Hyers-Ulam Stability for Linear Differences with Time Dependent and Periodic Coefﬁcients: The Case When the Monodromy Matrix Has Simple Eigenvalues

: Let q ≥ 2 be a positive integer and let ( a j ) , ( b j ) and ( c j ) (with j nonnegative integer) be three given C -valued and q -periodic sequences. Let A ( q ) : = A q − 1 · · · A 0 , where A j is deﬁned below. Assume that the eigenvalues x , y , z of the “monodromy matrix” A ( q ) verify the condition ( x − y )( y − z )( z − x ) (cid:54) = 0. We prove that the linear recurrence in C x n + 3 = a n x n + 2 + b n x n + 1 + c n x n , n ∈ Z + is Hyers–Ulam stable if and only if ( | x | − 1 )( | y | − 1 )( | z | − 1 ) (cid:54) = 0, i.e., the spectrum of A ( q ) does not intersect the unit circle Γ : = { w ∈ C : | w | = 1 } .


Introduction
Exponential dichotomy and its links with the unconditional stability of differential dynamics systems were first highlighted by O. Perron in 1930 [1]. The reader can find details on the subsequent evolution of this topic in Coppel's monograph [2]. The history of the Ulam problem (concerning the stability of a functional equation) and of stability in the sense of Hyers-Ulam is well known. In particular, Hyers-Ulam stability for linear recurrences and for systems of linear recurrences is considered in [3][4][5][6][7][8][9][10][11][12][13][14][15][16], and the references therein.
The relationship between exponential stability and Hyers-Ulam stability has been studied in the articles [3,8,9,17,18], and this article continues these studies.

Notations and Definitions
By C, we denote the set complex numbers and Z + is the set of all nonnegative integers. Now, C m (with m a given positive integer) is the set of all vectors v = (ξ 1 , · · · , ξ m ) T with ξ j ∈ C for every integers 1 ≤ j ≤ m; here and in as follows T denotes the transposition. The norm on C m is the well-known Euclidean norm defined by v := (|ξ 1 | 2 + · · · + |ξ m | 2 ) 1 2 . In addition, C m×n (with m and n given positive integers) denotes the set of all m by n matrices with complex entries. In particular, C m×m becomes a Banach algebra when it is endowed with the (Euclidean) matrix norm defined by M := sup v ≤1 Mv , v ∈ C m , M ∈ C m×m . As is usual, the rows and columns of a matrix M ∈ C m×n are identified by vectors of the corresponding dimensions and in that case its norm is the vector norm. The entry m ij of a matrix M (i.e., the entry in M located at the intersection between the ith row and the jth column) is denoted by [M] ij . As is usual, the uniform norm of a C m -valued and bounded sequence g = (g n ) is defined and denoted by g ∞ := sup j∈Z + g j .
Let ε > 0 be given. We recall (see also [8] for the two-dimensional case) that a scalar valued sequence (y j ) is an ε-approximative solution of the linear recurrence x n+3 = a n x n+2 + b n x n+1 + c n x n , n ∈ Z + if |y n+3 − a n y n+2 − b n y n+1 − c n y n | ≤ ε, ∀n ∈ Z + .
The recurrence in Equation (1) is Hyers-Ulam stable if there exists a positive constant L such that for every ε > 0 and every ε-approximative solution y = (y j ) of Equation (1) there exists an exact solution θ = (θ j ) of Equation (1) such that y − θ ∞ ≤ Lε.

Remark 1.
Since any ε-approximative solution of the recurrence in Equation (1) can be seen as a solution of the nonhomogeneous equation for some scalar valued sequence ( f n ) with f 0 = 0 and ( f k ) ∞ ≤ ε, one has that Equation (1) is Hyers-Ulam stable if and only if there exists a positive constant L such that for every ε > 0, every sequence as above, and every initial condition Here, and in what follows, (φ(n, Y 0 , ( f k )) denotes the solution of the nonhomogeneous linear recurrence in Equation (3) initiated from Y 0 .
Proof. See the proof of Proposition 3.1 in [9].

Background, Previous Results and the Main Result
Proposition 1. ( [19]) Let A be a 3 by 3 matrix whose spectrum (i.e., the set of its eigenvalues σ(A) := {x, y, z}) satisfies the condition Then, for every nonnegative integer n, one has and (6) are orthogonal projections, that is BC = BD = CD = 0 3 ; the null matrix of order three,

Remark 2. (i) The matrices B, C and D in Equation
and B 2 = B, C 2 = C, and D 2 = D.
(ii) In addition, B, C, and D are nonzero matrices. (5), the characteristic polynomial P A and the minimal polynomial m A of A coincide and P A (λ) = (λ − x)(λ − y)(λ − z). Thus, from the Hamilton-Cayley (9) becomes clear.

and Equation
To prove Equation (10), it is enough to see that the details are clear thus omitted. Then, we apply the Hamilton-Cayley theorem and obtain Equation (10). Finally, assuming that B = 0 3 , the polynomial is annulated by A and its degree is equal 2 and is a contradiction with the minimality of the degree of m A .
Let q, (a j ), (b j ), (c j ) be as above. Recall that Our main result reads as follows.
Theorem 1. Assume that the eigenvalues x, y, z (of A(q)) satisfy the condition in Equation (5). Then, the following two statements are equivalent: 1. The linear recurrence in C x n+3 = a n x n+2 + b n x n+1 + c n x n , n ∈ Z + (12) is Hyers-Ulam stable.

The eigenvalues of A(q) verify the condition
The proof of the implication 2 ⇒ 1 is covered (for the most part) in the existing literature. We present the ideas and complete the details. For unexplained terminology, we refer the reader to [8,9]. The following result is taken directly from the second section of [9].
Let X be a complex, finite dimensional Banach space and let B = {B n } n∈Z + and P = {P n } n∈Z + be two families of linear operators acting on X. Assume that: [A1] B n+q = B n and P n+q = P n , for all n ∈ Z + and some positive integer q.
[A2] P 2 n = P n , for all n ∈ Z + , that is, P is a family of projections.
[A3] B n P n = P n+1 B n , for all n ∈ Z + . In particular, this yields that B n x ∈ ker(P n+1 ) for each x ∈ ker(P n ).
(3) For each bounded sequence (G n ) n∈Z + , G 0 = 0 (of X-valued functions) there exists a unique bounded solution (starting from ker(P 0 )) of the difference equation.
We mention that the equivalence between (2) and (3) still works when X is an infinite dimensional Banach space (see [20]). We use Theorem 2 to prove 2⇒1 in Theorem 1.
The main ingredient in the proof of the implication 1 ⇒ 2 in Theorem 1 is the following Lemma. With A we denote the set of all matrices A j (with j ∈ Z + ), where A j is given in Equation (11)) and the matrix U A (n, k) is defined above.

Lemma 1.
If the spectrum of A(q) intersects the unit circle then for each ε > 0 there exists a C-valued sequence ( f j ) j∈Z + with f 0 = 0 and ( f j ) ∞ ≤ ε such that for every initial condition Z 0 = (x 0 , y 0 , z 0 ) T ∈ C 3 , the C-valued sequence (with F k = (0, 0, f k ) T ), is unbounded.

Proofs
Proof of Lemma 1. We first use Proposition 1 with A(q) instead of A. Assume that the eigenvalue x has modulus 1. Let P x be the Riesz projection associated to A(q) and x; that is where C(x, r) is the circle centered at x of radius r, and r is small enough that y and z are located outside of the circle. Using Dunford calculus (see [21]), it is easy to see that P x A(q) n = x n B, for each n ∈ Z + . Consider the matrix B from Equation (6), of the form: The solution of the system X n+1 = A n X n + F n+1 , n ∈ Z + , (15) initiated from Z 0 , where X n := z n v n w n T ∈ C 3 , F n = 0 0 f n T and A n :=    0 1 0 0 0 1 c n b n a n    is given by Denote by (ϕ(n, Z 0 , ( f k )) the solution of Equation (1). An obvious calculation yields In fact, one has Φ n = ϕ n ϕ n+1 ϕ n+2 where u 0 := 0 0 c 0 T and c 0 is a randomly chosen nonzero complex scalar with |c 0 | < ε. Successively, one has x j x n−j Bu 0 . = x n BZ 0 + nx n Bu 0 .
Since the sequence (x n BZ 0 ) n is bounded, it is enough to prove that the sequence ([nx n Bu 0 ] 11 ) n is unbounded, and note |[nx n Bu 0 ] 11 | = n |b 13 c 0 | → ∞ as n → ∞.

Case 1.2.
Let b 23 = 0. Arguing as above we can show that (ϕ n+1 ) is unbounded, that is that (ϕ n ) is unbounded as well.
where u 0 and c 0 are taken as above. We obtain Φ nq = A(q) n Z 0 + ∑ n j=1 x j A(q) n−j A q−1 u 0 which leads to ϕ nq = ∑ n j=1 x j P x A(q) n−j A q−1 u 0 11 = ∑ n j=1 x j x n−j BA q−1 u 0 11 = ∑ n j=1 x n b 12 c 0 = |nx n b 12 c 0 | = n |b 12 c 0 | → ∞ as n → ∞. Case 2.2. Let b 22 = 0. Similar to the previous case, we can show that (ϕ n+1 ) is unbounded, that is that (ϕ n ) is unbounded as well.
As in the previous cases, we obtain ∑ n j=1 x j P x A n−j A q−2 A q−1 u 0 11 = |nx n b 11 c 0 | = n |b 11 c 0 | → ∞ as n → ∞, therefore (ϕ n ) is again unbounded.
Finally, we remark that the matrix B cannot be of the form Proof of Theorem 1. 1⇒2. We argue by contradiction. Suppose that σ(A(q)) intersects the unit circle. Without loss of generality, assume that x is an eigenvalue of A(q) and |x| = 1. Let Y 0 and X 0 be as in the Remark 1. From Lemma 1, it follows that the sequence in Equation (14) with (Y 0 − X 0 ) instead of Z 0 is unbounded and this contradicts Equation (4). 2⇒1. From the assumption and Theorem 2, it follows that the system X n+1 = A n X n is Hyers-Ulam stable. Thus, for a certain positive constant L, every ε > 0, every sequence ( f n ), every Y 0 and some X 0 one has |φ(n, Y 0 , ( f k )) − φ(n, X 0 , (0))| for all n ∈ Z + . Now, the assertion follows from Remark 1.

An Example
The following example illustrates our theoretical result.

Example 1.
The linear recurrence of order three x n+3 = sin 2nπ 3 x n+2 + cos 2nπ 3 x n+1 + c n x n , n ∈ Z (with c n = 1, if n is a multiple of 3 0, elsewhere ) is Hyers-Ulam stable. Indeed, with the above notation one has Now, the monodromy matrix associated to Equation (21) is The characteristic equation associated to A(3) is and the absolute value of each of its solutions is different to 1.

Remark 3.
Reading [22], we note that an interesting question is if the spectral condition (|x| − 1)(|y| − 1)(|z| − 1) = 0 is equivalent to Hyers-Ulam stability of the recurrence in Equation (12) with Z + replaced by Z. We thank the anonymous reviewer who made us aware of the work in [22].