Consider $f \in \mathbb{R}^{(N-1)\cdot (M - 1)}$. It's a sequence of some numbers, for which we know : $f_{n,m} = \displaystyle \sum_{j=1}^{M-1}\sum_{i=1}^{N-1}2c_{i,j}\sin \left(\frac{\pi in}{N}\right) \sin \left(\frac{\pi j m}{M}\right)$ (discrete Fourier transform).
Now let $A$ be Laplace transform operator it works so: $\displaystyle \triangle v =(Av)_{n,m} =-\left(\frac{v_{n+1,m} -2v_{n,m}+v_{n-1,m}}{h_x^2} + \frac{v_{n,m+1} -2v_{n,m}+v_{n,m+1}}{h_y^2} \right)$.
We know that eigenvalues are $\pi^2 (n^2+m^2)$ correspond to $\displaystyle \psi_{n,m} = C\sin \left(\frac{\pi x n}{N}\right) \sin \left(\frac{\pi y m}{M}\right)$.
So if we represent $u_{n,m} = \displaystyle \sum_{j=1}^{M-1}\sum_{i=1}^{N-1}2d_{i,j}\sin \left(\frac{\pi in}{N}\right) \sin \left(\frac{\pi j m}{M}\right)$ and consider $Au = f$, we should have that $\displaystyle d_{i,j} = \frac{c_{i,j}}{\pi^2(i^2+j^2)}$.
So it seems that we can solve system of equations $Au = f$ :
1) Find $\{c_{i,j}\}$.
2) Divide by $\lambda_i +\lambda_j$ and summarize them.
But there is a problem with it. I've made a Fourier transform on PC , check that $\bar{f}_{n,m} \approx f_{n,m}$ , so it should be that $\{c_{i,j}\}$ are correct. Next i've considered $\bar{u}_{n,m} = \displaystyle \sum_{j=1}^{M-1}\sum_{i=1}^{N-1} \frac{2}{\pi^2(i^2+j^2)} \sin \left(\frac{\pi in}{N}\right) \sin \left(\frac{\pi j m}{M}\right)$ and get the wrong result. Simply checking $\|f-A\bar{u}\|$ gives me the problem.
I'm interested in does there some problem in mathematics, or not? Maybe there should be other eigenvalues or not?