APM346-2015S > HA2

HA2 problem 1

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Victor Ivrii:
Consider equation with the initial conditions
\begin{align}
& u_{tt}-9u_{xx}=0,\qquad &&t>0, x>vt,  \label{eq-HA2.1}\\\\
&u|_{t=0}= \cos (x), \qquad &&x>0, \label{eq-HA2.2}\\\\
&u_t|_{t=0}= \sin(x), \qquad &&x>0, \label{eq-HA2.3}
\end{align}


a.  Let $v=4$. Find which of these conditions (a)-(c) at $x=vt$, $t>0$ could be added to (\ref{eq-HA2.1})-(\ref{eq-HA2.3}) so that the  resulting problem would have a unique solution and solve the problem you deemed as a good one:
   1.  None,
   2.  $u|_{x=vt}=0$ ($t>0$),
   3.  $u|_{x=vt}=u_x|_{x=vt}=0$ ($t>0$).


b.  Let $v=2$. Find which of these conditions (a)-(c) at $x=vt$, $t>0$ could be added to (\ref{eq-HA2.1})-(\ref{eq-HA2.3}) so that the  resulting problem would have a unique solution and solve the problem you deemed as a good one:
   1. None
   2. $u|_{x=vt}=0$ ($t>0$),
   3. $u|_{x=vt}=u_x|_{x=vt}=0$ ($t>0$).

c.  Let $v=-4$. Find which of these conditions (a)-(c) at $x=vt$,
    $t>0$ could be added to (\ref{eq-HA2.1})-(\ref{eq-HA2.3}) so that the
    resulting problem would have a unique solution and solve the problem you deemed as a good one:
   1. None
   2. $u|_{x=vt}=0$ ($t>0$),
   3. $u|_{x=vt}=u_x|_{x=vt}=0$ ($t>0$).
   
d.  Let $v=-3$. Find which of these conditions (a)-(c) at $x=vt$,  $t>0$ could be added to (\ref{eq-HA2.1})-(\ref{eq-HA2.3}) so that the  resulting problem would have a unique solution and solve the problem you deemed as a good one:
   1. None
   2. $u|_{x=vt}=0$ ($t>0$),
   3. $u|_{x=vt}=u_x|_{x=vt}=0$ ($t>0$).

Solve the problem you deemed as a good one.

Ping Wei:
A) The problem always has  unique solution. No extra consitions are necessary. Since x>4t, we are confident that x−3t is always positive. I.e. initial value functions are defined everywhere in domain of u(t,x). Using d'Alembert's formula we write:
u(t,x)=1/3cos(x+3t)+2/3cos(x-3t)

Jessica Chen:
b. When $v=2$, we need to impose $u|_{x=2t}=0, t>0$

First when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)

Then when $2t<x<3t$, $x+3t>0$, but $x-3t<0$.

\begin{align}
\phi(x) = \frac{1}{2}\cos(x)+\frac{1}{6}\int_0^x \! \sin(y) dx.\\
\phi(x) = \frac{1}{2}\cos(x) - \frac{1}{6}(\cos(x)-1)\\
\phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}\\
\end{align}
We need to impose the extra condition such that
\begin{align}
u = \phi(5t) + \psi(-t) = 0\\
\phi(5t) = -\psi(-t)\\
\psi(t) = \phi(5t)\\
\psi(x-3t) = \frac{1}{3}\cos(5x-15t) + \frac{1}{6}\\
\end{align}
Then we get $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x-15t)) + \frac{1}{3}$ for $2t<x<3t$.

c. When $v=-4$, we need to impose $u|_{x=-4t}=ux|_{x=-4t}=0, t>0$

Firstly, when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)

Then when $-4t<x<-3t$, $x+3t<0$ and $x-3t<0$.

\begin{align}
\phi(5t) + \psi(-t) = 0\\
\phi'(5t) + \psi'(-t) = 0\\
\end{align}
Then we get $\phi'(5t)=0$ and $\psi(-t) = 0 \implies u(x,t) = 0$ for $-4t<x<-3t$.
Lastly, -3t<x<3t, the situation is similar to part b) $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x-15t)) + \frac{1}{3}$ .

d. When $v=-3$, we need to impose $u|_{x=-3t}=0, t>0$

First, when $x>3t$ then the domain satisfies both $\phi(x+3t)$ and $\psi(x-3t)$, thus $u(x, t)$ is same as part a)

Then when $ -3t<x<3t$, $x+3t>0$ and $x-3t<0$.

Then $\phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}$

We need to find $\psi(x)$ by imposing the condition:

\begin{align}
\phi(0) + \psi(-6t) = 0\\
0 + \psi(-6t) = 0\\
\psi(x-3t) = 0
\end{align}

Then we get $u(x,t) = \phi(x+3t) = \frac{1}{3}\cos(x+3t) + \frac{1}{6}$ for  $ -3t<x<3t$.

Aladdin Seaifan:
Hey I'm looking over your solutions for #2 and either I spot a mistake or I'm misunderstanding something.

I agree with what you have for ( 8 )
I follow up to step ( 10 ) .
But from ( 10 ) to ( 11 ) , you got rid of the negative sign for some reason. I don't think the negative signs simply cancel out.

When I leave things as is, my answer is $u(x,t) = \frac{1}{3}(cos(x + 3t) - 5cos(5x - 15t))$

Bethany Garnett:
Im very confused on part c) you have used the initial conditions with x=2t, shouldn't it be x=-4t? since in c) we are told v=-4 so then the set of equations would change?

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