Mathematical Physics Vol 1

5.3 Legendre: equation, function, polynomial

229

or

∞ ∑ i = 2

∞ ∑ i = 2

∞ ∑ i = 1

∞ ∑ i = 0

i − 2 −

i − 2

i + k ( k + 1 )

i = 0 .

i ( i − 1 ) a i x

i ( i − 1 ) a i x

ia i x

a i x

(5.25)

This relation can be expanded as follows

2 + ··· +( s + 2 )( s + 1 ) a 2 −···− s ( s − 1 ) a s x s −··· 2 −···− 2 · s · a 2 + ··· + k ( k + 1 ) · a s · x s −··· s · x

s + ···

2 · 1 · a 2 + 3 · 2 · a 3 · x + 4 · 3 · a 4 · x

s + 2 · x

− 2 · 1 · a 2 · x

− 2 · 1 · a 1 · x − 2 · 2 · a 2 · x

k ( k + 1 ) a 0 + k ( k + 1 ) a 1 · x + k ( k + 1 ) a 2 · x s + ··· = 0 . From the condition that the above relation must be an identity, according to Theorem 12, we obtain 2 a 2 + k ( k + 1 ) a 0 = 0 , coefficient next to x 0 6 a 3 +[ − 2 + k ( k + 1 )] a 1 = 0 , coefficient next to x 1 ( s + 2 )( s + 1 ) a s + 2 +[ − s ( s − 1 ) − 2 s + k ( k + 1 )] a s = 0 , coefficient next to x s . From the last relation we obtain the so called recurrence formulas for determining the coefficients a i a s + 2 = − ( k − s )( k + s + 1 ) ( s + 2 )( s + 1 ) a s , s = 0 , 1 , 2 ,... (5.26) It is visible from these relations that all coefficients are determined except a 0 and a 1 , which remain arbitrary. Thus, all other coefficients ca be expressed in terms of these two. For example a 2 = − k ( k + 1 ) 2 · 1 a 0 = − k ( k + 1 ) 2! a 0 a 3 = − ( k − 1 )( k + 2 ) 3 · 2 a 1 = − ( k − 1 )( k + 2 ) 3! a 1 a 4 = − ( k − 2 )( k + 3 ) 4 · 3 a 2 = ( k − 2 ) k ( k + 1 )( k + 3 ) 4! a 0 a 5 = − ( k − 3 )( k + 4 ) 5 · 4 a 3 = ( k − 3 )( k − 1 )( k + 2 )( k + 4 ) 5! a 1 etc. The above examples show that the even coefficients can be expressed in terms of a 0 , and the odd ones in terms of a 1 , and thus the initial solution can be expressed in the form y = a 0 y 1 ( x )+ a 1 y 2 ( x ) , (5.27) where y 1 ( x )= 1 − k ( k + 1 ) 2! x 2 + ( k − 2 ) k ( k + 1 )( k + 3 ) 4! x 4 + ···

( k − 1 )( k + 2 ) 3!

( k − 3 )( k − 1 )( k + 2 )( k + 4 ) 5!

x 3 +

x 5 + ···

y 2 ( x )= x −

This series converges for | x | < 1.

R Note that y 1 contains only even powers, and y 2 only odd ones, and thus their quotient is not constant. It follows that y 1 and y 2 are not proportional, that is, these two functions are linearly independent. Thus, the function y = a 0 y 1 + a 1 y 2 represents the general solution of the initial (Legendre) equation in the interval − 1 < x < 1.