Math 2.0

Ordinary differential equations Complex domain and special equations Example

Legendre's equation

Example

Let λ∈ℂ\lambda \in \C. Legendre's equation is (1−z2) y′′−2z y′+λ y=0,(1 - z^2)\, y'' - 2 z\, y' + \lambda\, y = 0, usually written with λ=n(n+1)\lambda = n (n + 1). In normalised form, p(z)=−2z1−z2p(z) = \frac{-2z}{1 - z^2} and q(z)=λ1−z2q(z) = \frac{\lambda}{1 - z^2}, which are rational. They have simple poles at z=±1z = \pm 1, so ±1\pm 1 are regular singular points, and every other finite point is ordinary. As z→∞z \to \infty, zp(z)→2z p(z) \to 2 and z2q(z)→−λz^2 q(z) \to -\lambda, so ∞\infty is ordinary or regular singular. Hence Legendre's equation is a Fuchsian equation, with singular points among −1-1, 11 and ∞\infty.

Remarks

At the ordinary point 00, a power series solution ∑akzk\sum a_k z^k satisfies (k+2)(k+1) ak+2=(k(k+1)−λ)ak(k + 2)(k + 1)\, a_{k+2} = \left( k (k + 1) - \lambda \right) a_k. The series converge for ∣z∣<1\abs{z} < 1, the distance from 00 to the nearest singular point. If λ=n(n+1)\lambda = n (n + 1) with n≥0n \ge 0 an integer, the recurrence stops, and one solution is the Legendre polynomial PnP_n of degree nn. For n=2n = 2, P2(z)=3z2−12P_2(z) = \frac{3 z^2 - 1}{2}: then y′′=3y'' = 3 and y′=3zy' = 3z, and (1−z2)⋅3−2z⋅3z+6⋅3z2−12=0(1 - z^2) \cdot 3 - 2z \cdot 3z + 6 \cdot \frac{3 z^2 - 1}{2} = 0. The equation arises from the Laplace equation in spherical coordinates.

Further reading

  1. Web https://en.wikipedia.org/wiki/Legendre_polynomials
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Instance ofFuchsian equation
Changed2 November 2026

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