程序代写代做代考 chain Homework 7 Solutions

Homework 7 Solutions

Chapter 5, Exercise 15

Part a: Let g be the same function from Exercise 2. By the Poisson summation
formula,

∞∑
n=−∞

sin2(πn+ πα)

π2(n+ α)2
=

∞∑
−∞

ĝ(n+ α) =

∞∑
−∞

g(n)e2πinα = 1.

Since sin(πn+ πα) = (−1)n sin(πα), the desired identity follows immediately.

Part b: It suffices to prove the claim when α ∈ (0, 1), since the sum on the
left hand side only depends on the fractional part of α. So let 0 < α < 1. First we note that for any N ∈ N N−1∑ n=−N 1 n+ 1 2 = 0, since the positive terms cancel the negative ones. Next note that for any α ∈ (0, 1), 1 n+ α − 1 n+ 1 2 = ∫ 1/2 α 1 (n+ x)2 dx, and thus by combining the previous two identities we get N−1∑ n=−N 1 n+ α = ∫ 1/2 α N−1∑ n=−N 1 (n+ x)2 dx. Letting N →∞ on both sides and using uniform convergence on [α, 1/2] of the series inside the integral, we get lim N→∞ N−1∑ n=−N 1 n+ α = ∫ 1/2 α π2 sin2(πx) dx = π tan(πα) , where the RHS is interpreted as 0 for α = 1/2, and we used the fact that the antiderivative of csc2 u is − cotu. 1 Chapter 5, Problem 1 Let us write U(y, t) = u(e−y, t). Then we see by the chain rule that Ut(y, t) = ut(e −y, t), Uy(y, t) = −e−yux(e−y, t), Uyy(y, t) = e −2yuxx(e −y, t) + e−yux(e −y, t). In particular, if ut = x 2uxx + axux, then Ut(y, t) = ut(e −y, t) = e−2yuxx(e −y, t) + ae−yux(e −y, t) = Uyy(y, t) + (1− a)Uy(y, t). Thus U satisfies the constant-coefficient equation Ut = Uyy + (1− a)Uy. Taking Fourier transforms, we see that Û satisfies the following ODE for fixed ξ: ∂tÛ(ξ, t) = −4π2ξ2Û(ξ, t) + 2πiξ(1− a)Û(ξ, t), where Û(ξ, t) = ∫ R U(y, t)e −2πiξydy is the Fourier transform in the y variable. Solving the ODE for Û gives the solution Û(ξ, t) = Û(ξ, 0)e(−4π 2ξ2+2(1−a)πiξ)t, so by inverting the Fourier transform, U(y, t) = ∫ R Û(ξ, 0)e(−4π 2ξ2+2(1−a)πiξ)te2πiξydξ. This last expression is simply the convolution of U(y, 0) = f(e−y) =: F (y) with the kernel pt(y) := ∫ R e (−4π2ξ2+2(1−a)πiξ)te2πiξydξ. By completing the square and then using the fact that ∫ R e −(cξ+d)2dξ = √ π/c2 for c ∈ R, we obtain that pt(y) = e −(y+(1−a)t)2/4t ∫ R e − ( 2πξt1/2− i 2t1/2 ( y+(1−a)t ))2 dξ = 1 √ 4πt e−(y+(1−a)t) 2/4t. Thus U(y, t) = pt ∗ F (y) = 1√4πt ∫ R F (z)e −(y−z+(1−a)t)2/4tdz, so that (trans- forming variables back to the original function), u(x, t) = U(− log x, t) = 1 √ 4πt ∫ R F (z)e−(− log x−z+(1−a)t) 2/4tdz = 1 √ 4πt ∫ ∞ 0 f(v)e−(log(v/x)+(1−a)t) 2 dv v , where we substituted z = − log v in the final line, so that dz became −dv v , F (z) = f(e−z) became f(v), and the limits changed from (−∞,∞) to (0,∞). 2 Chapter 5, Problem 5 Part a: Proceed by showing the contrapositive (i.e., prove that if u solves the heat equation but violates the stated maximum principle, then we can find a solution to the heat equation which violates the statement made in part a). To this end, let u solve the heat equation, then let A := max∂′R u, and sup- pose u has a strict maximum at (x0, t0) ∈ R\∂′R. Now consider the function v := A− u, and check that this also solves the heat equation. Clearly, v ≥ 0 on ∂′R, but v(x0, t0) < 0, since u(x0, t0) > A.

Part b: Suppose that v (as defined in the book) has a minimum at (x1, t1) ∈ R,
where x1 /∈ {a, b} and t1 6= 0. Since v has a minimum at (x1, t1),

vt(x1, t1) ≤ 0 and vxx(x1, t1) ≥ 0.

Here we are using the fact that t1 > 0 which ensures that v is actually differ-
entiable at (x1, t1), and we are also using x1 /∈ {a, b} (since boundary minima
need not satisfy the second derivative test). The above expression implies that
vxx(x1, t1)−vt(x1, t1) ≥ 0. On the other hand, since u solves the heat equation,
vt = ut + � = uxx + � = vxx + �, and thus vxx(x1, t1) − vt(x1, t1) = −� < 0, contradiction. Part c: Assume (see part a) that u ≥ 0 on ∂′R. For � > 0, we know from
part b that v� := u + �t achieves its minimum at some point (x�1, t


1) ∈ ∂′R,

which means that

u(x, t) + �t = v�(x, t) ≥ v�(x�1, t

1) = u(x


1, t


1) + �t


1 ≥ �t


1,

where we used the fact that u ≥ 0 on ∂′R in the final inequality. Consequently,
we find that u(t, x) ≥ �(t�1−t) for every � > 0, as desired. The claim now follows
by letting �→ 0.

3

Chapter 6, Problem 8

Part a: We set β = 2π|x| on both sides of the given expression. In order to
prove the expression, we need to take the Fourier transform (in the x variable)
of both sides and check they are equal. First the left side:∫

R
e−2π|x|e−2πiξxdx =

∫ ∞
0

e−2πx(1+iξ)dx+

∫ 0
−∞

e2πx(1−iξ)dx

=
1

[
1

1 + iξ
+

1

1− iξ

]
=

1

π(1 + ξ2)
.

Now the right side: we recall that

R e
−cx2e−2πiξxdx =


π
c
e−π

2ξ2/c, and so by
changing the order of integration, we get∫

R

[ ∫ ∞
0

e−u

πu
e−4π

2×2/4udu

]
e−2πiξxdx =

∫ ∞
0

e−u

πu

[ ∫
R
e−π

2×2/ue−2πiξxdx

]
du

=

∫ ∞
0

e−u

πu

[√
u

π
e−uξ

2

]
du

=
1

π

∫ ∞
0

e−(1+ξ
2)udu

=
1

π(1 + ξ2)
,

thus proving the claim.

Part b: Using the subordination principle, we may write

e−2π|ξ|y =

∫ ∞
0

e−u

πu
e−π

2|ξ|2y2/udu,

where ξ ∈ Rd and |ξ| = (ξ21 + …+ ξ2d)
1/2. Thus we find that

(continued on next page)

4

P (d)y (x) =


Rd
e2πix·ξe−2π|ξ|ydξ

=


Rd
e2πix·ξ

[ ∫ ∞
0

e−u

πu
e−π

2|ξ|2y2/udu

]

=

∫ ∞
0

e−u

πu

[ ∫
Rd
e2πix·ξe−π

2|ξ|2y2/udξ

]
du

=

∫ ∞
0

e−u

πu

[
ud/2

πd/2yd
e
−u |x|

2

y2

]
du

= π−(d+1)/2y−d
∫ ∞
0

u(d−1)/2e
−u(1+ |x|

2

y2
)
du

=
π−(d+1)/2y−d

(1 +
|x|2
y2

)(d+1)/2

∫ ∞
0

v(d−1)/2e−vdv

=
Γ((d+ 1)/2) y

π(d+1)/2(y2 + |x|2)(d+1)/2
.

In the second-to-last equality, we made the substitution v = u(1 +
|x|2
y2

), and in

the final line we used the definition of the gamma function: Γ(z) =
∫∞
0
vz−1e−vdv,

and also multiplied the numerator and denominator by yd+1.

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