I am trying to solve the following Neumann problem on $H=\{(x,y)|-\infty<x<\infty, \, y>0\}$:
$\begin{align}v_{xx}(x,y)+v_{yy}(x,y)=0 &\, \text{in}\,H \\ v_{y}(x,0)=g(x), &\,\, -\infty <x < \infty \\ v(x,y) & \text{bounded in}\, H\end{align}$
Earlier, I solved the related Dirichlet problem on the same region $H$:
$\begin{align}u_{xx}(x,y) + u_{yy}(x,y) = 0 &\, \text{in}\, H \\ u(x,0)=f(x), &\,\, -\infty < x < \infty \\ u(x,y) & \text{bounded in}\, H \end{align}$
and I know that I can find the solution to the Neumann problem by letting $w(x,y) = v_{y}(x,y)$, and noticing that $w$ satisfies the Dirichlet problem. I found that $\displaystyle u(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{y}{z^{2}+y^{2}}f(x-z)dz$, so here we have that $\displaystyle w(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{y}{y^{2}+z^{2}}g(x-z)dz = v_{y}(x,y)$. Integrating this with respect to $y$, and keeping in mind that the constant function $C(x)$ introduced from the integration must satisfy both $C^{\prime\prime}(x) =0$ and $C(x)$ bounded $\forall x$, we see that $C(x) = C$, an arbitrary constant. Therefore, the solution to the Neumann problem should be $\displaystyle v(x,y) = \frac{1}{2\pi}\int_{-\infty}^{\infty}g(x-z)\log(y^{2}+z^{2})dz + C $
However, I also need to be able to solve the Neumann problem by directly applying the Fourier transform to it. I have been trying to do this, but I am getting stuck on applying the boundary conditions to the transformed problem.
The transformed Neumann problem is the following:
$\begin{align}-\alpha^{2} V(\alpha, y) + V^{\prime \prime}(\alpha, y) = 0, &\,y>0 \\ V^{\prime}(\alpha,0)=G(\alpha) \\ V(\alpha,y),\, V^{\prime}(\alpha, y)\, & \text{bounded for}\,y>0 \end{align}$
The general solution to this problem is $V(\alpha, y) = A \exp(-|\alpha|y) + B\exp (|\alpha|y)$, $y>0$.
Then, since as $y \to \infty$, $\exp(|\alpha|y)$ grows without bound, we need $B = 0$.
To take care of $A$, we look at $V^{\prime}(\alpha, y) = -|\alpha|A\exp(-|\alpha|y)$.
Then, applying the B.C.: $V^{\prime}(\alpha,0)=-|\alpha|A = G(\alpha)$. So, $\displaystyle A = -\frac{G(\alpha)}{|\alpha|}$, $\alpha \neq 0$.
But then, $\displaystyle V(\alpha, y) = -\frac{1}{|\alpha|}G(\alpha)\exp(-|\alpha|y)$.
Then, using the fact that the Fourier transform of the convolution is the product of the Fourier transforms: $\displaystyle \mathcal{F}(g * f) = \mathcal{F}(g(\alpha)) \cdot F(\alpha) = \mathcal{F}(g(\alpha))\cdot \left(-\frac{1}{|\alpha|}\exp(-|\alpha|y) \right)$, all I would need to do is find the function $f(y)$ whose image under the Fourier transform is $\displaystyle F(\alpha) =-\frac{1}{|\alpha|}\exp(-|\alpha|y) $. However, I am having a lot of trouble doing that. In order to do so, I need to integrate $\displaystyle \int_{-\infty}^{\infty}-\frac{1}{|\alpha|}\exp(-|\alpha|y)\exp(iy\alpha)d\alpha$. Letting $y = x-z$, this becomes $\begin{align}\displaystyle -\int_{-\infty}^{\infty}\frac{1}{|\alpha|}\exp(-(x-z)|\alpha|)\exp(i y \alpha) d\alpha \\ \mathbf{= \int_{-\infty}^{0} \frac{1}{\alpha}\exp([(x-z)+iy]\alpha)\,d\alpha - \int_{0}^{\infty}\frac{1}{\alpha}\exp(-[(x-z)-iy]\alpha)\,d\alpha}\end{align}$.
Now, evaluating each of these integrals is not fun. I attempted to solve the first integral by parts, which needed to be performed twice, and then I needed to add $\int \frac{1}{\alpha}\exp([(x-z)+iy]\alpha)\,d\alpha$ to both sides and wound up getting $0 = 0$.
Should I try to split this up into a convolution product as well? Or should I proceed with integration by parts on both pieces? If the former, how would I do that, and then put everything back together? If the latter, do you have any pointers on how I can do these integrations in the least painful manner?
Thank you