TITLE: Laplace equation Fourier transform
QUESTION [8 upvotes]: Suppose
$$
\begin{cases}
u_{xx} + u_{yy} = 0, & (x,y) \in \mathbb R \times (0, \infty)
\\
u_{y}(x,0) = f(x).
\end{cases}
$$
Show that
$$u(x,y)= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \ln((x-z)^2+y^2)f(z)dz+cx+d $$
where $c$,$d$ are constants.


Hint: if $ w(x,y)=\ln[(x-z)^2+y^2] $ then $ w_{xx}+w_{yy}=0  $

Thank you for your help.
Edit:
I know that
$$ u(x,y)= \frac{y}{\pi} \int_{-\infty}^{\infty}\frac{f(z)}{(x-z)^2+y^2}dz $$
with the $ u(x,0)=f(x)  $ boundary condition, but this questions boundary condition is different you know that $ u_{y}(x,0)=f(x)$ I tried something but it doesn't work.

REPLY [11 votes]: I think where you are stuck is in the FT solution.  Note that the FT equation is
$$\hat{u}(k,y) = \int_{-\infty}^{\infty} dx \, u(x,y) e^{-i k x}$$
so that
$$-k^2 \hat{u} + \frac{\partial^2 \hat{u}}{\partial y^2} = 0$$
which means that,
$$\hat{u}(k,y) = A(k) e^{k y} + B(k) e^{-k y}$$
subject to ,
$$\frac{\partial \hat{u}}{\partial y}(k,0) = \hat{f}(k)$$
$\hat{f}$ being the FT of $f$.  There is another condition, which is implied, that the solution must die away as $y \to \infty$.  This requires some care because $k$ can be positive or negative.  The way to do this while requiring continuity at $k=0$ is for ,
$$\hat{u}(k,y) = C(k) e^{-|k| y}$$
The boundary condition at $y=0$ requires that
$$C(k) = -\frac{\hat{f}(k)}{|k|}$$
That all worked out, we may now write
$$u(x,y) = -\frac{1}{2 \pi} \int_{-\infty}^{\infty} dk  \frac{\hat{f}(k)}{|k|} e^{-|k| y} e^{i k x}$$
This is a little nasty with the $|k|$ in the denominator, so I will take a derivative in $y$:
$$\frac{\partial u}{\partial y} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk  \, \hat{f}(k)\, e^{-|k| y} e^{-i k x}$$
By the convolution theorem, we may rewrite this integral.  Note that the inverse FT of $e^{-|k| y}$ is $(y/\pi)/(x^2+y^2)$.  Then we have,
$$\frac{\partial u}{\partial y} = \frac{y}{\pi} \int_{-\infty}^{\infty} dx' \frac{f(x')}{(x-x')^2+y^2}$$
Now integrate with respect to $y$:
$$u(x,y) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dx' \, f(x') \, \ln{[(x-x')^2+y^2]} + P x+Q$$
where $P$ and $Q$ are constants of integration.