Legendre s Equation PHYS 500 - Southern Illinois University October 18, 2016 PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 1 / 11
Legendre s Equation Recall We are trying to find solutions to the equation (1 x 2 )P 2xP + l(l + 1)P = 0. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 2 / 11
Legendre s Equation Recall We are trying to find solutions to the equation (1 x 2 )P 2xP + l(l + 1)P = 0. We have taken P(x) = k=0 a kx k and derived a recursion relation on the coefficients: (k + 1)(k + 2)a k+2 = [(k + 1)k l(l + 1)]a k. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 2 / 11
Legendre s Equation Recall We are trying to find solutions to the equation (1 x 2 )P 2xP + l(l + 1)P = 0. We have taken P(x) = k=0 a kx k and derived a recursion relation on the coefficients: (k + 1)(k + 2)a k+2 = [(k + 1)k l(l + 1)]a k. This leads to two classes of bounded solutions. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 2 / 11
Legendre s Equation Even Solutions { 0 for odd k a k = k 1 j=1 [(2j + 1)2j l(l + 1)] a 0 (2k)! for even k. For a finite solution l = 2n for some non-negative integer n. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 3 / 11
Legendre s Equation Even Solutions { 0 for odd k a k = k 1 j=1 [(2j + 1)2j l(l + 1)] a 0 (2k)! for even k. For a finite solution l = 2n for some non-negative integer n. Odd Solutions a k = { 0 for even k k j=1 [2j(2j 1) l(l + 1)] a 1 (2k+1)! for odd k. For a finite solution l = 2n 1 for some positive integer n. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 3 / 11
Legendre s Equation Even Solutions : P n (x) = Odd Solutions : P n (x) = a k x k = a k x k = a 2k x 2k k=0 even k a k x k = a k x k = k=0 k=0 odd k k=0 a 2k+1 x 2k+1. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 4 / 11
Legendre s Equation Even Solutions : P n (x) = Odd Solutions : P n (x) = a k x k = a k x k = a 2k x 2k k=0 even k a k x k = a k x k = k=0 k=0 odd k k=0 Consider again the recurrence relation (k + 1)(k + 2)a k+2 = [(k + 1)k l(l + 1)]a k. a 2k+1 x 2k+1. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 4 / 11
Legendre s Equation Even Solutions : P n (x) = Odd Solutions : P n (x) = a k x k = a k x k = a 2k x 2k k=0 even k a k x k = a k x k = k=0 k=0 odd k k=0 Consider again the recurrence relation (k + 1)(k + 2)a k+2 = [(k + 1)k l(l + 1)]a k. a 2k+1 x 2k+1. l = n : (k + 1)(k + 2)a k+2 = [(k + 1)k n(n + 1)]a k. a k+2 = 0 for k > n. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 4 / 11
Legendre s Equation Therefore, we just focus on k < n. Let k = n 2j. Then (n 2j + 1)(n 2j + 2) a n 2j = (n 2j + 1)(n 2j) n(n + 1) a n 2j+2, = (n 2j + 1)(n 2j + 2) a n 2j+2, j = 1, 2,, n (2j)(2n 2j + 1) 2. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 5 / 11
Legendre Polynomials (n 1)(n) a n 2 = (2)(2n 1) a n (n 3)(n 2) a n 4 = (4)(2n 3) a n 2 2 (n 3)(n 2) (n 1)(n) = ( 1) (4)(2n 3) (2)(2n 1) a n a n 2j = ( 1)j 2 j j! n(n 1)(n 2) (n 2j + 1) (2n 1)(2n 3) (2n 2j + 1) a n j = 1, 2,, n 2. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 6 / 11
Legendre Polynomials Therefore P n (x) := n 2 j=0 ( 1) j 2 j j! n(n 1)(n 2) (n 2j + 1) (2n 1)(2n 3) (2n 2j + 1) a nx n 2j. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 7 / 11
Legendre Polynomials Therefore P n (x) := n 2 j=0 ( 1) j 2 j j! n(n 1)(n 2) (n 2j + 1) (2n 1)(2n 3) (2n 2j + 1) a nx n 2j. The polynomials P n are called the Legendre Polynomials. The factor a n is customarily chosen so that P n (1) = 1. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 7 / 11
Legendre Polynomials Therefore P n (x) := n 2 j=0 ( 1) j 2 j j! n(n 1)(n 2) (n 2j + 1) (2n 1)(2n 3) (2n 2j + 1) a nx n 2j. The polynomials P n are called the Legendre Polynomials. The factor a n is customarily chosen so that P n (1) = 1. The n th Legendre Polynomial P n (x) is the only bounded polynomial solution to Legendre s Equation (1 x 2 )P 2xP + n(n + 1)P = 0 that satisfies P n (1) = 1. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 7 / 11
Rodrigues Formula Theorem. The Legendre Polynomials are equivalently given by the formula d n P n (x) = 1 2 n n! dx n (x 2 1) n. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 8 / 11
Rodrigues Formula Theorem. The Legendre Polynomials are equivalently given by the formula d n P n (x) = 1 2 n n! dx n (x 2 1) n. Proof Idea: Let y = (x 2 1) n. Show that y (n) satisfies Legendre s Equation (1 x 2 )P 2xP + n(n + 1)P = 0. We will also need to use the following Lemma, which is a higher-order form of the product rule. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 8 / 11
Rodrigues Formula Theorem. The Legendre Polynomials are equivalently given by the formula d n P n (x) = 1 2 n n! dx n (x 2 1) n. Proof Idea: Let y = (x 2 1) n. Show that y (n) satisfies Legendre s Equation (1 x 2 )P 2xP + n(n + 1)P = 0. We will also need to use the following Lemma, which is a higher-order form of the product rule. Lemma. For any functions a(x) and b(x) and any integer k, d k dx k [ab] = k ( ) k a (l) b (k l). l l=0 PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 8 / 11
Rodrigues Formula Proof of Lemma Prove by induction and use the relations ( ( k l 1) + k ) ( l = k+1 ) l, ( k ( k) = ( k+1), and k ) ( 0 = k+1 ) 0. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 9 / 11
Rodrigues Formula Proof of Lemma Prove by induction and use the relations ( ( k l 1) + k ) ( l = k+1 ) l, ( k ( k) = ( k+1), and k ) ( 0 = k+1 ) 0. Proof of Rodrigues Formula First note the equality (x 2 1)y = 2nxy. Then differentiate both sides n + 1 times. This gives us (1 x 2 )[y (n) ] 2x[y (n) ] + n(n + 1)y (n) = 0. PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 9 / 11
Rodrigues Formula Proof of Lemma Prove by induction and use the relations ( ( k l 1) + k ) ( l = k+1 ) l, ( k ( k) = ( k+1), and k ) ( 0 = k+1 ) 0. Proof of Rodrigues Formula First note the equality (x 2 1)y = 2nxy. Then differentiate both sides n + 1 times. This gives us (1 x 2 )[y (n) ] 2x[y (n) ] + n(n + 1)y (n) = 0. 1 The last step is to verify that 2 n n! y (n) (1) = 1. Since P n (x) is the only bounded polynomial solution to Legendre s equation with P n (1) = 1, we have that P n (x) = 1 2 n n! y (n) (x) = 1 d n 2 n n! dx (x 2 1) n. n PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 9 / 11
Legendre Polynomials We have the equality P n (x) = n 2 j=0 ( 1) j 2 j j! d n = 1 2 n n! dx n (x 2 1) n. n(n 1)(n 2) (n 2j + 1) (2n 1)(2n 3) (2n 2j + 1) a nx n 2j PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 10 / 11
Legendre Polynomials We have the equality P n (x) = n 2 j=0 ( 1) j 2 j j! d n = 1 2 n n! dx n (x 2 1) n. n(n 1)(n 2) (n 2j + 1) (2n 1)(2n 3) (2n 2j + 1) a nx n 2j Use this to find the constant a n. Compare the coefficients of x n : d n a n = 1 2 n n! dx n x 2n = 1 2 n n! (2n)!. n! PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 10 / 11
Legendre Polynomials We have the equality P n (x) = n 2 j=0 ( 1) j 2 j j! d n = 1 2 n n! dx n (x 2 1) n. n(n 1)(n 2) (n 2j + 1) (2n 1)(2n 3) (2n 2j + 1) a nx n 2j Use this to find the constant a n. Compare the coefficients of x n : d n a n = 1 2 n n! dx n x 2n = 1 2 n n! a n = (2n)! 2 n (n!) 2. (2n)!. n! PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 10 / 11
Legendre Polynomials P 0 (x) = 1 P 1 (x) = x P 2 (x) = 1 2 (3x 2 1) P 3 (x) = 1 2 (5x 3 3x) P 4 (x) = 1 8 (35x 4 30x 2 + 3) PHYS 500 - Southern Illinois University Legendre s Equation October 18, 2016 11 / 11