Legendre, Hermite and UNIT 3 LEGENDRE, HERMITE AND Laguerre Polynomials LAGUERRE POLYNOMIALS
Structure Page No.
3.1 Introduction 83 Objectives 3.2 Legendre Polynomials 84 Rodrigue’s Formula Generating Function Recurrence Relations Orthogonality Property Legendre Function of the Second Kind 3.3 Hermite and Laguerre Polynomials 100 3.4 Applications to Physical Situations 106 3.5 Summary 111 3.6 Solutions/Answers 112
3.1 INTRODUCTION
In Unit 2, we discussed methods of finding series solutions of second order, linear, homogeneous differential equations with variable coefficients. In this unit and the unit to follow, we shall apply these methods to a few special second order equations occurring frequently in applied mathematics, engineering and physics. These equations are Legendre’s, Bessel’s, Hermite’s and Laguerre’s equations. The solution to these equations, that occur in applications, are referred to as special functions . They are called ‘special’ as they are different from the standard functions like sine, cosine, exponential, logarithmic etc. In this unit, we shall concentrate on Legendre, Hermite and Laguerre polynomials which are polynomial solutions to Legendre’s, Hermite’s and Laguerre’s differential equations.
Legendre polynomials first arose in the problem of expressing the Newtonian potential of a conservative force field in an infinite series involving the distance variable of two points and their included central angle. Other similar problems dealing with either gravitational potential or electrostatic potentials and steady-state heat conduction problems in spherical solids, also lead to Legendre polynomials. Other polynomials which commonly occur in applications are the Hermite and Laguerre. They play an important role in quantum mechanics, and in probability theory.
We have started the unit by obtaining the power series solutions of Legendre’s differential equation in Sec.3.2 and introduced Legendre polynomials and Legendre functions of both first and second kind. Various properties of Legendre polynomials are also discussed in this section. Polynomials solutions of Hermite’s and Laguerre’s equations and their properties are discussed in Sec.3.3. Applications of the Legendre and Hermite polynomials to physical situtations are discussed in Sec.3.4. Objectives After studying this unit you should be able to • obtain the power series solutions of Legendre’s differential equation; • derive Rodrigue’s formula, for Legendre polynomials; • obtain Legendre polynomials through generating function; • use recurrence relations for Legendre polynomials and its orthogonality property in various applications; • derive Rodrigue’s formula, generating function, recurrence relations and orthogonal property of Hermite and Laguerre polynomials and use them in various applications. 83
Ordinary Differential Equations 3.2 LEGENDRE POLYNOMIALS The differential equation of the form d 2 y dy 1( − x 2 ) − 2x + n(n + y)1 = 0 (1) dx 2 dx is called Legendre’s differential equation or simply Legendre’s equation , where ‘n’ is a real constant. We can also write Legendre’s equation in the form d dy 1( − )x 2 + n(n + y)1 = 0 (2) dx dx In Eqn.(1), if we divide by 1( − x 2 ) throughout then we can write the equation as d 2 y 2x dy 1 − + n(n + y)1 = ,0 x ≠ ± .1 (3) dx 2 1− x 2 dx 1− x 2 2x n(n + )1 Here − and are analytic functions for x = 0 . Also if we use 1− x 2 1− x 2 Binomial expressions, then both these converge for |x| <1 . Hence x = 0 is an ordinary point of Legendre’s Eqn.(1) and this suggests that Eqn.(1) has a power series solution about x = 0 . Assume the series solution ∞ k )x(y = ∑ c k x (4) k=0 Differentiting Eqn.(4) w.r. to x , we get ∞ k−1 y′ )x( = ∑ c k k x k=1 ∞ k−2 and, y ′′ )x( = ∑ c k k(k − )1 x k=2 Substituting for ),x(y y′ )x( and y′′ )x( in Eqn.(1), we get ∞ ∞ ∞ 2 k−2 k−1 k 1( − x )∑c k k(k − x)1 − 2x ∑c k k x + n(n + )1 ∑ c k x = 0 k=2 k= 1 k= 0
∞ ∞ ∞ ∞ k−2 k k k i.e., ∑c k k(k − x)1 − ∑c k k(k − x)1 − 2 ∑c k kx + n(n + )1 ∑ c k x = 0 k=2 k= 2 k= 1 k= 0
∞ ∞ ∞ k k+2 k+1 i.e., ∑c k+2 k( + k()2 + x)1 − ∑c k+2 k( + k()2 + x)1 − 2 ∑ c k+1 k( + x)1 k=0 k=0 k=0 ∞ k + n(n + )1 ∑ c k x = 0 (5) k=0 Equating the coefficients of various powers of x on both sides of Eqn.(5), we get n(n + )1 For x 0 : c )1.2( + n(n + c)1 = 0 .,e.i c = − c (6) 2 0 2 2 0 2 − n(n + )1 For x1 : c )2.3( − 2 c + n(n + c)1 = 0 .,e.i c = c (7) 3 1 1 3 3.2 1 k For x : c k+2 k( + k()2 + )1 − c k k(k − )1 − 2 c k k + n(n + c)1 k = 0 k(k − )1 + 2k − n(n + )1 .,e.i c = c , for k ≥ 0 k +2 k( + k()2 + )1 k k( − k()2 − )1 − n(n + )1 .,e.i c = c , for k ≥ 2 k k(k − )1 k −2 − n[ − k( − n[])2 + k( − ])1 = c , for k ≥ 2 (8) k k( − )1 k−2 84
From recurrence relation (8), alongwith Eqns.(6) and (7), we observe that coefficients Legendre, Hermite and Laguerre Polynomials c k for k even are multiples of c 0 while those for k odd are multiples of c1 . Thus, in general, we have − n( − 2k + n()2 + 2k − )1 c = c , for k ≥ 1 2k 2k 2( k − )1 2k−2 n( − 2k + n()2 − 2k + )4 K n( − ).2 n(n + n()1 + )3 L n( + 2k − )1 = (− )1 k c , 2( )!k 0 for k ≥1 (9) − n( − 2k + n()1 + 2 )k and, c = c , for k ≥ 1 2k+1 2( k + 2()1 )k 2k−1 n( − 2k + n()1 − 2k + )3 K n( − n()1 + n()2 + )4 K n( + 2 )k = (− )1 k c , 2( k + )!1 1 for k ≥ 1 (10) Substituting from relations (9) and (10) in relation (4), the power series solution of Eqn.(1) can be written as n(n + )1 2 n( + n()3 + n)1 n( − )2 4 L )x(y = c0 1 − x + x − )!n( )!4(
n( + n()2 − )1 3 n( + n()4 + n()2 − n()1 − )3 5 L + c1 x − x + x − )!3( )!5(
∞ L K k n( − 2k + n()2 − 2k + )4 n( − ).2 n(n + n()1 + )3 n( + 2k − )1 2k = c 0 1 + ∑ (− )1 x k=1 2( )!k
∞ L K k n( − 2k + n()1 − 2k + )3 n( −1).( n + n()2 + )4 n( + 2 )k 2k+1 = c1 x + ∑ (− )1 x k=1 2( k + )!1
= c0 y1 )x( + c1 y 2 )x( (11) where, ∞ n( − 2k + n()2 − 2k + )4 K n( − ).2 n(n + n()1 + )3 K n( + 2k − )1 k 2k 12) y1 )x( = 1+ ∑ (− )1 x k=1 2( )!k and ∞ K K k n( − 2k + n()1 − 2k + )3 n( − ).1 n(n + n()2 + )4 n( + 2 )k 2k+1 (13) y2 )x( = x + ∑(− )1 x k=1 2( k + )!1 when n is not an integer both y1 )x( and y 2 )x( converge for |x| < 1.
Since c 0 and c1 are arbitrary constants, and y1 and y 2 are linearly independent solutions of Eqn.(1), solution (11) can be considered as the general solution of Legendre’s Eqn.(1). The functions defined by Eqns.(12) and (13) are called Legendre Functions.
In case, n is a non-negative integer, then y1 )x( reduce to a polynomial of degree n , if n even; and the same is true for y 2 )x( if n is odd. For example, for n even , we have
For n = 0 : y1 )x( =1 2 For n = 2 : y1 )x( =1 − 3x 35 For n = 4 : y )x( =1 −10 x 2 + x 4 1 3 and so on. Thus y1 )x( reduces to a polynomial of even powers. For these values of
y,n 2 )x( remains an infinite series. 85
Ordinary Differential In general, for n = 2m , where m is an integer, Equations m K K k 2( m − 2k + 2()2 m − 2k + )4 2m 2( m + 2()1 m + )3 2( m + 2k − )1 2k (14) y1 )x( = 1+ ∑(− )1 x k=1 2( )!k and for k = m + k,1 = m + ,2 K etc. the terms in the summation on the R.H.S. of Eqn.(14) are zero.
When n is odd , y1 )x( remains an infinite power series and y 2 )x( reduces to polynomial of odd powers. For example,
For n = 1 : y 2 )x( = ,x 5 For n = 3: y )x( = x − x 3 . 2 3 14 21 For n = 5 : y )x( = x − x 3 + x 5 . 2 3 5 In general, for n = 2m +1 , we may obtain
∞ K K k 2( m − 2k + 2()2 m − 2k + )4 2m.( 2m + 2()3 m + )5 2( m + 2k + )1 2k+1 . (15) y2 )x( = x + ∑(− )1 x k=1 2( k + )!1
These polynomials multiplied by suitable constants are called Legendre polynomials.
The Legendre polynomials are denoted by Pn )x( where n denotes the order of the polynomial. Therefore, when n takes positive integral values, one of the linearly independent solutions of Eqn.(1) is a Legendre polynomial and the second solution is an infinite series. In order to explicitly write the expressions for the Legendre polynomials, we need to evaluate the multiplicative constants. The values of the
multiplicative constants are obtained by setting Pn )1( =1. It is too cumbersome to use the recurrence relation given by Eqn.(8) or the polynomials given above to determine the multiplicative constants. It is easy to use the Rodrigue’s formula which we shall discuss now to find the expressions for the Legendre Polynomials. 3.2.1 Rodrigue’s Formula We shall now show that a polynomial of degree n , obtained by applying the binomial theorem to x( 2 − )1 n and differentiating it n times, is a solution of the Legendre’s
differential equation. The expression for Pn )x( can then be obtained by using
Pn )1( =1 . Consider the function )x(f = x( 2 − )1 n . (16) Differentiating it w.r. to x, we get f ′ )x( = x(n 2 − )1 n−1 2. x ⇒ x( 2 − f)1 ′ )x( − 2n )x(f x = 0 . (17) Differentiating Eqn.(17), n( + )1 times by using the Leibnitz rule, we get
2 n( + )2 n( + )1 n(n + )1 )n( x( − f)1 )x( + n( + 2()1 f)x )x( + 2. f )x( 2 − 2 [n x f n+1 )x( + n( + ).1 f.1 )n( ])x( = ,0 or, x( 2 − f)1 n( + )2 )x( + 2x f n( + )1 )x( − n(n + f)1 )n( )x( = 0 , d 2 d or, 1( − x 2 ) f[ )n( ])x( − 2x f[ )n( ])x( + n(n + f[)1 )n( ])x( = 0 . dx 2 dx d n d n This shows that f )n( )x( = )]x(f[ = x([ 2 − )1 n ] satisfies Lagendre’s Eqn.(1). dx n dx n 86
Legendre, Hermite and d n Thus Legendre polynomial P )x( must be a constant multiple of [( x 2 − )1 n ] , i.e., Laguerre Polynomials n dx n d n P )x( = A [( x 2 − )1 n ] , where A is a constant. The constant A is determined by n dx n setting Pn )1( = .1 d Now P )x( = A D n x([ 2 − )1 n ] = A D n [( x − )1 n x( + )1 n ], where D = n dx n n n−r n r n = A ∑ C r [D x({ − )1 [].} D x({ + )1 ] r=0 = )!n( A x( + )1 n + terms with x( − )1 as factor .
n Thus, Pn )1( =1⇒ n(.A )!. 2 =1 1 i.e., A = . )!n( 2n n 1 d 2 n Hence, Pn (x) = (x{ − 1) }. (18) (n)!2 n dx n
It is called Rodrigue’s Formula for Legendre Polynomials. Using the Binomial theorem we can write n n r 2n− r2 2 n n 2 n−r r (− )1 x!n x( − )1 = ∑Cr x( ) (− )1 = ∑ r=0 r =0 n(!r − )!r Differentiating the above expression n times and substituting in Eqn.(18), we get 1 n (− )1 r !n d n x( 2n− r2 ) P )x( = n n ∑ n 2!n r=0 n(!r − )!r dx 1 N (−1) r (2n − 2r)! xn−2r (19) = n ∑ 2 r=0 r! (n − r)! (n − 2r)! where, N = 2/n or n( − 2/)1 whichever is an integer. In notation we usually write ,2/n if n is even n N = ]2/n[ , where = n −1 . 2 , if n is odd 2
This function Pn )x( as given by Eqn.(19) is called Legendre function of the first kind or Legendre polynomial of degree n. Example 1: Show that .5.3.1 K n( − )1 (− )1 /n 2 , if isn even Pn )0( = .4.2 Kn . 0 , if n is odd Solution: From Rodrigue’s formula for Legendre polynomials , n 1 d 2 n Pn )x( = ()x −1 2n !n dx n 1 d n = []()()x +1 n x −1 n 2n !n dx n
1 n n n−1 n n−2 !n 2 = x(!n + )1 + C1n x( + )1 x( − )1 !n + C 2 n(n − x()1 + )1 x( − )1 2n !n 2 + L + x(!n − )1 n ]
1 n n 2 n−1 n 2 n−2 2 = [ x(!n + )1 + ( C1 ) x( + )1 x( − !n)1 + ( C 2 ) x( + )1 x( − )1 !n 2n !n + L + x(!n − )1 n ] 87
Ordinary Differential Putting x = 0 , we get Equations 1 n 2 n 2 n 2 L n n 2 Pn )0( = [(!n C0 ) − ( C1 ) + ( C 2 ) + + (− )1 ( C n ) ] 2n !n 1 0 , if n is odd n n = !n [Q C = C ] n n 2/ n k n−k 2 !n (− )1 . C n 2/ , if isn even
∴ Pn )0( = 0 if n is odd 1 and P )0( = (− )1 n 2/ n C , if n is even n 2n /n 2 1 !n = (− )1 n 2/ 2n n n ! ! 2 2 .5.3.1 K n( − )1 = (− )1 n 2/ 2.4.6.Kn *** You may now try the following exercises. E1) Using Rodrigue’s Formula, show that 1 (− )1 n 1 a) P)x(f dx)x( = x( 2 − )1 n f )n( dx)x( . ∫n 2n !n ∫ −1 −1 1 2n+1 )!n( 2 b) x n P dx)x( = . ∫ n 2( n + )!1 −1
E2) Prove that all the roots of Pn )x( = 0 are real, distinct and lie between −1 and +1.
1 E3) Show that ∫ Pn dx)x( = 0 , except when n = 0 , in which case, the value of the −1 integral is 2. Thus you have seen that Legendre polynomials are obtained either as polynomial solutions of Legendre’s equation or by Rodrigue’s formula. We shall now discuss the historical discovery of the Legendre polynomials through expansion of the generating function into a particular type of series. We shall show that these polynomials can also be obtained as coefficients of t n in the expansion of 1( − 2xt + t 2 ) − 2/1 . 3.2.2 Generating Function The subject of potential theory is concerned with the forces of attraction due to the presence of a gravitational field. Central to the discussion of problems of gravitational attraction is Newton’s law of universal gravitation. Every particle of matter in the universe attracts every other particle with a force whose direction is that of the line joining the two, and whose magnitude is directly as the product of their masses and inversely as the square of their distance from each other. The force field generated by a single particle is usually considered to be conservative. That is, there exists a potential function V such that the gravitational force F at a point of free space (i.e., free of point masses) is related to the potential function according to F = −∇ V (20) where the minus sign is conventional. If r denotes the distance between a point mass and a point of free space, the potential function can be shown to have the form 88
k Legendre, Hermite and V )r( = (21) Laguerre Polynomials r where k is a constant whose numerical value does not concern us. Because of spherical symmetry of the gravitational field, the potential function V depends on only the radial distance r . To obtain Legendre’s results, let us suppose that a particle of mass m is located at point P , which is ‘a’ units from the origin of our coordinate system (see Fig.1). P r
a Q
φ b
O Fig.1 Let the point Q represent a point of free space r units from P and b units from the origin O . Let us assume that b > a , then, from the law of cosines, we find the relation r 2 = a 2 + b 2 − ab2 cos φ (22) where φ is the central angle between segments OP and OQ . By rearranging the terms and factoring out b 2 , it follows that a a 2 r 2 = b 2 1 − 2 cos φ + , a < b (23) b b We introduce the parameters a t = , x = cos φ (24) b and taking the square root of Eqn.(23), we obtain
2/1 r = b (1 − 2xt + t 2 ) (25) Finally, substituting Eqn.(25) into Eqn.(21), we obtain the potential function k − 2/1 V = (1 − 2xt + t 2 ) , 0 < t <1 (26) b We refer to the function w )t,x( = 1( − 2xt + t 2 )− 2/1 as the generating function of the Legendre polynomials. We shall now develop w )t,x( in a power series in the variable t and show that ∞ 1 n = ∑ t Pn ),x( t ≠ 1 (27) 2 1− 2xt + t n=0 We write 1( − 2xt + t 2 ) − /1 2 = 1[ − 2( x − ]t)t − /1 2 = 1( − )u − /1 2 where, u = 2( x − t)t Expanding in a binomial series, we get 1 )2/3()2/1( )2/5()2/3()2/1( 1( − )u − /1 2 = 1+ u + u 2 + u 3 +L 2 !2 !3
1 !2 !4 !6 2( )!n = + u + u 2 + u 3 +L+ u n +L 2 2)!1( 2 )!2( 2 2 4 )!3( 2 26 )!n( 2 2 2n
Substituting u = 2( x − t)t , we get 89
Ordinary Differential 2 − 2/1 !2 !4 2 2 Equations 1( − 2xt + t ) =1 + 2( x − t)t + 2( x − )t t + L )1( 2 22 )!2( 2 24 2( n − )!r2 2( )!n + 2( x − )t n−r t n−r + L + 2( x − )t n t n + L (28) n([ − )!r ]2 22n− r2 )!n( 2 22n Since we are interested in the coefficients of t n in the above series, we consider the n( − )r th term of the series and expand the product 2( x − )t n−r t n−r in powers of t .
We have n( − 2!)r n− r2 t n−r [ n−r C (− )t r 2( )x n− r2 ] = (− )1 r t n x n− r2 r n(!r − !)r2 The n( − )r th term of series (28) then becomes 2( n − !)r2 n( − 2!)r n− r2 (− )1 r 2( n − )!r2 t n x n− r2 . (− )1 r t n x n− r2 = L (29) n([ − ]!)r 2 22n− r2 n(!r − !)r2 2n n( − n(!r!)r − !)r2
n n( − )1 Summing the r.h.s of Eqn(29) for r = ,1,0 K, N , where N = or , whichever 2 2 is an integer, we get on using Eqn.(19) N (− )1 r 2( n − x!)r2 n− r2 t n = P t)x( n (30) ∑ n n r=0 2 n(!r − n(!)r − )!r2 ∞ 2 − 2/1 n Therefore, 1( − 2xt + t ) = ∑ Pn t)x( t, ≠1 n=0
Thus Pn )x( can be obtained by the generating function. You may observe here that
when n is an even number the polynomial Pn )x( is an even function; and when n is odd, the polynomial is an odd function and the relation
n Pn (− )x = (− )1 Pn ,)x( (31) holds for n = ,2,1,0 K .
In Fig.2 below, we have given the graphs of Pn ),x( n = ,4,3,2,1,0 over the interval −1≤ x ≤1 . Pn )x(
1 P0
P1 P3 P4
x −1 1
P2
−1
Fig.2: Graph of Pn ),x( n = 4,3,2,1,0 Further, considering Eqn.(26) with x = cos φ and t = b/a , we find that the potential function has the series expansion 90
∞ n Legendre, Hermite and k a Laguerre Polynomials V = ∑ Pn (cos φ) , a < b (32) b n=0 b In terms of the argument cos φ , the Legendre polynomials can be expressed as trigonometric polynomials of the form shown below
P0 (cos φ) =1
P1 (cos φ) = cos φ 1 1 P (cos φ) = 3( cos 2 φ − )1 = 3( cos 2φ + )1 2 2 4 1 1 P (cos φ) = 5( cos 3 φ − 3cos φ) = 5( cos 3φ + 3cos φ) 3 2 8 Let us now take up the following examples. Example 2: Show that n (a) Pn (− )x = (− )1 Pn )x(
(b) Pn )1( =1 n (c) Pn (− )1 = (− )1
Solution: From Generating Function for Legendre Polynomials, we get ∞ −1 n 2 2 (a) ∑ Pn (− )x h = 1( + 2xh + h ) n=0 −1 2 = []1 − 2x (− )h + (− )h 2 ∞ n = ∑ Pn ()x( − )h n=0 ∞ n n = ∑ (− )1 Pn )x( h n=0 n ∴Pn (− )x = (− )1 Pn )x( 1 ∞ − n 2 2 (b) ∑ Pn )1( h = 1( − 2h + h ) n=0 = 1( − )h −1 =1 + h + h 2 + h 3 + L + h n + L ∞ = ∑ h n n=0
∴ Pn )1( =1. 1 ∞ − n 2 2 (c) ∑ Pn (− h)1 = 1( + 2h + h ) n=0 = 1( + )h −1 ∞ = ∑ (− )1 n h n n=0 n ∴ Pn (− )1 = (− )1 ***
Example 3: Using the generating function for the Legendre polynomial show that 2 3x −1 1 3 P0 )x( = ,1 P1 )x( = ,x P2 )x( = , P3 )x( = 5( x − 3 )x 2 2 1 4 2 P4 )x( = 35( x − 30 x + ).3 8 91
Ordinary Differential Solution: From generating function for Legendre polynomials we know that Equations 1 ∞ − n 2 2 − 2/1 ∑ h Pn )x( = 1( − 2hx + h ) = 1[ − h 2( x − ])h n=0 1 3.1 5.3.1 =1 + h 2( x − )h + h 2 2( x − )h 2 + h 3 2( x − )h 3 2 4.2 6.4.2 7.5.3.1 + h 4 2( x − )h 4 +L 8.6.4.2 2 3 4 L ⇒ P0 )x( + h P1 )x( + h P2 )x( + h P3 )x( + h P4 )x( + 1 1 1 =1 + h.x + 3( x 2 − h)1 2 + 5( x 3 − 3 h)x 3 + 35( x 4 − 30 x + h)3 4 + L 2 2 8 Equating the coefficients of like powers of h , we get 1 1 P )x( = P,1 )x( = P,x )x( = 3( x 2 − ),1 P )x( = 5( x 3 − 3 )x 0 1 2 2 3 2
1 P )x( = 35( x 4 − 30 x 2 + ).3 4 8 *** Example 4: Express )x(f = x 4 + 3x 3 + 4x 2 − x + 2 in terms of Legendre polynomials. Solution: We have shown in Example 3, that 2 3x −1 1 3 P0 )x( = P,1 1 )x( = P,x 2 )x( = , P3 )x( = 5( x − 3 )x 3 2 1 P )x( = 35( x 4 − 30 x 2 + ).3 4 8 The above results can also be obtained by substituting n = ,3,2,1,0 and 4 in relation (19). From these relations we can write P 2 2 3 2 3 x 2 = 0 + P. ),x( x 3 = P )x( + x = P )x( + P )x( and 3 3 2 5 3 5 5 3 5 1 8 6 3 x 4 = P )x( + x 2 − 35 4 7 35 8 4 1 = P )x( + P )x( + P 35 4 7 2 5 0 Thus, x 4 + 3x 3 + 4x 2 − x + 2 8 4 P 2 3 2 P = P )x( + P )x( + 0 + 3 P )x( + P )x( + 4 P )x( + 0 − P )x( + 2P )x( 35 4 7 2 5 5 3 5 1 3 2 3 1 0 8 6 68 4 53 = P )x( + P )x( + P )x( + P )x( + P ).x( 35 4 5 3 21 2 5 1 15 0 *** You may now attempt the following exercises. E4) Prove that 1 1 1 1 1 1 1 P − = P − P + P − P +L + P − P . n 2 0 2 2n 2 1 2 2n−1 2 2n 2 0 2 E5) Show that 1 a) P′ )1( = n n( + )1 n 2 1 b) P′ (− )1 = (− )1 n+1 n n( + )1 n 2 E6) Show that a) P )0( = 0 , n = ,2,1,0 K 92 2n+1
n Legendre, Hermite and (− )1 2 !n K Laguerre Polynomials b) P2n )0( = , n = ,2,1,0 22n )!n( 2 E7) Verify that
a) P0 (cos φ) =1
b) P1 (cos φ) = cos φ 1 c) P (cos φ) = 3( cos 2φ + )1 2 4 1 d) P (cos φ) = 5( cos 3φ + 3cos φ) 3 8 We now give some relations between Legendre polynomials of different order. 3.2.3 Recurrence Relations We shall now establish a few useful relations involving Legendre polynomials of different order or between Legendre polynomials and their derivatives, which are called recurrence relations. From Rodrigue’s formula, we have
1 n( + )1 2 n+1 d Pn+1 )x( = D x([ − )1 ,] where D = (33) 2 n+1 n( + !)1 dx Differentiating Eqn.(33) w.r. to x , we get 1 P′ )x( = D n+1 n([ + x()1 2 − )1 n 2. x ] (34) n+1 2n+1 n( + !)1 1 = D n x([ 2 − )1 n + 2nx 2 x( 2 − )1 n−1 ] 2n !)n( 1 = D n x([ 2 − )1 n + 2n x( 2 −1 + x()1 2 − )1 n−1 ] 2n !)n( 1 1 = D n 2([ n + x()1 2 − )1 n + 2n x( 2 − )1 n−1 ] 2n !n 2n + 1 1 = D n x( 2 − )1 n + D{D n−1 x( 2 − )1 n−1 } 2n !n 2n−1 n( − !)1
= 2( n + )1 Pn )x( + Pn′−1 )x(
Thus Pn′+1 (x) = (2n + 1) Pn (x) + Pn′−1 (x) (35) Now we use Leibnitz rule on the right hand side of Eqn.(34) to obtain 1 P′ )x( = {[ D n+1 x( 2 − )1 n }x + n( + ){1 D n x( 2 − )1 n ]1.} n+1 2n !)n( 1 1 = .x D D n x({ 2 − )1 n } + n( + ).1 {D n x( 2 − )1 n } 2n )!n( 2n !)n(
= x Pn′ )x( + n( + P)1 n )x(
⇒ Pn′+1 (x) = x Pn′ (x) + (n + 1) Pn (x) (36)
Subtracting Eqn.(36) from Eqn.(35), we get
n Pn (x) = xPn′ (x) − Pn′−1 (x) (37) Changing n( + )1 to n in relation (35) and using it in relation (37), we get
n Pn )x( = 2([x n − )1 Pn−1 )x( + Pn′−2 ])x( − Pn′−1 )x(
= 2( n − )1 x Pn−1 )x( +[ x Pn′−2 )x( − Pn′−1 ])x(
= 2( n − )1 x Pn−1 )x( − n( − )1 Pn−2 )x( ( using relation (37)) 93
Ordinary Differential We can thus write Equations (n + 1) Pn+1 (x) = (2n + 1)x Pn (x) − n Pn−1 (x) (38) which is a three-term recurrence relation.
From Eqn.(36) and (37) following relations can also be obtained
2 (1 − x ) Pn′ (x) = n{ Pn−1 (x) − x Pn (x) } (39) 2 and (1 − x ) Pn (x) = (n + 1) {x Pn (x) − Pn+1 (x) } (40) We are leaving them here for you to do it yourself. E8) Establish relations (39) and (40). E9) Derive the identity 2 K 1( − x ) Pn′ )x( = n( + x[)1 Pn )x( − Pn+1 ])x( , n = ,2,1,0 .
Legendre polynomials belong to an important class of orthogonal polynomials which we shall discuss now. 3.2.4 Orthogonality Property
We shall now show that the Legendre polynomials Pn )x( satisfy the following orthogonal property +1 0 if, m ≠ n P )x( P dx)x( = 2 ' (41) ∫ m n if, m = n −1 2n +1 Consider the Legendre equation d 2 y dy 1( − x 2 ) − 2x + n(n + )1 y = .0 dx 2 dx This can be written as
d 2 dy 1( − x ) + n(n + y)1 = 0 (42) dx dx
Legendre Polynomials Pm )x( and Pn )x( satisfy Eqn.(42), thus
d 2 [ 1( − x ) Pn′ ] = − n(n + )1 Pn (43) dx
d 2 and [ 1( − x ) Pm′ ]= −m(m + )1 Pm (44) dx
Multiplying Eqn.(43) by Pm and Eqn.(44) by Pn and substracting the resulting expressions, we get
d 2 [ 1( − x P() m Pn′ − Pn Pm′ )]= []m(m + )1 − n(n + )1 Pn Pm dx Integrating both sides w.r. to x from x = −1 to x = 1, we get 1 2 +1 1( − x P() m Pn′ − Pn Pm′ ) = [m(m + )1 − n(n + )]1 Pn )x( Pm dx)x( −1 ∫ −1 The left hand term vanishes at x = ±1 +1
∴ n( + m + ()1 m − )n ∫ Pm )x( Pn dx)x( = 0 −1 Since m and n are non-negative integers, ∴ n + m +1 ≠ 0 . Thus if m ≠ n , i.e., m − n ≠ 0 , then from above
+1
∫ Pm P)x( n dx)x( = ,0 for m ≠ .n −1 To prove the property for m = n , we use Rodrigue’s formula 94
n Legendre, Hermite and 1 d 2 n Laguerre Polynomials Pn )x( = x({ − )1 } )!n( 2 n dx n Let )x(f be any function with at least n continuous derivatives on the interval −1≤ x ≤1 and consider the integral 1
I = ∫ )x(f Pn dx)x( . −1 Using Rodrigue’s formula 1 1 d n I = )x(f x( 2 − )1 n dx n ∫ n 2 !n −1 dx Integrating by parts, we get 1 1 d n−1 x( 2 − )1 n 1 1 d n−1 x( 2 − )1 n I = )x(f − f ′ )x( dx n n−1 n ∫ n−1 2 !n dx −1 2 !n −1 dx The first expression in the above equation on the right hand side, vanishes at both limits, so 1 1 d n−1 x( 2 − )1 n I = − f ′ )x( dx n ∫ n−1 2 !n −1 dx and continuing to integrate by parts, we obtain 1 (− )1 n I = f )n( x()x( 2 − )1 n dx . n ∫ 2 !n −1 2 !n Now if we put )x(f = P )x( in the above integral then since P )n( )x( = , it n n 2n !n follows that 1 2 !n 1 I = P 2 dx)x( = 1( − x 2 ) n dx ∫n 2 2n )!n( 2 ∫ −1 −1 2(2 )!n 1 = 1( − x 2 ) n dx 2n 2 ∫ 2 )!n( 0 Using the substitution x = sin θ , and recalling the reduction formula, 1 2n cos 2n+1 θ dθ = cos 2n θsin θ + cos 2n−1 θ dθ . ∫2n +1 2n +1 ∫
Then the definite integral in I becomes 1 π/ 2 2n π / 2 1( − x 2 ) n dx = cos 2n+1 θ dθ = cos 2n−1 θdθ ∫ ∫2n +1 ∫ 0 0 0 2n 2n − 2 2 π / 2 = L cos θ dθ 2n +1 2n −1 3 ∫ 0 2 n !n 2 2n )!n( 2 = = 3.1 K 2( n − 2()1 n + )1 2( )!n 2( n + )1 Thus, we conclude that
2(2 )!n 1 I = 1( − x 2 ) n dx 2n 2 ∫ 2 )!n( 0 2(2 )!n 22n )!n( 2 2 = = 22n )!n( 2 2( )!n 2( n + )1 2n +1 1 2 or, P 2 dx)x( = (45) ∫ n 2n +1 −1 95
Ordinary Differential Example 5: Assuming the orthogonality property of Legendre polynomials, show Equations that 1 0 , if m ≠ n 2 )1( )1( ∫ 1( − x P) n )x( Pm dx)x( = 2 n(n + )1 , if m = n −1 2n +1 where m and n are integers. 1 2 )1( )1( Solution: L.H.S. = ∫ 1( − x ) Pn )x( Pm dx)x( −1 1 1 2 )1( d 2 )1( = 1( − x ) Pm ).x( Pn )x( − ∫ Pn )x( {}1( − x ) Pm )x( dx −1 dx −1
Since Pm )x( is a solution of Legendre’s equation, thus
d 2 )1( [ 1( − x ) Pm )x( ]= −m (m + )1 Pm )x( dx
1 0 , if n ≠ m 1 ∴ .S.H.L = m(m + )1 ∫ Pn )x( Pm dx)x( = 2 n(n + )1 ∫ Pn dx)x( , if n = m −1 −1 0 , if n ≠ m = 2 n(n + )1 us[ ing result (41)] , if n = m 2n +1 *** Many problems of potential theory depend on the possibility of expanding a given function in a series of Legendre polynomials. It is easy to see that this can always be done when the given function is itself a polynomial. For example, any third-degree 2 3 polynomials )x(p = b 0 + b1x + b 2 x + b3 x can be written as 1 2 3 2 (see Example 4) )x(p = b 0 P0 )x( + b1P1 )x( + b 2 P0 )x( + P2 )x( + b3 P1 )x( + P3 )x( 3 3 5 3 b b3 2b 2b = b + 2 P )x( + b + 3 P )x( + 2 P )x( + 3 P )x( 0 3 0 1 5 1 3 2 5 3 3 = ∑ a n Pn )x( n=0
More generally, since Pn )x( is a polynomial of degree n for every positive integer n , it can easily be seen that x n can always be expressed as a linear combination of K P0 ),x( P1 ),x( P, n )x( , so any polynomial )x(p of degree k has an expansion of the form k )x(p = ∑ a n Pn )x( (46) n=0 But when the given function is not a polynomial then it may not be easy to expand an ‘arbitrary’ function )x(f in series of the form: ∞ )x(f = ∑ a n Pn )x( (47) n=0
We then need a new procedure for calculating the coefficients a n in Eqn.(47), and the
key lies in formula (41). We multiply Eqn.(47) by Pm )x( and integrate term by term from −1 to 1 and obtain 1 ∞ 1 ∫ )x(f Pm dx)x( = ∑ a n ∫ Pm )x( Pn dx)x( n=0 −1 −1 Using formula (41), it reduces to 96
Legendre, Hermite and 1 a2 )x(f P dx)x( = m Laguerre Polynomials ∫ m 2m +1 −1
Coefficients a n in Eqn.(47) are then obtained as 1 1 a = n + )x(f P dx)x( (48) n 2 ∫ n −1 The above manipulations are easy to justify if )x(f is known in advance to have a series expansion of the form (47) and this series is integrable term by term on the interval −1≤ x ≤1 . Both the conditions are obviously satisfied when )x(f is a polynomial. However, we shall not discuss here the conditions under which the representation of form (47) and (48) is valid for any other type of functions. For now it suffices to say that for certain functions the series (47) will converge throughout the interval −1≤ x ≤1 , even at points of finite discontinuities of the given function. Series of this type are called Legendre series. In practice, the evaluation of integrals like (48) is performed numerically. However, if the function f is not too complicated, we can sometimes use various properties of Legendre polynomials to evaluate such integrals in closed form. We illustrate this through the following example. Example 6: Find the Legendre series for − ,1 −1≤ x < 0 )x(f = . 1 , 0 < x ≤1
Solution: The given function f is an odd function. Hence )x(f Pn )x( is an odd function when n is even, and in this case 1 1 a = n + )x(f P dx)x( = ,0 n = ,4,2,0 K n 2 ∫ n −1
For n odd, the product )x(f Pn )x( is an even function, and therefore 1 1 a = n + )x(f P dx)x( n 2 ∫ n −1 1 1 K = 2( n + )1 ∫)x(f Pn dx)x( = 2( n + )1 ∫ Pn dx)x( , n = ,5,3,1 0 0 Now let us use the identity (35) 1 P )x( = []P′ )x( − P′ )x( n 2n +1 n+1 n−1 and set n = 2k +1 for k = ,2,1,0 K 1
a 2k+1 = 4( k + )3 ∫ P2k+1 dx)x( 0 1 ′ ′ = ∫ [ P2k+2 )x( − P2k )]x( dx 0 1 = []P )x( − P )x( 2k+2 2k 0 Q = P2k )0( − P2k+2 )0( [ Pn )1( = 1 for all n ]
Using the result of E6), we have
(− )1 k 2( !)k (− )1 k+1 2( k + !)2 a 2k+1 = − 22k )!k( 2 22k+2 []k( + !)1 2 97
Ordinary Differential (− )1 k 2( )!k 2( k + 2()2 k + )1 Equations = 1 + 22k )!k( 2 22 k( + )1 2
(− )1 k 2( )!k 2k +1 (− )1 k 2( )!k 4( k + )3 = 1 + = 22k )!k( 2 2k + 2 22k+1 k(!k + )!1 and thus ∞ (− )1 k 2( )!k 4( k + )3 )x(f = P ,)x( −1≤ x ≤1 . ∑ 2k+1 2k+1 k=0 2 k(!k + )!1 *** You may try a few exercises now.
E10) If )x(f is a polynomial of degree less then n , prove that 1 ∫ )x(f Pn dx)x( = 0 . −1
E11) Express )x(f = x 3 + a x 2 + b x + c as a linear combination of Legendre polynomials.
2 E12) Show that 2( n + x()1 − )1 Pn′ = n(n + )1 P( n+1 − Pn−1 ) .
1 P )x( t2 n E13) Prove that n dx = . ∫ 2 2n +1 −1 1 − 2 tx + t 1 1 L θ θ E14) Prove that 1 + P1 (cos θ) + P2 (cos θ) + = ln 1 + sin / sin . 2 3 2 2 E15) Make the change of variable x = cos φ in the DE 1 d dy sin φ + n n( + )1 y = 0 sin φ dφ dφ and show that it reduces to Legendre’s Eqn.(1).
E16) Determine the values of n for which y = Pn )x( is a solution of a) 1( − x 2 y) ′′ − 2xy′ + n n( + )1 y = ,0 )0(y = ,0 )1(y =1 b) 1( − x 2 y) ′′ − 2xy′ + n n( + )1 y = ,0 y′ )0( = ,0 )1(y =1.
3.2.5 Legendre Function of the Second Kind
We have shown that for positive integral values of n the Legendre polynomials Pn )x( represents one of the linearly independent solutions of Legendre’s Eqn.(1), viz., 1( − x 2 ) y ′′ − 2xy′ + n n( + y)1 = 0 . Because the equation is of second order, we know from the theory of differential
equations that there exists a second linearly independent solution Q n )x( (say ) , such that
y = c1 Pn )x( + c 2Q n )x( (49)
where c1 and c 2 are arbitrary constants, is a general solution of Eqn.(1). Also from the theory of second-order linear differential equations it is well known that
if y1 )x( is a non-trivial solution of y′′ + )x(a y′ + )x(b y = 0 (50) then a second linearly independent solution can be defined by exp [−∫ dx)x(a ] y )x( = y )x( dx [ref. Unit 7, MTE-08]. (51) 2 1 ∫ 2 y1 )x( 98
Hence, if we express Eqn.(1) in the form Legendre, Hermite and Laguerre Polynomials 2x n n( + )1 y ′′ − y′ + y = 0 1 − x 2 1( − x 2 ) and let y1 )x( = Pn )x( , it follows that dx y )x( = P )x( (52) 2 n ∫ 2 2 1( − x P[) n )]x( is a second solution, linearly independent of Pn )x( . Because any linear combination of solutions is also a solution of a homogeneous differential equation, we defined the second solution of Eqn.(1) as follows: dx Q )x( = P )x( A + B , (53) n n n n ∫ 2 2 1( − x P[) n )]x( where A n and Bn are constants to be chosen for each n . We refer to Q n )x( as the Legendre function of the second kind of integral order.
Accordingly, when n = 0 , we choose A 0 = 0 and B0 =1, and hence dx Q0 )x( = ∫ 1 − x 2 which leads to 1 1 + x Q )x( = ln , |x| <1 (54) 0 2 1 − x
For n =1, we set A1 = 0 and B1 =1, from which we obtain dx Q1 )x( = x ∫ 1( − x 2 x) 2
1 1 = x + dx ∫ 1 − x 2 x 2 1 1 + x = lnx −1 (55) 2 1 − x or, Q1 )x( = x Q 0 )x( −1 |x|, < 1 (56) If we continue in this fashion then we will have to evaluate more difficult integrals. Since it is assumed that all solutions of Legendre’s equation satisfy the recurrence formulas for Pn )x( , hence we select the Legendre functions Q n )x( so that, 2n +1 n Q )x( = x Q )x( − Q )x( (57) n+1 n +1 n n +1 n−1 K for n = ,3,2,1 with Q0 )x( and Q1 )x( already defined. The substitution n =1 into Eqn.(57) yields 3 1 Q 2 )x( = x Q1 )x( − Q 0 )x( 2 2 1 3 = 3( x 2 − )1 Q )x( − x 2 0 2 3 1 Q 2 or, Q 2 )x( = P2 )x( Q0 )x( − x |x|, < 1 P2 )x( = 3( x − )1 (58) 2 2 For n = 2 , we find 5 2 Q )x( = P )x( Q )x( − x 2 + |x|, < 1 (59) 3 3 0 2 3 whereas in general, we state the result [ n( − 2/)1 ] 2( n − 4k − )1 Q n )x( = Pn )x( Q0 )x( − ∑ Pn−2k−1 |x|,)x( < 1 (60) k=0 2( k + n()1 − )k for n = ,3,2,1 K .
Because of the logarithmic term in Q 0 ),x( Q n )x( has infinite discontinuities at x = ±1. However, within the interval −1< x <1 , these functions are well defined. 99
Ordinary Differential Without giving the details, we state that the Legendre functions Q )x( satisfy all Equations n recurrence relations given in Sec.3.2.3 for Pn )x( . In addition, there are several
relations that directly involve both Pn )x( and Q n )x( . For example, if |t| < |x| , then 1 ∞ = ∑ 2( n + )1 Pn )t( Q n )x( (61) x − t n=0 From this result, it can be easily shown that 1 1 P )t( Q )x( = n dt , n = ,2,1,0 K (62) n 2 ∫ x − t −1 which is called the Neumann formula. You may now try the following exercises. E17) Verify result (62). E18) Find the general solution of the following differential equations in terms of
Pn )x( and Q n )x( a) 1( − x 2 ) y ′′ − 2xy′ = 0 b) 1( − x 2 ) y ′′ − 2xy′ +12 y = 0 c) 1( − x 2 ) y ′′ − 2xy′ + 2y = 0 d) 1( − x 2 ) y ′′ − 2xy′ + 30 y = 0 . 1 1 + x E19) Given P0 )x( =1 and Q 0 )x( = ln , find W P[ 0 ),x( Q0 )]x( where , W is 2 1 − x 2 the wronskian.
E20) Using Eqn.(53), show that the wronskian of Pn )x( and Q n )x( is given by Bn K W ()Pn ,)x( Q n )x( = , n = ,2,1,0 . 1 − x 2 Other polynomials which commonly occur in applications are the Hermite and Laguerre polynomials which we shall be discussing now. 3.3 HERMITE AND LAGUERRE POLYNOMIALS Let us first consider the Hermite polynomials. Hermite Polynomials The Hermite polynomials play an important role in problems involving Laplace’s equations in cylinderical coordinates, in various problems in quantum mechanics and in probability theory. The Hermite polynomials are polynomial solutions to Hermite’s equation y′′ − 2xy′ + 2n y = 0 (63) where n is a constant.
Using the power series method it can be easily shown that )x(y = a 0 y1 )x( + a1y 2 )x( is the general solution of Eqn.(63), where 2n n( − )2 23 n(n − )2 n( − )4 y )x( =1 − x 2 + 22 n x 4 − x 6 + L (64) 1 2! 4! 6! and n(2 − )1 2 2 n( − )1 n( − )3 23 n( − n()1 − n()3 − )5 y )x( = x − x 3 + x 5 − x 7 + L (65) 2 3! 5! 7! and both the series converge for all x . We shall not be going into the details of the series solution here and leave it for you to verify it yourself. 100
If n is a non-negative integer, then one of these series terminates and is thus a Legendre, Hermite and Laguerre Polynomials polynomial- y1 )x( if n is even, and y 2 )x( if n is odd, while the other remains an infinite series. It can be easily verified that for n = ,1,0 ,4,3,2 ,5 these polynomials are 2 4 4 4 1,x,1 − 2x 2 x, − x 3 1, − 4x 2 + x 4 x, − x 3 + x 5 , respectively. 3 3 3 15 The polynomial solutions of Hermite’s Eqn.(63) are constant multiples of these polynomials. The constant multiples with the property that the terms containing the n n highest powers of x are of the form 2 x are denoted by H n )x( and called the Hermite polynomials. Let us now define the Hermite polynomials Generating Function
We define the Hermite polynomials H n )x( by means of the relation ∞ n (2xt −t 2 ) t e = ∑ H n (x) , for all finite x and t . (66) n=0 n! The function on the left hand side is called the generating function of the Hermite polynomials. This definition of Hermite polynomials is used often in statistical applications. We have ∞ m ∞ 2 k 2 2 2( xt ) (−t ) e 2( xt −t ) = e 2xt e. −t = ∑ ∑ m=0 m! k=0 !k ∞ /n[ ]2 (− )1 k 2( )x n−2k = ∑ ∑ t n (67) n=0 k =0 n(!k − 2 !)k where the last step follows from the index change m = n − 2k and ]2/n[ is the standard notation to the greatest integer ≤ 2/n . From Eqns.(66) and (67) it follows that [n/2] k (−1) n! n−2k H n (x) = ∑ (2x) (68) k=0 k! (n − 2k) ! n n , if n is even where = 2 n −1 2 , if n is odd 2
Eqn.(68) shows that H n )x( is a polynomials of degree n in x and that
n n H n )x( = 2 x + π n−2 )x( , where π n−2 )x( is a polynomial of degree n( − )2 in x . Further, it follows that it is an even function of x for even n and an odd function of x for odd n . Thus, it follows that n H n (−x) = (−1) H n (x) (69) The first few Hermite polynomials are listed in Table-1. Table 1
H )x( 1 0 = H1 )x( = 2x H )x( = 4x 2 − 2 2 3 H 3 )x( = 8x −12 H )x( =16 x 4 − 48 x 2 +12 4 5 3 H 5 )x( = 32 x −160 x +120 x 101
Ordinary Differential In addition to series (68), the Hermite polynomials can be defined by the Rodrigues Equations formula also. The Rodrigue’s Formula n 2 d 2 We prove the formula H )x( = (− )1 n e x e( −x ) (70) n dx n We have the relation ∞ 2 2 2 H )x( e tx2 −t = e x − t( − )x = ∑ n t n n=0 !n If we examine this relation in the light of Maclaurin’s theorem, we see that n n ∂ 2 2 ∂ 2 e( tx2 −t ) = e x e( − t( − )x ) = H )x( n n n ∂t t=0 ∂t t=0
n ∂ 2 2 or, e[ − t( − )x ] = e −x H )x( n n ∂t t=0 ∂ ∂ If we introduce a new variable z = x − t and use the fact that = − , then since ∂t ∂z t = 0 corresponds to z = x , the above expression reduces to n n −x2 d 2 d e( ) 2 (− )1 n e( −z ) = (− )1 n = e −x H )x( n n n dz z=x dx n 2 d 2 ∴H (x) = (−1) n e x (e −x ) n dx n which proves the Rodrigue’s formula (70). Hermite polynomials satisfy a number of properties, we list some of these here without proving them. You can verify them yourself. Recurrence Relations Using the generating function for Hermite polynomials it can be shown that
x H′n (x) = n H′n−1 (x) + n H n (x) (71) The relation ∞ n 2 H t)x( e 2( xt −t ) = ∑ n (72) n=0 !n on differentiating with respect to x , yields ∞ n 2 H′ t)x( et2 2( xt −t ) = ∑ n (73) n=0 !n From Eqns.(72) and (73), we get ∞ 2H t)x( n+1 ∞ H′ t)x( n ∑ n = ∑ n n=0 !n n=0 !n
which with a shift of index on the left, yields H′0 (x) = 0 , and for n ≥1,
H′n (x) = 2n H n−1 (x) (74) Combining relations (71) and (74), we obtain
H n (x) = 2x H n−1 (x) − H′n−1 (x) (75) Further, using relations (74) and (75), we obtain the relations
H n (x) = 2x H n−1 (x) − (n2 − 1) H n−2 (x) (76) and Hermite’s differential equation
H′n′ (x) − 2x H′n (x) + 2n H n (x) = 0 (77) 102 Relation (77) can be used to prove the orthogonality property of Hermite polynomials.
Orthogonality Legendre, Hermite and Laguerre Polynomials
The Hermite polynomials H n )x( satisfy the following orthogonality property ∞ 2 0 , if m ≠ n e −x H (x) H (x) dx = (78) ∫ n m 2n !n π if, m = n −∞ Eqn.(77) can be written in the form
−x2 ′ −x2 e[ H′n ])x( + 2n e H n )x( = 0 (79)
Similarly, for a polynomial H m )x( , we can write
−x2 ′ −x2 e[ H′m )]x( + 2m e H m )x( = 0 (80)
Multiplying Eqn.(79) by H m )x( and Eqn.(80) by H n )x( and adding the two, we get −x2 −x2 ′ −x2 ′ n(2 − m e) H n )x( H m )x( = H n )[x( e H′m )]x( − H m )[x( e H′n )]x( −x2 = e[ {}H n )x( H′m )x( − H′n )x( H m ])x( ′ Therefore, it follows that ∞ 2 2 −x −x ′ ′ ∞ n(2 − m) ∫ e H n )x( H m dx)x( = e[ {H n )x( H m )x( − H n )x( H m ]})x( −∞ (81) −∞ 2 Since the product of any polynomial in x by e −x → 0 as x → ∞ or as x → −∞ , we conclude that ∞ −x2 ∫ e H n )x( H m dx)x( = ,0 m ≠ n (82) −∞ This proves the first part of formula, that is, the Hermite polynomials form an 2 orthogonal set over the interval ] − ∞, ∞[ with weight function e −x . To establish the second part of (78), we use Rodrigue’s formula (70) and integrate ∞ ∞ n −x2 2 d e e −x H )x( H )x( dx = (− )1 n H )x( dx ∫n n ∫ n n −∞ −∞ dx by parts with u = H n ),x( du = H′n dx)x( n n−1 d 2 d 2 dv = e( −x dx) , v = e( −x ) dx n dx n−1 2 Since uv is the product of e −x and a polynomial, it vanishes at both limits and ∞ ∞ n−1 2 d 2 e −x [H ])x( 2 dx = (− )1 n+1 H′ )x( e( −x ) dx ∫n ∫ n n−1 −∞ −∞ dx ∞ n−2 d 2 = (− )1 n+2 H ′′ )x( e( −x dx) ∫ n n−2 −∞ dx ∞ LL 2n )n( −x2 = = (− )1 ∫ H n )x( e dx −∞ n n )n( n Now the term containing the highest power of x in H n )x( is 2 x , so H n )x( = 2 !n and the last integral is
∞ ∞ 2 2 2 n !n ∫e −x dx = 2 n 2!n ∫ e −x dx = 2n !n π −∞ 0 which gives the desired result ∞ −x2 2 n ∫ e H n )x( dx = 2 !n π −∞ 103
Ordinary Differential These orthogonality properties can be used to expand an arbitrary function )x(f in a Equations series ∞ f(x) = ∑ an H n (x) (83) n=0 −x2 The coefficients a n can be found by multiplying Eqn.(83) by e H m )x( and integrating term by term from − ∞ to ∞ . Using formula (78), this gives ∞ ∞ ∞ −x2 −x2 m ∫ e H m )x( )x(f dx = ∑ a n ∫ e H m )x( H n dx)x( = a m 2 m! π −∞ n=0 −∞ Replacing m by n , in the above equation, we get
∞ 1 2 a = e −x H (x) f(x) dx (84) n n π ∫ n 2 n! −∞ The series (83), with its coefficients calculated by formula (84), is called the Hermite series and is useful in the theory of orthogonal functions. Let us consider the following example. Example 7: Express )x(f = e 2bx in a Hermite series, and use this result to deduce the value of the integral ∞ −x2 +2bx ∫ e H n )x( dx . −∞ Solution: Set t = b in the generating function (66) to obtain ∞ n 2bx −b2 b e = ∑ H n )x( n=0 !n We can then write the series ∞ n ∞ n 2bx b2 b b b2 )x(f = e = e ∑H n )x( = ∑ a n H n )x( , where, a n = e n=0 1n n= 0 n! Also, from formula (84), we obtain ∞ 1 2 a = e −x +2bx H dx)x( , n = ,2,1,0 K n n ∫ n 2 !n π −∞ Thus, it follows that
∞ n b2 2 b!n e 2 e −x +2bx H dx)x( = π 2n = π 2( )b n e b , n = ,2,1,0 K ∫ n !n −∞ ***
You may now try the following exercises.
E21) Verify that y1 )x( and y 2 )x( given by Eqns.(64) and (65) respectively, are two linearly independent solutions of Hermite’s Eqn.(63). E22) Use the generating function for the Hermite polynomials to find
H 0 ),x( H1 ),x( H 2 )x( and H 3 )x( .
E23) Prove the following relations for Hermite polynomials H n )x( .
a) x H′n )x( = n H′n−1 )x( + n H n )x(
b) H n )x( = 2x H n−1 )x( − n(2 − )1 H n−2 )x(
c) H′n′ )x( − 2x H′n )x( + 2n H n )x( = 0 104
E24) Show that Legendre, Hermite and Laguerre Polynomials 2 !n a) H )0( = (− )1 n b) H )0( = 0 2n n! 2n+1 2( n + !)2 c) H′ )0( = 0 d) H′ )0( = (− )1 n 2n 2n+1 n( + !)1 E25) Use the generating function (66) to derive the relation n[ ]2/ !n H )x( x n = n−2k , n = ,2,1,0 K ∑ n k=0 2 n(!k − 2 !)k
∞ 3 2 E26) Expand )x(f = x − 3x + 2x in a series of the form ∑ a n H n )x( . n=0
Let us now consider the Laguerre polynomials. Laguerre Polynomials
The Laguerre polynomials L n )x( are polynomial solutions to Laguerre’s equation xy′′ + 1( − )x y′ + ny = 0 (85) where n is a constant. These polynomial solutions are obtained when n is a non- negative integer. An important application involving Laguerre polnomials in quantum mechanics is to find the wave function associated with the electron in a hydrogen atom. We now define the Laguerre polynomials and list some of their useful properties. We shall not be proving these properties here. We shall leave them for you to verify yourself. Generating Function
We define the Laguerre polynomials L n )x( by means of the relations −xt ∞ −1 1−t n (1 − t) e = ∑ Ln (x) t |t|, < 1, 0 ≤ x ≤ ∞ (86) n=0 The function on the left hand side is called the generating function of the Laguerre polynomials. We have −xt ∞ (− )1 k 1( − )t −1 e 1−t = ∑ xt( ) k 1( − )t −k−1 k=0 !k ∞ k ∞ (− )1 k − k −1 m m = ∑xt( ) ∑ (− )1 t (87) k=0 !k m= 0 m
− k −1 (−k − ()1 −k − ()2 −k − )3 K(−k − m) But since = m m!
m k( + k()1 + )2 K k( + m) m k + m = (− )1 = (− )1 , m! m Eqn.(87) becomes −xt ∞ ∞ (− )1 k k( + m)! x k 1( − )t −1 e 1−t = t k+m (88) ∑ ∑ 2 m=0 k=0 )!k( m! where we have reversed the order of summation.
Using the change of index m = n − k in Eqn.(88) and comparing with Eqn.(86), the Laguerre polynomials are defined by 105
Ordinary Differential ∞ (−1) k n! xk Equations L (x) (89) n = ∑ 2 k=0 (k! ) (n − k)!
In Table-2, we have listed the first few polynomials L n )x( Table 2
L 0 )x( =1
L1 )x( = −x + 1 1 L )x( = x( 2 − 4x + )2 2 !2 1 L )x( = (−x 3 + 9x 2 −18 x + )6 3 !3
1 4 3 2 L 4 )x( = x( −16 x + 72 x − 96 x + 24 ) !4 The Rodrigue’s Formula
The Rodrigue’s formula for Laguerre polynomials L n )x( is given by e x d n L (x) = (x ne −x ,) n = 0, 1, 2, K (90) n n! dx n which can be verified by the application of the Leibniz formula d n n n d n−k f d k g fg( ) = , n = ,3,2,1 K (91) n ∑ n−k k dx k=0 k dx dx Recurrence Relations We now list a few important properties satisfied by Laguerre polynomials
(n + 1) L n+1 (x) + (x − 1 − 2n) Ln (x) + n L n−1 (x) = 0 (92)
L′n (x) − L′n−1 (x) + Ln−1 (x) = 0 (93)
x L′n (x) = nL n (x) − n Ln−1 (x) (94)
and x L′n′ (x) + (1 − x) L′n (x) + n Ln (x) = 0 (95) Relation (95) can be used to prove the following orthogonality property of Laguerre polynomials. Orthogonality
The Laguerre polynomials L n )x( satisfy the orthogonality property ∞ −x ∫ e Ln (x) Lm (x) dx = 0, if m ≠ n (96) 0 You may now try the following exercises.
E27) Determine the Laguerre polynomials L 0 ),x( L1 ),x( L 2 )x( and L3 )x( . E28) Prove the recurrence relations (92) – (95).
∞ −x E29) Prove that ∫ e L n )x( L m dx)x( = 0 , if m ≠ n . 0
We shall now discuss some applications of these polynomials to physical situations. 3.4 APPLICATIONS TO PHYSICAL SITUATIONS We shall first illustrate the role played by Legendre polynomials in solving certain boundary value problems of mathematical physics. 106
Steady-state temperature in a sphere Legendre, Hermite and Laguerre Polynomials Consider a solid sphere of radius a . Let origin be its centre. Let the surface of the sphere be held at a prescribed temperature say, (g φ) , which is assumed to be independent of coordinate θ , until within the sphere a steady state for temperature (T ρ,φ) is produced by the flow of heat. Here (ρ,θ, φ) are the spherical polar coordinates and we wish to find the steady-state temperature (T ρ,φ) which satisfies Laplace equation. You may recall that Laplace equation for any function )z,y,x(T is given by ∂ 2T ∂ 2T ∂ 2T + + = .0 ∂x 2 ∂y 2 ∂z 2 and in the spherical polar coordinates (ρ,θ, φ) where , x = ρsin φcos θ, y = ρ sin φsin θ, z = ρ cos φ , it reduces to
2 1 ∂ 2 ∂T 1 ∂ ∂T 1 ∂ T ρ + sin φ + = 0 (97) ρ2 ∂ρ ∂ρ ρ 2 sin φ ∂φ ∂φ ρ 2 sin 2 φ ∂θ2 In the given situation since T does not depend on θ , Eqn.(97) reduces to
∂ 2 ∂T 1 ∂ ∂T ρ + sin φ = 0 (98) ∂ρ ∂ρ sin φ ∂φ ∂φ We have to find the solution of Eqn(98) subject to the boundary condition ,a(T φ) = g (φ) (99) To solve Eqn.(98) , we use the method of separation of variable and take the product solution of the form (T ρ,φ) = R (ρ) (P φ) (100) Using Eqn.(100), Eqn.(98) becomes
d 2 dR 1 d dP P ρ + R sin φ = 0 dρ dρ sin φ dφ dφ
1 d 2 dR 1 1 d dP ⇒ ρ = − sin φ (101) R dρ dρ P sin φ dφ dφ The left side of Eqn.(101) is a function of ρ only whereas right hand side of Eqn.(101) is a function of φ only. This is possible only when each side is equal to the same constant, sayλ . Then Eqn.(101) is equivalent to the equations. d 2 R dR ρ2 + 2ρ −λ R = 0 (102) dρ 2 dρ 1 d dP and, sin φ +λ P = 0 (103) sin φ dφ dφ Eqn.(102) is an Euler equation, its auxiliary equation is
m(m − )1 + 2m −λ = ,0 or , m 2 + m −λ = .0 −1 ± 1 + 4λ ⇒ m = . 2 General solution of Eqn.(102) can be written as
−1 1 −1 1 + λ+ − λ+ 2 4 2 4 R = c1 ρ + c 2ρ (104) 107
Ordinary Differential In order that R is single-valued and bounded near ρ = 0 , we put in Eqn.(104) Equations 1 1 c = 0 and − + λ + = n , where n is a non-negative integer. It then follows that 2 2 4 λ = n(n + )1 and Eqn.(104) reduces to n R = c1ρ (105) d 2 P cos φ dP and Eqn.(103) becomes + + n(n + P)1 = 0 . dφ2 sin φ dφ In this equation if we let x = cos φ, then we get d 2P dP 1( − x 2 ) − 2x + n(n + )1 P = 0 (106) dx 2 dx which is Legendre’s equation. By the physics of the problem, the function P must be bounded for 0 ≤ φ ≤ π i.e., −1≤ x ≤1 and we know that, the solution of Eqn.(106) are constant multiples of
Legendre polynomials, Pn )x( . If we combine this result with Eqn.(105), then we have particular solution of Eqn.(98) of the form n K α n ρ Pn (cos φ) , for each n = 3,2,1,0
where α n are arbitrary constants. The general solution of Eqn.(98) can be written as ∞ n (T ρ,φ) = ∑ α n ρ Pn (cos φ) (107) n=0 The boundary condition (99) now requires ∞ n (g φ) = ∑ α n a Pn (cos φ) n=0 or equivalently, ∞ −1 n g(cos )x = ∑ α n a Pn )x( n=0
Multiplying both sides of the above equation by Pm )x( and integrating w.r. to x from x = −1 to x = 1and using orthogonality property of Legendre polynomials, we get P 1 1 1 α = n + g(cos −1 P)x dx)x( (108) n n 2 ∫ n a −1 Hence relation (107) gives the required solution where coefficients are given by r2 r1 relation (108). r Let us look at another application of Legendre polynomials .
θ Electrostatic Dipole Potential Consider two point charges of equal magnitude, say q , but of opposite sign. − q a 0 a q Fig.3 Let these charges be placed in a polar coordinate system (see Fig.3). With suitable units of measurement, the potential U at the point P is q q U = − (109) r1 r2 2 2 where, r1 = r + a − ra2 cos θ
2 2 and, r2 = r + a + a2 r cos θ From generating function relation for the Legendre’s polynomials, we have ∞ 2 − 2/1 n 1( − 2xt + t ) = ∑ t Pn )x( (110) n=0 108
when r > a , we can use Eqn.(110) to write Legendre, Hermite and n Laguerre Polynomials 1 1 1 1 ∞ a = = ∑ Pn (cos θ) (111) 2 r1 r 1− )r/a(2 cos θ + )r/a( r n=0 r and, similarly, n 1 1 1 1 ∞ a = = ∑ Pn (− cos θ) (112) 2 r2 r 1 − )r/a(2 cos (−θ) + )r/a( r n=0 r From Eqns.(109), (110) and (111), we get n q ∞ a U = ∑[ Pn (cos θ) − Pn (− cos θ ]) (113) r n=0 r
From the properties of Legendre’s polynomial Pn )x( , we know that Pn )x( is even if n is even and it is odd if n is odd. Thus P[ n (cos θ) − Pn (− cos θ ]) equals 0 or, 2Pn (cos θ) according as n is even or odd. Thus relation (113) becomes 2n 1 2q ∞ a + U = ∑ P2n+1 (cos θ) r n=0 r 3 2q a a L = P1 (cos θ) + P3 (cos θ) + . (114) r r r When r is large compared with a , we assume that all terms except the first can be neglected and since P1 )x( = x , relation (114) becomes cos θ U = a2 q r 2 Physicists use this approximation for the dipole potential. We shall now consider the application of Hermite polynomials and understand the role played by the Hermite polynomials H n )x( and the corresponding Hermite function −x2 2/ e H n )x( in the theory of the linear harmonic oscillator in quantum mechanics. Simple Harmonic Oscillator In wave mechanics, the basic equation for describing the location of a particle attracted by an energy potential V )z( is the Schr Ödinger’s (time-independent) equation d 2ψ 8mπ2 + [V )z( − E ] ψ = 0 (115) dz 2 h 2 where m is the mass of the particle, E is the total energy, and h is Plank’s constant. The unknown quantity ψ is called the wave function i.e. the amplitude of the wave whose intensity gives the probability of finding the particle at any given point in space. A fundamental problem in wave mechanics concerns the motion of a particle bounded in a potential well. It has been established that bounded solutions of Schrodinger’s equation for such problems are obtained only for certain discrete energy levels of the particle within the well. A particular example of this important class of problems is the linear oscillator (also called the simple harmonic oscillator), the solution of which lead to Hermite polynomials. If the restoring force on a particle displaced a distance z from its equilibrium position is − kz , where k can be associated with the spring constant of the classical oscillator, 1 then its potential energy is V )z( = kz 2 . Substituting this expression into Eqn.(115) 2 and introducing the dimensionless parameters 4/1 4m kπ2 4πE m 4πE x = z , λ = = 2 h h k hω 109
Ordinary Differential where ω = /k m is the angular frequency of the classical oscillator, we find that Equations Eqn.(115) becomes ψ ′′ + (λ − x 2 ) ψ = 0 , − ∞ < x < ∞ (116) where the primes denote differentiation with respect to x . In addition to Eqn.(116), the wave function ψ must satisfy the boundary condition lim ψ )x( = 0 (117) |x| →∞ In order to look for bounded solutions of Eqn.(116), we start with the observation that λ becomes negligible compared with x 2 for large values of x . Thus, asymptotically we expect the solution of Eqn.(116) to behave like 2 ψ )x( ~ e ±x / 2 , |x| → ∞ where only the negative sign in the exponent is appropriate in order that Eqn.(117) be satisfied. Based on this observation, we make the assumption that Eqn.(116) has solutions of the form 2 ψ )x( = )x(y e −x 2/ (118) for suitable y . The substitution of Eqn.(118) into Eqn.(116) yields the differential equation y′′ − 2xy′ + (λ − )1 y = 0 (119) The boundary condition (117) suggests that whatever functional form y assumes, it 2 must either be finite for all x or approach infinity at a rate slower than e −x 2/ approaches zero. It is seen that the only solution of Eqn.(119) satisfying this condition are those for which λ −1 = 2 ,n or K λ = λ n = 2n + ,1 n = ,2,1,0 (120) called eigenvalues . In terms of E , we find 1 n + hω 2 E = , n = ,2,1,0 K (121) n 2π
with E 0 = hω 4/ π being the lowest or minimum energy level. With λ so restricted, we see that Eqn.(119) becomes y′′ − 2xy′ + 2ny = 0
which is Hermite’s equation with solution y = H n )x( . Hence, we conclude that to
each eigenvalue λ n given by Eqn.(120) there corresponds the solution of Eqn.(116) (called an eigenfunction) given by −x2 2/ K ψ n )x( = H n )x( e , n = ,2,1,0 (122) You may now try the following exercises.
−x2 2/ E30) Show that the functions ψ n )x( = e H n )x( satisfy the relations.
a) 2n ψ n−1 )x( = x ψ n )x( + ψ′n )x(
b) 2x ψ n )x( − 2n ψ n−1 )x( = ψ n+1 )x(
c) ψ′n )x( = x ψ n )x( − ψ n+1 )x( −x2 2/ E31) By writing ψ n )x( = c n e H n )x( , show that the normalized eigen function, ∞ 2 that is ∫ [ ψ n ])x( dx =1, assumes the form −∞ − 2/1 n −x2 2/ K ψ n )x( = [ π 2 !n ] e H n ),x( n = ,2,1,0 . We now end this unit by giving a summary of what we have covered in it. 110
Legendre, Hermite and 3.5 SUMMARY Laguerre Polynomials In this unit, we have learnt the following points: 1. The equation 1( − x 2 ) y ′′ − 2xy′ + n(n + y)1 = 0 is called Legendre’s Equation where n is a real constant. The polynomial solutions of this equation for integral values of n are Legendre polynomials . 2. Legendre polynomial of degree n , having value 1 at x =1, is denoted by
Pn )x( and is called Legendre function of the first kind and is given by ]2/n[ k (− )1 2( n − 2 !)k n−2k Pn )x( = ∑ n x k=0 2 n(!)k( − n(!)k − 2 !)k 2/n , if n is even where ]2/n[ = n −1 , if n is odd . 2
3. Legendre function of the second kind is denoted by Qn )x( , when n is an integer and is given by dx Q )x( = P )x( A + B n n n n ∫ 2 2 1( − x P[) n )]x(
where A n and Bn are constants. 4. Rodrigue’s formula for Legendre’s polynomials is given by 1 d n P )x( = x([ 2 − )1 n ]. n 2n !)n( dx n
5. Orthogonality property of the Legendre’s polynomials Pn )x( is 0 , if m ≠ n 1 ∫ Pn )x( Pm )x( dx = 2 , if m = n −1 2n +1 6. The function on the left side of
1 2 L n L = P0 )x( + t P1 )x( + t P2 )x( + + t Pn )x( + |x| < 1 1− 2xt + t 2 is called the generating function of the Legendre polynomials. 7. Some of the recurrence relations for Legendre polynomials are
Pn′+1 )x( = 2( n + )1 Pn )x( + Pn′−1 )x(
Pn′+1 )x( = x Pn′ )x( + n( + P)1 n )x(
nP n )x( = x Pn′ )x( − Pn′−1 )x(
n( + P)1 n+1 )x( = 2( n + xP)1 n )x( − nP n−1 )x( 2 1( − x ) Pn′ )x( = P[n n−1 )x( − x Pn ])x( 2 1( − x ) Pn )x( = n( + x[)1 Pn )x( − Pn+1 ])x(
8. The Hermite polynomials H n )x( are polynomial solutions to Hermite’s equation y′′ − 2xy′ + 2ny = 0 , where n is a constant, and are given by
n[ ]2/ k (− )1 !n n−2k H n )x( = ∑ 2( )x k=0 n(!k − 2 )!k where ]2/n[ is the greatest integer ≤ 2/n . ∞ n 2( xt −t2 ) t 9. The function on the left side of equation e = ∑ H n )x( , for all n=0 !n
finite x and t is the generating function of the Hermite polynomials H n )x( . 10. Some of the recurrence relations of Hermite polynomials are 111
Ordinary Differential x H′ (x) = n H′ (x) + n H (x) Equations n n−1 n H′ (x) = 2n H (x) n n−1 . H n (x) = 2x H n−1 (x) − H n−1 (x)
H n (x) = 2x H n−1 (x) − 2(n − 1) H n−2 (x) 11. The Rodrigues formula for Hermite polynomials is given by n 2 d 2 H )x( = (− )1 n e x e( −x ) n dx n
12. Orthogonality property of the Hermite polynomials H n )x( is ∞ 2 0 , if m ≠ n e −x H )x( H dx)x( = ∫ n m 2n !n π if, m = n −∞
13. The Laguerre polynomials L n )x( are polynomial solutions to Laguerre’s equation x y′′ + 1( − )x y′ + ny = 0 , where n is a constant, and are given by ∞ (− )1 k x!n k L )x( = n ∑ 2 k=0 )!k( n( − )!k 14. The function on the left hand side of the equation −xt ∞ −1 1−t n 1( − )t e = ∑ L n t)x( |t|, < ,1 0 ≤ x < ∞ n=0 is the generating function of the Laguerre polynomials.
15. The Rodrigues formula for the Laguerre polynomials L n )x( is e x d n L )x( = x( n e −x ) n !n dx n 16. Some of the recurrence relations of Laguerre polynomials are
(n + 1) Ln+1 (x) + (x − 1 − 2n) L n (x) + n Ln−1 (x) = 0
L′n (x) − L′n−1 (x) + Ln−1 (x) = 0
x L′n (x) = n Ln (x) − n Ln−1 (x)
17. Orthogonality property of the Laguerre polynomials L n )x( is ∞ −x ∫ e L n )x( L m )x( dx = 0 , if m ≠ n . 0 3.6 SOLUTIONS/ANSWERS
1 E1) a) Let I = ∫ )x(f Pn dx)x( −1 1 1 d n = )x(f x([ 2 − dx])1 2n !n ∫ dx n −1 1 1 1 d n−1 d n−1 = )x(f {( x 2 − )1 n } − f )1( )x( {( x 2 − )1 n }dx 2n !n dx n−1 ∫ dx n−1 −1 −1 (− )1 1 d n−1 = f )1( )x( x({ 2 − )1 n }dx 2n !n ∫ dx n−1 −1 Proceeding like this
(− )1 n 1 I = f )n( x().x( 2 − )1 n dx . 2n !n ∫ −1
b) Here )x(f = x n , f )n( )x( = !n 112
Legendre, Hermite and 1 (− )1 n 1 ∴ x n P dx)x( = x(!.n 2 − )1 n dx Laguerre Polynomials ∫n 2n !n ∫ −1 −1 (− )1 n 1 = x( 2 − )1 n dx 2 n ∫ −1 2n+1 )!n( 2 = 2( n + )!1
E2) By Rodrigue’s formula 1 d n P )x( = {( x 2 − )1 n } n 2n !n dx n since x( 2 − )1 n vanishes n times at x =1 and x = −1, thus its n th derivative
and hence , Pn )x( have all real zeroes lying between −1 and +1.
∴ the roots of Pn )x( = 0 are real and lie between −1 and +1. Let, if possible, the equation
Pn )x( = 0 has a repeated root, say 'α' . Then
Pn (α) = 0 and Pn′ (α) = 0 (123)
since Pn )x( is a solution of Legendre’s equation, therefore, 2 )2( )1( 1( − x ) Pn )x( − 2xP n + n(n + P)1 n )x( = 0 (124) Differentiating m times, by Leibnitz’s theorem, we get 2 (m+ )2 (m+ )1 (m) 1( − x P) n )x( − 2 (x m + P)1 n )x( + n( − m n() − m + P)1 n )x( = ,0 (125) Putting x = α in Eqn.(124) and using Eqn.(123), we get )2( Pn (α) = 0 Putting m = ,3,2,1 K in Eqn.(125), we get )3( )4( )5( L )n( Pn (α) = 0 = Pn (α) = Pn (α) = = Pn (α) 2n But from definition, P )n( (α) = ≠ 0 . n 2 n !n
Thus Eqn.(123) is not valid and hence Pn )x( = 0 cannot have repeated roots.
Hence all the roots of Pn )x( = 0 are distinct.
E3) From Rodrigue’s formula, we have 1 d n P )x( = {( x 2 − )1 n } n 2n !n dx n 1 1 1 d n ∴ P dx)x( = x( 2 − )1 n dx ∫n 2 n !n ∫ dx n −1 −1 +1 1 d n−1 = {( x 2 − )1 n } 2n !n dx n−1 −1 +1 1 d n−1 = []x( − )1 n x( + )1 n 2n !n dx n−1 −1 1 = 0[ − ]0 = 0 if n ≠ 0 2n n!
Now when n = 0 , then P0 )x( =1 1 1 +1 ∴ P0 dx)x( = dx.1 = x = 2 . ∫ ∫ −1 −1 −1 113
Ordinary Differential E4) We know that Equations ∞ 2 − 2/1 n 1( − 2xh + h ) = ∑ Pn )x( h (126) n=0 1 Putting x = in Eqn.(126), we get 2 ∞ 2 − /1 2 1 n 1( − h + h ) = ∑ Pn h (127) n=0 2 1 Again, putting x = − in Eqn.(126), we get 2 ∞ 2 − 2/1 1 n 1( + h + h ) = ∑ Pn − h (128) n=0 2 Now put h 2 for h in Eqn.(128), we get ∞ 2 4 − 2/1 2n 1 1( + h + h ) = ∑ h Pn − (129) n=0 2 But 1( + h 2 + h 4 ) − 2/1 ={( 1 + h 2 ) 2 − h 2 }− 2/1 = 1( + h + h 2 ) − 2/1 1( − h + h 2 ) − 2/1 Using Eqns.(127), (128) and (129) in the above, we get ∞ ∞ ∞ 2n n n ∑h Pn (− )2/1 = ∑ h Pn (− )2/1 ∑ h Pn )2/1( n=0 n=0 n=0 2 L 2 L = P[ 0 (− )2/1 + h P1 (− )2/1 + h P2 (− )2/1 + ]× P[ 0 )2/1( + h P1 )2/1( + h P2 )2/1( + ] Now equating the coefficient of h 2n on both sides, we get 1 1 1 1 1 1 1 P − = P − P + P − P + L + P − P n 2 0 2 2n 2 1 2 2n−1 2 2n−1 2 1 2 1 1 + P − P . 2n 2 0 2
E5) Since Pn )x( satisfies Legendre’s equation, we have 2 1( − x P) n′′ )x( − 2xPn′ )x( + n(n + P)1 n )x( = 0 (130) a) putting x =1 in Eqn.(130), we get
− 2Pn′ )1( + n(n + )1 Pn )1( = 0 1 i.e., P′ )1( = n(n + )1 P )1( n 2 n 1 = n n( + )1 [QP )1( = ]1 2 n b) putting x = −1 in Eqn.(130), we get
2Pn′ (− )1 + n n( + )1 Pn (− )1 = 0 1 i.e., P′ (− )1 = − n n( + )1 P (− )1 n 2 n 1 = (− )1 n−1 n()n( + )1 [Q P (− )1 = (− )1 n ] 2 n E6) a) Putting x = 0 in Eqn.(27), we obtain ∞ 2 − /1 2 n 1( + t ) = ∑ Pn t)0( (131) n=0 Expending left hand side of Eqn.(131) in binomial series ∞ 2 − /1 2 − 2/1 2n 1( + t ) = ∑ ()n t (132) n=0 114
Comparing terms on the right in series (131) and (132), we see that series Legendre, Hermite and K Laguerre Polynomials (132) has only even powers of t . Thus, Pn )0( = 0 for n = ,5,3,1 or, K equivalently, P2n+1 )0( = n,0 = ,2,1,0 . b) Since all odd terms in series (131) are zero, replace n by 2n in the series and compare with (132), we obtain −1 −1 −1 −1K − n +1 2 2 2 P )0( = ()− 2/1 = 2n n n! 1 1 1 (− )1 n +1K + n −1 2 2 2 = !n
1 1 1 1 (− )1 n + n −1 + n − 2K +1 2 2 2 2 = n!
n 1 1 (− )1 Γn + n + n −1 n n− 2/1 2 = (− )1 2 = (− )1 ()n = 1 n !n Γ 2 (− )1 n 2( )!n 1 2 !n 1 = . QΓ n + = Γ . n 2 2n 2 )!n( 2 2 !n 2
E7) In the expressions for P0 ),x( P1 ),x( P2 ),x( P3 )x( in terms of x , as obtained in Example 3, put x = cos φ and obtain the required results.
E8) a) Multiplying Eqn.(37) by x and then subtracting it from Eqn.(36), we get 2 1( − x P) n′ )x( = {n Pn−1 )x( − xP n })x( . b) Multiplying (36) by x and then subtracting it from (37), we get 2 1( − x ) Pn′ )x( = n( + xP{)1 n )x( − Pn+1 })x( .
E9) From relation (39), we have 1( − x 2 ) P′ )x( = {n P )x( − x P )}x( n n−1 n = 2( n + )1 x Pn )x( − n( + )1 Pn+1 )x( − nx Pn )x( [using (38)]
= n( + x[)1 Pn )x( − Pn+1 )]x(
E10) We know that K .3.1 2( n − )1 n n(n − )1 n−2 L Pn )x( = x − x + !n 2(2 n − )1
Thus we see that P0 )x( is a constant, P1 )x( is a polynomial of degree 1, P2 )x( is a polynomial of degree 2 and so on. m K Thus we can express x in terms of Legendre forms P0 ),x( P1 ),x( Pm )x( , and hence a polynomial )x(f of degree k < n can be expressed as L )x(f = A 0 P0 )x( + A1P1 )x( + + A k Pk )x( 1 k 1 ∴ ∫ P)x(f n dx)x( = ∑ A r ∫ Pr )x( Pn dx)x( r=0 −1 −1 k = A 0. , as r ≠ n ∑ r r=0 = 0.
E11) Proceeding as in Example 4. 115
Ordinary Differential )x(f = x 3 + ax 2 + bx + c Equations 2 3 P0 )x( 2 = P3 )x( + P1 )x( + a + P2 )x( + b P1 )x( + c P0 )x( 3 5 3 3
2 2 3 a = P )x( + a P )x( + + b P )x( + + c P ).x( 3 3 3 2 5 1 3 0
E12) From E8) (a) and (b), we have 2 1( − x ) Pn′ )x( = n{Pn−1 )x( − x Pn })x( 2 and 1( − x ) Pn′ )x( = n( + )1 x{ Pn )x( − Pn+1 )}x(
Substituting for xP n )x( from the first relation in second relation, we get 2 2 1( − x P) n′ )x( 1( − x P) n′ )x( = n( + )1 Pn−1 )x( − − Pn+1 )x( n
2 n +1 or, 1( − x ) 1 + Pn′ )x( = n( + P[)1 n−1 )x( − Pn+1 )]x( n 2 or, 2( n + 1()1 − x ) Pn′ )x( = n n( + P[)1 n−1 )x( − Pn+1 ])x( .
E13) From generating function of Legendre Polynomials, we have ∞ 2 − 2/1 m 1[ − 2xt + t ] = ∑ t Pm )x( m=0 1 1 ∞ Pn )x( m ∴ dx = Pn )x( ∑ t Pm )x( dx ∫2 ∫ m=0 −1 1 − 2xt + t −1 1 n 2 = t ∫ [ Pn )]x( dx , [using formula (41)] −1 2 = t n . . 2n +1
E14) We know that ∞ 2 − 2/1 n 1( − 2xt + t ) = ∑ t Pn )x( n=0 Integrating both sides w.r.t t between the limit 0 to 1, we get 1 ∞ 1 2 − 2/1 n ∫ 1( − 2xt + t ) dt = ∑ Pn )x( ∫ t dt n=0 0 0 Putting x = cos θ , we get 1 1 dt ∞ t n+1 = ∑ Pn (cos θ). ∫ 2 n=0 n + 1 0 1 − 2cos θ t. + t 0 1 dt ∞ 1 or, = ∑ Pn (cos θ). ∫ 2 2 n=0 n +1 0 t( − cos θ) + sin θ 1 ∞ P (cos θ) or, ln t( − cos θ) + sin 2 θ + t( − cos θ) 2 = n ∑ 0 n=0 n +1 1( − cos θ) + sin 2 θ + 1( − cos θ)2 ∞ P (cos θ) or, ln = ∑ n 1 − cos θ n +1 n=0 2 + 1 − cos θ ∞ P (cos θ) or, ln = ∑ n 1 − cos θ n=0 n +1 116
Legendre, Hermite and θ 2 + 2sin 2 Laguerre Polynomials 2 ∞ P (cos θ) or, ln = ∑ n θ n +1 2sin 2 n=0 2 θ 1 + sin 1 1 1 or, ln 2 =1 + P (cos θ) + P (cos θ) + P (cos θ) + L . θ 2 1 3 2 4 3 sin 2
dx dy dy E15) x = cos φ , therefore, = −sin φ = − 1 − x 2 and = −sin φ dφ dφ dx
1 d dy −1 2 dy ∴ sin φ + n(n + y)1 = 2x 1 − x sin φ dφ dφ 1 − x 2 dx
2 2 d y 2 + 1( − x ) − 1 − x + n(n + y)1 dx 2 d 2 y dy = 1( − x 2 ) − 2x + n(n + y)1 . dx 2 dx
E16) a) y = Pn )x( satisfies the given equation and the boundary conditions are
)0(y = 0 ⇒ Pn )0( = 0 and )1(y =1⇒ Pn )1( =1, (which is true).
Now Pn )0( = 0 , for n odd (see Example 1) K ∴ y = Pn )x( is a solution for n = ,5,3,1 . Q b) y = Pn )x( satisfies the given equation and the condition )1(y =1 [ Pn )1( = ]1
y′ )0( = 0 ⇒ Pn′ )0( = 0 , but Pn′ )0( = n Pn−1 )0( [using (39)] K Pn′ )0( = 0 ⇒ Pn−1 )0( ⇒ n = ,4,2,0 . K ∴ y = Pn )x( is a solution for n = ,4,2,0 .
E17) Multiplying both sides of Eqn.(61) with Pm )t( and integrating from −1 to 1 w.r.t. t , we get 1 1 ∞ 1 Pm dt)t( = ∑ Qn 2()x( n + )1 Pn )t( Pm dt)t( ∫ x − t ∫ −1 n=0 −1 1 P )t( 2 ∴ m dt = Q )x( 2( n + )1 ∫ x − t n 2n + 1 −1 1 1 P )t( or, Q )x( = n dt . n 2 ∫ x − t −1
E18) a) Comparing the given equation with Eqn.(1), we see that given equations is a Legendre’s equation with n = 0 , so its general solution is
y = C1P0 )x( + C 2Q 0 )x( . b) Comparing the given equation with Eqn.(1), we find that n(n + )1 =12 , therefore, n = ,3 − 4 . Since n is non-negative, the general solution of the
given equation is y = C1P3 )x( + C 2Q3 )x( .
c) y = C1P1 )x( + C 2Q1 )x( .
d) y = C1P5 )x( + C 2Q5 )x( .
P0 Q 0 1 E19) W P( 0 ,Q 0 ) = = P0Q′0 − Q 0 P0′ = . P′ Q′ 1 − x 2 0 0 117
Ordinary Differential E20) W P[ ),x( Q ])x( = P Q′ − Q P′ Equations n n n n n n Substituting in the above equation for Q n and Q′n from Eqn.(53), we get dx 1 W P[ ),x( Q )]x( = P P′ A + B + P[ )]x( 2 B n n n n n n ∫ 2 2 n n 2 2 1( − x P[) n )]x( 1( − x P[) n )]x( dx − P P′ A + B n n n n ∫ 2 2 1( − x P[) n )]x( B = n , n = ,2,1,0 L. 1 − x 2 ∞ m E21) Consider y = ∑ c m x , c0 ≠ 0 . Substitute for y,y ′ and y′′ in Eqn.(63), m=0 simplify and equate the coefficients of equal power of x to get two linearly
independent solutions y1 )x( and y 2 )x( [ref. E9) Unit 2].
∞ n 2 H t)x( H )x( H )x( tx2 −t n 2 2 3 3 L E22) e = ∑ = H 0 )x( + H1 t)x( + t + t + n=0 !n !2 !3 2 2 2 3 2 tx2( − t ) tx2( − t ) Now, e tx2 −t =1 + tx2( − t 2 ) + + + L !2 !3 3 2 2 4x − 6x 3 L =1 + 2( t)x + 2( x − t)1 + t + 3 Comparing the two series, we have 2 3 H 0 )x( = ,1 H1 )x( = 2 ,x H 2 )x( = 4x − ,2 H 3 )x( = 8x −12 x .
∞ 2 H )x( E23) a) Let w = e 2( xt −t ) = ∑ n t n n=0 !n ∂w ∂w then, x( − )t − t = 0 ∂x ∂t ∞ H′ )x( ∞ n H t)x( n ∞ n H′ )x( or, ∑x n t n − ∑n = ∑ n−1 t n n=0 !n n= 0 !n n= 0 !n Comparing coefficients of t n on both the sides, weget
x H′n )x( − nH n )x( = n H n−1 ).x(
b) From relation (75), we get H n )x( = 2x H n−1 )x( − n(2[ − )1 H n−2 ])x( [using relation (74)].
c) H′n′ )x( = 2n H′n−1 )x( [ using Eqn.(74)]
= 2 2[n x H n−1 )x( − H n ])x( [using Eqn.(75)]
= 2x H′n )x( − 2n H n )x( [ using Eqn.(74)].
E24) a) We have from Eqn.(68) n (− )1 k 2 2(!n )x 2n−2k H 2n )x( = ∑ k=0 2(!k n − 2 )!k The last term in the above summation is a constant and therefore, we have (− )1 n 2 !n H )0( = . 2n n! b) From Eqn.(68), we have n (− )1 k 2n + 2(!1 )x 2n+1−2k H 2n+1 )x( = ∑ k=0 2(!k n +1 − 2 )!k The last term in the above summation is a function of x , hence we get H )0( = 0 . 118 2n+1
c) Differentiating both sides of Eqn.(68), we obtain Legendre, Hermite and Laguerre Polynomials n−1 2 (− )1 k 2(!n )x n−2k−1 2. H′n )x( = ∑ k=0 n(!k − 2k − )!1 from which it follows that n−1 (− )1 k 2( n)!. .(2 2 )x 2n−2k−1 H′2n )x( = ∑ k=0 2(!k n − 2k − )!1 Putting x = 0 , we get
H′2n )0( = 0 n (− )1 k 2( n +1)!. .(2 2 )x 2n−2k d) H′2n+1 )x( = ∑ k=0 2(!k n − 2 )!k (− )1 n 2( n +1)!. 2 (− )1 n 2( n + )!2 H′ )0( = = . 2n+1 !n n( + )!1
E25) From Eqn.(66), we have ∞ k 2( xt −t2 ) t e = ∑ H k )x( !k k=0 ∞ k 2xt t2 t e = e ∑ H k )x( k=0 !k ∞ 2( )x n t n ∞ t 2m ∞ t k ∴∑= ∑ ∑ H k )x( n=0 !n m= 0 m! k= 0 !k ∞ n[ ]2/ H t)x( n [by interchang ing m and k = ∑ ∑ n−2k and setting k = n − 2 ].k m=0 k =0 n(!k − 2 !)k
Comparing the coefficients of t n on both the sides, we have n[ ]2/ !n H )x( x n = n−2k ∑ n k=0 2 n(!k − 2 !)k
− 3 7 3 1 E26) Using E22) required result is, H + H − H + H . 2 0 4 1 4 2 8 3
d n E27) We have, L )x( = e x x( n e −x ) , then n dx n d L )x( = ,1 L )x( = e x x( e −x ) =1 − ,x 0 1 dx d 2 L )x( = e x x( 2e−x ) = 2 − 4x + x 2 2 dx 2
3 x d 3 −x 2 3 L3 )x( = e x( e ) = 6 −18 x + 9x − x dx 3
− xt E28) Let w )t,x( = 1( − )t −1 exp . It satisfy the identity 1 − t ∂w 1( − )t 2 + x( −1 + )t w = 0 ∂t Using Eqn.(86), the above equation reduces to ∞ n ∑ [( n + )1 L n+1 )x( + x( −1 − 2 )n L n )x( + n L n−1 t])x( = 0 n=1 Equating the coefficients of t n to zero, relation (89) is proved. 119
Ordinary Differential ∂w Equations Similarly, using Eqn.(86) into the identity 1( − )t + tw = 0 , relation (90) is ∂x obtained.
Differentiating relation (89), using relation (90) in it and then eliminating
L′n+1 )x( and L′n−1 )x( from it relation (91) is obtained.
Differentiating relation (91) and using relation (90), relation (92) is obtained.
E29) From Laguerre’s differential equation we have for any two Laguerre
polynomials L m )x( and L n )x( x L′′ + 1( − )x L′ + mL = 0 m m m x L n′′ + 1( − )x L′n + n L n = 0
Multiplying these equations by L n and L m respectively and subtracting, we obtain
L[x n L′m′ − L m L′n′ ] + 1( − L[)x n L′m − L m L′n ]= n( − m) L m L n d 1( − )x n( − m) or, L[ L′ − L L′ ] + L[ L′ − L L′ ] = L L dx n m m n x n m m n x m n d or, x{ e −x L[ L′ − L L′ }] = n( − m e) −x L L , where we have multiplied dx n m m n m n
∫ 1( − /)x x dx ln x−x −x by I.F. e = e = x e Integrating from 0 to ∞ ∞ ∞ n( − m) e −x L L)x( dx)x( = x e −x L[ L′ − L L′ ] = 0 ∫ m n {}n m m n 0 0 ∞ −x If m ≠ ,n ∫ e L m )x( L n dx)x( = 0 . 0 −x2 2/ −x2 2/ −x2 2/ E30) a) Putting ψ n )x( = e H n )x( and ψ′n )x( = e H′n )x( − x e H n )x( , in the given equation, it reduces to
2nH n−1 )x( = H′n )x( which is relation (74) b) The given equation reduces to
2xH n )x( − 2n H n−1 )x( = H n+1 )x( replacing n by n( − )1 we get relation (76).
c) Substituting for ψ n )x( and its derivatives in the given equation, relation (75) is obtained.
∞ ∞ 2 ⇒ 2 −x2 2 E31) ∫[]ψ n )x( dx =1 ∫ c n e [ H n )x( ] dx =1 −∞ −∞
2 n 1 ⇒ c n 2 !n π =1 or c, n = 2/1 [using result (78)] []2n !n π 1 − n 2 −x2 K ∴ψ n )x( = []π2 !n e H n ),x( n = ,2,1,0 . —x—
120