# Toronto Math Forum

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

Title: Problem 2
Post 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.
$$\begin{equation*} \phi (r) = v(r, 0) = ru(r, 0) = r \Phi (r) \\ \psi (r) = v_t(r, 0) = ru_t(r, 0) = r \Psi (r) \\ u(r, t) = \frac{1}{2r} \left[(r+ct) \Phi (r+ct) + (r - ct) \Phi (r - ct) \right] + \frac{1}{2cr} \int_{r-ct}^{r+ct}s \Psi (s) ds \\ \end{equation*}$$

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 $0$.
Title: Re: Problem 2
Post by: Peishan Wang on October 01, 2012, 09:08:02 PM
Used a different way to do part (d)
Title: Re: Problem 2
Post by: Jinlong Fu on October 01, 2012, 09:55:17 PM
for part d), I calculated the general form of solution u(0,t)
Title: Re: Problem 2
Post by: Victor Ivrii on October 02, 2012, 07:13:01 AM
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/APM346-2011F-forum/index.php?topic=22.msg64#msg64 (http://weyl.math.toronto.edu:8888/APM346-2011F-forum/index.php?topic=22.msg64#msg64)

$\frac{1}{r}f (ct+r)$ is an expanding spherical wave

$\frac{1}{r}g (ct-r)$ 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

$\frac{1}{r}\bigl[f (ct+r)-f(ct-r0\bigr])$ both source and sink cancel one another at original 3D wave equation holds in the origin as well.