BSJ Bessel J

BSJ.1 Introduction

Let x be a complex variable of ℂ \ {0, ∞} and let ν denote a parameter (independent of x).The function Bessel J (noted J_{ν}) is defined by the following second order differential equation

\begin{aligned}& (x^{2} − ν^{2})y(x) + x\frac{∂y(x)}{∂x} + x^{2} \frac{∂^{2}y(x)}{∂x^{2}} & = 0. \\ (BSJ.1.1) \end{aligned}

Although 0 is a singularity of BSJ.1.1, the initial conditions can be given by

\begin{aligned}& \frac{∂ \frac{J_{ν}(x)}{x^{ν}}}{∂x} & = \frac{1}{Γ(ν + 1)2^{ν}} . \\ (BSJ.1.2) \end{aligned}

The formulae of this document are valid for −2ν ∉ ℤ. Related functions: Hankel H1,Hankel H2,Bessel Y

BSJ.2 Series and asymptotic expansions

BSJ.2.1 Asymptotic expansion at 0. BSJ.2.1.1 First terms.

J_{ν}(x) ≈ x^{ν}(\frac{1}{Γ(ν + 1)2^{ν}} − \frac{x^{2}}{4Γ(ν + 1)2^{ν}(ν + 1)}+

\frac{x^{4}}{32Γ(ν + 1)2^{ν}(ν + 1)(ν + 2)}−

\frac{x^{6}}{384Γ(ν + 1)2^{ν}(ν + 1)(ν + 2)(ν + 3)} …).

(BSJ.2.1.1.1)

BSJ.2.1.2 General form.

\begin{aligned}& J_{ν}(x) ≈ x^{ν} ∑_{n=0}^{∞} u(n)x^{n}. \\ (BSJ.2.1.2.1) \end{aligned}

The coefficients u(n) satisfy the recurrence

\begin{aligned}& u(n)(−ν^{2} + (ν + n)^{2}) + u(n − 2) & = 0. \\ (BSJ.2.1.2.2) \end{aligned}

Initial conditions of BSJ.2.1.2.2 are given by

\begin{aligned}u(1) & = 0, \\ u(0) & = \frac{1}{Γ(ν + 1)2^{ν}} . \end{aligned} ; (BSJ.2.1.2.3)

The recurrence BSJ.2.1.2.2 has the closed form solution

\begin{aligned}u(2n + 1) & = 0, \\ u(2n) & = \frac{(−1)^{n}}{2^{ν}4^{n}Γ(n + 1)Γ(n + ν + 1)}. \end{aligned} ; (BSJ.2.1.2.4)

BSJ.2.2 Asymptotic expansion at ∞.

BSJ.2.2.1 First terms.

J_{ν}(x) ≈ e^{(− \frac{RootOf_{ξ,2}(1+ξ^{2})}{x})} \sqrt{x}y_{0}(x)+

e^{(− \frac{RootOf_{ξ,1}(1+ξ^{2})}{x})} \sqrt{x}y_{1}(x),

where

y_{0}(x) = \frac{\sqrt{2} e^{(− \frac{i}{4} π(2ν+1))}}{2\sqrt{π}} − \frac{−(4ν^{2} − 1)\sqrt{2} e^{(− \frac{i}{4} π(2ν+1))} x}{16\sqrt{π} RootOf_{ξ,2}(1 + ξ^{2})}+

\frac{(4ν^{2} − 9)(4ν^{2} − 1)\sqrt{2} e^{(− \frac{i}{4} π(2ν+1))} x^{2}}{256\sqrt{π} RootOf_{ξ,2}(1 + ξ^{2})^{2}} −

\frac{−(4ν^{2} − 25)(4ν^{2} − 9)(4ν^{2} − 1)\sqrt{2} e^{(− \frac{i}{4} π(2ν+1))} x^{3}}{6144\sqrt{π} RootOf_{ξ,2}(1 + ξ^{2})^{3}} +

2 …

y_{1}(x) = \frac{\sqrt{2} e^{\frac{i}{4} π(2ν+1)}}{2\sqrt{π}} − \frac{−(4ν^{2} − 1)\sqrt{2} e^{\frac{i}{4} π(2ν+1)} x}{16\sqrt{π} RootOf_{ξ,1}(1 + ξ^{2})}+

\frac{(4ν^{2} − 9)(4ν^{2} − 1)\sqrt{2} e^{\frac{i}{4} π(2ν+1)} x^{2}}{256\sqrt{π} RootOf_{ξ,1}(1 + ξ^{2})^{2}} −

\frac{−(4ν^{2} − 25)(4ν^{2} − 9)(4ν^{2} − 1)\sqrt{2} e^{\frac{i}{4} π(2ν+1)} x^{3}}{6144\sqrt{π} RootOf_{ξ,1}(1 + ξ^{2})^{3}} + 2 …

BSJ.2.2.2 General form.

BSJ.2.2.2.1 Auxiliary function y_{0}(x). The coefficients u(n) of y_{0}(x) satisfy the following recurrence

8u(n)n RootOf_{ξ,2}(1 + ξ^{2}) + u(n − 1)(−4ν^{2} − 3 + 4n + 4(n − 1)^{2}) = 0

whose initial conditions are given by

u(0) = \frac{\sqrt{2} e^{(− \frac{i}{4} π(2ν+1))}}{2\sqrt{π}}

BSJ.2 SERIES AND ASYMPTOTIC EXPANSIONS 3

This recurrence has the closed form solution

u(n) = (2^{(n+ \frac{1}{2})} RootOf_{ξ,2}(1 + ξ^{2})^{n}Γ(n − ν + \frac{1}{2})Γ(ν + \frac{1}{2} + n)

sin(\frac{π(2ν + 1)}{2})(−2)^{n}(−1)^{n} e^{(− \frac{i}{4} π(2ν+1))})/

(2Γ(n + 1)8^{n}π^{\frac{3}{2}} ).

BSJ.2.2.2.2 Auxiliary function y_{1}(x). The coefficients u(n) of y_{1}(x) satisfy the following recurrence

8u(n)n RootOf_{ξ,1}(1 + ξ^{2}) + u(n − 1)(−4ν^{2} − 3 + 4n + 4(n − 1)^{2}) = 0 whose initial conditions are given by

u(0) = \frac{\sqrt{2} e^{\frac{i}{4} π(2ν+1)}}{2\sqrt{π}}

This recurrence has the closed form solution

u(n) = (2^{(n+ \frac{1}{2})} RootOf_{ξ,1}(1 + ξ^{2})^{n}Γ(n − ν + \frac{1}{2})Γ(ν + \frac{1}{2} + n)

sin(\frac{π(2ν + 1)}{2})(−2)^{n}(−1)^{n} e^{\frac{i}{4} π(2ν+1)})/

(2Γ(n + 1)8^{n}π^{\frac{3}{2}} ).