1. No category

Existence and Attractors of Solutions
for Nonlinear Parabolic Systems
By A. El Hachimi and H. El Ouardi
Abstract: We prove existence and asymptotic behaviour results for weak
solutions of a mixed problem (S). We also obtain the existence of the global at
2
tractor and the regularity for this attractor in H 2 (Ω) and we derive estimates
of its Haussdorf and fractal dimensions.
—————————————–
Keywords: Nonlinear parabolic systems; existence of solutions; global attractor; asymptotic behaviour; Haussdorf and fractal dimensions.
AMS subject classifications : 35K55, 35K57, 35K65, 35B40
0. Introduction
We consider the
nonlinear system
 following
∂b1 (u1 )
− 4u1 + f1 (x, u1 , u2 ) = 0
in Ω × (0, T )

∂t


∂b2 (u2 )
−
4u
+
f
(x,
u
,
u
)
=
0
in Ω × (0, T )
2
2
1
2
∂t
(S )

u
=
u
=
0
in
∂Ω × (0, T )
1
2


(b1 (u1 (x, 0), b2 (u2 (x, 0)) = (b1 (ϕ0 (x)), b2 (ψ 0 (x))) in Ω
where Ω is a bounded open subset in RN , N ≥ 1, with a smooth boundary
∂Ω. (S) is an example of nonlinear parabolic systems modelling a reaction diffusion process for which many results on existence, uniqueness and regularity
have been obtained in the case where bi (s) = s ( see, for instance [6, 7, 18]).
The case of a single equation of the type (S) is studied in [1, 2, 3, 4, 5, 8, 9, 19] .
The purpose of this paper is the natural extension to system (S) of the results
by [8], which concerns the single equation ∂β(u)
∂t − ∆u + f (x, t, u) = 0.
Actually, our work generalizes the question of existence and regularity of the
global attractor obtained therein.
In the first section of this paper, we give some assumptions and preliminaries
and in section 2, we prove the existence of absorbing sets and the existence of
the gobal attractor; while in section 3, we present the regularity of the attractor
and show stabilization property. Finally, section 4 is devoted to estimates of
the Haussdorf and fractal dimensions.
1. Preliminaries, Existence and Uniqueness
1.1 Notations and Assumptions
Let bi , (i = 1, 2) be continuous functions with bi (0) = 0. We define for t ∈ R
Rt
Ψi (t) = 0 bi (τ )dτ . Then the Legendre transform Ψ∗ of Ψ is defined by
Ψ∗i (τ ) = sup {τ s − Ψi (s)} . Ω stands for a regular open bounded subset of
s∈R
EJQTDE, 2001 No. 5, p. 1
RN and for any T > 0, we set QT = Ω × (0, T ) and ST = ∂Ω × (0, T ), where
∂Ω is the boundary of Ω . The norm in a space X will be denoted by : k.kr
if X = Lr (Ω) for all r : 1≤ r ≤ +∞ , k.kX otherwise and h., .iX,X 0 will denote
0
the duality product between X and its dual X .
We start by introducing our assumptions and making precise the meaning
of a solution of (S). Consider the system ( S) under the following assumptions:
(H1) (ϕ0 , ψ 0 ) ∈ L2 (Ω) × L2 (Ω).
(H2) bi is an increasing continuous function from R into R , bi (0) = 0, and
there exists cij > 0 such that : |bi (s)| ≤ ci1 |s| + ci2 , for all s ∈ R, i = 1, 2.
( H3) fi ∈ C 1 (Ω × R × R).
( H4) ∀x ∈ Ω, ∀ξ
∈ R, ∃c1 > 0, c2 > 0 :
sign(ξ)f1 (x, ξ, 0) ≥ −c1
sign(ξ)f2 (x, 0, ξ) ≥ −c2 .
( H5) For any
N > 0, ∃c3 > 0, c4 > 0, c5 > 0 :
p −1
sign(ξ)f1 (x, ξ, v) ≥ c3 |ξ| 1 − c4



p1 −1
|f1 (x, ξ, v)| ≤ c5 (|ξ|
+ 1)
p1 > 2
+
+

|f
(x,
u,
v)|
≤
a
(|u|),
where
a
:
R
→
R
is
increasing
1


for any v : |v| ≤ N.
( H6) For any
M > 0, ∃c6 > 0, c7 > 0, c8 > 0 :
p −1
sign(ξ)f2 (x, u, ξ) ≥ c6 |ξ| 2 − c7



p −1
|f2 (x, u, ξ)| ≤ c8 (|ξ| 2 + 1)
p2 > 2

|f2 (x, u, v)| ≤ a2 (|v|), where a2 : R+ → R+ is increasing


for any u : |u| ≤ M.
(H7) 0 < γ i ≤ b0i (s) for all s ∈ R.
Definition By a weak solution of (S), we mean an element
ui ∈ Lpi (0, T ; Lpi (Ω)) ∩ L2 (0, T ; H01 (Ω)) ∩ L∞ (t0 , T ; L∞ (Ω)), for all t0 > 0
such that
∗
∂bi (ui )
p∗
i
T ; Lpi (Ω)) + L2 (0, T ; H −1(Ω)) and ∀φi ∈ L2 (0, T ; H −1(Ω)) :
∂tD ∈ L (0,
E
R T ∂bi (ui )
RT R
RT R
dt + 0 Ω ∇ui ∇φi dxdt + 0 Ω fi (x, u1 , u2 )dxdt = 0,
∂t , φi
0
∗
Vi ,Vi
and if (φ )t ∈ L2 (0, T ; L2(Ω)), φi (T ) = 0
RT R
R T D ∂bi (uii ) E
dt = − 0 Ω (bi (ui (t) − bi (ui (x, 0)) (φi )t dxdt,
0
∂t , φi
0
Vi ,Vi
0
0
where Vi = Lpi (Ω) ∩ H01 (Ω), Vi = Lpi (Ω) + H −1 (Ω), p10 + p1i = 1, i = 1, 2.
i
1.2. Existence theorem.
Theorem 1 Let (H1) to ( H6) be satisfied. Then there exists a solution
(u1 , u2 ) of problem (S) such that for i = 1, 2 , we have
ui ∈ Lpi (0, T ; Lpi (Ω)) ∩ L2 (0, T ; H01(Ω)) ∩ L∞ (t0 , T ; L∞(Ω)), ∀t0 > 0
Proof: By theorem 3.2 in [8] ,we can choose u0i ∈ Lpi (QT )∩L2 (0, T ; H01(Ω))∩
∞
L (τ , T ; L∞(Ω)), for any τ > 0 such that :
 ∂b (u0 )
1
1

− 4u01 + f1 (x, u01 , 0) = 0
in QT
∂t
0
u1 = 0
in ST

b1 (u01 )t=0 = b1 (ϕ0 )
in Ω
EJQTDE, 2001 No. 5, p. 2


∂b2 (u02 )
∂t
u02 = 0
− 4u02 + f2 (x, 0, u02 ) = 0
in QT
in ST

b1 (u02 )t=0 = b1 (ψ 0 )
in Ω
and we construct
sequences of functions (un1 ) and (un2 ), such that :
 ∂b (untwo
1
1)

− 4un1 + f (x, un1 , u2n−1 ) = 0
in QT
(1.1)
∂t
n
u1 = 0
in ST
(1.2)

b1 (un1 )t=0 = b1 (ϕ0 )
in Ω
(1.3)
 ∂b (un )
 2∂t 2 − 4un2 + f2 (x, u1n−1 , un2 ) = 0
in QT
(1.4)
un2 = 0
in ST
(1.5)

b2 (un2 )t=0 = b2 (ψ 0 )
in Ω
(1.6)
We need lemma 1 and lemma 2 below to complete the proof of theorem 1.
From now on we denote by ci various positive constants independent of n.
Lemma 1
∀τ > 0, ∃cτ > 0 such that kuni kL∞ (τ ,T ;L∞ (Ω)) ≤ cτ .
(1.7)
Proof : For n = 0, (1.7) is proved in [7]. So, suppose (1.7) for (n − 1).
Multiplying (1.1) by |b1 (un1 )|k b1 (un1 ) and using (H2),(H5) , we obtain :
Z
Z
Z
1
n k+p1
n k+2
|b1 (un1 )|k+1 dx.
|b1 (u1 )|
dx ≤ c10
|b1 (u1 )|
dx + c9
k+2 Ω
Ω
Ω
Setting yk,n (t) = kb1 (un1 )kLk+2 (Ω) and using Holder’s inequality on both
sides, we have the existence of two constants λ > 0 and δ > 0 such that
dyk,n (t)
p1 −1
+ λyk,n
(t) ≤ δ,
dt
which implies from lemma 5.1([22]) that ∀t ≥ τ > 0
δ 1
1
yk,n (t) ≤ ( ) p1 −1 +
.
1
λ
[λ(p1 − 2)t] p1 −2
As k → ∞, we obtain
|un1 (t)| ≤ cτ
∀t ≥ τ > 0 .
The same holds also for un2 .
Lemma 2 ∀τ > 0, ∃ci = ci (τ , ϕ0 , ψ 0 ) > 0 :
kuni kL2 (0,T ;H 1 (Ω))
≤ c11,
kuni kL∞ (τ ,T ;H 1 (Ω)∩L∞ (Ω))
0
≤ c12
0
and
"Z
2
X
i=1
T
0
Z
Ω
2
|∇uni |
dx + c13
Z
T
0
Z
Ω
p
|uni | i
#
dx ≤ c14 .
EJQTDE, 2001 No. 5, p. 3
Proof of lemma 2 : Multiplying (1.1) by un1 and (1.4) by un2 , and adding ,
we get :
2
dX
dt i=1
Z
Ω
Ψ∗i (bi (uni ))dx
+
2 Z
X
i=1
2
Ω
|∇uni | dx + c15
2 Z
X
i=1
p
Ω
|uni | i dx ≤ c16 . (1.8)
But
R
R
|ϕ0 |L2 (Ω) + |ψ 0 |L2 (Ω) ≤ c ⇒ Ω Ψ∗1 (b1 (ϕ0 ))dx + Ω Ψ∗2 (b2 (ψ 0 ))dx ≤ c,
2 R R
2 R R
P
P
T
T
n pi
n 2
so we deduce that
|∇u
|
dx
+
c
17
i
0
Ω |ui | dx ≤ c18 .
0
Ω
i=1
i=1
Whence lemma 2.
From lemma 2 and Lemma 1, there is a subsequence uni (i = 1, 2) with the
following properties:
uni → ui weakly in L2 (0, T ; H01 (Ω))∩ Lpi (0, T ; Lpi (Ω))
bi (uni ) → χi weakly in L2 (0, T ; L2(Ω))
bi (uni ) → χi strongly in L2 (τ , T ; H −1 (Ω)) ( by the compactness result of
Aubin ( see [22])). By lemma 7([9]), we have χi = bi (ui ). Moreover,
f1 (., un1 , u2n−1 ) converges to f1 (., u1 , u2 ) in Lr (τ , T ; Lr (Ω)), ∀r ≥ 1, ∀τ ≥ 1
and f2 (., u1n−1 , un2 ) converges to f2 (., u1 , u2 ) in Lr (τ , T ; Lr (Ω)), ∀r ≥ 1;
taking the limit as n goes to ∞ , we deduce that ( u1 , u2 ) is a weak solution
of (S).
1.3. Uniqueness
Theorem 2.
Assume that f
1 and f2 verify :
∀M > 0, ∀N > 0, ∃cM > 0, cN > 0 :


 ∀u, u, v, v : |u| + |u| ≤ M and |v| + |v| ≤ N, we have
(H8)
|f (x, u, v) − f1 (x, u, v)|2 + |f2 (x, u, v) − f2 (x, u, v)|2 ≤


 1
cM (b1 (u) − b1 (u))(u − u) + cN (b2 (v) − b2 (v))(v − v).
Then (S) has a unique solutions (u, v) in QT .
Proof : Let (u, v) and ( u, v) be solutions of (S); then we have :
∂(b1 (u) − b1 (u))
− 4(u − u) = f1 (x, u, v) − f1 (x, u, v)
∂t
(1.9)
∂(b2 (v) − b2 (v))
− 4(v − v) = f2 (x, u, v) − f2 (x, u, v).
∂t
(1.10)
and
Multiplying (1.9) by w1 = (−4)−1 (b1 (u) − b1 (u)) and (1.10) by
w2 =h (−4)−1 (b2 (v) − b2 (v)) and adding, we get i
2
2
1 d
2 dt kb1 (u) − b1 (u)kH −1 (Ω) + kb2 (v) − b2 (v)kH −1 (Ω) +
b1 (u) − b1 (u), u − u)L2 (Ω) + (b2 (v) − b2 (v), v − v L2 (Ω) ≤
c kf1 (x, u, v) − f1 (x, u, v)kH −1 (Ω) kb1 (u) − b1 (u)kH −1 (Ω) +
EJQTDE, 2001 No. 5, p. 4
c kf2 (x, u, v) − f2 (x, u, v)kH −1 (Ω) kb2 (v) − b2 (v)kH −1 (Ω) .
(1.11)
From hypothesis (H8) we obtain
2
2
kf1 (x, u, v) − f1 (x, u, v)kH −1 (Ω) + kf2 (x, u, v) − f2 (x, u, v)kH −1 (Ω)
h
i
2
2
≤ c kf1 (x, u, v) − f1 (x, u, v)kL2 (Ω) + kf2 (x, u, v) − f2 (x, u, v)kL2 (Ω)
≤ ccM (b1 (u) − b1 (u), u − u)L2 (Ω) + ccN (b2 (v) − b2 (v), v − v)L2 (Ω) ,
(1.12)
where M = kukL∞ (0,T ;L∞ (Ω)) +kukL∞ (0,T ;L∞ (Ω)) and N = kvkL∞ (0,T ;L∞(Ω)) +
kvkL∞ (0,T ;L∞ (Ω)) .
Therefore, using Schwartz inequality in (1.11), the fact that ( bi , i = 1, 2 ) is
increassing andh (1.12), we deduce that
i
2
d
dt
2
kb1 (u) − b1 (u)kH −1 (Ω) + kb2 (v) − b2 (v)kH −1 (Ω) ≤
i
h
d
c dt kb1 (u) − b1 (u)k2H −1 (Ω) + kb2 (v) − b2 (v)k2H −1 (Ω) .
Thus, we deduce that b1 (u) = b1 (u) and b2 (v) = b2 (v), hence u = u and
v = v.
Remark 1. Theorem 1 establishes the existence of dynamical system
{S(t)}t≥0 which maps L2 (Ω)×L2 (Ω) into L2 (Ω)×L2 (Ω) such that S(t)(ϕ0 , ψ 0 ) =
(u1 (t), u2 (t)).
2. Global attractor
Proposition 1 Assume that (H1)-( H8) hold; then the solution (u1 , u2 ) of
system (S) satisfies :
|u1 (t)|L∞ (Ω) + |u2 (t)|L∞ (Ω) ≤ c(t0 )
∀t ≥ t0
(2.0)
and
2 Z
X
i=1
|∇ui |2 dx ≤ c
∀t ≥ t0 + r
(2.1)
Ω
Proof : Reasoning as the proof of lemma 1, we also have(2.0).
Multiplying the first equation of (S) by u1 and the second by u2 , by (H2)
and (2.5), we get :
EJQTDE, 2001 No. 5, p. 5
2
dX
dt i=1
Z
Ω
Ψ∗i (bi (ui ))dx +
Z
2
X
1
i=1
2
2
|∇ui | dx = −
Ω
2 Z
X
i=1
fi (x, u)ui dx ≤ c. (2.2)
Ω
For fixed r > 0 and τ > 0, integrate (2.2) on ]t, t + r[
∀t ≥ τ > 0
2 Z
X
i=1
t+r
t
Z
2
|∇ui | dxds ≤ c(τ ).
(2.3)
Ω
Multiplying the first equation of (S) by (u1 )t and the second by (u2 )t , we get
2 Z
2 Z
X
1 d X
∂ui 2
2
(
) dx =
|∇ui | dx) +
(b0i (ui )(
2 dt i=1 Ω
∂t
i=1 Ω
2 Z
X
i=1
2
fi (x, u)(ui )t ≤
Ω
1X
2
i=1
Z
Ω
(2.4)
2 Z
1X
∂ui 2
fi2 (x, u)
dx +
) dx.
(b0 (ui )(
b0i (ui )
2 i=1 Ω i
∂t
By (H7) and the properties of functions fi , we obtain:
#
" 2 Z
d X
|∇ui |2 dx ≤ c(τ ).
dt i=1 Ω
(2.5)
From the uniform Gronwall’s lemma see [22], we get (2.1).
Remark 2.
By proposition1 we deduce that there exist absorbing sets
in Lσ1 (Ω) × Lσ1 (Ω) for any σ i : 1 ≤ σ i ≤ +∞ and absorbing sets in H01 (Ω) ×
H01 (Ω); then assumptions (1.1) ,(1.4) and (1.12) in theorem 1.1 [22, p.23] are
2
satisfied with U = L2 (Ω) , so we have the following :
Theorem 2. Assume that (H1) - (H7) are satisfied. Then the semi-group
S(t) associated with the boundary
value problem (S)
possesses a maximal
attractor A, which is bounded in H01 (Ω) ∩ L∞ (Ω) × H01 (Ω) ∩ L∞ (Ω) , com
2
pact and connected in L2 (Ω) . Its domain of attraction is the whole space
2
2
L (Ω) .
3. A regularity property of the attractor
In this section we shall show supplementary regularity estimates on the solution of problem (S) and by use of them, we shall obtain more regularity on
the attractor obtained in section 3. We shall assume that
(H9) N ≤ 3 and bi is of class C 3 .
EJQTDE, 2001 No. 5, p. 6
Hereafter, we shall assume that there exist positive constants δ i > 0 and a
function Φ from RN +2 to R such that :
∂Φ
fi (x, u) = fi (u) − hi (x) = δ i ∂u
i
(H10)
fi satisfying (H3) to (H6) and hi ∈ L∞ (Ω).
We shall denote : r(t) =
2 R
P
i=1
0
Ω bi (ui )
0 2
ui dx.
Theorem 3 Let fi and bi satisfies hypothesis (H1) to (H10). Then the
solution ( u1 , u2 ) of problem (S) satisfies the following regularity estimates:
∂ bi (ui )
∈ L2 (t0 , +∞; L2 (Ω) ),
∂t
(3.0)
∂ ∇ui
∈ L2 (t0 , +∞; L2 (Ω) ),
∂t
(3.1)
and
ui ∈ H 2 (Ω) .
(3.2)
To prove this theorem, we need the following lemma:
Lemma 3 Under the assumptions of theorem 3, there exist constants C =
C(ϕ0 , ψ 0 ), such that for any T > 0 :
kui kL∞ (0,T,H 1 (Ω)) ≤ C < ∞
(3.3)
0
and
∂ui ∂t ≤ C < ∞.
(3.4)
L2 (QT )
(ui )
∂Φ
Proof of lemma 3 : Multiplying the equation ∂bi∂t
=0
−div [∇ui ]+δ i ∂u
i
1
by δi (ui )t and adding the two equations, we obtain :
Z
2
X
1
δ
i=1 i
Z
b0i (ui )(
QT
Z
2
X
∂ui 2
1
2
) dxdt +
|∇ui (., T )| dx =
∂t
2δ
i
Ω
i=1
1
[−Φ(., u1 (T ), u2 (T )) + Φ(., ϕ0 , ψ 0 ] dx =
2δ1
Ω
Z
1
|∇ϕ0 | dx +
2δ2
Ω
2
Z
2
Ω
|∇ψ 0 | dx.
(3.5)
Φ being continuous and (u1 , u2 ) bounded, we then obtain:
Z
2
X
γ
i
i=1
δi
Z
2
X
1
∂ui 2
2
) dxdt +
|∇ui (., T )| dx ≤ C(ϕ0 , ψ 0 ) ,
(
∂t
2δ
i Ω
QT
i=1
(3.6)
EJQTDE, 2001 No. 5, p. 7
whence (3.3) and (3.4).
Proof of theorem 3 :
fi (u1 , u2 ) = hi we obtain
Differentiating equation
0
b0i (ui )u00i + b00i (ui )(u0i )2 − div (∇ui )) +
∂bi (ui )
∂t
2
X
∂fi (u)
j=1
∂uj
− div [∇ui ] +
u0j = 0 .
(3.7)
Now multiplying (3.7) by u0i , and integrating over Ω gives


Z
2 X
2
2
2 Z
X
X
X
∂fi (u) 0  0
1
1 0

kui 0 kH 1 (Ω) +
b00 (ui )(u0i )3 dx +
r (t) +
u ui dx = 0 .
0
2
2 i=1 Ω i
∂uj j
Ω
i=1 j=1
i=1
(3.8)
The L∞ estimate and hypothesis ( H9) imply successively :


Z
2 Z
2
2 X
X
X
∂fi (u) 0  0
2

(u0i ) dx,
uj ui dx ≤ M
∂u
j
Ω
i=1 Ω
i=1 j=1
γ
2 Z
X
i=1
2
(u0i ) d ≤ r(t) ≤ M
Ω
2
X
2
kui 0 kH 1 (Ω) ,
0
(3.9)
(3.10)
i=1
and
2
1X
−
2 i=1
Z
Ω
b00 (ui )(u0i )3 dx ≤
2
X
Mi
i=1
2
3
|u0i |L3 (Ω) .
(3.11)
Since for N ≤ 3, H01 (Ω) is continuously imbedded in L6 (Ω), we then obtain
by Young’s inequality that :
9
3
18
3
|ui 0 |L3 (Ω) ≤ c |u0i |L4 2 (Ω) |u0i |L3 (Ω) ≤ c |u0i |L52 (Ω) +
1
2
kui 0 kH 1 (Ω) .
0
2
(3.12)
By (3.9),(3.10),(3.11) and (3.12), (3.7) becomes :
2
r0 (t) +
9
1X
2
kui 0 kH 1 (Ω) + cr(t) ≤ cr(t) 5 + cr(t) ≤ cr(t)2 + c.
0
2 i=1
(3.13)
On the other hand, using (2.4) we obtain :
2 Z
X
i=1
τ +r
τ
Z
2
Ω
b0i (ui ) (u0i ) dxdt ≤ cτ , for any τ ≥ t0 .
(3.14)
EJQTDE, 2001 No. 5, p. 8
Estimates (3.13) and (3.14) and the use of the uniform Gronwall’ lemma
thus gives
r(t) ≤ c(t0 ), for any
∀t ≥ t0 .
(3.15)
Now, by (3.15) and hypothesis (H1) , we get :
2
2 Z X
∂bi (ui )
dx ≤ M r(t) ≤ c(t0 ) f or any t ≥ t0 .
∂t
i=1 Ω
Then (3.0) is satisfied. Now, as we have :
−4ui = −fi (x, u) − bi (ui )t ∈ L∞ (t0 , +∞; L2 (Ω)),
then by (S) ui (., t) is in bounded subset of H 2 (Ω). Hence estimate (3.2)
follows.
For a solution
the ω − limit set by :
(u1 , u2 ) of (S), we define
w = (w1 , w2 ) ∈ H01 (Ω) × L∞ (Ω) ∩ H01 (Ω) × L∞ (Ω)
ω(ϕ0 , ψ 0 ) =
.
∃tn → +∞ u1 (., tn ) → w1 in L2 (Ω), u2 (., tn ) → w2 in L2 (Ω)
Corollary 1. Under the assumptions (H1) to (H10), we have ω(ϕ 0 , ψ 0 ) 6= ∅
and any (w1 , w2 ) ∈ ω(ϕ0 , ψ0 ) is a bounded weak solution of the stationary
problem
−∆ui + fi (x, u1 , u2 ) = 0 in Ω
ui = 0
on
∂Ω
Proof : From (3.3) we obtain ω(ϕ0 , ψ 0 ) 6= ∅. Setting wi = Lim ui (., tn )
n→∞
and w = (w1 , w2 ) ∈ ω(ϕ0 , ψ 0 ), we get that w = (w1 , w2 ) is a solution of the
Dirichlet problem for elliptic system. The proof is analogous to El Ouardi and
de Th´elin [12] and is omitted here.
Corollary 2. Under the assumptions (H1) to (H10), we have
2
A ⊂ W 2,6 (Ω) if N = 3
and
2
A ⊂ W 2,r (Ω) for all r < ∞ if N ≤ 2.
Proof: Taking the inner product of (4.7) with u00i , we get
2
2
2
P
P
P
2
4
2
d
ku0i k 1 ) ≤ c( |u0i | 1 +
|u0i | 2 ).
dt (
H (Ω)
0
i=1
i=1
H (Ω)
0
L (Ω)
i=1
By uniform Gronwall’s lemma, we get
2
P
i=1
Then
2
P
i=1
ku0i k
2
Lαi (Ω)
ku0i k
2
1 (Ω)
H0
≤ c, ∀t ≥ T.
≤ c, ∀t ≥ T for all t ≥ τ and αi = 6 if N = 3 or
1 ≤ αi < ∞ if N ≤ 2.
4. Dimension of the attractor A
EJQTDE, 2001 No. 5, p. 9
4.1 Linearized problem
Let (ϕ0 , ψ 0 ) ∈ A; then by theorem3, u(t) = (u1 (t), u2 (t)) belongs to a
2
bounded subset of H 2 (Ω) . This fact allows us to linearize the system ( S)
along u(t). Formally, the candidate for the linearized problem is

2
U = (U1 , U2 ) ∈ L2 (0, T ; H01(Ω)



2
P
∂fi
∂
0
(SL )
(b
(u
U
)
−
4U
+
i
i
i
i
∂t
∂uj Uj = 0

j=1


U (0) = (U1 (0), U2 (0)) = U0
The existence and uniqueness of solution can be deduced from (4.0) below
Ui ∈ L∞ (0, T ; L2(Ω)) ∩ L2 (0, T ; H01 (Ω)).
(4.0)
To deduce (4.0), Multiply the equation in (SL ) by b0i (ui )Ui , we obtain
2
2
X
1 d X 0
2
(
|b (ui )Ui |L2 (Ω) ) +
(∇ui , ∇(b0i (ui )Ui ))L2 (Ω)
2 dt i=1 i
i=1
=
2 X
2
X
∂fi
Uj , b0i (ui )Ui )L2 (Ω)
(
∂u
j
i=1 j=1
(4.1)
By the hypothesis on bi , we have
∇(b0i (ui )Ui ) = b0i (ui )∇Ui + b00i (ui )∇ui .Ui ,
(4.2)
(∇Ui , b00i (ui )∇ui .Ui )L2 (Ω) ≤ c |∇ui |L4 (Ω) |Ui |L4 (Ω) |∇Ui |L2 (Ω) ,
(4.3)
and
2 X
2
2
2 X
2
X
X
X
∂fi
2
0
Uj , bi (ui )Ui )L2 (Ω) ≤ M
|Uj Ui | 1 ≤ c
|Ui | 2 .
(
L (Ω)
L (Ω)
∂u
j
i=1 j=1
i=1
i=1 j=1
(4.4)
From (4.2) to (4.4), (4.1) becomes
2
2
X
1 d X 0
2
2
|Ui | 1 ≤
(
|b (ui )Ui |L2 (Ω) ) + γ
H0 (Ω)
2 dt i=1 i
i=1
c
X2
|Ui |
2
L2 (Ω)
≤c
i=1
2
X
2
|b0i (ui )Ui |L2 (Ω) .
i=1
By standard application of Gronwall’s inequality, we get (4.0).
4.2 Differentiability of the Semigroup
We assume that fi ∈ C 2 (R × R) (∀i = 1, 2). Let u0 =(ϕ0 , ψ 0 ), v0 =(ϕ0 , ψ 0 ),
S(t) be the solution of (S) and S 0 (t, u0 ) the solution of (SL ). The results of [6]
imply the following proposition :
EJQTDE, 2001 No. 5, p. 10
Proposition 3.
2
Assume (H1) to (H10), then for any (u0 , v0 ) ∈ L∞ (Ω) × H 2 (Ω) , we have
|S(t)v0 − S(t)u0 − S 0 (t, u0 )(v0 − u0 )| 2 2 ≤ c(t)o(|v0 − u0 | 2 2 )
(L (Ω))
(L (Ω))
We need the lemma 3 for the proof of proposition 3:
Let (u1 , u2 ) and (v1 , v2 ) be two solutions of (S) in A with (u1 (0), u2 (0)) =
(ϕ0 , ψ 0 ) and (v1 (0), v2 (0)) =(ϕ1 , ψ 1 ). Setting w1 = u1 −v1 and w2 = u2 −v2 , we
have
Lemme 3 Assume (H1) to (H10). For all T > 0, there exists c(T ) > 0 such
that for all t ∈ [0, T ] ,
2
X
2
|wi (t)|H 1 (Ω) ≤ c(T )
0
2
X
2
|wi (0)|H 1 (Ω) ,
0
(4.7)
i=1
i=1
t
2
X
|wi (t)|2H 1 (Ω) ≤ c(T )
0
i=1
2
X
|wi (0)|2L2 (Ω) ,
(4.8)
i=1
and
2
X
2
|wi0 (t)|L2 (0,T ;L2 (Ω)) ≤ c(T )
2
X
2
|wi (0)|H 1 (Ω)
0
(4.9)
i=1
i=1
Proof : We have



∂bi (ui )
∂t


− 4ui + fi (x, u) = 0
u(0) = (u1 (0), u2 (0)) = (ϕ0 ,ψ 0 )

∂bi (vi )
∂t
− 4vi + fi (x, v) = 0
v(0) = (v1 (0), v2 (0)) = (ϕ1 ,ψ 1 ) .
Thus, the difference wi = ui − vi satisfies
b0i (ui )wi0 − 4wi = [b0i (vi ) − b0i (ui )] vi0 + fi (v) − fi (u).
(4.10)
Setting
F11 =
R1
R 1 ∂f1
∂f1
0 ∂u1 (x, u1 +θ(u2 −u1 ), u2 )dθ, F21 = 0 ∂u2 (x, u1 , u2 +θ(u2 −u1 ))dθ,
R 1 ∂f2
R 1 ∂f2
0 ∂u1 (x, u1 + θ(u2 − u1 ), u2 )dθ and F22 = 0 ∂u2 (x, u1 , u2 + θ(u2 −
F12 =
u1 ))dθ,
(4.10) becomes
b0i (ui )wi0
− 4wi =
[b0i (vi )
−
b0i (ui )] vi0
2
X
+
Fij wj .
(4.11)
i=1
EJQTDE, 2001 No. 5, p. 11
We multiply (4.11) by
wi
b0i (ui )
2
2
X
1 dX
wi
2
))L2 (Ω) =
|wi |L2 (Ω) +
(∇wi , ∇( 0
2 dt i=1
b
i (ui )
i=1
2
2 X
2
X
X
b0 (vi ) − b0i (ui ) 0
wi
2
( i
v
,
w
)
+
(Fij wj , 0
)L2 (Ω)
i L (Ω)
i
0 (u )
b
b
i
i
i (ui )
i=1
i=1 j=1
(4.12)
And parallel to lemma 16 in [8], it is easy to see that (4.7), (4.8) and (4.9)
hold.
Proof of proposition 3 : It is similar to the proof for the lemma 15 in
[8, p.125] and is omited.
4.3 Dimension Estimates
Consider the linearized problem

Ui0 = Fi0 (ui (τ ))Ui

Ui = 0
(SL )

Ui (0) = ξ i
in Ω × R+
on ∂Ω × R
00
where Fi0 (ui (τ ))Ui =
1
b0i (ui (τ )) 4Ui
−
bi (ui (τ )) 0
b0i (ui (τ )) ui Ui
−
1
b0i (ui (τ ))
This problem can be rewritten as

 U 0 = F 0 (u(τ ))U
U =0
(L)

U (0) = ξ
where U =
U1
U2
,
0
F (u(τ )) =
F10 (u1 (τ ))
0
2
P
j=1
0
F10 (u1 (τ ))
∂fi
∂uj Uj .
.
Let U 1 , ......., U m be m solutions of (L) corresponding to the initial data
ξ , , ........, ξ m and Qm (τ ) be the orthogonal projector in H = L2 (Ω)×L2 (Ω) such
m
that Qm H ⊂ V = H01 (Ω) × H01 (Ω). If W k = (w1k , w2k k=1 is an orthonormal
basis of Qm (τ )H; then
2 P
m
m
P
P
(Fi0 (ui (τ ))Wik , Wik )L2 (Ω)
(F 0 (u(τ ))W k , W k )H =
T r(F 0 (u(τ ))◦Qm (τ ) =
k=1
i=1k=1
and
00
wk
b (u (τ ))
(Fi0 (ui (τ ))Wik , Wik )L2 (Ω) = (4wik , b0 (uii(τ )) )L2 (Ω) − ( bi0 (uii(τ )) u0i wik , wik )L2 (Ω) −
i
i
2
P
∂fi k
1
k
( b0 (ui (τ ))
∂uj wj , wi )L2 (Ω) .
i
j=1
Since u = (u1 , u2 ) ∈ (L∞ (0, +∞, L∞ (Ω)))2 , we have bi (ui ), b0i (ui ),
EJQTDE, 2001 No. 5, p. 12
00
bi (ui ) ∈ (L∞ (0, +∞, L∞ (Ω)))2 , so
2
X
(Fi0 (ui (τ ))Wik , Wik )L2 (Ω)
≤
i=1
2
X
(4wik ,
i=1
2 Z
X
2
wik
)L2 (Ω) + c
|u0i | wik dx +
b0i (ui (τ ))
i=1 Ω
Z X
2 X
2
∂fi (x, u)
Ω i=1 j=1
∂uj
wjk wik )dx.
(4.13)
Now, we have
2
X
(4wik ,
i=1
where
J=
2
X
k 2
wik
w 1
2 (Ω) ≤ −c
)
L
i H (Ω) + cJ
0
0
bi (ui (τ ))
i=1
Z X
2
2
2 Z
X
X
k 2
k 2
wi 1 ) 12 (
wi dx) 21
∇wik wik |∇ui | dx ≤ c(
H (Ω)
i=1
Ω
0
i=1
2
≤
and
cX
w k 2 1
i H (Ω) + c(
0
2 i=1
Z X
2 X
2
∂fi (u)
Ω i=1 j=1
∂uj
Ω i=1
Z X
2
k 2
w dx)
i
(4.14)
Ω i=1
wjk wik )dx ≤ c(
Z X
2
k 2
wi dx).
(4.15)
Ω i=1
According to (4.14) and (4.15), relation (4.13) becomes
2
X
(Fi0 (ui (τ ))Wik , Wik )L2 (Ω) ≤ −c
i=1
2
X
k 2
w i +c
H 1 (Ω)
i=1
c(
T r(F 0 (u(τ )) ◦ Qm (τ ) ≤ −c105
H01 (Ω)
+
2 X
m
X
k 2
wi +c
H 1 (Ω)
0
2 X
m
X
k 2
w i i=1 k=1
Z X
2 X
m
2 X
m Z
X
k 2
2
w dx) + c
|u0i | wik dx.
i
Ω i=1 k=1
(4.16)
Ω
i=1
i=1 k=1
c(
i=1
Z X
2
2 Z
X
k 2
2
w dx) + c
|u0i | wik dx
i
Ω i=1
Which leads to
0
2
X
k 2
wi i=1 k=1
H01 (Ω)
+
(4.17)
Ω
EJQTDE, 2001 No. 5, p. 13
m 2 P
m P
P
wk (x)2 =
wk (x)2 and γ(t) = max(|u01 (t)| , |u02 (t)| ,
We set ρ(x) =
i
i=1k=1
k=1
R
5
θ(t) = Ω γ(t) 2 dx. So, we get
2 X
m Z
X
i=1 k=1
2
|u0i | wik dx ≤
Ω
Z
γ(t)ρ(x)dx ≤ c
Ω
Z
5
γ(t) 2 dx
Ω
52 Z
5
ρ 3 dx
Ω
35
Therefore by theorem 4.1 in [22] , there exists c01 > 0 such that :
Z
m
X
k 2
5
w (x) 1 .
ρ 3 dx ≤ c01
H (Ω)
Ω
0
k=1
.
(4.18)
(4.19)
So, we get
m Z
2 X
X
i=1 k=1
Ω
2
|u0i | wik dx ≤ c
Z
5
γ(t) 2 dx +
Ω
m
cX
wk (x)2 1 .
H0 (Ω)
5
k=1
From (4.18) to (4.19), (4.16) becomes
Z
Z
m
X
k 2
5
ρdx + c
γ(t) 2 dx,
T r(F (u(τ )) ◦ Qm (τ ) ≤ −c
w H 1 (Ω) + c
0
0
k=1
Ω
Ω
and as in [22] , we obtain :
2
Setting
T r(F 0 (u(τ )) ◦ Qm (τ ) ≤ −cm1+ N + c0 m + θ(t).
(4.20)
n R
o
1 t
0
qm (t) = sup
sup
t 0 T r(F (S(τ )u0 )) ◦ Qm (τ )dτ
u0 ∈A
and
ξ i ∈H, |ξ i |≤1
i=1,....,m
qm = Lim sup qm (t).
t→+∞
Rη
Then, by lemma 15 in [8, p.119] , we have 0 θdτ ≤ c(η) and u0i ∈ L∞ (η, +∞, L∞ (Ω)).
2
2
1+ N
+ c0 m + c0 (η) and qm ≤ −cm1+ N + c0 m + c0 (η)
Thus, qm (t) ≤ c(η)
t − cm
2
and for all integers j > 0, we get µ1 + µ2 ......... + µj ≤ qj ≤ −cj 1+ N +
0
0
c j + c (η).
Hence
µ1 + µ2 ......... + µm < 0
for any
m < c00 .
(4.21)
This shows that the fractal dimension of the attractor A is finite and arguing
as for theorem V3.3 in [22], we conclude to the following :
Theorem 4. Assume (H1) to (H10) and let m be an integer satisfying
(4.21). Then
EJQTDE, 2001 No. 5, p. 14
(i)
dimF ractale (A) ≤ 2m
( ii )
dimH (A) ≤ m.
Acknowledgement. The authors would like to thank the referee for his
refereeing and helpful comments.
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A.El Hachimi: UFR Math´ematiques Appliqu´ees et Industrielles
Facult´e des Sciences, B.P 20, El Jadida - Maroc
e-mail adress: [email protected]
H. El Ouardi: UFR Math´ematiques Appliqu´ees et Industrielles
Facult´e des Sciences, B.P 20, El Jadida - Maroc
and
Ecole Nationale Sup´erieure d’Electricit´e et de M´ecanique
B.P 8118-Casablanca-Oasis - Maroc
e mail adress: [email protected]
EJQTDE, 2001 No. 5, p. 16