6.4 Differential Equations with Discontinuous Forcing Functions
Chapter 5. Series Solutions of Second Order Linear Equations
n
∞
∞
n
n
∞
is satisfied. Try to identify the function represented by the series n
5.2 Series Solutions near an Ordinary Point, Part I In Chapter 3 we described methods of solving second order linear differential equations with constant coefficients. We now consider methods of solving second order linear
ODE
equations when the coefficients are functions of the independent variable. In this chapter we will denote the independent variable by x. It is sufficient to consider the homogeneous equation P(x) d2 y + Q(x) dy + R(x)y = 0, (1)
d x2
d x since the procedure for the corresponding nonhomogeneous equation is similar.
A wide class of problems in mathematical physics leads to equations of the form (1) having polynomial coefficients; for example, the Bessel equation x2 y + x y + (x2 − ν2)y = 0, where ν is a constant, and the Legendre equation (1 − x2)y − 2xy + α(α + 1)y = 0, where α is a constant. For this reason, as well as to simplify the algebraic computations, we primarily consider the case in which the functions P, Q, and R are polynomials.
However, as we will see, the method of solution is also applicable when P, Q, and R
are general analytic functions.
For the present, then, suppose that P, Q, and R are polynomials, and that they have no common factors. Suppose also that we wish to solve Eq. (1) in the neighborhood of a point x . The solution of Eq. (1) in an interval containing x is closely associated
0
0
with the behavior of P in that interval.
A point x such that P(x ) = 0 is called aince P is continuous, 0
0
it follows that there is an interval about x in which P(x) is never zero. In that interval
0
we can divide Eq. (1) by P(x) to obtain y + p(x)y + q(x)y = 0, (2)
where p(x) = Q(x)/P(x) and q(x) = R(x)/P(x) are continuous functions. Hence, according to the existence and uniqueness Theorem 3.2.1, there exists in that interval a unique solution of Eq. (1) that also satisfies the initial conditions y(x ) = y , y(x ) =
0
0
0
y for arbitrary values of y and y . In this and the following section we discuss the
0
0
0
solution of Eq. (1) in the neighborhood of an ordinary point.
On the other hand, if P(x ) = 0, then x is called a singular point of Eq. (1). In
0
0
this case at least one of Q(x ) and R(x ) is not zero. Consequently, at least one of the
0
0
coefficients p and q in Eq. (2) becomes unbounded as x → x , and therefore Theorem
0
3.2.1 does not apply in this case. Sections 5.4 through 5.8 deal with finding solutions of Eq. (1) in the neighborhood of a singular point.
We now take up the problem of solving Eq. (1) in the neighborhood of an ordinary point x . We look for solutions of the form
0
∞
y = a + a (x − x ) + · · · + a (x − x )n + · · · = a (x − x )n, (3)
0
1
0
n
0
n
0
n=0
and assume that the series converges in the interval |x − x | < ρ for some ρ > 0.
0
While at first sight it may appear unattractive to seek a solution in the form of a power series, this is actually a convenient and useful form for a solution. Within their intervals of convergence power series behave very much like polynomials and are easy to manipulate both analytically and numerically. Indeed, even if we can obtain a solution in terms of elementary functions, such as exponential or trigonometric functions, we are likely to need a power series or some equivalent expression if we want to evaluate them numerically or to plot their graphs.
The most practical way to determine the coefficients a is to substitute the series n
(3) and its derivatives for y, y, and y in Eq. (1). The following examples illustrate this process. The operations, such as differentiation, that are involved in the procedure are justified so long as we stay within the interval of convergence. The differential equations in these examples are also of considerable importance in their own right.
Find a series solution of the equation E X A M P L E
1
y + y = 0, −∞ < x < ∞.
(4)
As we know, two linearly independent solutions of this equation are sin x and cos x, so series methods are not needed to solve this equation. However, this example illustrates the use of power series in a relatively simple case. For Eq. (4), P(x) = 1, Q(x) = 0, and R(x) = 1; hence every point is an ordinary point.
We look for a solution in the form of a power series about x = 0, 0
∞
y = a + a x + a x2 + · · · + a xn + · · · = a xn,
(5)
0
1
2
n
n
n=0
and assume that the series converges in some interval |x| < ρ.
Differentiating term by term yields
∞
y = a + 2a x + · · · + na xn−1 + · · · = na xn−1,
(6)
1
2
n
n
n=1
∞
y = 2a + · · · + n(n − 1)a xn−2 + · · · = n(n − 1)a xn−2.
(7)
2
n
n
n=2
Substituting the series (5) and (7) for y and y in Eq. (4) gives
∞
∞
n(n − 1)a xn−2 + a xn = 0.
n
n
n=2
n=0
Chapter 5. Series Solutions of Second Order Linear Equations
To combine the two series we need to rewrite at least one of them so that both series display the same generic term. Thus, in the first sum, we shift the index of summation by replacing n by n + 2 and starting the sum at 0 rather than 2. We obtain
∞
∞
(n + 2)(n + 1)axn + a xn = 0
n+2
n
n=0
n=0
or
∞
[(n +2)(n +1)a +a ]xn = 0.
n+2
n
n=0
For this equation to be satisfied for all x, the coefficient of each power of x must be zero; hence, we conclude that (n + 2)(n + 1)a + a = 0,n = 0, 1, 2, 3, . . . .
(8)
n+2
n
Equation (8) is referred to as a The successive coefficients can be evaluated one by one by writing the recurrence relation first for n = 0, then for n = 1, and so forth. In this example Eq. (8) relates each coefficient to the second one before it. Thus the even-numbered coefficients (a , a , a , . . .) and the odd-numbered
0
2
4
ones (a , a , a , . . .) are determined separately. For the even-numbered coefficients we
1
3
5
have
a
a
a
a
a
aa = − 0 = − 0 ,a = − 2 = + 0 ,a = − 4 = − 0 , . . . .
2
2 · 1 2!
4
4 · 3 4!
6
6 · 5 6!
These results suggest that in general, if n = 2k, then (−1)ka = a = a ,k = 1, 2, 3, . . . .
(9)
n
2k(2k)! 0
We can prove Eq. (9) by mathematical induction. First observe that it is true for k = 1.
Next, assume that it is true for an arbitrary value of k and consider the case k + 1. We have a(−1)k(−1)k+1
a
= −
2k
= −
a = a .
2k+2 (2k + 2)(2k + 1)(2k + 2)(2k + 1)(2k)! 0 (2k + 2)! 0
Hence Eq. (9) is also true for k + 1 and consequently it is true for all positive integers k.
Similarly, for the odd-numbered coefficients a
a
a
a
a
aa = − 1 = − 1 ,a = − 3 = + 1 ,a = − 5 = − 1 , . . . ,
3
2 · 3 3!
5
5 · 4 5!
7
7 · 6 7!
and in general, if n = 2k + 1, then2 (−1)ka = a
=
a ,k = 1, 2, 3, . . . .
(10)
n
2k+1 (2k + 1)! 1
Substituting these coefficients into Eq. (5), we have
a
a
a
ay = a + a x − 0 x2 − 1 x3 + 0 x4 + 1 x5 +
0
1
2!
3!
4!
5!
2The result given in Eq. (10) and other similar formulas in this chapter can be proved by an induction argument resembling the one just given for Eq. (9). We assume that the results are plausible and omit the inductive argument hereafter.
5.2Series Solutions near an Ordinary Point, Part I(−1)na(−1)na + · · · +
0
1
(
x2n + x2n+1 + · · · 2n)!
(2n + 1)!
(−
=
1)n
a 1 − x2 + x4 + · · · + x2n + · · ·
0
2!
4!
(2n)!
(−
+
1)n
ax − x3 + x5 + · · · + x2n+1 + · · ·
1
3!
5!
(2n + 1)! ∞
(−
∞
(−
=
1)n 1)n
ax2n + ax2n+1.
(11)
0
(2n)!
1 (2n + 1)! n=0
n=0
Now that we have formally obtained two series solutions of Eq. (4), we can test them for convergence. Using the ratio test, it is easy to show that each of the series in Eq. (11)
converges for all x, and this justifies retroactively all the steps used in obtaining the solutions. Indeed, we recognize that the first series in Eq. (11) is exactly the Taylor series for cos x about x = 0 and the second is the Taylor series for sin x about x = 0.
Thus, as expected, we obtain the solution y = a cos x + a sin x.
0
1
Notice that no conditions are imposed on a and a ; hence they are arbitrary. From
0
1
Eqs. (5) and (6) we see that y and y evaluated at x = 0 are a and a , respectively.
0
1
Since the initial conditions y(0) and y(0) can be chosen arbitrarily, it follows that a0 and a should be arbitrary until specific initial conditions are stated.
1
Figures 5.2.1 and 5.2.2 show how the partial sums of the series in Eq. (11) ap proximate cos x and sin x. As the number of terms increases, the interval over which the approximation is satisfactory becomes longer, and for each x in this interval the accuracy of the approximation improves. However, you should always remember that a truncated power series provides only a local approximation of the solution in a neighborhood of the initial point x = 0; it cannot adequately represent the solution for large |x|.
y
n = 4 n = 8 n = 12 n = 16 n = 20
2
1
2
4
6
8
10
x
–1
y = cos x
–2
n = 2 n = 6 n = 10 n = 14 n = 18 FIGURE 5.2.1 Polynomial approximations to cos x. The value of n is the degree of the approximating polynomial.
Chapter 5. Series Solutions of Second Order Linear Equations
y
n = 5 n = 9 n = 13 n = 17 n = 21
2
1
2
4
6
8
10
x
–1
y = sin x
–2
n = 3 n = 7 n = 11 n = 15 n = 19
FIGURE 5.2.2 Polynomial approximations to sin x. The value of n is the degree of the approximating polynomial.
In Example 1 we knew from the start that sin x and cos x form a fundamental set of solutions of Eq. (4). However, if we had not known this and had simply solvusing series methods, we would still have obtained the solution (11). In recognition of the fact that the differential equation (4) often occurs in applications we might decide to give the two solutions of Eq. (11) special names; perhaps,
∞ (−1)n
∞ (−1)nC(x) =
(
x2n,S(x) = x2n+1.
2n)!
(2n + 1)! n=0
n=0
Then we might ask what properties these functions have. For instance, it follows at once from the series expansions that C(0) = 1, S(0) = 0, C(−x) = C(x), and S(−x) = −S(x). It is also easy to show that S(x) = C(x),C(x) = −S(x).
Moreover, by calculating with the infinite series3 we can show that the functions C(x)
and S(x) have all the usual analytical and algebraic properties of the cosine and sine functions, respectively.
Although you probably first saw the sine and cosine functions defined in a more elementary manner in terms of right triangles, it is interesting that these functions can be defined as solutions of a certain simple second order linear differential equation.
To be precise, the function sin x can be defined as the unique solution of the initial value problem y + y = 0, y(0) = 0, y(0) = 1; similarly, cos x can be defined as the unique solution of the initial value problem y + y = 0, y(0) = 1, y(0) = 0. Many other functions that are important in mathematical physics are also defined as solutions of certain initial value problems. For most of these functions there is no simpler or more elementary way to approach them.
3Such an analysis is given in Section 24 of K. Knopp, Theory and Applications of Infinite Series (New York: Hafner, 1951).
5.2Series Solutions near an Ordinary Point, Part I
Find a series solution in powers of x of Airy’s4 equation E X A M P L E
2
y − x y = 0, −∞ < x < ∞.
(12)
For this equation P(x) = 1, Q(x) = 0, and R(x) = −x; hence every point is an ordinary point. We assume that
∞
y = a xn,
(13)
n
n=0
and that the series converges in some interval |x| < ρ. The series for y is given by Eq. (7); as explained in the preceding example, we can rewrite it as
∞
y = (n + 2)(n + 1)axn.
(14)
n+2
n=0
Substituting the series (13) and (14) for y and y in Eq. (12), we obtain
∞
∞
∞
(n + 2)(n + 1)axn = xa xn = a xn+1.
(15)
n+2
n
n
n=0
n=0
n=0
Next, we shift the index of summation in the series on the right side of this equation by replacing n by n − 1 and starting the summation at 1 rather than zero. Thus we have
∞
∞
2 · 1a + (n + 2)(n + 1)axn =
axn.
2
n+2
n−1
n=1
n=1
Again, for this equation to be satisfied for all x it is necessary that the coefficients of like powers of x be equal; hence a = 0, and we obtain the recurrence relation
2
(n + 2)(n + 1)a = a for n = 1, 2, 3, . . . .
(16)
n+2
n−1
Since a is given in terms of a , the a’s are determined in steps of three. Thus
n+2
n−1
a determines a , which in turn determines a , . . . ; a determines a , which in turn
0
3
6
1
4
determines a , . . . ; and a determines a , which in turn determines a , . . . . Since
7
2
5
8
a = 0, we immediately conclude that a = a = a = · · · = 0.
2
5
8
11
For the sequence a , a , a , a , . . . we set n = 1, 4, 7, 10, . . . in the recurrence rela 0
3
6
9
tion:
a
a
a
a
aa = 0 ,a = 3
=
0
,a = 6
=
0
, . . . .
3
2 · 3 6
5 · 6 2 · 3 · 5 · 6
9
8 · 9 2 · 3 · 5 · 6 · 8 · 9
For this sequence of coefficients it is convenient to write a formula for a , n =
3n
1, 2, 3, . . . . The preceding results suggest the general formula a
a
=
0
,n = 1, 2, . . . .
3n
2 · 3 · 5 · 6 · · · (3n − 4)(3n − 3)(3n − 1)(3n) For the sequence a , a , a , a , . . . , we set n = 2, 5, 8, 11, . . . in the recurrence
1
4
7
10
relation:
a
a
a
a
aa = 1 , a =
4
=
1
, a =
7
=
1
, . . . .
4
3 · 4 7
6 · 7 3 · 4 · 6 · 7
10
9 · 10 3 · 4 · 6 · 7 · 9 · 10 4Sir George Airy (1801–1892), an English astronomer and mathematician, was director of the Greenwich Observatory from 1835 to 1881. One reason Airy’s equation is of interest is that for x negative the solutions are oscillatory, similar to trigonometric functions, and for x positive they are monotonic, similar to hyperbolic functions. Can you explain why it is reasonable to expect such behavior?
Chapter 5. Series Solutions of Second Order Linear Equations
In the same way as before we find that a
a
=
1
,n = 1, 2, 3, . . . .
3n+1 3 · 4 · 6 · 7 · · · (3n − 3)(3n − 2)(3n)(3n + 1) The general solution of Airy’s equation is y = a 1 + x3 +
x6
+ · · · +
x3n
+ · · ·
0
2 · 3 2 · 3 · 5 · 6 2 · 3 · · · (3n − 1)(3n)
+ a x + x4 +
x7
+ · · · +
x3n+1
+ · · ·
1
3 · 4 3 · 4 · 6 · 7 3 · 4 · · · (3n)(3n + 1)
∞
∞
=
x3n
x3n+1
a
1 + + a x + .
0
2 · 3 · · · (3n − 1)(3n)
1
3 · 4 · · · (3n)(3n + 1)
n=1
n=1
(17)
Having obtained these two series solutions, we can now investigate their convergence.
Because of the rapid growth of the denominators of the terms in the series (17), we might expect these series to have a large radius of convergence. Indeed, it is easy to use the ratio test to show that both these series converge for all x; see Problem 20.
Assuming for the moment that the series do converge for all x, let y and y denote
1
2
the functions defined by the expressions in the first and second sets of brackets, respectively, in Eq. (17). Then, by first choosing a = 1, a = 0 and then a = 0, a =
0
1
0
1
1, it follows that y and y are individually solutions of Eq. (12). Notice that y satisfies
1
2
1
the initial conditions y (0) = 1, y (0) = 0 and that y satisfies the initial conditions
1
1
2
y (0) = 0, y (0) = 1. Thus W (y , y )(0) = 1 = 0, and consequently y and y are
2
2
1
2
1
2
linearly independent. Hence the general solution of Airy’s equation is y = a y (x) + a y (x), −∞ < x < ∞.
0 1
1 2
In Figures 5.2.3 and 5.2.4, respectively, we show the graphs of the solutions y and
1
y of Airy’s equation, as well as graphs of several partial sums of the two series in
2
Eq. (17). Again, the partial sums provide local approximations to the solutions in a neighborhood of the origin. While the quality of the approximation improves as the number of terms increases, no polynomial can adequately represent y and y for large
1
2
|x|. A practical way to estimate the interval in which a given partial sum is reasonably accurate is to compare the graphs of that partial sum and the next one, obtained by including one more term. As soon as the graphs begin to separate noticeably, one can be confident that the original partial sum is no longer accurate. For example, in Figure 5.2.3 the graphs for n = 24 and n = 27 begin to separate at about x = −9/2. Thus, beyond this point, the partial sum of degree 24 is worthless as an approximation to the solution.
Observe that both y and y are monotone for x > 0 and oscillatory for x < 0.
1
2
One can also see from the figures that the oscillations are not uniform, but decay in amplitude and increase in frequency as the distance from the origin increases. In contrast to Example 1, the solutions y and y of Airy’s equation are not elementary functions
1
2
that you have already encountered in calculus. However, because of their importance in some physical applications, these functions have been extensively studied and their properties are well known to applied mathematicians and scientists.
5.2Series Solutions near an Ordinary Point, Part I
y
2
n = 3 n = 48
36
24
12
n ≥ 6
42
30
18
6
y = y1(x)
–10
– 8 – 6 –4
–2
2
x
39
27
15
n = 45
33
21
9
3
–2
FIGURE 5.2.3
Polynomial approximations to the solution y (x) of Airy’s equation. The value
1
of n is the degree of the approximating polynomial.
y
n ≥ 4
34
22
10
2
n = 46
40
28
16
4
y = y2(x)
–10
–8
–6
–4
–2
2
x
n = 49
37
25
13
–2
43
31
19
7
FIGURE 5.2.4
Polynomial approximations to the solution y (x) of Airy’s equation. The value
2
of n is the degree of the approximating polynomial.
Find a solution of Airy’s equation in powers of x − 1.
E X A M P L E
The point x = 1 is an ordinary point of Eq. (12), and thus we look for a solution of
3
the form
∞
y = a (x − 1)n,
n
n=0
where we assume that the series converges in some interval |x − 1| < ρ. Then
∞
∞
y = na (x − 1)n−1 = (n + 1)a(x − 1)n,
n
n+1
n=1
n=0
Chapter 5. Series Solutions of Second Order Linear Equations and
∞
∞
y = n(n − 1)a (x − 1)n−2 = (n + 2)(n + 1)a(x − 1)n.
n
n+2
n=2
n=0
Substituting for y and y in Eq. (12), we obtain
∞
∞
(n + 2)(n + 1)a(x − 1)n = xa (x − 1)n.
(18)
n+2
n
n=0
n=0
Now to equate the coefficients of like powers of (x − 1) we must express x, the coefficient of y in Eq. (12), in powers of x − 1; that is, we write x = 1 + (x − 1). Note that this is precisely the Taylor series for x about x = 1. Then Eq. (18) takes the form
∞
∞
(n + 2)(n + 1)a(x − 1)n = [1 + (x − 1)] a (x − 1)n
n+2
n
n=0
n=0
∞
∞
=
a (x − 1)n + a (x − 1)n+1.
n
n
n=0
n=0
Shifting the index of summation in the second series on the right gives
∞
∞
∞
(n + 2)(n + 1)a(x − 1)n = a (x − 1)n + a(x − 1)n.
n+2
n
n−1
n=0
n=0
n=1
Equating coefficients of like powers of x − 1, we obtain 2a = a ,
2
0 (3 · 2)a = a + a ,
3
1
0 (4 · 3)a = a + a ,
4
2
1 (5 · 4)a = a + a ,
5
3
2
...
The general recurrence relation is (n + 2)(n + 1)a = a + a for n ≥ 1.
(19)
n+2
n
n−1
Solving for the first few coefficients a in terms of a and a , we find that
n
0
1
a
a
a
a
a
a
a
a
a
a
aa = 0 , a = 1 + 0 , a = 2 + 1 = 0 + 1 , a = 3 + 2 = 0 + 1 .
2
2
3
6
6
4
12
12
24
12
5
20
20
30
120
Hence
(x − 1)2 (x − 1)3 (x − 1)4 (x − 1)5 y = a
1 +
+
+
+
+ · · ·
0
2
6
24
30
(
(
(
+
x − 1)3 x − 1)4 x − 1)5 a(x − 1) +
+
+
+ · · · .
(20)
1
6
12
120
In general, when the recurrence relation has more than two terms, as in Eq. (19), the determination of a formula for a in terms a and a will be fairly complicated, if
n
0
1
not impossible. In this example such a formula is not readily apparent. Lacking such a formula, we cannot test the two series in Eq. (20) for convergence by direct methods such as the ratio test. However, even without knowing the formula for a we shall see
n
in Section 5.3 that it is possible to establish that the series in Eq. (20) converge for all x, and further define functions y and y that are linearly independent solutions of the
3
4
Airy equation (12). Thus y = a y (x) + a y (x) 0 3
1 4 is the general solution of Airy’s equation for −∞ < x < ∞.
It is worth emphasizing, as we saw in Example 3, that if we look for a solution of
∞
Eq. (1) of the form y = a (x − x )n, then the coefficients P(x), Q(x), and R(x)n
0
n=0
in Eq. (1) must also be expressed in powers of x − x . Alternatively, we can make the
0
change of variable x − x = t, obtaining a new differential equation for y as a function
0
∞
of t, and then look for solutions of this new equation of the form a tn. When we n
n=0
have finished the calculations, we replace t by x − x (see Problem 19).
0
In Examples 2 and 3 we have found two sets of solutions of Airy’s equation. The functions y and y defined by the series in Eq. (17) are linearly independent solutions
1
2
of Eq. (12) for all x, and this is also true for the functions y and y defined by the
3
4
series in Eq. (20). According to the general theory of second order linear equations each of the first two functions can be expressed as a linear combination of the latter two functions and vice versa—a result that is certainly not obvious from an examination of the series alone.
Finally, we emphasize that it is not particularly important if, as in Example 3, we are unable to determine the general coefficient a in terms of a and a . What is essential n
0
1
is that we can determine as many coefficients as we want. Thus we can find as many terms in the two series solutions as we want, even if we cannot determine the general term. While the task of calculating several coefficients in a power series solution is not difficult, it can be tedious. A symbolic manipulation package can be very helpful here; some are able to find a specified number of terms in a power series solution in response to a single command. With a suitable graphics package one can also produce plots such as those shown in the figures in this section.
PROBLEMS
In each of Problems 1 through 14 solve the given differential equation by means of a power series about the given point x . Find the recurrence relation; also find the first four terms in each
0
of two linearly independent solutions (unless the series terminates sooner). If possible, find the general term in each solution.
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Chapter 5. Series Solutions of Second Order Linear Equations In each of Problems 15 through 18: (a) Find the first five nonzero terms in the solution of the given initial value problem.
(b) Plot the four-term and the five-term approximations to the solution on the same axes.
(c) From the plot in part (b) estimate the interval in which the four-term approximation is reasonably accurate.
䉴
see Problem 2
䉴
see Problem 6
䉴
see Problem 7
䉴
see Problem 12 find two linearly independent series solutions of
20. Show directly, using the ratio test, that the two series solutions of Airy’s equation about
The equation
n
0
5
(b) Look for a solution of the initial value problem in the form of a power series about
䉴
see Problem 2 䉴 24. (2 + x2)y − xy + 4y = 0,y(0) = 1,y(0) = 0; see Problem 6 䉴 25. y + xy + 2y = 0,y(0) = 0,y(0) = 1; see Problem 7
䉴
see Problem 10 䉴 27. y + x2y = 0,y(0) = 1,y(0) = 0; see Problem 4 䉴 28. (1 − x)y + xy − 2y = 0,y(0) = 0,y(0) = 1 5Charles Hermite (1822–1901) was an influential French analyst and algebraist. He introduced the Hermite functions in 1864, and showed in 1873 that e is a transcendental number (that is, e is not a root of any polynomial equation with rational coefficients). His name is also associated with Hermitian matrices (see Section 7.3), some of whose properties he discovered.