Functionals

A functional is a rule that associates a number to a function, where a function is a rule that associates a number to a variable. That is, given a function $f$, the functional $F[f]$ is a mapping between the entirety of the values of $f$ on its domain of definition to a number. A general way to write $F[f]$ is:

$$F[f] = \int g[f(x)] dx,$$ (14.1)

where $g$ is some function that determines $F$, and the integral is extended to the entire domain on which $f$ is defined. We are interested to obtain an expression for the linear order variation of $F[f]$ as $f$ is varied, i.e. writing $F[f+\delta f] - F[f] = \delta F + \delta^2 F + \dots$, with $\delta^i F$ the part that depends to $i^{th}$ order on $\delta f$, we are interested in $\delta F$. The reason is that we are often interested in the function $f$ that minimises $F$, which is the one for which $\delta F = 0$. Any variation $\delta f(x) dx$ can contribute to the total variation $\delta f$. Using  we have:

$$F[f+\delta f] - F[f] = \int \left \{ g[f(x) + \delta f(x)] - g[f(x)] \right\} dx.$$ (14.2)

Let us approximate the integral  with the sum:

$$F[f+\delta f] - F[f] \simeq \sum_i \{g[f(x_i) + \delta f(x_i)] - g[f(x_i)]\} \Delta x_i,$$ (14.3)

which becomes exact in the limit in which $f(x_i)$ and $\delta f(x_i)$ are constant over the interval $\Delta x_i$. Let us now choose $\delta f(x)$ in such a way that it is equal to a small quantity $\epsilon$ over some interval $\Delta x_j$ and zero otherwise. With this choice, all the elements in the sum in  are zero, a part from the term with $i=j$, so that we have:

$$F[f+\delta f] - F[f] \simeq \{g[f(x_j) + \epsilon] - g[f(x_j)]\} \Delta x_j.$$ (14.4)

Since the approximation  is equivalent to approximate $f(x_j)$ to be constant over the interval $\Delta x_j$, the term in curly brackets on the r.h.s. of  is equal to:

$$g[f(x_j) + \epsilon] - g[f(x_j)] = \frac{\delta g[f(x_j)]}{\delta f(x_j)} \epsilon + o(\epsilon^2),$$ (14.5)

where $\frac{\delta g[f(x_j)]}{\delta f(x_j)}$ is the derivative of the function $g$ w.r.t. to its argument– $\delta f(x_j)$ in this case. Therefore, for the choice of $\delta f(x_j) = \epsilon$ for $x_j \in \Delta x_j$ and zero otherwise, we have:

$$\delta F = \frac{\delta g[f(x_j)]}{\delta f(x_j)} \epsilon \Delta x_j = \frac{\delta g[f(x_j)]}{\delta f(x_j)}\delta f(x_j) \Delta x_j,$$ (14.6)

which allows us to define:

$$\frac{\delta F}{\delta f(x)} = \frac{\delta g[f(x)]}{\delta f(x)}.$$ (14.7)

If other terms $\delta f(x_i)$ are contributing to the variation of $f$, they all contribute to the variation of $F$, the linear order of which becomes:

$$\delta F = \sum_i \frac{\delta g[f(x_i)]}{\delta f(x_i)}\delta f(x_i) \Delta x_i,$$ (14.8)

or, coming back to the limit $\Delta x_i \rightarrow 0$,

$$\delta F = \int\frac{\delta g[f(x)]}{\delta f(x)}\delta f(x) dx = \int\frac{\delta F} {\delta f(x)}\delta f(x) dx.$$ (14.9)

Note that $\delta F$ in  depends on the choice of the displacement function $\delta f$. If we are interested in the extreme $\delta F = 0$, since $\delta f(x)$ is arbitrary, we must have $\frac{\delta F} {\delta f(x)} = 0$ for every value of $x$.