Hyers-Ulam Stability for Linear Differences with Time Dependent and Periodic Coefﬁcients

: Let q ≥ 2 be a positive integer and let ( a j ) , ( b j ) , and ( c j ) (with j a non-negative integer) be three given C - valued and q -periodic sequences. Let A ( q ) : = A q − 1 · · · A 0 , where A j is as is given below. Assuming that the "monodromy matrix" A ( q ) has at least one multiple eigenvalue, we prove that the linear scalar recurrence 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 the spectrum of A ( q ) does not intersect the unit circle Γ : = { w ∈ C : | w | = 1 } . Connecting this result with a recently obtained one it follows that the above linear recurrence is Hyers-Ulam stable if and only if the spectrum of A ( q ) does not intersect the unit circle. 34D09, 39B82, 34K20


Introduction
An open problem, arising naturally in [1], is a problem referring to the relationship between the Hyers-Ulam stability of a certain linear recurrence of order n with periodic coefficients and the exponential dichotomy of the monodromy matrix associated to the recurrence. The corresponding problem for second-order recurrences was completed in [2], where second-order linear differential equations were also analyzed.
Here, we continue the analysis started in [3] and, finally, we complete the discussion raised in [1] for periodic linear recurrences of order three. Thus, this article can be seen as a new link in the chain of articles [1][2][3][4][5] which address the Hyers-Ulam stability of linear scalar recurrences. The connections of this topic to those existing in the literature was already presented in [3], so we do not present them again here.
It seems that the methods used here can be extended to recurrences of higher order in Banach spaces and, hopefully, this will be considered in the future; the autonomous case was analyzed in [6][7][8][9]. For developments concerning differential equations with impulses see, [10][11][12][13], and the references therein.

Definitions and Notations
We use the same notation as in [3]. Recall that the entry m ij (of a matrix M) is denoted by [M] ij , and 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 [3]):
Obviously, any ε-approximative solution of the recurrence (1) can be seen as a solution of the non-homogeneous equation for some scalar valued sequence We denote the solution of the nonhomogeneous linear recurrence (3) initiated from Y 0 by (φ(n, Y 0 , ( f k )).
The solution of the system initiated from Z 0 , where X n := z n v n w n T ∈ C 3 , F n = 0 0 f n T , and is given by Denoting by (ϕ(n, Z 0 , ( f k )) the solution of (1), obviously we have ϕ n := ϕ(n, Z 0 , ( f k )) = U A (n, 0)Z 0 + ∑ n k=1 U A (n, k)F k 11 (7) and Φ n = ϕ n ϕ n+1 ϕ n+2 T . Here, A is the family of all matrices A j (with j ∈ Z + , where A j is given in (5)) and the matrix U A (n, k) is given by U A (n, k) = A n−1 · · · A k , n > k and U A (n, n) = I 3 . The family U A := {U A (n, k) : n ≥ k ∈ Z + } will be called the evolution family associated to A.

Background and the Main Result
The next two propositions appear (in a slightly different form) in [14]. Proposition 1. Suppose that the eigenvalues x, y, and z of the matrix A ∈ M(3, C) verify the condition Then, A n = x n (n 2 B + nC + I 3 ) for all n ∈ Z + , and Proposition 2. If the characteristic polynomial of the matrix A is , with x = y and x = 0, then its natural powers are given by where B, C, and D are given by Remark 1. Taking into account that x and y are different roots of the minimal polynomial m A (of A), the matrices C in (15) and D in (16) are not the zero matrix.
Let q, (a j ), (b j ), and (c j ) be as above. Recall that Our main result reads as follows.
Theorem 1. Assume that either of the conditions (8) or (12) (concerning the spectrum of A(q)) are fulfilled. The linear recurrence 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 the spectrum of A(q) does not intersect the unit circle.
Combining this result with ([3], Theorem 3.1) we get the following Corollary that completes an open problem, raised in [1] for the particular case n = 3.   [15], we are also interested in studying the Hyers-Ulam stability of the linear recurrence in (18), but with Z instead of Z + . This can be seen as a symmetrization of the result in Corollary 1. Next, we summarize some ideas, but do not give all the details. For simplicity, we assume that c j = 0 for all j ∈ Z.
It is well-known that the equivalent statements of Corollary 1 are also equivalent to the fact that the system of recurrences in C 3 possesses a discrete dichotomy on Z + ; see ( [1], Proposition 1.2, Theorem 2.1) for a more general framework of this result.
A new challenge for us is to see if the following three statements (presented in formal terms) are equivalent.
1. The linear recurrence is Hyers-Ulam stable on Z − .

The "symmetric" linear system
possesses a discrete dichotomy on Z − .
3. The spectrum of the monodromy matrix associated with (21) does not intersect the unit circle. It seems that all these statements are also equivalent to the fact that the spectrum of the monodromy matrix associated with (19) does not intersect the unit circle.
The main ingredient in the proof in Section 3 of the "if" part of the Theorem 1 is the following technical Lemma, whose proof is presented in the next section.

Lemma 1.
Assume that either of the conditions (8) or (12) (concerning the spectrum of A(q)) are fulfilled. 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. Proof of Lemma 1. We use Propositions 1 and 2, with A(q) instead of A. Denote by x, y, and z the eigenvalues of A(q). Case I. Let x = y = z and |x| = 1. We use the notation of the previous sections. I.1. When B = (b rs ) r,s∈{1,2,3} = 0 3 , there exists a pair (i, j) with i, j ∈ {1, 2, 3} such that b ij = 0. We analyze three cases: I.1.1. Let j = 3 and b 13 = 0. Set where u 0 := 0 0 c 0 T and c 0 is a given nonzero complex scalar with |c 0 | < ε. Successively, we have On the other hand, Thus, (24) and (25) yield the unboundedness of the sequence (φ n ). When b 23 = 0 or b 33 = 0, arguing as above, we can show that the sequences (ϕ n+1 ) and (ϕ n+2 ) are unbounded, and that then (ϕ n ) is unbounded as well.

11
(27) + x n n 2 BZ 0 + x n C nZ 0 + (n − 1)n 2 A q−1 u 0 + x n Z 0 + nA q−1 u 0 11 . For our purposes, it is enough to prove that the sequence (whose general term is given in (27)) is unbounded. Indeed, we have as n → ∞. The cases when b 22 = 0 and b 23 = 0 can be treated in a similar manner, and we omit the details. I.1.3. Let j = 1 and b 11 = 0. Set with u 0 and c 0 as above. As in the previous cases, we obtain that φ nq = x n B n 2 Z 0 + (n − 1)n(2n − 1) 6 A q−2 A q−1 u 0 11 , and so (φ nq ) is unbounded, as Let c ij = 0, for some pair (i, j) with i, j ∈ {1, 2, 3}. Then, We have to consider the following three steps: I.2.1. Let j = 3 and c 13 = 0. Set F k = F 1 k . As above, we have Thus, The sequence (φ n ) is unbounded, since x n (n − 1)n 2 Cu 0 11 = (n − 1)n 2 |c 13 c 0 | → ∞ as n → ∞.
When c 23 = 0 or c 33 = 0, we can argue as in the previous cases. I.2.2. Let j = 2 and c 12 = 0. Set F k = F 2 k . Then, 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 − 1)n 2 CA q−1 u 0 11 + x n CZ 0 + x n Z 0 + nA q−1 u 0 11 , and the sequence (φ nq ) is unbounded because When c 22 = 0 or c 23 = 0, we can proceed in a similar manner.

I.2.3.
let j = 1 and c 11 = 0. Set F k = F 3 k . As in the previous cases, we obtain Therefore, (ϕ n ) is unbounded. I.3. When B = 0 3 and C = 0 3 , then A(q) n = x n I 3 for all n ∈ Z + . Set F k = F 1 k . Then, It is enough to prove that the sequence ([nx n u 0 ] 11 ) n is unbounded, and note that |[nx n u 0 ] 11 | = n |c 0 | → ∞ as n → ∞.
Case II. When the characteristic polynomial p A(q) (λ) is given by Let λ be an eigenvalue of A(q) and let P λ be the Riesz projection associated to A(q) and λ; that is, where C(λ, r) is the circle centered at λ of radius r, and r is small enough such that all other eigenvalues of A(q) are located outside of the circle. Using the Dunford integral calculus (see [16]) and the Cauchy formula (see, e.g., [17], Theorem 10.15) it is easy to show that P 2 x = P x and P x P y = P y P x = 0 3 . On the other hand, by Proposition 2 and the Spectral Decomposition Theorem (see, e.g., [18], Theorem 1), for every v ∈ C 3 , one has A(q) n v = x n (nB + C) v + y n Dv = x n (nB + C) P x v + y n DP y v , and so P x A(q) n = x n (nB + C)P x = x n (nB + C) and P y A(q) n = y n DP y = y n D for every n ∈ Z + . In the following, we will analyze three cases: II.1. When |x| = 1 and |y| < 1. II.1.1. When B = 0 3 , let us first assume that b 13 = 0 and set F k = F 1 k . Then, x n Cu 0 = (n − 1)n 2 x n Bu 0 + x n (nB + C) Z 0 + nx n Cu 0 .
The cases b 23 = 0 and b 33 = 0 can be treated in a similar manner, and we omit the details. II.1.2. When b 12 = 0, set F k = F 2 k . Then, Therefore, the sequence P x Φ nq 11 n is unbounded, as Note that the last three terms in (29) are bounded (as functions of n). Now, if [C] 13 = 0 then |[x n nCu 0 ] 11 | = n|[C] 13 c 0 | → ∞ as n → ∞, and (29) yields the unboundedness of the sequence (ϕ nq ). As C = 0 3 , at least one of its entries is nonzero and we arrive at the same conclusion by arguing in a similar manner (such arguments were given above a few times, so we omit the details).

II.2.2.
When |x| = |y| = 1 and B = 0 3 , it can be treated like in Case II.1.1.. II. 3. When |x| = 1 and |y| > 1. Taking into account that D is not the zero matrix (of order 3), we can choose a sequence (F k ) and a pair (i, j) such that the sequence ([P y Φ(nq)] i,j ) n is unbounded (we omit the details).
The proof of Lemma 1 is now complete.
Proof. Proof of Theorem 1. Necessity: We argue by contradiction. Suppose that σ(A(q)) intersects the unit circle. Let Y 0 and X 0 be as in [ [3], Remark 1]. From Lemma 1, it follows that the sequence in (22) with (Y 0 − X 0 ) instead of Z 0 is unbounded and this contradicts the Hyers-Ulam stability property of the recurrence given in (18). Sufficiency: Can be done exactly as in the proof of the implication 2 ⇒ 1 in ([3], Theorem 1); we omit the details.
We find the values of the real parameter a, such that the recurrence in (30) is Hyers-Ulam stable. With the above notation, we have: