# Representing solutions of $-\Delta u+au=f$ when $a\leq 0$

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).
1. 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?
2. 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?

Take the eigenbasis $$\phi_n$$ you exhibited for $$a=0$$. It is still an eigenbasis for $$-\Delta + a$$, with shifted eigenvalues $$\lambda_n + a$$. Write $$E_a=\{n\in\mathbb N : \lambda_n =- a\},$$ which is either an empty set or a finite set. If $$E_a=\emptyset$$ (case 1.) then $$u = \sum_{n\in\mathbb N} \frac{1}{\lambda_n + a} \langle\phi_n,f\rangle \phi_n,$$ and this sum is well defined. Note that only finitely many terms correspond to the case $$\lambda_n+a<0$$, since $$\lambda_n\to\infty$$.
If $$E_a\neq\emptyset$$ (case 2.) note that if $$\langle\phi_n,f\rangle\neq0$$ for some $$n\in E_a$$, there isn't a solution, as an integration by parts against $$\phi_n$$ will show.
If $$\langle\phi_n,f\rangle=0$$ for all $$n\in E_a$$, then any solution writes $$u=\sum_{n\in E_a} \alpha_n \phi_n + \sum_{n\not\in E_a} \frac{1}{\lambda_n + a} \langle\phi_n,f\rangle \phi_n,$$ where the $$\alpha_n$$ are arbitrary. So to obtain uniqueness, you need to set $$\langle\phi_n,u\rangle$$ for all $$n \in E_a$$. When $$a=0$$, there is a unique eigensolution, corresponding to $$\lambda=0$$, which is the constant eigensolution $$\phi_0=1$$. So usually one sets $$\langle\phi_0,u\rangle=\int_{[0,1]^2} u dxdy=0$$ to ensure uniqueness (but you could choose another constant if you so wished).