On the controllability of the fifth order Korteweg-de Vries equation

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1 On the controllability of the fifth order Korteweg-de Vries equation O. Glass & S. Guerrero January 22, 29 Abstract In this paper, we consider the fifth-order Korteweg-de Vries equation in a bounded interval. We prove that this equation is locally well-posed when endowed with suitable boundary conditions, and establish a result of local controllability to the trajectories. Résumé Dans ce papier, nous considérons l équation de Korteweg-de Vries du cinquième ordre en domaine borné. Nous montrons que l équation est localement bien posée lorsque l on impose certaines données au bord, et établissons un résultat de contrôlabilité locale aux trajectoires. Keywords. Controllability, Fifth order Korteweg-de Vries equation, initial-boundary value problem. 1 Introduction In this paper, we study the controllability of the fifth order Korteweg-de Vries equation: u t + αu 5x + µu xxx + βuu xxx + δu x u xx + P (uu x =, (1 where α, µ, β and δ are real constants and P is a cubic polynomial: P (u = pu + qu 2 + ru 3. (2 This class of equations was introduced by Kichenassamy and Olver [16]. It contains in particular the Kawahara equation [14] introduced to model magneto-acoustic waves, the various models derived by Olver [18] for the unidirectional propagation of waves in shallow water when the third order term appearing in the Korteweg-de Vries equation is small, and many other models. See [16] for a discussion of them. In this paper, we are interested in studying this equation in a bounded domain. We will both consider the Cauchy problem with boundary conditions and the boundary controllability problem. Note that there is an important literature concerning the Cauchy problem in the real line, see for instance [6, 9, 16, 15, 17, 2] and references therein. But for what concerns the boundary value problem, the problem was still completely open as far as we know. Note that the initial boundary value problem for the (third-order Korteweg-de Vries equation has drained much attention (see in particular [1, 4, 5, 1, 13]. The controllability problem was also, up to our knowledge, completely open. The equivalent for the Korteweg-de Vries equation has also known many developments lately [2, 3, 7, 12, 21, 22, 23, 24, 25]. To be more precise, we will consider in the sequel that α > : this is not a restriction since it suffices to make the change of variable x = 1 x and to invert the role of the left and right boundaries. The spatial domain will be [, 1], which is not a restriction either in the present paper, since it will suffice to rescale in space to obtain a result on an interval of arbitrary length. (Note that this is not necessarily the case for the Korteweg-de Vries equation with Neumann boundary control, see [21]. The boundary conditions that we will consider are the following: u x= = v 1, u x=1 = v 2, u x x= = v 3, u x x=1 = v 4, u xx x= = v 5. (3 The authors are partially supported by the Agence Nationale de la Recherche, Project ContrôleFlux. They would like to thank Jean-Cluade Saut for useful discussions. They also thank the anonymous referee for helpful suggestions. 1

2 The first and main result of this paper concerns a boundary controllability result for equation (1. To be more precise, we will control the system from the right endpoint (by using only v 2 and v 4 while maintaining v 1, v 3, v 5 to zero, and the type of controllability that we consider is the local controllability to trajectories. That is to say, we consider T > and a fixed trajectory u of (1, and prove that for any initial state u sufficiently close to u t=, there exist controls (v 2, v 4 which steer the system from u to u t=t. The precise result is the following. Theorem 1. Let T >. Let u L (, T ; W 3, (, 1 be a trajectory of (1 with boundary conditions u x= = u x x= = u xx x= =. There exists ε > such that for any u L 2 (, 1 such that u u(, L 2 (,1 ε, (4 there exist two controls v 2, v 4 in L 2 (, T such that the solution u of (1 with initial condition u t= = u, (5 and boundary condition (3 with controls (, v 2,, v 4,, belongs to C ([, T ]; L 2 (, 1 L 2 (, T ; H 2 (, 1 and satisfies u t=t = u(t,. (6 As we will see, the solution to the controllability problem that we construct is in fact more regular than C ([, T ]; L 2 (, 1 L 2 (, T ; H 2 (, 1. Indeed, in order to prove Theorem 1, we will first take the controls (v 2, v 4 as zero and we will prove, thanks to Theorem 3 below, that the state becomes H 2. Then we will work with more regular solutions (belonging to L 2 (ɛ, T ; H 4 (, 1 C ([ɛ, T ]; H 2 (, 1. Remark 1. Using the reversible character of this system stated on the whole real line, it is not difficult to deduce that equation (1 is locally exactly controllable near when using the five controls (for sufficiently regular states. Remark 2. As we will see in Section 4.3, a solution of (1 with boundary condition (3 with v 1 = v 3 = v 5 = is regularized away from the right endpoint and the initial condition. This involves that equation (1 cannot be locally exactly controllable by means of v 2 and v 4 only. The next results of this paper concerns the Cauchy problem. We prove that the problem is well posed locally in time, and regularizes the state of the system when the boundary conditions are homogeneous. Theorem 2. Given u L 2 (, 1, T >, v 1, v 2 H 2 5 (, T, v 3, v 4 H 1 5 (, T, v 5 L 2 (, T there exists T (, T ] such that the nonlinear problem (1 with boundary conditions (3, admits a unique solution u L 2 (, T ; H 2 (, 1 C ([, T ]; L 2 (, 1 satisfying ( u L 2 (,T ;H 2 (,1 C ([,T ];L 2 (,1 C u L 2 (,1 + 5 k=1 v k H 2 (k 1/2 5 (,T Theorem 3. Moreover, if v 1 = v 2 = v 3 = v 4 = v 5 =, this solution is regularized in the sense that for any τ (, T ], u C ([τ, T ] [, 1] with, for all k, u H k (τ,t ;H k (,1 C(τ, k u L2 (,1. (7. We conclude this introduction by a remark concerning the choice of the controls among v 1,..., v 5. The controllability to the trajectories described in Theorem 1 may not take place if one chooses another set of controls, for instance when acting through v 5 only and keeping v 1 = v 2 = v 3 = v 4 =. This is given in the next example. Proposition 1. Let L > be a solution of tan(l = L. The system ϕ t ϕ 5x ϕ xxx = in (, T ( L, L, ϕ x= L = ϕ x=l = ϕ x x= L = ϕ x x=l = ϕ xx x=l = in (, T, ϕ t=t = ϕ T on ( L, L, (8 2

3 has solutions satisfying ϕ(, and ϕ xx (t, L = in (, T, (9 for all T >. As a consequence the system u t + u 5x + u xxx = in (, T ( L, L, u x= L = u x=l = u x x= L = u x x=l = in (, T, u xx x= L = v 5 (t in (, T, is not null approximately controllable by the control v 5. (1 Proof of Proposition 1. We introduce the following (time-independent function where g(x = cos(x + µx 2 (cos(l + µl 2, µ := sin(l 2L = cos(l. 2 It is elementary to check that { g5x + g xxx = in (, T ( L, L, g x= L = g x=l = g x x= L = g x x=l = g xx x= L = g xx x=l = in (, T, hence g satisfies (8 and (9. Now the equivalence between the unique continuation of (8 and the approximate controllability of (1 is an application of the standard duality in PDE control theory, see for instance [8]. Let us note that this phenomenon of critical values of the length of the domain was raised by Rosier [21] for the linearized KdV equation (see [2, 3, 7] for further developments on this subject. Hence, according to the values of the length of the domain and of the coefficients, a similar behaviour can take place here. We believe that this leads to many open and challenging problems. The structure of the paper is the following. In Section 2, we study the initial boundary value problem for a linearized equation. This requires proving a regularizing effect on the equation ζ t + ζ 5x = g. In Section 3, we study the controllability of a linearized equation. In Section 4, we use a fixed point argument to establish Theorems 2 and 3 and an inverse mapping theorem to establish Theorem 1. Finally Sections 5 and 6 are devoted to the most technical parts of the paper, namely, the proof of a Carleman estimate and a proof of the regularizing effect to the left. 2 Cauchy problem for the linearized equation In this section, we study the well posedness of the following linearized equation: where the functions a k satisfy 3 y t + αy 5x = a k (t, x xy k + h, (11 k= a k L (, T ; W k, (, 1 for k = (12 Recall that we consider α >. We will state the corresponding result in Paragraph 2.3. For this, we will first study the adjoint system of (11: ψ t + αψ 5x = 3 k= ( 1k+1 x(a k k (t, xψ + f, in (, T (, 1, ψ x= = ψ x x= = in (, T, ψ x=1 = ψ x x=1 = ψ xx x=1 = in (, T, ψ t=t = in (, 1. (13 3

4 2.1 The equation ζ t + αζ 5x = g We begin with the following proposition. Proposition 2. Consider α >. Given ζ T (H 5 H 2 (, 1 satisfying ζ T,xx (1 = and g C 1 ([, T ]; L 2 (, 1 there exists a unique solution ζ C ([, T ]; H 5 (, 1 C 1 ([, T ]; L 2 (, 1 of ζ t + αζ 5x = g in (, T (, 1, ζ x= = ζ x x= = in (, T, ζ x=1 = ζ x x=1 = ζ xx x=1 = in (, T, ζ t=t = ζ T in (, 1. (14 Proof of Proposition 2. This follows from the standard Lumer-Phillips theory. We can introduce the operator A : D(A L 2 (, 1: D(A = {ϑ (H 5 H 2 (, 1/ϑ xx (1 = } and Aϑ = αϑ 5x. Then one can see that its adjoint is defined via D(A = {h (H 5 H 2 (, 1/h xx ( = } and A h = αh 5x. Then it is elementary to check that Aϑ, ϑ L 2 and A h, h L 2 so that Proposition 2 follows from standard operator theory [19]. Now we prove some estimates for the solutions of (14. We define the spaces, for k N, X k := {y L 2 (, T ; H k+2 (, 1 C ([, T ]; H k (, 1, y xx x= H k/5 (, T }, endowed with their natural norm. Proposition 3. One has the following estimates on the solutions of (14: and ζ Xs C g L 2 (,T ;H s 2 (,1, for s [, 1], (15 ζ Xs C g L1 (,T ;H s (,1, for s [, 1], (16 ζ xxx x= H 1/5 (,T + ζ xxx x=1 H 1/5 (,T C g L 2 ((,T (,1, (17 ζ 4x x= L 2 (,T + ζ 4x x=1 L 2 (,T C g L 2 ((,T (,1. (18 Moreover in (17 and (18 one can replace g L 2 ((,T (,1 by g L 1 (,T ;H 2 (,1. Remark 3. If we interpolate (15 and (16, we also deduce ζ Xs C g L 4/3 (,T ; H s 1 (,1, for s [, 1]. (19 Proof of Proposition 3. We consider a smooth solution of (14 and establish several estimates on it. 1. Proof of (15-(16. Estimate in X. We multiply (14 by (1 + xζ: 1 d 2 dt (1 + xζ 2 dx + αζxx x= α ζxx 2 dx = 4 (1 + xgζ dx. (2

5 It follows that It is also clear that from (2 it follows ζ X C g L 2 (,T ;H 2 (,1. (21 ζ X C g L 1 (,T ;L 2 (,1. (22 Estimate in X 5. Now we consider g L 2 (, T ; H 3 (, 1. Observe that due to (14, for such a g, the traces of ζ 5x and ζ 6x on both sides, and the trace of ζ 7x on the right, vanish. We apply the operator 5x to the equation and we apply (21: Using the equation, this gives ζ 5x X C g 5x L 2 (,T ;H 2 (,1. ζ xx x= H1 (,T C g L2 (,T ;H 3 (,1. This yields also In the same way, we have ζ X5 C g L 2 (,T ;H 3 (,1. (23 ζ X5 C g L 1 (,T ;H 5 (,1. (24 Estimate in X 1. Here we consider g L 2 (, T ; H 8 (, 1. We apply the operator 5x to the equation and we apply (23 (since g 5x L 2 (, T ; H 3 (, 1: ζ 5x X5 C g 5x L 2 (,T ;H 3 (,1. This yields as previously Also we have ζ X1 C g L 2 (,T ;H 8 (,1. (25 ζ X1 C g L 1 (,T ;H 1 (,1. (26 Interpolation argument. By an interpolation argument, we deduce (15 and (16 for every s [, 1]. 2. Proof of (17. Let ρ C 4 ([, 1]; R satisfying ρ(x = for x [, 1/2] and ρ(x = 1 for x [3/4, 1]. We use estimate (15 for s = 1: ζ X1 C g L2 (,T ;H 1 (,1. (27 Multiplying (14 with ρζ xx, integrating in space and integrating by parts, we get, for almost any t [, T ], α 2 ζ xxx x=1 2 = 3α 2 ρ x ζ xxx 2 dx α 2 ρ xxx ζ xx 2 dx 1 d 2 dt ρ ζ x 2 dx ρ x ζ x, ζ t H 2 (,1 H 2 (,1 g, ρζ xx H 1 (,1 H 1 (,1. (28 Integrating in time and thanks to (14-(27, we get ζ X1 + ζ xxx x=1 L 2 (,T + ζ xxx x= L 2 (,T C g L 2 (,T ;H 1 (,1. (29 An estimate for ζ xxx x= can be done in the same way, by employing the weight 1 ρ. Now, we use estimate (15 for s = 6: ζ X6 C g L2 (,T ;H 4 (,1. (3 5

6 In order to prove that ζ txxx x=1 L 2 (, T, we multiply (14 by ρ t 7 xζ, we integrate in space and we integrate by parts (using again what we know on the traces of ζ 5x, ζ 6x and ζ 7x : 1 2 ζ txxx x=1 2 = 7 2 α 2 d dt ρ x ζ txxx 2 dx + ζ txxx (6ρ xx ζ txx + 4ρ xxx ζ tx + ρ 4x ζ t dx (ρ ζ 6x 2 ρ xx ζ 5x 2 dx + α ρ x ζ 6x, ζ t5x H 2 H 2 + ζ txxx x=1g xxx x=1 4 ζ txxx j= ( ( 4 j xg j x 4 j ρ dx. (31 As previously, the same can be done for ζ txxx x=. Integrating in time, using Cauchy-Schwarz inequality to estimate the last term in the second line and using (14-(3, we get ζ X6 + ζ txxx x=1 L2 (,T + ζ txxx x= L 2 (,T C g L 2 (,T ;H 4 (,1. (32 An interpolation argument applied to (29 and (32 provides ( Proof of inequality (18. We multiply the equation of ζ by ρζ 4x and we integrate in space. After some integration by parts, we obtain: α 2 ζ 4x x=1 2 = α 2 ρ x ζ 4x 2 dx 1 d 2 dt Integrating in time this identity, we have ρ ζ xx 2 dx ρ xx ζ xx, ζ t H 1 H 1 2 ρ x ζ xx, ζ tx H 2 H + ρζ 2 4x g dx. (33 ζ 4x x=1 L2 (,T C( ζ L (,T ;H 2 (,1 + ζ t L 2 (,T ;H 1 (,1 + ζ L 2 (,T ;H 4 (,1 + g L 2 ((,T (,1. Using (15, we have estimated ζ 4x x=1 as in (18. (1 ρζ 4x. The estimate ζ 4x x= is similar by multiplying by Finally, the proof of (17 and (18 with g L 1 (, T ; H 2 (, 1 is completely analogous. 2.2 Well posedness for the adjoint equation Now we can state the following existence and regularity result for (13. Proposition 4. Given a k satisfying (12 and f L 2 (, T ; L 2 (, 1, there exists a unique solution ψ L 2 (, T ; H 4 (, 1 C ([, T ]; H 2 (, 1 of (13. Proof of Proposition 4. We use a fixed point scheme. Given ˆψ L 2 (, T ; H 4 (, 1 C ([, T ]; H 2 (, 1, we consider the solution ψ := T ˆψ of where ˆT (, T is to be fixed later. ψ t + αψ 5x = 3 k= ( 1k+1 x(a k k (t, x ˆψ + f in ( ˆT, T (, 1, ψ x= = ψ x x= = in ( ˆT, T, ψ x=1 = ψ x x=1 = ψ xx x=1 = in ( ˆT, T, ψ t=t = in (, 1, (34 6

7 Using Proposition 3 (precisely (15 for s = 2, we infer that T ˆψ 1 T ˆψ 2 X2 C( a k L (,T ;W k, ˆψ 1 ˆψ 2 L2 (,T ;H 3 Note that the constant in (15 is independent of ˆT (, T. Then by interpolation we deduce that C( a k L (,T ;W k, ˆT 1/4 ˆψ 1 ˆψ 2 L4 (,T ;H 3. (35 T ˆψ 1 T ˆψ 2 X2 C( a k L (,T ;W k, ˆT 1/4 ˆψ 1 ˆψ 2 1/2 L (,T ;H 2 ˆψ 1 ˆψ 2 1/2 L 2 (,T ;H 4 C( a k L (,T ;W k, ˆT 1/4 ˆψ 1 ˆψ 2 X2. (36 It follows that T is contracting for sufficiently small time ˆT. Then extending the solution obtained in ( ˆT, T to a solution in (, T is standard using the linear character of the equation. Furthermore, the solutions described in Proposition 4 possess the following regularity property. Proposition 5. Under the assumptions of Proposition 4, the solution ψ has the following hidden regularity: ψ X2 + ψ xx x=,1 H 2/5 (,T + ψ xxx x=,1 H 1/5 (,T + ψ 4x x=,1 L2 (,T C f L2 ((,T (,1. (37 Proof of Proposition 5. This is a consequence of Propositions 3 and 4. Note that due to the contracting character of T and using Proposition 3, we have Now we can use ψ X2 T ( X2 f L2 ((,T (,1. g := 3 ( 1 k+1 x(a k k (t, xψ + f, k= as a right hand side in (14 to deduce (37 from Proposition Well posedness for the initial boundary value problem In this paragraph we give the notion of solution of y t + αy 5x = 3 k= a k(t, x xy k + h in (, T (, 1, y x= = v 1, y x=1 = v 2, y x x= = v 3 in (, T, y x x=1 = v 4, y xx x= = v 5 in (, T, y t= = y in (, 1, where y, h, v 1,..., v 5 are given function. The solution of (38 for homogeneous boundary conditions and h L 2 (, T ; L 2 (, 1 is granted by Proposition 4 (replace t by T t and x by 1 x. Hence we can suppose without loss of generality that h =. Definition 1. Let y H 2 (, 1, v 1, v 2 L 2 (, T, v 3, v 4 H 1/5 (, T and v 5 H 2/5 (, T. We call y a solution by transposition of (38 with h =, a function y L 2 ((, T (, 1 such that y f dx dt = u, ψ t= H 2 (,1 H 2 (,1 + α v 1 ψ 4x x= dt α α v 3, ψ xxx x= H 1/5 (,T H 1/5 (,T + α v 4, ψ xxx x=1 H 1/5 (,T H 1/5 (,T +α v 5, ψ xx x= H 2/5 (,T H 2/5 (,T + where ψ is the solution of (13 associated to f. v 2 ψ 4x x=1 dt a 3 x= v 1 ψ xx x= dt, f L 2 ((, T (, 1, (38 (39 7

8 Proposition 6. Let a k satisfy (12. There exists a unique solution by transposition of system (38 with h =. Moreover, there exists C > such that y L 2 ((,T (,1 C( y H 2 (,1+ v 1 L 2 (,T + v 2 L 2 (,T + v 3 H 1/5 (,T + v 4 H 1/5 (,T + v 5 H 2/5 (,T. Proof: All comes to prove ψ C ([, T ]; H 2 (, 1, ψ 4x x=,1 L 2 (, T, ψ xxx x=,1 H 1/5 (, T, ψ xx x= H 2/5 (, T and the following inequality: ψ L (,T ;H 2 (,1 + ψ 4x x=,1 L 2 (,T + ψ xxx x=,1 H 1/5 (,T This was established in Proposition 5. + ψ xx x= H 2/5 (,T C h L2 ((,T (,T. (4 Corollary 1. Suppose that a k = for k = For any T > there exists C > (non decreasing in T such that, if v 1, v 2 H 2/5 (, T, v 3, v 4 H 1/5 (, T and v 5 L 2 (, T, the above solution of system (38 with h = and y = belongs to L 2 (, T ; H 2 (, 1 and satisfies that y L2 (,T ;H 2 (,1 C ([,T ];L 2 (,1 C( v 1 H 2/5 (,T + v 2 H 2/5 (,T + v 3 H 1/5 (,T + v 4 H 1/5 (,T + v 5 L2 (,T. Proof. Suppose that v 1, v 2 H 1 (, T, v 3, v 4 H 4/5 (, T and v 5 H 3/5 (, T. We apply Proposition 6 to initial state and controls v 1t, v 2t, v 3t, v 4t and v 5t, and deduce a solution z in L 2 ((, T (, 1. Using Proposition 3 (for ψ in the L 1 (, T ; H 2 (, 1 framework, we deduce as previously that z is moreover in C ([, T ]; H 2 (, 1. Next we apply Proposition 6 to initial state and controls v 1, v 2, v 3, v 4 and v 5, and deduce a solution y. Using Definition 1 with smooth test functions f, it is not difficult to see that z = y t. Using the equation, we infer that y L 2 (, T ; H 5 (, 1 C ([, T ]; H 3 (, 1. Now the conclusion follows by interpolation. The fact that the constant C can be chosen non decreasing in time is simple: given < T < T and v 1, v 2 H 2/5 (, T, v 3, v 4 H 1/5 (, T and v 5 L 2 (, T, one extends these boundary values to [, T ] by. One applies the inequality for time [, T ], and one sees that one can choose a constant for T which is not larger. 3 Controllability of the linearized equation 3.1 Carleman estimate We consider the following dual system ϕ t + αϕ 5x = 3 k= ( 1k+1 x(a k k (t, xϕ + f in (, T (, 1, ϕ x= = ϕ x=1 = ϕ x x= = ϕ x x=1 = ϕ xx x=1 = in (, T, ϕ t=t = ϕ T on (, 1, (41 where the functions a k satisfy (12. A central argument in this paper consists in establishing a Carleman inequality for (41. For this let us set β(x α(t, x =, (42 t 1/4 (T t 1/4 for (t, x. Weight functions of this kind were first introduced by A. V. Fursikov and O. Yu. Imanuvilov; see [11]. In the above equation β is a positive, strictly decreasing and concave polynomial of degree 2 in [, 1]. Observe that the function α satisfies C T 1/2 α, C α α x C 1 α, C α α xx C 1 α in (, T [, 1], (43 α t + α xt + α xxt CT α 5, α tt C(T 2 α 9 + α 5 CT 2 α 9 in (, T [, 1], (44 where C, C and C 1 are positive constants independent of T. We have: 8

9 Proposition 7. Suppose that (12 applies. There exists a positive constant C independent of T such that, for any ϕ T L 2 (, 1 and f L 2 (, T ; L 2 (, 1, we have e 2sα α( ϕ 4x 2 + s 2 α 2 ϕ xxx 2 + s 4 α 4 ϕ xx 2 + s 6 α 6 ϕ x 2 + s 8 α 8 ϕ 2 dt dx ( C α x=1 e 2sα x=1 ( ϕ 4x x=1 2 + s 2 α x=1 2 ϕ xxx x=1 2 dt + s 1 e 2sα f 2 dt dx, (45 for any s C(T 1/4 + T 1/2, where ϕ is the solution of (41. The proof of this inequality is postponed to Section 5. Remark 4. We will also require β to satisfy max β(x < 2 min β(x. (46 x [,1] x [,1] This is not needed for Proposition 7 (nor to Proposition 8 below, but will be useful later. 3.2 Weighted observability estimate Now let us deduce from Proposition 7 a slightly modified inequality, with a weight function not vanishing at t =. We begin by introducing a new weight. Set l on [, T ] by T 2 if t T l(t := 4 2, (47 t(t t otherwise. Now introduce γ(t, x = β(x. (48 l(t 1/4 Proposition 8. Suppose that (12 applies. There exist two positive constants s and C > depending on T such that, for any ϕ T L 2 (, 1 and any f L 2 (, T ; L 2 (, 1, we have e 2sγ γ 9 ϕ 2 dx dt + where ϕ is the solution of (41. ( ϕ(, x 2 dx C e 2sγ f 2 dt dx + γ x=1 e 2sγ x=1 ( ϕ 4x x=1 2 + γ x=1 2 ϕ xxx x=1 2 dt, (49 Proof of Proposition 8. We use the following energy estimate: ϕ L (,T/2;L 2 (,1 C exp{c a k L (,T ;W k, (,1}( f L 2 (,3T/4;L 2 (,1 + ϕ L 2 (T/2,3T/4;L 2 (,1. (5 To get (5, introduce η C ([, T ]; R such that η = 1 in [, T/2] and η = in [3T/4, T ], multiply equation (41 by η(t(1 + xϕ, and perform several integration by parts as in (21. Let us notice that the weight functions γ and e 2sγ are positive for t [, T/2]. Hence there is a constant C such that e sγ γ 9/2 ϕ L (,T/2;L 2 (,1 C exp{c a k L (,T ;W k, (,1} ( e sγ f L2 (,3T/4;L 2 (,1 + e sγ γ 9/2 ϕ L2 (T/2,3T/4;L 2 (,1. (51 9

10 Next, we use (45 and the choice of γ to deduce T 2 (,1 ( e 2sγ γ 9 ϕ 2 dx dt C e 2sγ f 2 dt dx + γ x=1 e 2sγ x=1 ( ϕ 4x x=1 2 + γ x=1 2 ϕ xxx x=1 2 dt, (52 for s large enough. Combining (51 and (52 we obtain (49. Let us consider s as in Proposition 8. We introduce κ := s 2 T 1/4 max β(x and κ 1 := s x [,1] T 1/4 min β(x. (53 x [,1] Corollary 2. Under the assumptions of Proposition 8, one has e 2κ (T t 1/4 (T t 9/4 ϕ 2 dx dt + + ( ϕ(, x 2 dx C e 2κ 1 (T t 1/4 f 2 dt dx (T t 1/4 e 2κ 1 (T t 1/4 ( ϕ 4x x=1 2 + (T t 1/2 ϕ xxx x=1 2 dt. ( Controllability We introduce the following space: κ 1 E = {y L 2 (, T ; L 2 (, 1 / e (T t 1/4 y L 2 (, T ; L 2 (, 1}. (55 We have the following controllability result. κ Proposition 9. Given h such that (T t 9/8 e (T t 1/4 h L 2 ((, T (, 1 and y L 2 (, 1, there exist controls v 2, v 4 L 2 (, T satisfying κ 1 (T t 1/8 e (T t 1/4 v 2 L 2 (, T and (T t 3/8 e (T t 1/4 v 4 L 2 (, T, (56 such that if we call y the solution of (38 starting from y with v 1 = v 3 = v 5 =, then y belongs to E. In particular y, which belongs to C ([, T ]; H 5 (, 1, satisfies Besides, there exists a constant C > such that κ 1 κ 1 y t=t = on (, 1. (57 e (T t 1/4 y L 2 (,T ;L 2 (,1 + (T t 1/8 e (T t 1/4 (v 2, (T t 1/4 v 4 L 2 (,T ( C y L 2 (,1 + (T t 9/8 e κ 1 κ (T t 1/4 h L 2 ((,T (,1. (58 Proof of Proposition 9. The proof is inspired by Fursikov and Imanuvilov s approach [11]. Define L Ly := y t + αy 5x 3 a k (t, x xy, k (59 k= 1

11 and L its dual operator: 3 L φ := φ t αφ 5x ( 1 k x(a k k (t, xφ. (6 k= Let us set F = {φ C ([, T ] [, 1]; R / φ x= = φ x=1 = φ x x= = φ x x=1 = φ xx x=1 = }. Consider the bilinear form a( ˆφ, φ = e 2κ 1 (T t 1/4 L ˆφ L φ dx dt + e 2κ 1 (T t 1/4 (T t 1/4[ ˆφ4x x=1 φ 4x x=1 + (T t 1/2 ] ˆφxxx x=1 φ xxx x=1 dt ˆφ, φ F. We also introduce the linear form l, φ = hφ dt dx + u φ t= dx. (61 Introduce F the completion of F for the norm φ a(φ, φ 1/2 (it is a norm from Corollary 2. The next step in this proof is to demonstrate that there exists exactly one ˆφ in the class F satisfying a( ˆφ, φ = l(φ, φ F. (62 Now F is a Hilbert space for the scalar product a(,, hence in order to get (62 it is sufficient to prove that l is a continuous linear form on F. From Cauchy-Schwarz inequality, we see that hφ dt dx (T t 9/8 κ e (T t 1/4 h L2 (,T ;L 2 (,1 (T t 9/8 e κ (T t 1/4 φ L2 (,T ;L 2 (,1. (63 Using the assumption on h and Corollary 2, one sees that l is indeed a continuous linear form on F. Hence there exists a unique ˆφ F satisfying (62. Let us set y = e 2κ 1 (T t 1/4 L ˆφ v2 = (T t 1/4 e 2κ 1 (T t 1/4 ˆφ4x x=1 and v 4 = (T t 3/4 e 2κ 1 (T t 1/4 ˆφxxx x=1. (64 Finally it is not difficult to see that y E, that (v 2, v 4 satisfies (56 and that y is a solution of (38 with v 1 = v 3 = v 5 =. This concludes the proof of Proposition 9. Now we define the space κ 1 E 1 = {y E / (T t 5/4 e (T t 1/4 y L 2 (, T ; H 4 (, 1 C ([, T ]; H 2 (, 1, y x= = y x x= = y xx x= =, (T t 9/8 e (T t 1/4 Ly L 2 (, T ; L 2 (, 1}. (65 κ κ Proposition 1. Given h such that (T t 9/8 e (T t 1/4 h L 2 ((, T (, 1 and y H 2 (, 1, there exist controls (v 2, v 4 L 2 (, T 2 such that the associated solution y of (38 with v 1 = v 3 = v 5 = belongs to E 1 and moreover satisfies (T t 5/4 κ 1 ( e (T t 1/4 y L 2 (,T ;H 4 (,1 C ([,T ];H 2 (,1 C y H 2 (,1+ (T t 9/8 κ e, for some C >. (T t 1/4 h L 2 ((,T (,1 (66 11

12 Proof of Proposition 1. We extend the problem to the interval [, 2]. We extend y (resp. h by in [1, 2] (resp. [, T ] [1, 2], we call ỹ (resp. h the resulting function. We also extend ak in [, T ] [1, 2] in a way that keeps the L (, T ; W k, regularity (in a continuous way, and in such a way that a k (t, x = in [, T ] [ 3 2, 2]. We now consider the following control problem ỹ t + αỹ 5x = 3 k= ãk(t, x xỹ k + h in (, T (, 2, ỹ x= = ỹ x x= = ỹ xx x= = in (, T, ỹ x=2 = ṽ 2, ỹ x x=2 = ṽ 4, in (, T, ỹ t= = ỹ in (, 2. (67 According to Proposition 9, there exist ṽ 2, ṽ 4 fulfilling (56 such that the corresponding solution ỹ belongs to E (adapted to the interval [, 2] of course. Now we claim that the restriction of ỹ to [, T ] [, 1] satisfies the required properties. We have to establish that κ 1 (T t 5/4 e (T t 1/4 y L 2 (, T ; H 4 (, 1 C ([, T ]; H 2 (, 1. For that, we introduce This function satisfies κ 1 y (t, x := (T t 5/4 e (T t 1/4 ỹ(t, x. (68 y t + αy 5x = 3 k= ãk(t, x k xy + h in (, T (, 2, y x= = y x x= = y xx x= = in (, T, y x=2 = v 2, y x x=2 = v 4 in (, T, y t= = y in (, 2, (69 where y = T 5/4 e κ 1 T 1/4 ỹ, (v2, v4 = (T t 5/4 e (T t 1/4 (ṽ 2, ṽ 4 κ 1 and h = (T t 5/4 κ 1 d e (T t 1/4 h + dt [(T t5/4 e κ 1 (T t 1/4 ]ỹ. These data are in H 2 (, 2, in L 2 (, T 2 and in L 2 (, T ; L 2 (, 2 respectively, thanks to Proposition 9. We will use the following lemma, whose proof is postponed to the Appendix. Lemma 1. For k large enough, one has (2 x k y L 2 (, T ; H 4 (, 2 C ([, T ]; H 2 (, 2 with the estimate (2 x k+ 1 2 y L (,T ;H 2 (,2 + for some positive constant C. 4 (2 x k+j 4 xy j L2 (,T ;L 2 (,2 j= C ( h L2 (,T ;L 2 (,2 + y H 2 (,2 + v 2 L 2 (,T + v 4 L 2 (,T, (7 Now we use (7 and the continuity of the previous extensions from (, 1 to (, 2 to deduce κ 1 (T t 5/4 e (T t 1/4 y L 2 (,T ;H 4 (,1 C ([,T ];H 2 (,1 ( C y H 2 (,1 + (T t 5/4 κ 1 e (T t 1/4 h L2 ((,T (,1 + d dt [(T t5/4 e κ 1 κ 1 (T t 1/4 ]y L2 ((,T (,1 + (T t 1/8 e (T t 1/4 (v 2, (T t 1/4 v 4 L2 (,T, 12

13 for some C >. Finally, we use κ 1 < κ to estimate the second term in the right hand side, and (58 to estimate the last two terms. We deduce (66. 4 Nonlinear problem 4.1 Proof of Theorem 2 We use a fixed point scheme to prove local existence and uniqueness in X := L 2 (, T ; H 2 (, 1 C ([, T ]; L 2 (, 1. Given z X, we introduce the solution of u t + αu 5x + µu xxx + βzu xxx + δz x u xx + P (zu x = in (, T (, 1, u x= = v 1, u x=1 = v 2, u x x= = v 3, u x x=1 = v 4, u xx x= = v 5 in (, T, (71 u t= = u in (, 1. Call T the operator which maps z to u. The existence and uniqueness of u is obtained as in Proposition 4: one associates to ˆψ X the solution of u t + αu 5x = µ ˆψ xxx βz ˆψ xxx δz x ˆψxx P (z ˆψ x in (, T (, 1, u x= = v 1, u x=1 = v 2, u x x= = v 3, u x x=1 = v 4, u xx x= = v 5 in (, T, (72 u t= = u in (, 1. Let us notice that ˆψ xxx L 2 (, T ; H 1 (, 1 while (by interpolation z L 4 (, T ; H 1 (, 1, and hence z ˆψ xxx L 4/3 (, T ; H 1 (, 1; on the other hand, one also sees that z x ˆψxx L 4/3 (, T ; H 1 (, 1. It follows then from Remark 3 and Corollary 1 that (72 defines a solution in L 2 (, T ; H 2 (, 1 C ([, T ]; L 2 (, 1. Now consider ˆψ 1 and ˆψ 2 in X, and their images u 1 and u 2 by the above mapping. Making the difference of the two equations, multiplying by (2 x(u 1 u 2 and performing the same operations as in Proposition 3, we infer [ u 1 u 2 2 X C + We deduce for small ε > (u 1 u 2 x ( ˆψ 1 ˆψ 2 xx dt dx + z(u 1 u 2 x ( ˆψ 1 ˆψ 2 xx dt dx + u 1 u 2 2 X C ˆψ 1 ˆψ 2 L 2 (,T ;H 2 (,1 [ u 1 u 2 L 2 (,T ;H 1 (,1 (u 1 u 2 z x ( ˆψ 1 ˆψ 2 xx dt dx (u 1 u 2 (1 + z 2 ( ˆψ 1 ˆψ 2 x dt dx ]. (73 + z x L 8 3+2ɛ (,T ;L (,1 u 1 u 2 L 8 1 2ε (,T ;L 2 (,1 + z L 8 1 2ε (,T ;L 2 (,1 u 1 u 2 L 8 + C ˆψ 1 ˆψ 2 L 4 (,T ;H 1 (,1 3+2ɛ (,T ;W 1, (,1 [ u 1 u 2 L 4/3 (,T ;L 2 (,1 + u 1 u 2 L 4 (,T ;L 6 (,1 z 2 L 4 (,T ;L 6 (,1 Note that by interpolation and Sobolev imbeddings we have ] ]. (74 L 2 (, T ; H 2 (, 1 L (, T ; L 2 (, 1 L 12 (, T ; L 6 (, 1 L 8 3+2ε (, T ; W 1, (, 1. (75 We infer that (at least if T 1 u 1 u 2 2 X CT 1 2ε 8 (1 + z 2 X ˆψ 1 ˆψ 2 X u 1 u 2 X. Hence the operator is contractive for sufficiently small time T, which proves the local well posedness of (71. 13

14 Now let us prove that T has a fixed point. We decompose u = û + ǔ with ǔ t + αǔ 5x = in (, T (, 1, ǔ x= = v 1, ǔ x=1 = v 2, ǔ x x= = v 3, ǔ x x=1 = v 4, ǔ xx x= = v 5 in (, T, ǔ t= = in (, 1, and û t + αû 5x + µû xxx + βzû xxx + δz x û xx + P (zû x = ȟ in (, T (, 1, û x= = û x=1 = û x x= = û x x=1 = û xx x= = in (, T, û t= = u in (, 1, (76 with ȟ = (µǔ xxx + βzǔ xxx + δz x ǔ xx + P (zǔ x. Let us prove that for some constant C > independent of T 1 and T 1/16 z X small enough, the solution u of (71 satisfies ( ( u X C exp CT 1/16 (1 + z 2 X u L 2 (,1 + v 1 H 2/5 (,T + v 2 H 2/5 (,T + v 3 H 1/5 (,T + v 4 H 1/5 (,T + v 5 L 2 (,T. (77 That ǔ satisfies (77 for some constant C > is a consequence of Corollary 1. For what concerns û, we multiply (76 by (2 xû; after some integration by parts, we can deduce d dt û 2 L 2 + 5α 2 û xx 2 dx C(1 + z x (t, 2 + z(t, 2 û 2 L + ν û 2 xx 2 dx + 2 (2 xûȟ dx, for arbitrarily small ν. We use (75 for z and deduce when putting ε = 1/4 that (1 + z x (t, 2 + z(t, 2 dt T 1/16 (1 + z 2 X. We choose ν small enough, use a Gronwall argument and deduce for all t [, T ] that û(t 2 L 2 + 2α t û xx 2 dx dτ exp{ct 1/16 (1 + z 2 X} ( u 2 L Now we take the supremum in t. The last term is treated as follows: (2 xûz x ǔ xx dx dt T 1 2ε 8 T 1 2ε 8 ûz x ǔ xx dx L 8 7+2ε (,T û L 2 (,1 z x L (,1 ǔ xx L 2 (,1 T 1 2ε 8 û X z X ǔ X, (2 xûȟ dx dt. L 8 7+2ε (,T using again (75. The other trilinear terms can be estimated analogously (up to an integration by parts. Then taking again ε = 1/4 and imposing T 1/16 z X 1 we obtain (77. Now let us show that T is contractive on { B := u X / u X (C + 1 ( u L2 (,1 + v 1 H 2/5 (,T + v 2 H 2/5 (,T for sufficiently small T, where C is the constant in (77. + v 3 H 1/5 (,T + v 4 H 1/5 (,T + v 5 L2 (,T }, 14

15 From (77, we see that, provided that T is suitably small, B is stable by T. We now consider that this is the case. Now consider z 1, z 2 B and denote u 1 := T z 1, u 2 := T z 2, z := z 1 z 2, u := u 1 u 2. We have u t + αu 5x + µu xxx + βz 1 u xxx + βzu 2,xxx + δz 1,x u xx + δz x u 2,xx + P (z 1 u x + [P (z 1 P (z 2 ]u 2,x = in (, T (, 1, (78 We multiply (78 by (2 xu, integrate in both time and space and perform the same reasoning as in (73-(74. After lengthy but straightforward computations, we deduce (if T 1 u 2 X CT 1/16 [ u 2 X u X z X (1 + z 1 X + z 2 X + (1 + z 1 2 X u 2 X]. (79 Using that both u 2 and z 1 belong to B, this establishes that T is contractive on B for sufficiently small time T. 4.2 Proof of Theorem 3 We consider a solution u L 2 (, T ; H 2 (, 1 C ([, T ]; L 2 (, 1 of (1 with homogeneous boundary conditions. We introduce ε (, 1 arbitrarily small and η C ([, T ]; R such that η = in [, ε T/4] and η = 1 in [ε T/2, T ]. We consider the equation satisfied by ηu (ηu t + α(ηu 5x = µηu xxx βηuu xxx δηu x u xx ηp (uu x + η u in (, T (, 1. Now let us look at the regularity of the right hand side. It is not difficult to see that the less regular terms are the second and third one. Concerning the term u x u xx, we have by interpolation u L 4 (, T ; H 1 (, 1, so that (u x 2 L 2 (, T ; L 1 (, 1, and hence u x u xx L 2 (, T ; W 1,1 (, 1 L 2 (, T ; H 3 2 ε (, 1. For what concerns uu xxx, we use that uu xxx = (uu xx x u x u xx. Now we use that by interpolation u L ε (, T ; H 2 +ε (, 1 and u xx L 8 3 2ε (, T ; H 1 2 ε (, 1, so that the product is in L 2 (, T ; H 1 2 ε (, 1. As a conclusion, the term uu xxx has the same regularity as u x u xx. Now we use Proposition 3, and infer that for arbitrary ε >, one has ηu L 2 (, T ; H 5 2 ε (, 1 C ([, T ]; H 1 2 ε (, 1. Then repeating the above steps we can show by a bootstrap argument that the solution u becomes C in time and space in arbitrary small time. 4.3 Proof of Remark 2 We start from a solution u L 2 (, T ; H 2 (, 1 C ([, T ]; L 2 (, 1. Then proceeding as in the proof of Theorem 3 we can prove that the right hand side is in L 2 (, T ; H 3 2 ε (, 1. Now using Lemmata 2 and 3 posed in [, 1] rather than in [, 2] (see in Section 6 below and interpolation arguments, we deduce that (1 x 7 u L 2 (, T ; H 5 2 ε (, 1. Then restricting to the space interval [, 1 ν] for ν > suitably small and applying a bootstrap procedure, we see that u is regular at positive distance of the right endpoint. 4.4 Proof of Theorem 1 We consider a trajectory u as indicated in the statement; then u = u + y satisfies (1 if and only if y satisfies y t + y 5x + µy 3x + β((u + yy 3x + u 3x y + δ((u + y x y xx + u xx y x + py x + 2q((u + yy x + u x y + 3r(y(2u + y(u + y x + u 2 y x =. (8 Conspicuously, the controllability of (1 to the trajectory u is equivalent to the null controllability of (8. Now we have the following result for (8. Proposition 11. Given y L 2 (, 1 and u L (, T ; W 3, (, 1, there exists T > such that the nonlinear problem (8 with homogenous boundary conditions (3 (v 1 = v 2 = v 3 = v 4 = v 5 = admits 15

16 a unique solution y L 2 (, T ; H 2 (, 1 C ([, T ]; L 2 (, 1, which regularizes in the sense that for any τ (, T ], u L 2 ([τ, T ]; H 4 (, 1 C ([τ, T ]; H 2 (, 1, with moreover y C ([τ,t ];H 2 (,1 C(τ, u u L 2 (,1. (81 The proof of Proposition 11 follows the steps of the proof of Theorems 2 and 3; all the computations are justified thanks to u L (, T ; W 3, (, 1. We omit the details. We now turn to the proof of Theorem 1. The solution of the controllability problem is obtained in two successive steps. In a first step, we set the controls (v 2, v 4 to (,. According to Proposition 11, this regularizes the state of the system, so that we may consider that the initial state y belongs to H 2 (, 1 and is small (see (81. From now, we consider that this is the case, and proceed to the proof of the null-controllability of system (8 with such an initial state, by using the inverse mapping theorem. We introduce the coefficients a k as follows Recall that L is expressed by (59. Define a = βu 3x + 2qu x + 6ru u x, a 1 = δu xx + p + 2qu + 3ru 2, a 2 = δu x, a 3 = µ + βu. Y 1 := {f L 2 (, T ; L 2 (, 1 / (T t 9/8 e (T t 1/4 f L 2 (, T ; L 2 (, 1}, (82 equipped with the clear corresponding norm. We consider the following map κ Λ : { E1 H 2 (, 1 Y 1 y (y(, Ly + βyy xxx + δy x y xx + (2q + 6ruyy x + 3ruy 2 + 3ry 2 y x. (83 Recall that the definition of E 1 was given in (65. Note that the mapping Λ is well defined and C 1. Indeed, from y E 1, we find out that Then, thanks to (46 and (53, we have that 2κ 1 β(t t 5/2 e (T t 1/4 yy xxx L 2 (, T ; L 2 (, 1. κ β(t t 9/8 e (T t 1/4 yy xxx L 2 (, T ; L 2 (, 1. The same can be done for all the other terms (since they are bilinear or trilinear. Now using Proposition 1, we see that Λ ( is a surjective map. Hence there exists a neighborhood of (, in H 2 (, 1 Y 1 on which Λ is onto. This gives the desired result. 5 Proof of Proposition 7 Let ψ := e sα ϕ, where α is given by (42 and ϕ fulfills system (41. We deduce that L 1 ψ + L 2 ψ = L 3 ψ, with and L 3 ψ = L 1 ψ = ψ t + ψ 5x + 1s 2 α 2 xψ xxx + 5s 4 α 4 xψ x, (84 L 2 ψ = 5sα x ψ 4x + 1s 3 α 3 xψ xx + s 5 α 5 xψ + sα t ψ + 1sα xx ψ xxx + 3s 3 α 2 xα xx ψ x, (85 { [e e sα sα (3sα xx ψ x + 3s 2 α x α xx ψ ] + [ e sα (3sα xx xx ψ xx + 6s 2 α x α xx ψ x ] { x } 6sα xx ψ xxx + 12s 2 α x α xx ψ xx 15s 3 αxα 2 xx ψ x + 7s 4 αxα 3 xx ψ + 6s 3 α x αxxψ 2 { +e sα f + } 3 k= ( 1k+1 x(a k k (t, xe sα ψ. } (86 16

17 (We recall that α xxx =. Then, we have L 1 ψ 2 L 2 ( + L 2ψ 2 L 2 ( + 2 L 1 ψ L 2 ψ dx dt = L 3 ψ 2 L 2 (. (87 The main part of what follows consists in evaluating the double product term. We will denote by (L i ψ j (1 i 4, 1 j 6 the j-th term in the expression of L i ψ. We recall that α >, α x <, α xx < and α xxx =. In the sequel we will repeatedly use that ψ x=,1 = ψ x x=,1 =. First, integrating by parts with respect to x and t, we have ((L 1 ψ 1, (L 2 ψ 1 L2 ( = 5s α x ψ tx ψ xxx dt dx 5s α xx ψ t ψ xxx dt dx = 5 2 s α x ( ψ xx 2 t dt dx + 1s α xx ψ tx ψ xx dt dx + 5s CsT α 5 ψ xx 2 dt dx + 1s α xx ψ tx ψ xx dt dx. For the second term, we get ((L 1 ψ 1, (L 2 ψ 2 L 2 ( = = 5s 3 αx( ψ 3 x 2 t dt dx 3s 3 Cs 3 T α 7 ψ x 2 dt dx 3s 3 α 2 xα xx ψ t ψ x dt dx α 2 xα xx ψ t ψ x dt dx. α xxx ψ t ψ xx dt dx (88 (89 For the third term, we obtain ((L 1 ψ 1, (L 2 ψ 3 L 2 ( = 1 2 s5 αx(ψ 5 2 t dt dx Cs 5 T α 9 ψ 2 dt dx. We consider now the fourth term of L 2 ψ and using (44 we readily get ((L 1 ψ 1, (L 2 ψ 4 L2 ( = s α t (ψ 2 t dt dx 2 CsT 2 α 9 ψ 2 dt dx. (9 (91 The next term gives ((L 1 ψ 1, (L 2 ψ 5 L 2 ( = 1s α xx ψ tx ψ xx dt dx. (92 The last term gives ((L 1 ψ 1, (L 2 ψ 6 L 2 ( = 3s 3 α 2 xα xx ψ x ψ t dt dx. (93 All these computations ((88-(93 show that ((L 1 ψ 1, (L 2 ψ L2 ( CsT α 5 ψ xx 2 dt dx Cs 3 T α 7 ψ x 2 dt dx C(s 5 T + st 2 α 9 ψ 2 dt dx ɛs 5 α 5 ψ xx 2 dt dx ɛs 7 α 7 ψ x 2 dt dx ɛs 9 α 9 ψ 2 dt dx, (94 17

18 for any ɛ >, provided that s CT 1/4, where C depends on ɛ. Now we consider the second term of L 1. The product with the first term of L 2 gives ((L 1 ψ 2, (L 2 ψ 1 L 2 ( = 5 2 s α xx ψ 4x 2 dt dx s α x x=1 ψ 4x x=1 2 dt 5 2 s α x x= ψ 4x x= 2 dt. (95 Similar computations give the following for the second term: ((L 1 ψ 2, (L 2 ψ 2 L2 ( = 5s 3 αx( ψ 3 xxx 2 x dt dx 1s 3 αx x= 3 ψ 4x x=ψ xx x= dt 3s 3 αxα 2 xx ψ 4x ψ xx dt dx 45s 3 αxα 2 xx ψ xxx 2 dt dx 5s 3 αx x=1 3 ψ xxx x=1 2 dt For the third one we have ((L 1 ψ 2, (L 2 ψ 3 L 2 ( = s 5 +5s 3 αx x= 3 ψ xxx x= 2 dt 1s 3 αx x= 3 ψ 4x x=ψ xx x= dt T +3s 3 αx x= 2 α xx x=ψ xxx x= ψ xx x= dt Cs 3 α 3 ψ xxx ψ xx dt dx. = s5 2 +5s 5 α 5 xψ 4x ψ x dt dx 5s 5 αx( ψ 5 xx 2 x dt dx + 1s 5 s5 2 Cs 5 (α 4 xα xx x ψ xxx ψ dt dx α 5 x x= ψ xx x= 2 dt 25 2 s5 α 5 ( ψ xxx ψ + ψ xx ψ x dt dx. α 4 xα xx ψ 4x ψ dt dx α 4 xα xx ψ xxx ψ x dt dx α 4 xα xx ψ xx 2 dt dx (96 (97 Then, we see that ((L 1 ψ 2, (L 2 ψ 4 L 2 ( = s α t ψ 4x ψ x dt dx s α tx ψ 4x ψ dt dx = s α t ( ψ xx 2 x dt dx + 2s α tx ψ 3x ψ x dt dx 2 +s α txx ψ xxx ψ dt dx CsT α 5 ( ψ xx 2 + ψ xxx ( ψ + ψ x dt dx Next, ((L 1 ψ 2, (L 2 ψ 5 L2 ( = 1s We used that α xxx =. 1s CsT α 5 x= ψ xx x= 2 dt. α xx ψ 4x 2 dt dx + 1s α xx x=1 ψ 4x x=1 ψ xxx x=1 dt α xx x= ψ 4x x= ψ xxx x= dt. (98 (99 18

19 Finally, ((L 1 ψ 2, (L 2 ψ 6 L 2 ( = 3s 3 αxα 2 xx ψ 4x ψ xx dt dx 3s 3 (αxα 2 xx x ψ 4x ψ x dt dx = 3s 3 αxα 2 xx ψ xxx 2 dt dx + 3s 3 (αxα 2 xx x= ψ xxx x= ψ xx x= dt Cs 3 α 3 ( ψ 4x ψ x + ψ xxx ψ xx dt dx. (1 Putting together all the computations concerning the second term of L 1 ψ ((95-(1, we obtain ((L 1 ψ 2, L 2 ψ L2 ( 25 2 s α xx ψ 4x 2 dt dx + 75s 3 αxα 2 xx ψ xxx 2 dt dx 25 2 s5 αxα 4 xx ψ xx 2 dt dx 5 2 s α x x= ψ 4x x= 2 dt s5 αx x= 5 2 ψ xx x= 2 dt 1s 3 αx x= 3 ψ 4x x=ψ xx x= dt T +5s 3 αx x= 3 ψ xxx x= 2 dt ɛs 9 α 9 ψ 2 dt dx ɛs 7 α 7 ψ x 2 dt dx ɛs 5 α 5 ψ xx 2 dt dx ɛs 3 α 3 ψ xxx 2 dt dx ɛs α ψ 4x 2 dt dx ɛs 5 α x= 5 ψ xx x= 2 dt ɛs 3 α x= 3 ψ xxx x= 2 dt ɛs T α x= ψ 4x x= 2 dt Cs Cs 3 α x=1 3 ψ xxx x=1 2 dt, T α x=1 ψ 4x x=1 2 dt for any ɛ >, provided that s C(T 1/4 + T 1/2, where C depends on ɛ. (We used that s C(ɛT 1/2 for appropriate C(ɛ and α CT α 3. We consider now the products concerning the third term of L 1 ψ. First, we have ((L 1 ψ 3, (L 2 ψ 1 L2 ( = 75s 3 αxα 2 xx ψ xxx 2 dt dx + 25s 3 αx x=1 3 ψ xxx x=1 2 dt Secondly ((L 1 ψ 3, (L 2 ψ 2 L 2 ( = 25s 5 25s 3 αx x= 3 ψ xxx x= 2 dt. (11 (12 αxα 4 xx ψ xx 2 dt dx 5s 5 αx x= 5 ψ xx x= 2 dt. (13 Third, ((L 1 ψ 3, (L 2 ψ 3 L2 ( = 5s 7 15s 7 α 7 x( ψ x 2 x dt dx 7s 7 α 6 xα xx ψ x 2 dt dx Cs 7 α 6 xα xx ψ xx ψ dt dx α 7 ψ x ψ dt dx. For the fourth term, we have ((L 1 ψ 3, (L 2 ψ 4 L2 ( = 1s 3 αxα 2 t ψ xx ψ x dt dx 1s 3 (αxα 2 t x ψ xx ψ dt dx Cs 3 T α 7 ( ψ x 2 + ψ ψ xx dt dx. (14 (15 19

20 We obtain the following for the fifth term: ((L 1 ψ 3, (L 2 ψ 5 L2 ( = 1s 3 Finally, ((L 1 ψ 3, (L 2 ψ 6 L2 ( = 3s 5 αxα 4 xx ψ xx 2 dx dt Cs 5 α 2 xα xx ψ xxx 2 dt dx. (16 α 5 ψ xx ψ x dt dx. (17 Consequently, we get the following for the third term of L 1 ψ ((12-(17: ((L 1 ψ 3, L 2 ψ L 2 ( 25s 3 αxα 2 xx ψ xxx 2 dt dx 55s 5 αxα 4 xx ψ xx 2 dt dx +15s 7 αxα 6 xx ψ x 2 dt dx 25s 3 αx x= 3 ψ xxx x= 2 dt 5s 5 αx x= 5 ψ xx x= 2 dt ɛs 9 α 9 ψ 2 dt dx ɛs 7 α 7 ψ x 2 dt dx ɛs 5 α 5 ψ xx 2 dt dx Cs 3 α x=1 3 ψ xxx x=1 2 dt, for any ɛ >, where again s C(T 1/4 + T 1/2 and C depends on ɛ. Now, we compute the fourth term. First, we have: ((L 1 ψ 4, (L 2 ψ 1 L 2 ( = 25 2 s5 αx( ψ 5 xx 2 x dx dt 125s 5 αxα 4 xx ψ xxx ψ x dx dt s5 αxα 4 xx ψ xx 2 dt dx s5 αx x= 5 ψ xx x= 2 dt Cs 5 α 5 ψ x ψ xx dx dt. (18 (19 Next, we obtain ((L 1 ψ 4, (L 2 ψ 2 L2 ( = 175s 7 α 6 xα xx ψ x 2 dt dx. (11 For the third term, we get ((L 1 ψ 4, (L 2 ψ 3 L2 ( = 45 2 s9 α 8 xα xx ψ 2 dt dx. (111 Then, ((L 1 ψ 4, (L 2 ψ 4 L 2 ( Cs 5 T α 9 ψ 2 dt dx. (112 The fifth term gives ((L 1 ψ 4, (L 2 ψ 5 L 2 ( = 5s 5 αxα 4 xx ψ xx 2 dt dx Cs 5 α 5 ψ xx ψ x dt dx. (113 Direct computations for the last term provides ((L 1 ψ 4, (L 2 ψ 6 L2 ( = 15s 7 αxα 6 xx ψ x 2 dt dx. (114 2

21 All these computations ((19-(114 gives ((L 1 ψ 4, L 2 ψ L2 ( s5 αxα 4 xx ψ xx 2 dt dx 25s 7 αxα 6 xx ψ x 2 dt dx 45 2 s9 αxα 8 xx ψ 2 dt dx s5 αx x= 5 ψ xx x= 2 dt ɛs 9 α 9 ψ 2 dt dx ɛs 7 α 7 ψ x 2 dt dx ɛs 5 α 5 ψ xx 2 dt dx, (115 for any ɛ >, where again s C(T 1/4 + T 1/2 and C depends on ɛ. Let us now gather all the product (L 1 ψ, L 2 ψ L2 ( coming from (94, (11, (18 and (115: (L 1 ψ, L 2 ψ L 2 ( 25 2 s α xx ψ 4x 2 dt dx + 1s 3 αxα 2 xx ψ xxx 2 dt dx 425s 5 αxα 4 xx ψ xx 2 dt dx + 8s 7 αxα 6 xx ψ x 2 dt dx 45 2 s9 αxα 8 xx ψ 2 dt dx 5 2 s α x x= ψ 4x x= 2 dt 2s 3 αx x= 3 ψ xxx x= 2 dt 38s 5 αx x= 5 ψ xx x= 2 dt 1s 3 αx x= 3 ψ 4x x=ψ xx x= dt ɛs 9 α 9 ψ 2 dt dx ɛs 7 α 7 ψ x 2 dt dx ɛs 5 α 5 ψ xx 2 dt dx ɛs 3 α 3 ψ xxx 2 dt dx ɛs α ψ 4x 2 dt dx ɛs 5 α x= 5 ψ xx x= 2 dt ɛs 3 α x= 3 ψ xxx x= 2 dt ɛs Cs T α x= ψ 4x x= 2 dt α x=1 ψ 4x x=1 2 dt Cs 3 α x=1 3 ψ xxx x=1 2 dt, for s C(T 1/4 + T 1/2. Let us explain how we handle the wrongly signed terms in ψ xxx 2 and ψ x 2. After integration by parts, we get 1s 3 αxα 2 xx ψ xxx 2 dt dx 1s 3 αxα 2 xx ψ xx ψ 4x dt dx ɛs 5 α 5 ψ xx 2 dt dx (117 ɛs 3 α x= 3 ψ xxx x= 2 ɛs 5 α x= 5 ψ xx x= 2 dt, by taking s CT 1/2. The last two terms in the right hand side are already in (116, while the first one is estimated as follows, by using Cauchy-Schwarz s inequality: 1s 3 αx α 2 xx ψ xx ψ 4x dt dx 12s α xx ψ 4x 2 dt dx s5 αx α 4 xx ψ xx 2 dt dx. (118 On the other hand, by integration by parts and Cauchy-Schwarz s inequality, we have for s CT 1/2 : 8s 7 αxα 6 xx ψ x 2 dt dx (22 + ɛs 9 αx α 8 xx ψ 2 dt dx s5 αx α 4 xx ψ xx 2 dt dx. (119 Observe that ( < 425. (

22 Finally, we have 1s 3 αx x= 3 ψ 4x x=ψ xx x= dt 2s α x x= ψ 4x x= 2 dt s5 α x x= 5 ψ xx x= 2 dt. (121 Now we observe that thanks to (43 we can absorb all the ɛ terms in (116 provided that s CT 1/2. Finally, using again (43 we deduce from (87 the following inequality for ψ: s α ψ 4x 2 dt dx + s 3 α 3 ψ xxx 2 dt dx + s 5 α 5 ψ xx 2 dt dx + s 7 α 7 ψ x 2 dt dx + s 9 α 9 ψ 2 dt dx C ( L 3 ψ 2 L 2 ( + s α x=1 ψ 4x x=1 2 dt + s 3 α x=1 3 ψ xxx x=1 2 dt. (122 Now it is not difficult to see that all the terms in L 3 ψ yield a L 2 -norm estimated by ( L 3 ψ 2 L 2 ( C s 2 α 2 ψ xxx 2 dt dx + s 4 α 4 ψ xx 2 dt dx + s 6 α 6 ψ x 2 dt dx + s 8 α 8 ψ 2 dt dx + e 2sα f 2 dt dx, (123 for s CT 1/2. Here we have used that a k L (, T ; W k, (, 1. Hence they can be absorbed by the left hand side of (122 provided that s CT 1/2. We deduce the Carleman inequality for ψ s α ψ 4x 2 dt dx + s 3 α 3 ψ xxx 2 dt dx + s 5 α 5 ψ xx 2 dt dx + s 7 α 7 ψ x 2 dt dx + s 9 α 9 ψ 2 dt dx ( C s α x=1 ψ 4x x=1 2 dt + s 3 α x=1 3 ψ xxx x=1 2 dt + It remains to replace ψ by ϕ, to use (43 and s CT 1/2 in order to deduce (45. 6 Proof of Lemma 1 We first establish two lemmas before turning to the core of the proof. e 2sα f 2 dt dx. (124 Lemma 2. Let p satisfy Then for k 2, one has p t + αp 5x = g in (, T (, 2, p x= = p x x= = p xx x= = in (, T, p x=1 = ˆv 2, p x x=2 = ˆv 4 in (, T, p t= = p in (, 2. (125 (2 x k+ 1 2 p L (,T ;L 2 (,2 + (2 x k p xx L 2 (,T ;L 2 (,2 + (2 x k 1 p x L 2 (,T ;L 2 (,2 (2 x k+1 g L2 (,T ;H 2 (,2 + (2 x k 2 p L2 (,T ;L 2 (,2 + (2 x k+ 1 2 p L2 (,2. (126 Proof of Lemma 2. As previously, we multiply by (2 x 2k+1 p; we get = 1 d 2 dt 2 2 (2 x 2k+1 p 2 dx + 5 2k + 1 α 2 (2 x 2k+1 pg dx+6k(2k+1α 2 2 (2 x 2k p xx 2 dx (2 x 2k 1 p x p xx dx 2k(2k+1(2k 1α 2 (2 x 2k 2 pp xx dx. 22

23 We utilize Young s inequality: (2 x 2k 2 pp xx ɛ (2 x 2k p xx 2 dx + 1 ɛ 2 (2 x 2k 1 p x p xx ɛ (2 x 2k p xx 2 dx + 1 ɛ Integrate by parts in the last term, we deduce ( (2 x 2k 4 p 2 dx, (2 x 2k 2 p x 2 dx. Lemma 3. Let p satisfy (125. Then for k 7, one has 4 (2 x k+ 1 2 p L (,T ;H 2 (,2 + (2 x k+j 4 xp j L 2 (,T ;L 2 (,2 j= g L2 (,T ;L 2 (,2 + p H2 (,2 + ˆv 2 L2 (,T + ˆv 4 L2 (,T. (127 Proof of Lemma 3. First step. Higher order estimates. Let g L 2 (, T ; H 3 (, 2. We apply Lemma 2 to p 5x (which satisfies the boundary conditions, and get (2 x k+ 1 2 p5x L (,T ;L 2 (,2 + (2 x k p 5x L2 (,T ;H 2 (,2 (2 x k+1 g 5x L2 (,T ;H 2 (,2 By an integration by parts, this inequality yields + (2 x k 2 p 5x L 2 (,T ;L 2 (,2 + (2 x k+ 1 2 p,5x L 2 (,2. (128 (2 x k+ 1 2 p5x L (,T ;L 2 (,2 + (2 x k p 7x L 2 (,T ;L 2 (,2 + (2 x k 1 p 6x L 2 (,T ;L 2 (,2 (2 x k+1 g 5x L2 (,T ;H 2 (,2 + (2 x k 2 p 5x L2 (,T ;L 2 (,2 + (2 x k+ 1 2 p,5x L2 (,2. (129 Now in order to estimate the term concerning p 5x in the right hand side, we observe that 2 2 (2 x 2k 4 p 5x p 5x dx dt = (2 x 2k 2 p 6x p 6x dx dt = 2 2 (2 x 2k 4 p 6x p 4x dx dt+(k 2(2k 5 (2 x 2k 2 p 5x p 7x dx dt+(k 1(2k (2 x 2k 6 p 4x 2 dx dt, (13 (2 x 2k 4 p 2 5x dx dt. (131 The identity (131 may be used to estimate the first integral in the right hand side of (13 (with ab ɛa 2 + b 2 /ɛ. Now injecting in (129, we obtain (2 x k+ 1 2 p5x L (,T ;L 2 (,2 + 2 j= (2 x k j 7 j x p L2 (,T ;L 2 (,2 (2 x k+1 g 5x L2 (,T ;H 2 (,2 + (2 x k 3 p 4x L2 (,T ;L 2 (,2 + (2 x k+ 1 2 p,5x L2 (,2. (132 Now to absorb the term concerning p 4x in the right hand side, we operate in the same way, but here a boundary term appears: 2 (2 x 2k 6 p 4x p 4x dx dt = + (k 3(2k (2 x 2k 6 p 5x p 3x dx dt (2 x 2k 8 p 3x 2 dx dt + 2 2k 6 p 3x x= p 4x x= dt. (133 23

24 This latter term is treated as follows: p 3x x= p 4x x= dt ɛ Now p 4x x= 2 dt = 1 2 2k 6 2 which can be treated as above, while p 3x x= 2 dt = 1 2 2k 8 which leads us to 2 (2 x k+ 1 2 p5x L (,T ;L 2 (,2 + p 4x x= 2 dt + 1 ɛ (2 x 2k 5 p 4x p 5x dt dx (2 x 2k 7 p 3x p 4x dt dx 3 j= (2k 5 2 2k 5 (2k 7 2 2k 7 (2 x k j 7 j x p L 2 (,T ;L 2 (,2 p 3x x= 2 dt. 2 2 (2 x 2k 6 p 4x 2 dt dx, (2 x 2k 8 p 3x 2 dt dx, (2 x k+1 g 5x L 2 (,T ;H 2 (,2 + (2 x k 4 p 3x L 2 (,T ;L 2 (,2 + (2 x k+ 1 2 p,5x L 2 (,2. (134 Then one follows the same steps as previously (note that p x= = p x x= = p xx x= = and finally get (2 x k+ 1 2 p5x L (,T ;L 2 (,2 + 6 (2 x k 7+j xp j L 2 (,T ;L 2 (,2 j= (2 x k+1 g 5x L 2 (,T ;H 2 (,2 + (2 x k 7 p L 2 (,T ;L 2 (,2 + (2 x k+ 1 2 p,5x L 2 (,2, (135 and consequently, using Proposition 6, (2 x k+ 1 2 p5x L (,T ;L 2 (,2 + 6 (2 x k 7+j xp j L 2 (,T ;L 2 (,2 j= g L2 (,T ;H 3 (,2 + (2 x k+ 1 2 p,5x L2 (,2 + ˆv 2 L2 (,T + ˆv 4 L2 (,T. (136 Second step. Interpolation. Now we consider the operator which maps (p, g, ˆv 2, ˆv 4 to (2 x k+1 p: it is continuous from respectively L 2 (, 2 L 2 (, T ; H 2 (, 2 L 2 (, T 2 to L 2 (, T ; H 2 (, 2 C ([, T ]; L 2 (, 2, H 5 (, 2 L 2 (, T ; H 3 (, 2 L 2 (, T 2 to L 2 (, T ; H 7 (, 2 C ([, T ]; H 5 (, 2. By interpolation, it is hence continuous from H 2 (, 2 L 2 (, T ; L 2 (, 2 L 2 (, T 2 to L 2 (, T ; H 4 (, 2 C ([, T ]; H 2 (, 2. This concludes the proof of Lemma 3. Proof of Lemma 1. We apply Lemma 3 with p = y and g = 3 j= ãj(t, x j xy + h. We infer (2 x k+ 1 2 y L (,T ;H 2 (, (2 x k+j 4 xy j L2 (,T ;L 2 (,2 h L2 (,T ;L 2 (,2 j= 3 ã j xy j L2 (,T ;L 2 (,2 + y H2 (,2 + v2 L2 (,T + v4 L2 (,T. (137 j= 24

25 Now, using that the supports of ã j are away from 2, we can estimate the terms 3 j= ã j j xy L2 (,T ;L 2 (,2 as follows 3 ã j xy j L 2 (,T ;L 2 (,2 ɛ j= exactly as in Lemma 3. We get (2 x k+ 1 2 y L (,T ;H 2 (,2 + 4 (2 x k+j 4 xy j L 2 (,T ;L 2 (,2 + C y L 2 (,T ;L 2 (,2, j= 4 (2 x k+j 4 xy j L 2 (,T ;L 2 (,2 j= h L2 (,T ;L 2 (,2 + y L2 (,T ;L 2 (,2 + y H2 (,2 + v 2 L 2 (,T + v 4 L 2 (,T. Using again Proposition 6, and thanks to (58, this gives (7, hence completing the argument. References [1] Bona J., Sun S. M., Zhang B.-Y., A nonhomogeneous boundary-value problem for the Korteweg-de Vries equation posed on a finite domain, Comm. Partial Differential Equations 28 (23, no. 7-8, pp [2] Cerpa E., Exact controllability of a nonlinear Korteweg-de Vries equation on a critical spatial domain, SIAM J. Control Optim. 46 (27, pp [3] Cerpa E., Crépeau E., Boundary controllability for the non linear Korteweg-de Vries equation on any critical domain, preprint 27, to appear in Ann. Inst. H. Poincaré Anal. Non Linéaire. [4] Colin T., Ghidaglia J.-M., An initial-boundary value problem for the Korteweg-de Vries equation posed on a finite interval, Adv. Differential Equations 6 (21, no. 12, pp [5] Colliander J. E., Kenig C. E., The generalized Korteweg-de Vries equation on the half line, Comm. Partial Diff. Eq. 27 (22, no , pp [6] Cui S. B., Deng D. G., Tao S. P., Global Existence of Solutions for the Cauchy Problem of the Kawahara Equation with L 2 Initial Data, Acta Math. Sin. (Engl. Ser. 22 (26, no. 5, pp [7] Coron J.-M., Crépeau E., Exact boundary controllability of a nonlinear KdV equation with critical lengths, J. Eur. Math. Soc, 6, 24, pp [8] Coron J.-M., Control and Nonlinearity, Mathematical Surveys and Monographs 136, American Mathematical Society, Providence, RI, 27. [9] Dawson L., Uniqueness properties of higher order dispersive equations, J. Differential Equations 236 (27, no. 1, pp [1] Faminskii A. V., On Two Initial Boundary Value Problems for the Generalized KdV Equation, Nonlinear Boundary Problems 14 (24, pp [11] Fursikov A., Imanuvilov O. Yu., Controllability of Evolution Equations, Lecture Notes #34, Seoul National University, Korea, [12] Glass O., Guerrero S., Some exact controllability results for the linear KdV equation and uniform controllability in the zero-dispersion limit, preprint 27, to appear in Asymp. Anal. [13] Holmer J., The initial-boundary value problem for the Korteweg-de Vries equation, Communications in Partial Differential Equations, 31 (26, pp [14] Kawahara R., Oscillatory solitary waves in dispersive media, J. Phys. Soc. Japan, 33 (1972, pp