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Soft Matter Physics$

Masao Doi

Print publication date: 2013

Print ISBN-13: 9780199652952

Published to Oxford Scholarship Online: December 2013

DOI: 10.1093/acprof:oso/9780199652952.001.0001

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(p.236) C Variational calculus

(p.236) C Variational calculus

Source:
Soft Matter Physics
Publisher:
Oxford University Press

The variational derivative of a functional is an extension of the partial derivative of a function of many variables. We explain this by taking a special problem as an example.

Consider a system consisting of ( N + 1 ) mass points (of mass m) each connected by a string of length ℓ. The string is under tension T, and the two end points are fixed at the same height (see Fig. C.1(a) ). We consider the mechanical equilibrium of the system under gravity g.


                     C Variational calculus

Fig. C.1 (a) A string connecting N + 1 mass points is hung with a tensile force T. (b) A string with mass density ρ is hung with a tensile force T.

The mechanical equilibrium is the state at which the potential energy of the system is a minimum. Let y i ( i = 0 , 1 , , N ) be the vertical displacement of the particle i from the line connecting the two end points. The potential energy is written as

(C.1)
f ( y 0 , y 1 , , y N ) = T i = 0 N 1 ( y i + 1 y i ) 2 + 2 i = 1 N 1 m g y i

For simplicity, we consider the case that the displacement | y i + 1 y i | is much less than ℓ. Equation ( C.1 ) is then written as

(C.2)
f = T 2 i = 0 N 1 ( y i + 1 y i ) 2 i = 1 N 1 m g y i

Let ( y 0 , 0 , , y N , 0 ) be the state of minimum energy. To find the state, we consider the variation of f when y i is changed slightly from y i , 0 to y i , 0 + δ y i . Since f becomes a minimum at the state ( y 0 , 0 , , y N , 0 ) , the following condition must be satisfied for any δ y i 1

(C.3)
δ f = f ( y 0 , 0 + δ y 0 , , y N , 0 + δ y N ) f ( y 0 , 0 , y 10 , , y N , 0 ) 0

The right-hand side can be expanded with respect to δ y i as

(C.4)
δ f = i = 0 N f y i δ y i

For this to be positive for any values of ( δ y 1 , , δ y N 1 ) the following conditions must hold:

(C.5)
f y i = 0 i = 1 , , N 1

(p.237) In the case that f is given by eq. ( C.2 ), eq. ( C.5 ) gives

(C.6)
T 2 ( y i + 1 2 y i + y i 1 ) m g = 0 , i = 1 , , N 1

This determines the equilibrium state.

Let us now consider that the string has a uniform mass density ρ . The state of the string is represented by a function y ( x ) as in Fig. C.1(b) . The potential energy of the system is now written as

(C.7)
f [ y ( x ) ] = T 0 L d x 1 + d y d x 2 1 0 L d x ρ g y ( x )

For the case ( d y / d x ) 2 1 , eq. ( C.7 ) is written as

(C.8)
f [ y ( x ) ] = T 2 0 L d x d y d x 2 0 L d x ρ g y ( x )

In eqs. ( C.7 ) and ( C.8 ), f [ y ( x ) ] represents a scalar, the value of which is determined by the function y ( x ) . Therefore f [ y ( x ) ] is regarded as a function of the function y ( x ) , and is called a functional.

The variational calculus is a method of finding the minimum (or maximum) of functionals. The calculation is essentially the same as that for usual functions. The only difference is that the set of variables y i ( i = 0 , 1 , , N ) is replaced by the function y ( x ) , i.e., the discrete index i is replaced by a continuous variable x.

Let y 0 ( x ) be the state which gives the minimum of f [ y ( x ) ] . Consider the change of f when y ( x ) is slightly changed from y 0 ( x ) to y 0 ( x ) + δ y ( x ) . The change in f is written as

(C.9)
δ f = f [ y 0 ( x ) + δ y ( x ) ] f [ y 0 ( x ) ]

For small δ y ( x ) , the right-hand side can be expanded with respect to δ y ( x ) , and can be written as

(C.10)
δ f = 0 L d x δ f δ y ( x ) δ y ( x )

This corresponds to eq. ( C.4 ), where the summation over i is replaced by the integral over x, and the partial derivative f / y i is replaced by δ f / δ y ( x ) . The quantity δ f / δ y ( x ) is called the functional derivative.

By the same argument as above, the condition that the state y 0 ( x ) is the state of minimum energy is written as

(C.11)
δ f δ y ( x ) = 0 , for y ( x ) = y 0 ( x )

(p.238) Calculation of the functional derivative δ f / δ y ( x ) is straightforward. Let us consider the functional ( C.8 ). For notational simplicity we write y 0 ( x ) as y ( x ) . The variation of f [ y ( x ) ] is written as

(C.12)
δ f [ y ( x ) ] = T 2 0 L d x d y d x + d δ y d x 2 d y d x 2 0 L d x ρ g δ y

Ignoring terms of second order in δ y , we have

(C.13)
δ f [ y ( x ) ] = T 0 L d x d y d x d δ y d x 0 L d x ρ g δ y

The first term on the right-hand side can be written by using integration by parts, as

(C.14)
δ f [ y ( x ) ] = T d y d x δ y ( x ) 0 L T 0 L d x d 2 y d x 2 δ y 0 L d x ρ g δ y

Since the position of the string is fixed at x = 0 and x = L , δ y ( x ) becomes zero at x = 0 and at x = L . Therefore the first term on the right-hand side of eq. ( C.14 ) becomes zero. Hence

(C.15)
δ f [ y ( x ) ] = 0 L d x T d 2 y d x 2 ρ g δ y ( x )

Therefore the functional derivative is given by

(C.16)
δ f δ y ( x ) = T d 2 y d x 2 ρ g

Let us consider the case that the functional f [ y ( x ) ] is written in the following form

(C.17)
f [ y ( x ) ] = a b d x F ( y ( x ) , y ( x ) , x )

where y ( x ) = d y / d x , and F ( y , y , x ) is a certain function which has three arguments, y, y , and x. The functional derivative of such a functional is calculated as

(C.18)
δ f δ y ( x ) = F y d d x F y