Research Article | Open Access

Tran Dinh Ke, Valeri Obukhovskii, Ngai-Ching Wong, Jen-Chih Yao, "On Semilinear Integro-Differential Equations with Nonlocal Conditions in Banach Spaces", *Abstract and Applied Analysis*, vol. 2012, Article ID 137576, 26 pages, 2012. https://doi.org/10.1155/2012/137576

# On Semilinear Integro-Differential Equations with Nonlocal Conditions in Banach Spaces

**Academic Editor:**Malisa R. Zizovic

#### Abstract

We study the abstract Cauchy problem for a class of integrodifferential equations in a Banach space with nonlinear perturbations and nonlocal conditions. By using MNC estimates, the existence and continuous dependence results are proved. Under some additional assumptions, we study the topological structure of the solution set.

#### 1. Introduction

In this paper, we investigate the following problem: Here takes values in a Banach space ; , for each , is a linear operator on ; maps and are given. In this model, is the generator of a strongly continuous semigroup on .

It is known that (1.1) with arises from some real applications. For example, the classical heat equation for medium with memory can be written as where and (for more details, see [1, 2]). In addition, if we replace the initial condition by the nonlocal condition (1.2), it allows to describe the model more effectively. As an example of , the following function can be considered: where are given constants and . As another example, one can take where are given linear operators. Regarding to (1.3), in the case , the operators can be given by where are continuous kernel functions.

Semilinear problem (1.1)-(1.2) with was studied extensively. In [3–5], the existence and uniqueness results were obtained by using the contraction mapping principle, under the Lipschitz conditions imposed on and . Supposing Carathéodory-type conditions on , the authors in [6] proved the global existence result with the assumption that the semigroup is compact. However, as it was indicated in [7], if the Lipschitz condition is relaxed, one may get difficulties in proving the compactness of the solution map since the map , in general, is not uniformly continuous in , even in case when is compact. Recently, Fan and Li [8] gave an asymptotical method to solve this problem for the case when is a compact strongly continuous semigroup and the nonlocal function is supposed to be continuous only.

It is known that, in the case , the mild solution of (1.1)-(1.2) on is defined via the integral equation Problem (1.1)-(1.2) involving integro-differential equations was introduced in [2]. The complete references to integro-differential equations can be found in [1, 9, 10]. For some additional problems on solvability and controllability of integro-differential equations, we refer the reader to [11–13]. In order to represent the mild solutions via the variation of constants formula for this case, the notion of so-called resolvent for the corresponding linear equation can be applied. More precisely, an operator-valued function is called the resolvent of (7) if it satisfies the following:(1), the identity operator on ,(2)for each , the map is continuous on ,(3)if is the Banach space formed from , the domain of , endowed with the graph norm, then for and For the existence of resolvent operators, we refer the reader to [14].

It is worth noting that, from definition of resolvent operator and the uniform boundedness principle, there exists such that Then the mild solution on can be represented as By a similar approach as in [3], the authors in [2] obtained the existence and uniqueness of solutions for (1.11) with the assumptions of the Lipschitz conditions on and .

In this work, instead of the Lipschitz conditions posed on and , we assume the regularity of and expressed in terms of the measure of noncompactness. The mentioned regularity can be considered as a generalization of the Lipschitz condition. We first prove the existence of solutions for (1.1)-(1.2) in Section 2. Our method is to find fixed points of a corresponding condensing map, which yields the existence but does not provide the uniqueness of solutions. The arguments in this work are mainly based on the estimates with measure of noncompactness (MNC estimates). It should be noted that this technique was developed in [15], and it has been employed widely for differential inclusions. In Section 3, we prove that the solution set of our problem is continuously dependent on initial data. Section 4 is devoted to a special case when is a Lipschitz function and is compact for . We show that, in this case, the solution set to (1.1)-(1.2) has the so-called -set structure. We end this paper with an example in Section 5.

#### 2. Existence Results

We start with the recalling of some notions and facts (see, e.g. [15, 16]).

*Definition 2.1. *Let be a Banach space with power set , and a partially ordered set. A function is called a measure of noncompactness (MNC) in if
where is the closure of convex hull of . An MNC is called (i)*monotone*, if such that , then ;(ii)*nonsingular*, if for any ;(iii)*invariant with respect to union with compact sets*, if for every relatively compact set and . If, in addition, is a cone in a normed space, we say that is (iv)*algebraically semiadditive*, if for any ;(v)*regular*, if is equivalent to the relative compactness of .

An important example of MNC is *the Hausdorff* MNC, which satisfies all properties given in the previous definition:
Other examples of MNC defined on the space of continuous functions on an interval with values in a Banach space are the following: (i)the modulus of fiber noncompactness:
where is the Hausdorff MNC on and ;(ii)the modulus of equicontinuity:
As indicated in [15], these MNCs satisfy all properties mentioned in Definition 2.1 except the regularity.

Let , that is, is a bounded linear operator from into . We recall the notion of -norm (see e.g., [16]) as follows:
The -norm of can be evaluated as
where and are the unit sphere and the unit ball in , respectively. It is easy to see that

*Definition 2.2. *A continuous map is said to be condensing with respect to a MNC (-condensing) if for every bounded set that is not relatively compact, we have

Let be a monotone nonsingular MNC in . The application of the topological degree theory for condensing maps (see, e.g., [15, 16]) yields the following fixed point principles.

Theorem 2.3 (cf. [15, Corollary 3.3.1]). *Let be a bounded convex closed subset of and a - condensing map. Then is a nonempty compact set.*

Theorem 2.4 (cf. [15, Corollary 3.3.3]). *Let be a bounded open neighborhood of zero, and a - condensing map satisfying the following boundary condition:
**
for all and . Then the fixed point set is nonempty and compact.*

Now, returning to problem (1.1)-(1.2), we impose the following assumptions for and : (G1)the map is continuous;(G2)there exist function and nondecreasing function such that

for a.e. and for all ;(G3)there exists a function such that for each nonempty, bounded set we have

for a.e. , where is the Hausdorff MNC in ;(H1) is a continuous function and there is a nondecreasing function such that

for all , where ;(H2)there is a constant such that

for any bounded subset , where is defined in (2.3).(H3)if is a bounded set, then

*Remark 2.5. *(1) If is a finite dimensional space, one can exclude the hypothesis (G3) since it can be deduced from (G2).

(2) It is known (see, e.g, [15, 16]) that condition (G3) is fulfilled if
where is Lipschitz with respect to the second argument:
for a.e. and with and is compact in second argument; that is, for each and bounded , the set is relatively compact in .

(3) If we assume that is completely continuous, that is, it is continuous and compact on bounded sets, then (H2)-(H3) will be satisfied. It is obvious that if the function in (1.4) obeys (H1)-(H2) and function is uniformly continuous, (H3) is also satisfied. It is worth noting that the function given by (1.5)-(1.6) obeys (H1)–(H3).

As in [2], we assume in the sequel that(F1) for and for continuous with values in ;(F2) for each , the function is continuously differentiable on .It is known that under conditions (F1)-(F2), the resolvent operator for (1.8) exists. We assume, in addition, that(HA) is uniformly norm continuous for . We define the following operator:
Before collecting some properties of , we recall the following definitions.

*Definition 2.6. *A subset of is said to be integrably bounded if there exists a function such that
for all .

*Definition 2.7. *The sequence is called semicompact if it is integrably bounded and the set is relatively compact in for a.e. .

By using hypothesis (HA) and the same arguments as those in [15, Lemma 4.2.1, Theorem 4.2.2, Proposition 4.2.1, and Theorem 5.1.1], one can verify the following properties for :

() the operator sends any integrably bounded set in to equicontinuous set in ;()the following inequality holds:
for every ;()for any compact and sequence such that for a.e. , the weak convergence implies the uniform convergence ;()if is an integrably bounded sequence and is a nonnegative function such that , then(5)if is a semicompact sequence, then is weakly compact in and is relatively compact in . Moreover, if , then . Denote
for and . By we denote the Nemytskii operator corresponding to the nonlinearity , that is,
We see that is a solution of (1.1)-(1.2) if and only if
Let
Then the solutions of (1.1)-(1.2) can be considered as the fixed points of , the operator defined on .

It follows from (G1) and (H1) that is continuous on . Consider the function
where and are defined in (2.3) and (2.4), respectively, denotes the collection of all countable subsets of , and the maximum is taken in the sense of the ordering in the cone . By applying the same arguments as in [15], we have that is well defined. That is, the maximum is archiving in and so is an MNC in the space , which satisfies all properties in Definition 2.1 (see [15, Example 2.1.3] for details).

Theorem 2.8. *Let satisfy (F1)-(F2). Assume that conditions (G1)–(G3) and (H1)–(H3) are fulfilled. If
**
then is -condensing.*

*Proof. *Let be such that
We will show that is relatively compact in . By the definition of , there exists a sequence such that
Following the construction of , one can take a sequence such that
where
Using assumption (G3), we have
for all . Then by (), we obtain
Noting that
we have
due to (2.5)-(2.7) and (H2). Combining (2.29), (2.31), and (2.32), we get
Combining the last inequality with (2.27), we have
and therefore
But then (2.35) implies
Putting (2.37) together with (2.31), we obtain that is semicompact. Hence, by () one that has is relatively compact. This yields
By (H3), we have
Taking (2.29) into account again, we obtain
Now it follows from (2.38)-(2.41) that
By regularity of , we come to the conclusion that is relatively compact.

*Remark 2.9. *If is compact for , we can drop assumption (G3) in the foregoing theorem. Indeed, for any bounded sequence , by setting , one sees that under hypothesis (G2), is an integrably bounded sequence in . In addition, since , is compact, we have
Then by [15, Corollary 4.2.5], we obtain
for each . By this reason, inequality (2.32) becomes
without the reference to (G3).

We now can formulate the existence result in the following way.

Theorem 2.10. *Under assumptions of Theorem 2.8, if one has
**
then the solution set to problem (1.1)-(1.2) is nonempty and compact.*

*Proof. *We will use Theorem 2.3. Applying the results of Theorem 2.8, we only need to verify the existence of a number such that
where is the closed ball in centered at origin with radius . Indeed, assume to the contrary that for each , there is such that
Recalling that
we have
due to (H1) and (G2). Then
Equivalently,
Passing in the last inequality to the limit as , one gets a contradiction due to assumption (2.46). Thus the proof is completed.

We have some special cases related to the growth of and .

Corollary 2.11. *Assume hypotheses of Theorem 2.8, in which (G2) and (H1) are replaced by*(G2′)*, for all ;*(H1′)* is continuous and
for all , respectively.Then the solution set to problem (1.1)-(1.2) is nonempty and compact.*

*Proof. *Since and , condition (2.46) in Theorem 2.10 is testified obviously. Then we get the conclusion.

Corollary 2.12. *Assume hypotheses of Theorem 2.8, in which (G2) and (H1) are replaced by*(G2′′)*, for all ;*(H1′′)* is continuous and
for all , respectively. If one has
then the solution set to problem (1)-(2) is nonempty and compact.*

*Proof. *Under (G2′′) and (H1′′), condition (2.55) is equivalent to (2.46) and the conclusion of Theorem 2.10 holds.

It should be mentioned that if in (H1′), that is, the nonlocal function is uniformly bounded, then one can relax the growth of , by the arguments similar to those in [17].

Theorem 2.13. *Assume the hypotheses of Theorem 2.8, in which (H1) is replaced by*(H1b)* is a continuous function and for all , where is a positive constant.**If one has
where , then the solution set to problem (1.1)-(1.2) is nonempty and compact.*

*Proof. *In this case we employ Theorem 2.4. It suffices to verify the boundary condition in Theorem 2.4. We show that if for , then must belong to a bounded set. Indeed, suppose
It follows that
Putting
we have *,* for all , and
due to the fact that is nondecreasing. Then, by using (2.56), we have
for all . The last inequalities imply that is bounded, so is .

#### 3. Continuous Dependence Result

We start with some notions from the theory of multivalued maps (see, e.g. [15] for details).

Let and be metric spaces; denotes the collection of all nonempty compact subsets of . A multivalued map (multimap) is said to be (i) *upper semicontinuous (u.s.c.)* if for each and there exists such that condition implies , where denotes the -neighborhood of the set induced by the metric (ii) *closed* if its graph is a closed subset of (iii) *compact* if is relatively compact in (iv) *quasicompact* if its restriction to any compact set is compact.

The following assertion gives a sufficient condition for upper semicontinuity.

Lemma 3.1 (see[15]). *Let be a closed quasicompact multimap. Then is u.s.c.*

Consider the solution multimap Notice that, as we demonstrated previously, under conditions of our existence theorems, the solution multimap has compact values. We will study the continuity properties of .

Theorem 3.2. *Under the assumptions of Theorem 2.10, the solution map defined in (3.1) is u.s.c.*

*Proof. *We first prove that is a quasicompact multimap. Let be a compact set. We will show that is relatively compact in . Suppose that . Then there exists a sequence such that
where .

Notice that the sequence is bounded. In fact, from (3.2) we have the estimate
Supposing to the contrary that the sequence is unbounded, by dividing the last inequality over and using condition (2.46) and the boundedness of the sequence , we get a contradiction.

Since is relatively compact, we obtain from (3.2) that
Using (G3) we have
for all . Referring to (), one gets
and then
On the other hand, by (H2) one has
Combining the last inequality with (3.4)-(3.7), we have
This leads to the conclusion that .

Now, condition (G2) implies that is integrably bounded in . Thus () ensures that is equicontinuous. Then applying condition (H3), we obtain
So we have and therefore is relatively compact in .

In order to prove that is u.s.c., it remains, according to Lemma 3.1, to show that is closed. Let in and in . We claim that . Indeed, one has
We first observe that
in in accordance with (H1). In addition, since is a continuous function, we have a.e. . The Lebesgue dominated convergence theorem implies that
due to the fact that is integrably bounded. Therefore, taking (3.11) into account, we arrive at
The proof is completed.

#### 4. Lipschitz Assumption for the Function from Nonlocal Condition

##### 4.1. Existence Result

In this section, we assume that is a Lipschitz function.(H2’) There exists a constant such that

This implies the growth of : and the last inequality covers (H1).

Let be the Hausdorff MNC in . We have for all , where is given in (2.21). Thus Then we know that (see [15]) condition (H2’) implies for any bounded set . We recall the following facts, which will be used in the sequel: for each bounded set , one has the following:(i), for all ;(ii)if is an equicontinuous set (), then

Theorem 4.1. *Assume that satisfies (G1)–(G3) and obeys (H2’). If the following relations
**
hold true, then problem (1.1)-(1.2) has at least one solution.*

*Proof. *As we know from the proof of Theorem 2.10, condition (4.8) implies that there exists a ball , , such that
To apply Theorem 2.3, we verify that is -condensing. Let be a bounded set satisfying the inequality
We will show that is relatively compact. Notice that
where
Then we have

The boundedness of in implies that is a bounded set in . By (), the set is equicontinuous and therefore we have
due to (G3) and (). Thus
Combining (4.5), (4.13), and (4.15), we obtain
Relations (4.7) and (4.10) yield
Since , we have . The regularity of ensures that is relatively compact.

*Remark 4.2. *(1) Assumption (H2^{’}) allows us to drop (H1)–(H3).(2) As indicated in Remark 2.9, in the case when is compact for , condition (G3) can be dropped and condition (4.7) is reduced to
which is covered by (4.8). Recall that in this case we have
for any bounded sequence and for all .

##### 4.2. The Structure of the Solution Set

We are in a position to study the structure of the solution set to (1.1)-(1.2) under the hypotheses of Theorem 4.1 and the assumption that , is compact. At first, let us recall some notions.

*Definition 4.3. *A subset of a metric space is said to be contractible in if the inclusion map is null-homotopic; that is, there exist and a continuous map such that and for every .

The following notion [18] is important for our purposes.

*Definition 4.4. *Let be a metric space; a subset is called an -set if can be represented as the intersection of a decreasing sequence of compact contractible sets.

The next lemma gives us a tool for the recognition of -set.

Lemma 4.5 (see [19]). *Let be a metric space, a Banach space, and a proper map; that is, is continuous and is compact for each compact set . Suppose that there exists a sequence of mappings from into such that *(1)* is proper and converges to uniformly on ;*(2)*for a given point and for all in a neighborhood of in , there exists exactly one solution of the equation .** Then is an -set.*

In order to use this lemma, we need the following result, which is called Lasota-Yorke Approximation Theorem (see, e.g. [20]).

Lemma 4.6. *Let be a normed space, a metric space, and a continuous map. Then for each , there is a locally Lipschitz map such that
*

We now can formulate the main result of this section.

Theorem 4.7. *Assume that satisfies (G1)-(G2) and obeys (H2′). If is compact for and
**
then the solution set of problem (1.1)-(1.2) is an -set.*

*Proof. *By Theorem 4.1 and Remark 4.2, the hypotheses of Theorem 4.7 provide the existence result, that is,
We will show that is an -set.

Consider the nonlinearity . By Lemma 4.6, there exists a sequence of functions such that(i) is continuous and locally Lipschitz map;(ii) for all , where as . One can assume, without loss of generality, that
for all and . Let us consider the following equation:
where is a given function. Define by

By applying the same arguments as in the proof of Theorem 4.1 and in Remark 4.2, we can see that is -condensing. In addition, using the similar estimates as in Theorem 2.10, one can find a ball , , such that
due to (4.23). Therefore, has a fixed point due to Theorem 2.3 and then (4.24) has at least one solution. Moreover, since is Lipschitzian and is a locally Lipschitz function, the solution to (4.24) is unique.

Now by setting
we see that converges to uniformly on . In addition, for a given the equation
has a unique solution, which is the fixed point of mentioned previously.

We now show that and are proper. We proceed with ; the proof for is similar. Obviously, is continuous. Let be a compact set and . We claim that is a compact set in . Since is continuous and is closed, we deduce that is closed. Assume that is a sequence in , then one can take a sequence such that
Equivalently,
We first show that the sequence is bounded. We have
due to (H2^{’}) and (4.23). Thus
If is unbounded, then there is a subsequence (still denoted by ) such that as . Now from the last inequality, it follows that
Passing in the last inequality to limits as , one gets a contradiction due to the hypotheses of the Theorem. Now from (4.30), we have
Using the same arguments as in the proof of Theorem 4.1 and Remark 4.2, we obtain that
Substituting the last inequalities into (4.34) and using the fact that is a compact sequence, we obtain
Noting that , we deduce . Therefore, is a relatively compact sequence in and we arrive at the conclusion that is compact and then is proper.

Finally, by the observation that , from Lemma 4.5, we obtain that is an -set.

*Remark 4.8. *The topological structure of the solution set of problem (1.1)-(1.2) for the case of a non-compact resolvent is an open problem.

*Further Remarks*

Some additional remarks can be given in the case when , is compact. Following the technique presented in [8], we can consider the following problem:
where . Since is continuous, is a completely continuous function. Then it satisfies (H2)-(H3). Therefore, under the assumptions (G1)–(G3), and (H1) and (2.46), problem (4.37)-(4.38) has at least one mild solution . Furthermore,
for all bounded subset . Thus, one can drop assumption (G3), and then the solution operator for (4.37)-(4.38)
is -condensing without assuming the condition
In fact, we have the following assertion.

Theorem 4.9. *Let be compact for . Under assumptions (G1)-(G2) and (H1) and (2.46), the solution set of problem (4.37)-(4.38) is a nonempty compact set.*

Using the same arguments as in [8], one can prove that the sequence of solutions to (4.37)-(4.38) is relatively compact. Finally, passing to the limit as in the equation we obtain the solution of problem (1.1)-(1.2).

#### 5. Example

We conclude this note with an example, in which we find a representation for the resolvent operator generated by the linear part and impose suitable conditions to demonstrate the existence result and the structure of the solution set. Precisely, consider the following system: