Boundary Controllability for the Nonlinear Korteweg-De Vries Equation on Any Critical Domain

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www.elsevier.com/locate/anihpc

Boundary controllability for the nonlinear Korteweg–de Vries equation on any critical domain

^✩

Eduardo Cerpa

^a

, Emmanuelle Crépeau

^b,c,^∗

aUniversité Paris-Sud, Laboratoire de Mathématiques d’Orsay, Bât. 425, 91405 Orsay Cedex, France bINRIA Rocquencourt, Domaine de Voluceau, 78150 Le Chesnay, France

cUniversité de Versailles Saint-Quentin en Yvelines, 78035 Versailles, France Received 27 April 2007; received in revised form 6 November 2007; accepted 6 November 2007

Available online 25 January 2008

Abstract

It is known that the linear Korteweg–de Vries (KdV) equation with homogeneous Dirichlet boundary conditions and Neumann boundary control is not controllable for some critical spatial domains. In this paper, we prove in these critical cases, that the nonlinear KdV equation is locally controllable around the origin provided that the time of control is large enough. It is done by performing a power series expansion of the solution and studying the cascade system resulting of this expansion.

MSC:93B05; 35Q53

Keywords:Controllability; Korteweg–de Vries equation; Critical domains; Power series expansion

1. Introduction and main result

LetL >0 be fixed. Let us consider the following Neumann boundary control system for the Korteweg–de Vries (KdV) equation with homogeneous Dirichlet boundary conditions

yt+yx+yxxx+yyx=0, y(t,0)=y(t, L)=0, yx(t, L)=κ(t ),

(1) where the state isy(t,·):[0, L] →Rand the control isκ(t )∈R. This equation has been introduced by Korteweg and de Vries in [16] to describe the propagation of small amplitude long waves in a uniform channel. The KdV equation also appears in the study of various physical phenomena like long internal waves in a density-stratified ocean, ionic- acoustic waves in a plasma, etc.

In this paper, we are concerned with the controllability of (1). More precisely, for a timeT >0, we want to prove the following local exact controllability property.

✩ This work has been partially supported by the ANR project C-QUID.

* Corresponding author at: Université de Versailles Saint-Quentin en Yvelines, 78035 Versailles, France.

E-mail addresses:eduardo.cerpa@math.u-psud.fr (E. Cerpa), crepeau@math.uvsq.fr (E. Crépeau).

doi:10.1016/j.anihpc.2007.11.003

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P(T ) There existsr >0 such that, for every(y₀, y_T)∈L²(0, L)²withy₀_L²_(0,L)< r andy_T_L²_(0,L)< r, there existκ∈L²(0, T )and

y∈C

[0, T], L²(0, L)

∩L²

0, T , H¹(0, L) satisfying(1),y(0,·)=y₀andy(T ,·)=y_T.

In order to deal with the nonlinear term in (1), one can perform a power series expansion of(y, κ)around 0. To find the different terms of this development, one can write, formally, for a parametersmall,

y=α+²β+³γ+ · · ·, κ=u+²v+³w+ · · ·

thus, the nonlinear term can be written as yy_x=²αα_x+³(αβ)_x+(higher terms) and therefore the three main orders are given by

α_t+α_x+α_xxx=0, α(t,0)=α(t, L)=0, α_x(t, L)=u(t ),

(2) β_t+β_x+β_xxx= −αα_x,

β(t,0)=β(t, L)=0, β_x(t, L)=v(t ),

(3) and γ_t+γ_x+γ_xxx= −(αβ)_x,

γ (t,0)=γ (t, L)=0, γx(t, L)=w(t ).

(4) In [18] Rosier studies the control system (1) by using a first order expansion, i.e. he considers the linear control system (2) where the state isα(t,·):[0, L] →Rand the control isu(t )∈R. First, by using multiplier technique and the HUM method (see [17]), he proves that (2) is exactly controllable if and only if

L /∈N:=

2π

k²+kl+l²

3 ; k, l∈N^∗

, (5)

and then, by means of a fixed point theorem, he gets the following result.

Theorem 1.1.(See [18, Theorem 1.3].) IfL /∈N, then propertyP(T )holds for everyT >0.

Remark 1.2.If one is allowed to use more than one boundary control input, there is no critical spatial domain and the exact controllability holds for anyL >0. More precisely, let us consider the nonlinear control system

yt+yx+yxxx+yyx=0,

y(t,0)=u1(t ), y(t, L)=u2(t ), yx(t, L)=u3(t ), (6)

where the controls areu₁(t ), u₂(t )andu₃(t ). As it has been pointed out by Rosier in [18], for everyL >0 the system (6) with u₁≡0 is locally exactly controllable inL²(0, L)around the origin. Moreover, using all the three control inputs, Zhang proves in [22] that for everyL >0, the system (6) is exactly controllable in the spaceH^s(0, L)for any s0, in a neighborhood of a given smooth solution of the KdV equation.

IfL∈N, one says thatLis a critical length since the linear control system (2) is no more controllable. Indeed, Rosier proves in [18] that there exists a finite-dimensional subspace ofL²(0, L), denoted byM=M(L), which is unreachable from 0 for the linear system. More precisely, for every nonzero stateψ∈M, for everyu∈L²(0, T )and for everyα∈C([0, T], L²(0, L))∩L²(0, T , H¹(0, L))satisfying (2) andα(0,·)=0, one hasα(T ,·)=ψ.

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Let us recall the description ofM given in [4]. LetL∈N. There exist ndistinct pairs(k_j, l_j)∈N^∗×N^∗ with k_jl_j such that

∀j∈ {1, . . . , n}, L=2π

k_j²+kjlj+l_j²

3 . (7)

L=2π

k²+kl+l²

3 , kl, k, l∈N^∗ ⇒ (∃j∈ {1, . . . , n}s.t.k=k_j andl=l_j). (8) Let us introduce the notation

J^>:=

j ∈ {1, . . . , n}; k_j> l_j

, J⁼:=

j∈ {1, . . . , n}; k_j=l_j

, n^>:=J^>. (9)

For everyj∈ {1, . . . , n}, we define the real number p_j:=(2kj+l_j)(k_j−l_j)(2lj+k_j)

2π 3L

3

. (10)

We have then (see [18]), ξ³−ξ+pj=

ξ−γ₁^j

ξ−γ₂^j ξ−γ₃^j with ⎧

⎪⎪

⎪⎨

⎪⎪

⎪⎩

γ₁^j= −1

3(2kj+lj)2π L , γ₂^j=γ₁^j+kj

2π L , γ₃^j=γ₂^j+l_j2π

L .

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Lemma 1.3.With the previous notations, we get 1. ifj∈J^>,pj=0,

2. ifj∈J⁼,p_j=0, 3. ifi=j,p_i=p_j.

Proof. Items 1. and 2. are obvious with (10). Leti, j∈J such thatpi=pj. Then,γ_kⁱ=γ_k^j fork=1,2,3. With the definitions ofγ_k^j, (11) we obtaink_i=k_j,l_i=l_j and hencei=j. 2

Remark 1.4.We can easily notice that|J⁼|1.

Thus we can reorganize the indexes such that p₁> p₂>· · ·> p_n0.

With this notation, we define,

• forj∈J^>, the subspace ofL²(0, L) M_j:=

λ₁ϕ₁^j+λ₂ϕ₂^j; λ₁, λ₂∈R

= ϕ^j₁, ϕ^j₂

, where the real-valued functionsϕ₁^j, ϕ₂^j are given by

ϕ^j₁(x):=C_j

cos γ₁^jx

−γ₁^j−γ₃^j γ₂^j−γ₃^jcos

γ₂^jx

+γ₁^j−γ₂^j γ₂^j−γ₃^j cos

γ₃^jx , ϕ^j₂(x):=C_j

sin

γ₁^jx

−γ₁^j−γ₃^j γ₂^j−γ₃^j

sin γ₂^jx

+γ₁^j−γ₂^j γ₂^j−γ₃^j

sin

γ₃^jx , (12)

whereC_j is a constant chosen so thatϕ₁^jL²(0,L)= ϕ₂^jL²(0,L)=1;

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• forj∈J⁼, the subspace ofL²(0, L) M_j:=

λ(1−cosx); λ∈R

=

1−cos(x) . Then, one can define the following subspaces ofL²(0, L)

M:=

n j=1

M_j and H:=M^⊥. Note that

n^>

j=1

ϕ^j₁, ϕ^j₂

(ifL=2π kfor anyk) or

1−cos(x)ⁿ^>

j=1

ϕ^j₁, ϕ^j₂

(ifL=2π kfor somek) is an orthogonal basis fromM.

The subspaceH is the space of reachable states for the linear control system. More precisely, from the work of Rosier one has the exact controllability inH for the control system (2).

Theorem 1.5. Let L >0 and T >0. For every (y₀, y_T)∈H ×H, there exist u∈L²(0, T ) and α∈C([0, T], L²(0, L))∩L²(0, T , H¹(0, L))satisfying(2),α(0,·)=y₀andα(T ,·)=y_T.

In [11], Coron and Crépeau study the first critical case:

n=1 and k1=l1=k.

In this case the subspace Mis one-dimensional. First, they prove that one can reach all the missed directions lying inM, i.e. (1−cos(x)) and (cos(x)−1), with a third order power series expansion.

Proposition 1.6. (See [11, Proposition 8].) Let L ∈N be such that dimM(L)=1. Let T >0. There exist (u^±, v^±, w^±)∈L²(0, T )³such that ifα^±, β^±, γ^±are the solutions of (2), (3)and(4)with initial conditions

α^±(0,·)=0, β^±(0,·)=0, γ^±(0,·)=0, then

α^±(T ,·)=0, β^±(T ,·)=0, γ^±(T ,·)= ±

1−cos(x) .

Then, using Theorem 1.5 and a fixed point theorem, they prove that property P(T ) holds for every T >0 [11, Theorem 2]. They also prove that for this first critical case, a second order expansion is not sufficient to enter into the subspaceM[11, Corollary 19].

Remark 1.7.The proof ofP(T )given in [11] requires that the subspaceMis one-dimensional, but this is not implied by the fact thatL=2kπ for somek∈N^∗. It is necessary to add a condition as the following one

m²+mn+n²=3k², m∈N^∗, n∈N^∗

⇒ (m=n=k). (13)

This condition, not explicitly given in [11], appears in [10]. In this book it is also proved that there are infinitely many positive integersksatisfying (13) and therefore there are infinitely many lengthsLsuch thatMis one-dimensional.

In [4], the same approach is used to treat the second critical case:

n=1 and k1> l1.

In this case, the spaceMis two-dimensional and a second order expansion allows to enter into the subspaceM.

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Proposition 1.8. (See [4, Proposition 3.1].) Let L∈ N be such that dimM(L) =2. Let T >0. There exist u, v∈L²(0, T )such that ifα, βare the solutions of (2)and(3)with initial conditions

α(0,·)=0, β(0,·)=0, then

α(T ,·)=0, β(T ,·)∈M\ {0}.

It is also proved that if the time of control is large enough, one can reach all the missed directions. Using this and a fixed point argument, one obtains propertyP(T )provided that the time of controlT is large enough [4, Theorem 1.4].

The aim of this paper is to proveP(T )in the critical cases for whichn >1, i.e. when the dimension of the subspace Mis higher than 2. We use an expansion to the second order ifL=2π kfor anyk∈N^∗and an expansion to the third order ifL=2π kfor somek∈N^∗. Our main result is the following.

Theorem 1.9.LetL∈N. Then, there existsT_L0such thatP(T )holds provided thatT > T_L.

The paper is organized as follows. First, in Section 2, we recall the well-posedness results for both linear and nonlinear KdV control systems. Next, in Section 3, we prove by using a second order power series expansion, that one can reach all the missed states in the subspacesM_j forj∈J^>. Then, in Section 4, we prove that ifL=2π k, one can reach the missed states±(1−cos(x))with a third order expansion and finally, in Section 5 we get Theorem 1.9 by using a fixed point argument.

Remark 1.10. From our proof of Theorem 1.9, it follows that there exists a constant C >0 such that for every y₀, y_T ∈L²(0, L)small enough, the solutiony and the controlκ given by propertyP(T )satisfy

yC([0,T],L²(0,L))+ yL²(0,T ,H¹(0,L))+ κL²(0,T )C

y0L²(0,L)+ yTL²(0,L)

1/3

ifL=2kπ for somek∈N^∗and

yC([0,T],L²(0,L))+ yL²(0,T ,H¹(0,L))+ κL²(0,T )C

y₀L²(0,L)+ y_TL²(0,L)

1/2

ifL=2kπ for anyk∈N^∗. The power 1/3 and 1/2 come from the order of the series expansion needed in each case.

Remark 1.11.One can find other results on the controllability of KdV control systems in [14,19–22] and the refer- ences therein.

Remark 1.12.The power series expansion method is a classical tool to study finite-dimensional control systems. It has been used for the first time in infinite dimension in [11]; see also [4] as well as [2] for a Schrödinger equation.

This method and others such as quasi-static deformations (see [1,12,13] and [10, Chapter 7]) and the return method (see [1,6–8] and [10, Chapter 6]) are very useful to deal with nonlinear systems and to get properties which are not a consequence of the linearized system behavior.

2. Well-posedness results

The aim of this section is to precise what we mean by “a solution” of the KdV equations appearing in this paper and to recall the existence and uniqueness results we will use.

Let us introduce the spaceB:=C([0, T], L²(0, L))∩L²(0, T , H¹(0, L))endowed with the norm y_B:= max

t∈[0,T]y(t )

L²(0,L)+ T

0

y(t )²

H¹(0,L)dt 1/2

. Let us begin with the linear case.

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Definition 2.1.LetT >0,f ∈L¹(0, T , L²(0, L)),y₀∈L²(0, L)andκ∈L²(0, T )be given. A solution of the Cauchy problem

⎧⎪

⎨

⎪⎩

yt+yx+yxxx=f, y(t,0)=y(t, L)=0, y_x(t, L)=κ(t ), y(0,·)=y₀,

(14)

is a functiony∈Bsuch that, for everyτ∈ [0, T]and for everyφ∈C³([0, τ] × [0, L])satisfying φ(t,0)=φ(t, L)=φ_x(t,0)=0, ∀t∈ [0, τ],

one has

− τ 0

L 0

(φ_t+φ_x+φ_xxx)y dx dt− τ 0

κ(t )φ_x(t, L) dt+ L 0

y(τ, x)φ(τ, x) dx− L 0

y₀(x)φ(0, x) dx

= τ 0

L 0

f φ dx dt.

With this definition and from the work of Rosier in [18], we have the following result.

Theorem 2.2.LetT >0,f ∈L¹(0, T , L²(0, L)),y₀∈L²(0, L)andκ∈L²(0, T ). Then, there exists one and only one solution of the Cauchy problem(14).

Let us now give the definition of a solution for the nonlinear equation.

Definition 2.3.LetT >0,g∈L¹(0, T , L²(0, L)),y₀∈L²(0, L)andκ∈L²(0, T )be given. A solution of the Cauchy problem

⎧⎪

⎨

⎪⎩

y_t+y_x+y_xxx+yy_x=g, y(t,0)=y(t, L)=0, y_x(t, L)=κ(t ), y(0,·)=y0,

(15)

is a functiony∈Bsatisfying (14) withf =g−yy_x.

Remark 2.4.Note that ify∈B, thenyy_x∈L¹(0, T , L²(0, L))and therefore(g−yy_x)as well.

Theorem 2.5.(See [11, Appendix].) LetT >0. Then there exists >0such that, for everyg∈L¹(0, T , L²(0, L)), y₀∈L²(0, L)andκ∈L²(0, T )satisfying

gL¹(0,T ,L²(0,L))+ y0L²(0,L)+ κL²(0,T ), (16)

the Cauchy problem (15)has one and only one solution. Furthermore, there exists a constant C >0such that this solution satisfies

y_BC

gL¹(0,T ,L²(0,L))+ y0L²(0,L)+ κL²(0,T )

. (17)

Remark 2.6.In [3] and [15], one can find some well-posedness results in the case where there are nonhomogeneous Dirichlet boundary conditions.

Remark 2.7.Recently, in [5] the author proved Theorem 2.5 with= ∞, that is, without a smallness condition on the data.

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3. Motion in the missed subspacesMj, forj∈∈∈J^>

Here and in the sequel, we denote by L a critical length such that dimM(L) >2 and by P_A the orthogonal projection on a subspaceAinL²(0, L). We also adopt the notations introduced in Section 1.

The first point is that for anyj∈J^>, we canenterinto the two-dimensional subspaceM_j. The strategy is the same as in [11] and [4]. We consider a power series expansion of(y, κ)with the same scaling on the statey and on the controlκ. One has the following result that can be proved in the same way as in [4, Proposition 3.1].

Proposition 3.1.LetT >0. For everyi=1, . . . , n^>, there exists(u_i, v_i)∈L²(0, T )²such that ifα_i =α_i(t, x)and β_i =β_i(t, x)are the solutions of

⎧⎪

⎨

⎪⎩

α_it+α_ix+α_ixxx=0, α_i(t,0)=α_i(t, L)=0, α_ix(t, L)=u_i(t ), α_i(0,·)=0,

(18)

and ⎧

⎪⎨

⎪⎩

β_it+β_ix+β_ixxx= −α_iα_ix, β_i(t,0)=β_i(t, L)=0, βix(t, L)=vi(t ), βi(0,·)=0,

(19)

then

α_i(T ,·)=0, P_H

β_i(T ,·)

=0 and P_M_i

β_i(T ,·)

=0.

Let us denote, forj=1, . . . , n^>, φ^j_i :=P_M_j

β_i(T ,·) .

From Proposition 3.1,φ_iⁱ=0. Consequently, using scaling on the controls, we can assume thatφⁱ_i_L²_(0,L)=1. Notice that the previous proposition says nothing aboutφ_i^j forj=i.

Now, we shall prove that we can reach all the states lying in the subspace M^>:=

i∈J^>

M_i, in any timeT > T^>, where

T^>:=π

n^>

i=1

n^>+1−i1 pi

.

In order to do that, we will strongly use the fact (proved in [4]) that if there is no control (i.e.κ=0) and if the initial condition lies inMj forj∈J^>(i.e.y0∈Mj), then the solutiony of the linear KdV equation only turns in the two- dimensional subspaceM_j with an angular velocity equal top_j (defined in (10)) and conserves itsL²-norm. More precisely, we have the following result.

Lemma 3.2.Letj∈J^>. Lety₀∈M_j. Letλ0andδ∈ [0,2π )be such that

y₀=λcos(δ)ϕ₁^j+λsin(δ)ϕ₂^j. (20)

Then the solution of y_t+y_x+y_xxx=0,

y(t,0)=y(t, L)=y_x(t, L)=0, y(0,·)=y₀

(21) is given by

y(t, x)=λcos(pjt+δ)ϕ₁^j+λsin(pjt+δ)ϕ₂^j. (22)

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For the sake of brevity we introduce, forj∈J^>,θ∈Randy₀∈M_j reading as (20), the notation

R^j(y₀, θ ):=λcos(θ+δ)ϕ^j₁+λsin(θ+δ)ϕ^j₂, (23)

i.e.R^j(·, θ )represents a rotation of an angleθin the subspaceMj. Thus, the solution of (21) can be written as y(t, x)=R^j(y₀, p_jt ).

Proposition 3.3. Let T > T^>. Let ψ∈M^>. There exists (u_ψ, v_ψ)∈L²(0, T )² such that if α_ψ =α_ψ(t, x) and β_ψ=β_ψ(t, x)are the solutions of (18)and(19), then

αψ(T ,·)=0, βψ(T ,·)=ψ.

Proof. First at all, let us notice that ifL=2kπ for somek∈N^∗, thenM_n= 1−cosxanda prioriP_M_n(β_ψ(T ,·)) may be non-null. However, we know from [11, Corollary 19] that a second order expansion is not sufficient to enter into the subspaceM_n and thereforeP_M_nβ_ψ(T .·)=0. That is the reason why we do not care about the projection on M_nof second-order trajectories.

The casen^>=1 has already been studied in [4]. Let us consider the casen^>=2, i.e. where we have 2 subspaces, M₁ and M₂ associated to (k₁, l₁)and (k₂, l₂) with p₁> p₂>0 (for instance, L=2π√

91 leads to the couples (k₁, l₁)=(16,1)and(k₂, l₂)=(11,8)).

LetT >^2π_p

1 +_p^π₂. LetT₁be such that T₁> π

p1

and T −T₁> π p1+ π

p2

. LetT_θ>0 andT_c>0 be such that

T_c< T_θ, T_c< π p1

, T_c+T_θ<min

T −T1− π p₁ − π

p₂, π p₂− π

p₁, T1− π p₁ .

Thanks to Proposition 3.1, there exist two pairs of controls,(u₁, v₁)and(u₂, v₂)inL²(0, Tc)such that the respec- tive solutions of (18) and (19), (α₁, β₁)and(α₂, β₂), satisfyP_M₁(β₁(T_c,·))=0 andP_M₂(β₂(T_c,·))=0. With the notations introduced before,

φ₁¹, φ₁²

= P_M₁

β₁(T ,·) , P_M₂

β₁(T ,·) ,

φ₂¹, φ₂²

= P_M₁

β₂(T ,·) , P_M₂

β₂(T ,·) .

We now use the rotation phenomena explained before and Proposition 3.1 to reach a basis for the missed directions lying inM^>. For the seek of clarity in our control strategy, we define for a timet₁, the following control inL²(0, T )

(Ut₁, Vt₁)(t ):=

(0,0) ift∈(0, t1), (u1(t−t1), v1(t−t1)) ift∈(t1, t1+Tc), (0,0) ift∈(t1+Tc, T ).

This control becomes active at timet=t1, betweent=t1andt=t2, it drives the system to enter into the space M1and aftert=t2, it becomes inactive, producing a rotation inM1.

Now, we define the controls u¹₁, v₁¹

:=(U_t₁, V_t₁) witht1=T −T_c, u²₁, v₁²

:=(Ut₁, Vt₁) witht1=T −Tc− π 2p1

, u³₁, v₁³

:=(U_t₁, V_t₁) witht₁=T −T_c− π p1

, u⁴₁, v₁⁴

:=(U_t₁, V_t₁) witht₁=T −T_c− π p1−T_θ.

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Letα₁^j, β₁^j∈Bbe the solutions of (18) and (19) with controlsu^j₁andv₁^j forj=1, . . . ,4 and let us denote ψ₁^j:=P_M₁β₁^j(T ,·) and ψ˜₂^j:=P_M₂β₁^j(T ,·).

It is easy to see that

ψ₁¹=φ¹₁, ψ˜₂¹=φ₁², ψ₁²=R¹

φ₁¹,π

2 , ψ˜₂²=R²

φ₁²,p₂π 2p1

, ψ₁³=R¹

φ₁¹, π

= −φ₁¹, ψ˜₂³=R²

φ₁²,p₂π p1

, ψ₁⁴=R¹

−φ₁¹, p₁T_θ

, ψ˜₂⁴=R²

φ₁², p₂

T_θ+ π p₁

.

Thus, we have constructed some controls allowing to reach the missed states ψ₁¹+ ˜ψ₂¹, ψ₁²+ ˜ψ₂², ψ₁³+ ˜ψ₂³, and ψ₁⁴+ ˜ψ₂⁴.

Now, we define for a timet₂, the following control inL²(0, T ) U^t², V^t²

(t ):=

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎩

(0,0) ift∈(0, t2),

(u₁(t−t₂), v₁(t−t₂)) ift∈(t₂, t₂+T_c), (0,0) ift∈(t₂+T_c, t₂+_p^π₁), (u₁(t−t₂−_p^π

1), v₁(t−t₂−_p^π

1)) ift∈(t₂+_p^π

1, t₂+_p^π

1 +T_c), (0,0) ift∈(t₂+_p^π₁ +T_c, T ), which is the superposition of two controls of type(Ut₁, Vt₁)

U^t², V^t²

(t )=(U_t₂₊^π

p1

, V_t₂₊^π

p1

)+(U_t₂, V_t₂).

This fact means that the solution corresponding to the controls(U^t², V^t²)is the addition of two trajectories which enter intoMand then turn during different times.

We define the following controls inL²(0, T ), u¹₂, v₂¹

=

U^t², V^t²

witht₂=T −T₁− π p1−T_c, u²₂, v₂²

=

U^t², V^t²

witht₂=T −T₁− π

p₁−T_c−T_θ, u³₂, v₂³

=

U^t², V^t²

witht2=T −T1− π p₁− π

p₂−Tc, u⁴₂, v₂⁴

=

U^t², V^t²

witht₂=T −T₁− π p1− π

p2−T_c−T_θ.

Letα₂^j, β₂^j∈Bbe the solutions of (18) and (19) with controlsu^j₂andv₂^j forj=1, . . . ,4 and let us denote ψ₂^j:=P_M₂β₂^j(T ,·).

Here, it is very important to note that, by construction and sincep₁> p₂, one has PM₁β₂¹(T ,·)=0 and ψ₂¹=R²

φ₁², p2T1

+R²

φ₁², p2(T1+π/p1)

=0.

Thus, we have constructed some controls allowing to reach the following missed states ψ₂¹, ψ₂², ψ₂³, and ψ₂⁴,

where

ψ₂²=R²

ψ₂¹, p2Tθ

, ψ₂³=R²

ψ₂¹, π

= −ψ₂¹, ψ₂⁴=R²

−ψ₂², p₂T_θ .

(10)

Furthermore, we have fork=1,2 M_k=

4 j=1

M_k^j (24)

where M_k¹:=

d_k¹ψ_k¹+d_k²ψ_k²; d_k¹>0, d_k²0 , M_k²:=

d_k¹ψ_k²+d_k²ψ_k³; d_k¹>0, d_k²0 , M_k³:=

d_k¹ψ_k³+d_k²ψ_k⁴; d_k¹>0, d_k²0 , M_k⁴:=

d_k¹ψ_k⁴+d_k²ψ_k¹; d_k¹>0, d_k²0 .

Letψ∈M^>. From (24), we know thatP_M₁(ψ )∈M₁ⁱ for somei∈ {1, . . . ,4}. Hence, there existd₁¹>0, d₁²0, such that

ψ=d₁¹ψ₁ⁱ+d₁²ψ₁ⁱ⁺¹+P_M₂(ψ ).

Let us writeψas follows

ψ=d₁¹ψ₁ⁱ+d₁²ψ₁ⁱ⁺¹+d₁¹ψ˜₂ⁱ+d₁²ψ˜₂ⁱ⁺¹+

PM₂(ψ )−d₁¹ψ˜₂ⁱ−d₁²ψ˜₂ⁱ⁺¹ . Since the statesψ˜₂ⁱ,ψ˜₂ⁱ⁺¹lie inM₂, there existsj∈ {1, . . . ,4}such that

P_M₂(ψ )−d₁¹ψ˜₂ⁱ−d₁²ψ˜₂ⁱ⁺¹∈M₂^j

and therefore there existd₂¹>0, d₂²0 such that ψ=d₁¹

ψ₁ⁱ+ ˜ψ₂ⁱ +d₁²

ψ₁ⁱ⁺¹+ ˜ψ₂ⁱ⁺¹

+d₂¹ψ₂^j+d₂²ψ₂^j⁺¹.

Thus, we have decomposed ψ in terms of reachable directions for the second-order expansion. Now, we take the controlsu_ψ, v_ψ defined by

(u_ψ, v_ψ)=

d₁¹uⁱ₁+

d₁²uⁱ₁⁺¹+

d₂¹u^j₂+

d₂²u^j₂⁺¹, d₁¹v₁ⁱ+d₁²v₁ⁱ⁺¹+d₂¹v₂^j+d₂²v₂^j⁺¹ ,

andα_ψ, β_ψ∈Bthe corresponding solutions of (18) and (19) respectively. Here, it is important to note that, with the choices ofT , T₁, T_c andT_θ, the supports of the trajectoriesα_k^j fork=1,2 andj=1, . . . ,4 are disjoint and that all these trajectories go from 0 att=0 to 0 att=T, i.e.

α^j_k(0,·)=α^j_k(T ,·)=0.

Thus, it is not difficult to verify that αψ(T ,·)=0 and βψ(T ,·)=ψ

which ends the proof in the casen^>=2. The previous method can be easily adapted to the case wheren^>>2. In order to construct the controls needed in the general case, our method requires a time of controlT greater thanT^>. 2 4. Motion in the missed directions±(1−cosx)ifL=2kπ

We assume in this section thatL=2kπ for somek∈N^∗. Let us recall that in this case we have

M_n= 1−cosx and n^>=n−1. (25)

Thanks to [11], we have the following result that one can prove in a similar way to [11, Proposition 8].

(11)

Proposition 4.1.LetT_c>0. There exists(u, v, w)inL²(0, Tc)³such that, ifα, β, γ are the mild solutions of

⎧⎪

⎨

⎪⎩

α_t+α_x+α_xxx=0, α(t,0)=α(t, L)=0, α_x(t, L)=u(t ), α(0,·)=0,

(26)

⎧⎪

⎨

⎪⎩

βt+βx+βxxx= −ααx, β(t,0)=β(t, L)=0, βx(t, L)=v(t ), β(0,·)=0,

(27)

⎧⎪

⎨

⎪⎩

γ_t+γ_x+γ_xxx= −(αβ)_x, γ (t,0)=γ (t, L)=0, γ_x(t, L)=w(t ), γ (0,·)=0,

(28)

then

α(T_c,·)=0, β(T_c,·)=0 and γ (T_c,·)=(1−cosx)+

n^>

i=1

P_M_i

γ (T_c,·) .

The idea to vanish the projections ofγ (T_c,·)onM_i, and thus to reach the direction (1−cos(x)), is the same as before, that is, to use the rotation phenomena given in Lemma 3.2. In addition, we use the fact that the function (1−cosx)satisfies

y_x+y_xxx=0,

y(0)=y(2kπ )=y_x(2kπ )=0.

The casen=1 has already been considered in [11]. We deal with the casen=2 (for example,L=14π leads to the couples(k₁, l₁)=(11,2)and(k₂, l₂)=(7,7)).

Let us define the following control lying inL²(0, T )³, whereT > π/p₁. (Here, we omit the time translation needed for the controlsu, vandwwhich are defined in(0, Tc).)

(u₊, v₊, w₊)(t )=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

(0,0,0) ift∈(0, T−T_c−_p^π₁), (u, v, w) ift∈(T−Tc−_p^π₁, T −_p^π₁), (0,0,0) ift∈(T−_p^π₁, T −T_c), (u, v, w) ift∈(T−Tc, T ).

By definingα₊, β₊, γ₊∈Bas the solutions of (26) with controlu₊, (27) with controlv₊and (28) with control w₊respectively, it is not difficult to see that

α₊(T ,·)=0, β₊(T ,·)=0, γ₊(T ,·)=2(1−cosx). (29) Now, if we consider the control(u₋, v₋, w₋):=(−u₊, v₊,−w₊)we get

α₋(T ,·)=0, β₋(T ,·)=0, γ₋(T ,·)= −2(1−cosx), (30) where obviouslyα₋, β₋, γ₋∈Bare the solutions of (26), (27) and (28) with controlsu₋, v₋andw₋respectively.

Thus we can reach all directions inM2in a timeT >_p^π

1.

We can easily deduce the same result in the case n >2. We just have to construct a control that vanishes the components in the other missed subspacesM_j, j∈J^>. In order to do that, a time of controlT, with

T > Tⁿ:=π

n−1

i=1

1

p_i, (31)

is sufficient.