Electronic Journal of Differential Equations, Vol. 2008(2008), No. 94, pp. 1–5.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
A CHARACTERIZATION OF DICHOTOMY IN TERMS OF
BOUNDEDNESS OF SOLUTIONS FOR SOME CAUCHY
PROBLEMS
AKBAR ZADA
Abstract. We prove that a quadratic matrix of order n having complex entries is dichotomic (i.e. its spectrum does not intersect the imaginary axis) if
and only if there exists a projection P on Cn such that P etA = etA P for all
t ≥ 0 and for each real number µ and each vector b ∈ Cn the solutions of the
following two Cauchy problems are bounded:
ẋ(t) = Ax(t) + eiµt P b,
t ≥ 0,
x(0) = 0
and
ẏ(t) = −Ay(t) + eiµt (I − P )b,
t ≥ 0,
y(0) = 0 .
1. Notation and Preliminary Results
It is well-known that if a nonzero solution of the scalar differential equation
ẋ(t) = ax(t), t ≥ 0 is asymptotically stable then each another solution has the
same property and this it happens if and only if the real part of the complex
number a is negative or if and only if for each real number µ and each complex
number b the solution of the Cauchy Problem
ż(t) = az(t) + eiµt b,
t ≥ 0,
z(0) = 0
is bounded.
This result can be extended with approximatively the same formulation for the
case of bounded linear operators acting on a Banach space X. See [1]. The result
can be also extended for strongly continuous bounded semigroups, see [3, 4, 5, 8].
For discrete systems see [6].
Under a slightly different assumption the result on stability is also preserved
for any strongly continuous semigroups acting on complex Hilbert spaces, see for
example [9, 10] and references therein. See also, [7, 2], for counter-examples. In
2000 Mathematics Subject Classification. 47D06, 35B35.
Key words and phrases. Stable and dichotomic matrices; Cauchy problem;
spectral decomposition theorem.
c
2008
Texas State University - San Marcos.
Submitted May 29, 2008. Published July 5, 2008.
1
2
A. ZADA
EJDE-2008/94
this paper the above result is extended for the case of dichotomic matrices. Our
proof uses the Spectral Decomposition Theorem which is reminded below.
It is well known that the solution of the linear Cauchy Problem
ẋ(t) = Ax(t),
tA
x(0) = x0 ,
t∈R
(1.1)
n
is given by x(t) = e x0 . Here x0 ∈ C and A is a quadratic matrix of order n having
complex numbers as entries. Let pA be the characteristic polynomial associated
with the matrix A and let σ(A) = {λ1 , λ2 , . . . , λk }, k ≤ n be its spectrum. There
exist the integer numbers n1 , n2 , . . . , nk ≥ 1 such that
pA (λ) = (λ − λ1 )n1 (λ − λ2 )n2 . . . (λ − λk )nk ,
n1 + n2 + · · · + nk = n.
Let j ∈ {1, 2, . . . , k} and Yj := ker(A − λj I)nj . Clearly Yj is an etA -invariant
subspace of Cn and dim(Yj ) ≥ 1. Moreover, the following Spectral Decomposition
Theorem holds.
Theorem 1.1. For each x ∈ Cn there exist yj ∈ Yj (j ∈ {1, 2, . . . , k}) such that
etA x = etA y1 + etA y2 + · · · + etA yk ,
t ∈ R.
Moreover, if yj (t) := etA yj then yj (t) ∈ Yj for all t ∈ R and there exist a Cn -valued
polynomials pj (t) with deg (pj ) ≤ nj − 1 such that
yj (t) = eλj t pj (t),
t ∈ R, j ∈ {1, 2, . . . , k}.
(1.2)
To make this article self contained, we give here a sketch of the proof. From the
Hamilton-Cayley theorem and using the fact that
ker[pq(A)] = ker[p(A)] ⊕ ker[q(A)]
whenever the complex valued polynomials p and q are relative prime follows that
Cn = Y1 ⊕ Y2 ⊕ · · · ⊕ Yk .
(1.3)
n
Let x ∈ C . For each j ∈ {1, 2, . . . , k} there exists a unique yj ∈ Yj such that
x = y1 + y 2 + · · · + y k
and then
−λj t
Let zj (t) = e
etA x = etA y1 + etA y2 + · · · + etA yk ,
yj (t). A simple calculus shows that
t ∈ R.
(1.4)
dnj zj (t)
= e−λj t (A − λj )nj yj (t) = 0.
dt
The last equality follows because yj (t) belongs to Yj for each t ∈ R. Then zj is a
Cn -valued polynomial having degree less then nj .
2. Dichotomy and Boundedness
Let us denote C+ := {z ∈ C : Re(z) > 0}, C− := {z ∈ C : Re(z) < 0} and iR =
{iη : η ∈ R}. Clearly C = C+ ∪ C− ∪ iR.
A matrix A is called:
(i) stable if σ(A) belongs to C− or, equivalently, if there exist two positive
constants N and ν such that ketA k ≤ N e−νt for all t ≥ 0,
(ii) expansive if σ(A) belongs to C+ and
(iii) dichotomic if σ(A) does not intersect the set iR.
Our first result reads as follows.
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A CHARACTERIZATION OF DICHOTOMY
3
Theorem 2.1. The matrix A is stable if and only if for each µ ∈ R and each
b ∈ Cn the solution of the Cauchy Problem
ẏ(t) = Ay(t) + eiµt b,
t ≥ 0,
y(0) = 0
(2.1)
is bounded.
Proof. Necessity: Let µ ∈ R and b ∈ Cn be given. The solution of (2.1) is
Z t
ψµ,b (t) =
e(t−s)A eiµs b ds t ≥ 0.
0
Then
Z
t
ke(t−s)A eiµs bkds
kψµ,b (t)k ≤
0
Z
t
ke(t−s)A kkbkds
=
0
Z
≤
t
N e−ν(t−s) kbkds
0
≤
N
kbk.
ν
Thus ψµ,b is bounded.
Sufficiency: The solution of (2.1) can be written as
Z t
ψµ,b (t) = eiµt
e(−iµI+A)r b dr, t ≥ 0.
0
It is clear that
σ(−iµI + A) = {−iµ + λ1 , −iµ + λ2 , . . . , −iµ + λk }.
Suppose for the contrary that the matrix A is not stable i.e. there exists ν ∈
{1, 2, . . . , k} such that Re(λν ) ≥ 0. Then Re(−iµ + λν ) ≥ 0. Denoting Bµ :=
(−iµI + A) may write
eBµ s b = eBµ s z1 + eBµ s z2 + · · · + eBµ s zk .
Let choose b = zν be a nonzero vector. Then e(−iµI+A)s b = eBµ s zν and by Spectral
Decomposition Theorem follows that eBµ s zν = eµν s p(s), where µν := −iµ + λν and
p is a Cn -valued polynomial with deg(p) ≤ nν − 1.
We are going to consider two cases.
Case 1: When Re(λν ) > 0. Applying again Theorem 1.1 above, obtain
Z t
iµt
ψµ,b (t) = e
eµν s p(s)ds = eiµt eµν t q(t), t ≥ 0.
0
Here q(t) is a Cn -valued, nonzero polynomial. Thus ψµ,b is an unbounded function
and we arrived at a contradiction.
Case 2: When Re(λν ) = 0. Let µ = λiν . Then µν t = 0 and
(
Z t
eiµt q(t)if deg(p) ≥ 1
iµt
ψµ,b (t) = e
p(s)ds =
teiµt
if deg(p) = 0.
0
Here q is a polynomial of degree greater than 1. In this case ψµ,b is also unbounded,
which is a contradiction.
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A. ZADA
EJDE-2008/94
Having in mind that A is expansive if and only if (−A) is stable we obtain the
following result.
Corollary 2.2. The matrix A is expansive if and only if for each µ ∈ R and each
b ∈ Cn the solution of (2.1), with −A instead of A, is bounded.
A linear map P acting on Cn is called projection if P 2 = P .
Theorem 2.3. The matrix A is dichotomic if and only if there exist a projection
P having the property etA P = P etA for all t ≥ 0 such that for each µ and each
b ∈ Cn the following two inequalities hold
Z t
e(−iµ+A)s P bdsk < ∞,
(2.2)
sup k
t≥0
0
Z
sup k
t≥0
t
e(−iµ−A)s (I − P )bdsk < ∞ .
(2.3)
0
Proof. Necessity: Working under the assumption that A ia a dichotomic matrix
we may suppose that there exists ν ∈ {1, 2, . . . , k} such that
Re(λ1 ) ≥ Re(λ1 ) ≥ . . . <(λν ) > 0 > <(λν+1 ) ≥ · · · ≥ <(λk ).
Having in mind the decomposition of Cn given by (1.3) let us consider
X0 = Y1 ⊕ Y2 ⊕ · · · ⊕ Yν ,
X1 = Yν+1 ⊕ Yν+2 ⊕ · · · ⊕ Yk .
Then Cn = X0 ⊕ X1 . Let us define P : Cn → Cn by P x = x0 , where x = x0 + x1 ,
x0 ∈ X0 and x1 ∈ X1 . It is clear that P is a projection. Moreover for all x ∈ Cn
and all t ≥ 0, this yields
P etA x = P (etA (x0 + x1 )) = P (etA (x0 ) + etA (x1 )) = etA (x0 ) = etA P x,
where the fact that X0 is an etA − invariant subspace, was used. Then P etA = etA P .
Now, we have
es(−iµI+A) P b = e−iµs P esA b
= e−iµs P eλ1 s p1 (s) + · · · + eλν s pν (s) + · · · + eλk s pk (s)
= e−iµs eλ1 s p1 (s) + eλ2 s p2 (s) + · · · + eλν s pk (s) ,
where p1 , pR2 , . . . , pν are polynomials as in Theorem 1.1. Now it is clear that the
t
map t 7→ 0 e(−iµ+A)s P bds is bounded. The condition (2.3) can be verified in a
similar manner.
Sufficiency: Suppose for a contradiction that A is not dichotomic. Then there
exists j ∈ {1, 2, . . . , k} such that λj = iη with η ∈ R. Let us take b = xj ∈ Yj ,
xj 6= 0 and consider xj0 =: P xj and xj1 =: (I − P )xj . We have
Z t
Z t
ψµ,P xj (t) =
e−iµs esA P xj ds =
ei(−µ+η)s pj (s)ds.
0
0
First consider the case when deg(pj ) ≥ 1. If xj0 6= 0 the map t 7→ ψµ,P xj (t) is
clearly unbounded and if xj1 6= 0 we may repeat the above argument in order to
Rt
arrive at the contradiction that the map t 7→ 0 e−iµs e−sA (I −P )xj ds is unbounded.
Next we consider the case when deg(pj ) = 0. If xj0 6= 0 choose µ = η and then
EJDE-2008/94
A CHARACTERIZATION OF DICHOTOMY
5
there exists pj ∈ Yj , pj 6= 0 such that ψµ,P xj (t) = tpj which is unbounded. If
xj0 = 0 choose µ = −η and then
Z t
Z t
Z t
−iµs −sA
−iµs −sA
e
e
(I − P )xj ds =
e
e
xj1 ds =
qj ds = qj t, qj ∈ Yj .
0
0
0
Here qj 6= 0 because xj1 6= 0. Then ψµ,(I−P )xj is an unbounded map. This is a
contradiction and the proof is complete.
It is clear that the Theorem 2.1 is a particular case (P = 1) of Theorem 2.3. We
present its proof because it is different from that in [1].
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Akbar Zada
Government College University, Abdus Salam School of Mathematical Sciences, (ASSMS),
Lahore, Pakistan
E-mail address: zadababo@yahoo.com