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Next: Borel Summation Up: No Title Previous: The Gamma function and

The Bessel function and the related Neumann function

In the book these subjects are discussed on pp. 16-19. Here we shall give a more detailed discussion of the two solutions in the case where the index is an integer. In particular, this leads to the Neumann function.

The power series solution of the Bessel equation (1-52) should satisfy the recursion relation (1-54). In the coefficients cn, n is an even integer, so we take n=2k. Then

(k+s)(k+s-1)}=(-1)^k\frac{c_0}{2^{2k}\Gamma (k+1)\Gamma (k+s+1)}.
\end{displaymath} (46)

To conform to standard notation we replace the index m2 in the Bessel equation by $\nu$, i.e. $m^2\rightarrow\nu^2$. The solutions for the two allowed cases $s=\pm \nu$ are thus

 \begin{displaymath}J_\nu (x)=\frac{x^\nu}{2^\nu}\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{2^{2k}
\Gamma (k+1)\Gamma (k+\nu+1)},
\end{displaymath} (47)


 \begin{displaymath}J_{-\nu} (x)=x^{-\nu}2^\nu\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{2^{2k}
\Gamma (k+1)\Gamma (k-\nu+1)},
\end{displaymath} (48)

These functions are called Bessel functions of index $\nu$ and $-\nu$, respectively. The general solution of the Bessel equation is therefore

\begin{displaymath}y(x)=C_1J_\nu (x)+C_2J_{-\nu}(x),
\end{displaymath} (49)

where the C's are arbitrary constants.

This solution breaks down when $\nu=n$, where n is an integer. The reason is that for k=0,1,2,...,n-1 the function $\Gamma (k-n+1)$ has poles, and therefore there are no contributions to the sum in (48) for these values of k. Hence

\begin{displaymath}J_{-n}(x)=2^nx^{-n}\sum_{k=n}^\infty (-1)^k\frac{x^{2k}}{2^{2k}
\Gamma (k+1)\Gamma (k-n+1)}.
\end{displaymath} (50)

Introducing the new summation variable l=k-n with $l\geq 0$ we get

\begin{displaymath}J_{-n}(x)=2^{-n}x^{-n}\sum_{l=0}^\infty (-1)^{l+n}\frac{x^{2l+2n}}{2^{2l}
\Gamma (l+n+1)\Gamma (l+1)}=(-1)^nJ_n(x).
\end{displaymath} (51)

Thus J-n and Jn are not linearly independent, and only one of these functions can be used in the complete solution. We therefore consider the Neumann function Nn defined by

 \begin{displaymath}\pi N_n(x)=\lim_{\nu\rightarrow n}\left[\frac{\partial}{\part...
...u (x)
\end{displaymath} (52)

The reader can easily check that this function satisfies the Bessel equation. Using the $\psi$-function introduced in the last section, we get

 \begin{displaymath}\frac{\partial}{\partial\nu}J_\nu (x)\vert _{\nu=n}=J_n(x) \l...
\Gamma(k+1)\Gamma (k+n+1)}\psi(k+n+1),
\end{displaymath} (53)


 \begin{displaymath}\frac{\partial}{\partial\nu}J_\nu (x)\vert _{\nu=-n}=-J_{-n}(...
\Gamma(k+1)\Gamma (k-n+1)}\psi(k-n+1),
\end{displaymath} (54)

Using the result (39) in the last section we get from (53)

\begin{displaymath}\frac{\partial}{\partial\nu}J_\nu (x)\vert _{\nu=n}=J_n(x) \l...
...k+1)\Gamma (k+n+1)}\left(\sum_{l=1}^{k+n}\frac{1}{l}-C\right),
\end{displaymath} (55)

where C is Euler's constant. Writing $\gamma=e^C=$1.781 we have

\begin{displaymath}\frac{\partial}{\partial\nu}J_\nu (x)\vert _{\nu=n}=J_n(x) \l...
...frac{1}{\Gamma(k+1)\Gamma (k+n+1)}\sum_{l=1}^{k+n}\frac{1}{l}.
\end{displaymath} (56)

To treat the case where $\nu=-n$ we need to use results derived in the previous section. By means of (34) we easily deduce that

 \begin{displaymath}\psi(x+1)\rightarrow -\frac{1}{x+n}~{\rm and}~~\frac{\psi (x+... (x+1)}
\rightarrow (-1)^n(n-1)!~{\rm for}~x\rightarrow -n.
\end{displaymath} (57)

To obtain the first of these equations we have computed $\Gamma'(x+1)/\Gamma (x+1)$ near the pole, using (34). The values k=0,1,2,...,n-1 can thus be treated by use of (57). Using this in (54) we get
$\displaystyle \frac{\partial}{\partial\nu}J_\nu (x)\vert _{\nu=-n}$ = $\displaystyle (-1)^{n+1}J_n(x)
\ln \frac{\gamma x}{2}+
\left(\frac{x}{2}\right)^{-n}~\sum_{k=0}^{n-1} (-1)^k\frac{x^{2k}}{2^{2k}
  + $\displaystyle (-1)^n\left(\frac{x}{2}\right)^{n}\sum_{k=1}^\infty (-1)^k\frac{x^{2k}}
{2^{2k}\Gamma(k+1)\Gamma (k+n+1)}\sum_{l=1}^k\frac {1}{l}.$ (58)

Collecting results we finally get by use of the definition (52)
$\displaystyle \pi N_n(x)$ = $\displaystyle 2 \ln\frac{\gamma x}{2}J_n(x)-\left(\frac{2}{x}\right)^n~
\sum_{k=0}^{n-1} \frac{(n-k-1)!}{k!}\left(\frac{x^2}{4}\right)^k$  
  - $\displaystyle \left(\frac{x}{2}\right)^{n}\frac{1}{n!}\sum_{l=1}^n\frac{1}{l}
...)!(k+n)!}\left(\sum_{l=1}^{n+k}\frac {1}{l}+\sum_{l=1}^{k}
\frac {1}{l}\right).$ (59)

We see that near x=0 the Neumann functions are singular like x-n. For n=0 there is only a logarithmic singularity,

\begin{displaymath}\pi N_0(x)=2J_0(x)\ln\frac{\gamma x}{2}-2\sum_{k=1}^\infty\fr...
\end{displaymath} (60)


\begin{displaymath}J_0(x)=\sum_{k=0}^\infty \frac{1}{(k!)^2}\left(\frac{-x^2}{4}\right)^k.
\end{displaymath} (61)

We mention that the Neumann function in general is defined by

 \begin{displaymath}N_\nu(x)=\frac{\cos (\nu\pi)J_\nu (x)-J_{-\nu}(x)}{\sin (\nu\pi)}.
\end{displaymath} (62)

By means of l'Hospital's rule it is easily seen that this definition agrees with (52) in the case $\nu\rightarrow n$.

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Next: Borel Summation Up: No Title Previous: The Gamma function and
Mette Lund