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I'd like to know if I have got the following ideas right:

1) $f(r,\theta,t)=\sum\limits_{n=-\infty}^\infty \sum\limits_{k=1}^\infty a_{nk}J_n(j_{nk}r)\exp[in\theta-j^2_{nk}t]$ subjected to initial condition $f_0(r,\theta)$ is just $\int\limits_0^1 f_0(r,0)J_n(j_{nk}r)rdr \over{1\over2}J^2_{n+1}(j_{nk})$, where $J_n$ is the $n$th order Bessel function and $j_{nk}$ its $k$th zero;

2) In order that $F(r,\theta,t)={f(r,\theta,t)\over f_0(r,\theta)}$ is purely a function of time, we must have $f_0(r,\theta )$ to have the form $a_{nk}J_n(j_{nk}r)\exp[in\theta]$ i.e. only a single term so that all the time-independent stuff cancels;

3) If $f_0(r,\theta)=F_0(r,\theta)$ for all $r<1$, then... what is the form of $f(r,\theta,t)$? (I really need enlightenment on this bit. I am not even sure that I fully understand what $F_0(r,\theta)$ is.)

Thanks.

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Re 1: You can write $f_0(r,\theta)$ as $$ f_0(r,\theta) = \sum_{n=-\infty}^\infty e^{in\theta} b_n(r) \qquad\text{with}\qquad b_n(r)= \sum_{k=1}^\infty a_{nk} J_n(j_{nk} r) $$ The coefficients $b_n(r)$ can be written in terms of $f_0(r,\theta)$ according to $$ b_n(r) = \frac{1}{2\pi} \int_{-\pi}^\pi f_0(r,\theta) e^{-i n\theta} \mathrm{d}\theta $$ Then $a_{nk}$ is equal to $$ a_{nk} = \frac{\int_0^1 b_n(r) J_n(j_{nk} r) r \mathrm{d}r}{\frac{1}{2}J^2_{n+1}(j_{nk})} = \frac{\frac{1}{2\pi}\int_0^1 \int_{-\pi}^\pi f_{0}(r,\theta) e^{-i n\theta} J_n(j_{nk} r) r \ \mathrm{d}\theta\ \mathrm{d}r}{\frac{1}{2}J^2_{n+1}(j_{nk})/2} $$

Re 2: Since $J_{-n}=(-1)^n J_n$, we have $j_{-nk}=j_{nk}$. This means that if $f_0(r,\theta)$ is of the form $$a_{-nk} J_{-n}(j_{-nk} r)\exp(-in\theta) + a_{nk} J_{n}(j_{nk} r)\exp(in\theta) $$ the non-time dependent behaviour cancels as well.

Re 3: What is $F_0(r,\theta)$? In the case that $F_0(r,\theta) = F(r,\theta,0)$ we would have $f_0(r,\theta) = 1$. Then $b_n(r)=\delta_{n0}$ and $$ a_{nk} = \frac{2 \delta_{n0}}{J_{1}(j_{0k})j_{0k}} $$

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Thank you so much! You made my day :-) –  needhelp Nov 5 '11 at 14:19
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