Exact boundary controllability of a nonlinear KdV equation with critical lengths

Transcription

1 J. Eur. Math. Soc. 6, c European Mathematical Society 24 Jean-Michel Coron Emmanuelle Crépeau Exact boundary controllability of a nonlinear KdV equation with critical lengths Received November 2, 23 Abstract. We study the boundary controllability of a nonlinear Korteweg de Vries equation with the Dirichlet boundary condition on an interval with a critical length for which it has been shown by Rosier that the linearized control system around the origin is not controllable. We prove that the nonlinear term gives the local controllability around the origin. Keywords. Controllability, nonlinearity, Korteweg de Vries. Introduction Let us consider the following Korteweg de Vries control system: { yt + y (KdV) x + y xxx + yy x =, y(t, ) = y(t, L) =. For this control system, L > is given, the state is y( ) : [, L] R, and for the control one can take, for example, u( ) = y x (, L) R. The Korteweg de Vries equation serves to model various physical phenomena (see e.g. [9]), for example the propagation of small amplitude long water waves in a uniform channel. Let us recall that Bona and Winther have pointed out in [3] that the term y x in (KdV) has to be added to model the water waves when x denotes the spatial coordinate in a fixed frame. We are interested in the local controllability of (KdV) around. Rosier has proved in [2] that the control system (KdV) is locally controllable around provided that the length of the spatial domain is not critical. Theorem ([2, Theorem.3]). Let T >, and assume that { j L / N := 2π 2 + l 2 + jl ; j, l N }. (.) 3 J.-M. Coron: Université de Paris-Sud, Laboratoire ANEDP, Mathématique, UMR 8628, Bât. 425, 945 Orsay Cedex, France; Jean-Michel.Coron@math.u-psud.fr E. Crépeau: INRIA, Projet SOSSO, Domaine de Voluceau-Rocquencourt, 7853 Le Chesnay Cedex; Emmanuelle.Crepeau@inria.fr Mathematics Subject Classification (2): 93B5, 76B5

2 368 Jean-Michel Coron, Emmanuelle Crépeau Then there exists r > such that, for every (y, y T ) L 2 (, L) 2 with y L 2 (,L) < r and y T L 2 (,L) < r, there exists y C([, T ], L 2 (, L)) L 2 (, T, H (, L)) satisfying (KdV) such that y(, ) = y and y(t, ) = y T. The aim of this paper is to study the local exact controllability around of the nonlinear KdV equation when L = 2kπ N (take j = l = k in (.)). Our main theorem is the following one. Theorem 2. Let k be a positive integer and let T >. There exists r > such that, for every (y, y T ) L 2 (, 2kπ) 2 with y L 2 (,2kπ) < r and y T L 2 (,2kπ) < r, there exists y C([, T ], L 2 (, 2kπ)) L 2 (, T, H (, 2kπ)) satisfying (KdV) with L = 2kπ such that y(, ) = y and y(t, ) = y T. When L = 2kπ the linearized control system of (KdV) around is { yt + y (KdVL) x + y xxx =, y(t, ) = y(t, 2kπ) =. It has been shown by Rosier in [2] that this linear control system is not controllable. To prove that the nonlinear term yy x gives the local controllability, a first approach could be to use the exact controllability of the nonlinear equation around nontrivial stationary solutions proved in [8] and to apply the method introduced in [6] (that is, the return method [4, 5] together with quasi-static deformations; see also [7] for this last point). But, with this method, it seems that one can only obtain the local exact controllability for large time. To prove Theorem 2 we use a different strategy that we briefly describe now. We first point out that in this theorem we may assume that y = : this follows easily from the invariance of (KdV) under the change of variables τ = T t, ξ = 2kπ x. Then we use the following result, due to Rosier, for the linearized control system (KdVL). Theorem 3 ([2, Remark 3.6]). Let T > and H = { y L 2 (, 2kπ); 2kπ } ( cos(x))y dx =. For (y, y T ) H H, there exists y C([, T ], L 2 (, 2kπ)) L 2 (, T, H (, 2kπ)) satisfying (KdVL) such that y(, ) = y and y(t, ) = y T. Then, as we shall prove in Section 2, the nonlinear term yy x allows us to go in the two directions ±( cos(x)) which are missed by the linearized control system (KdVL). Finally, in Section 3 we derive Theorem 2 from Section 2 by means of a fixed point theorem.

3 Boundary controllability of KdV equation 369 Remark 4. For the other critical lengths, we believe that the same result holds. Note that the situation is more complicated in these other cases: there are now four noncontrollable directions for the linearized control system around (see [2, proof of Lemma 3.5 and Remark 3.6]). Remark 5. The method we use here (try to move in the directions which are missed by the linearized control system) is classical to study the local controllability of a control system in finite dimensions. Here we fix the time and perform a power series expansion, with the same scaling on the state and on the control. In finite dimensions, much more subtle tools have been introduced: for example different scalings on the components of the control and of the state as well as scaling on time. See e.g. [, 2, 9,, 7, 8, 2] and the references therein. Remark 6. One can find other results on the controllability of KdV control systems in [, 3 6] and the references therein. 2. Motion in the ±( cos(x)) directions Let L >. We first recall some properties proved by Rosier in [2] for the following linear KdV Cauchy problem: y t + y x + y xxx = f, (2.) y(t, ) = y(t, L) =, (2.2) y x (t, L) = h(t), (2.3) y(t, x) = y (x). (2.4) We adopt the notations of [2]. Let A denote the operator Aw = w w defined on D(A) := {w H 3 (, L); w() = w(l) = w x (L) = } and let (S(t) t ) denote the semigroup of contractions associated with A (see [2, Proposition 3.]). For T < T, let endowed with the norm B T,T := C([T, T ], L 2 (, L)) L 2 (T, T, H (, L)) y BT,T = max{ y(t, ) L 2 (,L) ; t [T, T ]} + Rosier has proved the following proposition. ( T T ) /2 y(t, ) 2 H (,L) dt. Proposition 7 ([2, (Proofs of) Propositions 3.2 and 3.7]). Let T < T. There exist unique continuous linear maps T,T and δ T,T T,T : L 2 (, L) L 2 (T, T ) L (T, T, L 2 (, L)) B T,T, (y, h, f ) T,T (y, h, f ), δ T,T : L 2 (, L) L 2 (T, T ) L (T, T, L 2 (, L)) L 2 (T, T ), (y, h, f ) δ T,T (y, h, f ),

4 37 Jean-Michel Coron, Emmanuelle Crépeau such that, for y D(A), h C 2 ([T, T ]) with h(t )= and f C ([T, T ], L 2 (, L)), T,T (y, h, f ) is the unique classical solution of (2.) to (2.4), δ T,T (y, h, f )(t) = ( T,T (y, h, f )) x (t, ). The function T,T (y, h, f ) is called the mild solution of (2.) to (2.4). For simplicity, we write B for B T,T and for T,T when (T, T ) = (, T ). Note that the existence of the continuous linear map δ T,T shows that, with y := T,T (y, h, f ), y x (t, ) makes sense in L 2 (T, T ). For simplicity we shall write y x (t, ) instead of δ T,T (y, h, f )(t). Let f L 2 (T, T, L 2 (, L)). We say that y : [T, T ] [, L] R is a mild solution of y t + y x + y xxx = f, y(t, ) = y(t, L) =, if there exists h L 2 (T, T ) such that y is the mild solution of (2.) to (2.4) with y (x) := y(t, x). Note that it follows from the proof of Theorem 3 given in [2] that this theorem holds for mild solutions of (KdVL). Until the end of this section we assume that The aim of this section is to prove the following result. L {2kπ; k N }. (2.5) Proposition 8. Let T >. There exists (u +, v +, w + ) in L 2 (, T ) 3 and (u, v, w ) in L 2 (, T ) 3 such that, if α ±, β ±, γ ± are the mild solutions of of and of then α ±t + α ±x + α ±xxx =, (2.6) α ± (t, ) = α ± (t, L) =, (2.7) α ±x (t, L) = u ± (t), (2.8) α ± (, x) =, (2.9) β ±t + β ±x + β ±xxx = α ± α ±x, (2.) β ± (t, ) = β ± (t, L) =, (2.) β ±x (t, L) = v ± (t), (2.2) β ± (, x) =, (2.3) γ ±t + γ ±x + γ ±xxx = (α ± β ± ) x, (2.4) γ ± (t, ) = γ ± (t, L) =, (2.5) γ ±x (t, L) = w ± (t), (2.6) γ ± (, x) =, (2.7) α ± (T, x) =, β ± (T, x) =, γ ± (T, x) = ±( cos(x)). (2.8)

5 Boundary controllability of KdV equation 37 Remark 9. It would have been quite natural to look for the existence of (u +, v + ) in L 2 (, T ) 2 and of (u, v ) in L 2 (, T ) 2 such that, if α ±, β ± are the mild solutions of (2.6) to (2.3), then α ± (T, x) =, β ± (T, x) = ±( cos(x)). The existence of such (u ±, v ± ) would have also implied Theorem 2. Unfortunately, as proved in Corollary 9 below, such (u ±, v ± ) do not exist. Roughly speaking, an expansion to the 2 nd order is not sufficient. We must go to the 3 rd order to get local controllability. In order to prove Proposition 8, let us first remark that (u +, v +, w + ) has the required properties if and only if (u, v, w ) := ( u +, v +, w + ) does. Moreover, in order to prove the existence of (u +, v +, w + ) it suffices to prove the existence of (u +, v +, w + ) in L 2 (, T ) 3 such that, if α +, β + and γ + are the mild solutions of (2.6) to (2.7) with γ + := γ + and w + := w +, then α + (T, ) =, β + (T, ) =, γ + (T, x)( cos(x)) dx = cos(x) 2 L 2 (,L). Indeed, by Theorem 3 (for mild solutions), there exists w + in L2 (, T ) such that the mild solution γ + of satisfies γ +t + γ +x + γ +xxx =, γ+ (t, ) = γ + (t, L) =, γ+x (t, L) = w + (t), γ+ (, x) =, γ + (T, ) = P H ( γ + (T, )), where P H denotes the orthogonal projection on H for the L 2 -scalar product. Then u +, v + and w + := w + +w+ have the properties required by Proposition 8 (with γ + := γ + +γ+ ). Similarly, in order to prove the existence of (u, v, w ) it suffices to prove the existence of (u, v, w ) in L 2 (, T ) 3 such that, if α, β and γ are the mild solutions of (2.6) to (2.7) with γ := γ and w := w, then α (T, ) =, β (T, ) =, γ (T, x)( cos(x)) dx = cos(x) 2 L 2 (,L). From (2.4), (2.5) and (2.7), one gets, using integration by parts (which can be easily justified by density arguments), γ ± (T, x)( cos(x)) dx = = = T T T γ ±t (t, x)( cos(x)) dx dt ( γ ±x γ ±xxx (α ± β ± ) x )( cos(x)) dx dt α ± β ± sin(x) dx dt. Hence Proposition 8 is a consequence of the following proposition.

6 372 Jean-Michel Coron, Emmanuelle Crépeau Proposition. Let T >. There exists (u, v) in L 2 (, T ) 2 such that, if α, β are the mild solutions of and of then α t + α x + α xxx =, (2.9) α(t, ) = α(t, L) =, (2.2) α x (t, L) = u(t), (2.2) α(, x) =, (2.22) β t + β x + β xxx = αα x, (2.23) β(t, ) = β(t, L) =, (2.24) β x (t, L) = v(t), (2.25) β(, x) =, (2.26) α(t, ) =, β(t, ) =, (2.27) T Let T >. Let α B be a mild solution of such that αβ sin(x) dx dt =. (2.28) α t + α x + α xxx =, (2.29) α (t, ) = α (t, L) =, (2.3) α (, x)( cos(x)) dx =. (2.3) Let us multiply (2.29) by cos(x) and integrate the resulting equality on [, T ] [, L]. Then, using integrations by parts together with (2.3) and (2.3), one gets α (T, x)( cos(x)) dx =. (2.32) By Theorem 3, (2.3) and (2.32), α can be extended to [ T, 2T ] [, L] in such a way that this extension is still a mild solution of (2.29) (2.3) and satisfies α ( T, x) = α (2T, x) =. Let β : [, T ] [, L] R be a mild solution of β t + β x + β xxx = α α x, (2.33) β (t, ) = β (t, L) =. (2.34)

7 Boundary controllability of KdV equation 373 By Theorem 3, there exists θ R and a mild solution β B T,2T of such that β t + β x + β xxx = α α x, (2.35) β (t, ) = β (t, L) =, (2.36) β (t, x) = β (t, x) + θ ( cos(x)), t [, T ], (2.37) β ( T, x) =, P H (β (2T, )) =. (2.38) By Corollary 9 below, it follows that β (2T, x) =. Let now α 2 : [, T ] [, L] R be a mild solution of such that α 2t + α 2x + α 2xxx =, (2.39) α 2 (t, ) = α 2 (t, L) =, (2.4) α 2 (, x) = α 2 (T, x) =. (2.4) By Theorem 3, there exists a mild solution β 2 : [, T ] [, L] R of such that By Corollary 9 again, β 2t + β 2x + β 2xxx = α 2 α 2x, β 2 (t, ) = β 2 (t, L) =, β 2 (, x) =, P H (β 2 (T, )) =. β 2 (T, x) =. Similarly, by Theorem 3 and Corollary 9, there exists a mild solution β 3 : [, T ] [, L] R of such that β 3t + β 3x + β 3xxx = (α α 2 ) x, (2.42) β 3 (t, ) = β 3 (t, L) =, (2.43) β 3 (, x) =, β 3 (T, x) =. (2.44) We extend α 2, β 2 and β 3 to [ T, 2T ] [, L] by requiring α 2 (t, x) = β 2 (t, x) = β 3 (t, x) =, t [ T, ] [T, 2T ].

8 374 Jean-Michel Coron, Emmanuelle Crépeau Let us consider, for (ρ, ρ 2 ) R 2, α := ρ α + ρ 2 α 2 and β := ρ 2 β + ρ 2 2 β 2 + ρ ρ 2 β 3. Let u(t) := α x (t, L), v(t) := β x (t, L) for t [ T, 2T ]. Then (2.9) to (2.2) and (2.23) to (2.25) hold in the mild sense and α( T, x) = β( T, x) = α(2t, x) = β(2t, x) =. Now, assume that Proposition is false. Then, for every (ρ, ρ 2 ) R 2, 2T T αβ sin(x) dx dt T = (ρ α + ρ 2 α 2 )(ρ 2 β + ρ 2 2 β 2 + ρ ρ 2 β 3 ) sin(x) dx dt =. (2.45) By looking at the coefficient of ρ 2 ρ 2 in (2.45), we get T (α β 3 + α 2 β ) sin(x) dx dt =. (2.46) For the time being the functions α, β, β 2 and β 3 have been assumed to be real-valued. But, looking at the real and imaginary parts and applying the same trick we have used to get (2.46), we see that (2.46) also holds if these four functions take their values in C. (Introduce arbitrary (ρ, ρ 2 ). If α = a +ib, consider α := ρ a +ρ 2 b etc.) Of course one says that y = y + iy 2 : [T, T ] [, L] C is a mild solution of y t + y x + y xxx = f, y(t, ) = y(t, L) =, with f = f + if 2 L (T, T, L 2 (, L, C)), if y and y 2 are mild solutions of y t + y x + y2 xxx = f, y 2 t + y 2 2x + y2 xxx = f 2, y (t, ) = y (t, L) =, y 2 (t, ) = y 2 (t, L) =. Let λ C. Let y λ C ([, L], C) be such that We take, for t [, T ] and x [, L], λy λ + y λx + y λxxx =, (2.47) y λ () = y λ (L) =. (2.48) α (t, x) = e λt y λ (x). (2.49) From (2.47), (2.48) and (2.49), we get (2.29) and (2.3). Multiplying (2.47) by cos(x), integrating the resulting equality on [, L] and using integrations by parts together with (2.48), we get By (2.49) and (2.5), if λ ( cos(x))y λ dx =. (2.5) λ =, (2.5)

9 Boundary controllability of KdV equation 375 which will be assumed until the end of this section, (2.3) holds. Let à denote the operator Aw = w w defined on D(Ã) := {w H 3 (, L); w() = w(l) =, w x () = w x (L)}. Then iã is a self-adjoint operator on L 2 (, L) with compact resolvent. Hence, the spectrum σ (Ã) of à is a discrete subset of ir. Assume that λ σ (Ã). (2.52) Then there exists one (and only one) φ λ C ([, L], C) such that λφ λ + φ λx + φ λxxx = y λ sin(x), (2.53) φ λ () = φ λ (L) =, φ λx () = φ λx (L). (2.54) We multiply (2.53) by y λ (L x), integrate on [, L], and use integrations by parts together with (2.5), (2.47), (2.48) and (2.54) to get From now on we assume that By (2.47), (2.48), (2.52) and (2.56), which, together with (2.55), gives φ λx (L)(y λx (L) y λx ()) =. (2.55) From (2.42), (2.43), (2.49), (2.53), (2.54) and (2.58), we get d dt We also assume that e λt φ λ β 3 dx = y λ =. (2.56) y λx (L) = y λx (), (2.57) φ λx (L) =. (2.58) α β 3 sin(x) dx + Then there also exists a unique z λ C ([, L], C) such that Define β by α α 2 φ λx e λt dx. (2.59) 2λ σ (Ã). (2.6) 2λz λ + z λx + z λxxx = y λ y λx, (2.6) z λ () = z λ (L) =, z λx () = z λx (L). (2.62) β (t, x) = e2λt z λ (x). (2.63)

10 376 Jean-Michel Coron, Emmanuelle Crépeau From (2.49) and (2.49) to (2.63), (2.33) and (2.34) hold. Let also (use (2.6) again) ψ λ C ([, L], C) be the unique solution of 2λψ λ + ψ λx + ψ λxxx = z λ sin(x), (2.64) ψ λ () = ψ λ (L) =, ψ λx () = ψ λx (L). (2.65) Let Then µ(x) = 2 x sin(x) + 2 ( cos(2x)). µ x + µ xxx = ( cos(x)) sin(x), (2.66) µ() = µ(l) = µ x () =, (2.67) µ x (L) = L/2. (2.68) From (2.37), (2.39), (2.4), (2.53), (2.54) and (2.63) to (2.68), we get d dt By (2.4) and (2.44), T ( d dt (e 2λt ψ λ + θ µ)α 2 dx = e 2λt ψ λx (L)(α 2x (t, L) α 2x (t, )) From (2.59), (2.69) and (2.7), we get T L 2 θ α 2x (t, L) + ((e 2λt ψ λ + θ µ)α 2 + e λt φ λ β 3 ) dx (α β 3 + α 2 β ) sin(x) dx dt = T T α 2 β sin(x) dx. (2.69) e λt α α 2 φ λx dx dt ) dt =. (2.7) ( e 2λt ψ λx (L)(α 2x (t, L) α 2x (t, )) L ) 2 θ α 2x (t, L) dt. (2.7) Let (see (2.6)) δ λ C ([, L], C) be the unique solution of 2λδ λ + δ λx + δ λxxx = y λ φ λx, (2.72) δ λ () = δ λ (L) =, δ λx () = δ λx (L). (2.73) From (2.39), (2.4), (2.49), (2.72) and (2.73), we get d dt e 2λt δ λ α 2 dx = e 2λt δ λx (L)(α 2x (t, L) α 2x (t, )) + which, together with (2.4), gives T e λt φ λx α α 2 dx dt + T e λt φ λx α α 2 dx, (2.74) e 2λt δ λx (L)(α 2x (t, L) α 2x (t, )) dt =. (2.75)

11 Boundary controllability of KdV equation 377 From (2.46), (2.7) and (2.75), we get T ( e 2λt (δ λx (L) ψ λx (L))(α 2x (t, L) α 2x (t, )) + L ) 2 θ α 2x (t, L) dt =. (2.76) Let us restrict ourselves to the case where α 2 = on [3T /4, T ] [, L]. (2.77) This allows us to perform a time translation of ɛ [, T /4]: if we define α 2ɛ (t, x) := α 2 (t ɛ, x), t [ɛ, T ], (2.78) α 2ɛ (t, x) :=, t [, ɛ], (2.79) then α 2ɛ also satisfies (2.39), (2.4) (in the mild sense) and (2.4). Hence, by (2.76) associated to α 2ɛ, T ( e 2λt (δ λx (L) ψ λx (L))(α 2x (t ɛ, L) α 2x (t ɛ, )) + L ) 2 θ α 2x (t ɛ, L) dt = for all ɛ [, T /4], which is equivalent to T ( e 2λ(t+ɛ) (δ λx (L) ψ λx (L))(α 2x (t, L) α 2x (t, )) + L ) 2 θ α 2x (t, L) dt = for all ɛ [, T /4]. This last property, together with (2.5), implies that (δ λx (L) ψ λx (L)) T Let a R \ [ / 3, / 3]. We take λ := 2ia(4a 2 ). Let with (e 2λt (α 2x (t, L) α 2x (t, ))) dt =. (2.8) y λ (x) := κe ( 3a 2 ia)x + ( κ)e ( 3a 2 ia)x e 2iax (2.8) κ := e 2iaL e 3a ( 2 ia)l. (2.82) e ( 3a 2 ia)l e 3a ( 2 ia)l One easily checks that such a y λ satisfies (2.53), (2.54) and (2.56). Let, with λ := 2ia(4a 2 ), := {a R \ [ / 3, / 3]; λ σ (Ã) and 2λ σ (Ã)}. Then the function S : C, S(a) = δ λx (L) ψ λx (L), is continuous (and even analytic). In Appendix C we prove the following lemma: Lemma. The function S is not identically equal to.

12 378 Jean-Michel Coron, Emmanuelle Crépeau This lemma and (2.8) imply that α 2x (t, L) α 2x (t, ) =. (2.83) Indeed, let L : C C, λ T (e2λt (α 2x (t, L) α 2x (t, ))) dt. The function L is holomorphic. Hence the zeros of L are isolated if L =. However, by Lemma, (2.8) and the continuity of S, there exists a nonempty open subset of ir on which L vanishes (let us recall that σ (Ã) is a discrete subset of ir). Hence L =, which implies (2.83). We multiply (2.39) by ᾱ 2, take the real part, and integrate on [, L]. Then, using integrations by parts and (2.4) together with (2.83), we get d dt which, combined with (2.4), implies that α 2 2 dx =, α 2 =. (2.84) But, by Theorem 3 (for mild solutions), there are mild solutions of (2.39) and (2.4) satisfying (2.4) and (2.77) such that (2.84) does not hold. This ends the proof of Proposition and therefore of Proposition Local exact controllability In this section we still assume that (2.5) holds and we end the proof of Theorem 2. As pointed out in Section, the invariance of the control system (KdV) under the change of variables τ = T t, ξ = L x allows us to prove only that, for every T >, there exists r > such that, for every y T L 2 (, L) with y T L 2 (,L) r, there exists u L 2 (, T ) such that the mild solution y of y t + y x + y xxx + yy x =, (3.) y(t, ) = y(t, L) =, (3.2) y x (t, L) = u(t), (3.3) y(, x) =, (3.4) satisfies y(t, ) = y T. Of course, by y is a mild solution of (3.) to (3.4), we mean that y is in B and is the mild solution of y t + y x + y xxx = f, y(t, ) = y(t, L) =, y x (t, L) = u(t), y(, x) =, with f := yy x (note that, if y is in B, then yy x L (, T, L 2 (, L))). We use similar natural conventions until the end of this paper. It follows from Propositions 4 and 5 below that, for a given u L 2 (, T ), there exists at most one mild solution of (3.) to (3.4) and that such a solution exists if u L 2 (,T ) is small enough (the smallness depending on T and L).

13 Boundary controllability of KdV equation 379 By (the proof of) Theorem 3, there exists a continuous linear map such that the mild solution of Ɣ : h H L 2 (, L) Ɣ(h) L 2 (, T ) (3.5) y t + y x + y xxx =, y(t, ) = y(t, L) =, y x (t, L) = Ɣ(h)(t), y(, x) =, satisfies y(t, x) = h(x). (One can take for Ɣ the control obtained by means of HUM; see [2, Remark 3.].) Let y T L 2 (, L) be such that y T L 2 (,L) r, r > to be chosen later, small enough so that the maps introduced below are well defined in a neighborhood of. Let T yt denote the map with T yt : L 2 (, L) L 2 (, L), z z + y T F (G(z)), F : L 2 (, T ) L 2 (, L), u y(t, ), where y is the mild solution of (3.) to (3.4) and G : L 2 (, L) L 2 (, T ) is defined as follows. We decompose z = P H (z) + ρ(z)( cos(x)). Then. If ρ(z), then G(z) = Ɣ(P H (z)) + ρ /3 (z)u + + ρ 2/3 (z)v + + ρ(z)w If ρ(z) <, then G(z) = Ɣ(P H (z)) + ρ(z) /3 u + ρ(z) 2/3 v + ρ(z) w. (The functions u ±, v ± and w ± are fixed as in Proposition 8.) Clearly, each fixed point z of T yt satisfies F (G(z )) = y T, and the control u = G(z ) is a solution to our problem. Until the end of this paper, we adopt the following notations: For z L 2 (, T, H (, L)), z L 2 (H ) = z L 2 (,T,H (,L)), For z L (, T, L 2 (, L)), z L (L 2 ) = z L (,T,L 2 (,L)), B R = {z L 2 (, L); z L 2 (,L) R}. First of all we prove a lemma about the map T. Lemma 2. There exist C = C (T, L) > and ɛ = ɛ (T, L) > such that, for every z B ɛ, T z L 2 (,L) C z 4/3 L 2 (,L). (3.6) Let z L 2 (, L). Let (ũ, ṽ, w)=(u +, v +, w + ) if ρ(z) and (ũ, ṽ, w)=(u, v, w ) if ρ(z) <. Let y be the mild solution of y t + y x + y xxx + yy x =, (3.7) y(t, ) = y(t, L) =, (3.8) y x (t, L) = Ɣ(P H (z))(t) + ρ(z) /3 ũ(t) + ρ(z) 2/3 ṽ(t) + ρ(z) w(t), (3.9) y(, x) =. (3.)

14 38 Jean-Michel Coron, Emmanuelle Crépeau By Propositions 4 and 5, there exist ɛ 2 = ɛ 2 (T, L) > and C 2 = C 2 (T, L) > such that, for every z L 2 (, L) with z L 2 (,L) ɛ 2, there exists a unique mild solution of (3.7) to (3.), and this mild solution satisfies Let ỹ, α, β, γ be the mild solutions of Let y B C 2 z /3 L 2 (,L). (3.) ỹ t + ỹ x + ỹ xxx =, (3.2) ỹ(t, ) = ỹ(t, L) =, (3.3) ỹ x (t, L) = Ɣ(P H (z))(t), (3.4) ỹ(, x) =, (3.5) α t + α x + α xxx =, (3.6) α(t, ) = α(t, L) =, (3.7) α x (t, L) = ũ(t), (3.8) α(, x) =, (3.9) 2 β t + β x + β xxx = α α x, (3.2) β(t, ) = β(t, L) =, (3.2) β x (t, L) = ṽ(t), (3.22) β(, x) =, (3.23) γ t + γ x + γ xxx = ( α β) x, (3.24) γ (t, ) = γ (t, L) =, (3.25) γ x (t, L) = w(t), (3.26) γ (, x) =. (3.27) φ := y ỹ ρ(z) /3 α ρ(z) 2/3 β ρ(z) γ, (3.28) a := ỹ + ρ(z) /3 α + ρ(z) 2/3 β + ρ(z) γ, (3.29) b := ỹỹ x + (ỹ( ρ(z) /3 α + ρ(z) 2/3 β + ρ(z) γ )) x + ρ(z) 4/3 ( α γ ) x + ρ(z) 4/3 β β x + ρ(z) 5/3 ( β γ ) x + ρ(z) 2 γ γ x. (3.3) By Proposition 7, (3.) to (3.3) and standard estimates, there exists C 3 = C 3 (T, L) > such that, for every z L 2 (, L) with z L 2 (,L) ɛ 2, a B C 3 z /3 L 2 (,L), (3.3) b L (L 2 ) C 3 z 4/3 L 2 (,L), (3.32) φ B C 3 z /3 L 2 (,L). (3.33)

15 Boundary controllability of KdV equation 38 Similarly standard estimates give the existence of C 4 = C 4 (T, L) > such that Note that, by (2.8), (3.7) to (3.) and (3.2) to (3.28), (φa) x L (L 2 ) C 4 φ B a B. (3.34) φ(t, ) = F (G(z)) z = T z. (3.35) Moreover, by (3.7) to (3.) and (3.2) to (3.3), φ is a mild solution of φ t + φ x + φ xxx + φφ x = (φa) x b, (3.36) φ(t, ) = φ(t, L) = φ x (t, L) =, (3.37) φ(, x) =. (3.38) From (3.32), (3.34) to (3.37), (3.38) and Proposition 5, there exists C 5 = C 5 (T, L) > such that φ 2 B ( z 8/3 L 2 (,L) + φ 2 B a 2 B )ec 5(+ φ 2 B ). (3.39) From (3.3), (3.33) and (3.39), one gets the existence of ɛ 3 = ɛ 3 (T, L) > and of C 6 = C 6 (T, L) > such that, for every z L 2 (, L) with z L 2 (,L) ɛ 3, which, together with (3.35), ends the proof of Lemma 2. φ B C 6 z 4/3 L 2 (,L), (3.4) We now study P H T yt : H H, on the space H. For ω R and y T L 2 (, L), let g g + P H (y T ) P H (F (G(g + ωe))), where e(x) := cos(x) and ω R. (In fact we should write, for example, yt,ω, but for simplicity we omit the indices y T and ω.) To prove the existence of a fixed point for, we apply the Banach fixed point theorem to the restriction of to the closed ball B R H, with y T L 2 (,L) + ω R/3 and where R > small enough. Let (y T, ω) L 2 (, L) R be such that y T L 2 (,L) + ω R/3. Let g, h H B R. With (3.6), we have, for R > small enough, (g) L 2 (,L) y T L 2 (,L) + g + ωe F (G(g + ωe)) L 2 (,L) R/3 + 2R/3 = R. (3.4) Hence, for R > small enough, (B R H ) B R H. (3.42)

16 382 Jean-Michel Coron, Emmanuelle Crépeau Let us now look at the contracting property of. Let ω R, g H and h H. Let (ũ, ṽ, w) = { (u+, v +, w + ) if ω, (u, v, w ) if ω <. Let y, z, ỹ, z be the mild solutions of the following problems: y t + y x + y xxx + yy x =, (3.43) y(t, ) = y(t, L) =, (3.44) y x (t, L) = Ɣ(g)(t) + ω /3 ũ(t) + ω 2/3 ṽ(t) + ω w(t), (3.45) y(, x) =, (3.46) z t + z x + z xxx + zz x =, (3.47) z(t, ) = z(t, L) =, (3.48) z x (t, L) = Ɣ(h)(t) + ω /3 ũ(t) + ω 2/3 ṽ(t) + ω w(t), (3.49) z(, x) =, (3.5) ỹ t + ỹ x + ỹ xxx =, (3.5) ỹ(t, ) = ỹ(t, L) =, (3.52) ỹ x (t, L) = Ɣ(g)(t), (3.53) ỹ(, x) =, (3.54) z t + z x + z xxx =, (3.55) z(t, ) = z(t, L) =, (3.56) z x (t, L) = Ɣ(h)(t), (3.57) z(, x) =. (3.58) Let φ = y ỹ and ψ = z z. Let γ = φ ψ. By (3.43) to (3.58), γ is a mild solution of with γ t + γ x + γ xxx + γ γ x = (γ a) x b, γ (t, ) = γ (t, L) =, γ x (t, L) =, γ (, x) =, a = ψ + ỹ, b = (ψ(ỹ z)) x + ỹỹ x z z x. Let us notice that there exists C 7 = C 7 (T ) > such that b L (L 2 ) C 7( z B + ỹ B + z B ) ỹ z B, (γ a) x L (L 2 ) C 7( z B + z B + ỹ B ) γ B.

17 Boundary controllability of KdV equation 383 Then, using again Propositions 4 and 5 as in the the proof of (3.4), we get the existence of ɛ 4 = ɛ 4 (T, L) > such that, for every (g, h, ω) H H R with one has Note that (3.59) implies that g L 2 (,L) ɛ 4, h L 2 (,L) ɛ 4, ω ɛ 4, γ B 2 g h L 2 (,L). (3.59) (g) (h) L 2 (,L) = γ (T, ) L 2 (,L) γ B /2 g h L 2 (,L). (3.6) Therefore, by (3.42) and (3.6), there exists ɛ 5 = ɛ 5 (T, L) > such that, for R ɛ 5, for every (y T, ω) L 2 (, L) R such that y T L 2 (,L) + ω R/3, has a unique fixed point h(y T, ω) in B R H. Note that the map h is continuous in a neighborhood of (, ) H R. We now apply the intermediate value theorem to the map τ : R R, ω ρ(ωe + h(y T, ω) + y T F (G(h(y T, ω) + ωe))). By (3.6), there exists ɛ 6 = ɛ 6 (T, L) > such that, if y T L 2 (,L) ɛ 6 3L/8, then τ([ ɛ 6, ɛ 6 ]) [ ɛ 6, ɛ 6 ]. Hence, if y T L 2 (,L) ɛ 6 3L/8, we deduce, by the intermediate value theorem, that τ has at least one fixed point ω. We have which ends the proof of Theorem 2. F (G(h(y T, ω ) + ω e)) = y T, Let us remark that it follows from our proof of Theorem 2 that the following theorem, slightly more precise than Theorem 2, also holds. Theorem 3. Let k be a positive integer and let T >. There exist r > and C > such that, for every (y, y T ) L 2 (, 2kπ) 2 with y L 2 (,2kπ) < r and y T L 2 (,2kπ) < r, there exists y C([, T ], L 2 (, 2kπ)) L 2 (, T, H (, 2kπ)) satisfying, in the mild sense, (KdV) with L = 2kπ such that y(, ) = y, y(t, ) = y T, 2kπ y B C ( P H (y ) L 2(,2kπ) + ( cos(x))y dx 2kπ + P H (y ) L 2 (,2kπ) + ( cos(x))y dx /3 /3).

18 384 Jean-Michel Coron, Emmanuelle Crépeau A. Appendix: Existence and uniqueness of solutions to Cauchy problems for KdV equations We first prove the existence of solutions to the Cauchy problem for nonlinear KdV equations (with small data). Proposition 4. Let L > and T >. There exist ɛ > and C > such that, for every f L (, T, L 2 (, L)), every u L 2 (, T ) and every y L 2 (, L) such that f L (L 2 ) + u L 2 (,T ) + y L 2 (,L) ɛ, there exists at least one mild solution y of y t + y x + y xxx + yy x = f, y(t, ) = y(t, L) =, y x (t, L) = u(t), y(, x) = y (x), which satisfies y B C( f L (L 2 ) + u L 2 (,T ) + y L 2 (,L) ). (A.) Proof. Let T >. For f L (, T, L 2 (, L)), u L 2 (, T ) and y L 2 (, L), define M f,u,y : B B, y (y, u, f yy x ). A fixed point of M f,u,y is a solution of (A.). One easily gets the existence of C 8 = C 8 (T ) > such that, for every (y, z) B 2, yy x L (L 2 ) C 8 y 2 B, zz x yy x L (L 2 ) C 8( z B + y B ) z y B. Hence, by continuity of (see Proposition 7), there exists a constant C 9 = C 9 (T, L) > such that, for every f L (, T, L 2 (, L)), every u L 2 (, T ), every y L 2 (, L), every y in B and every z in B, M f,u,y (y) B C 9 ( f L (L 2 ) + u L 2 (,T ) + y L 2 (,L) + y 2 B ), M f,u,y (z) M f,u,y (y) B C 9 ( z B + y B ) z y B. From these two inequalities and the Banach fixed point theorem, one sees that the assertion of Proposition 4 holds with ɛ := 9C9 2, C := 3C 9 2. (Note that M f,u,y ( B) B if B :={y B; y B /(3C 9 )}, M f,u,y (z) M f,u,y (y) B (2/3) z y B for every (y, z) B 2, and y L 2 (,L) ( C 9 y L 2 (,L) ) C 9( f L (L 2 ) + u L 2 (,T ) + y L 2 (,L) ) if y is a fixed point of M f,u,y.) We now prove the uniqueness of the mild solution of the Cauchy problem for our nonlinear KdV equation, together with estimates of this solution.

19 Boundary controllability of KdV equation 385 Proposition 5. Let T > and let L >. There exists C = C (T, L) > such that for every (y, z ) L 2 (, L) 2, (u, v) L 2 (, T ) 2 and (f, g) L (, T, L 2 (, L)) 2 for which there exist mild solutions y and z of y t + y x + y xxx + yy x = f, y(t, ) = y(t, L) =, y x (t, L) = u(t), y(, x) = y (x), (A.2) (A.3) (A.4) (A.5) and of z t + z x + z xxx + zz x = g, z(t, ) = z(t, L) =, z x (t, L) = v(t), z(, x) = z (x), (A.6) (A.7) (A.8) (A.9) one has the following inequalities: T (z x (t, x) y x (t, x)) 2 dx dt ( ) (z y ) 2 dx + u v 2 L 2 (,T ) + f g 2 L (L 2 ) (z(t, x) y(t, x)) 2 dx e C (+ y 2 L 2 (H ) + z 2 L 2 (H ) ), (A.) ( ) (z y ) 2 dx + u v 2 L 2 (,T ) + f g 2 L (L 2 ) e C (+ y 2 L 2 (H ) + z 2 L 2 (H ) ), (A.) for all t [, T ]. Proof. Let := z y. (A.2) Then is a mild solution of t + x + xxx = y x z x (f g), (t, ) = (t, L) =, x (t, L) = v(t) u(t), (, x) = z (x) y (x). (A.3) (A.4) (A.5) (A.6)

20 386 Jean-Michel Coron, Emmanuelle Crépeau Formally, by integrating by parts in 2x ( t + x + xxx + y x + z x + (f g)) dx =, using (A.4) and (A.5), we readily get d dt x 2 dx x = 2 dx + L(v(t) u(t)) z 2 dx + 4 x (f g) dx. xz x dx xy x dx (A.7) (A.8) (Note that the multiplier x for (A.3) has been introduced by Rosier in [2, p. 48].) Even if (A.7) is only formal, using standard approximation arguments, one easily sees that (A.8) always holds (in the sense of distributions). By (A.3) and the continuous Sobolev embedding H (, L) C ([, T ]), there exists C = C (L) > such that 2 xy x dx C y x L 2 (,L) x x dx. Thus, 2 Similarly, 4 xy x dx 2 xz x dx 2 We have the following lemma: 2 x dx + C2 2 y x 2 L 2 (,L) L 2 x dx + 2C2 z x 2 L 2 (,L) x 2 dx. x 2 dx. Lemma 6. For every φ H (, L) with φ() =, and every a [, L], Indeed, φ 2 dx = a a a2 2 φ 2 dx + ( φ 2 dx a2 2 a φ 2 dx φ 2 x dx + a a φ 2 x (s) ds ) x dx + a φ 2 x dx + a xφ 2 dx. ( x xφ 2 dx. ) 2 φ x (s) ds dx + a xφ 2 dx xφ 2 dx (A.9) (A.2) (A.2)

21 Boundary controllability of KdV equation 387 Thanks to Lemma 6, there exists C 2 > such that 2 dx 2 2 x dx + C 2 x 2 dx. (A.22) Moreover, by (A.7) and the Sobolev embedding H (, L) C ([, T ]), there exists C 3 = C 3 (L) > such that 2 z 2 L dx C 3 z x L 2 (,L) 2 dx. Hence, using (A.4) and Lemma 6 with a := min{c /2 3 z x /2 C 4 = C 4 (L) > such that 2 Moreover, 2 z 2 dx 2 L 2 (,L) 2 x dx + C 4( + z x 3/2 L 2 (,L) ) x 2 dx., L}, there exists (A.23) ( /2 x (f g) dx 2 L f g L 2 (,L) x dx) 2. (A.24) Thus, by (A.2), (A.22), (A.23), and (A.24), there exists C 5 = C 5 (L) > such that d dt In particular, d dt x 2 dx + 2 x dx L(v(t) u(t)) L f g L 2 (,L) ( + C 5 ( + y x 2 L 2 (,L) + z x 2 L 2 (,L) ) x 2 dx. x 2 dx L(v(t) u(t)) L f g L 2 (,L) ) /2 x 2 dx ( + C 5 ( + y x 2 L 2 (,L) + z x 2 L 2 (,L) ) x 2 dx. Let us assume for the time being that the following lemma holds: ) /2 x 2 dx (A.25) (A.26) Lemma 7. Let T >. Let a, b and c be three nonnegative functions in L (, T ). Let w C ([, T ]) be a nonnegative function such that, in the sense of distributions, ẇ a(t) + b(t) w(t) + c(t)w(t). Then ( t ( t ) 2 ) t w(t) 3 w() + a(s) ds + b(s) ds e c(s) ds.

22 388 Jean-Michel Coron, Emmanuelle Crépeau From (A.6), (A.26) and Lemma 7, we get, for every t [, T ], ( ) x 2 (t, x) dx 3 x(z y ) 2 dx + L u v 2 L 2 (,T ) + 4L f g 2 L (L 2 ) e C 5(T + y 2 L 2 (H ) + z 2 L 2 (H ) ). (A.27) Using (A.2), (A.25) and (A.27), we get the existence of C 6 = C 6 (T, L) > such that T (z x (t, x) y x (t, x)) 2 dx dt ( ) (z y ) 2 dx + u v 2 L 2 (,T ) + f g 2 L (L 2 ) e C 6(+ y 2 L 2 (H ) + z 2 L 2 (H ) ), (A.28) which gives (A.). Finally, in order to get (A.), we multiply (A.3) by and integrate on [, L]. Using (A.4), (A.5) and integrations by parts, we get Moreover d 2 dt 2 dx x (t, ) = (v(t) u(t))2 2 (y x 2z x ) dx (y x 2z x + f g) dx. ( ) 2 x dx + 2 y2 + 2z 2 2 dx. (A.29) (A.3) By (A.3), (A.7), (A.29), (A.3), and the continuous Sobolev embedding H (, L) C ([, L]) there exists C 7 = C 7 (L) > such that d 2 dt 2 dx 2 (v(t) u(t))2 + 2 x dx + f (t, ) g(t, ) L 2 (,L) ( ) /2 2 dx + C 7 ( y x 2 L 2 (,L) + z x 2 L 2 (,L) ) 2 dx, which, combined with (A.6), (A.28) and Lemma 7, gives (A.) for C = C (T, L) > large enough. It remains to prove Lemma 7. Considering w(t) := w(t)e t c(s) ds,

23 Boundary controllability of KdV equation 389 we easily see that, without loss of generality, we may assume that c =. Moreover, we may also assume that there exists ɛ > such that, for almost every s in [, T ], a(s) ɛ. Let ( t ) w(t) := max w(t) a(s) ds,. Then, in the sense of distributions, ( d t ) w(t) + a(s) ds b(t) dt 2 + a(t). t 2 a(s) ds Integrating this inequality on [, t], we get t w(t) + a(s) ds w() + t t b(s) ds + a(s) ds, 2 which gives ( w(t) 3 w() + ( t 2 t ) b(s) ds) + a(s) ds, 4 and ends the proof of Lemma 7. B. Motion failure in the ±( cos(x)) directions for a 2 nd order power series expansion Throughout all this section, we again assume that (2.5) holds. We now denote by L 2 (, L) the space of measurable complex-valued functions such that f 2 dx <. We use the similar convention for C ([, L]), L 2 (, T ), H, D(A), mild solutions etc. We also still denote by P H the orthogonal projection on H L 2 (, L) for the hermitian product on L 2 (, L). The main result of this section is the following. Proposition 8. Let z be a mild solution of z t + z x + z xxx =, z(t, ) = z(t, L) =, z(, x) = z(t, x) =, (B.) (B.2) (B.3) and let y be a mild solution of y t + y x + y xxx =, y(t, ) = y(t, L) =, (B.4) (B.5) such that y(t, x)( cos(x)) dx =, t [, T ]. (B.6)

24 39 Jean-Michel Coron, Emmanuelle Crépeau Then T yz sin(x) dx dt =. (B.7) Before giving the proof of this proposition, let us first mention a corollary: Corollary 9. Let T, L, y and z be as in Proposition 8. Let w B be a mild solution of w t + w x + w xxx = (yz) x, w(t, ) = w(t, L) =, (B.8) (B.9) such that w(, x) =, P H (w(t, )) =. (B.) (B.) Then w(t, x) =. (B.2) Indeed, from (B.8) and (B.9), we get d dt ( cos(x))w dx = which, together with (B.7) and (B.), gives yz sin(x) dx, ( cos(x))w(t, x) dx =. (B.3) Finally, (B.2) follows from (B.) and (B.3). Let us now prove Proposition 8 in two special cases, from which we will then deduce the general case. Let λ C \ σ (A). Then there exists a unique y λ C ([, L]) such that λy λ + y λx + y λxxx =, y λ () = y λ (L) =, y λx (L) =. (B.4) (B.5) (B.6) Let Y λ (t, x) := e λt y λ (x). (B.7) Then the following lemma holds: Lemma 2. For λ C \ σ (A), the assertion of Theorem 8 holds if y(t, x) := Y λ (t, x). (B.8)

25 Boundary controllability of KdV equation 39 (Note that (B.4), (B.5), (B.7) and (B.8) imply (B.4) and (B.5).) Let us prove this lemma. Since λ C \ σ (A), there exists a unique φ C ([, L]) such that λφ + φ x + φ xxx = y λ sin(x), φ() = φ(l) =, φ x (L) =. (B.9) (B.2) (B.2) After some integrations by parts, we get, using (B.), (B.2), (B.9), (B.2), (B.7) and (B.8), yz sin(x) dx = d dt which, together with (B.3) and (B.2), gives (B.7) if zφe λt e λt [z x φ x ] L, φ x () =. (B.22) We multiply (B.9) by y λ (L x) to get, in view of (2.5), λφy λ (L x) + (φ x + φ xxx )y λ (L x) dx = y λ (L x)y λ (x) sin(x) dx =. (B.23) Performing integrations by parts in (B.23), we get, using (B.4), (B.5) and (B.2), φ x ()y λx (L) =, which, together with (B.6), gives (B.22) and ends the proof of Lemma 2. Let now µ σ (Ã) and ξ C ([, L]) be such that If µ =, we assume that µξ + ξ x + ξ xxx =, (B.24) ξ() = ξ(l) =, ξ x (L) = ξ x (). (B.25) (B.26) ( cos(x))ξ dx =. (B.27) Note that (B.27) is implied by (B.24) and (B.25) if µ =. One has the following lemma. Lemma 2. The assertion of Proposition 8 holds for y(t, x) := e µt ξ(x). (B.28)

26 392 Jean-Michel Coron, Emmanuelle Crépeau (Again (B.24) to (B.28) imply (B.4) to (B.6).) Let us prove this lemma. Let us first deal with the case where µ =. Then, since, by [2, Remark 3.6 (ii)], it follows that µ σ (A). Hence σ (A) σ (Ã) = {}, and there exists a unique φ C ([, L]) such that (B.29) (ξ x (L) = ) (ξ = ) (B.3) µφ + φ x + φ xxx = ξ sin(x), φ() = φ(l) =, φ x (L) =. We then proceed as in the proof of Lemma 2 to get the desired property. Let us now turn to the case where µ =. Then (B.24) to (B.27) imply the existence of c C such that We now define φ C ([, L]) by We have ξ(x) = c sin(x). φ(x) := 6 (x x cos(x) + 2 sin(2x) sin(x)). φ x + φ xxx = sin 2 (x) 3 ( cos(x)), φ() = φ(l) = φ x () = φ x (L) =, which, together with (B.) and (B.2), leads to z sin 2 (x) dx = 3 But, from (B.) and (B.2), we get d dt which, together with (B.3), gives ( cos(x))z dx + d dt ( cos(x))z dx =, zφ. (B.3) (B.32) ( cos(x))z dx =. (B.33) Equality (B.7) follows from (B.3), (B.28), (B.3), (B.32) and (B.33). This ends the proof of Lemma 2. Let E be the subspace of L 2 (, T ) L 2 (, L) spanned by the following pairs: (Y λx (, L), Y λ (, )) with λ C \ σ (A), (e µt ξ x (L), ξ) with µ σ (Ã) and ξ C ([, L]) satisfying (B.24) to (B.27).

27 Boundary controllability of KdV equation 393 Let us point out that, as in Section 2, we still have (2.5). Therefore E L 2 (, T ) H. We have Lemma 22. For every T >, E is dense in L 2 (, T ) H. Let us prove this lemma. Let v L 2 (, T ) and h L 2 (, L) be such that (v, h) E. Thus, for all λ C \ σ (A), T v(t)e λt dt + h(x)y λ (x) dx =. (B.34) Let us give an estimate on y λ in order to prove that v =. We multiply (B.4) by ȳ λ, take the real part and integrate on [, L]. Then, using integrations by parts together with (B.5) and (B.6), we get 2 Re(λ) From this equality we deduce that y λ L 2 (,L) 2Re(λ) y λ 2 dx = y λx () 2. (B.35) if Re(λ) >. (B.36) The same computations show that, if w D(A) satisfies Aw = λw with λ C, then 2Re(λ) w 2 dx = w x () 2. Therefore σ (A) {λ C; Re(λ) }. Hence by (B.34) and (B.36) the holomorphic function λ C T v(t)eλt is bounded and converges to as Re(λ) tends to +. Therefore this holomorphic function is identically equal to, which implies that v =. Let us now prove that h H. We have T hξ dx = (B.37) for every ξ C ([, L]) satisfying (B.24) for some µ σ (Ã) and (B.25) to (B.27). But, since iã is selfadjoint with compact resolvent, it follows from the spectral decomposition of such operators that the vector space spanned by such ξ is dense in H. Hence, by (B.37), h H. Let us now end the proof of Proposition 8 by a density argument on y. Let z be as in the hypotheses of Proposition 8. Let F : L 2 (, T ) H C, where y is the mild solution of y t + y x + y xxx =, (u, φ) T yz dx dt, y(t, ) = y(t, L) =, y x (t, L) = u(t), y(, x) = φ(x). By Lemmas 2 and 2, this linear map F vanishes on E. By Proposition 7, F is continuous. Hence F vanishes on the closure of E, which, by Lemma 22, is equal to L 2 (, T ) H. This ends the proof of Proposition 8.

28 394 Jean-Michel Coron, Emmanuelle Crépeau C. Proof of Lemma Let b R \ [ / 3, / 3] be such that 2λ = 2ib(4b 2 ). We have y 2λ = ηe ( 3b 2 ib)x + ( η)e ( 3b 2 ib)x e 2ibx with e 2ibL e 3b ( 2 ib)l η :=. e ( 3b 2 ib)l e 3b ( 2 ib)l After some integrations by parts, we get, using (2.47) for 2λ in place of λ, (2.65) and (2.73), (2λ(δ λ ψ λ ) + (δ λ ψ λ ) x + (δ λxxx ψ λxxx ))y 2λ dx Thus, by (2.64) and (2.72), = (δ λx (L) ψ λx (L))(y 2λx (L) y 2λx ()). (z λ sin(x) y λ φ λx )y 2λ dx = (δ λx (L) ψ λx (L))(y 2λx (L) y 2λx ()). Then, in order to prove Lemma, one just just needs to check that a (z λ sin(x) y λ φ λx )y 2λ dx is not identically on. (C.) Straightforward computations give φ λ (x) = 2a(2a + )(2a ) [κ[(a + i 3a 2 )e ( 3a 2 ia+i)x + (a i 3a 2 )e ( 3a 2 ia i)x ] + ( κ)[(a + + i 3a 2 )e ( 3a 2 ia+i)x + (a + i 3a 2 )e ( 3a 2 ia i)x ] + [(2a )e (2ia+i)x + (2a + )e (2ia i)x ]] + K e 2iax + K 2 e ( 3a 2 ia)x + K 3 e ( 3a 2 ia)x, with K := 3(2a + )(2a ), K 2 := A B, K 3 := C D, where A := (6ia 2 i + 2a 3a 2 )e 2iaL + ( 6ia 2 + i 2a 3a 2 )e L( + (6ia 2 2i 2a 3a 2 )e 4iaL + ( i 4a 3a 2 )e 2L( + (3i 6ia 2 + 6a 3a 2 )e L( 3a 2 +ia), 3a 2 ia) (C.2) 3a 2 +ia)

29 Boundary controllability of KdV equation 395 B := 6a(2a + )(2a )(e L( 3a 2 ia) e L( 3a 2 ia) ) (( 3ia + 3a 2 )e L( 3a 2 +ia) 2 3a 2 e 2iaL + (3ia + 3a 2 )e L( 3a 2 ia) ), C := ( i + 4a 3a 2 )e L( 3a 2 ia) + (6ia 2 i 2a 3a 2 )e L( 3a 2 )e 2iaL ][e L( 3a 2 ia) e 2iaL ], + ( 6ia 2 + 2i 2a D := 6a(2a + )(2a )(e L( 3a 2 ia) e L( 3a 2 ia) ) We also have 3a 2 )e L( 3a 2 +ia) (( 3ia + 2 3a 2 e 2iaL + (3ia + z λ (x) = k (x) + ( k () + 3a 2 )e L( 3a 2 ia) ). k (L) k () e ( 3b 2 +ib)l e 3b ( 2 +ib)l )e ( 3b 2 +ib)x k (L) k () e ( 3b 2 +ib)l e 3b ( 2 +ib)l + M λ ( e 2ibx + ηe ( 3b 2 +ib)x + ( η)e ( 3b 2 +ib)x ), where M λ C is a constant and Since k (x) := κ2 i( 3a 2 ia) 2(24a 3 9a + 3i 3a 2 ) e( 2 ( κ)2 i( 3a 2 ia) 2(24a 3 9a 3i 3a 2 ) e(2 + 3a 2 2ia)x 3a 2 2ia)x 3a 2 + ia) κ( κ) 2a 2 3 e 2iax + e4iax 24a 2 κi( 2(2a 3 3a) ) 3a 2 ia) e ( 3b 2 +ib)x e ( 3a 2 +ia)x (C.3) (C.4) i( κ)( 3a 2 + ia) 3a 2(2a 3 e ( 2 +ia)x. (C.5) 3a) ( e 2ibx + ηe ( 3b 2 +ib)x + ( η)e ( 3b 2 +ib)x )y 2λ sin(x) dx = ȳ 2λ y 2λ sin(x) dx =, one does not need to know M λ to compute the integral in (C.). Let Q := E (z λ sin(x) y λ φ λx )y 2λ dx, (C.6)

30 396 Jean-Michel Coron, Emmanuelle Crépeau with E = (e L( 3a 2 ia) e L( 3a 2 ia) )(e L( 3b 2 +ib) e L( 3b 2 +ib) ) (( 3ia + 3a 2 )e L( 3a 2 +ia) 2 + (3ia + 3a 2 )e L( 3a 2 +ia) ). 3a 2 e 2iaL Then Q is a finite sum of terms of the form P c (a)e (c 3a 2 +c 2 ia+c 3 3b 2 +c 4 ib)l where P c (a) is a rational function of a, b, 3a 2 and 3b 2 and c := (c, c 2, c 3, c 4 ) Z 4. (There are such terms.) It is easy to prove that (C.) is identically (if and) only if every P c is identically. We look at P (3,,2,2). After lengthy but straightforward computations, we get with P (3,,2,2) (a) = (3ia + 3a 2 )(f + f 2 + f 3 + f 4 + f 5 + f 6 + f 7 + f 8 + f 9 ) + 6ia2 i 2a 3a 2 (f + f ), (C.7) 6a(2a + )(2a ) ( 3a f = 2 ( ia)i 48a 3 8a 6i 3a 2 (2 3a 2 2ia + 3b 2 ib) 2 + ) (2, 3a 2 2ia + 2ib) 2 + f 2 = ( ) 24a 2 (4ia + 3b 2 ib) 2 + (4ia + 2ib) 2, + f 3 = ( 3a 2 ( + ia)i 24a 3 6a ( 3a 2 + ia + 3b 2 ib) 2 + ) (, 3a 2 + ia + 2ib) 2 + f 4 = 2a(2a + )(2a ) [( a + + i 3a 2 2 3a 2 2ia + i + 3b 2 ib a + i 3a + 2 ) 2 3a 2 2ia i + 3b 2 ib ( a + + i 3a 2 2 3a 2 2ia + i + 2ib + a + i 3a 2 2 3a 2 2ia i + 2ib )],

31 Boundary controllability of KdV equation 397 f 5 = 2a(2a + )(2a ) [( f 6 = + a + + i 3a 2 3a 2 + ia + i + 3b 2 ib a + i 3a 2 ) 3a 2 + ia i + 3b 2 ib ( a + + i 3a 2 3a 2 + ia + i + 2ib + a + i 3a 2 3a 2 + ia i + 2ib i [( 2a + ( f 7 = i 2a ( 3a 2 + ia + i + 3b 2 ib ) 3a 2 + ia i + 3b 2 ib 3a 2 + ia + i + 2ib + 3a 2 + ia i + 2ib [( 4ia + i + 3b 2 ib + )] 4ia + i + 2ib +, 4ia i + 2ib 2ia f 8 = 3(2a + )(2a ) [ )], )], 4ia i + 3b 2 ib 3a 2 + ia + 3b 2 ib 3a 2 + ia + 2ib [ 2ia f 9 = 3(2a + )(2a ) 4ia + 3b 2 ib 4ia + 2ib f = ( 3a 2 ia) [ ] 2 3a 2 2ia + 3b 2 ib 2, 3a 2 2ia + 2ib [ f = ( 3a 2 ia) 3a 2 + ia + 3b 2 ib 3a 2 + ia + 2ib As a +, one concludes that P (3,,2,2) (a) p/a with p := ( 3 i)(6i 2 ( 3) i + 2 /3 3 + i2 + / i i2 ) 4/ i + 2 /3 3 + i2 =. /3 3 + i i2 4/3 In particular P (3,,2,2) is not identically equal to. ], ) ], ].

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