### Author Topic: Q3  (Read 3844 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 $x > 3t$,
$$u = \phi ( x + 3t )$$

when $0 < x < 3t$,
$$u = \phi ( x + 3t ) - \frac{7}{5} \phi (3t - x) + c$$ 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