Author Topic: TT1-P3  (Read 421 times)

Victor Ivrii

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TT1-P3
« on: October 19, 2016, 10:27:53 PM »
Find  solution to
\begin{align}
&u_{tt}-9u_{xx}=0, \qquad&& t>0, \ \ 0<x< t,\label{eq-1}\\
&u|_{t=0}=\sin (x), && x>0,\label{eq-2}\\
&u_t|_{t=0}=3\cos (x), && x>0,\label{eq-3}\\
&u|_{x=t}= 0, &&t>0.\label{eq-4}
\end{align}

XinYu Zheng

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Re: TT1-P3
« Reply #1 on: October 19, 2016, 10:47:13 PM »
Let $u(x,t)=\phi(x+3t)+\psi(x-3t)$. Applying D'Alembert's formula, for $x>0$ we have
$$\phi(x)=\frac{1}{2}\sin(x)+\frac{1}{6}\int_0^x 3\cos x'\,dx'=\sin(x)\\
\psi(x)=\frac{1}{2}\sin(x)-\frac{1}{6}\int_0^x 3\cos x'\,dx'=0$$
We need to find $\psi(x)$ for $x<0$. To do this, we apply boundary condition:
$$0=u|_{x=t}=\phi(4t)+\psi(-2t)\,\,t>0$$
Therefore, we have
$$\psi(t)=-\phi(-2t)\,\,t<0$$
Thus we have the solution
$$u(x,t)=\sin(x+3t)-\sin(6t-2x)$$
which is valid for $0<x<3t$. But the original equation is defined on a domain that is a subset of this (since $x<t\implies x<3t$), so this is the solution to the original problem.
« Last Edit: October 19, 2016, 10:50:43 PM by XinYu Zheng »

Victor Ivrii

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Re: TT1-P3
« Reply #2 on: October 20, 2016, 04:52:18 AM »
Please draw a picture and clarify where solution is given by this expression.

Roro Sihui Yap

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Re: TT1-P3
« Reply #3 on: October 20, 2016, 09:17:09 AM »
In domain $ 0< x < t,$ $x - 3t < 0 $

Victor Ivrii

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Re: TT1-P3
« Reply #4 on: October 20, 2016, 10:30:46 AM »
Crap: it was a misprint nobody noticed: domain should be $\{0<t < x\}$. Then it would be a proper problem. As stated on TT problem cannot be recovered. So, I instruct TAs not to grade P3 but to grade any other problem out of 5.

Also: please as bonus solve a correct problem

Roro Sihui Yap

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Re: TT1-P3
« Reply #5 on: October 20, 2016, 12:21:08 PM »
Using the functions found above,

$\phi(x) = \sin(x)$ for $x > 0$
$\psi(x) = 0$ for $x > 0$

and

$\psi(x)=-\phi(-2x)=-sin(-2x)=sin(2x)$ for $x < 0$

In the region, $ 0<t<x<3t $, we have $x+3t > 0$,$x-3t < 0$
$$u(x,t) = \sin (x + 3t ) + \sin (2x - 6t)$$

In the region $ x > 3t > 0 $, we have $x+3t > 0$,$x-3t > 0$
$$u(x,t) = \sin (x + 3t ) $$
« Last Edit: October 20, 2016, 12:23:30 PM by Roro Sihui Yap »

Victor Ivrii

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Re: TT1-P3
« Reply #6 on: October 20, 2016, 01:50:12 PM »
OK, модифицированная (и та, которую я и имел в виду) проблема решена правильно