An Instructor’s Solutions Manual to Accompany
ADVANCED ENGINEERING MATHEMATICS, 8TH EDITION
PETER V. O’NEIL
© 2018, 2012 Cengage Learning WCN: 01-100-101
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READ IMPORTANT LICENSE INFORMATION
INSTRUCTOR’S SOLUTIONS MANUAL TO ACCOMPANY
ADVANCED ENGINEERNG MATHEMATICS
8th EDITION
PETER V. O’NEIL UNIVERSITY OF ALABAMA AT BIRMINGHAM
Contents
1
2
3
First-Order Differential Equations 1
1.1 Terminology and Separable Equations 1 1.2 The Linear First-Order Equation 12 1.3 Exact Equations 19 1.4 Homogeneous, Bernoulli and Riccati Equations 28
Second-Order Differential Equations 37
2.1 The Linear Second-Order Equation 37 2.2 The Constant Coefficient Homogeneous Equation 41 2.3 Particular Solutions of the Nonhomogeneous Equation 46 2.4 The Euler Differential Equation 53 2.5 Series Solutions 58
The Laplace Transform 69
3.1 Definition and Notation 69 3.2 Solution of Initial Value Problems 72 3.3 The Heaviside Function and Shifting Theorems 77 3.4 Convolution 86 3.5 Impulses and the Dirac Delta Function 92 3.6 Systems of Linear Differential Equations 93
iii
iv
4
5
6
7
8
9
10
11
CONTENTS
Sturm-Liouville Problems and Eigenfunction Expansions 101
4.1 Eigenvalues and Eigenfunctions and Sturm-Liouville Problems 101 4.2 Eigenfunction Expansions 107 4.3 Fourier Series 114
The Heat Equation 137
5.1 Diffusion Problems on a Bounded Medium 137 5.2 The Heat Equation With a Forcing Term F (x, t) 147
5.3 The Heat Equation on the Real Line 5.4 The Heat Equation on a Half-Line 5.5 The Two-Dimensional Heat Equation
The Wave Equation
6.1 Wave Motion on a Bounded Interval
6.2 Wave Motion in an Unbounded Medium 6.3 d’Alembert’s Solution and Characteristics 6.4 The Wave Equation With a Forcing Term 6.5 The Wave Equation in Higher Dimensions
150 153 155 157 157 167 173
K (x, t) 190 192 197 197 202 205 7.4 The Dirichlet Problem for Unbounded Regions 205 7.5 A Dirichlet Problem in 3 Dimensions 208 7.6 The Neumann Problem 211 7.7 Poisson’s Equation 217
Laplace’s Equation
7.1 The Dirichlet Problem for a Rectangle 7.2 The Dirichlet Problem for a Disk
7.3 The Poisson Integral Formula
Special Functions and Applications 221
8.1 Legendre Polynomials 221 8.2 Bessel Functions 235 8.3 Some Applications of Bessel Functions 251
Transform Methods of Solution 263
9.1 Laplace Transform Methods 263 9.2 Fourier Transform Methods 268 9.3 Fourier Sine and Cosine Transforms 271
Vectors and the Vector Space Rn 275 10.1 Vectors in the Plane and 3− Space 275 10.2 The Dot Product 277 10.3 The Cross Product 278 10.4 n− Vectors and the Algebraic Structure of Rn 280 10.5 Orthogonal Sets and Orthogonalization 284 10.6 Orthogonal Complements and Pro jections 287
Matrices, Determinants and Linear Systems 291
11.1 Matrices and Matrix Algebra 291 11.2. Row Operations and Reduced Matrices 295 11.3 Solution of Homogeneous Linear Systems 299 11.4 Nonhomogeneous Systems 306 11.5 Matrix Inverses 313 11.6 Determinants 315 11.7 Cramer’s Rule 318 11.8 The Matrix Tree Theorem 320
12
13
14
15
16
17
Eigenvalues, Diagonalization and Special Matrices
12.1 Eigenvalues and Eigenvectors
12.2 Diagonalization
12.3 Special Matrices and Their Eigenvalues and Eigenvectors 12.4 Quadratic Forms
Systems of Linear Differential Equations
13.1 Linear Systems
13.2 Solution of X′ = AX When A Is Constant 13.3 Exponential Matrix Solutions
13.4 Solution of X′ = AX + G for Constant A 13.5 Solution by Diagonalization
Nonlinear Systems and Qualitative Analysis
14.1 Nonlinear Systems and Phase Portraits 14.2 Critical Points and Stability
14.3 Almost Linear Systems
14.4 Linearization
Vector Differential Calculus
15.1 Vector Functions of One Variable 15.2 Velocity, Acceleration and Curvature 15.3 The Gradient Field
15.4 Divergence and Curl
15.5 Streamlines of a Vector Field
Vector Integral Calculus
16.1 Line Integrals
16.2 Green’s Theorem
16.3 Independence of Path and Potential Theory 16.4 Surface Integrals
16.5 Applications of Surface Integrals
16.6 Gauss’s Divergence Theorem
16.7 Stokes’s Theorem
Fourier Series
17.1 Fourier Series on [−L, L]
17.2 Sine and Cosine Series
17.3 Integration and Differentiation of Fourier Series 17.4 Properties of Fourier Coefficients
17.5 Phase Angle Form
17.6 Complex Fourier Series
17.7 Filtering of Signals
323
323 327 332 336 339 339 341 348 350 353 359 359 363 364 369 373 373 376 381 385 387 391 391 393 398 405 408 412 414 419 419 423 428 430 432 435 438
v
vi
18
19
20
21
22
23
CONTENTS
Fourier Transforms 441
18.1 The Fourier Transform 441 18.2 Fourier sine and Cosine Transforms 448 Complex Numbers and Functions 451 19.1 Geometry and Arithmetic of Complex Numbers 451 19.2 Complex Functions 455 19.3 The Exponential and Trigonometric Functions 461 19.4 The Complex Logarithm 467 19.5 Powers 468
Complex Integration 473
20.1 The Integral of a Complex Function 473 20.2 Cauchy’s Theorem 477 20.3 Consequences of Cauchy’s Theorem 479
Series Representations of Functions 485
21.1 Power Series 485 21.2 The Laurent Expansion 492 Singularities and the Residue Theorem 497 22.1 Classification of Singularities 497 22.2 The Residue Theorem 499 22.3 Evaluation of Real Integrals 505
Conformal Mappings 515
23.1 The Idea of a Conformal Mapping 515 23.2 Construction of Conformal Mappings 533
Chapter 1
First-Order Differential Equations
1.1 Terminology and Separable Equations
1. The differential equation is separable because it can be written 3y2 dy = 4x,
or, in differential form, Integrate to obtain
dx
3y2 dy = 4x dx.
y3 = 2×2 + k.
This implicitly defines a general solution, which can be written explicitly
as
y = (2×2 + k)1/3, with k an arbitrary constant.
2. Write the differential equation as
xdy =−y,
which separates as
dx
1 dy = − 1 dx yx
ifx̸=0andy̸=0. Integratetoget
ln |y| = − ln |x| + k.
Then ln |xy| = k, so
xy = c 1
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2
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
with c constant (c = ek). y = 0 is a singular solution, satisfying the original differential equation.
3. If cos(y) ̸= 0, the differential equation is
y =sin(x+y) dx cos(y)
= sin(x) cos(y) + cos(x) sin(y) cos(y)
= sin(x) + cos(x) tan(y).
There is no way to separate the variables in this equation, so the differen-
tial equation is not separable.
4. Write the differential equation as
exey dy = 3x, dx
which separates in differential form as
ey dy = 3xe−x dx.
Integrate to get
ey =−3e−x(x+1)+c,
with c constant. This implicitly defines a general solution.
5. The differential equation can be written xdy =y2 −y,
dx
1 dy = 1 dx,
and is therefore separable. Separating the variables assumes that y ̸= 0 and y ̸= 1. We can further write
1 1 1 y−1−y dy=xdx.
Integrate to obtain
ln |y − 1| − ln |y| = ln |x| + k. Using properties of the logarithm, this is
y − 1
ln xy =k.
or
y(y − 1) x
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1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 3
Then
y − 1 = c, xy
with c = ek constant. Solve this for y to obtain the general solution y=1.
1−cx
y = 0 and y = 1 are singular solutions because these satisfy the differential
equation, but were excluded in the algebra of separating the variables.
6. The differential equation is not separable.
7. The equation is separable because it can be written in differential form as
sin(y) dy = 1 dx.
cos(y) x
This assumes that x ̸= 0 and cos(y) ̸= 0. Integrate this equation to obtain − ln | cos(y)| = ln |x| + k.
This implicitly defines a general solution. From this we can also write sec(y) = cx
with c constant.
The algebra of separating the variables required that cos(y) ̸= 0. Now cos(y) = 0 if y = (2n+1)π/2, with n any integer. Now y = (2n+1)π/2 also satisfies the original differential equation, so these are singular solutions.
8. The differential equation itself requires that y ̸= 0 and x ̸= −1. Write the equation as
x dy = 2y2 + 1 y dx x
and separate the variables to get
1 dy= 1 dx.
y(2y2 + 1) x(x + 1)
Use a partial fractions decomposition to write this as
1 2y 1 1 y−2y2+1 dy= x−x+1 dx.
Integrate to obtain
ln |y| − 1 ln(1 + 2y2) = ln |x| − ln |x + 1| + c 2
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4
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
with c constant. This implicitly defines a general solution. We can go a step further by writing this equation as
yx
ln 1+2y2 =ln x+1 +c
and take the exponential of both sides to get
y x 1+2y2 =k x+1 ,
which also defines a general solution. The differential equation is
dy = ex − y + sin(y), dx
and this is not separable. It is not possible to separate all terms involving x on one side of the equation and all terms involving y on the other.
9.
10.
Substitute
11.
into the differential equation to get the separated differential form (cos(y) − sin(y)) dy = (cos(x) − sin(x)) dx. Integrate to obtain the implicitly defined general solution
cos(y) + sin(y) = cos(x) + sin(x) + c. If y ̸= −1 and x ̸= 0, we obtain the separated equation
y2 1
y + 1 dy = x dx.
sin(x − y) = sin(x) cos(y) − cos(x) sin(y), cos(x + y) = cos(x) cos(y) − sin(x) sin(y),
and
cos(2x) = cos2(x) − sin2(x)
To make the integration easier, write this as 11
y−1+1+y dy=xdx. 1y2 −y+ln|1+y|=ln|x|+c.
Integrate to obtain
2
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1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 5 This implicitly defines a general solution. The initial condition is y(3e2) =
2,soputy=2andx=3e2 toobtain 2−2+ln(3) = ln(3e2)+c.
Now so
ln(3e2) = ln(3) + ln(e2) = ln(3) + 2, ln(3) = ln(3) + 2 + c.
Then c = −2 and the solution of the initial value problem is implicitly defined by
12. Integrate
1y2 −y+ln|1+y|=ln|x|−2. 2
1 dy=3x2dx, y+2
assuming that y ̸= −2, to obtain
ln |2 + y| = x3 + c.
This implicitly defines a general solution. To have y(2) = 8, let x = 2 and y = 8 to obtain
ln(10) = 8 + c.
The solution of the initial value problem is implicitly defined by
ln|2+y|=x3 +ln(10)−8. We can take this a step further and write
2+y 3
ln 10 =x−8.
By taking the exponential of both sides of this equation we obtain the explicit solution
y = 10ex3−8 − 2.
13. With ln(yx) = x ln(y), we obtain the separated equation
Integrate to obtain For y(2) = e3, we need
ln(y) dy = 3x dx. y
(ln(y))2 = 3×2 + c. (ln(e3))2 = 3(4) + c,
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6
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
or 9 = 12 + c. Then c = −3 and the solution of the initial value problem is defined by
(ln(y))2 = 3×2 − 3. Solve this to obtain the explicit solution
√
y=e 3(x2−1)
Because ex−y2 = exe−y2 , the variables can be separated to obtain
14.
if |x| > 1.
Integrate to get
To satisfy y(4) = −2 we need
2yey2 dy = ex dx. ey2 =ex+c.
e4 = e4 + c
so c = 0 and the solution of the initial value problem is implicitly defined
15.
Because y(4) = −2, the explicit solution is y = −√x for x > 0. Separate the variables to obtain
by
which reduces to the simpler equation
Integrate to get
y cos(3y) dy = 2x dx.
1 y sin(3y) + 1 cos(3y) = x2 + c, 39
ey2 = ex, x = y2.
which implicitly defines a general solution. For y(2/3) = π/3, we need 1 π sin(π) + 1 cos(π) = 4 + c.
This reduces to
3399
− 1 = 4 + c, 99
so c = −5/9 and the solution of the initial value problem is implicitly defined by
or
1 y sin(3y) + 1 cos(3y) = x2 − 5 , 399
3y sin(3y) + cos(3y) = 9×2 − 1.
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1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 7
16. Let T(t) be the temperature function. By Newton’s law of cooling, T′(t) = k(T −60) for some constant k to be determined. This equation is separable and is easily solved to obtain:
T(t) = 60 + 30ekt. To determine k, use the fact that T (10) = 88:
Then
so
T(10) = 60 + 30e10k = 88. e10k = 88−60 = 14,
30 15
k = 1 ln(14/15). 10
Now we know the temperature function completely:
T(t)=60+30ekt =60+30e10kt/10 14 t/10
=60+30 15 . We want to know T (20), so compute
142 T(20)=60+30 15 ≈86.13
degrees Fahrenheit. To see how long it will take for the object to reach 65 degrees, solve for t in
Then
so
14 t/10 T(t)=65=60+30 15 .
14t/10 65−60 1 15 =30=6,
t ln(14/15)=ln(1/6)=−ln(6). 10
The object reaches 65 degrees at time
t = − 10ln(6) ≈ 259.7
ln(14/15)
minutes.
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8 17.
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
Suppose the thermometer was removed from the house at time t = 0, and let T(t) be the temperature function. Let A be the ambient temperature outside the house (assumed constant). By Newton’s law,
T′(t) = k(t − A).
We are also given that T (0) = 70 and T (5) = 60. Further, fifteen minutes
after being removed from the house, the thermometer reads 50.4, so T (15) = 50.4.
We want to determine A, the constant outside temperature. From the differential equation for T,
1 dT=kdt. T−A
Integrate this, as we have done before, to get T(t) = A + cekt.
Now,
so c = 70 − A and
T (0) = 70 = A + c, T(t) = A + (70 − A)ekt.
Now use the other two conditions:
T(5)=A+(70−A)e5k =15.5andT(15)=A+(70−A)e15k =50.4. From the equation for T (5), solve for e5k to get
Then
e5k = 60−A. 70−A
15k 5k3 60−A3 e = e = 70−A .
Substitute this into the equation T (15) to get 60 − A3
Then
(70−A) 70−A = 50.4−A. (60−A)3 =(50.4−A)(70−A)2.
The cubic terms cancel and this reduces to the quadratic equation 10.4A2 − 1156A + 30960 = 0,
with roots 45 and (approximately) 66.15385. Clearly the outside temper- ature must be less than 50, and must therefore equal 45 degree.
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1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 9
18. The amount A(t) of radioactive material at time t is modeled by A′(t) = kA, A(0) = e3
together with the given half-life of the material, A(ln(2)) = 1e3.
2 Solve this (as in the text) to obtain
Then
1 t/ ln(2) A(t)= 2 e3.
3 1 3/ ln(2)
A(3) = e 2 = 1 tonne.
19. The problem is like Problem 18, and we find that the amount of Uranium-
235 at time t is
1 t/(4.5(109 )) U(t)=10 2 ,
with t in years. Then
U(10)=10 2 ≈8.57kg.
this for k to get
1 9.1 k=4ln 12 .
The half-life of this element is the time t∗ it will take for there to be 6
9 1 1/4.5
20. At time t there will be A(t) = 12ekt grams, and A(4) = 12e4k = 9.1. Solve
grams, so
Solve this to get
21. Let
A(t∗) = 6 = 12eln(9.1/12)t∗/4. t∗ = 4 ln(1/2) ≈ 10.02 minutes.
ln(9.1/12)
∞22 e−t −(x/t) dt.
0
I(x) =
The integral we want is I(3). Compute
′ ∞ 1 −t2−(x/t)2
I (x) = −2x
t2 e dt.
0
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10
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
Let u = x/t, so t = x/u and
dt=−x du.
Then
∞ √π
′
I(x)=−2x
= −2I(x).
u2
0 u2 −(x/u)2−u2 −x
x2 e u2 du
satisfies the separable differential equation I ′ = −2I , with
∞
Then I (x)
general solution of the form I(x) = ce−2x. Now observe that
e−t2 dt= 2 =c, √
I(0)=
in which we used a standard integral that arises often in statistics. Then
0
I(x)= πe−2x. 2
Finally, put x = 3 for the particular integral of interest:
22.
Begin with the logistic equation
P′(t) = aP(t) − bP(t)2,
in which a and b are positive constants. Then dP =P(a−bP)
so
1 dP = dt
and the variables are separated. To make the integration easier, write this
I(3) =
∞ √π
e−t2−(9/t)2 dt = 2 e−6.
0
dt
P(a − bP)
equation as
Integrate to obtain
11 b 1 aP+aa−bP dP=dt.
1 ln(P ) − b ln(a − bP ) = t + c. aa
if P (t) > − and a − bP (t) > 0. Using properties of the logarithm, we can write this equation as
P
ln a−bP =at+k,
© 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 11 in which k = ac is still constant. Then
P = eat+k = ekeat = Keat, a−bP
in which K = ek is a positive constant. Now suppose the initial population (say at time zero) is p0. Then P(0) = p0 and
We now have
p0 = K. a−bp0
P = p0 eat. a−bP a−bp0
It is a straightforward algebraic manipulation to solve this for P (t): P(t) = ap0 eat.
a−bp0 +bp0
This is the solution of the logistic equation with P(0) = p0.
Because a − bp0 > 0 by assumption, then
bp0eat < a − bp0 + bpeat,
so
This means that this population function is bounded above. Further, by
P(t)< ap0 eat=a. bp0 eat b
multiplying the numerator and denominator of P (t) by e(−at, we have lim P(t)= lim ap0
t→∞ t→∞ (a − bp0)e−at + bp0 = lim ap0 = a.
t→∞ bp0 b
23. With a and b as given, and p0 = 3, 929, 214 (the population in 1790), the
logistic population function for the United States is
P(t) = 123,141.5668 e0.03134t.
0.03071576577 + 0.0006242342283e0.03134t
If we attempt an exponential model Q(t) = Aekt, then take A = Q(0) =
3, 929, 214, the population in 1790. To find k, use the fact that Q(10) = 5308483 = 3929214e10k
and we can solve for k to get
1 5308483
k = 10 ln 3929214 ≈ 0.03008667012.
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12 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
year population
1790 3,929,213 3,929,214 0 3,929,214 0 1800 5,308,483 5,336,313 0.52 5,308,483 0 1810 7,239,881 7,228,171 -0.16 7,179,158 -0.94 1820 9,638,453 9,757,448 1.23 7,179,158 0.53 1830 12,886,020 13,110,174 1.90 13,000,754 1.75 1840 17,169,453 17,507,365 2.57 17,685,992 3.61 1850 23,191,876 23.193,639 0.008 23,894,292 3.03 1860 31,443,321 30,414,301 -3.27 32,281,888 2.67 1870 38,558,371 39,374,437 2.12 43,613,774 13.11 1880 50,189,209 50,180,383 -0.018 58,923,484 17.40 1890 62,979766 62,772,907 -0.33 79,073,491 26.40 1900 76,212,168 76,873,907 0.87 107,551,857 41.12 1910 92,228,496 91,976,297 -0.27 145,303,703 57.55 1920 106,021,537 107,398,941 1.30 196,312,254 83.16 1930 123,202,624 122,401,360 -0.65
1940 132,164,569 136,329,577 3.15
1950 151,325,798 148,679,224 -1.75
1960 179,323,175 150,231,097 -11.2
1970 203,302,031 167,943,428 -17.39
1980 226,547,042 174,940,040 -22.78
P (t) percent error
Q(t) percent error
1.2
1.
Table 1.1: Census data for Problem 23
The exponential model, using these two data points (1790 and 1800 pop- ulations), is
Q(t) = 3929214e0.03008667012t.
Table 1.1 uses Q(t) and P(t) to predict later populations from these two initial figures. The logistic model remains quite accurate until about 1960, at which time it loses accuracy quickly. The exponential model becomes quite inaccurate by 1870, after which the error becomes so large that it is not worth computing further. Exponential models do not work well over time with complex populations, such as fish in the ocean or countries throughout the world.
The Linear First-Order Equation
With p(x) = −3/x, and integrating factor is
e (−3/x) dx = e−3 ln(x) = x−3
for x > 0. Multiply the differential equation by x−3 to get x−3y′ − 3x−4 = 2x−1.
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1.2. THE LINEAR FIRST-ORDER EQUATION 13 or
d (x−3y)= 2. dx x
Integrate to get
x−3y = 2 ln(x) + c,
with c an arbitrary constant. For x > 0 we have a general solution
y = 2×3 ln(x) + cx3.
In the last integration, we can allow x < 0 by replacing ln(x) with ln |x|
ex to get
Then
to derive the solution for x ̸= 0.
y = 2x3 ln |x| + cx3
2. e dx = ex is an integrating factor. Multiply the differential equation by
and an integration gives us
y′ex +yex = 1(e2x −1). 2
(exy)′ = 1(e2x −1) 2
Then
exy = 1e2x − 1x + c. 42
y = 1ex − 1xe−x +ce−x 42
is a general solution, with c an arbitrary constant.
3. e 2 dx = e2x is an integrating factor. Multiply the differential equation by
e2x :
or
Integrate to get
y′e2x + 2ye2x = xe2x, (e2xy)′ = xe2x.
e2xy = 1xe2x − 1e2x + c. 24
giving us the general solution
y = 1x − 1 + ce−2x. 24
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14 4.
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
For an integrating factor, compute
e sec(x) dx = eln | sec(x)+tan(x)| = sec(x) + tan(x).
Multiply the differential equation by this integrating factor:
y′(sec(x) + tan(x)) + sec(x)(sec(x) + tan(x))y
= y′(sec(x) + tan(x)) + (sec(x) tan(x) + sec2(x))y
= ((sec(x) + tan(x))y)′
= cos(x)(sec(x) + tan(x)) = 1 + sin(x).
We therefore have
Integrate to get
Then
((sec(x) + tan(x))y)′ = 1 + sin(x). y(sec(x) + tan(x)) = x − cos(x) + c.
y = x−cos(x)+c . sec(x) + tan(x)
This is a general solution. If we wish, we can also observe that 1 = cos(x)
to obtain
1 + sin(x) First determine the integrating factor
e −2dx =e−2x. Multiply the differential equation by e−2x to get
Integrate to get
(e−2xy)′ = −8x2e−2x.
−8x2e−2x dx = 4x2e−2x + 4xe−2x + 2e−2x + c.
e−2xy =
This yields the general solution
sec(x) + tan(x) 1 + sin(x)
cos(x)
y = (x−cos(x)+c) 1+sin(x) = xcos(x)−cos2(x)+ccos(x).
5.
y=4x2 +4x+2+ce2x.
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1.2. THE LINEAR FIRST-ORDER EQUATION 15 6. e 3 dx = e3x is an integrating factor. Multiply the differential equation by
e3x to get
Integrate this equation:
(e3xy)′ = 5e5x − 6e3x.
e3xy=e5x −2e3x +c. Now we have a general solution
y = e2x − 2 + ce−3x. y(0) = 2 = 1 − 2 + c,
We need
so c = 3. The unique solution of the initial value problem is
y=e2x +3e−3x −2.
7. x − 2 is an integrating factor for the differential equation because
e (1/(x−2)) dx = eln(x−2) = x − 2. Multiply the differential equation by x − 2 to get
Integrate to get
This gives us the general solution
((x−2)y)′ =3x(x−2). (x−2)y=x3 −3x2 +c.
y= 1 (x3−3x2+c). x−2
Now we need
so c = 4 and the solution of the initial value problem is
y= 1 (x3−3x2+4). x−2
8. e (−1) dx = e−x is an integrating factor. Multiply the differential equation by e−x to get:
y(3) = 27 − 27 + c = 4,
Integrate to get
(ye−x)′ = 2e3x. ye−x = 2e3x +c,
3
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16
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
and we have the general solution
y = 2e4x + cex.
3
We need
so c = −11/3 and the initial value problem has the solution
y = 2e4x − 11ex. 33
First derive the integrating factor
e (2/(x+1)) dx = e2 ln(x+1) = eln((x+1)2 ) = (x + 1)2 .
Multiply the differential equation by (x + 1)2 to obtain (x + 1)2y′ = 3(x + 1)2.
Integrate to obtain Then
y=x+1+ 4 . (x+1)2
An integrating factor is
e (5/9x) dx = e(5/9) ln(x) = eln(x5/9) = x5/9.
Multiply the differential equation by x5/9 to get (yx5/9)′ = 3x32/9 + x14/9.
y(0) = 2 + c = −3, 3
9.
(x+1)2y = (x+1)3 +c. y=x+1+ c .
Now
so c = 4 and the initial value problem has the solution
(x+1)2 y(0) = 1 + c = 5
10.
Integrate to get Then
Finally, we need
yx5/9 = 27x41/9 + 9 x23/9 + c. 41 23
y=27x4+ 9x2+cx−5/9. 41 23
y(−1)=27+ 9 −c=4. 41 23
Then c = −2782/943, so the initial value problem has the solution y=23x4+ 9x2−2782x−5/9.
41 23 943
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1.2. THE LINEAR FIRST-ORDER EQUATION 17 11. Let (x,y) be a point on the curve. The tangent line at (x,y) must pass
through (0, 2x2), and so has slope
′ y − 2x2
y=x. This is the linear differential equation
y′ − 1y = −2x. x
An integrating factor is
e−(1/x)dx = e−ln(x) = eln(1/x) = 1,
x so multiply the differential equation by 1/x to get
This is
Integrate to get
1y′− 1y=−2. x x2
1 ′
xy =−2.
1 y = −2x + c. x
Then
in which c can be any number.
y = −2x2 + cx,
12. Let A(t) be the number of pounds of salt in the tank at time t ≥ 0. Then
dA = rate salt is added − rate salt is removed dt
A(t) =6−2 50+t .
We must solve this subject to the initial condition A(0) = 25. The differ- ential equation is
A′+ 2 A=6, 50+t
which is linear with integrating factor
e 2/(50+t) dt = e2 ln(50+t) = (50 + t)2.
Multiply the differential equation by (50 + t)2 to get (50+t)2A′ +2(50+t)A=6(50+t)2.
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18
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
which we will write as
A(t) = 2(50 + t) + c . (50 + t)2
We need c so that
so c = 187, 500. The number of pounds of salt in the tank at time t is
A(t)=2(50+t)− 187,500. (50 + t)2
Let A1(t) and A2(t) be the number of pounds of salt in tanks 1 and 2, respectively, at time t. Then
This is
Integrate this equation to get
(50 + t)2A′ = 6(50 + t)2. (50+t)2A=2(50+t)3 +c,
A(0)=100+ c =25, 2500
13.
and
A′1(t) = 5 − 5A1(t);A1(0) = 20 2 100
A′2(t) = 5A1(t) − 5A2(t);A2(0) = 90. 100 150
Solve the linear initial value problem for A1(t) to get A1(t) = 50 − 30e−t/20.
Substitute this into the differential equation for A2(t) to get A′2+ 1A2=5−3e−t/20;A2(0)=90.
30 2 2 Solve this linear problem to obtain
A2(t) = 75 + 90e−t/20 − 75e−t/30.
Tank 2 has its minimum when A′2(t) = 0, and this occurs when
2.5e−t/30 − 4.5e−t/20 = 0. This occurs when et/60 = 9/5, or t = 60 ln(9/5). Then
A2(t)min = A2(60 ln(9/5)) = 5450 81
pounds.
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1.3. EXACT EQUATIONS 19 1.3 Exact Equations
In these problems it is assumed that the differential equation has the form M(x,y)+N(x,y)y′ =0,or,indifferentialform,M(x,y)dx+N(x,y)dy=0.
1.WithM(x,y)=2y2+yexy andN(x,y)=4xy+xexy+2y.Then ∂N = 4y + exy + xyexy = ∂M
∂x ∂y
for all (x,y), so the differential equation is exact on the entire plane. A
potential function φ(x, y) must satisfy
∂φ =M(x,y)=2y2 +yexy
and
∂x
∂φ =N(x,y)=4xy+xexy +2y. ∂y
Choose one to integrate. If we begin with ∂φ/∂x = M, then integrate with respect to x to get
φ(x, y) = 2xy2 + exy + α(y),
with α(y) the “constant” of integration with respect to x. Then we must
have
∂φ =4xy+xexy +α′(y)=4xy+xexy +2y. ∂y
This requires that α′(y) = 2y, so we can choose α(y) = y2 to obtain the potential function
φ(x,y)=2xy2 +exy +y2.
The general solution is defined implicitly by the equation
2xy2 + exy + y2 = c, , with c an arbitrary constant.
2. ∂M/∂y = 4x = ∂N/∂x for all (x, y), so this equation is exact on the entire plane. For a potential function, we can begin by integrating
to get Then
∂φ = 2x2 + 3y2 ∂y
φ(x,y)=2x2y+y3 +c(x). ∂φ =4xy+2x=4xy+c′(x).
∂x
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20
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
Then c′(x) = 2x so we can choose c(x) = x2 to obtain the potential function
φ(x,y)=2x2y+y3 +x2. The general solution is defined implicitly by
2x2y + y3 + x2 = k, with k an arbitrary constant.
3. ∂M/∂y = 4x + 2x2 and ∂N/∂x = 4x, so this equation is not exact (on any rectangle).
4.
∂M =−2sin(x+y)+2xcos(x+y)= ∂N, ∂y ∂x
for all (x,y), so this equation is exact on the entire plane. Integrate ∂φ/∂x = M or ∂φ/∂y = N to obtain the potential function
φ(x, y) = 2x cos(x + y). The general solution is defined implicitly by
2x cos(x + y) = k with k an arbitrary constant.
5. ∂M/∂y = 1 = ∂N/∂x, for x ̸= 0, so this equation is exact on the plane except at points (0, y). Integrate ∂φ/∂x = M or ∂φ/∂y = N to find the potential function
φ(x, y) = ln |x| + xy + y3
for x ̸= 0. The general solution is defined by an equation
ln|x|+xy+y3 =k. 6. For the equation to be exact, we need
∂M = αxyα−1 = ∂N = −2xyα−1. ∂y ∂x
This will hold if α = −2. With this choice of α, the (exact) equation is 3x2 + xy−2 − x2y−3y′ = 0.
Routine integrations produce a potential function
3 x2 φ(x,y)=x +2y2.
The general solution is defined by the equation
3 x2
x +2y2 =k,
for y ̸= 0.
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1.3. EXACT EQUATIONS 21 7. For this equation to be exact, we need
∂M =6xy2 −3= ∂N =−3−2αxy2. ∂y ∂x
This will be true if α = −3. By integrating, we find a potential function φ(x,y)=x2y3 −3xy−3y2
and a general solution is defined implicitly by x2y3 − 3xy − 3y2 = k.
8. We have
∂M = 2 − 2y sec2(xy2) − 2xy3 sec2(xy2) tan(xy2) = ∂N ,
∂y ∂x
for all (x, y), so this equation is exact over the entire plane. By integrating
∂φ/∂x = 2y − y2 sec2(xy2) with respect to x, we find that φ(x, y) = 2xy − tan(xy2) + c(y).
Then
∂φ = 2x − 2xy sec2(xy2) ∂y
= 2x − 2xy sec2(xy2) + c′(y).
Then c′(y) = 0 and we can choose c(y) = 0 to obtain the potential function
φ(x, y) = 2xy − tan(xy2). A general solution is defined implicitly by
2xy − tan(xy2) = k.
For the solution satisfying y(1) = 2, put x = 1 and y = 2 into this
implicitly defined solution to get
4 − tan(4) = k.
The solution of the initial value problem is defined implicitly by 2xy − tan(xy2) = 4 − tan(4).
9. Because ∂M/∂y = 12y2 = ∂N/∂x, this equation is exact for all (x,y). Straightforward integrations yield the potential function
φ(x, y) = 3xy4 − x.
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22
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
A general solution is defined implicitly by 3xy4 − x = k.
To satisfy the condition y(1) = 2, we must choose k so that 48 − 1 = k,
so k = 47 and the solution of the initial value problem is specified by the equation
3xy4 − x = 47.
In this case we can actually write this solution explicitly with y in terms
10.
of x. First,
∂M = 1ey/x − 1ey/x − y ey/x ∂y x x x2
= − y ey/x = ∂N , x2 ∂x
so the equation is exact for all (x, y) with x ̸= 0. For a potential function, we can begin with
∂φ = ey/x ∂y
and integrate with respect to y to get
φ(x, y) = xey/x + c(x).
Then we need
∂φ = 1+ey/x − yey/x = ey/x − yey/x +c′(x).
∂x x x
This requires that c′(x) = 1 and we can choose c(x) = x. Then
φ(x, y) = xey/x + x.
The general solution of the differential equation is implicitly defined by
xey/x + x = k. To have y(1) = −5, we must choose k so that
e−5 + 1 = k.
The solution of the initial value problem is given by
xey/x +x=1+e−5.
This can be solve for y to obtain the explicit solution
1 + e−5 y = x ln x + 1
for x + 1 > 0.
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1.3. EXACT EQUATIONS 23 11. First,
∂M =−2xsin(2y−x)−2cos(2y−x)= ∂N, ∂y ∂x
so the differential equation is exact for all (x, y). For a potential function, integrate
∂φ =−2xcos(2y−x) ∂y
with respect to y to get
φ(x, y) = −x sin(2y − x) + c(x).
Then we must have
∂φ =xcos(2y−x)−sin(2y−x)
∂x
= x cos(2y − x) − sin(2y − x) + c′(x).
Then c′(x) = 0 and we can take c(x) to be any constant. Choosing c(x) = 0 yields
φ(x, y) = −x sin(2y − x). The general solution is defined implicitly by
−x sin(2y − x) = k. To satisfy y(π/12) = π/8, we need
− π sin(π/6) = k, 12
so choose k = −π/24 to obtain the solution defined by −x sin(2y − x) = − π
which of course is the same as
24
We can also write
for x ̸= 0. 12.
xsin(2y−x)= π. 24
y=1x+arcsin π 2 24x
∂M = ey = ∂N ∂y ∂x
so the differential equation is exact. Integrate ∂φ = ey
∂x
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24
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
with respect to x to get
Then
φ(x, y) = xey + c(y).
∂φ =xey +c′(y)=xey −1,
∂y
so c′(y) = −1 and we can let c(y) = −y. This gives us the potential
function
The general solution is given by
xey − y = 5.
13. φ + c is also a potential function if φ is because
φ(x,y)=xey −y. xey − y = k.
For y(5) = 0 we need
so k = 5 and the solution of the initial value problem is given by
5−0=k
and
∂φ = ∂(φ+c) ∂x ∂x
∂φ = ∂(φ+c).
∂y ∂y
The function defined implicitly by
φ(x, y) = k
is the same as that defined by
φ(x, y) + c = k
if k is arbitrary. 14. (a)
∂M = 1 and ∂N = −1 ∂y ∂x
so this equation is not exact over any rectangle in the plane. (b) Multiply the differential equation by x−2 to obtain
This is exact because
yx−2 − x−1y′ = 0. ∂M∗ −2 ∂N∗
∂y=x =∂x.
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1.3. EXACT EQUATIONS 25 This new equation has potential function φ(x,y) = −yx−1 and so has
general solution defined implicitly by
−y =k. x
This also defines a general solution of the original differential equation. (c) Multiply the differential equation by y−2 to obtain
y−1 − xy−2y′ = 0.
This is exact on any region of the plane not containing y = 0, because
∂M∗∗ −2 ∂N∗∗ ∂y =−y = ∂x .
The new equation has potential function φ(x,y) = xy−1, so its general solution is defined implicitly by
xy−1 = k.
It is easy to check that this also defines a solution of the original differential
equation.
(d) Multiply the differential equation by xy−2 to obtain
xy−2 − x2y−3y′ = 0.
This is exact (on any region not containing y = 0) because
∂M∗∗∗ −3 ∂N∗∗∗ ∂y =−2xy = ∂x .
Integrate ∂φ/∂x = xy−2 with respect to x to obtain φ(x,y)= 1x2y−2 +c(y).
Then
2
∂φ = −x−2y−3 + c′(y) = −x2y−3,
∂y
soc′ =0andwecanchoosec(y)=0. Then
φ(x,y)= 1x2y−2 2
and we can define a general solution of this differential equation as x2y−2 = k.
Here we absorbed the factor of 1/2 into the arbitrary constant c. This again defines a solution of the original differential equation.
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26
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
(e) The original differential equation can be written as the linear equation y′ − 1y = 0.
x
This has integrating factor
e −(1/x) dx = e− ln(x) = eln(x−1) = x−1.
Multiply the differential equation by x−1 to write this equation as (x−1y)′ = 0,
so x−1y = c implicitly defines the general solution.
(f) The methods of (b) through (e) yield the same general solution. For example, in (b) we obtained −yx−1 = c, which we can write as y = −cx. Because c is an arbitrary constant, this general solution can be written y = kx. And in (d) we obtained x2y−2 = c, and this gives the same solutions as y2 = cx2, or y = kx.
First,
∂M =x−3y−5/2 and ∂N =2x. ∂y 2 ∂x
and these are not equal on any rectangle in the plane. In differential form, the differential equation is
(xy + y−3/2) dx + x2 dy = 0. Multiply this equation by xayb to get
(xa+1yb+1 +xayb−3/2)dx+xa+2ybdy=0=M∗dx+N∗dy. For this to be exact, we need
∂M∗ 3
∂y = (b + 1)xa+1yb + (b − 2 xayb−5/2
∂N∗
= ∂x = (a + 2)xa+1yb.
15.
Divide this equation by xayb to get
3 −5/2
(b+1)x+ b−2 y =(a+2)x.
This will hold for all x and y if we let b = 3/2 and then choose a and b so
thatb+1=a+2. Thuschoose
a = 1 and b = 3
22
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1.3. EXACT EQUATIONS 27 to get the integrating factor μ(x,y) = x1/2b3/2. Multiply the original
differential equation by this to get
(x3/2y5/2 + x1/2) dx + x5/2y1/2 dy = 0.
To find a potential function, integrate
∂φ = x5/2y3/2 ∂y
with respect to y to get
φ(x,y)= 2×5/2y5/2 +c(x).
Then we need
5
∂φ = x3/2y5/2 + c′(x) = x3/2y5/2 + x1/2. ∂x
Therefore c′(x) = x1/2, so c(x) = 2×3/2/3 and φ(x) = 2×5/2y5/2 + 2×3/2.
53
The general solution of the original differential equation is given implicitly by
2(xy)5/2 + 2×3/2 = k. 53
In this we must have x ̸= 0 and y ̸= 0 to ensure that the integrating factor μ(x, y) ̸= 0.
16. It is routine to verify that the differential equation is not exact. To find an integrating factor, first multiply by xayb to get
(2xayb+2 − 9xa+1yb+1) dx + (3xa+1yb+1 − 6xa+2yb) dy = 0. For this to be exact, we must have
∂M = 2(b + 2)xayb+1 − 9(b + 1)xa+1yb ∂y
= ∂N = 3(a + 1)xayb+1 − 6(a + 2)x+1yb. ∂x
Divide by xayb and rearrange terms to obtain (2(b+2)−3(a+1))y = (9(b+1)−6(a+2))x.
Because x and y are independent, both coefficients must be zero: 2(b + 2) − 3(a + 1) = 0 and 9(b + 1) − 6(a + 2) = 0.
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28
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
Solve these to get a = b = 1, so μ(x,y) = xy is an integrating factor. Multiply the differential equation by xy to obtain, in differential form,
(2xy3 − 9x2y2) dx + (3x2y2 − 6x3y) dy = 0. This equation is exact. For a potential function, integrate
∂φ = 2xy3 − 9x2y2 ∂x
with respect to x to get
φ(x, y) = x2y3 − 3x3y2 + c(y).
Then
∂φ = 3x2y2 − 6x3y + c′(y) = 3x2y2 − 6x3y. ∂y
1.4
1.
Then c′(y) = 0 and we can choose c(y) = 0 got s potential function φ(x,y)=x2y3 −3x3y2.
A general solution of this equation, and also the original equation, is given by
x2y3 − 3x3y2 = k. This requires that μ(x, y) ̸= 0.
Homogeneous, Bernoulli and Riccati Equa- tions
This is a Riccati equation and one solution (by inspection) is S(x) = x. Let y = x + 1/z to obtain
1′ 1 12 1 1 2−z2z=x2 x+z −x x+z +1.
This simplifies to
z′ + 1z = − 1 , x x2
a linear equation with integrating factor
e (1/x) dx = eln(x) = x.
The differential equation for z can therefore be written (xz)′ =−1.
Integrate to get
x
xz = − ln(x) + c,
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1.4. HOMOGENEOUS, BERNOULLI AND RICCATI EQUATIONS 29
so
for x > 0. Then
z = − ln(x) + c = c − ln(x) . xxx
y=x+1=x+ x
z c − ln(x)
for x > 0.
2. This is a Bernoulli equation with α = −4/3. Put v = y7/3, so y = v7/3.
Substitute this into the differential equation to get 3v−4/7v′ + 7 v3/7 = 14 .
7 3x 3×2 This simplifies to the linear differential equation
v′ + 7 v = 14 3x 3×2
which has integrating factor
e 7/3x dx = e(7/3) ln(x) = eln(x7/3 ) = x7/3
for x > 0. Multiply the differential equation by x7/3 to get (x7/3v)′ = 14×1/3.
Integrate to get
3
vx7/3 = 7×4/3 +c. 2
Because v = y7/3, this gives us
2y7/3×7/3 − 7×4/3 = k,
in which k = 2c is an arbitrary constant. This equation implicitly defines the general solution.
3. ThisisaBernoulliequationwithα=2,soletv=y1−α =y−1 fory̸=0 and y = 1/v. Compute
y′ = dy dv = − 1 xv′. dv dx v2
The differential equation becomes −1v′+x= x.
This is
v2 vv2 v′ − xv = −x,
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30
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
a linear equation with integrating factor e−x2/2. We can therefore write (e−x2/2v)′ = −xe−x2/2.
Integrate to get
so
e−x2/2v = e−x2/2 + c, v = 1 + ce−x2/2.
4.
The original differential equation has the general solution
y=1=1, v 1+ce−x2/2
in which c is an arbitrary constant.
This equation is homogeneous. With y = ux we obtain
u + xu′ = u + 1 . u
Then
a separable equation. In differential form, this is
xdu = 1, dx u
Integrate to get
Then
u du = 1 dx. x
1u2 =ln|x|+c. 2
1 x2 = 2 ln |x| + k, y2
5.
where k = 2c is an arbitrary constant. This implicitly defines the general solution.
This differential equation is homogeneous and setting y = ux gives us u+xu′= u .
1+u
This is the separable equation
xdu= u −u
dx 1+u
which, in terms of x and y, is 111
u2+u du=−xdx.
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1.4. HOMOGENEOUS, BERNOULLI AND RICCATI EQUATIONS 31 Integrate to get
1 +ln|u|=−ln|x|+c. u
With u = y/x this reduces to
−x + y ln |y| = cy,
with c an arbitrary constant.
6. This is a Riccati equation and one solution (by inspection) is S(x) = 4.
After some routine computation we obtain the general solution 6×3
y = 4 + c − x3 .
7. The differential equation is exact, with general solution defined implicitly
by
8. The differential equation is homogeneous, and y = ux yields the general
solution defined by
sec y + tan y = cx. xx
xy − x2 − y2 = c.
9. The differential equation is of Bernoulli type with α = −3/4. The general solution is defined by
5(xy)7/4 + 7x−5/4 = c.
10. The differential equation is homogeneous and y = ux leads to the separable
differential equation
1−u+u2 1 du= xdx.
Integrate and set u = y/x to obtain the general solution implicitly defined by
2 2y − x
√ arctan √ =ln|x|+c.
3 3x
11. The equation is Bernoulli with α = 2 and the change of variables v = y−1
leads to the general solution y=2+2.
defined by
1 x2
2y2 =ln|x|+c.
cx2 − 1
12. The equation is homogeneous and y = ux leads to the general solution
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32 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS 13. The differential equation is Riccati and one solution is S(x) = ex. A
general solution is given explicitly by 2ex
y = ce2x − 1 .
14. The equation is Bernoulli with α = 2 and a general solution is given by
y=2. 3+cx2
15. For the first part,
ax+by+c a+b(y/x)c/x y
F dx+py+r =F d+p(y/x)+r/x =f x
if and only if c = r = 0. Next,supposex=X+handy=Y +k. Then
dY
dX =F
=F
a(X +h)+b(Y +k)+c d(x+h)+p(Y +k)+r
aX+bY +c+ah+bk+c dX+pY +r+dh+pk+r .
This equation is homogeneous exactly when h and k can be chosen so that
ah + bk = −c and dh + pk = −r.
This 2 × 2 system of algebraic equations has a solution exactly when the determinant of the coefficients is nonzero, and this is the condition that
a b=ap−bd̸=0. d p
16. Comparing this with problem 15, we have
a = 0, b = 1, c = −3, d = p = 1 and r = −1.
The system to solve for h and k is
k = 3, h + k = 1.
Thenk=3andh=−2. LetX=x−2,Y =y+3toobtain dY= Y .
dX X+Y
This is a homogeneous equation solved in problem 5. The general of the
current problem is defined by
(y − 3) ln |y − 3| − (x + 2) = c(y − 3),
with c an arbitrary constant.
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1.4. HOMOGENEOUS, BERNOULLI AND RICCATI EQUATIONS 33 17. LetX=x−2,Y =y+3togetthehomogeneousequation
dY =3X−Y. dX X+Y
The general solution of the original equation (in terms of x and y) is defined by
3(x−2)2 −2(x−2)(y+3)−(y+3)2 =c, with c an arbitrary constant.
18. Set X = x + 5, Y = y + 1 to obtain the implicitly defined general solution (x+5)2 +4(x+5)(y+1)−(y+1)2 =c.
19. LetX=x−2,Y =y+1toobtainthegeneralsolutiongivenby (2x+y−3)2 =c(y−x+3).
20. Suppose at time t = 0 the dog is at the origin of an x,y− coordinate system, and the person is at (A,0). The person moves directly upward and at time t is at (A, vt), while the dog is at (x, y) and runs toward the person at a speed 2v. The tangent to the dog’s path joins these two points, and so has slope
y′ = vt−y. A−x
To find the equation of the dog’s path, we will first eliminate t from this equation. In the time the person has moved vt units upward, the dog has run 2vt units along its path of motion, so
x dy21/2 2vt= 1+dξ dξ.
Then
Then
dy 1 x
0
vt=y+(A−x)dx = 2
′ x dy2
dy2
dξ dξ.
1+ dξ dξ−2y.
−2y′,
1+
2(A−x)y = Differentiate this equation to get
0
0
2(A−x)y′′ −2y′ =
2(A − x)y′′ = 1 + (y′)21/2 ,
dy2 1+ dx
so
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34
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
together with the conditions y(0) = y′(0) = 0.
Now let u = y′ and rewrite the resulting equation to get
1 du= 1 dx. (1 + u2)1/2 2(A − x)
This has the general solution
ln(u + 1 + u2) = −1 ln(A − x) + c. 2
Use the condition that y′(0) = u(0) = 0 to obtain A 1/2
u+ 1+u2 = A−x . In terms of y, we now have
But so
Letw=y′ toget
√ y′+1+(y′)2=√ A ;y(0)=0.
A−x
1 + (y′)2 = 2(A − x)y′′,
√
y′ +2(A−x)y′′ = √ A .
A−x √
w′+ 1 w= A . 2(A − x) 2(A − x)3/2
√
This linear differential equation has integrating factor 1/ w′ A
A − x, so
Integrate this to get
√A−x = 2(A−x)2. √A 1 √
w= 2 √A−x+c A−x. Use the fact that w(0) = 0 to get
√A1 1√ ′ w= 2 √A−x−2√A A−x=y.
Integrate this to get
y=− A A−x+ √ (A−x) +c.
√√ 1 3/2 3A
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1.4. HOMOGENEOUS, BERNOULLI AND RICCATI EQUATIONS 35 Because y(0) = 0,
√√ 1 3/22 y=− A A−x+3√A(A−x) +3A.
Now the dog catches the person at x = A, so they meet at (A, 2A/3). This is also the point (A, vt), so vt = 2A/3 and they meet at time
t = 2A. 3v
21. It is convenient to use polar coordinates to formulate a model for this problem. Put the origin at the submarine at the time of sighting, and the polar axis the line from there to the destroyer at this time (the point (9, 0)). Initially the destroyer should steam at speed 2v directly toward the origin, until it reaches (3, 0). During this time the submarine, moving at speed v, will have moved three units from the point where it was sighted. Let θ = φ be the ray (half-line) along which the submarine is moving.
Upon reaching (3, 0), the destroyer should execute a search pattern along a path r = f(θ). The object is to choose this path so that the sub and the destroyer both reach (f(φ,φ) at the same time T after the sighting.
From sighting to interception, the destroyer travels a distance
so
φ
(f(θ))2 + (f′(θ))2 dθ, 1φ
6 +
0
T = 2v 6+ For the submarine,
(f(θ))2 +(f′(θ))2 dθ . T = 1f(φ).
0
v
Equate these two expressions for T and differentiate with respect to φ to
get
1(f(φ))2 + (f′(φ))2 = f′(φ). 2
Denote the variable as θ and rearrange the last equation to obtain f′(θ) 1
The positive sign here indicates that the destroyer should execute a star- board (left) turn, while the negative sign is for a portside turn. Taking the positive sign, solve for f(θ) to get
√
f(θ) = keθ/ 3.
f(θ) =±√3.
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36
CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
Now f(0) = k = 3, so the path of the destroyer is part of the graph of √
f(θ) = 3seθ/ 3.
After sailing directly to (3,0), the destroyer should execute this spiral
pattern. A similar conclusion follows if the negative sign of 1/ 3 is used.
This shows that the destroyer can carry out a maneuver that will take it directly over the submarine at some time. However, there is no way to solve for the specific time, so it is unknown when this will occur.
(a) Observe that each bug follows the same curve of pursuit relative to the center from which it starts. Place a polar coordinate system as suggested
and determine the pursuit curve for the bug starting at θ = 0, r = a/ 2. At t > 0, the bug will be at (f(θ),θ) and its target is at (f(θ),θ + π/2).
√
22.
√
Show that
dy = dy/dθ = f′(θ)sin(θ)+f(θ)cos(θ).
dx dx/dθ f′(θ)cos(θ)−f(θ)sin(θ)
At the same time, the direction of the tangent must be from the position
(f (θ), θ) to the target location (f (θ), θ + π/2), so we also have
dy = f(θ)sin(θ+π/2)−f(θ)sin(θ) dx f (θ) cos(θ + π/2) − f (θ) cos(θ)
= cos(θ) − sin(θ) = sin(θ) − cos(θ) . − sin(θ) − cos(θ) sin(θ) + cos(θ)
Equate these two expressions for dy/dx and rearrange terms to get f′(θ) + f(θ) = 0.
√
Further, f(0) = a/ 2. This is a separable, and also linear, differential equation, and the initial value problem has the solution
a −θ r=f(θ)=√e .
2
This is the pursuit curve (in polar coordinates).
(b) The distance traveled is
∞
r2 + (r′)2 dθ
0
∞a −θ2 a −θ21/2
= √e +−√e dθ 022
∞ 0
e−θ dθ = a.
The actual distance between pursuer and quarry is ae−θ.
= a
(c) Because r = ae−θ/√2 > 0 for all θ, no bug actually catches its quarry.
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Chapter 2
Second-Order Differential Equations
2.1
1.
The Linear Second-Order Equation
It is a routine exercise in differentiation to show that y1(x) and y2(x) are solutions of the homogeneous equation, while yp(x) is a solution of the nonhomogeneous equation. The Wronskian of y1(x) and y2(x) is
sin(6x) cos(6x) 2 2 W(x)=6cos(6x) −6sin(6x)=−6sin (x)−6sin (x)=−6,
and this is nonzero for all x, so these solutions are linearly independent on the real line. The general solution of the nonhomogeneous differential equation is
y=c1sin(6x)+c2cos(6x)+ 1(x−1). 36
For the initial value problem, we need
so c2 = −179/36. And
y(0)=c2 − 1 =−5 36
y′(0)=2=6c1 + 1 36
so c1 = 71/216. The unique solution of the initial value problem is y(x) = 71 sin(6x) − 179 cos(6x) + 1 (x − 1).
2.
216
The Wronskian of e4x and e−4x is
e 4 x W(x) = 4e4x
36 36
e − 4 x
−4e−4x = −8 ̸= 0
37
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38
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
so these solutions of the associated homogeneous equation are indepen- dent. With the particular solution yp(x) of the nonhomogeneous equation, this equation has general solution
y(x)=c1e4x+c2e−4x−1×2− 1. 4 32
From the initial conditions we obtain
y(0)=c1+c2− 1 =12 32
and
Solve these to obtain c1 = 409/64 and c2 = 361/64 to obtain the solution
y′(0) = 4c1 − 4c2 = 3.
y(x)= 409e4x + 361e−4x − 1×2 − 1 .
3.
64 64 4 32
The associated homogeneous equation has solutions e−2x and e−x. Their
Wronskian is
e−2x e−x −3x W (x) = −2e−2x −e−x = e
and this is nonzero for all x. The general solution of the nonhomogeneous differential equation is
y(x) = c1e−2x + c2e−x + 15. 2
For the initial value problem, solve
y(0) = −3 = c1 + c2 + 15 2
and
to get c1 = 23/2, c2 = −22. The initial value problem has solution
y′(0)=−1=−2c1 −c2 y(x) = 23e−2x − 22e−x + 15.
4.
22 The associated homogeneous equation has solutions
y1(x) = e3x cos(2x), y2(x) = e3x sin(2x). The Wronskian of these solutions is
e3x cos(2x) e3x sin(2x)
6x W (x) = 3e3x cos(2x) − 2e3x sin(2x) 3e3x sin(2x) + 2e3x cos(2x) = e
̸= 0
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2.1. THE LINEAR SECOND-ORDER EQUATION 39 for all x. The general solution of the nonhomogeneous equation is
y(x) = c1e3x cos(2x) + c2e3x sin(2x) − 1ex. 8
To satisfy the initial conditions, it is required that
and
y(0) = −1 = c1 − 1 8
3c1 + 2c2 − 1 = 1. 8
Solve these to obtain c1 = −7/8 and c2 = 15/8. The solution of the initial value problem is
y(x) = −7e3x cos(2x) + 15e3x sin(2x) − 1ex. 888
5. The associated homogeneous equation has solutions y1(x) = ex cos(x), y2(x) = ex sin(x).
These have Wronskian
ex cos(x) ex sin(x) 2x
̸= 0
so these solutions are independent. The general solution of the nonhomo-
W(x) = ex cos(x) − ex sin(x) ex sin(x) + ex cos(x) = e
geneous differential equation is
We need and
y(x) = c1ex cos(x) + c2ex sin(x) − 5c2 − 5x − 5. 22
y(0) = c1 − 5 = 6 2
y′(0)=1=c1 +c2 −5.
Solve these to get c1 = 17/2 and c2 = −5/2 to get the solution
y(x) = 17ex cos(x) − 5ex sin(x) − 5×2 − 5x − 5. 2222
6. Suppose y1 and y2 are solutions of the homogeneous equation (2.2). Then y′′+py′ +qy =0
and
y′′+py′ +qy =0. 222
111
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40
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
Multiply the first equation by y2 and the second by −y1 and add the resulting equations to obtain
y′′y −y′′y +p(y′y −y′y )=0. 12211221
We want to relate this equation to the Wronskian of these solutions, which is
W = y 1 y 2′ − y 2 y 1′ . W′ =y y′′ −y y′′.
W ′ + pW = 0.
This is a linear first-order differential equation for W . Multiply this equa-
tion by the integrating factor
e p(x) dx
We p(x)dx +pWe p(x)dx = 0,
Now Then
12 21
to obtain
which we can write as
Integrate this to obtain with k constant. Then
p(x)dx′
W e = 0.
We p(x)dx =k, W(x)=ke− p(x)dx.
This shows that W(x) = 0 for all x (if k = 0), and W(x) ̸= 0 for all x (if k ̸= 0).
Now suppose that y1 and y2 are independent and observe that d y2 y1y2′ −y2y1′ 1
dx y = y2 = y2W(x). 111
If k = 0, then W(x) = 0 for all x and the quotient y2/y1 has zero derivative
and so is constant:
y2 =c y1
for some constant c. But then y2(x) = cy1(x), contradicting the assump- tion that these solutions are linearly independent. Therefore k ̸= 0 and W(x) ̸= − for all x, as was to be shown.
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2.2. THE CONSTANT COEFFICIENT HOMOGENEOUS EQUATION 41
7. The Wronskian of x2 and x3 is
x2 x3 4 W(x)=2x 3×2=x .
Then W(0) = 0, while W(x) ̸= 0 if x ̸= 0. This is impossible if x2 and x3 are solutions of equation (2.2) for some functions p(x) and q(x). We conclude that these functions are not solutions of equation (2.2).
8. It is routine to verify that y1(x) and y2(x) are solutions of the differential equation. Compute
x x2 2 W(x)=1 2x=x.
Then W (0) = 0 but W (x) > 0 if x ̸= 0. However, to write the differential equation in the standard form of equation (2.2), we must divide by x2 to
obtain
y′′−2y′+ 2y=0. x x2
This is undefined at x = 0, which is in the interval −1 << 1, so the theorem does not apply.
9. If y2(x) and y2(x) both have a relative extremum (max or min) at some
x0 within (a, b), then
But then the Wronskian of these functions vanishes at 0, and these solu-
tions must be independent.
10. By assumption, φ(x) is the unique solution of the initial value problem
y′′ + py′ + qy = 0; y(x0) = 0.
But the function that is identically zero on I is also a solution of this initial value problem. Therefore these solutions are the same, and φ(x) = 0 for all x in I.
11. If y1(x0) = y2(x0) = 0, then the Wronskian of y1(x) and y2(x) is zero at x0, and these two functions must be linearly dependent.
2.2 The Constant Coefficient Homogeneous Equa- tion
1. From the differential equation we read the characteristic equation λ2 −λ−6=0,
which has roots −2 and 3. The general solution is y(x) = c1e−2x + c2e3x.
y′(x0) = y2′ (x0) = 0.
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42 2.
3.
4.
5.
6.
7.
8.
9.
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
The characteristic equation is
λ2 −2λ+10=0
with roots 1 ± 3i. We can write a general solution y(x) = c1ex cos(3x) + c2ex sin(3x).
The characteristic equation is
λ2 +6λ+9=0
with repeated roots −3, −3. Then
y(x) = c1e−3x + c2xe−3x
is a general solution.
The characteristic equation is
with roots 0, 3, and is a general solution.
λ2 − 3λ = 0 y(x) = c1 + c2e3x
characteristic equation λ2 + 10λ + 26 = 0, with roots −5 ± i; general solution
y(x) = c1e−5x cos(x) + c2e−5x sin(x).
characteristic equation λ2 +6λ−40 = 0, with roots 4, −10; general solution y(x) = c1e4x + c2e−10x.
characteristic equation λ2 +3λ+18 = 0, with roots −3/2±3√7i/2; general solution
3√7x 3√7x y(x) = c2e−3x/2 cos 2 + c2e−3x/2 sin 2 .
characteristic equation λ2 + 16λ + 64 = 0, with repeated roots −8, −8; general solution
y(x) = e−8x(c1 + c2x).
characteristic equation λ2 − 14λ + 49 = 0, with repeated roots 7, 7; general
solution
y(x) = e7x(c1 + c2x).
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2.2. THE CONSTANT COEFFICIENT HOMOGENEOUS EQUATION 43 10. characteristic equation λ2 −6λ+7 = 0, with roots 3±√2i; general solution
y(x) = c1e3x cos(√2x) + c2e3x sin(√2x).
In each of Problems 11–20 the solution is found by finding a general solution of the differential equation and then using the initial conditions to find the particular solution of the initial value problem.
11. The differential equation has characteristic equation λ2 + 3λ = 0, with roots 0, −3. The general solution is
y(x) = c1 + c2e−3x. Choose c1 and c2 to satisfy:
y(0) = c1 + c2 = 3, y′(0) = −3c2 = 6.
Then c2 = −2 and c1 = 5, so the unique solution of the initial value problem is
y(x) = 5 − 2e−3x.
12. characteristic equation λ2 + 2λ − 3 = 0, with roots 1, −3; general solution
y(x) = c1ex + c2e−3x. y(0)=c1 +c2 =6,y′(0)=c1 −3c2 =−2
Solve
togetc1 =4andc2 =2. Thesolutionis
y(x) = 4ex + 2e−3x.
13. The initial value problem has the solution y(x) = 0 for all x. This can be seen by inspection or by finding the general solution of the differential equation and then solving for the constants to satisfy the initial conditions.
14. y(x) = e2x(3 − x)
15. characteristic equation λ2 + λ − 12 = 0, with roots 3, −4. The general
solution is
We need
and
Solve these to obtain
y(x) = c1e3x + c2e−4x. y(2) = c1e6 + c2e−8 = 2 y′(2) = 3c1e6 − 4c2e−8 = −1. c1 = e−6,c2 = e8.
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44
16.
17. 18.
19.
20.
where
root. The general solution is
y(x) = (c1 + c2x)eαx.
(b) The characteristic equation is λ2 − 2αλ + (α2 − ε2) = 0, with roots α + ε, α − ε. The general solution is
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
The solution of the initial value problem is
y(x) = e−6e3x + e8e−4x.
This can also be written
y(x) = e3(x−2) + e−4(x−2).
√
6 x √6x y(x)=4ee −e
−√6x
y(x) = ex−1(29 − 17x) y(x) =
8 √ 5x/2 √
√ sin( 23)e cos( 23x/2)
e5 23
8 √ 5x/2 √
− √ cos( 23)e sin( 23x/2) e5 23
y(x) = e(x+2)/2 cos(√15(x + 2)/2)
5 + √
√
sin( 15(x + 2)/2)
15
y(x) = ae(−1 + √5)x/2 + be(−1−√5)x/2,
9+7√5 √ 7√5−9 √ √ e−2+ 5andb= √ e−2− 5
21.
a=
(a) The characteristic equation is λ2 − 2αλ + α2 = 0, with α as a repeated
We can also write
25 25
yε(x) = c1e(α+ε)x + c2e(α−ε)x. yε(x) = c1eεx + c2e−εx eαx.
In general,
Note, however, that the coefficients in the differential equations in (a) and
(b) can be made arbitrarily close by choosing ε sufficiently small.
lim yε(x) = (c1 + c2)eαx ̸= y(x). ε→0
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2.2. THE CONSTANT COEFFICIENT HOMOGENEOUS EQUATION 45 22. With a2 = 4b, one solution is y1(x) = e−ax/2. Attempt a second solution
y2(x) = u(x)e−ax/2. Substitute this into the differential equation to get ′′ ′ a2 ′ a −ax/2
u −au + 4u+a u −2u +bu e =0.
Because a2 − 4b = 0, this reduces to u′′(x) = 0.
Then u(x) = cx + d, with c and d arbitrary constants, and the functions (cx + d)e−ax/2 are also solutions of the differential equation. If we choose c = 1 and d = 0, we obtain y2(x) = xe−ax/2 as a second solution. Further, this solution is independent from y1(x), because the Wronskian of these solutions is
e−ax/2 W (x) =
xe−ax/2
= e
−ax
,
e−ax/2 − (a/2)xe−ax/2 23. The roots of the characteristic equation are
√√ λ1=−a+ a2−4bandλ2=−a− a2−4b.
22
−(a/2)e−ax/2 and this is nonzero.
Because a2 − 4b < a2 by assumption, λ1 and λ2 are both negative (if a2 − 4b ≥ 0), or complex conjugates (if a2 − 4b < 0). There are three cases.
Case 1 - Suppose λ1 and λ2 are real and unequal. Then the general solution is
y(x) = c1eλ1x + c2eλ2x
and this has limit zero as x → ∞ because λ1 and λ2 are negative.
Case 2 - Suppose λ1 = λ2. Now the general solution is y(x) = (c1 + c2x)eλ1x,
and this also has limit zero as x → ∞.
Case 3 - Suppose λ1 and λ2 are complex. Now the general solution is
y(x) = c1 cos(4b − a2x/2) + c2 sin(4b − a2x/2) e−ax/2,
and this has limit zero as x → ∞ because a > 0.
If, for example, a = 1 and b = −1, then one solution is e(−1+ 5)x/2, and
this tends to ∞ as x → ∞.
√
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46
2.3
1.
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
Particular Solutions of the Nonhomogeneous Equation
Two independent solutions of y′′ + y = 0 are y1(x) = cos(x) and y2(x) = sin(x), with Wronskian
u1(x) = −
tan(x) sin(x) dx
= −
sin2 (x) cos(x) dx
cos(x) sin(x)
W (x) = − sin(x) cos(x) = 1.
Let f(x) = tan(x) and use equations (2.7) and (2.8). First,
y2(x)f(x) W (x) = −
1 − cos2(x) =− cos(x) dx
= cos(x) dx − sec(x) dx = sin(x) − ln | sec(x) + tan(x)|.
The general solution is
y(x) = c1 cos(x) + c2 sin(x) + u1(x)y1(x) + u2(x)y2(x)
= c1 cos(x) + c2 sin(x) − cos(x) ln | sec(x) + tan(x)|.
Two independent solutions of the associated homogeneous equation are y1(x) = e3x and y2(x) = ex. Their Wronskian is W(x) = −2e4x. Compute
Next,
y1(x)f(x) u2 (x) = W (x) dx =
cos(x) tan(x) dx = sin(x) dx = − cos(x).
2.
u1(x) = −
2ex cos(x + 3) −2e4x dx
e−3x cos(x + 3) dx =−3e−3xcos(x+3)+ 1e−3xsin(x+3)
=
10 10
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2.3. PARTICULARSOLUTIONSOFTHENONHOMOGENEOUSEQUATION47 and
u2(x) = =
2e3x cos(x + 3) −2e4x dx
e−x cos(x + 3) dx
= 1 e−x cos(x + 3) − 1 e−x cos(x + 3).
The general solution is
y(x) = c1e3x + c2ex
− 3 cos(x+3)+ 1 sin(x+3) 10 10
+ 1 cos(x + 3) − 1 sin(x + 3). 22
More compactly, the general solution is
y(x) = c1e3x + c2ex + 1 cos(x + 3) − 2 sin(x + 3). 55
For Problems 3–6, some details of the calculations are omitted.
3. The associated homogeneous equation has independent solutions y1(x) =
22
cos(3x) and y2(x) = sin(3x), with Wronskian 3. The general solution is y(x) = c1 cos(3x) + c2 sin(3x) + 4x sin(3x) + 4 cos(3x) ln | cos(3x)|.
3
4. y1(x) = e3x and y2(x) = e−x, with W(x) = −4e−2x. With
f (x) = 2 sin2 (x) = 1 − cos(2x) we find the general solution
y(x) = c1e3x + c2e−x − 1 + 7 cos(2x) + 4 sin(2x). 365 65
5. y1(x) = ex and y2(x) = e2x, with Wronskian W(x) = e3x. With f(x) = cos(e−x), we find the general solution
y(x) = c1ex + c2e2x − e2x cos(e−x).
6. y1(x) = e3x and y2(x) = e2x, with Wronskian W(x) = e−5x. Use the
identity
in determining u1(x) and u2(x) to write the general solution
8 sin2(4x) = 4 cos(8x) − 1
y(x) = c1e3x + c2e2x + 2 + 58 cos(8x) + 40 sin(8x).
3 1241 1241
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48 CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
In Problems 7–16 the method of undetermined coefficients is used to find a particular solution of the nonhomogeneous equation. Details are included for Problems 7 and 8, and solutions are outlined for the remainder of these problems.
7. The associated homogeneous equation has independent solutions y1(x) = e2x and e−x. Because 2×2 +5 is a polynomial of degree 2, attempt a second degree polynomial
yp(x)=Ax2 +Bx+C
for the nonhomogeneous equation. Substitute yp(x) into this nonhomoge-
neous equation to obtain
2A−(2Ax+B)−2(Ax2 +Bx+C)=2×2 +5.
Equating coefficients of like powers of x on the left and right, we have the equations
−2A = 2( coefficients of x2) −2A − 2B − 0( coefficients of x 2A − 2B − 2C = 5( constant term.)
ThenA=−1,B=1andC=−4. Then yp(x)=−x2 +x−4
and a general solution of the (nonhomogeneous) equation is y = c1e2x +c2e−x −x2 +x−4.
8. y1(x) = e3x and y2(x) = e−2x are independent solutions of the associated homogeneous equation. Because e2x is not a solution of the homogeneous equation, attempt a particular solution yp(x) = Ae2x of the nonhomoge- neous equation. Substitute this into the differential equation to get
4A − 2A − 6A = 8, so A = −2 and a general solution is
y(x) = c1e3x + c2e−2x − 2e2x.
9. y1(x) = ex cos(3x) and y2(x) = ex sin(3x) are independent solutions of the associated homogeneous equation. Try a particular solution yp(x) = Ax2 + Bx + C to obtain the general solution
y(x) = c1ex cos(3x) + c2ex sin(3x) + 2×2 + x − 1.
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2.3. PARTICULARSOLUTIONSOFTHENONHOMOGENEOUSEQUATION49
10. For the associated homogeneous equation, y1(x) = e2x cos(x) and y2(x) =
e2x sin(x). Try yp(x) = Ae2x to get A = 21 and obtain the general solution y(x) = c1e2x cos(x) + c2e2x sin(x) + 21e2x.
11. For the associated homogeneous equation, y1(x) = e2x and y2(x) = e4x. Because ex is not a solution of the homogeneous equation, attempt a particular solution of the nonhomogeneous equation of the form yp(x) = Aex. We get A = 1, so a general solution is
y(x) = c1e2x + c2e4x + ex.
12. y1(x) = e−3x and y2(x) = e−3x. Because f(x) = 9cos(3x) (which is not a solution of the associated homogeneous equation), attempt a particular solution
yp (x) = A cos(3x) + B sin(3x).
This attempt includes both a sine and cosine term even though f(x) has only a cosine term, because both terms may be needed to find a particular solution. Substitute this into the nonhomogeneous equation to obtain A = 0 and B = 1/2, so a general solution is
y(x) = (c1 + c2x)e−3x + 1 sin(3x). 2
In this case yp(x) contains only a sine term, although f(x) has only the cosine term.
13. y1(x) = ex and y2(x) = e2x. Because f(x) = 10sin(x), attempt yp (x) = A cos(x) + B sin(x).
Substitute this into the (nonhomogeneous) equation to find that A = 3 and B = 1. A general solution is
y(x) = c1ex + c2e2x + 3 cos(x) + sin(x).
14. y1(x) = 1 and y2(x) = e−4x. Finding a particular solution yp(x) for this problem requires some care. First, f(x) contains a polynomial term and an exponential term, so we are tempted to try yp(x) as a second degree polynomial Ax2 + Bx + C plus an exponential term De3x to account for the exponential term in the equation. However, note that y1(x) = 1, a constant solution, is one term of the proposed polynomial part, so multiply this part by x to try
yp(x)=Ax3 +Bx2 +Cx+De3x.
Substitute this into the nonhomogeneous differential equation to get
6Ax+2B+9De3x −4(3Ax2 +2Bx+C+3De3x)=8×2 +2e3x.
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50
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
Matching coefficients of like terms, we conclude that
2B − 4C = 0( from the constant terms) 6A − 8B = 0( from the x terms),
−12A = 8( from the x2 terms)
−3D = 2. ThenD=−2,A=−2,B=−1 andC=−1. Then
3324
yp(x) = −2×3 − 1×2 − 1x − 2e3x
3243
and a general solution of the nonhomogeneous equation is
y(x) = c1 + c2e−4x − 2×3 − 1×2 − 1x − 2e3x. 3243
15. y1(x) = e2x cos(3x) and y2(x) = e2x sin(3x). Try ypx = Ae2x + Be3x.
This will work because neither e2x nor e3x is a solution of the associated homogeneous equation. Substitute yp(x) into the differential equation and obtain A = 1/3, B = −1/2. The differential equation has general solution
y(x) = [c1 cos(3x) + c2 sin(3x)]e2x + 1e2x − 1e3x. 32
16. y1(x) = ex and y2(x) = xex. Try
yp(x) = Ax + B + C cos(3x) + D sin(3x).
This leads to the general solution
y(x) = (c1 + c2x)ex + 3x + 6 + 3 cos(3x) − 2 sin(3x). 2
In Problems 17–24 the strategy is to first find a general solution of the dif- ferential equation, then solve for the constants to find a solution satisfying the initial conditions. Problems 17–22 are well suited to the use of undetermined co- efficients, while Problems 23 and 24 can be solved fairly directly using variation of parameters.
17. y1 (x) = e2x and y2 (x) = e−2x . Because e2x is a solution of the asso- ciated homogeneous equation, use xe2x in the method of undetermined coefficients, attempting
yp(x) = Axe2x + Bx + C.
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2.3. PARTICULARSOLUTIONSOFTHENONHOMOGENEOUSEQUATION51
Substitute this into the nonhomogeneous differential equation to obtain 4Axe2x + 4Axe2x − 4Axe2x − 4Bx − 4C = −7e2x + x.
Then A = −7/4, B = −1/4 and C = 0, so the differential equation has the general solution
We need and
y(x) = c1e2x + c2e−2x − 7xe2x − 1x. 44
y(0) = c1 + c2 = 1
y′(0) = 2c1 − 2c2 − 7 − 1 = 3.
44
Then c1 = 7/4 and c2 = −3/4. The initial value problem has the unique
solution
y(x) = 7e2x − 3e−2x − 7xe2x − 1x. 4444
18. y1 = 1 and y2(x) = e−4x are independent solutions of the associated homogeneous equation. For yp(x), try
yp(x) = Ax + B cos(x) + C sin(x),
with the term Ax because 1 is a solution of the homogeneous equation.
This leads to the general solution
y(x)=c1 +c2e−4x +2x−2cos(x)+8sin(x).
Now we need
and
y(0)=c1 +c2 −2=3 y′(0)=−4c2 +2+8=2.
Then c1 = 3 and c2 = 2, so the initial value problem has the solution
y(x)=3+2e−4x +2x−2cos(x)+8sin(x). 19. We find the general solution
y(x)=c1e−2x+c2e−6x+1e−x+ 7. 5 12
The solution of the initial value problem is
y(x)=3e−2x− 19e−6x+1e−x+ 7. 8 120 5 12
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52 CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS 20. 1 and e3x are independent solutions of the associated homogeneous equa-
tion. Attempt a particular solution
yp(x) = Ae2x cos(x) + Be2x sin(x)
of the nonhomogeneous equation to find the general solution y(x)=c1 +c2e3x − 1e2x(cos(x)+3sin(x)).
5
The solution of the initial value problem is
y(x) = 1 + e3x − 1 (cos(x) + 3 sin(x)). 55
21. e4x and e−2x are independent solutions of the associated homogeneous equation. The nonhomogeneous equation has general solution
y(x) = c1e4x + c2e−2x − 2e−x − e2x. The solution of the initial value problem is
y(x) = 2e4x + 2e−2x − 2e−x − e2x. 22. The general solution of the differential equation is
√3 √3 y(x)=ex/2 c1cos 2 x +c2sin 2 x +1.
It is easier to fit the initial conditions specified at x = 1 if we write this general solution as
√3 √3 y(x)=ex/2 d1cos 2 (x−1) +d2sin 2 (x−1) +1.
Now
√
y(1) = e1/2d1 + 1 = 4 and y′(1) = 1e1/2d1 + 3e1/2d2 = −2.
Solve these to obtain d1 = 3e−1/2 and d2 = −7e−1/2/√3. The solution of the initial value problem is
√3 7 √3 y(x)=e(x−1)/2 3cos 2 (x−1) −√3sin 2 (x−1) +1.
23. The differential equation has general solution
y(x) = c1ex + c2e−x − sin2(x) − 2.
The solution of the initial value problem is
y(x) = 4e−x − sin2(x) − 2.
22
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2.4. THE EULER DIFFERENTIAL EQUATION 53
24. The general solution is
y(x) = c1 cos(x) + c2 sin(x) − cos(x) ln | sec(x) + tan(x)|,
and the solution of the initial value problem is
y(x) = 4 cos(x) + 4 sin(x) − cos(x) ln | sec(x) + tan(x)|.
2.4 The Euler Differential Equation
Details are included with solutions for Problems 1–2, while just the solutions are given for Problems 3–10. These solutions are for x > 0.
1. Read from the differential equation that the characteristic equation is r2 + r − 6 = 0
with roots 2, −3. The general solution is
y(x) = c1x2 + c2x−3.
2. The characteristic equation is
r2 + 2r + 1 = 0
with repeated root −1, −1. The general solution is y(x) = c1x−1 + c2x−1 ln(x).
We can also write
for x > 0. 3.
4.
5.
6.
7.
y(x) = 1 (c1 + c2 ln(x)) x
y(x) = c1 cos(2 ln(x)) + c2 sin(2 ln(x)) y(x)=cx2+c 1
1 1 x2 y(x)=cx2+c 1
1 1 x4
y(x)= 1 (c2cos(3ln(x))+c2sin(3ln(x)))
x2
y(x)=c 1 +c 1
1 x2 2 x3
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54 CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS 8.
9.
10.
y(x) = x2(c1 cos(7 ln(x)) + c2 sin(7 ln(x))) y(x)= 1(c1+c2ln(x))
x12
y(x) = c1x7 + c2x5
11. The general solution of the differential equation is
y(x) = c1x3 + c2x−7. From the initial conditions, we need
y(2)=8c1+2−7c2 =1andy′(2)=3c122−7c22−8 =0. Solve for c1 and c2 to obtain the solution of the initial value problem
7 x3 3 x−7 y(x)=10 2 +10 2 .
12. The initial value problem has the solution y(x) = −3 + 2×2.
13. y(x) = x2(4 − 3 ln(x))
14. y(x) = −4x−12(1 + 12 ln(x))
15. y(x) = 3×6 − 2×4
16.
y(x) = 11×2 + 17x−2 44
17. With Y (t) = y(et), use the chain rule to get y′(x)= dY dt = 1Y′(t)
and then
dtdx x ′′ d1′
y(x)=dx xY(t)
=− 1 Y′(t)+ 1 d (Y′(t))
x2 x dx =−1Y′(t)+1dY′ dt
x2 x dt dx = − 1 Y ′(t) + 1 1 Y ′′(t)
x2 xx
= 1(Y′′(t)−Y′(t)).
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x2
2.4. THE EULER DIFFERENTIAL EQUATION 55 Then
x2y′′(x) = Y ′′(t) − Y ′(t). Substitute these into Euler’s equation to get
Y ′′(t) + (A − 1)Y ′(t) + BY (t) = 0.
This is a constant coefficient second-order homogeneous differential equa-
tion for Y (t), which we know how to solve.
18. Ifx<0,lett=ln(−x)=ln|x|. Wecanalsowritex=−et. Notethat
dt = 1 (−1) = 1 dx−x x
just as in the case x > 0. Now let y(x) = y(−et) = Y (t) and proceed with chain rule differentiations as in the solution of Problem 17. First,
and
Then
y′(x)= dY dt = 1Y′(t) dtdx x
′′ d1′ y(x)=dx xY(t)
=− 1 Y′(t)+ 1 dtY′′(t) x2 x dx
=− 1 Y′(t)+ 1 Y′′(t). x2 x2
x2y′′(x) = Y ′′(t) − Y ′(t)
just as we saw with x > 0. Now Euler’s equation transforms to
Y′′ +(A−1)Y′ +BY =0.
We obtain the solution in all cases by solving this linear constant coefficient second-order equation. Omitting all the details, we obtain the solution of Euler’s equation for negative x by replacing x with |x| in the solution for positive x. For example, suppose we want to solve
x2y′′ + xy′ + y = 0 forx<0. Solvethisforx>0toget
y(x) = c1 cos(ln(x)) + c2 sin(ln(x)). The solution for x < 0 is
y(x) = c1 cos(ln |x|) + c2 sin(ln |x|).
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56 19.
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
The problem to solve is
x2y′′ −5dxy′ +10y=0;y(1)=4,y′(1)=−6.
We know how to solve this problem. Here is an alternative method, us- ing the transformation x = et, or t = ln(x) for x > 0 (since the initial conditions are specified at x = 1). Euler’s equation transforms to
Y′′ −6Y′ +10Y =0. However, also transform the initial conditions:
Y (0) = y(1) = 4, Y ′(0) = (1)y′(1) = −6. This differential equation for Y (t) has general solution
20.
so c2 = −18. The solution of the transformed initial value problem is
Y (t) = 4e3t cos(t) − 18e3t sin(t).
The original initial value problem therefore has the solution
y(x) = 4×3 cos(ln(x)) − 19×3 sin(ln(x))
for x > 0. The new twist here is that the entire initial value problem (including initial conditions) was transformed in terms of t and solved for Y (t), then this solution Y (t) in terms of t was transformed back to the solution y(x) in terms of x.
Suppose
x2y′′ + Axy′ + By = 0
has repeated roots. Then the characteristic equation
r2 + (A − 1)r + B = 0
has (1 − A)/2 as a repeated root, and we have only one solution y1(x) = x(1−A)/2 so far. For another solution, independent from y1, look for a solution of the form y2(x) = u(x)y1(x). Then
Now
and
Y (t) = c1e3t cos(t) + c2e3t sin(t). Y(0)=c2 =4
Y′(0)=3c1 +c2 =−6,
and
y2′ =u′y1+uy1′
y′′ = u′′y + 2u′y′ + uy′′.
2111
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2.4. THE EULER DIFFERENTIAL EQUATION 57 Substitute y2 into the differential equation to get
x2(u′′y +2u′y′ +uy′′)+Ax(u′y +uy′)+Buy =0. 111111
Three terms in this equation cancel, because u(x2y′′ + Axy′ + By ) = 0
111 by virtue of y1 being a solution. This leaves
x2u′′y1 + 2x2u′y1′ + Axu′y1 = 0. Assuming that x > 0, divide by x to get
xu′′y1 + 2xu′y1′ + Au′y1 = 0. Substitute y1(x) = x(1−A)/2 into this to obtain
′′ (1−A)/2 ′ 1 − A (−1−A)/2 ′ (1−A)/2 xu x +2xu 2 x +Aux
Divide this by x(1−A)/2 to get
xu′′ +(1−A)u′ +Au′ = 0,
=0.
and this reduces to
Let z = u′ to obtain
or
Then xz = c, constant, so
xu′′ + u′ = 0. xz′ + z = 0, (xz)′ = 0.
z = u′ = c . x
Then u(x) = c ln(x) + d. We only need one second solution, so let c = 1 and d = 0 to get u(x) = ln(x). A second solution, independent from y1(x), is
y2(x) = y1(x) ln(x), as given without derivation in the chapter.
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58
2.5
2.5.1
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
Series Solutions
Power Series Solutions
1. Put y(x) = ∞n=0 anxn into the differential equation to obtain
∞∞
y′ − xy = nanxn−1 − anxn+1
n=1 n=0 ∞∞
= nanxn−1 − an−2xn−1 n=1 n=2
∞
= a1 + (2a2 − a0)x + (nan − an−2)xn−1
n=3
= 1 − x.
Then a0 is arbitrary, a1 = 1, 2a2 − a0 = −1, and
an = 1 an−2 for n = 3, 4, · · · . n
This is the recurrence relation. If we set a0 = c0 + 1, we obtain the
coefficients
and so on. Further,
a2=1c0,a4= 1 c0,a6= 1 c0, 2 2·4 2·4·6
a1=1,a3=1,a5= 1,a7= 1
3 3·5 3·5·7
and so on. The solution can be written
y(x) = 1 + ∞ 1 x2n+1
+ c0
3·5···2n+1 ∞1
n=0
1 +
x2n .
= a1 + 2a2x + 3a3x2 + (nan − an−4)xn−1 = 4. n=4
The recurrence relation is
an = 1 an−4 for n = 4, 5, · · · ,
n
2. Write
y′ − x3y = nanxn−1 − anxn+3
n=1 n=0
∞
∞∞
2·4···2n
n=1
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2.5. SERIES SOLUTIONS 59 with a0 arbitrary, a1 = 4 and a2 = a3 = 0. This yields the solution
y(x) = 4 ∞ 1 x4n+1
1·5·9···(4n+1) ∞1
n=0
+ a0 1 +
n=1
3. Write
y′ +(1−x2)y=nanxn−1 +anxn −anxn+2
n=1 n=0 n=0 ∞
= (a1 + a0) + (2a2 + a1)x + (nan + an−1 − an−3)xn−1 n=3
= x.
The recurrence relation is
nan+an−1−an−3 =0forn=3,4,···.
Here a0 is arbitrary, a1 +a0 = 0 and 2a2 +a1 = 1. This gives us the
4. Begin with
∞∞∞
y′′ +2y′ −xy=n(n−1)anxn−2 +2nanxn−1 +anxn+1
n=2 n=1 n=0 =(2a2 +2a1)+(3·2a3 +2·2a2 +a0)x
∞
+ (n(n − 1)an + 2(n − 1)an−1 + an−2)xn−2 = 0.
n=1
The recurrence relation is
n(n−1)an+2(n−1)an−1+an−2 =0forn=4,5,···. Further, a0 and a1 are arbitrary, a2 = −A1 and
6a3 +4a2 +a0 =0. Taking a0 = 1, a1 = 0, we obtain the solution
y1(x)=1−1×3+ 1×4− 1×5+ 1×6+···. 6 12 30 60
∞∞∞
solution
y(x)=a0 1−x+2!x +3!x −4!x + 5!x +···
4·8·12···4n
x4n .
121374195 + 1×2− 1×3+ 1×4+11×5−31×6+···.
2! 3! 4! 5! 6!
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60
5.
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
With a0 = 0, a1 = 1 we get a second, linearly independent solution y2(x)=x−x2+2×3− 5×4+ 7×5+···.
is
an+2 = n−1 forn=1,2,···. (n+2)(n+1)
3 12 60
Write
y′′ −xy′ +y = n(n−1)nxn−2 −nanxn +anxn
n=2 n=1 n=0 ∞∞∞
=2a2 +a0 +(n+2)(n+1)an+2xn −nanxn +anxn =3. n=0 n=1 n=0
Here a0 and a1 are arbitrary and a2 = (3−a0)/2. The recurrence relation
∞∞∞
This yields the general solution
y(x) = a0 + a1x + 3 − a0 x2 + 3 − a0 x4 2 4!
+ 3(3−a0)x6 + 3·5(3−a0)x8 +··· . 6! 8!
6.
Begin with
y′′ +xy′ +xy = n(n−1)axnxn−2 +nanxn +anxn+1
n=2 n=1 n=0 =2a2 +(n(n−1)an +(n−2)an−2 +an−3)xn−2 =0.
n=3
Here a0 and a1 are arbitrary and a2 = 0. The recurrence relation is
an=−(n−2)an−2−an−3 forn=3,4,···. n(n−1)
∞
∞∞∞
With a0 = 1 and a1 = 0, we obtain one solution y1(x) = 1 − 2×3 + 3 x5
3 2·3·4·5
+ 1 x6 − 3 · 5 x7 + · · · .
2·3·5·6 2·3·4·5·6·7
With a0 = 0 and a1 = 1, we obtain a second, linearly independent solution
y2(x)=x− 1 x3− 1 x4 2·3 3·4
+ 3 x5 + 3 · 5 x6 + · · · . 2·3·4·5 2·3·5·6
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2.5. SERIES SOLUTIONS
61
7. We have
∞
y′′ −x2y′ +2y = n(n−1)anxn−2
n=2 ∞∞
− nanxn+1 + 2anxn n=1 n=0
= 2a2 +2a0 +(6a3 +2a1)x ∞
+ (n(n − 1)an − (n − 3)an−3 + 2an−2)xn−2 = x. n=1
Then a0 and a1 are arbitrary, a2 = −a0, and 6a3+2a1 = 1. The recurrence relation is
an = (n − 3)an−3 − 2an−2 n(n−1)
for n = 4,5,···. The general solution has the form 2141516
y(x)=a0 1−x +6x −10x −90x +··· 13141576
+a1 x−3x +12x +30z −180x +··· +1×3−1×5+ 1×6+ 1 x7− 1 x8+···.
6 6 60 1260 480
Note that a0 = y(0) and a1 = y′(0). The third series represents the
solution obtained subject to y(0) = y′(0) = 0.
8. Using the Maclaurin expansion for cos(x), we have
∞∞
y′ + xy = nanxn − 1 + anxn+1
n=1
= a1 + (2na2n + a2n−2)x2n−1
n=0 ∞
+ ((2n + 1)a2n+1 + a2n−1)x2n n=1
∞ (−1)n
∞
(2n)! x2n.
a2n = − 1 a2n−2 and a2n+1 = −a2n−1 + (−1)n/((2n)!)
= cos(x) =
a0 is arbitrary and a1 = 1. The recurrence relation is
n=0
n=0
2n 2n+1
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62
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
for n = 1,2,···. This yields the solution
12 142·4·66
y(x)=a0 1−2x +2·4x − x +··· 331357976339
+ x−3!x +5!x −7!x + 9!x +··· .
9. We have
y′′ +(1−x)y′ +2y = n(n−1)anxn−2
n=2 ∞∞∞
+nanxn−1 −nanxn +2anxn
∞
n=1 n=1
=(2a2 +a1 +2a0)+(n(n−1)an +(n−1)an−1 −(n−4)a2n−2)xn−2
n=3
= 1 − x2.
Then a0 and a1 are arbitrary, 2a2 +a1 +2a0 = 1, 6a3 +2a2 +a1 = 0, and
12a4 + 3a3 = −1. The recurrence relation is
an = −(n − 1)an−1 + an−4an−2
n(n−1) for n = 5,6,···. The general solution is
2131415 y(x)=a0 1−x +3x −12x +30x −···
1 2 1 2 1 3 1 4 1 6 1 7
+a1 x−2x +2x −6x −24x −360x +2520x +···.
∞
n=0
Here a0 = y(0) and a1 = y′(0).
10. Using the Maclaurin expansion of ex, we have
∞∞
y′′ +xy′ =n(n−1)anxn−2 +nanxn
n=2
= 2a2 + (n(n − 1)an + (n − 2)an−2)xn−2
n=3
=−∞ 1xn−2.
(n−2)!
Then a0 and a1 are arbitrary, a2 = 0 and
an = −(n−2)an−2 −1/(n−2)! n(n−1)
n=3
∞
n=1
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2.5. SERIES SOLUTIONS
63
for n = 3,4,···. This leads to the solution
13 35 157 1059
y(x)=a0+a1 x−3!x +5!x −7!x + 9!x +··· 13 14 25 36 117 198
+ −3!x −4!x +5!x +6!x −7!x +8!x +··· . Note that a0 = y(0) and a1 = y′(0).
2.5.2 Frobenius Solutions
1. Substitute y(x) = ∞n=0 cnxn+r into the differential equation to get
∞
xy′′ +(1−x)y′ +y = (n+r)(n+r−1)cnxn+r−2
n=0 ∞∞∞
+ (n + r)cnxn+r−1 − (n + r)cnxn+r + cnxn+r
n=0
= r2c0xr−1 + ((n + r)2cn − (n + r − 2)cn−1)xn+r−1
n=1
∞
n=0 n=0
= 0.
Because c0 is assumed to be nonzero, r must satisfy the indicial equation
r2 =0,sor1 =r2 =0. Onesolutionhastheform
∞
y1(x) = cnxn, n=0
while a second solution has the form
y2(x) = y1(x) ln(x) + c∗nxn.
n=0
For the first solution, choose the coefficients to satisfy c0 = 1 and
cn = n − 2 cn−1 for n = 1, 2, · · · . n2
This yields the solution y1(x) = 1 − x. The second solution is therefore
∞
y2(x) = (1 − x) ln(x) + c∗nxn.
n=0
∞
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64
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
Substitute this into the differential equation to obtain 2 1−x 1−x
x −x− x2 +(1−x) −ln(x)+ x
∞∞ +(1−x)ln(x)+n(n−1)c∗nxn−1 +(1−x)c∗nxn−1
∞
+ c ∗n x n
n=2
n=1
n=1
= (−3 + c∗1) + (1 + 4c∗2)x + (n2c∗n − (n − 2)c∗n−1)xn−2
n=3
= 0.
The coefficients are determined by c∗1 = 3, c∗2 = −1/4, and
c∗n = n − 2 for n = 3, 4, · · · . n2
r1 = 1 and r2 = 0. There are solutions
∞∞ y1(x)=cnxn+1 andy2(x)=ky1(x)ln(x)c∗nxn.
n=0 n=0 For y1, the recurrence relation is
cn = 2(n + r − 2) cn−1 (n + r)(n + r − 1)
forn=1,2,···. Withr=1andc0 =1,thisyields y1(x) = x,
a solution that can be seen by inspection from the differential equation. For the second solution, substitute y2(x) into the differential equation to get
(2c∗0 + k) + 2(c∗2 − k)x ∞
+ (n(n − 1)c∗n − 2(n − 2)c∗n−1)xn−1 = 0. n=1
A second solution is
y2(x) = (1 − x) ln(x) + 3x − ∞ 1 xn.
n(n−1)
2. Omitting some routine details, the indicial equation is r(r − 1) = 0, so
∞
n=2
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2.5. SERIES SOLUTIONS 65 Choose c∗0 = 1 to obtain k = −2. c∗1 is arbitrary, and we will take c∗1 = 0.
n(n−1) This yields the second solution
y2 (x) = −2x ln(x) + 1 −
3.Theindicialequationisr2−4r=0,sor1 =4andr2 =0. Thereare solutions of the form
∞∞
y1(x) = cnxn+4 and y2(x) = ky1(x) ln(x) + c∗nxn.
n=0 n=0 With r = 4 the recurrence relation is
cn = n + 1 cn−1 for n = 1, 2, · · · . n
Finally, c∗2 = −2 and
c∗n = 2(n−2)c∗n−1 forn=3,4,···.
Then
Using the geometric series, we can observe that
y1(x)=x4(1+2x+3×2 +4×3 +···). y1(x)=x4 d (1+x+x2 +x3 +···)
∞ 2n
n=2
n!(n − 1) xn .
=x dx 1−x =(1−x)2. This gives us the second solution
dx
4d1 x4
y2(x)= 3−4x. (1−x)2
4. The indicial equation is 4r2 − 9 = 0, with roots r1 = 3/2 and r2 = −3/2. There are solutions
∞∞
y1(x) = cnxn+3/2 and y2(x) = ky1(x) ln(x) + c∗nxn−3/2.
n=0 n=0
Upon substituting these into the differential equation, we obtain
∞
y1 (x) = x3/2 y2 (x) = x−3/2
(−1)n
2n n!(5 · 7 · 9 · · · (2n + 3)) x2n ∞
1 +
1 +
and
n=1
(−1)n+1
n=1
2n+1 n!(3) · · · (2n − 3) x2n .
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66 5.
CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS
The indicial equation is 4r2 − 2r = 0, with roots r1 = 1/2 and r2 = 0. There are solutions of the form
∞∞
y1(x) = cnxn+1/2 and y2(x) = c∗nxn.
n=0 n=0
Substitute these into the differential equation to get
∞
n=1
y2 (x) = 1 +
=1−1x+1×2−1×3+ 1 x4+···.
(−1)n
2n n!(3 · 5 · 7 · · · (2n + 1)) xn
=x 1−6x+120x −5040x +362880x +···
y1 (x) = x1/2 1 +
1/2 1 1 2 1 3 1 4
and
∞ (−1)n
2n n!(1 · 3 · 5 · · · (2n − 1)) xn 2 24 720 40320
n=1
6.
The indicial equation is 4r2 − 1 = 0, with roots r1 = 1/2 and r2 = −1/2. There are solutions
∞∞
y1(x) = cnxn+1/2 and y2(x) = ky1(x) ln(x) + c∗nxn−1/2.
n=0 n=0
After substituting these into the differential equation, we obtain the simple
solutions
y1(x) = x1/2 and y2(x) = x−1/2.
These solutions are consistent with the observation that, upon division by
4, the differential equation is an Euler equation.
The indicial equation is r2 −3r+2 = 0, with roots r1 = 2 and r2 = 1. There are solutions
∞∞
y(x) = cnxn+2 and c∗nxn−2. n=0 n=0
Substitute these in turn into the differential equation to obtain the solu-
7.
tions
and
y1(x)=x2 + 1×4 + 1×6 + 1×8 +··· 3! 5! 7!
y2(x) = x − x2 + 1 x3 − 1 x4 + 1 x5 − · · · . 2! 3! 4!
We can recognize these series as
y1(x) = x sinh(x) and y2(x) = xe−x.
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2.5. SERIES SOLUTIONS 67 8. The indicial equation is r2 − 2r = 0, with roots r1 = 2, r2 = 0. There are
solutions
y1(x) = cnxn+2 and y2(x) = ky1(x) ln(x) + c∗nxn.
n=0 n=0 The recurrence relation for the c′ns is
cn = −2 for n = 3,4,··· n(n−2)
and we obtain, with c0 = 1,
∞ (−1)n 2n+1
∞∞
n(n+2) xn+2
3 6 45 540
y1(x)=
=x2−2×3+1×4− 1×5+ 1 x6+···.
n=0
For the second solution, substitute y2(x) into the differential equation to get
∗ ∗ ∞ ∗ ∗ (−1)n2nk n−1
n(n − 2)cn + cn−1 + n((n − 2)!)2 x = 0. take this to be zero), and the recurrence relation
∗ 1 ∗ (−1)n2n+1 cn =−n(n−2) 2cn−1+n((n−2)!)2
for n = 3,4,···. We obtain the second solution y2(x)=−2y1ln(x)+1+2x+16×3−25×4+ 157×6−···.
9. The indicial equation is 2r2 = 0, with roots r1 = r2 = 0. There are solutions
∞∞
y1(x) = cnxn and y2(x) = y1(x) ln(x) + c∗nxn.
n=0 n=1
Upon substituting these into the differential equation, we obtain the in-
2c0 − c1 +
Setting c∗0 = 1 for simplicity, we obtain c∗1 = 2, k = −2, c∗2 arbitrary (we
n=1
dependent solutions and
y1(x) = 1 − x x
y2(x)=(1−x)ln x−2 −2.
9 36 1350
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68 CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS 10. The indicial equation is r2 − 1 = 0, with roots r1 = 1 and r2 = −1. There
are solutions
y1(x) = cnxn+1 and y2(x) = ky1(x) ln(x) + c∗nxn−1.
n=0 n=0 Substitute each of these into the differential equation to get
and
∞∞
∞ y1(x)=x 1+(−1)n(1·4·7···(3n−2)) x3n
3nn!(5·8·11···(3n+2)) ∞
y2(x)= 1 1+(−1)n+1(1·2·5···(3n−1)) x3n. x 3nn!(4·7···(3n−2))
n=1
n=1
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Chapter 3
The Laplace Transform
3.1
1.
2.
3.
4.
5.
6.
7. 8.
9.
Definition and Notation
F (s) = 3(s2 − 4) (s2 + 4)2
G(s) = 8 (s+4)2 +64
H(s) = 14 − 7
s2 s2 + 49
W(s)= s − s
s2 + 9 K(s) = − 10
s2 + 49 + 3
q(t) = cos(8t)
s2 + 9
√√ g(t)= √ sin( 12t)−4cos( 8t)
(s + 4)2
r(t) = 7 sinh(3t)
5 12
3
p(t) = e−42t − 1t3e−3t 6
69
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70 10.
11.
CHAPTER 3. THE LAPLACE TRANSFORM
f(t) = −5tsin(t) 2
(a) From the definition,
F(s) = L[f](s) = lim e−stf(t)dt.
write
∞ 0
∞ (n+1)T n=0 nT
because f(u + nT) = f(u).
(c) Use the results of (a) and (b) to write
∞ (n+1)T n=0 nT
∞ T e−snT
0
n=0 nT
e−st f (t) dt =
= e−snT
e−suf(u)du,
L[f](s) =
=
e−st f (t) dt
R R→∞ 0
For each R, let N be the largest integer such that (N + 1)T ≤ R to write
N (n+1)T
R 0
R (N+1)T
e−st f (t) dt =
By choosing R sufficiently large, we can make the last integral on the right
e−st f (t) dt +
as small as we like. Further, R → ∞ as N → ∞, so
e−st f (t) dt.
e−stf(t) dt =
(b) Use the periodicity of f (t) and the change of variables u = t − nT to
T nT 0
T 0
(n+1)T
e−s(u+nT ) f (u + nT ) dt
f (t) dt
n=0
∞ T
∞∞n1 e−snt = e−sT = 1−e−sT .
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e−snT n=0
e−st f (t) dt,
(d) For s > 0, 0 < e−st < 1 and we can use the geometric series to obtain
=
because the summation is independent of t.
0
0
e−stf(t) dt.
Therefore
n=0 n=0 1T
L[f](s)= 1−e−sT
e−stf(t)dt.
3.1. DEFINITION AND NOTATION 71 12. f has period T = 6, and
Then
635 e−stf(t)dx= 5e−st = s(1−e−3s).
00
L[F ](s) = 5 1 − e−3s s 1 − e−6s
13. f has period T = π/ω and
T
= 5 1 − e−3s
s (1 − e−3s)(1 + e−3s)
=5. s(1 − e−3s)
π/ω
e−stf(t) dt = 00
Ee−st sin(ωt) dt = Eω (1+e−πs/ω).
s2 +ω2 1−e−πs/ω 14. f has period T = 25 and, from the graph,
0 for0
)H(t−4).
1t √
=√ τ2sin( 5(t−τ))dτ
50
= 1t− 2 + 2 cos(√5t).
8.
2 sin(√5t) L s2s2+5(t)=t∗ √5
s+2 −1 2 1
s
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5 25 25
3.4. CONVOLUTION 89 9. Take the transform of the initial value problem and solve for Y (s) to get
F(s) 1 1 Y(s)=s2−5s+6= s−3−s−2 F(s).
By the convolution theorem,
y(t)=e3t ∗f(t)−e2t ∗f(t).
10. Take the transform of the initial value problem to obtain Y (s) = F (s) + s
Then
(s+6)(s+4) (s+6)(s+4)
11131 =2F(s) s+4−s+6 + s+6−2s+4 .
y(t) = 1e−4t ∗ f(t) − 1e−6t ∗ f(t) + 3e−6t − 2e−4t. 22
For Problems 11–16 the solution is given, but the details (similar to those of Problems 9 and 10) are omitted.
11.
12.
13.
14.
15.
y(t) = 1e6t ∗ f(t) − 1e2t ∗ f(t) + 2e6t − 5e2t 44
y(t) = 1e5t ∗f(t)− 1e−t ∗f(t)+ 1e5t + 3e−t 6622
y(t) = 1 sin(3t) ∗ f(t) − cos(3t) + 1 sin(3t) 33
y(t) = 1 sinh(kt) ∗ f (t) − 2 cosh(kt) − 4 sinh(kt) kk
y(t) = 1e2t ∗f(t)+ 1 e−2t ∗f(t)− 1et ∗f(t)− 1e2t − 1 e−2t + 4et 4 12 3 4 12 3
16.
y(t)=42e3t∗f(t)−42e−3t∗f(t)− 28e 2t∗f(t)− 28e− 2t∗f(t)
1 1 √2√ √2√
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90 CHAPTER 3. THE LAPLACE TRANSFORM
17. The integral equation can be expressed as
f(t) = −1 + f(t) ∗ e−3t.
Take the transform of this equation to obtain F (s) = − 1 + F (s) .
s s+3 F(s)=− s+3 =1 1 −31.
Then
Invert to obtain the solution
f(t) = 1e−2t − 3. 22
18. The equation is f(t) = −1 + f(t) ∗ sin(t). Take the transform of this equation and solve for F (s) to obtain
s(s + 2) 2 s + 2 2 s
Then
F(s)=−(s2+1)=−1 − 1. s4 s2s4
f(t) = −t − 1t3. 6
19. The equation is f (t) = e−t + f (t) ∗ 1. Take the transform and solve for F(s) to get
Then
F(s)= s =1 1 +1 1 . (s − 1)(s + 1) 2 s + 1 2 s − 1
f(t) = 1e−t + 1et = cosh(t). 22
20. The equation is f (t) = −1 + t − 2f (t) ∗ sin(t). From this we obtain
F(s) = (1−s)(s2 +1) s2(s2 + 3)
11112s21 =3s2 −3s−3 s2+3 +3 s2+3 .
Invert this to get
f(t) = 1t − 1 − 2 cos(√3t) + 2√3 sin(√3t). 333 9
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3.4. CONVOLUTION 91 21. The equation is f (t) = 3 + f (t) ∗ cos(2t). From this we obtain
F (t) = 3(s2 + 4) s(s2 −s+4)
= 3 + 3 .
The inverse of this is
f(t)=3+ 5 et/2sin 2 t .
22. Write the equation as
t 0
2√15
f(τ)e−2(t−τ) dτ = cos(t) + f(t) ∗ e−2t.
s
s2 −s+4 √15
f(t) = cos(t) +
Transform this and solve for F (s) to obtain
F (s) = s(s + 2)
(s + 1)(s2 + 1)
1 3s11 =−2(s+1)+2 s2+1 +2 s2+1 .
Invert this to get the solution
f(t) = 1e−t + 3 cos(t) + 1 sin(t). 222
23. LetF =L[f]andG=L[g]. Then
∞
Now recall that
e−sτg(τ)dτ. e−sτ F (s) = L[H (t − τ )f (t − τ )](s).
F(s)G(s) = F(s)
Substitute this into the expression for F (s)G(s) to get ∞
F(s)G(s) =
But, from the definition of the Laplace transform,
Then
L[H (t − τ )f (t − τ )] = ∞ ∞
0
e−st H (t − τ )f (t − τ ) dt.
F(s)G(s)= =
00
∞∞ 00
e−stH(t−τ)f(t−τ)dt g(τ)dτ e−stg(τ)H(t−τ)f(t−τ)dτ dt.
0
0
L[H(t − τ)f(t − τ)](s)g(τ)dτ. ∞
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92
CHAPTER 3. THE LAPLACE TRANSFORM
ButH(t−τ)=0if0≤t<τ,whileH(t−τ)=1ift≥τ. Therefore
∞∞ 0τ
The last integration is over the wedge in the t,τ− plane consisting of points (t,τ) with 0 ≤ τ ≤ t < ∞. Reverse the order of integration to write
e−st g(τ)f(t−τ)dτ dt 00
∞ 0
3.5
This is what we wanted to show.
Impulses and the Dirac Delta Function
F(s)G(s) =
e−stg(τ)f(t−τ)dtdτ.
F(s)G(s) =
=
e−stg(τ)f(t−τ)dτ dt ∞t
∞t 00
=
= L[f ∗ g](s).
e−st(f ∗g)(t)dt
In Problem 1 details of the solution are given. For Problems 2–5, the details are similar and only the solution is given.
1. Transform the initial value problem to obtain
(s2 +5s+6)Y(s) = 3e−2s −4e−5s.
Using a partial fractions decomposition, this gives us
1 1 −3s 1 1 −5s
Y(s)=3 s+2−s+3 e −4 s+2−s+3 e . Invert this to obtain the solution
2.
3. 4.
y(t) = 4e2(t−3) sin(3(t − 3))H(t − 3) 3
y(t) = 6(e−2t − e−t + te−t)
y(t) = 3 cos(4t) + 3 sin(4(t − 5π/8))H(t − 5π/8)
y(t) = 3 e−2(t−2) − e−3(t−2) H(t − 2) − 4 e−2(t−5) − e−3(t−5) H(t − 5).
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3.6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 93 5.
y(t) = (B + 9)e−2t − (B + 6)e−3t ∞ ∞1
f (t)δε (t − a) dt = ε [H (t − a) − H (t − a − ε)]f (t) dt 00
a+ε 1
f(t)dt= εf(tε). a
∞
f (t)δε (t − a) dt = f (tε ).
0
3.6
1.
Systems of Linear Differential Equations
Take the transform of the system:
sX − 2sY = 1 , sX − X + Y = 0.
6. Begin with
t + ε such that
Then
a+ε a
f (t) dt.
By the mean value theorem for integrals, there is some tε between t and
==
Now let ε → 0+. Then a+ε → a, so tε → a also. Because f(t) is assumed to be continuous, f(tε) → f(a). Finally, we have
∞ ∞ lim f(t)δε(t−a)dt=
ε→0+ 0
f(t) lim δε(t−a)dt ε→0+
f(t)δ(t−a) lim f(tε) = f(a).
ε→0+
= =
∞ 0
0
Then
s
X(s)= 1
s2(2s−1) s2 s 2s−1
1−s =−1 −1+ 2 s2(2s−1) s2 s 2s−1
x(t) = −t − 2 + 2et/2, y(t) = −t − 1 + et/2.
=−1 −2+ 4 ,
Y(s)=
Apply the inverse transform to get the solution
.
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94
CHAPTER 3. THE LAPLACE TRANSFORM
2. Take the transform of the system to obtain 2sX+(s−3)Y =0,sX+sY = 1.
Then
X(s) = − s − 3 , Y (s) = s2(s + 3)
s2
2 .
s2(s + 3)
Using partial fractions, we can write X(s)=21−21+1−2 1 ,
9 s 3 s2 s3 9 s + 3 Y(s)=−21+21+2 1 .
x(t) = 2 − 2t + 1t2 − 2e−3t, 9329
y(t) = −2 + 2t + 2e−3t. 939
3. After transforming the system, we obtain sX+(2s−1)Y = 1,2sX+Y =0.
9 s 3 s2 9 s + 3 Inverting, we obtain the solution
Then
s
X(s)=− 1 =4− 16 +1, s2(4s − 3) 9s 9(4s − 3) 3s2
Y(s)= 2 =−2+ 8 . s(4s−3) 3s 3(4s−3)
Invert these to obtain
x(t) = 4(1 − e3t/4) + 1t, 93
y(t)= 2(−1+e3t/4). 3
4. The transformed system is
sX+sY−X= s ,sX+2sY=0.
Then
X(s) = 2s
s3 −2s2 +s−2
= −1 4s − 2 + 4 , 5s2 +1 5(s−2)
s2 + 1
Y (s) = − s
s3 −2s2 +s−2
= 1 2s − 1 − 2 . 5s2 +1 5(s−2)
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3.6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
95
The solution is
Then
Then
X(s)= 2(s−1) = 3 + 1 + 1 − Y(s)=− 2 = 3 + 1 −92(3s−2)
x(t) = −4 cos(t) + 2 sin(t) + 4e2t, 555
y(t) = 2 cos(t) − 1 sin(t) − 2e2t. 555
5. The system transforms to
3sX − Y = 2 sX + sY − Y = 0.
s2
9 s2(3s−2) 4s 2s2 s3 4(3s−2)
,
s2(3s−2) 2s s2 .
x(t) = 3 + 1t + 1t2 − 3e2t/3, 4224
y(t) = 3 + t − 3e2t/3. 22
6. Apply the transform to get
sX+4sY −Y =0sX+2Y = 1 .
Then
Then
X(s)= 4s−1 =1+ 32 − 5 s(4s2 +s−3) 3s 21(4s−3) 7(s+1)
Y(s)=− 1 =− 4 + 1 . 4s2 +s−3 7(4s−3) 7(s+1)
x(t)= 1 + 8 e3t/4 − 5e−t, 321 7
y(t) = −1e3t/4 + 1e−t. 77
s+1
7. The transform of the system is
sX + 2X − sY = 0, sX + Y + X = 2 .
s3
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96
CHAPTER 3. THE LAPLACE TRANSFORM
X(s)= 2 =1+ s+1 −1,
Then
Y(s)= The solution is
2(s+2) s3(s2 +2s+2)
=−1 + 1
s2 s2 +2s+2
s3
s2(s2 +2s+2) s2 s2 +2s+2
s
+ 2.
8.
In inverting X(s) and Y (s), the terms involving s2 +2s+2 can be treated by using a shifting theorem, expressing these as functions of s + 1.
The system transforms to
sX+4X−Y =0,sX+sY = 1.
s2
Then
X(s)= 1 =1−1− 1 +1,
x(t) = t + e−t cos(t) − 1 y(t) = −t + e−t sin(t) + t2.
s2(s + 5) 125s 25s2 125(s + 5) Y(s)= s+4 =− 1 + 1 + 1
5s2
+ 4 .
9.
The solution is
First,
Then
s3(s + 5) 125s 25s2 125(s + 5)
x(t)= 1 − 1t− 1 e−5t+ 1t2 125 25 125 10
y(t)=− 1 + 1t+ 1 e−3t+2t2. 12525125 5
5s3
sX + sY + X − Y = 0, sX + 2sY + X = 1 . s
X(s)= 1−s = −2 − 1 +1, s(s+1)2 s+1)2 s+1 s
Y(s)= 1 =1−1. s(s+1) s s+1
10.
Then
The transform of the system is
x(t) = 1 − e−t(2t + 1), y(t) = 1 − e−t.
sX + 2sY − X = 0, 4sX + 3sY + Y = − 6 . s
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3.6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
Then,
X(s) = − 12 , Y (s) = 6(s − 1) 5s2 +2s+1 s(5s2 +2s+1)
By using a shifting theorem, invert these to get x(t) = −6e−t/5 sin(2t/5),
y(t) = −6 + 6e−t/5 (cos(2t/5) + sin(2t/5)) . 11. The system transforms to
97
.
sX−2sY +3X=0,X−4sY +3sZ= 1,X−2sY +3sZ=−1. s2 s
Then
X(s)=s+1=2− 2 −1,
s2(s+3) 9s 9(s+3) 3s2 Y(s)=−1s+1=−1− 1,
The solution is
− 16 . 3 s3(s+3) 9s 81s 81(s+3) 27s2
x(t) = 2 + 1t − 2e−3t, 939
y(t) = −1t(t + 2) 4
z(t)=−16t−1t2− 2 + 2e−3t. 27 9 81 81
2s3 2s3 2s2 Z(s)=−2s2+3s+1=−2 − 2 + 2
12. The loop currents in the circuit satisfy
5i′1 +5i1 −5i′2 =1−H(t−4)sin(2(t−4)),
− 5 i ′1 + 5 i ′2 + 5 i 2 = 0 .
Apply the Laplace transform to these equations and solve for I1(s) and
I2(s) to get
I1(s)= s+1 1−2e−4s 5(2s+1) s s2+4
11122 s 9−4s =5 s−2s+1 −85 2s+1−s2+4+s2+4 e
1 2e−4s I2(s)=2s+1 1−s2+4
1 2 2 s 8 −4s =5(2s+1)+85 2s+1−s2+4−s2+4 e .
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98
CHAPTER 3. THE LAPLACE TRANSFORM
Apply the inverse Laplace transform to obtain the solution for the currents: 1 1 −t/2
i1(t)=5 1−2e
2 −(t−4)/2 9
− 85 e −cos(2(t−4))+ 2 sin(2(t−4)) H(t−4), i2(t) = 1 e−t/2
13.
10
+ 2 e−(t−4)/2 − cos(2(t − 4)) − 4 sin(2(t − 4)) H(t − 4). 85
The equations for the loop currents are
20i′1 +10(i1 −i2)=E(t)=5H(t−5),
30i′2 +10i2 +10(i2 −i1) = 0. Initial conditions are
i1(0) = i2(0) = 0. Transform the system to obtain
I1(s) = 5(30s + 20)e−5s s(600s2 + 700s + 100)
1 1 27 1 −5s = s−10(s+1)−56s+1 e ,
I2(s) = 50e−5s s(600s2 + 700s + 100)
1 10 181−5s =2s+s+1−56s+1e .
Invert these to obtain the current functions:
1 i1(t)= 1−10e
respectively, at time t. Now
x′1(t) = rate of change of salt in tank 1
−(t−5)
H(t−5),
9 −(t−5)/6 −10e
−(t−5)/6
Let x1(t) and x2(t) be the amounts of salt (in pounds) in tanks 1 and 2,
1 1 i2(t)= 2+10e
−(t−5) 3 −10e
H(t−5).
14.
Then
= (rate salt is added ) − ( rate salt is removed). x′1(t)=1+ 3x2− 5x1.
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3 18 60
3.6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 99 Similarly,
x′2 = 5 x1 − 5 x2 +11(H(t−4)−H(t−6)). 60 18
Initial conditions are x1 (0) = 11, x2 (0) = 7. Transform this system to obtain
(12s+1)X1 −2X2 = 4 +132, s
−3X1 + (36s + 10)X2 = 396(e−4s − e−6s) + 252. s
Then
X1(s) = 4752s2 + 1968s + 40 + 792(e−4s − e−6s)
s(432s2 + 156s + 4)
10 6 108 99 27 3888 −4s −6s
= s −3s+1+36s+1+2 s +3s+1−36s+1 (e −e ), X2(s) = 3024s2 + 648s + 12 + 396(12s + 1)(e−4s − e−6s)
s(432s2 + 156s + 4)
3 9 36 99 81 2592 −4s −6s
=s+3s+1+36s+1+ s −3s+1−36s+1 (e −e ). Apply the inverse transform to obtain the solution:
x1(t) = 10 − 2e−t/3 + 3e−t/36 + 2(99 + 9e−(t−4)/3 − 108e−(t−4)/6)H(t − 4)
− 2(99 + 9e−(t−6)/3 − 108e−(t−6)/36)H(t − 6),
x2(t) = 3 + 3e−t/3 + e−t/36 + (99 − 27e−(t−4)/3 − 72e−(t−4)/36)H(t − 4)
− (99 − 27e−(t−6)/3 − 72e−(t−6)/36)H(t − 6).
15. Using the notation of the preceding problem, we can write
x ′1 = − 6 x 1 + 3 x 2 , 200 100
x′2= 4x1− 4x2+5H(t−3). 200 200
Initial conditions are x1(0) = 10,x2(0) = 5. Apply the transform to this initial value problem and rearrange terms to obtain
(100s + 3)X1 − 3X2 = 1000, −2X1 + (100s + 4)X2 = 500 + 500e−3s.
Solve these to get
X1(s) = 100000s + 5500 + 1500e−3s 10000s2 + 700s + 6
50 900 300 150 −3s = 50s+3 + 100s+1 + 100s+1 − 50s+3 e
,
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100
CHAPTER 3. THE LAPLACE TRANSFORM
and
Apply the inverse transform to obtain the solution:
i1(t) = e−3t/50 + 9e−t/100 + 3(e−(t−3)/100 − e−3(t−3)/50)H(t − 3), i2(t) = −e−3t/50 + 6e−t/100 + (3e−3(t−3)/50 + 2e−(t−3)/100)H(t − 3).
X2(s) = 50000s + 3500 + (50000s + 1500)e−3s 10000s2 + 700s + 6
50 600 150 200 −3s =−50s+3+100s+1+ 50s+3+100s+1 e .
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Chapter 4
Sturm-Liouville Problems and Eigenfunction Expansions
4.1 Eigenvalues, Eigenfunctions and Sturm-Liouville Problems
For these problems, an eigenfunction is found for each eigenvalue, and it is understood that nonzero constant multiples of eigenfunctions are also eigen- functions.
1. The problem is regular on [0, L]. To find the eigenvalues and eigenfunc- tions, take cases on λ.
Case 1. If λ = 0, the differential equation is y′′ = 0, with solutions y=a+bx. Nowy(0)=a=0,soy=bx. Buttheny′(L)=b=0also, so this case has only the trivial solution and 0 is not an eigenvalue of this problem.
Case 2. If λ is negative, say λ = −α2 with α > 0, then the differential equation is
with general solution Now
so c2 = −c1 and
y′′ − α2y = 0
y = c1eαx + c2e−αx. y(0) = c1 + c2 = 0,
y(x) = c1 eαx − e−αx = 2c1 sinh(αx). 101
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102CHAPTER4. STURM-LIOUVILLEPROBLEMSANDEIGENFUNCTIONEXPANSIONS
From the other boundary condition,
y′(L) = 2c1α cosh(αL) = 0.
Butcosh(αL)>0andα>0,sowemusthavec1 =0andthiscasealso has only the trivial solution. This problem has no positive eigenvalue.
Case 3. Suppose λ is positive, write λ = α2, with α > 0. Now the differential equation is
y′′ + α2y = 0,
with general solution
y(x) = c1 cos(αx) + c2 sin(αx). Immediately y(0) = c1 = 0, so
y(x) = c2 sin(αx). From the other boundary condition, we must have
y′(L) = c2α cos(αL) = 0.
We need to be able to choose c2 ̸= 0 to have nontrivial solutions. This
requires that we α must be chosen to satisfy cos(αL) = 0.
We know that the zeros of the cosine function have the form (2n − 1)π/2 for integer π, so let
αL = (2n − 1)π , 2
with n = 1,2,···. Then acceptable values of α are α = (2n − 1)π .
2L
Because λ = α2, the eigenvalues of this problem, indexed by n, are
(2n − 1)π2 λn = 2L
for n = 1, 2, · · · . Corresponding eigenfunctions are (2n−1)π
φn(x) = sin 2L x , or any nonzero constant multiple of this function.
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4.1. EIGENVALUES,EIGENFUNCTIONSANDSTURM-LIOUVILLEPROBLEMS103
2. The problem is regular on [0, L]. The eigenvalues are λ0 =0,λn =n2 forn=1,2,···.
Corresponding eigenfunctions are
φn(x) = cos(nπx) for n = 0,1,2,··· .
Any nonzero constant multiple of an eigenfunction is also an eigenfunction (for the same eigenvalue). Notice in this example that 0 is an eigenvalue. The number zero can be an eigenvalue, but the trivial function (identically zero) cannot be an eigenfunction.
3. The problem is regular on [0, L];
1 π2
λn= n−2 4
is an eigenvalue for n = 1, 2, · · · , with eigenfunctions
(2n − 1)π φn(x) = cos 8 .
4. The problem is periodic on [0,π]. Eigenvalues are λn = 4n2 for n = 0,1,2,··· andeigenfunctionsare
φn(x) = an cos(2nx) + bn sin(2nx) with an and bn constant and not both zero.
5. The problem is periodic on [−3π, 3π]. Eigenvalues are n2
λ0 =0andλn = 9 forn=1,2,···.
Eigenfunctions are
φn(x) = an cos(nx/3) + bn sin(nx/3)
for n = 0,1,2,···, with an and bn constant and not both zero.
6. The problem is regular on [0,π]. To find the eigenvalues and eigenfunc-
tions, take cases on λ.
Case 1. If λ = 0, then y = ax+b for constants a and b. Because
y(0)=b=0,theny=ax. Butthen
y(π) + 2y′(π) = 0 = aπ,
so a = 0 and this case has only the trivial solution. 0 is not an eigenvalue of this problem.
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104CHAPTER4. STURM-LIOUVILLEPROBLEMSANDEIGENFUNCTIONEXPANSIONS
Case2. Ifλ=−α2,withα>0,then
y(x) = c1eαx + c2e−αx.
Now y(0) = c1 + c2 = 0, so c2 = −c1 and
y(x) = c1(eαx − e−αx).
From the boundary condition at π, we must have
y(π) + 2y′(π) = c1(eαπ − e−απ) = 0.
But e−απ < eαπ because the exponential function is strictly increasing. Therefore c1 = 0 and this case admits only the trivial solution. The problem has no negative eigenvalue.
Case3. Ifλ=α2 withα>0,then
y(x) = c1 cos(αx) + c2 sin(αx).
Immediately, y(0) = c1 = 0, so y(x) = c2 sin(αx). The second boundary condition is
y(π) + 2y′(π)0 = c2 sin(απ) + 2c2α cos(απ).
To look for nontrivial solutions, suppose c2 ̸= 0. Then this equation is
or
sin(απ) = −2α cos(απ) tan(απ) = −2α.
This is a transcendental equation, which cannot be solved by algebraic manipulations. There are infinitely many positive solutions, however, be- cause the graphs of y = tan(απ) and y = −2α intersect infinitely often in the right half-plane. Let the first coordinates of these points of inter- section be α1, α2, · · · , in increasing order. Then the eigenvalues of this problem are
λ n = α n2 .
Using a numerical approximation program, the first four eigenvalues are
approximately
λ1 ≈ 0.48705, λ2 ≈ 2.54914, λ2 ≈ 6.56059, λ4 ≈ 12.56423.
Corresponding eigenfunctions are
φn(x) = sin(λnx).
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4.1. EIGENVALUES,EIGENFUNCTIONSANDSTURM-LIOUVILLEPROBLEMS105
7. The problem is regular on [0,1]. The analysis to find eigenvalues and eigenfunctions is similar to that done for Problem 6. Take cases on λ. It is routine to check that λ = 0 or λ < 0 lead only to the trivial solution, so the problem has no negative eigenvalue and zero is not an eigenvalue. Ifλ=α2 forα>0,then
y(x) = c1 cos(αx) + c2 sin(αx). Using the first boundary condition,
y(0)−2y′(0)=0=c1 −2c2α
so c1 = 2αc2 and
y(x) = 2αc2 cos(αx) + c2 sin(αx). Now use the boundary condition at 1:
y′(1) = −2α2c2 sin(α) + c2α cos(α) = 0.
For a nontrivial solution we need to be able to choose c2 nonzero. This
requires that or
−2α sin(α) + cos(α) = 0, tan(α)= 1.
2α
Solutions of this equation must be numerically approximated. There are infinitely many positive solutions α1 < α2 < · · · , and the eigenvalues are λj = αj2. The first four eigenvalues are
λ1 ≈ 0.42676, λ2 ≈ 10.8393, λ3 ≈ 40.4702, λ4 ≈ 89.8227. Corresponding eigenfunctions are
√ φn(x) = 2 λ cos( λnx) + sin( λnx).
8. The problem is regular on [0, 1]. The differential equation has character- istic equation
with roots r = −1 ± is
r2 +2r+(1+λ)=0, √
λi. The general solution of the differential equation −x √ −x √
y(x) = c1e cos( λx) + c2e sin( λx). Because y(0) = 0 = c1, we have just
−x √ y(x) = c2e sin( λx).
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106CHAPTER4. STURM-LIOUVILLEPROBLEMSANDEIGENFUNCTIONEXPANSIONS The boundary condition at 1 requires that
−1 √ y(1) = 0 = c2e sin( λ).
This can be satisfied with c2 ̸= 0 if √
λ = nπ,
a positive zero of the sine function, for n = 1, 2, · · · . The eigenvalues are
therefore
and the eigenfunctions are
λn = n2π2 φn(x) = e−x sin(nπx).
9. The problem is regular on [0, π]. The differential equation can be written y′′ + 2y′ + λy = 0.
and the characteristic equation has roots
√
−1± 1−λ.
Here it is convenient to take cases on 1 − λ.
Case1. If1−λ=0,thenλ=1andthegeneralsolutionis
y(x) = c1e−x + c2xe−x. Now y(0) = c1 = 0, so y(x) = c2xe−x. And
y(π)=0=c2πe−π =0
forces c2 = 0, so this case has only the trivial solution and 0 is not an
eigenvalue.
Case2. If1−λispositive,say1−λ=α2 withα>0,then
Now
so c2 = −c1 and
Then
y(x) = c1e(−1+α)x + c2e(−1−α)x.
y(0) = c1 + c2 = 0
y(x)=c e(−1+α)x−e(−1−α)x. 1
(−1+α)π (−1−α)π y(π)=c1 e −e
.
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4.2. EIGENFUNCTION EXPANSIONS 107 If c1 ̸= 0, this requires that
eαπ =e−απ,
which is impossible if α > 0. The problem has no negative eigenvalue.
Case3. If1−λisnegative,write1−λ=−α2. Now y(x) = c1e−x cos(αx) + c2e−x sin(αx).
Immediately y(0) = c1 = 0. Next,
y(π) = c2e−π sin(απ) = 0.
To have c2 ̸= 0, we must choose
√
α= λ−1=n,
any positive integer. Then λ = 1 + n2, so the eigenvalues are
Eigenfunctions are
λn =1+n2 forn=1,2,···. φn(x) = e−x sin(nx).
10. The problem is regular on [0, 8]. Details are similar to those of Problem 9 and we find eigenvalues
λn =8+n2π2 for n = 1, 2, · · · , and eigenfunctions
φn(x) = e3x sin(nπx).
11. If λn = 1 − 1/n, then the eigenvalues are listed in increasing order, and
lim λn = 0. n→∞
This is impossible by Theorem 4.1.
4.2 Eigenfunction Expansions
In Problems 1–5, the weight function is p(x) = 1 (read from the differential equation). In Problem 6 the differential equation must be put into standard Sturm-Liouville form to read the weight function p(x) as the coefficient of λ.
In graphing partial sums of eigenfunctions and comparing them to the func- tion, note the differences in the number of terms that must be taken to have the partial sum fit reasonably close to the function. The convergence theorem does not give any information about how fast an eigenfunction expansion converges to the function.
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108CHAPTER4. STURM-LIOUVILLEPROBLEMSANDEIGENFUNCTIONEXPANSIONS 1. It is routine to find the eigenfunctions
φn(x) = sinnπx 2
for this problem. The expansion has the form
where
These integrals are
and
2(1 − ξ) sin(nπξ/2) dξ 0
∞
cn sin(nπx/2), n=1
cn= 2 2 . 0 sin (nπξ/2)dξ
2
sin2(nπξ/2) dξ = 1
0
2 2(1 + (−1)n)
(1 − ξ) sin(nπξ/2) dξ = nπ . 0
The eigenfunction expansion on [0, 2] is
∞ 2(1 + (−1)n) sin(nπx/2).
nπ
Figure 4.1 shows a graph of f (x) = 1 − x and the fortieth partial sum of this expansion. By the convergence theorem, this expansion converges to 1−x for 0 < x < 2. Clearly the expansion converges to 0 at both x = 0 and x = 2 because the eigenfunctions vanish there.
2. The problem has eigenfunctions
φn(x) = sin((2n − 1)x/2).
The eigenfunction expansion on [0, π] has the coefficients cn = 8 (−1)n+1
and the expansion is
n=1
∞
π (2n − 1)2
cn sin((2n − 1)x/2). n=1
Figure 4.2 shows a graph of the function and the fifth partial sum of this eigenfunction expansion. Unlike Problem 1, this expansion converges very rapidly to the function. By the convergence theorem, it converges to x for 0 < x < π. The graph suggests that it converges to x at the endpoints as well, but this is not given by the theorem.
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4.2. EIGENFUNCTION EXPANSIONS 109
Figure 4.1: Comparison of 1−x and the fortieth partial sum of its eigenfunction expansion on [0, 2].
Figure 4.2: Comparison of x and the fifth partial sum of its eigenfunction ex- pansion on [0, π].
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110CHAPTER4. STURM-LIOUVILLEPROBLEMSANDEIGENFUNCTIONEXPANSIONS
Figure 4.3: Comparison of f(x) and the sixtieth partial sum of its eigenfunction expansion on [0, 4].
3. The eigenfunctions are
φn(x) = cos((2n − 1)πx/8). The coefficients in the expansion of f(x) on [0,4] are
2 − cos((2n − 1)πx/8) dx + 4 cos((2n − 1)πx/8) dx 02
cn = 4 cos2((2n − 1)πx/8) dx 0
= 4 (−1)n+1 + √2(cos(nπ/2) − sin(nπ/2)) . (2n − 1)π
The expansion has the form
∞
cn cos((2n − 1)πx/8). n=1
Figure 4.3 compares f(x) with the sixtieth partial sum of this eigenfunc- tion expansion. The theorem tells us that the expansion converges to f(x) on (0,2) and on(2,4), as well at to 0 at x = 0 (average of left and right limits there).
4. The eigenfunctions are
φ0(x) = 1,φn = cos(nx) for n = 1,2,··· .
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4.2. EIGENFUNCTION EXPANSIONS 111
Figure 4.4: Comparison of f(x) and the tenth partial sum of its eigenfunction expansion in Problem 4.
The coefficients in the eigenfunction expansion of f(x) = sin(2x) on [0,π]
are
1π
cn = π
The eigenfunction expansion is
c0 = π 2π
c2 = π
sin(1ξ) dξ = 0, sin(2ξ) cos(2ξ) dξ = 0,
0
0
and, for n = 1,3,4,5,···, 2 π
4 (−1)n − 1 sin(2ξ)cos(nξ)dξ = π n2 −4 .
0
4 ∞ 4 ((−1)n − 1) cos(nx).
π n=1,n̸=2 π n2 − 4
Figure 4.4 compares a graph of f(x) with the tenth partial sum of this eigenfunction expansion. The theorem tells us that this expansion con- verges to sin(2x) for 0 < x < π.
5. The eigenfunctions are
φ0(x) = 1, φn(x) = an cos(nx/3) + bn sin(nx/3)
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112CHAPTER4. STURM-LIOUVILLEPROBLEMSANDEIGENFUNCTIONEXPANSIONS
Figure 4.5: Comparison of f(x) and the fifth partial sum of its eigenfunction expansion in Problem 5.
for n = 1, 2, · · · . The coefficients in the eigenfunction expansion of x2 on [−3π, 3π] are
1 3π
a0 = 6π
an = 1 ξ2 cos(nξ/3) dξ = 36 (−1)n, 3π n2
1 3π
bn = 3π
−3π The eigenfunction expansion is
ξ2 dξ = 3π2,
−3π
ξ2 sin(nξ/3) dξ = 0.
2 ∞(−1)n
n2 cos(nx/3).
Figure 4.5 shows f (x) and the fifth partial sum of this eigenfunction expan-
sion. By the theorem, the expansion converges to x2 for −3π < x < 3π. 6. The eigenfunctions are
φn(x) = e−x sin(nπx)
3π + 36
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n=1
4.2. EIGENFUNCTION EXPANSIONS 113
Figure 4.6: Comparison of f (x) and the eightieth partial sum of its eigenfunction expansion in Problem 6.
for n = 1, 2, · · · . Notice that the differential equation of the problem can written in standard Sturm-Liouville form as
e2xy′′ + (1 + λ)e2xy = 0,
with the coefficient of λ equal to e2x. Therefore, in the expansion of a function in terms of these eigenfunctions, the weight function p(x) = e2x must be used. With this in mind, the coefficients in this expansion are
1 e2ξe−ξ sin(nπξ)dξ 1/2
cn = 1 e2ξe−2ξ sin2(nπξ)dξ 0
=
1 eξ sin(nπξ) dξ 1/2
1 sin2(nπξ) dξ 0
2e1/2(nπ cos(nπ/2) − sin(nπ/2)) − 2enπ(−1)n = 1+n2π2 .
Figure 4.6 shows a graph of the function and the eightieth partial sum of this eigenfunction expansion.
7. Recall that the complex conjugate of z = a+ib is z = a−ib. Suppose λ is an eigenvalue of a Sturm-Liouville problem, with eigenfunction φ(x). By
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114CHAPTER4. STURM-LIOUVILLEPROBLEMSANDEIGENFUNCTIONEXPANSIONS
taking the complex conjugate of the Sturm-Liouville differential equation and appropriate boundary conditions, it is routine to check that λ is also an eigenvalue with eigenfunction φ(x). If λ is complex and not real, then λ ̸= λ, so the eigenfunctions must be orthogonal with respect to the weight function p, and
Now, so
b
p(x)φ(x)φ(x) dx = 0.
a
b
p(x)|φ(x)|2| dx = 0.
a
φ(x)φ(x) = |φ(x)|2,
4.3
1.
This is impossible because p(x) > 0 on (a, b) and φ(x) is continuous and not identically zero on the interval. This contradiction shows that λ = λ, so λ is real.
Fourier Series
The Fourier series of f (x) = 4 on [−3, 3] has the form 1 ∞
2 a0 + [an cos(nπx/3) + bn sin(nπx/3)]. n=1
All that is left is to compute the coefficients. First, because f(x) is an even function, each bn = 0. Compute
23
a0 = 3
23
4 dx = 8
and, for n = 1,2,···,
−3
4 cos(nπx/3) dξ = 0. With each an = 0 for n = 1,2,···, the Fourier series is
1a0 =4. 2
This series consists of a single term, namely the constant term (which seems obvious by hindsight, if not noticed immediately). This one-term series converges to 4 on [−3, 3].
Because f(x) is an odd function on [−1,1], each an = 0. Compute the Fourier coefficients
bn = 2
an = 3
0
2.
12n
0
−xsin(nπx)dx = nπ(−1) .
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4.3. FOURIER SERIES 115
Figure 4.7: Comparison of f(x) and the thirtieth partial sum of the Fourier series in Problem 2.
The Fourier series of −x on [−1, 1] is 2∞ (−1)n
n sin(nπx).
This series converges to −x for −1 < x < 1, and to 0 at x = −1 and at
a0 = 2
π
n=1
x = 1. The latter is consistent with the convergence theorem because lim (−x)+ lim (−x) = 1−1 = 0.
x→−1+ x→1−
Figure 4.7 is a graph of f(x) and the thirtieth partial sum of the Fourier
series.
3. Because cosh(πx) is an even function, each bn = 0 in the Fourier series, which will have the appearance
1 ∞
2a0 +
an cos(nπx).
12
cosh(πx) dx π sinh(π)
Compute
n=1
0
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116CHAPTER4. STURM-LIOUVILLEPROBLEMSANDEIGENFUNCTIONEXPANSIONS
Figure 4.8: Comparison of f(x) and the eighth partial sum of the Fourier series in Problem 3.
and, for n = 1,2,···, 1
an =2 The Fourier series is
2sinh(π) (−1)n cosh(πx)cos(nπx)dx= π 1+n2.
0
1
π sinh(π) +
∞ 2sinh(π) (−1)n
π 1 + n2 cos(nπx).
This series converges to cosh(πx) for −1 ≤ x ≤ 1. Figure 4.7 shows a
graph of f(x) and the eighth partial sum of this Fourier series.
4. Omitting the routine integrations, the Fourier series of 1 − |x| on [−2, 2]
is
8 ∞ 1 cos((2n−1)πx/2).
n=1
5. The Fourier series of f(x) on [−π,π] is
16 ∞ 1 sin((2n − 1)x).
n=1
π2
This series converges to 1 − |x| for −2 ≤ x ≤ 2. Figure 4.9 shows a graph
(2n − 1)2
of f(x) and the fifth partial sum of this Fourier series.
π
n=1
(2n − 1)2
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4.3. FOURIER SERIES 117
Figure 4.9: Comparison of f(x) and the fifth partial sum of the Fourier series in Problem 4.
This series converges to
+ (−1)n n=1
cos(nπx/2)+ sin(nπx/2) . nπ
−4 for−π
u(x, 0) = B.
The coefficients in the series solution are
2L
c0 = L
B dξ = 2B
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0
5.1. DIFFUSION PROBLEMS ON A BOUNDED MEDIUM 141 and, for n = 1,2,···,
2L
cn = L
This is consistent with intuition – with no energy loss, the bar maintains
a constant temperature.
9. The initial-boundary value problem for the temperature function is
ut =uxx for0
u(x,0)= Bx. L
Separate variables by putting u(x, t) = X (x)T (t) to obtain X′′ + λX = 0; X(0) = X′(L) = 0
and
T′′ +λkT =0.
By taking cases on λ, we find the eigenvalues and corresponding eigen-
Further,
The solution has the form
∞
u(x, t) = cn sin((2n − 1)πx/2L)e−k(2n−1)2 π2 t/4L2 .
n=1 The coefficients are
The solution is
B cos(nπξ/L) dξ = 0.
0
u(x, t) = B.
functions:
λn = 2L and Xn(x) = sin((2n − 1)πx/2L).
(2n − 1)π2
Tn(x) = e−k(2n−1)2π2t/4L2 .
The solution is u(x, t) = − π2
cn = 2 L Bξsin((2n−1)πξ/2L)dξ L0L
= −8B (−1)n. π2(2n − 1)2
(−1)n
(2n − 1)2 sin((2n − 1)πx/2L)e .
8B ∞ n=1
−k(2n−1)2π2t/4L2
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142 10.
CHAPTER 5. THE HEAT EQUATION
The initial-boundary value problem for the temperature function is
ut =9uxx for0
u(x,0) = x2.
A routine separation of variables leads to eigenvalues and eigenfunctions
n2π2
λn = 4 , Xn(x) = sin((2n − 1)πx/4)
and solutions for the time-dependent parts as Tn(t) = e−9n2π2t/4.
The solution has the form
where
∞
u(x, t) = cn sin((2n − 1)πx/4)e−9n2π2t/4,
n=1
2 0
ξ2 sin((2n − 1)πξ/4) dξ 64 2 + (2n − 1)π(−1)n
cn =
= −π3 (2n − 1)3 .
11.
Make the transformation u(x, t) = eαx+βtv(x, t). Following the discussion of the text, let α = −A/2 = −4/2 = −2 and β = k(B − A2/4) = −2 also, so
u(x, t) = e−2x−2tv(x, t) and v is the solution of the problem
vt =vxx for0
v(x, 0) = e2xu(x, 0) = xe2x(π − x). This has the solution
where
2π
∞
v(x,t) = cn sin(nx)e−ntt,
n=1
cn = π ξe2ξ(π−ξ)sin(nξ)dξ 0
= − 4 24n − 2n3 + 16nπ + 4n3π − 24ne2π(−1)n π(4 + n2)3
+2e2πn3(−1)n +16nπe2π(−1)n +4n3πe2π(−1)n.
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5.1. DIFFUSION PROBLEMS ON A BOUNDED MEDIUM 143 The solution of the original problem is
u(x, t) = e−2x−2tv(x, t).
12. Again following the discussion in the text, we have A = 6,B = 0,k =
1,L=4andu(x,0)=f(x)=1. Findthatα=−3andβ=−9tolet u(x, t) = e−3x−9tv(x, t).
The v(x, t) is the solution of the standard problem
vt =vxx for0
v(x, 0) = e3xf(x) = e3x. This problem has the solution
∞
v(x, t) = cn sin(nπx/4)e−n2π2t/16,
n=1
14
where
∞
u(x, t) = e−3x−9tv(x, t) = e−3x−9t cn sin(nπx/4)e−n2π2t/16.
n=1
13. HerewehaveA=6,B=0,k=1,L=πandu(x,0)=f(x)=x(π−x).
Let α = 3 and β = −9 and let
u(x, t) = e3x−9tv(x, t).
The v(x, t) satisfies
vt =vxx for0
v(x, 0) = e−3xf(x) = x(π − x)e−3x. The solution of this problem is
∞
v(x,t) = cn sin(nx)e−n2t,
n=1
=
The original problem has the solution
cn = 2
e3ξ sin(nπξ/4) dξ 2nπ (1 − e12(−1)n).
0
144 + n2π2
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144
CHAPTER 5. THE HEAT EQUATION
where
cn = π
e−3ξξ(π−ξ)sin(nξ)dξ
4n (1 − (−1)ne−3π)(3π(n2 + 9) + n2 − 27).
2π
0
=
The original problem has the solution
14.
π(n2 + 9)3
v(x, t) = e3x−9tv(x, t).
Because of the nonhomogeneous boundary conditions, we cannot solve this problem by a standard separation of variables. Transform the problem by letting
u(x, t) = v(x, t) + ψ(x),
where Lψ(x) must be chosen so that we have a problem for v(x,t) that we know how to solve. Substitute u(x,t) into the initial-boundary value problem to obtain
vt = 16vxx + 16ψ′′(x) for 0 < x < 1, t > 0, v(0, t) + ψ(0) = u(0, t) = 2, v(1, t) + ψ(1) = u(x, t) = 5,
v(x, 0) + ψ(x) = u(x, 0) = x(1 − x)2.
First, simplify the differential equation for v by setting ψ′′(x) = 0, so
ψ(x) = cx + d.
Now, we will have v(0,t) = 0 if ψ(0) = 2, and v(1,t) = 0 if ψ(1) = 5.
Therefore choose d = 2 and c = 3, so
ψ(x) = 3x+2.
The problem for v is
vt =16vxx for0
v(x, 0) = u(x, 0) − ψ(x) = sin(πx) − 3x − 2. This problem has the solution
where
∞
v(x, t) = cn sin(nπx)e−16n2π2t,
n=1
1 0
(sin(πξ) − 3ξ − 2) sin(nπξ) dξ nπ
cn = 2
= −1 + 10(−1)n
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5.1. DIFFUSION PROBLEMS ON A BOUNDED MEDIUM
145
if n = 2, 3, · · · , and 1
−14+π (sin(πξ)−3ξ−2)sin(πξ)dξ= π
c1 =2
This determines v(x, t), and u(x, t) = v(x, t) + 3x + 2.
.
Then
The problem for v is
ψ(0) = T,ψ(L) = 0.
ψ(x)= T(L−x) L
0
15. Let u(x, t) = v(x, t) + ψ(x). To get a standard problem for v(x, t), choose ψ(x) so that ψ′′ = 0 and
vt =kvxx for0
v(x, 0) = u(x, 0) − ψ(x) = x(L − x)2 − T (L − x). L
This has the solution
v(x, t) = cn sin(nπx/L)e−kn2 π2 t/L2 ,
where
∞
n=1
2L 2T
cn = L ξ(1−ξ) − L(L−ξ) sin(nπx/L)dξ
0
= 2 −n2π2T + 4L3 + 2L3(−1)n .
n3π3
16. Herewehavek=4,L=9andu(x,0)=xsin((9−x)π). Let
v(x, t) = ekAtv(x, t) and the problem for v(x, t) is
vt =4vxx for0
v(x, 0) = u(x, 0) = x sin((9 − x)π). The solution for v(x, t) is
∞
v(x, t) = cn sin(nπx/9)e−4n2π2t/81,
n=1
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146
CHAPTER 5. THE HEAT EQUATION
29
cn = 9 ξ sin((9 − ξ)π) sin(nπξ/9) dξ
0
= − 324n(1 + (−1)n) π2(n − 9)2(n + 9)2
17.
where
for n ̸= 9, and
c9 = 9
0
29 9
ξsin((9−ξ)π)sin(πξ)dξ = 2. The solution of the original problem is
u(x, t) = ekAtv(x, t).
Let u(x, t) = v(x, t) + h(x) and substitute this into the initial-boundary value problem to choose h(x) and obtain a standard problem for v(x,t). We find that x
h(x)=T 1−L and the problem for v(x, t) is
vt =9vxx for0
v(x, 0) = −T 1 − x . L
This has the solution
v(x, t) = cn sin(nπx/L)e−9n2 π2 t/L2 ,
where
∞
n=1
2L ξ 2T cn=L −T 1−L sin(nπξ/L)dξ=−nπ.
0
Then
u(x,t)=T1−x−2T∞ 1sin(nπx/L)e−9n2π2t/L2.
n=1 The diffusion equation in this case is
ut = duxx − dηux. L
Let u(x,t) = eαx+βtv(x,t) and solve for α and β to obtain a standard problem for v(x, t). We find that
dη d3η2 α = 2L and β = − 4L2 .
18.
Lπn
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5.2. THE HEAT EQUATION WITH A FORCING TERM F(X,T) 147 The problem for v(x, t) is
vt =dvxx for0
v(x,0)=e−dηx/2Lu(x,0)= kLe−dηx/2L1−e−η(1−x/L). dη
The solution for v(x, t) is
where
∞
v(x, t) = cn sin(nπx/L)e−dn2 π2 t/L2 ,
n=1
cn = 2 L kLe−dηξ/2L 1 − e−η(1−ξ/L) dξ L 0 dη
= 4Lk d−2−de−η +2e−dη/2 d2η2(d − 2)
ifd̸=2. Ifd=2,then
cn =−Lk(e−η +1+η). η2
This determines v(x, t) and therefore u(x, t).
5.2 The Heat Equation With a Forcing Term
F(x,t)
In Problems 1–5, notation of the text is used for Bn(t), bn and Tn(t). Note that the second term in the solution for u(x, t) is the solution to the problem without the forcing.
1. Herek=4,L=π,f(x)=x(π−x)andF(x,t)=t. Weneed 2π 2tn
and
tsin(nξ)dξ=nπ(1−(−1) ), f(ξ)sin(nξ)dξ = nπ3(1−(−1) ),
t22 e−4n (t−τ)Bn(τ)dτ+bne−4n t
0
1 (1 − (−1)n)(−1 + 4n2t + e−4n2t).
8πn5
Bn(t)=π 2π4n
bn = π Tn(t)=
=
0
0
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148
CHAPTER 5. THE HEAT EQUATION
3.
First,
and
The solution is
u(x, t) = ∞ 1 (1 − (−1)n)(−1 + 4n2t + e−4n2t) sin(nx)
n=1 + ∞
n=1
Compute
Bn(t) = 2
8πn5
4 (1 − (−1)n ) sin(nx)e−4n2 t .
πn3
2.
and
and Tn(t)=
8 n+1 ξ sin(t) sin(nπξ/4) dξ = nπ (−1)
142n bn = 2 sin(nπξ/4)dξ = nπ(1−(−1) )
0
sin(t)
22
14 0
∞ 128(−1)n
n=1 + ∞
n=1
nπ(n4π4 +256)(16cos(t)−n2π2sin(t)−16e−n π t/16)sin(nπx/4) 2 (1 − (−1)n) sin(nπx/4)e−n2π2t/16.
nπ
Bn(t) = 5
t cos(ξ) sin(nπξ/5) dξ
2t ((−1)n+1(nπ + 5) + nπ),
25
0
n2π2 −25
=
25 500
ξ2(5 − ξ) sin(nπξ/5) dξ = n3π3 ((−1)n+1 − 1) Tn(t)= 50(1−cos(5)(−1)n)n2π2t−25+25e−n2π2t/25.
bn = 5
0
n3π3(n2π2 − 25)
The solution is
u(x,t)=∞ 50(1−cos(5)(−1)n)(n2π2t−25+25e−n2π2t/25)sin(nπx/5)
n3π3(n2π2 − 25)
+ ∞ 500 ((−1)n+1 − 1) sin(nπx/5)e−n2π2t/25.
n3π3
n=1
n=1
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5.2. THE HEAT EQUATION WITH A FORCING TERM F(X,T) 149
Figure 5.1: Solution surface for Problem 3, without effects of the forcing term included.
Figure 5.2: Solution surface for Problem 3, including effects of the forcing term.
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150 CHAPTER 5. THE HEAT EQUATION
Sometimes a graphic can display features of a solution. This is done for Problem 3 (Section 5.2). Figure 5.1 shows part of a surface plot of the solution without the forcing term, and Figure 5.2 shows the solution with the forcing term. In Figure 5.1, the temperature decreases quickly to zero, without the introduction of new energy, while in Figure 5.2 this does not occur.
4. First we need Bn(t) =
and
bn = 3 Tn(t) =
K sin(nπξ/3) dξ = nπ (1 − (−1) ),
1 2K
K sin(nπξ/2) dξ = nπ (1 − cos(nπ/2)), b1 =1andbn =0forn=2,3,···,
Tn(t) = 2K (1 − cos(nπ/2))(1 − e−n2π2t). n3π3
0
The solution is
u(x,t)=∞ 2K (1−cos(nπ/2))(1−e−n2π2t)sin(nπx/2)
5.
First compute
Bn(t) = 3
n3π3
+ sin(πx/2)e−π2t.
n=1
23 6t n+1
ξt sin(nπξ/3) dξ = nπ (−1) , 23 2Kn
0
and
0 27(−1)n+1
128n5π5 ∞ 27(−1)n+1
(16n2π2 − 9 + 9e−16n π t/9). 2 2
The solution is
u(x, t) = 128n5π5 (16n2π2 − 9 + 9e−16n π t/9) sin(nπx/3)
2 2
5.3
1.
The Heat Equation on the Real Line
aω = π
n=1
+ ∞ 2K (1 − (−1)n) sin(nπx/5)e−16n2π2t/9.
nπ
n=1
With f(x) = e−|x|, compute
1∞ 81
−∞
e−|ξ| cos(ωξ) dξ = π 16 + ω2
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5.3. THE HEAT EQUATION ON THE REAL LINE 151 and bω = 0 because f(x) is an even function on the real line. The solution
is
u(x,t) = 8 ∞ 1 cos(ωx)e−ω2kt. π 0 16+ω2
The solution can also be written in the form 1∞2
u(x, t) = √ e−|ξ|e−(x−ξ) /4kt dξ. 2 πkt −∞
2. The coefficients in the Fourier integral solution are aω = 0 because f(x) is an odd function, and
The solution is
π0 ω2−1 Alternatively, we can write the solution as
1π2
u(x, t) = √ sin(ξ)e−(x−ξ) /4kt dξ.
2 πkt −π
bω =π
−π
1π 2sin(ωπ)
sin(ξ)sin(ωξ)dξ=π(ω2−1). u(x, t) = 2 ∞ sin(ωπ) sin(ωx)e−ω2kt.
3. The coefficients are
14 1
ξ cos(ωξ) dξ = πω2 (4ω sin(4ω) + cos(4ω) − 1) 14 1
(aω cos(ωx) + bω sin(ωx))e−ω kt dω.
and
aω = π bω = π
0
0
The solution is
u(x, t) =
We can also write
ξ sin(ωξ) dξ = πω2 (sin(4ω) − 4ω cos(4ω)). ∞2
1 1 −1
2 cos(ω) sinh(1) + ω sin(ω) cosh(1) e−ξcos(ωξ)dξ=π ω2+1
0
u(x, t) =
142 √ ξe−(x−ξ)
2 πkt 0
/4kt dξ.
4. We need
aω =π
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152
CHAPTER 5. THE HEAT EQUATION
−1
(aω cos(ωx) + bω sin(ωx))e−ω kt dω. The solution can also be written
and
bω = π
11 2
e−ξ sin(ωξ) dξ = π ω cos(ω) sinh() − sin(ω) cosh(1)ω2 + 1. The solution is
2∞ 2
u(x, t) = π
u(x,t) = √ e−ξe−(x−ξ) /4kt dξ.
0
112 2 πkt −1
In each of Problems 5–8, the solution has the form
∞2 [aωcos(ωx)+bωsin(ωx)]e−ω kt
0
and just aω and bω are given
5. aω = 0 because f(x) is an even function, and
6.
and
bω = 4(1 − cos(ω)). πω
aω =− 2 (2sin(ω)+3sin(3ω)−7sin(9ω)) πω
bω = 2 (2cos(ω)+3cos(3ω)−7cos(9ω)+2) πω
u(x,t)=
7. Each bω = 0, while
8.
and
9. Let
aω = 2 πω2
aω = 2cos(πω/2). π(1 − ω2)
(cos(ω) + 2ω sin(ω) − 2 + cos(2ω) + 3ω sin(2ω)),
bω = 2 πω2
(− sin(ω) + 2ω cos(ω) + sin(2ω) − 3ω cos(2ω))
F(x) =
∞2
e−ζ cos(xζ)dζ.
0
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5.4.
THE HEAT EQUATION ON A HALF-LINE 153 By differentiating under the integral sign and then integrating by parts,
we obtain
Therefore
∞2
−ζe−ζ sin(xζ)dζ
F′(x) =
= 2ζe−ζ sin(xζ) − 2x e−ζ cos(xζ)dζ
0
1 2 ∞ 1∞ 2 00
1∞2
e−ζ cos(xζ)dζ = −1xF(x).
=−2x 2
0
The linear differential equation
F′(x) + 1xF(x) = 0 2
has the general solution
with k an arbitrary constant. However, we also know that
F(x) = ke−x2/4,
∞2 1√
e−ζ dζ=2 π, ∞ 2 1√ 2
F(0)=
an integral that can be found in tables and is widely used in statistics.
0
F(x)= Now let x = α/β to obtain
e−ζ cos(xζ)dζ=2 πe−x /4.
0
∞ 2 αζ 1√ 2 2
e−ζcos β dζ=2πe−α/4β. 0
Finally, this integral is half the value of the integral of the same function from −∞ to ∞, so
∞ 2 αζ √ 2 2
e−ζ cos β dζ = πe−α /4β . −∞
5.4
This is equation (5.17).
The Heat Equation on a Half-Line
In each of Problems 1–4, the initial condition is u(x, 0) = f (x) and the solution
has the form
where
u(x, t) =
∞2
bω sin(ωx)e−kω t dω,
bω = π
0
0
2∞
f(ξ)sin(ωξ)dξ.
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154 CHAPTER 5. THE HEAT EQUATION 1. Compute
bω = π so the solution is
2∞ 2ω
e−αξ sin(ωξ) dξ = π ω2 + α2 , u(x, t) = 2 ∞ ω sin(ωx)e−kω2 t dω.
0
2. First,
π0 ω2+α2
2∞ 2
bω = π so the solution is
0
ξe−αξ sin(ωξ) dξ = π αω(α2 + ω2)2,
u(x, t) = 4 ∞ αω π0 (α2+ω2)2
3. The coefficients are
sin(ωx)e−kω2 t dω.
bω = π and the solution is
0
2 h
2 1 − cos(hω) sin(ωξ)dξ = π ω
,
u(x,t)= 2 ∞ 1−cos(hω)sin(ωx)e−kω2tdω. π0ω
4. The coefficients are
2 2
bω = π so the solution is
2 sin(2ω) − 2ω cos(2ω)
ω2 , −kω2t
In Problems 5–8, the heat equation is to be solved on the half-line x > 0, but the initial condition is now the insulation condition ux(x,0) = f(x). Now
ξ sin(ωξ) dξ = π
2 ∞ sin(2ω) − 2ω cos(2ω)
0
u(x,t)=π 0 ω2
sin(ωx)e dω.
the solution is
where
u(x, t) =
∞2
aω cos(ωx)e−ω kt dω,
0
2∞
aω = π
f(ξ)cos(ωξ)dξ. We will just give aω for each problem.
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0
5.5.
5.
6.
7.
8.
5.5
THE TWO-DIMENSIONAL HEAT EQUATION 155
24
aω = π = 2
ξ(ξ+1)cos(ωξ)dξ
(20ω2 sin(4ω) − ω − 2 sin(4ω) + 9ω cos(4ω))
aω = π = 2
ξ2 cos(ωξ)dξ
(−2 sin(πω) + ω2π2 sin(πω) + 2ωπ cos(πω)
29
2∞
0 πω3
2π
0 πω3
aω = π
4cos(ωξ)dξ
= 8(sin(9ω) − sin(5ω))
πω
5
where
and
∞∞
cnm sin(nπx/L) sin(mπy/K)e−αnmkt, n=1 n=1
n2π2 m2π2 αnm = L2 + K2
4LK
e−ξ sin(ξ) cos(ωξ) dξ = 4−2ω2
aω = π
π(2 + 2ω + ω2)(2 − 2ω + ω2)
0
The Two-Dimensional Heat Equation
With the condition of zero initial temperature on the sides of the rectangle, and initial temperature u(x, y, 0) = f (x, y), the solution is
cnm = LK temperature function.
f(ξ,η)sin(nπξ/L)sin(mπη/K)dηdξ.
In the problems we will give the values of αnm and cnm for the particular initial
00
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156 CHAPTER 5. THE HEAT EQUATION
1. Here k = 1, L and K are positive numbers, and f(x, y) = x(L − x)y2(K − y).
Because f(x,y) is a product of a function of x and a function of y, we have
4 L K cnm = LK ξ(L − ξ) sin(nπξ/L) dξ η2(K − η) sin(mπη/K) dη
= −
00 16 n3m3π6(1 − (−1)n)(1 + 2(−1)m))
L2K3 and
n2π2 m2π2 αnm = L2 + K2 .
2. Nowk=4,L=2andK=3,and
f(x, y) = x2(2 − x)(3 − y) sin(y).
Now
cnm = π2 ξ2(2 − ξ) sin(nπξ/2) dξ
4 π π 00
16L2K3
= n3m3π6 (−1 + (−1)n)(1 + 2(−1)m).
(3 − η) sin(η) sin(mη/3) dη
And
n2π2 m2π2 αnm= 4 + 9 .
3. Nowk=1andL=K=π,and
4 π π
cnm = π2 Now,
π 0
sin(ξ) sin(nξ) dξ 00
η cos(η/2) cos(mη) dη
.
sin(ξ) sin(nξ) dξ =
π/2 ifn=1,
0 for n = 2,3,···.
Therefore, in the double summation for u(x, y, t), we have only c1m terms and the summation is for m = 1 to ∞. Completing the computation of the integrations with respect to η, we obtain
Further,
The solution is
u(x, y, t) =
32m(−1)m+1 c1,m = (4m2 − 1)2 .
m=1
α1m =1+m2. ∞ 32m(−1)m+1
−(1+m2 )t (4m2 − 1)2 sin(x) sin(my)e .
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Chapter 6
The Wave Equation
6.1 Wave Motion on an Interval
For each of Problems 1–8, the problem involves the wave equation on [0, L], with fixed ends, initial position y(x,0) = f(x), and initial velocity yt(x,0) = g(x). The solution is
where
and
∞
y(x, t) = [an cos(nπct/L) + bn sin(nπct/L)] sin(nπx/L), n=1
2L
an = L 2L
f(ξ)sin(nπξ/L)dξ
bn = nπc g(x) =
0
g(ξ)sin(nπξ/L)dξ.
1. Here c = 1,L = 2, the initial position is given by f(x) = 0, and the initial
velocity is
0
2x for0≤x≤1, 0 for 1 < x ≤ 2.
In the general expression for the solution, then, we have an = 0 for n = 1,2,··· and
22
bn = nπ 21
= nπ = 8
n3π3
0
0
g(ξ)sin(nπξ/2) 2ξ sin(nπξ/2) dξ
[2 sin(nπ/2) − nπ cos(nπ/2)]. 157
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158
CHAPTER 6. THE WAVE EQUATION
1 [2 sin(nπ/2) − nπ cos(nπ/2)] sin(nπx/2) sin(nπt/2). n3
The solution is y(x, t) = 8 ∞
π3
2. Now c = 3,L = 4,f(x) = 2sin(πx) and g(x) = 0, so the string is lifted to
n=1
its initial position and released from rest. Now each bn = 0, and 14 2 ifn=4,
an = 2 The solution is
1sin(πξ)sin(nπξ/4)dξ = 0 for n = 1,2,3,5,6,···. y(x, t) = 2 sin(πx) cos(3πt).
0
3. Eachan =0and
13 54n
ξ(3−ξ)sin(nπx/3)dξ = n4π4(1−(−1) ). y(x,t)=∞ 54 (1−(−1)n)sin(nπx/3)sin(2nπt/3).
bb = nπ The solution is
n=1
Because (1−(−1)n) is 2 if n is odd, and zero if n is even, we can also write the solution by summing only over the odd positive integers. This is achieved by replacing n with 2n − 1 in the expression being summed, and replacing each (1 − (−1)n) with 2:
y(x,t)=sin(x)cos(3t)+ 4 ∞ 1 sin((2n−1)x)sin(3(2n−1)t). 3π (2n − 1)2
0
6. The solution is y(x, t) = 5 ∞
π3
n=1
24 ∞ n=1
n4π4
y(x, t) = ∞ 108 sin((2n − 1)πx/3) sin(2(2n − 1)πt/3).
(2n − 1)4π4
4. After carrying out the integrations for the coefficients, we get
5. The solution is y(x, t) = π
n=1
(−1)n+1 √ (2n − 1)2 sin((2n − 1)x/2) cos((2n − 1) 2t).
1 [5 sin(4nπ/5) + nπ cos(4nπ/5)] sin(nπx/5) sin(2nπt/5). n3
n=1
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6.1. WAVE MOTION ON AN INTERVAL 159 7. The solution is
y(x, t) = − 32 ∞ 1 sin((2n − 1)πx/2) cos((3(2n − 1)πt/2)
π3
+ 4 ∞ 1 [cos(nπ/4) − cos(nπ/2)] sin(nπx/2) sin(3nπt/2).
(2n − 1)3 π2 n2
n=1
n=1 8. The solution is
y(x,t)=sin(2x)cos(10t)+2∞ 1 sin(nx)sin(5nt). 5 n2
n=1
9. Let y(x, t) = Y (x, t) + ψ(x) and substitute into the wave equation
ytt = Ytt = 3yxx +2x = 3Yxx +3ψ′′(x)+2x. Choose ψ(x) so that 3ψ′′(x) + 2x = 0. This means that
ψ(x)=−1x3 +cx+d. 9
Now,
soletd=0tohaveψ(0)=0. ThenY(0,t)=0.
y(0, t) = Y (0, t) + ψ(0) = 0
Next
WewillhaveY(2,t)=0ifc=4/9. Thismeansthat
y(2, t) = Y (2, t) + ψ(2) = Y (2, t) − 8 + 2c = 0. 9
ψ(x) = −1x3 + 4x = 1x(4 − x2). 999
The problem for Y (x, t) is
Ytt =3Yxx for0
Y (0, t) = Y (2, t) = 0, Y(x,0)=y(x,0)−ψ(x)= 1x(x2 −4).
9
The solution for Y (x, t) is
Y (x, t) =
The original problem has the solution
9
∞ 32(−1)n √
3 n3π3 sin(nπx/2) cos(nπ y(x,t)=Y(x,t)+ 1x(4−x2).
t/2).
n=1
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160 10.
CHAPTER 6. THE WAVE EQUATION
Here L = 4 and c = 3. Let y(x,t) = Y(x,t)+ψ(x) and substitute this into the wave equation to get
Ytt = 9(Yxx + ψ′′(x)) + x2. Set 9ψ′′(x) + x2 = 0. By two integrations,
ψ(x)=− 1 x4 +cx+d. 108
Now,y(0,t)=Y(0,t)+ψ(0)=0willgiveusY(0,t)=0ifψ(0)=0. Thus choose d = 0.
Next, y(4,t) = Y(4,t)+ψ(4) = 0 will give us Y(4,t) = 0 if ψ(4) = 0. Then c = 16/27, so
ψ(x)= 1×4+16x. 108 27
The problem for Y (x, t) is
Ytt =9Yxx for0
Y (0, t) = Y (4, t) = 0,
Y (x, 0) = y(x, 0) − ψ(x) = sin(πx) + 108 − 27 x,
Yt(x,0) = 0.
This has a solution of the form
x4 16
where
if n ̸= 4, and
∞
Y (x, t) = an cos(3nπt/4) sin(nπx/4), n=1
1 4
an = 2 = 8
ξ4 16
sin(πξ) + 108 − 27 ξ sin(nπξ/4) dξ
1 ξ4 16
a4=2 sin(πξ)+108−27ξ sin(πξ)dξ
1 8 + 9π3 =9 π3 .
0
9 n5π5(n2 − 16)
1 128n2 − 128n1(−1)n − 2048 +2048(−1)n + 64n4π2(−1)n − 1024n2π2(−1)n ,
With these, the solution of the original problem is y(x, t) = Y (x, t) + ψ(x).
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6.1. WAVE MOTION ON AN INTERVAL 161
11. Let y(x, t) = Y (x, t) + ψ(x). Substitute this into the wave equation to get ytt = Ytt = yxx = Yxx + ψ′′(x) − cos(x).
This will give us Ytt = Yxx if ψ′′(x) = cos(x), which means that ψ(x) = − cos(x) + cx + d.
Now
y(0, t) = 0 = Y (0, t) + ψ(0) = −1 + d. ThiswillgiveusY(0,t)=0ifd=1. Next,
y(2π, t) = 0 = Y (2π, t) − cos(2π) + 2πc + 1. ThiswillgiveusY(2π,0)=0ifc=0. Then
Finally,
ψ(x) = − cos(x) + 1.
y(x, 0) = Y (x, 0) − cos(x) + 1 = 0
implies that Y (x, 0) = cos(x) − 1. And
yt(x,0) = Yt(x,0) = x.
The problem for Y (x, t) is:
Ytt =Yxx for0
Y (x, 0) = cos(x) − 1, Yt(x, 0) = x. This has a solution of the form
where
and
∞
Y (x, t) = [an cos(nt/2) + bn sin(nt/2)] sin(nx/2), n=1
1 2π
an = π
16
(cos(ξ) − 1) sin(nξ/2) dξ
0
nπ(n2 −4)
bn = π
0
=
if n is odd, 0 if n is even,
2 2π 8 n+1
ξ sin(nξ/2) dξ = n2 (−1) .
These coefficients determine Y (x, t), and then y(x, t) = Y (x, t)+1−cos(x).
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162 12.
CHAPTER 6. THE WAVE EQUATION
Let y(x, t) = Y (x, t) + ψ(x) and substitute into the wave equation to get Ytt = 9(Yxx + ψ′′(x)) + 5×3.
Choose ψ(x) so that 9ψ′′(x) = −5×3. This requires that ψ(x)=− 1 x5 +cx+d.
by choosing d = 0. And
ψ(4)=− 1 44 +4c=0 36
requires that c = 64/9. Then
ψ(x) = − 1 x5 + 64x.
36 9
Now Y satisfies:
Ytt =9Yxx for0
Y (0, t) = Y (4, t) = 0,
Y(x,0)=f(x)−ψ(x)=1−cos(πx)+ 1×5−64x,yt(x,0)=0.
36 9
36 TohaveY(0,t)=Y(4,t)=0,weneedψ(0)=ψ(4)=0. Wegetψ(0)=0
This has the solution
Y (x, t) = an cos(3nπt/4) sin(nπx/4),
where
∞
n=1
15 64
1−cos(πξ)+ 36ξ − 9 x sin(nπξ/4)dξ.
a4 = 20 8π2 − 3 9 π5
an = 2 We find that
while, for n ̸= 4,
0
14
an =
32 9n6π6(n2 − 16)
−9n5π5 + 30720nπ(−1)n + 9n5π5(−1)n +320n5π3(−1)n − 5120n3π3(−1)n − 1920n3π(−1)n .
The original problem has the solution
y(x, t) = Y (x, t) + ψ(x).
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6.1. WAVE MOTION ON AN INTERVAL 163 13. Let y(x, t) = Y (x, t) + ψ(x) and substitute this into the wave equation of
the problem to choose ψ(x) so that
7ψ′′(x)+e−x =0,ψ(0)=ψ(2)=0.
This leads to
Now
ψ(x)=−1e−x+ 1(e−2−1)x+1. 7 14 7
Ytt =7Yxx for0
Y (x, 0) = −ψ(x), Yt(x, 0) = 5x. This has the solution
Y (x, t) = ∞ an cos(nπ√7t/2) + bn sin(nπ√7t/2) sin(nπx/2), n=1
where
an= =
and
7e −14(e −1)ξ−7 sin(nπξ/2)dξ
21−ξ 1−2 1
0
7nπ(4 + n2π2)
bn =
nπ70
These coefficients determine Y (x, t), and then y(x, t) = Y (x, t) + ψ(x).
14. Let y(x, t) = Y (x, t) + ψ(x). Solve
4ψ′′ +cos(πx)=0;ψ(0)=ψ(4)=0
2
(−4 − n2π2e−2(−1)n + e−1n2π2(−1)n + 4e−1(−1)n),
2 2 √
5ξsin(nπξ/2)dξ=
40(−1)n+1 2 2√ .
to get
Then
ψ(x) = 1 (cos(πx) − 1). 4π2
Ytt =4Yxx for0
Y (x, 0) = x(π − x) − ψ(x), Yt(x, 0) = x2.
nπ 7
Then
Y (x, t) = [an cos(nπt/2) + bn sin(nπt/2)] sin(nπx/4) dx,
n=1
∞
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164
CHAPTER 6. THE WAVE EQUATION
141
where
and
n4π4 Finally, y(x, t) = Y (x, t) + ψ(x).
(a) Substitute y(x, t) = X(x)T (t) into the fourth-order differential equa- tion to get
X(4) −λX =0,T′′ +λa4λT =0,
with λ the separation constant. Note – by rearranging terms differently, we can reach different separated equations for X and T . For example, we could have kept the a4 factor with the X terms.
(b) Consider cases on λ, noting that the boundary conditions are X′′(0) = X′′(π) = X(3)(0) = X(3)(π) = 0.
Case 1 – Suppose λ = 0. Then X(4)(x) = 0 and four integrations give us X(x)=A+Bx+Cx2 +DX3.
The boundary conditions force C = D = 0, while A and B are arbitrary. Therefore 0 is an eigenvalue of this problem, with eigenfunctions X0(x) = A + Bx, with A and B not both zero. In this case solutions for T are T (t) = α + βt.
Case 2 – Suppose λ < 0. The notation is simplified if we set λ = −4α4, with α > 0. The differential equation for X is
X(4) + 4α4X = 0,
with characteristic equation r4 + 4α4 = 0. This has roots
(1 + i)α,(1 − i)α,(−1 + i)α and (−1 − i)α. In this case solutions are
X(x) = eαx(A cos(αx) + B sin(αx)) + e−αx(C cos(αx) + D sin(αx)).
an = 2 = −
ξ(4 − ξ) − 4π2 (cos(πξ) − 1) sin(nπξ/4) dξ
0
n3π3(n2 − 16)
14
8
(128 − 7n2 + 7n2(−1)n − 128(−1)n),
ξ2 sin(nπξ/4) dξ
=− 64 (2(1−(−1n)+n2π2(−1)n).
bn = πn
0
15.
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6.1. WAVE MOTION ON AN INTERVAL
Apply the boundary conditions to this general solution to obtain:
B − D = 0,
A − B − C − D = 0,
− Aeαπ sin(απ) + Beαπ cos(απ) + Ce−απ sin(απ) − De−απ cos(απ) − Aeαπ (cos(απ) + sin(απ)) + Beαπ (cos(απ) − sin(απ))
− Ce−απ(cos(απ) − sin(απ)) − De−απ(cos(απ) + sin(απ)) = 0.
165
= 0,
This is a 4 × 4 homogeneous system of linear algebraic equations. This system has a nontrivial solution if and only if the determinant of the coefficients is zero:
cosh(2απ) − cos(2απ) = 0.
But this equation is satisfied only by α = 0, and in this case α > 0. Therefore the system has only the trivial solution A = B = C = D = 0, and this problem has no negative eigenvalue.
Case3-Supposeλ>0,sayλ=α4 withα>0. NowX(4)−α4X=0, and the characteristic equation has roots
α, −α, αi, −αi.
The general solution is
X (x) = A cos(αx) + B sin(αx) + C cosh(αx) + D sinh(αx).
The boundary conditions give us four equations:
− A + C = 0,
− A cos(απ) − B sin(απ) + C cosh(απ) + D sinh(απ) = 0, − B + D = 0,
A sin(απ) + B cos(απ) + C sinh(απ + D cosh(απ) = 0.
From the first and third equations, A = C and B = D. This reduces the
system to the second and fourth equations in two unknowns: C(cosh(απ) − cos(απ)) + D(cosh(απ) − sin(απ)) = 0,
C(sinh(απ) + sin(απ)) + D(cosh(απ) − cos(απ)) = 0.
This 2 × 2 system has a nontrivial solution if and only if the determinant
of the system is nonzero. This requires that cos(απ) cosh(απ) = 1.
It may not be obvious, but this equation has infinitely many positive solutions for α (two are 2.499752670 and 0.000000207171091). If these solutions for α are listed α1 < α2 < ···, then λn = αn4 is an eigenvalue of the problem. Eigenfunctions then have the form
Xn(x) = A cos(αnx) + B sin(αnx) + C cosh(αnx) + D sinh(αnx).
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166 16.
CHAPTER 6. THE WAVE EQUATION
Separation of variables yields
X′′ +λX =0;X(0)=X(L)=0
and
T′′ +AT′ +(B+c2λ)T =0;T′(0)=0.
The problem for X(x) is a familiar one, with eigenvalues and eigenfunc-
tions
n2π2
λn = L2 , Xn(t) = sin(nπx/L).
With these eigenvalues, the characteristic equation for the differential equation for T is
with roots
r2 +Ar+(B+c2n2π2/L2)=0
A 1 c2n2π2
r=−2±2 A2−4 B+ L2 .
The given condition that A2L2 < 4(V L2 + c2n2π2) ensures that these
roots for r are complex. Let
rn = 4(BL2 + c2n2π2) − A2L2.
The roots are then
r = −A ± rn i. 2 2L
Then
Now T′(0) = 0 gives us −Aan/2 + bnrn/2L = 0, so
bn = ALan. rn
By superposition,
Tn(t) = e−At/2[an cos(rnt/2L) + bn sin(rnt/2L)].
−At/2∞ AL
u(x, t) = e
To satisfy u(x, 0) = f (x), choose
an cos(rnt/2L) + r sin(rnt/2L) n
sin(nπx/L).
n=1
an = L
2L
f(ξ)sin(nπξ/L)dξ.
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0
6.2. WAVE MOTION IN AN UNBOUNDED MEDIUM 167
6.2 Wave Motion in an Unbounded Medium
For the wave equation on a the real line,
ytt =c2yxx for −∞
y(x, 0) = f(x), yt(x, 0) = 0,
specifying an initial position but zero initial velocity, a solution can be found very much like the problem for an interval [0, L], with Fourier integrals replacing the Fourier series seen in the bounded interval case. The solution is
where
and
f(ξ)cos(ωξ)dξ
y(x, t) =
∞ 0
1∞
aω = π 1∞
bω = π
but with initial velocity g(x)), then the solution is
[aω cos(ωx) + bω sin(ωx)] cos(ωct) dω,
−∞
f(ξ)sin(ωξ)dξ.
If y(x,0) = 0 and yt(x,0) = g(x) (string released without initial displacement,
where
and
1∞
g(ξ)cos(ωξ)dξ
g(ξ)sin(ωξ)dξ.
y(x, t) =
∞ 0
αω = πωc 1∞
βω = πωc
−∞
[αω cos(ωx) + βω sin(ωx)] sin(ωct) dω,
−∞
−∞
If the problem has f(x) and g(x) both nonzero, then the solution is the sum of the solution with zero initial velocity, and the solution with no initial displace- ment.
1. With c = 5, f(x) = e−5|x| and g(x) = 0, compute 1∞ 10
e−5|ξ| cos(ωξ) dξ = (25 + ω2)π and bω = 0 because f(x) is an even function. The solution is
aω = π
10 ∞ 1
y(x, t) = π 0 25 + ω2
−∞
cos(ωx) cos(12ωt) dω.
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168 CHAPTER 6. 2. The coefficients are
THE WAVE EQUATION
1 − cos(8ω) πω2
aω = π
1 8
1 8
(8−ξ)cos(ωξ)dξ =
and
0
0
bω = π The solution is
(8 − ξ) sin(ωξ) dξ =
8ω − sin(8ω) πω2 .
y(x, t) = +
0
∞1−cos(8ω)
πω2 cos(ωx) cos(8ωt) dω ∞8ω−sin(8ω)
πω2 sin(ωx) cos(8ωt) dω. 3. Compute the coefficients to determine the solution
y(x, t) = 4. The solution is
0
− 2πω(ω2 − 1) sin(ωx) sin(4ωt) dω.
0
∞ sin(πω)
y(x, t) = 5. The solution is
∞2
πω2 (2 − cos(2ω))
[αω cos(ωx) + βω sin(ωx)] sin(3ωt) dω,
where
and
1 ∞
αω = 3πω
1 ∞
1 2cos(ω)−ωsin(ω) e−2ξ cos(ωξ)dξ = 3e2πω 4+ω2
cos(ωx) cos(ωt) dω.
y(x, t) =
0
∞ 0
βω = 3πω 6. The solution is
1
1
1 2sin(ω)+ωcos(ω) e−2ξ sin(ωξ)dξ = 3e2πω 4+ω2
.
∞1−cos(2ω)
y(x, t) =
7. The solution for the problem with the given displacement and zero initial
velocity is
∞ 0
y1(x, t) =
[aω cos(ωx) + bω sin(ωx)] cos(7ωt) dω,
0
πω2 sin(ωx) sin(2ωt) dω.
© 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6.2. WAVE MOTION IN AN UNBOUNDED MEDIUM 169 where
15
f(ξ)cos(ωξ)dξ
= sin(ω) − 2 sin(2ω) + 3 sin(5ω)
aω = π 15
−1
πω
and
placement, is
y2(x, t) =
f(ξ)sin(ωξ)dξ
= cos(ω) + 2 cos(2ω) − 3 cos(5ω) .
bω = π
The solution for the problem with the given velocity, but zero initial dis-
where
and
αω = 7πω
−1 2
e−|ξ| cos(ωξ) dξ
(e−1 cos(ω) − ωe−1 sin(ω) − 1)
∞ 0
11
−1
πω
[αω cos(ωx) + βω sin(ωx)] sin(7ωt) dω,
Now,
where
y(x, t) = y1(x, t) + y2(x, t). ∞
√
= −
7π(1 + ω2) 11
e−|ξ| sin(ωξ) dξ = 0.
The solution of the problem with initial displacement f(x) and initial
βω = 7πω
y(x, t) = y1(x, t) + y2(x, t).
velocity g(x) is
−1
8. Let y1(x,t) be the solution of the problem with initial displacement f(x) and zero initial velocity, and y2(x,t) the solution with no initial displace- ment, but initial velocity g(x). Then
y1(x, t) =
[aω cos(ωx) + bω sin(ωx)] cos(
1 3π
sin(ξ)cos(ωξ)dξ = 4cos(πω)sin2(πω)
π(ω2 − 1)
7ωt).
0
aω = π
−π
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170
CHAPTER 6. THE WAVE EQUATION
and
And
where
and
Then
1 3π
bω = π ∞
sin(ξ)sin(ωξ)dξ = 4sin(πω)cos2(πω).
−π
αω = √ g(ξ)cos(ωξ)dξ 7πω −∞
y2(x, t) = 1∞
√
7ωt),
π(1 − ω2)
[αω cos(ωx) + βω sin(ωx)] sin(
0
=
2√ (8ω2sin(4ω)−sin(4ω)+4ωcos(4ω)) πω2 7
7πω4
y(x, t) = y1(x, t) + y2(x, t).
1∞ 7πω −∞
βω = √
= √−2 (1 + 8ω2 cos(4ω) − cos(4ω) − 4ω sin(4ω)).
g(ξ)sin(ωξ)dξ
9.
Following the notation of Problems 7 and 8, form y1(x, t) with coefficients
and
43
αω = πω ξ2 cos(ωξ)dξ
bω = π And y2(x, t) has coefficients
−2
πω2
aω = π 12
12
ξ cos(ωξ) dξ = 0
−2
ξsin(ωξ)dξ
= 2sin(2ω)−4cos(2ω).
−3
= 8 (9ω2 sin(3ω) − 2 sin(3ω) + 6ω cos(3ω))
πω4
and βω = 0.
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6.2. WAVE MOTION IN AN UNBOUNDED MEDIUM 171 The solution is
∞ 0
∞ 0
The the problem on a half-line [0, ∞), there is a boundary condition which we will take to be
y(0, t) = 0
along with initial conditions
y(x, 0) = f(x), yt(x, 0) = g(x)
for x > 0. The solution has the form ∞
where
and
2∞
Aω = π 2∞
f(ξ)sin(ωξ)dξ
y(x, t) =
[Aω cos(ωct) + Bω sin(ωct)] sin(nωx) dω,
0
y(x, t) = +
an cos(ωx) cos(ωt/4) βω sin(ωx) sin(ωt/4).
0
Bω = π 21
ξ(1−ξ)sin(ωξ)dξ 22 sin(ω)
g(ξ)sin(ωξ)dξ. 10. With zero initial velocity, we need only compute
0
Aω = π
=π ω3(1−cos(ω))− ω2 .
0
The solution is
2∞2 sin(ω)
y(x, t) = π 11. Here Aω = 0 and
0 ω3 (1 − cos(ω)) − ω2
2 11
sin(ωx) cos(3ωt) dω.
2sin(ωξ)dξ = 4(cos(4ω) − cos(11ω)) .
3πω2
4 ∞ cos(4ω) − cos(11ω)
Bω = 3πω
4
The solution is
y(x, t) = 3π 0 ω2 sin(ωx) sin(3ωt) dω.
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172
12. Aω =0and
The solution is y(x, t) =
CHAPTER 6. THE WAVE EQUATION
2 5π/2
cos(ξ)sin(ωξ)dξ = sin(ωπ/2) − sin(5ωπ/2) .
13. Withg(x)=0,Bω =0and
2∞
14. NowAω =0and
Bω = 14πω
= The solution is
15. Compute
and
ξ2(3 − ξ) sin(ωξ) dξ
(2 sin(3ω) − 4ω cos(3ω) − 3ω2 sin(3ω) − 2ω).
Bω = 2πω
π/2
πω(ω2 − 1)
∞ sin(ωπ/2) − sin(5ωπ/2)
sin(ωx) sin(2ωt) dω.
The solution is
y(x, t) = − 4 ∞ ω sin(ωx) cos(6ωt) dω.
0
πω(ω2 − 1)
−2e−ξ sin(ωξ) dξ =− 4ω .
Aω = π
0
π(1 + ω2)
π0 1+ω2 23
0 7πω5
3
y(x, t) =
∞ 0
21
f(ξ)sin(ωξ)dξ = 2sin(ω)
Aω = π
π2 − ω2
Bω sin(ωx) sin(14ωt) dω.
0
24
Bω = √ g(ξ)sin(ωξ)dξ
13πω 0
= √ 2 (1−2cos(ω)+cos(4ω)).
13πω2
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6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS 173 The solution is
y(x, t) =
16. Compute the coefficients
∞ 0
√√
[Aω cos( 13ωt) + Bω sin( 13ωt)] sin(ωx) dω.
2∞ 12ω
Aω = π
2 2π
ξe−3ξ sin(ωξ) dξ = π(9 + ω2)2
0
and
Bω = 5πω The solution is
1. The characteristics are the lines x−t = k1 and x+t = k2, with k1 and k2 arbitrary real numbers. d’Alembert’s solution of the problem is
12 21x+t y(x,t)=2[(x−t) +(x+t)]+2 −ξdξ
∞ 0
0
4sin2(πω) cos(ξ) sin(ωξ) dξ = (w2 − 1)π .
y(x, t) =
6.3 d’Alembert’s Solution and Characteristics
[Aω cos(5ωt) + Bω sin(5ωt)] sin(ωx) dω.
x−t
12 2 2 2 12x+t
= 2[x −2xt+t +x +2xt+t − 2ξ
x−t
= x2 − xt + t2.
2. characteristics: x − 4t = k1, x + 4t = k2;
y(x,t)= 1(x−4t)2 −2(x−4t)+(x+4t)2 −2(x+4t) 2
1 x+4t
+8 (cos(ξ)dξ
x−4t
= x2 + 16t2 − 2x + 1 cos(x) sin(4t)
4
3. characteristics: x − 7t = k1, x + 7t = k2;
y(x,t)= 1[cos(π(x−7t))+cos(π(x+7t))]+t−x2t− 49t3 23
This solution can also be written
y(x,t) = 1 cos(πx)cos(7πt)+t−x2t− 49t3. 23
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174 CHAPTER 6. THE WAVE EQUATION
4. characteristics: x − 5t = k1, x + 5t = k2;
y(x,t)= 1[sin(2(x−5t))+sin(2(x+5t))]+x3t+25xt3 2
= sin(2x) cos(10t) + x3 t + 25xt3
5. characteristics: x − 14t = k1, x + 14t = k2;
y(x, t) = 1 ex−14t + ex+14t + xt 2
= ex cosh(14t) + xt.
6. characteristics: x − 12t = k1, x + 12t = k2;
y(x,t)=x2 +144t2 −5x+3t. √√
7. characteristics x − 3t = k1, x + 3t = k2;
y(x, t) = 1 e−3|x−√3t| + e−3|x+√3t|
2√√
1 x+ 3t x− 3t
+√3 sin 2 −sin 2 8. characteristics: x − 9t = k1, x + 9t = k2;
y(x,t)= 1(2−cos(x−9t)−cos(x+9t)) 2
+ 1e−x+9t(cos(−x+9t)−sin(−x+9t)) 36
− 1e−x−9t(cos(x+9t)+sin(x+9t)) 36
9. The solution with y(x, 0) = f (x) = sin(x) is
y(x,t)= 1(sin(x−t)+sin(x+t)).
10. Let
yε(x,t)= 1sin(x−t)+ε+sin(x+t)+ε=y(x,t)+ε. 2
y(x, t) = F (x − ct) + G(x + ct)
=(x−ct)2 −(x−ct)+(x+ct)cos(x+ct).
x
With y(x, 0) = sin(x) + ε, the solution is
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6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS 175
Compute
yx(x, t) = 2(x − ct) − 1 + cos(x + ct) − (x + ct) sin(x + xt),
yxx(x, t) = 2 − sin(x + ct) − sin(x + ct) − (x + ct) cos(x + ct), yt(x, t) = 2(x − ct)(−c) + c + c cos(x + ct) − c(x + ct) sin(x + ct),
and
ytt(x,t)=2c2 −c2sin(x+ct)−c2sin(x+ct)−c2(x+ct)cos(x+ct).
Now it is routine to check that
ytt = c2yxx.
11. Now
Then
y(x, t) = e−3(x−ct) + sin(4(x + ct) = e−3xe3ct + sin(4(x + ct)).
yx = −3e−3x e3ct + 4 cos(4(x + ct)), yxx = 9e−3x e3ct − 16 sin(4(c + ct)),
yt = 3ce−3x e3ct + 4c cos(4(x + ct)), ytt = 9c2e−3xe3ct − 16c2 sin(4(x + ct)).
It is easy to check that ytt = c2yxx.
In each of Problems 12–17, with c = 1 and g(x) = 0, the forward wave is
F(x,t) = 1f(x−t) and the backward wave is B(x,t) = 1f(x+t). The solution 22
is
y(x, t) = F (x, t) + B(x, t). 12. Figures 6.1–6.7 are graphs of y(x, t) for
t = 0, 1, 3,2,3,4,6 24
respectively.
13. Figures 6.8–6.13 show graphs of y(x, t) for times
t = 0, 1 , 3 , 7 , 1, 3 . 248 2
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176 CHAPTER 6. THE WAVE EQUATION
Figure 6.1: y(x, 0) in Problem 12.
Figure 6.2: y(x, 1/2), Problem 12.
Figure 6.3: y(x, 3/4), Problem 12.
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6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS 177
Figure 6.4: y(x, 2), Problem 12.
Figure 6.5: y(x, 3), Problem 12.
Figure 6.6: y(x, 4), Problem 12.
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178 CHAPTER 6. THE WAVE EQUATION
Figure 6.7: y(x, 6), Problem 12.
Figure 6.8: y(x, 0), Problem 13.
Figure 6.9: y(x, 1/2), Problem 13.
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6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS 179
Figure 6.10: y(x, 3/4), Problem 13.
Figure 6.11: y(x, 7/8), Problem 13.
Figure 6.12: y(x, 1), Problem 13.
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180 CHAPTER 6. THE WAVE EQUATION
Figure 6.13: y(x, 3/2), Problem 13.
Figure 6.14: y(x, 0), Problem 14.
Figure 6.15: y(x, 1/4), Problem 14.
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6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS 181
Figure 6.16: y(x, 1/2), Problem 14.
Figure 6.17: y(x, 3/4), Problem 14.
Figure 6.18: y(x, 1), Problem 14.
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182 CHAPTER 6. THE WAVE EQUATION
Figure 6.19: y(x, 5/4), Problem 14.
14. Figures 6.14–6.21 show the graphs of y(x, t) for times t = 0, 1 , 1 , 3 , 1, 5 , 3 , 2.
424 42 15. Figures 6.22–6.27 show graphs of y(x, t) for
t = 0, 1 , 1 , 3 , 1, 3 . 424 2
16. Figures 6.28–6.34 show graphs of y(x, t) for
t = 0, 1 , 1 , 3 , 1, 5 , 7 .
424 44 17. Figures 6.35–6.41 show graphs of y(x, t) for
t = 0, 1, 1, 3, 5/4 7, 5. 424,42
18. Derive d’Alembert’s solution as follows. Begin with the fact that any solution of the wave equation on the line must look like
y(x, t) = F (x + ct) + G(x − ct),
for some twice differentiable functions F and G. To satisfy the initial
conditions, we need
y(x, 0) = F (x) + G(x) = f (x) and, by chain rule differentiations,
yt(x, 0) = −cF ′(x) + cG′(x) = g(x).
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6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS 183
Figure 6.20: y(x, 3/2), Problem 14.
Figure 6.21: y(x, 2), Problem 14.
Figure 6.22: y(x, 0), Problem 15.
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184 CHAPTER 6. THE WAVE EQUATION
Figure 6.23: y(x, 1/4), Problem 15.
Figure 6.24: y(x, 1/2), Problem 15.
Figure 6.25: y(x, 3/4), Problem 15.
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6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS 185
Figure 6.26: y(x, 1), Problem 15.
Integrate the last equation to obtain 1x
−F(x)+G(x)= c
Add this to the equation for y(x, 0) to obtain
1x
2G(x)=f(x)+ c
Solve this for G(x) to obtain 11x11
g(ξ)dξ− 2F(0)+ 2G(0). But then, from the equation for y(x, 0),
G(x)= 2f(x)+ 2c
0
g(ξ)dξ−F(0)+G(0).
0
g(ξ)dξ−F(0)+G(0).
0
F(x) = f(x) − G(x) 11x11
= 2f(x)− 2c From the last two equations,
y(x, t) = F (x − ct) + G(x + ct) 1 1 x−ct
= 2f(x−ct)− 2c 11x1
0
g(ξ)dξ+ 2F(0)− 2G(0).
1
g(ξ)dξ+ 2(F(0)−G(0))
0
+ 2f(x+ct)+ 2c
Upon combining the integrals and canceling terms where possible, this
g(ξ)dξ− 2(F(0)−G(0)).
0
1 1 x+ct
y(x,t)= 2(f(x−ct)+f(x+ct))+ 2c
yields d’Alembert’s formula. 19. We know that
g(ξ)dξ
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x−ct
186 CHAPTER 6. THE WAVE EQUATION
Figure 6.27: y(x, 3/2), Problem 15.
Figure 6.28: y(x, 0), Problem 16.
Figure 6.29: y(x, 1/4), Problem 16.
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6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS 187
and
Then
Figure 6.30: y(x, 1/2), Problem 16.
1 ̃ ̃1 y ̃(x,t)= 2(f(x−ct)+f(x+ct))+ 2cg ̃(ξ)dξ.
y(x, t) − y ̃(x, t)
1 1 x+ct
= 2 (f(x−ct)+f(x+ct))+ 2c
1 ̃ ̃ 1x+ct
1 x+ct
−2 f(x−ct)+f(x+ct) −2 1 ̃1 ̃
x−ct
g(ξ)dξ g ̃(ξ)dξ
Then
x−ct
= 2(f(x−ct)−f(x−ct))+ 2(f(x+ct)−f(x+ct))
x−ct
(g(ξ) − g ̃(ξ)) dξ.
+ 2c
|y(x, t) − y ̃(x, t)|
1 ̃1 ̃
≤ 2|f(x−ct)−f(x−ct)|+ 2|f(x+ct)−f(x+ct)|
1 x+ct
+ 2c |g(ξ) − g ̃(ξ)| dξ x−ct
1 1 1 x+ct
≤ 2ε1 + 2ε1 + 2c 2c
ε2 dξ ≤ε1+ 1ε2((x+ct)−(x−ct))
x−ct
= ε1 + ε2t.
20. For the problem on the real line, with initial position fo(x) and initial
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188 CHAPTER 6. THE WAVE EQUATION
Figure 6.31: y(x, 3/4), Problem 16.
Figure 6.32: y(x, 1), Problem 16.
Figure 6.33: y(x, 5/4), Problem 16.
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6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS 189
Figure 6.34: y(x, 7/4), Problem 16.
Figure 6.35: y(x, 0), Problem 17.
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190
CHAPTER 6. THE WAVE EQUATION
velocity go(x), the d’Alembert solution is
1 1 x+ct
y(x,t)= 2(fo(x−ct)+fo(x+ct))+ 2c allx>0. Further,ifx≥0,
g(ξ)dξ.
This function satisfies the wave equation for all x, and therefore also for
y(x,0)= 1(fo(x)+fo(x))= 1(f(x)+f(x))=f(x). 22
And,
yt(x,0)=1(−cfo′(x)+cfo′(x))+ 1(cgo(x)−(−c)go(x))
Therefore y(x, t) is also a solution of the problem on the half-line x ≥ 0.
The Wave Equation With a Forcing Term
K(x,t)
Herec=4,f(x)=x,g(x)=e−x andK(x,t)=x+t. Thesolutionis
x−ct
6.4
1.
2 2c = go(x) = g(x).
1 1 x+4t
e−ξ dξ
y(x, t) = 2 ((x − 4t) + (x + 4t)) + 8 1 t x+4t−4T
x−4t + 8 (X + T ) dX dT
0 x−4t+4T
= x + 1 e−x+4t − e−x−4t
8
t
+ (xt−xT +tT −T2)dT
2.
0
= x + 1e−x e4t − e−4t + 1t3 + 1xt2.
862 We can also write this solution as
y(x,t) = x + 1e−x sinh(4t) + 1t3 + 1xt2. 462
We are given c = 2,f(x) = sin(x),g(x) = 2x and K(x,t) = 2xt. The
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6.4. THE WAVE EQUATION WITH A FORCING TERM K(X, T ) 191 solution is
1 1 x+2t
y(x, t) = 2 (sin(x − 2t) + sin(x + 2t)) + 4 1 t x+2t−2T
8 (X + T ) dX dT 0 x−2t+2T
= 1(sin(x − 2t) + sin(x + 2t)) + 1 (x + 2t)2 − (x − 2t)2 24
1t
T (x+2t−2T)2 −(x−2t+2T)2 dT.
After carrying out the last integration, this expression reduces to
y(x,t)= 1(sin(x−2t)+sin(x+2t))+2xt+ 1xt3. 23
x−2t
2ξ dξ
+ 4
0
3. The solution is
1 1 x+8t
y(x,t)= 2(f(x−8t)+f(x+8t))+ 16 1 t x+8t−8T
cos(2ξ)dξ
=x2 +64t2 −x+ 1 (sin(−2x+16t)+sin(2x+16t)) 32
t
+ −xT2(−t+T)dT
0
=x2 +64t2 −x+ 1 (sin(−2x+16t)+sin(2x+16t))+ 1 xt4.
+ 16 XT2 dX dT 0 x−8t+8T
x−8t
4.
1
y(x,t)= (x−4t)2+(x+4t)2+
32
32
1 x+4t
ξe−ξdξ
0 x−4t+4T
= x2 + 16t2 + 1e−x+4t + 1xe−x+4t − 1te−x+4t
882
− 1e−x−4t − 1xe−x+4t − 1te−x−4t 882
t
+ x(t−T)sin(T)dT
2
1 t x+4t−4T
8 x−4t + 8 X sin(T ) dX dT
0
= x2 + 16t2 + 1 (1 + x)e−x+4t − 1 t e−x+4t + e−x−4t
82
− 1(1+x)e−x−4t +xt−xsin(t). 8
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192 5.
CHAPTER 6. THE WAVE EQUATION
1
y(x, t) = 2 (cosh(x − 3t) + cosh(x + 3t)) + 6
1 t x+3t−3T
+ 6 3XT3 dX dT 0 x−3t+3T
1t
dξ
1 x+3t
x−3t
= 2(cosh(x−3t)+cosh(x+3t))+t+
= 1(cosh(x−3t)+cosh(x+3t))+t+ 3 xt5.
y(x,t)= 1((1+x−7t)+(x+x+7t)) 2
1 t x+7t−7T
−3xT3(T −t)dT
0
6.
2 20
+ 14
= 1 + x +
(X − cos(T )) dX dT
(xt − xT − t cos(T ) + T cos(T )) dT
6.5
0
= x + 1 xt2 + cos(t).
2
The Wave Equation in Higher Dimensions
0
x−7t+7T t
For Problems 1–3 the solution has the form
∞∞
z(x, y, t) = anm sin(nπx/L) sin(mπy/K) cos(αnmπct), n=1 m=1
where
and
αnm =
4LK
n2 m2 L2 + K2
anm = LK
we can compute the coefficients as a product of integrals:
f(ξ,η)sin(nπξ/L)sin(mπη/K)dηdξ.
1. Because f(x,y) = x2y is a product of a function of x and a function of y,
00
1 2π K anm = π2 ξ2 sin(nξ/2) dξ
η sin(mη/2) dη 2(1 − (−1)n) + n2π2(−1)n .
αnm= 1n2+m2. 2π
=
32(−1)m mn3 π
00
Further,
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6.5. THE WAVE EQUATION IN HIGHER DIMENSIONS 193 Because c = 1, the solution is
Now
And
anm = 1
for n = 2,3,···,
z(x, y, t) =
4 0
0
− 4m(−1+(−1)m )
if m = 2, if m ̸= 2.
∞∞
32(−1)m
mn3π (2(1 − (−1)n)
+ n2π2(−1)n sin(nx/2) sin(my/2) cos(n2 + m2t/2).
2. Now c = 3,L = 1 and K = 4, and f(x,y) = sin(πx)cos(πy/2). The coefficients in the solution are
n=1 m=1
1 4 sin(πξ)sin(nπξ)dξ
00
cos(πη/2)sin(mπη/4)dη.
0
1/2 forn=1.
π(m2−4)
single summation over m, with m = 2 excluded. Further, m2
sin(πξ) sin(nπξ) dξ = 0
cos(πη/2) sin(mπη/4) dη =
This means that n will only assume the value n = 1 and the solution is a
α1m= 1+16 4m(−1 + (−1)m)
for m = 1,3,4,5,···. Then
−
anm = π2 ξ sin(nξ) dξ eη sin(mη) dη 00
∞ m=1,m̸=2
2π(m2 − 4) sin(nπx) sin(mπy/4) cos(3α1mπt). 4π π
z(x, y, t) =
3. We have c = 1,L = K = π and f(x,y) = xey. A routine integration yields
=
4(−1)n+1m (1 − eπ(−1)m). πn2(m2 + 1)
αnm = 1n2 +m2. π
Further,
The solution is
z(x, y, t) = anm sin(nx) sin(my) cos(2n2 + m2t).
n=1 m=1
∞∞
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194 4.
CHAPTER 6. THE WAVE EQUATION
Now suppose that the membrane is initially unmoved and that an initial velocity is given by zt(x, y, 0) = g(x, y). Separate variables in the two- dimensional wave equation. Because the boundary conditions are the same as in the zero initial velocity case, we obtain
and
n2π2 λn = L2
m2π2 μm = K2
, Xn(x) = sin(nπx/L)
, Ym(y) = sin(mπy/K).
The problem for T is
T+c L2+K2 T=0,
2 n2π2 m2π2 but now we have T(0) = 0, not T′(0) = 0. Then
where
Tnm(t) = cos(αnmπct), n2 m2
′′
αm = L2 + K2 . We therefore are led to attempt a solution
∞∞
z(x, y, t) = bnm sin(nπx/L) sin(mπy/K) cos(αnmπct). n=1 m=1
We need
zt(x, y, 0) = αnmπc sin(nπx/L) sin(mπy/K) = g(x, y).
n=1 m=1
This is a Fourier double series expansion of g(x,y), and the coefficients
5.
are
Then
4LK
∞∞
bnmαnmπc = LK
4 LK
g(ξ, η) sin(nπξ/L) sin(mπη/K).
00
g(ξ,η)sin(nπξ/L)sin(mπη/K). Supposec=3,L=K=π,f(x,y)=0andg(x,y)=xy. Now
αnm = 1n2 +m2. π
bnm = LKα πc
nm 0 0
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6.5. THE WAVE EQUATION IN HIGHER DIMENSIONS 195 Compute
4 (−1)n+m bnm = 3√n2 + m2 nm .
The solution is
z(x, y, t) = bnm sin(nx) sin(mx) cos(3n2 + m2t).
n=1 m=1
6. With L = K = 2π, we have αnm = 1 √n2 + m2. Compute 2
bnm = =
1 √ 2 2 π2 n2+m2nm
sin(mη)dη
∞∞
2π (2π)(2π)2π n+m2π0
The solution is
2π 0
4
sin(nξ)dξ √ 1 4 (1−(−1)n)(1−(−1)m).
∞
z(x, y, t) = sum∞m=1bnm sin(nx/2) sin(my/2) cos(n2 + m2t)
n=1
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196 CHAPTER 6. THE WAVE EQUATION
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Chapter 7
Laplace’s Equation
7.1
1.
The Dirichlet Problem for a Rectangle
Substitute u(x, y) = X(x)Y (y) into Laplace’s equation to obtain X′′ +λX =0;X(0)=X(1)=0
and
Solutions for X are
Y′′ −λY =0;Y(π)=0.
λ = n2π2, Xn(x) = sin(nπx).
With these values of λ, the problem for Y (y) has solutions that are con- stant multiples of sinh(nπ(π − y). To find a solution satisfying the bound- ary condition u(x, 0) = sin(πx), use a superposition
We need
∞
u(x, y) = an sin(nπx) sinh(nπ(π − y)). n=1
∞
u(x, 0) = an sin(πx) sinh(nπ2) = sin(πx). n=1
We can compute the Fourier coefficients of this sine expansion, or simply observe that we can take an = 0 for n = 2,3,··· and
The solution is
a1= 1 . sinh(π2)
1 (sin(πx) sinh(π(π − y))). sinh(π2)
197
u(x, y) =
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198 2.
CHAPTER 7. LAPLACE’S EQUATION
Separate variables and use the boundary conditions u(x, 0) = u(x, 2) = 0 to obtain the problem
Y′′ +λY =0;Y(0)=Y(2)=0, with eigenvalues and eigenfunctions
n2π2
λn = L2 , Yn(y) = sin(nπy/2).
The problem for X is
′′ n2π2
X − L2 X=0;X(3)=0.
Solutions are constant multiples of sinh(nπ(3 − x)/2). Now attempt a
solution of the form
u(x, y) = bn sinh(nπ(3 − x)/2) sin(nπy/2).
n=1
To satisfy u(3, y) = 0, we need
∞
u(0, y) = y(2 − y) = bn sinh(3nπ/2) sin(nπy/2). n=1
This is the Fourier sine expansion of y(2 − y) on [0, 2], and the coefficients are
12
η(2 − η) sin(nπη/2) dη 1 − (−1)n
bn = sinh(3nπ/2) 16
0
∞
n3π3 .
u(x, y) = 16 ∞ 1 − (−1)n sinh(nπ(3 − x)/2) sin(nπy/2).
u(x, y) =
The coefficients must be determined so that
∞
u(x, 4) = an sin(nπx) = x cos(πx/2). n=1
= sinh(3nπ/2)
3.
The solution is
n3 sinh(3nπ/2) find that the solution has the form
π3
After separating the variables and applying the boundary conditions, we
n=1
∞ sinh(nπy)
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n=1
an sinh(4nπ) sin(nπx).
7.1. THE DIRICHLET PROBLEM FOR A RECTANGLE 199 This is a Fourier sine expansion of x cos(πx/2) on [0, 1]. Choose the coef-
ficients as
an = 2
1 32n(−1)n+1
ξ cos(πξ/2) sin(nπξ) dξ = π2(4n2 − 1)2 . These determine the solution.
5. There are nonhomogeneous boundary conditions on two sides of the rect- angle, so write
u(x, y) = v(x, y) + w(x, y)
where
∇2v = 0;v(0,y) = v(π,y) = v(x,0) = 0,v(x,π) = xsin(πx)
and
∇2w = 0;w(x,0) = w(x,π) = w(0,y) = 0,w(w,y) = sin(y).
These are defined on 0 < x < 2,0 < y < π. Solve these problems inde- pendently.
First, separate variables in the problem for w to find that it has a solution
of the form
0
∞ sinh(nx)
w(x, y) =
Observe that we can solve this problem for w by taking b1 = 1 and all
other bn = 0, so
bn sin(ny) sinh(2n) . w(x, y) = sin(y) sinh(x) .
n=1
sinh(2) The problem for v has a solution of the form
v(x, y) =
an sin(nπx/2) sinh(nπ2/2) .
∞ sinh(nπy/2)
We need
∞
n=1
v(x, π) = x sin(πx) = an sin(nπx/2). n=1
This is a Fourier sine expansion of x sin(πx) on [0, 2], so choose 2
bn = =
ξsin(nξ)sin(nπξ/2)dξ 16n ((−1)n − 1)
0
π2 ((n2 −4)2 )
for n = 1, 3, 4, 5, · · · , for n = 2.
1
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200 CHAPTER 7. LAPLACE’S EQUATION Then
v(x, y) = sin(πx) sinh(πy) sinh(π2)
+ 10 ∞ n
π2 (n2 − 4)2
((−1)n −1)sin(nπx/2)sinh(nπy/2). sinh(nπ2/2)
n=1,n̸=2
6. A routine separation of variables, together with the zero boundary condi-
tions on sides y = 0, y = b and x = 0, lead to a solution of the form
∞ sinh((2n − 1)πx/2b)
Then
an sin((2n − 1)πy/2) sinh((2n − 1)πa/2b) .
u(x, y) =
Now we need to choose the coefficients so that
n=1
u(a, y) = g(y) = an sin((2n − 1)πy/2b). n=1
2b
∞
g(η)sin((2n−1)πη/2b)dη.
7. Separation of variables and the zero boundary conditions on x = 0, x = a
an = b
and y = 0 yield a general form of the solution:
u(x, y) = Now we need
n=1
an sin((2n − 1)πx/2a) sinh((2n − 1)πb/2a) .
an = a the form of the solution:
0
0
∞ sinh((2n − 1)πy/2a)
∞
u(x, 0) = f (x) = an sin((2n − 1)πx/2a). n=1
This leads us to choose 2a
f(ξ)sin((2n−1)πξ/2a)dξ.
8. The homogeneous conditions on the sides x = 0, x = a and y = 0 lead to
Then
∞
u(x, y) = ∞ sin(nπx/a) sinh(nπy/a) .
n=1
u(x, a) = x(x − a)2 = an sin(nπx/a). n=1
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sinh(nπb/a)
7.1. THE DIRICHLET PROBLEM FOR A RECTANGLE 201 The coefficients in this eigenfunction expansion are
2a
an = a ξ(ξ − a)2 sin(nπξ/a) dξ
0
= 4 (1 + 2nπ(−1)n). n3π3
9. Write the solution as u(x, y) = v(x, y) + w(x, y), where v is the solution of the problem
∇2v = 0,v(x,0) = v(x,1) = v(4,y) = 0,v(0,y) = sin(πy), and w is the solution of
∇2(w) = 0, w(x, 0) = w(x, 1) = w(0, y) = 0, w(4, y) = y(1 − y). These problems are defined on 0 ≤ x ≤ 4,0 ≤ y ≤ 1. A separation of
variables yields a general form of the solution of the problem for v:
We need
∞
v(x, y) =
an sin(nπy) sinh(4nπ) .
∞ sinh(nπ(4 − x))
n=1
v(0, y) = sin(πy) = an sin(nπy), n=1
sobyobservationwecanleta1 =1andan =0forn=2,3,···. Then v(x, y) = sin(πy) sinh(π(4 − x)) .
sinh(4π)
Another separation of variables leads to a general form of the solution for
w:
Then
so
∞
w(x, y) = bn sin(nπy) sinh(nπx). n=1
∞
w(4, y) = y(1 − y) = bn sinh(4nπ) sin(nπy), n=1
21
bn = sinh(4nπ)
= 4(1 − (−1)n)
n3π3 sinh(4nπ)
ξ(1 − ξ) sin(nπξ) dξ
0
.
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202 CHAPTER 7. LAPLACE’S EQUATION
7.2 The Dirichlet Problem for a Disk
For each of Problems 1–8, a solution
1 ∞rn
u(r, θ) = 2 a0 +
an = πRn
[an cos(nθ) + bn sin(nθ)],
where
n=1 1π
for n = 0,1,2,··· and
for n = 1,2,···.
f(ξ)cos(nξ)dξ
f(ξ)sin(nξ)dξ
R
−π
1π
bn = πRn
−π
1. We can see by observation that u(r, θ) = 1 is a solution. This can also be obtained the long way by carrying out the integrations, obtaining a0 = 2 andan =bn =0forn=1,2,···.
2. As with Problem 1, we can obtain the solution by inspection by matching coefficients of the proposed series solution to settle on only one nonzero coefficient, namely a4. If we carry out the integrations, we get
1π
an = π3n for n = 0,1,2,3,5,6,···, and
1π
bn = π3n for n = 1,2,···, while
8 cos(4ξ) cos(4ξ) dξ = 34 . The solution therefore consists of a single term:
3. Calculate
−π
8 cos(4ξ) cos(nξ) dξ = 0
8 cos(4ξ) sin(nξ) dξ = 0
−π
1π 8
a4 = π34
u(r, θ) = 8 3 cos(4θ).
−π
r4
1π 2π2
(ξ2−ξ)dξ= 3, 1π 4(−1)n
an=2nπ
a0=π
−π
(ξ2−ξ)cos(nξ)dξ= n22n
−π
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7.2. THE DIRICHLET PROBLEM FOR A DISK
and
203
The solution is
u(r,θ)= 3 +2
bn = 2nπ π2
(ξ2 −ξ)sin(nξ)dξ = n2n ∞ rn (−1)n
.
1π 2(−1)n
−π
2 n2 [2cos(nθ)+nsin(nθ)]. 4. After carrying out the integrations, the solution is
n=1
1 ∞ (−1)nn rn u(r,θ)=−10rsin(θ)+2 n2 −1 5
n=2
sin(nθ).
5. The solution is
sinh(π) 2∞ (−1)nrn
u(r, θ) = π + π
6. If we write sin2(θ) = (1 − cos(2θ))/2, we can identify the only nonzero coefficients in the series solution as a0 = 1/2 and a2 = −1. The solution
is
u(r, θ) = 1 − r cos(2θ). 22
n=1
n2 + 1 4 sinh(π)[cos(nθ) + n sin(nθ)].
7. After the integrations, we obtain the solution
1 2 ∞ 4(−1)n+1 rn
u(r, θ) = 2 cosh(2π) − sinh(2π) 4π
2∞ (−1)n rn
u(r,θ)=1−3π +
8. After integrating to find the coefficients, the solution is
n=1
n2 8 cos(nθ).
+ π
2∞ (−1)n rn
[cosh(2π)(8π + 2nπ ) + sinh(2π)(n + 4)] cos(nθ) 3 3
2 2
n=1
(n2 + 4)2 4 (n2 + 4)2 4
[cosh(2π)(4nπ + n π ) − 4n sinh(2π)] sin(nθ).
9. Let U (r, θ) = u(r cos(θ), r sin(θ). The problem given in rectangular coor-
+ π
n=1
dinates converts to the following problem in polar coordinates: ∇2U(r,θ) = 0 for 0 ≤ r < 4,U(4,θ) = 16cos2(θ).
If we write 16 cos2(θ) = 8(1 + cos(2θ), we can recognize by inspection that 1a0 =8,a2(42)=8,
2
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204 CHAPTER 7. LAPLACE’S EQUATION and all other an = 0. The solution in polar coordinates is
r2 U(r,θ)=8+8 4 cos(2θ).
Because the original problem was posed in rectangular coordinates, con- vert this to rectangular coordinates by using x = rcos(θ),y = rsin(θ), and the identity cos(2θ) = 2 cos2(θ) − 1, to obtain
u(x,y)=8+ 1(x2 −y2). 2
10. In polar coordinates this Dirichlet problem is
∇2U(r, θ) = 0 for 0 ≤ r < 3, U(3, θ) = 3(cos(θ) − sin(θ)).
In the series solution for this problem, identify 3a1 = 3 and 3b1 = 3, with all other coefficients zero. Then
U (r, θ) = r(cos(θ) − sin(θ)). In rectangular coordinates, this solution is
u(x, y) = x − y.
Perhaps by hindsight, this could have been seen with no computation
because x − y is harmonic on the entire plane.
11. In polar coordinates this problem is
∇2U(r, θ) = 0 for 0 ≤ r < 2, U(2, θ) = 4(cos2(θ) − sin2(θ)) = 4 cos(2θ). Identify a222 = 4, with all other coefficients zero, so
U(r,θ) = r2 cos(2θ). In rectangular coordinates the solution is
u(x,y)=x2 −y2.
12. In polar coordinates the problem is
∇2U(r,θ) = 0 for 0 ≤ r < 5,U(5,θ) = 25sin(θ)cos(θ).
If we write
U (5, θ) = 25 sin(2θ) 2
we can identify the coefficients in the series solution: a2(52) = 25/2, with all other coefficients zero. The solution is
U(r,θ) = 1r2 sin(2θ). 2
To convert this to rectangular coordinates, use
to obtain
1 r2 sin(2θ) = r sin(θ)r cos(θ) 2
u(x, y) = xy.
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7.3. THE POISSON INTEGRAL FORMULA 205 7.3 The Poisson Integral Formula
In Problems 1–4 the idea is to use the Poisson integral formula to write the re- quested solution value as an integral which can be approximated by a numerical technique. This assumes the availability of software that will do this.
For Problem 1, the Poisson integrals are given for each approximate value calculated. For Problems 2, 3 and 4, we give just the approximate values.
1. WithR=1andf(θ)=θ,thesolutionis
Then
U(1/2,π)= 1 π 3ξ
2π −π 5−4cos(ξ−π)
1 pi 1−r2
U(r,θ)= 2π −π 1+r2 −2rcos(ξ−θ)ξdξ.
Next,
U (3/4, π/3) = 1 π 7ξ dξ ≈ 0.8826128645.
0.132145.
3. U(4,π)≈−16.4654,U(12,π/6)≈0.0694,U(8,π/4)≈1.5281
4. U(5.5,3π/5)≈0.8230986392,U(4,2π/7)≈1.609925382,U(1,π)≈3.986586385, U (4, 9π/4) ≈ .461163821
7.4 The Dirichlet Problem for Unbounded Re- gions
1. If we put f(ξ) = K in equation (7.7), we get the solution
dξ=0.
2π −π 25−14cos(ξ−π/3) U (1/5, π/4) = 1 π 24ξ
And
dξ ≈ 0.2465422.
2. U(1,π/6) ≈ 0.0033829, U(3,7π/2) ≈ 0, U(1,π/4) ≈ 0, U(2.5,π/12) ≈
2π −π 26−10cos(ξ−π/4)
u(x,y) = Ky ∞ 1
π −∞y2+(ξ−x)2
dξ =lim2 2dξ
Ky L 1 πL→∞ −Ly+(ξ−x)
K L−x −L−x = lim arctan −arctan
πL→∞ y y = K π + π = K.
π22
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206 CHAPTER 7. LAPLACE’S EQUATION
Or, we can avoid this computation by observing that u(x, y) = K is har- monic on the entire plane, and equals K on the real line (the boundary of the upper half-plane). Therefore the solution is u(x, y) = K.
2. By equation (7.7), the solution of this problem for the upper half-plane is u(x, y) = y ∞ 1 e−|ξ| dξ.
π −∞y2+(ξ−x)2 3. By equation (7.7), the solution is
u(x, y) = y ∞ ξ
π −∞y2+(ξ−x)2
dξ.
4. The solution is
u(x, y) = y k 1
1 ξ − x k = arctan
dξ
π −ky2+(ξ−x)2
π y−k
1
= π arctan
1
=π arctan
k − x y
k − x y
− arctan
+arctan
−k − x y
k + x y ,
in which the last line makes use of the fact that the arctangent is an odd function.
5. Suppose u(x,y) is harmonic on the upper half-plane and u(x,0) = f(x). Then the function v(x, y) defined by v(x, y) = u(x, −y) on the lower half- plane is harmonic, and v(x,0) = f(x). But we know an integral formula for u(x, y). Therefore the problem for the lower half-plane has the solution
v(x,y) = u(x,−y) = −y ∞ f(ξ)
π −∞y2+(ξ−x)2
dξ.
for all x and for y < 0.
6. By the result of Problem 5, the solution of this problem for the lower
half-plane is
u(x,y) = −y ∞ f(ξ)
π −∞y2+(ξ−x)2
dξ
y0 −1 y1 1
=−π −1 y2 +(ξ−x)2 dξ− π 0 y2 +(ξ−x)2 dξ
1 x x + 1 x − 1 = − π 2 arctan y − arctan y − arctan y .
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7.4. THE DIRICHLET PROBLEM FOR UNBOUNDED REGIONS 207
7. The boundary of the right quarter-plane consists of the nonnegative hor- izontal and vertical axes. Define two Dirichlet problems, in each of which boundary data is nonzero on just one part of the boundary:
Problem 1 ∇2v = 0 for x > 0,y > 0 and v(x,0) = f(x),v(0,y) = 0, and Problem 2 ∇2w = 0 for x > 0,y > 0 and w(x,0) = 0,w(0,y) = g(y).
Both problems can be solved by separation of variables and Fourier inte- grals, obtaining
and
−ωx sin(ωy)e dω.
2 ∞ ∞
v(x, y) = π
2 ∞ ∞
−ωy sin(ωx)e dω
∞ω
e−3ξ sin(ωξ) dξ = 9 + ω2 .
f (ξ) sin(ωξ) dξ
w(x, y) = π
The solution of the original problem is u(x, y) = v(x, y) + w(x, y).
8. Using the result of Problem 7, compute
00
g(η) sin(ωη) dη
The solution is
9.
u(x, y) = 2 ∞ ω π0 9+ω2
sin(ωx)e−ωy dω.
x − 8 y
bω =
00
0
u(x, y) = Ay 8 1
π 4 y2+(ξ−x)2
dξ
= π arctan y − arctan
A x − 4
10. The two-dimensional heat equation is
ut = k(uxx + uyy) = k∇2u.
In the steady-state case, ut = 0 and u satisfies Laplace’s equation. Use the result of Problem 7. The solution is
where
2∞
Bω = π
Bω sin(ωy)e−ωx dω, 2∞ ω
u(x, y) = π
0
0
e−η sin(ωη) dη = 1 + ω2 .
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208 11.
CHAPTER 7. LAPLACE’S EQUATION
Using the results of Problem 7, the solution is
where
and
bω = π
0
∞ 0
2∞
ξsin(ωξ)dξ
= 2 sin(πω) − 2ωπ cos(πω)
πω2
2∞
η2 sin(ωη)dη
= 4(cos(πω) − 1) − 2ω2π2 cos(πω) + 4π sin(πω).
πω3
u(x, y) =
[bω sin(ωx)e−ωy + Bω sin(ωy)e−ωx] dω,
Bω = π
0
7.5
1.
A Dirichlet Problem in 3 Dimensions
Let u(x, y, z) = X(x)Y (y)Z(z) to separate variables, obtaining first that X′′ +λX =0;X(0)=X(1)=0
and
Then
and
Further,
Y′′ +μY −0;Y(0)=Y(1)=0. λn = n2π2, Xn(x) = sin(nπx) μm = m2π2, Ym(y) = sin(mπy). Z′′ − (n2 + m2)π2Z = 0; Z(0) = 0.
This leads to functions
unm(x, y, z) = cnm sin(nπx) sin(mπy) sinh(αnmπz),
√
where αnm = superposition
n+m2. To satisfy the condition u(x,y,1) = xy, use a
∞∞
u(x, y, z) = cnm sin(nπx) sin(mπy) sinh(αnmπz). n=1 m=1
We must choose the coefficients so that
∞∞
u(x, y, 1) = xy = sin(nπx) sin(mπy) sinh(αnmπ). n=1 m=1
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7.5. A DIRICHLET PROBLEM IN 3 DIMENSIONS 209 This is a double Fourier series on the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and
we know from experience with the heat and wave equations that 411
cnm = sinh(α π) ξ sinh(nπξ) dξ η sin(mπη) dη nm0 0
4(−1)n+m
= nmπ2 sinh(αnmπ).
2. Again separate variables, except now exploit the zero boundary conditions on sides z = 0, z = 1, y = 0, y = 2π and x = 0 to obtain Sturm-Liouville problems for Y (y) and Z(z):
Y′′ +λY =0;Y(0)=Y(2π)=0 Z′′ + μZ = 0; Z(0) = Z(1) = 0.
n2
λn = 4 , Yn(y) = sin(ny/2)
μm = m2π2, Zm(z) = sin(nπz). Xnm(x) = sinh(αnmx/2)
√
and Then
and Further
where αnm =
u(x, y, z) = cnm sin(ny/2) sin(mπz) sinh(αnmx/2).
n=1 m=1
For the solution, choose the coefficients so that
u(2π, y, z) = z.
n2 + 4m2π2. Therefore use the superposition ∞∞
Thus choose
cnm = sinh(α π)π 2sin(nη/2)dη 2sin(mπτ)dτ
1 1 2π 1 nm0 0
= 4 (1 − (−1)n)(1 − (−1)m). sinh(αnmπ)
3. The solution is the sum of the solutions of the following two problems:
∇2w = 0,
w(0,y,z) = w(1,y,z) = w(x,0,z) = w(x,2π,z) = w(x,y,0) = 0, w(x,y,π) = 1
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210
CHAPTER 7. LAPLACE’S EQUATION
and
Each of these problems is solved by a separation of variables. For the first, we obtain
∞∞
w(x, yz) = anm sin(nπx) sin(my/2) sinh(4n2π2 + m2z/2), n=1 m=1
∇2v = 0,
v(0,y,z) = v(1,y,z) = v(x,y,0) = v(x,y,π) = v(x,0,z) = 0, v(x, 2π, z) = xz2.
in which anm =
√
1 1
2sin(nπξ)dξ
2π 1
sin(nπη)dη
.
0 π
1 1 − (−1)n 1 − (−1)m
sinh( 4n2π2+m2π/2) 0
= sinh(√n2π2 + m2π/2) nπ mπ
For the second problem, obtain
∞∞
v(x, y, z) = bnm sin(nπx) sin(mz) sinh(n2π2 + m2y), n=1 m=1
in which
bnm = 2 √ 2 2 2 ξsin(nπξ)dξ τ2sin(mτ)dτ)
41π π sinh( n π +m 2π) 0 0
4 (−1)n+12−2(−1)m +m2π2(−1)m
4.
nπ m3 . u(x, y, z) = w(x, y, z) + v(x, y, z).
The solution u(x, y, z) is a sum of the solutions of the following two prob- lems:
= π2 sinh(√n2π2 + m22π) The original problem has solution
and
∇2(v) = 0,
v(0, y, z) = 0, v(1, y, z) = sin(πy) sin(z), v(x,0,z) = v(x,2,z) = 0,
v(x,y,0) = v(x,y,π) = 0
∇2w = 0,
w(0,y,z) = w(1,y,z) = 0,
w(x,0,z) = w(x,2,z) = 0,
w(x, y, 0) = x2(1 − x)y(2 − y), w(x, y, π) = 0.
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7.6.
THE NEUMANN PROBLEM 211
Solve these problems by separation of variables. The first problem has a solution of the form
∞∞
v(x, y, z) = anm sinh(m2 + (mπ/2)2) sin(nπy/2) sin(mz), n=1 m=1
while the second problem has a solution of the form
∞∞
w(x, y, z) = bnm sin(nπx) ∼ (mπy/2) sinh((nπ)2 + (mπ/2)2(π−z)). n=1 m=1
√
π2 + 1 =
The coefficients anm can be obtained by inspection. Observe that a21 1 and all other anm = 0. The coefficients bnm require integrations:
bnm =
sinh(π 4n2 + m2/2) 0
212 √ ξ2(1 − ξ) sin(nπξ) dξ
η(2 − η) sin(mπη/2) dη 0
= − 64 (1 + 2(−1)n)(1 − (−1)m). π6
’
7.6
1.
The Neumann Problem
First,
and
These have solutions of the form
1
4 cos(πx) dx = 0,
0
so this problem may have a solution. (If this integral were nonzero, we could conclude that the problem has no solution).
A separation of variables, making use of the three homogeneous boundary conditions, leads to the problems
X′′ +λX =0;X′(0)=X′(1)=0 Y′′−λY =0;Y′(1)=0.
λn = n2π2, Xn(x) = cos(nπy)
and
Thus attempt a solution of the Neumann problem of the form
∞
u(x, y) = c0 + cn cosh(nπ(1 − y)) cos(nπy). n=1
Yn(x) = cosh(nπ(1 − y)).
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212
CHAPTER 7. LAPLACE’S EQUATION
The condition uy (0) = 4 cos(πx) requires that
∞
−cnnπ sinh(nπ) cos(nπx) = 4 cos(πx). n=1
This is satisfied if we put cn = 0 for n = 2,3,4,···, and choose c1 so that −c1π sinh(1) cos(πx) = 4 cos(πx).
Therefore −c1π sinh(π) = 4, and
c1=− 4 .
π sinh(π)
The solution is
u(x, y) = c0 − 4 cosh(π(1 − y)) cos(πx).
2.
First check that
π y − π dy = 0, 02
π sinh(π)
Here c0 is an arbitrary constant, so this solution is not unique.
so it is worthwhile to look for a solution. From the zero boundary condi- tions at y = 0 and y = π, separation of variables gives us a solution of the form
∞
u(x, y) = c0 + [cn cosh(nx) + dn cosh(1 − x)] cos(ny). n=1
The boundary conditions on the edges x = 0 and x = 1 give us
∂u π∞
∂x(0,y) = y − 2 =
∂u ∞
∂x (1, y) = cos(y) = The coefficients are given by
ncn sinh(n) cos(ny).
−ndn sinh(n)cos(ny)
and
n=1
and
if n = 1
−1 π π 2(1−(−1)n)
cn = n sinh(n) π
cos(y) cos(ny) dy if n ̸= 1,
dn=nsinh(n)
0
y−2 dy= πn2sinh(n)
=
0
1/ sinh(1)
1 2π
n=1
0
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7.6. THE NEUMANN PROBLEM 213 for n = 1,2,···. The solution is
u(x, y) = c0 + 1 cosh(x) sinh(1)
0
conditions on edges x = 0 and x = π, separation of variables yields a solution of the form
∞
u(x, y) = c0 + [cn cosh(ny) + dn cosh(n(π − y))] cos(nx). n=1
Now
+ ∞ 2((−1)n − 1) cosh(n(1 − x)) cos(ny).
πn2 sinh(n)
3. A solution may exist because π cos(3x) dx = 0. From the zero boundary
n=1
∂u ∞
−ndn sinh(nπ) cos(nx) d3=− 1
3 sinh(3π)
∂y (x, 0) = cos(3x) =
anddn =0ifn̸=3. Next,theboundaryconditionaty=πgivesus
so
n=1
∂u ∞
∂u (x, π) = 6x − 3π =
1 2π
ncn sinh(nπ) cos(nx).
Then
3 sinh(3π)
+ ∞ 12((−1)n − 1) cosh(ny) cos(nx).
n=1
(6x − 3π) cos(nx) dx = 1 12 ((−1)n − 1).
cn = n sinh(nπ) π
n sinh(nπ) n2π
0
The solution is
u(x, y) = c0 − cosh(3(π − y)) cos(3x)
n3π sinh(nπ)
4. Let u(x,y) = X(x)Y(y) and use the boundary conditions to obtain the
problems and
X′′ − λX = 0; X′(0) = X′(π) = 0 Y ′′ + λY = 0.
n=1
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214
CHAPTER 7. LAPLACE’S EQUATION
We find that X0(x) = 1 and Xn(x) = cos(nx) for n = 1,2,···, while Yn(y) = cn cosh(ny) + dn cosh(n(π − y))
for n = 1,2,···. We therefore seek a solution of the form
∞
u(x, y) = c0 + [cn cosh(ny) + dn cosh(n(π − y))] cos(nx). n=1
At the edge y = π, we have
u(x, y) = 0 = c0 + [cn cosh(nπ) + dn] cos(nx).
n=1 Along the edge y = 0, we have
∞
u(x, 0) = f (x) = c0 + [cn + dn cosh(nπ)] cos(nx). n=1
In order to have a solution, the coefficients must be chosen to satisfy the equations
c0 = 0,
cn cosh(nπ) + dn = 0, 2π
cn +dncosh(nπ)= π
for n = 1,2,···. For n ≥ 1, the determinant of the coefficients of this
system of equations is
cosh(nπ) 1 = cosh2(nπ) − 1 = sinh2(nπ) ̸= 0.
5.
Therefore there is a unique solution of these algebraic equations for the coefficients cn and dn for n = 1, 2, · · · . The problem therefore has a unique solution (recalling that c0 = 0).
With u(x, y) = X(x)Y (y), we obtain:
X′′ − λX = 0
and Then
1 cosh(nπ)
∞
Y′′ +λY =0;Y(0)=Y(1)=0.
Yn(y) = sin(nπy) and Xn(x) = cn cosh(nπx) + dn cosh(nπ(1 − x))
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0
f(ξ)cos(nξ)dξ
7.6. THE NEUMANN PROBLEM 215
for n = 1,2,···. Look for a solution of the form ∞
u(x, y) = [cn cosh(nπx) + dn cosh(nπ(1 − x))] sin(nπy). n=1
To solve for the constants, use the other two boundary conditions. First,
nπcn sinh(nπ) sin(nπy)
∂x (1, y) = 0 = so we each cn = 0. Next,
∂u ∞
n=1
∂u 2 ∞
∂x(0,y)=3y −2y= −2 1
−nπdnsinh(nπ)sin(nπy).
Then
n4π4 sinh(nπ) for n = 1,2,···. The solution is
(3η2 − 2η) sin(nπη) dη [n2π2(−1)n + 6(1 − (−1)n)]
dn = nπ sinh(nπ) = 2
0
n=1
u(x, y) =
∞ 2 [n2π2(−1)n + 6(1 − (−1)n)] cosh(nπ(1 − x)) sin(nπy).
n4π4 sinh(nπ)
6. Because π sin(3θ)dθ = 0, this problem may have a solution. Such a
1 ∞
u(r,θ)= 2a0 + [anrncos(nθ)+bnrnsin(nθ)].
n=1 The boundary conditions give us
∂u(R,θ) = sin(3θ) ∂r
∞
= [nanRn−1 cos(nθ) + nbnRn−1 sin(nθ)].
n=1
By inspection, we see that we can choose each an = 0 and bn = 0 for
n=1
−pi
solution will have the form
n ̸= 3. Further, 3b3R2 = 1, so
b3= 1
and the solution is
3R2
1 R r 3
u(r,θ)=2a0+3 R sin(nθ).
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216 CHAPTER 7. LAPLACE’S EQUATION 6. First check that π cos(2θ) dθ = 0, a necessary condition for a solution
1 ∞
u(r, θ) = 2 a0 + [an cos(nθ) + bn sin(nθ)].
n=1
From the boundary condition at r = R, we have
∂u(R,θ) = cos(2θ) ∂r
∞
= [nanRn−1 cos(nθ) + nbnRn−1 sin(nθ)].
n=1
As in the preceding problem, compare coefficients on both sides of this equation to choose each bn = 0 and an = 0 except for n = 2. Further, 2a2R = 1. The solution is
−π
to exist. A solution must have the form
8. Because
1 R r 2 u(r,θ)=2a0+2 R cos(2θ).
∞
xe−|x| dx = 0
−∞
this problem may have a solution. Using the general solution developed in Section 7.6.3, we can immediately write the solution
1∞
u(x, y) = 2π ∞
−∞
ln(y2 + (ξ − x)2)ξe−|ξ| dξ + c. e−|ξ| sin(ξ) dξ = 0,
9. Because
−∞
a necessary condition for a solution to exist is satisfied. The solution is 1∞2 2−|ξ|
u(x,y)= 2π −∞ ln(y +(ξ−x) )e sin(ξ)dξ.
10. A solution of the Dirichlet problem for the lowr half-plane was requested in Problem 5, Section 7.4. Use this result and the line of argument used to solve the Neumann problem for the upper half-plane to obtain the solution
1∞
u(x,y)=−2π
11. Problem 7, Section 7.4 requested a solution of the Dirichlet problem for the right quarter-plane. Using this, we are led to attempt a solution for the Neumann problem for the right quarter-plane of the form
∞ 0
u(x, y) =
aω cos(ωx)e−ωy dω.
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−∞
ln(y2 +(ξ−x)2)f(ξ)dξ.
7.7. POISSON’S EQUATION
Now
∂y (x, 0) = This tells us to choose
2∞
aω =−πω 12. Attempt a solution of the form
217
∂u ∞
−ωaω cos(ωx) dω.
∞ 0
bω e−ωy sin(ωx) dω.
0
u(x, y) =
This is harmonic and has been chosen to satisfy u(0, y) = 0. Now
0
f(ξ)cos(ωξ)dξ.
where
∞
v(x, y) = an sin(nπ(1 − x)) sin(nπy), n=1
2(−1)n+1 an = nπsinh(nπ).
∂u ∞
−ωbω sin(ωx)dω,
∂y(x,0) = f(x) = 2∞
0
so choose
bω =−πω 7.7 Poisson’s Equation
f(ξ)sin(ωξ)dξ.
0
1. Write u(x, y) = v(x, y) + w(x, y), where v is the solution of the Dirichlet problem
∇2v = 0 for 0 < x < 1,0 < y < 1, v(x,0) = v(x,1) = 0,
v(1, y) = 0,
v(0, y) = y
and w is the solution of the problem
∇2w = 0 for 0 < x < 1,0 < y < 1, w(0,y) = w(1,y) = w(x,0) = w(x,1) = 0, w(x, y) = xy for 0 < x < 1, 0 < y < 1.
For the first problem, for v(x, y), separate variables to obtain the solution:
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218
CHAPTER 7. LAPLACE’S EQUATION
The problem for w(x, y) has the solution
where
∞∞
w(x, y) = knm sin(nπx) sin(mπy), n=1 m=1
00 4(−1)n+m+1
1 1
−4
knm = π2(n2 + m2) ξ sin(nπξ dξ η sin(mπη) dη
2.
= (n2 + m2)nmπ4 .
Split the Poisson problem into two problems, as in Figure 7.6. The first
is a Dirichlet problem:
∇2v = 0 for 0 < x < 1,0 < y < 2,
v(0,y = v(1,y) = v(x,2) = 0,v(x,0) = x2. This has the solution
where
∞
v(x, y) = cn sin(nπx) sinh(nπ(2 − y)), n=1
21
cn = sinh(2nπ) ξ2 sin(nπξ) dξ
0
2 −2 + 2(−1)n − n2π2(−1)n
= sinh(2nπ) n3π3 . The second problem is
∇2w = 0,
w(x, y) = 0 on the boundary of the rectangle, w(x, y) = x sin(y) for 0 < x < 1, 0 < y < 2.
This problem has the solution
∞∞
w(x, y) = knm sin(nπx) sin(mπy/2) n=1 m=1
where
knm =−π2(4n2 +m2) ξsin(nπξ)dξ η2sin(mπη/2)dη
.
812
00
16 (−1)n+1 mπ sin(2)(−1)m
= π2(4n2 +m2) nπ −4+m2π2
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7.7. POISSON’S EQUATION 219
3. Split the problem into two problems, as we have been doing. However, the first problem (see Figure 7.6) must itself be broken up into two problems, in the first of which v(0,y) = 1 and v(π,y) = 0, and in the second of which v(0, y) = 0 and v(π, y) = 0. Applying straightforward separation of variables to these problems, we obtain
and v(π, y) = y, we obtain
v2(x, y) = ∞ 2 (−1)n+1 sin(ny) sinh(nx).
where knm =
v1(x, y) = ∞ 2 (1 − (−1)n) sin(my) sinh(n(π − x))
nπ sinh(nπ)
for the Dirichlet problem with v(0, y) = 1 and v(π, y) = 0. If v(0, y) = 0
n=1
n sinh(nπ)
For the problem for w in Figure 7.6, we have
n=1
w(x, y) = knm sin(nx) sin(my), n=1 m=1
∞∞
4 ππ π
− π2 n2π2 + m2π2 ξ2 sin(nξ) dξ η2 sin(mη) dη
000
= 4 (2 − 2(−1)n + n2π2(−1)n)(−2 + 2(−1)m − m2π2(−1)m).
π2(n2 + m2)
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220 CHAPTER 7. LAPLACE’S EQUATION
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Chapter 8
Special Functions and Applications
8.1 Legendre Polynomials
For Problems 1–4 and 6, graphs of the function and the sixth partial sum of its Fourier-Legendre expansion on[−1, 1] appear nearly indistinguishable within the scale of the graph. The “most“ functions many terms of this expansion are needed to achieve a good fit between the partial sum and the function. This is seen in Problem 5, where the sixth partial sum is a poor fit to the function, while the fiftieth partial sum is much closer (though still a poor fit).
1. The coefficients are
cn = 2
2n + 1 1 −1
sin(πx/2)Pn(x)dx. Carrying out these integrations, we obtain
c0 =c2 =c4 =0,c1 = 12, π2
c3 = 168(π2 −10),c5 = 660(π4 −112π2 +1008). π4 π6
Figure 1 shows a graph of f(x) and 5n=0 cnPn(x). 2. The coefficients are
cn = 2
−1 221
2n + 1 1
e−xPn(x)dx.
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222 CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
Figure 8.1: Graph of sin(πx/2) and the sixth partial sum of its Fourier-Legendre expansion.
The first six coefficients are
c0 = 1 (e − e−1 ), c1 = − 3 , c2 = 5 e − 35 e−1 ,
2e22 c3 = 35 e − 259 e−1 , c4 = 162e − 259 e−1 ,
The first six are
222
c5 = 3619e − 26741e−1. 22
Figure 2 is a graph of f(x) = e−x and the sixth partial sum of its Fourier- Legendre expansion.
3. The coefficients are
2n + 1 1
cn = 2
c0 =−1sin(1)cos(1)+1,c1 =c3 =c5 =0,
c2 =−5sin(1)cos(1)+15−15cos2(1), 884
c4 = − 585 + 585 cos2 (1) + 531 sin(1) cos(1). 3216 32
22
sin2(x)Pn(x)dx.
−1
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8.1. LEGENDRE POLYNOMIALS 223
Figure 8.2: Graph of e−x and the sixth partial sum of its Fourier-Legendre expansion.
Figure 8.3 shows the function and the sixth partial sum of this Fourier- Legendre expansion.
4. The coefficients are
cn = 2
2n + 1 1 −1
(cos(x) − sin(x))Pn(x) dx.
The first six coefficients are
c0 =sin(1),c1 =−3sin(1)+3cos(1),c2 =15cos(1)−10sin(1),
c4 =549sin(1)−855cos(1),c5 =−5940sin(1)+9250cos(1). Figure 8.4 is a graph of the function and the sixth partial sum.
5. The coefficients are
cn = 2 The first six coefficients are
−1
2n + 1 1
f(x)Pn(x)dx.
c0 =c2 =c4 =0,c1 = 3,c3 =−7,c5 = 11. 2 8 16
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224 CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
Figure 8.3: Graph of sin2(x) and the sixth partial sum of its Fourier-Legendre expansion.
Figure 8.4: Graph of cos(x) − sin(x) and the sixth partial sum of its Fourier- Legendre expansion.
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8.1. LEGENDRE POLYNOMIALS 225
Figure 8.5: Graph of f(x) and the sixth partial sum of its Fourier-Legendre expansion.
Figure 8.6: Graph of f(x) and the fiftieth partial sum of its Fourier-Legendre expansion.
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226 CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
Figure 8.7: Graph of (x + 1) cos(x) and the sixth partial sum of its Fourier- Legendre expansion.
Figure 8.5 shows a graph of the function and this partial sum. For this function the sixth partial sum does not fit the function well at all on [−1, 1]. Figure 8.6 shows the fiftieth partial sum, a better fit to the function.
6. The coefficients are
2n + 1 1
n
(x + 1) cos(x)Pn(x) dx.
−1 The first six coefficients are
c0 =sin(1),c1 =−3sin(1)+6cos(1),c2 =15cos(1)−10sin(1) c3 =238sin(1)−378cos(1),c4 =549sin(1)−855cos(1)
c5 =−34969sin(1)+54461cos(1).
Figure 8.7 shows a graph of (x + 1) cos(x) and the sixth partial sum of the Fourier-Legendre expansion.
7. Forn=7,[n/2]=[7/2]=3,so
3 (14 − 2k)!
xn−2k = 429x7 − 693x5 + 315x3 − 35x.
P7 = (−1)k k=0
27k!(7 − k)!(7 − 2k)! 16 16 16 16
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8.1. LEGENDRE POLYNOMIALS
For n = 8, [n/2] = [4] = 4 and 4
P8(x) = (−1)k k=0
227
(16 − 2k)! 28k!(8 − k)!(8 − 2k)!
x8−2k
= 6435x8 − 3003x6 + 3465x4
128 32
− 315x2 + 35 . 33 128
64
For n = 9, [n/2] = [9/2] = 4 and 4
P9(x) = (−1)k k=0
(18 − 2k)! 29k!(9 − k)!(9 − 2k)!
x9−2k
= 12155x9 − 6435x7 + 9009x5 128 32 64
− 1155x3 + 315x. 32 128
For n = 10, [n/2] = [5] = 5 and 5
256 256 128 − 15015x4 + 3465x2 − 63 .
128 256 256 8. Put n = 0 into the integral formula to obtain
(20 − 2k)! 210k!(10 − k)!(10 − 2k)
x10−2k = 46189x10 − 109395x8 + 45045x6
P10(x) = (−1)k k=0
1π 1
dθ= π(π)=1=P0(x).
1 π π x+x2 −1cos(θ) dθ π00
= 1 xθ + x2 − 1 sin(θ)π π0
= 1(xπ)=x=P1(x). π
π Next put n = 1 to get
0
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228
CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
With n = 2, we have
1π 2 2
π x+ x −1cos(θ) dθ
0
= 1 π x2 +2x2 −1cos(θ)+(x2 −1)cos2(θ) dθ
π0
1π22 12
x +2 x −1cos(θ)+ 2(x −1)(1+cos(2θ)) dθ 122 12 π
= π
= π x θ+2 x −1sin(θ)+ 2(x −1)(θ+sin(2θ)/2)
0
1212 =π xπ+2(x −−1)π
= 3x2 − 1 = P2(x). 22
Finally, wit n = 3, we have 1π 2 3
0
x+ x −1cos(θ) dθ
= 1 π x3 +3x2x2 −1cos(θ)−3x3 cos2(θ) dθ
π
0
π0
+ 1 π −3xcos2(θ)+x2x2 −1cos3(θ)−x2 −1cos3(θ) dθ
π0 13333
1π 2 n
=π xπ+2πx−2πx = 5x3 − 3x = P3(x).
9.
Let
22
Qn(x)=π
x+ x −1cos(θ) dθ
0
for n = 0, 1, 2, · · · . The strategy is to show that Qn(x) satisfies the same recurrence relation (8.7) that the Legendre polynomials do. From Prob- lem 8, we also have that Q0(x) = P0(x) and Q1(x) = P1(x). Then the recurrence relation will give us Q2(x) = P2(x), and then Q4(x) = P4(x), and so on.
To show that Qn(x) satisfies equation (8.7), first substitute the integral for Qn(x) into this equation and rearrange terms to obtain
1 π −n(x2 −1)sin2(θ)+x2 −1cos(θ)[x+x2 −1cos(θ)] π0
2 n−1
× x+ x −1cos(θ) dθ.
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8.1. LEGENDRE POLYNOMIALS 229 Now integrate this by parts, with
and
to obtain
u=x+x2 −1cos(θ)n dv = cos(θ) dθ
1 π x+x2 −1cos(θ)n x2 −1cos(θ)dθ π0
12 n−12 2
= π x+ x −1cos(θ) n(x −1)sin (θ).
Use this in the substitution of Qn(x) into equation (8.7) to show that Qn(x) satisfies this recurrence relation. This shows that Qn(x) = Pn(x).
10. We want to show that
12
Pn2(x)dx= 2n+1 −1
for n = 0,1,2,···. Let An be the coefficient of xn in Pn(x). We have an explicit formula for An in equation (8.8). For convenience, denote
1 −1
Consider the polynomial
q(x) = Pn(x) −
pn =
Pn2(x)dx.
An xPn−1(x). An−1
Both Pn(x) and xPn−1(x) are of degree n, and all terms involving xn in q(x) cancel (by design), so q(x) has degree ≤ n − 1. Further,
Then
Pn(x) = q(x) + An xPn−1(x). An−1
Pn2(x) = q(x)Pn(x) + An xPn−1(x)Pn(x). An−1
Integrate this equation, using Theorem 8.3, to get
An 1
pn = A xPn−1(x)Pn(x)dx.
n−1 −1
Now use the recurrence relation to write
xPn(x) = n + 1 Pn+1(x) + n Pn−1(x). 2n+1 2n+1
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230
CHAPTER 8.
SPECIAL FUNCTIONS AND APPLICATIONS
n + 1 Pn+1(x)Pn−1(x) + n Pn2−1(x). 2n+1 2n+1
Then
xPn(x)Pn−1(x) = Finally, we have
An 1
pn = A xPn(x)Pn−1(x)dx
n−1 −1
An+11 n1
= n Pn+1(x)Pn−1(x)dx+ Pn2−1(x)dx . An−1 2n+1 −1 2n+1 −1
Connecting the first and last expressions in this equation, we have pn= An n pn−1.
An−1 2n + 1
This is a recurrence relation for the numbers pn. Because we know An,
this is
pn = 1·3·5···(2n−3)(2n−1)
n!
(n−1)! n pn−1 1 · 3 · 5 · · · (2n − 3) 2n + 1
= 2n−1pn. 2n+1
From this we deduce pn:
1n 11 2
p1 = 3
−1
P0(x)2 dx = 3 3p1 = 32 = 2,
dx = 3,
p2 = p3 = p4 =
−1 5535
5p3 = 2, 77
7p3 = 2, 99
and so on. A straightforward induction now yields pn= 2 .
to get
Then
11.
2n+1
Put x = t = 1/2 into the generating formula for the Legendre polynomials
1 ∞ 1 1 n 3/4= Pn 2 2 .
n=0
2 ∞ 1 1
√3= 2nPn 2 . n=0
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8.1. LEGENDRE POLYNOMIALS 231
Then
∞ 1 1 1 n=02n+1Pn 2 =√3.
12. One way to derive these results is to use the recurrence relation. Another way is to use the formula given in Problem 7:
xn−2k.
[n/2] (2n − 2k)!
Pn(x) = (−1)k k=0
2nk!(n − k)!(n − 2k)!
To do this, first observe that
2n+1 1
so
2 =n+2=n, (4n + 2 − 2k)!
n k=0
x2n+1−2k.
(−1)k
Each term of this sum involves a positive power of x, so
Next,
so
P2n+1(0) = 0. 2n
P2n+1(x) =
22n+1k!(2n + 1 − k)!(2n + 1 − 2k)!
n P2n(x) = (−1)k
2 =[n]=n, (4n − 2k)!
x2n−2k. The constant term in this polynomial occurs when k = n, so
k=0
13. Apply the law of cosines to the triangle in the diagram to get R2 = r2 + d2 − 2rd cos(θ).
φ(x, y, z) = = R
dR
= . d 1−2r cos(θ)+ r2
22nk!(2n − k)!(2n − 2k)!
P2n(0) = (−1)n(2n)!. 22n (n!)2
Then
Then
R2 r r2 d2 =1−2dcos(θ)+d2.
11d11
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d d2
232
CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
For the remainder of the problem, consider two cases on r/d. First, sup- poser/d<1,sor
1 Pn(cos(θ))rn. dn+1
R2 d d2 r2 =1−2rcos(θ)+r2.
Then
Again, comparing this with the generating function, we have 1 ∞ d n
φ(r) = r and this is equivalent to
1 ∞
φ(r) = r With f(φ) = φ2, the solution is
r1 =.
R 1−2d cos(θ)+ d2 r r2
u(ρ, φ) =
n=0
ρ n ∞cn R
Pn(cos(φ)),
n=0
Pn(cos(θ)) r ,
dnPn(cos(θ))e−n.
14.
n=0
in which
2n + 1 1
f(arccos(ξ))Pn(ξ),dξ (arccos(ξ))2Pn(ξ) dξ.
cn = 2
2n + 1 1
−1
= 2
Many of these coefficients can be explicitly computed. For example,
−1
c0 = 1π2 −2,c1 =−3π2,c2 = 10, 289
c3 =− 7 π2,c4 = 8,c5 =−11π2. 128 25 512
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8.1. LEGENDRE POLYNOMIALS 233 With R = 1, use of the twenty-first partial sum of the solution series yields
the following approximations:
u(1, π/4) ≈ 0.615257, u(1, π/6) ≈ 0.272881, u(1, π/8) ≈ 0.154213.
These can be compared to f(φ) as a check on their accuracy. 15. With f(φ) = sin(φ), we have
2n + 1 1
cn = 2 u(ρ, φ) =
sin(arccos(ξ))Pn(ξ)dξ ∞ ρn
cn R Pn(cos(φ)).
and
−1
n=0
In computing the coefficients, use can be made of the identity
sin(arccos(ξ)) = 1 − ξ2.
With R = 1, and using the twenty-first partial sum of the solution, we
obtain the approximations
u(1, π/4) ≈ 0.707274, u(1, π/6) ≈ 0.500761, u(1, π/8) ≈ 0.382683.
16. With f(φ) = φ3, let
cn = 2
and
2n + 1 1 −1
(arccos(ξ))3Pn(ξ)dξ ∞ ρn
cn R Pn(n, cos(φ)).
With R = 1, the twenty-first partial sum of this series yields the following
approximate values:
u(1, π/4) ≈ 0.476973, u(1, π/6) ≈ 0.143548, u(1, π/8) ≈ 0.060559.
17. Withf(φ)=2−φ2,let
2n + 1 1
and
cn = 2 (2−arccos2(ξ))Pn(ξ)dξ −1
∞ ρn
u(ρ, φ) =
n=0
cn R Pn(cos(φ)).
u(1, π/4) ≈ 1.384743, u(1, π/6) ≈ 1.725844, u(1, π/8) ≈ 1.845787.
u(ρ, φ) =
With R = 1, use the twenty-first partial sum to approximate:
n=0
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234 CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS 18. Now f(φ) = φ − cos(φ). Let
and
2n + 1 1
cn = 2 u(ρ, φ) =
(arccos(ξ) − ξ)Pn(ξ) dξ ∞ ρn
−1
cn R Pn(cos(φ)).
With R = 1, the twenty-first partial sum of this series gives us the following
approximate values:
u(1, π/4) ≈ 0.077951, u(1, π/6) ≈ −0.342074, u(1, π/8) ≈ −0.531180.
n=0
19. In spherical coordinates, the Dirichlet problem to be solved is:
uρρ + 2uφφ + cot(φ)uφ =0,R1 <ρ
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0
0
tn+1(1 − t2)m−n−1Jn(xt) dt.
8.2. BESSEL FUNCTIONS 249 14. Use a result of part (g) of Problem 13, with n = −1/2 and m > −1/2 to
write
Jm(x) = m+1/2
2 1/2 πxt
= 2m−1Γ(m + 1/2)
This is the requested result with m used in place of n.
15. Start with the following result from Problem 14:
xm 1
2xm+1/2 1 2 Γ(m+1/2) 0
xm 1
t1/2(1 − t2)m−1/2
(1 − t2)m−1/2 cos(xt) dt.
cos(xt) dt
Mm(x) = 2m−1Γ(m + 1/2)
0
(1 − t2)m−1/2 cos(xt) dt. Make the change of variables t = sin(θ) to get
0
π/2
(cos2(t))m−1/2 cos(x sin(θ)) dθ
0
π/2
cos2m(θ) cos(x sin(θ)) dθ. 16. Let y = xaJν(bxc) and compute the derivatives:
xm
Jm(x) = 2m−1Γ(m + 1/2)
y′ = axa−1Jν (bxc) + xabcxc−1Jν′ (bxc)
y′′ = a(a − 1)xa−2Jν (bxc)
+ [2axa−1bcxc−1 + xabc(c − 1)xc−2Jν′ (bxc)]
+ xab2c2x2c−2J′′(bxc). ν
xm
= 2m−1Γ(m + 1/2)
0
and
Substitute these into the differential equation and simplify to obtain c2xa−2[(bxc)2J′′ + bxcJ′ (bxc) + ((bxc)2 − ν2)J (bxc)] = 0.
ννν
In Problems 17–24, the strategy is to match the given differential equation to the differential equation of Problem 16 by choosing a, b, c and ν. This makes it possible to write a general solution in terms of Bessel functions
17. The differential equation matches that of Problem 16 if 2a−1=−1,2c−2=0,b2c2=1,anda2−ν2c2= 7.
Then
We can write a general solution
3
144
a = 1 , b = c = 1, and ν = 1 . 34
y(x) = c1x1/3J1/4(x) + c2x1/3J−1/4(x).
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250 CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS 18. We need
−(2a−1)=1,2c−2=2,b2c2 =4, anda2−ν2c2 =−4. 9
Then
Then
a = 0, c = 2, b = 1 and ν = 1 . 3
y(x) = c1J1/3(x2) + c2J−1/3(x2) is a general solution.
19. Choosea=3,c=4,b=2,ν=1/2toget
y(x) = c1x3J1/2(2×4) + c2J−1/2(2×4).
20. Leta=−1,c=b=2,ν=3/4toobtain
y(x)=c 1J (2×2)+c 1J (2×2).
1x 3/4 2x −3/4 21. Leta=2,c=3,b=1,ν=2/3toget
y(x) = c1x2J2/3(x3) + c2x2J−2/3(x3).
22. Choose a = 4,c = 3,b = 2,ν = 3/4 for a general solution
y(x) = c1x4J3/4(2×3) + c2x4J−3/4(2×3).
23. Here we get a = b = 0, so this method produces only the trivial solution.
However, if the differential equation is multiplied by x2, we obtain x2y′′+xy′− 1y=0,
16
which is an Euler equation with general solution
y(x) = c1x1/4 + c2x−1/4.
24. Witha=−2,c=b=3andν=1/2,thegeneralsolutionis
y(x) = c1x−2J1/2(3×3) + c2x−2J−1/2(3×3).
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8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS 251
Figure 8.20: First normal mode in Problem 1 at times t = 1/4, 1/2, 3/4.
8.3 Some Applications of Bessel Functions
1. With f(r) = r(1 − r) and g(r) = r2, the coefficients in the solution are 21
where
n=1
jnct jnct jn an cos R + bn sin R J0 R r .
zn(r, t) =
an = J2(j ) Rs2(1 − RS)J0(jns) ds 1n0
and
The solution is
R21 bn = j cJ2(j )
R2s3J0(jns)ds. z(r, t) = zn(r, t),
n1n0 ∞
Figures 8.20 through 8.23 show graphs of the first four normal modes of the solution times t = 1/2,1,2 and 4, for R = 1, c = 2, f(r) = r(1−r) and g(r) = 0.
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252 CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
Figure 8.21: Second normal mode in Problem 1.
Figure 8.22: Third normal mode in Problem 1.
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8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS 253
Figure 8.23: Fourth normal mode in Problem 1.
2. With f (r) = r cos(πr/2) and g(r) = e−r , the coefficients an and bn in the nth normal mode are
and
21
an = J2(j ) Rs2 cos(πRs/2)J0(Rs) ds
1n0
R21
bn = j cJ2(j ) se−RsJ0(jns)ds.
n1n0
With R = 1, c = 2, f(r) = rcos(πr/2 and g(r) = 0, Figures 8.24–8.27
show the first four normal modes for times t = 1/4, 1/2, 3/4. 3. With f(r) = r2(1 − r) and g(r) = r + r2, the coefficients are
and
21
an = J2(j ) R2s3(1−Rs)J0(jns)ds
1n0
R21
bn = j cJ2(j ) Rs2(1+Rs)J0(jns)ds.
n1n0
With R = 1, c = 2, f(r) = r2(1−r) and g(r) = 0, Figures 8.28–8.31 show
the first four normal modes for times t = 1/4, 1/2, 3/4.
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254 CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
Figure 8.24: First normal mode in Problem 2.
Figure 8.25: Second normal mode in Problem 2.
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8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS 255
Figure 8.26: Third normal mode in Problem 2.
Figure 8.27: Fourth partial sum in Problem 2.
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256 CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
Figure 8.28: First normal mode in Problem 3.
Figure 8.29: Second normal mode in Problem 3.
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8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS 257
Figure 8.30: Third normal mode in Problem 3.
Figure 8.31: Fourth normal mode in Problem 3.
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258 CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
Figure 8.32: First normal mode in Problem 4.
4. The coefficients in the solution are given by 21
an = J2(j ) ssin(πs)J0(jns)ds. 1n0
Figures 8.32–8.35 show the first four normal modes, with R = 1,c = 2, g(r) = 0 and f(r) = sin(πr).
In each of Problems 5–8, the solution has the form
∞
u(r, t) = anJ0(jnr)e−2jn2 t.
n=1
For each, the expression for the coefficients is given.
5.
6.
7.
21
an = J2(j ) ξ(1 + cos(πξ))J0(jnξ) dξ.
1n0 21
an = J2(j ) 1n0
21 an = J2(j )
ξ(1 − ξ2)J0(jnξ) dξ. ξ2 cos(3πξ/2)J0(jnξ).
1n0
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8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS 259
Figure 8.33: Second normal mode in Problem 4.
Figure 8.34: Third normal mode in Problem 4.
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260
CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
8.
and 2/5.
Figure 8.35: Fourth normal mode in Problem 4.
21
an = J2(j ) ξ(2ξ3 −ξ−1)J0(jnξ)dξ.
1n0
Figures 8.36–8.39 show graphs of u(r,t0) for t0 equal to 1/10, 1/5, 1/7
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8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS 261
Figure 8.36: Temperature profiles at specific times in Problem 5.
Figure 8.37: Temperature profiles in Problem 6.
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262 CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS
Figure 8.38: Temperature profiles in Problem 7.
Figure 8.39: Temperature profiles in Problem 8.
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Chapter 9
Transform Methods of Solution
9.1
1.
Laplace Transform Methods
Apply the Laplace transform with respect to t to the partial differential equation to get
s2Y (x, s) = c2Y ′′(x, s) + K . s
Here primes denote differentiation with respect to x and the initial condi- tions have been inserted in using the operational rule for the derivatives. Write this equation as
′′s2 K Y −c2Y=−c2s.
Think of this as a second-order differential equation in x, with s carried along as a parameter. The general solution is
Y (x, s) = c1esx/c + c2e−sx/c + K . s3
Here c1 and c2 are “constants”, but may involve s, because x is the variable of the differential equation. Now
Y (0, s) = [y(0, t)](s) = F (s) = c1 + c2 + K . s3
We need limx→∞y(0,t) = 0, so lims→∞Y(x,s) = 0. Therefore c1 = 0
and
We now have
c2 = F (s) − K . s3
K −sx/c F (s) − s3 e
263
K + s3 .
Y (x, s) =
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264
2.
CHAPTER 9. TRANSFORM METHODS OF SOLUTION
The solution is the inverse Laplace transform of Y (x, s). Recalling the formula for the inverse Laplace transform of e−asF(s), we obtain
x K x2 x 1 2 y(x,t)= f t−c −2 t−c H t−c +2Kt,
in which H is the Heaviside function.
Apply the Laplace transform with respect to t to the partial differential
equation to get
Then
9s2Y (x, s) + Y ′′(x, s) − 6sY ′(x, s) = 0. Y′′ −6sY′ +9s2Y =0.
This second-order differential equation has characteristic equation r2 − 6sr + 9s2 = 0,
with repeated root r = 3s. Then
Y (x, s) = c1e3sx + c2xe3sx.
Now, so Next, so
L[y(0, t)](s) = Y (0, s) = 0 = c1, Y (x, s) = c2xe3xs.
L[y(2, t)](s) = F (s) = 2c2e6s,
c2 = 1e−6sF(s). 2
Then
The solution is obtained by inverting this to get
Y (x, s) = 1 e−6s F (s)xe3xs = 1 xF (s)e(3x−6)s . 22
y(x,t)= 1xf(t−(6−3x))H(t−(6−3x)). 2
3.
From the partial differential equation and the initial conditions, s2Y(x,s)=c2Y′′ − A.
Then
s2 ′′ s2 A
Y −c2=s2,
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9.1. LAPLACE TRANSFORM METHODS 265 with general solution
Y(x,s)=c1esx/c +c2e−sx/c − A. s4
Becauselimx→∞y(x,t)=0,wemusthavelims→∞Y(x,s)=0,soc1 =0
and
Next, y(0, t) = 0, so
and then
Then
Y(x,s)=c2e−sx/c − A. s4
Y (0, s) = c2 − A s4
c2 = A. s4
Y (x, s) = A e−sx/c − A . s4 s4
The solution is the inverse of this,
A x3 x A3 y(x,t)=6t−c Ht−c−6t.
5. Transform the partial differential equation to get s2 Y (x, s) = c2 Y ′′ (x, s) − Ax .
s2 ′′ s2 Ax
s4
The condition that limx→∞ y(x, t) = 0 forces lims→∞ Y (x, s) = 0, so
Then
This has general solution
Then
Invert this for the solution
Y −c2Y=c2s2.
Y (x, s) = c1 esx/c + c2 e−sx/c − Ax .
c1 = 0 and Next,
Y (x, t) = c2 e−sx/c − Ax . s4
L[y(0, t)](s) = F (s) = Y (0, s) = c2. Y (x, s) = F (s)e−sx/c − Ax .
s4 y(x,t)=ft− xHt− x− 1Axt3.
cc6
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266 6.
CHAPTER 9. TRANSFORM METHODS OF SOLUTION
Take the Laplace transform (in t) in the heat equation and use the initial
conditions to get
with general solution
U(x,s)=c1e s/kx+c2e− s/kx.
Because limx→∞ u(x,t) = 0, we have lims→∞ U(x,s) = 0, so c1 = 0 and √
U(x,s)=c2e− s/kx. Next take the transform of u(0, t) = t2 to get
U(0,s)= 2 =c2, s3
U′′ − sU = 0, k
√√
so
As preparation for using the convolution theorem, think of this as the
U(x,s)= 2e−√s/kx. s3
product
2 1 −√s/kx U(x,s)= s2 se
.
7.
By the convolution theorem, the inverse of this is the solution
x u(x, t) = 2t ∗ erfc √ .
2 kt
Take the transform with respect to t of the heat equation to get
or
sU(x,s)−e−x =kU′′(x,s), U′′ − sU = −1e−x.
kk
The associated homogeneous equation of this nonhomogeneous equation has the general solution
√√
Uh(x, s) = c1e s/kx + c2e− s/kx.
For a particular solution, use undetermined coefficients, trying
Up(x, s) = Ae−x.
Substitute this into the nonhomogeneous differential equation to get
A−sA=−1, kk
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9.1. LAPLACE TRANSFORM METHODS
267
1 +s−ke .
so
A=s−k .
Then, U(x,s)=Uh(x,s)+Up(x,s)=c1e
−√s/kx U(x,s) = c2e−√s/kx + 1 e−x.
s−k
Take the transform of u(0, t) = 0 to get U(0,s)=c2+ 1 .
√s/kx Because limx→∞ u(x, t) = 0, choose c1 = 0, so
−x
+c2e
Then so
s−k
c2=− 1 s−k
U(x,s) = − 1 e−√s/kx + 1 e−x. s−k s−k
Using the convolution theorem, write
u(x, t) = −e−kt ∗ L−1 e−√s/kx (t) + ekte−x.
By consulting a table, we obtain
L−1 e−(x/√k)s (t) = √x e−x2/4kt. 2 πkt3
Then,
e−x2/4kt +ekt−x. 8. Apply the transform (in t) to the heat equation to get
u(x,t)=−e−kt ∗ √x
2 πkt3
with general solution
U′′ − sU = 0, k
√√
U(x,s)=c1e s/kx+c2e− x/kx.
Now u(x, 0) forces c1 = −c2 and the transformed solution has the form
U (x, s) = c sinh(s/kx).
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268
CHAPTER 9. TRANSFORM METHODS OF SOLUTION
With u(1,t) = f(t), we get U(1,s) = F(s), so F (s)
Then
and the solution is
c = sinh(s/k). sinh(s/kx)
U(x,s)= sinh(s/k)F(s) u(x, t) = L−1 [U (x, s)](t).
9.2
Fourier Transform Methods
For the first four problems, the solution is given by
1∞2 u(x, t) = √ f(ξ)e−(x−ξ) /4kt.
2 πkt −∞ 1. With f(x) = e−4|x|, the solution is
1∞2
√ e−4|ξ|e−(x−ξ) /4kt dξ.
2 πkt −∞
2. The solution is
3.
4.
u(x, t) =
5. Take the Fourier transform of the wave equation with respect to x to get
u(x, t) =
1π2 √ sin(ξ)e−(x−ξ)
2 πkt −π
/4kt dξ.
u(x, t) =
142
√ ξe−(x−ξ) /4kt dξ.
2 πkt 0 112
√ e−ξ−(x−ξ) /4kt dξ. 2 πkt −1
or Then
′′ 2 2
y = 144(iω) y = −144ω y,
′′ 2
y + 144ω y = 0.
y(ω, t) = c1 cos(12ωt) + c2 sin(12ωt),
in which primes denote differentiation with respect to t. Because yt(x, 0) =
0,thenc1 =0and
y(ω, t) = c1 cos(12ωt).
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9.2. FOURIER TRANSFORM METHODS
269
Next, y(x,0) = f(x), so
Then
y(ω, t) = 10 cos(144ωt). 25+ω2
y(x,t)=Re 2π −∞ 25+ω2 cos(12ωt)e dω .
Of course y(x,t) is a real quantity, so in the last line we have taken the
real part of the integral. If we replace
eiωx = cos(ωx) + i sin(ωx)
in this integral, we can write more explicitly
y(x, t) = 1 ∞ 10 cos(ωx) cos(12ωt) dω.
y(ω,0) = f(ω).
Then c1 = f(ω), so
It is routine to compute (or use a software routine to find)
y(ω, t) = f (ω) cos(12ωt).
f(ω) = 10 . 25+ω2
Finally, use the integral formula for the inverse Fourier transform to obtain 1∞ 10 iωx
2π −∞25+ω2
6. Apply the Fourier transform (in x) to the initial-boundary value problem
to get
(8 − ξ)e−iωξ dξ = ω2 , This problem has the solution
1−8ωi−e−8ωi
y(ω, t) = ω2 cos(8ωt).
′′ 2
y + 64ω y = 0,
0
y(ω, 0) = ′
8 1−8iω−e−8ωi
y (ω, 0) = 0.
Invert this to obtain
y(x, t) = Re 2π −∞ ω2 cos(8ωt)eiωx dω
1 ∞ 1−8ωi−e−8ωi
.
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270 CHAPTER 9. TRANSFORM METHODS OF SOLUTION 7. Apply the Fourier transform to the initial-boundary value problem to get
′′ 2
y + 16ω y = 0,
y(ω, 0) = 0,
′
Invert this to obtain the solution
1∞ isin(πω)
sin(ξ)e
The solution of the transformed problem is
y (ω, 0) =
dξ =
π −π
−iωξ
2i sin(πω) ω2 − 1 .
y(ω,t)= 2isin(πω) sin(4ωt). 4ω(ω2 − 1)
iωx y(x, t) = Re 2π −∞ 2ω(ω2 − 1) sin(4ωt)e dω .
8. The initial-boundary value problem transforms (in x) to ′′ 2
y + ω y = 0 ,
y(ω, 0) = ′
22
(2 − |ξ|)e−iωξ dξ = ω2 (1 − cos(2ω)), 2 (1 − cos(2ω)) cos(ωt)eiωx dω.
−2 y (ω, 0) = 0.
This has solution
y(ω, t) =
ω2
9. The problem transforms (in x) to
′′ 2
y + 9 ω y = 0 ,
y(ω, 0) = 0,
′ −2x
(2 − iω)e−(2+iω) y(ω, t) = 3ω(4 + ω2) .
1 ∞ (2−iω)e−(2+iω)
y(x, t) = 2π −∞ 3ω(4 + ω2) sin(3ωt)eiωx dω
y (ω, 0) = F[e This problem has the solution
.
H(x − 1)](ω) =
(2 − iω)e−(2+iω) 4 + ω2
Then
.
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9.3. FOURIER SINE AND COSINE TRANSFORM METHODS
271
10. Transform the problem to get
′′ 2
y + 4 ω y = 0 ,
y(ω, 0) = 0,
y (ω, 0) = This has the solution
−iωξ g(ξ)e dξ =
2 −2
2(1 − cos(2ω)) ω .
′
Invert this to get
y(x, t) = Re 2π −∞ ω2 sin(2ωt)e
y(ω, t) = 1 − cos(2ω) sin(2ωt). ω2
1 ∞ 1−cos(2ω) iωx
11. This problem was solved by the Fourier transform in the text, so we can
dω
.
use the result to immediately write
y0 −1 4 1
u(x,y)= π y2 +(ξ−x)2 dξ+ y2 +(ξ−x)2 dξ −4 0
y x+4 x−4 =−π arctan y +arctan y .
12. The solution is
u(x, y) = y 1 ξ dξ
π 0 y2+(ξ−x)2
− 1 ln(x2+y2)−ln((x−1)2+y2)−2xarctan(x/y)+2xarctan((x−1)/y).
2π
9.3 Fourier Sine and Cosine Transform Methods
1. Take a Fourier sine transform (in x) of the problem to get ′′ 2
y + 9 ω y = 0 , SS
yS (ω, 0) =
′
1 0
ξ(1 − ξ) sin(ωξ) dξ =
2(1 − cos(ω) − ω sin(ω) ω3
cos(3ωt).
,
y (ω,0)=0. S
The solution of the transformed problem is 2(1−cos(ω))−ωsin(ω)
yS(ω,t) = ω3 Invert this to obtain the solution
2 ∞2(1−cos(ω))−ωsin(ω)
y(x, t) = π 0 ω3 sin(ωx) cos(3ωt) dω.
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272 CHAPTER 9. TRANSFORM METHODS OF SOLUTION 2. Apply the sine transform to the problem to get
′′ 2
y + 9 ω y = 0 ,
SS yS(ω,0) = 0,
′
11
2(cos(4ω) − cos(11ω)) ω
2sin(ωξ)dξ= This problem has the solution
y = S
.
4
yS (ω, t) = 2(cos(4ω) − cos(11ω)) sin(3ωt). 3ω2
Invert this to obtain the solution
4 ∞ cos(4ω) − cos(11ω)
y(x, t) = 3π 0 ω2 3. The sine transformed problem is
′′ 2
y + 4 ω y = 0 ,
sin(ωx) sin(3ωt).
sin(ωπ/2) − sin(5ωπ/2) ω2 − 1
SS yS(ω,0) = 0,
′
5π/2 π/2
.
y (ω,0)= S
cos(ξ)sin(ωξ)dξ=
This has solution
yS (ω, t) = sin(ωπ/2) − sin(5ωπ/2) sin(2ωt).
2ω(ω2 − 1) Invert this to obtain the solution
2 ∞ sin(ωπ/2) − sin(5ωπ/2)
y(x, t) = π 0 2ω(ω2 − 1) sin(ωx) sin(2ωt).
4. Upon taking the sine transform (in x) of the problem, we get ′′ 2
This has the solution
Invert this to obtain
y(x, t) = − 4 ∞ ω
sin(ωx) cos(6ωt).
y + 36ω y = 0, SS
yS (ω, 0) =
′
y (ω,0)=0. S
∞ 0
−2e−ξ sin(ωξ) dξ
yS(ω,t)=− 2ω cos(6ωt). 1+ω2
π0 1+ω2
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9.3. FOURIER SINE AND COSINE TRANSFORM METHODS 273 5. After applying the sine transform to the initial-boundary value problem,
we obtain
yS(ω,0) = 0,
0
′′ 2
y + 196ω y = 0,
SS
3 ′2
ξ (3−ξ)sin(ωξ)dξ
= 3 (2 sin(3ω) − 4ω cos(3ω) − 3ω2 sin(3ω) − 2ω).
y (ω,0)= S
ω4
The transformed problem has the solution
y S ( ω , t ) =
3 (2 sin(3ω) − 4ω cos(3ω) − 3ω2 sin(3ω) − 2ω) sin(14ωt).
14ω5 Then
2∞
yS (ω, t) sin(ωx) dω.
6. Apply the Fourier sine transform with respect to x to the problem to get
y(x, t) = π ′2
0
uS +ω uS +tuS =0,
uS(ω,0) = FS xe−x = 2ω .
(1 + ω2)2
This problem (involving a linear first-order differential equation) has the
solution
uS(ω,t) = 2ω e−(ω2t−t2/2). (1 + ω2)2
Invert this to solve the problem:
u(x, t) = 4 ∞ ω e−ω2t−t2/2 sin(ωx) dω.
π 0 (1+ω2)2
7. Apply the Fourier cosine transform in x to the problem to get
′2
uC +(1+ω )uC =−f(t);uC(ω,0)=0.
This has the solution uC(ω,t)=e−(1+ω )t
Invert this to obtain u(x,t)=−π
2t2 f(τ)e(1+ω )τ dτ
0
= −f(τ) ∗ e−(1+ω2)t.
2∞ 2
f(t)∗e−(1+ω )tcos(ωx)dω.
0
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274 CHAPTER 9. TRANSFORM METHODS OF SOLUTION
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Chapter 10
Vectors and the Vector
Space Rn
10.1 Vectors in the Plane and 3−Space
1.
2.
3.
4.
5.
√√
F+G=(2+ 2)i+3j,F−G=(2− 2)i−9j+10k,
√√ 2F=4i−6j+10k,3G=3 2i+18j−15k,∥F∥= 38
F + G = i + 4j − 3k, F − G = i − 4j − 3k, √
2F=2i−6k,3G=12j,∥F∥= 10
F + G = 3i − k, F − G = i − 10j + k,
√
2F=2i−6k,3G=3i+15j−3k,∥F∥= 29
√√
F+G=( 2+8)i+j−4k,F−G=( 2−9)i+j−8k,
√ 2F=4i−6j+10k,3G=24i+6k,∥F∥= 39
F + G = 3i − j + 3k, F − G = −i + 3j − k,
√
2F=2i+2j+2k,3G=6i−6j+6k,∥F∥= 3 275
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276 CHAPTER 10. VECTORS AND THE VECTOR SPACE RN 6. The vector
F = −5i + j − 2k
is in the direction from the first point to the second, but this vector has
√
length ∥ F ∥= 30. The vector
√F 30
has length 5 and extends from the first point to the second.
5
7.
8.
9.
9
√ (−5i−4j+2k)
45
12
√ (10i−3j−4k)
125
4(−4i+7j+4k) 9
10. The vector M = i+j−3k is oriented from (1,0,4) to (2,1,1). The vector V = (x − 1)i + yj + (z − 4)k
is parallel to M exactly when (x, y, z) is on the line through the two given points. Then, for some t, V = tM, so
Then
Then
(x − 1)i + yj + (z − 4)k = t(i + j − 3k). x − 1 = t, y = t, z − 4 = −3t.
x = 1 + t, y = t, z = 4 − 3t
varies over all real values. As a check, notice that these points are on the
line given by these parametric equations for t = 0 and t = 1 respectively.
11. x=3−6t,y=1−t,z=0
12. x = 2,y = 1,z = 1−3t This line is parallel to the z− axis, and also passes through (2, 1, 0) when t = 0.
13. x=0,y=1−t,z=3−2t
14. x=1−3t,y=−2t,z=−4+9t
15. x=2−3t,y=−3+9t,z=6−2t
are parametric equations of the line through (1, 0, 4) and (2, 1, 1). Here t
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10.2. THE DOT PRODUCT 277 10.2 The Dot Product
In Problems 1–6, F is the first vector given in the problem G is the second, and θ is the angle between F and G.
1. F·G=2and
F·G 2 cos(θ)= ∥F∥∥G∥ = √14.
F and G are not orthogonal. √
2. F · G = 8, cos(θ) = 8/ 82 and the vectors are not orthogonal. √√
3. F · G = −23, cos(θ) = −23/ 29 41 and the vectors are not orthogonal. √√
4. F · G = −63, cos(θ) = −63/ 75 74 and the vectors are not orthogonal.
5. F · G = −18, cos(θ) = −9/10 and the vectors are not orthogonal.
6. F · G = 4, cos(θ) = 2/3 and the vectors are not orthogonal.
In Problems 7–12, if the given point is (x0,y0,z0) and the normal to the proposed plane is N = ai+bj+ck, then the plane through the point and having N as normal vector has equation
a(x − x0) + b(y − y0) + c(z − z0) = 0.
The constant terms are usually collected to write this equation in the form
ax + by + cz = k.
7. The plane has equation
3(x + 1) − (y − 1) + 4(z − 2) = 0,
or
8. x−2y=−1
9. 4x−3y+2z=25
10. −3x+2y=1
11. 7x+6y−5z=−26
12. 4x+3y+z=−6
3x − y + 4z = 4.
In each of Problems 13–17, the projection of v onto u is the vector u·v u.
∥ u ∥2
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278 13.
14.
15.
16.
17.
10.3
CHAPTER 10.
VECTORS AND THE VECTOR SPACE RN
projuv = − 9 u 14
−11u 30
1u 62
73 u 101
15u 53
1.
and
2.
3.
4.
The Cross Product
i j k
F × G = −3 6 1 = 8i + 2j + 12k
−1 −2 1 i j k
G×F=−1 −2 1=−8i−2j−12k=−F×G −3 6 1
F × G = i + 12j + 6k = −G × F
F × G = −8i − 12j − 5k = −G × F
F × G = 112k = −G × F
In Problems 5–9, the three given points are used to find two nonparallel vectors in the plane that is sought (assuming that the points are not collinear). The cross product of these vectors is a normal to the plane. We than have a point on the plane and a normal to the plane, so we can find an equation of the plane.
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10.3. THE CROSS PRODUCT 279 5. Vectors from the first point to the second and third points are F = 4i −
j−6kandG=i−k. Compute
N = F × G = i − 2j + k.
Because this cross product is not the zero vector, the given points are not collinear. The plane containing these points has equation
This is or
((x + 1)i + (y − 1)j + (z − 6)k) · N = 0. x + 1 − 2(y − 1) + z − 6 = 0,
x − 2y + z = 3.
6. The points are not collinear and the plane containing them has equation
x + 2y + 6z = 12.
7. 2x−11y+z=0
8. 5x+16y−2z=−1
9. 29x+37y−12z=30
In Problems 10–12, recall that a plane ax + by + cz = k has normal vector ai + bj + ck, or any nonzero multiple of this vector.
10. N = 8i − j + k, or any nonzero scalar multiple of this vector.
11. N=i−j+2k
12. n=i−3j+2k
13. If two sides of a parallelogram meet at a point, with an angle θ between the sides, then the area of the parallelogram is the product of the lengths of these sides, times the cosine of θ. Now suppose F and G are vectors drawn from a corner of the parallelogram, along two incident sides, and θ the angle between these vectors. Then
∥ F ∥∥ G ∥ cos(θ)
is the product of the lengths of incident sides, times the angle between
them, hence is the area of the parallelogram. But
∥ F ∥∥ G ∥ cos(θ) =∥ F × G ∥
so the magnitude of this cross product is the area of the parallelogram.
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CHAPTER 10. VECTORS AND THE VECTOR SPACE RN
The vector N = G × H is normal to the base of the parallelopiped (box) having sides along F,G,H (drawn as arrows from a corner of the box). From Problem 13, the area of the base of this box is N. Then
(|G × F) · F = F · N =∥ N ∥∥ F ∥ cos(θ)
is the magnitude of the volume of the box having incident sides F, G, H,
because
∥ F ∥ cos(θ) = ± altitude of the box.
10.4 n− Vectors and the Algebraic Structure of
Rn
1. Let F =< −2,1,−1,4 >. Then O =< 0,0,0,0 > is in S (the scalar
multiple of 0 with F). Further, if a and b are real numbers, then aF + bF = (a + b)F,
so a sum of scalar multiples of F is again a scalar multiple of F, hence is in S. Finally,
a(bF) = (ab)F
so a scalar multiple of a vector in S is in S. Therefore S is a subspace of
R4.
2. < 0, 0, 0, 0, 0, 0 > is in S, having zero third and fifth components. Further, if F and G have zero third and fifth components, so does any scalar multiple of these vectors, as well as any sum of scalar multiples of these vectors. Therefore S is a subspace of R6.
3. S is not a subspace of R5. The zero vector does not have fourth component 1, and a sum of vectors with fourth component 1 does not have fourth component 1. S fails on scalar multiples as well.
4. S is not a subspace of R8. The zero vector is in S, but a sum of vectors of length less than 1 need not have length less than 1, and scalar multiples could also fail to have length less than 1.
5. S is not a subspace of R4. For example, < 1,1,1,0 > and < 0,1,1,1 > are both in S, but their sum is not, having no zero component.
6. The zero vector is in S, and a linear combination of two vectors with equal first and fourth components will have equal first and fourth components. S is a subspace of R5.
In Problems 7–16, keep in mind that a set of vectors is linearly dependent if some linear combination of these vectors, with at least one nonzero coefficient, is equal to the zero vector. And the vectors are linearly independent exactly when the only linear combination of these vectors adding up to the zero vector is the trivial one, with all zero coefficients.
280 14.
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10.4. N− VECTORS AND THE ALGEBRAIC STRUCTURE OF RN 281 7. If
then
a(3i + 2j) + b(i − j) =< 0, 0 >, 3a + b = 0 and 2a − b = 0.
From the second equation, b = 2a, and then the first equation is 5a = 0, so a = 0 and then b = 0. The only linear combination of these vectors that equals the zero vector is the trivial combination, so the vectors are linearly independent.
We could also observe that neither vector is a linear combination of the other, which would be the case of these two vectors were linearly depen- dent.
8. These vectors are linearly dependent because no one of them is a linear combination of the others. It is also easy to check that the only linear combination of these vectors that equals < 0, 0, 0, 0 > is the trivial linear combination (all coefficients zero).
9. These two vectors are linearly independent because neither is a scalar mul- tiple of the other (which would occur for two linearly dependent vectors).
10. These vectors are linearly dependent because
4 < 1,0,0,0 > −6 < 0,1,1,0 > + < −4,6,6,0 >=< 0,0,0,0 > .
11. The vectors are linearly dependent because
2 < 1,2,−3,1 > + < 4,0,0,2 > − < 6,4,−6,4 >=< 0,0,0,0 > .
12. Suppose that
α < 0,1,1,1 > +β < −3,2,4,4 > +γ < −2,2,34,2 >
+ δ < 1, 1, −6, 2 >=< 0, 0, 0, 0 > .
Looking at the components of the sum on the left, we obtain
−3β − 2γ + δ = 0, α + 2β + 2γ + δ = 0, α + 4β + 34γ − 6δ = 0, α + 4β + 2γ + 2δ = 0.
It is routine to solve these equations and show that α = β = γ = δ = 0.
Therefore the vectors are linearly independent.
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282 CHAPTER 10. VECTORS AND THE VECTOR SPACE RN
13. The vectors are linearly dependent because
2 < 1, −2 > −2 < 4, 1 > + < 6, 6 >=< 0, 0 > .
14. The vectors are linearly independent (each has a component of 1 where
the other two have components of 0).
15. The vectors are linearly independent.
16. The vectors are linearly independent.
In each of Problems 17–21 it is routine to show that S is not empty, and that a linear combination of vectors of S is in S, showing that S is a subspace of the appropriate Rn. We will show how to find a basis for S.
17. Every vector in S has the form
x < 1, 0, 0, −1 > +y < 0, 1, −1, 0 > .
Therefore the two vectors < 1,0,0,−1 > and < 0,1,−1,0 > span S. These vectors are also linearly independent, and so form a basis for S, which has dimension 2.
18. Every vector in S is of the form
x < 1, 0, 2, 0 > +y < 0, 1, 0, 3 > .
The independent vectors < 1, 0, 2, 0 > and < 0, 1, 0, 3 > therefore span S and form a basis. S has dimension 2.
19. S consists of all vectors in Rn of the form
< x1,0,x2,··· ,xn−1 > .
The n − 1 independent vectors
< 1,0,0,··· ,0 >,< 0,0,1,0,··· ,0 >,··· ,< 0,0,0,··· ,0,1 >
span S, so S has dimension n − 1.
20. S is spanned by the independent vectors
< 1,1,0,0,0,0 >,< 0,0,1,1,0,0 >,< 0,0,0,0,0,1 > and so has dimension 3.
21. Every vector in S is a scalar multiple of 0,1,0,2,0,3,0 >
so this vector forms a basis for S and S has dimension 1.
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10.4. N− VECTORS AND THE ALGEBRAIC STRUCTURE OF RN 283
22. The spanning vectors are independent, so they form a basis for the sub-
space S of R3 that they span. Further, by inspection,
< 3, 1, 0 >= 3 < 1, −1, 0 > +2 < 0, 1, 0 > .
23. The spanning vectors are independent and form a basis for the subspace S of R3 that they span. Further, by inspection,
< −5, −3, −3 >= −5 < 1, 1, 1 > +2 < 0, 1, 1 > .
24. It is routine to check that the three spanning vectors are independent,
hence form a basis for the subspace S of R4 that they span. By inspection, < 4, 5, 9, 4 >= 2 < 2, 1, 1, 0 > +3 < 0, 1, 1, 0 > +4 < 0, 0, 1, 1 > .
25. The spanning vectors are independent and so form a basis for the subspace S of R4 that they span. This subspace has dimension 3. For X to be in S, we need numbers a,b and c so that
a < 1,0,−3,2 > +b < 1,0,−1,1 >=< −4,0,10,−5 > . This requires that
a+b=−4,−3a+b=10, and2a+b=−5. Then a = −3, b = −1, so
< −4,0,10,−7 >= −3 < 1,0,−3,2 > − < 1,0,−1,1 > and < −4,0,10,−4 > is in S.
26. All that is needed is to show that the dot product of the two vectors is zero:
(X − Y) · (X + Y) = X · X + X · Y − Y · X − Y · Y =∥X∥2 −∥Y∥2=0.
27. Because U is in S, which is spanned by V1,··· ,Vk, we must have U=c1V1 +···+ckVk,
so V1, · · · , Vk, U are linearly independent.
28. If U1, · · · , Uk are linearly independent, then they form a basis for the subspace of Rn spanned by S. If these vectors are not independent, then at least one of the vectors must be a linear combination of the others. Suppose Uk is a linear combination of U1, · · · , Uk−1. Then
U1,··· ,Uk−1
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284
CHAPTER 10. VECTORS AND THE VECTOR SPACE RN
span S. If these vectors are linearly independent, they form a basis for S. If not, one of these vectors, say Uk−1 is a linear combination of the others. Now
U1,··· ,Uk−2
span S. Continue in this way. If these vectors are linearly independent, they form a basis for S. If not, one must be a linear combination of the others, and then we obtain a set of k − 3 of the original set of vectors that spans S. Continue in this way until a linearly independent set of the original spanning vectors spans S. This set is a basis for S.
Suppose the set of vectors is V1, · · · , Vk, O. Then 0V1 +0V2 +···+0Vk +1O=O
is a linear combination of the vectors adding up to the zero vector, and with a nonzero coefficient. This makes the original set of vectors (including the zero vector) linearly dependent.
29.
10.5 Orthogonal Sets and Orthogonalization
1.
2. Let
∥V1 +···+Vk ∥2=(V1 +···+Vk)·(V1 +···+Vk) =V1 ·(V1 +···+Vk)+V2 ·(V1 +···+Vk)+··· +Vk ·(V1 +···+Vk)
=V1 ·V1 +V2 ·V2 +···+Vk ·Vk
=∥V1 ∥2 +∥V2 ∥2 +···∥Vk ∥2, inwhichwehaveusedthefactthatVi·Vj =0ifi̸=j.
k
Y = X − (X · Vj )Vj .
j=1
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10.5.
ORTHOGONAL SETS AND ORTHOGONALIZATION 285 Compute
0 ≤∥ Y ∥2= Y · Y kk
= X − (X · Vj )Vj · X − (X · Vj )Vj
3.
Reason as in Problem 2, except now use the fact that
n
(X·Vj)Vj =X. j=1
Then
kk
0≤∥X∥2 −2(X·Vj)2 +(X·Vj)2
j=1 j=1 k
=∥X∥2 −(X·Vj)2. j=1
k
(X·Vj)2 ≤∥X∥2 . j=1
j=1
k
= X · X − 2X · (X · Vj )Vj
j=1
j=1 kk
+ (X · Vj)Vj · (X · Vj)Vj
j=1
j=1
k
= X · X − 2X (X · Vj )Vj
j=1 +(X·Vj)(X·Vr)Vj ·Vr.
kk j=1 r=1
Because Vj ·Vr = 0 if j ̸= r, and Vj ·Vj = 1, the double sum collapses to just terms in which j = r and we have
In each of Problem 4–11, the given vectors are denoted X1 , · · · , Xk in the given order.
4. LetV1 =X1 and
V2 = X2 − X2 · X1 X1 X1 · X1
= X2 + 18X1 17
=< 52/17, −13/17, 0 > .
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286 CHAPTER 10. VECTORS AND THE VECTOR SPACE RN 5. Take V1 = X1 and
V2 = X2 + 11X1 =< 0,4/5,2/5,0 > . 5
6. Let V1 = X1. Next,
V2 = X2 + 7X1 =< 0,4/3,13/6,29/6 > .
Finally,
7. LetV1 =X1 and
Finally, let
8. V1 = X1,
6
V3 = X3 − 3V1 − 43/2 V2 6 179/6
=X3 − 1V1 − 129V2 2 179
= 1 < 0,7,−11,3 > . 179
V2 = X2 + 5 X1 26
= 1 < 109,0,−41,58 > . 26
V3 = X3 − 17X1 − 331/26V1 26 651/26
= X3 − 17V1 − 331V2 26 651
= 1 < −962,0,−1406,0,814 > . 651
V2 = X2 − 5X1 7
= 1 < 0,0,−1,−19,40 >, 9
V3 = X3 + 2V1 + 17V2 99
= 1 < 0,218,−341,279,62 >, 218
V4=6X1+13V2− 435V3 9 3 1179
= 1 < 0,248,88,−24,−32 > . 393
and
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10.6. ORTHOGONAL COMPLEMENTS AND PROJECTIONS 287 9. V1 = X1,
and
V2 = X2 − 1 X1 10
= 1 < 21,−8,−60,−31,−18,0 >, 10
V3 =X3− 3X1−163/10V2 10 269/10
= 1 < −423, −300, 489, −759, 132, 0 >, 269
V4 = X4 + 15X1 − 13/2 V2 − 4455/269V4 10 260/10 4095/269
= 1 < 337,−145,250,29,−9,0 > . 91
V2 =X2 + 3X1 = 1 <0,0,−3,3,0,0>. 22
10. V1 = X1 and
11. V1 = X1 and V2 = X2 because X2 and X1 are orthogonal. Finally, V3 =X3 + 4 V1 + 4V2 = 1 <0,−8,0,−8,0,16>.
12 2 3
10.6 Orthogonal Complements and Projections
1. Let V1 =< 1,−1,0,0 > and V2 =< 1,1,0,0 >. These form an orthogonal basis for S. Then
uS = u·V1 V1+ u·V2 V2 V1 · V1 V2 · V2
=−4V1 +2V2 =<−2,6,0,0> u⊥ =u−uS =<0,0,1,7>.
and
The distance between u and S is
2. Let
∥ u⊥ ∥= √50.
V1 =< 1,0,0,2,0 > and V2 =< −2,0,0,1,0 > .
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288 CHAPTER 10. VECTORS AND THE VECTOR SPACE RN
We find that
uS = 2 V1 + 1 V2 =< 0, 0, 0, 1, 0 > 55
and
The distance between u and S is
u⊥ =< 0,−4,−4,0,3 > . ∥ u⊥ ∥= √41.
3. Let
V1 =< 1,−1,0,1,−1 >,V2 =< 1,0,0,−1,0 >,V3 =< 0,−1,0,0,1 > .
Then
and
uS = 7V1 +V2 −3V3 = 1 <9,−1,0,5,−13>. 22
u⊥ = 1 < −1,−1,6,−1,−1 > . 2
The distance between u and S is
4. Let Then
∥ u⊥ ∥= √10.
V1 =< 1,−1,0,0 > and V2 =< 1,1,6,1 > .
uS = −3V1 + 31 V2 =< −86/39, 148/39, 62/13, 31/39 > 39
and
The distance between u and S is
u⊥ = 1 < 203,203,−230,−226 > . 309
∥u⊥∥= 1186,394. 309
V1 =< 1,0,1,0,1,0,0 > and V2 =< 0,1,0,1,0,0,0 > . uS = 3V1 + 1 V2 = 1 < 6, 1, 6, 1, 6, 0, 0 >
22
5. Let
We find that
and
The distance between u and S is
u⊥ = 1 < 10,1,−4,−1,−6,−6,8 > . 2
∥ u⊥ ∥= 1√254. 2
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10.6. ORTHOGONAL COMPLEMENTS AND PROJECTIONS 289
6. If every vector in S⊥ is orthogonal to every vector in S, then, going the
other way, every vector in S is orthogonal to every vector in S⊥. Then (S⊥⊥) = S.
7. Let v1,··· ,vk be an orthogonal basis for S, and u1,···ur an orthogonal basis for S⊥. Then the vectors
u1,··· ,uk,v1,···vr
span Rn because every n− vector is a sum of a vector in S (a linear combination of v1 , · · · , vk ) and a vector in S ⊥ (a linear combination of u1,··· ,ur). Further, the set of vectors
u1,··· ,uk,v1,···vr
are linearly independent, because no one of them is a linear combination of others in this set. These k + r vectors therefore form a basis for Rn. Because Rn has dimension n,
n = k + r.
That is,
dimension(S) + dimension (S⊥) = dimension(Rn).
8. Let u =< 1,−1,3,−3 >. The idea is to use an orthogonal basis for S to
produce uS, which is the vector we seek. The given vectors V1 =< 1,0,1,0 > and V2 =< −2,0,2,1 >
form an orthogonal basis for S. Then
uS = 2V1 + 1 V2 = 1 < 16, 0, 20, 1 > .
99 This is the vector in S closest to u.
9. Denote
V1 =< 2,1,−1,0,0 >,V2 =< −1,2,0,1,0 >,V3 =< 0,1,1,−2,0 > .
These form an orthogonal basis for S. With u =< 4, 3, −3, 4, 7 >, compute uS =7V1+V2−4V3
33
= 1 < 11,9,−11,11,0 > . 3
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290 CHAPTER 10. VECTORS AND THE VECTOR SPACE RN
10. Denote
V1 =< 0,1,1,0,0,1 >,V2 =< 0,0,3,0,0,−3 >,V3 =< 6,0,0,−2,0,0 > . With u =< 0,1,1,−2,−2,6 >, compute
uS =8V1−5V2+ 1V3 3 6 10
=< 3/5, 8/3, 1/6, −1/5, 0, 31/6 > . This is the vector in S closest to u.
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Chapter 11
Matrices, Determinants and Linear Systems
11.1
1.
2.
Matrices and Matrix Algebra
14 −2 6 2A − 3B = 10 −5 −6
−26 −43 −8 19 2
−5A+3B= 6 −2 −28 38
−27 35
3.
A +2AB= 4+2x+2ex +2xex −22−2x+e2x +2excos(x)
2+2x−x2 12x+(1−x)(x+ex +2cos(x))
2
4. −3A−4B = (18) This is a 1×1 matrix, which we can think of as just
the single matrix entry, in this case 18. 5.
6.
−36 0 68 196 20 4A+8B = 128 −40 −36 −8 72
2 2−1718−4082710 A −B = 6 1 − −5 −39 = 11 40
291
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292CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS 7. BA is not defined;
8.
9. AB = (115);
10.
−16 6 6 184 11. AB is not defined;
12 −32
−10 −34 −16 −30 −14
AB= 10 −2 −5 1
−16 0
AB= 17 28 ;BA= −14 0
3 −18−6−42 66
−2 12 4 28 −44 BA = −6 36 12 84 −132 0 0 0 0 0
4 −24 −8 −56 88
48 1 1 −58
−96 2 2 220 76 152
AB = −288 −22 −22 −68;BA = 50 136
410 36 −56 227 BA= 17 253 40 −1
−22 30 −10 −4 AB= −42 45 30 6
13. AB is not defined;
14. Neither AB nor BA is defined.
15. BA is not defined;
39 −84 21 AB= −23 38 3
12. BA is not defined;
16. AB is not defined;
17. AB is 14×14, BA is 21×21.
BA = 28 30 18. Neither AB nor BA is defined.
BA = −16 −13 −5
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−11 −8 −45 15 61 −63
11.1. MATRICES AND MATRIX ALGEBRA 293
19. AB is not defined, BA is 4×2.
20. AB is 1×3, BA is not defined.
21. AB is not defined, BA is 7×6.
22. There are infinitely many examples. Here is one. Let
2 1 2 1 6 0 A= 8 4 ,B= −1 1 ,C= −1 1 .
Then B ̸= C, but
12 6 BA=CA= 6 3 .
23. The adjacency matrix of G is
Compute
0 1 1 0 0
1 1 0 1 1 A=1 1 0 1 1.
0 1 1 0 1 01110
2 7 7 4 4 14 17 17 18 18
7 8 9 9 9 17 34 33 26 26 A3=7 9 8 9 9andA4=17 33 34 26 26.
4 9 9 6 7 18 26 26 25 24
4 9 9 7 6 18 26 26 24 25
The number of v1 − −v4 walks of length 3 is (A)314 = 4 and the number of v1 − −v4 walks of length 4 is (A4)14 = 18. The number of v2 − −v3 walks of length 3 is 9 and the number of v2 −−v4 walks of length 4 is 26.
24. The adjacency matrix of H is
0 1 1 0 1
1 0 1 0 1 A=1 1 0 1 0.
0 0 1 0 1 11010
3 2 1 2 1 19 18 11 14 11
Then
The number of v1 − −v4 walks of length 4 is (A4)14 = 14, the number of v2 − −v3 walks of length 2 is (A2)23 = 1.
2 3 1 2 1 18 19 11 11 11 A2=1 1 3 0 3andA4=11 11 20 4 20.
22020 1414 4 12 4 11303 111120 4 20
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294CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS
25. The adjacency matrix of K is 0
1 1 1 1 0 1 1 0
Then
and
1 0110
4 2 3 3 2 10 10 11 11 10
23223 10 6 1010 6
1
A = 11 01 1
. 1 1 1 0 1
A2=3 2 4 3 2,A3=11 10 10 11 10
3 2 3 4 2 11 10 11 10 10
23223
42
32 A4 = 41 41
32 41
30 32
32 42
32 31
10 6 10 10 6 41 32
32 30 41 32 . 42 32
32 30 32 32 30
The number of v4 −−v5 walks of length 2 is 2 and the number of v2 −−v3 walks of length 3 is 10. The number of v1 − −v2 walks of length 4 is 32 and the number of v4 − −v5 walks of length 4 is 32.
26. (a) The i, i element of A2 is the number of vi − −vi walks of length 2 in the graph. Each such walk has the form vi − −vj − −vi for some j ̸= i, hence corresponds to a vertex vj adjacent to vi in the graph. Therefore A2ii counts the number of vertices adjacent to vi in the graph, and this is the degree of vi.
(b) The i−−i element of A3 is the number of vi −−vi walks of length 3 in the graph. Any such walk has the form vi −−vj −−vk −−vi with j ̸= k and neither j nor k equal to i. These three vertices therefore form the vertices of a triangle in the graph. However, each such triangle is counted twice in the i, i element of A3 because this triangle actually represents two vI −−vi walks of length 3, namely vi −−vj −−vk −−vi and vi −−vk −−vj −−vi. Therefore
(A3)ii = 2(number of triangles in G).
27. Let M be the set of all n × n real matrices. Each such matrix has nm elements in its n rows and m columns (some possibly the same number, but differentiated by location). If we string out the rows, with row 2 following row 1, then row 3 following row 2, and so on, we have a string of nm elements, which we can think of as a vector in Rnm. Each nm − −vector corresponds to an n×m matrix, and each n×m matrix produces a unique nm−− vector. In this way the n×m matrices correspond in a one-to-one fashion with the vectors in Rnm.
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11.2. ROW OPERATIONS AND REDUCED MATRICES 295 11.2 Row Operations and Reduced Matrices
1. A is 3×4, so multiply A on the left by the matrix Ω formed by performing this elementary row operation on I3. Thus form
1 0 0
2.
1 0 0 0 Ω=0 1 0 0 0 6 1 0
0001
Ω = 0 001
3. In this case more than one elementary operation is performed, so Ω is a product of elementary operations, the right-most factor performing the first operation, then the next to right-most, the second operation, and so on, with the left factor performing the last operation:
5 0 00 1 01 0 √13 Ω=01010001 0.
001001001 If these products are carried out, we get
√
3 0 .
This choice of Ω can be checked by verifying that ΩA = B.
4.
5.
6.
Then
05 0 √
1 0 13 001
1 00100 1 00 Ω=−1 1 00 0 1=−1 0 1
001010 010 0 11 01 √3 0 15
Ω= 1 0 0 15 0 1 = 1 √3
1 0 01 0 01 0 01 0 0
Ω=0 1 00 1 0 3 1 00 1 0. 011 004 001 101
√
1 0 0 √
√ Ω=310.
4+ 31 4
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296CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS
7.
8.
500
In these and later problems, the delta notation is sometimes useful:
1 ifi=j, δij= 0 ifi̸=j.
9.LetA=[aij]beann×mmatrix. Now,B=[bij]andE=[eij]are obtained, respectively, by interchanging rows s and t of A and In. Then, for i ̸= s and i ̸= t,
bij =aij andeij =δij.
Ifi=s,bsj =atj andesj =δtj. Andifi=t,bij =asj andeij =δsj. Nowconsiderthei,j−−elementofEA. Fori̸=sandi̸=t,
1001 00100 1 00 Ω=0 0 114 1 00 1 0=0 0 4
0 1 0 0 0 1 0 0 4 14 1 0
1 0 00 0 11 0 01 0 0 Ω=0 1 00 1 00 1 00 0 1.
005 100 031 010 If these are multiplied out, we get
0 1 3 Ω=0 0 1.
For i = s,
And for i = t,
n
(EA)sj = eikakj = aij = bij.
k=1
nn
(EA)sj = eskakj = δtkakj = atj = bsj.
k=1 k=1
nn
(EA)tj =eikakj =δskakj =asj =btj
k=1 k=1 for j = 1,··· ,m. Therefore EA = B.
10. Let A be n × m. B and E are formed, respectively, by multiplying row sofAandIn byα. Then,fori̸=s,bij =aij andeij =δij. Fori=s, bsj =αasj andesj =αδsj.
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11.2.
ROW OPERATIONS AND REDUCED MATRICES 297
Now consider the i, j − −element of EA. For i ̸= s, nn
11.
k=1 k=1 for j = 1,2,··· ,m. This shows that EA = B.
Let A be n × m. Now B and E are formed, respectively, from A and In byaddingαtimesrowstorowt. Fori̸=t,bij =aij andeij =δij,while fori=t,btj =atj +αasj andetj =δtj +αδsj.
Nowconsiderthei,j−−elementofofEA. Fori̸=t, nn
and
(EA)ij = eikakj = αδikakj = aij = bij k=1 k=1
nn
(EA)sj =eskakj =αδskakj =bsj
For i = t,
(EA)ij = eikakj = δikakj = aij. k=1 k=1
nn
(EA)tj = etkakj = (δtk +αakj)
k=1 k=1 =atj +αasj =bsj.
This shows that EA = B.
In Problems 12–23, keep in mind that a given matrix can be reduced by different sequences of row operations, but the end result must be the same – a matrix has only one reduced form. There may therefore be different matrices Ω1 and Ω2 such that
Ω1A = AR = Ω2A. We give one such matrix for each problem.
12. A is reduced by simply adding row two to row one:
1 1 0 1 0 5 Ω=0 1 0 andAR =0 1 2.
001 000
13. We can reduce A by first subtracting row two from row one of I2, then
multiplying row one of the resulting matrix by 1/3:
1 0 1 −1 1/3 −1/3
Then
I2= 0 1 → 0 1 → 0 1 =Ω. 1 0 1/3 4/3
ΩA=AR= 0 1 0 0 .
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298CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS
14. We can reduce A by first interchanging rows two and four of I4, then (on the resulting matrix), multiplying row one by −1, then adding row two to row one:
1 0 0 0 1 0 0 0
I4=0 1 0 0→0 0 0 1
0 0 1 0 0 0 1 0
0001 0100
−1 0 0 0 −1 0 0 1 →0 0 0 1→0 0 0 1=Ω.
0 0 1 0 0 0 1 0 0100 0100
15. A is in reduced form, so A = AR.
16. Starting with I4, subtract row one from row four, subtract 3 times row one from row three, interchange rows two and three, and then interchange rows one and two:
17.
18.
01 11 Ω= 1 −2 ,AR= 0 0 .
Then
1 −4 −1 0
ΩA = 0 0 0 1 = AR.
1 I4→0
0 0 0 1 100→0 0 1 0 − 3
0 0 0 100
0 1 0 0 0 1
0 1 0
0
−1 0
0 0
1 −1 0 −3
1 0 → −3 0
0 0 0 0 0000
0 0 1.
0→1
0 1 0 0 − 1 0 0 −1 0 0 1 −1 0 0 1
Then
For Problems 17–23, just Ω and AR are given.
1 0 ΩA=0 1=AR.
0 0 00
−8 −2 38 1 0 0 270
Ω= 1 37 43 −7 ,AR= 0 1 0 =I3. 19 −29 11 0 0 1
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11.3. SOLUTION OF HOMOGENEOUS LINEAR SYSTEMS 299
19.
20.
21.
22.
23.
−1/3 0 1 −4/3 −4/3 Ω=01,AR=00 0
01 1000 Ω= 1/2 1/2 ,AR= 0 1 3/2 1/2
0 0 1 1 0 0 −3/4 4
Ω=1 4 −4 −8 ,AR= 0 1 0 3 −4 8 8 0 0 1 0
0 1/2 −1 1 Ω= 0 0 1 ,AR=0 −1/7 2/7 −3/7 0
0 0 1 0 1 0 1 3 0,AR=0 1 0 −6 0 0
0 0 1 0 0 1
0 0 −1 1 0
11.3 Solution of Homogeneous Linear Systems
In Problems 1–12, we use the facts that (1) the system AX = O has the same solutions as the reduced system ARX = O, and (2) the solution of the reduced system can be read by inspection from the reduced matrix AR.
1. The coefficient matrix and its reduced form are 12−11 10 1 −1
A= 0 1 −1 1 ,AR= 0 1 −1 1 .
Because AR has two nonzero rows and m = 4, the solution space has dimension m − 2 = 2, which means that the general solution is in terms of two of the unknowns, which can be given any values. Specifically, from the reduced matrix,
x1 = −x3 + x4 x2 = x3 − x4.
All solutions are given by
x1 −x3 +x4 −1 1 X=x2= x3 −x4 =x3 1 +x4−1.
x 3 x 3 1 0 x4 x4 0 1
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300CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS It looks a little neater to write the general solution
−1 1 X=α 1 +β−1
1 0
01
in which α and β are arbitrary real numbers. The solution space of this
system is the subspace of R4 having basis vectors
< −1,1,1,0 >,< 1,−1,0,1 > .
2. The coefficient matrix is
A=0 1 1 0 4
and
−3 1 −1 1 1 0 0 −3 2 1
1 0 0 1/9 11/9 AR=0 1 0 2/3 13/3.
0 0 1 −2/3 −1/3 With x4 = α and x5 = β, the general solution is
−1/9 −11/9
−2/3 −13/3 X=α 2/3 +β 1/3 .
1 0 01
The solution space is a subspace of R5 having dimension 2 and basis < −1/9,−2/3,2/3,1,0 >,< −11/9,−13/3,1/3,0,1 > .
3. The coefficient matrix and its reduced form are
−2 1 2 1 0 0 A=1 −1 0,AR=0 1 0=I3. 110 001
Here A has rank 3, the number of nonzero rows of A, and m− rank (A) = 3 − 3 = 0, so the solution space has dimension zero, consisting just of the
trivial solution
0 X = 0 .
0
This is consistent with the reduced system being the system
x1 =0,x2 =0,x3 =0.
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11.3. SOLUTION OF HOMOGENEOUS LINEAR SYSTEMS 301
4. The
coefficient matrix and its reduced form are
4 1 −3 1 1 0 −1/2 0
A= 2 0 −1 0 ,AR= 0 1 −1 1 . The general solution can be written
1/2 1
X=α 1 +β0−101. 0
The solution space is the subspace of R4 having basis vectors < 1/2,1,1,0 >,< 0,−1,0,1 > .
5. The
and
coefficient matrix is
1 −1 3 −1 4 A=2 −2 1 1 0 1 0 −2 0 1
0 0 1 1 −1
1 0 0 0 9/4 AR=0100 7/4.
0 0 1 0 5 / 8 0 0 0 1 −13/8
The solution space has dimension 5 − 4 = 1 and the general solution is
−9/4
−7/4 X = α −5/8 .
13/8 1
The single vector < −9/4, −7/4, −5/8, 13/8, 1 > is a basis for the solution space.
6. The coefficient matrix is
A=1 0 0 −1 2
and
6 −1 1 0 0 1 0 0 1 −2
1 0 0 0 −2 AR=0 1 −1 0 −12.
0 0 0 1 −4
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302CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS The general solution is
0 2 1 12
X = α 1 + β 0 . 0 4
01
The solution space is the subspace of R5 of dimension 2, having basis
< 0,1,1,0,0 >,< 2,12,0,4,1 > . 7. The coefficient matrix is
and
−10 −1 4 1 1 −1 A=0 1 −1 3 0 0 2 − 1 0 0 1 0
0 1 0 −1 0 1
1 0 0 05/6 5/9 AR=010 0 2/310/9.
0 0 1 0 8/3 13/9 0 0 0 1 2/3 1/9
From the reduced system read the general solution
−5/6 −5/9
−2/3 −10/9
−8/3 −13/9 X = α −2/3 + β −1/9 .
1 0 01
The solution space is a subspace of R6 having dimension 2, with basis vectors
< −5/6, −2/3, −8/3, −2/3, 1, 0 >, < −5/9, −10/9, −13/9, −1/9, 0, 1 > . 8. The coefficient matrix is
1 0 0 0 7/6 −5/4 A=0100−20/3 9/2.
0 0 0 1 14/3 −11/2 0001−3 2
From the reduced matrix we can write the general solution
−7/6
20/3
5/4
−9/2
−14/3
X=α 3 +β −2 .
11/2
1 0 01
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11.3.
9.
SOLUTION OF HOMOGENEOUS LINEAR SYSTEMS 303 The solution space is the subspace of R6 of dimension 2, having basis
vectors
< −7/6,20/3,−14/3,3,1,0 >,< 5/4,−9/2,11/2,−2,0,1 > .
There is no x3 in the equations, so we actually have a system of three equations in the four unknowns x1, x2, x4 and x5. The coefficient matrix
10.
The coefficient matrix is
is
and
Now x1, x2 and x4 depend on x5 and
5/14 X = α11/7.
6/7 1
and
1 0 0 0 −41/6 −2/3 AR =0 1 0 0 −49/6 −1/3.
0 0 1 0 −13/6 −7/3 0001 23/6 −4/3
1 0
AR= 0 0 1−6/7
.
The general solution is
0 1 −31 A=2 −1 1 0
4 −3 A=0 2
0 1 1 −3 0 1−1−6 0 0 4 −1
2 −3 0 4
0 −5/140 1 0 −11/17
The solution space is the subspace of R4 of dimension 1, having basis vector < 5/14, 11/7, 6/7, 1 >.
3 −2
2 1 −3 4 0 0
41/6
49/6
2/3 1/3
13/6
X = α −23/6 + β 4/3 .
10
01
The solution space is a subspace of R6 of dimension 2 and having basis vectors
< 41/6, 49/6, 13/6, −23/6, 1, 0 >, < 2/3, 1/3, 7/3, 4/3, 0, 1 > .
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7/3
304CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS 11. The coefficient matrix is
and
1 −2 0 0 1 −1 1 A=0 0 1−1 1 −23 1 0 0 0 −1 2 0
2 0 0 −3 1 0 0 1000−1 2 0
The general solution is
and
1 00
0 1 0 AR=0 0 1 0 0 0 0 0 0
00 −3 0 0 −1/2 0 0 −2/3 1 0 −1/6 0 1 −3/2
−7/2
−1/2
7/2 −1/2 1/2 −1/2 2/3 −1. 1/6 −1/2 3/2 −1/2
1/2 1/2
The general solution is
AR=0 1 0 0 0 1 0 0 0
0 −1 3/2 −1/2. 0 0 −2/3 3 1 −1 4/3 0
−2 0 −3/2 1/2
1 1
2/3 −3 X=α1+β−4/3+γ 0 .
1 0 0 0 1 0
001
The solution space is the subspace of R7 of dimension 3, with basis vectors
< 1,1,0,1,1,0,0 >,< −2,−3/2,2/3,−4/3,0,1,0 >,< 0,1/2,−3,0,0,0,1 > . 12. The coefficient matrix is
0
2 0 0 0 −4 0 1
0 2 0 0 0 −1 1 A=0 0 1 −4 0 0 0 0 1 − 1 1 0 0 0 0 1 0 0−1 1 −1 0
1 −1
1 0
3 1/2
2/3
1/6 1 0 0 0 1 0
001
−2/3
1 1/2
−1/6
X = α 3/2 + β −3/2 + γ 1/2 .
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11.3.
13.
SOLUTION OF HOMOGENEOUS LINEAR SYSTEMS 305 The solution space is the subspace of R8 having basis vectors
< 3, 1/2, 2/3, 1/6, 3/2, 1, 0, 0 >, < −7/2, −1/2, −2/3, −1/6, −3/2, 0, 1, 0 >, < 1/2,1/2,1,1/2,1/2,0,0,1 > .
The answer is yes. All that is required is that m − rank(A) > 0, so that the solution space has positive dimension, hence non-zero vectors, which are solutions of the system.
As a specific example, consider the system AX = O, where
1 0 3 A=0 1 −1.
309
This is a homogeneous system with three equations in three unknowns.
14.
Then AR has two nonzero rows, so the solution space of AX = O has dimension 3 − 2 = 1 > 0. In fact, the general solution is
3 X = α1.
1
Suppose A is n × m. Let the columns of A be C1,··· ,Cm, written as
n × 1 column matrices. Now notice that, if a1
a2 X= .
. am
then AX = O is equivalent to
a1C1 +···+amCm =O,
the n × 1 zero matrix.
With these preliminaries, we can prove the proposition. If the columns of A are linearly dependent, then there are numbersa1, · · · , am not all zero such that
a1C1 +···+amCm =O,
and then X, formed as a column with these coefficients as elements, is a
nontrivial solution of AX = O.
We find that
1 0 3 AR=0 1 −1.
000
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306CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS Conversely, suppose the system has a nontrivial solution
Then
a1
a2 X=X= . .
. am
a1C1 +···+amCm =O,
and at least one of these coefficients is nonzero, so the columns of A are
linearly dependent.
15. Let the rows of A be R1,··· ,Rn. These are vectors in Rm. Let R be the
row space of A, which is the subspace of Rm spanned by the row vectors.
Now, X is in the solution space S(A) of the homogeneous system with coefficient matrix A exactly when AX = O. This is true exactly when the dot product Rj · X = O for j = 1,··· ,n, which is true exactly when each row is orthogonal to X. But this is equivalent to X being orthogonal to every linear combination of these rows, hence to every vector in the row space R of A. This makes the solution space of the system the orthogonal complement of the row space:
R⊥ = S(A).
Because the columns of At are the rows of A, similar reasoning shows that the solution space S(At) of the system AtX = O has the column space C of A as its orthogonal complement:
C⊥ = S(At). 11.4 Nonhomogeneous Systems
1. The augmented matrix is
with reduced form
. 3 −2 1 . 6
. . [A.B]=1 10 −1 . 2
. −3−2 1 .0
. 100.1
. . [A.B]R = 0 1 0 . 1/2.
. 001.4
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11.4.
NONHOMOGENEOUS SYSTEMS 307 The reduced forms of the matrix of coefficients of the system and the
augmented matrix have the same number of nonzero rows, so both A
and [A.B] have the same rank (in this case 3). Therefore this system is consistent. We can read the solution from the reduced form of the augmented matrix:
1 X = 1/2 .
4
In this case the solution is unique because m minus the rank of A is 3 − 3 = 0, so the associated homogeneous system has only the trivial solution.
2.
The augmented matrix is
3.
We have
and
83
X= 0 +α 7/3 . −31/3 −52/9
01
2 −3 0 1 0 −1 [A.B]=3 0 −2 0 1 0
. . 0 . . 1 . . 3
and
. . 8
. . 0 .
. . −31/3
. 4 −2 3 10 .1
. . [A.B]=1 0 0 −3 .8 ,
. 2 −3 0 1 . 16
1 0 0 −3 .
[A.B]R =0 1 0 −7/3
0 0 1 52/9
Because A and [A.B] have the same rank 3, this system is consistent. Read from the reduced augmented matrix that
.
. 0 0 1 −3/2 −1/2 51/4 . 25/4
0 1 0 −1 0 6
.
1 0 0 −1 0 17/2 . 9/2
.
.
[A.B]R=010−1 0 6.3.
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308CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS A and [A.B] have the same rank 3, so the system has solutions. Read
from the reduced augmented matrix that
9/2 1 0 −17/2
3 1 0 −6 25/4 3/2 1/2 −51/4
X= 0 +α1+β0+γ 0 .
0 0 1 0 0001
4. From the system, we have
. [A.B] = −1
0 3 0 −4 0 0
[A.B]=1 −3 0 0 4 −1
0 1 1−60 1
1 −1 0 0 0 1
. . 10
. . 8 . .−9 . . 0
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2 −3
. . 1 . . 0 . . 3
3 1 −4
and
The left two columns of this reduced augmented matrix form AR, and this has two nonzero rows, while the reduced form of the augmented matrix has three nonzero rows. Because the rank of A does not equal the rank
of [A.B], this system is inconsistent.
In this case the absence of solutions can be seen easily from the reduced
augmented matrix, which is the system
x1 = 0,
x2 = 0, 0x1 + 0x2 = 1.
Clearly the third of these equations can have no solution. 5. The augmented matrix is
. 10.0
.
[A.B]R =0 1 . 0.
. .
00.1
11.4.
NONHOMOGENEOUS SYSTEMS 309 and its reduced form is
. 1 0 0 0 −2 2 . −4
. [A.B]R=0100 −2 1 . −4.
. 0 0 1 0 −7 9/2 . −38
. 0 0 0 1 −3/2 3/4 . −11/2
Both A and [A.B] have the same rank 4 (number of nonzero rows in their reduced forms), so this system has a solution. From the reduced augmented matrix we read the general solution
6.
We have
and
. . 1
. . 0 . .0
.
.11/20
. . 1/30 .
. . −1/10
7.
The augmented matrix is
.
[A.B]=0 1 0 1 −1
0 0 1 −3 2
−4 −4
2 2
−2
−1
−9/2 X = −11/2 + α 3/2 + β −3/4 .
7
001
−38
0 1 0
2 −3 0 1
A=0 3 1 −1
2−310 0
100 1/20
. [A.B]R = 0
1
0 0 1 −1/10
0
Because A has rank 3, the same as the rank of the augmented matrix, this system is consistent. Read the general solution from the reduced augmented matrix:
11/20 −1/20 X= 1/30 +α 9/30 .
−9/30
−1/10 1/10 01
8 −4 0 1/2 10
. . 1
. . 2 . . 0
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310CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS with reduced form
. . 9/8
. . 2.
augmented matrix), and
9/8 −1/2 −3/4
2 −1 1 X= 0 +α 3 +β −2 .
0 1 0 001
8. The augmented matrix is
with reduced form
. . 1
. .1
1 0 0 1/2 3/4 [AR.C]=0 1 0 1 −1
.
. . 0
0 0 1 −3 2
The system is consistent (same number of zero rows in AR and the reduced
2 0 −3
1 −1
100. 3/4
. . 0 1 0 . −1/12
. 001. 1/6
2 −4 1
.
Both the matrix of coefficients and the augmented matrix have rank 3, so the system is consistent. The system has the unique solution
3/4 X = −1/12 .
1/6
This unique solution could have been foreseen from the fact that the num- ber of columns of A minus its rank is 3 − 3 = 0, so the associated homo- geneous system has only the trivial solution.
9. The augmented matrix is
.
0014 0 −301. 2
.
1 1 1 −1 0 1 0 . −4
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. . 2
11.4.
NONHOMOGENEOUS SYSTEMS 311 with reduced form
. 1101 3/14 1−1/14.−29/7.
. 0 0 1 0 −3/14 0 1/14 . 1/7
The matrix of coefficients and its augmented matrix have rank 2, so the system is consistent. The general solution is
−29/7 −1 1 −3/14 −1 1/14 0
0 1 0 0 0 −1/14
0 .
0 0
0 0 0 0 1 000001
10.
The augmented matrix of coefficients is
.
11.
The augmented matrix is
1/7 0 0 3/14 0 X= 0 +α 0 +β1+γ 0 +δ 0 +ε
0 0 0 1 0
with reduced form
43.4
.
and the reduced augmented matrix is
1 0 0 19/15
0 1 0 −3
0 0 1 −67/13
. .
. .
. .
3 −2 . −1 .
10. 5/17 . .
0 1 . 16/17
The rank of the coefficient matrix equals 2, the same as the rank of the
augmented matrix, so the system is consistent. The solution is unique:
5/17 X = 16/17
.
. . −7
. . 6
. . −5
22/15
.
−5
−121/15
7 −3 4 0
2 1 −1 4
0 1 0 −3
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312CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS The rank of A and of the augmented matrix is 3, so the system is consis-
tent. The general solution is
22/15 −19/15 X= −5 +α 3 .
−121/15 67/15 01
12. The augmented matrix of coefficients is
with reduced form
. 1 0 0 .
. 010.
. 001.
−4 5 −6
−616−11. 1
2 −6 1
.
The coefficient matrix and its augmented matrix have the same rank 3, so the system is consistent. Because the number of unknowns is also 3, this system has a unique solution
13. The augmented matrix is
and its reduced form is
−2 1 7
.
. . 1
. . 0
−137/48 X= 1/6 .
41/24
4 −1 4
1 1 −5
1 0 0 . 16/57
. . 0 1 0 . 99/57
. 0 0 1 . 23/57
A and the augmented matrix both have rank 3, which is also the number of unknowns, so the system has a unique solution
16/57 X = 99/57 .
23/57
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. . 2
. . −5
−137/48 .
1/6
41/24
. . 4
11.5.
14.
MATRIX INVERSES 313 The augmented matrix is
. −6 2 −1 1 . 0
. 1 4 0 −1 . −5
. 1 1 1 −7 . 0
15.
Write
−15/23 −21/23 X=−25/23+α 11/23 .
40/23 171/23
01
α1 α2
X= . .
and its reduced form is
.
1 0 0 21/23 . −15/23
. . 0 1 0 −11/23 . −25/23
. 0 0 1 −171/23 . 40/23
A and its augmented matrix have the same rank 3, so the system is con- sistent. The solution is not unique because m − 3 = 4 − 3 = 1, so the associated homogeneous system as a solution space of dimension 1. The general solution is
αm
Let the columns of A be C1,··· ,Cm. Then AX = B if and only if
α1C1 +···+αmCm =B.
This means that X can be a solution if and only if X is a linear combination
of the columns of A, hence is in the column space of A. 11.5 Matrix Inverses
Problem 1 shows all of the row operations used to reduce the matrix and find the inverse, or show that the matrix is singular. For Problems 2–10, just the inverse is given if the matrix is nonsingular.
1. Reduce
. .
−1 2 . 1 0 → add2timesrowonetorowtwo −1 2 . 1 0 . .
21.01 05.21
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314CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS .
→ multiplyrowoneby−1 1 −2 . −1 0 .
→ multiplyrowtwoby1/5 1 −2 . −1 0 .
→ add2timesrowtwotorowone 1 0 . −1/5 2/5 . .
01. 2/5 1/5 Because I2 has appeared on the left, the given matrix is nonsingular and
the right two columns of this augmented matrix form the inverse matrix: −1 1 −1 2
A=521.
2. The matrix is singular (has no inverse) because we find that
05.21
.
0 1 .2/5 1/5
.
3.
4.
5.
6.
7.
1 1/4 AR= 0 0 ̸=I2.
A−1=1−2 2 12 1 5
A−1 = 1 4 0 4 −4−1
A−1=13 −2
6
49
A−1 = 1 −8 4
12 −3 64 −4
−7 0 0 14
31 8. The matrix is singular because
1 −7
−1 10
56
−6 11
A−1=1 3 10
2
1 0 3
AR =0 1 1̸=I3. 000
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11.6. DETERMINANTS
9.
11.
12.
13.
14.
15.
315
10. A has no inverse because
AR=0 1 14/9.
6 −6 0
A−1=−1 −3 −9 2 12
3 −3 −2
1 0 28/27 000
−1 −1 8 4 1 X=A−1B=1−9 2 −5 142=1−75
−23
11 2 2 −5 3 0 11 −9 3 3 −2 −1 −5 14
5 5 54 9 55 11
X=A−1B=1 −10 34 23 0 =1 15 56175 13
X = A−1B
11 12 9 −4 22
28 7 =13165 5=127
82448 30
4 4 0 4 −1
52 52 X=A−1B=17−639 −5=1 58
111130 −66 5 −15−150 −21
5 X = A−1B = − 1 −10 15 10 0 = 1 14
25 11.6 Determinants
−5100−7 0
In Problems 1–6 we provide a sequence of row and/or column operations leading to a determinant that is easily evaluated. Other sequences of operations can also be used.
1. Add 2 times row two to row one and then −7 times row two to row three to obtain
−241016 7 3+1 16 7
1 6 3 = 1 6 3 = (−1) (1) = −22.
7 0 4 0 −42 −17 −42 −17
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316CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS 2. Add 3 times row two to row one, then add row two to row three to obtain
2 −3 7 44 0 10 2+2 44 10
14 1 1 = 14 1 1 = (−1) (1) = 254.
−4 5 6 1 5 21 3+2 −2 3 5 = 1 3 14 = (−1) 2 −2 6 0 −2 0
−13 −1 5 1 0 6 1 6
3. Add column two to column one, then 2 times column two to column three:
1 14
4. Add 2 times row three to row one and 2 times row three to row two to
obtain
2 −5 8 28 −5 0 3+328 −5
4 3 8 =30 3 0 =(−1) =−936. 13 0 −4 13 0 −4 30 3
5. Add 2 times column three to column one and then add column three to column two to obtain
17 −2
1 12
14 7 −7 0
27 (−7)
3 =−2,247.
7 3 Putting the pieces together,
=−249.
5 27
3 5 3+3 12 0 =(−1)
0 =1
6. Add column one to column two, then 3 times column two to column three,
0 −7
then 2 times column one to column four to obtain
1
12
−3 3 9 6 −3 0 0 0 1 −2 15 6 1 −1 18 8
7 1 1 5=7 8 22 19
2 1 −1 3 2 3 5 7 −1 18 8 −1 18
= (−1)1+1(−3) 8 22 19 = −3 8 22 357 357
Now the problem of evaluating a 4 × 4 determinant has been reduced to one of a 3 × 3 determinant. In this determinant, add 18 times column one to column two and then 8 times column one to column three to obtain
−1 18 8 22 3 5
8 −1 0 19= 8 166
0 166 83
−3 3 9
1 −2 15
7 1 1 5 2 1 −1 3
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83=(−1)
59 31 59 31
6
6 = (−3)(−249) = 747.
1 21
(−2) = −14.
8 19.
11.6. DETERMINANTS 317
For Problems 7–10, we just give the value of the determinant.
7. −122
8. 293
9. 72
10. −2,667
11. −15,698
12. 1,693
13. 3,372
14.
1a a2 1aa2 =0 b−a (b−a)(b+a)=(b−a)(c−a)0 1 b+a
1 a a2 1 a
1 b b2=0 b−a b2−a2
a2 1 c c 2 0 c − a c 2 − a 2
0 c−a (c−a)(c+a) 0 1 c+a 1 b+a
=(b−a)(c−a)1 c+a=(b−a)(c−a)(c−b).
15. Add columns two, three and four to column one, then factor (a+b+c+d) out of column one to obtain
a b c d 1 b c d b c d a 1 c d a
c d a b=(a+b+c+d)1 d a b
d a b c 1 a b c (−1)row two + row three − row four
Now add
to row one and factor out b−a+d−c from the new row one to obtain
1 b c d 1 c d a
1 a b c
16. Define the function
1 z y
L(x,y)=1 x2 y2=(y2 −y3)x+(x3 −x2)y+x2y3 −x3y2. 1 x3 y3
(a+b+c+d)(b−a+d−c)1 d a b.
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318CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS
Notice that L(x,y) has the form ax+by+c, where a = y2−y3, b = x3−x2 and c = x2y3 − x3y2 are given numbers. The graph of L(x,y) = 0 is a straight line in the x, y− plane. Because
L(x2,y2) = 0 and L(x3,y3) = 0, both of these points are on the line L(x, y) = 0.
Finally, L(x1,y1) = 0 (x1,y1) is also on this line, and this occurs only if 1 x1 y1
1 x2 y2=0. 1 x3 y3
17. Use induction to prove that, for n = 2,3,···, the determinant of an n×n upper triangular matrix is the product of the main diagonal elements.
For n = 2, this is obvious because
a11 a12 = a11a22.
0 a22
Now suppose the statement is true for some n ≥ 2. We want to prove that it is true for n+1. Let A be an n+1×n+1 upper triangular matrix. Then ai1 = 0 for i = 2,··· ,n+1, so by expanding by column one, we have
|A| = a11|B|,
where B is the n × n upper triangular matrix obtained by deleting row
one and column one of A. By the inductive hypothesis, |B| = a22a33 · · · an+1,n+1.
Then
This completes the proof by induction.
11.7 Cramer’s Rule
1. |A| = 47 ̸= 0 so Cramer’s rule applies:
x1 = 1 5 −4=−11,×2 = 1 15
47 −4 1 47 47 8 2. |A| = −3 and the solution is
|A| = a11a22 · · · an+1,n+1.
5 =−100. −1 47
x1 =−13 4=−1,×2 =−11 3=1. 3 0 1 3 1 0
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11.7. CRAMER’S RULE 319 3. |A| = 132 and the solution is
and
4. |A|
5. |A|
6. |A|
7. |A|
8. |A|
9. |A|
10. |A|
0 −4 3
x1=1−5 5 −1=−66=−1,
132−4 6 1 132 2
8 0 3
x2= 1 1 −5 −1=−114=−19,
132−2−41 132 22
8 −4 0
x3=11 5 −5=24=2.
132 −2 6 −4 132 11
= 108 and the solution is
x1 =− 63 =− 7 ,x2 =−165 =−55,×3 = −943 =−9. 108 12 108 36 108 4
= −6 and
x1 = 5,×2 =−10,×3 =−5. 636
130 130 130 130 130 = 4 and the solution is
x1 = −172 = −86,×2 = −109,×3 = −43,×4 = 37. 2222
= −130 and
x1 = 197,×2 = 255,×3 = 1260,×4 = 42 ,x5 = 173.
= 12 and
= 93 and
= 42,
x1 = 117 , 63 , x3 = 3 , x4 = 21 . 12 12 2 12
x1 = 33,×2 =−409,×3 =− 1 ,x4 = 116. 93 93 93 93
x1 = 69,×2 = 162,×3 = 24,×4 = −54. 21 21 21 21
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320CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS 11.8 The Matrix Tree Theorem
1. The tree matrix for this graph is
2 0 −1 0 −1
0 2 −1 −1 0 T=−1 −1 4 −1 −1.
0 − 1 − 1 3 − 1 −1 0 −1 −1 3
Evaluate any 4 × 4 cofactor of T to obtain 21 as the number of spanning trees in the labeled graph.
2.
3.
4.
5.
4 −1−1−1 0 −1 −1 3 −1 −1 0 0 −1 −1 2 0 0 0
T=−1 −1 0 4 −1 −1 0 0 0 −1 2 −1
−1 0 0 −1 −1 3 and each cofactor yields 55 spanning trees in G.
4 −1 0 −1−1−1 −1 2 −1 0 0 0 0 −1 3 −1−1 0
T=−1 0 −1 4 −1 −1 − 1 0 − 1 − 1 3 0
−1 0 0 −1 0 2
and each cofactor gives 61 as the number of spanning trees
4 −1−1 0 −1−1 −1 3 −1 −1 0 0 −1 −1 3 −1 0 0
T=0 −1 −1 3 −1 0 −1 0 0 −1 3 0
−1 0 0 0 −1 2
and each cofactor gives 64 for the number of spanning trees.
3 −1 0 0 −1 −1 −1 3 −1 0 −1 0 0 −1 4 −1 −1 −1
T=0 0 −1 2 −1 0 − 1 − 1 − 1 − 1 4 0
−1 0 −1 0 0 2 and the number of spanning trees is 61.
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11.8.
6.
THE MATRIX TREE THEOREM 321 The tree matrix for Kn is the n × n matrix
n−1 −1 −1 ··· −1
−1 n−1 −1 ··· −1
−1 −1 n−1··· −1
.
T=
. . . ··· .
−1 −1 −1 ··· n−1
We will compute −1)1+1M11, which is the n−1×n−1 determinant formed by deleting row one and column one of T. In M11, add the last n−2 rows to row one to obtain the new n − 1 × n − 1 determinant still equal to M11:
1 1 1 ··· 1
−1n−1 −1 ··· −1
−1 −1 n−1··· −1 M11 = .
. . . ··· .
−1 −1 −1 ··· n−1
Subtract column one of this determinant from each other column. This
does not change the value of the determinant, and we now have
1 0 0 0 ··· 0
−1 n 0 0 ··· 0
−1 −1 n 0 ··· 0 M11 = .
. . . . . .
−1 0 0 0 ··· 0
This is a lower triangular n − 1 × n − 1 determinant, which is therefore equal to the product of the diagonal elements, which is nn−2. Therefore the number of spanning trees in Kn is nn−2.
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322CHAPTER11. MATRICES,DETERMINANTSANDLINEARSYSTEMS
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Chapter 12
Eigenvalues and Diagonalization
12.1 Eigenvalues and Eigenvectors
1.
so the eigenvalues of A are 1 + 6 and 1 − 6, with eigenvectors, respec-
tively,
√6 −√6 V1= 2 ,V2= 1 .
pA(λ)=|λI2 −A|=λ2 −2λ−5 √√
The Gerschgorin circles are of radius 3 about (1,0) and radius 2 about (1, 0).
2. The characteristic equation of A is
λ2 −2λ−8=0.
Eigenvalues and eigenvectors are
0 6
λ1=4,V1= 4 ,λ2=−2,V2= −1 .
Gerschgorin circles are of radius 1 about (4,0) and radius 0 about −2. The circle of radius 0 is a “degenerate” circle, consisting of just the center point.
3. The characteristic equation is
λ2 +3λ−10=0
323
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324 CHAPTER 12. EIGENVALUES AND DIAGONALIZATION and eigenvalues and eigenvectors are
7 0 λ1=−5,V1= −1 ,λ2=2,V2= 1 .
One Gerschgorin circle has radius 1 and center (2, 0), and the other is the degenerate circle of radius 0 about (−5, 0).
4. pA(λ)=λ2 −10λ+18,
√ 2√ √ 2√
λ1=5+ 7,V1= 1− 7 ,λ2=5− 7,V2= 1+ 7 Gerschgorin circles have radius 2, center (6, 0), and radius 3, center (4, 0).
5. pA(λ)=λ2 −3λ+14,
λ1 =(3+ 47i)/2,V1 =
−1 + √47i 4
(2, 0).
6. pA(λ) = λ2, and eigenvalues are λ1 = λ2 = 0. All engenvectors are
nonzero scalar multiples of
1 V=0.
The lone Gerschgorin circle has radius 1 and center the origin.
7. pA(λ)=λ3 −5λ2 +6λ.
0 2 0 λ1 =0,V1 =1,λ2 =2,V2 =1,λ3 =3,V3 =2
003 The Gershgorin circle has radius 3, center (0, 0).
8. pA(λ) = (λ+1)(λ2 −λ−7)
0 2
√ √
−1 − √47i 4
λ2 =(3− 47i)/2,V2 =
The Gerschgorin circles have radius 6, center (1, 0), and radius 2, center
and
√√
λ1 =1,V1 =0,λ2 =(1+ 29)/2,V2 =5+ 29 10
2
0
√√ λ3 =(1− 29)/2,V3 =5−
29
Gershgorin circles have radius 1, center (−2, 0), and radius 1, center (3, 0).
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12.1. EIGENVALUES AND EIGENVECTORS 325
9. pA(λ)=λ2(λ+3)
1 1 λ1 =−3,V1 =0,λ2 =λ3 =0,V2 =0
03
All eigenvectors associated with the double eigenvalue 0 are constant mul-
tiples of V2. The Gershgorin circle has radius 2, center (−3,0).
10. pA(λ) = λ(λ2 + 2)
0 √ 1 √ 1 λ=0,V= 1 ,λ= 2i,V= −1 ,,λ=−2i,V= −1
1 12 2√3 3√ 0 −2i 2i
The Gershgorin circles have center (0, 0) and radii 1 and 2.
11. pA(λ) = (λ + 14)(λ − 2)2,
−16 0 λ1 =−14,V1 = 0 λ2 =λ3 =2,V2 =0.
11
All eigenvectors associated with λ2 are constant multiples of V2. The
Gershgorin circles have radius 1, center (−14, 0) and radius 3, center (2, 0).
12. pA(λ) = (λ−3)(λ2 +λ−42),
0 30 0 λ1 =6,V1 = 1 ,λ2 =3,V2 =−2,λ3 =−7,V3 =8
−1 5 5
The Gerschgorin circles have radius 9, center (−2, 0) and radius 5, center
(1, 0).
13. pA(λ) = λ(λ2 − 8λ + 7),
14 6 0 λ1 =0,V1 =7,λ2 =1,V2 =0,λ3 =7,V3 =0
10 5 1
The Gershgorin circles have radius 2, center (1,0), and radius 5, center
(7, 0).
14. pA(λ) = λ2(λ2 + 2λ − 1),
1 0 λ1 =λ2 =0,V1 =2,V2 =0
0 1 10
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326
CHAPTER 12. EIGENVALUES AND DIAGONALIZATION
so the repeated eigenvalue 0 has two linearly independent eigenvectors associated with it. For the other two eigenvalues,
1√ 1√
0 0 λ3 =−1+√2,V3 =1+ 2,λ4 =−1−√2,V4 =1− 2
00
The Gershgorin circles have radius 1, center (−2, 0) and radius 2, center
(0, 0).
pA(λ) = (λ−1)(λ−2)(λ2 +λ−13),
λ3 = −1+ 2
15.
−2 0 λ1 = 1,V1 = −11,λ2 = 2,V2 = 0
0 1 10
√ √53−7 √ −√53−7 53,V3 = 0 ,λ4 = −1− 53,V4 = 0
0 2 0 22
The Gershgorin circles have radius 2, center (−4, 0) and radius 1, center (3, 0).
16. pA(λ) = λ2(λ − 1)(λ − 5),
λ1 =1,V1 =−4,λ2 =5,V2 =0
1 1
0 0 00
0 ,λ3 =λ4 =0,V3 =0
All eigenvectors associated with the double eigenvalue 0 are constant mul- tiples of just the one eigenvector V3. The Gershgorin circles have radius 10, center (5, 0), radius 9, center (1, 0), radius 9, center (0, 0).
17. We know that AE = λE. Then
A(AE) = A2E = A(λE)
= λAE = λ(λE) = λ2E.
This says that λ2 is an eigenvalue of A2 with eigenvector E. It is not a routine inductive argument to show that λn is an eigenvalue of An with eigenvector E, for any positive integer n.
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1 0
12.2. DIAGONALIZATION 327 18. The characteristic polynomial of A is
The
pA(λ) = |λIn − A|. pA(0) = | − A|
is the constant term in this polynomial. But | − A| = (−1n)|A|
because −A is A with each of the n rows multiplied by −1, and each such multiplication introduces a factor of −1 into the determinant of the resulting matrix.
Now, if A is singular, then |A| = 0. But then pA(0) = (−1)n|A| = 0,
so 0 is a root of the characteristic polynomial, and is therefore an eigen- value of A.
12.2 Diagonalization
1. The characteristic equation is λ2 − 3λ + 4 = 0, so the eigenvalues are √√
λ1=3+ 7iandλ2=3− 7i. 22
Corresponding eigenvectors are
2√ 2√
V1= −3+ 7i andV2= −3− 7i . 2√2√
The matrix
If we wrote the eigenvectors in different order in defining the columns of P, then the columns of P−1AP would be reversed.
2. The characteristic equation is λ2 − 8λ + 12 = 0 so the eigenvalues are λ1 = 2 and λ2 = 6. Corresponding eigenvectors are
−1 3 V1 = 1 and 1 .
P= −3+ 7i −3− 7i
−1 (3+√7i)/2 0
diagonalizes A, and
P AP= 0 (3−√7i)/2 .
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328 CHAPTER 12. The matrix
diagonalizes A, and
EIGENVALUES AND DIAGONALIZATION
−1 3 P=11
−1 2 0 P AP= 0 6 .
3. The characteristic equation is λ2 −2λ+1 = 0, with repeated root 1. Every eigenvector is a scalar multiple of
0 V1 = 1 .
Because A does not have two independent eigenvectors, A is not diago- nalizable.
4. Eigenvalues and corresponding eigenvalues are
1 3
The matrix
diagonalizes A, and
P= 0 14
−1 −5 0
λ1=−5,V1= 0 ,λ2=9,V2= 14 . 1 3
A AP= 0 9 .
5. The eigenvalues and corresponding eigenvectors are
0 5 0 λ1 = 0,V1 = 1,λ2 = 5,V2 = 1,λ3 = −2,V3 = −3.
The matrix
diagonalizes A and
002
P=0 5 0 1 1−3
002
0 0 0 P−1AP=0 5 0.
0 0 −2
6. Eigenvalues and corresponding eigenvectors are
−2 √ 0
λ=0,V= −3 ,λ=(3+ 17)/2,V= 4
1 12 2√
1 3+ 17
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12.2. DIAGONALIZATION
and
The matrix
329
√
λ =(3− 17)/2,V =
0 √
33
4 . 3− 17
−2 0 0 P=−34 4
√√ 1 3+ 17 3− 17
diagonalizes A, and
P−1AP= 0 (3+√17)/2 0
0 0 0 √
0 0 (3− 17)/2 7. Eigenvalues and corresponding eigenvectors are
.
0 −3 λ1 =1,V1 =1,λ2 =λ3 =−2,V2 = 1 .
00
All eigenvectors associated with the repeated eigenvalue −2 are scalar multiples of V2. Because A does not have three linearly independent eigenvectors, A is not diagonalizable.
8. Eigenvalues and corresponding eigenvectors are
1 0 0
λ1 =2,V1 =0,λ2 =2+i,V2 =−i,λ3 =2−i,V3 =i. 011
Let
1 0 0 P=0 −i i.
0 1 −1
Then P diagonalizes A and
P−1AP=0 2+i 0 .
9. The characteristic equation is
2 0 0 002−i
(λ−1)(λ−4)(λ2 +5λ+5) = 0. √√
Eigenvalues are λ1 = 1, λ2 = 4, λ3 = (−5+ 5)/2, and λ4 = (−5− Corresponding eigenvectors are
5)/2).
1 0 V1 = 0,V2 = 1,
0 0 00
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330
CHAPTER 12. EIGENVALUES AND DIAGONALIZATION
00 √√
(2−3 5)/41 (2+3 5)/41
V3 = √ ,V4 = (−1 + 5)/2 (−1 −
√ . 5)/2
11 100 0
Let
0011 Then P diagonalizes A:
√√
0 1 (2−3 5)/41 (2+3 5)/41
P = √ √ . 0 0 (−1 + 5)/2 (−1 − 5)/2
1000 √ √
P−1AP= 0 1 0 00 0 (−5+ 5)/2 0 . 0 0 0 (−5− 5)/2
10.
11.
The characteristic equation is (λ + 2)4 = 0, so A has a repeated eigen- value −2 of multiplicity 4. We find that A has only three independent eigenvectors,
0 0 0 1,0, and 0.
0 1 0 001
Because A does not have four independent eigenvectors, A is not diago- nalizable.
Let
Then so Then
λ1 0 0 ··· 0
0 λ2 0 0 ··· 0
00λ0···0 D = 3 .
. . . . ··· .
0 0 0 0 ··· λn D = P−1AP,
A = PDP−1.
Ak = (PDP−1)k
= (PDP−1)(PDP−1 · · · (PDP−1) = PDkP−1,
with the interior pairings of P and P−1 canceling.
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12.2. DIAGONALIZATION 331
Problems 12–15 can be solved using the idea of Problem 11, coupled with the fact that the kth power of a diagonal matrix is the diagonal matrix formed by raising each diagonal element to the kth power.
12. Eigenvalues of A are −1,−6. Make respective eigenvectors the columns
of a matrix
Then
Then
Then
−3 1 P=21.
−1 −1/5 2/5 P = 1/5 3/5 .
−1 −1 0 P AP=D= 0 −6 .
5 −1 −3111 −4665 A=PDP =−3110−4666.
5
13. Eigenvalues of A are −1, −5, and the matrix of respective eigenvectors, 4 0
diagonalizes A to
Now,
so compute
P=11 −1 0
D= 0 −5 . −1 1/4 0
P =−1/41
6 6−110
A=PDP =−390615625.
14. Eigenvalues of A are −3 + (10), −3 − √10. Form P from respective
eigenvectors:
1+√10 1−√10 P=33.
Then P diagonalizes A to
−3+√10 0√
Then
D= 0 −3− 10 . 4 4 −1 493 −694
A=PDP = −684 949 .
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332 CHAPTER 12. EIGENVALUES AND DIAGONALIZATION
15. Eigenvalues of A are
Then
Let
Then
√√
2, − 2. Form
√2 −√2 P=11.
−1 √2/4 1/2 P = −√2/4 1/2 .
√2 0 B= 0 −√2 .
6 6−180 A=PDP =08.
16. Suppose A2 is an n × n diagonalizable matrix. Then A must have n lin- early independent eigenvectors V1, · · · , Vn, corresponding, respectively, to eigenvalues λ1, · · · , λn. We know that
pA(λj) = (A2 − λjIn)X = O for j = 1,··· ,n. But then
Then
pA(λj) = (A − λjIn)(A + λjIn) = pA(λj )pA(−λj ) = 0.
pA(λj) = 0 or pA(−λj) = 0.
Then λj or −λj is an eigenvalue of A with eigenvector Vj. This means that A has n linearly independent eigenvectors, and is therefore diagonalizable.
12.3 Special Matrices and Their Eigenvalues and Eigenvectors
In Problems 1–12, find independent eigenvectors for the given matrix. Normalize these by dividing each eigenvector by its length. These normalized eigenvectors form columns of an orthogonal matrix that diagonalizes the given matrix.
It is routine to show that eigenvectors are orthogonal by taking their dot product. We will omit the arithmetic of this verification.
1. We find eigenvectors
1 −2 V1= 2 ,V2= 1 .
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12.3. SPECIALMATRICESANDTHEIREIGENVALUESANDEIGENVECTORS333 Divide each by its norm to get unit eigenvectors and form the orthogonal
matrix
1/√5 −2/√5 Q= 2/√5 1/√5 .
This is an orthogonal matrix that diagonalizes A.
2.
3. Eigenvectors are
10 10 V1 = 7+√149 ,V2 = 7−√149 .
Normalize these to form columns of the orthogonal matrix
√10√ √10√ 298+14 149 298−14 149
Q = √ 7+√149 √ 7−√149 . √√
298+14 149 298−14 149
1+√2 1−√2 V1= 1 ,V2= 1 .
Normalize these to form
1+√2 1−√2 √√√√
Q= 4+2 2 4−2 2. √1√ √1√
4. Eigenvalues are
√√
√17 − 4 −√17 − 4
4+2 2 4−2 2
17 and − 17, with eigenvectors, respectively,
V1= 1 ,V2= 1 .
Normalize these vectors to form an orthogonal matrix Q that diagonalizes
A:
1 √17−4 −√17−4 Q = 34 − 8√17 1 1
.
√√
5. Eigenvalues of A are 3, 2 − 1 and − 2 − 1, with corresponding eigen-
vectors
0 √1 √1 V1 =0,V2 = 2−1V3 =− 2−1.
100 For an orthogonal matrix that diagonalizes A, let
0√1√ √1√
Q=
2−1 − 2−1 0√√√√
4−2 2 4+2 2 √ √
4−2 2 4+2 2. √2−1 −√2−1
1√√√√ 4−2 2 4+2 2
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334 CHAPTER 12. EIGENVALUES AND DIAGONALIZATION 6. Eigenvalues of A are
1, 1 (5 + √5), 1 (5 − √5) 22
with eigenvectors, respectively,
1 0√ 0√
V1 =0,V2 =−1+ 5,V3 =−1− 5. 022
Normalize these eigenvectors and use them as columns of an orthogonal matrix that diagonalizes A:
10 0 √√
−1+5 −1−5 Q=0√ √ √ √.
10−2 5 10+2 5 0√2√ √2√
7. Eigenvalues are
10−2 5 10+2 5
7, 1(5 + √41), 1(5 − √41) 22
with corresponding eigenvectors
0 5 + √41 5 − √41 V1 =1,V2 = 0 ,V3 = 0 .
044 The following orthogonal matrix diagonalizes A:
5+√41 5−√41 1√√√√
82+10 41 82−10 41 Q=10 0.
0 √ 4 √ √ 4 √ 82+10 41 82−10 41
√√
8. Eigenvalues are 0, 1 + 17, 1 − 17, with corresponding eigenvectors
0 1 + √17 1 − √17 V1 =0,V2 = −4 ,V3 = −4 .
166
For an orthogonal matrix that diagonalizes A, use the normalized eigen- vectors as columns:
√√
1+ 17 1− 17
0 √ √ √ √
Q=0√√ 70−2 17
1√6√ 70−2 17
√√. 70+2 17
70+2 17 −4
70−2 17 −4
√6√ 70+2 17
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12.3. SPECIALMATRICESANDTHEIREIGENVALUESANDEIGENVECTORS335
9. The matrix is not hermitian, skew-hermitian or unitary. Eigenvalues are
2, 2.
10. The matrix is not hermitian, skew-hermitian or unitary. Eigenvalues are
−1, −1.
11. The matrix is skew-hermitian because St = −S. Eigenvalues are 0, √3i, −√3i.
12. The matrix is unitary. Eigenvalues are
1, 1(1 + i)(√2 + √6i), 1(−1 − i)(−√2 + √6i). 44
13. The matrix is not unitary, hermitian or skew-hermitian. Eigenvalues are 2, i, −i.
14. The matrix is hermitian, because Ht = H. Eigenvalues are √√
1,−1− 41,−1+ 41. 22
15. Suppose H is hermitian. Then
H = Ht.
Then
HHt =HHt =HH=HH.
16. If H is hermitian, then Ht = H, so the diagonal elements satisfy hjj = hjj,
so these diagonal elements must be real.
17. Suppose S is skew-hermitian. Then St = −S, so
sjj = −sjj for j = 1,2,··· ,n. Writesjj =ajj +ibjj. Then
sjj =ajj +ibjj =−ajj +ibjj =−ajj +ibjj.
But then each ajj = −ajj, so ajj = 0 for j = 1,2,··· ,n. This makes each
diagonal element of S either pure imaginary (if bjj ̸= 0) or zero.
18. Suppose U and V are unitary n × n matrices. Then
But then
U−1 = Ut and V−1 = Vt.
(UV)−1 = V−1U−1 = V tUt = ((U)(V ))t = UVt,
so UV satisfies the equation defining a matrix to be unitary.
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336 CHAPTER 12. EIGENVALUES AND DIAGONALIZATION
12.4 Quadratic Forms
1. The matrix of the quadratic form is
−5 2
A=23. √√
This matrix has eigenvalues −1 + 2 5, −1 − 2 5 and the quadratic form has standard form
2. The matrix is
with eigenvalues
(−1 + 2√5)y12 + (−1 − 2√5)y2. 4 −6
A= −6 1
1(5 + √153) and 1(5 − √153).
22 The standard form of this quadratic form is
1(5 + √153)y12 + 1(5 − √153)y2. 22
−3 2 27
√
29. The standard form is (2 + √29)y12 + (2 − √29)y2.
4 −2 −2 1
√
22 0 −3
3. The matrix is
with eigenvalues 2 ±
4. The matrix is
with eigenvalues 3 ±
5. The matrix is
with eigenvalues 2 ±
17)/2. The standard form is 1(3 + √17)y12 + 1(3 − √17)y2.
−3 4
13. The standard form is
√
(2 + √13)y12 + (2 − √13)y2.
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12.4. QUADRATIC FORMS
6. The matrix is
337
with eigenvalues 1, 6. The standard form is y 12 + 6 y 2 2 .
7. The matrix is
with eigenvalues 1 ±
0 −1 −1 2
√
5 2 22
2. the standard form is
(1 + √2)y12 + (1 − √2)y2.
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338 CHAPTER 12. EIGENVALUES AND DIAGONALIZATION
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Chapter 13
Systems of Linear Differential Equations
13.1 Linear Systems
1. The two given solutions are linearly independent because neither is a con- stant multiple of the other. Use them as columns of a fundamental matrix
−e2t 3e6t Ω(t) = e2t e6t .
Notice that Ω(0) has a nonzero determinant, which is also an indicator that the columns are independent.
Now we have a general solution
X(t) = Ω(t)C,
in which C is a 2 × 1 matrix of constants. To satisfy the initial condition, we need to choose C so that
0 Ω(0)C = 4 .
Then c1 = 3, c2 = 1, so the solution of the initial value problem is −e2t 3e6t 3 −3e2t + 3e6t
X(t)= e2t e6t 1 = 3e2t+e6t . 339
This requires that
0 11c=4.
−1 3 c1 2
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340 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
2. The two solutions are independent because Φ1(0) and Φ2(0) are indepen- dent in R2. We can form a fundamental matrix
e5t e3t Ω(t) = 3e5t e3t .
We now have a general solution
X(t) = Ω(t)C,
in which C is an arbitrary 2 × 1 matrix of real numbers. To solve the initial value problem, we need
X(0)= 1 =Ω(0)C= 3 1 c
Then c1 = 3/2 and c2 = −7/2, so the solution of the initial value problem
is
−2 1 1 c1 −2
= 1 .
e5t e3t 3/2 1 3e5t −7e3t
.
3. Because Φ(0) and Φ(0) are independent in R2, these solutions are inde-
X(t)=Ω(t)C= 3e5t e3t −7/2 = 2 9e5t −7e3t pendent. Form the fundamental matrix
(2 + 2√3)e(1+2√3t) (2 − 2√3)e(1−2√3)t Ω(t) = e(1+2√3)t e(1−2√3)t .
Then X(t) = Ω(t)C is a general solution. To solve the initial value prob- lem, we need
2 Ω(0)C = 2
2
and we find that we must choose
c1 =1−1√3,c2 =1+1√3.
66
4. From the given solutions we can write the fundamental matrix
Ω(t) =
− cos(√15t/2) − √15 sin(√15t/2) √15 cos(√15t/2) − sin(√15t/2) √√
e3t/2
The general solution is Ω(t)C. For the solution of the initial value prob-
8 cos( 15t/2) 8 cos( 15t/2) lem, solve for C so that
We find that
0 Ω(0)C = 7 .
7 √15 C = 8(1 + √15) 1 .
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13.2. SOLUTION OF X′ = AX WHEN A IS CONSTANT 341 5. Form the fundamental matrix
1.
2.
3.
4.
5.
6.
Ω(t) =
Ω(t) =
7e3t 0 5e3t e−4t
2et e6t −3et e6t
1 e2t
et −et Ω(t) = et 0 0 et
e−3t 3e−3t .
e−3t
The general solution is X(t) = Ω(t)C. To solve the initial value problem,
we need C such that
Solve this equation to get
1 Ω(0)C = −3 .
5
24 C = 14.
−9
13.2 Solution of X′ = AX When A Is Constant
In these problems, different fundamental matrices may be found, depending on the choice of eigenvectors used corresponding to eigenvalues of the coefficient matrix.
In each problem, the general solution has the form X = Ω(t)C, where Ω(t) is a fundamental matrix. We will give one choice for Ω(t).
Ω(t) = et
Ω(t) = et et
−1 e−t
e−t 2e−t
2e3t 3e3t
e2t
e2t
2e2t e2t
−e−4t 2e−4t
1 Ω(t)= 6
−13 Ω(t) =
e−4t
−2e3t 2et e−t
et e−t
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342 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 7. A has eigenvalues and eigenvectors
Write
−2i 2+2i, 2i 1 ,2−2i, 1 .
2i 0 2 1=1+i0
to form two independent solutions
Φ1(t) = e2t −2 sin(2t)
and
cos(2t) 2t 2 cos(2t)
Φ2(t) = e sin(2t) . Use these as columns of a fundamental matrix
−2 sin(2t) 2 cos(2t) Ω(t) = cos(2t) sin(2t) .
√
8. A has complex eigenvalues, with the eigenvalue 3+ 2 + √5i 2 √5
3=3+i0. Two independent solutions are
5i having eigenvector
2 cos(√5t) − √5 sin(√5t) Φ1(t) = e2t √
√5 cos(√5t) − √5 sin(√5t) Φ2(t) = e2t √ .
and
3 cos( 5t)
3 sin( 5t) These form the columns of a fundamental matrix.
9. A has eigenvalues 2, 5, 5, with corresponding eigenvectors 1 −3
0,−3. 01
All eigenvectors associated with 5 are scalar multiples of this eigenvector. Immediately we can write two independent solutions
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Φ1(t) = e2t 0 0
1
13.2.
SOLUTION OF X′ = AX WHEN A IS CONSTANT 343
and
For a third solution, denote the eigenvector associated with 5 as E and let Φ3(t) = Ete5t + Ke5t.
Substitute this into X′ = AX and use the fact that AE = E to obtain E+5K = AK.
10.
11.
−b − 3c = 1. Then a = −2/3, b = −1, c = 0, so
−2/3 K= −10
and we obtain the third solution
−3te5t − (2/3)e5t Φ3(t)= −3te5t −e5t .
te5t
The three solutions obtained are linearly independent and can be used to
form the columns of a fundamental matrix.
The coefficient matrix has eigenvalues 2, −2 − 3 and we can use the inde- pendent eigenvectors to form the fundamental matrix
If we let
we obtain
a K = b ,
c
−3a + 5b + 6c = −3 3b + 9c = −3
Φ2(t) = e5t −3 . 1
e−3t −3e−3t .
eigenvectors
2i −2i 1,0.
From these form two independent solutions which for the columns of the fundamental matrix
−2e2t sin(2t) 2e2t cos(2t) Ω(t) = e2t cos(2t) e2t sin(2t) .
4e2t 0 Ω(t) = 3e2t e−t 3e2t e−2t
2e−3t
The coefficient matrix has eigenvalues 2+2i and 2−2i, with corresponding
−3
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344 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 12. The coefficient matrix has eigenvalues −1 ± 2i, and the eigenvector asso-
ciated with −1 + 2i is
Form the fundamental matrix
t 5 cos(2t)
− cos(2t) − 2 sin(2t)
5 sin(2t) 2 cos(2t) − sin(2t) .
5 −1+2i .
Ω(t) = e
13. The coefficient matrix has eigenvalues 1 ± i. An eigenvector associated
with 1+i is
2+i 1.
Form the fundamental matrix
t 2 cos(t) − sin(t)
0 2 1,1.
11
The second eigenvector is associated with 2, and every eigenvector asso- ciated with −1 is a scalar multiple of the first. Immediately we have two independent solutions
Ω(t) = e cos(t)
14. Eigenvalues of the coefficient matrix are −1, −1, 2. Eigenvectors are
and
0 Φ1(t) = 1 e−t
1 2
Φ2(t) = 1 e2t. 1
Try
where E is the eigenvector associated with the eigenvalue −1, and
Φ3(t) = Ete−t + Ke−t, a
K = b c
is to be determined. Substitute Φ3(t) into X′ = AX and use the fact that AE = −E to get
E − K = AK.
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cos(t) + 2 sin(t) sin(t) .
13.2.
15.
SOLUTION OF X′ = AX WHEN A IS CONSTANT 345 Solve the equations for a, b and c to get a = 1, b = 2, c = 0, in which c is
arbitrary and has been chosen to be zero. Then
0 1 1
Φ3(t)=1te−t +2e−t =t+2e−t. 10t
These solutions form the columns of a fundamental matrix for the system. The coefficient matrix has eigenvalues −2, −1 + 2i, −1 − 2i. From −2 we
16.
obtain the solution
An eigenvalue for −1 + 2i is
0
30 This gives us two more solutions:
and
Φ(t)=e−2t 0 . 1
1 0 1+i2.
cos(2t)
Φ2(t) = e−t cos(2t) − 2 sin(2t) 3 cos(2t)
sin(2t)
Φ3(t) = e−t
These solutions form the columns of a fundamental matrix.
The coefficient matrix has eigenvalues 2, 2, and all eigenvectors are scalar multiples of
One solution is
0 E=1.
0 Φ1(t)= e2t .
For a second solution, try
Φ2(t) = Ete2t + Ke2t.
2 cos(2t) + sin(2t) 3 sin(2t)
.
Substitute this into the differential equation to find that we can take 1/5
K=0. This yields a second solution
e2t /5 Φ2(t) = te2t .
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346 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 17. Thecoefficientmatrixhaseigenvalues3,3,andeveryeigenvectorisascalar
multiple of
One solution is
1 E=0.
e3t Φ1(t)= 0 .
Attempt a second solution
Φ2(t) = Ete3t + Ke3t.
Solve for K to get
Then
0 K= 1/2 .
te3t Φ2(t) = e3t/2 .
These solutions form columns of a fundamental matrix.
18. The eigenvalues of A are 1, 1, 1 and all eigenvectors are scalar multiples of
One solution is
For a second solution, let
and obtain
Then
For a third solution, let
0 E = 0 .
1 0
Φ1(t) = 0. et
Φ2(t) = Etet + Ket
1/4 K=0.
0
et /4
Φ2(t) = tet . Φ3(t) = 1Et2 + Ktet + Met.
2
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13.2. SOLUTION OF X′ = AX WHEN A IS CONSTANT Substitute this into the system and solve for
347
1/2 M=0.
−2
1 t2 + 1 t + 1
242
Φ3(t) = et
19. The coefficient matrix has eigenvalues 4 + 29i and 4 −
sponding eigenvectors
−2 −√29 3+i0.
This gives us two independent solutions:
This gives us
0 .
1 t2 − 2 2
√√
29i, with corre-
−2 cos(√29t) + √29 sin(√29t) √
and
Φ1(t) = e4t
3 cos( 29t)
−√29 cos(√29t) − 2 sin(√29t)
Φ2(t) = e4t
20. The coefficient matrix has eigenvalues −1, −3 +
.
√
3 sin( 29t) √√
responding eigenvectors
3 −4 − √10
10, −3 − 10, with cor- −4 + √10
√√
V1 =1,V2 =10+3 10,V3 =10−3 10.
644 The general solution is
√√ X(t) = V1e−t + V2e(−3+ 10)t + V3e(−3− 10)t.
These solutions can also be used as columns of a fundamental matrix.
21. The coefficient matrix has eigenvalues 1, 1, 3, 0, with corresponding eigen- vectors
0 1 3 2 V1 = −2,V2 = 0,V3 = 2,V4 = 0.
−2 0 2 1 1000
We can write a general solution
X(t) = V1et + V2et + V3e3t + V4.
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348 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
13.3 Exponential Matrix Solutions
In Problems 1–8, the exponential matrix can be obtained using the Putzer algorithm or a software program.
1.
2.
3.
4.
5.
6.
7.
e=e
At cos(2t) − 1 sin(2t) 1 sin(2t) e=5221
− 2 sin(2t) cos(2t) + 2 sin(2t)
At 2e−3t + 1 1 − 1e−3t e=3 333
2 − 2e−3t 1e−3t + 2 3333
eAt = e3t cos(2t) + sin(2t) 2 sin(2t)
cos(√7t/2) + 1 sin(√7t/2)
At3t/2 7 7
− sin(2t) cos(2t) − sin(2t)
− 2 sin(√7t/2)
√
e=e
4 sin(√7t/2) cos(√7t/2) − 1 77
sin(√7t/2)
√√
eAt = et cos(2t) −1 sin(2t)
2
2 sin(2t) cos(2t)
−2e5t + 2 −2e5t + 2 4 + 1e5t
5555
At 1 + 4e5t e=5555
cos(3√3t/2) + 1 sin(3√3t/2) − 2 sin(3√3t/2) At−t/2 3 3
√√
2 sin(3√3t/2) cos(3√3t/2) − 1 sin(3√3t/2) 33
√√
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13.3. EXPONENTIAL MATRIX SOLUTIONS 349 8. eAt isthe3×3matrix[eij],where
e11 =
e12 =
e13 =
e21 =
e22 =
e23 =
3e2t + 2 cos(t) − 1 sin(t), 555
2 sin(t) + 1 cos(t) − 1 e2t 555
1 sin(t) − 1 cos(t) + 1 e2t 555
3 cos(t) − 3 e2t − 4 sin(t) 555
1 e2t + 4 cos(t) + 3 sin(t) 555
7 sin(t) + 1 cos(t) − 1 e2t 555
e31 = −3 cos(t) + 3e2t − 1 sin(t)
555
e32 = 1 cos(t) − 1 e2t − 3 sin(t) 555
e33 = 1 e2t + 4 cos(t) − 2 sin(t) 555
9. First, because D is a diagonal matrix, Dn is a diagonal matrix for any positive integer n, and the diagonal element of Dn is dnj . Then
eDt=∞ 1Dntn
n! and the diagonal element of eDt is
which is edj t. 10. Notice that
n=0
∞ 1 ( d j ) n t n ,
n!
n=0
Bn = (P−1AB)n
= (P−1AP)(P−1AP) · · · (P−1AP) = P−1AnP.
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350 CHAPTER 13. Then
SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS
eBt = ∞ n=0
= ∞ n=0
∞1
Antn P
1 (P−1AP)ntn n!
1 P−1AnPtn n!
= P−1 n=0
n! = P−1eAtP.
11. From the result of Problem 10,
eAt = PeDtP−1,
where P−1 AP = D, the diagonal matrix having the eigenvalues λ1 , · · · , λn of A down its diagonal. But from the result of Problem 9, eDt is the di- agonal matrix having diagonal elements eλj t .
13.4 Solution of X′ = AX + G for Constant A
1. The coefficient matrix A has the repeated eigenvalue 3, 3, and every eigen-
vector is a scalar multiple of
1 −1 .
One solution of the homogeneous system X′ = AX is 3t 1
e −1 .
Using methods from Section 13.2, find a second, independent solution of
this homogeneous system to write the fundamental matrix 3t1+2t 2t
Ω(t) = e −2t 1 − 2t .
For a particular solution of the nonhomogeneous system, first compute
−1 −3t 1 − 2t −2t Ω (t)=e 2t 1+2t .
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13.4.
SOLUTION OF X′ = AX + G FOR CONSTANT A 351
Now use variation of parameters to compute a solution of the nonhomo- geneous system. For this method, we need
u(t) = Ω−1(t)G(t) dt
3t1−2t −2t−3et
= e 2t 1+2t e3t dt 6te−2t − 3e−2t − 2t
= −6te−2t +1+2t dt= The general solution is
X(t) = Ω(t)C + Ω(t)u(t) =e3t1+2t 2t c1
−2t 1−2t c2 + e3t 1 + 2t 2t
=
2. A has eigenvalues 0, 0, and every eigenvector is a scalar multiple of
2 1.
The associated homogeneous system has this eigenvector as a constant solution. Apply the method of Section 13.2 to find a second, linearly independent solution and produce the fundamental matrix
2 1+2t Ω(t)=1 t .
Use this fundamental matrix and variation of parameters to find the gen- eral solution of the nonhomogeneous system
2c1 +c2(1+2t)+t+t2 −2t3 X(t)= c1 +c2t+2t2 −t3 .
3. A has eigenvalues 6, 6 and eigenvectors are scalar multiples of 1
1. We find the fundamental matrix
1 1+t Ω(t)=1 t
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−3te−2t − t2 (3/2)(1+2t)e−2t +t+t2
.
−3te−2t − t2 −2t 1−2t (3/2)(1+2t)e−2t +t+t2
e3t(c1(1 + 2t) + 2c2t) + t2e3t e3t(−2c1t + c2(1 − 2t)) + (t − t2)e3t + 3et/2
.
352 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS for the associated homogeneous system. Use this and variation of param-
eters to find the general solution of the nonhomogeneous system: 6t c1 +c2(1+t)+2t+t2 −t3
X(t)=e c1 +c2t+4t2 −t3 .
4. A has eigenvalues 2, 2, 2, and there are two linearly independent eigenvec-
tors
1 0 0,1.
01 We find the fundamental matrix
vectors
and 1 has the eigenvector
0 0 1,−9,
0 2 10
0 0 ,
0 1
10 0
Ω(t)=e2t 0 1 −4t 0 1 1−4t
of the associated homogeneous system. A general solution for the nonho- mogeneous system is
c1e2t X(t)= (c2 −4c3t)e2t +1 .
(c2 +c3(1−4t))e2t +1
5. A has eigenvalues 1, 1, 3, 3. The eigenvalue 3 has two independent eigen-
with all other eigenvectors associated with 1 scalar multiples of this one. A fundamental matrix for the homogeneous system X′ = AX is
0et 00
Ω(t) = 0 −2etet −9e3t . 0 0 0 2e3t
et −5tet e3t 0 The nonhomogeneous system has general solution
c2et X(t)= −2c2et +(c3 −9c4)e3t +et .
2 c 4 e 3 t (c1 − 5c2t)et + c3e3t + (1 + 3t)et
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13.5. SOLUTION BY DIAGONALIZATION 353
For Problems 6–9, the idea is to find a general solution for the system, then solve for the constants to obtain the solution satisfying the initial condition. For these problems only the solution of the initial value problem is given.
6.
7.
8.
9.
−1+e2t X(t) = −5t + (3 + 5t)e2t
(−1 − 14t)et X(t) = (3 − 14t)et
13t − (8 + 12t + 3t2)e2t
X(t)=
4et +(7+2t)e2t −et − e2t
X(t) =
(6 + 12t + (1/)t2)e−2t
(2 + 12t + (1/2)t2)e−2t
(3 + 38t + 66t2 + (13/6)t3)e−2t
13.5 Solution by Diagonalization
1. The coefficient matrix A is diagonalized by 1 1
and we find that
P=14
−1 14 −1
P =3 −1 1 .
′ −1 0 14 −1 0
The system for Z is
Z=02Z+3−11 10cos(t). This is the system
′ z1′ −z1 1−10cos(t) Z(t)= z′ = 2z +3 10cos(t) .
Then
Z(t) = c2 e2t − (4/3) cos(t) + (2/3) sin(t) .
c1e−t + c2e2t − 3 cos(t) − sin(t)
22
Solve these two independent differential equations to get c1 e−t − (5/3) cos(t) − (5/3) sin(t)
X(t) = PZ(t) = c1e−t + 4c2e2t − 7 cos(t) + sin(t) .
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354 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 2. The coefficient matrix has eigenvalues 2, 6 and is diagonalized by
We find that
3 1 P= −1 1 .
−1 1 1 −1 P=413.
With X = PZ, the transformed system has the solution c1e2t − 1 − e3t
Z(t)= c2e6t −1/3−et . The original system has the general solution
3c1e2t + c2e6t − 4e3t − 10/3 X(t) = −c1e2t + c2e6t + 2/3 .
3. The coefficient matrix has eigenvalues 0, 2 and is diagonalized by 1 1
which has inverse
P= −1 1
−1 1 1 −1
P=211. The system for Z(t) has the solution
Then
c1−2t+e3t Z(t)= c2e2t −1+3e3t .
c1 +2c2e2t −1−2t+4e3t X(t)= −c1 +c2e2t −1+2t+2e3t .
4. A has eigenvalues 1, 7, and we find
15−1 11−5
We obtain Z(t) =
P= −1 1 ,P =6 1 1 .
c1 et + (1/15) cos(3t) − (3/15) sin(3t) + 20/3
c2 e7t + (7/87) cos(3t) − (2/58) sin(3t) − 4/21 . c1et + 5c2e7t + (68/145) cos(3t) − (54/145) sin(3t) + 40/7
Then
X(t) = −c1 et + c2 e7t + (2/145) cos(3t) + (24/145) sin(3t) − 48/7 .
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13.5. SOLUTION BY DIAGONALIZATION
355
5. A has eigenvalues 3i, −3i and is diagonalized by 1+i 1−i
We find that
P=33. −1 1 −3i 1 + i
P =6 3i 1−i . The transformed problem for Z has the solution
d1e3it + ((2 − i)/6)e2t Z(t) = d2e−3it + ((2 + i)/6)e2t
.
If Euler’s formula is used on the complex exponential terms, we obtain the real solution
c1(cos(3t) − sin(3t)) − c2(cos(3t) + sin(3t)) + e2t X(t) = 3c1 cos(3t)03c2 sin(3t) + 2e2t
.
6. The coefficient matrix is
1 1 A=11
with eigenvalues 0, 2. A is diagonalized by 1 1
and we find that
The uncoupled system is
′ 0 0 1/2
P= −1 1
−1 1/2 −1/2 P = 1/2 1/2 .
−1/2 6e2t Z= 0 2 Z+ 1/2 1/2 2e2t
.
The initial conditions convert to
Z(0)=P X(0)= 3 .
Solve for Z to obtain
2+e2t Z(t) = 3e2t + 4te2t .
X(t)=PZ(t)= −2+2(1+t)e2t .
The solution of the initial value problem for X is
2+4(1+t)e2t
−1 3
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356 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 7. The coefficient matrix has eigenvalues 0, 3 and is diagonalized by
Compute
2 −1 P=11.
−1 11 1 P =3 −1 2 .
The uncoupled system is
′ 0 0 1/2 1/32t
Z= 0 3 Z+ −1/3 2/3 The initial condition is
5 .
Then
(1/3)t2 + (5/3)t + 25/3 (127/27)e3t + (2/9)t − 28/27 .
−1 25/3 Z(0)=P X(0)= 11/3 .
Z(t) =
The solution of the initial value problem for X is
−(127/27)e3t + (2/3)t2 + (28/9)t + 478/27 X(t) = PZ(t) = (127/27)e3t + (1/3)t2 + (17/9)t + 197/27 .
8. The coefficient matrix has eigenvalues i, −i and is diagonalized by 5 5
We find that
P=2−i2+i. −1 1 1 − 2i 5i
P =10 1+2i −5i . With X = PZ, the uncoupled system is
′ i 0 (1 − 2i)/10 i/2 5 sin(t) Z= 0 −i Z+ (1+2i)/10 −i/2 0
with initial conditions
Z(0)=P X(0)= 1−i/2 .
,
−1 1 + i/2
Solve these uncoupled equations for z1 and z2 and then obtain
10 cos(t) − (5/2)t sin(t) − 5t cos(t) X(t) = PZ(t) = 5 cos(t) + (5/2) sin(t) − (5/2)t sin(t) .
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13.5.
9.
SOLUTION BY DIAGONALIZATION 357
The coefficient matrix has eigenvalues 1,1,−3, but there are two inde- pendent eigenvectors associated with the repeated eigenvalue 1 and A is diagonalized by
10.
We find that
1 −1 1 P=1 0 3.
011
3 −2 3 P−1=1 −1 2.
−1 1 −1 With X = PZ we obtain the uncoupled system
1 0 0 3 = 2 3 −3e−3t Z′=0 1 0Z+1 −1 2 t .
00−3 −11−1 0 The initial conditions are
11 Z(0)=P−1X(0)= 6 .
−4
Solve this uncoupled system to obtain
(5/2)et − (8/3)e−3t + 3te−3t + (8/9) + (4/3)t X(t) = PZ(t) = (27/4)et − (113/12)e−3t + 9te−3t + (5/3) + 3t
(17/4)et − (113/36)e−3t + 3te−3t + (8/9) + (4/3)t The coefficient matrix has eigenvalues 1, 2, 2 and is diagonalized by
.
and we find that
1 1 1 P=1 1 0
101
−1 1 1 P−1 = 1 0 −1 .
1 −1 0
The uncoupled system is
1 0 0 −1 1 1− Z′=0 2 0Z+1 0 −1t.
002 1−102et The initial conditions are
Z(0) = P−1X(0).
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358 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS Solve this initial value problem for Z(t) and then obtain
−(1/4)e2t + (2 + 2t)et − (3/4) − (1/2)t X(t)=PZ(t)= e2t +(2+2t)et −1−t .
(5/4)e2t + 2tet − (3/4) − (1/2)t
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Chapter 14
Nonlinear Systems and Qualitative Analysis
14.1 Nonlinear Systems and Phase Portraits
1. The coefficient matrix is
3 −5 A= 5 −7 ,
with eigenvalues −2, −2. Every eigenvector is a scalar multiple of 1
1. The origin is an improper nodal sink.
2. The coefficient matrix has eigenvalues −3,4 and the origin is a saddle point.
For Problems 3 − −16, only the eigenvalues of the coefficient matrix, and the classification of the origin, are given. As typical cases, phase portraits are drawn for the systems of Problems 3, 5, 6, 7 and 11.
Phase portraits are included for the systems of Problems 3, 4, 5 and 11.
3. eigenvalues 2i, −2i; center
4. 2, 3, nodal source
5. 4+5i,4−5i, spiral point
6. −3, −5, nodal sink
7. 3,3,andthecoefficientmatrixdoesnothavetwoindependenteigenvectors; improper node
359
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360CHAPTER14. NONLINEARSYSTEMSANDQUALITATIVEANALYSIS
Figure 14.1: Center of Problem 3.
Figure 14.2: Nodal source of Problem 4.
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14.1.
NONLINEAR SYSTEMS AND PHASE PORTRAITS 361
8.
9. 10.
11. 12. 13. 14.
15. 16. 17.
Figure 14.3: Spiral source of Problem 5.
√√
31i, − 31i, center
√√
eigenvalues −2 + 3i, −2 − 3i, spiral sink
−13, −13, and the coefficient matrix does not have two independent eigen-
vectors; improper node
√√
5, − 5, saddle point
√√
3 + 5i, 3 − 5i, spiral source
√√
−3 + 7, −3 − 7, both eigenvalues negative, nodal sink
11, 11, and the coefficient matrix does not have two independent eigenvec- tors, improper node
√√
2+ 3,2− 3, nodal source
√√
13i, − 13i, center
(a) First, write
as
Integrate to get
dx = 1x dt t
1 dx = 1 dt. xt
ln(x) = ln(t) + c
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362CHAPTER14. NONLINEARSYSTEMSANDQUALITATIVEANALYSIS
Figure 14.4: Saddle point of Problem 11.
for x > 0,t > 0. Then x = ct with c constant. Put this into the second equation to get
Write this as
or
Then
Integrate to get
Then
y′ =ct−1y. t
y′ + 1y = ct, t
ty′ + y = ct2. (ty)′ = ct2.
ty = c t3 + d. 3
y = ct2 + d. 3t
(b) Suppose x(t0) = 1 and y(t0) = 0. Then it is routine to solve for c and d from part (a) to obtain
x(t)=1t,y(t)= 1t2−1t20. t0 3t0 3t
(c) In part (b), we have trajectories through (1,0) at any time t0 ̸= 0. However, these translations are not trajectories of each other.
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14.2. CRITICAL POINTS AND STABILITY 363 14.2 Critical Points and Stability
In Problems 1–16, the stability type of the origin is given, based on information in Problems 1–16 of Section 14.1.
1. The origin is an improper node that is both stable and asymptotically stable
2. The origin is an unstable saddle point
3. stable but not asymptotically stable center
4. unstable nodal source
5. unstable spiral source
6. stable and asymptotically stable nodal sink
7. unstable improper node
8. stable but not asymptotically stable center
9. stable and asymptotically stable spiral sink
10. stable and asymptotically stable improper node
11. unstable saddle point
12. unstable spiral point
13. stable and asymptotically stable nodal sink
14. unstable improper node
15. unstable nodal source
16. stable but not asymptotically stable center
√√
17. If ε = 0, the eigenvalues are 5i, − 5i and the origin is a center, which
is stable but not asymptotically stable. If ε > 0, the eigenvalues are
1 ε + 1 (ε − 2)2 − 24, 1 ε(ε − 2)2 − 24. 222
√
These have positive real part. If 0 < ε < 2(1 + 6), then the origin is an √
unstable spiral point. If ε > 2(1 + 6), the origin is an unstable saddle
point. If ε = 2(1 +
√
6), the origin is an unstable improper node.
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364CHAPTER14. NONLINEARSYSTEMSANDQUALITATIVEANALYSIS 18. If ε = 0, the eigenvalues are −3,−3 and every eigenvector is a scalar
multiple of
1
−1 .
In this case the origin is a stable and asymptotically stable improper node.
If ε > 0, the eigenvalues are
ε − 6 + 1 (ε + 10)2 − 100, ε − 6 − 1 (ε + 10)2 − 100.
These eigenvalues are real and distinct. Further, if 0 < ε < 9/8, these eigenvalues are both negative, so the origin is a stable and asymptotically stable nodal sink. If ε > 9/8, then the eigenvalues are of opposite sign and the origin is an unstable saddle point. If ε = 9/8, then the origin is not an isolated singularity.
14.3 Almost Linear Systems
In Problem 1, the details of showing that the system is almost linear are in- cluded. Problems 2–10 omit this demonstration and concentrate on analyzing the critical point (0, 0).
1. To show that the system is almost linear, consider
x2 lim
(x,y)→(0,0) x2 + y2
r→0
The origin is a critical point and the matrix of coefficients of the linear
22 22
r2 cos2(θ) r→0 r
= lim
= lim r cos2(θ) = 0.
part is
which has eigenvalues
1 −1 12,
1(3 + √3i) and 1(3 − √3)i. 22
The origin is an unstable spiral point of the linear part, hence also of the given system.
2. The matrix of coefficients of the linear part is
1 −1 11
with eigenvalues 1 + i, 1 − i. The origin is an unstable spiral source.
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14.3. ALMOST LINEAR SYSTEMS 365
3. The linear part has matrix
−2 2 A(0,0) = 1 4
√√
with eigenvalues 1+ 11, 1− 11. These are of opposite sign, so the origin
is an unstable saddle point.
4. The matrix of the linear part is
−2 −3 14
√
with eigenvalues 1 ± 6, which are of opposite sign. The origin is an
unstable saddle point.
5. The linear part has matrix
3 12 −1 −3 ,
√√
with eigenvalues 3i, − 3i. The origin is a center for the linear part of
the system, so the nonlinear system could have a center or a spiral point there.
6. The linear part has matrix
2 −4 11,
√
with eigenvalues (3 ± 15i)/2. The origin is a spiral point and is unstable
because the real part of the eigenvalues is positive.
7. The linear part has matrix
−3 −4 11.
This matrix has eigenvalues −1, −1, and all eigenvectors are scalar multi- ples of
−2 1.
The origin is a stable improper nodal sink.
8. The linear part has matrix
−3 −4 −1 1
√
with eigenvalues −1 ± 2 2. Both eigenvalues are positive, so the origin is
an unstable nodal source.
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366CHAPTER14. NONLINEARSYSTEMSANDQUALITATIVEANALYSIS
9. The linear part has matrix
−2 −1 −4 1
with eigenvalues 2, −3, so the origin is an unstable saddle point.
10. The linear part has matrix
1 0 −2 1
with eigenvalues 1, 1. All eigenvectors associated with this eigenvalue are scalar multiples of
0 1,
so the origin is an unstable (positive eigenvalue) improper node.
11. Refer to these as systems I and II, in the order given.
(a) For each system the linear part has coefficient matrix
0 1 A= −1 0
with eigenvalues i, −i. Therefore the origin is a center for each system. (b) For The first system, use polar coordinates, with
and
Then
x = rcos(θ),y = rsin(θ) x2 + y2 = r.
lim
(x,y)→(0,0)
−xx2 + y2
x2 + y2
= lim
−r2 cos(θ) r
independent of θ. And, similarly
−yx2 + y2 lim
= lim−rsin(θ)=0. r→0
r→0
= lim −r cos(θ) = 0,
r→0
x2 + y2 the same for the second system.
(x,y)→(0,0)
Therefore the first system is almost linear. The argument is essentially
(c) With r2 = x2 + y2, we have
rr′ = xx′ + yy′,
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14.3.
ALMOST LINEAR SYSTEMS 367
where primes denote differentiation with respect to t. Now insert the expressions for x′ and y′ from the differential equations of system I to obtain
Then
rr′ =x(y−xx2 +y2)+y(−x−yx2 +y2) = −(x2 + y2)x2 + y2
= −r3.
r′ = dr = −r2 for system I.
dt
Similarly, if we insert the expressions for x′ and y′ from system II, we
obtain
Then
(d) For system I,
rr′ =x(y+xx2 +y2)+y(−x+yx2 +y2) = (x2 + y2)x2 + y2
= r3.
dr = r2 for system II. dt
dr = −r2. dt
This is the separable equation
−1 dr=dt.
r2
This shows that, for system I, r′(t) < 0, so the distance between the point
and the origin is decreasing with time.
Now integrate the differential equation for r(t) to get
1 = t + c. r
To satisfy the initial condition r(t0) = r0, we need 1 = t0 + c,
so
Then
r0
c= 1 −t0.
r0 r(t) = 1
t−t0 +1/r0
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368CHAPTER14. NONLINEARSYSTEMSANDQUALITATIVEANALYSIS for system I. Then
r(t) → 0 as t → ∞.
The first system is asymptotically stable at the origin.
(e) By an entirely analogous derivation, for system II, r′ =r2 >0
so r(t) is increasing with time for system II. Solve this separable differential equation subject to r(t0) = r0 to get
r(t) = 1 . 1/r0 +t0 −t
Then for system II, r(t) → ∞ as t → t0 + 1/r0 from the left. We conclude from this that the second system is unstable at the origin.
Parts (d) and (e) show that, in the case of a center at the origin, behavior of the the linear part of an almost linear system at the origin does not provide definitive information about the stability of the origin for the nonlinear system.
12. The linear part of the system has coefficient matrix α −1
A= 1 −α . √√
The eigenvalues are α2 − 1 and − α2 − 1. There are the following cases:
1. If 0 < |α < 1, the eigenvalues are pure imaginary. In this case the origin is a center of the linear part and may be a center or spiral point of the almost linear system.
2. If |α| > 1, then the eigenvalues are real and of opposite sign, so the origin is an unstable saddle point of the linear part, hence also of the nonlinear system.
3. If α = ±1, then A is a singular matrix and the theory developed for almost linear systems does not apply.
The constants h and k will influence quantitative aspects of the phase portrait of the system, but not the qualitative behavior.
13. Using results from Problem 11, we have
rr′ = xx′ + yy′
=x(y+εx(x2 +y2))+y(−x+εy(x2 +y2))
= ε(x2 + y2)(x2 + y2) = εr4.
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14.4. LINEARIZATION
Then
This is separable
Integrate to get
Then
369
dr = εr3. dt
1 dr=εdt. r3
−1r−2 =εt+c. 2
r(t)=√1 , k − 2εt
where k = 2c is an arbitrary constant which is determined by specifying a point that the trajectory is to pass through at some positive time.
If ε < 0, then
1
r(t)= k+2|ε|t →0
as t → ∞. In this case trajectories approach the origin as t → ∞ and the nonlinear system is asymptotically stable.
However, something different happens if ε > 0. Suppose r(0) = ρ, so a trajectory starts at a point ρ units from the origin at time zero. Then k = 1/ρ2 and
1
r(t) = (1/ρ)2 − 2εt .
Now, as t starts at zero and increases toward 1/(2ερ2), r(t) → ∞, so there is a time close to which the point is arbitrarily far from the origin. This makes the origin unstable.
14.4 Linearization
1. For critical points other than the origin, solve x−y+x2 =0,x+2y=0
to get (−3/2, 3/4). We find that
A(−3/2,3/4) = 1 2 ,
−2 −1
which has eigenvalues 3 and − 3. This critical point is an unstable
√√
saddle point of the linear part, and therefore of the given system.
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370CHAPTER14. NONLINEARSYSTEMSANDQUALITATIVEANALYSIS 2. The critical points are
We have
with eigenvalues
−1+√17 −1−√17 2,4,2,4.
√17 −1 A((−1+√17)/2,4) = −7 2 − √17
√√ 1− 25−2 17,1+ 25−2 17.
,
These are of opposite sign, being approximately equal to −3.093 · · · and 5.093 · · · . This critical point is an unstable saddle point.
Next,
−√17 −1 A((−1−√17)/2,4) = −7 2 + √17
,
with eigenvalues 1 − 25 + 2√17 and 1 + 25 + 2√17. These are also of opposite sign, so this critical point is also an unstable saddle point.
3. The critical point other than the origin is (−5, −5). We find that −2 2
A(−5,−5) = 1 −6 . √√
This has eigenvalues −4 + 6, −4 − 6, which are unequal and both negative. This critical point is a stable and asymptotically stable nodal sink of the nonlinear system.
4. The only critical point other than the origin is (−20, 5). We find that −2 −13
A(−20,5) = 1 4 ,
having eigenvalues 1 + 2i, 1 − 2i. These are complex with positive real
part, so (−20, 5) is an unstable spiral source.
5. The system has one critical point other than the origin, (−1/2, 1/8). Cal-
culate
with eigenvalues are point.
3 12 A(−1/2,1/8) = −1/4 −3 ,
√√
6, − 6, so this critical point is an unstable saddle
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14.4. LINEARIZATION 371 6. The system has two critical points other than the origin, (1,−2) and
(−2/3,2/9). First,
−4 −1 A(1,−2)= 3 1
√
and the eigenvalues are (−3± 13)/2. These have opposite sign, so (1, −2)
√
is an unstable saddle point.
A(−2/3,2/9) = −1/2 1 ,
8/3 −6
with eigenvalues (11 ± 97)/2. These are both positive, so (−2/3, 2/9) is
an unstable nodal source.
7. Aside from the origin, the system has critical points (1/2, −1/2). We have
−2 −3 A(1/2,−1/2) = 1 1
√
with eigenvalues (−1 ±
asymptotically stable because the real part of the eigenvalues is negative.
3i)/2, so (1/2, −1/2) is a spiral point, stable and 8. The critical points other than the origin are
First,
7 √21 7 √21 3,− 4 and − 3, 4 .
−3 −4 A(√7/3,−√21/4) = −5/2 −4/3
√
with eigenvalues (−13 ± 385)/6, both positive, so this critical point is
an unstable nodal source. Next,
−3 −4 A(−√7/3,√21/4) = 9/2 −4/3 .
√
This has eigenvalues (−13 ± 623i)/6. This critical point is a stable and
asymptotically stable spiral point.
9. The critical point other than the origin is (−3/8, −3/2). We find that
−2 −2 A(−3/8,−3/2) = −4 1
√
with eigenvalues (−1 ± 23i)/2, so (−3/8, −3/2) is a stable and asymp-
totically stable spiral point.
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372CHAPTER14. NONLINEARSYSTEMSANDQUALITATIVEANALYSIS
10. The system has two critical points other than the origin:
√ 2+√2 √ 2−√2
3+2 2, 2
, 3−2 2, 2
1 −4−2√2
.
First,
A(3+2√2,(2+√2)/2) = with eigenvalues
(−1 + 2√2)/2 11 + 7√2 7 1 √
7 1 √
6+√2−2 128 2+90,6+√2+2 128 2+190.
These are approximately 1.3188 · · · and 20.5806 · · · , both positive, so this critical point is an unstable proper nodal source,
1 −4+2√2 A(3−2√2,(2−√2)/2) = (−1 − 2√2)/2 11 − 7√2
with eigenvalues
7 1 √ 7 1 √
7− √2 − 2 190−128 2,6− √2 + 2 190−128 2.
These are approximately −0.4481 · · · and 2.5486 · · · , having opposite signs,
so this critical point is an unstable saddle point.
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Chapter 15
Vector Differential Calculus
15.1 Vector Functions of One Variable
In Problems 1 and 2 the details of the differentiation are carried out both ways. For Problems 3–8 just the derivative is given.
1. First use the product rule:
(f(t)F(t))′ = f′(t)F(t) + f(t)F′(t)
= (−12 sin(3t))F(t) + 4 cos(3t)(6tj + 2k)
= −12 sin(3t)i + (24t cos(3t) − 36t2 sin(3t))j + (8 cos(3t) − 24t sin(3t))k.
If we first carry out the product, we have
f (t)F(t) = 4 cos(3t)i + 12t2 cos(3t)j + 8t cos(3t)k,
so
2.
(f(t)F(t))′ =−12sin(3t)i
+ (24t cos(3t) − 36t2 sin(3t))j + (8 cos(3t) − 24t sin(3t))k.
(F(t) · G(t))
= (i − 6tk) · (i + cos(t)k) + (ti − 3t2k) · (− sin(t)k) = 1 − 6t cos(t) + 3t2 sin(t)
373
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374 3.
4.
CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS
Apply the product rule for cross products:
i j k i j k (F(t)×G(t))′ =1 0 0+t 1 4
1 − cos(t) t 0 sin(t) 1
= −tj − cos(t)k + (1 − 4 sin(t))i − tj + t sin(t)k = (1 − 4 sin(t))i − 2tj − (cos(t) − t sin(t))k.
(F(t) × G(t))′ = (3t2 − 2t sinh(t) − t2 cosh(t))i − 2tj − (sinh(t) + t cosh(t))k
5.
(f(t)F(t))′ = (1 − 8t2)i
6. 7. 8. 9.
+ (6t2 cosh(t) − (1 − 2t3) sinh(t))j + (et − 6t2et − 2t3et)k
(F(t)·G(t))′ =sin(t)+tcos(t)+4+5t4
(F(t) × G(t))′ = tet(2 + t)(j − k)
(F(t) · G(t))′ = −16 cos2(t) + 16 sin2(t)
F(t) = sin(t)i + cos(t)j + 45tk is a position vector for the curve, and
F′(t) = cos(t)i − sin(t)j + 45k is a tangent vector. The distance function along C is
tt√√
s(t) = ∥ F′(τ) ∥ dτ = 2026dτ = 2026t.
00
√
Then t = s/ 2026 and we can write a position vector in terms of s:
s s 45s G(s) = F(t(s)) = sin √ i + cos √ j + √ k.
2026 2026 2026
Now we can write a tangent vector in terms of s:
′ 1s s
G(s)=√ cos √ i−sin √ j+45k . 2026 2026 2026
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15.1. VECTOR FUNCTIONS OF ONE VARIABLE 375
10. F(t) = t3(i + j + k) is a position vector, and a tangent vector is F′(t) = 3t2(i + j + k).
A distance function along C is given by
t t√
Then
Let
s(t)= ∥F′(ξ)∥dξ= 3 3(t3+1). −1 −1
s 1/3 t=√−1.
3
s G(s)=F(t(s))= √ −1 (i+j+k).
3 This gives us the unit tangent vector
′1
G(s)= √ (i+j+k).
3
11. F = t2(2i + 3j + 4k) is a position vector for the curve, and F′(t) = 2t(2i + 3j + 4k)
is a tangent vector. The distance function along the curve is given by
Then Let
Then
t√t√
s(t)= ∥F′(ξ)∥dξ=2 29 ξdξ= 29(t2−1).
11
√
t= 1+s/ 29. s
G(s)=F(t(s))= √ +1 (2i+3j+4k). 29
′1
G(s)= √ (2i+3j+k)
29
is a unit tangent vector.
12. Because F(t) × F′(t) = O, we must have
∥ F(t) ∥∥ F′(t) ∥ sin(θ) = 0
where θ is the angle between these two vectors. Then at least one of F(t) and F′(t) is the zero vector, or the angle between these vectors is zero (so the vectors are parallel). Consider cases.
If F(t) = O then the particle simply remains at the origin for all time.
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376
CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS
If F′(t) = O, then F(t) is a constant vector and the particle does not move.
In the case that F(t) and F′(t) are parallel, then the position and tangent vectors are parallel, so the velocity vector is directed along the path of motion and the motion is in a straight line.
We could also argue as follows in the last case. If the position and tangent vectors are parallel, then for some number c,
Then
Then
Then
F′(t) = cF(t).
x′(t)i + y′(t)j + z′(t)k = c(x(t)i + y(t)j + z(t)k). x′(t) = cx(t), y′(t) = cy(t), z′(t) = cz(t).
x = x0ect,y = y0ect,z = z0ect,
where F(0) = x0i + y0j + z0k. But these are parametric equations of a
straight line.
15.2 Velocity, Acceleration and Curvature
In Problems 1–10, we can compute
v = F′(t), a(t) = F′′(t), v(t) =∥ v(t) ∥
by straightforward calculations. Next,
T(t)= 1 v(t)= 1 F′(t).
v(t) ∥ F′(t) ∥
Tangential and normal components of the acceleration can be obtained as
The unit normal is
aT = dv and aN = ∥ a ∥2 −a2T . dt
N(t) = 1 (a(t) − aT T(t)). aN
In this way it is not necessary to compute s(t) and write vectors in terms of s, which is often awkward. We can also compute
N(t) = 1 dT. ∥ dT/dt ∥ dt
Curvature is often easily computed as
κ = aN .
v2
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15.2. VELOCITY, ACCELERATION AND CURVATURE 377 We can also compute curvature as
κ = ∥ T′(t) ∥. ∥ F′(t) ∥
κ can also be obtained from a formula requested in Problem 13: κ= ∥F′(t)×F′′(t)∥.
1. The velocity is and the speed is The acceleration is A unit tangent is
The curvature is
∥ F′(t) ∥3
v(t) = F′(t) = 3i + 2tk
v(t) =∥ v(t) ∥= 9 + 4t2. a(t) = F′′(t) = 2k.
T(t)= √ 1 (3i+2tk). 9+4t2
κ = ∥ T′(t) ∥ = 6
∥ F′(t) ∥ (9 + 4t2)3/2
aT =dv=√ 4t
dt 9+4t2
Finally,
and
2.
a N = ∥ a ∥ 2 − a 2T = √ 6 9+4t2
.
.
v(t) = (sin(t) + t cos(t))i + (cos(t) − t sin(t))j, a(t) = (2 cos(t) − t sin(t))i − (2 sin(t) + t cos(t))j,
T(t) = √ 1 v, 1+t2
v(t) = 1 + t2,
t 2+t2
aT = √1+t2,aN = 1+t2, κ= 2+t2
(+t2 )3/2
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378 3.
4.
5.
6.
7.
CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS
v(t) = 2i − 2j + k, v = 3, T= 1(2i−2j+k),
3
aT =aN =κ=0
v(t) = et(sin(t) + cos(t))i + et(cos(t) − sin(t))k, v(t) = √2et, a(t) = 2et(cos(t)i − sin(t)k),
1
T(t)=√ ((sin(t)+cos(t))i+(cos(t)−sin(t))k),
2
√ t 1 −t aT= 2e=aN,κ=√e
2
v(t) = −3e−t(i + j − 2k), a(t) = 3e−t(i + j − 2k),
√−t 1
v(t)=3 6e ,T(t)= √ (−i−j+2k),
6
aT = −3√6e−t,aN = 0,κ = 0
v(t)=−αsin(t)i+βj+αcos(t)k,v(t)=α2 +β2, a(t) = −α cos(t)i − α sin(t)k,
1
T(t) = α2 + β2 (−α sin(t)i + βj + α cos(t)k),
aT =0,aN =α,κ= α
α2 + β2
v(t) = 2 cosh(t)j − 2 sinh(t)k, v(t) = 2cosh(2t), a(t) = 2 sinh(t)j − 2 cosh(t)k,
1
T(t) = cosh(t) (cosh(t)j − sinh(t)k),
2 sinh(2t) 2
a(t) = cosh(2t) , aN = cosh(2t) ,
κ=1 2(cosh(2t))3/2
Here we have used the identity
cosh(2t) = cosh2(t) + sinh2(t).
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15.2.
8.
9.
and where
10.
VELOCITY, ACCELERATION AND CURVATURE 379
v(t)= 1(i−j+2k),a(t)=−1(i−j+2k), t t2
√
61
v(t)= t ,T(t)=√6(i−j+2k),
√
aT =− 6,aN =0,κ=0
t2
v(t) = 2t(αi + βj + γk),
a(t) = 2(αi + βj + γk)
v(t) = 2|t|α2 + β2 + γ2, 1
T(t)= α2 +β2 +γ2(αi+βj+γk) aN =0,κ=0,
aN = 2(sgn(t))α2 + β2 + γ2,
1 if t > 0, −1 ift<0.
v(t) = (3 cos(t) − 3t sin(t))j − (3 sin(t) + 3t cos(t))k, v(t) = 31 + t2
sgn(t) =
a(t) = (−6 sin(t) − 3t cos(t))j − (6 cos(t) − 3t sin(t))k, T(t)=√ 1 ((cos(t)−tsin(t))j−(sin(t)+tcos(t))k,
9(1 + t2)3/2
The position vector for a straight line has the form
F(t) = (a + bt)i + (c + dt)j + (p + ht)k. The tangent vector is the constant vector
T(t) = bi + dj + hk.
Then T′(t) = O, so κ = 0.
1+t2
aT = √1+t2,aN = √1+t2 ,
3t (3t2 + 6)2 κ= (3t2+6)2
11.
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380
CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS
Conversely, suppose C is a smooth curve having zero curvature. Then κ =∥ T′(s) ∥=∥ F′′(s) ∥= 0.
If we write
this means that
F(s) = f(s)i + g(s)j + h(s)k, f′′(s) = g′′(s) = h′′(s) = 0.
But then f(s) = a + bs,g(s) = c + ds,h(s) = p + hs for some constants a, b, c, d, p, h. This makes F(s) the position vector of a straight line.
12. We may suppose that C is a circle of radius r about the origin in the x,y−plane (translations and rotations will not affect the curvature). A position vector for C in polar coordinates has the form
F(t) = r cos(t)i + r sin(t)j
with0≤t≤2π. Then
F′(t) == r sin(t)i + r cos(t)j
is a tangent vector, which has length r. A unit tangent is T = − sin(t)i + cos(t)j.
Then Then
13. First write
vT×F′′ =vT(aTT+aNN)
= vaT T × T + vaN T × N
= vaN T × N
= v(v2κ)T × N.
But T and N are orthogonal unit vectors, so ∥ T × N ∥= 1.
T′(t) = − cos(t)i − sin(t)j. κ= ∥T′(t)∥ = 1.
∥ F′(t) ∥ r
T(t)= 1 f′(t)= 1 F′(t).
∥ F′(t) ∥ v(t) F′ = vT.
This enables us to write
Now, F′′(t) is the acceleration a(t), and T × T = O, so
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15.3.
THE GRADIENT FIELD
Then Finally, so
381
15.3
∥ F′ × F′′ ∥= v3κ. v =∥ F′ ∥
∥ F′(t) × F′′(t) ∥3 κ = ∥ F′(t) ∥
The Gradient Field
.
1.
∇φ(x,y,z)= ∂ (xyz)i+ ∂ (xyz)j+ ∂ (xyz)k=yzi+xzj+xyk,
∂x ∂y ∂z ∇φ(1, 1, 1) = i + j + k
The maximum value of Du (1, 1, 1) is
∥ ∇φ(1, 1, 1) ∥= 3.
√ The minimum value is − 3.
2.
√
∇φ(x, y, z) = (2xy − z cos(xz))i + x2j − x cos(xz)k, √2π √2
∇φ(1,−1,π/4)= −2− 8 i+j− 2 k. The maximum value of Du(1, −1, π/4) is
√
3.
11/2 + π2/32 + π/ 2.
The minimum value is the negative of this maximum value.
∇φ(x, y, z) = (2y + ez )i + 2xj + xez k, ∇φ(−2, 1, 6) = (2 + e6)i − 4j − 2e6k.
The maximum value of Du (−2, 1, 6) is
20 + 4e6 + 5e12
and the minimum value is the negative of this.
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382 4.
CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS
∇φ(x, y, z) = −yz sin(xyz)i − xz sin(xyz)j − xy sin(xyz)k, ∇φ(−1, 1, π/2) = π i − π j − k.
22 The maximum value of Du(−1, 1, π/2) is
π2 1+4.
The minimum value is the negative of this radical. 5.
∇φ(x, y, z) = 2y sinh(2xy)i + 2x sinh(2xy)j − cosh(z)k, ∇φ(0, 1, 1) = − cosh(1)k.
The maximum − cosh(1).
6.
value of Du(0, 1, 1) is cosh(1). The minimum value is
1
∇φ(x,y,z)= x2 +y2 +z2[xi+yj+zk]
1 ∇φ(2,2,2)= √ (i+j+k).
3
maxDu =∥ ∇φ(2, 2, 2) ∥= 1,
and the minimum is −1. 7.
8.
Duφ(x, y, z) = ∇φ(x, y, z) · u 21
= ((8y −z)i+16xyj−xk)· √ (i+j+k) 3
12
= √ (8y −z+16xy−x)
3
Duφ(x, y, z)
=(−sin(x−y)i+sin(x−y)je k)·√ (i−j+2k)
1z = √ (−2sin(x−y)+2e )
6
z1 6
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15.3.
9.
10.
11.
THE GRADIENT FIELD
383
Du(x,y,z)
= (2xyz i+x z j+3x yz k)· √ (2j+k)
323221 5
12322 = √ (2x z + 3x yz )
5
Du(x,y,z) =((z+y)i+(z+x)j+(y+x)k)· √
1
=√ (z−3y−4x)
17
1 17
(i−4k)
Let φ(x,y,z) = x2 +y2 +z2 so the level surface is given by φ(x,y,z) = 4. The gradient provides a normal vector
Then
N(x, y, z) = ∇φ(x, y, z) = 2xi + 2yj + 2zk.
√√
N(1,1, 2)=2i+2j+2 2k √
is normal to the surface at (1, 1, 2). The tangent plane at this point has the equation
√√ 2(x−1)+2(y−1)+2 2(z− 2)=0,
or
The normal line at (1, 1,
√
x+y+ 2z=4. √
2) has parametric equations
√
12.
x=y=1+2t,z= 2(1+2t)
for all real t.
Write φ(x,y,z) = x2 +y−z, so the level surface is given by φ(x,y,z) = 0.
A normal vector at (−1, 1, 2) is given by
N(−1, 1, 2) = ∇φ(−1, 1, 2) = −2i + j − k.
The tangent plane at (−1, 1, 2) has the equation −2x + y − z = 1.
The normal line at (−1, 1, 2) is given by
x = −1 − 2t, t = 1 + t, z = 2 − t
in which t takes on all real values.
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384 CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS
13. Let φ(x, y, z) = x2 − y2 − z2. The normal vector at (1, 1, 0) is N(1, 1, 0) = ∇φ(1, 1, 0) = 2i − 2j.
The tangent plane at (1, 1, 0) has equation 2x − 2y = 0
or x = y. The normal line at (1, 1, 0) has parametric equations x = 1 + 2t, y = 1 − 2t, z = 0.
14. Let φ(x, y, z) = x2 − y2 + z2. A normal vector at (1, 1, 0) is given by N = 2i − 2j.
The tangent plane at (1, 1, 0) has equation y = x. The normal line at this point has parametric equations
x = 1 + 2t, y = 1 − 2t, z = 0.
15. A normal vector is given by
has parametric equations
x = 1 + 2t, y = π, z = 1.
N = ∇(2x − cos(xyz))
= 2i.
The tangent plane has equation x = 1 and the normal line at the point
16. A normal vector at the point is
444 N=∇(3x +3y +6z )
(1,1,1) The tangent plane at (1, 1, 1) has equation
=12(i+j+2k).
x + y + 2z = 4 and the normal line has parametric equations
x = 1 + 12t, y = 1 + 12t, z = 1 + 24t.
17. Because ∇φ(x,y,z) = i+k, the normal to the surface φ(x,y,z) = c is the
constant vector
The surface must therefore be the plane x + z = c.
N(x, y, z) = i + k.
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(1,π,1)
15.4. DIVERGENCE AND CURL 385 15.4 Divergence and Curl
1.
2.
3.
4.
5.
6.
∇·F= ∂ (x)+ ∂ (y)+ ∂ (2z)=4, ∂x ∂y ∂z
i j k
∇×F=∂/∂x ∂/∂y ∂/∂z=0i+0j+0k=O,
x y 2z
∇ · (∇ × F) = 0.
∇ · F = xz cosh(xyz),
∇ × F = −xy cosh(xyz)i + yz cosh(xyz)k
∇ · (∇ × F) = ∂ (−xy cosh(xyz)) + ∂ (yz cosh(xyz)) ∂x ∂z
= cosh(xyz)(y − y) + sinh(xyz)(−xy2z + xy2z) = 0. ∇·F=2y+xey +2,
∇×F=(ey −2x)k, ∇·(∇×F) = ∂ (ey −2x) = 0.
∂x
∇·F=xzexyz +3z2, ∇×F=−xyexyzi+(yexyz −1)k, ∇·(∇×F)=−yexyz −xy2exyz +yexyz +xy2exyz =0.
∇ · F = cosh(x) + xz sinh(xyz) − 1,
∇ × F = (−1 − xy sinh(xyz))i − j + yz sinh(xyz)k,
∇ · (∇ × F) = (−y + y) sinh(xyz)
+ ((−xy2z + xy2z) cosh(xyz) = 0.
∇ · F = cosh(x − z) + 2 + 1 = 3 + cosh(x − z), ∇ × F = −2yi − cosh(x − z)j,
∇·(∇×F)= ∂ (2y)+ ∂ (−cosh(x−z))=0. ∂x ∂y
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386 7.
8.
9.
10.
11.
CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS
∇φ = i − j + 4zk,
i j k
∇ × (∇φ) = ∂/∂x ∂/∂y ∂/∂z = 0. 1 −1 4z
∇φ = (18yz + ex)i + 18xzj + 18xyk,
∇ × (∇φ) = (18x − 18x)i + (18y − 18y)j + (18z − 18z)k = O.
∇φ = −6x2yz2i − 2x3z2j − 4x3yzk, ∇ × (∇φ) = (−4x3z + 4x3z)i
+ (−12x2yzi + 12x2yz)j + (6x2z2 − 6x2z2)k = O. ∇φ = z cos(xz)i + x cos(xz)k,
i j k ∇×(∇φ)= ∂/∂x ∂/∂y ∂/∂z
z cos(xz) 0 x cos(xz)
= 0i + (cos(xz) − xz sin(xz) − cos(xz) + xz sin(xz))j + 0k = O.
∇φ = (cos(x + y + z) − x sin(x + y + z))i − x sin(x + y + z)j − x sin(x + y + z)k,
12.
∇ × (∇φ) = (−x cos(x + y + z) + x cos(x + y + z))i
+ (− sin(x + y + z) − x cos(x + y + z) + sin(x + y + z) + x cos(x + y + z))j + (− sin(x + y + z) − x cos(x + y + z) + sin(x + y + z) + x cos(x + y + z))k = O.
∇φ = ex+y+z(i + j + k), ∇×(∇φ)=(ex+y+z −ex+h+z)(i+j+k)=O.
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15.5. STREAMLINES OF A VECTOR FIELD
13. Let F = fi + gj + hk. Then
∇ · (φF) = ∇ · (φfi + φgj + φhk)
387
Next,
i j ∇ × (φF) = ∂/∂x ∂/∂y
k ∂/∂z
φh
φf φg ∂∂
= ∂y(φh)−∂z(φg) i ∂∂
= ∂ (φf)+ ∂ (φg)+ ∂ (φh) ∂x ∂y ∂z
= φxf + φyg + φzh +φ(fx +gy +hz)
= ∇φ · F + φ(∇ · F).
+ ∂z(φf)−∂x(φh) j ∂∂
+ ∂x(φg)−∂y(φf k ∂φ ∂φ ∂φ
∂φ ∂φ = ∂yh−∂zg i+ ∂zf−∂xh j+ ∂xg−∂yf k
∂φ
∂h ∂g ∂f ∂h ∂g ∂f +φ ∂y−∂z i+ ∂z−∂x j+ ∂x−∂y k = ∇φ × F + φ(∇ × F).
15.5 Streamlines of a Vector Field
1. The streamlines satisfy
dx=−dy = dz. y2 z
y Next integrate dx = (1/z) dz to get
Integrate dx = −(1/y2) dy to obtain
x = 1 + c1.
x = ln |z| + c2.
Using x as the parameter, we can write equations of the streamline for
this vector field:
x=x,y= 1 ,z=ex−c2. x−c1
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388 CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS For the streamline through (2, 1, 1), let x = 2. Then
1 = 1 and 1 = e2−c2 . 2−c1
Thenc1 = 1 and c2 = 2, so this streamline has parametric equations x=x,y= 1 ,z=ex−2.
2. Streamlines satisfy
x−1
dx = −1 dy = dz
2
and two integrations give us
y = −2x + c1, z = x + c2.
For the streamline through (0, 1, 1), choose c1 = c2 = 1.
3. We have
x dx = dy = dz . ex −1
Integrate xex dx = dy to obtain
y = xex − ex + c1.
Integrate x dx = −dz to get
Using x as parameter, streamlines are given by
4. Streamlines satisfy
2 dx = dy ,dz=0.
x2 =−2z+c2.
y = xex − ex + c1 , z = 1 (c2 − x2 ).
2 For the streamline through (2, 0, 4), we need
e2 + c1 = 0 and 4 = 1 (c2 − 4). 2
Then c1 = −e2 and c2 = 12, so this streamline has parametric equations x = x, y = xex − ex − e2 , z = 1 (12 − x2 ).
cos(y) sin(x) Integrate sin(x) dx = cos(y) dy and dz = 0 to get
−cos(x)+c1 =sin(y)andz=c2.
To pass through (π/2, 0, −4) we need c1 = 0 and c2 = −4. This streamline,
with x as parameter, is
x = x, y = arcsin(− cos(x)), z = −4.
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15.5. STREAMLINES OF A VECTOR FIELD
5. Streamlines satisfy
dy =− dz . 2ez cos(y)
This is the separable equation
cos(y) dy = −2ez dz.
and c2 = 2 + 2/2. With y as parameter, this streamline is given by √2 1
x=3,y=y,z=ln 4 +1−2sin(y) . 6. Streamlines satisfy
dx =−dy=dz. 3x2 yz3
389
Integrate this to get
We also have x = c . For the streamline through 3, π/4, 0), we need c
√
sin(y)=c2 −2ez. 11
= 3
Integrate the equations
1 dx = −3 dy and 1 dy = − 1 dz
to obtain
x2 y y z3
1 = −3 ln |y| + c1 and 2 ln |y| + c2 = 1 .
x z2
For the streamline passing through (2, 1, 6), we need c1 = 1/2 and c2 =
1/36. With y as parameter, this streamline has parametric equations 26
x = 1 + 6 ln(y) , y = y, z = 1 + 72 ln(y) .
7. Circular streamlines about the origin in the x, y − −plane can be written
as x2 + y2 = r2, so Then
A vector field having these streamlines is F(x, y) = 1 i − 1 j.
xy
x dx + y dy = 0. dx = − dy , dz = 0.
yx
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390 CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS
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Chapter 16
Vector Integral Calculus
16.1 Line Integrals
1. OnC,x=t,y=tandz=t3,so 1
(t(1)−(1)+t3(3t2))dt (t − 1 + 3t5) dt = 0
2.
3.
=
(−4(−t2)(−2t)02 − 0) dt −8t3 dt = −2
xdx−dy+zdz = C0
1 0
1
−4x dx + y2 dy − yz dz = C0
1 0
22 (x+y)ds= (2t 1+1+4t dt
C0
=
2 2 1 23/22 262 = 2t2+4tdt=(2+4t)=
0
√
4. Parametric equations of C are
x = t, y = 1 + t, z = 1 − 2t for 0 ≤ t ≤ 1.
Then
1√ x2zds= t2(1−2t) 6dt
C0
√1 1 = 6 (t2−2t3)dt=−√.
06 391
603
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392 CHAPTER 16. VECTOR INTEGRAL CALCULUS 5.
3
F · dR = (cos(t)i + t2j + tk) · (i − 2tj + 0k) dt
C0
=
2√28√6 4xy ds = 4t2 6 dt = 3
3 81
6.
0
(cos(t)−2t3)dt=sin(3)− 2
C1
7. Parametrize C as x = 2cos(t),y = 2sin(t),z = 0 for 0 ≤ t ≤ 2π. Then
(2 cos(t)i + 2 sin(t)j) · (−2 sin(t)i + 2 cos(t)j) dt
(−4 cos(t) sin(t) + 4 cos(t) sin(t)) dt = 0.
8. ParametrizeC byx=1,y=t,z=t2 for0≤t≤2toget
2
yz ds = t(t2) 1 + 4t2 dt
C0
21√
2π C0
F · dR = =
2π 0
t3 1 + 4t2 dt = 120 (391 17 + 1).
The last integration can be carried out by an integration by parts.
9.
10.
=
0
9
√ −xyzdz= −z zdz
C4
2 5/29 422 = − z = −
C1 11. Parametrize the line segment as
3 xz dy =
545
t(−4t2) dt = −80
x = y = z = 1 + 3t for 0 ≤ t ≤ 1. 1
The work done is
F·dR= ((1+3t)2 −2(1+3t)2 +1+3t)(3)dt
C0
(1 + 3t)2 (1 + 3t)3 1 27
=2−3=−2. 0
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16.2.
12.
GREEN’S THEOREM 393 Parametrize the shape and location of the wire by x = y = z = t for
0≤t≤3. ThemassM is
3√27√3 M= δ(x,y,z)ds= 3t 3dt= 2 .
C0
Because the density function and the position of the wire are symmetric
in the first octant, we will have x = y = z, so we need only compute
2 x = √
xδ(x, y, z) ds √
t(3t) 3 dt = 2.
3 27 3 0
3 27 3 0
2 = √
13.
The centroid is (2, 2, 2).
TakeF(x)=f(x)iandR(t)=tjfora≤t≤b. Thegraphofthecurveis
defined by this position vector is the interval [a, b], and b
F · dR = f (x) dx. Ca
16.2 Green’s Theorem
1. The work done by F is
∂∂
2.
01
C
work = xydx+xdy= ∂x(x)−∂y(xy) dA CΩ
1 6x 4 8−2x
= (1−x)dydx+ (1−x)dydx 00 10
work = = = =
F·dR
(ex −y+xcosh(x))dx+(y3/2 +x)dy
=
1 4 6x(1 − x) dx +
(8 − 2x)(1 − x) dx = −8
C
∂ 3/2 ∂ x
D
D
∂x(y +x)−∂y(e −y+xcosh(x)) dA 2dA = 2(area of D) = 2(62)π = 72π
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394 3.
CHAPTER 16. VECTOR INTEGRAL CALCULUS
4.
5.
6(1 − x) dx = −12
∂ ∂
work = = = =
Fd ̇R C
(−cosh(4x4)+xy)dx+(e−y +x)dy ∂−y ∂ 4
C
∂x(e +x)− ∂y(−cosh(4x )+xy) dA 37
D
(1 − x) dA = D11
F·dR= ∂x(−x)−∂y(2y) dA CD
3 1
(1 − x) dy dx
(−3) dA = −3(area of D)
∂ ∂ 2
∂x(−2xy)−∂y(x) dA 6 (22−2y)/5
=
= −3(16π) = −48π
D
= = =
F·dR= ∂x(x−y)−∂y(x+y) dA CD
D
(−2y)dA = −2ydxdy D 1 (y+4)/5
6 2y
(3y−18)5 dy=−40
∂ ∂
6.
7.
=
F·dR= ∂x(8xy )dA= 8y dA. CDD
1
0 dA = 0
∂2 2
D
To evaluate this integral, change to polar coordinates x = r cos(θ), y = rsin(θ),with0≤θ≤2πand0≤r≤4. Weget
2π 4 D00
8r2 sin2(θ)r dr dθ
8y2 dA =
= sin2(θ) dθ
4 00
2π
8r3 dr = 512π.
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16.2.
8.
9.
10.
11.
GREEN’S THEOREM 395
∂ 3y ∂2
F·dR= CD
D
∂x(cos(2y)−e +4x)−∂y(x −y) dA 5 dA = 125
=
∂x ∂x
F·dR= CD
D
∂x(−e sin(y))−∂y(e cos(y)) dA (−ex sin(y) + ex sin(y)) dA = 0
=
∂2∂2
F·dR= CD
∂x(−xy)−∂y(xy) dA π/2 2
= = 2
(−y2 −x2)dA= D00
(−r2)rdrdθ
π2 0
−r3 dr = −2π
̇ ∂ 2 cos(y) ∂
FdR= CD
∂x(xy −e
)−∂y(xy) dA 3 5−5x/3
=
(y2 −x)dydx 31 5x 3 5x
(y2 −x)dA= D00
= 3 5− 3 dx− x 5− 3 ],dx 00
12.
= 95 4
(a) By Green’s theorem with F = −yi,
∂∂
−ydx= ∂x(0)−∂y(−y) dA= CDD
(b) Now apply Green’s theorem with F = xj to get ∂∂
F·dR= ∂x(x)−∂y(0) dA= CDD
(c) Add the results of parts (a) and (b).
dA= areaofD.
dA= areaofD.
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396 CHAPTER 16. VECTOR INTEGRAL CALCULUS 13. By Green’s theorem,
∂ ∂ ∂∂u∂∂u −∂ydx+∂xdy= ∂x ∂x −∂y −∂y dA
CD
∂g
∂x dA = D
c α(y) d
=
∂2u ∂2u
∂x2 + ∂y2 dA.
D
14. Assume that C is the join of two curves in two ways. First, C has an upper piece, the graph of y = p(x), and a lower piece, the graph of y = q(x), for a ≤ x ≤ b. Then D consists of the points (x, y) with
a ≤ x ≤ b, q(x) ≤ y ≤ p(x).
And C also has a left piece, the graph of x = α(y), and a right piece, the graph of x = β(y), for c ≤ y ≤ d. Then D also consists of the points (x,y) such that
c ≤ y ≤ d, α(y) ≤ x ≤ β(y).
Now use both of these descriptions of D as follows. Using the second (looking at D from left to right),
dc
g(x, y) dy = g(β(y), y) dy + g(α(y), y) dy.
Ccd
Note that, on the right part of C, we take y varying from d to c to maintain a counterclockwise orientation around C. Further,
dβ(y)∂g
∂y dx dy
g(x, y) dy = ∂x dA. CD
This is “half” of the conclusion of Green’s theorem. For the rest, use the first description of C. Now, looking from the lower to the upper piece of C, and keeping in mind the counterclockwise orientation on C, we have
=
Then ∂
and
(f (x, p(x)) − f (x, q(x))) dx bp(x)∂f
∂y dA = D
∂y dA
(f (x, p(x)) − f (x, q(x))) dx.
c
(g(β(y), y) − g(α(y), y)) dy.
ab
f (x, y) dx = f (x, p(x)) dx + f (x, q(x)) dx Cba
∂f
= −
b a
=
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a q(x) b
a
16.2.
GREEN’S THEOREM
397
Then
Upon adding these two results, we have
∂f
f(x,y)dx = − CD
∂x−∂y dA.
If C does not enclose the origin, then Green’s theorem applies and
F·dR C
f(x,y)dx+g(x,y)dy= CD
∂y dA.
∂g ∂f
15.
= =
∂x ∂−y 2
∂x x2+y2−2y −∂y x2+y2+x dA 0 dA = 0.
D
D
If C does enclose the origin, let K be a circle about the origin of sufficiently small radius r that K is in the region enclosed by C. Then, using the extended Green’s theorem and polar coordinates, we have
F · dR = F · dR CK
2π −rsin(θ)
2 2 +r cos (θ) (−rsin(θ)) dθ
=
+
=
= θ +
If C does not enclose the origin, then
r2
2π rcos(θ)
0
0
2π
r2
− 2r sin(θ) (r cos(θ)) dθ
(1 − r2 cos2(θ) sin(θ) − 2r2 sin(θ) cos(θ)) dθ
0
r3 2π
cos2(θ) − r2 sin2(θ) = 2π. 30
16.
∂ x 2 ∂ −y
∂x x2+y2 −y −∂y x2+y2 +3x dA=0, because all terms in the integrand cancel. If C does enclose the origin, let
K be a circle about the origin and entirely in the region enclosed by C.
F·dR= CD
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398 CHAPTER 16. VECTOR INTEGRAL CALCULUS Then
F · dR = F · dR CK
ory
1. First observe that
∂(y3)= ∂(3xy2−4) ∂y ∂x
2π −rsin(θ)
+ 3r cos(θ)
F · dR = 0 C
if C does not enclose the origin, and
F · dR = F · dR = 0 CK
if C does enclose the origin, and K is a circle about the origin entirely in the region enclosed by C.
18. If C does not enclose the origin, then by Green’s theorem we get
F · dR = 0. C
If C encloses the origin, let K be a circle of radius r about the origin and entirely in the region enclosed by C. Using polar coordinates for K, a straightforward calculation yields
F · dR = F · dR = 0. CK
19. By arguing as in the last four problems, obtain
F · dR = 0 C
whether or not C encloses the origin.
16.3 Independence of Path and Potential The-
=
+
=
17. By a calculation like those of Problems 15 and 16, obtain
0
− r sin(θ)
(1 − 4r2 sin(θ) cos(θ)) dθ = 2π.
(−r sin(θ)) dθ (r cos(θ)) dθ
r2
2π rcos(θ)
0
2π
0
r2
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16.3.
INDEPENDENCE OF PATH AND POTENTIAL THEORY 399 on the entire plane, so this vector function is conservative. To find a
potential function, we can begin with
∂φ = y3 ∂x
and integrate with respect to x to get
φ(x, y) = xy3 + k(y).
Then we must have
∂φ = 3xy2 +k′(y) = 3xy2 −4 ∂y
to conclude that k′(y) = −4, so we can choose k(y) = −4y. Then φ(x, y) = xy3 − 4y
is a potential function for F. First,
∂ (6y+yexy)=6+exy +xyexy = ∂ (6x+xexy) ∂y ∂x
for all (x,y). Then F has a potential function defined for all (x,y). To find a potential function, we can begin with
∂φ = 6y + yexy ∂x
and integrate with respect to x to get φ(x,y)=6xy+exy +k(y).
2.
Then we must have
∂φ =6y+xexy +k′(y)=6y+yexy.
∂y
Then k′(y) = 0 and we can choose k(y) to be any constant, say k(y) = 0
for convenience, to get
φ(x, y) = 6xy + exy . 3. F is conservative over the entire plane because
∂ (16x)= ∂ (2−y2)=0 ∂y ∂x
for all (x, y). To find a potential function, we can begin with ∂φ = 16x
∂x
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400
CHAPTER 16. VECTOR INTEGRAL CALCULUS
and integrate with respect to y to get
φ(x, y) = 8x2 + k(y).
Then
∂φ =k′(y)=2−y2 ∂y
4.
so k(y) = 2y − y3/3 and
φ(x, y) = 8x2 + 2y − 1 y3
3
is a potential function. First,
∂ (2xy cos(x2)) = 2x cos(x2) = ∂ (sin(x2)) ∂y ∂x
for all (x,y). This vector field is conservative. For a potential function, we can begin with
to conclude that Then we need
∂φ = sin(x2) ∂y
φ(x, y) = y sin(x2) + k(x).
∂φ = 2x cos(x2) = 2xy cos(x2) + k′(x).
5.
φ(x, y) = y sin(x2). First, if (x, y) ̸= (0, 0), then
∂ 2x 4xy ∂ 2y ∂y x2+y2 =−(x2+y2)2 =∂y x2+y2 .
Then F is conservative on the plane with the origin removed. For a po- tential function, we can begin with
∂φ= 2x ∂x x2 +y2
and integrate with respect to x to get
φ(x, y) = ln(x2 + y2) + k(y).
∂x
Then k′ = 0 and we can choose k(y) = 0 to get the potential function
Then we need
∂φ = 2y = 2y +k′(y). ∂y x2 +y2 x2 +y2
We can choose k(y) = 0 to obtain the potential function φ(x,y)=ln(x2 +y2).
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16.3. INDEPENDENCE OF PATH AND POTENTIAL THEORY 401
6. In this simple case, we can see a potential function by inspection:
φ(x,y,z)=x2 −y2 +z2
for all (x, y, z).
7. By inspection,
φ(x, y, z) = x − 2y + z is a potential function for F, for all (x, y, z).
8. A routine calculation shows that ∇ × F = O for all (x,y,z), so F is conservative. A potential function must satisfy
∂φ = yzcos(x), ∂φ = zsin(x)+1, ∂φ = ysin(x). ∂x ∂y ∂z
Integrate the first of these equations with respect to x to get φ(x, y) = yz sin(x) + k(y, z).
Next, we need
Then
∂φ = z sin(x) + 1 = z sin(x) + ∂k .
∂y
∂y
∂k = 1 ∂y
Integrate this with respect to y to get
k(y, z) = y + c(z).
So far we have
Finally, using the last equation, we have
∂φ = y sin(x) + c′(z). ∂z
Then c′(z) = 0, so c(z) can be any constant. For convenience, choose c(z) = 0. Then
φ(x, y, z) = yz sin(x) + y is a potential function for F.
9. We find that ∇ × F ̸= O, so this vector field is not conservative.
10. ∇ × F ̸= O, so F is not conservative.
In Problems 11–20 we provide a potential function to use in evaluating the line integral, but do not include the details of finding this potential function.
φ(x, y, z) = yz sin(x) + y + c(z).
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402 CHAPTER 16. VECTOR INTEGRAL CALCULUS 11. By integrating, we find a potential function
Then
φ(x,y)=x3(y2 −4y).
F · dR = φ(2, 3) − φ(1, 1) = −24 − 3 = −27.
C
12. φ(x, y) = ex cos(y) and
e2 F·dR=φ(2,π/4)−φ(0,0)= √ −1.
C2
13. In any region not containing points of the y− axis, we can use the potential
function
φ(x, y) = x2y − ln |y|.
If C does not cross the x− axis, then
F · dR = φ(2, 2) − φ(1, 3) C
= 8−ln(2)−3+ln(3) = 5+ln(3/2).
14. With φ(x, y) = x + 3y2 − cos(y), we have
F · dR = φ(1, 3) − φ(0, 0) = 29 − cos(3). C
15. φ(x,y)=x3y2−6xy3,so
F · dR = φ(1, 1) − φ(0, 0) = −5. C
16. φ(x, y, z) = xy cos(xz), so
F · dR = φ(1, 1, 7) − φ(1, 0, π) = cos(7). C
17. φ(x,y,z)=x−3y3z,so
F · dR = φ(0, 3, 5) − φ(1, 1, 1) = −403. C
18. φ(x, y, z) = −8xy2 − 4zy, so
F · dR = φ(1, 3, 2) − φ(−2, 1, 1) = −108. C
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16.3. INDEPENDENCE OF PATH AND POTENTIAL THEORY 403
19. φ(x,y,z) = 2x3eyz, so
F · dR = φ(2, 1, −1) − φ(0, 0, 0) = 2e−2. C
20. φ(x,y,z)=xy−2x2z+z3,so
F · dR = φ(3, 1, 4) − φ(1, 1, 1) = −5. C
21. Let C be a smooth path of motion having position vector R(t) = x(t)i + y(t)j + z(t)k. Let L(t) be the sum of the potential and kinetic energies. Then
L(t) = m ∥ R′(t) ∥ −φ((xt), y(t), z(t)) 2
= mR′(t)·R′(t)−φ(x(t),y(t),z(t)). 2
Then
L′(t) = m(2R′′(t) · R′(t)) − ∂φx′(t) − ∂φy′(t) − ∂φz′(t)
2 ∂x ∂y ∂z = (mR′′(t) · R′(t) − ∇φ · R′(t)
= (mR′′(t) − ∇φ) · R′(t).
Now, ∇φ is the force acting on the particle, so by Newton’s second law, mR′′ = ∇φ.
Therefore L′(t) = 0.
22. Wewanttoshowthat,intheplane,apotentialfunctionexistsforF(x,y)=
f(x,y)i+g(x,y)j if
∂g = ∂f . ∂x ∂y
We will use this condition to construct a potential function for F (on the relevant region D). First observe that, if K is a closed path in D, then by Green’s theorem,
∂g ∂f
∂x−∂y dA=0,
where M is the region enclosed by K. This means that F · dR is
C
independent of path in D. Fix a point P0 : (a, b) in D. Then, for any P in D, define
C
F·dR= KM
φ(x, y) =
F · dR
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404
CHAPTER 16. VECTOR INTEGRAL CALCULUS
inwhichCisanysmoothpathinDfromP0 toP. Weclaimthat∇φ=F. To show this, we will first show that
∂φ =f(x,y). ∂x
Choose ∆x small enough that (x + ∆x, y) is also in D. Now φ(x+∆x,y)−φ(x,y)
= = = =
(x+∆x,y) P0
(x,y) P0
(x,y) F·dR− F·dR
P0
(x+∆x,y) (x,y)
(x+∆x,y) (x,y)
(x+∆x,y) (x,y)
F·dR
f (ξ, η) dξ + g(ξ, η) dη.
F·dR+ F·dR− F·dR (x,y) P0
integral is over the horizontal segment from (x, y) to (x + ξ = x + t∆x, η = y for 0 ≤ t ≤ 1.
On this segment, dξ = (∆x) dt and dη = 0, because y is constant on this segment. Then
The last line
∆x, y), which can be parametrized by
Then
φ(x + ∆x, y) − φ(x, y) = ∆x
f (x + t∆x, y) dt.
φ(x+∆x,y)−φ(x,y) ∆x
1 0
1
= f(x+t∆x,y)dt.
0
By the mean value theorem for integrals, there is some t0 in (0, 1) such
that
1
f(x + t∆x, y) dt = f(x + t0∆x, y).
0
Then
Inthelimitas∆x→0,wemusthavet0 →0,sox+t0∆x→xand
φ(x+∆x,y)−φ(x,y) =f(x+t0∆x,y). ∆x
f(x+t0∆x,y)→f(x,y). Then
∂φ = lim φ(x+∆x,y)−φ(x,y) ∂x ∆x→0 ∆x
= lim f(x+t0∆x,y)=f(x,y). ∆x→0
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16.4. SURFACE INTEGRALS 405 A similar argument, using a vertical segment from (x, y) to (x, y + ∆y),
shows that
∂φ =g(x,y). ∂y
16.4 Surface Integrals
1. On the surface, z = 10−x−4y, so
dσ = 1 + (∂z/∂x)2 + (∂z/∂y)2 dA = 3√2 dA.
Then
xdσ= ΣD√
√ 5/210−4y 3 25/2
=3 2 xdxdy= 2 (10−4y)2dy
√00 0 2 35/2 √
= (10−4y) =1252. 80
2. On the surface, z = x, so
dσ = 1 + 12 + 02 dA = √2 dA
√
3 2xdA
and
3. On Σ,
and D is the annulus 2 ≤ x2 + y2 ≤ 7. Then, using polar coordinates,
√
2π7 2
dσ= √ r 1+4rrdrdθ
Σ02
1 23/27π3/2
2 √2
y dσ= 2y dA ΣD√
√ 2 4 128 2
y2dxdy= 3 . dσ = 1+4x2 +4y2 dA,
= 2
00
√
=2π 12(1+4r ) √ = 6(29 −27). 2
4. On the surface, z = (25 − 4x − 8y)/10, so 3
dσ = 1+(2/5)2 +(4/5)2 dA = √
dA.
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5
406 CHAPTER 16. Then
VECTOR INTEGRAL CALCULUS
3 (x+y)√ dA
(x+y)dσ= ΣD5
31x
=√ (x+y)dydx
=√5 0 2xdx=2 5. 5. On the surface, z2 = x2 + y2, so
500
3132 3√
Then
Then
Then
6. OnΣ,dσ=
2z∂z =2xand2z∂z =2y. ∂x ∂y
∂z = x and ∂z = y. ∂x z ∂y z
x2 y2 √ dσ= 1+ z + z dA= 2dA.
√22 zdσ= 2 x +y dA
ΣD
√ π/24
= 2
02
3dAandz=x+y,so √
28π√ r2drdθ= 3 2.
√
xyzdσ = 3 xy(x+y)dA ΣD
√11 1 = 3 (x2y+xy2)dydx=√.
003
√
7. On the surface, dσ = 1+4x2 dA, so 2
ydσ= y 1+4xdA ΣD
23 2 92 2
= y 1+4xdydx=2 1+4xdx
000
= 9 ln(4 + √17 + 4√17). 8
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16.4. SURFACE INTEGRALS 407 8. Onthesurface,dσ=1+4(x2+y2)dA,so
=
For the θ−integration, use the identity
cos2(θ) = 1(1 + cos(2θ)) 2
and for the r−integration, use the substitution u = 1 + 4r2.
These give us
x2 dσ = ΣD
x21+4(x2 +y2)dA
=
2π 2 00
r2 cos2(θ) 1 + 4r2r dr dθ
2π 00
cos2(θ) dθ
2
r3 1 + 4r2 dr.
1 2π 1 17 x2 dσ = 2 (1+cos(2θ)dθ 32
Σ0√1 = π (782 17+2).
240
√ 3dAandz=x−y,so
(u3/2 −u1/2)du
9. OnΣ,dσ=
z dσ = ΣD
√
3(x − y) dA
√15 √
= 3
(x−y)dydx=−10 3.
10. OnΣ,dσ=1+4y2dAandz=1+y2,so
22
xyzdσ= xy(1+y) 1+4ydA ΣD
00
= = 2
xy(1+y2)1+4y2 dydx y(1+y2)1+4y2 dy.
11 00
11 0
For this integral, let u = 1 + 4y2 to get
115 1√ 3
xyzdσ=232 (u3/2+3u1/2)du=16 5 5−5 . Σ1
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408 CHAPTER 16. VECTOR INTEGRAL CALCULUS 16.5 Applications of Surface Integrals
1. The triangular shell is on the plane 6x + 2y + 3z = 6, which is the plane through the three given points. The projection of Σ onto the x,y−plane isthesetDofpoints(x,y)suchthat0≤y≤3−2x. OnΣ,
Then
and
z = 2 − 2 y − 2x. 3
4
dσ= 1+9+4dA
m = =
(xz + 1) dσ 2
Σ
x 2−3y−2x +1 dA 713−3x 2
D
=3
The first coordinate of the center of mass is
49 x 2−3y−2x +1 dydx=12.
x= 49
Σ
00
12
x(xz+1)dσ 1272
x x 2−3y−2x +1 dydx 413−3x2
=493
=7 x x 2−3y−2x +1 dx
D
00
1 24 2 12 4 12 12
−7x+7x+7x dx=35. The second coordinate is
=
12
0
y = 49
=7 y y 2−3y−2x +1 dydx
y(xz + 1) dσ 413−3x2
Σ
00
1 24 18 2 36 3 12 4 18
= −7x−7x+7x−7x+7 dx 0
= 33. 35
And, without all the details, the third coordinate is
12 z = 49
Σ
24 z(xz + 1) dσ = 35 .
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16.5. APPLICATIONS OF SURFACE INTEGRALS 409 2. On Σ,
dσ = 1+(x/z)2 +(y/z)2 dA = 3 dA. z
The part of the sphere above the plane z = 1 projects onto the x, y−−plane √
onto the disk D of radius 2 2 about the origin. The mass of the shell is
m=
Use polar coordinates:
3 1
KzdA=3K 2 2 dA.
Σ
D
9−(x +y )
dr dθ
√
2π22 r m = 3K √
By symmetry, x = y = 0. Finally,
0 0 9−r2 √
222 =6πK − 9−r
=12Kπ.
=3K(areaofD)= 1(24Kπ)=2. mm
The center of mass is (0, 0, 2). 3. On the surface,
0
1 z = m
Kz dσ
Then
Σ
3 =m KzzdA
D
1
dσ=1+(x/z)2 +(y/z)2dA=√2dA.
mass =
√ 2π3 Kdσ=K 2
ΣD00
√
rdrdθ=9πK 2.
By symmetry, x = y = 0, and 1
z = m z dσ
√Σ√ 2K2π3 18 2Kπ
= m
The center of mass is (0, 0, 2).
4. On the surface,
r2 dr dθ = m = 2. dσ = 1 + 4(x2 + y2) dA
00
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410
CHAPTER 16. VECTOR INTEGRAL CALCULUS
and Σ projects onto the x, y − −plane to the quarter annulus D consisting of points (x, y) with
The mass is
x ≥ 0, y ≥ 0 and 1 ≤ x2 + y2 ≤ 9.
xy
m= 2 2dσ
= =
xy dA = D00
1+4(x +y ) π/2 3
r3 sin(θ) cos(θ) dr dθ 1 2 π/21 43
sin (θ) r 2041
Σ
= 10.
Because the region is symmetric about the line y = x, and
δ(x,y,z) = δ(r,θ) = √ 1 1+4r2
is independent of θ, we must have x = y. Compute
1 x = m
1 = 10
Σ
xδ(x, y, z) dσ
1 π/2 3
x2ydA = 10 D01
121 r4 cos2(θ)sin(θ)drdθ = 75 .
Finally,
1 122
z = m zδ(x,y,z)dσ = 10 (16−x −y )xydA
ΣD
1 π/2 3
331 (16−r2)r3 sin(θ)cos(θ)drdθ = 48 .
= 10
The center of mass is (121/75, 121/75, 331/48).
By symmetry of the surface and the density function, x = y = 0. Further,
5.
01
dσ = 1+4x2 +4y2 dA.
Then
2 2 1+4x +4y dσ
m= =
= (1+4r2)rdrdθ 00
= 2π(39) = 78π.
Σ
(1−4x2 +4y2)dA √
D 2π 6
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16.5. APPLICATIONS OF SURFACE INTEGRALS 411 Finally,
1 x = m
zδ(x, y, z) dσ
1 2 2 2 2
= m
12π 6
Σ
(6−x −y )(1+4x +4y )dA √
(6−r2)(1+4r2)rdrdθ
D
= m
= 162π = 27.
00 m 13
6. By symmetry, x = y = 0. On the surface,
dσ = 1+(x/z)2 +(y/z)2 dA = 1 dA.
The mass is m =
z
K 1 1
Kzdσ = Kπ z(1/z)dA = πarea of D = 4. ΣD
7. Aunitnormaltotheplanex+2y+z=8is
1
n= √ (i+2j+k).
6
1
F·n= √ (x+2y−z).
6
Further,dσ= 1+4+1dA= 6dA,sothefluxofFacrossthesurface is
Then
On Σ, z = 8 − x − 2y, so √√
F · n = 2x + 4y − 8.
4 8−2y F·n dσ = (2x+4y−8) dA =
ΣD00 8. Thespherex2+y2+z2 =4hasunitnormal
(2x+4y−8) dx dy =
128 3 .
Then Further,
n= 1(xi+yj+zk). 2
F·n= 1(x2z−yz). 2
dσ = 1 + (−x/z)2 + (−y/z)2 dA.
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412
CHAPTER 16. VECTOR INTEGRAL CALCULUS
Therefore the flux of the force across Σ is
F · n dσ = ΣD
(x2 − y) dA.
√ UsepolarcoordinatesonD,whichisthedisk0≤r≤ 3,0≤θ≤2π.
Then the flux is
√
2π 3 92π (r2 cos2(θ)−rsin(θ))rdrdθ = 4
000
9π cos2(θ)dθ = 4 .
16.6 Gauss’s Divergence Theorem
1. ∇·F=1,so
Σ
M
F · n dσ =
= volume of V = 4π43 = 256π.
33
∇ · F dV
2. ∇·F=4−6=−2,so
3. ∇·F=0,so
∇·FdV =−2(volumeofM)=−2π(22)(2)=−16π. M
4. ∇ · F = 3(x2 + y2 + z2), so
(x2+y2+z2)dV.
∇·FdV =3 MM
Using spherical coordinates, we have
M
∇·FdV = =
2π π 1
000 2π π
3ρ4 sin(φ)dρdφdθ 1
∇·FdV =0. M
sin(φ) dφ 000
3ρ4 dρ
dθ
3 12π =(2π)(2) 5 = 5 .
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16.6. GAUSS’S DIVERGENCE THEOREM 413
5. With∇·F=4,compute
8π ∇·FdV=4(volumeof)V= 3.
M
6. ∇·F=2+x,so
324 4
∇ · F = (2 + x) dx dy dz = 6 000 0
(2 + x) dx = 96. 7. ∇ · F = 2(x + y + z), so, using cylindrical coordinates, we have
M
√√ 2π 2 2
∇·FdV = 2
00r
Do these integrations in turn. First,
√
(rcos(θ)+rsin(θ)+z)rdzdrdθ.
M
2 √1
(r2(cos(θ)+sin(θ))+rz)dz = r2(cos(θ)+sin(θ))( 2−r)+2r(2−r2). r
Next,
√
22√1211
r (cos(θ) + sin(θ))( 2 − r) + 2r(2 − r ) and finally,
dr = 3(cos(θ)+sin(θ))+2, dθ = π.
0
Therefore
2π 1 1 0 3(cos(θ) + sin(θ)) + 2
8. ∇·F=1+2x,socompute
∇·FdV =2π. M
∇·FdV = MM
(1+2x)dV 2
= dz (1+2x)dA, 0D
where D is the plane region given in polar coordinates by 0 ≤ r ≤ 0 ≤ θ ≤ 2π. Now compute
√
2,
M
(1+2x)dV = 2
(1+2rcos(θ))rdrdθ 4√2 2π
D
=2 2π+ 3
cos(θ)dθ =4π.
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0
414 CHAPTER 16. VECTOR INTEGRAL CALCULUS
9. With the given conditions on F, Σ and M, we have
Σ
M
because ∇ · (∇ × F) = 0.
10. Apply the divergence theorem to get
1 3
1 F · n dσ = 3
M
1 = 3
(∇×F)·ndσ=
∇·(∇×F)dV =0
(∇ · R) dV
3dV = volume of M.
1. The surface is a function of θ and φ, and (θ, φ) varies over its parameter domain0≤θ≤2π,0≤φ≤π. Thisisarectangleintheθ,φ−plane, with lower side L1, upper right side L2, top L3 and left side L4. For orientation, imagine (θ, φ) moves around this rectangle counterclockwise, starting along L1 from the origin. We want to know what each side maps to on the surface.
On L1, the point (θ, 0) moves from (0, 0) to (2π, 0) as θ increases from 0 to 2π. The image point
Σ(θ, 0) = (R cos(θ), R sin(θ), 0)
moves from (R,0,0) along the circle x2 + y2 = R2 in the plane z = 0, all
the way around to end at (R, 0, 0).
Then the point (2π,φ) moves up along L2 as φ increases from 0 to π.
Image points of points (2π, φ) on L2 are
Σ(2π, φ) = (R cos(φ), 0, R sin(φ))
which is a half-circle x2 + z2 = R2 in the y = 0 plane, starting at (R, 0, 0) and ending at (−R, 0, 0).
From (2π,π), (θ,φ) now moves left along L3. The points are (θ,π), but θ varies from 2π to 0 to maintain counterclockwise orientation on the rectangle. The image points of L3 are
Σ(θ, π) = (−R cos(θ), −R sin(θ), 0)
asθ varies from2π to 0. The image of L3 on the surface consists of the
points
Σ(θ, π) = (−R cos(θ), −R sin(θ), 0),
Σ
16.7 Stokes’s Theorem
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M
16.7.
STOKES’S THEOREM 415
and this point moves along the half-circle x2 + y2 = R2
from (−R,0,0) to (−R,0,0) in the z = 0 plane.
Finally, on L4, θ = 0 and φ varies from π to 0. Image points are
Σ(0, φ) = (R cos(φ), 0, R sin(φ)) from (−R, 0, 0) to (R, 0, 0).
Now trace out the image point on the surface as (θ,φ) moves over all four sides of the rectangle. This curve on the graph of the surface is the boundary of Σ.
The boundary curve C of the surface can be parametrized by x = 2 cos(t), y = 2 sin(t), z = 0 for 0 ≤ t ≤ 2π.
On C, R(t) = 2 cos(t)i + 2 sin(t)j + 0k. Then
F · dR = (−16 cos2(t) sin2(t) − 16 cos2(t) sin2(t)) dt = −32 cos2(t) sin2(t).
2.
Then
2π C0
−32 cos2(t) sin2(t) dt = −8π.
F dR =
Now evaluate (∇×F)·ndσ. First,
Σ
∇×F=−(x2 +y2)k. Further, we can use the gradient to find a normal to Σ:
∇(x2 +y2 +z2) = 2(xi+yj+zk). This gives us the normal vector
Then
Then
n= 1(xi+yj+zk). 2
dσ = 1+(x/z)2 +(y/z)2 dA = 2 dA. z
r3 dr dθ = −8π.
00
(x2 + y2) dA 2π 2
(∇ × F) · n dσ = − ΣD
= −
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416 CHAPTER 16. VECTOR INTEGRAL CALCULUS In Problems 3–8, one side of Stokes’s theorem is computed in detail, with the
choice being determined by which side appears to be the easiest computation. 3. The boundary curve C of the surface is the top of the parabolic bowl.
This is the circle of radius 3 about (0, 0, 9). Parametrize C by x = 3 cos(t), y = 3 sin(t), z = 9 for 0 ≤ t ≤ 2π.
On C,
Further,
Then
F(t) = 9 cos(t) sin(t)i + 27 sin(t)j + 27 cos(t)k. dR = (−3 sin(t)i + 3 cos(t)j) dt.
F · dR = (−27 cos(t) sin2 (t) + 81 cos(t) sin(t)) dt.
A routine integration gives
(−27 cos(r) sin2(t) + 81 cos(t) sin(t)) dt = 0.
2π C0
F · dR =
Evaluation of (∇ × F) dσ involves considerably more labor.
Σ
4. Compute∇×F=i+j+k. AunitnormaltoΣis 1
n= x2 +y2 +z2(xi+yj+zk) 1
dσ= 1+x2+y2+x2+y2dA= 2dA.
(∇·F)dσ=
Σ Dx+y
This is
Further,
Then
Here we used the fact that z = x2 + y2 on σ. This integral is easily evaluated using polar coordinates to get
n= √2x2 +y2(xi+yj−2k). x2 y2 √
x+y−z
2 2 dA
x+y Dx+y
= 2 2−1dA.
(∇ × F) dσ = −16π. Σ
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16.7. STOKES’S THEOREM 417
For the line integral around the boundary C of Σ, parametrize C by x = 4 sin(t), y = 4 cos(t), z = 4 for 0 ≤ t ≤ 2π.
This orientation is determined by n. Notice that the normal in this prob- lem has negative k component, unlike the normal in the preceding prob- lem, where the surface was a parabolic bow. Parametrize C by
and
Finally, dσ =
1 n= √
(2i+4j+k).
(−16 cos(t) sin(t) − 16 cos2(t)) dt = −16π.
2π C0
F · dR =
5. The boundary curve of Σ is the circle x2 + y2 = 6 in the x,y− plane. Parametrize C by
√√
x= 6cos(t),y= 6sin(t),z=0for0≤t≤2π.
Then
2π C0
6 cos2(t)6 sin2(t) dt = 0.
F · dR =
6. The boundary C of the surface must be parametrized by three smooth curves. This is not difficult, but it is tedious. This leads us to try the surface integral side of Stokes’s theorem. First,
∇ × F = (x − y)i − yj − xk
√
21 dA, so
C
F · dR. Take Σ to be the disk 0 ≤ x2 + y2 ≤ 1
=
(x − 6y) dx dy = − 3 .
21
(∇×F)dσ = ΣD
(x−6y)dA
2 4−2y 32
00
7. The circulation is
with boundary C parametrized by
x = cos(t), y = sin(t), z = 0 for 0 ≤ t ≤ 2π.
The proper unit normal to Σ (a disk in the x, y− plane) is n = k. Now, ∇ × F = −azj + (2xy + 1)k.
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418
CHAPTER 16. VECTOR INTEGRAL CALCULUS
8.
First,
A normal to the surface is
Then
Further, dσ = dA, so
F · dR = CΣ
(∇ × F) · n = 2xy + 1.
=
(2r3 cos(θ) sin(θ) + r) dr dθ = π. ∇ × F = −i − j − k.
n = i + 4j + k
2π 1 D00
(∇ × F) · n dσ (2xy + 1) dA =
so we have the unit normal
n= √ (i+4j+k).
18
C is the boundary of the part of the plane x+4y+z = 12 lying in the first octant, and consists of three line segments: the line from (0,0,12) to (12,0,0), then from (12,0,0) to (0,3,0), and then from (0,3,0) to (0, 0, 12). We can think of this portion of the plane in the first octant as having equation
z = 12 − x − 4y
with (x, y) varying over the triangle D bounded by the segment [0, 12] on the x− axis, the segment [0, 3] on the y− axis, and the line x + 4y = 12.
1
D has area
By Stokes’s theorem the circulation is
area of D = 1(12)(3) = 18. 2
Now
And Then
6 (∇×F)·n=−√ .
18
√ dσ=∥N∥dxdy= 18.
F · dT ds = CD
(∇ × F) · n dσ.
6
(∇×F)·ndσ =
D D18
−√ dxdy
= −6(area of D) = −6(18) = −108.
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Chapter 17 Fourier Series
17.1 Fourier Series on [−L,L] 1. The Fourier coefficients are
13
a0 = 3 13
4 dx = 8,
4 cos(nπξ/3) dξ = 0,
an = 3 13
−3
and
−3
4 sin(nπξ/3) dξ = 0. The Fourier series of 4 on [−3, 3] is just
1a0 2
or 4, as we might expect. This converges to 4 on [−3, 3].
2. The Fourier coefficients an are zero for n = 0,1,2,··· because f is an odd
function on [−1, 1]. Next,
bn = 3
−3
bn =
The Fourier series of f(x) on [−1,1] is
12n
f(ξ)sin(nπξ)dξ = nπ(−1) . 2 ∞ 1 (−1)n sin(nπx).
−1
This series converges to
πn
n=1
−x for−1
points are the points on the parabola, which are the points x + i x for x ≥ 0. W does not contain any of its boundary points, and is open. W is not closed.
34. One way to envision R is to think of certain points on vertical segments of length 1. Begin with n = 1. The points
1+1i m
are the points 1+i, 1+(1/2)i, 1+(1/3)i,···, going down the line x = 1 and approaching arbitrarily lose to (but not reaching) 1 on the horizontal axis. When n = 2, we have points
1 + 1 i, 2m
moving down from x = 1/2 and approaching arbitrarily close to x = 1/2 as m is chosen larger. And so on – as n increases, these points remain on parallel horizontal lines, but move down vertical segments closer to the imaginary axis.
The boundary points are all points (1/m)i and all points 1/n for positive integer n and m. None of these boundary points belong to R, so R is not closed. But R is open, because every point of R is an interior point.
19.2 Complex Functions
1.
f(z) = z − i = x + iy − i = x + (y − 1)i
√
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456
CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS
so u(x, y) = x and v(x, y) = y − 1. The Cauchy-Riemann equations for
this function are
and
∂u =1= ∂v ∂x ∂y
∂u =0=−∂v. ∂y ∂x
Because u and v are continuous with continuous first partial derivatives, and the Cauchy-Riemann equations are satisfied for all z, f(z) is differen- tiable for all z.
2.
f(z)=z2 −iz=x2 −y2 +2ixy−ix+y=x2 −y2 +y+(2xy−x)i.
Let u(x, y) = x2 − y2 + y and v(x, y) = 2xy − x. Then ∂u = 2x = ∂v
and
∂x ∂y
∂v = −2y + 1 = ∂u ∂x ∂y
3.
so the Cauchy-Riemann equations are satisfied at every point. Because u and v are continuous with continuous partial derivatives at all points, f(z) is differentiable for all z.
f(z) = |x + iy| = x2 + y2, so
u(x,y)=x2+y2 andv(x,y)=0.
If z ̸= 0, then
and
∂u x ∂u y
∂x = x2 +y2,∂y = x2 +y2
∂v = ∂v = 0. ∂x ∂y
The Cauchy-Riemann equations are not satisfied at any nonzero z. To check what happens at z = 0, compute
∂u(0,0)= lim u(h,0)−u(0,0)
∂x
h→0 √ h
h2 h→0 h
= lim
= lim |h|.
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h→0 h
19.2.
COMPLEX FUNCTIONS 457 This limit does not exist, because
|h| 1 if h > 0, h = −1 ifh<0.
Similarly, (∂u/∂y)(0,0) does not exist. Therefore the Cauchy-Riemann equations are not satisfied at any point, including the origin, and f(z) is not differential for any z.
4. f(z) is defined for all nonzero z. For z ̸= 0,
f(z) = 2z + 1 = 2 + 1.
Then
Then
zz
f(z) = 2 + 1 x+iy
=2+ x−iy x2 + y2
x2 y =2+x2+y2 −x2+y2i.
x2 y u(x,y)=2+x2+y2 andv(x,y)=−x2+y2.
For z ̸= 0 compute the partial derivatives
and
∂u y2−x2 ∂u ∂x = (x2 +y2)2, ∂y
∂v 2xy ∂v ∂x = (x2 +y2)2,∂y
−2xy
= (x2 +y2)2,
y2 − x2
= (x2 +y2)2.
The Cauchy-Riemann equations hold for all nonzero z. Because u,v and its partial derivatives are continuous for all nonzero z, f(z) is differentiable for all z ̸= 0.
5. f(z) = i|z|2 = (x2 + y2)i, so
u(x, y) = 0 and v(x, y) = x2 + y2.
Then
and
∂u = ∂u = 0 ∂x ∂y
∂v =2x,∂v =2y. ∂x ∂y
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458
CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS
The Cauchy-Riemann equations are satisfied only at z = 0, so f(z) is certainly not differentiable at any nonzero z. To check at z = 0, fall back on the definition of the derivative:
z(h) − f (0) h→0 h
i|h|2 h→0 h
ihh
= lim = lim ih = 0.
lim
= lim
h→0 h h→0
Therefore f′(0) = 0. 0 is the only point at which this function is differen-
tiable.
6. f(z)=x+iy+y=(x+y)+yisolet
u(x, y) = x + y and v(x, y) = y.
The Cauchy-Riemann equations fail to hold at any point, because
7. First,
∂u = 1 but ∂v = 0. ∂y ∂x
f(z)= x+iy =1+ yi xx
for x ̸= 0. This function is defined for all z except for points on the imaginary axis. For x ̸= 0, we can let
Now
and
u(x,y) = 1,v(x,y) = y. x
∂u = ∂u = 0 ∂x ∂y
∂v = − y , ∂v = 1. ∂x x2 ∂y x
These are not satisfied at any z at which the function is defined. Therefore f(z) is not differentiable at any point at which it is defined.
8. Write
f(z)=(x+iy)3 −8(x+iy)+2
=x3 −3xy2 −8x+2+(3x2y−y3 −8y)i.
Let
u(x, y) = x3 − 3xy2 − 8x + 2 and v(x, y) = 3x2y − y3 − 8y.
Then
∂u =3x2 −3y2 −8= ∂v ∂x ∂y
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19.2.
9.
COMPLEX FUNCTIONS 459 and
∂u = −6xy = −∂v. ∂y ∂x
The Cauchy-Riemann equations are satisfied at every z. Further, u,v and their first partial derivatives are continuous for all (x,y), so f(z) is differentiable for all z.
First, so let Then
while
f(z)=(z)2 =(x−iy)2 =x2 −y2 −2xyi, u(x,y)=x2−y2 andv(x,y)=−2xy.
∂u = 2x but ∂v = −2x, ∂x ∂y
∂v = −2y = ∂u. ∂x ∂y
The Cauchy-Riemann equations hold only a z = 0, so this is the only point at which f(z) might have a derivative. To check this, look at
f(h) − f(0) h→0 h
(h)2 h→0 h
h
= lim h = 0
lim
because h/h has magnitude 1 and h → 0 if h → 0.
Therefore f′(0) = 0, and 0 is the only point at which the function has a derivative.
Ifz=x+iy,then
f(z)=iz+|z|=x2 +y2 −y+xi.
Let Then
while
= lim
h→0 h
10.
u(x, y) = x2 + y2 − y and v(x, y) = x. ∂ux∂u y
∂x = x2 + y2 , ∂y = −1 + x2 + y2 , ∂v =1,∂v =0.
∂x ∂y
Using these, it is routine to check that f(z) is not differentiable at any z.
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460 CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS 11. For z ̸= 0, write
f(z) = −4z + 1 = −4x − 4iy + 1
z x+iy
=−4x−4yi+ x−iy x2 + y2
for (x, y) ̸= (0, 0). Let
u(x,y)=−4x+x2+y2 andv(x,y)= −4y−x2+y2 .
xy
Then
and
The Cauchy-Riemann equations are satisfied at each nonzero z. Because u,v and the partial derivatives are continuous for (x,y) ̸= (0,0), f(z) is differentiable for all nonzero z.
∂u y2 −x2 ∂u −2xy
∂x = −4 + (x2 + y2)2 , ∂y = (x2 + y2)2 ,
∂v 2xy ∂v y2 −x2
∂x = (x2 + y2)2 , ∂y = −4 + (x2 + y2)2 .
12. First,
Let
f(z)= z−i = x+(y−1)i z + i x + (y + 1)i
= x2 +y2 −1−2xi. x2 +(y+1)2
u(x,y)= x2+y2−1 andv(x,y)=− 2x . x2 +(y+1)2 x2 +(y+1)2
for all (x, y) with x2 + (y + 1)2 ̸= 0. The partial derivatives are
∂u= 4x(y+1) ,
∂x x2 + (y + 1)2
∂u 2(y+1)2 −2x2
∂y= x2+(y+1)2 ,
∂v 2x2 −2(y+1)2 ∂v 4x(y+1) ∂x = x2 +(y+1)2 ,∂y = x2 +(y+1)2.
For (x, y) with x2 + (y − 1)2 ̸= 0, the Cauchy-Riemann equations are sat- isfied. Further, u(x, y) and v(x, y) are continuous with continuous partial derivatives. Therefore f(z) is differentiable at all z with z ̸= i.
13. Let zn = xn +iyn and z0 = x0 +iy0. Write f(z) = u(x,y)+iv(x,y). Because u and v are continuous at (x0,y0), then
f(zn) = u(xn, yn) + iv(xn, yn) → u(x0, y0) + iv(x0, y0) = f(z0).
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19.3. THE EXPONENTIAL AND TRIGONOMETRIC FUNCTIONS 461 19.3 The Exponential and Trigonometric Func-
tions
ei = e0+i = e0 (cos(1) + i sin(1)) = cos(1) + i sin(1)
2. There are several ways to proceed. One is to use
sin(x + iy) = sin(x) cosh(y) + i cos(x) sinh(y)
to get
sin(1 − 4i) = sin(1) cosh(4) − i cos(1) sinh(4).
We could also have begun with the definition of sin(z) and used Euler’s
formula.
3. Use the fact that
cos(x + iy) = cos(x) cosh(y) − i sin(x) sinh(y)
to get
cos(3 + 2i) = cos(3) cosh(2) − i sin(3) sinh(2).
4. From their definitions, the trigonometric and hyperbolic functions are re-
1.
lated by
Then
5.
6.
sin(z) = −i sinh(iz) and cos(z) = cosh(iz).
tan(3i) = sin(3i) = −i sinh(−3) cos(3i) cosh(−3)
= isinh(3) . cosh(3) = i tanh(3)
e5+2i = e3e2i = e3 cos(2) + ie3 sin(2).
cot(1 − πi/4) = cos(1 − πi/4) sin(1 − πi/4)
= cos(1)cosh(π/4)+isin(1)sinh(π/4). sin(1) cosh(π/4) − i cos(1) sinh(π/4)
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462
CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS
To identify the real and imaginary parts of this quotient, multiply the numerator and denominator by the conjugate of the denominator to obtain
cot(1 − iπ/4)
= (cos(1) cosh(π/4) + i sin(1) sinh(π/4))(sin(1) cosh(π/4) + i cos(1) sinh(π/4))
sin2(1) cosh2(π/4) + cos2(1) sinh2(π/4)
= sin(1)cos(1)(cosh2(π/4) − sinh2(π/4)) + i sinh(π/4) cosh(π/4)(sin2 + cos2(1))
=
sin2(1) cosh2(π/4) + cos2(1) sinh2(π/4) sin(1) cos(1) + i sinh(π/4) cosh(π/4) .
sin2(1) cosh2(π/4) + cos2(1) sinh2(π/4) 7.
sin2(1 + i) = 1(1 − cos(2(1 + i))) 2
= 1 [1 − cos(2) cosh(2) + i sin(2) sinh(2)]. 2
8.
9.
10.
11.
cos(2−i)−sin(2−i)
= cos(2) cosh(1) + i sin(2) sinh(1) − sin(2) cosh(1) − i cos(2) sinh(1) = cosh(1)[cos(2) − sin(2)] − i sinh(1)[cos(2) − sin(2)]
Begin with
Then
eiπ/2 = cos(π/2) + i sin(π/2) = i
sin(ei) = sin(cos(1) + i sin(1))
= sin(cos(1)) cosh(sin(1))i cos(cos(1)) sinh(sin(1))
ez2 = ex2−y2+2ixy
= ex2−y2 [cos(2xy) + i sin(2xy)].
u(x,y) = ex2−y2 cos(2xy) and v(x,y) = ex2−y2 sin(2xy).
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19.3.
THE EXPONENTIAL AND TRIGONOMETRIC FUNCTIONS 463 Now compute
∂u = ex2−y2 [2x cos(2xy) − 2y sin(2xy)], ∂x
∂u = ex2−y2 [−2y cos(2xy) − 2x sin(2xy)], ∂y
∂v = ex2−y2 [2x sin(2xy) + 2y sin(2xy)], ∂x
∂v = ex2−y2 [−2y sin(2x) + 2x cos(2xy)]. ∂y
Then u and v satisfy the Cauchy-Riemann equations for all (x, y). Begin by writing
12.
Then
1= 1 = x − y i. z x + iy x2 + y2 x2 + y2
1/z x/(x2+y2)y y e =e cos x2+y2 −isin x2+y2
= u(x, y) + iv(x, y). Take the partial derivatives:
∂u ex/(x2+y2)2 2 y y ∂x = (x2 +y2)2 (y −x )cos x2 +y2 +2xysin x2 +y2 ,
∂u ex/(x2+y2)
∂y = (x2 +y2)2 −2xycos
y 2 2 y x2 +y2 −(x −y )sin x2 +y2
∂v ex/(x2+y2)2 2
∂v ex/(x2+y2) y 2
y x2 +y2
, ,
y ∂x = (x2 +y2)2 (x −y )sin x2 +y2
+2xycos 2
∂y = (x2 +y2)2 2xycos x2 +y2 −(x −y )cos
It is easy to verify that u and v satisfy the Cauchy-Riemann equations for
y x2 +y2
.
all (x, y). 13.
f(z)=zez =(x+iy)ex(cos(y)+isin(y))
= xex cos(y) − yex sin(y) + (yex cos(y) + xex sin(y))i = u(x, y) + iv(x, y).
Then
∂u = ex[cos(y) + x cos(y) − y sin(y)] = ∂v ∂x ∂y
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464
CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS
∂u =ex[−xsin(y)−sin(y)−ycos(y)]=−∂v. ∂y ∂x
and
The Cauchy-Riemann equations are satisfied for all z. 14.
f(z) = cos2(z) = 1(1 − cos(2z)) 2
= 1 − 1 (cos(2x) cosh(2y) − i sin(2x) sinh(2y)). 22
Then
Now,
u(x, y) = 1 − 1 cos(2x) cosh(2y), v(x, y) = 1 sin(2x) sinh(2y). 222
∂u = sin(2x) cosh(2y), ∂x
∂u = − cos(2x) sinh(2y), ∂y
∂v = cos(2x) sinh(2y), ∂x
∂v = sin(2) cosh(2y). ∂y
15.
The Cauchy-Riemann equations are satisfied at every (x, y). Supposeez =2i. Withz=x+iy,then
Then
Because ex ̸= 0, cos(y) = 0, so
ex cos(y) + iex sin(y) = 2i. ex cos(y) = 0 and ex sin(y) = 2.
in which n can be any integer. Now we have x 2n+1
esin 2π=2. Nowex >0forrealx,sosin((2n+1)π/2)>0. But
y = (2n + 1)π 2
2n+1 1 ifniseven, sin 2 π = −1 ifnisodd.
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19.3. THE EXPONENTIAL AND TRIGONOMETRIC FUNCTIONS 465 Therefore n must be even, say n = 2m. Now we have
y = 4m + 1π. 2
Nowweareleftwithex =2,sox=ln(2). Allthesolutionsofez =2iare ln(2) + 4m + 1 π
2
with m any integer.
16. For the first identity, write
sin(z) cos(w) + cos(z) sin(w)
= 1 (eiz − e−iz)(eiw + e−iw) + (eiz + e−iz)(eiw − e−iw)
4i
= 1 ei(z+w) − e−i(z+w) 2i
= sin(z + w).
The second identity is proved by a similar manipulation.
17. Use the polar form of the given equation. If z = reiθ, the equation is
ez =ereiθ =−2. Because θ is real, |eiθ| = 1, so
|ez|=er =|−2|=2. Then r = ln(2). Next, we must also have
eiθ = −1 = cos(θ) + i sin(θ).
Then sin(θ) = 0, so θ = nπ, in which (so far) n can be any integer. But
cos(θ) = −1 means that n must be odd, so θ = (2m+1)π
in which m can be an y integer. Then
z = ln(2)+(2m+1)π,
with m any integer, are all the solutions for z.
18. We want all z such that sin(z) = i. We need, for z = x + iy,
Then
sin(x) cosh(y) + i cos(x) sinh(y) = i. sin(x) cosh(y) = 0 and cos(x) sinh(y) = 1.
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466
CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS
Because cosh(y) > 0 for y real, we must have sin 9x) = 0, so x = nπ, with n as yet any integer.
The second equation now becomes
cos(nπ) sinh(y) = 1.
Then
Then
Write this equation as
sinh(y) = 1 = (−1)n. cos(nπ)
ey − e−y = 2(−1)n. e2y −2(−1)ney −1=0
and consider cases. First, if n is even, then (−1)n = 1 and we have a quadratic equation for ey:
e2y −2ey −1=0 ey =1±√2.
with roots
Because1−√2<0andey >0,discardthisrootandset
Then y = ln(1 +
ey =1+√2. √
2) and solutions for z are √
z=2mπ+ln(1+ 2)i,
with m any integer. In the case that n is odd, write n = 2m+1. Now
(−1)n = −1 and we have
with roots −1 ± so in this case
e2y + 2ey − 1 = 0, √√
2. Again, we can only use the positive root −1 + 2, √
y=ln(−1+ 2)
√
and solutions for z are
z=(2m+1)π+ln(−1+ 2)i,
with m any integer.
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19.4. THE COMPLEX LOGARITHM
19.4 The Complex Logarithm
467
1. In polar form,
so π
log(−4i) = ln(4) + 2 + 2nπ
2. First, 2 − 2i = 2√2e7πi/4, so
z = −4i = 4e3nπi/2
√ 7π log(2−2i)=ln(2 2)+ 4 +2nπ i.
i.
3. −5 = 5eπi, so
4. 1 + 5i = √26earctan(5)i, so
log(1 + 5i) = 1 ln(26) + (arctan(5) + 2nπ)i. 2
log(−5) = ln(5) + (2n + 1)πi.
5. −9 + 2i = √85e(arctan(−2/9)+π)i, so
log(−9 + 2i) = 1 ln(85) + (− arctan(2/9) + (2n + 1)π)i.
2
6. 5 is its own polar form (argument zero), so
log(5) = ln(5) + 2nπi.
7. Note that log(zw),log(z) and log(w) all have infinitely many different values, so we cannot expect to write the complex logarithm of the product as the sum of the logarithms of the factors. What we can show is that every value of log(zw) is the sum of a value of log(z) and a value of log(w). Suppose that z and w are nonzero. Let θz be any argument of z, and θw any argument of w. Then
Then while
z = |z|e(θz+2nπ)i and w = |w|e(θw+2mπ)i. zw = |z||w|e(θz+θw+2kπ)i,
log(z) + log(w) = ln(|z|) + ln(|w|) + (θz + θw + 2(n + m)π)i.
This means that for any choice of n and m, we can choose k = n+m to
obtain a value of log(zw) that is equal to log(z) + log(w).
8. The argument is nearly identical to that used in Problem 7, except now
z = |z| e(θz−θw+2(n−m)π)i. w |w|
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468 CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS
19.5 Powers
In these problems, n denotes an arbitrary integer. 1.
i1+i = e(1+i) log(i) = e(1+i)(π/2+2nπ)i =e−(π/2+2nπ)cosπ +2nπ+isinπ +2nπ
= ie−(π/2+2nπ).
2.
3.
√
22
(1 + i)2i = e2i log(1+i) = e2i(ln( 2)+i(π/4+2nπ))
= e−(π/2+4nπ)[cos(ln(2)) + i sin(ln(2))].
ii = ei log(i) = ei(i(π/2+2nπ)) = e−π/2+2nπ.
This is consistent with Problem 1, because i1+i = iii. 4.
(1 + i)2−i = e(2−i) log(1+i) √
5.
6.
= e(2−i)(ln( 2)+i(π/4+2nπ))
= eln(2)+π/2+2nπ cos π + 4nπ − ln(√2) + i sin π + 4nπ − ln(√2)
π/4+2nπ √2 √ 2 = 2 [sin(ln( 2)) + i cos(ln( 2))].
(−1 + i)−3i = e−3i log(−1+i) √
= e−3i ln( 2)+i(3π/4+2nπ)
= e9π/4+6nπ [cos(3 ln(√2)) + i sin(3 ln(√2))].
(1 − i)1/3 = √2e−i(π/4+2nπ)1/3 = 21/6e−i(π/12+2nπ).
We obtain the three distinct cube roots by choosing n = 0,1,2. Other choices of n repeat those obtained for these values.
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19.5.
7.
8.
POWERS
469
10.
11.
12.
(−4)2−i = e(2−i) log(−4) = e(2−i)(ln(4)+i(π+2nπ))
= e2 ln(4)+π+2nπ [cos(ln(4)) − i sin(ln(4))].
6−2−3i = e(−2−3i) log(6)
= e(−2−3i)(ln(6)+2nπi)
= e−2 ln(6)+6nπ e−(3 ln(6)+4nπ)i
= 1e6nπ[cos(3ln(6))−isin(3ln(6))].
36
i1/4 = ei(π/2+2nπ)1/4 = ei(π/8+nπ/2),
the the four fourth roots obtained for n = 0,1,2,3. Other choices of n repeat these roots.
161/4 = (16e2nπi)1/4 = 2enπi/2
= 2[cos(nπ/2) + i sin(nπ/2)]
with the distinct fourth roots obtained by using n = 0, 1, 2, 3. 9.
i(π+2nπ)1/4 i(π/4+nπ/2)
= 2 cos π + nπ + i sin π + nπ .
42 42
the four fourth roots by taking n = 0, 1, 2, 3. These fourth
√√√√ 2(1+i), 2(−1+i), 2(−1−i), 2(1−i).
1/4
(−16) = 16e =2e
We obtain roots are
First compute
1+i = (1+i)(1+i) =i. 1−i (1−i)(1+i)
We therefore want i1/3. These roots are
1/3 i(π/2+2nπ)1/3 i(π/6+2nπ/3)
i=e=e
π 2nπ π 2nπ
=cos 6+ 3 +isin 6+ 3 .
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470
13.
14.
15.
CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS
The three roots are obtained for n = 0, 1, 2 and are 1 (√3 + i), 1 (−√3 + i), −i.
22 These are the sixth roots of unity:
11/6 = e2nπi1/6 = enπi/3
= cos(nπ/3) + i sin(nπ/3).
These sixth roots are obtained for n = 0, 1, 2, 3, 4, 5, and are
1, 1 (1 + √3i), 1 (−1 + √3i), −1, 1 (−1 − √3i), 1 (1 − √3i).
2222
(7i)3i = e3i log(7i) = e3i(ln(7)+i(π/2+2nπ))
= e−2(π/2+2nπ)[cos(3 ln(7)) + i sin(3 ln(7))].
Let ω be any nth root of 1 different from 1. The numbers ωj, for j = 0,1,··· ,n − 1 are distinct, hence are all of the nth roots of 1. It is therefore enough to show that
n−1
ωj =0. j=0
But this is a finite geometric series, whose sum is known:
n−1
j 1−ωn
ω=1−ω=0 j=0
because ωn = 1.
The conclusion can also be proved as follows. Let ω1, · · · , ωn be the nth
roots of unity. Let S = nj=1 ωj.
Now, one of the nth roots of unity is 1, but the other n − 1 roots are different from 1. Pick one root that does not equal 1, say, possibly by relabeling, ω1 ̸= 1. The numbers
ω1ω1,ω1ω2,··· ,ω1ωn
are also nth roots of unity and are distinct, so this list includes all the nth roots of unity. The sum of these numbers is therefore S:
n
S=ω1ωj =ω1S.
j=1
S(1 − ω1) = 0.
But then
Because ω1 ̸= 1, then S = 0.
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19.5. POWERS
16. First recall that, if a ̸= 1, then n−1
471
j 1−an
a=1−a. j=0
Further, replacing a with −a, n−1 n−1
j (−a) =
j j 1−(−1)nan
. Apply this result with a = e2π/n, so an = e2πi = 1, to obtain
(−1)a = 1+a
n−1
j
1−(−1)n 0 if n is even, = 1+e2πi/n = 2/(1+e2πi/n) if n is odd.
j=0
(−1) e
j=0
2πij/n
j=0
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472 CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS
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Chapter 20
Complex Integration
20.1 The Integral of a Complex Function
1. In this problem f(z) = 1 is differentiable for all z and we can write an antiderivative F (z) = z. The curve has initial point γ(1) = 1 − i and terminal point γ(3) = 9 − 3i, so
f (x) dz = F (9 − 3i) − F (1 − i) = 9 − 3i − (1 − i) = 8 − 2i. γ
We can also evaluate the integral by using the parametric equations of the curve. On γ, z = γ(t) = t2 − it, so dz = 2t − i and
f(z)dz = dz γγ
3
= (2t−i)dt 1
2 3
=t −it =(9−3i)−(1−i)
1
= 8 − 2i.
2. f(z) = z2 − iz is differentiable for all z, with antiderivative
F(z)= 1z3 − iz2. 32
Because the curve extends from 2 to 2i, then 84
f (z) dz = F (2i) − F (1) = − 3 + 3 i. γ
If we want to use a parametrization of the curve, we can use polar coor- 473
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474 CHAPTER 20. COMPLEX INTEGRATION dinates and write γ(t) = 2eit for 0 ≤ t ≤ π/2. Then
(4e2it −2ieit)(2ieit dt (8ie3it + 4e2it) dt
π/2 γ0
3. f(z) = Re(z) is not differentiable, so there is no antiderivative. There are many ways to parametrize the curve. One is by setting
γ(t) = 1 + (1 + i)t for 0 ≤ t ≤ 1. On γ, f(z) = 1 + t and dz = (1 + i) dt, so
1
f(z)dz = (1+t)(1+i)dt γ0
1
= (1+i+(1+i)t)dt
0
= (1 + i)t + = 3(1+i).
2
4. Parametrize γ(t) = 4eit for π/2 ≤ t ≤ 3π/2. Then
1 3π/21 it
f(z)dz = =
π/2 0
8 3it 2itπ/2
− 3e −2ie
0
= −8 + 4i. 33
1+i 21 t
20
γ
zdz= 4eit4ie dt=πi. π/2
5. F(z) = (z − 1)2/2 is an antiderivative of f(z), which is differentiable for all z, so
13
f(z)dz=F(1−4i)−F(2i)=−2 +2i. γ
6. F(z) = iz3/3 is an antiderivative of f(z), which is differentiable for all z,
so 1
f(z)dz=F(3+i)−F(1+2i)= 3(−28+29i). γ
7. f(z) is differentiable for all z and has antiderivative F(z) = −cos(2z)/2, so
f (z) dz = F (−4i) − F (−i) γ
= − 1 (cos(−8i) − cos(−2i)) = − 1 [cosh(8) − cosh(2)]. 22
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20.1. THE INTEGRAL OF A COMPLEX FUNCTION 475
8. f(z) is differentiable for all z and has antiderivative F(z) = z + z3/3, so
f (z) dz = F (3i) − F (−3i) = −12i. γ
9. f(z) is differentiable for all z and has antiderivative F(z) = −isin(z), so
f (z) dz = F (2 + i) − F (0) = −i sin(2 + i) γ
= −i[sin(2) cosh(1) + i cos(2) sinh(1)] = − cos(2) sinh(1) − i sin(2) cosh(1).
10. f(z) has no antiderivative, so parametrize the curve, say by γ(t) = −4 + (4+i)tfor0≤t≤1. Onthecurve,
and Then
z(t) = −4 + (4 + i)t and dz = (4 + i) dt |z|2 =|4(t−1)+it|2 =16(t−1)2 +t2
1
|z|2 dz = (16(t−1)2 +t2)(4+i)dt
γ0
= 4+i[16(t−1)3 +t3]dt
3
= 17(4+i). 3
11. Use the antiderivative F (z) = (z − i)4/4 to get
f (z) dz = F (2 − 4i) − F (0) = 10 + 210i. γ
12. Use the antiderivative F(z) = −ieiz to get
f (z) dz = F (−4 − i) − F (−2) γ
=−ie1−4i −e−2i
= −e sin(4) + sin(2) + [cos(2) − e cos(1)]i.
13. f(z) has no antiderivative because this function is not differentiable. Parametrize thecurvebyγ(t)=(−4+3i)tfor0≤t≤1. Then
1
iz dz = −i(4t − 3ti)(−4 + 3i) dt
γ0
=(−4+3i) 2−2i = 2i.
3 25
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476 CHAPTER 20. COMPLEX INTEGRATION 14. f(z) = Im(z) is not differentiable, so parametrize γ(t) = 4eit for 0 ≤ t ≤
2π. Then
16(− sin2(t) + i cos(t) sin(t)) dt = −16π.
15. f(z) = |z|2 has no antiderivative, so write γ(t) = (1+i)t−i for 0 ≤ t ≤ 1
to get
16. Begin with
2π γ0
12 |z|2dz= (t2 +(t−1)2)(1+i)dt= 3(1+i).
γ0
2
Im(z) dz = =
4 sin(t)4ieit dt
2π 0
cos(z )dz≤8πM,
where 8π is the length of the circle γ, and M is a bound for cos(z2) on γ,
so
|cos(z2)| ≤ M for z on γ.
There are of course many possible choices for M. Indeed, if we find a bound, then any larger number is also a bound. To find one such number, let z = x + iy so
γ
z2 = x2 − y2 + 2ixy
and
in which we have used the facts that the real hyperbolic cosine is positive, and that the real sine and cosine functions are bounded by 1.
Now, for points on γ, x = 4cos(t) and y = 4sin(t) for 0 ≤ t ≤ 2π, so e2xy = e2(4 cos(t))(4 sin(t)) = e32 sin(t) cos(t)
= e16 sin(2t) ≤ e16. We can choose M = e16 to obtain
2 16 cos(z )dz≤8πe .
γ
This is a massively huge bound for this integral, and is not claimed to be an approximation of the integral in any sense. But this is what we get with the quick and crude bounds made along the way. With more effort we may get a smaller bound. However, in some applications, it is enough to know that a certain term is bounded, and the size of the bound may not matter.
| cos(z2)| = | cos(x2 − y2 + 2ixy)|
= | cos(x2 − y2) cosh(2xy) − i sin(x2 − y2) sinh(2xy)| ≤ cosh(2xy) + | sinh(2xy)| = e2xy .
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20.2.
17.
CAUCHY’S THEOREM 477 √
The length of γ is 5. Now we need a number M such that 1
1+z≤M onγ.
Notice that the point on γ closest to z = −1 is 2+i, so for z on the curve,
√ |z+i|=|z−(−1)|≥|2+i+i|= 10.
Then
We can therefore use M = 1/ 10 to get the bound √
111
1+z= |1+z| ≤ √10. √
151
dz≤√ =√. 10 2
1. sin(z) is differentiable for all z, hence on an open set containing the curve and all points enclosed by the curve. By Cauchy’s theorem,
sin(z) dz = 0. γ
2. The circle encloses i, at which f(z) is not defined (hence not differen- tiable), so Cauchy’s theorem does not apply. Evaluate the integral by parametrizing γ(t) = 2i + 2eit for 0 ≤ t ≤ 2π. Then
2i 2π 2i+6eit
γ 1+z 20.2 Cauchy’s Theorem
dt
Then
z − i dz = γ0
3eit 3ieit
(−2 + 6ieit) dt = −4π.
=
3. γ encloses 2i, at which f(z) is not defined. Parametrize
2π 0
γ(t)=2i+2eit for0≤t≤2π. 1 2π 1 it
(z − 2i)3 dt = (2eit)3 2ie dt γ0
i 2π
e−2itdt=0.
This integral happens to be zero, but we could not conclude this from
= 4 Cauchy’s theorem, which does not apply here.
0
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478 CHAPTER 20. COMPLEX INTEGRATION
4. Because z2 sin(z) is differentiable everywhere, and in particular on the curve and throughout the points it encloses, then Cauchy’s theorem applies
and γ
z2 sin(z) dz = 0.
5. f(z) = z is not differentiable. Write γ(t) = eit for 0 ≤ t ≤ 2π. Then
2π
z dz = e−itieit dt = 2πi.
γ0
6. f(z) = 1/z is differentiable except at the origin, which is enclosed by the
curve. Parametrize γ(t) = 5eit for 0 ≤ t ≤ 2π. Then 1 2π2π 1 it
z dz = 5e−it 5ie dt γ00
region it encloses, then by Cauchy’s theorem,
zez dz = 0. γ
8. A polynomial is differentiable on the entire complex plane, so
(z2 − 4z + i) dz = 0. γ
9. f(z) = |z|2 is not differentiable at any point other than 0, Cauchy’s theo- rem does not apply. Write γ(t) = 7eit for 0 ≤ t ≤ 2π to obtain
γ, so Cauchy’s theorem applies and
sin(1/z) dz = 0. γ
11. f(z) = Re(z) is not differentiable, so write γ(t) = 2eit for 0 ≤ t ≤ 2π. Then
2 cos(t)(2ieit) dt
[4i cos2(t) − 4 cos(t) sin(t)] dt = 4πi.
=
2π γ0
2π 0
ie2it dt = 0.
7. Because f(z) = zez is differentiable on the curve and throughout the
|z|2 dz =
10. f(z) = sin(1/z) is not differentiable at 0, which is not on or enclosed by
2π γ0
Re(z) dz = =
2π 0
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49(7ieit) dt = 0.
20.3.
12.
CONSEQUENCES OF CAUCHY’S THEOREM 479 f(z) = z2 is differentiable for ll z, so by Cauchy’s theorem,
z2 dz = 0 γ
for any simple closed path in the plane. We can therefore concentrate on the integral of just Im(z) around the square. Let S1 be the left side, S2 the lower side, S3 the right side, and S4 the top side, and orient the rectangle counterclockwise. We can parametrize each side:
S1 : γ1(t) = −2it,
S2 :γ2(t)=2t−2i,
S3 :γ3(t)=2−2i(1−t), S4 :γ4(t)=2(1−t).
For each side, t varies from 0 to 1. Then 1
Im(z) dz = (−2t)(−2i) dt γ0
In total, then,
111
+ (−2)2dt+ (−2)(1−t)(2i)dt+ 0dt
000
= 2i − 4 − 2i = −4.
(z2 + Im(z)) dz = −4. γ
20.3 Consequences of Cauchy’s Theorem
1. Because 2i is the center of the circle γ, we can apply Cauchy’s integral formula with f(z) = z4 to obtain
z4
γ z − 2i dz = 2πif(2i) = 2πi(2i)4 = 32πi.
2. By Cauchy’s integral formula with f(z) = sin(z2),
sin(z2) dz = 2πif(5) = 2πi sin(25).
γ z−5
3. Use Cauchy’s integral formula, with f(z) = z2 − 4z + i, to obtain
z2−4z+i
γ z−1+2i dz=2πif(1−2i)
=2πi[(1−2i)2 −4(1−2i)+i]=2πi(−8+7i) = −14π − 16πi.
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480 CHAPTER 20. COMPLEX INTEGRATION 4. Apply the Cauchy integral formula for derivatives with n = 1 and f(z) =
2z3 to obtain
2z3 ′
γ (z − 2)2 dz = 2πif (2) = 48πi.
5. We can use the Cauchy integral formula for derivatives with n = 1 and f(z) = iez:
iez ′
γ (z − 2 + i)2 dz = 2πif (2 − i)
= 2πi(ie2−i) = −2πe2[cos(1) − i sin(1)].
6. Apply Cauchy’s formula for derivatives with n = 2 and f(z) = cos(z − i)
to obtain
cos(z−i)dz=2πif′′(−2i) γ (z+2i)3 2
= −πi cos(−3i) = −πi cosh(3).
7. With f(z) = zsin(3z) and n = 2, Cauchy’s formula for derivatives gives
us
zsin(3z)dz=2πif′′(−4) γ (z+4)3 2
= πi[6 cos(12) − 36 sin(12)].
8. γ is not a closed curve, so parametrize γ(t) = 1 − t − it for 0 ≤ t ≤ 1. On
the curve,
f(γ(t)) = 2iγ(t)|γ(t)| = 2i[(1 − t) + it]1 − 2t + 2t2
so
12 2iz|z|dz = 2i[(1−t)+it] 1−2t+2t dt
γ0 1212
= 2 1−2t+2t dt+2i (2t−1) 1−2t+2t 0√0
√ 2+1 =1+ 24ln √ .
dt
2−1
These definite integrals can be evaluated using
1−2t+2t2dt= −1+4t1−2t+4t2 8
7√2 −14 1 + 32 sinh √7 t−4
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20.3. CONSEQUENCES OF CAUCHY’S THEOREM 481
and 2 1 23/2 (2t−1) 1−2t+2t dt= 3(1−2t+2t ) .
9.
−(2+i)sin(z4) γ (z + 4)2
d 4
dz = −2πi(2 + i) dz (sin(z ))z=−4
= 2πi(1 − 2i) 4z3 cos(z4) = −512π(1 − 2i) cos(256).
z=−4
10. γ is not a closed curve. An antiderivative of (z − i)2 is (z − i)3/3, so
2 1 3−i (z−i)dz= (z−i)
γ
11. Parametrizeγ(t)=3−t+(1−6t)ifor0≤t≤1. Then 1
3i
= 1(−2i)3 = 8i.
12.
= (−1 − 6i) 13 = − 13 − 39i. 22
dz = 2πi dz (3z
= 2πi 6 cosh(z) − 3z2 sinh(z)
13. First evaluate
ez
γ z dz
Re(z + 4) dz = (7 − t)(−1 − 6i) dt γ0
3z2cosh(z)
d 2
33
γ
(z + 2i)2
cosh(z))z=−2i
= 2πi[−12i cosh(2i) + 12 sinh(2i)]
= 24π[cos(2) − sin(2)].
z=−2i
by Cauchy’s integral formula to obtain
ez
γz z=0
dz = 2πie
z
= 2πi.
Now evaluate this integral by parametrizing γ(t) = eit for 0 ≤ t ≤ 2π: ez 2π ecos(t)+i sin(t)
z dz = γ0
eit ieit dt
2π
ecos(t) cos(sin(t)) dt −
= i
= 2πi.
2π 00
ecos(t) sin(sin(t)) dt
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482
CHAPTER 20. COMPLEX INTEGRATION
By equating the real parts of both sides of this equation, and then the imaginary parts, we obtain
14.
First write
f(z)= z−4i = z−4i . z3 +4z z(z−2i)(z+2i)
and
2π
ecos(t) cos(sin(t)) dt = 2π
0
2π
ecos(t) sin(sin(t)) dt = 0. 0
The first integral is not obvious. The second could be done without com- plex analysis by observing that the integral from 0 to π is the negative of the integral from π to 2π.
Now let γ1, γ2 and γ3 be nonintersecting circles in the region bounded by γ and having centers, respectively, 0, 2i, −2i. By the extended deformation theorem,
3
f(z)dz = f(z)dz.
γ
f(z) = (z − 4i)/(z − 2i)(z + 2i). z
n=1 γj
Consider each integral on the right. On and in the interior of γ1, think of
Apply the Cauchy integral formula to obtain
z−4i
f(z)dz=2πi (z−2i)(z+2i)
γ1 z=0
−4i
= 2πi (−2i)(2i) = 2π.
For the integral over γ2, we can apply Cauchy’s integral formula to
(z − 4i)/z(z + 2i) z − 2i
−2i π (2i)(4i) = −2.
f(z) = (z − 4i)/z(z − 2i) z + 2i
f(z)dz =
γ2 γ2
= 2πi And, for the integral over γ3, think of
and apply Cauchy’s integral formula in this region to obtain −6i 3π
f(z)dz=2πi (−2i)(−4i) = 2. γ3
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20.3. CONSEQUENCES OF CAUCHY’S THEOREM 483 Finally, we have
π3π
f(z)dz=2π−2− 2 =0. γ
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484 CHAPTER 20. COMPLEX INTEGRATION
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Chapter 21
Series Representations of Functions
21.1 Power Series
In each of Problems 1–6, the strategy is to take the limit of the magnitude of the ratio of successive terms of the series. The series converges when this limit (if it exists) is less than 1.
1. Take the limit of the magnitude of successive terms: (n + 2)/2n+1 1 n + 2
2
The series converges (absolutely) if
1 |z + 3i| < 1 2
or
The power series has radius of convergence 2 and open disk of convergence
2.
(n+1)/2n |z+3i|= 2n+1(z+3i) → 1|z+3i|.
|z + 3i| < 2.
|z + 3i| < 2, the open disk of radius 2 about the center −3i.
1/(2n+3)2 (z−i)n+1 2n+12 1/(2n+1)2 (z−i)n = 2n+3 |z−i|2
→ |(z − i)2|.
if |z − i| < 1. This power series has radius of convergence 1, and open disk
of convergence |z − i| < 1, the open disk of radius 1 about the center i. 485
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486 3.
CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS
(n + 1)n+1/(n + 2)n+1 nn/(n + 1)n (z − 1 + 3i)
(n + 1)2n+1
= nn(n+2)n+1|z−1+3i|
n + 1 n n + 1 n+1
= n n + 2 |z − 1 + 3i|
1n 1+1/nn 1+1/n = 1+ n 1+2/n 1+2/n
|z−1+3i|
and the limit of this quantity is less than 1 if |z − 1 + 3i| < 1. The radius of convergence is 1 and the disk of convergence is the open disk of radius 1 centered at 1 − 3i.
In this limit, we have used (several times) the fact that
xn x lim1+ =e.
4.
n→∞ n (2i/(5 + i))n+1
(2i/(5+i)n)n |z+3−4i| 2i
→ 5 + i ( z + 3 − 4 i )
2
=√ |z+3−4i|,
26 and this limit is less than 1 when
√
|z+3−4i|< 26.
2
√
This power series has radius of convergence 26/2, and the open disk of √
convergence is the disk of radius 26/2 centered at −3 + 4i.
5.
6.
in+1 /2n+2 1
in/2n (z+8i)→ 2|z+8i|
This ratio has limit < 1 if |z + 8i| < 2. The power series has radius of convergence 2 and the open disk of convergence is the open disk of radius 2 centered at −8i.
(1−i)n+1/(n+3) n+2√
(1−i)n/(n+2) (z−3)= n+1 2|z−3|
√
→ 2|z−3|
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21.1.
7. 8.
POWER SERIES 487 √
and this limit is less than 1 if |z − 3| < 1/ 2. This power series has radius √
of convergence 1/ 2 and the open disk of convergence is the open disk of √
radius 1/ 2 centered at 3.
No. The power series has center 2i. If the series converges at 0, it must
also converge at the point i that is closer to the center 2i than 0 is.
The center of this power series is 4−2i. Now, 1+i is closer to 4−2i than
i is, so if the series converges at i, it must also converge at 1 + i.
In each of Problems 9–14, we attempt to use known series to derive the
requested series.
9. Assuming that we know the series for cos(z):
cos(z) =
This converges for all z. Replace z with 2z to obtain
(2n)! ̸= 2n!. 10. We know that
for all z. Then
n! ∞∞
∞ (−1)n
(2n)! z2n. ∞ (−1)n 22n
n=0
(2n)! z2n. e z = ∞ 1 z n
cos(2z) =
In writing these series, be careful with factorials. For example, in general
n=0
1
e−z = n!(−z)n =
(−1)n
n=0
The expansion of e−z about −3i will have powers of z + 3i, so write
converging for all z.
n=0
e−z = e−z−3i+3i = e3ie−(z+3i)
∞ (−1)n
n!
= e3i
(z + 3i)n,
n! zn. n=0
n=0
11. This is just a rearrangement of the given polynomial into powers of z−2+i. This can be done algebraically, or we can write the Maclaurin series of this polynomial about 2 − i. This series will be
f(z)=z2 −3z+i=c0 +c1(z−2+i)+z2(z−2+i)2,
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488 CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS where
and
c0 =f(2−i)=−3,c1 =f′(2−i)=1−2i c2 = 1f′′(2−i)=1.
2 The expansion of f (z) about 2 − i is
z2 −3z+i=−3+(1−2i)(z−2+i)+(z−2+i)2. 12. Use the series for ez and sin(z):
to write
and
c0 =f(1+i)=63−16i,c1 =f′(1+i)=−16+2i c2 = 1f′′(1+i)=1.
2
ez =
n!zn and sin(z) = ∞∞
∞∞
1
(−1)n
ez −isin(z)=
1
(2n + 1)!z2n+1 (−1)n
n=0
n=0
n=0
n=0
n!zn −i
(2n+1)!z2n+1.
13. Like Problem 11, this can be done as an algebraic rearrangement of terms in f(z) = (z−9)2, or as a power series about 1+i, which will be in powers of z − i − i. Using the latter approach, compute the coefficients
Then
(z−9)2 =63−16i+(−16+2i)(z−1−i)+(z−1−i)2.
14. Replace z with z + i in the Maclaurin expansion of sin(z) to obtain ∞ (−1)n
(2n + 1)!(z + i)2n+1. n=0
This converges for all z.
15. We know that f(0) = 1, f′(0) = i, and f′′(z) = 2f(z) + 1. Compute
f′′(0) = 2f(0) + 1 = 3, f(3)(0) = 2f′′(0) = 2i, f(4)(0) = 2f′′(0) = 6, f(5)(0) = 2f(3)(0) = 4i.
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21.1.
POWER SERIES 489 Now use Taylor’s formula for the coefficients (in this case, about 0,
cn = 1 f(n)(0) n!
to write the first six terms of the expansion:
1 + iz + 3z2 + 2iz3 + 6 z4 + 4iz5. 2 3! 4! 5!
In this problem it is not difficult to write the entire Maclaurin expansion, because an inductive argument shows that
f(2n)(0) = 2n + 2n−1 and f(2n+1)(0) = 2ni.
With f(z) = sin2(z), we will need
f′(z) = 2 sin(z) cos(z) = sin(2z), f′′(z) = 2 cos(2z),
f(3)(z) = −4 sin(2z), f(4)(z) = −8 cos(2z), f(5)(z) = 16 sin(2z), f(6)(z) = 32 cos(2z).
(a) Using these derivatives, we can find the first seven terms of the power series expansion of f(z) about 0:
sin2(z)=z2−1z4+ 2z6+···. 3 45
16.
(b) Multiply 213151315
sin (z)= z−6z +120z +··· z−6z +120z +··· 2 1 14 1 1 16
=z − 6+6 z + 120+36+120 z +··· =z2−1z4= 2z6+···,
3 45
(c) Use the definition of sin(z) to write
sin2(z) = −1 eiz − e−iz2 4
=−1e2iz +e−2iz −2 4
1 1 (2iz)2 (2iz)7 =2−4 1+2iz+ 2! +···+ 7! +···
1 (2iz)2 (2iz)7 − 4 1−2iz+ 2! +···− 7! +···
1124486 =2−4 1−4z +3z −45z +···
=z2−1z4+ 2z6+··· 3 45
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490
CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS
(d) We can also write
sin2(z) = 1 − 1 cos(2z)
22
1 1 (2z)2 (2z)4 (2z)6
==2−2 1− 2! + 4! − 6! +··· 1122446
=2−2 1−z +3z −45z +··· =z2−1z4+ 2z6+···
17.
3 45
Let z be a complex number and consider the integral 1 zn
2πi γ n!wn+1 ezw dw.
Here γ is the unit circle about the origin, oriented counterclockwise as usual. Expand ezw in its Maclaurin series and parametrize γ(t) = eit for 0 ≤ t ≤ π to write
z n 1 z n ∞ ( z w ) k
γ n!wn+1 ezw dw = 2πi γ n!wn+1
1 ∞ zn+kwk−n−1
=2πi γk=0 n!k!
1 2π ∞ zn+k i(k−n−1)t
k=0
k! dw dw
= 2πi 0 k=0 e n!k!ieit dt
∞ 1 2 π z n + k
2π 0 n!k! ei(k−n)t dt.
Now,
We therefore have
2π 0
0 if k ̸= n, 2π ifk=n.
=
ei(k−n)t dt =
(zn)2 2πi γ n!wn+1 ezw dw = (n!)2 .
1 zn
k=0
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21.1. POWER SERIES
Finally, we can write
491
∞∞
1 1zn
n=0
(n!)2 z2n = 2πi n!wn+1 ezw dw n=0
12π∞ (ze−it)n
12π∞ zn it = 2πi 0 n=0 n!ei(n+1)t eze
= 2π 0 n=0
n!
eit dt ezeit dt
1 2π −it
=2π eze eze dt
0
1 2π
= 2π
1 2π
it
it −it ez(e +e ) dt
= 2π
0
0
e2z cos(t) dt.
18. f(z) has a zero of order 3 at 0 because z3 has a zero of order 3 there and
cos(0) ̸= 0.
19. f(z)hasazerooforder4at0becausez2 hasazerooforder2at0and sin2(z) also has a zero of 2 there (because sin(z) has a zero of order 1 at 0).
20. f(z) = (z − π/2)2 cos(z) has a zero of order 3 at π/2.
21. f(z) has a zero of order 3 at 3π/2 because cos(z) has a simple zero there.
22. f(z) has a zero of order 4 at 0 because cos(z) has a simple zero at π/2.
23. f(z) is not defined at z = 0, so we cannot really speak of it having a zero there. However, notice something interesting. Using the Maclaurin expansion of sin(z), with z4 in place of z, and divided by z2, we can write
n=0
This power series converges for all z, and has the value 0 at 0. We can therefore extend f(z) by giving it the value 0 at z = 0, and obtain a differentiable function. This extended function has a zero of order 2 at 0.
24. As in Problem 23, f (z) is not defined at π. However, (z − π)5 has a fifth order zero at π, and sin2(z) has a second order zero at π, and if we extend f(z) by giving it the value 0 at π, this extended function has a third order zero at π.
1 ∞(−1)n
z2 sin(z4) =
(2n + 1)!z8n+2.
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492 CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS 5. Compute the kth derivative of f(z) at z0 using each series, obtaining
∞
f k (z0 ) = an (n)(n − 1) · · · (n − k + 1)(z − z0 )n−k
n=0 ∞
= bn (n)(n − 1) · · · (n − k + 1)(z − z0 )n−k . n=0
Then
so for each k = 0,1,2,···,
k!
21.2 The Laurent Expansion
fk(z0) = k!ak = k!bk, ak = 1 f(k)(z0) = bk.
Problems 1–10 are solved using manipulations of known series, such as geomet- ric series and power series for exponential and trigonometric functions. It is sometimes best, in seeking an expansion about z0, to focus on getting an ex- pression involving powers of z−z0, using algebraic manipulations, or sometimes integration and differentiation.
In particular, it is useful to know the geometric series
1 ∞
1−r = 1 ∞
1 + r = (−1)nrn, n=0
rn
and
n=0
for |r| < 1.
1. We want an expansion in powers of z − i. To this end, begin with
2z = 1 + 1 . 1 + z2 z − i z + i
The first term is already an expansion in powers of z − i (having only one term). For the second term, write
111
z + i = 2i + (z − i) = 2i 1 − z−i
= =
(−)n n=0
1 ∞
z − i n 2i
2i
2i
∞ (−1)n
n=0
(2i)n+1 (z − i)n.
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21.2. THE LAURENT EXPANSION 493 This expansion is valid for
z − i 1
2i =2|z−i|<1,
or
The Laurent expansion of f(z) about i is therefore
=
plane with the origin removed. 3. Ifz̸=0,then
|z − i| < 2. 1 ∞ (−1)n
(2i)n+1(z−i)n. This represents f(z) in the annulus 0 < |z − i| < 2.
z−i +
n=0
2. For z ̸= 0, write
sin(z) z2
1 ∞ (−1)n
= z2
n=0
(2n + 1)!z2n+1 ∞ (−1)n
(2n + 1)!z2n−1.
This series represents sin(z)/z2 in the annulus 0 < |z < ∞, which is the
1 − cos(2z) z2
∞ 1 (−1)n
n=0
(2n)! (2z)2n = (2n)! z2n−2.
= z2 1 −
∞ (−1)n+1 4n
4. Write
2
n=1
i 2 ∞ (−1)n i 2n z cos z =z (2n)! z
n=0
n=0
=∞ 1z2−2n,
(2n)!
in which we have used the fact that i2n = (−1)n. This expansion repre-
sents the function in the annulus 0 < |z| < ∞.
5. The denominator is already in terms of z − 1, so concentrate on the nu-
merator:
n=0
z2 ((z − 1) + 1)2 1 + 2(z − 1) + (z − 1)2 1−z = 1−z =− z−1
=− 1 −2−(z−1). z−1
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494 CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS This represents the function for 0 < |z − 1| < ∞, the complex plane with
1 removed.
6. As with Problem 5, algebraic manipulation will be enough here. Write
z2 +1 1 z2 +1 11+[(z−1/2)+1/2]2 2z−1 = 2 z−1/2 = 2 z−1/2
5 1 1 1 1 =8z−1/2+2+2 z−2
1
for 0 < z − 2 < ∞.
7. Use the exponential series to obtain
1ez2=1∞ 1z2n=∞ 1z2n−2,
z2 z2 n! n!
n=0 n=0
for 0 < |z| < ∞.
8. Use the Maclaurin expansion of the sine function to get
(−1)n ∞ (−1)n
(2n + 1)!z2n+1 = (2n + 1)!z2n−1,
n=0
for 0 < |z| < ∞.
9. The denominator is already a power of z − i, so we can write
z+i = 2i+(z−i) =1+ 2i z−i z−i z−i
for 0 < |z − i| < ∞.
10. Use the Maclaurin expansion of sinh(z) (or write sinh(z) = (ez − e−z )/2
and use the exponential series) to obtain
1 ∞ 1 12n+1 ∞ 1
1 1 ∞
z sin(4z) = z
n=0
sinh =
(2n+1)! z3
= z−6n−3
(2n+1)!
z3
n=0
n=0
for 0 < |z| < ∞.
11. By Cauchy’s integral formula, for any z enclosed by Γ1,
f(z)=1 f(w)dw. 2πi Γ1 w−z
Because Γ2 does not enclose z,
1 f(w)dw=0
2πi Γ2 w−z
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21.2.
THE LAURENT EXPANSION 495
by Cauchy’s theorem. The factor of 1/2πi was included in the last equation so we can add these two integrals to get
1
f(z)=2πi
Γ2
f(w) w−zdw+
f(w)
Orientation on both curves is counterclockwise. In this sum of integrals, L1 and L2 are traversed in both directions, so the integrals over these segments are zero. The integrals in square brackets therefore give us the integrals over γ1 and γ2, but counterclockwise on γ1 and clockwise on γ2. Reversing this orientation on γ1 so that all integrals are over counterclock- wise curves, we have
f(z)=2πi
γ2
1
f(w) f(w) w−zdw− w−zdw .
Γ1
w−zdw .
γ1
Now manipulate the 1/(w − z) factor in each integral to derive the result
we want. For the integral over γ2, write 1=1
w−z
w−z0 −(z−z0) =11
w−z0 1−(z−z0)/(w−w0) = 1 ∞ z − z 0 n
w−z0 n=0 w−z0
= ∞ 1
n=0 (w − z0)n+1
( z − z 0 ) n . This geometric expansion is valid because, for w on γ2,
w − z 0 z−z0 <1.
For the integral over γ1, use the fact that, for w on this curve, w − z 0
Now we have
z−z0 <1. 1=1
w−z
w−z0 −(z−z0) =−1 1
z−z0 1−(w−z0)/(z−z0) = − 1 ∞ w − z 0
z − z0 n=0 z − z0 ∞ 1
= − (w − z0)n . n=0 (z − z0)n+1
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496
CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS
Substitute these expressions into the sum of integrals representing f(z) and interchange the integrals with the summation to obtain
1 ∞ f(w) f(z) =
dw (z−z0)n
+ = +
f(w)(w − z0)n dw ∞1 f(w) n
2πi γ2 n=0 (w − z0)n+1 1 ∞
1 n+1 2πi γ1 n=0 z−z0
2πi γ2 (w−z0)n+1 dw (z−z0)
n=0
∞1 n1
2πi f(w)(w−z0) dw (z−z )n+1. n=0 γ1 0
Finally, use the deformation theorem to replace these integrals over γ1 and γ2 with integrals over Γ, which is any simple closed path in the annulus and enclosing z0. This gives us
∞
f(z)= cn(z−z0)n, n=−∞
with the integral expressions given for the coefficients cn.
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Chapter 22
Singularities and the Residue Theorem
22.1 Singularities
1. cos(z)/z has one singularity, a double pole at z = 0.
2. f(z) has a double pole at −i and a simple pole at 1.
3. e1/z (z + 2i) has an essential singularity at 0.
4. Note that
so
sin(z) = − sin(z − π) z−π z−π
lim sin(z) = − lim sin(z − π) z→π z−π z→π z−π
=−lim sin(z) =1̸=0. z→0 z
Because this limit is nonzero, the function has a removable singularity at z = π.
5. The function has a double pole at 1 and simple poles at i and −i.
6. The function has a double pole at −1.
7. Write
z−i=z−i=1, z2 +1 (z+i)(z−i) z+i
so the function has a simple pole at −i. 497
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498 CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM
8. We need to know if sinh(z) has any zeros other than z = 0. For this, solve
sinh(z)= 1ez −e−z=0. 2
For this to be true, we must have
e2z = 1
so 2z = 2nπi, or z = nπi, with n any integer. These are simple zeros of sinh(z) because cosh(nπi) ̸= 0 for any integer n.
For n = 0, observe that 0 is a simple zero of both sin(z) and sinh(z), so 0 is a removable singularity of the function.
9. The denominator has simple zeros at 1,−1,i,−i and these are simple poles of the function because the numerator does not vanish at any of these numbers.
10. tan(z) = sin(z)/ cos(z) has simple poles at the zeros of cos(z), which are simple and occur at points z = (2n + 1)π/2, in which n any integer.
11. sec(z) = 1/cos(z) has simple poles at the zeros of cos(z), which are the simple zeros (2n + 1)π/2 with n any integer.
12. e1/z(z+1) has essential singularities at 0 and −1. One way to see this is to
write
so
1=1−1 z+1 z z+1
e1/z(z+1) = e1/ze−1/(z+1).
13. Suppose f is differentiable at z0 and f(z0) ̸= 0, while g has a pole of order m at z0. We want to show that the product fg has a pole of order m at z0.
Because g has a pole of order m at z0, the Laurent expansion in some annulus about z0 has the form
k ∞
g(z)= (z−z0)m + with k ̸= 0. Then
∞
(z − z0)mg(z) = k +
If we denote the power series on the right as h(z), then
(z − z0)mg(z) = h(z),
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n=−m+1
cn(z−z0)n.
cn(z − z0)n+m. n=−m+1
22.2. THE RESIDUE THEOREM 499 where h(z0) = k ̸= 0. Further, in some annulus about z0,
f(z)g(z) = f(z)h(z) . (z − z0)m
Because f(z0)h(z0) ̸= 0, f(z)g(z) has a pole of order m at z0. 22.2 The Residue Theorem
1. The function has simple poles at 1 and −2i, both enclosed by γ. Keep in mind that only singularities enclosed by the curve are relevant in evaluat- ing the integral by the residue theorem.
Compute
and
Then
Res(f,1)=lim d 1+z2 z→1dz z+2i
= lim (z+2i)(2z)−(1+z2)
(z + 2i)2
γ (z−1)2(z+2i)dz=2πi −3+4i−−3+4i =2πi.
2. γ encloses i, which is a double pole and the only singularity of the function.
Then 2z dz = 2πiRes(f,i) = lim d (2z) = 4πi. γ (z − i)2 z→i dz
3. The only singularity of ez/z is a simple pole at 0, and this is not enclosed
−3+4i Res(f,−2i)= lim
1+z2 −3
2 = .
z→1
= 4i ,
z→−2i (z−1) −3+4i
1+z2 4i 3
by γ, so
ez
γ zdz=0
by Cauchy’s theorem.
4. f(z) has singularities at 2i and −2i. These are both simple poles, and
both are enclosed by γ. Then
cos(z) dz = 2πiRes(f, 2i) + 2πiRes(f, −2i)
4i −4i
= 0.
γ 4+z2
= 2πi cos(2i) + 2πi cos(−2i)
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500 CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM √√
5. The function has simple poles at 6i and − 6i, both enclosed by γ. Then z + i dz = 2πi Res(6, √6i) + Res(f, −√6i)
γ z2 + 6
=2πi √ + √ =2πi.
√6+1 √6−1 26 26
6. f(z) has a simple pole at −1/2. This is the only singularity, and it is enclosed by γ. Then
z−i 11 γ 2z+1dz=2πiRes(f,01/2)=−2 2+i .
7. z/sinh2(z) has a simple pole at 0 and double poles at nπi, for every nonzero integer n. The only singularity enclosed by γ is 0, so
f (z) dz = 2πiRes(f, 0). γ
Compute this residue as
Res(f,0) = lim zf(z) = lim 2
z2 z→0 z→0 sinh (z)
= lim z2
z→0 z2 + 1 z4 + · · ·
6
= lim 1
z→0 1 + 1 z2 + · · ·
6
= 1.
Then
z dz = 2πi. γ sinh2(z)
8. cos(z)/zez has only one singularity, a simple pole at 0, and this is enclosed by γ. Therefore
cos(z) dz = 2πiRes(f, 0) γ zez
= 2πi lim cos(z) = 2πi. z→0 ez
9. f(z) has simple poles at i,3i and −3i. Only the pole at −3i is enclosed by the curve, so
iz dz = 2πiRes(f, −3i) γ (z2 +9)(z−i)
iz 1πi =2πilim =2πi− =−.
z→−3i (z−3i)(z−i) 8 4
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22.2. THE RESIDUE THEOREM 501
10. e3/z2 has an essential singularity at 0 and no other singularities. 0 also lies in the region bounded by the curve. We need the residue of f(z) at 0. Here we do not have a formula, but we can look at the Laurent expansion about 0:
∞ 13n
3/z2
e dz = 0.
γ
11. f(z) has only one singularity, a simple pole at −4i, and this is outside the region bounded by the curve. By Cauchy’s theorem,
8z−4i+1
dz = 0.
12. f(z) has a simple pole at 1 − 2i, which is enclosed by γ, so
z2
γ z−1+2idz=2πiRes(f,1−2i)
= 2πi((−1 + 2i)2) = 2π(4 − 3i).
13. The singularities of coth(z) = cosh(z)/sinh(h) are the zeros of sinh(z). This means that coth(z) has simple poles at nπi, with n any integer. Only the simple pole at 0 is enclosed by the curve, so
e3/z2 = n=0
.
The coefficient of 1/z in this expansion is zero, so the residue is zero and
n! z2
γ
z + 4i
cosh(0)
coth(z) dz = Res(f, 0) = 2πi cosh(0) = 2πi. γ
14. f(z) has simple poles at 2,2e2πi/3 and 2e4πi/3. Only 2 is enclosed by γ, so (1−z)2
γ z3−8 dz=2πiRes(f,2) =2πilim (1−z)2
=2πi 12 =6.
15. 0 and 4i are simple poles of f(z) and both are enclosed by γ, so
e2z
γ z(z−4i)dz=2πi[Res(f,0)+Res(f,4i)]
1 e4i =2πi −4i+ 4i
= π[cos(8)−1+isin(8)]. 2
z→2 3z2 1 πi
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502 CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM 16. The function has a double pole at 1 and γ is assumed to be a closed path
that enclosed 1, so
z2
γ (z−1)2 dz=2πiRes(f,1)
=2πilim d(z2)=4πi. z→1 dz
17. z0 is a zero of order 2 of h(z), but g(z0) ̸= 0. We want to show that Res(g/h,z0) = 2g′(z0) − 2 g(z0)h(3)(z0).
To do this, first write
with φ(z0) ̸= 0. Then Res(g/h, z0) =
Now, ,
and Then
h′′ (z0 ) 3 (h′′ (z0 ))2 h(z) = (z − z0)2φ(z),
d 2g(z) lim dz (z − z0) h(z)
z→z0
= lim dz (z−z0) φ(z)(z−z )2φ(z)
d2 g(z) z→z0 0
= lim d g(z) z→z0 dz φ(z)
= φ(z0)g′(z0) − φ′(z0)g(z0). (φ(z0 ))2
h′(z) = 2(z − z0)φ(z) + (z − z0)2φ′(z).
h′′(z) = 2φ(z) + 4(z − z0)φ′(z) + (z − z0)2φ′′(z), h(3)(z0) = 6φ′(z) + 6(z − z0)φ′′(z) + (z − z0)2φ(3)(z).
φ(z0) = 1h′′(z0) and φ′(z0) = 1h(3)(z0). 26
Substituting these into the expression for the residue, we have Res(g/h,z0) = 2g′(z0) − 2 g(z0)h(3)(z0).
h′′ (z0 ) 3 (h′′ (z0 ))2
18. Suppose first that f has a zero of order k at z0 in G. Consider the residue
of f′/f at z0. First, write
f(z) = (z − z0)kg(z),
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22.2.
THE RESIDUE THEOREM 503 for some function g(z) that is differentiable on an open disk about z0, and
g(z0) ̸= 0. Then
f′(z) = k(z − z0)k−1g(z) + (z − z0)kg′(z)
f(z)
= k
(z − z0)kg(z) + g′(z).
z − z0
Now g′/g is differentiable at z0 so the last equation implies that f′/f has
a simple pole at z0, and
Res(f′/f,z0) = k.
Now see what happens if f has a pole of order m at z1. In some annulus
about z1,
∞
f(z)= dn(z−z1)n n=−m
g(z)
with d−m ̸= 0. Then
(z−z1)mf(z)= dn(z−z1)n =dn−m(z−z1)n =h(z),
∞∞ n=−m n=0
with h(z) differentiable at z1 and h(z1) = d−m ̸= 0. Then f(z) = 1 h(z).
(z − z1)m
Now, in some disk about z1,
f′(z) = −m(z − z1)−m−1h(z) + (z − z1)−mh′(z) f(z) (z − z1)−mh(z)
This implies that
= −m +h′(z). z − z1 h(z)
Res(f′/f,z1) = −m.
Therefore, the sum of the residues of f′/f at poles of f enclosed by γ in G counts each zero of f enclosed by γ, according to its multiplicity, and each pole of f enclosed by γ, according to the negative of its multiplicity. If Z is the number of zeros of f enclosed by γ, counting multiplicities, and P the number of poles of f enclosed by γ, also counting multiplicities, then by the residue theorem,
f′(z)dz=2πi(Z−P). γ f(z)
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504 CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM 19. By the residue theorem, with g(z) = z/(2 + z2),
z
γ 2+z2
dz = 2πi Res(g, √2i) + Res(g, −√2i)
√2i −√2i =2πi √ + √
2 2i −2 2i = 2πi.
To use the argument principle, write g(z)=f′(z)=1 2z ,
f(z) 2 2 + z2
with f(z) = 2 + z2. Then f′/f = 2g. Now f(z) has two simple zeros
enclosed by γ, and no poles, so Z = 2, P = 0, and z dz=1 2z dz
γ 2+z2 2 γ 1+z2
= 1(2πi)(Z − P) = 2πi.
2
20. Let g(z) = tan(z). Then g has simple poles enclosed by γ at ±π/2. By the residue theorem,
tan(z) dz = 2πi[Res(g, π/2) + Res(g, −π/2)] γ
sin(π/2) sin(−π/2) = 2πi − sin(π/2) + − sin(−π/2) = 2πi(−1 + (−1)] = −4πi.
To use the argument principle, write g(z) = −f′(z)/f(z) with f(z) = cos(z). Now f(z) has no poles, and simple zeros at ±π/2 enclosed by γ. Then Z = 2 and P = 0, so
f′(z)
g(z)dz=− γγ
f(z) =−2πi(Z−P)=−4πi.
21. g(z)=(z+1)/(z2+2z+4)hassimplepolesat−1±√3ienclosedbyγ.
By the residue theorem
z + 1
γ z2 + 2z + 4
dz = 2πi Res(g, −1 − √3) + Res(g, −1 + √3)
1−1−√3i −1+√3i+1 =2πi √ + √
2(−1− 3i)+2 2(−1+ 3i)+2 1 1
=2πi 2+2 =2πi.
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22.3.
EVALUATION OF REAL INTEGRALS 505 To use the argument principle, write
z+1 = 1f′(z) z2 + 2z + 4 2 f(z)
where f(z) = z2 +2z+4. f(z) has z = 2 zeros enclosed by γ and no poles (P = 0), so
2z + 2 dz = πi(Z − P ) = 2πi. γ z2 + 2z + 4
Because p(z) has exactly n simple zeros enclosed by γ, then p′(z)/p(z) has simple poles at z1,··· ,zn and
Res(p′/p, zj ) = p′(zj ) = 1. p′(zj)
22.
By the residue theorem,
p ′ ( z ) n
Res(p′/p, zj ) = 2nπi. enclosed by γ, and a polynomial has no poles, so
γ p(z) dz = 2πi
To use the argument principle, note that p(z) has exactly n simple zeros
j=1
p′(z) dz = 2πi(n−0) = 2nπi. γ p(z)
22.3 Evaluation of Real Integrals
1. With z = eiθ, so
11 1 cos(θ)=2 z+z anddθ=izdz
2π 1 1 1
2 − cos(θ) dθ = 2 − 1 (z + 1/z) iz dz
0γ2
= 2i
The integrand has simple poles at z1 = 2− 3 and z2 = 2+ 3. Only z1
1
z2 − 4z + 1 dz. √√
is enclosed by γ, and
√
γ
11 √ =− √ .
Res(f,2− 3)=
Then
2π
2(2− 3)−4 2 3 1 −12π
2 − cos(θ) dθ = 2i(2πi)2√3 = √3. 0
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506 CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM 2. f(z) = 1/(z4 + 1) has two simple poles in the upper half-plane:
11
z1=√ (1+i)andz2=√ (−1+i).
22 We need the residues of f(z) at these points:
Res(f,z1) = 1 and Res(f,z2) = −1z2. 4z12 4
Then
∞
−∞x4+1dx=−4 (z1+z2)=√2.
1 2πi π
3. f(z) = 1/(1 + z6) has simple poles in the upper half-plane at z = i,
poles are
so
Then
4.
Res(f,zj) = 1 = −1zj, 6zj5 6
∞1 1 2π −∞1+x6dx=2πi 6(z1+z2+z3) = 3.
∞ 1 dx=π. 01+x6 3
2π 1 1 1
6 + sin(θ) dθ = 1 + 1 (z − 1/z) iz dz
√√
1
z2 = ( 3+i)/2, and z3 = (− 3+i)/2. The residues of f(z) at these
0γ2
=2
The integrand has simple poles at z1 = (−6+ 37)i and z2 = (−6− 37)i.
1
z2 +12iz+1dz. √√
γ
Of these, only z1 is enclosed by γ. Further,
√
Res(f,(−6+ 37)i)= 2z1 +12i = 2√37i.
11
Then, recalling the factor of 2 in front of the previously written integral,
we have
5. Let
2π1 12π
6+sin(θ)dθ=2πi √37i = √37. ze2iz
f(z)= z4 +16.
0
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22.3.
507 f has simple poles in the upper half-plane at z = (1+i)/ 2 and z =
EVALUATION OF REAL INTEGRALS
√
√
(−1 + i)/ 2. Compute the residues:
12
√√
Res(f,z1)= to obtain
e2
2i(−1+i) e2 2(−1 − i) 16i and Res(f,z2)= −16i
e−2√2e2√2i −e−2√2i −∞ x4 + 16 dx = Im 2πi 8 2i
√
πe−2 2 √
∞ xsin(2x)
= 4 sin(2 2).
6. f (z) = 1/(z2 − 2z + 6) has one simple pole in the upper half-plane, and it
is z1 = 1 +
so
7. First use the identity
to write
Let
√ Res(f,1+ 5i)= √
√
5i. The residue there is
1
2 5i
∞
−∞ x2 − 2x + 6 dx = √5 .
cos2(x) = 1(1 + cos(2x)) 2
1π
∞
−∞(x2+4)2 2 −∞ (x2+4)2
cos2(x) dx= 1 ∞ 1+cos(2x)dx. 1+e2iz
f(z) = (z2 + 4)2 .
Then f has a pole of order 2 in the upper half-plane at 2i. Compute
Then
d 1+e2iz Res(f,2i)= lim 2 =
1+5e−1 32i
.
z→2idx (z+2i)
∞ cos2(x) 1 1+5e−1 −∞(x2+4)2dx=2Re 2πi 32i
= π(1+5e−4). 32
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508 CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM 8. Complex methods will work with this integral, but it is easier to observe
that, with the change of variables θ = 2π − φ,
2π sin(θ) π sin(φ)
Then
2+sin(θ) dθ = − 2+sin(φ) dφ. π0
2π sin(θ) dθ = 0. 0 2 + sin(θ)
9. Let f(z) = z2/(z2 +4)2. The only singularity of f in the upper half-plane is 2i, which is a double pole. Compute
d z2 i Res(f,2i)= lim 2 =− .
Then
10. Let
−∞(x2+4)2dx=2πi −8 =4. eiβx
z→2idz (z+2i)
∞ x2 iπ
8
f(z) = (z2 + α2)2 .
The only singularity of f in the upper half-plane is a double pole at αi.
We need
Then
11. Let
Res(f,αi)= lim d eiαz z→αi dz (z + αi)2
(αβ + 1)e−αβ =− 4α3 i.
∞
−∞(x2+α2)2dx=2πi − 4α3
cos(βx) (αβ + 1)e−αβ = (αβ+1)eαβπ.
i
f(z)= z2 +1.
The only singularity f has in the upper half-plane is a simple pole at i.
Compute
Then
e−α Res(f,i)= 2i.
∞ cos(αx) e−α −α −∞ x2+1dx=2πi 2i =πe .
2α3 eiαz
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22.3. EVALUATION OF REAL INTEGRALS 509
12. Let
z2eiαz f(z) = (z2 + β2)2 .
This function has a double pole at βi in the upper half-plane. The residue there is
Res(f,βi)= lim d z2eiαz z→βi dz (z + βi)2
= lim (2zeiαz(z + βi)−2 + iz2αz(z + βi)−2 − 2z2eiαz(z + βi)−3) z→βi
= 2βi(2βi)−2 + iα(−β)2e−αβ(2βi)−2 − 2(−β2)e−αβ(2βi)−3 e−αβ
= 4β i(αβ−1). Then
∞ x2 cos(αx) e−αβ −∞ (x2 +β2)2 dx=2πi 4β i(αβ−1
= π e−αβ(1−αβ). αβ
2π 1 dθ 0 α2 cos2(θ) + β2 sin2(θ)
i γ (α2 −β2)z4 +2(α2 +β2)z2 +(α2 −β2) Singularities of the integrand satisfy
z2 = β − α or z2 = β − α. β+α β+α
13. Begin with
11
=
=4 z dz.
γ
α2(z + 1/z)2/4 − β2(z − 1/z)2/4 iz dz
Because α and β are positive,
β − α β + α
β + α < 1 and β − α > 1.
The simple poles enclosed by the unit circle are the square roots z1 and z2 of (β − α)/(β + α). The residue of the integrand at each of these poles can be computed using Corollary 22.1. Omitting the arithmetic, we obtain
Res(f,zj)= 1 8αβ
for j = 1,2. Then
2π 1 dθ=4(2πi)2 =2π.
0 α2 cos2(θ) + β2 sin2(θ) i 8αβ αβ
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510 CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM
14. Let
Write
g(θ) = 1 . α + sin2(θ)
π/2 π/2 π
g(θ) dθ = g(θ) dθ +
0 0 π/2
3π/2 +
g(θ) dθ 2π
g(θ) dθ.
In the second, third and fourth integrals on the right, put, respectively,
3π/2
θ = π − u, θ = π + u, θ = 2π − u.
π/2 1 2π 1
g(θ) dθ = 4 α + sin2(θ) dθ
00
π
g(θ) dθ +
This leads to
fying
zj2 =(1+2α)−2α(α+1).
Omitting the arithmetic, the residues at each of these poles is the same,
= 1 1
4 γ α−(z−1/z)2/4iz
1 dz z4 −(2+4α)z2 +1 dz.
π/2 0
z γ
= i
The integrand has two simple poles bounded by the unit circle, and satis-
given by
Then
−1 Res(f,zj)= 8α(α+1).
1 −2 α + sin2(θ) dθ = i(2πi) 8α(α + 1)
π
= 2α(α+1).
15. Let Γ be the given rectangular path. The four sides are:
Γ1 :z=t,−R≤t≤R,
Γ2 :z=R+it,0≤t≤β, Γ3 :z=t+iβ,−R≤t≤R, Γ4 :z=−R+it,0≤t≤β.
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22.3.
EVALUATION OF REAL INTEGRALS 511
These are, respectively, the lower side, right side, top side and left side of the rectangle. In carrying out the integrations, limits of integration must be consistent with counterclockwise orientation of Γ.
Because e−z2 is differentiable for all z, then by Cauchy’s theorem, 4
e−z2 dz=0= e−z2 dz. Γ j=1 Γj
Evaluate each of the four integrals in the sum on the right as follows.
2R2 e−z dz = e−t
−R 2β22
dt,
2β 2
et [cos(2Rt)−isin(2Rt)]dt,
0
2R22 e−z dz = e−(t +2βti−β ) dt
Γ3 −R
−R
Γ1
e−z dz = e−(r +2Rti−t )i dt
Γ2 0 =ie−R
=e−β 2022
2R 2
e−t [cos(2βt)−isin(2βt)]dt,
and
Now let R → ∞. The terms having a factor of e−R2 go to zero in the limit, and upon adding these integrals over the sides of the rectangle, we obtain, using x as the variable of integration on the line,
∞22∞
e−x dx − e−β [cos(2βx) − i sin(2βx)] dx = 0.
−∞
Now, e−x2 sin(2βx) is an odd function on the real line, so
∞2
e−x sin(2βx) dx = 0.
−∞
We are therefore lift with
2∞ ∞2
eβ cos(2βx)dx= e−x dx.
−∞ −∞
e−z dz = e−(R −2Rti−t Γ4 β
=ie−R
2β 2 0
−∞
)i dt
[et [−cos(2βt)−isin(2βt)]dt.
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512
CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM
Finally, use the known result that
to conclude that
∞2√ e−x dx= π
−∞
∞2 √2 e−x cos(2βx)dx= πe−β .
16.
−∞
Finally, because e−x2 cos(2βx) is an even function on the real line, then
∞ √π
e−x2 cos(2βx) dx = 2 e−β2 . 0
Denote the given closed curve as Γ. By Cauchy’s theorem,
iz2
e dz=0.
Γ
Γ consists of the segment Γ1 on the x− axis, the circular arc Γ2, and the
segment Γ3 from the end of this arc to the origin. On Γ1, z = x and
2R eiz dz =
Γ1 0 OnΓ2,z=Reiθ and
2
eiz dz =
[cos(x2) + i sin(x2)] dx.
π/4 2
eiR e2iθ dθ.
Γ2 0
On Γ3, z = reiπ/4 and 202
eiz dz = e−r eiπ/4 dr, Γ3 R
with the limits of integration set to maintain counterclockwise orientation around Γ.
Now we want to take the limit as R → ∞. The integral over Γ3 has limit zero because of the factor e−r2 . The integral over Γ1 only has R in its upper limit of integration and becomes an integral over the half-line. The integral over Γ2 is not as obvious. Make the change of variable u = 2θ to get
2 1π/22 2
eiz dz = 2 eiR cos(u)−R sin(u)iReiu/2 du.
Γ2 0
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22.3.
EVALUATION OF REAL INTEGRALS 513 Then
iz2 R π/2 iR2 cos(u) iu/2 −R2 sin(u) edz≤2 e ee du
Γ2 0
R π/2 2
e−R sin(u) du
≤Rπ 2 2πR2
=π→0 4R
as R → ∞. In the limit, we therefore obtain
∞ ∞2
= 2
0
Because
and
we finally have
e
iπ/4
1
=√ (1+i),
2
[cos(x2) + i sin(x2)] dx − eiπ/4 00
∞ √π
e−x2dx= 2, 0
e−x
dx = 0.
17.
∞ ∞ √π cos(x2)dx+i sin(x2)dx= √ (1+i).
0022
From the real and imaginary parts of both sides of this equation, we have
∞ ∞ 1π cos(x2)dx = sin(x2)dx = 2 2.
00
First observe that, because the integrand is an even function, ∞xsin(αx)dx=1∞ xsin(αx)dx.
Now,
0 x4+β4 2 −∞ x4+β4 f(z) = zeiαz
z4 + β4
has simple poles in the upper half-plane at z1 = βeiπ/4 and z2 = βe3πi/4.
Compute the residues of f at these poles: zeiαz
eiαβzk = 4z2 .
Res(f,zk)= 4z3 1 eiαβeiπ/4 and
4β2i
z=zk
k
1 eiαβe3iπ/4 . −4β2i
In particular, Res(f, z1) =
Res(f, z2) =
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514
CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM
Then
∞ xsin(αx) 1 2πi iαβ(1+i)/√2
0 x4+β4 dx=2Im 4β2 e −e
√
πe−αβ/ 2 αβ
iαβ(−1+i)/√2 1 i
= 2β2 sin √2 .
2π 1 π 1 2π 1
18. First write
(α+βcos(θ))2 dθ = (α+βcos(θ))2 dθ+ (α+βcos(θ))2 dθ.
00π
In the last integral on the right, put θ = 2π − u to show that the two
integrals on the right are equal. Therefore
π 1
0 (α+βcos(θ))2
The function
f(z) =
has double poles at the zeros of βz2 + 2αz + β, which are
−α+α2 −β2 −α−α2 −β2 z1= β andz2= β .
dθ = 1 2π 1 dθ 2 0 (α+βcos(θ))2
= 1 1
2 γ (α+β(z+1/z)/2)2 iz
=2 z dz. i γβz2+2αz+β)2
1 dz
z
(βz2 + 2αz + β)2
Because z2 lies outside the unit disk, we need only the residue at z1: Res(f,z1)= lim d z
Then
π 0
z→z1 dz β2(z − z1)2 =1 αβ2
β2 4(α2 − β2)3/2 =α.
1 dθ = 2(2πi) α
(α + β cos(θ))2 i 4(α2 − β2)3/2
= πα . (α2 − β2)3/2
4(α2 − β2)3/2
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Chapter 23
Conformal Mappings and Applications
23.1 Conformal Mappings
For Problems 1–3, the image of the given rectangle is given as a graph for each part of the problem.
1. (a)Therectangledefinedby0≤x≤π,0≤y≤πmapstothesector 1 ≤ r ≤ eπ,0 ≤ θ ≤ π.
See Figure 23.1.
(b) This rectangle maps to the sector
1 ≤ r ≤ e, − π ≤ θ ≤ π . e22
See Figure 23.2.
(c) The rectangle maps to the sector
1 ≤ r ≤ e, 0 ≤ θ ≤ π . 4
See Figure 23.3.
(d) The rectangle maps to the sector
e ≤ r ≤ e2,0 ≤ θ ≤ π. (e) The rectangle maps to the sector
See Figure 23.4.
See Figure 23.5.
1 ≤ r ≤ e2,−π ≤ θ ≤ π. e22
515
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516 CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS
Figure23.1: Imageoftherectangle0≤x≤π,0≤y≤πunderw=ez.
Figure 23.2: Image of −1 ≤ x ≤ 1, −π/2 ≤ y ≤ π/2.
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23.1. CONFORMAL MAPPINGS 517
Figure 23.3: Image of the rectangle 0 ≤ x ≤ 1,0 ≤ y ≤ π/4.
Figure23.4: Imageoftherectangle1≤x≤2,0≤y≤π.
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518
CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS
2.
Figure 23.5: Image of the rectangle −1 ≤ x ≤ 2, −π/2 ≤ y ≤ π/2.
To determine the image of a rectangular area with sides parallel to the axes, we need to know the images of horizontal and vertical lines under this mapping. First write
w = cos(z) = cos(x + iy) = u + iv = cos(x) cosh(y) − i sin(x) sinh(y), so
u = cos(x) cosh(y) and v = − sin(x) sinh(y).
This means that points x, y) map to points (cos(x) cosh(y), − sin(x) sinh(y))
in the w−plane.
Consider the images of vertical and horizontal lines, beginning with a
vertical line x = a. If cos(a) and sin(a) are nonzero, then u2 v2
cos2(a) − sin2(a) = 1.
This is the image of one branch of a hyperbola in the w−plane, because
cosh(y) > 0, forcing u > 0 if cos(a) > 0 and u < 0 if cos(a) < 0.
Next consider the case that sin(a) = 0 or cos(a) = 0.
If sin(a) = 0, then a = nπ, with n an integer, and cos(a) = (−1)n. If n is even, say n = 2m, then cos(a) = 1. Because cosh(y) ≥ 1, the vertical line x = 2mπ maps to the interval u ≥ 1,v = 0 on the real line in the
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23.1. CONFORMAL MAPPINGS 519
Figure 23.6: Image of the rectangle 0 ≤ x ≤ 1,1 ≤ y ≤ 2 under w = cos(z).
Figure 23.7: Image of the rectangle π/2 ≤ x ≤ π,1 ≤ y ≤ 3.
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520
CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS
3.
Figure 23.8: Image of the rectangle 0 ≤ x ≤ π,−π/2 ≤ y ≤ π.
w−plane. If n is odd, say n = 2m + 1, then cos(a) = −1 and a vertical line x = (2m + 1)π maps to the interval u ≤ −1, v = 0 in the w−plane.
For the case cos(a) = 0, suppose a = (2n + 1)π/2, with n any integer. The image of a point on the line x = a is (0, − sin((2n + 1)π/2) sinh(y), varying over the entire imaginary axis in the w−plane as y varies over all real values.
This takes care of images of horizontal lines. Now look at a horizontal line y = b. The image of a point on this line is a point (u,v) with
u = cos(x) cosh(b), y = − sin(x) sinh(b). If b ̸= 0, these points lie on the ellipse
u2 v2 cosh2(b) + sinh2(b) = 1.
The image of a horizontal x = b ̸= 0 is an ellipse. Ifb=0,theny=0istherealline. Noww=u=cos(x),andtheimage
of the real line is the interval −1 ≤ u ≤ 1 on the real line in the w−plane. For the images of the rectangles of parts (a) through (e), see Figures 23.6–
23.10, respectively.
For the images of the given rectangles of parts (a) through (e), see Figures 23.11–23.15, respectively.
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23.1. CONFORMAL MAPPINGS 521
Figure 23.9: Image of the rectangle π ≤ x ≤ 2π,1 ≤ y ≤ 2.
Figure 23.10: Image of the rectangle 0 ≤ x ≤ π/2,0 ≤ y ≤ 1.
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522 CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS
Figure 23.11: Image of the rectangle 0 ≤ x ≤ π/2, 0 ≤ y ≤ π/2.
Figure 23.12: Image of the rectangle π/4 ≤ x ≤ π, 0 ≤ y ≤ π/2.
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23.1. CONFORMAL MAPPINGS 523
Figure 23.13: Image of the rectangle 0 ≤ x ≤ 1,0 ≤ y ≤ π/6.
Figure 23.14: Image of the rectangle π/2 ≤ x ≤ 3π, 0 ≤ y ≤ π/2.
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524 CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS
Figure 23.15: Image of the rectangle 0 ≤ x ≤ π/2,0 ≤ y ≤ 1.
Figure 23.16: Image of the rectangle 1 ≤ x ≤ 2,1 ≤ y ≤ 2.
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23.1. CONFORMAL MAPPINGS 525
4. Writez=reiθ sow=z2 =r2e2iθ. Ifrvariesfrom0to∞,sodoesr2. And as θ varies from π/4 to 5π/4, 2θ varies from π/2 to 5π/2. This is an interval of length 2π. Therefore the image of the sector π/4 ≤ θ ≤ 5π/4 is the entire plane.
5. The analysis proceeds like that of Problem 4. Let z = reiθ. If π/6 ≤ θ ≤ π/3, then π/2 ≤ 3θ ≤ π, so image points under this mapping lie in the second quadrant. It is routine to verify that this mapping is onto the second quadrant.
6. Let u = reiθ. Then
1iθ 1−iθ w=u+iv=2 re +ee .
u=2 r+r cos(θ)andv=2 r−r sin(θ). It is routine to check that
u 2 v 2
1 (r + 1/r) + 1 (r − 1/r) = 1
assuming that r ̸= 1. This is an ellipse in the w−plane. Because r+1/r > r − 1/r, this ellipse has foci at (±c, 0), where
2 1 12 12 c=4 r+r −r−r =1.
This means that a circle of radius r ̸= 1 maps to an ellipse with foci (±1, 0) in the w−plane.
If r = 1, then v = 0 and u = 2 cos(θ), so the image of the unit circle about the origin in the x,y−plane is the interval [−2,2] on the real axis in the w−plane.
7. Using some of the analysis from Problem 6, a half-line θ = k maps to points u + iv with
1 1 1 1
u=2 r+r cos(k),v=2 r−r sin(k).
If sin(k) ̸= 0 and cos(k) ̸= 0, then a little algebraic manipulation shows
Using Euler’s formula, we obtain
1 1 1 1
22
that
u2 v2 cos2(k) − sin2(k) = 1.
This is the equation of a hyperbola with foci (±c, 0), where c2 = cos2(k) + sin(k) = 1.
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526
CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS
Figure 23.17: Image of the rectangle in Problem 8(a) with α = 2.
Finally, the cases cos(k) = 0 and sin(k) = 0 must be considered separately. Ifcos(k)=0,thenk=(2n+1)π/2. Nowu=0and−∞
w = iz − 2i = (z − 2)i.
5. We can map the line Re(z) = 0 to the circle by |w| = 4 by a bilinear transformation. The domain Re(z) < 0 consists of numbers to the left of the imaginary axis, which is the boundary of this domain. Choose three points on this axis, ordered upward so the region Re(z) < 0 is on the left as we walk up the line. Next choose three points on the image circle |w| = 4, counterclockwise so the interior of the circle is on the left as we walk counterclockwise around the circle. Convenient choices are
z1 =−i,z2 =0,z3 =iandw1 =−4i,w2 =4,w3 =−4i.
Using equation (23.1), we find the bilinear transformation mapping zj →
wj:
1+z w=T(z)=4 1−z .
As a check, z = −1, which has negative real part, maps to 0, interior to thedisk|w|<4. ThusT mapsRe(z)<1to|w|<4,ratherthantothe exterior |w| > 4.
z + 2i z + 2i
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23.2. CONSTRUCTION OF CONFORMAL MAPPINGS 535
6. The domain Im(z) > −4 consists of all x + iy lying above the horizontal line y = −4. The boundary of this domain is this line. We want to map this domain onto |w − i| > 2, the exterior of the circle of radius 2 and center i in the w−plane. The circle |w − i| = 2 is the boundary of this domain. Choose three points on the line y = −4 in the z−plane, ordered from left to right so the domain is on the left. Choose three points on the circle in the w−plane, clockwise so as we walk around this circle, the exterior is on the left. Convenient choices are
z1 =−1−4i,z2 =−4i,z3 =1−4iandw1 =3i,w2 =2+i,w3 =−i. The bilinear transformation mapping zj → wj (use equation (23.1)) is
w= (−2+i)z−(3+10i). z + 3i
7. Because the boundary of the wedge in the w−plane is not a circle or line, a bilinear transformation will not work here. However, wedges suggest polar representations. Let z = reiθ for 0 < θ < π. These are points in the upper half-plane. Let
w = z1/3 = r1/3eiθ/3 = ρeiφ.
Here ρ > 0 and 0 ≤ φ ≤ π/3. This mapping is conformal because
dw = 1z−2/3 ̸= 0 dz 3
for z in the upper half-plane, and the mapping takes the open upper half- plane onto the open wedge 0 < θ < π/2.
8. Let z = x+iy = reiθ with y > 0. Then arg(z) = θ is unique (restricted to 0 ≤ θ < 2π). Further,
w = ln(r) + iθ.
Because r can be any positive number, ln(r) varies over all the real num-
bers. Further,
Therefore log(z) is in the strip 0 < Im(z) < π.
Im(z) = θ in (0, π).
To show that this mapping is onto, choose any w = u + iv in this strip.
Let z = rew. Then
and
Im(z) = eu sin(v) > 0 log(z) = u + iv = w.
The mapping is therefore onto. Finally,
d (log(z)) = 1 ̸= 0, dz z
so the mapping is conformal.
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536 9.
CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS
The solution of this problem requires some familiarity with the gamma and beta functions.
To show that f maps the upper half-plane onto the given rectangle, we will evaluate the function at −1, 0, 1 and ∞ and then show that these are the vertices of that rectangle.
First, it is obvious that f(0) = 0. Next,
1 0
Let ξ = u1/2 to obtain
1
f(1) = 2i
(ξ2 − 1)−1/2ξ−1/2 dξ 1 (1 − ξ2)−1/2 −1/2
= 2
(1 − ξ2)−1/2ξ−1/2 dξ.
1 0
=2i i ξ dξ 0
(1 − u)−1/2u−3/4 du Γ(3/4)
in which B(x,y) is the beta function and Γ(x) is the gamma function. Next, write
1 0
∞ 0
1 0
f(1) =
= B(1/4, 1/2) = Γ(1/4)Γ(1/2) = c,
0
f(−1) = 2i
(ξ2 − 1)−1/2ξ−1/2 dξ.
f(−1) = 2i Let ξ = −u to obtain
Finally,
(ξ + 1)−1/2(ξ − 1)−1/2ξ−1/2 dξ.
The first integral in the last line of the last equation is B(1/4,1/2). In
1 0
(1 − u2)−1/2u−1/2 du
= iB(1/4, 1/2) = i Γ(1/4)Γ(1/2) = ic.
f(∞) = 2i = 2i + 2i
(ξ + 1)−1/2(ξ − 1)−1/2ξ−1/2 dξ (ξ + 1)−1/2(ξ − 1)−1/2ξ−1/2 dξ
∞ 1
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Γ(3/4)
23.2.
CONSTRUCTION OF CONFORMAL MAPPINGS
the second integral, put ξ = 1/u to get
0 1+u−1/2 1−u−1/2
f(∞) = c + 2i u1/2
537
du
1 1 u u u2
= c + 2i
1 0
(1 − u2)−1/2u−1/2 du = (1 + i)u.
© 2018 Cengage Learning. All Rights reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.