Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 25, pp. 1–19. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu SUB-SUPER SOLUTION METHOD FOR NONLOCAL SYSTEMS INVOLVING THE p(x)-LAPLACIAN OPERATOR GELSON C. G. DOS SANTOS, GIOVANY M. FIGUEIREDO, LEANDRO S. TAVARES Abstract. In this article we study the existence of solutions for nonlocal systems involving the p(x)-Laplacian operator. The approach is based on a new sub-super solution method. 1. Introduction In this work we are interested in the nonlocal system −A(x, |v|Lr1(x))∆p1(x)u = f1(x, u, v)|v|α1(x) Lq1(x) + g1(x, u, v)|v|γ1(x) Ls1(x) in Ω, −A(x, |u|Lr2(x))∆p2(x)v = f2(x, u, v)|u|α2(x) Lq2(x) + g2(x, u, v)|u|γ2(x) Ls2(x) in Ω, u = v = 0 on ∂Ω, (1.1) where Ω is a bounded domain in RN (N > 1) with C2 boundary, | · |Lm(x) is the norm of the space Lm(x)(Ω), −∆p(x)u := −div(|∇u|p(x)−2∇u) is the p(x)-Laplacian operator, ri, pi, qi, si, αi, γi : Ω → [0,∞), i = 1, 2 are measurable functions and A, f1, f2, g1, g2 : Ω× R→ R are continuous functions satisfying certain conditions. In the previous decades there have been several works related to the p and p(x) Laplacian operator; see for example [1, 4, 9, 12, 25, 26, 27, 28, 29, 34, 35, 38, 39] and the references therein. Partial differential equations involving the p(x)-Laplacian arise in several areas of Science and Technology such as nonlinear elasticity, fluid mechanics, non-Newtonian fluids and image processing. Regarding the mentioned applications we point out [1, 14, 36, 41, 42]. The nonlocal term | · |Lm(x) with the condition p(x) = r(x) ≡ 2 was considered in the well known Carrier’s equation ρutt − a(x, t, |u|2L2)∆u = 0 which models the vibrations of a elastic string under certain contidions. See [11] for more details. We also quote the applicability of such nonlocal term in Population Dynamics, see [15, 17]. Several works related to (1.1) in the p-Laplacian case, that is, with p(x) = p (a constant) can be found, see [10, 13, 19, 20, 23, 43] and the references provided in such manuscripts. For example Corrêa & Lopes [20] studied the system −∆um = a|v|αLp in Ω, 2010 Mathematics Subject Classification. 35J60. Key words and phrases. Fixed point argument; nonlocal problem; p(x)-Laplacian; sub-super solutions. c©2020 Texas State University. Submitted April 9, 2019. Published March 19, 2020. 1 2 G. C. G. DOS SANTOS, G. M. FIGUEIREDO, L. S. TAVARES EJDE-2020/25 −∆vn = b|u|βLq in Ω, u = v = 0 on ∂Ω, and in [13] a related system was considered using the Galerkin method. In [19] the authors used a theorem due to Rabinowitz [40] to study the problem −∆p1u = |v|α1 Lq1 in Ω, −∆p2v = |u|α2 Lq2 in Ω, u = v = 0 on ∂Ω. The system −A(x, |v|Lr1(x))∆u = f1(x, u, v)|v|α1(x) Lq1(x) + g1(x, u, v)|v|γ1(x) Ls1(x) in Ω, −A(x, |u|Lr2(x))∆u = f2(x, u, v)|u|α2(x) Lq2(x) + g2(x, u, v)|u|γ2(x) Ls2(x) in Ω, u = v = 0 on ∂Ω, where A : Ω × R → R is a function satisfying some conditions, was considered in [43]. The approach in such paper consists in use an abstract result involving sub and supersolutions, whose proof is based on the Schaefer’s fixed point theorem. Specifically, it was considered a sublinear system, a concave-convex problem and a system of logistic equations. The scalar version of (1.1), −A(x, |u|Lr(x))∆p(x)u = f(x, u)|u|α(x) Lq(x) + g(x, u)|u|γ(x) Ls(x) in Ω, u = 0 on ∂Ω, (1.2) was considered in [44]. The authors obtained an abstract result involving sub and super solutions for (1.1) that generalizes [43, Theorem 1]. As an application of such result the authors generalized for the p(x)-Laplacian operator the three applications of [43, Theorem 1]. The goal of this work is to prove [43, Theorem 2] for the p(x)-Laplacian operator and use it in three applications of the mentioned paper. Thus, we provide a gener- alization of [43] with respect to systems with variable exponents. Next we describe the main differences and difficulties of this work when compared with [43]. (i) The homogeneity of the Laplacian operator (−∆, H1 0 (Ω)) and the eigenfunc- tion associated to the first eigenvalue were used in [43] for constructing a subso- lution. Differently from the p-Laplacian (p(x) ≡ p constant) the p(x)-Lapalcian is not homogeneous. Besides that, it can occurs that the first eigenvalue and the first eigenfunction of the p(x)-Laplacian operator (−∆p(x),W 1,p(x) 0 (Ω)) do not exist. Even if the first eigenvalue and the associated eigenfunction exist the homogeneity, in general, does not allows to use the first eigenfunction to construct a subsolution. In order to avoid such difficulties we explore some arguments of [44]. (ii) Some arguments of [43] were improved and weaker conditions on ri, qi, si, αi, γi, i = 1, 2 are considered here. (iii) We generalize [43, Theorem 2] and as an application it is considered some nonlocal problems that generalizes the three systems studied in [43]. (iv) As in [43, Theorem 2] and differently from several works that consider the nonlocal term A(x, |u|Lr(x)) satisfying A(x, t) ≥ a0 > 0 (where a0 is a constant), Theorem 1.1 permits us to study (1.1) in the mentioned case and in situations where A(x, 0) = 0. EJDE-2020/25 SUB-SUPER SOLUTION METHOD 3 (v) The abstract result involving sub and super solutions is proved by using a different argument. It is used a theorem due to Rabinowitz that can be found in [40] and some arguments of [43] are improved. In this work we assume that ri, pi, qi, si, αi, γi satisfy (H1) pi ∈ C1(Ω), ri, qi, si ∈ L∞+ (Ω), where L∞+ (Ω) = { m ∈ L∞(Ω) with ess inf m(x) ≥ 1 } and for i = 1, 2, αi, γi ∈ L∞(Ω) and satisfy 1 < p−i : = inf Ω pi(x) ≤ p+ i : = sup Ω pi(x) < N, αi(x), γi(x) ≥ 0 a.e in Ω . Some definitions are needed to present the main results. We say that the pair (u1, u2) is a weak solution of (1.1), if ui ∈W 1,pi(x) 0 (Ω) ∩ L∞(Ω) and∫ Ω |∇ui|pi(x)−2∇ui∇ϕ = ∫ Ω (fi(x, u1, u2)|uj |αi(x) Lqi(x) A(x, |uj |Lri(x)) + gi(x, u1, u2)|uj |γi(x) Lsi(x) A(x, |uj |Lri(x)) ) ϕ, for all ϕ ∈W 1,pi(x) 0 (Ω) and i 6= j with i, j = 1, 2. Given u, v ∈ S(Ω) we write u ≤ v if u(x) ≤ v(x) a.e. in Ω. If u ≤ v we define [u, v] := { w ∈ S(Ω) : u(x) ≤ w(x) ≤ v(x) a.e. in Ω } . To simplify the next definition we denote f̃1(x, t, s) = f1(x, t, s), g̃1(x, t, s) = g1(x, t, s), f̃2(x, t, s) = f2(x, s, t), g̃2(x, t, s) = g2(x, s, t). We say that the pairs (ui, ui), i = 1, 2 are a sub-super solutions for (1.1) if ui ∈ W 1,pi(x) 0 (Ω) ∩ L∞(Ω), ui ∈ W 1,pi(x)(Ω) ∩ L∞(Ω) with ui ≤ ui, ui = 0 ≤ ui on ∂Ω and for all ϕ ∈W 1,pi(x) 0 (Ω) with ϕ ≥ 0 the following inequalities hold∫ Ω |∇ui|pi(x)−2∇ui∇ϕ ≤ ∫ Ω ( f̃i(x, ui, w)|uj | αi(x) Lqi(x) A(x, |w|Lri(x)) + g̃i(x, ui, w)|uj | γi(x) Lsi(x) A(x, |w|Lri(x)) ) ϕ, ∫ Ω |∇ui|pi(x)−2∇ui∇ϕ ≥ ∫ Ω ( f̃i(x, ui, w)|uj |αi(x) Lqi(x) A(x, |w|Lri(x)) + g̃i(x, ui, w)|uj |γi(x) Lsi(x) A(x, |w|Lri(x)) ) ϕ, (1.3) for all w ∈ [ujuj ] where i, j = 1, 2 with i 6= j. Our main result reads as follows. Theorem 1.1. Suppose that ri, pi, qi, si, αi and γi satisfy (H1), that (ui, ui) is a sub-super solution for (1.1) with ui > 0 a.e. in Ω, that fi(x, t, s), gi(x, t, s) ≥ 0 in Ω × [0, |u1|L∞ ] × [0, |u2|L∞ ] and that A : Ω × (0,∞) → R is a continuous function with A(x, t) > 0 in Ω × [ σ, σ ] , where σ := min { |w|Lri(x) , i = 1, 2 } , σ := max { |w|Lri(x) , i = 1, 2 } , w := min{ui, i = 1, 2} and w := max{ui, i = 1, 2}. Then (1.1) has a weak positive solution (u1, u2) with ui ∈ [ui, ui], i = 1, 2. 2. Preliminaries In this section, we present some facts regarding the spaces Lp(x)(Ω), W 1,p(x)(Ω) and W 1,p(x) 0 (Ω) that will be often used in this work. For more details see Fan-Zhang [27] and the references therein. 4 G. C. G. DOS SANTOS, G. M. FIGUEIREDO, L. S. TAVARES EJDE-2020/25 Let Ω ⊂ RN (N ≥ 1) be a bounded domain. Given p ∈ L∞+ (Ω), we define the generalized Lebesgue space Lp(x)(Ω) = { u ∈ S(Ω) : ∫ Ω |u(x)|p(x)dx <∞ } , where S(Ω) := { u : Ω → R : u is measurable } . Then Lp(x)(Ω) is a Banach space with the norm |u|p(x) := inf { λ > 0 : ∫ Ω |u(x) λ |p(x)dx ≤ 1 } . Given m ∈ L∞(Ω), we define m+ := ess supΩm(x), m− := ess infΩm(x). Proposition 2.1. Let ρ(u) := ∫ Ω |u|p(x)dx. Then for u, un ∈ Lp(x)(Ω), and n ∈ N, the following assertions hold (i) Let u 6= 0 in Lp(x)(Ω), then |u|Lp(x) = λ⇔ ρ(uλ ) = 1. (ii) If |u|Lp(x) < 1 (= 1, > 1), then ρ(u) < 1 (= 1, > 1). (iii) If |u|Lp(x) > 1, then |u|p − Lp(x) ≤ ρ(u) ≤ |u|p + Lp(x) . (iv) If |u|Lp(x) < 1, then |u|p + Lp(x) ≤ ρ(u) ≤ |u|p − Lp(x) . (v) |un|Lp(x) → 0⇔ ρ(un)→ 0, and |un|Lp(x) →∞⇔ ρ(un)→∞. Theorem 2.2. Let p, q ∈ L∞+ (Ω). Then the following statements hold (i) If p− > 1 and 1 q(x) + 1 p(x) = 1 a.e. in Ω, then∣∣ ∫ Ω uvdx ∣∣ ≤ ( 1 p− + 1 q− ) |u|Lp(x) |v|Lq(x) . (ii) If q(x) ≤ p(x) a.e. in Ω and |Ω| <∞, then Lp(x)(Ω) ↪→ Lq(x)(Ω). We define the generalized Sobolev space as W 1,p(x)(Ω) := { u ∈ Lp(x)(Ω) : ∂u ∂xj ∈ Lp(x)(Ω), j = 1, . . . , N } with the norm ‖u‖∗ = |u|Lp(x) + N∑ j=1 ∣∣ ∂u ∂xj ∣∣ Lp(x) . The space W 1,p(x) 0 (Ω) is defined as the closure of C∞0 (Ω) with respect to the norm ‖ · ‖∗. Theorem 2.3. If p− > 1, then W 1,p(x)(Ω) is a Banach, separable and reflexive space. Proposition 2.4. Let Ω ⊂ RN be a bounded domain and p, q ∈ C(Ω). Define the function p∗(x) = Np(x) N−p(x) if p(x) < N and p∗(x) = ∞ if N ≥ p(x). Then the following statements hold. (i) (Poincaré inequality) If p− > 1, then there is a constant C > 0 such that |u|Lp(x) ≤ C|∇u|Lp(x) for all u ∈W 1,p(x) 0 (Ω). (ii) If p−, q− > 1 and q(x) < p∗(x) for all x ∈ Ω, then the embedding W 1,p(x)(Ω) ↪→ Lq(x)(Ω) is continuous and compact. From (i) of Proposition 2.4, we have that ‖u‖ := |∇u|Lp(x) defines a norm in W 1,p(x) 0 (Ω) which is equivalent to the norm ‖ · ‖∗. EJDE-2020/25 SUB-SUPER SOLUTION METHOD 5 Definition 2.5. For u, v ∈W 1,p(x)(Ω), we say that −∆p(x)u ≤ −∆p(x)v, if∫ Ω |∇u|p(x)−2∇u∇ϕ ≤ ∫ Ω |∇v|p(x)−2∇v∇ϕ, for all ϕ ∈W 1,p(x) 0 (Ω) with ϕ ≥ 0. The following result appears in [29, Lemma 2.2] and [26, Proposition 2.3]. Proposition 2.6. Let u, v ∈ W 1,p(x)(Ω). If −∆p(x)u ≤ −∆p(x)v and u ≤ v on ∂Ω, (i.e., (u − v)+ ∈ W 1,p(x) 0 (Ω)) then u ≤ v in Ω. If u, v ∈ C(Ω) and S = { x ∈ Ω : u(x) = v(x) } is a compact set of Ω, then S = ∅. Lemma 2.7 ([26, Lemma 2.1]). Let λ > 0 be the unique solution of the problem −∆p(x)zλ = λ in Ω, u = 0 on ∂Ω. (2.1) Define ρ0 = p− 2|Ω| 1 N C0 . If λ ≥ ρ0 then |zλ|L∞ ≤ C∗λ 1 p−−1 , and |zλ|L∞ ≤ C∗λ 1 p+−1 if λ < ρ0. Here C∗ and C∗ are positive constants depending only on p+, p−, N, |Ω| and C0, where C0 is the best constant of the embedding W 1,1 0 (Ω) ↪→ L N N−1 (Ω). Regarding the function zλ of the previous result, it follows from [25, Theorem 1.2] and [29, Theorem 1] that zλ ∈ C1(Ω) with zλ > 0 in Ω. The proof of Theorem 1.1 is mainly based on the following result by Rabinowitz: Theorem 2.8 ([40]). Let E be a Banach space and Φ : R+ × E → E a compact map such that Φ(0, u) = 0 for all u ∈ E. Then the equation u = Φ(λ, u) possesses an unbounded continuum C ⊂ R+ × E of solutions with (0, 0) ∈ C. We point out that a mapping Φ : E → E is compact if it is continuous and for each bounded subset U ⊂ E, the set Φ(U) is compact. 3. Proof of main results Proof of Theorem 1.1. For i = 1, 2 consider the operators Ti : Lpi(x)(Ω)→ L∞(Ω) defined by Tiz(x) =  ui(x), if z(x) ≤ ui(x), z(x), if ui(x) ≤ z(x) ≤ ui(x), ui(x), if z(x) ≥ ui(x). Since Tiz ∈ [ui, ui] and ui, ui ∈ L∞(Ω) it follows that the operators Ti are well- defined. We define p′i(x) = pi(x)/ ( pi(x) − 1 ) and consider the operators Hi : [u1, u1] × [u2, u2]→ Lp ′ i(x)(Ω) given by Hi(u1, u2)(x) = fi(x, u1(x), u2(x))|uj |αi(x) Lqi(x) A(x, |uj |Lri(x)) + gi(x, u1(x), u2(x))|uj |γi(x) Lsi(x) A(x, |uj |Lri(x)) where i 6= j with i, j = 1, 2, and | · |Lm(x) denotes the norm of the space Lm(x)(Ω). We consider in the space Lp1(x)(Ω)× Lp2(x)(Ω) with the norm |(u, v)|1,2 = |u|Lp1(x) + |v|Lp2(x) . 6 G. C. G. DOS SANTOS, G. M. FIGUEIREDO, L. S. TAVARES EJDE-2020/25 Since fi, gi,A are continuous functions, A(x, t) > 0 in the compact set Ω× [ σ, σ], Tizi ∈ [ui, ui] for all zi ∈ Lpi(x)(Ω), ui, ui ∈ L∞(Ω), and |w|θ(x) Lm(x) ≤ |w|θ − Lm(x) + |w|θ+ Lm(x) for all w ∈ Lm(x)(Ω) with θ ∈ L∞(Ω), it follows that there are constants Ki > 0 such that |Hi(T1z1, T2z2)| ≤ Ki (3.1) for all (z1, z2) ∈ Lp1(x)(Ω)× Lp2(x)(Ω). By the Lebesgue Dominated Convergence Theorem, the mappings (z1, z2) 7→ Hi(T1z1, T2z2) are continuous from Lp1(x)(Ω)× Lp2(x)(Ω) in Lp ′ i(x)(Ω), i = 1, 2. From [27, Theorem 4.1] the operator Φ : R+×Lp1(x)(Ω)×Lp2(x)(Ω)→ Lp1(x)(Ω)× Lp2(x)(Ω) given by Φ(λ, z1, z2) = (u1, u2), where (u1, u2) ∈W 1,p1(x) 0 (Ω)×W 1,p2(x) 0 (Ω) is the unique solution of −∆p1(x)u1 = λH1(T1z1, T2z2) in Ω, −∆p2(x)u2 = λH2(T1z1, T2z2) in Ω, u = v = 0 on ∂Ω, (3.2) is well-defined. Claim 1: Φ is compact. Let (λn, z 1 n, z 2 n) ⊂ R+ × Lp1(x)(Ω) × Lp2(x)(Ω) be a bounded sequence and consider (u1 n, u 2 n) = Φ(λn, z 1 n, z 2 n). The definition of Φ imply that ∫ Ω |∇uin|pi(x)−2∇un∇ϕ = λn ∫ Ω Hi(T1z 1 n, T2z 2 n)ϕ, ∀ϕ ∈ W 1,pi(x) 0 (Ω), where i, j = 1, 2 blue with i 6= j. Considering the test function ϕ = uin, the boundness of (λn) and inequality (3.1), we obtain ∫ Ω |∇uin|pi(x) ≤ λKi ∫ Ω |uin| for all n ∈ N. Here λ is a constant that does not depend on n ∈ N. Since p−i > 1, the embedding Lpi(x)(Ω) ↪→ L1(Ω) holds. Combining such em- bedding with the Poincaré inequality we obtain∫ Ω |∇uin|pi(x) ≤ CKi‖uin‖, for all n ∈ N. Suppose that |∇uin|Lpi(x) > 1. Thus by Proposition 2.1 we have ‖uin‖p −−1 ≤ CKi for all n ∈ N where C is a constant that does not depend on n. Then we conclude that (uin) is bounded in W 1,pi(x) 0 (Ω). The reflexivity of W 1,pi(x) 0 (Ω) and the compact embedding W 1,pi(x) 0 (Ω) ↪→ Lpi(x)(Ω) provides the result. Claim 2: Φ is continuous. Consider a sequence (λn, z 1 n, z 2 n) in R+×Lp1(x)(Ω)× Lp2(x)(Ω) converging to (λ, z1, z2) in R+×Lp1(x)(Ω)×Lp2(x)(Ω). Define (u1 n, u 2 n) = Φ(λn, z 1 n, z 2 n) and (u1, u2) = Φ(λ, z1, z2). Using the definition of Φ we obtain∫ Ω |∇uin|pi(x)−2∇uin∇ϕ = λn ∫ Ω Hi(T1z 1 n, T2z 2 n)ϕ, (3.3)∫ Ω |∇ui|pi(x)−2∇ui∇ϕ = λ ∫ Ω Hi(T1z 1, T2z 2)ϕ (3.4) EJDE-2020/25 SUB-SUPER SOLUTION METHOD 7 for all ϕ ∈W 1,pi(x) 0 (Ω) where i, j = 1, 2 and i 6= j. Considering ϕ = (uin−ui) in (3.3) and (3.4) and subtracting (3.4) from (3.3) we obtain ∫ Ω 〈 |∇uin|pi(x)−2∇uin − |∇ui|pi(x)−2∇ui,∇(uin − ui) 〉 = ∫ Ω λnH(T1z 1 n, T2z 2 n)(uin − ui)− ∫ Ω λH(T1z 1, T2z 2) ] (uin − ui). Using Hölder’s inequality we have∣∣ ∫ Ω 〈 |∇uin|pi(x)−2∇uin − |∇u|pi(x)−2∇ui,∇(uin − u) 〉∣∣ ≤ |uin − ui|pi(x)|λnHi(T1z 1 n, T2z 2 n)− λHi(T1z 1, T2z 2)|p′i(x) The arguments above ensures that (uin) is bounded in W 1,pi(x) 0 (Ω). Since λn → λ and Hi(T1z 1 n, T2z 2 n)→ Hi(T1z 1, T2z 2) in Lp ′ i(x)(Ω) for i = 1, 2 we have∣∣ ∫ Ω 〈 |∇uin|pi(x)−2∇uin − |∇u|pi(x)−2∇ui,∇(uin − u) 〉∣∣→ 0. Therefore uin → ui in Lpi(x)(Ω) for i = 1, 2 which proves the continuity of Φ. Combining the fact that Φ(0, z1, z2) = (0, 0, 0) for all (z1, z2) ∈ Lp1(x)(Ω) × Lp2(x)(Ω) with the previous claims we have by Theorem 2.8 that the equation Φ(λ, u, v) = (u, v) possesses an unbounded continuum C ⊂ R+ × Lp1(x)(Ω) × Lp2(x)(Ω) of solutions with (0, 0, 0) ∈ C. Claim 3: C is bounded with respect to the parameter λ. Suppose that there exists λ∗ > 0 such that λ ≤ λ∗ for all (λ, u1, u2) ∈ C. For (λ, u1, u2) ∈ C the definition of Φ imply that −∆p1(x)u1 = λH1(T1u1, T2u2) in Ω, −∆p2(x)u2 = λH2(T1u1, T2u2) in Ω, u1 = u2 = 0 on ∂Ω. (3.5) Using the test function ui in (3.5) and considering (3.1) we obtain∫ Ω |∇ui|pi(x) ≤ λ∗C|ui|Lp(x) . Suppose that |∇ui|Lp(x) > 1. Then using Proposition 2.1 and the Poincaré inequal- ity we obtain that |ui|pi−1 Lpi(x) ≤ λ∗C. Thus C is bounded in R+ × Lp1(x)(Ω)× Lp2(x)(Ω), which is a contradiction. Considering λ = 1, by (3.5) we have∫ Ω |∇ui|pi(x)−2∇ui∇ϕ = ∫ Ω (fi(x, T1u1, T2u2)|Tjuj |αi(x) Lqi(x) A(x, |Tjuj |Lri(x)) ) ϕ + ∫ Ω (gi(x, T1u1, T2u2)|Tjuj |γi(x) Lsi(x) A(x, |Tjuj |Lri(x)) ) ϕ, (3.6) for all ϕ ∈W 1,pi(x) 0 (Ω) where i, j = 1, 2 with i 6= j. 8 G. C. G. DOS SANTOS, G. M. FIGUEIREDO, L. S. TAVARES EJDE-2020/25 Now we claim that ui ∈ [ui, ui] for i = 1, 2. To prove the claim we define L1(u1 − u1)+ := ∫ {u1≥u1} 〈 |∇u1|p1(x)−2∇u1 − |∇u1|p1(x)−2∇u1,∇(u1 − u1) 〉 . Using the facts that T2u2 ∈ [u2, u2], ui(x) > 0 a.e. in Ω, i = 1, j = 2, considering w = T2u2 and ϕ = (u1 − u1)+ in the first inequality of (1.3) and combining with equation (3.6) we obtain L1(u1 − u1)+ ≤ ∫ {u1≥u1} f1(x, u1, T2u2)(|u2| α1(x) Lq1(x) − |T2u2|α1(x) Lq1(x)) A(x, |T2u2|Lr1(x)) (u1 − u1) + ∫ {u1≥u1} g1(x, u1, T2u2)(|u2| γ1(x) Ls1(x) − |T2u2|γ1(x) Ls1(x)) A(x, |T2u2|Lr1(x)) (u1 − u1), which implies that∫ {u1≥u1} 〈 |∇u1|p1(x)−2∇u1 − |∇u1|p1(x)−2∇u1,∇(u1 − u1) 〉 ≤ 0. Therefore u1 ≤ u1. The same reasoning imply the other inequalities. Since ui ∈ [ui, ui], we have Tiui = ui. Therefore the pair (u1, u2) is a weak positive solution of (S). � 4. Applications In this section we apply Theorem 1.1 to some nonlocal problems. 4.1. A sublinear problem: In this section, we use Theorem 1.1 to study the nonlocal problem −A(x, |v|Lr1(x))∆p1(x)u = (uβ1(x) + vγ1(x))|v|α1(x) Lq1(x) in Ω, −A(x, |u|Lr2(x))∆p2(x)v = (uβ2(x) + vγ2(x))|u|α2(x) Lq2(x) in Ω, u = v = 0 on ∂Ω. (4.1) This problem with p1(x) ≡ p1(x) ≡ 2, was considered in [43]. The result in this section generalizes [43, Theorem 6]. Theorem 4.1. Suppose that pi, qi, ri, si, i = 1, 2 satisfy (H1) and αi, βi ∈ L∞(Ω), i = 1, 2. Assume also that 0 < α+ 1 + γ+ 1 < p−i − 1, 0 < α+ 1 p−2 − 1 + β+ 1 p−1 − 1 < 1, 0 < α+ 2 + γ+ 2 < p−i − 1, 0 < α+ 2 p−1 − 1 + β+ 2 p−2 − 1 < 1 for i = 1, 2. Let a0 > 0 be a positive constant. Suppose that one of the following two sets of conditions holds A(x, t) ≥ a0 in Ω× [0,∞), (4.2) or 0 < A(x, t) ≤ a0 in Ω× (0,∞) and lim t→+∞ A(x, t) = a∞ > 0 uniformly in Ω. (4.3) Then (4.1) has a positive solution. EJDE-2020/25 SUB-SUPER SOLUTION METHOD 9 Proof. Suppose that (4.2) holds. We will start by constructing (u, v). Let λ > 0 be a positive number, which will be chosen later and denote by zλ ∈W 1,p1(x) 0 (Ω)∩L∞(Ω) and yλ ∈W 1,p2(x) 0 (Ω) ∩ L∞(Ω) the unique solutions of (2.1) respectively. For λ > 0 sufficiently large it follows from Lemma 2.7 that there is a constant K > 1 that does not depend on λ such that 0 < zλ(x) ≤ Kλ 1 p − 1 −1 in Ω, (4.4) 0 < yλ(x) ≤ Kλ 1 p − 2 −1 in Ω. (4.5) Since α+ 1 + γ+ 1 < p−2 − 1 and α+ 1 p−2 −1 + β+ 1 p−1 −1 < 1, it is possible to choose λ > 1 such that (4.4), (4.5) and 1 a0 (Kβ+ 1 λ β + 1 p − 1 −1 + α + 1 p − 2 −1 +Kγ+ 1 λ α + 1 +γ + 1 p − 2 −1 ) max{|K|α − Lq1(x) , |K|α + Lq1(x)} ≤ λ (4.6) hold. By (4.4), (4.5) and (4.6), we obtain 1 a0 (z β1(x) λ + wγ1(x))|yλ|α1(x) Lq1(x) ≤ λ,w ∈ [0, yλ]. Thus for w ∈ [0, yλ] we obtain −∆p1(x)zλ ≥ 1 A(x, |w|Lr1(x)) (z β1(x) λ + wγ1(x))|yλ|α1(x) Lq1(x) in Ω, zλ = 0 on ∂Ω. Considering, if necessary, a larger λ > 0, the previous reasoning imply that −∆p2(x)yλ ≥ 1 A(x, |w|Lr2(x)) (wβ2(x) + yλ γ2(x))|zλ|α2(x) Lq2(x) in Ω, yλ = 0 on ∂Ω, for all w ∈ [0, zλ]. Now we construct (ui, vi), i = 1, 2. Since ∂Ω is C2, there is a constant δ > 0 such that d ∈ C2(Ω3δ) and |∇d(x)| ≡ 1, where d(x) := dist(x, ∂Ω) and Ω3δ := {x ∈ Ω; d(x) ≤ 3δ}. From [34, Page 12], we have that, for σ ∈ (0, δ) sufficiently small, the function φi = φi(k, σ), i = 1, 2 defined by φi(x) =  ekd(x) − 1 if d(x) < σ, ekσ − 1 + ∫ d(x) σ kekσ ( 2δ−t 2δ−σ ) 2 p − i −1 dt if σ ≤ d(x) < 2δ, ekσ − 1 + ∫ 2δ σ kekσ ( 2δ−t 2δ−σ ) 2 p − i −1 dt if 2δ ≤ d(x), belongs to C1 0 (Ω), where k > 0 is an arbitrary number and that −∆pi(x)(µφi) 10 G. C. G. DOS SANTOS, G. M. FIGUEIREDO, L. S. TAVARES EJDE-2020/25 =  −k(kµekd(x))pi(x)−1 [ (pi(x)− 1) + (d(x) + ln kµ k )∇pi(x)∇d(x) + ∆d(x) k ] if d(x) < σ,{ 1 2δ−σ 2(pi(x)−1) p−i −1 − ( 2δ−d(x) 2δ−σ )[ ln kµekσ ( 2δ−d(x) 2δ−σ ) 2 p − i −1∇pi(x)∇d(x) +∆d(x) ]} (kµekσ)pi(x)−1 ( 2δ−d(x) 2δ−σ ) 2(pi(x)−1) p − i −1 −1 if σ < d(x) < 2δ, 0 if 2δ < d(x), for all µ > 0 and i = 1, 2. Define Aλ := max { A(x, t) : (x, t) ∈ Ω × [ 0,max{|yλ|Lr1(x) |zλ|Lr2(x)} ]} . Then we have a0 ≤ A(x, |w|Lr1(x)) ≤ Aλ in Ω for all w ∈ [0, yλ]. Let σ = 1 k ln 2 and µ = e−ak where a = min{p−1 − 1, p−2 − 1} max{maxΩ |∇p1|+ 1,maxΩ |∇p2|+ 1} . Then ekσ = 2 and kµ ≤ 1 if k > 0 is sufficiently large. Let x ∈ Ω with d(x) < σ. If k > 0 is large enough we have |∇d(x)| = 1 and then∣∣d(x) + ln(kµ) k ∣∣|∇p1(x)||∇d(x)| ≤ ( |d(x)|+ | ln(kµ)| k ) |∇p1(x)| ≤ ( σ − ln(kµ) k ) |∇p1(x)| = ( ln 2 k − ln k k ) |∇p1(x)|+ a|∇p1(x)| < p−1 − 1. (4.7) Note also that there exists a constant A > 0, that does not depend on k, such that |∆d(x)| < A for all x ∈ ∂Ω3δ. Using the last inequality and the expression of −∆p1(x)(µφ), we obtain −∆p1(x)(µφ1) ≤ 0 for x ∈ Ω with d(x) < σ or d(x) > 2δ for k > 0 large enough. Therefore −∆p1(x)(µφ1) ≤ 0 ≤ 1 Aλ (µφ1)β1(x)|µφ2|α1(x) Lq1(x) ≤ 1 Aλ ((µφ1)β1(x) + wγ1(x))|µφ2|α1(x) Lq1(x) for all w ∈ L∞(Ω) with w ≥ µφ2 and d(x) < σ or 2δ < d(x). Using the idea in the proof of [34, estimate (3.10)] we obtain −∆p1(x)(µφ1) ≤ C̃(kµ)p − 1 −1| ln kµ| = C̃(kµ)p − 1 −1 ∣∣ ln k eak ∣∣ if σ < d(x) < 2δ. (4.8) From the proof of [44, Theorem 2] and the fact that α+ 1 + γ+ 1 < p−1 − 1 we obtain lim k→+∞ C̃kp − 1 −1 eak(p−1 −1−(α+ 1 +γ+ 1 )) ∣∣ ln k eak ∣∣ = 0. (4.9) Note that φ1(x) ≥ 1 if σ ≤ d(x) < 2δ because φ1(x) ≥ ekσ − 1 and ekσ = 2 for all k > 0. Thus, there is a constant C0 > 0 that does not depend on k such that EJDE-2020/25 SUB-SUPER SOLUTION METHOD 11 |φ2|α1(x) Lq1(x)(Ω) ≥ C0 if σ < d(x) < 2δ. By (4.9), we can choose k > 0 large enough such that C̃kp − 1 −1 eak[(p−1 −1)−(α+ 1 +β+ 1 )] ∣∣ ln k eak ∣∣ ≤ C0 Aλ . (4.10) Therefore from (4.8) and (4.10) we have −∆p1(x)(µφ1) ≤ 1 Aλ ((µφ1)β1(x) + wγ1(x))|µφ2|α1(x) Lq1(x) , for all w ∈ L∞(Ω) with w ≥ µφ2 and σ < d(x) < 2δ for k > 0 large enough. Thus it is possible to conclude that −∆p1(x)(µφ1) ≤ 1 Aλ ((µφ1)β1(x) + wγ1(x))|µφ2|α1(x) Lq1(x) in Ω. Fix k > 0 satisfying the above property and −∆p1(x)(µφ1) ≤ 1. For λ > 1 we have −∆p1(x)(µφ1) ≤ −∆p1(x)zλ. Therefore µφ1 ≤ zλ. Since α+ 2 + γ+ 2 < p−2 − 1, a similar reasoning imply that there is µ > 0 small enough such that −∆p2(x)(µφ2) ≤ 1 A(x, |w|Lr2 (x)) (wβ2 + (µφ2)γ2)|µφ1|α2(x) Lq2(x)(Ω) in Ω for all w ∈ L∞(Ω) with w ≥ µφ1 and that µ2φ ≤ yλ. The first part of the result is proved. Now suppose that 0 < A(x, t) ≤ a0 in Ω× (0,∞). Let δ, σ, µ, a, λ, zλ, yλ and φi for i = 1, 2 as before. From the previous arguments there exist k > 0 large enough and µ > 0 small such that −∆p1(x)(µφ1) ≤ 1, −∆p1(x)(µφ) ≤ 1 a0 ((µφ1)β1(x) + wγ1(x))|µφ2|α1(x) Lq1(x) (4.11) in Ω for all w ∈ [µφ2, yλ], and −∆p2(x)(µφ2) ≤ 1, −∆p2(x)(µφ2) ≤ 1 a0 (wβ2(x) + (µφ2)γ2(x))|µφ1|α2(x) Lq2(x) (4.12) in Ω for all w ∈ [µφ1, zλ]. Since limt→∞A(x, t) = a∞ > 0 uniformly in Ω there is a large constant a1 > 0 such that A(x, t) ≥ a∞ 2 on Ω× (a1,∞). Let mk := min { A(x, t) : (x, t) ∈ Ω× [min{|µφ1|Lr1(x) , |µφ2|Lr2(x)}, a1] } > 0 and Ak := min { mk, a∞ 2 } . Then we have A(x, t) ≥ Ak in Ω× [min{|µφ1|Lr1(x) , |µφ2|Lr2(x)},∞). Fix k > 0 satisfying (4.11) and (4.12). Consider λ > 1 such that (4.4), (4.5) and 1 Ak ( Kβ+ 1 λ β + 1 p − 1 −1 + α + 1 p − 2 −1 +Kγ+ 1 λ α + 1 +γ + 1 p − 2 −1 ) max{|K|α − 1 Lq1(x) , |K| α+ 1 Lq1(x)} ≤ λ, 1 Ak ( Kβ+ 2 λ β + 2 +α + 2 p − 1 −1 +Kγ+ 2 λ γ + 2 p − 2 −1 + α + 2 p − 1 −1 ) max{|K|α + 2 Lq2(x) , |K| α−2 Lq2(x)} ≤ λ, where K > 1 is a constant that does not depend on k or λ (see Lemma 2.7). Therefore, −∆p1(x)zλ ≤ 1 A(x, |w|Lr1(x)) (z β1(x) λ + wγ1(x))|yλ|α1(x) Lq1(x) in Ω, w ∈ [µφ2, yλ]. 12 G. C. G. DOS SANTOS, G. M. FIGUEIREDO, L. S. TAVARES EJDE-2020/25 Arguing as before and considering a suitable choice for λ and k we obtain −∆p2(x)yλ ≤ 1 A(x, |w|Lr2(x)) (wβ2(x) + y β2(x) λ )|zλ|α2(x) Lq2(x) in Ω, w ∈ [µφ1, zλ]. The comparison principle implies that µφ1 ≤ zλ and µφ2 ≤ yλ if µ is small. The proof is complete. � 4.2. A concave-convex problem. In this section we consider the following non- local problem with concave-convex nonlinearities −A(x, |v|Lr1(x))∆p1(x)u = λ|u|β1(x)−1u|v|α1(x) Lq1(x) + θ|v|η1(x)−1v|v|γ1(x) Ls1(x) in Ω, −A(x, |u|Lr2(x))∆p2(x)v = λ|v|β2(x)−1v|u|α2(x) Lq2(x) + θ|u|η2(x)−1u|u|γ2(x) Ls2(x) in Ω, u = v = 0 on ∂Ω. (4.13) The scalar and local version of (4.13) with p(x) ≡ 2 and constant exponents was considered in the famous paper by Ambrosetti-Brezis-Cerami [5] in which a sub- supersolution argument is used. In [43], problem (4.13) was studied with p(x) ≡ 2. The following result generalizes [43, Theorem 7]. Theorem 4.2. Suppose that ri, pi, qi, si, αi, ηi satisfy (H1) for i = 1, 2 and that βi ∈ L∞(Ω), i = 1, 2 are nonnegative functions with 0 < α−i + β−i ≤ α+ i + β+ i < p−i − 1, i = 1, 2. Let a0, b0 > 0 be positive numbers. Then the following assertions hold (1) If p+ 2 −1 < η−1 +γ−1 , p+ 1 −1 < η−2 +γ−2 and A(x, t) ≥ a0 in Ω× [0, b0], then for each θ > 0 there exists λ0 > 0 such that for each λ ∈ (0, λ0), problem (4.13) has a positive solution uλ,θ. (2) p+ 2 − 1 < η−1 + γ−1 , p+ 1 − 1 < η−2 + γ−2 and β+ 1 p−1 − 1 + α+ 1 p−2 − 1 < 1, β+ 2 p−2 − 1 + α+ 2 p−1 − 1 < 1 . Suppose that 0 < A(x, t) ≤ a0 in Ω× (0,∞) and limt→∞A(x, t) = b0 uniformly in Ω. Then given a λ > 0, there exists θ0 > 0 such that for each θ ∈ (0, θ0), problem (4.13) has a positive solution uλ,θ. Proof. Suppose that (1) occurs. Consider zλ ∈ W 1,p1(x) 0 (Ω) ∩ L∞(Ω) and yλ ∈ W 1,p2(x) 0 (Ω) ∩ L∞(Ω) the unique solutions of (2.1) respectively, where λ ∈ (0, 1) will be chosen later. Lemma 2.7 imply that for λ > 0 small enough there exists a constant K > 1 that does not depend on λ such that 0 < zλ(x) ≤ Kλ 1 p + 1 −1 in Ω, (4.14) 0 < yλ(x) ≤ Kλ 1 p + 2 −1 in Ω. (4.15) To construct ui we will prove, for each θ > 0, that there exists λ0 > 0 such that 1 a0 ( λ|zλ|β1(x)−1zλ|yλ|α1(x) Lq1(x) + θ|w|η1(x)−1w|yλ|γ1(x) Ls1(x) ) ≤ λ, ∀w ∈ [0, yλ], (4.16) 1 a0 ( λ|yλ|β2(x)−1yλ|zλ|α2(x) Lq2(x) + θ|w|η2(x)−1w|zλ|γ2(x) Ls2(x) ) ≤ λ, ∀w ∈ [0, zλ]. (4.17) EJDE-2020/25 SUB-SUPER SOLUTION METHOD 13 Let K := max i=1,2 { Kβ+ i |K|α + i Lqi(x) ,Kβ+ i |K|α − i Lqi(x) ,Kη+i |K|γ + i Lsi(x) ,Kη+i |K|γ − i Lsi(x) } . (4.18) Since 0 < α−1 + β−1 and p+ 2 − 1 < η−1 + γ−1 , there exists λ0 > 0 such that 1 a0 ( λ p + 1 −1+β − 1 p + 1 −1 + α − 1 p + 2 −1K + θλ η − 1 +γ − 1 p + 2 −1 K ) ≤ λ, (4.19) for all λ ∈ (0, λ0). If necessary, we consider small λ0 > 0 such that |yλ|Lr1(x) ≤ |K|Lr1(x)λ 1 p + 2 −1 ≤ b0 for all λ ∈ (0, λ0). Therefore A(x, |w|Lr1(x)) ≥ a0, w ∈ [0, yλ]. It follows from (4.14), (4.15) and (4.19) that (4.16) holds. Then we can conclude that −∆p1(x)zλ ≥ 1 A(x, |w|Lr1(x)) ( λzλ β1(x)|yλ|α1(x) Lq1(x) + θwη1(x)|yλ|γ1(x) Ls1(x) ) , (4.20) for all w ∈ [0, yλ]. Assume also that λ0 satisfies 1 a0 ( λ p + 2 −1+β − 2 p + 2 −1 + α − 2 p + 1 −1K + θλ η − 2 +γ − 2 p + 1 −1 K ) ≤ λ (4.21) and |zλ|Lr2(x) ≤ |K|Lr2(x)λ 1 p + 1 −1 ≤ b0 for all λ ∈ (0, λ0). Therefore A(x, |w|Lr2(x)) ≥ a0, w ∈ [0, zλ]. Thus from (4.14), (4.15) and (4.21) we have that (4.17) holds. Then we can conclude that −∆p2(x)yλ ≥ 1 A(x, |w|Lr2(x)) ( λzλ β2(x)|zλ|α2(x) Lq2(x) + θwη2(x)|zλ|γ2(x) Ls2(x) ) (4.22) for all w ∈ [0, zλ]. To construct ui consider φi, δ, σ, µ as in the proof of Theorem 4.1. Using the inequalities α+ i + β+ i < p−i − 1, i = 1, 2 and repeating the arguments of Theorem 4.1, we have that exists a number µ > 0 such that µφ1 ≤ zλ, µφ2 ≤ yλ, −∆p1(x)(µφ1) ≤ λ, −∆p1(x)(µφ1) ≤ 1 A(x, |w|Lr1(x)) ( λ(µφ1)β1(x)|µφ1|α1(x) Lq1(x) + θwη1(x)|µφ2|γ1(x) Ls1(x) ) , for all w ∈ [µφ2, yλ] and −∆p2(x)(µφ2) ≤ λ, −∆p2(x)(µφ2) ≤ 1 A(x, |w|Lr2(x)) ( λ(µφ2)β2(x)|µφ1|α2(x) Lq2(x) + θwη2(x)|µφ1|γ2(x) Ls2(x) ) , for all w ∈ [µφ2, zλ]. Then by Theorem 1.1 we have the desired result. Now we consider the condition (2). Let φi, δ and σi, i = 1, 2 as in the first part of the result and let λ > 0 fixed. Since α+ i + β+ i < p−i − 1, i = 1, 2 there exists µ > 0 depending only on λ such that −∆pi(x)(µφi) ≤ 1, −∆pi(x)(µφ) ≤ 1 a0 λ(µφi) βi(x)|µφj |αi(x), for w ∈ L∞(Ω) with w ≥ µφj , i 6= j and i, j = 1, 2. Let M > 0 that will be chosen later and assume zM ∈W 1,p1(x) 0 (Ω)∩L∞(Ω) is a solution of −∆p1(x)zM = M in Ω, 14 G. C. G. DOS SANTOS, G. M. FIGUEIREDO, L. S. TAVARES EJDE-2020/25 zM = 0 on ∂Ω, and yM ∈W 1,p2(x) 0 (Ω) ∩ L∞(Ω) is a solutions of −∆p2(x)yM = M in Ω, yM = 0 on ∂Ω . For M large enough from Lemma 2.7, there exists a constant K > 1 that does not depend on M such that 0 < zM (x) ≤ KM 1 p − 1 −1 in Ω, (4.23) 0 < yM (x) ≤ KM 1 p − 2 −1 in Ω. (4.24) To construct ui we will show that exist θ0 > 0 depending on λ with the following property: if we assume θ ∈ (0, θ0) then there is a constant M depending only on λ and θ satisfying M ≥ 1 A(x, |w|Lr1(x)) ( λzM β1(x)|yM |α1(x) Lq1(x) + θwη1(x)|yM |γ1(x) Ls1(x) ) , (4.25) for w ∈ [µφ2, yM ], and M ≥ 1 A(x, |w|Lr2(x)) ( λyM β2(x)|zM |α2(x) Lq2(x) + θwη2(x)|zM |γ2(x) Ls2(x) ) , (4.26) for w ∈ [µφ1, zM ]. Since A is continuous and limt→+∞A(x, t) = b0 > 0 uniformly in Ω, there exists a1 > 0 large enough such that A(x, t) ≥ b0 2 in Ω× (a1,+∞). Define mλ := {A(x, t) : (x, t) ∈ Ω× [min{|µφ1|Lr1(x) , |µφ2|Lr2(x)}, a1]} andAλ := min{mλ, b0 2 }. ThenA(x, t) ≥ Aλ in Ω×[min{|µφ1|Lr1(x) , |µφ2|Lr2(x)},∞). Thus Aλ ≤ A(x, |w|Lr1(x)) ≤ a0 for all w ∈ L∞(Ω) with µφ1 ≤ w or µφ2 ≤ w. Note that from (4.23) and (4.24) the inequalities (4.25) and (4.26) hold if we have simul- taneously the inequalities 1 Aλ ( λKM β + 1 p − 1 −1 + α + 1 p − 2 −1 + θKM η + 1 +γ + 1 p − 2 −1 ) ≤M, 1 Aλ ( λKM β + 2 p − 2 −1 + α + 2 p − 1 −1 + θKM η + 2 +γ + 2 p − 1 −1 ) ≤M, where K is given by (4.18). To obtain such inequalities we will study the inequality 1 Aλ ( λKMρ−1 + θKMτ−1 ) ≤ 1 (4.27) where ρ := max { β+ 1 p−1 − 1 + α+ 1 p−2 − 1 , β+ 2 p−2 − 1 + α+ 2 p−1 − 1 } , τ := max {η+ 1 + γ+ 1 p−2 − 1 , η+ 2 + γ+ 2 p−1 − 1 } . Define Ψλ,θ(M) := λK Aλ Mρ−1 + θK Aλ Mτ−1, M > 0. EJDE-2020/25 SUB-SUPER SOLUTION METHOD 15 Since 0 < ρ < 1 and τ > 1 we have limM→0+ Ψλ,θ(M) = limM→+∞Ψλ,θ(M) = +∞. Note that Ψλ,θ ′(M) = 0 if, and only if M = Mλ,θ := (λ θ ) 1 τ−ρ c, c := (1− ρ τ − 1 ) 1 τ−ρ . (4.28) From the above properties of Ψλ,µ we have that the global minimum of Ψλ,θ is attained at Mλ,θ. The inequality (4.27) is equivalent to finding Mλ,θ > 0 such that Ψλ,θ(Mλ,θ) ≤ 1. By (4.28), we have that Ψλ,θ(Mλ,θ) ≤ 1, if and only if λK Aλ (λ θ ) ρ−1 τ−ρ cρ−1 + θ1−( τ−1 τ−ρ ) K Aλ λ τ−1 τ−ρ cτ−1 ≤ 1. (4.29) Thus from (4.28) and (4.29), we have that given λ > 0 there exists θ0 > 0 such that for each θ ∈ (0, θ0) there exists Mλ,θ satisfying Mλ,θ ≥ 1 and 1 Aλ ( λKMλ,θ ρ−1 + θKMλ,θ τ−1 ) ≤ 1. Therefore, −∆p1(x)zM ≥ 1 Aλ ( λzM β1(x)|yM |α1(x) Lq1(x) + θwη1(x)|yM |γ1(x) Ls1(x) ) in Ω, for all w ∈ [µφ2, yM ], and −∆p2(x)yM ≥ 1 Aλ ( λyM β2(x)|zM |α2(x) Lq2(x) + µwη2(x)|zM |γw(x) Lsw(x) ) in Ω, for all w ∈ [µφ1, zM ]. Since Mλ,θ → +∞ as θ → 0+ and the map θ 7−→Mλ,θ is decreasing we have −∆p1(x)(µφ1) ≤ 1 ≤Mλ,θ0 ≤Mλ,θ, θ ∈ (0, θ0) for θ0 small enough. Similarly, we have −∆p2(x)(µφ2) ≤ Mλ,θ0 ≤ Mλ,θ for all θ ∈ (0, θ0), for θ0 small. The weak maximum principle imply that µφ1 ≤ zM and µφ2 ≤ yM . The proof is complete. � 4.3. A generalization of the logistic equation. In the previous sections, we considered at least one of the conditions A(x, t) ≥ a0 > 0 or 0 < A(x, t) ≤ a∞, t > 0. In this section we study a generalization of the classic logistic equation where the function A(x, t) satisfies A(x, 0) ≥ 0, lim t→0+ A(x, t) =∞, and lim t→+∞ A(x, t) = ±∞. We consider the problem −A(x, |v|Lr1(x))∆p1(x)u = λf1(u)|v|α1(x) Lq1(x) in Ω, −A(x, |u|Lr2(x))∆p2(x)v = λf2(v)|u|α2(x) Lq2(x) in Ω, u = v = 0 on ∂Ω. (4.30) We suppose that there are numbers θi > 0, i = 1, 2 such that the functions fi : [0,∞)→ R satisfy the following conditions: (H2) fi ∈ C0([0, θi],R), i = 1, 2; (H3) fi(0) = fi(θi) = 0, fi(t) > 0 in (0, θi) for i = 1, 2. Problem (4.30) is a generalization of the problemes studied in [16, 18, 43]. The next result generalizes [43, Theorem 8]. 16 G. C. G. DOS SANTOS, G. M. FIGUEIREDO, L. S. TAVARES EJDE-2020/25 Theorem 4.3. Suppose that ri, pi, qi, αi satisfy (H1). Also that fi, i = 1, 2 satisfies (H2), (H3) and that A(x, t) > 0 in Ω × ( 0,max{|θ1|Lr2(x) , |θ2|Lr1(x)} ] . Then there exists λ0 > 0 such that (4.30) has a positive solution for λ ≥ λ0. Proof. Consider the functions f̃i(t) = fi(t) for t ∈ [0, θi], and f̃i(t) = 0 for t ∈ R \ [0, θi], i = 1, 2. The functional Jλ(u, v) = ∫ Ω 1 p1(x) |∇u|p1(x)dx− λ ∫ Ω F̃1(u)dx+ ∫ Ω 1 p2(x) |∇v|p2(x)dx− λ ∫ Ω F̃2(v)dx := J1,λ(u) + J2,λ(v), where F̃i(t) = ∫ t 0 f̃i(s)ds is of class C1(W 1,p1(x) 0 ×W 1,p2(x) 0 (Ω),R) and W 1,p1(x) 0 (Ω)× W 1,p2(x) 0 (Ω) is a Banach space endowed with the norm |(u, v)| := max{|∇u|p1(x), |∇v|p2(x)}. Since |f̃i(t)| ≤ C, t ∈ R for some constant which does not depends on i = 1, 2 we have that J is coercive. Thus J has a minimum (zλ, wλ) ∈W 1,p1(x) 0 (Ω)×W 1,p2(x) 0 (Ω) with −∆p1(x)zλ = λf̃1(zλ) in Ω, zλ = 0 on ∂Ω, (4.31) and −∆p2(x)wλ = λf̃2(wλ) in Ω, wλ = 0 on ∂Ω. (4.32) Note that the unique solutions of (4.31) and (4.32) are given by the minimizers of functionals J1,λ and J2,λ respectively. Consider a function ϕ0 ∈ W 1,pi(x) 0 (Ω), i = 1, 2 with F̃i(ϕ0) > 0, i = 1, 2. Define (z0, w0) := (zλ̃0 , wλ̃0 ), where λ̃0 satisfies∫ Ω 1 pi(x) |∇ϕ0|pi(x)dx < λ̃0 ∫ Ω F̃i(ϕ0)dx, i = 1, 2. We have J1,λ̃0 (z0) ≤ J1,λ̃0 (ϕ0) < 0 and that J2,λ̃0 (z0) < 0. Therefore z0 6= 0 and w0 6= 0. Since −∆p1(x)z0 and −∆p2(x)w0 are nonnegative, we have z0, w0 > 0 in Ω. Note that by [28, Theorem 4.1] and [25, Theorem 1.2], we obtain that z0, w0 ∈ C1,α(Ω) for some α ∈ (0, 1]. Using the test function ϕ = (z0 − θ1)+ ∈W 1,p1(x) 0 (Ω) in (4.31) we obtain∫ Ω |∇z0|p1(x)−2∇z0∇(z0 − θ1)+dx = λ̃0 ∫ {z0>θ} f̃1(z0)(z0 − θ1)dx = 0. Therefore,∫ {z0>θ} 〈 |∇z0|p(x)−2∇z0 − |∇θ1|p1(x)−2∇θ1,∇(z0 − θ1) 〉 dx = 0, which imply (z0 − θ1)+ = 0 in Ω. Thus 0 < z0 ≤ θ1. A similar reasoning provides 0 < w0 ≤ θ2. Note that there is a constant C > 0 such that |z0|α1(x) Lq1(x) , |w0|α2(x) Lq2(x) ≥ C. We define A0 = max { A(x, t) : (x, t) ∈ Ω× [min{|z0|Lr2(x) , |w0|Lr1(x)}, EJDE-2020/25 SUB-SUPER SOLUTION METHOD 17 max{|θ1|Lr2(x) , |θ2|Lr1(x)} } and µ0 = A0 C . Then, we have −∆p1(x)z0 = λ̃0f1(z0) = 1 A0 λ̃0µ0f1(z0)|w0|α1(x) Lq1(x) A0 µ0|z0|α1(x) Lq1(x) ≤ 1 A0 λ̃0µ0f1(z0)|w0|α1(x) Lq1(x) . Thus for each λ ≥ λ0 := λ̃0µ0 and w ∈ [w0, θ2], we obtain −∆p1(x)z0 ≤ 1 A(x, |w|Lr1(x)) λf1(z0)|w0|α1(x) Lq1(x) . If necessary, we can consider a larger λ0 > 0 such that −∆p2(x)w0 ≤ 1 A(x, |w|Lr2(x)) λf2(w0)|z0|α2(x) Lq2(x) , for all λ ≥ λ0 and w ∈ [z0, θ1]. Since fi(θi) = 0, i = 1, 2, we have that (z0, θ1) and (w0, θ2) are sub-super solutions pairs for (4.30). The proof is complete. � We remark that is possible to use the functions φi from the proof of Theorem 4.1 for problem (4.30). However, more restrictions on the functions pi, fi, i = 1, 2 are needed. Acknowledgements. G. M. 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Tavares Universidade Federal do Cariri, Centro de Ciências e Tecnologia, CEP: 63048-080, Juazeiro do Norte - CE, Brazil Email address: leandro.tavares@ufca.edu.br 1. Introduction 2. Preliminaries 3. Proof of main results 4. Applications 4.1. A sublinear problem: 4.2. A concave-convex problem 4.3. A generalization of the logistic equation Acknowledgements References