Attractors of reaction-diffusion systems via a preconditioning …lmb.univ-fcomte.fr/IMG/pdf/Andreianov-talkBordeaux2-2011... · 2011. 7. 13. · Attractors of reaction-diffusion

Concrete examples and questions Results and Assumptions Ingredients of the Proof Steps of the Proof

Attractors of reaction-diffusion systemsvia a preconditioning technique

B. Andreianov1

based on joint work withHalima Labani (El-Jadida, Maroc)

1Laboratoire de Mathématiques CNRS UMR6623Université de Franche-Comté

Besançon, France

Seminar at Victor Segalen (Bordeaux 2) UniversityBordeaux, March 2011


Plan of the talk

1 Concrete examples and questions

2 Results and Assumptions

3 Ingredients of the Proof

4 Steps of the Proof


CONCRETE EXAMPLES

AND GOAL OF THE STUDY


Concrete examples to be treated

The motivation comes from the concrete example of 3× 3 “facilitateddiffusion” system with different boundary conditions (BC) modellingthe blood oxigenation reaction Hb + O2 HbO2 , or of 5× 5 systemmodelling coupled reactions Hb + O2 HBO2, Hb + CO2 HbCO2 .Let Ω be a bounded, smooth enough domain of Rn. Consider ∂tu1 − d1∆u1 = u3 − u1u2

∂tu2 − d2∆u2 = u3 − u1u2∂tu3 − d3∆u3 = u1u2 − u3,

(1)

∂tu1 − d1∆u1 = −K1u1u5 + K2u2∂tu2 − d2∆u2 = K1u1u5 − K2u2∂tu3 − d3∆u3 = −K3u3u5 + K4u4∂tu4 − d4∆u4 = K3u3u5 − K4u4∂tu5 − d5∆u5 = (−K1u1u5 + K2u2) + (−K3u3u5 + K4u4)

(2)

with non-homogeneous BC of the following general form:

λi∂nui + (1−λi )ui =αi on ∂Ω, αi > 0, i = 1..3 or i = 1..5. (3)

Here 0 6 λi 6 1, and: λi 6≡ λ for all i is a difficulty.




∂tu2 − d2∆u2 = u3 − u1u2∂tu3 − d3∆u3 = u1u2 − u3,

(1)


(2)







∂tu2 − d2∆u2 = u3 − u1u2∂tu3 − d3∆u3 = u1u2 − u3,

(1)


(2)







∂tu2 − d2∆u2 = u3 − u1u2∂tu3 − d3∆u3 = u1u2 − u3,

(1)


(2)





Issues of interest. Estimates and (E.A.T) estimates.

local in time existence of solutions : standard,one writes a fixed-point formulation in L∞ from the Duhamel formula(solution := mild solution ≡ strong solution in L2 ≡ ...)global existence requires uniform L∞ bounds:

‖U(t)‖∞ 6 Ψ(‖U0‖∞)

where Ψ denotes a generic non-decreasing function on R+

existence of a maximal attractor in L∞ relies upon:

— compactness of the linear semigroups e−tdi ∆

— a bounded absorbing set that is obtained via“estimates of attractor type” (E.A.T.) in L∞:

‖U(t)‖∞ 6 Φ(‖U0‖∞ , t

)where for all t , Ψ(·, t) is non-decreasing and

supr>0

limt→∞

Φ(r , t) 6 const .

NB: without loss of generality, Φ(r , ·) can be assumed non-increasing.Additional dependence is in subscripts. With this notation, we have e.g.supt∈R Φ(r , t) = Ψ(r), C + e−δt Ψp(r) + Φ(r , t) = Φδ,p(r , t),

Ψ(r)1l[0,2δ)(t) + Φ(r , t − 2δ)1l[2δ,+∞)(t) = Φδ(r , t).




‖U(t)‖∞ 6 Ψ(‖U0‖∞)





‖U(t)‖∞ 6 Φ(‖U0‖∞ , t


supr>0

limt→∞

Φ(r , t) 6 const .






‖U(t)‖∞ 6 Ψ(‖U0‖∞)





‖U(t)‖∞ 6 Φ(‖U0‖∞ , t


supr>0

limt→∞

Φ(r , t) 6 const .






‖U(t)‖∞ 6 Ψ(‖U0‖∞)





‖U(t)‖∞ 6 Φ(‖U0‖∞ , t


supr>0

limt→∞

Φ(r , t) 6 const .






‖U(t)‖∞ 6 Ψ(‖U0‖∞)





‖U(t)‖∞ 6 Φ(‖U0‖∞ , t


supr>0

limt→∞

Φ(r , t) 6 const .






‖U(t)‖∞ 6 Ψ(‖U0‖∞)





‖U(t)‖∞ 6 Φ(‖U0‖∞ , t


supr>0

limt→∞

Φ(r , t) 6 const .




MAIN RESULTS.ABSTRACT FRAMEWORK

AND ASSUMPTIONS


Results for the concrete systems above...

Assume the following :

U0 = (u01 ,u

02 ,u

03) in

((L∞)+

)3

αi ∈ L∞(∂Ω) are > 0 and belong to the ad hoc trace spaceAssume one of the three following situations occurs:

either λi ∈ (0,1), i = 1..3 , or λ1 = λ2 = λ3 = 0 ,or λi ∈ [0,1) with αi = 0 for i such that λi = 0 .

[ λi = 0 : Dirichlet ; λi = 1 : Neumann ; 0 < λi < 1 : Robin ]

Results :global existence of solutions in L∞ (⇒ there exists a nonlinearsemigroup S(t)t>0 on ((L∞(Ω))+)3 of solutions of the system)

existence of a maximal attractor in L∞ :∃M a compact set that is invariant for the semigroup S(t)t>0on ((L∞(Ω))+)3 and satisfies

∀ r > 0 limt→∞

supU0∈((L∞(Ω))+)3, ‖U0‖∞6r dist(S(t)U0,M

)= 0.

if Neumann BC also allowed : only global existence is proved.




U0 = (u01 ,u

02 ,u

03) in

((L∞)+

)3






∀ r > 0 limt→∞


)= 0.





U0 = (u01 ,u

02 ,u

03) in

((L∞)+

)3






∀ r > 0 limt→∞


)= 0.





U0 = (u01 ,u

02 ,u

03) in

((L∞)+

)3






∀ r > 0 limt→∞


)= 0.





U0 = (u01 ,u

02 ,u

03) in

((L∞)+

)3






∀ r > 0 limt→∞


)= 0.





U0 = (u01 ,u

02 ,u

03) in

((L∞)+

)3






∀ r > 0 limt→∞


)= 0.





U0 = (u01 ,u

02 ,u

03) in

((L∞)+

)3






∀ r > 0 limt→∞


)= 0.





U0 = (u01 ,u

02 ,u

03) in

((L∞)+

)3






∀ r > 0 limt→∞


)= 0.





U0 = (u01 ,u

02 ,u

03) in

((L∞)+

)3






∀ r > 0 limt→∞


)= 0.



Abstract framework and assumptions...

Following Bénilan and Labani, we recast the 3× 3 problem under theabstract form :

(S)

ddt

ui + Ai (ui−αi ) = fi (u1,u2,u3),

ui (0) = u0i , i = 1..3,

where for i = 1..3,(−Ai ) is the infinitesimal generator of an analytic exponentiallystable semigroup of positive linear operators e−tAi on L2(Ω) ;we assume that these semigroups are Lp-nonexpansive ;we assume that these semigroups are hypercontractive .

Further, in (S) we assume

αi ∈ (L∞(Ω))+ with e−tAi αi 6 αi , i = 1..3

To get from the concrete system to (S) one takes for αi the solution ofthe appropriately defined elliptic problem with BC given by αi :−di ∆αi = 0 in Ωλi∂nαi + (1− λi )αi = αi on ∂Ω.




(S)

ddt

ui + Ai (ui−αi ) = fi (u1,u2,u3),

ui (0) = u0i , i = 1..3,








(S)

ddt

ui + Ai (ui−αi ) = fi (u1,u2,u3),

ui (0) = u0i , i = 1..3,








(S)

ddt

ui + Ai (ui−αi ) = fi (u1,u2,u3),

ui (0) = u0i , i = 1..3,






...Abstract framework and assumptions...

The source terms fi , i = 1..3 in (S) are assumed to be locallyLipschitz continuous on (R+)3 and verify for all u1,u2,u3 ∈ R+:

[ “preservation of the positive cone” ]

f1(0,u2,u3) > 0, f2(u1,0,u3) > 0, f3(u1,u2,0) > 0; (4)

[ “some amount of compensation” ]

f1(u1,u2,u3) + f3(u1,u2,u3) 6 0; (5)

[ “linear growth in u3 for «good components» u1 and u2” ]

∃a > 0 f1(u1,u2,u3) 6 a(1 + u3), f2(u1,u2,u3) 6 a(1 + u3); (6)

[ “polynomial growth in u1,u2 for «bad component» u3” ]

∃b > 0, β > 0, γ > 0 f3(u1,u2,u3) 6 b(1 + uβ1 + uγ2 ). (7)





f1(0,u2,u3) > 0, f2(u1,0,u3) > 0, f3(u1,u2,0) > 0; (4)


f1(u1,u2,u3) + f3(u1,u2,u3) 6 0; (5)


∃a > 0 f1(u1,u2,u3) 6 a(1 + u3), f2(u1,u2,u3) 6 a(1 + u3); (6)


∃b > 0, β > 0, γ > 0 f3(u1,u2,u3) 6 b(1 + uβ1 + uγ2 ). (7)





f1(0,u2,u3) > 0, f2(u1,0,u3) > 0, f3(u1,u2,0) > 0; (4)


f1(u1,u2,u3) + f3(u1,u2,u3) 6 0; (5)


∃a > 0 f1(u1,u2,u3) 6 a(1 + u3), f2(u1,u2,u3) 6 a(1 + u3); (6)


∃b > 0, β > 0, γ > 0 f3(u1,u2,u3) 6 b(1 + uβ1 + uγ2 ). (7)





f1(0,u2,u3) > 0, f2(u1,0,u3) > 0, f3(u1,u2,0) > 0; (4)


f1(u1,u2,u3) + f3(u1,u2,u3) 6 0; (5)


∃a > 0 f1(u1,u2,u3) 6 a(1 + u3), f2(u1,u2,u3) 6 a(1 + u3); (6)


∃b > 0, β > 0, γ > 0 f3(u1,u2,u3) 6 b(1 + uβ1 + uγ2 ). (7)





f1(0,u2,u3) > 0, f2(u1,0,u3) > 0, f3(u1,u2,0) > 0; (4)


f1(u1,u2,u3) + f3(u1,u2,u3) 6 0; (5)


∃a > 0 f1(u1,u2,u3) 6 a(1 + u3), f2(u1,u2,u3) 6 a(1 + u3); (6)


∃b > 0, β > 0, γ > 0 f3(u1,u2,u3) 6 b(1 + uβ1 + uγ2 ). (7)


Preconditioning operator...

Finally, assume there exists a “preconditioning operator” B on L2(Ω)satisfying the same requirements as those imposed on Ai , i = 1..3(infinitesimal generator of a positive, analytic, exponentially stablesemigroup on L2, non-expansive in all Lp spaces, hypercontractive) ;and such that, for A = Ai , i = 1..3, the two properties hold:

[ “lower bound on B−1A” ](I−B−1A) 6 0 in the sense thatfor all u ∈ D(A) ∩ L∞(Ω), u > 0, one has u 6 B−1A u

[ “upper bound on B−1A” ](only needed for i such that αi 6≡ 0)

for all p < +∞ there exists Cp > 0 such thatfor all u ∈ D(A) ∩ L∞(Ω), u > 0, one has ‖B−1A u‖Lp(Ω) 6 Cp‖u‖L∞(Ω) .


























...Preconditioning operator...

Idea of the preconditioning:roughly speaking, the study of solutions of

( ddt + A

)(u − α) = f is

reduced to the study of solutions to( ddt

+ B)

(B−1(u − α)) = B−1f + (I−B−1Ai ) (u − α).

If αi = 0, the negativity of (I − B−1A) permits to upper bound the rhs.In general, also some bound for control of B−1A (u − α)− is needed.In practice, when does a preconditioner operator exist ?

Proposition

Let A be the operator associated with −d∆ on Ω with the BCλ∂nu + (1−λ)u = 0 on ∂Ω with parameter λ ∈ [0,1).Let B be another such operator with parameters e and µ. Then

(I − B−1A) 6 0 if 0 < e 6 d and λ 6 µ < 1B−1A is L∞-Lp bounded if either λ = µ = 0 or λ > 0.

Proof: maximum principle for (i); duality + Calderón-Zygmund for (ii).




( ddt + A

)(u − α) = f is


+ B)

(B−1(u − α)) = B−1f + (I−B−1Ai ) (u − α).


Proposition







( ddt + A

)(u − α) = f is


+ B)

(B−1(u − α)) = B−1f + (I−B−1Ai ) (u − α).


Proposition







( ddt + A

)(u − α) = f is


+ B)

(B−1(u − α)) = B−1f + (I−B−1Ai ) (u − α).


Proposition







( ddt + A

)(u − α) = f is


+ B)

(B−1(u − α)) = B−1f + (I−B−1Ai ) (u − α).


Proposition







( ddt + A

)(u − α) = f is


+ B)

(B−1(u − α)) = B−1f + (I−B−1Ai ) (u − α).


Proposition





MAIN INGREDIENTS


Properties and arguments used in the proof...

Definition

We say that A is an operator of class A if the following holds:• −A is the inf. generator of an analytic semigroup e−tA on L2(Ω)

• the semigr. e−tA is positive , in the sense e−tAu > 0 for u > 0;• e−tA is non-expansive on all spaces Lp(Ω) , i.e., for all t > 0,

∀p ∈ [1,+∞] ‖e−tAu‖p 6 ‖u‖p for u ∈ L∞(Ω);

• e−tA is exponentially stable on L2(Ω) , i.e. there exists ω > 0 st

for all t > 0 ‖e−tA‖L(L2) 6 e−ωt ;

• e−tA is hypercontractive , i.e., there exist σ > 0 and c > 0 such that

‖e−tAu‖L∞ 6 ctσ ‖u‖L1 .

Remark: The different Laplace operators Ai in our concrete examplesare of class A , provided λi < 1, i = 1..3 (Dirichlet or Robin BC case) .If λi = 1 (Neumann BC case), then for all c > 0, (Ai + cI) is of class A.



Definition



∀p ∈ [1,+∞] ‖e−tAu‖p 6 ‖u‖p for u ∈ L∞(Ω);








Definition



∀p ∈ [1,+∞] ‖e−tAu‖p 6 ‖u‖p for u ∈ L∞(Ω);








Definition



∀p ∈ [1,+∞] ‖e−tAu‖p 6 ‖u‖p for u ∈ L∞(Ω);








Definition



∀p ∈ [1,+∞] ‖e−tAu‖p 6 ‖u‖p for u ∈ L∞(Ω);








Definition



∀p ∈ [1,+∞] ‖e−tAu‖p 6 ‖u‖p for u ∈ L∞(Ω);








Definition



∀p ∈ [1,+∞] ‖e−tAu‖p 6 ‖u‖p for u ∈ L∞(Ω);







...Properties and arguments used in the proof...

Assume that A is of class A. Then(i) A−1 is bounded, and for all u > 0, one has A−1u > 0;

(ii) there exists c > 0 such that for all t > 0, ‖Ae−tA‖L(L2(Ω)) 6ct .

(iii) the unique mild solution of the evolution problemddt u + Au = 0, u(0) = u0 ∈ L2(Ω)

verifies the equation in the classical sense in L2(Ω), for all t > 0.Namely, u − u0 ∈ C0([0,∞); L2(Ω)) ∩ C1((0,∞); L2(Ω)), both d

dt u andAu are in C((0,+∞), L2(Ω)), and the equality holds in L2(Ω) for t > 0.

(iv) there exist σ > 0 and C > 0 such that for all p > 2 there exists λp > 0such that

‖e−tAu‖Lp(Ω) 6 C e−λp t t−σp ‖u‖Lp/2(Ω);

in addition, for all q > 1‖e−tAu‖L∞(Ω) 6 C t−

σq ‖u‖Lq (Ω).

(v) for all δ > 0 there exists a constant Cδ such that

∀ p ∈ [2,+∞) ∀t > δ ‖Ae−tA‖L(Lp(Ω),L∞(Ω)) 6 Cδ.

(vi) A−11 ∈ Lp(Ω) for all p < +∞.











σq ‖u‖Lq (Ω).



(vi) A−11 ∈ Lp(Ω) for all p < +∞.











σq ‖u‖Lq (Ω).



(vi) A−11 ∈ Lp(Ω) for all p < +∞.











σq ‖u‖Lq (Ω).



(vi) A−11 ∈ Lp(Ω) for all p < +∞.











σq ‖u‖Lq (Ω).



(vi) A−11 ∈ Lp(Ω) for all p < +∞.











σq ‖u‖Lq (Ω).



(vi) A−11 ∈ Lp(Ω) for all p < +∞.



Finally, recall the following maximal regularity statement (Lamberton).

Theorem

Assume that −A is the infinitesimal generator of an analyticsemigroup on L2(Ω), non-expansive in Lp. Let p ∈ (1,+∞).Then the unique mild solution of the evolution problem

ddt

u + Au = f ∈ Lploc([0,+∞)× Ω), u(0) = 0

verifies the equation in the strong sense in Lp(Ω). Namely,u ∈W 1,p

0 ([0,∞); Lp(Ω)), both ddt u and Au belong to Lp(Ω), and the

equality holds in Lp(Ω) for a.e. t > 0.Moreover, there exists Cp > 0 such that for all T > 0, the maximalregularity estimate holds:∥∥ d

dtu∥∥

Lp([0,T ]×Ω)+ ‖Au‖Lp([0,T ]×Ω) 6 Cp ‖f‖Lp([0,T ]×Ω).


STEPS OF THE PROOF


Step 1...

The proof of global existence + (E.A.T) consists in four Steps.Step 1 The following (E.A.T) hold:

∀i ,j =1..3 ∀p∈ [1,+∞) ∀t<Tmax

‖A−1j ui (t)‖Lp(Ω) 6 Φp(‖U0‖L∞(Ω), t);

∀i =1..3 ∀δ > 0 ∀t > δ ∀τ <Tmax

‖e−tAi ui (τ)‖L∞(Ω) 6 Φδ(‖U0‖L∞(Ω), τ).

Principle: apply ( ddt + B) to

w(t) := B−1((u1−α1) + (u3−α3)).

Because (S) is verified in the strong sense for t < Tmax , we getddt

w + Bw = B−1(f1 + f3) +∑

i=1,3

(I − B−1Ai

)(ui−αi )

w(0) = B−1((u01−α1) + (u0

3−α3))

in the strong sense (thus in the mild sense: Duhamel formula).


Step 1...


∀i ,j =1..3 ∀p∈ [1,+∞) ∀t<Tmax

‖A−1j ui (t)‖Lp(Ω) 6 Φp(‖U0‖L∞(Ω), t);

∀i =1..3 ∀δ > 0 ∀t > δ ∀τ <Tmax



w(t) := B−1((u1−α1) + (u3−α3)).


w + Bw = B−1(f1 + f3) +∑

i=1,3

(I − B−1Ai

)(ui−αi )

w(0) = B−1((u01−α1) + (u0

3−α3))



Step 1...


∀i ,j =1..3 ∀p∈ [1,+∞) ∀t<Tmax

‖A−1j ui (t)‖Lp(Ω) 6 Φp(‖U0‖L∞(Ω), t);

∀i =1..3 ∀δ > 0 ∀t > δ ∀τ <Tmax



w(t) := B−1((u1−α1) + (u3−α3)).


w + Bw = B−1(f1 + f3) +∑

i=1,3

(I − B−1Ai

)(ui−αi )

w(0) = B−1((u01−α1) + (u0

3−α3))



Step 1...


∀i ,j =1..3 ∀p∈ [1,+∞) ∀t<Tmax

‖A−1j ui (t)‖Lp(Ω) 6 Φp(‖U0‖L∞(Ω), t);

∀i =1..3 ∀δ > 0 ∀t > δ ∀τ <Tmax



w(t) := B−1((u1−α1) + (u3−α3)).


w + Bw = B−1(f1 + f3) +∑

i=1,3

(I − B−1Ai

)(ui−αi )

w(0) = B−1((u01−α1) + (u0

3−α3))



...Step 1...

From the compensation assumption on f1 + f3 and the assumptions onthe preconditioning operator B, we infer

ddt

w + Bw 6 g(t), ‖g(t)‖Lq (Ω) 6 Cq for all q < +∞,

(in the strong sense); by the Duhamel formula and the positivity of e−tB ,

w(t) 6 e−tBw(0) +

∫ t

0e−(t−s)Bg(s) ds, ‖g(t)‖Lq (Ω) 6 Cq .

The first term admits an (E.A.T.) (boundedness in terms of ‖U0‖∞ andexponential decay to zero in all Lp, p <∞).For p large , we have σ

p < 1 and take q = p/2. Then

‖w+(t)‖Lp(Ω) 6 Φp(‖U0‖L∞(Ω), t) + C∫ t

0eλp(t−s)(t−s)

−σp ‖g(s)‖Lp/2(Ω) ds

6 Φp(‖U0‖L∞(Ω), t) + Cp

∫ t

0e−λpzz−

σp dz

6 Φp(‖U0‖L∞(Ω), t) + Cp.

This yields the (E.A.T.) ‖w+(t)‖Lp(Ω) 6 Φp(‖U0‖L∞(Ω), t).The negative part is bounded by B−1(α1 + α2), whence the (E .A.T .)estimate on w(t) and then on B−1u1(t) and B−1u2(t) in Lp.


...Step 1...


ddt



w(t) 6 e−tBw(0) +

∫ t




‖w+(t)‖Lp(Ω) 6 Φp(‖U0‖L∞(Ω), t) + C∫ t

0eλp(t−s)(t−s)


6 Φp(‖U0‖L∞(Ω), t) + Cp

∫ t

0e−λpzz−

σp dz

6 Φp(‖U0‖L∞(Ω), t) + Cp.



...Step 1...


ddt



w(t) 6 e−tBw(0) +

∫ t




‖w+(t)‖Lp(Ω) 6 Φp(‖U0‖L∞(Ω), t) + C∫ t

0eλp(t−s)(t−s)


6 Φp(‖U0‖L∞(Ω), t) + Cp

∫ t

0e−λpzz−

σp dz

6 Φp(‖U0‖L∞(Ω), t) + Cp.



...Step 1...


ddt



w(t) 6 e−tBw(0) +

∫ t




‖w+(t)‖Lp(Ω) 6 Φp(‖U0‖L∞(Ω), t) + C∫ t

0eλp(t−s)(t−s)


6 Φp(‖U0‖L∞(Ω), t) + Cp

∫ t

0e−λpzz−

σp dz

6 Φp(‖U0‖L∞(Ω), t) + Cp.



...Step 1.

But then, because I − B−1Aj 6 0 and solution and semigroup are positive, wefind 0 6 A−1

j ui 6 B−1ui (for all j but i = 1, 3).It remains to bound B−1u2 (then we deduce bounds on A−1

j u2 as above). Wesimply apply ( d

dt + B) to B−1(u2 − α2) and argue as above.

The equation we get is

ddt

w + Bw = B−1f2 +(I − B−1A2

)(u2−α2)

6 a B−1(1 + u3) +(I − B−1A2

)(u2−α2),

taking into account the growth assumption on f2.

Because B−1u3 is already estimated, [by an (E.A.T)], we can conclude in thesame way.NB here sums, products, "chain rules" for (E.A.T) etc. are useful !

Finally, we can estimate e−tAi ui (τ) writing (Aie−tAi )A−1i :

we have A−1i ui estimated, and Aie−tAi is uniformly bounded for t > δ.


...Step 1.






ddt

w + Bw = B−1f2 +(I − B−1A2

)(u2−α2)

6 a B−1(1 + u3) +(I − B−1A2

)(u2−α2),






...Step 1.






ddt

w + Bw = B−1f2 +(I − B−1A2

)(u2−α2)

6 a B−1(1 + u3) +(I − B−1A2

)(u2−α2),






...Step 1.






ddt

w + Bw = B−1f2 +(I − B−1A2

)(u2−α2)

6 a B−1(1 + u3) +(I − B−1A2

)(u2−α2),






...Step 1.






ddt

w + Bw = B−1f2 +(I − B−1A2

)(u2−α2)

6 a B−1(1 + u3) +(I − B−1A2

)(u2−α2),






...Step 1.






ddt

w + Bw = B−1f2 +(I − B−1A2

)(u2−α2)

6 a B−1(1 + u3) +(I − B−1A2

)(u2−α2),






Step 2.

Step 2 The following (E.A.T) of «good components» hold:

∀i =1,2 ∀p∈ [1,+∞) ∀δ>0 ∀τ <Tmax−2δ

‖ui (τ + · )‖Lp((δ,2δ)×Ω) 6 Φδ,p(‖U0‖L∞(Ω), τ).

Moreover, if Tmax > 2δ, then

∀i =1,2 ∀p∈ [1,+∞) ∀δ>0 ∀t62δ

‖ui (·)‖Lp((0,t)×Ω) 6 Ψδ,p(‖U0‖L∞(Ω)).

Indeed, because we already have a bound on e−tA1u1(τ) , wecan fix the initial time at τ (i.e., consider u1(·+ τ)), assume theinitial datum to be zero, and write:

(ddt

+A1)[A−11 u1(·+τ)] = A−1

1 f1(U(·+τ) 6 A−11 a(1+u3)(·+τ).

(we applied A−11 to each term).The rhs is Lp bounded because of

the previous bound on A−11 u3; but then the maximum regularity

yields an Lp((δ, 2δ)× Ω) bound on A1[A−11 u1(·+ τ)] ≡ u1(·+ τ) .


Step 2.


∀i =1,2 ∀p∈ [1,+∞) ∀δ>0 ∀τ <Tmax−2δ



∀i =1,2 ∀p∈ [1,+∞) ∀δ>0 ∀t62δ

‖ui (·)‖Lp((0,t)×Ω) 6 Ψδ,p(‖U0‖L∞(Ω)).


(ddt

+A1)[A−11 u1(·+τ)] = A−1

1 f1(U(·+τ) 6 A−11 a(1+u3)(·+τ).





Step 2.


∀i =1,2 ∀p∈ [1,+∞) ∀δ>0 ∀τ <Tmax−2δ



∀i =1,2 ∀p∈ [1,+∞) ∀δ>0 ∀t62δ

‖ui (·)‖Lp((0,t)×Ω) 6 Ψδ,p(‖U0‖L∞(Ω)).


(ddt

+A1)[A−11 u1(·+τ)] = A−1

1 f1(U(·+τ) 6 A−11 a(1+u3)(·+τ).





Step 2.


∀i =1,2 ∀p∈ [1,+∞) ∀δ>0 ∀τ <Tmax−2δ



∀i =1,2 ∀p∈ [1,+∞) ∀δ>0 ∀t62δ

‖ui (·)‖Lp((0,t)×Ω) 6 Ψδ,p(‖U0‖L∞(Ω)).


(ddt

+A1)[A−11 u1(·+τ)] = A−1

1 f1(U(·+τ) 6 A−11 a(1+u3)(·+τ).





Step 3.

Step 3 The required (E.A.T) hold :

∀i =1..3 ∀δ > 0 ∀τ ∈ [2δ,Tmax ) ‖ui (τ)‖L∞(Ω) 6 Φδ(‖U0‖L∞(Ω), τ).

We first estimate the «bad component» u3; at the very end, with thisestimate in hand one easily bounds u1 and u2.Again, we can shift the initial moment (now at t = τ − δ) and drop the IC.Then

(ddt

+ A3)u3(·+τ−δ) = gτ , w(0) = 0,

with gτ (·) := f3(U(·+τ−δ)) 6 b(1 + |u1|β(·+τ−δ) + |u2|γ(·+τ−δ)

).

Use the Duhamel formula + Lp − L∞ regularizing effect + the (E.A.T)

∀p ∈ [1,+∞) ∀τ ∈ [2δ,Tmax ) ‖g+τ ‖Lp((0,δ)×Ω) 6 Φδ,p(‖U0‖L∞(Ω), τ −δ) :

in this way, for p large enough (i.e., for σ p′

p < 1) we get at t = δ:

‖u3(δ +τ−δ)‖L∞(Ω) 6∫ δ

0Cp(δ − s)

−σp ‖g+τ (s)‖Lp(Ω) ds

6 Cp

(∫ δ

0

((δ − s)

−σp)p′)1/p′

‖g+τ ‖Lp((0,δ)×Ω)

6 Cδ,pΦδ,p(‖U0‖L∞(Ω), τ − δ) = Φδ,p(‖U0‖L∞(Ω), τ).


Step 3.




(ddt

+ A3)u3(·+τ−δ) = gτ , w(0) = 0,


).





‖u3(δ +τ−δ)‖L∞(Ω) 6∫ δ

0Cp(δ − s)


6 Cp

(∫ δ

0

((δ − s)

−σp)p′)1/p′

‖g+τ ‖Lp((0,δ)×Ω)



Step 3.




(ddt

+ A3)u3(·+τ−δ) = gτ , w(0) = 0,


).





‖u3(δ +τ−δ)‖L∞(Ω) 6∫ δ

0Cp(δ − s)


6 Cp

(∫ δ

0

((δ − s)

−σp)p′)1/p′

‖g+τ ‖Lp((0,δ)×Ω)



Step 3.




(ddt

+ A3)u3(·+τ−δ) = gτ , w(0) = 0,


).





‖u3(δ +τ−δ)‖L∞(Ω) 6∫ δ

0Cp(δ − s)


6 Cp

(∫ δ

0

((δ − s)

−σp)p′)1/p′

‖g+τ ‖Lp((0,δ)×Ω)



Step 3.




(ddt

+ A3)u3(·+τ−δ) = gτ , w(0) = 0,


).





‖u3(δ +τ−δ)‖L∞(Ω) 6∫ δ

0Cp(δ − s)


6 Cp

(∫ δ

0

((δ − s)

−σp)p′)1/p′

‖g+τ ‖Lp((0,δ)×Ω)



Attractor.

Denote by E an absorbing set obtained from the (E.A.T).By the general result (see Bénilan and Labani; cf. Temam), under theadditional assumption of asymptotic compactness of the solution semigroup,

M =⋂t>0

⋃δ>0

S(t + δ)E is the maximal attractor in L∞.

[ maximal attractor = compact invariant setthat attracts the images of bounded sets as t →∞ ]

Indeed, it is not difficult to show thatM is invariant for S(t) and

∀ r > 0 limt→∞


)= 0.

By construction,M is bounded and closed. Because S(t)M =M, thecompactness of S(t) is enough to infer the compactness ofM.Thus it remains to deduce compactness of the nonlinear semigroup (S(t))t>0

from the compactness of the linear semigroups e−tAi . The only delicate pointis to bypass the continuity of the semigroup in L∞: indeed, S(t)t>0 is notcontinuous in the topology (L∞(Ω))3, thus the L2-continuity and the L2 − L∞

regularizing effect are used instead.


Attractor.


M =⋂t>0

⋃δ>0




∀ r > 0 limt→∞


)= 0.





Attractor.


M =⋂t>0

⋃δ>0




∀ r > 0 limt→∞


)= 0.





Attractor.


M =⋂t>0

⋃δ>0




∀ r > 0 limt→∞


)= 0.





Attractor.


M =⋂t>0

⋃δ>0




∀ r > 0 limt→∞


)= 0.





Oufff !!!

MERCI !

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Attractors of reaction-diffusion systems via a preconditioning …lmb.univ-fcomte.fr/IMG/pdf/Andreianov-talkBordeaux2-2011... · 2011. 7. 13. · Attractors of reaction-diffusion