ACM 30020 Advanced Mathematical Methods
Solution of the 2nd order linear homogeneous ODEs using power series
Equations with variable coefficients
Consider the ODE:
L[y] = y” + p(x)y 0 + q(x)y = 0
(1)
where p(x) and q(x) are continuous function of x. This problem formulated as IVP (by adding initial
conditions for both y and y 0 ) has a unique solution on the interval of continuity of p and q. If p and q are
polynomials it has unique solution for all x.
As continuous functions p and q could be expanded into converging infinite power series:
p = p0 + p1 x + p2 x2 + ... =
∞
X
ps xs
q = q0 + q1 x + q2 x2 + ... =
s=0
y =
∞
X
sas x
qs x s ;
s=0
Let look for solution of Eq. (1) in the form y = a0 + a1 x + a2 x2 + ... =
0
∞
X
00
s−1
y =
s=1
∞
X
P∞
s=0
as xs . We could compute
s(s − 1)as xs−2 ;
s=2
Replacing y,y 0 ,y 00 p and q by their expansions we get:
L[y] =
∞
X
s=2
s(s − 1)as xs−2 +
∞
X
ps xs
s=0
∞
X
s=1
sas xs−1 +
∞
X
s=0
qs x s
∞
X
as xs = 0 .
(2)
s=0
Let collect the coefficients corresponding the same powers of x and pay attention that Eq. (2) is satisfied
only when all these terms are equal to 0.
x0
x
1
2 · 1 · a2 + p0 · 1 · a1 + q0 a0 = 0
3 · 2 · a3 + p0 · 2 · a2 + p1 · 1 · a1 + q0 a1 + q1 a0 = 0
x2
4 · 3 · a4 + p0 · 3 · a3 + p1 · 2 · a2 + p2 · 1 · a1 + q0 a2 + q1 a1 + q2 a0 = 0
.
...
xs
(s + 2)(s + 1)as+2 + f (s, q0 , ..., qs , p0 , ..., ps , a0 , ..., as s + 2) = 0
where f (s, q0 , ..., qs , p0 , ..., ps , a0 , ..., as s + 2) is a multi-variable polynomial.
To find the coefficient corresponding to x0 we need {a0 , a1 , a2 }, x1 involves {a0 , a1 , a2 , a3 }. Extending
this to as one can deduce that we have s + 1 equations for s + 1 coefficients, but s + 3 unknowns. The
system of linear equations we have derived to find coefficients as has two more unknowns than equations
or by other words two degrees of freedom. We could choose these coefficients arbitrary under conditions
that they yield linearly independent solutions.
One can easily check by computing the Wronskian that choosing sets:
{a0 = 1, a1 = 0}
{a0 = 0, a1 = 1}
we get two linearly independent solutions:
y1 = 1 +
∞
X
as xs
y2 =
s=2
∞
X
as xs ;
s=1
and the general solution of Eq. (1) is their linear combination: y = C1 y1 + C2 y2
Example: Airy Equation
y 00 − xy = 0
This equation raises in analysis of wave propagation and scattering. By substitution the series expansions
for y and y 00 into Eq. (2) we get
(2 · 1 · a2 + 3 · 2 · a3 x + 4 · 3 · a4 x2 + . . . ) − x(a0 + a1 x + a2 x2 + . . . ) = 0
Next we construct the table: The general formula for as is
x0
2 · 1 · a2 = 0
x1
3 · 2 · a3 − a0 = 0
2
4 · 3 · a4 − a1 = 0
x
.
x
...
s
(s + 2)(s + 1)as+2 − as−1 = 0
as+2 =
as−1
(s + 2)(s + 1)
The first of the linearly independent solutions corresponds to the set {a0 = 1, a1 = 0}. The rest of
parameters:
a2 = 0, a3 = 1/(2 · 3), a4 = 0, a5 = 0, a6 = 1/(2 · 3 · 5 · 6) .
The second solution is given by:
a0 = 0, a1 = 1, a2 = 0, a3 = 0, a4 = 1/(3 · 4), a5 = 0, a6 = 0, a7 = 1/(3 · 4 · 6 · 7) .
The general solution could be written as
x4
x3
x6
x7
y = C1 y1 + C2 y2 = C1 1 +
+
+ . . . + C2 x +
+
+ ...
6
180
12 504
Convergence To test convergence we apply ratio test: If an 6= 0 we compute
an+1 an+1 (x − x0 )n+1 = |x − x0 | lim = K
lim n→∞ an n→∞
an (x − x0 )n If K < 1 then the series converges absolutely; if K > 1 then the series diverges; if K = 1 then
additional tests are needed. Singular points Consider the differential equation:
A(x)y 00 + B(x)y 0 + C(x)y = 0
(3)
If we are interested to solve it (using series expansion, in general case) we need first to divide it by A(x)
to bring to the canonical form. However, A(x) could be equal to 0 at one or more points in our interval
of interest. In such a case we need to look for the solution in the vicinity of singular point(s). Let x0 be
a point s.t. A(x0 ) = 0. We divide the Eq. (8) by A(x) and multiply it by (x − x0 )2 :
B(x 0
C(x)
2 00
(x − x0 ) y + (x − x0 ) (x − x0 )
y + (x − x0 )2
y=0
(4)
A(x)
A(x)
If
B(x
lim (x − x0 )
= lim p(x) = p0 < ∞
x→x0
x→x0
A(x)
and
lim (x − x0 )2
x→x0
C(x)
= lim q(x) = q0 < ∞
A(x) x→x0
then x0 is called a regular singular point. Otherwise, x0 is an irregular singular point.
Example
2x4 (1 − x2 )y 00 + 3x2 y = 0
1
3
C(x)
B(x)
= 3
, =
2
2
A(x)
x (1 − x )
A(x) 2x (1 − x2 )
Suspicious points: {−1, 0, 1}.
x0 = −1 – regular singular point
(x + 1)
1
=−
2
−x )
3
2
3(x + 1)
=0
lim
x→−1 2x2 (1 − x2 )
lim
x→−1
x3 (1
(5)
x0 = 0 – irregular singular point
x
=∞
− x2 )
3x2
3
lim
=
2
2
x→−1 2x (1 − x )
2
lim
x→0 x3 (1
Easy to show that x = 1 is the regular singular point as well
Solution of 2nd order linear equations in the vicinity of the regular singular point In vicinity of regular singular points we could compute solution of Eq. (4) using the power series expansion. Conditions
(5) let us to do so, as both p(x) and q(x) are continuous functions everywhere on the interval of interest
including x0 . To simplify our computations we consider that x0 = 0. If this is not the case we could
always change the variable X = x − x0 , which will make the shift. As we already wrote:
2
p = p0 + p1 x + p2 x + ... =
∞
X
ps x
s
2
q = q0 + q1 x + q2 x + ... =
Now let look for solution y(x) in the form: y = xr (a0 + a1 x + a2 x2 + ...) = xr
we could compute
y =
∞
X
(r + s)as x
qs x s ;
s=0
s=0
0
∞
X
00
r+s−1
y =
∞
X
P∞
s=0
as xs . As before
(r + s)(r + s − 1)as xr+s−2 ;
s=0
s=0
Replacing y,y 0 ,y 00 p and q by their power series expansions we get:
L[y] =
∞
X
(s + r)(r + s − 1)as x
s=0
s+r
+
∞
X
s=0
ps x
s
∞
X
s=0
(r + s)as x
s+r
+
∞
X
qs x
s=0
s
∞
X
as xs+r = 0 .
(6)
s=0
Let collect the coefficients corresponding the same powers of xr and pay attention that Eq. (6) is satisfied
only when all these terms are equal to 0.
xr
r(r − 1)a0 + p0 ra1 + q0 a0 = [r(r − 1) + p0 r + q0 ]a0 = 0
xr+1
r(r + 1)a1 + p0 (r + 1)a1 + p1 ra0 + q0 a1 + q1 a0 = 0
.
...
xr+s
(s + r)(s + r − 1)as + f (s, r, q0 , ..., qs , p0 , ..., ps , a0 , ..., as ) = 0
where f (s, r, q0 , ..., qs , p0 , ..., ps , a0 , ..., as ) is a multi-variable polynomial. The top of equations has nontrivial solution if r(r − 1) + p0 r + q0 = 0. In such a case a0 is a free parameter, so so we could choose
a0 = 1 and find a1 , a2 , ....
The equation r(r − 1) + p0 r + q0 = 0 is called a characteristic equation, which, in fact corresponds to
the homogeneous Euler equation: x2 y 00 + P xy 0 + Qy = 0, where P and Q are constants. The Euler
equation (on any interval that doesn’t include a regular singular point x = 0) is converted to the equation
with the constant coefficients by the substitution x = et .
Solving the characteristic equation we get three options:
1. r1 6= r2 and r1 − r2 6= ±1, ±2, .... In such a case we have two linearly independent solutions:
y1 = xr1 (1 +
∞
X
as x s )
y2 = xr2 (1 +
s=1
∞
X
as xs );
s=1
One can easily check that these solutions are linearly independent by computing their Wronskian.
If r1,2 = α ± iβ then the solution is converted to the real form using the Euler formula.
2. r1 = r2 = r. In this case we have one solution
r
y1 = x (1 +
∞
X
as xs ) ,
s=1
and use it to find y by reduction of order (substitution y = vy1 . The solution y2 will have a form:
y2 = y1 ln x + xr
∞
X
bs x s
s=0
3. r2 − r1 = N = 1, 2, ... (we set r2 > r1 . This is the most complicated case. To explain why we
define:
F (r) = r(r − 1) + p0 r + q0
We know that the only condition for Eq. (4) to have nontrivial solution (when a0 6= 0) is to require
F (r1 ) = 0. However, in such a case F (r2 ) = F (r1 +N ) = (r1 +N )(r1 +N −1)+p0 (r1 +N )+q0 =
0 This means (from the expression we derive for computing the coefficients as ) that starting from
the term s = N + 1 the coefficients in the expansion corresponding to r2 will be equal to 0 (or the
series will be finite), which is true only for polynomials (but we don’t know the form of solution
without finding it). Therefore in this case we need again to use the reduction of order
The general form of solution is similar to the previous case:
r
y1 = x (1 +
∞
X
as xs ) ,
s=1
and
y2 = αy1 ln x + x
r
∞
X
s=0
bs x s
Example: Bessel’s Equation: The equation
x2 y 00 + 2xy 0 + (x2 − ν 2 )y = 0
(7)
is called Bessel’s equation of order ν. We consider ν = 1/3. So the problem has a form
x2 y 00 + 2xy 0 + (x2 − 1/9)y = 0
(8)
We build the corresponding table: The recursive formula for the coefficients:
xr
r(r − 1)a0 + ra0 − (1/9)a0 = (r2 − 1/9)a0 = 0
xr+1
r(r + 1)a1 + (r + 1)a1 − (1/9)a1 = 0
xr+2
(r + 2)(r + 1)a2 + (r + 2)a2 + a0 − (1/9)a2 = 0
.
...
xr+s
(s + r)(s + r − 1)as + (r + s)as + as−2 − (1/9)as = 0
as = −
as−2
(r + s)2 ) − (1/9)
The roots of characteristic equation are r1,2 = ±1/3 = ±ν. The general solution has a form:
1/3
y = C1 y1 + C2 y2 = C1 |x|
∞
X
s
−1/3
as (1/3)x + C2 |x|
s=0
∞
X
as (−1/3)xs
s=0
The first few coefficients (a0 = 1):
r = 1/3: a1 = 0, a2 = −1/[(2 + 1/3)2 − 1/9] = −3/16, a3 = 0
r = −1/3: a1 = 0, a2 = −1/[(2 − 1/3)2 − 1/9] = −3/8, a3 = 0
Few more words about Bessel’s functions.
Depending on values of ν (or more precisely 2ν) we have different types of solution for Bessel’s equation,
which correspond to the different types of Bessel’s functions:
• 2ν is not an integer
y = C1 J−ν (x) + C2 Jν (x)
where Jν (x) is called Bessel’s function of the 1st kind. This is the function from the example we’ve
solved.
• 2ν = 2n. It corresponds to the ν = n being integer number and the difference between the roots
of the characteristic equation is integer number.In this case
y = C1 Jν (x) + C2 Kν (x) ,
where Kν (x) is Bessel’s function of the 2nd kind.
Kn (x) = αJn (x) ln x + x
−n
∞
X
bs x s ,
s=0
where
Jn (x) =
∞
X
s=0
n+2s
x
(−1)2
s!(n + s)! 2
• 2ν = 2n + 1 or ν = (2n + 1)/2. In this case the solution has the following form:
r 2
1
1
y=
C1 Pn
sin x + C2 Qn
cos x .
πx
x
x