Introductory lecture notes on Partial Differential Equations c Anthony Peirce.
Not to be copied, used, or revised without explicit written permission from the copyright owner.
( Compiled 3 March 2014 )
In this lecture we will define a Regular Singular Point about which a Taylor series will not work. We will also introduce the concept of the radius of convergence of the series and how it relates to the coefficient of the highest derivative of the
ODE.
Key Concepts: Ordinary Points and Regular Singular Points, radius of convergence of power series.
3 Radius of Convergence and Nearest Singular Points
Example 1: (1 + x 2 ) y 00 + 2 xy 0 + 4 x 2 y = 0 .
(1) If we were given y (0) = 0 and y 0 (0) = 1 then we would want a power series expansion of the form y = n =0 c n x n about x
0
= 0 .
Roots of 1 + x 2 = 0 are x = ± i , so we expect the radius of convergence of the TS for 1
1+ x 2 to be 1 since
(3.1)
1
1 + x 2
= 1 − x 2 + x 4 − · · · lim a n +2 a n
= 1 ρ = 1
(2) If we were given y (1) = 1, y
√ case ρ = 2.
0 (1) = 0 then a power series expansion of the form
P c n
( x − 1) n
(3.2) is required. In this
Example 2: ( x − 1)(2 x − 1) y 00 + 2 xy 0 − 2 y = 0.
x = 0 is an ordinary point.
x = 1 and x = 1
2 are singular points. One solution of this equation is y ( x ) =
1 x − 1
= − (1 + x + x
2
+ · · · ) ρ = 1 .
This Taylor Series solution about the ordinary point x = 0 converges beyond the singular point x = 1
2
.
Example 3: ( x 2 − 2 x ) y 00 + 5( x − 1) y 0 + 3 y = 0 y (1) = 7 y
£ x = 1 is an ordinary point.
x = 0 is a singular point ( x
0
−
(1) = 3.
1) 2 − 1
¤ y 00 + 5( x − 1) y 0 + 3 y = 0.
(3.3)
1

2
Let t = x − 1 so that d dt
= d dx and the equation is transformed to n =2 y = n =0 n ( n − 1) c c n n t t n n
− n =2
( t
, ˙ = n
2
(
− n =1 n
1) ˙˙
− c n
1)
+ 5 nt c n t y n − 1 t
˙
, n − 2
+ 3 y
˙˙ y
+ 5
= 0
=
P
P
=2 nc n =1 c n n n t n
( n −
+ 3
1) t n =0 n − 2 c n t n m = n − 2 n = m + 2 n = 2 = m = 0
= 0 m =2
[ − c m +2
( m + 2)( m + 1) + { m ( m − 1) + 5 m + 3 } c m
] t m
− 2 c
2
+ 3 c
0
+ [ − c
3
3 .
2 + 5 c
1
+ 3 c
1
] t = 0 c
0
: t t
0
1
>
> c
2
= 3
2 c
0 c
3
= 8
6 c
1
= 4
3 c
1 t m > c m +2
= c m
( m +1)( m +3)
( m +1)( m +2) m ≥ 2 .
c
4
= y
0
( x ) =
5 c
2
P n =0
= 5
4
¡
3
2
¢ c
0
,
357 ...
(2 n +1)
246 ...
(2 n )
( x − 1) 2 n c
6
= 7
6 c
4
= 7
6
5
4
3
2 c
0 c
1
: c
5
= y
1
( x ) =
6
5 c
P
3
=
46 ...
2 n +2
35 ...
2 n +1
( x − 1) 2 n +1 n =0
6
5
4
3 c
1 c
2 n +1
= lim n →∞ c m +2 c m
= m + 3 m + 1
= 1 ρ = 1
46 ...
2 n +2
35 ...
2 n +1 c
1 y y ( x ) = c
0 y
0
( x ) + c
1 y
1
( x )
(1) = c
0
= 7 y 0 (1) = c
1
= 3 .
(3.4)
(3.5)
(3.6)
(3.7)
3.1 Singular Points:
Consider
P ( x ) y 00 + Q ( x ) y 0 + R ( x ) y = 0 .
If P , Q and R are polynomials without common factors then singular points are points x
0 at which P ( x
0
) = 0.
Note: At singular points the solution is not necessarily analytic.
Examples:
(3.8)
(1)
(2)
(3) x 2 y 00 + xy 0 y = x r → r
= 0
( r − 1) + r = 0 → y = c
1
+ c
2 ln x
The x 2 y 00 admits wild behaviour.
x 2 y 00 − 2 y = 0 y = x r → r ( r − 1) − 2 = 0 r = 2 , − 1 y = c
1 x 2 + c
2 x − 1
Again the x 2 y 00 admits wild behaviour.
x 2 y 00 − 2 xy 0 y = x r → r (
+ 2 r − y
1)
= 0
− 2 r + 2 = 0 r = 1 , 2 y = c
1 x + c
2 x 2

Series solutions to ODE with variable coefficients
In this case both solutions are analytic.
3
3.2 Regular Singular Points - polynomial coefficients:
Notice that all these cases are equidimensional equations for which we can identify solutions of the form x r or x r log x .
There is a special class of singular points called regular singular points in which the singularities are no worse than those in the equidimensional equations.
x 2 y 00 + αxy 0 + βy = 0 .
(3.9)
If P , Q and R are polynomials and suppose P ( x
0
) = 0 then x
0 is a regular singular point if lim x → x
0
( x − x
0
)
Q ( x )
P ( x ) and lim x → x
0
( x − x
0
)
2
R ( x )
P ( x ) are finite.
(3.10)
I.E. ( x − x
0
)
Q ( x )
P ( x )
= p
0
+ p
1
( x − x
0
) + p
2
( x − x
0
)
2
+ · · ·
→ singularity no worse than
( x − x
0
) 2
R ( x )
P ( x )
= q
0
+ q
1
( x − x
0
) + q
2
( x − x
0
) 2 + · · ·
1 x − x
0
→ singularity no worse than 1
( x − x
0
) 2
(3.11)
Examples:
(1)
(1 − x 2 ) y 00 − 2 xy 0 + 4 y = 0
P = 1 − x 2 lim x → 1
( x − 1)
P ( ± 1) = 0 Q = − 2 x R = 4
( − 2 x )
(1 − x )(1+ x )
= 1 lim x → 1
( x − 1) 2 4
(1+ x )(1 − x )
= 0 x = 1 is a R.S.P. (similarly for x = − 1).
(3.12)
(2) x 3 y 00
P ( x ) = x 3 lim x → 0 x 2
− y = 0
Q
¢
R = − 1
− 1 x 3
= ∞
(3.13)
Thus x = 0 is an irregular singular point . Actually y ∼ x 3 / 4 e ± 2 /x
1 / 2 the simple power law x r or x r log x .
as x → 0+ which is much wilder than
Note: Any singular point that is not a regular singular point is called an irregular singular point.
(3) 2( x − 2) 2 xy 00 + 3 xy 0 + ( x − 2) y = 0. Singular points at x = 0 , 2.
x = 0 is a regular singular point.
x = 2 is an irregular singular point.

4
3.3 More General Definition of a Regular Singular Point:
If P , Q , and R are not limited to polynomials then consider x = 0 is a regular singular point if
P ( x ) y 00 + Q ( x ) y 0 + R ( x ) y = 0 or
µ xQ
P x 2 y
¶
00 + x
³ xQ
P
´ y 0 and
µ x 2 R
¶
P
+
³ x
2
P
R
´ y = 0 are analytic at x = 0, i.e., xQ
P
= p ( x ) = p
0
+ p
1 x + · · · and x 2 R
P
= q ( x ) = q
0
+ q
1 x + · · · .
In this case
Ly = x 2 y 00 + xp
0 y 0 + q
0 y + z x { p
1 small as xy 0 + q
1 x → 0
{ y + · · · } = 0 .
Then as x → 0 x 2 y 00 + xp
0 y 0 + q
0 y ≈ 0 which is an Euler Equation which has solutions of the form y = x r .
Thus about a regular singular point we look for solutions of the form y = x r n =0 a n x n
=
Our task is to determine: n =0 a n x n + r
.
(3.14)
(3.15)
(3.16)
(i) r
(ii) the coefficients a n
(iii) the radius of convergence.
Example: x 2 y 00 + 2(e x − 1) y 0 + e − x cos xy = 0 , P = x 2 , Q = 2(e x − 1), R = e − x cos x .
x = 0 is a singular point.
lim x → 0 xQ x 2
P
R lim x → 0
P
= lim k → 0
= lim x → 0 x
2(e x − x 2
1) x 2 e − x x cos
2 x
= lim x → 0
= 1 <
2(e
∞ .
x x
− 1) 0
= lim x → 0
2e x
1
= 2 L’Hopital
(3.17)
Since the quotient functions p = xQ/P and q = x 2 R/P have Taylor Expansions about x = 0, x = 0 is a regular singular point.