Special Functions – Bessel functions

The Bessel function \(J_{\nu}(z)\) of the first kind of order \(\nu\) is defined by

\[J_{\nu}(z)=\frac{(z/2)^{\nu}}{\Gamma(\nu+1)}\,{}_0F_1\left(\genfrac{}{}{0pt}{}{-}{\nu+1}\,;\,-\frac{z^2}{4}\right) =\left(\frac{z}{2}\right)^{\nu}\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(\nu+k+1)\,k!}\left(\frac{z}{2}\right)^{2k}.\tag1\]

For \(\nu\geq0\) this is a solution of the Bessel differential equation

\[z^2y''(z)+zy'(z)+\left(z^2-\nu^2\right)y(z)=0,\quad\nu\geq0.\tag2\]

For \(\nu\notin\{0,1,2,\ldots\}\) we have that \(J_{-\nu}(z)\) is a second solution of the differential equation (2) and the two solutions \(J_{\nu}(z)\) and \(J_{-\nu}(z)\) are clearly linearly independent. For \(\nu=n\in\{0,1,2,\ldots\}\) we have

\[J_{-n}(z)=\left(\frac{z}{2}\right)^{-n}\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(-n+k+1)\,k!} \left(\frac{z}{2}\right)^{2k}=\left(\frac{z}{2}\right)^{-n}\sum_{k=n}^{\infty} \frac{(-1)^k}{\Gamma(-n+k+1)\,k!}\left(\frac{z}{2}\right)^{2k},\]

since

\[\frac{1}{\Gamma(-n+k+1)}=0\quad\text{for}\quad k=0,1,2,\ldots,n-1.\]

This implies that

\begin{align*} J_{-n}(z)&=\left(\frac{z}{2}\right)^{-n}\sum_{k=n}^{\infty}\frac{(-1)^k}{\Gamma(-n+k+1)\,k!}\left(\frac{z}{2}\right)^{2k}= \left(\frac{z}{2}\right)^{-n}\sum_{k=0}^{\infty}\frac{(-1)^{n+k}}{\Gamma(k+1)\,(n+k)!}\left(\frac{z}{2}\right)^{2(n+k)}\\[2.5mm] &=(-1)^n\left(\frac{z}{2}\right)^n\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(n+k+1)\,k!}\left(\frac{z}{2}\right)^{2k}=(-1)^nJ_n(z). \end{align*}

This implies that \(J_n(z)\) and \(J_{-n}(z)\) are linearly dependent for \(n\in\{0,1,2,\ldots\}\).

A second linearly independent solution can be found as follows. Since \((-1)^n=\cos n\pi\), we see that \(J_{\nu}(z)\cos\nu\pi-J_{-\nu}(z)\) is a solution of the differential equation (2) which vanishes when \(\nu=n\in\{0,1,2,\ldots\}\). Now we define

\[Y_{\nu}(z):=\frac{J_{\nu}(z)\cos\nu\pi-J_{-\nu}(z)}{\sin\nu\pi},\tag3\]

where the case that \(\nu=n\in\{0,1,2,\ldots\}\) should be regarded as a limit case. By l'Hopital's rule we have

\[Y_n(z):=\lim\limits_{\nu\rightarrow n}Y_{\nu}(z)=\frac{1}{\pi}\left[\frac{\partial J_{\nu}(z)}{\partial\nu}\right]_{\nu=n} -\frac{(-1)^n}{\pi}\left[\frac{\partial J_{-\nu}(z)}{\partial\nu}\right]_{\nu=n}.\]

This implies that \(Y_{-n}(z)=(-1)^nY_n(z)\) for \(n\in\{0,1,2,\ldots\}\). The function \(Y_{\nu}(z)\) is called the Bessel function of the second kind of order \(\nu\).

Using the definition (3) we find that

\[\left[\frac{\partial J_{\nu}(z)}{\partial\nu}\right]_{\nu=n}=J_n(z)\ln\left(\frac{z}{2}\right) -\left(\frac{z}{2}\right)^n\sum_{k=0}^{\infty}\frac{(-1)^k\psi(n+k+1)}{(n+k)!\,k!}\left(\frac{z}{2}\right)^{2k},\]

where

\[\psi(z)=\frac{d}{dz}\ln\Gamma(z)=\frac{\Gamma'(z)}{\Gamma(z)}.\]

For \(\nu\notin\{0,1,2,\ldots\}\) we have

\[\frac{\partial J_{-\nu}(z)}{\partial\nu}=-J_{\nu}(z)\ln\left(\frac{z}{2}\right) +\left(\frac{z}{2}\right)^{-\nu}\sum_{k=0}^{\infty}\frac{(-1)^k\psi(-\nu+k+1)}{\Gamma(-\nu+k+1)\,k!}\left(\frac{z}{2}\right)^{2k}.\]

Now we use

\[\lim\limits_{z\rightarrow -n}(z+n)\Gamma(z)=\frac{(-1)^n}{n!}\quad\Longrightarrow\quad \Gamma(z)\sim\frac{(-1)^n}{(z+n)\,n!}\quad\text{for}\quad z\rightarrow -n\]

and

\[\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}\sim\frac{-\displaystyle\frac{(-1)^n}{(z+n)^2\,n!}}{\displaystyle\frac{(-1)^n}{(z+n)\,n!}} =-\frac{1}{z+n}\quad\text{for}\quad z\rightarrow -n.\]

This implies that

\[\lim\limits_{z\rightarrow -n}\frac{\psi(z)}{\Gamma(z)}=(-1)^{n+1}n!\quad\text{for}\quad n=0,1,2,\ldots.\]

Hence

\begin{align*} \lim\limits_{\nu\rightarrow n}\;\sum_{k=0}^{\infty}\frac{(-1)^k\psi(-\nu+k+1)}{\Gamma(-\nu+k+1)\,k!} \left(\frac{z}{2}\right)^{2k}&=(-1)^n\sum_{k=0}^{n-1}\frac{(n-k-1)!}{k!}\left(\frac{z}{2}\right)^{2k} +\sum_{k=n}^{\infty}\frac{(-1)^k\psi(-n+k+1)}{\Gamma(-n+k+1)\,k!}\left(\frac{z}{2}\right)^{2k}\\[2.5mm] &=(-1)^n\sum_{k=0}^{n-1}\frac{(n-k-1)!}{k!}\left(\frac{z}{2}\right)^{2k}+(-1)^n\left(\frac{z}{2}\right)^{2n} \sum_{k=0}^{\infty}\frac{(-1)^k\psi(k+1)}{\Gamma(k+1)\,(n+k)!}\left(\frac{z}{2}\right)^{2k}. \end{align*}

This implies that

\[\left[\frac{\partial J_{-\nu}(z)}{\partial\nu}\right]_{\nu=n}=-J_{-n}(z)\ln\left(\frac{z}{2}\right)+(-1)^n\left(\frac{z}{2}\right)^{-n} \sum_{k=0}^{n-1}\frac{(n-k-1)!}{k!}\left(\frac{z}{2}\right)^{2k} +(-1)^n\left(\frac{z}{2}\right)^n\sum_{k=0}^{\infty}\frac{(-1)^k\psi(k+1)}{(n+k)!\,k!}\left(\frac{z}{2}\right)^{2k}.\]

Finally we use the fact that \(J_{-n}(z)=(-1)^nJ_n(z)\) to conclude that

\[Y_n(z)=\frac{2}{\pi}J_n(z)\ln\left(\frac{z}{2}\right)-\frac{1}{\pi}\left(\frac{z}{2}\right)^{-n} \sum_{k=0}^{n-1}\frac{(n-k-1)!}{k!}\left(\frac{z}{2}\right)^{2k} -\frac{1}{\pi}\left(\frac{z}{2}\right)^n\sum_{k=0}^{\infty}\frac{(-1)^k}{(n+k)!\,k!} \left[\psi(n+k+1)+\psi(k+1)\right]\left(\frac{z}{2}\right)^{2k}\]

for \(n\in\{0,1,2,\ldots\}\). Compare with the theory of Frobenius for linear second-order differential equations.

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Last modified on September 29, 2021