📂Odinary Differential Equations

Series Solution of the Bessel Equation: Bessel Functions of the First Kind

Definition¹

For $\nu \in \mathbb{R}$, a differential equation of the following form is called a $\nu$ order Bessel equation.

$$\begin{align*} && x^{2} y^{\prime \prime} +xy^{\prime}+(x^{2}-\nu^{2})y &= 0 \\ \text{or} && y^{\prime \prime}+\frac{1}{x} y^{\prime} + \left( 1-\frac{\nu^{2}}{x^{2}} \right)y &= 0 \end{align*}$$

Explanation

The Bessel equation emerges when solving the wave equation in spherical coordinates. The coefficients are not constant but depend on the independent variable $x$. Since, at $x=0$, the following equation is satisfied, $x=0$ is a regular singular point.

$$\lim \limits_{x\rightarrow 0} x \frac{x}{x^{2}}=1<\infty,\quad \lim\limits_{x\rightarrow 0}x^{2}\frac{x^{2}-\nu^{2}}{x^{2}}=-\nu^{2} < \infty$$

Therefore, the solution can be found using the Frobenius method, and the series solution is as follows.

$$\begin{align*} J_{\nu}(x) &= \sum \limits_{n=0}^{\infty} \frac{(-1)^{n} }{\Gamma (n+1) \Gamma (n+\nu+1)} \left(\frac{x}{2} \right)^{2n+\nu} \\ J_{-\nu}(x) & =\sum \limits_{n=0}^{\infty}\frac{(-1)^{n}}{\Gamma (n+1)\Gamma (n-\nu+1)} \left( \frac{x}{2} \right)^{2n-\nu} \end{align*}$$

This is called the Bessel function of the first kind of order nu. Here, $\Gamma (x)$ is the gamma function. By looking at the order of both series, we can see that they are linearly independent. Therefore, the general solution to the $\nu$ order Bessel equation is as follows.

$$y(x)=AJ_{\nu}(x)+BJ_{-\nu}(x)$$

However, this only applies when $\nu$ is not an integer. If $\nu$ is an integer, since $J_{\nu}$ and $J_{-\nu}$ are not independent, we need to find the second solution, also known as the Neumann function.

Solution

Let’s assume the solution to the Bessel equation is the following power series.

$$\begin{equation} y=\sum \limits_{n=0}^{\infty}a_{n}x^{n+r}=x^{r}(a_{0}+a_{1}x+a_{2}x^{2}+\cdots)=a_{0}x^{r}+a_{1}x^{r+1}+a_{2}x^{r+2}+\cdots \label{1} \end{equation}$$

First, let’s slightly modify the form of the Bessel equation. Since $x(xy^{\prime})^{\prime}=x^{2}y^{\prime \prime}+xy^{\prime}$, the Bessel equation can be expressed as follows.

$$x(xy^{\prime})+(x^{2}-\nu^{2})y=0$$

To substitute into the Bessel equation, let’s find $x(xy^{\prime})^{\prime}, x^{2}y$ from $\eqref{1}$, as follows.

$$\begin{align*} y^{\prime} &=ra_{0}x^{r-1}+(r+1)a_{1}x^{r}+(r+2)a_{2}x^{r+1}+\cdots \\ xy^{\prime}&=ra_{0}x^{r}+(r+1)a_{1}x^{r+1}+(r+2)a_{2}x^{r+2}+\cdots \\ (xy^{\prime})^{\prime}&=r^{2}a_{0}x^{r-1}+(r+1)^{2}a_{1}x^{r}+(r+2)^{2}a_{2}x^{r+1}+\cdots \\ x(xy^{\prime})^{\prime}&=r^{2}a_{0}x^{r}+(r+1)^{2}a_{1}x^{r+1}+(r+2)^{2}a_{2}x^{r+2}+\cdots \\ &= \sum \limits_{n=0}^{\infty} a_{n}(r+n)^{2}x^{n+r} \end{align*}$$

Substituting this into the Bessel equation yields the following.

$$\begin{align*} &&&\left( r^{2}a_{0}x^{r}+(r+1)^{2}a_{1}x^{r+1}+(r+2)^{2}a_{2}x^{r+2}+\cdots \right) +(x^{2}-\nu^{2})\left( a_{0}x^{r}+a_{1}x^{r+1}+a_{2}x^{r+2}+\cdots \right) \\ \implies&&& \left( r^{2}a_{0}x^{r}+(r+1)^{2}a_{1}x^{r+1}+(r+2)^{2}a_{2}x^{r+2}+\cdots \right) +\left( a_{0}x^{r+2}+a_{1}x^{r+3}+a_{2}x^{r+4}+\cdots \right) \\ &&&+ \left( -\nu^{2}a_{0}x^{r}-\nu^{2}a_{1}x^{r+1}-\nu^{2}a_{2}x^{r+2}+\cdots \right) \end{align*}$$

Rearranging this to match the order of $x$,

$$\begin{align*} a_{0}(r^{2}-\nu^{2})x^{r}+a_{1}\left( (r+1)^{2}-\nu^{2} \right)x^{r+1} +\left(a_{2}(r+2)^{2}-a_{2}\nu^{2} +a_{0} \right)x^{r+2}& \\ +\cdots + (a_{n}(r+n)^{2}-a_{n}\nu^{2}+a_{n-2})x^{n+r}+\cdots &= 0 \end{align*}$$

For the equation to always hold for any $x$, all coefficients must be $0$. Let’s examine the first term.

$$a_{0}(r^{2}-\nu^{2})=0$$

Since $a_{0}\ne 0$, we have $r=\pm\nu$. Now, let’s look at the second term.

$$a_{1}\left( (r+1)^{2}-\nu^{2} \right)=0$$

Given the condition that $r=\pm \nu$, the expression inside the parentheses can never be $0$. Therefore, $a_{1}=0$. From the third term onwards, the coefficient is generally expressed as follows.

$$a_{n}(r+n)^{2}-a_{n}\nu^{2}+a_{n-2}=0$$

Arranging this,

$$\begin{equation} a_{n}=\frac{-a_{n-2}}{(r+n)^{2}-\nu^{2}} \label{2} \end{equation}$$

Combining the previously obtained $a_{1}=0$ with the above condition, one can see that for all odd $n$, $a_{n}=0$ applies. Therefore, we only need to derive when $n$ is even.

Case 1. $r=\nu$

In this case, $\eqref{2}$ is

$$a_{n}=\frac{- a_{n-2}}{n^{2}+2n\nu}=\frac{-a_{n-2}}{n(n+2\nu)}$$

Since we are only interested in when $n$ is even, let’s denote $n$ as $2n$. Then,

$$a_{2n}=\frac{-a_{2n-2}}{2n(2n+2\nu)}=\frac{-a_{2n-2}}{2^{2}n(n+\nu)}$$

Now, starting from $a_{2}$, we can derive as follows.

$$\begin{align*} a_{2} & =\frac{-a_{0}}{2^{2}\cdot 1(\nu+1)} \\ a_{4} &= \frac{-a_{2}}{2^{2}\cdot 2(\nu+2)}=\frac{a_{0}}{2^{4}\cdot2\cdot1(\nu+1)(\nu+2)} \\ a_{6} &=\frac{-a_{4}}{2^{2}\cdot3 (\nu+3)}=\frac{-a_{0}}{2^{6}\cdot3\cdot2\cdot1(\nu+1)(\nu+2)(\nu+3)} \\ \vdots \\ a_{2n}&=\frac{(-1)^{n}a_{0}}{2^{2n}n!(\nu+1)(\nu+2)\cdots(\nu+n)} \end{align*}$$ Using the gamma function simplifies the expression. The gamma function has the following properties.

$$\Gamma (\nu+1)=\nu\Gamma (\nu) \implies \dfrac{1}{\nu} = \dfrac{\Gamma (\nu)}{\Gamma (\nu+1)}$$

Using this, we obtain the following.

$$\begin{align*} && \frac{1}{\nu+1}&=\frac{\Gamma (\nu+1)}{\Gamma (\nu+2)}& \\ \implies && \frac{1}{(\nu+1)(\nu+2)}&=\frac{\Gamma (\nu+1)}{(\nu+2)\Gamma (\nu+2)}=\frac{\Gamma (\nu+1)}{\Gamma (\nu+3)} \\ \implies && \frac{1}{(\nu+1)(\nu+2)(\nu+3)}&=\frac{\Gamma (\nu+1)}{(\nu+3)\Gamma (\nu+3)} =\frac{\Gamma (\nu+1)}{\Gamma (\nu+4)} \\ \implies && &\vdots \\ \implies && \frac{1}{(\nu+1)\cdots(\nu+n)}&=\frac{\Gamma (\nu+1)}{\Gamma (\nu+n+1)} \end{align*}$$

Substituting this back into the coefficients we derived earlier, and expressing the factorial as a gamma function, each coefficient becomes as follows.

$$\begin{align*} a_{2} & =\frac{-a_{0}\Gamma (\nu +1)}{2^{2} 1! \Gamma (\nu+2)} \\ a_{4} &= \frac{a_{0} \Gamma (\nu+1)}{2^{4} 2!\Gamma (\nu+3)} \\ a_{6} &=\frac{-a_{0}\Gamma (\nu+1)}{2^{6}3!\Gamma (\nu+4)} \\ \vdots \\ a_{2n}&=\frac{(-1)^{n}\Gamma (\nu+1)}{2^{2n}\Gamma (n+1)\Gamma (\nu+n+1)}a_{0} \end{align*}$$

Substituting this into $\eqref{1}$,

$$\begin{align*} y &= \sum\limits_{n=0}^{\infty}a_{2n}x^{2n+\nu} \\ &= \sum \limits_{n=0}^{\infty}\frac{(-1)^{n}a_{0}\Gamma (\nu+1)}{2^{2n}\Gamma (n+1)\Gamma (\nu+n+1)}x^{2n+\nu} \end{align*}$$

Arranging the form,

$$y=\sum \limits_{n=0}^{\infty}\frac{(-1)^{n}2^{\nu}a_{0}\Gamma (\nu+1)}{\Gamma (n+1)\Gamma (n+\nu+1)} \left(\frac{x}{2}\right)^{2n+\nu}$$

If we let $a_{0}=\frac{1}{2^{\nu} \Gamma (\nu+1)}$,

$$J_{\nu}(x)=\sum \limits_{n=0}^{\infty}\frac{(-1)^{n}}{\Gamma (n+1)\Gamma (n+\nu+1)} \left(\frac{x}{2}\right)^{2n+\nu}$$

This is referred to as the Bessel function of the first kind of order $\nu$.

Case 2. $r=-\nu$

This case is simply the result of Case 1. with $\nu$ changed to $-\nu$. The solution process is completely the same, so we will only list the key results without detailed calculation steps and explanations.

$$a_{n}=\frac{- a_{n-2}}{n(n-2\nu)}$$

$$\begin{align*} a_{2n}&=\frac{ -a_{2n-2}}{ 2^{2}n(n-\nu) } \\ &=\frac{(1-)^{n}a_{0}}{2^{2n}n!(1-\nu)(2-\nu)\cdots(n-\nu)} \end{align*}$$

$$\frac{1}{(1-\nu)(2-\nu)\cdots(n-\nu)}=\frac{\Gamma (1-\nu)}{\Gamma (n-\nu+1)}$$

$$a_{2n}=\frac{(-1)^{n}\Gamma (1-\nu)}{2^{2n}\Gamma (n+1)\Gamma (n-\nu+1)}$$

$$\begin{align*} y &=\sum \limits _{n=0} ^{\infty} a_{2n}x^{2n-\nu} \\ &=\sum \limits_{n=0}^{\infty}\frac{(-1)^{n}a_{0}\Gamma (1-\nu)}{2^{2n}\Gamma (n+1)\Gamma (n-\nu+1)}x^{2n-\nu} \\ &=\sum \limits_{n=0}^{\infty}\frac{(-1)^{n}2^{-\nu}a_{0}\Gamma (1-\nu)}{\Gamma (n+1)\Gamma (n-\nu+1)} \left( \frac{x}{2} \right)^{2n-\nu} \end{align*}$$

$$a_{0}=\frac{2^{\nu}}{\Gamma (1-\nu)}$$

$$J_{-\nu}(x)=\sum \limits_{n=0}^{\infty}\frac{(-1)^{n}}{\Gamma (n+1)\Gamma (n-\nu+1)} \left( \frac{x}{2} \right)^{2n-\nu}$$

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