Toronto Math Forum
APM3462016F => APM346Tests => TT2 => Topic started by: Victor Ivrii on November 17, 2016, 03:24:00 AM

Solve
\begin{align}
&u_{xx}+u_{yy}=0\qquad &\infty<x<\infty, \ 0<y<\infty,\label{21}\\
&u_{y=0}=g(x)=\left\{\begin{aligned} &1 &&x<1,\\ &0 &&x>1,\end{aligned}\right.\label{22}\\
&\maxu<\infty. \label{23}\end{align}
Hint: Use partial Fourier transform with respect to $x$. Write solution as a Fourier integral without calculating it.

Applying fourier transform with respect to $x$ so that $u(x,y)\to \hat{u}(k,y)$ the PDE becomes
$$\begin{cases}
k^2\hat{u}+\hat{u}_{yy}=0\\
\hat{u}_{y=0}=\hat{g}(k)
\end{cases}$$
This PDE has general solution $\hat{u}=A(k)e^{ky}+B(k)e^{ky}$. We drop the second term because it goes unbounded. Now applying the B.C. we see that $\hat{g}(k)=A(k)$. So we compute $\hat{g}(k)$:
$$\hat{g}(k)=\frac{1}{2\pi}\int_{1}^1e^{ikx}\,dx=\frac{1}{2\pi}\int_0^1 e^{ikx}+e^{ikx}\,dx=\frac{1}{\pi}\int_0^1 \cos(kx)\,dx=\frac{\sin k}{k\pi}$$
Where in the middle we have split the integral in two parts and did a change of variables $x\to x$ in the second one. Now we just need to apply IFT for the solution:
$$u(x,y)=\frac{1}{\pi}\int_{\infty}^\infty \frac{\sin k}{k}e^{ky+ikx}\,dk$$