Differential Equations: Special Functions

Lesson 2 of 3

Module 2: The Bessel Equation & Functions

Core140 min

The Bessel Equation

1. Definition and Solution

For the equation:

$x^2 y'' + x y' + (x^2 - k^2)y = 0$

The general solution is:

$y = c_1 J_k(x) + c_2 Y_k(x)$

Where $J_k(x)$ is the Bessel function of the first kind (finite at $x=0$ for $k \geq 0$ ) and $Y_k(x)$ is the Bessel function of the second kind (singular at $x=0$ ).

2. Series Definition

The Bessel function of the first kind is defined as:

$J_k(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \, \Gamma(n+k+1)} \left( \frac{x}{2} \right)^{2n+k}$

3. Important Identities (Recurrence Relations)

We must master these relationships:

1. $\frac{d}{dx} [x^k J_k(x)] = x^k J_{k-1}(x)$

2. $\frac{d}{dx} [x^{-k} J_k(x)] = -x^{-k} J_{k+1}(x)$

3. $J_{k-1}(x) - J_{k+1}(x) = 2J'_k(x)$

4. $J_{k-1}(x) + J_{k+1}(x) = \frac{2k}{x} J_k(x)$

Watch: Bessel Functions in Physics

15 min · watch

Visualizing the modes of vibration on a circular drum head, which correspond to the zeros of Bessel functions.

Solved Problem: Proving Identities

30 min · read

Problem

Prove that $\frac{d}{dx} J_0(x) = -J_1(x)$ .

Proof

Step 1: Write the series for $J_0(x)$

$J_0(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2} \left(\frac{x}{2}\right)^{2n}$

Step 2: Differentiate term-by-term

$J_0'(x) = \sum_{n=1}^{\infty} \frac{(-1)^n}{(n!)^2} \cdot 2n \left(\frac{x}{2}\right)^{2n-1} \cdot \frac{1}{2}$

Note: The sum starts at $n=1$ because the derivative of the constant term ( $n=0$ ) is 0.

Step 3: Simplify coefficients

$J_0'(x) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n!(n-1)!} \left(\frac{x}{2}\right)^{2n-1}$

Step 4: Re-index the sum

Let $m = n-1$ . Then $n = m+1$ . When $n=1$ , we have $m=0$ .

$J_0'(x) = \sum_{m=0}^{\infty} \frac{(-1)^{m+1}}{(m+1)! \, m!} \left(\frac{x}{2}\right)^{2m+1}$

$J_0'(x) = - \sum_{m=0}^{\infty} \frac{(-1)^m}{m! (m+1)!} \left(\frac{x}{2}\right)^{2m+1}$

Step 5: Compare with $J_1(x)$

The series for $J_1(x)$ is exactly $\sum \frac{(-1)^m}{m! (m+1)!} (x/2)^{2m+1}$ .

Therefore:

$J_0'(x) = -J_1(x)$

Q.E.D.

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1. Definition and Solution

For the equation:

x^2 y'' + x y' + (x^2 - k^2)y = 0

The general solution is:

y = c_1 J_k(x) + c_2 Y_k(x)

Where

J_k(x)

is the Bessel function of the first kind (finite at

x=0

for

k \geq 0

) and

Y_k(x)

is the Bessel function of the second kind (singular at

x=0

3. Important Identities (Recurrence Relations)

We must master these relationships:

\frac{d}{dx} [x^k J_k(x)] = x^k J_{k-1}(x)

\frac{d}{dx} [x^{-k} J_k(x)] = -x^{-k} J_{k+1}(x)

J_{k-1}(x) - J_{k+1}(x) = 2J'_k(x)

J_{k-1}(x) + J_{k+1}(x) = \frac{2k}{x} J_k(x)

Problem

Prove that $\frac{d}{dx} J_0(x) = -J_1(x)$ .

Proof

Step 1: Write the series for $J_0(x)$

$J_0(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2} \left(\frac{x}{2}\right)^{2n}$

Step 2: Differentiate term-by-term

$J_0'(x) = \sum_{n=1}^{\infty} \frac{(-1)^n}{(n!)^2} \cdot 2n \left(\frac{x}{2}\right)^{2n-1} \cdot \frac{1}{2}$

Note: The sum starts at $n=1$ because the derivative of the constant term ( $n=0$ ) is 0.

Step 3: Simplify coefficients

$J_0'(x) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n!(n-1)!} \left(\frac{x}{2}\right)^{2n-1}$

Step 4: Re-index the sum

Let $m = n-1$ . Then $n = m+1$ . When $n=1$ , we have $m=0$ .

$J_0'(x) = \sum_{m=0}^{\infty} \frac{(-1)^{m+1}}{(m+1)! \, m!} \left(\frac{x}{2}\right)^{2m+1}$

$J_0'(x) = - \sum_{m=0}^{\infty} \frac{(-1)^m}{m! (m+1)!} \left(\frac{x}{2}\right)^{2m+1}$

Step 5: Compare with $J_1(x)$

The series for $J_1(x)$ is exactly $\sum \frac{(-1)^m}{m! (m+1)!} (x/2)^{2m+1}$ .

Therefore:

$J_0'(x) = -J_1(x)$

Q.E.D.