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
d 
∫ − 1 1 Px()   dx   (1− x ) dP x ()  + ll ( + 1 ) ( ) 0 (7.100)
P x dx =
2
l
dx 
l
m

Integrating the first term by parts, we obtain:


∫ − 1 1  (x − ) 1 dP dx () x dP x + ( ll 1+ )P x P x dx = 0 (7.101)
()
m

2
() () 
l
m
l

dx
Similarly, we can write the ODE for P (x), and multiply on the left
m
by P (x); this results in the equation:
l


∫ − 1 1 (x − ) 1 dP dx dx ()x + mm 1+ ) ( )P x P m ( )x dx = 0 (7.102)
()x dP
m
l


2
(
l

Now, subtracting Eq. (7.102) from Eq. (7.101), we obtain:
[ (mm + 1 ) ll− ( + 1 )] ∫ 1 P l ()x P m ()x dx = 0 (7.103)
−1
But because l ≠ m, this can only be satisfied if the integral is zero,
which is the result that we are after.
6. Finally, we compute the normalization of the basis functions; that
is, compute:
∫ 1 Px Px dx() () = N 2 l (7.104)
l
l
−1
From Eq. (7.90), we can write:
Px() (− 2 l − 1 ) xP () (+ l − 1 P ) x () = 0 (7.105)
x
l l−1 l−2
If we multiply this equation by (2l + 1)P (x) and subtract from it
l
Eq. (7.90), which we multiplied by (2l + 1)P (x), we obtain:
l–1
ll(2 + ) 1 Px() (2+ l 1− l )( − P ) 1 () x
2
x P ()
−
−
l l 1 l 2
(7.106)
x P () −
− l ( + 12 l 1− P ) () x l l(2 − P ) 1 2 x () = 0
)(
−
+
−
l 1 l 1 l 1
Now integrate over the interval [–1, 1] and using Eq. (7.103), we
obtain, for l = 2, 3, …:
© 2001 by CRC Press LLC