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# 10.2: Bessel’s Equation

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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

$\sum_n(n+\nu)(n+\nu-1) a_\nu x^{m+\nu} +\sum_n(n+\nu)a_\nu x^{m+\nu} +\sum_n(x^2-\nu^2)a_\nu = 0$

which leads to

$[(m+\nu)^2-\nu^2] a_m= -a_{m-2}$ or $a_m= -\frac{1}{m(m+2\nu)}a_{m-2}.$

If we take $$\nu=n>0$$, we have

$a_m= -\frac{1}{m(m+2n)}a_{m-2}.$

This can be solved by iteration,

\begin{aligned} a_{2k} &= -\frac{1}{4}\frac{1}{k(k+n)}a_{2(k-1)}\nonumber\\ &= \left(\frac{1}{4}\right)^2\frac{1}{k(k-1)(k+n)(k+n-1)}a_{2(k-2)} \nonumber\\ &= \left(-\frac{1}{4}\right)^k\frac{n!}{k!(k+n)!}a_{0}.\end{aligned}

If we choose1 $$a_0 = \frac{1}{n!2^n}$$ 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 $$Y$$, $y(x) = A J_n(x)+B Y_n(x).$

1. 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.