Chapter 4 • Theory & Derivations

Series Solutions Near Ordinary Points & Legendre Differential Equation

Analytic function theory, power series expansions near ordinary points, recurrence relations, Legendre polynomials P_n(x), Rodrigues' formula, and complete orthogonality.

§4.1 Real Analytic Functions, Ordinary vs Singular Points & Complex Radius of Convergence

1. Classification of Points for Linear ODEs

Consider the second-order homogeneous linear differential equation written in standard normalized form:

$$y'' + P(x) y' + Q(x) y = 0$$

Definition 4.1: A point $x_0 \in \mathbb{R}$ is called an ordinary point of the ODE if both coefficient functions $P(x)$ and $Q(x)$ are real analytic at $x_0$; that is, both possess Taylor series expansions with non-zero radius of convergence: $$P(x) = \sum_{n=0}^\infty p_n (x - x_0)^n, \quad Q(x) = \sum_{n=0}^\infty q_n (x - x_0)^n$$ If either $P(x)$ or $Q(x)$ fails to be analytic at $x_0$, then $x_0$ is called a singular point.

2. The Fuchs-Frobenius Radius of Convergence Theorem

Theorem 4.1: If $x_0$ is an ordinary point of $y'' + P(x)y' + Q(x)y = 0$, then every solution $y(x)$ is analytic at $x_0$ and can be expanded as a power series: $$y(x) = \sum_{n=0}^\infty a_n (x - x_0)^n = a_0 y_1(x) + a_1 y_2(x)$$ where $y_1(x)$ and $y_2(x)$ are linearly independent analytic solutions. The radius of convergence $R$ of these series is at least equal to the distance from $x_0$ to the nearest singularity of $P(z)$ or $Q(z)$ in the complex plane $\mathbb{C}$.

§4.2 The Power Series Method Near an Ordinary Point & Recurrence Relations

1. The Systematic Power Series Algorithm

To solve $y'' + P(x)y' + Q(x)y = 0$ about $x_0 = 0$, we substitute:

$$y(x) = \sum_{n=0}^\infty a_n x^n, \quad y'(x) = \sum_{n=1}^\infty n a_n x^{n-1}, \quad y''(x) = \sum_{n=2}^\infty n(n - 1) a_n x^{n-2}$$

Shift summation indices so that every term involves $x^k$. Factoring $x^k$, linear independence of powers $\{1, x, x^2, \dots\}$ requires that each coefficient bracket vanishes independently, yielding the recurrence relation for $a_{k+2}$ in terms of preceding coefficients.

The constants $a_0 = y(0)$ and $a_1 = y'(0)$ remain completely arbitrary, parametrizing the fundamental solution pair:

$$y_1(x) = 1 + \sum_{n=2}^\infty c_n^{(1)} x^n \quad (a_0 = 1, a_1 = 0), \qquad y_2(x) = x + \sum_{n=2}^\infty c_n^{(2)} x^n \quad (a_0 = 0, a_1 = 1)$$

§4.3 Legendre's Differential Equation, Legendre Polynomials P_n(x), Rodrigues' Formula & Orthogonality

1. Legendre's Differential Equation

Legendre's differential equation arises ubiquitously in electrostatics, quantum mechanics, and gravitational potential theory when solving Laplace's equation in spherical coordinates:

$$(1 - x^2) y'' - 2x y' + \alpha(\alpha + 1) y = 0$$

Dividing by $1 - x^2$, the singular points are $x = \pm 1$. The origin $x_0 = 0$ is an ordinary point. Substituting $y = \sum a_n x^n$ yields the two-step recurrence relation:

$$a_{n+2} = -\frac{(\alpha - n)(\alpha + n + 1)}{(n + 1)(n + 2)} a_n$$

When $\alpha = n$ is a non-negative integer, the series terminates after the term $x^n$, producing a polynomial of degree $n$. Normalized such that $P_n(1) = 1$, these are the Legendre Polynomials $P_n(x)$:

$$P_0(x) = 1, \quad P_1(x) = x, \quad P_2(x) = \frac{1}{2}(3x^2 - 1), \quad P_3(x) = \frac{1}{2}(5x^3 - 3x), \quad P_4(x) = \frac{1}{8}(35x^4 - 30x^2 + 3)$$

2. Rodrigues' Formula & Orthogonality

Rodrigues' Formula: $$P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n$$ Orthogonality Relation: The Legendre polynomials form a complete orthogonal set in $L^2[-1, 1]$: $$\int_{-1}^1 P_n(x) P_m(x)\,dx = \frac{2}{2n + 1} \delta_{nm}$$

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 1 • Foundational Example 4.1: Power Series Solution of Airy-Type Equation about x = 0

Find the general power series solution about the ordinary point $x_0 = 0$ for:

$$y'' - x y = 0$$

Step 1: Series Substitution
Let $y = \sum_{n=0}^\infty a_n x^n$. Then:

y'' = \sum_{n=2}^\infty n(n - 1) a_n x^{n-2}, \quad x y = \sum_{n=0}^\infty a_n x^{n+1}

Step 2: Align Indices
Let $k$ be the power of $x$: In $y''$: $k = n - 2 \implies n = k + 2 \implies \sum_{k=0}^\infty (k + 2)(k + 1) a_{k+2} x^k$. For $k = 0$: $2(1) a_2 = 0 \implies a_2 = 0$. For $k \ge 1$: let $j = k - 1 \ge 0$, equating coefficients gives:

(k + 2)(k + 1) a_{k+2} = a_{k-1} \implies a_{k+2} = \frac{a_{k-1}}{(k + 2)(k + 1)}

Setting $k + 2 = n$:

a_n = \frac{a_{n-3}}{n(n - 1)} \quad \text{for } n \ge 3

Step 3: Compute Coefficients
Since $a_2 = 0$, all terms $a_{3m+2} = 0$.
Multiples of 3 (governed by $a_0$):

a_3 = \frac{a_0}{3 \cdot 2}, \quad a_6 = \frac{a_3}{6 \cdot 5} = \frac{a_0}{6 \cdot 5 \cdot 3 \cdot 2}

Terms of form $3m+1$ (governed by $a_1$):

a_4 = \frac{a_1}{4 \cdot 3}, \quad a_7 = \frac{a_4}{7 \cdot 6} = \frac{a_1}{7 \cdot 6 \cdot 4 \cdot 3}

Step 4: Solution

y(x) = a_0 \left( 1 + \frac{x^3}{6} + \frac{x^6}{180} + \dots \right) + a_1 \left( x + \frac{x^4}{12} + \frac{x^7}{504} + \dots \right)

Final Answer & Physical Insight

$y(x) = a_0 \left(1 + \frac{x^3}{6} + \frac{x^6}{180} + \dots\right) + a_1 \left(x + \frac{x^4}{12} + \frac{x^7}{504} + \dots\right)$

Tier 1 • Foundational Example 4.2: Legendre Polynomial P_3(x) and Orthogonality Verification

Derive $P_3(x)$ via Rodrigues' formula and verify the orthogonality $\int_{-1}^1 P_1(x) P_3(x)\,dx = 0$.

Step 1: Rodrigues' Formula for $n = 3$

P_3(x) = \frac{1}{2^3 \cdot 3!} \frac{d^3}{dx^3} (x^2 - 1)^3 = \frac{1}{48} \frac{d^3}{dx^3} (x^6 - 3x^4 + 3x^2 - 1)

First derivative:

\frac{d}{dx}(x^6 - 3x^4 + 3x^2 - 1) = 6x^5 - 12x^3 + 6x

Second derivative:

\frac{d^2}{dx^2} = 30x^4 - 36x^2 + 6

Third derivative:

\frac{d^3}{dx^3} = 120x^3 - 72x

Dividing by 48:

P_3(x) = \frac{120x^3 - 72x}{48} = \frac{5}{2}x^3 - \frac{3}{2}x = \frac{1}{2}(5x^3 - 3x)

Step 2: Orthogonality integral with $P_1(x) = x$

\int_{-1}^1 P_1(x) P_3(x)\,dx = \int_{-1}^1 x \cdot \frac{1}{2}(5x^3 - 3x)\,dx = \frac{1}{2} \int_{-1}^1 (5x^4 - 3x^2)\,dx

Because the integrand is even:

= \int_0^1 (5x^4 - 3x^2)\,dx = \left[ x^5 - x^3 \right]_0^1 = (1 - 1) - 0 = 0

Orthogonality is verified identically!

Final Answer & Physical Insight

$P_3(x) = \frac{1}{2}(5x^3 - 3x)$ and $\int_{-1}^1 P_1(x)P_3(x)\,dx = 0$.

Tier 3 • Honors Challenge Example 4.3: Complete Legendre Series Expansion & Parseval's Identity

Expand $f(x) = x(1 - x^2)$ in terms of Legendre polynomials on $[-1, 1]$ and verify Parseval's identity.

Step 1: Algebraic decomposition
$f(x) = x - x^3$. We express $f(x)$ as a linear combination of Legendre polynomials:

P_1(x) = x \implies x = P_1(x)

P_3(x) = \frac{1}{2}(5x^3 - 3x) \implies 5x^3 = 2P_3(x) + 3x = 2P_3(x) + 3P_1(x) \implies x^3 = \frac{2}{5}P_3(x) + \frac{3}{5}P_1(x)

Substituting $x^3$:

f(x) = P_1(x) - \left( \frac{2}{5}P_3(x) + \frac{3}{5}P_1(x) \right) = \frac{2}{5}P_1(x) - \frac{2}{5}P_3(x)

All other coefficients $c_n = 0$.

Step 2: Parseval's Identity Verification
LHS: Directly compute $\int_{-1}^1 [f(x)]^2\,dx$:

$$[f(x)]^2 = x^2(1 - x^2)^2 = x^2(1 - 2x^2 + x^4) = x^2 - 2x^4 + x^6$$

\int_{-1}^1 (x^2 - 2x^4 + x^6)\,dx = 2 \left[ \frac{1}{3} - \frac{2}{5} + \frac{1}{7} \right] = 2 \left( \frac{35 - 42 + 15}{105} \right) = 2 \left(\frac{8}{105}\right) = \frac{16}{105}

RHS: Compute $\sum c_n^2 \|P_n\|^2$ where $\|P_n\|^2 = \frac{2}{2n + 1}$:

c_1^2 \|P_1\|^2 + c_3^2 \|P_3\|^2 = \left(\frac{2}{5}\right)^2 \left(\frac{2}{3}\right) + \left(-\frac{2}{5}\right)^2 \left(\frac{2}{7}\right)

= \frac{4}{25} \left( \frac{2}{3} + \frac{2}{7} \right) = \frac{4}{25} \left( \frac{20}{21} \right) = \frac{80}{525} = \frac{16}{105}

LHS = RHS = $\frac{16}{105}$. Parseval's identity holds with 100% precision!

Final Answer & Physical Insight

$f(x) = \frac{2}{5}P_1(x) - \frac{2}{5}P_3(x)$ and $\|f\|^2 = \sum c_n^2 \|P_n\|^2 = \frac{16}{105}$.