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Legendre Polynomials and Applications

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Legendre Polynomials and Applications LEGENDRE POLYNOMIALS AND APPLICATIONS We construct Legendre polynomials and apply them to solve Dirichlet problems in spherical coordinates. 1. Legendre equation: series solutions The Legendre equation is the second order differential equation (1) (1 − x2)y00 − 2xy0 + λy = 0 which can also be written in self-adjoint form as [ ]0 (2) (1 − x2)y0 + λy = 0 : This equation has regular singular points at x = −1 and at x = 1 while x = 0 is an ordinary point. We can find power series solutions centered at x = 0 (the radius of convergence will be at least 1). Now we construct such series solutions. Assume that X1 j y = cjx j=0 be a solution of (1). By substituting such a series in (1), we get X1 X1 X1 2 j−2 j−1 j (1 − x ) j(j − 1)cjx − 2x jcjx + λ cjx = 0 j=2 j=1 j=0 X1 X1 X1 X1 j−2 j j j j(j − 1)cjx − j(j − 1)cjx − 2jcjx + λ cjx = 0 j=2 j=2 j=1 j=0 After re-indexing the first series and grouping the other series, we get X1 X1 j 2 j (j + 2)(j + 1)cj+2x − (j + j − λ)cjx = 0 j=0 j=0 and then X1 [ ] 2 j (j + 2)(j + 1)cj+2 − (j + j − λ)cj x = 0 : j=0 By equating each coefficient to 0, we obtain the recurrence relations (j + 1)j − λ c = c ; j = 0; 1; 2; ··· j+2 (j + 2)(j + 1) j We can obtain two independent solutions as follows. For the first solution we make c0 =6 0 and c1 = 0. In this case the recurrence relation gives 2 − λ 12 − λ c = c = 0; c = c = 0; ··· ; c = 0 : 3 6 1 5 20 3 odd The coefficients with even index can be written in terms of c0: −λ (2 · 3 − λ) (2 · 3 − λ)(−λ) c = c ; c = c = c 2 2 0 4 4 · 3 2 4! 0 Date: April 19, 2016. 1 2 LEGENDRE POLYNOMIALS AND APPLICATIONS we prove by induction that [(2k − 1)(2k − 2) − λ][(2k − 3)(2k − 4) − λ] ··· [3 · 2 − λ][−λ] c = c 2k (2k)! 0 We can write this in compact form as ! − c kY1 c = 0 [(2i + 1)(2i) − λ] 2k (2k)! i=1 This give a solution ! 1 − X kY1 x2k y (x) = c [(2i + 1)(2i) − λ] 1 0 (2k)! k=0 i=0 A second series solution (independent from the first) can be obtained by making c = 0 and c =6 0. In this case c = 0 and 0 1 even ! c Yk c = 1 [2i(2i − 1) − λ] 2k+1 (2k + 1)! i=1 The corresponding solution is ! 1 X Yk x2k+1 y (x) = c [(2i)(2i − 1) − λ] 2 1 (2k + 1)! k=0 i=1 Remark 1. It can be proved by using the ratio test that the series defining y1 and y2 converge on the interval (−1; 1) (check this as an exercise). 2. It is also proved that for every λ either y1 or y2 is unbounded on (−1; 1). That is, as x ! 1 or as x ! −1, one of the following holds, either jy1(x)j ! 1 or jy2(x)j ! 1. 3. The only case in which Legendre equation has a bounded solution on [−1; 1] is when the parameter λ has the form λ = n(n + 1) with n = 0 or n 2 Z+. In this case either y1 or y2 is a polynomial (the series terminates). This case is considered below. 2. Legendre polynomials Consider the following problem Problem. Find the parameters λ 2 R so that the Legendre equation [ ]0 (3) (1 − x2)y0 + λy = 0; −1 ≤ x ≤ 1 : has a bounded solution. This is a singular Sturm-Liouville problem. It is singular because the function (1 − x2) equals 0 when x = ±1. For such a problem, we don't need boundary con- ditions. The boundary conditions are replaced by the boundedness of the solution. As was pointed out in the above remark, the only values of λ for which we have bounded solutions are λ = n(n + 1) with n = 0; 1; 2; ··· . These values of λ are the eigenvalues of the SL-problem. To understand why this is so, we go back to the construction of the series solu- tions and look again at the recurrence relations giving the coefficients (j + 1)j − λ c = c ; j = 0; 1; 2; ··· j+2 (j + 2)(j + 1) j LEGENDRE POLYNOMIALS AND APPLICATIONS 3 If λ = n(n + 1), then (n + 1)n − λ c = c = 0: n+2 (n + 2)(n + 1) n By repeating the argument, we get cn+4 = 0 and in general cn+2k = 0 for k ≥ 1. This means • if n = 2p (even), the series for y1 terminates at c2p and y1 is a polynomial of degree 2p. The series for y2 is infinite and has radius of convergence equal to 1 and y2 is unbounded. • If n = 2p + 1 (odd), then the series for y2 terminates at c2p+1 and y2 is a polynomial of degree 2p + 1 while the solution y1 is unbounded. For λ = n(n+1), we can rewrite the recurrence relation for a polynomial solution in terms of cn. We have, (j + 2)(j + 1) (j + 2)(j + 1) c = c = − c ; j j(j + 1) − n(n + 1) j+2 (n − j)(n + j + 1) j+2 for j = n − 2; n − 4; ··· 1 or 0. Equivalently, (n − 2k + 2)(n − 2k + 1) c − = − c − ; k = 1; 2; ··· ; [n=2]: n 2k (2k)(2n − 2k + 1) n 2k+2 I will leave it as an exercise to verify the following formula for cn−2k in terms of cn: (−1)k n(n − 1) ··· (n − 2k + 1) c − = c : n 2k 2k k! (2n − 1)(2n − 3) ··· (2n − 2k + 1) n The polynomial solution is therefore [n=2] X (−1)k n(n − 1) ··· (n − 2k + 1) y(x) = c xn−2k 2k k! (2n − 1)(2n − 3) ··· (2n − 2k + 1) n k=0 where cn is an arbitrary constant. The n-th Legendre polynomial Pn(x) is the above polynomial of degree n for the particular value of cn (2n)! c = : n 2n(n!)2 This particular value of cn is chosen to make Pn(1) = 1. We have then (after simplification) [n=2] 1 X (−1)k(2n − 2k)! P (x) = xn−2k : n 2n k!(n − k)!(n − 2k)! k=0 Note that if n is even (resp. odd), then the only powers of x involved in Pn are even (resp. odd) and so Pn is an even (resp. odd). The first six Legendre polynomials are. P0(x) = 1 P1(x) = x 1 1 P (x) = (3x2 − 1) P (x) = (5x3 − 3x) 2 2 3 2 1 1 P (x) = (35x4 − 30x2 + 3) P (x) = (63x5 − 70x3 + 15x) 4 8 5 8 We have the following proposition. 4 LEGENDRE POLYNOMIALS AND APPLICATIONS P P 0 1 P P 2 3 P P 4 5 P P 6 7 Proposition. If y(x) is a bounded solution on the interval (−1; 1) of the Legendre equation (1) with λ = n(n + 1), then there exists a constant K such that y(x) = KPn(x) where Pn is the n-th Legendre polynomial. Remark. When λ = n(n + 1) a second solution of the Legendre equation, inde- pendent from Pn, can be found in the form 1 1 + x Q (x) = P (x) ln + R (x) n 2 n 1 − x n where Rn is a polynomial of degree n − 1. The construction of Qn can be achieved by the method of reduction of order. Note that jQn(x)j ! 1 as x ! ±1. The general solution of the Legendre equation is then y(x) = APn(x) + BnQn(x) and such a function is bounded on the interval (−1; 1) if and only if B = 0. 3. Rodrigues' formula The Legendre polynomials can be expressed in a more compact form. Theorem 1. (Rodrigues' Formula) The n-th Legendre polynomial Pn is given by the following 1 dn [ ] (4) P (x) = (x2 − 1)n n 2n n! dxn (thus expression (4) gives a solution of (3) with λ = n(n + 1)). Proof. Let y = (x2 − 1)n. We have following Claim. The k-th derivative y(k)(x) of y satisfies the following: d2y(k) dy(k) (5) (1 − x2) + 2(n − k − 1)x + (2n − k)(k + 1)y(k) = 0: dx2 dx Proof of the Claim. By induction. For k = 0, y = y(0). We have y0 = 2nx(x2 − 1)n−1 ) (1 − x2)y0 + 2nxy = 0 and after differentiation, we get (1 − x2)y00 + 2(n − 1)xy0 + 2ny = 0 LEGENDRE POLYNOMIALS AND APPLICATIONS 5 So formula (5) holds when k = 0. By induction suppose the (5) holds up to order k − 1. We can rewrite (5) for k − 1 as dy(k) (1 − x2) + 2(n − k)xy(k) + (2n − k + 1)ky(k−1) = 0: dx We differentiate to obtain d2y(k) dy(k) (1 − x2) + [2(n − k) − 2]x + [2(n − k) + k(2n − k + 1)]y(k) = 0: dx2 dx which is precisely (5). Now if we let k = n in (5), we obtain d2y(n) dy(n) (1 − x2) − 2x + n(n + 1)y(k) = 0: dx2 dx Hence y(n) solves the Legendre equation with λ = n(n + 1). Since y(n) is a polyno- mial of degree 2n, then by Proposition 1, it is a multiple of Pn.
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