Functional - Aetherwisp

Given a space of functions $U$ , a functional defined on $U$ is a map $F$ that maps a function $u\in U$ to a Scalar number:

\begin{align} F:\;&U\mapsto \mathbb{R} \\ &u\mapsto F[u]\in \mathbb{R} \end{align}

where we chose real numbers. Functionals can also be dependent on multiple functions:

\begin{align} F:\;&U\times U\times\ldots\times U\mapsto \mathbb{R} \\ &(u_{1},\ldots,u_{n})\mapsto F[u_{1},\ldots,u_{n}]\in \mathbb{R} \end{align}

Properties#

A functional is linear if $F[\alpha u+\beta v]=\alpha F[u]+\beta F[v]$ .

Variation#

The variation $\delta F$ of a functional $F$ in some point $u_{0}$ relative to the variation $\delta u$ is defined as

\delta F[u_{0},\delta u]\equiv \left.{\frac{d}{d\alpha} F[u_{0}+\alpha u]}\right|_{\alpha=0}

It is itself a functional, which takes two functions as an argument. It is fundamentally an extension of the directional derivative for a functional. In fact, for a function $F:U\subset \mathbb{R}^{N}\to \mathbb{R}$ , the directional derivative $\delta F$ on a direction vector $\delta \mathbf{x} \in \mathbb{R}^{N}$ is:

\delta F(\mathbf{x},\delta \mathbf{x})=\left.{\frac{d}{d\alpha}F(\mathbf{x}+\delta \mathbf{x})}\right|_{\alpha=0}=\left.{\sum_{i=1}^{N} \frac{ \partial F }{ \partial x_{i} } (\mathbf{x}+\delta \mathbf{x})\delta \mathbf{x}_{i}}\right|_{\alpha=0}=\sum_{i=1}^{N} \frac{ \partial F }{ \partial x_{i} } (\mathbf{x})\delta \mathbf{x}_{i}=\nabla F\cdot \delta \mathbf{x}

Similarly to usual derivatives, the functional $F$ is said to be stationary in $u_{0}$ if $\delta F[u_{0},\delta u]=0$ . More formally, we say that $u_{0}$ is a stationary point for $F$ for variations $\delta u$ null at the boundaries ( $\delta u(x_\text{start})=\delta u(x_\text{end})=0$ ) if and only if $u_{0}$ satisfies the equation

f(x)=\frac{d}{dx}\frac{ \partial L }{ \partial u' } (u_{0}(x),u_{0}'(x),x)-\frac{ \partial L }{ \partial u } (u_{0}(x),u_{0}'(x),x)=0\tag{1}

This equation is the Lagrange equation.

Proof

$(\Leftarrow)$ : Immediate. $(\Rightarrow)$ : We need to prove that if $\delta F[u_{0},\delta u]=0$ for all $\delta u$ , then $(1)$ is true. We'll claim that $f(x)$ is actually $\neq 0$ for some $\tilde{x}\in[x_\text{start},x_\text{end}]$ . If that's true, by continuity $f(x)\neq 0$ for all $x$ in a neighborhood of $\tilde{x}$ , which we'll call $I_{\tilde{x}}$ . If $I_{\tilde{x}}$ is sufficiently small, $f(x)$ always has the same sign in it, and thus there exists some $\delta u$ such that $\delta F[u_{0},\delta u]\neq 0$ . This is because $\delta u(x)=0$ if $x\not\in I_{\tilde{x}}$ and $\delta u(x)\geq 0$ if $x \in I_{\tilde{x}}$ . But this contradicts the hypothesis that $\delta F[u_{0},\delta u]=0$ for all $\delta u$ . Thus, $f(x)$ cannot be nonzero, concluding the proof.

Furthermore, if we define $\Delta F=F[u+\delta u]-F[u]$ for some small $\delta u$ , $\delta F$ is the linear part of $\delta u$ of the increase $\Delta F$ .

Examples#

Let $U=\{ \text{smooth functions}:[0,1]\to \mathbb{R} \}$ .

> If for instance $u(t)=\sin \pi t$, the functional would give us > $$F[\sin \pi t]=\int_{0}^{1}\sin \pi t\ dt=\frac{1}{\pi}\int_{0}^{\pi}\sin \xi\ d\xi=\frac{2}{\pi}

> which gives us the value $t_{0}$ for which the function is equal to $1$. With our previous $u$: > $$F[\sin \pi t]=\sin(\pi t_{0})=1\quad\to \quad t_{0}= \frac{1}{2}

One could also do the same thing to find the zeros of a derivative:

> This statement is in essence a compact way of expressing the search for maxima and minima of $u$. > [!example]- Curve length > An example of a functional of several functions is the length of a [[Curve]]. Given some curve in $\mathbb{R}^{3}$, $\gamma(t)=(u(t),v(t),w(t))$, its length is the functional > $$F[u,v,w]=\int_{0}^{1}\sqrt{ u'(t)^{2}+v'(t)^{2}+w'(t)^{2} }\ dt

> The increment $\Delta F$ is > $$\Delta F=\int_{0}^{1}(u+\delta u)dt-\int_{0}^{1}udt=\int_{0}^{1}\delta u\ dt

If we now take the integral of the square of the function $F[u]=\int_{0}^{1}u(t)^{2}dt$ we get

> and the increment is > $$\Delta F=\int_{0}^{1}(u^{2}+2u\delta u+\delta u^{2})dx-\int_{0}^{1}u^{2}dx=2\int_{0}^{1} u\delta udx+\int_{0}^{1} \delta u^{2}dx