Toronto Math Forum
APM3462018S => APM346Tests => Term Test 2 => Topic started by: Victor Ivrii on March 23, 2018, 06:09:19 AM

Solve
\begin{align}
&u_{xx}+u_{yy}=0\qquad &\infty<x<\infty, \ 0<y<\infty,\label{21}\\
&(u_y+\alpha 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.
Find restriction to $\alpha$ , so that there will be no singularities.

PFT$x\mapsto \omega$ \eqref{21} becomes
\begin{equation}\label{24}\hat{u}_{yy}\omega^2\hat{u}=0.\end{equation}
Due to \eqref{23} general solution for \eqref{24} is
$$\hat{u}(y)=Ae^{\omegay}.$$
Plugging in \eqref{22}
$$\hat{g}(\omega)=A(\omega)\omega+A(\omega)\alpha\implies A(\omega)=\frac{\hat{g}(\omega)}{\alpha\omega}$$
and so
\begin{equation}\label{25}\hat{u}(\omega, y)=\hat{g}(\omega)\frac{e^{\omegay}}{\alpha\omega}.\end{equation}
Regularity of \eqref{25} is guaranteed whenever $\alpha\neq 0$. Now taking IFT we have
$$u(x,y)=\int_{\infty}^\infty \frac{\sin\omega}{\pi\omega} \frac{e^{\omegay+i\omega x}}{\alpha\omega}\,d\omega.$$

http://forum.math.toronto.edu/index.php?topic=1122.0
why the singularity condition is α=0？

Same question. The TA says $\alpha<0$ or $\alpha =n\pi$ which I don't really quite understand.

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 $g$.