Roberto Camporesi a a Dipartimento di Scienze Matematiche, Politecnico di Torino,

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1 This article was downloaded by: [Roberto Camporesi] On: 9 June 25, At: :8 Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: Registered office: Mortimer House, 37-4 Mortimer Street, London WT 3JH, UK Click for updates International Journal of Mathematical Education in Science and Technology Publication details, including instructions for authors and subscription information: A fresh look at linear ordinary differential equations with constant coefficients. Revisiting the impulsive response method using factorization Roberto Camporesi a a Dipartimento di Scienze Matematiche, Politecnico di Torino, Corso Duca degli Abruzzi 24, 29 Torino, Italy Published online: 9 Jun 25. To cite this article: Roberto Camporesi (25): A fresh look at linear ordinary differential equations with constant coefficients. Revisiting the impulsive response method using factorization, International Journal of Mathematical Education in Science and Technology, DOI:.8/2739X To link to this article: PLEASE SCROLL DOWN FOR ARTICLE Taylor & Francis makes every effort to ensure the accuracy of all the information (the Content ) contained in the publications on our platform. However, Taylor & Francis, our agents, and our licensors make no representations or warranties whatsoever as to the accuracy, completeness, or suitability for any purpose of the Content. Any opinions and views expressed in this publication are the opinions and views of the authors, and are not the views of or endorsed by Taylor & Francis. The accuracy of the Content should not be relied upon and should be independently verified with primary sources of information. Taylor and Francis shall not be liable for any losses, actions, claims, proceedings, demands, costs, expenses, damages, and other liabilities whatsoever or howsoever caused arising directly or indirectly in connection with, in relation to or arising out of the use of the Content. This article may be used for research, teaching, and private study purposes. Any substantial or systematic reproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in any form to anyone is expressly forbidden. Terms &

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3 International Journal of Mathematical Education in Science and Technology, 25 A fresh look at linear ordinary differential equations with constant coefficients. Revisiting the impulsive response method using factorization Roberto Camporesi Dipartimento di Scienze Matematiche, Politecnico di Torino, Corso Duca degli Abruzzi 24, 29 Torino, Italy (Received 3 September 24) Downloaded by [Roberto Camporesi] at :8 9 June 25 We present an approach to the impulsive response method for solving linear constantcoefficient ordinary differential equations of any order 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 variation of parameters. The approach presented here can be used in a first course on differential equations for science and engineering majors. Keywords: ordinary differential equations; Green functions; instructional exposition. Introduction In view of their many applications in various fields of science and engineering, linear constant-coefficient differential equations constitute an important chapter in the theory and application of ordinary differential equations. 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 [] 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). 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 be implemented also in the case of constant coefficients ([4], chapter 2). However, camporesi@polito.it C 25 Taylor & Francis

4 2 R. Camporesi 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, for example, [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 write the nth order equation in the form y (n) + a y (n ) + a 2 y (n 2) + +a n y + a n y = f (x), (.) Downloaded by [Roberto Camporesi] at :8 9 June 25 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, a 2,..., a n are complex constants and the forcing term f is a complex-valued continuous function in an interval I. When f, the equation is called non-homogeneous. When f =, we get the associated homogeneous equation y (n) + a y (n ) + a 2 y (n 2) + +a n y + a n y =. (.2) The basic tool of our investigation is the so-called impulsive response. This is the function defined as follows. Let p(λ) = λ n + a λ n + + a n λ + a n (λ C) be the characteristic polynomial, and let λ,...,λ n C be the roots of p(λ) (not necessarily distinct, each counted with its multiplicity). We define the impulsive response g = g λ λ n recursively by the following formulas: for n =, we set g λ (x) = e λ x,forn 2weset g λ λ n (x) = e λ nx e λ nt g λ λ n (t) dt (x R). (.3) It turns out that g solves the homogeneous Equation (.2) with the initial conditions y() = y () = =y (n 2) () =, y (n ) () =. 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 g denotes the impulsive response of order n and if I, we shall see that the general solution of (.) in the interval I can be written as y = y p + y h, where the function y p is given by the convolution integral y p (x) = g(x t)f (t) dt, (.4) and solves (.) with trivial initial conditions at the point x = (i.e., y (k) p () = fork =,,..., n ), whereas the function n y h (x) = c k g (k) (x) (.5) k=

5 International Journal of Mathematical Education in Science and Technology 3 Downloaded by [Roberto Camporesi] at :8 9 June 25 gives the general solution of the associated homogeneous Equation (.2) as the coefficients c k vary in C. In other words, the n functions g, g, g,..., g (n ) are linearly independent solutions of this equation and form a basis of the vector space of its solutions. For n 2, we will not assume, a priori, existence, uniqueness, and extendability of the solutions of a linear initial value problem (homogeneous or not). We shall rather prove these facts directly, in the case of constant coefficients, by obtaining explicit formulas for the solutions. The proof of (.4) that we give is by induction on n and is based on the factorization of the differential operator acting on y in (.) 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 =, we get that the unique solution of the homogeneous initial value problem with all vanishing initial data is the zero function. This implies immediately the uniqueness for the initial value problem (homogeneous or not) with arbitrary initial data. We remark that in the usual approach to linear equations with constant coefficients, the proof of uniqueness relies on an estimate for the rate of growth of any solution y of the homogeneous equation, together with its derivatives y, y,..., y (n ), in terms of the coefficients a, a 2,..., a n in (.2). (See, e.g., [4], Theorems 3 and 4, chapter 2.) As regards (.5), one can give a similar proof by induction on n using factorization. We shall rather prove (.5) by computing the linear relationship between the coefficients c k and the initial data b k = y (k) () = y (k) h () for any given n. Ifx is any point of I, we can just replace with x in (.4), and g (k) (x) with g (k) (x x ) in (.5). The function y p satisfies then y p (k) (x ) = for k n, and the relationship between c k and b k = y (k) (x ) = y (k) h (x ) remains the same as before. We will then look for explicit formulas for the impulsive response g. Forn =, g is an exponential, for n = 2, 3 it is easily computed. For generic n, one can use the recursive formula (.3) to compute g at any prescribed order. A simple formula is obtained if the roots of p(λ) are all equal or all different. In the general case, one can prove by induction on k that if λ,..., λ k are the distinct roots of p(λ), of multiplicities m,..., m k, then there exist polynomials G,..., G k, of degrees m,..., m k, such that g(x) = k G j (x)e λ j x. (.6) j= An explicit formula for the polynomials G j will be given in the Appendix. We also give the formula for the polynomials G j based on the partial fraction expansion of /p(λ). This formula can be proved by induction, but it is most easily proved using the Laplace transform or distribution theory. Using (.6) in (.5) gives the well-known formula for the general solution of (.2) in terms of the complex exponentials e λ j x. The case of real coefficients follows easily from this result. Using (.6) in (.4) gives a simple proof of the method of undetermined coefficients for the case in which f (x) = P (x)e λ x, where P is a complex polynomial and λ C. This paper is an outgrowth of [5], where the method of factorization was illustrated for second-order equations. Here, we treat equations of arbitrary order n. Since the methodology used here is the same as that of [5], a reading of [5] enables the understanding of the current

6 4 R. Camporesi paper. We also refer to [6] and the references therein, for the approach by factorization to linear constant-coefficient differential equations. 2. Linear equations of order n Consider the linear constant-coefficient non-homogeneous differential equation of order n y (n) + a y (n ) + a 2 y (n 2) + +a n y + a n y = f (x), (2.) where y (k) = dk y dx k, a, a 2,..., a n are complex constants, and the forcing term f is a complexvalued continuous function in the interval I, i.e., f C (I). When f =, we obtain the associated homogeneous equation y (n) + a y (n ) + a 2 y (n 2) + +a n y + a n y =. (2.2) Downloaded by [Roberto Camporesi] at :8 9 June 25 We will write (2.) and (2.2) in operator form as Ly = f(x) and Ly =, where L is the linear differential operator of order n with constant coefficients defined by Ly = y (n) + a y (n ) + a 2 y (n 2) + +a n y + a n y for any function y at least n times differentiable. Denoting by d the differentiation operator, dx we have L = ( ) d n ( ) d n d + a + +a n dx dx dx + a n. (2.3) L defines a map C n (R) C (R) that to each complex-valued function y at least n-times differentiable over R with continuous nth derivative associates the continuous function Ly. The fundamental property of L is its linearity; that is, L(c y + c 2 y 2 ) = c Ly + c 2 Ly 2, (2.4) c,c 2 C, y,y 2 C n (R). This formula implies some important facts. First, if y and y 2 are any two solutions of the homogeneous equation on R, then any linear combination c y + c 2 y 2 is also a solution of this equation on R. In other words, the set V ={y C n (R) : Ly = } is a complex vector space. We will prove that dim V = n, and will see how to get a basis of V. Note that if y V, then y C (R), as follows from (2.2) by isolating y (n) and differentiating any number of times. Second, if y and y 2 are two solutions of (2.) on a given interval I I, then their difference y y 2 solves (2.2). It follows that if we know a particular solution y p of the non-homogeneous equation on an interval I, then any other solution of (2.) on I is given by y p + y h, where y h is a solution of the associated homogeneous equation. We will prove that the solutions of (2.) are defined on the whole of the interval I in which f is continuous (and are of course of class C n there). The fact that L has constant coefficients allows one to find explicit formulas for the solutions of (2.) (2.2). For the moment, we only observe (by the formula d dx eλx = λe λx,

7 International Journal of Mathematical Education in Science and Technology 5 λ C) that the complex exponential e λx is a solution of (2.2) if and only if λ is a root of the characteristic polynomial p(λ) = λ n + a λ n + +a n λ + a n. (2.5) Let λ,λ 2,...,λ n C be the roots of p(λ), not necessarily all distinct, each counted with its multiplicity. The polynomial p(λ) factors as p(λ) = (λ λ )(λ λ 2 ) (λ λ n ). The differential operator L can be factored in a similar way as Downloaded by [Roberto Camporesi] at :8 9 June 25 L = ( )( ) ( ) d d d dx λ dx λ 2 dx λ n, (2.6) where the order of composition of the factors is unimportant as they all commute. For example, for n = 2wehaveLy = y + a y + a 2 y, and ( )( ) ( ) d d d dx λ dx λ 2 y = dx λ (y λ 2 y) = y (λ + λ 2 )y + λ λ 2 y, which coincides with Ly since λ + λ 2 = a and λ λ 2 = a 2. Moreover, it is clear that ( )( ) ( )( ) d d d d dx λ dx λ 2 = dx λ 2 dx λ. The idea is now to use (2.6) to reduce the problem to first-order differential equations. The following result gives a particular solution of (2.). Theorem 2.: Let f C (I), and suppose that I. Let λ, λ 2,..., λ n be n complex numbers (not necessarily all distinct), and let L be the differential operator (2.6). Then, the initial value problem { Ly = f (x) y() = y () = =y (n ) () = (2.7) has a unique solution, defined on the whole of I, and given by the formula y(x) = g(x t)f (t) dt (x I), (2.8) where g = g λ λ n is the function defined recursively as follows: for n =, weset g λ (x) = e λx (λ C),forn 2 we set g λ λ n (x) = g λn (x t) g λ λ n (t) dt (x R). (2.9)

8 6 R. Camporesi The function g λ λ n is of class C on the whole of R. In particular, if we take f =,weget that the unique solution of the homogeneous problem { Ly = y() = y () = =y (n ) () = (2.) is the zero function y =. Proof: We proceed by induction on n. First, we prove that g λ λ n C (R). Indeed, this holds for n =. Suppose it holds for n. Then, it holds also for n, since by (2.9) g λ λ n (x) = e λ nx e λ nt g λ λ n (t) dt, Downloaded by [Roberto Camporesi] at :8 9 June 25 so g λ λ n is the product of two functions of class C on R. (The integral function of a C function is C.) We now prove (2.8). The result holds for n =, in view of the formula y(x) = for the unique solution to the first-order initial value problem e λ(x t) f (t) dt (2.) { y λy = f (x) y() =. Assuming the result holds for n, let us prove it for n. Consider then the problem (2.7) with L given by (2.6) { ( d dx λ )( d dx λ ) 2 ( d dx λ n) y = f (x) y() = y () = =y (n ) () =. Letting h = ( d dx λ n) y, we see that y solves the problem (2.7) if and only if h solves { ( d dx λ )( d dx λ ) 2 ( d dx λ n ) h = f (x) h() = h () = =h (n 2) () =, and y solves { y λ n y = h(x) y() =. (The initial conditions for h = y λ n y follow by computing h, h,..., h (n 2) and setting x =.) By the inductive hypothesis and by (2.), we have h(x) = g λ λ n (x t)f (t) dt,

9 International Journal of Mathematical Education in Science and Technology 7 y(x) = e λ n(x t) h(t) dt. Substituting h from the first formula into the second, we obtain y(x) as a repeated integral ( x I): y(x) = ( t ) g λn (x t) g λ λ n (t s)f (s) ds dt. (2.2) Downloaded by [Roberto Camporesi] at :8 9 June 25 To fix ideas, suppose x >. 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) asa double integral y(x) = g λn (x t)g λ λ n (t s) f (s) ds 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), (x, x). In (2.2), 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 F (s, t) = g λn (x t) g λ λ n (t s) f (s) 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, seethe picture below. t s t x s t x x s We thus obtain y(x) = ( s ) g λn (x t)g λ λ n (t s) dt f (s) ds.

10 8 R. Camporesi By substituting t with t + s in the integral with respect to t, we obtain ( s ) y(x) = g λn (x s t)g λ λ n (t) dt f (s) ds = g λ λ n (x s)f (s) ds, which is (2.8). For x <, we can reason in a similar way and we get the same result. By induction, one can verify that the function g λ λ n solves the following homogeneous initial value problem: Downloaded by [Roberto Camporesi] at :8 9 June 25 { Ly = y() = y () = =y (n 2) () =, y (n ) () =. (2.3) Conversely, if y solves (2.3), then y is given by (2.9). In other words, the solution of the initial value problem (2.3) is unique and is calculable for n 2 by the recursive formula (2.9). This is proved by induction, reasoning as in the proof of Theorem 2.. For example, take n = 2 and suppose y solves the problem { ( d dx λ )( d dx λ 2) y = y() =, y () =. Letting h = ( d dx λ 2) y, the function h solves { ( d dx λ ) h = h() =, whence h = g λ. Therefore, y solves { y λ 2 y = g λ (x) y() =, and by (2.), we get y(x) = g λ2 (x t) g λ (t) dt = g λ λ 2 (x). The function g = g λ λ n is called the impulsive response of the differential operator L.The name comes from the initial conditions in (2.3) for the case n = 2, namely y() =, y () =. We will see later how to compute it explicitly using (2.9). Note that as a function of λ,..., λ n, g λ λ n is symmetric in the interchange λ i λ j, i, j =,..., n. This follows from the fact that g λ λ n solves the problem (2.3) with L symmetric in λ i λ j, i, j. It is interesting to verify directly that the function y given by (2.8) solves the nonhomogeneous problem (2.7). Indeed, using repeatedly the formula d x F F (x,t) dt = F (x,x) + (x,t) dt, dx x

11 International Journal of Mathematical Education in Science and Technology 9 we get from (2.8) y (x) = g()f (x) + y (x) = g ()f (x) +. y (n ) (x) = g (n 2) ()f (x) + y (n) (x) = g (n ) ()f (x) + g (x t)f (t) dt = g (x t)f (t) dt = g (n ) (x t)f (t) dt = g (x t)f (t) dt g (x t)f (t) dt g (n) (x t)f (t) dt = f (x) + g (n ) (x t)f (t) dt g (n) (x t)f (t) dt, Downloaded by [Roberto Camporesi] at :8 9 June 25 where we used (2.3). It follows that (Ly)(x) = f (x) + (Lg)(x t)f (t) dt = f (x), x I, being Lg =. The initial conditions y (k) () = for k n are immediately verified from these formulas. For the initial value problem with arbitrary (complex) initial data at x =, we have the following result. Theorem 2.2: Let f C (I), I, and let b,b,...,b n be n complex numbers. Let L be the differential operator in (2.3), (2.6), and let g be the impulsive response of L. Then, the initial value problem { Ly = f (x) y() = b, y () = b,..., y (n ) (2.4) () = b n has a unique solution, defined on the whole of I, and given for any x Iby y(x) = g(x t)f (t) dt + c g(x) + c g (x) + +c n g (n ) (x), (2.5) where c = b n + a b n 2 + +a n 2 b + a n b c = b n 2 + a b n 3 + +a n 3 b + a n 2 b. c n 3 = b 2 + a b + a 2 b c n 2 = b + a b c n = b. (2.6) In particular (taking f = ), the solution of the homogeneous problem { Ly = y() = b, y () = b,..., y (n ) (2.7) () = b n

12 R. Camporesi is unique, of class C on the whole of R, and is given by n y h (x) = c k g (k) (x) (x R). (2.8) k= Proof: The uniqueness follows from the fact that if y, y 2 both solve the problem (2.4), then their difference ỹ = y y 2 solves the problem (2.), whence ỹ = by Theorem 2.. To show that y(x) given by (2.5) satisfies Ly(x) = f(x), just observe that any derivative g (k) of g satisfies the homogeneous equation, since Downloaded by [Roberto Camporesi] at :8 9 June 25 ( ) d k ( ) d k Lg (k) = L g = Lg =. dx dx By Theorem 2. and the linearity of L, itfollowsthat (Ly)(x) = f (x) + L ( n ) n c k g (k) (x) = f (x) + c k (Lg (k) )(x) = f (x). k= There remain to prove the relations (2.6) between the coefficients c k and b k = y (k) () ( k n ). By computing y, y,..., y (n ) from (2.5) and imposing the initial conditions in (2.4), we obtain the linear system c n = b k= c n 2 + c n g (n) () = b c n 3 + c n 2 g (n) () + c n g (n+) () = b 2 c n 4 + c n 3 g (n) () + c n 2 g (n+) () + c n g (n+2) () = b 3. c + c g (n) () + c 2 g (n+) () + +c n g (2n 2) () = b n, (2.9) whereweusedg (k) () = for k n 2, g (n ) () =. The derivatives g (k) () with k n can be computed from the differential equation Lg = by isolating g (n) (x), differentiating and letting x =. One gets g (n) () = a g (n ) () = a g (n+) () = a g (n) () a 2 g (n ) () = a 2 a 2 g (n+2) () = a g (n+) () a 2 g (n) () a 3 g (n ) () = a 3 + 2a a 2 a 3, and so on. These relations, substituted in the system (2.9), allow one to solve it recursively and we get exactly the relations (2.6). Remark : By imposing the initial conditions at an arbitrary point x of the interval I,we get that the unique solution to the problem { Ly = f (x) y(x ) = b, y (x ) = b,..., y (n ) (x ) = b n

13 International Journal of Mathematical Education in Science and Technology is given, x I, by y(x) = n g(x t)f (t) dt + c k g (k) (x x ), (2.2) x where the coefficients c k are given by (2.6). The proof of this is analogous to that of [5], Theorem 3.3. Corollary 2.3: The set V of solutions of the homogeneous equation Ly = on R is a complex vector space of dimension n, with a basis given by {g, g,g,...,g (n ) }.IfL has real coefficients (i.e., a,...,a n in (2.3) are all real numbers), then the set V R of real solutions of Ly = on R is a real vector space of dimension n, with a basis given by {g, g, g,...,g (n ) }. k= Downloaded by [Roberto Camporesi] at :8 9 June 25 Proof: Let y be any solution of Ly = onr.letb k = y (k) () ( k n ), and define y h by formula (2.8), with the c k given by (2.6). Then, y and y h both satisfy (2.7), whence y = y h by Theorem 2.2. We conclude by (2.8) that every solution of Ly = onr is in C (R) and can be written as a linear combination of g, g, g,..., g (n ) in a unique way. (The coefficients c k in this combination are uniquely determined by the initial data b k at x = through (2.6).) In particular, the n functions g, g, g,..., g (n ) are linearly independent in V and form a basis of V.IfL has real coefficients, then the solutions of Ly = with real initial data at x = (i.e., b k = y (k) () R for k n ) are real-valued. Indeed by writing y = Re y + iim y, one gets that Im y solves the problem (2.), whence Im y =. In particular, the impulsive response g is real-valued. The set V R of real solutions of Ly = onr is then a real vector space of dimension n, with a basis given by {g, g, g,..., g (n ) }. 3. Explicit formulas for the impulsive response We now consider the problem of explicitly computing g for an equation of order n. By iterating (2.9), we obtain the following formula for the impulsive response of order n as an (n )-times repeated integral of exponential functions ( tn g λ λ n (x) = e λ n(x t n ) e λ n (t n t n 2 ) ( t3 ( t2 ) ) ) e λ 3(t 3 t 2 ) e λ 2(t 2 t ) e λ t dt dt 2 dt n 2 dt n. (3.) For n = 2, we have g λ λ 2 (x) = e λ 2(x t) e λ t dt, and we obtain e λx e λ 2x if λ g λ λ 2 (x) = λ 2, λ λ 2 xe λ x if λ = λ 2.

14 2 R. Camporesi For n = 3, we have g λ λ 2 λ 3 (x) = e λ 3(x t) g λ λ 2 (t) dt, and we easily obtain the following formulas for g = g λ λ 2 λ 3. () If λ λ 2 λ 3 (all distinct roots), then g(x) = (2) If λ = λ 2 λ 3, then e λ x (λ λ 2 )(λ λ 3 ) + e λ2x (λ 2 λ )(λ 2 λ 3 ) + e λ3x (λ 3 λ )(λ 3 λ 2 ). Downloaded by [Roberto Camporesi] at :8 9 June 25 g(x) = (3) If λ = λ 2 = λ 3, then (λ λ 3 ) 2 [ e λ 3 x e λ x + (λ λ 3 ) xe λ x ]. g(x) = 2 x2 e λ x. Note that if λ R and λ 2, 3 = α ± iβ with α, β R, β, then one computes g(x) = [ e λx + ] (λ α) 2 + β 2 β (α λ ) e αx sin βx e αx cos βx. (3.2) Example : Solve the initial value problem { y + y = sin x y( π 2 ) = y ( π 2 ) = y ( π 2 ) =. Solution. The characteristic polynomial p(λ) = λ 3 + λ = λ(λ 2 + ) has the roots λ =, λ 2 = i, λ 3 = i. Using (3.2), we compute the impulsive response g(x) = cos x. The forcing term /sin x is continuous in the interval I = (, π) x = π/2. Using (2.2) with c k = k, we get the following solution of the proposed problem in the interval I: y(x) = = = π/2 π/2 cos(x t) sin t dt cos x sin t [ log tan t 2 ] x π/2 dt π/2 cos t sin t dt sin x dt π/2 cos x [log sin t] x π/2 (x π/2) sin x = log tan x cos x log sin x (x π/2) sin x. 2

15 International Journal of Mathematical Education in Science and Technology 3 For generic n, we get simple formulas for g = g λ λ n when the roots of p(λ) areall distinct or all equal. Proposition 3.: () If λ i λ j for i j (all distinct roots), then g(x) = c e λx + c 2 e λ2x + +c n e λnx, where c j = i j (λ j λ i ) = p (λ j ) ( j n). (2) If λ = λ 2 = = λ n, then Downloaded by [Roberto Camporesi] at :8 9 June 25 g(x) = (n )! xn e λ x. (3.3) Proof: This is readily proved by induction on n using (2.9). For (), it is useful to remind that the function g = g λ λ n is symmetric in the interchange λ i λ j, i, j. A method for computing g in the general case is given by the following result. Theorem 3.2: Let λ, λ 2,..., λ k be the distinct roots of the characteristic polynomial (2.5), of multiplicities m,m 2,...,m k, and let k j= (λ λ j ) m j = k j= be the partial fraction expansion of differential operator (2.3) is given by ( cj, + c j,2 λ λ j (λ λ j ) + + c ) j,m j 2 (λ λ j ) m j p(λ) (3.4) over C. Then, the impulsive response of the g(x) = G (x)e λ x + G 2 (x)e λ 2x + +G k (x)e λ kx, (3.5) where G,...,G k are the polynomials given by G j (x) = c j, + c j,2 x + c j,3 x 2 2! + +c j,m j x mj (m j )! ( j k). Proof: This theorem can be proved by induction, though it is most easily proved using the Laplace transform or distribution theory. Indeed, one can show that the Laplace transform L of the impulsive response g is precisely the reciprocal of the characteristic polynomial: (Lg)(λ) = p(λ). Decomposing /p(λ) according to (3.4) and taking the inverse Laplace transform gives the result (see [], Equation (2) p. 8). See also [2], p. 4 42, for another proof using distribution theory in the convolution algebra D +.

16 4 R. Camporesi We can now use the impulsive response to compute the general solution of the homogeneous equation Ly = in terms of the complex exponentials e λ j x. Theorem 3.3: Every solution of Ly = can be written in the form y h (x) = P (x)e λ x + P 2 (x)e λ 2x + +P k (x)e λ kx, (3.6) Downloaded by [Roberto Camporesi] at :8 9 June 25 where P j ( j k) is a complex polynomial of degree m j. Conversely, any function of this form is a solution of Ly =. Proof: Let y h be a solution of Ly =. By Corollary 2.3, y h can be written as a linear combination of g, g,..., g (n ), i.e., it is given by (2.8) for some complex numbers c, c,..., c n. Substituting (3.5) in (2.8), we get an expression of the form (3.6), where the degrees of the polynomials P j can be less than or equal to m j (depending on the coefficients c, c,..., c n ). For the converse, observe that since λ, λ 2,..., λ k are the distinct roots of p(λ), of multiplicities m, m 2,..., m k, p(λ) can be factored according to the formula p(λ) = (λ λ ) m (λ λ 2 ) m2 (λ λ k ) m k. Similarly, the differential operator L factors as L = ( ) d m ( ) d m2 ( ) d mk dx λ dx λ 2 dx λ k, (3.7) where the order of composition of the various factors is irrelevant. We now observe that if λ C and m N +, then ( ) d m dx λ y = y(x) = P (x)e λx, (3.8) where P is a complex polynomial of degree m. Indeed, given a function y and letting h(x) = y(x)e λx, we compute [( ) ] d h (x) = y (x)e λx λy(x)e λx = dx λ y (x) e λx, h (x) = ( y (x) 2λy (x) + λ 2 y(x) ) [ ( ) d 2 e λx = y] dx λ (x) e λx,. [( ) d m h (m) (x) = y] dx λ (x) e λx. It follows that ( ) d m dx λ y = h (m) = h is a polynomial of degree m, as claimed. Note that the implication from left to right also follows by substituting (3.3) in (2.8), as already seen in the first part of the proof for the case of k distinct roots. By writing

17 International Journal of Mathematical Education in Science and Technology 5 now L in the form (3.7) and using the commutativity of the various factors and (3.8), we see that any function of the form (3.6) satisfies Ly h =. Theorem 3.3 identifies another basis of the vector space V of solutions of Ly = onr, namely the well-known basis given explicitly by the n functions e λ j x, xe λ j x, x 2 e λ j x,..., x m j e λ j x ( j k). Indeed, these functions belong to V (by the converse part of the theorem), and every element of V is a linear combination of them (by (3.6)). Since V has dimension n (by Corollary 2.3), these functions must be linearly independent. When L has real coefficients, we get the well-known basis of V R given by the n functions Downloaded by [Roberto Camporesi] at :8 9 June 25 e λ j x, xe λ j x,..., x n j e λ j x ( j p), e α j x cos β j x, xe α j x cos β j x,..., x m j e α j x cos β j x ( j q), e α j x sin β j x, xe α j x sin β j x,..., x m j e α j x sin β j x ( j q), where λ j ( j p) is a real root of the characteristic polynomial p(λ) of multiplicity n j, and α j ± iβ j ( j q) is a pair of complex conjugate roots of p(λ) both of multiplicity m j, with n + + n p + 2m + + 2m q = n. Example 2: Find the general real solution of the differential equation y (4) 2y + y = ch 3 x. (3.9) Solution. The characteristic polynomial p(λ) = λ 4 2λ 2 + = (λ 2 ) 2 has the roots λ =, λ 2 =, both of multiplicity m = 2. The general solution of the homogeneous equation is then y h (x) = (ax + b) e x + (cx + d) e x (a, b, c, d R). To this, we must add any particular solution of (3.9); for example, the solution y p of the initial value problem { y (4) 2y + y = ch 3 x y() = y () = y () = y () =. From (A-3), (A-4) and (A-5), we find the impulsive response g(x) = ( x)ex + ( x)e x = 2 x ch x sh x. 2 Alternatively, use the partial fraction expansion of /p(λ) (λ 2 ) = a 2 λ + b (λ ) + c 2 λ + + d (λ + ) 2, compute a = /4, b = c = d = /4, and use Theorem 3.2 to get the same result.

18 6 R. Camporesi The forcing term ch 3 x y p (x) = 2 is continuous on R. By (2.8), we get x R (x t)ch(x t) ch 3 dt t 2 sh(x t) ch 3 dt. t Now, use the formulas ch(x t) = ch x ch t sh x sh t, sh(x t) = sh x ch t ch x sh t. The following integrals are immediate: ch 2 dt = th t + c, t sh t ch 3 t dt = 2ch 2 t + c. Integrating by parts, we compute Downloaded by [Roberto Camporesi] at :8 9 June 25 Thus, t ch 2 dt = t th t log(ch t) + c, t The final result can be written as t sh t ch 3 t y p (x) = 2 ch x log(ch x) x sh x. 4 y(x) = y h (x) + ch x log(ch x), 2 dt = t 2ch 2 t + th t + c. 2 by noting that the term x sh x is a solution of the homogeneous equation. 4 Disclosure statement No potential conflict of interest was reported by the author. References [] Doetsch G. Introduction to the theory and application of the Laplace transformation. New York (NY): Springer-Verlag; 974. [2] Schwartz L. Methodes mathematiques pour les sciences physiques [Mathematical methods for physical sciences]. Enseignement des Sciences. Paris: Hermann; 96. [3] Coddington EA, Levinson N. Theory of ordinary differential equations. New York (NY): McGraw-Hill Book Company, Inc.; 955. [4] Coddington EA. An introduction to ordinary differential equations. Prentice-Hall Mathematics Series. Englewood Cliffs (NJ): Prentice-Hall, Inc.; 96. [5] Camporesi R. Linear ordinary differential equations with constant coefficients. Revisiting the impulsive response method using factorization. Internat J Math Ed Sci Tech. 2;42: [6] Robin WA. Solution of differential equations by quadratures: Cauchy integral method. Internat J Math Ed Sci Tech. 994;25: Appendix. An explicit formula for the impulsive response In this appendix, we give an explicit formula for the impulsive response g in the general case of k distinct roots λ, λ 2,..., λ k of respective multiplicities m, m 2,..., m k, with m + m m k = n. The following result is independent of Theorem 3.2 and can be proved without using that theorem.

19 International Journal of Mathematical Education in Science and Technology 7 Theorem A.: There exist polynomials G,G 2,...,G k, of degrees m, m 2,...,m k, respectively, such that g(x) = G (x)e λ x + G 2 (x)e λ 2x + +G k (x)e λ kx. (A-) More precisely, G is the polynomial of degree m given, for k 2,by G (x) = m r r 2 r = r 2 = r 3 = r k 2 r k = ( m +m 2 2 r m 2 ( ) m r k r k! x r k )( m3 +r r 2 )( m4 +r 2 r 3 ) m 3 m 4 ( mk +r k 2 r k ) m k. (λ λ 2 ) m +m 2 r (λ λ 3 ) m 3+r r 2(λ λ 4 ) m 4+r 2 r 3 (λ λ k ) m k+r k 2 r k (A-2) Downloaded by [Roberto Camporesi] at :8 9 June 25 To get G j,j 2, just exchange λ λ j and m m j in (A-2). The proof of this result will be given elsewhere. However, the main ideas of the proof using our previous theory are indicated below for the case k = 2. Let us make explicit the formula for g given by Theorem A. in the cases k = 2, 3. (For k =, formula (A-2) is somewhat degenerate. The result g = G e λ, with G (x) = xm (m, follows from )! (3.3).) For k = 2, i.e., for two distinct roots λ, λ 2 of multiplicities m, m 2, we get from (A-) (A-2) the impulsive response where m G (x) = r= m 2 G 2 (x) = r= g(x) = G (x)e λ x + G 2 (x)e λ 2x, ( ) m r r! ( ) m 2 r r! (A-3) ( ) m + m 2 r 2 x r m 2 (λ λ 2 ), (A-4) m +m 2 r ( ) m + m 2 r 2 x r m (λ 2 λ ). (A-5) m +m 2 r Note that G, G 2 have degrees m, m 2, respectively, and that G 2 is obtained from G by interchanging λ λ 2 and m m 2,sothatg is symmetric under this interchange. The simplest way to prove this is to go back to formula (3.). Suppose we label our roots as {λ,..., λ n } = {λ,..., λ, λ 2,..., λ 2 },whereλ (resp. λ 2 ) is repeated m (resp. m 2 ) times, with m + m 2 = n. By interchanging the order of integration in some of the nested integrals in (3.), we arrive easily at the following formula: g(x) = = g λ2 λ 2 (x t)g λ λ (t) dt (m 2 )! (x t)m 2 e λ 2(x t) (m )! t m e λt dt. (A-6) At this point, we use the following result, that can be proved by elementary calculus. Lemma A.2: Let p, q N and b, c C, b c. Then, x R, we have p!q! ebx (x t) p t q e (c b)t dt = P (x,p,q,b c) e bx + P (x,q,p,c b) e cx, (A-7)

20 8 R. Camporesi where P(x, p, q, a), a C \{}, is the polynomial of degree p in x given by P (x,p,q,a) = p ( ) ( ) p r p + q r x r r! q a. p+q+ r r= (A-8) Downloaded by [Roberto Camporesi] at :8 9 June 25 Using (A-7) (A-8) in (A-6) with p = m 2, q = m gives immediately the result (A-3) (A-5). For k 3, we can reason in a similar way to prove Theorem A. by formula (3.) and Lemma A.2. For example, for k = 3wegetg = G e λ + G 2 e λ 2 + G 3 e λ 3,where G (x) = = m r= m s= r s= m r=s ( ) m s and analogous formulae for G 2 and G 3. s! ( ) m s s! ( m +m 2 2 r m 2 )( m3 +r s m 3 (λ λ 2 ) m +m 2 r (λ λ 3 ) m 3+r s xs ( m +m 2 2 r m 2 ) )( m3 +r s m 3 (λ λ 2 ) m +m 2 r (λ λ 3 ) m 3+r s xs, )