### Author Topic: HA4-P4  (Read 2082 times)

#### Victor Ivrii ##### HA4-P4
« on: September 28, 2015, 01:04:56 PM »

#### Zaihao Zhou

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• Karma: 0 ##### Re: HA4-P4
« Reply #1 on: October 03, 2015, 02:48:39 PM »
a)
\begin{equation} u_r = \frac{v_rr-v}{r^2} \end{equation}
\begin{equation} u_{rr} = \frac{v_{rr}r^2-2v_rr+2v}{r^3}\end{equation}
\begin{equation} u_{tt} = \frac{v_{tt}}{r} \end{equation}

Thus, the original equation can be written as \begin{equation} \frac{v_{tt}}{r} = c^2 [\frac{v_{rr}r^2-2v_rr+2v}{r^3} + \frac{2}{r}\frac{v_rr-v}{r^2}] \end{equation}
Then we get \begin{equation} v_{tt} = c^2v_{rr}\end{equation}
b) \begin{equation} u = \frac{1}{r}[f(r+ct)+g(r-ct)] \end{equation}
c)\begin{equation}v_{t=0} = ru_{t=0} = \phi(r) \end{equation}, thus \begin{equation} \phi(r) = r\Phi(r)\end{equation}
similarly, \begin{equation} \psi(r) = r\Psi(r)\end{equation}
Substitute, we get\begin{equation} u = \frac{1}{2r}[(r+ct)\Phi(r+ct)+(r-ct)\Phi(r-ct)]+\frac{1}{2cr}\int^{r+ct}_{r-ct}s\Psi(s)ds \end{equation}

d) not sure I understand the question. I tried to use L'Hopital rule to determine the limit of u, this seems to be well defined as long as \begin{equation} \Phi'(x)\end{equation} is well defined. However u(0,t) is infinite. That means \begin{equation} \Phi'(x)\end{equation} should be infinite? Not sure if I'm doing the right thing.

#### Yumeng Wang

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« Reply #2 on: October 04, 2015, 03:15:05 PM »
For (d). The following is my answer.

Given u is continuous when r=0, which means limrâ†’0 u(r,t) exists.
Then limrâ†’0 [f(r+ct) + g(r-ct)] = 0. Because otherwise, limrâ†’0 [f(r+ct) + g(r-ct)] â‰  0 implies limrâ†’0 u(r,t) tends to be infinity.

So f(ct) + g(-ct) = 0
f(ct) = - g(-ct)
-f(x) = g(-x)
So g(r-ct) = -f(ct-r). As a result, u = r-1 [f(r+ct) - f(ct-r)]

#### Victor Ivrii ##### Re: HA4-P4
« Reply #3 on: October 04, 2015, 04:24:35 PM »
You did it collectively

Yumeng Wangâ€”learn LaTeX and avoid html tags for math