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# Calculus of Variations

Module by: Narayan C.R.. E-mail the author

In physics, apart from problems which involve determining the maximal and minimal values of a certain function y=f(x)y=f(x), we often come across problems where one has to find the maximal and minimal values of special quantities called Functionals.

A Functional is a kind of function, where the indepenent variable is itself a function i.e., its a "function of function". A Functional could also involve several functions.

The regular function specified by f(x,y,,,)f(x,y,,,) is a mathematical relation which maps(corresponds) one or more numbers(variable) into one number, which is the value of the function. But here a Functional is a mathematical relation which maps one or more functions into a number.

For example, the arc length L of a plane or space curve connecting two given points A(x1,y1)A(x1,y1) and B(x2,y2)B(x2,y2) is a functional . The arc length L can be computed if the equation of the curve y=f(x)y=f(x) is given

L = x 1 x 2 1 + d y d x 2 d x = x 1 x 2 1 + y ' 2 d x L = x 1 x 2 1 + d y d x 2 d x = x 1 x 2 1 + y ' 2 d x
(1)

The co-ordinates of centre of gravity of a homogeneous curve is also a functional.

x = L x d L L d L = x 1 + d y d x 2 d x 1 + d y d x 2 d x x = L x d L L d L = x 1 + d y d x 2 d x 1 + d y d x 2 d x
(2)

Similarly y and z can be expressed and are functionals.

Similarly moment of inertia, the co-ordinates of the centre of gravity of a homogeneous surface are all functionals since their values are determined by the choice of a curve or a surface.

Calculus of variations deals with the methods of finding Extremal values(maximal and minimal) of Functionals.

We will consider three simple examples which are typical of the problems encountered and solved in the calculus of variations.

1. To find the equation of the shortest plane curve joining two given points, which is also called the problem of geodesic on a plane. A Geodesic is a curve along a surface which marks the shortest distance between two neighbouring points.

Given two points (x1,y1)(x1,y1) , (x2,y2)(x2,y2) we are supposed to find the curve joining these points whose arc length is a minimum.

To find the the arc length of any curve we have

d s = d x 2 + d y 2 = 1 + d y d x 2 d x d s = d x 2 + d y 2 = 1 + d y d x 2 d x
(3)

So, the arc length between the two given points, denoted here by II, is

I = x 1 x 2 d s = x 1 x 2 1 + d y d x 2 d x I = x 1 x 2 d s = x 1 x 2 1 + d y d x 2 d x
(4)

which is equal to

I = x 1 x 2 1 + ( y ' ) 2 d x I = x 1 x 2 1 + ( y ' ) 2 d x
(5)

Our problem is to find the function y=f(x)y=f(x) which will make the above integral as small as possible. The function which does this is called an extremal.

2. The Curve of Quickest Descent:

Here we are given two points (x1,y1)(x1,y1) and (x2,y2)(x2,y2) laying in a vertical plane. The problen is to find the curve joining the two points, down which a bead will slide from rest in the least time.

We shall use rectangular co-ordinates . Let one of the points be the origin (see fig 1). Now the problem is to minimize dtdt Let y=f(x)y=f(x) be a curve joining (0,0)(0,0) and (x2,y2)(x2,y2)

The velocity of the bead at any point on the curve is found by using the law of conservation of energy, to be equal to 2gx2gx

Therefore we have

d s d t = 2 g y d s d t = 2 g y
(6)

so

d t = d s 2 g y d t = d s 2 g y
(7)

but we have

d s = 1 + y ' 2 d x d s = 1 + y ' 2 d x
(8)

so it follows that

d t = 1 + y ' 2 2 g y d x d t = 1 + y ' 2 2 g y d x
(9)

Therefore =

t = 0 x 2 1 + y ' 2 2 g y d x t = 0 x 2 1 + y ' 2 2 g y d x
(10)

Here again we have to find the function y=f(x)y=f(x) , so that the integral shall be minimum.

3. The Minimum Surface of Revolution

This is another representative problem whose solution requires the techniques of the calculus of variation.

Here in this problem, given two points and a curve which are in the same plane. By revolving the curve y(x)y(x) whose extremities are the given two points, about the x-axis, we get a surface of revolution. Now the problem is to find the curve y(x)y(x) so that the area of the resulting surface will be a minimum.

Considering rectangular axis , for the element of area shown in the fig. 3

d A = 2 π d s = 2 π y 1 + y ' 2 d x d A = 2 π d s = 2 π y 1 + y ' 2 d x
(11)

So area

A = x 1 x 2 2 π y 1 + y ' 2 d x A = x 1 x 2 2 π y 1 + y ' 2 d x
(12)

Here the variational problem is to find the curve y(x)y(x) for which the integral is minimum.

All the above considered problems involve functionals which can be written in the form

I = a b F ( x , y , y ' ) d x I = a b F ( x , y , y ' ) d x
(13)

Here F will be a given function , as are the constants aa and bb. The corresponding values of y can also be found. Now the problem is to find the function y=f(x)y=f(x) for which the Functional(Integral) attain a stationary(extremum) value, i.e., a maximum or a minimum. But it should be noted here that in most cases of physical interest the extremum would be a minimum.

In ordinary problems of finding extremum values of a given function y=f(x)y=f(x) at x=ax=a we find f'(x)f'(x) and equate it to zero. The equation is solved for real values of xx. They may correspond to maximum points, minimum points or points of inflection. We next find f''(x)f''(x) put the values of x in it one by one. If f''(x)f''(x) is negative when x=ax=a, then f(x)f(x) is maximum at x=ax=a and corresponding maximum value of f(x)f(x) is f(a)f(a). If f''(x)f''(x) is positive when x=ax=a then f(x)f(x) is minimum at x=ax=a and the corresponding minimum value of f(x)f(x) is f(a)f(a). Here f'(x)=0f'(x)=0 is a necessary condition and not a sufficient condition.

In the calculus of variations, we however have to find a curve for which a functional(Integral)is greater or less than for any neighbouring curve having the same end points. We actually do something similar to equating f'(x)f'(x) to 0, as we did in ordinary calculus. As for as the question of whether we have found a maximum or a minimum or neither is generally difficult mathematical problem. But these difficulties can be overcome by relying on the physics of the problem. As a matter of fact in most of the problems of physical interest "stationary" is all that is required.

We now consider the general problem as stated above. The problem is to find yy for which the integral

I = x 1 x 2 F ( y , y ' , x ) , d s I = x 1 x 2 F ( y , y ' , x ) , d s
(14)

will have a stationary value, where FF is a given function. y(x)y(x) is the extremal. We will now algebraically represent all the possible curves which have the given end points, but differing from the extremal , which is yet to determined, by small amounts. There are infinitely many curves which we can make as close to the extremal as possible. Here we assume the existence of an extremal which is continuous and twice differentiable. Also it can be shown that the conditions imposed are not too restrictive and that the result is not changed when the conditions are slackened.

We construct a function representing the family of curves having the same extremities in the following way. Let η(x)η(x) represent the deformation of the path and a scale factor ϵϵ, represent the magnitude of the variation. Here η(x)η(x) is a function of xx and is zero at x1x1 and x2x2 . But for these restrictions it is completely arbitrary function. Now we define

Y ( x ) = y ( x ) + ϵ η ( x ) Y ( x ) = y ( x ) + ϵ η ( x )
(15)

Here Y(x)Y(x) represents any curve which joins the given extremities(fig.2). Now, out of all these curves we want to pick the one which minimizes the (functional) integral. We form an integral by replacing yy and y'y' in the above integral by YY and Y'Y', which is clearly a function of ϵϵ.

I ( ϵ ) = x 1 x 2 F ( Y , Y ' , x ) d x I ( ϵ ) = x 1 x 2 F ( Y , Y ' , x ) d x
(16)

Y'Y' is given by

Y ' ( x ) = y ' ( x ) + ϵ η ' ( x ) Y ' ( x ) = y ' ( x ) + ϵ η ' ( x )
(17)

It can be easily seen that by making ϵϵ equal to zero we can replace YY and Y'Y' by yy and y'y'. Its clearly seen that the integral is a minimum with respect to ϵϵ when ϵϵ = 0, since we are assuming y(x)y(x) as the extremal. This holds irrespective of the choice of η(x)η(x).

Now the problem is reduced to finding the minimum of a function, as we do in ordinary calculus, with respect to the variable ϵϵ. That is we now will find dI(ϵ)dϵdI(ϵ)dϵ and equate it to 0, when ϵ=0ϵ=0. Since YY and Y'Y' are functions of ϵϵ , differentiating under the integral sign with respect to ϵϵ, we get

I ' ( ϵ ) = d I d ϵ = x 1 x 2 F Y d Y d ϵ + F Y ' d Y ' d ϵ d x I ' ( ϵ ) = d I d ϵ = x 1 x 2 F Y d Y d ϵ + F Y ' d Y ' d ϵ d x
(18)

substituing the value of Y(x)Y(x) and Y'(x)Y'(x) in to the above equation, we get

d I d ϵ = x 1 x 2 F Y η ( x ) + F Y ' η ' ( x ) d x d I d ϵ = x 1 x 2 F Y η ( x ) + F Y ' η ' ( x ) d x
(19)

Since we need to find dIdϵdIdϵ at ϵ=0ϵ=0 we have to replace YY and Y'Y' with yy and y'y'. That is

d I d ϵ ϵ = 0 = x 1 x 2 F y η ( x ) + F y ' η ' ( x ) d x = 0 d I d ϵ ϵ = 0 = x 1 x 2 F y η ( x ) + F y ' η ' ( x ) d x = 0
(20)

Since we have assumed y''y'' to be continuous , we can go ahead with the integration of the second term. Integrating it by parts, we get

x 1 x 2 F y ' η ' ( x ) d x = F y ' η ( x ) x 1 x 2 - x 1 x 2 d d x F y ' η ( x ) d x . x 1 x 2 F y ' η ' ( x ) d x = F y ' η ( x ) x 1 x 2 - x 1 x 2 d d x F y ' η ( x ) d x .
(21)

Since η(x)η(x) is zero at both x1x1 and x2x2 , the integrated term in the above expression is zero. Now we can write

d I d ϵ ϵ = 0 = x 1 x 2 F y - d d x F y ' η ( x ) d x = 0 d I d ϵ ϵ = 0 = x 1 x 2 F y - d d x F y ' η ( x ) d x = 0
(22)

Since η(x)η(x) is arbitrary we can assume the bracketed term in the integrand to be zero. That is

F y - d d x F y ' = 0 . F y - d d x F y ' = 0 .
(23)

This is called Euler-LagrangeEuler-Lagrange equation. This is a necessary condition for II to have an extremum value.

Alternate Forms of Euler-Lagrange equation

1.Expanding the total derivative with respect to xx ,

d d x F y ' = x F y ' + y F y ' d y d x + y ' F y ' d y ' d x d d x F y ' = x F y ' + y F y ' d y d x + y ' F y ' d y ' d x
(24)
= 2 F x y ' + 2 F y y ' d y d x + 2 F y ' 2 d y ' d x . = 2 F x y ' + 2 F y y ' d y d x + 2 F y ' 2 d y ' d x .
(25)

the Euler-LagrangeEuler-Lagrange equation becomes

F y - 2 F x y ' - 2 F y y ' y ' - 2 F y ' 2 y ' ' = 0 F y - 2 F x y ' - 2 F y y ' y ' - 2 F y ' 2 y ' ' = 0
(26)

a second order differential equation.

2.Integrand does not depend on y explicitly

In this case

F y ' = 0 F y ' = 0
(27)

Therefore the Euler-LagrangeEuler-Lagrange equation is reduced to

d d x F y ' = 0 d d x F y ' = 0
(28)

Implies

F y ' = c . F y ' = c .
(29)

where cc a constant. This is a first order differential equation which does not depend on yy. Solving for y'y', we obtain an equation of the form

y ' = f ( x , c ) y ' = f ( x , c )
(30)

from which y(x)y(x) can be found.

3. Integrand does not depend on xx explicitly

Here

F x = 0 F x = 0
(31)

FF is a function of yy and y'y', and since y=y(x)y=y(x), F is implicitly depend on xx.

d F d x = F x + F y d y d x + F y ' d y ' d x d F d x = F x + F y d y d x + F y ' d y ' d x
(32)
d F d x = F y y ' + F y ' y ' ' d F d x = F y y ' + F y ' y ' '
(33)

Consider

d d x y ' F y ' = y ' ' F y ' + y ' d d x F y ' d d x y ' F y ' = y ' ' F y ' + y ' d d x F y '
(34)

Subtracting equation (4)(4) from (3)(3) , we have

d F d x - d d x y ' F y ' = F y y ' - y ' d d x F y ' d F d x - d d x y ' F y ' = F y y ' - y ' d d x F y '
(35)

Or

d d x F - y ' F y ' = [ F y - d d x F y ' ] y ' d d x F - y ' F y ' = [ F y - d d x F y ' ] y '
(36)
d d x F - y ' F y ' = 0 d d x F - y ' F y ' = 0
(37)

The above equation implies that

F - y ' F y ' = c F - y ' F y ' = c
(38)

This is a first order differential equation which can be solved for y(x)y(x).

Example . To find the shortest between two points in a plane or finding geodesicsgeodesics in a plane.

As stated earlier the arc length of any curve between the two given points (x1,y1)(x1,y1) and (x2,y2)(x2,y2) is

I = x 1 x 2 1 + ( y ' ) 2 d x I = x 1 x 2 1 + ( y ' ) 2 d x
(39)

Here the above equation is of the form of equation(14)equation(14) with

F = 1 + ( y ' ) 2 F = 1 + ( y ' ) 2
(40)

The condition for the curve to be the shortest path is that the integralIintegralI should be minimum. Thus FF must satisfy Euler-LagrangeEuler-Lagrange equation. Substituting it into equation(23)equation(23) with

F y = 0 , F y = 0 ,
(41)

and

F y ' = y ' 1 + ( y ' ) 2 F y ' = y ' 1 + ( y ' ) 2
(42)

we have

d d x y ' 1 + ( y ' ) 2 = 0 , d d x y ' 1 + ( y ' ) 2 = 0 ,
(43)

The above equation implies

y ' 1 + ( y ' ) 2 = c . y ' 1 + ( y ' ) 2 = c .
(44)

Here cc is integration constant. Solving for y'y' we get,

y ' = c 1 + c 2 y ' = c 1 + c 2
(45)

This is nothing but

y ' = m y ' = m
(46)

where m is another constant. Now again integrating this equation we get

y = m x + c y = m x + c
(47)

Where cc is another constant of integration. This clearly is the equation of a straight line. The constants mm and cc are determined by the two end points (x1,y1)(x1,y1) and (x2,y2)(x2,y2).

Example. The curve of quickest descent: As stated earlier we need to find shape of the path of the bead , so that it takes least amount of time to reach B (fig1) . We have from the earlier discussion (page2 ) , the time tt which the beat takes to travel from point A to point B is given by

t = 0 x 2 1 + y ' 2 2 g y d x = 1 2 g 0 x 2 1 + y ' 2 y d x t = 0 x 2 1 + y ' 2 2 g y d x = 1 2 g 0 x 2 1 + y ' 2 y d x
(48)

The curve can found by minimizing the above integral. Here the integrand does not explicitly depend on x, so the Euler-LagrangeEuler-Lagrange equation can be written in the modified form of equation (38)(38)

F - y ' F y ' = c F - y ' F y ' = c
(49)

with

F = 1 + y ' 2 y F = 1 + y ' 2 y
(50)

Thus

F = 1 + y ' 2 y - y ' 2 y 1 + y ' 2 = c F = 1 + y ' 2 y - y ' 2 y 1 + y ' 2 = c
(51)

Squaring both sides and solving for y'y' we have

y ' = 1 - c 2 y c 2 y y ' = 1 - c 2 y c 2 y
(52)

Since

y ' = d y d x y ' = d y d x
(53)

this equation can be rewritten as

c 2 y 1 - c 2 y d y = d x c 2 y 1 - c 2 y d y = d x
(54)

To find y(x)y(x) we have to integrate the above equation. We will carry out the integration by a change of variable . Let

c 2 y = 1 2 ( 1 - c o s θ ) c 2 y = 1 2 ( 1 - c o s θ )
(55)

so

1 - c 2 y = 1 2 ( 1 + c o s θ ) 1 - c 2 y = 1 2 ( 1 + c o s θ )
(56)

and so

d y = 1 2 c 2 s i n θ d θ d y = 1 2 c 2 s i n θ d θ
(57)

We have

2 s i n 2 θ 2 = 1 - c o s θ 2 s i n 2 θ 2 = 1 - c o s θ
(58)
2 c o s 2 θ 2 = 1 + c o s θ 2 c o s 2 θ 2 = 1 + c o s θ
(59)
s i n θ = s i n θ 2 + θ 2 = 2 s i n θ 2 c o s θ 2 s i n θ = s i n θ 2 + θ 2 = 2 s i n θ 2 c o s θ 2
(60)

So

c 2 y 1 - c 2 y d y = 1 c 2 sin ( θ / 2 ) ) cos ( θ / 2 ) sin θ 2 cos θ 2 d θ = 1 c 2 sin 2 θ 2 d θ c 2 y 1 - c 2 y d y = 1 c 2 sin ( θ / 2 ) ) cos ( θ / 2 ) sin θ 2 cos θ 2 d θ = 1 c 2 sin 2 θ 2 d θ
(61)
= 1 2 c 2 ( 1 - cos θ ) d θ = 1 2 c 2 ( 1 - cos θ ) d θ
(62)

So that

c 2 y 1 - c 2 y d y = d x c 2 y 1 - c 2 y d y = d x
(63)

becomes

1 2 c 2 ( 1 - c o s θ ) d θ = d x 1 2 c 2 ( 1 - c o s θ ) d θ = d x
(64)

which gives

1 2 c 2 ( θ - s i n θ ) = x + k 1 2 c 2 ( θ - s i n θ ) = x + k
(65)

Since the curve passes through (0,0) we have when x=0x=0, y=0y=0. By equation(55)equation(55), we have that when y=0y=0θθ must be equal to zero. So we have x=θ=0x=θ=0. Putting these valus in the above equation we get k=0k=0. The remaining constants can determined by the fact that the curve the other end point. So we get

x = 1 2 c 2 ( θ - sin θ ) x = 1 2 c 2 ( θ - sin θ )
(66)
y = 1 2 c 2 ( 1 - cos θ ) y = 1 2 c 2 ( 1 - cos θ )
(67)

These are the parametric the equations of the curve along which the particle moves in least possible time. The curve is called cycloidcycloid. As the shape of the curve suggests, the bead may travel uphill along the cycloidcycloid, for some distance,but nonetheless it travels faster than it would have done along a straight line or any other path. The problem is traditionally called The BrachistochroneBrachistochrone problem.

Example. The minimum surface of revolution.

As stated earlier the total surface area is given by

x 1 x 2 2 π y 1 + y ' 2 d x x 1 x 2 2 π y 1 + y ' 2 d x
(68)

Here the minimizing function is

f = y 1 + y ' 2 f = y 1 + y ' 2
(69)

since fx=0fx=0 we can directly apply equation(23)equation(23) By which we obtain

y 1 + y ' 2 - y ' y y ' 1 + y ' 2 = a y 1 + y ' 2 - y ' y y ' 1 + y ' 2 = a
(70)
y ( 1 + y ' 2 ) - y y ' 2 = a 1 + y ' 2 y ( 1 + y ' 2 ) - y y ' 2 = a 1 + y ' 2
(71)
y = a 1 + y ' 2 y = a 1 + y ' 2
(72)
y 1 + y ' 2 = a y 1 + y ' 2 = a
(73)
y 2 a 2 - 1 = y ' 2 y 2 a 2 - 1 = y ' 2
(74)
d x d y = 1 y ' = a y 2 - a 2 d x d y = 1 y ' = a y 2 - a 2
(75)
x = a d y y 2 - a 2 = a cosh - 1 y a + b x = a d y y 2 - a 2 = a cosh - 1 y a + b
(76)

and manipulating the above equation ,

y = a c o s h x - b a y = a c o s h x - b a
(77)

Again, here as in the previous problem, the constants aa and bb are determined by making the curve to pass through the given points. The curve represented by the above parametric equations is called a catenarycatenary. It is a curve of a flexible cord hanging freely between two points of support. The minimun surface area generated by its rotation is called a catenoidcatenoid.

Application to Mechanics

Variational Calculus leads to a new and elegant way to formulate mechanics. By using it, mechanics can be reduced to the investigation of a scalar integral. The condition for the stationary value of this scalar integral gives all the equations of motion. As a simple example consider a particle of mass mm moving under gravity. Considering recatangular co-ordinates, the equations of motion are

m y ' ' = - m g , m x ' ' = 0 m y ' ' = - m g , m x ' ' = 0
(78)

where the differentiation is carried out with respect to time. Denoting the kinetic energy 12m(x'2+y'2)12m(x'2+y'2) of the particle by TT and the potential energy, mgymgy by VV we construct an equation

L = T - V = 1 2 m ( x ' 2 + y ' 2 ) - m g y L = T - V = 1 2 m ( x ' 2 + y ' 2 ) - m g y
(79)

LL is called the LagrangianLagrangian of the system. Then

L y - d d t L y ' = - m g - m y ' ' = 0 L y - d d t L y ' = - m g - m y ' ' = 0
(80)

by using equation (78). Here we know that x'x' is constant, and taking tt as independent variable and yy as dependent variable, LL now resembles the standard functional equation(14)equation(14) . Therefore it follows that equation(78)equation(78) is the Euler-LagrangeEuler-Lagrange equation for the integral(functional)

I = a b L d t I = a b L d t
(81)

Thus it follows that for the motion of a particle under gravity, II is stationary. We can conclude, therefore, that the time integral of the Lagrangian is smallest if the motion obeys Newton's Laws. This in fact is a special case of Hamilton's principle which states that if L=T-VL=T-V, where TT is kinetic and VV is the potential energy of a system then

t 0 t 1 L d t t 0 t 1 L d t
(82)

for any fixed time interval, is stationary for a physical path.

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