The objective, in this work, is to study the alpha-norm, the existence, the continuity dependence in initial data, the regularity, and the compactness of solutions of mild solution for some semi-linear partial functional integrodifferential equations in abstract Banach space. Our main tools are the fractional power of linear operator theory and the operator resolvent theory. We suppose that the linear part has a resolvent operator in the sense of Grimmer. The nonlinear part is assumed to be continuous with respect to a fractional power of the linear part in the second variable. An application is provided to illustrate our results.
- mild solution
- resolvent operator
- fractional power operator
We consider, in this manuscript, partial functional equations of retarded type with deviating arguments in terms of involving spatial partial derivatives in the following form :
where is the infinitesimal generator of an analytic semigroup on a Banach space . is a closed linear operator with domain time-independent. For , is the fractional power of which will be precise in the sequel. The domain is endowed with the norm called norm. is the Banach space of continuous functions from to endowed with the following norm:
is a continuous function, and as usual, the history function is defined by
As a model for this class, one may take the following Lotka-Volterra equation:
Here and are appropriate functions.
In the particular case where , many results are obtained in the literature under various hypotheses concerning , , and (see, for instance, [2, 3, 4, 5, 6] and the references therein). For example, in , Ezzinbi et al. investigated the existence and regularity of solutions of the following equation:
The authors obtained also the uniqueness and the representation of solutions via a variation of constant formula, and other properties of the resolvent operator were studied. In , Ezzinbi et al. studied a local existence and regularity of Eq. (3). To achieve their goal, the authors used the variation of constant formula, the theory of resolvent operator, and the principle contraction method. Ezzinbi et al. in  studied the local existence and global continuation for Eq. (3). Recall that the resolvent operator plays an important role in solving Eq. (3); in the weak and strict sense, it replaces the role of the semigroup theory. For more details in this topic, here are the papers of Chen and Grimmer , Hannsgen , Smart , Miller [12, 13], and Miller and Wheeler [14, 15]. In the case where the nonlinear part involves spatial derivative, the above obtained results become invalid. To overcome this difficulty, we shall restrict our problem in a Banach space , to obtain our main results for Eq. (1).
Considering the case where , Travis and Webb in  obtained results on the existence, stability, regularity, and compactness of Eq. (1). To achieve their goal, the authors assumed that is the infinitesimal generator of a compact analytic semigroup and is only continuous with respect to a fractional power of A in the second variable. The present paper is motivated by the paper of Travis and Webb in .
The paper is organized as follows. In Section 2, we recall some fundamental properties of the resolvent operator and fractional powers of closed operators. The global existence, uniqueness, and continuous dependence with respect to the initial data are studied in Section 3. In Section 4, we study the local existence and bowing up phenomena. In Section 5 we prove, under some conditions, the regularity of the mild solutions. And finally, we illustrate our main results in Section 6 by examining an example.
2. Fractional power of closed operators and resolvent operator for integrodifferential equations
We shall write for endowed with the graph norm for and will denote the space of bounded linear operators from to , and for , we write with norm . We also frequently use the Laplace transform of which is denoted by . If we assume that generates an analytic semigroup and, without loss of generality, that , then one can define the fractional power for , as a closed linear operator on its domain with its inverse given by
where is the gamma function
We have the following known results.
Theorem 2.1.  The following properties are true.
is a Banach space with the norm for .
is a closed linear operator with domain and .
is a bounded linear operator in .
If then . Moreover the injection is compact if is compact for .
and for some and .
For all , is continuous for .
for . For , , and for we have
What follows is we assume the hypothesis taken from  which implies the existence of an analytic resolvent operator .
(V1) generates an analytic semigroup on . is a closed operator on with domain at least a.e with strongly measurable for each and with absolutely convergent for .
(V2) exists as a bounded operator on which is analytic for where . In if , there exists so that .
(V3) for and is analytic from to . and for . Given , there exists a positive constant so that for and with and as in . In addition, for some with . Further, there exists which is dense in such that and are contained in and is bounded for each and with .
Theorem 2.3.  Assume that conditions (V1)–(V3) are satisfied. Then there exists an analytic resolvent operator . Moreover, there exist positive constants such that for and .
We take the following hypothesis.
(H0) The semigroup is compact for .
Theorem 2.4.  Under the conditions (V1)–(V3) and (H0), the corresponding resolvent operator is compact for .
3. Global existence, uniqueness, and continuous dependence with respect to the initial data
Definition 3.1. A function is called a strict solution of Eq. (1), if:
is continuously differentiable on .
satisfies Eq. (1) on .
Definition 3.2. A continuous function is called a mild solution of Eq. (1) if
Now to obtain our first result, we take the following assumption.
(H1) There exists a constant such that
Theorem 3.3. Assume that (V1)–(V3) and (H1) hold. Then for , Eq. (1) has a unique mild solution which is defined for all
Proof. Let . For , we define the set by
The set is a closed subset of where is the space of continuous functions from to equipped with the uniform norm topology
For , we introduce the extension of on defined by for and for . We consider the operator defined on by
We claim that In fact for , we have , and by continuity of and for , we deduce that In order to obtain our result, we apply the strict contraction principle. In fact, let and . Then
Using the norm, we have
Now we choose such that
Then is a strict contraction on , and it has a unique fixed point which is the unique mild solution of Eq. (1) on . To extend the solution of Eq. (1) in , we show that the following equation has a unique mild solution:
Notice that the solution of Eq. (6) is given by
Let be the function defined by for and for . Consider now again the set defined by
provided with the induced topological norm. We define the operator on by
We have and is continuous. Then it follows that Moreover, for , one has
Since in we deduce that
Then we deduce that has a unique fixed point in which extends the solution in . Proceeding inductively, is uniquely and continuously extended to for all , and this ends the proof.
Now we show the continuous dependence of the mild solutions with respect to the initial data.
Theorem 3.4. Assume that (V1)–(V3) and (H1) hold. Then the mild solution of Eq. (1) defines a continuous Lipschitz operator in by . That is, is continuous from to for each fixed . Moreover there exist a real number and a scalar function such that for and we have
Proof. We use the gamma formula
where (see , p. 265). The continuity is obvious that the map is continuous. Now, let If we pose then we have
Let a real number be such that
We define the function by
Fix and let . If , then from Eq. (8), we have
If , we have
For , we have
which implies that
Then the result follows.
4. Local existence, blowing up phenomena, and the compactness of the flow
We start by generalizing a result, obtained in  in the case of the usual norm on , in the case where . We take the following assumption.
(H2) for some , a.e and for with where
Theorem 4.1. Assume that (V1)–(V3) and (H2) hold. Then for any there exists a positive constant such that for we have
Proof. Let and . Then
We deduce that satisfies the equation of the form
Then by the variation o constante formula, it follows that
Which yields that
Taking the norm, we obtain that
Let be such that , so Then it follows that
And the proof is complete.
The local existence result is given by the following Theorem.
Theorem 4.2. Suppose that (V1)–(V3), (H0), and (H2) hold. Moreover, assume that defined from into is continuous where is an open set in . Then for each , Eq. (1) has at least one mild solution which is defined on some interval .
Proof. Let . For any real and we define the following sets:
For , we choose and such that and By continuity of , there exists such that for in . We consider as the function defined by for and . Suppose that and choose such that
Let Then we have for and since Consider the mapping from to defined by
Notice that finding a fixed point of in is equivalent to finding a mild solution of Eq. (1) in . Furthermore, is a mapping from to , since if we have and
which implies that . We claim that is compact in for fixed In fact, let be such that . The above estimate show that is bounded in . Since is compact operator, we infer that is compact in hence is compact in . Next, we show that is equicontinuous. The equicontinuity of at follows from the above estimation of . Now let with be fixed. Then we have
Using Theorem 4.1, it follows that
As the set is compact in , we have
We obtain the same results by taking be fixed with Then we claim that uniformly in which means that is equicontinuous. Then by Ascoli-Arzela theorem, is relatively compact in . Finally, we prove that is continuous. Since is continuous, given , there exists , such that
Then for , we have
This yields the continuity of , and using Schauder’s fixed point theorem, we deduce that has a fixed point. Then the proof of the theorem is complete.
The following result gives the blowing up phenomena of the mild solution in finite times.
Theorem 4.3. Assume that (V1)–(V3), (H0), and (H2) hold and is a continuous and bounded mapping. Then for each , Eq. (1) has a mild solution on a maximal interval of existence . Moreover if , then .
Proof. Let be the mild solution of Eq. (1) defined on . Similar arguments used in the local existence result can be used for the existence of and a function defined from to satisfying
By a similar proceeding, we show that the mild solution can be extended to a maximal interval of existence . Assume that and . There exists such that for . We claim that is uniformly continuous. In fact, let . Then
By continuity of , we claim that is uniformly continuous on each compact set. Moreover, Theorem 4.1 implies that uniformly in when In fact we have
Then using Theorem 4.1, we obtain that
We claim that the set is relatively compact. In fact, let be a sequence of . Then there exist a subsequence and a real number such that . Using the dominated convergence theorem, we deduce that
This implies that is relatively compact. Now using Banach-Steinhaus’ theorem, we deduce that
uniformly when with respect to . Moreover we have
Consequently uniformly in . If , that is, for , we have
one can show similar results by using the same reasoning. This implies that is uniformly continuous. Therefore And consequently, can be extended to which contradicts the maximality of .
The next result gives the global existence of the mild solutions under weak conditions of . To achieve our goal, we introduce a following necessary result which is a consequence of Lemma 7.1.1 given in (, p. 197, Exo 4).
Lemma 4.4.  Let and Also assume that is nonnegative and locally integrable on with
Then there exists a constant such that on
Theorem 4.5. Assume that (V1)–(V3), (H0), and (H2) hold and is a completely continuous function on . Moreover suppose that there exist continuous nonnegative functions and such that for and Then Eq. (1) has a mild solution which is defined for .
Proof. Let be the maximal interval of existence of a mild solution . Assume that By Theorem 4.3 we have . Recall that the solution of Eq. (1) is given by and
Then taking the norm, we obtain
where and . Then we deduce that
Now we claim that the function
is nondecreasing. In fact, let . Then
which yields the result. Then it follows from Eq. (15) that
Then using Lemma 4.4, we deduce that is bounded in . Then we obtain that which contradicts our hypothesis. Then the mild solution is global.
We focus now to the compactness of the flow defined by the mild solutions.
Theorem 4.6. Assume that (V1)–(V3) and (H0)–(H2) hold. Then the flow defined from to by is compact for , where denotes the mild solution starting from .
Proof. We use Ascoli-Arzela’s theorem. Let be a bounded subset of and let be fixed, but arbitrary. We will prove that is compact. It follows from (H1) and inequality Eq. (7) that there exists such that
For each we define by . We show now that for fixed the set is precompact in . For any , we have
As is compact for , we need only to prove that the set
is compact. Also we have
where is the measure of non-compactness. Moreover, using Theorem 4.1, we have
We deduce that
On the other hand, for , we have
Thus is a bounded subset of . The precompactness in now follows from the compactness of . Then the set is precompacted in . We prove that the family is equicontinuous. Let in and with be fixed and . Then
Then it follows that
Using the compactness of the set and the continuity of for , the right side of the above inequality can be made sufficiently small for small enough. Then we conclude that is equicontinuous. Consequently, by Ascoli-Arzela’s theorem, we conclude that the set is compact, which means that the operator is compact for .
5. Regularity of the mild solutions
We define the set by as the set of continuously differentiable functions from to . We assume the following hypothesis.
(H3) is continuously differentiable, and the partial derivatives and are locally Lipschitz in the classical sense with respect to the second argument.
Theorem 5.1. Assume that (V1)–(V3), (H1), and (H3) hold. Let in be such that and Then the corresponding mild solution becomes a strict solution of Eq. (1).
Proof. Let . Take such that and , and let be the mild solution of Eq. (1) which is defined on . Consider the following equation:
Using the strict contraction principle, we can show that there exists a unique continuous function solution in of Eq. (16). We introduce the function defined by
Then it follows
Consequently, the maps and are continuously differentiable, and the following formula holds
This implies that
On the other hand, from equality Eq. (4), we have
We rewrite as follows:
Then it follows that
We deduce, for that
The set is compact in . Since the partial derivatives of are locally Lipschitz with respect to the second argument, it is well-known that they are globally Lipschitz on . Then we deduce that
where with and the Lipschitz constant of and , respectively, which implies that
If we choose such that
then in . Now we will prove that in Assume that there exists such that . Let By continuity, one has for , and there exists such that for . Then it follows that for ,
Now choosing such that
then in which gives a contradiction. Consequently, for . We conclude that from to and from to are continuously differentiable. Thus, we claim that is a strict solution of Eq. (1) on [22, 23, 24, 25, 26, 27, 28, 29, 30, 31].
For illustration, we propose to study the model Eq. (2) given in the Introduction. We recall that this is defined by
where and are appropriate functions. To study this equation, we choose , with its usual norm . We define the operator by
and For , we define where for each . We define = equipped with norm and the functions and and by for a.e and , and finally
Then Eq. (18) takes the abstract form
The is a closed operator and generates an analytic compact semigroup on . Thus, there exists in and such that is contained in , the resolvent set of , and for . The operator is closed and for , . The operator has a discrete spectrum, the eigenvalues are , and the corresponding normalized eigenvectors are . Moreover the following formula holds:
One also has the following result.
Lemma 6.1  LetThenis absolutely continuous,and
We assume the following assumptions.
(H4) The scalar function and satisfies the Laplace transform of h) and for . Further, as for and .
(H5) The function is continuous and Lipschitz with respect to the second variable.
By assumption (H4), the operator exists as a bounded operator on , which is analytic in and satisfies On the other hand, for , we have
Since is bounded because , then has the growth properties of which tends to 1 if goes to infinity. Then we deduce that . Moreover, it is analytic from to . Now, for , one has
Then it follows that
We deduce that , , and Considering , we see that the conditions (V1)–(V3) and (H0) are verified. Hence the homogeneous linear equation of Eq. (18) has an analytic compact resolvent operator . The function is continuous in the first variable from the fact that is continuous in the first variable. Moreover from Lemma 6.1 and the continuity of , we deduce that is continuous with respect to the second argument. This yields the continuity of in . In addition, by assumption (H5) we deduce that
Then is a continuous globally Lipschitz function with respect to the second argument. We obtain the following important result.
Proposition 6.2. Suppose that the assumptions (H4)–(H5) hold. Then Eq. (19) has a mild solution which is defined for .