Throughout the paper will denote absolute value, and let
Then E is a Banach space endowed with the norm .
We first establish some preliminary results to prove our main result. An easy but useful property of ϕ and is the following one.
Lemma 2.1 Let . Then is an odd, increasing homeomorphism with . Moreover, ϕ has the following properties:
ϕ is convex up on and is concave up on .
For each , there exists such that , with and for each , there exists such that , with .
For each , there exists such that , . For each with , there exists such that , .
Proof By a simple computation, it follows that and . So ϕ is an odd, increasing homeomorphism with . Moreover, from , we get that ϕ is convex up on . Notice that is also an odd, increasing homeomorphism with . It is easy to verify that is concave up on .
For each , there exists such that
and for each , there exists such that
By a similar argument, it is not difficult to compute that for each , there exists such that , . For each with , there exists such that , . □
Lemma 2.2 Let with . Assume that w is the solution of
Then on and , where .
Proof By integrating, it follows that (2.1) has the unique solution given by
where C is such that . Hence we must have . Further, since , by using , we obtain , and follows, here .
Since , there exists such that and it follows from that is decreasing on . Then for and for . Hence, on . □
Note that from Lemma 2.2, there exists , , such that with depending on . Define the cone P in E by
and for , let .
Lemma 2.3 ([, Lemma 4.1 and Lemma 4.2])
For each , (2.1) has a unique solution given by
where C is such that with . Moreover, the operator is continuous and sends equicontinuous sets in into a relatively compact set in E.
We next state the fixed point index theorem which will be used to prove our results.
Lemma 2.4 ([, Chapter 6])
Let E be a Banach space and P be a cone in E. Assume that Ω is a bounded open subset of E with , and let be a completely continuous operator such that , .
If , , then .
If , , then .
From Lemma 2.2, problem (1.3) is equivalent to the fixed point problem
in the space E, where is such that with , since otherwise,
which is a contradiction. This together with Lemma 2.3 implies that is a completely continuous mapping. Moreover, for any fixed , we have
and . In addition, from Lemma 2.2, it follows that on and there exist , , such that . So is a completely continuous operator.
Lemma 2.5 Let be given. If there exists small enough with such that , then
where is defined as in Lemma 2.1(iii).
Proof From the definition of , for any , we have
Lemma 2.6 Let be given. If and for , then
where and .
Proof From problem (1.3), since , it follows that there exists such that . Let . Then u satisfies the following boundary value problem:
Let v be the solution of the problem
Then we have
and by a comparison argument, we get that on . In fact, from , , there exists such that , i.e., . Thus, by a simple computation, we have that
This together with ϕ is an increasing homeomorphism implies that
Integrating from 0 to x for (2.6) and integrating from x to for (2.7), respectively, we have that for .
where is such that , and hence . If , then it follows that
which is a contradiction. Thus, . Moreover, we have
where . Consequently, for . Obviously,
where satisfies . It follows from that . Therefore,
If , then
If , let w be the solution of
where satisfies , i.e., . If , then
which is contradiction. Hence .
By a similar argument as before, it follows that on . Moreover,
where . So , . Therefore, we have
Let . Then
Lemma 2.7 Let be given. If , then
Proof Obviously, for any , it follows that for . So we have
Lemma 2.8 Let be given. If , then
where and .
Proof By using a similar argument of the proof of Lemma 2.6, we have that for . Meanwhile, (2.8) is true. If , then we can get that
If , then (2.10) holds and it follows that
Let , then