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I have the series

$\psi(r,\theta;p)=\sum_{n=0}^{\infty} a_n J_{\pi n/\Phi}(p r)\cos(\pi n \theta/\Phi)$

and the boundary conditions

$\psi(r,\pm\Phi;p)=\sum_{n=0}^{\infty} a_n J_{\pi n/\Phi}(p r)\cos(\pi n)=1.$

Here $J_\nu$ stands for a Bessel function and the coefficients $a_n$ are unknown. It can be readily found from the above BC at $r=0$ that

$a_0=1.$

A simple solution can be found for $\Phi=\pi/2$, namely

$a_0=1$ and $a_{n}=2\times(-1)^n$ for $n>1$.

However, I want to solve this problem for $\Phi=3\pi/4$. It is not even evident that a solution exists for $\Phi=3\pi/4$. Indeed, if one expands the Bessel functions into the Taylor series and equate to $0$ the coefficients before different powers of $x$ then one obtains that $a_n=0$ for all $n$. In any case, these problems seems to be ill-posed. I computed a numerical solution for this problem using different mehods (see e.g. Inverse problems for an asymptotic series which depends on a parameter) but still not sure that I did that well. Could someone direct me to a correct way?

Finally I need to compute the Laplace inverse for

$\Gamma(7/3)p^{-7/3}\psi(r,\theta;p)$

which is an electrostatic potential outside a rectangular corner of a homogeneous beam of charged particles provided that the so called Pierce electrodes compensate an electrostatic repulsion of the particles in the directions transversal to direction of the beam propagation.

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  • $\begingroup$ Are you interpreting the functions $f_n(z) = J_\nu(z)$, with $\nu=\pi n /\Phi$, as a complete system of eigenfunctions of a Sturm-Liouville type problem? If so, why not use the corresponding orthogonality relation for the $f_n(z)$ to decompose the function $1$ and obtain the coefficients $a_n$ directly? $\endgroup$
    – Igor Khavkine
    Commented Oct 24, 2013 at 23:37
  • $\begingroup$ In any case, there seems to be an orthogonality relation (dlmf.nist.gov/10.22.E55) for the system of functions $g_n(z) = J_{\nu + 2n + 1}(z)$. Your $f_n(z)$ don't fit the same pattern unless $\Phi = \pi/2$. But perhaps for your preferred $\Phi$ you can set the $a_n=0$ coefficients of those functions that don't fit the pattern and the use the orthogonality relation together with the integral identity (dlmf.nist.gov/10.22.E43). $\endgroup$
    – Igor Khavkine
    Commented Oct 24, 2013 at 23:41
  • $\begingroup$ @Igor Khavkine: Yes, $J_\nu n$ is a complete system of eigenfunctions of a Sturm-Liouville type problem. I tried to drop some terms as you proposed. Then $a_n$ can be found in analytic form, but the sum (in the boundary condition) does not converge to $1$. $\endgroup$
    – Igor Kotelnikov
    Commented Oct 25, 2013 at 7:08
  • $\begingroup$ If that is so, there should be an orthogonality relation like $\int w(z) f_n(z) f_m(z) = C_n \delta_{n,m}$ for some $w(z)$ determined by the Sturm-Liouville problem. It may be different than the one that I found at DLMF. Well, the $a_n$ can be obtained by replacing $f_m(z)$ in the integral by $1$. If the system of eigenfunctions $f_n(z)$ is complete, it would then be the unique solution. Would that method work, or do you know that it fails at some point? $\endgroup$
    – Igor Khavkine
    Commented Oct 25, 2013 at 11:49

1 Answer 1

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Let us just use your second identity, $$ J_0(pr)+\sum_{n=1}^\infty b_n J_{\pi n/ \phi} (pr)=1 \mbox{ for all }r\in(0,r_0) $$ Since $J_{\nu}(x)\approx \left(\frac{z}{2}\right)^{\nu}\frac{1}{\Gamma(\nu+1)}$, and $$ 1-J_0(x)=\sum_{k=1}^\infty \left(\frac{z^{k}}{2 (k!)}\right)^2(-1)^{k+1} $$ you obtain at first order $$ b_1 \left(\frac{pr}{2}\right)^{\pi/\phi}\frac{1}{\Gamma(\pi/\phi+1)} \approx \left(\frac{pr}{2}\right)^{2} $$
so if $\pi/\phi\neq 2$ there are no solutions. If $2\phi=\pi$, $b_1=2$ (and as you pointed out all other $b_i$ as well). Added after comments: from the growth of $J_{\nu}$ given above, you can see that this is also the answer provided $$ b_n < (C)^{\pi n/\phi}\Gamma(\frac{\pi n}{\phi}+1) $$ for some positive $C$, provided that $2Cr_0\leq 1$. Now if you want an approximate solution, that's a completely different question, but then you have to explain which type (in which norm) you want the approximation to hold...and forget uniqueness.

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  • $\begingroup$ Thank you, @user400001. Your arguments are quite reasonable, at least for a converging series. But this is not a sort of converging series as it is obtained from ill-posed problem. At least, it is possible to compute the coefficients $b_n$ which provide a sufficiently good approximation for the sum=1. $\endgroup$
    – Igor Kotelnikov
    Commented Oct 31, 2013 at 5:16

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