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.