Series Solutions of Differential Equations

Transcription

1 Chapter 6 Series Solutions of Differential Equations In this chapter we consider methods for solving differential equations using power series. Sequences and infinite series are also involved in this treatment. Sequences, infinite series, and power series are topics which are typically studied in Calculus II. See the Appendices for a brief review of that material. Section 6.1. The Taylor Series Method. Recall that if a function g has derivatives of all orders on some open interval I and a I, then the Taylor series expansion of g in powers of (x a) is: g(x) = g(a) + g (a)(x a) + g (a) 2! (x a) 2 + g (a) (x a) 3 + = g (n) (a) (x a) n. Suppose we are given the initial-value problem y + p(x)y = f(x), y(a) = α. We assume that the solution y = y(x) has the Taylor series expansion y(x) = y(a) + y (a)(x a) + y (a) 2! (x a) 2 + y (a) (x a) 3 + = y (n) (a) (x a) n. on some interval I = (a R, a+ R). The initial condition, y(a) = α, determines the first term in the series expansion. From the differential equation, we have y (x) = f(x) p(x)y(x) and so y (a) = f(a) p(a)y(a) 217

2 and this determines the second term of the expansion. Differentiating the differential equation (assuming that p and f are differentiable functions on I), we get y (x) = f (x) p(x)y (x) p (x)y(x) and so y (a) = f (a) p(a)y (a) p (a)y(a). We continue by calculating successive derivatives, followed by evaluating at x = a. Note that the functions p and f will have to be infinitely differentiable on I in order to carry this method out completely. Example 1. Given the initial-value problem y y = 2e x, y(1) = 0. Here p(x) 1 and q(x) = 2e x are infinitely differentiable on I = (, ). Assuming that the solution y = y(x) has the Taylor series expansion y(x) = y(1) + y (1)(x 1) + y (1) 2! we have y(1) = 0 and y (1) = 2e 1 + y(1) = 2e 1. From the differential equation, (x 1) 2 + y (1) (x 1) 3 +, y (x) = 2e x + y (x) and so y (1) = 2e 1 + y (1) = 2e 1 + 2e 1 = 0. Continuing this process, and so on. y (x) = 2e x + y (x) and y (1) = 2e 1 y (4) (x) = 2e 1 + y (x) and y (4) (1) = 0 y (5) (x) = 2e 1 + y (4) (x) and y (5) (1) = 2e 1 In general, we get y (n) (x) = ( 1) n+1 2e x + y (n 1) (x) and 0, n even y (n) (1) = 2e 1, n odd Therefore, y(x) = 2e 1 (x 1) + 2e 1 (x 1) 3 + 2e 1 (x 1) 5 + = 5! 2e 1 (2n + 1)! (x 1)2n+1. Of course, we could have solved this initial-value problem using the method of Section 2.1. Using that method, the solution is y(x) = e x 2 e x. You can verify that the Taylor series expansion in powers of x 1 of this function is the series given above. 218

3 This method can be applied equally well to higher order equations. For example, suppose we are given the second order linear initial-value problem y + p(x)y + q(x)y = f(x), y(a) = α, y (a) = β. As above, we assume that the solution y = y(x) has the Taylor series expansion y(x) = y(a) + y (a)(x a) + y (a) 2! (x a) 2 + y (a) (x a) 3 + = y (n) (a) (x a) n. on some interval I = (a R, a+r). The initial conditions, y(a) = α, y (a) = β, determine the first two terms in the series expansion, and from the differential equation, we have y (x) = f(x) p(x)y (x) q(x)y(x) and so y (a) = f(a) p(a)y (a) q(a)y(a) This determines the third term of the expansion. Differentiating the differential equation (again, assuming that p, q and f are differentiable functions on I), we get y (x) = f (x) p(x)y (x) p (x)y (x) q(x)y (x) q (x)y(x) and so y (a) = f (a) p(a)y (a) p (a)y (a) q(a)y (a) q (a)y(a). We obtain the higher ordered derivatives by successive differentiation followed by evaluation at x = a. Example 2. Given the initial-value problem y + x 2 y 2y = 0, y(0) = 1, y (0) = 1. Here p(x) x 2, q(x) 2, f(x) 0 are infinitely differentiable on I = (, ). Assuming that the solution y = y(x) has the Taylor series expansion y(x) = y(0) + y (0)x + y (0) 2! x 2 + y (0) x 3 +, we have y(0) = 1, y (0) = 1, and from the differential equation, y (x) = x 2 y (x) + 2y(x) and so y (0) = 2. Continuing this process, y (x) = x 2 y 2xy + 2y and y (0) = 2 y (4) (x) = x 2 y 4xy 2y + 2y and y (4) (1) = 6 y (5) (x) = x 2 y (4) 6xy 6y + 2y and y (5) (0) = 16 and so on. 219

4 Therefore, we find that y(x) = 1 x + 2 2! x2 2 x ! x4 16 5! x5 + = 1 x + x x x x5 + The Taylor Series method can also be applied to nonlinear initial-value problems. Example 3. Consider the initial-value problem y = 1 + y 2, y(0) = 0. Assume that the solution y has a Taylor series expansion in powers of x which converges for x ( R, R) for some R > 0. Then y(x) = y(0) + y (0)x + y (0) 2! x 2 + y (0) x 3 +. We will calculate the first several terms of this series. From the initial condition, y(0) = 0, and from the differential equation y (0) = 1. We calculate successive derivatives of y using the differential equation, and we evaluate the derivatives at x = 0. y = 2yy, y (0) = 2y(0)y (0) = 0 y = 2yy + 2(y ) 2, y (0) = 2 y (4) = 2yy + 2y y + 4y y = 2yy + 6y y, y (4) (0) = 0 y (5) = 2yy (4) + 2y y + 6y y + 6(y ) 2 = 2yy (4) + 8y y + 6(y ) 2, y (5) (0) = 16 and so on. Therefore, we have y(x) = x + 2 x ! x5 + = x x x5 + Exercises 6.1. Use the Taylor series method to find a power series expansion of the solution of the initialvalue problem. Find at least three non-zero terms beyond those given by the initial conditions. 1. (1 x 2 )y xy = 1, y(0) = 0 2. y + xy = ln x y(1) = 1 3. y + xy 2y = 0, y(0) = 1, y (0) = 0 4. y + xy 2y = 0, y(0) = 0, y (0) = 1 220

5 5. y + e x y xy = 0, y(0) = 1, y (0) = 0 6. y 2xy + x 2 y = 0, y(0) = 1, y (0) = 1 7. y 2xy = 0, y(2) = 1, y (2) = 0 8. y 2xy = x 2, y(1) = 0, y (1) = 2 The equations in Exercises 9 14 are nonlinear. Use the Taylor series method to find a power series expansion of the solution of the initial-value problem. Find at least three non-zero terms beyond those given by the initial conditions. 9. y = x 2 + y 2, y(1) = y = 1 x 2 y 2, y(0) = y = sin y, y(1) = π/2 12. y = ln(y x), y(1) = yy + 3y 2 = 0, y(0) = 1, y (0) = Find a Taylor series solution of yy = 1, y(1) = 1. The differential equation is separable. Solve this initial-value problem using the method of Section 2.2. Compare the two results. Section 6.2. The Power Series Method; Ordinary Points In this section we discuss a method for finding power series solutions of linear differential equations. In Section 3.7 we stated that an n th -order linear differential equation is an equation that can be written in the form y (n) + p n 1 (x)y (n 1) + p n 2 (x)y (n 2) + + p 1 (x)y + p 0 (x)y = f(x) (L) where the functions p 0, p 1,..., p n 1 (called the coefficients) and f (called the forcing function or nonhomogeneous term) are continuous on some interval I. We will be seeking solutions of (L) having the form y(x) = c n (x a) n (T) for some real number a I. 221

6 In the preceding section we saw that we could find a series solution provided the functions involved in the equation were infinitely differentiable on the interval I. We will need to elaborate on this restriction in this section. Recall that a function g defined on an interval I containing a point x = a is analytic at x = a if g has a Taylor series expansion in powers of (x a) with radius of convergence R > 0. For the purposes of this section, we assume that the coefficients p 0, p 1,..., p n 1 and f are analytic functions. That is, if we are seeking a power series solution in the form (T), then we assume that the coefficient functions P, Q, R and G each have a Taylor series expansion in powers of (x a) with a positive radius of convergence. Before we present the power series methods, we need to introduce some concepts which relate the coefficients of a linear equation with the existence, uniqueness and behavior of solutions. Singular Points, Ordinary Points In Section 2.1 we studied first order linear differential equations written in the standard form y + p(x)y = f(x) (S) where p and f were assumed to be continuous functions on an interval I. In many examples and exercises in that section, we started with an equation in the form P(x)y + Q(x)y = G(x) (L) Our first step, then, was to divide the equation (L) by P(x) to obtain an equation in the form (S). We then considered (S) on intervals which did not contain points at which P(x) = 0. A point c is a singular point of (L) if P(c) = 0. Example 1. Consider the differential equations: (a) (1 + x 2 )y + xy = 2x Here P(x) = 1 + x 2, Q(x) = x, G(x) = 2x are continuous functions on the interval J = (, ). Since 1 + x 2 0 for all x J, this equation has no singular points. Using the solution method presented in Section 2.1, the general solution of this equation is C y(x) = x 2 (b) (x 1)y 2y = 2 Here P(x) = x 1, Q(x) = 2, G(x) 0 are continuous functions on the interval J = (, ). Since x 1 = 0 when x = 1, 1 is a singular point. From Section 2.1, the general solution of this equation is y(x) = 1 + C(x 1)

7 (c) (4 x 2 )y 4xy = 4x Here q(x) = 4 x 2, r(x) = x, g(x) = 2x are continuous functions on the interval J = (, ). Since 4 x 2 = 0 when x = ±2, 2 and 2 are singular points. The general solution of this equation is y(x) = 4 + C 4 x 2. Remarks. Equations (b) and (c) in this example illustrate possible effects on the solutions of an equation with singular points. In (b), the one-parameter family of solutions is defined for all x, but every member of the family has the value 1 at x = 1. Therefore, every member of the family is a solution of the initial-value problem but none of the initial-value problems has a solution. (x 1)y 2y = 2, y(1) = 1 (x 1)y 2y = 2, y(1) = α, α 1 On the other hand, in (c), the only member of the one-parameter family of solutions which is defined at either of the singular points is y 4 when C = 0. All other members of the family are undefined at the singular points. The ideas presented above for first order linear equations also apply to second order linear equations. The general second order linear differential equation is an equation of the form q 0 (x)y + q 1 (x)y + q 2 (x)y = f(x) The points at which q 0 (x) = 0, if any, are called singular points of the equation Example 2. Consider the first-order linear differential equation y y = 0 As we saw in Chapter 1 and in Section 2.1, the one-parameter family of solutions is y = Ce x. Assume that the differential equation has a power series solution of the form y = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + = a n x n (1) which converges on some interval I = ( R, R), 0 < R. Since a power series can be differentiated term-by-term to produce a new series with the same radius of convergence, we have y = a 1 + 2a 2 x + 3a 3 x 2 + = na n x n 1 (2) 223

8 Substituting (1) and (2) into the differential equation, we get or na n x n 1 a n x n = 0 (3) (a 1 + 2a 2 x + 3a 3 x 2 + ) (a 0 + a 1 x + a 2 x 2 + a 3 x 3 + ) = 0 which can be written a 1 a 0 + (2a 2 a 1 )x + (3a 3 a 2 )x 2 + = 0 Since the powers of x are linearly independent functions, it follows that a 1 a 0 = 0, 2a 2 a 1 = 0, 3a 3 a 2 = 0, and, in general, na n a n 1, n = 1, 2, 3,... We can use this formula to express each coefficient a n, n 1 in terms of a 0. a 1 = a 0, a 2 = a 1 2 = a 0 2, a 3 = a 2 3 = a 0 2 3,... a n = a n 1 n = a 0,... Setting a 0 = C, an arbitrary constant, and substituting these equations into (1), we have y(x) = C xn = C x n. (4) Of course, we recognize the power series in (4) as the Maclaurin series expansion of e x so we get, as expected, the solution y = Ce x. The calculations involved in substituting the power series for y and y into the differential equation can be organized using summation notation. In the example we obtained equation (3) na n x n 1 a n x n = 0 The index of summation, n, is a dummy variable, comparable to the variable of integration in a definite integral, and so changes of indices are possible. In particular, if we let m = n 1, (which means n = m + 1), then the first summation in (4) can be written na n x n 1 = (m + 1)a m+1 x m. The second summation can, of course be, written as a n x n = a m x m Thus (3) becomes (m + 1)a m+1 x m a m x m = 0 224

9 or from which it follows that [(m + 1)a m+1 a m ]x m = 0 (m + 1)a m+1 a m = 0, m = 0, 1, 2,... (5) Thus, the coefficients in the power series expansion of the solution can be obtained by solving (5), a so-called first order recursion formula whose solution is a m = a 0 m!. The form of the power series in Example 1, that is, powers of x, was chosen for convenience. In general, the form of the assumed power series is usually dictated by the particular equation or by an initial-value problem whose solution is sought. Example 3. Consider the initial-value problem which we solved in Section 6.1. y y = 2e x, y(1) = 0. Assume that the solution has a power series expansion of the form y = a n (x 1) n. (1) Note that we chose an expansion about x = 1 since the initial value is specified at 1. The derivative of y is: y = na n (x 1) n 1. (2) The Taylor series expansion of e x in powers of (x 1) is: e x = ( 1) n e 1 (x 1) n. (3) Substituting (1), (2), and (3) into the differential equation, we get na n (x 1) n 1 a n (x 1) n = ( 1) n e 1 (x 1) n. (4) The first summation on the left can be written equivalently as Thus, (4) becomes (n + 1)a n+1 (x 1) n. (n + 1)a n+1 (x 1) n a n (x 1) n = 225 ( 1) n e 1 (x 1) n

10 which can be written as [(n + 1)a n+1 a n ](x 1) n = From this equation we get the recursion formula ( 1) n e 1 (x 1) n. (n + 1)a n+1 a n = 2e 1( 1)n, n = 0, 1, 2, Successively setting n = 0, 1, 2,... in the recursion formula, we have a 0 a 1 = 2e 1, Therefore, 2a 2 a 1 = 2e 1, 3a 3 a 2 = 2 12!, 4a 4 a 3 = 2e 1,... 1! e a 1 = a 0 + 2e 1, a 2 = a 0 2!, a 3 = a 0 + 2e 1 Assuming that the pattern continues, we find that a 0, n even a n = a 0 + 2e 1 n odd Applying the initial condition y(1) = 0 in (1), we get 0 = y(1) = so a 0 = 0. We can now conclude that a n = and y(x) = a n (1 1) n = a 0 2e 1 0, n even n odd 2e 1 (x 1)2n+1 (2n + 1)!, a 4 = a 0 4!,... which is the result we obtained using the Taylor series method in Section 6.1. The radius of convergence of this series is R = ; the series converges for all x (, ). In the next example we find a powers series solution of a second order equation. Example 4. Consider the second order equation y + y = 0. As you know from Chapter 3, the general solution of this equation is y = C 1 cos x+c 2 sin x. 226

11 Assume that the differential equation has a power series solution of the form y = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + = a n x n (1) which converges on some interval I = ( R, R), 0 < R. Differentiating twice, we have y = a 1 + 2a 2 x + 3a 3 x 2 + 4a 4 x 3 + = a n x n 1, y = 2a a 3 x + 3 4a 4 x 2 + = n(n 1)a n 2 x n 2, (2) Substituting (1) and (2) into the differential equation, we get n(n 1)a n 2 x n 2 + a n x n = 0 (3) n=2 To add these two series we change the index of summation in the first series by setting m = n 2, and we set and m = n in the second series. The result is (m + 2)(m + 1)a m+2 x m + a m x m = 0 or n=2 [(m + 2)(m + 1)a m+2 + a m ]x m = 0. From this equation we get the second order recursion formula (m + 2)(m + 1)(n + 1)a m+2 + a m = 0, m = 0, 1, 2, Successively setting m = 0, 1, 2,... in the recursion formula, we have a 2 = a 0 2 1, a 3 = a a 4 = a = a , a 5 = a = a , Assuming that the pattern continuous (and returning to the original variable n), we find that ( 1) n a 0, n even (2n)! a n = ( 1) n a 1 (2n + 1)!, n odd Therefore, y = a 0 + a 1 x a 0 2! x2 a 1 x3 + a 0 4! x4 + a 1 5! x5.... Rearranging the terms and factoring out a 0 and a 1, we get [ ] ] y = a 0 1 x2 2! + x4 4!... + a 1 [x x3 + x5 5!... which is ( 1) n 1) n y = a 0 (2n)! x2n + a 1 (2n + 1)! x2n+1 = a 0 cos x + a 1 sin x. 227

12 In the preceding examples, the differential equation had constant coefficients. The power series method can also be applied to equations with nonconstant coefficients. Example 5. Find power series solutions of in powers of (x 1). Set y(x) = y xy = 0 a n (x 1) n. (1) We differentiate twice and substitute into the differential equation to get n=2 n(n 1)a n (x 1) n 2 x a n (x 1) n = 0. Before we can manipulate the series and combine the coefficients of the powers of (x 1), we must write the coefficient x in terms of x 1; x = (x 1) + 1. We now have n=2 n=2 n(n 1)a n (x 1) n 2 [(x 1) + 1] a n (x 1) n = n(n 1)a n (x 1) n 2 (x 1) a n (x 1) n a n (x 1) n = n(n 1)a n (x 1) n 2 a n (x 1) n+1 a n (x 1) n = 0 n=2 By changing indices, this equation can be written (2a 2 a 0 ) + [(n + 2)(n + 1)a n+2 a n a n 1 ](x 1) n = 0. Equating the coefficients of corresponding powers of (x 1), we get a 2 = a 0 2, a 3 = a a 0 6, a 4 = a a 1 12, a 5 = a a 1 120, Substituting these coefficients into (1), we have [ y(x) = a (x 1) (x 1) (x 1)4 + 1 ] 30 (x 1)5 + + (2) [ a 1 (x 1) (x 1) (x 1)5 + 1 ] 120 (x 1)5 + It can be verified that the two series y ( x) = (x 1) (x 1) (x 1) (x 1)

13 and y 2 (x) = (x 1) (x 1) (x 1) (x 1)5 + are linearly independent. Therefore (2), with a 0 and a 1 being two independent arbitrary constants, is the general solution of the given differential equation. Exercises 6.2 Find power series representations of the general solution of each of the following differential equations. Express your powers series solutions in terms of elementary functions whenever possible. 1. y + 4y = 0 in powers of x. 2. y + xy = 0 in powers of x. 3. y x 2 y y = 0 in powers of x. 4. y + (x 1)y + 3y = 0 in powers of (x 1). 5. y (x + 1)y y = 0 in powers of (x + 1). 6. (1 + x 2 )y + 2xy = 0 in powers of x. 7. (x 2 1)y xy + y = 0 in powers of x. 8. x 2 y xy + 2y = 0 in powers of (x 1). 9. (1 + x 2 )y 4xy + 6y = 0 in powers of x. 10. (1 x 2 )y 4xy 2y = 0 in powers of x. Find power series solutions of the following initial-value problems. 11. y 3y + 2y = 0, y(0) = 0, y (0) = y + xy 2y = 0, y(0) = 1, y (0) = y + xy 2y = 0, y(0) = 0, y (0) = x 2 y + xy 4y = 0, y(1) = 0, y (1) = 1. Find power series solution of the following nonhomogeneous differential equations 15. y y = e x in powers of x. 16. x 2 y 2xy + 2y = x + 1 in powers of (x + 1). 229

14 17. y 2xy + 2y = 1 1 x in powers of x. 18. y xy = ln(1 + x) in powers of x. 19. Consider the linear homogeneous equation where a, b and c are constants. (ax 2 + 1)y + bxy + cy = 0. (a) Give a necessary and sufficient condition for the equation to have a polynomial solution. (b) Give a necessary and sufficient condition for the equation to have exactly one polynomial solution and one power series solution. (c) Give a necessary and sufficient condition for the equation to have only polynomial solutions. 230