# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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What is the radius of convergence of the Taylor series for (1 + x2) 1 about x = 0? One way to proceed is to find the Taylor series in question, namely,

= 1 - x2 + x4 - x6 + ... + (1)nx2n + ....

1 + x2

Then it can be verified by the ratio test that p = 1. Another approach is to note that the zeros of 1 + x2 are x = ąz'. Since the distance from 0 to i or to i in the complex plane is 1, the radius of convergence of the power series about x = 0 is 1.

What is the radius of convergence of the Taylor series for (x2 2x + 2)1 about x = 0? About x = 1?

First notice that

x2 2x + 2 = 0

has solutions x = 1 ą i. The distance from x = 0 to either x = 1 + i or x = 1 i in the complex plane is ë/2; hence the radius of convergence of the Taylor series expansion

TO

anxn about x = 0 is V2.

n=0

The distance from x = 1 to either x = 1 + i or x = 1 i is 1; hence the radius of

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convergence of the Taylor series expansion bn (x 1)n about x = 1 is 1.

n=0

6Immanuel Lazarus Fuchs (1833-1902) was a student and later a professor at the University of Berlin. He proved the result of Theorem 5.3.1 in 1866. His most important research was on singular points of linear differential equations. He recognized the significance of regular singular points (Section 5.4), and equations whose only singularities, including the point at infinity, are regular singular points are known as Fuchsian equations.

252

Chapter 5. Series Solutions of Second Order Linear Equations

EXAMPLE

3

According to Theorem 5.3.1 the series solutions of the Airy equation in Examples

2 and 3 of the preceding section converge for all values of x and x - 1, respectively, since in each problem P (x) 1 and hence is never zero.

A series solution may converge for a wider range of x than indicated by Theorem

5.3.1, so the theorem actually gives only a lower bound on the radius of convergence of the series solution. This is illustrated by the Legendre polynomial solution of the Legendre equation given in the next example.

Determine a lower bound for the radius of convergence of series solutions about x 0 for the Legendre equation

(1 - x2)ó" - 2xy/ + a(a + 1)ó 0,

where a is a constant.

Note that P(x) 1 - x2, Q(x) 2x, and R(x) a(a + 1) are polynomials, and that the zeros of P, namely, x ą1, are a distance 1 from x 0. Hence a series

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solution of the form Y anxn converges for |x| < 1 at least, and possibly for larger

ď0 ď

values of x. Indeed, it can be shown that if a is a positive integer, one of the series solutions terminates after a finite number of terms and hence converges not just for |x | < 1 but for all x. For example, if a 1, the polynomial solution is ó x. See Problems 22 through 29 at the end of this section for a more complete discussion of the Legendre equation.

EXAMPLE

4

Determine a lower bound for the radius of convergence of series solutions of the differential equation

(1 + x 2)y" + 2xy' + 4x2 ó 0 (9)

about the point x 0; about the point x -1.

Again P, Q, and R are polynomials, and P has zeros at x ąz'. The distance in the

complex plane from 0 to ąz' is 1, and from - 2 to ąz' is ^1 + | ë/5/2. Hence in the

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first case the series Y anxn converges at least for |x| < 1, and in the second case the

ď0 ď

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series E bn (x + 2)n converges at least for |x + 11 < V5/2.

ď0 ď 2 2

An interesting observation that we can make about Eq. (9) follows from Theorems

3.2.1 and 5.3.1. Suppose that initial conditions y(0) y0 and y'(0) y0 are given.

Since 1 + x2 0 for all x, we know from Theorem 3.2.1 that there exists a unique

solution of the initial value problem on -to < x < to. On the other hand, Theorem

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5.3.1 only guarantees a series solution of the form Y anxn (with a0 y0, a1 y0)

n0

for - 1 < x < 1. The unique solution on the interval -to < x < to may not have a power series about x 0 that converges for all x.

I

EXAMPLE

5

Can we determine a series solution about x 0 for the differential equation

y" + (sin x) y' + (1 + x2) ó 0, and if so, what is the radius of convergence?

5.3 Series Solutions near an Ordinary Point, Part II

253

For this differential equation, p(x) = sin x and q (x) = 1 + x2. Recall from calculus that sin x has a Taylor series expansion about x = 0 that converges for all x. Further, q also has a Taylor series expansion about x = 0, namely, q (x) = 1 + x2, that converges

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