The Bessel Equation

In this note we investigate series solutions to the Bessel equation

\begin{equation} \label{eq:bessel} z^2 u'' + zu' + (z^2 - \nu^2)u = 0 \end{equation}

Solutions to the Bessel equation are transcendental functions called Bessel functions, and they have wide application. For instance they appear in the classical theory of modular forms on the upper half plane, and the solutions of the heat equation and Schrodinger equation in cylindrical coordinates. There are many different kinds of Bessel function. Each of these tends to be best suited to a specific type of application. In this note we treat the most common kinds of Bessel function.

For convenience, we introduce the notation

\begin{equation*} L_\nu = z^2 \frac{\mathrm{d}^2}{\mathrm{d}z^2} + z \frac{\mathrm{d}}{\mathrm{d}z} + (z^2 - \nu^2) \end{equation*}

for the differential operator in \ref{eq:bessel}. The Bessel equation can then be rewritten as

\begin{equation*} L_\nu[v] = 0 \end{equation*}

Bessel Functions of the First Kind

Derivation of Solutions

Note that our equation has a regular singular point at \(z = 0\). In this section we follow the usual approach for finding series solutions to ODE with regular singularities. Namely, we look for solutions of the form

\begin{equation*} u(z) = z^\alpha \sum_{n=0}^{\infty} a_n z^n = \sum_{n=0}^{\infty} a_n z^{n+\alpha} \end{equation*}

where \(a_0 \ne 0\). Substituting this into equation \ref{eq:bessel} yields

\begin{align*} 0 &= \sum_{n=0}^{\infty} (n+\alpha)(n+\alpha -1)z^{n+\alpha} + \sum_{n=0}^{\infty} (n+\alpha) z^{n+\alpha} + \sum_{n=0}^{\infty} a_n z^{n+2+\alpha} - \sum_{n=0}^{\infty} \nu^2 a_n z^{n+\alpha} \\ &= \sum_{n=0}^{\infty} [((n+\alpha -1) + 1)(n+\alpha) - \nu^2]a_nz^{n+\alpha} + \sum_{n=2}^{\infty} a_{n-2}z^{n+\alpha} \\ &= \sum_{n=0}^{\infty} ((n+\alpha)^2 - \nu^2)a_nz^{n+\alpha} + \sum_{n=2}^{\infty} a_{n-2}z^{n+\alpha} \\ &= (\alpha^2 - \nu^2)a_0z^\alpha + ((1 + \alpha)^2 - \nu^2)a_1z^{1+\alpha} + \sum_{n=2}^{\infty} [((n + \alpha)^2 - \nu^2)a_n + a_{n-2}]z^{n+\alpha} \end{align*}

Comparing \(z^{n+\alpha}\)-coefficients in the equation above produces a series of identities involving the \(a_n\) terms. We consider the coefficients on \(z^\alpha\), \(z^{1+\alpha}\), and \(z^{n+\alpha}\) separately. To keep matters simple for the moment, it will be useful to assume that the terms \((n + \alpha)^2 - \nu^2\) appearing above do not vanish for any integer \(n \geq 1\); we treat the remaining cases later. Since

\begin{equation*} (n + \alpha)^2 - \nu^2 = (\alpha + \nu + n)(\alpha - \nu + n), \end{equation*}

it will be enough to assume \(\alpha \pm \nu\) is not a negative integer.

Case \(z^{\alpha}\): Comparing these coefficients shows that

\begin{equation*} (\alpha^2 - \nu^2)a_0 = 0 \end{equation*}

Since \(a _ 0 \ne 0\), this implies \(\alpha = \pm \nu\). The restriction \(\alpha \pm \nu \not\in \mathbb{Z} _ {<0}\) thus reduces to requiring \(-2\nu \not\in \mathbb{Z} _ {<0}\) when \(\alpha = -\nu\) (resp. \(2\nu \not\in \mathbb{Z}_{<0}\) when \(\alpha = \nu\)).

Case \(z^{\alpha + 1}\): From this coefficient we obtain

\begin{equation*} ((1 - \alpha)^2 - \nu^2)a_1 = (\alpha - \nu + 1)(\alpha + \nu + 1)a_1 = 0 \end{equation*}

We have assumed \((\alpha \pm \nu) \ne - 1\), so the above implies \(a_1 = 0\).

Case \(z^{\alpha + n}\): This time we obtain a recurrence of the form

\begin{equation*} [(n + \alpha)^2 - \nu^2]a_n + a_{n-2} = (\alpha - \nu + n)(\alpha + \nu + n)a_n + a_{n-2} = 0 \end{equation*}

Rearranging, this is the same as

\begin{equation*} a_n = - \frac{a_{n-2}}{(\alpha - \nu + n)(\alpha - \nu + n)} \end{equation*}

This allows us to quickly compute the odd-order terms \(a_n = a_{2k - 1}\). In particular, bootstrapping from the above case,

\begin{equation*} a_3 = - \frac{a_{1}}{(\alpha - \nu + 3)(\alpha - \nu + 3)} = 0 \end{equation*}

Repeating this for \(n = 5, 7, \dots\) demonstrates that \(a _ {2k-1} = 0\) for all \(k \in \mathbb{Z} _ {>0}\). Induction also allows us to compute the even-order terms \(a_n = a_{2k}\). Indeed, when \(n = 2\) we have

\begin{equation*} a_2 = \frac{-a_{0}}{(\alpha - \nu + 2)(\alpha - \nu + 2)} \end{equation*}

and in general

\begin{equation*} a_{2k} = \frac{(-1)^k a_{0}}{(\alpha - \nu + 2)(\alpha - \nu + 4) \cdots (\alpha - \nu + 2k)(\alpha + \nu + 2) \cdots (\alpha - \nu + 2k)} \end{equation*}

If we take \(\alpha = \nu\), this reduces to

\begin{equation*} a_{2k} = - \frac{(-1)^k a_{0}}{2^{2k} k! (\nu + 1)(\nu + 2) \cdots (\nu + k)} \end{equation*}

Notice that there is no restriction on \(a_0\).

Combining all of our work so far, we arrive at our first candidate solution for the Bessel equation:

\begin{equation} \label{eq:candidate1} u(z) = z^\nu \left[a_0 + \sum_{k=1}^{\infty} \frac{a_0 (-1)^k (\frac{1}{2}z)^{2k}}{k! (\nu +1)(\nu +2)\cdots (\nu + m)} \right] \end{equation}

Since we have no restriction on \(a_0\), it is convenient to choose \(a_0 = \frac{1}{2^\nu \Gamma(\nu + 1)}\). The solution \(u\) in equation \ref{eq:candidate1} then simplifies to

\begin{equation*} J_\nu(z) = u(z) = \sum_{k=0}^{\infty} \frac{(-1)^k (\frac{1}{2}z)^{\nu + 2k}}{k! \Gamma(\nu + k + 1)} \end{equation*}

If we had instead choosen \(\alpha = -\nu\) above, we could have set \(a_0 = \frac{1}{2^{-\nu} \Gamma(-\nu + 1)}\). This leads to a solution of the form

\begin{equation*} J_{-\nu}(z) = \sum_{k=0}^{\infty} \frac{(-1)^k (\frac{1}{2}z)^{\nu + 2k}}{k! \Gamma(\nu + k + 1)} \end{equation*}

In all situations considered so far (ie. when \(2\nu\) is not an integer) the sums defining \(J_{\pm \nu}\) converge for all \(z \in \mathbb{C}\setminus \{0\}\).

Definition 1: The functions \(J_\nu\) and \(J_{-\nu}\) defined above are called Bessel functions of the first kind. ∎

It can be shown, eg. with the Weierstrass M-test, that the sum defining \(J_\nu\) converges absolutely and uniformly in a neighbourhood of any \(z \in \mathbb{C} \setminus \{ 0 \}\) and in any bounded domain of values of \(\nu\).

Fundamental System of \(1^{st}\)-kind solutions.

In this section we show that, provided \(\nu \not\in \mathbb{Z}\), the system \(\{ J_\nu, J_{-\nu}\}\) forms a linearly independent basis of solutions for the Bessel equation \ref{eq:bessel}. To do so, it suffices to show the Wronskian determinant \(\mathcal{W}(J_\nu, J_{-\nu}) \) does not vanish. In other words, we want to show

\begin{equation*} \mathcal{W}(J_{\nu}(z), J_{-\nu}(z)) = \det \begin{pmatrix} J_\nu & J_{-\nu} \\ \frac{\partial J_\nu}{\partial z} & \frac{\partial J_{-\nu}}{\partial z} \end{pmatrix} = 0 \end{equation*}

First, we attempt to extract a Wronskian-like quantity from the Bessel equation by computing

\begin{align*} J_\nu \cdot L_\nu[J_{-\nu}] &- J_{-\nu} \cdot L_\nu[J_\nu] \\ &= z^2 J_\nu \frac{\partial^2 J_{-\nu}}{\partial z^2} - z^2 J_{-\nu} \frac{\partial^2 J_\nu}{\partial z^2} + z J_\nu \frac{\partial J_{-\nu}}{\partial z} - z J_{-\nu} \frac{\partial J_\nu}{\partial z} \\ &= z \left[z J_\nu \frac{\partial^2 J_{-\nu}}{\partial z^2} - z J_{-\nu} \frac{\partial^2 J_\nu}{\partial z^2} + J_\nu \frac{\partial J_{-\nu}}{\partial z} - J_{-\nu} \frac{\partial J_\nu}{\partial z} \right] \\ &= 0 \end{align*}

The whole expression equals zero since \(L_\nu[J_{-\nu}] = 0 = L_\nu[J_\nu]\). The bracketed term on the last line above somewhat resembles the product \(z \mathcal{W}(J_\nu, J_{-\nu})\). The main differences between the two are the second-order derivatives and factor of \(z\). Both of these differences can be accounted for by taking a derivative. We compute:

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}z} [ zW(J_\nu &, J_{-\nu})] \\ &= z J_\nu \frac{\partial^2 J_{-\nu}}{\partial z^2} + \left(z \frac{\partial J_{\nu}}{\partial z} + J_\nu\right)\frac{\partial J_{-\nu}}{\partial z} - z J_{-\nu} \frac{\partial^2 J_\nu}{\partial z^2} - \left(z \frac{\partial J_{-\nu}}{\partial z} + J_{-\nu}\right) \frac{\partial J_\nu}{\partial z} \\ &= z J_\nu \frac{\partial^2 J_{-\nu}}{\partial z^2} J_{-\nu} - z J_{-\nu} \frac{\partial^2 J_\nu}{\partial z^2} + J_\nu \frac{\partial J_{-\nu}}{\partial z} - J_{-\nu} \frac{\partial J_\nu}{\partial z} \\ &= \frac{J_\nu \cdot L_\nu[J_{-\nu}] - J_{-\nu} \cdot L_\nu[J_\nu]}{z} \\ &= 0 \end{align*}

In other words, the function \(z \mapsto z\cdot \mathcal{W}(J_\nu(z), J_{-\nu}(z))\) is constant. Solving for the Wronskian, we have

\begin{equation} \label{eq:frac} \mathcal{W}(J_\nu, J_{-\nu}) = \frac{C}{z} \end{equation}

In order to evaluate the constant \(C\), first let us note the easily obtained formulae, valid when \(\nu \not\in \mathbb{Z}\) and for small \(|z|\):

\begin{equation*} J_\nu(z) = \frac{\left(\frac{1}{2} z\right)^\nu}{\Gamma(\nu + 1)} (1 + O(z^2)), \quad\quad \frac{\partial J_{\nu}(z)}{\partial z} = \frac{\left(\frac{1}{2} z\right)^{\nu-1}}{2\Gamma(\nu)} (1 + O(z^2)) \end{equation*}

Similar formulae exist for \(J_{-\nu}\) and \(\frac{\partial J_{-nu}}{\partial z}\). This allows us to compute

\begin{align*} \mathcal{W}(J_\nu, J_{-\nu}) = J_\nu \frac{\partial J_\nu}{\partial z} - J_{-\nu} \frac{\partial J_{-\nu}}{\partial z} &= \frac{1}{z} \left[\frac{1}{\Gamma(\nu + 1)\Gamma(-\nu)} - \frac{1}{\Gamma(\nu)\Gamma(-\nu + 1)} \right] + O(z) \\ &= -\frac{2\sin(\nu\pi)}{\pi z} + O(z) \end{align*}

In light of equation \ref{eq:frac} the \(O(z)\) term above must vanish. We conclude

\begin{equation*} \mathcal{W}(J _ \nu, J _ {-\nu}) = \frac{2\sin(\nu\pi)}{\pi z} \end{equation*}

Since \(\nu \not\in \mathbb{Z}\), \(\sin(\nu z\) does not vanish. We arrive at the following theorem:

Theorem 1: When \(\nu \not\in \mathbb{Z}\), the \(1^{st}\)-kind functions \(J_\nu\) and \(J_{-\nu}\) form a fundamental system of solutions to Bessel’s equation. ∎

On account of the relation \(J_n(z) = (-1)^n J_{-n}(z)\), these functions do not form a fundamental system when \(\nu = n \in \mathbb{Z}\).

Bessel Functions of the Second Kind

Derivation of Solutions

In the previous section we derived a fundamental system of solutions to the Bessel equation \ref{eq:bessel} subject to the condition \(\pm 2\nu \ne n\) for any integer \(n\in \mathbb{Z}\). That is, we required \(\nu\) not be an even integer, nor half of an odd integer. In this section we shall derive a solution that is valid even in the case \(\nu \in \mathbb{Z}\).

For the remainder of this section let us fix some \(n \in \mathbb{Z}\). Now, since \(J_\nu\) and \(J_{-\nu}\) solve the Bessel equation for non-integer \(\nu\), so does the linear combination

\begin{equation*} \frac{J_\nu(z) - (-1)^n J_{-\nu}(z)}{\nu - n} \end{equation*}

If we then set

\begin{equation*} \mathbf{Y}(z) = \lim_{\nu \rightarrow n} \frac{J_\nu(z) - (-1)^n J_{-\nu}(z)}{\nu - n} \end{equation*}

it seems reasonable to believe that \(\mathbf{Y}\) is also a solution to Bessel’s equation. In order to check this, we recall \(J_\nu\) is analytic in \(\nu\) and compute

\begin{align*} \mathbf{Y}(z) &= \lim_{\nu \rightarrow n} \frac{J_\nu(z) - (-1)^n J_{-\nu}(z)}{\nu - n} \\ &= \lim_{\nu \rightarrow n} \frac{J_\nu(z) - J_n(z)}{\nu - n} - (-1)^n \frac{J_{-nu}(z) - J_{-n}(z)}{\nu - n} \\ &= \left. \frac{\partial}{\partial\nu}J_\nu(z) \right|_{\nu = n} - \left. (-1)^n \frac{\partial}{\partial\nu} J_{-\nu} \right|_{\nu = n} \end{align*}

We will show this expression solves equation \ref{eq:bessel}. First we exploit the equality of mixed partials to write

\begin{equation*} \frac{\partial}{\partial\nu} L_\nu[J_{\pm\nu}] \end{equation*}