### Author Topic: HA3-P2  (Read 5180 times)

#### Victor Ivrii

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##### HA3-P2
« on: September 28, 2015, 01:04:20 PM »

#### Zaihao Zhou

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##### Re: HA3-P2
« Reply #1 on: October 03, 2015, 01:31:48 PM »
Not sure if I'm right. Since involve graphs I uploaded a picture. Skipped 4 and 6 since I don't see the point and whether I am doing the right thing. Please correct me if I'm wrong.

#### Yumeng Wang

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##### Re: HA3-P2
« Reply #2 on: October 03, 2015, 07:41:12 PM »
For (5) I think you miss the last case : x<-ct and xâ‰¥ct
In these conditions, I get (-x+ct)\2c

#### Victor Ivrii

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##### Re: HA3-P2
« Reply #3 on: October 04, 2015, 11:13:44 AM »

I give example of the proper solution to one of the problems.

\begin{align}
&u_{tt}-c^2u_{xx}=0,\\
\end{align}
with

h(x)=\left\{\begin{aligned}
&1 &&|x| < 1,\\
&0 && |x| \ge 1.
\end{aligned}\right.

Observe that  $g$, $h$ are even with respect to $x$. Then $u$ is even with respect to $x$. Since $g=0$ we conclude that $u$ is odd with respect to $t$. So basically we need to consider only $x>0,t>0$. Here teal and orange are characteristics passing through the ends of the segment $(-1,1)$ on $\{t=0\}$

« Last Edit: October 04, 2015, 01:14:29 PM by Victor Ivrii »

#### Bruce Wu

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##### Re: HA3-P2
« Reply #4 on: October 08, 2015, 01:11:58 AM »
The yellow diamond is the only place where both initial conditions are satisfied. Will we have to include in our answer that it is where the solution is valid?

#### Victor Ivrii

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##### Re: HA3-P2
« Reply #5 on: October 08, 2015, 08:02:34 AM »
The yellow diamond is the only place where both initial conditions are satisfied. Will we have to include in our answer that it is where the solution is valid?

Initial conditions are satisfied where they are given: as $t=0$, $-\infty<x<\infty$. Yellow diamond is just a zone like any other, or if you want where both $\phi(x+ct)$ and $\psi(x-ct)$ are not $0$

#### Bruce Wu

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##### Re: HA3-P2
« Reply #6 on: October 08, 2015, 11:44:58 AM »
I will do the last part.
$$g(x) = 0, h(x) = \begin{cases} \cos(x) &\mbox{if } |x| < \frac{\pi}{2} \\ 0 & \mbox{if } |x| \geq \frac{\pi}{2}. \end{cases}$$

The solution is $$u=\frac{1}{2c}\int^{x+ct}_{x-ct}h(y)dy$$
Let $H$ be the primitive of $h$, i.e $H'(y)=h(y)$
Then $$H(y)=\begin{cases} 0 &\mbox{if } y \leq -\frac{\pi}{2} \\ \sin(y)+1 & \mbox{if } -\frac{\pi}{2} < y < \frac{\pi}{2} \\ 2 & \mbox{if } y \geq \frac{\pi}{2}. \end{cases}$$
And we have $$u=\frac{1}{2c}[H(x+ct)-H(x-ct)]$$
Since g and h here are also even with respect to x, we apply the same logic as the professor and only worry about $x>0$ and $t>0$.
So we have $$u(x,t)=\begin{cases} \frac{1}{c} &\mbox{if } x-ct \leq -\frac{\pi}{2}, x+ct \geq \frac{\pi}{2} \\ \frac{1}{2c}[1-\sin(x-ct)] & \mbox{if } -\frac{\pi}{2} < x-ct < \frac{\pi}{2}, x+ct \geq \frac{\pi}{2} \\ 0 & \mbox{if } x-ct \geq \frac{\pi}{2}, x+ct \geq \frac{\pi}{2} \\ \frac{1}{2c}[\sin(x+ct)-\sin(x-ct)] & \mbox{if } -\frac{\pi}{2} < x-ct < \frac{\pi}{2}, -\frac{\pi}{2} < x+ct < \frac{\pi}{2} \end{cases}$$
Then we can use the symmetry of u to find its value in other regions.
« Last Edit: October 08, 2015, 11:30:07 PM by Fei Fan Wu »

#### Victor Ivrii

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##### Re: HA3-P2
« Reply #7 on: October 08, 2015, 12:54:38 PM »
I was not completely clear here, see below
« Last Edit: October 08, 2015, 02:53:22 PM by Victor Ivrii »

#### Bruce Wu

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##### Re: HA3-P2
« Reply #8 on: October 08, 2015, 01:32:57 PM »
Sorry I forgot about the yellow diamond. My solution above has been fixed.
« Last Edit: October 08, 2015, 11:25:27 PM by Fei Fan Wu »

#### Emily Deibert

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##### Re: HA3-P2
« Reply #9 on: October 08, 2015, 01:41:44 PM »
Professor, I am a little confused about the different regions. Can you please explain this?

#### Victor Ivrii

1) Functions $g$, $h$ are supported in some interval $[a,b]$ (which means they are $0$ outside of it). We draw characteristics through this interval ends, we get two orange and two teal lines. They divide plane into 9 regions.
2) Observe that $u(x,t)$ is even with respect to $x$ (indeed, both $g$ and $h$ are) and odd with respect to $t$ (indeed, $g=0$; if $h=0$ we would have $u(x,t)$ even with respect to $t$).   So we need to consider only $x>0$, $t>0$ and continue into 3 other quadrants by a symmetry. Instead of 9 regions we have only 4.