In this section, we construct Green’s function associated with BVP (1.1), (1.2).
Let , be roots of the polynomial , namely,
Then we have
where , , . Under the basic assumption on α, β, it is easy to see that .
Consider the two initial value problems:
(4.1)
(4.2)
By the direct computing, we get (4.1) has a unique solution
and (4.2) has a unique solution
Let
Lemma 4.1 Let and be fixed. Then the problem
(4.3)
has a unique solution
where
is given by
Remark 4.1 Green’s function defined by Lemma 4.1 is positive on .
Define operators , respectively, by
(4.4)
(4.5)
Now, from Lemma 4.1, it is easy to see that BVP (1.1), (1.2) has a solution if and only if u is a fixed point of the operator A. It follows from the continuity of f that is completely continuous.
Lemma 4.2 Let . Then the linear discrete fourth-order boundary value problem
(4.6)
has a unique solution
and
Proof The conclusion is obvious, so we omit it. □
We will use the following assumptions.
(H1) α, β are real parameters and satisfy
(H2) is continuous, for any , ; for any and , .
(H3) There exists an even number such that
(4.7)
where uniformly for , is defined in Lemma 3.1 and .
(H4) There exists an even number such that
(4.8)
where uniformly for , and are given in the condition (H3).
(H5) There exists a constant such that for any ,
(4.9)
where .
Lemma 4.3 Suppose that (H2) holds and is a solution of BVP (1.1), (1.2). Then .
Proof According to (H2) and the positivity of Green’s function defined in Lemma 4.1, we can easily get the desired conclusion. □
Remark 4.2 Similarly to Lemma 4.3, we also know that if (H2) holds and is a solution of BVP (1.1), (1.2), then .
Lemma 4.4 Suppose that (H2)-(H4) hold. Then the operator A is Fréchet differentiable at θ and ∞, where the operator A is defined by (4.5). Moreover, and .
Proof By (H3), for any , there exists such that for any , . Hence, noticing that for any , we have
(4.10)
for any with , where . Consequently,
(4.11)
This means that the nonlinear operator A is Fréchet differentiable at θ and .
By (H4), for any , there exists such that for any , . Let . By the continuity of with respect to x, we have . Then, for any , . Thus
(4.12)
for any . Consequently,
(4.13)
which implies that operator A is Fréchet differentiable at ∞ and . The proof is completed. □
Lemma 4.5 Let M be given in the condition (H5). Suppose that (H1)-(H4) hold. Then , . Moreover, one has the following.
(i) There exists an such that for any ,
(4.14)
(ii) There exists an such that for any ,
(4.15)
Proof By (H2) and the fact that is positive on , we get that for any , , , and , . Then and .
We only need to prove conclusion (i). The proof of conclusion (ii) is similar and will be omitted here. Let . The condition (H3) yields . It follows from (4.12) that there exists such that
(4.16)
where . Setting , then is completely continuous. For any and , we obtain that
(4.17)
According to the homotopy invariance of the fixed point index, for any , we have
(4.18)
(4.19)
Let . Then and (see Lemma 3.1 and the proof of Lemma 4.2). We claim
(4.20)
Indeed, we assume that there exist and such that . Obviously, . Since , , then . Set . It is clear that and . Then
(4.21)
Since , then , which contradicts the definition of . This proves (4.21).
It follows from Lemma 2.1 and (4.20) that
(4.22)
Similarly to (4.22), we know also that
(4.23)
By (4.18), (4.19), (4.22), and (4.23), we conclude
(4.24)
□