Nodal solutions of second-order two-point boundary value problems

AbstractWe shall study the existence and multiplicity of nodal solutions of the nonlinear second-order two-point boundary value problems,
 u″+f(t,u)=0,t∈(0,1),u(0)=u(1)=0.
 The proof of our main results is based upon bifurcation techniques.Mathematics Subject Classifications: 34B07; 34C10; 34C23.

From above literature, we can see that the existence and multiplicity results are largely based on the assumption that t and u are separated in nonlinearity term. It is interesting to know what will happen if t and u are not separated in nonlinearity term? We shall give a confirm answer for this question.
In this article, we consider the existence and multiplicity of nodal solutions for the nonlinear BVP under the following assumptions: 1], and the inequality is strict on some subset of positive measure in (0,1), where l k denotes the kth eigenvalue of 1], and all the inequalities are strict on some subset of positive measure in (0, 1), where l k denotes the kth eigenvalue of (1.5); (H 3 ) f(t, s)s > 0 for t (0, 1) and s ≠ 0.  [5,6] and López-Gómez [7], the proofs of these theorems contain gaps, the original statement of Theorem 1.40 of [4] is not correct, the original statement of Theorem 1.27 of [4] is stronger than what one can actually prove so far. Although there exist some gaps in the proofs of Rabinowitz's Theorems 1.27, 1.40, and 1.27 has been used several times in the literature to analyze the global behavior of the component of nodal solutions emanating from u = 0 in wide classes of boundary value problems for equations and systems [1,2,8,9]. Fortunately, López-Gómez gave a corrected version of unilateral bifurcation theorem in [7].
By possesses two solutions u + k and u − k , such that u + k has exactly k -1 zeros in (0, 1) and is positive near 0, and u − k has exactly k -1 zeros in (0,1) and is negative near 0. Similarly, we also have the following: 1], and all the inequalities are strict on some subset of positive measure in (0, 1), where l k denotes the kth eigenvalue of (1.5); 1], and the inequality is strict on some subset of positive measure in (0, 1), where l k denotes the kth eigenvalue of (1.5), then problem (1.4) possesses two solutions u + k and u − k , such that u + k has exactly k -1 zeros in (0,1) and is positive near 0, and u − k has exactly k -1 zeros in (0,1) and is negative near 0. Remark 1.2. We would like to point out that the assumptions (H 1 ) and (H 2 ) are weaker than the corresponding conditions of Theorem A. In fact, if we let By the strict decreasing of μ k (f) with respect to weight function f (see [10]), where μ k (f) denotes the kth eigenvalue of (1.2) corresponding to weight function f, we can show that our con- and the inequalities are strict on some subset of positive measure in (0,1), where l k denotes the kth eigenvalue of (1.5); 1], and the inequality is strict on some subset of positive measure in (0, 1), where l k denotes the kth eigenvalue of (1.5).
Using Sturm Comparison Theorem, we also can get a non-existence result when f satisfies a non-resonance condition.
Theorem 1.4. Let (H 3 ) hold. Assume that there exists an integer k N such that for any t [0, 1], where l k denotes the kth eigenvalue of (1.5).
By the strict decreasing of μ k (f) with respect to weight function f (see [11]), where μ k (f) denotes the kth eigenvalue of (1.2) corresponding to weight function f, we can show that our condition c(t) ≤ l k < ... <l k+j ≤ a(t) is equivalent to the condition f 0 <μ k < · · · <μ k+j <f ∞ . Similarly, our condition a(t) ≤ l k < · · · <l k+j ≤ c(t) is equivalent to the condition f ∞ <μ k < ... <μ k+j <f 0 . Therefore, Theorem B is the corollary of Theorem 1.3. Similar, we get Theorem C is also the corollary of Theorem 1.4.

Preliminary results
To where r ℝ is a parameter, u X, X is a Banach space, θ is the zero element of X, and G: X = R × X → X is completely continuous. ind(0, l 0 T) changes sign as l crosses l 0 , then each of the components C ν λ 0 , ν ∈ {+, −} satisfies (λ 0 , θ ) ∈ C ν λ 0 , and either (i) meets infinity in X, where V is the complement of span{ϕ λ 0 }, ϕ λ 0 denotes the eigenfunction corresponding to eigenvalue l 0 . Lemma 2.2 [[7], Theorem 6.5.1]. Under the assumptions: (A) X is an order Banach space, whose positive cone, denoted by P, is normal and has a nonempty interior; (B) The family ϒ(r) has the special form where T is a compact strongly positive operator, i.e., T(P\{0}) ⊂ int P; (C) The solutions of u = rTu + H(r, u) satisfy the strong maximum principle. Then the following assertions are true: (1) Spr (T) is a simple eigenvalue of T, having a positive eigenfunction denoted by ψ 0 > 0, i.e., ψ 0 int P, and there is no other eigenvalue of T with a positive eigenfunction; (2) For every y int P, the equation thenς is nondecreasing with respect to u and lim u→0 +ς (t, u) |u| = 0.
If u E, it follows from (2.3) that as a bifurcation problem from the trivial solution u ≡ 0. Equation (2.4) can be converted to the equivalent equation Further we note that ||L -1 [ζ(t, u(t))] || E = o(||u|| E ) for u near 0 in E.
where l k denotes the kth eigenvalue of (1.5). Proof. It is easy to see that the problem (2.4) is of the form considered in [7], and satisfies the general hypotheses imposed in that article.
Combining Lemma 2.1 with Lemma 2.3, we know that there exists a continuum C ν k ⊂ E of solutions of (2.4) such that: where j N, l j is another eigenvalue of (1.5) and different from l k ; (c) or C ν k contains a point where V is the complement of span{ k }, k denotes the eigenfunction corresponding to eigenvalue l k .
We finally prove that the first choice of the (a) is the only possibility.
In fact, all functions belong to the continuum sets C ν k have exactly k -1 simple zeros, this implies that it is impossible to exist (λ j , θ ) ∈ C ν k , j ∈ N. Next, we shall prove (c) is impossible, suppose (c) occurs, then C ν k is bounded and without loss of generality, suppose there exists a point (ι, y) ∈ R × (V\{θ }) ∩ C + k . Moreover, it follows from Lemma 2.1 that Note that as the complement V of span{ k } in E, we can take Thus, for this choice of V, the component C + k cannot contain a point then y > 0 in (0, a 0 ), where a 0 denotes the first zero point of y, and there exists u E for which u − λ k Lu = y > 0, in (0, a 0 ).
Lemma 2.5. If (μ, u) ∈ E is a non-trivial solution of (2.4), then u ∈ S ν k for ν and some k N.
Proof. Taking into account Lemma 2.4, we only need to prove that , then c j should be a solution of problem, u j E .

Proof of main results
Proof of Theorems 1.1 and 1.2. We only prove Theorem 1.1 since the proof of Theorem 1.2 is similar. It is clear that any solution of (2.4) of the form (1, u) yields a solution u of (1.4). We shall show C ν k crosses the hyperplane {1} × E in ℝ × E. By the strict decreasing of μ k (c(t)) with respect to c(t) (see [11]), where μ k (c(t)) is the kth eigenvalue of (1.2) corresponding to the weight function c(t), we have μ k (c(t)) >μ k (l k ) = 1.
We note that μ j > 0 for all j N, since (0,0) is the only solution of (2.4) for μ = 0 Step 1: We show that if there exists a constant M > 0, such that for j N large enough, then C ν k crosses the hyperplane {1} × E in ℝ × E. In this case it follows that We divide the equation , after taking a subsequence if necessary, we have thatūj →ū for someū ∈ E with ||u|| E = 1. By (3.1), using the similar proof of (2.3), we have that lim j→+∞ ξ (t, u j (t)) By the compactness of L we obtain It is clear thatū ∈ C ν k ⊆ C ν k since C ν k is closed in ℝ × E. Therefore,μ(a(t)) is the kth eigenvalue of u (t) + μa(t)u(t) = 0, t ∈ (0, 1), u(0) = u(1) = 0.
Step 2: We show that there exists a constant M such that μ j (0, M] for j N large enough.
On the other hand, we note that −u j = μ j f (t, u j ) u j u j .
In view of Remark 1.1, we have μ j f (t, u j ) u j > λ k on [a, b] and for j large enough and all t [0, 1]. By Lemma 3.2 of [12], we get u j must change its sign more than k times on [a, b] for j large enough, which contradicts the act that u j ∈ S μ k . Therefore,  The results follows.