Unique weak solution to Helmholtz equation on a square

I've recently started studying the modern theory of PDEs. I studied some basic properties of Sobolev space and then started with linear elliptic PDEs.

I consider the following problem:

For which $\lambda\in\mathbb{R}$ there is a unique weak solution in $W_0^{1,2}((0,1)^2)$ to \begin{align*} -\Delta u=\lambda u\quad &\mathrm{in}\ (0,1)^2,\\ u = 0\quad &\mathrm{on}\ \partial (0,1)^2 \end{align*} ?

The only tool I know for proving the existence and uniqueness is Lax-Milgram theorem.

My thoughts:

If $\lambda\leq 0$ then I can prove the uniqueness easily using Lax-Milgram. So I was wondering, what if $\lambda > 0$. I know Poincaré inequality and the following inequality, holding under some assumptations for every $u\in W^{1,2}(\Omega)$ and $\alpha > 0$, that there exists $C > 0$ such that: $$C||u||_2^2\leq||\nabla u ||_2^2 + \alpha\int\limits_{\partial\Omega}{|u|^2\ \mathrm{dS}}.$$ So, from this inequality it follows, using Lax-Milgram, that there is a unique weak solution for every $\lambda < C$, doesn't it?

However, the problem is that I can prove the inequality just by contradiction, so I have no clue how big the constant $C$ is.

Is it possible to find a constant $C > 0$ (as big as possible) explicitly for this particular choice of $\Omega$? I tried googling it but I didn't find anything useful (I probably just don't know what I should look for)...

• Well, it seems that I'm looking for smallest possible constant in Poincaré inequality. Using Hölder inequality, I can show that there is a unique weak solution for all $\lambda < 8$. Dec 5 '15 at 12:15
As stated, existence is trivial: $u\equiv 0$ is a solution. When a nontrivial solution exists, it is not unique since one can multiply it by a constant. But this happens only for some specific $\lambda$: those that are in the spectrum of the Dirichlet Laplacian.
You can't infer this kind of structure from Lax-Milgram. Rather, sine Fourier series should be used: expand $u(x,y)=\sum_{m,n}\sin \pi m x\sin \pi n y$ and note that $$-\Delta u = \pi^2 \sum_{m,n} (m^2+n^2)\sin \pi m x\sin \pi n y$$ This gives eigenvalues $\lambda_{m,n} = \pi^2(m^2+n^2)$, the lowest of them being $2\pi^2\approx 19.7$. This set is the entire spectrum of Dirichlet Laplacian on this domain: for $\lambda \notin \{\pi^2(m^2+n^2) : m,n\in\mathbb{Z}\}$, the zero solution is the only one.