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# 10.5: Properties of Bessel functions

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Bessel functions have many interesting properties: \begin{aligned} J_{0}(0) &= 1,\\ J_{\nu}(x) &= 0\quad\text{(if \nu>0),}\\ J_{-n}(x) &= (-1)^{n }J_{n}(x),\\ \frac{d}{dx} \left[x^{-\nu}J_{\nu}(x) \right] &= -x^{-\nu}J_{\nu+1}(x),\\ \frac{d}{dx} \left[x^{\nu}J_{\nu}(x) \right] &= x^{\nu}J_{\nu-1}(x),\\ \frac{d}{dx} \left[J_{\nu}(x) \right] &=\frac{1}{2}\left[J_{\nu-1}(x)-J_{\nu+1}(x)\right],\\ x J_{\nu+1}(x) &= 2 \nu J_{\nu}(x) -x J_{\nu-1}(x),\\ \int x^{-\nu}J_{\nu+1}(x)\,dx &= -x^{-\nu}J_{\nu}(x)+C,\\ \int x^{\nu}J_{\nu-1}(x)\,dx &= x^{\nu}J_{\nu}(x)+C.\end{aligned}

Let me prove a few of these. First notice from the definition that $$J_{n}(x)$$ is even or odd if $$n$$ is even or odd,

$J_{n}(x) = \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!(n+k)!} \left(\frac{x}{2}\right)^{n+2k}.$

Substituting $$x=0$$ in the definition of the Bessel function gives $$0$$ if $$\nu >0$$, since in that case we have the sum of positive powers of $$0$$, which are all equally zero.

Let’s look at $$J_{-n}$$:

\begin{aligned} J_{-n}(x) &= \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!\Gamma(-n+k+1)!} \left(\frac{x}{2}\right)^{n+2k}\nonumber\\ &= \sum_{k=n}^{\infty}\frac{(-1)^{k}}{k!\Gamma(-n+k+1)!} \left(\frac{x}{2}\right)^{-n+2k}\nonumber\\ &= \sum_{l=0}^{\infty}\frac{(-1)^{l+n}}{(l+n)!l!} \left(\frac{x}{2}\right)^{n+2l}\nonumber\\ &= (-1)^{n} J_{n}(x).\end{aligned}

Here we have used the fact that since $$\Gamma(-l) = \pm \infty$$, $$1/\Gamma(-l) = 0$$ [this can also be proven by defining a recurrence relation for $$1/\Gamma(l)$$]. Furthermore we changed summation variables to $$l=-n+k$$.

The next one:

\begin{aligned} \frac{d}{dx} \left[x^{-\nu}J_{\nu}(x) \right] &= 2^{-\nu}\frac{d}{dx} \left\{ \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!\Gamma(\nu+k+1)} \left(\frac{x}{2}\right)^{2k} \right\} \nonumber\\&= 2^{-\nu} \sum_{k=1}^{\infty}\frac{(-1)^{k}}{(k-1)!\Gamma(\nu+k+1)} \left(\frac{x}{2}\right)^{2k-1} \nonumber\\&= -2^{-\nu} \sum_{l=0}^{\infty}\frac{(-1)^{l}}{(l)!\Gamma(\nu+l+2)} \left(\frac{x}{2}\right)^{2l+1} \nonumber\\&= -2^{-\nu} \sum_{l=0}^{\infty}\frac{(-1)^{l}}{(l)!\Gamma(\nu+1+l+1)} \left(\frac{x}{2}\right)^{2l+1} \nonumber\\&= -x^{-\nu} \sum_{l=0}^{\infty}\frac{(-1)^{l}}{(l)!\Gamma(\nu+1+l+1)} \left(\frac{x}{2}\right)^{2l+\nu+1} \nonumber\\&= -x^{-\nu}J_{\nu+1}(x).\end{aligned} Similarly \begin{aligned} \frac{d}{dx} \left[x^{\nu}J_{\nu}(x) \right] &=x^{\nu}J_{\nu-1}(x).\end{aligned}

The next relation can be obtained by evaluating the derivatives in the two equations above, and solving for $$J_{\nu}(x)$$:

\begin{aligned} x^{-\nu}J'_{\nu}(x)-\nu x^{-\nu-1}J_{\nu}(x)&= -x^{-\nu}J_{\nu+1}(x),\\ x^{\nu}J_{\nu}(x)+\nu x^{\nu-1}J_{\nu}(x)&=x^{\nu}J_{\nu-1}(x).\end{aligned}

Multiply the first equation by $$x^{\nu}$$ and the second one by $$x^{-\nu}$$ and add:

\begin{aligned} -2\nu \frac{1}{x}J_{\nu}(x) = -J_{\nu+1}(x)+J_{\nu-1}(x).\end{aligned}

After rearrangement of terms this leads to the desired expression.

Eliminating $$J_{\nu}$$ between the equations gives (same multiplication, take difference instead) \begin{aligned} 2 J'_{\nu}(x) &=J_{\nu+1}(x)+J_{\nu-1}(x).\end{aligned}

Integrating the differential relations leads to the integral relations.

Bessel function are an inexhaustible subject – there are always more useful properties than one knows. In mathematical physics one often uses specialist books.