Differential Equations: Special Functions

Lesson 1 of 3

Module 1: The Legendre Equation & Polynomials

Core120 min

The Legendre Equation

1. Differential Equation & Series Solution

The Legendre equation arises in problems with spherical symmetry (e.g., electrostatics, gravitation):

$(1-x^2) \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + n(n+1)y = 0$

Using the Frobenius method near the ordinary point $x=0$ , we assume a solution $y = \sum a_k x^k$ . The recurrence relation for coefficients is:

$a_{k+2} = -\frac{(n-k)(n+k+1)}{(k+1)(k+2)} a_k$

If $n$ is an integer, the series terminates, yielding the Legendre Polynomials $P_n(x)$ .

2. Rodrigues' Formula

A compact closed-form expression for $P_n(x)$ is:

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

3. Orthogonality

Legendre polynomials form a complete orthogonal set on $[-1, 1]$ :

$\int_{-1}^{1} P_m(x) P_n(x) \, dx = \frac{2}{2n+1} \delta_{mn}$

where $\delta_{mn}$ is the Kronecker delta (1 if $m=n$ , 0 otherwise).

Watch: Deriving Legendre Polynomials

20 min · watch

A step-by-step walkthrough of the power series substitution and the derivation of the recurrence relation.

Solved Problem: Rodrigues' Formula

25 min · read

Problem

Verify that $P_2(x) = \frac{1}{2}(3x^2 - 1)$ using Rodrigues' Formula.

Solution

From Rodrigues' formula with $n=2$ :

$P_2(x) = \frac{1}{2^2 \cdot 2!} \frac{d^2}{dx^2} (x^2 - 1)^2$

Step 1: Simplify the term inside

$(x^2 - 1)^2 = x^4 - 2x^2 + 1$

Step 2: First Derivative

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

Step 3: Second Derivative

$\frac{d}{dx} (4x^3 - 4x) = 12x^2 - 4$

Step 4: Final Calculation

$P_2(x) = \frac{1}{8} (12x^2 - 4) = \frac{4(3x^2 - 1)}{8} = \frac{1}{2}(3x^2 - 1)$

Q.E.D.

Solved Problem: Legendre Expansion

30 min · practice

Problem

Express $f(x) = x^3$ in terms of Legendre polynomials.

Solution Strategy

We know:

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

Rearrange the $P_3$ equation to solve for $x^3$ :

$2P_3(x) = 5x^3 - 3x$

$5x^3 = 2P_3(x) + 3x$

Since $x = P_1(x)$ :

$5x^3 = 2P_3(x) + 3P_1(x)$

Therefore:

$x^3 = \frac{2}{5}P_3(x) + \frac{3}{5}P_1(x)$

Next Lesson

1. Differential Equation & Series Solution

The Legendre equation arises in problems with spherical symmetry (e.g., electrostatics, gravitation):

(1-x^2) \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + n(n+1)y = 0

Using the Frobenius method near the ordinary point

x=0

, we assume a solution

y = \sum a_k x^k

. The recurrence relation for coefficients is:

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

n

is an integer, the series terminates, yielding the Legendre Polynomials

P_n(x)

3. Orthogonality

Legendre polynomials form a complete orthogonal set on

[-1, 1]

\int_{-1}^{1} P_m(x) P_n(x) \, dx = \frac{2}{2n+1} \delta_{mn}

where

\delta_{mn}

is the Kronecker delta (1 if

m=n

, 0 otherwise).