Definition of Bessel Equation
The Bessel Equation is a type of second-order linear ODE that takes the form of
\begin{align}
x^2y^{\prime\prime} + xy’ + (x^2 -\nu^2)y = 0 \tag{1}
\end{align}
where \( \nu \) is a given parameter. It has a regular singularity at \( x = 0 \), and so we may want to apply the method of Frobenius to obtain two linearly independent solutions for (1). The indicial equation ((8) of this tutorial) in this case is
\begin{align}
m^2 + (1-1)m -\nu^2 &= 0 \\
m^2 -\nu^2 &= 0 \\
m &= \pm \nu \tag{2}
\end{align}
With \( m = \pm \nu \) as the exponents, plugging (4)-(6) of the referred tutorial into (1) produces
\begin{align}
\begin{aligned} x^2\sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{m+n-2} \\
+ x\sum_{n=0}^{\infty} (m+n)a_nx^{m+n-1} + (x^2 -\nu^2)\sum_{n=0}^{\infty} a_nx^{m+n}\end{aligned} &= 0 \\
\begin{aligned} \sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{m+n} \\
+ \sum_{n=0}^{\infty} (m+n)a_nx^{m+n} + \sum_{n=0}^{\infty} a_nx^{m+n+2} -\nu^2\sum_{n=0}^{\infty} a_nx^{m+n} \end{aligned} &= 0 \\
\begin{aligned} \sum_{n=0}^{\infty} (m+n)(m+n-1)a_nx^{m+n} \\
+ \sum_{n=0}^{\infty} (m+n)a_nx^{m+n} + \sum_{n=2}^{\infty} a_{n-2}x^{m+n} -\nu^2\sum_{n=0}^{\infty} a_nx^{m+n} \end{aligned} &= 0 \tag{3}
\end{align}
while adjusting the index in the third summation by \( -2 \). Comparing the coefficients yields the recurrence relation:
\begin{align}
(m+n)(m+n-1)a_n + (m+n)a_n + a_{n-2} -\nu^2a_n &= 0 \\
((m+n)^2 -\nu^2) a_n + a_{n-2} &= 0 \\
\Rightarrow a_n = -\frac{1}{(m+n)^2 -\nu^2}a_{n-2} \tag{4}
\end{align}
This is a two-term recurrence relation. Note that when \(n=1\) (and \( m = \pm \nu \)), (3) gives
\begin{align}
((1\pm \nu)^2 -\nu^2) a_1 &= 0 \\
(1 \pm 2\nu) a_1 &= 0 \tag{5}
\end{align}
and so all the coefficients for odd indices are zeros (as long as \(\nu \neq 1/2\)). Only the even indices \( n = 0,2,4,\ldots \) remain and are generated via
\begin{align}
a_n = -\frac{1}{n(n \pm 2\nu)}a_{n-2} \tag{6}
\end{align}
Bessel Functions of the First Kind
When \(\nu\) is not an Integer
When \( \nu \) is not an integer and also not a half-integer, the roots of the indicial equation \( \pm \nu \) will not differ by an integer and the Frobenius method has no problem straightforwardly giving two linearly independent solutions via (5) and (6). The more interesting scenario is when \( \nu = p+1/2 \) is a half integer, e.g. \( \nu = 1/2, 3/2, \ldots \), so that the exponents differ by an odd integer \( m_1 -m_2 = 2\nu = 2p+1 \). According to this tutorial, only the Frobenius series for the larger exponent is guaranteed, and we now consider whether the second Frobenius series for the smaller exponent is possible. In this case, the part \( n(n \pm 2\nu) \) in the denominator of (6) becomes \( n(n \pm (2p+1)) \). Notice that \(n\) is always an even integer, so this quantity is never zero, \( a_{n} \) will remain finite, and the computation can proceed.
The corresponding solutions are then
\begin{align}
y_{\pm \nu} &= x^{\pm \nu} (1 -\frac{x^2}{2(2\pm2\nu)} + \frac{x^4}{(2)(4)(2\pm2\nu)(4\pm2\nu)} -\cdots) \\
&= x^{\pm \nu} \sum_{m=0}^{\infty} \frac{(-1)^mx^{2m}\Gamma(1\pm \nu)}{(2^mm!)(2^m\Gamma(m+1\pm \nu))} \tag{7}
\end{align}
The normalization convention is to require \( a_0 = 1/(2^{\pm \nu} \Gamma(1\pm \nu))\), so that (7) becomes
\begin{align}
y_{\pm \nu} &= \sum_{m=0}^{\infty} \frac{(-1)^m}{m!\Gamma(m+1\pm \nu)} (\frac{x}{2})^{\pm\nu+2m}\tag{8}
\end{align}
This defines the Bessel functions of the first kind of order \(\pm \nu\): \(J_{\nu}(x)\) and \(J_{-\nu}(x)\). It is clear that if \( \nu \) is not an integer, they are linearly independent as they cannot be a constant multiple of each other. So the general solution in this case will simply be \( y = c_1 J_{\nu}(x) + c_2 J_{-\nu}(x) \).
Two particular relevant values are \(J_{1/2}(x)\) and \(J_{-1/2}(x)\). They can be shown to be equal
\begin{align}
J_{1/2}(x) &= \sqrt{\frac{2}{\pi x}} \sin(x) & J_{-1/2}(x) &= \sqrt{\frac{2}{\pi x}} \cos(x) \tag{9}
\end{align}
We can directly analyze the form of (8), but alternatively we may consider the Bessel equation where \( \nu = \pm 1/2 \):
\begin{align}
x^2y^{\prime\prime} + xy’ + (x^2 -\frac{1}{4})y = 0 \tag{10}
\end{align}
The two solutions are then \(J_{1/2}(x)\) and \(J_{-1/2}(x)\). But with the technique of this tutorial, we rewrite (10) as
\begin{align}
y^{\prime\prime} + \frac{1}{x}y’ + (1 -\frac{1}{4x^2})y = 0 \tag{11}
\end{align}
where
\begin{align}
u(x) &= \exp(-\frac{1}{2} \int^x \frac{1}{x’} dx’) \\
&= \exp(-\frac{1}{2} \ln x) = \frac{1}{\sqrt{x}} \tag{12} \\
g(x) &= (1 -\frac{1}{4x^2}) -\frac{1}{4}(\frac{1}{x})^2 -\frac{1}{2}(-\frac{1}{x^2}) \\
&= 1 -\frac{1}{4x^2} -\frac{1}{4x^2} +\frac{1}{2x^2} = 1 \tag{13}
\end{align}
and thus the transformed equation will be
\begin{align}
\frac{d^2v}{dx^2} + v = 0 \tag{14}
\end{align}
where we can easily conclude that \( v(x) = c_1\sin x + c_2\cos x \) and \( y = uv = \frac{c_1}{\sqrt{x}}\sin x + \frac{c_2}{\sqrt{x}}\cos x \). By observing and matching the terms, they can be seen to correspond to \(J_{1/2}(x)\) and \(J_{-1/2}(x)\) within a multiplicative factor.
When \(\nu\) is an Integer
When \( \nu \) is an integer, difficulties arise for the Frobenius method. First of all, if \( \nu = 0 \), it is a repeated root for the indicial equation (2) so there is only one Frobenius solution. It is derived in the same way as (7) and (8):
\begin{align}
J_0(x) &= (1 -\frac{x^2}{2^2} + \frac{x^4}{(2^2)(4^2)} -\cdots) \\
&= \sum_{m=0}^{\infty} \frac{(-1)^m}{(m!)^2} (\frac{x}{2})^{2m} \tag{15}
\end{align}
Meanwhile, when \( \nu = p = 1, 2, \ldots \) is a positive integer, the exponents now differ by an even integer \( m_1 -m_2 = 2p \). The same argument as before means that for the negative exponent, the part \( n(n-2\nu) \) in the denominator of (6) becomes \( n(n-2p) \). As \(n\) runs through the even numbers, at some point the denominator will be \( 0 \) and the coefficients will become infinite, thus there is no second Frobenius solution.
Another perspective to see the problem is that, the general solution derived above \( y = c_1 J_{\nu}(x) + c_2 J_{-\nu}(x) \) for a non-integer \( \nu \) will not be complete if \( \nu \) is instead an integer, because
\begin{align}
J_{-\nu}(x) &= \sum_{m=0}^{\infty} \frac{(-1)^m}{m!\Gamma(m+1-\nu)} (\frac{x}{2})^{-\nu+2m} \\
&= \sum_{m=\nu}^{\infty} \frac{(-1)^m}{m!\Gamma(m+1-\nu)} (\frac{x}{2})^{-\nu+2m} \\
&= \sum_{m=0}^{\infty} \frac{(-1)^{(m+\nu)}}{(m+\nu)!\Gamma((m+\nu)+1-\nu)} (\frac{x}{2})^{-\nu+2(m+\nu)} \\
&= (-1)^\nu \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(m+\nu)!\Gamma(m+1)} (\frac{x}{2})^{\nu+2m} \\
&= (-1)^\nu \sum_{m=0}^{\infty} \frac{(-1)^{m}}{\Gamma(m+1+\nu)m!} (\frac{x}{2})^{\nu+2m} \\
&= (-1)^{\nu} J_{\nu}(x) \tag{16}
\end{align}
The first-to-second line is due to the fact that \(\Gamma(n) = \pm \infty \) for a non-positive integer \(n = 0,-1,-2,\ldots\) so the relevant terms become \( 0 \) by the denominator. The second-to-third line is doing a change of variable \( m \rightarrow m+\nu \). This shows that \( J_{\nu}(x) \) and \( J_{-\nu}(x) \) are linearly dependent when \( \nu \) takes an integer value.
To find a second linearly independent solution we must resort to other ways. One possible approach is to utilize the already obtained complementary solution \( J_{\nu}(x) \) and apply this tutorial. The required form is then given by
\begin{align}
v’ &= \frac{1}{J_{\nu}(x)^2} e^{-\int (1/x) dx} = \frac{1}{xJ_{\nu}(x)^2} \tag{17} \\
y_2 &= y_1v = J_{\nu}(x)\int \frac{dx}{xJ_{\nu}(x)^2} \tag{18}
\end{align}
However, this is quite difficult to compute explicitly, and so we need a more tactful method. For this, let’s evaluate the Wronskian of \( J_{\nu}(x) \) and \( J_{-\nu}(x) \):
\begin{align}
&\quad W[J_{\nu}(x), J_{-\nu}(x)] \\
&= J_{\nu}(x)J_{-\nu}'(x) -J_{\nu}'(x)J_{-\nu}(x) \tag{19}
\end{align}
We have
\begin{align}
&\quad J_{\nu}(x)J_{-\nu}'(x) \\
&= (\sum_{m=0}^{\infty} \frac{(-1)^m}{m!\Gamma(m+1+\nu)} (\frac{x}{2})^{\nu+2m}) (\frac{1}{2}\sum_{l=0}^{\infty} \frac{(-1)^l(-\nu+2l)}{l!\Gamma(l+1-\nu)} (\frac{x}{2})^{-\nu+2l-1}) \\
&= \frac{1}{2}\sum_{m=0}^{\infty}\sum_{l=0}^{\infty} \frac{(-1)^{l+m}(-\nu+2l)}{l!m!\Gamma(m+1+\nu)\Gamma(l+1-\nu)} (\frac{x}{2})^{2m+2l-1} \\
&= \frac{1}{2}\sum_{l=0}^{\infty}\sum_{m=0}^{l} \frac{(-1)^{l}(-\nu+2(l-m))}{(l-m)!m!\Gamma(m+1+\nu)\Gamma(l-m+1-\nu)} (\frac{x}{2})^{2l-1} \tag{20}
\end{align}
where we have used the triangular summation, and similarly
\begin{align}
&\quad J_{-\nu}(x)J_{\nu}'(x) \\
&= \frac{1}{2}\sum_{l=0}^{\infty}\sum_{m=0}^{l} \frac{(-1)^{l}(\nu+2(l-m))}{(l-m)!m!\Gamma(m+1-\nu)\Gamma(l-m+1+\nu)} (\frac{x}{2})^{2l-1} \tag{21}
\end{align}
Now we may make use of Euler’s reflection formula:
\begin{align}
\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)} \tag{22}
\end{align}
and note that for \( l \geq 1 \), (20) and (21) cancel out each other nicely, so what remains is
\begin{align}
&\quad J_{\nu}(x)J_{-\nu}'(x) -J_{\nu}'(x)J_{-\nu}(x) \\
&= \frac{1}{2}\frac{-\nu}{\Gamma(1+\nu)\Gamma(1-\nu)} (\frac{2}{x}) -\frac{1}{2}\frac{\nu}{\Gamma(1-\nu)\Gamma(1+\nu)} (\frac{2}{x}) \\
&= \frac{-1}{\Gamma(\nu)\Gamma(1-\nu)} \frac{2}{x} \\
&= -\frac{2}{\pi x} \sin(\pi \nu) \tag{23}
\end{align}
This form indicates that \( J_{\nu}(x) \) and \( J_{-\nu}(x) \) are linearly dependent because the \( \sin(\pi \nu) \) part vanishes whenever \( \nu \) is an integer. So, a natural remedy is to divide some specific linear combination of \( J_{\nu}(x) \) and \( J_{-\nu}(x) \) by \( \sin(\pi \nu) \) so that the new Wronskian may then become non-zero so it yields a new linearly independent solution. Such an educated guess is not too far from the actual expression used, which is
\begin{align}
y_2 &= \frac{J_{\nu}(x) \cos(\nu \pi) -J_{-\nu}(x)}{\sin(\nu \pi)} \tag{24}
\end{align}
This form is further motivated by the fact that, if the denominator \( \sin(\pi \nu) \) takes the value of \( 0 \) when \( \nu \) is an integer, then the only hope to compute something well-defined is to have the numerator also \( 0 \) (due to (16)) so that it is an indeterminate form and we can proceed in the sense of the L’Hôpital’s rule. (24) is then known as the Bessel functions of the second kind of order \(\nu\), denoted by \( Y_{\nu}(x) \), and for an integer \( \nu \) we take the limit \(Y_{\nu}(x) = \lim_{\mu \to \nu}Y_{\mu}(x) \). Notice that even when \( \nu \) is not an integer, \( Y_{\nu}(x) \) is valid and linearly independent from \( J_{\nu}(x) \). So for all values of \( \nu \), the general solution to the Bessel equation (1) can be written as
\begin{align}
y = c_1 J_{\nu}(x) + c_2 Y_{\nu}(x) \tag{25}
\end{align}
We note that in this case, the corresponding Wronskian \( W[J_{\nu}(x), Y_{\nu}(x)] \) is nicely \( 2/\pi x\).
Exercise
Show that an equivalent definition of \( Y_{p}(x) \) with \( p \) being an integer is
\begin{align}
Y_{p}(x) = \frac{1}{\pi} [\frac{\partial}{\partial \nu}J_\nu(x) -(-1)^p\frac{\partial}{\partial \nu}J_{-\nu}(x)]_{\nu = p}\tag{26}
\end{align}
by L’Hôpital’s rule.
Answer
It is a straightforward use of L’Hôpital’s rule on (24):
\begin{align}
Y_{p}(x) &= \frac{\frac{\partial}{\partial \nu}[J_{\nu}(x) \cos(\nu \pi) -J_{-\nu}(x)]_{\nu = p}}{[\frac{\partial}{\partial \nu}\sin(\nu \pi)]_{\nu = p}} \\
&= \frac{[(\frac{\partial}{\partial \nu}J_{\nu}(x)) \cos(\nu \pi) -\pi J_{\nu}(x) \sin(\nu \pi) -\frac{\partial}{\partial \nu}J_{-\nu}(x)]_{\nu = p}}{[\pi\cos(\nu \pi)]_{\nu = p}} \\
&= \frac{[\frac{\partial}{\partial \nu}J_\nu(x) -\frac{1}{\cos(\nu \pi)}\frac{\partial}{\partial \nu}J_{-\nu}(x)]_{\nu = p}}{\pi} \\
&= \frac{1}{\pi} [\frac{\partial}{\partial \nu}J_\nu(x) -(-1)^p\frac{\partial}{\partial \nu}J_{-\nu}(x)]_{\nu = p}
\end{align}
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