Mathematical Tools for Physics - Department of Physics - University

4—Differential Equations 88

Assume a Frobenius solutions aboutx= 0

y=

∑∞

0

akxk+s

and substitute into (4.55). Could you use an ordinary Taylor series instead? Yes, the pointx= 0is

not a singular point at all, but it is just as easy (and more systematic and less prone to error) to use
the same method in all cases.

(1−x^2 )

∑∞

0

ak(k+s)(k+s−1)xk+s−^2 − 2 x

∑∞

0

ak(k+s)xk+s−^1 +C

∑∞

0

akxk+s= 0

∑∞

0

ak(k+s)(k+s−1)xk+s−^2 +

∑∞

0

ak

[

−2(k+s)−(k+s)(k+s−1)

]

xk+s+C

∑∞

0

akxk+s= 0

∑∞

n=− 2

an+2(n+s+ 2)(n+s+ 1)xn+s−

∑∞

n=0

an

[

(n+s)^2 + (n+s)

]

xn+s+C

∑∞

n=0

anxn+s= 0

In the last equation you see the usual substitutionk=n+ 2for the first sum andk=nfor the rest.

That makes the exponents match across the equation. In the process, I simplified some of the algebraic
expressions.

The indicial equation comes from then=− 2 term, which appears only once.

a 0 s(s−1) = 0, so s= 0, 1

Now set the coefficient ofxn+sto zero, and solve for an+2in terms ofan. Also note thatsis a

non-negative integer, which says that the solution is non-singular atx= 0, consistent with the fact

that zero is a regular point of the differential equation.

an+2=an

(n+s)(n+s+ 1)−C

(n+s+ 2)(n+s+ 1)

(4.56)

a 2 =a 0

s(s+ 1)−C

(s+ 2)(s+ 1)

, then a 4 =a 2

(s+ 2)(s+ 3)−C

(s+ 4)(s+ 3)

, etc. (4.57)

This looks messier than it is. Notice that the only combination of indices that shows up isn+s. The

indexsis 0 or 1, andnis an even number, so the combinationn+scovers the non-negative integers:

0 , 1 , 2 ,...

The two solutions to the Legendre differential equation come from the two cases,s= 0, 1.

s= 0 : a 0

[

1 +

(

−C

2

)

x^2 +

(

−C

2

)(

2. 3 −C

4. 3

)

x^4 +

(

−C

2

)(

2. 3 −C

4. 3

)(

4. 5 −C

6. 5

)

x^6 ···

]

s= 1 : a′ 0

[

x+

(

1. 2 −C

3. 2

)

x^3 +

(

1. 2 −C

3. 2

)(

3. 4 −C

5. 4

)

x^5 +···

]

(4.58)