Poisson equation - Eigenvalue expansion method; only trivial solutions I have a problem with the following question (it can be found in the book Partial Differential Equations by Asmar as well if any of you have that; exercise 3.9.6).
So I need to solve this equation using the eigenfunction expansion method:
$\nabla^2u=u_{xx}+u$
So I started using the method described in the book, which gives me the following:
Try as a solution
$u(x,y)=\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{m=1}^{\infty}E_{mn}\sin{m\pi x}\sin{n\pi y}$
Filling this into the equation gives:
$\displaystyle\sum_{n,m}E_{mn}\nabla^2(\sin{m\pi x}\sin{n\pi y})=-\displaystyle\sum_{n,m}E_{mn}(m\pi)^2\sin{m\pi x}\sin{n\pi y}+\displaystyle\sum_{n,m}E_{mn}\sin{m\pi x}\sin{n\pi y}$
$\displaystyle\sum_{n,m}E_{mn}(-\lambda_{mn}+(m\pi)^2-1)\sin{m\pi x}\sin{n\pi y}=0$
Where $\lambda_{mn}$ are the eigenvalues: $\lambda_{mn}=(m\pi)^2+(n\pi)^2$
Unlike the example they provide in the book (I'll write it out at the bottom), because there is no term in this problem without a u, this equals zero. Because of this, if I use the double sine expansion Taylor series, I get:
$E_{mn}(-\lambda_{mn}+(m\pi)^2-1)=0$
Which gives me only trivial solutions for $E_{mn}$. So you could you help me out how to find a nontrivial solution for this problem? 
Many thanks!!  

Example from the book with nontrivial solutions:
Here they had the equation $\nabla^2u=u+3$.
Filling the same equation for $u$ then gives:
$\displaystyle\sum_{n,m}-E_{mn}\lambda_{mn}\sin{m\pi x}\sin{n\pi y}=\displaystyle\sum_{n,m}E_{mn}\sin{m\pi x}\sin{n\pi y}+3$
$\displaystyle\sum_{n,m}E_{mn}(-1-\lambda_{mn})\sin{m\pi x}\sin{n\pi y}=3$
Using the double sine fourier expansion now gives
$E_{mn}(-1-\lambda_{mn})=4\int_0^1\int_0^1f(x)\sin{m\pi x}\sin{n\pi y}\,\mathrm{d}x \mathrm{d}y$.
Where in this case $f(x)=3$, whereas in the case described above we unfortunately have $f(x)=0$, which is why we get trivial solutions.
 A: I think the clue lies in the boundary conditions, which you didn't specify. If you try to expand the solution in eigenfunctions of the Laplacian $\nabla^2$, and you write $u(x,y) = \sum_{m,n} E_{mn} \sin(m\pi x) \sin(n\pi y)$, you use the eigenfunctions of the Laplacian which are associated to homogeneous Dirichlet boundary conditions. In other words, for an eigenfunction $\phi_{mn}(x,y)$ which solves $-\nabla^2 \phi = \lambda_{mn} \phi$, you impose the conditions that $\phi(x,0) = 0$, $\phi(x,1) = 0$, $\phi(0,y) = 0$ and $\phi(1,y) = 0$.
Now, taking the comment of @Baron Mingus into consideration, your equation indeed simplifies to $u_{yy} = u$, which has the general solution $u(x,y) = A(x) \exp(y) + B(x) \exp(-y)$. When we now impose homogeneous Dirichlet boundary conditions on this, what would we get? Is it then so strange that you are not able to find a nontrivial solution with the eigenfunction expansion method?
If, perchance, the boundary conditions in your problem are different, you would obtain another 'basis of eigenfunctions' $\phi(x,y)$ for the Laplacian. Then, the problem $\nabla^2 u = u_{xx} + u$ might have a nontrivial solution after all! 
