In this section, by using the techniques of approximate solutions and fixed points, we establish a result on the existence of mild solutions for the nonlocal impulsive problem (1.1) when the nonlocal item
and the impulsive functions
are only assumed to be continuous in
and
, respectively.
In practical applications, the values of
for
near zero often do not affect
. For example, it is the case when
So, to prove our main results, we introduce the following assumptions.
(H2)
is a continuous function, and there is a
such that
for any
with
,
. Moreover, there exist
such that
for any
.
(H3)There exists a
such that
is a continuous function, and
for any
with
,
. Moreover, there exist
such that
for any
,
, and
for any
,
.
(H4)The function
is continuous a.e.
; the function
is strongly measurable for all
. Moreover, for each
, there exists a function
such that
for a.e.
and all
, and
(H5)
is continuous for every
, and there exist positive numbers
such that
for any
and
.
We note that, by Theorem 2.1, there exist
and
such that
and
For simplicity, in the following we set
and will substitute
by
below.
Theorem 3.1.
Let (H1)–(H5) hold. Then the nonlocal impulsive Cauchy problem (1.1) has at least one mild solution on
, provided
To prove the theorem, we need some lemmas. Next, for
, we denote by
the maps
defined by
In addition, we introduce the decomposition
, where
for
and
.
Lemma 3.2.
Assume that all the conditions in Theorem 3.1 are satisfied. Then for any
, the map
defined by (3.7) has at least one fixed point
.
Proof.
To prove the existence of a fixed point for
, we will use Darbu-Sadovskii's fixed point theorem.
Firstly, we prove that the map
is a contraction on
. For this purpose, let
. Then for each
and by condition (H3), we have
Thus,
which implies that
is a contraction by condition (3.6).
Secondly, we prove that
,
,
are completely continuous operators. Let
be a sequence in
with
in
. By the continuity of
with respect to the second argument, we deduce that for each
,
converges to
in
, and we have
Then by the continuity of
,
,
, and using the dominated convergence theorem, we get
in
, which implies that
are continuous on
.
Next, for the compactness of
we refer to the proof of [4, Theorem 3.1].
For
and any bounded subset
of
, we have
which implies that
is relatively compact in
for every
by the compactness of
. On the other hand, for
, we have
Since
is relatively compact in
, we conclude that
which implies that
is equicontinuous on
. Therefore,
is a compact operator.
Now, we prove the compactness of
. For this purpose, let
Note that
Thus according to Lemma 2.3, we only need to prove that
is precompact in
, as the remaining cases for
,
, can be dealt with in the same way; here
is any bounded subset in
. And, we recall that
,
, which means that
Thus, by the compactness of
, we know that
is relatively compact in
for every
.
Next, for
, we have
Thus, the set
is equicontinuous due to the compactness of
and the strong continuity of operator
. By the Arzela-Ascoli theorem, we conclude that
is precompact in
. The same idea can be used to prove that
is precompact for each
. Therefore,
is precompact in
, that is, the operator
is compact.
Thus, for any bounded subset
, we have by Lemma 2.4,
Hence, the map
is an
-contraction in
.
Now, in order to apply Lemma 2.5, it remains to prove that there exists a constant
such that
. Suppose this is not true; then for each positive integer
, there are
and
such that
. Then
Dividing on both sides by
and taking the lower limit as
, we obtain that
This is a contradiction with inequality (3.6). Therefore, there exists
such that the mapping
maps
into itself. By Darbu-Sadovskii's fixed point theorem, the operator
has at least one fixed point in
. This completes the proof.
Lemma 3.3.
Assume that all the conditions in Theorem 3.1 are satisfied. Then the set
is precompact in
for all
, where
and
is the constant in (H2).
Proof.
The proof will be given in several steps. In the following
is a number in
.
Step 1.
is precompact in
.
For
, define
by
For
, let
,
,
,
, and we define
by
By condition (H3),
is well defined and for
, we have
On the other hand, for
,
, we have
,
. So,
Now, for
, we have
By the compactness of
,
, we get that
is relatively compact in
for every
and
is equicontinuous on
, which implies that
is precompact in
.
By the same reasoning,
is precompact in
.
For
, we claim that
is Lipschitz continuous with constant
. In fact, (H3) implies that for every
and
,
that is,
Therefore,
is Lipschitz continuous with constant
.
Clearly,
is precompact in
, and so is
in
.
Thus, by (3.29) and Lemma 2.4, we obtain
By (3.6),
, which implies
. Consequently,
is precompact in
.
Step 2.
is precompact in
.
For
, let
and define
by
By (H3),
is well defined and for
, we have
So, for
,
, we have
where
According to the proof of Step 1, we know that
are all precompact in
and
is Lipschitz continuous with constant
.
Next, we will show that
is precompact in
. Firstly, it is easy to see that
is precompact in
. Thus according to Lemma 2.3, it remains to prove that
is precompact in
. And, we recall that
,
, which means that
By Step 1,
is precompact in
. Without loss of generality, we may suppose that
Therefore,
, as
in
. Thus, by the continuity of
and
, we get
as
, which implies that
is relatively compact in
. And, for
, by the compactness of
,
,
is also relatively compact in
. Therefore,
is relatively compact in
for every
.
Next, for
, we have
Thus, the set
is equicontinuous on
due to the compactness of
and the strong continuity of operator
,
. By the Arzela-Ascoli theorem, we conclude that
is precompact in
. Therefore,
is precompact in
.
Thus, by Lemma 2.4, we obtain
By (3.6),
, which implies
. Consequently,
is precompact in
.
Step 3.
The same idea can be used to prove the compactness of
in
for
, where
. This completes the proof.
Proof of Theorem 3.1.
For
,
, let
where
comes from the condition (H2). Then, by condition (H2),
.
By Lemma 3.3, without loss of generality, we may suppose that
, as
. Thus, by the continuity of
and
, we get
as
. Thus,
is precompact in
. Moreover,
and
are both precompact in
. And
is Lipschitz continuous with constant
. Note that
Therefore, by Lemma 2.4, we know that the set
is precompact in
. Without loss of generality, we may suppose that
in
. On the other hand, we also have
Letting
in both sides, we obtain
which implies that
is a mild solution of the nonlocal impulsive problem (1.1). This completes the proof.
Remark 3.4.
From Lemma 3.3 and the above proof, it is easy to see that we can also prove Theorem 3.1 by showing that
is precompact in
.
The following results are immediate consequences of Theorem 3.5.
Theorem 3.5.
Assume (H1), (H3)–(H5) hold. If
, then the impulsive Cauchy problem (1.1) has at least one mild solution on
, provided
Theorem 3.6.
Assume (H1), (H2), (H4), and (H5) hold. If
, then the nonlocal impulsive problem (1.1) has at least one mild solution on
, provided
.
Theorem 3.7.
Assume (H1), (H4), and (H5) hold. If
, then the impulsive problem (1.1) has at least one mild solution on
, provided
.
Remark 3.8.
Theorems 3.5-3.6 are new even for many special cases discussed before, since neither the Lipschitz continuity nor compactness assumption on the impulsive functions is required.