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APM346-2016F => APM346--Tests => Q3 => Topic started by: Roro Sihui Yap on October 13, 2016, 08:48:28 PM

Title: Q3
Post by: Roro Sihui Yap 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,
$u = \begin{cases}\phi ( x + 3t ) && x > 3t \\\phi ( x + 3t ) - \frac{7}{5} \phi (3t - x) + \frac{7}{5} \phi (0) && 0 < x < 3t \end{cases}$
Title: Re: Q3
Post by: Tianyi Zhang 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.
Title: Re: Q3
Post by: Victor Ivrii on October 13, 2016, 09:48:16 PM
Good job!
Title: Re: Q3
Post by: Roro Sihui Yap 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 $+\frac{7}{5} \phi (0)$
Title: Re: Q3
Post by: Victor Ivrii 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!
Title: Re: Q3
Post by: Roro Sihui Yap 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 ?
Title: Re: Q3
Post by: Victor Ivrii 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
Title: Re: Q3
Post by: ziyao hu 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?
Title: Re: Q3
Post by: Victor Ivrii 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.
Title: Re: Q3
Post by: zion on October 19, 2016, 12:55:52 PM
How do we discuss the reflected wave in this case