AN EXTENSION OF A LYAPUNOV APPROACH TO THE STABILIZATION OF SECOND ORDER COUPLED SYSTEMS

https://doi.org/10.1051/cocv/2019075 www.esaim-cocv.org

AN EXTENSION OF A LYAPUNOV APPROACH TO THE STABILIZATION OF SECOND ORDER COUPLED SYSTEMS

Thierry Horsin

and Mohamed Ali Jendoubi

Abstract. This paper deals with the convergence to zero of the energy of the solutions of a second order linear coupled system. It revisits some previous results on the stabilization of such systems by exhibiting Lyapunov functions. The ones used are constructed according to some scalar cases situations.

These simpler situations explicitely show that the assumptions made on the operators in the coupled systems seem, first, natural and, second, give insight on their forms.

Mathematics Subject Classification.35B40, 49J15, 49J20.

Received March 27, 2019. Accepted November 27, 2019.

1. Introduction, functional framework

Let us consider a quite general coupled system in abstract form ( u⁰⁰+Bu⁰+A₁u+αCv= 0

v⁰⁰+A2v+αC^∗u= 0, (1.1)

whereA1andA2andCare, in general, unbounded operators. F. Alabauet al.considered in [4], the case when C =Id, and A1 and A2 are densely closed linear self-adjoint coercive operator and B is a coercive bounded self-adjoint operator. They proved that if |α|kCk<1 then the energy of the solution (u, v) in polynomially decreasing under quite large assumption on A1 and A2. In this paper our main concern is the caseA2 =A²₁ which is a special case of the aforementioned paper.

When A₁=A₂ and |α|kCk<1, A. Haraux and M.A. Jendoubi proved in [9] (see also [8]) the polynomial convergence to 0 of the energy by means of a Lyapunov method.

∗The first author wishes to thank the Tunisian Mathematical Society (SMT) for its kind invitation to its annual congress during which this work was completed.

∗∗The second author wishes to thank the department of mathematics and statistics EPN6 and the research department M2N (EA7340) of the CNAM where this work was initiated.

∗∗∗Both authors are grateful to the reviewers for their helpful comments and suggestions.

Keywords and phrases:damping, linear evolution equations, dissipative hyperbolic equation, decay rates, Lyapunov function.

1 Laboratoire M2N, EA7340, CNAM, 292 rue Saint-Martin, 75003 Paris France.

2 Universit´e de Carthage, Institut Pr´eparatoire aux Etudes Scientifiques et Techniques, B.P. 51 2070 La Marsa, Tunisia.

3 Laboratoire ´equations aux d´eriv´ees partielles, Facult´e des sicences de Tunis, Universit´e Tunis El Manar, Campus Universitaire El Manar, 2092 El Manar, Tunisia.

*∗∗∗Corresponding author:thierry.horsin@lecnam.net

c

The authors. Published by EDP Sciences, SMAI 2020

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

As we previously said, in this paper, we investigate such a method in the case when A2 =A²₁ and C=A^β₁ with β∈[0,³₂]. The main result of this paper is Theorem3.3 which also proves the polynomial convergence to 0 of the solution (u, v). Compared to the result in ([4], p. 144, Prop. 5.3), the convergence that we obtain is in weaker norms, but requires less regularity on the initial data.

Let us mention that the stabilization of such systems settled in abstract form has been widely studied and many results are connected to those in our paper. The case when A1 andA2are unbounded operators, andC is a bounded was considered by Alabau in [3] and by Alabau et al. in [5] where the polynomial statibility is proven. The optimal energy decay was proven in [11] in the caseA1=A2=A,C=IandB =A^γ whereγ <0.

The indirect stabilization of abstract coupled equations is also considered in [1,2,6,7].

Nonlinear damping are also currently investigated. See e.g.[12].

In order to motivate the Lyapunov function that we construct in the proof of our main result, we explain the strategy in Section2in the framework of a coupled scalar differential system.

In Section3 we introduce the functional framework and an existence theorem that lead to state and prove our main result, namely Theorem3.3.

2. A Lyapunov function for the scalar case

As mentioned in the preceding section, we consider the (real) scalar coupled system ( u⁰⁰+u⁰+λu+cv= 0

v⁰⁰+µv+cu= 0 (2.1)

where λ, µ >0, andc are such that 0< c²< λµ. The damping coefficient is set to 1 for simplicity but a time scale change reduces general damping terms bu⁰ to this case. In order to shorten the formulas, let us introduce for each solution (u, v) of (2.1), its total energy

E(u, u⁰, v, v⁰) =1 2

u⁰²+v⁰²+λu²+µv² +cuv.

Then we have for all t≥0

d

dtE(u, u⁰, v, v⁰) =−u⁰². Now we introduce

K(t) =1 2

u⁰²+v⁰²+λu²+µv² . Our first result is the following

Proposition 2.1. There are some constantsη >0,δ >0 such that

∀t≥0 K(t)≤ηe^−δtK(0).

Proof. For allε >0 we define the function

Hε=E −εvv⁰+ 2εuu⁰+3ε

2c(µu⁰v−λuv⁰). (2.2)

It is easy to check that

C1K(t)≤Hε(t)≤C2K(t) (2.3)

where

C₁=





√λµ− |c|

√λµ −ε





2 min√

λ,√

µ+ 3

2|c|max√ λ,√

µ







 (2.4)

and

C₂=





√λµ+|c|

√λµ +ε





2 min√

λ,√

µ+ 3

2|c|max√ λ,√

µ







. (2.5)

In fact, using Young’s inequality, we get

|cuv|= c

√λ√ µ

√λ|u|√

µ|v| ≤ |c|

√λ√ µ

λ 2u²+µ

2v²

|−εvv⁰|= ε

õ

√µ|v||v⁰| ≤ ε

√µ µ

2v²+v⁰² 2

|2εuu⁰|= 2ε

√λ

√λ|u||u⁰| ≤ 2ε

√λ λ

2u²+u⁰² 2

3ε 2cµu⁰v

= 3ε 2|c|

√µ|u⁰|√

µ|v| ≤ 3ε 2|c|

√µ u⁰²

2 +µ 2v²

− 3ε 2|c|λuv⁰

= 3ε 2|c|

√ λ√

λuv⁰ ≤ 3ε 2|c|

√ λ

λu²

2 +v⁰² 2

Then we deduce H_ε≤

1 +ε

2

√

λ+3√ µ 2|c|

| {z }

A₁

u⁰² 2 +

"

1 +ε 1

√µ+3√ λ 2|c|

!#

| {z }

A2

v⁰² 2 +

"

1 + |c|

√ λ√

µ+ε 2

√

λ+3√ λ 2|c|

!#

| {z }

A3

λ 2u²+

"

1 + |c|

√ λ√

µ+ε 1

√µ+3√ λ 2|c|

!#

| {z }

A4

µ 2v².

It is clear that for alli∈ {1,2,3,4}

Ai≤C2,

where C2 is given in (2.5). A similar proof gives similar inequalities for C1.

Letε1>0 the value ofεsuch thatC1 defined in (2.4) is equal to 0. Letε∈(0, ε1).In this caseC1 becomes positive.

A straightforward computation gives

H_ε⁰ =−u⁰²−εv⁰²−εvv⁰⁰+ 2εu⁰²+ 2εuu⁰⁰+3ε

2c[(µ−λ)u⁰v⁰+µu⁰⁰v−λuv⁰⁰]

=−(1−2ε)u⁰²−εv⁰²+εcuv+εµv²−2εuu⁰−pελu²−2εcuv +3ε

2c[(µ−λ)u⁰v⁰−µu⁰v−µλuv−µcv²+λcu²+λµuv]

=−(1−2ε)u⁰²−εv⁰²−λε1

2u²−µε1

2v²−2εuu⁰−εcuv

−µε3

2cu⁰v+ (µ−λ)ε3 2cu⁰v⁰. Now we have

λu²+ 2cuv+µv²=λ

u²+2c λuv+µ

λv²

=λ

u+ c λv²

+ µ

λ− c² λ²

v²

≥ λµ−c² λ v². Similarly, we get

λu²+ 2cuv+µv²≥ λµ−c² µ u², and then

λu²+ 2cuv+µv²≥ λµ−c² 2

1 µu²+ 1

λv²

Using Young’s inequality, we can find some constantsc₁, c₂, c₃>0 such that

|2uu⁰| ≤ 1

8µ(λµ−c²)u²+c1u⁰²; 3

2c|µu⁰v| ≤ 1

8λ(λµ−c²)v²+c2u⁰²; 3

2c|µ−λ||u⁰v⁰| ≤ 1

2v⁰²+c3u⁰². Finally we obtain

H_ε⁰ ≤ −(1−(2 +c1+c2+c3)ε)u⁰²−ε

2v⁰²− ε

8µ(λµ−c²)u²− ε

8µ(λµ−c²)v².

Now by choosing ε∈(0, ε₁) such that 1−(2 +c₁+c₂+c₃)ε >0,some constant C₃>0 can be found such that

H_ε⁰ ≤ −C3K(t).

By combining this with the inequality (2.3), we get for allt≥0 H_ε⁰(t)≤ −C3

C₂Hε(t).

We conclude the proof by integrating this last inequality and using (2.3) again.

3. The case A

= A

, and C = A

with β ∈ [0,

]

This section is devoted to the proof of Theorem 3.3. In order to proceed we first introduce the functional framework and give an existence theorem.

3.1. Functional framework

LetH be a separable Hilbert space, whose norm and scalar product will be denotedk · kandh·,·irespectively.

We consider A:H→H an unbounded closed self-adjoint operator such that the injection D(A)⊂H is dense and compact. We assume moreover throughout the paper that there exists a >0 such that

∀u∈D(A), hAu, ui ≥ahu, ui. (3.1)

Following for example the exposition given in [10], by denoting (λn)n∈N^∗the increasing sequence of eigenvalues ofA, the largestafor which (3.1) is true is λ₁.

Besides, let us consider (e_n)_n∈N^∗an orthonormal basis ofH constituted by eigenvectors ofA. For anyβ >0, we consider u=

∞

X

n=1

hu, e_iie_i,∈H (u∈H ⇐⇒

∞

X

n=1

hu, e_ii²<∞), and we setA^β :H →H given by

A^βu=

∞

X

i=1

λ^β_ihu, eiiei (3.2)

then (seee.g.[10])

D(A^β) ={u∈H,

∞

X

i=1

λ^2β_i hu, eii²<∞}

and A^β is an unbounded self-adjoint operator such that the inclusion D(A^β)⊂H is dense and compact. We also have

∀u∈D(A^β), hA^βu, ui ≥ahu, ui, (3.3)

for some a >0. The largestafor which this inequality is true beingλ^β₁.

As usual we write A⁰= Id. In this case of course the operatorA^αis a continuous linear operator onH. We will denote V =D(A^1/2) and W =D(A). Thus V and W are Hilbert spaces whose norms k · k_V and k · kW are given respectively by

kukV =kA^1/2uk, kukW =kAuk.

We have, if we identify H with its dual

W ⊂V ⊂H ⊂V⁰ ⊂W⁰ (3.4)

with dense and compact injections when the norms on the Hilbert spaces V⁰ andW⁰ are given by

∀u∈V⁰, kukV⁰ =hu, A⁻¹ui^1/2_V0,V

and

∀u∈W⁰, kukW⁰ =hu, A⁻²ui^1/2_W0,W

where h·,·iV⁰,V denotes the action ofV⁰ onV (with a similar notation forW). Of course whenu∈H one has kukV⁰ =hA⁻¹u, ui^1/2,kukW⁰ =hA⁻²u, ui^1/2.

Let us remark that with these definitions Amaps continuouslyV toV⁰ andA² mapsW to W⁰.

3.2. Existence result

Letαandβ two reals numbers withβ≥0. We recall that we consider the problem ( u⁰⁰+u⁰+Au+αA^βv= 0

v⁰⁰+A²v+αA^βu= 0 (3.5)

which can be rewritten as the first order system















u⁰−w= 0 v⁰−z= 0

w⁰+Au+w+αA^βv= 0 z⁰+A²v+αA^βu= 0.

(3.6)

Let us first establish an existence and uniqueness result for (3.5).

We concentrate on the caseβ ∈[¹₂,³₂], the caseβ ∈[0,¹₂) being easier.

Let us consider

H:=V ×W×H×H.

For two elements of HU_i= (u_i, v_i, w_i, z_i),i= 1,2, we define

hU1, U2iH :=hAu2, u1iV⁰,V +hA²v2, v1iW⁰,W+hw1, w2i+hz1, z2i +αhA^βv2, u1iV⁰,V +αhA^βu2, v1iW⁰,W,

where h·,·iV⁰,V denotes the usual duality pairing between V and V⁰, with similar notation for W, while h·,·i denotes the scalar product on H for which it is a Hilbert space.

It is straightforward to prove thath·,·i_Hdefines a scalar product onHfor which it is a Hilbert space provided that

|α|< λ

3−2β 2

1 . (3.7)

From now on we assume that (3.7) holds true.

We now consider the unbounded operator A:H −→ Hdefined by

D(A) :={U = (u, v, w, z)∈ H,(−w,−z, Au+αA^βv+w, A²v+αA^βu)∈ H}

and forU = (u, v, w, z)∈D(A)

AU = (−w,−z, Au+αA^βv+w, A²v+αA^βu).

It is clear that Ahas a dense domain inH.

Let us remark that for anyU = (u, v, w, z)∈D(A) one has

hAU, UiH =−hAu, wiV⁰,V − hA²v, ziW⁰,W +hAu+αA^βv+w, wi+hA²v+αA^βu, zi

−αhA^βv, wiV⁰,V −αhA^βu, ziW⁰,W, and thereforehAU, Ui=kwk²≥0. Indeed

hA²v+αA^βu, zi=hA²v+αA^βu, ziW⁰,W, sinceA^βu∈W⁰ forβ ∈[1,3/2] andu∈V =D(A^1/2).

Let us show that I+Ais onto. For this we take (f, g, h, k)∈ H. We want to find (u, v, w, z)∈D(A) such that

u−w=f v−z=g

Au+αA^βv+ 2u=h+ 2f A²+αA^βu+v=k+g.

We define Φ : (V ×W)²→Rby

Φ(u₁, v₁, u₂, v₂) :=hA^1/2u₁, A^1/2u₂i+hAv1, Av₂i+αhA^βv₁, u₂iV⁰,V + αhAu1, A^β−1v2iV⁰,V + 2hu1, u2i+hv1, v2i.

Clearly Φ is continuous on V ×W. It is also clear that Φ is coercive if we assume|α|< λ

3−2β 2

1 . By the Lax-Milgram theorem, there exists a unique (u, v)∈V ×W such that

∀(δu, δv)∈V ×W, Φ(u, v, δu, δv) =hh+ 2f, δui+hk+g, δvi.

We therefore get

A²v+αA^βu+v=g+k

Au+αA^βv+ 2u=h+ 2f.

Now if we denotew=u−f andz=v−gthenw∈V sinceu, f ∈V andz∈W sincev, g∈W. We have thus proven thatAis maximal monotone. By classical theory, we get that

Theorem 3.1. For any (u0, v0, u1, v1) ∈ H, there exists a unique solution to (3.5) in C([0, T],H)× C¹([0, T], D(A)⁰).

Remark 3.2. It is also well known that if (u0, v0, u1, v1) ∈ D(A) then the solution to (3.5) belongs to C([0, T], D(A))∩C¹(0, T,H).

3.3. Main result of the paper

Theorem 3.3. Assumeβ ∈[0,³₂]andα6= 0. Let(u, v)be a solution of (3.5)such that(u(0), u⁰(0), v(0), v⁰(0))∈ H, then there exists a constantc >0 such that

∀t >0, kA^β2−1u⁰(t)k²_W0+kA^β2−1v⁰(t)k²_W0 +kA^β−12 u(t)k²_W0+kA^β2v(t)k²_W0

≤c t

ku⁰(0)k²+kv⁰(0)k²+ku(0)k²_V +kv(0)k²_W

ifβ ∈[0,1];

∀t >0, kA^−β2 u⁰(t)k²_W0+kA^−β2 v⁰(t)k²_W0+kA^1−β2 u(t)k²_W0 +kA^1−β2v(t)k²_W0

≤c t

ku⁰(0)k²+kv⁰(0)k²+ku(0)k²_V +kv(0)k²_W

ifβ ∈[1,3 2].

Remark 3.4. If we replace (3.5) by

( u⁰⁰+Bu⁰+Au+αA^βv= 0 v⁰⁰+A²v+αA^βu= 0

(3.8)

where, as mentionned in the introduction, B is a bounded self-adjoint operator on H for which there exists µ >0 such that

hBu, ui ≥µkuk², the results of Theorem 3.1and Remark3.4remain true.

Remark 3.5. If we replace (3.5) by

( u⁰⁰+Bu⁰+Au+αA^βv= 0 v⁰⁰+A₂v+αA^βu= 0

where A₂ is a self-adjoint unbounded operator such thatD(A₂) =D(A²) and there exist ν₁, ν₂>0 such that

∀u∈D(A2), ν1hA²u, ui ≤ hA2u, ui ≤ν2hA²u, ui,

and ifBis as in the remark3.4, the result of Theorem3.3remains true provided|α|is small enough (depending onλ₁,ν₁ andν₂).

Remark 3.6. In the caseβ= 0, in order to obtain the decay of the energy, we must assume A²u(0)∈V, A²v(0)∈W, A²u⁰(0)∈H, A²v⁰(0)∈H.

In the paper [4], the authors obtain such a decay with merely

Au(0)∈V, A²v(0)∈W, Au⁰(0)∈H, A²v⁰(0)∈H.

We, of course, would expect that the energy decay also holds with (u(0), v(0), u⁰(0), v⁰(0))∈D(A) but unfortunately we are not able to prove it for the moment being.

Proof of Theorem 3.3. All the computations below will be made assuming that (u0, v0, u1, v1)∈D(A) which ascertains them. By density and continuity the inequalities stated in Theorem3.3remain true.

Let us also recall that αsatisfies|α|< λ

3−2β 2

1 . We introduce the energy of the system by

E(t) = 1 2

ku⁰(t)k²+kv⁰(t)k²+ku(t)k²_V +kv(t)k²_W

+αhA^βv, ui.

Then we have

E⁰(t) =−ku⁰(t)k². Letp >1 andε >0 two real numbers to be fixed later and let

H_ε=E−ελ^2−β₁ hA^β−2v, v⁰i_W⁰+pελ^−a₁ hA^au, u⁰i_W⁰+ρε

hu⁰, vi_W⁰ − hu, A⁻¹v⁰i_W⁰ where ρ=^p+1_2αλ^2−β₁ anda= min(0,1−β). We find easily

H_ε⁰ =−ku⁰k²−ελ^2−β₁ kA^β2−1v⁰k²_W0−ελ^2−β₁ hA^β−2v, v⁰⁰iW⁰+pελ^−a₁ kA^a2u⁰k²_W0

−pελ^−a₁ hA^au, u⁰+Au+αA^βviW⁰+ρεhu⁰, v⁰iW⁰−ρεhu⁰, A⁻¹v⁰iW⁰

−ρεhu⁰+Au+αA^βv, viW⁰+ρεhu, A⁻¹(A²v+αA^βu)iW⁰

=−ku⁰k²−ελ^2−β₁ kA^β2−1v⁰k²_W0+ελ^2−β₁ hA^β−2v, A²v+αA^βuiW⁰+pελ^−a₁ kA^a2u⁰k²_W0

−pελ^−a₁ hA^au, u⁰i_W⁰−pελ^−a₁ kA^a+12 uk²_W0−pελ^−a₁ αhA^au, A^βvi_W⁰+ρεhu⁰, v⁰i_W⁰

−ρεhu⁰, A⁻¹v⁰i_W⁰−ρεhu⁰, vi_W⁰−ρεαkA^β2vk_W⁰+ρεαkA^β−12 uk²_W0

=−ku⁰k²−ελ^2−β₁ kA^β2−1v⁰k²_W0+ελ^2−β₁ kA^β2vk²_W0+ελ^2−β₁ αhA^β−2v, A^βui_W⁰+pελ^−a₁ kA^a2u⁰k²_W0

−pελ^−a₁ hA^au, u⁰iW⁰−pελ^−a₁ kA^a+12 uk²_W0−pελ^−a₁ αhA^au, A^βviW⁰+ρεhu⁰, v⁰iW⁰

−ρεhu⁰, A⁻¹v⁰iW⁰−ρεhu⁰, viW⁰−ρεαkA^β2vk²_W0+ρεαkA^β−12 uk²_W0

=−ku⁰k²+pελ^−a₁ kA^a2u⁰k²_W0−ελ^2−β₁ kA^β2−1v⁰k²_W0−pελ^−a₁ kA^a+12 uk²_W0−εp−1

2 λ^2−β₁ kA^β2vk²_W0

+ρεαkA^β−12 uk²_W0+ελ^2−β₁ αhA^β−2v, A^βuiW⁰−pελ^−a₁ hA^au, u⁰iW⁰ −pελ^−a₁ αhA^au, A^βviW⁰

+ρεhu⁰, v⁰iW⁰−ρεhu⁰, A⁻¹v⁰iW⁰ −ρεhu⁰, viW⁰. First case :β ∈[0,1]. In this casea= 0. We have

H_ε⁰ =−ku⁰k²+pεku⁰k²_W0−ελ^2−β₁ kA^β2−1v⁰k²_W0−pεkA¹²uk²_W0−εp−1

2 λ^2−β₁ kA^β2vk²_W0

+ρεαkA^β−12 uk²_W0+ελ^2−β₁ αhA^β−2v, A^βuiW⁰−pεhu, u⁰iW⁰−pεαhu, A^βviW⁰

+ρεhu⁰, v⁰iW⁰−ρεhu⁰, A⁻¹v⁰iW⁰−ρεhu⁰, viW⁰.

Now since

kA^β−12 uk²_W0 ≤ 1

λ^2−β₁ hAu, ui_W⁰ = 1

λ^2−β₁ kA¹²uk²_W0, we get

H_ε⁰ ≤ −ku⁰k²+pεku⁰k²_W0−ελ^2−β₁ kA^β2−1v⁰k²_W0−p−1

2 εkA¹²uk²_W0−εp−1

2 λ^2−β₁ kA^β2vk²_W0

+ελ^2−β₁ αhA^β−2v, A^βuiW⁰ −pεhu, u⁰iW⁰ −pεαhu, A^βviW⁰

+ρεhu⁰, v⁰iW⁰−ρεhu⁰, A⁻¹v⁰iW⁰−ρεhu⁰, viW⁰. Let us remark that

ελ^2−β₁ αhA^β−2v, A^βuiW⁰ ≤ελ^2−β₁ |α|kA^β2vkW⁰kA^3β2−2ukW⁰

≤ελ^2−β₁ |α|kA^β2vkW⁰

1 λ

5−3β 2

1

kA¹²ukW⁰

≤ελ

β−1 2

1 |α|kA^β2vkW⁰kA¹²ukW⁰

and that

−pεαhu, A^βviW⁰ ≤pε|α|kA^β2vkW⁰kA^β2ukW⁰

≤pελ

β−1 2

1 |α|kA^β2vk_W⁰kA¹²uk_W⁰. Thus

ελ^2−β₁ αhA^β−2v, A^βuiW⁰−pεαhu, A^βviW⁰

≤ελ

β−1 2

1 |α|kA^β2vkW⁰kA¹²ukW⁰ +pελ

β−1 2

1 |α|kA^β2vkW⁰kA¹²ukW⁰

≤ελ

β−1 2

1 |α|(p+ 1)kA^β2vk_W⁰kA¹²uk_W⁰

≤ελ

β−1 2

1 (p+ 1)|α|

2

γkA¹²uk²_W0+1

γkA^β2vk²_W0

where we chooseγ >0 such that δ:= p−1

2 −λ

β−1 2

1 (p+ 1)|α|

2 γ >0, ζ:= p−1

2 λ^2−β₁ −λ

β−1 2

1 (p+ 1)|α|

2γ >0 (3.9)

which is equivalent to

λ

β−1 2

1 (p+ 1)|α|

(p−1)λ^2−β₁ < γ < p−1 λ

β−1 2

1 (p+ 1)|α|

.

This choice is possible provided that

p+ 1 p−1

2

< λ^3−2β₁

|α|² . (3.10)

Now we choose p >1 such that (3.10) is satisfied. Then we have

H_ε⁰ =−ku⁰k²+pεku⁰k²_W0−ελ^2−β₁ kA^β2−1v⁰k²_W0−εδkA¹²uk²_W0−εζkA^β2vk²_W0

−pεhu, u⁰iW⁰+ρεhu⁰, v⁰iW⁰ −ρεhu⁰, A⁻¹v⁰iW⁰−ρεhu⁰, viW⁰. Now let us observe first that if one considers someU ∈W⁰ one has

kUk²_W0 =hU, A⁻²UiW⁰,W =hA^−1/2U, A^−3/2UiV⁰,V =kA^−1/2Uk²_V0. (3.11) Second, since u⁰ ∈V, one hasAu⁰∈V⁰. Therefore according to (3.11) we have

kAu⁰k²_W0 =kA^1/2u⁰k²_V0. (3.12) MoreoverA^1/2u⁰∈H, thus, one has

kAu⁰k²_W0 =kA^1/2u⁰k²_V0 =hA⁻¹A^1/2u⁰, A^1/2u⁰i=ku⁰k². (3.13) According to (3.13), since

hu⁰, v⁰iW⁰ =hAu⁰, A⁻¹v⁰iW⁰, we have

|hu⁰, v⁰iW⁰| ≤ |hAu⁰, A⁻¹v⁰iW⁰|

≤ kAu⁰k_W⁰kA⁻¹v⁰k_W⁰

=ku⁰kkA⁻¹v⁰kW⁰. Yet we have

kA⁻¹v⁰k²_W0 ≤ 1

λ^β₁hA^βA⁻¹v⁰, A⁻¹v⁰i_W⁰ = 1

λ^β₁kA^β2−1v⁰k²_W0. Using Young’s inequality, we find some constants c1, c2, c3, c4>0 such that

−phu, u⁰i_W⁰ ≤c₁ku⁰k²+δ

2kA¹²uk²_W0; (δdefined by (3.9)) ρhu⁰, v⁰iW⁰ ≤c2ku⁰k²+λ^2−β₁

3 kA^β2−1v⁰k²_W0

ρhu⁰, A⁻¹v⁰iW⁰ ≤c3ku⁰k²+λ^2−β₁

3 kA^β2−1v⁰k²_W0

−ρhu⁰, viW⁰ ≤c₄ku⁰k²+ζ

2kA^β2vk²_W0 (ζ defined by (3.9)).

By choosing εsmall enough, we find a constantη=η(p, ε)>0 such that for all t≥0 H_ε⁰ ≤ −η

ku⁰k²+kA^β2−1v⁰k²_W0+kA¹²uk²_W0+kA^β2vk²_W0

.

Let

E˜= 1 2

hkA^β2−1u⁰(t)k²_W0+kA^β2−1v⁰(t)k²_W0 +kA^β−12 u(t)k²_W0+kA^β2v(t)k²_W0

i

+αhA^2β−2v, uiW⁰

and

K(t) =kA^β2−1u⁰(t)k²_W0+kA^β2−1v⁰(t)k²_W0+kA^β−12 u(t)k²_W0+kA^β2v(t)k²_W0. For allt≥0, we have

E˜⁰=−kA^β2−1u⁰(t)k²_W0. Then ˜E is nonincreasing. Observe that

αhA^2β−2v, uiW⁰

≤ |α|

λ

3−2β 2

1

kA^β−12 u(t)kW⁰kA^β2v(t)kW⁰,

from which we deduce that

λ

3−2β 2

1 − |α|

2λ

3−2β 2

1

K(t)≤E˜ ≤λ

3−2β 2

1 +|α|

2λ

3−2β 2

1

K(t). (3.14)

Now since

kA^β2−1u⁰(t)kW⁰ ≤ 1 λ²⁻

β 2

1

ku⁰k, kA^β−12 u(t)kW⁰≤ 1 λ¹⁻

β 2

1

kA¹²ukW⁰,

then there exists a constant γ >0 such that for allt≥0

H_ε⁰ ≤ −γK(u, v, u⁰, v⁰). (3.15) From (3.15), assumingεpossibly smaller in order to achieve positivity of the quadratic formHε, we get

Z t 0

K(u(s), v(s), u⁰(s), v⁰(s))ds≤ 1

γHε(u(0), v(0), u⁰(0), v⁰(0)).

Using inequality (3.14), we obtain 2λ

3−2β 2

1

λ

3−2β 2

1 +|α|

Z t 0

E(u(s), v(s), u˜ ⁰(s), v⁰(s))ds≤ 1

γHε(u(0), v(0), u⁰(0), v⁰(0)).

Now since ˜E is nonincreasing, it follows

E(u(t), v(t), u˜ ⁰(t), v⁰(t))≤ λ

3−2β 2

1 +|α|

2λ

3−2β 2

1 γ

1

tHε(u(0), v(0), u⁰(0), v⁰(0)).

Using inequality (3.14) we get

K(u(t), v(t), u⁰(t), v⁰(t))≤ λ

3−2β 2

1 +|α|

(λ

3−2β 2

1 − |α|)γ 1

tHε(u(0), v(0), u⁰(0), v⁰(0)).

Second case :β∈(1,³₂]. In this casea= 1−β.

H_ε⁰ =−ku⁰k²+pελ^β−1₁ kA^1−β2 u⁰k²_W0−ελ^2−β₁ kA^β2−1v⁰k²_W0−pελ^β−1₁ kA^1−β2uk²_W0−εp−1

2 λ^2−β₁ kA^β2vk²_W0

+ρεαkA^β−12 uk²_W0+ελ^2−β₁ αhA^β−2v, A^βuiW⁰−pελ^β−1₁ hA^1−βu, u⁰i

−pελ^β−1₁ αhA^1−βu, A^βvi_W⁰+ρεhu⁰, v⁰i_W⁰−ρεhu⁰, A⁻¹v⁰i_W⁰−ρεhu⁰, vi_W⁰. Now since

kA^β−12 uk²_W0 ≤ 1

λ^3−2β₁ hA^2−βu, uiW⁰ = 1

λ^3−2β₁ kA^1−β2uk²_W0, we get

H_ε⁰ ≤ −ku⁰k²+pελ^β−1₁ kA^1−β2 u⁰k²_W0−ελ^2−β₁ kA^β2−1v⁰k²_W0−p−1

2 ελ^β−1₁ kA^1−β2uk²_W0

−εp−1

2 λ^2−β₁ kA^β2vk²_W0+ελ^2−β₁ αhA^β−2v, A^βuiW⁰−pελ^β−1₁ hA^1−βu, u⁰i

−pελ^β−1₁ αhA^1−βu, A^βviW⁰+ρεhu⁰, v⁰iW⁰−ρεhu⁰, A⁻¹v⁰iW⁰−ρεhu⁰, viW⁰

Let us remark that

ελ^2−β₁ αhA^β−2v, A^βuiW⁰ ≤ελ^2−β₁ |α|kA^β2vkW⁰kA^3β2−2ukW⁰

≤ελ^2−β₁ |α|kA^β2vkW⁰

1

λ^3−2β₁ kA^1−β2ukW⁰

≤ελ^β−1₁ |α|kA^β2vkW⁰kA^1−β2ukW⁰

and that

−pελ^β−1₁ αhA^1−βu, A^βvi_W⁰ ≤pελ^β−1₁ |α|kA^β2vk_W⁰kA^1−β2uk_W⁰, we therefore get

ελ^2−β₁ αhA^β−2v, A^βui_W⁰−pελ^β−1₁ αhA^1−βu, A^βvi_W⁰

≤ελ^β−1₁ |α|kA^β2vkW⁰kA^1−β2ukW⁰ +pελ^β−1₁ |α|kA^β2vkW⁰kA^1−β2ukW⁰

=ε(p+ 1)λ^β−1₁ |α|kA^β2vkW⁰kA^1−β2ukW⁰

≤ελ^β−1₁ (p+ 1)|α|

2

γkA^1−β2uk²_W0+1

γkA^β2vk²_W0

where we chooseγ >0 such that δ:=p−1

2 λ^β−1₁ −λ^β−1₁ (p+ 1)|α|

2 γ >0, ζ:= p−1

2 λ^2−β₁ −λ^β−1₁ (p+ 1)|α|

2γ >0 which is equivalent to

λ^β−1₁ (p+ 1)|α|

(p−1)λ^2−β₁ < γ < p−1 (p+ 1)|α|. This choice is possible provided that

p+ 1 p−1

²

< λ^3−2β₁

|α|² . (3.16)

We choose p >1 such that (3.16) is satisfied. Then we have

H_ε⁰ ≤ −ku⁰k²+pελ^β−1₁ kA^1−β2 u⁰k²_W0−ελ^2−β₁ kA^β2−1v⁰k²_W0−εδkA^1−β2uk²_W0−εζkA^β2vk²_W0

−pελ^β−1₁ hA^1−βu, u⁰i+ρεhu⁰, v⁰iW⁰−ρεhu⁰, A⁻¹v⁰iW⁰ −ρεhu⁰, viW⁰. By use of Young’s inequality we deduce that there arec1, c2, c3, c4>0 such that

−pλ^β−1₁ hA^1−βu, u⁰i ≤c1ku⁰k²+δ

2kA^1−β2uk²_W0, ρhu⁰, v⁰iW⁰ ≤c2ku⁰k²+λ^2−β₁

3 kA^β2−1v⁰k²_W0, ρεhu⁰, A⁻¹v⁰iW⁰ ≤c3ku⁰k²+λ^2−β₁

3 kA^β2−1v⁰k²_W0,

−ρhu⁰, vi_W⁰ ≤c₄ku⁰k²+ζ

2kA^β2vk²_W0.

By choosingεsmall enough, we find a constant η=η(p, ε)>0 such that for allt≥0 H_ε⁰ ≤ −η

ku⁰k²+kA^β2−1v⁰k²_W0+kA^1−β2uk²_W0+kA^β2vk²_W0

Let

E˜ = 1 2

hkA^−β2 u⁰(t)k²_W0+kA^−β2 v⁰(t)k²_W0+kA^1−β2 u(t)k²_W0+kA^1−β2v(t)k²_W0

i+αhv, uiW⁰

and

K(t) =kA^−β2 u⁰(t)k²_W0 +kA^−β2 v⁰(t)k²_W0+kA^1−β2 u(t)k²_W0+kA^1−β2v(t)k²_W0.

For allt≥0, we have

E˜⁰=−kA^−β2u⁰(t)k²_W0. Then ˜E is nonincreasing.

Since

|αhv, uiW⁰| ≤ |α|

λ

3−2β 2

1

kA^1−β2 u(t)kW⁰kA^1−β2v(t)kW⁰

then we get

λ

3−2β 2

1 − |α|

2λ

3−2β 2

1

K(t)≤E˜ ≤λ

3−2β 2

1 +|α|

2λ

3−2β 2

1

K(t). (3.17)

Now since

kA^−β2v⁰(t)kW⁰ ≤ 1

λ^β−1₁ kA^β2−1v⁰kW⁰, kA^1−β2 u(t)kW⁰ ≤ 1 λ₁¹²

kA^1−β2ukW⁰,

then there exists a constant γ >0 such that for allt≥0

H_ε⁰ ≤ −γK(u, v, u⁰, v⁰). (3.18) From (3.18), assumingεpossibly smaller in order to achieve positivity of the quadratic formHε, we get

Z t 0

K(u(s), v(s), u⁰(s), v⁰(s))ds≤ 1

γHε(u(0), v(0), u⁰(0), v⁰(0)).

Using inequality (3.17), we obtain 2λ

3−2β 2

1

λ

3−2β 2

1 +|α|

Z t 0

E(u(s), v(s), u˜ ⁰(s), v⁰(s))ds≤ 1

γH_ε(u(0), v(0), u⁰(0), v⁰(0)).

Now since ˜E is nonincreasing, it follows

E(u(t), v(t), u˜ ⁰(t), v⁰(t))≤ λ

3−2β 2

1 +|α|

2λ

3−2β 2

1 γ

1

tH_ε(u(0), v(0), u⁰(0), v⁰(0)).

Using inequality (3.17) we get

K(u(t), v(t), u⁰(t), v⁰(t))≤ λ

3−2β 2

1 +|α|

(λ

3−2β 2

1 − |α|)γ 1

tH_ε(u(0), v(0), u⁰(0), v⁰(0)).