Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 think I am stuck in solving this problem. It involves a wave equation in a circular membrane, so polar coordinates must be used:

$u_{tt}=c²(u_{rr}+{1\over r}u_r+{1\over r²}u_{\theta\theta})$, r<1, t>0




where $u_{tt}={\delta²u(r,\theta,t)\over\delta t²}$, being r the radius and $\theta$ the angle. The only extra assumption I can make is that $\Theta(\theta)=\Theta(\theta+2\pi)$.

By separation of variables, I get $u_t(r,\theta,t)=R(r)\Theta(\theta)T(t)$. The only extra assumption I can make is that $\Theta(\theta)=\Theta(\theta+2\pi)$. By substituting in the original PDE:

${T''\over c²T}={R''\over R}+{R'\over rR}+{\Theta''\over r²\Theta}$

And now I can separate:

${\Theta''\over \Theta}=r²{T''\over c²T}-r²{R''\over R}-r{R'\over R}=-\lambda$

$r²{R''\over R}+r{R'\over R}=r²{T''\over c²T}+\lambda=-\mu$

${T''\over c²T}=-{(\mu+\lambda) \over r²}$

So it's time to solve for the angle eigenvalue problem. I get these solutions, which I assume correct as I got them in previous exercises:

${\lambda_0=1; \Theta_0=1}$

${\lambda_n=n²; \Theta_n=A\cos(n\theta)+B\sin(n\theta})$

So now, it's time to solve for the radius. I can't find why, according to my class notes, I must solve this equation:

${r²R''+rR'+(\mu r²-\lambda)}=0$

I don't know how to get to that equation from the previous steps. This may be really simple but, after some time looking into it, I feel like my head is totally biased and I won't find the mistake...

Thank you very much for reading.

share|cite|improve this question
The more accurate procedure can be found in…. – doraemonpaul Aug 16 '12 at 0:01
Hi @doraemonpaul, I alreade read that before posting but, even understanding it, I can't find why the solution in my notes works. In fact, when I try to do it I end up with something more similar to wikipedia's. Even so, I can't find the way to the radius equation here, while they should be equivalent. – Serge Aug 16 '12 at 8:16

Your Answer


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

Browse other questions tagged or ask your own question.