Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:, 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|cite|improve this question
Didn't an extremely similar question get migrated to Maths last month? – Willie Wong Feb 28 '13 at 0:14
up vote 3 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|cite|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


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.