# Bessel function recursion relation

+ 1 like - 0 dislike
41 views

I'm reading a paper and the following set of radial equations is derived:

$-i \lbrack \partial_r + \frac{1}{r} \left( \frac{1}{2} - \nu \right) \rbrack u(r) = \pm k v(r)$

$-i \lbrack \partial_r + \frac{1}{r} \left( \frac{1}{2} + \nu \right) \rbrack v(r) = \pm k u(r)$

Where $\nu$ is some constant whose meaning doesn't really matter here. The authors state that the left-hand sides of these equations are a version of the recursion relations for Bessel functions, so the solutions have the form:

$\left( \begin{array} &u(r) \\ v(r) \end{array} \right) = \left( \begin{array} &(i \epsilon)^{-1/2} J_{\epsilon (\nu - 1/2)}(kr) \\ \pm (i \epsilon)^{-1/2} J_{\epsilon (\nu + 1/2)}(kr) \end{array} \right)$

and $\epsilon = \pm 1$.

No derivation is provided and I don't understand this step. Could someone briefly outline the method used here or provide some helpful reading material? I haven't had any luck so far.

This post imported from StackExchange Mathematics at 2014-06-01 19:31 (UCT), posted by SE-user Calavera
Have you checked the book of Watson, "A Treatise on the theory of Bessel functions"?

This post imported from StackExchange Mathematics at 2014-06-01 19:31 (UCT), posted by SE-user Norbert Schuch
@NorbertSchuch Thanks, I'll see if I can find it. Part of the issue is that I'm not on my University campus for the next few days so I don't have access to the library.

This post imported from StackExchange Mathematics at 2014-06-01 19:31 (UCT), posted by SE-user Calavera
You might be interested in this.

This post imported from StackExchange Mathematics at 2014-06-01 19:31 (UCT), posted by SE-user Ｊ. Ｍ.

+ 0 like - 0 dislike

You can verify the correctness of the solutions by using the two recursion relations $$J_n(z) = \frac{z}{2n}(J_{n-1}(z)+J_{n+1}(z))$$ and $$\frac{\mathrm d}{\mathrm dz}J_n(z) = \frac{J_{n-1}(z)-J_{n+1}(z)}{2}$$ which you can find e.g. on Wikipedia.

This post imported from StackExchange Mathematics at 2014-06-01 19:31 (UCT), posted by SE-user Norbert Schuch
answered Apr 14, 2013 by (290 points)
Thanks, I'll give it a try.

This post imported from StackExchange Mathematics at 2014-06-01 19:31 (UCT), posted by SE-user Calavera
I get that the relative phase between $u$ and $v$ in the solution should be $\pm i$ rather than $\pm 1$, but except for that, it works out. (Indeed, using the fact that everything else can be chosen real, the relative phase must be $\pm i$.)

This post imported from StackExchange Mathematics at 2014-06-01 19:31 (UCT), posted by SE-user Norbert Schuch
+ 0 like - 0 dislike

It is in fact just a systems of linear first-order ODEs.

$\begin{cases}-i\left[\partial_r+\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)\right]u(r)=\pm kv(r)\\-i\left[\partial_r+\dfrac{1}{r}\left(\dfrac{1}{2}+\nu\right)\right]v(r)=\pm ku(r)\end{cases}$

$\begin{cases}\partial_ru(r)+\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)u(r)=\pm kiv(r)\\\partial_rv(r)+\dfrac{1}{r}\left(\dfrac{1}{2}+\nu\right)v(r)=\pm kiu(r)\end{cases}$

$\therefore\partial_{rr}u(r)+\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)\partial_ru(r)-\dfrac{1}{r^2}\left(\dfrac{1}{2}-\nu\right)u(r)=\pm ki\partial_rv(r)$

$\partial_{rr}u(r)+\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)\partial_ru(r)-\dfrac{1}{r^2}\left(\dfrac{1}{2}-\nu\right)u(r)=\pm ki\left(-\dfrac{1}{r}\left(\dfrac{1}{2}+\nu\right)v(r)\pm kiu(r)\right)$

$\partial_{rr}u(r)+\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)\partial_ru(r)-\dfrac{1}{r^2}\left(\dfrac{1}{2}-\nu\right)u(r)=\mp\dfrac{ki}{r}\left(\dfrac{1}{2}+\nu\right)v(r)-k^2u(r)$

$\partial_{rr}u(r)+\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)\partial_ru(r)-\dfrac{1}{r^2}\left(\dfrac{1}{2}-\nu\right)u(r)=-\dfrac{1}{r}\left(\dfrac{1}{2}+\nu\right)\left(\partial_ru(r)+\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)u(r)\right)-k^2u(r)$

$\partial_{rr}u(r)+\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)\partial_ru(r)-\dfrac{1}{r^2}\left(\dfrac{1}{2}-\nu\right)u(r)=-\dfrac{1}{r}\left(\dfrac{1}{2}+\nu\right)\partial_ru(r)-\dfrac{1}{r^2}\left(\dfrac{1}{4}-\nu^2\right)u(r)-k^2u(r)$

$\partial_{rr}u(r)+\dfrac{1}{r}\partial_ru(r)+\left(k^2-\dfrac{1}{r^2}\left(\nu^2-\nu+\dfrac{1}{4}\right)\right)u(r)=0$

$r^2\partial_{rr}u(r)+r\partial_ru(r)+\biggl(k^2r^2-\left(\nu-\dfrac{1}{2}\right)^2\biggr)u(r)=0$

$u(r)=\begin{cases}C_1J_{\nu-\frac{1}{2}}(kr)+C_2Y_{\nu-\frac{1}{2}}(kr)&\text{when}~\nu-\dfrac{1}{2}\text{is an integer}\\C_1J_{\nu-\frac{1}{2}}(kr)+C_2J_{\frac{1}{2}-\nu}(kr)&\text{when}~\nu-\dfrac{1}{2}\text{is not an integer}\end{cases}$

According to http://people.math.sfu.ca/~cbm/aands/page_361.htm,

$\partial_ru(r)=\begin{cases}-C_1\left(J_{\nu+\frac{1}{2}}(kr)-\dfrac{1}{r}\left(\nu-\dfrac{1}{2}\right)J_{\nu-\frac{1}{2}}(kr)\right)-C_2\left(Y_{\nu+\frac{1}{2}}(kr)-\dfrac{1}{r}\left(\nu-\dfrac{1}{2}\right)Y_{\nu-\frac{1}{2}}(kr)\right)&\text{when}~\nu-\dfrac{1}{2}\text{is an integer}\\-C_1\left(J_{\nu+\frac{1}{2}}(kr)-\dfrac{1}{r}\left(\nu-\dfrac{1}{2}\right)J_{\nu-\frac{1}{2}}(kr)\right)+C_2\left(J_{-\nu-\frac{1}{2}}(kr)-\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)J_{\frac{1}{2}-\nu}(kr)\right)&\text{when}~\nu-\dfrac{1}{2}\text{is not an integer}\end{cases}$

$\therefore v(r)=\begin{cases}\mp\dfrac{i}{k}\left(-C_1\left(J_{\nu+\frac{1}{2}}(kr)-\dfrac{1}{r}\left(\nu-\dfrac{1}{2}\right)J_{\nu-\frac{1}{2}}(kr)\right)-C_2\left(Y_{\nu+\frac{1}{2}}(kr)-\dfrac{1}{r}\left(\nu-\dfrac{1}{2}\right)Y_{\nu-\frac{1}{2}}(kr)\right)+\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)(C_1J_{\nu-\frac{1}{2}}(kr)+C_2Y_{\nu-\frac{1}{2}}(kr))\right)&\text{when}~\nu-\dfrac{1}{2}\text{is an integer}\\\mp\dfrac{i}{k}\left(-C_1\left(J_{\nu+\frac{1}{2}}(kr)-\dfrac{1}{r}\left(\nu-\dfrac{1}{2}\right)J_{\nu-\frac{1}{2}}(kr)\right)+C_2\left(J_{-\nu-\frac{1}{2}}(kr)-\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)J_{\frac{1}{2}-\nu}(kr)\right)+\dfrac{1}{r}\left(\dfrac{1}{2}-\nu\right)(C_1J_{\nu-\frac{1}{2}}(kr)+C_2J_{\frac{1}{2}-\nu}(kr))\right)&\text{when}~\nu-\dfrac{1}{2}\text{is not an integer}\end{cases}$

$v(r)=\begin{cases}\pm\dfrac{i}{k}\left(C_1J_{\nu+\frac{1}{2}}(kr)+C_2Y_{\nu+\frac{1}{2}}(kr)\right)&\text{when}~\nu-\dfrac{1}{2}\text{is an integer}\\\pm\dfrac{i}{k}\left(C_1J_{\nu+\frac{1}{2}}(kr)-C_2J_{-\nu-\frac{1}{2}}(kr)\right)&\text{when}~\nu-\dfrac{1}{2}\text{is not an integer}\end{cases}$

This post imported from StackExchange Mathematics at 2014-06-01 19:31 (UCT), posted by SE-user doraemonpaul
answered Apr 14, 2013 by (20 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOve$\varnothing$flowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.