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Q3

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Roro Sihui Yap:
\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, 
$u = \begin{cases}\phi ( x + 3t ) && x > 3t \\\phi ( x + 3t )  - \frac{7}{5} \phi (3t - x) + \frac{7}{5} \phi (0) && 0 < x < 3t \end{cases}$ 

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

Victor Ivrii:
Good job!

Roro Sihui Yap:
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:
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!

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