Legendre's equation has a power-series solution about whose coefficients obey the…

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Ordinary differential equations · Special functions and orthogonal polynomials

Conditions of Validity, Ch Series solutions, singular points and where the special functions come from

Problem

Legendre's equation (1−x2)y′′−2xy′+λy=0(1-x^2)y'' - 2xy' + \lambda y = 0 has a power-series solution about x=0x=0 whose coefficients obey the recursion ak+2=ak k(k+1)−λ(k+1)(k+2),a_{k+2} = a_k\,\frac{k(k+1)-\lambda}{(k+1)(k+2)}, splitting into an even series built on a0a_0 (powers x0,x2,x4,…x^0, x^2, x^4,\ldots) and an odd series built on a1a_1 (powers x1,x3,x5,…x^1, x^3, x^5,\ldots). Only a solution whose series terminates after finitely many terms stays bounded on [−1,1][-1,1]; an unterminated series of either parity diverges logarithmically as x→±1x\to\pm1. ((a)) Show that if the series of a given parity terminates, so that its coefficients vanish for all kk from some index onward, then λ=ℓ(ℓ+1)\lambda = \ell(\ell+1) for a non-negative integer ℓ\ell, and the surviving terms form a polynomial of degree ℓ\ell. ((b)) Take λ=12\lambda = 12. Find the integer ℓ\ell, state which of the two series (even or odd) terminates, and explain why a bounded solution must have the leading coefficient of the other series equal to zero.

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