8 - Series Solutions of Differential
Equations
8.2 Power Series and Analytic Functions
Homework: p. 434 - 436 #
ü Introduction
Our earlier technques allowed us to write our solutions in terms of elementary functions (such as polynomial functions,
exponential functions, trigonometric functions, etc). In most cases we are not able to do so. We can approximate the
solution using numerical methods (Euler’s method, etc.), or we can use power series.
ü Power Series
A power series about the point x0 is an expression of the form
⁄ an Hx - x0 Ln = a0 + a1 Hx - x0 L + a2 Hx - x0 L2 + ∫,
¶
(1)
n=0
where x is a variable and the an ’s are constants. We say that (1) converges at the point x = c if the infinite series (of
real numbers) ⁄ an Hc - x0 Ln converges; that is, the limit of the partial sums limNض ⁄ an Hc - x0 Ln exists (as a finite
¶
N
n=0
n=0
number). If this dimit does not exist, the power series is said to diverge at x = c. Note that (1) converges at x = x0 .
Theorem 1 - Radius of Convegence
For each power series of the form (1), there is a number r (0 § r § ¶), called the radius of convergence of the
power series, such that (1) converges absolutely for †x - x0 § < r and diverges for †x - x0 § > r.
If the series (1) converges for all values of x, then r = ¶. When the series (1) converges only at x0 , then r = 0.
Note that at the interior points of the interval, the power series converges absolutely, that is, the series
⁄ †an Hx - x0 Ln § converges for every x in the interior.
¶
n=0
Recall that we had to test the endpoints of the interval individually. At the endpoints, the power series could diverge,
converge conditionally, or converge absolutely.
For the geometric series ⁄ xn , the radius of convergence is r = 1, since
¶
n=0
1
1-x
= 1 + x + x2 + ∫ = ⁄ xn for -1 < x < 1.
¶
n=0
So the radius of convergence is r = 1 and the interval of convergence is -1 < x < 1. Recall that if the series was not a
geometric series, we used the ratio test to find the interval of convergence.
2
CH_08_notes.nb
Theorem 2 - Ratio Test for Power Series
If, for n large, the coefficients an are nonzero and satisfy
limnض ¢ a n ¶ = L
a
(0 § L § ¶)
then the radius of convergence of the power series ⁄ an Hx - x0 Ln is r = L
n+1
¶
n=0
ü Example 8.2.1
Determine the convergence set of the given power series.
⁄
¶
n=0
3n
n!
xn
Theorem 3 - Power Series Vanishing on an Interval
If ⁄ an Hx - x0 Ln = 0 for all x in some open interval, then each coefficient an equals zero.
¶
n=0
Given two power series f HxL = ⁄ an Hx - x0 Ln and gHxL = ⁄ bn Hx - x0 Ln with nonzero radii of convergence.
¶
n=0
¶
n=0
f HxL + gHxL = ⁄ Han + bn L Hx - x0 Ln for all x in the common interval of
¶
• The sum is given by
n=0
convergence for both power series.
n
• The product, given by f HxL gHxL = ⁄ cn Hx - x0 Ln where cn = ⁄k=0
ak bn-k , for all x in the
¶
n=0
common open interval of convergence for both power series.
Theorem 4 - Differentiation and Integration of Power Series
If the series f HxL = ⁄ an Hx - x0 Ln has a positive radius of convergence r, then f is differentiable in the interval
¶
†x - x0 § < r and termwise differentiation gives the power series for the derivative:
n=0
f ' HxL = ⁄ n an Hx - x0 Ln-1 for †x - x0 § < r
¶
n=1
Furthermore, termwise integration gives the power series for the integral of f:
Ÿ f HxL „ x = ⁄
¶
n=0
an
n+1
Hx - x0 Ln+1 + C for †x - x0 § < r
CH_08_notes.nb
ü Shifting the Summation Index
Note that
⁄ an Hx - x0 Ln = ⁄ ak Hx - x0 Lk = ⁄ a j Hx - x0 L j
¶
¶
¶
n=0
k=0
j=0
Occasionally we will need to shift the summation index to determine the terms of a power series.
ü Example 8.2.2
Express the given power series as a series with generic term xk .
⁄ nHn - 1L an xn+2
¶
(a)
n=2
(b)
⁄ n an Hx - aLn-1 + 3 ⁄ an Hx - aLn
¶
¶
n=1
n=0
ü Analytic Functions
Definition 1 - Analytic Function
A function f is said to be analytic at x0 if, in an open interval about x0 , this funciton is the sum of a power series
⁄ an Hx - x0 Ln that has a positive radius of convergene.
¶
n=0
3
4
CH_08_notes.nb
Elementary functions such as ‰x , sin x, and cos x are analytice for all x, while ln x is analytic for x > 0. A rational
function
P HxL
QHxL
is an analytic funtion except at those x0 for which QHx0 L = 0. Recall the representations of some of these
functions:
(10)
(11)
(12)
(13)
x2
2!
‰x = 1 + x +
sin x = x cos x = 1 -
3
+
x2
2!
+
+∫= ⁄
¶
n=0
x
5!
+∫= ⁄
x4
4!
+∫= ⁄
5
x
3!
ln x = Hx - 1L -
x3
3!
+
1
2
¶
n=0
¶
n=0
Hx - 1L2 +
1
3
xn
n!
H-1Ln
H2 n+1L!
H-1Ln
H2 nL!
x2 n+1
x2 n
Hx - 1L3 + ∫ = ⁄
¶
H-1Ln-1
n
Hx - 1Ln
The inportance of f being analytice is the fact that if f HxL i analytic at x0 , then it is the sum of some power series that
converges in a neighborhood of x0 :
n=1
f HxL = ⁄ an Hx - x0 Ln
¶
n=0
Any power series—regardless of how it is derived—that converges in some neighborhood of x0 to a function has to be
the Taylor series of that function.
CH_08_notes.nb
5
8.3 Power Series Solutions to Linear Differential Equations
Homework: p. 445 - 446 #
ü Introduction
We start by considering the linear differential equation
a2 HxL y '' + a1 HxL y ' + a0 HxL y = 0
(1)
that has been written in the standard form
y '' + pHxL y ' + qHxL y = 0
(2)
Definition 2 - Ordinary and Singular Points
A point x0 is called an ordinary point of equation (1) if both p =
a1
a2
and q =
a0
a2
are analytic at x0 . If x0 is not an
ordinary point, it is called a singular point of the equation.
ü Example 8.3.1
Determine all singularpoints of the given differential equation.
Ix2 + xM y '' + 3 y' - 6 x y = 0
At an ordinary point x0 of equation (1) or (2), the coefficient functions pHxL and qHxL are analytic. It runs out, in a
eighborhood of an ordinary point x0 , the solutions to (1) or (2) can be expressed as a power series about x0 . Thus, we
want to find a power series solution of the form
yHxL = a0 + a1 Hx - x0 L + a2 Hx - x0 L2 + ∫ = ⁄ an Hx - x0 Ln
¶
(4)
n=0
Note that
y' HxL = a1 + 2 a2 Hx - x0 L + 3 a3 Hx - x0 L2 + ∫ = ⁄ n an Hx - x0 Ln-1
¶
n=1
Note that
y '' HxL = 2 a2 + 6 a3 Hx - x0 L + 12 a4 Hx - x0 L2 + ∫ = ⁄ nHn - 1L an Hx - x0 Ln-2
¶
n=2
6
CH_08_notes.nb
ü Example 8.3.2
Find at least the first four nonzero terms in a power series expansion about x = 0 for a general solution to the given
differential equation:
Ix2 + 1M y '' - x y' + y = 0
CH_08_notes.nb
ü Example 8.3.3
Find at least the first four nonzero terms in a power series expansion about x = 0 for the solution to the given initial
value problem.
y '' + Hx - 2L y' - y = 0;
yH0L = -1, y ' H0L = 0
7
8
CH_08_notes.nb
8.4 Equations with Analytic Coefficients
Homework: p. 450 - 451 #
ü Introduction
We have intrduced a procedure for finding the power series solution of
(1)
y '' + pHxL y ' + qHxL y = 0
Theorem 5 - Existence of Analytic Solutions
Suppose x0 is an ordinary point for equation (1). Then (1) has two linearly independent analytic solutions of the
form
yHxL = ⁄ an Hx - x0 Ln
¶
(2)
n=0
Moreover, the radius of convergence of any power series solutionfhe form given by (2) is at least as large as the
distance from x0 to the nearest singular pont (real or complex-valued) of equation (1)
ü Example 8.4.1
Find a minimum value for the radius of convergence of a power series solution about
x0 = 0.
Ix2 - 5 x + 6M y'' - 3 x y ' - y = 0
CH_08_notes.nb
ü Example 8.4.2
Find at least the first four nonzero terms in a power series expansion about x = 2 for a general solution to the given
differential equation:
x2 y '' - x y' + 2 y = 0;
yH0L = -1, y ' H0L = 0
9
10
CH_08_notes.nb
8.6 Method of Frobenius
Homework: p. 465 - 466 #
ü Introduction
Definition 3 - Regular Singular Points
A singular point x0 of
(7)
y'' HxL + pHxL y' HxL + qHxL yHxL = 0
is said to be a regular singular point if both Hx - x0 L pHxL and Hx - x0 L2 qHxL are analytic at x0 . Otherwise x0 is
called an irregular singular point.
ü Example 8.6.1
Classify each singular point (real or complex) of the given equation as regular or irregular.
Ix2 - 4M y '' + 3 Hx - 2L y ' + 5 y = 0
2
Let us assume that x = 0 is a regular singular point of
y '' HxL + pHxL y' HxL + qHxL yHxL = 0.
Then x pHxL and x qHxL are analytic functions at x = 0. Which means that in some open interval about x = 0
2
x pHxL = p0 + p1 x + p2 x2 + ∫ = ⁄ pn xn
¶
(4)
n=0
x2 qHxL = q0 + q1 x + q2 x2 + ∫ = ⁄ qn xn
¶
(5)
n=0
One result of (4) and (5) is
(6)
limxØ0 x pHxL = p0
and
Note that (4) and (5) can be rewritten as follows:
pHxL = ⁄ pn xn-1
¶
(9)
n=0
limxØ0 x2 qHxL = q0
qHxL = ⁄ qn xn-2
¶
and
n=0
CH_08_notes.nb
11
Frobenius (inspired by the solutions to Cauchy-Euler equations) considered solutions of
y '' HxL + pHxL y' HxL + qHxL yHxL = 0
at the regular singular point x = 0 to have the form:
wHr, xL = xr ⁄ an xn = ⁄ an xn+r ,
(10)
¶
¶
n=0
n=0
x>0
Differentiating w with respect to x gives
w ' Hr, xL = ⁄ Hn + rL an xn+r-1
¶
(11)
n=0
w '' Hr, xL = ⁄ Hn + rL Hn + r - 1L an xn+r-2
¶
(12)
Substituting these (9), (10), (11), and (12) into y '' HxL + pHxL y' HxL + qHxL yHxL = 0 are regrouping terms by powers of x
we obtain
n=0
(14)
@rHr - 1L + p0 r + q0 D a0 xr-2 + @Hr + 1L r a1 + p0 Hr + 1L a1 + p1 r a0 + q0 a1 + q1 a0 D xr-1 + ∫ = 0
Since each coefficient is equal to zero, we consider the equation
(15)
@rHr - 1L + p0 r + q0 D a0 = 0
Definition 4 - Indicial Equation
If x0 is a regular sinular point of y '' + p y ' + q y = 0, then the indicial equation for this point is
(16)
rHr - 1L + p0 r + q0
where
and
q0 := limxØx0 Hx - x0 L2 qHxL
p0 := limxØx0 Hx - x0 L pHxL
The roots of the indicial equation are called the exponents (indices) of the singularity x0 .
ü Example 8.6.2
Find the indicial equation and the exponents for the singularity x = 0 of 3 x y '' + y ' - y = 0.
To find the coefficients we will be looking for a recursive relationship as before. To do this, we will use the larger root
of the indicial equation.
12
CH_08_notes.nb
ü Example 8.6.3
Find a series expansion about the regular singular point x = 0 for a solution to 3 x y '' + y ' - y = 0.
CH_08_notes.nb
Method of Frobenius
To derive a series solution about the singular point x0 of
a2 HxL y'' HxL + a1 HxL y ' HxL + a0 HxL yHxL = 0,
(29)
a1 HxL
,
a2 HxL
a0 HxL
.
a2 HxL
Set pHxL :=
(b)
singular point and the remaining steps apply.
Let
(c)
(d)
(e)
(f)
x > x0
If both Hx - x0 L pHxL and Hx - x0 L2 qHxL are analytic at x0 , then x0 is a regular
(a)
(30)
qHxL :=
13
wHr, xL = Hx - x0 Lr ⁄ an Hx - x0 Ln = ⁄ an Hx - x0 Ln+r
¶
¶
n=0
n=0
and, using termwise differentiation, substitute wHr, xL into equation (29) to obtain an equation of the form
A0 Hx - x0 Lr+J + A1 Hx - x0 Lr+J +1 + ∫ = 0
Set the coefficients A0 , A1 , A2 , . . . equal to zero. [Notice that the equation A0 = 0 is just a constant multiple
of the indicial equation rHr - 1L + p0 r + q0 = 0.]
Use the system of equations
A0 = 0, A1 = 0, A2 = 0, . . .
to find the recurrence relation involving ak and a0 , a1 , . . ., ak-1 .
Take r = r1 , the larger root of the indicial equation and use the relation obtained in step (d) to determine a1 ,
a2 , . . . , recursively in terms of a0 and r1 .
A series expansion of a solution to (29) is
wHr1 , xL = Hx - x0 Lr1 ⁄ an Hx - x0 Ln ,
¶
(31)
x > x0
n=0
where a0 is arbitrary and the an ’s are defined in terms of a0 and r1 .
What is the radius of convergence of the power series in (31)?
Theorem 6 - Frobenious’s Theorem
If x0 is a regular sinular point of equation (29), then there exists at least one series solution of the form (30), where
r = r1 is the larger root of the associated indicial equation. Moreover, this series conrges for all x such that
0 < x - x0 < R, where R is the distance from x0 to the nearest other singular point (real or complex) of (29).
For simplicity, will consider only expansions about the regular singular point x = 0 and only those equations for which
the associated indicial equation has real roots.
14
CH_08_notes.nb
ü Example 8.6.4
Find a series expansion about the regular singular point x = 0 for a solution to x2 y '' - x y ' + H1 - xL y = 0, x > 0.
CH_08_notes.nb
15
8.7 Finding a Second Linearly Independent Solution
Homework: p. 474 - 476 #
ü Introduction
In the last section, we used the method of Frobenius to find one series solution, how do we find the second one, if it
exists?
Theorem 7 - Form a Second Linearly Independent Solution
Let x0 be a regular singular point for y '' + p y' + q y = 0 and let r1 and r2 be the roots of the associated indicial
equation, where Re r1 ¥ Re r2 .
(a)
If r1 - r2 is not an integer, then there exist two linearly independent solutions of the form
y1 HxL = ⁄ an Hx - x0 Ln+r1 ,
¶
(11)
a0 ∫ 0
y2 HxL = ⁄ bn Hx - x0 Ln+r2 ,
n=0
¶
(12)
b0 ∫ 0
n=0
(b)
If r1 = r2 , then there exist two linearly independent solutions of the form
y1 HxL = ⁄ an Hx - x0 Ln+r1 ,
¶
(13)
a0 ∫ 0
y2 HxL = y1 HxL lnHx - x0 L + ⁄ bn Hx - x0 Ln+r1
n=0
¶
(14)
n=1
(c)
If r1 - r2 is a positive integer, then there exist two linearly independent solutions of the form
y1 HxL = ⁄ an Hx - x0 Ln+r1 ,
¶
(15)
a0 ∫ 0
y2 HxL = C y1 HxL lnHx - x0 L + ⁄ bn Hx - x0 Ln+r2 ,
n=0
¶
(16)
b0 ∫ 0
n=0
where C is a constant that could be zero.
ü Example 8.7.1
Find the first few terms in the series expansion about the regular singular point x = 0 for two linearly independent
solutions to x2 y'' - x y ' + H1 - xL y = 0, x > 0.