HA4-P3

[Victor Ivrii]

Problem 7: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.7

Reflects new numeration of the problems

[Bruce Wu]

Below is what I have so far. Please let me know if I am correct before I continue.
a) If $v>c$, then $x>ct$. In this region both $\phi(x+ct)$ and $\psi(x-ct)$ are fully defined. We do not need any boundary conditions, which is condition 1.
If $-c<v<c$, then $-ct<x<ct$. Here $\phi$ is fully defined but we need to define $\psi$ for negative arguments. Therefore we need 1 boundary condition, either 2. or 3. (need to work out which one)
If $v<-c$, then $x<-ct$. We need to define both $\phi$ and $\psi$ for negative arguments. We need 2 boundary conditions, so condition 4.

[Victor Ivrii]

Right

[Zaihao Zhou]

Sorry I'm confused about what form of boundary conditions should take.

Why do we need here condition Ux=vt but not Ux=0? If x = vt means "location of the wave is its velocity times t" then why don't we use Ux=ct boundary condition for cases without v mentioned? Shouldn't c be the velocity? If we have v, what is c?

Could someone or professor please explain? Thanks!

[Victor Ivrii]

Fei Fan Wu provided almost full explanations. Solution is given by $u=\phi(x+ct)+\psi(x-ct)$ and initial data provide us with $\phi(x)$ and $\psi(x)$ as $x>0$. So $\phi(x+ct)$ is defined as $x+ct>0$ and $\psi(x-ct)$ is defined as $x-ct>0$.

a) $v>c$. Then $x+ct > x-ct >0$ as $x>vt$ ($t>0$).  completely defined!

b) $c>v> -c$.  Then $x+ct$ as $x>vt$ but $x-ct$ could be less than $0$ so we need to find $\psi(x)$ for negative $x$. For this we need 1 boundary condition. Plug $u$ to it:
\begin{equation}
(\alpha + \beta c) \phi' ((v+c)t) + (\alpha - \beta c )\psi' ((v-c)t) =0,\qquad t>0.
\end{equation}
We can always find $\psi$ out of it except one special case which should be a mentioned.


c) $v<-c$. Then both $x-ct$ and $x+ct$ could be negative and we need to find $\phi(x)$ and $\psi(x)$ as $x<0$. We need two boundary conditions. Plug $u$ into boundary conditions:
\begin{align}
&\phi((v+c)t) + \phi'(v-c)t)=0,\\
$c\phi '( (v+c)t)-c\psi'((v-c)t)=0.
\end{align}
 Integrating the second one (never mind, constant $C$ could be safely taken $0$) and comparing with the first one we find that $\phi(x)=\psi(x)=0$ as $x<0$.

QED

[Zaihao Zhou]

Fei Fan Wu provided almost full explanations. Solution is given by $u=\phi(x+ct)+\psi(x-ct)$ and initial data provide us with $\phi(x)$ and $\psi(x)$ as $x>0$. So $\phi(x+ct)$ is defined as $x+ct>0$ and $\psi(x-ct)$ is defined as $x-ct>0$.

a) $v>c$. Then $x+ct > x-ct >0$ as $x>vt$ ($t>0$).  completely defined!

b) $c>v> -c$.  Then $x+ct$ as $x>vt$ but $x-ct$ could be less than $0$ so we need to find $\psi(x)$ for negative $x$. For this we need 1 boundary condition. Plug $u$ to it:
\begin{equation}
(\alpha + \beta c) \phi' ((v+c)t) + (\alpha - \beta c )\psi' ((v-c)t) =0,\qquad t>0.
\end{equation}
We can always find $\psi$ out of it except one special case which should be a mentioned.


c) $v<-c$. Then both $x-ct$ and $x+ct$ could be negative and we need to find $\phi(x)$ and $\psi(x)$ as $x<0$. We need two boundary conditions. Plug $u$ into boundary conditions:
\begin{align}
&\phi((v+c)t) + \phi'(v-c)t)=0,\\
$c\phi '( (v+c)t)-c\psi'((v-c)t)=0.
\end{align}
 Integrating the second one (never mind, constant $C$ could be safely taken $0$) and comparing with the first one we find that $\phi(x)=\psi(x)=0$ as $x<0$.

QED

Thanks professor but I understand the process of doing this but not why we need initial condition Ux=vt here but not Ux=0, or similarly when with problems without v, why not use boundary condition Ux=ct?

[Victor Ivrii]

Thanks professor but I understand the process of doing this but not why we need initial condition Ux=vt here but not Ux=0, or similarly when with problems without v, why not use boundary condition Ux=ct?

This is completely incomprehensible. What do you really mean by this charade?

[Zaihao Zhou]

Thanks professor but I understand the process of doing this but not why we need initial condition Ux=vt here but not Ux=0, or similarly when with problems without v, why not use boundary condition Ux=ct?

This is completely incomprehensible. What do you really mean by this charade?

My questions is about boundary condition. For the general form of 1D wave, there is no v present. We used boundary condition\begin{equation} u_{x=0} = 0 \end{equation}. Here you are asking v. I understand that we need conditions, but why here we need:\begin{equation} u_{x=vt} = something \end{equation} but not \begin{equation} u_{x=ct} = something \end{equation} or \begin{equation} u_{x=0} = something \end{equation}

And what is v? where does it come from and how it relates to c here?

Hope that clarifies my questions. Sorry for the confusion.

[Emily Deibert]

From my understanding v is just some constant that could have any value. We consider what conditions will be necessary for the various cases of v in relation to c (as mentioned in the problem).

[Victor Ivrii]

The domain is $\{t>0, x>vt\}$ and the b.c. is on its border, not inside!

This problem could be interpreted in the following way: we have an air pipe but its left end is moving with a constant speed $v$. If this movement is subsonic $|v|<c$ then there should be one condition on it, if the movement is supersonic $|v|>c$ then there should be either $2$ conditions as this end goes to the left or none as it goes to the right. It is a toy-model for 3D or 2D theory where we consider airflow which has different speeds at different points and the most difficult case is transonic when at some points it is subsonic but in others supersonic.