4 variation

7/31/2019 4 Variation

1/37

Calculus of Variations


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contdyExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle

Paul Lim Calculus of Variations 1 / 37


2/37

The Euler-Lagrange Equation


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3


ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


I DZba

F.y;y0; x/ dx; (1)

where a, b and the form of F are fixed by given considerations, but

y.x/ will be chosen to make I stationary.

Suppose y.x/ is the function required to make I stationary and

consider making the replacement

y.x/ ! y.x/ C .x/; (2)

where the parameter is small and .x/ is an arbitrary function.

For I to be stationary with respect to these variations, we require

dI

d

D0

D 0 for all .x/: (3)


3/37

The Euler-Lagrange Equation 2


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3


ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


Substituting (2) into (1) and expanding as a Taylor series in weget

I.y;/ D Zb

a F .y C ;y0

C 0

; x/d

x

DZba

F.y;y0; x/ dx

CZ

b

a

@F@y

C @F@y0

0

dx C O.2/:

With this form for I.x;y/ the condition (3) implies that for all .x/

we require

dI

d

D0

DZba

@F

@y C @F

@y00

dx D 0:

Integrating second term by parts, this becomes


4/37

The Euler-Lagrange Equation 3


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3


ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


@F

@y0

ba

CZba

@F

@y d

dx

@F

@y0

.x/ dx D 0: (4)

If the end-points are fixed, then .a/

D.b/

D0. From (4), we

thus have@F

@yD d

dx

@F

@y0

: (5)

This is known as the Euler-Lagrange (EL) equation.

F does not contain y explicitly

In this case, @F=@y

D0, and (5) can be integrated giving

@F

@y0D constant: (6)


5/37

Example 1


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3


ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


ExampleShow that the shortest curve joining two points is a straight line.

Solution

FIG. 1: An arbitrary path between two fixed points.


6/37

Example 1 contd


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contd

yExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


Let the points be labeled A and B , and have coordinates .a; y.a// and.b; y.b// respectively (Fig. 1). The length of an element path ds is

ds D .dx/2 C .dy/2

1=2 D .1 C y02/1=2 dx:

Thus the total length along curve is

L D Zb

a

.1 C y02/1=2 dx: (7)

Hence using (6), we have

k D @F@y0

D y0

.1C

y02/1=2

where k is a constant. Integrating we get

y

D

k

.1 k2

/1=2

x

Cc: (straight line)


7/37

F does not containx explicitly


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contd

yExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


In this case, multiplying Eq. (5) by y0 and using

d

dx y0

@F

@y0 D y0d

dx @F

@y0C y00

@F

@y0

we obtain

y0@F

@yC y00 @F

@y0D d

dx y0@F

@y0 ;)

@F

@x d

dx

F y0 @F

@y0

D 0:

Hence, integrating,

F y0 @F@y0

D constant: (8)


8/37

Example 2


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contd

yExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


ExampleFind the closed convex curve of length l that encloses the greatest

possible area.

Solution

FIG. 2: A convex closed curve that is symmetrical about thex-axis.

Without loss of generality, assume that the curve passes through

the origin and suppose also that it is symmetric about the x-axis.


9/37

Example 2 contd


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contd

yExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


Using the distance s along the curve, measured from the origin, asthe independent variable, then y.0/ D y.l=2/ D 0, and

dA

Dy dx

Dy .ds/2 .dy/

21=2 :The total area is

AD

2 Zl=20

y.1

y02/1=2 ds: (9)

Here y0 stands for dy=ds. Thus, using (8) we have

y.1 y02

/

1=2

C yy02

.1 y02

/

1=2

D kor ky0 D .k2 y2/1=2; which using y.0/ D 0, integrates to

y=k D sin.s=k/: (10)


10/37

Example 2 contd


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contd

yExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


The other endpoint, y.l=2/ D 0, fixes the value of k as l=2 togive

y D l2

sin2s

l:

For this we obtain dy D cos.2s=l/ ds and since.ds/2 D .dx/2 C .dy/2, we also find that dx D sin.2s=l/ ds.This in turn can be integrated and, using x.0/ D 0, gives x interms of s as

x l2

D l2

cos2s

l:

Thus, x and y lie on a circle of radius l=2

x l

2

2

C y2 D l2

42:

Substituting the solution (10) into the expression for the total area

(9), it can be verified that A D l2=4 .


11/37

Example 3


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contd

yExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


ExampleTwo rings, each of radius a, are placed parallel with their centres

2b apart and on a common normal. An axially symmetric soap film

is formed between them but does not cover the ends of the rings.

Find the shape assumed by the film.

Solution

FIG. 3: Possible soap films between two parallel circular rings.


12/37

Example 3 contd


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contd

yExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


We take cylindrical polar coordinates as in Figure 3(c), and let theradius of the soap film at height z be .z/ with .b/ D a.Counting only one side of the film, the element of surface area

between z and z

Cdz is

dSD 2 .dz/2 C .d/21=2 ;so the total surface area is given by

SD 2Zbb

.1 C 02/1=2 dz: (11)

Since the integrand does not contain z explicitly, we can use (8) toobtain an equation for that minimises S, i.e.

.1 C 02/1=2 02.1C 02/1=2 D k;

where k is a constant.


13/37

Example 3 contd


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contd

yExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


Multiplying through by .1C 02/1=2, rearranging to find an explicitexpression for 0 and integrating we find

cosh1

k Dz

k Cc;

where c is the constant of integration.

Using the boundary conditions .

b/

Da, we require c

D0 and

k such that a= k D coshb=k (if b=a is too large, no such k can befound). Thus the curve that minimises the surface area is

= k

Dcosh.z=k/;

with the minimum distance from the axis equal to k.


14/37

Several Dependent & Independent Variables


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contd

yExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


Several dependent variables

We have F D F .y1; y01; y2; y02; : : : yn; y0n; x/, and

@F@yi

D ddx

@F@y0i

for i D 1 ; 2 ; : : : ; n : (12)

Several independent variables

I DZ

: : :

ZF

y;

@y

@x1; : : :

@y

@xn; x1; : : : xn

dx1 : : : dxn:

The extremising function y D y.x1; : : : ; xn/ must satisfy@F

@yD

n

XiD1@

@xi

@F

@yxi

; where yxi stands for @y=@xi (13)


15/37

Higher-order derivatives


yEL Equation

yEL Equation 2

yEL Equation 3

yExample 1

yExample 1 contd

yF not contain x

yExample 2

yExample 2 contd

yExample 2 contd

yExample 3

yExample 3 contd

yExample 3 contd

ySeveral Variables

yHigher-order

derivatives

Variable end-points

Fermats Principle

Hamiltons Principle


If in (1), F D F.y;y0; y00; : : : y.n/; x/, then using the same methodas before and performing repeated integration by parts, we have

@F

@y d

dx @F

@y0C d

2

dx2 @F

@y00 : : :

C.1/n dn

dxn

@F

@y.n/

D 0; (14)

provided that y D y0 D D y.n/ D 0 at both end-points.


16/37

Variable end-points


Variable end-points

yVariable end-points

yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle



17/37

Variable end-points


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


Suppose we wish to find the function y.x/ that extremises theintegral

I D Zb

a

F.y;y0; x/ dx;

and that only the lower end-point is fixed and allow y.b/ to be

arbitrary. Repeating the earlier analysis, we find from (4) that

@F

@y0ba

C Zba

@F@y

ddx

@F@y0

.x/ dx D 0: (15)

Since EL (5) must still hold for the second term on the LHS to

vanish and that .a/ D 0, we must require that@F

@y0

xDb

D 0: (16)

If both end-points may vary, then @F=@y0 must vanish at both ends.


18/37

Example 4


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


FIG. 4: A frictionless wire along which a small bead of massm slides. We

seek the shape of the wire that allows the bead to travel from the origin O

to the linex

Dx0 in the least possible time.


19/37

Example 4 contd


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


ExampleA frictionless wire in a vertical plane connects two points A and B,

A being higher than B. Let the position A be fixed at the origin of an

xy-coordinate system, but allow B to lie anywhere on the vertical

line x D x0 (see Fig. 4). Find the shape of the wire such that abead of mass m placed on it at A will slide under gravity to B in the

shortest possible time.

SolutionConservation of energy tells us that the particle speed is given by

v

D

ds

dt Dp2gy;

where s is the path length along the wire and g is the acceleration

due to gravity. Since the element of path length is

ds

D.1

Cy02/1=2dx, the total time taken to travel to the line

x D x0 is given by


20/37

Example 4 contd


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


t DZxDx0xD0

ds

vD 1p

2g

Zx00

s1C y02

ydx:

Using (8) with FD p1 C y02=py, we have

y.1C y02/1=2 D k;where k is a constant. Letting a D k

2

and solving for y

0

, we find

y0 D dydx

Dr

a yy

;

which on substituting y D a sin2 integrates to give

x D a2

.2 sin 2 / C c:


21/37

Example 4 contd


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


Thus the parametric equation of the curve is given by

x D b. sin /C c; y D b.1 cos /;

where b D a=2 and D 2 and define a cycloid.Since the curve passes through the origin, c D 0. Now since y.x0/is arbitrary, i.e. the upper endpoint can lie anywhere on the curve

x D x0, (16) gives@F

@y0

xDx0

D y0

py.1 C y02/

xDx0

D 0;

which implies that y0 D 0 at x D x0, or that the tangent to thecycloid at B must be parallel to the x-axis; this requires b D x0.


22/37

Constrained Variation


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


Suppose that we wish to find the stationary values of

I D Zb

a

F.y;y0; x/ dx;

subject to the constraint that the value of

J

D Zb

a

G.y;y0; x/ dx

is held constant. Following the method of Lagrange undetermined

multipliers, let us define the new functional

KD IC J DZ

b

a

.FC G/ dx;

and find its unconstrained stationary values.


23/37

Constrained Variation 2


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


Repeating earlier analysis leads to

@F

@y d

dx

@F

@y0

C

@G

@y d

dx

@G

@y0

D 0;

which, together with the original constraint J D constant, will yieldthe required solution y.x/.

In general, if the integral I is subject to multiple constraints Ji ,i D 1 ; : : : ; n, then we simply find the unconstrained stationaryvalues of the new integral

KD ICnX

iD1

iJi :


24/37

Example 5


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


FIG. 5: A uniform rope with fixed endpoints suspended under gravity.

Example

Find the shape assumed by a uniform rope when suspended by its

ends from two points at equal heights.


25/37

Example 5 contd


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


SolutionLet the rope of length 2L be suspended between the points

x D a, y D 0 .L > a/, and have uniform density per unitlength. We need to find the stationary value of the ropes

gravitational potential energy

I D gZ

y ds D gZ

a

a

y.1C y02/1=2 dx;

with respect to small changes in the form of the rope, but subject to

the constraint that the total length of the rope remains constant, i.e.

J D Zds D Zaa

.1C y02/1=2 dx D 2L

Define a new integral

KD IC J DZ

a

a

.gy C /.1C y02/1=2 dx:


26/37

Example 5 contd


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


Using (8) to find the first integral:

.gy C /

1 C y02

1=2 .gy C /

1 C y02

1=2

y02 D k;

where k is a constant. This reduces to

y02 D

gy C k

2 1:

Making the substitution gy C D k cosh z and integrating, givesk

gcosh

1 gy C

k D x C cwhere c is the constant of integration. Three unknowns , k and c

will be evaluated using the two end conditions y.a/ D 0 and theconstraint J

D2L.


27/37

Example 5 contd


Variable end-points


yExample 4

yExample 4 contd

yExample 4 contd

yExample 4 contd

yConstraint

yConstrained 2

yExample 5

yExample 5 contd

yExample 5 contd

yExample 5 contd

Fermats Principle

Hamiltons Principle


The end conditions give

coshg.a C c/

kD

kD cosh g.a C c/

k:

Since a 0, these imply c D 0 and = k D cosh.ga=k/. Puttingc D 0 into the constraint, in which y0 D sinh.gx=k/, we obtain

2L D Za

a1C sinh2

.gx=k/1=2

dx D2k

g sinh.ga=k/:

Collecting together the values for the constants, the form adopted

by the rope is therefore

y.x/ D kg

hcosh

gxk

cosh

gak

i;

where k is the solution ofsinh

.ga=k/ D gL=k.


28/37

Fermats Principle in Optics


Variable end-points

Fermats Principle

yFermats Principle

yExample 6

yExample 6 contd

yExample 6 contd

Hamiltons Principle



29/37

Fermats Principle in Optics


Variable end-points

Fermats Principle

yFermats Principle

yExample 6

yExample 6 contd

yExample 6 contd

Hamiltons Principle


It states that a ray of light traveling in a region of variable refractive

index follows a path such that the total optical path length (physical

length refractive index) is stationary.


30/37

Example 6


Variable end-points

Fermats Principle

yFermats Principle

yExample 6

yExample 6 contd

yExample 6 contd

Hamiltons Principle


ExampleFrom Fermats principle deduce Snells law of refraction at an

interface.

FIG. 6: Path of a light ray at the plane interface between media with

refractive indicesn1 andn2, wheren2 < n1.

Solution

Let the interface be y D constant (see Fig. 6).


31/37

Example 6 contd


Variable end-points

Fermats Principle

yFermats Principle

yExample 6

yExample 6 contd

yExample 6 contd

Hamiltons Principle


For a ray that passes through points A and B, its element ofphysical path length is ds D .1 C y02/1=2 dx; thus, total opticalpath length is

P DZ

B

A

n.y/.1C y02/1=2 dx:

Integrand doesnt explicitly contain the independent variable x, so

(8) givesn.y/

1 C y021=2 D k;

where k is a constant.

Now y0 is the tangent of the angle between the instantaneous

direction of the ray and the x-axis; thus, along a ray, we have

n cos D

constant


32/37

Example 6 contd


Variable end-points

Fermats Principle

yFermats Principle

yExample 6

yExample 6 contd

yExample 6 contd

Hamiltons Principle


Since n is constant in each medium, y0 is also a constant. Thus

the rays travel in straight lines in each medium, and we have

n1 cos1 D n2 cos 2

or

n1 sin 1 D n2 sin 2


33/37

Hamiltons Principle in Mechanics


Variable end-points

Fermats Principle

Hamiltons Principle

yHamiltons Principle

yExample 7

yExample 7 contd

yExample 7 contd



34/37

Hamiltons Principle in Mechanics


Variable end-points

Fermats Principle

Hamiltons Principle


yExample 7

yExample 7 contd

yExample 7 contd


Consider a mechanical system whose configuration is defined by anumber of coordinates qi and time t , and which only experiences

forces derivable from a potential.

Hamiltons principle states that in moving from one configuration attime t0 to another at time t1 the motion is such as to make

D Zt1

t0

L.q1; : : : ; qn;Pq1; : : : ;

Pqn; t / dt (17)

stationary. The Lagrangian L is defined as L D T V, where T iskinetic energy and V (which is a function of qi only) is potential

energy. Applying the EL equation to we obtain Lagrangesequations

@L

@qiD d

dt

@L

@ Pqi

; i D 1 ; : : : ; n :

E l


35/37

Example 7


Variable end-points

Fermats Principle

Hamiltons Principle


yExample 7

yExample 7 contd

yExample 7 contd


ExampleUsing Hamiltons principle derive the wave equation for small

transverse oscillations of a taut string.

FIG. 7: Transverse displacement on a taut string that has an unstretched

lengthl .

E l 7 d


36/37

Example 7 contd


Variable end-points

Fermats Principle

Hamiltons Principle


yExample 7

yExample 7 contd

yExample 7 contd


SolutionIf and are the local density and tension of the string, both of

which may depend on x, then referring to Fig. 7, the kinetic and

potential energies of the string are given by

T DZl0

2

@y

@t

2dx; V D

Zl0

2

@y

@x

2dx;

and (17) becomes

D 12

Zt1t0

dt

Zl0

"

@y

@t

2

@y

@x

2#dx:

Using (13) and the fact that y does not appear explicitly, we obtain

@

@t @y

@t

@

@x@y

@x D 0:

E l 7 td


37/37

Example 7 contd


Variable end-points

Fermats Principle

Hamiltons Principle


yExample 7

yExample 7 contd

yExample 7 contd


If and do not depend on x or t , then

@2y

@x2D 1

c2@2y

@t2;

where c2 D =.

4 variation

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