Take the 2-minute tour ×

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.

share|improve this question
Didn't an extremely similar question get migrated to Maths last month? –  Willie Wong Feb 28 '13 at 0:14

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
Thank you very much, that's what I had tried, to try to find a variable change of the form $u=vr^\alpha$, but I didn't know how to justify a change of that form and not other. –  MyUserIsThis Feb 28 '13 at 0:31
One can justify this by using a little bit of differential geometry. The Laplacian term in the wave equation, using the curvilinear coordinate expression of the Laplace-Beltrami operator can be written in spherical/cylindrical symmetry in the form $r^{-\beta} \partial_r(r^{\beta} \partial_r u))$, which contains a term that looks like the logarithmic derivative $D \log f = f^{-1} Df$, and differs from the LHS of the identity I wrote down in where the "inner" $\partial_r$ sits. –  Willie Wong Feb 28 '13 at 0:41
Ok thank you, that's good enough –  MyUserIsThis Feb 28 '13 at 9:19

Your Answer


