In this section, we shall state some necessary definitions and preliminaries results.
be a real Banach space. A nonempty closed set
is called a cone if it satisfies the following two conditions:
A cone is said solid if it contains interior points, . A cone P is called to be generating if , i.e., every element can be represented in the form , where . A cone P in E induces a partial ordering in E given by if . If and , we write ; if cone P is solid and , we write .
Definition 2.2 A cone is said to be normal if there exists a positive constant N such that , , , .
be a metric space and S
be a bounded subset of E
. The measure of non-compactness
is defined by
Definition 2.4 An operator is said to be completely continuous if it is continuous and compact. B is called a k-set-contraction () if it is continuous, bounded and for any bounded set , where denotes the measure of noncompactness of S.
A k-set-contraction is called a strict-set contraction if . An operator B is said to be condensing if it is continuous, bounded, and for any bounded set with .
Obviously, if B is a strict-set contraction, then B is a condensing mapping, and if operator B is completely continuous, then B is a strict-set contraction.
It is well known that
is a solution of the problem (1.1) if and only if
is a solution of the following nonlinear integral equation:
where . In what follows, we write , (), . By making use of (2.1), we can prove that has the following properties.
Proposition 2.1, .
Let . It is easy to verify is a Banach space with norm . Obviously, is a cone in Banach space .
Let , . It is easy to see that is a Banach space with the norm . Evidently, and is a cone in Banach space . For any , by making use of the mean value theorem (), obviously we see that exists and .
Let . For any , let , , .
A map is called a nonnegative solution of problem (1.1) if , for and satisfies problem (1.1). An operator is called a positive solution of problem (1.1) if y is a nonnegative solution of problem (1.1) and .
For convenience and simplicity in the following discussion, we denote
where ν denote 0 or ∞.
To establish the existence of multiple positive solutions in E of problem (1.1), let us list the following assumptions, which will stand throughout the paper:
are Lebesgue integrable functionals on J
) and satisfying
and there exist positive constants
(H3) for any bounded set , , and together with are relatively compact sets,
We shall reduce problem (1.1) to an integral equation in E
. To this end, we first consider operator
Lemma 2.1is a solution of problem
(1.1) if and only ifis a solution of the following impulsive integral equation
i.e., y is a fixed point of operator A defined by (2.6) in.
First suppose that
is a solution of problem (1.1). It is easy to see by the integration of problem (1.1) that
Integrate again, we get
in (2.8) and (2.9), we find that
Substituting (2.12) and (2.13) into (2.9), we obtain
is a solution of the integral equation (2
, direct differentiation of the integral equation (2
and . So and (). It is easy to verify that and . The proof is complete. □
Thanks to (2.1), we know that
In the following, let . For , we denote , and ().
Lemma 2.2 ()
Letbe a bounded set
. Suppose thatis equi
-continuous on each
where ϒ anddenote the Kuratowski measures of noncompactness of bounded sets in E and, respectively.
Lemma 2.3 ()
Lemma 2.4 ()
is relatively compact if and only if each elementandare uniformly bounded and equicontinuous on each ().
Lemma 2.5 ()
, from Proposition 2.1 and (2.6), we obtain
On the other hand, for any
, by (2.6) and Proposition 2.2, we know that
Hence, . □
Lemma 2.7 Suppose that (H1) and (H3) hold. Thenis completely continuous.
Firstly, we show that
is continuous. Assume that
Thus, for any
, from the Lebesgue dominated convergence theorem together with (2.14) and (2.15), we know that
Hence, is continuous.
be any bounded set, then there exists a positive constant
. Thus, for any
, we know that
from 0 to 1 and exchanging integral sequence, then
Thus, by (H1
) and (2.17), we have
. Hence, for any
and for all
, from (2.16), we know that
From (2.17), (2.18), and the absolutely continuity of integral function, we see that is equicontinuous.
On the other hand, for any
, we know that
is uniformly bounded. By virtue of Lemma 2.3 and (H3
), we know that
So, . Therefore, A is compact. To sum up, the conclusion of Lemma 2.7 follows. □
The main tools of the paper are the following well-known fixed-point index theorems (see [2–4]).
Lemma 2.8 Letbe a completely continuous mapping andfor
, we have the following conclusions