Linear Algebra, Theory And Applications, 2012a

428 APPLICATIONS TO DIFFERENTIAL EQUATIONS
Recall r 0 (t) ≡ 0andr 1 (t) =e (−1+i)t . Then
r ′ 2 =(−1 − i) r 2 + e (−1+i)t ,r 2 (0) = 0
and so
r 2 (t) = e(−1+i)t − e (−1−i)t
= e −t sin (t)
2i
Putzer’s method yields the fundamental matrix as
( )
( )
Φ(t) = e (−1+i)t 1 0
+ e −t 1 − i 1
sin (t)
0 1
−2 −1 − i
( )
e
=
−t (cos (t)+sin(t)) e −t sin t
−2e −t sin t e −t (cos (t) − sin (t))
From variation of constants formula the desired solution is
( ) ( )( )
x1
e
(t) =
−t (cos (t)+sin(t)) e −t sin t
1
x 2 −2e −t sin t e −t (cos (t) − sin (t)) 0
∫ t
( )(
)
e
+
−s (cos (s)+sin(s)) e −s sin s
0
0
−2e −s sin s e −s (cos (s) − sin (s)) cos (t − s)
( )
e
=
−t ∫ t
(
)
(cos (t)+sin(t))
e
−2e −t +
−s sin (s) cos (t − s)
sin t
0
e −s ds
(cos s − sin s) cos (t − s)
( ) (
e
=
−t (cos (t)+sin(t)) −
1
−2e −t + 5 (cos t) e−t − 3 5 e−t sin t + 1 5 cos t + 2 5 sin t )
sin t
− 2 5 (cos t) e−t + 4 5 e−t sin t + 2 5 cos t − 1 5 sin t ( 4
= 5 (cos t) e−t + 2 5 e−t sin t + 1 5 cos t + 2 5 sin t )
− 6 5 e−t sin t − 2 5 (cos t) e−t + 2 5 cos t − 1 5 sin t
Thus y (t) =x 1 (t) = 4 5 (cos t) e−t + 2 5 e−t sin t + 1 5 cos t + 2 5
sin t.
C.5 Geometric Theory Of Autonomous Systems
Here a sufficient condition is given for stability of a first order system. First of all, here is
a fundamental estimate for the entries of a fundamental matrix.
Lemma C.5.1 Let the functions, r k be given in the statement of Theorem C.4.8 and suppose
that A is an n × n matrix whose eigenvalues are {λ 1 , ··· ,λ n } . Suppose that these
eigenvalues are ordered such that
Re (λ 1 ) ≤ Re (λ 2 ) ≤···≤Re (λ n ) < 0.
Then if 0 > −δ > Re (λ n ) is given, there exists a constant, C such that for each k =
0, 1, ··· ,n,
|r k (t)| ≤Ce −δt (3.29)
for all t>0.
Proof: This is obvious for r 0 (t) because it is identically equal to 0. From the definition
of the r k ,r ′ 1 = λ 1 r 1 ,r 1 (0) = 1 and so r 1 (t) =e λ1t which implies
|r 1 (t)| ≤e Re(λ 1)t .