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.
$\endgroup$ 32 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$Fourier transform the original equation and he boundary condition twice in x, get an ODE, solve it using the boundary condition, then inverse transform the solution to get the desired. Add the linear term at the end to satisfy the original equation. Use, e.g., these lecture notes as guide. Hope this helps.
