Transcript Math 260
Ch 5.5: Series Solutions Near a Regular Singular Point, Part I
We now consider solving the general second order linear equation in the neighborhood of a regular singular point
x
0 . For convenience, will will take
x
0 = 0.
The point
x
0 = 0 is a regular singular point of 2
P
(
x
)
d dx
2
y
Q
(
x
)
dy dx
R
(
x
)
y
0 iff
x Q
(
P
(
x
)
x
)
xp
(
x
) and
x
2
R
(
x
)
P
(
x
)
x
2
q
(
x
) are analytic at
x
0 iff
xp
(
x
)
n
0
p n x n
and
x
2
q
(
x
)
n
0
q n x n
, convergent on
x
Transforming Differential Equation
Our differential equation has the form
P
(
x
)
y
Q
(
x
)
y
R
(
x
)
y
0 Dividing by
P
(
x
) and multiplying by
x
2 , we obtain
x
2
y
x
xp
(
x
)
y
x
2
q
(
x
)
y
0 Substituting in the power series representations of
p
and
q
,
xp
(
x
)
n
0
p n x n
,
x
2
q
(
x
)
n
0
q n x n
, we obtain
x
2
y
x
p
0
p
1
x
p
2
x
2
y q
0
q
1
x
q
2
x
2
y
0
Comparison with Euler Equations
Our differential equation now has the form
x
2
y
x
p
0
p
1
x
p
2
x
2
y q
0
q
1
x
q
2
x
2
y
0 Note that if
p
1
p
2
q
1
q
2 0 then our differential equation reduces to the Euler Equation
x
2
y
p
0
x y
q
0
y
0 In any case, our equation is similar to an Euler Equation but with power series coefficients. Thus our solution method: assume solutions have the form
y
(
x
)
x r
a
0
a
1
x
a
2
x
2
n
0
a n x r
n
, for
a
0 0 ,
x
0
Example 1: Regular Singular Point
(1 of 13) Consider the differential equation 2
x
2
y
x y
1
x
y
0 This equation can be rewritten as
x
2
y
x
2
y
1 2
x y
0 Since the coefficients are polynomials, it follows that
x
= 0 is a regular singular point, since both limits below are finite: lim
x
0
x x
2
x
2 1 2 and
x
lim 0
x
2 1 2
x
2
x
1 2
2
x
2
y
x y
1
x
y
0
Example 1: Euler Equation
(2 of 13) Now
xp
(
x
) = -1/2 and
x
2
q
(
x
) = (1 +
x
)/2, and thus for
xp
(
x
)
n
0
p n x n
,
x
2
q
(
x
)
n
0
q n x n
, it follows that
p
0 1 / 2 ,
q
0 1 / 2 ,
q
1 1 / 2 ,
p
1
p
2
q
2
q
3 0 Thus the corresponding Euler Equation is
x
2
y
p
0
x y
q
0
y
0 2
x
2
y
x y
y
0 As in Section 5.5, we obtain
x r
2
r
(
r
1 )
r
1 0 2
r
1
r
1 0
r
1 ,
r
1 / 2 We will refer to this result later.
2
x
2
y
x y
1
x
y
0
Example 1: Differential Equation
(3 of 13) For our differential equation, we assume a solution of the form
y
(
x
)
n
0
a n x r
n
,
y
(
x
)
n
0
a n
r y
(
x
)
n
0
a n
r
n
r
n
1
x r
n
2
n
x r
n
1 , By substitution, our differential equation becomes
n
0 2
a n
r
n
r
n
1
x r
n
n
0
a n
r
n
x r
n
n
0
a n x r
n
n
0
a n x r
n
1 0 or
n
0 2
a n
r
n
r
n
1
x r
n
n
0
a n
r
n
x r
n
n
0
a n x r
n
n
1
a n
1
x r
n
0
Example 1: Combining Series
(4 of 13) Our equation
n
0 2
a n
r
n
r
n
1
x r
n
n
0
a n
r
n
x r
n
n
0
a n x r
n
n
1
a n
1
x r
n
0 can next be written as
a
0 2
r
(
r
1 )
r
1
x r
n
1
a n
2
r
n
r
n
1 (
r
n
) 1
a n
1
x r
n
0 It follows that
a
0 2
r
(
r
1 )
r
1 0 and
a n
2
r
n
r
n
1 (
r
n
) 1
a n
1 0 ,
n
1 , 2 ,
Example 1: Indicial Equation
(5 of 13) From the previous slide, we have
a
0 2
r
(
r
1 )
r
1
x r
n
1
a n
2
r
n
r
n
1 (
r
n
) 1
a n
1
x r
n
The equation
a
0 2
r
(
r
1 )
r
1 0
a
0 0 2
r
2 3
r
1 ( 2
r
1 )(
r
1 ) 0 0 is called the
indicial equation
, and was obtained earlier when we examined the corresponding Euler Equation. The roots
r
1 = 1,
r
2 = ½, of the indicial equation are called the
exponents of the singularity,
for regular singular point
x
= 0. The exponents of the singularity determine the qualitative behavior of solution in neighborhood of regular singular point.
Example 1: Recursion Relation
(6 of 13) Recall that
a
0 2
r
(
r
1 )
r
1
x r
n
1
a n
2
r
n
r
n
1 (
r
n
) 1
a n
1
x r
n
0 We now work with the coefficient on
x r+n
:
a n
2
r
n
r
n
1 (
r
n
) 1
a n
1 0 It follows that
a n
2
r
n
r
n a n
1 1 (
r
n
) 1 2
r
n
2
a n
1 3 (
r
n
) 1 2
r
n
a n
1 1
r
n
1 ,
n
1
Example 1: First Root
(7 of 13) We have
a n
2
r
n
a n
1 1
r
n
1 , for
n
1 ,
r
1 1 and
r
2 1 / 2 Starting with
r
1 = 1, this recursion becomes
a n
2 1
n
a
1
n
1 1
n
1 2
n a n
1 1
n
,
n
1 Thus
a
1
a
0 3 1
a
2 5
a
1 2
a
0
a
3 7
a
2 3 3 5 7
a
0 1 2 3 ,
a n
3 5 7 ( 1 )
n a
0 2
n
1
n
!
,
n
1 etc
Example 1: First Solution
(8 of 13) Thus we have an expression for the
n
-th term:
a n
3 5 7 ( 1 )
n
2
a n
0 1
n
!
,
n
1 Hence for
x
> 0, one solution to our differential equation is
y
1 (
x
)
n
0
a n x n
r a
0
x
n
1 3 5 ( 1 )
n
7
a
0 2
n x n
1 1
n
!
a
0
x
1
n
1 3 5 7 ( 1 )
n x
2
n n
1
n
!
Example 1: Second Root
(10 of 13) Recall that
a n
2
r
n
a n
1 1
r
n
1 , for
n
1 ,
r
1 1 and
r
2 1 / 2 When
r
2 = 1/2, this recursion becomes
a n
2 1 / 2
n
a n
1 1 1 / 2
n
1 2
n
n a n
1 1 / 2
n
2
a n n
1 1 ,
n
1 Thus
a
1
a
0 1 1
a
2 2
a
1 3
a
0
a
3
a n
3
a
2 5 1 2 3
a
0 1 3 5 , 1 3 ( 1 ) 5
n a
0 2
n
1
n
!
, etc
n
1
Example 1: Second Solution
(11 of 13) Thus we have an expression for the
n
-th term:
a n
1 3 5 ( 1 )
n
2
a n
0 1
n
!
,
n
1 Hence for
x
> 0, a second solution to our equation is
y
2 (
x
)
n
0
a n x n
r
a
0
x
1 / 2
n
1 1 ( 3 1 ) 5
n
a
0
x
2
n n
1 / 2 1
n
!
a
0
x
1 / 2 1
n
1 1 3 ( 1 ) 5
n
2
n x n
1
n
!
Example 1: General Solution
(13 of 13) The two solutions to our differential equation are
y
1 (
x
)
x
1
n
1 3 5 7 ( 1 )
n
2
x n n
1
n
!
y
2 (
x
)
x
1 / 2 1
n
1 1 3 ( 1 ) 5
n
2
n x n
1
n
!
Since the leading terms of
y
1 it follows that
y
1 and
y
2 and
y
2 are
x
and
x
1/2 , respectively, are linearly independent, and hence form a fundamental set of solutions for differential equation.
Therefore the general solution of the differential equation is
y
(
x
)
c
1
y
1 (
x
)
c
2
y
2 (
x
),
x
0 , where
y
1 and
y
2 are as given above.
Shifted Expansions
For the analysis given in this section, we focused on
x
= 0 as the regular singular point. In the more general case of a singular point at
x
=
x
0 , our series solution will have the form
y
(
x
)
x
x
0
r n
0
a n
x
x
0
n