Section 5.3, Exercise 22

Transcription

1 The Legendre equation is where α is a constant. Section 5.3, Exercise 22 (1 x 2 ) 2x + α(α + 1) 0 Determine two linearl independent solutions in powers of x for x < 1. Assume (x) a n x n and substitute this into Legendre s equation. 0 (1 x 2 ) n(n 1)a n x n 2 2x na n x n + α(α + 1) a n x n n2 n(n 1)a n x n 2 n2 n1 n(n 1)a n x n n2 (n + 2)(n + 1)a n+2 x n n1 n(n 1)a n x n 2na n x n + α(α + 1) a n x n [(n + 2)(n + 1)a n+2 (n(n + 1) α(α + 1))a n ] x n We obtain the recurrence relation: a n+2 (n(n + 1) α(α + 1))a n. (n + 2)(n + 1) 2na n x n + α(α + 1) a n x n If we let a 0 1 and a 1 0 then we can calculate the following coefficients. α(α + 1) a 2 2! a 4 (2(3) α(α + 1))a 2 (4)(3) α(α + 1)[α2 + α 6] 4! α(α 2)(α + 1)(α + 3) 4! a 6 (4(5) α(α + 1))a 4 (6)(5) α(α 2)(α + 1)(α + 3)(α2 + α 20) 6! α(α 2)(α 4)(α + 1)(α + 3)(α + 5) 6!

2 Hence we have one solution. a 2m () mα (α 2m + 2)(α + 1) (α + 2m 1) (2m)! α(α + 1) 1 (x) 1 x 2 α(α 2)(α + 1)(α + 3) + x 4 2! 4! α (α 2m + 2)(α + 1) (α + 2m 1) + x 2m. (2m)! m3 Using the Ratio Test for absolute convergence we see that lim a 2(m+1) x 2(m+1) m a 2m x 2m lim m lim m x 2. α (α 2m)(α+1) (α+2m+1) x 2m+2 (2m+2)! α (α 2m+2)(α+1) (α+2m) (2m)! (α 2m)(α + 2m + 1) x 2 (2m + 2)(2m + 1) Thus the power series converges for x < 1. If we let a 0 0 and a 1 1 then we can calculate the following coefficients. 1(2) α(α + 1) a 3 3! (α 1)(α + 2) 3! a 5 (3(4) α(α + 1))a 3 (5)(4) (α 1)(α + 2)(α2 + α 12) 5! (α 1)(α 3)(α + 2)(α + 4) 5! a 7 (5(6) α(α + 1))a 5 (7)(6) (α 1)(α 3)(α + 2)(α + 4)(α2 + α 30) 7! (α 1)(α 3)(α 5)(α + 2)(α + 4)(α + 6) 7! x 2m. a 2m+1 () m(α 1) (α 2m + 1)(α + 2) (α + 2m) (2m + 1)!

3 Hence we have another solution (α 1)(α + 2) 2 (x) x x 3 (α 1)(α 3)(α + 2)(α + 4) + x 5 3! 5! + () m(α 1) (α 2m + 1)(α + 2) (α + 2m) x 2m+1. (2m + 1)! m3 Using the Ratio Test for absolute convergence we see that lim a 2(m+1)+1 x 2(m+1)+1 (α) (α 2m)(α+2) (α+2m+2) x 2m+3 (2m+3)! m a 2m+1 x 2m+1 lim m (α) (α 2m+1)(α+2) (α+2m) x (2m+1)! 2m+1 lim (α 2m 1)(α + 2m + 2) m x 2 (2m + 3)(2m + 2) x 2. Thus the power series converges for x < 1. We ma verif the solutions are linearl independent using the Wronskian. Section 5.3, Exercise 23 W( 1, 2 )(0) 1 (0) 2(0) 1(0) 2 (0) 1 0 Show that if α is zero or a positive even integer 2n, the series solution 1 reduces to a polnomial of degree 2n containing onl even powers of x. Find the polnomials corresponding to α 0, 2, and 4. Show that if α is a positive odd integer 2n+1, the series solution 2 reduces to a polnomial of degree 2n + 1 containing onl odd powers of x. Find the polnomials corresponding to α 1, 3, and 5. If α 0 then 1 (x) 1 which is a polnomial of degree zero. If α 2n for some n N then according to the recurrence relation a 2n+2k 0 for all k N which implies that α(α + 1) 1 (x) 1 x 2 + 2! + 2n m3 α(α 2)(α + 1)(α + 3) x 4 4! α (α 2m + 2)(α + 1) (α + 2m 1) x 2m, (2m)! a polnomial of degree 2n containing onl even powers of x. α 0: (x) 1 α 2: (x) 1 3x 2 α 4: (x) 1 10x x4

4 If α 1 then 2 (x) x which is a polnomial of degree 1 containing onl odd powers of x. If α 2n + 1 for some n N then according to the recurrence relation a 2n+1+2k 0 for all k N which implies that (α 1)(α + 2) 2 (x) x x 3 (α 1)(α 3)(α + 2)(α + 4) + x 5 3! 5! 2n+1 + () m(α 1) (α 2m + 1)(α + 2) (α + 2m) x 2m+1, (2m + 1)! m3 a polnomial of degree 2n + 1 containing onl odd powers of x. α 1: α 3: Section 5.3, Exercise 24 (x) x (x) x 5 3 x3 α 5: (x) x 14 3 x x5 The Legendre polnomial P n (x) is defined as the polnomial solution of the Legendre equation with α n that also satisfies the condition P n (1) 1. (a) Using the results of Problem 23, find the Legendre polnomials P 0 (x),..., P 5 (x). We can easil see that P 0 (x) 1 and P 1 (x) x. Since then a /2 and we have P 2 (x) a(1 3x 2 ) P 2 (1) a(1 3) 1 P 2 (x) 1 2 (1 3x2 ). In the same wa we see that P 3 (x) a (x 53 ) x3 ( P 3 (1) a 1 5 ) 1 3 a 3 2 P 4 (x) a (1 10x ) x4 ( P 4 (1) a ) 1 3

5 To summarize a 3 8( P 5 (x) a x 14 3 x ) 5 x5 ( P 5 (1) a ) 1 5 a Legendre Polnomials P 0 (x) 1 P 1 (x) x P 2 (x) 1 2 (3x2 1) P 3 (x) 1 2 (5x3 3x) P 4 (x) 1 8 (35x4 30x 2 + 3) P 5 (x) 1 8 (63x5 70x x) (b) Plot the graphs of P 0 (x),..., P 5 (x) for x 1. P 0 (x) x

6 P 1 (x) x x 0.5 P 2 (x) 1 2 (3x2 1) x P 3 (x) 1 2 (5x3 3x) x 0.5

7 P 4 (x) 1 8 (35x4 30x 2 + 3) x P 5 (x) 1 8 (63x5 70x x) x 0.5 (c) Find the zeros of P 0 (x),..., P 5 (x). P 0 (x) 1 has no zeros. P 1 (x) x has a zero at x 0. P 2 (x) 1 2 (3x2 1) has zeros at x ± P 3 (x) 1 2 (5x3 3x) has zeros at x 0 and at x ± 5. P 4 (x) 1 8 (35x4 30x 2 15 ± ) has zeros at x ± (63x5 70x x) has zeros at x 0 and at x ± 35 ±

8 Section 5.3, Exercise 25 It can be shown that the general formula for P n (x) is P n (x) 1 2 n [n/2] k0 () k (2n 2k)! k!(n k)!(n 2k)! xn 2k, where [n/2] denotes the greatest integer less than or equal to n/2. B observing the form of P n (x) for n even and n odd, show that P n () () n. B exercise 23 we know P n (x) is a polnomial of even (odd) degree when n is even (odd) and contains onl even (odd) powers of x. Thus P n (x) is an even (odd) function when n is even (odd). B exercise 24, P n (1) 1, thus when n is even P n () 1 () n and when n is odd P n () P n (1) () n. Section 5.3, Exercise 26 The Legendre polnomials pla an important role in mathematical phsics. For example in solving Laplace s equation (the potential equation) in spherical coordinates, we encounter the equation d 2 F(ϕ) dϕ 2 + cotϕ df(ϕ) dϕ + n(n + 1)F(ϕ) 0, 0 < ϕ < π, where n is a positive integer. Show that the change of variable x cosϕ leads to the Legendre equation with α n for f(x) F(arccosx). Using the chain rule for derivatives we see that df(ϕ) dϕ df(ϕ) dx dx dϕ df(arccosx) ( sin ϕ) df dx dx ( sin ϕ) sin ϕ. Differentiating once more we have d 2 F(ϕ) dϕ 2 d dϕ [ sin ϕ ] cos ϕ + sin 2 ϕ. Substituting into the equation above we have 0 0 cosϕ + sin 2 ϕ + cot ϕ ( sin ϕ ) + n(n + 1) sin 2 ϕ 2 cosϕ + n(n + 1) (1 cos 2 ϕ) 2 cosϕ + n(n + 1) (1 x 2 ) 2x + n(n + 1).

9 Section 5.3, Exercise 27 Show that for n 0, 1, 2, 3, the corresponding Legendre polnomial is given b P n (x) 1 d n 1) n. 2 n n! dx n(x2 This formula, known as Rodrigues formula, is true for all positive integers n. For the case n 0 P 0 (x) 1 d 0 1) ! dx 0(x2 For the case n 1 P 1 (x) 1 d 2 1 1! dx (x2 1) 1 (2x) x. 2 For the case n 2 P 2 (x) 1 d 2 1) 2 1 d 2 2x 2 + 1) 1 d 2 2 2! dx 2(x2 8 dx 2(x4 8 dx (4x3 4x) 1 8 (12x2 4) 3 2 x For the case n 3 P 3 (x) 1 d 3 1) 3 1 d 3 3x 4 + 3x 2 1) 1 d 2 12x 3 + 6x) 2 3 3! dx 3(x2 48 dx 3(x6 48 dx 2(6x5 1 d 48 dx (30x4 36x 2 + 6) 1 48 (120x3 72x) 5 2 x3 3 2 x. Section 5.3, Exercise 28 Show that the Legendre equation can also be written as Then it follows that [(1 x 2 ) ] α(α + 1). [(1 x 2 )P n (x)] n(n + 1)P n (x) and [(1 x 2 )P m (x)] m(m + 1)P m (x). B multipling the first equation b P m (x) and the second equation b P n (x), integrating b parts, and then subtracting one equation from the other, show that P n (x)p m (x) dx 0 if n m. This propert of the Legendre polnomials is known as the orthogonalit propert. If m n, it can be show that the value of the preceding integral is 2/(2n + 1).

10 Starting with the Legendre equation 0 (1 x 2 ) 2x + α(α + 1) (1 x 2 )( ) + (1 x 2 ) + α(α + 1) α(α + 1) [(1 x 2 ) ] Thus when (x) P n (x) we have (product rule). [(1 x 2 )P n (x)] n(n + 1)P n (x). Suppose we multipl this equation b P m (x) where m n, then [(1 x 2 )P n (x)] P m (x) n(n + 1)P n (x)p m (x) [(1 x 2 )P n(x)] P m (x) dx n(n + 1) P n (x)p m (x) dx We can appl integration b parts to the integral on the left-hand side of this equation. Thus u P m (x) du P m (x) v (1 x 2 )P n(x) dv [(1 x 2 )P n (x)] dx [(1 x 2 )P n (x)] P m (x) dx P m (x)[(1 x 2 )P n (x)] 1 Similarl when (x) P m (x) we have (1 x 2 )P n(x)p m(x) dx. [(1 x 2 )P m(x)] m(m + 1)P m (x). Suppose we multipl this equation b P n (x), then [(1 x 2 )P m(x)] P n (x) m(m + 1)P m (x)p n (x) [(1 x 2 )P m (x)] P n (x) dx m(m + 1) (1 x 2 )P n (x)p m (x) dx P m (x)p n (x) dx We can appl integration b parts to the integral on the left-hand side of this equation. Thus u P n (x) du P n(x) v (1 x 2 )P m (x) dv [(1 x 2 )P m(x)] dx [(1 x 2 )P m(x)] P n (x) dx P n (x)[(1 x 2 )P m(x)] 1 (1 x 2 )P m(x)p n(x) dx (1 x 2 )P m (x)p n (x) dx.

11 Combining the equations we see that m(m + 1) [m(m + 1) n(n + 1)] (1 x 2 )P m (x)p n (x) dx (1 x 2 )P n (x)p m (x) dx P m (x)p n (x) dx n(n + 1) P m (x)p n (x) dx 0. Thus either P m (x)p n (x) dx 0 which is the orthogonalit propert, or 0 [m(m + 1) n(n + 1)] (m n)(m + n + 1) P n (x)p m (x) dx Since m, n N and we have assumed m n then this equation is not satisfied. Section 5.3, Exercise 29 Given a polnomial f of degree n, it is possible to express f as a linear combination of P 0, P 1, P 2,..., P n : n f(x) a k P k (x). Using the result of Problem 28, show that then Suppose 2m a k 2k f(x) k0 f(x)p k (x) dx. n a k P k (x). k0 f(x)p m (x) f(x)p m (x) dx n a k P k (x)p m (x) k0 n a k P k (x)p m (x) dx k0 f(x)p m (x) dx a m 2 2m + 1 a m