I'm trying to solve the Dirichlet problem for the Helmholtz equation \begin{aligned}-\triangle u & = & \lambda u, & x\in\Omega,\\ u & = & 0, & x\in\partial\Omega, \end{aligned}
on the Hilbert cube $\Omega=\left(0,\pi\right)^{n}=\underbrace{\left(0,\pi\right)\times\cdots\times\left(0,\pi\right)}_{n\ \textrm{times}}\subset\mathbb{R}^{n}$. Namely, I'm looking to find an expression for $\lambda$.
I can get so far but then I get bogged down by indices and get confused about what to do next. Can anyone help? This is my attempt so far.
Separating the variables, we set $$ u\left(x_{1}x_{2},\ldots,x_{n}\right):=\prod_{i=1}^{n}X_{i}\left(x_{i}\right) $$ so that \begin{eqnarray*} \triangle u\left(x_{1},x_{2},\ldots,x_{n}\right) & = & X_{1}''\left(x_{1}\right)X_{2}\left(x_{2}\right)\ldots X_{n}\left(x_{n}\right)+X_{1}\left(x_{1}\right)X_{2}''\left(x_{2}\right)\ldots X_{n}\left(x_{n}\right)+\cdots+X_{1}\left(x_{1}\right)X_{2}\left(x_{2}\right)\ldots X_{n}''\left(x_{n}\right)\\ & = & -\lambda u\left(x_{1},x_{2},\ldots,x_{n}\right)\\ & = & -\lambda X_{1}\left(x_{1}\right)X_{2}\left(x_{2}\right)\ldots X_{n}\left(x_{n}\right). \end{eqnarray*} We are looking for non-trivial solutions, so we are safe to divide each side by $\prod_{i=1}^{n}X_{i}\left(x_{i}\right)$ to yield $$ \sum_{i=1}^{n}\frac{X_{i}''\left(x_{i}\right)}{X_{i}\left(x_{i}\right)}=-\lambda. $$ Now each of the terms on the LHS is a function of a single variable, independent of each of the other terms; hence $$ \frac{X_{i}''\left(x_{i}\right)}{X_{i}\left(x_{i}\right)}=c_{i} $$ for some constants $c_{i}$. Now $u$ must satisfy the boundary condition, $u=0$ on $\partial\Omega$. Hence, for non-trivial solutions, we have for each $i\in\left[1:n\right]$, $$ X_{i}\left(0\right)=0=X_{i}\left(\pi\right). $$ Hence for each $i\in\left[1:n\right]$, we must solve the system $$ \begin{cases} X_{i}''\left(x_{i}\right)-c_{i}X_{i}\left(x_{i}\right)=0, & x_{i}\in\left(0,\pi\right),\\ X_{i}\left(0\right)=0=X_{i}\left(\pi\right). \end{cases} $$
Now this is just a Sturm-Liouville system, so doing the usual separation of variables again, we find that $c_i > 0$ and
$$ c_{i}=\left(\frac{n_{i}\pi}{\pi}\right)^{2}=n_{i}^{2},\ n_{i}\in\mathbb{N}. $$
I'm not absolutely sure about the last assertion. I feel that it should be something like
$$ c_{i,i_j} = \left(\frac{n_{i,i_j}\pi}{\pi}\right)^{2}=n_{i,i_j}^{2},\ n_{i_j}\in\mathbb{N} $$
because for each fixed $i$, we have a set of solutions $c_i$ indexed by the natural numbers, and "the $n$'s aren't the same for each $c_i$" (I'm finding it difficult to explain what I think I mean).
Can anyone clear the fog, help me out a bit with the notation and show me how to proceed from here?
Edit: This is what I know I should be concluding, but I'm not sure how to get there.
Thus the mutually-orthogonal eigenfunctions can be described by the multiply-indexed family of products $$ u_{k}\left(x_{1},\ldots,x_{n}\right)=\prod_{j=1}^{n}\sin k_{j}x_{j} $$ where $k=\left(k_{1},\ldots,k_{n}\right)\in\mathbb{N}^{n}$, and with corresponding eigenvalues $$ \lambda_{k}=\sum_{j=1}^{n}k_{j}^{2}. $$