Ammari H., Benkirane A., Touzani A. (editors) Recent Developments in Nonlinear Analysis
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December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
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Consider the operator B
n
: W
1,p
0
(Ω, w) −→ W
−1,p
0
(Ω, w
∗
) defined as
hB
n
v, wi =
Z
Ω
g
n
(x, v, ∇v)w dx +
Z
Ω
a(x, v, ∇v)∇w dx. (27)
Lemma 4.2. The operator B
n
from K
ψ
into W
−1,p
0
(Ω, w
∗
) is pseu-
domonotone. Moreover, B
n
is coercive in the following sense:
hB
n
v, v − v
0
i
kvk
−→ +∞ if kvk −→ +∞, v ∈ K
ψ
,
where v
0
is associated to a through (7).
The proof of this Lemma will be given in Appendix 1.
Lemma 4.2 implies that the problem (25) has a solution, applying the
classical result for nonlinear equations associated to pseudo-monotone op-
erators (see [,
15
Theorem 8.2]).
STEP 2: A priori estimates on the approximate solution and con-
vergence.
Let k ≥ kv
0
k
∞
and let ϕ
k
(s) = se
γs
2
, where γ = (
2b(k)
α
)
2
.
An essential role will be played the following property enjoyed by ϕ
k
(s)
(the proof is trivial):
ϕ
0
k
(s) −
2b(k)
α
|ϕ
k
(s)| ≥
1
2
, ∀s ∈ IR. (28)
Taking u
n
− ηϕ
k
(T
l
(u
n
− v
0
)) (η small enough) as test function in (25),
where l = k + kv
0
k
∞
, we obtain
Z
Ω
a(x, u
n
, ∇u
n
)∇T
l
(u
n
− v
0
)ϕ
0
k
(T
l
(u
n
− v
0
)) dx
+
Z
Ω
g
n
(x, u
n
, ∇u
n
)ϕ
k
(T
l
(u
n
− v
0
)) dx ≤ hf, ϕ
k
(T
l
(u
n
− v
0
))i.
Using the fact that g
n
(x, u
n
, ∇u
n
)ϕ
k
(T
l
(u
n
− v
0
)) ≥ 0 on the subset {x ∈
Ω : |u
n
(x)| > k}, we get
Z
{|u
n
−v
0
|≤l}
a(x, u
n
, ∇u
n
)∇(u
n
− v
0
)ϕ
0
k
(T
l
(u
n
− v
0
)) dx
≤
Z
{|u
n
|≤k}
|g
n
(x, u
n
, ∇u
n
)||ϕ
k
(T
l
(u
n
− v
0
))| dx
+
Z
Ω
F ∇T
l
(u
n
− v
0
)ϕ
0
k
(T
l
(u
n
− v
0
)) dx.
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
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Now we use (7) and (23), we can write
α
Z
{|u
n
−v
0
|≤l}
N
X
i=1
w
i
(x)|
∂u
n
∂x
i
|
p
ϕ
0
k
(T
l
(u
n
− v
0
)) dx
≤ b(|k|)
Z
Ω
c(x) +
N
X
i=1
w
i
(x)|
∂T
k
(u
n
)
∂x
i
|
p
!
|ϕ
k
(T
l
(u
n
− v
0
))|dx
+
Z
Ω
δ(x)ϕ
0
k
(T
l
(u
n
− v
0
)) dx +
Z
{|u
n
−v
0
|≤l}
F.∇u
n
ϕ
0
k
(T
l
(u
n
− v
0
)) dx
+
Z
{|u
n
−v
0
|≤l}
F.∇v
0
ϕ
0
k
(T
l
(u
n
− v
0
)) dx.
Applying the Young’s inequality and the fact that F ∈
N
Y
i=1
L
p
0
(Ω, w
1−p
0
i
),
we get
α
Z
{|u
n
−v
0
|≤l}
N
X
i=1
w
i
(x)|
∂u
n
∂x
i
|
p
ϕ
0
k
(T
l
(u
n
− v
0
)) dx
≤ b(|k|)
Z
Ω
c(x) +
N
X
i=1
w
i
(x)|
∂T
k
(u
n
)
∂x
i
|
p
!
|ϕ
k
(T
l
(u
n
− v
0
))|dx
+
α
2
Z
{|u
n
−v
0
|≤l}
N
X
i=1
w
i
(x)|
∂u
n
∂x
i
|
p
ϕ
0
k
(T
l
(u
n
− v
0
)) dx
+
Z
Ω
δ(x)ϕ
0
k
(T
l
(u
n
− v
0
)) + C
1
(k)
where C
1
(k) is a positive constant depending on k.
Thanks to {x ∈ Ω, |u
n
(x)| ≤ k} ⊆ {x ∈ Ω : |u
n
−v
0
| ≤ l} and the fact that
c, δ ∈ L
1
(Ω), we have
Z
Ω
N
X
i=1
w
i
(x)|
∂T
k
(u
n
)
∂x
i
|
p
ϕ
0
k
(T
l
(u
n
−v
0
))−
2b(k)
α
|ϕ
k
(T
l
(u
n
−v
0
))|
dx ≤ C
2
(k).
Thus by (28), we deduce
Z
Ω
N
X
i=1
w
i
(x)|
∂T
k
(u
n
)
∂x
i
|
p
dx ≤ 2C
2
(k). (29)
Now, we claim that,
Z
Ω
N
X
i=1
w
i
(x)|
∂u
n
∂x
i
|
p
dx ≤ C.
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
192
Let k ≥ kv
0
k
∞
. Taking v = v
0
as a test function in (25), we get
Z
Ω
a(x, u
n
, ∇u
n
)∇(u
n
− v
0
) dx +
Z
Ω
g
n
(x, u
n
, ∇u
n
)(u
n
− v
0
) dx
≤
Z
Ω
F ∇(u
n
− v
0
) dx.
(30)
Furthermore, since g
n
(x, u
n
, ∇u
n
)(u
n
− v
0
) ≥ 0 on the subset {x ∈
Ω, |u
n
(x)| > k}, the inequality (30) implies that
Z
Ω
a(x, u
n
, ∇u
n
)∇(u
n
− v
0
) dx
≤ 2k
Z
{|u
n
|≤k}
|g
n
(x, u
n
, ∇u
n
)|dx +
Z
Ω
F ∇(u
n
− v
0
) dx,
which gives by using (23) and Young’s inequality
Z
Ω
a(x, u
n
, ∇u
n
)∇(u
n
− v
0
) dx
≤ 2kb(k)
"
Z
Ω
c(x) dx +
Z
Ω
N
X
i=1
w
i
(x)|
∂T
k
(u
n
)
∂x
i
|
p
dx
#
+
α
2
Z
Ω
N
X
i=1
w
i
(x)|
∂u
n
∂x
i
|
p
dx + C.
(31)
Using (29) and (31), we obtain
Z
Ω
a(x, u
n
, ∇u
n
)∇(u
n
− v
0
) dx ≤
α
2
Z
Ω
N
X
i=1
w
i
(x)|
∂u
n
∂x
i
|
p
dx + C
3
(k).
Thanks to (7), we have
Z
Ω
N
X
i=1
w
i
(x)|
∂u
n
∂x
i
|
p
dx ≤ C. (32)
Then, we conclude that there exists a function u ∈ W
1,p
0
(Ω, w) such that
u
n
* u weakly in W
1,p
0
(Ω, w),
u
n
→ u strongly in L
q
(Ω, σ) and a.e. in Ω.
(33)
This yields, by using (20), the existence of a function h ∈
N
Y
i=1
L
p
0
(Ω, w
1−p
0
i
),
such that
a(x, u
n
, ∇u
n
) * h weakly in
N
Y
i=1
L
p
0
(Ω, w
1−p
0
i
). (34)
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STEP 3: Almost everywhere convergence of the gradient.
We fix k > kv
0
k
∞
, and let w
n,k
= T
k
(u
n
) − T
k
(u).
For η > 0, we consider the following function:
v
n
= u
n
− ηϕ
k
(w
n,k
). (35)
We choose η such that v
n
∈ K
ψ
(see Appendix 2). As in,
7
we take v
n
as a
test functions in (25), we get
Z
Ω
a(x, u
n
, ∇u
n
)∇w
n,k
ϕ
0
k
(w
n,k
) dx +
Z
Ω
g
n
(x, u
n
, ∇u
n
)ϕ
k
(w
n,k
) dx
≤ hf, ϕ
k
(w
n,k
)i.
(36)
Remark that, we have
T
k
(u
n
) − T
k
(u) =
k − T
k
(u) ≥ 0 if u
n
≥ k
u
n
− T
k
(u) ≤ 0 if |u
n
| ≤ k
−k − T
k
(u) ≤ 0 if u
n
≤ −k
which implies, by using (22), g(x, u
n
, ∇u
n
)ϕ
k
(w
n,k
) ≥ 0 on the set {x ∈
Ω, |u
n
(x)| > k}. So by (36),
Z
Ω
a(x, u
n
, ∇u
n
)∇w
n,h
ϕ
0
k
(w
n,k
) dx +
Z
{|u
n
|≤k}
g
n
(x, u
n
, ∇u
n
)ϕ
k
(w
n,k
) dx
≤ hf, ϕ
k
(w
n,k
)i.
(37)
Splitting the first integral on the left hand side of (37) where |u
n
| ≤ k and
|u
n
| > k, we can write,
Z
Ω
a(x, u
n
, ∇u
n
)∇w
n,k
ϕ
0
k
(w
n,k
) dx
=
Z
{|u
n
|≤k}
a(x, u
n
, ∇u
n
)[∇T
k
(u
n
) − ∇T
k
(u)]ϕ
0
k
(w
n,k
) dx
+
Z
{|u
n
|>k
a(x, u
n
, ∇u
n
)∇w
n,k
ϕ
0
k
(w
n,k
) dx.
(38)
Now we estimate the first term in the right hand side of (38) as follows
Z
{|u
n
|≤k}
a(x, u
n
, ∇u
n
)∇w
n,k
ϕ
0
k
(w
n,k
) dx
≥
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u
n
))[∇T
k
(u
n
) − ∇T
k
(u)]ϕ
0
k
(w
n,k
) dx
−ϕ
0
k
(2k)
Z
{|u
n
|>k}
N
X
i=1
|a
i
(x, T
k
(u
n
), 0)||
∂T
k
(u)
∂x
i
| dx.
(39)
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We claim that last term of the above inequality goes to zero as n tends to
zero.
Indeed we have, for that i = 1, . . . , N |a
i
(x, T
k
(u
n
), 0)|χ
{|u
n
|>k}
con-
verges to |a(x, T
k
(u), 0)|χ
{|u|>k}
strongly in L
p
0
(Ω, w
1−p
0
i
), moreover, since
|
∂T
k
(u)
∂x
i
| ∈ L
p
(Ω, w
i
), then
−ϕ
0
k
(2k)
Z
{|u
n
|>k}
N
X
i=1
|a
i
(x, T
k
(u
n
), 0)||
∂T
k
(u)
∂x
i
| dx = ε(n).
Where ε(n) is a quantity (which is possible to be changed from a line to
another) such that lim
n→+∞
ε(n) = 0 We now investigate the behaviour of the
second term in the right hand side of (38) we get,
Z
{|u
n
|>k}
a(x, u
n
, ∇u
n
)∇w
n,k
ϕ
0
k
(w
n,k
) dx
≥ −ϕ
0
k
(2k)
Z
{|u
n
|>k}
N
X
i=1
|a
i
(x, u
n
, ∇u
n
)||
∂T
k
(u)
∂x
i
| dx.
(40)
By (34) and the fact that
∂T
k
(u)
∂x
i
χ
{|u
n
|>k}
→
∂T
k
(u)
∂x
i
χ
{|u|>k}
= 0
in L
p
(Ω, w
i
), we have by (34)
−ϕ
0
k
(2k)
Z
{|u
n
|>k}
N
X
i=1
|a
i
(x, u
n
, ∇u
n
)||
∂T
k
(u)
∂x
i
| dx = ε(n) (41)
Gathering (39), (40) and (41) we have that (38) implies:
Z
Ω
a(x, u
n
, ∇u
n
)∇w
n,k
ϕ
0
k
(w
n,k
) dx
≥
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u
n
))[∇T
k
(u
n
) − ∇T
k
(u)]ϕ
0
k
(w
n,k
) dx
+ ε(n).
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Which implies that
Z
Ω
a(x, u
n
, ∇u
n
)∇w
n,k
ϕ
0
k
(w
n,k
) dx
≥
Z
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)]ϕ
0
k
(w
n,k
) dx
+
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u))[∇T
k
(u
n
) − ∇T
k
(u)]ϕ
0
k
(w
n,k
) dx
+ ε(n).
(42)
By the continuity of the Nemitskii operator (see
11
), we have for all i =
1, . . . , N,
a
i
(x, T
k
(u
n
), ∇T
k
(u))ϕ
0
(T
k
(u
n
) − T
k
(u)) → a
i
(x, T
k
(u), ∇T
k
(u))ϕ
0
(0)
strongly in L
p
0
(Ω, w
1−p
0
i
), while
∂(T
k
(u
n
))
∂x
i
*
∂(T
k
(u))
∂x
i
weakly in L
p
(Ω, w
i
).
Which yields
lim
n→∞
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u))[∇T
k
(u
n
) − ∇T
k
(u))]ϕ
0
k
(w
n,k
) dx = 0. (43)
Du to (42) and (43), we conclude that
Z
Ω
a(x, u
n
, ∇u
n
)[∇T
k
(u
n
) − ∇T
k
(u)]ϕ
0
(w
n,k
) dx
≥
Z
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)]ϕ
0
(w
n,k
) dx + ε(n).
(44)
We now, discus the behaviour of the second integral of the left hand side
of (37).
Using (23), we have
Z
{|u
n
|≤k}
g
n
(x, u
n
, ∇u
n
)ϕ
k
(w
n,k
) dx
≤ b(k)
Z
Ω
(c(x) +
N
X
i=1
w
i
|
∂T
k
(u
n
)
∂x
i
|
p
|ϕ
k
(w
n,k
)| dx
≤ b(k)
Z
Ω
c(x)|ϕ
k
(w
n,k
)| dx +
b(k)
α
Z
Ω
δ(x)|ϕ
k
(w
n,k
)|
+
b(k)
α
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u
n
))∇T
k
(u
n
)|ϕ
k
(w
n,k
)| dx
−
b(k)
α
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u
n
))∇v
0
|ϕ
k
(w
n,k
)| dx.
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From (20) and (29), there exists a function h
k
∈
N
Y
i=1
L
p
0
(Ω, w
1−p
0
i
) such that
a(x, T
k
(u
n
), ∇T
k
(u
n
)) * h
k
weakly in
N
Y
i=1
L
p
0
(Ω, w
1−p
0
i
) (45)
Moreover, since ∇v
0
|ϕ
k
(w
n,k
)| → 0 in
N
Y
i=1
L
p
(Ω, w
i
) as n tend to infinity,
the third term of the right hand side of the last inequality tend to zero as
n tend to infinity, hence
Z
{|u
n
|≤k}
g
n
(x, u
n
, ∇u
n
)ϕ
k
(w
n,k
) dx
≤
b(k)
α
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u
n
))∇T
k
(u
n
)|ϕ
k
(w
n,k
)| dx
+ b(k)
Z
Ω
c(x)|ϕ
k
(w
n,k
| dx +
b(k)
α
Z
Ω
δ(x)|ϕ
k
(w
n,k
)| dx + ε(n).
(46)
Next we estimate the first term in the right hand side of (46) as follows:
b(k)
α
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u
n
))∇T
k
(u
n
)|ϕ
k
(w
n,k
)| dx
=
b(k)
α
Z
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)]|ϕ
k
(w
n,k
)| dx
+
b(k)
α
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u))[∇T
k
(u
n
) − ∇T
k
(u))]|ϕ
k
(w
n,k
)| dx
+
b(k)
α
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u
n
))∇T
k
(u)|ϕ
k
(w
n,k
)| dx.
(47)
By Lebesgue’s Theorem, we deduce that
∇T
k
(u)|ϕ
k
(w
n,k
)| → ∇T
k
(u)|ϕ
k
(0)| = 0 strongly in
N
Y
i=1
L
p
(Ω, w
i
).
Which and using (45) implies that the third term of (47) tends to 0 as
n → ∞.
On the other side reasoning as in (43), the second term of (47) tends to 0
as n → ∞.
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From (46) and (47), we obtain
Z
{|u
n
|≤k}
g
n
(x, u
n
, ∇u
n
)ϕ
k
(w
n,k
) dx
≤
Z
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)]|ϕ
k
(w
n,k
)| dx
+b(k)
Z
Ω
c(x)|ϕ
k
(w
n,k
)| dx +
b(k)
α
Z
Ω
δ(x)|ϕ
k
(w
n,k
)| dx + ε(n).
(48)
Combining (37), (44) and (48), we obtain
Z
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)](ϕ
0
k
(w
n,k
) −
b(k)
α
|ϕ
k
(w
n,k
)|) dx
≤ b(k)
Z
Ω
c(x)|ϕ
k
(w
n,k
)| dx +
b(k)
α
Z
Ω
δ(x)|ϕ
k
(w
n,k
)| dx
+hf, ϕ
k
(w
n,k
)i + ε(n).
(49)
Therefore (28) implies
Z
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)] dx
≤ 2b(k)
Z
Ω
c(x)|ϕ
k
(w
n,k
)| dx + 2
b(k)
α
Z
Ω
δ(x)|ϕ
k
(w
n,k
)| dx
+hf, ϕ
k
(w
n,k
)i + ε(n).
(50)
Now, since c, δ ∈ L
1
(Ω)), ϕ
k
(w
n,k
) * 0 weakly in W
1,p
0
(Ω, w) as n → ∞,
all the terms of the right hand side of the last inequality tends to 0 as
n → +∞.
This implies that
lim
n→∞
Z
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)] dx = 0.
Finally, Lemma 4.1 implies that
T
k
(u
n
) → T
k
(u) strongly in W
1,p
0
(Ω, w) ∀k > 0. (51)
Since k arbitrary, we have for a subsequence
∇u
n
→ ∇u a.e. in Ω. (52)
Indeed, we show that (as in
8
) ∇u
n
converge to ∇u in measure (and there-
fore, we can always assume that the convergence is a.e. after passing to a
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
198
suitable subsequence).
Let k > 0 large enough, we have
kmeas({|u
n
| > k} ∩ B
R
) =
Z
{|u
n
|>k}∩B
R
|T
k
(u
n
)|dx ≤
Z
B
R
|T
k
(u
n
)|dx
≤
Z
Ω
|u
n
|
p
w
0
dx
1
p
Z
B
R
w
1−p
0
0
dx
1
q
0
≤ c
0
Z
Ω
N
X
i=1
|
∂u
n
∂x
i
|
p
w
i
(x) dx
!
1
p
≤ c
1
where B
R
= {x ∈ Ω; |x| ≤ R}. Which implies that
meas({|u
n
| > k} ∩ B
R
) ≤
c
1
k
∀k > 1. (53)
same, since u ∈ W
1,p
0
(Ω, w), we get
meas({|u| > k}∩ B
R
) ≤
c
2
k
∀k > 1. (54)
We have, for every δ > 0,
meas({|∇u
n
− ∇u| > δ}) ≤ meas({|u
n
| > k}) + meas({|u| > k})
+meas({|∇T
k
(u
n
) − ∇T
k
(u)| > δ})
≤ meas({|u
n
| > k} ∩ B
R
) + meas({|u| > k} ∩ B
R
)
+2meas({|x| > R}) + meas({|∇T
k
(u
n
) − ∇T
k
(u)| > δ}).
(55)
Since T
k
(u
n
) converge strongly in W
1,p
0
(Ω, w), we can assume that ∇T
k
(u
n
)
converge to ∇T
k
(u) in measure in Ω.
Let ε > 0, for R large enough, by (53), (54) and (55), there exists some
n
0
(k, R, δ, ε) > 0 such that meas({|∇u
n
− ∇u| > δ}) < ε for all n, m ≥
n
0
(k, R, δ, ε). This concludes the proof of (52).
Which yields
a(x, u
n
, ∇u
n
) → a(x, u, ∇u) a.e. in Ω
g
n
(x, u
n
, ∇u
n
) → g(x, u, ∇u) a.e. in Ω.
(56)
Step 4: Equi-integrability of the nonlinearities.
We need to prove that
g
n
(x, u
n
, ∇u
n
) → g(x, u, ∇u) strongly in L
1
(Ω), (57)
in particular it is enough to prove the equi-integrable of g
n
(x, u
n
, ∇u
n
). To
this purpose. We consider the function T
1
(u
n
− v
0
− T
h
(u
n
− v
0
)) (with
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
199
h ≥ kv
0
k
∞
). This function can be write as follows
T
1
(u
n
− v
0
− T
h
(u
n
− v
0
)) =
0 if |u
n
− v
0
| ≤ h
u
n
− v
0
− h if h ≤ u
n
− v
0
≤ h + 1
u
n
− v
0
+ h if −h −1 ≤ u
n
− v
0
≤ −h
1 if u
n
− v
0
≥ h + 1
−1 if u
n
− v
0
≤ −h − 1
which implies that
g
n
(x, u
n
, ∇u
n
)T
1
(u
n
− v
0
− T
h
(u
n
− v
0
)) ≥ 0 (58)
and
g
n
(x, u
n
, ∇u
n
)T
1
(u
n
− v
0
− T
h
(u
n
− v
0
)) = |g
n
(x, u
n
, ∇u
n
)| (59)
on the set {x ∈ Ω; |u
n
− v
0
| ≥ h + 1}.
Now, we take u
n
−T
1
(u
n
−v
0
−T
h
(u
n
−v
0
)) as a test function in (25),
we obtain by using (58) and (59)
Z
{h≤|u
n
−v
0
|≤h+1}
a(x, u
n
, ∇u
n
)∇(u
n
− v
0
) dx
+
Z
{|u
n
−v
0
|>h+1}
|g
n
(x, u
n
, ∇u
n
)| dx
≤
Z
{h≤|u
n
−v
0
|≤h+1}
F ∇(u
n
− v
0
) dx
using the Young’s inequality and (7), we deduce
Z
{|u
n
−v
0
|>h+1}
|g
n
(x, u
n
, ∇u
n
)| dx ≤ c
N
X
i=1
Z
{|u
n
−v
0
|>h}
|F
i
|
p
0
w
1−p
0
i
dx
+
Z
{|u
n
−v
0
|>h}
(|F ∇v
0
| + δ(x)) dx
(60)
Let ε > 0, since the functions |F
i
|
p
0
w
1−p
0
i
, |F ∇v
0
| and δ belongs to L
1
(Ω)
then there exists h(ε) ≥ 1 such that
Z
{|u
n
−v
0
|>h(ε)}
|g(x, u
n
, ∇u
n
)| dx < ε/2. (61)
For any measurable subset E ⊂ Ω, we have
Z
E
|g
n
(x, u
n
, ∇u
n
)| dx ≤
Z
E
b(h(ε) + kv
0
k
∞
)(c(x)
+
N
X
i=1
w
i
|
∂T
h(ε)+kv
0
k
∞
(u
n
)
∂x
i
|
p
) dx +
Z
{|u
n
−v
0
|>h(ε)}
|g(x, u
n
, ∇u
n
)| dx.
(62)