Ammari H., Benkirane A., Touzani A. (editors) Recent Developments in Nonlinear Analysis
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Some remarks on a sign condition for perturbations of
nonlinear problems
A. Benkirane, J. Benouna, and M. Rhoudaf
∗
Facult´e des Sciences Dhar-Mahraz, B.P 1796 Atlas F`es, Morocco
E-mail:
∗
rhoudafmohamed@gmail.fr
In this paper, we shall be concerned with the existence result of the quasilinear
elliptic equations of the form,
Au + g(x, u, ∇u) = f,
where A is a Leray-Lions operator from W
1,p
0
(Ω) into its dual. On the nonlin-
ear lower order term g(x, u, ∇u), we assume that it is a Carath´eodory function
having natural growth with respect to |∇u|, but without assuming the sign con-
dition. The main novelty of our work is a new technique based on a Poincar´e’s
inequality. The right hand side f belongs to W
−1,p
0
(Ω) .
Keywords: Quasilinear elliptic equation; Leray-Lions operator; Existence.
1. Introduction
Let Ω be a bounded open subset of IR
N
, N ≥ 2. Let p be a real number, with
1 < p < +∞, and let p
0
be its conjugate H¨older exponent (i.e.
1
p
+
1
p
0
= 1).
Let us consider the following nonlinear elliptic problem
Au + g(x, u, ∇u) = f. (1)
Where A is a Leray-Lions operator from W
1,p
0
(Ω) into its dual W
−1,p
0
(Ω)
and g(x, u, ∇u) is a nonlinearity which satisfies the following growth con-
dition
|g(x, s, ξ)| ≤ b(|s|)γ(x) + h(s)|ξ|
p
with γ ∈ L
p
p−r
(Ω), h ∈ L
1
(IR), b(|s|) ≤ β|s|
r−1
where 0 ≤ r < p and
h, b ≥ 0.
More precisely, this paper deals with the existence of solutions to the
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following problem
u ∈ W
1,p
0
(Ω), g(x, u, ∇u) ∈ L
1
(Ω),
Z
Ω
a(x, u, ∇u)T
k
(u −v) dx +
Z
Ω
g(x, u, ∇u)T
k
(u −v) dx
≤
Z
Ω
fT
k
(u −v) dx
∀ v ∈ W
1,p
0
(Ω) ∩ L
∞
(Ω)
(2)
where f ∈ W
−1,p
0
(Ω).
Our principal goal in this paper is to prove the existence result for the
problem (2) without assuming any sign condition on g.
Recently Porreta has proved in
5
the existence result for the problem (2)
but the result is restricted to b(.) ≡ 1.
A different approach (without sign condition) was used in,
2
under the
assumption g(x, s, ξ) = λs −|ξ|
2
with λ > 0.
We recall also that the authors used in
2
the methods of lower and upper-
solutions. For the case of sign condition, many important works have ap-
peared during these last decades. Namely.
1
In the literature of the same problems, the sign condition play a crucial
role in the proof of main result, to overcame this difficultly we use a new test
functions (see lemma below) and a new technique based under inequality
of poincar´e.
2. Main results
Let A be the nonlinear operator from W
1,p
0
(Ω) into its dual W
−1,p
0
(Ω)
defined as
Au = −div(a(x, u, ∇u))
where a : Ω ×IR × IR
N
is a Carath´eodory function satisfying the following
assumptions: (H
1
)
|a(x, s, ξ)| ≤ [k(x) + |s|
p−1
+ |ξ|
p−1
] (3)
[a(x, s, ξ) − a(x, s, η)](ξ − η) > 0. for all ξ 6= η ∈ IR
N
, (4)
a(x, s, ξ)ξ ≥ α|ξ|
p
(5)
where k(x) is a positive function in L
p
0
(Ω) and α is a positive constant.
Let g(x, s, ξ) be a Carath´eodory function satisfying the following as-
sumptions: (H
2
)
|g(x, s, ξ)| ≤ b(|s|)γ(x) + h(s)|ξ|
p
(6)
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
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where
b(|s|) ≤ β|s|
r−1
where 0 ≤ r < p, γ ∈ L
p
p−r
(Ω) (7)
and h : IR → IR
+
with h ∈ L
1
(Ω)
Lemma 2.1.
Let ϕ(t) = c
0
te
R
|t|
0
h(|s|)
α
ds
, then ϕ
0
(t) −
h(|t|)
α
|ϕ(t)| ≥ c
0
.
Proof. We can tucking c
0
= 1 if t ≥ 0
ϕ
0
(t) = e
R
|t|
0
h(|s|)
α
ds
+ t
h(|t|)
α
e
R
|t|
0
h(|s|)
α
ds
= e
R
|t|
0
h(|s|)
α
ds
+
h(|t|)
α
|ϕ(t)|
(8)
if t ≤ 0
ϕ
0
(t) = e
R
|t|
0
h(|s|)
α
ds
+ t
h(|t|)
α
e
R
|t|
0
h(|s|)
α
ds
= e
R
|t|
0
h(|s|)
α
ds
+
h(|t|)
α
|ϕ(t)|
which implies that
ϕ
0
(t) −
h(|t|)
α
|ϕ(t)| = e
R
|t|
0
h(|s|)
α
ds
≥ 1.
Theorem 2.1. Assume that the assumption (H
1
) and (H
2
) hold and let f
belongs to W
−1,p
0
(Ω). Then, there exists a measurable function u solution
of the following problem:
(P )
u ∈ W
1,p
0
(Ω), g(x, u, ∇u) ∈ L
1
(Ω)
Z
Ω
a(x, u, ∇u)∇T
k
(u −ϕ) dx +
Z
Ω
g(x, u, ∇u)T
k
(u −ϕ) dx
≤
Z
Ω
fT
k
(u −ϕ) dx, ∀ϕ ∈ W
1,p
0
(Ω) ∩ L
∞
(Ω) ∀k > 0.
Step (2) Approximate problem
Let us consider the sequence of approximate problem
(P
n
)
u
n
∈ W
1,p
0
(Ω)
Z
Ω
a(x, u
n
, ∇u
n
)∇(u
n
− v) dx +
Z
Ω
g
n
(x, u
n
, ∇u
n
)(u
n
− v) dx
≤
Z
Ω
f
n
(u
n
− v) dx
∀ v ∈ W
1,p
0
(Ω) ∩ L
∞
(Ω)
where
g
n
(x, s, ξ) =
g(x, s, ξ)
1 +
1
n
|g(x, s, ξ)|
. (9)
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Note that |g
n
(x, s, ξ)| ≤ |g(x, s, ξ)| et |g
n
(x, s, ξ)| ≤ n, then for fixed n ∈ IN,
the approximate problem (P
n
) has at least one solution.
4
Lemma 2.2. Let u
n
be a solution of the problem (P
n
), then we have
Z
Ω
|∇u
n
|
p
1
p
≤ c, (10)
where c is a positive constant not depending on n.
Proof. Let v = u
n
− ϕ(u
n
) where ϕ(t) = te
R
|t|
0
h(|s|)
α
ds
(the function h ap-
pears in (H
2
)). Since v ∈ W
1,p
0
(Ω)) v is admissible test function in (P
n
),
then
Z
Ω
a(x, u
n
, ∇u
n
)∇(ϕ(u
n
)) dx +
Z
Ω
g
n
(x, u
n
, ∇u
n
)ϕ(u
n
) dx ≤
Z
Ω
fϕ(u
n
) dx
which implies that,
Z
Ω
a(x, u
n
, ∇u
n
)∇u
n
ϕ
0
(u
n
) dx ≤
Z
Ω
b(|u
n
|)γ(x)|ϕ(u
n
)| dx
+
Z
Ω
h(|u
n
|)|ϕ(u
n
)||∇u
n
|
p
dx +
Z
Ω
|f||ϕ(u
n
)| dx
since ϕ
0
≥ 0, then
Z
Ω
|∇u
n
|
p
(ϕ
0
(u
n
) −
h(|u
n
|)
α
|ϕ(|u
n
|)|) ≤ βc
0
Z
Ω
|u
n
|
r
γ(x) + c
0
Z
Ω
|f||u
n
|
with c
0
= e
R
+∞
0
h(|s|)
α
ds
and by using poincar´e inequality and Lemma 2.1,
we deduce that
Z
Ω
|∇u
n
|
p
≤ c
Z
Ω
|∇u
n
|
p
r
p
+ c
Z
Ω
|∇u
n
|
p
1
p
. (11)
Consequently, since r < p we have
Z
Ω
|∇u
n
|
p
1
p
≤ c (12)
then, we can extracts a subsequence still denote by u
n
such that,
u
n
* u weakly in W
1,p
0
(Ω) (13)
u
n
→ u strongly in L
p
(Ω) (14)
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this yields, by (3), the existence of a function h ∈
N
Y
i=1
L
p
0
(Ω) such that
a(x, u
n
, ∇u
n
) * h weakly in
N
Y
i=1
L
p
0
(Ω). (15)
Lemma 2.3. Let u
n
be a solution of the problem (P
n
), then we have the
following assertions:
Assertion (i)
lim
j→∞
lim
n→∞
Z
{j≤|u
n
|≤j+1}
a(x, u
n
, ∇u
n
)∇u
n
dx = 0. (16)
Assertion (ii)
lim
j→∞
lim
n→∞
Z
Ω
(a(x, T
k
(u
n
), ∇T
k
(u
n
)) − a(x, T
k
(u), ∇T
k
(u)))
∇(T
k
(u
n
) − T
k
(u))h
j
(u
n
) = 0.
(17)
Assertion (iii)
T
k
(u
n
) → T
k
(u) strongly in W
1,p
0
(Ω) as n → ∞. (18)
Proof assertion (i). Consider the following function v = u
n
−
e
R
|u
n
|
0
h(|s|)
α
ds
T
1
(u
n
− T
j
(u
n
))
+
, for j > 1, then we obtain,
Z
Ω
a(x, u
n
, ∇u
n
)∇u
n
h(|u
n
|)
α
e
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
+
+
Z
Ω
a(x, u
n
, ∇u
n
)∇T
1
(u
n
− T
j
(u
n
))
+
e
R
|u
n
|
0
h(s)
α
ds
+
Z
Ω
g
n
(x, u
n
, ∇u
n
)e
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
+
≤
Z
Ω
fe
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
+
.
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From the growth condition (H
2
), we have
Z
Ω
a(x, u
n
, ∇u
n
)∇u
n
h(|u
n
|)
α
e
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
+
+
Z
Ω
a(x, u
n
, ∇u
n
)∇T
1
(u
n
− T
j
(u
n
))
+
e
R
|u
n
|
0
h(s)
α
ds
≤
Z
Ω
|γ(x)|b(|u
n
|)e
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
+
+
Z
Ω
h(|u
n
|)|∇u
n
|
p
e
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
+
+
Z
Ω
|f|e
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
+
which thanks to (5) and 1 ≤ e
R
|u
n
|
0
h(s)
α
ds
≤ e
R
+∞
0
h(s)
α
ds
= c
0
, gives:
Z
Ω
a(x, u
n
, ∇u
n
)∇T
1
(u
n
− T
j
(u
n
))
+
e
R
|u
n
|
0
h(s)
α
ds
≤ c
0
Z
Ω
|γ(x)|b(|u
n
|)T
1
(u
n
− T
j
(u
n
))
+
+ c
0
Z
Ω
|f|T
1
(u
n
− T
j
(u
n
))
+
(19)
which implies that,
Z
{j≤u
n
≤j+1}
a(x, u
n
, ∇u
n
)∇u
n
≤ c
0
β
Z
{|u
n
|>j}
|u
n
|
r−1
|γ(x)|T
1
(u
n
− T
j
(u
n
))
+
+c
0
Z
Ω
|f|T
1
(u
n
− T
j
(u
n
))
+
(20)
we deduce that
Z
{j≤u
n
≤j+1}
a(x, u
n
, ∇u
n
)∇u
n
≤ c
0
β
Z
Ω
|u
n
|
r−1
|γ(x)|T
1
(u
n
− T
j
(u
n
))
+
+
Z
Ω
|f|T
1
(u
n
− T
j
(u
n
))
+
≤ c
0
β
Z
Ω
|u
n
|
p
r
p
Z
Ω
(γ(x)|T
1
(u
n
− T
j
(u
n
))
+
|)
p
p−r
p−r
p
+
Z
Ω
c
0
|f|T
1
(u
n
− T
j
(u
n
))
+
(21)
by using the inequality poincar´e and (10) we get,
Z
{j≤u
n
≤j+1}
a(x, u
n
, ∇u
n
)∇u
n
≤ c
Z
Ω
(γ(x)|T
1
(u
n
− T
j
(u
n
))
+
|)
p
p−r
p−r
p
+
Z
Ω
c
0
|f|T
1
(u
n
− T
j
(u
n
))
+
(22)
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then, by Lebesgue’s theorem the right hand side goes to zero as n and j
tends to infinity.
Therefore, passing to the limit first in n, then in j, we obtain from (22)
lim
j→∞
lim
n→∞
Z
{j≤u
n
≤j+1}
a(x, u
n
, ∇u
n
)∇u
n
dx = 0 (23)
on the other hand, consider the test function v = u
n
+e
−
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
−
T
j
(u
n
))
−
in (P
n
), then
Z
Ω
a(x, u
n
, ∇u
n
)∇
−e
−
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
−
dx
+
Z
Ω
g
n
(x, u
n
, ∇u
n
)
−e
−
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
−
dx
≤
Z
Ω
f
−e
−
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
−
dx
which implies according to (H
2
) that,
Z
Ω
a(x, u
n
, ∇u
n
)∇u
n
h(|u
n
|)
α
e
−
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
−
dx
−
Z
Ω
a(x, u
n
, ∇u
n
)e
−
R
|u
n
|
0
h(s)
α
ds
∇T
1
(u
n
− T
j
(u
n
))
−
dx
≤
Z
Ω
b(|u
n
|)e
−
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
−
γ(x) dx
+
Z
Ω
h(|u
n
|)e
−
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
−
|∇u
n
|
p
dx
+
Z
Ω
|f|e
−
R
|u
n
|
0
h(s)
α
ds
T
1
(u
n
− T
j
(u
n
))
−
dx.
From (5) and since 0 ≤ e
−
R
|u
n
|
0
h(s)
α
ds
≤ 1 it is possible to conclude that,
−
Z
Ω
a(x, u
n
, ∇u
n
)e
−
R
|u
n
|
0
h(s)
α
ds
∇T
1
(u
n
− T
j
(u
n
))
−
dx
≤
Z
Ω
b(|u
n
|)T
1
(u
n
− T
j
(u
n
))
−
γ(x) dx +
Z
Ω
|f|T
1
(u
n
− T
j
(u
n
))
−
dx
then,
−
Z
{−j−1≤u
n
≤−j}
a(x, u
n
, ∇u
n
)∇u
−
n
e
−
R
|u
n
|
0
h(s)
α
ds
dx
≤
Z
{u
n
≤−j}
β|u
n
|
p−1
T
1
(u
n
− T
j
(u
n
))
−
γ(x) dx
+
Z
Ω
|f|T
1
(u
n
− T
j
(u
n
))
−
dx
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
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Consequently, we have
Z
{−j−1≤u
n
≤−j}
a(x, u
n
, ∇u
n
)∇u
n
≤ β
Z
{u
n
≤−j}
|u
n
|
p
!
r
p
Z
{u
n
≤−j}
(γ(x)T
1
(u
n
− T
j
(u
n
))
−
)
p
p−r
!
p−r
p
+
Z
Ω
|f|T
1
(u
n
− T
j
(u
n
))
−
dx
(24)
since (u
n
)
n∈IN
is bounded in L
p
(Ω), we deduce by Lebesgue’s theorem that
the terms of the right-hand side of the last inequality goes to zero as n and
j tends to infinity. Then, (24) becomes
lim
j→∞
lim
n→∞
Z
{−j−1≤u
n
≤−j}
a(x, u
n
, ∇u
n
)∇u
n
dx = 0. (25)
Finally, combining (23) and (25), we have
lim
j→∞
lim
n→∞
Z
{j≤|u
n
|≤j+1}
a(x, u
n
, ∇u
n
)∇u
n
dx = 0. (26)
Proof of assertion (ii). We will use the following function of one real
variable, which is defined as follow:
h
j
(s) =
1 if |s| ≤ j
0 if |s| ≥ j + 1
j + 1 − s if j ≤ s ≤ j + 1
s + j + 1 if − j − 1 ≤ s ≤ −j.
(27)
Let v = u
n
−e
R
|u
n
|
0
h(s)
α
ds
(T
k
(u
n
) −T
k
(u))
+
h
j
(u
n
) as test function in (P
n
),
we obtain
Z
Ω
a(x, u
n
, ∇u
n
)∇(e
R
|u
n
|
0
h(s)
α
ds
(T
k
(u
n
) − T
k
(u))
+
h
j
(u
n
)) dx
+
Z
Ω
g
n
(x, u
n
, ∇u
n
)e
R
|u
n
|
0
h(s)
α
ds
(T
k
(u
n
) −T
k
(u))
+
h
j
(u
n
) dx
≤
Z
Ω
fe
R
|u
n
|
0
h(s)
α
ds
(T
k
(u
n
) − T
k
(u))
+
h
j
(u
n
) dx
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
38
using (5) and (H
2
), we obtain
Z
Ω
a(x, u
n
, ∇u
n
)∇(T
k
(u
n
) −T
k
(u))
+
e
R
|u
n
|
0
h(s)
α
ds
h
j
(u
n
) dx
≤
Z
Ω
b(|u
n
|)γ(x)e
R
|u
n
|
0
h(s)
α
ds
(T
k
(u
n
) − T
k
(u))
+
h
j
(u
n
) dx
+
Z
{j≤|u
n
|≤j+1}
a(x, u
n
, ∇u
n
)∇u
n
e
R
|u
n
|
0
h(s)
α
ds
(T
k
(u
n
) − T
k
(u))
+
dx
+
Z
Ω
|f|e
R
|u
n
|
0
h(s)
α
ds
(T
k
(u
n
) − T
k
(u))
+
h
j
(u
n
) dx
(28)
since h ∈ L
1
(Ω), we have e
R
|u
n
|
0
h(s)
α
ds
≤ e
R
+∞
0
h(s)
α
ds
= c
0
< +∞, which
implies that,
Z
Ω
a(x, u
n
, ∇u
n
)∇(T
k
(u
n
) −T
k
(u))
+
e
R
|u
n
|
0
h(s)
α
ds
h
j
(u
n
) dx
≤ c
0
β
Z
Ω
|u
n
|
r−1
γ(x)(T
k
(u
n
) − T
k
(u))
+
h
j
(u
n
) dx
+c
2
Z
{j≤|u
n
|≤j+1}
a(x, u
n
, ∇u
n
)∇u
n
dx + c
1
Z
Ω
|f|(T
k
(u
n
)
−T
k
(u))
+
h
j
(u
n
) dx
By the Holder’s inequality, we have,
Z
Ω
a(x, u
n
, ∇u
n
)∇(T
k
(u
n
) − T
k
(u))
+
e
R
|u
n
|
0
h(s)
α
ds
h
j
(u
n
) dx
≤ c
Z
Ω
|u
n
|
p
dx
r
p
Z
Ω
(γ(x)(T
k
(u
n
) −T
k
(u))
+
h
j
(u
n
))
p
p−r
dx
p−r
p
+c
2
Z
{j≤|u
n
|≤j+1}
a(x, u
n
, ∇u
n
)∇u
n
dx
+c
1
Z
Ω
|f|(T
k
(u
n
) − T
k
(u))
+
h
j
(u
n
) dx
applying again (16) and Lebesgue’s theorem the terms of the right hand
side of last inequality goes to zero as n and j tend to infinity, then,
lim
j→∞
lim
n→∞
Z
Ω
a(x, u
n
, ∇u
n
)∇(T
k
(u
n
) − T
k
(u))
+
e
R
|u
n
|
0
h(s)
α
ds
h
j
(u
n
) dx = 0.
(29)
Moreover, (29) becomes,
Z
{T
k
(u
n
)−T
k
(u)≥0}
a(x, T
k
(u
n
), ∇T
k
(u
n
))∇(T
k
(u
n
) − T
k
(u))
×e
R
|u
n
|
0
h(s)
α
ds
h
j
(u
n
) dx
−
Z
{T
k
(u
n
)−T
k
(u)≥0,|u
n
|≥k}
a(x, u
n
, ∇u
n
)∇T
k
(u)e
R
|u
n
|
0
h(s)
α
ds
h
j
(u
n
) dx
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
39
which gives
lim
j→∞
lim
n→∞
Z
{T
k
(u
n
)−T
k
(u)≥0}
a(x, T
k
(u
n
), ∇T
k
(u
n
))(∇T
k
(u
n
) − ∇T
k
(u))
×e
R
|u
n
|
0
h(s)
α
ds
h
j
(u
n
) dx = 0.
Since e
R
|u
n
|
0
h(s)
α
ds
≥ 1, the consequently we can write ,
lim
j→∞
lim
n→∞
Z
{T
k
(u
n
)−T
k
(u)≥0}
[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)]h
j
(u
n
) dx = 0
(30)
on the other hand, taking
v = u
n
+ e
−
R
|u
n
|
0
h(s)
α
ds
(T
k
(u
n
) − T
k
(u))
−
h
j
(u
n
)
as test function in (P
n
) and reasoning as in (30) it is possible to conclude
that,
lim
j→∞
lim
n→∞
Z
{T
k
(u
n
)−T
k
(u)≤0}
[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)]h
j
(u
n
) dx = 0
(31)
Combining (30) and (31), we deduce (17).
Proof of assertion (iii). First we have
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
=
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)]h
j
(u
n
) 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)](1 − h
j
(u
n
)) dx
thanks to (17) the first integral of the right hand side converges to zero as
n and j tend to infinity, for the second term, we have for j large enough
(j > k)
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)](1 − h
j
(u
n
)) dx
=
Z
Ω
a(x, T
k
(u
n
), ∇T
k
(u))∇T
k
(u)(1 − h
j
(u
n
)) dx