Linear Algebra, Theory And Applications, 2012a

426 APPLICATIONS TO DIFFERENTIAL EQUATIONS
and so by the uniqueness theorem, z (t) =y (t) for all t showing that Φ (t) v =Ψ(t) v for
all t. Since v is arbitrary, this shows that Φ (t) =Ψ(t) for every t.
It is useful to consider the differential equations for the r k for k ≥ 1. As noted above,
r 0 (t) =0andr 1 (t) =e λ1t .
Thus
Therefore,
r ′ k+1 = λ k+1 r k+1 + r k ,r k+1 (0) = 0.
r k+1 (t) =
r 2 (t) =
∫ t
0
∫ t
0
e λ k+1(t−s) r k (s) ds.
e λ2(t−s) e λ1s ds = eλ1t − e λ2t
−λ 2 + λ 1
assuming λ 1 ≠ λ 2 .
Sometimes people define a fundamental matrix to be a matrix Φ (t) such that Φ ′ (t) =
AΦ(t) and det (Φ (t)) ≠0forallt. Thus this avoids the initial condition, Φ (0) = I. The
next proposition has to do with this situation.
Proposition C.4.9 Suppose A is an n × n matrix and suppose Φ(t) is an n × n matrix for
each t ∈ R with the property that
Φ ′ (t) =AΦ(t) . (3.25)
Then either Φ(t) −1 exists for all t ∈ R or Φ(t) −1 fails to exist for all t ∈ R.
Proof: Suppose Φ (0) −1 exists and (3.25) holds. Let Ψ (t) ≡ Φ(t)Φ(0) −1 . Then Ψ (0) =
I and
Ψ ′ (t) =Φ ′ (t)Φ(0) −1 = AΦ(t)Φ(0) −1 = AΨ(t)
so by Theorem C.4.8, Ψ (t) −1 exists for all t. Therefore, Φ (t) −1 also exists for all t.
Next suppose Φ (0) −1 does not exist. I need to show Φ (t) −1 does not exist for any t.
Suppose then that Φ (t 0 ) −1 does exist. Then letΨ (t) ≡ Φ(t 0 + t)Φ(t 0 ) −1 . Then Ψ (0) =
I and Ψ ′ = AΨ so by Theorem C.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 (3.25) 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 C.4.10 Let f be continuous on [0,T] and let A be an n × n matrix and x 0 a
vector in C n . Then there exists a unique solution to (3.17), x, given by the variation of
constants formula,
∫ t
x (t) =Φ(t) x 0 +Φ(t) Φ(s) −1 f (s) ds (3.26)
for Φ(t) the fundamental matrix for A. Also, Φ(t) −1 =Φ(−t) and Φ(t + s) =Φ(t)Φ(s)
for all t, s and the above variation of constants formula can also be written as
x (t) = Φ(t) x 0 +
= Φ(t) x 0 +
∫ t
0
0
∫ t
0
Φ(t − s) f (s) ds (3.27)
Φ(s) f (t − s) ds (3.28)