Toronto Math Forum
APM3462012 => APM346 Math => Home Assignment 2 => Topic started by: Zarak Mahmud on October 01, 2012, 09:01:03 PM

Part (a): Make the variable change and use product rule to find $u_{rr}$, $u_r$, and $u_{tt}$.
$$
\begin{equation*}
u = vr \\
\frac{\partial u}{\partial t} = \frac{1}{r}\frac{\partial v}{\partial t} \\
\frac{\partial^2 u}{\partial t^2} = \frac{1}{r} \frac{\partial^2 v}{\partial t^2} \\
u_{tt} = \frac{1}{r}v_{tt} \\
\frac{\partial u}{\partial r} = \frac{\partial}{\partial r}\frac{v}{r} = \frac{1}{r}\frac{\partial v}{\partial r}  \frac{v}{r^2}\\
\frac{\partial^2 u}{\partial r^2} = \frac{\partial}{\partial r}\frac{\partial u}{\partial v} \\
= \frac{\partial}{\partial r} \left(\frac{1}{r}\frac{\partial v}{\partial r}  \frac{v}{r^2} \right) \\
= \frac{1}{r^2}\frac{\partial v}{\partial r} + \frac{1}{r}\frac{\partial^2 v}{\partial r^2}  \frac{1}{r^2} + \frac{2v}{r^3} \\
u_{rr} + \frac{2}{r}u_r = \frac{v_{rr}}{r} \\
u_{tt} = \frac{v_{tt}}{r} \\
\frac{v_{tt}}{r} = c^2\frac{v_{rr}}{r} \\
v_{tt} = c^2v_{rr}
\end{equation*}
$$
Part (b):
$$\begin{equation*}
v_{tt} = c^2v_{rr} \\
u(r, t) = \frac{f(r + ct) + g(r  ct)}{r} \\
\end{equation*}$$
Part (c): Take initial conditions with change of variable and then plug into D'Alambert formula.
$$
\begin{equation*}
\phi (r) = v(r, 0) = ru(r, 0) = r \Phi (r) \\
\psi (r) = v_t(r, 0) = ru_t(r, 0) = r \Psi (r) \\
u(r, t) = \frac{1}{2r} \left[(r+ct) \Phi (r+ct) + (r  ct) \Phi (r  ct) \right] + \frac{1}{2cr} \int_{rct}^{r+ct}s \Psi (s) ds \\
\end{equation*}
$$
Part (d): For function to be continuous at $r = 0$, we must have
$$
\begin{equation*}
\lim\limits_{r \to 0} \left[(r+ct) \Phi (r+ct) + (r  ct) \Phi (r  ct) \right] = 0 \\
ct \lim\limits_{r \to 0} \left[ \Phi (r+ct)  \Phi (r  ct) \right] = 0 \\
\Phi (ct)  \Phi (ct) = 0 \\
\implies \Phi (ct) = \Phi(ct)
\end{equation*}
$$
Thus, $\Phi$ is an even function. Similarly, $\Psi$ must be odd since the integral of an odd function between symmetric bounds, i.e., $[ct, ct]$ is equal to $0$.

Used a different way to do part (d)

for part d), I calculated the general form of solution u(0,t)

MZ solution is OK except (d), PW solution of (d) is correct
There are my comments in 1 y.a. forum
http://weyl.math.toronto.edu:8888/APM3462011Fforum/index.php?topic=22.msg64#msg64 (http://weyl.math.toronto.edu:8888/APM3462011Fforum/index.php?topic=22.msg64#msg64)
$\frac{1}{r}f (ct+r)$ is an expanding spherical wave
$\frac{1}{r}g (ctr)$ is a collapsing spherical wave (note my slightly different notations)
however both of them violate an original 3D wave equation as $r=0$ as an expanding spherical wave requires a source and a collapsing spherical wave require a sink and only for
$\frac{1}{r}\bigl[f (ct+r)f(ctr0\bigr])$ both source and sink cancel one another at original 3D wave equation holds in the origin as well.