# A primer on the Euler-Lagrange equations

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

$L[u] = \int_{\Omega} F(Du(x),u(x),x)~dx,$

where $u: \Omega \subset \mathbb{R}^n \to \mathbb{R}$, and (as usual) Du is the vector of partial derivatives of u.  For example, perhaps $L[u] = \int_0^1 |u(x)|~dx,$ which would be minimized by choosing u(x) identically equal to zero.  Or maybe we want to minimize $L[u] = \int_0^1 u'(x)-u(x)~dx$, 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 $L[u] \leq L[u+tv]$, 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

$f(t) = L[u+tv] = \int_{\Omega} F(Du(x)+tDv(x),u(x)+tv(x),x)dx.$

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, $F: \mathbb{R}^n \times \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}$, since x and Du are vectors.  I will denote the derivative of F with respect to its first n variables by $D_pF$, and the derivative of F with respect to u by $D_uF$.  The p in the first derivative is convention, and notice that $D_uF$ 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

$f'(t) = \int_{\Omega}Dv\cdot D_pF(Du+tDv,u+tv,x) + vD_uF(Du+tDv,u+tv,x)~dx,$

so

$f'(0) = \int_{\Omega} Dv \cdot D_pF(Du, u,x) + vD_uF(Du,u ,x)~dx.$

Now if you’re really sharp with integrating by parts (and remember that v has compact support, and so vanishes on the boundary of $\Omega$), you would probably know that the left hand summand may be transformed (and f'(0) swapped with 0) to get

$0=\int_{\Omega}-v(div_x(D_pF(Du,u,x)) + vD_uF(Du,u,x)~dx,$

which can be rewritten as

$0 = \int_{\Omega}(D_uF(Du,u,x)-div_x(D_pF(Du,u,x)))v~dx.$

This equality must hold for all smooth functions v.  Hence if the big ugly guy is nonzero anywhere in $\Omega$, 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!