# Are these critical points minima to the variational problem?

Let $\Omega\equiv (0, 1)\times(0, 1)\subset\mathbb{R}^2$ and consider the variational integral \begin{equation*} I[u]\equiv\int_{\Omega}\frac{1}{2}|Du|^2\ \mathrm{d}x-\frac{5\pi^2}{2}|u|^2\ \mathrm{d}x, \end{equation*} taken over the admissible class \begin{equation*} \mathcal{A}\equiv\{u\in C^2(\Omega)\cap C(\overline{\Omega})\ |\ u=0\text{ on }\partial\Omega\}. \end{equation*}The corresponding Euler-Lagrange equation is $$\tag{1} -\triangle u=5\pi^2u\quad\text{in }\Omega.$$

The Dirichlet eigenvalue problem on $\Omega$ is to solve $$\left\{\begin{array}{l r} -\triangle w=\lambda w &\text{in }\Omega\\ \ \ \ \quad w=0&\text{on }\partial\Omega\end{array}\right.$$

Via the method of separation of variables, it can be shown that \begin{equation*} \mu_{m, n}\equiv\pi^2(m^2+n^2)\quad(m, n\in\mathbb{N}), \end{equation*}and \begin{equation*} \phi_{m, n}(x, y)\equiv\sin(m\pi x)\sin(n\pi y)\quad (x, y)\in \Omega,\ (m, n\in\mathbb{N}) \end{equation*}are the eigenvalues and their corresponding eigenfunctions.

So we see that the only non-zero solutions to (1) are $$u=\sin(\pi x)\sin(2\pi y)\quad\text{and}\quad v=\sin(2\pi x)\sin(\pi y),$$with $I[u]=I[v]=0$.

Clearly $u$ and $v$ are critical points of $I$; are they minima?

I tried to mimic the proof of the Rayleigh quotient optimisation problem to try and show that $I[w]\geq 0$ for all $w\in\mathcal{A}$ but I couldn't make progress.

First, all functions in the span of $u$ and $v$ are critical points.
Second, consider $u_0 = \sin(\pi x)\sin(\pi y)\ge 0$, $u=t\cdot u_0$, i.e., eigenfunctions to the smallest eigenvalue of the Laplacian. Then $$I[u] = \int_\Omega \frac12|Du|^2 - \frac{5\pi^2}2|u|^2 = \frac12\int_\Omega u(-\Delta u - 5\pi^2u) = \frac12t^2 \int_\Omega u_0(2\pi^2 - 5\pi^2)u_0\\= -\frac{3\pi^2}{2} t^2 \int_\Omega u_0^2,$$ and hence $I[u]\to -\infty$ for $t\to+\infty$.
• I agree with your first point...I subconsciously had considered equivalence upto multiplicative constant. In your working I think there should be the factor of half as in $$I[tu_0]=\frac{1}{2}\int_{\Omega}-t^2u_0\triangle u_0-5\pi^2t^2u_0^2\ \mathrm{d}x\mathrm{d}y=\cdots=-\frac{3}{2}\pi^2t^2\int_{\Omega}u_0^2\ \mathrm{d}x\mathrm{d}y.$$ So since there are $u_0\in\mathcal{A}$ with $I[u_0]<0$ it implies that $u$ and $v$ are not minima...correct? – Nirav Nov 9 '15 at 18:37