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Calculus of Variations

Lagrangian formalism is the main tool of theoretical classical mechanics. Calculus of Variations is apart of Mathematics which Lagrangian formalism is based on. In this section, we discuss the basics

of the Calculus of Variations and, in particular, consider some simple applications. This will provideus with the mathematical language necessary for formulating the Lagrangian Mechanics.The key object of the Calculus of Variations is a functional, a real-valued function defined on some

class of functions. That is the argument of a functional is a function of one or many variables. Thefunctional assigns a real value to a function of the given class. Typically, one deals with functionalsof the form

F[f] =

xbxa

g(f, fx, x) dx , (1)

where f(x) is any function of a given class, fx f(x) is its derivative, and g is a given function of

three variables. Apart from being differentiable, the functions of the given class can be subject tosome constraint. Say,

f(xa) = ya , f(xb) = yb . (2)

The definition of functional is straightforwardly generalized to the cases of more than one function f,more than one variable x, and higher-order (partial) derivatives.

Example 1. Length of a line in a plane. Suppose we have a smooth line in the xy-plane passing throughthe two given points (xa, ya) and (xb, yb). The line is supposed to be specified by the equation

y = f(x) . (3)

The length of the line is a functional of f. Let us make sure that it can be represented in the form(1). We have

l[f] =

dl =

(dx)2 + (dy)2 =

(dx)2 + (fxdx)2 =xbxa

1 + (fx)2 dx . (4)

We see that l[f] has the form (1) with

g(f, fx, x) =

1 + (fx)2 . (5)

Note that in this example the function g does not depend explicitly on f and x.

Example 2. Potential energy of a flexible cable suspended from two fixed points. The cable is supposedto be at rest. Choose the y axis perpendicular to the surface of the Earth, the cable being representedby a curve (9) the xy plane, with the constraint (2). Then its potential energy is given by (we do not

care about the units, and thus neglect the dimensional proportionality coefficient)

U[f] =

y dl =

xbxa

f

1 + (fx)2 dx . (6)

(The potential energy of the element dl is proportional to its hight y and its length dl.) We see thatl[f] has the form (1) with

g(f, fx, x) = f

1 + (fx)2 . (7)

Here the function g explicitly depends on f and fx, but not on x.

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Example 3. Length of a line in a surface. Suppose we have a smooth line L in the surface S, passingthrough the two given points (xa, ya, za) and (xb, yb, zb). Let the surface S and the line L be specifiedby the equations

z = s(x, y) , (8)

y = f(x) , z = s(x, f(x)) , (9)

respectively. Let us show that the length of the line is a functional of the form (1). First we introduceconvenient notation

sx =s(x, y)

x, sy =

s(x, y)

y, (10)

and note that

dy = fx dx , dz = sx dx + sy dy = sx dx + sy fx dx . (11)

Then,

l[f] =

dl =

(dx)2 + (dy)2 + (dz)2 =

xbxa

g(f, fx, x) dx , (12)

where

g(f, fx, x) =

1 + f2

x + (sx(x, f) + sy(x, f) fx)2

. (13)A typical problem which arises in connection with a functional F[f] is the problem of finding such

a function f that minimizes the functional. In the context of above examples, this problem comesfrom natural questions like: What is the shortest distance between two points in a given surface?What is the shape of the suspended cable (corresponding to the minimum of the potential energy at agiven length)? Calculus of Variations provides mathematical tools for solving the problem. Supposethe function f is a (local) minimum/maximum of the functional F. Then, for any small variationof the function f variation of the functional is supposed to be sign-definite. Let us see what are theimplications of this fact. We introduce the symbol f(x) for an infinitesimal variation of the functionf and the symbol F for corresponding variation of the functional F. That is

F = F[f + f] F[f] . (14)

With a simple observation that

f f + f fx fx + (f) (15)

we write

F =

xbxa

g( f + f, fx + (f), x) dx

xbxa

g(f, fx, x) dx . (16)

Then we take into account the smallness of f:

g( f + f, fx + (f), x) g(f, fx, x) +

g

ff +

g

fx(f) . (17)

F =

xb

xa

gf

f dx +

xb

xa

gfx

(f) dx . (18)

Doing by parts the second integral, with Eq. (2) taken into account [meaning f(xa) = f(xb) = 0 ],we have xb

xa

g

fx(f) dx =

xbxa

dx (f)d

dx

g

fx. (19)

That is

F =

xbxa

A(x) f dx , (20)

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where

A(x) =g

f

d

dx

g

fx. (21)

We see that the variation F is a linear functional of f. The linearity means that the sign of Fchanges if f f. On the other hand, the condition of minimum/maximum requires that thesign of variation do not change. This is only possible if A(x) 0, and we arrive at the celebrated

Eulers equation, the fundamental equation of the Calculus of Variations:g

f

d

dx

g

fx= 0 . (22)

As an easy example, let us make sure that the shortest distance between two points is a straight line.Applying Eq. (22) to the function g of the Example 1, see Eq. (5), we readily get

f(x) = 0 f(x) = C1x + C2 , (23)

where C1 and C2 are some constants. And this is the equation of a straight line. The constants arefixed by the boundary conditions (2).

Alternate form of Eulers equation

If fx = 0, then Euler equation is equivalent to

g

x

d

dx

g fx

g

fx

= 0 . (24)

Indeed, by the chain rule we have

dg

dx=

g

ffx +

g

fxfxx +

g

x. (25)

Taking into account that

ddx

fx gfx

= fxx gfx

+ fx ddx

gfx

, (26)

we then getd

dx

g fx

g

fx

=

g

x+ fx

g

f

d

dx

g

fx

. (27)

This means that Eq. (24) is equivalent to

fx

g

f

d

dx

g

fx

= 0 , (28)

which, in its turn, is equivalent to the original Eulers equation (22) if fx = 0. To put it different, anysolution of Eq. (22) is simultaneously a solution of Eq. (24), but, in contrast to Eq. (22), equation

(24) has also a trivial solution f = const.Now if g = g(f, fx), that is g does not depend explicitly on x, Eq. (24) simplifies to

d

dx

g fx

g

fx

= 0 , (29)

and immediately yields the first integral

g fxg

fx= const . (30)

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Soap film. The equilibrium shape of a soap film corresponds to a (local) minimum of its surfaceenergy. The latter is just proportional to the surface area. Hence, to find the equilibrium shape ofa soap film one has to find the (local) minimum of the surface area under the appropriate boundaryconditions. Consider simplest example when the film is axially symmetric being stretched between

two parallel coaxial wire circles. Let the center of the first circle be at x = 0 and the center of thesecond circle be at x = h; the two radii being Ra and Rb, respectively; x being the symmetry axis.The surface is parameterized by the radius r at a given x:

r = f(x) , f(0) = Ra , f(h) = Rb . (31)

The surface area, A, is then a functional of f:

A[f] =

2rdl =

2r

(dx)2 + (dr)2 = 2

h0

f

1 + f2x dx . (32)

We see thatg = 2 f1 + f2x (33)

does not depend explicitly on x and we can use the Euler equation in the form of its first integral(30). This yields

f

1 + f2x f f2x1 + f2x

= C0 , (34)

where C0 is a constant. A simple algebra then leads to

f = C0

1 + f2x f

2 = C20 (1 + f2x) fx =

(f /C0)2 1 . (35)

We get a simple differential equation and easily solve it:

df

dx =

(f /C0)2 1 . (36)df

(f /C0)2 1=

dx . (37)

C0 cosh1(f /C0) = x + const . (38)

f(x) = C0 cosh

x x0

C0

. (39)

Note that the sign is absorbed by the constant x0. The solution (39) has two free constants whichare fixed by two boundary conditions:

cosh(x0/C0) = Ra/C0 ,cosh((h x0)/C0) = Rb/C0 .

(40)

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Functional of many functions

The generalization of the theory to the case of more than one function, f f(1), f(2), . . . , f (m) , isquite straightforward. Now we have

F[f(1), f(2), . . . , f (m)] = xb

xa

g(f(1), f(2), . . . , f (m), f(1)x , f(2)x , . . . , f

(m)x , x) dx , (41)

f(j)(xa) = y(j)a , f

(j)(xb) = y(j)b (j = 1, 2, . . . , m) . (42)

Introducing the variationsf(j) f(j) + f(j) , (43)

and evaluating F, we get

F =m

j=1

xbxa

Aj(x) f(j) dx , (44)

Aj(x) =g

f(j)

d

dx

g

f(j)x

. (45)

Now if we are looking for minimum/maximum of the functional F, then by definition we have F 0for a minimum, or F 0 for a maximum. But this is only possible if

Aj(x) 0 (j = 1, 2, . . . , m) . (46)

Indeed, if for some j = j0 we had Aj0(x) = 0, then we could set f(j) 0 for all js but j0 and write

F =

xbxa

Aj0(x) f(j0) dx . (47)

And then we just repeat the reasoning we used to prove A(x) 0 in the case of just one function.Hence, for a set of functions {f(1), f(2), . . . , f (m)} to correspond to a minimum/maximum of the

functional (41) the following Eulers equations have to be satisfied:

g

f(j)

d

dx

g

f(j)x

= 0 (j = 1, 2, . . . , m) . (48)

This system of m differential equations defines the functions {f(1), f(2), . . . , f (m)} up to 2m freeconstants. The constants are then fixed by the boundary conditions (42).

Important Theorem

What is the generalization, if any, of the alternative Eulers equation (24) for the case of manyfunctions? In the case of m functions, we cannot replace the whole system of m equations (48)with alternative ones, equivalent to (24). The generalization of Eq. (24) comes in the form of atheorem that may seem too abstract, but actually is extremely important being directly relevant tothe conservation of energy in Lagrangian mechanics. Consider the quantity (we use the same letterwe use for energy, and do it on purpose)

E(f(1)(x), . . . , f (m)(x), f(1)x (x), . . . , f (m)x (x), x) =

mj=1

f(j)xg

f(j)x

g . (49)

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The theorem says that if the functions {f(1), f(2), . . . , f (m)} minimize/maximise the functional (41),

then for the quantity E(f(1)(x), . . . , f (m)(x), f(1)x (x), . . . , f

(m)x (x), x) the following relation takes

placedE

dx=

g

x. (50)

In particular, if the function g does not explicitly depend on x, then the right-hand side is zero and

the quantity E is just a constant.The proof is a straightforward generalization of Eqs. (25)-(27):

d

dx

mj=1

f(j)xg

f(j)x

=m

j=1

f(j)xxg

f(j)x

+m

j=1

f(j)xd

dx

g

f(j)x

. (51)

dg

dx=

mj=1

g

f(j)f(j)x +

mj=1

g

f(j)x

f(j)xx +g

x. (52)

Hence,dE

dx=

g

x+

m

j=1

f(j)x

d

dx

g

f

(j)

x

g

f(j)

. (53)

So far we did just identical transformations. Now we take into account that the functions{f(1), f(2), . . . , f (m)} minimize/maximise the functional (41) and are supposed thus to satisfy Eulersequations (48). Eqs. (48) state that each term in the brackets in the r.h.s. of (53) is identically equalto zero, which completes the proof.

Lagrange Multipliers

Let us recall the problem of the flexible cable (Example 2). This problem involves an extra feature,a constraint that the total length of the curve be fixed. To solve problems like that we have to un-derstand how to treat the constraints. Actually, the idea is the same as in the calculus of functions:

one can use Lagrange multipliers.Suppose we want to find an extremum of the functional (1) on the set of functions subject to the

boundary conditions (2) and also the constraint that

Q[f] = C , (54)

where C is a given constant and

Q[f] =

xbxa

q(f, fx, x) dx (55)

is a given functional. (In our Example 2, Q[f] is the length of the curve.)Consider a new functional

F[f] = F[f] Q[f] , (56)

where is a fixed real number. Under the constraint (54), an extremum of the functional F issimultaneously an extremum of the functional F, and vice versa, since the two differ by the fixedconstant C. Now consider a genuine (i.e. unconstrained) extremum of the functional (56). At anarbitrary , this genuine extremum is not supposed to have anything to do with the extremum underthe constraint. However, normally their exists a special choice of = at which the function f

(),the genuine extremum of the functional (56), just satisfies the condition

Q[f()] = C . (57)

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Clearly, the function f() is the solution to the original problem, since the genuine extremum auto-matically implies the extremum under given (and satisfied) constraint. Once we know the value of, the problem is solved. And to find we simply solve the problem of the genuine extremum ofthe functional (56) with an arbitrary (undefined) , and then a posteriori fix the proper value = by requiring that the condition (57) be satisfied.

Solution to the problem of the flexible cable. We have (see Example 2)

g(f, fx, x) = f

1 + (fx)2 , (58)

q(f, fx, x) =

1 + (fx)2 . (59)

Hence,

F[f] =

xbxa

[g(f, fx, x) q(f, fx, x)] dx =

xbxa

(f )

1 + (fx)2 dx . (60)

With this particular form of the functional F it is very convenient to introduce the new function

f = f , (61)

because then we will have

F[f] =

xbxa

f

1 + (fx)2 dx , (62)

which is (up to a numeric factor) is the functional which we have minimized already in the contextof the problem of the soap film. We just write the answer, Eq. (39):

f(x) = C0 cosh

x x0

C0

f(x) = C0 cosh

x x0

C0

+ . (63)

The values of the three constants, C0, x0, , should be chosen to satisfy two boundary conditions,Eq. (2), and the constraint that the total length of the curve is equal to a given value l.

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