Let $\Omega=[0,1]\times [0,1]$ be the square. We say a function $f\in H^1(\Omega)$ is periodic on $\Omega$ if $f(x,0)=f(x,1)$ and $f(0,y)=f(1,y)$ (in the sense of traces of course). Now consider the problem $$\begin{cases}-\Delta u+au=f,\\ u,u_x,u_y\text{ are periodic on }\Omega.\end{cases} $$ It is easily seen that the variational formulation of the above equation is $$(\nabla u,\nabla v)_{L^2}+(u,v)_{L^2}=(f,v)_{L^2},\;\;\;\;\text{ for all } v\in H^1_{per}(\Omega):=\{v\in H^1(\Omega):v\text{ is periodic on }\Omega\}. $$

- When $a>0$, we have the ideal situation where the solution operator $T_a:f\mapsto u$ is well defined (Lax-Milgram on the Hilbert space $H^1_{per}$), self-adjoint (since $(\nabla u,\nabla v)_{L^2}+a(u,v)_{L^2}$ is an inner product) and compact on $L^2(\Omega)$ (by Sobolev embeddings). Thus one can find a Hilbert basis for $L^2(\Omega)$ composed from eigenfunctions $\{\phi_n\}$ of $T_a$. Thus we have that $u=\sum_n (\lambda_n+a)^{-1}(f,\phi_n)_{L^2}\phi_n$ where the $\lambda_n$'s are the eigenvalues of $-\Delta$.
- When $a\leq 0$, we have two situations given by the Fredholm alternative applied to the inverse of the operator of $L_\epsilon u=-\Delta u+au+\epsilon u$ (where $\epsilon$ is chosen so that the problem is coercive).
- If $-a$ is not an eigenvalue of $-\Delta$, then a unique solution $u$ exists for the given $f$. Thus we can also define a solution operator $T_a:f\mapsto u$ which is not necessarily continuous from $L^2$ to $H^1$ (a priori). Do the eigenvalues of $T_a$ still form a basis for $L^2(\Omega)$? If not how do we represent a solution in this case?
- When $-a$ is an eigenvalue, either we have no solutions or $f \perp V$ where $V$ is a subspace of dimension $n\in\mathbb{N}$. (I believe it's some eigenspace). How to represent solution in this case?