Bessel’s equation is given by
$$ x^2 \frac{\text d^2 y}{\text dx^2}+x\frac{\text dy}{\text dx}+(x^2-m^2)y=0 $$
subject to boundary conditions $y$ is finite at $x=0$
We are solving using it using the method of Frobenius which uses a series solution with a leading factor of $x^s$ to account for non-constant behaviour at small $x$
$$ y(x)=x^s\sum^\infin_{n=0}a_n x^n $$
🗒️ Note: $s$ could be $s<0$ and could be complex in general
Take the derivative
$$ \begin{aligned} x\frac{\text dy}{\text dx}&=\sum^\infin_{n=0}a_n(n+s)x^{n+s} \\ x^2 \frac{\text d^2 y}{\text dx^2}&=\sum^\infin_{n=0} a_n(n+s)(n+s-1)x^{n+s} \\ (x^2-m^2)y&=\sum^\infin_{n=0}a_n x^{n+s+2}-m^2 \sum^\infin_{n=0} a_n x^{n+s} \end{aligned} $$
To help equating coefficients we write this first term on the right hand side as
$$ \sum_{n=2}^\infin a_{n-2}x^{n+s} $$
Subbing in we get
$$ \begin{aligned} &\sum^\infin_{n=0} a_n(n+s)(n+s-1)x^{n+s}+\sum^\infin_{n=0} a_n(n+s)x^{n+s} \\&+\sum^\infin_{n=2}a_{n-2} x^{n+s}-m^2 \sum^\infin_{n=0} a_n x^{n+s}=0 \end{aligned} $$
We can now equate coefficients
when $n=0$
$$ \begin{aligned} a_0 s(s-1)+a_0s-m^2a_0&=0 \\ a_0(s^2-m^2)&=0 \\ \Rightarrow \qquad s&=\pm m
\end{aligned} $$
when $n=1$
$$ \begin{aligned} a_1s(s+1)+(1+s)a_1-m^2a_1&=0 \\ \left [ (s+1)^2-m^2\right ]a_1&=0 \\ a_1&=0 \end{aligned} $$
when $n\ge2$
$$ \begin{aligned} [(n+s)(n+s-1)+(n+s)-m^2]a_n+a_{n-2}=0 \\ \left [ (n+s)^2-m^2 \right ]a_n+a_{n-2}=0
\end{aligned} $$
For solutions that are regular at the origin we take $s=+m>0$ we have
$$ \begin{aligned} a_2&=-\frac{a_0}{(m+2)^2-m^2} \\ a_4&=-\frac{a_2}{(m+4)^2-m^2} \\ \vdots\;&\qquad \qquad \quad \vdots
\end{aligned} $$
We can expand these solutions as
$$ \begin{aligned} a_2&= -\frac{a_0}{2^2(1+m)}=-\frac{a_0}{2(2+2m)} \\ a_4&=-\frac{a_2}{2^3(2+m)}=\frac{a_0}{2.4(2+2m)(4+2m)}=\frac{a_0}{2^{2.2}2!(1+m)(2+m)} \end{aligned} $$
We can also evaluate $a_6$ by a similar procedure
$$ a_6=-\frac{a_0}{2^{2.3}3!(1+m)(2+m)(3+m)} $$
We also have $a_n=0$ is $n=$ odd. The series converges for all values $x$ since the ratio of the $(n+2)$th to the $n$th term in the series is
$$ \left | \frac{a_{n+2}x^{n+2}}{a_nx^n} \right |=\frac{x^2}{(n+2+m)^2-m^2} $$
which vanishes as $n\to\infin$ for all $x$
For positive integer values of $m$ we can write the expression in terms of factorials
$$ \begin{aligned} &=\frac{1}{(1+m)(2+m)\ldots(j+m)}\\ &=\frac{1.2.3\ldots (m-1)m}{1.2.3\ldots (m-1)(m)(1+m)\ldots (m+j)}\\&=\frac{m!}{(m+j)!} \end{aligned} $$
Thus in general for integer $m$ we have
$$ a_{2j}=\frac{(-1)^ja_0 m!}{2^{2j}j!(m+j)!} $$
$m=0,1,2,\ldots$
🗒️ Note: this can be generalised to non-integer $m$ using the gamma function
This gives a series of coefficients in the original power series and a solution to Bessel’s equation
$$ \begin{aligned} y(x)&=\sum^\infin_{n=0}a_nx^{m+n} \\ &=\sum^\infin_{j=0}a_{2j}x^{m+2j} \\ &=\sum^\infin_{j=0}\frac{(-1)^ja_0 m!}{2^{2j}j!(m+j)!}x^{m+2j} \\ &=a_0m! 2^m\sum^\infin_{j=0}\frac{(-1)^j}{2^{2j}2^mj!(m+j)!}x^{m+2j} \\ &=a_0m! 2^m\sum^\infin_{j=0}\frac{(-1)^j}{j!(m+j)!}\left ( \frac x2 \right )^{m+2j}
\end{aligned} $$
If $m$ is an integer it is customary to write
$$ a_0=\frac{1}{2^mm!} $$
Thus we have
$$ y(x)=a_02^mm! J_m(x) $$
where
$$ J_m(x)=\sum^\infin_{j=0}(-1)^j\frac{1}{j!(m+j)!}\left ( \frac x2 \right ) ^{m+2j} $$
This series solution corresponds to Bessel’s equation of the first kind
🗒️ Notes:
-
$y(x)\propto J_m(x)$
-
All others are zero at the origin $x=0$
-
$J_0(0)=1$
The orthogonality relation for these Bessel functions
$$ \int^\ell_0xJ_p\left ( \frac{\chi_n}{\ell}x\right ) J_p\left ( \frac{\chi_m}{\ell}x\right ) \,\text dx=\left \{ \begin{aligned} &\frac{\ell^2}{2}J^2_{p+1}(\chi_m) \; &n=m \\ &0 \quad &n\ne m \end{aligned}\right . $$
where we have
$$ J_p(\chi_n)=0 \qquad n\in \N^* $$
which corresponds to locating the zeros of the Bessel function $J_p$
