Author Topic: TT2--P2  (Read 1352 times)

Victor Ivrii

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TT2--P2
« on: March 23, 2018, 06:09:19 AM »
Solve
\begin{align}
&u_{xx}+u_{yy}=0\qquad &-\infty<x<\infty, \ 0<y<\infty,\label{2-1}\\
&(u_y+\alpha 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.

Find restriction to $\alpha$ , so that there will be no singularities.

Jingxuan Zhang

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Re: TT2--P2
« Reply #1 on: March 24, 2018, 08:16:29 AM »
PFT$x\mapsto \omega$ \eqref{2-1} becomes
\begin{equation}\label{2-4}\hat{u}_{yy}-\omega^2\hat{u}=0.\end{equation}
Due to \eqref{2-3} general solution for \eqref{2-4} is
$$\hat{u}(y)=Ae^{-|\omega|y}.$$
Plugging in \eqref{2-2}
$$\hat{g}(\omega)=-A(\omega)|\omega|+A(\omega)\alpha\implies A(\omega)=\frac{\hat{g}(\omega)}{\alpha-|\omega|}$$
and so
\begin{equation}\label{2-5}\hat{u}(\omega, y)=\hat{g}(\omega)\frac{e^{-|\omega|y}}{\alpha-|\omega|}.\end{equation}
Regularity of \eqref{2-5} is guaranteed whenever $\alpha\neq 0$. Now taking IFT we have
$$u(x,y)=\int_{-\infty}^\infty \frac{\sin\omega}{\pi\omega} \frac{e^{-|\omega|y+i\omega x}}{\alpha-|\omega|}\,d\omega.$$
« Last Edit: March 24, 2018, 08:19:48 AM by Jingxuan Zhang »

Qiaochu Yang

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tt2-Q2 problem
« Reply #2 on: April 05, 2018, 04:08:10 PM »
http://forum.math.toronto.edu/index.php?topic=1122.0
why the singularity condition is α=0?

Jingxuan Zhang

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Re: tt2-Q2 problem
« Reply #3 on: April 05, 2018, 04:36:45 PM »
Same question. The TA says $\alpha<0$ or $\alpha =n\pi$ which I don't really quite understand.

Victor Ivrii

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Re: TT2--P2
« Reply #4 on: April 05, 2018, 05:25:30 PM »
Observe that the denominator is $0$ as $|\omega|=\alpha$ and to avoid it we need to assume that $\alpha<0$ (since $|\omega|$ runs from $0$ to $\infty$.

If $\alpha=0$, $\sin (\omega)\sim \omega$, while denominator is $\omega|\omega|$, so still singularity.

Nathan observed that as $\alpha=n\pi$ with $n=1,2,\ldots$ singularity does not appear since $\sin (\omega)/(\omega -2\pi n)\sim 1$. However $\sin(\omega)$ appears only because of the special $g(x)$, and it saves only some $g$, not all of them, so while indeed we have a solution for such $\alpha$, it is only for some $g$.