Series Solutions of Second Order Linear Equations
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Euler Equations; Regular Singular Points
Example 5.14
Determine the singular points of the Legendre equation
(1 − x 2 )y 00 − 2xy 0 + α(α − 1)y = 0
and determine whether they are regular or irregular.
Solution
In this case P(x) = 1 − x 2 , so the singular points are x = 1 and
x = −1. Therefore, we have for x = 1
limx→1 (x − 1)
−2x
−2x(x − 1)
2x
= limx→1
= limx→1
=1
2
1−x
(1 − x)(1 + x)
1+x
and
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Euler Equations; Regular Singular Points
limx→1 (x − 1)2
limx→1
(x − 1)2 α(α − 1)
α(α − 1)
=
lim
=
x→1
1 − x2
1 − x2
(x − 1)2 α(α − 1)
(1 − x)α(α − 1)
= limx→1
=0
(1 − x)(1 + x)
(1 + x)
Since these limits are finite, the point x = 1 is a regular singular
point. It can be shown in a similar manner that x = −1 is also a
regular singular point.
Example 5.15
Determine the singular points of the equation
2x(x − 2)2 y 00 + 3xy 0 + (x − 2)y = 0
and determine whether they are regular or irregular.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Euler Equations; Regular Singular Points
Solution
In this case P(x) = 2x(x − 2)2 , so the singular points are x = 0
and x = 2. Therefore, we have for x = 0
limx→0 (x − 0)
3x
3x
= limx→0
=0
2
2x(x − 2)
2(x − 2)2
and
limx→0 (x − 0)2
x2
(x − 2)
=
lim
=0
x→1
2x(x − 2)2
2x(x − 2)
Since these limits are finite, the point x = 0 is a regular singular
point.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Euler Equations; Regular Singular Points
For x = 2 we have
limx→2 (x − 2)
3x
3(x − 2)
3
= limx→2
= limx→2
=
2
2
2x(x − 2)
2(x − 2)
2(x − 2)2
so the limit does not exist, hence x = 2 is an irregular singular
point.
Example 5.16
Determine the singular points of the equation
π 2 00
) y + cos(x)y 0 + sin(x)y = 0
2
and determine whether they are regular or irregular.
(x −
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Euler Equations; Regular Singular Points
Solution
In this case P(x) = (x − π2 )2 , so the only singular point is x = π2 .
Therefore, we have for x = π2
limx→ π2 (x −
cos(x)
π cos(x)
π
)
= −1
π 2 = limx→ 2
2 (x − 2 )
x − π2
and
limx→ π2 (x −
π 2 sin(x)
)
= limx→ π2 sin(x) = 1
2 (x − π2 )2
Since these limits are finite, the point x =
point.
Dr. Marco A Roque Sol
π
2
is a regular singular
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
We now consider the question of solving the general second order
linear equation
P(x)y 00 + Q(x)y 0 + R(x)y = 0
in the neighborhood of a regular singular point x = x0 . For
convenience we assume that x0 = 0. If x0 6= 0, we can transform
the equation into one for which the regular singular point is at the
origin by letting x − x0 equal t.
The assumption that x = 0 is a regular singular point means that
the functions p(x) = Q(x)/P(x) and q(x) = R(x)/P(x) have
convergent power series expansions of the form
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
xp(x) =
∞
X
pn x n ;
n=0
x 2 q(x) =
∞
X
qn x n
n=0
on some interval |x| < ρ about the origin, where ρ > 0. Assuming
that we have
P(x)y 00 + Q(x)y 0 + R(x)y = 0
y 00 + p(x)y 0 + q(x)y = 0
x 2 y 00 + x (xp(x)) y 0 + x 2 q(x)y = 0
x 2 y 00 + x (p0 + p1 x + ... + pn x n + ...) y 0
+x 2 (q0 + q1 x + ... + qn x n + ...) y = 0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
Now, if pn = qn = 0 except possibly at
p0 = limx→0
xQ(x)
P(x)
and
q0 = limx→0
x 2 R(x)
P(x)
the equation can reduced to
x 2 y 00 + p0 xy 0 + q0 y = 0
the Euler equation associate to the ODE, which was discussed
previously. Thus we can see the differential equation as
x 2 y 00 + x (xp(x)) y 0 + x 2 q(x)y = 0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
x 2 y 00 + xp0 (1 + (p1 /p0 )x + ... + (pn x n /p0 ) + ...) y 0 +
q0 (1 + (q1 /q0 )x + ... + (qn /q0 )x n + ...) y = 0
Thus, the coefficients in the differential equation can be viewed as
”Euler coefficients” times power series. Thus it may seem natural
to seek solutions of the form of ”Euler solutions” times power
series. Hence we assume that
r
n
y = x (a0 + a1 x + ... + an x + ...) = x
r
∞
X
n
an x =
n=0
Dr. Marco A Roque Sol
Ordinary Differential Equations
∞
X
n=0
an x r +n
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
where a0 6= 0. In other words, r is the exponent of the first
nonzero term in the series, and a0 is its coefficient. Of course, as
part of the solution, we have to determine:
1. The values of r for which the equation has a solution of the
form given above.
2. The recurrence relation for the coefficients an .
P
n
3. The radius of convergence of the series ∞
n=0 an x
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
The general theory was constructed by Ferdinand Georg Frobenius
(1849-1917) ( https:
//en.wikipedia.org/wiki/Ferdinand_Georg_Frobenius )
and is fairly complicated. A this point we are going to assume it
and show how to use it.
Example 5.17
Solve the differential equation
2x 2 y 00 − xy 0 + (1 + x)y = 0
Solution
It is easy to show that x = 0 is a regular singular point.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
Further, xp(x) = −1/2 and x 2 q(x) = (1 + x)/2. Thus
p0 = −1/2, q0 = 1/2, q1 = 1/2, and all other p 0 s and q 0 s are zero.
We assume that there is a solution of the form
y=
∞
X
an x r +n
n=0
Then, y 0 and y 00 are given by
0
y =
∞
X
an (r + n)x r +n−1
n=0
y 00 =
∞
X
an (r + n)(r + n − 1)x r +n−2
n=0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions of Second Order Linear Equations
Series Solutions Near a Regular Singular Point, Part
I
By substituting the expressions for y , y 0 , and y 00 in the ODE, we
obtain
"∞
#
X
2x 2 y 00 − xy 0 + (1 + x)y = 2x 2
an (r + n)(r + n − 1)x r +n−2 −
n=0
"
x
∞
X
#
"
an (r + n)x r +n−1 + (1 + x)
n=0
∞
X
#
an x r +n = 0
n=0
"
2x 2 y 00 − xy 0 + (1 + x)y =
∞
X
#
2an (r + n)(r + n − 1)x r +n −
n=0
"∞
X
n=0
#
an (r + n)x r +n +
"
∞
X
n=0
Dr. Marco A Roque Sol
#
an x r +n +
"
∞
X
#
an x r +n+1 = 0
n=0
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
2 00
0
2x y − xy + (1 + x)y = x
r
(" ∞
X
#
2an (r + n)(r + n − 1)x
n
n=0
"
∞
X
n=0
#
an (r + n)x n +
"
∞
X
n=0
#
an x n +
"∞
X
#)
an x n+1
=0
n=0
2x 2 y 00 − xy 0 + (1 + x)y = a0 [2r (r − 1) − r + 1]x r +
∞
X
([2(r + n)(r + n − 1) − (r + n) + 1] an + an−1 ) x r +n = 0
n=1
Dr. Marco A Roque Sol
Ordinary Differential Equations
−
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
If the above equation is to be satisfied for all x, the coefficient of
each power of x must be zero. From the coefficient of x r we
obtain, since a0 6= 0,
F (r ) = 2r (r − 1) − r + 1 = 2r 2 − 3r + 1 = (r − 1)(2r − 1) = 0
this equation is called the indicial equation . Note that it is
exactly the characteristic equation we would obtain for the Euler
equation associated with the ODE. The roots of the indicial
equation are
r1 = 1;
Dr. Marco A Roque Sol
r2 = 1/2
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
These values of r are called the exponents at the singularity for
the regular singular point x = 0. They determine the qualitative
behavior of the solution in the neighborhood of the singular point.
Now, the recurrence relation is
[2(r + n)(r + n − 1) − (r + n) + 1] an + an−1 = 0;
an−1
;
− 3(r + n) + 1]
n≥1
an−1
;
[(r + n) − 1] [2(r + n) − 1]
n≥1
an = −
an = −
[2(r +
n)2
Dr. Marco A Roque Sol
Ordinary Differential Equations
n≥1
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
For each root r1 and r2 of the indicial equation, we use the
recurrence relation to determine a set of coefficients a1 , a2 , .... For
r = r1 = 1, we have
an = −
an−1
;
(2n + 1)n
a1 = −
a2 = −
a3 = −
n≥1
a0
3·1
a1
a0
=
5·2
(3 · 5)(1 · 2)
a2
a0
=
7·3
(3 · 5 · 7)(1 · 2 · 3)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
In general we have
an =
(−1)n
a0 ;
[3 · 5 · 7 . . . (2n + 1)]n!
n≥1
If we multiply both the numerator and denominator of the right
side of the equation by 2 · 4 · 6 · · · 2n (= 2[1 · 4 · 6 . . . 2n] = 2 · 2
[1 · 2 · 6 · · · 2n] = 2 · 2 · 2[1 · 2 · 3 . . . 2n] = · · · = 2 · 2 · 2 · · · 2
[1 · 2 · 3 · · · n] = 2n n!), we can rewrite an as
(−1)n 2n
a0 ; n ≥ 1
(2n + 1)!
Hence, if we omit the constant multiplier a0 , one solution is
"
#
∞
X
(−1)n 2n n
y1 = x 1 +
x ; x >0
(2n + 1)!
an =
n=0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
To determine the radius of convergence of the series, we use the
ratio test
an+1 x n+1 2|x|
= limn→∞
=0
limn→∞ n
an x
(2n + 2)(2n + 3)
Corresponding to the second root r = r2 = 1/2 , we proceed
similarly. We have
an = −
an−1
an−1
=−
; n≥1
1
n(2n − 1)
2n(n − 2 )
a0
a1 = −
1·1
a1
a0
=
a2 = −
2·3
(1 · 2)(1 · 3)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
a3 = −
a0
a2
=
3·5
(1 · 2 · 3)(1 · 3 · 5)
In general we have
an =
(−1)n
a0 ; n ≥ 1
[1 · 3 · 5 . . . (2n − 1)]n!
If we multiply both the numerator and denominator of the right
side of the equation by 2 · 4 · 6 . . . 2n = 2n n! , we can rewrite an as
an =
(−1)n 2n
a0 ; n ≥ 1
(2n)!
Dr. Marco A Roque Sol
Ordinary Differential Equations
Series Solutions of Second Order Linear Equations
Series Solutions Near an Ordinary Point, Part II
Series Solutions Near a Regular Singular Point, Part I
Series Solutions Near a Regular Singular Point, Part
I
Hence, if we omit the constant multiplier a0 , one solution is
#
"
∞
X
(−1)n 2n n
1/2
x ; x >0
y2 = x
1+
(2n)!
n=0
To determine the radius of convergence of the series, we can use
the ratio test and to show that converges for all x. The functiones
y1 and y2 form a fundamental set of solutions. Hence the general
solution is
y = c1 y1 (x) + c2 y2 (x); x > 0
The preceding example illustrates that if x = 0 is a regular singular
point, then sometimes there are two solutions of the form
y = xr
∞
X
an x n
n=0
Dr. Marco A Roque Sol
Ordinary Differential Equations