 Author Topic: TT2-P2  (Read 1797 times)

Victor Ivrii TT2-P2
« on: November 17, 2016, 03:24:00 AM »
Solve
\begin{align}
&u|_{y=0}=g(x)=\left\{\begin{aligned} &1 &&|x|<1,\\ &0 &&|x|>1,\end{aligned}\right.\label{2-2}\\
&\max|u|<\infty. \label{2-3}\end{align}

Hint: Use partial Fourier transform with respect to $x$. Write solution as a Fourier integral without calculating it.

XinYu Zheng

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• Karma: 0 Re: TT2-P2
« Reply #1 on: November 17, 2016, 08:07:03 AM »
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 $\hat{u}=A(k)e^{-|k|y}+B(k)e^{|k|y}$. 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:
$$u(x,y)=\frac{1}{\pi}\int_{-\infty}^\infty \frac{\sin k}{k}e^{-|k|y+ikx}\,dk$$