So I have the following exercise :

1) Solve the following differential equation : $$\frac{\partial^2}{\partial x^2}u(x,y)+\frac{\partial^2}{\partial y^2}u(x,y)=0$$

where $x\in \mathbb{R}$, $y > 0$

knowing that $u(x,0)=h(x)$ where $h: \mathbb{R}\rightarrow\mathbb{C}$ is an absolute, integrable, continuous function.

Hint: use the Fourier transform and the convolution theorem. Keep in mind for the solution of the system that $u(\cdot,y)$ should be bounded.

My attempt: I found on the internet how to use the Fourier Transformation with PDE. So I transformed the left side "with respect to x" and rewrote the equation as $$\frac{\partial^2}{\partial y^2}û(\omega,y)=\omega^2û(\omega,y)$$

The solution of that would be $$û(\omega,y)= c_1e^{\omega y}$$ I suppose.

Now my question is : If I plug in the initial condition right now, I would get that $c_1=h(x)$, but I'm not quite sure if it's correct because the initial condition is about the function (u), not the function transformed (û).

So I could use the convolution theorem, and get : $$u(x,y)= F^{-1}(c_1)*F^{-1}(e^{\omega y})$$ where * is the convolution of both functions. But now, how can I isolate $c_1$ to be able to "plug in" my initial condition and find the solution ? And is my attempt even correct ?

I could plug it in now and get $$ u(x,0)=F^{-1}(c_1)*F^{-1}(1)$$ but I'm not quite sure how this could help me.

Thanks for your help !

Edit : I found the solution with help from the comments. Thank you.

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    $\begingroup$ Hint: Since $u(x,0) = h(x)$, we have $\hat u(x,0) = \hat h(x)$ on the Fourier transform side, so $\hat h(x) = \hat u(x,0) = c_1 e^{\omega \cdot 0} = c_1$. $\endgroup$ – MisterRiemann Jul 10 '18 at 12:18
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    $\begingroup$ It is a lot easier using change of variables: $s=x+iy,z=x-iy, u(s,z)=u(x,y)$ $\endgroup$ – Holo Jul 10 '18 at 12:21
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    $\begingroup$ Yes, $c_1$ is a function of $x$. $\endgroup$ – MisterRiemann Jul 10 '18 at 12:21
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    $\begingroup$ Also, your notation is kinda messed up. You should write $\hat u (\omega, y)$ on the Fourier transform side. $\endgroup$ – MisterRiemann Jul 10 '18 at 12:24
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    $\begingroup$ @Poujh Since $h$ is absolutely integrable and continuous, the Inversion theorem can be applied, so that $\mathcal{F}^{-1} \hat h = h$ at every point. (In particular, $\mathcal{F}^{-1} c_1 = h$.) $\endgroup$ – MisterRiemann Jul 10 '18 at 12:27

I will assume this convention for the Fourier transform

\begin{align} \hat u(\omega) &= \int_{\Bbb R} e^{-i\omega x}u(x)\ dx \\ u(x) &= \frac{1}{2\pi}\int_{\Bbb R} e^{i\omega x}\hat{u}(\omega)d\omega \end{align}

Denote $\hat{u}(\omega,y)$ as the Fourier transform with respect to $x$, then we obtain the ODE in $y$

$$ \frac{\partial^2}{\partial y^2}\hat{u} + \omega^2 \hat{u} = 0 $$

with boundary condition

$$ \hat{u}(\omega,0) = \hat{h} (\omega) $$

where $\hat{h}(\omega)$ is the Fourier transform of $h(x)$

The general solution to the above equation is

$$ \hat{u}(\omega,y) = c_1(\omega)e^{\omega y} + c_2(\omega)e^{-\omega y} $$

where the coefficients $c_1,c_2$ are constant w.r.t to $y$

In order for the Fourier transform to exist, we require this solution to be bounded (also per the problem statement) for $y > 0$. Therefore $c_1 = 0$ if $\omega > 0$ and $c_2 = 0$ if $\omega < 0$. We can rewrite the above as

$$ \hat{u}(\omega,y) = \begin{cases} c_1(\omega)e^{\omega y}, & \omega < 0, y > 0 \\ c_2(\omega)e^{-\omega y}, & \omega > 0, y > 0 \end{cases} $$

The boundary condition implies $c_1 = c_2 = \hat{h}(\omega)$, so finally

$$ \hat{u}(\omega,y) = \hat{h}(\omega) e^{-|\omega|y}, \quad y > 0 $$

By the convolution theorem

$$ u(x,y) = h(x) * \mathcal{F}^{-1}\{e^{-|\omega| y}\} $$

You can compute the inverse Fourier transform of the $y$ function to find

$$ \mathcal{F}^{-1}\{e^{-|\omega|y}\} = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega x}e^{-|\omega| y}d\omega = \frac{1}{\pi}\frac{y}{x^2 + y^2} $$

Hence, the final solution is

$$ u(x,y) = \frac{y}{\pi}\int_{-\infty}^\infty \frac{h(\xi)}{(x-\xi)^2 + y^2}d\xi $$


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