269
Chapter 8
First-Order Linear Systems
As we have seen on numerous occasions, a differential equation of order two
or higher can be converted to a first-order system for the purpose of applying
a numerical approximation procedure. First-order systems that arise in this
manner have a particular form. First-order systems also arise naturally in various
applications that involve two or more quantities. In this chapter, we begin to
look at first-order systems in general, concentrating on linear systems.
The reader who is not readily familiar with the basic manipulations of linear
algebra—such as matrix-vector and matrix-matrix multiplication, transposes,
determinants, etc.—is encouraged to study Appendix I before going on here.
8.1 Introduction
Recall that in Chapter 2 we studied linear, first-order (scalar) differential equations, which were of the form
y 0 − a y = f,
where a and f are functions of the independent variable.
In this chapter we will study a more general version the same problem. Instead of scalar-valued functions y and f , the problem will involve n-component
vector-valued functions y and f :
 
 
f1
y1
..  ,
.



..
, f=
y=
.
fn
yn
and the scalar a will instead be an n × n matrix A. The system of equations
y10 − y1 + 2t y2 = sin t
y20 − t y1 + y2 = 0,
for example, will be viewed as the single equation
0 y1
1 −2t
y1
sin t
−
=
y2
t −1
y2
0
for the (column) vector quantity y = (y1 , y2 )T .
∗
* The superscript T signifies the transpose of a vector or matrix. (See Appendix I.) The
transpose of a row vector is simply the corresponding column vector, and vice versa.
270
Chapter 8. First-Order Linear Systems
The general form of the systems we shall study is
y0 − A y = f ,
(1)
where y and f are vector-valued functions with n components, A is an n × n
matrix-valued function, and the entries of f and A are continuous on an interval
I. When f = 0, the system is said to be homogeneous, and when f 6= 0, the
homogeneous equation associated with (1) is
y0 − A y = 0.
(2)
When A and f have constant entries, the system is said to be autonomous.
Initial-value problems associated with (1) take the form
y0 − A y = f , y(t0 ) = y0 ,
(3)
where t0 in I and y0 in Rn . Some of our subsequent arguments will rely upon
the following existence and uniqueness theorem. Its proof is in Appendix III.
THEOREM 8.1.1. If the entries of the matrix A and the vector f are continuous
on an interval I, then for any t0 in I and y0 in Rn , the initial-value problem (3)
has a unique solution on I.
J
The usual consequences of linearity are in force for solutions of (1):
i) If y1 and y2 are two solutions of (2), then so is c1 y1 + c2 y2 for any scalar
constants c1 and c2 ;
ii) If ỹ and y are solutions of (1) and (2), respectively, then ỹ + c y is a
solution of (1) for any scalar constant c.
These important facts follow easily from the linearity of matrix multiplication.
• Example 1 Easy calculations reveal that the vector-valued functions
3e t
e−t
y1 =
and y2 =
−e t
−e−t
each satisfy the (autonomous) homogeneous equation
2
3
0
y −
y = 0.
−1 −2
Consequently, so does c1 y1 +c2 y2 for any constants c1 , c2 . Also, ỹ = (t+1, t−1)T
is a particular solution of the nonhomogeneous system
2
3
2 − 5t
0
y −
y=
;
−1 −2
3t
therefore, c1 y1 + c2 y2 + ỹ is a solution for any constants c1 , c2 .
8.1. Introduction
271
• Example 2 Easy calculations reveal that the vector-valued functions
t
t3
y1 =
and y2 =
−t
t3
each satisfy the (nonautonomous) homogeneous system
1 2 1
0
y = 0.
y −
t 1 2
Consequently, so does c1 y1 + c2 y2 for any constants c1 , c2 . Also, ỹ = (−t, −t2 )T
is a particular solution of the nonhomogeneous system
1 2 1
t+1
0
y −
y=
;
1
t 1 2
therefore, c1 y1 + c2 y2 + ỹ is a solution for any constants c1 , c2 .
Applications
To motivate the investigation of linear systems of differential equations, we
will present here a couple of standard elementary applications—a two-container
mixing problem and a two-loop RLC circuit.
• Example 3 The following mixing problem involving two well-stirred containers is depicted in Figure 1. (The reader may wish to review Section 2.2.2 at
this point.) Suppose that the volume of mixture in each of the two containers
is a constant V , and the amounts of impurity in the two container are a1 and
a2 , respectively. Thus the respective impurity concentrations are a1 /V and a2 /V.
Figure 1
Suppose further that a mixture with an impurity concentration of c 0 is being
pumped into the first container at a rate of r0 units of volume per unit time, and
that the well-stirred contents of the second container is being pumped back into
the first container at a rate of r1 units of volume per unit time, while the wellstirred contents of the first container flows into the second container at a rate
272
Chapter 8. First-Order Linear Systems
of r0 + r1 units of volume per unit time. The contents of the second container
is pumped out at a rate of r0 units of volume per unit time. The differential
equations for the amounts a1 and a2 each take the form
dai
= rate in − rate out.
dt
Thus we have the pair of equations
da1
r1
r0 + r1
= r0 c 0 + a2 −
a1
dt
V
V
da2
r0 + r1
r1
r0
=
a1 − a2 − a2 ,
dt
V
V
V
or in vector form as in (1), the single equation
da
1 −(r0 + r1 )
r1
r0 c 0
a=
,
−
r0 + r1
−(r0 + r1 )
0
dt
V
where a = (a1 , a2 )T .
• Example 4 Consider the simple two-loop RLC circuit shown in Figure 2, in
which a capacitor discharges through a resistor and an inductor connected in
parallel. (The reader may wish to review Section 2.2.3 at this point.)
Figure 2
Let the currents i1 , i2 , and i3 be defined as shown in Figure 2. The system of
equations will be obtained by applying Kirchhoff’s current law—which says that
the sum of the currents entering a node at any instant is zero—and Kirchhoff’s
voltage law (see Section 2.2.3) to each of the closed loops in the circuit. The
current law applied to the top node in this circuit results in
−i1 − i2 + i3 = 0, or i3 = i1 + i2 .
The voltage law applied to the RL loop on the right results in
L i03 + R i2 = 0.
The voltage law applied to the RC loop on the left results in R i2 − Q/C = 0,
where Q is the charge on the capacitor. Since Q0 = i1 , differentiation of that
8.1. Introduction
273
equation produces
1
i1 = 0.
C
Combining these three relationships among the currents yields a pair of differential equations for i1 and i2 :
R i02 −
R
1
i1 + i2 = 0
RC
L
1
0
i2 −
i1 = 0.
RC
The current law then relates i3 trivially to i1 and i2 . In vector form as in (1),
this pair of differential equations becomes the single equation
di
1
−L −R2 C
−
i = 0,
L
0
dt
RLC
i01 +
where i = (i1 , i2 )T and the zero on the right side is the zero vector (0, 0)T .
Problems
For each of the linear systems in Problems 1 through 3, find the matrix A such that
the system can be written as y0 − A y = f . Identify each system as homogeneous or
nonhomogeneous.
1. y10 = y1 − y2 + sin t, y20 − y2 = t y1
2. y10 = e−t y1 − y2 , y20 = y1 + 2 y2
3. y10 = y1 − y2 + y3 , y20 + y2 = −y3 + y1 , y30 + y3 = y1 + y2
In Problems 4 through 6, write down the differential equation for each component of
the solution of y0 − A y = 0.


−1 −3 2
1 −3
t −1
4. A =
5. A =
6. A =  3 −2 −2 
2 −2
1 −t2
1
1 −1
7. Find a second-order differential equation for x(t) (that does not involve y(t)), given
that x(t) and y(t) together satisfy
x0 − x + y = 0,
y 0 + x − 2y = 0.
8. Find a 2 × 2 matrix A such that the functions y1 = e−t + e−2t and y2 = e−t − e−2t
satisfy y0 − A y = 0. Is the answer unique?
9. Verify each of the solutions y1 , y2 , and ỹ in Example 1.
10. Verify each of the solutions y1 , y2 , and ỹ in Example 2.
11. Verify that y1 = cos t and y2 = sin t − cos t together satisfy the system
−1 −1
0
y −
y = 0.
2
1
274
Chapter 8. First-Order Linear Systems
Determine a system of three first-order differential equations that governs each of the
systems represented schematically in the following figures. Each is an extension of either
Example 3 or Example 4. (Each of the two circuits requires conversion of a second-order
equation to a pair of first-order equations.)
12.
13.
14.
15.
8.2. Two Ad Hoc Methods
275
8.2 Two Ad Hoc Methods
In this section we will look at two methods for solving linear systems of differential equations with constant coefficients. These methods are ad hoc in the
sense that they are based upon techniques we are already familiar with in the
context of a single equation and provide little insight into the general structure
of solutions.
The Method of Elimination
This method amounts essentially to replacing a linear first-order system with
a single higher-order differential equation. For a system of two equations, the
result is a single second-order equation, which can be solved by the method of
Chapter 6.∗ We will simply illustrate the method with an example.
• Example 1 Consider the system of equations for the two-tank mixing problem
in Example 3 of Section 8.1. With parameter values V = 4 m3 , r0 = 3 m3 /h,
r1 = 1 m3 /h, and c0 = 1 kg/m3 , and zero initial concentrations in both tanks,
we obtain the initial-value problem
−1 1/4
3
0
a −
a=
,
1 −1
0
0
a(0) =
.
0
Rewriting the differential equation as a pair of equations in operator form, we
have
1
(D + I )a1 − a2 = 3
4
−a1 + (D + I )a2 = 0.
Now we apply D + I to each side of the second equation and add the result
to the first equation. This eliminates a1 and produces a single, second-order
equation for a2 :
1
(D + I )2 a2 − a2 = 3.
4
Simplification results in
3
(D2 + 2D + I )a2 = 3.
4
At this point it is easy to see that a particular solution is ã2 = 4. Also, since
D2 + 2D + 34 I = (D + 12 )(D + 32 ), we see that e−t/2 and e−3t/2 are solutions of
the associated homogeneous equation. Therefore, the general solution is
a2 = c1 e−t/2 + c2 e−3t/2 + 4.
* Higher-order equations are discussed in Appendix IV.
276
Chapter 8. First-Order Linear Systems
Returning now to the second of the two original differential equations, we simply
compute a1 :
a1 = (D + I )a2
c1
c2
= e−t/2 − e−3t/2 + 4.
2
2
The zero initial values give us the following system of equations for c1 and c2 :
c1 + c2 = −4,
c1 − c2 = −8.
Solving these equations gives us c1 = −6
and c2 = 2; thus the solution of the initialvalue problem is
−t/2 − e−3t/2 + 4
−3e
a=
.
−6e−t/2 + 2e−3t/2 + 4
The graphs of a1 and a2 versus t are seen
in Figure 1. Decide for yourself which is
which.
Figure 1
Laplace Transforms
The Laplace transform method extends naturally to systems of linear differential
equations. As you might expect, the algebraic step that produces the transform
of the solution will involve solving a system of equations in which the coefficients
depend on the variable s. In fact, given an initial-value problem of the form
y0 − A y = f ,
y(0) = y0 ,
the vector-valued Laplace transform Y(s) of the solution satisfies
s Y(s) − y0 − A Y(s) = F(s),
which becomes the linear system
(s I − A)Y(s) = F(s) + y0 .
Solving this equation produces the transform Y(s). The inverse transform of
each component of Y(s) is then found by the usual means.
• Example 2 Consider the system of equations for the two-loop circuit in Example
4 of Section 8.1. With parameter values R = 1 ohm, L = .01 henry, and C = .125
farad, and initial currents i1 (0) = 1 and i2 (0) = −1 ampere, we obtain the
initial-value problem
−8 −100
1
0
i −
i = 0, i (0) =
.
8
0
−1
8.2. Two Ad Hoc Methods
277
Using Ψ(s) = (Ψ1 (s), Ψ2 (s))T to denote the Laplace transform of i (t), we have
1
−8 −100
s Ψ(s) −
−
Ψ(s) = 0,
−1
8
0
which becomes
s+8
−8
100
s
Ψ(s) =
1
−1
.
The solution of this system is
Ψ(s) =
1
s2 + 8 s + 800
s + 100
−s
.
After a bit of processing, we arrive at


24
28
s+4
+
 (s + 4)2 + 282
7 (s + 4)2 + 282 
Ψ(s) = 
,
s+4
1
28
−
+
(s + 4)2 + 282
7 (s + 4)2 + 282
which tells us that the solution of the original system is


24
 cos 28t + 7 sin 28t 
i (t) = e−4t 
.
1
− cos 28t + sin 28t
7
From this we find that the current through the inductor is
i3 (t) = i1 (t) + i2 (t) =
25 −4t
e
sin 28t.
7
Figure 2 shows the graph of this function.
Figure 2
Problems
1. Rework Example 1 with the parameter values V = 10 m3 , r0 = 5 m3 /h, r1 = 4 m3 /h,
and c0 = 1.2 kg/m3 and zero initial concentrations in both tanks.
278
Chapter 8. First-Order Linear Systems
2. Rework Example 2 with the parameter values R = 1 ohm, L = 1 henry, and C = 2/9
farad and initial currents i1 (0) = 1 and i2 (0) = −1 ampere.
3. (ab) Use Laplace transforms to rework Example 1 and Problem 1.
4. (ab) Use the method of elimination to rework Example 2 and Problem 2.
In each of Problems 5 through 12, solve the initial-value problem y0 −Ay = 0, y(0) = y0 ,
(a) by the method of elimination;
(b) using the Laplace transform.
(Solving the 3 × 3 systems in Problems 9 through 12 by elimination will require the
solution of a cubic characteristic equation.)
−2 1
1
−2 −1
1
5. A =
, y0 =
6. A =
, y0 =
1 −2
1
1 −2
0
−1 1
0
0 4
1
7. A =
, y0 =
8. A =
, y0 =
−1 −1
1
−1 0
1

 


 

1
−1 5 0
0
−3 1 0
10. A =  −5 1 −5  , y0 =  0 
9. A =  0 −3 1  , y0 =  0 
1
0 5 −1
1
−4 0 0


 


 
−1 1
0
0
0
0 1
1
12. A =  0 −2 1  , y0 =  0 
11. A =  −1 1 −1  , y0 =  1 
0 −1 −1
0
−1 0 0
0
8.3 Vector-Valued Functions and Linear Independence
Just as in the case of differential equations of order two or higher, the notion
of linear independence is crucial to understanding the structure of solutions to
linear first-order systems of differential equations. Since we view such solutions
as vector-valued functions, certain basic concepts from linear algebra play an
important role. We begin with a very brief review of the the notion of linear
independence for numeric vectors, after which we will develop the notion of linear
independence for vector-valued functions and ultimately describe the general
solution of of y0 − A y = 0.
Numeric Vectors and Linear Independence
A basic definition from linear algebra is that a set of m vectors x1 , x2 , . . . , xm
in Rn is linearly dependent if there are scalars c1 , c2 , . . . , cm , with at least
one ci 6= 0, such that
c1 x1 + c2 x2 + · · · + cm xm = 0,
where 0 represents the zero vector in Rn . When the only such scalars that exist
are all zero, then we say that x1 , x2 , . . . , xm are linearly independent.
8.3. Vector-Valued Functions and Linear Independence
279
Let X be the constant n × m matrix whose j th column is xj ; that is, the ij th
entry of X is xi,j :


x1,1 x1,2
x1,m
 x2,1 x2,2 · · · x2,m 
X = (x1 x2 · · · xm ) = 
.
..
.. 
 ...
.
. 
xn,1
xn,2
xn,m
Note that multiplication of a vector c = (c1 , c2 , . . . , cm )T by the matrix X results
in a linear combination of the columns x1 , x2 , . . . , xm of X:
Xc = c1 x1 + c2 x2 + · · · + cm xm .
Therefore, x1 , x2 , . . . , xm are linearly independent if the equation Xc = 0 has
only the trivial solution c = 0, and they are linearly dependent if Xc = 0 has
at least one nontrivial solution c.
Consider now the case where m = n, in which X is an n × n (square) matrix,
and let the symbol |X| denote the determinant of X. (See Appendix IV.) The
following theorem from linear algebra states the simple connection between |X|
and the linear dependence/independence of the columns of X.
THEOREM 8.3.1. Let X be an n × n matrix. The columns of X are linearly
dependent if |X| = 0, and they are linearly independent if |X| =
6 0.
J
• Example 1a Consider the 3 × 3 matrix
X=
1
3
1
2
1
1
1
2
2
!
.
By a Laplace expansion on the first row (see Appendix IV), the determinant of
X is
1 2
3 2
3 1
|X| = (1) − (2) + (1) 1 2
1 2
1 1
= (1) (1)(2) − (2)(1) − (2) (3)(2) − (2)(1) + (1) (3)(1) − (1)(1)
= 0 − 8 + 2 = −6.
Since |X| =
6 0, we conclude that the columns of X are linearly independent.
• Example 1b Consider the 3 × 3 matrix
X=
1
3
1
2
1
1
3
4
2
!
.
280
Chapter 8. First-Order Linear Systems
By a Laplace expansion on the first row, the determinant of X is
1 4
3 4
3 1
|X| = (1) − (2) + (3) 1 2
1 2
1 1
= (1) (1)(2) − (4)(1) − (2) (3)(2) − (4)(1) + (3) (3)(1) − (1)(1)
= −2 − 4 + 6 = 0,
from which we conclude that the columns of X are linearly dependent. Note
that the third column of X is equal to the sum of the first two.
Vector-valued Functions and Linear Independence
Let y1 , y2 , . . . , ym be vector-valued functions from an interval I into Rn :


y1,j (t)
 y2,j (t) 

yj (t) = 
 ...  , j = 1, 2, . . . , m.
yn,j (t)
We say that these functions are linearly dependent on I if there are scalars
c1 , c2 , . . . , cm , with at least one ci 6= 0, such that
c1 y1 (t) + c2 y2 (t) + · · · + cm ym (t) = 0 for all t in I.
When the only such scalars that exist are all zero, then y1 , y2 , . . . , ym are said
to be linearly independent on I.
It is crucial to understand that the values y1 (t), y2 (t), . . . , ym (t) at certain
t in I may form a linearly dependent set of numeric vectors, while the functions
y1 , y2 , . . . , ym are linearly independent on I.
sin t
cos t
• Example 2 Let y1 (t) =
, y2 (t) =
. Then
cos t
sin t
y1 (π/4) − y2 (π/4) = 0,
which shows that y1 (π/4) and y2 (π/4) are linearly dependent vectors, but it is
easy to see that y1 and y2 are linearly independent functions on I = R, since
there are no scalars c1 and c2 such that c1 y1 (t) + c2 y2 (t) = 0 for all t in R,
other than c1 = c2 = 0.
Now suppose that m = n, and let Y be the n × n matrix-valued function
whose j th column is yj ; that is, the ij th entry of Y is yi,j . Then y1 , y2 , . . . ,
yn are linearly dependent on an interval I, if and only if there is a nonzero,
constant vector c = (c1 , c2 , . . . , cn )T such that
Y (t)c = 0 for all t in I.
From this we can conclude that the following theorem is true.
8.3. Vector-Valued Functions and Linear Independence
281
THEOREM 8.3.2. If y1 , y2 , . . . , yn are linearly dependent functions on an interval I, then |Y (t)| = 0 for all t in I, where |Y (t)| is the determinant of Y (t).
We must be very careful here. The converse of Theorem 2 is false, as the
following example shows.
t
1
• Example 3 Let y1 (t) =
, y2 (t) =
. The determinant of the matrix
t
1
t 1
Y (t) =
t 1
is zero for all t. Let c = (c1 , c2 )T , where c1 , c2 are constants. Then
c1 t + c2
Y (t)c =
,
c1 t + c2
which clearly cannot equal (0, 0)T for all t unless c1 = c2 = 0. So the columns of
Y (t) are linearly independent functions on R, in spite of the fact that |Y (t)| = 0
for all t.
While the converse of Theorem 8.3.2 is false in general, we are about to see
that it is true if y1 , y2 , . . . , yn are solutions of a linear system y0 − Ay = 0 on I.
Solutions of Homogeneous Linear Systems
Let A be an n×n matrix with entries that are continuous functions of t (perhaps
constants) on an interval I.
Definition. Suppose that the vector-valued functions y1 , y2 , . . . , yn each satisfy
the homogeneous differential equation
y0 − A y = 0
on an interval I, and let Y (t) be the matrix whose columns are y1 , y2 , . . . , yn .
Such a matrix is said to be a solution matrix for the equation y0 − A y = 0
on I, or more simply, a solution matrix generated by A.
Suppose that Y (t) is a solution matrix generated by A and that |Y (t0 )| = 0,
where t0 is some number in I. By Theorem 1, since |Y (t0 )| = 0, there are
numbers c1 , c2 , . . . , cn , with at least one ci 6= 0, such that
c1 y1 (t0 ) + c2 y2 (t0 ) + · · · + cn yn (t0 ) = 0.
Given this particular collection of numbers c1 , c2 , . . . , cn , let
yc (t) = c1 y1 (t) + c2 y2 (t) + · · · + cn yn (t)
for all t in I. Because of linearity, yc satisfies the differential equation. Therefore,
since yc (t0 ) = 0, uniqueness of solutions implies that yc (t) = 0 for all t in I.
282
Chapter 8. First-Order Linear Systems
Thus y1 , y2 , . . . , yn are linearly dependent, and so |Y (t)| = 0 for all t in I. This
argument is the basis of the proof of the following theorem.
THEOREM 8.3.3. Let A be an n × n matrix with entries that are continuous
functions of t on I, and let Y (t) be a solution matrix generated by A. Then the
following statements are true.
i) Either |Y (t)| = 0 for all t in I or else |Y (t)| =
6 0 for all t in I.
ii) The columns y1 , y2 , . . . , yn of Y (t) are linearly independent on I if and
only if |Y (t0 )| =
6 0 for some t0 in I.
iii) The columns y1 , y2 , . . . , yn of Y (t) are linearly independent on I if
and only if the numeric vectors y1 (t0 ), y2 (t0 ), . . . , yn (t0 ) are linearly
independent for some t0 in I.
J
• Example 4 The functions
y1 (t) =
et
3e t
and y2 (t) =
e−t
2e−t
satisfy the differential equation
0
y −
−5 2
−12 5
y = 0,
and the entries of the matrix are continuous on any interval. Note that
−t et
e
= −1,
|Y (t)| = t
3e 2e−t which is nonzero for all t, thus illustrating part (i) of Theorem 8.3.3. Part (ii)
tells us that y1 and y2 are linearly independent on any interval. By part (iii),
we could have arrived at the same conclusion from the linear independence of
the vectors y1 (0) = (1, 3)T and y2 (0) = (1, 2)T , for example.
• Example 5 Consider the differential equation
1
t −1
0
y = 0,
y − 2
t − 1 −1 t
in which the entries of the matrix are continuous on any interval that does not
contain t = ±1. It is easy to check that the functions
t
1
y1 (t) =
and y2 (t) =
1
t
satisfy the differential equation on any interval that does not contain t = ±1.
Now notice that
t 1
= t2 − 1,
|Y (t)| = 1 t
8.3. Vector-Valued Functions and Linear Independence
283
and so |Y (t)| = 0 at t = ±1 and is nonzero elsewhere. This is consistent with
part (i) of the Theorem 3, because y1 and y2 are solutions of the differential
equation only on intervals that do not contain t = ±1.
A simple consequence of linearity is that, if y1 , y2 , . . . , yn each satisfy the
differential equation y0 − A y = 0 on an interval I, then any linear combination
of y1 , y2 , . . . , yn is also a solution. It turns out, moreover, that every solution
must be of that form, provided that y1 , y2 , . . . , yn are linearly independent on I.
To see why, first suppose that y1 , y2 , . . . , yn are linearly independent solutions
of y0 − A y = 0 on I and that ỹ is any solution whatsoever of y0 − A y = 0 on I.
Now pick any number t0 in I. We know that |Y (t0 )| 6= 0; so by Theorem 8.3.1
there are scalars c1 , c2 , . . . , cm , with at least one ci 6= 0, such that
c1 y1 (t0 ) + c2 y2 (t0 ) + · · · + cm yn (t0 ) = ỹ(t0 ).
With this particular choice of c1 , c2 , . . . , cm , let
yc = c1 y1 + c2 y2 + · · · + cm yn .
Then yc has the same value as ỹ at t0 and satisfies the same differential equation.
Therefore, uniqueness of solutions implies that yc (t) = ỹ(t) for all t in I, which
tells us that ỹ is a linear combination of y1 , y2 , . . . , yn .
We summarize the result of this argument in the following theorem.
THEOREM 8.3.4. Suppose that the n × n matrix A has entries that are continuous functions on an interval I and that y1 , y2 , . . . , yn are linearly independent
solutions of y0 − A y = 0 on I. Then every solution of y0 − A y = 0 on I can
be expressed as a linear combination of y1 , y2 , . . . , yn ; that is, every solution
is described by the general solution
y = c1 y1 + c2 y2 + · · · + cn yn .
Stated in terms of the solution matrix Y (t) = (y1 y2 · · · yn ), every solution
of y0 − A y = 0 on I is of the form y = Y (t)c for some constant vector
c = (c1 , c2 , . . . , cn )T .
J
Problems
1. Prove Theorem 1 for 2 × 2 matrices X =
p
k
h
.
q
Problems 2 through 4: Compute the determinant in order to determine
columns of the matrix are linearly dependent or linearly independent.





0
1 1
1 2 0
2
1
0
−2
1
1

2.  −1 3 1 
3.  −1 3
7 
4. 
1 −1 0
0 2 2
2 −1 −4
1
0 2
whether the

1
0 

2
−2
284
Chapter 8. First-Order Linear Systems
Problems 5 through 7: Compute each product as a linear combination of the columns
of the matrix (rather than using dot products).




 
2
1 −1
3
5 1 3
0
1 2
3
5.
6.  −1 3
1  −1 
7.  2 −3 1  1 
−1 3
2
2 −1 3
2
3 −1 2
0
t
1
8. Consider the functions u and v defined on R by u(t) =
and v(t) =
.
3
t+4
(a) Show that u and v are linearly independent functions on R.
(b) Find all t in R at which the numeric vectors u(t) and v(t) are linearly dependent.
9. Consider the functions u, v, and w defined on R by






t
0
−t
u(t) =  e−t  , v(t) =  e−t  , w(t) =  e−t  .
2 et
et
et
Use Theorem 2 to show that u, v, w are linearly independent on every interval I.
In each of Problems 10 through 13,
(a) verify the statement;
(b) explain the choice of the interval I;
(c) verify statement (i) in Theorem 3, and then use statement (ii) to conclude that
the columns of Y (t) are linearly independent on I;
(d) write the general solution on I in the form y = c1 y1 + c2 y2 .
t t−1
is a solution matrix on I = (0, 1) for the equation
10. Y (t) =
t t−2
1
t + 1 −2t
y = 0.
y0 −
3
−t − 2
t (1 − t)
11. Y (t) =
sin t cos t
cos t sin t
is a solution matrix on I = (− π4 , π4 ) for the equation
− tan 2t sec 2t
y −
sec 2t − tan 2t
0
12. Y (t) = tan t
1
sec t
sec t
1
y + sec t csc t
13. Y (t) =
e−t
2 e−t
3 e−2t
e−2t
y = 0.
is a solution matrix on I = (0, π2 ) for the equation
0
−2 cos t
cos t −2
y = 0.
is a solution matrix on I = R for the equation
y0 −
1
5
−11 3
−2 −4
y = 0.
8.4. Homogeneous Autonomous Systems: The Matrix Exponential
285
14. Use either the method of elimination or Laplace transforms to find a pair of linearly
independent solutions on R of the equation
−3 1
0
y −
y = 0.
2 −2
Then form the corresponding solution matrix Y (t) and verify that |Y (t)| =
6 0 on R.
−t
te
t
15. Using Theorem 8.3.3, explain why Y (t) =
cannot be a solution matrix
e−t e−t
0
for y − A y = 0 on I if I contains 0.
t/2
e
2 e−t
16. Find the matrix A, if Y (t) =
is a solution matrix generated by A.
2 et/2 −e−t
17. Suppose that u and v are solutions of the second-order equation x00 +p x0 +q x = 0 and
that we convert this equation to a first-order system y0 −A y = 0, where y = (x, x0 )T .
(a) What solution matrix Y (t) generated by A corresponds to u and v?
(b) Show that |Y (t)| = W (t), where W (t) is the Wronskian of u and v. (See §6.?.)
8.4 Homogeneous Autonomous Systems and eAt
We will assume throughout this section that the equation
y0 − A y = 0
(1)
is autonomous; that is, the entries of A are constants. Thus we may assume that
I = R, and therefore we need make no further qualifying statements regarding I.
Definition. Suppose that Y (t) is a solution matrix generated by A. If the
columns of Y (t) are linearly independent on R, then Y (t) is said to be a fundamental matrix generated by A.
Note that, by Theorem 8.3.3, a solution matrix Y (t) is a fundamental matrix
generated by A if and only if |Y (t0 )| =
6 0 for some t0 .
The following theorem states three important properties of any fundamental
matrix. Part (i) follows from the fact that the columns of Y (t) satisfy (1), and
parts (ii) and (iii) follow from Theorem 8.3.4 and Theorem 8.1.1.
THEOREM 8.4.1. Let Y (t) be a fundamental matrix generated by A. Then
i) Y 0 (t) − A Y (t) = 0 for all t;
ii) Every solution of (1) has the form Y (t)c for some constant vector c in Rn ;
iii) For any y0 in Rn , y(t) = Y (t)Y (0)−1 y0 is the unique solution of (1) that
satisfies y(0) = y0 .
J
286
Chapter 8. First-Order Linear Systems
It is clear that the matrix Y (t)Y (0)−1 in part (iii) of Theorem 8.4.1 plays
a very special role. It is itself a fundamental matrix, and it is the fundamental
matrix that “maps” the initial vector y0 onto the solution at time t. Moreover,
it is independent of the choice of Y (t). Because of properties that we will list
below, this matrix is called the matrix exponential generated by A.
Definition. Given any fundamental matrix generated by a square matrix A
with constant entries , the matrix exponential eAt is defined by
eAt = Y (t)Y (0)−1 .
• Example 1 Consider the differential equation
1
y0 −
2
−3
1
1
−1
−5
1
2
2
−4
!
y = 0.
It turns out that a simple fundamental matrix is
Y (t) =
e−t
e−t
e−t
−e−2t
e−2t
e−2t
e−3t
e−3t
−e−3t
!
,
as is straightforward to verify. Consequently, the the matrix exponential is
!
!
e−t −e−2t e−3t
1 −1 1 −1
At
−1
−t
−2t
−3t
e = Y (t)Y (0) = e
e
e
1 1
1
e−t e−2t −e−3t
1 1 −1
!
!
1 0 1
e−t −e−2t e−3t
1
−1 1 0
= e−t e−2t
e−3t
−t
−2t
0 1 −1
e
e
−e−3t 2
!
−2t
−t
−3t
−2t
e
+e
e
−e
−e−3t + e−t
1
=
−e−2t + e−t e−3t + e−2t −e−3t + e−t .
2 −e−2t + e−t −e−3t + e−2t e−3t + e−t
Part (iii) of Theorem 8.4.1 may be restated in terms of eAt as
iii 0 ) For any y0 in Rn , y(t) = eAt y0 is the unique solution of (1) that satisfies
y(0) = y0 .
Note that this is a natural extension of the fact that y(t) = y0 eat is the unique
solution of the “scalar equation” y 0 − ay = 0 that satisfies y(0) = y0 .
The following corollary represents a somewhat different way of thinking about
eAt . Not only does it map the solution from time 0 to time t, it also maps the
solution from any time s to time s + t.
COROLLARY 8.4.1. For any solution y of (1) and any s and t,
y(s + t) = eAt y(s) = eAs y(t) = eA(s+t) y(0).
J
8.4. Homogeneous Autonomous Systems: The Matrix Exponential
287
Proof. We need only show that eAt y(s) = eA(s+t) y(0), since we already know
that y(s+t) = eA(s+t) y(0). Once that is shown, it will follow by simply switching
s and t that eAs y(t) = eA(s+t) y(0). So let u(t) = eAt y(s). We will prove that
u(t) = y(t + s). We know that, for all t,
u0 (t) − Au(t) = 0, u(0) = y(s).
Now, because the entries of A are constant, it follows that
u0 (t − s) − Au(t − s) = 0,
and so, if we set w(t) = u(t − s), we have
w0 (t) − Aw(t) = 0 and w(s) = u(0) = y(s),
which tells us that w and y satisfy the same initial-value problem. Thus, by
uniqueness, w(t) = u(t − s) = y(t) for all t, and now by replacing t with t + s,
we have u(t) = y(t + s) as desired.
The next theorem lists properties of the matrix exponential that are interestingly analogous to properties of the exponential function e at .
THEOREM 8.4.2. For any matrix A with constant entries, the matrix exponential eAt has the following properties:
i) eA0 = I ;
ii) eA(s+t) = eAt eAs = eAs eAt for all t and s;
iii) (eAt )−1 = e−At for all t;
iv)
v)
d At
=
dt e
At
e =I
A eAt for all t;
+ tA +
1 2 2
2! t A
+
1 3 3
3! t A
+ · · · for all t.
J
Parts (i) and (iv) here follow directly from the definition of eAt and Theorem
8.4.1. Part (ii) is essentially the same as Corollary 8.4.1. Part (iii) follows from
(i) and (ii) with s = −t. Part (v) may be verified by showing that (iv) and (i)
are true for the given series.
It is possible to use property (v) of Theorem 8.4.2 to compute eAt . When
A has the property that An = 0 for some n, this is an especially convenient
method. (Such matrices are called idempotent.)
• Example 2 Consider the matrix
A=
−1
−1
1
−1
0
1
−1
−1
1
!
.
288
Chapter 8. First-Order Linear Systems
Straightforward computation reveals that
!
1 0 1
2
A =
0 0 0
and A3 =
−1 0 −1
0
0
0
0
0
0
0
0
0
!
.
Consequently all higher powers of A are also zero, and thus
eAt
1
= I + t A + t2 A2 =
2!
1 − t + t2 /2 −t −t + t2 /2
−t
1
−t
t − t2 /2
t 1 + t − t2 /2
!
.
Eigenvalues and Eigenvectors
Let us begin by recalling the “scalar case.” If a is a constant, then every solution
of y 0 − ay = 0 is of the form y = c e at , where c is a constant. So we will look
for solutions of the system y0 − A y = 0 in the form y = eλt p, where λ is a
scalar constant and p is a nonzero constant vector. Substituting y = eλt p into
y0 − A y = 0 results in
λ eλt p − eλt A p = 0,
which, after we divide by eλt and rearrange, becomes
A p = λ p.
This leads us to the following definition. (See also Appendix A.)
Definition. Let A be an n × n with constant entries. A complex (possibly real)
scalar λ is an eigenvalue of A if there is a nonzero vector p such that A p = λ p.
Such a vector p is called an eigenvector of A corresponding to the eigenvalue
λ. An eigenvalue λ and a corresponding eigenvector p are collectively called an
eigenpair (λ, p) of A.
So any eigenpair (λ, p) of A provides us with a nontrivial solution eλt p of
− A y = 0. An important issue now is whether we can get enough solutions
in this way to construct a fundamental matrix.
We point out that any nonzero scalar multiple of an eigenvector is still an
eigenvector. More precisely, given any eigenpair (λ, p) and any scalar constant
c 6= 0, (λ, c p) is also an eigenpair of A. This is due to the fact that if A p = λ p,
then A c p = λ c p. Thus, eigenpairs (λ, p) and (λ, cp) are considered equivalent.
Also, it is possible for a given eigenvalue to have two (or more) linearly independent corresponding eigenvectors, resulting in eigenpairs (λ, p1 ) and (λ, p2 )
in which p1 and p2 are linearly independent.
The following theorem, which we state for reference here without proof, describes one important circumstance under which eAt can be constructed from
the eigenpairs of A.
y0
8.4. Homogeneous Autonomous Systems: The Matrix Exponential
289
THEOREM 8.4.5. Let A be an n×n matrix with constant entries. If (λi , pi ) and
(λj , pj ) are eigenpairs with λi 6= λj , then pi and pj are linearly independent.
Moreover, any set of eigenvectors corresponding to distinct eigenvalues of A is
linearly independent. In particular, if A has n distinct eigenvalues, then A has
n linearly independent eigenvectors.
J
It follows directly from Theorem 8.4.5 that, given a collection of n distinct
eigenvalues of A, the corresponding eigenvectors will be linearly independent,
and so the corresponding solutions eλt p of y0 − A y = 0 make up the columns
a fundamental matrix Y (t), from which we can then compute the matrix exponential as e At = Y (t)Y (0)−1 .
So how can we compute eigenpairs? Since λ is an eigenvalue of A if and only
if the equation (A − λ I )x = 0 has a nonzero solution, it follows that λ is an
eigenvalue of A if and only if |A − λ I | = 0. The determinant |A − λ I | is a
polynomial in the variable λ and is called the characteristic polynomial of
A. Its roots are precisely the eigenvalues of A. Once an eigenvalue λ of A is
known, any nonzero solution of (A − λ I )x = 0 is a corresponding eigenvector.
1 −2
• Example 3 The characteristic polynomial of A =
is
1 4
1 − λ
−2
= (1 − λ)(4 − λ) + 2
|A − λ I | = 1
4 − λ
= λ2 − 5λ + 6 = (λ − 2)(λ − 3).
Therefore, the eigenvalues of A are λ1 = 2 and λ2 = 3. To find an eigenvector
corresponding to λ1 = 2, we first form the matrix
−1 −2
A − λ1 I =
1
2
and then simply observe ∗ that (2, −1)T is a solution of (A − λ1 I )x = 0. To find
an eigenvector corresponding to λ2 = 3, we form the matrix
−2 −2
A − λ2 I =
1
1
and then observe that (1, −1)T is a solution of (A − λ2 I )x = 0. Thus we have
found eigenpairs
2
1
.
2,
and
3,
−1
−1
* Note that we are merely looking for some x1 and x2 so that x1 + 2x2 = 0.
290
Chapter 8. First-Order Linear Systems
We now have the following linearly independent solutions of y0 − A y = 0:
2
1
2t
3t
y1 = e
and y2 = e
,
−1
−1
from which it follows that
Y (t) =
2 e 2t
−e 2t
e 3t
−e 3t
is a fundamental matrix generated by A. The matrix exponential is therefore
−1
2 e 2t e 3t
2
1
e At = Y (t)Y (0)−1 =
−e 2t −e 3t
−1 −1
2t
3t
1 1 −1
2e
e
=
−e 2t −e 3t 3 1 2
2 e 2t − e 3t 2 e 2t − 2 e 3t
=
.
−e 2t + e 3t −e 2t + 2e 3t
Complex Eigenvalues
A matrix with real entries can have nonreal complex eigenvalues. An eigenvector
corresponding to a nonreal complex eigenvalue must have at least one nonreal
complex entry. Since the eigenvalues of A are the roots of a polynomial, nonreal complex eigenvalues must occur as conjugate pairs. It also turns out that
the corresponding eigenvectors will be conjugate pairs as well; in particular, if
(α + β i, r + i q ) is an eigenpair, then so is (α − β i, r − i q ).
The following theorem summarizes the relationship between complex eigenpairs of A and solutions of y0 − A y = 0.
THEOREM 8.4.6. Let A be an n × n matrix with constant entries, and suppose
that (λ, p) is an eigenpair of A. Then eλt p is a solution of y0 − A y = 0. A complex conjugate pair of eigenvalues λ± = α ± β i with corresponding eigenvectors
p± = r ± i q gives rise to a pair of linearly independent real solutions:
e αt (cos β t r − sin β t q ) and e αt (sin β t r + cos β t q ).
(2)
These are the real and imaginary parts, respectively, of the complex-valued
solution eλ+ p+ = e(α+β i)t (r + i q).
J
Remark. The solutions given in Theorem 8.4.6 are the columns of an n×2 matrix
eαt P R(t), where P = (p q) and R(t) is a standard 2 × 2 “rotation” matrix:
cos β t sin β t
R(t) =
.
− sin β t cos β t
8.4. Homogeneous Autonomous Systems: The Matrix Exponential
• Example 4 The characteristic polynomial of A =
−2 − λ
|A − λ I | = −3
−2
−3
6
4
291
is
6 = (−2 − λ)(4 − λ) + 18 = λ2 − 2λ + 10.
4 − λ
By the quadratic formula, the roots of |A − λ I | are the complex conjugate pair
λ1 = 1 + 3i, λ2 = 1 − 3i.
To find an eigenvector corresponding to λ1 = 1 + 3i, we first form the matrix
−3 − 3i
6
−1 − i
2
A − λ1 I =
=3
−3
3 − 3i
−1
1−i
and then observe (because of the entries in the second row of the matrix) that
(1 − i, 1)T is a solution of (A − λ1 I ) x = 0. Thus we have the eigenpair
1−i
1
−1
1 + 3i,
= 1 + 3i,
+i
,
1
1
0
from which we use (2) to obtain the following pair of linearly independent solutions of y0 − A y = 0:
1
−1
cos 3t + sin 3t
t
t
e cos 3t
− sin 3t
=e
1
0
cos 3t
and
e
t
1
−1
sin 3t − cos 3t
t
sin 3t
+ cos 3t
=e
.
1
0
sin 3t
These produce the solution matrix
cos 3t + sin 3t sin 3t − cos 3t
t
Y (t) = e
,
cos 3t
sin 3t
from which we compute the matrix exponential as follows:
−1
cos 3t + sin 3t sin 3t − cos 3t
1 −1
e At = e t
cos 3t
sin 3t
1 0
cos 3t + sin 3t sin 3t − cos 3t
0 1
t
=e
cos 3t
sin 3t
−1 1
cos 3t − sin 3t
2 sin 3t
= et
.
− sin 3t
cos 3t + sin 3t
Deficient Eigenvalues and Generalized Eigenvectors
It should be clear by now that if an n × n matrix A with constant entries has
n linearly independent eigenvectors, then once the eigenvectors are computed,
it is a relatively simple matter to form e At . According to Theorem 3, this is
292
Chapter 8. First-Order Linear Systems
the case any time A has n distinct eigenvalues. It is also possible, though not
guaranteed, when A does not have n distinct eigenvalues. Our purpose now is
to understand the solutions of y0 − A y = 0 when A fails to have n linearly
independent eigenvectors.
Definition. An eigenvalue λi of A is said to have multiplicity m if λi is a root
of multiplicity m of the characteristic polynomial |A − λ I |—that is, (λ − λi )m
is a factor of |A − λ I | while (λ − λi )m+1 is not.
According to the fundamental theorem of algebra, the sum of the multiplicities of all of the eigenvalues of A is n. If A fails to have n distinct eigenvalues,
it is because one (or more) of its eigenvalues has multiplicity greater than 1.
Often an eigenvalue with multiplicity m > 1 will have m corresponding linearly
independent eigenvectors. If that is the case for all “repeated” eigenvalues of
A, then A has n linearly independent eigenvectors. So our concern lies with
the repeated eigenvalues of A for which the number of corresponding linearly
independent eigenvectors is less than the eigenvalue’s multiplicity.
Definition. If λi is an eigenvalue of A with multiplicity m and A has only
m − k linearly independent eigenvectors corresponding to λi , then λi is said to
be deficient with a deficiency of order k.
Though its proof requires a deeper foray into linear algebra than is appropriate here, we state the following theorem that describes the generalized eigenvectors of A and the associated solutions of y0 − A y = 0. (See also Problem
31.)
THEOREM 8.4.7. Suppose that λ is a deficient eigenvalue of A with multiplicity m and deficiency k. Then A has m linearly independent generalized
eigenvectors corresponding to λ. These include the m − k eigenvectors of A
corresponding to λ, plus k additional vectors v1 , v2 , . . . , vk that satisfy
(A − λ I )j+1 vj = 0, (A − λ I )j vj 6= 0
for j = 1, . . . , k. Moreover, each vj produces a solution of y0 − A y = 0 in the
form
tj
t2
2
j
λi t
(3)
y=e
I + t (A − λ I ) + (A − λ I ) + · · · + (A − λ I ) vj .
2!
j!
• Example 5 Consider the matrix
A=
−2
1
−1
1
−2
−5
−1
1
1
!
,
8.4. Homogeneous Autonomous Systems: The Matrix Exponential
293
whose characteristic polynomial is
−2 − λ
1
−1 |A − λ I | = 1
−2 − λ
1 −1
−5
1 − λ
= (−2 − λ)((−2 − λ)(1 − λ) + 5) − ((1 − λ) + 1) − (−5 − (2 + λ))
= −1 − 3 λ − 3 λ2 − λ3 = −(λ + 1)3 .
Thus A has only one eigenvalue, λ = −1, which has multiplicity m = 3. To
determine corresponding eigenvectors, we form the matrix
!
−1 1 −1
A+I =
1 −1 1
,
−1 −5 2
which turns out to have the row-echelon form:
!
1 0 1/2
0 1 −1/2 .
0 0
0
Therefore, p = (−1, 1, 2)T is an eigenvector corresponding to λ = −1. All other
eigenvectors are simply a scalar multiples of p, so there can be no other eigenvector linearly independent from p. This tells us that λ has a deficiency of order
k = 2; so we need to find two generalized eigenvectors. We begin by computing
!
3
3 0
(A + I )2 = −3 −3 0 .
−6 −6 0
By simple observation, a solution of (A+I )2 x = 0 is the vector v1 = (1, −1, 0)T ,
and it is easily verified that (A + I )v1 6= 0. Now after computing
!
1−t
t
−t
I + t (A + I ) =
t
1−t
t
,
−t
−5t 1 + 2t
we find the solution
e
−t
(I + t (A + I )) v1 = e
−t
1 − 2t
2t − 1
4t
!
.
Computation now shows that (A + I )3 = 0. Consequently, any vector x for
which (A + I )2 x 6= 0 may serve as v2 . So we make the simple choice v2 =
(1, 0, 0)T . We now compute the matrix


1 − t + 32 t2
t + 32 t2
−t
2
t
I + t (A + I ) + (A + I )2 =  t − 32 t2
1 − t − 32 t2
t 
2
2
2
−t − 3t
−5t − 3t
1 + 2t
294
Chapter 8. First-Order Linear Systems
followed by the solution
e−t
I + t (A + I ) +
t2
2
(A + I )2

1 − t + 32 t2
v2 = e−t  t − 32 t2  .
−t − 3t2

We now have three linearly independent solutions of y0 − A y = 0 :


!
!
1 − t + 32 t2
−1
1 − 2t
e−t
1
, e−t 2t − 1 , e−t  t − 32 t2  .
2
4t
−t − 3t2
The resulting fundamental matrix generated by A is


−1 1 − 2t 1 − t + 23 t2
Y (t) = e−t  1 2t − 1
t − 32 t2  ,
2
4t
−t − 3t2
which produces the matrix exponential as

−1 1 − 2t
e At = Y (t)Y (0)−1 = e−t  1 2t − 1
2
4t

−1 1 − 2t
= e−t  1 2t − 1
2
4t

1 − t + 23 t2
−t 
=e
t − 23 t2
−t − 3t2
follows:

!
1 − t + 23 t2
−1 1 1 −1
1 −1 0
t − 32 t2 
2
2
0 0
−t − 3t

!
1 − t + 23 t2
0 0 1
1
0 −2 1
t − 32 t2 
2 2 2 0
2
−t − 3t

t + 32 t2
−t
1 − t − 32 t2
t .
2
−5 t − 3t
1 + 2t
Problems
1. Assume that parts (i) and (ii) of Theorem 8.4.2 are true, and prove part (iii) by
showing that eAt eA(−t) = eA(−t) eAt = I . (Hint: Use s = −t in part (ii).)
2. (abc) For each of the matrix exponentials found in Examples 1, 2, and 3, verify parts
(i) through (iv) of Theorem 8.4.2.
−3 9
3. Let A =
. Show that A2 = 0, and use (v) in Theorem 8.4.2 to find e At .
−1 3


0 1 0
4. Let A =  2 0 1 . Show that A3 = 0, and use (v) in Theorem 8.4.2 to find e At .
0 −2 0
In Problems 5 through 8, assume that A satisfies the given equation. Compute e At in
terms of A by means of (v) in Theorem 8.4.2.
5. A2 = A
6. A2 = −A
7. A2 = −I
8. A3 = A
8.4. Homogeneous Autonomous Systems: The Matrix Exponential
9. Find e At given A =
295
0 1
. What trigonometric identities follow from part (ii)
−1 0
of Theorem 8.4.2?
10. The trace of a square matrix A, written
Pn tr(A), is defined to be the sum of the
diagonal entries of A; that is, tr(A) = i=1 aii .
(a) Show that the characteristic polynomial of a 2 × 2 matrix A is
|A − λ I | = λ2 − tr(A)λ + |A| .
(b) Show that the characteristic polynomial of a 3 × 3 matrix A is
|A − λ I | = −λ3 + tr(A)λ2 − c1 λ + |A| ,
where
a
c1 = 11
a21
a12 a11
+
a22 a31
a13 a22
+
a33 a32
a23 .
a33 In Problems 11 through 17,
(a) find the eigenvalues of the given matrix A, using one of the results of Problem 10
to write down the characteristic polynomial;
(b) find a fundamental matrix generated by A;
(c) compute e At .
−1 3
−1 −1
1
3
−2 1
11.
12.
13.
14.
1 1
5
1
−2 −4
1 −2






−3 −1 −1
1
2 −2
−4 2 2
17.  −3 1 2 
15.  −2 −2 2 
16.  −2 −3 2 
−1 1 −3
−2 −2 1
−2 1 0
Compute
−1
18.
2
−4
21.
4

−1
24.  1
−1
e At for each of the matrices A in Problems 18 through 26.
2
−1 2
0
19.
20.
−1
2 −4
−1
2 −9
−10
2 −2 − π2
22.
23.
−1
−4
2
−2




0 1
0 −1 1
0
0 −1 
25.  1 −1 0 
26.  −2
1 1
0 −1 0
0
3
−4
33
4
−1
0
−2

0
−1 
0
In Problems 27 through 31, compute e At for the given matrix A. In each case, A has at
least one deficient eigenvalue.
9
8 −9
5 − 16
27.
28.
16 −16
36 −4






0
1 −1 0
−2 −1 1
−2 2 2
0
1 
 −1 1
30.  2 1 1 
31. 
29.  −1 2 −1 

−1 1
0
0
−2 6 −3
−4 1 1
0 −1 1 −2
296
Chapter 8. First-Order Linear Systems
32. This problem provides a partial proof of Theorem 8.4.7. Suppose that (λ, p) is an
eigenpair of A.
a) Show that, if v satisfies (A − λ I )2 v = 0, then y = eλt (I + t (A − λ I )) v is a
solution of y0 − A y = 0.
b) Show that, if v satisfies (A − λ I )3 v = 0, then
1
y = eλt I + t (A − λ I ) + t2 (A − λ I )2 v
2
is a solution of y0 − A y = 0.
33. This problem describes a technique for generating the generalized eigenvectors described in Theorem 8.4.7. It is a technique that avoids the computation of the matrix
powers (A−λ I )j . Let (λ, p) be an eigenpair of A in which λ is a deficient eigenvalue,
and suppose that v1 , v2 , v3 , . . . , vk are nonzero vectors satisfying
(A − λ I )v1 = p and (A − λ I )vj = vj−1 for j = 2, 3, . . . , k.
a) Show that, for j = 1, . . . , k, vj satisfies
(A − λ I )j vj 6= 0, (A − λ I )j+1 vj = 0.
b) Show that the solution produced by vj (see Theorem 8.4.7) can be written as
j
t
t j−1
λt
y=e
p+
v1 + · · · + t vj−1 + vj .
j!
(j − 1)!
Each of the matrices A in Problems 34 through 36 has only a single eigenvalue with multiplicity 3 and an order 2 deficiency. Use the method of Problem 35 to find a fundamental
matrix generated by A.






1
6
3
−2 2
1
5 −1 −3
36.  2 −2 2 
35.  2 −1 −2 
34.  −2 0
2 
−3 −6 −5
−1 2
0
1 −3 1
In Problems 37 through 40, you are given a matrix A along with its characteristic
polynomial. Find a fundamental matrix generated by A.




0 0 −2 3
1 −3 3 −4
 2 1 1 −2 
 −2 0 −1 2 
37. 
38. 
 ; (λ − 1)4
 ; λ (λ + 1)3
−1 0 0
2
0 −1 0 −1
−1 0 −1 3
2 −2 2 −4




−1 1 −1 1
1 −1 −1 −1
0
0 −1 
 −1 1 −2 1 
1
39. 
40. 
 ; λ2 (λ + 1)2
 ; (λ2 + 1)2
−1 0 −1 0
0 −1 0
1
0 −1 1 −1
1
0 −1 −1
8.5. Homogeneous Autonomous Systems: The Matrix Exponential
297
8.5 Nonhomogeneous Autonomous Systems
We now consider the nonhomogeneous system
y0 − A y = f ,
(1)
where A has constant entries and f is a continuous, vector-valued function on
R. Let Y (t) be any fundamental matrix generated by A—possibly but not necessarily eAt .
Since every solution of the associated homogeneous system is of the form
y = Y (t)c, where c is a constant vector, we will look at a particular solution ỹ in
the form ỹ = Y (t)u(t), where u(t) is a vector-valued function to be determined.
Recall that such a technique is called variation of constants.
Substitution of ỹ = Y (t)u(t) into (1) produces
Y 0 u + Y u0 − A Y u = f
by the product rule for matrix-vector products. Since Y 0 = A Y , this becomes
Y u0 = f .
Since |Y (t)| =
6 0 for all t, Y (t) is invertible for all t; so we can express u0 as
0
−1
u = Y f . Thus integration over [0, t] results in
Z t
u(t) =
Y (s)−1 f (s) ds.
0
So a particular solution of (1) is given by
Z t
ỹ(t) = Y (t)
Y (s)−1 f (s) ds.
(2)
0
Note that this particular solution satisfies ỹ(0) = 0.
Now consider the nonhomogeneous initial-value problem
y0 − A y = f , y(0) = y0 ,
(3)
under the same assumptions on A and f as stated previously. The solution
satisfying y(t0 ) = 0 is given by (2). Thanks to linearity, we can superimpose ỹ
with the solution Y (t)Y (0)−1 y0 of the corresponding homogeneous problem to
arrive at
Z t
−1
−1
y(t) = Y (t) Y (0) y0 +
Y (s) f (s) ds .
(4)
0
If we take the fundamental matrix Y (t) to be the matrix exponential eAt , this
becomes the simpler formula
Z t
At
−As
y(t) = e
y0 +
e
f (s) ds .
0
298
Chapter 8. First-Order Linear Systems
Since eAt e−As = eA(t−s) by Theorem 8.4.2, this becomes
Z t
y(t) = eAt y0 +
eA(t−s) f (s) ds.
(5)
0
We will refer to (4) and (5) as variation of constants formulas.
• Example 1 Let’s solve the initial-value problem
−2 1
0
0
y −
y=
, y(0) = 0.
1 −2
e−2t
We begin by computing eAt . The eigenvalues of the coefficient matrix are the
roots of
−2 − λ
1 = (2 + λ)2 − 1 = λ2 + 4λ + 3 = (λ + 1)(λ + 3),
1
−2 − λ which are λ = −1, −3. Corresponding eigenvectors turn out to be, respectively,
1
1
and
.
1
−1
So a fundamental matrix is
Y (t) =
e−t
e−t
e−3t
−e−3t
.
Therefore the matrix exponential is
−1
e−t e−3t
1 1
At
−1
e = Y (t)Y (0) =
e−t −e−3t
1 −1
1
−1 −1
e−t e−3t
=
e−t −e−3t −2 −1 1
1 e−t + e−3t e−t − e−3t
=
.
−t
−3t e−t + e−3t
2 e −e
So the vector function we need to integrate is
1 e−(t−s) + e−3(t−s) e−(t−s) − e−3(t−s)
0
A(t−s)
e
f (s) =
e−2s
2 e−(t−s) − e−3(t−s) e−(t−s) + e−3(t−s)
1 e−(t+s) − e−3t+s
=
,
2 e−(t+s) + e−3t+s
and so the solution of the initial-value problem is
Z 1 t e−(t+s) − e−3t+s
1 −e−(t+s) − e−3t+s t
y(t) =
ds =
2 0 e−(t+s) + e−3t+s
2 −e−(t+s) + e−3t+s 0
1 −2e−2t + e−t + e−3t
.
=
e−t − e−3t
2
8.5. Homogeneous Autonomous Systems: The Matrix Exponential
299
• Example 2 Let’s solve the following initial-value problem, in which the coefficient matrix is the same as in Example 1 of Section 8.4:
!
!
t e−t
1 −3 −1 2
0
y −
1 −5 2
y=
0
, y(0) = 0.
2
1
1 −4
0
In this case, we will work with the fundamental matrix
!
e−t −e−2t e−3t
−t
−2t
−3t
Y (t) = e
e
e
e−t e−2t −e−3t
and use formula (4), rather computing the more complicated matrix exponential
needed for (5). A (computer-aided) calculation reveals that
!
0
et
et
1
−1
2t
2t
−e
e
0
.
Y (t) =
2
0
e 3t −e 3t
The vector function we need to integrate is then
!
!
es
0
es
s e−s
1
1
−1
2
s
2
s
Y (s) f (s) =
−e
e
0
0
=
2
2
3
s
3
s
0
e
−e
0
s
−s e−s
0
!
.
The resulting integral is
t
Z
−1
Y (s)
0
1
f (s) ds =
2
Z
0
t
s
−s e−s
0
!
1
ds =
4
t2
2(1 − t)e t − 2
0
!
.
So the solution we seek is
Z t
Y (t)
Y (s)−1 f (s) ds =
0
!
!
t2
e−t −e−2t e−3t
1
e−t e−2t
e−3t
2(1 − t)e t − 2
4
−t
−2t
−3t
0
e
e
−e
!
−2t
2
−t
2e
+ (t + 2t − 2)e
1
=
−2e−2t + (t2 − 2t + 2)e−t .
4 −2e−2t + (t2 − 2t + 2)e−t
Problems
1. Use the variation of constants formula (5) to solve the following initial-value problem,
in which the coefficient matrix is the same as in Problem 3 of §8.4:
−3 9
t
0
y0 −
y=
, y(0) =
.
−1 3
1
1
300
Chapter 8. First-Order Linear Systems
2. Use the variation of constants formula (5) to solve the following initial-value problem,
in which the coefficient matrix is the same as in Problem 4 of §8.4:


 
0 1 0
0
y0 −  2 0 1  y =  1  , y(0) = 0.
0
0 −2 0
3. Let A be the coefficient matrix in Example 1. Use the variation of constants formula
(5) to find the solution of
0
1
0
y − Ay =
, y(0) =
.
e−t
0
1 2 1
1
4. Suppose that A is a 2 × 2 matrix with eigenpairs − ,
and − ,
.
3 2
3 −1
Solve
−t/3 6e
0
y − Ay =
, y(0) = 0.
3e−2t/3
1
1
.
and − i,
5. Suppose that A is a 2 × 2 matrix with eigenpairs i,
1+i
1−i
Solve
cos t
0
y − Ay =
, y(0) = 0.
sin t
±i
6. Suppose that A is a 2 × 2 matrix with eigenpairs − 1 ± i,
. Solve
1±i
0
1
0
y − Ay =
, y(0) =
.
e−t
0
7. Solve
y0 −
−2
2
1
−1
y=
3
, y(0) = 0.
0
8. Solve



0
1
1  y =  2e−2t  , y(0) = 0.
0
0
q11 q12
x1
9. a) Let the entries of the matrix Q =
and the vector x =
be
q21 q22
x2
differentiable functions of t. Prove the product rule: (Q x)0 = Q0 x + Q x0 . Then
explain why essentially the same proof works when Q is m × n and x has length n.

0
y0 −  1
1
−1
−2
−1
b) Let the entries of the matrices Q and M be differentiable functions of t. Explain
why the product rule (QM )0 = Q0 M + Q M 0 follows directly from the product rule
in part (a).
8.6. e at and the Cayley-Hamilton Theorem
301
8.6 e At and the Cayley-Hamilton Theorem
In this section we present a method for computing e At that does not require
the computation of eigenvectors—although we will still need to compute the
eigenvalues of A. At the heart of this method lies the following centerpiece from
linear algebra known as the Cayley-Hamilton theorem.
THEOREM 8.6.1. Let A be a constant n × n matrix and let χ(λ) = |A − λ I |
be the characteristic polynomial of A. Then
χ(A) = 0,
where the zero on the right side represents the n × n zero matrix.
J
The characteristic polynomial χ always has degree n, so suppose that
χ(λ) = λn + cn−1 λn−1 + · · · + c1 λ + c0 .
Theorem 1 says then that
An + cn−1 An−1 + · · · + c1 A + c0 I = 0,
which we may rewrite as
An = −cn−1 An−1 − · · · − c1 A − c0 I .
The upshot of this is that An (and therefore any higher power of A) can always
be expressed as a linear combination of the lesser powers An−1 , An−2 , . . . , A, I .
(Note that I = A0 .) As a result we have the following corollary.
COROLLARY 8.6.1 Any linear combination of powers of A can be expressed as
a linear combination of An−1 , An−2 , . . . , A, I .
J
• Example 1 Let’s compute A4 + A3 + A2 + A for the 2 × 2 matrix
1 2
A=
.
3 −2
The characteristic polynomial is
1 − λ
2 χ(λ) = = λ2 + λ − 8.
3
−2 − λ By Theorem 1,
A2 + A − 8I = 0,
and so A2 = −A + 8I , which is easily verified. Furthermore,
A3 = −A2 + 8A = −(−A + 8I ) + 8A = 9A − 8I ,
302
Chapter 8. First-Order Linear Systems
and
A4 = 9A2 − 8A = 9(−A + 8I ) − 8 A = −17A + 72I .
Therefore,
4
3
2
A + A + A + A = −8A + 72I =
64
−24
−16
88
.
A limiting argument based on the Cayley-Hamilton theorem can be made in
order to establish the following result.
THEOREM 8.6.2. Let A be an n × n matrix and suppose that ϕ is a function
on R whose Maclaurin series expansion
ϕ00 (0) 2 ϕ000 (0) 3
x +
x + ···
2!
3!
converges for all x in R. Then for each t in R the matrix series
ϕ(x) = ϕ(0) + ϕ0 (0)x +
ϕ00 (0) 2 2 ϕ000 (0) 3 3
t A +
t A + ···
2!
3!
converges ∗ to an n × n matrix, to which we assign the symbol ϕ(tA). Moreover,
there is a polynomial ψ of degree at most n − 1, whose coefficients depend on t,
such that ϕ(tA) = ψ(A) for all t.
J
ϕ(0)I + ϕ0 (0) tA +
For our purposes here, the most important implication of Theorem 2 is the
following:
COROLLARY 8.6.2. There is a polynomial γ of degree at most n − 1, with
coefficients depending upon t, such that e At = γ(A).
J
So we now focus our efforts on finding the polynomial γ whose existence is
asserted by Corollary 2. Let’s begin by writing
γ(A) = c0 (t) + c1 (t) A + c2 (t) A2 + · · · + cn−1 (t) An−1 = e At
(1)
and assuming that A has eigenpairs (λ1 , p1 ), (λ2 , p2 ), . . . , (λn , pn ), where the λi
are distinct. For each i, we know that
e At pi = eλi t pi ,
(2)
because each side of the equation is the solution of y0 −A y = 0 with y(0) = pi .
Furthermore, for k = 1, 2, 3, . . . ,
Ak pi = λki pi ,
* This means that the individual entries in the partial sums of the series each converge.
(3)
8.6. e at and the Cayley-Hamilton Theorem
303
because Ak pi = Ak−1 λi pi = Ak−2 λ2i pi = · · · = λki pi . We now equate e At pi
with γ(A) pi , using (1), (2), and (3), to obtain n linear equations of the form
c0 (t) + c1 (t) λi + c2 (t) λ2i + · · · + cn−1 (t) λin−1 = eλi t , i = 1, 2, . . . , n
for the coefficients
tions becomes

1
1

1

.
 ..
1
(4)
c0 (t), c1 (t), . . . , cn−1 (t). In matrix form, this system of equaλ1
λ2
λ3
..
.
λn
λ21
λ22
λ23
..
.
λ2n
···
···
···
···

 

λ1n−1
c0 (t)
e λ1 t
 λ2 t 
λ2n−1 
  c1 (t) 
  e λ3 t 
n−1  
c2 (t)  =  e  .
λ3  

  . 
..
.. 

  .. 

.
.
e λn t
cn−1 (t)
λnn−1
The coefficient matrix, an example of a Vandermonde matrix, is guaranteed to
be invertible, provided that the λi ’s are distinct. Therefore, the system has a
unique solution (c0 (t), c1 (t), . . . , cn−1 (t))T for each t.
• Example 2 Suppose that A is a 2 × 2 matrix with characteristic polynomial
χ(λ) = λ2 + 3 λ + 2. The eigenvalues of A are λ1 = −1 and λ2 = −2; so the
system of equations of interest is
1 −1
c0 (t)
e−t
=
.
1 −2
c1 (t)
e−2t
Inversion of the matrix shows that
c0 (t)
2 −1
e−t
2e−t − e−2t
=
=
.
c1 (t)
1 −1
e−2t
e−t − e−2t
Therefore, the polynomial γ is
γ(x) = 2e−t − e−2t − (e−t − e−2t ) x,
and the matrix exponential is
e At = γ(A) = (2e−t − e−2t )I − (e−t − e−2t )A,
which could be carried further, were we given the entries of A. (Note that all
of this was done with no knowledge of A other than its eigenvalues.) With the
help of Theorem 1, which tells us that A2 = −3A − 2 I , this result can be easily
verified by showing that Y (t) = e At satisfies Y 0 − A Y = 0 and Y (0) = I .
The argument preceding Example 2 gives a method for finding γ that amounts
to solving an n × n system of equations. More important, however, it casts the
problem as one of finding an interpolating polynomial. We summarize this result
as follows:
304
Chapter 8. First-Order Linear Systems
THEOREM 8.6.3. If A has distinct eigenvalues λ1 , λ2 , . . . , λn , then the coefficients of γ are uniquely determined by the equations
γ(λ1 ) = eλ1 t , γ(λ2 ) = eλ2 t , . . . , γ(λn ) = eλn t .
That is, for each fixed t, γ is the unique polynomial in x that agrees with ext at
x = λ1 , . . . , λ n .
J
There are techniques for polynomial interpolation that are far more efficient
in general than solving an n × n system of equations involving a Vandermonde
matrix. A simple technique that will serve our present needs nicely is interpolation by Lagrange polynomials.
Lagrange Interpolation
Consider for a moment the general problem of finding a polynomial ψ of degree
n − 1 (or less) whose graph contains the points (x1 , y1 ), (x2 , y2 ), . . . , (xn , yn ) in
which x1 , x2 , . . . , xn are distinct. For each i = 1, 2, . . . , n define a degree n − 1
polynomial `i by
`i (x) =
(x − x1 ) · · · (x − xi−1 )(x − xi+1 ) · · · (x − xn )
.
(xi − x1 ) · · · (xi − xi−1 )(xi − xi+1 ) · · · (xi − xn )
These are the fundamental polynomials associated with x1 , x2 , . . . , xn . Notice that the numerator of `i (x) contains each of the factors (x − xj ) except
(x − xi ) and that the denominator is just the numerator evaluated at xi . Figure
1 shows the graph of `3 , based on x1 = 1, x2 = 2, x3 = 3, x4 = 4, x5 = 5:
`3 (x) =
1
(x − 1)(x − 2)(x − 4)(x − 5)
= (x − 1)(x − 2)(x − 4)(x − 5).
(3 − 1)(3 − 2)(3 − 4)(3 − 5)
4
Figure 1
Lagrange observed that such polynomials are useful in the interpolation problem, because each `i has the property that
`i (xi ) = 1, and `i (xj ) = 0 if j 6= i.
In fact, the polynomial we seek is precisely
ψ(x) = y1 `1 (x) + y2 `2 (x) + · · · + yn `n (x).
This is the so-called Lagrange form of the interpolating polynomial.
8.6. e at and the Cayley-Hamilton Theorem
305
• Example 3 Let’s find the cubic polynomial ψ that interpolates the points (0, 1),
(1, 0), (3, 2), and (4, −1). We first build fundamental polynomials:
(x − 1)(x − 3)(x − 4)
x (x − 3)(x − 4)
, `2 (x) =
,
−12
6
x (x − 1)(x − 3)
x (x − 1)(x − 4)
,
`4 (x) =
,
`3 (x) =
−6
12
and then compute
`1 (x) =
ψ(x) = 1 `1 (x) + 0 `2 (x) + 2 `3 (x) − 1 `4 (x)
1
1
1
= − (x − 1)(x − 3)(x − 4) − x(x − 1)(x − 4) −
x(x − 1)(x − 3)
12
3
12
1
= − 3 x3 − 16 x2 + 19 x − 6 .
6
In the next two examples, we’ll use Lagrange interpolation to help us compute
a matrix exponential.
• Example 4 The 3 × 3 matrix
1
A=
4
−8 −4 4
1 −3 1
5
5 −7
!
has the characteristic polynomial χ(λ) = − 12 (2 λ3 + 9 λ2 + 10 λ + 3) and therefore eigenvalues λ1 = −1, λ2 = − 12 , and λ3 = −3. To compute e At we need a
quadratic polynomial γ(x) that satisfies
γ(−1) = e−t , γ(−1/2) = e−t/2 , γ(−3) = e−3t .
So we construct fundamental polynomials:
(x + 1/2)(x + 3)
1
= − (2 x + 1)(x + 3),
−1
2
(x + 1)(x + 3)
4
`2 (x) =
= (x + 1)(x + 3),
5/4
5
(x + 1)(x + 1/2)
1
`3 (x) =
=
(x + 1)(2 x + 1),
5
10
and then compute
`1 (x) =
γ(x) = e−t `1 (x) + e−t/2 `2 (x) + e−3t `3 (x)
4
1 −3t
1
e (x + 1)(2x + 1).
= − e−t (2x + 1)(x + 3) + e−t/2 (x + 1)(x + 3) +
2
5
10
Therefore,
e At = −
e−t
4e−t/2
e−3t
(2A+ I )(A+3 I )+
(A+ I )(A+3 I )+
(A+ I )(2A+ I ),
2
5
10
306
Chapter 8. First-Order Linear Systems
which after some (computer-aided) computation becomes
!
!
1 −1 1
0 0 0
−t
−t/2
e
e
e−3t
e At =
−1 1 −1 +
1 1 1 +
2
2
2
0
0
0
1 1 1
!
e−t + e−3t −e−t + e−3t
e−t − e−3t
1
=
−e−t + e−t/2 e−t + e−t/2 −e−t + e−t/2 .
2 e−t/2 − e−3t e−t/2 − e−3t e−t/2 + e−3t
1
1 −1
0
0
0
−1 −1 1
!
1 −6
• Example 5 The 2 × 2 matrix A =
has the characteristic polynomial
3 −5
χ(λ) = λ2 +4 λ+13 and therefore nonreal complex eigenvalues λ1 = −2+3i and
λ2 = −2 − 3i. To compute e At we need a linear polynomial γ(x) that satisfies
γ(−2 + 3i) = e(−2+3i) t , γ(−2 − 3i) = e(−2−3i) t .
So we construct the fundamental polynomials
1
x − (−2 − 3i)
= − i (x + 2 + 3i),
6i
6
x − (−2 + 3i)
1
`2 (x) =
= i (x + 2 − 3i),
−6i
6
`1 (x) =
and then compute
γ(x) = e(−2+3i)t `1 (x) + e(−2−3i)t `2 (x)
1
1
= − e(−2+3i)t i (x + 2 + 3i) + e(−2−3i)t i (x + 2 − 3i).
6
6
We arrive at the following real form for γ(x) by applying Euler’s formula and
simplifying:
i e−2t γ(x) = −
(cos 3t + i sin 3t)(x + 2 + 3i)
6
− (cos 3t − i sin 3t)(x + 2 − 3i)
e−2t =
3 cos 3t + (x + 2) sin 3t
3
Therefore,
e−2t 3 cos 3t I + (A + 2 I ) sin 3t ,
3
which after some computation becomes
cos 3t + sin 3t
−2 sin 3t
At
−2t
e =e
.
sin 3t
cos 3t − sin 3t
e At =
8.6. e at and the Cayley-Hamilton Theorem
307
Repeated Eigenvalues
When A has repeated eigenvalues (deficient or not), the preceding method fails,
because there are not enough equations to determine the required polynomial.
This can be remedied by including appropriate derivative information.
THEOREM 8.6.4. Suppose that A has eigenvalues λ1 , λ2 , . . . , λk , with multiplicities m1 , m2 , . . . , mk , respectively. Then the coefficients of γ are uniquely
determined by n equations consisting, for i = 1, 2, . . . , k, of
γ(λi ) = eλi t , γ 0 (λi ) = t eλi t , γ 00 (λi ) = t2 eλi t , . . . , γ (mi −1) (λi ) = tmi −1 eλi t . J
• Example 6 Consider the 3 × 3 matrix
A=
−2
−1
−2
−1
2
6
1
−1
−3
!
,
whose characteristic polynomial turns out to be χ(λ) = (λ + 1)3 . So the only
eigenvalue of A is λ1 = −1, with multiplicity m1 = 3. Therefore, the polynomial
γ will be determined by
γ(−1) = e−t , γ 0 (−1) = t e−t , γ 00 (−1) = t2 e−t .
If we let c0 , c1 , c2 be the (t-dependent) coefficients of γ, that is,
γ(x) = c0 + c1 x + c2 x2 ,
then those equations become
c0 − c1 + c2 = e−t , c1 − 2c2 = t e−t , 2c2 = t2 e−t ,
which are easily solved by back-substitution to yield
c0 =
e−t 2
e−t 2
(t + 2t + 2), c1 = e−t (t2 + t), c2 =
t .
2
2
Therefore,
γ(x) =
e−t
2
(t2 + 2t + 2) + 2 (t2 + t)x + t2 x2 ,
and so
e−t
(t2 + 2t + 2)I + 2 (t2 + t)A + t2 A2 .
2
Further computation reveals, finally, that
!
1−t
−t + 2 t2
t − t2
At
−t
2
2
e =e
−t 1 + 3t + 2 t
−t − t
.
−2t
6t + 4 t2
1 − 2t − 2 t2
e At =
308
Chapter 8. First-Order Linear Systems
Problems
1. Let A be a 2 × 2 matrix with eigenvalues λ1 and λ2 .
a) Assuming that λ1 6= λ2 , derive the formula
e At = (λ1 − λ2 )−1 eλ1 t (A − λ2 I ) − eλ2 t (A − λ1 I ) .
b) Assuming that λ1 and λ2 are complex conjugates α ± β i, show that the formula
in part (a) becomes
e At = β −1 eαt β cos β t I + sin β t (A − α I ) .
c) Assuming that λ1 = λ2 , derive the formula e At = eλ1 t ( t A + (1 − λ1 t) I ) .
2. Let A be a 2 × 2 matrix with eigenvalues λ1 and λ2 .
a) Use the Cayley-Hamilton theorem to show that A2 = (λ1 + λ2 )A − λ1 λ2 I .
b) For each of the formulas in Problem 1, verify that eA0 = I and, with the help of
d
part (a), verify that dt
e At = A e At .
For each matrix A in Problems 3 through 6, use the appropriate formula from Problem
1 to compute e At .
1 3
0 −1
−4 1
1 −3
3.
4.
5.
6.
3 1
2 −2
1 −4
3 −5
−1 2
7. Let A =
. Use the appropriate formula from Problem 1 to compute e At ,
−2 −1
and then use the variation of constants formula (1) in Section 8.5 to solve
sin 2t
1
0
−t
y − Ay = e
, y(0) =
.
cos 2t
0
In Problems 8 through 11, find e At in terms of A, where A is a 3 × 3 matrix with the
given eigenvalues.
8. 0, −1, −2
9. −1, −2 ± i
10. −1, −2, −2
11. −2, −2, −2
In Problems 12 and 13, use the Cayley-Hamilton theorem to express A5 in terms of A2 ,
A, and I , where A is a 3 × 3 matrix with the given eigenvalues.
12. −1, −1, −2
13. −1 ± i, 0
8.7 Nonautonomous Systems
Throughout this section, A is assumed to be an n × n matrix whose entries are
continuous functions of t (possibly constants) on an interval I. We will concern
ourselves first with the homogeneous problem
y0 − A y = 0 on I.
(1)
The development here will parallel that of autonomous case in §8.4. The definition of fundamental matrix is exactly the same. Theorem 8.4.1 remains true;
8.7. Nonautonomous Systems
309
however, for our purposes here, we state a more general version of part (iii) of
that theorem as follows:
THEOREM 8.7.1. Let Y (t) be a fundamental matrix generated by A, and let
y be any solution of (1) on I. Then
y(t) = Y (t)Y (s)−1 y(s) for any s, t in I.
J
Thus, given a fundamental matrix Y (t), multiplication by Y (t)Y (s)−1 maps,
or “evolves,” any solution of (1) from time s to time t. We will call this matrix
the evolution matrix generated by A.
Definition. Let Y (t) be any fundamental matrix generated by A. The evolution matrix Φ(t, s) generated by A is
Φ(t, s) = Y (t)Y (s)−1 .
(2)
Note that Theorem 8.7.1 now says that, if y is a solution of (1), then
y(t) = Φ(t, s)y(s) for any s, t in I;
in particular, given t0 in I and y0 in Rn , y(t) = Φ(t, t0 )y0 is the unique solution
of (1) that satisfies y(t0 ) = y0 .
We should also make note of the fact that, if (1) is autonomous (i.e., A has
constant entries), then Φ(t, s) = eA(t−s) and, in particular, Φ(t, 0) = eAt . When
(1) is not autonomous, Φ(t, s) cannot be expressed in terms of t−s. For instance,
if y1 and y2 are solutions of (1) for which y1 (0) = y2 (1), there is no guarantee
that y1 (1) will equal y2 (2). In other words, solutions can evolve differently, from
the same initial state, over different time intervals of the same length. This is
the crucial difference between autonomous and nonautonomous systems.
In the following example, a fundamental matrix is given with no indication
of how it was obtained. Do not worry about this. In fact, the system was computed from the solution—rather than the other way around. There is no general
method for finding a fundamental matrix in the nonautonomous case.
• Example 1 Consider the differential equation
1
t
1
0
y −
y = 0,
(t + 1)2 −1 2 + t
and note that the entries of the matrix are continuous on I = (−1, ∞). It turns
out that a simple fundamental matrix is
2+t
t
Y (t) =
,
1
1 + 2t
310
Chapter 8. First-Order Linear Systems
as is readily verified. From this we compute the inverse, finding that
1
1 + 2s −s
Y (s)−1 =
.
−1
2+s
2 (1 + s)2
Consequently, the evolution matrix is
Φ(t, s) = Y (t)Y (s)
−1
1
=
2 (1 + s)2
2+t
1
t
1 + 2t
1 + 2s −s
−1
2+s
,
which after some simplification becomes
1
1 + (2 + t)s
t−s
.
Φ(t, s) =
s−t
1 + (2 + s)t
(1 + s)2
So the solution satisfying y(0) = y0 , for instance, is
1
t
y(t) = Φ(t, 0)y0 =
y0 ,
−t 1 + 2t
while the solution satisfying y(1) = y0 , on the other hand, is
1 t+3 t−1
y0 .
y(t) = Φ(t, 1)y0 =
4 1 − t 1 + 3t
We point out that Φ(t, 0) and Φ(t, 1) could have been computed directly using
Y (0)−1 and Y (1)−1 , avoiding the computation of Y (s)−1 .
Properties of the evolution matrix
The following theorem states properties of Φ(t, s) that are analogous to properties of the matrix exponential eAt listed in Theorem 8.4.2.
THEOREM 8.7.2. If the entries of the matrix A are continuous on I, then the
evolution matrix generated by A has the following properties:
i) Φ(t, t) = I for all t in I;
ii) Φ(t, s) = Φ(t, r) Φ(r, s) for all r, s, t in I;
iii) Φ(t, s)−1 = Φ(s, t) for all s, t in I;
iv)
∂
∂t Φ(t, s)
= A(t)Φ(t, s) for all s, t in I.
J
Nonhomogeneous Systems
Consider the nonhomogeneous initial-value problem
y0 − A y = f , y(t0 ) = y0 ,
(3)
where the entries of A and f are continuous on I, and let Y (t) be any fundamental matrix generated by A—possibly some evolution matrix Φ(t, s).
8.7. Nonautonomous Systems
311
Precisely the same calculations as in §8.5 lead to the following solution of
(3):
Z t
−1
−1
Y (s) f (s) ds .
(4)
y(t) = Y (t) Y (t0 ) y0 +
t0
In terms of the evolution matrix Φ(t, s), this becomes
Z t
Φ(t, s)f (s) ds.
y(t) = Φ(t, t0 )y0 +
(5)
t0
These are the variation of constants formulas for the nonautonomous case.
• Example 2 The following system is a nonhomogeneous version of the equation
from Example 1:
1
t
1
1
0
y −
y=
.
t2
(t + 1)2 −1 2 + t
Let’s compute the particular solution that satisfies y(0) = 0. The evolution
matrix turned out to be
1
1 + (2 + t)s
t−s
Φ(t, s) =
.
s−t
1 + (2 + s)t
(1 + s)2
Thus the solution we seek is
Z t
1
1 + (2 + t)s
t−s
1
y(t) =
ds
2
2
s
−
t
1
+
(2
+
s)t
s
0 (1 + s)
Z t
1
s2 (t − s)
ds
=
2
s2 + s2 (2 + s) t
0 (1 + s)
Z t
1+t 2 − s + t − 1+s
1 4t + t2 − 2(t + 1) ln(t + 1)
= ··· =
ds =
.
3
1+t
1 + st − 1+s
2 2t + t − 2(t + 1) ln(t + 1)
0
Problems
In Problems 1 through 3,
a) use a solution matrix from one of Problems 10 through 12 in Section 8.3 to find
the evolution matrix Φ(t, s) generated by the coefficient matrix A;
b) use the variation of constants formula to solve the initial-value problem. Also
state the domain I of the solution.
1
t + 1 −2t
8
y=
, y(1/2) = 0
1. y0 −
3 −t − 2
8/t
t (1 − t)
− tan 2t sec 2t
sin t
0
0
2. y −
y=
, y(0) =
sec 2t − tan 2t
cos t
1
√ −2 cos t
2
3. y + sec t csc t
y = 12 sin4 t
, y(π/4) = 0
cos t −2
−1
312
Chapter 8. First-Order Linear Systems
In each of Problems 4 through 6, the given y1 and y2 are linearly independent solutions
of y0 − A y = 0 on an interval I containing t = 0. For the given function f , find the
solution on I of the initial-value problem y0 − A y = f , y(0) = 0.
t
−2t e
e
8
4. y1 =
, y2 =
, f (t) =
et
−e−2t
4
t
t+1
1
5. y1 =
, y2 =
, f (t) =
t+1
0
(1 + t)2
(t + 1)3/2
(t + 1)5/2
(t + 1)1/2
6. y1 =
, y2 =
, f (t) =
−(t + 1)3/2
(t + 1)5/2
−(t + 1)1/2
8.8 Asymptotic Stability
We now turn our attention to questions concerning the asymptotic behavior —
that is, the behavior as t → ∞ —of solutions of
y0 − A y = f ,
(1)
where the entries of A and f are continuous for all t in [0, ∞). We allow the case
n = 1, in which y = y, A = a, and f = f are scalar valued.
In order to state a definition of asymptotic stability, we first need to recall
the notation kx − ak for the distance from x to a, which is defined by
kx − ak =
n
X
1/2
(xi − ai )2
.
i=1
Note that in R (i.e., when n = 1) kx − ak reduces to |x − a|. In R2 and R3 ,
kx − ak is the usual distance formula.
Definition. Let ỹ be a particular solution of (1). If there is some number ε > 0
such that
lim ky(t) − ỹ(t)k = 0
t→∞
for all solutions y with ky(0) − ỹ(0)k < ε, then ỹ is said to be asymptotically
stable.
This definition essentially says that an asymptotically stable solution is one
with the property that solutions sufficiently “close” to it at t = 0 eventually
become closer and closer as t → ∞. Because the system is linear , it turns out
that if a solution ỹ is asymptotically stable, then
lim ky(t) − ỹ(t)k = 0
t→∞
8.8. Asymptotic Stability
313
for all solutions y, not just solutions whose intial values are close to ỹ(0). Moreover, if the system has one asymptotically stable solution, then all of its solutions are asymptotically stable. These facts are results of the following theorem,
whose proof is the subject of Problem 15.
THEOREM 1 If every solution of the homogeneous equation y0 − A y = 0
approaches 0 as t → ∞, then all solutions of equation (1) are asymptotically
stable; otherwise no solution of (1) is asymptotically stable.
J
• Example 1 Consider the equation
y 0 − (cos t − 1/4) y = −4 sin t.
Let ỹ be a given solution, and let y be any other solution. Their difference
u = y − ỹ will satisfy
u0 − (cos t − 1/4) u = 0, u(0) = u0 = y(0) − ỹ(0),
whose solutions is
u = u0 e sin t−t/4 = u0 e sin t e−t/4 .
It follows that
lim |y(t) − ỹ(t)| = lim |u(t)| = 0;
t→∞
t→∞
therefore, ỹ is asymptotically stable. Since
ỹ represents an arbitrary solution, we conclude that every solution is asymptotically
stable. This fact is illustrated by the solutions plotted in Figure 1, all of which apparently approach a certain nonconstant,
periodic solution, which begs description
as a periodic attractor.
Figure 1
The behavior that we observed in Example 1 is actually typical of linear,
scalar, first-order equations with asymptotically stable solutions. Consider
y 0 − py = f ,
(2)
where p and f are continuous functions on [0, ∞). Given any two solutions y
and ỹ, their difference u = y − ỹ will satisfy the homogeneous equation
u0 − pu = 0.
Therefore, every pair of solutions y and ỹ of (2) will have the property that
lim |y(t) − ỹ(t)| = 0,
t→∞
(3)
314
Chapter 8. First-Order Linear Systems
if and only if every solution of the homogeneous equation (3) approaches zero
as t → ∞. Since every solution of (2) is of the form
Z t
P (t)
p(s)ds,
u = u(0)e
, where P (t) =
0
it follows that every solution u of (3) approaches 0 as t → ∞, if and only if
P (t) → −∞ as t → ∞. So we conclude that every solution of (2) is asymptotically stable, if
Z t
lim
p(s)ds = −∞;
t→∞ 0
otherwise no solution of (2) is asymptotically stable.
• Example 2 Consider the system
0
y − Ay =
cos t
sin 2t
,
where
A=
1
1 + t2
t − t2 − 1
−1
1
t − t2 − 1
.
It is simple to verify that the evolution matrix generated by A (at s = 0) is
e−t −t e−t
Φ(t, 0) =
.
t e−t
e−t
Since each entry in Φ0 (t) approaches 0 as
t → ∞, it follows that every solution of
the homogeneous equation y0 − Ay = 0
approaches 0 as t → ∞. (Why?) Therefore, all solutions of the nonhomogeneous
system are asymptotically stable. Figure 2
shows phase-plane orbits of two solutions
with initial points (0, 0.5) and (0.5, −0.5).
Notice that, as t → ∞, they both approach the same closed curve (which is not
the orbit of any solution of the system). Figure 2
We now turn our attention to equation (1) under the assumption that A is a
constant n×n matrix and f is continuous on [0, ∞). We have seen in this chapter
that solutions of the homogeneous equation y0 − A y = 0 can be constructed
from the eigenvalues of A, which are the roots of the characteristic polynomial
χ(λ) = |A − λ I |.
8.8. Asymptotic Stability
315
Since the solution produced by each eigenvalue λi has eλi t as a factor that is
dominant as t → ∞, it is a relatively simple matter to prove the following:
COROLLARY 1 Let A have constant entries. If the real part of every eigenvalue
of A is negative, then all solutions of (1) are asymptotically stable; otherwise
no solution of (1) is asymptotically stable.
J
• Example 4 Consider the system y0 − A y = f , where
1 8 −18
A=
.
3 9 −10
The eigenvalues of A are − 31 ± 3i; hence each has negative real part. Therefore,
solutions of y0 − A y = f are asymptotically stable for any choice of f . Figure 4
shows the graphs of y1 and y2 corresponding to the constant nonhomogeneous
term f (t) = (2, 1)T (and y(0) = (−1, 0)T ). Note that y1 and y2 approach constants as t → ∞. Figure 5 shows the phase-plane orbit of the same solution
y = (y1 , y2 )T . Here we see that the orbit spirals in toward an equilibrium point
as t → ∞. Similar behavior would be seen with any initial point y(0).
Figure 4
Figure 5
When the nonhomogeneous term f is constant, as was the case in Example 4,
an isolated equilibrium point—if one exists—clearly must be the unique constant
solution y∗ satisfying
−A y∗ = f .
(Note that if solutions are asymptotically stable, then A is invertible. Why?) In
the case of a homogeneous system (i.e., f = 0), this gives us y∗ = 0. When f is
not constant, one does not expect any equilibrium points to exist.
• Example 5 Consider the system
1 8 −18
cos 3t
0
y −
y=
,
sin 3t
3 9 −10
y(0) = 0.
The matrix here is the same as in Example 4. Figure 6 shows the graphs of
y1 and y2 corresponding to y(0) = (0, 0)T . Here, y1 and y2 approach periodic
functions as t → ∞. Figure 7 shows phase-plane orbit corresponding to the
316
Chapter 8. First-Order Linear Systems
same solution y = (y1 , y2 )T . As t → ∞ the orbit spirals out toward an ellipse
that corresponds to a periodic solution of the system. Similar behavior would
be seen with any initial point y(0), although if y(0) were outside the ellipse,
the orbit would spiral inward. Thus the particular solution corresponding to
the elliptical orbit could be viewed as a periodic attractor of all other solutions.
Figure 6
Figure 7
Some (computer-aided) computation reveals that this solution is
3
496 1 − e−t/3 cos 3t + 153 − 172 e−t/3 sin 3t
y=
.
325 162 1 − e−t/3 cos 3t + 2 158 − 167 e−t/3 sin 3t
Since the eigenvalues of A are − 13 ± 3i, we know that solutions of the homogeneous system y0 − A y = 0 will have e−t/3 as a factor. So we separate the
solution into two terms as follows:
3 e−t/3 496 cos 3t + 172 sin 3t
3
496 cos 3t + 153 sin 3t
y=−
+
.
162 cos 3t + 334 sin 3t
325
325 162 cos 3t + 316 sin 3t
The first term, which satisfies the homogeneous system y0 − A y = 0, is sometimes called the transient part of the solution, and the second term is precisely
the periodic attractor:
3
496 cos 3t + 153 sin 3t
.
y∗ =
325 162 cos 3t + 316 sin 3t
Figure 8 shows the graphs of y1∗ and y2∗ versus t, and Figure 9 shows the phaseplane orbit for y∗ .
Figure 8
Figure 9
8.8. Asymptotic Stability
317
Problems
For each of the linear, first-order equations in Problems 1 through 6,
(a) find the general solution, and
(b) determine whether solutions are asymptotically stable. If so, describe their behavior as t → ∞.
1
1
1. y 0 +
y=2
2. y 0 −
y=2
t+1
t+1
1
1
y=2
4. y 0 − √
y=2
3. y 0 + √
t+1
t+1
sin t
y
5. y 0 −
y = cos t
6. (t + 1) y 0 +
=1
2 + cos t
ln(t + 1)
For each matrix A in Problems 7 through 9, determine all values of γ for which solutions
of y0 − Ay = f are asymptotically stable.


−1 1
γ
−1 γ
γ b
7.
8.
9.  1 −1 1 
4 −1
c γ
−γ 1 −1
a b
10. Show that solutions of y0 − Ay = f , where A =
, are asymptotically stable
c d
if and only if a + d < 0 and ad − bc > 0.
−2 1
10 sin t
11. For the equation y0 −
y=
,
1 −2
0
(a) show that all solutions are asymptotically stable;
(b) find a periodic solution.
In Problems 12 and 13, show that every solution of the given equation approaches a
certain periodic function w, and show that w is not a solution of the differential equation.
t+1
12. y 0 + y = e−t + cos t
13. y 0 +
y = cos t
t
14. (a) Let a and b be continuous functions on [0, ∞) (possibly constants). Show that if
solutions of y 0 − a y = f and y 0 − b y = f are asymptotically stable, then solutions
of y 0 − (a + b)y = f are also asymptotically stable.
(b) Find a pair of 2 × 2 matrices A and B for which solutions of y0 − A y = f and
y0 − B y = f are asymptotically stable, yet solutions of y0 − (A + B)y = f are
not asymptotically stable.
15. Prove Theorem 1 as follows, noting first that for any initial point the existence of a
unique solution on [0, ∞) is guaranteed.
(i) Suppose that every solution of y0 − A y = 0 approaches 0 as t → ∞. Show that,
for any two solutions u and v of y0 −A y = f , the difference w = u−v approaches
0 as t → ∞. Conclude that every solution of y0 −A y = f is asymptotically stable.
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Chapter 8. First-Order Linear Systems
(ii) Suppose that there is a solution of y0 −A y = 0—call it w—that does not approach
0 as t → ∞. Let ỹ be any solution of y0 − A y = f . Show that, for any number
c 6= 0, y = ỹ + c w is also a solution of y0 − A y = f and that y − ỹ does not
approach 0 as t → ∞. Conclude that ỹ is not asymptotically stable.
Problems 16 through 20 concern the stability of solutions of nonlinear, scalar equations.
The definition of stability stated in this section applies to nonlinear problems as well as
linear ones, provided we make an assumption about the existence of solutions for t ≥ 0.
Consider a single equation y 0 = f (t, y), where f is continuous on all of R2 . A solution ỹ
is asymptotically stable if (1) it exists for all t ≥ 0 and (2) there is some number ε > 0
such that every solution y with |y(0) − ỹ(0)| < ε exists for all t ≥ 0 and
lim (y(t) − ỹ(t)) = 0.
t→∞
In other words, ỹ is asymptotically stable if for every solution y with a “nearby” initial
value, the difference y(t) − ỹ(t) approaches zero as t → ∞.
16. Show that each positive solution of the (nonlinear) logistic equation
y y 0 = ky 1 −
, with k, M > 0,
M
is asymptotically stable and that nonpositive solutions are not asymptotically stable.
For each equation in Problems 17 and 18, make a sketch of the solution curves (cf.
Section 4.3), and describe the asymptotically stable solutions.
17. y 0 = y (1 − y)(2 − y)
18. y 0 = −y (1 − y)(2 − y)
In Problems 19 and 20, the given Bernoulli/logistic equation has a positive, asymptotically stable, periodic solution. Find it.
cos t y = −y 2
20. y 0 − y = −(2 + cos t) y 2 .
19. y 0 − 1 +
2 + sin t