Cylindrical wave

Question

I know that a wave dependent of the radius (cylindrical symmetry), has a good a approximations as $$u(r,t)=\frac{a}{\sqrt{r}}[f(x-vt)+f(x+vt)]$$ when $r$ is big. I would like to know how to deduce that approximation from the wave equation, which is this (after making symmetry simplifications): $$u_{tt}-v^2\left(u_{rr}+\frac{1}{r}u_r\right)=0$$

Proving that's a good approximation is easy (just plug it in the the equation), I want to know how to deduce that from the above equation.

I've been searching and I found this: http://vixra.org/abs/0908.0045, which actually solved me a couple of problems, but the way they do it looks a bit clumsy to me, saying for example that "assuming the function $g$ depends on $r$ so some terms just go away..."

Thanks in advance.

Answer 1

Use the following identity:

$$ r^{-\alpha} \partial^2_{rr} \left( r^\alpha f(r) \right) = f_{rr} + \frac{2\alpha}{r} f_r + \frac{\alpha(\alpha - 1)}{r^2} f $$

Now, by inspection and comparing the above equation to the cylindrical wave equation you have that

$$ u_{tt} - \nu^2(u_{rr} + \frac1r u_r) = u_{tt} - \nu^2 \left( \frac{1}{\sqrt{r}}\partial^2_{rr} \left[ \sqrt{r} u\right] + \frac{1}{4r^{5/2}} \sqrt{r} u\right) $$

So writing $U = \sqrt{r} u$ we have that

$$ U_{tt} - \nu^2 U_{rr}+ \frac{\nu^2}{4r^2} U = 0 $$

So that $U = \sqrt{r} u$ solves the 1 dimensional wave equation up to a term that decays quickly (as inverse square). Hence $u$ is approximated by $1/\sqrt{r}$ times a solution of the 1 dimensional wave equation when $r$ is large.