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Bessel Function's Recursive Relations 📂Functions

Bessel Function's Recursive Relations

Theorem

Jν(x)=∑n=0∞(−1)nΓ(n+1)Γ(n+ν+1)(x2)2n+ν(1) J_{\nu}(x)=\sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu+1)} \left(\frac{x}{2} \right)^{2n+\nu} \tag{1}

The function above is called the first kind Bessel function of order ν\nu. The first kind Bessel function Jν(x)J_{\nu}(x) satisfies the following equation.

ddx[xνJν(x)]=xνJν−1(x)ddx[x−νJν(x)]=−x−νJν+1(x)Jν−1(x)+Jν+1(x)=2νxJν(x)Jν−1(x)−Jν+1(x)=2Jν′(x)Jν′(x)=−νxJν(x)+Jν−1(x)=νxJν(x)−Jν+1(x) \begin{align*} \frac{d}{dx}[x^{\nu} J_{\nu}(x)] &= x^{\nu}J_{\nu-1}(x) \tag{a} \\ \frac{d}{dx}[x^{-\nu}J_{\nu}(x)] &= -x^{-\nu}J_{\nu+1}(x) \tag{b} \\ J_{\nu-1}(x)+J_{\nu+1}(x) &= \frac{2\nu}{x}J_{\nu}(x) \tag{c} \\ J_{\nu-1}(x)-J_{\nu+1}(x) &= 2J^{\prime}_{\nu}(x) \tag{d} \\ J_{\nu}^{\prime}(x) = -\frac{\nu}{x}J_{\nu}(x)+J_{\nu-1}(x) &= \frac{\nu}{x}J_{\nu}(x)-J_{\nu+1}(x) \tag{e} \end{align*}

Proof

(a)(a)

By multiplying xνx^{\nu} to (1)(1) and then differentiating, it can be easily obtained.

ddx[xνJν(x)]=ddx[xν∑n=0∞(−1)nΓ(n+1)Γ(n+ν+1)(x2)2n+ν)=ddx∑n=0∞(−1)nΓ(n+1)Γ(n+ν+1)x2n+2ν22n+ν=∑n=0∞(−1)n(2n+2ν)Γ(n+1)Γ(n+ν+1)x2n+2ν−122n+ν \begin{align*} \frac{d}{dx}[x^{\nu} J_{\nu}(x)] &= \frac{d}{dx} \left[ x^{\nu} \sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu+1)} \left(\frac{x}{2} \right)^{2n+\nu} \right) \\ &= \frac{d}{dx} \sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu+1)} \frac{x^{2n+2\nu}}{2^{2n+\nu}} \\ &= \sum \limits_{n=0}^{\infty} \frac{(-1)^{n}(2n+2\nu)}{\Gamma (n+1) \Gamma (n+\nu+1)} \frac{x^{2n+2\nu-1}}{2^{2n+\nu}} \tag{2} \end{align*}

Since the gamma function satisfies the relation Γ(n+ν+1)=(n+ν)Γ(n+ν)\Gamma (n+\nu+1)=(n+\nu)\Gamma (n+\nu), simplifying the 2(n+ν)2(n+\nu) in the denominator of (2)(2) yields

ddx[xνJν(x)]=∑n=0∞(−1)nΓ(n+1)Γ(n+ν)x2n+2ν−122n+ν−1=xν∑n=0∞(−1)nΓ(n+1)Γ(n+ν)x2n+ν−122n+ν−1=xνJν−1(x) \begin{align*} \frac{d}{dx}[x^{\nu} J_{\nu}(x)] &= \sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu)} \frac{x^{2n+2\nu-1}}{2^{2n+\nu-1}} \\ &= x^{\nu}\sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu)} \frac{x^{2n+\nu-1}}{2^{2n+\nu-1}} \\ &= x^{\nu}J_{\nu-1}(x) \end{align*}

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(b)(b)

The general approach to the proof is the same as (a)(a), but there are subtle parts that are easy to miss, so it is explained without omission. By multiplying x−νx^{-\nu} to (1)(1) like (a)(a) and then differentiating

ddx[x−νJν(x)]=ddx∑n=0∞(−1)nΓ(n+1)Γ(n+ν+1)x2n22n+ν=∑n=0∞(−1)n2nΓ(n+1)Γ(n+ν+1)x2n−122n+ν \begin{align*} \frac{d}{dx}[x^{-\nu} J_{\nu}(x)] &= \frac{d}{dx}\sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu+1)} \frac{x^{2n}}{2^{2n+\nu}} \\ &= \sum \limits_{n=0}^{\infty} \frac{(-1)^{n}2n }{\Gamma (n+1) \Gamma (n+\nu+1)} \frac{x^{2n-1}}{2^{2n+\nu}} \end{align*}

Since it is Γ(n+1)=nΓ(n)\Gamma (n+1)=n\Gamma (n), after simplifying the 2n2n in the denominator and adjusting the order of x2\frac{x}{2}

ddx[x−νJν(x)]=x−ν∑n=0∞(−1)nΓ(n)Γ(n+ν+1)x2n+ν−122n+ν−1 \begin{align*} \frac{d}{dx}[x^{-\nu} J_{\nu}(x)] &= x^{-\nu}\sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n) \Gamma (n+\nu+1)} \frac{x^{2n+\nu-1}}{2^{2n+\nu-1}} \end{align*}

Here, let’s substitute the index with n=k+1n=k+1. Then

ddx[x−νJν(x)]=x−ν∑k=−1∞(−1)k+1Γ(k+1)Γ(k+ν+2)x2k+ν+122k+ν+1 \begin{align*} \frac{d}{dx}[x^{-\nu} J_{\nu}(x)] &= x^{-\nu}\sum \limits_{k=-1}^{\infty} \frac{(-1)^{k+1} }{\Gamma (k+1) \Gamma (k+\nu+2)} \frac{x^{2k+\nu+1}}{2^{2k+\nu+1}} \end{align*}

If it is k=−1k=-1, the denominator diverges to Γ(k+1)=Γ(0)=∞\Gamma (k+1)=\Gamma (0)=\infty, so it is 00. Therefore, it is okay to start the index from k=0k=0.

ddx[x−νJν(x)]=x−ν∑k=0∞(−1)k+1Γ(k+1)Γ(k+ν+2)x2k+ν+122k+ν+1=−x−νJν+1(x) \begin{align*} \frac{d}{dx}[x^{-\nu} J_{\nu}(x)] &= x^{-\nu}\sum \limits_{k=0}^{\infty} \frac{(-1)^{k+1} }{\Gamma (k+1) \Gamma (k+\nu+2)} \frac{x^{2k+\nu+1}}{2^{2k+\nu+1}} \\ &= -x^{-\nu}J_{\nu+1}(x) \end{align*}

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(c)(c), (d)(d)

By organizing (a)(a) for Jν−1(x)J_{\nu-1}(x) and organizing (b)(b) for Jν+1(x)J_{\nu+1}(x), then adding and subtracting, it can be immediately obtained.

Jν−1(x)=x−νddx[xνJν(x)]=x−ννxν−1Jν(x)+x−νxνJν′(x)=νx−1Jν(x)+Jν′(x)(3) \begin{align*} J_{\nu-1}(x) &= x^{-\nu} \frac{d}{dx}\left[x^{\nu} J_{\nu}(x)\right] \\ &= x^{-\nu} \nu x^{\nu-1}J_{\nu}(x)+x^{-\nu}x^{\nu}J_{\nu}^{\prime}(x) \\ &= \nu x^{-1}J_{\nu}(x) + J_{\nu}^{\prime}(x) \end{align*} \tag{3}

Jν+1(x)=−xνddx[x−νJν(x)]=−xν(−ν)x−ν−1Jν(x)−xνx−νJν′(x)=νx−1Jν(x)−Jν′(x)(4) \begin{align*} J_{\nu+1}(x) &= -x^{\nu} \frac{d}{dx}\left[x^{-\nu} J_{\nu}(x) \right] \\ &= -x^{\nu}(-\nu) x^{-\nu-1}J_{\nu}(x)-x^{\nu}x^{-\nu}J_{\nu}^{\prime}(x) \\ &= \nu x^{-1}J_{\nu}(x) - J_{\nu}^{\prime}(x) \end{align*} \tag{4}

If you calculate (3)+(4)(3)+(4)

Jν−1+Jν+1(x)=2νxJν(x) J_{\nu-1}+J_{\nu+1}(x)=\frac{2\nu}{x}J_{\nu}(x)

If you calculate (3)−(4)(3)-(4)

Jν−1−Jν+1(x)=2Jν′(x) J_{\nu-1}-J_{\nu+1}(x)=2J_{\nu}^{\prime}(x)

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(e)(e)

By expanding and organizing the left sides of (a)(a) and (b)(b), it can be easily obtained. First, by expanding the left side of (a)(a) and organizing for Jν′(x)J_{\nu}^{\prime}(x)

ddx[xνJν(x)]=xνJν−1(x)νxν−1Jν(x)+xνJν′(x)=xνJν−1(x)Jν′(x)=−νxJν(x)+Jν−1(x) \begin{align*} \frac{d}{dx}[x^{\nu} J_{\nu}(x)] &=x^{\nu}J_{\nu-1}(x) \\ \nu x^{\nu-1}J_{\nu}(x) + x^{\nu} J_{\nu}^{\prime}(x) &= x^{\nu}J_{\nu-1}(x) \\ J_{\nu}^{\prime}(x) &= -\frac{\nu}{x}J_{\nu}(x)+J_{\nu-1}(x) \end{align*}

The same process is done for (b)(b)

ddx[x−νJν(x)]=−x−νJν+1(x)−νx−ν−1Jν(x)+x−νJν′(x)=−x−νJν+1(x)Jν′(x)=νxJν(x)−Jν+1(x) \begin{align*} \frac{d}{dx}[x^{-\nu} J_{\nu}(x)] &=-x^{-\nu}J_{\nu+1}(x) \\ -\nu x^{-\nu-1}J_{\nu}(x) + x^{-\nu} J_{\nu}^{\prime}(x) &=-x^{-\nu}J_{\nu+1}(x) \\ J_{\nu}^{\prime}(x) &= \frac{\nu}{x}J_{\nu}(x)-J_{\nu+1}(x) \end{align*}

Therefore

Jν′(x)=−νxJν(x)+Jν−1(x)=νxJν(x)−Jν+1(x) J_{\nu}^{\prime}(x)=-\frac{\nu}{x}J_{\nu}(x)+J_{\nu-1}(x)=\frac{\nu}{x}J_{\nu}(x)-J_{\nu+1}(x)

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