Why is it interesting?

Variational calculus is the alternative to the usual calculus methods when we want to find functions that minimize something. As an analogy, usually when we search for the extrema of a function, we differentiate the function, set the derivative of the function to zero and find the point that yields the extrema. Similar results can be computed by using variational calculus.In variational calculus we find extrema of functionals which are functions of functions with respect some function (instead of variable). This is extremely important for the Lagrangian formalism.

Layman

Another way of saying a thing is least is to say that if you move the path a little bit at first it does not make any difference. Suppose you were walking around on hills – but smooth hills, since the mathematical things involved correspond to smooth things – and you come to a place where you are lowest, then I say that if you take a small step forward you will not change your height. When you are at the lowest or at the highest point, a step does not make any difference in the altitude in first approximation, whereas if you are on a slope you can walk down the slope with a step and then if you take the step in the opposite direction you walk up. That is the key to the reason why, when you are at the lowest place, taking a step does not make much difference, because if it did make any difference then if you took a step in the opposite direction you would go down. Since this is the lowest point and you cannot go down, your first approximation is that the step does not make any difference. We therefore know that if we move a path a little bit it does not make any difference to the action on a first approximation."The Character of Physical Law" by R. Feynman

Student

Variational calculus is a powerful mathetmatical tool to find the extremums (maxima, minima and saddle points) of functions and even functionals ( we can call them functions of functions).

=== What we know from Calculus ===

On calculus, when we want to find the extremum of a function, we use the derivative:

We differentiate the function $f(x)$, then demand that the resulting derivative vanishes: $$\frac{d f(x)}{dx} \stackrel{!}{=} 0 .$$

While this mathematical tool is great for functions it does not help us if we want to calculate thee extrema of different objects, like functionals. A functional is a function of a function. This means, a functional $S[f(x)]$ has as an argument a function $f(x)$ and spits out a number for each function that we put into it. This is to be contrasted with what a function is: A function $f(x)$ eats a number $x$ and spits our a number.

(Functionals are especially important for the Lagrangian framework.)

We will see in a moment that the variational calculus enables us to calculate the extrema of functions and functionals.

To "invent" this new theory that is capable of finding the minima of functionals, we need to take a step back and think about what characterizes a mathematical minimum.

The answer of variational calculus is that a minimum is characterised by the neighbourhood of the minimum. For example, let's find the minimum $x_{\mathrm{min }}$ of an ordinary function $f(x)=3x^2+x$. We start by looking at one specific $x=a$ and take a close look at its neighborhood. Mathematically this means $a+ \epsilon$, where $\epsilon$ denotes an infinitesimal (positive or negative) variation. We put this variation of $a$ into our function $f(x)$:

$$f(a+\epsilon)=3(a+\epsilon)^2+(a+\epsilon)=3(a^2+2a \epsilon + \epsilon^2)+a+\epsilon.$$

If $a$ is a minimum, first order variations in $\epsilon$ must vanish, because otherwise we can choose $\epsilon$ to be negative $\epsilon<0$ and then $f(a+\epsilon)$ is smaller than $f(a)$. Therefore, we collect all terms linear in $\epsilon$ and demand this to be zero

$$3\cdot 2a \epsilon+ \epsilon \stackrel{!}{=}0 \rightarrow 6a+1 \stackrel{!}{=} 0.$$

So we find the minimum

$x_{\mathrm{min }}= a=\frac{-1}{6},$$

which is of course exactly the same result we get if we take the derivative $f(x)=3x^2+x \rightarrow f'(x)= 6x+1$ and demand this to be zero. In terms of ordinary functions this is just another way of doing the same thing, but varitional calculus is in addition able to find the extrema of functionals.

== Mathematics ==

Fundamental Lemma of Variational Calculus Assume $f\in\mathcal{C}[a,b]$ and that for all $h\in\mathcal{C}[a,b]$ wich is zero at the endpoints it holds that $$\int f(x) h(x) d x =0$$ . Then $f(x)=0$ for all $x \in [a,b]$ Proof

This is analogous to the vector algebra proposition, let $v,w\in V$, where $V$ is a vector space. If $v\cdot w = 0$ for all $w\in V$ then $v=0$. In fact, if you dig deeper, it is the same result: the space of continuous functions form $a$ to $b$ is a vector space, and we can define the integral of the multiplication as the inner product!

Now let's say that $h$ is in fact a derivative of another function, or to simplify notation, we have $\dot h$ instead of $h$ inside the integral?

Corollary for the Fundamental Lemma TODO

Definition [Functional] Let $\Omega(\mathcal{Q})$ be the set of functions $q:\mathbb{R} \to \mathcal{Q}$, then a functional F is a map $$F:\Omega \to \mathbb{R}; F[q] \mapsto \alpha \in\mathbb{R}$$

So we can see how a functional is a function of functions as we said before.

References

Gelfand and Fomin, Calculus of Variations

Researcher

Common Question 1

Common Question 2

Examples

Example1

Example2:

History