Electronic Journal of Qualitative Theory of Differential Equations2015, No. 58, 1–12; doi: 10.14232/ejqtde.2015.1.58 http://www.math.u-szeged.hu/ejqtde/

Hyers–Ulam stability and exponential dichotomy oflinear differential periodic systems are equivalent

Dorel Barbu1, Constantin Bus, eB 1 and Afshan Tabassum2

1West University of Timis, oara, Department of Mathematics,Bd. V. Pârvan No. 4, Timis, oara – 300223, România

2Government College University, Abdus Salam School of Mathematical Sciences, (ASSMS),Lahore, Pakistan

Received 7 March 2015, appeared 17 September 2015

Communicated by Nickolai Kosmatov

Abstract. Let m be a positive integer and q be a positive real number. We prove that them-dimensional and q-periodic system

x(t) = A(t)x(t), t ∈ R+, x(t) ∈ Cm (∗)

is Hyers–Ulam stable if and only if the monodromy matrix associated to the family{A(t)}t≥0 possesses a discrete dichotomy, i.e. its spectrum does not intersect the unitcircle.

Keywords: differential equations, dichotomy, Hyers–Ulam stability.

2010 Mathematics Subject Classification: 12H20, 34D09, 39B82.

1 Introduction

The notion of exponential dichotomy comes from a paper published in 1930 by Oscar Perron[25]. Over the years this concept has proven to be very useful in investigating properties ofthe solutions of ordinary and functional differential equations. In particular, the existence ofbounded and periodic solutions of several families of semi-linear systems has been studiedusing the Green matrix G(t, s) of the system (∗) and concluding that for any bounded f , theconvolution G ∗ f is a bounded solution of the non-homogeneous linear system

x(t) = A(t)x(t) + f (t). (1.1)

In 1940 S. M. Ulam has tackled some open problems (see [30] and [31]), one of those problemsconcerns the stability of a certain functional equation. The first answer to that problem wasprovided by D. H. Hyers in 1941, see [15]. Later on, this was coined as the Hyers–Ulamproblem and its study became an extensive object for many mathematicians. See for example[1, 3–7, 12, 13, 16–24, 27–29, 32] and the references therein.

BCorresponding author. Email: buse@math.uvt.ro, buse1960@gmail.com

2 D. Barbu, C. Bus,e and A. Tabassum

The set of all m×m matrices having complex entries will be denoted by Cm×m. Denote byIm the identity matrix in Cm×m. Assume that the map t 7→ A(t) : R 7→ Cm×m is continuousand then the Cauchy problem{

X(t) = A(t)X(t), t ∈ R, X(t) ∈ Cm×m

X(0) = Im,(1.2)

has a unique solution denoted by ΦA(t). It is well known that ΦA(t) is an invertible matrixand that its inverse is the unique solution of the Cauchy problem{

X(t) = −X(t)A(t), t ∈ R

X(0) = Im.

The evolution family UA = {UA(t, s) : t, s ∈ R}, where

UA(t, s) := ΦA(t)Φ−1A (s),

has the following properties:

(i) UA(t, t) = Im, for all t ∈ R;

(ii) UA(t, s) = UA(t, r)UA(r, s) for all t, s, r ∈ R;

(iii) ∂∂t UA(t, s) = A(t)UA(t, s) for all t, s ∈ R;

(iv) ∂∂s UA(t, s) = −UA(t, s)A(s) for all t, s ∈ R;

(v) the map (t, s) 7→ UA(t, s) : R2 → Cm×m is continuous.

If, in addition, the map A(·) is q- periodic, for some positive number q, then:

(vi) UA(t + q, s + q) = UA(t, s) for all t, s ∈ R;

(vii) there exist ω > 0 and Mω ≥ 1 such that

‖UA(t, s)‖ ≤ Mωeω(t−s), t ≥ s;

(viii) ΦA(t + q) = ΦA(t) ·ΦA(q) for all t ∈ R.

To prove the latter statement, we remark that the map t 7→ ΦA(t + q)(ΦA(q))−1 is asolution of (1.2). Now, by using the uniqueness it must be ΦA(·). The matrix Tq := UA(q, 0) isthe matrix of monodromy associated with the family A. Having in mind that Tq is invertiblethere exists a matrix B ∈ Cm×m such that Tq = eqB. Thus there is a periodic (period q) matrixfunction t 7→ R(t) such that ΦA(t) = R(t)etB for all t ∈ R. This will be used to show thatcertain family of projections described below is periodic.

The complex unit circle is denoted by Γ := {z ∈ C : |z| = 1}. Recall that the matrix Ais said to be dichotomic (or that it possesses a discrete dichotomy) if its spectrum does notintersect the unit circle, i.e. σ(A) ∩ Γ = ∅. An m × m complex matrix P, verifying P2 = Pis called projection. The circle and closed disk centered in the eigenvalue λj ∈ σ(A) arerespectively denoted by

Cr(λj) = {z ∈ C : |z− λj| = r}

Hyers-Ulam stability and exponential dichotomy 3

andDr(λj) = {z ∈ C : |z− λj| ≤ r}.

Here r is any positive real number, small enough such that σ(A) ∩ Dr(λj) = {λj}, for every1 ≤ j ≤ k. The projection Eλj(A) := Ej(A) : Cm → Cm, defined by

Ej(A) =1

2πi

∮Cr(λj)

(zIm − A)−1 dz,

is called spectral projection associated to the eigenvalue λj, [10, Chap. 7]. Obviously, Im =

Eλ1(A) + Eλ2(A) + · · ·+ Eλk(A). The stable spectral projection of A is given by

Π−(A) :=1

2πi

∮Cr(0)

(zIm − A)−1 dz,

where 0 < r < 1 is large enough such that

{λ ∈ σ(A) : |λ| < 1} ⊂ {λ ∈ C : |λ| < r}.

Clearly, Π−(A) commutes with any natural power of A.Coming back to the non-autonomous case let Π− := Π−(Tq) and let

Π−(t) := ΦA(t)Π−(ΦA(t))−1 and Π+(t) := Im −Π−(t)

for t ∈ R. Next, we list the main properties of this family of projections.

(i) Π2−(t) = Π−(t) and Π2

+(t) = Π+(t) for all t ∈ R.

(ii) Π±(t)U(t, s) = U(t, s)Π±(s) for all t, s ∈ R, (the signs correspond).

(iii) The maps t 7→ Π±(t) are continuous on R and q-periodic.

(iv) Π−(t) + Π+(t) = Im and Π−(t) ·Π+(t) = 0 for all t ∈ R.

(v) For each t, s ∈ R, U(t, s) is an isomorphism from ker(Π−(s)) to ker(Π−(t)).

Proposition 1.1. The following two statements, concerning an invertible m×m matrix A, are equiv-alent.

(1) A possesses a discrete dichotomy.

(2) There exist four positive constants N1 = N1(A), N2 = N2(A), ν1 = ν1(A), ν2 = ν2(A) suchthat

(i) ‖AnΠ−(A)x‖ ≤ N1e−ν1n‖Π−(A)x‖, for all x ∈ Cm and all n ∈ Z+.

(ii) ‖AnΠ+(A)x‖ ≤ N2eν2n‖Π+(A)x‖, for all x ∈ Cm and all n ∈ Z− := {0,−1,−2, . . . }.

The argument is standard and the details are omitted. Mention that the above result canbe stated in a more general form with any projection P, commuting with A, instead of Π−(A).Moreover, the assumption of invertibility can be removed. See, for example, Proposition 2.1from [2]. For further details about the concept of dichotomy see for example [8, 26].

4 D. Barbu, C. Bus,e and A. Tabassum

Let t 7→ f (t) be a Cm-valued locally Riemann integrable function on R+ and let x ∈ Cm bea given vector. Consider the Cauchy problem{

x(t) = A(t)x(t) + f (t), t ≥ 0

x(0) = x.(1.3)

The solution of (1.3) is given by

φ f ,x(t) = UA(t, 0)x +∫ t

0UA(t, s) f (s) ds.

In order to prove Theorem 1.3 below, we need the following proposition, which containsequivalent characterizations for exponential dichotomy.

Proposition 1.2. The following three statements concerning the matrix family A are equivalent.

(1) Tq is dichotomic.

(2) There exist the positive constants N′1, N′2, ν′1, ν′2 such that

(i) ‖UA(t, s)Π−(s)‖ ≤ N′1e−ν′1(t−s), for all t ≥ s ≥ 0, and

(ii) ‖UA(t, s)Π+(s)‖ ≤ N′2eν′2(t−s), for all 0 ≤ t ≤ s.

(3) For each locally Riemann integrable and bounded function f : R+ → Cm there exists a uniquex ∈ ker(Π−), such that φ f ,x(·) is bounded on R+.

Proof. (1) ⇒ (2) Let t ≥ s ∈ R+ and let n and k be the integer parts of tq and s

q respectively,i.e., n = [ t

q ] and k = [ sq ]. Therefore t = nq + µ and s = kq + ρ, with n, k ∈ Z+ and µ, ρ ∈ [0, q).

We analyze the following cases.Case 1. When n > k, then

UA(t, s)Π−(s) = UA(nq + µ, nq)UA(nq, (k + 1)q)UA((k + 1)q, kq + ρ)Π−(kq + ρ)

= UA(µ, 0)UA((n− k− 1)q, 0)UA(q, ρ)Π−(ρ)

= UA(µ, 0)Tn−k−1q Π−UA(q, ρ).

In the view of Proposition 1.1 and taking into account that Π−(0) = Π−(q) = Π−, we get

‖UA(t, s)Π−(s)‖ ≤ MeωqN1e−ν1(n−k−1)Meωq‖Π−‖≤ N′1e−ν′1(t−s),

where N′1 = N1M2e2ωqe2ν1‖Π−‖ and ν′1 = ν1q .

Case 2. When n = k, then µ ≥ ρ and

‖UA(s, t)Π−(s)‖ = ‖UA(µ, ρ)Π−(ρ)‖.

By using supρ∈[0,q] ‖Π−(ρ)‖ ≤ c < ∞ and letting ν be an arbitrary positive number, wemay choose N ∈ R+ large enough, such that

‖UA(t, s)Π−(s)‖ ≤ Meω(µ−ρ)‖Π−(ρ)‖ ≤ cNe−ν(µ−ρ) = N′1e−ν′1(t−s).

Similar estimations can be obtained in order to prove (2) (ii). We omit the details.

Hyers-Ulam stability and exponential dichotomy 5

(2) ⇒ (1). Put s = 0 and t = nq in (2) (i), (ii) and apply Proposition 1.1 with Tq insteadof A.

(2)⇒ (3). The map

t 7→ y(t) :=∫ t

0UA(t, s)Π−(s) f (s) ds−

∫ ∞

tUA(t, s)Π+(s) f (s) ds

is a solution of (1.1), [8, Chap. 3]. Indeed, the second integral is well defined because, from(2) (ii), have that ∫ ∞

t‖UA(t, s)Π+(s) f (s)‖ ds ≤

∫ ∞

tN′2eν′2(t−s)‖ f ‖∞ ds

=N′2ν′2‖ f ‖∞.

Also from (2), the solution is bounded, and

supt≥0|y(t)| ≤

(N′1ν′1

+N′2ν′2

)supt≥0| f (t)|.

Moreover, since ker(Π−) is a closed subspace, the initial value

y(0) = −∫ ∞

0UA(0, s)Π+(s) f (s) ds ∈ ker(Π−).

Let us suppose that there exist two bounded solutions of the differential equation x(t) =A(t)x(t) + f (t), t ≥ 0 having their start in ker(Π−). Denote them by y1(·) and y2(·). Then

y1(t) = UA(t, 0)x1 +∫ t

0UA(t, s) f (s) ds, x1 ∈ ker(Π−)

and

y2(t) = UA(t, 0)x2 +∫ t

0UA(t, s) f (s) ds, x2 ∈ ker(Π−).

Their difference is bounded and y1(t)− y2(t) = UA(t, 0)(x1 − x2). Since the map y1(·)− y2(·)is bounded on R+, and because Tq is dichotomic it follows that x1 − x2 ∈ Range(Π−). On theother hand, x1, x2 ∈ ker(Π−) yields x1 − x2 ∈ ker(Π−) and therefore x1 = x2.

(3)⇒ (1). Suppose that there exists λ ∈ σ(T), with |λ| = 1. Then, there exists x0 6= 0 suchthat Tqx0 = λx0, and therefore UA(nq, 0) = λnx0, for all n ∈ Z+.

Set

f (t) :=

{UA(s, 0)x0, if s ∈ [0, q)

x0, if s = q,

and let us denote also by f its continuation by periodicity on R+. By assumption there existsa unique y0 ∈ ker(Π−) such that the map

t 7→ ψ(t) := UA(t, 0)y0 +∫ t

0UA(t, s) f (s) ds

is bounded on R+. Next we analyze two cases.

6 D. Barbu, C. Bus,e and A. Tabassum

Case 1. When λ = 1. The sequence (ψ(nq))n∈Z+ should be bounded. But,

ψ(nq) := UA(nq, 0)y0 +∫ nq

0UA(nq, s) f (s) ds

= UA(nq, 0)y0 +n−1

∑k=0

∫ (k+1)q

kqUA(nq, s) f (s) ds

= UA(nq, 0)y0 +n−1

∑k=0

UA(nq, (k + 1)q)∫ q

0UA(q, r) f (r) dr

= UA(nq, 0)y0 +n−1

∑k=0

UA(nq, kq)x0 = UA(nq, 0)y0 + nx0.

If y0 = 0, obviously we arrive at a contradiction, since the map n 7→ nx0 is unbounded,and if y0 6= 0, let denote y0(n) := UA(nq, 0)y0. Then, one has

‖y0‖ = ‖UA(0, nq)y0(n)‖= ‖UA(0, nq)Π+y0(n)‖ ≤ N′2e−ν′2nq‖UA(nq, 0)y0‖,

where the fact that

Π+y0(n) = UA(nq, 0)Π+y0 = UA(nq, 0)(y0 −Π−y0) = y0(n),

was used. This yields

‖UA(nq, 0)y0‖ ≥1

N′2eν′2nq‖y0‖,

and a contradiction arises again.Case 2. When λ = eiuq 6= 1, u ∈ R, i2 = −1. Then 1 ∈ σ(e−iuqT), Tu(q) := e−iuqTq is themonodromy matrix of the evolution family

{UA,u(t, s) := e−iu(t−s)UA(t, s) : t, s ∈ R}

and, as before, we obtain that the sequence

(e−iunqψ(nq))n∈Z+ = (UA,u(nq, 0)y0 + qnx0)n∈Z+ ,

is unbounded, which is a contradiction.

In the present paper we assume that the matrix-valued map t 7→ A(t) is continuous andq-periodic for some positive q. Next we outline the Hyers–Ulam problem for a family of m×mmatrices A = {A(t)}t≥0, m being a positive integer. Let R+ be the set of all nonnegative realnumbers and let ρ(·) be a Cm-valued function defined on R+. Consider the systems

x(t) = A(t)x(t), t ∈ R, x(t) ∈ Cm (1.4)

andx(t) = A(t)x(t) + ρ(t), t ∈ R+, x(t) ∈ Cm. (1.5)

Let ε be a positive real number. A continuous Cm-valued function y(·) defined on R+ := [0, ∞)

is called ε-approximate solution for (1.4) if it is continuously differentiable on R+ \ (qZ+) and

‖y(t)− A(t)y(t)‖ ≤ ε, ∀t ∈ R+ \ (qZ+). (1.6)

Hyers-Ulam stability and exponential dichotomy 7

The family A is said to be Hyers–Ulam stable if there exists a nonnegative constant L suchthat, for every ε-approximate solution φ(·) of (1.4), there exists an exact solution θ(·) of (1.4)such that

supt∈R+

‖φ(t)− θ(t)‖ ≤ Lε. (1.7)

The result of this paper reads as follows.

Theorem 1.3. The family A = {A(t)}t≥0 is Hyers–Ulam stable if and only if its monodromy matrixTq possesses a discrete dichotomy.

2 Hyers–Ulam stability and exponential dichotomy for linear differ-ential systems

We can see an ε-approximate solution of (1.4) as an exact solution of (1.5) corresponding to aforced term ρ(·) which is bounded by ε.

Remark 2.1. Let ε be a given positive number. The following two statements are equivalent:

1. The matrix family A (or the system (1.4)) is Hyers–Ulam stable.

2. There exists a nonnegative constant L such that for every function ρ(·), continuous onR+ \ (qZ+), with supt≥0 ‖ρ(t)‖ ≤ ε, and every x ∈ Cm there exists x0 ∈ Cm and

supt≥0

∥∥∥∥UA(t, 0)(x− x0) +∫ t

0UA(t, s)ρ(s) ds

∥∥∥∥ ≤ Lε. (2.1)

Proof. Let ε be a given positive number. Assume first that the system (1.4) is Hyers–Ulamstable and let L be a positive constant verifying (1.7). Let ρ(·) be as assumed in the secondstatement and x ∈ Cm. Obviously, the solution φ(·) of the Cauchy problem

x(t) = A(t)x(t) + ρ(t), x(0) = x

is an ε-approximative solution for (1.4). Thus, by assumption, there exists an exact solutionθ(·) of (1.4) such that (1.7) holds true. Let x0 := θ(0). Now, in view of (1.6) the inequality in(2.1) holds true as well.

Now assume that the second statement is true and let L be a positive constant verifying(2.1) and φ(·) be an ε-approximative solution of (1.4). Set ρ(t) := φ(t) − A(t)φ(t) for t ∈R+ \ (qZ+) and ρ(t) := ε in the rest, and let x := φ(0). Thus ‖ρ‖∞ ≤ ε and, by assumption(2.1) holds true for a certain x0 ∈ Cm. The required exact solution of (1.4), verifying (1.7), isdefined by θ(t) := U(t, 0)x0.

Proof of Theorem 1.3.Necessity. Suppose that Tq is not dichotomic. Then, there exist an integer j with 1 ≤ j ≤ k andλj = eiµjq ∈ σ(Tq), where µj is a certain real number. Let ε > 0 be fixed and let

ρ(t) :=

{UA(s, 0)u0, if s ∈ [0, q)

u0, if s = q,

where u0 ∈ Cm and ‖u0‖ ≤ (Mωeωq)−1ε. Let us denote also by ρ the continuation by period-icity of the previous function. Obviously, the function ρ(·) is locally Riemann integrable on

8 D. Barbu, C. Bus,e and A. Tabassum

R+ and bounded by ε. By assumption, the family matrix A is Hyers–Ulam stable. Hence, thesolution

φ(t) = UA(t, 0)(x− x0) +∫ t

0UA(t, s)ρ(s) ds,

of the Cauchy problem {y(t) = A(t)y(t) + ρ(t), t ≥ 0

y(0) = x− x0,

is bounded by Lε for certain x− x0 ∈ Cm. Then, the sequence

n 7→ Ej(Tq)φ(nq) = Ej(Tq)

[UA(nq, 0)(x− x0) +

∫ nq

0UA(nq, s)ρ(s) ds

]should be also bounded by Lε. On the other hand, see for example [2, Lemma 4.5], [9, 11, 14],there exists an m×m matrix-valued polynomial Pj = Pj(Tq) (in n) having the degree at mostmj − 1, such that

Ej(Tq)UA(nq, 0) = eiµjqnPj(n), ∀n ∈ Z+.

But,

Ej(Tq)

[UA(nq, 0)(x− x0) +

∫ nq

0UA(nq, s)ρ(s) ds

]= eiµjnqPj(n)(x− x0) +

∫ nq

0Ej(Tq)UA(nq, s)ρ(s) ds

= eiµjnqPj(n)(x− x0) +n−1

∑k=0

∫ (k+1)q

kqEj(Tq)UA(nq, s)ρ(s) ds

= eiµjnqPj(n)(x− x0) +n−1

∑k=0

∫ q

0Ej(Tq)UA(nq, (k + 1)q)UA(q, s)ρ(s) ds

= eiµjnqPj(n)(x− x0) +n−1

∑k=0

λn−kj Pj(n− k)u0.

Now, if λj = 1 then by choosing an appropriate u0 6= 0, we have that

deg[Pj(n)(x− x0)] ≤ deg[Pj(n)] = deg[Pj(n)u0] < 1 + deg[Pj(n)]

= deg[qj(n)],

where qj(n) := ∑n−1k=0 Pj(n− k)u0 and the fact that the degree of the polynomial in n, p(n) =

1k + 2k + · · · + nk, is equal to k + 1 was used. Therefore, the sequence (Pj(n)(x − x0) +

qj(n))n∈Z+ , is unbounded and a contradiction arises.When λj 6= 1, then 1 ∈ σ(Tµj(q)) and the map t 7→ e−iµjtφ(t) should be bounded on R+.

Then the sequencen 7→ e−iµjnqφ(nq), n ∈ Z+,

is bounded as well. On the other hand

e−iµjnqEj(Tµj(q))φ(nq) = Ej(Tµj(q))[

UA,µj(nq, 0)(x− x0) +∫ nq

0UA,µj(nq, s)e−iµjsρ(s) ds

].

Again, as above, there exists a matrix valued polynomial Qj(n) = Qj(Tµj(q)) (in n) having thedegree at most mj − 1 such that

Ej(Tµj(q))UA,µj(nq, 0) = Qj(n) for every n ∈ Z+.

Hyers-Ulam stability and exponential dichotomy 9

Thus after a standard calculation

e−iµjnqEj(Tµj(q))φ(nq) = Qj(n)(x− x0) +n−1

∑k=0

Qj(n− k)u0.

For an appropriate u0 ∈ Cm, the last expression is a vector valued polynomial of degree atleast one and so it is unbounded and a contradiction is provided again.

Sufficiency. The absolute constant L will be settled later. Let ρ : R+ → Cm be a boundedlocally Riemann integrable function on R+, with ‖ρ‖∞ ≤ ε and let x ∈ Cm. By Proposition 1.1,there exists a unique bounded solution y(·) of the equation (1.5) starting from the subspaceker(Π−). Let denote u0 := y(0). Then

‖y(t)‖ =∥∥∥∥UA(t, 0)u0 +

∫ t

0UA(t, s)ρ(s) ds

∥∥∥∥=

∥∥∥∥∫ t

0UA(t, s)Π−(s)ρ(s)ds−

∫ ∞

tUA(t, s)Π+(s)ρ(s) ds

∥∥∥∥≤(

N′1ν′1

+N′2ν′2

)ε.

The desired assertion follows by choosing L =(

N′1ν′1

+N′2ν′2

)and setting x0 = x− u0.

A more general result, described in the following, can be stated. Its proof is very similarto that given before and we omit the details.

Let X be a complex, finite dimensional Banach space and let A = {A(t)}t∈R+ and P =

{P(t)}t∈R+ be two families of linear operators acting on X. Assume the following.

H1. A(t + q) = A(t) and P(t + q) = P(t), for all t ∈ R+ and some positive q.

H2. P(t)2 = P(t), for all t ∈ R+, i.e., P is a family of projections.

H3. UA(t, s)P(s) = P(t)UA(t, s), for any t ≥ s ∈ R+. In particular, this yields that UA(t, s)x ∈ker(P(t)) for each x ∈ ker(P(s)).

H4. For each t ≥ s ∈ R+, the map

x 7→ UA(t, s)x : ker(P(s))→ ker(P(t))

is invertible. Denote by UA|(s, t) its inverse.

We say that the family A is P-dichotomic if there exist four positive constants N1, N2, ν1

and ν2 such that

(i) ‖UA(t, s)P(s)‖ ≤ N1e−ν1(t−s) for all t ≥ s ≥ 0;

(ii) ‖UA|(t, s)(I − P(s))‖ ≤ N2eν2(t−s) for all 0 ≤ t < s.

Proposition 2.2. Assume that the families A and P satisfy H1–H4 above. Thus the following threestatements are equivalent.

(1) Tq possesses a discrete dichotomy.

(2) The family A is P-dichotomic.

10 D. Barbu, C. Bus,e and A. Tabassum

(3) The family A is Hyers–Ulam stable.

We conclude this note with the one-dimensional version of our result.

Corollary 2.3. Let t 7→ a(t) : R+ → C be a given continuous and q-periodic function (for somepositive q). The scalar differential equation

x(t) = a(t)x(t), t ∈ R+, x(t) ∈ C (2.2)

is Hyers–Ulam stable if and only if ∫ q

0<[a(r)] dr 6= 0.

Proof. Indeed, we have

Tq = e∫ q

0 a(r) dr, σ(Tq) = {Tq} and |Tq| = e∫ q

0 <[a(r)] dr.

From Theorem 1.3 follows that (2.2) is Hyers–Ulam stable if and only if |Tq| 6= 1 or equivalentlyif and only if

∫ q0 <[a(r)] dr 6= 0.

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