Solve the BVP:
$$u_{xx}+u_{yy}=0,~~x\in \mathbb{R},~y>0,~~~u(x,0)=0,~~~u_y(x,0)=a\sin(x/a),~x\in \mathbb{R},$$
where $a$ is a positive constant.
Attempt. By separation of variables we get $u=XY$, where $X''Y+XY''=0$, i.e. $$X''-cX=0, ~Y''+cY=0.$$ The boundary condition $u(x,0)=0$ gives $Y(0)=0$ and the only bounded solutions for $y>0$ are $Y(y)=\sin(ky),$ for $c=k^2>0,~k>0$. Then $X=Ae^{-k|x|}+Be^{k|x|}$ and in order to have bounded solutions for $x$ we get $X=Ae^{-k|x|}$. Thus, our solutions are $u_k=X_kY_k$, for $k>0$.
Since $k>0$, $u_k's$ are uncountable. How can we get a sum of the form $\sum_kE_ke^{-k|x|}\sin(ky)$ in order to use the boundary condition $u_y(x,0)=a\sin(x/a)$?
Thanks in advance.