For convenience, we list the following conditions:
(H1) There exist , (), () and () such that
where is nonincreasing, () and , are nondecreasing. And there exist , (, ) such that
(H2) There exists a ( denotes the dual cone of ) such that . And for any , there exists a such that
(H3) There exists a such that
where b, (), (), φ, (), (), (, ) and are defined as in conditions (H1) and (H2), and , .
(H4) There exist (), such that
(), . There exist (, ) such that
Remark Obviously, condition (H4) is satisfied automatically when E is finite dimensional.
Lemma 3.1 Suppose conditions (H1), (H2) and (H3) are satisfied. Then Q defined by
is a nonempty, convex and closed subset of .
It is clear that . Since , for , by (11), one can see that
which implies () for .
By conditions (H1), (H2), (H3), and (11), we have
and . Therefore, and Q is a nonempty set.
Now, we check that Q is a convex subset of . In fact, for any , , we write , which means . It is clear that
By virtue of the characters of elements of Q and the characters of φ, we have
In the same way,
Therefore, . Thus, Q is a convex subset of . It is clear that Q is a closed subset of . So the conclusion holds. □
Lemma 3.2 Assume that conditions (H1), (H2) and (H3) are satisfied. Then , where the operator A is defined by
Proof For any , i.e.,
For any and ,
Because of , and , , we know
Analogously, for , it is easy to see
Differentiating (15) times, we get
Obviously, () exist and
where is understood as θ for . Similarly, () exist. Hence,
and (; ), it follows from (15), (16), (17) and (18) that
It is clear that
Since , by (22), (26) and condition (H2), we have
Now, we show that
By (15), (19), (21), conditions (H1) and (H3) imply
which implies that (29) is true. By (24), (26) to (29), the conclusion holds. □
Lemma 3.3 Suppose conditions (H1) to (H4) are satisfied. Let
in which , . Then
in which ().
Proof In order to avoid the singularity, given , let
By conditions (H1), (H2) and (H3), for any , , we have
By virtue of absolute continuity of the Lebesgue integrable function, we have
in which, denotes the Hausdorff distance between and . Therefore,
Now, we show
In fact, by Lemma 2.2, we have
where , .
On the other hand, for , it follows from (19) and (21) that
Taking , , by (16), (17) and (18), one can see that
Therefore, by condition (H4) and (36), for , it is easy to get
Since B is a bounded set of and is a bounded set, is equicontinuous on each (). By Lemma 2.3, it is easy to get
Substituting (41) into (40), we get (31).
Similarly, we obtain (32) and our conclusion holds. □
Lemma 3.4 Let conditions (H1) to (H3) be satisfied. is a solution of SBVP (1), if and only if is a fixed point of the operator A defined by (15).
Proof First of all, by mathematical induction, for , Taylor’s formula with the integral remainder term holds,
In fact, as , for , let , it is easy to see that
Adding these together, we get
This proves that (42) is true for .
Suppose (42) is true for , i.e., for , the next formula holds:
Now we check that (42) is also true for n. In fact, suppose . Then , by (43), one can see
Substituting the above equation into (44), we get
So, (42) is also true for n. By mathematical induction, (42) holds.
Suppose is a solution of SBVP (1). By (42), we can see that
into (47), by (15), we get . So u is a fixed point of the operator A defined by (15) in Q.
Conversely, if is a fixed point of the operator A, i.e., u is a solution of the following impulsive integro-differential equation:
Then, by (15), similar to (26), from the derivative of both sides of the above equation one can draw the following conclusions:
So, we have
It follows from (48) and (50) that . By (48), (49) and (50), we have
It is easy to see by (48) and (50)
By (51) to (54), u is a solution of SBVP (1). □
Theorem 3.1 Let conditions (H1) to (H4) be satisfied. Assume that
Then SBVP (1) has at least a solution .
Proof We will use Lemma 2.5 to prove our conclusion. By (H1)-(H3), from Lemma 3.2, we know .
We affirm that is continuous. In fact, let , , (as ). From the continuity of f and (, ) and the definition of A, by virtue of the Lebesgue dominated convergence theorem, we see that
For (fixed), we have (). We also see that is bounded and is equicontinuous on each . By Lemma 2.1, it is easy to get
i.e., is a relatively compact set in . The reduction to absurdity is used to prove that A is continuous. Suppose . Then , such that
On the other hand, since is a relatively compact set in , there exists a subsequence of which converges to . Without loss of generality, we may assume itself converges to y, that is,
By virtue of (56), we see that . Obviously, this is in contradiction to (58). Hence,
Consequently, is continuous.
By Lemma 2.4, for any countable , which satisfies , one can see