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
130
By choosing T
γ
(u
− T
β
(u
)), β ≥ kψk
∞
as test function in (P
), we get
h
∂u
∂t
, T
γ
(u
− T
β
(u
))i +
Z
β≤|u
|≤β+γ
a(x, t, u
, ∇u
)∇u
dx dt
−
Z
Q
1
T
1
(u
− ψ)
−
T
γ
(u
− T
β
(u
)) dx dt =
Z
Q
f
T
γ
(u
− T
β
(u
)) dx dt
(14)
On the one hand, we have
h
∂u
∂t
, T
γ
(u
− T
β
(u
))i =
Z
Ω
S
β
γ
(u
(T )) dx −
Z
Ω
S
β
γ
(u
0
) dx (15)
where S
β
γ
(s) =
Z
s
0
T
γ
(t − T
β
(t)) dt, and by using the fact that
Z
Ω
S
β
γ
(u
(T )) dx ≥ 0 and |
Z
Ω
S
β
γ
(u
0
) dx| ≤ γku
0
k, we get
α
Z
β≤|u
|≤β+γ
N
X
i=1
∂u
∂x
i
p
w
i
(x) dx dt
−
1
Z
Q
T
1
(u
− ψ)
−
T
γ
(u
− T
β
(u
)) dx dt ≤ cγ, ∀ > 0
(16)
so that
−
Z
Q
1
T
1
(u
− ψ)
−
T
γ
(u
− T
β
(u
))
γ
dx dt ≤ c
since −
Z
Q
1
T
1
(u
− ψ)
−
T
γ
(u
− T
β
(u
))
γ
dx dt ≥ 0, for every β ≥ kψk
∞
,
we deduce by Fatou’s lemma as γ → 0 that
Z
Q
1
T
1
(u
− ψ)
−
≤ c. (17)
Using in (P
) the test function T
(u
)
χ
(0,τ )
, we get for every τ ∈ (0, T ),
Z
Ω
S
k
(u
(τ)) dx +
Z
Q
τ
a(x, t, T
k
(u
), ∇T
k
(u
))∇T
k
(u
) dx dt
−
1
Z
Q
T
1
((u
− ψ)
−
)T
k
(u
) dx dt ≤ ck
which gives thanks to (17)
Z
Ω
S
k
(u
(τ)) dx +
Z
Q
τ
a(x, t, T
k
(u
), ∇T
k
(u
)) dx dt ≤ ck. (18)
Then,
α
Z
Q
N
X
i=1
∂T
k
(u
)
∂x
i
p
w
i
(x) dx dt ≤ ck, ∀ k ≥ 1. (19)
December 29, 2009 9:37 WSPC - Proceedings Trim Size: 9in x 6in recent
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Hence, T
k
(u
) is bounded in L
p
(0, T, W
1,p
0
(Ω, w)).
Let k > 0 large enough and B
R
be a ball of Ω, we have,
k meas({|u
| > k} ∩ B
R
× [0, T ]) =
Z
T
0
Z
{|u
|>k}∩B
R
|T
k
(u
)| dx dt
≤
Z
T
0
Z
B
R
|T
k
(u
)| dx dt
≤
Z
Q
|T
k
(u
)|
p
w
0
dx dt
1
p
×
Z
T
0
Z
B
R
w
1−p
0
0
dx dt
!
(20)
then, thanks to (H
1
), we deduce that,
k meas({|u
| > k} ∩ B
R
× [0, T ]) ≤ c
Z
Q
N
X
i=1
∂T
k
(u
)
∂x
i
p
w
i
(x) dx dt
!
1
p
≤ ck
1
p
(21)
which implies that,
meas({|u
| > k} ∩ B
R
× [0, T ]) ≤
c
1
k
1−
1
p
, ∀ k ≥ 1.
So, we have,
lim
k→+∞
(meas({(x, t) ∈ Q : |u
| > k} ∩ B
R
× [0, T ]) = 0 (22)
uniformly with respect to . Consider now a function nondecreasing ξ
k
∈
C
2
(IR) such that
ξ
k
(s) = s for |s| ≤
k
2
ξ
k
(s) = k for |s| ≥ k.
Multiplying the approximate equation by ξ
0
k
(u
), we get
∂
∂t
(ξ
k
(u
)) − div(a(x, t, u
, ∇u
)ξ
0
k
(u
)) + a(x, t, u
, ∇u
)ξ
00
k
(u
)
−
1
T
1
((u
− ψ)
−
)ξ
0
k
(u
) = f
ξ
0
k
(u
),
in the sense of distribution.
This implies, thanks to (19) and the fact that ξ
0
k
has compact support,
that ξ
k
(u
) is bounded in L
p
(0, T, W
1,p
0
(Ω, w)), while it’s time derivative
∂
∂t
(ξ
k
(u
)) is bounded in L
p
0
(0, T, W
−1,p
0
(Ω, w
∗
)) + L
1
(Q
T
), hence lemma
3.3 allows us to conclude that ξ
k
(u
) is compact in L
p
loc
(Q
T
, σ).
Thus, for a subsequence, it also converges in measure and almost every
where in Q
T
since we have, for every λ > 0
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
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meas({|u
− u
η
| > λ} ∩ B
R
× [0, T ]) ≤ meas({|u
| >
k
2
} ∩ B
R
× [0, T ])
+meas({|u
η
| >
k
2
} ∩ B
R
× [0, T ])
+meas({|ξ
k
(u
) − ξ
k
(u
η
)| > λ} ∩ B
R
× [0, T ]).
(23)
Let σ > 0, then, by (22) and the fact that ξ
k
(u
) is compact in L
p
loc
(Q
T
, σ),
there exists k(σ) > 0 such that,
meas({|u
− u
η
| > λ} ∩ B
R
× [0, T ]) ≤ σ for all , η ≤
0
(k(σ), λ, R).
This proves that (u
) is a Cauchy sequence in measure in B
R
×[0, T ], thus
converges almost everywhere to some measurable function u. Then for a
subsequence denoted again u
, we can deduce from (19) that,
T
k
(u
) * T
k
(u) weakly in L
p
(0, T, W
1,p
0
(Ω, w)). (24)
and then, the compact imbedding (7) gives,
T
k
(u
) → T
k
(u) strongly in L
p
(Q
T
, σ) and a.e. in Q
T
. (25)
Step 2: About the gradient of approximate solutions
In the sequel and throughout the paper, we will denote α(, µ, s) all quan-
tities (possibly different) such that,
lim
s→∞
lim
µ→∞
lim
→+0
α(, µ, s) = 0.
Taking now T
η
(u
− (T
k
(u))
µ
), η > 0 as test function in (P
), we get
h
∂u
∂t
, T
η
(u
− (T
k
(u))
µ
)i +
Z
Q
a(x, t, u
, ∇u
)∇T
η
(u
− (T
k
(u))
µ
)
−
1
Z
Q
T
T
1
((u
− ψ)
−
)T
η
(u
− (T
k
(u))
µ
) dx dt ≤ cη,
which implies that,
h
∂u
∂t
, T
η
(u
− (T
k
(u))
µ
)i +
Z
Q
T
a(x, t, u
, ∇u
)∇T
η
(u
− (T
k
(u))
µ
)
≤
η
Z
Q
T
T
1
((u
− ψ)
−
) dx dt + cη
and by (17)
h
∂u
∂t
, T
η
(u
− T
k
(u)
µ
)i +
Z
Q
T
a(x, t, u
, ∇u
)∇T
η
(u
− (T
k
(u)
µ
)
≤ cη.
(26)
The first term of the left-hand side of the last inequality reads as,
h
∂u
∂t
, T
η
(u
− T
k
(u)
µ
)i = h
∂u
∂t
−
∂T
k
(u)
µ
∂t
, T
η
(u
− T
k
(u)
µ
)i
+h
∂T
k
(u)
µ
∂t
, T
η
(u
− T
k
(u)
µ
)i.
(27)
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The second term of the last equality can be written as,
h
∂u
∂t
−
∂T
k
(u)
µ
∂t
, T
η
(u
− T
k
(u)
µ
)i =
Z
Ω
S
η
(u
(T ) − T
k
(u)
µ
(T )) dx
−
Z
Ω
S
η
(u
0
) dx ≥ −η
Z
Ω
|u
0
| dx
≥ −ηc.
(28)
The third term can be written as,
h
∂T
k
(u)
µ
∂t
, T
η
(u
−T
k
(u)
µ
)i = µ
Z
Q
T
(T
k
(u)−T
k
(u)
µ
)(T
η
(u
−T
k
(u)
µ
)) dx dt
(29)
thus by letting → 0 and by using Lebesgue theorem,
Z
Q
T
(T
k
(u) − T
k
(u)
µ
)(T
η
(u
− T
k
(u)
µ
)) dx dt
=
Z
Q
T
(T
k
(u) −T
k
(u)
µ
)(T
η
(u −T
k
(u)
µ
)) dx dt.
Consequently,
h
∂u
∂t
, T
η
(u
− T
k
(u)
µ
)i ≥ α(, µ) − ηc (30)
on the other hand,
Z
Q
T
a(x, t, u
, ∇u
)∇T
η
(u
− T
k
(u)
µ
) dx dt
=
Z
{|u
−T
k
(u)
µ
)|<η}
a(x, t, u
, ∇u
)(∇u
− ∇T
k
(u)
µ
) dx dt
=
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
a(x, t, T
k
(u
), ∇T
k
(u
))(∇T
k
(u
) − ∇T
k
(u)
µ
) dx dt
+
Z
{|u
|>k}∩{|u
−T
k
(u)
µ
)|<η}
a(x, t, u
, ∇u
)(∇u
− ∇T
k
(u)
µ
) dx dt
which implies, by using the fact that
Z
{|u
|>k}∩{|u
−T
k
(u)
µ
)|<η}
a(x, t, u
, ∇u
)∇u
dx dt ≥ 0,
that
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
a(x, t, T
k
(u
), ∇T
k
(u
))(∇T
k
(u
) −∇T
k
(u)
µ
) dx dt
≤ cη +
Z
{|u
|>k}∩{|u
−T
k
(u)
µ
)|<η}
a(x, t, u
, ∇u
)|∇T
k
(u)
µ
| dx dt.
(31)
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Since a(x, t, T
k+η
(u
), ∇T
k+η
(u
)) is bounded
N
Y
i=1
L
p
0
(Q
T
, w
∗
i
), there exists
some h
k+η
∈
N
Y
i=1
L
p
0
(Q
T
, w
∗
i
) such that,
a(x, t, T
k+η
(u
), ∇T
k+η
(u
)) * h
k+η
weakly in
N
Y
i=1
L
p
0
(Q
T
, w
∗
i
).
Consequently,
Z
{|u
|>k}∩{|u
−T
k
(u)
µ
)|<η}
a(x, t, u
, ∇u
)|∇T
k
(u)
µ
| dx dt
=
Z
{|u|>k}∩{|u−T
k
(u)
µ
)|<η}
h
k+η
|∇T
k
(u)
µ
| dx dt + α()
thanks to proposition 3.1, one easily has,
Z
{|u|>k}∩{|u−T
k
(u)
µ
)|<η}
h
k+η
|∇T
k
(u)
µ
| dx dt + α() = α(, µ).
Hence,
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
a(x, t, T
k
(u
), ∇T
k
(u
))(∇T
k
(u
) − ∇T
k
(u)
µ
) dx dt
≤ cη + α(, µ).
(32)
On the other hand, note that
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
a(x, t, T
k
(u
), ∇T
k
(u
))(∇T
k
(u
) − ∇T
k
(u)
µ
) dx dt
=
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
a(x, t, T
k
(u
), ∇T
k
(u
))(∇T
k
(u
) − ∇T
k
(u)) dx dt
+
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
a(x, t, T
k
(u
), ∇T
k
(u
))(∇T
k
(u) − ∇T
k
(u)
µ
) dx dt
(33)
the last integral tends to 0 as → 0 and µ → ∞.
Indeed, we have that
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
a(x, t, T
k
(u
), ∇T
k
(u
))(∇T
k
(u) − ∇T
k
(u)
µ
) dx dt →
Z
{|T
k
(u)−T
k
(u)
µ
)|<η}
h
k
(∇T
k
(u) −∇T
k
(u)
µ
) dx dt as → 0.
It is obviously that,
Z
{|T
k
(u)−T
k
(u)
µ
)|<η}
h
k
(∇T
k
(u) − ∇T
k
(u)
µ
) dx dt → 0 as µ → ∞.
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
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We deduce then that,
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
a(x, t, T
k
(u
), ∇T
k
(u
))(∇T
k
(u
) − ∇T
k
(u)) dx dt
≤ cη + α(, µ).
(34)
Let A
be expression in brace above, then for any 0 < η < 1
I
=
Z
{|T
k
(u
)−T
k
(u)
µ
)|≤η}
A
θ
dx dt +
Z
{|T
k
(u
)−T
k
(u)
µ
)|>η}
A
θ
dx dt
since, a(x, T
k
(u
), ∇T
k
(u
)) is bounded in
N
Y
i=1
L
p
0
(Q
T
, w
1−p
0
i
), while
∇T
k
(u
)
L
∞
bounded in
N
Y
i=1
L
p
(Q
T
, w
i
), then by applying the H¨older’s in-
equality, we obtain,
I
≤ c
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
A
dx dt
!
θ
+c
2
meas{(x, t) ∈ Q
T
: |T
k
(u
) −T
k
(u)
µ
)| > η}
1−θ
,
(35)
on the other hand, we have,
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
A
dx dt
=
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
a(x, t, T
k
(u
), ∇T
k
(u
))(∇T
k
(u
) −∇T
k
(u)) dx dt
−
Z
{|T
k
(u
)−T
k
(u)
µ
)|<η}
a(x, t, T
k
(u
), ∇T
k
(u))(∇T
k
(u
) −∇T
k
(u)) dx dt
= I
1
+ I
2
(36)
using (34), we have,
I
1
≤ cη + α(, µ). (37)
Concerning I
2
the second term of the right hand side of the (36), it is easy
to see that
I
2
= α() (38)
because for all i = 1, . . . , N, we have, a
i
(x, t, T
k
(u
), ∇T
k
(u)) →
a
i
(x, t, T
k
(u), ∇T
k
(u)) strongly in L
p
0
(Q
T
, w
1−p
0
i
), while
∂T
k
(u
)
∂x
i
*
∂T
k
(u)
∂x
i
weakly in L
p
(Q
T
, w
i
).
Combining (35), (36), (37) and (38) we get,
I
≤ c meas{|T
k
(u
) − T
k
(u)
µ
)| < η}
θ
+ c(α(, µ, η))
1−θ
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and by passing to the limit sup over , µ and η
lim
→0
Z
Q
T
[a(x, t, T
k
(u
), ∇T
k
(u
))
−a(x, t, T
k
(u
), ∇T
k
(u))][∇T
k
(u
) − ∇T
k
(u)]
θ
dx dt = 0
(39)
and thus there exist a subsequence also denoted by u
such that,
∇u
−→ ∇u a.e. in Q
T
. (40)
Step 3: Passage to the limit
Let ϕ ∈ K
ψ
∩L
∞
(
¯
Q), choosing T
k
(u
−ϕ)
χ
(0,τ )
as test function in (P
), we
get,
h
∂u
∂t
, T
k
(u
− ϕ)i
Q
τ
+
Z
Q
τ
a(x, t, u
, ∇u
)∇T
k
(u
− ϕ) dx dt
−
1
Z
Q
τ
T
1
(u
− ψ)
−
T
(u
− ϕ) dx dt
=
Z
Q
τ
f
T
(u
− ϕ) dx dt
(41)
since, −
1
Z
Q
τ
T
1
(u
−ψ)
−
T
(u
−ϕ) dx dt ≥ 0 and
∂u
∂t
=
∂
∂t
(u
−ϕ) +
∂ϕ
∂t
,
we get
Z
Ω
S
k
(u
(τ) − ϕ(τ)) dx + h
∂ϕ
∂t
, T
k
(u
− ϕ)i
Q
τ
+
Z
Q
τ
a(x, t, u
, ∇u
)∇T
k
(u
− ϕ) dx dt
≤
Z
Q
τ
f
T
(u
− ϕ) dx dt +
Z
Ω
S
k
(u
(0) −ϕ(0)) dx.
(42)
Lemma 3.7. The sequence (u
) is a Cauchy sequence in C([0, T ], L
1
(Ω)),
moreover, u ∈ C([0, T ], L
1
(Ω)) and (u
) converges to u in C([0, T ], L
1
(Ω)).
This lemma will be proved below.
Because of u
→ u in C([0, T ], L
1
(Ω)), then ∀ τ ≤ T , u
(t) → u(t) in L
1
(Ω),
thus
Z
Ω
S
k
(u
(τ) − ϕ(τ)) dx →
Z
Ω
S
k
(u −ϕ) dx (43)
and
Z
Ω
S
k
(u
(0) − ϕ(0)) dx →
Z
Ω
S
k
(u
0
− ϕ(0)) dx.
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137
Let M = k + kϕk
∞
, then, we can write,
Z
Q
T
a(x, t, u
n
, ∇u
n
)∇T
k
(u
n
− ϕ) dx dt
=
Z
Q
T
a(x, t, T
M
(u
n
), ∇T
M
(u
n
))∇T
k
(u
n
− ϕ) dx dt,
which implies by Fatou’s lemma (and the fact
a(x, t, T
M
(u
n
), ∇T
M
(u
n
)) * a(x, t, T
M
(u), ∇T
M
(u))
weakly in
N
Y
i=1
L
p
0
(Q, w
1−p
0
i
), that we can deduce,
Z
Q
T
a(x, t, T
M
(u), ∇T
M
(u))∇T
k
(u −ϕ)
≤ lim inf
→0
Z
Ω
a(x, t, T
M
(u
), ∇T
M
(u
))∇T
k
(u
− ϕ) dx dt.
(44)
Moreover, since
∂ϕ
∂t
∈ L
p
0
(0, T, W
−1,p
0
(Ω, w
∗
)) and ∇T
k
(u
−ϕ) * ∇T
k
(u−
ϕ) weakly in
N
Y
i=1
L
p
(Q
T
, w
i
), we get,
Z
Q
τ
∂ϕ
∂t
T
k
(u
− ϕ) dx dt →
Z
Q
τ
∂ϕ
∂t
T
k
(u − ϕ) dx dt (45)
Z
Q
f
T
k
(u
− ϕ) dx dt →
Z
Q
fT
k
(u −ϕ) dx dt. (46)
Finally, by (42)-(46) we get,
Z
Ω
S
k
(u
(τ) − ϕ(τ)) dx + h
∂ϕ
∂t
, T
k
(u −ϕ)i
Q
τ
+
Z
Q
τ
a(x, t, u, ∇u)∇T
k
(u − ϕ) dx dt
≤
Z
Q
τ
fT
k
(u − ϕ) dx dt +
Z
Ω
S
k
(u(0) −ϕ(0)) dx,
which completes the proof of the theorem.
Proof of Lemma 3.7:
Since T
l
(u) ∈ K
ψ
, for every l ≥ kψk
∞
.
Let v
i,l
µ
= (T
l
(u))
µ
+ e
−µt
T
l
(η
i
) with η
i
≥ 0 converges to u
0
in L
1
(Ω), as
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
138
test function in (46),
h
∂u
∂t
, T
k
(u
− v
i,l
µ
)i
Q
τ
+
Z
Q
τ
a(x, t, u
, ∇u
)∇T
k
(u
− v
i,l
µ
) dx dt
−
1
Z
Q
τ
T
1
(u
− ψ)
−
T
k
(u
− v
i,l
µ
) dx dt
=
Z
Q
τ
f
T
k
(u
− v
i,l
µ
) dx dt
(47)
since,
∂u
∂t
=
∂
∂t
(u
− v
i,l
µ
) +
∂
∂t
(v
i,l
µ
)
=
∂
∂t
(u
− v
i,l
µ
) + µ(T
l
(u) −v
i,l
µ
)
we deduce,
h
∂u
∂t
, T
k
(u
− v
i,l
µ
)i
Q
τ
= h
∂
∂t
(u
− v
i,l
µ
), T
k
(u
− v
i,l
µ
)i + α(, µ, l)
which implies that,
h
∂
∂t
(u
− v
i,l
µ
), T
k
(u
− v
i,l
µ
)i = h
∂u
∂t
, T
k
(u
− v
i,l
µ
)i
Q
τ
+ α(, µ, l) (48)
on the other hand, by using the monotonicity of a and the fact that,
−
Z
Q
τ
1
T
1
(u
− ψ)
−
T
k
(u
− v
i,l
µ
) dx dt ≥ 0, we deduce that by (47),
h
∂u
∂t
, T
k
(u
− v
i,l
µ
)i
Q
τ
+
Z
Q
τ
a(x, t, u
, ∇v
i,l
µ
)∇T
k
(u
− v
i,l
µ
) dx dt
≤
Z
Q
τ
f
T
k
(u
− v
i,l
µ
) dx dt
since for all i = 1, ..., N a
i
(x, t, T
2k+l
(u
), ∇v
i,l
µ
) → a
i
(x, t, T
k+2l
(u), ∇v
i,l
µ
)
strongly in L
p
0
(Q, w
1−p
0
i
) while
∂
∂x
i
T
k
(u
−v
i,l
µ
) *
∂
∂x
i
T
k
(u −v
i,l
µ
) weakly
in L
p
(Q, w
i
), we have,
h
∂u
∂t
, T
k
(u
− v
i,l
µ
)i ≤ α(, µ, i, l). (49)
Therefore, by writing
Z
Ω
S
k
(u
(τ) − v
i,l
µ
(τ)) dx = h
∂u
∂t
, T
k
(u
− v
i,l
µ
)i
Q
τ
− h
∂v
i,l
µ
∂t
, T
k
(u
− v
i,l
µ
)i
Q
τ
+
Z
Ω
S
k
(u
0
− T
l
(η
i
)) dx
(50)
December 28, 2009 12:15 WSPC - Proceedings Trim Size: 9in x 6in recent
139
and using (48) and (49) we get,
Z
Ω
S
k
(u
(τ) − v
i,l
µ
(τ)) dx ≤ α(, µ, i, l) (51)
which implies, by writing,
2
Z
Ω
S
k
(
u
− u
λ
2
) dx ≤
Z
Ω
S
k
(u
(τ) − v
i,l
µ
(τ)) + S
k
(u
λ
(τ) − v
i,l
µ
(τ))
dx,
that
Z
Ω
S
k
(
u
− u
λ
2
) dx ≤ α(, λ). (52)
Finally, by H¨older’s inequality, we have,
Z
Ω
|u
− u
λ
| dx =
Z
{|u
−u
λ
|≤1}
|u
− u
λ
| dx +
Z
{|u
−u
λ
|>1}
|u
− u
λ
| dx
≤
Z
{|u
−u
λ
|≤1}
|u
− u
λ
|
2
dx
!
1
2
meas(Ω)
1
2
+
Z
{|u
−u
λ
|>1}
|u
− u
λ
| dx
≤ meas(Ω)
1
2
Z
{|u
−u
λ
|≤1}
2S
1
|u
− u
λ
|
2
dx
!
1
2
+
Z
{|u
−u
λ
|>1}
2S
1
(u
− u
λ
) dx
since
(
|y|
2
)
χ
{|y|>1}
≤ (
|y|
2
+
|y|− 1
2
)
χ
{|y|>1}
= S
1
(y)
χ
{|y|>1}
and (
|y|
2
2
)
χ
{|y|≤1}
= S
1
(y)
χ
{|y|≤1}
then by (52) we deduce that,
Z
Ω
|u
(τ) −
u
λ
(τ)| dx ≤ α(, λ), not depending on τ. And thus (u
) is a Cauchy sequence
in C([0, T ], L
1
(Ω)), and since u
→ u, a.e. in Q, we deduce that u
→ u,
a.e. in C([0, T ], L
1
(Ω)).
References
1. R. Adams, Sobolev spaces, AC, Press, New York, (1975)
2. L. Aharouch, E. Azroul and M. Rhoudaf Strongly nonlinear variational
parabolic problems in Weighted Sobolev spaces to appear in AJMAA.
3. Y. Akdim, E. Azroul and A. Benkirane, Existence Results for Quasilin-
ear Degenerated Equations Via Strong Convergence of Truncations, Revista
Matematica Complutense 17, , N.2, (2004) pp 359-379.
4. Y. Akdim, E. Azroul and A. Benkirane, Existence of Solution for quasi-
linear degenerated Elliptic Unilateral Problems, Ann. Math. Blaise pascal vol
10 (2003) pp 1-20.
5. Y. Akdim, E. Azroul and A. Benkirane, Existence of solution for quasi-
linear degenerated elliptic equation, Electronic J. Diff. Equ. Vol 2001 , N71,
(2001) pp 1-19.