Bessel function recursion relation

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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
asked Apr 13, 2013
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.

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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
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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)

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