Calculate the Fourier expansion of $u$ for $$\nabla^2u=u_{xx}+u_{yy}=0,\\ y\ge0,0\le x\le L\\ u(0,y)=0=u_x(L,y), u(x,0)=g(x)=x/L$$.
Solution:
By separation of variables, we propose the solution $u(x,y)=X(x)Y(y)$.
After calculations and after applying the conditions I get that the general solution is $u(x,y)=\sum_{n=0}^\infty A_ne^{\sqrt{(\lambda_n)}y}D_n\sin(\sqrt \lambda_nx),$ where $A_n=\frac{(g(x),X_n(x))}{\Vert X_n\Vert^2}, X_n(x)=\sin(\frac{(n+1/2)\pi}{L}x),\lambda_n=(\frac{n\pi}{L})^2$
I found $A_n$ when I applied the boundary condition $u(x,0)$. Is there a way to find $D_n$ so the solution would be complete?
Or the solution it's fine as it is?
Thanks in advance for your time and help.
Separation of variables gives
$$ X(x) = \sin\left(\beta_n x\right) $$
Where $ \beta_n = \sqrt{\lambda_n} = (n+\frac12)\frac{\pi }{L}$
Then
$$ Y'' - \beta^2 Y = 0 $$ $$ \implies Y = Ae^{-\beta y} + Be^{\beta y} $$
Since there are two constants, you need two boundary conditions here to determine a solution. Due to the domain, we will assume that $Y(y)$ remains bounded as $y\to\infty$. Hence, $B=0$ and the general solution is
$$ u(x,y) = \sum_{n=0}^\infty A_ne^{-\beta_n y} \sin(\beta_n x) $$
Plugging in the B.C at $y=0$ results in a Fourier series in $x$, which you can use to solve for $A_n$
$$ A_n = \frac{2}{L}\int_0^L g(x)\sin(\beta_n x)\ dx $$