Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 36, pp. 1–18.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

WEAK SOLUTION BY THE SUB-SUPERSOLUTION METHOD

FOR A NONLOCAL SYSTEM INVOLVING LEBESGUE

GENERALIZED SPACES

ABDOLRAHMAN RAZANI, GIOVANY M. FIGUEIREDO

Abstract. We consider a system of nonlocal elliptic equations

−A(x, |v|
Lr1(x) ) div(a1(|∇u|p1(x))|∇u|p1(x)−2∇u)

= f1(x, u, v)|∇v|α1(x)

Lq1(x) + g1(x, u, v)|∇v|γ1(x)
Ls1(x) ,

−A(x, |u|
Lr2(x) ) div(a2(|∇v|p2(x))|∇u|p2(x)−2∇u)

= f2(x, u, v)|∇u|α2(x)

Lq2(x) + g2(x, u, v)|∇u|γ2(x)
Ls2(x) ,

with Dirichlet boundary condition, where Ω is a bounded domain in RN (N >
1) with C2 boundary. Using sub-supersolution method, we prove the existence

of at least one positive weak solution. Also, we study a generalized logistic

equation and a sublinear system.

1. Introduction

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 (see [8, 31, 37, 38]). In the previous decades there have
been several works related to the p and p(x)-Laplacian operators; see [1, 2, 6, 16,
17, 19, 20, 21, 22, 23, 29, 30, 34, 35, 40] and the references therein.

Nonlocal problems including Laplace operator have been intensively studied since
their first appearance in the work of Kirchhoff [27] who studied a wave equation
which is a generalization of the D’Alembert equation. On this subject the reader
may also consult Carrier [5] and Lions [28].

However, non-local problems are not restricted to mechanical motivations as in
the aforementioned works. They also appear in a wide variety of applications as
population dynamics [9, 10, 12], Ohmic heating [26], the formation of shear bands in
materials [32], heat transfer in thermistors [25], combustion theory [33], microwave
heating of ceramic materials [3].

2020 Mathematics Subject Classification. 35J91, 35J60, 35D30.
Key words and phrases. Nonlocal problem, p(x)-Laplacian, sub-supersolution;

minimal wave speed.
©2022. This work is licensed under a CC BY 4.0 license.

Submitted June 18, 2021. Published May 1, 2022.

1



2 A. RAZANI, G. M. FIGUEIREDO EJDE-2022/36

Here, we consider the nonlocal system

−A(x, |v|Lr1(x)) div
(
a(|∇u|p1(x))|∇u|p1(x)−2∇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)) div
(
a(|∇v|p2(x))|∇v|p2(x)−2∇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.
To be more specific about the structure of the operator in (1.1), we consider func-
tions a : R+ → R+ of class C1 satisfying the following conditions:

(A1) There exist constants k1, k2, k3, k4 ≥ 0, 2 < pi ≤ oi < N such that

k1t
pi + k2t

oi ≤ a(tpi)tpi ≤ k3t
pi + k4t

oi , for all t ≥ 0.

(A2) The function t 7→ Ai(t
pi) is strictly convex, where Ai(t) =

∫ t
0
ai(s)ds.

(A3) The function t 7→ ai(t
pi)tpi−2 is increasing.

Various operators occurring in applications are included in models for the bound-
ary value problem (1.1) as one can see from next examples. The following operators
satisfy (A1)–(A3):

(i) If ai(t) = 1 for i = 1, 2, we obtain the p-Laplacian and problem (1.1) becomes

−A(x, |v|Lr1(x))∆p1u = f1(x, u, v)|∇v|α1(x)

Lq1(x) + g1(x, u, v)|∇v1|γ1(x)

Ls1(x) in Ω,

−A(x, |u|Lr2(x))∆p2v = 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,

with qi = pi, k1 + k2 = 1 and k3 + k4 = 1.

(ii) If ai(t) = 1 + t
oi−pi
pi for i = 1, 2, we obtain the (p, o)-Laplacian or p&o-

Laplacian and problem (1.1) becomes

−A(x, |v|Lr1(x)) (∆p1u+ ∆o1u) = f1(x, u, v)|∇v|α1(x)

Lq1(x) + g1(x, u, v)|∇v1|γ1(x)

Ls1(x) in Ω,

−A(x, |u|Lr2(x)) (∆p2v + ∆o2v) = 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,

with k1 = k2 = k3 = k4 = 1.
(iii) If ai(t) = 1 + 1

(1+t)
pi−2
pi

for i = 1, 2, we obtain

−A(x, |v|Lr1(x)) div
∣∣∣∇u|p1−2∇u+

|∇u|p1−2∇u

(1 + |∇u|p1)
p1−2
p1

)
= 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)) div
(
|∇v|p2−2∇v +

|∇v|p2−2∇v

(1 + |∇v|p2)
p2−2
p2

)
= f2(x, u, v)|∇u|α2(x)

Lq2(x) + g2(x, u, v)|∇u|γ2(x)

Ls2(x) in Ω,



EJDE-2022/36 WEAK SOLUTION BY THE SUB-SUPERSOLUTION METHOD 3

u = v = 0 on ∂Ω,

with qi = pi, k1 + k2 = 1, and k3 + k4 = 2.

(iv) If ai(t) = 1 + t
oi−pi
pi + 1

(1+t)
pi−2
pi

for i = 1, 2, we obtain

−A(x, |v|Lr1(x))
(

∆p1u+ ∆o1u+ div
( |∇u|p1−2∇u

(1 + |∇u|p1)
p1−2
p1

))
= 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))
(

∆p2v + ∆o2v + div
( |∇v|p2−2∇v

(1 + |∇v|p2)
p2−2
p2

))
= 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,

with k1 = k2 = k4 = 1 and k3 = 2.
Several works related to (1.1) in the p-Laplacian case, that is, with p(x) = p (a

constant) can be found in [4, 7, 14, 15, 18, 24, 39] and their references. Chen et
al. [7] proved the existence of positive solutions for a class of nonvariational elliptic
system with nonlocal source

−∆um = f1(x, u)|v|αLp in Ω,

−∆vn = f2(x, v)|u|βLq in Ω,

u > 0, v > 0 in Ω,

u = v = 0 on ∂Ω.

using the Galerkin method, a fixed point theorem in finite dimensions, and sub-
supersolution technique. Corrêa et al [14] studied the existence of positive solutions
for the nonlocal problem

−∆p1u = |v|α1

Lq1 in Ω,

−∆p2v = |u|α2

Lq2 in Ω, u = v = 0 on ∂Ω,

by using Rabinowitz’s theorem [36]. Santos et al [39] studiedthe 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 certain conditions. They use an
abstract result involving sub-supersolution, 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 [40]. The authors obtained an abstract result involving sub and
super solutions for (1.1) that generalizes [39, Theorem 1]. As an application of



4 A. RAZANI, G. M. FIGUEIREDO EJDE-2022/36

such result the authors presented three applications of [39, Theorem 1] for the
p(x)-Laplacian operator.

The main result of this paper proves the existence of at least one weak posi-
tive solution for (1.1) via sub-supersolution method. This is an extension of [39,
Theorem 2] and [41, Theorem 1.1] for the p(x)-Laplacian operator.

2. Function spaces

Here, we introduce a suitable function space, where the solution of problem (1.1)
make sense. Next we recall some facts about the known spaces Lp(x)(Ω), W 1,p(x)(Ω)

and W
1,p(x)
0 (Ω) (see [21] and the references therein for more details).

Let Ω ⊂ RN , (N ≥ 1) be a bounded domain. Given p ∈ L∞+ (Ω), the generalized
Lebesgue space is

Lp(x)(Ω) :=
{
u ∈ S(Ω) :

∫
Ω

|u(x)|p(x)dx <∞
}
,

where S(Ω) :=
{
u : Ω → R : u is measurable

}
. The 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. Assume p, q ∈ L∞+ (Ω). 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).

One can define the generalized Sobolev space

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 separable and reflexive Banach
space.



EJDE-2022/36 WEAK SOLUTION BY THE SUB-SUPERSOLUTION METHOD 5

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). The following

statements are 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.

By part (i) of Proposition 2.4, ‖u‖ := |∇u|Lp(x) defines a norm in W
1,p(x)
0 (Ω)

which is equivalent to the norm ‖ · ‖∗.

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 [23, Lemma 2.2] and [20, 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 = ∅.

Next we recall [20, Lemma 2.1].

Lemma 2.7. 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 [19, Theorem
1.2] and [23, Theorem 1] that zλ ∈ C1(Ω) with zλ > 0 in Ω. The proof of Theorem
3.5 is mainly based on the following result by Rabinowitz [36].

Theorem 2.8. 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. Existence of solutions

In this section, we prove Theorem 3.5 which shows the existence of at least one
weak solution for system (1.1), via new sub-supersolution method. For this, we
recall preliminaries.



6 A. RAZANI, G. M. FIGUEIREDO EJDE-2022/36

Definition 3.1. The pair (u1, u2) is called a weak solution of (1.1), if ui ∈
W

1,pi(x)
0 (Ω) ∩ L∞(Ω) and∫

Ω

a1(|∇u1|p1(x))|∇u1|p1(x)−2∇u1∇ϕdx

=

∫
Ω

(f1(x, u1, u2)|u2|α1(x)

Lq1(x)

A(x, |u2|Lr1(x))
+
g1(x, u1, u2)|u2|γ1(x)

Ls1(x)

A(x, |u2|Lr1(x))

)
ϕdx,∫

Ω

a2(|∇u2|p2(x))|∇u2|p2(x)−2∇u2∇ϕdx

=

∫
Ω

(f2(x, u1, u2)|u1|α2(x)

Lq2(x)

A(x, |u1|Lr2(x))
+
g2(x, u1, u2)|u1|γ2(x)

Ls2(x)

A(x, |u1|Lr2(x))

)
ϕdx,

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 notation in 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).

Definition 3.2. The pairs (ui, ui), i = 1, 2 are called sub-supersolutions 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

ai(|∇ui|pi(x))|∇ui|pi(x)−2∇ui∇ϕdx

≤
∫

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))

)
ϕdx,∫

Ω

|ai(|∇ui|pi(x))|∇ui|pi(x)−2∇ui∇ϕdx

≥
∫

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))

)
ϕdx,

(3.1)

for all w ∈ [ujuj ] where i, j = 1, 2 with i 6= j.

Remark 3.3. The space Lp1(x)(Ω)× Lp2(x)(Ω) with the norm

|(u, v)|1,2 = |u|Lp1(x) + |v|Lp2(x) .

is a Banach space.

In what follows, we study the existence and uniqueness of solution for

−div(a1(|∇u|p1(x))|∇u|p1(x)−2∇u) = G1(z1, z2) in Ω,

−div(a2(|∇v|p2(x))|∇v|p2(x)−2∇v) = G2(z1, z2) in Ω,

u = v = 0 on ∂Ω,

(3.2)

where (z1, z2) ∈ Lp1(x)(Ω)× Lp2(x)(Ω).



EJDE-2022/36 WEAK SOLUTION BY THE SUB-SUPERSOLUTION METHOD 7

Lemma 3.4. Let Ω ⊂ RN , N ≥ 2, be a smooth bounded domain and ai : R+ →
R+ be a C1 function satisfying (A1), (A2) and (a3). Assume Gi : Lp1(x)(Ω) ×
Lp2(x)(Ω) → Lp

′
i(x)(Ω), where p′i(x) = pi(x)

(pi(x)−1) , Gi(z1, z2) are continuous and

|Gi(z1, z2)| ≤ Ki for all (z1, z2) ∈ Lp1(x)(Ω)×Lp2(x)(Ω). Then problem (3.2) has a

unique solution (u, v) ∈W 1,l1(x)
0 (Ω)×W 1,l2(x)

0 (Ω).

Proof. Consider the functional I : W
1,l1(x)
0 (Ω)×W 1,l2(x)

0 (Ω)→ R defined as

I(u, v) =
1

p1

∫
Ω

A(|∇u|p1(x))dx+
1

p2

∫
Ω

A(|∇v|p2(x))dx

−
∫

Ω

G1(z, z)udx−
∫

Ω

G2(z, z)vdx.

(3.3)

From (A1) the functional (3.3) is well defined and so I ∈ C1(W 1,q1
0 (Ω)×W 1,q2

0 (Ω),R).
Notice that, (A2) implies that I is strictly convex and weakly lower semicontinuous.
Also, (A1), |Gi(z1, z2)| ≤ Ki and Hölder’s inequality imply

I(u, v) ≥ k1

p−1
‖u‖p1(x)

W
1,p1(x)
0 (Ω)

+
k1

p−2
‖v‖p2(x)

W
1,p2(x)
0 (Ω)

+
k2

l−1
‖u‖l1(x)

W
1,l1(x)
0 (Ω)

+
k2

l−2
‖v‖l2(x)

W
1,l2(x)
0 (Ω)

−K0C
(
‖u‖

W
1,l1(x)
0 (Ω)

+ ‖v‖
W

1,l2(x)
0 (Ω)

)
for C > 0 and all (u, v) ∈W 1,l1(x)

0 (Ω)×W 1,l2(x)
0 (Ω) with ρ(|∇u|), ρ(|∇v|) ≥ 1, what

shows that I is coercive. Hence I has a unique critical point (a global minimizer),
which is the unique solution to (3.2). �

To state the main result of this article we need the following assumptions on
ri, pi, qi, si, αi, γi in (1.1):

(A4) 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, we have αi, γi ∈ L∞(Ω) and

1 < p−i : = inf
Ω
pi(x) ≤ p+

i := sup
Ω
pi(x) < N, αi(x), γi(x) ≥ 0 a.e. in Ω .

We set

σ := min
{
|w|Lri(x) , fori = 1, 2

}
, σ := max

{
|w|Lri(x) , for i = 1, 2

}
,

w := min{ui, for i = 1, 2}, w := max{ui, for i = 1, 2}.
(3.4)

Theorem 3.5. Assume

• ri, pi, qi, si, αi and γi satisfy (A4),
• (ui, ui) is a sub-supersolution for (1.1) with ui > 0 a.e. in Ω,
• fi(x, t, s), gi(x, t, s) ≥ 0 in Ω× [0, |u1|L∞ ]× [0, |u2|L∞ ],
• A : Ω× (0,∞)→ R is a continuous function with A(x, t) > 0 in Ω×

[
σ, σ

]
,

• ai : R+ → R+ is a C1 function satisfying (A1), (A2) and (a3).

Then (1.1) has at least one weak positive solution (u1, u2) with ui ∈ [ui, ui], i = 1, 2.

Proof. 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).



8 A. RAZANI, G. M. FIGUEIREDO EJDE-2022/36

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).

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 (3.5)

for all (z1, z2) ∈ Lp1(x)(Ω)× Lp2(x)(Ω).
By the Lebesgue Dominated Convergence Theorem, the mappings (z1, z2) 7→

Hi(T1z1, T2z2) from Lp1(x)(Ω)× Lp2(x)(Ω) to Lp
′
i(x)(Ω), i = 1, 2, are continuous.

The operator Φ : R+ × Lp1(x)(Ω)× Lp2(x)(Ω)→ Lp1(x)(Ω)× Lp2(x)(Ω) given by

Φ(λ, z1, z2) = (u1, u2),

is well-defined, by [21, Theorem 4.1], where (u1, u2) ∈W 1,p1(x)
0 (Ω)×W 1,p2(x)

0 (Ω) is
the unique solution of

− div(a1(|∇u1|p1(x))|∇u1|p1(x)−2∇u1) = λH1(T1z1, T2z2) in Ω,

−div(a2(|∇u2|p2(x))|∇u2|p2(x)−2∇u2) = λH2(T1z1, T2z2) in Ω,

u1 = u2 = 0 on ∂Ω,

(3.6)

by Lemma 3.4, where (z1, z2) ∈ Lp1(x)(Ω)× Lp2(x)(Ω).

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

implies that∫
Ω

ai(|∇uin|pi(x)−2)|∇uin|pi(x)−2∇uin∇ϕ = λn

∫
Ω

Hi(T1z
1
n, T2z

2
n)ϕ,

for all 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 (3.5),
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) is hold. Combining such

embedding 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 (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.



EJDE-2022/36 WEAK SOLUTION BY THE SUB-SUPERSOLUTION METHOD 9

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

ai(|∇uin|pi(x)−2)|∇uin|pi(x)−2∇uin∇ϕ = λn

∫
Ω

Hi(T1z
1
n, T2z

2
n)ϕ, (3.7)∫

Ω

ai(|∇ui|pi(x)−2)|∇ui|pi(x)−2∇ui∇ϕ = λ

∫
Ω

Hi(T1z
1, T2z

2)ϕ (3.8)

for all ϕ ∈W 1,pi(x)
0 (Ω) where i, j = 1, 2, and i 6= j.

Considering ϕ = (uin−ui) in (3.7) and (3.8) and subtracting (3.8) from (3.7) we
obtain ∫

Ω

〈
a(|∇uin|pi(x)−2)|∇uin|pi(x)−2∇uin

− a(|∇ui|pi(x)−2)|∇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

〈
a(|∇uin|pi(x)−2)|∇uin|pi(x)−2∇uin

− a(|∇ui|pi(x)−2)|∇u|pi(x)−2∇ui,∇(uin − ui)
〉∣∣

≤ |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

〈
a(|∇uin|pi(x)−2)|∇uin|pi(x)−2∇uin

− a(|∇ui|pi(x)−2)|∇ui|pi(x)−2∇ui,∇(uin − ui)
〉∣∣→ 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

−div(a(|∇u1|p1(x)−2)|∇u1|p1(x)−2∇u1) = λH1(T1u1, T2u2) in Ω,

−div(a(|∇u2|p2(x)−2)|∇u2|p2(x)−2∇u2) = λH2(T1u1, T2u2) in Ω,

u1 = u2 = 0 on ∂Ω.

(3.9)

Using the test function ui in (3.9) and considering (3.5), we obtain∫
Ω

|∇ui|pi(x) ≤ λ∗C|ui|Lp(x) .



10 A. RAZANI, G. M. FIGUEIREDO EJDE-2022/36

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 (3.9) we have∫

Ω

a(|∇ui|pi(x)−2)|∇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,

(3.10)

for all ϕ ∈W 1,pi(x)
0 (Ω) where i, j = 1, 2 with i 6= j.

Now we claim that ui ∈ [ui, ui] for i = 1, 2. To prove this claim we define

L1(u1 − u1)+ :=

∫
{u1≥u1}

〈
a(|∇u1|p1(x)−2)|∇u1|p1(x)−2∇u1

− a(|∇u1|p1(x)−2)|∇u1|p1(x)−2∇u1,∇(u1 − u1)
〉
dx.

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 (3.1) and combining with
equation (3.10) 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}

〈
a(|∇u1|p1(x)−2)|∇u1|p1(x)−2∇u1

− a(|∇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

The main goal of this section is to apply Theorem 3.5 to some nonlocal problems.

4.1. A generalized logistic equation. Here we present a generalization of the
classic logistic equation studied in [11, 13, 39] and [39, Theorem 8]. We consider

−A(x, |v|Lr1(x)) div(a1(|∇u|p1(x))|∇u|p1(x)−2∇u) = λf1(u)|v|α1(x)

Lq1(x) in Ω,

−A(x, |u|Lr2(x)) div(a1(|∇v|p2(x))|∇v|p2(x)−2∇v) = λf2(v)|u|α2(x)

Lq2(x) in Ω,

u = v = 0 on ∂Ω.

(4.1)



EJDE-2022/36 WEAK SOLUTION BY THE SUB-SUPERSOLUTION METHOD 11

where the function A(x, t) satisfies

A(x, 0) ≥ 0, lim
t→0+

A(x, t) =∞, lim
t→+∞

A(x, t) = ±∞.

Assume that there are numbers θi > 0, for i = 1, 2 such that the functions fi :
[0,∞)→ R satisfy the conditions:

(A5) fi ∈ C0([0, θi],R), for i = 1, 2,
(A6) fi(0) = fi(θi) = 0, fi(t) > 0 in (0, θi) for i = 1, 2.

Remark 4.1. Notice that 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)

}
.

Theorem 4.2. Suppose that ri, pi, qi, αi satisfy (A4). Assume fi satisfies (A5),
(A6) and ai : R+ → R+ is a C1 function satisfying (A1)–(A3) for i = 1, 2. If
A(x, t) > 0 in Ω×

(
0,max{|θ1|Lr2(x) , |θ2|Lr1(x)}

]
, then there exists λ0 > 0 such that

(4.1) has a positive weak 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)
A(|∇u|p1(x))dx+

∫
Ω

1

p2(x)
A(|∇v|p2(x))dx

− λ
∫

Ω

F̃1(u)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).

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

−div(a1(|∇zλ|p1(x))|∇zλ|p1(x)−2∇zλ) = λf̃1(zλ) in Ω,

zλ = 0 on ∂Ω,
(4.2)

and

−div(a2(|∇wλ|p2(x))|∇wλ|p2(x)−2∇wλ) = λf̃2(wλ) in Ω,

wλ = 0 on ∂Ω.
(4.3)

Note that the unique solutions of (4.2) and (4.3) are given by the minimum of
functionals J1,λ and J2,λ respectively.

Consider a function ϕ0 ∈ W
1,pi(x)
0 (Ω) for for i = 1, 2, with F̃i(ϕ0) > 0, for

i = 1, 2. Define (z0, w0) := (zλ̃0
, wλ̃0

), where λ̃0 satisfies∫
Ω

1

pi(x)
A(|∇ϕ0|pi(x))dx < λ̃0

∫
Ω

F̃i(ϕ0)dx,

for i = 1, 2. We have J1,λ̃0
(z0) ≤ J1,λ̃0

(ϕ0) < 0 and also that J2,λ̃0
(z0) < 0.

Therefore z0 6= 0 and w0 6= 0. Since −div(a1(|∇z0|p1(x))|∇z0|p1(x)−2∇z0) and
−div(a2(|∇w0|p2(x))|∇w0|p2(x)−2∇w0) are nonnegative, we have z0, w0 > 0 in Ω.
Note that by [22, Theorem 4.1] and [19, Theorem 1.2], we obtain that z0, w0 ∈
C1,α(Ω) for some α ∈ (0, 1].



12 A. RAZANI, G. M. FIGUEIREDO EJDE-2022/36

Using the test function ϕ = (z0 − θ1)+ ∈W 1,p1(x)
0 (Ω) in (4.2) we obtain∫

Ω

a1(|∇z0|p1(x))|∇z0|p1(x)−2∇z0∇(z0 − θ1)+dx

= λ̃0

∫
{z0>θ}

f̃1(z0)(z0 − θ1)dx = 0.

Therefore ∫
{z0>θ}

〈
a1(|∇z0|p1(x))|∇z0|p1(x)−2∇z0

− a1(|∇θ1|p1(x))|∇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. Define

A0 := max
{
A(x, t) : (x, t) ∈ Ω× [min{|z0|Lr2(x) , |w0|Lr1(x)},
max{|θ1|Lr2(x) , |θ2|Lr1(x)}]

}
and µ0 = A0/C. Then

−div(a1(|∇zλ|p1(x))|∇zλ|p1(x)−2∇zλ) = 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

−div(a1(|∇zλ|p1(x))|∇zλ|p1(x)−2∇zλ) ≤ 1

A(x, |w|Lr1(x))
λf1(z0)|w0|α1(x)

Lq1(x) .

If necessary, we can consider a larger λ0 > 0 such that

−div(a2(|∇w0|p2(x))|∇w0|p2(x)−2∇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 for i = 1, 2, we have that (z0, θ1)
and (w0, θ2) are sub-supersolutions pairs for (4.1). �

4.2. Sublinear problem. Here, we study a nonlocal problem to generalize [39,
Theorem 6]. We prove the following theorem.

Theorem 4.3. Assume that

• pi, qi, ri, si for i = 1, 2 satisfy (A4);
• αi, βi ∈ L∞(Ω), for i = 1, 2;
• for i = 1, 2, we have

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;

• a0 > 0 is a positive constant;
• One of the following two conditions holds



EJDE-2022/36 WEAK SOLUTION BY THE SUB-SUPERSOLUTION METHOD 13

(A7) A(x, t) ≥ a0 on Ω× [0,∞),
(A8) 0 < A(x, t) ≤ a0 on Ω × (0,∞), and limt→+∞A(x, t) = a∞ > 0

uniformly on Ω.

Then the problem

−A(x, |v|Lr1(x))
(
∆p1(x)u−∆u

)
= (uβ1(x) + vγ1(x))|v|α1(x)

Lq1(x) in Ω,

−A(x, |u|Lr2(x))
(
∆p2(x)v −∆v

)
= (uβ2(x) + vγ2(x))|u|α2(x)

Lq2(x) in Ω,

u = v = 0 on ∂Ω,

(4.4)

has a positive solution.

Proof. Suppose that (A7) is hold, that is, A(x, t) ≥ a0 in Ω× [0,+∞). We 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, (4.5)

0 < yλ(x) ≤ Kλ
1

p
−
2 −1 in Ω. (4.6)

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 (4.5), (4.6) 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 (4.7)

occur. By (4.5), (4.6) and (4.7), 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 −∆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 implies that

−∆p2(x)yλ −∆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 [29, Page 12], 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),



14 A. RAZANI, G. M. FIGUEIREDO EJDE-2022/36

belongs to C1
0 (Ω), where k > 0 is an arbitrary number and that

−∆pi(x)(µφi) =



−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),

and

−∆(µφi) =


−k(kµekd(x))

[
1 + ∆d(x)

k

]
if d(x) < σ,{

2
2δ−σ −

( 2δ−d(x)
2δ−σ

)
∆d(x)

}
(kµekσ)

( 2δ−d(x)
2δ−σ

)
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)}

]}
. 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

we have∣∣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.

(4.8)

Note that there exists a constant A > 0, that does not depend on k, such that
|∆d(x)| < A for all x ∈ ∂Ω3δ. Using the above inequality and the expression of
−∆p1(x)(µφ) and −∆(µφ), we obtain −∆p1(x)(µφ1) −∆(µφ1) ≤ 0 for x ∈ Ω with
d(x) < σ or d(x) > 2δ for k > 0 large enough. Therefore

−∆p1(x)(µφ1)−∆(µφ1) ≤ 0 ≤ 1

Aλ
(µφ1)β1(x)|µφ2|α1(x)

Lq1(x)

≤ 1

Aλ
((µφ1)β1(x) + wγ1(x))|µφ2|α1(x)

Lq1(x)



EJDE-2022/36 WEAK SOLUTION BY THE SUB-SUPERSOLUTION METHOD 15

for all w ∈ L∞(Ω) with w ≥ µφ2 and d(x) < σ or 2δ < d(x). Using an idea in [29,
estimate (3.10)], if σ < d(x) < 2δ, then

−∆p1(x)(µφ1)−∆(µφ1) ≤ C̃(kµ)p
−
1 −1| ln kµ|+ C̃(kµ)| ln kµ|

= C̃((kµ)p
−
1 −1 + kµ)

∣∣ ln k

eak
∣∣ . (4.9)

[40, Theorem 2] and α+
1 + γ+

1 < p−1 − 1 imply

lim
k→+∞

C̃kp
−
1 −1 + C̃k

eak(p−1 −1−(α+
1 +γ+

1 ))

∣∣ ln k

eak
∣∣ = 0. (4.10)

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

|φ2|α1(x)

Lq1(x)(Ω)
≥ C0 if σ < d(x) < 2δ. By (4.10), we can choose k > 0 large enough

such that

C̃kp
−
1 −1C̃k

eak[(p−1 −1)−(α+
1 +β+

1 )]

∣∣ ln k

eak
∣∣ ≤ C0

Aλ
. (4.11)

Therefore from (4.9) and (4.11), we have

−∆p1(x)(µφ1)−∆(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)−∆(µφ1) ≤ 1

Aλ
((µφ1)β1(x) + wγ1(x))|µφ2|α1(x)

Lq1(x) in Ω.

Fix k > 0 satisfying the above property and the inequality−∆p1(x)(µφ1)−∆(µφ1) ≤
1. For λ > 1 we have −∆p1(x)(µφ1) − ∆(µφ1) ≤ −∆p1(x)zλ − ∆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)−∆(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, i = 1, 2 as before. From the previous arguments there exist k > 0 large enough
and µ > 0 small such that

−∆p1(x)(µφ1)−∆(µφ1) ≤ 1 in Ω,

−∆p1(x)(µφ)−∆(µφ) ≤ 1

a0
((µφ1)β1(x) + wγ1(x))|µφ2|α1(x)

Lq1(x) in Ω
(4.12)

for all w ∈ [µφ2, yλ], also that

−∆p2(x)(µφ2)−∆(µφ2) ≤ 1 in Ω

−∆p2(x)(µφ2)−∆(µφ2) ≤ 1

a0
(wβ2(x) + (µφ2)γ2(x))|µφ1|α2(x)

Lq2(x) in Ω
(4.13)



16 A. RAZANI, G. M. FIGUEIREDO EJDE-2022/36

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 in Ω× (a1,∞). Let

mk := min
{
A(x, t) : (x, t) ∈ Ω× [min{|µφ1|Lr1(x) , |µφ2|Lr2(x)}, a1]

}
> 0 ,

Ak := min
{
mk,

a∞
2

}
.

We have A(x, t) ≥ Ak in Ω× [min{|µφ1|Lr1(x) , |µφ2|Lr2(x)},∞).
Fix k > 0 satisfying (4.12) and (4.13). Consider λ > 1 such that (4.5), (4.6),

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 and λ (see Lemma 2.7).
Therefore,

−∆p1(x)zλ −∆zλ ≤
1

A(x, |w|Lr1(x))
(z
β1(x)
λ + wγ1(x))|yλ|α1(x)

Lq1(x)

in Ω, w ∈ [µφ2, yλ]. Arguing as before and considering a suitable choice for λ and
k, we obtain

−∆p2(x)yλ −∆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 imply that µφ1 ≤ zλ and µφ2 ≤ yλ if
µ is small. �

Acknowledgments. The authors would like to express their gratitude to the
anonymous referees for their constructive comments and suggestions that allowed
to improve this manuscript.

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Abdolrahman Razani
Department of Pure Mathematics, Faculty of Science, Imam Khomeini International

University, Postal code: 34148-96818, Qazvin, Iran

Email address: razani@sci.ikiu.ac.ir

Giovany M. Figueiredo

Departamento de Matemática, Universidade de Brasilia, 70.910-900, Brasilia (DF), Brazil
Email address: giovany@unb.br


	1. Introduction
	2. Function spaces
	3. Existence of solutions
	4. Applications
	4.1. A generalized logistic equation
	4.2. Sublinear problem
	Acknowledgments

	References