18.03 Power Series Techniques
Jeremy Orloff
This is taken from chapter 3 of Edwards and Penney. I recommend reading that chapter if
your want to learn more about this.
We will consider the second order linear DE:
y 00 + P (x)y 0 + Q(x) = 0
We will do all our work around x = 0, it is easy to translate this to x = a. Also, we will
not worry about radius of convergence or other analytic issues.
Ordinary points: If P (x) and Q(x) are analytic (have a convergent power series) around
x = a then a is called an ordinary point
Example 1: Solve y 00 + y = 0.
Try a power series solution:
y(x) =
∞
X
an xn .
0
⇒ y 00 =
∞
X
n(n − 1)an xn−2 =
2
So, y 00 + y = 0 ⇒
∞
X
(n + 2)(n + 1)an+2 xn
0
∞
X
[(n + 2)(n + 1)an+2 + an ]xn = 0.
0
⇒ the coefficient of xn is 0
⇒ (n + 2)(n + 1)an+2 + an = 0 ⇒ an+2 = −
an
(This last equation is called
(n + 2)(n + 1)
the recurrence relation for the coefficients an ).
Pick an arbitrary a0 :
a0
n = 0 ⇒ a2 = −
2·1
a2
a0
a0
n = 2 ⇒ a4 = −
=
=
4·3
4·3·2·1
4!
a4
a0
n = 4 ⇒ a6 = −
=−
6·5
6!
n a0
.
⇒ a2n = (−1)
(2n)!
Likewise, pick an arbitrary a1 :
a1
n = 1 ⇒ a3 = −
3·2
a1
n
. . . a2n+1 = (−1)
.
(2n + 1)!
⇒ y = a0 (1 − x2 /2! + x4 /4! − . . .) + a1 (x − x3 /3! + x5 /5! − . . .) = a0 cos(x) + a1 sin(x) (as
expected).
(continued)
1
18.03 Power Series Techniques
2
Example 2: Solve y 0 = x2 y.
∞
X
Try y =
an xn .
0
⇒ y0 =
∞
X
nan xn−1 = a1 + 2a2 x +
0
∞
X
(n + 1)an+1 xn .
2
y 0 − x2 y = 0 ⇒ a1 + 2a2 x +
∞
X
[(n + 1)an+1 − an−2 ]xn = 0.
2
All the coefficients are 0 ⇒ a1 = 0, a2 = 0, (n+1)an+1 −an−2 = 0 ⇒ an+1 = an−2 /(n+1).
a3
a0
a0
a0
=
= 2
Take arbitrary a0 ⇒ a3 = , a6 =
3
6
6·3
3 · 2!
a6
a0
a0
a0
Likewise a9 =
= 3
, a12 = 4
, a3n = n
.
9
3 · 3!
3 · 4!
3 · n!
a1 = 0 ⇒ a4 = 0 ⇒ a7 = 0 . . .
a2 = 0 ⇒ a5 = 0 ⇒ a8 = 0 . . .
∞
X
x3n
3
= a0 ex /3 (as expected).
⇒ y = a0
n
3 · n!
0
Regular Singular Points
p(x) 0 q(x)
y + 2 y = 0. Because the coefficients are discontinuous at
Consider the DE y 00 +
x
x
x = 0 we call 0 a singular point. If p(x) and q(x) are analytic (in particular if they are
polynomials) then x = 0 is a regular singular point.
In this case we only look for solutions for x > 0.
Example 3: Solve y 00 + (p0 /x)y 0 + (q0 /x2 )y = 0 (p0 , q0 are constants)
Try y = xr ⇒ r(r − 1)xr−2 + p0 rxr−2 + q0 xr−2 = 0.
⇒ r(r − 1) + p0 r + q0 = 0 (this is called the indicial equation).
If the roots are real and different we have two solutions.
If the roots are complex, say r = a ± ib, we have the solution
y = xa+ib = xa eib log x = xa (cos(b log x) + i sin(b log x)).
Likewise for xa−ib . And, just like CC homogemeous linear DE’s, this means we have two
real solutions
y1 = xa cos(b log x) and y2 = xa sin(b log x).
From now on we will focus on equations with real roots.
(continued)
18.03 Power Series Techniques
3
Example 4: Solve x2 y 00 + xy 0 + (x2 − m2 )y = 0 (Bessel equation of order m).
(Note, x = 0 is a regular singular point.)
A simple y r won’t work, instead we need the Frobenius Solution
y = xr
∞
X
an xn =
∞
X
0
an xn+r
0
Note, this is not a power series if r is not an integer. Also, without loss of generality we
can require a0 6= 0. We get:
∞
∞
X
X
x2 y 00 =
(n + r)(n + r − 1)an xn+r = r(r − 1)a0 xr + (r + 1)ra1 xr+1 +
(n + r)(n + r − 1)an xn+r
0
xy 0 =
∞
X
2
(n + r)an xn+r = ra0 xr + (r + 1)a1 xr+1 +
0
x2 y =
m y=
(n + r)an xn+r
2
∞
X
an xn+r+2 =
0
2
∞
X
∞
X
∞
X
an−2 xn+r
2
2
m an x
n+r
2
r
2
= m a0 x + m a1 x
r+1
+
0
∞
X
m2 an xn+r
2
Substituting in the DE we get
[r(r − 1) + r − m2 ]a0 xr + [(r + 1)r + (r + 1) − m2 ]a1 xr+1
∞
X
[((n + r)(n + r − 1) + (n + r) − m2 )an + an−2 ]xn+r = 0
+
2
Cleaning things up and setting coefficients to 0 we get
(r2 − m2 )a0 = 0,
((r + 1)2 − m2 )a1 = 0,
((n + r)2 − m2 )an + an−2 = 0.
Example 5: Take m = 1/3 in the above example.
We take a0 arbitrary. The indicial equation is r2 − m2 = 0 ⇒ r = ±1/3.
Take r = 1/3: the recurrence equations become a1 = 0 and an = −an−2 /((n + r)2 − m2 ).
a0
⇒ a3 = a5 = a7 = . . . = 0 and a2 = −
etc.
(2 + 1/3)2 − 1/9
Likewise for r = −1/3.
If the roots differ by an integer the recurrence relations can sometimes run into trouble.
You can read section 3.3 and 3.4 of E&P for details.
Example 6: Take m = 1/2: indicial equation r2 − 1/4 = 0 ⇒ r = ±1/2.
4a0
4a2
Take r = −1/2, a0 arbitrary ⇒ a2 = − 2
, a4 = − 2
...
3 −1
7 −1
2
2 )a = 0 shows a can also be arbitrary then a = −4a /(52 −1) . . ..
The equation ((r+1)
1
1
3
1
P −mn+r
The solution y =
an x
has two arbitrary constants. (Note if we take r = 1/2 we get
the same solution as above with a0 = 0 and a1 arbitrary
(continued)
18.03 Power Series Techniques
4
Example 7: Take m = 1: indicial equation r2 − 1/4 = 0 ⇒ r = ±1.
a0
a2
Take r = 1, a0 6= 0: ⇒ a2 = − 2
, a4 = − 2
...
3 −1
5 −1
The equation ((r + 1)2 − m2 )a1 = 0 ⇒ a1 = 0 ⇒ a3 = a5 = . . . = 0.
∞
X
⇒ y1 =
a2n x2n+1 .
0
a0
doesn’t work.
(2 + r)2 − 1
Trick (reduction of order): try y2 (x) = v(x)y1 (x) ⇒ y1 v 00 + (2y10 + 1/x)v 0 = 0. Let u = v 0
⇒ y1 u0 + (2y10 + 1/x)u = 0, a first order linear DE.
∞
X
This leads to, y2 = C1 y1 ln x + x−1
bn xn . Substitution gives relations for C1 and bn .
Take r = −1, a0 6= 0: the equation a2 = −
0
Example 8: Take m = 0: indicial equation r2 = 0 ⇒ repeated roots.
Take r = 0, a0 6= 0 ⇒ solution y1 as in example 7.
∞
X
Reduction of order leads to y2 = y1 ln x +
bn xn . Substitution gives relations for bn .
0