Let the Banach space be endowed with the norm . Let τ be a real constant with and define the cone by
where , obviously, .
Define the operator by
In order to get the integrated and rigorous theory, we make the following assumptions.
(H1) There are constants such that the functional φ satisfies the inequality
(3.1)
(H2) For each given , there are and such that
(H3) There exists a function satisfying the growth condition
for some . For each given and , there is such that
The following result plays an important role in the coming discussion.
Lemma 3.1 is completely continuous.
Proof It is easy to see that the operator F is continuous in view of the continuity of G and f.
By Lemmas 2.1 and 2.6, we have
where . Thus, .
Now let be bounded, i.e., there exists a positive constant such that for all . By the continuity of , , and φ, we easily see that and are bounded, so there exist constants and such that and . Let . Then for , from Lemma 2.6, we have
(3.3)
we rearrange (3.3) as follows:
Hence, is bounded.
On the other hand, for any given , there exists small enough, such that holds for each and with , that is to say, is equicontinuous. In fact,
(3.4)
Now, we estimate (we discuss in the same way, the proof here is omitted):
-
(1)
for , , ;
-
(2)
for , ;
-
(3)
for , from the mean value theorem of differentiation, we have .
Thus, we have
By means of the Arzela-Ascoli theorem, is completely continuous. The proof is complete. □
Theorem 3.1 Assume that the nonlinearity splits in the sense that , for continuous functions and such that and . Suppose conditions (H1)-(H3) and
(3.5)
hold, where is a closed subinterval. Then the boundary value problem (1.1)-(1.2) has at least one solution.
Proof Begin by selecting the number such that
(3.6)
Now, there exists a number such that for . Then take the open set
By Remark 2.2, Lemma 2.6, and (3.6), for , we find
(3.7)
Hence, , that is, F is a cone expansion on .
On the other hand, we consider two cases:
Case 1. Suppose that g is bounded for . We may find sufficiently large such that
(3.8)
Condition (3.5) implies the existence of such that
(3.9)
Now if , then by (H1) we get . According to condition (H2), it follows that
(3.10)
for all with .
Next, since is a closed subinterval, we may select such that . Then for each ,
(3.11)
Thus combining condition (H3) we see that
(3.12)
Take . Set .
From (3.8) we may assume without loss of generality that
(3.13)
Then by (3.9), (3.10), (3.12), and (3.13), for each , we have
(3.14)
Case 2. Suppose that g is unbounded at +∞. By condition , there exists a number such that for we find that , where meets
(3.15)
Noting that g is unbounded at +∞, we may find such that
(3.16)
Now, put . Then for each we find that by (3.9), (3.10), (3.12), (3.15), and (3.16),
(3.17)
Hence, .
To summarize, we conclude from (3.14) and (3.17) that F is a cone compression on .
With the help of Lemma 2.4 we can now deduce the existence of function such that . Hence, the problem (1.1)-(1.2) has at least one solution. The proof is completed. □
Next, by choosing suitable forms of and φ, we present the corresponding boundary value problems with separated boundary conditions, integral boundary conditions and multi-point boundary conditions as corollaries of Theorem 3.1 to illustrate the universality and generalization of our results.
Corollary 3.1 Assume is defined as in Theorem 3.1 and (H1)-(H3) hold. If , . Then the boundary value problem
(3.18)
(3.19)
has at least one solution.
This existence result of positive solution for the boundary value problem (3.18)-(3.19) has been studied by Ferreira in [20] and Yang et al. in [27].
Corollary 3.2 Assume is defined as in Theorem 3.1, . Then φ is linear functional, is a closed subinterval, just need , where m is Lebesgue measure. If (H 1)-(H 3) hold, in addition
then the boundary value problem (3.20)-(3.21) with integral boundary conditions
(3.20)
(3.21)
has at least one solution.
Corollary 3.3 Assume is defined as in Theorem 3.1, . Then φ is linear functional, just need , is a closed subinterval. If (H1)-(H3) hold, moreover,
then the boundary value problem (3.22)-(3.23) with multi-point boundary conditions
(3.22)
(3.23)
has at least one solution.