### Author Topic: HA2 problem 1  (Read 3255 times)

#### Victor Ivrii

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##### HA2 problem 1
« on: January 27, 2015, 09:23:27 PM »
Consider equation with the initial conditions
\begin{align}
&u|_{t=0}= \cos (x), \qquad &&x>0, \label{eq-HA2.2}\\\\
\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

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##### Re: HA2 problem 1
« Reply #1 on: January 29, 2015, 09:42:54 PM »
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(x-3t)

#### Jessica Chen

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##### Re: HA2 problem 1
« Reply #2 on: January 29, 2015, 10:35:13 PM »
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  $-3t<x<3t$.
« Last Edit: March 05, 2015, 09:33:32 PM by Jessica Chen »

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##### Re: HA2 problem 1
« Reply #3 on: February 11, 2015, 08:51:29 PM »
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 $u(x,t) = \frac{1}{3}(cos(x + 3t) - 5cos(5x - 15t))$
« Last Edit: February 11, 2015, 08:53:40 PM by Aladdin Seaifan »

#### Bethany Garnett

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##### Re: HA2 problem 1
« Reply #4 on: February 11, 2015, 09:53:45 PM »
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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##### Re: HA2 problem 1
« Reply #5 on: February 12, 2015, 01:28:04 PM »
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.

#### Victor Ivrii

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