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Purpose

In this paper, we study a fourth-order singular problems of type Leray–Lions with singular weight and with no-flux boundary conditions.

Design/methodology/approach

We use the Nehari manifold to establish our result for our problem.

Findings

We prove that there exist at least two nontrivial positive weak solutions.

Originality/value

In particular, it investigates the problem in the presence of singular weights and variable exponents, a case that has not been extensively studied in the existing literature.

In this article, we study the following problem.

where ΩRN(N>2) is a bounded domain containing the origin with sufficiently smooth boundary and Δ(a(x, Δu)) is the fourth-order Leray–Lions operator, b(.)C(Ω¯) is a positive weight function and λ is a real parameter. The variable exponents p(.),q(.),r(.),α(.),β(.)C+(Ω¯)=hC(Ω¯) and h(x)>1 on Ω¯.

The Leray–Lions-type operator is a more quiet general in the fourth-order, we can take for example a(x, t) = h(x)|t|p(x)−2t, when h(x) = 1 and t = Δu the above operator turns into the operator Δp()2u (the p(x) − biharmonic operator), for mor details about Leray–Lions type operators one can see Refs. [1–4].

It should be noted that this type of problem is very active in several domains, such electrorheological fluids [5], elastic materials [6] and stationary thermo-rheological viscous flows [7].

In the last few years, various authors have studied this kind of problem; in the case of exponents constant, one can see Refs. [8, 9]. In 2020, Mokhtar, M. E. [10] studied the existence of solution by using variational approaches and utilizing the Pohozaev identity to show the nonexistence of solution for the following p-Laplace singular problem

Moreover, many authors have studied problems similar to our problem; in the case of variable exponents, see for example [11–15]. In 2023, Repovš, D. D. and Saoudi, K. [16]. studied this p(x)-Laplace singular problem

They used the Nehari manifold technique to demonstrate their results for the problem (PN).

We start by giving the assumptions that we will consider for our problem.

  • (H1) 0<s<s+<1<2<p<p+<qq(x)<Nα(x)Np2(x).

  •      where hminxΩ¯h(x),h+maxxΩ¯h(x) and p2(x)=Np(x)Np(x), for all xΩ¯.

  • (H2) Ns(x) < β(x) < N.

  •      (b) b ∈ L(Ω) and strictement positive in Ω.

  • (A1) a:Ω¯×RR is a Carathéodory function and a(x, 0) = 0 for a.e. xΩ¯.

  • (A2) there exists c1 > 0 such that

such that A(x,t)=0ta(x,s)ds

  • (A3) [a(x, s) − a(x, t)](s − t) ≥ 0 holds for all s,tR and a.e. x ∈ Ω with equality if and only if s = t.

  • (A4) There exist 0<c0<c1q1p+1 and a non-negative function d(x) ∈ Lp(x)(Ω), such that

for all tR and d(x) that verifies |d|p(x)<12min{qc1c01p,qp+c0}.

  • (A5) A:Ω¯×RR is homogeneous of degre p(x), that is A(x,tu)=tp(x)A(x,u)(t>0)xΩ¯, uR.

Under the above hypotheses, we give our main result.

Theorem 1.

Suppose that (A1) − (A5), (H1), (H2) and (b) hold. Then there exists a positive constants λ such that the problem (PV) has at least two positive solutions for each λ0,λ.

The structure of this article is as follows: Section 2 presents key background results concerning variable exponent Lebesgue and Sobolev spaces, along with a compact embedding theorem of the Sobolev–Hardy type. Section 3 introduces the essential lemmas required for our analysis. We then establish that the energy functional achieves its minimum on both Nλ+ and Nλ. We conclude with the proof of our main result.

In this part, we review several fundamental properties and definitions of Lebesgue and Sobolev spaces with variable exponents. For further details, readers are encouraged to consult the comprehensive studies in Refs. [17–20].

In this study, we denote by M(Ω) the set of all real-valued functions that are measurable with respect to the Lebesgue measure on Ω, and we suppose that:

  • (H) p is log-Hölder continuous, satisfying 1 < p ≤ p+ < .

That is, there exists k > 0 such that

Let b ∈ M(Ω) satisfies the condition (b), we define the space:

equipped with the norm,

Lb(x)p(x)(Ω),||Lb(x)p(x) is a Banach space. For more details on this norm, (see for example [20], Theorem 2.5 and Corollary 2.7).

Additionally, we can establish the following inequality,

(1)

for all u ∈ Lp(x)(Ω) and v ∈ Lp(x)(Ω), (see Ref. [20], Theorem 2.1).

We denote by W01,p(x)(Ω) the closure of C0(Ω) in W1,p(x)(Ω).

Taking into account the log-Hölder countinuity of the exponent p(x), we have

As a consequence of the Poincaré inequality, uW1,p(x)(Ω) and |▽u|p(x) are equivalent norms on W01,p(x)(Ω). Therefore, for any uW01,p(x)(Ω), we can define an equivalent norm uW01,p(x)(Ω) such that

and which makes W01,p(x)(Ω) a separable and reflexive Banach space (see Ref. [21], Proposition 2.1).

Under condition (H), we can see that C(Ω¯) is dense in Wk,p(x)(Ω) (see Refs. [22, 17]).

Theorem 2.

(see ([22]) and ([19], Theorem 2.3)). If qC+(Ω¯) and q(x)<p2*(x) for all xΩ¯. Then W2,p(x)(Ω)↪↪Lq(x)(Ω). , where

The following lemma is very important.

Lemma 1.

Let us consider r(x), γ(x)C+(Ω¯) and

(2)

Then W2,p(x)(Ω)L|x|γ(x)r(x)(Ω).

Proof. We notice |x|γ(x)LNϵγ(x)(Ω) such that ϵ > 0 is a constant small enough (see Ref. [21]). Let u ∈ W2,p(x)(Ω). Set h(x)=Nϵγ(x)r(x)=(Nϵ)r(x)Nϵγ(x).

Then (2) implies h(x)<p2*(x) and by Theorem 2.3 (in Ref. [19]) we have W2,p(x)(Ω)↪↪Lh(x)(Ω), then for u ∈ W2,p(x)(Ω) we have |u(x)|r(x)LNϵNγ(x)ϵ(Ω) and, from inequality 1, from inequality, one can see that

This proves W2,p(x)(Ω)L|x|γ(x)r(x)(Ω).

Now let (un) ⊂ W2,p(x)(Ω) and un ⇀ 0 in W2,p(x)(Ω). Using Theorem 2, we find that un → 0 in Lh(x)(Ω), and from this, we get |unr(x)NϵNγ(x)ϵ0. Then

that implies un(r(x),|x|γ(x))0. Therefore, we have W2,p(x)(Ω)L|x|γ(x)r(x)(Ω).

Remark 1.

when r(x)=γ(x)=p(x)xΩ¯. Then W2,p(x)(Ω)L|x|p(x)p(x)(Ω).◽

This result indicates the existence of a positive constant note Cp such that,

(3)

where ± = + if uW2,p(x)(Ω)>1, and ± = − if uW2,p(x)(Ω)1.

Throughout this paper, we opted for the norm:

where b(x) satisfies (b), this norm represents on both W2,p(x)(Ω) and W2,p(x)(Ω)W01,p(x)(Ω) and we can see by relation (3) that is equivalent to W2,p(x)(Ω). We will search for the solutions of problem (PV) in this space.

Y,W2,p(x)(Ω) is a reflexive and separable Banach space ([23]). As a result, we take into account the modular ρ:YR such that

The following inequalities have a significant connection to the norm ‖.‖b, (see for example [23], Proposition 1). If u ∈ W2,p(x)(Ω) then:

(4)
(5)

We define the functional ϝ:YR such that

To see the properties of this functional, refer to ([1], Proposition 3; 5 and Theorem 3.2).

Definition 1.

u ∈ Y is a weak solution of problem (PV) if

for all υ ∈ Y.

For the problem (PV), the energy functional T:YR is defined as

where

The weakly lower semi-continuous function T has a Gâteaux derivative that is defined by

for all υ ∈ Y.

Remark 2.

T is not in C1(Y,R) because of the singular term in I1.

For some problems, like (PV), T may not be bounded below on Y, but it does possess a bounded below on Nλ, such that

u ∈ Nλ if and only if

We see that all solutions to the problem (PV) are included in Nλ.

Defined ϕu(t) = T(tu),

As a result, it makes sense to divide Nλ into the following three parts.

ϕu(t)=Ωtp(x)A(x,Δu)dx+Ωtp(x)b(x)p(x)|u|p(x)|x|p(x)dxΩtq(x)q(x)|u|q(x)|x|α(x)dxλΩt1s(x)1s(x)|u|2s(x)|x|β(x)dx.

ϕu(t)=Ωp(x)tp(x)1A(x,Δu)dx+Ωb(x)tp(x)1|u|p(x)|x|p(x)dxΩtq(x)1|u|q(x)|x|α(x)dxλΩts(x)|u|1s(x)|x|β(x)dx

ϕu(t)=Ωp(x)(p(x)1)tp(x)2A(x,Δu)dx+Ω(p(x)1)b(x)|u|p(x)|x|p(x)dxΩ(q(x)1)tq(x)2|u|q(x)|x|α(x)dx+λΩs(x)t1s(x)|u|1s(x)|x|β(x)dx.

Lemma 2.

T is bounded below on Nλ and is coercive.

Proof. Lemma 1 and (A2) provide us, if u ∈ Nλ such that ‖ub > 1.

With C1 = min{1, c1}, since 1 − s+ < p, then T(u) → + when ‖ub → +.◽

Lemma 3.

Given u as a local minimum of T on subsets Nλ+ or Nλ, where u is not an element of Nλ0, it follows that u represents a critical point of T.

Proof. Assume that, under the given restriction, u is a local minimizer of T.

(6)

Thus, we use the theory of Lagrange multipliers to prove that there exists μR that satisfies

Consequently,

From this uNλ0, it follows that Φu(1)0. As a result, μ = 0.◽

Lemma 4.

There is λ0 such that, Nλ± and Nλ0=, for every 0 < λ < λ0.

Proof. Initially, applying Lemma 3, we can infer that Nλ± for λ ∈ (0, λ0). Now since u ∈ Nλ, we deduce the following

(7)

Assuming uNλ0 with ‖ub > 1. By combining the above equality with (6), under the conditions (A1) and (A4), we get

such that C2 = min{c1, 1}. From lemma 1 and (4), we get

Hence

(8)

Since u ∈ Nλ and since uNλ0, we get

Since ‖ub > 1, by lemma 1 and (4) we get

such that C3 = min{(q − 1)c1 − c0(p+ − 1), q − p+}, hence

(9)

Using the above inequality, we find

(10)

we get

Then, if λ small enough, we obtain ‖ub < 1, which is impossible. Therefore, Nλ0= for all λ ∈ (0, λ0). Hence, this proof is complete.◽

Now let us show that the energy functional T in Nλ+ has a minimum. Furthermore, we will demonstrate that this minimum corresponds to a solution for the problem (PV).

Theorem 3.

There is u0Nλ+ for any λ ∈ (0, λ0), which satisfies

Proof. Consider λ ∈ (0, λ0), it follows that T is bounded below on Nλ and, consequently on Nλ+. Hence, we can establish the existence of a sequence {un}Nλ+, satisfying T(un)infuNλ+T(u) as n. T is coercive on Nλ and {un} is bounded in V; by utilizing lemma 1, we deduce the following

First we will show that infuNλ+T(u)>0

Let u0Nλ+, then ϕu0(1)>0, which gives

(11)

We can write

(12)

We multiply (6) with s+, we obtain

(13)

invoking (13) and (11), we obtain

(14)

However, we multiply (6) with 11s+, and from (12), we get

Replacing the inequality (14) in the previous inequality, we get

and by the conditions A2 and A4, we have

such that 0<C4=min{c0p(1p+1q+(1s+))c1(11s++s+q+(s+1)),1s+pp(1s+)+p+1+s+q+(1s+)}, then by the definition of the function d and the above inequality, we get

Assume that unu0 strongly in Y. Then

We use the compact embedding to derive the following

Now, by (6) and Theorem 2.3 in Ref. [24].

and by (A2), we get

which C5 = min{c1, 1}, now passing n, we obtain

using Theorem 2.3 in Ref. [24], we get

Since p > 1 − s+, which give a contradiction. Therefore, unu0 in Y and infuNλ+T(un)=T(u0)

Second, we shall now prove that the functional energy T in Nλ is at a minimum.◽

Theorem 4.

for every λ ∈ (0, λ0), there exists u1Nλ satisfying

Proof. Consider that λ ∈ (0, λ0); thus, it follows T, which is bounded below on Nλ. Consequently, there exists {un}Nλ, with T(un)infuNλT(u) as n. T is coercive in Nλ, {un} is bounded in V; therefore, we can assume that using unu1 in Y by lemma 1, we obtain

Currently, prove that unu1 in Y. First, we will show that infuNλT(u)>0

Let u1Nλ, then from (6), we give

(15)

Then

(16)
(17)

As we have p > 1 − β+, we can make the following choice.

We obtain T(u1) > 0. Moreover, since Nλ+Nλ= and infuNλ+T(u)<0, then u1Nλ. The same, if u1Nλ; hence, there is t1, satisfying t1u1Nλ and so T(t1u1) ≤ T(u1).

we get

Since 1 − s+ < p+ < q, Iλ(t1u1)<0; so by definition of Nλ, we have t1u1Nλ

Now, let us assume that unu1 in Y, we know that

which gives us

This contradicts with t1u1Nλ; hence, unu1 in Y and T(u1)=infuNλT(u).

Proof. Invoking Theorems 3 and 4, we can conclude that for every λ ∈ (0, λ0), there exist u0Nλ+ and u1Nλ such that

On the one hand, considering the properties of the function T, we can assume that u0 and u1 are non-negative, and by applying Lemma 2, we can show that they are critical points of T on the set Y, which makes them weak solutions of the problem (PV). Using the Harnack inequality and the results of Zhang-Liu [25], we can then demonstrate that u0 and u1 are non-negative solutions of (PV).

Since Nλ+Nλ=, we conclude that u0, u1 must be distinct. Thus, we have demonstrated that the solutions u0 and u1 obtained from Theorem 3 and Theorem 4 are indeed different. Consequently, Theorem 1 has now been completely proved.◽

The anonymous editors and referees who thoroughly examined this work and offered helpful criticism are acknowledged by the authors.

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