Linear Operators and the General Solution of Elementary Linear Ordinary Differential Equations

Transcription

1 CODEE Journal Volume 9 Article Linear Operators and the General Solution of Elementary Linear Ordinary Differential Equations Norbert Euler Follow this and additional works at: Recommended Citation Euler, Norbert (2012 "Linear Operators and the General Solution of Elementary Linear Ordinary Differential Equations," CODEE Journal: Vol. 9, Article 11. Available at: This Article is brought to you for free and open access by the Journals at Claremont at Claremont. It has been accepted for inclusion in CODEE Journal by an authorized administrator of Claremont. For more information, please contact scholarship@cuc.claremont.edu.

2 Linear Operators and the General Solution of Elementary Linear Ordinary Differential Equations Norbert Euler Luleå University of Technology Keywords: General solutions of linear scalar differential equations, linear operators Manuscript received on February 8, 2012; published on May 12, Abstract: We make use of linear operators to derive the formulae for the general solution of elementary linear scalar ordinary differential equations of order n. The key lies in the factorization of the linear operators in terms of first-order operators. These first-order operators are then integrated by applying their corresponding integral operators. This leads to the solution formulae for both homogeneous- and nonhomogeneous linear differential equations in a natural way without the need for any ansatz (or educated guess. For second-order linear equations with nonconstant coefficients, the condition of the factorization is given in terms of Riccati equations. 1 Introduction In this paper we introduce linear differential operators and show that a linear differential equation can be solved by factoring these operators in terms of linear first-order operators. The factorization and integration of these operators then leads to a direct method and solution formulae to integrate any linear differential equation with constant coefficients without the need of an ansatz for its solutions. This method provides an alternative to the method of variation of parameters or the method of judicious guessing which one finds in most standard textbooks on elementary differential equations, e.g [1]. A more detailed discussion of the method of linear operators for differential equations is given in [2]. 2 Definitions In this section we introduce linear operators and introduce a integral operator that corresponds to a general first-order linear differential operator. This integral operator is the key to the integration of the linear equations. Let C(I denote the vector space of all continuous functions on some domain I R and let C n (I denote that vector subspace of C(I that consist of all functions with continuous nth order derivatives on I. CODEE Journal

3 Definition 2.1. We define the linear mapping for all f (x C n (I on the interval I R as L : C n (I C(I where L is the linear differential operator of order n L : f (x Lf (x, (2.1 L := p n (xd n x + p n 1 (xd (n 1 x + + p 1 (xd x + p 0 (x. (2.2 Here n N and p j (x (j = 0, 1,..., n are differentiable functions with so that D (k x := dk dx k, D x D x (1 Lf (x = p n (xf (n (x + p n 1 (xf (n 1 (x + + p 1 (xf (x + p 0 (xf (x. Example 2.1. Consider the second-order linear operator L = cos xdx 2 + e x D x + x 2. As an example, let L act on both e 2x and on u(x e x : L e 2x = 4e 2x cos x + 2e 3x + x 2 e 2x and L u(x e x = e x cos x u + 1 2e x cos x u + e x cos x + x 2 e x 1 u Lu(x, where L is another linear operator given by L := e x cos xdx e x cos x D x + e x cos x + x 2 e x 1. 2

4 Next we define the composite linear operator and the integral operator D 1 x. Definition 2.2. a Consider two linear differential operators of the form (2.2, namely L 1 of order m and L 2 of order n, and consider a function f (x C m+n (I with I R. The composite operator L 1 L 2 is defined by L 1 L 2 f (x := L 1 (L 2 f (x (2.3 where L 1 L 2 is a linear differential operator of order m + n. b The integral operator, D 1 x, is defined by the linear mapping D 1 x : C(I C 1 (I for all f (x C(O on the interval I R as D 1 x : f (x D 1 x f (x (2.4 where Dx 1 f (x := f (xdx (2.5 Note that, in general, L 1 L 2 f (x L 2 L 1 f (x. Example 2.2. We consider the two linear differential operators L 1 = D x + x 2, L 2 = xd 2 x + 1. Then L 2 L 1 f (x = xd 2 x + 1 f + x 2 f = x f (3 + x 3 f + (4x 2 + 1f + (x 2 + 2xf L 3 f (x, where L 3 := xd 3 x + x 3 D 2 x + (4x 2 + 1D x + x 2 + 2x. Furthermore L 1 L 2 f (x = x f (3 + (x 3 + 1f + f + x 2 f L 4 f (x where L 4 := xd 3 x + (x 3 + 1D 2 x + D x + x 2. Clearly L 2 L 1 f (x L 1 L 2 f (x. Let f be a differentiable function. Then, following Definition 2.2 b, we have D 1 x D x f (x = D 1 x f (x = f (x + c, (2.6 3

5 where c is an arbitrary constant of integration and, furthermore, D x Dx 1 f (x = d ( f (xdx + c = f (x. (2.7 dx In the next section we shall introduce a first-order linear operator and a corresponding integral operator which is essential for our application. In terms of the linear operator (2.2, the nth-order linear nonhomogeneous differential equation, p n (xu (n + p n 1 (xu (n p 1 (xu + p 0 (xu = f (x (2.8 takes the form Lu(x = f (x, where L is the linear operator (2.2. In the next section we introduce a method to solve Lu = f by factoring L into firstorder linear operators and then use the corresponding integral operators to eliminate all derivatives. For this purpose the following integral operator plays a central role. Definition 2.3. Let a and b be continuous real-valued functions on some interval I R, such that a(x 0 for all x I and consider the first-order linear operator L = a(xd x + b(x. (2.9 We define ˆL as the integral operator corresponding to L, as follows: ˆL := e ξ (x D 1 x 1 a(x eξ (x, (2.10 where dξ (x dx = a(x b(x. (2.11 By Definition 2.3, this proposition follows. Proposition 2.1. For any continuous function, f (x, we have [ ] f (x ˆLf (x = e ξ (x a(x eξ (x dx + c (2.12 and where c is an arbitrary constant of integration. ˆL L f (x = f (x + e ξ (x c, (2.13 4

6 We show that relation (2.13 holds: ˆL L f (x = ˆL (a(xf (x + b(xf (x ( a(xf = e ξ (x (x + b(xf (x a(x ( = e ξ (x f (xe ξ (x dx + e ξ (x dx + c b(x a(x f (x eξ (x dx + c Note that f (x e ξ (x + f (x e ξ (x = f (xe ξ (x, so that (2.13 follows.. Example 2.3. We consider L = xd x + x 2 and f (x = e x2 /2. Then the corresponding integral operator is ˆL = e x2 /2 Dx 1 e x2 /2 /x so that ˆL e x2 /2 = e x2 /2 (ln x + c and ˆL L e x2 /2 = e x2 /2 (1 + c. 3 Application to linear differential equations 3.1 First-order linear equations Consider the first-order linear differential equation in the form u + д(xu = h(x. (3.1 In terms of a linear differential operator (2.2 we can write (3.1 in the form where L is the first-order linear operator Lu(x = h(x, (3.2 L = D x + д(x. (3.3 Following Definition 2.3 we now apply the corresponding integral operator ˆL to both sides of (3.2 to gain the general solution of (3.1. We illustrate this explicitly in the next example. Example 3.1. We find the general solution of u + д(xu = h(x. Following Definition 2.3 we apply ˆL, given by ˆL = e ξ (x D 1 x e ξ (x with ξ (x = д(xdx, (3.4 to the left-hand side and the right-hand side of (3.2. For the left-hand side we obtain ˆL Lu(x = u(x + c 1 e ξ (x, (c 1 is an arbitrary constant 5

7 and for the right-hand side ˆLh(x = e ξ (x [h(xe ξ (x dx + c 2 ] (c 2 is an arbitrary constant. Thus u(x + c 1 e ξ (x = e ξ (x [h(xe ξ (x dx + c 2 ], or, equivalently u(x = e ξ (x [ ] h(xe ξ (x dx + c, where c = c 2 c 1 is an arbitrary constant and ξ (x = д(xdx. 3.2 Linear differential equations with constant coefficients In order to solve higher-order linear differential equations, we need to factorize the differential equatons into first-order factors. This is in principle always possible, but in practise there are some obstacles. We first look at the constant-coefficient case and then the more general case which allows nonconstant coefficient functions in the linear operator. It should be clear that the linear constant-coefficient homogeneous equation factors in terms of first-order differential operators in the same manner as the corresponding characteristic equation does. Take, for example, the second-order equation u + pu + qu = f (x, p R, q R, (3.5 with characteristic equation (following the substitution u = e zx z 2 + pz + q = 0 (3.6 and roots z 1 = 1 2 ( p + p 2 4q, z 2 = 1 ( p p 2 2 4q. (3.7 Equation (3.5 then takes the form Lu(x = f (x, where L = Dx 2 + pd x + q. (3.8 It is now easy to show that the second-order operator (3.8 factors as L = (D x z 1 (D x z 2 L 1 L 2, (3.9 where L 1 = D x z 1 and L 2 = D x z 2. Since L 1 L 2 u(x = (D x z 1 (u z 2 u = u (z 1 + z 2 u + z 1 z 2 u 6

8 and, by (3.7, z 1 + z 2 = p and z 1 z 2 = q. Thus Lu(x = L 1 L 2 u(x = u + pu + qu = f (x. This easily extends to linear constant-coefficient equations of order n. Proposition 3.1. Consider a constant coefficient nth-order nonhomogeneous differential equation of the form where Lu(x = f (x, (3.10 L = a n D n x + a n 1 D n 1 x + + a 1 D x + a 0 (3.11 a j R, j = 0, 1,..., n. The characteristic equation of (3.10, following the substitution u(x = e zx, is the nth degree polynomial a n z n + a n 1 z n a 1 z + a 0 = 0 (3.12 with n roots, {z 1, z 2,..., z n }, so that (3.12 can be factorized as Then equation (3.10 factors in the form where (z z 1 (z z 2 (z z n = 0. L 1 L 2 L n u(x = f (x, (3.13 L j = D x z j, j = 1, 2,..., n. (3.14 The general solution of (3.10 then follows by applying, successively, the corresponding integral operators ˆL 1, ˆL 2,..., ˆL n to (3.13. Applying Proposition 3.1 to second-order equations leads to Proposition

9 Proposition 3.2. Consider the 2nd-order equation with characteristic equation u + pu + qu = f (x, p, q R (3.15 z 2 + pz + q = 0. (3.16 a If the two roots, z 1 and z 2, of (3.16 are real and distinct, then the general solution of (3.15 is u(x = c 1 e z 1x + c 2 e z 2x + u p (x, (3.17 where u p (x is a particular solution of (3.15 given by ( ( 1 u p (x = e z 1x z 1 z 2 and c 1 and c 2 are arbitrary constants. f (xe z1x dx e z 2x f (xe z2x dx (3.18 b If the two roots, z 1 and z 2, of (3.16 are complex numbers, then z 2 = z 1 (z 2 is the complex conjugate of z 1 and the general solution of (3.15 is u(x = c 1 Re {e z 1x } + c 2 Im {e z 1x } + u p (x, (3.19 where u p (x is a particular solution of (3.15 given by ( ( 1 u p (x = e z 1x z 1 z 1 f (xe z1x dx e z 1x f (xe z 1x dx (3.20 and c 1 and c 2 are arbitrary constants. Note that although z 1 and z 1 are complex numbers, the solution u(x is always a real-valued function. c If the two roots for (3.16 are equal, i.e. z 1 = z 2 R, then the general solution of (3.15 is u(x = c 1 e z 1x + c 2 x e z 1x + u p (x (3.21 where c 1 and c 2 are arbitrary constants and a particular solution, u p (x, of (3.15 is ( u p (x = e z 1x f (xe z1x dx dx. (3.22 Proof: Equation (3.15 can be written in the form L 1 L 2 u(x = f (x, L 1 = D x z 1, L 2 = D x z 2, (3.23 where z 1 and z 2 are the roots of the characteristic equation (3.16. The corresponding 8

10 integral operators for L 1 and L 2 are respectively. From (3.23 we obtain L 2 u(x + e z1x k 1 = e z 1x ˆL 1 = e z 1x D 1 x e z 1x, ˆL 2 = e z 2x D 1 x e z 2x, (3.24 or L 2 u(x = e z 1x f (xe z 1x dx + k 2 e z 1x f (xe z 1x dx + k 3 e z 1x, (3.25 where k 3 k 2 k 1 is another constant of integration. Applying now ˆL 2 to each side of (3.25 we have (k3 u(x + e z2x k 4 = e z 2x e (z 1 z 2 x + e (z 1 z 2 x F(x dx + k 5 e z 2x or u(x = e z 2x (k3 e (z 1 z 2 x + e (z 1 z 2 x F(x dx + k 6 e z 2x, (3.26 where k 6 k 5 k 4 is a constant of integration and F(x := f (xe z1x dx. (3.27 If z 1 z 2, then (3.26 reduces (after integration by parts to (3.17 for z 1 R and z 2 R and to (3.19 for complex roots z 1 and z 2 = z 1, or to (3.21 for equal roots z 1 = z 2 R. Note that c 1 k 3 and c 2 k 6 are the two arbitrary constants for the general solution. Example 3.2. We find the general solution of u + 4u = 8x 2. (3.28 The characteristic and its roots are z = 0, z 1 = 2i, z 2 = 2i so the equation can be presented in factorized form L 1 L 2 u(x = 8x 2, where L 1 = D x 2i, L 2 = D x + 2i. Following Proposition (3.2, the general solution, ϕ h (x, of the homogeneous part of (3.28 is ϕ h (x;c 1,c 2 = c 1 cos(2x + c 2 sin(2x. (3.29 For a particular solution, u p (x, we use formula (3.20 and calculate the integrals: u p (x = 1 4i (e 2ix 8x 2 e 2ix dx e 2ix 8x 2 e 2ix dx = 2x 2 1. The general solution of (3.28 is thus u(x = c 1 cos(2x + c 2 sin(2x + 2x 2 1, where c 1 and c 2 are arbitrary constants. 9

11 For an nth-order linear constant-coefficient equations we can obtain this result. Proposition 3.3. Consider a constant coefficient nth-order nonhomogeneous differential equation with n > 2 of the form where Equation (3.30 factors in the form Lu(x = f (x, (3.30 L = a n D n x + a n 1 D n 1 x + + a 1 D x + a 0 (3.31 a j R, j = 0, 1,..., n. L 1 L 2 L n u(x = f (x, (3.32 where L j = D x z j (j = 1, 2,..., n and {z 1, z 2,..., z n } are the roots of its characteristic equation. The general solution of (3.30 is then u(x = ϕ h (x;c 1,c 2,...,c n + u p (x, (3.33 where ϕ h (x is the general solution of the homogeneous part of (3.30, namely ( ϕ h (x;c 1,c 2,...,c n = c 1 e z nx e (z n 1 z n x G {n 2} 12 (x dx + c 2 e znx G {n 2} 23 (x +c 3 e z nx G {n 3} 34 (x + + c n 1 e z nx G {1} (n 1n (x + c ne z nx (c 1, c 2,..., c n are arbitrary constants and u p (x is a particular solution, (3.34 u p (x = e z nx (e (z n 1 z n x F n 1 (xdx. (3.35 Here G {1} kl (x := e (z k z l x dx, G {j} kl (x := e (z l+j 2 z l+j 1 x G {j 1} (x dx, kl j = 2, 3,... ; k = 1, 2,... ; l = 2, 3,.... F 1 (x := f (xe z1x dx, F i (x := e (z i 1 z i x F i 1 (xdx i = 2, 3,.... To prove Proposition 3.3 we apply the corresponding integral operators on (3.32 and identify the patterns. The details are ommitted. Remark: For complex numbers, {z 1, z 2,..., z n }, the expression for ϕ h (x;c 1,c 2,...,c n, (3.34, will be a complex-valued solution for (3.30 for which the real- and the imaginary 10

12 parts are real-valued solutions of (3.30. We therefore need to combine the real- and imaginary parts of ϕ h such that we have n linear independent real-valued solutions. We remark further that (3.35 is a real particular solution for (3.30, even for the case where {z 1, z 2,..., z n } are complex roots of the characteristic equation. Example 3.3. We find the general solution for u (3 u + u u = e 2x cos x. (3.36 The characteristic equation is z 3 z 2 + z 1 = 0, with roots z 1 = 1, z 2 = i, z 3 = i. Thus (3.36 can be presented in the factorized form L 1 L 2 L 3 u(x = e 2x cos x, where L 1 = D x 1, L 2 = D x i, L 3 = D x + i. Following Proposition 3.3 the general solution of the homogeneous part of (3.36 is ( ϕ h (x;c 1,c 2,c 3 = c 1 e z 3x e (z 2 z 3 x G {1} 12 dx + c 2 e z3x G {1} 23 + c 3e z3x, where G {1} 12 = G {1} 23 = e (z 1 z 2 x dx = e (z 2 z 3 x dx = ( e (1 ix 1 dx = 1 i ( 1 e 2ix dx = e 2ix. 2i e (1 ix We find where ϕ h (x;c 1,c 2,c 3 = ( 1 ( 1 c 1 e 2 x + c 2 e ix + c 3 e ix, 2i Re {ϕ h } = 1 2 c 1e x c 2 sin x + c 3 cos x Im {ϕ h } = 1 2 c 2 cos x c 3 sin x are real-valued solutions of the homogeneous part of (3.36. Since {e x, sin x, cos x} is a linearly independent set of functions for all x R, the general solution of the homogeneous part of (3.36 is ϕ h (x; a 1, a 2, a 3 = a 1 e x + a 2 sin x + a 3 cos x, where a 1, a 2 and a 3 are arbitrary constants. Following Proposition 3.3, a particular solution for (3.36 is of the form ( u p (x = e ix e 2ix F 2 (xdx, 11

13 where F 2 (x = e (1 ix F 1 (xdx, F 1 (x = e x cos x dx. Calculating the integrals we obtain F 1 (x = 1 2 ex (cos x + sin x F 2 (x = 1 8 e(2 ix (cos x + 2 sin x + i sin x so that the particular solution becomes [ ] 1 u p (x = e ix 8 e2x e ix (cos x + 2 sin x + i sin x dx = 1 8 e2x sin x. The general solution of (3.36 is thus u(x = a 1 e x + a 2 sin x + a 3 cos x e2x sin x. 3.3 Linear differential equations with nonconstant coefficients First we consider second-order linear nonhomogeneous equations of the form u + д(xu + h(xu = f (x, (3.37 where f, д and h are differentiable functions on some common interval I R. Assume a factorization in terms of two linear first-order differential operators, such that (3.37 is equivalent to Now (3.39 takes the form L 1 = D x + q 1 (x, L 2 = D x + q 2 (x, (3.38 L 1 L 2 u(x = f (x. (3.39 u + (q 1 + q 2 u + (q 1 q 2 + q 2 u = f (x (3.40 and, comparing (3.40 to (3.37 leads to the condition q 1 + q 2 = д(x, q 2 + q 1q 2 = h(x, or, equivalently, q 2 = q2 2 д(xq 2 + h(x, q 1 (x = д(x q 2 (x. To find the general solution of (3.39 we apply the corresponding integral operators ˆL 1 and ˆL 2, successively. This leads to Proposition

14 Proposition 3.4. The 2nd-order linear equation can be written in the factorized form u + д(xu + h(xu = f (x (3.41 L 1 L 2 u(x = f (x, (3.42 where L 1 = D x + q 1 (x and L 2 = D x + q 2 (x, if and only if q 2 (x satisfies the Riccati equation q 2 = q2 2 д(xq 2 + h(x. (3.43 Then q 1 (x = д(x q 2 (x. Applying the corresponding integral operators, ˆL 1 and ˆL 2 successively on (3.42, leads to the general solution of (3.41 namely u(x = c 1 e ξ2(x + c 2 e ξ 2(x e ξ2(x e ξ1(x dx + u p (x (3.44 where u p (x is a particular solution of (3.41 given by u p (x = e ξ 2(x e ξ2(x e ξ1(x F(xdx. (3.45 Here c 1 and c 2 are arbitrary constants and F(x := f (xe ξ1(x dx, ξ 1 (x := q 1 (xdx, ξ 2 (x := q 2 (xdx. 13

15 Example 3.4. (Second-order Cauchy-Euler equation We consider the second-order nonhomogeneous Cauchy-Euler equation ax 2 u + bxu + cu = f (x, x 0, (3.46 where a 0, b and c are real constants and f is a continuous function on some interval I R. Equation (3.46 can equivalently be presented in the form ( b ( c u + u + ax ax 2 u = f (x ax 2. (3.47 Comparing (3.47 and (3.41 we identify д(x = b ax, h(x = c ax 2, so that, by Proposition 3.4, equation (3.46 can be factorized in the form (3.42 if q 2 satisfies the following Riccati equation: 14

16 ( b q 2 = q2 2 q 2 + c ax ax 2. (3.48 A solution of (3.48 is of the form q 2 (x = αx β with β = 1 and α satisfying the quadratic equation ( α b α + c = 0. (3.49 a a As an explicit example we consider a = 1, b = 1 and c = 1. This corresponds to the equation A solution for (3.49 is then α = 1, so that x 2 u xu + u = x 3. (3.50 q 2 (x = 1 x, q 1(x = 0, ξ 1 (x = 0, ξ 2 (x = ln x. Thus the equation factors in the form ( 1 u x ( 1 u + x 2 u = x ( (D x D x 1 u(x = x x so that, by the solution formula (3.44, the general solution of (3.50 becomes where c 1 and c 2 are arbitrary constants. u(x = c 1 x + c 2 x ln x x3, Example 3.5. Consider the equation Comparing (3.51 and (3.41 we identify u + 2xu + (x 2 + 1u = e x2 /2. (3.51 д(x = 2x, h(x = x 2 + 1, so that, by Proposition 3.4, equation (3.51 can be factorized in the form (3.42 if q 2 satisfies the following Riccati equation: q 2 = q2 2 2xq 2 + x (

17 A solution for (3.52 is q 2 (x = x so that q 1 (x = x. Hence (3.51 takes the factorized form (D x + x(d x + xu(x = e x2 /2 so that, by the solution formula (3.44, the general solution of (3.51 becomes where c 1 and c 2 are arbitrary constants. u(x = c 1 e x2 /2 + c 2 xe x2 / x2 e x2 /2 An extension to higher-order nonconstant-coefficient linear equations is possible, but the general condition for the factorization into first-order linear operators becomes rather complicated. 4 Conclusion We have shown that the factorization of linear operators in terms of first-order linear operators can be applied to find the general solution of an nth-order linear differential equation. This provides an alternate approach to the ansatz-based methods that are often taught to students in an introductory course on ordinary differential equations. The method is based on the integral operator as defined in Definition 2.3. and also applies to linear differential equations with nonconstant coefficients. This paper has been written on a basic level and with many examples to make it accessible to a broad audience. Acknowledgements I thank Marianna Euler, Thomas Gunnarsson and Lech Maligranda for a careful reading of the manuscript and for suggestions. I also thank the referee for pointing out some corrections. References [1] Braun M, Differential Equations and Their Applications, Springer, 1986 [2] Euler M and Euler N, An Invitation to Differential Equations (in preparation,