The associated Legendre equation, again, is

[(1 x2)y ] +

k2 1 x2

y = 0, 1

(1 x2)y 2xy + y = 0,

differentiate this equation k times to show that y(k) satisfies the equation (1 x2)y(k+2) 2(k + 1)xy(k+1) + [ k(k + 1)]y = 0.

292 An Introduction to Partial Differential Equations with MATLABR b) Next, make the change of dependent variable y(x) = (1 x2) k/2z(x) in the associated Legendre equation. c) Conclude that if f(x) is a Legendre function, that is, a solution of Legendre's equation

(1 x2)y 2xy + y = 0,

then

gk (x) = (1 x2) k/2f(k) (x)

is a solution of the associated Legendre equation. In particular, show that the associated Legendre function of degree n and order k

Pk n (x) = (1 x2) k/2P(k) n (x)

is a solution of (1 x2)y 2xy +

n(n + 1) k2 1 x2

y = 0.

(It turns out that, as with Legendre's equation, the associated Leg- endre's equation has bounded solutions if and only if = n(n+ 1),

n = 0, 1, 2,... . These bounded solutions are the associated Leg- endre functions.)