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Examples : Euler-Lagrange Equation (Part - 2) - Classical Mechanics, CSIR-NET Physical Sciences Physics Notes | EduRev

Physics for IIT JAM, UGC - NET, CSIR NET

Created by: Akhilesh Thakur

Physics : Examples : Euler-Lagrange Equation (Part - 2) - Classical Mechanics, CSIR-NET Physical Sciences Physics Notes | EduRev

The document Examples : Euler-Lagrange Equation (Part - 2) - Classical Mechanics, CSIR-NET Physical Sciences Physics Notes | EduRev is a part of the Physics Course Physics for IIT JAM, UGC - NET, CSIR NET.
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Example 8 : Show that the time taken by a particle moving along a curve y = y ( x )
with velocity om 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

Therefore the total time required for the particle to move from the point (0, 0) to the point (1, 1) is given by

. . . (1)
where the infinitesimal distance between two points on the path is given by

Hence the equation (1) becomes

. . . (2)

Time t is minimum if the integrand

. . . (3)
must satisfy the Euler-Lagrange’s equation

. . . (4)

Solving for y′ we get

Integrating we get

Put . . . (5)

Therefore,

. . . (6)

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

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

. . . (1)

The parametric equations of the right circular cylinder are

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

on the surface of the cylinder becomes

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

. . . (2)
For s to be extremum, the integrand

. . . (3)

must satisfy the Euler-Lagrange’s equation

. . . (4)

Integrating the equation and solving for z′ we get

(constant).
Integrating we get

. . . (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

. . . (1)

The parametric equations of the cone are given by

. . . (2)

where for a = sinθ , b = cosθ are constant. Thus the metric

on the surface of the cone becomes

. . . (3)

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

. . . (4)

The length s is stationary if the integrand

. . . (5)

must satisfy the Euler-Lagrange’s equation

. . . (6)

. . . (7)

where  c1 = 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

Example 11 : Find the curve for which the functional

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

Solution:  To find the extremal curve of the functional

. . . (1)
we must solve Euler’s equation

. . . (2)
where

. . . (3)

. . . (4)

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

. . . (5)

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

. . . (6)
The second boundary point moves along the line

where from equation (6) we have y′ = b cos x . Thus  gives

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

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

Then show that f is the total derivative of some function of x and y and conversely.

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

. . . (1)

We claim that

where

As

we write equation (1) explicitly as

. . . (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.

Integrating w. r. t. y′ we get

Integrating once again we get

. . (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

and

Therefore equation (1) becomes

.(4)

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

Therefore,

. . . (5)
This proves the necessary part.

Conversely, assume that  We prove that f satisfies the Euler-Lagrange’s equation

Since

Therefore, we find

Consider now

satisfies Euler-Lagrange’s equation.

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

has the first integral   if the integrand does not depend on x.

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

to be extremum is given by

. . . (1)

If f does not involve x explicitly, then

Therefore, we have

. . . (2)

Multiply equation (2) by y′ we get

. . . (3)
But we know that

. . . (4)

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

. . . (5)
This is the first integral of Euler-Lagrange’s equation, when the functional

Example 14 : Find the minimum of the functional

if the values at the end points are not given.

Solution: For the minimum of the functional

. . . (1)
the integrand

. . . (2)
must satisfy the Euler-Lagrange’s equation

. . . (3)

. . . (4)
Integrating we get

. . . (5)
Further integrating we get

. . . (6)

where c, c2 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.

. . . (7)
These two conditions will determine the values of the constants.

. . . (8)
where from equation (5) and (6) we have

similarly,

Thus the equations (8) become

Solving these equations for cand c2 we obtain

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

. . . (9)

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