Section 4.7 Bessel’s Differential Equation
The differential equation
\begin{equation}
x^{2}+\frac{d^{2}y}{dx^{2}}+x\frac{\,dy}{\,dx}+(x^{2}-n^{2})y =0 \tag{4.7.1}
\end{equation}
is called the Bessel’s differential equation. Such type of equation yields on solving the Laplace’s equation in cylindrical harmonics. This equation has a regular singularity at \(x=0\) and its solution can be obtained as a power series developed about this point.
Let the solution of this equation is in the form
\begin{equation}
y=\sum\limits_{k=0}^{\infty}a_{k}x^{m+k}\tag{4.7.2}
\end{equation}
so that
\begin{equation*}
\frac{dy}{dx}=\sum a_{k}(m+k)x^{m+k-1}
\end{equation*}
and
\begin{equation*}
\frac{d^{2}y}{dx^{2}}=\sum a_{k}(m+k)(m+k-1)x^{m+k-2}
\end{equation*}
substitutuing these values in equation (4.7.1), we get -
\begin{equation*}
\sum a_{k}(m+k)(m+k-1)x^{m+k}+\sum a_{k}(m+k)x^{m+k}
\end{equation*}
\begin{equation*}
+\sum a_{k}x^{m+k+2}-n^{2}\sum a_{k}x^{m+k}=0
\end{equation*}
or,
\begin{equation*}
\sum a_{k}\left[(m+k)(m+k-1)+(m+k)-n^{2}\right]x^{m+k}+\sum a_{k}x^{m+k+2}=0
\end{equation*}
or,
\begin{equation}
\sum a_{k}\left[(m+k)^{2}-n^{2}\right]x^{m+k}+\sum a_{k}x^{m+k+2}=0\tag{4.7.3}
\end{equation}
Equating the coefficient of lowest of power of \(x\text{.}\) i.e., \(x^{m}\) to zero, by putting \(k=0\) in equation (4.7.3), we get the indicial equation as \(a_{o}(m^{2}-n^{2})=0 \text{.}\) Since, \(a_{o} \neq 0\text{,}\) we have \(m=\pm n \text{.}\)
Equating the coefficient of \(x^{m+1}\) to zero, we get -
\begin{equation*}
a_{1}[(m+1)^{2}-n^{2}]=0
\end{equation*}
for \(k=1\text{,}\) i.e., \(a_{1} =0\text{.}\) Since \((m+1)^{2}-n^{2} \neq 0\text{,}\) for the given value of \(m =\pm n \text{.}\)
Equating the coefficient of general term \(x^{m+k+2}\) to zero, we get the recurrence relation for the coefficients as
\begin{equation*}
a_{k+2}[(m+k+2)^{2}-n^{2}]+a_{k}=0
\end{equation*}
or,
\begin{equation}
a_{k+2}=-\frac{a_{k}}{(m+k+2)^{2}-n^{2}}\tag{4.7.4}
\end{equation}
for
\begin{equation*}
m=+n, \quad a_{k+2}=-\frac{a_{k}}{(n+k+2)^{2}-n^{2}}
\end{equation*}
and
\begin{equation*}
a_{2}=-\frac{a_{o}}{(n+2)^{2}-n^{2}}=-\frac{a_{o}}{4(n+1)};
\end{equation*}
\begin{equation*}
a_{3}=-\frac{a_{1}}{(n+3)^{2}-n^{2}}=0 \quad [\because a_{1}=0]
\end{equation*}
\begin{equation*}
\therefore \quad a_{5} =a_{7} =\cdots=0,
\end{equation*}
\begin{equation*}
a_{4}=-\frac{a_{2}}{(n+4)^{2}-n^{2}} =+\frac{a_{o}}{4(n+1)8(n+2)}
\end{equation*}
\begin{equation*}
=\frac{a_{o}}{4^{2}2!(n+1)(n+2)}
\end{equation*}
\begin{equation*}
a_{6}=-\frac{a_{4}}{(n+6)^{2}-n^{2}}=- \frac{a_{0}}{4^{2}2!(n+1)(n+2)(n+3)12}
\end{equation*}
\begin{equation*}
=-\frac{a_{0}}{4^{3}3!(n+1)(n+2)(n+3)}
\end{equation*}
and so on.
Substituting these values in equation (4.7.2), we have
\begin{equation*}
y=a_{o}x^{n}\left[1-\frac{1}{4(n+1)}x^{2}+\frac{1}{4^{2}2!(n+1)(n+2)}x^{4}\right.
\end{equation*}
\begin{equation*}
\left.-\frac{1}{4^{3}3!(n+1)(n+2)(n+3)}x^{6}+\cdots\right.
\end{equation*}
\begin{equation*}
\left.+\cdots+\frac{(-1)^{r}x^{2r}}{4^{r}r!(n+1)(n+2)\cdots(n+r)}+\cdots\right]
\end{equation*}
\begin{equation}
\therefore \quad y=a_{o}\sum\limits_{r=0}^{\infty}(-1)^{r}\frac{x^{n+2r}}{2^{2r}r!(n+1)(n+2)\cdots(n+r)}\tag{4.7.5}
\end{equation}
where \(a_{o}\) is an arbitrary constant. If we choose \(a_{o} = \frac{1}{2^{n}n!}\) then the solution (4.7.5) is called Bessel’s function of the first kind and is denoted by \(J_{n}(x)\text{.}\) Thus
\begin{equation*}
J_{n}(x) = \frac{1}{2^{n}n!} \sum\limits_{r=0}^{\infty}(-1)^{r}\frac{x^{n+2r}}{2^{2r}r!(n+1)(n+2)\cdots(n+r)}
\end{equation*}
\begin{equation}
= \sum\limits_{r=0}^{\infty}\frac{(-1)^{r}}{r!(n+r)!}\left(\frac{x}{2}\right)^{n+2r}\tag{4.7.6}
\end{equation}
for \(m=-n\text{,}\) we have
\begin{equation*}
a_{k+2}=-\frac{a_{k}}{(-n+k+2)^{2}-n^{2}}
\end{equation*}
\begin{equation}
J_{-n}(x)=\sum\limits_{r=0}^{\infty}\frac{(-1)^{r}}{r!(-n+r)!}\left(\frac{x}{2}\right)^{-n+2r}\tag{4.7.7}
\end{equation}
Therefore, the general solution of Bessel’s differential equation (4.7.1) is
\begin{equation}
y=AJ_{n}(x)+BJ_{-n}(x)\tag{4.7.8}
\end{equation}
Where \(A\) and \(B\) are two arbitrary constants.
