### 3.1. Introduction

Functional differential and partial differential equations arise in many areas of applied mathematics and such equations have received much attention in recent years. A good guide to the literature for functional differential equations is the books by Hale [43] and Hale and Verduyn Lunel [44], Kolmanovskii and Myshkis [45], and Wu [46] and the references therein.

In a series of papers (see [47–50]), the authors considered some classes of initial value problems for functional differential equations involving the Riemann-Liouville and Caputo fractional derivatives of order In [51, 52] some classes of semilinear functional differential equations involving the Riemann-Liouville have been considered. For more details on the geometric and physical interpretation for fractional derivatives of both the Riemann-Liouville and Caputo types see [53, 54].

In the following, we consider the semilinear functional differential equation of fractional order of the form

where is the standard Riemann-Liouville fractional derivative, is a continuous function, is a closed linear operator (possibly unbounded), a given continuous function with , and a real Banach space. For any function defined on and any we denote by the element of defined by

Here represents the history of the state from time , up to the present time .

The reason for studying (3.1) is that it appears in mathematical models of viscoelasticity [55], and in other fields of science [54, 56]. Equation (3.1) is equivalent to solve an integral equation of convolution type. It is also of interest to explore the neighborhood of the diffusion (). In this survey paper, we use the fractional derivative in the Riemann-Liouville sense. The problems considered in the survey are subject to zero data, which in this case, the Riemann-Liouville and Caputo fractional derivatives coincide. From a practical point of view, in some mathematical models it is more appropriate to consider traditional initial or boundary data. This is what we are considering in this survey.

In all our paper we suppose that the operator is the infinitesimal generator of a -semigroup . Denote by

Before stating our main results in this section for problem (3.1) and (3.2) we give the definition of the mild solution.

Definition 3.1 (see [23]).

One says that a continuous function is a mild solution of problem (3.1) and (3.2) if and

### 3.2. Existence Results for Finite Delay

By using the Banach's contraction principle, we get the following existence result for problem (3.1) and (3.2).

Theorem 3.2.

Let continuous. Assume the following.

Then there exists a unique mild solution of problem (3.1) and (3.2) on

Proof.

Transform the IVP (3.1) and (3.2) into a fixed point problem. Consider the operator defined by

Let us define the iterates of operator by

It will be sufficient to prove that is a contraction operator for sufficiently large. For every we have

Indeed,

Therefore (3.9) is proved for . Assuming by induction that (3.9) is valid for , then

and then (3.9) follows for .

Now, taking sufficiently large in (3.9) yield the contraction of operator .

Consequently has a unique fixed point by the Banach's contraction principle, which gives rise to a unique mild solution to the problem (3.1) and (3.2).

The following existence result is based upon Theorem 2.9.

Theorem 3.3.

Assume that the following hypotheses hold.

Then the problem (3.1) and (3.2) has at least one mild solution on

Proof.

Transform the IVP (3.1) and (3.2) into a fixed point problem. Consider the operator as defined in Theorem 3.2. To show that is continuous, let us consider a sequence such that in . Then

Since is a continuous function, then we have

Thus is continuous. Now for any , and each we have for each

Thus maps bounded sets into bounded sets in .

Now, let , Thus if and we have for any

As and sufficiently small, the right-hand side of the above inequality tends to zero, since is a strongly continuous operator and the compactness of for implies the continuity in the uniform operator topology [29]. By the Arzelá-Ascoli theorem it suffices to show that maps into a precompact set in .

Let be fixed and let be a real number satisfying . For we define

Since is a compact operator for , the set

is precompact in for every Moreover

Therefore, the set is precompact in . Hence the operator is completely continuous. Now, it remains to show that the set

is bounded.

Let be any element. Then, for each ,

Then

We consider the function defined by

Let such that , if then by (3.22) we have, for (note )

If then and the previous inequality holds.

By Lemma 2.7 we have

Hence

This shows that the set is bounded. As a consequence of Theorem 2.9, we deduce that the operator has a fixed point which is a mild solution of the problem (3.1) and (3.2).

### 3.3. An Example

As an application of our results we consider the following partial functional differential equation of the form

where is continuous and is a given function.

Let

Take and define by with domain

Then

where is the inner product in and is the orthogonal set of eigenvectors in It is well known (see [29]) that is the infinitesimal generator of an analytic semigroup in and is given by

Since the analytic semigroup is compact, there exists a constant such that

Also assume that there exist continuous functions such that

We can show that problem (3.1) and (3.2) is an abstract formulation of problem (3.27). Since all the conditions of Theorem 3.3 are satisfied, the problem (3.27) has a solution on

### 3.4. Existence Results for Infinite Delay

In the following we will extend the previous results to the case when the delay is infinite. More precisely we consider the following problem

where is the standard Riemann-Liouville fractional derivative, is a continuous function, the phase space [41], is the infinitesimal generator of a strongly continuous semigroup , a continuous function with and a real Banach space. For any the function is defined by

Consider the following space:

where is the restriction of to Let be the seminorm in defined by

Definition 3.4.

One says that a function is a mild solution of problem (3.34) if and

The first existence result is based on Banach's contraction principle.

Theorem 3.5.

Assume the following.

Then there exists a unique mild solution of problem (3.34) on

Proof.

Transform the IVP (3.34) into a fixed point problem. Consider the operator defined by

For , we define the function

Then . Set

It is obvious that satisfies (3.38) if and only if satisfies and

Let

For any , we have

Thus is a Banach space. Let the operator defined by

It is obvious that has a fixed point is equivalent to has a fixed point, and so we turn to proving that has a fixed point. As in Theorem 3.2, we show by induction that satisfy for any , the following inequality:

which yields the contraction of for sufficiently large values of . Therefore, by the Banach's contraction principle has a unique fixed point . Then is a fixed point of the operator , which gives rise to a unique mild solution of the problem (3.34).

Next we give an existence result based upon the nonlinear alternative of Leray-Schauder type.

Theorem 3.6.

Assume that the following hypotheses hold.

Then, the problem (3.34) has at least one mild solution on

Proof.

Transform the IVP (3.34) into a fixed point problem. Consider the operator defined as in Theorem 3.5. We will show that the operator is continuous and completely continuous. Let be a sequence such that in . Then

Since is a continuous function, then we have

Thus is continuous. To show that maps bounded sets into bounded sets in it is enough to show that for any there exists a positive constant such that for each we have Let , then

Then we have for each

Taking the supremum over we have

Now let , thus if and we have for each

As and sufficiently small, the right-hand side of the above inequality tends to zero, since is a strongly continuous operator and the compactness of for implies the continuity in the uniform operator topology (see [29]). By the Arzelá-Ascoli theorem it suffices to show that maps into a precompact set in . Let be fixed and let be a real number satisfying . For we define

Since is a compact operator for , the set

is precompact in for every Moreover

Therefore, the set is precompact in . Hence the operator is completely continuous. Now, it remains to show that the set

is bounded. Let be any element. Then, for each ,

Then

but

Take the right-hand side of the above inequality as , then by (3.60) we have

Using the above inequality and the definition of we have

By Lemma 2.7, there exists a constant such that we have

Then there exists a constant such that This shows that the set is bounded. As a consequence of the Leray-Schauder Theorem, we deduce that the operator has a fixed point, then has one which gives rise to a mild solution of the problem (3.34).

### 3.5. An Example

To illustrate the previous results, we consider in this section the following model:

where are continuous functions.

Consider and define by with domain

Then generates a semigroup (see [29]).

For the phase space , we choose the well-known space : the space of uniformly bounded continuous functions endowed with the following norm:

If we put for and

Then, problem (3.65) takes the abstract neutral evolution form (3.34).