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Linear ordinary differential equations with constant coefficients. Revisiting
the impulsive response method using factorization
Roberto Camporesia
a
Dipartimento di Matematica, Politecnico di Torino, Torino 10129, Italy
First published on: 08 April 2011
To cite this Article Camporesi, Roberto(2011) '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, 497 — 514, First published on: 08 April 2011 (iFirst)
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International Journal of Mathematical Education in
Science and Technology, Vol. 42, No. 4, 15 June 2011, 497–514
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 10129, Italy
Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 2011
(Received 14 July 2010)
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].
*Email: [email protected]
ISSN 0020–739X print/ISSN 1464–5211 online
ß 2011 Taylor & Francis
DOI: 10.1080/0020739X.2010.543162
http://www.informaworld.com
498
R. Camporesi
Downloaded By: [Camporesi, Roberto] At: 13:48 19 May 2011
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
y00 þ ay0 þ by ¼ f ðxÞ,
ð1:1Þ
(k)
where we use the notation y(x) in place of x(t) or y(t), y 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 0 the equation is called nonhomogeneous. When f ¼ 0 we get the associated homogeneous equation
y00 þ ay0 þ by ¼ 0:
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 g1 2 by the following formula:
Zx
g1 2 ðxÞ ¼ e2 x
eð1 2 Þt dt ðx 2 RÞ:
0
It turns out that g solves the homogeneous equation (1.2) with the initial conditions
yð0Þ ¼ 0,
y0 ð0Þ ¼ 1:
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 0 2 I, we shall see that the general solution of (1.1) in the
interval I can be written as
y ¼ yp þ y h ,
where the function yp is given by the convolution integral
Zx
yp ðxÞ ¼
gðx tÞ f ðtÞdt,
ð1:3Þ
ð1:4Þ
0
and solves (1.1) with trivial initial conditions at the point x ¼ 0, i.e.
yp ð0Þ ¼ y0p ð0Þ ¼ 0,
whereas the function
yh ðxÞ ¼ c0 gðxÞ þ c1 g0 ðxÞ
ð1:5Þ
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International Journal of Mathematical Education in Science and Technology 499
gives the general solution of the associated homogeneous equation (1.2) as the
coefficients c0, c1 vary in R. In other words, the two functions g, g0 are linearly
independent solutions of this equation and form a basis of the vector space of its
solutions.
The linear relation between the coefficients c0, c1 in (1.5) and the initial data
the initial
b0 ¼ y(0) ¼ yh(0) and b1 ¼ y0 ð0Þ ¼ y0h ð0Þ is easily obtained. RIf we impose
Rx
x
conditions at an arbitrary point x0 2 I, we can just replace 0 with x0 in (1.4), and
g(x), g0 (x) with g(x x0), g0 (x x0) in (1.5). The function yp then satisfies yp ðx0 Þ ¼
y0p ðx0 Þ ¼ 0, and the relation between ck and bk ¼ yðkÞ ðx0 Þ ¼ yhðkÞ ðx0 Þ ðk ¼ 0, 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 yp as
a convolution integral between the impulsive response g and the forcing term f. In
particular if we take f ¼ 0, 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
y0 þ ay ¼ f ðxÞ,
ð2:1Þ
dy
where y0 ¼ 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
ð2:2Þ
yðxÞ ¼ eax eax f ðxÞdx,
500
R. Camporesi
R
where eaxf (x)dx denotes the set of all primitives of the function eaxf (x) in the
interval I (i.e. its indefinite integral).
Rx
Suppose that 0 2 I, and consider the integral function 0 eat f ðtÞdt. By the
Fundamental Theorem of Calculus, this is the primitive of eaxf (x) that vanishes at 0.
The theorem of the additive constant for primitives implies that
Z
Zx
eax f ðxÞdx ¼
eat f ðtÞdt þ k ðk 2 RÞ,
0
and we can rewrite (2.2) in the form
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Zx
yðxÞ ¼ eax
eat f ðtÞdt þ keax
Zx 0
¼
eaðxtÞ f ðtÞdt þ keax
0
Zx
¼
gðx tÞ f ðtÞdt þ kgðxÞ,
ð2:3Þ
0
where g(x) ¼ eax. The function g is called the impulsive response of the differential
equation y0 þ ay ¼ 0. It is the unique solution of the initial value problem
0
y þ ay ¼ 0,
yð0Þ ¼ 1:
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 y0 þ ay ¼ 0 and of any particular solution of
(2.1). In (2.3) the function
Zx
yp ðxÞ ¼
gðx tÞ f ðtÞdt
ð2:4Þ
0
is the particular solution of (2.1) that vanishes at x ¼ 0.
If x0 is any point of I, it is easy to verify that
Zx
yðxÞ ¼
gðx tÞ f ðtÞdt þ y0 gðx x0 Þ
ðx 2 I Þ
x0
is the unique solution of (2.1) in the interval I that satisfies y(x0) ¼ y0 ( y0 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
y00 þ ay0 þ by ¼ f ðxÞ,
ð3:1Þ
2
d y
where y00 ¼ dx
2 , a, b 2 R, and the forcing term f : I ! R is a continuous function in the
interval I R, i.e. f 2 C0(I ). For f ¼ 0 we get the associated homogeneous equation
y00 þ ay0 þ by ¼ 0:
ð3:2Þ
International Journal of Mathematical Education in Science and Technology 501
We will write (3.1) and (3.2) in operator form as Ly ¼ f (x) and Ly ¼ 0, where L is the
linear second-order differential operator with constant coefficients defined by
Ly ¼ y00 þ ay0 þ by,
for any function y at least twice differentiable.
d
Denoting by dx
the differentiation operator, we have
2
d
d
þ b:
L¼
þa
dx
dx
ð3:3Þ
L defines a map C2(R) ! C 0(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,
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Lðc1 y1 þ c2 y2 Þ ¼ c1 Ly1 þ c2 Ly2
8c1, c2 2 R, 8y1, y2 2 C2(R). This formula implies some important facts. First of all, if
y1 and y2 are any two solutions of the homogeneous equation, then any linear
combination c1y1 þ c2 y2 is also a solution of this equation. In other words, the set
V ¼ y 2 C2 ðRÞ : Ly ¼ 0
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 C1 there.
Second, if y1 and y2 are two solutions of (3.1) (in a given interval I0 I ), then
their difference y1 y2 solves (3.2). It follows that if we know a particular solution yp
of the non-homogeneous equation (in an interval I 0 ), then any other solution of (3.1)
in I0 is given by yp þ yh, where yh 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 C2 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 y1 þ iy2 (with y1,
y2 : R ! R) is such a function, the derivative y0 may be defined by linearity as
y0 ¼ y01 þ iy02 . It follows that L( y1 þ iy2) ¼ Ly1 þ iLy2. In a similar way one defines
the integral of y as
Z
Z
Z
Zd
Zd
Zd
yðxÞdx ¼ y1 ðxÞdx þ i y2 ðxÞdx,
yðxÞdx ¼
y1 ðxÞdx þ i
y2 ðxÞdx:
c
c
c
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 x
e ¼ ex
dx
8 2 C:
It follows that the complex exponential ex 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
pðÞ ¼ ð 1 Þð 2 Þ:
502
R. Camporesi
The operator L factors in a similar way as a product (composition) of first-order
differential operators
d
d
1
2 :
ð3:4Þ
L¼
dx
dx
Indeed we have
d
1
dx
d
d
2 y ¼
1 ð y0 2 yÞ
dx
dx
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¼ y00 ð1 þ 2 Þ y0 þ 1 2 y,
which coincides with Ly since 1 þ 2 ¼ a, and 12 ¼ b. Note that in (3.4) the order
with which the two factors are composed is unimportant. In other words, the two
d
d
1 Þ and ðdx
2 Þ commute
operators ðdx
d
d
d
d
1
2 ¼
2
1 :
dx
dx
dx
dx
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 y0 y ¼ 0 with 2 C has the general solution
yðxÞ ¼ kex ðk 2 CÞ:
d
(Indeed if y0 ¼ y, then dx
yðxÞex ¼ 0, whence y(x)ex ¼ k.) The first-order nonhomogeneous equation
d
y ¼ f ðxÞ ð 2 CÞ,
y0 y ¼
dx
with complex forcing term f : I R ! C continuous in I 3 0, has the general solution
Z
yðxÞ ¼ ex ex f ðxÞdx
Zx
x
et f ðtÞdt þ kex
¼e
0
Zx
¼
g ðx tÞ f ðtÞdt þ kg ðxÞ ðx 2 I, k ¼ yð0Þ 2 CÞ:
ð3:5Þ
0
x
d
Þ. It is
Here g(x) ¼ e is the impulsive response of the differential operator ðdx
0
the (unique) solution of y y ¼ 0, y(0) ¼ 1. Formula (3.5) can be proved as (2.3) in
the real case. In particular, the solution of the first-order problem
0
y y ¼ f ðxÞ,
yð0Þ ¼ 0
( 2 C) is unique and is given by
Z
x
eðxtÞ f ðtÞdt:
yðxÞ ¼
0
ð3:6Þ
International Journal of Mathematical Education in Science and Technology 503
The following result gives a particular solution of (3.1) as a convolution integral.
Theorem 3.1:
Let f 2 C0(I ), and suppose that 0 2 I. Then the initial value problem
00
y þ ay0 þ by ¼ f ðxÞ,
ð3:7Þ
yð0Þ ¼ 0, y0 ð0Þ ¼ 0
has a unique solution, defined on the whole of I, and given by the formula
Zx
gðx tÞ f ðtÞdt ðx 2 I Þ,
yðxÞ ¼
ð3:8Þ
0
where g is the function defined by
Zx
gðxÞ ¼
e2 ðxtÞ e1 t dt
ðx 2 RÞ:
ð3:9Þ
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0
In particular, if we take f ¼ 0, we get that the only solution of the homogeneous
problem
00
y þ ay0 þ by ¼ 0,
ð3:10Þ
yð0Þ ¼ 0, y0 ð0Þ ¼ 0
is the zero function y ¼ 0.
Proof: We rewrite the differential equation (3.1) in the form
d
d
1
2 y ¼ f ðxÞ:
dx
dx
Letting
h¼
d
2 y ¼ y0 2 y,
dx
we see that y solves the problem (3.7) if and only if
0
0
h 1 h ¼ f ðxÞ,
y 2 y ¼ hðxÞ,
h solves
and y solves
hð0Þ ¼ 0,
yð0Þ ¼ 0:
From (3.6) we get
Z
x
e1 ðxtÞ f ðtÞdt,
hðxÞ ¼
0
Z
x
e2 ðxtÞ hðtÞdt:
yðxÞ ¼
0
Substituting h from the first formula into the second, we obtain y(x) as a repeated
integral (for any x 2 I ):
Z t
Zx
2 ðxtÞ
1 ðtsÞ
e
e
f ðsÞds dt:
ð3:11Þ
yðxÞ ¼
0
0
504
R. Camporesi
To fix ideas, let us suppose that x 4 0. Then in the integral with respect to t we have
0 t x, whereas in the integral with respect to s we have 0 s t. We can then
rewrite y(x) as a double integral:
Z
2 x
eð1 2 Þt e1 s f ðsÞds dt,
yðxÞ ¼ e
Tx
where Tx is the triangle in the (s, t) plane defined by 0 s t x, with vertices at the
points (0, 0), (0, x) and (x, x). In (3.11) we first integrate with respect to s and then
with respect to t. Since the triangle Tx is convex both horizontally and vertically, and
since the integrand function
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Fðs, tÞ ¼ eð1 2 Þt e1 s f ðsÞ
is continuous in Tx, we can interchange the order of integration and integrate with
respect to t first. Given s (between 0 and x) the variable t in Tx varies between s and
x, see the figure below.
t
s =t
x
s≤t≤x
0
x
s
We thus obtain
Z x Z
x
yðxÞ ¼
0
e2 ðxtÞ e1 ðtsÞ dt f ðsÞds:
s
By substituting t with t þ s in the integral with respect to t we finally get
Zx
Z x Z xs
e2 ðxstÞ e1 t dt f ðsÞds ¼
gðx sÞ f ðsÞds,
yðxÞ ¼
0
0
ð3:12Þ
0
which is (3.8). For x 5 0 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 a2 4b 6¼ 0) then
gðxÞ ¼
1 1 x
e e 2 x ;
1 2
ð3:13Þ
gðxÞ ¼ xe1 x :
ð3:14Þ
(2) if 1 ¼ 2 (,D ¼ 0) then
International Journal of Mathematical Education in Science and Technology 505
Note that g is always a real function. Letting ¼ a/2 and
(
( pffiffiffiffiffiffiffi
i if D 5 0
D=2 if D 5 0
¼ pffiffiffiffi
so that 1,2 ¼
if D 4 0,
D=2
if D 4 0,
we have
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8
1
1 ðþiÞx
>
>
eðiÞx ¼ ex sinðxÞ if D 5 0,
>
< 2i e
gðxÞ ¼
>
1 ðþÞx
1
>
>
e
eðÞx ¼ ex sinhðxÞ if D 4 0:
:
2
Also note that g 2 C1(R).
It is easy to check that g solves the following homogeneous initial value problem:
( 00
y þ ay0 þ by ¼ 0,
ð3:15Þ
yð0Þ ¼ 0, y0 ð0Þ ¼ 1:
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 y0 :
Z
Zx
d x
0
y ðxÞ ¼
gðx tÞ f ðtÞdt ¼ gð0Þ f ðxÞ þ
g0 ðx tÞ f ðtÞdt ðx 2 I Þ:
ð3:16Þ
dx 0
0
Indeed, given h such that x þ h 2 I, we have
Z xþh
Zx
yðx þ hÞ yðxÞ 1
¼
gðx þ h tÞ f ðtÞdt gðx tÞ f ðtÞdt :
h
h 0
0
ð3:17Þ
As g 2 C2(R), we can apply Taylor’s formula with the Lagrange remainder
gðx0 þ hÞ ¼ gðx0 Þ þ g0 ðx0 Þh þ
1 00
g ðÞh2
2
at the point x0 ¼ x t, where is some point between x0 and x0 þ h. Substituting this
in (3.17) and using
Zx
Z xþh
Z xþh
gðx tÞ f ðtÞdt ¼
gðx tÞ f ðtÞdt þ
gðx tÞ f ðtÞdt
0
0
x
gives
1
1
ð yðx þ hÞ yðxÞÞ ¼
h
h
þ
Z
Z
xþh
xþh
g0 ðx tÞ f ðtÞdt
gðx tÞ f ðtÞdt þ
0
x
1
h
2
Z
xþh
g00 ðÞ f ðtÞdt
ð3:18Þ
0
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(0)f
R x (x), by the Fundamental Theorem of Calculus.
The second term in (3.18) tends to 0 g0 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
506
R. Camporesi
bounded function of h in a neighbourhood of h ¼ 0 (this is easy to prove). We thus
obtain formula (3.16). Recalling that g(0) ¼ 0, we finally get
Zx
0
y ðxÞ ¼
g0 ðx tÞ f ðtÞdt ðx 2 I Þ:
0
In the same way we compute the second derivative
Z
d 2 x
gðx tÞ f ðtÞdt
dx
0
Z
d x 0
¼
g ðx tÞ f ðtÞdt
dx 0
Zx
¼ g0 ð0Þ f ðxÞ þ
g00 ðx tÞ f ðtÞdt
0
Zx
¼ f ðxÞ þ
g00 ðx tÞ f ðtÞdt,
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y00 ðxÞ ¼
0
0
where we used g (0) ¼ 1. It follows that
00
Z
0
x
ð g00 þ ag0 þ bgÞðx tÞ f ðtÞdt
y ðxÞ þ ay ðxÞ þ byðxÞ ¼ f ðxÞ þ
0
¼ f ðxÞ 8x 2 I,
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(0) ¼ 0 ¼ y0 (0) are
immediately verified.
We now come to the solution of the initial value problem with an arbitrary initial
data at the point x ¼ 0.
Theorem 3.2: Let f 2 C0(I ), 0 2 I, and let y0 , y00 be two arbitrary real numbers. Then
the initial value problem
00
y þ ay0 þ by ¼ f ðxÞ,
ð3:19Þ
yð0Þ ¼ y0 , y0 ð0Þ ¼ y00
has a unique solution, defined on the whole of I, and given by
Zx
gðx tÞ f ðtÞdt þ y00 þ ay0 gðxÞ þ y0 g0 ðxÞ
yðxÞ ¼
ðx 2 I Þ:
ð3:20Þ
0
In particular (taking f ¼ 0), the solution of the homogeneous problem
00
y þ ay0 þ by ¼ 0,
yð0Þ ¼ y0 ,
y0 ð0Þ ¼ y00
is unique, of class C1 on the whole of R, and is given by
yh ðxÞ ¼ y00 þ ay0 gðxÞ þ y0 g0 ðxÞ ðx 2 RÞ:
ð3:21Þ
Proof: The uniqueness of the solutions of the problem (3.19) follows from the
fact that if y1 and y2 both solve (3.19), then their difference y~ ¼ y1 y2 solves
the problem (3.10), whence y~ ¼ 0 by Theorem 3.1. Now note that the function g0
International Journal of Mathematical Education in Science and Technology 507
satisfies the homogeneous equation (like g). Indeed, since L has constant coefficients,
we have
"
d
Lg ¼ L g ¼
dx
d
dx
0
2
#
d
d
d
þb
g¼
Lg ¼ 0:
þa
dx
dx
dx
By the linearity of L and by Theorem 3.1 it follows that the function y given
by (3.20) satisfies (Ly)(x) ¼ f (x) 8x 2 I. It is immediate that y(0) ¼ y0. Finally, since
Zx
y0 ðxÞ ¼
g0 ðx tÞ f ðtÞdt þ y00 þ ay0 g0 ðxÞ þ y0 g00 ðxÞ,
0
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we have
y0 ð0Þ ¼ y00 þ ay0 þ y0 g00 ð0Þ
¼ y00 þ ay0 þ y0 ða g0 ð0Þ b gð0ÞÞ
¼ y00 :
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
Z
yðxÞ ¼
0
x
gðx sÞ f ðsÞds þ y00 2 y0 gðxÞ þ y0 e2 x :
This formula agrees with (3.20) in view of the equality e2 x ¼ g0 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 0(I ), x0 2 I, y0 , y00 2 R. The solution of the initial value problem
y00 þ ay0 þ by ¼ f ðxÞ,
yðx0 Þ ¼ y0 , y0 ðx0 Þ ¼ y00
is unique, it is defined on the whole of I, and is given by
Zx
yðxÞ ¼
gðx tÞ f ðtÞdt þ y00 þ ay0 gðx x0 Þ þ y0 g0 ðx x0 Þ
ð3:22Þ
ðx 2 I Þ:
ð3:23Þ
x0
In particular (taking f ¼ 0), the solution of the homogeneous problem
y00 þ ay0 þ by ¼ 0,
yðx0 Þ ¼ y0 ,
y0 ðx0 Þ ¼ y00 ,
with x0 2 R arbitrary, is unique, of class C1 on the whole of R, and is given by
yh ðxÞ ¼ y00 þ ay0 gðx x0 Þ þ y0 g0 ðx x0 Þ ðx 2 RÞ:
ð3:24Þ
508
R. Camporesi
Proof: Let y be a solution of the problem (3.22), and let x0 denote the translation
map defined by x0 ðxÞ ¼ x þ x0 . Set
~
yðxÞ
¼ y x0 ðxÞ ¼ yðx þ x0 Þ:
Since the differential operator L has constant coefficients, it is invariant under
translations, that is,
Lð y x0 Þ ¼ ðLyÞ x0 :
It follows that
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~
ðLyÞðxÞ
¼ ðLyÞðx þ x0 Þ ¼ f ðx þ x0 Þ:
~ ¼ yðx0 Þ ¼ y0 , y~ 0 ð0Þ ¼ y0 ðx0 Þ ¼ y00 , we see that y solves the
Since, moreover yð0Þ
problem (3.22) in the interval I if and only if y~ solves the initial value problem
(
~
y~ 00 þ ay~0 þ by~ ¼ fðxÞ,
0
~ ¼ y0 , y~ ð0Þ ¼ y00
yð0Þ
in the translated interval I x0 ¼ {t x0 : t 2 I} and with the translated forcing term
~ ¼ f ðx þ x0 Þ. By Theorem 3.2 we have that y~ is unique and is given by
fðxÞ
Zx
~
~
yðxÞ
¼
gðx sÞ fðsÞds
þ y00 þ ay0 gðxÞ þ y0 g0 ðxÞ:
0
Therefore y is also unique and is given by
Z xx0
~ x0 Þ ¼
gðx x0 sÞ f ðs þ x0 Þds
yðxÞ ¼ yðx
0
0
þ y0 þ ay0 gðx x0 Þ þ y0 g0 ðx x0 Þ:
Formula (3.23) follows immediately from this by making the change of variable
g
s þ x0 ¼ t in the integral with respect to s.
We observe that
R x the solution in (3.23) can be written as y ¼ yp þ yh, where the
function yp ðxÞ ¼ x0 gðx tÞ f ðtÞdt solves (3.1) with the initial conditions yp ðx0 Þ ¼
y0p ðx0 Þ ¼ 0, whereas the function yh, 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, g0 form a basis of V.
Proof: Let y be a real solution of (3.2). Let x0 be any point at which y is defined,
and let y0 ¼ y(x0), y00 ¼ y0 ðx0 Þ. Define yh by formula (3.24). Then y and yh both satisfy
(3.2) with the same initial conditions, whence y ¼ yh. In particular, y 2 C1(R).
Repeating this with x0 ¼ 0, we conclude by (3.21) that every element of the
vector space V can be written as a linear combination of g and g0 in a unique way
(the coefficients in this combination are uniquely determined by the initial data
International Journal of Mathematical Education in Science and Technology 509
at the point x ¼ 0). In particular, g and g0 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 0 can be
written in the form y ¼ c1y1 þ c2y2 (c1, c2 2 C), where y1 and y2 are given by
(
e 1 x , e 2 x
if 1 6¼ 2 ,
ð y1 ðxÞ, y2 ðxÞÞ ¼
e1 x , xe1 x
if 1 ¼ 2 :
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Conversely, any function of this form is a solution of Ly ¼ 0.
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 0 can be
written as a complex linear combination of g and g0 (indeed if y ¼ Re y þ iIm y solves
Ly ¼ 0, then L(Re y) ¼ 0 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 g0 , so y is a complex linear
combination of g and g0 ). Our claim then follows immediately from formulas (3.13)
and (3.14). The last statement is easily verified.
g
The two functions y1, y2 (like g, g0 ) form then a basis of the complex vector space
VC of complex solutions of Ly ¼ 0. If 1 and 2 are real (,D 0), then y1 and y2 are
real functions and form a basis of V as well. If instead 1,2 ¼ i with 6¼ 0
(,D 5 0), then
y1,2 ðxÞ ¼ eðiÞx ¼ ex ðcos x i sin xÞ,
and a basis of V is given by the functions
Re y1 ðxÞ ¼ ex cos x,
Im y1 ðxÞ ¼ ex sin x:
ð3:25Þ
Indeed if c1 ¼ c þ id and c2 ¼ c0 þ id 0 , with c, d, c0 , d 0 2 R, then
c1 e1 x þ c2 e2 x ¼ ðc þ id ÞeðþiÞx þ ðc0 þ id 0 ÞeðiÞx
¼ ex ðc þ c0 Þ cos x þ ðd 0 d Þ sin x þ i½ðd þ d 0 Þ cos x þ ðc c0 Þ sin x :
It follows that the function y ¼ c1y1 þ c2 y2 is real-valued if and only if c ¼ c0 and
d ¼ d 0 , i.e. iff c1 ¼ c2 . In this case y is a real linear combination of the functions
in (3.25).
Example 1: Solve the initial value problem
8
ex
< 00
,
y 2y0 þ y ¼
xþ2
:
yð0Þ ¼ 0, y0 ð0Þ ¼ 0:
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 ¼ xex :
510
R. Camporesi
x
e
The forcing term f ðxÞ ¼ xþ2
is continuous for x 4 2 and for x 5 2. Since the
initial conditions are posed at x ¼ 0, 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:
Zx
Zx
t
xt
xt e
x
dt
dt ¼ e
ðx tÞe
yðxÞ ¼
t
þ
2
0
0 tþ2
Z x
xþ2
¼ ex
1 dt ¼ ex ½ðx þ 2Þ logðt þ 2Þ tx0
t
þ
2
0
¼ ex ½ðx þ 2Þ logðx þ 2Þ x ðx þ 2Þ log 2
xþ2
xex :
¼ ex ðx þ 2Þ log
2
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Example 2: Solve the initial value problem
8
<
1
,
cos x
:
yð0Þ ¼ 0, y0 ð0Þ ¼ 0:
y00 þ y ¼
Solution: We have that p() ¼ 2 þ 1, whence 1 ¼ 2 ¼ i and the impulsive
response is
gðxÞ ¼ sin x:
The initial data are given at x ¼ 0 and we can work in the interval I ¼ (/2, /2),
where the forcing term f ðxÞ ¼ cos1 x is continuous. By formula (3.8) we get
Zx
1
dt
yðxÞ ¼
sinðx tÞ
cos
t
0
Zx
1
¼
dt
ðsin x cos t cos x sin tÞ
cos t
0
Zx
Zx
sin t
¼ sin x
dt
dt cos x
0
0 cos t
¼ x sin x þ cos x logðcos xÞ:
Example 3: Solve the initial value problem
8
<
1
,
ch x
:
yð0Þ ¼ 0, y0 ð0Þ ¼ 0:
y00 y0 ¼
Solution: We have that p() ¼ 2 ¼ ( 1), thus 1 ¼ 1, 2 ¼ 0, and the
impulsive response is
gðxÞ ¼ ex 1:
The forcing term f ðxÞ ¼ 1 is continuous on the whole of R. Formula (3.8) gives
chx
Zx
Zx
Z x t
xt
1
e
1
dt ¼ ex
dt:
dt e 1
yðxÞ ¼
ch t
0
0 ch t
0 ch t
International Journal of Mathematical Education in Science and Technology 511
The two integrals are easily computed:
Z
Z
Z
1
1
et
dt ¼ 2
dt
dt
¼
2
t
t
2t
ch t
e þe
e þ1
¼ 2 arctan et þ C,
Z
Z
Z
et
1
1 þ e2t e2t
dt ¼ 2
dt
¼
2
dt
e2t þ 1
ch t
1 þ e2t
¼ 2t log 1 þ e2t þ C0 :
We finally get
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x x
yðxÞ ¼ ex 2t log 1 þ e2t 0 2 arctan et 0
¼ ex 2x log 1 þ e2x þ ex log 2 2 arctanðex Þ þ :
2
4. Discussion
Remark 1 (The convolution): To better understand the structure of the particular
solution (2.4)–(3.8), let us recall that if h1 and h2 are suitable functions (e.g. if h1 and
h2 are piecewise continuous on R with h1 bounded and h2 absolutely integrable over
R, or if h1 and h2 are two signals, that is, they are piecewise continuous on R and
vanish for x 5 0), then their convolution h1 * h2 is defined as
Z þ1
h1 ðx tÞh2 ðtÞdt:
h1 h2 ðxÞ ¼
1
The change of variables x t ¼ s shows that the convolution is commutative
Z þ1
h1 ðsÞh2 ðx sÞds:
h1 h2 ðxÞ ¼ h2 h1 ðxÞ ¼
1
Moreover, one can show that the convolution is associative: (h1 * h2) * h3 ¼
h1 * (h2 * h3).
If h1 and h2 are two signals, it is easy to verify that h1 * h2 is also a signal and that
for any x 2 R one has
Zx
h1 h2 ðxÞ ¼
h1 ðx tÞh2 ðtÞdt:
0
In a similar way, if h1 and h2 vanish for x 4 0, the same holds for h1 * h2 and we
have that
Z0
h1 h2 ðxÞ ¼
h1 ðx tÞh2 ðtÞdt:
x
If denotes the Heaviside step function, given by
1 if x 0,
ðxÞ ¼
0 if x 5 0,
512
R. Camporesi
and if we let
~ ¼ ðxÞ ¼
ðxÞ
0
if x 4 0,
1
if x 0,
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then for any two functions g and f we have
Zx
g f ðxÞ
if x 0,
gðx tÞ f ðtÞdt ¼
ð4:1Þ
~ f
~ ðxÞ if x 0.
g
0
Rx
In other words, the integral 0 gðx tÞ f ðtÞdt is the convolution of g and f for
~ and f
~ for x 0.
x 0, it is the opposite of the convolution of g
Furthermore, if we denote g in (3.9) by g1 2 , we can rewrite (3.9) in the suggestive
form
Zx
g2 g1 ðxÞ
if x 0,
g2 ðx tÞ g1 ðtÞdt ¼
g1 2 ðxÞ ¼
ð4:2Þ
~ 2 g
~ 1 ðxÞ if x 0:
g
0
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 0) and then using
the associativity of this
yðxÞ ¼ g2 ðg1 f Þ ðxÞ ¼ ðg2 g1 Þ f ðxÞ:
ð4:3Þ
This is just the equality (3.11) ¼ (3.12) for x 4 0. 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 0.
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 D0þ of distributions with support in [0, þ1), which generalizes the algebra of
signals. One can show that the elementary solution in D0þ , 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. If yp solves
Ly ¼ f (x) with trivial initial conditions at x ¼ 0, then L(yp) ¼ f, whence
d
þ a. This approach requires,
yp ¼ g * f. This is just (2.4) for x 4 0 and L ¼ dx
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. 104–105. 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
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, y0 , y00 2 C and f : I R ! C continuous in I 3 x0, is given by (3.23). Finally, the
set VC of complex solutions of (3.2) is a complex vector space of dimension 2, with a
basis given by {g, g0 }.
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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
ð4:4Þ
yðnÞ þ a1 yðn1Þ þ a2 yðn2Þ þ þ an1 y0 þ an y ¼ f ðxÞ,
k
k
d y
d
where yðkÞ ¼ dx
y, a1, a2, . . . , an 2 C and the forcing term f : I R ! C is a
k ¼ dx
continuous complex-valued function in the interval I. When f ¼ 0 we obtain the
associated homogeneous equation
yðnÞ þ a1 yðn1Þ þ a2 yðn2Þ þ þ an1 y0 þ an y ¼ 0:
ð4:5Þ
We can write (4.4) and (4.5) in the form Ly ¼ f (x) and Ly ¼ 0, where L is the linear
differential operator of order n with constant coefficients given by
n
n1
d
d
d
þ an :
L¼
þ a1
þ þ an1
dx
dx
dx
As in the case of n ¼ 2, one easily proves that the complex exponential ex, 2 C, is a
solution of (4.5) if and only if is a root of the characteristic polynomial
pðÞ ¼ n þ a1 n1 þ þ an1 þ an :
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 ¼ ð 1 Þð 2 Þ ð n Þ:
The differential operator L can be factored in a similar way as
d
d
d
1
2 n ,
L¼
dx
dx
dx
where the order of composition of the factors is unimportant as they all commute.
We define the impulsive response of L, g ¼ g1 ... n , recursively by the following
formulas: for n ¼ 1 we set g1ðxÞ ¼ e1 x , for n 2 we set
Zx
g1 ...n ðxÞ ¼
gn ðx tÞ g1 ...n1 ðtÞdt ðx 2 RÞ:
ð4:6Þ
0
The impulsive response g solves the homogeneous equation (4.5) with the initial
conditions
yð0Þ ¼ y0 ð0Þ ¼ ¼ yðn2Þ ð0Þ ¼ 0,
yðn1Þ ð0Þ ¼ 1:
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 0 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 yp is given by (1.4) and solves (4.4) with trivial initial conditions at
the point x ¼ 0 (i.e. yðkÞ
p ð0Þ ¼ 0 for k ¼ 0, 1, . . . , n 1), whereas the function
yh ðxÞ ¼
n1
X
ck gðkÞ ðxÞ
ð4:7Þ
k¼0
gives the general solution of (4.5) as the coefficients ck vary in C. Thus the n
functions g, g0 , g00 , . . . , 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 0 is given by (4.7) with ck 2 R.
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References
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[2] L. Schwartz, Methodes Mathematiques Pour les Sciences Physiques, Enseignement des
Sciences, Hermann, Paris, 1961.
[3] E.A. Coddington and N. Levinson, Theory of Ordinary Differential Equations, McGrawHill Book Company, Inc., New York–Toronto–London, 1955.
[4] E.A. Coddington, An Introduction to Ordinary Differential Equations, Prentice-Hall
Mathematics Series, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1961.
[5] W.A. Robin, Solution of differential equations by quadratures: Cauchy integral method,
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[6] D.H. Parsons, Linear ordinary differential equations with constant coefficients: identification
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