Bessel函数

在这篇文章中，我们将会罗列Bessel函数的一些基本性质。

A. Definition and Basic Properties

We define the Bessel function $J_{\nu}$ of order $\nu$ by its Poisson representation formula $$J_{\nu}(t) = \frac{(t/2)^{\nu}}{\Gamma(\nu + 1/2)\Gamma(1/2)}\int_{-1}^1e^{its}(1 - s^2)^{\nu}\frac{{\rm d}s}{\sqrt{1 - s^2}},$$ where ${\rm Re}\ \nu > -1/2$ and $t \geq 0$. Note that in this case $J_{\nu}(t)$ is a real number.

Here are some basic properties of Bessel functions.

1. (Recurrence formula)$$\frac{{\rm d}}{{\rm d}t}(t^{-\nu}J_{\nu}(t)) = -t^{-\nu}J_{\nu + 1}(t), \qquad {\rm Re} \ \nu > -\frac{1}{2}; $$
2. (Companion recurrence formula) $$\frac{{\rm d}}{{\rm d}t}(t^{\nu}J_{\nu}(t)) = t^{\nu}J_{\nu - 1}(t), \qquad {\rm Re} \ \nu > \frac{1}{2}; $$
3. $J_{\nu}(t)$ satisfies the Bessel equation $$t^2f''(t) + tf'(t) + (t^2 - \nu^2)f(t) = 0;$$
4. If $\nu \in \mathbb{Z}_+$, then we have the following identity, which was taken by Bessel as the definition of $J_{\nu}$ for integer $\nu$: $$J_{\nu}(t) = \frac{1}{2\pi}\int_0^{2\pi}e^{it\sin\theta}e^{-i\nu\theta} \,{\rm d}\theta =  \frac{1}{2\pi}\int_0^{2\pi}\cos(t\sin\theta - \nu\theta) \,{\rm d}\theta;$$
5. For ${\rm Re}\ \nu > -1/2$ we have the following identity: $$J_{\nu}(t) = \frac{(t/2)^{\nu}}{\Gamma(1/2)}\sum\limits_{j = 0}^{\infty}(-1)^j\frac{\Gamma(j + 1/2)}{\Gamma(j + \nu + 1)}\frac{t^{2j}}{(2j)!} = \sum\limits_{j = 0}^{\infty}\frac{(-1)^j}{j!}\frac{(t/2)^{2j + \nu}}{\Gamma(j + \nu + 1)}; $$
6. For ${\rm Re}\ \nu > 1/2$ we have the following identity: $$\frac{{\rm d}}{{\rm d}t}(J_{\nu}(t)) = \frac{1}{2}(J_{\nu - 1}(t) - J_{\nu + 1}(t)).$$

Remark. The property 1, 2, 3 and 4 still hold when $t \in \mathbb{C}$.

Proposition.  Let ${\rm Re} \ \mu > -1/2, {\rm Re} \ \nu > -1$ and $t > 0$. Then the following identity is valid: \begin{equation}\label{1}\int_0^1J_{\mu}(ts)s^{\mu + 1}(1 - s^2)^{\nu} \,{\rm d}s = \frac{\Gamma(\nu + 1)2^{\nu}}{t^{\nu + 1}}J_{\mu + \nu + 1}(t).\end{equation}

B. The Fourier Transform of Surface Measure on $\mathbb{S}^{n - 1}$

Let ${\rm d}\sigma$ denote the surface on $\mathbb{S}^{n - 1}$ for $n \geq 2$. Then the following is true: $$\widehat{{\rm d}\sigma}(\xi) = \int_{\mathbb{S}^{n - 1}}e^{-2\pi i\xi \cdot \theta} \,{\rm d}\theta = \frac{2\pi}{|\xi|^{\frac{n - 2}{2}}}J_{\frac{n - 2}{2}}(2\pi|\xi|).$$ Using polar coordinates, we can also obtain the Fourier transform of a radial function on $\mathbb{R}^n$: $$\widehat{f}(\xi) = \frac{2\pi}{|\xi|^{\frac{n - 2}{2}}}\int_0^{\infty}J_{\frac{n - 2}{2}}(2\pi r|\xi|)r^{\frac{n}{2}} \,{\rm d}r,$$ where $f(x) = f_0(|x|)$.

Example. Consider the radial function $f(x) = \chi_{B_1(0)}(x)$ on $\mathbb{R}^n$. It follows that $$(\chi_{B_1(0)})^{\wedge}(\xi) = \frac{2\pi}{|\xi|^{\frac{n - 2}{2}}}\int_0^1J_{\frac{n - 2}{2}}(2\pi|\xi|r)r^{\frac{n}{2}} \,{\rm d}r = \frac{J_{\frac{n}{2}}(2\pi|\xi|)}{|\xi|^{\frac{n}{2}}},$$ where we use identity \eqref{1}. More generally, for ${\rm Re}\ \lambda > -1$, let \begin{equation*}m_{\lambda}(\xi) = \begin{cases}(1 - |\xi|^2)^{\lambda} \quad &|\xi| \leq 1, \\ 0 \quad &|\xi| > 1. \end{cases}\end{equation*} Then $$m_{\lambda}^{\vee}(x) = \frac{\Gamma(\lambda + 1)}{\pi^{\lambda}}\frac{J_{\frac{n}{2} + \lambda}(2\pi|x|)}{|x|^{\frac{n}{2} + \lambda}}.$$

C. Asymtotics of Bessel Funtions

Let ${\rm Re}\ \nu > -1/2$. We have the following results:

\begin{equation*}J_{\nu}(t) = \begin{cases}\frac{t^{\nu}}{2^{\nu}\Gamma(\nu + 1)} + O(t^{{\rm Re}\ \nu + 1}) \qquad &t \rightarrow 0^+, \\ \sqrt{\frac{2}{t\pi}}\cos\left(t - \frac{\nu\pi}{2} - \frac{\pi}{4}\right) + O(t^{-\frac{3}{2}}) \qquad &t \rightarrow \infty. \\ \end{cases}\end{equation*} In particular, for fixed $\nu$, we have $J_{\nu}(t) = O(t^{-\frac{1}{2}})$ as $t \rightarrow \infty$.

 

Ref.  Grafakos, L. Classical Fourier Analysis.

苗长兴, 现代调和分析及其应用讲义.

 

标签：xi,frac,函数,Re,rm,pi,Bessel,nu
来源： https://www.cnblogs.com/tron-math/p/16663112.html