Toronto Math Forum
APM3462015S => APM346Home Assignments => HA2 => Topic started by: Thierry Serafin Nadeau on January 27, 2015, 08:57:40 PM

I was wondering whether there might be a typo in problem 2.2 as otherwise, we are asked to solve for v from its general solution but without being given any boundary or initial conditions.
No  just find the general solution. V.I.

A spherical wave is a solution of the threedimensional wave equation of the form $u(r, t)$, where r is the distance to the origin (the spherical coordinate). The wave equation takes the form
\begin{equation}
u_{tt} = c^2 \bigl(u_{rr}+\frac{2}{r}u_r\bigr)
\qquad\text{(â€œspherical wave equationâ€).}
\label{eqHA2.4}
\end{equation}
a. Change variables $v = ru$ to get the equation for $v$: $v_{tt} = c^2 v_{rr}$.
b. Solve for $v$ using
\begin{equation}
v = f(r+ct)+g(rct)
\label{eqHA2.5}
\end{equation}
and thereby solve the spherical wave equation.
c. Use
\begin{equation}
v(r,t)=\frac{1}{2}\bigl[
\phi (r+ct)+\phi (rct)\bigr]+\frac{1}{2c}\int_{rct}^{r+ct}\psi (s) \,ds
\label{eqHA2.6}
\end{equation}
with $\phi(r)=v(r,0)$, $\psi(r)=v_t(r,0)$ to solve it with initial conditions $u(r, 0) = \Phi (r)$, $u_t(r, 0) = \Psi(r)$.
4. Find the general form of solution $u$ to (\ref{eqHA2.4}) which is continuous as $r=0$.

But in that case it asks us to find the general solution of v by using the general solution of v, which doesn't make ant sense.
No, it asks to find the general solution $u$ of the original equation by using the general solution $v$ of the reduced equation. V.I.

(a): $v = ur$, then $u = \frac{v}{r}$ and
\begin{align}
&u_t = \frac{ v_t }{r},\notag \\
&u_{tt} = \frac{1}{r}{v_{tt}}\label{A}\\
&u_r =  \frac{v}{{{r^2}}} + \frac{{{v_r}}}{r},\label{B}\\
&u_{rr} = \frac{{2v}}{{{r^3}}}  \frac{{2{v_r}}}{{{r^2}}} + \frac{{{v_{rr}}}}{r}.\label{C}
\end{align}
Plugging (\ref{A})(\ref{C}) into the equation we have
$$
\frac{{{v_{tt}}}}{r} = {c^2}(\frac{{2v}}{{{r^3}}}  \frac{{2{v_r}}}{{{r^2}}} + \frac{{{v_{rr}}}}{r} + \frac{{2{v_r}}}{{{r^2}}}  \frac{{2v}}{{{r^3}}})
\frac{{{v_{tt}}}}{r} = {c^2}(\frac{{{v_{rr}}}}{r})$$
Hence, ${v_{tt}} = {c^2}{v_{rr}}$.
(b) From (a), known that ${v_{tt}} = {c^2}{v_{rr}}$:
$$v(x,t) = f(x + ct) + g(x  ct)$, $u =r^{1} \left[ f(r + ct) + g(r  ct) \right]$$
where $f$ and $g$ are arbitrary functions.
(c) $\phi (r) = v(r,0) = ru(r,0) = r\Phi ( r ) $, $\varphi (r) = {v_t}(r,0) = r{u_t}(r,0) = r\Psi (r )$,
$$
u(r,t) = \frac{1}{2}\left[ {(r + ct)\Phi (r + ct) + (r  ct)\Psi (r  ct)} \right] + \frac{1}{{2cr}}\int\limits_{r  ct}^{r + ct} {s\Phi (s)ds} .$$
(d) If this solution is continuous at $r=0$, $\lim_{r \to 0} u(r,t)$ should exist; so, $\lim_{r \to 0} \left[ {f(r + ct) + g(r  ct)} \right] = 0$. If $\lim_{r \to 0} \left[ {f(r + ct) + g(r  ct)} \right] \ne 0$, then $\lim_{r \to 0} u(r,t)=\infty$.
Therefore, $f(ct) + g(  ct) = 0$, that is, $f(ct) =  g(  ct)$. So $g(r  ct) =  f(ct  r)$,
$$u(r,t) = r^{1}\left[ f(r + ct)  f(ct  r)\right].$$
I fixed LaTeX. Please do not use the crapware you used to produce that code! It was produced by some aliens and is not for human use! V.I.

It is correct but virtually impossible to read. Fix it (see as I did with you Problem 3)