APM346-2015F > Test 2

TT2-P2

(1/4) > >>

Victor Ivrii:
Solve
\begin{align}
&u_{xx}+u_{yy}=0\qquad -\infty<x<\infty, \ 0<y<\infty,\label{2-1}\\
&u_{y=0}=\frac{1}{x^2+1},\label{2-2}\\
&\max |u |<\infty.\label{2-3}
\end{align}

Hint: Use partial Fourier transform with respect to $x$, and formula
\begin{equation}
F (x^2+a^2)^{-1}=\frac{1}{2a}e^{-|k|a}\qquad \text{as\ \ } a>0.
\label{2-4}
\end{equation}

Xi Yue Wang:
Use Fourier Transform，$u(x,y) \rightarrow \hat{u} (k,y)$ $$-k^2\hat{u}+\hat{u}_{yy}=0\\\hat{u}(k,y)=A(k)e^{-|k|y}+B(k)e^{|k|y}$$
We discard second term because it is unbounded.$$\hat{u}(k,0) = A(k) = \hat{f}(k) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{x^2 +1}e^{-ikx} dx = \frac{1}{2}e^{-|k|} \\u(x,y)=\frac{1}{2}\int_{-\infty}^{\infty} e^{-|k|y}e^{|-k|}e^{ikx} dk $$ Then the answer is same as Emily.

But for this one, from lecture notes, suppose we don't have a hint for $\hat{f}(k)$
We get IFT of $e^{-|k|y}$ is $ \frac{2y}{x^2+y^2}$ $$u(x,y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x')\frac{2y}{((x-x')^2+y^2)}dx'\\= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{(x'^2 +1)}\frac{y}{((x-x')^2+y^2)}dx'$$
This interval is supposed to be same as the correct answer. I am confused about this part.

Emily Deibert:
I believe there may be a mistake in this one. As we did in HA7, we must change the given condition at $y=0$ by Fourier transform as well. So we will have (making use of the hint):

\begin{equation}
u|_{y=0} \longrightarrow \hat{u}|_{y=0} = \hat{\frac{1}{x^2 + 1}} = \frac{1}{2}e^{-|k|}
\end{equation}

We should be able to proceed similarly to what Xi Yue Wang has done, with the integral now being:

\begin{equation}
u(x,y)=\frac{1}{2} \int_{-\infty}^{\infty} e^{-|k|y}e^{-|k|}e^{ikx} dk
\end{equation}

It's actually a simple integral to solve; we split it up so that we have:

\begin{equation}
u(x,y)=\frac{1}{2} \left( \int_{-\infty}^{0} e^{ky}e^{k}e^{ikx} dk +  \int_{0}^{\infty} e^{-ky}e^{-k}e^{ikx} dk \right)
\end{equation}

So we get:

\begin{equation}
u(x,y) = \frac{1}{2} \left( \frac{e^{ky+k+ikx}|_{-\infty}^0}{y + 1 + ix} - \frac{e^{-ky-k+ikx}|_{0}^{\infty}}{y + 1 - ix} \right)
\end{equation}

Evaluating, we get:

\begin{equation}
u(x,y) = \frac{1}{2} \left( \frac{1}{y + 1 + ix} + \frac{1}{y + 1 - ix} \right)
\end{equation}

The final answer is then:

\begin{equation}
u(x,y) = \frac{y+1}{(y+1)^2 + x^2}
\end{equation}

Rong Wei:
I believe Xi Yue Wang is right  ;D

Bruce Wu:
Xi Yue's solution makes no sense. By her answer $u(x,0)=0$, which clearly does not satisfy the boundary condition.

Navigation

[0] Message Index

[#] Next page