Mathematics 22: Lecture 19

Transcription

1 Mathematics 22: Lecture 19 Legendre s Equation Dan Sloughter Furman University February 5, 2008 Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

2 Example: Legendre s equation The equation (1 t 2 ) d 2 u du 2t + α(α + 1)u = 0 dt2 dt is known as Legendre s equation. Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

3 Example: Legendre s equation The equation (1 t 2 ) d 2 u du 2t + α(α + 1)u = 0 dt2 dt is known as Legendre s equation. Standard form: d 2 u dt 2 2t du α(α + 1) 1 t 2 + dt 1 t 2 u = 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

4 Example: Legendre s equation The equation (1 t 2 ) d 2 u du 2t + α(α + 1)u = 0 dt2 dt is known as Legendre s equation. Standard form: Thus the coefficients d 2 u dt 2 2t du α(α + 1) 1 t 2 + dt 1 t 2 u = 0. p(t) = 2t α(α + 1) and q(t) = 1 t2 1 t 2 are analytic at t = 0 with power series converging on ( 1, 1). Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

5 As usual, let u(t) = a n t n. n=0 Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

6 As usual, let Then u(t) = a n t n. n=0 du dt = na n t n 1 and d 2 u dt 2 = n(n 1)a n t n 2. n=1 Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

7 As usual, let Then u(t) = a n t n. n=0 du dt = na n t n 1 and d 2 u dt 2 = n(n 1)a n t n 2. n=1 So we want 0 = (1 t 2 ) n(n 1)a n t n 2 2t na n t n 1 + α(α + 1) a n t n. n=0 n=1 Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

8 That is, 0 = n(n 1)a n t n 2 n(n 1)a n t n 2na n t n + n=0 α(α + 1)a n t n n=0 n=1 = (n + 2)(n + 1)a n+2 t n n(n 1)a n t n 2na n t n + α(α + 1)a n t n. n=0 n=1 Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

9 Hence 0 = 2a 2 + 6a 3 t + (n + 2)(n + 1)a n+2 t n n(n 1)a n t n 2a 1 t 2na n t n + α(α + 1)a 0 + α(α + 1)a 1 t + α(α + 1)a n t n = α(α + 1)a 0 + 2a 2 + (α(α + 1)a 1 2a 1 + 6a 3 )t + ((n + 2)(n + 1)a n+2 + ( n(n 1) 2n + α(α + 1))a n )t n = α(α + 1)a 0 + 2a 2 + ((α(α + 1) 2)a 1 + 6a 3 )t + ((n + 2)(n + 1)a n+2 + (α(α + 1) n(n + 1))a n )t n. Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

10 Thus we need α(α + 1) a 2 = a 0, 2 α(α + 1) 2 (α 1)(α + 2) a 3 = a 1 = a 1, 6 6 and, for n = 2, 3, 4,..., a n+2 = n(n + 1) α(α + 1) a n = (n + 2)(n + 1) (n α)(n + α + 1) a n. (n + 2)(n + 1) Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

11 Thus we need α(α + 1) a 2 = a 0, 2 α(α + 1) 2 (α 1)(α + 2) a 3 = a 1 = a 1, 6 6 and, for n = 2, 3, 4,..., a n+2 = n(n + 1) α(α + 1) a n = (n + 2)(n + 1) (n α)(n + α + 1) a n. (n + 2)(n + 1) In fact, we have, for n = 0, 1, 2,..., a n+2 = (α n)(α + n + 1) a n. (n + 2)(n + 1) Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

12 For example, α(α + 1) a 2 = a 0, 2 (α 2)(α + 3) a 4 = a 2 = 4 3 (α 4)(α + 5) a 6 = a = (α 2)α(α + 1)(α + 3) a 0, 4! (α 4)(α 2)α(α + 1)(α + 3)(α + 5) a 0. 6! Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

13 And (α 1)(α + 2) a 3 = a 1, 6 (α 3)(α + 4) a 5 = a 3 = 5 4 (α 5)(α + 6) a 7 = a = (α 3)(α 1)(α + 2)(α + 4) a 1 5! (α 5)(α 3)(α 1)(α + 2)(α + 4)(α + 6) a 1. 7! Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

14 Thus we have ( α(α + 1) u(t) = a 0 1 t 2 (α 2)α(α + 1)(α + 3) + t 4 2 4! (α 4)(α 2)α(α + 1)(α + 3)(α + 5) ) t 6 + 6! ( (α 1)(α + 2) + a 1 t t 3 3! (α 3)(α 1)(α + 2)(α + 4) + t 5 5! (α 5)(α 3)(α 1)(α + 2)(α + 4)(α + 6) ) t 7 + 7! Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

15 Thus we have ( α(α + 1) u(t) = a 0 1 t 2 (α 2)α(α + 1)(α + 3) + t 4 2 4! (α 4)(α 2)α(α + 1)(α + 3)(α + 5) ) t 6 + 6! ( (α 1)(α + 2) + a 1 t t 3 3! (α 3)(α 1)(α + 2)(α + 4) + t 5 5! (α 5)(α 3)(α 1)(α + 2)(α + 4)(α + 6) ) t 7 + 7! Note: the solution is of the form u(t) = a 0 u 1 (t) + a 1 u 2 (t). Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

16 Legendre polynomials Note: when α is an integer, one of the solutions reduces to a polynomial. Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

17 Legendre polynomials Note: when α is an integer, one of the solutions reduces to a polynomial. Example: when α = 4, u 1 (t) = 1 10t t4. Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

18 Legendre polynomials Note: when α is an integer, one of the solutions reduces to a polynomial. Example: when α = 4, The polynomial u 1 (t) = 1 10t t4. P 4 (t) = u 1(t) u 1 (1) = 3 8 u 1(t) = t t4 is an example of a Legendre polynomial. Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

19 Convergence We are guaranteed that the series solutions will converge on at least ( 1, 1). Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

20 Convergence We are guaranteed that the series solutions will converge on at least ( 1, 1). This is in fact the interval of convergence since ρ(t) = lim a n+2 t n+2 n a n t n = t 2 lim (n α)(n + α + 1) n (n + 2)(n + 1) = t 2, showing that ρ(t) < 1 if and only if 1 < t < 1. Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11

21 Convergence We are guaranteed that the series solutions will converge on at least ( 1, 1). This is in fact the interval of convergence since ρ(t) = lim a n+2 t n+2 n a n t n = t 2 lim (n α)(n + α + 1) n (n + 2)(n + 1) = t 2, showing that ρ(t) < 1 if and only if 1 < t < 1. Note: of course, the polynomial solutions are defined for all t in (, ). Dan Sloughter (Furman University) Mathematics 22: Lecture 19 February 5, / 11