To set the framework for our main existence results, we will make use of the following lemma.
Lemma 3.1.
Let
. Assume that
is a sectorial operator of type
and
is a solution operator generated by
. Suppose in addition that
is a continuous function. If
is a mild solution of the Cauchy problem (2.14) in the sense of Definition 2.1, then,
satisfying the following impulsive integral equation:
is a mild solution of problem (1.3), where
Proof.
Assume that
is a mild solution of (2.14) in the sense of Definition 2.1. Obviously, if
, then one sees from Definition 2.1, that the assertion of theorem remains true. Thus, the rest proof of the theorem is done under
.
By Definition 2.1, note that
for all
. Taking
, then we get
Hence, it follows form
that
If
, then combining Definition 2.1 and the result above, we deduce that
This proves, for the case
, that the conclusion of theorem holds.
Now taking
in (3.7), one has
which implies that
provided that
. Then, again making use of Definition 2.1, we get for all
,
here
and
are given by (3.2) and (3.3) with
, respectively. A continuation of the same process shows that for any
, the assertion of theorem holds.
In this work, we adopt the following concept of mild solution for the problem (1.3).
Definition 3.2.
Let
. Assume that
is a sectorial operator of type
,
is a solution operator generated by
, and
and
are given by (3.2) and (3.3), respectively. A solution
of the integral equation
here
, is called a mild solution of the Cauchy problem (1.3).
Remark 3.3.
Note that if there is no discontinuity, that is, if
,
, then Definition 2.1 is equivalent to Definition 3.2.
Now we present and prove our main results.
Theorem 3.4.
Let
. Assume that
is a sectorial operator of type
and
is a solution operator generated by
. Suppose in addition that assumptions
are fulfilled. Then the Cauchy problem (1.3) admits at least one mild solution, provided
Proof.
Consider the mapping
, which is defined for each
by
Then it is clear that
is well defined.
To prove the theorem, it is sufficient to prove that
has a fixed point in
.
Put
for
as selected below.
We first show that there exists an integer
such that
maps
into
. For the case
, by assumption
and the estimate (2.7), a straightforward calculation yields that
We claim that there exists an integer
such that
provided that
. In fact, if this is not the case, then for each
, there would exist
and
such that
. Thus, by (3.15) and assumption
we obtain
Dividing on both sides by
and taking the lower limit as
, we get
which contradicts (3.12).
Since the interval
is divided into finite subintervals by
, we only need to prove that for a fixed
,
maps
into
, here
is a positive number yet to be determined, as the cases for other subintervals are the same.
From the Hypotheses
, we infer for any
,
Now, an application of the same idea with above discussion yields that there exists a
such that
. Indeed, if this is not the case, then we would deduce that
This is a contradiction to (3.12). Thus, we prove that there exists an integer
such that
.
For
, we decompose the mapping
as follows:
Next, we show that for each
,
is completely continuous, while
is a contraction. In fact, it follows from assumption
and the estimate (2.7) that
,
is completely continuous. Note also that
For the case
, it is clear that the conclusion holds in view of (3.12). For
, by
,
and (2.7) we get
provided that
. Hence, we deduce that
which means that
is a contraction due to (3.12).
Thus,
is a condensing map on
. Then, it follows from Lemma 2.2 that the Cauchy problem (1.3) admits at least one mild solution. This completes the proof.
Theorem 3.5.
Let
. Assume that
is a sectorial operator of type
,
is a solution operator generated by
, and the Hypotheses
are satisfied. Then the Cauchy problem (1.3) admits at least one mild solution, provided
Proof.
Assume that the map
and the set
are defined the same as in Theorem 3.4. First we claim that there exists an positive number
such that
. For the case
, the proof of the assertion follows from Theorem 3.4. For the case
, if the conclusion is not true, then for each positive integer
, there would exist
and
such that
with
, where
denotes
depending upon
. Thus, by assumptions
,
,
, we have
Dividing on both sides by
and taking the lower limit as
, we have
This is a contradiction to (3.25).
For
, decompose the mapping
as follows:
Next, we will verify that for each
,
is a completely continuous operator, while,
is a contraction. Obviously, by assumptions
, it easily seen that
is a completely continuous operator. Moreover, by a similar proof with that in Theorem 3.4, we can prove that
is a contraction.
As a consequence of the above discussion and Lemma 2.2, we can conclude that the problem (1.3) admits at least one mild solution. The proof is completed.
Theorem 3.6.
Let
. Assume that
is a sectorial operator of type
and
is a solution operator generated by
. Then, under assumptions
,
,
, the Cauchy problem (1.3) has a unique mild solution, provided
Proof.
Assume that the map
is defined the same as in Theorem 3.4. Now, we prove that
is a contraction. Take any
. For the case
, the conclusion follows from assumptions
,
, and (3.29). For
, a direct calculation yields
in view of assumptions
,
,
. Hence, we deduce that
which implies
is a contractive mapping on
due to (3.29). Thus
has a unique fixed point
, this means that
is a mild solution of (1.3). This completes the proof of the theorem.