Solvability of a nonlinear second order m -point boundary value problem with p -Laplacian at resonance

We study the existence of solutions of the nonlinear second order m -point boundary value problem with p -Laplacian at resonance

It is well known that the research of m-point boundary value problems is significant in the theory of ordinary differential equations and practical applications, see [4,13].
We specifically mention that the existence of solutions of problem (1.1) with p = 2 has been studied by Gupta [12], Feng and Webb [3], Infante and Webb [14], Ma [20], Infante and Zima [15], Liang and Lin [18], in which the Leray-Schauder continuation theorem, the Mawhin's continuation theorem, the theory of fixed point index, the nonlinear alternative of Leray-Schauder, the Leggett-Williams norm-type theorem due to O'Regan and Zima [21], and the fixed point theorem in cones are used, respectively.In the case that p > 1, problem (1.1) has been studied by García-Huidobro et al. [5][6][7] via some continuation lemma of Leray-Schauder type due to themselves, Zhu and Wang [24] via Mawhin's continuation theorem.For other types of works regarding the second order nonlocal boundary value problems involving p-Laplacian, we refer the reader to [1, 2, 8-10, 16, 23].
Inspired by the above works and [17,22], the aim of this paper is to establish the new existence results of solutions to problem (1.1) by using the topological transversality method together with the barrier strip technique and the cut-off technique.
We would like to emphasize that the results of this paper are new even when p = 2, and the main research tool used is the topological transversality theorem, which is different from those in [5-7, 20, 24].The advantage of the topological transversality theorem is that it can transform the resonance problem into a nonresonance problem by appropriately selecting a convex subset U of X. Besides, in [5][6][7], the order of growth of f with respect to the derivative term is less than or equal to p.However, in our results, we just impose some local sign conditions on the nonlinear term f and do not require the growth constraints, so the degree of the derivative term in f can exceed p.
This work is organized as follows: In Sect.2, we first briefly introduce the topological transversality theory.And then, a modified boundary value problem is constructed by the cut-off technique.Finally, the barrier strip technique is used to estimate a prior bound of solutions of the modified boundary value problems in C 1 [0, 1].In Sect.3, the topological transversality method is used to establish some existence theorems of solutions of problem (1.1).As applications of our main results, some examples are given in the last section.
Throughout this paper, the following local conditions on f will be used.

Preliminaries
Firstly, we briefly review some concepts and results of topological transversality theory.
Let U be a convex subset of a Banach space X and D ⊂ U be an open set.Denote by H ∂D (D, U) the set of compact operators F : D → U that are fixed point free on ∂D.We say that F ∈ H ∂D (D, U) is essential if every operator in H ∂D (D, U) that agrees with F on ∂D has a fixed point in D.
The next two lemmas can be found in [11].
Lemma 2.1 If q ∈ D and F ∈ H ∂D (D, U) is a constant operator, F(x) = q for x ∈ D, then F is essential.
) is essential and therefore it has a fixed point in D.
According to the intermediate value property of continuous functions, it is not difficult to obtain the following.

Lemma 2.3 Assume that
Let the constants r 1 , r 2 , R 1 , R 2 be such that We define the modification g of f as follows: and define the modification h of g by setting Obviously, both g(t, x, y) and h(t, x, y) are continuous on [0, 1] × R 2 .
We note here that in the following discussion, we agree that when (H 1 ) is true, the constants r 1 and r 2 in function g are the ones in (H 1 ), while for i = 2, 3, 4, 5, when (H i ) is true, the constants R 1 and R 2 in function h are the ones in (H i ).
Consider the family of the following modified boundary value problem: where λ ∈ (0, 1].
The following lemma is a prior estimate of the possible solutions of the modified problem (2.1), (2.2) in C 1 [0, 1].Lemma 2.4 Assume that (H 1 ) and (H 2 ) hold.Let x be a solution of the modified problem (2.1), (2.2) for some λ ∈ (0, 1].Then (2.3) Proof We divided the proof into two steps.
Similar to the proof of Lemma 2.5, we can easily show the following two results.
Obviously, U is a convex subset of X and D is an open subset of U.

Lemma 2.8 Let the operator H
(2.10) Proof Notice that to prove H(x, r, λ) is a compact operator, it is sufficient to show the second component of H(x, r, λ), that is, We now divided the proof into two steps.
Step 1.We prove that H 2 (x, r, λ) is continuous.To do this, we let From the continuity of h and φ -1 p , we have Hence Step 2. We show that H 2 (D × [0, 1]) is a relatively compact set in R. Notice that for all (x, r, λ) ∈ D × [0, 1], we have where Thus, H 2 (D × [0, 1]) is a relatively compact set in R.This completes the proof of the lemma.
Lemma 2.9 Assume that (H 1 ) holds.Let the operator F : D → U be defined by Then F is essential.

Lemma 2.10 Let the operator G
, where the symbol " * " denotes the transpose of vector.Then G is a compact operator.
Proof Firstly, we define operators Next, we divided the proof into three steps.
Step 1.We prove that The continuity of h and φ -1 p imply that uniformly on [0, 1] as n → ∞, and and This shows that G 1 : Step 2. We show that the set where M is defined by (2.11).Since φ -1 p (•) is uniformly continuous on [-M, M], for any ε > 0, there exists μ > 0 such that when |s 2 - Now, we choose δ = μ/M.Then, for all t 1 , t 2 ∈ [0, 1] with |t 2 -t 1 | < δ and for all (x, r) ∈ D, we have Step 3. We prove that The compactness of the operator G 2 is clear, the proof is similar to the one of H 2 in Lemma 2.8.
In summary, G : D × [0, 1] → U is a compact operator.This completes the proof of the lemma.

Main results
With the preparatory work in Sect.2, we can now establish the existence results of solutions of problem (1.1).Theorem 3.1 Assume that (H 1 ) and (H 2 ) hold.Then boundary value problem (1.1) has at least one solution x = x(t) satisfying (2.3) and (2.4).
Proof We note that to obtain the existence of solutions satisfying (2.3) and (2.4) of problem (1.1), it is sufficient to prove that the modified problem (2.1), (2.2) has a solution x = x(t) satisfying (2.3) and (2.4).
Below, we will prove in two steps.
Suppose that (x 1 , r 1 ) is a fixed point of G(•, •, 1).It follows from the definition of the operator G that Furthermore, we have and thus , it is easy to see that x 2 is a solution of the modified problem (2.1), (2.2) with λ = 1, and the validity of (2.3) and (2.4) follows from Lemma 2.4.
The following conclusion can be obtained by Theorem 3.1 immediately.
The following theorems can be proved by using similar arguments to those of Theorem 3.1.

Some examples
As applications of our results, this section will provide four illustrative examples.
where φ p (s) = |s| p-2 s, p > 1; n is an even number, c i ∈ R (i = 0, 1, . . ., n); the polynomial n i=0 c i x i has at least two distinct real roots with different signs; P l (y) is a polynomial of degree l ∈ N, P l (0) = 0; and a i > 0 Obviously, f ∈ C([0, 1] × R 2 ).Let r 1 and r 2 be the minimum and maximum real root of the polynomial n i=0 c i x i , respectively.Then from the assumption it follows that r 1 ≤ 0 ≤ r 2 , r 2 1 + r 2 2 > 0 and Hence, condition (H 1 ) of Theorem 3.1 is satisfied.Let Then, for x ∈ [r 1 , r 2 ], y ∈ R, we have Thus, if there exists y 1 < 0 such that P l (y 1 ) -F > 0 and there exists y 2 > 0 such that P l (y 2 ) + F < 0, then condition (H 2 ) of Theorem 3.1 is satisfied.In summary, problem (4.1) has at least one solution x = x(t) provided inf y>0 P l (y) + < 0 < sup y<0 P l (y) -F.
Note that if l is an odd number and the coefficient of y l in P l (y) is negative, then the above inequality holds.