Author Topic: Q3  (Read 676 times)

Roro Sihui Yap

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Q3
« on: October 13, 2016, 08:48:28 PM »
\begin{align*}
& u_{tt}-9u_{xx}=0, &&&t>0, x>0,  \\
&u|_{t=0}= \phi (x),   &&u_t|_{t=0}= 3\phi'(x) &x>0, \\
&(u_x+2u_{t})|_{x=0}=0,  &&&t>0
\end{align*}

$ u = f(x+3t) + g(x-3t) $

From $u|_{t=0}= \phi (x)$, we get $f(x) + g(x) = \phi (x)$
From $u_t|_{t=0}= 3\phi'(x)$, we get $3f'(x) - 3g'(x) = 3\phi'(x)$, and thus $f(x) - g(x) = \phi (x) - \phi (0) $

Solving the equations, $ f(x) = \phi (x) - \frac{\phi (0)}{2}$ and $g(x) = \frac{\phi (0)}{2}$ for $x>0$ only

From $(u_x+2u_{t})|_{x=0}=0$, we get $f'(3t) + g'(-3t) + 6f'(3t) - 6g'(-3t)= 0 $
let $x = -3t$, since $t > 0$, we have $x < 0$
$7f'(-x) - 5g'(x) = 0 $
$-7f(-x) - 5g(x) = -k$ where k is some constant
$g(x) = \frac{k}{5} - \frac{7\phi(-x)}{5} +  \frac{7\phi (0)}{10} $ for $x < 0 $

when $ x > 3t$,
\begin{equation}u = \phi ( x + 3t ) \end{equation}

when $ 0 < x < 3t$,
\begin{equation}u = \phi ( x + 3t )  - \frac{7}{5} \phi (3t - x) + c\end{equation} where c is some constant

If we want the function to be continuous, as $ x \rightarrow 3t $, both of the above functions have to be equal.
when $x = 3t$,
(1) $u = \phi (6t)$                     
(2) $u = \phi (6t) - \frac{7}{5} \phi (0) + c $
In order for them to be equal $c = \frac{7}{5} \phi (0)  $

Thus, 
$u = \begin{cases}\phi ( x + 3t ) && x > 3t \\\phi ( x + 3t )  - \frac{7}{5} \phi (3t - x) + \frac{7}{5} \phi (0) && 0 < x < 3t \end{cases}$ 
« Last Edit: October 13, 2016, 09:02:19 PM by Roro Sihui Yap »

Tianyi Zhang

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Re: Q3
« Reply #1 on: October 13, 2016, 09:39:13 PM »
Do we have to get the constant right? I don't remember u has to be continuous in the question.
« Last Edit: October 13, 2016, 09:54:37 PM by Tianyi Zhang »

Victor Ivrii

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Re: Q3
« Reply #2 on: October 13, 2016, 09:48:16 PM »
Good job!

Roro Sihui Yap

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Re: Q3
« Reply #3 on: October 13, 2016, 10:35:12 PM »
Even if we do not consider u being continuous at the line $x = 3t $
Since $u|_{t=0}= \phi (x)$, then $u(0,0) = \phi (0)$
If we do not have the constant, $u(0,0) = \phi ( 0)  - \frac{7}{5} \phi (0) \neq \phi (0)$
we need the constant $+\frac{7}{5} \phi (0) $ 

Victor Ivrii

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Re: Q3
« Reply #4 on: October 14, 2016, 06:27:23 AM »
Roro, if we do not assume $u$ to be continuous, then $u(0,0)$ would not be defined , so we really need a continuity condition to define a constant!

Roro Sihui Yap

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Re: Q3
« Reply #5 on: October 14, 2016, 09:18:33 AM »
In an exam, should we always assume that U is continuous ?
or are we allowed to drop the constant term if it is not stated that U is continuous ?

Victor Ivrii

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Re: Q3
« Reply #6 on: October 14, 2016, 09:32:39 AM »
If the boundary condition was $u|_{x=0}=k(t)$ then there would be no integration, thus no constant and in general no continuity.

Otherwise yes: you need to choose a correct constant but if so, it will be explicitly said so

ziyao hu

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Re: Q3
« Reply #7 on: October 17, 2016, 12:49:32 PM »
If the boundary condition was $u|_{x=0}=k(t)$ then there would be no integration, thus no constant and in general no continuity.

Otherwise yes: you need to choose a correct constant but if so, it will be explicitly said so

Does it mean that if the boundary condition was $u|_{x=0}=k(t)$ then there would be in general no continuity. We do not necessarily need to make it continuous.
But in this problem, we need to select constant to make it continuous?

Victor Ivrii

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Re: Q3
« Reply #8 on: October 17, 2016, 01:15:22 PM »

Does it mean that if the boundary condition was $u|_{x=0}=k(t)$ then there would be in general no continuity.
Indeed: assuming that $k,g$ are continuous,  if $u|_{x=0}=k(t)$ and $u|_{t=0}=g(x)$, $u_t|_{t=0}=h(x)$ then $u(x,t)$ is continuous iff $g(0)=k(0)$. There are more conditions to make it continuously differentiable $k'(0)=g(0)$, twice continuously differentiable and so on.

On the other hand, if $u_x|_{x=0}=k(t)$, then the correct choice of the constant makes $u(x,t)$ continuous, but it make it continuously differentiable we need $k(0)=g'(0)$ and so on.

zion

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Re: Q3
« Reply #9 on: October 19, 2016, 12:55:52 PM »
How do we discuss the reflected wave in this case