📂Mathematical Physics

Derivation of Bessel's Equation

Definition

The differential equation below is called the $\nu$th order Bessel’s equation.

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

Description

The solution to the Bessel’s equation is called the Bessel function.

Bessel functions are often seen in physics, engineering, and more, especially in problems involving cylindrical symmetry. For this reason, Bessel functions are also known as cylinder functions, though this term is less commonly used.

Derivation

In two-dimensional polar coordinates, the wave equation is given as follows.

$$\begin{equation} \dfrac{\partial ^2 u}{\partial t^2} = c^2 \left( \dfrac{\partial ^2 u}{\partial r^2}+\frac{1}{r}\dfrac{\partial u}{\partial r}+\frac{1}{r^2} \dfrac{\partial ^2 u}{\partial \theta ^2}\right) \end{equation}$$

$c$ is a constant. Let’s assume that the solution $u$ to the above equation is a function that can be separated into variables.

$$u(t, r, \theta)=T(t)R(r)\Theta (\theta)$$

Substituting $(1)$ gives

$$T^{\prime \prime}R\Theta=c^2\left( TR^{\prime \prime}\Theta + \dfrac{1}{r}TR^{\prime}\Theta + \frac{1}{r^2}TR\Theta^{\prime \prime} \right)$$

Dividing both sides by $c^2TR\Theta$ gives

$$\dfrac{T^{\prime \prime}}{c^2T}=\dfrac{R^{\prime \prime}}{R}+\dfrac{R^{\prime}}{rR}+\dfrac{\Theta^{\prime \prime}}{r^2\Theta}$$

The left-hand side is a function solely of $t$, and the right-hand side is a function of $r$ and $\theta$, thus both sides of the equation must be constant. If the left-hand side were not constant with respect to $t$, changing the value of $t$ would change the left-hand side without altering the right, breaking the equality. Therefore, for all $t$, $r$, and $\theta$, both sides must be constant. Let’s call this constant $-\mu ^2$. Then,

$$\begin{equation} \dfrac{T^{\prime \prime}}{c^2T}=\dfrac{R^{\prime \prime}}{R}+\dfrac{R^{\prime}}{rR}+\dfrac{\Theta^{\prime \prime}}{r^2\Theta}=-\mu^2 \end{equation}$$

First, let’s examine the equation for $r$ and $\theta$.

$$\dfrac{R^{\prime \prime}}{R}+\dfrac{R^{\prime}}{rR}+\dfrac{\Theta^{\prime \prime}}{r^2\Theta}=-\mu^2$$

Multiply both sides by $r^2$ and separate the equation into terms for $r$ and $\theta$,

$$\dfrac{r^2R^{\prime \prime}}{R}+\dfrac{rR^{\prime}}{R}+r^2\mu^2=-\dfrac{\Theta^{\prime \prime}}{\Theta}$$

Both sides of the equation, for reasons mentioned earlier, must also be constant. Let’s call this constant $\nu^2$. Then, we get the following equation.

$$\begin{equation} -\dfrac{\Theta^{\prime \prime}}{\Theta}=\nu^2 \quad \implies \quad \Theta^{\prime \prime} =-\nu^2 \Theta \quad \end{equation}$$

Returning to $(2)$ and organizing the equation for $t$ yields

$$\begin{equation} T^{\prime \prime}=-c^2\mu^2T \end{equation}$$

Substituting $(3)$ and $(4)$ into $(2)$ and organizing appropriately gives the following.

$$\begin{align*} &&\dfrac{-c^2 \mu^2 T}{c^2T}=\dfrac{R^{\prime \prime}}{R}+\frac{R^{\prime}}{rR}+\dfrac{-\nu^2\Theta}{r^2\Theta} \\ \implies &&\frac{1}{R}R^{\prime \prime}+\dfrac{1}{rR}R^{\prime}+\left(\mu^2-\frac{\nu^2}{r^2}\right) =0 \\ \implies && r^2R^{\prime \prime}(r)+rR^{\prime}(r)+(\mu^2r^2-\nu^2)R(r)=0 \end{align*}$$

Now, let’s introduce a substitution $\mu r=x$. And then let it be as follows.

$$R(r)=f(\mu r)=f(x),\quad R^{\prime}(r)=\mu f^{\prime}(\mu r)=\mu f^{\prime}(x),\quad R^{\prime \prime}(r)=\mu^2 f^{\prime \prime}(\mu r)=\mu^2 f^{\prime \prime}(x)$$

Substituting these equations into the ones we obtained earlier gives

$$\begin{align*} && \frac{x^2}{\mu^2}\mu^2f^{\prime \prime}(x) + \dfrac{x}{\mu}\mu f(x)+(x^2-\nu^2)f(x)&= 0 \\ \implies && x^2f^{\prime \prime}(x) + x f(x)+(x^2-\nu^2)f(x)&= 0 \end{align*}$$

The above equation is known as the $\nu$th order Bessel’s equation. It is commonly found in the following form.

$$\begin{align*} x^2 y^{\prime \prime} +xy^{\prime} +(x^2-\nu^2)y&= 0 \\ x(xy^{\prime})^{\prime}+(x^2 \nu ^2) y&= 0 \end{align*}$$

The first solution to this equation is as follows, and it is called the first kind Bessel function.

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

The second solution is as follows, and it is called the second kind Bessel function.​​​

$$N_{\nu}(x)=Y_{\nu}(x)=\frac{\cos (\nu \pi)J_{\nu}(x)-J_{-\nu}(x)}{\sin (\nu\pi)}$$

Therefore, the general solution to the Bessel’s equation is as follows.

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

In this case, $A$ and $B$ are constants.