Multiple solutions for a perturbed mixed boundary value problem involving the one-dimensional p-Laplacian

The existence of three distinct weak solutions for a perturbed mixed boundary value problem involving the one-dimensional p-Laplacian operator is established under suitable assumptions on the nonlinear term. Our approach is based on recent variational methods for smooth functionals defined on reflexive Banach spaces.


Introduction
Consider the following perturbed mixed boundary value problem Using two kinds of three critical points theorems obtained in [4,8] which we recall in the next section (Theorems 2.1 and 2.2), we ensure the existence of at least three weak solutions for the problem (1); see Theorems 3.1 and 3.2.These theorems have been successfully employed to establish the existence of at least three solutions for perturbed boundary value problems in the papers [5,6,14,16,17].
Existence and multiplicity of solutions for mixed boundary value problems have been studied by several authors and, for an overview on this subject, we refer the reader to the papers [2,3,12,15,18].We also refer the reader to the papers [7,9,10,11] in which the existence of multiple solutions is ensured.
A special case of Theorem 3.1 is the following theorem.Assume that F (η) > 0 for some η > 0 and F (ξ) ≥ 0 in [0, η] and Then, there is λ * > 0 such that for each λ > λ * and for every L 1 -Carathéodory function g : Moreover, the following result is a consequence of Theorem 3.2.
Theorem 1.2.Let f : R → R be a nonnegative continuous function such that admits at least three weak solutions.
The present paper is arranged as follows.In Section 2 we recall some basic definitions and preliminary results, while Section 3 is devoted to the existence of multiple weak solutions for the eigenvalue problem (1).

Preliminaries
Our main tools are the following three critical points theorems.In the first one the coercivity of the functional Φ − λΨ is required, in the second one a suitable sign hypothesis is assumed.
Then, for each λ ∈ Λ r the functional Φ − λΨ has at least three distinct critical points in X.
Theorem 2.2 ([4], Theorem 3.3).Let X be a reflexive real Banach space, Φ : X −→ R be a convex, coercive and continuously Gâteaux differentiable functional whose derivative admits a continuous inverse on X * , Ψ : X −→ R be a continuously Gâteaux differentiable functional whose derivative is compact, such that 1. inf X Φ = Φ(0) = Ψ(0) = 0; 2. for each λ > 0 and for every u 1 , u 2 ∈ X which are local minima for the functional Φ − λΨ and such that Ψ(u 1 ) ≥ 0 and Ψ(u 2 ) ≥ 0, one has Assume that there are two positive constants r 1 , r 2 and v ∈ X, with 2r 1 < Φ(v) < r 2 2 , such that Then, for each λ ∈ 3 2 , the functional Φ − λΨ has at least three distinct critical points which lie in In order to study the problem (1), the variational setting is the space We observe that the norm • is equivalent to the usual one.
It is well known that (X, for every u ∈ X.
We need the following proposition in the proof of Theorem 3.1.
Proposition 2.3.Let T : X → X * be the operator defined by for every u, v ∈ X.Then T admits a continuous inverse on X * .
Proof.In the proof, we use C 1 , C 2 , . . ., C 9 to denote suitable positive constants.For any u ∈ X \ {0}, Thus, the map T is coercive.Now, taking into account (2.2) in [19], we see that ( At this point, if p ≥ 2, then it follows that so T is uniformly monotone.By [20,Theorem 26.A (d)], T −1 exists and is continuous on X * .On the other hand, if 1 < p < 2, by Hölder's inequality, we obtain Similarly, one has Then, relation (3) together with ( 4) and ( 5), yields Thus, T is strictly monotone.By [20,Theorem 26.A (d)], T −1 exists and is bounded.Moreover, given g 1 , g 2 ∈ X * , by the inequality So T −1 is continuous.This completes the proof.
We use the following notations: Corresponding to f and g we introduce the functions We mean by a (weak) solution of problem (1), any function EJQTDE, 2013 No. 24, p. 5 Following the construction given in [6], in order to introduce our first result, fixing two positive constants θ and η such that and taking and where we read ρ/0 = +∞, so that, for instance, δ λ,g = +∞ when lim sup and G η = G θ = 0. Now, we formulate our main result.
Theorem 3.1.Assume that there exist two positive constants θ and η with θ < η such that Then, for each λ ∈ Λ and for every L there exists δ λ,g > 0 given by (8) such that, for each µ ∈ [0, δ λ,g [, the problem (1) admits at least three distinct weak solutions in X.
Proof.In order to apply Theorem 2.1 to our problem, we introduce the functionals Φ, Ψ : X → R for each u ∈ X, as follows Let us prove that the functionals Φ and Ψ satisfy the required conditions.It is well known that Ψ is a differentiable functional whose differential at the point u ∈ X is for every v ∈ X as well as is sequentially weakly upper semicontinuous.Furthermore, Ψ ′ : X → X * is a compact operator.Indeed, it is enough to show that Ψ ′ is strongly continuous on X.For this end, for fixed u ∈ X, let u n → u weakly in X as n → ∞, then u n converges uniformly to u on [a, b] as n → ∞; see [20].Since f, g are L 1 -Carathéodory functions, f, g are continuous in R for every Thus we proved that Ψ ′ is strongly continuous on X, which implies that Ψ ′ is a compact operator by Proposition 26.2 of [20].Moreover, Φ is continuously differentiable whose differential at the point u ∈ X is for every v ∈ X, while Proposition 2.3 gives that Φ ′ admits a continuous inverse on X * .Furthermore, Φ is sequentially weakly lower semicontinuous.Clearly, the weak solutions of the problem (1) are exactly the solutions of the equation Φ ′ (u) − λΨ ′ (u) = 0.
Put r := ρ 0 θ p p(b − a) p−1 , and It is easy to see that w ∈ X and, in particular, one has Taking into account 0 < θ < η, using (6), we observe that Bearing in mind relation (2), we see that and it follows that On the other hand, by using condition (A 1 ), since 0 ≤ w(x) ≤ η for each x ∈ [a, b], we infer Therefore, we have and Since µ < δ λ,g , one has Hence from ( 10)-( 12), we observe that the condition (a 1 ) of Theorem 2.1 is satisfied.Finally, since µ < δ λ,g , we can fix l > 0 such that lim sup and µl < ρ 0 p(b − a) p .Therefore, there exists a function h for every x ∈ [a, b] and t ∈ R.
for every x ∈ [a, b] and t ∈ R. Taking (2) into account, it follows that, for each u ∈ X, and thus lim which means the functional Φ − λΨ is coercive, and the condition (a 2 ) of Theorem 2.1 is verified.By using relations (10) and ( 12) one also has .
Finally, Theorem 2.1 (with x = w) ensures the conclusion.Now, we present a variant of Theorem 3.1 in which no asymptotic condition on the nonlinear term is requested.In such a case f and g are supposed to be nonnegative.EJQTDE, 2013 No. 24, p. 9 For our goal, let us fix positive constants θ 1 , θ 2 and η such that 3 2 , and taking With the above notations we have the following multiplicity result.
Assume that there exist three positive constants θ 1 , θ 2 and η with 2 1/p θ 1 < η < θ 2 2 1/p such that assumption (A 1 ) in Theorem 3.1 holds.Furthermore, suppose that Then, for each λ ∈ Λ and for every nonnegative L 1 -Carathéodory function g : [a, b] × R → R, there exists δ * λ,g > 0 given by min such that, for each µ ∈ [0, δ * λ,g [, the problem (1) admits at least three distinct weak solutions u i for i = 1, 2, 3, such that Proof.Fix λ, g and µ as in the conclusion and take Φ and Ψ as in the proof of Theorem 3.1.We observe that the regularity assumptions of Theorem 2.2 on Φ and Ψ are satisfied.Then, our aim is to verify (b 1 ) and (b 2 ).
Since µ < δ * λ,g and G η = 0, one has sup Therefore, (b 1 ) and (b 2 ) of Theorem 2.2 are verified.Finally, we verify that Φ − λΨ satisfies the assumption 2. of Theorem 2.2.Let u 1 and u 2 be two local minima for Φ − λΨ.Then u 1 and u 2 are critical points for Φ − λΨ, and so, they are weak solutions for the problem (1).We want to prove that they are nonnegative.
Let u 0 be a weak solution of problem (1).Arguing by a contradiction, assume that the set for every v ∈ X.Thus, from our sign assumptions on the data, we have Finally, a simple computation shows that all assumptions of Theorem 3.2 are fulfilled.The desired conclusion follows.