Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

272

Multiplying the first equation by P m ( x ) , and the second equation by P n ( x ) , we obtain P m h 1 − x 2 P ′ n i ′ + n ( n + 1 ) P n P m = 0 , P n h 1 − x 2 P ′ m i ′ + m ( m + 1 ) P m P n = 0 . By subtracting the resulting equations we further obtain [ n ( n + 1 ) − m ( m + 1 )] P m P n + P m h 1 − x 2 P ′ n i ′ − P n h 1 − x 2 P ′ m i ′ = 0 . Let us note that P m h 1 − x 2 P ′ n i ′ − P n h 1 − x 2 P ′ m i ′ = = h P m 1 − x 2 P ′ n i ′ − P ′ m 1 − x 2 P ′ n − h P n 1 − x 2 P ′ m i ′ + P ′ n 1 − x 2 P ′ m = = h P n 1 − x 2 P ′ m − P m 1 − x 2 P ′ n i ′ − 1 − x 2 P ′ n P ′ m − P ′ m P ′ n = = h P n 1 − x 2 P ′ m − P m 1 − x 2 P ′ n i ′ . Thus, the previous relation can be rewritten as [ n ( n + 1 ) − m ( m + 1 )] P m P n + h 1 − x 2 P n P ′ m − P m P ′ n i ′ = 0 . Integrating this relation from − 1 to + 1yields [ n ( n + 1 ) − m ( m + 1 )] + 1 Z − 1 P m ( x ) P n ( x ) d x +

+ 1 Z

d d x h

n i d x = 0 , for n̸ = m .

1 − x 2 P

′ m − P m P ′

n P

+

− 1

Given that

+ 1

+ 1 Z

d h 1 − x 2 P

n i = 1 − x

n

2 P

n P m ′ − P m P ′

′ m − P m P ′

n P

= 0 ,

− 1

− 1

because, for x = − 1 and x =+ 1, 1 − x 2 = 0, and P bounded functions, it follows that P n ( ± 1 ) < ∞ , P m ( ± 1 ) < ∞ , P ′ ∞ , and, as [ n ( n + 1 ) − m ( m + 1 )]̸= 0, we finally obtain + 1 Z − 1 P m ( x ) P n ( x ) d x = 0 , za n̸ = m , which is the condition of orthogonality. For the norm we obtain n ( x ) , P ′

n ( x ) , P m ( x ) and P ′ n ( ± 1 ) < ∞ i P ′

m ( x ) are m ( ± 1 ) <

d n u

n ( x ) d x n

+ 1 Z

2

1 ( 2 n n ! )

2

∥ P n ( x ) ∥

d x ,

=

2

− 1