Q:

Laplace's equation via Fourier transformation

I have a Laplace equation with some data along the $y$-axis:
$$
\begin{cases}
u_{xx} + u_{yy} &= 0 \\
u(0,y) &= f(y) \\
u_x(0,y) &= g(y).
\end{cases}
$$
There is no information of any region, properties of the solution or properties of $f$ and $g$ other than that we may assume $f$ and $g$ to have well-defined Fourier transforms.
So, Fourier transforming everything with respect to the $y$-variable we obtain: 
$$
\begin{cases}
\hat{u}_{xx} -\xi ^2 \hat{u} &= 0 \\
\hat{u}(0,\xi ) &= \hat{f}(\xi ) \\
\hat{u}_x(0,\xi ) &= \hat{g}(\xi ).
\end{cases}
$$
Thus $\hat{u}(x,\xi ) = A(\xi )e^{-\xi x} + B(\xi ) e^{\xi x}$. 
I suppose this is where I become puzzled. Clearly $\hat{u}$ behaves very badly as $\xi \to \pm \infty $, depending on the sign of $x$. To fix this I first restricted to the case $x>0$ and simply assumed that $A(\xi ) = 0$ for $\xi <0$ and $B(\xi ) = 0$ for $\xi > 0$ so that 
$$
\hat{u}(x,\xi ) = C(\xi )e^{-|\xi | x} = \hat{f}(\xi )e^{-|\xi | x}. 
$$
This is easy to invert since $\mathcal{F}^{-1}[e^{-|\xi |x}](x,y)$ is well-known. But now it seems there is no room for $g$ (or $\hat{g}$) to fit in anymore. 
I suppose I must have gone wrong when I restricted to $x>0$ and singled out certain solutions, but how can I avoid this? 

A:

You cannot put $A(\xi)=0$ given that boundary conditions. So,
$$
   \hat u(0,\xi)=A(\xi)+B(\xi)=\hat f(\xi)
$$
and
$$
   \hat u_x(0,\xi)=-\xi A(\xi)+\xi B(\xi)=\hat g(\xi).
$$
This is a simple system you have to solve to get $A$ and $B$.

