Consider $$u_{xx} + u_{yy} = 0 $$ on the bounds: $$o < x < L$$ and $$-\infty<y<\infty$$
The initial condition is: $$u(0,y) = f(y)$$ and $$u(L,y)=g(y)$$ I've tried performing fourier transform on both sides but looking at the solutions, I knew it wasn't right.
Could someone give me a leg up?
Break this into two problems--one with $f=0$ and $g$ general and the other with $g=0$ and $f$ general. Then add the solutions together.
Separation of variables: $$ \frac{X''}{X}= \lambda^{2} = -\frac{Y''}{Y}. $$ The solutions in $Y$ should be chosen so that they are bounded, which is why I've written the separation paramter as $\lambda^{2}$, and assume $\lambda$ is real. For the first problem where $f=0$ and $g$ is general, the separated solutions are $$ A(\lambda)e^{i\lambda y}\sinh(\lambda x) $$ Summing these gives $$ F(x,y) = \int_{-\infty}^{\infty}A(\lambda)e^{i\lambda y}\sinh(\lambda x)dx $$ The requirement that $F(L,y)=g(y)$ requires finding the coefficient function $A(\lambda)$ such that $$ g(y) = \int_{-\infty}^{\infty}A(\lambda)\sinh(\lambda L)e^{i\lambda x}d\lambda \\ \implies A(\lambda)\sinh(\lambda L) = \frac{1}{2\pi}\int_{-\infty}^{\infty}g(y)e^{-i\lambda y}dy $$ So a solution of this subproblem where $f=0$ is $$ F(x,y) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \left(\int_{-\infty}^{\infty}g(y)e^{-i\lambda y}dy\right)\frac{\sinh(\lambda x)}{\sinh(\lambda L)}e^{i\lambda y}d\lambda. $$ Similarly, a solution where $g=0$ is $$ G(x,y) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \left(\int_{-\infty}^{\infty}f(y)e^{-i\lambda y}dy\right) \frac{\sinh(\lambda(L-x))}{\sinh(\lambda L)}d\lambda. $$ A solution of your problem is $$ u(x,y)=F(x,y)+G(x,y). $$ Note about Uniqueness: For constants $A$ and $B$, the functions $$ u_{n}(x,y) = \sin(n\pi x/L)\{ Ae^{n\pi y/L}+Be^{-n\pi y/L}\}, \;\;\; n=1,2,3,\cdots, $$ are solutions of $\nabla^{2}u_{n} = 0$ that vanish at $x=0$ and $x=L$. So you can add any linear combination of these solutions to the above solution $u$ and still have a solution of the original problem. Some assumption of boundedness in $y$ is required in order to have a unique solution of your stated problem, which is often the case on semi-infinite or infinite regions.