Hyers–ulam Stability and Exponential Dichotomy of Linear Differential Periodic Systems Are Equivalent

Let m be a positive integer and q be a positive real number. We prove that the m-dimensional and q-periodic system ˙ x(t) = A(t)x(t), t ∈ R + , x(t) ∈ C m (*) 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 unit circle.


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 of the solutions of ordinary and functional differential equations.In particular, the existence of bounded and periodic solutions of several families of semi-linear systems has been studied using the Green matrix G(t, s) of the system ( * ) and concluding that for any bounded f , the convolution G * f is a bounded solution of the non-homogeneous linear system ẋ(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 problems concerns the stability of a certain functional equation.The first answer to that problem was provided by D. H. Hyers in 1941, see [15].Later on, this was coined as the Hyers-Ulam problem 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.
The set of all m × m matrices having complex entries will be denoted by C m×m .Denote by I m the identity matrix in C m×m .Assume that the map t → A(t) : R → C m×m is continuous and then the Cauchy problem has a unique solution denoted by Φ A (t).It is well known that Φ A (t) is an invertible matrix and that its inverse is the unique solution of the Cauchy problem The evolution family U A = {U A (t, s) : t, s ∈ R}, where has the following properties: If, in addition, the map A(•) is qperiodic, for some positive number q, then: (vi) U A (t + q, s + q) = U A (t, s) for all t, s ∈ R; (vii) there exist ω > 0 and M ω ≥ 1 such that To prove the latter statement, we remark that the map t → Φ A (t + q)(Φ A (q)) −1 is a solution of (1.2).Now, by using the uniqueness it must be Φ A (•).The matrix T q := U A (q, 0) is the matrix of monodromy associated with the family A. Having in mind that T q is invertible there exists a matrix B ∈ C m×m such that T q = e qB .Thus there is a periodic (period q) matrix function t → R(t) such that Φ A (t) = R(t)e tB for all t ∈ R.This will be used to show that certain family of projections described below is periodic.
The complex unit circle is denoted by Γ := {z ∈ C : |z| = 1}.Recall that the matrix A is said to be dichotomic (or that it possesses a discrete dichotomy) if its spectrum does not intersect the unit circle, i.e. σ(A) ∩ Γ = ∅.An m × m complex matrix P, verifying P 2 = P is called projection.The circle and closed disk centered in the eigenvalue λ j ∈ σ(A) are respectively denoted by Here r is any positive real number, small enough such that σ(A) ∩ D r (λ j ) = {λ j }, for every 1 ≤ j ≤ k.The projection E λ j (A) := E j (A) : C m → C m , defined by is called spectral projection associated to the eigenvalue λ j , [10,Chap. 7].Obviously, The stable spectral projection of A is given by Clearly, Π − (A) commutes with any natural power of A.
Coming back to the non-autonomous case let Π − := Π − (T q ) and let for t ∈ R. Next, we list the main properties of this family of projections.
(iii) The maps t → Π ± (t) are continuous on R and q-periodic.
Proposition 1.1.The following two statements, concerning an invertible m × m matrix A, are equivalent.
(2) There exist four positive constants The argument is standard and the details are omitted.Mention that the above result can be 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 2.1 from [2].For further details about the concept of dichotomy see for example [8,26].
Let t → f (t) be a C m -valued locally Riemann integrable function on R + and let x ∈ C m be a given vector.Consider the Cauchy problem (1. 3) The solution of (1.3) is given by In order to prove Theorem 1.3 below, we need the following proposition, which contains equivalent characterizations for exponential dichotomy.
Proposition 1.2.The following three statements concerning the matrix family A are equivalent.
(2) There exist the positive constants N 1 , N 2 , ν 1 , ν 2 such that , for all t ≥ s ≥ 0, and (3) For each locally Riemann integrable and bounded function f : R + → C m there exists a unique x Proof.(1) ⇒ (2) Let t ≥ s ∈ R + and let n and k be the integer parts of t q and s q respectively, i.e., n = [ t q ] and k = [ s q ].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 In the view of Proposition 1.1 and taking into account that Π − (0) = Π − (q) = Π − , we get where By using sup ρ∈[0,q] Π − (ρ) ≤ c < ∞ and letting ν be an arbitrary positive number, we may choose N ∈ R + large enough, such that Similar estimations can be obtained in order to prove (2) (ii).We omit the details.
(2) ⇒ (3).The map is a solution of (1.1), [8,Chap. 3].Indeed, the second integral is well defined because, from Also from (2), the solution is bounded, and Moreover, since ker(Π − ) is a closed subspace, the initial value Let us suppose that there exist two bounded solutions of the differential equation ẋ(t) = A(t)x(t) + f (t), t ≥ 0 having their start in ker(Π − ).Denote them by y 1 (•) and y 2 (•).Then Their difference is bounded and y is bounded on R + , and because T q is dichotomic it follows that x 1 − x 2 ∈ Range(Π − ).On the other hand, (3) ⇒ (1).Suppose that there exists λ ∈ σ(T), with |λ| = 1.Then, there exists x 0 = 0 such that T q x 0 = λx 0 , and therefore U A (nq, 0) = λ n x 0 , for all n ∈ Z + .Set and let us denote also by f its continuation by periodicity on R + .By assumption there exists a unique y 0 ∈ ker(Π − ) such that the map Next we analyze two cases.
Case 1.When λ = 1.The sequence (ψ(nq)) n∈Z + should be bounded.But, If y 0 = 0, obviously we arrive at a contradiction, since the map n → nx 0 is unbounded, and if y 0 = 0, let denote y 0 (n) := U A (nq, 0)y 0 .Then, one has where the fact that was used.This yields and a contradiction arises again.Case 2. When λ = e iuq = 1, u ∈ R, i 2 = −1.Then 1 ∈ σ(e −iuq T), T u (q) := e −iuq T q is the monodromy matrix of the evolution family {U A,u (t, s) := e −iu(t−s) U A (t, s) : t, s ∈ R} and, as before, we obtain that the sequence is unbounded, which is a contradiction.
In the present paper we assume that the matrix-valued map t → A(t) is continuous and q-periodic for some positive q.Next we outline the Hyers-Ulam problem for a family of m × m matrices A = {A(t)} t≥0 , m being a positive integer.Let R + be the set of all nonnegative real numbers and let ρ(•) be a C m -valued function defined on R + .Consider the systems Let ε be a positive real number.A continuous C m -valued function y(•) defined on The family A is said to be Hyers-Ulam stable if there exists a nonnegative constant L such that, for every ε-approximate solution φ(•) of (1.4), there exists an exact solution θ(•) of (1.4) such that sup 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 matrix T q possesses a discrete dichotomy.

Hyers-Ulam stability and exponential dichotomy for linear differential systems
We can see an ε-approximate solution of (1.4) as an exact solution of (1.5) corresponding to a forced 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.
Proof of Theorem 1.3.Necessity.Suppose that T q is not dichotomic.Then, there exist an integer j with 1 ≤ j ≤ k and λ j = e iµ j q ∈ σ(T q ), where µ j is a certain real number.Let ε > 0 be fixed and let where u 0 ∈ C m and u 0 ≤ (M ω e ωq ) −1 ε.Let us denote also by ρ the continuation by periodicity of the previous function.Obviously, the function ρ(•) is locally Riemann integrable on R + and bounded by ε.By assumption, the family matrix A is Hyers-Ulam stable.Hence, the solution 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 P j = P j (T q ) (in n) having the degree at most m j − 1, such that E j (T q )U A (nq, 0) = e iµ j qn P j (n), ∀n ∈ Z + . But, Now, if λ j = 1 then by choosing an appropriate u 0 = 0, we have that where q j (n) := ∑ n−1 k=0 P j (n − k)u 0 and the fact that the degree of the polynomial in n, p(n) = 1 k + 2 k + • • • + n k , is equal to k + 1 was used.Therefore, the sequence (P j (n)(x − x 0 ) + q j (n)) n∈Z + , is unbounded and a contradiction arises.
Again, as above, there exists a matrix valued polynomial Q j (n) = Q j (T µ j (q)) (in n) having the degree at most m j − 1 such that E j (T µ j (q))U A,µ j (nq, 0) = Q j (n) for every n ∈ Z + .
We conclude this note with the one-dimensional version of our result.
Corollary 2.3.Let t → a(t) : R + → C be a given continuous and q-periodic function (for some positive q).The scalar differential equation [a(r)] dr = 0.
Theorem 1.3 follows that (2.2) is Hyers-Ulam stable if and only if |T q | = 1 or equivalently if and only if