6
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • $\begingroup$ 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
    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
    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
    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
    Oct 25, 2013 at 11:49

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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
    Oct 31, 2013 at 5:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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