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Applied Mathematics

Section 7.7 Bessel’s equation and Bessel Functions

Bessel’s equation is
\begin{equation} x^2 y'' + x y' + (x^2-\nu^2)y = 0\tag{7.7.1} \end{equation}
A solution can be obtained by a power series solution and represented as
\begin{equation*} J_{\nu} (x) = \frac{x^{\nu}}{2^{\nu}} \sum_{m=0}^{\infty} \frac{(-1)^m}{2^m m! \Gamma(m+\nu+1)} x^{2m} \end{equation*}
where \(\Gamma\) is the gamma function, a generalized factorial. The function \(J_{\nu}(x)\) is called the Bessel Function of the first kind. We are often interested in solutions of (7.7.1) in which \(\nu=n\) is an integer. If this is the case, then \(J_n(x)\) and \(J_{-n}(x)\) are two linearly independent solutions. The power series representation in this case is
\begin{equation} J_{n} (x) = \frac{x^{n}}{2^{n}} \sum_{m=0}^{\infty} \frac{(-1)^m}{2^m m! (m+n)!} x^{2m}\tag{7.7.2} \end{equation}

Subsection 7.7.1 Propeties of \(J_n(x)\)

The following is a plot of \(J_0(x)\) (solid line) and \(J_1(x)\) (dashed line) on \(0 \leq x \leq 10\text{.}\) Each of the Bessel functions have osciallatory behavior with decay and an infinite number of roots for \(x \geq 0\text{.}\) Also note that the roots of \(J_1\) are between the roots of \(J_0\text{.}\)
Figure 7.7.1. A plot of the bessel functions \(J_0\) and \(J_1\) on \([0,10]\text{.}\)
Using (7.7.2), it can be shown that
\begin{equation*} J_n(0) = \begin{cases} 1 \amp \text{if $n=0$} \\ 0 \amp \text{otherwise} \end{cases} \end{equation*}
In addition, using the power series representation, one can show that the other solution of (7.7.1) can be written:
\begin{equation*} J_{-n}(x) = (-1)^n J_n(x) \end{equation*}
However for \(n < 0\text{,}\)\(J_n(x)\) has a term \(x^{-n}\) which means that it is undefined at \(x=0\text{,}\) which is generally why it not relevant as we will show later. There are a number of identities that are useful for understanding Bessel functions. Two of these are shown in the follow lemma.

Proof.

First we will prove (7.7.3). Using (7.7.2), we can write
\begin{align*} x^nJ_n(x) \amp = x^n \biggl( \frac{x^{n}}{2^{n}} \sum_{m=0}^{\infty} \frac{(-1)^m}{2^m m! (m+n)!} x^{2m} \biggr) \\ \amp = \sum_{m=0}^{\infty} \frac{(-1)^m}{2^{m+n} m! (m+n)!} x^{2m+2n} \end{align*}
and differentiating,
\begin{align*} \frac{d}{dx} \bigl(x^nJ_n(x)\bigr) \amp = \sum_{m=0}^{\infty} \frac{(-1)^m(2m+2n)}{2^{m+n} m! (m+n)!} x^{2m+2n-1} \\ \amp = \frac{x^{n-1}}{2^{n-1}} \sum_{m=0}^{\infty} \frac{(-1)^m2(m+n)}{2^{m+1} m! (m+n)!} x^{2m+n} \\ \amp = x^n \biggl(\frac{x^{n-1}}{2^{n-1}} \sum_{m=0}^{\infty} \frac{(-1)^m}{2^{m} m! (m+n-1)!} x^{2m}) \\ \amp = x^{n} J_{n-1}(x) \end{align*}
The proof for (7.7.4) is very similar and is not shown.
In addition, there are another two identities for Bessel functions that are often called recurrence relationships.

Proof.

If we use the product rule to expand (7.7.3) and (7.7.4), we get
\begin{align*} n x^{n-1} J_n(x) + x^n J'_n(x) \amp = x^n J_{n-1} (x) \\ -nx^{-n-1} J_n(x) +x^{-n} J'_n(x) \amp = - x^{-n} J_{n+1}(x) \end{align*}
and multiply the first equation by \(x^{1-n}\) and the second by \(x^{1+n}\text{,}\) one gets
\begin{align*} n J_n + x J'_n \amp = xJ_{n-1} \\ -nJ_n+xJ'_n \amp = -xJ_{n+1} \end{align*}
Adding the two above equations and dividing through by \(x\) results in (7.7.5) whereas subtracting the bottom equation from the top results in (7.7.6).
These properties can now be used to find higher order Bessel functions, the derivatives of Bessel functions as well as the closed form of some integrals as shown in the next three examples.

Example 7.7.4.

Use the identities in Lemma 7.7.2 and (7.7.5) to find \(J_3\) in terms of \(J_0\) and \(J_1\text{.}\)
Solution.
Let \(n=2\) in (7.7.5) or
\begin{align*} x (J_1 + J_3) \amp = 4J_2 \amp\amp\text{solving for $J_3$} \\ J_3 \amp = \frac{4}{x} J_2 - J_1 \end{align*}
use (7.7.5) again with \(n=1\) or \(x(J_0 + J_2) = 2J_1\) which can be written \(J_2 = (2/x)J_1 -J_0\)
\begin{align*} \amp = \frac{4}{x} \biggl( \frac{2}{x} J_1 - J_0 \biggr) - J_1 \\ J_3 \amp = \biggl( \frac{8}{x^2} -1\biggr) J_1 - \frac{2}{x} J_0 \end{align*}
The above technique can be used to find \(J_n\) where \(n\) is an integer in terms of \(J_0\) and \(J_1\text{,}\) showing the importance of the first two Bessel functions. The next example shows how to calculate the derivatives of the first two Bessel functions.

Example 7.7.5.

Use the identities in lemmas Lemma 7.7.2 and Lemma 7.7.3 to find \(J'_0\) and \(J'_1\) in terms of \(J_0\) and \(J_1\text{.}\)
Solution.
First, differentiate (7.7.5) with \(n=1\) to get
\begin{align*} x (J'_0 +J'_2) + J_0 + J_2 \amp = 2 J' _1 \quad\text{solving for $xJ'_0$} \\ xJ'_0 \amp = 2J'_1 -J_0 - J_2 - xJ'_2 \end{align*}
using (7.7.6) with \(n=1\) and \(n=2\text{,}\)
\begin{align*} xJ'_0 \amp = (J_0 - J_2) -J_0 -J_2 - \frac{xJ_1 - xJ_3}{2} \\ \amp = -2J_2 -\frac{xJ_1 - xJ_3}{2} \end{align*}
Using (7.7.5) with \(n=1\)
\begin{align*} \amp = \frac{-x(J_1+J_3)}{2} - x\frac{J_1 -J_3}{2} = -xJ_1 \end{align*}
and finally dividing through by \(x\)
\begin{equation*} J'_0 = -J_1 \end{equation*}

Example 7.7.6.

Evaluate \(\int x^4 J_1 (x) \, dx\text{.}\)
Solution.
Integrating this by parts with \(u=x^2\) and \(du = x^2 J_1\, dx\) results in
\begin{equation*} \int x^4 J_1 \, dx = \int x^2 (x^2 J_1) \,dx = x^2 (x^2 J_2) - \int x^2 J_2 (2x) \,dx - 2 \int x^3 J_2 \, dx \end{equation*}
where \((x^2 J_2)'=x^2 J_1\) is used from (7.7.3). Next, if we again apply (7.7.3) with \(n=3\text{,}\) to the last integral, we get
\begin{equation*} \int x^4 J_1 \, dx = x^4 J_2 -2x^3 J_3 + C \end{equation*}

Subsection 7.7.2 Roots of the Bessel functions

There is not an analytic way to find the roots of any of the bessel functions, so we will resort to numerical approximation. Many Computer Algebra Systems and scientific computing languages have bessel functions built in and roots can be found with techniques such as Newton’s method or bisection.
In general, the \(i\)th root of \(J_0(x)\) is between \((i-1)\pi\) and \(i\pi\text{,}\) so the following Maple code will find the first 50.
\begin{equation*} \sigma := \text{[seq(fsolve(BesselJ(0,x)=0,x,(i-1)\cdot\pi,i\cdot\pi))]} \end{equation*}
The first ten values are: \(\{2.404825558, 5.520078110, 8.653727913, 11.79153444, \\14.93091771, 18.07106397, 21.21163663, 24.35247153, 27.49347913, 30.63460647\}\text{.}\)