Control of a Boussinesq system of KdV-KdV type on a bounded interval

We consider a Boussinesq system of KdV-KdV type introduced by J. Bona, M. Chen and J.-C. Saut as a model for the motion of small amplitude long waves on the surface of an ideal fluid. This system of two equations can describe the propagation of waves in both directions, while the single KdV equation is limited to unidirectional waves. We are concerned here with the exact controllability of the Boussinesq system by using some boundary controls. By reducing the controllability problem to a spectral problem which is solved by using the Paley-Wiener method introduced by the third author for KdV, we determine explicitly all the critical lengths for which the exact controllability fails for the linearized system, and give a complete picture of the controllability results with one or two boundary controls of Dirichlet or Neumann type. The extension of the exact controllability to the full Boussinesq system is derived in the energy space in the case of a control of Neumann type. It is obtained by incorporating a boundary feedback in the control in order to ensure a global Kato smoothing effect.


Introduction
J. L. Boussinesq introduced in [5,6] several simple nonlinear systems of PDEs (including the Korteweg-de Vries equation) to explain certain physical observations concerning the water waves, e.g. the emergence and stability of solitons. Unfortunately, several systems derived by Boussinesq proved to be ill-posed, so that there was a need to propose other systems similar to Boussinesq's ones but with better mathematical properties.
The four-parameter family of Boussinesq systems η t + v x + (ηv) x + av xxx − bη xxt = 0, v t + η x + vv x + cη xxx − dv xxt = 0 (1.1) was introduced by J. J. Bona, M. Chen and J.-C. Saut in [3] to describe the motion of small amplitude long waves on the surface of an ideal fluid under the gravity force and in situations when the motion is sensibly two-dimensional. In (1.1), η is the elevation of the fluid surface from the equilibrium position, and v = v θ is the horizontal velocity in the flow at height θh, where h is the undisturbed depth of the liquid. The parameters a, b, c, d are required to fulfill the relations where θ ∈ [0, 1] specifies which horizontal velocity the variable v represents. As it has been proved in [3], the initial value problem for the linear system associated with (1.1) is well posed on R if and only if the parameters a, b, c, d fall in one of the following cases The wellposedness of the full system (1.1) on the line (x ∈ R) was investigated in [4].
Recently, a rather complete picture of the control properties of (1.1) on a periodic domain with a locally supported forcing term was given in [26]. According to the values of the four parameters a, b, c, d, the linearized system may be either controllable in any positive time, or solely in large time, or may not be controllable at all. These results were also extended in [26] to the generic nonlinear system (1.1); that is, when all the parameters are different from 0.
When b = d = 0 and (C2) is satisfied, then necessarily a = c = 1/6. Nevertheless, the scaling x → x/ √ 6, t → t/ √ 6 gives a system equivalent to (1.1) for which a = c = 1, namely 3) The above system will be referred to as a Boussinesq system of KdV-KdV type, or as a KdV-KdV system.
The KdV-KdV system is expected to admit global solutions on R [4], and it also possesses good control properties on the torus [26].
To our best knowledge, the boundary control of the Boussinesq system of KdV-KdV type on a bounded domain (0, L) is completely open. The aim of this paper is to investigate the control properties of the following system with the boundary conditions η(t, 0) = h 0 (t), η(t, L) = h 1 (t), η x (t, 0) = h 2 (t) in (0, T ), v(t, 0) = g 0 (t), v(t, L) = g 1 (t), v x (t, L) = g 2 (t) in (0, T ) (1. 7) and the initial conditions η(0, x) = η 0 (x), v(0, x) = v 0 (x) in (0, L). (1.8) It is of course desirable to obtain control results (or stabilization results) with a few number of controls inputs. Here, we will provide a complete picture of the exact controllability of the linearized system with one or two controls among h 0 , h 1 , h 2 , g 0 , g 1 , g 2 .
A similar study was performed for the Korteweg -de Vries (KdV) equation y t + y xxx + y x + yy x = 0, (1.9) with the boundary conditions y(t, L) = h 0 (t), y(t, L) = h 1 (t), y x (t, L) = h 2 (t). (1.10) More precisely, the exact controllability of (1.9)-(1.10) was established in [30] when h 0 = h 1 = 0 (h 2 being the only effective control) for a length L which is not critical, in [8,9,13] for a length L which is critical, and in [19] when h 0 = h 2 = 0 for a length L not critical. The null controllability of (1.9)-(1.10) was proved in [31] (see also [18]) when h 1 = h 2 = 0 for any length L > 0. Note that in that case there is no critical length, and that solely the null controllability holds (say in L 2 (0, L)), because the terminal state y(·, T ) is C ∞ smooth for x > 0. A length L is said to be critical when the linearized equation y t + y xxx + y x = 0, (1.11) fails to be controllable. This phenomenon was first noticed in [30]. It is due to the influence of the first order derivative ∂ x on the spectrum of the operator ∂ 3 x with the boundary conditions y(0) = y(L) = y x (L) = 0. If the high frequencies are asymptotically preserved, the low frequencies may be strongly modified, and some of the corresponding eigenfunctions may become uncontrollable for (1.11) for certain values of L (the critical ones).
On the other hand, the set of critical lengths for the linear control system y t + y xxx + y x = 0, (1.14) y(t, 0) = y x (t, L) = 0, y(t, L) = h 1 (t) (1.15) was proved in [19] to be discrete, countable and given bỹ The determination of the critical lengths for system (1.12)-(1.13) in [30] was based on a series of reductions (as in [1]), first to a unique continuation property for the adjoint system, next to a spectral problem with an extra condition: − y − y = λy, y(0) = y(L) = y (0) = 0 and y (L) = 0.
(1. 16) This spectral problem was then solved by extending the function y by 0 outside (0, L), by taking its Fourier transform and by using Paley-Wiener theorem. The length L is then critical if and only if there exist some numbers p ∈ C and (α, β) ∈ C 2 \ {(0, 0)} such that the function is analytic on C; that is, the roots of ξ 3 − ξ + p are also roots of α − βe −iLξ with at least the same multiplicity. The determination of N follows then with some algebra. By contrast, the critical lengths for (1.14)- (1.15), that is the elements ofÑ, are not explicitly known. This is likely due to the lack of symmetries in the corresponding spectral problem − y − y = λy, y(0) = y(L) = y (0) = 0 and y (L) = 0. (1.18) Note that the boundary conditions in (1.16) are preserved by the transformation x → L − x. This symmetry yields a more tractable function f in (1.17). Boussinesq system is more convenient than KdV as a model for the propagation of water waves, for it is adapted to the propagation of waves in both directions, and it is still valid after bounces of waves at the boundary. It is striking that the control theory for Boussinesq system (exposed in this paper) is better understood than for KdV as far as the critical lengths are concerned: indeed, the critical lengths for Boussinesq system are explicitly given for any set of boundary controls (except in Case 3 where only g 0 is used, and in Case 12 where h 1 and g 0 are used, see below Table 1), which is not the case for KdV with a control as in (1.15). We believe that this is due to the numerous symmetries of Boussinesq system: for instance, x = 0 and x = L (resp. η and v) play a symmetric role for the linearized Boussinesq system. The price to be paid is the lack of any Kato smoothing effect (the system being conservative), which makes the extension of the control results to the nonlinear Boussinesq system more delicate than for KdV. We refer the reader to [7,8,9,10,11,12,13,16,18,19,20,21,27,29,30,31,32,33] for the control and stabilization of KdV, and [14,15,17] for the critical lengths concerning some other dispersive equations.
To investigate the control properties of the linearized system, we proceed as in [30]. To prove the observability inequality, we use the reduction to a spectral problem with an extra condition and the Paley-Wiener method. Here, we have to see for which value of L > 0 two functions are entire for a set of parameters. For instance, in Case 1 where only g 2 is used, the two functions read for some p ∈ C and (α, β, γ, α , γ ) ∈ C 5 \ {0}. Using the symmetries of the system, we can prove that the set of critical lengths is N.
In the reduction to the Unique Continuation Property (UCP), we use in most cases an identity obtained by using Morawetz multiplier, namely (2.10) (see below). However, for Case 3 (see below Table 1), the identity (2.10) proved to be useless. We replace the observability inequality by a weaker estimate which in turn is established by performing a careful investigation of the spectral reduction of We show that the space [L 2 (0, L)] 2 admits an orthonormal basis constituted of eigenfunctions of A. To do that, we introduce the selfadjoint operator (By . Then the spectral reduction of B yields at once those for A. We give in Table 1 (see below) a complete picture of the exact controllability results for the linearized Boussinesq system. It is not difficult to extend those exact controllability results to the nonlinear Boussinesq system in spaces of sufficiently regular functions (e.g. some subspaces of [H 2 (0, L)] 2 ), but here we will not do it for the sake of shortness. Rather, we shall explain how to obtain exact controllability/stabilization results in the energy space [L 2 (0, L)] 2 by incorporating a feedback law in the control that yields a smoothing effect, as it was done in [22] for the Benjamin-Ono equation. We shall limit ourserves to the case of a single Neumann boundary control (Case 1), namely to the system x ∈ (0, L), (1.20) The first main result in this paper is a local exact controllability result for (1.20).
one can find a control g 2 ∈ L 2 (0, T ) and a solution ( The second main result is a local exponential stability result for (1.20).
Theorem 1.2. Let T > 0, L ∈ (0, +∞) \ N, and α > 0. Then there exist some positive numbers δ, µ, C such that for all initial data and it holds The paper is outlined as follows. The wellposedness of the linearized Boussinesq system with the boundary conditions (1.7) is studied in Section 2. Section 3 is concerned with the exact controllability of the linearized Boussinesq system with one or two boundary control inputs. The proof of the main results, namely Theorems 1.1 and 1.2, is provided in Section 4. Finally, the proof of the weak observability estimate (1.19) based on some spectral reduction is given in appendix.

2.1.
Wellposedness of the homogeneous problem. Let L > 0 be a fixed number. Introduce the spaces denotes the Banach space obtained by the complex interpolation method (see e.g. [2]). The space X 0 (resp. X 3 ) is endowed with the norm It is easily seen that (2.5) and that in the space X 1 (resp. X 2 ), the norm (η, v) ). We shall use at some place the space endowed with its natural norm, and for s ∈ {1, 2}, the space X −s = (X s ) which is the dual of X s with respect to the pivot space X 0 = [L 2 (0, L)] 2 . Note that The bracket ·, · X −s ,Xs stands for the duality bracket. We first investigate the wellposedness of the initial value problem We introduce the operator Proposition 2.1. The operator A is skew-adjoint in X 0 , and thus it generates a group of isometries (e tA ) t∈R in X 0 .
for some constant C = C(T ).

Controllability of the linearized system
In this section, we are interested in the exact controllability of the linear system (2.17) using at most two control inputs (the other control inputs being replaced by 0). The exact controllability in X −2 is defined as follows.
As it is well known, the exact controllability of a control system is equivalent to the observability of its adjoint system (see e.g. [12,23,24]). Here, using (2.20)-(2.21), we see that the exact controllability of (2.17) in X −2 with the six controls g 0 , where (θ, u) denotes the solution of (2.18). The exact controllabity in X −2 with less control inputs is equivalent to the observability inequality (3.2) in which we remove the traces corresponding to the missing controls. Similarly, the exact controllability of (2.17) in X −1 with the two controls g 2 , h 2 ∈ L 2 (0, T ) is equivalent to the existence of a constant C > 0 such that for all (θ 0 , u 0 ) ∈ X 1 , The aim of this section is to provide a complete picture of the exact controllability results for system (2.17) with one or two control inputs among g 0 , g 1 , We introduce some subsets of (0, +∞) that are needed to give a complete picture of the critical lengths for the control of the linearized KdV-KdV system: Then the control results for the linearized Boussinesq system with one or two boundary controls of Dirichlet or Neumann type are outlined in the following theorem.
Theorem 3.2. Consider system (2.17) with one or two control inputs among g 0 , g 1 , g 2 , h 0 , h 1 , h 2 (the other being set to 0). Then the exact controllability of (2.17) holds (in any time T > 0) if and only if L is not a critical length, a concept depending on the set of control inputs that are available. The precise results are reported in Table 1.

Controls
Properties Table 1. Controllability results for the linearized system Remark 3.3. Using the symmetries, we notice that all the possibilities with one or two control inputs are really considered in Table 1: indeed, h 2 alone is similar to g 2 alone (case 1); h 0 alone is similar to g 1 alone (case 2); h 1 alone is similar to g 0 alone (case 3); the pair (h 0 , g 0 ) is similar to the pair (h 1 , g 1 ) (case 5); the pair (h 2 , g 1 ) is similar to the pair (h 0 , g 2 ) (case 8); the pair (h 1 , h 2 ) is similar to the pair (g 0 , g 2 ) (case 9); the pair (g 0 , h 2 ) is similar to the pair (h 1 , g 2 ) (case 6); the pair (h 0 , h 2 ) is similar to the pair (g 1 , g 2 ) (case 7); the pair (h 0 , h 1 ) is similar to the pair (g 0 , g 1 ) (case 11).
The proof of Theorem 3.2 is detailed in a series of propositions or theorems displayed in several subsections.
3.1. Neumann controls. We consider first two controls (case 4) and next only one control (case 1). We shall use repeatedly the following results from [30].

4)
where y = y(t, x) denotes the solution to the linearized KdV system x ∈ (0, L). (3.5) is an entire function, i.e. a complex analytic function on C.
The controllability result for the case 4 is a consequence of the following observability inequality.
Proposition 3.6. Let L ∈ N and T > 0. Then there exists a constant C = C(L, T ) > 0 such that where (θ, u) denotes the solution of (2.18).
Let us investigate the exact controllability of (2.17) in case 4. If L ∈ N, then by Proposition 3.6, system (2.17) is exactly controllable in X −1 . If L ∈ N, then by [30, Remark 3.6 (i)] there exists a nontrivial solution y of (3.5) such that y x (·, 0) = 0. Then (θ(t, x), u(t, x)) : We now turn into the problem of the controllability of (2.17) with only one Neumann control (case 1). The following observability inequality improves Proposition 3.6, for only one boundary measurement is needed to estimate the initial data.
Theorem 3.7. Let L ∈ N and T > 0. Then there exists a constant C = C(L, T ) > 0 such that where (θ, u) denotes the solution of (2.18).
Proof. We follow closely [30]. In the first step, we transform the problem into a spectral problem.
In the second step, we solve the spectral problem by using Paley-Wiener theorem combined with complex analysis. Introduce some notations. For any T > 0, let Step 1. Reduction to a spectral problem.
Pick any L ∈ (0, +∞) \ N and any T > 0. If (3.9) is not true, one can find a sequence Extracting a subsequence if necessary, one can assume that (θ 0 n , u 0 n ) → (θ 0 , u 0 ) weakly in X 1 , and hence strongly in X 0 . Let (θ n , u n ) and (θ, u) denote the solutions of (2.18) associated with (θ 0 n , u 0 n ) and (θ 0 , u 0 ), respectively. Using (3.10), we infer that θ n,x (·, L) → 0 strongly in L 2 (0, T ), and hence θ x (·, L) = 0. Using (2.10), (3.10) and the fact that (θ n , u n ) → (θ, u) in C([0, T ], X 0 ), we see that (θ n , u n ) → (θ, u) strongly in L 2 (0, T, X 1 ), so that (θ 0 n , u 0 n ) → (θ 0 , u 0 ) strongly in X 1 , and hence (θ 0 , u 0 ) X 1 = 1. Thus (θ 0 , u 0 ) ∈ N T \ {(0, 0)}. To obtain a contradiction, it is sufficient to prove the following On the other hand, the vector space N T has a finite dimension, for its unit ball is compact. Indeed, the same argument as above shows that any sequence (θ 0 n , u 0 n ) n∈N in the unit ball of N T has a convergent subsequence for the norm · X 1 . If Claim 1 is not true, one may find T > 0 such that dim N T > 0. Since the map T ∈ (0, +∞) → dim N T ∈ N is nonincreasing, one may pick T ∈ (0, T ) and ε > 0 such that T + ε < T and and let (θ, u) denote the solution of (2.18). By the semigroup property, we have In Let N C T denote the complexification of N T , and let = d/dx. Since it was assumed that dim N T = 0, we infer that the linear map (3.14) We show in the next step that (3.11)-(3.14) has no nontrivial solution when L ∈ N.
Step 2. Study of the spectral problem.
Let (θ, u) be as in Claim 2, and extend θ and u to R by setting where δ ζ denotes the Dirac measure at x = ζ and the derivatives u (0), u (0), u (L), θ (0), θ (L) are those of the functions u and θ when restricted to [0, L]. Taking the Fourier transform of 2 We removed the superscript 0 to simplify the notations. We still used the notation X3 to denote the space of complex-valued functions in (2.2).
each term in the above system, we obtain Setting λ = −ip, f (ξ) :=θ(ξ) +û(ξ), and g(ξ) :=θ(ξ) −û(ξ), we arrive to Since both θ and u have a compact support, it follows from the Paley-Wiener theorem that the functions f and g have to be entire (i.e. holomorphic in the whole plane C).
(ii) If β = 0, then since g(−ξ) is entire, each root µ k of Q has to satisfy α − βiµ k + γ e iLµ k = 0, i.e. µ k = (α + γ e iLµ k )/(iβ). Since e iLµ k is a root of P , we arrive to the conclusion that We infer that Q cannot have three distinct roots.
-If γ = 0, then taking the module of both terms in (3.18) yields L = 0, which is impossible. Assume now that µ 0 = − 1 √ 3 . Then, proceeding as above, we obtain that β = Lγe iL/ √ 3 and . Again, we see that we obtain a contradiction for γ = 0 or for γ = 0. The proof of Theorem 3.7 is complete.
With Theorem 3.7 at hand, we deduce that (2.17) is exactly controllable in X −1 with g 2 as unique control (case 1) when L ∈ N. On the other hand, the exact controllability fails when L ∈ N according to case 4.
Let us turn our attention to the cases 6, 7, 8 and 9 for which we have added one Dirichlet control to the Neumann control g 2 . First, we notice that in case 1, we have for L ∈ N the exact controllability in X −2 by picking the controls g 2 in H − 1 3 (0, T ). Indeed, from (3.9), we have for By interpolation between (3.9) and (3.19), we infer that for all (θ 0 , u 0 ) ∈ X 2 , . (3.20) The observability inequality (3.20) implies the exact controllability in X −2 in case 1 (with g 2 ∈ H − 1 3 (0, T )) as well as in cases 6 to 9 for L ∈ N. It remains to show that any L ∈ N is still a critical length in cases 6 to 9. We claim that in each of those cases, the corresponding spectral problem has for L ∈ N a nontrivial solution (θ, u), so that the corresponding observability inequality fails for the exponential solution Re[e λt (θ, u)].
Proof. Let L ∈ N. Then we can write L = 2π k 2 +kl+l 2 3 with k, l ∈ N * . Following [30], we introduce the numbers We use again the notations in the proof of Theorem 3.7 (Claim 2). Let . We know that we can find coefficients α, β, γ, α , γ not all zero so that the two functions f (ξ) and g(−ξ) are entire, the roots of Q(ξ) = ξ 3 −ξ+p being µ 0 , µ 1 , µ 2 . Furthermore, it follows from Paley-Wiener theorem (see [30]) that the spectral problem (3.11)-(3.14) has indeed a nontrivial solution. Our concern is to prove that we can as well impose the addition condition in each case. From our choice of the µ k 's, we have that the quantity e −iLµ k is independent of k. Set We shall pick β = 0 in all the cases. 1. The additional condition u (L) = 0 is equivalent to γ = −γ. We can pick γ = 1, γ = −1, α = −C, and α = C.

Dirichlet controls.
We consider the cases in which only Dirichlet controls are involved. We start with a preparatory result which is an observability inequality with the measurement of three traces.
Proof. First, we introduce the space which is a Banach space when endowed with the norm Pick any (θ 0 , u 0 ) ∈ X 4 and let (θ, u) denote the solution of (2.18). Taking the derivative w.r.t.
We are in a position to investigate the case 5. The following observability inequality improves those in Proposition 3.9.
Proof. We proceed in two steps as in the proof of Theorem 3.7.
Step 2. Study of the spectral problem.
We now turn our attention to cases 2 and 3. We need the following estimate, whose (long) proof is postponed in an appendix.
Proof of Corollary 3.12: We proceed in two steps as in the proof of Theorem 3.7.
Step 1. Reduction to a spectral problem.
Step 2. Study of the spectral problem.
and hence (3.64) Set λ = −ip. By Paley-Wiener theorem, we conclude that (3.58) is equivalent to the existence of numbers p ∈ C and (α, α , β, γ, γ ) ∈ C 5 \ {(0, 0, 0, 0, 0)} such that the two functions defined for ξ ∈ R by fulfill the conditions f and g are entire; (3.67) are entire. Let Q(ξ) = ξ 3 − ξ + p and let µ 0 , µ 1 , µ 2 denote its roots. The polynomial function Q cannot have a triple root (see the proof of Corollary 3.10), so either the roots µ k , k = 0, 1, 2, are simple, or there are a double root µ 0 = µ 1 and a simple root µ 2 = µ 0 . 1. Assume that the roots µ 0 , µ 1 , µ 2 of Q are simple. If ξ ∈ {µ 0 , µ 1 , µ 2 }, then from the fact that the functions in (3.70)-(3.71) are entire, we infer that and hence The polynomial functions in the two sides of the above equation are of degree two and they take the same values on the numbers µ k , k = 0, 1, 2. Therefore, they must have the same coefficients; that is,  • Assume that − γ iγ ∈ {µ 0 , µ 1 , µ 2 }, then each root µ j of Q should also be a root of 1 − sin(Lξ) by (3.81), and hence it could be written as We arrive to the conclusion that We arrive to the system
2. Assume that Q has a double root µ 0 = µ 1 and a simple root . We shall consider the case (µ 0 , 3 ) being similar. As µ 0 is a double root of Q, it should be root of the numerators of the functions f (ξ) + g(ξ) and f (ξ) − g(−ξ) (see (3.70) and (3.71)) and of their first derivatives, so that As µ 2 is a simple root of Q, it has to be a root of the numerators of the functions f (ξ) + g(−ξ) and f (ξ) − g(−ξ), so that Note that (3.92)-(3.93) can be rewritten as In particular, (iγ , γ) ∈ R 2 . From (3.90) and (3.94), we infer that Combined with (3.96) and (3.97), this yields As the set of solutions of (3.98) cannot have a limit point, we conclude that on any segment [0, R] there are at most finitely many L satisfying (3.98). From (3.91) and (3.95), we infer that Combined with (3.96) and (3.97), this yields (with β = 1) We claim that cos( Indeed, otherwise we would have either cos( L On the other hand, it follows from (3.98) that .
and hence L = 0, which is impossible. The proof of Proposition 3.13 and of Corollary 3.12 is complete.
We now turn our attention to case 3.
Proof of Theorem 3.16: We proceed in two steps as in the proof of Theorem 3.7.
Step 1. Reduction to a spectral problem.
It is similar to Case 3. We find, instead of (3.130)-(3.131), the system Introducing again a := iLµ 0 , b = iLµ 1 , we see that the existence of p ∈ C, γ 2 ∈ C * and µ 0 , µ 1 , µ 2 (the roots of Q) holds if and only if L ∈ G .
2. Assume that Q has a double root µ 0 = µ 1 and a simple root µ 2 = µ 0 . Then (µ 0 , . We shall consider the case (µ 0 , 3 ) being similar. As µ 0 is a double root of Q, it should be a root of the numerators of the functions f (ξ) + g(−ξ) and f (ξ) − g(−ξ) (see (3.121) and (3.122)) and of their first derivatives, so that As µ 2 is a simple root of Q, it has to be a root of the numerators of the functions f (ξ) + g(−ξ) and f (ξ) − g(−ξ), so that Note that (3.134)-(3.135) can be rewritten as If β = 0, then we obtain −γ 2 iL + iγ = γ 1 + i γ Assume now that β = 0. Picking β = 1, we obtain We can compute all the coefficients in (3.132)-(3.137) in terms of L. Indeed, we infer from (3.133) and (3.139) that cos(L/ √ 3) = 0 and · Combined with (3.138), this yields · It follows that 132) and (3.136), we infer that Substituting the values of γ 2 , γ 1 + γ i √ 3 and γ 1 − γ 2i √ 3 in (3.140), we obtain the equation Simplifying the above equation, we arrive to From (3.133) and (3.137), we infer that Substituting the values of γ 2 , γ 1 + γ i √ 3 and γ 1 − γ 2i √ 3 in (3.141), we obtain the equation After some simplifications, we arrive to Next, we investigate the spectral problem associated with case 12. Proof. We use the same notations as in the proof of Proposition 3.17. The extra condition u (L) = 0 means that γ 1 = 0. As for Proposition 3.17, the case when Q has multiple roots is impossible. We therefore assume that Q has three different roots. a. If γ = 0, then β = 0 and for each root ξ of Q, we have that We assume γ 2 = 0 (otherwise all the coefficients are zero), and we arrive to e 2iLξ = −1 and 2Lξ = π + 2πk, k ∈ Z. Since the sum of the three roots of Q is 0, we arrive to a contradiction. b. If γ = 0, then we have (still) γ 1 = 0, β = ±iγ = 0 and α = ±iγ 2 , a case already considered in the proof of Proposition 3.17. The cases 1 and 2 have to be excluded, while the cases 3 and 4 are valid and they lead to L ∈ G ∪ G .

Proof of Theorems 1.1 and 1.2.
We shall first investigate the wellposedness of system (1.20) when picking g 2 (t) = −αη x (t, L) as control input, with α > 0. We first notice that a global Kato smoothing effect holds, thanks to which we can derive the wellposedness of the system in the energy space Let us first pay our attention to the linearized system x ∈ (0, L). (4.1) We introduce the operatorÃ Proposition 4.1. Assume that α > 0. Then the operatorÃ generates a semigroup of contractions (e tÃ ) t≥0 in X 0 . Furthermore, for any T > 0 and any (η 0 , v 0 ) ∈ X 0 , the solution (η, v) := e tÃ (η 0 , v 0 ) of (4.1) belongs to L 2 (0, T, [H 1 (0, L)] 2 ) and we have Proof. It is clear that D(Ã) is dense in X 0 and thatÃ is closed. Let (η, v) ∈ D(Ã). Then we readily obtain that (Ã(η, v), (η, v)) X 0 = −αη x (L) 2 ≤ 0, so thatÃ is a dissipative operator. Introduce the operator B(θ, u) := (u x + u xxx , θ x + θ xxx ) with domain We easily obtain that (B(θ, u), (θ, u)) X 0 = −αθ x (L) 2 ≤ 0, so that B is also a dissipative operator. We claim that B =Ã * . First, for all (η, v) ∈ D(Ã) and all (θ, u) ∈ D(B), we have by integration by parts that Let us now check thatÃ * ⊂ B. Pick any (θ, u) ∈ D(Ã * ). Then, we have for some constant C > 0 that Picking v = 0 and η ∈ C ∞ c (0, L), we infer from (2.7) that u x + u xxx ∈ L 2 (0, L), and hence that u ∈ H 3 (0, L). Similarly, we obtain that θ ∈ H 3 (0, L). Integrating by parts in the left hand side of (4.3), we obtain that It follows that so that (θ, u) ∈ D(B) andÃ * = B. We conclude that the operator A is m−dissipative in X 0 and that it generates a semigroup of contractions in X 0 . Pick any (η 0 , v 0 ) ∈ D(Ã) and let (η, v)(t) := e tÃ (η 0 , v 0 ). Scaling the two first equations in (4.1) by η and v respectively, and summing the two obtained equations we arrive to the energy identity Scaling the two first equations in (4.1) by xv and xη, respectively, and summing the obtained equations, we arrive to Combined with (4.4), it yields We search a solution of x ∈ (0, L), as a solution of the integral equation Introducing the map Γ defined by we will show that for any (sufficiently small) time T > 0, the map Γ has a unique fixed-point in some closed ball B(0, R) in the space endowed with the norm Such a fixed-point yields a local solution of (4.5).
Proof. The proof is very similar to those of the wellposedness of the KdV equation with the boundary conditions u(t, 0) = u(t, L) = u x (t, L) = 0 (see [29,30]). First, we can prove as in [30] that for some constant C 1 > 0, we have for all T ∈ (0, 1] and all (f, g) Following [29], one can find a constant C(L) > 0 such that for all T ∈ (0, 1] and all (η, v) ∈ E T , it holds + v x L 2 (0,T,L 2 (0,L)) η 1 2 L ∞ (0,T,L 2 (0,L)) η x 1 2 L 2 (0,T,L 2 (0,L)) . (4.8) It follows that there are some positive constants C 2 (L), C 3 (L) such that for all T > 0 and all Picking R = 2C 2 (η 0 , v 0 ) X 0 and T > 0 such that 2C 3 T 1 4 R = 1/2, we see that the map Γ is a contraction in the closed ball B(0, R) of E T , and hence that it has a unique fixed point by the contraction mapping theorem.
Let us now proceed to the proof of Theorems 1.1 and 1.2. Assume that L ∈ (0, +∞)\N. First, we show that the semigroup (e tÃ ) t≥0 is exponentially stable in X 0 ; that is, for some constants C, µ > 0 it holds (4.11) It is actually sufficient to prove that for some T > 0 and some C = C(T ) > 0, Indeed, combining (4.12) with (4.4), we obtain (η(T ), v(T )) 2 which yields (4.11) by the semigroup property.
Thus ( Let us now proceed to the proof of Theorem 1.1. We first notice that the linear system x ∈ (0, L), is well posed for (η 0 , v 0 ) ∈ X 0 and h ∈ L 2 (0, T ). Clearly, the wellposedness can be derived for h ∈ C 2 ([0, T ]) by performing the change of unknownsη(t, x) := η(t, x),ṽ(t, L) := v(t, L) + h(t)g(x), where the function g ∈ C ∞ ([0, L]) is such that g(0) = g(L) = 0 and g (L) = −1. To extend the result from C 2 ([0, T ]) to L 2 (0, T ), we need to derive some a priori estimates. Scaling in the first (resp. second) equation of (4.15) by η (resp. v), we obtain after some integrations by parts This yields for all T ≥ 0 and thus (η, v) ∈ C([0, T ], X 0 ) if (η 0 , v 0 ) ∈ X 0 and h ∈ L 2 (0, T ). Scaling now in the first (resp. second) equation of (4.15) by xv (resp. xη) yields for some constant The same computations as in the proof of Theorem 1.2 show that the operator generates a semigroup of contractions in X 0 . Next, performing the change of variables t → T − t and x → L − x, we infer that for any (θ 1 , u 1 ) ∈ X 0 , the backward system x ∈ (0, L), We now prove the exact controllability in X 0 of the linear system (4.15). Scaling in the first (resp. second) equation of (4.15) by θ (resp. u), we obtain after some integrations by parts By the Hilbert Uniqueness Method (see [23,24]), the exact controllability of (4.15) holds in X 0 with control inputs h ∈ L 2 (0, T ) if and only if the following observability inequality holds for the solutions of the backward system (4.17). Again the backward system enjoys the global Kato smoothing property: Proceeding as in [30], we easily see that if (4.18) is false, then we can find a pair of data (θ 1 , u 1 ) in X 0 such that (θ 1 , u 1 ) X 0 = 1 and θ x (·, L) ≡ 0. But this is impossible by Theorem 3.7, for L ∈ N.
We have established the exact controllability of the linear system (4.15). The (local) exact controllability of the nonlinear system (1.20) in X 0 follows at once by applying the contraction mapping theorem as in [30] for KdV. (Note that η x (·, L) ∈ L 2 (0, T ) by (4.16), and hence g 2 = −αη x (·, L) + h ∈ L 2 (0, T ) as well.) We omit the details for the sake of shortness. The proof of Theorem 1.1 is complete.
The proof of Theorem 3.11 is not based on the multiplier method, but on the analysis of the spectral properties of the operator A. More precisely, we estimate the asymptotic behavior of the eigenvalues of A and use it to establish the observability inequalities (3.50)-(3.51). The proof of Theorem 3.11 is outlined as follows. In Step 1, we introduce the operator By = −y (L − x) − y (L − x) with domain D(B) = {y ∈ H 3 (0, L) ∩ H 1 0 (0, L); y (L) = 0}, which is closely related to the operator A (but more easy to handle). We prove that it is selfadjoint and that it has a compact resolvent, so that it can be diagonalized in an orthonormal basis in L 2 (0, L). In Step 2, we estimate the asymptotic behavior of the eigenvalues of B. Finally, in Step 3 we show that A can be diagonalized in an orthonormal basis of [L 2 (0, L)] 2 and use the expansions of the solutions in terms of the eigenfunctions to prove (3.50) and (3.51).
Step 1 (Study of the operator B) Let Picking any y, z ∈ D(B), we readily obtain by integration by parts that i.e. (By, z) L 2 = (y, Bz) L 2 where (., .) L 2 stands for the scalar product in L 2 (0, L). This means that D(B) ⊂ D(B * ) and that B * = B on D(B). Conversely, pick any z ∈ D(B * ). Then Setting w(x) = z(L − x), we arrive to |(y + y , w) L 2 | ≤ C y L 2 , or equivalently for y ∈ D = D(0, L) w + w , y D ,D ≤ C y L 2 · It follows that w + w ∈ L 2 (0, L), and hence w ∈ H 3 (0, L) and z ∈ H 3 (0, L). Integrating by parts in (By, z) L 2 and using the fact that y + y ∈ L 2 (0, L), we arrive to We search for the function y in the form y(x) = y 1 (x) + y 2 (x) with We see at once that y 1 ∈ H 3 (0, L) with y 1 (x) + y 1 (x) = −z(L − x) a.e. in (0, L) and y 1 (L) = y 1 (L) = y 1 (L) = 0. Thus, it remains to find y 2 ∈ H 3 (0, L) such that y 2 + y 2 = 0 and y 2 (L) = y 2 (L) = 0, y 2 (0) = −y 1 (0). By linearity, there is no loss of generality in assuming that y 2 (0) = 1. The three roots of the equation r 3 + r = 0 are r 1 = i, r 2 = −i and r 3 = 0. We search for y 2 in the form where the coefficients a 1 , a 2 , a 3 have still to be found. It is clear that y 2 solves y 2 + y 2 = 0, and the boundary conditions y 2 (L) = y 2 (L) = 0 and y 2 (0) = 1 give the following conditions  Using (4.19), (4.20) and the precise values of r 1 , r 2 , r 3 , we obtain that a 3 = 1 2 − a 1 − a 2 and (Note that L ∈ 2πZ, for L ∈ N.) Plugging these expressions of a 2 and a 3 in (4.21), we arrive to We readily see that the coefficient behind a 1 is not zero, for L ∈ 2πZ, so that the last equation for a 1 can be solved. Claim A.2 is proved.
There is an orthonormal basis (v n ) n∈N in L 2 (0, L) composed of eigenvectors of B: for all n ∈ N, v n ∈ D(B) and Bv n = λ n v n for some λ n ∈ R.
It is a direct consequence of Claims A.1 and A.2, since B −1 is a bounded compact selfadjoint operator in L 2 (0, L). Thus B −1 is diagonalizable in an orthonormal basis in L 2 (0, L), and the same is true for B.
Step 2 (Asymptotics of the eigenvalues of B) Claim A.4 Using a convenient relabeling, the sequence of eigenvalues of B can be written (λ n ) n∈Z , with λ n ≤ λ n+1 for all n ∈ Z and λ n = for some numbers k 1 , k 2 ∈ Z. The eigenvalues of B as given in Claim A.3 satisfy |λ n | → +∞ as n → +∞. We shall show that they can be separated into two subsequences, one with the asymptotics (4.22) and another one with the asymptotics (4.23).
Thus if e r 2 L = e r 3 L , then a 2 can be expressed in terms of a 1 , and the system has a solution (a 1 , a 2 , a 3 ) = (0, 0, 0) if and only if r 1 (e r 1 L + i)(e r 2 L − e r 3 L ) + r 2 (e r 2 L + i)(e r 3 L − e r 1 L ) + r 3 (e r 3 L + i)(e r 1 L − e r 2 L ) = 0. (4.32) The case e r 2 L = e r 3 L is actually impossible for |λ| large enough, see below (4.33) and (4.37). We shall use several times the following classical result.
Theorem 4.3. (Rouché's theorem, see e.g. [35, 3.42]) Let f and g be analytic inside and on a closed contour Γ, and such that |f (z) − g(z)| < |f (z)| on Γ. Then f and g have the same number of zeros inside Γ.
• Assume that λ → −∞. We still choose r 1 , r 2 , r 3 so that r 1 ∼ −iλ Then we obtain that Clearly, |e r 1 L | → 1, |e r 2 L | → 0 and |e r 3 L | → +∞. On the other, we have that  On the other hand, we can prove as above that for all n ∈ Z with n < 0 and |n| large enough, there is indeed an eigenvalue and that it is simple.
• Finally, we can relabel the λ n 's so that the eigenvalues of B form a sequence (λ n ) n∈Z as in Claim A.4.
Pick any T > 0. It follows then from Ingham's lemma that there exist an integer N ∈ N and a constant K > 0 such that for all (c + n ) n∈Z , (c − n ) n∈Z ∈ l 2 (Z), we have Next, we compare |v n (0)| to |v n (L)| as |n| → +∞.
It follows from (4.27) that v n (0) = partially supported by CNPq (Brazil) and the International Cooperation Agreement Brazil-France. LR was partially supported by the ANR project Finite4SoS (ANR-15-CE23-0007) and the project ICoPS Math-AmSud.