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1 This article was downloaded by: [Camporesi, Roberto] On: 19 May 211 Access details: Access Details: [subscription number ] Publisher Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: Registered office: Mortimer House, Mortimer Street, London W1T 3JH, UK International Journal of Mathematical Education in Science and Technology Publication details, including instructions for authors and subscription information: Linear ordinary differential equations with constant coefficients. Revisiting the impulsive response method using factorization Roberto Camporesi a a Dipartimento di Matematica, Politecnico di Torino, Torino 1129, Italy First published on: 8 April 211 To cite this Article Camporesi, Roberto(211) 'Linear ordinary differential equations with constant coefficients. Revisiting the impulsive response method using factorization', International Journal of Mathematical Education in Science and Technology, 42: 4, , First published on: 8 April 211 (ifirst) To link to this Article: DOI: 1.18/2739X URL: PLEASE SCROLL DOWN FOR ARTICLE Full terms and conditions of use: This article may be used for research, teaching and private study purposes. Any substantial or systematic reproduction, re-distribution, re-selling, loan or sub-licensing, systematic supply or distribution in any form to anyone is expressly forbidden. The publisher does not give any warranty express or implied or make any representation that the contents will be complete or accurate or up to date. The accuracy of any instructions, formulae and drug doses should be independently verified with primary sources. The publisher shall not be liable for any loss, actions, claims, proceedings, demand or costs or damages whatsoever or howsoever caused arising directly or indirectly in connection with or arising out of the use of this material.

2 International Journal of Mathematical Education in Science and Technology, Vol. 42, No. 4, 15 June 211, Linear ordinary differential equations with constant coefficients. Revisiting the impulsive response method using factorization Roberto Camporesi* Dipartimento di Matematica, Politecnico di Torino, Corso Duca degli Abruzzi 24, Torino 1129, Italy (Received 14 July 21) Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 We present an approach to the impulsive response method for solving linear constant-coefficient ordinary differential equations based on the factorization of the differential operator. The approach is elementary, we only assume a basic knowledge of calculus and linear algebra. In particular, we avoid the use of distribution theory, as well as of the other more advanced approaches: Laplace transform, linear systems, the general theory of linear equations with variable coefficients and the variation of constants method. The approach presented here can be used in a first course on differential equations for science and engineering majors. Keywords: linear differential equations; factorization; Green functions 1. Introduction Linear constant-coefficient differential equations constitute an important chapter in the theory of ordinary differential equations, also in view of their many applications in various fields of science. In introductory courses on differential equations, the treatment of second or higher order non-homogeneous equations is usually limited to illustrating the method of undetermined coefficients. Using this, one finds a particular solution when the forcing term is a polynomial, an exponential, a sine or a cosine, or a product of terms of this kind. It is well known that the impulsive response method gives an explicit formula for a particular solution in the more general case in which the forcing term is an arbitrary continuous function. This method is generally regarded as too difficult to implement in a first course on differential equations. Students become aware of it only later, as an application of the theory of the Laplace transform [1] or of distribution theory [2]. An alternative approach, which is sometimes used consists in developing the theory of linear systems first, considering then linear equations of order n as a particular case of this theory. The problem with this approach is that one needs to digest the theory of linear systems, with all the issues related to the diagonalization of matrices and the Jordan form [3, Chapter 3]. * camporesi@polito.it ISSN 2 739X print/issn online ß 211 Taylor & Francis DOI: 1.18/2739X

3 498 R. Camporesi Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 Another approach is by the general theory of linear equations with variable coefficients, with the notion of Wronskian and the method of the variation of constants. This approach can also be implemented in the case of constant coefficients [4, Chapter 2]. However, in introductory courses, the variation of constants method is often limited to first-order equations. Moreover, this method may be very long to implement in specific calculations, even for rather simple equations. Finally, within this approach, the occurrence of the particular solution as a convolution integral is rather indirect, and appears only at the end of the theory (see, e.g. [4, Exercise 4, p. 89]). The purpose of these notes is to give an elementary presentation of the impulsive response method using only very basic results from linear algebra and calculus in one or many variables. We discuss in detail the case of second-order equations, but our approach can easily be generalized to linear equations of any order (see Remark 5 in Section 4). Consider the second-order equation y þ ay þ by ¼ f ðxþ, ð1:1þ where we use the notation y(x) in place of x(t) ory(t), y (k) denotes as usual the derivative of order k of y, a, b are real constants and the forcing term f : I R! R is a continuous function in an interval I. When f 6¼ the equation is called nonhomogeneous. When f ¼ we get the associated homogeneous equation y þ ay þ by ¼ : ð1:2þ The basic tool of our investigation is the so called impulsive response. This is the function defined as follows. Let p() ¼ 2 þ a þ b ( 2 C) be the characteristic polynomial, and let 1, 2 2 C be the roots of p() (not necessarily distinct). We define the impulsive response g ¼ g 1 2 by the following formula: g 1 2 ðxþ ¼e 2x e ð 1 2 Þt dt ðx 2 RÞ: It turns out that g solves the homogeneous equation (1.2) with the initial conditions yðþ ¼, y ðþ ¼1: Moreover, the impulsive response allows one to solve the non-homogeneous equation with an arbitrary continuous forcing term and with arbitrary initial conditions. Indeed, if 2 I, we shall see that the general solution of (1.1) in the interval I can be written as y ¼ y p þ y h, ð1:3þ where the function y p is given by the convolution integral y p ðxþ ¼ gðx tþ f ðtþdt, and solves (1.1) with trivial initial conditions at the point x ¼, i.e. y p ðþ ¼y pðþ ¼, ð1:4þ whereas the function y h ðxþ ¼c gðxþþc 1 g ðxþ ð1:5þ

4 International Journal of Mathematical Education in Science and Technology 499 Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 gives the general solution of the associated homogeneous equation (1.2) as the coefficients c, c 1 vary in R. In other words, the two functions g, g are linearly independent solutions of this equation and form a basis of the vector space of its solutions. The linear relation between the coefficients c, c 1 in (1.5) and the initial data b ¼ y() ¼ y h () and b 1 ¼ y ðþ ¼y hðþ is easily obtained. If we impose the initial conditions at an arbitrary point x 2 I, we can just replace R x with R x x in (1.4), and g(x), g (x) with g(x x ), g (x x ) in (1.5). The function y p then satisfies y p ðx Þ¼ y p ðx Þ¼, and the relation between c k and b k ¼ y ðkþ ðx Þ¼y ðkþ h ðx Þ ðk ¼, 1Þ remains the same as before. The proof of (1.4) that we give is based on the factorization of the differential operator acting on y in (1.1) into first-order factors, along with the formula for solving first-order linear equations. It requires, moreover, the interchange of the order of integration in a double integral, that is, the Fubini theorem. The proof is constructive in that it directly produces the particular solution y p as a convolution integral between the impulsive response g and the forcing term f. In particular if we take f ¼, we get that the unique solution of the homogeneous initial value problem with all vanishing initial data is the zero function. By linearity, this implies the uniqueness of the solutions of the initial value problem (homogeneous or not) with arbitrary initial data. In general, the factorization method provides an elementary proof of existence, uniqueness and extendability of the solutions of a linear initial value problem with constant coefficients (homogeneous or not). We thus obtain a foundation for a complete theory of linear constant-coefficient differential equations. We would like to mention that the approach by factorization to linear constantcoefficient differential equations is certainly not new (see, e.g. [5], and the references therein). The use of formula (1.4) for finding a particular solution of the nonhomogeneous equation is also known as the Cauchy or Boole integral method, and has been discussed in general in [6] (see also [7], for an approach using factorization). The plan of this article is as follows. In Section 2 we briefly review the case of first-order linear equations within the framework of the impulsive response method. We then proceed, in Section 3, with the basic case of second-order equations. Some examples are given to illustrate the method. Finally, in Section 4, we collect some general remarks, including a brief discussion of higher-order linear equations with constant coefficients. 2. First-order equations Consider the first-order linear differential equation y þ ay ¼ f ðxþ, ð2:1þ where y ¼ dy dx, a is a real constant, and the forcing term f is a continuous function in an interval I R. It is well known that the general solution of (2.1) is given by Z yðxþ ¼e ax e ax f ðxþdx, ð2:2þ

5 5 R. Camporesi Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 where R e ax f (x)dx denotes the set of all primitives of the function e ax f (x) in the interval I (i.e. its indefinite integral). Suppose that 2 I, and consider the integral function R x eat f ðtþdt. By the Fundamental Theorem of Calculus, this is the primitive of e ax f (x) that vanishes at. The theorem of the additive constant for primitives implies that Z e ax f ðxþdx ¼ e at f ðtþdt þ k ðk 2 RÞ, and we can rewrite (2.2) in the form yðxþ ¼ e ax e at f ðtþdt þ ke ax ¼ ¼ e aðxtþ f ðtþdt þ ke ax gðx tþ f ðtþdt þ kgðxþ, ð2:3þ where g(x) ¼ e ax. The function g is called the impulsive response of the differential equation y þ ay ¼. It is the unique solution of the initial value problem y þ ay ¼, yðþ ¼1: Formula (2.3) illustrates a well-known result in the theory of linear differential equations. Namely, the general solution of (2.1) is the sum of the general solution of the associated homogeneous equation y þ ay ¼ and of any particular solution of (2.1). In (2.3) the function y p ðxþ ¼ gðx tþ f ðtþdt is the particular solution of (2.1) that vanishes at x ¼. If x is any point of I, it is easy to verify that ð2:4þ yðxþ ¼ gðx tþ f ðtþdt þ y gðx x Þ ðx 2 I Þ x is the unique solution of (2.1) in the interval I that satisfies y(x ) ¼ y ( y 2 R). We shall now see that the formula (2.4) gives a particular solution of the nonhomogeneous equation also in the case of second-order linear constant-coefficient differential equations, by suitably defining the impulsive response g. 3. Second-order equations Consider the second-order non-homogeneous linear differential equation y þ ay þ by ¼ f ðxþ, ð3:1þ where y ¼ d2 y, a, b 2 R, and the forcing term f : I! R is a continuous function in the dx 2 interval I R, i.e. f 2 C (I ). For f ¼ we get the associated homogeneous equation y þ ay þ by ¼ : ð3:2þ

6 International Journal of Mathematical Education in Science and Technology 51 We will write (3.1) and (3.2) in operator form as Ly ¼ f (x) and Ly ¼, where L is the linear second-order differential operator with constant coefficients defined by Ly ¼ y þ ay þ by, for any function y at least twice differentiable. Denoting by d dx the differentiation operator, we have L ¼ d 2 þ a d dx dx þ b: ð3:3þ L defines a map C 2 (R)! C (R) that to each function y at least twice differentiable over R with continuous second derivative associates the continuous function Ly. The fundamental property of L is its linearity, that is, Lðc 1 y 1 þ c 2 y 2 Þ¼c 1 Ly 1 þ c 2 Ly 2 Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 8c 1, c 2 2 R, 8y 1, y 2 2 C 2 (R). This formula implies some important facts. First of all, if y 1 and y 2 are any two solutions of the homogeneous equation, then any linear combination c 1 y 1 þ c 2 y 2 is also a solution of this equation. In other words, the set V ¼ y 2 C 2 ðrþ : Ly ¼ is a vector space over R. We shall see that this vector space has dimension two, and that the solutions of (3.2) are defined in fact on the whole of R and are of class C 1 there. Second, if y 1 and y 2 are two solutions of (3.1) (in a given interval I I ), then their difference y 1 y 2 solves (3.2). It follows that if we know a particular solution y p of the non-homogeneous equation (in an interval I ), then any other solution of (3.1) in I is given by y p þ y h, where y h is a solution of the associated homogeneous equation. We shall see that the solutions of (3.1) are defined on the whole of the interval I in which f is continuous (and are of course of class C 2 there). The fact that L has constant coefficients (i.e. a and b in (3.3) are constants) allows one to find explicit formulas for the solutions of (3.1) and (3.2). To this end, it is useful to consider complex-valued functions, y : R! C. If y ¼ y 1 þ iy 2 (with y 1, y 2 : R! R) is such a function, the derivative y may be defined by linearity as y ¼ y 1 þ iy 2. It follows that L( y 1 þ iy 2 ) ¼ Ly 1 þ ily 2. In a similar way one defines the integral of y as Z Z Z Z d Z d Z d yðxþdx ¼ y 1 ðxþdx þ i y 2 ðxþdx, yðxþdx ¼ y 1 ðxþdx þ i y 2 ðxþdx: The theorem of the additive constant for primitives and the Fundamental Theorem of Calculus extend to complex-valued functions. It is then easy to verify that d dx ex ¼ e x 8 2 C: It follows that the complex exponential e x is a solution of (3.2) if and only if is a root of the characteristic polynomial p() ¼ 2 þ a þ b. Let 1, 2 2 C be the roots of p(), so that c c c pðþ ¼ð 1 Þð 2 Þ:

7 52 R. Camporesi The operator L factors in a similar way as a product (composition) of first-order differential operators L ¼ d d dx 1 dx 2 : ð3:4þ Indeed we have d d dx 1 dx 2 y ¼ d dx 1 ð y 2 yþ ¼ y ð 1 þ 2 Þ y þ 1 2 y, Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 which coincides with Ly since 1 þ 2 ¼a, and 1 2 ¼ b. Note that in (3.4) the order with which the two factors are composed is unimportant. In other words, the two operators ðdx d 1Þ and ðdx d 2Þ commute d d dx 1 dx 2 ¼ d d dx 2 dx 1 : The idea is now to use (3.4) to reduce the problem to first-order differential equations. It is useful to consider linear differential equations with complex coefficients, whose solutions will be, in general, complex-valued. For example, the first-order homogeneous equation y y ¼ with 2 C has the general solution yðxþ ¼ke x ðk 2 CÞ: (Indeed if y ¼ y, then d dx yðxþex ¼, whence y(x)e x ¼ k.) The first-order nonhomogeneous equation y y ¼ d dx y ¼ f ðxþ ð 2 CÞ, with complex forcing term f : I R! C continuous in I 3, has the general solution Z yðxþ ¼e x e x f ðxþdx ¼ e x e t f ðtþdt þ ke x ¼ g ðx tþ f ðtþdt þ kg ðxþ ðx 2 I, k ¼ yðþ2cþ: ð3:5þ Here g (x) ¼ e x is the impulsive response of the differential operator ð d dx Þ. Itis the (unique) solution of y y ¼, y() ¼ 1. Formula (3.5) can be proved as (2.3) in the real case. In particular, the solution of the first-order problem y y ¼ f ðxþ, yðþ ¼ ( 2 C) is unique and is given by yðxþ ¼ e ðxtþ f ðtþdt: ð3:6þ

8 International Journal of Mathematical Education in Science and Technology 53 The following result gives a particular solution of (3.1) as a convolution integral. Theorem 3.1: Let f 2 C (I ), and suppose that 2 I. Then the initial value problem y þ ay þ by ¼ f ðxþ, yðþ ¼, y ð3:7þ ðþ ¼ has a unique solution, defined on the whole of I, and given by the formula yðxþ ¼ where g is the function defined by gðx tþ f ðtþdt ðx 2 I Þ, ð3:8þ Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 gðxþ ¼ e 2ðxtÞ e 1t dt ðx 2 RÞ: ð3:9þ In particular, if we take f ¼, we get that the only solution of the homogeneous problem y þ ay þ by ¼, yðþ ¼, y ð3:1þ ðþ ¼ is the zero function y ¼. Proof: We rewrite the differential equation (3.1) in the form d d dx 1 dx 2 y ¼ f ðxþ: Letting h ¼ d dx 2 y ¼ y 2 y, we see that y solves the problem (3.7) if and only if h 1 h ¼ f ðxþ, y 2 y ¼ hðxþ, h solves and y solves hðþ ¼, yðþ ¼: From (3.6) we get hðxþ ¼ yðxþ ¼ e 1ðxtÞ f ðtþdt, e 2ðxtÞ hðtþdt: Substituting h from the first formula into the second, we obtain y(x) as a repeated integral (for any x 2 I): Z t yðxþ ¼ e 2ðxtÞ e 1ðtsÞ f ðsþds dt: ð3:11þ

9 54 R. Camporesi To fix ideas, let us suppose that x 4. Then in the integral with respect to t we have t x, whereas in the integral with respect to s we have s t. We can then rewrite y(x) as a double integral: Z yðxþ ¼e 2x e ð 1 2 Þt e 1s f ðsþds dt, T x where T x is the triangle in the (s, t) plane defined by s t x, with vertices at the points (, ), (, x) and (x, x). In (3.11) we first integrate with respect to s and then with respect to t. Since the triangle T x is convex both horizontally and vertically, and since the integrand function Fðs, tþ ¼e ð 1 2 Þt e 1s f ðsþ Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 is continuous in T x, we can interchange the order of integration and integrate with respect to t first. Given s (between and x) the variable t in T x varies between s and x, see the figure below. We thus obtain t x yðxþ ¼ s t x s x s =t e 2ðxtÞ e 1ðtsÞ dt f ðsþds: By substituting t with t þ s in the integral with respect to t we finally get s yðxþ ¼ e 2ðxstÞ e 1t dt f ðsþds ¼ gðx sþ f ðsþds, s ð3:12þ which is (3.8). For x 5 we can reason in a similar way and we get the same result. g The integral in formula (3.9) can be computed exactly as in the real field. We obtain the following expression of the function g: (1) if 1 6¼ 2 (,D ¼ a 2 4b 6¼ ) then 1 gðxþ ¼ e 1x e 2x ; ð3:13þ 1 2 (2) if 1 ¼ 2 (,D ¼ ) then gðxþ ¼xe 1x : ð3:14þ

10 International Journal of Mathematical Education in Science and Technology 55 Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 Note that g is always a real function. Letting ¼a/2 and ( p ffiffiffiffiffiffiffi ( D =2 if D5 i if D5 ¼ p ffiffiffi so that 1,2 ¼ D =2 if D4, if D4, we have 8 1 2i eðþiþx e ðiþx 1 >< ¼ ex sinðxþ if D5, gðxþ ¼ 1 2 eðþþx e ðþx 1 >: ¼ ex sinhðxþ if D4: Also note that g 2 C 1 (R). It is easy to check that g solves the following homogeneous initial value problem: ( y þ ay þ by ¼, ð3:15þ yðþ ¼, y ðþ ¼1: The function g is called the impulsive response of the differential operator L. It is interesting to verify directly that the function y given by (3.8) solves (3.1). First, let us prove the following formula for the derivative y : y ðxþ ¼ d gðx tþ f ðtþdt ¼ gðþ f ðxþþ dx g ðx tþ f ðtþdt ðx 2 I Þ: ð3:16þ Indeed, given h such that x þ h 2 I, we have yðx þ hþyðxþ ¼ 1 þh gðx þ h tþ f ðtþdt gðx tþ f ðtþdt : h h ð3:17þ As g 2 C 2 (R), we can apply Taylor s formula with the Lagrange remainder gðx þ hþ ¼gðx Þþg ðx Þh þ 1 2 g ðþh 2 at the point x ¼ x t, where is some point between x and x þ h. Substituting this in (3.17) and using gives þh gðx tþ f ðtþdt ¼ gðx tþ f ðtþdt þ þh x gðx tþ f ðtþdt 1 yðx þ hþyðxþ h ð Þ ¼ 1 h þh x gðx tþ f ðtþdt þ þ 1 þh 2 h g ðþ f ðtþdt þh g ðx tþ f ðtþdt ð3:18þ for some between x t and x t þ h. When h tends to zero, the first term in the right-hand side of (3.18) tends to g()f (x), by the Fundamental Theorem of Calculus. The second term in (3.18) tends to R x g ðx tþ f ðtþdt, by the continuity of the integral function. Finally, the third term tends to zero, since the integral that occurs in it is a

11 56 R. Camporesi bounded function of h in a neighbourhood of h ¼ (this is easy to prove). We thus obtain formula (3.16). Recalling that g() ¼, we finally get Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 y ðxþ ¼ g ðx tþ f ðtþdt ðx 2 I Þ: In the same way we compute the second derivative y ðxþ ¼ d 2 gðx tþ f ðtþdt dx ¼ d g ðx tþ f ðtþdt dx ¼ g ðþ f ðxþþ ¼ f ðxþþ where we used g () ¼ 1. It follows that y ðxþþay ðxþþbyðxþ ¼f ðxþþ g ðx tþ f ðtþdt g ðx tþ f ðtþdt, ¼ f ðxþ 8x 2 I, ðg þ ag þ bgþðx tþ f ðtþdt g being a solution of the homogeneous equation. Therefore the function y given by (3.8) solves (3.1) in the interval I. The initial conditions y() ¼ ¼ y () are immediately verified. We now come to the solution of the initial value problem with an arbitrary initial data at the point x ¼. Theorem 3.2: Let f 2 C (I ), 2 I, and let y, y be two arbitrary real numbers. Then the initial value problem y þ ay þ by ¼ f ðxþ, yðþ ¼y, y ðþ ¼y ð3:19þ has a unique solution, defined on the whole of I, and given by yðxþ ¼ gðx tþ f ðtþdt þ y þ ay gðxþþy g ðxþ ðx 2 I Þ: ð3:2þ In particular (taking f ¼ ), the solution of the homogeneous problem y þ ay þ by ¼, yðþ ¼y, y ðþ ¼y is unique, of class C 1 on the whole of R, and is given by y h ðxþ ¼ y þ ay gðxþþy g ðxþ ðx 2 RÞ: ð3:21þ Proof: The uniqueness of the solutions of the problem (3.19) follows from the fact that if y 1 and y 2 both solve (3.19), then their difference ~y ¼ y 1 y 2 solves the problem (3.1), whence ~y ¼ by Theorem 3.1. Now note that the function g

12 satisfies the homogeneous equation (like g). Indeed, since L has constant coefficients, we have Lg ¼ L d dx g ¼ " # d 2 þ a d dx dx þ b d dx g ¼ d Lg ¼ : dx By the linearity of L and by Theorem 3.1 it follows that the function y given by (3.2) satisfies (Ly)(x) ¼ f (x) 8x 2 I. It is immediate that y() ¼ y. Finally, since y ðxþ ¼ g ðx tþ f ðtþdt þ y þ ay g ðxþþy g ðxþ, we have International Journal of Mathematical Education in Science and Technology 57 Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 y ðþ ¼y þ ay þ y g ðþ ¼ y þ ay þ y ðag ðþbgðþþ ¼ y : It is also possible to give a constructive proof, analogous to that of Theorem 3.1. Indeed, by proceeding as in the proof of this theorem and using (3.5), we find that y solves the problem (3.19) if and only if y is given by yðxþ ¼ gðx sþ f ðsþds þ y 2 y gðxþþy e 2x : This formula agrees with (3.2) in view of the equality e 2x ¼ g ðxþ 1 gðxþ, which follows from (3.9) by interchanging 1 and 2 (to see that formula (3.9) is symmetric in 1 $ 2, just make the change of variables x t ¼ s in the integral with respect to t). g By imposing the initial conditions at an arbitrary point of the interval I, we get the following result. Theorem 3.3: Let f 2 C (I ), x 2 I, y, y 2 R. The solution of the initial value problem y þ ay þ by ¼ f ðxþ, yðx Þ¼y, y ðx Þ¼y ð3:22þ is unique, it is defined on the whole of I, and is given by yðxþ ¼ gðx tþ f ðtþdt þ y þ ay gðx x Þþy g ðx x Þ ðx 2 I Þ: ð3:23þ x In particular (taking f ¼ ), the solution of the homogeneous problem y þ ay þ by ¼, yðx Þ¼y, y ðx Þ¼y, with x 2 R arbitrary, is unique, of class C 1 on the whole of R, and is given by y h ðxþ ¼ y þ ay gðx x Þþy g ðx x Þ ðx 2 RÞ: ð3:24þ

13 58 R. Camporesi Proof: Let y be a solution of the problem (3.22), and let x denote the translation map defined by x ðxþ ¼x þ x. Set ~yðxþ ¼y x ðxþ ¼yðx þ x Þ: Since the differential operator L has constant coefficients, it is invariant under translations, that is, It follows that Lð y x Þ¼ðLyÞ x : ðl ~yþðxþ ¼ðLyÞðx þ x Þ¼f ðx þ x Þ: Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 Since, moreover ~yðþ ¼yðx Þ¼y, ~y ðþ ¼y ðx Þ¼y, we see that y solves the problem (3.22) in the interval I if and only if ~y solves the initial value problem ( ~y þ a ~y þ b ~y ¼ fðxþ, ~ ~yðþ ¼y, ~y ðþ ¼y in the translated interval I x ¼ {t x : t 2 I} and with the translated forcing term ~fðxþ ¼f ðx þ x Þ. By Theorem 3.2 we have that ~y is unique and is given by ~yðxþ ¼ gðx sþ fðsþds ~ þ y þ ay gðxþþy g ðxþ: Therefore y is also unique and is given by x yðxþ ¼ ~yðx x Þ¼ gðx x sþ f ðs þ x Þds þ y þ ay gðx x Þþy g ðx x Þ: Formula (3.23) follows immediately from this by making the change of variable s þ x ¼ t in the integral with respect to s. g We observe that the solution in (3.23) can be written as y ¼ y p þ y h, where the function y p ðxþ ¼ R x x gðx tþ f ðtþdt solves (3.1) with the initial conditions y p ðx Þ¼ y p ðx Þ¼, whereas the function y h, given by (3.24), solves the homogeneous equation with the same initial conditions as y. We can describe this fact by saying that the non-homogeneity in the differential equation (3.1) can be dealt with separately from the initial conditions. Corollary 3.4: The set V of real solutions of the homogeneous equation (3.2) is a vector space of dimension 2 over R, and the two functions g, g form a basis of V. Proof: Let y be a real solution of (3.2). Let x be any point at which y is defined, and let y ¼ y(x ), y ¼ y ðx Þ. Define y h by formula (3.24). Then y and y h both satisfy (3.2) with the same initial conditions, whence y ¼ y h. In particular, y 2 C 1 (R). Repeating this with x ¼, we conclude by (3.21) that every element of the vector space V can be written as a linear combination of g and g in a unique way (the coefficients in this combination are uniquely determined by the initial data

14 International Journal of Mathematical Education in Science and Technology 59 at the point x ¼ ). In particular, g and g are linearly independent and form a basis of V. g Another basis of V can be obtained from the following result. Theorem 3.5: Every complex solution of the homogeneous equation Ly ¼ can be written in the form y ¼ c 1 y 1 þ c 2 y 2 (c 1, c 2 2 C), where y 1 and y 2 are given by ( ðy 1 ðxþ, y 2 ðxþþ ¼ e1x, e 2x if 1 6¼ 2, e 1x, xe 1x if 1 ¼ 2 : Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 Conversely, any function of this form is a solution of Ly ¼. Proof: This can be shown directly using (3.4) and arguing as in the proof of Theorem 3.1. Alternatively, we observe that any complex solution of Ly ¼ can be written as a complex linear combination of g and g (indeed if y ¼ Re y þ iim y solves Ly ¼, then L(Re y) ¼ ¼ L(Im y), since L has real coefficients; by Corollary 3.4, Re y and Im y are real linear combinations of g and g,soyis a complex linear combination of g and g ). Our claim then follows immediately from formulas (3.13) and (3.14). The last statement is easily verified. g The two functions y 1, y 2 (like g, g ) form then a basis of the complex vector space V C of complex solutions of Ly ¼. If 1 and 2 are real (,D ), then y 1 and y 2 are real functions and form a basis of V as well. If instead 1,2 ¼ i with 6¼ (,D5), then y 1,2 ðxþ ¼e ðiþx ¼ e x ðcos x i sin xþ, and a basis of V is given by the functions Re y 1 ðxþ ¼e x cos x, Imy 1 ðxþ ¼e x sin x: ð3:25þ Indeed if c 1 ¼ c þ id and c 2 ¼ c þ id, with c, d, c, d 2 R, then c 1 e 1x þ c 2 e 2x ¼ðcþidÞe ðþiþx þðc þ id Þe ðiþx ¼ e x ðc þ c Þcosx þðd dþsinx þ i½ðd þ d Þcosx þðcc ÞsinxŠ : It follows that the function y ¼ c 1 y 1 þ c 2 y 2 is real-valued if and only if c ¼ c and d ¼d, i.e. iff c 1 ¼ c 2. In this case y is a real linear combination of the functions in (3.25). Example 1: Solve the initial value problem 8 < y 2y þ y ¼ ex x þ 2, : yðþ ¼, y ðþ ¼: Solution: The characteristic polynomial is p() ¼ 2 2 þ 1 ¼ ( 1) 2, so that 1 ¼ 2 ¼ 1, and the impulsive response is gðxþ ¼xe x :

15 51 R. Camporesi The forcing term f ðxþ ¼ ex xþ2 is continuous for x 4 2 and for x 5 2. Since the initial conditions are posed at x ¼, we can work in the interval I ¼ (2, þ1). By Theorem 3.1 we get the following solution of the proposed problem in I: Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 Example 2: yðxþ ¼ ðx tþe xt x t dt ¼ ex t þ 2 t þ 2 dt ¼ e x x þ 2 t þ 2 1 dt ¼ e x ½ðx þ 2Þ logðt þ 2ÞtŠ x ¼ e x ½ðx þ 2Þ logðx þ 2Þxðxþ2Þlog 2Š ¼ e x ðx þ 2Þ log x þ 2 xe x : 2 e t Solve the initial value problem 8 < y þ y ¼ 1 cos x, : yðþ ¼, y ðþ ¼: Solution: We have that p() ¼ 2 þ 1, whence 1 ¼ 2 ¼ i and the impulsive response is gðxþ ¼sin x: The initial data are given at x ¼ and we can work in the interval I ¼ (/2, /2), where the forcing term f ðxþ ¼ 1 cos x is continuous. By formula (3.8) we get yðxþ ¼ ¼ ¼ sin x sinðx tþ 1 cos t dt ðsin x cos t cos x sin tþ dt cos x ¼ x sin x þ cos x logðcos xþ: 1 cos t dt sin t cos t dt Example 3: Solve the initial value problem 8 < y y ¼ 1 ch x, : yðþ ¼, y ðþ ¼: Solution: We have that p() ¼ 2 ¼ ( 1), thus 1 ¼ 1, 2 ¼, and the impulsive response is gðxþ ¼e x 1: The forcing term f ðxþ ¼ 1 is continuous on the whole of R. Formula (3.8) gives chx Z yðxþ ¼ e xt 1 x e t 1 1 dt ¼ ex dt ch t ch t ch t dt:

16 International Journal of Mathematical Education in Science and Technology 511 The two integrals are easily computed: Z Z Z 1 ch t dt ¼ 2 1 e t dt ¼ 2 þ et ¼ 2 arctan e t þ C, e t e 2t þ 1 dt Z e t Z Z ch t dt ¼ þ e 2t e 2t þ 1 dt ¼ 2 e 2t 1 þ e 2t dt ¼ 2t log 1 þ e 2t þ C : We finally get yðxþ ¼e x 2t log 1 þ e 2t x x 2 arctan et ¼ e x 2x log 1 þ e 2x þ e x log 2 2 arctanðe x Þþ 2 : Downloaded By: [Camporesi, Roberto] At: 13:48 19 May Discussion Remark 1 (The convolution): To better understand the structure of the particular solution (2.4) (3.8), let us recall that if h 1 and h 2 are suitable functions (e.g. if h 1 and h 2 are piecewise continuous on R with h 1 bounded and h 2 absolutely integrable over R, orifh 1 and h 2 are two signals, that is, they are piecewise continuous on R and vanish for x 5 ), then their convolution h 1 * h 2 is defined as h 1 h 2 ðxþ ¼ Z þ1 1 h 1 ðx tþh 2 ðtþdt: The change of variables x t ¼ s shows that the convolution is commutative h 1 h 2 ðxþ ¼h 2 h 1 ðxþ ¼ Z þ1 1 h 1 ðsþh 2 ðx sþds: Moreover, one can show that the convolution is associative: (h 1 * h 2 ) * h 3 ¼ h 1 * (h 2 * h 3 ). If h 1 and h 2 are two signals, it is easy to verify that h 1 * h 2 is also a signal and that for any x 2 R one has h 1 h 2 ðxþ ¼ h 1 ðx tþh 2 ðtþdt: In a similar way, if h 1 and h 2 vanish for x 4, the same holds for h 1 * h 2 and we have that h 1 h 2 ðxþ ¼ Z x h 1 ðx tþh 2 ðtþdt: If denotes the Heaviside step function, given by 1 if x, ðxþ ¼ if x 5,

17 512 R. Camporesi Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 and if we let if x 4, ~ðxþ ¼ðxÞ ¼ 1 if x, then for any two functions g and f we have g f ðxþ if x, gðx tþ f ðtþdt ¼ ð4:1þ ~g ~f ðxþ if x. In other words, the integral R x gðx tþ f ðtþdt is the convolution of g and f for x, it is the opposite of the convolution of ~g and ~f for x. Furthermore, if we denote g in (3.9) by g 1 2, we can rewrite (3.9) in the suggestive form g 1 2 ðxþ ¼ g 2 ðx tþ g 1 ðtþdt ¼ g 2 g 1 ðxþ if x, ð4:2þ ~g 2 ~g 1 ðxþ if x : This formula relates the (complex) first-order impulsive responses to the impulsive response of order 2, and admits a generalization to linear equations of order n. Finally, we observe that the proof of Theorem 3.1 can be simplified, at least at a formal level, by rewriting (3.11) in convolution form (e.g. for x 4 ) and then using the associativity of this yðxþ ¼g 2 ðg 1 f ÞðxÞ ¼ðg 2 g 1 ÞfðxÞ: ð4:3þ This is just the equality (3.11) ¼ (3.12) for x 4. The interchange of the order of integration in the double integral considered above is then equivalent to the associative property of convolution for signals. Now (4.3) and (4.2) imply (3.8) for x 4. Remark 2 (The approach by distributions): The origin of formulas (2.4) (4.1) (or (3.8) (4.1)) is best appreciated in the context of distribution theory. Given a linear constant-coefficient differential equation, written in the symbolic form Ly ¼ f (x), we can study the equation LT ¼ S in a suitable convolution algebra, for example, the algebra D þ of distributions with support in [, þ1), which generalizes the algebra of signals. One can show that the elementary solution in D þ, that is, the solution of LT ¼, or equivalently, the inverse of L ( the Dirac distribution), is given precisely by g, where g is the impulsive response of the differential operator L. Ify p solves Ly ¼ f (x) with trivial initial conditions at x ¼, then L(y p ) ¼ f, whence y p ¼ g * f. This is just (2.4) for x 4 and L ¼ d dx þ a. This approach requires, however, a basic knowledge of distribution theory (see [2], Chapters II and III for a nice presentation). Remark 3 (Existence, uniqueness and extendability of the solutions): It is well known that a linear initial value problem (with constant or variable coefficients) has a unique solution defined on the whole of the interval I in which the coefficients and the forcing term are continuous (on the whole of R in the homogeneous constantcoefficient case); see, e.g. [4], Theorems 1 and 3, pp Theorems 3.1, 3.2 and 3.3 give an elementary proof of this fact (together with an explicit formula for the solutions) for a linear second-order problem with constant coefficients. Remark 4 (Complex coefficients): Theorems 3.1, 3.2, 3.3 and 3.5 remain valid if L is a linear second-order differential operator with constant complex coefficients, i.e. if

18 International Journal of Mathematical Education in Science and Technology 513 a, b 2 C in (3.3). In this case 1 and 2 in (3.4) are arbitrary complex numbers. The impulsive response g of L is defined again by (3.9) (4.2) (or explicitly by formulas (3.13) (3.14)), and solves (3.15). The solution of the initial value problem (3.22), with a, b, y, y 2 C and f : I R! C continuous in I 3 x, is given by (3.23). Finally, the set V C of complex solutions of (3.2) is a complex vector space of dimension 2, with a basis given by {g, g }. Remark 5 (Linear equations of order n): The results obtained in Section 3 for second-order equations can be generalized to linear equations of order n, by working directly with complex coefficients and using induction on n. We give only a brief outline here. Consider the linear constant-coefficient non-homogeneous differential equation of order n Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 y ðnþ þ a 1 y ðn1þ þ a 2 y ðn2þ þþa n1 y þ a n y ¼ f ðxþ, ð4:4þ where y ðkþ ¼ dk y ky, ¼ d dx k dx a1, a 2,..., a n 2 C and the forcing term f : I R! C is a continuous complex-valued function in the interval I. When f ¼ we obtain the associated homogeneous equation y ðnþ þ a 1 y ðn1þ þ a 2 y ðn2þ þþa n1 y þ a n y ¼ : ð4:5þ We can write (4.4) and (4.5) in the form Ly ¼ f (x) and Ly ¼, where L is the linear differential operator of order n with constant coefficients given by L ¼ d n d n1 d þ a 1 þþa n1 dx dx dx þ a n: As in the case of n ¼ 2, one easily proves that the complex exponential e x, 2 C, isa solution of (4.5) if and only if is a root of the characteristic polynomial pðþ ¼ n þ a 1 n1 þþa n1 þ a n : Let 1, 2,..., n 2 C be the roots of p(), not necessarily all distinct, each counted with its multiplicity. The polynomial p() factors as pðþ ¼ð 1 Þð 2 Þð n Þ: The differential operator L can be factored in a similar way as L ¼ d d dx 1 dx d 2 dx n, where the order of composition of the factors is unimportant as they all commute. We define the impulsive response of L, g ¼ g 1... n, recursively by the following formulas: for n ¼ 1 we set g 1 ðxþ ¼e 1x, for n 2 we set g 1... n ðxþ ¼ g n ðx tþ g 1... n1 ðtþdt ðx 2 RÞ: ð4:6þ The impulsive response g solves the homogeneous equation (4.5) with the initial conditions yðþ ¼y ðþ ¼¼y ðn2þ ðþ ¼, y ðn1þ ðþ ¼1:

19 514 R. Camporesi It can be explicitly computed by iterating the recursive formula (4.6). It is then easy to prove by induction on n that if 2 I and if g is the impulsive response of L, then the general solution of (4.4) in the interval I can be written in the form (1.3), where y p is given by (1.4) and solves (4.4) with trivial initial conditions at the point x ¼ (i.e. y ðkþ p ðþ ¼ for k ¼, 1,..., n 1), whereas the function y h ðxþ ¼ Xn1 c k g ðkþ ðxþ ð4:7þ k¼ gives the general solution of (4.5) as the coefficients c k vary in C. Thus the n functions g, g, g,..., g (n1) are linearly independent solutions of this equation and form a basis of the vector space of its solutions. If L has real coefficients then g is real, and the general real solution of Ly ¼ is given by (4.7) with c k 2 R. Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 211 References [1] G. Doetsch, Introduction to the Theory and Application of the Laplace Transformation, Springer-Verlag, New York, Heidelberg, [2] L. Schwartz, Methodes Mathematiques Pour les Sciences Physiques, Enseignement des Sciences, Hermann, Paris, [3] E.A. Coddington and N. Levinson, Theory of Ordinary Differential Equations, McGraw- Hill Book Company, Inc., New York Toronto London, [4] E.A. Coddington, An Introduction to Ordinary Differential Equations, Prentice-Hall Mathematics Series, Prentice-Hall, Inc., Englewood Cliffs, NJ, [5] W.A. Robin, Solution of differential equations by quadratures: Cauchy integral method, Int. J. Math. Edu. Sci. Tech. 25 (1994), pp [6] D.H. Parsons, Linear ordinary differential equations with constant coefficients: identification of Boole s integral with that of Cauchy, Proc. Edinb. Math. Soc. 12(2) (196), pp [7] R.R. Burnside, Convolution and linear differential equations, Int. J. Math. Edu. Sci. Tech. 5 (1974), pp