Let be a Banach space with the maximum norm for . Define and . Then P is a total cone in E, that is, . denotes the dual cone of P, namely, . Let denote the dual space of E, then by Riesz representation theorem, is given by
We assume that the following condition holds throughout this paper.
(H1) is the unique solution of the linear boundary value problem
Let solve the following inhomogeneous boundary value problems, respectively:
Let , , , .
(H2) , , .
Lemma 2.1 ()
If (H1) and (H2) hold, then BVP (1.1) is equivalent to
where is the Green function for (1.1).
Define an operator as follows:
It is easy to show that is a completely continuous nonlinear operator, and if is a fixed point of A, then u is a solution of BVP (1.1) by Lemma 2.1.
For any , define a linear operator as follows:
It is easy to show that is a completely continuous nonlinear operator and holds. By , the spectral radius of K is positive. The Krein-Rutman theorem  asserts that there are and corresponding to the first eigenvalue of K such that
Here is the dual operator of K given by:
The representation of , the continuity of G and the integrability of h imply that . Let . Then , and (2.4) can be rewritten equivalently as
Lemma 2.2 ()
If (H1) holds, then there is such that is a subcone of P and .
Lemma 2.3 ()
Let E be a real Banach space and be a bounded open set with . Suppose that is a completely continuous operator. (1) If there is with such that for all and , then . (2) If for all and , then . Here deg stands for the Leray-Schauder topological degree in E.
Lemma 2.4 Assume that (H1), (H2) and the following assumptions are satisfied:
(C1) There exist , and such that (2.3), (2.4) hold and K maps P into .
(C2) There exists a continuous operator such that
(C3) There exist a bounded continuous operator and such that for all .
(C4) There exist and such that for all .
Let , then there exists such that
Proof Choose a constant . From (C2), for , there exists such that implies
Now we shall show
provided that R is sufficiently large.
In fact, if (2.7) is not true, then there exist and satisfying
Since , , . Multiply (2.8) by on both sides and integrate on . Then, by (C4), (2.5), we get
By (2.9), holds. Then (2.3), (2.6) and (2.10) imply
where is a constant.
(C3) shows and (C1) implies . Then (C1), (2.8) and Lemma 2.2 tell us that
The definition of yields
It follows from (2.6), (2.11) and (2.12) that
where is a constant.
Since , then (2.13) deduces that (2.7) holds provided that R is sufficiently large such that . By (2.13) and Lemma 2.3, we have