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The propagation of sound waves produced by a moving sphere of radius $a$ in an uniform, infinite fluid, is given by the following equation: $$ r^2 \frac{\partial^2{u}}{\partial t^2} = c^2 \frac{\partial}{\partial r} \left (r^2 \frac{\partial u}{\partial r} \right) \qquad \qquad (1) $$

Where $t$ represents time, $r$ the distance from the center of the sphere and $c$ the wave propagation speed. The following conditions are known:

$$ u(r,0) = \frac{\partial u(r,0)}{\partial t} = 0 \qquad \qquad (2)$$ and $$\frac{\partial u(a,t)}{\partial r} = f(t) \qquad \qquad (3)$$ and its also known that $u\to 0$ when $r\to \infty$.


The first step of the solution is to consider a variable $v = ru$ and, from that, rewrite the first equation as

$$ \frac{\partial^2 v}{\partial t^2} = c^2 \frac{\partial^2 v}{\partial r^2} \qquad \qquad (4) $$

then using Laplace's Transform...

I could not see how this rewriting of the first equation is valid because if $v=ru$ and assuming $$\frac{\partial r}{\partial t} = \frac{\partial t}{\partial r} = 0 \qquad \qquad (5)$$

then

$$\frac{\partial v^2}{\partial t^2} = r \frac{\partial u^2}{\partial t^2} \qquad \qquad (6)$$

and

$$c^2 \frac{\partial v^2}{\partial r^2} = c^2 \frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r} + u \right) = c^2 \left( r \frac{\partial^2 u}{\partial r^2} + 2 \frac{\partial u}{\partial r} \right ) \qquad \qquad (7)$$


I tried to consider $v = r^2u$ and got:

$$\frac{\partial v^2}{\partial t^2} = r^2 \frac{\partial u^2}{\partial t^2} \qquad \qquad (8)$$

and

$$c^2 \frac{\partial v^2}{\partial r^2} = c^2 \frac{\partial}{\partial r} \left( r^2 \frac{\partial u}{\partial r} + 2ru \right) = c^2 \left( r^2 \frac{\partial^2 u}{\partial r^2} + 2r + 2u + 2r \frac{\partial u}{\partial r} \right ) \qquad \qquad (9)$$

The equality of the first and fourth equations does not hold yet. Am I missing something?

1 Answers 1

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When you did $v=ru$, you have $ru_{tt} = c^2( ru_{rr}+2u_r)$. Multiply both sides by $r$ to get $r^2u_{tt}=c^2(r^2u_{rr}+2ru_{r})$. Note that, by product rule, $(r^2u_r)_r = r^2u_{rr}+2ru_{r}$, and so you've already got the desired equality, $r^2u_{tt}=c^2(r^2u_r)_r$.

(Note the subscript notation is just a shorthand, e.g. $u_{r} = \frac{\partial u}{\partial r}$.)

  • 0
    So, I just needed to multiply (6) and (7) by $r$.2017-02-24
  • 0
    Basically, yeah. That and noting that r*(7) is the same as the RHS of (1).2017-02-27