Fourier transform of PDE on finite and infinite bound simultaneously. 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?
 A: 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.
