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CALCULUS OF VARIATIONS

MA SOLUTION MANUAL

B Neta

Department of Mathematics

Naval Postgraduate Scho ol

Co de MANd

Monterey California

June

c

Professor B Neta

Contents

Functions of n Variables

Examples Notation

First Results

Variable EndPoint Problems

Higher Dimensional Problems and Another Pro of of the Second Euler

Equation

Integrals Involving More Than One Indep endentVariable

Examples of Numerical Techniques

The RayleighRitz Metho d

Hamiltons Principle

Degrees of Freedom Generalized Co ordinates

Integrals Involving Higher Derivatives i

List of Figures

tan sec Plot of y and y

ii

Credits

Thanks to Lt William K Co oke USN Lt Thomas A Hamrick USN Ma jor Michael

R Hub er USA Lt Gerald N Miranda USN Lt Coley R Myers USN Ma jor Tim A

Pastva USMC Capt Michael L Shenk USA who worked out the solution to some of the

problems iii

CHAPTER

Functions of n Variables

Problems

Use the metho d of Lagrange Multipliers to solve the problem

minimize f x y z

sub ject to xy z

Show that

max

cosh cosh

where is the p ositiverootof

cosh sinh

Sketchtoshow

Of all rectangular parallelepip eds whichhave sides parallel to the co ordinate planes and

which are inscrib ed in the ellipsoid

y z x

a b c

determine the dimensions of that one which has the largest volume

Of all parab olas which pass through the p oints and determine that one

which when rotated ab out the xaxis generates a solid of revolution with least p ossible

volume b etween x and x Notice that the equation may b e taken in the form

y x cx x when c is to b e determined

a If x x x x isarealvector and A is a real symmetric matrix of order n

n

show that the requirementthat

T T

F x Ax x x

b e stationary for a prescib ed Atakes the form

Ax x

Deduce that the requirement that the quadratic form

T

x Ax

b e stationary sub ject to the constraint

T

leads to the requirement

Ax x

where is a constant to b e determined Notice that the same is true of the requirement

that is stationary sub ject to the constraintthat constant with a suitable denition

of

b Show that if we write

T

x Ax

T

x x

the requirement that b e stationary leads again to the matrix equation

Ax x

Notice that the requirement d can b e written as

d d

or

d d

Deduce that stationary values of the ratio

T

x Ax

T

x x

are characteristic numb ers of the symmetric matrix A

f x y z

xy z

F f x y z xy z

F

x y

x

F

y x

y

F

z

z

F

xy z

z

z xy

and xy

Substitute in

x xy y

y xy x

x xy y

y x y x

xy y x

x or y or x y

x z y by

z x by y

x y z xy

x

Not p ossible

So the only p ossibility

x y z

f

Find max

cosh

Dierentiate

d cosh sinh

d cosh

cosh

Since cosh

cosh sinh

The p ositiverootis

Thus the function at b ecomes

cosh

No need for absolute value since

5

4

3

2

1

0 λ 0 −1

−2

−3

−4

−5

−5 −4 −3 −2 −1 0 1 2 3 4 5

Figure

x y z

max xy z st

a b c

x y z

Write F xy z then

a b c

x

F yz

x

a

y

F xz

y

b

z

F xy

z

c

y z x

F

a b c

If anyofx y or z are zero then the volume is zero and not the max Therefore x y

z so

x y y x

xF yF

x y

a b b a

Also

x y y z

zF yF

z y

a b b c

y b b

p

Then by y taking only the square ro ot length y

b

a c

p p

x z by resp ectfully

y z x

has dimension The largest volume parallelepip ed inside the ellipsoid

a b c

b c a

p p p

y x cx x

Z

Volume V y dx

Z

min V x cx x dx

Z

dV c

x cx x x xdx

dc

Z Z

x xdx c x x dx

c x x x x x

c

c

c

y x x x

Z

h i

V c x cx xc x x dx

e c

V c

T T

F x Ax x x

X X

A x x x

ij i j

i

i j i

X X

F

A x A x x k n

kj j ik i k

x

k

j i

T

Ax A x x

Since A is symmetric

Ax x

T T

min F x Ax x x c

implies by dierentiating with resp ect to x kn

k

Ax x

T

x Ax

b

T

x x

To minimize we require

d d

d

Divide by

d d

or

d d

CHAPTER

Examples Notation

Problems

For the integral

Z

x

f x y y dx I

x

with

y f y

write the rst and second variations I and I

Consider the functional

Z

J y xy dx

where y is twice continuously dierentiable and y and y Of all functions of

the form

y xx c x xc x x

where c and c are constants nd the one that minimizes J

f y y

y y f

y

y y f

y

Z

x

I dx y y y y

x

y y f

yy

y y f

yy

f y

y y

Z

x

y y y y y dx I

x

We are given thatafter expanding y x c x c c x c x Then wealso

have that

y x c c c x c x

and that

y x c xc c c x c c

x c c x c c c c x

Therefore wenow can integrate J y and get a solution in terms of c and c

R

c c c c xy dx

c c c c

c c c c

To get the minimumwewanttosolve J andJ After taking these partial

c c

derivatives and simplifying we get

J c c

c

and

c J c

c

Putting this in matrix form wewanttosolve

c

c

Using Cramers rule wehavethat

c

and

c

Therefore wehavethatthey xwhich minimizes J y is

y x x x x

x x x

Using a technique found in Chapter it can b e shown that the extremal of the J y is

ln x y x

ln

which after expanding ab out x is represented as

y x x x x Rx

ln ln ln

x x x Rx

So we can see that the form for y xgiven in the problem is similar to the series representation

gotten using a dierent metho d

CHAPTER

First Results

Problems

Find the extremals of

Z

x

F x y y dx I

x

for each case

a F y k y k constant

b F y y

c F y xy

d F y yy y

e F x y yy y

f F axy bxy

g F y k cos y

Z

h i

b

Solve the problem minimize I y y dx

a

with

y a y y b y

a b

What happ ens if b a n

ShowthatifF y xy y thenI has the same value for all curves joining the

endp oints

A geo desic on a given surface is a curve lying on that surface along which distance

between two p oints is as small as p ossible On a plane a geo desic is a straight line Determine

equations of geo desics on the following surfaces

v

u

Z

u

dz

t

d a a Right circular cylinder Take ds a d dz and minimize

d

v

u

Z

u

d

t

or a dz

dz

b Right circular cone Use spherical co ordinates with ds dr r sin d

c Sphere Use spherical co ordinates with ds a sin d a d

d Surface of revolution Write x r cos y r sin z f r Express the desired

relation b etween r and in terms of an integral

Determine the stationary function asso ciated with the integral

Z

y f x ds I

when y and y where

x

f x

x

Find the extremals

Z

dx y y a J y y

Z

b J y yy dx y y

Z

xy y dx y y c J y

Find extremals for

Z

y

a J y dx

x

Z

x

y y ye dx b J y

Obtain the necessary condition for a function y to b e a lo cal minimum of the functional

Z ZZ Z

b b

y tf tdt y dt J y K s ty sy tdsdt

a a R

where K s tisa given continuous function of s and t on the square R for which a s t b

K s t is symmetric and f t is continuous

Hint the answer is a Fredholm integral equation

Find the extremal for

Z

J y xy dx y y

What is the extremal if the b oundary condition at x is changed to y

Find the extremals

Z

b

J y x y y dx

a

Z

x

F x y y dx Find the externals

a F y y y k y k constant

by Eulers equation F y F c so

y

y ky y y y ky c

y ky c y ky c

dy dy

dx idx

ky c ky c

Z

p

du

p

u a j we get Using the fact that lnju

u a

Z Z

q

dy

lnjky ky cj idx ix

ky c

q

ix

ky ky c e

Z

du u

p

Lets try another way using sin

a

a u

Z Z

dy

dx

p

c ky

ky ky

p p

sin x sinx sin x

c c

p

c

y c sin x

k

b F y y y y

y y y y y y c F y F

y

dy

p

y y c dx

y c

y c x y c x

x c y

c F x y xy y

use F c x y c

y

c x xc x y y

d F y yy y

c see F y F

y

y y F

y

y yy y y y y F y F

y

y y

y y c

y y c

q

y c y

Z Z

dy

p

dx

c y

y

p

c x ar c cosh

c

q

can also b e written as ln j y y c j

y

p

x c ar c cosh

c

y

p

cosh x c

c

p

y c cosh x c

e F xy yy y

d

F see F

y y

dx

F y

y

xy y F

y

d

xy y y

dx

y xy y y

xy y

xy

xy c

c

y

x

c dx

dy

x

c

y ln j x j c

f F axy bxy

F bxy

y

F axy

y

d

F axy by F

y y

dx

axy a y by

Linear nonconstant co ecients Can b e Solved

g F y k cos y

F y Fy c

Fy y

y k cos y y c

y k cos y c

y c k cos y

q

dy

c k cos y

dx

dy

p

dx

c k cos y

Z

dy

p

x c

c k cos y

F y y

From problem a with k wehave

p

c sin x c y

p

c sin a y a y y

a a

p

y b y y c sin b

b b

y sin b

b

to get a solution

sin a y

a

y y

b a

The solution is not unique y sin x sin x

sin b sin a

p

c sina n If b a n then y

b

p

c sin a

then y y for a solution

b a

otherwise no solution

If F y xy y show I has the same values for all curves joining the endp oints

Using Eulers equation in Chapter we need only show

d

xF x x x x F

y y

dx

F xy F y xy

y y

d d

x F xy xy y

y

dx dx

whichisF

y

Note that

d

xy F

dx

x

R

x

F xy indep endentofcurve

x

x

a Right circular cylinder

v

u

Z

u

dz

t

min a d

d

p

a z F z z

F z F c

z

F a z z

z

p

p

a z z c

a z

p

a z a z z c

p

a

a z

c

a

a z

c

a

a z

c

v

u

u

a

t

z a

c

v

u

u

a

t

a c z

c

parameter family of helical lines

b Right circular cone

v

u

Z

u

dr

t

r sin d

d

q

F rr r r sin

No dep endence on thus we can use

F r r sin r

r

c F r F

r

q

r

p

r r sin c

r r sin

q

r r sin r c r r sin

sin

r r r sin

c

r sin

r r sin

c

q

sin

r sin c r r

c

dr d

q

c

r sin r sin c

Let r sin

Z Z

d sin d

q

c

c

r sin

sec c c

sin c

r sin c sec c c sin

c Sphere

v

u

Z

u

d

t

a a sin d

d

q

F a sin a

F F c

q

a

q

a sin a c

a sin a

q

a sin a a c a sin a

a sin

sin

c

s

a

sin sin

c

d

q

d

a

sin sin

c

d Surface is given as

r

in parametric form

x cos

y sin

z f

The length

Z

q

t

r r r r r r dt L

t

r cos i sin j f k

r sin i cos j

r r cos sin f f

r r

r r

Z

q

t

L f dt

t

or

v

u

Z

u

d

t

L f d

d

d

So F is a function of and

d

F F c

d

f f F

d

q

q

f f c

f

q

f c

f

c

v

u

u

u

c

t

f

F f x y

d

F F Using

y y

dx

F

y

f x y F

y

d

f x y

dx

f x y c

c

y

f x

Z Z

c

dy dx

f x

Z

c

dx k y

f x

using y

Z

x

c

d y x

f

Z

c

y d

f

Z Z

Substituting for f cd cd

c c

c

c

Z

x

d y x

f

Z

a J y y dx y y

Eulers equation in this case is

d

dx

which is satised for all y Clearly that y should also satisfy the b oundary conditions ie

y x

Lo oking at this problem from another p oint of view notice that J y can b e computed

directly and wehave after using the b oundary condition

J y

Since this value is constant the functional is minimzed byany y that satises the b oundary

conditions

Z

b J y yy dx y y

Eulers equation in this case is

d

y y

dx

which is the identity y y which is satised for all y Clearly that y should also satisfy

the b oundary conditions ie y x

Lo oking at this problem from another p oint of view notice that J y can b e computed

directly and wehave after using the b oundary condition

J y

Since this value is constant the functional is minimzed byany y that satises the b oundary

conditions

Z

xy y dx y y c J y

Eulers equation in this case is

d

xy xy

dx

whichis

y xy xy

or

y

Clearly that y could NOT satisfy the b oundary conditions

Z

y

a J y dx

x

y

F

x

d

Euler equation F F

y y

dx

y

F

y

x

F

y

y

c Integrate Eulers equation F c

y

x

y cx

cx

y

cx

b y

Z

x

b J y y y ye dx

x

F y y ye

x

F ye

x

x

F y e

y

y F

y

d

Euler equation F F

y y

dx

d

x

y y e

dx

x

y y e

Solve the homogeneous

p p

x x

c e y y y c e

Find a particular solution of the nonhomogeneous

x x

y y e y e

Therefore the general solution of the nonhomogeneous is

p p

x x x

y c e c e e

Obtain the necessary condition for a function y to b e a lo cal minimum of the functional

b b b b

Z Z Z Z

y tf tdt y t dt K s ty sy tdsdt J y

a a a a

Find the rst variation of J

b b b

Z Z Z

J y K s ty s sy t tdsdt y t t dt

a a a

b

Z

y t tf tdt

a

Then

b b

Z Z

d

J y fK s ty s s t K s ty t t sg dsdt

d

a a

b b

Z Z

y t t tdt tf tdt

a a

Now letting wehave

b b b b

Z Z Z Z

d

J y K s ty sds tdt K s ty tdt sds

d

a a a a

b b

Z Z

y t tdt f t tdt

a a

Since the limits of s and t are constants wecaninterchange s for t and vice versa in the

second of four terms ab ove

b b b b b b

Z Z Z Z Z Z

K s ty sds tdt K t sy sds tdt y t tdt f t tdt

a a a a a a

Combining the rst two terms and factoring out an tdt yields

b b

Z Z

K s tK t s y sds y t f t tdt

a a

Setting this equal to implies

b

Z

K s tK t s y sds y t f t

a

WhichisaFredholm equation

Given F xy It is easy to nd that

F y x

y

F

y

d d

F y x Therefore

y

dx dx

Integrating b oth sides we obtain

c

y x c y

x

Integrating again leads to

y c ln xc

Now applying the b oundary conditions

y c ln c c

y c ln c

ln

Therefore the nal solution is

ln x

y

ln

It is easy to show that in that case the functional J y is

ln

If our b oundary condition at x was y then

c

y c ln x and y

x

c

Then y c

and we get the trivial solution

R

b

Find the extremal J y x y y dx

a

x y F x y y F

y

F y

y

d

F x y xy

y

dx

d

F F Eulers Equation

y y

dx

x y xy y

x y xy y

This is an Euler equation Thus a bmust not contain the origin

r

Let y x

r

y rx

r

y r r x

Substituting

r r r

r r x rx x

r

r r r x

r r r

r r

p

p

r

p p

x x y y

y x c x c x for a x b

CHAPTER

Variable EndPoint Problems

Problems

Z

h i

x

y y Solve the problem minimize I dx

with left end p oint xed and y x is along the curve

x

Find the extremals for

Z

I y yy y y dx

where end values of y are free

Solve the EulerLagrange equation for

Z

q

b

y I y dx

a

where

y a A y b B

b Investigate the sp ecial case when

a b A B

and show that dep ending up on the relative size of b B there may b e none one or two

candidate curves that satisfy the requisite endp oints conditions

Solve the EulerLagrange equation asso ciated with

Z

h i

b

y yy y dx I

a

What is the relevant EulerLagrange equation asso ciated with

Z

h i

I y xy y dx

Investigate all p ossibilities with regard to tranversality for the problem

Z

q

b

min y dx

a

Determine the stationary functions asso ciated with the integral

Z

i h

dx y y y y I

where and are constants in each of the following situations

a The end conditions y and y are preassigned

b Only the end conditions y is preassigned

c Only the end conditions y is preassigned

d No end conditions are preassigned

Determine the natural b oundary conditions asso ciated with the determination of ex

tremals in each of the cases considered in Problem of Chapter

Find the curves for which the functional

p

Z

x

y

dx I

y

with y canhave extrema if

a The p ointx y can vary along the line y x

b The p ointx y can vary along the circle x y

If F dep ends up on x show that the transversality condition must b e replaced by

Z

x

F F

F y dx

y x

x

x x

Find an extremal for

Z

e

x y y dx y ye is unsp ecied J y

Find an extremal for

Z

J y y dx y y y is unsp ecied

F y y

d

F F

y y

dx

d

y y

dx

y y

y A cos x B sin x

using y

y B sin x

Now for the transversity condition

F y F

y

x

slop e of curve

Since the curveis x vertical line slop e is innite we should rewrite the condition

y

F F

y

z z

F

y

x

B y B cos

x

y

F y yy y y

d

F F

y y

dx

F y

y

F y y

y

d

y y y F F

y y

dx

y y y

y

y Ax B

H

x y

P

x y Ax B

Free ends at x x

F y F

y

x

F y F

y

x

The free ends are on vertical lines x x

y y F

y

x

A B

F y y

y

x

A B

A B

A B

A A

B A

y x x

q

F y y

q

F y

y

y y y F

y

yy

p

F

y

y

F y F c

y

q

yy

p

y y c

y

q

y y yy c y

y

y

c

s

y

y

c

Z Z

c dy

q

dx

y c

y

c x c ar c cosh

c

q

OR c ln y y c c x

x c y

ar c cosh

c c

x c

c cosh y

c

a c

A y a A c cosh

c

b c

y b B c cosh B

c

A

a c c ar c cosh

c

B

b c c ar c cosh

c

A B

ar c cosh a b c ar c cosh

c c

This gives c then

A

c a c ar c cosh

c

If a b A B

zero on left zero on right

of of

Thus we cannot sp ecify c based on that free c we can get c using Thus wehavea

one parameter family

F y yy y

c F y F

y

y yy y y y y c

y y c

y y c

q

y y c

dy

p

dx

y c

y

c x ar c cosh

p

c

F y xy y

d

F F

y y

dx

d

y x y

dx

y y x

y y x

x x

y y y Ae Be

h

y Cx D

p

Cx D x

C D

y x

p

x x

y Ae Be x

q

F y

y

q

F yAxB

y

y

y A

F y F

y

x a

F y F

y

x b

q

y

q

y y

x a b

y

y y y

y A

x a

y A

x b

Therefore if b oth end p oints are free then the slop es are the same

a b

A

F y yy y

a y

y

d

F F

y y

dx

d

y y y

dx

y y y

y

y Ax B

y B

y A

y x

b If only y y Ax

Transversality condition at x

F

y

x

imples

y y

Substituting for y

A A

A

Thus the solution is

y x

c y only

y Ax B

y A B B A

y Ax A

F

y

x

y y

A A

A

A

d No end conditions

y Ax B

y A

y y

y y

A B

A A B

A A

B A

B

Natural Boundary conditions are

F

y

x a b

a For

F y k y a of chapter

y F

y

y a

y b

b For F y y exactly the same

c For F y xy

F y x

y

y aa

y bb

d F y yy y

y y F

y

y ay a

y by b

e F xy yy y

F xy y

y

ay a y a

by b y b

f F ax y bxy

ax y F

y

a y

a y

Divide bya or a to get same as part a

g F y k cos y

F y

y

same as part a

p

y

F

y

d

F F

y y

dx

p

y

F

y

y

y

p

F

y

y y

p p

y y

p

y y y y y y y

d

y

F

y

dx y y

p

y y y y y yy y d y

p

F F

y y

dx y

y y y

y y y y y

y y y y y y

yy y

yy

yy x c

ydy x c dx

x

y c x c

y x c x c

y c

a Transversality condition

F y F

y

x x

p

y y y

p

y

y y

x x

y y y

x x

y x y x

Since y x c x

yy x c

at x x

y x x c y x

z z

x

on

the line

x x c

c

p

y x x or y x x

b On x y

The slop e is computed

x yy

yy x

x

At x x x

y x

Remember that at x y x from the solution

y x x c x

is the same as from the circle

y x x

x c x x

x c x x x

c x x

Substituting in the transversality condition

F y F

y

x

wehave

x y x

z

x c

y

x

x x c

y x y x

x x c

y x

y x x x c

z

x

x x x c

x x x c

x c

Solve this with

c x x

to get

c

x

x

x

x

x c x

y x x

In this case equation will have another term resulting from the dep endence of F on

x that is

Z

x

F

dx

x

x

In this problem one b oundary is variable and the line along which this variable p oint

moves is given by y ey which implies that is the line x e First we satisfy Eulers

rst equation Since F x y y wehave

d

F F

y y

dx

and so

d

y x y y xy x y

dx

x y xy y

Therefore

y x y xy

r

This is a CauchyEuler equation with assumed solution of the form y x Plugging this in

and simplifying results in the following equation for r

r r

whichhastwoidentical real ro ots r r and therefore the solution to the dierential

equation is

y x c x c x ln x

The initial condition y implies that c The solution is then

y x c x ln x

To get the other constant wehave to consider the transversality condition Therefore we

need to solve

j F y F

y

x e

Which means we solve the following note that is a vertical line

y

F

j F F

y y

x e

x e

x y j

x e

which implies that y e isour natural boundary condition

y x c x ln x c x

With this natural b oundary condition we get that c and therefore the solution is

y x x ln x

Z

Find an extremal for J y y dx y where y y is unsp ecied

F y y

F F y

y y

Notice that since y is unsp ecied the righthandvalue is on the vertical line x

By the Fundamental Lemma an extremal solution y must satisfy the Euler equation

d

F F

y y

dx

d

y

dx

y

y

Solving this ordinary dierential equation via standard integration results in the follow

ing

y Ax B

Given the xed left endp oint equation y this extremal solution can b e further

rened to the following

y Ax

Additionally y must satisfy a natural b oundary condition at y In this case where

y is part of the functional to minimize we substitute the solution y Ax into the

functional to get

Z

A dx A A A I A

Dierentiating I with resp ect to A and setting the derivative to zero necessary condition

for a minimum wehave

A A

Therefore

A

and the solution is

x y

CHAPTER

Higher Dimensional Problems and Another Pro of

of the Second Euler Equation

Problems

A particle moves on the surface x y z from the p ointx y z to the p oint

x y z in the time T Show that if it moves in suchaway that the integral of its kinetic

energy over that time is a minimum its co ordinates must also satisfy the equations

y z x

x y z

Sp ecialize problem in the case when the particle moves on the unit sphere from

to in time T

Determine the equation of the shortest arc in the rst quadrant which passes through

the p oints and and encloses a prescrib ed area A with the xaxis where A

Finish the example on page What if L

Solve the following variational problem by nding extremals satisfying the conditions

Z

y dx y y y J y y

y y y y

Solve the isoparametric problem

Z

J y y x dx y y

and

Z

y dx

Derive a necessary condition for the isoparametric problem

Minimize

Z

b

dx y I y y Lx y y y

a

sub ject to

Z

b

Gx y y y y dx C

a

and

y a A y a A y bB y bB

where C A A B and B are constants

Use the results of the previous problem to maximize

Z

t

xy y x dt I x y

t

sub ject to

Z

q

t

x y dt

t

Show that I represents the area enclosed by a curve with parametric equations x xt

y y y and the contraint xes the length of the curve

Find extremals of the isoparametric problem

Z

I y y dx y y

sub ject to

Z

y dx

Kinetic energy E is given by

Z

T

x y z dt E

The problem is to minimize E sub ject to

x y z

Let F x y z x y z x y z

Using

d

F F j

y

y

j

j

dt

x d

x

x

dt

x

d y

y

y

dt

y

d z

z

z

dt

z

y z x

x y z

If x y z

x y z

then

x y z

x x

y y

z z

p p

t B sin t Solving x A cos

p p

y C cos t D sin t

p p

z E cos t G sin t

Use the b oundary condition at t

x y z

A

C

E

Therefore the solution b ecomes

p

t x B sin

p

y D sin t

p p

z cos t G sin t

The b oundary condition at t T

xT

y T

z T

p p

B sin t t n

p

n

T

same conclusion for y

n

x B sin t

T

n

y D sin t

T

n n

z cos t G sin t

T T

Nowusez T

cosn G sin n

z

nisodd

n

x B sin t

T

n

t y D sin

n odd

T

n n

z cos t G sin t

T T

Now substitute in the kinetic energy integral

Z

T

E x y z dt

Z

T

n n n n

B cos t D cos t

T T T T

n n n

sin t G cos t dt

T T T

Z

T

n n

t B D G cos

T T

n n

sin t

T T

n n n

G sin t cos t dt

T T T

Z

T

T

n T n

sin tdt cos t

T n T

Z

T

T

T n T n

dt t sin t t sin

T n T

z

n

t cos

T

Z

T

T n

dt cos t

T

z

n

t cos

T

n T

B D G E

T

Clearly E increases with nthus the minimum is for n

Therefore the solution is

t x B sin

T

y D sin t

T

z cos sin t G sin t

T T

Z

q

Min L y dx

Z

sub ject to A ydx

q

F y y

F y F c

y

q

y

p

y c y y

y

q q

y y y y c y

q

y y c

y

c y

s

y i

c y

dy

q

idx

c y

Z Z

c y dy

q

i dx

c y

Use substitution

u c y

du c y dy

du

c y dy

Z Z

du

i dx

u

u

ix c

substitute for u

q

c y ix c

square b oth sides

c y ix c

c

y x ix c c

c

x ic x c y

z

x D

c

y x D

We need the curve to go thru and

c

x y D

c

D x y

D D

D D D

D

c

D y x

Let

c

k

then the equation is

x y k k

To nd k we use the area A

s

Z Z

A ydx k x k dx

use

Z

p p

u a u

a u du a u ar c sin

a

where

a k

u x

s

k

x

x

q

A x ar c sin k kx

k

s

k

q

k k ar c sin

k

s

k

q

k ar c sin

k

p

k k k ar c sin A

k

k

p

A k ar c sin

k

p

A k k ar c sin k ar c cot k

k

So

A k k ar c cot k

and

y k k x

at c c

c c at

subtract

c c

c c c

c

Now use

Since y tan

Z

sec dx L

since

x c

sin

dx cos d

c

x sin

c

x sin

Z

sec cos d arc sin L

L

ar c sin

Supp ose wesketch the two sides as a function of

is the value suchthat

L

ar c sin

is a function of L

c L

c L

1.5

y=1.2/(2λ) 1

0.5

0 λ 1/(2 0)

−0.5

−1

−1.5

−1.5 −1 −0.5 0 0.5 1 1.5

Figure

L

c

The curve is then y x

1.8

1.6

1.4

1.2

1

0.8

0.6 L=π /2

0.4

0.2 A=π /4

0

−0.2

−0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

Figure

Solve the following variational problem by nding the extremals satisfying the conditions

Z

y y dx y J y y y

y y y y

Vary eachvariable indep endently bycho osing and in C satisfying

Form a one parameter admissible pair of functions

y and y

Yielding two Euler equations of the form

d d

F F and F F

y y

y y

dx dx

For our problem

F y y y y

Taking the partials of F yields

F y

y

F y

y

F y

y

F y

y

Substituting the partials with resp ect to y into the Euler equation

d

y y

dx

y y

Substituting the partials with resp ect to y into the Euler equation

d

y y

dx

y y

Solving for y and substituting into the rst second order equation

y y y y

th

Since this is a order homogeneous constant co ecient dierential equation wecan

assume a solution of the form

rx

y e

Now substituting into y y gives

rx rx

r e e

r

r

r i

This yields a homogeneous solution of

ix ix x x

C e C e y C e C e

x x

C e C e C cos x C sin x

Now using the result from ab ove

y y

d

x x

C e C e C sin x C cos x

dx

x x

C e C e C cos x C sin x

x x

C e C e C cos x C sin x

Applying the initial conditions

y C C C

C e C e C y

y C C C

C e y C e C

Wenowhave equations with unknowns

C

C

C e e

e C e

Performing Gaussian Elimination on the augmented matrix

e e e e

e e

The augmented matrix yields

C C C

C C

C C

C e C C e

Substituting C and C into the rst and fourth equations gives

C C C C

e

C e and C C C C C e

e

Finally

x x

y e e cos x sin x

x x

y e e cos x sin x

The problem is solved using the Lagrangian technique

Z Z

L y x dx y dx

L F G y x y

where

F y x and G y

L y and L y

y y

Nowwe use Eulers Equation to obtain

d

y y

dx

y y

Solving for y

p p

y A cos xB sin x

Applying the initial conditions

p p

B sin A y A cos

p

y B sin

p

If B then we get the trivial solution Therefore wewant sin

p

This implies that n n

Nowwesolve for B using our constraint

y B sinn x

Z Z

y dx B sin n xdx

sin x x

B B

B or B

Therfore our nal solution is

y sin n x n

Derive a necessary condition for the isop erimetric problem

Z

b

Minimize I y y L x y y y y dx

a

Z

b

sub ject to G x y y y y dx C

a

and y aA ya A ybB ybB

where A A B B and C are constants

Assume L and G are twice continuously dierentiable functions The fact that

Z

b

dx C is called an isop erimetric constraint y G x y y y

a

Z

b

Let W G x y y y y dx

a

Wemust emb ed an assumed lo cal minimum y x in a family of admissible functions with

resp ect to whichwe carry out the extremization Intro duce a twoparameter family

z y x x i

i i i i

where C a b and

a b i

i i

and are real parameters ranging over intervals containing the orign Assume W do es

not have an extremum at y then for anychoice of and there will b e values of and

i

in the neighb orho o d of for which W z C

Evaluating I and W at z gives

Z Z

b b

J L x z z z z dx and V G x z z z z dx

a a

Since y is a lo cal minimum sub ject to V the p oint must b e a lo cal minimum

for J sub ject to the constraint V C

This is just a dierential calculus problem and so the Lagrange multiplier rule maybe

applied There must exist a constant such that

J J

at

where J is dened by

Z

b

dx z J J V L x z z z

a

with

L L G

Wenow calculate the derivatives in afterward setting Accordingly

Z

i h

b

J

x y y y y dx i L x y y y y L

i

i y y

a

i

Integrating the second term by parts as in the notes and applying the conditions of

gives

Z

b

d J

L x y y y L y x y y y y dx i

y y i

dx

a

i

Therefore from and b ecause of the arbitrary character of or the Fundamental

Lemma implies

d

y x y y y L y x y y y L

y y

dx

Which is a necessary condition for an extremum

Let the two dimensional p osition vector R be R xi y j then the velo cityvector

v xi y j From vector calculus it is known that the triple a b c gives the volume of the

parallelepip ed whose edges are these three vectors If one of the vectors is of length unity

then the volume is the same as the area of the parallelogram whose edges are the other

vectors Now lets take a k b R and c v Computing the triple wehave xy x y

whichistheintegrand in I The second integral gives the length of the curve from t to t

see denition of arc length in any Calculus b o ok

To use the previous problem let

Lt x y x y xy xy

q

x y Gt x y x y

then

L y L x

x y

G G

x y

L y L x

x y

y x

p p

G G

y x

x y x y

Substituting in the Euler equations we end up with the two equations

xy x y

y

x y

x y x y

x

x y

Case y

Substituting this in the second equation yieldsx

Thus the solution is x c y c

Case x then the rst one yieldsy andwehave the same solution

Case x andy

In this case the term in the braces is zero or

x y xy x y

d x

The right hand side can b e written asy

dt y

x

Now let u weget

y

du

dy

u

For this we use the trigonometric substitution u tan This gives the following

s

x x

y c

y y

Simplifying we get

y c

q

dx dy

y c

Substitute v y c and we get

k c

x y

which is the equation of a circle

Let F F G y y Then Eulers rst equation gives

d

y y y y

dx

y y

r

p

r

rx

Where we are substituting the assumed solution form of y e into the dierential equation

to get an equation for r Note that and b oth lead to trivial solutions for y x

R

and there would b e no way to satisfy the condition that y dx Therefore assume

o

that Wethenhave that the solution has the form

p p

xc sin x y x c cos

p

The initial conditions result in c and c sin Since c would giveusthe

p

trivial solution again it must b e that n where n This implies that

n or eqivalently n n

R

Wenow use this solution and the requirement y dx to solve for the constant c

o

Therefore wehave

Z Z

n

c

sin udu c sin nxdx

n

n

c sinu u

n

sinn c

c

for n

After solving for the constantwehave that

s

y x sinnx n

Z

If wenow plug this solution into the equation y dx we get that I y n whichimplies

we should cho ose n to minimize I y Therefore our nal solution is

s

sinx y x

CHAPTER

Integrals Involving More Than One Indep endent

Variable

Problem

Find all minimal surfaces whose equations have the form z x y

Derive the Euler equation and obtain the natural b oundary conditions of the problem

ZZ

h i

x y u x y u x y u dxdy

x y

R

In particular showthatif x y x y the natural b oundary condition takes the form

u

u

n

u

is the normal derivativeof u where

n

Determine the natural b oundary condition for the multiple integral problem

ZZ

Lx y u u u dxdy uC R u unsp ecied on the b oundary of R I u

x y

R

Find the Euler equations corresp onding to the following functionals

ZZ

a I u x u y u dxdy

x y

R

ZZ

dxdt c is constant c u b I u u

x t

R

z x y

Z Z

q

S z z dx dy

x y

R

Z Z

q

x y dx dy

R

x y

p p

x y

Dierentiate and multiply by

q q

x

q q

y

Expand and collect terms

q q

y y x x

Separate the variables

x y

y x

One p ossibilityis

x y xAx

y By

z Ax By C whichisaplane

The other p ossibilityisthateach side is a constant left hand side is a function of only x

and the right hand side dep ends only on y

y x

x y

Let x then

d

dx

ar c tan x c

tanx c

Integrate again

Z

x tan x c dx

ln cos x c c x

c x

e cosx c

Similarly for y sign is dierent

D cos y D ln y

y D

e cosy D

Divide equation by equation

cosy D

c D y x

e

cosx c

using z x y wehave

cosy D

c D z

e e

cosx c

If we let x y z b e on the surface wend

cos x c cosy D

z z

e

cosx c cosy D

x y u x y u F x y u

y x

F F F see equation

u u u

x y

x y

F x y u

u x

x

F x y u

u y

y

F x y u

u

x y u x y u x y u

x y

x y

The natural b oundary conditions come from the b oundary integral

F cos F sin

u u

x y

x y u cos x y u sin

x y

If x y x y then

x y u cos u sin

x y

z

ru n

z

u

n

u

n

Determine the natural b oundary condition for the muliple integral problem

ZZ

I u Lx y u u u dxdy u C R

x y

R

u unsp ecied on the b oundary of R

Let ux y b e a minimizing function among the admissible functions for I u Consider

the oneparameter family of functions uux y x y where C over R and

x y on the b oundary of RThenif

ZZ

Lx y u u u dxdy I

x x y y

R

a necessary condition for a minimum is I

ZZ

Now I L L L dxdy where the arguments in the partial deriva

u x u y u

x y

R

tives of L are the elements x y u u u of the minimizing function u Thus

x y

ZZ ZZ

L L L L dxdy dxdy I L

u u u u u

x y x y

x y x y

R R

The second integral in this equation is equal to by Greens Theorem

I

L mL ds

u u

x y

R

where and m are the direction cosines of the outward normal to R and ds is the arc

length of the R But since x y on R this integral vanishes Thus the condition

I which holds for all admissible x y reduces to

ZZ

L L dxdy L

u u u

x y

x y

R

Therefore L L L at all p oints of R This is the EulerLagrange equation

u u u

x y

x y

for the two dimensional problem

Now consider the problem

Z ZZ Z

d b

Lx y u u u dxdy I u Lx y u u u dxdy

x y x y

R c a

where all or or a p ortion of the R is unsp ecied This condition is analogous to the single

integral variable endp oint problem discussed previously Recall the line integral presented

ab ove

I

L mL ds where and m are the direction cosines of the outward normal to

u u

x y

R

R and ds is the arc length of the R Recall that in the case where u is given on R

analogous to xed endp oint this integral vanishes since x y on RHowever in

the case where on all or a p ortion of R u is unsp ecied x y Therefore the natural

b oundary condition whichmust hold on R is L mL where and m are the

u u

x y

direction cosines of the outward normal to R

Eulers equation

F F F

u u u

x y

x y

y u a F x u

y x

Dierentiate and substitute in Eulers equation wehave

xu x u yu y u

x xx y yy

b F u c u

t x

Dierentiate and substitute in Eulers equation wehave

u c u

tt xx

whichisthewave equation

CHAPTER

Examples of Numerical Techniques

Problems

Z

h i

x

y y dx Find the minimal arc y x that solves minimize I

a Using the indirect xed end p oint metho d when x

b Using the indirect variable end p oint metho d with y and y x Y x

Z

Find the minimal arc y xthatsolves minimize I y yy y y dx

where y and y

Z

h i

x

y yy y dx Solve the problem minimze I

a Using the indirect xed end p oint metho d when x

b Using the indirect variable end p oint metho d with y and y x Y x

Solve for the minimal arc y x

Z

h i

I y xy y dx

where y and y

a Here is the Matlab function dening all the derivatives required

odefm

function xdotodeftx

fyfy fyy nd partial wrt y y

fyy fyy nd partial wrt y y

fy fy st partial wrt y

fyx fyx nd partial wrt y x

fyy

fyy

fy x

fyx

rhsfyyfyyfyfyxfyy

xdotxrhsxrhs

The graph of the solution is given in the following gure

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure

First we give the mo died nputm

function VALUE FINPUTxyyprimenum returns the value of the

functions Fxyy Fyxyy Fyxyy for a given num

num defines which function you want to evaluate

for F for Fy for Fy

if nargin errorFour arguments are required break end

if num num

errornum must be between and break

end

if num value ypypyypy end F

if num value yp end Fy

if num value ypy end Fy

The b oundary conditions are given in the main program dmetho dm see lecture notes

The graph of the solution using direct metho d follows

Solution y(x) using the direct method 2

1.8

1.6

1.4

1.2

y 1

0.8

0.6

0.4

0.2

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x

Figure

a Here is the Matlab function dening all the derivatives required

odefm

function xdotodeftx

fyfy fyy nd partial wrt y y

fyy fyy nd partial wrt y y

fy fy st partial wrt y

fyx fyx nd partial wrt y x

fyy

fyy

fy xx

fyx

rhsfyyfyyfyfyxfyy

xdotxrhsxrhs

The graph of the solution is given in the following gure

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure

First we give the mo died nputm

function VALUE FINPUTxyyprimenum returns the value of the

functions Fxyy Fyxyy Fyxyy for a given num

num defines which function you want to evaluate

for F for Fy for Fy

if nargin errorFour arguments are required break end

if num num

errornum must be between and break

end

if num value yxyypend F

if num value yx end Fy

if num value end Fy

The b oundary conditions are given in the main program dmetho dm see lecture notes

The graph of the solution using direct metho d follows

Solution y(x) using the direct method 1

0.8

0.6

y 0.4

0.2

0

-0.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x

Figure

CHAPTER

The RayleighRitz Metho d

Problems

Write a MAPLE program for the RayleighRitz approximation to minimize the integral

Z

h i

y y xy dx I

y

y

Plot the graph of y y y and the exact solution

Solve the same problem using nite dierences

withplots

phi x

y phi

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phi xphi axx

y phi phi

dy diffyx

f dy y xy

w intfx

dw diffwa

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y phi phi phi

dy diffyx

f dy y xy

w intfx

dw diffwb

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dw diffwb

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b cc

bsolvebb

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phi x

phi cxx

phi cxxx

phi cxxxx

y phi phi phi phi

dy diffyx

f dy y xy

w intfx

dw diffwc

csolvedwc

dw diffwc

csolvedwc

dw diffwc

csolvedwc

a c c

asolveac

a c c

asolveac

b a a

csolvebc

ca

cc

pplotyxcolorbluestylepoint

y cosx cossinsinx x

pplotyxcolorredstyleline

displayppppp

Note Delete p or p or b oth if you wanttomake the True versus Approximations more noticable

2

1.8

1.6

1.4

1.2

1 0 0.2 0.4 0.6 0.8 1

x

Figure

F dyyyx

withplots

f yiyidelx yi xiyi

phi sumyiyidelx yi xiyidelxi

dy diffphiy

dy diffphiy

dy diffphiy

x

x

x

delx

y

y

ysolvedyy

parrayxyxyxy

pplotp

phi sumyiyidelx yi xiyidelxi

dy diffphiy

dy diffphiy

dy diffphiy

dy diffphiy

x

x

x

x

delx

y

y

dsolvedyy

dsolvedyy

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Figure

CHAPTER

Hamiltons Principle

Problems

If is not preassigned show that the stationary functions corresp onding to the problem

Z

y dx

sub ject to

y y sin

are of the form y x cos where satises the transcendental equation

cos sin

and Also verify that the smallest p ositivevalue of is b etween

If is not preassigned show that the stationary functions corresp onding to the problem

Z

h i

y y dx

sub ject to

y y

x

are of the form y x where isoneofthetwo real ro ots of the quartic equation

A particle of mass m is falling vertically under the action of gravityIfy is distance

measured downward and no resistive forces are present

a Show that the Lagrangian function is

L T V m y gy constant

and verify that the Euler equation of the problem

Z

t

Ldt

t

is the prop er equation of motion of the particle

b Use the momentum p my to write the Hamiltonian of the system

c Show that

H y

p

H p

y

A particle of mass m is moving vertically under the action of gravity and a resistive

force numerically equal to k times the displacement y from an equilibrium p osition Show

that the equation of Hamiltons principle is of the form

Z

t

dt my mg y ky

t

and obtain the Euler equation

A particle of mass m is moving vertically under the action of gravity and a resistive force

numerically equal to c times its velo cityy Show that the equation of Hamiltons principle

is of the form

Z Z

t t

cy y dt my mg y dt

t t

Three masses are connected in series to a xed supp ort by linear springs Assuming

that only the spring forces are present show that the Lagrangian function of the system is

h i

L m x m x m x k x k x x k x x constant

where the x represent displacements from equilibrium and k are the spring constants

i i

If is not preassigned show that the stationary functions corresp onding to the problem

Z

y dx

Sub ject to

y and y sin

Are equal to

y x cos

d

Using the Euler equation L L with

y y

dt

L y

L

y

L y

y

We get the nd order ODE

y

y

Integrating twice wehave

y Ax B

Using our initial conditions to solve for for A and B

y A B B

sin

y sin A A

Substituting A and B into our original equation gives

sin

x y

Now b ecause wehaveavariable right hand end p oint wemust satisfy the following transver

sality condition

F y F j

y x

Where

F y

sin

F

y

cos

sin

y

Therefore

sin

y y cos

sin sin

y cos

sin sin sin

cos

sin sin

cos

sin cos sin

cos sin

Which is our transversality condition Since satises the transcendental equation ab ove

wehave

sin

cos

Substituting this backinto the equation for y yields

y x cos

Whichiswhatwewanted to show

Toverify that the smallest p ositivevalue of is b etween and wemust rst solvethe

transcendental equation for

cos sin

sin

cos cos

tan sec

20

15

10

5

0 y −5

−10

−15

−20 pi/2 −25 0 0.5 1 1.5 2 2.5 3

l

Figure Plot of y and y tan sec

Then plot the curves

y

y tan sec

between and Pi to see where they intersect

Since they app ear to intersect at approximately lets verify the limits of y tan sec

analytically

tan sec lim

l

sin

cos cos

sin

cos

is the smallest value of Which agrees with the plot Therefore

Z

h i

y y dx

sub ject to

y y

Since L y y

wehave L and L y

y

y

d d

L y b ecomes Thus Eulers equation L

y

y

dx dx

Integrating leads to

c

y x

c

Integrating again y x x c

Now use the left end condition y c

c

c At x wehave y

Thus the solution is y x x

Lets dierentiate y for the transversality condition y x

Nowwe apply the transversality condition

L y L where and

y

x

y and y and evaluating at x we obtain Now substituting for L L

y

Therefore the nal solution is

y x x

where is one of the two real ro ots of

First using Newtons Second Law of Motion a particle with mass m with p osition

vector y is acted on by a force of gravity Summing the forces gives

my F

Taking the downward direction of y to b e p ositive F mgy Thus

my mg y

my we obtain From Eqn and the denition of T

Z

t

T F dy dt

t

From Eqn

Z

t

my y F y dt

t

Dening the p otential energy as

F y V mg y y

gives

Z

t

T V dt

t

or

Z

t

my mg y dt

t

If we dene the Lagrangian L as L T V we obtain the result

L m y gyconstant

Note The constant is arbitrary and dep endent on the initial conditions

Toshow the Euler Equation holds recall

L m y gyconstant

d

L mg L L my my

y y y

dt

Thus

d

L L mg my mg y

y y

dt

Since the particle falls under gravity no initial velo cityy g and

d

L L

y y

dt

The Euler Equation holds

b Let p my The Hamiltonian of the system is

H t x p Lt x t x p pt x p

m y gyconstant my t x p

H c

p

H y by denition

p

H mg my p

y

Newtons second law mR F Note that F mg kRsowehave

Z

t

mR R mgR kRR dt

t

This can also b e written as

Z

t

mR mg R kR dt

t

To obtain Eulers equation we let

mR mg R kR L

Therefore

L mg kR

R

L mR

R

d

L mg kR mR L

R

R

dt

The rst two terms are as b efore coming from ma and the gravity The second

integral gives the resistive force contribution which is prop ortional toy with a constantof

prop ortionality c Note that the same is negative b ecause it acts opp osite to other forces

Here we notice that the rst spring moves a distance of x relativetorest The

second spring in the series moves a distance x relative to its original p osition but x was

the contribution of the rst spring therefore the total is x x Similarly the third moves

x x units

CHAPTER

Degrees of Freedom Generalized Co ordinates

Problems

Consider the functional

Z

h i

b

r ty q ty dt I y

a

Find the Hamiltonian and write the canonical equations for the problem

Give Hamiltons equations for

Z

q

b

I y t y y dt

a

Solve these equations and plot the solution curves in the yp plane

A particle of unit mass moves along the y axis under the inuence of a p otential

f y y ay

where and a are p ositive constants

a What is the p otential energy V y Determine the Lagrangian and write down the

equations of motion

b Find the Hamiltonian H y p and show it coincides with the total energy Write

down Hamiltons equations Is energy conserved Is momentum conserved

and y what is the initial velo city c If the total energy E is

d Sketch the p ossible phase tra jectories in phase space when the total energy in the

system is given by E

a

q

p

E V y Hint Note that p

What is the value of E ab ove which oscillatory solution is not p ossible

A particle of mass m moves in one dimension under the inuence of the force F y t

t

ky e where y t is the p osition at time tandk is a constant Formulate Hamiltons

principle for this system and derive the equations of motion Determine the Hamiltonian

and compare it with the total energy

A Lagrangian has the form

a

Lx y y y ay Gy Gy

where G is a given dierentaible function Find Eulers equation and a rst integral

If the Lagrangian L do es not dep end explicitly on time t prove that H constant and

if L do esnt dep end explicitly on a generalized co ordinate y provethat p constant

Consider the dierential equations

k

r C r r r

m

governing the motion of a mass in an inversely square central force eld

a Showbythechain rule that

dr dr d r

r Cr C r r C r

d d d

and therefore the dierential equations may b e written

k dr d r

r r r

d d C m

b Let r u and showthat

d u k

u

d C m

c Solve the dierential equation in part b to obtain

k

cos u r

C m

where and are constants of integration

d Show that elliptical orbits are obtained when

CHAPTER

Integrals Involving Higher Derivatives

Problems

Derive the Euler equation of the problem

Z

x

F x y y y dx

x

in the form

F d F F d

dx y dx y y

and show that the asso ciated natural b oundary conditions are

x

d F F

y

dx y y

x

and

x

F

y

y

x

Derive the Euler equation of the problem

Z Z

y x

F x y u u u u u u dxdy

x y xx xy yy

y x

where x x y and y are constants in the form

F F F F F F

x u xy u y u x u y u u

xx xy yy x y

and show that the asso ciated natural b oundary conditions are then

x

F F F

u

x u y u u

x

xx xy x

x

F

u

x

u

x

xx

and

y

F F F

u

y u x u u

y

yy xy y

y

F

u

y

u

y

yy

Sp ecialize the results of problem in the case of the problem

Z Z

x y

u u u u u dxdy

xx yy

xx yy xy

x y

where is a constant

Hint Show that the Euler equation is r u regardless of the value of butthe

natural b oundary conditions dep end on

Sp ecialize the results of problem in the case

F axy bxy cxy

Find the extremals

Z

a I y yy y dx y y y y

Z

y y y y dx y y y y b I y

Find the extremals for the functional

Z

b

y y y dt I y

a

Solve the following variational problem by nding extremals satisfying the given condi

tions

Z

y dx y y y y I y

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