Here is a discussion of linear systems of the form
x′ = Ax + f (t)
where A is a constant n×n matrix and f is a vector valued function having all entries continuous.
Of course the existence theory is a very special case of the general considerations above but I will
give a self contained presentation based on elementary first order scalar differential equations and
linear algebra.
Definition D.4.1Suppose t → M
(t)
is a matrix valued function of t. Thus M
(t)
=
(mij(t))
.Then define
′ ( ′ )
M (t) ≡ m ij(t) .
In words, the derivative of M
(t)
is the matrix whose entries consist of the derivatives of theentries of M
(t)
. Integrals of matrices are defined the same way. Thus
∫ b (∫ b )
M (t)di ≡ mij(t)dt .
a a
In words, the integral of M
(t)
is the matrix obtained by replacing each entry of M
(t)
by theintegral of that entry.
With this definition, it is easy to prove the following theorem.
Theorem D.4.2Suppose M
(t)
and N
(t)
are matrices for which M
(t)
N
(t)
makes sense. Thenif M^{′}
(t)
and N^{′}
(t)
both exist, it follows that
′ ′ ′
(M (t)N (t)) = M (t)N (t)+ M (t)N (t).
Proof:
( )
( ′) ( )′ ∑ ′
(M (t)N (t)) ij ≡ (M (t)N (t))ij = M (t)ikN (t)kj
k ( )
= ∑ (M (t) )′N (t) + M (t) N (t) ′
k ik kj ik kj
∑ ( ′) ( ′)
≡ M (t) ikN (t)kj + M (t)ik N (t) kj
k ′ ′
≡ (M (t)N (t)+ M (t)N (t))ij ■
In the study of differential equations, one of the most important theorems is Gronwall’s
inequality which is next.
Theorem D.4.3Suppose u
(t)
≥ 0 and for all t ∈
[0,T]
,
∫ t
u(t) ≤ u0 + Ku (s)ds. (4.10)
0
(4.10)
where K is some nonnegative constant. Then
u(t) ≤ u0eKt. (4.11)
(4.11)
Proof:Let w
(t)
= ∫_{0}^{t}u
(s)
ds. Then using the fundamental theorem of calculus, 4.10w
and you can
repeat the argument which was just given to conclude that x
(t)
= y
(t)
for all t ≤ a. ■
Definition D.4.5Let A be an n × n matrix. We say Φ
(t)
is a fundamental matrix for Aif
′
Φ (t) = AΦ (t),Φ (0) = I, (4.16)
(4.16)
and Φ
(t)
^{−1}exists for all t ∈ ℝ.
Why should anyone care about a fundamental matrix? The reason is that such a matrix valued
function makes possible a convenient description of the solution of the initial value
problem,
x′ = Ax + f (t),x(0) = x0, (4.17)
(4.17)
on the interval,
[0,T ]
. First consider the special case where n = 1. This is the first order linear
differential equation,
′
r = λr + g,r (0) = r0, (4.18)
(4.18)
where g is a continuous scalar valued function. First consider the case where g = 0.
Lemma D.4.6There exists a unique solution to the initial value problem,
′
r = λr,r(0) = 1, (4.19)
(4.19)
and the solution for λ = a + ib is given by
at
r (t) = e (cos bt+ isinbt). (4.20)
(4.20)
This solution to the initial value problem is denoted as e^{λt}. (If λ is real, e^{λt}as defined here reducesto the usual exponential function so there is no contradiction between this and earlier notationseen in calculus.)
Proof: From the uniqueness theorem presented above, Theorem D.4.4, applied to the case
where n = 1, there can be no more than one solution to the initial value problem, 4.19. Therefore,
it only remains to verify 4.20 is a solution to 4.19. However, this is an easy calculus exercise.
■
With this lemma, it becomes possible to easily solve the case in which g≠0.
Theorem D.4.7There exists a unique solution to 4.18and this solution is given by theformula,
∫ t
r(t) = eλtr + eλt e−λsg (s)ds. (4.22)
0 0
(4.22)
Proof:By the uniqueness theorem, Theorem D.4.4, there is no more than one solution. It
only remains to verify that 4.22 is a solution. But r
(0)
= e^{λ0}r_{0} + ∫_{0}^{0}e^{−λs}g
(s)
ds = r_{0} and so the
initial condition is satisfied. Next differentiate this expression to verify the differential
equation is also satisfied. Using 4.21, the product rule and the fundamental theorem of
calculus,
∫
′ λt λt t −λs λt −λt
r (t) = λe r0 + λe 0 e g (s)ds + e e g(t) = λr(t)+ g(t). ■
Now consider the question of finding a fundamental matrix for A. When this is done, it will be
easy to give a formula for the general solution to 4.17 known as the variation of constants formula,
arguably the most important result in differential equations.
The next theorem gives a formula for the fundamental matrix 4.16. It is known as Putzer’s
method [1],[22].
Theorem D.4.8Let A be an n×n matrix whose eigenvalues are
{λ1,⋅⋅⋅,λn }
listed accordingto multiplicity as roots of the characteristic equation. Define
k
Pk (A ) ≡ ∏ (A − λmI ),P0(A) ≡ I,
m=1
and let the scalar valued functions, r_{k}
(t)
be defined as the solutions to the following initial valueproblem
^{−1} does not exist for any
t. Suppose then that Φ
(t)
0
^{−1} does exist. Then letΨ
(t)
≡ Φ
(t + t)
0
Φ
(t)
0
^{−1}. Then
Ψ
(0)
= I and Ψ^{′} = AΨ so by Theorem D.4.8 it follows Ψ
(t)
^{−1} exists for all t and so for all
t,Φ
(t+ t )
0
^{−1} must also exist, even for t = −t_{0} which implies Φ
(0)
^{−1} exists after all.
■
The conclusion of this proposition is usually referred to as the Wronskian alternative and
another way to say it is that if 4.24 holds, then either det
(Φ(t))
= 0 for all t or det
(Φ (t))
is
never equal to 0. The Wronskian is the usual name of the function, t → det
(Φ(t))
.
The following theorem gives the variation of constants formula,.
Theorem D.4.10Let f be continuous on
[0,T]
and let A be an n×n matrix and x_{0}a vector inℂ^{n}. Then there exists a unique solution to 4.17,x,given by the variation of constantsformula,
∫
t −1
x(t) = Φ (t)x0 + Φ (t) 0 Φ (s) f (s)ds (4.25)
(4.25)
for Φ
(t)
the fundamental matrix for A. Also, Φ
(t)
^{−1} = Φ
(− t)
and Φ
(t+ s)
= Φ
(t)
Φ
(s)
for allt,s and the above variation of constants formula can also be written as