But what does $\delta F = 0$ mean?

The following discussion is meant to give some more formalism to students of the Variational Principles course and perhaps tie it to material that they might have come across in analysis or linear algebra courses.

This page is a work in progress and may not be accurate; formulae may have mistakes or be improperly formatted. Use with care.

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A functional is a function of a function: $$ F[y] = \int_a^b f(x, y, y'), \mathrm{d}x $$ We are concerned with scalar-valued functionals where $f$ and $F$ take real values.

The idea is that the function space $y$ lives in some function space $U$, e.g. differentiable functions on $(a, b)$. This $U$ is a vector space over $\mathbb{R}$ meaning that it is legal to write down expressions like $y = \lambda y_1 + \mu y_2$ where $\lambda, \mu \in \mathbb{R}$ are scalars. The expression $f(x, y, y')$ is arbitrary, and in general nonlinear.

The functional $F: U \rightarrow \mathbb{R}$ is a function from a function space $U$ to $\mathbb{R}$. Our goal is to understand what it means to differentiate such functions in $U$, "with respect to $y$".

Differentiation in finite-dimensional vector spaces

We shall begin by considering a finite-dimensional vector space $V = \mathbb{R}^n$ and a scalar-valued function $\phi: \mathbb{R}^n \rightarrow \mathbb{R}$, and trying to understand what it means to differentiate $\phi$.

When we consider the derivative of $\phi$ at a point $x$ we are really looking at what happens when the argument is changed by a small amount: the difference betwen $\phi(x + \epsilon n)$ and $\phi(x)$ in the limit $\epsilon\rightarrow0$. Here, $n$ is some vector that specifies the direction in which we are differentating.

If $\phi$ is differentiable at $x$ in the direction of $n$, then we would expect that as $\epsilon\rightarrow0$, $$ \phi(x + \epsilon n) = \phi(x) + g_n \epsilon + o(\epsilon) $$ or equivalently (expanding the definition of the $o(\epsilon)$ notation) $$ \lim_{\epsilon \rightarrow 0} \frac{\phi(x + \epsilon n) - \phi(x)}{\epsilon} = g_n $$ for some quantity $g_n$ that depends on $n$. That is, the variation in $\phi$ should be linearly proportional to $\epsilon$ at leading order. This is reminiscent of the definition of differentiability in one dimension: we are in fact looking at the function $\varphi(\epsilon) = \phi(x + \epsilon n)$ which is a function over the scalar variable $\epsilon$.

For $\phi: \mathbb{R}^n \rightarrow \mathbb{R}$ it turns out we need a stronger condition: we actually need $g_n$ to be a linear function of the direction vector $n$ as well. Thus (by some linear algebra) there needs to be a vector $g$ such that $g_n = g \cdot n$ , and $g$ doesn't depend on $n$ (but may depend on $x$).

(Note that it is not enough for the partial derivatives along the coordinate axes to merely exist for $\phi$ to be differentiable.)

Absorbing $\epsilon$ into the magnitude of $n$, the definition of $\phi: \mathbb{R}^n \rightarrow \mathbb{R}$ being differentiable is that there needs to exist a vector $g$ such that for any direction $n$ we have $$ \phi(x + n) = \phi(x) + g \cdot n + o\left(||n||\right) $$ If such a $g$ exists, then we give it the notation $\nabla\phi$ and call it the gradient of $\phi$ at $x$. Its components are given by the partial derivatives $\nabla \phi_i = \partial \phi / \partial x_i$ along the coordinate axes.

To look at stationary points of $\phi$, we look for places where the vector $\nabla\phi = 0$: that is, the change in $\phi$ along any direction $n$ needs to be sublinear, $o\left(||n||\right)$.

Function spaces

Back to our functional $F[y] = \int_a^b f(x, y, y'), \mathrm{d}x$. Now the finite dimensional $V$ is replaced by an infinite dimensional function space $U$. But this is still a vector space, so notions such as "linear map" and "dot product" still apply, in disguised form.

Inner products

The key idea is that we can define an inner product on $U$, between two functions $r, s \in U$: $$ r \cdot s = \int_a^b r(x) , s(x) , \mathrm{d}x $$ It can be shown that this satisfies the requirements to be an inner product over a vector space. Importantly, it is linear in each of $r$ and $s$.

Aside: Linear functionals

The converse is trickier. In the finite-dimensional case, we used the fact that any linear function of a vector $n$ must take the form $g \cdot n$ for some vector $g$. This is not the case in the space $U$: a scalar-valued linear functional $L[s]$ does not necessarily need to be $r \cdot s$ for some $r \in U$. Simple counterexample, function evaluation $L[s] = s(c)$ at some $c \in (a, b)$.

In the finite dimensional case a vector space $V$ is isomorphic to $V^*$ and so we can identify $g$ as a member of $V$ corresponding to the map $n \mapsto g\cdot n$. The problem is that this is not the case in the function space $U$, where the concept of a "dual space" requires a bit more care to define. See Wikipedia for further details.

It turns out that such difficulties can be overcome by introducing the concept of a distribution – including the ever-popular Dirac delta function, which is not actually a function but a distribution. In the example above, we could take $r(x) = \delta(x-c$) to get our functional for evaluating at $c$.

Norms and limits

The inner product also defines a norm on $U$, by $$ ||r|| = \left( \int_a^b r(x)^2 ,\mathrm{d}x \right)^{1/2} $$ The point of the norm is that it now makes sense to talk about the "size" of $r$, and in particular, what it means to take a limit $r \rightarrow 0$ in the function space $U$. This also lets us write down a term like $o\left( ||r|| \right)$ to denote a quantity that is sublinear in $||r||$.

The variation $\delta F$

Now we want to see what the "derivative" of $F[y]$ is at the "point" $y$, where a "point" $y \in U$ is a function. By analogy with the finite-dimensional case, we first consider the variation of $F$ along a certain "direction" $\eta$ – this direction is also a member of $U$. So we consider the behaviour of $$ \Delta_y [\eta] = F[y + \eta] - F[y] $$ in the limit $||\eta|| \rightarrow 0$ and we hope that this is, to leading order, linear in $\eta$. We consider this to be a functional in $\eta$, although it also depends on the $y$ where we are evaluating it.

By standard derivations (expand and IBP) we get $$ \begin{align} \Delta_y [\eta] &= \int_a^b \left( \frac{\partial f}{\partial y} - \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\partial f}{\partial y'} \right) \right) , \eta , \mathrm{d}x &+ o\left(||\eta||\right) \ &= \int_a^b EL(x, y, y') , \eta(x) ,\mathrm{d}x &+ o\left(||\eta||\right) \ &= EL \cdot \eta &+ o\left(||\eta||\right) \end{align} $$ We observe that this is an inner product: the expression $EL(x, y, y')$ does not depend on $\eta$; and when we evaluate it at the "point" $y = y(x)$, this itself becomes a function of just $x$, written $EL(x)$. Thus $\Delta_y [\eta]$ is an inner product of $EL$ with $\eta$.

We write $$ F[y + \eta] = F[y] + \delta F [\eta] + o\left(||\eta||\right) $$ where $\delta F$ is that first-order approximation, and call it the variation. Note that it does depend on the "direction" of $\eta$. It also depends on $y$ but we usually don't write this explicitly.

Stationary points

Again a stationary point of $F$ is defined to be a "point" $y$ where the change in $F$ is sublinear to the change in the argument.

If we require that $\Delta \eta = o\left(||\eta||\right)$ for all "directions" $\eta$ in the limit $||\eta||\rightarrow0$, this is equivalent (subject to caveats!) to requiring that $EL = 0$, which recovers us the Euler–Lagrange equation: $$ \frac{\partial f}{\partial y} - \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\partial f}{\partial y'} \right) = 0 $$

The second variation

In finite dimensions

Going back to the finite-dimensional case, for our function $\phi$ over a finite dimensional domain to be twice differentiable, we need the correction to $\phi$ to be second-order in $\epsilon$, i.e. $$ \phi(x + \epsilon n) = \phi(x) + (g \cdot n),\epsilon + S_n \epsilon^2 + o(\epsilon^2) $$ for some quantity $S_n$ that, again, in general, depends on the direction $n$. It can be shown that $S_n$ actually needs to be a quadratic form, i.e. $$ \phi(x + \epsilon n) = \phi(x) + (g \cdot n),\epsilon + ( n \cdot H \cdot n ) , \epsilon^2 + o(\epsilon^2) $$ for some symmetric matrix $H$ which we call the Hessian. Its values are given by the second partial derivatives along the coordinate axes, $H_{ij} = \partial^2f / (\partial x_i \partial x_j)$.

One sometimes writes $H = \nabla\nabla \phi$, especially in continuum mechanics. However it is incorrect to write $H = \nabla^2 \phi$ as this notation denotes the Laplacian, which is a scalar value; in fact $\nabla^2 \phi = \mathrm{trace}, H$.

For functionals

As for the second variation $\delta^2 F$, this means expanding to a second order approximation. That is, we suppose that $$ F[y + \eta] = F[y] + \delta F [\eta] + \delta^2 F[\eta] + o\left(||\eta||^2 \right) $$ for some functional $\delta^2 F[\eta]$ (again this implicitly depends on the $y$ where we are evaluating); and we insist that the size of $\delta^2 F[\eta]$ be quadratic in $\eta2$. This is a generalisation of Taylor's theorem.

Expanding $F[y+\eta]$ to second order in $\eta$, we find that the second order correction gives us $$ \delta^2 F[\eta] = \int_a^b (P \eta^2 + Q \eta'^2) , \mathrm{d}x $$ for some functions $P$ and $Q$. We estimate the size of this functional by: $$ ||\delta^2 F[\eta]|| \leq (a-b)\left[||P|| \times||\eta||^2 + ||Q|| \times ||\eta'||^2 \right] $$ This almost gives us what we want, but there is the problem that differentiation is an "unbounded" operation: the size of $||\eta'||$ can be large even in the limit $||\eta|| \rightarrow 0$. (Consider a very wriggly function.) So it is not necessarily the case that $||\eta'||^2$ is proportional to $||\eta||^2$.

We ignore such pathologies, but the way to avoid this issue is to backtrack and work not in the limit of $||\eta||\rightarrow0$, but instead to let $\eta$ be a fixed "direction" and consider $F[y + \epsilon \eta]$ and take the limit $\epsilon\rightarrow0$; in that way one finds that $$ F[y + \epsilon\eta] = F[y] + \delta F [\eta] \epsilon + \delta^2 F[\eta] \epsilon^2 + o\left(\epsilon^2 \right) $$

Local minimum

The second variation can now be used to show that $F$ takes a local minimum at a point $y$, if $\delta F = 0$.

It is not easy and depends on the magnitude and sign of $P$ and $Q$ over the interval $(a, b)$, but if it can be shown that $\delta ^2 F[\eta] \geq 0$ for all $\eta$ then it follows that $$ F[y + \epsilon\eta] = F[y] + \delta^2 F[\eta] \epsilon^2 + o(\epsilon^2) \geq F[y] $$ for sufficiently small values of $\epsilon$. So $F[y]$ is a local minimum. Note that this does not tell us anything about the global behaviour.