In this section, we introduce the existence and uniqueness theorems of mild solutions of Eq. (1). The discussions are based on fractional calculus and Krasnoselskii’s fixed point theorem. Our main results are as follows.
Theorem 1 Suppose that the assumptions ()-() hold. If and the following inequality holds:
then Eq. (1) has at least one mild solution on J.
Proof Define two operators and as follows:
Obviously, u is a mild solution of Eq. (1) if and only if u is a solution of the operator equation on J. To prove the operator equation has solutions, we first show that there is a positive number r such that for every pair . If this were not the case, then for each , there would exist and such that . Thus, from Lemma 3, () and (), we have
Dividing on both sides by r and taking the lower limit as , we have
which contradicts (3). Hence, for some , for every pair .
The next proof will be given in two steps.
Step 1. is a contraction on .
For any and , according to Lemma 3 and assumption (), we have
which implies that . It follows from (3) that , hence is a contraction on .
Step 2. is a completely continuous operator on .
We first prove that is continuous on . Let with as . Then for any , , by assumption (), we have
as , and from assumption (), we have
This together with the Lebesgue dominated convergence theorem gives that
as . Hence, . This means that is continuous on .
Next, we will show that the set is relatively compact. It suffices to show that the family of functions is uniformly bounded and equicontinuous, and for any , the set is relatively compact.
For any , we have for some , which means that is uniformly bounded. In what follows, we show that is a family of equicontinuous functions. For , we have
Hence, it is only necessary to consider . For , from Lemma 3 and assumption (), we have
For any , we have
It follows from Lemma 3 that as and independently of . From the expressions of and , it is clear that and as independently of . Therefore, we prove that is a family of equicontinuous functions.
It remains to prove that for any , the set is relatively compact.
Obviously, is relatively compact in . Let be fixed. For each , and , we define an operator by
Then the sets are relatively compact in since by Lemma 2, the operator is compact for in . Moreover, for every , from Lemma 3 and assumption (), we have
Therefore, there are relatively compact sets arbitrarily close to the set for and since it is compact at , we have the relative compactness of in for all .
Therefore, the set is relatively compact by the Ascoli-Arzela theorem. Thus, the continuity of and relative compactness of the set imply that is a completely continuous operator. Hence, Krasnoselskii’s fixed point theorem shows that the operator equation has a solution on . Therefore, Eq. (1) has at least one mild solution. The proof is completed. □
The following existence and uniqueness theorem for Eq. (1) is based on the Banach contraction principle. We will also need the following assumptions.
() There exists a constant such that the functions are strongly measurable.
() For , there exist functions such that
for any and .
Theorem 2 Let the assumptions ()-() be satisfied. If and the inequalities (3) and
hold, then Eq. (1) has a unique mild solution.
Proof From Lemma 4 and assumption (), it is easy to see that is Bochner integrable with respect to for all . For any , we define an operator Q by
According to the proof of Theorem 1, we know that for some . For any and , from Lemma 3, assumptions () and (), we have
which means that Q is a contraction according to (4). By applying the Banach contraction principle, we know that Q has a unique fixed point on , which is the unique mild solution of Eq. (1). This completes the proof. □