Examples : Euler-Lagrange Equation - 2 | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

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 c₁ , 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 c₁ and c₂ we obtain

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

 . . . (9)