Mo Chen

Lionel Rosier
email: lionel.rosier@univ-littoral.fr

Exact controllability of the linear Zakharov-Kuznetsov equation

Keywords: Zakharov-Kuznetsov equation, null controllability, reachable states, exact controllability, flatness approach, Gevrey functions. 2010 Mathematics Subject Classification: 37L50, 93B05

We consider the linear Zakharov-Kuznetsov equation on a rectangle with a left Dirichlet boundary control. Using the flatness approach, we prove the null controllability of this equation and provide a space of analytic reachable states.

Introduction

The Zakharov-Kuznetsov (ZK) equation

u t + au x + ∆u x + uu x = 0, (1.1)
provides a model for the propagation of nonlinear ionic-sonic waves in a plasma. In (1.1), x, t ∈ R and y ∈ R d (with d ∈ {1, 2}) are the independent variables, u = u(x, y, t) is the unknown, u t = ∂u/∂t, u x = ∂u/∂x, ∆u = ∂ 2 u/∂x 2 + d i=1 ∂ 2 u/∂y 2 i , and the constant a > 0 stands for the sound velocity. The ZK equation is, from the mathematical point of view, a natural extension to R d+1 of the famous Korteweg-de Vries equation

z t + az x + z xxx + zz x = 0, (1.2)
which has been extensively studied from the control point of view (see e.g. the surveys [START_REF] Cerpa | Control of a Korteweg-de Vries equation: a tutorial[END_REF][START_REF] Rosier | Control and stabilization of the Korteweg-de Vries equation: recent progresses[END_REF]).

If we focus on the situation where (1.2) is supplemented with the following boundary conditions z(0, t) = h(t), z(L, t) = z x (L, t) = 0, (1.3) where L > 0 is a given number and h is the control input, then it was proved in [START_REF] Glass | Some exact controllability results for the linear KdV equation and uniform controllability in the zero-dispersion limit[END_REF][START_REF] Rosier | Control of the surface of a fluid by a wavemaker[END_REF] that (1.2)-(1.3) was null controllable on the domain (0, L). Due to the smoothing effect, with such a control at the left endpoint the exact controllability can only hold in a space of analytic functions.

More recently, a space of analytic reachable states was provided in [START_REF] Martin | Exact controllability of a linear Korteweg-de Vries equation by the flatness approach[END_REF] for the linearized KdV equation

z t + z x + z xxx = 0
with the same boundary conditions as in (1.3). The method of proof was based on the flatness approach, as introduced in [START_REF] Martin | On the reachable states for the boundary control of the heat equation[END_REF] to study the reachable states of the heat equation. The aim of the paper is to extend the results given in [START_REF] Martin | Exact controllability of a linear Korteweg-de Vries equation by the flatness approach[END_REF] to the ZK equation.

The wellposedness of various initial boundary value problems for ZK were studied in [START_REF] Faminski | The Cauchy problem for the Zakharov?Kuznetsov equation[END_REF][START_REF] Faminskii | Initial-boundary value problems in a rectangle for two-dimensional Zakharov-Kuznetsov equation[END_REF][START_REF] Linares | The Cauchy problem for the 3D Zakharov-Kuznetsov equation[END_REF][START_REF] Linares | Well-posedness for the ZK equation in a cylinder and on the background of a KdV Soliton[END_REF][START_REF] Saut | An initial boundary-value problem for the Zakharov-Kuznetsov equation[END_REF][START_REF] Saut | An initial and boundary-value problem for the Zakharov-Kuznestov equation in a bounded domain[END_REF]. Some unique continuation property for ZK derived with a Carleman estimate was done in [START_REF] Chen | Unique continuation property for the Zakharov-Kuznetsov equation[END_REF]. Exact controllability results for ZK in the same spirit as those for KdV in [START_REF] Rosier | Exact boundary controllability for the Korteweg-de Vries equation on a bounded domain[END_REF] are given in [START_REF] Faminskii | Initial-boundary value problems in a rectangle for two-dimensional Zakharov-Kuznetsov equation[END_REF][START_REF] Perla-Menzala | Boundary control of the Zakharov-Kuznetsov equation[END_REF].

Here, we limit ourselves to the case d = 1, so that y ∈ R. By a translation, we can assume without loss of generality that x ∈ (-1, 0) (this will be more convenient when using series to represent the solutions). We set Ω := (-1, 0) × (0, 1). The paper is concerned with the control properties of the system: u t + u xxx + u xyy + au x = 0, (x, y) ∈ Ω, t ∈ (0, T ), (1.4) u(0, y, t) = u x (0, y, t) = 0, y ∈ (0, 1), t ∈ (0, T ), (1.5) u(-1, y, t) = h(y, t), y ∈ (0, 1), t ∈ (0, T ), (1.6) u(x, 0, t) = u(x, 1, t) = 0, x ∈ (-1, 0), t ∈ (0, T ), (1.7) u(x, y, 0) = u 0 (x, y), (x, y) ∈ Ω, (1.8) where u 0 = u 0 (x, y) is the initial data and h = h(y, t) is the control input.

We shall address the following issues:

1. (Null controllability) Given any u 0 ∈ L 2 (Ω), can we find a control h such that the solution u of (1.4)-(1.8) satisfies u(., T ) = 0? 2. (Reachable states) Given any u 1 ∈ R (a subspace of L 2 (Ω) defined thereafter), can we find a control h such that the solution u of (1.4)-(1.8) with u 0 = 0 satisfies u(., T ) = u 1 ?

We shall investigate both issues by the flatness approach and derive an exact controllability in R by combining our results.

To state our result, we need introduce notations. A function u ∈ C ∞ ([t 1 , t 2 ]) is said to be Gevrey of order s ≥ 0 on [t 1 , t 2 ] if there exist some constant C, R ≥ 0 such that

|∂ n t u(t)| ≤ C (n!) s R n ∀n ∈ N, ∀t ∈ [t 1 , t 2 ].
The set of functions Gevrey of order s on [t 1 , t 2 ] is denoted by

G s ([t 1 , t 2 ]). A function u ∈ C ∞ ([x 1 , x 2 ] × [y 1 , y 2 ] × [t 1 , t 2 ]
) is said to be Gevrey of order s 1 in x, s 2 in y and s 3 in t on [x 1 , x 2 ] × [y 1 , y 2 ] × [t 1 , t 2 ] if there exist some constants C, R 1 , R 2 , R 3 > 0 such that

|∂ n 1 x ∂ n 2 y ∂ n 3 t u(x, y, t)| ≤ C (n 1 !) s 1 (n 2 !) s 2 (n 3 !) s 3 R n 1 1 R n 2 2 R n 3 3 ∀n 1 , n 2 , n 3 ∈ N, ∀(x, y, t) ∈ [x 1 , x 2 ]×[y 1 , y 2 ]×[t 1 , t 2 ].
The set of functions Gevrey of order s 1 in x, s 2 in y and s

3 in t on [x 1 , x 2 ] × [y 1 , y 2 ] × [t 1 , t 2 ] is denoted by G s 1 ,s 2 ,s 3 ([x 1 , x 2 ] × [y 1 , y 2 ] × [t 1 , t 2 ]).
The first main result in this paper is a null controllability result with a control input in a Gevrey class.

Theorem 1.1. Let u 0 ∈ L 2 (Ω) and s ∈ [ 3 2 , 2). Then there exists a control input h ∈ G s 2 ,s ([0, 1]× [0, T ]) such that the solution u of (1.4)-(1.8) satisfies u(•, •, T )=0. Furthermore, it holds that u ∈ C([0, T ]; L 2 (Ω)) ∩ G s 2 , s 2 ,s ([-1, 0] × [0, 1] × [ε, T ]), ∀ ε ∈ (0, T ).
Introduce the differential operator P u := u x + au x and the following space

R R 1 ,R 2 := {u ∈ C ∞ ([-1, 0] × [0, 1]); ∃C > 0, |∂ p x ∂ q y u(x, y)| ≤ C (p!) 2 3 (q!) 2 3 R p 1 R q 2 ∀p, q ∈ N, ∀(x, y) ∈ Ω,
and

P n u(0, y) = ∂ x P n u(0, y) = P n u(x, 0) = P n u(x, 1) = 0, ∀n ∈ N, ∀x ∈ [-1, 0], ∀y ∈ [0, 1]}.
Our second main result provides a set of reachable states for system (1.4)-(1.8).

Theorem 1.2. Let R 0 := 3 9(a + 2)e (3e) -1 , and let R 1 , R 2 ∈ (R 0 , +∞). Then for any

u 1 ∈ R R 1 ,R 2 , there exists a control input h ∈ G 1,2 ([0, 1] × [0, T ]) such that the solution u of (1.4)-(1.8) with u 0 = 0 satisfies u(•, •, T ) = u 1 . Furthermore, u ∈ G 1,1,2 ([-1, 0] × [0, 1] × [0, T ])
, and the trajectory u = u(x, y, t) and the control h = h(y, t) can be expanded as series:

u(x, y, t) = ∞ j=1 ∞ i=0 g i,j (x)z (i) j (t)e j (y), (1.9) h(y, t) = ∞ j=1 ∞ i=0 g i,j (-1)z (i) j (t)e j (y). (1.10)
We refer the reader to Section 2 for the definitions of the functions g i,j (i ≥ 0, j ≥ 1) and of the functions e j (j ≥ 1).

Combining Theorem 1.1 and Theorem 1.2, we obtain the following result which implies the exact controllability of (1.4)- (1.8)

in R R 1 ,R 2 for R 1 > R 0 and R 2 > R 0 . Corollary 1.1. Let R 0 := 3 9(a + 2)e (3e) -1 , and let R 1 , R 2 ∈ (R 0 , +∞). Let u 0 ∈ L 2 (Ω) and u 1 ∈ R R 1 ,R 2 . Then there exists h ∈ G 1,2 ([0, 1] × [0, T ]) such that the solution of (1.4)-(1.8) satisfies u(., T ) = u 1 .
The paper is outlined as follows. Section 2 introduces the eigenfunctions e j , the generating functions g i,j , and provide some estimates needed in the sequel. The null controllability of ZK is established in Section 3, while the reachable states of ZK are investigated in Section 4.

Preliminaries

First we introduce the operator

Au := -P u = -u x -au x with domain D(A) = {u ∈ L 2 (Ω); P u ∈ L 2 (Ω), u(-1, y) = u(0, y) = u x (
0, y) = 0 for a.e. y ∈ (0, 1) and u(x, 0) = u(x, 1) = 0 for a.e. x ∈ (-1, 0)}.

It is well-known (see e.g. [START_REF] Saut | An initial and boundary-value problem for the Zakharov-Kuznestov equation in a bounded domain[END_REF]) that the operator A generates a semigroup of contractions in

L 2 (Ω). In what follows, we denote f D(A) = f L 2 (Ω) + A f L 2 (Ω) for all f ∈ D(A).
It would be natural to expect, as for KdV, that the domain D(A) coincide with the set {u ∈ H 3 (Ω) ∩ H 1 0 (Ω); u x (0, y) = 0 for a.e. y ∈ (0, 1)}, but this is not the case. The best description (up to date) of D(A) is given in the following lemma.

Lemma 2.1. We have the following inclusions:

{u ∈ H 2 (Ω) ∩ H 1 0 (Ω); u x ∈ H 2 (Ω) and u x (0, y) = 0 for a.e. y ∈ (0, 1)} ⊂ D(A), (2.1) D(A) ⊂ {u ∈ H 2 (Ω) ∩ H 1 0 (Ω); (x + 1)u x ∈ H 2 (Ω) ∩ H 1 0 (Ω)}. (2.2) Proof. The inclusion (2.1) is obvious. For (2.2), it follows from [19, Proposition 2] that D(A) ⊂ H 2 (Ω) ∩ H 1 0 (Ω). If u ∈ D(A), then f := ∆u x + au x ∈ L 2 (Ω) and hence ∆((x + 1)u x ) = (x + 1)∆u x + 2u xx = (x + 1)(f -au x ) + 2u xx ∈ L 2 (Ω).
On the other hand, we claim that (x + 1)u x ∈ H 1 0 (Ω). Indeed, u x ∈ H 1 (Ω) and hence (x + 1)u x ∈ H 1 (Ω). Moreover, u(., 0) = u(., 1) = 0 in H 3 2 (-1, 0) gives u x (., 0) = u x (., 1) = 0 in H 1 2 (-1, 0), and finally ((x + 1)u x )(-1, .) = u x (0, .) = 0 in H 1 2 (0, 1). By the classical boundary H 2 regularity result for the Dirichlet problem on a Lipschitz domain, we infer that (x + 1)

u x ∈ H 2 (Ω) ∩ H 1 0 (Ω).
Remark 2.1. It can be shown that the inclusion (2.1) is strict.

The following lemmas will be used several times thereafter.

Lemma 2.2. For any n ∈ N * and any

f ∈ D(A n ) with A i f ∈ H 2(n-i) (Ω) for i = 0, 1, ..., n, we have ∂ 2p y f (x, 0) = ∂ 2p y f (x, 1) = 0, ∀x ∈ [-1, 0], ∀p ∈ {0, ..., n -1}. (2.3)
Proof. We proceed by induction on n. For n = 1, the property (2.3) is obvious since f ∈ D(A). Assume now that (2.3) is true for n

-1 ≥ 1. If f ∈ D(A n ) with A i f ∈ H 2(n-i) (Ω) for i = 0, 1, ..., n, then P f = -Af ∈ D(A n-1 ) with A i P f = -A i+1 f ∈ H 2(n-i-1)
(Ω) for i = 0, 1, ..., n -1, , so that by (2.3) applied to P f and p = n -2

∂ 2n-4 y P f (x, 0) = ∂ 2n-4 y P f (x, 1) = 0.
This implies

∂ 3 x ∂ 2n-4 y f (x, 0) + ∂ x ∂ 2n-2 y f (x, 0) + a∂ x ∂ 2n-4 y f (x, 0) = 0, (2.4) ∂ 3 x ∂ 2n-4 y f (x, 1) + ∂ x ∂ 2n-2 y f (x, 1) + a∂ x ∂ 2n-4 y f (x, 1) = 0. (2.5) Since (2.
3) is true for n -1, we obtain that ∂ 2p y f (x, 0) = ∂ 2p y f (x, 1) = 0 for p = 0, 1, ..., n -2, and hence (taking p = n -2 and using (2.4)-(2.5))

∂ x ∂ 2n-2 y f (x, 0) = ∂ x ∂ 2n-2 y f (x, 1) = 0.
This means that we have for some constants

C 1 and C 2 ∂ 2n-2 y f (x, 0) = C 1 , ∂ 2n-2 y f (x, 1) = C 2 ∀x ∈ [-1, 0]. Note that ∂ 2n-2 y f ∈ H 2 (Ω) ⊂ C(Ω). On the other hand, it follows from the assumption f ∈ D(A) that ∂ 2n-2 y f (0, y) = 0 ∀y ∈ [0, 1].
Taking y = 0 and next y = 1, we see that C 1 = C 2 = 0. The proof of Lemma 2.2 is complete.

Remark 2.2. It will be proved in Proposition 2.1 (see below) that D(A n ) ⊂ H 2n (Ω) for all n ∈ N, so that the conclusion of Lemma 2.2 will be still valid when assuming solely that f ∈ D(A n ).

The following lemma is classical. Its proof is omitted.

Lemma 2.3. Let A = ∂ 2 y with domain D(A ) = H 2 (0, 1) ∩ H 1 0 (0, 1). Then for any m ∈ N * , it holds D(A m 2 ) = {g ∈ H m (0, 1); g (2p) (0) = g (2p) (1) = 0 for 0 ≤ p ≤ m -1 2 }• Let h ∈ L 2 (0, 1) be decomposed as h(y) = ∞ j=1
c j e j (y), and let m ∈ N * . Then

h ∈ D(A m 2 ) ⇐⇒ ∞ j=1 |λ m 2 j c j | 2 < ∞.
Furthermore, for any h ∈ D(A m

2 ), we have

h (q) 2 L 2 (0,1) = ∞ j=1 λ q j |c j | 2 ∀q ∈ {0, ..., m}.
We are in a position to state the main result in this section.

Proposition 2.1. For any n ∈ N, it holds D(A n ) ⊂ H 2n (Ω). Furthermore, there exists a constant B ≥ 1 such that

u H 2n (Ω) ≤ B n n i=0 P i u L 2 (Ω) , ∀n ∈ N, ∀u ∈ D(A n ). (2.6)
Proof. Let {e j } j≥1 be an orthonormal basis in L 2 (0, 1) such that e j is an eigenfunction for the Dirichlet Laplacian on (0, 1), λ j being the corresponding eigenvalue; that is -e j (y) = λ k e j (y), e j (0) = e j (1) = 0.

A classical choice is e j (y) = √ 2 sin(jπy) and λ j = (jπ) 2 for j ≥ 1. Following [START_REF] Saut | An initial and boundary-value problem for the Zakharov-Kuznestov equation in a bounded domain[END_REF], we decompose any function u ∈ L 2 (Ω) as

u(x, y) = ∞ j=1 ûj (x)e j (y). Note that u 2 L 2 (Ω) = ∞ j=1 ûj 2
, where we denote h = h L 2 (-1,0) for all h ∈ L 2 (-1, 0) for the sake of simplicity. If u ∈ D(A) and g := ∆u x + au x , then for any j ≥ 1

û j + (a -λ j )û j = ĝj in L 2 (-1, 0) (2.7)
where = d/dx. For n = 0, (2.6) is obvious if we pick C 0 ≥ 1. Let us assume first that n = 1. Note that ûj ∈ H 3 (-1, 0) by (2.7). Multiplying (2.7) by λ j (x + 1)û j , we obtain

3 2 λ j 0 -1 |û j | 2 dx -(a -λ j ) λ j 2 0 -1 |û j | 2 dx = λ j 0 -1 (x + 1)û j ĝj dx. Let j 0 := √ 2a
π . Then for j > j 0 , we have a ≤ λ j /2 and hence |a -λ j |λ j /2 ≥ λ 2 j /4. Using

λ j 0 -1 (x + 1)û j ĝj dx ≤ λ 2 j 8 0 -1 |û j | 2 dx + 2 0 -1 |ĝ j | 2 dx,
we infer that for j > j 0

3 2 λ j 0 -1 |û j | 2 dx + λ 2 j 8 0 -1 |û j | 2 dx ≤ 2 0 -1 |ĝ j | 2 dx,
and that for 1

≤ j ≤ j 0 3 2 λ j 0 -1 |û j | 2 dx + λ 2 j 8 0 -1 |û j | 2 dx ≤ Λ 0 -1 |û j | 2 dx + 2 0 -1 |ĝ j | 2 dx, (2.8)
where Λ := max

1≤j≤j 0 λ 2 j 8 + (a -λ j ) λ j 2 .
Obviously, (2.8) is valid for any j ≥ 1. Summing in j, we

obtain 3 2 Ω |u xy | 2 dxdy + 1 8 Ω |u yy | 2 dxdy ≤ Λ u 2 + 2 g 2 .
(2.9)

Dividing in (2.8) by λ j ≥ π 2 and summing in j, we obtain

3 2 Ω |u x | 2 dxdy + 1 8 Ω |u y | 2 dxdy ≤ Λ π 2 u 2 + 2 π 2 g 2 . ( 2
|û j | 2 dx + û j û j | 0 -1 + (a -λ j ) 0 -1 |û j | 2 dx = 0 -1 ĝj û j dx,
and hence

0 -1 |û j | 2 dx ≤ |λ j -a| 0 -1 |û j | 2 dx + ĝj û j + |û j (-1)û j (-1)|.
We are let to estimate û j (-1) and û j (-1). Multiplying in (2.7) by λ j ûj results in

λ j û j (-1) 2 2 = λ j 0 -1 ûj ĝj dx.
Combined with (2.8), this yields

λ j 2 û j (-1) 2 ≤ λ j ûj • ĝj ≤ 1 4 λ j ûj 2 + ĝj 2 ≤ 5 ĝj 2 + 2Λ ûj 2 . (2.11)
Multiplying in (2.7) by x, we obtain

- 0 -1 û j dx + xû j | 0 -1 + (a -λ j ) - 0 -1 ûj dx + xû j | 0 -1 = 0 -1 xĝ j dx which yields û j (-1) = -û j (-1) + (a -λ j ) 0 -1 ûj dx + 0 -1 xĝ j dx, so that |û j (-1)| 2 ≤ 3 |û j (-1)| 2 + 2 a 2 + |λ j | 2 ûj 2 + ĝj 2 .
Using (2.8) and (2.11), we conclude that |û j (-1)| 2 = O( ûj 2 + ĝj 2 ). The same is true for û j 2 . Gathering together the above estimates, we arrive at

u 2 H 2 (Ω) ≤ C 1 u 2 L 2 (Ω) + P u 2 L 2 (Ω)
for some constant

C 1 = C 1 (a) > 0. Let us check that D(A n ) ⊂ H 2n (Ω) for n ≥ 2. We proceed by induction on n. Assume that D(A p ) ⊂ H 2p (Ω) for p = 0, 1, ..., n -1 (with n -1 ≥ 1), and pick any u ∈ D(A n ). Then g = Au ∈ D(A n-1 ) ⊂ H 2(n-1) (Ω). Let h := (-1) n-1 ∂ 2(n-1) y g ∈ L 2 (Ω).
Then, using Lemmas 2.2 and 2.3, we have that for all j ≥ 1

λ n-1 j (û j + (a -λ j )û j ) = ĥj .
(2.12)

Multiplying in (2.12) by λ n j (x + 1)û j , we obtain

3 2 λ 2n-1 j 0 -1 |û j | 2 dx -(a -λ j ) λ 2n-1 j 2 0 -1 |û j | 2 dx = λ n j 0 -1 (x + 1)û j ĥj dx.
This yields

λ 2n-1 j û j 2 + λ 2n j ûj 2 = O( ûj 2 + ĥj 2 ), (2.13)
Multiplying in (2.12) by λ n j ûj gives

λ 2n-1 j û j (-1) 2 2 = λ n j 0 -1 ûj ĥj dx and λ 2n-1 j |û j (-1)| 2 = O( ûj 2 + ĥj 2 ).
From

λ n-1 j û j (-1) = -λ n-1 j û j (-1) + (a -λ j )λ n-1 j 0 -1 ûj dx + 0 -1
x ĥj dx,

we infer that λ 2n-2 j |û j (-1)| 2 = O( ûj 2 + ĥj 2 ).
It follows from

-λ 2n-2 j 0 -1 |û j | 2 dx + λ 2n-2 j û j û j | 0 -1 + (a -λ j )λ 2n-2 j 0 -1 |û j | 2 dx = λ n-1 j 0 -1 ĥj û j dx, that λ 2n-2 j û j 2 = O( ûj 2 + ĥj 2 ). (2.14)
So far, we have proved that

∞ j=1 λ 2n j ûj 2 + λ 2n-1 j û j 2 + λ 2n-2 j û j 2 < +∞. Using Lemma 2.3, this gives that ∂ 2n y u, ∂ 2n-1 y ∂ x u, and ∂ 2n-2 y ∂ 2 x u belong to L 2 (Ω).
For the other derivatives of order 2n, we apply the operator ∂ 2k

x (for k ∈ N with 2k + 3 ≤ 2n) to each term in (2.7) to obtain û(2k+3

) j + (a -λ j )û (2k+1) j = ĝ(2k) j .
This yields λ

(2n-3-2k) j û(2k+3) j 2 = O( ûj 2 + ĥj 2 + λ 2n-3-2k j ĝ(2k) j 2 ).
On the other hand, (2.7) gives by differentiation with respect to x that û(4) j + (a -λ j )û j = ĝ j , and we obtain in a similar way that

λ (2n-4-2k) j û(2k+4) j 2 = O( ûj 2 + ĥj 2 + λ 2n-4-2k j ĝ(2k+1) j 2 ).
for k ∈ N with 2k + 4 ≤ 2n. Thus we conclude that

2n q=0 ∞ j=1 λ (2n-q) j û(q) j 2 < +∞.
Using Lemma 2.3, we infer that for q ∈ {0, ..., 2n}, ∂ q x u ∈ L 2 (-1, 0, H 2n-q (0, 1)), and hence that

∂ 2n-q y ∂ q x u ∈ L 2 (Ω). We also have that ∂ 2n-1-q y ∂ q x u ∈ L 2 (Ω) for q ∈ {0, ..., 2n -1}. Taking into account the fact that u ∈ D(A n-1 ) ⊂ H 2(n-1) (Ω), we conclude that u ∈ H 2n (Ω). The proof of the inclusion D(A n ) ⊂ H 2n (Ω) is complete.
It remains to prove that the constant in the r.h.s. of (2.6) is indeed of the form B n . This will require a series of lemmas.

Lemma 2.4. For any ε 0 > 0, there exists a constant K = K(ε 0 ) > 0 such that for all ε ∈ (0, ε 0 ) and all f ∈ H 2 (-1, 0),

0 -1 |f (t)| 2 dt ≤ Kε 0 -1 |f (t)| 2 dt + Kε -1 0 -1 |f (t)| 2 dt.
(2.15)

Lemma 2.4 is a direct consequence of [1, Lemma 4.10] (which is concerned with twice continuously functions) by density of

C 2 ([-1, 0]) in H 2 (-1, 0).
For any j ∈ N * , we define the operator P j by

P j f := f -(λ j -a)f , ∀f ∈ H 3 (-1, 0). Lemma 2.5. There exists a constant C 1 ≥ 1 such that f 2 H 2n (-1,0) ≤ C n 1 n i=0 λ 2n-2i j P i j f 2 , ∀n ∈ N, ∀j ∈ N * , ∀f ∈ H 3n (-1, 0). (2.16)
Proof. For n = 0, (2.16) is obvious. For n = 1, it follows from the definition of P j and Lemma 2.4 that

f 2 H 2 (-1,0) = f 2 + f 2 + f 2 ≤C( f 2 + f 2 ) ≤C( f 2 + 1 λ j f 2 + λ j f 2 ) ≤C( f 2 + 1 λ j P j f 2 + λ j f 2 ) ≤C( f 2 + P j f 2 ) + 1 2 f 2 + Cλ 2 j f 2 ≤C(λ 2 j f 2 + P j f 2 ) + 1 2 f 2 .
This shows that we can find a constant C 2 ≥ 1 such that

f 2 H 2 (-1,0) ≤ C 2 (λ 2 j f 2 + P j f 2 ).
Let us prove (2.16) for n ≥ 2 by induction on n. Assume (2.16) to be true for n -1 ≥ 0. It follows that

f 2 H 2n (-1,0) = f 2 H 2n-2 (-1,0) + f (2n-1) 2 + f (2n) 2 ≤ f 2 H 2n-2 (-1,0) + f (2n-2) 2 H 2 (-1,0) ≤ f 2 H 2n-2 (-1,0) + C 2 (λ 2 j f (2n-2) 2 + P j f (2n-2) 2 ) ≤2C 2 λ 2 j f 2 H 2n-2 (-1,0) + C 2 P j f 2 H 2n-2 (-1,0) ≤2C 2 λ 2 j C n-1 1 n-1 i=0 λ 2n-2-2i j P i j f 2 + C 2 C n-1 1 n-1 i=0 λ 2n-2-2i j P i+1 j f 2 ≤2C 2 C n-1 1 n-1 i=0 λ 2n-2i j P i j f 2 + C 2 C n-1 1 n i=1 λ 2n-2i j P i j f 2 ≤3C 2 C n-1 1 n i=0 λ 2n-2i j P i j f 2 . If we pick C 1 = 3C 2 , (2.16
) is true for n.

Lemma 2.6. There exists a positive constant C 3 such that

u 2 H 2n (Ω) ≤ C 3 n m=0 m k=0 ∂ 2k x ∂ 2m-2k y u 2 L 2 (Ω) , ∀n ∈ N, ∀u ∈ D(A n ).
Proof. For any p ∈ N, we set

I p := a,b∈N, a+b=p ∂ a x ∂ b y u 2 L 2 (Ω) .
Decompose u as u(x, y) = ∞ j=1 ûj (x)e j (y).

(2.17)

Let us go back to the proof of Lemma 2.6. Pick any u ∈ D(A n ), for some n ∈ N. Using Lemma 2.2 and applying Lemma 2.3 to the functions ∂ 2m+1-k x u(x, .) for 0 ≤ m ≤ n -1, 10 0 ≤ k ≤ 2m + 1, and x ∈ (-1, 0), we obtain that

I 2m+1 = 2m+1 k=0 ∂ 2m+1-k x ∂ k y u 2 L 2 (Ω) = 2m+1 k=0 ∞ j=1 λ k j û(2m+1-k) j 2 = ∞ j=1 û(2m+1) j 2 + 2m+1 k=1 ∞ j=1 λ k j û(2m+1-k) j 2 ≤ ∞ j=1 λ j û(2m+1) j 2 + 1 2 2m+1 k=1 ∞ j=1 λ k-1 j û(2m+1-k) j 2 + 1 2 2m+1 k=1 ∞ j=1 λ k+1 j û(2m+1-k) j 2 = ∂ 2m+1 x ∂ y u 2 L 2 (Ω) + 1 2 2m+1 k=1 ∂ 2m+1-k x ∂ k-1 y u 2 L 2 (Ω) + 1 2 2m+1 k=1 ∂ 2m+1-k x ∂ k+1 y u 2 L 2 (Ω) ≤ 3 2 I 2m+2 + 1 2 I 2m ,
where we used Young's estimate. Thus, we have

u 2 H 2n (Ω) = n m=0 I 2m + n-1 m=0 I 2m+1 ≤ n m=0 I 2m + n-1 m=0 ( 3 2 I 2m+2 + 1 2 I 2m ) ≤3 n m=0 I 2m .
(2.18)

Next, we consider I 2m . For m = 0, I 0 = u 2 L 2 (Ω) . For m ≥ 1, we have

I 2m = m k=0 ∂ 2k x ∂ 2m-2k y u 2 L 2 (Ω) + m-1 k=0 ∂ 2k+1 x ∂ 2m-2k-1 y u 2 L 2 (Ω) , (2.19)
and it remains to estimate the second term in the r.h.s. of (2.19). Applying Lemma 2.4, we obtain

m-1 k=0 ∂ 2k+1 x ∂ 2m-2k-1 y u 2 L 2 (Ω) = m-1 k=0 ∞ j=1 λ 2m-2k-1 j û(2k+1) j 2 ≤C m-1 k=0 ∞ j=1 λ 2m-2k-2 j û(2k+2) j 2 + m-1 k=0 ∞ j=1 λ 2m-2k j û(2k) j 2 =C m k=1 ∞ j=1 λ 2m-2k j û(2k) j 2 + m-1 k=0 ∞ j=1 λ 2m-2k j û(2k) j 2 ≤C m k=0 ∞ j=1 λ 2m-2k j û(2k) j 2 =C m k=0 ∂ 2k x ∂ 2m-2k y u 2 L 2 (Ω) .
(2.20) Combining (2.18)-(2.20), the conclusion of Lemma 2.6 follows.

Lemma 2.7. There exists a constant C 4 ≥ 1 such that

λ 2m j P i j ûj 2 ≤ C m 4 m l=0 m l P i+l j ûj 2 , ∀m, i ∈ N, ∀j ∈ N * , ∀u ∈ D(A m+i ), (2.21)
where ûj is the Fourier coefficients of u as in (2.17).

Proof. The proof is by induction on m. For m = 0, (2.21) is obvious for any C 4 ≥ 1.

For m = 1 and u ∈ D(A 1+i ), we have that P i u ∈ D(A) and, by [19, Lemma 4.1],

(P i u)(x, y) = ∞ j=1 (P i j ûj )(x)e j (y),
where the function P i j ûj satisfies for each j ∈ N * (P i j ûj ) -(λ j -a)(P i j ûj ) = P i+1 j ûj , (P i j ûj )(-1) = (P i j ûj )(0) = (P i j ûj ) (0) = 0.

x ∈ (-1, 0), (

Multiplying the first equation in (2.22) by λ j (x + 1)P i j ûj and integrating over (-1, 0) results in

3 2 λ j 0 -1 |(P i j ûj ) | 2 dx + (λ j -a) λ j 2 0 -1 |P i j ûj | 2 dx = λ j 0 -1 (x + 1)(P i j ûj )(P i+1 j ûj )dx.
After some elementary calculations, we can find a constant C 4 = C 4 (a) ≥ 1 such that

λ j (P i j ûj ) 2 + λ 2 j P i j ûj 2 ≤ C 4 ( P i j ûj 2 + P i+1 j ûj 2 ).
Therefore, (2.21) holds for m = 1. Pick now any m ≥ 2, and assume that (2.21) is true for m -1 ≥ 0. For any u ∈ D(A m+i ), we have

λ 2m j P i j ûj 2 = λ 2 j λ 2m-2 j P i j ûj 2 ≤ λ 2 j C m-1 4 m-1 l=0 m -1 l P i+l j ûj 2 .
Since u ∈ D(A m+i ), for any l = 0, 1, ..., m -1, system (2.22) is satisfied with P i+l j ûj substituted to P i j ûj , and it follows as above that

λ 2 j P i+l j ûj 2 ≤ C 4 ( P i+l j ûj 2 + P i+l+1 j ûj 2 ).
We infer that

λ 2m j P i j ûj 2 ≤C m 4 m-1 l=0 m -1 l ( P i+l j ûj 2 + P i+l+1 j ûj 2 ) =C m 4 ( P i j ûj 2 + m-1 l=1 m -1 l P i+l j ûj 2 + m-1 l=1 m -1 l -1 P i+l j ûj 2 + P i+m j ûj 2 ) =C m 4 ( P i j ûj 2 + m-1 l=1 m l P i+l j ûj 2 + P i+m j ûj 2 ) =C m 4 m l=0 m l P i+l j ûj 2
where we used Pascal's Rule. The proof of Lemma 2.7 is achieved.

We are in a position to complete the proof of Proposition 2.1. The estimate (2.6) is obvious for n = 0. Let n ≥ 1. Using Lemmas 2.5, 2.6, and 2.7, we obtain that

u 2 H 2n (Ω) ≤C 3 n m=0 m k=0 ∂ 2k x ∂ 2m-2k y u 2 L 2 (Ω) =C 3 n m=0 m k=0 ∞ j=1 λ 2m-2k j û(2k) j 2 ≤C 3 n m=0 m k=0 ∞ j=1 λ 2m-2k j C k 1 k i=0 λ 2k-2i j P i j ûj 2 ≤C 3 C n 1 n m=0 m k=0 ∞ j=1 k i=0 λ 2m-2i j P i j ûj 2 .
Using the fact that i ≤ k ≤ m ≤ n in the sum above, we obtain

u 2 H 2n (Ω) ≤C 3 C n 1 n m=0 n k=0 ∞ j=1 n i=0 λ 2n-2i j P i j ûj 2 ≤C 3 C n 1 (n + 1) 2 ∞ j=1 n i=0 λ 2n-2i j P i j ûj 2 ≤C 3 C n 1 (n + 1) 2 ∞ j=1 n i=0 C n-i 4 n-i l=0 n -i l P i+l j ûj 2 ≤C 3 C n 1 (n + 1) 2 C n 4 2 n ∞ j=1 n i=0 n l=i P l j ûj 2 ≤C 3 C n 1 (n + 1) 2 C n 4 2 n ∞ j=1 n i=0 n l=0 P l j ûj 2 ≤C 3 C n 1 (n + 1) 3 C n 4 2 n n l=0 ∞ j=1 P l j ûj 2 ≤B n n l=0 P l u 2 L 2 (Ω) with B := 16C 1 C 3 C 4 .
Indeed, it is easy to see that (n + 1) 3 ≤ 8 n for all n ∈ N. The proof of Proposition 2.1 is achieved.

Recall that λ j = (jπ) 2 for j ≥ 1. For any j ≥ 1, we consider a sequence of generating functions g i,j (i ≥ 0), where g 0,j is the solution of the Cauchy problem g 0,j (x) -(λ j -a)g 0,j (x) = 0, g 0,j (0) = g 0,j (0) = 0, g 0,j (0) = 1,

x ∈ (-1, 0), (2.23) while g i,j for i ≥ 1 is defined inductively as the solution of the Cauchy problem

g i,j (x) -(λ j -a)g i,j (x) = -g i-1,j (x), g i,j (0) = g i,j (0) = g i,j (0) = 0.
x ∈ (-1, 0), (2.24) Proposition 2.2. For any i ≥ 0, j ≥ 1 and x ∈ [-1, 0], we have

|g i,j (x)| ≤ e √ λ j 3 i i! (3i + 2)! . ( 2

.25)

Proof. It follows from (2.23) and (2.24) that

g i,j (x) = - x 0 g 0,j (x -ξ)g i-1,j (ξ)dξ = - x 0 g 0,j (x -ξ) ξ 0 ( ζ 0 g i-1,j (σ)dσ)dζ dξ, i, j ≥ 1.
(1) if λ j ≤ a, it is not difficult to obtain that

g 0,j (x) =      1 a -λ j (1 -cos( a -λ j x)), λ j < a; 1 2 x 2 , λ j = a, this implies 0 ≤ g 0,j (x) ≤ x 2 2 , ∀ j ≥ 1, x ∈ [-1, 0].
Then it follows from [13, Lemma 2.1] that

|g i,j (x)| ≤ |x| 3i+2 (3i + 2)! ≤ e √ λ j 3 i i! (3i + 2)! , ∀i ≥ 0, ∀j ≥ 1, ∀x ∈ [-1, 0].
(2) if λ j > a, we claim that

g i,j (x) ≤ cosh( λ j -ax) (-x) 3i+2 3 i i! (3i + 2)! , ∀i ≥ 0, ∀j ≥ 1, ∀x ∈ [-1, 0] (2.26)
which implies (2.25). Let us prove (2.26) by induction on i.

For i = 0, 0 ≤ g 0,j (x) = 1 λ j -a (cosh( λ j -ax) -1) = ∞ q=1 (λ j -a) q-1 x 2q (2q)! ≤ ∞ q=1 (λ j -a) q-1 x 2q-2 (2q -2)! x 2 2! = cosh( λ j -ax) x 2
2! , so that (2.26) is true for i = 0. Assume now that (2.26) is true for i -1 ≥ 0. We can deduce that for x ∈ [-1, 0]

|g i,j (x)| ≤ - x 0 g 0,j (x -ξ) ξ 0 ζ 0 |g i-1,j (σ)|dσ dζ dξ ≤ - x 0 ∞ p=0 (λ j -a) p (x -ξ) 2p (2p)! ξ 0 ζ 0 3 i-1 (i -1)! ∞ q=0 (λ j -a) q (-σ) 3i-1+2q (2q)!(3i -1)! dσ dζ dξ = -3 i-1 (i -1)! x 0 ∞ p=0 (λ j -a) p (x -ξ) 2p (2p)! ∞ q=0 (λ j -a) q (-ξ) 3i+1+2q (2q)!(3i -1)!(3i + 2q)(3i + 2q + 1) dξ = -3 i-1 (i -1)! x 0 ∞ p=0 ∞ q=0 (λ j -a) p+q (x -ξ) 2p (-ξ) 3i+1+2q (2p)!(2q)!(3i -1)!(3i + 2q)(3i + 2q + 1)
dξ.

Then, integrating by parts 2p times, we obtain

|g i,j (x)| ≤ -3 i-1 (i -1)! x 0 ∞ p=0 ∞ q=0 (λ j -a) p+q (-ξ) 3i+1+2q+2p (3i + 2q -1)! (2q)!(3i -1)!(3i + 1 + 2q + 2p)! dξ =3 i-1 (i -1)! ∞ p=0 ∞ q=0 (λ j -a) p+q (-x) 3i+2+2q+2p (3i + 2q -1)! (2q)!(3i -1)!(3i + 2 + 2q + 2p)! •
Next, we will show that

3 i-1 (i -1)!(3i + 2q -1)! (2q)!(3i -1)!(3i + 2 + 2q + 2p)! ≤ 3 i i! p + q + 1 1 (2p + 2q)!(3i + 2)! ∀ p, q ≥ 0, i ≥ 1. (2.27)
It is easy to see that (2.27) is equivalent to

(3i + 2q -1)! (2q)!(3i -1)! ≤ 3i p + q + 1 (3i + 2 + 2q + 2p)! (2p + 2q)!(3i + 2)! =6i (2p + 2q + 1)(2p + 2q + 3)(2p + 2q + 4) • • • (2p + 2q + 3i + 2) (3i + 2)! • (2.28)
Since the left hand side of (2.28) is independent of p and the right hand side of (2.28) is increasing in p, we only need to prove (2.27) for p = 0, namely, we need to show that

(3i + 2q -1)! (3i -1)! ≤ 3i q + 1 (3i + 2 + 2q)! (3i + 2)! ∀q ≥ 0, ∀i ≥ 1,
this is obvious due to the fact that (3i + 2)!(3i + 2q -1)! 3i(3i -1)!(3i + 2 + 2q)! = (3i + 1)(3i + 2) (3i + 2q)(3i + 2q + 1)(3i + 2q + 2)

≤ 1 3i + 2q ≤ 1 q + 1 • Applying (2.27), we infer that |g i,j (x)| ≤ (-x) 3i+2 3 i i! (3i + 2)! ∞ p=0 ∞ q=0 (λ j -a) p+q x 2p+2q (p + q + 1)(2p + 2q)! = (-x) 3i+2 3 i i! (3i + 2)! ∞ k=0 (λ j -a) k x 2k (2k)! = cosh( λ j -ax) (-x) 3i+2 3 i i! (3i + 2)! ,
where we have used the fact that for any function f :

N → R + , it holds ∞ p=0 ∞ q=0 f (p + q) = ∞ k=0 (k + 1)f (k).
This ends the proof of Proposition 2.2.

Remark 2.3. Compared with the result in [13, Lemma 2.1], it seems that a more natural estimate of g i,j is

|g i,j (x)| ≤ cosh( λ j -a) R i (-x) 3i+2
(3i + 2)! for some constant R > 0. According to the proof of Proposition 2.2, to prove this result, we need to obtain that

(3i + 2q -1)! (3i -1)! ≤ R q + 1 (3i + 2 + 2q)! (3i + 2)! ∀q ≥ 0, ∀i ≥ 1.
This is equivalent to

(q + 1)(3i)(3i + 1)(3i + 2) (3i + 2q)(3i + 2q + 1)(3i + 2q + 2) ≤ R ∀q ≥ 0, ∀i ≥ 1.
However, this is impossible if we pick q = 3i.

Using Proposition 2.2, we can obtain the following corollary which will be used in the proof of the main results.

Corollary 2.1. For any i ≥ 0, j ≥ 1 and x ∈ [-1, 0], we have

|g i,j (x)| ≤ Ce √ λ j 1 (2i)! , (2.29)
where the constant C is independent of i and j.

Proof. By Stirling's formula i! ∼ (i/e) i √ 2πi, and it follows from (2.25) that for i ≥ 1 and j ≥ 1 we have

|g i,j (x)| ≤e √ λ j 3 i i! (3i + 2)! ≤Ce √ λ j 3 i i! (3i + 1)(3i + 2) 3 3i 2 2i √ 6πi √ 2πi √ 4πi (2i)!i! ≤Ce √ λ j 1 (2i)! • 3 Null controllability Proposition 3.1. Let s ∈ [0, 2), 0 < t 1 < t 2 ≤ T and z j ∈ G s ([t 1 , t 2 ]) satisfy |z (i) j (t)| ≤ M j (i!) s R i
, where R is a positive constant and the positive constants M j are such that

∞ j=1 M j e √ λ j < ∞. (3.1)
Then the function u defined by (1.9) solves system (1.4)-(1.8) and u ∈ G

s 2 , s 2 ,s ([-1, 0] × [0, 1] × [t 1 , t 2 ]).
Proof. As the proof is similar to that of [13, Proposition 2.1], it is only sketched.

Let m, p, q ∈ N. By applying Proposition 2.1 and (1.9), we obtain that

|∂ m t ∂ p x ∂ q y u(x, y, t)| ≤C ∂ m t u(•, •, t) H p+q+2 (Ω) ≤CB [ p+q+2 2 ]+1 [ p+q+2 2 ]+1 n=0 P n ∂ m t u(•, •, t) L 2 (Ω) ≤CB [ p+q+2 2 ]+1 [ p+q+2 2 ]+1 n=0 sup (x,y)∈Ω |∂ m t P n u(x, y, t)| ≤CB [ p+q+2 2 ]+1 [ p+q+2 2 ]+1 n=0 sup (x,y)∈Ω ∞ j=1 ∞ i=0 |∂ m t P n (g i,j (x)z (i) j (t)e j (y))|.
By the definitions of g i,j and e j , it is clear that

∂ m t P n (g i,j (x)z (i) j (t)e i (y)) = z (i+m) j (t)(-1) n g i-n,j (x)e j (y), i ≥ n; 0, i < n.
Setting k = i -n and N = n + m, arguing as in [13, Proposition 2.1], we infer from Corollary 2.1 that

∞ j=1 ∞ i=0 |∂ m t P n (g i,j (x)z (i) j (t)e i (y))| = ∞ j=1 ∞ i=n |z (i+m) j (t)g i-n,j (x)e j (y)| ≤C ∞ j=1 ∞ k=0 M j (k + N )! s R k+N e √ λ j 1 (2k)! ≤C (N !) s ( R 2 s ) N ≤C (n!) s (m!) s R n 1 R m 2 ,
where

R 1 = R 2 = R/4 s .
Gathering the above estimates together, we obtain that

|∂ m t ∂ p x ∂ q y u(x, y, t)| ≤CB [ p+q+2 2 ]+1 [ p+q+2 2 ]+1 n=0 sup (x,y)∈Ω ∞ j=1 ∞ i=0 |∂ m t P n (g i,j (x)z (i) j (t)e i (y))| ≤CB [ p+q+2 2 ]+1 [ p+q+2 2 ]+1 n=0 (n!) s (m!) s R n 1 R m 2 ≤C (p!) s 2 (q!) s 2 (m!) s R p 1 R q 2 R m 3
for some positive constants R 1 , R 2 , R 3 . Finally, it is easily seen that u is indeed a solution of the ZK system.

Let u denote the solution of the free evolution for the ZK system:

       u t + au x + u x = 0, u(-1, y, t) = u(0, y, t) = u x (0, y, t) = 0, u(x, 0, t) = u(x, 1, t) = 0, u(x, y, 0) = u 0 (x, y), (x, y) ∈ (-1, 0) × (0, 1), t ∈ (0, T ), y ∈ (0, 1), t ∈ (0, T ), x ∈ (-1, 0), t ∈ (0, T ), x ∈ (-1, 0), y ∈ (0, 1). (3.2)
As for KdV, we have a Kato smoothing effect.

Proposition 3.2. Let u 0 ∈ L 2 (Ω). 1. System (3.2) admits a unique solution u ∈ C([0, T ]; L 2 (Ω)) ∩ L 2 (0, T ; H 1 0 (Ω)) and we have sup t∈[0,T ] u(•, •, t) 2 L 2 (Ω) + T 0 u(•, •, t) 2 H 1 (Ω) dt ≤ C u 0 2 L 2 (Ω) . (3.3) 2. If, in addition, u 0 ∈ D(A) ∩ H 3 (Ω), then u ∈ C([0, T ]; H 3 (Ω)) ∩ L 2 (0, T ; H 4 (Ω)
) and we have

sup t∈[0,T ] u(•, •, t) 2 H 3 (Ω) + T 0 u(•, •, t) 2 H 4 (Ω) dt ≤ C u 0 2 H 3 (Ω) . (3.4)
Proof. (i) comes from [START_REF] Saut | An initial and boundary-value problem for the Zakharov-Kuznestov equation in a bounded domain[END_REF]. Let us proceed with the proof of (ii). For any u 0 ∈ D(A) ∩ H 3 (Ω), we have that u ∈ C([0, T ]; D(A)) by the semigroup theory, and hence u ∈ C([0, T ]; H 2 (Ω) ∩ H 1 0 (Ω)). Let w 0 = Au 0 and w = Au. It is well known that w is the solution of (3.2) with initial value w 0 ∈ L 2 (Ω). According to (i), we have

-u x -au x = Au = w ∈ C([0, T ]; L 2 (Ω)) ∩ L 2 (0, T ; H 1 0 (Ω)).
Therefore u x ∈ C([0, T ]; L 2 (Ω)) ∩ L 2 (0, T ; H 1 (Ω)). Assume finally that u 0 ∈ D(A) ∩ H 3 (Ω), and let us prove that u ∈ C([0, T ], H 3 (Ω)) ∩ L 2 (0, T, H 4 (Ω)). Decompose u as u(x, y, t) = ∞ j=1 ûj (x, t)e j (y). Then for j ≥ 1, ûj solves

dû j dt + û j + (a -λ j )û j = 0, (3.5) ûj (-1, t) = ûj (0, t) = û j (0, t) = 0, (3.6) ûj (., 0) = û0 j , (3.7)
where u 0 (x, y) = ∞ j=1 û0 j (x)e j (y). Multiplying in (3.5) by ûj (resp. by (x + 1)û j ) and integrating over (-1, 0) x × (0, T ) t , we obtain respectively

0 -1 |û j (x, T )| 2 dx + T 0 |û j (-1, t)| 2 dt = 0 -1 |û 0 j (x)| 2 dx, (3.8) 0 -1
(x + 1)|û j (x, T )| 2 dx + 3 T 0 0 -1 |û j | 2 dxdt + (λ j -a) T 0 0 -1 |û j | 2 dxdt = 0 -1 (x + 1)|û 0 j (x)| 2 dx. (3.9) It follows from (3.8) that for any k ∈ N ∞ j=1 λ k j ûj (., t) 2 ≤ ∞ j=1 λ k j û0 j 2 , ∀t ∈ R + (3.10) (that is, ∂ k y u(., ., t) 2 L 2 (Ω) ≤ ∂ k y u 0 2 L 2 (Ω)
for all t ∈ R + ), and from (3.9) that

T 0 ∞ j=1 (λ k j û j (., t) 2 + λ k+1 j ûj (., t) 2 )dt ≤ (1 + aT ) ∞ j=1 λ k j û0 j 2 , ∀T > 0 (3.11) (that is, T 0 ∇∂ k y u(., ., t) 2 L 2 (Ω) dt ≤ (1 + aT ) ∂ k y u 0 2 L 2 (Ω)
for all T > 0). We need the following lemma.

Lemma 3.1. Let a ≥ 0 and λ > 0 be given. Let H k (k ∈ N) denote the Sobolev space H k (-1, 0), and let

H 3 := {u ∈ H 3 (-1, 0); u(-1) = u(0) = u (0) = 0}. Let • denote the norm • L 2 (-1,0) . 1. There exists a constant C > 0 such that 3 k=0 λ k ∂ 3-k x y 2 ≤ C y + (a -λ)y 2 + λ 3 y 2 ∀y ∈ H 3 , ∀λ ≥ λ 0 . (3.12)
2. There exists a constant C > 0 such that

4 k=0 λ k ∂ 4-k x y 2 ≤ C y (4) + (a -λ)y 2 + y + (a -λ)y 2 + λ 4 y 2 ∀y ∈ H 3 ∩ H 4 , ∀λ ≥ λ 0 . (3.13)
Proof of Lemma 3.1: 1. Pick any y ∈ H 3 and any λ ≥ 0. By the Interpolation Theorem and Young inequality, we have that

λ 2 y 2 ≤ Cλ 2 y 4 3 y 2 3 ≤ ε y 2 + C ε λ 3 y 2 , λ y 2 ≤ Cλ y 2 3 y 4 3 ≤ ε y 2 + C ε λ 3 y 2 .
We infer that if λ ≥ λ 0 > 0

y 2 ≤ 2 y + (a -λ)y 2 + 2(a -λ) 2 y 2 ≤ 2 y + (a -λ)y 2 + 2ε y 2 + 2C ε |a -λ| 3 y 2 ≤ 2 y + (a -λ)y 2 + 2ε y 2 + C ε λ 3 y 2
and (3.12) follows by picking ε < 1/4. 2. Pick now any y ∈ H 3 ∩ H 4 and any λ ≥ 0. Then we have

λ 3 y 2 ≤ Cλ 3 y 3 2 y 1 2 H 4 ≤ ε( y (4) 2 + y 2 ) + C ε λ 4 y 2 , λ 2 y 2 ≤ Cλ 2 y y H 4 ≤ ε( y (4) 2 + y 2 ) + C ε λ 4 y 2 , λ y 2 ≤ Cλ y 1 2 y 3 2 H 4 ≤ ε( y (4) 2 + y 2 ) + C ε λ 4 y 2 .
On the other hand, we have that for λ ≥ λ 0 > 0

y (4) 2 ≤ 2 y (4) + (a -λ)y 2 + 2(a -λ) 2 y 2 ≤ 2 y (4) + (a -λ)y 2 + 2ε( y (4) 2 + y 2 ) + C ε λ 4 y 2 ,
and (3.13) follows by picking ε < 1/4 and by using (3.12).

Assuming that u 0 ∈ D(A) ∩ H 3 (Ω) and using (3.10) and (3.12), we obtain that for any t ∈ [0, T ] (with a constant C that may vary from line to line)

u(., ., t) 2 H 3 (Ω) = u(., ., t) 2 H 2 (Ω) + 3 k=0 ∂ k y ∂ 3-k x u(., ., t) 2 L 2 (Ω) ≤ C u 0 2 D(A) + 3 k=0 ∞ j=1 λ k j ∂ 3-k x ûj (., t) 2 ≤ C u 0 2 D(A) + C 3 k=0 ∞ j=1 û j (., t) + (a -λ j )û j (., t) 2 + λ 3 j ûj (., t) 2 ≤ C u 0 2 D(A) + C ∂ 3 y u 0 2 L 2 (Ω) ≤ C u 0 2 H 3 (Ω) .
On the other hand u(., ., t) 2 H 4 (Ω) = u(., ., t) 2

H 3 (Ω) + 4 k=0 ∂ k y ∂ 4-k x u(., ., t) 2 L 2 (Ω) and it is clear that T 0 u(., ., t) 2 H 3 (Ω) dt ≤ C u 0 2 H 3 (Ω)
. Using (3.13), we obtain

T 0 4 k=0 ∂ k y ∂ 4-k x u(., ., t) 2 L 2 (Ω) dt = T 0 4 k=0 ∞ j=1 λ k j ∂ 4-k x ûj (., t) 2 dt ≤ C T 0 ∞ j=1 û(4) j + (a -λ j )û j 2 + û j + (a -λ j )û j 2 + λ 4 j ûj 2 dt ≤ C T 0 Au(., ., t) 2 H 1 (Ω) + ∂ 4 y u(., ., t) 2 L 2 (Ω) dt ≤ C u 0 2 H 3 (Ω)
where we used (3.11) with k = 3. This completes the proof of the proposition.

Interpolating between (3.3) and (3.4), we obtain

sup t∈[0,T ] u(•, •, t) 2 H 1 (Ω) + T 0 u(•, •, t) 2 H 2 (Ω) dt ≤ C u 0 2 H 1 (Ω) , sup t∈[0,T ] u(•, •, t) 2 H 2 (Ω) + T 0 u(•, •, t) 2 H 3 (Ω) dt ≤ C u 0 2 H 2 (Ω) .
This gives

u(•, •, t) H n+1 (Ω) ≤ C √ t u 0 H n (Ω) , for n ∈ {0, 1, 2, 3}.
Proceeding as in [13, Proposition 2.2], we can show that if u 0 ∈ L 2 (Ω), then u(t) ∈ D(A n ) for any t ∈ (0, T ] and n ∈ N, and it holds

A n u(•, •, t) L 2 (Ω) ≤ C n t 3n 2 n 3n 2 u 0 L 2 (Ω) . (3.14)
Without loss of generality, we assume that T = 1. Then for any p, q ∈ N, we infer from Proposition 2.1 that

|∂ p x ∂ q y u(x, y, t)| ≤ u(•, •, t) H p+q+2 (Ω) ≤C 0 B [ p+q+2 2 ]+1 [ p+q+2 2 ]+1 n=0 P n u(•, •, t) L 2 (Ω) ≤C 0 B [ p+q+2 2 ]+1 [ p+q+2 2 ]+1 n=0 C n t 3n 2 n 3n 2 u 0 L 2 (Ω) ≤Ct -3 2 [ p+q 2 ]-3 (p!) 3 4 (q!) 3 4 R p 1 R q 2 for some R 1 , R 2 > 0. This means that u(•, •, t) ∈ G 3 4 , 3 4 ([-1, 0] × [0, 1]) for any t ∈ (0, T ]. Let f j (t) := 1 0 e j (y)∂ 2
x u(0, y, t)dy.

Lemma 3.2. For any j ≥ 1 and n ≥ 0, there exist positive constants R 1 , R 2 and C such that

|f (n) j (t)| ≤ C (jπ) j t -3 2 (n+[ j 2 ]+3) (n!) 3 2 (j!) 3 4 R n 1 R j 2 •
Proof. Without loss of generality, we can assume that T = 1. Since u(•, •, t) ∈ D(A n ) for any t ∈ (0, T ] and n ∈ N, it follows from Lemma 2.2 that

∂ 2 x ∂ 2n y u(x, 0, t) = ∂ 2 x ∂ 2n y u(x, 1, t) = 0, ∀x ∈ [-1, 0], ∀t ∈ (0, T ], ∀n ∈ N.
Then, integrating by parts j-times, we deduce that

f j (t) = √ 2 1 0 sin(jπy)∂ 2 x u(0, y, t)dy = √ 2 jπ 1 0 cos(jπy)∂ 2
x ∂ y u(0, y, t)dy

= √ 2 (jπ) 2 1 0 sin(jπy)∂ 2 x ∂ 2 y u(0, y, t)dy =        √ 2 (jπ) j 1 0 sin(jπy)∂ 2
x ∂ j y u(0, y, t)dy, if j is even;

- √ 2 (jπ) j 1 0 cos(jπy)∂ 2
x ∂ j y u(0, y, t)dy, if j is odd.

(3.15)

To estimate |f

(n) j (t)|(n ∈ N), it remains to estimate |∂ n t ∂ 2 x ∂ j y u(0, y, t)|. Let l = [ j + 4 2 ] + 1.
Taking (2.6) (with u = P i u) and (3.14) into account, we obtain that

|∂ n t ∂ 2 x ∂ j y u(x, y, t)| =|P n ∂ 2 x ∂ j y u(x, y, t)| ≤C P n u(•, •, t) H j+4 (Ω) ≤CB l l k=0 P n+k u(•, •, t) L 2 (Ω) ≤CB l n+l k=0 P k u(•, •, t) L 2 (Ω) ≤CB l n+l k=0 C k k 3 2 k t 3 2 k u 0 L 2 (Ω) ≤CB l C n+l (n + l + 1)(n + l) 3 2 (n+l) t 3 2 (n+l) u 0 L 2 (Ω) ≤Ct -3 2 (n+[ j 2 ]+3) (n!) 3 2 (j!) 3 4 R n 1 R j 2 u 0 L 2 (Ω) (3.16)
for some R 1 , R 2 > 0. Combining (3.15) and (3.16), we obtain

|f (n) j (t)| ≤ C (jπ) j sup y∈[0,1] |∂ n t ∂ 2 x ∂ j y u(0, y, t)| ≤ C (jπ) j t -3 2 (n+[ j 2 ]+3) (n!) 3 2 (j!) 3 4 R n 1 R j 2 •
Now, we can prove the first main result in this paper.

Proof of Theorem 1.1. Pick any τ ∈ (0, T ), s ∈ [3/2, 2) and let

z j (t) = φ s t -τ T -τ f j (t), 0 ≤ t ≤ T,
where

φ s (ρ) =          1 if ρ ≤ 0, 0 if ρ ≥ 1, e - M (1-ρ) σ e -M ρ σ +e - M (1-ρ) σ if ρ ∈ (0, 1)
with M > 0 and σ = (s -1) -1 . As φ s is Gevrey of order s, there exist R φ > 0 such that

|φ (p) s (ρ)| ≤ C (p!) s R p φ ∀ p ∈ N, ρ ∈ R.
Then, applying Lemma 3.2, for any ε ∈ (0, T ) and t ∈ [ε, T ], we have

|z (i) j (t)| ≤ i n=0 i n ∂ i-n t [φ s t -τ T -τ ] |f (n) j (t)| ≤C i n=0 i n (i -n)! s R i-n φ ( 1 T -τ ) i-n 1 (jπ) j t -3 2 (n+[ j 2 ]+3) (n!) 3 2 (j!) 3 4 R n 1 R j 2 ≤C 1 (jπ) j ε -3 2 ([ j 2 ]+3) (j!) 3 4 R j 2 i n=0 i n (i -n)! s R i-n φ ( 1 T -τ ) i-n ε -3 2 n (n!) 3 2 R n 1 ≤C 1 (jπ) j ε -3 2 ([ j 2 ]+3) (j!) 3 4 R j 2 (i!) s min{R φ , R 1 } i i n=0 i n ( 1 T -τ ) i-n ε -3 2 n ≤M j (i!) s R i ,
where M j satisfies (3.1). Let

u(x, y, t) =      u 0 (x, y) if x ∈ [-1, 0], y ∈ [0, 1], t = 0, ∞ j=1 ∞ i=0 g i,j (x)z (i) j (t)e j (y) if x ∈ [-1, 0], y ∈ [0, 1], t ∈ (0, T ].
Then, it is easy to see that u(•,

•, T ) = 0. By Proposition 3.1, u ∈ G s 2 , s 2 ,s ([-1, 0] × [0, 1] × [ε, T ]
) for any ε ∈ (0, T ). Furthermore, we have u t + au x + ∆u x = 0 = u t + au x + ∆u x in Ω × (0, T ), u(0, y, t) = 0 = u(0, y, t), ∀y ∈ [0, 1], ∀t ∈ (0, τ ),

∂ x u(0, y, t) = 0 = ∂ x u(0, y, t), ∀y ∈ [0, 1], ∀t ∈ (0, τ ), ∂ 2
x u(0, y, t) = ∞ j=1 z j (t)e j (y) = ∂ 2 x u(0, y, t), ∀y ∈ [0, 1], ∀t ∈ (0, τ ).

It follows from Holmgren theorem that u(x, y, t) = u(x, y, t) for any (x, y, t) ∈

[-1, 0] × [0, 1] × (0, τ ). In particular, u ∈ C([0, T ]; L 2 (Ω)) and h = 0 for t ∈ [0, τ ), so that h ∈ G s 2 ,s ([0, 1] × [0, T ]). The proof of Theorem 1.1 is complete. 4 Reachable states Proposition 4.1. For any j ≥ 1, assume that z j ∈ G 2 ([0, T ]) is such that |z (i) j (t)| ≤ M j (2i)! R 2i , ∀ i ≥ 0, t ∈ [0, T ],
where R > 1 and M j satisfies (3.1). Then the function u defined by (1.9) solves system (1.4)-

(1.8) and u ∈ G 1,1,2 ([-1, 0] × [0, 1] × [0, T ]).
Proof. According to the proof of Proposition 3.1, for any m, p, q ∈ N, we have

|∂ m t ∂ p x ∂ q y u(x, y, t)| ≤ CB [ p+q+2 2 ]+1 [ p+q+2 2 ]+1 n=0 sup (x,y)∈Ω ∞ j=1 ∞ i=n |z (i+m) j (t)g i-n,j (x)e j (y)|.
Let k = 2i -2n and N = 2n + 2m. We can obtain by the same arguments as in [13, Proposition 3

.1] that ∞ j=1 ∞ i=n |z (i+m) j (t)g i-n,j (x)e j (y)| ≤ ∞ j=1 ∞ i=n M j (2i + 2m)! R 2i+2m Ce √ λ j (2i -2n)! = ∞ j=1 CM j e √ λ j ∞ k=0 (k + N )! R k+N k! ≤C ∞ k=0 (k + N )! R k+N k! =C ∞ k=0 (k + 1) • • • (k + N ) R k+N ≤C( αe R σ ) N N ! √ N ≤C (2n)!(2m)! R n 1 R m 2 ,
where R 1 , R 2 are two positive constants, σ ∈ (0, 1) and

α = sup k≥0 k + 2 (R 1-σ ) k+1 .

It follows from the above estimates that

|∂ m t ∂ p x ∂ q y u(x, y, t)| ≤CB [ p+q+2 2 ]+1 [ p+q+2 2 ]+1 n=0 (2n)!(2m)! R n 1 R m 2 ≤C p!q!(m!) 2 R p 1 R q 2 R m 3
for some positive constants R 1 , R 2 and R 3 . This ends the proof of Proposition 4.1.

As a particular case of [12, Proposition 3.6] (with a 0 = 1, a p = [2p(2p -1)] -1 for p ≥ 1), we have the following result. Proposition 4.2. Let {d q } q≥0 be a sequence of real numbers such that |d q | ≤ CH q (2q)! ∀ q ≥ 0 for some H > 0 and C > 0. Then for all H > e e -1 H, there exists a function f ∈ C ∞ (R) such that

f (q) (0) = d q ∀ q ≥ 0, |f (q) (x)| ≤ C H q (2q)! ∀ q ≥ 0, x ∈ R. Let X := {u ∈ C ∞ ([-1, 0] × [0, 1]); P n u(0, y) = ∂ x P n u(0, y) = P n u(x, 0) = P n u(x, 1) = 0, ∀n ∈ N, ∀x ∈ [-1, 0], ∀y ∈ [0, 1]}.
A result similar to Lemma 2.2 can be derived. Lemma 4.1. For any n ∈ N, we have

∂ 2n y f (x, 0) = ∂ 2n y f (x, 1) = 0, ∀f ∈ X , ∀x ∈ [-1, 0]. (4.1)
Proof. We proceed by induction on n. For n = 0, (4.1) is obvious since f ∈ X . Assume now that (4.1) is true for n -1 ≥ 0. If f ∈ X , then P f ∈ X , so that by the induction hypothesis

∂ 2n-2 y P f (x, 0) = ∂ 2n-2 y P f (x, 1) = 0.
This implies

∂ 3 x ∂ 2n-2 y f (x, 0) + ∂ x ∂ 2n y f (x, 0) + a∂ x ∂ 2n-2 y f (x, 0) = 0, ∂ 3 x ∂ 2n-2 y f (x, 1) + ∂ x ∂ 2n y f (x, 1) + a∂ x ∂ 2n-2 y f (x, 1) = 0.
Since (4.1) is true for n -1, we obtain that

∂ x ∂ 2n y f (x, 0) = ∂ x ∂ 2n y f (x, 1) = 0.
This means that for some constants C 1 and C 2 ,

∂ 2n y f (x, 0) = C 1 , ∂ 2n y f (x, 1) = C 2 ∀x ∈ [-1, 0].
On the other hand, we infer from the assumption f ∈ X that

∂ 2n y f (0, y) = 0 ∀y ∈ [0, 1].
Taking y = 0 and next y = 1, we see that C 1 = C 2 = 0. The proof of Lemma 4.1 is complete. Proof. To prove that (4.3) holds for any l ≥ 1 and any m ≥ 0, it is sufficient to show that for any M ∈ N, (4.3) holds for any l ≥ 1 and any m ≤ 3M + 2. We proceed by induction on M . For M = 0, we can take n = 0 in (4.2) to see that (4.3) holds for any l ≥ 1 and m ≤ 2.

Assume that (4.3) is true for any l ≥ 1 and any m ≤ 3M -1. We claim that (4.3) holds for any l ≥ 1 and m = 3M, 3M + 1, 3M + 2. Indeed, taking n = M in (4. In the last step, we used the fact that 2k + M ≤ 3M -1. Thus, we infer from (4.4) that 1 0 e l (y)∂ 3M x f (0, y)dy = 0, ∀ l ≥ 1.

We can show in the same way that (4.3) is true for m = 3M + 1, 3M + 2 by using the fact that The proof of Lemma 4.2 is complete. Now, we are in a position to prove the second main result in this paper.

Proof of Theorem 1.2. Assume that R := min{R 1 , R 2 } > R 0 = 3 9(a + 2)e (3e) -1 and pick any u 1 ∈ R R 1 ,R 2 . We intend to expand u 1 in the following form:

u 1 (x, y) = ∞ j=1 ∞ i=0
b i,j g i,j (x)e j (y), where b i,j = (-1) i 1 0 e j (y)∂ 2 x P i u 1 (0, y)dy.

Since u 1 ∈ R R 1 ,R 2 ⊂ X , we have that P i u 1 ∈ X for any i ∈ N. By Lemma 4.1, we infer that ∂ 2n y P i u 1 (x, 0) = ∂ 2n y P i u 1 (x, 1) = 0, ∀x ∈ [-1, 0]. Then, by integration by parts, we have

|b i,j | = |
R 2n+i+j+2 ≤ C (jπ) j i n=0 i n
2 n a i-n (3i + j + 2)! (jπ) j R j+2 2 2i (3i)! (jπ) j R j+2 3 2i (6πi)

2 3 R 3i+j+2 = C (jπ) j (3i + j + 2)!
1 3 (4πi) -1 2 (2i)!(2 + a) i R 3i ≤ C2 2 3 (j+2) (j + 2)! 2 3
(jπ) j R j+2 Since the map x → u(x, y, T ) -u 1 (x, y) is in G 1 ([-1, 0]) (i.e. is analytic) for any y ∈ [0, 1], we infer that u(x, y, T ) = u 1 (x, y) ∀(x, y) ∈ [-1, 0] × [0, 1].

The proof of Theorem 1.2 is complete.

Lemma 4 . 2 . 1 0 1 0 1 0 1 0

 421111 If f ∈ X is such that e l (y)P n f (0, y)dy = e l (y)∂ x P n f (0, y)dy = e l (y)∂ 2x P n f (0, y)dy = 0 (4.2)for any l ≥ 1 and any n ≥ 0, then e l (y)∂ m x f (0, y)dy = 0 (4.3)holds for any l ≥ 1 and any m ≥ 0.

1 0 4 ) 1 0 1 0

 1411 2), we have0 = (-1) M 1 0 e l (y)P M f (0, y)dy = e l (y)(∂ 2 x + ∂ 2 y + a) M ∂ M x f (0, y)dy = Since f ∈ X , it follows from Lemma 4.1 that ∂ 2n y f (x, 0) = ∂ 2n y f (x, 1) = 0, ∀x ∈ [-1, 0], ∀n ∈ N.Then, we obtain by integrations by parts that for k ∈ {0, ..., M -1} and i ∈ {0, ..., M -k} e l (y)∂ 2k+Mx ∂ 2i y f (0, y)dy = (-1) i (lπ) 2i e l (y)∂ 2k+Mx f (0, y)dy = 0.

1 0e 1 0

 11 l (y)∂ x P M f (0, y)dy = e l (y)∂ 2x P M f (0, y)dy = 0, ∀ l ≥ 1.

1 0e 2 y 2 3

 122 j (y)∂ 2 x P i u 1 (0, y)dy| ≤ C (jπ) j sup (x,y)∈Ω |∂ 2 x ∂ j y P i u 1 (x, y)|. Next, we estimate |∂ 2 x ∂ j y P i u 1 (x, y)|. |∂ 2 x ∂ j y P i u 1 (x, y)| =|∂ 2 x ) n ∂ n x (a∂ x ) i-n u 1 (x, (2n -2m + j)!

[ 9 ( 2 +

 92 a)] i (2i)! R 3i =M j [9(2 + a)] i (2i)! R 3i ,where M j satisfies (3.1). By Proposition 4.2, for any j ≥ 1, there exists a functionh j ∈ G 2 ([0, T ]) and a number R > 1 such that h (i) j (T ) = b i,j ∀ i ≥ 0,

  |h

  0, t ∈ [0, T ].

(4. 5 )

 5 Pick any τ ∈ (0, T ), s ∈ (1, 2) and letg(t) = 1 -φ s t -τ T -τ for t ∈ [0, T ].

Acknowledgements

Lionel Rosier was partially supported by the ANR project Finite4SoS (ANR-15-CE23-0007). Mo Chen was supported by NSFC Grant (11701078) and China Scholarship Council (No. 201806625055).

This author is supported by NSFC Grant (11701078) and China Scholarship Council. † This author is supported by the ANR project Finite4SoS (ANR-15-CE23-0007)

Setting z j (t) = h j (t)g(t) ∀ t ∈ [0, T ], following the method developed in [START_REF] Martin | On the reachable states for the boundary control of the heat equation[END_REF]Theorem 3.2], and taking into account the fact that s < 2, we see that z j satisfies

where R is the same as in (4.5) and C is a positive constant independent of i and j.

Let u be as in (1.9). According to (4.6), we have u 0 = 0 and

By Proposition 4.1,

Finally, for any l ≥ 1 and n ≥ 0, we have


D Roudil

P Peerani

S Boden

B Russel

M Herranz

M Crozet

L Aldave De

La Heras

M Bazin-Retours

The European R&D project INSIDER Acting on the upstream stages

des établissements d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

The radiological characterization of a facility prior to performing any D&D and remediation operations is a crucial step in the definition of a viable decommissioning scenario. This scenario must be robust and optimized not only with regards to technical issues, produced waste amount and workers dose, but also with regards to costs, deadlines and safety. INSIDER contributes to building confidence that the nuclear community is fit for taking responsibility for retiring a range of facilities and sites without imposing undue economic, social or radiological burdens. Today, the reconstitution of a 3D vision of a facility or components is partly based on off-site laboratory analyses, leading to technical and economic difficulties in the realization of representative samples. INSIDER aims at developing and qualify integrated approaches to enhance and support characterization methodologies.

The heart of the technical innovative part of the project is to improve the sampling strategy taking into account different types of measurement data, with different but known levels of accuracy. In-situ analysis techniques are of paramount importance for initial cartography. In constrained environment they must be complemented by more detailed off-site measurements. Realistic knowledge of their performance is essential, without systematic optimization.

The INSIDER project addresses 3 use cases: a fuel cycle facility, a nuclear power plant and a post accidental remediation. Each one is representative for different kind of constraints and analytical challenges. The project-integrated approach will be tested and validated by applying the methodologies on three full-scale D&D sites offered by JRC ISPRA, SCK-CEN and CEA, representative of the use cases.

A benchmarking supplemented by interlaboratory comparisons (ILC) on synthetic reference samples produced within the project will structure the validation and refinement of the INSIDER methodology.

The practical project implementation includes documenting, testing, verifying and assessing characterisation methodologies for the selected test cases. Thereby, a key contributor is effective knowledge sharing within the scientific community.

Final assessment of the outcome will strengthen the recommendations and guidance, and promote and share European expertise through guides and pre-normative texts.

The European R&D project INSIDER: Acting on the upstream stages

  . Roudil 1* , P. Peerani 2 , S. Boden 3 , B. Russell 4 , M. Herranz 5 , M. Crozet 1 , L. Aldave de la Heras 2 , M. Main Author, danielle.roudil@cea.frThe main objective of the European project INSIDER (Improved Nuclear Site Characterization for Waste minimization in Decommissioning under Constrained Environment) is to improve the management of waste coming from nuclear sites under D&D operations, in particular to direct D&D operations, so that the production of waste is optimized on the basis of criteria such as minimization of generated volumes, limitation of over-categorized waste, generated waste and effluents as engaged doses on sites during the D&D operations. INSIDER is a 4 years project launched on June 2017. It is coordinated by CETAMA (CEA Nuclear Energy Division) and includes 18 European partners from 10 member states.

	Bazin-Retours 6
	1 CEA Nuclear Energy division, Research Department on Mining and fuel Recycling Processes, BP 17171 F-
	30207 Bagnols sur cèze, France
	2 European Commission Joint Research Centre, Rue de la Loi 200, Brussels 1049, Belgium….
	3 SCK-CEN, Belgian Nuclear Research Centre, Boeretang 200, B-2400 Mol, Belgium
	4 NPL National Physical Laboratory, Hampton Road,Teddington, Middlesex, TW11 0LW
	5 UPV/EHU, Univ. of the Basque Country, EI Bilbao. Pza. Ing. Torres Quevedo 1, 48013 Bilbao, Spain Nuclear
	6 LGI-Consulting, 6 Cité de l'ameublement-75011 Paris, France

D*

Acknowledgment

The project INSIDER has received funding from the Euratom research and training program 2014-2018, under grant agreement No 755554.