8.1 EXERCISES. 426 Chapter 8 Series Solutions of Differential Equations

Transcription

1 426 Chapter 8 Series Solutions of Differential Euations to the solution yaxb. Now for nonlinear euations such as (4), the factor f (n ) AjB in the Lagrange error formula may grow too rapidly with n, and the convergence can be thwarted. But if the differential euation is linear and its coefficients and nonhomogeneous term enjoy a feature known as analyticity, our wish is granted; the error does indeed diminish to zero as the degree n goes to infinity, and the seuence of Taylor polynomials can be guaranteed to converge to the actual solution on a certain (known) interval. For instance, the exponential, sine, and cosine functions in Example all satisfy linear differential euations with constant coefficients, and their Taylor series converge to the corresponding function values. (Indeed, all their derivatives are bounded on any interval of finite length, so the Lagrange formula for their approximation errors approaches zero as n increases, for each value of x.) This topic is the theme for the early sections of this chapter. In Sections 8.5 and 8.6 we ll see that solutions to second-order linear euations can exhibit very wild behavior near points x 0 where the coefficient of y is zero; so wild, in fact, that no Euler or Runge Kutta algorithm could hope to keep up with them. But a clever modification of the Taylor polynomial method, due to Frobenius, provides very accurate approximations to the solutions in such regions. It is this latter feature, perhaps, that underscores the value of the Taylor methodology in the current practice of applied mathematics. 8. EXERCISES In Problems 8, determine the first three nonzero terms in the Taylor polynomial approximations for the given initial value problem y x 2 y 2 ; ya0b y y 2 ; ya0b 2 y sin y e x ; ya0b 0 y sinax yb ; ya0b 0 x tx 0 ; xa0b, x A0B 0 y y 0 ; ya0b 0, y A0B y AuB yaub 3 sin u ; ya0b 0, y A0B 0 y sin y 0 ; ya0b, y A0B 0 9. (a) Construct the Taylor polynomial p 3 AxB of degree 3 for the function f AxB ln x around x. (b) Using the error formula (6), show lna.5b p 3 A0.5B (c) Compare the estimate in part (b) with the actual error by lna.5b p 3 (d) Sketch the graphs of ln x and p 3 AxB (on the same axes) for 0 < x < (a) Construct the Taylor polynomial p 3 AxB of degree 3 for the function f AxB / A2 xb around x 0. (b) Using the error formula (6), show that ` f a 2 b p 3 a 2 b ` ` 2 3 p 3 a 2 b ` (c) Compare the estimate in part (b) with the actual error ` 2 3 p 3 a 2 b `. (d) Sketch the graphs of A2 xb and p 3 AxB (on the same axes) for 2 6 x Argue that if y faxb is a solution to the differential euation y p AxBy AxBy g AxB on the interval Aa, bb,wherep,,andg are each twice-differentiable, then the fourth derivative of faxb exists on Aa, bb. 2. Argue that if y faxb is a solution to the differential euation y p AxBy AxBy g AxB on the interval Aa, bb, where p,, and g possess derivatives of all orders, then f has derivatives of all orders on Aa, bb. 3. Duffing s Euation. In the study of a nonlinear spring with periodic forcing, the following euation arises: y ky ry 3 A cos vt. Let k r A and v 0. Find the first three nonzero terms in the Taylor polynomial approximations to the solution with initial values ya0b 0, y A0B. /

2 Section 8.2 Power Series and Analytic Functions Soft versus Hard Springs. For Duffing s euation given in Problem 3, the behavior of the solutions changes as r changes sign. When r > 0, the restoring force ky ry 3 becomes stronger than for the linear spring Ar 0B. Such a spring is called hard. When r < 0, the restoring force becomes weaker than the linear spring and the spring is called soft. Pendulums act like soft springs. (a) Redo Problem 3 with r. Notice that for the initial conditions ya0b 0, y A0B, the soft and hard springs appear to respond in the same way for t small. (b) Keeping k A and v 0, change the initial conditions to ya0b and y A0B 0. Now redo Problem 3 with r. (c) Based on the results of part (b), is there a difference between the behavior of soft and hard springs for t small? Describe. 5. The solution to the initial value problem xy AxB 2y AxB xyaxb 0 ; ya0b, y A0B 0 has derivatives of all orders at x 0 (although this is far from obvious). Use L Hôpital s rule to compute the Taylor polynomial of degree 2 approximating this solution. 6. van der Pol Euation. In the study of the vacuum tube, the following euation is encountered: y A0.BAy 2 By y 0. Find the Taylor polynomial of degree 4 approximating the solution with the initial values ya0b, y A0B POWER SERIES AND ANALYTIC FUNCTIONS The differential euations studied in earlier sections often possessed solutions yaxb that could be written in terms of elementary functions such as polynomials, exponentials, sines, and cosines. However, many important euations arise whose solutions cannot be so expressed. In the previous chapters, when we encountered such an euation we either settled for expressing the solution as an integral (see Exercises 2.2, Problem 27, page 43) or as a numerical approximation (Sections 3.6, 3.7, and 5.3). However, the Taylor polynomial approximation scheme of the preceding section suggests another possibility. Suppose the differential euation (and initial conditions) permit the computation of every derivative y AnB at the expansion point x 0. Are there any conditions that would guarantee that the seuence of Taylor polynomials would converge to the solution yaxb as the degree of the polynomials tends to infinity: n lim a y A jb Ax 0 B ns j 0 j! In other words, when can we be sure that a solution to a differential euation is represented by its Taylor series? As we ll see, the answer is uite favorable, and it enables a powerful new techniue for solving euations. The most efficient way to begin an exploration of this issue is by investigating the algebraic and convergence properties of generic expressions that include Taylor series long polynomials so to speak, or more conventionally, power series. Power Series A power series about the point Ax x 0 B j y A jb Ax 0 B a Ax x j 0 j! 0 B j yaxb? x 0 is an expression of the form () a a n Ax x 0 Bn a 0 a Ax x 0 B a 2 Ax x 0 B 2 p,

3 434 Chapter 8 Series Solutions of Differential Euations By the reasoning in the previous paragraph, the derivatives of f have convergent power series representations f AxB a na n Ax x 0 Bn 0 a na n Ax x 0 Bn, n f AxB a nan Ba n Ax x 0 Bn a nan Ba n Ax x 0 Bn 2, n 2 o f A jb AxB a nan B p An 3 j 4 B a n Ax x 0 B n j, o But if we evaluate these series at x x 0, we learn that f Ax 0 B a 0, f Ax 0 B # a, f Ax 0 B 2 # # a2, o f A jb Ax 0 B j! # aj, o that is, a and the power series must coincide with the Taylor series j f A jb Ax 0 B /j! f A jb Ax 0 B a Ax x j 0 j! 0 B j a nan B p An 3 j 4 B a n Ax x 0 B n j, n j about x 0. Any power series regardless of how it is derived that converges in some neighborhood of x 0 to a function has to be the Taylor series of that function. For example, the expansion for arctan x given in (9) of Example 2 must be its Taylor expansion. With these facts in mind we are ready to turn to the study of the effectiveness of power series techniues for solving differential euations. In the next sections, you will find it helpful to keep in mind that if f and g are analytic at x 0, then so are f g, cf, fg, and f/g if gax 0 B 0. These facts follow from the algebraic properties of power series discussed earlier. 8.2 EXERCISES In Problems 6, determine the convergence set of the given power series. 2 n 3 n. a Ax Bn 2. a n! xn n a n 2 3. Ax 2Bn 4. n 2 a n 4 Ax n 2 3Bn 2n 3 An 2B! 5. a Ax 2Bn 6. a Ax 2B n n 3 n n! When the expansion point x 0 is zero, the Taylor series is also known as the Maclaurin series.

4 Section 8.2 Power Series and Analytic Functions Sometimes the ratio test (Theorem 2) can be applied to a power series containing an infinite number of zero coefficients, provided the zero pattern is regular. Use Theorem 2 to show, for example, that the series and that a x a 3 x 3 a 5 x 5 a 7 x 7 p [Hint: Let z x Determine the convergence set of the given power series. 0. (a) (b) (c) sin x [euation ()] (d) cos x [euation (2)] (e) (f) In Problems 9 and 0, find the power series expansion a n x n for f AxB g AxB, given the expansions for f AxB and gaxb. 9. f AxB a g AxB a x n n In Problems 4, find the first three nonzero terms in the power series expansion for the product f AxBg AxB.. 2. a 0 a 2 x 2 a 4 x 4 a 6 x 6 p a has a radius of convergence r 2L, if lim ` a 2k ` L, ks a 2k 2 has a radius of convergence r 2M, if lim ` a 2k ` M. ks a 2k 3 a 22k x 2k Asin xb /x a a 22k x 4k f AxB a n 3 g AxB a n f AxB e x a a a 2k x 2k n xn, g AxB sin x a n 2 n Ax Bn 2 f AxB sin x a 2 n n! Ax Bn 2, g AxB cos x a A B n x 2n / A2n B! n! xn, A B k A2k B! x2k A B k A2k B! x2k, A B k A2kB! x2k a 22k x 2k a 2kx 2k Find the first few terms of the power series for the uotient by completing the following: (a) Let AxB a n x n, where the coefficients a n are to be determined. Argue that is the Cauchy product of and x n /2 n AxB x n /n!. (b) Use formula (6) of the Cauchy product to deduce the euations (c) Solve the euations in part (b) to determine the constants a 0, a, a 2, a To find the first few terms in the power series for the uotient AxB in Problem 5, treat the power series in the numerator and denominator as long polynomials and carry out long division. That is, perform In Problems 7 20, find a power series expansion for f AxB, given the expansion for f AxB f AxB e x a g AxB A xb a f AxB e x a g AxB e x a AxB a a 2 0 a 0, 2 a 0 a, 2 2 a 0 2 a a 2, 2 3 a 0 6 a 2 a 2 a 3,.... x 2 x2 p ` 2 x 4 x2 p. f AxB A xb a f AxB sin x a f AxB a a kx 2k 20. f AxB a na nx n n A B n x n! n, n! xn, A B n n! 2 n xn b ~ a a A B n x n x n A B k A B n x n n! xn b A2k B! x2k

5 436 Chapter 8 Series Solutions of Differential Euations In Problems 2 and 22, find a power series expansion for g AxB J x 0 f AtB dt, given the expansion for f AxB In Problems 23 26, express the given power series as a series with generic term x k. 23. a na nx n 24. n An Ba nx n 2 n n a a nx n Show that 28. Show that 2 a a nx n a nb nx n n In Problems 29 34, determine the Taylor series about the point x 0 for the given functions and values of x f AxB cos x, x 0 p f AxB A xb a f AxB sin x x a x 2 a n An Ba nx n b a 3 2a n An Bb n 4 x n. n f AxB x, x 0 f AxB x x, x 0 0 f AxB lna xb, x 0 0 f AxB x 3 3x 4, x f AxB x, x 0 A B n x n A B k A2k B! x2k a a n a n n 3 x n 3 a An 2BAn Ba n 2 x n. n The Taylor series for f AxB ln x about x 0 given in euation (3) can also be obtained as follows: (a) Starting with the expansion / A sb s n and observing that obtain the Taylor series for /x about x 0. (b) Since ln x x /tdt, use the result of part (a) and termwise integration to obtain the Taylor series for f AxB ln x about x Let f AxB and g AxB be analytic at x 0. Determine whether the following statements are always true or sometimes false: (a) 3f AxB g AxB is analytic at (b) f AxB /g AxB is analytic at (c) f AxB is analytic at (d) 3 f AxB 4 3 x x 0 g AtB dt is analytic at 37. Let x Ax B f AxB e e / x 2 x 0, x 0, 0, x 0. Show that f AnB A0B 0 for n 0,, 2,... and hence that the Maclaurin series for is p f AxB, which converges for all x but is eual to f AxB only when x 0. This is an example of a function possessing derivatives of all orders (at x 0 0), whose Taylor series converges, but the Taylor series (about x 0 0) does not converge to the original function! Conseuently, this function is not analytic at x Compute the Taylor series for f AxB lna x 2 B about x 0 0. [Hint: Multiply the series for A x 2 B by 2x and integrate.], x 0 x 0 x POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS In this section we demonstrate a method for obtaining a power series solution to a linear differential euation with polynomial coefficients. This method is easier to use than the Taylor series method discussed in Section 8. and sometimes gives a nice expression for the general term in the power series expansion. Knowing the form of the general term also allows us to test for the radius of convergence of the power series.

6 Section 8.3 Power Series Solutions to Linear Differential Euations EXERCISES In Problems 0, determine all the singular points of the given differential euation.. Ax By x 2 y 3y 0 2. x 2 y 3y xy 0 3. Au 2 2By 2y Asin uby 0 4. Ax 2 xby 3y 6xy 0 5. At 2 t 2Bx At Bx At 2Bx 0 6. Ax 2 By A xby Ax 2 2x By 0 7. Asin xby Acos xby 0 8. e x y Ax 2 By 2xy 0 9. Asin uby Aln uby lnax B 4 y Asin 2xBy e x y 0 In Problems 8, find at least the first four nonzero terms in a power series expansion about x 0 for a general solution to the given differential euation.. y Ax 2By 0 2. y y 0 3. z x 2 z 0 4. Ax 2 By y In Problems 9 24, find a power series expansion about x 0 for a general solution to the given differential euation. Your answer should include a general formula for the coefficients. 9. y 2xy y y 0 2. y xy 4y y xy y Ax By y 0 y 2y y 0 w x 2 w w 0 A2x 3By xy y 0 z x 2 z xz 0 Ax 2 By xy y 0 In Problems 25 28, 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. 25. w 3xw w 0 ; w A0B 2, w A0B Ax 2 x By y y 0 ; ya0b 0, y A0B 27. Ax By y 0 ; ya0b 0, y A0B 28. y Ax 2By y 0 ; ya0b, y A0B 0 In Problems 29 3, use the first few terms of the power series expansion to find a cubic polynomial approximation for the solution to the given initial value problem. Graph the linear, uadratic, and cubic polynomial approximations for 5 x y y xy 0 ; ya0b, y A0B y 4xy 5y 0 ; ya0b, y A0B 3. Ax 2 2By 2xy 3y 0 ; ya0b, y A0B Consider the initial value problem y 2xy 2y 0 ; ya0b a 0, y A0B a, a 0 a where and are constants. (a) Show that if a 0 0, then the solution will be an odd function [that is, ya xb yaxb for all x]. What happens when a 0? (b) Show that if a 0 and a are positive, then the solution is increasing on A0, B. (c) Show that if a 0 is negative and a is positive, then the solution is increasing on A, 0B. (d) What conditions on a 0 and a would guarantee that the solution is increasing on A, B? 33. Use the ratio test to show that the radius of convergence of the series in euation (3) is infinite. [Hint: See Problem 7, Exercises 8.2, page 435.] 34. Emden s Euation. A classical nonlinear euation that occurs in the study of the thermal behavior of a spherical cloud is Emden s euation y 2 x y yn 0, with initial conditions ya0b, y A0B 0. Even though x 0 is not an ordinary point for this euation (which is nonlinear for n ), it turns out that there does exist a solution analytic at x 0. Assuming that n is a positive integer, show that the first few terms in a power series solution are y x2 3! n x 4 5! p. [Hint: Substitute c 5 x 5 p y c 2 x 2 c 3 x 3 c 4 x 4 into the euation and carefully compute the first few terms in the expansion for y n. 4

7 446 Chapter 8 Series Solutions of Differential Euations 0. henrys R(t) = + t/0 ohms 2 farads (0) = 0 coulombs (0) = 0 amps k(t) = 6 t N/m 2 kg N-sec/m x(t) x(0) = 3 m x'(0) = 0 m/sec Figure 8.6 A mass spring system whose spring is being heated Figure 8.5 An RLC circuit whose resistor is being heated 35. Variable Resistor. In Section 5.7, we showed that the charge on the capacitor in a simple RLC circuit is governed by the euation L AtB R AtB AtB E AtB, C where L is the inductance, R the resistance, C the capacitance, and E the voltage source. Since the resistance of a resistor increases with temperature, let s assume that the resistor is heated so that the resistance at time t is R AtB t/0 Ω (see Figure 8.5). If L 0. H, C 2 F, E AtB 0, A0B 0 C, and A0B 0 A, find at least the first four nonzero terms in a power series expansion about t 0 for the charge on the capacitor. 36. Variable Spring Constant. As a spring is heated, its spring constant decreases. Suppose the spring is heated so that the spring constant at time t is k(t) 6 t N/m (see Figure 8.6). If the unforced mass spring system has mass m 2 kg and a damping constant b N-sec/m with initial conditions xa0b 3 m and x A0B 0 m/sec, then the displacement xatb is governed by the initial value problem 2x AtB x AtB A6 tbx(t) 0 ; xa0b 3, x A0B 0. Find at least the first four nonzero terms in a power series expansion about t 0 for the displacement. 8.4 EQUATIONS WITH ANALYTIC COEFFICIENTS In Section 8.3 we introduced a method for obtaining a power series solution about an ordinary point. In this section we continue the discussion of this procedure. We begin by stating a basic existence theorem for the euation () y AxB paxby AxB AxByAxB 0, which justifies the power series method. Existence of Analytic Solutions Theorem 5. Suppose x 0 is an ordinary point for euation (). Then () has two linearly independent analytic solutions of the form (2) yaxb a a n Ax x 0 Bn. Moreover, the radius of convergence of any power series solution of the form given by (2) is at least as large as the distance from x 0 to the nearest singular point (real or complexvalued) of euation ().

8 450 Chapter 8 Series Solutions of Differential Euations Thus, the solution to the initial value problem in (0) is (4) yaxb 2 x2 6 x3 20 x5 720 x6 p. Thus far we have used the power series method only for homogeneous euations. But the same method applies, with obvious modifications, to nonhomogeneous euations of the form (5) provided the forcing term gaxb and the coefficient functions are analytic at x 0. For example, to find a power series about x 0 for a general solution to (6) we use the substitution yaxb a n x n to obtain a power series expansion for the left-hand side of (6). We then euate the coefficients of this series with the corresponding coefficients of the Maclaurin expansion for sin x: Carrying out the details (see Problem 20), we ultimately find that an expansion for a general solution to (6) is (7) where y AxB paxby AxB AxByAxB gaxb, y AxB xy AxB yaxb sin x, sin x a A B n A2n B! x2n. yaxb a 0 y AxB a y 2 AxB y p AxB, (8) y AxB 2 x2 8 x4 48 x6 p, (9) y 2 AxB x 3 x3 5 x5 05 x7 p are the solutions to the homogeneous euation associated with euation (6) and (20) y p AxB 6 x3 40 x x7 p is a particular solution to euation (6). 8.4 EXERCISES In Problems 6, find a minimum value for the radius of convergence of a power series solution about x 0.. Ax By 3xy 2y 0 ; x 0 2. y xy 3y 0 ; x A x x 2 By 3y 0 ; x 0 4. Ax 2 5x 6By 3xy y 0 ; x y Atan xby y 0 ; x A x 3 By xy 3x 2 y 0 ; x 0 In Problems 7 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 euation with the given value for x y 2 Ax By 0 ; x 0 y 2xy 0 ; x 0 Ax 2 2xBy 2y 0 ; x 0 x 2 y xy 2y 0 ; x 0 2 x 2 y y y 0 ; x 0 2 y A3x By y 0 ; x 0 In Problems 3 9, find at least the first four nonzero terms in a power series expansion of the solution to the given initial value problem. 3. x Asin tbx 0 ; xa0b 4. y e x y 0 ; ya0b

9 Section 8.4 Euations with Analytic Coefficients To derive the general solution given by euations (7) (20) for the nonhomogeneous euation (6), complete the following steps: (a) Substitute yaxb a n x n and the Maclaurin series for sin x into euation (6) to obtain A2a 2 a 0 B a 3 Ak 2BAk B a k 2 Ak B a k 4 x k k (b) Euate the coefficients of like powers of x on both sides of the euation in part (a) and thereby deduce the euations (c) Show that the relations in part (b) yield the general solution to (6) given in euations (7) (20). In Problems 2 28, use the procedure illustrated in Problem 20 to find at least the first four nonzero terms in a power series expansion about x 0 of a general solution to the given differential euation Ax 2 By e x y y 0 ; ya0b, y A0B y ty e t y 0 ; ya0b 0, y A0B y Asin xby 0 ; yapb, y ApB 0 y Acos xby y 0 ; yap/2b, y Ap/2B y e 2x y Acos xby 0 ; ya0b, y A0B a A B n A2n B! x2n. a 2 a 0 2, a 3 6 a 3, a 4 a 0 8, a 5 40 a 5, a 6 a 0 48, a a 05. y xy sin x w xw e x z xz z x 2 2x y 2xy 3y x 2 A x 2 By xy y e x 26. y xy 2y cos x A x 2 By y y tan x y Asin xby cos x 29. The euation A x 2 By 2xy nan By 0, where n is an unspecified parameter, is called Legendre s euation. This euation occurs in applications of differential euations to engineering systems in spherical coordinates. (a) Find a power series expansion about x 0 for a solution to Legendre s euation. (b) Show that for n a nonnegative integer, there exists an nth-degree polynomial that is a solution to Legendre s euation. These polynomials, up to a constant multiple, are called Legendre polynomials. (c) Determine the first three Legendre polynomials (up to a constant multiple). 30. Aging Spring. As a spring ages, its spring constant decreases in value. One such model for a mass spring system with an aging spring is mx AtB bx AtB ke ht xatb 0, where m is the mass, b the damping constant, k and h positive constants, and xatb the displacement of the spring from its euilibrium position. Let m kg, b 2 N-sec/m, k N/m, and h AsecB. The system is set in motion by displacing the mass m from its euilibrium position and then releasing it AxA0B, x A0B 0B. Find at least the first four nonzero terms in a power series expansion about t 0 for the displacement. 3. Aging Spring without Damping. In the mass spring system for an aging spring discussed in Problem 30, assume that there is no damping (i.e., b 0), m, and k. To see the effect of aging, consider h as a positive parameter. (a) Redo Problem 30 with b 0 and h arbitrary but fixed. (b) Set h 0 in the expansion obtained in part (a). Does this expansion agree with the expansion for the solution to the problem with h 0? [Hint: When h 0, the solution is xatb cos t. 4

10 454 Chapter 8 Series Solutions of Differential Euations y x 0.5 Figure 8.7 Graph of x 0.3 cosa2 ln xb yields the repeated root r 0 /2. Thus, a general solution to () is obtained from euations (8) and (0) by setting r 0 /2. That is, yaxb C x C 2 x ln x, x 7 0. In closing we note that the solutions to Cauchy Euler euations can exhibit some extraordinary features near their singular points features unlike anything we have encountered in the case of constant coefficient euations. Observe that:. and x r ln x, for 0 < r <, have infinite slope at the origin; 2. x r, for r < 0, grows without bound near the origin; and 3. x a cosab ln xb and x a sinab ln xb oscillate infinitely rapidly near the origin. (See Figure 8.7.) x r Numerical solvers based on Euler or Runge Kutta procedures would be completely frustrated in trying to chart such behavior; analytic methods are essential for characterizing solutions near singular points. In Section 8.7 we discuss the problem of finding a second linearly independent series solution to certain differential euations. As we will see, operator methods similar to those described in this section will lead to the desired second solution. 8.5 EXERCISES In Problems 0, use the substitution y x r to find a 7. general solution to the given euation for x x 2 y AxB 6xy AxB 6yAxB 0 2x 2 y AxB 3xy AxB 5yAxB 0 x 2 y AxB xy AxB 7yAxB 0 x 2 y AxB 2xy AxB 3yAxB 0 d 2 y dx 2 5 x dx d 2 y dx 2 x dy 3 x 2 y dy dx 4 x 2 y x 3 y AxB 4x 2 y AxB 0xy AxB 0yAxB 0 x 3 y AxB 4x 2 y AxB xy AxB 0 x 3 y AxB 3x 2 y AxB 5xy AxB 5yAxB 0 x 3 y AxB 9x 2 y AxB 9xy AxB 8yAxB 0 In Problems and 2, use a substitution of the form y Ax cb r to find a general solution to the given euation for x 7 c.. 2 Ax 3B 2 y AxB 5 Ax 3By AxB 2yAxB Ax 2B 2 y AxB 5yAxB 0

11 Section 8.6 Method of Frobenius 455 In Problems 3 and 4, use variation of parameters to find a general solution to the given euation for x x 2 y AxB 2xy AxB 2yAxB x / 2 4. x 2 y AxB 2xy AxB 2yAxB 6x 2 3x In Problems 5 7, solve the given initial value problem. 5. t 2 x AtB 2xAtB 0 ; xab 3, x AB 5 6. x 2 y AxB 5xy AxB 4yAxB 0 ; yab 3, y AB 7 7. x 3 y AxB 6x 2 y AxB 29xy AxB 29yAxB 0 ; yab 2, y AB 3, y AB 9 8. Suppose r 0 is a repeated root of the auxiliary euation ar 2 br c 0. Then, as we well know, y AtB e r 0t is a solution to the euation ay by cy 0, where a, b, and c are constants. Use a derivation similar to the one given in this section for the case when the indicial euation has a repeated root to show that a second linearly independent solution is y 2 AtB te r0t. 9. Let L 3 y 4 AxB J x 3 y AxB xy AxB yaxb. (a) Show that L 3 x r 4 AxB Ar B 3 x r. (b) Using an extension of the argument given in this section for the case when the indicial euation has a double root, show that L 3 y 4 0 has the general solution yaxb C x C 2 x ln x C 3 xaln xb METHOD OF FROBENIUS In the previous section we showed that a homogeneous Cauchy Euler euation has a solution of the form yaxb x r, x 7 0, where r is a certain constant. Cauchy Euler euations have, of course, a very special form with only one singular point (at x 0). In this section we show how the theory for Cauchy Euler euations generalizes to other euations that have a special type of singularity. To motivate the procedure, let s rewrite the Cauchy Euler euation, () in the standard form (2) where and p are the constants and, respectively. When we substitute war, xb x r 0, 0 b/a c/a for y into euation (2), we get which yields the indicial euation (3) rar B p 0 r 0 0. Thus, if is a root of (3), then war, xb x r r is a solution to euations () and (2). Let s now assume, more generally, that (2) is an euation for which xpaxb and x 2 AxB, instead of being constants, are analytic functions. That is, in some open interval about x 0, (4) ax 2 y AxB bxy AxB cyaxb 0, x 7 0, y AxB paxby AxB AxByAxB 0, x 7 0, paxb p 0 x, AxB 0 x 2, 3 rar B p 0 r 0 4 x r 2 0, xpaxb p 0 p x p 2 x 2 p a p nx n, (5) x 2 AxB 0 x 2 x 2 p a nx n.

12 Section 8.6 Method of Frobenius 465 into (54) ultimately gives (57) Setting the coefficients eual to zero, we have (58) (59) and, for k, the recurrence relation (60) With r r 0, these euations lead to the following formulas: a 2k 0, k 0,,..., and (6) 3 rar B 3r 4 a 0 x r 3 Ar Br 3Ar B 4 a x r 3 rar B 3r 4 a 0 0, 3 Ar Br 3Ar B 4 a 0, Ak r BAk r 3Ba k a k 0. a 2k a 3 Ak r BAk r 3Ba k a k 4 x k r 0. k 3 2 # 4 p A2kB 434 # 6 p A2k 2B 4 a 0 Hence euation (54) has the power series solution (62) wa0, xb a 0 a 2 2k k!ak B! x2k, x k k!ak B! a 0, k 0. Unlike in Example 5, if we work with the second root r r 2 2 in Example 6, then we do not obtain a second linearly independent solution (see Problem 46). In the preceding examples we were able to use the method of Frobenius to find a series solution valid to the right (x > 0) of the regular singular point x 0. For x < 0, we can use the change of variables x t and then solve the resulting euation for t > 0. The method of Frobenius also applies to higher-order linear euations (see Problems 35 38). 8.6 EXERCISES In Problems 0, classify each singular point (real or complex) of the given euation as regular or irregular.. Ax 2 By xy 3y 0 2. x 2 y 8xy 3xy 0 3. Ax 2 Bz 7x 2 z 3xz 0 4. x 2 y 5xy 7y 0 5. Ax 2 B 2 y Ax By 3y 0 6. Ax 2 4By Ax 2By 3y 0 7. At 2 t 2B 2 x At 2 4Bx tx 0 8. Ax 2 xby xy 7y 0 9. Ax 2 2x 8B 2 y A3x 2By x 2 y 0 0. x 3 Ax By Ax 2 3xBAsin xby xy 0 In Problems 8, find the indicial euation and the exponents for the specified singularity of the given differential euations.. x 2 y 2xy 0y 0, at x 0 2. x 2 y 4xy 2y 0, at x 0 3. Ax 2 x 2B 2 z Ax 2 4Bz 6xz 0, at x 2 4. Ax 2 4By Ax 2By 3y 0, at x 2 5. u 3 y uasin uby Atan uby 0, at u 0 6. Ax 2 By Ax By 3y 0, at x 7. Ax B 2 y Ax 2 By 2y 0, at x 8. 4xAsin xby 3y 0, at x 0 In Problems 9 24, use the method of Frobenius to find at least the first four nonzero terms in the series expansion about x 0 for a solution to the given euation for x x 2 y 9x 2 y 2y 0

13 466 Chapter 8 Series Solutions of Differential Euations In Problems 25 30, use the method of Frobenius to find a general formula for the coefficient a n in a series expansion about x 0 for a solution to the given euation for x In Problems 3 34, first determine a recurrence formula for the coefficients in the (Frobenius) series expansion of the solution about x 0. Use this recurrence formula to determine if there exists a solution to the differential euation that is decreasing for x In Problems 35 38, use the method of Frobenius to find at least the first four nonzero terms in the series expansion about x 0 for a solution to the given linear thirdorder euation for x In Problems 39 and 40, try to use the method of Frobenius to find a series expansion about the irregular singular point x 0 for a solution to the given differential euation. If the method works, give at least the first four nonzero terms in the expansion. If the method does not work, explain what went wrong xAx By 3 Ax By y 0 x 2 y xy x 2 y 0 xy y 4y 0 x 2 z Ax 2 xbz z 0 3xy A2 xby y 0 4x 2 y 2x 2 y Ax 3By 0 x 2 y Ax 2 xby y 0 xw w xw 0 3x 2 y 8xy Ax 2By 0 xy Ax By 2y 0 xax By Ax 5By 4y 0 xy A xby y 0 x 2 y xa xby y 0 3xy 2 A xby 4y 0 xy Ax 2By y 0 6x 3 y 3x 2 y Ax x 2 By xy 0 6x 3 y x 2 y 2xy Ax 2By 0 6x 3 y 3x 2 y Ax 2 3xBy xy 0 6x 3 y A3x 2 x 3 By xy xy 0 x 2 y A3x By y 0 x 2 y y 2y 0 In certain applications, it is desirable to have an expansion about the point at infinity. To obtain such an expansion, we use the change of variables z /x and expand about z 0. In Problems 4 and 42, show that infinity is a regular singular point of the given differential euation by showing that z 0 is a regular singular point for the transformed euation in z. Also find at least the first four nonzero terms in the series expansion about infinity of a solution to the original euation in x x 3 y x 2 y y 0 8Ax 4B 2 Ax 6By 9xAx 4By 32y 0 r 43. Show that if and r 2 are roots of the indicial euation (6), with r the larger root (Re r Re r 2 B, then the coefficient of a in euation (9) is not zero when r r. 44. To obtain a second linearly independent solution to euation (20): (a) Substitute war, xb given in (2) into (20) and conclude that the coefficients a k, k, must satisfy the recurrence relation Ak r BA2k 2r Ba k 3 Ak r BAk r 2B 4 a k 0. (b) Use the recurrence relation with r /2 to derive the second series solution w a, xb 2 a 0 ax / x3 / x5 / x7 / 2 p b. (c) Use the recurrence relation with r to obtain wa, xb in (28). 45. In Example 5, show that if we choose r r 2 3, then we obtain two linearly independent solutions to euation (45). [Hint: a 0 and a 3 are arbitrary constants.] 46. In Example 6, show that if we choose r r 2 2, then we obtain a solution that is a constant multiple of the solution given in (62). [Hint: Show that a 0 and a must be zero while a 2 is arbitrary.] 47. In applying the method of Frobenius, the following recurrence relation arose: a k 5 7 a k/ Ak B9, k 0,, 2,.... (a) Show that the coefficients are given by the formula a k 5 7k a 0/ Ak!B9, k 0,, 2,.... (b) Use the formula obtained in part (a) with a 0 to compute a 5, a 0, a 5, a 20, and a 25 on your computer or calculator. What goes wrong? (c) Now use the recurrence relation to compute a k for k, 2, 3,..., 25, assuming a 0. (d) What advantage does the recurrence relation have over the formula?

14 474 Chapter 8 Series Solutions of Differential Euations 4 3 Euation (44) 2 Euation (38) Substituting these values for the x Figure 8.2 Partial sum approximations to the solutions of Example 4 b n s back into (39), we obtain the solution (43) y 2 AxB C e y AxB ln x 2x x x4 p f b 2 e 8 x2 92 x4 p f, where C and b 2 are arbitrary constants. Since the factor multiplying b 2 is the first solution y AxB, we can obtain a second linearly independent solution by choosing C and b 2 0: (44) y 2 AxB y AxB ln x 2x x x4 p. See Figure 8.2. In closing we note that if the roots r and r 2 of the indicial euation associated with a differential euation are complex, then they are complex conjugates. Thus, the difference r r 2 is imaginary and, hence, not an integer, and we are in case (a) of Theorem 7. However, rather than employing the display () (2) for the linearly independent solutions, one usually takes the real and imaginary parts of (); Problem 26 provides an elaboration of this situation. 8.7 EXERCISES In Problems 4, find at least the first three nonzero terms in the series expansion about x 0 for a general solution to the given euation for x 7 0. (These are the same euations as in Problems 9 32 of Exercises 8.6.). 9x 2 y 9x 2 y 2y xAx By 3 Ax By y 0 3. x 2 y xy x 2 y xy y 4y 0 x 2 z Ax 2 xbz z 0 3xy A2 xby y 0 4x 2 y 2x 2 y Ax 3By 0 x 2 y Ax 2 xby y 0 xw w xw x 2 y 8xy Ax 2By 0

15 Section 8.7 Finding a Second Linearly Independent Solution In Problems 5 and 6, determine whether the given euation has a solution that is bounded near the origin, all solutions are bounded near the origin, or none of the solutions are bounded near the origin. (These are the same euations as in Problems 33 and 34 of Exercises 8.6.) Note that you need to analyze only the indicial euation in order to answer the uestion xy Ax By 2y 0 xax By Ax 5By 4y 0 xy A xby y 0 x 2 y xa xby y 0 3xy 2 A xby 4y 0 xy Ax 2By y 0 (a) Truncated cone (b) Truncated pyramid with fixed thickness T and varying width T In Problems 7 20, find at least the first three nonzero terms in the series expansion about x 0 for a general solution to the given linear third-order euation for x 7 0. (These are the same euations as in Problems in Exercises 8.6.) x 3 y 3x 2 y Ax x 2 By xy 0 6x 3 y x 2 y 2xy Ax 2By 0 6x 3 y 3x 2 y Ax 2 3xBy xy 0 6x 3 y A3x 2 x 3 By xy xy 0 2. Buckling Columns. In the study of the buckling of a column whose cross section varies, one encounters the euation (45) x n y AxB A 2 yaxb 0, x 7 0, where x is related to the height above the ground and y is the deflection away from the vertical. The positive constant a depends on the rigidity of the column, its moment of inertia at the top, and the load. The positive integer n depends on the type of column. For example, when the column is a truncated cone [see Figure 8.3(a)], we have n 4. (a) Use the substitution x t to reduce (45) with n 4 to the form d 2 y dt 2 2 t dy dt a2 y 0, t 7 0. (b) Find at least the first six nonzero terms in the series expansion about t 0 for a general solution to the euation obtained in part (a). (c) Use the result of part (b) to give an expansion about x for a general solution to (45). 22. In Problem 2 consider a column with a rectangular cross section with two sides constant and the other two changing linearly [see Figure 8.3(b)]. In this case, n. Find at least the first four nonzero terms in the series expansion about x 0 for a general solution to euation (45) when n. 23. Use the method of Frobenius and a reduction of order procedure (see page 98) to find at least the first three nonzero terms in the series expansion about the irregular singular point x 0 for a general solution to the differential euation x 2 y y 2y The euation xy AxB A xby AxB nyaxb 0, where n is a nonnegative integer, is called Laguerre s differential euation. Show that for each n, this euation has a polynomial solution of degree n. These polynomials are denoted by L n AxB and are called Laguerre polynomials. The first few Laguerre polynomials are L 0 AxB, L AxB x. L 2 AxB x 2 4x Use the results of Problem 24 to obtain the first few terms in a series expansion about x 0 for a general solution for x > 0 to Laguerre s differential euation for n 0 and. 26. To obtain two linearly independent solutions to (46) Figure 8.3 Buckling columns x 2 y Ax x 2 By y 0, x 7 0, complete the following steps. (a) Verify that (46) has a regular singular point at x 0 and that the associated indicial euation has complex roots i.

16 476 Chapter 8 Series Solutions of Differential Euations (b) As discussed in Section 8.5, we can express x a ib x a x ib x a cosab ln xb ix a sinab ln xb. Deduce from this formula that Aa ibbx a ib. d dx xa ib (c) Set yaxb a n x n i, where the coefficients now are complex constants, and substitute this series into euation (46) using the result of part (b). (d) Setting the coefficients of like powers eual to zero, derive the recurrence relation n i a n An ib 2 a n, for n. (e) Taking a 0, compute the coefficients a and a 2 and thereby obtain the first few terms of a complex solution to (46). (f) By computing the real and imaginary parts of the solution obtained in part (e), derive the following linearly independent real solutions to (46): y AxB 3 cosaln xb 4 e 2 5 x 0 x2 p f 3 sinaln xb 4 e 5 x 20 x2 p f, y 2 AxB 3 cosaln xb 4 e 5 x 20 x2 p f 3 sinaln xb 4 e 2 5 x 0 x2 p f. 8.8 SPECIAL FUNCTIONS In advanced work in applied mathematics, engineering, and physics, a few special second-order euations arise very freuently. These euations have been extensively studied, and volumes have been written on the properties of their solutions, which are often referred to as special functions. For reference purposes we include a brief survey of three of these euations: the hypergeometric euation, Bessel s euation, and Legendre s euation. Bessel s euation governs the radial dependence of the solutions to the classical partial differential euations of physics in spherical coordinate systems (see Section 0.7, page 625); Legendre s euation governs their latitudinal dependence. C. F. Gauss formulated the hypergeometric euation as a generic euation whose solutions include the special functions of Legendre, Chebyshev, Gegenbauer, and Jacobi. For a more detailed study of special functions, we refer you to Basic Hypergeometric Series, 2nd ed., by G. Gasper and M. Rahman (Cambridge University Press, Cambridge, 2004); Special Functions, by E. D. Rainville (Chelsea, New York, 97); and Higher Transcendental Functions, by A. Erdelyi (ed.) (McGraw-Hill, New York, 953), 3 volumes. Hypergeometric Euation The linear second-order differential euation () xa xby 3 G AA B Bx 4 y ABy 0, where a, b, and g are fixed parameters, is called the hypergeometric euation. This euation has singular points at x 0 and, both of which are regular. Thus a series expansion about x 0 for a solution to () obtained by the method of Frobenius will converge at least for 0 < x < (see Theorem 6, page 46). To find this expansion, observe that the indicial euation associated with x 0 is rar B gr r 3 r A gb 4 0, which has roots 0 and g. Let us assume that g is not an integer and use the root r 0 to obtain a solution to () of the form (2) y AxB a a nx n.

17 Section 8.8 Special Functions EXERCISES In Problems 4, express a general solution to the given euation using Gaussian hypergeometric functions In Problems 5 8, verify the following formulas by expanding each function in a power series about x In Problems 9 and 0, use the method in Section 8.7 to obtain two linearly independent solutions to the given hypergeometric euation. 9. xa xby A 3xBy y Show that the confluent hypergeometric euation xy Ag xby ay 0, where a and g are fixed parameters and g is not an integer, has two linearly independent solutions AaB n y AxB F Aa; g; xb J a x n n n!agb n and y 2 AxB x g F Aa g; 2 g; xb. 2. Use the property of the gamma function given in (9) to derive relation (20). In Problems 3 8, express a general solution to the given euation using Bessel functions of either the first or second kind xa xby a 4xb y 2y 0 2 3xA xby A 27xBy 45y 0 2xA xby A 6xBy 2y 0 2xA xby A3 0xBy 6y 0 F A, ; 2; xb x lna xb F Aa, b; b; xb A xb a F a 2, ; 3 2 ; x2 b 2 x ln a x x b F a 2, ; 3 2 ; x2 b x arctan x xa xby A2 2xBy 4 y 0 4x 2 y 4xy A4x 2 By 0 9x 2 y 9xy A9x 2 6By 0 x 2 y xy Ax 2 By 0 x 2 y xy x 2 y 0 9t 2 x 9tx A9t 2 4Bx 0 8. x 2 z xz Ax 2 6Bz 0 In Problems 9 and 20, a Bessel euation is given. For the appropriate choice of n, the Bessel function J n AxB is one solution. Use the method in Section 8.7 to obtain a second linearly independent solution x 2 y xy Ax 2 By 0 x 2 y xy Ax 2 4By 0 2. Show that x n J n AxB satisfies the euation xy A 2nBy xy 0, x 7 0, and use this result to find a solution for the euation xy 2y xy 0, x 7 0. In Problems 22 through 24, derive the indicated recurrence formulas. 22. Formula (32) 23. Formula (33) 24. Formula (34) 25. Show that J / 2 AxB A2 / pxb / 2 sin x and J / 2 AxB A2 / pxb / 2 cos x. n n /2, n 26. The Bessel functions of order any integer, are related to the spherical Bessel functions. Use relation (33) and the results of Problem 25 to show that such Bessel functions can be represented in terms of sin x, cosx, and powers of x. Demonstrate this by determining a closed form for J 3 / 2 AxB and J 5/ 2 AxB. 27. Use Theorem 7 in Section 8.7 to determine a second linearly independent solution to Bessel s euation of order 0 in terms of the Bessel function J 0 AxB. 28. Show that between two consecutive positive roots (zeros) of J AxB, there is a root of J 0 AxB. This interlacing property of the roots of Bessel functions is illustrated in Figure 8.4. [Hint: Use relation (3) and Rolle s theorem from calculus.] 29. Use formula (43) to determine the first five Legendre polynomials. 30. Show that the Legendre polynomials of even degree are even functions of x, while those of odd degree are odd functions. 3. (a) Show that the orthogonality condition (44) for Legendre polynomials implies that P n AxBAxBdx 0 for any polynomial AxB of degree at most n. [Hint: The polynomials P 0, P,..., P n are

18 488 Chapter 8 Series Solutions of Differential Euations linearly independent and hence span the space of all polynomials of degree at most Thus, AxB a 0 P 0 AxB p n. a n P n AxB for suitable constants a k. 4 (b) Prove that if Q n AxB is a polynomial of degree n such that Q n AxBP k AxBdx 0 for k 0,,..., n, then Q n AxB cp n AxB for some constant c. [Hint: Select c so that the coefficient of x n for Q n AxB cp n AxB is zero. Then, since P 0,..., P n is a basis, Q n AxB cp n AxB a 0 P 0 AxB p a n P n AxB. Multiply the last euation by P k AxB A0 k n B and integrate from x to x to show that each is zero.] 32. Deduce the recurrence formula (5) for Legendre polynomials by completing the following steps. (a) Show that the function Q n AxB J An BP n AxB A2n BxP n AxB is a polynomial of degree n. [Hint: Compute the coefficient of the x n term using the representation (42). The coefficient of x n is also zero because P n AxB and xp n AxB are both odd or both even functions, a conseuence of Problem 30.] (b) Using the result of Problem 3(a), show that Q n AxBP k AxBdx 0 for (c) From Problem 3(b), conclude that Q n AxB cp n AxB and, by taking x, show that c n. [Hint: Recall that P m AB for all m.] From the definition of Q n AxB in part (a), the recurrence formula now follows. 33. To prove Rodrigues s formula (52) for Legendre polynomials, complete the following steps. (a) Let y n J Ad n /dx n BE Ax 2 B n F and show that y n AxB is a polynomial of degree n with the coefficient of x n eual to A2nB!/n!. (b) Use integration by parts n times to show that, for any polynomial AxB of degree less than n, y n AxBAxBdx 0. a k k 0,,..., n 2. Hint: For example, when n 2, Since n 2, the degree of AxB is at most, and so AxB 0. Thus d 2 dx 2 E Ax2 B 2 FAxBdx d AxB dx E Ax2 B 2 F 0 E AxBAx2 B 2 F 0 AxBAx 2 B 2 dx. d 2 dx 2 E Ax2 B 2 FAxBdx 0. (c) Use the result of Problem 3(b) to conclude that P and show that c /2 n n AxB cy n AxB n! by comparing the coefficients of x n in P n AxB and y n AxB. 34. Use Rodrigues s formula (52) to obtain the representation (43) for the Legendre polynomials P n AxB. Hint: From the binomial formula, P n AxB 2 n n! 2 n n! d n dx n e a 35. The generating function in (53) for Legendre polynomials can be derived from the recurrence formula (5) as follows. Let x be fixed and set f AzB J P n AxBz n. The goal is to determine an explicit formula for f AzB. (a) Show that multiplying each term in the recurrence formula (5) by z n and summing the terms from n to leads to the differential euation df dz x z 2xz z 2 f. Hint: a An BP n AxBz n n a An BP n AxBz n P AxB df dz x. d n dx n E Ax 2 B n F n m 0 n!a B m An mb!m! x2n 2m f. (b) Solve the differential euation derived in part (a) and use the initial conditions f A0B P 0 AxB to obtain f AzB A 2xz z 2 B / 2.

19 Section 8.8 Special Functions Find a general solution about x 0 for the euation A x 2 By 2xy 2y 0 by first finding a polynomial solution and then using the reduction of order formula given in Exercises 6., Problem 3 to find a second (series) solution. 37. The Hermite polynomials H n AxB are polynomial solutions to Hermite s euation y 2xy 2ny 0. The Hermite polynomials are generated by e 2tx t2 a Use this euation to determine the first four Hermite polynomials. 38. The Chebyshev (Tchebichef) polynomials T n AxB are polynomial solutions to Chebyshev s euation A x 2 By xy n 2 y 0. The Chebyshev polynomials satisfy the recurrence relation T n AxB 2xT n AxB T n AxB, with T 0 AxB and T AxB x. Use this recurrence relation to determine the next three Chebyshev polynomials. 39. The Laguerre polynomials L n AxB are polynomial solutions to Laguerre s euation xy A xby ny 0. The Laguerre polynomials satisfy Rodrigues s formula, d n H n AxB t n! n. L n AxB ex n! dx n Axn e x B. Use this formula to determine the first four Laguerre polynomials. 40. Reduction to Bessel s Euation. The class of euations of the form (54) y AxB cx n yaxb 0, x 7 0, where c and n are positive constants, can be solved by transforming the euation into Bessel s euation. (a) First, use the substitution y x / 2 z to transform (54) into an euation involving x and z. (b) Second, use the substitution s 2c n 2 xn / 2 to transform the euation obtained in part (a) into the Bessel euation s 2 d 2 z ds 2 s dz ds as2 (c) A general solution to the euation in part (b) can be given in terms of Bessel functions of the first and second kind. Substituting back in for s and z, obtain a general solution for euation (54). 4. (a) Show that the substitution zaxb x yaxb renders Bessel s euation (22) in the form (55) z a An 2B2b z 0, s n2 4x 2 b z 0. (b) For x W, euation (55) would apparently be approximated by the euation (56) z z 0. Write down the general solution to (56), reset y AxB zaxb / x, and argue the plausibility of formula (36). (c) For n /2, euation (55) reduces to euation (56) exactly. Relate this observation to Problem 25. Chapter Summary Initial value problems that do not fall into the solvable categories that have been studied (such as constant-coefficient or euidimensional euations) can often be analyzed by interpreting the differential euation as a prescription for computing the higher derivatives of the unknown function. The Taylor polynomial method uses the euation to construct a polynomial approximation matching the initial values of a finite number of derivatives of the unknown. If the euation permits the extrapolation of this procedure to polynomials of arbitrarily high degree, power series representations of the solution can be constructed.