Toronto Math Forum
APM3462016F => APM346Tests => TT1 => Topic started by: Victor Ivrii 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{eq1}\\
&u_{t=0}=\sin (x), && x>0,\label{eq2}\\
&u_t_{t=0}=3\cos (x), && x>0,\label{eq3}\\
&u_{x=t}= 0, &&t>0.\label{eq4}
\end{align}

Let $u(x,t)=\phi(x+3t)+\psi(x3t)$. 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
$$u(x,t)=\sin(x+3t)\sin(6t2x)$$
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.

Please draw a picture and clarify where solution is given by this expression.

In domain $ 0< x < t,$ $x  3t < 0 $

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

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 $x+3t > 0$,$x3t < 0$
$$u(x,t) = \sin (x + 3t ) + \sin (2x  6t)$$
In the region $ x > 3t > 0 $, we have $x+3t > 0$,$x3t > 0$
$$u(x,t) = \sin (x + 3t ) $$

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