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Consider the second order wave equation $u_{tt}=u_{rr}+1/ru_{r}+1/r^2u_{\theta\theta}$ with the boundary data

  • $u(r, \theta, 0)=f(r)$
  • $u_t(r, \theta, 0)=0$
  • $u(1, \theta, t)=0.$ Assuming that $u(r, \theta, t)=R(r)T(t)$ (by radial symmetry), we can conclude that

$$T''/T=(R''+1/rR')/R=c.$$ My question is about determining the sign of $c$. Wikipedia says that since we expect $T$ to be periodic rather than decaying or growing exponentially, $c$ has to be a negative number. However I am trying to understand the following alternative argument given in Knapp's advanced analysis.

"The indicial equation is $s^2 = 0$. One solution is given by a power series in $r$, while another involves $\log r$. We discard the solution with the logarithm because it would represent a singularity at the middle of the drum. To get at the sign of $c$, we use the condition $R(1) = 0$ and argue as follows: Without loss of generality, $R(0)$ is positive. Suppose $c > 0$, and let $r_1 \le 1$ be the first value of $r > 0$ where $R(r_1 ) = 0$. From the equation $r^{−1} (r R' )' = c R$ and the inequality $R(r) > 0$ for $0 < r < r_1$ , we see that $r R'$ is strictly increasing for $0 < r < r_1$. Examining the power series expansion for $R(r)$, we see that $R' (0) = 0$. Thus $R' (r ) > 0$ for $0 < r < r_1$. But $R(0) > 0$ and $R(r_1 ) = 0$ imply, by the Mean Value Theorem, that $R' (r)< 0$ somewhere in between, and we have a contradiction."

I have not been able to understand the part where he says "Examining the power series expansion for $R(r)$, we see that $R' (0) = 0$." Can someone please explain why $R'(0)$ has to be zero?

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There's a simpler explanation of why $R'(0) = 0$:

You assumed your solution is radially symmetric, and performed a separation of variables. If $R'(0) \neq 0$ you have a cusp at the origin, where the function $u$ is in fact not differentiable. So you require $R'(0) = 0$.

The power series explanation is basically the same: you need

$$ R'' + R'/r = c R $$

and you assume that $R$ is finite at the origin. If $R'\neq 0$ you will have that $R'' $ is infinite, and so your power series expansion cannot be taken.

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  • $\begingroup$ The second explanation is perhaps what Knapp has in mind, but I don't see why he is referring to the power series expansion of $R$; maybe just to insist that $R$ is smooth enough and that $R''$ cannot be infinite... $\endgroup$
    – EPS
    Commented Aug 20, 2014 at 16:14
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    $\begingroup$ @sam: if you write $R = \sum_{n = 0}^\infty R_n r^n$ as a power series expansion, you have $R'' = \sum_0^\infty (n+2)(n+1) R_{n+2} r^n$ and $R'/r = \sum_{-1}^\infty (n+2) R_{n+2} r^{n}$. Matching the coefficients you get a recurrence relation for $R_{n+2}$ in terms of $R_{n}$, and you must have $R_1 = 0$ since it is the only term that contributes to the coefficient of $r^{-1}$. $\endgroup$ Commented Aug 21, 2014 at 9:19

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