Here I will consider the case where the matrix A has both positive and negative eigenvalues. First
here is a useful lemma.
Lemma D.6.1Suppose A is an n × n matrix and there exists δ > 0 such that
0 < δ < Re (λ1) ≤ ⋅⋅⋅ ≤ Re(λn)
where
{λ1,⋅⋅⋅,λn}
are the eigenvalues of A, with possibly some repeated. Then there exists aconstant, C such that for all t < 0,
|Φ (t)x| ≤ Ceδt|x |
Proof: I want an estimate on the solutions to the system
Φ′(t) = AΦ (t),Φ (0) = I.
for t < 0. Let s = −t and let Ψ
(s)
= Φ
(t)
. Then writing this in terms of Ψ,
Ψ′(s) = − A Ψ (s),Ψ (0) = I.
Now the eigenvalues of −A have real parts less than −δ because these eigenvalues are obtained
from the eigenvalues of A by multiplying by −1. Then by Theorem D.5.3 there exists a constant,
C such that for any x,
|Ψ (s)x| ≤ Ce− δs|x|.
Therefore, from the definition of Ψ,
|Φ (t) x| ≤ Ceδt|x |.■
Here is another essential lemma which is found in Coddington and Levinson [6]
Lemma D.6.2Let pj
(t)
be polynomials with complex coefficients and let
∑m λjt
f (t) = pj (t)e
j=1
where m ≥ 1,λj≠λkfor j≠k, and none of the pj
(t)
vanish identically. Let
σ = max (Re (λ1),⋅⋅⋅,Re (λm )).
Then there exists a positive number, r and arbitrarily large positive values of t suchthat
e−σt|f (t)| > r.
In particular,
|f (t)|
is unbounded.
Proof: Suppose the largest exponent of any of the pj is M and let λj = aj + ibj. First assume
each aj = 0. This is convenient because σ = 0 in this case and the largest of the Re
(λj)
occurs in
every λj.
Then arranging the above sum as a sum of decreasing powers of t,
f (t) = tM fM (t)+ ⋅⋅⋅+ tf1(t)+ f0(t).
Then
( )
t−Mf (t) = fM (t)+ O 1
t
where the last term means that tO
(1)
t
is bounded. Then
∑m ibjt
fM (t) = cje
j=1
It can’t be the case that all the cj are equal to 0 because then M would not be the highest power
exponent. Suppose ck≠0. Then
1 ∫ T ∑m 1 ∫ T
lim -- t−M f (t)e−ibktdt = cj-- ei(bj−bk)tdt = ck ⁄= 0.
T→∞ T 0 j=1 T 0
Letting r =
|ck∕2|
, it follows
| |
|t−Mf (t)e− ibkt|
> r for arbitrarily large values of t. Thus it is also
true that
|f (t)|
> r for arbitrarily large values of t.
Next consider the general case in which σ is given above. Thus
−σt ∑ bjt
e f (t) = pj(t)e + g(t)
j:aj=σ
where limt→∞g
(t)
= 0, g
(t)
being of the form ∑sps
(t)
e
(as−σ+ibs)
t where as− σ < 0.
Then this reduces to the case above in which σ = 0. Therefore, there exists r > 0 such
that
||−σt ||
e f (t) > r
for arbitrarily large values of t. ■
Next here is a Banach space which will be useful.
Lemma D.6.3For γ > 0,let
{ }
Eγ = x ∈ BC ([0,∞),Fn) : t → eγtx(t) is also in BC ([0,∞ ),Fn)