Int. J. Anal. Appl. (2022), 20:52

On the Frictional Contact Problem of p(x)-Kirchhoff Type

Eugenio Cabanillas Lapa∗, Willy Barahona Martinez

Instituto de Investigación, Facultad de Ciencias Matemáticas-UNMSM, Lima, Perú

∗Corresponding author: cleugenio@yahoo.com

Abstract. In this article we consider a class of frictional contact problem of p(x)-Kirchhoff type, on

a bounded domain Ω ⊆ R2 . Using an abstract Lagrange multiplier technique and the Schauder’s
fixed point theorem we establish the existence of weak solutions. Furthermore, we also obtain the

uniqueness of the solution assuming that the datum f1 satisfies a suitable monotonicity condition.

1. Introduction

The purpose of this work is to investigate the existence of weak solutions for the boundary value

problem

−M
(∫

Ω

1

p(x)
|∇u|p(x)dx

)
div
(
|∇u|p(x)−2∇u

)
= f1(x,u) in Ω

u = 0 on Γ1

M
(∫

Ω

1

p(x)
|∇u|p(x)dx

)
|∇u|p(x)−2

∂u

∂ν
= f2(x) on Γ2∣∣∣M(∫

Ω

1

p(x)
|∇u|p(x)dx

)
|∇u|p(x)−2

∂u

∂ν

∣∣∣ ≤ g(x),
M
(∫

Ω

1

p(x)
|∇u|p(x)dx

)
|∇u|p(x)−2

∂u

∂ν
= −g

u

|u|
, if u 6= 0 on Γ3

(1.1)

where Ω ⊆ R2 is a bounded domain with smooth enough boundary Γ, partitioned in three parts
Γ1, Γ2, Γ3 such that meas (Γi ) > 0, (i = 1, 2, 3); f1 : Ω ×R → R, f2 : Γ2 → R, g : Γ3 → R and
M : [0, +∞[→ [m0, +∞[ are given functions, p ∈ C(Ω).
The study of the p(x)- Kirchhoff type equations with nonlinear boundary conditions of different class

Received: Aug. 16, 2022.

2010 Mathematics Subject Classification. 35J25, 46E35, 74G25.
Key words and phrases. p(x)-Kirchhoff type equation; frictional contact condition; Schauder fixed point theorem;

uniqueness of solutions.

https://doi.org/10.28924/2291-8639-20-2022-52
ISSN: 2291-8639

© 2022 the author(s).

https://doi.org/10.28924/2291-8639-20-2022-52


2 Int. J. Anal. Appl. (2022), 20:52

have attracted expensive attention in recent years, we refer to some interesting works [1,6,13,16] and

references therein. One reason of such interest is that various real fields require PDE problems with

variable exponent, for example, electrorheological fluids and image restoration. The other reason is that

the nonlocal problems with variable exponent, in addition to their contributions to the modeling of many

physical and biological phenomena, raise greater mathematical difficulties due to their nonlinearities;

see for example [2,15,19]. Cojocaru-Matei [5] studied the unique solvability of problem (1.1) in the

case M(s) = 1, f1(x,u) ≡ f1(x),p = constant ≥ 2, which models the antiplane shear deformation of
a nonlinearly elastic cylindrical body in frictional contact on Γ3 with a rigid foundation; see, e.g. [18].

They used a technique involving dual Lagrange multipliers, this allow to write efficient algorithms to

approximate the weak solutions; see [14]. For our situation, the behavior of the material is described

by the Hencky-type constitutive law:

σ(x) = ktrε(u(x))I3 + µ(x)‖εD(u(x))‖
p(x)−2

2 εD(u(x))

where σ is the Cauchy stress tensor, tr is the trace of a Cartesian tensor of second order,σ(x) ε is

the infinitesimal strain tensor, u is the displacement vector,I3 is the identity tensor, k,µ are material

parameters, p is a given function;εD is the desviator of the tensor ε defined by εD = ε − 1
3

(trε)I3

where trε =
3∑
i=1

εii; see for instance [12]. If, the Lamé coefficient is given by

µ = M
(∫

Ω

1

p(x)
|∇u|p(x)dx

)
we obtain our mechanical problem (1.1).

Thanks to the above mentioned research articles, we consider the problem (1.1), under appropriate

assumptions on M and f1, and establish the existence of a unique weak solution of this problem via

Lagrange multipliers and the Schauder fixed point theorem. In this sense, we generalize the main

result in [5]. Also, we state a simple uniqueness result under suitable monotonicity condition on f1.

The paper is designed as follows. In Section 2, we introduce the mathematical preliminaries and

give several important properties of p(x)-Kirchhoff-Laplace operator. We deliver a weak variational

formulation with Lagrange multipliers in a dual space. Section 3, is devoted to the proofs of main

results.

2. Preliminaries

For the reader’s convenience, we point out some basic results on the theory of Lebesgue-Sobolev

spaces with variable exponent. In this context we refer the reader to [8,17] for details. Firstly we state

some basic properties of spaces W 1,p(x)(Ω) which will be used later. Denote by S(Ω) the set of all

measurable real functions defined on Ω. Two functions in S(Ω) are considered as the same element



Int. J. Anal. Appl. (2022), 20:52 3

of S(Ω) when they are equal almost everywhere. Write

C+(Ω) = {h : h ∈ C(Ω),h(x) > 1 for any x ∈ Ω},

h− := min
Ω
h(x), h+ := max

Ω
h(x) for every h ∈ C+(Ω).

Define

Lp(x)(Ω) = {u ∈S(Ω) :
∫

Ω

|u(x)|p(x) dx < +∞ for p ∈ C+(Ω)}

with the norm

|u|Lp(x)(Ω) = |u|p(x) = inf{λ > 0 :
∫

Ω

|
u(x)

λ
|p(x) dx ≤ 1},

and

W 1,p(x)(Ω) = {u ∈ Lp(x)(Ω) : |∇u| ∈ Lp(x)(Ω)}

with the norm

‖u‖1,p(x) = |u|Lp(x)(Ω) + |∇u|Lp(x)(Ω).

Proposition 2.1 ( [11]). The spaces Lp(x)(Ω) and W 1,p(x)(Ω) are separable reflexive Banach spaces.

Proposition 2.2 ( [11]). Set ρ(u) =
∫

Ω
|u(x)|p(x) dx. For any u ∈ Lp(x)(Ω), then

(1) for u 6= 0, |u|p(x) = λ if and only if ρ( uλ) = 1;
(2) |u|p(x) < 1 (= 1; > 1) if and only if ρ(u) < 1 (= 1; > 1);
(3) if |u|p(x) > 1, then |u|

p−

p(x)
≤ ρ(u) ≤ |u|p

+

p(x)
;

(4) if |u|p(x) < 1, then |u|
p+

p(x)
≤ ρ(u) ≤ |u|p

−

p(x)
;

(5) limk→+∞ |uk|p(x) = 0 if and only if limk→+∞ρ(uk) = 0;
(6) limk→+∞ |uk|p(x) = +∞ if and only if limk→+∞ρ(uk) = +∞.

Proposition 2.3 ( [9,11]). If q ∈ C+(Ω) and q(x) ≤ p∗(x) (q(x) < p∗(x)) for x ∈ Ω, then there is a
continuous (compact) embedding W 1,p(x)(Ω) ↪→ Lq(x)(Ω), where

p∗(x) =




Np(x)
N−p(x) if p(x) < N,

+∞ if p(x) ≥ N.

Proposition 2.4 ( [11]). The conjugate space of Lp(x)(Ω) is Lq(x)(Ω), where 1
q(x)

+ 1
p(x)

= 1 holds

a.e. in Ω. For any u ∈ Lp(x)(Ω) and v ∈ Lq(x)(Ω), we have the following Hölder-type inequality∣∣∫
Ω

uv dx
∣∣ ≤ ( 1

p−
+

1

q−
)|u|p(x)|v|q(x).

We introduce the following closed space of W 1,p(x)(Ω)

X = {v ∈ W 1,p(x)(Ω) : γu = 0 a. e. on Γ1} (2.1)

where γ denotes the Sobolev trace operator and Γ1 ⊆ Γ, meas (Γ1) > 0, therefore X is a separable
reflexive Banach space. Now, we denote

‖u‖X = |∇u|p(x), u ∈ X.



4 Int. J. Anal. Appl. (2022), 20:52

This functional represents a norm on X.

Proposition 2.5 ( [3]). There exists c > 0 such that

‖u‖1,p(x) ≤ C‖u‖X for all u ∈ X.

Then, the norms ‖.‖X and ‖.‖1,p(x) are equivalent on X.
We write

L(u) =

∫
Ω

1

p(x)
|∇u|p(x) dx.

Proposition 2.6. The functional L : X → R is convex. The mapping L′ : X → X′ is a strictly
monotone, bounded homeomorphism, and is of (S+) type, namely

un ⇀ u and lim sup
n→+∞

L′(un)(un −u) ≤ 0 implies un → u,

where X′ is the dual space of X.

Proof. This result is obtained in a similar manner as the one given in [10], thus we omit the details. �

Now, we define the spaces

S =
{
u ∈ W

1
p′(x) ,p(x)(Γ) : ∃v ∈ X such that u = γv a.e on Γ

}
(2.2)

which is a real reflexive Banach space,
1

p(x)
+

1

p′(x)
= 1 for all x ∈ Ω, and

Y = S′, the dual of the space S. (2.3)

Let us introduce a bilinear form

b : X ×Y −→R : b(v,µ) = 〈 µ,γv 〉Y×S , (2.4)

a Lagrange multiplier λ ∈ Y ,

〈 λ,z 〉 = −
∫

Γ3

M
(∫

Ω

1

p(x)
|∇u|p(x)dx

)
|∇u|p(x)−2

∂u

∂ν
zdΓ , ∀z ∈ S

and the set of Lagrange multipliers

Λ =
{
u ∈ Y : 〈 µ,z 〉6

∫
Γ3

g(x)|z(x)| , ∀z ∈ S
}
. (2.5)

From (1.1)4 we deduce that λ ∈ Λ.
Let u be a regular enough function satisfying problem (1.1). After some computations we get (by

using density results)

M
(∫

Ω

1

p(x)
|∇u|p(x)dx

)∫
Ω

|∇u|p(x)−2∇u.∇vdx =
∫

Ω

f1(x,u)vdx

+

∫
Γ2

f2(x)γvdΓ + M
(∫

Ω

1

p(x)
|∇u|p(x)dx

)∫
Γ3

|∇u|p(x)−2
∂u

∂ν
γvdΓ (2.6)



Int. J. Anal. Appl. (2022), 20:52 5

for allv ∈ X , where u satisfies (1.1)5 on Γ3
Now, we write problem (2.6) as an abstract mixed variational problem (by means a Lagrange

multipliers technique)

We define the following operators:

i) A : X → X′,given by

〈 Au,v 〉 = M
(∫

Ω

1

p(x)
|∇u|p(x)dx

)∫
Ω

|∇u|p(x)−2∇u.∇vdx, u,v ∈ X.

ii) F : X → X′,given by

〈 F (u),v 〉 =
∫

Ω

f1(x,u)vdx +

∫
Γ2

f2(x)γvdx , u,v ∈ X.

(2.7)

So, we are led to the following variational formulation of problem (1.1)

Problem 1. Find u ∈ X and λ ∈ Λ such that

〈 Au,v 〉 + b(v,λ) = 〈 F (u),v 〉 , ∀v ∈ X (2.8)

b(u,µ−λ) ≤ 0 ∀µ ∈ Λ ⊆ Y

To solve this problem, we will apply the Schauder fixed point theorem.

Firstly, we "freeze" the state variable u on the function F, that is we fix w ∈ X such that
f = F (w) ∈ X′.

So, we are led to the following abstract mixed variational problem.

Problem 2. Given f ∈ X′ find u ∈ X and λ ∈ Λ such that

〈 Au,v 〉 + b(v,λ) = 〈 f ,v 〉 , ∀v ∈ X

b(u,µ−λ) ≤ 0 ∀µ ∈ Λ ⊆ Y. (2.9)

The unique solvability of Problem 2 is given under the following generalized assumptions.

Let (X,‖‖X) and (Y,‖‖Y ) be two real reflexive Banach space.

(B1): A : X → X′ is hemicontinuous;
(B2): ∃h : X →R such that
(a) h(tw) = tγh(w) with γ > 1 , ∀t > 0,w ∈ X;
(b) 〈 Au −Av,u −v 〉X×X ≥ h(v −u), ∀u,v ∈ X;
(c) ∀(xν) ⊆ X : xν ⇀ x inX =⇒ h(x) ≤ lim

ν→∞
sup h(xν)

(B3): A is coercive.

(B4): The form b : X ×Y es bilinear, and
(i) ∀(uν) ⊆ X : uν ⇀ u in X =⇒ b(uν,λν) → b(u,λ)
(ii) ∀(λν) ⊆ Y : λν ⇀ y in Y =⇒ b(vν,λν) → b(v,λ)

(iii) ∃α̂ > 0 : inf
µ∈I
u 6=0

sup
v∈X
v 6=0

b(v,µ)

|v|X|µ|Y
≥ α̂



6 Int. J. Anal. Appl. (2022), 20:52

(B5): Λ is a bounded closed convex subset of Y such that 0Y ∈ Λ.
(B6): ∃C1 > 0,q > 0 : h(v) ≥ C1‖v‖

q
X

, ∀v ∈ X.

Theorem 2.1. Assume (B1) - (B6). Then there exists a unique solution (u,λ) ∈ X × Λ of Problem
2.

Proof. See [5].

To solve Problem 1, we start by stating the following assumptions on M , f1 , f2 and g

(A1) M : [0, +∞[→ [m0, +∞[ is a locally Lipschitz-continuous and nondecreasing function; m0 > 0.
(A2) f1 : Ω ×R→R is a Caratheodory function satisfying

|f1(x,t)| ≤ c1 + c2|t|α(x)−1 , ∀(x,t) ∈ Ω ×R,

α ∈ C+(Ω)with α(x) < p∗(x), α+ < p−.
(A3) f2 ∈ Lp

′(x)(Γ2), g ∈ Lp
′(x)(Γ3), g(x) ≥ 0 a.e on Γ3.

We have the following properties about the operator A.

Proposition 2.7. If (A1) holds, then

(i) A is locally Lipschitz continuous.

(ii) A is bounded, strictly monotone. Furthermore

〈Au −Av,u −v〉≥ kp‖u −v‖
p̂
X

where

p̂ =


p
− if ‖u −v‖X > 1,

p+ if ‖u −v‖X ≤ 1.

So, we can take h(v) = kp‖v‖
p̂
X
.

(iii) 〈Au,u〉‖u‖X → +∞ as ‖u‖X → +∞.

Proof. (i) Assume that M is Lipschitz in [0,R1] with Lipschitz constant LM, R1 > 0. We have, for

u,v,w ∈ B(0,R1)

〈Au −Av,w〉 = [M(L(u)) −M(L(v))]
∫

Ω

|∇u|p(x)−2∇u.∇v dx

+ M(L(v))

∫
Ω

(
|∇u|p(x)−2∇u −|∇v|p(x)−2∇v

)
.∇w dx.

Using the Lipschitz continuity of M, the Holder inequality and the inequality 〈||x|α−2x−|y|α−2y,x−
y〉| ≤ c|x −y|(|x| + |y|)α−2 , ∀x,y ∈Rn, 2 ≤ α < +∞, we get

|〈Au −Av,w〉| ≤ C‖u −v‖X‖w‖X,

which implies ‖Au −Av‖X′ ≤ C‖u −v‖X.



Int. J. Anal. Appl. (2022), 20:52 7

ii)The functional S : X → X′ defined by

〈Su,v〉 =
∫

Ω

|∇u|p(x)−2∇u.∇v dx, ∀u,v ∈ X, (2.10)

is bounded (See [10]). Then

〈Su,v〉 = M(L(u))〈Su,v〉 ∀u,v ∈ X. (2.11)

Hence, since M is continuous and L is bounded (see Proposition 2.6), A is bounded.

To obtain that A is strictly monotone we develop the same arguments to those in [7], we omit it.

To establish the inequality in ii), we apply Lemma 3 in [4] to obtain

〈Au −Av,u −v〉≥
∫

Ω

(
M(L(u))|∇u|p(x)−2∇u −M(L(v))|∇v|p(x)−2∇v

)
.(∇v −∇u) dx

≥m0
∫

Ω

1

p(x)
(|∇u −∇u|p(x)) dx ≥

m0
p+

∫
Ω

|∇u −∇u|p(x) dx

≥
m0
p+
‖u −v‖p̂

X
.

iii)For u ∈ X with ‖u‖X > 1 we have

〈Au,u〉
‖u‖X

=

M
(∫

Ω

1

p(x)
|∇u|p(x) dx

)∫
Ω

|∇u|p(x) dx

‖u‖

≥m0‖u‖
p−−1
X

→ +∞ as ‖u‖X → +∞.

�

Proposition 2.8. The form b : X ×Y →R defined in (2.4) is bilinear and, it verifies i), ii) and iii) in
assumption (B4). Moreover

b(u,µ) ≤
∫

Γ3

g(x)|u(x)|dΓ for all µ ∈ Λ; (2.12)

b(u,λ) =

∫
Γ3

g(x)|u(x)|dΓ (2.13)

b(u,µ−λ) ≤0 for all µ ∈ Λ. (2.14)

Moreover, Λ is a bounded closed convex subset of Y such that 0Y ∈ Λ.

Proof. The assertions i), ii), iii) and Λ bounded are word for word as [5], Theorem 3, pags 138-139.

It is obvious to check (2.12). To justify (2.13), we have to show that, a.e. x ∈ Ω

−M
(∫

Ω

1

p(x)
|∇u|p(x)dx

)
|∇u(x)|p(x)−2

∂u(x)

∂ν
u(x) = g(x)|u(x)|

In fact, let x ∈ Ω . If |u(x)| = 0, then

−M
(∫

Ω

1

p(x)
|∇u|p(x) dx

)
|∇u(x)|p(x)−2

∂u(x)

∂ν
u(x) = 0 = g(x)|u(x)| on Γ3.



8 Int. J. Anal. Appl. (2022), 20:52

Otherwise, if |u(x)| 6= 0,then

−M(
∫

Ω

1

p(x)
|∇u|p(x) dx)|∇u(x)|p(x)−2

∂u(x)

∂ν
u(x) =g(x)

(u(x))2

|u(x)|

=g(x)|u(x)| on Γ3

Furthermore, for all µ ∈ Λ :

b(u,µ−λ) = b(u,µ) −b(u,λ) = 〈 µ,γu 〉Y×S −〈 λ,γu 〉Y×S . (2.15)

Hence, thanks to (2.12), (2.13) and (2.15), we obtain (2.14). �

3. Existence and uniqueness of solutions

We are ready to solve problem 1. For this, we consider the Banach spaces X and Y given in (2.1)

and (2.3) respectively, the form b : X ×Y →R defined in (2.4) and the set Λ in (2.5).

Theorem 3.1. Suppose (A1) − (A3) hold. Then problem 1 admits a solution (u,λ) ∈ X × Λ.

Proof. We apply the Schauder fixed point theorem.

As has been said before, we "freeze" the state variable u on the function F , that is, we fix w ∈ X
and consider the problem:

Find u ∈ X and λ ∈ Λ such that

〈 Au,v 〉 + b(v,λ) = 〈 f ,v 〉 , ∀v ∈ X (3.1)

b(u,µ−λ) ≤ 0 ∀µ ∈ Λ ⊆ Y. (3.2)

with f = F (w) ∈ X′. Note that by the hypotheses on α and f1, given in (A2), we have f1(w) ∈
Lα
′(x)(Ω) ↪→ X′.
By theorem (2.1), problem (3.1)-(3.2) has a unique solution (uw,λw ) ∈ X × Λ.
Here we drop the subscript w for simplicity. Setting v = u in (3.1) and µ = 0Y in (3.2), using

proposition 2.7 ii), we get

kp‖u‖
p̂
X
≤ (2C1Cα‖w‖σX + 2C2Cα|Ω| + cp|f2|p′(x),Γ2 )‖u‖X (3.3)

where

σ =


α

− if ‖w‖X > 1,

α+ if ‖w‖X ≤ 1,

and Cχ is the embedding constant of X ↪→ Lχ(x)(Ω).
Then

‖u‖X ≤ [C(1 + ‖w‖X)]
1
p̂−1 .

Therefore, either ‖u‖X ≤ 1 or

‖u‖X ≤ [C(1 + ‖w‖X)]
1

p−−1 . (3.4)



Int. J. Anal. Appl. (2022), 20:52 9

Since p− > α+ + 1, we have

tp
−−1 −Ctσ −C → +∞ as t → +∞

Hence, there is some R̄1 > 0 such that

R̄1
p−−1 −CR̄1

σ −C ≥ 0 (3.5)

From (3.4) and (3.5) we infer that if ‖w‖X ≤ R̄1 then ‖u‖X ≤ R̄1.
Thus there exists R1 = min{1, R̄1} such that

‖u‖X ≤ R1 for all u ∈ X. (3.6)

For this constant, define K as

K = {v : v ∈ Lα(x)(Ω),‖v‖X ≤ R1}

which is a nonempty, closed, convex subset of Lα(x)(Ω). We can define the operator

T : K → Lα(x)(Ω), Tw = uw

where uw is the first component of the unique pair solution of the problem (3.1)-(3.2), (uw,λw ) ∈
X × Λ

From (3.6) ‖Tw‖X ≤ R1, for every w ∈ K, so that T (K) ⊆ K.
Moreover, if (uν)ν≥1 (uwν ≡ uν) is a bounded sequence in K, then from (3.6) is also bounded in

X. Consequently, from the compact embedding X ↪→ Lα(x)(Ω), (Twν)ν≥1 is relatively compact in
Lα(x)(Ω) and hence, in K.

To prove the continuity of T , let (wν)ν≥1 be a sequence in K such that

wν → w strongly inLα(x)(Ω) (3.7)

and suppose uν = Twν. The sequence {(uν,λν)}ν≥1 satisfies

〈 Auν,v 〉 + b(v,λν) = 〈 F (wν),v 〉 , ∀v ∈ X

b(uν,µ−λν) ≤ 0 ∀µ ∈ Λ.

Using (3.6)-(3.7) we can extract a subsequence (uνk ) of (uν) and a subsequence (wνk ) of (wν) such

that
uνk → u

∗weakly inX,

uνk → u
∗ strongly in Lα(x)(Ω) and a.e. in Ω,

wνk → w a.e. in Ω,

L(uνk ) → t0, for some t0 ≥ 0,

(3.8)

and in view of continuity of M

M(L(uνk )) → M(t0). (3.9)



10 Int. J. Anal. Appl. (2022), 20:52

We shall show that u∗ = Tw. To this end, by choosing uνk −u
∗ as a test function, we have〈

Auνk,uνk −u
∗ 〉 + b(uνk −u∗,λν) = 〈 F (wνk ),uνk −u∗ 〉〈

Au∗,uνk −u
∗ 〉 + b(uνk −u∗,λ∗) = 〈 F (w),uνk −u∗ 〉 . (3.10)

Then

[M(L(u∗) −M(L(uνk )]
∫

Ω

|∇u∗|p(x)−2∇u∗.(∇uνk −∇u
∗) dx+

M(L(uνk ))

∫
Ω

(|∇u∗|p(x)−2∇u∗ −|∇uνk |
p(x)−2∇uνk ).(∇uνk −∇u

∗) dx+

b(uνk −u
∗,λ∗ −λνk ) =

〈
F (w) −F (wνk ),uνk −u

∗ 〉 .
(3.11)

Since b(uνk −u
∗,λ∗ −λνk ) ≥ 0, by the inequality |x|

p−2x −|y|p−2y ≥ C|x −y|p,
p ≥ 2, we obtain

m0Cp

∫
Ω

|∇uνk −∇u
∗|p(x) dx + [M(L(u∗) −M(L(uνk )]

∫
Ω

|∇u∗|p(x)−2∇u∗.(∇uνk −∇u
∗) dx

≤ |
〈
F (wνk ) −F (w),uνk −u

∗ 〉 |
(3.12)

But, using (3.8) we get

|[M(L(u∗) −M(L(uνk )]
∫

Ω

|∇u∗|p(x)−2∇u∗.(∇uνk −∇u
∗) dx|

≤
ϑνk
p−
|
∫

Ω

|∇u∗|p(x)−2∇u∗.(∇uνk −∇u
∗) dx|→ 0 as k →∞,

(3.13)

where ϑνk = max{‖uνk‖
p−

X
,‖uνk‖

p+

X
} + max{‖u∗‖p

−

X
,‖u∗‖p

+

X
} is bounded.

Also, by (A2), (3.8) and the compact embedding of X ↪→ Lα(x)(Ω) we deduce, thanks to the
Krasnoselki theorem, the continuity of the Nemytskii operator

Nf1 : L
α(x)(Ω) → Lα

′(x)(Ω)

w 7−→ Nf1 (w),
(3.14)

given by (Nf1 (w))(x) = f1(x,w(x)), x ∈ Ω.
Hence

‖f1(wνk ) − f1(w)‖α′(x) → 0

It follows from the definition of F and the above convergence that

|
〈
F (wνk ) −F (w),uνk −u

∗ 〉 |→ 0 (3.15)
Thus, from (3.12)-(3.15) we conclude that

uνk → u
∗ in X

Since the possible limit of the sequence (uν)ν≥1 is uniquely determined, the whole sequence converges

toward u∗ ∈ X
Therefore, from (3.7) and the continuous embedding X ↪→ Lα(x)(Ω), we get u∗ = Tw ≡ uw.



Int. J. Anal. Appl. (2022), 20:52 11

On the other hand

b(v,λ)

‖v‖X
=
〈F (w),v〉−〈Au,v〉

‖v‖X
≤
〈F (w),v〉
‖v‖X

+ ‖Au‖X′

≤
1

‖v‖X

[∫
Ω

f1(x,w)v dx +

∫
Γ2

f2(x)γv dΓ

]
+ LA‖u‖X + ‖A0‖X′

≤ C(‖f1(w)‖α′(x) + ‖f2‖p′(x),Γ2 + ‖A0‖X′ + 1)

(3.16)

Next, using the boundedness of the operator Nf1 and the sequence (uν)ν≥1, and the inf-sup property

of the form b, we get ‖λν‖Y ≤ C. It follows that up to a subsequence

λν → λ0 weakly in Y

for some λ0 ∈ Y .
So (u∗,λ∗) and (u∗,λ0) are solutions of problem (3.1)-(3.2).Then, by the uniqueness λ0 = λ∗ ≡

λw. This shows the continuity of T.

To prove that T is compact, let (wν)ν≥1 ⊆ K be bounded in Lα(x)(Ω) and uν = T (wν). Since
(wν)ν≥1 ⊆ K, ‖wν‖X ≤ C and then, up to a subsequence again denoted by (wν)ν≥1 we have

wν → w weakly in X

By the compact embedding Xinto Lα(x)(Ω), it follows that

wν → w strongly in Lα(x)(Ω).

Now, following the same arguments as in the proof of the continuity of T we obtain

uν = T (wν) → T (w) = u strongly in X

Thus

T (wν) → T (w) strongly in Lα(x)(Ω).

Hence, we can apply the Schauder fixed point theorem to obtain that T possesses a fixed point.

This gives us a solution of (u,λ0) ∈ X × Λ of Problem 1, which concludes the proof. �

Next, we consider the uniqueness of solutions of (2.8). To this end, we also need the following

hypothesis on the nonlinear term f1.

(A4) There exists b0 ≥ 0 such that

(f1(x,t) − f1(x,s))(t − s) ≤ b0|t − s|p(x) a.e. x ∈ Ω,∀t,s ∈R.

Our uniqueness result reads as follows.

Theorem 3.2. Assume that (A1) − (A4) hold. If, in addition 2 ≤ p for all x ∈ Ω̄, then (2.8) has a
unique weak solution provided that

kp

b0λ
−1
∗

< 1,



12 Int. J. Anal. Appl. (2022), 20:52

where

λ∗ = inf
u∈X\{0}

∫
Ω
|∇u|p(x) dx∫

Ω
|u|p(x) dx

> 0.

Proof. Theorem 3.1 gives a weak solution (u,λ) ∈ X × Λ. Let (u1,λ1), (u2,λ2) be two solutions of
(2.8). Considering the weak formulation of u1 and u2 we have

〈 Aui,v 〉 + b(v,λi ) = 〈 F (ui ),v 〉 , ∀v ∈ X (3.17)

b(ui,µ−λi ) ≤ 0 ∀µ ∈ Λ ⊆ Y i = 1, 2.

By choosing v = u1 −u2, µ = λ2 if i = 1 and µ = λ1 if i = 2, we have

〈Au1 −Au2,u1 −u2 〉 + b(u1 −u2,λ1 −λ2) = 〈F (u1) −F (u2),u1 −u2 〉 ,∀v ∈ X

b(u1 −u2,λ2 −λ1) ≤ 0 ∀µ ∈ Λ ⊆ Y. (3.18)

It gives

〈Au1 −Au2,u1 −u2 〉 = 〈F (u1) −F (u2),u1 −u2 〉 + b(u1 −u2,λ2 −λ1)

Thus, using (3.18) and repeating the argument used in the proof of Proposition 2.7, ii) we get

kp

∫
Ω

|∇u1 −∇u2|p(x) dx ≤ |〈 f1(u1) − f1(u2),u1 −u2 〉 |

≤ |
∫

Ω

(f1(x,u1) − f1(x,u2))(u1 −u2) dx|

≤ |
∫

Ω

|u1 −u2|p(x) dx ≤ b0λ−1∗
∫

Ω

|∇u1 −∇u2|p(x) dx

Consequently when kp
b0λ

−1
∗
< 1, it follows that u1 = u2. This completes the proof. �

Acknowledgements: This research was partially supported by the Vicerectorado de Investigación-

UNMSM-Perú and is part of the doctoral thesis of the second author.

Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the publi-

cation of this paper.

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	1. Introduction
	2. Preliminaries
	3. Existence and uniqueness of solutions
	Acknowledgements:

	References