Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 17, 1-11; http://www.math.u-szeged.hu/ejqtde/

Ψ− BOUNDED SOLUTIONS FOR A LYAPUNOV MATRIX DIFFERENTIAL

EQUATION

Aurel Diamandescu University of Craiova

Abstract It is proved a necessary and sufficient condition for the existence of at least one Ψ− bounded solution of a linear nonhomogeneous Lyapunov matrix differential equation.

Key Words: Ψ− bounded solution, Lyapunov matrix differential equation. 2000 Mathematics Subject Classification: 34C11, 34D05.

1. Introduction.

This work is concerned with linear nonhomogeneous Lyapunov matrix differential equation

\frac{dZ}{dt} = A(t)Z + ZB(t) + F(t)  (1)

where A, B and F are continuous n×n matrix-valued function on ℝ_{+} = [0,∞).

The basic problem under consideration is the determination of necessary and sufficient conditions for the existence of a solution with some specified boundedness condition.

The problem of Ψ− boundedness of the solutions for systems of ordinary differential equations was studied in many papers, as e.q. [1], [3], [5], [6]. In these papers, the function Ψ is a scalar continuous function (and increasing, differentiable and bounded in [1], nondecreasing and such that Ψ(t) ≥ 1 on ℝ_{+} in [3]).

Recently, in [7] was studied the existence of Ψ− bounded solutions for the corresponding Kronecker product system (6) associated with (1) (i.e. a linear nonhomogeneous differential system of the form \frac{dx}{dt} = G(t)x + f(t)) in the hypothesis that the free term f of the system is a Lebesgue Ψ− integrable function defined on ℝ_{+}. But the obtained results in [7] are particular cases of our general results stated in [4]. Indeed, if in Theorems 2.1 and 2.2 ([4]), the fundamental matrix Y is replaced with the fundamental matrix Z⊗Y of the linear system (3) of [7], the Theorems 1 and 2 ([7]) follow. In addition, in Theorems 1 and 2 ([7]) there are a few mistakes in connection with the matrix Ψ.

The purpose of present paper is to give a necessary and sufficient conditions so that the linear nonhomogeneous Lyapunov matrix differential equation (1) have at least one Ψ− bounded solution on ℝ_{+} for every Lebesgue Ψ− integrable matrixvalued function F on ℝ_{+}.

Here, as in [4], Ψ will be a continuous matrix function. The introduction of the matrix function Ψ permits to obtain a mixed asymptotic behavior of the components of the solutions.

The results of this paper include our results [4] as a particular case, when B = O_{n}.

2. Preliminaries. In this section we present some basic definitions, notations, hypotheses and

results which are useful later on.

Let ℝ^{n} be the Euclidean n - space. For x = (x_{1}, x_{2}, x_{3}, ...,x_{n})^{T} ∈ ℝ^{n}, let ∥x∥ = max{|x_{1}|, |x_{2}|, |x_{3}|, ..., |x_{n}|} be the norm of x ( ^{T} denotes transpose).

Let 𝕄_{m×n} be the linear space of all m×n real valued matrices. For a n×n real matrix A = (a_{ij}), we define the norm |A| by |A| = sup_{∥x∥≤1} ∥Ax∥ .

It is well-known that |A| = max_{1≤i≤n}{∑_{j=1}^{n} |a_{ij}|}.

Def inition 1. ([2]) Let A = (a_{ij}) ∈ 𝕄_{m×n} and B = (b_{ij}) ∈ 𝕄_{p×q}. The Kronecker product of A and B written A⊗B is defined to be the partitioned matrix

A ⊗ B = ( \begin{matrix}a_{11}B & a_{12}B & ⋯ & a_{1n}B \\ a_{21}B & a_{22}B & ⋯ & a_{2n}B \\ ⋮ & ⋮ & ⋮ & ⋮ \\ a_{m1}B & a_{m2}B & ⋯ & a_{mn}B \end{matrix} )

Obviously, A⊗B ∈ 𝕄_{mp×nq}.

Lemma 1. The Kronecker product has the following properties and rules, provided that the dimension of the matrices are such that the various expressions exist:

1). A⊗(B⊗C) = (A⊗B)⊗C; 2). (A⊗B)^{T} = A^{T}⊗B^{T}; 3). (A⊗B)^{−1} = A^{−1}⊗B^{−1}; 4). (A⊗B)·(C⊗D) = AC⊗BD; 5). A⊗(B + C) = A⊗B + A⊗C; 6). (A + B)⊗C = A⊗C + B⊗C;

7). I_{p}⊗A = ( \begin{matrix}A & O & ⋯ & O \\ O & A & ⋯ & O \\ ⋮ & ⋮ & ⋮ & ⋮ \\ O & O & ⋯ & A \end{matrix} );

8). \frac{d}{dt}(A(t)⊗B(t)) = \frac{dA(t)}{dt}⊗B(t) + A(t)⊗\frac{dB(t)}{dt}. (here, \frac{dX}{dt} denotes the derivative of X with respect to t). Proof. See in [2].

Def inition 2. The application Vec : 𝕄_{m×n} ⟶ ℝ^{mn}, defined by

Vec(A) = (a_{11},a_{21}, ⋯, a_{m1},a_{12},a_{22}, ⋯, a_{m2}, ⋯, a_{1n},a_{2n}, ⋯, a_{mn})^{T},

where A = (a_{ij}) ∈ 𝕄_{m×n}, is called the vectorization operator.

Lemma 2. The vectorization operator Vec : 𝕄_{n×n} ⟶ ℝ^{n^{2}}, is a linear and one-to-one operator. In addition, Vec and Vec^{−1} are continuous operators.

Proof. The fact that the vectorization operator is linear and one-to-one is immediate. Now, for A = (a_{ij}) ∈ 𝕄_{n×n}, we have

∥ Vec(A) ∥ = max_{1≤i,j≤n} {| a_{ij}|} ≤ max_{1≤i≤n}{∑_{j=1}^{n} | a_{ij}|} = | A | .

Thus, the vectorization operator is continuous and ∥ Vec∥ ≤ 1. In addition, for A = I_{n} (identity n×n matrix) we have ∥ Vec(I_{n})∥ = 1 = |I_{n}| and

then, ∥ Vec∥ = 1.

Obviously, the inverse of the vectorization operator, Vec^{−1} : ℝ^{n^{2}} ⟶ 𝕄_{n×n}, is defined by

Vec^{−1}(u) = ( \begin{matrix}u_{1} & u_{n+1} & ⋯ & u_{n^{2}−n+1} \\ u_{2} & u_{n+2} & ⋯ & u_{n^{2}−n+2} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ u_{n} & u_{2n} & ⋯ & u_{n}^{2} \end{matrix} ) ,

where u = (u_{1},u_{2},⋯,u_{n}^{2})^{T} ∈ ℝ^{n^{2}}.

We have

| Vec^{−1}u | = max_{1≤i≤n}{∑_{j=0}^{n−1} | u_{nj + i} |} ≤ n · max_{1≤i≤n^{2}} {| u_{i} |} = n· ∥ u ∥ .

Thus, Vec^{−1} is a continuous operator.

Remark. Obviously, if F is a continuous matrix function on ℝ_{+}, then f = Vec(F) is a continuous vector function on ℝ_{+} and reciprocally.

The next Lemma gives some basic properties of vectorization operator Vec.

Lemma 3. If A, B, M ∈ 𝕄_{n×n}, then 1). Vec(AMB) = (B^{T}⊗A)·Vec(M); 2). Vec(MB) = (B^{T}⊗I_{n})·Vec(M); 3). Vec(AM) = (I_{n}⊗A)·Vec(M); 4). Vec(AM) = (M^{T}⊗A)·Vec(I_{n}). Proof. It is a simple exercise.

Let Ψ_{i} : ℝ_{+} ⟶ (0,∞), i = 1,2,...,n, be continuous functions and

Ψ = diag [Ψ_{1}, Ψ_{2}, ⋯ Ψ_{n}].

Def inition 3. ([4]). A function f : ℝ_{+} ⟶ ℝ^{n} is said to be Ψ− bounded on ℝ_{+} if Ψf is bounded on ℝ_{+} (i.e. sup_{t≥0} ∥ Ψ(t)f(t)∥ < + ∞).

Extend this definition for matrix functions.

Def inition 4. A matrix function M : ℝ_{+} ⟶ 𝕄_{n×n} is said to be Ψ− bounded on ℝ_{+} if the matrix function ΨM is bounded on ℝ_{+} (i.e. sup_{t≥0} | Ψ(t)M(t)| < + ∞).

Def inition 5. ([4]). A function f : ℝ_{+} ⟶ ℝ^{n} is said to be Lebesgue Ψ− integrable on ℝ_{+} if f is measurable and Ψf is Lebesgue integrable on ℝ_{+}

(i.e. ∫ _{0}^{∞} ∥ Ψ(t)f(t)∥dt < ∞).

Extend this definition for matrix functions.

Def inition 6. A matrix function M : ℝ_{+} ⟶ 𝕄_{n×n} is said to be Lebesgue Ψ− integrable on ℝ_{+} if M is measurable and ΨM is Lebesgue integrable on ℝ_{+}

(i.e. ∫ _{0}^{∞} | Ψ(t)M(t)|dt < ∞).

Now, we shall assume that A and B are continuous n×n - matrices on ℝ_{+} and F is a Lebesgue Ψ− integrable matrix function on ℝ_{+}.

By a solution of (1), we mean an absolutely continuous matrix function Z(t) satisfying the equation (1) for almost all t ≥ 0.

The following lemmas play a vital role in the proof of main result.

Lemma 4. The matrix function M : ℝ_{+} ⟶ 𝕄_{n×n} is Lebesgue Ψ− integrable on ℝ_{+} if and only if the vector function VecM(t) is Lebesgue I_{n} ⊗ Ψ− integrable on ℝ_{+}.

Proof. From the proof of Lemma 2, it results that

\frac{1}{n} | A | ≤ ∥ VecA ∥_{ℝ^{^{n}^{2}}} ≤ | A |,

for every A ∈ 𝕄_{n×n}.

Setting A = Ψ(t)M(t), t ≥ 0 and using Lemma 3, we have the inequality

\frac{1}{n} | Ψ(t)M(t) | ≤ ∥ (I_{n} ⊗ Ψ(t)) · VecM(t) ∥_{ℝ^{^{n}^{2}}} ≤ | Ψ(t)M(t) | , t ≥ 0 (2)

for all matrix function M(t).

Now, the Lemma follows immediately.

Lemma 5. The matrix function M(t) is Ψ− bounded on ℝ_{+} if and only if the vector function VecM(t) is I_{n} ⊗ Ψ − bounded on ℝ_{+}.

Proof. It follows from the above inequality (2).

The next Lemma is Lemma 1 of [7]. Because the proof is incomplete, we present it with a complete proof.

Lemma 6. Let X(t) and Y(t) be the fundamental matrices for the equations

\frac{dX}{dt} = A(t)X  (3)

and

\frac{dY}{dt} = YB(t)  (4)

respectively.

Then, the matrix Z(t) = Y^{T}(t)⊗X(t) is a fundamental matrix for the equation

\frac{dZ}{dt} = (I_{n} ⊗ A(t) + B^{T}(t) ⊗ I_{n})Z.  (5)

If, in addition, X(0) = I_{n} and Y(0) = I_{n}, then Z(0) = I_{n}^{2}. Proof. Using the above properties of the Kronecker product, one obtains

\frac{dZ(t)}{dt} = \frac{d}{dt}(Y^{T}(t) ⊗ X(t)) =

= \frac{dY^{T}(t)}{dt} ⊗ X(t) + Y^{T}(t) ⊗ \frac{dX(t)}{dt} =
= (\frac{dY(t)}{dt})^{T} ⊗ X(t) + Y^{T}(t) ⊗ \frac{dX(t)}{dt} =
= (Y(t)B(t))^{T} ⊗ X(t) + Y^{T}(t) ⊗ A(t)X(t) =
= B^{T}(t)Y^{T}(t) ⊗ I_{n}X(t) + I_{n}Y^{T}(t) ⊗ A(t)X(t) =
= (B^{T}(t) ⊗ I_{n}) · (Y^{T}(t) ⊗ X(t)) + (I_{n} ⊗ A(t)) · (Y^{T}(t) ⊗ X(t)) =
= (I_{n} ⊗ A(t) + B^{T}(t) ⊗ I_{n})Z(t),

for all t ∈ ℝ_{+}.

On the other hand, the matrix Z(t) is a nonsingular matrix for all t ≥ 0 (because X(t) and Y(t) are nonsingular matrices for all t ≥ 0).

Thus, the matrix Z is a fundamental matrix on ℝ_{+} for the equation (5). Obviously, Z(0) = Y(0)⊗X(0) = I_{n}⊗I_{n} = I_{n}^{2}. The proof is now complete.

Lemma 7. The matrix function Z(t) is a solution of (1) if and only if the vector valued function z(t) = VecZ(t) is a solution of the differential system

\frac{dz}{dt} = (I_{n} ⊗ A(t) + B^{T}(t) ⊗ I_{n})z + f(t),  (6)

where f(t) = VecF(t).

Proof. Using Kronecker product notation, the vectorization operator Vec and the above properties, we can rewrite the equality

\frac{dZ(t)}{dt} = A(t)Z(t) + Z(t)B(t) + F(t), for almost all t ≥ 0,

in the equivalent form

Vec\frac{dZ(t)}{dt} = (I_{n} ⊗ A(t) + B^{T}(t) ⊗ I_{n})VecZ(t) + VecF(t), for almost all t ≥ 0.

If we denote VecZ(t) = z(t), VecF(t) = f(t), we have \frac{dz(t)}{dt} = Vec\frac{dZ(t)}{dt} and then, the above equality becomes

\frac{dz(t)}{dt} = (I_{n} ⊗ A(t) + B^{T}(t) ⊗ I_{n})z(t) + f(t), for almost all t ≥ 0.

The proof is now complete.

Def inition 7. The above system (6) is called “corresponding Kronecker product system associated with (1)”

Now, let Z(t) the above fundamental matrix for the system (5) with Z(0) = I_{n}^{2}. Let X̃_{1} denote the subspace of ℝ^{n^{2}} consisting of all vectors which are values

of I_{n} ⊗ Ψ− bounded solutions of (5) on ℝ_{+} for t = 0 and let X̃_{2} an arbitrary fixed subspace of ℝ^{n^{2}}, supplementary to X̃_{1}. Let P̃_{1}, P̃_{2} denote the corresponding projections of ℝ^{n^{2}} onto X̃_{1}, X̃_{2} respectively.

Finally, we remaind of two theorems which be useful in the proofs of our main results.

Theorem (2.1). ([4]). Let A be a continuous n×n real matrix function on ℝ_{+} and let Y the fundamental matrix of the homogeneous linear system \frac{dx}{dt} = A(t)x for which Y(0) = I_{n}.

Then, the system \frac{dx}{dt} = A(t)x + f(t) has at least one Ψ− bounded solution on ℝ_{+} for every Lebesgue Ψ− integrable function f on ℝ_{+} if and only if there is a positive constant K such that

\{ \begin{matrix}| Ψ(t)Y(t)P_{1}Y^{−1}(s)Ψ^{−1}(s) | & ≤ K, & for 0 ≤ s ≤ t \\ | Ψ(t)Y(t)P_{2}Y^{−1}(s)Ψ^{−1}(s) | & ≤ K, & for 0 ≤ t ≤ s \end{matrix}  (7)

Theorem (2.2). ([4]). Suppose that: 1^{∘}. The fundamental matrix Y(t) of the system \frac{dx}{dt} = A(t)x satisfies:

a). the above condition (7), where K is a positive constant; b). the condition lim_{t→∞} Ψ(t)Y(t)P_{1} = 0;

2^{∘}. The function f : ℝ_{+} ⟶ ℝ^{n} is Lebesgue Ψ− integrable on ℝ_{+}.

Then, every Ψ− bounded solution x of the system \frac{dx}{dt} = A(t)x + f(t) is such that

lim_{t→+∞} ∥ Ψ(t)x(t) ∥ = 0.

3. The main result. The main result of this paper is the following:

Theorem 1. Let A and B be continuous n×n real matrix functions on ℝ_{+} and let X and Y be the fundamental matrices of the homogeneous linear equations (3) and (4) respectively for which X(0) = Y(0) = I_{n}.

Then, the equation (1) has at least one Ψ− bounded solution on ℝ_{+} for every Lebesgue Ψ− integrable matrix function F : ℝ_{+} ⟶ 𝕄_{n×n} if and only if there is a positive constant K such that

| (Y^{T}(t) ⊗ (Ψ(t)X(t)))P̃_{1}((Y^{T}(s))^{−1} ⊗ (X^{−1}(s)Ψ^{−1}(s))) | ≤ K, 0 ≤ s ≤ t
| (Y^{T}(t) ⊗ (Ψ(t)X(t)))P̃_{2}((Y^{T}(s))^{−1} ⊗ (X^{−1}(s)Ψ^{−1}(s))) | ≤ K, 0 ≤ t ≤ s

(8)

Proof. First, we prove the “only if” part. Suppose that the equation (1) has at least one Ψ− bounded solution on ℝ_{+} for

every Lebesgue Ψ− integrable matrix function F : ℝ_{+} ⟶ 𝕄_{n×n}.

Let f : ℝ_{+} ⟶ ℝ^{n^{2}} a Lebesgue I_{n} ⊗Ψ− integrable function on ℝ_{+}. From Lemma 4, it follows that the matrix function F(t) = Vec^{−1}f(t) is Lebesgue Ψ− integrable matrix function on ℝ_{+}. From the hypothesis, the equation

\frac{dZ}{dt} = A(t)Z + ZB(t) + Vec^{−1}f(t)

has at least one Ψ− bounded solution Z(t) on ℝ_{+}.

From Lemma 7 and Lemma 5, it follows that the vector valued function z(t) = VecZ(t) is a I_{n} ⊗ Ψ− bounded solution on ℝ_{+} of the differential system

\frac{dz}{dt} = (I_{n} ⊗ A(t) + B^{T}(t) ⊗ I_{n})z + f(t).

Thus, this system has at least one I_{n} ⊗ Ψ− bounded solution on ℝ_{+} for every Lebesgue I_{n} ⊗ Ψ− integrable function f on ℝ_{+}.

From the Theorem 2.1 ([4]), there is a positive constant K such that the fundamental matrix Z(t) of the equation (5) satisfies the condition

| (I_{n} ⊗ Ψ(t))Z(t)P̃_{1}Z^{−1}(s)(I_{n} ⊗ Ψ(s))^{−1} | ≤ K, 0 ≤ s ≤ t | (I_{n} ⊗ Ψ(t))Z(t)P̃_{2}Z^{−1}(s)(I_{n} ⊗ Ψ(s))^{−1} | ≤ K, 0 ≤ t ≤ s .

We replace Z(t) = Y^{T}(t)⊗X(t). After computation, it follows that (8) holds.

Now, we prove the “if” part. Suppose that (8) holds for some K > 0.

Let F : ℝ_{+} ⟶ 𝕄_{n×n} a Lebesgue Ψ− integrable matrix function on ℝ_{+}. From Lemma 4, it follows that the vector valued function f(t) = VecF(t) is a Lebesgue I_{n} ⊗ Ψ− integrable function on ℝ_{+}.

From Theorem 2.1 ([4]), it follows that the differential system

\frac{dz}{dt} = (I_{n} ⊗ A(t) + B^{T}(t) ⊗ I_{n})z + f(t)

has at least one I_{n} ⊗ Ψ− bounded solution on ℝ_{+}. Let z(t) be this solution.

From Lemma 5 and Lemma 7, it follows that the matrix function Z(t) = Vec^{−1}z(t) is a Ψ− bounded solution of the equation (1) on ℝ_{+} (because F(t) = Vec^{−1}f(t)).

Thus, the differential equation (1) has at least one Ψ− bounded solution on ℝ_{+} for every Lebesgue Ψ− integrable matrix function F on ℝ_{+}.

The proof is now complete.

Remark. 1. The Theorem generalizes the Theorem 2.1, [4].

Indeed, if X and Y are fundamental matrices for the equations (3) and (4) respectively, then, the unique solution Z of the nonhomogeneous equation (1) which takes the value Z_{0} for t = t_{0} is given by

Z(t) = X(t)X^{−1}(t_{0})Z_{0}Y^{−1}(t_{0})Y(t) + ∫ _{t_{0}}^{t} X(t)X^{−1}(s)F(s)Y^{−1}(s)Y(t)ds, t ∈ ℝ_{+}.

(The proof is similar to the proof of well-known Variation of constants formula for the linear differential system \frac{dx}{dt} = A(t)x + f(t)).

In particular case B(t) = O_{n}, we have Y = I_{n} and then, the solution Z of the nonhomogeneous equation \frac{dZ}{dt} = A(t)Z + F(t) which takes the value Z_{0} for t = t_{0} is given by

Z(t) = X(t)X^{−1}(t_{0})Z_{0} + ∫ _{t_{0}}^{t} X(t)X^{−1}(s)F(s)ds, t ∈ ℝ_{+}.

If, in addition

F(t) = ( \begin{matrix}f_{1}(t) & f_{1}(t) & ⋯ & f_{1}(t) \\ f_{2}(t) & f_{2}(t) & ⋯ & f_{2}(t) \\ ⋮ & ⋮ & ⋮ & ⋮ \\ f_{n}(t) & f_{n}(t) & ⋯ & f_{n}(t) \end{matrix} ) and Z_{0} = ( \begin{matrix}z_{1}^{0} & z_{1}^{0} & ⋯ & z_{1}^{0} \\ z_{2}^{0} & z_{2}^{0} & ⋯ & z_{2}^{0} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ z_{n}^{0} & z_{n}^{0} & ⋯ & z_{n}^{0} \end{matrix} ) ,

it is easy to see that the solution of the equation \frac{dZ}{dt} = A(t)Z + F(t) is

Z(t) = ( \begin{matrix}x_{1}(t) & x_{1}(t) & ⋯ & x_{1}(t) \\ x_{2}(t) & x_{2}(t) & ⋯ & x_{2}(t) \\ ⋮ & ⋮ & ⋮ & ⋮ \\ x_{n}(t) & x_{n}(t) & ⋯ & x_{n}(t) \end{matrix} ) ,

where x = (x_{1}(t),x_{2}(t),⋯,x_{n}(t))^{T} is the solution of the problem

\frac{dx}{dt} = A(t)x + f(t), x(t_{0}) = (z_{1}^{0},z_{2}^{0}, ⋯,z_{n}^{0})^{T},

with f(t) = (f_{1}(t),f_{2}(t),⋯,f_{n}(t))^{T}.

In this case, the condition (8) becomes the condition (7), because the solution x(t) is Ψ− bounded on ℝ_{+} if and only if the solution Z(t) is Ψ− bounded on ℝ_{+}.

Thus, the Theorem generalizes the result from [4].

2. The particular case A = O_{n} come to the above case. 3). Other particular cases: A = I_{n} or B = I_{n} or A = B or A = B = O_{n}, etc.

In the end, we prove a theorem which shows that the asymptotic behavior of solutions of (1) is determined completely by the asymptotic behavior of the fundamental matrices X and Y of (3) and (4) respectively as t ⟶ ∞.

Theorem 2. Suppose that: 1). The fundamental matrices X(t) and Y(t) of (3) and (4) respectively

(X(0) = Y(0) = I_{n}) satisfy:

a). the condition (8) for some K > 0; b). the condition

lim_{t→∞} (Y^{T}(t) ⊗ (Ψ(t)X(t)))P̃_{1} = 0;

2). The matrix function F : ℝ_{+} ⟶ 𝕄_{n×n} is Lebesgue Ψ− integrable on ℝ_{+}. Then, every Ψ− bounded solution Z(t) of (1) satisfies the condition

lim_{t→∞} | Ψ(t)Z(t) | = 0.

Proof. Let Z(t) be a Ψ− bounded solution of (1). From Lemma 7 and Lemma 5, it follows that the function z(t) = VecZ(t) is a I_{n} ⊗ Ψ− bounded solution on ℝ_{+} of the differential system

\frac{dz}{dt} = (I_{n} ⊗ A(t) + B^{T}(t) ⊗ I_{n})z + f(t),

where f(t) = VecF(t).

Also, from Lemma 4, the function f(t) is Lebesgue I_{n} ⊗ Ψ− integrable on ℝ_{+}. From the Theorem 2.2 ([4]), it follows that

lim_{t→∞} ∥ (I_{n} ⊗ Ψ(t)) · z(t) ∥_{ℝ^{^{n}^{2}}} = 0.

Now, from the inequality (2) we have

| Ψ(t)Z(t) | ≤ n ∥ (I_{n} ⊗ Ψ(t)) · z(t) ∥_{ℝ^{^{n}^{2}}} , t ≥ 0

and then

lim_{t→∞} | Ψ(t)Z(t) | = 0.

The proof is now complete.

Remark. The Theorem generalizes the Theorem 2.2, [4].

Note that Theorem 2 is no longer true if we require that the matrix function F is Ψ− bounded on ℝ_{+}, instead of the condition 2) of the Theorem. Even if the matrix function F is such that lim_{t→∞} | Ψ(t)F(t)| = 0.

This is shown by the next

Example. Consider the equation (1) with

A(t) = ( \begin{matrix}0 & 0 \\ 0 & a \end{matrix} ) , B(t) = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) and F(t) = ( \begin{matrix}t & 0 \\ 0 & 0 \end{matrix} ) ,

where a ∈ ℝ, a < 0.

The fundamental matrices for the equations (3) and (4) are

X(t) = ( \begin{matrix}1 & 0 \\ 0 & e^{at} \end{matrix} ) , Y(t) = ( \begin{matrix}cos t & sin t \\ − sin t & cos t \end{matrix} )

respectively.

Consider

Ψ(t) = ( \begin{matrix}\frac{1}{t + 1} & 0 \\ 0 & \frac{e^{−at}}{t + 1} \end{matrix} ) , t ≥ 0.

It is easy to see that the conditions of the Theorem 2 are satisfied with

P̃_{1} = I_{4}, P̃_{2} = O_{4} and K = 2.

In addition, we have that the function matrix F is Ψ− bounded on ℝ_{+}, but it is not Lebesgue Ψ− integrable on ℝ_{+}.

On the other hand, the solutions of the equation (1) are

Z(t) = ( \begin{matrix}(c_{1} − 1) cos t − c_{2} sin t + 1 & (c_{1} − 1) sin t + c_{2} cos t + t \\ e^{at}(c_{3} cos t − c_{4} sin t) & e^{at}(c_{3} sin t + c_{4} cos t) \end{matrix} ) ,

where c_{1}, c_{2}, c_{3}, c_{4} ∈ ℝ.

It is easy to see that each solution Z is Ψ− bounded on ℝ_{+}, but

lim_{t→∞} | Ψ(t)Z(t) | = 1.

Note that the asymptotic properties of the components of the solutions are not the same. On the other hand, we see that the asymptotic properties of the components of the solutions are the same, via matrix function Ψ. This is obtained by using a matrix function Ψ rather than a scalar function.

Acknowledgement. The author would like to thank the anonymous referee for careful reading, valuable comments and suggestions.

References

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[7] Murty M.S.N. and Suresh Kumar G. − On Ψ− boundedness and Ψ− stability of matrix Lyapunov systems, J. Appl. Math. Comput(2008) 26: 67 - 84.

(Received November 12, 2008)

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