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Classical Mechanics Page No. 95

VARIATIONAL PRINCIPLES

Unit 1: Euler-Lagranges’s Differential Equations:

• Introduction:

We have seen that co-ordinates are the tools in the hands of a mathematician.

With the help of these co-ordinates the motion of a particle and also the path

followed by the particle can be discussed. The piece wise information of the

path ( )y f x= , whether it is minimum or maximum at a point can be obtained from

differential calculus by putting 0y =� . The function is either maximum or minimum at

the point depends upon the value of second derivative of the function at that point.

The function is maximum at a point if its second derivative is negative at the point,

and is minimum at the point if its second derivative is positive at that point.

However, if we want to know the information about the whole path, we use

integral calculus. i.e., the techniques of calculus of variation and are called

variational principles. Thus the calculus of variation has its origin in the

generalization of the elementary theory of maxima and minima of function of a

single variable or more variables. The history of calculus of variations can be traced

back to the year 1696, when John Bernoulli advanced the problem of the

brachistochrone. In this problem one has to find the curve connecting two given

points A and B that do not lie on a vertical line, such that a particle sliding down this

curve under gravity from A reaches point B in the shortest time.

Apart from the problem of brachistochrone, there are three other problems

exerted great influence on the development of the subject and are:

1. the problem of geodesic,

2. the problem of minimum surface of revolution and

CHAPTER - II


3. the isoperimetric problem.

Thus in calculus of variation we consider the motion of a particle or system

of particles along a curve y = f(x) joining two points ( )1 1,P x y and ( )2 2,Q x y . The

infinitesimal distance between two points on the curve is given by

( )1

2 2 2ds dx dy= + .

Hence the total distance between two point P and Q

along the curve is given by

( )( ) ( )2

1

12 21 ,

x

x

dyI y x y dx y

dx′ ′= + =∫

In general the integrand is a function of the independent variable x, the

dependent variable y and its derivative y′ . Thus the most general form of the integral

is given by

( )( ) ( )2

1

, , .

x

x

I y x f x y y dx′= ∫ . . . (1)

This integral may represents the total path between two given points, the surface

area of revolution of a curve, the time for quickest decent etc. depending upon the

situation of the problem. The functional I in general depends upon the starting point

( )1 1,x y , the end point ( )2 2,x y and the curve between two points. The question is

what function y is of x so that the functional ( )( )I y x has stationary value. Thus in

this chapter we first find the condition to be satisfied by y(x) such that the functional

( )( )I y x defined in (1) must have extremum value. The fascinating principle in

calculus of variation paves the way to find the curve of extreme distance between

two points. Its object is to extremize the values of the functional. This is one of the

most fundamental and beautiful principles in applied mathematics. Because from this

principle one can determine the

Q(x , y )2 2

P(x , y )2 2

ds


(a) Newton’s equations of motion,

(b) Lagrange’s equations of motion,

(c) Hamilton’s equations of motion,

(d) Schrödinger’s equations of motion,

(e) Einstein’s field equations for gravitation,

(f) Hoyle-Narlikar’s equations for gravitation and so on and so forth by

slightly modifying the integrand.

Note : A functional means a quantity whose values are determined by one or several

functions. i.e., domain of a functional is a set of all admissible functions.

e.g. The length of the path l between two points is a function of curves y(x),

which it self is a function of x. Such functions are called functional.

• Basic Lemma :

If 1x and ( )2 1x x> are fixed constants and ( )G x is a particular continuous

function for 1 2x x x≤ ≤ and if ( ) ( )2

1

0

x

x

G x x dxη =∫ for every choice of continuous

differentiable function ( )xη such that ( ) ( )1 20x xη η= = , then ( ) 0G x = identically

in 1 2x x x≤ ≤ .

Proof : Let the lemma be not true. Let us assume that there is a particular value x′ of

x in the interval such that ( ) 0G x′ ≠ . Let us assume that ( ) 0.G x′ >

[ ]x1 x’1 x’2 x2

x’( )

Since ( )G x is continuous function in 1 2x x x≤ ≤ and in particular it is

continuous at x x′= . Hence there must exist an interval surrounding x′ say

1 2x x x′ ′≤ ≤ in which ( ) 0G x > everywhere.


Let us now see whether the integral ( ) ( )2

1

0

x

x

G x x dxη = ∀∫ permissible choice

of ( )xη .

We choose ( )xη such that

( )

( ) ( )

1 1

2 2

1 2 1 2

2 2

0

0

x for x x x

x x x x for x x x

for x x x

η ′= ≤ ≤

′ ′ ′ ′= − − ≤ ≤

′= ≤ ≤

. . . (1)

For this choice of ( )xη which also satisfies

( ) ( )1 2 0x xη η′ ′= = ,

the integral ( ) ( )2

1

x

x

G x x dxη∫ becomes

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 1 2 2

1 1 1 2

x x x x

x x x x

G x x dx G x x dx G x x dx G x x dxη η η η′ ′

′ ′

= + +∫ ∫ ∫ ∫

( ) ( ) ( ) ( ) ( )2 2

1 1

2 2

1 2

x x

x x

G x x dx x x x x G x dxη′

′

′ ′⇒ = − −∫ ∫ . . . (2)

Since

( ) 0G x > in 1 2x x x′ ′≤ ≤ ,

⇒ R. H. S. of equation (2) is definitely positive.

⇒ ( ) ( )2

1

0,

x

x

x G x dxη >∫

This is contradiction to the hypothesis

( ) ( )2

1

0

x

x

G x x dxη =∫ .

If ( ) 0G x′ < ,


we obtain the similar contradiction. This contradiction arises because of our

assumption that ( ) 0G x ≠ for x in 1 2x x x≤ ≤ .

This implies that ( ) 0G x = identically in 1 2x x x≤ ≤ .

This completes the proof.

Theorem 1 : Find the Euler- Lagrange differential equation satisfied by twice

differentiable function y(x) which extremizes the functional

( )( ) ( )2

1

, ,

x

x

I y x f x y y dx′= ∫

where y is prescribed at the end points.

Proof: Let ( )1 1,P x y and ( )2 2,Q x y be two fixed

points in xy plane. The points P and Q can be joined

by infinitely many curves. Accordingly the value of

the integral I will be different for different paths. We

shall look for a curve along which the functional I

has an extremum value. Let c be a curve between P

and Q whose equation is given by ( ),0y y x= .

Let also the value of the functional along the curve c be extremum and is given by

( )( ) ( )2

1

, ,

x

x

I y x f x y y dx′= ∫ . . . (1)

We can label all possible paths starting from P and ending at Q by the family of

equations

( ) ( ) ( ), , 0y x y x xα αη= + , . . . (2)

where α is a parameter and ( )xη is any differentiable function of x.

P(x , y )1 1

Q(x , y )2 2

c : y

= y (x

, 0)

c : y (x

, ) =

y (x, 0) +

(x)

α

α ηy

xO


For different values of α we get different curves. Accordingly the value of the

integral I will be different for different paths. Since y is prescribed at the end points,

this implies that there is no variation in y at the end points. i.e., all the curves of the

family must be identical at fixed points P and Q.

( ) ( )1 20x xη η⇒ = = . . . (3)

Conversely, the condition (3) ensures us that the curves of the family that all pass

through the points P and Q. Let the value of the functional along the neighboring

curve be given by

( )( ) ( ) ( )( )2

1

, , , , ,

x

x

I y x f x y x y x dxα α α′= ∫ . . . (4)

From differential calculus, we know the integral I is extremum if 0

0I

αα =

∂ =

∂ ,

since for 0α = the neighboring curve coincides with the curve which gives

extremum values of I .

Thus 0

0I

αα =

∂ =

∂ , ⇒ ( ) ( )

2

1

0

x

x

f fx x dx

y yη η

∂ ∂′+ = ′∂ ∂

∫ .

Integrating the second integration by parts, we get

( ) ( ) ( )22 2

11

0

xx x

x xx

f f d fx dx x x dx

y y dx yη η η

∂ ∂ ∂+ − = ′ ′∂ ∂ ∂

∫ ∫ . . . (5)

As y is prescribed at the end points, hence on using equations (3) we obtain

( )2

1

0

x

x

f d fx dx

y dx yη

∂ ∂− = ′∂ ∂

∫ .

By using the basic lemma of calculus of variation we get

0f d f

y dx y

∂ ∂− = ′∂ ∂

. . . . (6)

This is required Euler- Lagrange differential equation to be satisfied by y(x) for

which the functional I has extremum value.


• Important Note :

If however, y is not prescribed at the end points then there is a difference in y

even at the end points and hence ( )( ) 2

1

0x

xxη ≠ . As the value of the functional I is

taken only on the extremal between two points and hence we must have the Euler-

Lagrange equation is true. Consequently, in this case we must have from equation (5)

that

2

1 1 2

0 0 0

x

x x x

f f fand

y y y

∂ ∂ ∂= ⇒ = = ′ ′ ′∂ ∂ ∂

. . . . (7)

We will prove this result a little latter in Theorem No. 2.

Aliter : (Proof of the above Theorem (1)):

Let ( )1 1,P x y and ( )2 2,Q x y be two fixed points in xy plane. Let c be the

curve between P and Q whose equation is given by y = y(x).Let the extremum value

of the functional along the curve c be given by

( )( ) ( )2

1

, ,

x

x

I y x f x y y dx′= ∫ . . . . (1)

To find the condition to be satisfied by y(x), let

the curve c be slightly deformed from the original

position such that any point y on the curve c is

displaced to y yδ+ , where yδ is the variation in

the path for an arbitrary choice of α , at any point

except at the end points, as y prescribed there.

Mathematically this means that

( ) ( )1 1 2 2,y x y y x y= =

( ) 2

1

0x

xyδ⇒ = . . . . (2)

P(x , y )1 1

Q(x , y )2 2

y

xO

δyy + δy

c : y = y(x, 0)


Thus the value of the functional along the varied path is given by

( )2

1

, ,

x

x

I f x y y y y dxδ δ′ ′ ′= + +∫ . . . . (3)

Hence the change in the value of the functional due to change in the path is given by

( ) ( )2

1

, , , ,

x

x

I I f x y y y y f x y y dxδ δ′ ′ ′ ′− = + + − ∫ ,

Let ( ) ( )2

1

, , , ,

x

x

I I I f x y y y y f x y y dxδ δ δ′ ′ ′ ′− = = + + − ∫ . . . . (4)

We recall the Taylor’s series expansion for the function of two variables’

( ) ( ), , , , ...f f

f x y y y y f x y y y yy y

δ δ δ δ ∂ ∂

′ ′ ′ ′+ + = + + + ′∂ ∂

Since yδ is very small, therefore by neglecting the higher order terms in yδ and

yδ ′ we have

( ) ( ), , , ,f f

f x y y y y f x y y y yy y

δ δ δ δ ∂ ∂

′ ′ ′ ′+ + − = + ′∂ ∂ .

Substituting this in the equation (4) we get

2

1

x

x

f fI y y dx

y yδ δ δ

∂ ∂′= + ′∂ ∂

∫

We know dy d

ydx dx

δ δ= ,

hence we have

2

1

( )

x

x

f f dI y y dx

y y dxδ δ δ

∂ ∂= + ′∂ ∂ ∫ .

Integrating the second integral by parts we get

22 2

1 11

xx x

x xx

f f d fI y dx y y dx

y y dx yδ δ δ δ

∂ ∂ ∂= + − ′ ′∂ ∂ ∂ ∫ ∫ .


On using equation (2) we get

2

1

x

x

f d fI y dx

y dx yδ δ

∂ ∂= − ′∂ ∂ ∫ . . . . (5)

If I is extremum along the curve y = y(x) then change in I is zero. ( ). ., 0i e Iδ = .

This resembles very closely with a similar condition of extremum of a function in

differential calculus.

⇒ 2

1

0

x

x

f d fy dx

y dx yδ

∂ ∂− = ′∂ ∂

∫ .

Since yδ is arbitrary, we have

0f d f

y dx y

∂ ∂− = ′∂ ∂

.

This is the Euler-Lagrange’s differential equation to be satisfied by y(x) for the

extremum of the functional between two points.

• Generalization of Theorem (1) : Euler-Lagrange’s equations for several

dependent variables.

Theorem 1a : Derive the Euler-Lagrange’s equations that are to be satisfied by twice

differential functions 1 2, ,...,n

y y y that extremize the integral

( )2

1

1 2 1 2, , ,..., , , ,...,

x

n n

x

I f x y y y y y y dx′ ′ ′= ∫

with respect to those functions 1 2, ,...,n

y y y which achieve prescribed values at the

fixed points 1 2, .x x

Proof: The functional which is to be extremized can be written as

( )2

1

, , , 1, 2,...,

x

i i

x

I f x y y dx i n′= =∫ .

Choose the family of neighboring curves as

( ) ( ) ( ), ,0i i iy x y x xα αη= +


and repeating the procedure delineated either in the Theorem (1) or in the alternate

proof we arrive the following set of Euler-Lagrange’s equations

0, 1, 2,...,i i

f d fi n

y dx y

∂ ∂− = =

′∂ ∂ .

Geodesic : Geodesic is defined as the curve of stationary (extremum) length between

two points.

Worked Examples •

Example 1 : Show that the geodesic (shortest distance between two points) in a

Euclidian plane is a straight line.

Solution: Take ( )1 1,P x y and ( )2 2,Q x y be two fixed points in a Euclidean plane.

Let ( )y f x= be the curve between P and Q. Then the element of distance between

two neighboring points on the curve ( )y f x= joining P and Q is given by

2 2 2ds dx dy= +

Hence the total distance between the point P and Q along the curve is given by

Q

P

I ds= ∫

( )2

1

12 21 ,

x

x

dyI y dx y

dx′ ′⇒ = + =∫ . . . (1)

Here the functional I is extremum if the integrand

( )1

2 21f y′= + . . . (2)

must satisfy the Euler-Lagrange’s differential equation

0f d f

y dx y

∂ ∂− = ′∂ ∂

. . . (3)

Now from equation (2) we find that


2

01

f f yand

y y y

′∂ ∂= =

′∂ ∂ ′+

2

01

d y

dx y

′ ⇒ = ′+

.

Integrating we get

21y c y′ ′= + .

Squaring we get

1 12

,1

cy c where c

c′ = =

+.

Integrating we get

1 2y c x c= + . . . . (4)

This is the required straight line. Thus the shortest distance between two points in a

Euclidean plane is a straight line.

Example 2 : Show that the shortest distance between two polar points in a plane is a

straight line.

Solution: Define a curve in a plane. If A ( ),x y and B ( ),x dx y dy+ + are

infinitesimal points on the curve, then an element of distance between A and B is

given by

2 2 2ds dx dy= + . . . . (1)

Let ( )rθ θ= be the polar equation of the curve and ( ) ( )1 1 2 2, ,P r and Q rθ θ be two

polar points on it. Recall the relations

cos ,

sin .

x r

y r

θ

θ

=

=

Hence equation (1) becomes

2 2 2 2ds dr r dθ= + . . . . (2)

Thus the total distance between the points P and Q becomes


( )2

1

12 2 21 ,

r

r

dI r dr

dr

θθ θ′ ′= + =∫ . . . . (3)

The functional I is shortest if the integrand

( )1

2 2 21f r θ ′= + . . . (4)

must satisfy the Euler-Lagrange’s differential equation

0f d f

drθ θ

∂ ∂ − = ′∂ ∂

, (5)

2

2 20

1

d r

dr r

θ

θ

′⇒ =

′+ ,

2 2 21r h rθ θ′ ′⇒ = + .

Squaring and solving for θ ′ we get

( )1

2 2 2

d h

drr r h

θ= ±

−

.

On integrating we get

1

0cosh

rθ θ−

= ± +

,

where 0θ is a constant of integration. We write this as

0cos( )h r θ θ= − . . . . (6)

This is the polar form of the equation of straight line. Hence the shortest distance

between two polar points is a straight line.

Note : If ( )r r θ= is the polar equation of the curve, then the length of the curve is

given by

1

0

2

2 drI r d

d

θ

θ

θθ

= +

∫ .

Since the integrand 2 2f r r′= + does not containθ , we therefore have


f

f r hr

∂′− =

′∂.

Solving this equation we readily obtain the same polar equation of straight line as the

geodesic.

Example 3 : Show that the geodesic ( )φ φ θ= on the surface of a sphere is an arc of

the great circle.

Solution : Consider a sphere of radius r described by the equations

sin cos ,

sin sin ,

cos .

x r

y r

z r

θ φ

θ φ

θ

=

=

=

. . . (1)

If A ( ), ,x y z and B ( ), ,x dx y dy z dz+ + + be two neighboring points on the curve

joining the points P and Q. Then the infinitesimal distance between A and B along

the curve is given by

2 2 2 2ds dx dy dz= + + , . . . (2)

where from equation (1) we find

cos cos sin sin ,

cos sin sin cos ,

sin .

dx r d r d

dy r d r d

dz r d

θ φ θ θ φ φ

θ φ θ θ φ φ

θ θ

= −

= +

= −

. . . (3)

Squaring and adding these equations we readily obtain

2 2 2 2 2 2sinds r d r dθ θ φ= + . . . . (4)

Hence the total distance between the points P and Q along the curve ( )φ φ θ= is

given by

( )2

1

12 2 21 sin ,

dI r d

d

θ

θ

φθ φ θ φ

θ′ ′= + =∫ . . . (5)

where

( )1

2 2 21 sinf r θ φ′= + . . . . (6)


The curve is geodesic if the functional I is stationary. This is true if the function f

must satisfy the Euler-Lagrange’s equations.

0f d f

dφ θ φ

∂ ∂− = ′∂ ∂

. . . (7)

2

2 2

sin0

1 sin

d r

d

θ φ

θ θ φ

′ ⇒ = ′+

.

Integrating we get

⇒2 2

2 2

sin,

1 sinc

θ φ

θ φ

′=

′+

Solving for φ′ we get

( )

2

12 2 2

cos

1 cos

c ec

c ec

θφ

θ

′ =

−

.

On simplifying we get

( )

2

12 2 2

cos

1 cot

d k ec

dk

φ θ

θθ

=

−

. . . . (8)

Put 2cot cosk t ec d dtθ θ θ= ⇒ − = ,

Therefore we have

21

dtd

tφ = −

−.

Integrating we get

1sin tφ α −= − ,

or ( )1sin cotkφ α θ−= − ,

( )cot sin ,

cos sin sin cos cos sin sin ,

k

k

θ α φ

θ α θ φ α θ φ

⇒ = −

⇒ = −

sin coskz x yα α⇒ = − . . . . (9)


This is the first-degree equation in x, y, z, which represents a plane. This plane

passes through the origin, hence cutting the sphere in a great circle. Hence the

geodesic on the surface of a sphere is an arc of a great circle.

Example 4 : Show that the curve is a catenary for which the area of surface of

revolution is minimum when revolved about y-axis.

Solution: Consider a curve between two points 1 1( , )x y and 2 2( , )x y in the xy plane

whose equation is ( )y y x= . We form a surface

by revolving the curve about y-axis. Our claim is

to find the nature of the curve for which the

surface area is minimum. Consider a small strip

at a point A formed by revolving the arc length

ds about y –axis. If the distance of the point A

on the curve from y-axis is x, then the surface

area of the strip is equal to 2 x dsπ .

But we know the element of arc ds is given by

21ds y dx′= + .

Thus the surface area of the strip ds is equal to

22 1x y dxπ ′+ .

Hence the total area of the surface of revolution of the curve ( )y y x= about y- axis

is given by

2

1

22 1

x

x

I x y dxπ ′= +∫ . . . . (1)

This surface area will be minimum if the integrand

22 1f x yπ ′= + . . . (2)

must satisfy Euler-Lagrange’s equation

xdsA

O

z

x

y


0f d f

y dx y

∂ ∂− = ′∂ ∂

, . . . (3)

2

20

1

d x y

dx y

π ′ ⇒ = ′+

,

2

01

d x y

dx y

′ ⇒ = ′+

.

Integrating we get

21xy a y′ ′= + .

Solving for y′ we get

2 2

dy a

dx x a=

−.

Integrating we get

1coshx

y a ba

− = +

.

Or coshy b

x aa

− =

. . . . (4)

This shows that the curve is the catenary.

• The Brachistochrone Problem :

The Brachistochrone is the curve joining two points not lie on the vertical

line, such that the particle falling from rest under the influence of gravity from higher

point to the lower point in minimum time. The curve is called the cycloid.

Example 5: Find the curve of quickest decent.

Or

A particle slides down a curve in the vertical plane under gravity. Find the curve

such that it reaches the lowest point in shortest time.


Solution: Let A and B be two points on the curve not lie on the vertical line. Let

dsv

dt= be the speed of the particle along the curve. Then the time required to fall an

arc length ds is given by

21

.

dsdt

v

ydt dx

v

=

′+⇒ =

Therefore the total time required for the particle to go

from A to B is given by

21

B

AB

A

yt dx

v

′+= ∫ . . . (1)

Since the particle falls freely under gravity, therefore its potential energy goes on

decreasing and is given by

V mgx= − ,

and the kinetic energy is given by

21

2T mv= .

Now from the principle of conservation of energy we have

T V+ = constant.

Initially at point A, we have 0x = and 0v = . Hence the constant is zero.

21

2mv mgx⇒ = ,

2v gx⇒ = . . . . (2)

Hence equation (1) becomes

2

1

21

2

x

AB

x

yt dx

gx

′+= ∫ . . . . (3)

Thus AB

t is minimum if the integrand

y

x

A

v

B


21

2

yf

gx

′+= , . . . (4)

must satisfy Euler-Lagrange’s equation

0f d f

y dx y

∂ ∂− = ′∂ ∂

. . . (5)

2

2

02 (1 )

0(1 )

d y

dx gx y

d y

dx x y

′ ⇒ = ′+

′ ⇒ = ′+

Integrating we get

2(1 )y c x y′ ′= + .

Solving it for y′ we get

dy x

dx a x=

−

Integrating we get

x

y dx ba x

= +−

∫ . . . (6)

Put

2sin ( / 2)

2 sin( / 2) cos( / 2)

x a

dx a d

θ

θ θ θ

=

⇒ =. . . . (7)

Hence

2sin ( / 2)y a d bθ θ= +∫ .

( )sin2

ay bθ θ⇒ = − + ,

If 0, 0 0y bθ= = ⇒ = ,

hence


( )sin2

ay θ θ= − . . . . (8)

Thus from equations (7) and (8) we have

( )

( )

1 cos ,

sin ,2

x b

ay b for b

θ

θ θ

= −

= − =

This is a cycloid. Thus the curve is a cycloid for which the time of decent is

minimum.

Example 6 : Find the extremal of the functional

( )2

2 2

0

2y y xy dx

π

′ − +∫

subject to the conditions that

( )0 0, 02

y yπ

= =

.

Solution: Let the functional be denoted by

( )2

2 2

0

2I y y xy dx

π

′= − +∫ . . . . (1)

The functional is extremum if the integrand

2 2 2f y y xy′= − + . . . (2)

must satisfy the Euler-Lagrange’s equation

0f d f

y dx y

∂ ∂− = ′∂ ∂

, . . . (3)

( ) ( )2 2 0d

x y ydx

′⇒ − − = ,

y y x′′⇒ + = . . . . (4)


This is second order differential equation, whose complementary function (C.F.) is

given by

1 2cos siny c x c x= + . . . (5)

where 1 2c and c are arbitrary constants.

The particular integral (P.I.) is

( )

121

.

y D x

y x

−

= +

=

Hence the general solution is given by

1 2cos siny c x c x x= + + . . . . (6)

This shows that the extremals of the functional are the two-parameter family of

curves. On using the boundary conditions we obtain

( ) 1

2

0 0 0,

0 .2 2

y c

y cπ π

= ⇒ =

= ⇒ = −

Hence the required extremal is

sin .2

y x xπ

= − . . . (7)

Example 7 : Find the extremal of the functional

2 3

2

1

xdx

y

′ ∫


( ) ( )1 0, 2 3y y= = .

Solution: Let the functional be denoted by

2 3

2

1

xI dx

y

= ′ ∫ . . . (1)

The functional is extremum if the integrand


3

2

xf

y=

′ . . . (2)


0f d f

y dx y

∂ ∂− = ′∂ ∂

. (3)

3

30

d x

dx y

⇒ = ′

.

Integrating we get

3 3x cy′=

or y ax′ = .

Integrating we get

2

2

ay x b= + . . . . (4)

Now using the boundary conditions we get

( )

( )

1 0 0,2

2 3 2 3.

ay b

y a b

= ⇒ + =

= ⇒ + =

Solving these two equations we obtain

2, 1a b= = − .

Hence the required functional becomes

2 1y x= − . . . . (5)

Example 8 : Show that the time taken by a particle moving along a curve ( )y y x=

with velocity ,ds

xdt

= from the point (0,0) to the point (1,1) is minimum if the curve

is a circle having its center on the x-axis.

Solution: Let a particle be moving along a curve y = y(x) from the point (0, 0) to the

point (1, 1) with velocity


ds

xdt

=

ds

dtx

⇒ = .

Therefore the total time required for the particle to move from the point (0, 0) to the

point (1, 1) is given by

1

0

dst

x= ∫ . . . (1)

where the infinitesimal distance between two points on the path is given by

21 ,dy

ds y dx ydx

′ ′= + = .

Hence the equation (1) becomes

1 2

0

1 yt dx

x

′+= ∫ . . . (2)

Time t is minimum if the integrand

21 y

fx

′+= . . . (3)


0f d f

y dx y

∂ ∂− = ′∂ ∂

. . . . (4)

2

2

0,1

1 .

d y

dx x y

y cx y

′ ⇒ = ′+

′ ′⇒ = +


2 2

1,

xy for a

ca x′ = =

−.

Integrating we get


2 2

xy dx b

a x= +

−∫

Put sin cosx a dx a dθ θ θ= ⇒ = . . . (5)

Therefore,

siny a d bθ θ= +∫ ,

cosy a bθ= − + . . . . (6)

Squaring and adding equations (5) and (6) we get

( )22 2

x y b a+ − = ,

which is the circle having center on y –axis.

Example 9 : Show that the geodesic on a right circular cylinder is a helix.

Solution: We know the right circular cylinder is characterized by the equations

2 2 2 ,x y a z z+ = = . . . (1)

The parametric equations of the right circular cylinder are

cos ,

sin ,

.

x a

y a

z z

θ

θ

=

=

=

where a is a constant. The element of the distance (metric)

2 2 2 2ds dx dy dz= + +

on the surface of the cylinder becomes

2 2 2 2ds a d dzθ= + ,

Hence the total length of the curve on the surface of the cylinder is given by

0

2 2 ,dz

s a z d for zd

θ

θ

θθ

′ ′= + =∫ . . . . (2)

For s to be extremum, the integrand

2 2f a z′= + . . . . (3)



0f d f

z d zθ

∂ ∂ − = ′∂ ∂

. . . . (4)

0d z

d fθ

′ ⇒ =

.

Integrating the equation and solving for z′ we get

z a′ = (constant).

Integrating we get

, 0z a b aθ= + ≠ , . . . (5)

where a, b are constants. Equation (5) gives the required equation of helix. Thus the

geodesic on the surface of a cylinder is a helix.

Example 10 : Find the differential equation of the geodesic on the surface of an

inverted cone with semi-vertical angle θ .

Solution: The surface of the cone is characterized by the equation

2 2 2 2tan , .x y z constθ θ+ = = . . . (1)

The parametric equations of the cone are given by

cos ,

sin ,

.

x ar

y ar

z br

φ

φ

=

=

=

. . . (2)

where for sin , cosa bθ θ= = are constant.

Thus the metric 2 2 2 2ds dx dy dz= + +

on the surface of the cone becomes

2 2 2 2 2ds dr a r dφ= + . . . (3)

Hence the total length of the curve ( )rφ φ= on the surface of the cone is given by

2 2 21 ,d

s a r drdr

φφ φ′ ′= + =∫ . . . (4)


The length s is stationary if the integrand

2 2 21f a r φ ′= + . . . (5)


0f d f

drφ φ

∂ ∂− = ′∂ ∂

. . . . (6)

2 2

0d a r

dr f

φ′ ⇒ =

,

Solving for φ′ we get

1

2 2 2

1

cd

dr ar a r c

φ=

−, . . . (7)

where 1c = constant.

This is the required differential equation of geodesic, and the geodesic on the surface

of the cone is obtained by integrating equation (7). This gives

1

2 2 2

1

cdr

ar a r cφ α= +

−∫ .

( )

1

1

1

1,

sec .

arsec

a c

cr a

a

φ α

φ α

− = +

⇒ = −

Example 11 : Find the curve for which the functional

( ) ( )4

2 2

0

I y x y y dx

π

′= − ∫

can have extrema, given that y(0)=0, while the right –hand end point can vary along

the line 4

xπ

= .


Solution: To find the extremal curve of the functional

( ) ( )4

2 2

0

I y x y y dx

π

′= − ∫ , . . . (1)

we must solve Euler’s equation

0f d f

y dx y

∂ ∂− = ′∂ ∂

, . . . (2)

where 2 2f y y′= − . . . (3)

0y y′′⇒ + = . . . . (4)

This is the second order differential equation, whose solution is given by

cos siny a x b x= + . . . . (5)

The boundary condition ( )0 0y = gives a = 0.

siny b x⇒ = . . . . (6)

The second boundary point moves along the line 4

xπ

= .

4

0x

f

y π=

∂⇒ = ′∂

4

( ) 0x

y π=

′⇒ = ,

where from equation (6) we have cosy b x′ = . Thus 4

y at xπ

′ = gives

b= 0. This implies the extremal is attained on the line y = 0.

Example 12 : If f satisfies Euler-Lagrange’s equation

0f d f

y dx y

∂ ∂− = ′∂ ∂

.

Then show that f is the total derivative dg

dx of some function of x and y and

conversely.


Solution: Given that f satisfies Euler-Lagrange’s equation

0f d f

y dx y

∂ ∂− = ′∂ ∂

. . . . (1)

We claim that ,dg

fdx

=

where ( ),g g x y= .

As ( ), ,f f x y y′= ,

we write equation (1) explicitly as

2 2 2

2 20

f f f fy y

y x y y y y

∂ ∂ ∂ ∂′ ′′− − − =

′ ′ ′∂ ∂ ∂ ∂ ∂ ∂. . . . (2)

We see from equation (2) that the first three terms on the l. h. s. of (2) contain at the

highest the first derivative of y. Therefore equation (2) is satisfied identically if the

coefficient of y′′ vanishes identically.

2

20.

f

y

∂⇒ =

′∂

Integrating w. r. t. y′ we get

( ),f

q x yy

∂=

′∂.

Integrating once again we get

( ) ( ), ,f q x y y p x y′= + , . . . (3)

where ( ),p x y and ( ),q x y are constants of integration and may be function of x and

y only. Then the function f so determined must satisfy the Euler –Lagrange’s

equation (1). From equation (3) we find

f q p

yy y y

∂ ∂ ∂′= +

∂ ∂ ∂,

and ( ),f

q x yy

∂=

′∂.


Therefore equation (1) becomes

( )( , ) 0q p d

y q x yy y dx

∂ ∂′ + − =∂ ∂

.

⇒ 0q p q q

y yy y x y

∂ ∂ ∂ ∂′ ′+ − − =∂ ∂ ∂ ∂

.

⇒ p q

y x

∂ ∂=

∂ ∂. .(4)

This is the condition that the equation pdx qdy+ is an exact differential

equation dg .

,

(3)

dg pdx qdy

dgp qy f by

dx

⇒ = +

′⇒ = + =

Therefore,

dg

fdx

= . . . . (5)

This proves the necessary part.

Conversely, assume that dg

fdx

= . We prove that f satisfies the Euler-Lagrange’s

equation

0f d f

y dx y

∂ ∂− = ′∂ ∂

.

Since dg g g

f ydx x y

∂ ∂′= = +

∂ ∂,

Therefore, we find

2 2

2,

.

f g gy

y x y y

g g

y y

∂ ∂ ∂′= +

∂ ∂ ∂ ∂

∂ ∂=

′∂ ∂


Consider now

2 2

2

f d f g g d gy

y dx y x y y dx y

∂ ∂ ∂ ∂ ∂′− = + − ′∂ ∂ ∂ ∂ ∂ ∂

.

2 2 2 2

2 2

f d f g g g gy y

y dx y x y y x y y

∂ ∂ ∂ ∂ ∂ ∂′ ′− = + − − ′∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

.

0f d f

y dx y

∂ ∂− = ′∂ ∂

dg

fdx

⇒ =

satisfies Euler-Lagrange’s equation.

Example 13 : Show that the Euler-Lagrange’s equation of the functional

( )( ) ( )2

1

, ,

x

x


has the first integral f

f y consty

∂′− =

′∂, if the integrand does not depend on x.

Solution: The Euler-Lagrange’s equation of the functional

( )( ) ( )2

1

, ,

x

x


to be extremum is given by

0f d f

y dx y

∂ ∂− = ′∂ ∂

. . . . (1)

2 2 2

2 20

ff f f

y yy x y y y y

∂ ∂ ∂ ∂′ ′′⇒ − − − =

′ ′ ′∂ ∂ ∂ ∂ ∂ ∂

If f does not involve x explicitly, then 0f

x

∂=

∂.

Therefore, we have


2 2

2 20

f f fy y

y y y y

∂ ∂ ∂′ ′′− − =

′ ′∂ ∂ ∂ ∂. . . . (2)

Multiply equation (2) by y′ we get

2 2

2

2 20

f f fy y y y

y y y y

∂ ∂ ∂′ ′ ′′ ′− − =

′ ′∂ ∂ ∂ ∂ . . . (3)

But we know that

2 2

2

2

d f f f f f ff y y y y y y y

dx y y y y y y y

∂ ∂ ∂ ∂ ∂ ∂′ ′ ′′ ′′ ′ ′ ′′− = + − − − ′ ′ ′ ′ ′∂ ∂ ∂ ∂ ∂ ∂ ∂

,

2 2

2

2

d f f f ff y y y y y

dx y y y y y

∂ ∂ ∂ ∂′ ′ ′ ′ ′′− = − − ′ ′ ′∂ ∂ ∂ ∂ ∂

. . . . (4)

From equations (3) and (4) we see that

0d f

f ydx y

∂′− = ′∂

,

.f

f y consty

∂′⇒ − =

′∂ . . . (5)

This is the first integral of Euler-Lagrange’s equation, when the functional

( ),f f y y′= .

Worked Examples •

Example 14 : Find the minimum of the functional

( )1

2

0

1( )

2I y x y yy y y dx

′ ′ ′= + + +

∫

if the values at the end points are not given.

Solution: For the minimum of the functional

( )1

2

0

1( )

2I y x y yy y y dx

′ ′ ′= + + +

∫ . . . (1)

the integrand


21

2f y yy y y′ ′ ′= + + + . . . (2)


0f d f

y dx y

∂ ∂− = ′∂ ∂

. . . . (3)

( )1 1 0d

y y ydx

′ ′⇒ + − + + = ,

1y′′⇒ = . . . . (4)

Integrating we get

1,y x c′ = + . . . (5)

Further integrating we get

2

1 22

xy c x c= + + , . . . (6)

where 1 2,c c are constants of integration and are to be determined.

However, note that the values of y at the end points are not prescribed. In this case

the constants are determined from the conditions.

0 1

0, 0x x

f fand

y y= =

∂ ∂= = ′ ′∂ ∂

. . . . (7)

These two conditions will determine the values of the constants.

( ) ( )0 1

1 0, 1 0x x

y y and y y= =

′ ′⇒ + + = + + = , . . . (8)

where from equation (5) and (6) we have

( ) ( )1 20 0y c and y c′ = = ,

similarly, ( ) ( )1 1 2

11 1 1

2y c and y c c′ = + = + + .

Thus the equations (8) become

1 2 1 2

51 0, 2 0

2c c c c+ + = + + = .


Solving these equations for 1c and 2c we obtain 1 2

3 1

2 2c and c= − =

Hence the required curve for which the functional given in (1) becomes minimum is

( )213 1 .

2y x x= − + . . . (9)

••

Theorem 3 : Find the Euler- Lagrange differential equation satisfied by four times

differentiable function y(x) which extremizes the functional

( )( ) ( )2

1

, , ,

x

x

I y x f x y y y dx′ ′′= ∫

under the conditions that both y and y′ are prescribed at the end points.

Proof : Let ( )1 1,P x y and ( )2 2,Q x y be two fixed points in xy plane. The points P

and Q can be joined by infinitely many curves. Accordingly the value of the integral

I will be different for different paths. We shall look for a curve along which the

functional I has an extremum value. Let c be a curve between P and Q whose

equation is given by y = y(x, 0). Let also the value of the functional along the curve c

be extremum and is given by

( )( ) ( )2

1

, , ,

x

x

I y x f x y y y dx′ ′′= ∫ . . . (1)

We can label all possible paths starting from P and ending at Q by the family of

equations

( ) ( ) ( ), , 0y x y x xα αη= + , . . . (2)

where α is a parameter and ( )xη is any differentiable function of x.

For different values of α we get different curves. Accordingly the value of

the integral I will be different for different paths. Since y and y′ are prescribed at

the end points, this implies that there is no variation in y and y′ at the end points.


i.e., all the curves of the family and their derivative must be identical at fixed points

P and Q.

This implies that

( ) ( )

( ) ( )1 1 1

2 2 2

, ,0 ,

, ,0 ,

y x y x y

y x y x y

α

α

= =

= =

and also

( ) ( )

( ) ( )1 1 1

2 2 2

, ,0 ,

, ,0

y x y x y

y x y x y

α

α

′ ′ ′= =

′ ′ ′= =

( ) ( )1 20 ,x xη η⇒ = = . . . (3a)

and ( ) ( )1 20x xη η′ ′= = . . . . (3b)

Conversely, the conditions (3) ensure us that the curves of the family that all pass

through the points P and Q. Let the value of the functional along the neighboring

curve be given by

( )( ) ( ) ( ) ( )( )2

1

, , , , , , ,

x

x

I y x f x y x y x y x dxα α α α′ ′′= ∫ . . . . (4)

From differential calculus, we know the integral I is extremum if 0

0I

αα =

∂ =

∂ ,

because for 0α = the neighboring curve coincides with the curve which gives

extremum values of I . Thus

0

0I

αα =

∂ =

∂ ,

⇒ ( ) ( ) ( )2

1

0

x

x

f f fx x x dx

y y yη η η

∂ ∂ ∂′ ′′+ + = ′ ′′∂ ∂ ∂

∫ .

Integrating the second and the third integrations by parts, we get


( ) ( ) ( ) ( )

( ) ( )

2 22 2

1 11 1

22

11

2

20,

x xx x

x xx x

x x

xx

f f d f fx dx x x dx x

y y dx y y

d f d fx x dx

dx y dx y

η η η η

η η

∂ ∂ ∂ ∂′+ − + − ′ ′ ′′∂ ∂ ∂ ∂

∂ ∂− + = ′′ ′′∂ ∂

∫ ∫

∫

. . . (5)

( ) ( ) ( )

( ) ( )

22 2

1 11

2 2

11

2

20.

xx x

x xx

x x

xx

f f d f d fx dx x x dx

y y dx y dx y

f d fx x dx

y dx y

η η η

η η

∂ ∂ ∂ ∂+ − − + ′ ′′ ′∂ ∂ ∂ ∂

∂ ∂′+ + = ′′ ′′∂ ∂

∫ ∫

∫

. . . (6)

As y and y′ are both prescribed at the end points, hence on using equations (3) we

obtain

( )2

1

2

20

x

x

f d f d fx dx

y dx y dx yη

∂ ∂ ∂− + = ′ ′′∂ ∂ ∂

∫ . . . . (7)


2

20

f d f d f

y dx y dx y

∂ ∂ ∂− + = ′ ′′∂ ∂ ∂

. . . . (8)

This is required Euler- Lagrange differential equation to be satisfied by y(x) for

which the functional I has extremum value.

Note : If the functions y and y′ are not prescribed at the end points then we must

have unlike the fixed end point problem, ( )xη and ( )xη′ need no longer vanish at

the points 1x and 2x . In order that the curve y =y(x) to be a solution of the variable

end point problem, y must be an extremal. i.e., y must be a solution of Euler’s

equation (8). Thus for the extremal we have from equation (6)

⇒ 0, 0.

x bx b

x a x a

f f d f

y y dx y

==

= =

∂ ∂ ∂= − = ′′ ′ ′′∂ ∂ ∂

. . . (9)


Thus to solve the variable end point problem, we must first solve Euler’s equation (8)

and then use the conditions (9) to determine the values of the arbitrary constants.

The above result can be summarized in the following theorem (3a).

Theorem (3a) : Derive the differential equation satisfied by four times differentiable

function ( )y x , which extremizes the integral

( )1

0

, , ,

x

x

I f x y y y dx′ ′′= ∫

under the condition that both ,y y′ are prescribed at both the ends. Show that if

neither y nor y′ is prescribed at either end points then

0 1

1

0

0

0

x x x x

x x

x x

f f

y y

f d f

y dx y

= =

=

=

∂ ∂= = ′′ ′′∂ ∂

∂ ∂− = ′ ′′∂ ∂

Remark: (General case of Theorem (3)) If the integrand in equation (1) of the

Theorem (3) is of the form ( ), , , ,..., nf f x y y y y′ ′′= with the boundary conditions

( ) ( ) ( )

( ) ( ) ( )

1 1

1 1 1 1 1 1

1 1

2 2 2 2 2 2

, ,....., ,

, ,....., ,

n n

n n

y x y y x y y x y

y x y y x y y x y

− −

− −

′ ′= = =

′ ′= = =

then the Euler-Lagrange’s differential equation is

( )2

2..... 1 0

nn

n n

f d f d f d f

y dx y dx y dx y

∂ ∂ ∂ ∂− + + + − = ′ ′′∂ ∂ ∂ ∂

.

Worked Examples •

Example 15 : Find the curve, which extremizes the functional

( )( ) ( )4

2 2 2

0

I y x y y x dx

π

′′= − +∫


under the conditions that

( ) ( )0 0, 0 1,

1

4 4 2

y y

y yπ π

′= =

′= =

Solution : For the extremization of the functional

( )( ) ( )4

2 2 2

0

I y x y y x dx

π

′′= − +∫ . . . (1)

the integrand

2 2 2f y y x′′= − + . . . (2)


2

20

f d f d f

y dx y dx y

∂ ∂ ∂− + = ′ ′′∂ ∂ ∂

. . . . (3)

( )2

22 2 0

dy y

dx′′⇒ − + = ,

i.e., 4

40

d yy

dx− = . . . . (4)

The solution of equation (4) is given by

cos sinx xy ae be c x d x−= + + + . . . (5)

where a, b, c, d are constants of integration and are to be determined.

Thus

( )

( )

4 4

4 4

0 0 0,

1 1 1 1,

4 2 2 2 2

0 1 1,

1 1 1 1

4 2 2 2 2

y a b c

y ae be c d

y a b d

y ae be c d

π π

π π

π

π

−

−

= ⇒ + + =

= ⇒ + + + =

′ = ⇒ − + =

′ = ⇒ − − + =


Solving these equations we get 0a b c= = = and 1d = .

Hence the required curve is siny x= .

Example 16 : Minimize the functional

( )2

2

0

1

2I x dt= ∫ �� ,

satisfies

( ) ( ) ( ) ( )0 1, 0 1, 2 1, 2 0x x x x= = = =� � .

Solution : To minimize the functional

( )2

2

0

1

2I x dt= ∫ �� , . . . (1)

the integral

21

2f x= �� . . . (2)


2

20

f d f d f

x dt x dt x

∂ ∂ ∂ − + =

∂ ∂ ∂ � ��. . . . (3)

This implies 4

40

d x

dt= . . . . (4)

Integrating we get

3 2

1 2 3 46 2

t tx c c c t c= + + + . . . . (5)

where x given in (5) must satisfy the conditions

( )

( )

( )

( )

4

3

1 2

1 2

0 1 1,

0 1 1,

2 1 4 6 3,

2 0 2 2 1.

x c

x c

x c c

x c c

= ⇒ =

= ⇒ =

= ⇒ + = −

= ⇒ + = −

�

�


Solving for 1c and 2c we get the required functional is

2

14

tx t= − + + .

Example 17 : Find the function on which the functional

( )( ) ( )1

2

0

2I y x y xy dx′′= −∫ ,

can be extremized such that

( ) ( ) ( )1

0 0 0, 1120

y y y′= = =

and ( )1y′ is not prescribed.

Solution : For the extremization of the functional

( )( ) ( )1

2

0

2I y x y xy dx′′= −∫ . . . (1)

the integrand

2 2f y xy′′= − . . . (2)


2

20

f d f d f

y dx y dx y

∂ ∂ ∂− + = ′ ′′∂ ∂ ∂

. . . . (3)

( )2

2

4

4

2 2 0,

.

dx y

dx

d yx

dx

′′⇒ − + =

⇒ =

Integrating we obtain

2

2

xy a′′′ = + ,

3

6

xy ax b′′⇒ = + + ,


4

2

24 2

x ay x bx c′⇒ = + + + ,

5

3 2

120 6 2

x a by x x cx d⇒ = + + + + , . . . (4)

where a, b, c, d are constants to be determined. Given that

( )

( )

( )

0 0 0,

0 0 0,

11 3 0.

120

y d

y c

y a b

= ⇒ =

′ = ⇒ =

= ⇒ + =

Since ( )1y′ is not prescribed. i.e., y′ at x= 1 is not given, then we have the

condition that

( )1

0 1 0x

fy

y=

∂′′= ⇒ = ′′∂

.

This gives from above equation that

6 6 1a b+ = − .

Solving the equations for a and b we get

1 1

,4 12

a and b= − = .

Substituting these values in equation (4) we get

( )5

2 31

120 24

xy x x= + − . . . . (5)

This is the required curve.

• When integrand is a function more than two dependent variables :

Example 18: Prove that the shortest distance between two points in a Euclidean 3-

space is a straight line.

Solution: Define the curve ( ) ( ),y y x z z x= = in the 3-dimentional Euclidean

space. Let P ( ), ,x y z and Q ( ), ,x dx y dy z dz+ + + be two neighboring points on the


curve joining the points ( )1 1 1, ,A x y z and ( )2 2 2, ,B x y z .Thus the infinitesimal

distance between P and Q along the curve is given by

2 2 2 2ds dx dy dz= + + .

Hence the total distance between the points A and B along the curve is given by

( )2

1

12 2 21 ,

x

x

dyI y z dx y

dx= + + =∫ � �� . . . (1)

Let ( )1

2 2 21f y z= + +� � . . . . (2)

We know the functional I is shortest if the function f must satisfy the Euler-

Lagrange’s equations.

0f d f

y dx y

∂ ∂− =

∂ ∂ �, . . . (3)

and 0f d f

z dx z

∂ ∂ − =

∂ ∂ �. . . . (4)

,d y

o y af adx f

⇒ = ⇒ = =

�� constant

and

⇒ ( )2 2 2 2 21y a a z a− − =� � . . . . (5)

Similarly, from equation (4) we obtain

( )2 2 2 2 21z b b y b− − =�� . . . . (6)

Solving equations (5) and (6) for y� and z� we get

2 2

,1

ay

a b= ±

− −� and

2 2,

1

az

a b= ±

− −�

i.e., 1,y c= ±� for ( )1

2 2 21 1c a a b

−

= − − . . . (7)

and 2z c= ±� for ( )1

2 2 22 1c b a b

−

= − − . . . . (8)


Integrating equations (7) and (8) we get

( )1 ,y c x zφ= ± + . . . (9)

( )2 ,z c x yψ= ± + . . . (10)

where ( )zφ and ( )yψ are constants of integration and may be functions of z and y

respectively. Thus the required curve is given by equations (9) and (10). But these

equations represent a pair of planes. The common point of intersection of these

planes is the straight line. Hence the shortest distance between two points in

Euclidean 3-space is a straight line.

Example 19 : Show that the geodesic defined in the 3-dimentional Euclidean space

by the equations ( ) ( ) ( ), ,x x t y y t z z t= = = is a straight line.

Solution : Let

( ) ( ) ( ), ,x x t y y t z z t= = = . . . (1)

be a curve in 3-dimentional Euclidean space, where t is a parameter of the curve. The

infinitesimal distance between two neighboring points on the curve (1) is given by

2 2 2 2ds dx dy dz= + + ,

where from equation (1) we have

, ,dx xdt dy ydt dz zdt= = =� � � .

Thus

( )2 2 2 2 2ds x y z dt= + +� � � .

Hence the total length of the curve between the points ( )0P t and ( )1P t is given by

( )1

0

1

22 2 2

t

t

I x y z dt= + +∫ � � � . . . (2)

The curve is geodesic if the length of the curve I is extremum. This is true if the

integrand


2 2 2f x y z= + +� � � . . . (3)

must satisfy the Euler-Lagrange’s equations.

0 1,2,3i i

f d fi

x dt x

∂ ∂− = ∀ =

∂ ∂ � with ( ), ,ix x y z= , . . . (4)

where

0i

f

x

∂=

∂ and 1,2,3i

i

xfi

x f

∂= ∀ =

∂

�

�.

, ,x af y bf z cf⇒ = = =� � � .

Thus the Euler-Lagrange’s equations become

( )

( )

( )

2 2 2 2 2 2

2 2 2 2 2 2

2 2 2 2 2 2

1 0,

1 0,

1 0.

a x a y a z

b x b y b z

c x c y c z

− + + =

+ − + =

+ + − =

� � �

� � �

� � �

. . . (5)

These equations are consistent provided

2 2 2

2 2 2

2 2 2

1

1 0

1

a a a

b b b

c c c

−

− =

−

2 2 2 1a b c⇒ + + = .

Solving equations (5) we obtain

2 2

,1

ax z

a b=

− −� � . . . (6)

2 2

, 01

by z z

a b= ≠

− −� � � . . . . (7)

Integrating equations (6) and (7) we obtain

( )1 12 2

,1

ax c z y c

a bφ= + =

− −, . . . (8)

( )2 22 2

,1

by c z x c

a bψ= + =

− −. . . . (9)


where ( )xψ and ( )yφ are constants of integration and may be functions of x and y

respectively. Equations (8) and (9) represent planes. The locus of the common points

of these planes is the straight line. Hence the geodesic in 3-dimentional Euclidean

space is the straight line.

Unit 2: Isoperimetric Problems:

The problems in which the function which is eligible for the extremization of

a given definite integral is required to confirm with certain restrictions that are given

as the boundary conditions. Such problems are called isoperimetric problems. The

method is exactly analogous to the method of finding stationary value of a function

under certain conditions by Lagrange’s multipliers method.

Theorem 4 : Obtain the differential equation, which is satisfied by the functional

( ), ,f x y y′ which extremizes the integral

( )( ) ( )2

1

, ,

x

x


subject to the conditions ( ) ( )1 1 2 2,y x y y x y= = , and the integral

( )2

1

, ,

x

x

J g x y y dx′= ∫ = constant.

Proof : Consider the functional between two points ( )1 1,P x y and ( )2 2,Q x y given

by

( )( ) ( )2

1

, ,

x

x

I y x f x y y dx′= ∫ . . . (1)

subject to the conditions

( ) ( )1 1 2 2,y x y y x y= = , . . . (2)


and ( )2

1

, ,

x

x

J g x y y dx′= ∫ = constant. . . . (3)

The points P and Q can be joined by infinitely

many curves.

Accordingly the value of the integral I will be

different for different paths. Let all possible paths

starting from P and ending at Q be given by two

parameters family of curves

( ) ( ) ( ) ( )1 1 2 2Y x y x x xε η ε η= + + . . . (4)

where 1 2,ε ε are parameters and ( ) ( )1 2,x xη η are arbitrary differentiable functions of

x such that

( ) ( )

( ) ( )1 1 1 2

2 1 2 2

0 ,

0

x x

x x

η η

η η

= =

= = . . . (5)

These conditions ensure us that the curves of the family that all pass through the

points P and Q.

Note that, we can not however, express Y(x) as merely a one parameter family of

curves, because any change in the value of the single parameter would in general

alter the value of J, whose constancy must be maintained as prescribed. For this

reason we introduce two parameter families of curves. We shall look for a curve

along which the functional I has an extremum value under the condition (3). Let c

be such a curve between P and Q whose equation is given by y = y (x) such that the

functional (1) along the curve c has extremum value. The values of the integrals (1)

and (3) along the neighboring curve (4) are obtained by replacing y by Y in both the

equations (1) and (3). Thus we have

( ) ( )2

1

1 2, , ,

x

x

I f x Y Y dxε ε ′= ∫ . . . (6)

P(x , y )1 1

Q(x , y )2 2

y = y (x)Y = y (x

) +

(x) +

ε η1

1

ε η2

2(x)

y

xO


and ( ) ( )2

1

1 2, , , .

x

x

J g x Y Y dx constε ε ′= =∫ . . . (7)

Equation (7) shows that 1ε and 2ε is not independent, but they are related by

( )1 2, .J constε ε = . . . (8)

Thus the changes in the value of the parameters are such that the constancy of (7) is

maintained. Thus our new problem is to extremizes (6) under the restriction (7). To

solve the problem we use the method of Lagrange’s multipliers.

Multiply equation (7) by λ and adding it to equation (6) we get

( ) ( )2

1

* *

1 2, , ,

x

x

I I J f x Y Y dxε ε λ ′= + = ∫ , . . . (9)

where λ is Lagrange’s undetermined multiplier and

*f f gλ= + . . . (10)

Thus extremization of (1) subject to the condition (3) is equivalent to the

extremization of (9). Thus from differential calculus, the integral *I is extremum if

1 2

*

0, 0

0j

I

ε εε

= =

∂= ∂

.

Thus from equation (9) we have

1 2

*

0, 0

0j

I

ε εε

= =

∂= ∂

⇒ ( ) ( )2

1

* *

0

x

j j

x

f fx x dx

y yη η

∂ ∂′+ = ′∂ ∂

∫ . J=1, 2 . . . (11)

Note here that by setting 1 2 0ε ε= = , we replace Y and Y ′ to ,y y′ .

Integrating the second integration by parts, we get

( ) ( ) ( )22 2

1 11

* * *

0

xx x

j j j

x xx

f f d fx dx x x dx

y y dx yη η η

∂ ∂ ∂+ − = ′ ′∂ ∂ ∂

∫ ∫ . . . . (12)

As any curve is prescribed at the end points, hence on using conditions (5) we obtain


( )2

1

* *

0, 1, 2

x

j

x

f d fx dx j

y dx yη

∂ ∂− = = ′∂ ∂

∫ .


* *

0f d f

y dx y

∂ ∂− = ′∂ ∂

. . . . (13)

This is required Euler- Lagrange’s differential equation to be satisfied by y (x) for

which the functional I has extremum value under the condition (3).

Remark : If y is not prescribed at either end point then from equation (12) we have

*

0f

y

∂=

′∂ at that end point.

• Generalization of Theorem 4 : Euler-Lagrange’s equations for several

dependent variables :

Theorem 4a : Obtain the differential equations which must be satisfied by the

function which extremize the integral

( )2

1

1 2 1 2, , ,..., , , ,...,

x

n n

x

I f x y y y y y y dx′ ′ ′= ∫

with respect to the twice differentiable functions 1 2, ,...,n

y y y for which

( )2

1

1 2 1 2, , ,..., , , ,..., .

x

n n

x

J g x y y y y y y dx const′ ′ ′= =∫

and with ,i

y y′ prescribed at points 1 2,x x .

Proof : The functional which is to be extremized can be written as

( )2

1

, , , 1, 2,...,

x

i i

x

I f x y y dx i n′= =∫

together with the conditions


( )2

1

, , . 1,2,...,

x

i i

x

J g x y y dx const i n′= = =∫

and ,i

y y′ prescribed at points 1 2,x x .

Repeating the procedure described in the Theorem (4) we arrive the following set of

Euler-Lagrange’s equations

* *

0, 1,2,...,i i

f d fi n

y dx y

∂ ∂− = =

′∂ ∂

where *f f gλ= + .

Theorem 5 : Obtain the differential equation, which is satisfied by four times

differential function y (x) which extremizes the functional

( )( ) ( )2

1

, , ,

x

x

I y x f x y y y dx′ ′′= ∫

subject to the conditions that the integral

( )2

1

, , ,

x

x

J g x y y y dx′ ′′= ∫ = constant.

and both y and y′ are prescribed at the end points.

Proof: Proof of the Theorem 5 runs exactly in the same manner as that of the proof

of Theorem 3 and Theorem 4. The required Euler-Lagrange’s differential equation in

this case is given by

* * 2 *

20

f d f d f

y dx y dx y

∂ ∂ ∂− + = ′ ′′∂ ∂ ∂

,

where *f f gλ= + .


Remarks :

1. If y is not prescribed at either end point, then we have the condition

* *

0f d f

y dx y

∂ ∂− = ′ ′′∂ ∂

at that end point.

2. If y′ is not prescribed at either end point then we have *

0f

y

∂=

′′∂ at that point.

3. In general if

( )

( )

, , , ,...,

, , , ,...,

n

n

f f x y y y y

g g x y y y y

′ ′′=

′ ′′=

with the boundary conditions that 1, , ,..., ny y y y −′ ′′ are prescribed at both the

ends, then in this case the Euler-Lagrange’s equation is

( )* * 2 * *

2... 1 0

nn

n n

f d f d f d f

y dx y dx y dx y

∂ ∂ ∂ ∂− + + + − = ′ ′′∂ ∂ ∂ ∂

.

Worked Examples •

Example 20 : Find the plane curve of fixed perimeter that encloses maximum area.

(The problem is supposed to have arisen from the gift of a king who was

happy with a person and promised to give him all the land that he could enclose by

running round in a day. The perimeter of his path was fixed.)

Solution: Let c : y = y (x) be a plane curve of fixed perimeter l .

2

1

x

x

l ds= ∫ , . . . (1)

where the infinitesimal distance between two points on the curve is given by

21 ,dy

ds y dx ydx

′ ′= + = .


Hence the total length of the curve between two points P and Q becomes

2

1

21

x

x

y dx l′+ =∫ . . . . (2)

The area bounded by the curve c and the x-axis is given by

( )( )2

1

x

x

A y x ydx= ∫ . . . . (3)

Thus we maximize (3) subject to the condition (2). Hence the required Euler-

Lagrange’s equation to be satisfied is

* *

0f d f

y dx y

∂ ∂− = ′∂ ∂

, . . . (4)

where

*

* 2

,

1 .

f f g

f y y

λ

λ

= +

′= + + . . . (5)

Solving equation (4) we get

2

1 0,1

d y

dx y

λ ′ − = ′+

21

yx a

y

λ ′⇒ − =

′+.


( )( )

2 22

21

yx a

y

λ ′− =

′+,

or

( )22

x ay

x aλ

−′ =

− −. . . . (6)

Integrating we get


( )

( )1

22 2

x ay dx b

x aλ

−= +

− −

∫ . . . (7)

Put

sin ,

cos

x a t

dx tdt

λ

λ

− =

⇒ = . . . (8)

sin ,y tdt bλ= +∫

cosy t bλ= − + . . . (9)


( ) ( )2 2 2

x a y b λ⇒ − + − = . . . . (10)

This is a circle centered at (a, b) and of radius λ and is to be determined. To

determine λ , we know that the circumference of the circle is 2 lπλ =2

lλ

π⇒ = .

Example 21 : Find the shape of the plane curve of fixed length l whose end points

lie on the x-axis and area enclosed by it and the x-axis is maximum.

Solution: Let c: y = y(x) be a plane curve of fixed length l whose end points lie on

the x-axis and the curve lies in the upper half

plane. The area bounded by the curve c and

the x-axis is given by

( )( )2

1

x

x

A y x ydx= ∫ . . . (1)

such that the length of the curve is fixed and

is given by

2 2

1 1

21

x x

x x

J ds y dx l′= = + =∫ ∫ . . . (2)

The area given in equation (1) is maximum under the condition (2), if

y

xO (x , 0)1 (x , 0)2

y = y(x)


* *

0f d f

y dx y

∂ ∂− = ′∂ ∂

, . . . (3)

where

*

* 2

,

1

f f g

f y y

λ

λ

= +

′= + + . . . (4)

where λ is Lagrange’s multiplier. Solving equation (3) we get

2

2

1 0,1

1

d y

dx y

yx a

y

λ

λ

′ − = ′+

′⇒ − =

′+


( )( )

2 22

21

yx a

y

λ ′− =

′+,

or

( )22

x ay

x aλ

−′ =

− −. . . . (5)

Integrating we get

( )

( )1

22 2

x ay dx

x aλ

−=

− −

∫ . . . . (6)

Put

sin ,

cos

x a t

dx tdt

λ

λ

− =

⇒ = . . . (7)

sin ,y tdt bλ= +∫

cosy t bλ= − + . . . . (8)


( ) ( )2 2 2

x a y b λ⇒ − + − = . . . . (9)


Thus the curve is a semi circle centered at (a, b) and of radius λ , and is to be

determined.

Since from the condition (2) the perimeter of the semi-circle is lπλ =l

λπ

⇒ = .

This is the radius of the curve of fixed length l which encloses maximum area.

Example 22 : Find the extremals for an isoperimetric problem

( )( ) ( )1

2 2

0

I y x y x dx′= +∫


( ) ( )1

2

0

2, 0 0, 1 0y dx y y= = =∫ .

Solution : The functional, which is to be extremized, is given by

( )( ) ( )1

2 2

0

I y x y x dx′= +∫ . . . (1)

such that

1

2

0

2y dx =∫ . . . (2)

and

( ) ( )0 0, 1 0y y= = . . . (3)

Thus for the extremizes of (1) under (2), we know the condition to be satisfied is

* *

0f d f

y dx y

∂ ∂− = ′∂ ∂

, . . . (4)

where

*

* 2 2 2

,

,

f f g

f y x y

λ

λ

= +

′= + + . . . (5)

where λ is Lagrange’s multiplier.


Solving equation (4) we get

0y yλ′′ − = . . . . (6)

This equation has roots λ± for 0λ > or i λ± for 0λ < .

Case 1 : Let 0λ > .

The solution of equation (6) is

xx

y ae beλλ −= + , . . . (7)

Conditions (3) give

( )

( )

0 0 0,

1 0 0.

y a b

y ae beλ λ−

= ⇒ + =

= ⇒ + =

Solving for a and b we have for

2

2 2

0, 1

1

2 2 0,

.

in

b e

e

in

in

λ

λ π

λ π

λ π

+

≠ =

⇒ =

⇒ + =

⇒ = −

This is contradictory to λ is positive, hence 0b = and consequently 0a = proving

that the equation has only trivial solution.

Case 2 : Let 0λ < .

The solution of equation (6) is

cos siny a x b xλ λ= + . . . . (8)

Boundary conditions (3) give

( )

( )

0 0 0

1 0 sin 0,

, 0.

y a

y b x

n for

λ

λ π λ

= ⇒ =

= ⇒ =

⇒ = ≠

Hence the required solution becomes we get

siny b n xπ= . . . . (9)


Condition (2) gives

2b = ± .

Therefore the required solution is 2siny n xπ= ± .

Example 23 : Find the extremals for an isoperimetric problem

( )( ) ( )2 2

0

I y x y y dx

π

′= −∫


( ) ( )0

1, 0 0, 1ydx y y

π

π= = =∫ .

Solution : It is given that

( )( ) ( )2 2

0

I y x y y dx

π

′= −∫ . . . (1)

where the functional I is to be extremized under the conditions

0

1,ydx

π

=∫ . . . (2)

and ( ) ( )0 0, 1y y π= = . . . (3)

The corresponding Euler-Lagrange’s equation is

* *

0f d f

y dx y

∂ ∂− = ′∂ ∂

, . . . (4)

where

*

* 2 2

,f f g

f y y y

λ

λ

= +

′= − + . . . (5)

Hence the equation (4) becomes

2

y yλ

′′ + = . . . . (6)

The C.F. of equation (6) is given by

cos siny a x b x= + ,


where as the P.I. is given by

2

yλ

= .

Hence the general solution becomes

cos sin2

y a x b xλ

= + + . . . (7)

To determine the arbitrary constants of integration we use the boundary

conditions (3)

( )0 0 ,2

y aλ

= ⇒ = −

( )1

1 1,2

y aπ λ= ⇒ = = −

To determine other constant of integration we use

0

1,ydx

π

=∫

0 0 0

1 1cos sin 1.

2 2xdx b xdx dx

π π π

⇒ − + + =∫ ∫ ∫

This gives the value of b as

2

4b

π− =

.

Hence the required curve is

( ) ( )1 1

1 cos 2 sin2 4

y x xπ= − + − . . . . (8)

Example 24 : Prove that the extremal of the isoperimetric problem

( ) ( )4

2

1

, 1 3, 4 24I y dx y y′= = =∫ ,

subject to the condition


4

1

36ydx =∫

is a parabola.

Solution: Here * 2f y yλ′= + .

The corresponding Euler-Lagrange’s differential equation is

2 0y λ′′ − = . . . . (1)

Integrating two times we get

2

4y x ax b

λ= + + , . . . (2)

where the constants of integration are to be determined. Now the boundary

conditions

( )

( )

1 3 4 4 12,

4 24 4 4 24.

y a b

y a b

λ

λ

= ⇒ + + =

= ⇒ + + = . . . (3)

Also the condition

4

1

36ydx =∫ ,

gives

4

2

1

364

x ax b dxλ

+ + = ∫ ,

21 30 12 144a bλ⇒ + + = . . . . (4)

Solving equations (3) and (4) we obtain

2, 0, 4.a b λ= = =

Thus the required curve is obtain by putting these values in equation (2) and is

2 2y x x= + .

We write this as ( )2

1 1x y+ = +


Or equivalently for X=(x+1), Y=(y+1), we have

2X Y= .

Hence the curve is a parabola.

Example 25 : Find the extremals for the isoperimetric problem

( )2

1

1

2

t

t

I xy yx dt= −∫ � � . . . (1)

with the conditions that

2

1

2 2

t

t

J x y dt l= + =∫ � � , . . . (2)

( ) ( )

( ) ( )1 2 0

1 2 0

,

.

x t x t x

y t y t y

= =

= = . . . (3)

Solution : We wand to find the function for which equation (1) is extremum w. r. t.

the functions x(t), y(t) satisfying the conditions (2) and (3). We know the conditions

that the integral (1) is extremum under (2) if the following Euler –Lagrange’s

equations are satisfied.

* *

0f d f

x dt x

∂ ∂− =

∂ ∂ �, . . . (4)

* *

0f d f

y dt y

∂ ∂− =

∂ ∂ �, . . . (5)

where

* ,f f gλ= +

( )* 2 21

2f xy yx x yλ= − + +� � � � . . . . (6)

Hence the equations (4) and (5) reduce to

2 2

02 2

y d y x

dt x y

λ − − + = +

� �

� �


2 2

02 2

x d x y

dt x y

λ − − + = +

� �

� �

.

Integrating these equations w. r. t. t we get

2 2

2 2

2 2

y y xa

x y

xy a

x y

λ

λ

− − + = +

⇒ − =+

�

� �

�

� �

and 2 2

2 2

,2 2

.

x x yb

x y

yx b

x y

λ

λ

+ + = +

⇒ − =+

�

� �

�

� �

Squaring and adding above equations we get

( ) ( )2 2 2

x b y a λ− + − = . . . . (7)

This is a circle of radius λ and centered at (b, a). Thus the closed curve for which

the enclosed area is maximum is a circle. The length of the circle is

2 lπλ =2

lλ

π⇒ = . This gives the radius of the curve.

Exercise:

1. Show that the shortest distance between two points along the curve

( ) ( ),x x t y y t= = in a Euclidean plane is a straight line.

2. Show that the geodesic defined by ( ) ( ),r r t tθ θ= = in a plane is a straight

line.

3. Show that the stationary (extremum) distance between two points along the

curve ( ) ( ),t tθ θ φ φ= = on the sphere

sin cos , sin sin , cosx r y r z rθ φ θ φ θ= = = is an arc of the great circle.


4. Find the geodesic on the surface obtained by generating the parabola

2 4y ax= about x-axis.

Ans.: The surface of revolution is 2 2 4y z ax+ = , whose parametric

representation is

2 , 2 sin , 2 cosx au y au v z au v= = =

The geodesic on this surface is obtain by solving the integral

2

1 22 2

1

1 uv c du c

u u c

+= +

−∫ .

5. Derive the Euler-Lagrange’s equations that are to be satisfied by twice

differential functions ( ) ( ) ( ), ,..., ,x t y t z t that extremize the integral

( )2

1

, ,..., , , ,..., , ,

t

t

dxI f x y z x y z t dt x

dt= =∫ � � ��

which achieve prescribed values at the fixed points 1 2,t t .

Ans : 0, 0,..., 0,f d f f d f f d f

x dt x y dt y z dt z

∂ ∂ ∂ ∂ ∂ ∂ − = − = − =

∂ ∂ ∂ ∂ ∂ ∂ � � �

6. Find the curve which generates a surface of revolution of minimum area

when it is revolved about x –axis.

Ans : Area of revolution of a curve about x-axis is

1

0

21

x

x

I y y dx′= +∫ .

The curve is a catenary given by sec , coshx b

y c or y aa

ψ−

= =

.

7. Find the function on which the functional can be extremized

( ) ( ) ( ) ( )1

2

0

12 , 0 0, 1 ,

120I y x y xy dx y y′′= − = = ∫

and y′ is not prescribed at both the ends.


Ans: 5 3

120 36 36

x x xy = − + .

8. Find the stationary function of ( )4

2

0

xy y dx′ ′−∫ , which is determined by the

boundary conditions ( ) ( )0 0, 4 3.y y= =

Ans : 2

4 4

x xy = − .

9. Find the extremum of ( ) ( ) ( )2 3

2

1

, 1 1, 2 4.x

I y x dx y yy

= = = ′∫

Ans: 2y x= .

10. Find the extremal of ( )1

0

2

3.

x

x

yI y x dx

x

′= ∫

Ans: 4

1 24

xy c c= + .

11. Find the extremal of ( ) ( ) ( )2 2

0

4 cos , 0 0, 0.y y y x dx y y

π

π′ − + = =∫

Ans: ( )2 siny c x x= + .

12. Find the extremal of the functional

( ) ( ) ( )1

2

0

12 , 0 1, 1 2.y xy dx y y′ − = =∫

Ans: 3 2 1y x x= − + + .

13. Determine the curve ( )z z x= for which the functional

( ) ( ) ( )2

2

1

2 , 1 0, 2 1I z xz dx z z′= − = = −∫ is extremal.

Ans: 3

3 6

x xz = − + .


14. Determine the curve ( )z z x= for which the functional

( ) ( )2

2 2

0

, 0 0, 12

I z z dx z z

π

π ′= − = =

∫

is extremal.

Ans: sinz x= .


( ) ( ) ( ) ( )2

2 2 2

0

, 0 1, 0, 0 0, 1.2 2

I y x y y x dx y y y y

π

π π ′′ ′ ′= − + = = = = −

∫

Ans : cos .y x= .

16. Obtain the differential equation in which the extremizing function makes the

integral

( )( ) ( )2

1

, ,

x

x


extremum subject to the conditions ( ) ( )1 1 2 2, ,y x y y x y= = and

( )2

1

, , . 1,2,...,

x

k k

x

J g x y y dx const k n′= = =∫

Ans : The problem would be extrmization of

( )2

1

* * , ,

x

x

I f x y y dx′= ∫ ,

where *

1

n

k k

k

I I Jλ=

= +∑ and *

1

n

k k

k

f f gλ=

= +∑ .

The required differential equation is * *

0f d f

y dx y

∂ ∂− = ′∂ ∂

.


17. Find the extremals of the isoperimetric problem

1 1

0 0

2 , . . .

x x

x x

I y dx s t ydx c′= =∫ ∫

Ans: 2

4

xy ax bλ= + + .


( ) ( ) ( ) ( ) ( ) ( )1

2

0

1 , 0 0, 1 1, 0 1, 1I y x y dx y y y y′′ ′ ′= + = = = ∫ .

Ans: y x= .


( ) ( ) ( )1

0

2 2 2, 2 2 .

x

x

I y x z x yz y y z dx′ ′= − + − ∫

Ans: ( ) ( )1 2 3 4cos sin , 2y c x c x c x c x z y y′′= + + + = + .


( ) ( )1

0

2 2 2 .

x

x

x

I y x y y ye dx′= + + ∫

Ans: 1 22

xx xxe

y c e c e−= + + .

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