Series Solutions of Linear ODEs

Transcription

1 Chapter 2 Series Solutions of Linear ODEs This Chapter is concerned with solutions of linear Ordinary Differential Equations (ODE). We will start by reviewing some basic concepts and solution methods for basic ODEs. For linear ODEs which cannot be solved in elementary functions, we will develop solutions in terms of power series. The latter that will converge in some neighbourhood of regular and regular singular points of those ODEs. 2.1 Ordinary Differential Equations. Solution Structure General and Particular Solutions of an ODE Definition 2.1. An Ordinary Differential Equation (ODE) of order N for the unknown function y(x) is an expression involving x, y(x) and derivatives of y(x) up to order N: E(x, y(x), y (x),..., y (N) (x)) = 0. (2.1) Here E(...) is a function of several variables. Definition 2.2. A function y(x) is a solution to the equation (2.1) if its substitution makes the equation an identity 0 = 0. Fact 2.3. The equation (2.1) has infinitely many solutions y(x). In many cases, all solutions can be described by a single formula, called the general solution: y gen (x) = F (x, C 1,..., C N ). (2.2) Here C 1,..., C N are free constants. Remark 2.4. A general solution for an ODE of order N involves exactly N arbitrary constants C 1,..., C N. Every particular solution y(x) of the equation (2.1) follows from the general solution (2.2) for some choice of constants C 1,..., C N. Example 2.5. Consider an equation y (x) + y(x) sin x = e x+cos x. (2.3) 1

2 It is a first-order equation. Its general solution is derived in the next subsection, and is given by y gen (x) = ( e x + C)e cos x. The equation is of order one, hence the general solution involves one free constant (C). This general solution can be used to produce infinitely many particular solutions, by choosing concrete values for C. For example, the following are particular solutions to (2.3): y 1 (x) = ( e x + 1)e cos x (with C = 1), y 2 (x) = ( e x 67)e cos x (with C = 67), y 3 (x) = ( e x + e π/2 )e cos x (with C = e π/2 ) Linear Homogeneous ODEs, Linear Non-Homogeneous ODEs All differential equations can be classified in the following categories: 1. Nonlinear; 2. Linear: a. Linear non-homogeneous; b. Linear homogeneous. The latter two have a simpler structure of general solution, as we will see. Definition 2.6. An ODE (2.1) is a linear homogeneous ODE if it has the form a N (x)y (N) (x) + a N 1 (x) y (N 1) (x) a 1 (x)y (x) + a 0 (x)y(x) = 0, (2.4) Fact 2.7. The general solution (2.2) has a particularly simple form: y gen (x) = C 1 y 1 (x) C N y N (x), (2.5) where y 1 (x),..., y N (x) is an arbitrary collection of N linearly independent particular solutions of (2.23). Example 2.8. Consider an ODE y (t) + 16y(t) = 0. (2.6) It is a second-order linear homogeneous equation. Its general solution (2.5) contains two arbitrary constants: y gen (x) = C 1 cos 4t + C 2 sin 4t. Here cos 4t, sin 4t are two linearly independent particular solutions. Definition 2.9. The ODE (2.1) is a linear non-homogeneous ODE if it has the form a N (x)y (N) (x) + a N 1 (x) y (N 1) (x) a 1 (x)y (x) + a 0 (x)y(x) = g(x), (2.7) with g(x) 0. 2

3 Fact The general solution of the linear non-homogeneous ODE (2.7) has the form y gen (x) = C 1 y 1 (x) C N y N (x) + y p (x), (2.8) where y 1 (x),..., y N (x) are any N different (linearly independent) particular solutions of the homogeneous equation (2.23), and y p is one particular solution of the non-homogeneous equation (2.8). In other words, y gen.non hom. (x) = y gen.hom. (x) + y partic.non hom. (x), (2.9) i.e., the general solution of the linear non-homogeneous ODE (2.7) is a sum of a general solution of its linear homogeneous version, and a particular solution of (2.7), which takes care of canceling the nonlinearity. Example Consider an ODE y (t) + 16y(t) = 32t. (2.10) It is a second-order linear non-homogeneous equation. We can guess a particular solution easily: for example, take y partic.non hom. (x) = 2t. The general solution of the homogeneous version of the equation has been derived in the previous example. Finally, the general solution of the ODE (2.10) is given by y gen.non hom. (x) = C 1 cos 4t + C 2 sin 4t + 2t, still containing two arbitrary constants, as needed. Keywords 2.1 Linear combination; General solution; Existence and Uniqueness. Problems 2.1 Problem 2.1a. Prove that and u(x) = C 1 cosh(ax) + C 2 sinh(ax) v(x) = D 1 e ax + D 2 e ax are both general solutions of the ODE y (x) a 2 y(x) = 0. [You have to show that (a) a given function is a solution for any values of arbitrary constants, and (b) it is indeed a general solution.] Thus the writing of the general solution is not unique (however, still any form of the general solution contains all particular solutions of a given ODE). 3

4 Problem 2.1b. Prove that and u(x) = C 1 cos(ax) + C 2 sin(ax) v(x) = D 1 e iax + D 2 e iax are both general solutions of the ODE y (x) + a 2 y(x) = 0. Problem 2.1c. Check that the function ( ) y(x) = x Aex/2 cos 2 x + Be x/2 sin is a solution of an ODE y (x) + y (x) + 4y(x) = x, for any constant A, B. Is it a general solution? ( ) 7 2 x 4

5 2.2 ODE Initial Value Problems IVP. Existence and Uniqueness Definition An Initial Value Problem (IVP) for the ODE (2.1) of order N is the ODE itself with N initial conditions (IC) given at some point x 0 : y (N) (x) = F (x, y(x), y (x),..., y (N 1) (x)), y(x 0 ) = y 0, y (x 0 ) = y 1, (2.11). y (N 1) (x 0 ) = y N 1. Fact The number of initial conditions of a well-posed ODE IVP equals the order of the ODE, N. Under mild conditions on the ODE form itself, the existence and uniqueness theorem guarantees the existence of a unique solution of each such IVP. The ODE itself (with no initial conditions specified) has infinitely many solutions. However only one of them satisfies a well-posed IVP. The following theorem holds. Theorem Consider an ODE IVP (2.11) in an interval I = {x : x x 0 ε}, ε > 0. Suppose that in I, the function F (...) is continuous in x, and satisfies the Lipschitz condition in y, y,...,y (N 1). Then there is a number ε 1 > 0 such that there exists a unique solution of the IVP (2.11) in the interval I 1 = {x : x x 0 ε 1 }. We note that the same holds under a stronger (but easier-to-verify) requirement that the function F (...) has continuous partial derivatives with respect to y, y,..., y (N 1) Solution of ODE IVPs In order to solve an IVP (2.11), one first finds the general solution (2.2) of the equation (2.11a), and then uses N initial conditions to find numeric values of all N constants C 1,..., C N in the general solution. As a result, one has a unique solution to the IVP (2.11). Example Consider an ODE y (x) (4x 2 )y (x) + (y (x)) 2 = sin(x). It is third-order, hence a well-posed IVP for it can have the form, for example, y (x) (4x 2 )y (x) + (y (x)) 2 = sin(x), y(3) = 1, y (3) = 1, y (3) = 42 with three initial conditions. This IVP has a unique solution y(x). (2.12) Example Consider an IVP { y (x) + y(x) sin x = e x+cos x, y(π/2) = 0. (2.13) 5

6 The equation order is 1, and the number of ICs is also 1. The solution sequence is as follows. First, find the general solution. As discussed above, the general solution is y gen (x) = ( e x + C)e cos x. Second, use IC to find the values of arbitrary constants, and thus specify the unique solution. We have y(π/2) = 0 = ( e π/2 + C)e 1, Hence C = e π/2. Finally, the unique solution to the IVP (2.13) is given by y(x) = ( e x + e π/2 )e cos x. Keywords 2.2 Lipschitz condition; Lipschitz-continuous functions. Problems 2.2 Problem 2.2a. Find the unique solution of the IVP y (x) = 2x, y(1) = 0, y (1) = 100. Problem 2.2b. Look at Theorem Consider an IVP y (x) = (y(x)) 2/3, y(0) = 0. Show that there is more than one solution y(x) satisfying this IVP (derive those solutions). Show that the Theorem 2.14 does not hold because the RHS of the ODE fails to satisfy the Lipschitz condition. 6

7 2.3 Finding Exact Solutions of Basic Classes of ODEs Solution of First Order Linear ODEs First Order Linear Homogeneous ODE We start from a simplest ODE y (x) = a(x)y(x), (2.14) where a(x) is a given integrable function. The general solution of (2.14) is done by separation of variables. Separating variables, one has dy y = a(x)dx. Integration yields ln y(x) = x x 0 a(q) dq + ln C, where C = const, and x 0 is any point in the solution domain. Hence the general solution of (2.14) is found: y gen (x) = Ce x x 0 a(q) dq. (2.15) First Order Linear Non-Homogeneous ODE For the non-homogeneous ODE y (x) = a(x)y(x) + b(x), (2.16) the solution structure is given by (2.9). In order to obtain both general homogeneous and particular non-homogeneous solution parts, one can apply the variation of parameters. Here one assumes that the solution of the non-homogeneous ODE (2.16) is given by (2.15) with C = C(x). This extra freedom will accommodate b(x) 0 in (2.16). Indeed, y (x) = d dx C(x) e x x 0 a(q) dq = (C (x) + a(x)) e x x 0 a(q) dq. Substitute it in the ODE (2.16) and cancel the a-term to get C (x) e x x 0 a(q) dq = b(x). Hence one finds x C(x) = b(p)e x x a(q) dq 0 dp + S, x 1 where x 0, x 1 are some fixed numbers and S is the integration constant. Finally, the general solution of the linear non-homogeneous ODE(2.16) is given by [ x x x a(q) y gen (x) = C(x)e dq 0 = b(p)e x x a(q) dq x 0 x a(q) dp + S] e dq 0, (2.17) x 1 and involves one free constant S, as required. 7

8 Example Find the general solution of the ODE y (x) + y(x) sin x = e x+cos x. (2.18) We start from the homogeneous version of this linear non-homogeneous 1st-order ODE: y (x) + y(x) sin x = 0. Separating variables, one has dy y = sin x dx. Integrating and solving for y(x), one gets y gen. hom. (x) = Ce cos x. Second, one varies a constant: y gen. non hom. (x) = C(x)e cos x. Substituting into the full ODE (2.18), one has (C (x) C(x) sin x)e cos x + C(x) sin xe cos x = e x+cos x. Upon the usual cancelation, one gets so Finally, C (x) = e x, C(x) = C 1 e x. y gen. non hom. (x) = C(x)e cos x = (C 1 e x )e cos x Separable First-Order ODEs A general first-order ODE y (x) = F (x, y(x)) is separable if it has the form y (x) = f(x)g(y(x)). Such equations are easily solved: separating variables, one has dy g(y) = f(x)dx, and the general solution y gen (x) is implicitly given by y(x) y 0 dq g(q) = x x 0 f(s) ds + C. 8

9 Example Find the general solution of the ODE y (x) = x2 y(x) 2. Solution: one has y 2 dy = x 2 dx, hence y 3 /3 = x 3 /3 + C, or y gen (x) = (3C + x 3 ) 1/ Basic Reduction of Order If an ODE of order N has the form, E(x, y (x),..., y (N) (x)) = 0, (2.19) where E(...) does not contain y(x) explicitly, then one denotes y (x) = z(x), and gets a reducedorder (order N 1) ODE for z(x): E(x, z(x),..., z (N 1) (x)) = 0. (2.20) The latter is generally easier to solve. One then reconstructs y gen (x) from the general solution z gen (x) by a single integration. Example Find the general solution of the ODE x 2 y (x) + xy (x) = 0. This is equivalent to x 2 z (x) + xz(x) = 0, z = y. The latter ODE is separable. One has dz z = dx x, and hence (show!) z(x) = C/x. To get y(x) we integrate: y gen (x) = z(x)dx = C ln x + D, which is indeed a general solution of the above 2nd-order ODE, since this solution contains two arbitrary constants. 9

10 Problems 2.3 Problem 2.3a. Find the general solution of the ODE y (x) + 3y(x) = 10x sin x. Problem 2.3b. Find the general solution of the ODE y (x) = y 2 (x) cos 2x. Problem 2.3c. Find the general solution of the ODE y (5) (x) + 2 x y(4) (x) = 0. 10

11 2.4 Solution of Some Linear Homogeneous ODEs Linear Dependence and Wronskian Determinants In order to solve linear homogeneous ODEs (2.23) (as well as linear non-homogeneous ODE (2.7)!) one needs to come up with a sufficient number of linearly independent particular solutions of a corresponding linear homogeneous ODE. For that purpose, one needs an effective linear dependence test, that would tell if some given set of functions {y 1 (x),..., y n (x)} is linearly dependent. The following objects are extremely helpful. Definition Consider an ordered set of n smooth functions {y i (x)} n i=1. The Wronskian matrix for that set is an n n matrix given by y 1 (x) y 2 (x) y n (x) y 1(x) y 2(x) y n(x) W[y 1 (x),..., y n (x)] = (2.21) y (n) 1 (x) y (n) 2 (x) y n (n) (x) Note that W[y 1 (x),..., y n (x)] can be called a matrix function of x, by analogy with the term vector function. Definition The Wronskian determinant of a set of n smooth functions {y i (x)} n i=1 is given by W [y 1 (x),..., y n (x)] = det W[y 1 (x),..., y n (x)]. (2.22) Then the desired linear dependence criterion is as follows. Theorem A set of n smooth functions {y i (x)} n i=1 is linearly dependent if and only if their Wronskian determinant W [y 1 (x),..., y n (x)] 0, i.e., is identically zero, for all x in its domain. The proof is rather straightforward; in particular, if functions are linearly dependent, and one y i (x) is a linear combination of the others, it is easy to see that a corresponding column of W is a linear combination of other columns, and hence W = 0. Example Show that x and x 3 are linearly independent. We compute W = x x3 1 3x 2 = 3x3 x 3 = 2x 3 0, hence x and x 3 are linearly independent. Example Show that y(x) = C 1 cos x+c 2 sin x is a general solution of the ODE y +y = 0. Solution: first, it is clear that for both y 1 (x) = cos x and y 2 (x) = sin x, we have y 1 = y 1, y 2 = 11

12 y 2, i.e., these are particular solutions of the indicated ODE. To show their linear independence, we compute the Wronskian W [cos x, sin x] = cos x sin x sin x cos x = 1 0. It follows that indeed, a linear combination y(x) = C 1 cos x + C 2 sin x is a general solution of the linear equation y + y = 0. Example The functions {sin 2 x, cos 2 x, cos 2x} are linearly dependent. It can be shown directly: cos 2 x sin 2 x = cos 2x, or algebraically through computation of the Wronskian Linear Homogeneous Constant-Coefficient ODE Definition The ODE (2.1) is a linear homogeneous ODE with constant coefficients is given by a N y (N) (x) + a N 1 y (N 1) (x) a 1 y (x) + a 0 y(x) = 0, (2.23) where a N,..., a 0 = const, a N 0. The general solution of the ODE (2.23) is given by (2.5). We only need to find N linearly independent particular solutions. For linear homogeneous constant-coefficient ODEs, it follows from the theory that one may use the solution ansatz y(x) = e λx, λ = const. (2.24) Substituting (2.24) into (2.23), one has where e λx P N (λ) = 0, P N (λ) = a N λ N a 1 λ + a 0 is the characteristic polynomial of the ODE. So each solution (2.24) arises from a root of a characteristic polynomial (which has N roots; some might repeat). Case 1. Distinct roots If all roots {λ 1, λ 2,..., λ N } of P N (λ) are different, then all functions y k (x) = e λ kx are linearly independent. (Prove!) Hence the general solution of the ODE (2.23) is given by y gen (x) = C 1 e λ 1x C N e λ N x. (2.25) Example Solve the ODE y (x) + 4y (x) + 3y(x) = 0. Solution: this is a linear homogeneous constant-coefficient ODE. Use the ansatz (2.24); get e λx (λ 2 + 4λ + 3) = 0, hence λ 1 = 3, λ 2 = 1, and y gen (x) = C 1 e 3x + C 2 e x. 12

13 Case 2. Repeating root(s) If a root λ k repeats n times, then instead of n linearly independent solutions, one has only one. One still is able to construct n independent solutions from λ k. It is done by reduction of order explained below. In this case, for the root λ k with multiplicity n, one has n linearly independent particular solutions given by e λ kx, xe λ kx,... x n 1 e λ kx. This should be done for any repeating root. Putting these all together, one obtains the general solutions which involves N linearly independent terms. Example Solve the ODE y (x) + 2y (x) + y(x) = 0. Solution: this is a linear homogeneous constant-coefficient ODE. Use the ansatz (2.24); get e λx (λ 2 + 2λ + 1) = e λx (λ 2 + 1) 2 = 0. The root is λ = 1 with multiplicity n = 2. Hence the general solution is y gen (x) = C 1 e x + C 2 xe x Linear Homogeneous Euler ODE Definition The linear homogeneous Euler ODE is given by a N x N y (N) (x) + a N 1 x N 1 y (N 1) (x) a 1 xy (x) + a 0 y(x) = 0, (2.26) where a N,..., a 0 = const, a N 0. It is a variable-coefficient ODE. The general solution of the linear ODE (2.26) is given by (2.5). We only need to find N linearly independent particular solutions. For Euler ODEs, one uses the ansatz y(x) = x r, r = const. (2.27) It only works for this special class of equations! Substituting (2.27) into (2.23), one has x r [a N r (r 1)... (r N) a 2 r(r 1) + a 1 r + a 0 ] = x r Q N (r) = 0, where Q N (r) is the degree r characteristic polynomial of the Euler ODE. So each solution (2.27) arises from a root of a characteristic polynomial (which, generally, again has N roots, but some may repeat). 13

14 Case 1. Distinct roots are linearly inde- If all roots {r 1,..., r N } of Q N (r) are different, then all functions y k (x) = x r k pendent. (Prove!) Hence the general solution of the ODE (2.26) is given by y gen (x) = C 1 x r C N x r N. (2.28) Example Solve the ODE x 2 y (x) + 3xy (x) + 5y(x) = 0. Solution: this is a linear homogeneous Euler ODE. Use the ansatz (2.27); get x r (r 2 + r + 6) = 0. The roots are r = 1 ± 2i. Hence the general solution is y gen (x) = C 1 x 1+2i + C 2 x 1 2i. The solution can be indeed written in a real form: y gen (x) = 1 x (C 1x 2i + C 2 x 2i ) = 1 x (C 1e 2i ln x + C 2 e 2i ln x ) = 1 x (D 1 cos(2 ln x) + D 2 (2 ln x)). Here Euler formulas were used, e iz = cos z + i sin z, and D 1 = C 1 + C 2, D 2 = i(c 1 + C 2 ). Case 2. Repeating root(s) If a root r k repeats n times, then instead of n linearly independent solutions, one has only one. It can be shown that to such root r k there still correspond n linearly independent particular solutions, given by x r, x r ln x,... x r (ln x) n. This should be done for any repeating root. Putting these all together, one obtains the general solutions which involves N linearly independent terms. Remark The transformation x = e z maps the Euler ODE (2.26) into the linear homogeneous ODE with constant coefficients (2.23); hence the similarity of the solution structure Reduction of Order in Linear Homogeneous ODEs Consider a linear homogeneous ODE of order two: y (x) + a(x)y (x) + b(x)y(x) = 0. (2.29) If one knows a formula for particular solution y 1 (x) satisfying (2.29), one is able to reduce the order of this ODE by one, and derive a formula for the second particular solution y 2 (x). It is sufficient to seek the second linearly independent solution in the form y 2 (x) = q(x)y 1 (x). (2.30) 14

15 Indeed, substituting (2.30) into the ODE (2.29), one gets q(x) [y 1(x) + a(x)y 1(x) + b(x)y 1 (x)]+y 1 (x) [q (x) + a(x)q (x)] = y 1 (x) [q (x) + a(x)q (x)]+2y 1(x)q (x) = 0 (the first bracket vanishes because y 1 (x) is a solution of the ODE (2.29)). Hence q satisfies an ODE y 1 (x) [q (x) + a(x)q (x)] + 2y 1(x)q (x) = 0, (2.31) which allows reduction of order by denoting z(x) = q (x). Hence q can be found from the solution of the 1st-order ODE (2.31), and the general solution of (2.29) can be found as a linear combiination of y 1 and y 2. Example Find the general solution of the ODE y 2y + y = 0. For this constant-coefficient equation, we use the ansatz y = e rx, and obtain the characteristic equation r 2 2r + 1 = (r 1) 2 = 0, and this yields only one solution r = 1, and so y 1 (x) = e x. To find y 2 (x), use reduction of order (2.30). Then (2.31) becomes e x (q (x) 2q (x)) + 2e x q (x) = e x q (x) = 0, or z = 0. Hence q(x) = Ax + B, and it is sufficient to take q = x to get a 2nd linearly independent solution y 2 (x) = xe x. Hence y gen (x) = C 1 e x + C 2 xe x. Remark Note that the same procedure can be used to reduce by one the order of any linear homogeneous ODE of order N. Remark The reduction of order procedures of Sections and are particular instances of the general method of reduction of order, based on symmetry analysis of differential equations. Keywords 2.4 Euler formulas; Reduction of order; Characteristic polynomial. Problems 2.4 Problem 2.4a. Find the general solution of the ODE y (x) + λy(x) = 0, λ R. Consider cases λ > 0, λ < 0, λ = 0 separately. 15

16 Problem 2.4b. Find the general solution of the ODE r 2 u (r) + ru (r) = a 2 u(r), a R. Consider cases a > 0, a = 0 separately. Problem 2.4c. Consider a linear ODE xy (x) (1 + x)y (x) + y(x) = 0. Show that y 1 (x) = e x is a solution. Using reduction of order, derive the general solution. 16

17 2.5 Solution of Linear Non-Homogeneous ODEs Recollect that linear non-homogeneous ODEs ar given by a N (x)y (N) (x) + a N 1 (x) y (N 1) (x) a 1 (x)y (x) + a 0 (x)y(x) = g(x) (2.32) with some nonzero g(x) 0. The general solution of the linear non-homogeneous ODE (2.7) has the form y gen.non hom. (x) = y gen.hom. (x) + y partic.non hom. (x) = (C 1 y 1 (x) C N y N (x)) + y p (x), i.e., the general solution of the linear non-homogeneous ODE (2.7) is a sum of a general solution of its linear homogeneous version, and a particular solution of (2.32), which takes care of canceling the nonlinearity. The solution of a linear non-homogeneous ODE (2.32) consists of two steps: 1. Find y gen.hom. (x) = C 1 y 1 (x) C N y N (x), the general solution of the linear nonhomogeneous ODE. 2. Find any particular solution of the non-homogeneous ODE (2.7). We suppose that part (a) is somehow accomplished (which is not always simple!), and we have formulas for a linearly independent set {y 1 (x),..., y N (x)} of N particular solutions of the linear non-homogeneous ODE. There are two main approaches to accomplish part (b). Construction of the particular solution of the non-homogeneous equation. The two basic methods are: Method of undetermined coefficients. This is not really a method, but rather a guessing framework. For some basic forms of the RHS g(x), for example, exponential, polynomial, etc., one seeks the solution y gen.hom. (x) as a linear combination of certain basic functions with coefficients to be determined. The method and examples are given in most basic textbooks. Method of variation of parameters. This is a systematic method explained below for a 2nd-order linear ODE. Consider a general second-order linear non-homogeneous ODE y (x) + p(x)y (x) + q(x)y(x) = g(x). (2.33) Suppose we know two solutions y 1,2 (x) of the corresponding linear homogeneous equation y (x) + p(x)y (x) + q(x)y(x) = 0. We are going to find y partic.non hom. (x), or in fact, the full general solution of the ODE (2.33). 17

18 Variation of Parameters: the Algorithm for a 2nd-order ODE. 1. Construct a linear combination Q(x) = C 1 (x)y 1 (x) + C 2 (x)y 2 (x), (2.34) for some unknown C 1 (x), C 2 (x). We will seek such C 1 (x), C 2 (x) that Q(x) will indeed solve (2.33). 2. In order to substitute Q(x) in (2.33), we need to differentiate it. We find Q (x) = C 1y 1 + C 2y 2 + C 1 y 1 + C 2 y 2. There is too much freedom in (2.34): two unknown functions, with only one requirement to satisfy one ODE (2.33). So we simplify our life by imposing a condition C 1y 1 +C 2y 2 = 0. It helps because Q will not involve second derivatives of C 1,2 (x). We therefore have Q (x) = C 1 y 1 + C 2 y 2, Q (x) = C 1y 1 + C 2y 2 + C 1 y 1 + C 2 y 2. Upon the substitution of y(x) = Q(x) into the ODE (2.33), and using the above expressions for Q (x), Q (x), we derive that the ODE (2.33) becomes as simple as C 1y 1 + C 2y 2 = g(x). 3. Finally, C 1(x), C 2(x) satisfy a linear system: [ ] [ ] [ ] y1 (x) y 2 (x) C 1 (x) 0 y 1(x) y 2(x) C 2(x) = g(x) 4. The system matrix is nondegenerate (since its determinant is the Wronskian W [y 1 (x), y 2 (x)] 0); hence one can solve for C 1(x), C 2(x). 5. Integration will yield C 1 (x), C 2 (x), with some arbitrary constants. 6. Finally, the general solution of (2.33) is given by Q(x) (2.34); it will already involve the particular solution of the linear non-homogeneous ODE, and the general solution of the linear homogeneous ODE (from those arbitrary integration constants). Problems 2.5 Problem 2.5a. Find the general solution of the non-homogeneous ODE y 4y y + 4y = 30e 2x. (Method of variation of parameters or undetermined coefficients can be used.) Clearly represent solution in the form (2.9). Problem 2.5b. Find the general solution of the non-homogeneous ODE 4x 3 y + 3xy 3y = 10. (Method of variation of parameters or undetermined coefficients can be used.) Clearly represent solution in the form (2.9). 18

19 2.6 Series and Power Series: a Brief Review Series Definition Let {a n } n=1 {a 1, a 2,..., a 100,...} R (C) be an infinite sequence (an ordered set) of real or complex numbers. A formal series is an expression a n = a 1 + a n=1 (2.35) A partial sum of N terms of a series is a number N s N = a n = a 1 + a a N. (2.36) n=1 The series is convergent if its sum exists: s = lim N s N. (2.37) If there is no limit of the partial sums, the series (2.35) is divergent. The remainder is the difference of the sum and a partial sum: r N = s s N = n=n+1 a n. (2.38) Theorem 2.36 (Series Convergence Criterion). A series (2.35) converges to s if and only if the remainder converges to zero: lim r N = 0. N Theorem 2.37 (A necessary condition of convergence). If a series (2.35) converges, then a n 0. Proof. If a series converges, then s = lim N s N, Also, clearly, s = lim N s N 1. Subtracting, one gets 0 = lim N (s N s N 1 ) = lim N a N. 1. n=1 (n!) diverges since it violates the necessary condition of conver- Example gence. 19

20 2. The geometric series n=1 qn converges if an only if q < 1. For instance, (0.5) n 1 = = 2. n=1 Definition If a n n=1 converges, then the series (2.35) is said to converge absolutely. (2.39) If the series (2.35) converges but the series of absolute values (2.39) diverges, then the series (2.35) is said to converge conditionally. Example The series 1 n n=1 diverges, whereas the series ( 1) n n=1 n converges conditionally. Remark Absolute convergence is a stronger condition than conditional convergence. It is easy to show that absolute convergence implies conditional convergence, but the opposite is generally false. According to the Riemann convergence theorem, if a series converges only conditionally, its terms can be rearranged to make it converge to any real number, as well as diverge to ±. Theorem 2.42 (d Alembert sufficient condition for convergence). For the sequence {a n } n=1, consider q = lim a n+1 n a n. If q exists and q < 1, the series converges absolutely. If q exists and q > 1, the series diverges, and a n. Theorem 2.43 (Cauchy sufficient condition for convergence). For the sequence {a n } n=1, consider q = lim n n a n. If q exists and q < 1, the series converges absolutely. If q exists and q > 1, the series diverges, and a n. Remark Both of the above criteria essentially compare the sequence to a geometric one with multiplier q. If the indicated limit q equals 1 or does not exist, then the corresponding criterion is inconclusive. The series still might converge or diverge. 20

21 2.6.2 Power Series and Taylor series Definition Let {a n } be a sequence of numbers. A formal power series about x = 0 is a function f(x) = c n x n. (2.40) A formal power series about x = x 0 is a function f(x) = c n (x x 0 ) n. (2.41) Now that x-dependence is present, the series may converge or diverge depending on x. The following holds. The power series (2.41) always converges at x = x 0 (and is zero there). The power series (2.41) may converge in I = (x 0 R, x 0 + R), R > 0. In this case, R is called the radius of convergence. For some series, the radius of convergence R = +. The following theorem holds. Theorem For any power series (2.41) there exists R 0, possibly R = +, such that (2.41) converges absolutely for x : x x 0 < R and diverges for x : x x 0 > R. Theorem 2.47 (Abel s convergence theorem). If the power series (2.41) converges at x 1, then it converges absolutely for all x : x x 0 < x 1 x 0. It follows that if a radius of convergence is known, then at x x 0 = R, the series might converge conditionally. But for x : x x 0 < R, it converges absolutely Formulas for the radius of convergence Let us analyze a power series around x 0 as just a series: c n (x x 0 ) n = a n, (2.42) where a n = a n (x) = c n (x x 0 ) n. Then from d Alembert sufficient condition for convergence (Theorem 2.42), it follows that the series converges when q(x) = lim a n+1 n a n = x x 0 lim c n+1 n c n < 1. It follows that the series converges at a given x, when x x 0 < R = lim c n. (2.43) n c n+1 21

22 (d Alembert formula for the radius of convergence). Similarly, from Cauchy sufficient condition for convergence (Theorem 2.43), we know that the series (2.48) converges when q(x) = lim n n a n = x x 0 lim n n a n < 1. It follows that the series converges at a given x, when x x 0 < R = 1/ lim n n a n. (2.44) (Cauchy formula for the radius of convergence). Remark The radius of convergence of Taylor series of an analytic function is given by the distance from x 0 to the closest singularity. Example Consider the power series x n n!. By d Alembert formula, one has R = lim c n n = lim n + 1 n 1 = +, c n+1 hence the series converges everywhere in R. Example Consider the power series x n 2. n By Cauchy formula, one has R = 1/ lim n n a n = 1/ lim n n 1/2 n = 2. hence the series converges in ( 2, 2) (here and in the previous example, x 0 = 0). Example Consider the power series n!(x 1) n. Here x 0 = 1. By d Alembert formula, one has R = lim c n n = lim 1 n n + 1 = 0, c n+1 hence the series converges nowhere but at x = 1 itself. Example For both series x n n, ( 1) n x n, n it is easy to see that the radius of convergence is one (show!). 22

23 2.6.4 Operations on Power Series Theorem 2.53 (Addition and multiplication of power series). Consider two power series about x 0, f(x) = c n (x x 0 ) n, g(x) = having radii of convergence R 1 and R 2, respectively. Then series f(x) + g(x) = f(x) g(x) = (c n + d n )(x x 0 ) n d n (x x 0 ) n, (2.45) q n (x x 0 ) n, q n = c 0 d n c n d 0, converge within R = min(r 1, R 2 ). Theorem 2.54 (Termwise integration and differentiation of power series). Consider a power series about x 0, f(x) = c n (x x 0 ) n n=1 (2.46) having the radius of convergence R. Then power series f (x) = nc n (x x 0 ) n 1 n=1 and F (x) = f(x) dx = c + n=1 c n 1 n (x x 0) n converge within (x 0 R, x 0 + R) Taylor series Theorem 2.55 (Taylor series). Suppose that f(x) is given by a power series f(x) = c n (x x 0 ) n (2.47) that converges in (x 0 R, x 0 + R), for some R > 0. Then f(x) C (x 0 R, x 0 + R), and the series is given by f(x) = f (n) (x 0 ) (x x 0 ) n. (2.48) n! 23

24 The proof proceeds by computing f (x) = nc n (x x 0 ) n 1, f (x) = n=1 n(n 1)c n (x x 0 ) n 2, n=2 etc., using Theorem 2.54, and finding derivatives at x = x 0. Definition A function f(x) is (real) analytic on an open set I R if for any x 0 I, f(x) is given by a Taylor series (2.48) convergent in some neighbourhood (x 0 R, x 0 + R) of x 0, with R > 0. Example Compute the Taylor series for f(x) = ln(1 x) about x = 0, and find its radius of convergence. We find and hence f(0) = 0, f (0) = 1, f (0) = 1/2,..., f (n) (0) = 1/n,..., ln(1 x) = x n n. By d Alembert s formula, R = lim n + 1 n n = 1. Hence the series converges in ( 1, 1) absolutely. Note that the series converges conditionally at x = 1 to ln 2, and diverges at x = 1 (verify!). Example Compute the Taylor series for f(x) = cos x about x = 0. Show that the radius of convergence is R =. Show that in any ( K, K), the series converges uniformly in x. Solution: first we note that any Taylor series can be represented exactly as a Taylor polynomial (partial sum) of any degree N plus the remainder: f(x) = N f (n) (0) x n + r N (x), n! where the remainder in the Lagrange form is given by r N (x) = f (N+1) (ξ) (N + 1)! xn+1, for some unknown ξ (0, x). a) the Taylor series for f(x) = cos x about x = 0 contains only even terms; it is given by cos x = ( 1) n (2n)! x2n. b) the above Taylor series written as a partial sum plus remainder is cos x = s 2N (x) + r 2N (x) = N ( 1) n (2n)! x2n + f (2N+1) (ξ) (2N + 1)! x2n+1. 24

25 For f(x) = cos x, f (2N+1) (ξ) 1. So within any ( K, K), r 2N Sup f (2N+1) 1 (x) ( K,K) (2N + 1)! K2N+1 = K2N+1 (2N + 1)!, which is a uniform estimate of error for all x in ( K, K). Indeed, function of N. Therefore for any ε > 0 N > 0, such that 2N +1 K (2N + 1)! < ε. If N > N, then for any x ( K, K), N cos(x) s 2N (x) = cos(x) ( 1) n (2n)! x2n = r 2N(x) < uniformly for all x. K 2N+1 (2N + 1)! KN+1 (N + 1)! < ε is a decreasing Keywords 2.6 Conditional convergence; Riemann convergence theorem. Problems 2.6 Problem 2.6a. Prove that if a series (2.35) converges absolutely, then it converges conditionally. Problem 2.6b. Find the Taylor series of f(x) = 1 1 x 2 about x 0 = 0, and compute its radius of convergence R. Show that R is indeed equal to the distance from x 0 to the closest singularity. Problem 2.6c. Find the Taylor series of f(x) = x about x 0 = 2, and compute its radius of convergence R. Show that R is indeed equal to the distance from x 0 to the closest singularity. Problem 2.6d. Consider a Taylor series for f(x) = ln x about x = 1. Compute the radius of convergence. Show that at x = 2, the series converges conditionally. Find a rearrangement of terms of the Taylor series representing ln 2 so that it converges to 2 ln 2. (Such a rearrangement must exist by Riemann convergence theorem.) Problem 2.6e. Show that f(x) = e 1/x2 is infinitely differentiable but is not analytic in any interval containing x = 0. In particular, show that it is not equal to its Taylor series about x = 0. Problem 2.6f. In a software package, plot partial sums s 2N (x) of Taylor series for f(x) = cos x for N = 1, 2,... in one plot with f(x), for x [ 10, 10] (see Example 2.58). Observe uniform convergence. 25

26 2.7 The Idea of the Power Series Method Consider a linear homogeneous ODE of order n which is NOT of a class for which we can construct an exact solution by any reduction method. The main idea of the power series method is to use linearity to seek Taylor-series type solutions around some point x 0 of interest, y i (x) = c i,n (x x 0 ) n, and if we can find n such linearly independent series-type solutions, then a general solution can be constructed. We will see later that not all ODEs have Taylor-series type solutions. But many do. Consider the following example: y (x) 2xy(x) = 0. (2.49) Clearly this ODE has the general solution y(x) = Ce x2. (2.50) But we will forget it and seek a solution of (2.49) using an unknown power series about x = 0: y(x) = c n x n. Substituting it into (2.49), we get y (x) 2xy(x) = (nc n x n 1 2c n x n+1 ) = (c 1 x 0 +2c 2 x 1 +3c 3 x ) 2(c 0 x 1 +c 1 x 2 +c 2 x ) = 0. (2.51) Now, powers of x are linearly independent. If the above series is zero, it means that the coefficient of every power of x must vanish separately. We have in (2.51) the following coefficients: x 0 : c 1 = 0 x 1 : 2c 2 2c 0 = 0 x 2 : 3c 3 2c 1 = 0 x 3 : 4c 4 2c 2 = Hence we get c 1 = c 3 =... = 0 (all odd terms), and c 0 R, c 2 = c 0, c 4 = c 0 /2, c 6 = c 0 /(2 3),..., c 2n = c 0 n!. 26

27 Therefore the solution is given by y(x) = c n x n = c 2n x 2n = c 0 x 2n n!, which is indeed the Taylor series of y(x) = c 0 e 2x. As a result, we have obtained the general solution of the ODE (2.49) using the power series method. Problems 2.7 Problem 2.7a. Use the above naive power series method to compute the general solutions of the ODEs (a) y (x) + y(x) = 0. (b) (1 + x)y = y. (c) x 2 y (x) 2xy (x) = 0. Compare with general solutions obtained by standard methods. 27

28 2.8 The Power Series Method for Second-Order Linear ODEs with Analytic Coefficients Big O notation Definition 2.59 (big O notation). Consider two functions f(x), g(x). 1. f(x) = O(g(x)) as x, if M > 0, x 0 : x > x 0, f(x) M g(x). 2. f(x) = O(g(x)) as x 0, if M, δ > 0, x : x < δ, f(x) M g(x). The big O notation is usually used in the sense that f(x) behaves like g(x), that is, f(x) is bounded by g(x). In particular, it is useful when the limit does not exist. Example sin x = O(x) as x 0. e x = O(1) as x 0. x = O(x 2 ) as x. cos x = O(1) as x The Power Series Method for ODEs with Analytic Coefficients Consider a linear second-order ODE y (x) + p(x)y (x) + q(x)y(x) = r(x). (2.52) The following theorem holds. Theorem If p(x), q(x), r(x) are analytic in some interval (x 0 A, x 0 + A), A > 0, then the general solution of the ODE IVP (2.11) can be represented as a power series y gen (x) = c n (x x 0 ) n convergent within (x 0 R, x 0 + R) for some R > 0. (Without proof). Example Compute several terms of the general solution of the ODE y y + x 2 y = 0 in terms of Taylor series about x 0 = 0. Solution: we compute derivatives and re-number infinite sums (the terms of the equation) to have the same powers of x, and to start from the same index. y(x) = c n x n, x 2 y(x) = c n x n+2 = n + 2 = k = c k 2 x k, 28 k=2

29 y (x) = y (x) = c n nx n 1 = n 1 = k = n=1 c n n(n 1)x n 2 = n 2 = k = n=2 These can be combined into the ODE to give (c 1 + 2c 2 x) + (2c 2 + 6c 3 x) + c k+1 (k + 1)x k = c 1 + 2c 2 x + k=0 c k+1 (k + 1)x k, k=2 c k+2 (k+1)(k+2)x k = 2c 2 +6c 3 x+ c k+2 (k+1)(k+2)x k. k=0 k=2 x k (c k+2 (k + 1)(k + 2) c k+1 (k + 1) + c k 2 ) = 0. k=2 Setting to zero the coefficients at all powers of x, we get x 0 : c 1 = 2c 2 ; x 1 : 2c 2 = 6c 3 ; x k (k 2) : c k+2 = c k+1(k + 1) c k 2. (k + 1)(k + 2) We observe that c 0, c 1 are free. The formulas yield: c 2 = c 1 2, c 3 = c 1 6, c 4 = 3c 3 c 0 = c c 0 12, c 5 = c 1 24 c 0 60, etc. Finally, several terms of the general solution are given by y gen (x) = c 0 y 1 (x) + c 1 y 2 (x) ( = c x4 1 ) ( 60 x5 + O(x 6 ) + c 1 x x x x4 1 ) 24 x5 + O(x 6 ). Example 2.63 (Legendre Equation). The Legendre equation is a linear homogeneous ODE given by (1 x 2 )y (x) 2xy (x) + ly(x) = 0, l = const R. (2.53) We will represent r as Clearly l = n(n + 1), n = const R. p(x) = 2x 1 x, q(x) = l, r(x) = 0 (2.54) 2 1 x2 are analytic within x ( 1, 1). We will seek the general series solution to (2.53) about x 0 = 0: y gen (x) = c m x m. m=0 29

30 We obtain, for the ODE, after the re-numbering in summations as appropriate, the following expression (check!) (m+2)(m+1)c m+2 x m + m(m 1)c m x m 2 mc m x m + n(n+1)c m x m = 0. (2.55) m=0 m=2 m=1 Setting to zero the coefficients at all powers of x, we get x 0 : 2c 2 + n(n + 1)c 0 = 0; x 1 : 6c 3 + [ 2 + n(n + 1)]c 1 = 0; x m (n m)(n + m + 1) (m 2) : c m+2 = c m. (2.56) (m + 1)(m + 2) As a result, we get independent even and odd coefficient families: c 0 free, c 2 = m=0 n(n + 1) (n 2)n(n + 1)(n + 3) c 0, c 4 = c 0,..., 2! 4! (n 1)(n + 2) (n 3)(n 1)(n + 2)(n + 4) c 1 free, c 3 = c 1, c 5 = c 1,..., 3! 5! etc. Finally, the general solution of the Legendre equation (2.53) is given by y gen (x) = c 0 y even (x) + c 1 y odd (x) (2.57) where y 1 (x) and y 2 (x) are the corresponding even and odd sums. Legendre polynomials arise from (2.57) when n is integer. (2.56), we have hence a n+2 = a n+4 = a n+6 =... = 0, y even (x) = p c 2m x 2m = P 2p (x) = P n (x) m=0 is a polynomial (Legendre polynomial) of degree n = 2p. solution y odd (x) = c 2m+1 x 2m+1 = Q n (x) m=0 Indeed, if n = 2p is even, from The second linearly independent is not a polynomial, but is given by an infinite series. It is called the Legendre function of the second kind. Similarly, when n = 2p + 1 is odd, we have y odd (x) = c 2m+1 x 2m+1 = P 2p+1 (x) = P n (x) m=0 is a Legendre polynomial of degree n = 2p+1. Now the even solution is still given by an infinite series: y even (x) = c 2m x 2m = Q n (x). m=0 30

31 Remark The series solutions to the Legendre equation (2.53) generally have the radius of convergence R = 1, as seen from (2.54). However, Legendre polynomials turn out to be regular everywhere in R. It is the Legendre functions Q n (x) that become singular at x = ±1. Remark The Legendre polynomials are usually normalized by choosing c n = (2n)! (2n 1) =. 2 n (n!) 2 n! and are subsequently called P n (x). Note that P n (±1) = ±1. The first few Legendre polynomials are given by P 0 (x) = 1, P 1 (x) = x, P 2 (x) = 1 2 (3x2 1), P 3 (x) = 1 2 (5x3 3x),... (2.58) Keywords 2.8 Associated Legendre functions; Families of orthogonal polynomials; Laguerre polynomials; Hermite polynomials; Chebyshev polynomials. Problems 2.8 Problem 2.8a. Use the power series about x 0 = 0 to compute the general solution of the ODE y + (1 + x 2 )y = 0. Problem 2.8b. Use the reduction of order for the Legendre ODE with n = 0 to derive a formula for the Legendre function of the second kind Q 0 (x) using the known Legendre polynomial P 0 (x) = 1. Problem 2.8c. Use the reduction of order for the Legendre ODE with n = 1 to derive a formula for the Legendre function of the second kind Q 1 (x) using the known Legendre polynomial P 1 (x) = x. Problem 2.8d. Prove the Rodriguez formula for the Legendre polynomials: P n (x) = 1 d n 2 n n! dx n [(x2 1) n ]. Hint: apply the binomial formula to the bracket, differentiate, and show that the result satisfies the Legendre equation. You also have to show that what the Rodriguez formula yields is actually a polynomial. Problem 2.8e. Use the Rodriguez formula to derive the explicit form of the Legendre polynomials P 6 (x), P 7 (x), P 8 (x). 31

32 2.9 Series Solutions of ODEs about Singular Points. Frobenius Method Classification of Regular and Singular Points of ODEs Consider a linear homogeneous second-order ODE y (x) + p(x)y (x) + q(x)y(x) = 0. (2.59) Definition The point x 0 is a regular point of the ODE (2.59) if p(x), q(x) are analytic in an open neighbourhood of x 0. Definition The point x 0 is a regular singular point of the ODE (2.59) if p(x) = b(x), q(x) = c(x) x x 0 (x x 0 ), (2.60) 2 where b(x), c(x) are analytic in an open neighbourhood of x 0. Definition The point x 0 is an irregular singular point of the ODE (2.59) if it is neither a regular nor a regular singular point. Example For the ODE y + 3x 2 y = 0, all points are regular points. For the Legendre ODE (1 x 2 )y (x) 2xy (x) + ly(x) = 0, all points x ±1 are regular points, and x = ±1 are regular singular points. Indeed, the ODE can be written as (2.59) with p(x) = 2x l(x 1)(x + 1), q(x) = (x 1)(x + 1) (x 1) 2 (x + 1), 2 where p(x), q(x) are analytic everywhere except for x = ±1. Then at x = 1, for example, they have the form (2.60) with b(x) = 2x l(x 1), c(x) = x + 1 x + 1, which are analytic at x = 1. So x = 1 is a regular singular point by definition. Similarly, x = 1 is a regular singular point. Theorem 2.70 (Frobenius theorem). If x 0 is a regular singular point of the ODE (2.59), then at least one of the two basis solutions is represented in the form of a generalized power series about x 0 : y 1 (x) = (x x 0 ) r a n (x x 0 ) n, (2.61) for some r R, convergent within (x 0 R, x 0 + R) for some R > 0. In order to study the form of the second basis solution, the following development is needed. 32

33 2.9.2 Indicial Equation. The Form of the Second Solution It is clear that by a transformation of variables z = x x 0, one can reduce the analysis of series solutions to that for the case x 0 = 0. We will assume x 0 = 0 from now on. Suppose that x 0 is a regular singular point of the ODE (2.59). Then the ODE can be rewritten as x 2 y (x) + xb(x)y (x) + c(x)y(x) = 0, (2.62) somewhat reminding the Euler equation. Here by analyticity, b(x) = c(x) = b n x n, b 0 = b(0); (2.63) c n x n, c 0 = c(0). (2.64) Further, the solution (2.61) will have the form, for a fixed number r, y(x) = x r a n x n = a n x n+r. (2.65) Substituting (2.65), (2.63) and (2.64) into the ODE (2.62), and collecting coefficients at x r, we get r(r 1) + b(0)r + c(0) = 0, (2.66) which is called the indicial equation. The meaning of the indicial equation (2.66) is clear: in order for the series (2.65) to solve the ODE (2.62), it is necessary that the constant r satisfies the indicial equation. Three important cases arise, that lead to three cases for the solution basis of the ODE (2.59) at a regular singular point. We now give the main result; examples are considered below. Theorem 2.71 (Frobenius method, solution basis at a regular singular point). If x 0 is a regular singular point of the ODE (2.59) (or equivalently (2.62)), and r 1, r 2 are the roots of the indicial equation (2.66), then: (a) Case 1: r 1 r 2, and r 1 r 2 Z. Then the solution basis is given by y 1 (x) = x r 1 a n x n, y 2 (x) = x r 2 A n x n. (b) Case 2: r 1 = r 2 = r. Then the solution basis is given by y 1 (x) = x r a n x n, y 2 (x) = y 1 (x) ln x + x r A n x n. n=1 33

34 (c) Case 3: r 1 > r 2, and r 1 r 2 Z. Then the solution basis is given by y 1 (x) = x r 1 a n x n, y 2 (x) = ky 1 (x) ln x + x r 2 where k R, possibly k = 0. A n x n, Remark In Case 3, note that the logarithmic term may be present for the smaller index r 2. In problems involving Case 3, one may initially assume k = 0. If it does not work, one proceeds with using the (k 0)-part of the second solution. Example Consider the Euler ODE x 2 y + b 0 xy + c 0 y = 0, b 0, c 0 = const. The usual Euler ansatz y = x r is nothing but a truncation of the series (2.65). The indicial equation (2.66) in this case is simply the Euler characteristic equation. If r 1 r 2, the general solution is y gen (x) = C 1 x r 1 + C 2 x r 2 (Frobenius Case 1 or 3); if r 1 = r 2 = r, the general solution is y gen (x) = C 1 x r + C 2 x r ln x (Frobenius Case 2). Example 2.74 (Lecture). Solve the ODE 6x 2 y (x) + xy (x) + e x y(x) = 0 (2.67) as a series about x = 0. We start with rewriting the ODE in the form (2.62): y (x) + 1 6x y (x) + ex y(x) = 0; 6x2 so b(x) = 1/6 and c(x) = e x /6, which are indeed analytic functions around x = 0. We have b(0) = c(0) = 1/6. The indicial equation is r(r 1) r = 0, which has the roots r 1 = 1/3, r 2 = 1/2, and we are in the Case 1 of Theorem We will work with general r, and then substitute r = r 1 to find y 1 (x), and r = r 2 to find y 2 (x). Substituting e x = x k k!, y(x) = a n x n+r k=0 into the given ODE (2.67), we get 6a n (n + r)(n + r 1)x n+r + ( a n (n + r)x n+r + 34 k=0 ) ( x k ) a n x n+r = 0. k!

35 Using series multiplication and rewriting, we have x [6a n+r n (n + r)(n + r 1) + a n (n + r) + The bracketed expression reveals the recurrence relation: ( 1 n! a )] 0! a n = 0. 1 a n = n! a ! a n 1 (n + r)(6(n + r) 5) + 1. (2.68) To find y 1 (x), we substitute r = 1/3, a 0 = 1 into (2.68), and find a 1 = 1 5, a 2 = 3 220,.... To find y 2 (x), we substitute r = 1/2, A 0 = 1 into (2.68), and find A 1 = 1 7, A 2 = 5 364,.... Finally the general solution is ( y gen = C 1 x 1/ x 3 ) ( 220 x C 2 x 1/ x 5 ) The same can be found, for example, using Maple routine dsolve(ode,y(x), series,x=0). Example Solve the ODE x(x 1)y + (3x 1)y + y = 0 as a series about x = 0. Solution: we insert the series (2.65) into the ODE, to get (n + r)(n + r 1)a n x n+r + n=2 +3 (n + r)a n x n+r n=1 (n + r)(n + r 1)a n x n+r 1 (2.69) n=2 (n + r)a n x n+r 1 + n=1 a n x n+r = 0, and set to zero coefficients of terms with different powers of x. The indicial equation is obtained at the lowest power, x r. It is hence (r(r 1) r)a 0 = 0, r 1 = r 2 = 0 is the double root, and we are in the Case 2 of Theorem To find the first solution, we insert r = 0 into (2.69), and set to zero the coefficient at each x m, m 0. We get (prove by re-numbering the sums!) m(m 1)a m (m + 1)ma m+1 + 3ma m (m + 1)a m+1 + a m = 0, 35 n=2

36 which simplifies to a m+1 = a m. Hence a 0 = a 1 =..., and the series solution is given by (upon putting a 0 = 1) y 1 (x) = n=1 x n = 1 1 x. Since this is a closed-form expression, in order to find the second solution, it is natural to use the reduction of order: y 2 (x) = q(x)y 1 (x). After some computation, one has q(x) = ln x, y 2 (x) = ln x 1 x, which concludes the example. (Compare the solution structure with Case 2 of Theorem 2.71.) Example Find the series solution of an ODE xy (x) + 5y (x) + xy(x) = 0 about x = 0. First, we rewrite equation in the canonical form x 2 y (x) + 5xy (x) + x 2 y(x) = 0, (2.70) and readily find b(x) = b(0) = 5; c(x) = x 2, c(0) = 0. Hence the indicial equation yields r 1 = 0, r 2 = 4, and 0 ( 4) = 4 is integer. We are in Case Finding the first solution. By Theorem 2.71, it is given by y 1 (x) = x r a n x n = a n x n+r, (2.71) where r = 0. We will however work with the form (2.71) with general r, hoping that both solutions (r = 0, r = 4) have the form (2.71). (It is indeed possible that in Case 3, k = 0.) Substituting (2.71) into the equation (2.70) and equating the coefficients at different powers of x to zero, one gets x r : r(r + 4)a 0 = 0; x r+1 : (r + 1)(r + 5)a 1 = 0; x r+m, m 2 : a m (r + m)(r + m + 4) + a m 2 = 0. (2.72) The first equation is indicial, hence a 0 is arbitrary. WLOG, set a 0 = 1. The second equation yields a 1 = 0, since (r + 1)(r + 5) 0 for both r = 0, r = 4. The third equation yields a recursion relation a m 2 a m =, m = 2, 3,... (2.73) (r + m)(r + m + 4) 36

37 Hence all a 1 = a 3 =... = 0 (odd terms). For the first solution corresponding to the maximal r (r = 0), we find a 2 = 1/(2 6), a 4 = 1/( ),..., thus y 1 (x) = x x x (2.74) 2. Finding the second solution. By the Frobenius Theorem, it is possible that in Case 3, for the second solution, k = 0. Then formulas (2.72) would still hold, now for r = r 2 = 4. In particular, from (2.73) we would have a m 2 a m =, m = 2, 4, 6... m(m 4) The latter breaks down at m = 4 and is not suitable. Therefore we expect the second solution to our ODE to have the form y 2 (x) = ky 1 (x) ln x + A n x n 4, (2.75) where the constants k 0 and A n are to be determined. Differentiating (2.75) yields y 2(x) = ky 1(x) ln x + k y 1(x) x + A n(n 4)x n 5, y 2(x) = ky 1(x) ln x + 2k y 1(x) k y 1(x) + (2.76) x x 2 A n(n 4)(n 5)x n 6. Substituting (2.75) and (2.76) into the ODE (2.70), we observe that log-terms cancel, since y 1 (x) itself solves the ODE (2.70). We are left with the following: 2kxy 1(x)+4ky 1 (x)+ A n (n 4)(n 5)x n 4 + 5A n (n 4)x n 4 + A n x n 2 = 0. (2.77) The lowest powers of x in (2.77) are -5, -4, which come from the first two sums: x 4 : A 0 ( 4)( 5) + 5A 0 ( 4) = 0 A 0 is arbitrary. x 3 : A 1 ( 3)( 4) + 5A 1 ( 3) = 0 A 1 = 0. For the terms x 3, x 2, the last sum in (2.77) starts playing: we get x 2 : A 2 ( 2)( 3) + 5A 2 ( 2) + A 0 = 0 A 2 = A 0 /4. x 1 : A 3 ( 1)( 2) + 5A 3 ( 1) + A 1 = 0 A 3 = 0. For the terms x 0, the term ky 1 (x) in (2.77) also starts playing: we get x 0 : 4ka A 2 = 0, where a 0 = 1 is the first coefficient of y 1 (x), hence k = A 2 /4. WLOG, assume A 0 = 1, hence A 2 = 1/4, k = 1/16. Coefficients at subsequent powers of x in (2.77) yield A odd = 0, and A even = respective expressions. Finally, one gets y 2 (x) = 1 ( 16 y 1(x) ln x + x x2 1 ) 288 x4 +..., which concludes the example. 37

38 Problems 2.9 Problem 2.9a. Classify regular, regular singular, and irregular singular points of the following ODEs. Specify b(x), c(x) for regular singular points where appropriate. 1. x 3 y (x) + 3x 2 y (x) + 4xy(x) = 0; 2. x 2 y (x) + 6e x y (x) = 0; 3. x 2 y (x) + (sin x)y (x) + (sin x cos x)y(x) = 0. Problem 2.9b. Use the Frobenius method to derive the power series form of the solution basis for an ODE. 2xy + (3 4x)y (x) + (2x 3)y(x) = 0. Write several first terms of the series for the basis solutions. If possible, combine the series solutions into known functions. Problem 2.9c. Use the Frobenius method to derive the power series form of the solution basis for an ODE x 2 y (x) + 4xy (x) (x 2 2)y(x) = 0. Write several first terms of the series for the basis solutions. Show that the two basis solutions can be written as y 1 (x) = sinh x x 2, y 2 (x) = cosh x x 2. Problem 2.9d. Use the Frobenius method to derive the power series form of the solution basis for the following ODEs. Write several first terms of the series for the two basis solutions. 1. xy (x) + y(x) = xy (x) + y (x) xy(x) = xy (x) + 2x 3 y (x) + (x 2 2)y(x) = 0. Problem 2.9e. Consider an ODE x 4 y (x)+2y (x)+y(x) = 0. Show that x 0 = 0 is an irregular singular point. What does one obtain if one seeks a solution of the above ODE in terms of a generalized power series y(x) = x r a nx n? Problem 2.9f. (Power series or Frobenius method?) Here are some practice ODEs. For each one, the task is to find the basis of solutions, in terms of series or generalized series about x = 0. Where possible, identify series expansions of known functions. Show all work. Compare with symbolic software solution, if desired. 1. y + 4y = 0 2. xy + (1 2x)y + (x 1)y = 0 3. (x 1) 2 y (x 1)y 35y = 0 38

39 4. 16(x + 1) 2 y + 3y = 0 5. x 2 y + xy + (x 2 5)y = 0 6. x 2 y + 2x 3 y + (x 2 2)y = 0 7. xy (x + 1)y + y = 0 8. xy + 3y + 4x 3 y = 0 9. y + 1 4x y = xy + y xy = 0 39

40 2.10 Bessel Functions The Gamma function The Gamma function is given by Γ(x) = 0 e t t x 1 dt. (2.78) Theorem a) Γ(1) = 1. b) Γ(x + 1) = xγ(x) for x > 0. Proof. a) Clearly Γ(1) = b) One has 0 e t dt = 1. Γ(x + 1) = e t t x dt = 0 0 t x de t = e t t x 0 + xt x 1 e t dt = 0 + xγ(x). 0 It follows that for an n N, Γ(n + 1) = nγ(n) = n(n 1)Γ(n 1) =... = n!γ(1) = n!, i.e., Γ(x + 1) is a continuous function generalizing the factorial. In fact, the Gamma function can be defined by analytical continuation for negative x and for complex numbers as well, except that it is singular at x = 0 and negative integers. See Figure 2.3. To compute the values of the Gamma function for x < 0, one may use the formula Γ(x + 1) = xγ(x) Bessel s equation and Bessel Functions J ±ν (x) The Bessel s equation is the linear homogeneous ODE with a real parameter ν, given by x 2 y (x) + xy (x) + (x 2 ν 2 )y(x) = 0. (2.79) The Bessel s equation arises in linear PDE problems involving cylindrical and polar coordinates, as well as nonlinear problems in many areas, including, for example, plasma physics. The two linearly independent solutions of the Bessel s equation cannot be expressed by elementary functions. They are referred to as y 1 = J ν (x) and y 2 = J ν (x). We are going to find them using their power series representation about x = 0. 40

41 Figure 2.1: The Gamma function. We proceed with the Frobenius method. First, we rewrite the ODE (2.79) in the canonical form y (x) + 1 x y (x) + x2 ν 2 The point x = 0 is a regular singular point, since b(x) = 1, c(x) = x 2 ν 2 are analytic in R. The indicial equation reads x 2 y(x) = 0. (2.80) r(r 1) + b(0)r + c(0) = r 2 r + r ν 2 = 0, hence the indices are r 1 = ν, r 2 = ν, (2.81) and r 1 r 2 = 2ν. Generally, we are in the first case of Theorem 2.71, unless ν is a half-integer. We proceed with this general situation, and assume y(x) = x r a n x n, (2.82) with y = y 1 for r = r 1, and y = y 2 for r = r 2. We substitute this solution form into the four different terms of Bessel s equation (2.79), to get (n + r)(n + r 1)a n x n+r + (n + r)a n x n+r + a n x n+r+2 ν 2 a n x n+r = 0. 41

42 In particular, in terms 1,2,4, no renumbering is required, and we assume m = n. In term 3, we need to renumber: m = n + 2. After re-numbering, the sum becomes a m x [ m+r (m + r)(m + r 1) + (m + r) ν 2] + m=0 a m 2 x m+r = 0. (2.83) Different powers of x are still linearly independent. We have the following coefficients at the two lowest powers: x r : (r 2 ν 2 )a 0 = 0; x r+1 : ((1 + r) 2 ν 2 )a 1 = 0. The first one is the indicial equation, do for our r = ±ν, a 0 remains free. The second one contradicts the first one, so we must have a 1 = 0. For higher-order terms involving x r+m, m 2 in (2.83), and using ν 2 = r 2, we obtain the recursion relation a m 2 a m =, m = 2, 4, 6... (2.84) m(m + 2r) In details, a 0 is free; a 0 m=2 a 2 = 2(2 + 2r), a 4 = 4(4 + 2r), a 6 = 6(6 + 2r),... or in terms of a 0, a 2 a 4 a 0 a 2 = 2(2 + 2r), a 4 = a (2 + 2r) (4 + 2r), a 6 = (2 + 2r) (4 + 2r) (6 + 2r),... a 0 We can write this in one expression: ( 1) m a 0 a 2m (r) = 2 2m m!(r + 1)(r + 2)...(r + m), m N; a 0 free; (2.85) a 2m 1 = 0, m N. In general, the solution of the Bessel s equation, for each r in (2.81), is given by (2.82). So we have the basis of two linearly independent solutions of the Bessel s ODE (2.79): y 1 (x) = x ν m=0 a 2m (ν)x 2m, y 2 (x) = x ν a 2m ( ν)x 2m. (2.86) m=0 Definition The Bessel functions of the first kind J ν (x), J ν (x) are given by y 1 (x) and y 2 (x) in (2.86) upon choosing a 0 (ν) = 1 2 ν Γ(ν + 1). 42

43 Then the coefficient form (2.85) simplifies: a 2m (ν) = ( 1) m 2 2m+ν m! Γ(m + ν + 1). So the Bessel functions of the first kind have the power series expansions J ν (x) = x ν m=0 ( 1) m 2 2m+ν m! Γ(m + ν + 1) x2m, (2.87) and J ν (x) is obtained from J ν (x) by ν ν: J ν (x) = x ν m=0 ( 1) m 2 2m ν m! Γ(m ν + 1) x2m. (2.88) The general solution of the Bessel s equation (2.79) is given by y gen (x) = C 1 J ν (x) + C 2 J ν (x). (2.89) Bessel Functions of Integer Order For non-integer order ν, the Bessel functions J ν (x) and J ν (x) are linearly independent. What happens for integer orders? Without loss of generality, let ν = n N 0 = {0, 1, 2,...}. Then in (2.87), Γ(m + ν + 1) = (m + n)!, and J n (x) = m=0 ( 1) m 2 2m+ν m! (m + n)! x2m+n. In order to compute J ν (x) at ν = n, we note that in (2.88), Γ(m ν + 1) may tend to infinity when ν n. This happens when or m n + 1 = 0, 1, 2,..., m = n 1, n 2,..., 0. In the corresponding expression, terms corresponding to these m vanish. The following turns out to be true. Theorem For n N 0, J n (x) = ( 1) n J n (x). 43

44 Figure 2.2: Bessel Functions J n (x) (plot from Wikipedia). So when ν = n is integer, we do not have a solution basis for the Bessel s equation (2.79) any more. Indeed, we are in Case 3 of the Frobenius Theorem The second linearly independent solution in this case is y 2 (x) = Y n (x), the Bessel functions of the second kind. All of them are singular at x = 0. Finally, the general solution of the Bessel s equation (2.79) with ν = n N 0 is given by y gen (x) = C 1 J n (x) + C 2 Y n (x). (2.90) Figure 2.3: Bessel Functions Y n (x) (plot from Wikipedia). 44