We make the following assumptions.
(H1) f : ℝ → ℝ is continuous and sf(s) > 0 for s ≠ 0.
(H2) .
Theorem 4.1. Suppose that (H1) and (H2) hold Assume that
(4.1)
or
(4.2)
Then (1.1) and (1.2) has two solutions u+ and u- such that u+ is positive on and u- is negative on .
Obviously, we can get the following lemma with ease.
Lemma 4.1. Suppose that u ∈ X and on satisfies (1.1) (or (1.3)) and there exists such that u(t0) = 0, then u(t0 - 1)u(t0 + 1) < 0.
Proof of Theorem 4.1. First, we deal with the case r > 0.
Let ζ, ξ ∈ C(ℝ, ℝ) such that
Clearly
(4.3)
Let
Then is nondecreasing and
(4.4)
Let us consider
(4.5)
as a bifurcation problem from the trivial solution u ≡ 0.
Equation (4.5) can be converted to the equivalent equation
(4.6)
It is easy to see that T : X → X is compact. Further we note that H(λ, u) = λL-1[m(·)ζ( u (·))] = o⊠u⊠near λ = 0 uniformly on bounded λ intervals, since
where and
Let under the product topology. Let . Set S- = -S+, S = S+ ∪ S-. Then S+ and S- are disjoint in X. Finally let Ψ± = ℝ × S± and Ψ = ℝ × S. Let Σ be the closure of the set of nontrivial solutions of (1.1) and (1.2).
It is easy to see that is simple. Now applying Theorems 2.1 and 2.2, we get the result as follows: Σ contains a maximum subcontinuum which is composed of two distinct connected set and such that and . Moreover, Lemma 4.1 guarantees the second case in Theorems 2.1 and 2.2 cannot happen. Otherwise, there will exist , such that y has a multiple zero point t0, (i.e., t0 satisfies y(t0) = 0 and y(t0 - 1)y(t0 + 1) > 0). However, this contradicts Lemma 4.1. Thus, for each joins to infinity in Ψvand .
It is obvious that any solution to (4.5) of the form (1, u) yields a solution u to (1.1) and (1.2). We will show that crosses the hyperplane {1} × X in ℝ × X. To achieve this goal, it will be enough to show that
(4.7)
or
(4.8)
where denotes the projection of on ℝ.
Let satisfy
We note that μ
n
> 0 for all n ∈ ℕ since (0,0) is the only solution of (4.5) for λ = 0 and .
Case 1. .
We divide the proof into two steps.
Step 1. We show that if there exists a constant number M > 0 such that
then (4.7) holds.
In this case it follows that
(4.10)
We divide the equation
(4.11)
by ⊠y
n
⊠
x
and set . Since is bounded in X and μ
n
is bounded in ℝ, after taking the subsequence if necessary, we have that for some with and for some μ ∈ ℝ. Moreover, from (4.4) and the fact that is nondecreasing, we have that
(4.12)
since . Thus,
which implies that
(4.13)
We claim that
We only prove that if , then . The other case that if , then can be treated similarly.
Obviously when , then on . Furthermore, on . In fact, if there exists a such that , then, by Lemma 4.1, we obtain which contradicts the fact that on . Thus, on . This together with the fact is a closed set in implies that . Moreover, , so that
Thus, (4.7) holds.
Step 2. We show that there exists a constant M > 0 such that μ
n
∈ (0, M] for all n.
Since {(μ
n
, y
n
)} are the solutions to (4.5), they follow that
(4.14)
where . From (H1) and (H2), there exist two positive constants ρ1 and ρ2, such that
(4.15)
Let η* > 0 be the positive principal eigenvalue of the following linear eigenvalue problem
(4.16)
and η* > 0 the positive principal eigenvalue of the following linear eigenvalue problem
(4.17)
where
By Theorem 3.2, (4.14), (4.15), (4.16), and (4.17), we get
Case 2. .
From Step 2 of Case 1, there exists M > 0 such that for all n ∈ ℕ,
Applying a similar argument to that used in Step 1 of Case 1 (after taking a subsequence and relabeling, if necessary), we get
which implies that (4.8) holds.
At last, we deal with the case r < 0.
Let us consider
(4.18)
as a bifurcation problem from the trivial solution u ≡ 0. Now, applying Theorems 2.1 and 2.2, we get the following results: Σ contains a maximum subcontinuum which is composed of two distinct connected set and such that and . Moreover, by Lemma 4.1, for each joins to infinity in Ψvand , where Σ and Ψvare defined as in the case r > 0.
It is clear that any solution to (4.18) of the form (-1, u) yields a solutions u of (1.1) and (1.2). We will show crosses the hyperplane {-1} × X in ℝ × X. To achieve this goal, it will be enough to show that
(4.19)
or
(4.20)
Let satisfy
We note that μ
n
< 0 for all n ∈ ℕ since (0, 0) is the only solution to (4.18) for λ = 0 and .
The rest of the proof is similar to the proof of the case r > 0, so we omit it.