I am again working through some old assignments for which I unfortunately have to solutions. The assignment ist:
Find the formal solution on $\Omega = ]0;1[^2$ for $-2 \frac{\partial^2u}{\partial x^2} - 3 \frac{\partial^2u}{\partial y^2} = \sin(x)\sin(3y)$ with the boundary condition $u|_{\partial\Omega}=0$.
First of all my approach (or basically how we solved them in class): We did express u as a sum of Fourier-basisfunctions, in this case, my $u(x,y)$ looks like that:
$u(x,y) = \sum_{k,l=1}^\infty u_{k,l} \cdot b_{k, l}(x,y)\text{ with }b_{k,l} = \sin(k\pi x)\sin(l\pi y)$
If you derive this term twice, insert it into the PDE you end up with:
$u_{k,l} = \frac{f_{k,l}}{2(k\pi)^2 + 3(l\pi)^2}$ where $f_{k,l}$ are the coefficents of the right hand side:
$f(x,y) = \sin(x)\sin(3y) = \sum_{k,l=1}^\infty f_{k,l} \cdot b_{k, l}(x,y)$
Up to this point I basically did anything just how we did it in the exercise, but now I wonder if there is a better way to express the $f_{k,l}$ (we only did that for generel right hand sides so far). My problem is: f also consists of 2 sine-functions, but since there is no $\pi$ as a factor in the argument, I can't easily express f by only one term of the infinity sum. So is there a way to express them, or is the sum really all I can do?
I hope my question got clear ;)