Toronto Math Forum
APM3462015F => APM346Home Assignments => HA3 => Topic started by: Victor Ivrii on September 28, 2015, 01:04:56 PM

http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter2/S2.3.P.html#problem2.3.P.4 (http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter2/S2.3.P.html#problem2.3.P.4)

a)
\begin{equation} u_r = \frac{v_rrv}{r^2} \end{equation}
\begin{equation} u_{rr} = \frac{v_{rr}r^22v_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^22v_rr+2v}{r^3} + \frac{2}{r}\frac{v_rrv}{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(rct)] \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)+(rct)\Phi(rct)]+\frac{1}{2cr}\int^{r+ct}_{rct}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.

For (d). The following is my answer.
Given u is continuous when r=0, which means lim_{râ†’0} u(r,t) exists.
Then lim_{râ†’0} [f(r+ct) + g(rct)] = 0. Because otherwise, lim_{râ†’0} [f(r+ct) + g(rct)] â‰ 0 implies lim_{râ†’0} u(r,t) tends to be infinity.
So f(ct) + g(ct) = 0
f(ct) =  g(ct)
f(x) = g(x)
So g(rct) = f(ctr). As a result, u = r^{1} [f(r+ct)  f(ctr)]

You did it collectively
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