# Existence and Nonexistence of a Solution for a Nonlinearp

Authors: Elhoussine Azroul; Abdelkrim Barbara; Hicham Redwane
Journal: International Journal of Analysis
 (2014)

---

## Abstract

We discuss the existence and nonexistence of solution of a nonlinear problemp
(
x
)-elliptic-div
(
a
(
x
,
∇
u
)
)
+
g
(
x
,
u
,
∇
u
)
=
μ, where μ is a Radon measure with bounded total variation, by considering the Sobolev spaces with variable exponents. This study is done in two cases: (i) μ is absolutely continuous with respect to p
(
x
)-capacity. and (ii) μ is concentrated on a Borel set of null p
(
x
)-capacity.


---

## Body

## 1. Introduction

LetΩ be a bounded domain of ℝ
N, let N
≥
2, and let p
(
·
)
∈
𝒞
+
(
Ω
¯
) (see definition below). Consider the following nonlinear p
(
x
)-elliptic problem:
(
𝒫
)
A
u
+
g
(
x
,
u
,
∇
u
)
=
μ
in
Ω
u
=
0
on
⁡
∂
Ω
,
where A
u
=
-
div
⁡
(
a
(
x
,
∇
u
)
) is a Leray-Lions operator from W
0
1
,
p
(
x
)
(
Ω
) into its dual W
-
1
,
p
′
(
x
)
(
Ω
). The nonlinear term g
(
x
,
u
,
∇
u
) satisfies a growth condition of the following form:
(1)
|
g
(
x
,
s
,
ξ
)
|
≤
b
(
|
s
|
)
(
d
(
x
)
+
|
ξ
|
p
(
x
)
)
,
where b
:
ℝ
+
→
ℝ
+ is a continuous increasing function and d
:
Ω
→
ℝ
+
,
d
∈
L
1
(
Ω
) and assuming also that g
(
x
,
s
,
ξ
)
sgn
⁡
(
s
)
≥
ρ
|
ξ
|
p
(
x
), a.e. x
∈
Ω, ∀
ξ
∈
ℝ
N, and ∀
s
∈
ℝ for |
s
|
≥
σ, where σ and ρ are two positive real constants. The second term μ is a Radon measure on Ω.We are interested in the existence and nonexistence of solution of the problem(
𝒫
). We prove the existence of solution (
𝒫
) if μ does not charge the sets E of null p
(
x
)-capacity, in other words, if and only if μ is absolutely continuous with respect to p
(
x
)-capacity. We give a necessary and sufficient condition that allows us the existence of solution (
𝒫
). Thus, if μ is concentrated on a Borel set E of Ω of null p
(
x
)-capacity, we will show that the problem (
𝒫
) admits no solution. Boccardo et al. [1] treated (
𝒫
) for p constant. We can also see other variations of this problem; for example, if the nonlinear term is independent of ∇
u and behaves as |
u
|
q
-
1
u, with q
>
1, the results of the existence and nonexistence of solution depend on the measure μ; see Baras and Pierre [2], Brezis [3], and Gallouët and Morel [4].


## 2. Preliminaries

LetΩ be a bounded open subset of ℝ
N
,
N
≥
2. The function p
(
·
) satisfies the log-Hölder continuity on Ω if
(2)
|
p
(
x
)
-
p
(
y
)
|
≤
C
|
log
⁡
|
x
-
y
|
|
,
∀
x
,
y
∈
Ω
¯
such
that
|
x
-
y
|
<
1
2
,
with C being a positive constant.LetC
+
(
Ω
¯
)
=
{
p
:
Ω
¯
→
ℝ be log-Hölder continuous such that 1
<
p
-
≤
p
+
<
N
}, where p
-
=
inf
⁡
x
∈
Ω
¯
(
p
(
x
)
) and p
+
=
sup
⁡
x
∈
Ω
¯
(
p
(
x
)
).Forp
∈
C
+
(
Ω
¯
), we define the variable exponent Lebesgue space
(3)
L
p
(
x
)
(
Ω
)
=
{
that
∫
Ω
d
x
<
∞
}
u
:
Ω
⟶
ℝ
as
measurable
such
that
∫
Ω
|
u
(
x
)
|
p
(
x
)
d
x
<
∞
}
,
normed by ∥
u
∥
p
(
x
)
=
inf
⁡
⁡
{
λ
>
0
,
∫
Ω
‍
|
u
(
x
)
/
λ
|
p
(
x
)
d
x
≤
1
}. The space L
p
(
x
)
(
Ω
) is a separable and reflexive Banach space, and its dual space is isomorphic to L
p
′
(
x
)
(
Ω
), where 1
/
p
(
x
)
+
1
/
p
′
(
x
)
=
1 (see [5, 6]).Proposition 1 (see [5, 6]).
(i) For anyu
∈
L
p
(
x
)
(
Ω
) and v
∈
L
p
′
(
x
)
(
Ω
), one has the following:
(4)
|
∫
Ω
u
v
d
x
|
≤
(
1
p
-
+
1
p
-
′
)
∥
u
∥
p
(
x
)
∥
v
∥
p
′
(
x
)
.
(ii) For allp
1
,
p
2
∈
C
+
(
Ω
¯
) such that p
1
(
x
)
≤
p
2
(
x
) a.e. in Ω; then L
p
2
(
x
)
(
Ω
)
↪
L
p
1
(
x
)
(
Ω
) and the embedding is continuous.Proposition 2 (see [5, 6]).
If one denotesρ
(
u
)
=
∫
Ω
|
u
(
x
)
|
p
(
x
)
d
x, ∀
u
∈
L
p
(
x
)
(
Ω
), then
(5)
min
⁡
⁡
{
∥
u
∥
p
(
x
)
p
-
,
∥
u
∥
p
(
x
)
p
+
}
≤
ρ
(
u
)
≤
max
⁡
{
∥
u
∥
p
(
x
)
p
-
,
∥
u
∥
p
(
x
)
p
+
}
.Define now the Sobolev space with variable exponent by(6)
W
1
,
p
(
x
)
(
Ω
)
=
{
u
∈
L
p
(
x
)
(
Ω
)
|
∇
u
|
∈
L
p
(
x
)
(
Ω
)
}
,
normed by
(7)
∥
u
∥
1
,
p
(
x
)
=
∥
u
∥
p
(
x
)
+
∥
∇
u
∥
p
(
x
)
,
∀
u
∈
W
1
,
p
(
x
)
(
Ω
)
.
Let W
0
1
,
p
(
x
)
(
Ω
) be the closure of C
0
∞
(
Ω
) in W
1
,
p
(
x
)
(
Ω
) and let p
*
(
x
)
=
N
p
(
x
)
/
(
N
-
p
(
x
)
) for p
(
x
)
<
N.Proposition 3 (see [5, 7]).
(i) Assuming that1
<
p
-
≤
p
+
<
∞, the spaces W
1
,
p
(
x
)
(
Ω
) and W
0
1
,
p
(
x
)
(
Ω
) are separable and reflexive Banach spaces.
(ii) Ifq
∈
C
+
(
Ω
-
) and q
(
x
)
<
p
*
(
x
), for any x
∈
Ω, then embedding W
0
1
,
p
(
x
)
(
Ω
)
↪
↪
L
q
(
x
)
(
Ω
) is continuous and compact.
(iii) The Poincaré inequality: there exists a constantC
>
0, such that
(8)
∥
u
∥
p
(
x
)
≤
C
∥
∇
u
∥
p
(
x
)
,
∀
u
∈
W
0
1
,
p
(
x
)
(
Ω
)
.
(iv) The Sobolev inequality: there exists another constantC
>
0, such that
(9)
∥
u
∥
p
*
(
x
)
≤
C
∥
∇
u
∥
p
(
x
)
,
∀
u
∈
W
0
1
,
p
(
x
)
(
Ω
)
.Remark 4.
By (iii) of the Proposition3, we deduce that ∥
∇
u
∥
p
(
x
) and ∥
u
∥
1
,
p
(
x
) are equivalent norms in W
0
1
,
p
(
x
)
(
Ω
).Proposition 5 (see [8]).
One denotes the dual space of the Sobolev spaceW
0
1
,
p
(
x
)
(
Ω
) by W
-
1
,
p
′
(
x
)
(
Ω
), and, for each F
∈
W
-
1
,
p
′
(
x
)
(
Ω
), there exist f
0
,
f
1
,
…
,
f
N
∈
L
p
′
(
x
)
(
Ω
) such that F
=
f
0
+
∑
i
=
1
N
(
∂
f
i
/
∂
x
i
), and, for all u
∈
W
0
1
,
p
(
x
)
(
Ω
), one has the following:
(10)
〈
F
,
u
〉
=
∫
Ω
f
0
u
d
x
-
∑
i
=
1
N
∫
Ω
f
i
∂
u
∂
x
i
d
x
.
Moreover,
(11)
∥
F
∥
-
1
,
p
′
(
x
)
≃
∑
i
=
0
N
∥
f
i
∥
p
′
(
x
)
.For allk
>
0 and s
∈
ℝ, the truncation function T
k
(
·
) can be defined by
(12)
T
k
(
s
)
=
{
s
,
if
|
s
|
≤
k
,
k
s
|
s
|
,
if
|
s
|
>
k
.
Let K be a compact Ω; the p
(
x
)-capacity of K over Ω is defined by
(13)
cap
p
(
x
)
(
K
,
Ω
)
=
inf
⁡
⁡
{
∫
Ω
|
∇
u
|
p
(
x
)
d
x
,
u
∈
C
0
∞
(
Ω
)
,
u
≥
χ
K
}
,
where χ
K is the characteristic function of K.IfU is open in Ω, the p
(
x
)-capacity of U from Ω is defined by
(14)
ca
p
p
(
x
)
(
U
,
Ω
)
=
sup
⁡
⁡
{
ca
p
p
(
x
)
(
K
,
Ω
)
,
K
compact
⊆
U
}
.
If B is any set of Ω, we define the p
(
x
)-capacity variational B over Ω by
(15)
ca
p
p
(
x
)
(
B
,
Ω
)
=
inf
⁡
⁡
{
ca
p
p
(
x
)
(
U
,
Ω
)
,
U
open
⊇
B
}
.
Denote by M
b
(
Ω
) the space of signed measures on Ω and
(16)
M
b
p
(
x
)
(
Ω
)
=
{
∀
E
⊂
Ω
;
ca
p
p
(
x
)
(
E
,
Ω
)
=
0
}
μ
∈
M
b
(
Ω
)
:
μ
(
E
)
=
0
,
∀
E
⊂
Ω
;
ca
p
p
(
x
)
(
E
,
Ω
)
=
0
}
.


## 3. Measures andp
(
x
)
-Variational Capacity

In this section we will give some results concerning the measures andp
(
x
)-capacity variational capacity of any set of Ω.Theorem 6.
Letμ
∈
M
b
(
Ω
)
,
μ
≥
0; then there exist γ
∈
W
-
1
,
p
′
(
x
)
(
Ω
) and h
∈
L
1
(
Ω
,
d
γ
) such that
(17)
μ
=
h
γ
.Proof.
Letu
∈
W
0
1
,
p
(
x
)
(
Ω
) and let u
~ be measurable representation ca
p
p
(
x
) quasicontinuous of u. Let F
:
W
0
1
,
p
(
x
)
(
Ω
)
→
[
0
,
∞
] and u
→
F
(
u
)
=
∫
Ω
u
~
+
d
μ.
F is convex and lower semicontinuous on W
0
1
,
p
(
x
)
(
Ω
) indeed; let u
n
→
u in W
0
1
,
p
(
x
)
(
Ω
); therefore, u
n
→
u a.e. in Ω, and hence u
n
~
+
→
u
~
+ a.e. in Ω. Then by Fatou’s lemma, we get
(18)
F
(
u
)
=
∫
Ω
u
~
+
d
μ
=
∫
Ω
lim
⁡
⁡
inf
⁡
n
u
n
~
+
d
μ
≤
lim
⁡
⁡
inf
⁡
n
∫
Ω
u
n
~
+
d
μ
≤
lim
⁡
⁡
inf
⁡
n
F
(
u
n
)
.
According to the separability of W
-
1
,
p
′
(
x
)
(
Ω
), there exist λ
n
∈
W
-
1
,
p
′
(
x
)
(
Ω
) and a
n
∈
ℝ such that F
(
u
)
=
sup
⁡
n
(
〈
λ
n
,
u
〉
+
a
n
), and we have the following:
(19)
t
F
(
u
)
=
F
(
t
u
)
≥
(
t
〈
λ
n
,
u
〉
+
a
n
)
,
∀
t
>
0
,
∀
n
,
F
(
u
)
≥
〈
λ
n
,
u
〉
+
a
n
t
,
∀
t
>
0
;
hence, consider the following:
(20)
F
(
u
)
≥
〈
λ
n
,
u
〉
,
∀
u
∈
W
0
1
,
p
(
x
)
(
Ω
)
.
Since F
(
0
)
=
sup
⁡
n
a
n
=
0, then a
n
≤
0 and we have the following:
(21)
F
(
u
)
≥
sup
⁡
n
⁡
〈
λ
n
,
u
〉
≥
sup
⁡
n
⁡
〈
λ
n
,
u
〉
+
a
n
=
F
(
u
)
.
We conclude that F
(
u
)
=
sup
⁡
n
〈
λ
n
,
u
〉.
Letφ
∈
C
c
∞
(
Ω
); we have the following:
(22)
〈
λ
n
,
φ
〉
≤
∫
Ω
φ
+
d
μ
≤
∥
μ
∥
M
b
(
Ω
)
∥
φ
∥
L
∞
(
Ω
)
,
〈
λ
n
,
-
φ
〉
≤
∫
Ω
(
-
φ
)
+
d
μ
=
-
∫
Ω
(
φ
)
+
d
μ
≤
∫
Ω
(
φ
)
+
d
μ
≤
∥
μ
∥
M
b
(
Ω
)
∥
φ
∥
L
∞
(
Ω
)
,
which implies that
(23)
-
∥
μ
∥
M
b
(
Ω
)
∥
φ
∥
L
∞
(
Ω
)
≤
〈
λ
n
,
φ
〉
.
Hence,
(24)
|
〈
λ
n
,
φ
〉
|
≤
∥
μ
∥
M
b
(
Ω
)
∥
φ
∥
L
∞
(
Ω
)
,
and then λ
n
∈
W
-
1
,
p
′
(
x
)
(
Ω
)
∩
M
b
(
Ω
).
SinceF
(
-
φ
)
=
0, ∀
φ
∈
C
c
∞
(
Ω
), φ
≥
0, then 0
≤
〈
λ
n
,
φ
〉
=
∫
Ω
φ
d
λ
n
. Hence,
(25)
∥
λ
n
∥
M
b
(
Ω
)
≤
∥
μ
∥
M
b
(
Ω
)
.
Let γ
=
∑
n
=
1
∞
‍
λ
n
/
2
n
(
∥
λ
n
∥
W
-
1
,
p
′
(
x
)
+
1
); we get that γ is absolutely convergent in W
-
1
,
p
′
(
x
)
(
Ω
). For all φ
∈
C
c
∞
(
Ω
), we have the following:
(26)
|
〈
γ
,
φ
〉
|
=
|
∑
n
=
1
∞
〈
λ
n
,
φ
〉
2
n
(
∥
λ
n
∥
W
-
1
,
p
′
(
x
)
+
1
)
|
≤
∑
n
=
1
∞
(
∥
λ
n
∥
M
b
(
Ω
)
∥
φ
∥
L
∞
(
Ω
)
)
2
n
≤
(
∑
n
=
1
∞
1
2
n
)
∥
μ
∥
M
b
(
Ω
)
∥
φ
∥
L
∞
(
Ω
)
≤
∥
μ
∥
M
b
(
Ω
)
∥
φ
∥
L
∞
(
Ω
)
.
This shows that γ
∈
W
-
1
,
p
′
(
x
)
(
Ω
). Let A be a Borel Ω, such that γ
(
A
)
=
0; we have the following:
(27)
∑
n
=
1
∞
‍
λ
n
(
A
)
2
n
(
∥
λ
n
∥
W
-
1
,
p
′
(
x
)
+
1
)
=
0
.
And we deduce that λ
n
(
A
)
=
0, for all n
≥
1, from which λ
n
≪
γ. Using the theorem of Radon-Nikodym, then there exists f
n
∈
L
1
(
Ω
,
d
γ
) such that, for all A
⊂
Ω, we have λ
n
(
A
)
=
∫
A
f
n
d
γ, and therefore λ
n
=
f
n
γ.
Letφ
∈
C
c
∞
(
Ω
) and φ
≥
0; we have, for any Borel B and for all n,
(28)
F
(
φ
)
=
∫
Ω
φ
d
μ
≥
sup
⁡
n
⁡
∫
Ω
φ
d
λ
n
=
sup
⁡
n
∫
Ω
φ
f
n
d
γ
,
then
f
n
γ
≤
μ
,
∫
B
‍
f
n
d
γ
≤
μ
(
B
)
.
Therefore,∫
B
sup
⁡
(
f
1
,
f
2
,
…
,
f
k
)
d
γ
≤
μ
(
B
), for all k
≥
1.
Fork
→
∞, we obtain the following, by monotone convergence theorem:
(29)
∫
B
f
d
γ
≤
μ
(
B
)
,
∀
B
⊂
Ω
,
where
f
=
sup
⁡
n
f
n
.
Remark that
(30)
∫
Ω
φ
d
μ
=
sup
⁡
n
⁡
∫
Ω
φ
f
n
d
γ
≤
∫
Ω
φ
f
d
γ
≤
∫
Ω
φ
d
μ
,
2222222222222222222222222221
∀
φ
∈
C
c
∞
(
Ω
)
;
hence μ
=
f
γ, and since μ
(
Ω
)
<
∞, then f
∈
L
1
(
Ω
,
d
γ
).Theorem 7.
Letp
∈
C
+
(
Ω
¯
) and let μ
∈
M
b
(
Ω
). Then,
(31)
μ
∈
L
1
(
Ω
)
+
W
-
1
,
p
′
(
x
)
(
Ω
)
⟺
μ
∈
M
b
p
(
x
)
(
Ω
)
.Proof.
Letμ
∈
L
1
(
Ω
)
+
W
-
1
,
p
′
(
x
)
(
Ω
), and show that μ
∈
M
b
p
(
x
)
(
Ω
). Indeed, we have μ
∈
L
1
(
Ω
)
+
W
-
1
,
p
′
(
x
)
(
Ω
) and then there exist f
∈
L
1
(
Ω
) and (
g
β
)
β
∈
L
p
′
(
x
)
(
Ω
) such that
(32)
μ
=
f
+
∑
|
β
|
≤
1
D
β
g
β
.
Let B
⊂
Ω such that ca
p
p
(
x
)
(
B
,
Ω
)
=
0, and we prove that μ
(
B
)
=
0. Indeed, for all ϵ
>
0, there exists u
ϵ
∈
C
0
∞
(
Ω
)
:
u
ϵ
=
1 in B, u
ϵ
≥
0 and ∫
Ω
|
∇
u
ϵ
|
p
(
x
)
d
x
<
ϵ.
Let(
f
n
)
n
∈
L
p
′
(
x
)
(
Ω
), such that f
n
→
f in L
1
(
Ω
); consider the sequence μ
n
=
f
n
+
∑
|
β
|
=
1
‍
D
β
g
β; we have μ
n
→
μ in M
b
(
Ω
). Indeed
(33)
μ
n
(
B
)
=
∫
B
d
μ
n
=
∫
B
(
f
n
+
∑
|
β
|
=
1
D
β
g
β
)
u
ϵ
d
x
.
We obtain the following:
(34)
|
μ
n
(
B
)
|
≤
∫
B
|
f
n
|
u
ϵ
d
x
+
∫
B
|
(
g
β
)
·
∇
u
ϵ
|
d
x
.
The Poincaré inequality leads to
(35)
|
μ
n
(
B
)
|
≤
C
(
∥
g
β
∥
L
p
(
x
)
(
Ω
)
∥
∇
u
ϵ
∥
L
p
(
x
)
(
Ω
)
)
∥
f
n
∥
L
p
′
(
x
)
(
Ω
)
∥
u
ϵ
∥
L
p
(
x
)
(
Ω
)
+
∥
g
β
∥
L
p
(
x
)
(
Ω
)
∥
∇
u
ϵ
∥
L
p
(
x
)
(
Ω
)
)
≤
C
ϵ
.
Since μ
n
→
μ in M
b
(
Ω
), then μ
n
(
B
)
→
μ
(
B
), and we deduce that μ
(
B
)
=
0.
Conversely, letμ
∈
M
b
p
(
x
)
(
Ω
), and we can always assume that μ
≥
0. And, by Theorem 6, we deduce that μ
=
h
γ where h
∈
L
1
(
Ω
,
γ
)
,
γ
∈
W
-
1
,
p
′
(
x
)
(
Ω
)
,
γ
≥
0, and h
≥
0. Let (
K
n
)
n be an increasing sequence of compact Ω such that ∪
n
=
1
∞
K
n
=
Ω and let μ
n
(
1
)
=
T
n
(
h
χ
K
n
)
γ
∈
W
-
1
,
p
′
(
x
)
(
Ω
)
,
(
μ
n
(
1
)
)
n be an increasing positive sequence in W
-
1
,
p
′
(
x
)
(
Ω
), with compact supports in Ω. Let
(36)
μ
0
=
μ
0
(
1
)
,
μ
n
=
μ
n
(
1
)
-
μ
n
-
1
(
1
)
.
Then, μ
=
∑
n
=
1
∞
‍
μ
n converges strongly in M
b
(
Ω
) and ∑
n
=
1
∞
‍
∥
μ
n
∥
M
b
(
Ω
)
<
∞.
Letρ
∈
𝒞
0
∞
(
ℝ
N
)
,
ρ
(
x
)
≥
0 a.e., and let ∫
ℝ
N
‍
ρ
(
x
)
d
x
=
1. Let ρ
n
(
x
)
=
n
N
ρ
(
n
x
). We have μ
n
*
ρ
m
→
ρ
n in W
-
1
,
p
′
(
x
)
(
Ω
) as m
→
∞ and μ
n
*
ρ
m
∈
𝒞
0
∞
(
ℝ
N
), for m is large enough.
Choosem
=
m
n such that ∥
μ
n
*
ρ
m
n
-
μ
n
∥
W
-
1
,
p
′
(
x
)
(
Ω
)
≤
2
-
n. Then μ
n
=
f
n
+
g
n, where
(37)
f
n
=
μ
n
*
ρ
m
n
,
g
n
=
μ
n
-
μ
n
*
ρ
m
n
.
We have ∑
g
n that converges in W
-
1
,
p
′
(
x
)
(
Ω
) and then g
=
∑
n
=
1
∞
g
n
∈
W
-
1
,
p
′
(
x
)
(
Ω
).
Moreover,∥
f
n
∥
L
1
(
Ω
)
=
∥
μ
n
*
ρ
m
n
∥
L
1
(
Ω
)
≤
∥
μ
n
∥
M
b
(
Ω
) and then ∑
f
n is absolutely convergent in L
1
(
Ω
). Then let f
=
∑
f
n
∈
L
1
(
Ω
); we conclude that
(38)
μ
=
∑
n
=
1
∞
μ
n
=
∑
n
=
1
∞
f
n
+
∑
n
=
1
∞
g
n
=
f
+
g
.Definition 8.
Letμ be a Radon measure and let E be a Borel set of Ω.(i)
The restriction ofμ to E is the measure λ
=
μ
|
E defined by
(39)
λ
(
B
)
=
μ
|
E
(
B
)
=
μ
(
E
∩
B
)
for any Borel setB of Ω.
(ii)
The measureλ is concentrated on E if λ
=
λ
|
E.Proposition 9.
Letμ
∈
M
b
(
Ω
) and let p
∈
𝒞
+
(
Ω
¯
) with p
+
≤
N.
Thenμ uniquely decomposes as μ
=
μ
0
+
λ, where
(40)
μ
0
∈
M
b
p
(
x
)
(
Ω
)
,
λ
=
μ
|
E
,
cap

p
(
x
)
(
E
,
Ω
)
=
0
.Proof.
We assume thatμ is positive (if not, consider μ
+ and μ
-). We first show the uniqueness. Assume that μ
=
μ
0
+
λ
=
μ
0
′
+
λ
′, where (
μ
0
,
μ
0
′
)
∈
(
M
b
p
(
x
)
(
Ω
)
)
2 and λ
=
μ
|
E with cap
p
(
x
)
(
E
,
Ω
)
=
0 and λ
′
=
μ
|
E
′ with cap
p
(
x
)
(
E
′
,
Ω
)
=
0. We have μ
0
-
μ
0
′
=
λ
′
-
λ. Let B be a Borel subset of Ω; if cap
p
(
x
)
(
B
,
Ω
)
=
0, then (
μ
0
-
μ
0
′
)
(
B
)
=
0, and therefore μ
0
=
μ
0
′ and λ
=
λ
′; if we consider cap
p
(
x
)
(
B
,
Ω
)
≠
0 and since λ
′
-
λ is focused on E
∩
E
′ such that cap
p
(
x
)
(
E
∩
E
′
,
Ω
)
=
0, then we have
(41)
(
λ
-
λ
′
)
(
B
)
=
(
λ
-
λ
′
)
(
B
∩
E
∩
E
′
)
=
(
μ
0
′
-
μ
0
)
(
B
∩
E
∩
E
′
)
≤
(
μ
0
′
-
μ
0
)
(
E
∩
E
′
)
=
0
.
Therefore, μ
0
=
μ
0
′, which proves the uniqueness.
Now we show the existence ofμ. Let
(42)
α
=
sup
⁡
{
and
cap
p
(
x
)
(
A
,
Ω
)
=
0
}
μ
(
A
)

:
A
be
a
Borel
subset
of
Ω
and
cap
p
(
x
)
(
A
,
Ω
)
=
0
}
<
∞
;
there exists a sequence (
A
n
)
n such that A
n is a Borel subset of Ω, lim
⁡
n
→
∞
μ
(
A
n
)
=
α, and cap
p
(
x
)
(
A
n
,
Ω
)
=
0. Let E
=
∪
n
A
n be a Borel subset of Ω; we have μ
(
E
)
=
α and cap
p
(
x
)
(
E
,
Ω
)
=
0.
Forλ
=
μ
|
E and μ
0
=
μ
-
λ, there exist E
∈
B
(
Ω
)
:
cap
p
(
x
)
(
E
,
Ω
)
=
0 and λ
=
μ
|
E.
It remains to show thatμ
0
∈
M
b
p
(
x
)
(
Ω
). By contradiction, assume that A is a Borel subset of Ω such that cap
p
(
x
)
(
E
,
Ω
)
=
0 and 0
<
μ
0
(
A
)
=
μ
(
A
)
-
λ
(
A
)
=
μ
(
A
)
-
μ
(
A
∩
E
)
=
μ
(
A
∖
E
). We have cap
p
(
x
)
(
A
∪
E
,
Ω
)
=
0 and μ
(
A
∪
E
)
=
μ
(
E
)
+
μ
(
A
∖
E
)
=
α
+
μ
(
A
∖
E
)
>
α, which contradicts the definition of α.Lemma 10.
Letλ be a positive measure in M
b
(
Ω
) concentrated on a set E such that cap

p
(
x
)
(
E
,
Ω
)
=
0. Then ∀
δ
>
0, ∃
ψ
δ
∈
C
0
∞
(
Ω
), and 0
≤
ψ
δ
≤
1, such that
(43)
∫
Ω
|
∇
ψ
δ
|
p
(
x
)
d
x
=
ε
δ
⟶
δ
→
0
+
0
,
∫
Ω
(
1
-
ψ
δ
)
d
λ
=
ε
δ
⟶
δ
→
0
+
0
.Proof.
Letλ
∈
M
b
(
Ω
); then for any δ
>
0, there exists K
δ which is a compact set in E with λ
(
E
∖
K
)
≤
δ.
Moreover,cap
p
(
x
)
(
E
,
Ω
)
=
0; hence, cap
p
(
x
)
(
K
δ
,
Ω
)
=
0, and then there exists ψ
δ
∈
C
0
∞
(
Ω
) such that ψ
δ
=
1 on a neighborhood of K
δ, 0
≤
ψ
δ
≤
1, and ∫
Ω
|
∇
ψ
δ
|
p
(
x
)
d
x
=
ε
δ converges to 0 as δ goes to 0
+.
We have0
≤
∫
Ω
(
1
-
ψ
δ
)
d
λ
=
∫
E
(
1
-
ψ
δ
)
d
λ
+
∫
Ω
∖
E
(
1
-
ψ
δ
)
d
λ, as λ is concentrated on E from which ∫
Ω
∖
E
‍
(
1
-
ψ
δ
)
d
λ
=
0; therefore ∫
Ω
(
1
-
ψ
δ
)
d
λ
=
∫
E
∖
K
δ
‍
(
1
-
ψ
δ
)
d
λ
+
∫
K
δ
(
1
-
ψ
δ
)
d
λ, and since ψ
δ
=
1 on a neighborhood of K
δ, we have then the following:
(44)
0
≤
∫
Ω
(
1
-
ψ
δ
)
d
λ
≤
λ
(
E
∖
K
δ
)
≤
δ
.Remark 11.
IfE is compact, we can choose K
δ
=
E, in view of Lemma 10, and we can deduce that ψ
δ converges to 0 strongly in W
0
1
,
p
(
x
)
(
Ω
) as δ goes to 0
+.


## 4. Essential Assumptions

LetΩ be a bounded open subset of ℝ
N
,
(
N
≥
2
) and p
∈
𝒞
+
(
Ω
-
). We consider a Leray-Lions operator a from W
0
1
,
p
(
x
)
(
Ω
) to dual W
-
1
,
p
′
(
x
)
(
Ω
) defined by
(45)
A
u
=
-
div
⁡
(
a
(
x
,
∇
u
)
)
,
where a
:
Ω
×
ℝ
N
→
ℝ
N is a Carathéodory operator, satisfying the following conditions:
(46)
|
a
(
x
,
ξ
)
|
≤
l
(
x
)
+
β
|
ξ
|
p
(
x
)
-
1
,
(47)
a
(
x
,
ξ
)
ξ
≥
α
|
ξ
|
p
(
x
)
,
(48)
[
a
(
x
,
ξ
)
-
a
(
x
,
ξ
¯
)
]
(
ξ
-
ξ
¯
)
>
0
,
∀
ξ
≠
ξ
¯
,
for a.e. x
∈
Ω, all ξ
∈
ℝ
N, where l
(
x
) is a positive function lying in L
p
′
(
x
)
(
Ω
) and β
>
0
,
α
>
0.The nonlinear termg
(
x
,
s
,
ξ
) satisfies the following:
(49)
|
g
(
x
,
s
,
ξ
)
|
≤
b
(
|
s
|
)
(
d
(
x
)
+
|
ξ
|
p
(
x
)
)
,
where b
:
ℝ
+
→
ℝ
+ is a continuous growth function and d
:
Ω
→
ℝ
+ with c
∈
L
1
(
Ω
). Consider the following:
(50)
g
(
x
,
s
,
ξ
)
sgn
⁡
(
s
)
≥
ρ
|
ξ
|
p
(
x
)
,
for a.e. x
∈
Ω, all ξ
∈
ℝ
N, and, for |
s
|
≥
σ, s
∈
ℝ, where σ
,
ρ
>
0. We consider the following problem:
(51)
-
div
⁡
(
a
(
x
,
∇
u
)
)
+
g
(
x
,
u
,
∇
u
)
=
μ
in
Ω
,
u
=
0
on
⁡
∂
Ω
,
where μ
∈
M
b
(
Ω
).


## 5. Approximate Problem

### 5.1. Existence of a Solution of the Approximate Problem

Consider the following approximate problem:(52)
u
n
∈
W
0
1
,
p
(
x
)
(
Ω
)
,
∫
Ω
a
(
x
,
∇
u
n
)
∇
v
d
x
+
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
v
d
x
=
∫
Ω
F
∇
v
d
x
,
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
,
where g
n
(
x
,
s
,
ξ
)
=
g
(
x
,
s
,
ξ
)
/
(
1
+
(
1
/
n
)
|
g
(
x
,
s
,
ξ
)
|
).Define the operatorsA and G
n of W
0
1
,
p
(
x
)
(
Ω
) to W
-
1
,
p
′
(
x
)
(
Ω
) by
(53)
〈
A
u
,
v
〉
=
∫
Ω
a
(
x
,
∇
u
)
∇
v
d
x
,
〈
G
n
u
,
v
〉
=
∫
Ω
g
n
(
x
,
u
,
∇
u
)
v
d
x
.Lemma 12.
The operatorG
n is bounded.Proof.
For allu
,
v in W
0
1
,
p
(
x
)
(
Ω
), and by the Hölder inequality,
(54)
|
〈
G
n
u
,
v
〉
|
=
|
∫
Ω
g
n
(
x
,
u
,
∇
u
)
v
d
x
|
≤
(
1
p
-
+
1
p
-
′
)
∥
g
n
(
x
,
u
,
∇
u
)
∥
p
′
(
x
)
∥
v
∥
p
(
x
)
≤
(
1
p
-
+
1
p
-
′
)
(
∫
Ω
∥
g
n
(
x
,
u
,
∇
u
)
∥
p
′
(
x
)
)
1
/
γ
∥
v
∥
p
(
x
)
≤
(
1
p
-
+
1
p
-
′
)
(
meas
⁡
(
Ω
)
)
1
/
γ
n
p
+
′
∥
v
∥
p
(
x
)
,
where
(55)
γ
=
{
p
-
′
if
∥
g
n
(
x
,
u
,
∇
u
)
∥
p
′
(
x
)
≥
1
,
p
+
′
if
∥
g
n
(
x
,
u
,
∇
u
)
∥
p
′
(
x
)
<
1
,
from which |

∫
Ω
g
n
(
x
,
u
,
∇
u
)
v
d
x
|
≤
C
n
∥
v
∥
p
(
x
), for all n fixed.Lemma 13.
The operatorT
n
=
A
+
G
n
:
W
0
1
,
p
(
x
)
(
Ω
)
→
W
-
1
,
p
′
(
x
)
(
Ω
) is pseudomonotone and coercive in the following sense: |
〈
T
n
v
,
v
〉
|
/
∥
v
∥
1
,
p
(
x
)
→
+
∞ when ∥
v
∥
1
,
p
(
x
)
→
+
∞.Proof.
Letu
,
v
∈
W
0
1
,
p
(
x
)
(
Ω
); we have
(56)
|
〈
A
u
,
v
〉
|
=
|
∫
Ω
a
(
x
,
∇
u
)
∇
v
d
x
|
≤
(
1
p
′
+
1
p
)
×
∥
a
(
x
,
∇
u
)
∥
p
′
(
x
)
∥
∇
v
∥
p
(
x
)
≤
(
1
p
′
+
1
p
)
(
∫
Ω
|
a
(
x
,
∇
u
)
|
p
′
(
x
)
d
x
)
1
/
γ
′
∥
∇
v
∥
p
(
x
)
≤
(
1
p
′
+
1
p
)
[
∫
Ω
2
p
+
′
(
l
p
′
(
x
)
(
x
)
+
|
∇
u
|
(
p
(
x
)
-
1
)
p
′
(
x
)
)
d
x
]
1
/
γ
′
×
∥
∇
v
∥
p
(
x
)
≤
(
1
p
′
+
1
p
)
2
p
+
′
/
γ
′
[
∫
Ω
(
l
p
′
(
x
)
(
x
)
+
|
∇
u
|
p
(
x
)
)
d
x
]
1
/
γ
′
×
∥
∇
v
∥
p
(
x
)
≤
(
1
p
′
+
1
p
)
2
p
+
′
/
γ
′
[
(
∫
Ω
l
p
′
(
x
)
(
x
)
d
x
)
1
/
γ
′
+
(
∫
Ω
|
∇
u
|
p
(
x
)
d
x
)
1
/
γ
′
]
∥
∇
v
∥
p
(
x
)
≤
C
1
(
C
+
|
∇
u
|
γ
0
/
γ
′
)
∥
∇
v
∥
p
(
x
)
,
where
(57)
γ
′
=
{
p
+
′
if
∥
a
(
x
,
∇
u
)
∥
p
′
(
x
)
≥
1
,
p
-
′
if
∥
a
(
x
,
∇
u
)
∥
p
′
(
x
)
<
1
,
γ
0
=
{
p
+
if
∥
∇
u
∥
p
(
x
)
≥
1
,
p
-
if
∥
∇
u
∥
p
(
x
)
<
1
.
Hence, operator A is bounded and, by Lemma 12, G
n is bounded; then it follows that T
n is bounded.
We show thatT
n is pseudomonotone: let (
u
k
)
k be a sequence in W
0
1
,
p
(
x
)
(
Ω
) such that
(58)
u
k
⇀
u
weakly
in
W
0
1
,
p
(
x
)
(
Ω
)
,
T
n
u
k
⇀
χ
weakly
in
W
-
1
,
p
′
(
x
)
(
Ω
)
,
lim
⁡
⁡
sup
⁡
k
〈
T
n
u
k
,
u
k
〉
≤
〈
χ
,
u
〉
.
We prove that χ
=
T
n
u and 〈
T
n
u
k
,
u
k
〉 converges to 〈
χ
,
u
〉 as k goes to +
∞. Indeed, we get that T
n
u
k weakly converges to χ in W
-
1
,
p
′
(
x
)
(
Ω
) as k tends to +
∞; then 〈
T
n
u
k
,
v
〉 converges to 〈
χ
,
v
〉, for all v
∈
W
0
1
,
p
(
x
)
(
Ω
), as k tends to +
∞, and we obtain
(59)
〈
χ
,
v
〉
=
lim
⁡
k
→
∞
(
∫
Ω
a
(
x
,
∇
u
k
)
∇
v
d
x
+
∫
Ω
g
n
(
x
,
u
k
,
∇
u
k
)
v
d
x
)
,
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
.
According to (46) and (49), we conclude that there exist φ
∈
(
L
p
′
(
x
)
(
Ω
)
)
N and ψ
n
∈
L
p
′
(
x
)
(
Ω
) such that a
(
x
,
∇
u
k
) converges to φ weakly in (
L
p
′
(
x
)
(
Ω
)
)
N and g
n
(
x
,
u
k
,
∇
u
k
) converges to ψ
n weakly in L
p
′
(
x
)
(
Ω
). Hence 〈
χ
,
v
〉
=
∫
Ω
‍
φ
∇
v
d
x
+
∫
Ω
‍
ψ
n
v
d
x. We have the following:
(60)
limsup
⁡
k
〈
T
n
u
k
,
u
k
〉
=
lim
⁡
⁡
sup
⁡
k
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
k
d
x
+
∫
Ω
g
n
(
x
,
u
k
,
∇
u
k
)
u
k
d
x
≤
∫
Ω
φ
∇
v
d
x
+
∫
Ω
ψ
n
v
d
x
.
Hence,
(61)
lim
⁡
⁡
sup
⁡
k
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
k
d
x
≤
∫
Ω
φ
∇
v
d
x
.
In view of (48), we have ∫
Ω
(
a
(
x
,
∇
u
k
)
-
a
(
x
,
∇
u
)
)
(
∇
u
k
-
∇
u
)
>
0, and then
(62)
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
k
d
x
>
-
∫
Ω
a
(
x
,
∇
u
)
∇
u
+
∫
Ω
a
(
x
,
∇
u
)
∇
u
k
d
x
+
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
d
x
.
We deduce that
(63)
lim
⁡
⁡
inf
⁡
k
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
d
x
≥
∫
Ω
φ
∇
u
d
x
.
In view of (61) we obtain the following:
(64)
lim
⁡
k
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
k
d
x
=
∫
Ω
φ
∇
u
d
x
.
Finally,
(65)
lim
⁡
k
〈
T
n
u
k
,
u
k
〉
=
〈
χ
,
u
〉
.
On the other hand, we use (64) and we deduce the following:
(66)
lim
⁡
k
∫
Ω
(
a
(
x
,
∇
u
k
)
-
a
(
x
,
∇
u
)
)
(
∇
u
k
-
∇
u
)
=
lim
⁡
k
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
k
d
x
-
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
d
x
-
∫
Ω
a
(
x
,
∇
u
)
∇
u
k
d
x
+
∫
Ω
a
(
x
,
∇
u
)
∇
u
d
x
=
0
.
So, in view of [9], we conclude that u
k converges to u a.e. in Ω and strongly in W
0
1
,
p
(
x
)
(
Ω
); hence a
(
x
,
∇
u
k
) converges to a
(
x
,
∇
u
) weakly in (
L
p
′
(
x
)
(
Ω
)
)
N and g
n
(
x
,
u
k
,
∇
u
k
) converges to g
n
(
x
,
u
,
∇
u
) weakly in L
p
′
(
x
)
(
Ω
). Since T
n
u
k converges weakly to χ, then
(67)
lim
⁡
k
(
∫
Ω
a
(
x
,
∇
u
k
)
∇
v
d
x
+
∫
Ω
g
n
(
x
,
u
k
,
∇
u
k
)
v
d
x
)
=
〈
χ
,
v
〉
,
〈
T
n
u
,
v
〉
=
〈
χ
,
v
〉
,
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
;
hence χ
=
T
n
u, and we conclude that the operator T
n is pseudomonotone.
We show thatT
n is coercive and we have the following:
(68)
〈
T
n
v
,
v
〉
=
∫
Ω
a
(
x
,
∇
v
)
∇
v
d
x
+
∫
Ω
g
n
(
x
,
v
,
∇
u
)
v
d
x
≥
α
∫
Ω
|
∇
v
|
p
(
x
)
d
x
+
∫
Ω
g
n
(
x
,
v
,
∇
u
)
sgn
⁡
(
v
)
|
v
|
d
x
≥
α
∫
Ω
|
∇
v
|
p
(
x
)
d
x
+
ρ
∫
Ω
|
∇
v
|
p
(
x
)
|
v
|
d
x
≥
α
∫
Ω
|
∇
v
|
p
(
x
)
d
x
from which
(69)
lim
⁡
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
→
∞
⁡
(
〈
T
n
v
,
v
〉
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
)
≥
lim
⁡
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
→
∞
⁡
α
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
γ
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
,
where
(70)
γ
=
{
p
+
if
∥
v
∥
1
,
p
(
x
)
≤
1
p
-
if
∥
v
∥
1
,
p
(
x
)
>
1
and γ
>
1, which implies that
(71)
lim
⁡
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
→
∞
⁡
α
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
γ
-
1
=
+
∞
;
hence T
n is coercive.
According to [10], there exists u
n
∈
W
0
1
,
p
(
x
)
(
Ω
) solution of (45).


### 5.2. Estimation of the Solution Approached

Consider the following approximate problem:(72)
A
u
n
+
g
n
(
x
,
u
n
,
∇
u
n
)
=
-
div
⁡
(
F
)
in
Ω
,
u
n
=
0
on
⁡
∂
Ω
,
where g
n
(
x
,
s
,
ξ
)
=
g
(
x
,
s
,
ξ
)
/
(
1
+
(
1
/
n
)
|
g
(
x
,
s
,
ξ
)
|
) and F
∈
(
L
s
(
x
)
(
Ω
)
)
N with s
(
·
)
>
N
/
(
p
(
·
)
-
1
).Lemma 14.
The sequence(
u
n
)
n is bounded in L
∞
(
Ω
).Proof.
Choosev
=
G
k
(
u
n
)
=
u
n
-
T
k
(
u
n
) as test function in (72); we obtain the following:
(73)
∫
Ω
a
(
x
,
∇
u
n
)
∇
G
k
(
u
n
)
d
x
+
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
G
k
(
u
n
)
d
x
=
∫
Ω
F
∇
G
k
(
u
n
)
d
x
.
For |
s
|
≥
k
≥
σ, we have G
k
(
s
)
=
s
-
k
sign
⁡
(
s
). Therefore, for |
s
|
≥
k
≥
σ, we have ∫
Ω
‍
g
n
(
x
,
u
n
,
∇
u
n
)
G
k
(
u
n
)
d
x
≥
0, and hence
(74)
∫
Ω
a
(
x
,
∇
u
n
)
∇
G
k
(
u
n
)
d
x
≤
∫
Ω
F
∇
G
k
(
u
n
)
d
x
.Assertion 1.
LetA
k
,
n
=
{
x
∈
Ω
/
|
u
n
(
x
)
|
≥
k
}. We have
(75)
α
2
∫
Ω
|
∇
G
k
(
u
n
)
|
p
(
x
)
d
x
≤
C
∫
A
k
,
n
|
F
|
p
′
(
x
)
d
x
.
Indeed
(76)
∫
Ω
a
(
x
,
∇
u
n
)
∇
G
k
(
u
n
)
d
x
≤
∫
Ω
F
∇
G
k
(
u
n
)
d
x
≤
C
∫
A
k
,
n
|
F
|
p
′
(
x
)
d
x
+
α
2
∫
A
k
,
n
|
∇
G
k
(
u
n
)
|
p
(
x
)
d
x
.
According to ∇
G
k
(
u
n
)
=
∇
u
n of A
k
,
n and to (47), we obtain Assertion 1.Assertion 2.
We have the following:(77)
(
∫
Ω
|
G
k
(
u
n
)
|
p
*
(
x
)
d
x
)
C
0
≤
C
(
meas
⁡
(
A
k
,
n
)
)
C
2
with p
*
(
x
)
=
N
p
(
x
)
/
(
N
-
p
(
x
)
), p
(
x
)
<
N, and C, C
0, and C
2 are positive constants.Proof.
In fact, we haveF
∈
(
L
s
(
x
)
(
Ω
)
)
N with s
(
·
)
>
N
/
(
p
(
·
)
-
1
), and since N
/
(
p
(
·
)
-
1
)
>
p
′
(
·
), then |
F
|
∈
L
p
′
(
x
)
(
Ω
). According to the inequality of Hölder, we obtain
(78)
∫
A
k
,
n
|
F
|
p
′
(
x
)
d
x
≤
(
1
(
s
/
p
′
)
-
+
1
(
s
/
(
s
-
p
′
)
)
-
)
×
∥
|
F
|
p
′
(
x
)
∥
s
(
x
)
/
p
′
(
x
)
∥
1
A
k
,
n
∥
s
(
x
)
/
(
s
(
x
)
-
p
′
(
x
)
)
≤
C
(
∫
Ω
|
F
|
s
(
x
)
d
x
)
1
/
γ
1
(
∫
Ω
|
1
A
k
,
n
|
d
x
)
1
/
γ
2
,
and therefore
(79)
∫
A
k
,
n
|
F
|
p
′
(
x
)
d
x
≤
C
(
∫
Ω
|
F
|
s
(
x
)
d
x
)
1
/
γ
1
(
meas
⁡
(
A
k
,
n
)
)
1
/
γ
2
,
where
(80)
γ
1
=
{
(
s
p
′
)
-
if
∥
|
F
|
p
′
(
x
)
∥
s
(
x
)
/
p
′
(
x
)
≥
1
,
(
s
p
′
)
+
if
∥
|
F
|
p
′
(
x
)
∥
s
(
x
)
/
p
′
(
x
)
<
1
,
γ
2
=
{
(
s
s
-
p
′
)
-
if
∥
1
A
k
,
n
∥
s
(
x
)
/
(
s
(
x
)
-
p
′
(
x
)
)
≥
1
,
(
s
p
′
)
+
if
∥
1
A
k
,
n
∥
s
(
x
)
/
(
s
(
x
)
-
p
′
(
x
)
)
<
1
,
since W
0
1
,
p
(
x
)
(
Ω
)
↪
L
p
*
(
x
) (Ω); hence ∥
G
k
(
u
n
)
∥
p
*
(
x
)
≤
C
∥
∇
G
k
(
u
n
)
∥
p
(
x
), and therefore
(81)
(
∫
Ω
|
G
k
(
u
n
)
|
p
*
(
x
)
d
x
)
1
/
γ
3
≤
C
(
∫
Ω
|
∇
G
k
(
u
n
)
|
p
(
x
)
d
x
)
1
/
γ
4
,
where
(82)
γ
3
=
{
(
p
*
)
-
if
∥
G
k
(
u
n
)
∥
p
*
(
x
)
≥
1
,
(
p
*
)
+
if
∥
G
k
(
u
n
)
∥
p
*
(
x
)
<
1
,
γ
4
=
{
p
-
if
∥
∇
G
k
(
u
n
)
∥
p
(
x
)
≥
1
,
p
+
if
∥
∇
G
k
(
u
n
)
∥
p
(
x
)
<
1
;
therefore
(83)
∫
Ω
|
∇
G
k
(
u
n
)
|
p
(
x
)
d
x
≥
C
(
∫
Ω
|
G
k
(
u
n
)
|
p
*
(
x
)
d
x
)
γ
4
/
γ
3
.
According to Assertion 1, (46), and (47), we obtain Assertion 2.Assertion 3.
We have, for alln,
(84)
∥
u
n
∥
L
∞
(
Ω
)
≤
C
1
.Proof.
Chooseh
>
k
+
1, and in A
h
,
n
=
{
x
∈
Ω
:
|
u
n
(
x
)
|
≥
h
} we have: |
G
k
(
u
n
)
|
≥
h
-
k. Hence, in view of Assertion 2, we obtain
(85)
∫
A
h
,
n
(
h
-
k
)
p
*
(
x
)
d
x
≤
∫
Ω
|
G
k
(
u
n
)
|
p
(
x
)
d
x
≤
C
(
meas
⁡
(
A
k
,
n
)
)
γ
3
/
γ
2
γ
4
and then (
h
-
k
)
p
-
*
(
meas
⁡
(
A
k
,
n
)
)
≤
C
(
meas
⁡
(
A
k
,
n
)
)
γ
3
/
γ
2
γ
4, for all h
>
k
>
σ, which gives
(86)
meas
⁡
(
A
k
,
n
)
≤
C
(
h
-
k
)
-
p
*
(
meas
⁡
(
A
k
,
n
)
)
γ
3
/
γ
2
γ
4
.
By a suitable choice of s
(
·
), we have γ
3
/
γ
2
γ
4
>
1. Now, we use the result of Stampacchia [11]; then there exists a constant M (independent of n) such that meas
⁡
(
A
k
,
n
)
=
0, for all k
≥
σ
+
M; then ∥
u
n
∥
L
∞
(
Ω
)
≤
σ
+
M, and we obtain Assertion 3.We prove the following theorem.Theorem 15.
There exists a subsequence of(
u
n
)
n still denoted by (
u
n
)
n and u such that u
n converges to u strongly in W
0
1
,
p
(
x
)
(
Ω
) and u is a weak solution of the following problem:
(87)
A
u
+
g
(
x
,
u
,
∇
u
)
=
-
div
⁡
(
F
)
i
n
Ω
,
u
=
0
o
n
∂
Ω
.Proof.
We need the following Lemmas.
Lemma 16.Let
c
and
d
be two positive constants, for
λ
=
c
2
/
4
d
2
, and let the function
φ
λ
(
t
)
=
t
e
λ
t
2.Then one has
d
φ
λ
′
(
t
)
-
c
|
φ
λ
(
t
)
|
≥
d
/
2.
Proof. We have φ
λ
′
(
t
)
=
e
λ
t
2
+
2
λ
t
2
e
λ
t
2; then
(88)
d
φ
λ
′
(
t
)
-
c
|
φ
λ
(
t
)
|
=
e
λ
t
2
(
2
λ
t
2
-
c
|
t
|
+
d
)
≥
e
λ
t
2
d
2
,
222222222222222222222222222222222222
∀
t
∈
ℝ
.
Lemma 17.There exists a positive constant
C
such that, for all
n
,
(89)
∥
u
n
∥
W
0
1
,
p
(
x
)
(
Ω
)
≤
C
.
Proof. In view of Assertion 2, we have ∥
u
n
∥
L
∞
(
Ω
)
≤
C
1, and we denote that C
2
=
b
(
C
1
).
Letv
n
=
φ
λ
(
u
n
)
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
), where λ
=
C
2
/
(
α
-
(
1
/
p
-
)
) (with α
>
1).
Choosev
n as test function in (52), and it follows that
(90)
∫
Ω
a
(
x
,
∇
u
n
)
∇
u
n
φ
λ
′
(
u
n
)
d
x
+
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
φ
λ
(
u
n
)
d
x
=
∫
Ω
‍
F
∇
u
n
φ
λ
′
(
u
n
)
d
x
.
Hence, consider
(91)
∫
Ω
a
(
x
,
∇
u
n
)
∇
u
n
φ
λ
′
(
u
n
)
d
x
=
∫
Ω
F
∇
u
n
φ
λ
′
(
u
n
)
d
x
-
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
φ
λ
(
u
n
)
d
x
;
this implies that
(92)
∫
Ω
a
(
x
,
∇
u
n
)
∇
u
n
φ
λ
′
(
u
n
)
d
x
≤
∫
Ω
|
F
|
|
∇
u
n
|
φ
λ
′
(
u
n
)
d
x
+
∫
Ω
|
g
n
(
x
,
u
n
,
∇
u
n
)
|
|
φ
λ
(
u
n
)
|
d
x
,
and we have
(93)
α
∫
Ω
|
∇
u
n
|
p
(
x
)
φ
λ
′
(
u
n
)
d
x
≤
∫
Ω
|
F
|
p
′
(
x
)
p
′
(
x
)
φ
λ
′
(
u
n
)
d
x
+
∫
Ω
|
∇
u
n
|
p
(
x
)
p
(
x
)
φ
λ
′
(
u
n
)
d
x
+
∫
Ω
b
(
|
u
n
|
)
(
|
∇
u
n
|
p
(
x
)
+
d
(
x
)
)
|
φ
λ
(
u
n
)
|
d
x
;
this implies that
(94)
(
α
-
1
p
-
)
∫
Ω
|
∇
u
n
|
p
(
x
)
φ
λ
′
(
u
n
)
d
x
-
∫
Ω
C
2
|
∇
u
n
|
p
(
x
)
|
φ
λ
(
u
n
)
|
d
x
≤
M
0
p
-
′
∫
Ω
|
F
|
p
′
(
x
)
d
x
+
C
2
∫
Ω
d
(
x
)
|
φ
λ
(
u
n
)
|
d
x
.
Hence,
(95)
∫
Ω
|
∇
u
n
|
p
(
x
)
[
(
α
-
1
p
-
)
φ
λ
′
(
u
n
)
-
C
2
|
φ
λ
(
u
n
)
|
]
d
x
≤
M
0
p
-
′
∫
Ω
|
F
|
p
′
(
x
)
d
x
+
C
2
M
1
∫
Ω
d
(
x
)
d
x
.
In view of Lemma 16 we have the following:
(96)
(
α
-
1
p
-
)
φ
λ
′
(
u
n
)
-
C
2
|
φ
λ
(
u
n
)
|
≥
(
α
-
(
1
/
p
-
)
)
2
.
Hence,
(97)
1
2
(
α
-
1
p
-
)
∫
Ω
|
∇
u
n
|
p
(
x
)
d
x
≤
M
0
p
-
′
∫
Ω
|
F
|
p
′
(
x
)
d
x
+
C
2
M
1
∫
Ω
d
(
x
)
d
x
.
Therefore, there exists a positive constant C such that
(98)
∥
u
n
∥
W
0
1
,
p
(
x
)
(
Ω
)
≤
C
.
There exists a subsequence still denoted by (
u
n
)
n and u such that
(99)
u
n
converges
to
u
a
.
e
.
in
Ω
,
weakly
in
W
0
1
,
p
(
x
)
(
Ω
)
and
in
L
∞
(
Ω
)
weakly-
*
.Lemma 18.
One has the following:(100)
lim
⁡
n
→
∞
∫
Ω
(
a
(
x
,
∇
u
n
)
-
a
(
x
,
∇
u
)
)
∇
(
u
n
-
u
)
d
x
=
0
,
and u
n converges to u strongly in W
0
1
,
p
(
x
)
(
Ω
) as n tends to +
∞.Proof.
We chooseu
n
-
u as a function test in the approximate problem (99) and we obtain
(101)
∫
Ω
a
(
x
,
∇
u
n
)
∇
(
u
n
-
u
)
d
x
+
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
(
u
n
-
u
)
d
x
=
∫
Ω
F
∇
(
u
n
-
u
)
d
x
.
This implies that
(102)
∫
Ω
(
a
(
x
,
∇
u
n
)
-
a
(
x
,
∇
u
)
)
∇
(
u
n
-
u
)
d
x
=
-
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
(
u
n
-
u
)
d
x
-
∫
Ω
a
(
x
,
∇
u
)
∇
(
u
n
-
u
)
d
x
+
∫
Ω
F
∇
(
u
n
-
u
)
d
x
.
Since F
∈
L
p
′
(
x
)
(
Ω
)
,
a
(
x
,
∇
u
)
∈
(
L
p
′
(
x
)
(
Ω
)
)
N, and u
n converges to u a.e. in Ω, weakly in W
0
1
,
p
(
x
)
(
Ω
), and in L
∞
(
Ω
) weakly-*, then
(103)
∫
Ω
F
∇
(
u
n
-
u
)
d
x
⟶
0
,
∫
Ω
a
(
x
,
∇
u
)
∇
(
u
n
-
u
)
d
x
⟶
0
.
We prove that ∫
Ω
‍
g
n
(
x
,
u
n
,
∇
u
n
)
(
u
n
-
u
)
d
x
→
0. Indeed, we have the following:
(104)
|
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
(
u
n
-
u
)
d
x
|
≤
∫
Ω
b
(
|
u
n
|
)
(
d
(
x
)
+
|
∇
u
n
|
p
(
x
)
)
|
u
n
-
u
|
d
x
≤
∫
Ω
b
(
c
)
|
∇
u
n
|
p
(
x
)
|
u
n
-
u
|
d
x
+
∫
Ω
b
(
c
)
d
(
x
)
|
u
n
-
u
|
d
x
≤
b
(
c
)
∫
Ω
‍
|
∇
u
n
|
p
(
x
)
|
u
n
-
u
|
d
x
+
∫
Ω
b
(
c
)
d
(
x
)
|
u
n
-
u
|
d
x
.
Since d
(
x
)
∈
L
1
(
Ω
)
,
|
∇
u
n
|
p
(
x
) is bounded in L
1
(
Ω
), and u
n
⇀
u in L
∞
(
Ω
) weakly-*, then
(105)
∫
Ω
|
∇
u
n
|
p
(
x
)
|
u
n
-
u
|
d
x
⟶
0
,
∫
Ω
b
(
c
)
d
(
x
)
|
u
n
-
u
|
d
x
⟶
0
.
Therefore, ∫
Ω
‍
g
n
(
x
,
u
n
,
∇
u
n
)
(
u
n
-
u
)
d
x
→
n
→
∞
0.
Finally, we obtain(106)
∫
Ω
‍
(
a
(
x
,
∇
u
n
)
-
a
(
x
,
∇
u
)
)
∇
(
u
n
-
u
)
d
x
⟶
0
.
We conclude, by [9], that u
n converges to u strongly in W
0
1
,
p
(
x
)
(
Ω
) as n tends to +
∞.


## 5.1. Existence of a Solution of the Approximate Problem

Consider the following approximate problem:(52)
u
n
∈
W
0
1
,
p
(
x
)
(
Ω
)
,
∫
Ω
a
(
x
,
∇
u
n
)
∇
v
d
x
+
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
v
d
x
=
∫
Ω
F
∇
v
d
x
,
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
,
where g
n
(
x
,
s
,
ξ
)
=
g
(
x
,
s
,
ξ
)
/
(
1
+
(
1
/
n
)
|
g
(
x
,
s
,
ξ
)
|
).Define the operatorsA and G
n of W
0
1
,
p
(
x
)
(
Ω
) to W
-
1
,
p
′
(
x
)
(
Ω
) by
(53)
〈
A
u
,
v
〉
=
∫
Ω
a
(
x
,
∇
u
)
∇
v
d
x
,
〈
G
n
u
,
v
〉
=
∫
Ω
g
n
(
x
,
u
,
∇
u
)
v
d
x
.Lemma 12.
The operatorG
n is bounded.Proof.
For allu
,
v in W
0
1
,
p
(
x
)
(
Ω
), and by the Hölder inequality,
(54)
|
〈
G
n
u
,
v
〉
|
=
|
∫
Ω
g
n
(
x
,
u
,
∇
u
)
v
d
x
|
≤
(
1
p
-
+
1
p
-
′
)
∥
g
n
(
x
,
u
,
∇
u
)
∥
p
′
(
x
)
∥
v
∥
p
(
x
)
≤
(
1
p
-
+
1
p
-
′
)
(
∫
Ω
∥
g
n
(
x
,
u
,
∇
u
)
∥
p
′
(
x
)
)
1
/
γ
∥
v
∥
p
(
x
)
≤
(
1
p
-
+
1
p
-
′
)
(
meas
⁡
(
Ω
)
)
1
/
γ
n
p
+
′
∥
v
∥
p
(
x
)
,
where
(55)
γ
=
{
p
-
′
if
∥
g
n
(
x
,
u
,
∇
u
)
∥
p
′
(
x
)
≥
1
,
p
+
′
if
∥
g
n
(
x
,
u
,
∇
u
)
∥
p
′
(
x
)
<
1
,
from which |

∫
Ω
g
n
(
x
,
u
,
∇
u
)
v
d
x
|
≤
C
n
∥
v
∥
p
(
x
), for all n fixed.Lemma 13.
The operatorT
n
=
A
+
G
n
:
W
0
1
,
p
(
x
)
(
Ω
)
→
W
-
1
,
p
′
(
x
)
(
Ω
) is pseudomonotone and coercive in the following sense: |
〈
T
n
v
,
v
〉
|
/
∥
v
∥
1
,
p
(
x
)
→
+
∞ when ∥
v
∥
1
,
p
(
x
)
→
+
∞.Proof.
Letu
,
v
∈
W
0
1
,
p
(
x
)
(
Ω
); we have
(56)
|
〈
A
u
,
v
〉
|
=
|
∫
Ω
a
(
x
,
∇
u
)
∇
v
d
x
|
≤
(
1
p
′
+
1
p
)
×
∥
a
(
x
,
∇
u
)
∥
p
′
(
x
)
∥
∇
v
∥
p
(
x
)
≤
(
1
p
′
+
1
p
)
(
∫
Ω
|
a
(
x
,
∇
u
)
|
p
′
(
x
)
d
x
)
1
/
γ
′
∥
∇
v
∥
p
(
x
)
≤
(
1
p
′
+
1
p
)
[
∫
Ω
2
p
+
′
(
l
p
′
(
x
)
(
x
)
+
|
∇
u
|
(
p
(
x
)
-
1
)
p
′
(
x
)
)
d
x
]
1
/
γ
′
×
∥
∇
v
∥
p
(
x
)
≤
(
1
p
′
+
1
p
)
2
p
+
′
/
γ
′
[
∫
Ω
(
l
p
′
(
x
)
(
x
)
+
|
∇
u
|
p
(
x
)
)
d
x
]
1
/
γ
′
×
∥
∇
v
∥
p
(
x
)
≤
(
1
p
′
+
1
p
)
2
p
+
′
/
γ
′
[
(
∫
Ω
l
p
′
(
x
)
(
x
)
d
x
)
1
/
γ
′
+
(
∫
Ω
|
∇
u
|
p
(
x
)
d
x
)
1
/
γ
′
]
∥
∇
v
∥
p
(
x
)
≤
C
1
(
C
+
|
∇
u
|
γ
0
/
γ
′
)
∥
∇
v
∥
p
(
x
)
,
where
(57)
γ
′
=
{
p
+
′
if
∥
a
(
x
,
∇
u
)
∥
p
′
(
x
)
≥
1
,
p
-
′
if
∥
a
(
x
,
∇
u
)
∥
p
′
(
x
)
<
1
,
γ
0
=
{
p
+
if
∥
∇
u
∥
p
(
x
)
≥
1
,
p
-
if
∥
∇
u
∥
p
(
x
)
<
1
.
Hence, operator A is bounded and, by Lemma 12, G
n is bounded; then it follows that T
n is bounded.
We show thatT
n is pseudomonotone: let (
u
k
)
k be a sequence in W
0
1
,
p
(
x
)
(
Ω
) such that
(58)
u
k
⇀
u
weakly
in
W
0
1
,
p
(
x
)
(
Ω
)
,
T
n
u
k
⇀
χ
weakly
in
W
-
1
,
p
′
(
x
)
(
Ω
)
,
lim
⁡
⁡
sup
⁡
k
〈
T
n
u
k
,
u
k
〉
≤
〈
χ
,
u
〉
.
We prove that χ
=
T
n
u and 〈
T
n
u
k
,
u
k
〉 converges to 〈
χ
,
u
〉 as k goes to +
∞. Indeed, we get that T
n
u
k weakly converges to χ in W
-
1
,
p
′
(
x
)
(
Ω
) as k tends to +
∞; then 〈
T
n
u
k
,
v
〉 converges to 〈
χ
,
v
〉, for all v
∈
W
0
1
,
p
(
x
)
(
Ω
), as k tends to +
∞, and we obtain
(59)
〈
χ
,
v
〉
=
lim
⁡
k
→
∞
(
∫
Ω
a
(
x
,
∇
u
k
)
∇
v
d
x
+
∫
Ω
g
n
(
x
,
u
k
,
∇
u
k
)
v
d
x
)
,
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
.
According to (46) and (49), we conclude that there exist φ
∈
(
L
p
′
(
x
)
(
Ω
)
)
N and ψ
n
∈
L
p
′
(
x
)
(
Ω
) such that a
(
x
,
∇
u
k
) converges to φ weakly in (
L
p
′
(
x
)
(
Ω
)
)
N and g
n
(
x
,
u
k
,
∇
u
k
) converges to ψ
n weakly in L
p
′
(
x
)
(
Ω
). Hence 〈
χ
,
v
〉
=
∫
Ω
‍
φ
∇
v
d
x
+
∫
Ω
‍
ψ
n
v
d
x. We have the following:
(60)
limsup
⁡
k
〈
T
n
u
k
,
u
k
〉
=
lim
⁡
⁡
sup
⁡
k
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
k
d
x
+
∫
Ω
g
n
(
x
,
u
k
,
∇
u
k
)
u
k
d
x
≤
∫
Ω
φ
∇
v
d
x
+
∫
Ω
ψ
n
v
d
x
.
Hence,
(61)
lim
⁡
⁡
sup
⁡
k
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
k
d
x
≤
∫
Ω
φ
∇
v
d
x
.
In view of (48), we have ∫
Ω
(
a
(
x
,
∇
u
k
)
-
a
(
x
,
∇
u
)
)
(
∇
u
k
-
∇
u
)
>
0, and then
(62)
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
k
d
x
>
-
∫
Ω
a
(
x
,
∇
u
)
∇
u
+
∫
Ω
a
(
x
,
∇
u
)
∇
u
k
d
x
+
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
d
x
.
We deduce that
(63)
lim
⁡
⁡
inf
⁡
k
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
d
x
≥
∫
Ω
φ
∇
u
d
x
.
In view of (61) we obtain the following:
(64)
lim
⁡
k
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
k
d
x
=
∫
Ω
φ
∇
u
d
x
.
Finally,
(65)
lim
⁡
k
〈
T
n
u
k
,
u
k
〉
=
〈
χ
,
u
〉
.
On the other hand, we use (64) and we deduce the following:
(66)
lim
⁡
k
∫
Ω
(
a
(
x
,
∇
u
k
)
-
a
(
x
,
∇
u
)
)
(
∇
u
k
-
∇
u
)
=
lim
⁡
k
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
k
d
x
-
∫
Ω
a
(
x
,
∇
u
k
)
∇
u
d
x
-
∫
Ω
a
(
x
,
∇
u
)
∇
u
k
d
x
+
∫
Ω
a
(
x
,
∇
u
)
∇
u
d
x
=
0
.
So, in view of [9], we conclude that u
k converges to u a.e. in Ω and strongly in W
0
1
,
p
(
x
)
(
Ω
); hence a
(
x
,
∇
u
k
) converges to a
(
x
,
∇
u
) weakly in (
L
p
′
(
x
)
(
Ω
)
)
N and g
n
(
x
,
u
k
,
∇
u
k
) converges to g
n
(
x
,
u
,
∇
u
) weakly in L
p
′
(
x
)
(
Ω
). Since T
n
u
k converges weakly to χ, then
(67)
lim
⁡
k
(
∫
Ω
a
(
x
,
∇
u
k
)
∇
v
d
x
+
∫
Ω
g
n
(
x
,
u
k
,
∇
u
k
)
v
d
x
)
=
〈
χ
,
v
〉
,
〈
T
n
u
,
v
〉
=
〈
χ
,
v
〉
,
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
;
hence χ
=
T
n
u, and we conclude that the operator T
n is pseudomonotone.
We show thatT
n is coercive and we have the following:
(68)
〈
T
n
v
,
v
〉
=
∫
Ω
a
(
x
,
∇
v
)
∇
v
d
x
+
∫
Ω
g
n
(
x
,
v
,
∇
u
)
v
d
x
≥
α
∫
Ω
|
∇
v
|
p
(
x
)
d
x
+
∫
Ω
g
n
(
x
,
v
,
∇
u
)
sgn
⁡
(
v
)
|
v
|
d
x
≥
α
∫
Ω
|
∇
v
|
p
(
x
)
d
x
+
ρ
∫
Ω
|
∇
v
|
p
(
x
)
|
v
|
d
x
≥
α
∫
Ω
|
∇
v
|
p
(
x
)
d
x
from which
(69)
lim
⁡
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
→
∞
⁡
(
〈
T
n
v
,
v
〉
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
)
≥
lim
⁡
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
→
∞
⁡
α
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
γ
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
,
where
(70)
γ
=
{
p
+
if
∥
v
∥
1
,
p
(
x
)
≤
1
p
-
if
∥
v
∥
1
,
p
(
x
)
>
1
and γ
>
1, which implies that
(71)
lim
⁡
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
→
∞
⁡
α
∥
v
∥
W
0
1
,
p
(
x
)
(
Ω
)
γ
-
1
=
+
∞
;
hence T
n is coercive.
According to [10], there exists u
n
∈
W
0
1
,
p
(
x
)
(
Ω
) solution of (45).


## 5.2. Estimation of the Solution Approached

Consider the following approximate problem:(72)
A
u
n
+
g
n
(
x
,
u
n
,
∇
u
n
)
=
-
div
⁡
(
F
)
in
Ω
,
u
n
=
0
on
⁡
∂
Ω
,
where g
n
(
x
,
s
,
ξ
)
=
g
(
x
,
s
,
ξ
)
/
(
1
+
(
1
/
n
)
|
g
(
x
,
s
,
ξ
)
|
) and F
∈
(
L
s
(
x
)
(
Ω
)
)
N with s
(
·
)
>
N
/
(
p
(
·
)
-
1
).Lemma 14.
The sequence(
u
n
)
n is bounded in L
∞
(
Ω
).Proof.
Choosev
=
G
k
(
u
n
)
=
u
n
-
T
k
(
u
n
) as test function in (72); we obtain the following:
(73)
∫
Ω
a
(
x
,
∇
u
n
)
∇
G
k
(
u
n
)
d
x
+
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
G
k
(
u
n
)
d
x
=
∫
Ω
F
∇
G
k
(
u
n
)
d
x
.
For |
s
|
≥
k
≥
σ, we have G
k
(
s
)
=
s
-
k
sign
⁡
(
s
). Therefore, for |
s
|
≥
k
≥
σ, we have ∫
Ω
‍
g
n
(
x
,
u
n
,
∇
u
n
)
G
k
(
u
n
)
d
x
≥
0, and hence
(74)
∫
Ω
a
(
x
,
∇
u
n
)
∇
G
k
(
u
n
)
d
x
≤
∫
Ω
F
∇
G
k
(
u
n
)
d
x
.Assertion 1.
LetA
k
,
n
=
{
x
∈
Ω
/
|
u
n
(
x
)
|
≥
k
}. We have
(75)
α
2
∫
Ω
|
∇
G
k
(
u
n
)
|
p
(
x
)
d
x
≤
C
∫
A
k
,
n
|
F
|
p
′
(
x
)
d
x
.
Indeed
(76)
∫
Ω
a
(
x
,
∇
u
n
)
∇
G
k
(
u
n
)
d
x
≤
∫
Ω
F
∇
G
k
(
u
n
)
d
x
≤
C
∫
A
k
,
n
|
F
|
p
′
(
x
)
d
x
+
α
2
∫
A
k
,
n
|
∇
G
k
(
u
n
)
|
p
(
x
)
d
x
.
According to ∇
G
k
(
u
n
)
=
∇
u
n of A
k
,
n and to (47), we obtain Assertion 1.Assertion 2.
We have the following:(77)
(
∫
Ω
|
G
k
(
u
n
)
|
p
*
(
x
)
d
x
)
C
0
≤
C
(
meas
⁡
(
A
k
,
n
)
)
C
2
with p
*
(
x
)
=
N
p
(
x
)
/
(
N
-
p
(
x
)
), p
(
x
)
<
N, and C, C
0, and C
2 are positive constants.Proof.
In fact, we haveF
∈
(
L
s
(
x
)
(
Ω
)
)
N with s
(
·
)
>
N
/
(
p
(
·
)
-
1
), and since N
/
(
p
(
·
)
-
1
)
>
p
′
(
·
), then |
F
|
∈
L
p
′
(
x
)
(
Ω
). According to the inequality of Hölder, we obtain
(78)
∫
A
k
,
n
|
F
|
p
′
(
x
)
d
x
≤
(
1
(
s
/
p
′
)
-
+
1
(
s
/
(
s
-
p
′
)
)
-
)
×
∥
|
F
|
p
′
(
x
)
∥
s
(
x
)
/
p
′
(
x
)
∥
1
A
k
,
n
∥
s
(
x
)
/
(
s
(
x
)
-
p
′
(
x
)
)
≤
C
(
∫
Ω
|
F
|
s
(
x
)
d
x
)
1
/
γ
1
(
∫
Ω
|
1
A
k
,
n
|
d
x
)
1
/
γ
2
,
and therefore
(79)
∫
A
k
,
n
|
F
|
p
′
(
x
)
d
x
≤
C
(
∫
Ω
|
F
|
s
(
x
)
d
x
)
1
/
γ
1
(
meas
⁡
(
A
k
,
n
)
)
1
/
γ
2
,
where
(80)
γ
1
=
{
(
s
p
′
)
-
if
∥
|
F
|
p
′
(
x
)
∥
s
(
x
)
/
p
′
(
x
)
≥
1
,
(
s
p
′
)
+
if
∥
|
F
|
p
′
(
x
)
∥
s
(
x
)
/
p
′
(
x
)
<
1
,
γ
2
=
{
(
s
s
-
p
′
)
-
if
∥
1
A
k
,
n
∥
s
(
x
)
/
(
s
(
x
)
-
p
′
(
x
)
)
≥
1
,
(
s
p
′
)
+
if
∥
1
A
k
,
n
∥
s
(
x
)
/
(
s
(
x
)
-
p
′
(
x
)
)
<
1
,
since W
0
1
,
p
(
x
)
(
Ω
)
↪
L
p
*
(
x
) (Ω); hence ∥
G
k
(
u
n
)
∥
p
*
(
x
)
≤
C
∥
∇
G
k
(
u
n
)
∥
p
(
x
), and therefore
(81)
(
∫
Ω
|
G
k
(
u
n
)
|
p
*
(
x
)
d
x
)
1
/
γ
3
≤
C
(
∫
Ω
|
∇
G
k
(
u
n
)
|
p
(
x
)
d
x
)
1
/
γ
4
,
where
(82)
γ
3
=
{
(
p
*
)
-
if
∥
G
k
(
u
n
)
∥
p
*
(
x
)
≥
1
,
(
p
*
)
+
if
∥
G
k
(
u
n
)
∥
p
*
(
x
)
<
1
,
γ
4
=
{
p
-
if
∥
∇
G
k
(
u
n
)
∥
p
(
x
)
≥
1
,
p
+
if
∥
∇
G
k
(
u
n
)
∥
p
(
x
)
<
1
;
therefore
(83)
∫
Ω
|
∇
G
k
(
u
n
)
|
p
(
x
)
d
x
≥
C
(
∫
Ω
|
G
k
(
u
n
)
|
p
*
(
x
)
d
x
)
γ
4
/
γ
3
.
According to Assertion 1, (46), and (47), we obtain Assertion 2.Assertion 3.
We have, for alln,
(84)
∥
u
n
∥
L
∞
(
Ω
)
≤
C
1
.Proof.
Chooseh
>
k
+
1, and in A
h
,
n
=
{
x
∈
Ω
:
|
u
n
(
x
)
|
≥
h
} we have: |
G
k
(
u
n
)
|
≥
h
-
k. Hence, in view of Assertion 2, we obtain
(85)
∫
A
h
,
n
(
h
-
k
)
p
*
(
x
)
d
x
≤
∫
Ω
|
G
k
(
u
n
)
|
p
(
x
)
d
x
≤
C
(
meas
⁡
(
A
k
,
n
)
)
γ
3
/
γ
2
γ
4
and then (
h
-
k
)
p
-
*
(
meas
⁡
(
A
k
,
n
)
)
≤
C
(
meas
⁡
(
A
k
,
n
)
)
γ
3
/
γ
2
γ
4, for all h
>
k
>
σ, which gives
(86)
meas
⁡
(
A
k
,
n
)
≤
C
(
h
-
k
)
-
p
*
(
meas
⁡
(
A
k
,
n
)
)
γ
3
/
γ
2
γ
4
.
By a suitable choice of s
(
·
), we have γ
3
/
γ
2
γ
4
>
1. Now, we use the result of Stampacchia [11]; then there exists a constant M (independent of n) such that meas
⁡
(
A
k
,
n
)
=
0, for all k
≥
σ
+
M; then ∥
u
n
∥
L
∞
(
Ω
)
≤
σ
+
M, and we obtain Assertion 3.We prove the following theorem.Theorem 15.
There exists a subsequence of(
u
n
)
n still denoted by (
u
n
)
n and u such that u
n converges to u strongly in W
0
1
,
p
(
x
)
(
Ω
) and u is a weak solution of the following problem:
(87)
A
u
+
g
(
x
,
u
,
∇
u
)
=
-
div
⁡
(
F
)
i
n
Ω
,
u
=
0
o
n
∂
Ω
.Proof.
We need the following Lemmas.
Lemma 16.Let
c
and
d
be two positive constants, for
λ
=
c
2
/
4
d
2
, and let the function
φ
λ
(
t
)
=
t
e
λ
t
2.Then one has
d
φ
λ
′
(
t
)
-
c
|
φ
λ
(
t
)
|
≥
d
/
2.
Proof. We have φ
λ
′
(
t
)
=
e
λ
t
2
+
2
λ
t
2
e
λ
t
2; then
(88)
d
φ
λ
′
(
t
)
-
c
|
φ
λ
(
t
)
|
=
e
λ
t
2
(
2
λ
t
2
-
c
|
t
|
+
d
)
≥
e
λ
t
2
d
2
,
222222222222222222222222222222222222
∀
t
∈
ℝ
.
Lemma 17.There exists a positive constant
C
such that, for all
n
,
(89)
∥
u
n
∥
W
0
1
,
p
(
x
)
(
Ω
)
≤
C
.
Proof. In view of Assertion 2, we have ∥
u
n
∥
L
∞
(
Ω
)
≤
C
1, and we denote that C
2
=
b
(
C
1
).
Letv
n
=
φ
λ
(
u
n
)
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
), where λ
=
C
2
/
(
α
-
(
1
/
p
-
)
) (with α
>
1).
Choosev
n as test function in (52), and it follows that
(90)
∫
Ω
a
(
x
,
∇
u
n
)
∇
u
n
φ
λ
′
(
u
n
)
d
x
+
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
φ
λ
(
u
n
)
d
x
=
∫
Ω
‍
F
∇
u
n
φ
λ
′
(
u
n
)
d
x
.
Hence, consider
(91)
∫
Ω
a
(
x
,
∇
u
n
)
∇
u
n
φ
λ
′
(
u
n
)
d
x
=
∫
Ω
F
∇
u
n
φ
λ
′
(
u
n
)
d
x
-
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
φ
λ
(
u
n
)
d
x
;
this implies that
(92)
∫
Ω
a
(
x
,
∇
u
n
)
∇
u
n
φ
λ
′
(
u
n
)
d
x
≤
∫
Ω
|
F
|
|
∇
u
n
|
φ
λ
′
(
u
n
)
d
x
+
∫
Ω
|
g
n
(
x
,
u
n
,
∇
u
n
)
|
|
φ
λ
(
u
n
)
|
d
x
,
and we have
(93)
α
∫
Ω
|
∇
u
n
|
p
(
x
)
φ
λ
′
(
u
n
)
d
x
≤
∫
Ω
|
F
|
p
′
(
x
)
p
′
(
x
)
φ
λ
′
(
u
n
)
d
x
+
∫
Ω
|
∇
u
n
|
p
(
x
)
p
(
x
)
φ
λ
′
(
u
n
)
d
x
+
∫
Ω
b
(
|
u
n
|
)
(
|
∇
u
n
|
p
(
x
)
+
d
(
x
)
)
|
φ
λ
(
u
n
)
|
d
x
;
this implies that
(94)
(
α
-
1
p
-
)
∫
Ω
|
∇
u
n
|
p
(
x
)
φ
λ
′
(
u
n
)
d
x
-
∫
Ω
C
2
|
∇
u
n
|
p
(
x
)
|
φ
λ
(
u
n
)
|
d
x
≤
M
0
p
-
′
∫
Ω
|
F
|
p
′
(
x
)
d
x
+
C
2
∫
Ω
d
(
x
)
|
φ
λ
(
u
n
)
|
d
x
.
Hence,
(95)
∫
Ω
|
∇
u
n
|
p
(
x
)
[
(
α
-
1
p
-
)
φ
λ
′
(
u
n
)
-
C
2
|
φ
λ
(
u
n
)
|
]
d
x
≤
M
0
p
-
′
∫
Ω
|
F
|
p
′
(
x
)
d
x
+
C
2
M
1
∫
Ω
d
(
x
)
d
x
.
In view of Lemma 16 we have the following:
(96)
(
α
-
1
p
-
)
φ
λ
′
(
u
n
)
-
C
2
|
φ
λ
(
u
n
)
|
≥
(
α
-
(
1
/
p
-
)
)
2
.
Hence,
(97)
1
2
(
α
-
1
p
-
)
∫
Ω
|
∇
u
n
|
p
(
x
)
d
x
≤
M
0
p
-
′
∫
Ω
|
F
|
p
′
(
x
)
d
x
+
C
2
M
1
∫
Ω
d
(
x
)
d
x
.
Therefore, there exists a positive constant C such that
(98)
∥
u
n
∥
W
0
1
,
p
(
x
)
(
Ω
)
≤
C
.
There exists a subsequence still denoted by (
u
n
)
n and u such that
(99)
u
n
converges
to
u
a
.
e
.
in
Ω
,
weakly
in
W
0
1
,
p
(
x
)
(
Ω
)
and
in
L
∞
(
Ω
)
weakly-
*
.Lemma 18.
One has the following:(100)
lim
⁡
n
→
∞
∫
Ω
(
a
(
x
,
∇
u
n
)
-
a
(
x
,
∇
u
)
)
∇
(
u
n
-
u
)
d
x
=
0
,
and u
n converges to u strongly in W
0
1
,
p
(
x
)
(
Ω
) as n tends to +
∞.Proof.
We chooseu
n
-
u as a function test in the approximate problem (99) and we obtain
(101)
∫
Ω
a
(
x
,
∇
u
n
)
∇
(
u
n
-
u
)
d
x
+
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
(
u
n
-
u
)
d
x
=
∫
Ω
F
∇
(
u
n
-
u
)
d
x
.
This implies that
(102)
∫
Ω
(
a
(
x
,
∇
u
n
)
-
a
(
x
,
∇
u
)
)
∇
(
u
n
-
u
)
d
x
=
-
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
(
u
n
-
u
)
d
x
-
∫
Ω
a
(
x
,
∇
u
)
∇
(
u
n
-
u
)
d
x
+
∫
Ω
F
∇
(
u
n
-
u
)
d
x
.
Since F
∈
L
p
′
(
x
)
(
Ω
)
,
a
(
x
,
∇
u
)
∈
(
L
p
′
(
x
)
(
Ω
)
)
N, and u
n converges to u a.e. in Ω, weakly in W
0
1
,
p
(
x
)
(
Ω
), and in L
∞
(
Ω
) weakly-*, then
(103)
∫
Ω
F
∇
(
u
n
-
u
)
d
x
⟶
0
,
∫
Ω
a
(
x
,
∇
u
)
∇
(
u
n
-
u
)
d
x
⟶
0
.
We prove that ∫
Ω
‍
g
n
(
x
,
u
n
,
∇
u
n
)
(
u
n
-
u
)
d
x
→
0. Indeed, we have the following:
(104)
|
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
(
u
n
-
u
)
d
x
|
≤
∫
Ω
b
(
|
u
n
|
)
(
d
(
x
)
+
|
∇
u
n
|
p
(
x
)
)
|
u
n
-
u
|
d
x
≤
∫
Ω
b
(
c
)
|
∇
u
n
|
p
(
x
)
|
u
n
-
u
|
d
x
+
∫
Ω
b
(
c
)
d
(
x
)
|
u
n
-
u
|
d
x
≤
b
(
c
)
∫
Ω
‍
|
∇
u
n
|
p
(
x
)
|
u
n
-
u
|
d
x
+
∫
Ω
b
(
c
)
d
(
x
)
|
u
n
-
u
|
d
x
.
Since d
(
x
)
∈
L
1
(
Ω
)
,
|
∇
u
n
|
p
(
x
) is bounded in L
1
(
Ω
), and u
n
⇀
u in L
∞
(
Ω
) weakly-*, then
(105)
∫
Ω
|
∇
u
n
|
p
(
x
)
|
u
n
-
u
|
d
x
⟶
0
,
∫
Ω
b
(
c
)
d
(
x
)
|
u
n
-
u
|
d
x
⟶
0
.
Therefore, ∫
Ω
‍
g
n
(
x
,
u
n
,
∇
u
n
)
(
u
n
-
u
)
d
x
→
n
→
∞
0.
Finally, we obtain(106)
∫
Ω
‍
(
a
(
x
,
∇
u
n
)
-
a
(
x
,
∇
u
)
)
∇
(
u
n
-
u
)
d
x
⟶
0
.
We conclude, by [9], that u
n converges to u strongly in W
0
1
,
p
(
x
)
(
Ω
) as n tends to +
∞.


## 6. Results of Existence

We state the following two theorems.Theorem 19.
Letf
∈
L
∞
(
Ω
) and let F
∈
(
L
s
(
x
)
(
Ω
)
)
N with s
(
·
)
>
N
/
(
p
(
·
)
-
1
).
Then there exists a solutionu
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
) of the following problem:
(107)
A
u
+
g
(
x
,
u
,
∇
u
)
=
f
-
div
⁡
(
F
)
i
n
Ω
,
u
=
0
o
n
∂
Ω
,
in the sense that
(108)
∫
Ω
a
(
x
,
∇
u
)
∇
v
d
x
+
∫
Ω
g
(
x
,
u
,
∇
u
)
v
d
x
=
∫
Ω
f
v
d
x
+
∫
Ω
F
∇
v
d
x
,
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
)
.Proof.
Without loss of generality, we can assume thatf
=
0. Let v
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
), and we have
(109)
∫
Ω
a
(
x
,
∇
u
n
)
∇
v
d
x
+
∫
Ω
g
n
(
x
,
u
n
,
∇
u
n
)
v
d
x
=
∫
Ω
F
∇
v
d
x
.
By Lemma 18, we have ∇
u
n
→
∇
u strongly in L
p
(
x
)
(
Ω
), and then ∇
u
n
→
∇
u a.e. in Ω, We use (46) and (49) and passing to the limit, we obtain the following:
(110)
∫
Ω
a
(
x
,
∇
u
)
∇
v
d
x
+
∫
Ω
g
(
x
,
u
,
∇
u
)
v
d
x
=
∫
Ω
F
∇
v
d
x
,
222222222222222222222
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
)
.Theorem 20.
Letμ
∈
M
b
(
Ω
). Under the assumptions (45)–(50), there exists a solution u of problem (51) in the sense that u
∈
W
0
1
,
p
(
x
)
(
Ω
)
,
g
(
x
,
u
,
∇
u
)
∈
L
1
(
Ω
), and
(111)
∫
Ω
a
(
x
,
∇
u
)
∇
v
d
x
+
∫
Ω
g
(
x
,
u
,
∇
u
)
v
d
x
=
∫
Ω
v
d
μ
,
222222222222222
∀
v
∈
C
0
∞
(
Ω
)
i
f
f
μ
∈
M
b
p
(
x
)
(
Ω
)
.Proof.
Suppose there exists a solutionu of problem (51), with u
∈
W
0
1
,
p
(
x
)
(
Ω
) and g
(
x
,
u
,
∇
u
)
∈
L
1
(
Ω
). According to (46), we have a
(
x
,
∇
u
)
∈
(
L
p
′
(
x
)
(
Ω
)
)
N; then -
div
⁡
(
a
(
x
,
∇
u
)
)
∈
W
-
1
,
p
′
(
x
)
(
Ω
), and μ
=
g
(
x
,
u
,
∇
u
)
-
div
⁡
(
a
(
x
,
∇
u
)
)
=
f
-
div
⁡
(
a
(
x
,
∇
u
)
) in L
1
(
Ω
)
+
W
-
1
,
p
′
(
x
)
(
Ω
); with f
=
g
(
x
,
u
,
∇
u
) and by Theorem 7, we have μ
∈
M
b
p
(
x
)
(
Ω
).
Conversely, ifμ
∈
M
b
p
(
x
)
(
Ω
), by Theorem 7, μ
∈
L
1
(
Ω
)
+
W
-
1
,
p
′
(
x
)
(
Ω
), and then μ
=
f
-
div
⁡
(
F
), where f
∈
L
1
(
Ω
) and F
∈
(
L
p
′
(
x
)
(
Ω
)
)
N. Let (
f
n
)
n be a sequence in L
∞
(
Ω
) such that f
n converges to f strongly in L
1
(
Ω
) and let (
F
n
)
n be a sequence in (
L
∞
(
Ω
)
)
N such that F
n converges to F strongly in (
L
p
′
(
x
)
(
Ω
)
)
N. By Theorem 19, there exists a solution u
n of
(112)
A
u
n
+
g
(
x
,
u
n
,
∇
u
n
)
=
f
n
-
div
⁡
(
F
n
)
in
Ω
,
u
n
=
0
on
⁡
∂
Ω
,
such that u
n
∈
W
0
1
,
p
(
x
)
(
Ω
)
,
g
(
x
,
u
n
,
∇
u
n
)
∈
L
1
(
Ω
), and, for all v
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
), we have
(113)
∫
Ω
a
(
x
,
∇
u
n
)
∇
v
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
v
d
x
=
∫
Ω
f
n
v
d
x
+
∫
Ω
F
n
∇
v
d
x
.
We choose v
=
φ
(
T
σ
(
u
n
)
), where φ is given by Lemma 16 and σ is given by (50); with d
=
α
/
2 and b
=
b
(
σ
), where b
(
s
) is given by (49), we obtain
(114)
∫
Ω
a
(
x
,
∇
u
n
)
∇
φ
(
T
σ
(
u
n
)
)
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
φ
(
T
σ
(
u
n
)
)
d
x
=
∫
Ω
f
n
φ
(
T
σ
(
u
n
)
)
d
x
+
∫
Ω
F
n
∇
φ
(
T
σ
(
u
n
)
)
d
x
,
which implies that
(115)
∫
Ω
a
(
x
,
∇
u
n
)
φ
σ
′
∇
T
σ
(
u
n
)
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
φ
σ
d
x
=
∫
Ω
f
n
φ
σ
d
x
+
∫
Ω
F
n
φ
σ
′
∇
T
σ
(
u
n
)
d
x
,
where φ
σ
=
φ
(
T
σ
(
u
n
)
), et φ
σ
′
=
φ
′
(
T
σ
(
u
n
)
).
According to (47) and Young’s inequality, we obtain
(116)
α
∫
Ω
|
∇
T
σ
(
u
n
)
|
p
(
x
)
φ
σ
′
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
φ
σ
d
x
≤
φ
(
σ
)
∫
Ω
|
f
n
|
d
x
+
φ
′
(
σ
)
∫
Ω
|
F
n
|
p
′
(
x
)
d
x
+
α
2
∫
Ω
|
∇
T
σ
(
u
n
)
|
p
(
x
)
φ
σ
′
d
x
.
Hence, consider
(117)
α
2
∫
Ω
|
∇
T
σ
(
u
n
)
|
p
(
x
)
φ
σ
′
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
φ
σ
d
x
≤
C
(
φ
(
σ
)
+
φ
′
(
σ
)
)
.
We can write the following:
(118)
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
φ
σ
d
x
=
∫
{
|
u
n
|
<
σ
}
g
(
x
,
u
n
,
∇
u
n
)
φ
σ
d
x
+
∫
{
|
u
n
|
≥
σ
}
g
(
x
,
u
n
,
∇
u
n
)
φ
σ
d
x
.
Using (49), we get
(119)
∫
{
|
u
n
|
<
σ
}
g
(
x
,
u
n
,
∇
u
n
)
φ
σ
d
x
≤
∫
{
|
u
n
|
<
σ
}
φ
σ
b
(
T
σ
(
u
n
)
)
(
|
∇
T
σ
(
u
n
)
|
p
(
x
)
+
d
(
x
)
)
d
x
and using (50), we obtain
(120)
∫
{
|
u
n
|
≥
σ
}
g
(
x
,
u
n
,
∇
u
n
)
φ
σ
d
x
≥
ρ
φ
(
σ
)
∫
{
|
u
n
|
≥
σ
}
|
∇
u
n
|
p
(
x
)
d
x
.
Now we use (119) and (120) in (117) and we write the following:
(121)
∫
Ω
|
∇
T
σ
(
u
n
)
|
p
(
x
)
[
α
2
φ
σ
′
-
b
(
σ
)
|
φ
(
σ
)
|
]
+
ρ
φ
(
σ
)
∫
{
|
u
n
|
≥
σ
}
|
∇
u
n
|
p
(
x
)
d
x
≤
C
.
In view of Lemma 16, we have (
α
/
2
)
φ
σ
′
-
b
(
σ
)
|
φ
(
σ
)
|
≥
α
/
4, and then
(122)
∫
Ω
|
∇
T
σ
(
u
n
)
|
p
(
x
)
d
x
+
∫
{
|
u
n
|
≥
σ
}
|
∇
u
n
|
p
(
x
)
d
x
≤
C
.
Hence, there is u
∈
W
0
1
,
p
(
x
)
(
Ω
) such that
(123)
u
n
converges
to
u
a
.
e
.
in
Ω
and
weakly
in
W
0
1
,
p
(
x
)
(
Ω
)
.Let us show that we haveu
n converging to u strongly in W
0
1
,
p
(
x
)
(
Ω
). We start by proving that
(124)
lim
⁡
k
→
∞
sup
⁡
n
∈
ℕ
∫
{
|
u
n
|
≥
k
}
|
∇
u
n
|
p
(
x
)
d
x
=
0
.
Indeed, for k
≥
2, choose v
=
T
1
(
u
n
-
T
k
-
1
(
u
n
)
)
≡
ψ
k
-
1
(
u
n
) as a test function in (113), and using (47) and Young’s inequality, we get
(125)
α
∫
Ω
|
∇
ψ
k
-
1
(
u
n
)
|
p
(
x
)
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
ψ
k
-
1
(
u
n
)
d
x
≤
∫
{
|
u
n
≥
k
-
1
|
}
|
f
n
|
d
x
+
∫
{
k
-
1
≤
u
n
≤
k
}
‍
|
F
n
|
p
′
(
x
)
d
x
+
α
2
∫
Ω
|
∇
ψ
k
-
1
(
u
n
)
|
p
(
x
)
d
x
.
We have ψ
k
-
1
(
s
). s
≥
0, for all s
∈
ℝ, and ψ
k
-
1
(
s
)
=
0, for all |
s
|
≤
k
-
1; hence,
(126)
|
g
(
x
,
u
n
,
∇
u
n
)
|
χ
{
u
n
≥
k
}
≤
g
(
x
,
u
n
,
∇
u
n
)
ψ
k
-
1
(
u
n
)
.
Therefore,
(127)
∫
(
|
u
n
|
≥
k
)
|
g
(
x
,
u
n
,
∇
u
n
)
|
d
x
≤
∫
(
|
u
n
|
≥
k
)
|
f
n
|
d
x
+
∫
(
k
-
1
≤
|
u
n
|
≤
k
)
|
F
|
p
′
(
x
)
d
x
,
and, by (125), we have
(128)
α
2
∫
Ω
|
∇
ψ
k
-
1
(
u
n
)
|
p
(
x
)
d
x
≤
∫
(
|
u
n
|
≥
k
)
|
f
n
|
d
x
+
∫
(
k
-
1
≤
|
u
n
|
≤
k
)
‍
|
F
|
p
′
(
x
)
d
x
.
Since (
u
n
)
n is bounded in L
1
(
Ω
), then lim
⁡
k
→
∞
sup
⁡
n
∈
ℕ
meas
⁡
{
x
∈
Ω
/
k
-
1
≤
|
u
n
|
}
=
0. We use (127) and (128) and since f
n strongly converges in L
1
(
Ω
) and |
F
n
| strongly converges in L
p
′
(
x
)
(
Ω
), we conclude that
(129)
lim
⁡
k
→
∞
lim
⁡
n
→
∞
∫
(
|
u
n
|
≥
k
)
|
g
(
x
,
u
n
,
∇
u
n
)
|
d
x
=
0
,
lim
⁡
k
→
∞
lim
⁡
n
→
∞
∫
Ω
|
∇
ψ
k
-
1
(
u
n
)
|
p
(
x
)
d
x
=
0
.
Now we show that ∀
k
≥
0 and
(130)
T
k
(
u
n
)
converges
to
T
k
(
u
)
strongly
in
W
0
1
,
p
(
x
)
(
Ω
)
.
Indeed, let k
≥
1 be fixed, and choose v
=
φ
(
T
k
(
u
n
)
-
T
k
(
u
)
) as test function in (113) where φ is the function of Lemma 16 with c
=
1 and d
=
b
(
k
)
/
α, and we obtain
(131)
∫
Ω
a
(
x
,
∇
u
n
)
φ
′
(
T
k
(
u
n
)
-
T
k
(
u
)
)
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
φ
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
=
∫
Ω
f
n
φ
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
+
∫
Ω
F
n
φ
′
(
T
k
(
u
n
)
-
T
k
(
u
)
)
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
.
To simplify writing, we put φ
n
′
=
φ
′
(
T
k
(
u
n
)
-
T
k
(
u
)
) and φ
n
=
φ
(
T
k
(
u
n
)
-
T
k
(
u
)
).We haveu
n which converges to u a.e. in Ω, φ
n which converges to φ
(
0
)
=
0 in L
∞
(
Ω
) weakly-*, and φ
n
′ which converges to φ
′
(
0
)
=
1, and since f
n strongly converges to f in L
1
(
Ω
), we obtain
(132)
lim
⁡
n
→
+
∞
∫
Ω
f
n
φ
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
=
0
.
Since T
k
(
u
n
) converges to T
k
(
u
) weakly in W
0
1
,
p
(
x
)
(
Ω
) and (
F
n
)
n converges to F strongly in (
L
p
′
(
x
)
(
Ω
)
)
N, we obtain
(133)
lim
⁡
n
→
+
∞
∫
Ω
F
n
φ
′
(
T
k
(
u
n
)
-
T
k
(
u
)
)
×
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
=
0
.
Write the first term on the left-hand side of (131) as
(134)
∫
Ω
a
(
x
,
∇
u
n
)
φ
′
(
T
k
(
u
n
)
-
T
k
(
u
)
)
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
=
∫
Ω
a
(
x
,
∇
T
k
(
u
n
)
)
φ
n
′
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
+
∫
Ω
a
(
x
,
∇
G
k
(
u
n
)
)
φ
n
′
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
,
where G
k
(
s
)
=
s
-
T
k
(
s
). Since a
(
x
,
∇
G
k
(
u
n
)
) is bounded in (
L
p
′
(
x
)
(
Ω
)
)
N and ∇
T
k
(
u
)
χ
(
|
u
n
|
≥
k
) converges to zero in (
L
∞
(
Ω
)
) weakly-* as n goes to +
∞, then
(135)
lim
⁡
n
→
+
∞
∫
Ω
a
(
x
,
∇
G
k
(
u
n
)
)
φ
n
′
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
=
0
.
The first term on the right-hand side of (134) is
(136)
∫
Ω
a
(
x
,
∇
T
k
(
u
n
)
)
φ
n
′
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
=
∫
Ω
(
a
(
x
,
∇
T
k
(
u
n
)
)
-
a
(
x
,
∇
T
k
(
u
)
)
)
×
φ
n
′
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
+
∫
Ω
a
(
x
,
∇
T
k
(
u
)
)
φ
n
′
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
.
Since ∇
T
k
(
u
n
) converges to ∇
T
k
(
u
) weakly in (
L
p
′
(
x
)
(
Ω
)
)
N and a
(
x
,
∇
T
k
(
u
)
)
∈
(
L
p
(
x
)
(
Ω
)
)
N, then
(137)
lim
⁡
n
→
+
∞
∫
Ω
a
(
x
,
∇
T
k
(
u
)
)
φ
n
′
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
=
0
.
We can write that
(138)
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
φ
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
=
∫
{
|
u
n
|
≤
k
}
g
(
x
,
u
n
,
∇
u
n
)
φ
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
+
∫
{
|
u
n
|
>
k
}
g
(
x
,
u
n
,
∇
u
n
)
φ
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
.
In view of (50), we have
(139)
∫
{
|
u
n
|
>
k
}
g
(
x
,
u
n
,
∇
u
n
)
φ
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
≥
0
.
On the other hand, we have
(140)
|
∫
{
|
u
n
|
≤
k
}
g
(
x
,
u
n
,
∇
u
n
)
φ
(
T
k
(
u
n
)
-
T
k
(
u
)
)
d
x
|
≤
b
(
k
)
∫
Ω
d
(
x
)
|
φ
n
|
d
x
+
b
(
k
)
∫
Ω
|
∇
T
k
(
u
n
)
|
p
(
x
)
|
φ
n
|
d
x
,
≤
b
(
k
)
∫
Ω
d
(
x
)
|
φ
n
|
d
x
+
b
(
k
)
α
×
[
∫
Ω
(
a
(
x
,
∇
T
k
(
u
n
)
)
-
a
(
x
,
∇
T
k
(
u
)
)
)
×
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
|
φ
n
|
d
x
+
∫
Ω
|
a
(
x
,
∇
T
k
(
u
n
)
)
∇
T
k
(
u
)
|
|
φ
n
|
d
x
+
∫
Ω
a
(
x
,
∇
T
k
(
u
)
)
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
|
φ
n
|
d
x
]
.
Since d
(
x
)
∈
L
1
(
Ω
), φ
n converges to 0 a.e. in Ω and in L
∞
(
Ω
) weakly-*, the sequence T
k
(
u
n
) converges to T
k
(
u
) weakly in W
0
1
,
p
(
x
)
(
Ω
), and a
(
x
,
∇
T
k
(
u
)
)
∈
(
L
p
′
(
x
)
(
Ω
)
)
N, then
(141)
lim
⁡
n
→
∞
∫
Ω
d
(
x
)
|
φ
n
|
d
x
=
0
,
lim
⁡
n
→
∞
∫
Ω
a
(
x
,
∇
T
k
(
u
n
)
)
∇
T
k
(
u
)
|
φ
n
|
d
x
=
0
.
Now since a
(
x
,
∇
T
k
(
u
n
)
) is bounded in (
L
p
′
(
x
)
(
Ω
)
)
N and ∇
T
k
(
u
)
φ
n converges to zero strongly in (
L
p
(
x
)
(
Ω
)
)
N, then we have
(142)
∫
Ω
|
a
(
x
,
∇
T
k
(
u
n
)
)
∇
T
k
(
u
)
|
|
φ
n
|
d
x
≤
(
1
p
-
+
1
p
′
-
)
∥
a
(
x
,
∇
T
k
(
u
n
)
)
∥
p
′
(
x
)
×
∥
∇
T
k
(
u
)
|
φ
n
|
∥
p
(
x
)
≤
C
(
1
p
-
+
1
p
′
-
)
∥
∇
T
k
(
u
)
|
φ
n
|
∥
p
(
x
)
≤
(
1
p
-
+
1
p
′
-
)
(
∫
Ω
|
∇
T
k
(
u
)
|
φ
n
|
|
p
(
x
)
d
x
)
1
/
γ
,
where
(143)
γ
=
{
p
-
if
∥
∇
T
k
(
u
n
)
|
φ
n
|
∥
p
(
x
)
≥
1
,
p
+
if
∥
∇
T
k
(
u
n
)
|
φ
n
|
∥
p
(
x
)
<
1
.
Therefore,
(144)
lim
⁡
n
→
∞
∫
Ω
|
a
(
x
,
∇
T
k
(
u
n
)
)
∇
T
k
(
u
)
|
|
φ
n
|
d
x
=
0
.
Now using (131) to (141), we conclude that
(145)
lim
⁡
n
→
∞
∫
Ω
[
a
(
x
,
∇
T
k
(
u
n
)
)
-
a
(
x
,
∇
T
k
(
u
)
)
]
×
∇
(
T
k
(
u
n
)
-
T
k
(
u
)
)
(
φ
n
′
-
b
(
k
)
α
|
φ
n
|
)
d
x
=
0
,
and in view of [9], we have (130).Now we prove that(146)
u
n
converges
to
u
strongly
in
W
0
1
,
p
(
x
)
(
Ω
)
.
Indeed, let E be a measurable subset of Ω, and then consider
(147)
∫
E
|
∇
u
n
|
p
(
x
)
d
x
=
∫
E
∩
(
|
u
n
|
≤
k
)
|
∇
u
n
|
p
(
x
)
d
x
+
∫
E
∩
(
|
u
n
|
>
k
)
|
∇
u
n
|
p
(
x
)
d
x
.
We have on the one hand
(148)
∫
E
∩
(
|
u
n
|
>
k
)
|
∇
u
n
|
p
(
x
)
d
x
≤
∫
(
|
u
n
|
>
k
)
|
∇
u
n
|
p
(
x
)
d
x
,
and since
(149)
lim
⁡
k
→
∞
sup
⁡
n
∈
ℕ
∫
(
|
u
n
|
>
k
)
|
∇
u
n
|
p
(
x
)
d
x
=
0
,
then, for ε
>
0, there exists k
≥
σ such that
(150)
∫
E
∩
{
|
u
n
|
>
k
}
|
∇
u
n
|
p
(
x
)
d
x
≤
ε
2
,
∀
n
∈
ℕ
.
Furthermore, ∫
E
∩
(
|
u
n
|
≤
k
)
|
∇
u
n
|
p
(
x
)
d
x
≤
∫
E
‍
|
∇
T
k
(
u
n
)
|
p
(
x
)
d
x, and since T
k
(
u
n
) converges to T
k
(
u
) strongly in W
0
1
,
p
(
x
)
(
Ω
), then there exists δ
>
0 such that, for mes
(
E
)
<
δ,
(151)
∫
E
∩
(
|
u
n
|
≤
k
)
|
∇
u
n
|
p
(
x
)
d
x
≤
ε
2
,
∀
n
∈
ℕ
.
In view of (150) and (151) we deduce that (
|
∇
u
n
|
p
(
x
)
)
n is equintegrable. Since ∇
u
n converges to ∇
u a.e., then by the Vitali theorem we obtain (146).Let us show that(152)
g
(
x
,
u
n
,
∇
u
n
)
converges
to
g
(
x
,
u
,
∇
u
)
strongly
in
L
1
(
Ω
)
.
Since g
(
x
,
u
n
,
∇
u
n
) converges to g
(
x
,
u
,
∇
u
), a.e. in Ω, it suffices to show that the sequence (
g
(
x
,
u
n
,
∇
u
n
)
)
n is equintegrable. Let E be a measurable subset of Ω; we have
(153)
∫
E
|
g
(
x
,
u
n
,
∇
u
n
)
|
d
x
≤
∫
E
|
g
(
x
,
u
n
,
∇
T
k
(
u
n
)
)
|
d
x
+
∫
{
|
u
n
|
>
k
}
|
g
(
x
,
u
n
,
∇
u
n
)
|
d
x
.
Since lim
⁡
k
→
∞
sup
⁡
n
∈
ℕ
∫
(
|
u
n
|
>
k
)
|
g
(
x
,
u
n
,
∇
u
n
)
|
d
x
=
0, then there exists k
≥
σ such that
(154)
∫
E
∩
(
|
u
n
|
>
k
)
|
g
(
x
,
u
n
,
∇
u
n
)
|
d
x
≤
ε
2
,
∀
n
∈
ℕ
.
In view of (49), we have
(155)
∫
E
∩
{
|
u
n
|
≤
k
}
|
g
(
x
,
u
n
,
∇
u
n
)
|
d
x
≤
b
(
k
)
∫
E
(
d
(
x
)
+
|
∇
T
k
(
u
n
)
|
p
(
x
)
)
d
x
,
since d
∈
L
1
(
Ω
) and T
k
(
u
n
) converges to T
k
(
u
) strongly in W
0
1
,
p
(
x
)
(
Ω
); therefore, there exists δ
>
0 such that meas
⁡
(
E
)
<
δ and ∫
E
∩
{
|
u
n
|
≤
k
}
‍
|
g
(
x
,
u
n
,
∇
u
n
)
|
d
x
≤
ε
/
2, ∀
n
∈
ℕ, and this shows that (
g
(
x
,
u
n
,
∇
u
n
)
)
n is equintegrable.Now, letu
n be a solution of the approximate problem; then we have
(156)
∫
Ω
a
(
x
,
∇
u
n
)
∇
v
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
v
d
x
=
∫
Ω
f
n
v
d
x
+
∫
Ω
F
n
∇
v
d
x
,
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
)
.
Letting n tend to +
∞, we conclude that u satisfies the following:
(157)
∫
Ω
a
(
x
,
∇
u
)
∇
v
d
x
+
∫
Ω
g
(
x
,
u
,
∇
u
)
v
d
x
=
∫
Ω
v
d
μ
,
2222222222222222222
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
)
.


## 7. A Result of Nonexistence

Theorem 21.
Letλ
∈
M
b
(
Ω
) be positive and concentrated in a subset E of Ω such that cap


p
(
·
)
(
E
)
=
0. Let (
f
n
)
n be a sequence of positive functions in L
1
(
Ω
) such that
(158)
lim
⁡
n
→
∞
∫
Ω
f
n
φ
d
x
=
∫
Ω
φ
d
μ
,
∀
φ
∈
C
0
(
Ω
¯
)
.
Assume (49), (50), and that g satisfies g
(
x
,
s
,
ξ
)
s
≥
0, a.e. x
∈
Ω, ∀
s
∈
ℝ, and ∀
ξ
∈
ℝ
N.
Letu
n be solution of
(159)
A
u
n
+
g
(
x
,
u
n
,
∇
u
)
=
f
n
i
n
Ω
,
u
n
=
0
o
n
∂
Ω
,
in the sense that
(160)
∫
Ω
a
(
x
,
∇
u
n
)
∇
v
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
)
v
d
x
=
∫
Ω
f
n
v
d
x
,
22222222222222222222
∀
v
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
)
.
Then, there exists k
>
0, such that T
k
(
u
n
) converges to 0 strongly in W
0
1
,
p
(
x
)
(
Ω
), the sequence u
n converges to 0 weakly in W
0
1
,
p
(
x
)
(
Ω
), and
(161)
lim
⁡
n
→
∞
∫
Ω
g
(
x
,
u
n
,
∇
u
)
φ
d
x
=
∫
Ω
φ
d
μ
,
∀
φ
∈
C
0
1
(
Ω
)
.Proof.
Consider the following.
Step
1. In view of the proof of Theorem 19 and the assumptions made about a and g, we have the sequence (
u
n
)
n which is bounded in W
0
1
,
p
(
x
)
(
Ω
) and then there exists a subsequence, still denoted by (
u
n
)
n, such that u
n converges to u weakly in W
0
1
,
p
(
x
)
(
Ω
) and a.e. in Ω. By (46), there exists G
∈
(
L
p
′
(
x
)
(
Ω
)
)
N such that a
(
x
,
∇
u
n
) converges to G weakly in (
L
p
′
(
x
)
(
Ω
)
)
N.
The functionb is continuous and let k
>
0 such that b
(
k
)
k
≤
α
/
2; choose v
=
(
k
-
T
k
(
u
n
)
)
ψ
δ
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
) as test function in (160), where ψ
δ is given by Lemma 16, and then we obtain
(162)
-
∫
Ω
(
x
,
∇
T
k
(
u
n
)
)
∇
T
k
(
u
n
)
ψ
δ
d
x
+
∫
Ω
a
(
x
,
∇
(
u
n
)
)
∇
ψ
δ
(
k
-
T
k
(
u
n
)
)
d
x
×
∫
Ω
g
(
x
,
u
n
,
∇
u
)
(
k
-
T
k
(
u
n
)
)
ψ
δ
d
x
=
∫
Ω
f
n
ψ
δ
(
k
-
T
k
(
u
n
)
)
d
x
.
Since ∇
ψ
δ
(
k
-
T
k
(
u
n
)
) converges to ∇
ψ
δ
(
k
-
T
k
(
u
)
) strongly in (
L
p
(
x
)
(
Ω
)
)
N as n tends to ∞ and ∇
ψ
δ
(
k
-
T
k
(
u
)
) converges to zero strongly in (
L
p
(
x
)
(
Ω
)
)
N as δ tends to 0, then
(163)
lim
⁡
δ
→
0
lim
⁡
n
→
+
∞
∫
Ω
a
(
x
,
∇
(
u
n
)
)
∇
ψ
δ
(
k
-
T
k
(
u
n
)
)
d
x
=
∫
Ω
G
∇
ψ
δ
(
k
-
T
k
(
u
)
)
d
x
=
0
,
lim
⁡
δ
→
0
lim
⁡
n
→
+
∞
∫
Ω
f
n
ψ
δ
(
k
-
T
k
(
u
n
)
)
d
x
=
0
.
In view of (49) we have k
-
T
k
(
u
n
)
=
0, for u
n
>
k, since u
n
≥
0, and we have
(164)
|
∫
Ω
g
(
x
,
u
n
,
∇
u
)
(
k
-
T
k
(
u
n
)
)
ψ
δ
d
x
|
≤
∫
(
0
≤
u
n
≤
k
)
b
(
u
n
)
(
k
-
T
k
(
u
n
)
)
ψ
δ
×
(
d
(
x
)
+
|
∇
u
n
|
p
(
x
)
)
d
x
.
≤
∫
Ω
b
(
k
)
k
ψ
δ
(
x
)
d
(
x
)
d
x
+
∫
Ω
b
(
k
)
k
c
|
∇
u
n
|
p
(
x
)
d
x
,
≤
α
2
∫
Ω
d
(
x
)
ψ
δ
(
x
)
d
x
+
α
2
∫
Ω
|
∇
T
k
(
u
n
)
|
p
(
x
)
d
x
.
Since ψ
δ converges to 0 weakly in L
∞
(
Ω
)
-
* and d
∈
L
1
(
Ω
), then
(165)
lim
⁡
δ
→
0
∫
Ω
d
(
x
)
ψ
δ
(
x
)
d
x
=
0
.
We use (47) and (163)–(165) in (162); we conclude that
(166)
lim
⁡
δ
→
0
∫
Ω
|
∇
T
k
(
u
n
)
|
p
(
x
)
ψ
δ
(
x
)
d
x
=
0
.
Step
2. Choose v
=
T
k
(
u
n
)
(
1
-
ψ
δ
)
∈
W
0
1
,
p
(
x
)
(
Ω
)
∩
L
∞
(
Ω
) as test function in (160); we have the following:
(167)
∫
Ω
a
(
x
,
∇
T
k
(
u
n
)
)
∇
T
k
(
u
n
)
(
1
-
ψ
δ
)
d
x
-
∫
Ω
a
(
x
,
∇
u
n
)
∇
ψ
δ
T
k
(
u
n
)
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
)
T
k
(
u
n
)
(
1
-
ψ
δ
)
d
x
=
∫
Ω
f
n
T
k
(
u
n
)
(
1
-
ψ
δ
)
d
x
.
We pass to the limit as n tends to +
∞ and δ tends to zero in (167), and using the same arguments of Step 1, we conclude that
(168)
lim
⁡
δ
→
0
lim
⁡
n
→
+
∞
α
∫
Ω
|
∇
T
k
(
u
n
)
|
p
(
x
)
(
1
-
ψ
δ
(
x
)
)
d
x
=
0
.
Using Step 1, we deduce that lim
⁡
n
→
+
∞
∫
Ω
|
∇
T
k
(
u
n
)
|
p
(
x
)
d
x
=
0, and we have T
k
(
u
n
) which converges to 0 strongly in W
0
1
,
p
(
x
)
(
Ω
). Since the limit is independent of the choice of the subsequence, the sequence (
T
k
(
u
n
)
)
n converges strongly in W
0
1
,
p
(
x
)
(
Ω
); hence u
n converges to u
=
0 weakly in W
0
1
,
p
(
x
)
(
Ω
) and ∇
u
n converges to 0 a.e. in Ω. Then, we obtain G
=
0.
Now, letφ
∈
C
0
1
(
Ω
), and we obtain
(169)
∫
Ω
a
(
x
,
∇
u
n
)
∇
φ
d
x
+
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
φ
d
x
=
∫
Ω
f
n
φ
d
x
.
Passing to the limit, we have
(170)
lim
⁡
n
→
∞
∫
Ω
a
(
x
,
∇
u
n
)
∇
φ
d
x
=
∫
Ω
G
∇
φ
d
x
=
0
,
lim
⁡
n
→
∞
∫
Ω
f
n
φ
d
x
=
lim
⁡
n
→
∞
∫
Ω
φ
d
μ
.
Therefore, consider
(171)
lim
⁡
n
→
∞
∫
Ω
g
(
x
,
u
n
,
∇
u
n
)
φ
d
x
=
∫
Ω
φ
d
μ
.
The proof of Theorem 21 is complete.



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*Source: 320527-2014-05-04.xml*