1 Introduction and main result

In this paper, we consider the following fourth-order problem:

where Ω is a nonempty bounded open subset of the Euclidean space (RN,│.│ ), N ≥ 5, with sufficient smooth boundary, 2∗ = 2N/(N 4), 1 < q < 2, λ and µ are positive parameters.

Bernis, Garcia-Azorero and Peral [3] study a fourth-order problem with a critical growth, which presents several difficulties. Indeed, the Palais–Smale condition, as well as the weak lower semi-continuity of the associated functional, may fail because the Sobolev embedding is not compact. To be precise, consider the problem where µ > 0 is a parameter. Bernis, Garcia-Azorero and Peral [3] study this problem following the idea of Ambrosetti, Brezis and Cerami [2]. They proved the following result.

Theorem 1. (See [3].) Fix 1 < s < 2. Then there is Λ > 0 such that for each µ ϵ ]0, Λ[, problem (D) admits at least two positive solutions.

Moreover, they also proved that if µ > Λ, the previous problem admits no solution (see [3, Thm. 2.1]). Their proof is combination of topological and variational methods. Precisely, they determine the existence of a first solution by using the method of sub- and super-solutions and then prove that this solution is the minimum of a suitable functional and apply the mountain pass theorem so ensuring the existence of a second solution. For other result of fourth-order problem and variational problem, we refer the reader to [1, 5, 8, 10,11,12,13,14,15,16] and references therein.

In this paper, we investigate a fourth-order problem with critical growth (Pλ). Our approach is due to Bonanno [4, 6]. Using the variational method, we will ensure that problem (Pλ) has one positive solution when the parameters λ and µ are in a suitable interval. Furthermore, when λ = 1, we can get another positive solution, where µ is in a suitable interval, and give the estimate of the parameter µ.

At first, we give the variational framework of this problem. As usual, put X :=

H01(Ω) ∩ H2(Ω) endowed with the norm

for all u ϵ X. Obviously, │ξ│ 2∗ /2∗ + µ │ξ│ q/q ≥ 0 for all ξ ϵ R.

By the Sobolev embedding,

and by Talenti [17] we obtain

Due to (2), by the Hölder inequality it follows that

where “ Ω ” denotes the Lebesgue measure of the set Ω and that the embedding X ‹ Ls(Ω) is not compact if s = 2∗.

where c2∗ , cq are given by (2) and (3).

Now, we give the first result of this paper.

Theorem 2. Fix q ∈]1, 2[. Then there exists µ∗ > 0, where

and cq, c2∗ are given by (3) and (2) such that for each λ ]0, λ¯r[ and µ ]0, µ∗[, problem (Pλ) admits at least one positive weak solution. Let λ = 1 and uµ be the positive solution. Then

Moreover, the mapping

is negative and strictly decreasing in ]0, µ∗[.

Next, we obtain the following existence result of two solutions. At the same time, an estimate of parameters is also obtained.

Theorem 3. Fix q ∈]1, 2[. Then there exists µ∗ > 0, where

and cq, c2∗ are given by (3) and (2) such that for each µ ∈]0, µ∗[, problem

admits at least two positive solutions uµ and wµ such that ǁuµǁ < (2∗/c22∗ )1/(2∗−2) and wµ > uµ.

We observe that the solution obtained in Theorem 2 is a local minimum for the considered functional. To obtain the second solution, we will use the mountain pass theorem of Ambrosetti and Rabinowitiz. This argument is the same in the part of [3, Thm. 1.1].

Example 1. Fix N = 5 and let Ω = {x ∈ R5: |x| < 1}. Then the problem

admits at least two positive solutions uµ and wµ such that wµ > uµ. In fact, it is enough to apply Theorem 3 by choosing q = 3/2 and taking into account that for which

2 Preliminaries

We present some definitions on differentiability of functionals and refer the reader to [4, Sect. 2]. Let X be a real Banach space. We denote the dual space of X by X∗, while ⟨·, ·⟩ stands for the duality pairing between X∗ and X. A functional I : X → R is called Gâteaux differentiable at u ∈ X if there is ϕ ∈ X∗ (denoted by I′(u)) such that

It is called continuously Gâteaux differentiable if it is Gâteaux differentiable for any u ∈ X and the functional u ›→ I(u) is a continuous map from X to its dual X∗.

Let Φ, Ψ : X → R be two continuously Gâteaux differentiable functionals and put

Fix r1, r2 ∈ [−∞, +∞] with r1 < r2. We say that the functional I verifies the Palais– Smale condition cut off lower at r1 and upper at r2 (in short (PS) ec r2 r1 -condition) if any sequence (un) such that

has a convergent subsequence.

When we fix r1 = −∞, that is, Φ(un) < r2 for all n ∈ N, we denote this type of Palais–Smale condition with (PS)[r2]. When, in addition, r2 = + ∞, it is the classical Palais–Smale condition.

Now, we recall the following local minimum theorem.

Theorem 4. (See [6, Thm. 3.3].) Let X be a real Banach space, and let Φ, Ψ : X → R be two continuously Gâteaux differentiable functionals such that infX Φ = Φ(0) = Ψ (0) = 0.

Assume that there are r ∈ R and u˜ ∈ X with 0 < Φ(u˜) < r such that

and, for each λ ∈]Φ(u˜)/Ψ (u˜), r/ supu∈Φ−1(]−∞,r[) Ψ (u)[, the functional Iλ = Φ − λΨ satisfies (PS)[r]-condition. Then, for each λ ∈]Φ(u˜)/Ψ (u˜), r/ supu∈Φ−1(]−∞,r[) Ψ (u)[, there is uλ ϵ Φ−1(]0, r[) (hence, uλ = 0) such that Iλ(uλ) ≤ Iλ(u) for all u ϵ Φ−1(]0, r[) and I′(uλ) = 0.

3 Proof of the main results

Firstly, we establish the following result.

Lemma 1. Let Φ and Ψ be the functional defined in (1) and fix r > 0. Then, for each λ ∈]0, λ¯r[, the functional Iλ = Φ − λΨ satisfies the (PS)[r]-condition.

Proof. Let (un) ⊆ X be a(PS)[r] sequence, that is,

From Φ(un) < r, for all ϵ n N, (un) is bounded in X. Going to a subsequence if necessary, we can assume

Taking (i) into account, for a constant c, limn→∞ Iλ(un) = c. Moreover, (un) is bounded in L2∗ (Ω). Now, we proof our result by many steps.

Step 1. u0 is a weak solution of problem (Pλ). Since (un) is bounded in L2∗ (Ω), we get that (un2∗−1) is bounded in L2∗ /(2∗−1)(Ω). Indeed, we have

Therefore, we get that

In fact, since un → u0 a.e. x ∈ Ω, we obtain un2∗−1 → uo2∗−1 a.e. x ∈ Ω, and that, together with the boundedness of (un2∗−1) in L2∗/(2∗−1), ensures the weak convergence of un2∗−1 to uo2∗−1 in L2∗ /(2∗−1) (see [7, Rem. (iii)]).

Moreover, since un → u0 in Lq(Ω), taking into account [18, Thm. A.2], one has that

In particular,

One has

for all v ∈ X, that is, u0 is a weak solution of (Pλ).

Step 2. We prove that

Let us consider the nonlinear term

So,

It follows that for all u ∈ X, ǁuǁ ≤ (2r)1/2, we obtained

Noting (iii) and Φ is sequentially weakly lower semicontinuous, we have

That is,

Step 3. Let vn = un − u0. We get that

In fact, ǁunǁ2 = ǁvn + u0ǁ2 = ǁvnǁ2 + ǁu0ǁ2 + 2⟨vn, u0⟩, so, we obtained

Moreover, by the Brezis–Lieb lemma (see [7, Thm. 1]) one has

Finally, since uΩ → ∫(1/q) │u│q dx is locally Lipschitz in Lq(Ω) (see, for example, [9, Thm. 7.2.1]) and un → u0 in Lq(Ω), we obtained

Hence,

that is,

We get (5).

Step 4. The following equality is satisfied:

From (ii) we have limn→∞ I’(un)(un) = 0. We get

Therefore, seen in the proof of (5) and

we get that │un│q-1 → │u0│q-1 in Lq=(q-1)( Ω) (see the first step) and un → u0 in Lq(Ω).

One has

that is,

Since u0 is a weak solution of (Pʎ), one has

We get,

that is, (6).

Conclusion. Finally, we observe that kvnk2 is bounded in R. Thus, there is a subsequence, still denoted by ǁvnǁ2, which converges to b ϵ R. That is, limn→∞ ǁvnǁ2 = b.

If b = 0, we have proved the lemma. In this situation, we have lim n→∞ ǁun - u0ǁ = 0.

We assume that b 6= 0, arguing by contradiction. From (6) we obtain

By the Sobolev embedding, ǁvnǁL2* (Ω) ≤ c2*ǁvnǁ, and passing to the limit, we obtained . Since b 6≠ 0, we get

Due to (4) and (5), one has

that is, c > −r + 2b/N . On the other hand, since

for all ξ ∈ R, we obtained

for all n ϵ N. That is, c ≤ r. Thus, -r + 2b/N < c ≤ r. It follows that 2b=N < 2r, that is, b < rN. Therefore, one has

so, it follows that 1/λ < (rNc2*N/2)4/(N−4). Hence, we get

and this is a contradiction.

Now, we give the proof of Theorem 2.

Proof of Theorem 2. Let

and

Fix 0 < µ < µ∗, and one has λ¯r > 1. Indeed,

and

Therefore, from Lemma 1 the functional Iλ = Φ − λΨ satisfies the (PS)[r]-condition for all λ ∈]0, λ¯r[.

Fix λ < λ¯r. We claim that there is a v0 ∈ X, with 0 < Φ(v0) < r, such that

Consider ǁuǁLs(Ω) ≤ csǁuǁ, u ∈ X, we get

Hence, we get

Let R = supx∈Ω d(x, ∂Ω), and let x0 ∈ Ω such that B(x, R) ⊆ Ω. Moreover, put

where l : Clearly, vδ ∈ X, and since

for every x ∈ B(x0, R) \ B(x0, R/2), we get

where Γ is the gamma function. Moreover, we get

and so,

From limt→0+ │t│q=t2 = +∞ we get that

So, by

there is a δ¯ > 0 such that

and Φ(vδ¯) < r. Therefore,

with 0 < Φ(vδ¯) < r. Hence, the claim is proved.

Finally, from Theorem 4 then functional Φ - λΨ admits a critical point uλ,µ such that ǁuλ,µǁ 2/2 > 0, which is a positive weak solution for problem (Pλ). In particular, by choosing λ = 1 a positive weak solution uµ for problem (Pλ) is obtained. Moreover, one has ǁuµǁ2/2 < r from which ǁuµǁ2/2 < (2∗/(2(2∗+2)/2c2*2∗ ))2/(2∗−2), that is,

Since uµ is a global minimum for I1 in Φ−1(]0, r[) again from Theorem 4, and vδ¯ Φ−1(]0, r[), one has I1(uµ) ≤ I1(vδ¯). So, by Ψ (vδ¯)/Φ(vδ¯) > 1/λ > 1 we get

Next, fix 0 < µ1 < µ2. We get

and the conclusion is achieved.

Proof of Theorem 3. Fix µ ϵ ]0, µ∗[. From Theorem 2 there exists a positive solution uµ of (Pλ) such that uµ is a local minimum for the functional

where F is the primitive of f (t), and

We consider a new problem

Clearly, if vµ is a positive weak solution to (7), then wµ = uµ + vµ is a weak solution of (Pλ) such that wµ > uµ > 0. Now, our aim is to prove that (7) admits at least one positive weak solution. Consider the functional J defined as

and

Clearly, nonzero critical points of J are positive weak solutions of (7). Since uµ is a local minimum of I, one has

for all v ∈ X such that ǁvǁ < δ for some δ > 0. So, taking into account that

for all v ϵ X (see [3]), we get J(v) ≥ 0 for all v ϵ X such that ǁvǁ < δ. That is, 0 is a local minimum of J.

By using the same proof in [3], the functional J admits a positive critical point vµ for which wµ = uµ + vµ is the second weak solution of (7), and the proof is complete.

Acknowledgments

The authors express their gratitude to the anonymous referees for useful comments and remarks.

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