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I seek the minimum of a certain functional which is always strictly greater than zero. The Euler-Lagrange equation is a "zeroth-order differential equation", that is, an implicit equation that a stationary function must satisfy. Assuming that there is only one such function, I seek to prove that it is in fact the minimizer of the functional.

Since the first variation doesn't depend on the derivative of the functional's argument, variations of all orders are zero at the stationary function, so I can't use the second variation test. I've looked a bit into direct methods, but I'm not a mathematician and I quickly got lost in the discussion of topologies of function spaces. The literature is so large and I am so unfamiliar with it that I'm having trouble sorting out exactly what assumptions are sufficient for the existence of a minimizer.

Here's my functional, $\mathcal{A}[f]$:

Let $w:\mathbb{R}\rightarrow\mathbb{R}^+$ be a weight function, let $g(x, y) \in C^2(\mathbb{R}^2)$, and let $u_{f} (x) = g(x, f(x))$. Then

$$\mathcal{A}[f] = \int^{x_1}_{x_0} f(x)[w(u_f(x))\frac{\mathrm{d}u}{\mathrm{d}x}-w(x)]\mathrm{d}x.$$

The Euler-Lagrange equation is

$$\frac{w(x)}{w(u_{f}(x))} = \left. \frac{\partial g}{\partial x} \right|_{y=f(x)}.$$

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@LiviuNicolaescu, since you wrote this answer --… -- I'm hoping you can help me with my question. – Cyan Oct 14 '12 at 17:05

I haven't verified your calculation of the E-L equation. But if the Jacobi-field equation (second variation) is degenerate then you have to show that $A[f+g] \leq A[f]$ for either all $g$ or, endowing your space of functions with a suitable norm for $g \leq \varepsilon$ for some $\varepsilon > 0$.

My feeling is to, assume your weight function is sufficiently smooth ($C^2$ or more), use Taylors theorem to expand out your nonlinear $w$ and $g$, bound out the highest order terms (the remainders) and look at the remaining problem.

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