# (p.236) C Variational calculus

# (p.236) C Variational calculus

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*.

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

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

Let $({y}_{0,0},\dots ,{y}_{N,0})$ 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 ${y}_{i,0}+\mathrm{\delta}{y}_{i}$. Since *f* becomes a minimum at the state $({y}_{0,0},\dots ,{y}_{N,0})$, the following condition must be satisfied for any $\mathrm{\delta}{y}_{i}$
^{1}

The right-hand side can be expanded with respect to $\mathrm{\delta}{y}_{i}$ as

For this to be positive for any values of $(\mathrm{\delta}{y}_{1},\dots ,\mathrm{\delta}{y}_{N-1})$ the following conditions must hold:

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

This determines the equilibrium state.

Let us now consider that the string has a uniform mass density $\rho $. 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

For the case $(dy/dx{)}^{2}\ll 1$, eq. ( C.7 ) is written as

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 ${y}_{i}$ ($i=0,1,\dots ,N$) 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 ${y}_{0}(x)+\mathrm{\delta}y(x)$. The change in *f* is written as

For small $\mathrm{\delta}y(x)$, the right-hand side can be expanded with respect to $\mathrm{\delta}y(x)$, and can be written as

This corresponds to eq. (
C.4
), where the summation over *i* is replaced by the integral over *x*, and the partial derivative $\mathrm{\partial}f/\mathrm{\partial}{y}_{i}$ is replaced by $\mathrm{\delta}f/\mathrm{\delta}y(x)$. The quantity $\mathrm{\delta}f/\mathrm{\delta}y(x)$ 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

(p.238) Calculation of the functional derivative $\mathrm{\delta}f/\mathrm{\delta}y(x)$ 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

Ignoring terms of second order in $\mathrm{\delta}y$, we have

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

Since the position of the string is fixed at $x=0$ and $x=L$, $\mathrm{\delta}y(x)$ 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

Therefore the functional derivative is given by

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

where ${y}^{\prime}(x)=dy/dx$, and $F(y,{y}^{\prime},x)$ is a certain function which has three arguments, *y*, ${y}^{\prime}$, and *x*. The functional derivative of such a functional is calculated as