jagomart
digital resources
picture1_Solved Problems Pdf 175470 | Legendre Equation


 126x       Filetype PDF       File size 0.09 MB       Source: jahandideh.iut.ac.ir


File: Solved Problems Pdf 175470 | Legendre Equation
sec 5 2 legendre s equation legendre polynomials pn x 175 5 2 legendre s equation legendre polynomials p x n 1 legendres differential equation 1 1 x2 ys 2xyr ...

icon picture PDF Filetype PDF | Posted on 28 Jan 2023 | 2 years ago
Partial capture of text on file.
                         SEC. 5.2  Legendre’s Equation. Legendre Polynomials Pn(x)                                                          175
                         5.2 Legendre’s Equation. 
                                    Legendre Polynomials P (x)
                                                                                             n
                                                                                  1
                                               Legendres differential equation
                                               (1)                         (1  x2)ys  2xyr  n(n  1)y  0                       (n constant)
                                               is one of the most important ODEs in physics. It arises in numerous problems, particularly
                                               in boundary value problems for spheres (take a quick look at Example 1 in Sec. 12.10).
                                                  The equation involves a parameter  n, whose value depends on the physical or
                                               engineering problem. So (1) is actually a whole family of ODEs. For n  1 we solved it
                                               in Example 3 of Sec. 5.1 (look back at it). Any solution of (1) is called a Legendre function.
                                               The study of these and other “higher” functions not occurring in calculus is called the
                                               theory of special functions. Further special functions will occur in the next sections.
                                                  Dividing (1) by 1  x2, we obtain the standard form needed in Theorem 1 of Sec. 5.1
                                                                                               2                        2
                                               and we see that the coefficients 2x>(1  x ) and n(n  1)>(1  x ) of the new equation
                                               are analytic at x  0, so that we may apply the power series method. Substituting
                                                                                              
                                               (2)                                     y        a  xm
                                                                                             a m
                                                                                            m0
                                               and its derivatives into (1), and denoting the constant n(n  1) simply by k, we obtain
                                                                                                                      
                                                               2                      m2                    m1                m
                                                       (1  x )       m(m  1)a  x         2x        ma  x       k       a  x    0.
                                                                  a               m               a m                  a m
                                                                 m2                             m1                  m0
                                               By writing the first expression as two separate series we have the equation
                                                                                                                          
                                                      m(m  1)a  xm2             m(m  1)a  xm             2ma  xm           ka  xm  0.
                                                 a                m            a               m         a        m         a m
                                                m2                           m2                       m1                m0
                                               It may help you to write out the first few terms of each series explicitly, as in Example 3
                                               of Sec. 5.1; or you may continue as follows. To obtain the same general power xs in all
                                               four series, set m  2  s (thus m  s  2) in the first series and simply write s instead
                                               of m in the other three series. This gives
                                                                                                                        
                                                         (s  2)(s  1)a      xs        s(s  1)a xs         2sa xs         ka xs  0.
                                                     a                    s2        a             s       a s            a s 
                                                     s0                             s2                   s1            s0
                                                 1
                                                  ADRIEN-MARIE LEGENDRE (1752–1833), French mathematician, who became a professor in Paris in
                                               1775 and made important contributions to special functions, elliptic integrals, number theory, and the calculus
                                               of variations. His book Éléments de géométrie (1794) became very famous and had 12 editions in less than
                                               30 years.
                                                                                                   [       ]    [         ]
                                                 Formulas on Legendre functions may be found in Refs.  GenRef1 and  GenRef10 .
                                176                              CHAP. 5          Series Solutions of ODEs. Special Functions
                                                                 (Note that in the first series the summation begins with s  0.) Since this equation with
                                                                 the right side 0 must be an identity in x if (2) is to be a solution of (1), the sum of the
                                                                 coefficients of each power of x on the left must be zero. Now x0 occurs in the first and
                                                                 fourth series only, and gives [remember that k  n(n  1)]
                                                                 (3a)                                               2 # 1a  n(n  1)a  0.
                                                                                                                             2                       0
                                                                 x1 occurs in the first, third, and fourth series and gives
                                                                 (3b)                                        3 # 2a  [2  n(n  1)]a  0.
                                                                                                                      3                                     1
                                                                                                   2     3 Á
                                                                 The higher powers x , x ,                         occur in all four series and give
                                                                 (3c)                   (s  2)(s  1)a                  [s(s  1)  2s  n(n  1)]a  0.
                                                                                                                  s2                                                           s
                                                                 The expression in the brackets [ Á ] can be written (n  s)(n  s  1), as you may
                                                                 readily verify. Solving (3a) for a and (3b) for a as well as (3c) for a                                                 , we obtain the
                                                                                                                       2                        3                                   s2
                                                                 general formula
                                                                                                                                (n  s)(n  s  1)                                                        Á
                                                                 (4)                                           a                                               a                        (s  0, 1,           ).
                                                                                                                 s2                (s  2)(s  1)                 s
                                                                 This is called a recurrence relation or recursion formula. (Its derivation you may verify
                                                                 with your CAS.) It gives each coefficient in terms of the second one preceding it, except
                                                                 for a and a , which are left as arbitrary constants. We find successively
                                                                         0           1
                                                                                      n(n  1)                                                          (n  1)(n  2)
                                                                          a                         a                                    a                                    a
                                                                          2                2!          0                                    3                      3!               1
                                                                                      (n  2)(n  3)                                                    (n  3)(n  4)
                                                                          a                                   a                          a                                    a
                                                                          4                   4 # 3               2                         5                   5 # 4               3
                                                                                  (n  2)n(n  1)(n  3)                                            (n  3)(n  1)(n  2)(n  4)
                                                                                                                         a                                                                           a
                                                                                                    4!                      0                                              5!                            1
                                                                 and so on. By inserting these expressions for the coefficients into (2) we obtain
                                                                 (5)                                                 y(x)  a y (x)  a y (x)
                                                                                                                                   0 1              1 2
                                                                 where
                                                                                                          n(n  1) 2               (n  2)n(n  1)(n  3) 4                               Á
                                                                 (6)                 y (x)  1                           x                                                x    
                                                                                      1                         2!                                   4!
                                                                                                 (n  1)(n  2) 3                    (n  3)(n  1)(n  2)(n  4) 5                                   Á
                                                                 (7)         y (x)  x                                    x                                                           x     .
                                                                              2                             3!                                              5!
                         SEC. 5.2  Legendre’s Equation. Legendre Polynomials P (x)                                                          177
                                                                                  n
                                               These series converge for ƒxƒ  1 (see Prob. 4; or they may terminate, see below). Since
                                               (6) contains even powers of x only, while (7) contains odd powers of x only, the ratio
                                               y >y   is not a constant, so that y      and y are not proportional and are thus linearly
                                                1   2                                1        2
                                               independent solutions. Hence (5) is a general solution of (1) on the interval 1  x  1.
                                                  Note that x 1 are the points at which 1  x2  0, so that the coefficients of the
                                               standardized ODE are no longer analytic. So it should not surprise you that we do not get
                                               a longer convergence interval of (6) and (7), unless these series terminate after finitely
                                               many powers. In that case, the series become polynomials.
                                               Polynomial Solutions. Legendre Polynomials P (x)
                                                                                                                                n
                                               The reduction of power series to polynomials is a great advantage because then we have
                                               solutions for all x, without convergence restrictions. For special functions arising as
                                               solutions of ODEs this happens quite frequently, leading to various important families of
                                               polynomials; see Refs. [GenRef1], [GenRef10] in App. 1. For Legendres equation this
                                               happens when the parameter n is a nonnegative integer because then the right side of (4)
                                               is zero for s  n, so that a       0, a       0, a       0,Á. Hence if n is even, y (x)
                                                                            n2          n4         n6                                    1
                                               reduces to a polynomial of degree n. If n is odd, the same is true for y (x). These
                                                                                                                                   2
                                               polynomials, multiplied by some constants, are called Legendre polynomials and are
                                               denoted by P (x). The standard choice of such constants is done as follows. We choose
                                                             n                              n
                                               the coefficient a   of the highest power x     as
                                                                 n
                                                                                   (2n)!     1 # 3 # 5 Á (2n  1)
                                               (8)                         a                                           (n a positive integer)
                                                                            n      n    2              n!
                                                                                 2 (n!)
                                               (and a  1 if n  0). Then we calculate the other coefficients from (4), solved for a in
                                                      n                                                                                     s
                                               terms of a     , that is,
                                                          s2
                                                                                         (s  2)(s  1)
                                               (9)                           a                            a                     (s  n  2).
                                                                               s      (n  s)(n  s  1) s2
                                               The choice (8) makes p (1)  1 for every n (see Fig. 107); this motivates (8). From (9)
                                                                         n
                                               with s  n  2 and (8) we obtain
                                                                                 n(n  1)             n(n  1) #    (2n)!
                                                                    a                     a   
                                                                     n2        2(2n  1) n          2(2n  1)      n    2
                                                                                                                   2 (n!)
                                               Using (2n)!  2n(2n  1)(2n  2)! in the numerator and n!  n(n  1)! and
                                               n!  n(n  1)(n  2)! in the denominator, we obtain
                                                                                     n(n  1)2n(2n  1)(2n  2)!
                                                                   a                                                      .
                                                                    n2                    n
                                                                               2(2n  1)2 n(n  1)! n(n  1)(n  2)!
                                               n(n  1)2n(2n  1) cancels, so that we get
                                                                                                (2n  2)!
                                                                              a                                .
                                                                               n2         n
                                                                                          2 (n  1)! (n  2)!
                     178                    CHAP. 5    Series Solutions of ODEs. Special Functions
                                            Similarly,
                                                                                       (n  2)(n  3)
                                                                            a                          a
                                                                            n4           4(2n  3)      n2
                                                                                           (2n  4)!
                                                                                  n
                                                                                    2 2! (n  2)! (n  4)!
                                            and so on, and in general, when n  2m  0,
                                                                                     m         (2n  2m)!
                                            (10)                    a       (1)                                   .
                                                                     n2m                n
                                                                                       2 m! (n  m)! (n  2m)!
                                            The resulting solution of Legendres differential equation (1) is called the Legendre
                                            polynomial of degree n and is denoted by P (x).
                                                                                            n
                                               From (10) we obtain
                                                                      M                   (2n  2m)!
                                                                                m                               n2m
                                                             P (x)        (1)                                x
                                                             n        a             n
                                                                     m0          2 m! (n  m)! (n  2m)!
                                            (11)                      (2n)!                 (2n  2)!
                                                                               n                               n2        Á
                                                                   n       2 x   n                         x      
                                                                     2 (n!)         2 1! (n  1)! (n  2)!
                                            where M  n>2 or (n  1)>2, whichever is an integer. The first few of these functions
                                            are (Fig. 107)
                                                        P(x)  1,                               P (x)  x
                                                         0                                     1
                                            (11)       P(x)  1 (3x2  1),                   P(x)  1 (5x3  3x)
                                                         2       2                             3       2
                                                        P(x)  1 (35x4  30x2  3),           P(x)  1 (63x5  70x3  15x)
                                                         4       8                             5       8
                                            and so on. You may now program (11) on your CAS and calculate P (x) as needed.
                                                                                                                       n
                                                                                       P(x)
                                                                                       n               P
                                                                                          1              0
                                                                                                       P
                                                                                                         1
                                                                                                  P
                                                                                                   4
                                                                     –1                       P                 1   x
                                                                                                3
                                                                                              P
                                                                                                2
                                                                                         –1
                                                                            Fig. 107. Legendre polynomials
The words contained in this file might help you see if this file matches what you are looking for:

...Sec legendre s equation polynomials pn x p n legendres differential ys xyr y constant is one of the most important odes in physics it arises numerous problems particularly boundary value for spheres take a quick look at example involves parameter whose depends on physical or engineering problem so actually whole family we solved back any solution called function study these and other higher functions not occurring calculus theory special further will occur next sections dividing by obtain standard form needed theorem see that coefficients new are analytic may apply power series method substituting xm m its derivatives into denoting simply k ma writing first expression as two separate have ka help you to write out few terms each explicitly continue follows same general xs all four set thus instead three this gives sa adrien marie french mathematician who became professor paris made contributions elliptic integrals number variations his book elements de geometrie very famous had editions...

no reviews yet
Please Login to review.