10.2: Bessel’s Equation
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 8332
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Bessel’s equation of order \(\nu\) is given by \[x^2 y'' + x y' + (x^2\nu^2) y = 0.\] Clearly \(x=0\) is a regular singular point, so we can solve by Frobenius’ method. The indicial equation is obtained from the lowest power after the substitution \(y=x^\gamma\), and is
\[\gamma^2\nu^2=0\]
So a generalized series solution gives two independent solutions if \(\nu \neq \frac{1}{2} n\). Now let us solve the problem and explicitly substitute the power series,
\[y = x^\nu \sum_n a_n x^n.\]
From Bessel’s equation we find
\[\sum_n(n+\nu)(n+\nu1) a_\nu x^{m+\nu} +\sum_n(n+\nu)a_\nu x^{m+\nu} +\sum_n(x^2\nu^2)a_\nu = 0\]
which leads to
\[[(m+\nu)^2\nu^2] a_m= a_{m2}\] or \[a_m= \frac{1}{m(m+2\nu)}a_{m2}.\]
If we take \(\nu=n>0\), we have
\[a_m= \frac{1}{m(m+2n)}a_{m2}.\]
This can be solved by iteration,
\[\begin{aligned} a_{2k} &= \frac{1}{4}\frac{1}{k(k+n)}a_{2(k1)}\nonumber\\ &= \left(\frac{1}{4}\right)^2\frac{1}{k(k1)(k+n)(k+n1)}a_{2(k2)} \nonumber\\ &= \left(\frac{1}{4}\right)^k\frac{n!}{k!(k+n)!}a_{0}.\end{aligned}\]
If we choose^{1} \(a_0 = \frac{1}{n!2^n}\) we find the Bessel function of order \(n\)
\[J_n(x) = \sum_{k=0}^\infty \frac{(1)^k}{k!(k+n)!} \left(\frac{x}{2}\right)^{2k+n}.\]
There is another second independent solution (which should have a logarithm in it) with goes to infinity at \(x=0\).
Figure \(\PageIndex{1}\): A plot of the first three Bessel functions \(J_n\) and \(Y_n\).
The general solution of Bessel’s equation of order \(n\) is a linear combination of \(J\) and \(Y\), \[y(x) = A J_n(x)+B Y_n(x).\]

This can be done since Bessel’s equation is linear, i.e., if \(g(x)\) is a solution \(C g(x)\) is also a solution.↩