6.4 Properties of Legendre, Hermite and Bessel Functions

6.4 Properties of Legendre, Hermite and Bessel Functions Legendre polynomial Pn (t) is a solution of the differential equation (1 − t2 ) d2 x dx − 2t + n(n + 1)x(t) = 0 2 dt dt (6.4.1) In terms of the notation of Chapter 6, this is eigenvalue problem of the form Lx = λx (6.4.2) where L is the second order differential operator   2 dx d 2 d x 2 dx Lx = (1 − t ) 2 − 2t = (1 − t ) dt dt dt dt (6.4.3) We can think of L as an operator defined on L2 [−1, 1] with D(L) as the set of all functions having second order derivatives. Then Green’s formula (to be done in Section 7.1) gives Z 1 −1  1 where J(x, y)|−1 = a0 y 1 (yLx − xL∗ y) dt = J(x, y) dx dy −x dt dt  −1 1 −1 with a0 (t) = (1 − t2 ). As a0 (1) = a0 (−1) = 0, it follows that J(x, y)|1−1 = 0 and hence Z 1 −1 (yLx − xL∗ y) dt = 0 ∀x, y ∈ D(L) = D(L∗ ) That is L∗ = L and D(L∗ ) = D(L). We appeal to Theorem 7.5.1, of Chapter 7 to claim that the eigenfunctions of L form a complete orthonormal set. We have already proved in Example 6.2.1 that Eq. (6.4.1) has sequence of eigenvalues {n(n + 1)} with {Pn (t)} corresponding eigenfunctions and hence they are orthogonal. However, for the sake of completeness we prove this result directly. Theorem 6.4.1 The set of Legendre polymonials {Pn (t)} satisfying Eq. (6.4.1) are orthogonal set of polynomials in L2 [−1, 1]. Proof : We have   d 2 dPn (1 − t ) = −n(n + 1)Pn (t) dt dt   d dPm (1 − t2 ) = −m(m + 1)Pm (t) dt dt (6.4.4) (6.4.5) 6.4. PROPERTIES OF LEGENDRE, HERMITE AND BESSEL FUNCTIONS Multiplying Eq. (6.4.4) by Pm (t) and Eq. (6.4.5) by Pn (t) and substracting the equations and integrating, we get Z 1 [m(m + 1) − n(n + 1)] Pm (t)Pn (t)dt −1 = = 1    dPn d 2 dPm (1 − t ) Pm (t) − (1 − t ) Pn (t) dt dt dt dt −1 h h ii1 (1 − t2 ) Ṗn (t)Pm (t) − Ṗm (t)Pn (t) −1 Z 1h h ii − (1 − t2 ) Ṗn (t)Ṗm (t) − Ṗm (t)Ṗn (t) Z  d dt  2 −1 = 0 Hence (Pm , Pn ) = R1 −1 Pm (t)Pn (t)dt = 0. That is Pm ⊥ Pn . Theorem 6.4.2 The Legendre polynomials Pn (t) satisfy the Rodrigues formula Pn (t) = Proof : We have  1 dn  2 (t − 1)n , n = 0, 1, 2, · · · n n 2 n! dt (6.4.6) M Pn (t) = 1 X (−1)k n! (2n − 2k)! n−2k t 2n n! k!(n − k)! (n − 2k)! k=0 n Since d  2n−2k  (2n − 2k)! n−2k t = t , it follows that dtn (n − 2k)! "M # 1 dn X (−1)k n! 2n−2k Pn (t) = n t 2 n! dtn k!(n − k)! k=0 It is now clear that the sum of (t2 − 1)n and hence  PM k=0 Pn (t) =  (−1)k n! 2n−2k t is the binomial expansion k!(n − k)! 1 dn 2 (t − 1)n 2n n! dtn Definition 6.4.1 Let {fn (t)} be a sequence of functions in an interval I. A function F (t, u) is said to be a generating function of this sequence if F (t, u) = ∞ X n=0 fn (t)un (6.4.7) We have the following theorem giving the generating function for the sequence {Pn (t)} of Legendre polynomials. Theorem 6.4.3 For Legendre polynomials {Pn (t)}, we have 2 − 12 (1 − 2tu + u ) = ∞ X Pn (t)un (6.4.8) n=0 1 Proof : Let |t| ≤ r (arbitary positive number) and |u| < (1 + r 2 ) 2 − r. Then |2tu − u2 | ≤ 2|t||u| + |u2 | 1 1 ≤ 2r(1 + r2 ) 2 − 2r2 + (1 + r2 ) + r2 − 2r(1 + r2 ) 2 = 1 1 So expanding (1 − 2tu + u2 )− 2 in a binomial series, we get [1 − u(2t − u)] − 12 1 13 2 = 1 + u(2t − u) + u (2t − u)2 + · · · 2 24 1.3......(2n − 1) n + u (2t − u)n + · · · 1.2.....(2n) The coefficient of un in this expression is 1.3......(2n − 1) 1.3......(2n − 3) (2t)n − (2t)n−2 + · · · 2.4....(2n) 2.4....(2n − 2)   1.3......(2n − 1) n n(n − 1) n−2 n(n − 1)(n − 2)(n − 3) n−4 = t − t + t + · · · n! (2n − 1)2 (2n − 1)(2n − 3)2.4 = Pn (t) Theorem 6.4.4 The Legendre polynomials satisfy the following recurrence relation (n + 1)Pn+1 (t) = (2n + 1)tPn (t) − nPn−1 (t), n = 1, 2, ... (6.4.9) Proof : Differentiating the relation given by Eq. (6.4.8) with respect to u, we get ∞ X 3 (t − u)(1 − 2tu + u2 )− 2 = nPn (t)un−1 n=1 This gives 1 (t − u)(1 − 2tu + u2 )− 2 = (1 − 2tu + u2 ) ∞ X n=0 nPn (t)un−1 6.4. PROPERTIES OF LEGENDRE, HERMITE AND BESSEL FUNCTIONS That is (t − u) Equivalently, ∞ X n=1 Pn (t)un = (1 − 2tu + u2 ) ∞ X = n=0 ∞ X tPn (t)un − ∞ X + nPn (t)un−1 n=0 Pn (t)un+1 n=0 nPn (t)un−1 − 2 n=1 ∞ X ∞ X ∞ X ntPn (t)un n=1 nPn (t)un+1 n=0 Rewriting, we get ∞ X = n=1 ∞ X tPn (t)un − ∞ X Pn−1 (t)un n=1 n (n + 1)Pn+1 (t)u − 2 n=1 ∞ X + n=1 ∞ X ntPn (t)un n=1 (n − 1)Pn−1 (t)un Comparing the coefficient of un , we get (n + 1)Pn+1 (t) = (2n + 1)tPn (t) − nPn−1 (t) Corollary 6.4.1 kPn k2 = Z 1 −1 Pn2 (t)dt = 2 2n + 1 (6.4.10) Proof : Recurrence relation gives (2n + 1)tPn (t) (2n − 1)tPn−1 (t) = (n + 1)Pn+1 (t) + nPn−1 (t) = nPn (t) + (n − 1)Pn−2 (t) These two equations give Z 1 0 = [tPn (t)Pn−1 (t) − tPn−1 (t)Pn (t)] dt −1 = (n + 1) (2n + 1) Z (n − 1) − (2n − 1) 1 n (2n + 1) Z n Pn−2 (t)Pn (t) − (2n − 1) −1 Z Pn+1 (t)Pn−1 (t) + −1 Z 1 1 −1 1 −1 2 Pn−1 (t)dt Pn2 (t)dt This implies that Z 1 −1 Pn2 (t)dt (2n − 1) = (2n + 1) Z 1 −1 2 Pn−1 (t)dt and hence Z 1 −1 Pn2 (t)dt = = (2n − 1) (2n − 3) 2 ··· (2n + 1) (2n − 1) 1 2 (2n + 1) Z 1 −1 P02 (t)dt We now examine the properties of the Hermite polynomials. Recall that Hermite polynomial Hn (t) satisfy the differential equation d2 x dx − 2t + 2nx = 0 dt2 dt and is given by Hn (t) = ∞ X (−1)n n! (2t)n−2k k!(n − 2k)! k=0 We have the following theorem concerning the orthogonality of {Hn (t)} in 2 L2 (−∞, ∞) with respect to the weight function e−t . Theorem 6.4.5 The Hermite Polynomials Hn (t) are orthogonal set of polyno2 mials in the space L2 (−∞, ∞) with respect to the weight function e−t . Theorem 6.4.6 For Hermite polynomials Hn (t), we have the following formulae. (i) Rodgriues formula Hn (t) = (−1)et 2 dn −t2 e dtn (ii) Generating function 2 e2tu−u = ∞ X n=0 Hn (t) un n! (iii) Recurrence relation Hn+1 (t) = 2tHn (t) − 2nHn−1 (t), H0 = 1, H1 = 2t We now enunciate the properties of the Bessel’s function Jn (t). 6.4. PROPERTIES OF LEGENDRE, HERMITE AND BESSEL FUNCTIONS Theorem 6.4.7 For each fixed nonnegative integer n, the sequence of Bessel functions Jn (km t), where km are the zeros of Jn (k), form an orthogonal set in L2 [0, 1] with respect to the weight function t. That is Z 1 0 tJn (kl t)Jn (km t)dt = 0, l 6= m (6.4.11) Proof : The Bessel function Jn (t) satisfies the differential equation (refer Eq. (6.3.6)) t2 J¨n (t) + tJ˙n (t) + (t2 − n2 )Jn (t) = 0 Set t = ks, then the above equation reduces to the differential equation  i  n2 d h ˙ 2 sJn (ks) + − + k s Jn (ks) = 0 ds s (6.4.12) Let km be the zeros of the Bessel function Jn (k), then we have and  i  n2 d h ˙ sJn (kl s) + − + kl2 s Jn (kl s) ds s = 0  i  n2 d h ˙ 2 sJn (km s) + − + km s Jn (km s) = 0 ds s (6.4.13) (6.4.14) Eq. (6.4.13) - Eq. (6.4.14) give Z 1 2 (kl2 − km ) sJn (kl s)Jn (km s)ds 0  Z 1 h i i d d h ˙ = sJ˙n (kl s) Jn (km s) − sJn (km s) Jn (kl s) ds ds ds 0 = 0, kl 6= km using the fact that Jn (kl ) = 0 = Jn (km ). Theorem 6.4.8 The generating function for the sequence {Jn (t)} of Bessel functions is given by exp    ∞ X 1 1 t u− = Jn (t)un 2 u n=−∞ and hence J−n (t) = (−1)n Jn (t). (6.4.15) Proof : Expanding exp exp  1 2t 1 u u−   1 1 t u− 2 u

= in powers of u, we get  =   1 −t exp 2 u ! ! ∞ ∞ X X (tu)k (−t)l 2k k! 2l l! ul 1 exp tu 2  k=0 = ∞ X l=0 cn (t)un −∞  2k+n t ∞ (−1) X 2 The coefficient cn (t) of un in the above expression is = Jn (t) k!(k + n)! k=0 and hence we get the generating function relation given by Eq. (6.4.15) for Jn (t). 1 If we replace u by − in Eq. (6.4.15), we get v k 1 1 exp( (v − )t) = 2 v = ∞ X n=−∞ ∞ X Jn (t)(−1)n v −n J−n (t)v n n=−∞ J−n (t) = (−1)n Jn (t) Hence, it follows that Theorem 6.4.9 The Bessel functions Jµ (t) satisfy the following recurrence relations Jµ−1 (t) + Jµ+1 (t) Jµ−1 (t) − Jµ+1 (t) 2µ Jµ (t) t = 2Jµ (t) = (6.4.16a) (6.4.16b) Proof : We have Jµ (t) = t µ ∞ X k=0 (−1)k t2k 2k+µ 2 (k!)Γ(µ + k + 1) ! Multiplying Jµ (t) by tµ and pulling t2µ under the summation, we have tµ Jµ (t) = ∞ X k=0 (−1)k t2k+2µ 22k+µ (k!)Γ(µ + k + 1) 6.4. PROPERTIES OF LEGENDRE, HERMITE AND BESSEL FUNCTIONS Differentiating the above expression we have d µ (t Jµ (t)) dt = ∞ X (−1)k 2(k + µ)t2k+2µ−1 k=0 22k+µ (k!)Γ(µ + k + 1) = tµ tµ−1 ∞ X k=0 (−1)k t2k 22k+µ−1 (k!)Γ(µ = tµ Jµ−1 (t) Thus, we have d µ (t Jµ (t)) = tµ Jµ−1 (t) dt and
d −µ t Jµ (t) = −t−µ Jµ+1 (t) dt Expanding the LHS of the above expressions, we have µtµ−1 Jµ + tµ J˙µ −µtµ−1 Jµ + tµ J˙µ = tµ Jµ−1 = −tµ Jµ+1 Adding and substracting the above relations, we get Jµ−1 + Jµ+1 Jµ−1 − Jµ+1 2µ Jµ (t) t = 2J˙µ (t) = Corollary 6.4.2 From the definition of Jµ (t), we have r r 2 2 J 12 (t) = sin t, J− 12 (t) = cos t πt πt and hence the above recurrence relations give r   2 sin t J 32 (t) = − cos t πt t r   2 cos t J− 32 (t) = − − sin t πt t We have the following graphics for the above functions. + k) 1 J-1\2 0.75 0.5 J1\2 0.25 5 10 15 20 t -0.25 Figure 6.4.2: Sketch of J−1/2 (t) and J1/2 (t) J3\2 0.5 0.25 5 10 15 20 t -0.25 -0.5 J-3\2 -0.75 -1 Figure 6.4.3: Sketch of J3/2 (t) and J−3/2 (t) For more details on varioius topics in this chapter, refer Agarwal and Gupta [1], Brown and Churchill [2], Hochstadt [3] and Kreyzig [4].