Advanced Engineering Mathematics, 7 th Edition Peter V. O’Neil © 2012 Cengage Learning Engineering. All Rights Reserved. CHAPTER 4 Series Solutions.

© 2012 Cengage Learning Engineering. All Rights Reserved.3 This solution is in closed form, which means that it is a finite algebraic combination of elementary functions (such as polynomials, exponentials, sines and cosines, and the like). CHAPTER 4 : Page 129

© 2012 Cengage Learning Engineering. All Rights Reserved.4 We may, however, encounter problems for which there is no closed form solution. For example, has the unique solution This solution (while explicit) has no elementary, closed form expression. CHAPTER 4 : Page 129

© 2012 Cengage Learning Engineering. All Rights Reserved.5 In such a case, we might try a numerical approximation. However, we may also be able to write a series solution that contains useful information. In this chapter, we will deal with two kinds of series solutions: power series (Section 4.1) and Frobenius series (Section 4.2). CHAPTER 4 : Page 129

© 2012 Cengage Learning Engineering. All Rights Reserved.6 4.1 Power Series Solutions CHAPTER 4 : Page 129 A function f is called analytic at x 0 if f (x) has a power series representation in some interval (x 0 − h, x 0 + h) about x 0. In this interval, where the a n ’s are the Taylor coefficients of f (x) at x 0 :

© 2012 Cengage Learning Engineering. All Rights Reserved.7 4.1 Power Series Solutions Here n! (n factorial) is the product of the integers from 1 through n if n is a positive integer, and 0! = 1 by definition. The symbol f (n) (x 0 ) denotes the nth derivative of f evaluated at x 0. CHAPTER 4 : Page 130

© 2012 Cengage Learning Engineering. All Rights Reserved.8 4.1 Power Series Solutions As examples of power series representations, sin(x) expanded about 0 is for all x, and the geometric series is for −1 < x < 1. CHAPTER 4 : Page 130

© 2012 Cengage Learning Engineering. All Rights Reserved.9 4.1 Power Series Solutions An initial value problem having analytic coefficients has analytic solutions. We will state this for the first- and second-order cases when the differential equation is linear. CHAPTER 4 : Page 130

© 2012 Cengage Learning Engineering. All Rights Reserved.10 THEOREM 4.1 1.1. If p and q are analytic at x 0, then the problem has a unique solution that is analytic at x 0. 2.If p, q, and f are analytic at x 0, then the problem has a unique solution that is analytic at x 0. CHAPTER 4 : Page 130

© 2012 Cengage Learning Engineering. All Rights Reserved.11 4.1 Power Series Solutions We are therefore justified in seeking power series solutions of linear equations having analytic coefficients. This strategy may be carried out by substituting into the differential equation and attempting to solve for the a n s. CHAPTER 4 : Page 130

© 2012 Cengage Learning Engineering. All Rights Reserved.12 EXAMPLE 4.1 We will solve We can solve this using an integrating factor, obtaining This is correct, but it involves an integral we cannot evaluate in closed form. CHAPTER 4 : Page 130

© 2012 Cengage Learning Engineering. All Rights Reserved.13 EXAMPLE 4.1 For a series solution, let Then with the summation starting at 1, because the derivative of the first term a 0 of the power series for y is zero. CHAPTER 4 : Page 130

© 2012 Cengage Learning Engineering. All Rights Reserved.15 EXAMPLE 4.1 We would like to combine these series and factor out a common power of x to solve for the a n ’s. To do this, write 1/(1−x) as a power series about 0 as for −1 < x < 1. Substitute this into equation (4.1) to obtain (4.2) CHAPTER 4 : Page 131

© 2012 Cengage Learning Engineering. All Rights Reserved.16 EXAMPLE 4.1 Now rewrite the series so that they all contain powers x n. This is like a change of variables in the summation index. First, and next, CHAPTER 4 : Page 131

© 2012 Cengage Learning Engineering. All Rights Reserved.17 EXAMPLE 4.1 Now equation (4.2) can be written as (4.3) These rearrangements allow us to combine these summations for n = 1, 2, ··· and to write the n = 0 terms separately to obtain (4.4) CHAPTER 4 : Page 131

© 2012 Cengage Learning Engineering. All Rights Reserved.18 EXAMPLE 4.1 Because the right side of equation (4.4) is zero for all x in (−1, 1), the coefficient of each power of x on the left, as well as the constant term a 1 − 1, must equal zero. This gives us and CHAPTER 4 : Page 131

© 2012 Cengage Learning Engineering. All Rights Reserved.19 EXAMPLE 4.1 Then a 1 = 1, and This is a recurrence relation for the coefficients, giving a n+1 in terms of a preceding coefficient a n−1. CHAPTER 4 : Page 131

© 2012 Cengage Learning Engineering. All Rights Reserved.22 EXAMPLE 4.1 With the coefficients computed thus far, the solution has the form This has one arbitrary constant, a 0, as expected. By continuing to use the recurrence relation, we can compute as many terms of the series as we like. CHAPTER 4 : Page 132

© 2012 Cengage Learning Engineering. All Rights Reserved.23 EXAMPLE 4.2 We will find a power series solution of expanded about x 0 = 0. Substitute into the differential equation. This will require that we compute CHAPTER 4 : Page 132

© 2012 Cengage Learning Engineering. All Rights Reserved.25 EXAMPLE 4.2 We will shift indices so that the power of x in both summations is the same, allowing us to combine terms from both summations. One way to do this is to write and CHAPTER 4 : Page 132

© 2012 Cengage Learning Engineering. All Rights Reserved.26 EXAMPLE 4.2 Now equation (4.5) is We can combine the terms for n ≥ 2 in one summation. This requires that we write the n = 0 and n = 1 terms in the last equation separately, or else we lose terms: CHAPTER 4 : Page 133

© 2012 Cengage Learning Engineering. All Rights Reserved.27 EXAMPLE 4.2 The left side can be zero for all x in some interval (−h, h) only if the coefficient of each power of x is zero: and The last equation gives us (4.6) CHAPTER 4 : Page 133

© 2012 Cengage Learning Engineering. All Rights Reserved.28 EXAMPLE 4.2 This is a recurrence relation for the coefficients of the series solution, giving us a 4 in terms of a 0, a 5 in terms of a 1, and so on. Recurrence relations always give a coefficient in terms of one or more previous coefficients, allowing us to generate as many terms of the series solution as we want. To illustrate, use n =2 in equation (4.6) to obtain CHAPTER 4 : Page 133

© 2012 Cengage Learning Engineering. All Rights Reserved.32 EXAMPLE 4.2 This is the general solution, since a 0 and a 1 are arbitrary constants. Because a 0 = y(0) and a 1 = y(0), a unique solution is determined by specifying these two constants. CHAPTER 4 : Page 134

© 2012 Cengage Learning Engineering. All Rights Reserved.34 4.2 Frobenius Solutions We will focus on the differential equation (4.7) If P(x)  0 on some interval, then we can divide by P(x) to obtain the standard form (4.8) CHAPTER 4 : Page 134

© 2012 Cengage Learning Engineering. All Rights Reserved.35 4.2 Frobenius Solutions CHAPTER 4 : Page 134 If P(x 0 ) = 0, we call x 0 a singular point of equation (4.7). This singular point regular if are analytic at x 0. A singular point that is not regular is an irregular singular point.

© 2012 Cengage Learning Engineering. All Rights Reserved.36 EXAMPLE 4.3 has singular points at 0 and 2. Now is not analytic (or even defined) at 0, so 0 is an irregular singular point. CHAPTER 4 : Page 134

© 2012 Cengage Learning Engineering. All Rights Reserved.38 4.2 Frobenius Solutions We will not treat the case of an irregular singular point. If equation (4.7) has a regular singular point at x 0, there may be no power series solution about x 0, but there will be a Frobenius series solution, which has the form with c 0  0. We must solve for the coefficients c n and a number r to make this series a solution. We will look at an example to get some feeling for how this works and then examine the method more critically. CHAPTER 4 : Page 135

© 2012 Cengage Learning Engineering. All Rights Reserved.40 EXAMPLE 4.4 Notice that the n = 0 term in the proposed series solution is c 0 x r, which is not constant if c 0  0, so the series for the derivatives begins with n = 0 (unlike what we saw with power series). Shift indices in the third summation to write this equation as CHAPTER 4 : Page 135

© 2012 Cengage Learning Engineering. All Rights Reserved.41 EXAMPLE 4.4 Combine terms to write Since we require that c 0  0, the coefficient of x r is zero only if This is called the indicial equation and is used to solve for r, obtaining the repeated root r = −2. CHAPTER 4 : Page 135

© 2012 Cengage Learning Engineering. All Rights Reserved.45 EXAMPLE 4.4 In general, for n =1, 2, 3, ···. We have found the Frobenius solution for x  0. This series converges for all nonzero x. CHAPTER 4 : Page 136

© 2012 Cengage Learning Engineering. All Rights Reserved.46 4.2 Frobenius Solutions Usually, we cannot expect the recurrence equation for c n to have such a simple form. Example 4.4 shows that an equation with a regular singular point may have only one Frobenius series solution about that point. A second, linearly independent solution is needed. The following theorem tells us how to produce two linearly independent solutions. For convenience, the statement is posed in terms of x 0 = 0. CHAPTER 4 : Page 136

© 2012 Cengage Learning Engineering. All Rights Reserved.47 THEOREM 4.2 Suppose 0 is a regular singular point of Then (1) The differential equation has a Frobenius solution with c 0  0. This series converges in some interval (0, h) or (−h, 0). CHAPTER 4 : Page 136

© 2012 Cengage Learning Engineering. All Rights Reserved.48 THEOREM 4.2 Suppose that the indicial equation has real roots r 1 and r 2 with r 1 ≥ r 2. Then the following conclusions hold. (2) If r 1 − r 2 is not a positive integer, then there are two linearly independent Frobenius solutions with c 0  0 and c ∗ 0  0. These solutions are valid at least in an interval (0, h) or (−h, 0). CHAPTER 4 : Page 137

© 2012 Cengage Learning Engineering. All Rights Reserved.49 THEOREM 4.2 (3) If r 1 − r 2 = 0, then there is a Frobenius solution with c 0  0, and there is a second solution These solutions are linearly independent on some interval (0, h). CHAPTER 4 : Page 137

© 2012 Cengage Learning Engineering. All Rights Reserved.50 THEOREM 4.2 (4) If r 1 − r 2 is a positive integer, then there is a Frobenius solution with c 0  0, and there is a second solution with c ∗ 0  0. y 1 and y 2 are linearly independent solutions on some interval (0, h). CHAPTER 4 : Page 137

© 2012 Cengage Learning Engineering. All Rights Reserved.51 4.2 Frobenius Solutions The method of Frobenius consists of using Frobenius series and Theorem 4.2 to solve equation (4.7) in some interval (−h, h), (0, h), or (−h, 0), assuming that 0 is a regular singular point. CHAPTER 4 : Page 137

© 2012 Cengage Learning Engineering. All Rights Reserved.52 4.2 Frobenius Solutions Proceed as follows: Step 1. –Substitute into the differential equation, and solve for the roots r 1 and r 2 of the indicial equation for r. This yields a Frobenius solution (which may or may not be a power series). CHAPTER 4 : Page 137

© 2012 Cengage Learning Engineering. All Rights Reserved.53 4.2 Frobenius Solutions Step 2. –Depending on which of Cases (2), (3), or (4) of Theorem 4.2 applies, the theorem provides a template for a second solution which is linearly independent from the first. Once we know what this second solution looks like, we can substitute its general form into the differential equation and solve for the coefficients and, in Case (4), the constant k. CHAPTER 4 : Page 137

© 2012 Cengage Learning Engineering. All Rights Reserved.54 4.2 Frobenius Solutions We will illustrate the Cases (2), (3), and (4) of the Frobenius theorem. For case (2), Example 4.5, we will provide all of the details. In Cases (3) and (4) (Examples 4.6, 4.7, and 4.8), we will omit some of the calculations and include just those that relate to the main point of that case. CHAPTER 4 : Page 137

© 2012 Cengage Learning Engineering. All Rights Reserved.56 EXAMPLE 4.5 Case 2 of the Frobenius Theorem It is routine to check that 0 is a regular singular point. Substitute the Frobenius series to obtain CHAPTER 4 : Page 138

© 2012 Cengage Learning Engineering. All Rights Reserved.57 EXAMPLE 4.5 Case 2 of the Frobenius Theorem In order to be able to factor x n+r from most terms, shift indices in the third and fourth summations to write this equation as CHAPTER 4 : Page 138

© 2012 Cengage Learning Engineering. All Rights Reserved.58 EXAMPLE 4.5 Case 2 of the Frobenius Theorem This equation will hold if the coefficient of each power of x is zero: (4.9) and for n =1, 2, 3, ···, (4.10) CHAPTER 4 : Page 138

© 2012 Cengage Learning Engineering. All Rights Reserved.59 EXAMPLE 4.5 Case 2 of the Frobenius Theorem Assuming that c 0  0, an essential requirement of the method, equation (4.9) implies that (4.11) CHAPTER 4 : Page 138

© 2012 Cengage Learning Engineering. All Rights Reserved.60 EXAMPLE 4.5 Case 2 of the Frobenius Theorem This is the indicial equation for this differential equation. It has the roots r 1 = 1 and r 2 = −1/2. This puts us in case 2 of the Frobenius theorem. From equation (4.10), we obtain the recurrence relation for n =1, 2, 3, ···. CHAPTER 4 : Page 138

© 2012 Cengage Learning Engineering. All Rights Reserved.61 EXAMPLE 4.5 Case 2 of the Frobenius Theorem First put r 1 = 1 into the recurrence relation to obtain for n = 1, 2, 3, ···. CHAPTER 4 : Page 138

© 2012 Cengage Learning Engineering. All Rights Reserved.63 EXAMPLE 4.5 Case 2 of the Frobenius Theorem One Frobenius solution is Because r 1 is a nonnegative integer, this first Frobenius series is actually a power series about 0. CHAPTER 4 : Page 139

© 2012 Cengage Learning Engineering. All Rights Reserved.64 EXAMPLE 4.5 Case 2 of the Frobenius Theorem For a second Frobenius solution, substitute r = r 2 = −1/2 into the recurrence relation. To avoid confusion with the first solution, we will denote the coefficients c ∗ n instead of c n.We obtain for n = 1, 2, 3, ···. CHAPTER 4 : Page 139

© 2012 Cengage Learning Engineering. All Rights Reserved.65 EXAMPLE 4.5 Case 2 of the Frobenius Theorem This simplifies to for n = 1, 2, 3, ···. It happens in this example that c ∗ 1 = 0, so each c ∗ n =0 for n = 1, 2, 3, ···, and the second Frobenius solution is for x > 0. CHAPTER 4 : Page 139

© 2012 Cengage Learning Engineering. All Rights Reserved.66 EXAMPLE 4.6 Case 3 of the Frobenius Theorem We will solve In Example 4.4, we found the indicial equation with repeated root r 1 = r 2 = −2 and the recurrence relation for n =1, 2, ···. CHAPTER 4 : Page 139

© 2012 Cengage Learning Engineering. All Rights Reserved.68 EXAMPLE 4.6 Case 3 of the Frobenius Theorem Conclusion (3) of Theorem 4.2 tells us the general form of a second solution that is linearly independent from y 1 (x). Set CHAPTER 4 : Page 139-140

© 2012 Cengage Learning Engineering. All Rights Reserved.69 EXAMPLE 4.6 Case 3 of the Frobenius Theorem Note that on the right, the series starts at n = 1, not n = 0. Substitute this series into the differential equation and find after some rearranging of terms that CHAPTER 4 : Page 140

© 2012 Cengage Learning Engineering. All Rights Reserved.70 EXAMPLE 4.6 Case 3 of the Frobenius Theorem The bracketed coefficient of ln(x) is zero because y 1 is a solution. Choose c ∗ 0 = 1 (we need only one second solution), shift the indices to write, and substitute the series for y 1 (x) to obtain CHAPTER 4 : Page 140

© 2012 Cengage Learning Engineering. All Rights Reserved.71 EXAMPLE 4.6 Case 3 of the Frobenius Theorem Set the coefficient of each power of x equal to 0. From the coefficient of x −1, we have c ∗ 1 = 2. From the coefficient of x n−2, we obtain (after some routine algebra) or for n =2, 3, 4, ···. CHAPTER 4 : Page 140

© 2012 Cengage Learning Engineering. All Rights Reserved.75 EXAMPLE 4.7 Case 4 of Theorem 4.2 with k = 0 The indicial equation is r 2 − r − 2 = 0 with roots r 1 = 2, r 2 = −1. Now r 1 − r 2 = 3, putting us in Case (4) of the theorem. From the coefficient of x n+r, we obtain the general recurrence relation for n =1, 2, 3, ···. CHAPTER 4 : Page 140

© 2012 Cengage Learning Engineering. All Rights Reserved.76 EXAMPLE 4.7 Case 4 of Theorem 4.2 with k = 0 For a first solution, use r = 2 to obtain the recurrence relation for n = 1, 2, ···. Using this, we obtain a first solution CHAPTER 4 : Page 141

© 2012 Cengage Learning Engineering. All Rights Reserved.77 EXAMPLE 4.7 Case 4 of Theorem 4.2 with k = 0 Now we need a second, linearly independent solution. Put r = −1 into the general recurrence relation to obtain for n = 1, 2, ···. CHAPTER 4 : Page 141

© 2012 Cengage Learning Engineering. All Rights Reserved.78 EXAMPLE 4.7 Case 4 of Theorem 4.2 with k = 0 When n = 3, this gives c ∗ 2 =0, which forces c ∗ n = 0 for n = 2, 3, ···. But then Substitute this into the differential equation to obtain CHAPTER 4 : Page 141

© 2012 Cengage Learning Engineering. All Rights Reserved.79 EXAMPLE 4.7 Case 4 of Theorem 4.2 with k = 0 Then c ∗ 1 = −c ∗ 0 /2, and a second solution is with c ∗ 0 arbitrary but nonzero. The functions y 1 and y 2 form a fundamental set of solutions. In these solutions, there is no y 1 (x) ln(x) term. CHAPTER 4 : Page 141

© 2012 Cengage Learning Engineering. All Rights Reserved.80 EXAMPLE 4.8 Case 4 of Theorem 4.2 with k  0 We will solve which has a regular singular point at 0. Substitute and rearrange terms to obtain CHAPTER 4 : Page 141

© 2012 Cengage Learning Engineering. All Rights Reserved.81 EXAMPLE 4.8 Case 4 of Theorem 4.2 with k  0 The indicial equation is r 2 − r = 0, with roots r 1 = 1, r 2 = 0. Here r 1 − r 2 = 1, a positive integer, putting us in Case (4) of the theorem. The general recurrence relation is for n = 1, 2, ···. CHAPTER 4 : Page 141

© 2012 Cengage Learning Engineering. All Rights Reserved.84 EXAMPLE 4.8 Case 4 of Theorem 4.2 with k  0 For a second solution, put r = 0 into the general recurrence relation to obtain for n=1, 2, ···. CHAPTER 4 : Page 142

© 2012 Cengage Learning Engineering. All Rights Reserved.85 EXAMPLE 4.8 Case 4 of Theorem 4.2 with k  0 If we put n = 1 into this, we obtain c 0 = 0, violating one of the conditions for the method of Frobenius. Here we cannot obtain a second solution as a Frobenius series. Theorem 4.2, Case (4), tells us to look for a second solution of the form CHAPTER 4 : Page 142

© 2012 Cengage Learning Engineering. All Rights Reserved.87 EXAMPLE 4.8 Case 4 of Theorem 4.2 with k  0 Now because y 1 is a solution. For the remaining terms, let c 0 = 1 in y 1 (x) for convenience (we need only one more solution) to obtain CHAPTER 4 : Page 142

© 2012 Cengage Learning Engineering. All Rights Reserved.90 EXAMPLE 4.8 Case 4 of Theorem 4.2 with k  0 Furthermore, the recurrence relation is for n =1, 2, ···.Since c ∗ 0 can be any nonzero number, we will for convenience let c ∗ 0 = 1. For a particular solution, we may also choose c ∗ 1 = 0. These give us CHAPTER 4 : Page 143

Advanced Engineering Mathematics, 7 th Edition Peter V. O’Neil © 2012 Cengage Learning Engineering. All Rights Reserved. CHAPTER 4 Series Solutions.

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Advanced Engineering Mathematics, 7 th Edition Peter V. O’Neil © 2012 Cengage Learning Engineering. All Rights Reserved. CHAPTER 4 Series Solutions.

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