# Existence of Solutions for Degenerate Elliptic Problems in Weighted Sobolev Space

Authors: Lili Dai; Wenjie Gao; Zhongqing Li
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

This paper is devoted to the study of the existence of solutions to a general elliptic problemA
u
+
g
(
x
,
u
,
∇
u
)
=
f
-
div
⁡
F, with f
∈
L
1
(
Ω
) and F
∈
∏
i
=
1
N
L
p
'
(
Ω
,
ω
i
*
), where A is a Leray-Lions operator from a weighted Sobolev space into its dual and g
(
x
,
s
,
ξ
) is a nonlinear term satisfying g
x
,
s
,
ξ
sgn
⁡
(
s
)
≥
ρ
∑
i
=
1
N
ω
i
|
ξ
i
|
p, |
s
|
≥
h
>
0, and a growth condition with respect to ξ. Here, ω
i, ω
i
* are weight functions that will be defined in the Preliminaries.


---

## Body

## 1. Introduction

LetΩ be a bounded domain in R
N
(
N
≥
1
) and let p be a real number with 1
<
p
<
∞. Denote by X the weighted Sobolev space W
0
1
,
p
(
Ω
,
ω
), associated with a vector of weight functions ω
=
{
ω
i
(
x
)
}
0
≤
i
≤
N, which is endowed with the usual norm ·
1
,
p
,
ω. In this paper, we consider a general class of degenerate elliptic problems:(1)
A
u
+
g
x
,
u
,
∇
u
=
μ
in
Ω
,
u
=
0
on
∂
Ω
,where A
u
=
-
div
⁡
(
a
(
x
,
u
,
∇
u
)
) and the right-hand side term μ
=
f
-
div
⁡
F, where f
∈
L
1
(
Ω
), F
∈
∏
i
=
1
N
L
p
′
(
Ω
,
ω
i
∗
). We also assume the following:(H
1)
The expression(2)
u
X
≔
∑
i
=
1
N
∫
Ω
∂
u
x
∂
x
i
p
ω
i
x
d
x
1
/
pis a norm defined on X and it is equivalent to ·
1
,
p
,
ω.
(H2)
There exist a weight functionσ
(
x
) and a parameter q, 1
<
q
<
∞, such that(3)
σ
1
-
q
′
∈
L
1
Ω
,with q
′
=
q
/
q
-
1. The Hardy inequality(4)
∫
Ω
u
x
q
σ
d
x
1
/
q
≤
C
∑
i
=
1
N
∫
Ω
∂
u
x
∂
x
i
p
ω
i
x
d
x
1
/
pholds for every u
∈
X with a constant C
>
0 independent of u. Moreover, the embedding(5)
X
↪
L
q
Ω
,
σis compact. Interested reader may refer to [1] for some examples of weights which satisfy the above Hardy inequality (see (4)).
(H3)
a
(
x
,
s
,
ξ
)
=
{
a
i
(
x
,
s
,
ξ
)
}
1
≤
i
≤
N
:
Ω
×
R
×
R
N
→
R
N is a Carathéodory vector-valued function, and for all i
=
1
,
…
,
N, there hold(6)
a
i
x
,
s
,
ξ
≤
c
1
ω
i
1
/
p
x
k
x
+
σ
1
/
p
′
s
q
/
p
′
+
∑
j
=
1
N
ω
j
1
/
p
′
x
ξ
j
p
-
1
,
(7)
a
x
,
s
,
ξ
·
ξ
≥
c
0
∑
i
=
1
N
ω
i
x
ξ
i
p
,
(8)
a
x
,
s
,
ξ
-
a
x
,
s
,
η
·
ξ
-
η
>
0
,
ξ
≠
η
∈
R
N
,where k
(
x
) is a positive function in L
p
′
(
Ω
), 1
/
p
+
1
/
p
′
=
1, and the constants c
0, c
1 are both positive.
(H4)
Letg
(
x
,
s
,
ξ
) be a Carathéodory function satisfying the following assumptions:(9)
g
x
,
s
,
ξ
sgn
⁡
s
≥
ρ
∑
i
=
1
N
ω
i
ξ
i
p
,
s
≥
h
,for some h
,
ρ
>
0, and(10)
g
x
,
s
,
ξ
≤
b
s
∑
i
=
1
N
ω
i
ξ
i
p
+
d
x
,with b
:
R
+
→
R
+, a continuous increasing function, and d
(
x
), a nonnegative function in L
1
(
Ω
).In the past decade, much attention has been devoted to nonlinear elliptic equations because of their wide application to physical models such as non-Newtonian fluids, boundary layer phenomena for viscous fluids, and chemical heterogenous model. When-
div
⁡
F
=
0, Akdim et al. [2] proved in the variational setting, under assumptions (H1)–(H4), that, for every f
∈
W
-
1
,
p
′
(
Ω
,
ω
∗
), with g satisfying the sign condition(11)
g
x
,
s
,
ξ
sgn
⁡
s
≥
0
.Problem (1) has a solution u
∈
W
0
1
,
p
(
Ω
,
ω
), where the authors used the approach based on the strong convergence of the positive part u
n
+ (negative part u
n
-) of u
n (the approximating sequence of u). Ammar [3] extended this existence result to problems with general data f
∈
L
1
(
Ω
), under hypotheses (
H
1
)
–
(
H
4
). They also used a similar approach to prove the existence of renormalized solutions. When -
div
⁡
F
≠
0, Aharouch et al. [4] proved the existence result for problem (1), by assuming the sign condition (11). For more details on weighted Sobolev spaces, the readers may refer to [5].Boccardo et al. [6] considered the nonlinear boundary value problem (12)
A
u
+
g
x
,
u
,
∇
u
=
μ
,where μ
∈
L
1
(
Ω
)
+
W
-
1
,
p
′
(
Ω
) and g
(
x
,
u
,
∇
u
)
∈
L
1
(
Ω
) with sign condition (9) for large values of s. By combining the truncation technique with some delicate test functions, the authors showed that the problem has a solution u
∈
W
0
1
,
p
(
Ω
). Mainly motivated by [4, 6], we investigate the elliptic problem (1) in weighted Sobolev space. By choosing test functions different from those employed in [4, 6], we show that problem (1) admits at least one weak solution with (9) instead of the sign condition (11). It is worth pointing out that (9) gives a sign condition on g
(
x
,
s
,
ξ
) only for large values of s, which brings about many difficulties. The essential one of those is that we have to construct some new test functions to obtain the a priori estimates of the approximation solutions u
n since the usual one T
k
(
u
n
) is not a proper test function for our problem. The outline of this paper is as follows. In Section 2, we give some preliminaries and some technical lemmas. The main results will be stated and proved in Section 3.


## 2. Preliminaries

In this section, we give some preliminaries (see [5]). Throughout this section, we assume that the vector field a
(
x
,
s
,
ξ
)
=
{
a
i
(
x
,
s
,
ξ
)
}
1
≤
i
≤
N
:
Ω
×
R
×
R
N
→
R
N satisfies assumptions (6)–(8) and g satisfies (9)-(10). Let ω
=
{
ω
i
(
x
)
}
0
≤
i
≤
N be a vector of measurable weight functions strictly positive a.e. in Ω, such that(13)
ω
i
∈
L
1
Ω
,
ω
i
-
1
/
p
-
1
∈
L
1
Ω
.We define the weighted space with weight γ on Ω as(14)
L
p
Ω
,
γ
=
u
=
u
x
:
u
γ
1
/
p
∈
L
p
Ω
.With this space, we equip the norm(15)
u
p
,
γ
=
∫
Ω
u
x
p
γ
x
d
x
1
/
p
.We denote by W
0
1
,
p
(
Ω
,
ω
) the space of all real-valued functions u
∈
L
p
(
Ω
,
ω
0
) such that the derivatives (see [5]) in the sense of distributions satisfy(16)
∂
u
∂
x
i
∈
L
p
Ω
,
ω
i
∀
i
=
1
,
…
,
N
,endowed with the norm(17)
u
1
,
p
,
ω
=
∫
Ω
u
x
p
ω
0
x
d
x
+
∑
i
=
1
N
∫
Ω
∂
u
x
∂
x
i
p
ω
i
x
d
x
1
/
p
.Let X
≔
W
0
1
,
p
(
Ω
,
ω
) be the closure of C
0
∞
(
Ω
) with respect to the norm ·
1
,
p
,
ω. Then, (
X
,
·
1
,
p
,
ω
) is a reflexive Banach space whose dual is equivalent to W
-
1
,
p
′
(
Ω
,
ω
∗
), where ω
∗
=
{
ω
i
∗
=
ω
i
1
-
p
′
=
ω
i
-
p
′
/
p
},  i
=
1
,
…
,
N, and p
′
=
p
/
p
-
1. As usual, for s, k in R, with k
≥
0, we denote T
k
(
s
)
=
max
⁡
(
-
k
,
min
⁡
(
k
,
s
)
) and G
k
(
s
)
=
s
-
T
k
(
s
).The following lemmas will be needed throughout this paper (refer to [2, 7]).Lemma 1.
Leta and b be two nonnegative real numbers, and let(18)
φ
s
=
s
e
θ
s
2
,with θ
=
b
2
/
4
a
2. Then,(19)
a
φ
′
s
-
b
φ
s
≥
a
2
,
s
∈
R
.Lemma 2.
Letg
∈
L
r
(
Ω
,
γ
) and g
n
∈
L
r
(
Ω
,
γ
), with g
n
L
r
(
Ω
,
γ
)
≤
c, 1
<
r
<
∞. If g
n
→
g
a
.
e
. in Ω, then g
n
⇀
g weakly in L
r
(
Ω
,
γ
), where γ is a weight function on Ω.Lemma 3 (assume(
H
1
)).
LetG
:
R
→
R be uniformly Lipschitzian, with G
(
0
)
=
0. Let u
∈
W
0
1
,
p
(
Ω
,
ω
). Moreover, if the set D of discontinuity points of G
′ is finite, then(20)
∂
G
∘
u
x
i
=
0
,
a.e.  in
x
∈
Ω
:
u
x
∈
D
,
G
′
u
∂
u
∂
x
i
,
a.e.  in
x
∈
Ω
:
u
x
∉
D
.Lemma 4 (assume(
H
1
)).
Letu
∈
W
0
1
,
p
(
Ω
,
ω
) and T
k
(
u
), k
∈
R
+, be the usual truncation. Then, T
k
(
u
)
∈
W
0
1
,
p
(
Ω
,
ω
). Moreover, one has(21)
T
k
u
⟶
u
s
t
r
o
n
g
l
y
i
n
W
0
1
,
p
Ω
,
ω
.Lemma 5 (assume(
H
1
) and (
H
2
)).
Let{
u
n
} be a sequence of functions in W
0
1
,
p
(
Ω
,
ω
) such that u
n
⇀
u weakly in W
0
1
,
p
(
Ω
,
ω
) and(22)
lim
n
→
∞
⁡
∫
Ω
a
x
,
u
n
,
∇
u
n
-
a
x
,
u
,
∇
u
·
∇
u
n
-
u
d
x
=
0
.Then, u
n
→
u strongly in W
0
1
,
p
(
Ω
,
ω
).


## 3. Main Results

Firstly, we give the definition of weak solution for problem (1).Definition 6.
One saysu
∈
W
0
1
,
p
(
Ω
,
ω
) is a weak solution to problem (1), provided that(23)
∫
Ω
a
x
,
u
,
∇
u
·
∇
v
d
x
+
∫
Ω
g
x
,
u
,
∇
u
v
d
x
=
∫
Ω
f
v
d
x
+
∫
Ω
F
·
∇
v
d
x
,
for  every
v
∈
W
0
1
,
p
Ω
,
ω
∩
L
∞
Ω
.Now, we will state and prove our main result on the existence of weak solutions to problem (1).Theorem 7.
Letf be in L
1
(
Ω
) and F
∈
∏
i
=
1
N
L
p
′
(
Ω
,
ω
i
∗
). Then, there exists at least one solution u to problem (1).Proof.
The proof will be divided into 5 steps.
Step 1  (the approximation equation). We introduce the following approximation equation of problem (1). Let f
n be a sequence of L
∞
(
Ω
) functions that converges to f strongly in L
1
(
Ω
) and let n
∈
N,(24)
g
n
x
,
s
,
ξ
=
g
x
,
s
,
ξ
1
+
1
/
n
g
x
,
s
,
ξ
;then g
n
(
x
,
s
,
ξ
) is bounded and satisfies (10) and(25)
g
n
x
,
s
,
ξ
·
sgn
⁡
s
≥
0
,for almost every x in Ω, for every ξ in R
N, and for every s in R with |
s
|
≥
h. By the results of [2], there exists a solution u
n
∈
W
0
1
,
p
(
Ω
,
ω
) of(26)
A
u
n
+
g
n
x
,
u
n
,
∇
u
n
=
f
n
-
div
⁡
F
in
Ω
,
u
n
=
0
on
∂
Ω
,which satisfies(27)
∫
Ω
a
x
,
u
n
,
∇
u
n
·
∇
v
d
x
+
∫
Ω
g
n
x
,
u
n
,
∇
u
n
v
d
x
=
∫
Ω
f
n
v
d
x
+
∫
Ω
F
·
∇
v
d
x
,for every v
∈
W
0
1
,
p
(
Ω
,
ω
)
∩
L
∞
(
Ω
).
Step 2  (the weak convergence
u
n
⇀
u
in
W
0
1
,
p
(
Ω
,
ω
)
). Take v
=
φ
(
T
h
(
u
n
)
) as a test function in (27), where h
>
0 is defined in (9) and φ
(
s
) is as in (19). Writing φ
h
′
=
φ
′
(
T
h
(
u
n
)
) and φ
h
=
φ
(
T
h
(
u
n
)
) for simplicity, we have(28)
∫
Ω
a
x
,
u
n
,
∇
u
n
·
∇
T
h
u
n
φ
h
′
d
x
+
∫
Ω
g
n
x
,
u
n
,
∇
u
n
φ
h
d
x
=
∫
Ω
f
n
φ
h
d
x
+
∫
Ω
F
·
∇
T
h
u
n
φ
h
′
d
x
.Thanks to Young’s inequality and (7), we have(29)
c
0
∫
Ω
∑
i
=
1
N
∂
T
h
u
n
∂
x
i
p
ω
i
x
φ
h
′
d
x
+
∫
Ω
g
n
x
,
u
n
,
∇
u
n
φ
h
d
x
≤
φ
h
∫
Ω
f
n
d
x
+
φ
′
h
∫
Ω
∑
i
=
1
N
F
i
·
ω
i
-
1
/
p
p
′
d
x
+
c
0
2
∫
Ω
∑
i
=
1
N
∂
T
h
u
n
∂
x
i
p
ω
i
x
φ
h
′
d
x
.Since {
f
n
} is bounded in L
1
(
Ω
) and F
∈
∏
i
=
1
N
L
p
′
(
Ω
,
ω
i
∗
), it follows from the above inequality that(30)
c
0
2
∫
Ω
∑
i
=
1
N
∂
T
h
u
n
∂
x
i
p
ω
i
x
φ
h
′
d
x
+
∫
Ω
g
n
x
,
u
n
,
∇
u
n
φ
h
d
x
≤
C
,where C is independent of n. Splitting the second term on the left-hand side where |
u
n
|
<
h and |
u
n
|
≥
h, we can write(31)
∫
Ω
g
n
x
,
u
n
,
∇
u
n
φ
h
d
x
=
∫
u
n
<
h
g
n
x
,
u
n
,
∇
u
n
φ
h
d
x
+
∫
u
n
≥
h
g
n
x
,
u
n
,
∇
u
n
φ
h
d
x
.Using (9) and (10), we get(32)
∫
u
n
≥
h
g
n
x
,
u
n
,
∇
u
n
φ
h
d
x
≥
ρ
φ
h
∫
u
n
≥
h
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
x
d
x
,
∫
u
n
<
h
g
n
x
,
u
n
,
∇
u
n
φ
h
d
x
≤
b
h
∫
Ω
∑
i
=
1
N
∂
T
h
u
n
∂
x
i
p
ω
i
x
φ
h
d
x
+
φ
h
d
L
1
Ω
.Hence,(33)
∫
Ω
∑
i
=
1
N
∂
T
h
u
n
∂
x
i
p
ω
i
x
c
0
2
φ
h
′
-
b
h
φ
h
d
x
+
ρ
φ
h
∫
u
n
≥
h
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
x
d
x
≤
C
.Recalling (19) in Lemma 1, let a
=
c
0
/
2, b
=
b
(
h
); we then obtain(34)
∫
Ω
∑
i
=
1
N
∂
T
h
u
n
∂
x
i
p
ω
i
x
d
x
+
∫
u
n
≥
h
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
x
d
x
≤
C
,which implies(35)
∫
Ω
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
x
d
x
≤
C
,or equivalently(36)
u
n
X
≤
C
,where C is some positive constant. Therefore, we can extract a subsequence, still denoted by itself, such that(37)
u
n
⇀
u
weakly  in
W
0
1
,
p
Ω
,
ω
.By (5) and (37), we have for a subsequence u
n
→
u strongly in L
q
(
Ω
,
ω
) and a.e. in Ω. Then, T
k
(
u
n
) is bounded in W
0
1
,
p
(
Ω
,
ω
). Hence, by the results of [8], we have(38)
T
k
u
n
⇀
T
k
u
weakly  in
W
0
1
,
p
Ω
,
ω
.
Step 3  (the strong convergence
u
n
→
u
in
W
0
1
,
p
(
Ω
,
ω
)
). For every k
≥
h, we will prove that T
k
(
u
n
) converges strongly to T
k
(
u
) in W
0
1
,
p
(
Ω
,
ω
). We first prove that(39)
lim
k
→
+
∞
⁡
sup
n
∈
N
∗
⁡
∫
u
n
≥
k
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
d
x
=
0
.Here, we denote by N
∗ the set of natural numbers. Choosing v
=
T
k
(
u
n
)
-
T
k
-
1
(
u
n
) as a test function in (27) with k
≥
h
+
1, using (7) and Young’s inequality, we obtain(40)
c
0
2
∫
Ω
∑
i
=
1
N
ω
i
x
∂
T
k
u
n
-
T
k
-
1
u
n
∂
x
i
p
d
x
+
∫
Ω
g
n
x
,
u
n
,
∇
u
n
T
k
u
n
-
T
k
-
1
u
n
d
x
≤
∫
u
n
≥
k
-
1
f
n
T
k
u
n
-
T
k
-
1
u
n
d
x
+
∫
k
-
1
≤
u
n
≤
k
∑
i
=
1
N
F
i
·
ω
i
-
1
/
p
p
′
d
x
.Noticing (9) and that T
k
(
u
n
)
-
T
k
-
1
(
u
n
) has the same sign as g
n
(
x
,
u
n
,
∇
u
n
) if |
u
n
|
>
h and is zero if |
u
n
|
≤
h, we get(41)
g
n
x
,
u
n
,
∇
u
n
T
k
u
n
-
T
k
-
1
u
n
≥
g
n
x
,
u
n
,
∇
u
n
χ
u
n
≥
k
.Dropping the nonnegative term, we have(42)
∫
u
n
≥
k
g
n
x
,
u
n
,
∇
u
n
d
x
≤
∫
u
n
≥
k
-
1
f
n
d
x
+
∫
k
-
1
≤
u
n
≤
k
∑
i
=
1
N
F
i
·
ω
i
-
1
/
p
p
′
d
x
.Since(43)
meas
⁡
u
n
≥
k
≤
∫
Ω
u
n
ω
0
1
/
p
k
ω
0
-
1
x
1
/
p
d
x
≤
1
k
u
n
L
p
Ω
,
ω
0
ω
0
-
1
1
/
p
L
p
′
Ω
≤
1
k
u
n
L
p
Ω
,
ω
0
ω
0
-
1
L
1
/
p
-
1
Ω
1
/
p
,we obtain(44)
lim
k
→
+
∞
⁡
sup
n
∈
N
∗
⁡
meas
⁡
u
n
≥
k
-
1
=
0
.Taking into account the fact that {
f
n
} is compact in L
1
(
Ω
) and F
∈
∏
i
=
1
N
L
p
′
(
Ω
,
ω
i
∗
), we deduce that(45)
lim
k
→
+
∞
⁡
sup
n
∈
N
∗
⁡
∫
u
n
≥
k
-
1
f
n
d
x
+
∫
k
-
1
≤
u
n
≤
k
∑
i
=
1
N
F
i
·
ω
i
-
1
/
p
p
′
d
x
=
0
.Hence,(46)
lim
k
→
+
∞
⁡
sup
n
∈
N
∗
⁡
∫
u
n
≥
k
g
n
x
,
u
n
,
∇
u
n
d
x
=
0
.Noticing that k
≥
h and (9), this completes the proof of assertion (39).
Letk
≥
h be fixed, 0
<
ɛ
<
k, and choose v
=
φ
(
T
ɛ
(
u
n
-
T
k
(
u
)
)
) as a test function in (27), where φ
(
s
) is defined in Lemma 1 (refer to [8–10]). We thus obtain(47)
∫
Ω
a
x
,
u
n
,
∇
u
n
·
∇
T
ɛ
u
n
-
T
k
u
φ
′
T
ɛ
u
n
-
T
k
u
d
x
︸
A
+
∫
Ω
g
n
x
,
u
n
,
∇
u
n
φ
T
ɛ
u
n
-
T
k
u
d
x
︸
B
=
∫
Ω
f
n
φ
T
ɛ
u
n
-
T
k
u
d
x
︸
C
+
∫
Ω
F
·
∇
T
ɛ
u
n
-
T
k
u
φ
′
T
ɛ
u
n
-
T
k
u
d
x
︸
D
.In the following, δ
(
ɛ
,
n
) represents a quantity which converges to zero as firstly n
→
∞ and secondly ɛ
→
0. For convenience, we write(48)
φ
ɛ
,
n
′
=
φ
′
T
ɛ
u
n
-
T
k
u
,
φ
ɛ
,
n
=
φ
T
ɛ
u
n
-
T
k
u
.Observe that, in the weak∗ topology of L
∞
(
Ω
) and almost everywhere in Ω, we have(49)
lim
ɛ
→
0
⁡
lim
n
→
∞
⁡
φ
ɛ
,
n
=
0
,
lim
ɛ
→
0
⁡
lim
n
→
∞
⁡
φ
ɛ
,
n
′
=
1
.Now, as {
f
n
} is compact in L
1
(
Ω
) and (49), we have(50)
C
=
δ
ɛ
,
n
.Thanks to T
k
+
ɛ
(
u
n
)
⇀
T
k
(
u
)
weakly in
W
0
1
,
p
(
Ω
,
ω
), F
∈
∏
i
=
1
N
L
p
′
(
Ω
,
ω
i
∗
), and (49), we obtain(51)
D
=
∫
Ω
F
·
∇
T
k
+
ɛ
u
n
-
T
k
u
χ
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
′
d
x
=
δ
ɛ
,
n
,where χ
ɛ
{
|
u
n
-
T
k
(
u
)
|
}
=
χ
{
|
u
n
-
T
k
(
u
)
|
≤
ɛ
}. We can decompose (
A
) as(52)
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
·
∇
T
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
′
d
x
︸
E
+
∫
Ω
a
x
,
u
n
,
∇
G
k
+
ɛ
u
n
χ
u
n
≥
k
+
ɛ
∇
T
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
′
d
x
︸
F
.Owing to {
|
u
n
-
T
k
(
u
)
|
≤
ɛ
}
⊂
{
|
u
n
|
≤
k
+
ɛ
}, we get(53)
F
=
∫
Ω
a
x
,
u
n
,
∇
G
k
+
ɛ
u
n
χ
u
n
≥
k
+
ɛ
·
∇
T
k
+
ɛ
u
n
-
T
k
u
χ
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
′
d
x
.Since ∇
T
k
+
ɛ
(
u
n
) is zero whenever ∇
G
k
+
ɛ
(
u
n
) is not zero, hence,(54)
F
=
-
∫
Ω
a
x
,
u
n
,
∇
G
k
+
ɛ
u
n
·
∇
T
k
u
χ
u
n
>
k
+
ɛ
χ
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
′
d
x
.Since ∇
T
k
(
u
)
≡
0 on the set {
|
u
|
>
k
+
ɛ
}, we see that, as n
→
∞,(55)
∇
T
k
u
χ
u
n
>
k
+
ɛ
⟶
0
,
a.e.  in
Ω
.As ∇
T
k
(
u
)
∈
∏
i
=
1
N
L
p
(
Ω
,
ω
i
), Lebesgue’s dominated convergence theorem guarantees that(56)
∇
T
k
u
χ
u
n
>
k
+
ɛ
⟶
0
strongly  in
∏
i
=
1
N
L
p
Ω
,
ω
i
,
n
⟶
∞
.By (6), (49), and the fact that a
(
x
,
u
n
,
∇
G
k
+
ɛ
(
u
n
)
) is bounded in ∏
i
=
1
N
L
p
′
(
Ω
,
ω
i
∗
), we obtain(57)
F
=
δ
ɛ
,
n
.Now we split (
E
) into(58)
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
-
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
·
∇
T
k
+
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
′
d
x
︸
G
+
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
·
∇
T
k
+
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
′
d
x
︸
H
.We will prove that(59)
a
i
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
⟶
a
i
x
,
T
k
+
ɛ
u
,
∇
T
k
u
strongly  in
L
p
′
Ω
,
ω
i
∗
.In fact,(60)
a
i
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
p
′
ω
i
-
p
′
/
p
≤
c
1
k
x
+
σ
1
/
p
′
T
k
+
ɛ
u
n
q
/
p
′
+
∑
i
=
1
N
ω
i
1
/
p
′
x
∂
T
k
u
∂
x
i
p
-
1
p
′
≤
c
2
k
x
p
′
+
σ
T
k
+
ɛ
u
n
q
+
∑
i
=
1
N
ω
i
x
∂
T
k
u
∂
x
i
p
,where c
1, c
2 are positive constants. Since T
k
+
ɛ
(
u
n
)
⇀
T
k
+
ɛ
(
u
) weakly in W
0
1
,
p
(
Ω
,
ω
) and W
0
1
,
p
(
Ω
,
ω
)
↪
L
q
(
Ω
,
σ
) is compact, then T
k
+
ɛ
(
u
n
)
→
T
k
+
ɛ
(
u
) strongly in L
q
(
Ω
,
σ
) and a.e. in Ω. Hence,(61)
a
i
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
p
′
ω
i
∗
⟶
a
i
x
,
T
k
+
ɛ
u
,
∇
T
k
u
p
′
ω
i
∗
a.e.  in
Ω
,
c
2
k
x
p
′
+
σ
T
k
+
ɛ
u
n
q
+
∑
i
=
1
N
ω
i
x
∂
T
k
u
∂
x
i
p
⟶
c
2
k
x
p
′
+
σ
T
k
+
ɛ
u
q
+
∑
o
=
1
N
ω
i
x
∂
T
k
u
∂
x
i
p
strongly  in
L
q
Ω
,
σ
.Then, by the generalized Lebesgue dominated convergence theorem, we deduce (59). By ∂
T
k
+
ɛ
(
u
n
)
/
∂
x
i
⇀
∂
T
k
(
u
)
/
∂
x
i weakly in L
p
(
Ω
,
ω
i
) and (49), we have(62)
H
=
δ
ɛ
,
n
.Using (57) and (62), we have(63)
A
=
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
-
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
·
∇
T
k
+
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
′
d
x
+
δ
ɛ
,
n
.As for the term (
B
), we decompose it as(64)
∫
u
n
≥
k
+
ɛ
g
n
x
,
u
n
,
∇
u
n
φ
ɛ
,
n
d
x
︸
I
+
∫
u
n
<
k
+
ɛ
g
n
x
,
u
n
,
∇
u
n
φ
ɛ
,
n
d
x
︸
J
.It is clear that on the set {
u
n
≥
k
+
ɛ
} we get(65)
φ
ɛ
,
n
=
φ
T
ɛ
u
n
-
T
k
u
=
φ
ɛ
≥
0
,while on the set {
u
n
<
-
k
-
ɛ
} we get(66)
φ
ɛ
,
n
=
φ
T
ɛ
u
n
-
T
k
u
=
φ
-
ɛ
≤
0
.By (9) and the fact that k
≥
h, we obtain(67)
I
≥
0
.Using (7) and (10) and noticing that ɛ
<
k, we have(68)
J
≤
∫
u
n
≤
k
+
ɛ
b
2
k
d
x
+
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
x
φ
ɛ
,
n
d
x
≤
b
2
k
∫
u
n
≤
k
+
ɛ
d
x
φ
ɛ
,
n
d
x
+
b
2
k
c
0
∫
u
n
≤
k
+
ɛ
a
x
,
u
n
,
∇
u
n
∇
u
n
φ
ɛ
,
n
d
x
=
δ
ɛ
,
n
+
b
2
k
c
0
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
·
∇
T
k
+
ɛ
u
n
φ
ɛ
,
n
d
x
≤
b
2
k
c
0
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
-
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
·
∇
T
k
+
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
d
x
+
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
·
∇
T
k
+
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
d
x
+
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
·
∇
T
k
u
φ
ɛ
,
n
d
x
+
δ
ɛ
,
n
.By the weak convergence of ∂
T
k
+
ɛ
(
u
n
)
/
∂
x
i
⇀
∂
T
k
(
u
)
/
∂
x
i in L
p
(
Ω
,
ω
i
), (49), and (59), we have(69)
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
·
∇
T
k
+
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
d
x
=
δ
ɛ
,
n
.Since a
(
x
,
T
k
+
ɛ
(
u
n
), ∇
T
k
+
ɛ
(
u
n
)
) is bounded in ∏
i
=
1
N
L
p
′
(
Ω
,
ω
i
∗
) and (49),(70)
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
·
∇
T
k
u
φ
ɛ
,
n
d
x
=
δ
ɛ
,
n
.We have(71)
J
≤
b
2
k
c
0
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
-
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
·
∇
T
k
+
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
d
x
+
δ
ɛ
,
n
.Invoking (63), we have(72)
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
-
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
·
∇
T
k
+
ɛ
u
n
-
T
k
u
φ
ɛ
,
n
′
-
b
2
k
c
0
φ
ɛ
,
n
d
x
=
δ
ɛ
,
n
.Consequently, by (19) and letting a
=
1 and b
=
b
(
2
k
)
/
c
0, it yields(73)
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
-
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
·
∇
T
k
+
ɛ
u
n
-
T
k
u
d
x
=
δ
ɛ
,
n
.Since {
|
u
n
|
≤
k
}
⊂
{
|
u
n
|
≤
k
+
ɛ
}, we have(74)
∫
Ω
a
x
,
T
k
u
n
,
∇
T
k
u
n
-
a
x
,
T
k
u
n
,
∇
T
k
u
·
∇
T
k
u
n
-
T
k
u
d
x
=
∫
u
n
≤
k
a
x
,
u
n
,
∇
u
n
-
a
x
,
u
n
,
∇
T
k
u
·
∇
u
n
-
T
k
u
d
x
≤
∫
u
n
≤
k
+
ɛ
a
x
,
u
n
,
∇
u
n
-
a
x
,
u
n
,
∇
T
k
u
·
∇
u
n
-
T
k
u
d
x
=
∫
Ω
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
+
ɛ
u
n
-
a
x
,
T
k
+
ɛ
u
n
,
∇
T
k
u
·
∇
T
k
+
ɛ
u
n
-
T
k
u
d
x
=
δ
ɛ
,
n
.Together with Lemma 5 and the assumptions on a, we obtain(75)
T
k
u
n
⟶
T
k
u
strongly  in
W
0
1
,
p
Ω
,
ω
,which in turn implies(76)
∇
u
n
⟶
∇
u
a.e.  in
Ω
.
For any measurable setE of Ω, we have(77)
∫
E
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
d
x
=
∫
E
∩
u
n
<
k
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
d
x
+
∫
E
∩
u
n
≥
k
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
d
x
.Let ɛ
>
0. Thanks to(78)
∫
E
∩
u
n
≥
k
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
d
x
≤
∫
u
n
≥
k
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
d
x
,by (39), there exists k
≥
h such that(79)
∫
E
∩
u
n
≥
k
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
d
x
≤
ɛ
2
,
n
∈
N
∗
.While k is fixed, we get(80)
∫
E
∩
u
n
<
k
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
d
x
≤
∫
E
∑
i
=
1
N
∂
T
k
u
n
∂
x
i
p
ω
i
d
x
.Owing to the strong compactness of {
T
k
(
u
n
)
} in W
0
1
,
p
(
Ω
,
ω
), there exists δ
′
>
0 such that if meas
⁡
(
E
)
<
δ
′, then(81)
∫
E
∩
u
n
<
k
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
d
x
≤
ɛ
2
,
n
∈
N
∗
.Hence,(82)
∫
E
∑
i
=
1
N
∂
u
n
∂
x
i
p
ω
i
d
x
≤
ɛ
,
n
∈
N
∗
.Thus, the sequence {
|
∇
u
n
|
p
} is equi-integrable. Thanks to Vitali theorem, the equi-integrability together with (76) implies that u
n converges strongly to u in W
0
1
,
p
(
Ω
,
ω
).
Step 4  (the strong convergence
g
n
(
x
,
u
n
,
∇
u
n
)
→
g
(
x
,
u
,
∇
u
)
in
L
1
(
Ω
)
). Note that (76) implies that(83)
g
n
x
,
u
n
,
∇
u
n
⟶
g
x
,
u
,
∇
u
a.e.  in
Ω
.On the other hand, for any measurable set E of Ω, we have(84)
∫
E
g
n
x
,
u
n
,
∇
u
n
d
x
=
∫
E
∩
u
n
<
k
g
n
x
,
u
n
,
∇
u
n
d
x
+
∫
E
∩
u
n
≥
k
g
n
x
,
u
n
,
∇
u
n
d
x
.Let ɛ
>
0 be fixed. We have(85)
∫
E
∩
u
n
≥
k
g
n
x
,
u
n
,
∇
u
n
d
x
≤
∫
u
n
≥
k
g
n
x
,
u
n
,
∇
u
n
d
x
.Choose k
≥
h in (46) such that(86)
∫
E
∩
u
n
≥
k
g
n
x
,
u
n
,
∇
u
n
d
x
≤
ɛ
2
,
n
∈
N
∗
.By (10), we have(87)
∫
E
∩
u
n
<
k
g
n
x
,
T
k
u
n
,
∇
T
k
u
n
d
x
≤
b
k
∫
E
∂
T
k
u
n
∂
x
i
p
ω
i
x
d
x
+
d
x
.Since d
(
x
) belongs to L
1
(
Ω
) and T
k
(
u
n
) is compact in W
0
1
,
p
(
Ω
,
ω
), there exists δ
′
′
>
0 such that if meas
⁡
(
E
)
<
δ
′
′, then(88)
∫
E
∩
u
n
<
k
g
n
x
,
T
k
u
n
,
∇
T
k
u
n
d
x
≤
ɛ
2
,
n
∈
N
∗
.Thus, we have proved that {
g
n
(
x
,
u
n
,
∇
u
n
)
} is equi-integrable. Invoking (86) and (88) and by Vitali theorem,(89)
g
n
x
,
u
n
,
∇
u
n
⟶
g
x
,
u
,
∇
u
strongly  in
L
1
Ω
.
Step 5  (passing to the limit). Now, by passing to the limit in (27), we obtain(90)
∫
Ω
a
x
,
u
,
∇
u
·
∇
v
d
x
+
∫
Ω
g
x
,
u
,
∇
u
v
d
x
=
∫
Ω
f
v
d
x
+
∫
Ω
F
·
∇
v
d
x
;that is, u is a weak solution to problem (1). The proof is complete.



---

*Source: 265127-2015-12-02.xml*