Toronto Math Forum
APM3462016F => APM346Tests => Q3 => Topic started 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(x3t) $
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,
$u = \begin{cases}\phi ( x + 3t ) && x > 3t \\\phi ( x + 3t )  \frac{7}{5} \phi (3t  x) + \frac{7}{5} \phi (0) && 0 < x < 3t \end{cases}$

Do we have to get the constant right? I don't remember u has to be continuous in the question.

Good job!

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) $

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!

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 ?

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

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?

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.

How do we discuss the reflected wave in this case