Positive solution for the (p, q)-Laplacian systems by a new version of sub-super solution method

Abstract

Purpose

In this paper, the authors give a new version of the sub-super solution method and prove the existence of positive solution for a (p, q)-Laplacian system under weak assumptions than usually made in such systems. In particular, nonlinearities need not be monotone or positive.

Design/methodology/approach

The authors prove that the sub-super solution method can be proved by the Shcauder fixed-point theorem and use the method to prove the existence of a positive solution in elliptic systems, which appear in some problems of population dynamics.

Findings

The results complement and generalize some results already published for similar problems.

Originality/value

The result is completely new and does not appear elsewhere and will be a reference for this line of research.

Citation

Moussa, M., Moussa, A. and Mazan, H. (2023), "Positive solution for the (p, q)-Laplacian systems by a new version of sub-super solution method", Arab Journal of Mathematical Sciences, Vol. 29 No. 2, pp. 145-153. https://doi.org/10.1108/AJMS-03-2021-0060

1. Introduction

Consider the following (p₁, p₂)-Laplacian system,

(1)−Δp1u1=μ1F1(x,u1,u2)   in  Ω−Δp2u2=μ2F2(x,u1,u2)   in  Ωu1=u2=0              on     ∂Ω

Ω is an open bounded domain of RN with smooth boundary ∂Ω. For i = 1, 2, Δpi=div(|∇ui|pi−2∇ui) is the p_i-Laplacian operator, p_i > 1, μ_i is a positive parameter and Fi:Ω¯×R×R→R is a continuous function.

Many authors have been interested by the problem (1) in different ways [1–4]. The sub-super solution method, given in [5] by using a monotony argument, is the principal tool used to prove the existence of solution of the problem (1) in [1, 3, 4]. Recently, a new version of the method of the sub-super solution is given to prove the existence of solution for the (p(x), q(x))-Laplacian systems by using the Schaefer’s fixed-point theorem [6].

Our main contribution in this article is, in first, to give a new version of the sub-super solution method based on Schauder’s famous fixed-point theorem and, in second, use the method to prove the existence of a positive solution of problem (1) under the continuity assumptions on functions F and G. The functions F and G need not to be nondecreasing as in [1, 4].

Recall that the sub-super solution method is a topological method, which does not require strong regularity assumptions as the variational method.

The paper is organized as follows: in Section 2, we present some preliminary results and our main results. In Section 3, we study some general problems studied previously. We end our paper by studying some concrete examples.

2. Preliminaries and main results

We start by the definition of sub-super solution of the problem (1).

Definition 2.1.

We say that (u̲1,u¯1),(u̲2,u¯2)∈W1,p1(Ω)∩L∞(Ω)×W1,p2(Ω)∩L∞(Ω) is a pair of sub-super solution of the problem (1) if they satisfy

H1.

u̲i≤u¯i a.e in Ω and u̲i≤0≤u¯i on ∂Ω for i = 1, 2.

H2.

−Δp1u̲1−F1(x,u̲1,v)≤0≤−Δp1u¯1−F1(x,u¯1,v), ∀v∈[u̲2,u¯2],

H3.

−Δp2u̲2−F2(x,u,u̲2)≤0≤−Δp2u¯2−F2(x,u,u¯2), ∀u∈[u̲1,u¯1].

Inequalities in H1 and H2 are in the weak sense. Where, for u ≤ v a.e. in Ω,

[u, v] = {z : u(x) ≤ z(x) ≤ v(x), a.e. ∈ Ω}.

Theorem 2.1.

For i = 1, 2, assume that F_i is continuous in Ω¯×R2. Then, if there exists a pair of sub-super solution of (1) in the sense of Definition (2.1), system (1) has a positive weak solution (u1,u2)∈[u̲1,u¯1]×[u̲2,u¯2].

Proof.

Consider, for i = 1, 2, the truncation operators, Ti:Lpi(Ω)→Lpi(Ω) defined by

Ti(zi)(x)=u̲i(x)ifzi(x)≤u̲i(x)zi(x)ifu̲i(x)≤zi(x)≤u¯i(x)u¯i(x)ifzi(x)≥u¯i(x)

then ∀zi∈Lpi(Ω), Ti(zi)∈[u̲i,u¯i] and ‖Ti(zi)‖∞∈[0,‖u¯i‖∞]. By the continuity of F_i, there exists a positive constant C_i such that

|Fix,T1(z1)(x),T2(z2)(x)|≤Ci.

Let Fi:Lpi(Ω)×Lpi(Ω)→Lpi(Ω) be the Nemytskii operator defined by

Fi(u1,u2)(x)=Fi(x,T1(u1)(x),T2(u2)(x)).

Then, Fi is Lpi(Ω)-bounded. By the dominate convergence theorem and the continuity of F_i, we conclude the continuity of Fi, and we have ∀(u1,u2)∈Lp1(Ω)×Lp2(Ω).

(2)‖Fi(u1,u2)‖∞≤Ci.

Now, fix (z1,z2)∈Lp1(Ω)×Lp2(Ω), there exists a unique pair (w1,w2)∈W01,p1(Ω)×W01,p2(Ω) solution of the problem,

(3)−Δp1w1=F1(T1(z1),T2(z2))   in  Ω−Δp2w2=F2(T1(z1),T2(z2))   in  Ωw1=w2=0                     on  ∂Ω

Therefore, we can define the operator S:Lp1(Ω)×Lp2(Ω)→Lp1(Ω)×Lp2(Ω) by S(z₁, z₂) = (w₁, w₂), where (w₁, w₂) is the unique solution of problem (3).

S is a compact operator. Indeed, let (z_1,n, z_2,n) be a bounded sequence in Lp1(Ω)×Lp2(Ω) and (w_1,n, w_2,n) = S(z_1,n, z_2,n), then ∀ϕi∈W01,pi(Ω)

∫Ω|∇wi,n|pi−2∇wi,n.∇ϕi=∫ΩFi(T1(z1,n),T2(z2,n))ϕi.

If the test function ϕ_i = w_i,n, by the Sobolev embedding theorem, there exists some constants K_i such tat

(4)‖wi,n‖W01,pi(Ω)pi=∫Ω|∇wi,n|pi=∫ΩF(T1(z1,n),T2(z2,n))wi,n≤Ci‖wi,n‖L1(Ω)≤Ki‖wi,n‖W01,pi(Ω).

Then, (w_1,n, w_2,n) is bounded in W01,p1(Ω)×W01,p2(Ω). By the compact embedding, there exists a convergent sub-sequence of (w_1,n, w_2,n) in Lp1(Ω)×Lp2(Ω). So, S is compact.

From (4), there exists L_i > 0 such that

(5)‖wi,n‖W01,pi(Ω)≤K¯i=Ki1pi−1⇒‖wi,n‖Lpi(Ω)≤Li.

Remark that in (2), (4), and (5), the constants are independent of the choice of (z₁, z₂). Then,

S(Lp1(Ω)×Lp2(Ω))⊂BLp1(Ω)×Lp2(Ω)(0,L¯).

for some L¯>0. By the Schauder fixed-point theorem, in BLp1(Ω)×Lp2(Ω)(0,L¯), there exists a unique (u1,u2)∈Lp1(Ω)×Lp2(Ω) such that S(u₁, u₂) = (u₁, u₂).

(6)−Δp1u1=F(T1(u1),T2(u2))in  Ω−Δp2u2=F2(T1(u1),T2(u2))in  Ωu1=u2=0on     ∂Ω

Finally, (u₁, u₂) is a solution of problem (1) if, and only if, T₁(u₁) = u₁ and T₂(u₂) = u_2, which means that u̲1≤u1≤u¯1 and u̲2≤u2≤u¯2. We need to prove that (u̲1−u1)+=0, (u1−u¯1)+=0, (u̲2−u2)+=0 and (u2−u¯2)+=0. Let us prove, for example, that (u̲1−u1)+=0. The same argument works for the others cases.

Let Ω+={x∈Ω,u̲1(x)>u1(x)}. Since (u̲1,u̲2) is a sub-solution, then for ϕ=(u̲1−u1)+ and v = T₂(u₂), we have,

∫Ω|∇u̲1|p1−2∇u̲1.∇(u̲1−u1)+≤∫ΩF1x,u̲1,T2(u2)(u̲1−u1)+

and as (u₁, u₂) is a solution of (6),

∫Ω|∇u1|p1−2∇u1.∇(u̲1−u1)+=∫ΩF1x,T1(u1),T2(u2)(u̲1−u1)+

Then,

∫Ω|∇u̲1|p1−2∇u̲1−|∇u1|p1−2∇u1.∇(u̲1−u1)+≤∫ΩF1(x,u̲1,T2(u2))−F1x,T1(u1),T2(u2)(u̲1−u1)+.

Remark that in Ω⁺, T1(u1)=u̲1. So,

∫Ω+|∇u̲1|p1−2∇u̲1−|∇u1|p1−2∇u1.∇(u̲1−u1)+=∫Ω+|∇u̲1|p1−2∇u̲1−|∇u1|p1−2∇u1.∇u̲1−∇u1≤ 0.

Therefore, by the monotonicity of the p₁-Laplacian, u̲1−u1=0 in Ω⁺. Then, u̲1≤u1 in Ω. In the same way, we get u1≤u¯1 in Ω. Then, u̲1≤u1≤u¯1 and u̲2≤u2≤u¯2⇒T1(u1)=u1 and T₂(u₂) = u₂. Finally, (u₁, u₂) is a solution of the problem (1). □

3. Applications

Theorem 3.1.

Consider system (1) and assume that for i = 1, 2,

A.1 ∃ C_i, α_i, β_i > 0, such that |Fi(x,s1,s2)|≤Ci1+|s1|αi+|s2|βi,

A.2 F₁(x, 0, 0) + F₂ (x, 0, 0) > 0 a.e. x ∈ Ω.

Then,

1. If max(α_i, β_i) < p_i − 1, for i = 1, 2, then ∀μ_i > 0, there exists a weak positive solution of problem (1).

2. If min(α_i, β_i) ≥ p_i − 1, for i = 1, 2, then, there exists positive numbers μ¯i such that ∀(μ1,μ2)∈]0,μ¯1]×]0,μ¯2] the problem (1) has at least a weak positive solution (u₁, u₂) such that ‖u_i‖_∞ ≤ ‖e_i‖_∞. e_i is the unique solution of problem (7).

3. If α_i + β_i < p_i − 1 and α_j + β_j ≥ p_j − 1. j = 2 if i = 1 and j = 1 if i = 2. Then there exists μ¯j such that problem (1) has at least a weak positive solution ∀μ_i > 0 and 0<μj≤μ¯j.

Proof.

By A.2, (0, 0) is a sub-solution, but not a solution, of problem (1). By Theorem 2.1, we need to find a super solution of problem. Let e_i be the unique positive solution of the Dirichlet boundary condition problem,

(7) −Δpiu=1  in     Ωu=0  on   ∂Ω

1. In the sub-linear case, as   max(α_i, β_i) < p_i − 1, for i = 1, 2, there exists K > 1, large enough, such that

   −Δp1(Ke1)=Kp1−1≥ μ1C11+Kα1‖e1‖∞α1+Kβ1‖e2‖∞β1≥ μ1C11+Kα1‖e1‖∞α1+‖v‖∞β1,≥ μ1F1x,Ke1,v,∀v∈[0,Ke2]−Δp2(Ke2)=Kp2−1≥ μ2C21+Kα2‖e1‖∞α2+Kβ2‖e2‖∞β2≥ μ2C21+‖u‖∞α2+Kβ2‖e2‖∞β2,≥ μ2F2x,u,Ke2,∀u∈[0,Ke1]

   in the weak sense. So, (u¯1,u¯2)=(Ke1,Ke2) is a super-solution of the problem (1). Therefore, there exists (u₁, u₂) ∈ [0, Ke₁] × [0, Ke₂] solution of system (1).

2. In the super-linear case, for i = 1, 2, min(α_i, β_i) ≥ p_i − 1. Put μ¯i=1Ci1+‖e1‖∞αi+‖e2‖∞βi and let 0<μi≤μ¯i. Then,

   −Δp1e1=1= μ¯1C11+‖e1‖∞α1+‖e2‖∞β1≥ μ1C11+‖e1‖∞α1+‖e2‖∞β1≥ μ1C11+‖e1‖∞α1+vβ1≥ μ1F1(x,e1,v),∀v∈[0,e2]−Δp2e2=1=μ¯2C21+‖e1‖∞α2+‖e2‖∞β2≥ μ2C21+‖e1‖∞α2+‖e2‖∞β2≥ μ2C21+uα2+‖e2‖∞β2≥ μ2F2(x,u,e2),∀u∈[0,e1].

   (e₁, e₂) is a super-solution, and we deduce that the problem (1) admits a weak positive solution (u₁, u₂), ∀(μ1,μ2)∈]0,μ¯1]×]0,μ¯2] such that 0 < ‖u_i‖_∞ ≤ ‖e_i‖_∞.

3. Consider the sub-super linear case, 0 < α₁ + β₁ < p₁ − 1 and α₂ + β₂ ≥ p₂ − 1 and put μ¯2=1C21+‖e1‖∞α2+‖e2‖∞β2. Let μ₁ > 0 and 0<μ2≤μ¯2. Then, there exists K, large enough, such that

   −Δp1(Ke1)=Kp1−1  ≥  μ1F1x,Ke1,v,  ∀v∈[0,Ke2]−Δp2e2=1       ≥  μ2F2(x,u,e2),   ∀u∈[0,e1].

So, (Ke₁, e₂) is as super solution of problem (P).

By the same argument, if α₁ + β₁ ≥ p₁ − 1 and 0 < α₂ + β₂ < p₂ − 1. Put μ¯1=1C11+‖e1‖∞α1+‖e2‖∞β1. For all 0<μ1≤μ¯1 and μ₂ > 0, there exists K, large enough, such that

−Δp1e1=1     ≥  μ1F1x,e1,v,  ∀v∈[0,e2]−Δp2(Ke2)=Kp2−1    ≥  μ2F2(x,u,e2),   ∀u∈[0,Ke1].

So, (e₁, Ke₂) is as super solution; hence, the problem (1) admits a weak positive solution (u₁, u₂). The proof of Theorem 3.1 is complete.

□

Remark 3.1.

The second point of Theorem 3.1 is of great importance because no restriction was made on the growth of nonlinearities but only on the parameter μ_i which must not be large.

The hypothesis A.2 plays an essential role in the way that we need only to find a super-solution. In what follows, we provide an example without the hypothesis A.2 of Theorem (3.1). We will see that the assumptions on nonlinearities become more restrictive.

Proposition 3.1.

For i = 1, 2, assume that ∃ 0 < c_i ≤ C_i, 0 < α_i, β_i < p_i − 1, ∀(s1,s2)∈R+×R+,

cisiαi≤Fi(x,s1,s2)≤Cis1αi+s2β1.

Then, ∀μ_i > 0, there exists a weak positive solution of the problem (1).

Proof.

According to the proof of Theorem 3.1, there exists K, large enough, such that (Ke₁, Ke₂) is a super-solution of the problem (1). Then, we need only to find a sub-solution. Let u̲i=εϕi, ϕ_i is the principal eigenfunction (positive) associated with the principal eigenvalue of p_i-Laplacian operator such that ‖ϕ_i‖_∞ = 1,

(8) −Δpiϕi=λ1,piϕipi−1 in    Ωϕi=0       on   ∂Ω

(u̲1,u̲2) is as sub-solution of the problem (1). We need to have

−Δp1u̲1=λ1,p1εp1−1ϕ1,p1p1−1≤ μ1c1εα1ϕ1,p1α1 ≤ μ1F1(x,u̲1,v),∀v∈[u̲2,u¯2]−Δp2u̲2=λ1,p2εp2−1ϕ1,p2p2−1≤ μ2c2εβ2ϕ1,p2β2 ≤ μ2F2(x,u,u̲2),∀u∈[u̲1,u¯1]

As ‖ϕ_i‖_∞ = 1, it is enough to have λ1,piεpi−1≤μiciεδi (δ₁ = α₁ and δ₂ = β₂). This is possible for ɛ, small enough, ɛ < 1 and for all μ_i > 0. To end the proof, we need to verify that εϕi=u̲i≤u¯i=Kei for a small ɛ and K large. By the maximum principle, we have

−Δpi(εϕi)=εpi−1λ1,piϕi≤εpi−1λ1,pi≤Kpi−1=−Δpi(Kei)⇒εϕi≤Kei.

□

4. Examples

In this section, we use Theorem 3.1 and solve some elliptic (p₁, p₂)-Laplacian systems studied in some published articles see [7].

Consider the following (p₁, p₂)-Laplacian system

(P)−Δp1u1=μ1a1(x)+b1(x)u1α1u2β1   in  Ω−Δp2u2=μ2a2(x)+b2(x)u1α2u2β2   in  Ωu1=u2=0                    on  ∂Ω

4.1 The case where (0, 0) is a sub-solution

Assume, for i = 1, 2, that a_i(x) and b_i(x) are continuous and nonnegative in Ω¯, a₁ or a₂ not identically null. Then, (0, 0) is a sub-solution of problem (P). Taking into account Theorem 3.1 and its proof, we get the following propositions,

Proposition 4.1.

Problem (P) has a positive weak solution provided that for i = 1, 2,

1. μ_i > 0 if 0 < α_i + β_i < p_i − 1 (The sub-linear case),

2. 0<μi≤μ¯i=1Ci1+‖e1‖∞αi‖e2‖∞βi if α_i + β_i ≥ p_i − 1. C_i =  max(‖a_i‖_∞, ‖b_i‖_∞) (The super linear case) and

3. μ_i > 0 and 0<μj≤μ¯j if α_i + β_i < p_i − 1 and α_j + β_j ≥ p_j − 1. j = 2 if i = 1 and j = 1 if i = 2 (The sub-super linear case).

Proof.

We have to look for a super solution of our problem.

• (1)

  Since 0 < α_i + β_i < p_i − 1, for ∀μ_i > 0, there exists K such that for i = 1, 2, Kpi−1≥μiCi1+Kαi+βi‖e1‖∞αi‖e2‖∞βi. Then, we have

  −Δp1(Ke1)=Kp1−1          ≥μ1C11+(K‖e1‖∞)α1(K‖e2‖∞)β1          ≥μ1a1(x)+b1(x)(Ke1)α1(Ke2)β1          ≥μ1a1(x)+b1(x)(Ke1)α1vβ1,∀v∈[0,Ke2].

In the same way, we show that −Δp2(Ke2)≥μ2a2(x)+b2(x)uα2(Ke2)β2, ∀u ∈ [0, Ke₁]. So, (Ke₁, Ke₂) is a super solution of problem (P), and then the problem has a weak positive solution.

• (2)

  Consider the super linear case, for i = 1, 2, α_i + β_i ≥ p_i − 1. Let 0<μi≤μ¯i, (e₁, e₂) is a super solution of problem (P). Indeed, we have

  −Δp1e1=1=μ¯1C11+‖e1‖∞α1‖e2‖∞β1          ≥ μ1C11+‖e1‖∞α1‖e2‖∞β1          ≥ μ1a1(x)+b1(x)e1α1e2β1          ≥ μ1a1(x)+b1(x)e1α1vβ1,∀v∈[0,e2].

In the same way, we obtain −Δp2e2≥μ2a2(x)+b2(x)uα2e2β2, ∀u ∈ [0, e₁]. We conclude that problem (P) has a positive weak solution in [0, e₁] × [0, e₂].

• (3)

  Finally, the third point, the sub-super linear case, follows from the two previous ones.

□

4.2 The case where (0, 0) is a trivial solution

We examine only the case where p₁ = p₂ = p. (0, 0) is a trivial solution of (P) if a_i, i = 1, 2, are identically null. Our goal is to find a positive solution of problem (P). To do this, we need to find a pair of positive sub-super solution. Nevertheless, we have to add some more assumptions. Assume that, for i = 1, 2, b_i is positive continuous in Ω¯. So, c_i ≤ ‖b_i‖_∞ ≤ C_i for some positive constants c_i and C_i. The problem becomes

(P1)−Δpu1=μ1b1(x)u1α1u2β1  in  Ω−Δpu2=μ2b2(x)u1α2u2β2  in  Ωu1=u2=0       on  ∂Ω

Proposition 4.2.

Assume that, for i = 1, 2, 0 < α_i + β_i < p − 1. Then, the problem (P₁) has a positive weak solution ∀μ_i > 0.

Proof.

According to Theorem 2.1, we need to find a pair of sub-super solution of the problem (P₁). Assume that for i = 1, 2, 0 < α_i + β_i < p − 1, then ∀μ_i > 0, we can choose 0 < ɛ < 1, such that λ1,pεp−1−(αi+βi)≤μici for i = 1, 2. Fix such ɛ and choose K≥max(λ1,p1p−1ε,1) such that Kp−1−(αi+βi)≥μiCi‖e1‖∞αi+βi for i = 1, 2. Then, (ɛϕ₁, ɛϕ₁) and (Ke₁, Ke₁) is a pair of sub-super linear solution of the problem (P₁) (‖ϕ₁‖_∞ = 1). Indeed, we have, by the maximum principle, ɛϕ₁ ≤ Ke₁ because

−Δp(εϕ1)=λ1,pεp−1ϕ1p−1≤λ1,pεp−1≤Kp−1=−Δp(Ke1).

∀(u, v) ∈ [ɛϕ₁, Ke₁] × [ɛϕ₁, Ke₁], we have

−Δp(εϕ1)=λ1,pεp−1ϕ1p−1≤  μ1c1εα1+β1ϕ1α1+β1≤  μ1c1(εϕ1)α1(εϕ1)β1≤  μ1b1(x)(εϕ1)α1vβ1.−Δp(εϕ1)=λ1,pεp−1ϕ1p−1≤  μ2c2εα2+β2ϕ1α2+β2≤  μ2c2(εϕ1)α2(εϕ1)β2≤  μ2b2(x)uα2(εϕ1)β2.

and

−Δp(Ke1)=Kp−1≥  μ1C1Kα1+β1‖e1‖∞α1+β1≥  μ1C1(Ke1)α1(Ke1)β1≥  μ1b1(x)(Ke1)α1vβ1.−Δp(Ke1)=Kp−1≥  μ2C2Kα2+β2‖e1‖∞α2+β2≥  μ2C2(Ke1)α2(Ke1)β2≥  μ2b2(x)uα2(Ke1)β2.

The problem (P₁) has a weak positive solution in the set [ɛϕ₁, Ke₁] × [ɛϕ₁, Ke₁]. The proof is complete. □