Chapter 5.3: Series solution near an ordinary point We continue to study ODE s with polynomial coefficients of the form: P (x)y + Q(x)y + R(x)y = 0. Recall that x 0 is an ordinary point if P (x 0 ) 0. We now consider convergence of the power series solution around an ordinary point. In particular, we want to find the radius of convergence of the power series solutions. 1
Series solution near an ordinary point We divide by P to put the equation into the form y + py + qy = 0, p = Q/P, q = R/P. Since x 0 is an ordinary point, the coefficients p, q are analytic functions near x 0. Indeed, if P is a polynomial and P (x 0 ) 0 then 1/P is analytic near x 0. 2
General fact about radius of convergence Theorem 1 If x 0 is an ordinary point of the equation P (x)y + Q(x)y + R(x)y = 0, then the general solution has the form y = a 0 y 1 (x)+ a 1 y 2 (x), where y 1, y 2 are analytic in an interval around x 0. The radius of convergence ρ of any solution is at least as large as the smaller of the radii of convergence of p = Q/P, q = R/P. This is very similar to the earlier existence and uniqueness theorem for solutions of second order linear equations (see Theorem 3.2.1 of section 3.2). It guarantees unique continuous solutions. We are now assuming the coefficients are polynomials, and we get analytic solutions, not just continuous solutions. 3
How to find the radius of convergence of Q/P, R/P Proposition 2 If P (x) is a polynomial, and if P (x 0 ) 0, then 1/P is an analytic function, i.e. has a convergence power series 1/P = n=0 a n (x x 0 ) n centered at x 0, and the radius of convergence of the series is the distance from x 0 to the nearest COMPLEX zero of P. Example: If P (x) = x 2 + 1, then every point is an ordinary point since P (x) 0 for any real number x (Note: it does have complex zeros). The radius of convergence of the power series 1 x 2 + 1 = 1 x2 + x 4 x 6 + + ( 1) n x n + equals 1. This reflects the fact that ±i are complex zeros, of distance 1 from 0. 4
Lower bound for radius of convergence of series solutions Consider Legendre s equation: (1 x 2 )y 2xy + α(α + 1)y = 0. Here, α is a constant. All coefficients are polynomials. The zeros of P (x) = 1 x 2. Hence, x = 0 is an ordinary point. Any solution of the equation is an analytic series solution y = n=0 a n x n and its radius of convergence is at least equal to 1, the distance from 0 to the nearest zero of P. 5
Lower bound for radius of convergence of series solutions We can also look at solutions of Legendre s equation with initial conditions at x = 10, say. The solutions will be analytic y = n=0 a n (x 10) n with radius of convergence at least 9. 6
More examples Find the radius of convergence of the power series solutions of at x = 0. y + (sin x)y + (1 + x 2 )y = 0 Answer: infinite, they converge for all x since P 1 has no zeros. 7
Hermite equation This is the equation y 2xy + λy = 0. All points are ordinary since P 1. Hermite s equation is one of the basic equations of quantum mechanics. It models the quantum harmonic oscillator. It is usually written in the form u + x 2 u = νu. But if you write u = ye x2 /2 you get the above equation for y, where λ = ν + 1. 8
Hermite polynomials If λ = 2n 0 for some positive integer n, then the equation has a polynomial solution. Indeed, the recurrence relation is (n + 2)(n + 1)a n+2 = (2n 2n 0 )a n. It is a two-step recurrence relation, so a 0 determines an even solution and a 1 determines an odd solution. There is a special polynomial solution H n0, which is even if n 0 is even and odd if n 0 is odd. 9
Hermite polynomials Indeed, the right side equals zero when n = n 0, and the recurrence relation says (n 0 + 2)(n 0 + 1)a n0 +2 = (2n 0 2n 0 )a n0 = 0. The solution is a n0 +2 = 0. The recurrence relation then implies that a n0 +4 = 0 = a n0 +8 =. All of the coefficients with the same parity as n 0 are zero. That is, suppose n 0 is an even integer. Then we want to consider the even solution. We have a n0 +2 = 0 and then a n0 +2k = 0 if k 1. Thus the even solution is a polynomial of degree n 0. 10
Hermite polynomials Suppose on the other hand that n 0 is an odd integer. Then again a n0 +2 = 0 and then all odd index coefficients are zero, a n0 +2k = 0 for k 1. The solution is an odd polynomial of degree n 0. Since the Hermite equation is linear, a constant multiple of a solution is another solution. The Hermite polynomials are the special solutions whose coefficient of x n is 2 n. 11