Lemma 8.6.2.
\begin{align}
\frac{d}{dx} \bigl( x^n J_n(x) \bigr) \amp = x^n J_{n-1} (x) \tag{8.6.8}\\
\frac{d}{dx} \bigl( x^{-n} J_n(x) \bigr) \amp = -x^{-n} J_{n+1} (x)\tag{8.6.9}
\end{align}
for all \(x > 0\text{.}\)
Section 8.6 Bessel’s equation and Bessel Functions
Bessel’s equation or
\begin{equation}
x^2 y'' + x y' + (x^2-\nu^2)y = 0\tag{8.6.1}
\end{equation}
for real constants \(\nu\) is related to solving partial differential equations in circular or cylindrical regions and with certain boundary conditions satisfy Sturm-Liouville problems seen in Section 8.5. In this section, we will provide solutions to (8.6.1) as well as a number of properties. In Chapter 9, we will use these solutions to solve partial differential equations.
Subsection 8.6.1 Solutions of Bessel’s Equation
We will use the Frobenius method from Subsection 8.4.5 to solve this. This means that we assume that the solution of the form:
\begin{equation}
y(x) = x^r \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{r+n}\tag{8.6.2}
\end{equation}
and recall that the first two derivatives of this is
\begin{align}
y'(x) \amp = \sum_{n=0}^{\infty} (r+n) a_n x^{r+n-1} \tag{8.6.3}\\
y''(x) \amp = \sum_{n=0}^{\infty} (r+n)(r+n-1) a_n x^{r+n-2} \tag{8.6.4}
\end{align}
\begin{equation*}
\begin{aligned} x^2 \sum_{n=0}^{\infty} (r+n)(r+n-1) a_n x^{r+n-2} + x \sum_{n=0}^{\infty} (r+n) a_n x^{r+n-1} + (x^2 - \nu^2) \sum_{n=0}^{\infty} a_n x^{r+n} \amp = 0 \\ \sum_{n=0}^{\infty} (r+n)(r+n-1) a_n x^{r+n} + \sum_{n=0}^{\infty} (r+n) a_n x^{r+n} + \sum_{n=0}^{\infty} a_n x^{r+n+2} - \nu^2 \sum_{n=0}^{\infty} a_n x^{r+n} \amp = 0 \end{aligned}
\end{equation*}
The first two terms can be combined and the third terms re-indexed to give
\begin{equation*}
\sum_{n=0}^{\infty} (r+n)^2 a_n x^{r+n} + \sum_{n=2}^{\infty} a_{n-2} x^{r+n} - \nu^2 \sum_{n=0}^{\infty} a_n x^{r+n} = 0
\end{equation*}
If the first two terms of the first and third series are written out then all series can be combined to give,
\begin{equation}
\begin{aligned} \left(r^2 - \nu^2 \right)a_0 x^r + \left((r+1)^2 - \nu^2\right)a_1 x^{r+1} \quad \quad\amp + \sum_{n=2}^{\infty} \left(\left( (r+n)^2 - \nu^2 \right)a_n + a_{n-2}\right) x^{r+n} \amp = 0 \end{aligned}\tag{8.6.5}
\end{equation}
For power series, each coefficient must be zero. The first term is zero when
\begin{equation*}
r^2 - \nu^2 = 0
\end{equation*}
which has two roots \(r=\nu\) and \(r=-\nu\) and consider the negative root later. The coefficient of the second term of (8.6.5) with \(r=\nu\) is
\begin{equation*}
\left((\nu+1)^2 - \nu^2\right)a_1 = (2\nu + 1)a_1
\end{equation*}
and this must be 0, so \(a_1=0\text{.}\)
The coefficient of the general \(x^{r+n}\) term of (8.6.5) is set to zero to give the following recurrence relationship for \(n\geq 2\) when \(r = \nu \) results in
\begin{equation}
a_n = - a_{n-2}\frac{1}{(\nu+n)^2 - \nu^2} = - a_{n-2} \frac{1}{n(n+2\nu)}.\tag{8.6.6}
\end{equation}
Because \(a_1=0\text{,}\) this relationship shows that all odd coefficients are also 0.
Writing out a few terms of the relationship above:
\begin{equation*}
\begin{aligned} a_2 \amp = - a_0 \frac{1}{2(2+2\nu)} = -a_0 \frac{1}{2^2(1+\nu)} \\ a_4 \amp = -a_2 \frac{1}{4(4+2\nu)} = a_0 \frac{1}{2^4 \cdot 2(1+\nu)(2+\nu)} \\ a_6 \amp = -a_4 \frac{1}{6(6+2\nu)} = -a_0 \frac{1}{2^6\cdot 3 \cdot 2 (3+\nu)(2+\nu)(1+\nu)} \\ a_8 \amp = -a_6 \frac{1}{8(8+2\nu)} = a_0 \frac{1}{2^8 \cdot 4 \cdot 3 \cdot 2 (4+\nu)(3+\nu)(2+\nu)(1+\nu) } \end{aligned}
\end{equation*}
Since the Gamma Function has the property that \(\Gamma(n+m+1) = (n+m)\Gamma(n+m) = (n+m)(n+m-1)\Gamma(n+m-1)\text{,}\) the terms
\begin{equation*}
(m+\nu)(m-1+\nu)(m-2+\nu) \cdots (1+\nu) = \frac{\Gamma(m+\nu+1)}{\Gamma(\nu+1)}
\end{equation*}
and the even coefficients can be written as
\begin{equation*}
a_{2m} = \frac{(-1)^ma_0}{2^{2m}m! \frac{\Gamma(m+\nu+1)}{\Gamma(\nu+1)}} = \frac{(-1)^m a_0 \Gamma(\nu+1)}{2^{2m}m!\Gamma(m+\nu+1)}.
\end{equation*}
Lastly, the coefficient \(a_0\) is often chosen such that \(a_0 \Gamma(\nu+1) = 1\) and this results in the power series solution of Bessel’s equation as
\begin{equation}
J_{\nu} (x) = \frac{x^{\nu}}{2^{\nu}} \sum_{m=0}^{\infty} \frac{(-1)^m}{4^m m! \Gamma(m+\nu+1)} x^{2m}\tag{8.6.7}
\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 (8.6.1) in which \(\nu=n\) is an integer. Recall that above, we found that \(r=\nu\) arose from the Frobenius method. The solution \(r = -\nu\) also satisfies this with the same steps and thus \(J_{-\nu}(x)\) is a second linearly independent solution to (8.6.1).
Subsection 8.6.2 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 \(J_0\text{.}\)
\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 (8.6.1) 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.
\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*}
In addition, there are another two identities for Bessel functions that are often called recurrence relationships.
Lemma 8.6.3.
\begin{align}
x(J_{n-1}+J_{n+1}) \amp = 2n J_n\tag{8.6.10}\\
J_{n-1} - J_{n+1} \amp = 2 J'_n \tag{8.6.11}
\end{align}
Proof.
\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*}
\begin{align*}
n J_n + x J'_n \amp = xJ_{n-1} \\
-nJ_n+xJ'_n \amp = -xJ_{n+1}
\end{align*}
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 8.6.4.
Use the identities in Lemma 8.6.2 and (8.6.10) to find \(J_3\) in terms of \(J_0\) and \(J_1\text{.}\)
Solution.
\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 (8.6.10) 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 8.6.5.
Use the identities in lemmas Lemma 8.6.2 and Lemma 8.6.3 to find \(J'_0\) and \(J'_1\) in terms of \(J_0\) and \(J_1\text{.}\)
Solution.
\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*}
\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*}
\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 8.6.6.
Evaluate \(\int x^4 J_1 (x) \, dx\text{.}\)
Solution.
\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 (8.6.8). Next, if we again apply (8.6.8) 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 8.6.3 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 one can use numerical techniques to find the roots in these intervals. However, also, there are often packages available to find these as well. In Julia, the package
FunctionZeros can be used as following
σ₀ = map(i -> besselj_zero(0, i), 1:10) σ₁ = map(i -> besselj_zero(1, i), 1:10) σ₂ = map(i -> besselj_zero(2, i), 1:10)
| Row | \(\sigma_0\) | \(\sigma_1\) | \(\sigma_2\) |
|---|---|---|---|
| 1 | 2.4048 | 3.8317 | 5.1356 |
| 2 | 5.5200 | 7.0155 | 8.4172 |
| 3 | 8.6537 | 10.1735 | 11.6198 |
| 4 | 11.7915 | 13.3237 | 14.7965 |
| 5 | 14.9309 | 16.4706 | 17.9598 |
| 6 | 18.0711 | 19.6159 | 21.1177 |
| 7 | 21.2116 | 22.7601 | 24.2701 |
| 8 | 24.3525 | 25.9037 | 27.4206 |
| 9 | 27.4935 | 29.0468 | 30.5692 |
| 10 | 30.6346 | 32.1897 | 33.7165 |
Subsection 8.6.4 Bessel Functions of the Second Kind
Above we mentioned that \(J_{-\nu}\) is a solution to (8.6.1), however, the function
\begin{equation}
Y_{\nu}(x) = \frac{J_{\nu}(x) \cos (\nu \pi) - J_{-\nu}(x)}{\sin(\nu \pi)}\tag{8.6.12}
\end{equation}
is generally used instead and is called the Bessel function of the second kind.
Example 8.6.8.
Solution.
We can write \(Y_{\nu}(x)\) as the power series
\begin{equation*}
Y_{\nu}(x) = \frac{\cos(\nu \pi)}{\sin(\nu \pi)} \frac{x^{\nu}}{2^{\nu}} \sum_{m=0}^{\infty} \frac{(-1)^m}{4^m m! \Gamma( m+\nu+1)} x^{2m} - \frac{1}{\sin(\nu \pi)}\frac{x^{-\nu}}{2^{-\nu}} \sum_{m=0}^{\infty} \frac{(-1)^m}{4^m m! \Gamma( m-\nu+1)} x^{2m}
\end{equation*}
