To me, one of the most beautiful pieces of mathematics is the Euler-Lagrange equations. The idea is this: you are seeking to minimize a functional, which is a function that eats functions and gives numbers. You won’t lose much by assuming that this functional is an integral. In particular, we will assume we are dealing with minimizing
where , and (as usual) Du is the vector of partial derivatives of u. For example, perhaps which would be minimized by choosing u(x) identically equal to zero. Or maybe we want to minimize , subject to the requirement that u(0) = u(1) = 1. Note that our desired “solution” will be a function on, in this case, the unit interval.
One could spend a while talking about classic problems in the calculus of variations (which the study of these problems is called), but I would like to move right to how to solve these problems. What we will do is extract a partial differential equation which will be zero at the minimizer (though it might be zero in other places too). The strategy is as follows: assuming we have a minimizer, “poking” the function in any direction will make L[u] larger. This is similar to the calculus argument that a relative extrema will have derivative zero. But in (single variable) calculus, we may “perturb” a point by moving it to the right or left on the real line. How do we “perturb” a function?
Well, we add a “small” function to it. Namely, we will suppose that u is a minimizer with the necessary boundary data (one typically specifies boundary data in these sorts of problems, to guarantee uniqueness), and let v be a smooth, compactly supported function. Then, for any real number t, we have , and that u+tv has the desired boundary data. So now we have a function of a single real variable, call it f(t), and have deduced that it has a minimum when t = 0.
Specifically, we have
In order to differentiate, the notation gets sort of bad. In practice, just remembering that you should differentiate now is pretty good, but let’s forge on. We notice that F is a map from a big vector space to the reals. Namely, , since x and Du are vectors. I will denote the derivative of F with respect to its first n variables by , and the derivative of F with respect to u by . The p in the first derivative is convention, and notice that is a number, not a vector. Also, I won’t need the derivative with respect to the last n variables, but if I did, I’d write $D_xF$.
Now then, differentiating with respect to t, we get
Now if you’re really sharp with integrating by parts (and remember that v has compact support, and so vanishes on the boundary of ), you would probably know that the left hand summand may be transformed (and f’(0) swapped with 0) to get
which can be rewritten as
This equality must hold for all
smooth functions v
. Hence if the big ugly guy is nonzero anywhere in
, we choose a v
that is supported right there, and break the equality. This sentence can be fleshed out into a full argument that the big ugly guy (which maybe I’ll call the Euler-Lagrange equation
instead) must vanish at any critical point of the functional. Once more for emphasis:
If u is a critical point of the functional L, then u satisfies the Euler-Lagrange equation
$latex D_uF(Du,u,x)-div_x(D_pF(Du,u,x)) = 0.$
I’ve had enough integral signs for one day, but at least I got to post another nice .gif I had lying around!