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).