Electronic Journal of Qualitative Theory of Differential Equations 2008, No. 26, 1-13; http://www.math.u-szeged.hu/ejqtde/

EXISTENCE OF Ψ− BOUNDED SOLUTIONS FOR LINEAR DIFFERENCE EQUATIONS ON ℤ

Aurel Diamandescu University of Craiova

Abstract

In this paper^{1}, we give a necessary and sufficient condition for the existence of Ψ− bounded solutions for the nonhomogeneous linear difference equation x(n + 1) = A(n)x(n) + f(n) on ℤ. In addition, we give a result in connection with the asymptotic behavior of the Ψ− bounded solutions of this equation.

1. Introduction

The problem of boundedness of the solutions for the system of ordinary differential equations x′ = A(t)x + f(t) was studied by Coppel in [2]. In [3], [4], [5], the author proposes a novel concept, Ψ− boundedness of solutions (Ψ being a matrix function), which is interesting and useful in some practical cases and presents the existence condition for such solutions. Also, in [1], the author associates this problem with the concept of Ψ− dichotomy on ℝ of the system x′ = A(t)x.

Naturally, one wonders whether there are any similar concepts and results on the solutions of difference equations, which can be seen as the discrete version of differential equations.

In [7], the authors extend the concept of Ψ− boundedness to the solutions of difference equation

x(n + 1) = A(n)x(n) + f(n)  (1)

(via Ψ− bounded sequence) and establish a necessary and sufficient condition for existence of Ψ− bounded solutions for the nonhomogeneous linear difference equation (1) in case f is a Ψ− summable sequence on ℕ.

In [6], the author proved a necessary and sufficient condition for the existence of Ψ− bounded solutions of (1) in case f is a Ψ− bounded sequence on ℕ.

Similarly, we can consider solutions of (1) which are bounded not only ℕ but on the ℤ.

In this case, the conditions for the existence of at least one Ψ−bounded solution are rather more complicated, as we will see below.

In this paper, we give a necessary and sufficient condition so that the nonhomogeneous linear difference equation (1) have at least one Ψ−bounded solution on ℤ for every Ψ−summable function f on ℤ

Here, Ψ is a matrix function. The introduction of the matrix function Ψ permits to obtain a mixed asymptotic behavior of the components of the solutions.

12000 Mathematics Subject Classification: 39A11, 39A10

Keywords: difference equations, Ψ− bounded solutions on ℤ.

2. Preliminaries

Let ℝ^{d} be the Euclidean d-space. For x = (x_{1}, x_{2},...,x_{d})^{T} ∈ ℝ^{d}, let ∥x∥ = max{|x_{1}|, |x_{2}|, ... |x_{d}|} be the norm of x. For a d × d real matrix A = (a_{ij}), the norm

|A| is defined by |A| = sup_{∥x∥≤1} ∥Ax∥ . It is well-known that |A| = max_{1≤i≤d} ∑_{j = 1}^{d} |a_{ij} | .

Let Ψ_{i} : ℤ ⟶ (0,∞), i = 1, 2, ...d and let the matrix function

Ψ = diag [Ψ_{1},Ψ_{2},...Ψ_{d}].

Then, Ψ(n) is invertible for each n ∈ ℤ.

Definition 1. A function φ : ℤ ⟶ ℝ^{d} is called Ψ− bounded iff the function Ψφ is bounded (i.e. there exists M > 0 such that ∥ Ψ(n)φ(n)∥ ≤ M for all n ∈ ℤ).

Definition 2. A function φ : ℤ ⟶ ℝ^{d} is called Ψ− summable on ℤ if

∑_{n =−∞}^{∞} ∥ Ψ(n)φ(n)∥ is convergent (i.e. lim_{{p→−∞ \atop q→+∞}} ∑_{n =p}^{q} ∥ Ψ(n)φ(n)∥ is finite).

Consider the nonautonomous difference linear equation

y(n + 1) = A(n)y(n)  (2)

where the d × d real matrix A(n) is invertible at n ∈ ℤ. Let Y be the fundamental matrix of (2) with Y(0) = I_{d} (identity d × d matrix). It is well-known that

i). Y(n) = \{ \begin{matrix}A(n − 1)A(n − 2) ⋯ A(1)A(0), & n > 0 \\ I_{d}, & n = 0 \\ [A(–1)A(–2) ⋯ A(n)]^{−1}, & n < 0 \end{matrix} ,

ii). Y(n + 1) = A(n)Y(n) for all n ∈ ℤ

iii). the solution of (2) with the initial condition y(0) = y_{0} is

y(n) = Y(n)y_{0}, n ∈ ℤ;

iv). Y is invertible for each n ∈ ℤ and

Y^{−1}(n) = \{ \begin{matrix}A^{−1}(0)A^{−1}(1) ⋯ A^{−1}(n − 1), & n > 0 \\ I_{d}, & n = 0 \\ A(–1)A(–2) ⋯ A(n), & n < 0 \end{matrix}

Let the vector space ℝ^{d} represented as a direct sum of three subspaces X_{−}, X_{0}, X_{+} such that a solution y of (2) is Ψ− bounded on ℤ if and only if y(0) ∈ X_{0} and Ψ− bounded on ℤ_{+} = {0,1,2,⋯ } if and only if y(0) ∈ X_{−} ⊕ X_{0}. Also let P_{−}, P_{0}, P_{+} denote the corresponding projection of ℝ^{d} onto X_{−}, X_{0}, X_{+} respectively.

3. Main result

The main result of this paper is the following.

Theorem 1. The equation (1) has at least one Ψ− bounded solution on ℤ for every Ψ− summable function f on ℤ if and only if there is a positive constant K such that

\{ \begin{matrix}| Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k) | & ≤ & K, & k+1 ≤ min\{0,n\} \\ | Ψ(n)Y(n)(P_{0} + P_{+})Y^{−1}(k+1)Ψ^{−1}(k) | & ≤ & K, & n < k+1 ≤ 0 \\ | Ψ(n)Y(n)(P_{0} + P_{−})Y^{−1}(k+1)Ψ^{−1}(k) | & ≤ & K, & 0 < k+1 ≤ n \\ | Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k) | & ≤ & K, & k + 1 > max\{0,n\} \end{matrix} (3)

Proof. First, we prove the ”only if” part. We define the sets:

B_{Ψ} = {x : ℤ ⟶ ℝ^{d} | x is Ψ− bounded},

B = {x : ℤ ⟶ ℝ^{d} | x is Ψ− summable on ℤ},

D = {x : ℤ ⟶ ℝ^{d} | x ∈ B_{Ψ}, x(0) ∈ X_{−} ⊕ X_{+}, (x(n + 1)−A(n)x(n)) ∈ B} Obviously, B_{Ψ}, B and D are vector spaces over ℝ and the functionals

x ⟼ ∥x∥_{B_{Ψ}} = sup_{n∈ℤ} ∥ Ψ(n)x(n)∥,

x ⟼ ∥x∥_{B} = ∑_{n =−∞}^{∞} ∥ Ψ(n)x(n)∥,
x ⟼ ∥x∥_{D} = ∥x∥_{B_{Ψ}} + ∥x(n + 1) − A(n)x(n)∥_{B}

are norms on B_{Ψ}, B and D respectively.

Step 1. It is a simple exercise that (B_{Ψ}, ∥ · ∥_{B_{Ψ}}) and (B, ∥ · ∥_{B}) are Banach spaces.

Step 2. (D, ∥ · ∥_{D}) is a Banach space. Let (x_{p})_{p∈ℕ} be a fundamental sequence in D. Then, (x_{p})_{p∈ℕ} is a fundamental

sequence in B_{Ψ}. Therefore, there exists a Ψ− bounded function x : ℤ ⟶ ℝ^{d} such that lim_{p→∞} Ψ(n)x_{p}(n) = Ψ(n)x(n), uniformly on ℤ. From

∥x_{p}(n) − x(n)∥ ≤ | Ψ^{−1}(n)|∥ Ψ(n)(x_{p}(n) − x(n))∥,

it follows that the sequence (x_{p})_{p∈ℕ} is almost uniformly convergent to function x on ℤ. Because x_{p}(0) ∈ X_{−} ⊕ X_{+}, p ∈ ℕ, it follows that x(0) ∈ X_{−} ⊕ X_{+}.

On the other hand, the sequence (f_{p})_{p∈ℕ}, f_{p}(n) = x_{p}(n + 1) − A(n)x_{p}(n), n ∈ ℤ, is a fundamental sequence in B. Therefore, there exists a function f ∈ B such that

∑_{n =−∞}^{∞} ∥ Ψ(n)f_{p}(n) − Ψ(n)f(n)∥ ⟶ 0 as p ⟶ ∞.

It follows that Ψ(n)f_{p}(n) ⟶ Ψ(n)f(n) and f_{p}(n) ⟶ f(n) for each n ∈ ℤ. For a fixed but arbitrary n ∈ ℤ, n > 0, we have

x(n + 1) − x(0) = lim_{p→∞} [x_{p}(n + 1) − x_{p}(0)] =

= lim_{p→∞} ∑_{i = 0}^{n} [x_{p}(i + 1)−x_{p}(i)] =
= lim_{p→∞} ∑_{i = 0}^{n} [x_{p}(i + 1) − A(i)x_{p}(i) + A(i)x_{p}(i) − x_{p}(i)] =
= lim_{p→∞} ∑_{i = 0}^{n} [f_{p}(i) − f(i) + f(i) + A(i)x_{p}(i) − x_{p}(i)] =

= ∑_{i = 0}^{n} [f(i) + A(i)x(i) − x(i)] =

= ∑_{i = 0}^{n − 1}[f(i) + A(i)x(i) − x(i)] + f(n) + A(n)x(n) − x(n) =

= x(n) − x(0) + f(n) + A(n)x(n) − x(n) = A(n)x(n) + f(n) − x(0). Similarly, we have

x(1) − x(0) = A(0)x(0) + f(0) − x(0)

and, for n ∈ ℤ, n < 0,

x(n) − x(0) = lim_{p→∞} [x_{p}(n) − x_{p}(0)] = lim_{p→∞} ∑_{i = n}^{−1} [x_{p}(i) − x_{p}(i + 1)] =

= lim_{p→∞} ∑_{i = n}^{−1} [x_{p}(i) − A(i)x_{p}(i) + A(i)x_{p}(i) − x_{p}(i + 1)] =

= lim_{p→∞} ∑_{i = n}^{−1} [x_{p}(i) − A(i)x_{p}(i) − f_{p}(i)] =

= ∑_{i = n}^{−1} [x(i) − A(i)x(i) − f(i)] =

= ∑_{i = n+1}^{−1} [x(i) − A(i)x(i) − f(i)] + x(n) − A(n)x(n) − f(n) =

= x(n + 1) − x(0) + x(n) − A(n)x(n) − f(n). By the above relations, we have that

x(n + 1) − A(n)x(n) = f(n), n ∈ ℤ.

It follows that x ∈ D. Now, from the relations

∑_{n =−∞}^{∞} ∥ Ψ(n)(x_{p} − x)(n+1) − Ψ(n)A(n)(x_{p} − x)(n)∥ ⟶ 0 as p ⟶ ∞,

∥x_{p} − x∥_{B_{Ψ}} ⟶ 0 as p ⟶ ∞, it follows that ∥x_{p} − x∥_{D} ⟶ 0 as p ⟶ +∞.

Thus, (D, ∥ · ∥_{D}) is a Banach space. Step 3. There exists a positive constant K such that, for every f ∈ B and for

corresponding solution x ∈ D of (1), we have

∥x∥_{B_{Ψ}} ≤ K· ∥f∥_{B}. (4)

We define the operator T : D ⟶ B, (Tx)(n) = x(n + 1) − A(n)x(n), n ∈ ℤ. Clearly, T is linear and bounded, with ∥T∥ ≤ 1. Let Tx = 0 be. Then,

x ∈ D and x(n + 1) = A(n)x(n). This shows that x is a Ψ− bounded solution of (2) with x(0) ∈ X_{−} ⊕ X_{+}. From the Definition of X_{0}, we have x(0) ∈ X_{0}. Thus, x(0) ∈ X_{0} ∩ (X_{−} ⊕ X_{+}) = {0}. It follows that x = 0. This means that the operator T is one-to-one.

Now, for f ∈ B, let x be a Ψ− bounded solution of the equation (1). Let z be the solution of the Cauchy problem

z(n + 1) = A(n)z(n) + f(n), z(0) = (P_{−} + P_{+})x(0).

Then, the function u = x − z is a solution of the equation (2) with

u(0) = x(0) − z(0) = P_{0}x(0) ∈ X_{0}.

It follows that the function u is Ψ− bounded on ℤ. Thus, the function z is Ψ− bounded on ℤ. It follows that z ∈ D and Tz = f. Consequently, T is onto.

From a fundamental result of Banach ”If T is a bounded one-to-one linear operator from a Banach space onto another, then the inverse operator T^{−1} is also bounded”, we have that

∥T^{−1}f∥_{D} ≤ ∥T^{−1}∥∥f∥_{B}, for f ∈ B.

Denoting T^{−1}f = x, we have ∥x∥_{D} = ∥x∥_{B_{Ψ}} + ∥f∥_{B} ≤ ∥T^{−1}∥∥f∥_{B} and then

∥x∥_{B_{Ψ}} ≤ (∥T^{−1}∥ − 1)∥f∥_{B} .

Thus, we have (4), where K = ∥T^{−1}∥ − 1. Step 4. The end of the proof. For a fixed but arbitrary k ∈ ℤ, ξ ∈ ℝ^{d}, we consider the function f : ℤ ⟶ ℝ^{d}

defined by

f(n) = \{ \begin{matrix}Ψ^{−1}(n)ξ, & if n = k \\ 0, & elsewhere \end{matrix} .

Obviously, f ∈ B and ∥f∥_{B} = ∥ ξ ∥ . The corresponding solution x ∈ D of (1) is x(n) = G(n,k+1)f(k), where

G(n,k) = \{ \begin{matrix}Y(n)P_{−}Y^{−1}(k) & k ≤ min\{0,n\} \\ −Y(n)(P_{0} + P_{+})Y^{−1}(k) & n < k ≤ 0 \\ Y(n)(P_{0} + P_{−})Y^{−1}(k) & 0 < k ≤ n \\ −Y(n)P_{+}Y^{−1}(k) & k > max\{0,n\} \end{matrix} .

Indeed, we prove this in more cases: Case k ≤ −1. a). for k + 1 ≤ n ≤ 0, x(n+1) = G(n+1,k+1)f(k) = Y(n+1)P_{−}Y^{−1}(k+1)f(k) = = A(n)Y(n)P_{−}Y^{−1}(k+1)f(k) = A(n)x(n) = A(n)x(n) + f(n) (because f(n) = 0); b). for n = k, x(n+1) = G(n+1,k+1)f(k) = Y(n+1)P_{−}Y^{−1}(k+1)f(k) = = Y(k+1)(I−P_{0}−P_{+})·Y^{−1}(k+1)f(k) = f(k)−A(k)Y(k)(P_{0}+P_{+})Y^{−1}(k+1)f(k) = = f(k) + A(k)G(k,k+1)f(k) = A(n)x(n) + f(n); c). for n < k, x(n+1) = G(n+1,k+1)f(k) = −Y(n+1)(P_{0} + P_{+})Y^{−1}(k+1)f(k) = = −A(n)Y(n)(P_{0} + P_{+})Y^{−1}(k+1)f(k) = A(n)x(n) = A(n)x(n) + f(n); d). for n > 0, x(n+1) = G(n+1,k+1)f(k) = Y(n+1)P_{−}Y^{−1}(k+1)f(k) = = A(n)Y(n)P_{−}Y^{−1}(k+1)f(k) = A(n)x(n) = A(n)x(n) + f(n); Case k > − 1. α). for n < 0, x(n+1) = G(n+1,k+1)f(k) = −Y(n+1)P_{+}Y^{−1}(k+1)f(k) = = − A(n)Y(n)P_{+}Y^{−1}(k+1)f(k) = A(n)x(n) = A(n)x(n) + f(n); β). for n = 0 and k = 0, x(1) = G(1,1)f(0) = Y(1)(P_{0} + P_{−})Y^{−1}(1)f(0) = Y(1)(I − P_{+})Y^{−1}(1)f(0) = = f(0) − A(0)Y(0)P_{+}Y^{−1}(1)f(0) = A(0)x(0) + f(0); γ). n = 0 and k > 0, x(1) = G(1,k+1)f(k) = −Y(1)P_{+}Y^{−1}(k+1)f(k) = −A(0)Y(0)P_{+}Y^{−1}(k+1)f(k) = = A(0)G(0,k+1)f(k) = A(0)x(0) + f(0); δ). for 0 < n = k, x(n+1) = G(k+1,k+1)f(k) = Y(k+1)(P_{0} + P_{−})Y^{−1}(k+1)f(k) = = Y(k+1)(I − P_{+})Y^{−1}(k+1)f(k) = f(k) − A(k)Y(k)P_{+}Y^{−1}(k+1)f(k) = = A(n)x(n) + f(n); ε). for 0 < n < k, x(n+1) = G(n+1,k+1)f(k) = −Y(n+1)P_{+}Y^{−1}(k+1)f(k) = = − A(n)Y(n)P_{+}Y^{−1}(k+1)f(k) = A(n)x(n) = A(n)x(n) + f(n); ζ). for n ≥ k + 1, x(n+1) = G(n+1,k+1)f(k) = Y(n+1)(P_{0} + P_{−})Y^{−1}(k+1)f(k) = = A(n)Y(n)(P_{0} + P_{−})Y^{−1}(k+1)f(k) = A(n)x(n) = A(n)x(n) + f(n). On the other hand, x(0) ∈ X_{−} ⊕ X_{+}, because

x(0) = G(0,k+1)f(k) = \{ \begin{matrix}+P_{−}Y^{−1}(k+1)f(k), & k + 1 ≤ 0 \\ −P_{+}Y^{−1}(k+1)f(k), & k + 1 > 0 \end{matrix} .

Finally, we have

x(n) = G(n,k+1)f(k) = \{ \begin{matrix}−Y(n)(P_{0} + P_{+})Y^{−1}(k+1)f(k), & n < k + 1 ≤ 0 \\ Y(n)(P_{0} + P_{−})Y^{−1}(k+1)f(k), & n ≥ k + 1 ≥ 0 \end{matrix} .

From the Definitions of X_{−}, X_{0} and X_{+}, it follows that the function x is Ψ− bounded on ℤ_{−} and ℕ. Thus, x is the solution of (1) in D.

Now, we have, ∥ Ψ(n)x(n)∥ = ∥ Ψ(n)G(n,k+1)f(k)∥ = ∥ Ψ(n)G(n,k+1)Ψ^{−1}(k)ξ ∥. The inequality (4) becomes

∥ Ψ(n)G(n,k+1)Ψ^{−1}(k)ξ ∥ ≤ K∥ ξ ∥, for all k, n ∈ ℤ, ξ ∈ ℝ.^{d}. It follows that | Ψ(n)G(n,k+1)Ψ^{−1}(k)| ≤ K, for all k, n ∈ ℤ, which is equivalent

with (3).

Now, we prove the ”if” part. For a given Ψ− summable function f : ℤ ⟶ ℝ^{d}, consider u : ℤ ⟶ ℝ^{d} defined

by

u(n) = \{ \begin{matrix}∑_{k =−∞}^{n−1} Y(n)P_{−}Y^{−1}(k+1)f(k) − ∑_{k = n}^{−1} Y(n)P_{0}Y^{−1}(k+1)f(k) − − ∑_{k = n}^{∞} Y(n)P_{+}Y^{−1}(k+1)f(k), & n < 0 \\ ∑_{k =−∞}^{−1} Y(0)P_{−}Y^{−1}(k+1)f(k) − ∑_{k = 0}^{∞} Y(0)P_{+}Y^{−1}(k+1)f(k), & n = 0 \\ ∑_{k =−∞}^{n−1} Y(n)P_{−}Y^{−1}(k+1)f(k) + ∑_{k = 0}^{n−1} Y(n)P_{0}Y^{−1}(k+1)f(k) − − ∑_{k = n}^{∞} Y(n)P_{+}Y^{−1}(k+1)f(k), & n > 0 \end{matrix}

Step 5. The function u is well-defined. For p, q ∈ ℤ, q < 0 < p, we have

∑_{k = q}^{−1} ∥Y(0)P_{−}Y^{−1}(k+1)f(k)∥ + ∑_{k = 0}^{p} ∥Y(0)P_{+}Y^{−1}(k+1)f(k)∥ ≤
≤ | Ψ^{−1}(0)| ∑_{k = q}^{−1} | Ψ(0)Y(0)P_{−}Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ +
+ | Ψ^{−1}(0)| ∑_{k = 0}^{p} | Ψ(0)Y(0)P_{+}Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ ≤
≤ K| Ψ^{−1}(0)| ( ∑_{k = q}^{p} ∥ Ψ(k)f(k) ∥) ,

and then, ∑_{k =−∞}^{−1}Y(0)P_{−}Y^{−1}(k+1)f(k) and ∑_{k = 0}^{∞}Y(0)P_{+}Y^{−1}(k+1)f(k) are absolutely

convergent series. Thus, u(0) is well-defined.

For m, n ∈ ℤ, m ≥ n > 0, we have

∑_{k=n}^{m} ∥Y(n)P_{+}Y^{−1}(k+1)f(k)∥ =

= ∑_{k=n}^{m} ∥ Ψ^{−1}(n)(Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k))(Ψ(k)f(k))∥ ≤

≤ | Ψ^{−1}(n)| ∑_{k = n}^{m} | Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ ≤

≤ K| Ψ^{−1}(n)| ( ∑_{k = n}^{m} ∥ Ψ(k)f(k) ∥) ,

and then, ∑_{k = n}^{∞}Y(n)P_{+}Y^{−1}(k+1)f(k) is an absolutely convergent series for n > 0.

For m ∈ ℤ, n ∈ ℕ, m < n − 1, we have

∑_{k = m}^{n−1} ∥Y(n)P_{−}Y^{−1}(k+1)f(k)∥ =

= ∑_{k = m}^{n−1} ∥ Ψ^{−1}(n)(Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k))(Ψ(k)f(k))∥

≤ | Ψ^{−1}(n)| ∑_{k = m}^{n−1} | Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ ≤

≤ K| Ψ^{−1}(n)| ∑_{k = m}^{n−1} ∥ Ψ(k)f(k)∥,

and then, ∑_{k =−∞}^{n−1}Y(n)P_{−}Y^{−1}(k+1)f(k) is an absolutely convergent series for n > 0.

Thus, the function u is well defined for n ≥ 0. Similarly, the function u is well defined for n < 0. Step 6. The function u is a solution of the equation (1). Indeed, using the expresion of the function u, we obtain:

• u(1) = ∑_{k =−∞}^{0}Y(1)P_{−}Y^{−1}(k+1)f(k) + Y(1)P_{0}Y^{−1}(1)f(0) −

− ∑_{k = 1}^{∞}Y(1)P_{+}Y^{−1}(k+1)f(k) = A(0)[ ∑_{k =−∞}^{0}Y(0)P_{−}Y^{−1}(k+1)f(k) +

+ Y(0)P_{0}Y^{−1}(1)f(0) − ∑_{k = 1}^{∞}Y(0)P_{+}Y^{−1}(k+1)f(k)] =

= A(0)[ ∑_{k =−∞}^{−1}Y(0)P_{−}Y^{−1}(k+1)f(k)+Y(0)P_{−}Y^{−1}(1)f(0)+Y(0)P_{0}Y^{−1}(1)f(0)

− ∑_{k = 0}^{∞}Y(0)P_{+}Y^{−1}(k+1)f(k) + Y(0)P_{+}Y^{−1}(1)f(0)] =

= A(0)u(0) + A(0)Y(0)(P_{−} + P_{0} + P_{+})Y^{−1}(1)f(0) = A(0)u(0) + f(0);

• for n > 0, u(n+1) = ∑_{k =−∞}^{n}Y(n+1)P_{−}Y^{−1}(k+1)f(k) +

+ ∑_{k = 0}^{n}Y(n+1)P_{0}Y^{−1}(k+1)f(k) − ∑_{k = n+1}^{∞}Y(n+1)P_{+}Y^{−1}(k+1)f(k) = = A(n)[ ∑_{k =−∞}^{n−1}Y(n)P_{−}Y^{−1}(k+1)f(k) + Y(n)P_{−}Y^{−1}(n+1)f(n) +

+ ∑_{k = 0}^{n−1}Y(n)P_{0}Y^{−1}(k+1)f(k) + Y(n)P_{0}Y^{−1}(n+1)f(n) −

− ∑_{k = n}^{∞}Y(n)P_{+}Y^{−1}(k+1)f(k) + Y(n)P_{+}Y^{−1}(n+1)f(n)] =

= A(n)u(n) + Y(n+1)(P_{−} + P_{0} + P_{+})Y^{−1}(n+1)f(n) = A(n)u(n) + f(n);

• u(0) = ∑_{k =−∞}^{−1}Y(0)P_{−}Y^{−1}(k+1)f(k) − ∑_{k = 0}^{∞}Y(0)P_{+}Y^{−1}(k+1)f(k) =

= A(−1)[ ∑_{k =−∞}^{−1}Y(−1)P_{−}Y^{−1}(k+1)f(k) − ∑_{k = 0}^{∞}Y(−1)P_{+}Y^{−1}(k+1)f(k)] =

= A(−1)[ ∑_{k =−∞}^{−2}Y(−1)P_{−}Y^{−1}(k+1)f(k) + Y(−1)P_{−}Y^{−1}(0)f(−1) −

− ∑_{k =−1}^{−1}Y(−1)P_{0}Y^{−1}(k+1)f(k) + Y(−1)P_{0}Y^{−1}(0)f(−1) −

− ∑_{k =−1}^{∞}Y(−1)P_{+}Y^{−1}(k+1)f(k) + Y(−1)P_{+}Y^{−1}(0)f(−1)] =

= A(−1)u(−1) + Y(0)(P_{−} + P_{0} + P_{+})Y^{−1}(0)f(−1) = = A(−1)u(−1) + f(−1);

• for n < −1, u(n+1) = ∑_{k =−∞}^{n}Y(n+1)P_{−}Y^{−1}(k+1)f(k) −

− ∑_{k = n+1}^{−1}Y(n+1)P_{0}Y^{−1}(k+1)f(k) − ∑_{k = n+1}^{∞}Y(n+1)P_{+}Y^{−1}(k+1)f(k) =

= A(n)[ ∑_{k =−∞}^{n−1}Y(n)P_{−}Y^{−1}(k+1)f(k) + Y(n)P_{−}Y^{−1}(n+1)f(n) −

− ∑_{k = n}^{−1}Y(n)P_{0}Y^{−1}(k+1)f(k) + Y(n)P_{0}Y^{−1}(n+1)f(n) −

− ∑_{k = n}^{∞}Y(n)P_{+}Y^{−1}(k+1)f(k) + Y(n)P_{+}Y^{−1}(n+1)f(n)] =

= A(n)u(n) + Y(n+1)(P_{−} + P_{0} + P_{+})Y^{−1}(n+1)f(n) = A(n)u(n) + f(n). These relations show that the function u is a solution of the equation (1). Step 7. The function u is Ψ− bounded on ℤ. Indeed, for n > 0 we have

∥ Ψ(n)u(n)∥ = ∥ ∑_{k =−∞}^{n−1} Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k) +

+ ∑_{k = 0}^{n−1} Ψ(n)Y(n)P_{0}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k) −

− ∑_{k = n}^{∞} Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k)∥ =

= ∥ ∑_{k =−∞}^{−1} Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k) +

+ ∑_{k = 0}^{n−1} Ψ(n)Y(n)(P_{0} + P_{−})Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k) − − ∑_{k = n}^{∞} Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k)∥ ≤

≤ ∑_{k =−∞}^{−1} | Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ +

+ ∑_{k = 0}^{n−1} | Ψ(n)Y(n)(P_{0} + P_{−})Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ +

+ ∑_{k = n}^{∞} | Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ ≤

≤ K( ∑_{k =−∞}^{−1} ∥ Ψ(k)f(k) ∥ + ∑_{k = 0}^{n−1} ∥ Ψ(k)f(k) ∥ + ∑_{k = n}^{∞} ∥ Ψ(k)f(k) ∥) =

= K ∑_{k =−∞}^{+∞} ∥ Ψ(k)f(k)∥ = K∥f∥_{B} .

For n < 0, we have

∥ Ψ(n)u(n)∥ = ∥ ∑_{k =−∞}^{n−1} Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k) −

− ∑_{k = n}^{−1} Ψ(n)Y(n)P_{0}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k) −

− ∑_{k = n}^{∞} Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k)∥ =

= ∥ ∑_{k =−∞}^{n−1} Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k) −

− ∑_{k = n}^{−1} Ψ(n)Y(n)(P_{0} + P_{+})Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k) −

− ∑_{k = 0}^{∞} Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k)∥ ≤

≤ ∑_{k =−∞}^{n−1} | Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ +

+ ∑_{k = n}^{−1} | Ψ(n)Y(n)(P_{0} + P_{+})Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ +

+ ∑_{k = 0}^{∞} | Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ ≤

≤ K( ∑_{k =−∞}^{n−1} ∥ Ψ(k)f(k) ∥ + ∑_{k = n}^{−1} ∥ Ψ(k)f(k) ∥ + ∑_{k = 0}^{∞} ∥ Ψ(k)f(k) ∥) =

= K ∑_{k =−∞}^{+∞} ∥ Ψ(k)f(k)∥ = K∥f∥_{B} .

Similarly, ∥ Ψ(0)u(0)∥ ≤ K∥f∥_{B} . Therefore, ∥ Ψ(n)u(n)∥ ≤ K∥f∥_{B}, for all n ∈ ℤ. Thus, the solution u of the equation (1) is Ψ− bounded on ℤ. The proof is now complete.

Corollary 1. If the homogeneous equation (2) has no nontrivial Ψ− bounded solution on ℤ, then, the equation (1) has a unique Ψ− bounded solution on ℤ for every Ψ− summable function f on ℤ if and only if there exists a positive constant K such that, for k, n ∈ ℤ,

\{ \begin{matrix}| Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k) | & ≤ & K, & for k+1 ≤ n \\ | Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k) | & ≤ & K, & for n < k+1 \end{matrix} (5)

Proof. Indeed, in this case, P_{0} = 0. Now, the Corollary follows from the above Theorem.

Finally, we give a result in which we will see that the asymptotic behavior of Ψ− bounded solutions of (1) is determined completely by the asymptotic behavior of the fundamental matrix Y of (2).

Theorem 2. Suppose that: 1^{∘}. the fundamental matrix Y of (2) satisfies the conditions (3) for some K > 0

and the conditions i). lim_{n→±∞} | Ψ(n)Y(n)P_{0} | = 0;

ii). lim_{n→+∞} | Ψ(n)Y(n)P_{−} | = 0;

iii). lim_{n→−∞} | Ψ(n)Y(n)P_{+} | = 0.

2^{∘}. the function f : ℤ ⟶ ℝ^{d} is Ψ− summable on ℤ. Then, every Ψ− bounded solution x of (1) satisfies the condition

lim_{n→±∞} ∥ Ψ(n)x(n)∥ = 0.

Proof. Let x be a Ψ− bounded solution of (1). Let u be the Ψ− bounded solution of (1) from the proof of Theorem 1 (”if” part).

Let the function y(n) = x(n) − u(n) − Y(n)P_{0}(x(0) − u(0)), n ∈ ℤ. It is easy to see that y is a Ψ− bounded solution of (2) and then y(0) ∈ X_{0}. On the other hand,

y(0) = (I − P_{0})(x(0) − u(0)) = (P_{−} + P_{+})(x(0) − u(0)) ∈ X_{−} ⊕ X_{+}. Thus, y(0) ∈ (X_{−} ⊕ X_{+}) ∩ X_{0} = {0}. It follows that y = 0 and then

x(n) = u(n) + Y(n)P_{0}(x(0) − u(0)), n ∈ ℤ.

Now, we prove that lim_{n→±∞} ∥ Ψ(n)x(n)∥ = 0. For n > 0, we have

x(n) = Y(n)P_{0}(x(0) − u(0)) + ∑_{k =−∞}^{n−1}Y(n)P_{−}Y^{−1}(k+1)f(k) +
+ ∑_{k = 0}^{n−1}Y(n)P_{0}Y^{−1}(k+1)f(k) − ∑_{k = n}^{∞}Y(n)P_{+}Y^{−1}(k+1)f(k)

and then Ψ(n)x(n) = Ψ(n)Y(n)P_{0}(x(0) − u(0)) +

+ ∑_{k =−∞}^{−1} Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k) +

+ ∑_{k=0}^{n−1} Ψ(n)Y(n)(P_{0} + P_{−})Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k) −

− ∑_{k = n}^{∞} Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k)Ψ(k)f(k).

By the hypotheses, for a given ε > 0, there exist:

• n_{1} ∈ ℕ such that, for n ≥ n_{1}, ∑_{k =−∞}^{−n} ∥ Ψ(k)f(k)∥ < \frac{ε}{5K} and ∑_{k = n}^{∞} ∥ Ψ(k)f(k)∥ < \frac{ε}{5K};

• n_{2} ∈ ℕ, n_{2} > n_{1}, such that, for n ≥ n_{2}, | Ψ(n)Y(n)P_{−} | < \frac{ε}{5} (1 + ∑_{k =−n_{1}+1}^{−1} ∥ Y^{−1}(k+1)f(k) ∥)^{−1} ;

• n_{3} ∈ ℕ, n_{3} > n_{2}, such that, for n ≥ n_{3}, | Ψ(n)Y(n)P_{0} | < \frac{ε}{5} (1 + ∥ x(0) − u(0) ∥)^{−1} ;

• n_{4} ∈ ℕ, n_{4} > n_{3}, such that, for n ≥ n_{4}, | Ψ(n)Y(n)(P_{0}+P_{−})| < \frac{ε}{5} (1 + ∑_{k = 0}^{n_{1}} ∥ Y^{−1}(k+1)f(k) ∥)^{−1} .

Then, for n ≥ n_{4} we have ∥ Ψ(n)x(n) ∥ ≤ | Ψ(n)Y(n)P_{0} |∥x(0) − u(0)∥ + + ∑_{k =−∞}^{−n_{1}} | Ψ(n)Y(n)P_{−}Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ +

+ ∑_{k =−n_{1}+1}^{−1} | Ψ(n)Y(n)P_{−} |∥Y^{−1}(k+1)f(k)∥ +

+ ∑_{k = 0}^{n_{1}} | Ψ(n)Y(n)(P_{0} + P_{−})|∥Y^{−1}(k+1)f(k)∥ +

+ ∑_{k = n_{1}+1}^{n−1} | Ψ(n)Y(n)(P_{0} + P_{−})Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ +

+ ∑_{k = n}^{∞} | Ψ(n)Y(n)P_{+}Y^{−1}(k+1)Ψ^{−1}(k)|∥ Ψ(k)f(k)∥ < < \frac{ε}{5} (1 + ∥ x(0) − u(0) ∥)^{−1} ∥x(0) − u(0)∥ + + K ∑_{k =−∞}^{−n_{1}} ∥ Ψ(k)f(k)∥ + | Ψ(n)Y(n)P_{−} | ∑_{k =−n_{1}+1}^{−1} ∥Y^{−1}(k+1)f(k)∥ +

+ | Ψ(n)Y(n)(P_{0} + P_{−})| ∑_{k = 0}^{n_{1}} ∥Y^{−1}(k+1)f(k)∥ +

+ K ∑_{k = n_{1}+1}^{n−1} ∥ Ψ(k)f(k)∥ + K ∑_{k = n}^{∞} ∥ Ψ(k)f(k)∥ < < \frac{ε}{5} + K\frac{ε}{5K} + \frac{ε}{5} + \frac{ε}{5} + K\frac{ε}{5K} = ε.

This shows that lim_{n→+∞} ∥ Ψ(n)x(n)∥ = 0. Similarly, lim_{n→−∞} ∥ Ψ(n)x(n)∥ = 0. The proof is now complete.

Corollary 2. Suppose that: 1^{∘}. the homogeneous equation (2) has no nontrivial Ψ− bounded solution on ℤ; 2^{∘}. the fundamental matrix Y of (2) satisfies:

a). the conditions (5) for some K > 0; b). the conditions:

i). lim_{n→+∞} | Ψ(n)Y(n)P_{−} | = 0

ii). lim_{n→−∞} | Ψ(n)Y(n)P_{+} | = 0.

2^{∘}. the function f : ℤ ⟶ ℝ^{d} is Ψ− summable on ℤ.

Then, the equation (1) has a unique solution x on ℤ such that lim_{n→±∞} ∥ Ψ(n)x(n)∥ = 0.

Proof. It results from the above Corollary and Theorem 2.

Note that the Theorem 2 (and the Corollary 2) is no longer true if we require that the function f be Ψ− bounded on ℤ, instead of the condition 2^{∘} of the Theorem. This is shown by the next

Example 1. Consider the system (1) with

A(n) = ( \begin{matrix}\frac{1}{2} & 0 \\ 0 & 2 \end{matrix} )

and

f(n) = \{ \begin{matrix}( \begin{matrix}1 \\ 0 \end{matrix} ) , & n = 0, 1, 2,... \\ ( \begin{matrix}0 \\ 0 \end{matrix} ) , & n = − 1, −2... \end{matrix} .

Then, Y(n) = ( \begin{matrix}2^{−n} & 0 \\ 0 & 2^{n} \end{matrix} ) is the fundamental matrix of (2) with Y(0) = I_{2}.

Consider Ψ(n) = ( \begin{matrix}1 & 0 \\ 0 & 3^{−n} \end{matrix} ) , n ∈ ℤ.

The first hypothesis of the Theorem 2 is satisfied with

P_{0} = O_{2}, P_{−} = I_{2}, P_{+} = O_{2} and K = 1.

In addition, we have ∥ Ψ(n)f(n)∥ = 1 for n ≥ 0 and ∥ Ψ(n)f(n)∥ = 0 for n < 0. The function f is not Ψ− summable on ℤ, but it is Ψ− bounded on ℤ.

On the other hand, the solutions on ℤ of the system (1) are

x(n) = \{ \begin{matrix}( \begin{matrix}2^{−n}c_{1} \\ 2^{n}c_{2} \end{matrix} ) , & for n < 0, \\ ( \begin{matrix}2^{−n}c_{1} + 2 − 2^{1 − n} \\ 2^{n}c_{2} \end{matrix} ) , & for n ≥ 0. \end{matrix} , c_{1}, c_{2} ∈ ℝ.

It results from this that there is no solution x for lim_{n→±∞} ∥ Ψ(n)x(n)∥ = 0. Acknowledgment. The author would like to thank the anonymous referee for

his/her valuable comments and suggestions.

References.

[1]. P. N. Boi − Existence of Ψ− bounded solutions on R for nonhomogeneous linear differential equations, Electronic Journal of Differential Equations, Vol. 2007(2007), No. 52, pp. 1-10.

[2]. W. A. Coppel − Stability and Asymptotic Behavior of Differential Equations, Heath, Boston, 1965.

[3]. A. Diamandescu − Existence of Ψ - bounded solutions for a system of differential equations, Electronic Journal of Differential Equations, Vol. 2004(2004), No. 63, pp. 1 - 6,

[4]. A. Diamandescu − Note on the Ψ - boundedness of the solutions of a system of differential equations, Acta. Math. Univ. Comenianae, Vol. LXXIII, 2(2004), pp. 223 - 233

[5]. A. Diamandescu − A Note on the Ψ− boundedness for differential systems, Bull. Math. Soc. Sc. math. Roumanie, Tome 48(96), No. 1, 2005, 33-43.

[6]. A. Diamandescu − Existence of Ψ− bounded solutions for nonhomogeneous linear difference equations, to appear.

[7]. Y. Han, J. Hong − Existence of Ψ− bounded solutions for linear difference equations, Applied Mathematics Letters, 20(2007) 301−305.

Author’s address Aurel Diamandescu University of Craiova, Department of Applied Mathematics, 13, ”Al. I. Cuza” st., 200585 Craiova, România. E-mail: adiamandescu@central.ucv.ro

(Received June 5, 2008)