### Author Topic: HA2 problem 2  (Read 3316 times)

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##### HA2 problem 2
« 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.
« Last Edit: January 27, 2015, 09:28:05 PM by Victor Ivrii »

#### Victor Ivrii

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##### Re: HA2 problem 2
« Reply #1 on: January 27, 2015, 09:25:26 PM »
A spherical wave is a solution of the three-dimensional 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

u_{tt} = c^2 \bigl(u_{rr}+\frac{2}{r}u_r\bigr)
\label{eq-HA2.4}

a.  Change variables $v = ru$ to get the equation for $v$: $v_{tt} = c^2 v_{rr}$.

b.  Solve for $v$ using

v = f(r+ct)+g(r-ct)
\label{eq-HA2.5}

and thereby solve the spherical wave equation.

c.  Use

v(r,t)=\frac{1}{2}\bigl[
\phi (r+ct)+\phi (r-ct)\bigr]+\frac{1}{2c}\int_{r-ct}^{r+ct}\psi (s) \,ds
\label{eq-HA2.6}

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{eq-HA2.4}) which is continuous as $r=0$.
« Last Edit: January 27, 2015, 09:50:44 PM by Victor Ivrii »

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##### Re: HA2 problem 2
« Reply #2 on: January 27, 2015, 09:40:25 PM »
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.
« Last Edit: January 27, 2015, 09:53:40 PM by Victor Ivrii »

#### Yiyun Liu

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##### Re: HA2 problem 2
« Reply #3 on: January 29, 2015, 09:01:08 PM »
(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.
« Last Edit: January 31, 2015, 05:53:23 AM by Victor Ivrii »