1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

5.9 Spherical Coordinates; Legendre Polynomials 337
Solutions of the differential equation are usually found by the power series
method. Assume thaty(x)=a 0 +a 1 x+ ··· +akxk+ ···.Thetermsofthe
differential equations are then

y′′= 2 a 2 + 3 · 2 a 3 x+ 4 · 3 a 4 x^2 +··· +(k+ 2 )(k+ 1 )ak+ 2 xk +···
−x^2 y′′=− 2 a 2 x^2 −··· −k(k− 1 )akxk +···
− 2 xy′=− 2 a 1 x − 2 a 2 x^2 −··· − 2 kakxk −···
μ^2 y=μ^2 a 0 +μ^2 a 1 x +μ^2 a 2 x^2 +··· +μ^2 akxk +···.
When this tableau is added vertically, the left-hand side is zero, according to
the differential equation. The right-hand side adds up to a power series, each
of whose coefficients must be zero. We therefore obtain the following relations:

2 a 2 +μ^2 a 0 = 0 ,

6 a 3 +

(

μ^2 − 2

)

a 1 = 0 ,

(k+ 2 )(k+ 1 )ak+ 2 +

[

μ^2 −k(k+ 1 )

]

ak= 0.

The last equation actually includes the first two, apparently special, cases. We
may write the general relation as

ak+ 2 =k(k+^1 )−μ

2

(k+ 2 )(k+ 1 )

ak,

valid fork= 0 , 1 , 2 ,....
Suppose for the moment thatμ^2 is given. A short calculation gives the first
few coefficients:

a 2 =−μ

2

2

a 0 , a 3 =^2 −μ

2

6

a 1 ,

a 4 =

6 −μ^2

12 a^2 , a^5 =

12 −μ^2

20 a^3 ,

=^6 −μ

2

12

·−μ

2

2

a 0 , =^12 −μ

2

20

·^2 −μ

2

6

a 1.

It is clear that all thea’s with even index will be multiples ofa 0 and those with
odd index will be multiples ofa 1 .Thusy(x)equalsa 0 times an even function
plusa 1 times an odd function, with botha 0 anda 1 arbitrary.
It is not difficult to prove that odd and even series produced by this process
diverge at bothx=±1, for generalμ^2 .However,whenμ^2 has one of the spe-
cial values

μ^2 =μ^2 n=n(n+ 1 ), n= 0 , 1 , 2 ,...,

one of the two series turns out to have all zero coefficients afteran.Forin-
stance, ifμ^2 = 3 ·4, thena 5 =0, and all subsequent coefficients with odd