Decay Mild Solutions of the Nonlocal Cauchy Problem for Second Order Evolution Equations with Memory

Our aim in this paper is to find decay mild solutions of the nonlocal Cauchy problem for a class of second order evolution equations with memory. By constructing a suitable measure of noncompactness on the space of solutions, we prove the existence of a compact set containing decay mild solutions to the mentioned problem.

Recently the Cauchy problem for Eq.(1.1) also has been studied in [2,4] with the replacement of f (t, u(t)) by ∇F(u(t)) or ∇F(u(t)) + g(t) and ∇F is Lipschitz in a neighborhood of 0. Motivated by the above work of [2,4], this paper also deals with problem (1.1)-(1.2).Here, the nonlinear f is more general than the one in [2,4] and the initial conditions are nonlocal.The concept of nonlocal initial conditions is introduced to extend the classical theory of initial value problems.This notion is more appropriate than the classical one in describing natural phenomena because it allows us to consider additional information (see, e.g., [5,7,11] and their references).
In this work, we will prove the existence of decay mild solutions for problem (1.1)-(1.2),basing on the fixed point theorem for condensing map for measure of noncompactness (MNC) in [6].
The rest of the paper is organized as follows.Section 2 introduces some useful preliminaries.In addition, we construct a regular MNC on BC(R + ; X) and give a fixed point principle.In Section 3, we prove the existence of mild solutions on [0, T], T > 0, for problem (1.1)-(1.2).Section 4 is devoted to show the decay mild solutions.In the last section, we give an example to illustrate the abtract results obtained in the paper.

Preliminaries
In this section, we introduce preliminary facts which are used throughout this paper.First, we consider the problem where F : R + → X is continuous, and g, h, x 0 , y 0 are given.

Definition 2.1 ([8]
).A family {S(t)} t≥0 ⊂ B(X) of bounded linear operators in X is called a resolvent for (2.1) if the following conditions are satisfied.
(S1) S(t) is strongly continuous on R + and S(0) = I; (S2) S(t) commutes with A, which means that S(t)D(A) ⊂ D(A) and AS(t)x = S(t)Ax for all x ∈ D(A) and t ≥ 0; (S3) the resolvent equation holds Integrating (2.1) twice we obtain the equivalent problem where and the star indicates the convolution.By a similar argument as in [8, p. 160], we get the mild solution given by the formula where S(t) is the resolvent of (2.1) and R(t) = t 0 S(s)ds is its integral.Moreover, since β(t) is real and A is selfadjoint, we obtain S(t) and R(t) are selfadjoint as well (cf.[ (ii) S(t), A 1/2 R(t) are strongly integrable, and converge strongly to 0 as t → ∞.
A typical example of kernel considered in [9] is as follows: where γ > 0, α ∈ (0, 1) and 0 < k 0 < γ α .Next, we recall the knowledge of the measure of noncompactness in Banach spaces.Among them the Hausdorff measure of noncompactness is important.Next, we mention the condensing map and the fixed point principle for condensing maps.We denote the collection of all nonempty bounded subsets in X by B X , and the norm of space C([0, T]; X) by • C , with u C = sup t∈[0,T] u(t) X .

Definition 2.3. A function
where co Ω is the closure of the convex hull of Ω.An MNC Φ in X is called (iii) invariant with respect to union with compact set if Φ(K ∪ Ω) = Φ(Ω) for every relatively compact K ⊂ X and Ω ∈ B X ; (iv) algebraically semi-additive if (v) regular if Φ(Ω) = 0 is equivalent to the relative compactness of Ω.
An important example of MNC is the Hausdorff MNC χ(•) which is defined as follows For T > 0, let χ T be the Hausdorff MNC in C([0, T]; X).We recall the following facts (see [6]): for each bounded D ⊂ C([0, T]; X), we have

χ(D(t)).
Consider the space BC(R + ; X) of bounded continuous functions on R + taking values on X. Denote by π T the restriction operator on [0 is an MNC.We give some measures of noncompactness as follows ) ) The regularity of MNC χ * is proved in [3,Lemma 2.6].
In the sequel, we need some basic MNC estimates.Recall that one can define the sequential MNC χ 0 as follows: where ∆(Ω) is the collection of all at-most-countable subsets of Ω (see [1]).We know that for all bounded set Ω ⊂ X.Then the following property is evident.
Proposition 2.4.Let χ be the Hausdorff MNC on Banach space X, Ω ∈ B X .Then there exists a sequence {x n } ∞ n=1 ⊂ Ω such that We have the following estimate whose proof can be found in [6].

Proposition 2.5 ([6]
).Let χ be the Hausdorff MNC on Banach space X, sequence ) To end this section, we recall a fixed point principle for condensing maps that will be used in the next sections.Definition 2.6.Let X be a Banach space, χ is an MNC on X, and ∅ = D ⊂ X.A continuous map Φ : D −→ X is said to be condensing with respect to χ (χ-condensing) if for all Ω ∈ B D , the relation implies the relative compactness of Ω.

Theorem 2.7 ([6]
).Let X be a Banach space, χ is an MNC on X, D is a bounded convex closed subset of X and let Φ : D −→ D be a χ-condensing map.Then the fixed point set of Φ

Existence result
It should be noted that is a Hilbert space equipped with the scalar product (x, y) ), respectively.In formulation of problem (1.1)-(1.2),we assume that obeys the following conditions: (i) g is continuous and ), where θ g : R + → R + is nondecreasing.
(ii) There exists a function for all v ∈ X 1 2 where m ∈ L 1 (0, T), θ f : R + → R + is continuous and nondecreasing function.
1.If g, h are Lipschitz, then (3.2) and (3.4) are satisfied.These conditions are also satisfied with η g = η h = 0 if g, h are completely continuous.
2. If f (t, •) satisfies the Lipschitzian condition respect to the second variable, i.e., for some k f ∈ L 1 (0, T), then (3.6) is satisfied.In fact, we have Here S(t) is the resolvent for the linear equation and where R > 0 given.We conclude that M is a bounded convex closed subset of C([0, T]; X 1 2 ).For each u ∈ M, we define the solution operator Φ : M → C([0, T]; X 1 2 ) as follows: Then u is a mild solution of problem (1.1)-(1.2) if it is a fixed point of solution operator Φ.
Thanks to the assumptions imposed of g, h, f , then Φ is continuous on M.

Lemma 3.3. Let the assumptions of Proposition 2.2 and the hypothesis (G)(i), (H)(i), (F)(i) be satisfied.
Then there exists R > 0 such that Φ(M) ⊂ M, provided that Proof.Assuming to the contrary that for each n ∈ N, there exists a sequence {u n } ∞ n=1 ⊂ M with u n C ≤ n and Φ(u n ) C > n.From the formulation of Φ, we have The above inequality leads to Passing to the limit as n → ∞, we get a contradiction to (3.9).
In order to apply the fixed point theory for condensing maps, we will establish the socalled MNC-estimate for the solution operator Φ.

Lemma 3.4. Let the assumptions of Proposition 2.2 and the hypothesis (G)(ii), (H)(ii), (F)(ii) be satisfied, then
for all bounded sets D ⊂ M, here η f = η f L 1 (0,T) . Proof.Setting We have We have It implies that Hence, (3.12) 2. By similar arguments as above, we get 3. Applying Proposition 2.4, there exists {u n } ∞ n=1 ⊂ D such that for every ε > 0, we obtain We invoke Proposition 2.5 to deduce that It is inferred that From (3.14) and (3.15), we obtain Combining (3.11), (3.12) and (3.16) yields The proof is completed.
Therefore χ T (D) = 0, and D is relatively compact.

Existence of decay mild solutions
In this section, we consider solution operator Φ on the following space: and its closed subspace We are going to prove Φ(M ∞ ) ⊂ M ∞ and using the MNC χ * defined by (2.7) to prove that Φ is a χ * -condensing map on M ∞ .In the hypothesis (G), (H), (F), we consider the conditions of g, h, f for any T > 0. The norm • C is replaced by the norm • BC .The conditions m, η f ∈ L 1 (0, T) are replaced by the m, η f ∈ L 1 (R + ).
We recall from [9, Lemma 6.1] the following result.

V. T. Luong
Lemma 4.1.Let X, Y be Banach spaces, and let {U(t)} t≥0 ⊂ B(X, Y) be a strongly continuous operator family which is such that U(•) as well as its adjoint operator U * (•) are strongly integrable.
Then the convolution T defined by is well-defined and bounded from L p (R + ; X) into L p (R + ; Y), for each p ∈ [1, ∞).T is also bounded from C 0 (R + ; X) into C 0 (R + ; Y), provided U(t) → 0 strongly as t → ∞.
We also definite the map Since R(t) is selfadjoint in X, by Proposition 2.2 and Lemma 4.1, Φ is well-defined.

Lemma 4.2.
Let A be a selfadjoint, positive definite operator in the Hilbert space X and let β ∈ L 1 (R + ) be a locally absolutely continuous, nonnegative, nonincreasing map such that b 0 = ∞ 0 β(t)dt < 1.Furthermore, the hypotheses (G), (H), (F) are satisfied.Then we have For every ε > 0, there exists T > 0 such that for any t > T, we get We find that for every Then for any t > T, we have From (4.1), (4.2), we obtain Φ(u)(t) → 0 as t → ∞.The proof is completed.
Lemma 4.3.Let the assumptions of Lemma 4.2 be satisfied.Then we have for all bounded sets D ⊂ M ∞ .
Proof.Let D ⊂ M ∞ be a bounded set.We have 1. Thanks to the Lemma 3.4, we obtain the following estimates: ) ) and From (4.5)-(4.8),we have 2. Next, we find that Applying Lemma 4.2, we obtain  Therefore, χ * (D) = 0, and so D is relatively compact.On the other hand, by condition (4.12) and the arguments in the proof of Lemma 3.3, one can find R > 0 such that Φ(B R ) ⊂ B R where B R is the ball in M ∞ with center at origin and radius R. Applying Theorem 2.7, the χ * -condensing map Φ defined by (3.8) has a nonempty and compact fixed point set Fix(Φ) ⊂ M ∞ .Hence, the problem (1.1)-(1.2) has at least a mild solution u(t), t ∈ R + described by (3.7) which satisfies lim t→∞ u(t) = 0.

An example
Let Ω be a bounded domain in R n with the boundary ∂Ω is smooth enough.Considering the operator L given by Lu

V. T. Luong
Let β ∈ L 1 (R + ) be the scalar memory kernel in previous section and let x 0 ∈ H 1 0 (Ω), y 0 ∈ L 2 (Ω).We consider the following problem: (5.1) It is well known that A is a selfadjoint, positive definite operator in X.Moreover, the fractional power √ A of A is well defined and )) (will be defined later), and Now we give the description for the functions g, h and f .(G) g : BC (ii) By Theorem 8.83 in [10], g is a compact operator, so χ g(D) = 0, for all bounded sets D ⊂ BC(R + ; X 1/2 ).