Solve the PDE $u_{xx}+u_{yy}+\lambda u=0, 0I am stuck with the following problem:
The PDE
 $u_{xx}+u_{yy}+\lambda u=0, 0<x,y<1$
$u(x,0)=u(x,1)=0; 0\leq x \leq 1$
$u(0,y)=u(1,y)=0; 0\leq y \leq 1$
has  

(a)a unique solution u for any $\lambda \in \mathbb R ,$
  (b)infinitely many solutions  for some  $\lambda \in \mathbb R ,$
  (c)a  solution  for countably many values of $\lambda \in \mathbb R ,$
  (d)infinitely many solutions  for all $\lambda \in \mathbb R .$ 

I do not know how to progress with it.Could someone point me in the right direction( e.g. a certain theorem or property i have to use?) Thanks in advance for your time.
 A: Consider the basis 
$$
B=\{\phi_{00},\ \theta_{0n},\ \xi_{n0},\ \phi_{mn},\ \psi_{mn},\ \theta_{mn},\xi_{mn}\}_{(m,n) \in \mathbb{N}^2}
$$ of $L^2([0,1]^2)$ given by
\begin{eqnarray}
\phi_{00}(x,y)&=&1,\ 
\theta_{0n}(x,y)=\sin(2\pi ny),\ \xi_{n0}(x)=\sin(2\pi nx),\\
\phi_{mn}(x,y)&=&\cos(2\pi mx)\cos(2\pi ny),\ \psi_{mn}(x,y)=\sin(2\pi mx)\sin(2\pi ny),\\
\theta_{mn}(x,y)&=&\cos(2\pi mx)\sin(2\pi ny),\ \xi_{mn}(x,y)=\sin(2\pi mx)\cos(2\pi ny).
\end{eqnarray}
Then for every $m,n \ge 0$ we have
$$
-\Delta\phi_{mn}=\lambda_{mn}\phi_{mn},\
-\Delta\psi_{mn}=\lambda_{mn}\psi_{mn},\
-\Delta\theta_{mn}=\lambda_{mn}\theta_{mn},\
-\Delta\xi_{mn}=\lambda_{mn}\xi_{mn}\ \forall\ m,n \ge 0,
$$
where
$$
\lambda_{mn}=4\pi^2(m^2+n^2) \quad \forall\ m, n\ge 0.
$$
This shows that the number $\lambda_{mn}$ is an eigenvalue of $-\Delta$ with corresponding eigenfunctions $\phi_{mn},\psi_{mn},\theta_{mn},\xi_{mn}$.
Because of the boundary conditions
\begin{eqnarray}
u(x,0)=u(x,1)&=&0 \quad \forall x \in [0,1],\\
u(0,y)=u(1,y)&=&0 \quad \forall y \in [0,1],
\end{eqnarray}
it is clear that a pair $(u,\lambda)$ satisfies the given PDE precisely when it belongs to $\{(c\psi_{mn},\lambda_{mn}): \ m,n \ge 1, \ c \in \mathbb{R}\setminus\{0\}\}$.
A: Hint: See the Dirichlet eigenvalues.
