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

Mohammed Moussa can be contacted at: mohammed.moussa@uit.ac.ma

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.

1. Introduction

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

\{\begin{cases} - Δ_{p_{1}} u_{1} & = μ_{1} F_{1} (x, u_{1}, u_{2}) i n Ω \\ - Δ_{p_{2}} u_{2} & = μ_{2} F_{2} (x, u_{1}, u_{2}) i n Ω \\ u_{1} = u_{2} & = 0 o n \partial Ω \end{cases}

(1)

Ω is an open bounded domain of $R^{N}$ with smooth boundary ∂Ω. For i = 1, 2, $Δ_{p_{i}} = div (| \nabla u_{i} |^{p_{i} - 2} \nabla u_{i})$ is the p_i-Laplacian operator, p_i > 1, μ_i is a positive parameter and $F_{i} : \bar{Ω} \times R \times R \to R$ 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 $({\underset{̲}{u}}_{1}, {\bar{u}}_{1}), ({\underset{̲}{u}}_{2}, {\bar{u}}_{2}) \in (W^{1, p_{1}} (Ω) \cap L^{\infty} (Ω)) \times (W^{1, p_{2}} (Ω) \cap L^{\infty} (Ω))$ is a pair of sub-super solution of the problem (1) if they satisfy

H1.

${\underset{̲}{u}}_{i} \leq {\bar{u}}_{i}$ a.e in Ω and ${\underset{̲}{u}}_{i} \leq 0 \leq {\bar{u}}_{i}$ on ∂Ω for i = 1, 2.

H2.

$- Δ_{p_{1}} {\underset{̲}{u}}_{1} - F_{1} (x, {\underset{̲}{u}}_{1}, v) \leq 0 \leq - Δ_{p_{1}} {\bar{u}}_{1} - F_{1} (x, {\bar{u}}_{1}, v)$ ⁠, $\forall v \in [{\underset{̲}{u}}_{2}, {\bar{u}}_{2}]$ ⁠,

H3.

$- Δ_{p_{2}} {\underset{̲}{u}}_{2} - F_{2} (x, u, {\underset{̲}{u}}_{2}) \leq 0 \leq - Δ_{p_{2}} {\bar{u}}_{2} - F_{2} (x, u, {\bar{u}}_{2})$ ⁠, $\forall u \in [{\underset{̲}{u}}_{1}, {\bar{u}}_{1}]$ ⁠.

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 $\bar{Ω} \times R^{2}$ ⁠. 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 $(u_{1}, u_{2}) \in [{\underset{̲}{u}}_{1}, {\bar{u}}_{1}] \times [{\underset{̲}{u}}_{2}, {\bar{u}}_{2}]$ ⁠.

Proof.

Consider, for i = 1, 2, the truncation operators, $T_{i} : L^{p_{i}} (Ω) \to L^{p_{i}} (Ω)$ defined by

T_{i} (z_{i}) (x) = \{\begin{cases} {\underset{̲}{u}}_{i} (x) i f z_{i} (x) \leq {\underset{̲}{u}}_{i} (x) \\ z_{i} (x) i f {\underset{̲}{u}}_{i} (x) \leq z_{i} (x) \leq {\bar{u}}_{i} (x) \\ {\bar{u}}_{i} (x) i f z_{i} (x) \geq {\bar{u}}_{i} (x) \end{cases}

then

\forall z_{i} \in L^{p_{i}} (Ω)

⁠,

T_{i} (z_{i}) \in [{\underset{̲}{u}}_{i}, {\bar{u}}_{i}]

and

‖ T_{i} (z_{i}) ‖_{\infty} \in [0, ‖ {\bar{u}}_{i} ‖_{\infty}]

⁠. By the continuity of F_i, there exists a positive constant C_i such that

| F_{i} (x, T_{1} (z_{1}) (x), T_{2} (z_{2}) (x)) | \leq C_{i} .

Let $F_{i} : L^{p_{i}} (Ω) \times L^{p_{i}} (Ω) \to L^{p_{i}} (Ω)$ be the Nemytskii operator defined by

F_{i} (u_{1}, u_{2}) (x) = F_{i} (x, T_{1} (u_{1}) (x), T_{2} (u_{2}) (x)) .

Then, $F_{i}$ is $L^{p_{i}} (Ω)$ -bounded. By the dominate convergence theorem and the continuity of F_i, we conclude the continuity of $F_{i}$ , and we have $\forall (u_{1}, u_{2}) \in L^{p_{1}} (Ω) \times L^{p_{2}} (Ω)$ ⁠.

‖ F_{i} (u_{1}, u_{2}) ‖_{\infty} \leq C_{i} .

(2)

Now, fix $(z_{1}, z_{2}) \in L^{p_{1}} (Ω) \times L^{p_{2}} (Ω)$ ⁠, there exists a unique pair $(w_{1}, w_{2}) \in W_{0}^{1, p_{1}} (Ω) \times W_{0}^{1, p_{2}} (Ω)$ solution of the problem,

\{\begin{cases} - Δ_{p_{1}} w_{1} = F_{1} (T_{1} (z_{1}), T_{2} (z_{2})) i n Ω \\ - Δ_{p_{2}} w_{2} = F_{2} (T_{1} (z_{1}), T_{2} (z_{2})) i n Ω \\ w_{1} = w_{2} = 0 o n \partial Ω \end{cases}

(3)

Therefore, we can define the operator $S : L^{p_{1}} (Ω) \times L^{p_{2}} (Ω) \to L^{p_{1}} (Ω) \times L^{p_{2}} (Ω)$ 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 $L^{p_{1}} (Ω) \times L^{p_{2}} (Ω)$ and (w_1,n, w_2,n) = S(z_1,n, z_2,n), then $\forall ϕ_{i} \in W_{0}^{1, p_{i}} (Ω)$

\int_{Ω} | \nabla w_{i, n} |^{p_{i} - 2} \nabla w_{i, n} . \nabla ϕ_{i} = \int_{Ω} F_{i} (T_{1} (z_{1, n}), T_{2} (z_{2, n})) ϕ_{i} .

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

‖ w_{i, n} ‖_{W_{0}^{1, p_{i}} (Ω)}^{p_{i}} = \int_{Ω} | \nabla w_{i, n} |^{p_{i}} = \int_{Ω} F (T_{1} (z_{1, n}), T_{2} (z_{2, n})) w_{i, n} \leq C_{i} ‖ w_{i, n} ‖_{L^{1} (Ω)} \leq K_{i} ‖ w_{i, n} ‖_{W_{0}^{1, p_{i}} (Ω)} .

(4)

Then, (w_1,n, w_2,n) is bounded in $W_{0}^{1, p_{1}} (Ω) \times W_{0}^{1, p_{2}} (Ω)$ ⁠. By the compact embedding, there exists a convergent sub-sequence of (w_1,n, w_2,n) in $L^{p_{1}} (Ω) \times L^{p_{2}} (Ω)$ ⁠. So, S is compact.

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

‖ w_{i, n} ‖_{W_{0}^{1, p_{i}} (Ω)} \leq {\bar{K}}_{i} = K_{i}^{\frac{1}{p_{i} - 1}} \Rightarrow ‖ w_{i, n} ‖_{L^{p_{i}} (Ω)} \leq L_{i} .

(5)

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

S (L^{p_{1}} (Ω) \times L^{p_{2}} (Ω)) \subset B_{L^{p_{1}} (Ω) \times L^{p_{2}} (Ω)} (0, \bar{L}) .

for some $\bar{L} > 0$ ⁠. By the Schauder fixed-point theorem, in $B_{L^{p_{1}} (Ω) \times L^{p_{2}} (Ω)} (0, \bar{L})$ ⁠, there exists a unique $(u_{1}, u_{2}) \in L^{p_{1}} (Ω) \times L^{p_{2}} (Ω)$ such that S(u₁, u₂) = (u₁, u₂).

\{\begin{cases} - Δ_{p_{1}} u_{1} = F (T_{1} (u_{1}), T_{2} (u_{2})) & i n Ω \\ - Δ_{p_{2}} u_{2} = F_{2} (T_{1} (u_{1}), T_{2} (u_{2})) & i n Ω \\ u_{1} = u_{2} = 0 & o n \partial Ω \end{cases}

(6)

Finally, (u₁, u₂) is a solution of problem (1) if, and only if, T₁(u₁) = u₁ and T₂(u₂) = u_2, which means that ${\underset{̲}{u}}_{1} \leq u_{1} \leq {\bar{u}}_{1}$ and ${\underset{̲}{u}}_{2} \leq u_{2} \leq {\bar{u}}_{2}$ ⁠. We need to prove that ${({\underset{̲}{u}}_{1} - u_{1})}^{+} = 0$ ⁠, ${(u_{1} - {\bar{u}}_{1})}^{+} = 0$ ⁠, ${({\underset{̲}{u}}_{2} - u_{2})}^{+} = 0$ and ${(u_{2} - {\bar{u}}_{2})}^{+} = 0$ ⁠. Let us prove, for example, that ${({\underset{̲}{u}}_{1} - u_{1})}^{+} = 0$ ⁠. The same argument works for the others cases.

Let $Ω^{+} = {x \in Ω, {\underset{̲}{u}}_{1} (x) > u_{1} (x)}$ ⁠. Since $({\underset{̲}{u}}_{1}, {\underset{̲}{u}}_{2})$ is a sub-solution, then for $ϕ = {({\underset{̲}{u}}_{1} - u_{1})}^{+}$ and v = T₂(u₂), we have,

\int_{Ω} | \nabla {\underset{̲}{u}}_{1} |^{p_{1} - 2} \nabla {\underset{̲}{u}}_{1} . \nabla {({\underset{̲}{u}}_{1} - u_{1})}^{+} \leq \int_{Ω} F_{1} (x, {\underset{̲}{u}}_{1}, T_{2} (u_{2})) {({\underset{̲}{u}}_{1} - u_{1})}^{+}

and as (u₁, u₂) is a solution of (6),

\int_{Ω} | \nabla u_{1} |^{p_{1} - 2} \nabla u_{1} . \nabla {({\underset{̲}{u}}_{1} - u_{1})}^{+} = \int_{Ω} F_{1} (x, T_{1} (u_{1}), T_{2} (u_{2})) {({\underset{̲}{u}}_{1} - u_{1})}^{+}

Then,

\begin{matrix} \int_{Ω} (| \nabla {\underset{̲}{u}}_{1} |^{p_{1} - 2} \nabla {\underset{̲}{u}}_{1} - | \nabla u_{1} |^{p_{1} - 2} \nabla u_{1}) . \nabla {({\underset{̲}{u}}_{1} - u_{1})}^{+} \leq \\ \int_{Ω} (F_{1} (x, {\underset{̲}{u}}_{1}, T_{2} (u_{2})) - F_{1} (x, T_{1} (u_{1}), T_{2} (u_{2})) {({\underset{̲}{u}}_{1} - u_{1})}^{+} . \end{matrix}

Remark that in Ω⁺, $T_{1} (u_{1}) = {\underset{̲}{u}}_{1}$ ⁠. So,

\begin{matrix} \int_{Ω^{+}} (| \nabla {\underset{̲}{u}}_{1} |^{p_{1} - 2} \nabla {\underset{̲}{u}}_{1} - | \nabla u_{1} |^{p_{1} - 2} \nabla u_{1}) . \nabla {({\underset{̲}{u}}_{1} - u_{1})}^{+} & = \\ \int_{Ω^{+}} (| \nabla {\underset{̲}{u}}_{1} |^{p_{1} - 2} \nabla {\underset{̲}{u}}_{1} - | \nabla u_{1} |^{p_{1} - 2} \nabla u_{1}) . (\nabla {\underset{̲}{u}}_{1} - \nabla u_{1}) & \leq 0 . \end{matrix}

Therefore, by the monotonicity of the p₁-Laplacian, ${\underset{̲}{u}}_{1} - u_{1} = 0$ in Ω⁺. Then, ${\underset{̲}{u}}_{1} \leq u_{1}$ in Ω. In the same way, we get $u_{1} \leq {\bar{u}}_{1}$ in Ω. Then, ${\underset{̲}{u}}_{1} \leq u_{1} \leq {\bar{u}}_{1}$ and ${\underset{̲}{u}}_{2} \leq u_{2} \leq {\bar{u}}_{2} \Rightarrow T_{1} (u_{1}) = u_{1}$ 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 $| F_{i} (x, s_{1}, s_{2}) | \leq C_{i} (1 + | s_{1} |^{α_{i}} + | s_{2} |^{β_{i}})$ ,

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

Then,

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).
If min(α_i, β_i) ≥ p_i − 1, for i = 1, 2, then, there exists positive numbers ${\bar{μ}}_{i}$ such that $\forall (μ_{1}, μ_{2}) \in] 0, {\bar{μ}}_{1}] \times] 0, {\bar{μ}}_{2}]$ 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).
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 ${\bar{μ}}_{j}$ such that problem (1) has at least a weak positive solution ∀μ_i > 0 and $0 < μ_{j} \leq {\bar{μ}}_{j}$ ⁠.

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,

\{\begin{cases} - Δ_{p_{i}} u & = 1 i n Ω \\ u & = 0 o n \partial Ω \end{cases}

(7)

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
$\{\begin{cases} - Δ_{p_{1}} (K e_{1}) = K^{p_{1} - 1} & \geq μ_{1} C_{1} (1 + K^{α_{1}} ‖ e_{1} ‖_{\infty}^{α_{1}} + K^{β_{1}} ‖ e_{2} ‖_{\infty}^{β_{1}}) \\ \geq μ_{1} C_{1} (1 + K^{α_{1}} ‖ e_{1} ‖_{\infty}^{α_{1}} + ‖ v ‖_{\infty}^{β_{1}}), \\ \geq μ_{1} F_{1} (x, K e_{1}, v), \forall v \in [0, K e_{2}] \\ - Δ_{p_{2}} (K e_{2}) = K^{p_{2} - 1} & \geq μ_{2} C_{2} (1 + K^{α_{2}} ‖ e_{1} ‖_{\infty}^{α_{2}} + K^{β_{2}} ‖ e_{2} ‖_{\infty}^{β_{2}}) \\ \geq μ_{2} C_{2} (1 + ‖ u ‖_{\infty}^{α_{2}} + K^{β_{2}} ‖ e_{2} ‖_{\infty}^{β_{2}}), \\ \geq μ_{2} F_{2} (x, u, K e_{2}), \forall u \in [0, K e_{1}] \end{cases}$
in the weak sense. So, $({\bar{u}}_{1}, {\bar{u}}_{2}) = (K e_{1}, K e_{2})$ is a super-solution of the problem (1). Therefore, there exists (u₁, u₂) ∈ [0, Ke₁] × [0, Ke₂] solution of system (1).
In the super-linear case, for i = 1, 2, min(α_i, β_i) ≥ p_i − 1. Put ${\bar{μ}}_{i} = \frac{1}{C_{i} (1 + ‖ e_{1} ‖_{\infty}^{α_{i}} + ‖ e_{2} ‖_{\infty}^{β_{i}})}$ and let $0 < μ_{i} \leq {\bar{μ}}_{i}$ ⁠. Then,
$\{\begin{cases} - Δ_{p_{1}} e_{1} = 1 & = {\bar{μ}}_{1} C_{1} (1 + ‖ e_{1} ‖_{\infty}^{α_{1}} + ‖ e_{2} ‖_{\infty}^{β_{1}}) \\ \geq μ_{1} C_{1} (1 + ‖ e_{1} ‖_{\infty}^{α_{1}} + ‖ e_{2} ‖_{\infty}^{β_{1}}) \\ \geq μ_{1} C_{1} (1 + ‖ e_{1} ‖_{\infty}^{α_{1}} + v^{β_{1}}) \\ \geq μ_{1} F_{1} (x, e_{1}, v), \forall v \in [0, e_{2}] \\ - Δ_{p_{2}} e_{2} = 1 & = {\bar{μ}}_{2} C_{2} (1 + ‖ e_{1} ‖_{\infty}^{α_{2}} + ‖ e_{2} ‖_{\infty}^{β_{2}}) \\ \geq μ_{2} C_{2} (1 + ‖ e_{1} ‖_{\infty}^{α_{2}} + ‖ e_{2} ‖_{\infty}^{β_{2}}) \\ \geq μ_{2} C_{2} (1 + u^{α_{2}} + ‖ e_{2} ‖_{\infty}^{β_{2}}) \\ \geq μ_{2} F_{2} (x, u, e_{2}), \forall u \in [0, e_{1}] . \end{cases}$
(e₁, e₂) is a super-solution, and we deduce that the problem (1) admits a weak positive solution (u₁, u₂), $\forall (μ_{1}, μ_{2}) \in] 0, {\bar{μ}}_{1}] \times] 0, {\bar{μ}}_{2}]$ such that 0 < ‖u_i‖_∞ ≤ ‖e_i‖_∞.
Consider the sub-super linear case, 0 < α₁ + β₁ < p₁ − 1 and α₂ + β₂ ≥ p₂ − 1 and put ${\bar{μ}}_{2} = \frac{1}{C_{2} (1 + ‖ e_{1} ‖_{\infty}^{α_{2}} + ‖ e_{2} ‖_{\infty}^{β_{2}})}$ ⁠. Let μ₁ > 0 and $0 < μ_{2} \leq {\bar{μ}}_{2}$ ⁠. Then, there exists K, large enough, such that
$\{\begin{cases} - Δ_{p_{1}} (K e_{1}) & = K^{p_{1} - 1} \geq μ_{1} F_{1} (x, K e_{1}, v), \forall v \in [0, K e_{2}] \\ - Δ_{p_{2}} e_{2} & = 1 \geq μ_{2} F_{2} (x, u, e_{2}), \forall u \in [0, e_{1}] . \end{cases}$

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 ${\bar{μ}}_{1} = \frac{1}{C_{1} (1 + ‖ e_{1} ‖_{\infty}^{α_{1}} + ‖ e_{2} ‖_{\infty}^{β_{1}})}$ ⁠. For all $0 < μ_{1} \leq {\bar{μ}}_{1}$ and μ₂ > 0, there exists K, large enough, such that

\{\begin{cases} - Δ_{p_{1}} e_{1} & = 1 \geq μ_{1} F_{1} (x, e_{1}, v), \forall v \in [0, e_{2}] \\ - Δ_{p_{2}} (K e_{2}) & = K^{p_{2} - 1} \geq μ_{2} F_{2} (x, u, e_{2}), \forall u \in [0, K e_{1}] . \end{cases}

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, $\forall (s_{1}, s_{2}) \in R^{+} \times R^{+}$ ,

c_{i} s_{i}^{α_{i}} \leq F_{i} (x, s_{1}, s_{2}) \leq C_{i} (s_{1}^{α_{i}} + s_{2}^{β_{1}}) .

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 ${\underset{̲}{u}}_{i} = ε ϕ_{i}$ ⁠, ϕ_i is the principal eigenfunction (positive) associated with the principal eigenvalue of p_i-Laplacian operator such that ‖ϕ_i‖_∞ = 1,

\{\begin{cases} - Δ_{p_{i}} ϕ_{i} & = λ_{1, p_{i}} ϕ_{i}^{p_{i} - 1} i n Ω \\ ϕ_{i} & = 0 o n \partial Ω \end{cases}

(8)

$({\underset{̲}{u}}_{1}, {\underset{̲}{u}}_{2})$ is as sub-solution of the problem (1). We need to have

\{\begin{cases} - Δ_{p_{1}} {\underset{̲}{u}}_{1} = λ_{1, p_{1}} ε^{p_{1} - 1} ϕ_{1, p_{1}}^{p_{1} - 1} & \leq μ_{1} c_{1} ε^{α_{1}} ϕ_{1, p_{1}}^{α_{1}} \leq μ_{1} F_{1} (x, {\underset{̲}{u}}_{1}, v), \forall v \in [{\underset{̲}{u}}_{2}, {\bar{u}}_{2}] \\ - Δ_{p_{2}} {\underset{̲}{u}}_{2} = λ_{1, p_{2}} ε^{p_{2} - 1} ϕ_{1, p_{2}}^{p_{2} - 1} & \leq μ_{2} c_{2} ε^{β_{2}} ϕ_{1, p_{2}}^{β_{2}} \leq μ_{2} F_{2} (x, u, {\underset{̲}{u}}_{2}), \forall u \in [{\underset{̲}{u}}_{1}, {\bar{u}}_{1}] \end{cases}

As ‖ϕ_i‖_∞ = 1, it is enough to have $λ_{1, p_{i}} ε^{p_{i} - 1} \leq μ_{i} c_{i} ε^{δ_{i}}$ (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 $ε ϕ_{i} = {\underset{̲}{u}}_{i} \leq {\bar{u}}_{i} = K e_{i}$ for a small ɛ and K large. By the maximum principle, we have

- Δ_{p_{i}} (ε ϕ_{i}) = ε^{p_{i} - 1} λ_{1, p_{i}} ϕ_{i} \leq ε^{p_{i} - 1} λ_{1, p_{i}} \leq K^{p_{i} - 1} = - Δ_{p_{i}} (K e_{i}) \Rightarrow ε ϕ_{i} \leq K e_{i} .

□

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) \{\begin{cases} - Δ_{p_{1}} u_{1} & = μ_{1} (a_{1} (x) + b_{1} (x) u_{1}^{α_{1}} u_{2}^{β_{1}}) i n Ω \\ - Δ_{p_{2}} u_{2} & = μ_{2} (a_{2} (x) + b_{2} (x) u_{1}^{α_{2}} u_{2}^{β_{2}}) i n Ω \\ u_{1} & = u_{2} = 0 o n \partial Ω \end{cases}

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 $\bar{Ω}$ ⁠, 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,

μ_i > 0 if 0 < α_i + β_i < p_i − 1 (The sub-linear case),
$0 < μ_{i} \leq {\bar{μ}}_{i} = \frac{1}{C_{i} (1 + ‖ e_{1} ‖_{\infty}^{α_{i}} ‖ e_{2} ‖_{\infty}^{β_{i}})}$ if α_i + β_i ≥ p_i − 1. C_i = max(‖a_i‖_∞, ‖b_i‖_∞) (The super linear case) and
μ_i > 0 and $0 < μ_{j} \leq {\bar{μ}}_{j}$ 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, $K^{p_{i} - 1} \geq μ_{i} C_{i} (1 + K^{α_{i} + β_{i}} ‖ e_{1} ‖_{\infty}^{α_{i}} ‖ e_{2} ‖_{\infty}^{β_{i}})$ ⁠. Then, we have
$\{\begin{cases} - Δ_{p_{1}} (K e_{1}) = K^{p_{1} - 1} \\ \geq μ_{1} C_{1} [1 + {(K ‖ e_{1} ‖_{\infty})}^{α_{1}} {(K ‖ e_{2} ‖_{\infty})}^{β_{1}}] \\ \geq μ_{1} [a_{1} (x) + b_{1} (x) {(K e_{1})}^{α_{1}} {(K e_{2})}^{β_{1}}] \\ \geq μ_{1} [a_{1} (x) + b_{1} (x) {(K e_{1})}^{α_{1}} v^{β_{1}}], \forall v \in [0, K e_{2}] . \end{cases}$

In the same way, we show that $- Δ_{p_{2}} (K e_{2}) \geq μ_{2} [a_{2} (x) + b_{2} (x) u^{α_{2}} {(K e_{2})}^{β_{2}}]$ ⁠, ∀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 $0 < μ_{i} \leq {\bar{μ}}_{i}$ ⁠, (e₁, e₂) is a super solution of problem (P). Indeed, we have
$\{\begin{cases} - Δ_{p_{1}} e_{1} = 1 = {\bar{μ}}_{1} C_{1} [1 + ‖ e_{1} ‖_{\infty}^{α_{1}} ‖ e_{2} ‖_{\infty}^{β_{1}}] \\ \geq μ_{1} C_{1} [1 + ‖ e_{1} ‖_{\infty}^{α_{1}} ‖ e_{2} ‖_{\infty}^{β_{1}}] \\ \geq μ_{1} [a_{1} (x) + b_{1} (x) e_{1}^{α_{1}} e_{2}^{β_{1}}] \\ \geq μ_{1} [a_{1} (x) + b_{1} (x) e_{1}^{α_{1}} v^{β_{1}}], \forall v \in [0, e_{2}] . \end{cases}$

In the same way, we obtain $- Δ_{p_{2}} e_{2} \geq μ_{2} [a_{2} (x) + b_{2} (x) u^{α_{2}} e_{2}^{β_{2}}]$ ⁠, ∀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 $\bar{Ω}$ ⁠. So, c_i ≤ ‖b_i‖_∞ ≤ C_i for some positive constants c_i and C_i. The problem becomes

(P_{1}) \{\begin{cases} - Δ_{p} u_{1} & = μ_{1} b_{1} (x) u_{1}^{α_{1}} u_{2}^{β_{1}} i n Ω \\ - Δ_{p} u_{2} & = μ_{2} b_{2} (x) u_{1}^{α_{2}} u_{2}^{β_{2}} i n Ω \\ u_{1} & = u_{2} = 0 o n \partial Ω \end{cases}

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 $λ_{1, p} ε^{p - 1 - (α_{i} + β_{i})} \leq μ_{i} c_{i}$ for i = 1, 2. Fix such ɛ and choose $K \geq \max (λ_{1, p}^{\frac{1}{p - 1}} ε, 1)$ such that $K^{p - 1 - (α_{i} + β_{i})} \geq μ_{i} C_{i} ‖ e_{1} ‖_{\infty}^{α_{i} + β_{i}}$ 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

- Δ_{p} (ε ϕ_{1}) = λ_{1, p} ε^{p - 1} ϕ_{1}^{p - 1} \leq λ_{1, p} ε^{p - 1} \leq K^{p - 1} = - Δ_{p} (K e_{1}) .

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

\{\begin{cases} - Δ_{p} (ε ϕ_{1}) = λ_{1, p} ε^{p - 1} ϕ_{1}^{p - 1} & \leq μ_{1} c_{1} ε^{α_{1} + β_{1}} ϕ_{1}^{α_{1} + β_{1}} \\ \leq μ_{1} c_{1} {(ε ϕ_{1})}^{α_{1}} {(ε ϕ_{1})}^{β_{1}} \\ \leq μ_{1} b_{1} (x) {(ε ϕ_{1})}^{α_{1}} v^{β_{1}} . \\ - Δ_{p} (ε ϕ_{1}) = λ_{1, p} ε^{p - 1} ϕ_{1}^{p - 1} & \leq μ_{2} c_{2} ε^{α_{2} + β_{2}} ϕ_{1}^{α_{2} + β_{2}} \\ \leq μ_{2} c_{2} {(ε ϕ_{1})}^{α_{2}} {(ε ϕ_{1})}^{β_{2}} \\ \leq μ_{2} b_{2} (x) u^{α_{2}} {(ε ϕ_{1})}^{β_{2}} . \end{cases}

and

\{\begin{cases} - Δ_{p} (K e_{1}) = K^{p - 1} & \geq μ_{1} C_{1} K^{α_{1} + β_{1}} ‖ e_{1} ‖_{\infty}^{α_{1} + β_{1}} \\ \geq μ_{1} C_{1} {(K e_{1})}^{α_{1}} {(K e_{1})}^{β_{1}} \\ \geq μ_{1} b_{1} (x) {(K e_{1})}^{α_{1}} v^{β_{1}} . \\ - Δ_{p} (K e_{1}) = K^{p - 1} & \geq μ_{2} C_{2} K^{α_{2} + β_{2}} ‖ e_{1} ‖_{\infty}^{α_{2} + β_{2}} \\ \geq μ_{2} C_{2} {(K e_{1})}^{α_{2}} {(K e_{1})}^{β_{2}} \\ \geq μ_{2} b_{2} (x) u^{α_{2}} {(K e_{1})}^{β_{2}} . \end{cases}

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

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Published in Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) licence. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this licence may be seen at http://creativecommons.org/licences/by/4.0/legalcode

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

1. Introduction

2. Preliminaries and main results

3. Applications

4. Examples

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

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

References

Data & Figures

Contents

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

1. Introduction

2. Preliminaries and main results

3. Applications

4. Examples

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

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

References

Data & Figures

Contents

Supplements

References

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New and popular articles

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