CS计算机代考程序代写 Basics of the Calculus of Variations

Basics of the Calculus of Variations
A List of Key Points:
• What is a variational problem?
• The problem of Lagrange
• Necessary conditions for the optimum
• Sufficient conditions for the optimum
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A Variational Problem
ConsiderafunctionalJ: F n, withF n denoting the set of functions from  to n.
Goal: Find, if possible, an optimal solution x* :  n
such that
J x*  min J(x)
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Examples of Functional
1)J(x)tb f(x,x,t)dt, (theproblemofLagrange) t
a
2) J x f(x(tb),tb) f(x(ta),ta)
 f (x(t),t) tb (the problem of Mayer)
ta
3)J (x) f (x(t),t)tb  tb f(x,x,t)dt tt
aa
(the problem of Bolza)
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John Bernoulli’s Brachistochrone Problem (1696)
Find the shortest time of descent of a bead along a wire joining two nearby fixed points.
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P0 g
(x0, y0)
P1 (x1, y1)

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Brachistochrone Problem
Let P  (x , y ), P  (x , y ) and  be 000111
the curve joining these points. Then,
the time T required for a bead to move
from P to P along  is: 01
T  T dt   ds 0v
with speed v  ds , ds  arc length along . dt
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Bernoulli’s Brachistochrone Problem (cont’d)
At location P  (x, y), the bead has: kinetic energy  1 mv2
2 potential energy  mgy
Without considering friction, according to conservation law, 1mv2 mgyconstantmgy
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2
0

Bernoulli’s Problem (cont’d)
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If one parameterizes  as :yY(x), xxx
01 then v 2gy0Y(x)
On the other hand, ds  1Y(x)2 dx Therefore,
 ds x 1Y(x)2
 1 dx J(Y)
v x0 2gy0Y(x)

Basic Definitions
 Strong relative minimum:
A function xa (t) is said to yield a strong relative minimum
J(xa)overtta,tb if 10s.t. 
J(xa)J(x)for xax1,tta,tb . 
 Weak relative minimum:
J ( xa ) is said to be a weak relative minimum
overtta,tbif 1,20s.t.
J(x)J(x)for x x, x x andtt,t .
a a 1a2 ab
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

Basic Definitions
 Strongly continuous:
J is strongly continous at xa if   0,   0 s.t.
xxa   J(x)J(xa).
 A variation x:
A (small) change to a function x  x(t), i.e.
x  x  x.
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Question
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So, how to solve the problem of Lagrange minJ(x)tb f(x,x,t)dt
t a
assuming x is at least continuous.

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Necessary Conditions
To this end, consider the increment
J  J(xx)J(x)
As a direct application of Taylor formula
around   0, it holds: JJ(xx)|0 2J(xx)
2  2!
 2  J  2J  3J 
0
2! 3!

J (x  x) |0 
 0,
 0. (minimization only)
2J(xx) 2
0
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Necessary Conditions
Therefore, we have necessary optimality conditions:

An Exercise
Let the functional J  J(x) be defined as J(x)  1x(t)2  x(t)3 dt
0
for all continuous functions x from [0,1] to ,
i.e. xC0 ([0,1];).
Give the first variation of J , around any given x.
1 
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Answer
J J(x;x)12x(t)3x(t)2x(t)dt 0
for any xC0 ([0,1];).
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Back to the Problem of Lagrange
As a particular case, consider J(x)tb f(x,x,t)dt
t a

J(xx)|
The Problem of Lagrange
ε=0

xf dttb  tb f f dtxdt
t xta axx
When the end conditions are fixed (e.g., in Bernoulli’s problem),
xfdttb 0.  x ta
Then,

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tb f f dtxdt0.
ta
xx

Proof:
 J  0  N  f x   f x d t  c o n s t a n t
tb f f dtxdttb N(t)xtdt
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Fundamental Lemma
t x x  t  aa
N(t)x t b  t
t dN(t)   b x t dt
tt 
t
dt
0
a
a

First‐order Necessary Condition
Any optimal solution must satisfy
f f dt  c xx
Curves satisfying the above equation are called “first-variational curves”.
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Consequences of the Necessary Condition
• A Corner Condition
• The Euler‐Lagrange Equation
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A Corner Condition
A corner point is a point at which dx/dt has a jump discontinuity.
At any corner point t  tc ,
lim f (x(t), x(t),t)  lim f (x(t), x(t),t).
ttc x  ttc x  thanks to the first-order necessary condition.
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The Euler‐Lagrange Equation
Differentiating both sides of the necessary condition yields:
ELequation: dfxfx0 dt
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Comment
Between corners of x(t), EL equation can be rewritten as
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d2xf dxf f f dt2 xx dt xx xt x
with boundary conditions x(ta )  ca , x(tb )  cb .

Application to the Shortest Distance Problem
For the purpose of illustration, consider the simple of problem of finding the shortest distance between two given points:
P  x , y  and P  x , y . 000111
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The Shortest Distance Problem (cont’d)
Consider curves parameterized as: y  (x), forx xx.Then,weintendtominimize
01
the arc length of such curve:
J()x x0
1(x) dx  x 1(t) dt x0
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1212

The Shortest Distance Problem (cont’d)
A direct application of EL equation yields:
d  dt
(t) 
0 
(t) 1 (t)2
constant
1 (t)2 
 
which, in turn, implies
(t)  A, or, (t)  At  B
where the constants A and B can be determined
from (x )  y , x  y . 0011
This is a straight – line segment, as wished !
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Another Application of EL Equation
Recall that EL equation is often called a two‐point boundary‐value problem:
d 2 x f x x  d x f x x  f x t  f x dt2 dt
with boundary conditions x(ta )  ca ,
x(t )  c . bb

Another Application (cont’d)
Consider the minimization problem
t
minJ(x) b ax2axxax2g(t)xg(t)xdt
ta where
 12312

a real constants, a , a  0, g (t)C1. i13i
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Solution
Applying EL equation to the problem yields:
2ax2axg (t)g(t) 1 3 2 1
D e f i n e  2  a 3 , g 0  g 2  g 1 a 2a
11
x(t)cet c et  1 tta e eg (t)d
1220 0 with the constants c , c and t determined by
12a the boundary conditions.
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Isoperimetric Constraints
Consider a constrained problem of the form
minJ(x)tb f(x,x,t)dt t
a
subjectto: K(x)tb f (x,x,t)dtK
ta 1  1 withK1 aconstant.

A Motivational Example
Goal: find the dimensions of the rectangle having the smallest perimeter among all rectangles with given fixed area A.
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Euler‐Lagrange Multiplier Theorem (Necessary Conditions)
Letx bealocalextremum.Assumethatbothvariations J and K be weakly continuous near x.
Then, at least one of the following conditions holds: 1. K(x;x)0, x.
2. There exists a constant  such that J(x;x)K(x;x), x.
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Problem formulation to the example
min J (x)  2(x  x ) (Perimeter) 12
subjectto K(x)xx A (Area) 12
x  0, x  0. 12
(Note: Of course, this particular problem can be also considered as a nonlinear optimization task and be solved using Cauchy-Schwarz inequality.)
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Solution to the example
First variations of J and K: J(x;x)2x 2x
K(x;x)xx xx 2112
Clearly,K(x; )0 x 0,x 0. 12
So, the 2nd possibility holds, i.e., , s.t. 2x 2x x*x x*x
122112
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12

Solution to the example (cont’d)
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(2x*)x (2x*)x 0 21 12
 (2x*)0, (2x*)0 21
 x*  x*  12
2

x*x* A,becausex*x*A.
1212 Finally, it is directly checked that
J(x*)J(x), xx*x.

Isoperimetric Theorem
Assume that the function x* (t) is a first-variational curve which results in the minimum of
J (x)tb f(x,x,t)h f (x,x,t)dt at11
a i.e.
t
J (x)J (x*) b f(x*,x*,t)hf(x*,x*,t) dt
 with h independent of x,t.
 aata 11
1
Further assume that the constraint is satisfied by x* (t).
Then, x*(t) is the minimal solution to the problem.
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