Chapter 6 • Theory & Derivations

Bessel Functions & Classical Orthogonal Systems (Laguerre & Hermite)

Cylindrical Bessel functions of the first and second kinds, circular drum vibrations, Laguerre polynomials in quantum mechanics, and Hermite polynomials of the quantum oscillator.

§6.1 Bessel's Differential Equation & Cylindrical Functions of the First Kind J_ν(x)

1. Bessel's Differential Equation

Bessel's equation of order $\nu \ge 0$ appears in physical problems possessing cylindrical symmetry (wave propagation in waveguides, heat conduction in cylinders, vibrations of circular membranes):

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

Here $x = 0$ is a regular singular point with indicial equation $r^2 - \nu^2 = 0 \implies r = \pm \nu$. Applying the method of Frobenius yields the Bessel functions of the first kind $J_\nu(x)$:

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

For non-integer $\nu$, $J_\nu(x)$ and $J_{-\nu}(x)$ are linearly independent.

§6.2 Bessel Functions of the Second Kind Y_ν(x), Generating Function & Recurrence Formulas

1. Bessel Functions of the Second Kind (Neumann Functions)

When $\nu = n$ is an integer, $J_{-n}(x) = (-1)^n J_n(x)$, so they are linearly dependent! The second independent solution is the Weber-Neumann function $Y_\nu(x)$:

$$Y_\nu(x) = \frac{J_\nu(x) \cos(\nu\pi) - J_{-\nu}(x)}{\sin(\nu\pi)}, \quad Y_n(x) = \lim_{\nu \to n} Y_\nu(x)$$

$Y_n(x)$ diverges logarithmically as $x \to 0^+$. The general solution for any order $\nu$ is $y(x) = c_1 J_\nu(x) + c_2 Y_\nu(x)$.

2. Master Differential Recurrence Relations

Bessel Recurrence Identities: $$\begin{aligned} \frac{d}{dx} \left[ x^\nu J_\nu(x) \right] &= x^\nu J_{\nu-1}(x) \\ \frac{d}{dx} \left[ x^{-\nu} J_\nu(x) \right] &= -x^{-\nu} J_{\nu+1}(x) \\ J_{\nu-1}(x) + J_{\nu+1}(x) &= \frac{2\nu}{x} J_\nu(x) \\ J_{\nu-1}(x) - J_{\nu+1}(x) &= 2 J_\nu'(x) \end{aligned}$$

§6.3 Laguerre's Differential Equation, Associated Laguerre Polynomials & Quantum Radial Eigenstates

1. Laguerre's Differential Equation

Laguerre's equation arises in the quantum mechanical description of the hydrogen atom:

$$x y'' + (1 - x) y' + n y = 0$$

For non-negative integers $n$, polynomial solutions are the Laguerre Polynomials $L_n(x)$:

$$L_n(x) = \frac{e^x}{n!} \frac{d^n}{dx^n} (x^n e^{-x}) = \sum_{k=0}^n \frac{(-1)^k}{k!} \binom{n}{k} x^k$$

They satisfy the orthogonality relation with weight $e^{-x}$ on $[0, \infty)$:

$$\int_0^\infty e^{-x} L_n(x) L_m(x)\,dx = \delta_{nm}$$

§6.4 Hermite's Differential Equation, Hermite Polynomials H_n(x) & Harmonic Oscillator Orthogonality

1. Hermite's Differential Equation

Hermite's equation governs the stationary wavefunctions of the quantum harmonic oscillator:

$$y'' - 2x y' + 2n y = 0$$

For integer $n \ge 0$, the solutions are the Hermite Polynomials $H_n(x)$:

$$H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} (e^{-x^2})$$

First few polynomials: $H_0(x) = 1$, $H_1(x) = 2x$, $H_2(x) = 4x^2 - 2$, $H_3(x) = 8x^3 - 12x$.

Orthogonality of Hermite Polynomials: $$\int_{-\infty}^\infty e^{-x^2} H_n(x) H_m(x)\,dx = 2^n n! \sqrt{\pi}\, \delta_{nm}$$

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 2 • Intermediate Exam Example 6.1: Bessel Differential Equation of Order Zero: J_0(x) and Y_0(x)

Obtain the series solution for Bessel's equation of order zero:

$$x y'' + y' + x y = 0$$

Step 1: Indicial equation
Multiply by $x$: $x^2 y'' + x y' + x^2 y = 0$. $p_0 = 1, q_0 = 0 \implies r^2 = 0 \implies r_1 = r_2 = 0$.

Step 2: Recurrence relation
Substitute $y = \sum a_n x^{n+r}$:

(n + r)^2 a_n = -a_{n-2} \quad (a_1 = 0)

Odd coefficients vanish: $a_{2m+1} = 0$. For even coefficients $n = 2m$:

a_{2m}(r) = \frac{(-1)^m a_0}{2^{2m} [(1 + r/2)(2 + r/2)\dots(m + r/2)]^2}

For $r = 0, a_0 = 1$:

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

Step 3: Second solution $Y_0(x)$
Differentiating with respect to $r$:

\left. \frac{\partial a_{2m}}{\partial r} \right|_{r=0} = -H_m a_{2m}(0)

y_2(x) = J_0(x) \ln x - \sum_{m=1}^\infty \frac{(-1)^m H_m}{(m!)^2} \left(\frac{x}{2}\right)^{2m}

The standard Neumann function is normalized as $Y_0(x) = \frac{2}{\pi}\left[ y_2(x) + (\gamma - \ln 2) J_0(x) \right]$.

Final Answer & Physical Insight

$J_0(x) = \sum_{m=0}^\infty \frac{(-1)^m}{(m!)^2} \left(\frac{x}{2}\right)^{2m}$ and $Y_0(x)$ contains logarithmic singularity $\frac{2}{\pi} J_0(x)\ln x$.

Tier 3 • Honors Challenge Example 6.2: Inhomogeneous Bessel Equation via Lommel's Integrals

Solve the inhomogeneous Bessel equation of order zero:

$$x^2 y'' + x y' + x^2 y = x$$

using Variation of Parameters.

Step 1: Normalized standard form
Divide by $x^2$:

y'' + \frac{1}{x} y' + y = \frac{1}{x}

The homogeneous solutions are $y_1(x) = J_0(x)$ and $y_2(x) = Y_0(x)$.

Step 2: Wronskian of Bessel functions
Abel's identity gives:

W(J_0, Y_0)(x) = \frac{2}{\pi x}

Step 3: Variation of Parameters formulas

u_1'(x) = -\frac{y_2(x) g(x)}{W(x)} = -\frac{Y_0(x) (1/x)}{2/(\pi x)} = -\frac{\pi}{2} Y_0(x)

u_2'(x) = \frac{y_1(x) g(x)}{W(x)} = \frac{J_0(x) (1/x)}{2/(\pi x)} = \frac{\pi}{2} J_0(x)

Step 4: Particular solution

y_p(x) = -\frac{\pi}{2} J_0(x) \int_0^x Y_0(s)\,ds + \frac{\pi}{2} Y_0(x) \int_0^x J_0(s)\,ds

This is the celebrated Struve function relation: $y_p(x) = \frac{\pi}{2} \mathbf{H}_0(x)$.

Final Answer & Physical Insight

$y(x) = c_1 J_0(x) + c_2 Y_0(x) + \frac{\pi}{2}\left[ Y_0(x)\int_0^x J_0(s)ds - J_0(x)\int_0^x Y_0(s)ds \right]$

Tier 2 • Intermediate Exam Example 6.3: Sturm Comparison Theorem & Interlacing Zeros of Bessel Functions

Prove using Sturm's Separation/Comparison Theorem that between any two consecutive positive zeros of $J_0(x)$, there is exactly one zero of $J_1(x)$.

Step 1: Recurrence relation between $J_0$ and $J_1$
Recall from Bessel recurrence relations:

\frac{d}{dx} J_0(x) = -J_1(x)

Step 2: Apply Rolle's Theorem
Let $0 < x_1 < x_2$ be two consecutive positive zeros of $J_0(x)$, so $J_0(x_1) = 0$ and $J_0(x_2) = 0$, and $J_0(x) \ne 0$ for all $x \in (x_1, x_2)$.
Since $J_0(x)$ is continuously differentiable on $[x_1, x_2]$, Rolle's theorem guarantees that there exists at least one point $\xi \in (x_1, x_2)$ such that $J_0'(\xi) = 0$.
Since $J_0'(x) = -J_1(x)$, this means $J_1(\xi) = 0$. Thus $J_1$ has at least one zero in $(x_1, x_2)$.

Step 3: Uniqueness via Sturm Separation
Now consider the identity:

\frac{d}{dx}[x J_1(x)] = x J_0(x)

Suppose $J_1(x)$ had two zeros $\xi_1 < \xi_2$ in $(x_1, x_2)$. Then applying Rolle's theorem to $g(x) = x J_1(x)$ on $[\xi_1, \xi_2]$ would imply $g'(\eta) = \eta J_0(\eta) = 0$ for some $\eta \in (\xi_1, \xi_2) \subset (x_1, x_2)$.
Since $\eta > 0$, this would force $J_0(\eta) = 0$, directly contradicting that $x_1$ and $x_2$ are consecutive zeros of $J_0(x)$!

Conclusion: There is strictly one zero of $J_1(x)$ between any two consecutive zeros of $J_0(x)$. The zeros strictly interlace: $0 < j_{0, 1} < j_{1, 1} < j_{0, 2} < j_{1, 2} < \dots$.

Final Answer & Physical Insight

Proved: zeros of $J_0(x)$ and $J_1(x)$ strictly interlace.