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Mathematics LibreTexts

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


    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 \(x=0\).

    A plot of the first three Bessel functions \(J_n\) and \(Y_n\).

    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.