Linear Algebra, Theory And Applications, 2012a

C.4. FIRST ORDER LINEAR SYSTEMS 427
Proof: From the uniqueness theorem there is at most one solution to (3.17). Therefore,
if (3.26) solves (3.17), the theorem is proved. The verification that the given formula works
is identical with the verification that the scalar formula given in Theorem C.4.7 solves the
initial value problem given there. Φ (s) −1 is continuous because of the formula for the inverse
of a matrix in terms of the transpose of the cofactor matrix. Therefore, the integrand in
(3.26) is continuous and the fundamental theorem of calculus applies. To verify the formula
for the inverse, fix s and consider x (t) =Φ(s + t) v, and y (t) =Φ(t)Φ(s) v. Then
x ′ (t) =AΦ(t + s) v = Ax (t) , x (0) = Φ (s) v
y ′ (t) =AΦ(t)Φ(s) v = Ay (t) , y (0) = Φ (s) v.
By the uniqueness theorem, x (t) =y (t) for all t. Since s and v are arbitrary, this shows
Φ(t + s) =Φ(t)Φ(s) for all t, s. Letting s = −t and using Φ (0) = I verifies Φ (t) −1 =
Φ(−t) .
Next, note that this also implies Φ (t − s)Φ(s) =Φ(t) andsoΦ(t − s) =Φ(t)Φ(s) −1 .
Therefore, this yields (3.27) and then (3.28)follows from changing the variable.
If Φ ′ = AΦ andΦ(t) −1 exists for all t, you should verify that the solution to the initial
value problem
x ′ = Ax + f, x (t 0 )=x 0
is given by
∫ t
x (t) =Φ(t − t 0 ) x 0 + Φ(t − s) f (s) ds.
t 0
Theorem C.4.10 is general enough to include all constant coefficient linear differential
equations or any order. Thus it includes as a special case the main topics of an entire
elementary differential equations class. This is illustrated in the following example. One
can reduce an arbitrary linear differential equation to a first order system and then apply the
above theory to solve the problem. The next example is a differential equation of damped
vibration.
Example C.4.11 The differential equation is y ′′ +2y ′ +2y =cost and initial conditions,
y (0) = 1 and y ′ (0) = 0.
To solve this equation, let x 1 = y and x 2 = x ′ 1 = y ′ . Then, writing this in terms of these
new variables, yields the following system.
x ′ 2 +2x 2 +2x 1 =cost
x ′ 1 = x 2
This system can be written in the above form as
( ) ′ (
x1
=
x 2
) (
x 2
0
+
−2x 2 − 2x 1 cos t
) (
0 1
=
−2 −2
)( ) (
x1 0
+
x 2 cos t
)
.
and the initial condition is of the form
( ) (
x1
1
(0) =
x 2 0
)
Now P 0 (A) ≡ I. The eigenvalues are −1+i, −1 − i and so
(( )
( )) ( 0 1
1 0 1 − i 1
P 1 (A) =
− (−1+i)
=
−2 −2
0 1 −2 −1 − i
)
.