### Author Topic: TT2-P2  (Read 6619 times)

#### Victor Ivrii ##### TT2-P2
« on: November 18, 2015, 08:38:04 PM »
Solve
\begin{align}
&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

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« Reply #1 on: November 18, 2015, 10:25:06 PM »
Use Fourier Transformï¼Œ$u(x,y) \rightarrow \hat{u} (k,y)$ $$-k^2\hat{u}+\hat{u}_{yy}=0\\\hat{u}(k,y)=A(k)e^{-|k|y}+B(k)e^{|k|y}$$
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 $\frac{2y}{x^2+y^2}$ $$u(x,y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x')\frac{2y}{((x-x')^2+y^2)}dx'\\= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{(x'^2 +1)}\frac{y}{((x-x')^2+y^2)}dx'$$
« Last Edit: November 19, 2015, 03:34:38 PM by Xi Yue Wang »

#### Emily Deibert ##### Re: TT2-P2
« Reply #2 on: November 19, 2015, 01:04:42 AM »
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}

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

#### Rong Wei

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« Reply #3 on: November 19, 2015, 01:08:57 AM »
I believe Xi Yue Wang is right #### Bruce Wu ##### Re: TT2-P2
« Reply #4 on: November 19, 2015, 01:13:48 AM »
Xi Yue's solution makes no sense. By her answer $u(x,0)=0$, which clearly does not satisfy the boundary condition.

#### Rong Wei

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« Reply #5 on: November 19, 2015, 01:14:48 AM »
as Xi Yue Wang did in homework 7 and lecture notes here: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-

#### Emily Deibert ##### Re: TT2-P2
« Reply #6 on: November 19, 2015, 01:17:00 AM »
as Xi Yue Wang did in homework 7 and lecture notes here: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-

Rong Wei, which section did you mean? The link does not work for me!

#### Rong Wei

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« Reply #7 on: November 19, 2015, 01:17:04 AM »
Xi Yue's solution makes no sense. By her answer $u(x,0)=0$, which clearly does not satisfy the boundary condition.
you are right Fei Fan, I think the process is not right, but the final answer is right based on Lecture Notes. #### Rong Wei

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« Reply #8 on: November 19, 2015, 01:18:02 AM »
as Xi Yue Wang did in homework 7 and lecture notes here: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-

Rong Wei, which section did you mean? The link does not work for me!
5.3! actually, solution in this section is quite confused for me #### Bruce Wu ##### Re: TT2-P2
« Reply #9 on: November 19, 2015, 01:18:37 AM »
The final answer cannot be right if it does not satisfy the boundary conditions.

#### Rong Wei

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« Reply #10 on: November 19, 2015, 01:20:51 AM »
The final answer cannot be right if it does not satisfy the boundary conditions.
I don't understand neither, but maybe you can check the lecture notes, http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter5/S5.3.html

#### Emily Deibert ##### Re: TT2-P2
« Reply #11 on: November 19, 2015, 01:25:06 AM »
The final answer cannot be right if it does not satisfy the boundary conditions.
I don't understand neither, but maybe you can check the lecture notes, http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter5/S5.3.html

Okay, thank you for the link! It did not work for me before. I am checking the textbook now, I hope I can understand.

#### Rong Wei

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« Reply #12 on: November 19, 2015, 01:26:25 AM »
The final answer cannot be right if it does not satisfy the boundary conditions.
I don't understand neither, but maybe you can check the lecture notes, http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter5/S5.3.html

Okay, thank you for the link! It did not work for me before. I am checking the textbook now, I hope I can understand.

welcome, Emily, #### Rong Wei

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« Reply #13 on: November 19, 2015, 01:28:28 AM »
my solution

#### Bruce Wu ##### Re: TT2-P2
« Reply #14 on: November 19, 2015, 01:42:47 AM »
Hopefully if one actually evaluates the integral we will get the same thing. Even wolfram alpha pro couldn't handle it, I tried. However as it stands right now I agree with Emily's answer.