Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 51, pp. 1–30.

ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu

DOI: 10.58997/ejde.2025.51

NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS IN A

DOMAIN WITH THE BOUNDARY EDGE

MIKHAIL BORSUK

Abstract. We investigate the behavior of weak solutions to the mixed problem with Neu-

mann and Robin boundary conditions for an elliptic quasi-linear second-order equation with the
variable p(x)-Laplacian in a neighborhood of the boundary edge.

1. Introduction

The aim of this article is the investigation of the behavior of the weak solutions to the mixed
problem with Neumann and Robin boundary conditions for quasi-linear elliptic second-order equa-
tions with the variable p(x)-Laplacian in a neighborhood of the boundary edge of 3-dimensional
cylindrical sector. Boundary value problems for elliptic second order equations with a non-standard
growth in function spaces with variable exponents have been an active investigations in recent
years. We refer to [10] for an overview. Differential equations with variable exponents-growth
conditions arise from the nonlinear elasticity theory, electrorheological fluids, etc. The Neumann -
Robin boundary conditions appear in the solving Sturm-Liouville problems which are used in many
contexts of science and engineering: for example, in electromagnetic problems, in heat transfer
problems and for convection-diffusion equations (Fick’s law of diffusion). These problems plays a
major role in the study of reflected shocks in transonic flow. Important applications of this prob-
lems is the capillary problem. There are many essential differences between the variable exponent
problems and the constant exponent problems. In the variable exponent problems, many singular
phenomena occurred and many special questions were raised. Zhikov [13, 14] has gave examples
of the Lavrentiev phenomenon for the variational problems with variable exponent.

Most of the works devoted to the quasi-linear elliptic second-order equations with the variable
p(x)-Laplacian refers to the Dirichlet problem in smooth bounded domains (see [10]). Concerning
the Robin problem for such equations we know only a few articles [1, 7, 8, 11], but in these works
a domain is smooth and lower order terms depend only on (x, u) and do not depend on |∇u|. Our
article [3] is dedicated to the Robin problem in a cone for such equations with a singular p(x)−
power gradient lower order term. The present article is a generalization of [3] and chapter 10 (with
χ(ω1) ≡ 0) of our monograph [2]. Here we describe qualitatively the behavior of the weak solution
to the mixed problem with the Neumann - Robin boundary conditions near a boundary edge of
3-dimensional cylindrical sector, namely we derive the sharp estimate of the type |u(x)| = O(|x|κ)
for the weak solution modulus (for the solution decrease rate) of our problem near a boundary
edge.

Our research methodology is based on:

• the investigation of the corresponding nonlinear eigenvalue problem;
• the maximum principle and the Stampacchia level method (see e.g. [12]);
• the comparison principle;
• the method of the barrier function.

2020 Mathematics Subject Classification. 35J20, 35J25, 35J70.

Key words and phrases. p(x)-Laplacian; boundary edge; mixed problem; Neumann problem; Robin problem.
©2025. This work is licensed under a CC BY 4.0 license.
Submitted January 20, 2025. Published May 24, 2025.

1



2 M. BORSUK EJDE-2025/51

Let R0 > 0, z0 > 0 be fixed, (r, ω, z) be the cylindrical coordinates of x = (x1, x2, x3) ∈ R3:

x1 = r cosω, x2 = r sinω, x3 = z; r ∈ (0,+∞), ω ∈ (−π, π), z ∈ (−∞,+∞)

and GR0
0 be an open bounded cylindrical sector in R3 with

∂GR0
0 = ΓR0

− ∪ ΓR0
+ ∪ Γ0 ∪ ΩR0

∪Π− ∪Π+,

• ΓR0
± = {(r, ω, z) : 0 < r < R0, ω = ±ω0

2 , z ∈ [−z0,+z0]}: 2-faces of a cylindrical sector;

• Γ0 = {(x1, x2, z) ∈ R3 : x1 = x2 = 0, z ∈ [−z0,+z0]} is the cylindrical sector edge;
• ΩR0 = {(R0, ω, z) : ω ∈ [−ω0

2 ,
ω0

2 ], z ∈ [−z0,+z0]}: lateral side of a cylindrical sector;
• Π± = {(r, ω, z) : r ∈ (0, R0), ω ∈ (−ω0

2 ,
ω0

2 ), z = ±z0}: upper/lower 2-face of a bounded
cylindrical sector.

Figure 1. Cylindrical sector

For cylindrical sectors, we use the following notation:

• dx = r dr dω dz: element of volume; ds
∣∣
Γ
R0
±

= drdz: area element of lateral 2-face ΓR0
± ;

dΩR = Rdω dz: area element of lateral 2-surface ΩR.
• △p(x)u ≡ div

(
|∇u|p(x)−2∇u

)
.

We investigate, in a neighborhood of the domain edge, the behavior of weak solutions to the mixed
problem with the Robin - Neumann boundary conditions on the lateral surface of the dihedral
cone:

−△p(x)u+ a(x)u|u|p(x)−1 + b(u,∇u) = f(x), x ∈ GR0
0 ,

∂u

∂−→n

∣∣∣
Γ
R0
− ∪Π−∪Π+

= 0, |∇u|p(x)−2 ∂u

∂−→n
+

γ(ω)

rp(x)−1
u|u|p(x)−2 = g(x), x ∈ ΓR0

+ ∪ ΩR0
.

(1.1)

We will work under the following assumptions:

(A1) 1 < p− ≤ p(x) ≤ p+ = p(0) < 3 for all x ∈ GR0
0 ;



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 3

(A2) p(x) ∈ C0,1(GR0
0 ) =⇒ 0 ≤ p(0) − p(x) ≤ L(z)r for all x ∈ GR0

0 , where L(z) is a nonde-
creasing function: L(0) = 1, L(z) ≤ L(z0) = L0 ≥ 1 =⇒

p+ − p(x) ≤ L0r, ∀x ∈ GR0
0 ;

(A3) |f(x)| ≤ f0r
β(x), f0 = const ≥ 0, β(x) = κ (p(x)− 1) − 2

s , s >
2

p−−1 > 1, κ = p+−1
p+−1+µλ,

0 ≤ µ < 2/3 for all x ∈ GR0
0 , where λ is the least positive eigenvalue of problem (2.1) (see

Section 2),

a(x) ≥

{
a0 = const > 0, if R0 ≤ 1,

a0r
−κ−p(x), if R0 > 1,

for all x ∈ GR0
0 . Also we suppose the validity of inequality (5.8) in section 5;

(A4) γ(ω) ∈ C0[−ω0

2 ,+
ω0

2 ], γ(ω) ≥ γ = const ≥ 1 for all ω ∈ [−ω0

2 ,+
ω0

2 ], γ(ω0

2 ) = γ0;

(A5) |g(x)| ≤ g0r
1+β(x), 0 ≤ g0 = const ≪ 1 (see (3.55)), g(x) ≤ 0 for all x ∈ ΓR0

+ ∪ ΩR0
;

(A6) the function b(u, ξ) is differentiable with respect to the u, ξ variables in M = R× Rn and
satisfy the following inequalities in M.

(A7) if µ = 0:

b(u, ξ) ≥ ν|u|−1|ξ|p(x) − b0|u|p(x)−1, ν > 0,

|b(u, ξ)| ≤ δ0|u|−1|ξ|p(x) + b0|u|p(x)−1, δ0 ≥ ν;

;
(A8) if µ > 0:

|b(u, ξ)| ≤ δ+|u|−1|ξ|p(x) + b0|u|p(x)−1, 0 < δ+ < µ;

(A9) ( n∑
i=1

∣∣∂b(u, ξ)
∂ξi

∣∣2)1/2

≤ b1|u|−1|ξ|p(x)−1,
∂b(u, ξ)

∂u
≥ b2|u|−2|ξ|p(x),

0 ≤ b0 ≤ a0, b1 ≥ 0, b2 ≥ 0.

We consider the class of functions

N
1,p(x)
−1,∞ (GR0

0 ) =
{
u : u(x) ∈ L∞(GR0

0 ) and

∫
G

R0
0

⟨r−p(x)|u|p(x) + |u|−1|∇u|p(x)}⟩ dx <∞
}
.

It is obvious that N
1,p(x)
−1,∞ (GR0

0 ) ⊂W 1,p(x)(GR0
0 ).

Definition 1.1. The function u is called a weak bounded solution of problem (1.1) provided that

u(x) ∈ N
1,p(x)
−1,∞ (GR0

0 ) and satisfies the integral identity

Q(u, η) :≡
∫
G

R0
0

〈
|∇u|p(x)−2uxi

ηxi
+ a(x)u|u|p(x)−1η + b (u,∇u) η

〉
dx

+

∫
Γ
R0
+

γ(ω)r1−p(x)u|u|p(x)−2η dS +

∫
ΩR0

γ(ω)R
1−p(x)
0 u|u|p(x)−2η dΩR0

−
∫
Γ
R0
+ ∪ΩR0

g(x)η dS

=

∫
G

R0
0

f(x)η(x) dx

(1.2)

for all η(x) ∈ N
1,p(x)
−1,∞ (GR0

0 ).

Remark 1.2. It is easy to verify that the assumptions (A1), (A3)–(A5) and (A6)–(A9) ensure
the existence of integrals in the identity (1.2). Therefore, the definition 1 is correct.

Our main result reads as follows.



4 M. BORSUK EJDE-2025/51

Theorem 1.3. Let u be a weak bounded solution of problem (1.1), M0 = sup
x∈Gd0

0
|u(x)| (see The-

orem3.4) and let λ be the least positive eigenvalue of problem (2.1). Suppose that
(A1)–(A9) hold. Then there exists a constant C0 > 0 depending only on λ,R0,M0, p+, p−, L0, n, (µ−
δ), ν, b0, f0, g0, and such that

|u(x)| ≤ C0r
κ , κ =

p+ − 1

p+ − 1 + µ
λ; ∀x ∈ GR0

0 . (1.3)

2. Nonlinear eigenvalue problem

To prove the main result we shall consider the nonlinear eigenvalue problem for the p+-Laplace-
Beltrami equation and for ψ(ω) ∈ C2

(
− ω0

2 ,
ω0

2

)
∩ C1[−ω0

2 ,
ω0

2 ]:

−
(
(λ2ψ2 + ψ′2)(p+−2)/2ψ′

)′
= λ (λ(p+ − 1) + 2− p+) (λ

2ψ2 + ψ′2)(p+−2)/2ψ, ω ∈
(
− ω0

2
,
ω0

2

)
,

ψ′(− ω0

2

)
= 0,

(λ2ψ2 + ψ′2)(p+−2)/2ψ′(ω) + γ ·
(p+ − 1 + µ

p+ − 1

)p+−1 × ψ(ω)|ψ(ω)|p+−2 = 0, ω = +
ω0

2
.

(2.1)
or (

λ2ψ2 + (p+ − 1)ψ′2
)
ψ′′(ω) + λ

(
λ(2p+ − 3) + 2− p+

)
ψ′2(ω)ψ(ω)

+ λ3
(
λ(p+ − 1) + 2− p+

)
ψ3(ω) = 0, ω ∈

(
−ω0

2
,
ω0

2

)
,

ψ′(− ω0

2

)
= 0,

(λ2ψ2 + ψ′2)(p+−2)/2ψ′(ω) + γ
(p+ − 1 + µ

p+ − 1

)p+−1

ψ(ω)|ψ(ω)|p+−2 = 0, ω = +
ω0

2
,

where γ = const ≥ 1 (see (A4)).
Note that if any two eigenfunctions solve the problem for the same value of λ, then they are

scalar multiplies of each other. Without loss of generality we can assume that ψ(−ω0

2 ) = 1. Let

us denote y(ω) = ψ′(ω)
ψ(ω) as well as y0 = y(ω0

2 ).

Lemma 2.1. We have

λ (λ(p+ − 1) + 2− p+) > 0. (2.2)

Moreover,

p+ − 1

p+
< λ <

1

2

{p+ − 2

p+ − 1
+

√(p+ − 2

p+ − 1

)2
+

(2τ∗(p+)
ω0

)2}
, if p+ > 2;

0 < λ <
π

2ω0
, if p+ = 2;

1

2

{√(2− p+
p+ − 1

)2
+
(2τ∗(p+)

ω0

)2 − 2− p+
p+ − 1

}
< λ <

1

2

{√(2− p+
p+ − 1

)2
+

( π
ω0

)2 − 2− p+
p+ − 1

}
,

if 1 < p+ < 2,

(2.3)

where τ∗(p+) is the least positive root of the equation

tan τ∗(p+) =
ω0γ̂(p+)

τ∗(p+)
,=⇒ τ∗(p+) ∈ (0,

π

2
). (2.4)

As well as

1 ≥ ψ(ω) ≥ ψ0 = exp(y0ω0) > 0, ∀ω ∈
[
− ω0

2
,+

ω0

2

]
, (2.5)



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 5

where y0 is the negative solution of problem (2.10) and transcendental equation (2.11) - see below;
in this connection y0 ̸= −∞. Moreover (see (2.9), (2.17), (2.16), (2.19)),

|y0| = γ(1 + µ), if p+ = 2,

ν(p+) tan
( ν(p+)ω0

1− λ+ ν(p+)

)
< |y0| < γ̂, if p+ > 2,

γ̂ < |y0| < ν(p+) tan
( ν(p+)ω0

1− λ+ ν(p+)

)
, if 1 < p+ < 2.

(2.6)

Here γ̂(p+) and ν(p+) are defined by (2.15).

Proof. We multiply (2.1) by ψ(ω) and integrate over Ω = (−ω0

2 ,+
ω0

2 ):

−
∫
Ω

ψ(ω)
(
(λ2ψ2(ω) + |ψ′(ω)|2)(p+−2)/2ψ′(ω)

)′
dω

= λ (λ(p+ − 1) + 2− p+)

∫
Ω

(λ2ψ2(ω) + |ψ′(ω)|2)(p+−2)/2ψ2(ω)dω.

Integrating by parts on the left integral, we obtain

−
∫
Ω

ψ(ω)
(
λ2ψ2(ω) + |ψ′(ω)|2)(p+−2)/2ψ′(ω)

)′
dω

=

∫
Ω

(
λ2ψ2(ω) + |ψ′(ω)|2

)(p+−2)/2 |ψ′(ω)|2dω

+ (λ2ψ2(ω) + |ψ′(ω)|2)(p+−2)/2ψ(ω)ψ′(ω)
∣∣∣
ω=−ω0

2

− (λ2ψ2(ω) + |ψ′(ω)|2)(p+−2)/2ψ(ω)ψ′(ω)
∣∣∣
ω=

ω0
2

=

∫
Ω

(
λ2ψ2(ω) + |ψ′(ω)|2

)(p+−2)/2 |ψ′(ω)|2dω + γ
(p+ − 1 + µ

p+ − 1

)p+−1

|ψ(ω0/2)|p+ ,

by the boundary conditions of (2.1). From above, we derive

λ (λ(p+ − 1) + 2− p+)

∫
Ω

(λ2ψ2(ω) + |ψ′(ω)|2)(p+−2)/2ψ2(ω)dΩ

=

∫
Ω

(
λ2ψ2(ω) + |ψ′(ω)|2

)(p+−2)/2 |ψ′(ω)|2dΩ+ γ ·
(p+ − 1 + µ

p+ − 1

)p+−1

|ψ(ω0/2)|p+ > 0,

(2.7)

since γ > 0. Because of ψ(ω) ̸≡ 0, the last inequality implies (2.2).

Case p+ = 2. Problem (2.1) takes the form

ψ′′ + λ2ψ = 0, ω ∈
(
− ω0

2
,
ω0

2

)
,

ψ′(− ω0

2

)
= 0;

ψ′(ω0

2

)
+ γ(1 + µ)ψ

(ω0

2

)
= 0.

(2.8)

We are interested in the least positive eigenvalue, therefore solving this problem we obtain

ψ(ω) = cos
[
λ
(
ω +

ω0

2

)]
, |y0| = γ(1 + µ),

tan(λω0) =
γ(1 + µ)

λ
=⇒ λ ∈ (0,

π

2ω0
),

(2.9)

(see Figure 2).



6 M. BORSUK EJDE-2025/51

Figure 2. Graphs of functions y = tan(x) and y = a/x, a > 0.

Case p+ ̸= 2. Now, we consider the problem (2.1) for the function y(ω). We obtain the Cauchy
problem (

(p+ − 1)y2 + λ2
)
y′ + (p+ − 1)y4 + λ

(
2λ(p+ − 1) + 2− p+

)
y2

+ λ3
(
λ(p+ − 1) + 2− p+

)
= 0, ω ∈

(
− ω0

2
,
ω0

2

)
,

y
(
− ω0

2

)
= 0

(2.10)

and the equation for λ,(
λ2 + y20

) p+−2

2 y0 = −γ
(p+ − 1 + µ

p+ − 1

)p+−1

, y0 = y
(ω0

2

)
< 0. (2.11)

From the Cauchy problem (2.10) we have y′(ω) < 0, therefore y(ω) is a decreasing function:

y
(ω0

2

)
= y0 ≤ y(ω) ≤ 0 = y

(
−ω0

2

)
, (2.12)

by (2.10) and (2.11), we have

y0 ̸= −∞, |y(ω)| ≤ |y0|, ∀ω ∈ [−ω0

2
,
ω0

2
].

From the definition of y(ω) and (2.12) it follows that

0 < ψ0 = exp(y0ω0) ≤ ψ(ω) = exp
(∫ ω

−ω0
2

y(ξ)dξ
)
≤ 1, ∀ω ∈ [−ω0

2
,
ω0

2
].

Thus, we derived (2.5).
By solving the Cauchy problem (2.10), we obtain

arctan
|y(ω)|
λ

+
1− λ√

λ2 + λ 2−p+
p+−1

arctan
|y(ω)|√

λ2 + λ 2−p+
p+−1

= ω +
ω0

2
. (2.13)



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 7

and consequently

arctan
|y0|
λ

+
1− λ√

λ2 + λ 2−p+
p+−1

arctan
|y0|√

λ2 + λ 2−p+
p+−1

= ω0. (2.14)

Now we set

p+ − 1 + µ

p+ − 1
γ

1
p+−1 ≡ γ̂(p+);

√
λ2 + λ

2− p+
p+ − 1

≡ ν(p+); ω0ν(p+) ≡ τ(p+). (2.15)

Case p+ > 2. From (2.2), (2.11) and (2.15) it follows that

|y0| < γ̂(p+), ν(p+) < λ =⇒ arctan
|y0|
λ

< arctan
|y0)|
ν(p+)

. (2.16)

Hence from (2.14) and (2.15) we obtain

ω0 <
1− λ+ ν(p+)

ν(p+)
arctan

|y0|
ν(p+)

=⇒

{
0 < 1− λ+ ν(p+) < 1;

ν(p+) tan
( ν(p+)ω0

1−λ+ν(p+)

)
< |y0| < γ̂.

(2.17)

Because of ν(p+)ω0 <
ν(p+)ω0

1−λ+ν(p+) , from (2.17) we obtain√
λ2 + λ

2− p+
p+ − 1

> λ− 1;

tan(ν(p+)ω0) <
γ̂(p+)

ν(p+)
=⇒ tan τ(p+) <

ω0γ̂(p+)

τ(p+)
=⇒

0 < τ(p+) < τ∗(p+) =⇒

√
λ2 + λ

2− p+
p+ − 1

<
τ∗(p+)

ω0
.

Solving these inequalities (see Figure 2), we derive (2.3) for p+ > 2.

Case 1 < p+ < 2. From (2.11) and (2.15) it follows that

|y0| > γ̂(p+), ν(p+) > λ > 0 =⇒ arctan
|y0|
λ

> arctan
|y0)|
ν(p+)

. (2.18)

Now, from (2.14), (2.18) it follows that

ω0 >
1− λ+ ν(p+)

ν(p+)
arctan

|y0|
ν(p+)

=⇒


1− λ+ ν(p+) > 1,

tan
( ν(p+)ω0

1−λ+ν(p+)

)
> |y0|

ν(p+) >
γ̂(p+)
ν(p+) ,

|y0| < ν(p+) tan
( ν(p+)ω0

1−λ+ν(p+)

)
.

(2.19)

Because of ν(p+)ω0 >
ν(p+)ω0

1−λ+ν(p+) , we obtain, by (2.19), that

tan(ν(p+)ω0) >
γ̂(p+)

ν(p+)
=⇒ tan τ(p+) >

ω0γ̂(p+)

τ(p+)
=⇒

τ∗ < τ(p+) <
π

2
=⇒ τ∗(p+)

ω0
<

√
λ2 + λ

2− p+
p+ − 1

<
π

2ω0
.

Solving this inequality (see Figure 2), we derive (2.3) for p+ < 2. □

Proposition 2.2. If assumption (A1) is satisfied and γ ≥ 1 (see assumption (A4)), then(κ
λ

√
λ2 + y20

)p(x)−p+
≤ 1, ∀x ∈ ΓR0

+ , (2.20)

where κ is defined by (1.3).



8 M. BORSUK EJDE-2025/51

Proof. We rewrite (2.11) for p = p+ with regard to (1.3):

|y0| =
γλ

κ
, if p+ = 2;√

λ2 + y20 =
( γ

|y0|

) 1
p+−2

( λ
κ

) p+−1

p+−2

, if p+ ̸= 2.

(2.21)

Case p+ = 2. Inequality (2.20) is true if p(x) ≡ 2. Now, let 1 < p(x) ≤ p+ = 2 for all x ∈ ΓR0
+ .

From (2.11) we have

|y0| = γ(1 + µ) =
γλ

κ
=⇒ κ

λ

√
λ2 + y20 ≥ κ

λ
|y0| = γ ≥ 1 =⇒

(p(x)− p+) ln
(κ
λ

√
λ2 + y20

)
≤ 0 =⇒ (2.20) is true.

Case p+ > 2. From (2.11) and (2.21) it follows that

|y0| ≤
λ

κ
γ

1
p+−1 and

√
λ2 + y20 ≥ λ

κ
γ

1
p+−1 =⇒

(p(x)− p+) ln
(κ
λ

√
λ2 + y20

)
≤ p(x)− p+

p+ − 1
ln γ ≤ 0 =⇒ (2.20) is true.

Case p+ < 2. From (2.11) and (2.21) we obtain that

|y0| = γ
( λ
κ

)p+−1(√
λ2 + y20

)2−p+
≥ γ

(
λ

κ

)p+−1

|y0|2−p+ =⇒ |y0| ≥
λ

κ
· γ

1
p+−1 and√

λ2 + y20 = γ
1

p+−2

( λ
κ

) p+−1

p+−2 |y0|
1

2−p+ ≥ λ

κ
· γ

1
p+−1 =⇒

(p(x)− p+) ln
(κ
λ

√
λ2 + y20

)
≤ p(x)− p+

p+ − 1
ln γ ≤ 0 =⇒ (2.20) is true.

□

3. Maximum principle

In this section we derive an L∞-a priori estimate of the weak bounded solution to problem
(1.1). First we formulate well known lemmas.

Lemma 3.1 (see [6, Lemma 2.1] and [5, Lemma 1.60]). Let us consider the function

η(x) =

{
eςx − 1, x ≥ 0,

−e−ςx + 1, x ≤ 0,

where ς > 0. Let a, b be positive constants, m > 1. If ς > (2b/a) +m, then we have

aη′(x)− bη(x) ≥ a

2
eςx, ∀x ≥ 0, (3.1)

η(x) ≥ [η(
x

m
)]m, ∀x ≥ 0. (3.2)

Moreover, there exist a q ≥ 0 and an M > 0 such that

η(x) ≤M
[
η
( x
m

)]m
and η′(x) ≤M [η(

x

m
)]m, ∀x ≥ q; (3.3)

|η(x)| ≥ x, ∀x ∈ R. (3.4)

Lemma 3.2 (Interpolation inequality, see [9, (7.9)]). Let 1 < p ≤ q ≤ r and 1
q = λ

p + 1−λ
r . Then

∥u∥Lq(G) ≤ ∥u∥λLp(G)∥u∥
1−λ
Lr(G) (3.5)

holds for all u ∈ Lr(G).



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 9

Lemma 3.3 (Stampacchia’s Lemma, see [12, Lemma 3.11]). Let φ : [k0 : ∞) → R be a nonnegative
and non-increasing function satisfying the condition

φ(l) ≤ C

(l − k)α
[φ(k)]β (3.6)

for l > k > k0, where C, α, β are positive constants and β > 1. Then φ(k0 + ϑ) = 0, where

ϑα = C|φ(k0)|β−12
αβ
β−1 .

Theorem 3.4. Let u(x) be a weak solution of (1.1). If assumptions (A1)–(A9) hold in GR0
0 , then

there exists a constant M0 > 0, depending only on measGR0
0 , p−, p+, s, µ, f0, g0, a0, γ, and such

that ∥u∥
L∞(G

R0
0 )

≤M0.

Proof. Let us define the set A(k) = {x ∈ GR0
0 : |u(x)| > k} with χA(k) being the character-

istic function of the set A(k). Then, for all q > 0, we have that A(k + q) ⊆ A(k). We take
η((|u| − k)+)χA(k) signu as the test function in the integral identity (1.2), where η is defined
by Lemma 3.1 and k ≥ k0 (without loss of generality we can assume k0 ≥ 1). Note that
η((|u| − k)+) ≥ 0 and η′((|u| − k)+) ≥ 0 on A(k). From the integral identity (1.2) it follows
that∫

A(k)

{
|∇u|p(x)η′((|u| − k)+) + a(x)|u|p(x)η((|u| − k)+) + b(u,∇u)η((|u| − k)+) signu

}
dx

+

∫
Γ
R0
+ ∩A(k)

γ(ω)r1−p(x)|u|p(x)−1η((|u| − k)+)ds

+

∫
ΩR0

∪∩A(k)

γ(ω)R
1−p(x)
0 |u|p(x)−1η((|u| − k)+)dΩR0

−
∫
(Γ

R0
+ ∪Ω

R0
+ )∩A(k)

|g(x)|η((|u| − k)+) dS

≤
∫
A(k)

|f(x)|η((|u| − k)+) dx.

(3.7)

By assumption (A3), for R0 > 1 we have

a(x) ≥ a0r
−κ−p(x) ≥ a0R

−κ−p(x)
0 ≥ a0R

−κ−p+
0 , ∀x ∈ GR0

0 .

Now we define the number

ã0 =

{
a0 = const > 0, for R0 ≤ 1,

a0R
−κ−p+
0 , for R0 > 1.

Then from assumption (A3) we obtain a(x) ≥ ã0, ∀x ∈ GR0
0 .

By assumptions (A7) and (A8), we have

b(u,∇u)η((|u| − k)+) signu ≥ −k−1
0

(
δ|∇u|p(x) + b0|u|p(x)

)
η((|u| − k)+),

where

δ =

{
δ0, if µ = 0,

δ+, if µ > 0.
(3.8)

Now, by assumption (A4) and γ(ω) ≥ 1 (see assumption (A4), from (3.7)–(3.8) it follows that∫
A(k)

{
|∇u|p(x)η′((|u| − k)+)− δk−1

0 |∇u|p(x)η((|u| − k)+)

+ (ã0 − b0k
−1
0 )|u|p(x)η((|u| − k)+)

}
dx

≤
∫
(
Γ
R0
0 ∪ΩR0

)
∩A(k)

g(x) signuη((|u| − k)+) dS +

∫
A(k)

|f(x)|η((|u| − k)+) dx.

(3.9)



10 M. BORSUK EJDE-2025/51

Since |u| > k ≥ k0 ≥ 1 and p(x) ≥ p− > 1, we have |u|p(x)−1 ≥ k
p−−1
0 and therefore from the

above it follows that∫
A(k)

{
|∇u|p(x)

(
η′((|u| − k)+)− δk−1

0 η((|u| − k)+)
)
+ (ã0k0 − b0)k

p−−1
0 η((|u| − k)+)

}
dx

≤
∫
(
Γ
R0
0 ∪ΩR0

)
∩A(k)

|g(x)|η((|u| − k)+) dS +

∫
A(k)

|f(x)| · η((|u| − k)+) dx.

Choosing

k0 >
b0
ã0
, (3.10)

we set â0 = ã0k0 − b0 > 0 and from the above we have∫
A(k)

{
|∇u|p(x)

(
η′((|u| − k)+)− δk−1

0 η((|u| − k)+)
)
+ â0k

p−−1
0 η((|u| − k)+)

}
dx

≤
∫
(
Γ
R0
0 ∪ΩR0

)
∩A(k)

|g(x)|η((|u| − k)+) dS +

∫
A(k)

|f(x)|η((|u| − k)+) dx.
(3.11)

Additionally, let us define the sets

A−(k) = A(k) ∩ {|∇u| ≤ 1}, A+(k) = A(k) ∩ {|∇u| ≥ 1}
=⇒ A(k) = A−(k) ∪A+(k).

(3.12)

and the functions

vk(x) := η
( (|u| − k)+

p−

)
, wk(x) := η

( (|u| − k)+
p+

)
. (3.13)

We note that by assumption (A1) the following inequalities hold

|∇u|p+ ≤ |∇u|p(x) ≤ |∇u|p− on A−(k); (3.14)

|∇u|p− ≤ |∇u|p(x) ≤ |∇u|p+ on A+(k). (3.15)

Direct calculations give

|∇vk| =
1

p−
|∇u|η′

( (|u| − k)+
p−

)
=

ς

p−
|∇u| exp

(
ς
(|u| − k)+

p−

)
, ς > 0

=⇒ |∇vk|p− =
( ς

p−

)p− |∇u|p−eς(|u|−k)+ . (3.16)

Putting a = 1 and b = δk−1
0 in (3.1) we have

η′((|u| − k)+)− δk−1
0 η((|u| − k)+) ≥

1

2
eς(|u|−k)+ , if ς > p− + 2δk−1

0 . (3.17)

From (3.15), (3.16), (3.17) it follows that∫
A+(k)

|∇u|p(x)
{
η′((|u| − k)+)− δk−1

0 η((|u| − k)+)
}
dx

≥ 1

2

(
p−
ς

)p− ∫
A+(k)

|∇vk|p− dx, if ς > p∗ + 2δk−1
0 .

(3.18)

Similarly, choosing ς > p+ + 2δk−1
0 and taking into account (3.13), (3.14), we obtain∫

A−(k)

|∇u|p(x){η′((|u| − k)+)+)− δk−1
0 η((|u| − k)+)+)} dx

≥ 1

2

(
p∗

ς

)p+ ∫
A−(k)

|∇wk|p+ dx.
(3.19)



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 11

Since p+ ≥ p−, adding the inequalities (3.18) and (3.19) we obtain

1

2

{(
p−
ς

)p− ∫
A+(k)

|∇vk|p− dx+

(
p+
ς

)p+ ∫
A−(k)

|∇wk|p+ dx
}

≤
∫
A(k)

|∇u|p(x)⟨η′((|u| − k)+)− δk−1
0 η((|u| − k)+)⟩ dx, if ς > p+ + 2δk−1

0 ,

(3.20)

by (3.12). Finally, from (3.11), (3.20) we derive

1

2

{(
p−
ς

)p− ∫
A+(k)

|∇vk|p− dx+

(
p+
ς

)p+ ∫
A−(k)

|∇wk|p+ dx
}

+ â0k
p−−1
0 ·

∫
A(k)

η((|u| − k)+) dx

≤
∫
(
Γ
R0
0 ∪ΩR0

)
∩A(k)

|g(x)|η((|u| − k)+) dS +

∫
A(k)

|f(x)| · η((|u| − k)+) dx.

Since, by (3.12),
∫
A(k)

=
∫
A+(k)

+
∫
A−(k)

, for ς > p+ + 2δk−1
0 , we have

1

2

{(
p−
ς

)p− ∫
A+(k)

|∇vk|p− dx+

(
p+
ς

)p+ ∫
A−(k)

|∇wk|p+ dx
}

+ â0k
p−−1
0

∫
A+(k)

η((|u| − k)+) dx+ â0k
p−−1
0

∫
A−(k)

η((|u| − k)+) dx

≤
∫
(
Γ
R0
0 ∪ΩR0

)
∩A+(k)

|g(x)|η((|u| − k)+) dS +

∫
(
Γ
R0
0 ∪ΩR0

)
∩A−(k)

|g(x)|η((|u| − k)+) dS

+

∫
A+(k)

|f(x)| · η((|u| − k)+) dx+

∫
A−(k)

|f(x)| · η((|u| − k)+) dx.

(3.21)

Now, by (3.2) and (3.13), we have

â0k
p−−1
0

∫
A+(k)

η((|u| − k)+) dx+ â0k
p−−1
0

∫
A−(k)

η((|u| − k)+) dx

≥ â0k
p−−1
0

(∫
A+(k)

v
p−
k dx +

∫
A−(k)

w
p+
k dx

)
.

(3.22)

From (3.21)–(3.22) it follows that

1

2

{(
p−
ς

)p− ∫
A+(k)

|∇vk|p− dx+

(
p+
ς

)p+ ∫
A−(k)

|∇wk|p+ dx
}

+ â0k
p−−1
0

(∫
A+(k)

v
p−
k dx +

∫
A−(k)

w
p+
k dx

)
≤

∫
(
Γ
R0
0 ∪ΩR0

)
∩A+(k)

|g(x)|η((|u| − k)+) dS +

∫
(
Γ
R0
0 ∪ΩR0

)
∩A−(k)

|g(x)|η((|u| − k)+) dS

+

∫
A+(k)

|f(x)| · η((|u| − k)+) dx+

∫
A−(k)

|f(x)| · η((|u| − k)+) dx, ς > p+ + 2δk−1
0 .

(3.23)

Next, we have∫
A±(k)

|f(x)| · η((|u| − k)+) dx

=

∫
A±(k+q)

|f(x)| · η((|u| − k)+) dx+

∫
A±(k)\A±(k+q)

|f(x)|η((|u| − k)+) dx, ∀q > 0.

(3.24)

By (3.3), we obtain

η((|u| − k)+)
∣∣∣
A+(k+q)

≤M
[
η
( (|u| − k)+

p−

)]p−
,



12 M. BORSUK EJDE-2025/51

η((|u| − k)+)
∣∣∣
A−(k+q)

≤M
[
η
( (|u| − k)+

p+

)]p+
.

Then (3.13) implies that∫
A+(k+q)

|f(x)| · η((|u| − k)+) dx ≤M

∫
A+(k+q)

|f(x)| · vp−k dx, (3.25)∫
A−(k+q)

|f(x)| · η((|u| − k)+) dx ≤M

∫
A−(k+q)

|f(x)| · wp+k dx. (3.26)

Using the definition of η from Lemma 3.1, we have

η((|u| − k)+)
∣∣∣
A±(k)\A±(k+q)

≤ eςq, ∀q > 0 =⇒∫
A±(k)\A±(k+q)

|f(x)| · η((|u| − k)+) dx ≤ eςq
∫
A±(k)\A±(k+q)

|f(x)| dx, ∀q > 0.
(3.27)

Now, we derive that f(x) ∈ Ls(G
R0
0 ), s > 2

p−−1 > 1. In fact, by assumptions (A1) and (A4), we

calculate:∫
G

R0
0

|f(x)|sdx

≤ fs0

∫
G

R0
0

rsβ(x) dx

= fs0

∫
G

R0
0

rsκ(p(x)−1)−2 dx = fs0

∫
G1

0

rsκ(p(x)−1)−2 dx+ fs0

∫
G

R0
1

rsκ(p(x)−1)−2 dx

≤ fs0 · 2z0ω0

{∫ 1

0

rsκ(p−−1)−1dr +

∫ R0

1

rsκ(p+−1)−1dr

}

=
2z0ω0f

s
0

sκ(p+ − 1)

{
R
sκ(p+−1)
0 +

p+ − p−
p− − 1

}
<∞

and therefore

∥f(x)∥
Ls(G

R0
+ )

≤ cf

{
R
sκ(p+−1)
0 +

p+ − p−
p− − 1

}1/s

, cf = f0

( 2z0ω0

sκ(p+ − 1)

)1/s

. (3.28)

Using the Hölder inequality with exponents s and s′ where 1
s +

1
s′ = 1, we obtain∫

A+(k+q)

|f(x)| · vp−k dx ≤ ∥f(x)∥
Ls(G

R0
0 )

(∫
A+(k)

v
p−s

′

k dx
)1/s′

. (3.29)

From the inequality 1
s <

p−−1
2 it follows that p−s

′ < p−
# = 2p−

3−p− and then the interpolation

inequality (3.5) gives(∫
A+(k)

v
p−s

′

k dx
)1/s′

≤
(∫

A+(k)

v
p−
k dx

)θ−(∫
A+(k)

v
p−

#

k dx
) (1−θ−)p−

p−#

,

where θ− ∈ (0, 1), which is defined by the equality 1
p−s′

= θ−
p−

+ 1−θ−
p−# =⇒ θ− = 1− 2

(p−−1)s . Now,

by using the Young inequality with exponents 1
θ−

and 1
1−θ− , from (3.29) we derive∫

A+(k+q)

|f(x)| · vp−k dx

≤ ε(1− θ−)
(∫

A+(k)

v
p−

#

k dx
)p−/p#−

+ θ−ε
θ−−1

θ− ∥f∥1/θ−
Ls(GR

0 )

∫
A+(k)

v
p−
k dx.

(3.30)



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 13

Similarly ∫
A−(k+q)

|f(x)| · wp+k dx

≤ ε(1− θ+)
(∫

A−(k)

w
p+

#

k dx
)p+/p#+

+ θ+ε
θ+−1

θ+ ∥f∥
1

θ+

Ls(GR
0 )

∫
A−(k)

w
p+
k dx

(3.31)

for

∀ε > 0, p−
# =

2p−
3− p−

, θ− = 1− 2

(p− − 1)s
;

p+
# =

2p+
3− p+

, θ+ = 1− 2

(p+ − 1)s
;

s > max
{ 2

p− − 1
,

2

p+ − 1

}
=

2

p− − 1
> 1, =⇒ 0 < θ− ≤ θ+ < 1,because of p− ≤ p+.

(3.32)

Then applying (3.30), (3.31) to (3.24) - (3.27), we obtain that∫
A+(k)

|f(x)| · η((|u| − k)+) dx

≤Mε(1− θ−)
(∫

A+(k)

v
p#−
k dx

)p−/p#−
+ eςq

∫
A+(k)

|f(x)| dx+Mθ−ε
θ−−1

θ− ∥f∥1/θ−
Ls(GR

0 )

∫
A+(k)

v
p−
k dx,∫

A−(k)

|f(x)| · η((|u| − k)+) dx

≤Mε(1− θ+)
(∫

A−(k)

w
p+

#

k dx
)p+/p#+

+ eςq
∫
A−(k)

|f(x)| dx+Mθ+ε
θ+−1

θ+ ∥f∥
1

θ+

Ls(GR
0 )

∫
A−(k)

w
p+
k dx.

(3.33)

By well known the Sobolev embedding theorem and taking into account the definition of p#−, p+
#

in (3.32), we obtain (∫
A+(k)

v
p−

#

k dx
)p−/p#− ≤ c−

∫
A+(k)

(
v
p−
k + |∇vk|p−

)
dx;

(∫
A−(k)

w
p+

#

k dx
)p+/p#+ ≤ c+

∫
A−(k)

(
w
p+
k + |∇wk|p+

)
dx,

(3.34)

where c−, c+ = const > 0. Finally, (3.23) - (3.34) imply that[1
2

(p−
ς

)p−
−Mc−(1− θ−)ε

] ∫
A+(k)

|∇vk|p− dx

+
[1
2

(p+
ς

)p+
−Mc+(1− θ+)ε

] ∫
A−(k)

|∇wk|p+ dx

+
[
â0k

p−−1
0 −Mc−(1− θ−)ε−Mθ−ε

θ−−1

θ− ∥f∥1/θ−
Ls(GR

0 )

] ∫
A+(k)

v
p−
k dx

+
[
â0k

p−−1
0 −Mc+(1− θ+)ε−Mθ+ε

θ+−1

θ+ ∥f∥
1

θ+

Ls(GR
0 )

] ∫
A−(k)

w
p+
k dx

≤
∫(

Γ
R0
+ ∪ΩR0

)
∩A+(k)

|g(x)|η((|u| − k)+) dS +

∫(
Γ
R0
+ ∪ΩR0

)
∩A−(k)

|g(x)|η((|u| − k)+) dS

+ eςq
∫
A(k)

|f(x)| dx, ς > p+ + 2δk−1
0 , ∀q > 0, ∀ε > 0.

(3.35)



14 M. BORSUK EJDE-2025/51

Similarly from (3.24)-(3.27) we obtain∫(
Γ
R0
+ ∪ΩR0

)
∩A+(k)

|g(x)|η((|u| − k)+) dS

≤M

∫(
Γ
R0
+ ∪ΩR0

)
∩A+(k+q)

|g(x)||vk|p− dS + eςq
∫(

Γ
R0
+ ∪ΩR0

)
∩{A+(k)\A+(k+q)}

|g(x)|dS,
(3.36)

∫(
Γ
R0
+ ∪ΩR0

)
∩A−(k)

|g(x)|η((|u| − k)+) dS

≤M

∫(
Γ
R0
+ ∪ΩR0

)
∩A−(k+q)

|g(x)||wk|p+ dS + eςq
∫(

Γ
R0
+ ∪ΩR0

)
∩{A−(k)\A−(k+q)}

|g(x)|dS.
(3.37)

Now we derive that

g(x) ∈ L j
j−1

(
ΓR0
+ ∪ ΩR0

)
, max{1, 1

κ
} < j <

2

3− p−
. (3.38)

By assumption (A5),∫
Γ
R0
+ ∪ΩR0

|g(x)|
j

j−1 dS ≤ g
j

j−1

0

∫
Γ
R0
+ ∪ΩR0

r
j

j−1 ·(1+β(x)) dS.

By assumptions (A1) and (A4),

(κ − 1)(p− − 1) + 1 ≤ 1 + β(x) ≤ κ(p+ − 1)− 2

s
+ 1, s >

2

p− − 1
,

we obtain:

Case 0 < R0 ≤ 1. In this case∫
Γ
R0
+

|g(x)|
j

j−1 dS ≤ 2z0g
j

j−1

0

∫ R0

0

r
j

j−1 ⟨(p−−1)(κ−1)+1⟩dr

= 2z0g
j

j−1

0

j − 1

j⟨(p− − 1)κ + 3− p−⟩ − 1
R

j⟨(p−−1)κ+3−p−⟩−1

j−1

0 <∞,

(3.39)

because of
j

j − 1
· ⟨(p− − 1)(κ − 1) + 1⟩+ 1 > 0.

In fact, if κ ≥ 1, this inequality is obvious. Let 0 < κ < 1. Then, since p− < 3 and j > 1
κ , we

have

j

j − 1
· ⟨(p− − 1)(κ − 1) + 1⟩+ 1 =

1

j − 1

(
j⟨(p− − 1)κ + 3− p−⟩ − 1

)
>

(p− − 1) + (2− p−)

j − 1
=

1

j − 1
> 0.

Similarly, ∫
ΩR0

|g(x)|
j

j−1 dΩR0
≤ 2z0ω0g

j
j−1

0 R
j⟨(p−−1)κ+3−p−⟩−1

j−1

0 . (3.40)

From (3.39) and (3.40) we derive

∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

) ≤ c−g0 ·R
(p−−1)κ+3−p−− 1

j

0 ,

c−g0 = g0(2z0)
j−1
j

{( j − 1

j⟨(p− − 1)κ + 3− p−⟩ − 1

) j−1
j

+ ω
j−1
j

0

}
.

(3.41)



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 15

Case R0 > 1. In this case∫
Γ
R0
+

|g(x)|
j

j−1 dS ≤ g
j

j−1

0

∫
Γ
R0
+

r
j

j−1 (1+β(x)) dS

≤ 2z0g
j

j−1

0

(∫ 1

0

r
j

j−1 ⟨(p−−1)(κ−1)+1⟩dr +

∫ R0

1

r
j

j−1 ⟨(p+−1)κ− 2
s+1⟩dr

)
.

Next we calculate∫ R0

1

r
j

j−1 ⟨(p+−1)κ− 2
s+1⟩dr

=
j − 1

j⟨(p+ − 1)κ − 2
s + 2⟩ − 1

{
R

j⟨(p+−1)κ− 2
s
+2⟩−1

j−1

0 − 1
}
> 0, R0 > 1,

(3.42)

because of j⟨(p+ − 1)κ − 2
s + 2⟩ − 1 > 0. In fact, since s > 2

p−−1 , j > max{1, 1
κ } we have:

(1) if κ > 1 then j⟨(p+ − 1)κ − 2
s + 2⟩ − 1 > (p+ − 1) + (1− p−) + 1 = p+ − p− + 1 ≥ 1;

(2) if 0 < κ < 1 then j⟨(p+−1)κ− 2
s+2⟩−1 > (p+−1)+j(3−p−)−1 > (p+−2)+(3−p−) =

p+ − p− + 1 ≥ 1.

Thus, from (3.39), (3.42) we obtain∫
Γ
R0
+

|g(x)|
j

j−1 dS

≤ 2z0g
j

j−1

0

j − 1

j⟨(p+ − 1)κ − 2
s + 2⟩ − 1

R
j⟨(p+−1)κ− 2

s
+2⟩−1

j−1

0

+ 2z0g
j

j−1

0 j(j − 1)
κ(p+ − p−) + (p− − 1)− 2

s(
j⟨(p− − 1)κ + 3− p−⟩ − 1

)(
j⟨(p+ − 1)κ − 2

s + 2⟩ − 1
) .

(3.43)

Similarly, ∫
ΩR0

|g(x)|
j

j−1 dΩR0 ≤ 2z0ω0g
j

j−1

0 R
j⟨(p+−1)κ− 2

s
+2⟩−1

j−1

0 . (3.44)

Applying the Jensen inequality

(a+ b)
j−1
j ≤ max

(
1, 2−

1
j

)(
a

j−1
j + b

j−1
j

)
= a

j−1
j + b

j−1
j ,

for j > 1, we derive from (3.43) - (3.44) that

∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

) ≤ g0

(
c+g0 ·R

(p+−1)κ− 2
s+2− 1

j

0 + cg+

)
, (3.45)

c+g0 = (2z0)
j−1
j

{( j − 1

j⟨(p+ − 1)κ − 2
s + 2− 1

j ⟩ − 1

) j−1
j

+ ω
j−1
j

0

}
; (3.46)

cg+ =
{ ⟨2z0j(j − 1)⟩ · ⟨κ(p+ − p−) + (p− − 1)− 2

s ⟩(
j⟨(p− − 1)κ + 3− p−⟩ − 1

)(
j⟨(p+ − 1)κ − 2

s + 2⟩ − 1
)} j

j−1

. (3.47)

At last, we proved that g(x) ∈ L j
j−1

(
ΓR0
+ ∪ ΩR0

)
for max{1, 1

κ } < j < 2
3−p− .

Now, by the Hölder inequality, we have∫(
Γ
R0
+ ∪ΩR0

)
∩A+(k+q)

|g(x)| |vk|p− dS ≤ ∥vp−k ∥
Lj

(
Γ
R0
+ ∪ΩR0

)
∩A+(k)

∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

)
= ∥vk∥p−

Ljp−

(
Γ
R0
+ ∪ΩR0

)
∩A+(k)

· ∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

) .
By the Sobolev boundary trace embedding theorem, we have∫(

Γ
R0
+ ∪ΩR0

)
∩A+(k+q)

|g(x)| |vk|p− dS ≤ Csob · ∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

) ∫
A+(k)

(|∇vk|p− + |vk|p−) dx,



16 M. BORSUK EJDE-2025/51

for j > max{1, 1
κ }. In a similar way we derive∫(

Γ
R0
+ ∪ΩR0

)
∩A−(k+q)

|g(x)| |wk|p+ dS ≤ Csob∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

) ∫
A−(k)

(|∇wk|p+ + |wk|p+) dx,

for j > max{1, 1
κ }. Hence from (3.36)-(3.37) it follows that∫(

Γ
R0
+ ∪ΩR0

)
∩A+(k)

|g(x)|η((|u| − k)+) dS

≤MCsob∥g(x)∥L j
j−1

(
Γ
R0
+ ∪ΩR0

) ∫
A+(k)

(|∇vk|p− + |vk|p−) dx+ eςq
∫(

Γ
R0
+ ∪ΩR0

)
∩A+(k)

|g(x)|dS,

(3.48)∫(
Γ
R0
+ ∪ΩR0

)
∩A−(k)

|g(x)|η((|u| − k)+) dS

≤MCsob∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

) ∫
A−(k)

(|∇wk|p+ + |wk|p+) dx+ eςq
∫(

Γ
R0
+ ∪ΩR0

)
∩A−(k)

|g(x)|dS.

(3.49)

Now, from (3.35), by (3.28), (3.40)-(3.49), it follows that[1
2

(
p−
ς

)p−
−Mc−(1− θ−)ε−MCsob∥g(x)∥

L j
j−1

(
Γ
R0
+ ∪ΩR0

)] ∫
A+(k)

|∇vk|p− dx

+
[1
2

(
p+
ς

)p+
−Mc+(1− θ+)ε−MCsob∥g(x)∥

L j
j−1

(
Γ
R0
+ ∪ΩR0

)] ∫
A−(k)

|∇wk|p+ dx

+
[
â0k

p−−1
0 −Mc−(1− θ−)ε−Mθ−ε

θ−−1

θ− ∥f∥1/θ−
Ls(GR

0 )

−MCsob∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

)] ∫
A+(k)

v
p−
k dx+

[
â0k

p−−1
0 −Mc+(1− θ+)ε

−Mθ+ε
θ+−1

θ+ ∥f∥
1

θ+

Ls(GR
0 )

−MCsob∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

)] ∫
A−(k)

w
p+
k dx

≤ eςq
{∫

A(k)

|f(x)| dx+

∫(
Γ
R0
+ ∪ΩR0

)
∩A(k)

|g(x)| dS
}
,

ς > p+ + 2δk−1
0 , ∀q > 0, ∀ε > 0, j > max{1, 1

κ
}.

(3.50)

Now we consider two cases according to assumption (A5):

(1) g(x) ≡ 0 ;
(2) g(x) ̸≡ 0 =⇒ 0 < g0 ≪ 1.

Case g0 = 0. In this case ∥g(x)∥
L j

j−1
(Γ

R0
+ ∪ΩR0

)
= 0 and inequality (3.50) takes the form

[1
2

(
p−
ς

)p−
−Mc−(1− θ−)ε

] ∫
A+(k)

|∇vk|p− dx

+
[1
2

(
p+
ς

)p+
−Mc+(1− θ+)ε

] ∫
A−(k)

|∇wk|p+ dx

+
[
â0k

p−−1
0 −Mc−(1− θ−)ε−Mθ−ε

θ−−1

θ− ∥f∥1/θ−
Ls(GR

0 )

] ∫
A+(k)

v
p−
k dx

+
[
â0k

p−−1
0 −Mc+(1− θ+)ε−Mθ+ε

θ+−1

θ+ ∥f∥
1

θ+

Ls(GR
0 )

] ∫
A−(k)

w
p+
k dx

≤ eςq
∫
A(k)

|f(x)| dx, ς > p+ + 2δk−1
0 , ∀q > 0, ∀ε > 0, j > max{1, 1

κ
}.

(3.51)



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 17

At first, we choose

ε =
1

4M
min

{ 1

c−(1− θ−)

(
p−
ς

)p−
,

1

c+(1− θ+)

(
p+
ς

)p+ }
(3.52)

and next

k0 ≥
(
2MF1

â0

) 1
p−−1

, (3.53)

where

F1 = max
{
c−(1− θ−)ε+ θ−ε

θ−−1

θ− c
1/θ−
f

{
R
s(p+−1)κ
0 +

p+ − p−
p− − 1

} 1
sθ− ,

c+(1− θ+)ε+ θ+ε
θ+−1

θ+ c
1

θ+

f

{
R
s(p+−1)κ
0 +

p+ − p−
p− − 1

} 1
sθ+

}
.

Thus, by the above arguments, we derive∫
A+(k)

(
|∇vk|p− + v

p−
k

)
dx+

∫
A−(k)

(
|∇wk|p+ + w

p+
k

)
dx ≤ C1

∫
A(k)

|f(x)| dx, (3.54)

where C1 = const(n, p−, p+, a0, b0, k0, δ, ν, µ, s, R0, λ, f0, c−, c+, ω0,M) > 0.

Case g0 ̸= 0. In this case we choose ε > 0 by (3.52), positive g0 ≪ 1 such

g0 ≤ 1

8

min
{(p−

ς

)p−
;
(p+
ς

)p+}
MCsob

(
c+g0R

(p+−1)κ− 2
s+2 1

j

0 + cg+
) (3.55)

and next

k0 ≥
(2MF2

â0

) 1
p−−1

, (3.56)

where

F2 = max
{
c−(1− θ−)ε+ θ−ε

θ−−1

θ− c
1/θ−
f

{
R
s(p+−1)κ
0 +

p+ − p−
p− − 1

} 1
sθ−

+ g0

(
c+g0R

(p+−1)κ− 2
s+2− 1

j

0 + cg+

)
,

c+(1− θ+)ε+ θ+ε
θ+−1

θ+ c
1

θ+

f

{
R
s(p+−1)κ
0 +

p+ − p−
p− − 1

} 1
sθ+

+ g0

(
c+g0 ·R

(p+−1)κ− 2
s+2− 1

j

0 + cg+

)}
.

Thus, by the above arguments,∫
A+(k)

(
|∇vk|p− + v

p−
k

)
dx+

∫
A−(k)

(
|∇wk|p+ + w

p+
k

)
dx

≤ C2

{∫
A(k)

|f(x)| dx+

∫(
Γ
R0
+ ∪ΩR0

)
∩A(k)

|g(x)|dS
}
,

(3.57)

where

C2 = const (n, p−, p+, a0, b0, k0, δ, ν, µ, s, R0, λ, f0, g0, Csob, c−, c+, ω0,M) > 0.

In both cases considered the inequalities (3.34) together with (3.54) or (3.57) gives(∫
A+(k)

v
p−

#

k dx
)p−/p#−

+
(∫

A−(k)

w
p+

#

k dx
)p+/p#+

≤ max{c−, c+}C3

{∫
A(k)

|f(x)| dx+

∫(
Γ
R0
+ ∪ΩR0

)
∩A(k)

|g(x)|dS
}
, ∀k ≥ k0.

(3.58)

Now we use again the Sobolev boundary trace embedding theorem,(∫(
Γ
R0
+ ∪ΩR0

)
∩A+(k)

v
p−

#

k dS
)p−/p#− ≤ Csob

∫
A+(k)

(
|∇vk|p− + v

p−
k

)
dx,



18 M. BORSUK EJDE-2025/51

(∫(
Γ
R0
+ ∪ΩR0

)
∩A−(k)

w
p+

#

k dS
)p+/p#+ ≤ Csob

∫
A−(k)

(
|∇wk|p+ + w

p+
k

)
dx.

Thus, from (3.57), (3.58) it follows that(∫
A+(k)

v
p−

#

k dx
)p−/p#−

+
(∫

A−(k)

w
p+

#

k dx
)p+/p#+

+
(∫(

Γ
R0
+ ∪ΩR0

)
∩A+(k)

v
p−

#

k dS
)p−/p#−

+
(∫(

Γ
R0
+ ∪ΩR0

)
∩A−(k)

w
p+

#

k dS
)p+/p#+

≤ C4

{∫
A(k)

|f(x)| dx+

∫(
Γ
R0
+ ∪ΩR0

)
∩A(k)

|g(x)|dS
}
, ∀k ≥ k0.

(3.59)

At last, by the Hölder inequality, we have∫
A(k)

|f(x)|dx ≤ |f(x)|Ls(GR
+) meas1−

1
s A(k); s >

2

p− − 1
> 1;∫(

Γ
R0
+ ∪ΩR0

)
∩A(k)

|g(x)|dS ≤ ∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

)[meas
{(

ΓR0
+ ∪ ΩR0

)
∩A(k)

}]1/j
,

for all k ≥ k0 and 3−p−
2 < 1

j < min {1,κ}. Next, from (3.59) it follows that

(∫
A+(k)

v
p−

#

k dx
)p−/p#−

+
(∫

A−(k)

w
p+

#

k dx
)p+/p#+

+
(∫(

Γ
R0
+ ∪ΩR0

)
∩A+(k)

v
p−

#

k dS
)p−/p#−

+
(∫(

Γ
R0
+ ∪ΩR0

)
∩A−(k)

w
p+

#

k dS
)p+/p#+

≤ C4

{
|f(x)|Ls(GR

0 ) meas1−
1
s A(k)

+ ∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

)[meas
{(

ΓR0
+ ∪ ΩR0

)
∩A(k)

}]1/j}
,

(3.60)

for all k ≥ k0,
3−p−

2 < 1
j < min{1,κ}, and s > 2

p−−1 > 1. Now from (3.31) we derive

p#−
p−

=
2

3− p−
> 1,

p+
#

p+
=

2

3− p+
> 1. (3.61)

Further, let l > k > k0. By (3.4) and the definition of the functions vk(x) and wk(x) we have

vk ≥ 1

p−
((|u| − k)+)+, wk ≥ 1

p+
((|u| − k)+)+.

Therefore, ∫
A+(l)

v
p−

#

k dx ≥
( l − k

p−

)p−#

measA+(l),∫
A−(l)

w
p+

#

k dx ≥
( l − k

p+

)p+#

measA−(l),∫(
Γ
R0
+ ∪ΩR0

)
∩A+(l)

v
p−

#

k dS ≥
( l − k

p−

)p−#

meas
[(
ΓR0
+ ∪ ΩR0

)
∩A+(l)

]
,∫(

Γ
R0
+ ∪ΩR0

)
∩A−(l)

w
p+

#

k dS ≥
( l − k

p∗

)p+#

meas
[(
ΓR0
+ ∪ ΩR0

)
∩A−(l)

]
.



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 19

Because of A±(l) ⊆ A±(k), the obtained inequalities can be rewritten as

measA+(l) ≤
( p−
l − k

)p−# ∫
A+(k)

v
p−

#

k dx, measA−(l) ≤
( p−
l − k

)p+# ∫
A−(k)

w
p+

#

k dx,

meas
[(
ΓR0
+ ∪ ΩR0

)
∩A+(l)

]
≤

( p−
l − k

)p−#(∫(
Γ
R0
+ ∪ΩR0

)
∩A+(k)

v
p−

#

k dS
)
,

meas
[(
ΓR0
+ ∪ ΩR0

)
∩A−(l)

]
≤

( p+
l − k

)p+#(∫(
Γ
R0
+ ∪ΩR0

)
∩A−(k)

w
p+

#

k dS
)
.

(3.62)

Let us introduce

ψ(k) = meas A(k) + meas
[(
ΓR0
+ ∪ ΩR0

)
∩A(k)

]
.

From A(l) = A+(l) ∪A−(l), we have

measA(l) = meas (A+(l) ∪A−(l)) ≤ measA+(l) + measA−(l).

Further, we use well known the Jensen inequality

(a+ b)
2

3−p ≤ 2
p−1
3−p

(
a

2
3−p + b

2
3−p

)
. (3.63)

Now, from (3.62), (3.60) with regard to (3.61), (3.63) we derive

measA(l) = meas (A+(l) ∪A−(l)) ≤ measA+(l) + measA−(l)

≤
( p−
l − k

)p−# ∫
A+(k)

v
p−

#

k dx+
( p+
l − k

)p+# ∫
A−(k)

w
p+

#

k dx

≤
c3−

(l − k)p−#

{
meas1−

1
s A(k) +

[
meas

(
ΓR0
+ ∪ ΩR0

)
∩A(k)

]1/j} 2
3−p−

+
c3+

(l − k)p+#

{
meas1−

1
s A(k) +

[
meas

(
ΓR0
+ ∪ ΩR0

)
∩A(k)

]1/j} 2
3−p+

≤ c3∗ · 2
p−−1

3−p−

(l − k)p−#

{
[measA(k)]

(1− 1
s )

2
3−p− +

[
meas

(
ΓR0
+ ∪ ΩR0

)
∩A(k)

] 1
j

2
3−p−

}
+
c3∗ · 2

p+
3−p+

(l − k)p+#

{
[meas A(k)]

(1− 1
s )

2
3−p+ +

[
meas

(
ΓR0
+ ∪ ΩR0

)
∩A(k)

] 1
j

2
3−p+

}
,

(3.64)

for all l > k ≥ k0. Similarly,

meas
[(
ΓR0
+ ∪ ΩR0

)
∩A(l)

]
≤ meas

[(
ΓR0
+ ∪ ΩR0

)
∩A+(l)

]
+meas

[(
ΓR0
+ ∪ ΩR0

)
∩A−(l)

]
≤ c3∗ · 2

p−−1

3−p−

(l − k)p−#

{
[measA(k)]

(1− 1
s )

2
3−p− +

[
meas

(
ΓR0
+ ∪ ΩR0

)
∩A(k)

] 1
j

2
3−p−

}
+
c3∗ · 2

p+
3−p+

(l − k)p+#

{
[measA(k)]

(1− 1
s )

2
3−p+ +

[
meas

(
ΓR0
+ ∪ ΩR0

)
∩A(k)

] 1
j

2
3−p+

}
,

(3.65)

for all l > k ≥ k0, where

c3∗ = const
(
p−, p−

#, ∥f(x)∥
Ls(G

R0
+ )

, ∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

)),
c3∗ = const

(
p+, p+

#, ∥f(x)∥
Ls(G

R0
+ )

, ∥g(x)∥
L j

j−1

(
Γ
R0
+ ∪ΩR0

)).



20 M. BORSUK EJDE-2025/51

Thus, by the definition of ψ(k), it follows that

ψ(l) ≤ c3∗ · 2
p−−1

3−p−

(l − k)p−#

{
[ψ(k)]

(1− 1
s )

2
3−p− + [ψ(k)]

1
j

2
3−p−

}
+
c3∗ · 2

p+
3−p+

(l − k)p+#

{
[ψ(k)]

(1− 1
s )

2
3−p+ + [ψ(k)]

1
j

2
3−p+

}
.

(3.66)

Since p− ≤ p+, we have

s >
2

p− − 1
≥ 2

p+ − 1
=⇒ 1− 1

s
>

3− p−
2

≥ 3− p+
2

=⇒(
1− 1

s

) 2

3− p−
> 1,(

1− 1

s

) 2

3− p+
> 1.

Similarly, from (3.60) we obtain

1

j

2

3− p−
> 1,

1

j

2

3− p+
> 1.

Therefore, from (3.66) it follows that for all l > k ≥ k0,

ψ(l) ≤ C̃ψβ(k)


1

(l−k)p∗# , if l − k ≥ 1;

1

(l−k)p∗# , if 0 < l − k < 1,

where β > 1 and C̃ = const(n, p∗, p
∗, â0, k0, δ, ν, s, C2, cf , cg0 , ch0

, R0, λ∗).
By the Stampacchia Lemma, we have that ψ(k0+ϑ) = 0 with ϑ depending only on the quantities

given in being proved Theorem. This fact means that |u(x)| ≤ k0+ϑ for almost all x ∈ GR0 . Thus,
we derive M0 = k0 + ϑ, where k0 > 1 is sufficiently large and is defined by (3.10), (3.53) with
(3.31) and (3.52). □

4. Comparison principle

In GR0
0 we consider the second-order quasi-linear degenerate operator T of the form

T (u, η) ≡
∫
G

R0
0

⟨Ai(x, ux)ηxi
+ b(x, u, ux)η(x)⟩ dx+

∫
Γ
R0
+

γ(ω)

rp(x)−1
u|u|p(x)−2η(x)ds

−
∫
ΩR0

Ai(x, ux) cos(r, xi)η(x)dΩR0
, γ(ω) ≥ γ∗ ≥ 0

(4.1)

for u(x) ∈ N
1,p(x)
−1,∞ (GR0

0 ) and for all non-negative η(x) belonging to N
1,p(x)
−1,∞ (GR0

0 ) under the follow-

ing assumptions: functions Ai(x, ξ), b(x, u, ξ) are Caratheodory, continuously differentiable with

respect to the u, ξ variables in M = GR0
0 × R× R3 and satisfy in M the following inequalities:

(i) ∂Ai(x,ξ)
∂ξj

ζiζj ≥ ςp|ξ|p(x)−2ζ2 for all ζ ∈ R3 \ {0} and ςp > 0;

(ii)
√∑n

i=1 |
∂b(x,u,ξ)
∂ξi

|2 ≤ b1|u|−1|ξ|p(x)−1, ∂b(x,u,ξ)∂u ≥ b2|u|−2|ξ|p(x) b1 ≥ 0, and b2 ≥ 0;

(iii) p(x) ≥ p− > 1.

Proposition 4.1. Let T satisfy assumptions (i)–(iii) and functions u,w ∈ N
1,p(x)
−1,∞ (GR0

0 ) satisfy

T (u, η) ≤ T (w, η) (4.2)

for all non-negative η ∈ N
1,p(x)
−1,∞ (GR0

0 ). Also assume that

u(x) ≤ w(x) on ΩR0
. (4.3)

Then u(x) ≤ w(x) in GR0
0 .



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 21

Proof. Let us define z = u− w and uτ = τu+ (1− τ)w, for τ ∈ [0, 1]. Then

0 ≥ T (u, η)− T (w, η)

=

∫∫
G

R0
0

〈
ηxi

zxj

∫ 1

0

∂Ai(x, u
τ
x)

∂uτxj

dτ + ηzxi

∫ 1

0

∂b(x, uτ , uτx)

∂uτxi

dτ

+ ηz

∫ 1

0

∂b(x, uτ , uτx)

∂uτ
dτ

〉
dx−

∫
ΩR0

(∫ 1

0

∂Ai(x, u
τ
x)

∂uτxj

dτ
)
cos(r, xi) · zxj

η(x)dΩR0

+

∫
Γ
R0
+

γ(ω)

rp(x)−1

(∫ 1

0

∂(uτ |uτ |p(x)−2)

∂uτ
dτ

)
z(x)η(x)ds

(4.4)

for all non-negative η ∈ N
1,p(x)
−1,∞ (GR0

0 ).
Now, we introduce the sets

(GR0
0 )+ := {x ∈ GR0

0 | u(x) > w(x)} ⊂ GR0
0 ,

(ΓR0
+ )+ := {x ∈ ΓR0

+ | u(x) > w(x)} ⊂ ΓR0
+

and assume that (GR0
0 )+ ̸= ∅ and (ΓR0

+ )+ ̸= ∅. Let k ≥ 1 be any an odd number. We choose

η = max{(u− w)k, 0} as a test function in the integral inequality (4.4). We have∫ 1

0

∂(uτ |uτ |p(x)−2)

∂uτ
dτ = (p(x)− 1)

∫ 1

0

|uτ |p(x)−2dτ > 0.

Then, by assumptions (i)–(iii) and η
∣∣∣
ΩR0

= 0, we obtain from (4.4) that∫
(G

R0
0 )+

{
kςpz

k−1
(∫ 1

0

|∇uτ |p(x)−2dτ
)
|∇z|2 + b2z

k+1
(∫ 1

0

|uτ |−2|∇uτ |p(x)dτ
)}

dx

≤ b1

∫
(G

R0
0 )+

zk
(∫ 1

0

|uτ |−1|∇uτ |p(x)−1dτ
)
|∇z|dx.

(4.5)

By the Cauchy inequality,

b1z
k|∇z||uτ |−1|∇uτ |p(x)−1 =

(
|uτ |−1z

k+1
2 |∇uτ |

p(x)
2

)(
b1z

k−1
2 |∇z||∇uτ |

p(x)
2 −1

)
≤ ε

2
|uτ |−2zk+1|∇uτ |p(x) + b21

2ε
zk−1|∇z|2|∇uτ |p(x)−2,

for all ε > 0. Taking ε = 2b2, we obtain from (4.5) that(
kςp −

b21
4b2

)∫
(G

R0
0 )+

zk−1|∇z|2
(∫ 1

0

|∇uτ |p(x)−2dτ
)
dx ≤ 0. (4.6)

Choosing the odd number k ≥ max
(
1;

b21
2b2ςp

)
, in view of z(x) ≡ 0 on ∂(GR0

0 )+, we obtain from

(4.6) that z(x) ≡ 0 in (GR0
0 )+. We have a contradiction to our definition of the set (GR0

0 )+. This
completes the proof. □

Remark 4.2. For the p(x)-Laplacian assumption (i) is satisfied with

ζp =

{
1, if p(x) ≥ 2;

p− − 1, if 1 < p− ≤ p(x) < 2.

4.1. Barrier function and eigenvalue problem (2.1). We shall study the barrier function
w(r, ω) ̸≡ 0 as a solution of the auxiliary problem

−△p+w = µw−1|∇w|p+ , x ∈ GR0
0 ,

∂w

∂−→n

∣∣∣
ΓR
−

= 0, |∇w|p+−2 ∂w

∂−→n
+

γ

rp+−1
w|w|p+−2 = 0, x ∈ ΓR0

+ ,

0 ≤ µ <
2

3
, γ ≥ 1.

(4.7)



22 M. BORSUK EJDE-2025/51

By direct calculations, we derive a solution of this problem in the form

w = w(r, ω) = rκψκ/λ(ω), κ =
p+ − 1

p+ − 1 + µ
λ, (4.8)

where (λ, ψ(ω)) is the solution of the eigenvalue problem (2.1). For this function we calculate with

regard to y(ω) = ψ′(ω)
ψ(ω) :

∂w

∂r
= κrκ−1ψκ/λ(ω),

∂w

∂ω
=

κ
λ
rκψ

κ
λ −1(ω)ψ′(ω),

|∇w| = κ
λ
rκ−1ψ

κ
λ −1(ω)

√
λ2ψ2(ω) + ψ′2(ω) =

κ
λ
rκ−1ψ

κ
λ (ω)

√
λ2 + y2(ω).

(4.9)

Proposition 4.3. w ∈ N
1,p(x)
−1,∞ (GR0

0 ).

Proof. From (4.8), (4.7) and (2.5) it follows that w ∈ L∞(GR0
0 ). Next,∫

G
R0
0

r−p(x)wp(x) dx =

∫
G

R0
0

r(κ−1)p(x)ψ
κ
λ p(x)(ω) dx.

By assumption (A1), we have

• for κ ≥ 1:

r(κ−1)p(x) ≤ r(κ−1)p− , if r ≤ 1;

r(κ−1)p(x) ≤ r(κ−1)p+ , if r ≥ 1;
(4.10)

• for 0 < κ ≤ 1:

r(κ−1)p(x) ≤ r(κ−1)p− , if r ≥ 1;

r(κ−1)p(x) ≤ r(κ−1)p+ , if r ≤ 1;
(4.11)

• ψ
κ
λ p(x)(ω) ≤ 1, by (2.5).

It follows that ∫
G

R0
0

r−p(x)wp(x) dx ≤
∫
G

R0
0

r(κ−1)p(x)ψ
κ
λ p(x)(ω) dx. (4.12)

If R0 > 1, then
∫
G

R0
0

=
∫
G1

0
+
∫
G

R0
1

. We consider two cases: (1) 0 < κ ≤ 1 and (2) κ > 1.

4.2. Case 0 < κ ≤ 1. By assumption (A1), (κ− 1)p+ ≤ (κ− 1)p(x) ≤ (κ− 1)p−, therefore with
regard to (2.5):∫

G1
0

r(κ−1)p(x)ψ
κ
λ p(x)(ω) dx ≤ 2z0ω0

∫ 1

0

r(κ−1)p++1dr =
2z0ω0

(κ − 1)p+ + 2
,

because (κ − 1)p+ + 2 > 0; in fact, it is obvious for p+ ≤ 2; if p+ > 2, then, by λ > p+−1
p+

(see

(2.3)) and µ ∈ [0, 2/3),

κ =
p+ − 1

p+ − 1 + µ
λ >

3(p+ − 1)2

p+(3p+ − 1)
=⇒ (κ − 1)p+ + 2 >

p+ + 1

3p+ − 1
> 0.

Similarly,∫
G

R0
1

r(κ−1)p(x)ψ
κ
λ p(x)(ω) dx ≤ 2z0ω0

∫ R0

1

r(κ−1)p−+1dr = 2z0ω0
R

(κ−1)p−+2
0 − 1

(κ − 1)p− + 2

and, because p− ≤ p+, by the above, we have

(κ − 1)p− + 2 ≥ (κ − 1)p+ + 2 > 0. (4.13)

From the inequalities obtained above, we derive∫
G

R0
0

r−p(x)wp(x) dx ≤ 2z0ω0 ·
{ (1− κ)(p+ − p−)

⟨(κ − 1)p+ + 2⟩⟨(κ − 1)p− + 2⟩
+

R
(κ−1)p−+2
0

(κ − 1)p− + 2

}
. (4.14)



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 23

Case κ > 1. Again, by assumption (A1), (κ − 1)p− ≤ (κ − 1)p(x) ≤ (κ − 1)p+; therefore,∫
G1

0

r(κ−1)p(x)ψ
κ
λ p(x)(ω) dx ≤ 2z0ω0

∫ 1

0

r(κ−1)p−+1dr =
1

(κ − 1)p− + 2
,

∫
G

R0
1

r(κ−1)p(x)ψ
κ
λ p(x)(ω) dx ≤ 2z0ω0

∫ R0

1

r(κ−1)p++1dr =
R

(κ−1)p++2
0 − 1

(κ − 1)p+ + 2
.

It follows that∫
G

R0
0

r−p(x)wp(x) dx ≤ 2z0ω0


(1−κ)(p+−p−)

⟨(κ−1)p++2⟩⟨(κ−1)p−+2⟩ +
R

(κ−1)p−+2

0

(κ−1)p−+2 , if 0 < κ ≤ 1,

(κ−1)(p+−p−)
⟨(κ−1)p++2⟩⟨(κ−1)p−+2⟩ +

R
(κ−1)p++2

0

(κ−1)p++2 , if κ > 1.
(4.15)

From (4.9) with regard to (2.12) we obtain that∫
G

R0
0

w−1|∇w|p(x) dx

=

∫
G

R0
0

(κ
λ

)p(x)
r(p(x)−1)κ−p(x)ψ(p(x)−1)κ

λ −p(x)(ω)
(
λ2ψ2(ω) + ψ′2(ω)

)p(x)/2
dx

≤ ψ
(p−−1)κ

λ −p+
0

∫
G

R0
0

r(p(x)−1)κ−p(x) (λ2 + y2(ω)
)p(x)/2

dx

≤ ψ
(p−−1)κ

λ −p+
0

∫
G

R0
0

r(p(x)−1)κ−p(x) (λ2 + y20
)p(x)/2

dx.

Since
√
λ2 + y20 = const(γ, µ, ω0, p+) ≡ Y0, (see (2.6)) and p(x) ∈ [p−, p+], we have(

λ2 + y20
)p(x)/2 ≤ C1 = const(γ, µ, ω0, p+, p−).

From the above inequality we obtain that∫
G

R0
0

w−1|∇w|p(x) dx ≤ C1ψ
(p−−1)κ

λ −p+
0

∫
GR

0

r(p(x)−1)κ−p(x) dx

= C1ψ
(p−−1)κ

λ −p+
0

∫
G

R0
0

r(κ−1)(p(x)−p+) · r(κ−1)p+−κ dx.

(4.16)

Now, by assumptions (Ai) and (A2), for 0 < r < 1 we derive that

r(κ−1)(p(x)−p+) ≤

{
r(1−κ)Lr, if κ > 1,

1, if κ ≤ 1.

Using the well known inequality

rα| ln r| ≤ 1

αe
, ∀α > 0, 0 < r < 1,

where e is the Euler number, for κ > 1 and α = 1, we establish the inequality

r(1−κ)Lr ≤ e
L(κ−1)

e , 0 < r < 1. (4.17)

Thus, from (4.16)-(4.17) it follows that∫
G1

0

w−1|∇w|p(x) dx ≤ 2C1z0ω0ψ
(p−−1)κ

λ −p+
0

∫ 1

0

r(κ−1)p+−κ+1dr

{
e

L(κ−1)
e , if κ > 1,

1, if κ ≤ 1.

□

Now we verify that (κ − 1)p+ − κ + 2 > 0 =⇒ (κ − 1)(p+ − 1) + 1 > 0. If κ ≥ 1, this
assertion is obvious. Let κ < 1. For 1 < p+ ≤ 2 we have (κ − 1)(p+ − 1) + 1 > κ > 0. Let us

κ < 1, 2 < p+ < 3. For this case we derived from the above that κ > 3(p+−1)2

p+(3p+−1) . It follows that

(κ − 1)(p+ − 1) + 1 >
−2p2+ + 7p+ − 3

p+(3p+ − 1)
> 0 for p+ ∈ (2, 3). (4.18)



24 M. BORSUK EJDE-2025/51

Therefore,∫
G1

0

w−1|∇w|p(x) dx ≤ C1ψ
(p−−1)κ

λ −p+
0

2z0ω0

(κ − 1)p+ − κ + 2

{
e

L(κ−1)
e , if κ > 1,

1, if κ ≤ 1.
(4.19)

Further, for r > 1 we have

(κ − 1)p(x)− κ ≤

{
r(κ−1)p− , if κ ≤ 1,

r(κ−1)p+−1, if κ ≥ 1;

therefore, with regard to (4.13),∫
G

R0
1

w−1|∇w|p(x) dx

≤ C1ψ
(p−−1)κ

λ −p+
0 2z0ω0


∫ R0

1
r(κ−1)p−+1dr =

R
(κ−1)p−+2

0 −1
(κ−1)p−+2 , if κ ≤ 1,∫ R0

1
r(κ−1)p+dr =

R
(κ−1)p++1

0 −1
(κ−1)p++1 , if κ > 1.

(4.20)

Adding (4.19) and (4.20) with regard to (4.18) we obtain∫
G

R0
0

w−1|∇w|p(x) dx

≤ C1ψ
(p−−1)κ

λ −p+
0 2z0ω0


(1−κ)(p+−p−)+κ

⟨(κ−1)p++2⟩⟨(κ−1)p+−κ+2⟩ +
R

(κ−1)p−+2

0

(κ−1)p−+2 , if 0 < κ ≤ 1;

κ−1
⟨(κ−1)p+−κ+2⟩⟨(κ−1)p++1⟩ +

R
(κ−1)p++1

0

(κ−1)p++1 , if κ > 1.

(4.21)

5. Proof of Theorem 1.3

Let A > 1, and let w(r, ω) be the barrier function defined above. By the definition of (1.2) we
consider the operator

Q(Aw, η) ≡
∫
G

R0
0

〈
Ap(x)−1|∇w|p(x)−2wxi

ηxi
+ a(x)Ap(x)wp(x)η(x)

+ b (Aw,A∇w) η(x)
〉
dx+

∫
Γ
R0
+

γ(ω)Ap(x)−1r1−p(x)wp(x)−1η(x) dS

−
∫
Γ
R0
+

g(x)η dS −
∫
ΩR0

Ap(x)−1|∇w|p(x)−2 ∂w

∂r
η(x)dΩR0

(5.1)

for all non-negative η ∈ N
1,p(x)
−1,∞ (GR0

0 ). Integrating by parts, we obtain∫
G

R0
0

Ap(x)−1|∇w|p(x)−2wxiηxi dx

= −
∫
G

R0
0

d

dxi
⟨Ap(x)−1|∇w|p(x)−2wxi

⟩η(x) dx+

∫
Γ
R0
−

Ap(x)−1|∇w|p(x)−2 dw

dn
η(x) dS

+

∫
Γ
R0
+

Ap(x)−1|∇w|p(x)−2 dw

dn
η(x) dS +

∫
ΩR0

Ap(x)−1|∇w|p(x)−2 ∂w

∂r
η(x)dΩR0 .

From (5.1), with regard to problem (4.7), it follows that

Q(Aw, η) = J
G

R0
0

+ J
Γ
R0
+

+ JΩR0
, (5.2)



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 25

where

J
G

R0
0

≡
∫
G

R0
0

〈
µAp(x)−1w−1|∇w|p(x) −Ap(x)−1|∇w|p+−2wxi

d|∇w|p(x)−p+
dxi

− ∂Ap(x)−1

∂xi
wxi

|∇w|p(x)−2 + a(x)Ap(x)wp(x) + b (Aw,A∇w)
〉
η(x) dx,

J
Γ
R0
+

≡
∫
Γ
R0
+

γ(ω)
(Aw
r

)p(x)−1

⟨1−
(r|∇w|

w

)p(x)−p+
⟩η(x) dS −

∫
Γ
R0
+

g(x)η dS,

JΩR0
≡

∫
ΩR0

Ap(x)−1|∇w|p(x)−2 ∂w

∂r
η(x)dΩR.

(5.3)

At first, we assert that J
Γ
R0
+

≥ 0. Indeed, by (4.9),(r|∇w|
w

)∣∣∣
Γ
R0
+

=
κ
λ

√
λ2 + y20

and the desired inequality follows from Proposition 2.2 and assumption (A5). And it is obvious
that JΩR0

≥ 0. Thus, from (5.2) it follows that

Q(Aw, η) ≥ J
G

R0
0
. (5.4)

Further, we proceed to the estimating of integral J
G

R0
0

. Setting W (x) = |∇w|p(x)−p+ , we calculate

lnW (x) = (p(x)− p+) ln |∇w|, =⇒
1

W (x)

∂W

∂xi
=

∂p

∂xi
ln |∇w|+ p(x)− p+

|∇w|
d|∇w|
dxi

=⇒ d

dxi

(
|∇w|p(x)−p+

)
= |∇w|p(x)−p+⟨ ∂p

∂xi
ln |∇w|+ p(x)− p+

|∇w|
d|∇w|
dxi

⟩.

Similarly,
d

dxi

(
Ap(x)−1

)
= Ap(x)−1 ∂p

∂xi
lnA.

By (5.3), we obtain that

J
G

R0
0

≥
∫
G

R0
0

{
Ap(x)−1|∇w|p(x)−2

〈
µw−1|∇w|2 − (∇p · ∇w)(lnA+ ln |∇w|)

− p(x)− p+
|∇w|

wxi

d|∇w|
dxi

〉
+ a(x)Ap(x)wp(x) + b (Aw,A∇w)

}
η(x) dx.

(5.5)

Passing to polar coordinates, we calculate

wxi

d|∇w|
dxi

=
∂w

∂r

∂|∇w|
∂r

+
1

r2
∂w

∂ω

∂|∇w|
∂ω

.

Now, by (5.4) and (5.5) with regard to assumption (A6), we obtain that

Q(Aw, η) ≥
∫
G

R0
0

Ap(x)−1
{
|∇w|p(x)−2

〈
σw−1|∇w|2 − (∇p · ∇w)(lnA+ ln |∇w|)

− p(x)− p+
|∇w|

(
∂w

∂r

∂|∇w|
∂r

+
1

r2
∂w

∂ω

∂|∇w|
∂ω

)〉
+Aa(x)wp(x)

}
η(x) dx,

(5.6)

with

σ =

{
µ− δ, if µ > 0;

ν, if µ = 0,

where δ is defined by (3.8).

Taking into account (4.9), condition (A2) for p(x), and ψ′(ω)
ψ(ω) = y(ω), we calculate the following

3 items:
(1)

|(∇p · ∇w)(lnA+ ln |∇w|)| ≤ |∇p| · |∇w|(lnA+ | ln |∇w||)



26 M. BORSUK EJDE-2025/51

≤ L0|∇w| · (lnA+ | ln |∇w||).

By (4.9), (2.5),and (2.12), we derive

| ln |∇w|| ≤ | ln κ
λ
|+ |κ − 1| | ln r|+ κ

λ
ψ

κ
λ −1 lnψ +

1

2
| ln(λ2 + y2(ω))|

≤ ln
λ

κ
+

1

2
| ln(λ2 + y20)|+ |κ − 1| · | ln r|

= lnC1(p+) + |κ − 1| | ln r|.

Note that C1(p+) =
p+−1+µ
p+−1

√
λ2 + y20 > 1: indeed,

* p+ = 2: from boundary condition (2.8) we have

C1(2) = (1 + µ)
√
λ2 + y20 > (1 + µ)|y0| = γ(1 + µ)2 ≥ 1,

since µ ≥ 0, γ ≥ 1, by assumption (A4);
* 1 < p+ < 2: by (2.16), we have

|y0| >
λ

κ
γ

1
p+−1 =⇒ C1(p+) >

λ

κ
|y0| >

(
λ

κ

)2

· γ
1

p+−1 > 1;

* p+ > 2: from (2.11) we obtain

|y0| <
λ

κ
γ

1
p+−1 =⇒

(
λ2 + y20

) p+−2

2 =
1

|y0|
γ

(
λ

κ

)p+−1

>

(
λ

κ

)p+−2

γ
p+−2

p+−1 > 1

=⇒ C1(p+) >
√
λ2 + y20 >

λ

κ
γ

1
p+−1 > 1.

Therefore,

|(∇p · ∇w)(lnA+ ln |∇w|)|

≤ L0
κ
λ
rκ−2ψ

κ
λ (ω)

√
λ2 + y2(ω)

(
R0 ln(AC1) + |κ − 1|R0 lnR0

)
, r < R0.

(2) ∂|∇w|
∂r = κ−1

r |∇w|,
∂|∇w|
∂ω

= |∇w|
(
κ
λ
+

y′(ω)

λ2 + y2(ω)

)
y(ω) = |∇w|

(
p+ − 1

p+ − 1 + µ
+

y′(ω)

λ2 + y2(ω)

)
y(ω),

by (4.8). Then

p+ − p(x)

|∇w|
·
(∂w
∂r

∂|∇w|
∂r

+
1

r2
∂w

∂ω

∂|∇w|
∂ω

)
= κ(p+ − p(x))

w

r2

〈
κ − 1 +

y2

λ

(
p+ − 1

p+ − 1 + µ
+

y′(ω)

λ2 + y2(ω)

)〉
.

From (2.10) we have that

y′(ω)

λ2 + y2(ω)
= − (p+ − 1)(y2 + λ2) + (2− p+)λ

(p+ − 1)y2 + λ2

implies

p+ − 1

p+ − 1 + µ
+

y′(ω)

λ2 + y2(ω)
≥ −λ 2− p+

(p+ − 1)y2 + λ2
− µ

p+ − 1

p+ − 1 + µ

y2 + λ2

(p+ − 1)y2 + λ2

≥ −λ 2− p+
(p+ − 1)y2 + λ2

− µ
y2 + λ2

(p+ − 1)y2 + λ2
.

It follows that

y2

λ

(
p+ − 1

p+ − 1 + µ
+

y′(ω)

λ2 + y2(ω)

)
≥ − 1

λ

y2

(p+ − 1)y2 + λ2
⟨(2− p+)λ+ µ(y2 + λ2)⟩



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 27

≥

−µ
λy

2
0 , if p+ ≥ 2, since y2+λ2

(p+−1)y2+λ2 ≤ 1,

− 2−p+
p+−1 − µ(y20+λ

2)
λ(p+−1) , if 1 < p+ < 2, since y2

(p+−1)y2+λ2 ≤ 1
p+−1 ,

and |y(ω)| ≤ |y0|.

It follows that

p+ − p(x)

|∇w|

(∂w
∂r

∂|∇w|
∂r

+
1

r2
∂w

∂ω

∂|∇w|
∂ω

)
≥ −L0C0(p+, λ, |y0|, µ)

w

r
;

(3) |w|−1|∇w|2 =
(κ
λ

)2
rκ−2ψ

κ
λ (ω)

(
λ2 + y2(ω)

)
:

From item (1)–(3) it follows that

σ
|∇w|2

w
− (∇p · ∇w)(lnA+ ln |∇w|)− p(x)− p+

|∇w|

(∂w
∂r

∂|∇w|
∂r

+
1

r2
∂w

∂ω

∂|∇w|
∂ω

)
≥

(κ
λ

)2
rκ−2ψ

κ
λ (ω)

(
λ2 + y2(ω)

) 〈
σ − L0R0

κ
lnA− L0R0

κ
lnC1 −

L0|κ − 1|
κ

R0 lnR0 −
L0C0R0

κ2

〉
.

(5.7)
From assumption (A3) and (4.8) we have

a0(x)Aw
p(x) ≥ Aa0r

(κ−1)p(x)−κψ
κ
λ p(x)(ω).

Then, using (4.8), (4.9), and (2.5), it follows that

|∇w|p(x)−2
〈
σw−1|∇w|2 − (∇p · ∇w)(lnA+ ln |∇w|)− p(x)− p+

|∇w|

×
(∂w
∂r

∂|∇w|
∂r

+
1

r2
∂w

∂ω

∂|∇w|
∂ω

)〉
+ a0(x)Aw

p(x)

≥
(κ
λ

)p(x)
rκ(p(x)−1)−p(x)ψ

κ
λ (p(x)−1)(ω)(λ2 + y2(ω))

p(x)
2

×
〈
σ +Aa0 −

L0R0

κ
lnA− L0R0

κ
lnC1 −

L0|κ − 1|
κ

R0 lnR0 −
L0C0R0

κ2

〉
.

Lemma 5.1. Let e be the Euler number. Then for all x > 1 and k ∈ (0, e] it holds x ≥ k lnx.

Figure 3. Graphs of functions y = x and y = e ln(x).



28 M. BORSUK EJDE-2025/51

Proof. It is obvious that for k = e and x = e the equality x = k lnx is fulfilled. From graphs of
functions y = x and y = e lnx (see Figure 3) it follows that

x ≥ e lnx, ∀x > 0 =⇒ x ≥ k lnx, ∀x > 1, ∀k ∈ (0, e],

see Figure 4. □

Figure 4. Graphs of functions y = x, y = k ln(x), 0 < k < e

At first, by Lemma 5.1, for all A > 1 the inequality A ≥ 2L0R0

κa0 lnA is valid. If 2L0R0

κa0 ≤ e, than

a0 ≥ 2L0R0

κe
. (5.8)

Next we can choose A > 1 as follows:

A ≥ 2L0R0

κa0

{
(lnC1 +

C0

κ ), for R0 ≤ 1;

(lnC1 + |κ − 1| lnR0 +
C0

κ ), for R0 > 1.
(5.9)

Therefore we can rewrite inequality (5.7) as

|∇w|p(x)−2
〈
σw−1|∇w|2 − (∇p · ∇w)(lnA+ ln |∇w|)

− p(x)− p+
|∇w|

(∂w
∂r

∂|∇w|
∂r

+
1

r2
∂w

∂ω

∂|∇w|
∂ω

)〉
+ a0(x)Aw

p(x)

≥ σκp(x)rκ(p(x)−1)−p(x) exp
(
y0ω0κ

p+ − 1

λ

)
≥ σκ0r

(κ−1)p(x)−κ ,

(5.10)

where

κ0 = exp
(
y0ω0κ

p+ − 1

λ

){
κp+ if κ < 1,

κp− if κ ≥ 1.

From (5.6) and (5.10) it follows that

Q(Aw, η) ≥ σκ0

∫
G

R0
0

Ap(x)−1r(κ−1)p(x)−κη(x) dx.



EJDE-2025/51 NEUMANN-ROBIN PROBLEM FOR p(x)-LAPLACIAN EQUATIONS 29

Since p(x) ≥ p− > 1 and A > 1, we have Ap(x)−1 ≥ Ap−−1. Therefore, taking into consideration
assumption (A3), the above inequality takes the form

Q(Aw, η) ≥ σκ0A
p−−1

∫
G

R0
0

r(κ−1)p(x)−κη(x) dx ≥ σκ0A
p−−1

∫
G

R0
0

rβ(x)η(x) dx

≥
∫
G

R0
0

f0r
β(x)η(x) dx ≥

∫
G

R0
0

|f(x)|η(x) dx

≥
∫
G

R0
0

f(x)η(x) dx = Q(u, η), by (1.2),

for all non-negative η ∈ N
1,p(x)
−1,∞ (GR0

0 ), if A > 1 satisfies

A ≥
( f0
σκ0

) 1
p−−1

. (5.11)

Further, we show that u(x) ≤ Aw(x) on ΩR0
. By (4.8) and (2.5),

w(x)|ΩR0
= R0

κψκ/λ(ω) ≥ R0
κψ

κ/λ
0 .

Because |u(x)| ≤M0 for all x ∈ GR0
0 , we can choose A such that

A ≥ M0

R0
κψ

κ/λ
0

(5.12)

and therefore,

Aw(x)|ΩR0
≥ AR0

κψ
κ/λ
0 ≥M0 ≥ u(x)|ΩR0

.

Thus, if we choose a large A > 1 according to (5.9), (5.11), and (5.12),

A ≥ max
{ M0

Rκ
0 ψ

κ/λ
0

,
( f0
σκ0

) 1
p−−1

,
2L0R0

κa0

{
(lnC1 +

C0

κ ), fo rR0 ≤ 1,

(lnC1 + |κ − 1| lnR0 +
C0

κ ), for R0 > 1,

}
,

then we come to the Comparison Principle

Q(u, η) ≤ Q(Aw, η) in GR0
0 ; u(x) ≤ Aw(x) on ΩR0

.

Thus, the Comparison Principle implies that

u(x) ≤ Aw(x) in GR0
0 .

Similarly, we derive the estimate u(x) ≥ −Aw(x) in GR0
0 when we replace u(x) with −u(x). By

this and (2.5), we obtain the required estimate

|u(x)| ≤ Aw(x) ≤ Arκ , in GR0
0 .

References

[1] S. Antontsev, L. Consiglieri; Elliptic boundary value problems with nonstandard growth conditions, Nonlinead
Analysis. 71, (2009), 891-902.

[2] M. Borsuk; Oblique Derivative Problems for Elliptic Equations in Conical Domains, A Springer Basel book
Series: Frontiers in Elliptic and Parabolic Problems. 1st ed. 2023, 327 p. ISBN: 978-3-031-28380-2.

[3] M. Borsuk; L∞ - estimate for the Robin pdoblem of a singulard variable p - Laplacian equation in a conical

domain, Electronic Journal of differential Equations, 2018, no. 49 (2018), 1 - 9.
[4] M. Borsuk, S. Jankowski; The Robin problem for singular p-Laplacian equation in a cone, Complex variables

and elliptic equations, 63, no. 3 (2018), 333 - 345. DOI: 10.1080/17476933.2017.1307837.

[5] M. Borsuk, V. Kondratiev; Elliptic Boundary Value Pdroblems of Second Orded in Piecewise Smooth domains,
North-Holland Mathematical Library, 69,531 pp., Elsevier, 531 pp., 2006

[6] G. R. Cirmi, M. M. Porzio; L∞− solutions for some nonlinear degenerate elliptic and parabolic equations,

Ann. mat. pura ed appl. (IV) 169 (1995), 67-86.
[7] Sh.-G. Deng; Positive solutions for Robin problem involving the p(x)−Laplacian, Journal of Mathematical

Anlysis and Applications, 360 (2009), 548-560.

[8] Sh.-G. Deng, Q. Wang, Sh. Cheng; On the p(x)-Laplacian Robin eigenvalue problem, Applied Mathematics
and Computation, 217 (2011), 5643-5649.

[9] D. Gilbarg, N.S. Trudinger; Elliptic Partial Differential Equations of Second Order, Springer-Verlag,
Berlin/Heidelberg/New York, 1977. Revised Third Printing, 1998.



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[10] P. Harjulehto, P. Hästö, Út Lê, M. Nuorito; Overview of differential equations with non-standard growth,

Nonlinear Analysis 72 (2010), 4551-4574.

[11] M. Mihǎilescu, C. Varga; Multiplicity results for some elliptic problems with nonlinear boundary conditions
involving variable exponents, Computeds and Mathematics with Applications 62 (2011), 3464-3471.

[12] M. K. V. Murthy, G. Stampacchia; Boundary value problem for some degenerate elliptic operators Ann. Math.

Pura Appl. Ser. IV, 80 (1968), 1-122.
[13] V. V. Zhikov; On Lavrentiev’s phenomenon, Russian J. Math. Phys. 13, no. 2 (1994), 249–269.

[14] V. Zhikov; On some variational problems, Russian J. Math. Phys. 5, no. 1 (1997), 105–116.

Mikhail Borsuk
Faculty of Mathematics and Computer Science, University of Warmia and Mazury in Olsztyn, 10-957

Olsztyn-Kortowo, Poland

Email address: mborsuk40@gmail.com


	1. Introduction
	2. Nonlinear eigenvalue problem
	Case p+=2
	Case p+=2
	Case p+> 2
	Case 1<p+<2
	Case p+ =2
	Case p+ >2
	Case p+ <2

	3. Maximum principle
	Case 0<R01
	Case R0 > 1
	Case g0=0
	Case g0 =0

	4. Comparison principle
	4.1. Barrier function and eigenvalue problem (??)
	4.2. Case 0<1
	Case > 1

	5. Proof of Theorem ??
	References