Toronto Math Forum
APM3462015S => APM346Home Assignments => HA2 => Topic started by: Victor Ivrii on January 27, 2015, 09:23:27 PM

Consider equation with the initial conditions
\begin{align}
& u_{tt}9u_{xx}=0,\qquad &&t>0, x>vt, \label{eqHA2.1}\\\\
&u_{t=0}= \cos (x), \qquad &&x>0, \label{eqHA2.2}\\\\
&u_t_{t=0}= \sin(x), \qquad &&x>0, \label{eqHA2.3}
\end{align}
a. Let $v=4$. Find which of these conditions (a)(c) at $x=vt$, $t>0$ could be added to (\ref{eqHA2.1})(\ref{eqHA2.3}) 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{eqHA2.1})(\ref{eqHA2.3}) 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{eqHA2.1})(\ref{eqHA2.3}) 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{eqHA2.1})(\ref{eqHA2.3}) 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.

A) The problem always has unique solution. No extra consitions are necessary. Since x>4t, we are confident that xâˆ’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(x3t)

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 $\psi(x3t)$, thus $u(x, t)$ is same as part a)
Then when $2t<x<3t$, $x+3t>0$, but $x3t<0$.
\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(x3t) = \frac{1}{3}\cos(5x15t) + \frac{1}{6}\\
\end{align}
Then we get $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x15t)) + \frac{1}{3}$ 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 $\psi(x3t)$, thus $u(x, t)$ is same as part a)
Then when $4t<x<3t$, $x+3t<0$ and $x3t<0$.
\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) $u(x, y) = \frac{1}{3}(\cos(x+3t) +\cos(5x15t)) + \frac{1}{3}$ .
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 $\psi(x3t)$, thus $u(x, t)$ is same as part a)
Then when $ 3t<x<3t$, $x+3t>0$ and $x3t<0$.
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(x3t) = 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$.

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 $u(x,t) = \frac{1}{3}(cos(x + 3t)  5cos(5x  15t))$

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?

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?
I'm pretty sure you're correct as well lol. And the question never told us to solve for it, although it would be good practice.
My answer to that question is c).
I'll post my attempt to the solutions after the midterm since I my answers are different.

Yes, we impose conditions as $x=v t$. Jessica, please fix your solution