Section 3.6 Calculus of Variations

III )

Section 3.6 Section 3.6 Calculus of Variations

For an extremum we want

dT / dx = ^O.

^Thus

-I dpi

⁺

-

^I

dP2

^{_ 0}

ci dx C2 dx -

^,

-I

¹

-I

¹

c -(x - xd I PI

⁺

c -(x - X2) = 2 P2

⁰

cll

^sin

4>1 = c2"1

^sin

4>2

This last equation is known as Snell's Law.

163

Now consider a medium in which the velocity of light is a function of

y;

^let

us say

c = c(y).

This would be the case in the Earth's atmosphere ^or in the ocean. Think of the medium as being composed of many thin layers, in each of which the velocity of light is constant. See Figure 3. 10, in which three layers are shown.

velocity c(

velocity c2

velocity c3

Figure 3.10 Snell's Law yields

sin

CI 4>1 =

^sin

C2 4>2 =

^sin

C3 4>3 = . . . = k

^constant

For a continuously varying speed

c(y),

the path of a ray of light should satisfy sin

4>(y) / c(y) = k.

Notice that the slope of the curve is

Hence

y'(x) =

^tan

(�

^-4»

=

^cot

4> =

^cos

4>/

^sin

4>

= J ¹

^-sin

2 4> /

^sin

4> = ^VI

^-

k2c2 / kc

--;::=::;=;;==;;:

_VI

_-

kc k2c2 dy = dx

^and

_{x -} ! ^{kc(y) dy}

- VI - k2c(y)2

Example 8. What. is the path of a light beam if the velocity of light in the medium is

c = ay

^(where

a

is a constant)?

Solution. The path is the graph of a function

y

such that The integration produces

! ^{kay d}

x = _VI - k2o.2y2 y

164

Chapter

³

Calculus in Banach Spaces

Here A ^and

k

are constants that can be adjusted so that the path passes through two given points. The equation can be written in the form

or in the standard form of a circle:

2 2

¹

(x

- A) +

Y

⁼

k2Q2

The analysis above can also be based directly on Fermat's Principle. The time taken to traverse a small piece of the path having length

tl.s

is

tl.s/c(y).

The total time elapsed is then

lb �-�..:...:.- dx

^{VI +}

Y'(X)2

a

c(y)

This is to be minimized under the constraint that

y(a)

⁼

Q

^and

^y (b)

^{= (3. Here}

the ray of light is to pass from

(a, Q)

^to

(b,

(3) in the shortest time. By Theorem 2, a necessary condition on

y

can be expressed (after some work) as

• In order to handle problems in which there are several unknown functions to be determined, one needs the following theorem.

(9)

Theorem 5.

that minimize the integral Suppose that Yh .. . , Yn are functions (of t) ⁱⁿ e2[a, b]

lb F(Yil .. . , Yn, y� , . . . , y�) dt

subject to endpoint constraints that prescribe values for all Then the Euler Equations hold: Yi(a), Yi(b).

d aF aF

₌

dt a ^y

:

a Y

ⁱ

(l � i � n)

Proof. Take functions 711 , _{. . .}

, 7Jn

in

e2[a, b]

that vanish at the endpoints. The expression

lb F(YI

^{+ 01 711 , . . .}

^,Yn

⁺

^{On7Jn) dt}

will have a minimum when (01 , • . •

, On)

= (0, 0, . . . ,0). Proceeding as in previous

proofs, one arrives at the given equations. •

Geodesic Problems. Find the shortest arc lying on a given surface and joining two points on the surface. Let the surface be defined by the two points be

( ^x

^o

,Y

^o,zo

)

^and

(XI,YI,ZI).

Arc length is ^z = z

( ^x

^,

^y).

^Let

Section

3.6

Calculus of Variations 165 If the curve is given parametrically as x

⁼

x(t), y = y(t), z = z(x(t), y(t)), o

�

t

�

1, then our problem is to minimize

( 10 11 ^{j x'2 + y,2}

⁺

^{(ZxX' + Zyy')2 dt}

subject to x

E

C2 [0, 1],

^Y E

C2[0, 1], x(O) = Xo, x(l) = XI , y(O)

=

Yo, y( l) =

YI . Example 9.

We search for geodesics on a cylinder. Let the surface be the cylinder x2 + z2 = 1 , or z = ( 1 - X2)1/2 (upper-half cylinder). In the general theory, F(x, y, x', y') = ..jX,2 + yl2 + (zxx' + Zyy')2. In this particular case this is F = [x,2

+

y,2 + z�x'2 f /2 = [( 1

^_

X2) -IX'2 + y'2 f ^/2

Then computations show that

of XX'2

oX ( 1 - x2)2 F of x' - = 0 of

[)y

of y' = oy'

^F

To simplify the work we take t to be arc length and drop the requirement that o

�

t

�

1. Since dt = ds = ..jX,2 + y,2 + Z'2 dt, we have X'2 + y'2 + Z'2

⁼

1 and F(x, y, x', y') = 1 along the minimizing curve. The Euler equations yield

( 1 - X2)X" + 2XX'2 xx,2

-'---'--=-:-7-- =

( 1 - x2)2 ( 1 - x2)2 and y" = 0

The first of these can be written x" = xx,2/(x2 - 1). The second one gives y = at +

b,

for appropriate constants a and

^b

that depend on the boundary conditions. The condition 1 = F2 leads to x,2/(1 - x2) + y,2 = 1 and then to x'2/ ( I - x2) = l - a2. The Euler equation for x then simplifies to x"

⁼

(a2 - 1)x.

There are three interesting cases:

Case 1: a = 1. Then x" = 0, and thus both x(t) and y(t) are linear expressions in t. The path is a straight line on the surface (necessarily parallel to the y-axis).

Case 2: a =

^o.

Then x" = -x, and x = c cos(t + d) for suitable constants

c

and d. The condition x'2/ ( 1 - x2) = 1 gives us

^c

= 1. It follows that x = ^{cos(t + d), y}

^{= b,}

and z = VI - x2 = sin(t + d). The curve is a circle parallel to the xz-plane.

Case 3: 0

^<

a

^<

1. Then x =

c

cos(

v'f=ll2

t + d), and as before, c

=

1. Again z = sin(

v'f=ll2

t + d) , and y = at +

b.

The curve is a spiral.

• Examples of Problems in the Calculus of Variations with No Solutions.

Some interesting examples are given in [CH], Vol. 1.

I. Minimize the integral fol \11 ⁺ y'2 dx subject to constraints y(O) = y(l) = 0, y'(O) = y'(I) = 1. An admissible curve is shown in Figure 3. 1 1 , but there is none of least length, since the infimum of the admissible lengths is 1 , but is not attained by an admissible y.

�

^I

o

�

Figure 3.1 1

166 Chapter 3 Calculus in Banach Spaces

II. Minimize

fl x2 y'(X)2 dx

subject to constraints that

y

be piecewise contin­

uously differentiable, continuous, and satisfy

y( -1) = -1, y(1)

⁼

1.

An admissible

y

is shown in Figure

3.12,

and the value of the integral for this function is

2£/3.

The infimum is

0

but is not attained by an admissible

y.

This example was given by Weierstrass himself!

· 1 + 1

· 1

Figure

3.12

Direct Methods in the Calculus of Variations. These are methods that proceed directly to the minimization of the given functional without first looking at necessary conditions. Such methods sometimes yield a constructive proof of existence of the solution. (Methods based solely on the use of necessary conditions never establish existence of the solution.)

The Rayleigh-Ritz Method. (We shall consider this again in Chapter 4.) Suppose that

U

is a set of "admissible" functions, and

<I>

is a functional on

U

that we desire to minimize. Put ^{p =} inf

{<I>( u)

^: U E

U}.

We assume ^p

>

-00, and seek a U E

U

such that

<I>(u) =

^p. The problem, of course, is that the infimum defining ^p need not be

attained.

In the Rayleigh-Ritz method, we start with a sequence of functions

W

I,

W2,

^•

.

^. such that every linear combination CI WI + ^{. .}. + CnWn is admissible. Also, we must assume that for each U E

U

and for each ^f

> 0

there is a linear combination v of the Wi such that <1>( ^v) ,:;; <1>(

u)

+ ^{f .} For each

n

we select Vn in the linear span of WI , ^{. . •}

,

Wn to minimize <1>( ^vn

)

^. This is an ordinary minimization problem for

n

real parameters CI , ^{• . .} , Cn . It can be attacked with the ordinary techniqueR of calculus.

Example 10. We wish to minimize the expression

I � ^I

^oQ

⁽ I/> ^;

⁺

I/>�) ^{dx dy}

^sub­

ject to the constraints that

I/>

be a continuously differentiable function on the rectangle R

= ^{{(x, y)}

^{: 0 ':;;}

^{x ,:;; a,}

^{0 ':;;}

y ':;; b},

that

I/>

⁼

⁰

on the perimeter of

R, and that

II R 1/>2 dx dy = 1.

A suitable set of base functions for this problem is the doubly indexed sequence

2 . n1rx . mrry

unm(x, y)

= rL

v ab

Sin -- Sin

a -b- (n, m � 1)

It turns out that this is an orthonormal set with respect to the inner prod­

uct

(u, v) = IIR u(x, y)v(x, y) dxdy.

We are looking for a function

I/> ⁼ L:�

m=1 CnmUnm that will solve the problem. Clearly, the function

I/>

^vanishes

on the perimeter of R. The condition

11 1/>2 = 1

means

L:�

^{m=1 c�m =}

¹

^{by the}

Parseval identity (page

73).

Now we compute

8

2 nrr nrrx . mrry

-

8

x u

nm

(x

,

y)

=

v ab a

rL - cos - sm -b-

a