6.3 Step Functions
C H A P T E R
5
Finding the general solution of a linear differential equation rests on determining a fundamental set of solutions of the homogeneous equation. So far, we have given a systematic procedure for constructing fundamental solutions only if the equation has constant coefficients. To deal with the much larger class of equations having variable coefficients it is necessary to extend our search for solutions beyond the familiar elementary functions of calculus. The principal tool that we need is the representation of a given function by a power series. The basic idea is similar to that in the method of undetermined coefficients: We assume that the solutions of a given differential equation have power series expansions, and then we attempt to determine the coefficients so as to satisfy the differential equation.
5.1 Review of Power Series
In this chapter we discuss the use of power series to construct fundamental sets of solutions of second order linear differential equations whose coefficients are functions of the independent variable. We begin by summarizing very briefly the pertinent results about power series that we need. Readers who are familiar with power series may go on to Section 5.2. Those who need more details than are presented here should consult a book on calculus.
231
Chapter 5. Series Solutions of Second Order Linear Equations
∞
1.
A power series a (x − x )n is said to converge at a point x if
n
0
n=0
m
lim
a (x − x )n
m→∞
n
0
n=0
exists for that x. The series certainly converges for x = x ; it may converge for all
0
x, or it may converge for some values of x and not for others.
∞
2.
The series a (x − x )n is said to converge absolutely at a point x if the series n
0
n=0
∞
∞
|a (x − x )n| = |a ||x − x |n
n
0
n
0
n=0
n=0
converges. It can be shown that if the series converges absolutely, then the series also converges; however, the converse is not necessarily true.
3.
One of the most useful tests for the absolute convergence of a power series is the ratio test. If a = 0, and if for a fixed value of x
n
a(x − x )n+1 a
lim n+1
0
= |x − x | lim n+1
= L|x − x |,
n→∞
a (x − x )n
0 n→∞
a
0
n
0
n
then the power series converges absolutely at that value of x if |x − x | < 1/L, 0
and diverges if |x − x | > 1/L. If |x − x | = 1/L, the test is inconclusive.
0
0
For which values of x does the power series E X A M P L E ∞
1
(−1)n+1n(x − 2)n
n=1
converge?
To test for convergence we use the ratio test. We have
(−
1)n+2(n + 1)(x − 2)n+1 n + 1
lim
= |x − 2| lim = |x − 2|.
n→∞ (−1)n+1n(x − 2)n
n→∞
n
According to statement 3 the series converges absolutely for |x − 2| < 1, or 1 <x < 3, and diverges for |x − 2| > 1. The values of x corresponding to |x − 2| = 1 are x = 1 and x = 3. The series diverges for each of these values of x since the nth term of the series does not approach zero as n → ∞.
∞
4.
If the power series a (x − x )n converges at x = x , it converges absolutely
n
0
1
n=0
for |x − x | < |x − x |; and if it diverges at x = x , it diverges for |x − x | >
0
1
0
1
0
|x − x |.
1
0
5.
There is a nonnegative number ρ, called the such that
∞
a (x − x )n convergesabsolutelyfor|x − x | < ρ anddivergesfor|x − x | >
n
0
0
0
n=0
ρ. For a series that converges only at x , we define ρ to be zero; for a series that
0
converges for all x, we say that ρ is infinite. If ρ > 0, then the interval |x − x | < ρ
0
is called the interval of convergence; it is indicated by the hatched lines in Figure 5.1.1. The series may either converge or diverge when |x − x | = ρ.
0
Series
Series
Series
converges
diverges
diverges absolutely
x
x
x
0 – ρ
ρ
0
x0 +
Series may converge or diverge FIGURE 5.1.1
The interval of convergence of a power series.
Determine the radius of convergence of the power series E X A M P L E ∞
2
(x + 1)n .
n2n
n=1
We apply the ratio test:
(
x + 1)n+1
n2n
|x + 1|
n
|x + 1|
lim
=
lim
=
.
n→∞ (n + 1)2n+1 (x + 1)n
2
n→∞ n + 1
2
Thus the series converges absolutely for |x + 1| < 2, or −3 < x < 1, and diverges for |x + 1| > 2. The radius of convergence of the power series is ρ = 2. Finally, we check the endpoints of the interval of convergence. At x = 1 the series becomes the harmonic series
∞
1,n
n=1
which diverges. At x = −3 we have
∞
(−3 + 1)n
∞
(−
=
1)n ,
n2n
n
n=1
n=1
which converges, but does not converge absolutely. The series is said to converge conditionally at x = −3. To summarize, the given power series converges for −3 ≤ x < 1, and diverges otherwise. It converges absolutely for −3 < x < 1, and has a radius of convergence 2.
∞
∞
If a (x − x )n and b (x − x )n converge to f (x) and g(x), respectively, n
0
n
0
n=0
n=0
for |x − x | < ρ, ρ > 0, then the following are true for |x − x | < ρ.
0
0
6.
The series can be added or subtracted termwise and
∞
f (x) ± g(x) = (a ± b )(x − x )n.
n
n
0
n=0
Chapter 5. Series Solutions of Second Order Linear Equations7.
The series can be formally multiplied and
∞
∞
∞
f (x)g(x) = a (x − x )nb (x − x )n = c (x − x )n,
n
0
n
0
n
0
n=0
n=0
n=0
where c = a b + a b + · · · + a b . Further, if g(x ) = 0, the series can be
n 0 n
1 n−1
n 0
0
formally divided and f (x)
∞
=
d (x − x )n.
g(x)n
0
n=0
In most cases the coefficients d can be most easily obtained by equating coeffi n
cients in the equivalent relation
∞
∞
∞
a (x − x )n = d (x − x )nb (x − x )n
n
0
n
0
n
0
n=0
n=0
n=0
∞
n
=
d b(x − x )n.
k n−k
0
n=0
k=0
Also, in the case of division, the radius of convergence of the resulting power series may be less than ρ.
8.
The function f is continuous and has derivatives of all orders for |x − x | < ρ.
0
Further, f , f , . . . can be computed by differentiating the series termwise; that is, f (x) = a + 2a (x − x ) + · · · + na (x − x )n−1 + · · ·
1
2
0
n
0
∞
=
na (x − x )n−1,
n
0
n=1
f (x) = 2a + 6a (x − x ) + · · · + n(n − 1)a (x − x )n−2 + · · ·
2
3
0
n
0
∞
=
n(n − 1)a (x − x )n−2,
n
0
n=2
and so forth, and each of the series converges absolutely for |x − x | < ρ.
0
9.
The value of a is given by
n
f (n)(x )a =
0 .
n
n!
The series is called the Taylor1 series for the function f about x = x .
0
∞
∞
10.
If a (x − x )n = b (x − x )n for each x, then a = b for n = 0, 1,
n
0
n
0
n
n
n=0 n=0 ∞
2, 3, . . . . In particular, if a (x − x )n = 0 for each x, then a = a = · · · =
n
0
0
1
n=0
a = · · · = 0.
n 1Brook Taylor (1685 –1731) was the leading English mathematician in the generation following Newton. In 1715 he published a general statement of the expansion theorem that is named for him, a result that is fundamental in all branches of analysis. He was also one of the founders of the calculus of finite differences, and was the first to recognize the existence of singular solutions of differential equations.
5.1Review of Power Series
A function f that has a Taylor series expansion about x = x , 0
∞
f (n)(x )f (x) = 0 (x − x )n,n!
0
n=0
with a radius of convergence ρ > 0, is said to be x = x . According to
0
statements 6 and 7, if f and g are analytic at x , then f ± g, f · g, and f /g [provided
0
that g(x ) = 0] are analytic at x = x .
0
0
Shift of Index of Summation.
The index of summation in an infinite series is a dummy parameter just as the integration variable in a definite integral is a dummy variable.
Thus it is immaterial which letter is used for the index of summation. For example,
∞
2nxn
∞
=
2 j x j .
n!
j !
n=0 j =0 Just as we make changes of the variable of integration in a definite integral, we find it convenient to make changes of summation indices in calculating series solutions of differential equations. We illustrate by several examples how to shift the summation index.
∞
Write a xn as a series whose first term corresponds to n = 0 rather than n = 2.
n
E X A M P L En=2
3
Let m = n − 2; then n = m + 2 and n = 2 corresponds to m = 0. Hence
∞
∞
a xn = a
xm+2.
(1)
n
m+2
n=2
m=0
By writing out the first few terms of each of these series, you can verify that they contain precisely the same terms. Finally, in the series on the right side of Eq. (1), we can replace the dummy index m by n, obtaining
∞
∞
a xn = a
xn+2.
(2)
n
n+2
n=2
n=0
In effect, we have shifted the index upward by 2, and compensated by starting to count at a level 2 lower than originally.
Write the series E X A M P L E ∞
4
(n + 2)(n + 1)a (x − x )n−2
(3)
n
0
n=2
as a series whose generic term involves (x − x )n rather than (x − x )n−2.
0
0
Again, we shift the index by 2 so that n is replaced by n + 2 and start counting 2 lower. We obtain
∞
(n +4)(n +3)a (x − x )n.
(4)
n+2
0
n=0
You can readily verify that the terms in the series (3) and (4) are exactly the same.
Chapter 5. Series Solutions of Second Order Linear Equations
Write the expression E X A M P L E ∞
5
x2 (r + n)a xr+n−1
(5)
n
n=0
as a series whose generic term involves xr+n.
First take the x2 inside the summation, obtaining
∞
(r +n)a xr+n+1.
(6)
n
n=0
Next, shift the index down by 1 and start counting 1 higher. Thus
∞
∞
(r + n)a xr+n+1 = (r + n − 1)axr+n.
(7)
n
n−1
n=0
n=1
Again, you can easily verify that the two series in Eq. (7) are identical, and that both are exactly the same as the expression (5).
Assume that E X A M P L E ∞
∞
6
na xn−1 = a xn (8)
n
n
n=1
n=0
for all x, and determine what this implies about the coefficients a .
n
We want to use statement 10 to equate corresponding coefficients in the two series.
In order to do this, we must first rewrite Eq. (8) so that the series display the same power of x in their generic terms. For instance, in the series on the left side of Eq. (8), we can replace n by n + 1 and start counting 1 lower. Thus Eq. (8) becomes
∞
∞
(n + 1)axn = a xn.
(9)
n+1
n
n=0
n=0
According to statement 10 we conclude that (n + 1)a = a ,n = 0, 1, 2, 3, . . .
n+1
n
or
a
a
=
n
,n = 0, 1, 2, 3, . . .
(10)
n+1
n + 1
Hence, choosing successive values of n in Eq. (10), we have
a
a
a
aa = a ,a = 1 = 0 ,a = 2 = 0 , 1
0
2
2
2
3
3
3!
and so forth. In general, aa = 0 ,n = 1, 2, 3, . . . .
(11)
n
n!
Thus the relation (8) determines all the following coefficients in terms of a . Finally,
0
using the coefficients given by Eq. (11), we obtain
∞
∞
xna xn = a = a ex,
n
0
n!
0
n=0
n=0
where we have followed the usual convention that 0! = 1.
5.1Review of Power Series
PROBLEMS
In each of Problems 1 through 8 determine the radius of convergence of the given power series.
∞
∞
1.
2.
xn
∞
∞
3.
4.
∞
∞
5.
6.
0
n
∞
∞
7.
8.
nn
0
function. Also determine the radius of convergence of the series.
0
0
0
0
1
14.
,
0
0
1
1
15.
,
16.
,
0
0
∞
∞
n
0
1
2
3
0
1
n
In each of Problems 19 and 20 verify the given equation.
∞
∞
19.
a
n
∞
∞
∞
20.
a
(a
)xk
k
1
∞
∞
21.
22.
n
n
∞
∞
∞
a xk
n
k
n
∞
∞
∞
∞
25.
26.
a xn m
k
n
n
∞
∞
a xn n
n
5
Finding the general solution of a linear differential equation rests on determining a fundamental set of solutions of the homogeneous equation. So far, we have given a systematic procedure for constructing fundamental solutions only if the equation has constant coefficients. To deal with the much larger class of equations having variable coefficients it is necessary to extend our search for solutions beyond the familiar elementary functions of calculus. The principal tool that we need is the representation of a given function by a power series. The basic idea is similar to that in the method of undetermined coefficients: We assume that the solutions of a given differential equation have power series expansions, and then we attempt to determine the coefficients so as to satisfy the differential equation.
5.1 Review of Power Series
In this chapter we discuss the use of power series to construct fundamental sets of solutions of second order linear differential equations whose coefficients are functions of the independent variable. We begin by summarizing very briefly the pertinent results about power series that we need. Readers who are familiar with power series may go on to Section 5.2. Those who need more details than are presented here should consult a book on calculus.
231
Chapter 5. Series Solutions of Second Order Linear Equations
∞
1.
A power series a (x − x )n is said to converge at a point x if
n
0
n=0
m
lim
a (x − x )n
m→∞
n
0
n=0
exists for that x. The series certainly converges for x = x ; it may converge for all
0
x, or it may converge for some values of x and not for others.
∞
2.
The series a (x − x )n is said to converge absolutely at a point x if the series n
0
n=0
∞
∞
|a (x − x )n| = |a ||x − x |n
n
0
n
0
n=0
n=0
converges. It can be shown that if the series converges absolutely, then the series also converges; however, the converse is not necessarily true.
3.
One of the most useful tests for the absolute convergence of a power series is the ratio test. If a = 0, and if for a fixed value of x
n
a(x − x )n+1 a
lim n+1
0
= |x − x | lim n+1
= L|x − x |,
n→∞
a (x − x )n
0 n→∞
a
0
n
0
n
then the power series converges absolutely at that value of x if |x − x | < 1/L, 0
and diverges if |x − x | > 1/L. If |x − x | = 1/L, the test is inconclusive.
0
0
For which values of x does the power series E X A M P L E ∞
1
(−1)n+1n(x − 2)n
n=1
converge?
To test for convergence we use the ratio test. We have
(−
1)n+2(n + 1)(x − 2)n+1 n + 1
lim
= |x − 2| lim = |x − 2|.
n→∞ (−1)n+1n(x − 2)n
n→∞
n
According to statement 3 the series converges absolutely for |x − 2| < 1, or 1 <x < 3, and diverges for |x − 2| > 1. The values of x corresponding to |x − 2| = 1 are x = 1 and x = 3. The series diverges for each of these values of x since the nth term of the series does not approach zero as n → ∞.
∞
4.
If the power series a (x − x )n converges at x = x , it converges absolutely
n
0
1
n=0
for |x − x | < |x − x |; and if it diverges at x = x , it diverges for |x − x | >
0
1
0
1
0
|x − x |.
1
0
5.
There is a nonnegative number ρ, called the such that
∞
a (x − x )n convergesabsolutelyfor|x − x | < ρ anddivergesfor|x − x | >
n
0
0
0
n=0
ρ. For a series that converges only at x , we define ρ to be zero; for a series that
0
converges for all x, we say that ρ is infinite. If ρ > 0, then the interval |x − x | < ρ
0
is called the interval of convergence; it is indicated by the hatched lines in Figure 5.1.1. The series may either converge or diverge when |x − x | = ρ.
0
Series
Series
Series
converges
diverges
diverges absolutely
x
x
x
0 – ρ
ρ
0
x0 +
Series may converge or diverge FIGURE 5.1.1
The interval of convergence of a power series.
Determine the radius of convergence of the power series E X A M P L E ∞
2
(x + 1)n .
n2n
n=1
We apply the ratio test:
(
x + 1)n+1
n2n
|x + 1|
n
|x + 1|
lim
=
lim
=
.
n→∞ (n + 1)2n+1 (x + 1)n
2
n→∞ n + 1
2
Thus the series converges absolutely for |x + 1| < 2, or −3 < x < 1, and diverges for |x + 1| > 2. The radius of convergence of the power series is ρ = 2. Finally, we check the endpoints of the interval of convergence. At x = 1 the series becomes the harmonic series
∞
1,n
n=1
which diverges. At x = −3 we have
∞
(−3 + 1)n
∞
(−
=
1)n ,
n2n
n
n=1
n=1
which converges, but does not converge absolutely. The series is said to converge conditionally at x = −3. To summarize, the given power series converges for −3 ≤ x < 1, and diverges otherwise. It converges absolutely for −3 < x < 1, and has a radius of convergence 2.
∞
∞
If a (x − x )n and b (x − x )n converge to f (x) and g(x), respectively, n
0
n
0
n=0
n=0
for |x − x | < ρ, ρ > 0, then the following are true for |x − x | < ρ.
0
0
6.
The series can be added or subtracted termwise and
∞
f (x) ± g(x) = (a ± b )(x − x )n.
n
n
0
n=0
Chapter 5. Series Solutions of Second Order Linear Equations7.
The series can be formally multiplied and
∞
∞
∞
f (x)g(x) = a (x − x )nb (x − x )n = c (x − x )n,
n
0
n
0
n
0
n=0
n=0
n=0
where c = a b + a b + · · · + a b . Further, if g(x ) = 0, the series can be
n 0 n
1 n−1
n 0
0
formally divided and f (x)
∞
=
d (x − x )n.
g(x)n
0
n=0
In most cases the coefficients d can be most easily obtained by equating coeffi n
cients in the equivalent relation
∞
∞
∞
a (x − x )n = d (x − x )nb (x − x )n
n
0
n
0
n
0
n=0
n=0
n=0
∞
n
=
d b(x − x )n.
k n−k
0
n=0
k=0
Also, in the case of division, the radius of convergence of the resulting power series may be less than ρ.
8.
The function f is continuous and has derivatives of all orders for |x − x | < ρ.
0
Further, f , f , . . . can be computed by differentiating the series termwise; that is, f (x) = a + 2a (x − x ) + · · · + na (x − x )n−1 + · · ·
1
2
0
n
0
∞
=
na (x − x )n−1,
n
0
n=1
f (x) = 2a + 6a (x − x ) + · · · + n(n − 1)a (x − x )n−2 + · · ·
2
3
0
n
0
∞
=
n(n − 1)a (x − x )n−2,
n
0
n=2
and so forth, and each of the series converges absolutely for |x − x | < ρ.
0
9.
The value of a is given by
n
f (n)(x )a =
0 .
n
n!
The series is called the Taylor1 series for the function f about x = x .
0
∞
∞
10.
If a (x − x )n = b (x − x )n for each x, then a = b for n = 0, 1,
n
0
n
0
n
n
n=0 n=0 ∞
2, 3, . . . . In particular, if a (x − x )n = 0 for each x, then a = a = · · · =
n
0
0
1
n=0
a = · · · = 0.
n 1Brook Taylor (1685 –1731) was the leading English mathematician in the generation following Newton. In 1715 he published a general statement of the expansion theorem that is named for him, a result that is fundamental in all branches of analysis. He was also one of the founders of the calculus of finite differences, and was the first to recognize the existence of singular solutions of differential equations.
5.1Review of Power Series
A function f that has a Taylor series expansion about x = x , 0
∞
f (n)(x )f (x) = 0 (x − x )n,n!
0
n=0
with a radius of convergence ρ > 0, is said to be x = x . According to
0
statements 6 and 7, if f and g are analytic at x , then f ± g, f · g, and f /g [provided
0
that g(x ) = 0] are analytic at x = x .
0
0
Shift of Index of Summation.
The index of summation in an infinite series is a dummy parameter just as the integration variable in a definite integral is a dummy variable.
Thus it is immaterial which letter is used for the index of summation. For example,
∞
2nxn
∞
=
2 j x j .
n!
j !
n=0 j =0 Just as we make changes of the variable of integration in a definite integral, we find it convenient to make changes of summation indices in calculating series solutions of differential equations. We illustrate by several examples how to shift the summation index.
∞
Write a xn as a series whose first term corresponds to n = 0 rather than n = 2.
n
E X A M P L En=2
3
Let m = n − 2; then n = m + 2 and n = 2 corresponds to m = 0. Hence
∞
∞
a xn = a
xm+2.
(1)
n
m+2
n=2
m=0
By writing out the first few terms of each of these series, you can verify that they contain precisely the same terms. Finally, in the series on the right side of Eq. (1), we can replace the dummy index m by n, obtaining
∞
∞
a xn = a
xn+2.
(2)
n
n+2
n=2
n=0
In effect, we have shifted the index upward by 2, and compensated by starting to count at a level 2 lower than originally.
Write the series E X A M P L E ∞
4
(n + 2)(n + 1)a (x − x )n−2
(3)
n
0
n=2
as a series whose generic term involves (x − x )n rather than (x − x )n−2.
0
0
Again, we shift the index by 2 so that n is replaced by n + 2 and start counting 2 lower. We obtain
∞
(n +4)(n +3)a (x − x )n.
(4)
n+2
0
n=0
You can readily verify that the terms in the series (3) and (4) are exactly the same.
Chapter 5. Series Solutions of Second Order Linear Equations
Write the expression E X A M P L E ∞
5
x2 (r + n)a xr+n−1
(5)
n
n=0
as a series whose generic term involves xr+n.
First take the x2 inside the summation, obtaining
∞
(r +n)a xr+n+1.
(6)
n
n=0
Next, shift the index down by 1 and start counting 1 higher. Thus
∞
∞
(r + n)a xr+n+1 = (r + n − 1)axr+n.
(7)
n
n−1
n=0
n=1
Again, you can easily verify that the two series in Eq. (7) are identical, and that both are exactly the same as the expression (5).
Assume that E X A M P L E ∞
∞
6
na xn−1 = a xn (8)
n
n
n=1
n=0
for all x, and determine what this implies about the coefficients a .
n
We want to use statement 10 to equate corresponding coefficients in the two series.
In order to do this, we must first rewrite Eq. (8) so that the series display the same power of x in their generic terms. For instance, in the series on the left side of Eq. (8), we can replace n by n + 1 and start counting 1 lower. Thus Eq. (8) becomes
∞
∞
(n + 1)axn = a xn.
(9)
n+1
n
n=0
n=0
According to statement 10 we conclude that (n + 1)a = a ,n = 0, 1, 2, 3, . . .
n+1
n
or
a
a
=
n
,n = 0, 1, 2, 3, . . .
(10)
n+1
n + 1
Hence, choosing successive values of n in Eq. (10), we have
a
a
a
aa = a ,a = 1 = 0 ,a = 2 = 0 , 1
0
2
2
2
3
3
3!
and so forth. In general, aa = 0 ,n = 1, 2, 3, . . . .
(11)
n
n!
Thus the relation (8) determines all the following coefficients in terms of a . Finally,
0
using the coefficients given by Eq. (11), we obtain
∞
∞
xna xn = a = a ex,
n
0
n!
0
n=0
n=0
where we have followed the usual convention that 0! = 1.
5.1Review of Power Series
PROBLEMS
In each of Problems 1 through 8 determine the radius of convergence of the given power series.
∞
∞
1.
2.
xn
∞
∞
3.
4.
∞
∞
5.
6.
0
n
∞
∞
7.
8.
nn
0
function. Also determine the radius of convergence of the series.
0
0
0
0
1
14.
,
0
0
1
1
15.
,
16.
,
0
0
∞
∞
n
0
1
2
3
0
1
n
In each of Problems 19 and 20 verify the given equation.
∞
∞
19.
a
n
∞
∞
∞
20.
a
(a
)xk
k
1
∞
∞
21.
22.
n
n
∞
∞
∞
a xk
n
k
n
∞
∞
∞
∞
25.
26.
a xn m
k
n
n
∞
∞
a xn n
n