Obtain a first order DE for a Bessel function of order $n$

Question

An exercise in my textbook is the following: Let $J_n (z)$ denote the Bessel function of order n. Set $$\theta(z,t)=\sum_{n=-\infty}^{+\infty}J_{n}(z)t^n$$ and assume that the series converges.

Obtain a first order DE relating $\frac{\partial \theta}{\partial z}$ and $\theta(z,t)$.

I hope anybody can help, because I am totally stuck at this moment.

score 3 · Accepted Answer · answered Dec 23 '17 at 21:00

3

$$\frac{\partial \theta}{\partial z}=\sum_m J'_m(z)t^m$$ Now using the derivative properties for the Bessel function, one has \begin{equation} \frac{\partial \theta}{\partial z}=\frac{1}{2}\sum \left[J_{m-1}(z)-J_{m+1}(z)\right]t^m \end{equation} It remains to shift the index $m$ in the summations to obtain \begin{align} \frac{\partial \theta}{\partial z}=\frac{1}{2}\left[t\sum_m J_m(z)t^m-\frac{1}{t}\sum_m J_m(z)t^m\right]\\ &=\frac{1}{2}\left( t-\frac{1}{t} \right)\theta(z,t) \end{align} Actually, $\theta$ is a generating function for the Bessel functions.

answered Dec 23 '17 at 21:00

Paul Enta

14,113

This is really helpful! and can we say that: $\theta(z,t)= e^{(1/2)z(t-1/t)}$ – mathpieuler Dec 23 '17 at 21:13
Yes, this is the expression of a generating function for the Bessel functions. – Paul Enta Dec 23 '17 at 21:16
If I may add a second part. if we set $t=e^{i\theta}$ and also $t=-e^{i\theta}$, how can we say that for complex z: $sin(z sin \theta) = 2\sum_{n=0}^{+\infty}J_{2n+1}(z) sin(2n+1)\theta$ – mathpieuler Dec 23 '17 at 21:26
Why don't you mention $\frac{\partial}{\partial z} \theta(z,t) = f(t) \theta(z,t)$ means $\theta(z,t) = \theta(0,t) e^{z f(t)}$ – reuns Dec 23 '17 at 22:24
Sorry, that was clear for me, but the question was to construct an ODE. The remark on the generating function in the comment was made to give a reference to the asker. – Paul Enta Dec 23 '17 at 22:36

Obtain a first order DE for a Bessel function of order $n$

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