Subrata Majumdar
email: subratamajumdar634@gmail.com.

Hugo Parada
email: hugo.parada@math.univ-toulouse.fr.

STABILIZATION OF KAWAHARA EQUATION WITH INPUT SATURATION AND SATURATED BOUNDARY FEEDBACK

Keywords: Mathematics Subject Classification. 93D15, 93B07, 35B40 Kawahara equation, input saturation, stabilization, unique continuation principle, Compactness uniqueness argument

In this paper, we study the stabilizability of the Kawahara equation in a bounded interval (0, L) by a saturated internal control and boundary feedback. We prove the exponential stabilization of the corresponding closed-loop system. We first show the well-posedness of the nonlinear system using linear, nonlinear semigroup theory and fixed-point arguments. Next, we prove two stabilization results by means of interior-saturating feedback and boundary-saturated feedback as well, using observability inequalities and certain unique continuation principle.

Introduction and main results

1.1. System under study. Let us consider the following nonlinear Kawahara equation

         ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u + u p ∂ x u + f 1 (t, x) = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, t ∈ (0, ∞), ∂ 2
x u(t, L) = f 2 (t), t ∈ (0, ∞), u(0, x) = u 0 (x),

x ∈ (0, L),

with p ∈ [1, 2), and the functions f 1 and f 2 are the controls acting internally and through the boundary conditions, respectively. The objective of this work is to show the exponential stabilization of (1.1) by acting with either internal or boundary controls, in the case where these controls are subject to saturation. The Kawahara equation

∂ t u + ∂ x u + ∂ 3 x u -∂ 5
x u + u p ∂ x u = 0 is fifth-order, nonlinear one dimensional equation and is typically used to model the propagation of small-amplitude long waves in fluid dynamics, water waves, plasma dynamics among other dispersive phenomena (see [START_REF] Cui | Global existence of solutions for the Cauchy problem of the Kawahara equation with L 2 initial data[END_REF][START_REF] Iguchi | A long wave approximation for capillary-gravity waves and the Kawahara equation[END_REF][START_REF] Kawahara | Oscillatory solitary waves in dispersive media[END_REF] and the references therein). This equation can be also considered as an extension of the Korteweg-de Vries equation (KdV), including high order dispersion effects. It has been studied as a fifth degree KdV equation and singularly perturbed KdV equation in [START_REF] Boyd | Weakly non-local solitons for capillary-gravity waves: fifth-degree korteweg-de vries equation[END_REF][START_REF] Pomeau | Structural stability of the korteweg-de vries solitons under a singular perturbation[END_REF] respectively. Discovered in 1972 by Kawahara [START_REF] Kawahara | Oscillatory solitary waves in dispersive media[END_REF], it was also obtained by Hasimoto for shallow water system in 1970 [START_REF] Hasimoto | Water waves[END_REF]. Since then, the Kawahara equation has been subject of study for different points of views. Recently, the control and stabilization properties have attracted the attention of the control community. Let us provide a brief overview about this subject.

As far as we know the first works dealing with this kind of control problems for Kawahara equation were [START_REF] Vasconcellos | Stabilization of the linear Kawahara equation with localized damping[END_REF][START_REF] Vasconcellos | Stabilization of the linear Kawahara equation with localized damping[END_REF]. In these works, the stabilization of global solutions of the linear Kawahara equation using localized damping was addressed following the ideas presented in the works [START_REF] Perla Menzala | Stabilization of the Korteweg-de Vries equation with localized damping[END_REF][START_REF] Rosier | Exact boundary controllability for the Korteweg-de Vries equation on a bounded domain[END_REF], where similar problems was analyzed for the KdV equation. These results were then extended to the nonlinear Kawahara equation, with nonlinearity u∂ x u. In [START_REF] Araruna | Energy decay for the modified Kawahara equation posed in a bounded domain[END_REF], a similar problem was studied for the generalized Kawahara equation with nonlinearity u p ∂ x u for p ∈ [START_REF] Araruna | Energy decay for the modified Kawahara equation posed in a bounded domain[END_REF][START_REF] Capistrano-Filho | Massera's theorems for a higher order dispersive system[END_REF]. It is important to remark that in [START_REF] Araruna | Energy decay for the modified Kawahara equation posed in a bounded domain[END_REF], the energy of solutions decay exponentially for arbitrary initial data in L 2 (0, L) for p ∈ [START_REF] Araruna | Energy decay for the modified Kawahara equation posed in a bounded domain[END_REF][START_REF] Boyd | Weakly non-local solitons for capillary-gravity waves: fifth-degree korteweg-de vries equation[END_REF]. However, this decay was only proved for small initial data if p ∈ [START_REF] Boyd | Weakly non-local solitons for capillary-gravity waves: fifth-degree korteweg-de vries equation[END_REF][START_REF] Capistrano-Filho | Massera's theorems for a higher order dispersive system[END_REF]. Internal and boundary delayed stabilization problems for Kawahara equation were considered in [START_REF] Capistrano-Filho | Two stability results for the Kawahara equation with a time-delayed boundary control[END_REF][START_REF] Capistrano-Filho | Stabilization results for delayed fifth-order KdV-type equation in a bounded domain[END_REF][START_REF] Chentouf | Well-posedness and exponential stability of the Kawahara equation with a time-delayed localized damping[END_REF]. The exponential stability was shown using multipliers, Lyapunov functionals and compactness ideas. We also mention [START_REF] Capistrano-Filho | Massera's theorems for a higher order dispersive system[END_REF], where Massera's theorems were obtained, [START_REF] Chen | Internal controllability of the Kawahara equation on a bounded domain[END_REF], where the internal null controllability was shown via Carleman estimates, [START_REF] Capistrano-Filho | Well-posedness and controllability of Kawahara equation in weighted Sobolev spaces[END_REF], where controllability in weighted Sobolev spaces was proved, and [START_REF] Capistrano-Filho | Control results with overdetermination condition for higher order dispersive system[END_REF][START_REF] Capistrano Filho | Control of Kawahara equation with overdetermination condition: The unbounded cases[END_REF], where the control with overdetermination condition was considered.

As we already mentioned, the focus of this paper is to analyze the case of saturated controllers. Saturation in control systems is related to the fact that in most real-life applications, the actuators are subject to constraints, limitations due to physical, economical, operational reasons, etc. It is wellknown that even in the finite-dimensional case the introduction of saturation in the controller could destabilize the system [START_REF] Tarbouriech | Stability and stabilization of linear systems with saturating actuators[END_REF]Example 1.1], making the saturated control problem a challenging task. In the finite-dimensional framework, the stability analysis could be done by studying an appropriate Lyapunov function and a sector condition related with the saturation map [START_REF] Sussmann | On the stabilizability of multiple integrators by means of bounded feedback controls[END_REF][START_REF] Tarbouriech | Stability and stabilization of linear systems with saturating actuators[END_REF][START_REF] Teel | Global stabilization and restricted tracking for multiple integrators with bounded controls[END_REF].

With respect to saturation in infinite-dimensional control systems there is not a general method to tackle these problems. Noteworthy works include [START_REF] Prieur | Wave equation with cone-bounded control laws[END_REF] where the stabilization of a one dimensional wave equation was considered with saturated internal and boundary controls. In this work, stability results were proved by using Lyapunov functions and LaSalle's invariance principle. In the case of parabolic equations in [START_REF] Lhachemi | Saturated boundary stabilization of partial differential equations using control-Lyapunov functions[END_REF], several results are collected in the one-dimensional case, the principal idea is to decompose the system into its stable and stable parts via a spectral approach. The unstable finite dimensional part is tackled as in [START_REF] Tarbouriech | Stability and stabilization of linear systems with saturating actuators[END_REF], and for the remaining part, an appropriated Lyapunov function is considered. In the case of KdV-type equations, the approach is to follow the ideas of the works [START_REF] Perla Menzala | Stabilization of the Korteweg-de Vries equation with localized damping[END_REF][START_REF] Rosier | Exact boundary controllability for the Korteweg-de Vries equation on a bounded domain[END_REF] and to prove the exponential stability via compactness arguments. This requires, in particular, the use of unique continuation results for nonlinear PDEs. In this direction we mention the works [START_REF] Marx | Stabilization of a linear Korteweg-de Vries equation with a saturated internal control[END_REF][START_REF]Global stabilization of a Korteweg-de Vries equation with saturating distributed control[END_REF][START_REF] Parada | Stability analysis of a Korteweg-de Vries equation with saturated boundary feedback[END_REF][START_REF]Global well-posedness of the KdV equation on a star-shaped network and stabilization by saturated controllers[END_REF].

1.2. Setting of the problem. This paper deals with the feedback stabilization of the system (1.1) by means of saturated control acting in-domain or through a boundary action. In this work, we consider two types of saturation function sat which can be given by the following. Let M > 0 be the saturation level. We define

sat loc (s) = s, if |s| ≤ M, M sgn(s), if |s| ≥ M, (1.2)
and the second one is

sat 2 (s) = s, if ∥s∥ L 2 (0,L) ≤ M, sM ∥s∥ L 2 (0,L) , if ∥s∥ L 2 (0,L) ≥ M.
(1.3) 1.2.1. Internal control. We start by describing the case where the control action is in-domain.

         ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u + u p ∂ x u + sat(a(x)u(t, x)) = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, t ∈ (0, ∞), ∂ 2 x u(t, L) = 0, t ∈ (0, ∞), u(0, x) = u 0 (x), x ∈ (0, L), (1.4)
where a ∈ L ∞ (0, L) is a localized function given by

a 1 ≥ a(x) ≥ a 0 > 0 on ω ⊂ [0, L], ω is a nonempty open subset of [0, L]. (1.5)
1.2.2. Boundary control. In this work, we also explore the case where the saturated actuator is at the boundary, consider the following equation:

         ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u + u p ∂ x u = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, t ∈ (0, ∞), ∂ 2 x u(t, L) = sat loc (α∂ 2 x u(t, 0)), t ∈ (0, ∞), u(0, x) = u 0 (x), x ∈ (0, L), (1.6)
where |α| < 1.

The main objective of this paper is to show the exponential stability of (1.4) and (1.6) in the presence of saturated control. More precisely, we investigate semiglobal exponential stabilization result (see Theorem 3.4) for the Kawahara equation (1.4) with internal saturation. That means, for any r > 0, we can find a positive number µ(r) > 0 such that the energy

E(t) = 1 2
local exponential stabilization result (see Theorem 3.3) for the system (1.6) with |α| < 1. That is, there exist positive constants r, µ, C > 0 such that whenever ∥u 0 ∥ L 2 (0,L) ≤ r, the solution of (1.6) satisfies

∥u(t, •)∥ L 2 (0,L) ≤ Ce -µt ∥u 0 ∥ L 2 (0,L) , ∀t ≥ 0. (1.7)
Our study is based on deriving appropriate observability inequalities for the saturated systems. These inequalities are obtained using compactness ideas and Carleman estimates. We also put special emphasis on the global well-posedness of the associated systems. Indeed, as we can observe from the definition of the saturation maps (1.2) and (1.3) if the argument of the map, it is small enough (in the appropriated norms) then the saturation is not really active. This was one of the main points for instance in [START_REF]Global well-posedness of the KdV equation on a star-shaped network and stabilization by saturated controllers[END_REF].

The structure of this paper is the following. In Section 2, we study the well-posedness results associated to (1.4) and (1.6). Then, in Section 3, we study the two exponential stabilization results for the case of boundary saturation and saturated interior feedback, respectively. At the end, we conclude the paper with some additional remarks and open questions.

well-posedness

This section is devoted to the existence and uniqueness of the system (1.4) and (1.6) with the saturated feedback. We will follow the works [START_REF] Marx | Stabilization of a linear Korteweg-de Vries equation with a saturated internal control[END_REF][START_REF]Global stabilization of a Korteweg-de Vries equation with saturating distributed control[END_REF][START_REF] Parada | Stability analysis of a Korteweg-de Vries equation with saturated boundary feedback[END_REF][START_REF]Global well-posedness of the KdV equation on a star-shaped network and stabilization by saturated controllers[END_REF] see also [START_REF] Parada | Control and stabilization of PDE on networks[END_REF].

2.1. Well-posedness with boundary feedback. We start by studying the well-posedness results for (1.6). First, we consider the system (1.6) without the internal nonlinearity, and we prove the wellposedness based on nonlinear semigroup theory. Then, we will add an internal source term in order to take in account the nonlinear term u p ∂ x u. Consider the system

         ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, t ∈ (0, ∞), ∂ 2 x u(t, L) = sat loc (α∂ 2 x u(t, 0)), t ∈ (0, ∞), u(0, x) = u 0 (x),
x ∈ (0, L).

(

Let A b : D(A b ) → L 2 (0, L) the nonlinear operator defined by

A b w := -w ′ -w ′′′ + w ′′′′′ (2.2) D(A b ) = {w ∈ H 5 (0, L) ∩ H 2 0 (0, L) : w ′′ (L) = sat loc (αw ′′ (0))}.
Note that the operator A b is nonlinear by the action of the boundary saturation. Following [START_REF] Marx | Stabilization of a linear Korteweg-de Vries equation with a saturated internal control[END_REF][START_REF] Parada | Stability analysis of a Korteweg-de Vries equation with saturated boundary feedback[END_REF], we show that A b is a closed, dissipative operator and that for some λ > 0, D(A b ) ⊂ R(I -λA b ), where I : L 2 (0, L) → L 2 (0, L), is the identity operator and R denotes the range of a operator. The following result, ensure the existence of mild and strong solutions for (2.1). Proposition 2.1. Let u 0 ∈ L 2 (0, L) and |α| < 1, then there exists a unique mild solution

u ∈ C([0, ∞), L 2 (0, L)) of (2.1). Moreover, if u 0 ∈ D(A b ), this solution is strong and u ∈ C([0, ∞), D(A b ))∩C 1 ((0, ∞), L 2 (0, L)). Proof. A b is a closed operator. Consider the operator A 0 b : D(A 0 b ) → L 2 (0, L) defined by A 0 b w := A b w, where D(A 0 b ) = {w ∈ H 5 (0, L) ∩ H 2 0 (0, L) : w ′′ (L) = 0}. Then, clearly A 0 b is a closed operator.
Similarly, by [17, Page 91] the operator sat loc satisfies

∀(s, s) ∈ R 2 , |sat loc (s) -sat loc (s)| ≤ |s -s| .
(2.3) Thus, sat loc is a Lipschitz continuous function and hence the operator

A b is closed. A b is dissipative. Let w, v ∈ D(A b )
, then after some integration by parts and (2.3) we have

⟨A b w -A b v, w -v⟩ L 2 (0,L) = L 0 (-w ′ -u ′′′ + w ′′′′′ + v ′ + v ′′′ -v ′′′′′ )(w -v)dx = 1 2 (w ′′ (L) -v ′′ (L)) 2 -(w ′′ (0) -v ′′ (0)) 2 = 1 2 (sat loc (αw ′′ (0)) -sat loc (αv ′′ (0))) 2 -(w ′′ (0) -v ′′ (0)) 2 ≤ α 2 -1 2 (w ′′ (0) -v ′′ (0)) 2 ≤ 0,
from where we obtain that A b is dissipative.

For λ > 0 small enough, D(A b ) ⊂ R(I -λA b ). Take λ > 0, then D(A b ) ⊂ R(I -λA b ), is equivalent to given w ∈ D(A b ) find v ∈ D(A b ) such that (I -λA b )v = w. Define λ = 1 λ , then we have to find solutions v ∈ D(A b ) of λv + v ′ + v ′′′ -v ′′′′′ = λw, v(0) = v(L) = v ′ (0) = v ′ (L) = 0, v ′′ (L) = sat loc (αv ′′ (0)). (2.4)
Consider the operator T : H 3 (0, L) → L 2 (0, L) defined by T (y) = z, where z is the solution of

λz + z ′ + z ′′′ -z ′′′′′ = λw, z(0) = z(L) = z ′ (0) = z ′ (L) = 0, z ′′ (L) = sat loc (αy ′′ (0)). (2.5)
Lemma 2.2. For all w ∈ D(A b ), the operator T is well-defined, i.e. for all y ∈ H 3 (0, L) there exists a unique solution to (2.5).

The proof of Lemma 2.2 is given in the Appendix A. Let us show now that T has a fixed point. Take C > 0 to be chosen later and define

K C = {z ∈ H 2 0 (0, L) : ∥z∥ H 2 0 (0,L) ≤ C}. (2.6)
As H 2 0 (0, L) is compactly embedded in L 2 (0, L), the set K C is a bounded set in H 2 0 (0, L) and relatively compact in L 2 (0, L). Furthermore, as K C is closed in L 2 (0, L) we can deduce that K C is compact in L 2 (0, L). Our idea now is to apply the Schauder fixed point theorem [START_REF] Coron | Control and nonlinearity[END_REF]Theorem B.19]. Thus, we have to prove the following lemma Lemma 2.3. There exists C > 0 such that T (H 3 (0, L)) ⊂ K C .

Proof. Multiplying (2.5) by z and integrating on (0, L) we get after some integration by parts

λ L 0 z 2 dx + 1 2 z ′′ (0) 2 = 1 2 sat loc (αy ′′ (0)) 2 + λ L 0 wzdx.
Then, using the definition of sat loc given in (1.2) and Cauchy-Schwartz, we obtain λ 2

∥z∥ 2 L 2 (0,L) ≤ M 2 2 + λ 2 ∥w∥ 2 L 2 (0,L) . (2.7)
Now, multiplying (2.5) by xz and integrating on (0, L) we get after some integration by parts

λ L 0 xz 2 dx + 3 2 L 0 (z ′ ) 2 dx + 5 2 L 0 (z ′′ ) 2 dx = 1 2 L 0 z 2 dx + L 2 sat(αy ′′ (0)) 2 + λ L 0 xwzdx, 3 2 ∥z ′ ∥ 2 L 2 (0,L) + 5 2 ∥z ′′ ∥ 2 L 2 (0,L) ≤ L λ 2 + 1 2 ∥z∥ 2 L 2 (0,L) + LM 2 2 + L λ 2 ∥w∥ 2 L 2 (0,L) , (2.8)
and hence using (2.7) and (2.8) we conclude, ∥z ′′ ∥ 2 L 2 (0,L) ≤ C, where C > 0 only depends on the level of saturation M , L, λ and w which ends the proof of Lemma 2.3. □

Using the Schauder fixed point theorem, we manage to find v ∈ H 3 (0, L) solution of (2.4), we get v ∈ D(A b ) using that the boundary conditions and w ∈ D(A b ). We conclude that for all λ > 0, D(A b ) ⊂ R(I -λA b ). Using [21, Theorem 4.2] A b is the generator of a semigroup of contractions S(t), t > 0. By [START_REF] Miyadera | Nonlinear semigroups[END_REF]Theorem 4.5] for all T > 0, (2.1). Moreover, by [START_REF] Miyadera | Nonlinear semigroups[END_REF]Theorem 4.10] this solution is unique. Also, we can deduce that for all T > 0, for all u 0 ∈ L 2 (0, L), there exists a unique mild solution of (2.1) u ∈ C([0, T ]; L 2 (0, L)) and then, the proof of Proposition 2.1 is finished. □

u 0 ∈ D(A b ), u = S(t)u 0 ∈ C([0, T ]; D(A b )) ∩ C 1 ((0, T ); L 2 (0, L)) is a strong solution to
Now we study the well-posedness of the system (2.1) with an extra source term f .

         ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u = f, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, t ∈ (0, ∞), ∂ 2 x u(t, L) = sat loc (α∂ 2 x u(t, 0)), t ∈ (0, ∞), u(0, x) = u 0 (x),
x ∈ (0, L).

(2.9) Formally, this system can be seen as ∂ t u = A b u + f . The following result can be obtained using [

∈ L 2 (0, L), |α| < 1, f ∈ L 1 ((0, ∞), L 2 (0, L)), then there exists a unique mild solution u ∈ C([0, ∞), L 2 (0, L)) of (2.9). Moreover, if u 0 ∈ D(A b ) and f ∈ W 1,1 ((0, ∞), L 2 (0, L)) this solution is strong and u ∈ C([0, ∞), L 2 (0, L)), u(t, •) ∈ D(A b ), for all t ≥ 0 and u ∈ C 1 ((0, ∞), L 2 (0, L)).
Now, in order to deal with the nonlinear term u p ∂ x u, we show that mild solutions of (2.9) have an extra regularity. Consider the space B T = C([0, T ], L 2 (0, L)) ∩ L 2 ((0, T ), H 2 (0, L)), endowed with the norm

∥u∥ 2 B T = ∥u∥ 2 C([0,T ],L 2 (0,L)) + ∥u∥ 2 L 2 ((0,T ),H 2 (0,L)) . Proposition 2.5. Let u 0 ∈ L 2 (0, L), |α| < 1, f ∈ L 1 ((0, T ), L 2 (0, L)) and T > 0, consider u ∈ C([0, T ], L 2 (0, L))
, the restriction of the solution of (2.9) on [0, T ], then u ∈ B T . Moreover, the following estimate holds,

∥u∥ B T ≤ C(1 + √ T ) ∥u 0 ∥ L 2 (0,L) + ∥f ∥ L 1 ((0,T ),L 2 (0,L)) .
(2.10)

For f = 0, the following estimate follows

∥u 0 ∥ 2 L 2 (0,L) ≤ 1 T ∥u∥ 2 L 2 ((0,T ),L 2 (0,L)) + ∥∂ 2 x u(•, 0)∥ 2 L 2 (0,T ) . (2.11)
Proof. Suppose u is a strong solution of (2.9), let T > 0 and 0 ≤ s ≤ T . Multiplying the first line of (2.9) by u and integrating on (0, L) × (0, s), we get

s 0 L 0 u∂ t udxdt + s 0 L 0 u∂ x udxdt + s 0 L 0 u∂ 3 x udxdt - s 0 L 0 u∂ 5 x udxdt = s 0 L 0 uf dxdt.
After some integration by parts, we observe

1 2 L 0 u(s, x) 2 dx + 1 2 s 0 ∂ 2 x u(t, 0) 2 -sat(α∂ 2 x u(t, 0)) 2 dt = 1 2 L 0 u 2 0 dx + s 0 L 0 f udxdt. (2.12)
Note that for all s ∈ [0, T ]

s 0 L 0 f udxdt ≤ T 0 ∥u(t, •)∥ L 2 (0,L) ∥f (t, •)∥ L 2 (0,L) dt ≤ ∥u∥ C([0,T ],L 2 (0,L)) ∥f ∥ L 1 ((0,T ),L 2 (0,L))
Now, using the above equation, (2.3), (2.12), we get for all s ∈ [0, T ],

1 2 L 0 u(s, x) 2 dx + C α s 0 ∂ 2 x u(t, 0) 2 dt ≤ 1 2 L 0 u 2 0 dx + ∥u∥ C([0,T ],L 2 (0,L)) ∥f ∥ L 1 ((0,T ),L 2 (0,L)) , (2.13)
where C α = 1 -|α| 2 2 > 0. Now, taking the supremum for s ∈ [0, T ], and with Young's inequality, we get

∥u∥ 2 C([0,T ],L 2 (0,L)) + T 0 ∂ 2 x u(t, 0) 2 dt ≤ C ∥u 0 ∥ 2 L 2 (0,L) + ∥f ∥ 2 L 1 ((0,T ),L 2 (0,L)) . (2.14)
Now, we multiply the first equation of (2.9) by xu and integrate on (0, L) × (0, T ) and we get

T 0 L 0 xu∂ t udxdt + T 0 L 0 xu∂ x udxdt + T 0 L 0 xu∂ 3 x udxdt - T 0 L 0 xu∂ 5 x udxdt = T 0 L 0 xuf dxdt.
(2.15) Thus, performing integration by parts

1 2 L 0 xu(T, •) 2 dx + 3 2 T 0 L 0 ∂ x u 2 dxdt + 5 2 T 0 L 0 ∂ 2 x u 2 dxdt = 1 2 L 0 xu 2 0 dx + 1 2 T 0 L 0 u 2 dxdt + L 2 T 0 sat(α∂ 2 x u(t, 0)) 2 dt + T 0 L 0 xuf dxdt. (2.16) Note that T 0 L 0 u 2 dxdt ≤ T ∥u∥ 2 C([0,T ],L 2 (0,L)) and T 0 L 0 xuf dxdt ≤ L∥u∥ C([0,T ],L 2 (0,L)) ∥f ∥ L 1 ((0,T ),L 2 (0,L)) .
Using these two inequalities, (2.14) we derive

∥∂ 2 x u∥ 2 L 2 ((0,T ),L 2 (0,L)) ≤ C(1 + T ) ∥u 0 ∥ 2 L 2 (0,L) + ∥f ∥ 2 L 1 ((0,T ),L 2 (0,L)) , (2.17)
where C > 0 does not depend on T . Thus u ∈ L 2 ((0, T ), H 2 (0, L)). Joining (2.14) and (2.17) we deduce (2.10). To deduce (2.11), consider f = 0 and multiply (2.1) by (T -t)u and integrate (0, T ) × (0, L), after some integration by parts we get

T L 0 u 2 0 dx + T 0 (T -t) sat loc (∂ 2 x u(t, 0)) 2 -∂ 2 x u(t, 0) 2 dt = T 0 L 0 u 2 dxdt, using (2.
3) we obtain (2.11). The result can be extended by density to mild solutions. □

Before going to prove the global well-posedness result for the nonlinear system (1.6), we first show some important estimates related to the nonlinear term:

Proposition 2.6. Let u ∈ B T . Then for 1 ≤ p ≤ 2, u p ∂ x u ∈ L 1 ((0, T ), L 2 (0, L)) and the map u ∈ B T → u p ∂ x u ∈ L 1 ((0, T ), L 2 (0, L)) is continuous. Moreover, there exists C > 0 such that, for any u, v ∈ B T ∥u p ∂ x v∥ L 1 ((0,T ),L 2 (0,L)) ≤ C(T (2-p)/4 + √ T )∥u∥ p B T ∥v∥ B T .
(2.18)

∥u p ∂ x u -v p ∂ x v∥ L 1 ((0,T ),L 2 (0,L)) ≤ C(T (2-p)/4 + √ T ) ∥u∥ B T ∥v∥ p-1 B T + ∥v∥ p B T + ∥u∥ p B T ∥u -v∥ B T . ( 2
∥u p ∂xu -v p ∂xv∥ L 2 (0,L) = T 0 ∥u p ∂x(u -v) + (u p -v p )∂xv∥ L 2 (0,L) ≤ C(T (2-p)/4 + √ T )∥u∥ p B T ∥u -v∥B T + T 0 ∥ |u| p-1 + |v| p-1 (u -v)∂xv∥ L 2 (0,L)
Using [31, Lemma 2.4], we estimate the last term of the above inequality as:

T 0 ∥ |u| p-1 + |v| p-1 (u -v)∂ x v∥ L 2 (0,L) ≤ CT (2-p)/4 ∥u -v∥ B T ∥u∥ B T ∥v∥ p-1 B T + ∥v∥ p B T .

□

We conclude this subsection by proving the global well-posedness result for nonlinear equation (1.6).

Theorem 2.7. Let u 0 ∈ L 2 (0, L) and |α| < 1, then there exists a unique mild solutions

u ∈ C([0, ∞), L 2 (0, L))∩ L 2 (0, ∞, H 2 (0, L)) of (1.6).
Proof. Let u 0 ∈ D(A b ) and R, θ > 0 to be chosen later. Consider the ball

B B θ (0, R) := {v ∈ B θ , ∥v∥ B θ ≤ R}. Then B B θ (0, R) it is a complete metric space. Let us introduce the map Φ : B θ → B θ , defined by Φ(v) = u, where u solves          ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u = -v p ∂ x v, (t, x) ∈ (0, θ) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, t ∈ (0, θ), ∂ 2 x u(t, L) = sat loc (α∂ 2 x u(t, 0)), t ∈ (0, θ), u(0, x) = u 0 (x),
x ∈ (0, L).

We observe first that Φ is well-defined, since by Proposition 2.6

v p ∂ x v ∈ L 1 ((0, θ), L 2 (0, L))
. By Proposition 2.5 and Proposition 2.6 we get

∥Φ(v)∥ B θ ≤C(1 + √ θ) ∥u 0 ∥ L 2 (0,L) + ∥v p ∂ x v∥ L 1 ((0,θ),L 2 (0,L)) ≤C 1 (1 + √ θ)∥u 0 ∥ L 2 (0,L) + C 2 (θ (2-p)/4 + √ θ)∥v∥ p+1 B T , for some C 1 , C 2 > 0. Then taking R = 3C 1 ∥u 0 ∥ L 2 (0,L) and θ > 0 small enough such that    C 2 (θ 2-p 4 + √ θ)R p < 1 3 , (1 + √ θ) < 2, we have that ∥Φ(v)∥ B θ ≤ R and thus Φ maps B B θ (0, R) into itself. Now take v 1 , v 2 ∈ B B θ (0, R), and consider w = u 1 -u 2 = Φ(v 1 ) -Φ(v 2 ). We observe that w solves          ∂ t w + ∂ x w + ∂ 3 x w -∂ 5 x w = f, (t, x) ∈ (0, θ) × (0, L), w(t, 0) = w(t, L) = ∂ x w(t, 0) = ∂ x w(t, L) = 0, t ∈ (0, θ), ∂ 2 x w(t, L) = h, t ∈ (0, θ), w(0, x) = 0, x ∈ (0, L), where f = -v p 1 ∂ x v 1 + v p 2 ∂ x v 2 and h = sat loc (α∂ 2 x u 1 (t, 0)) -sat loc (α∂ 2
x u 2 (t, 0)). Similar integration by parts as in Proposition 2.5 allows to write

∥w∥ B θ ≤ C ∥h∥ L 2 (0,θ) + ∥f ∥ L 1 ((0,θ),L 2 (0,L) .
(2.20)

Our idea now is to obtain an estimate of ∥h∥ L 2 (0,θ) . Proceeding as (2.12) we obtain

1 2 L 0 w(θ, •) 2 dx + 1 2 θ 0 (∂ 2 x w(t, 0) 2 -h(t) 2 )dt = θ 0 L 0 f wdxdt. Using (2.3) we get ∥h∥ L 2 (0,θ) = sat loc (α∂ 2 x u 1 (•, 0)) -sat loc (α∂ 2 x u 2 (•, 0)) L 2 (0,θ) ≤ ∥α∂ 2 x u 1 (•, 0) -α∂ 2 x u 2 (•, 0)∥ L 2 (0,θ) ≤ |α|∥∂ 2 x w(•, 0)∥ L 2 (0,θ) , which implies 1 2 L 0 w(θ, •) 2 dx + C α θ 0 ∂ 2 x w(t, 0) 2 dt ≤ θ 0 L 0 f wdxdt.
Thus by Young's inequality ∥w∥ C([0,θ],L 2 (0,L)) ≤ C∥f ∥ L 1 ((0,θ),L 2 (0,L) , then by the above equation

∥h∥ L 2 (0,θ) ≤ |α|∥∂ 2 x w(•, 0)∥ L 2 (0,θ) ≤ C∥f ∥ L 1 ((0,θ),L 2 (0,L) . Using this estimate in (2.20) we obtain ∥Φ(v 1 ) -Φ(v 2 )∥ B θ = ∥w∥ B θ ≤ C∥f ∥ L 1 ((0,θ),L 2 (0,L) = C∥v p 1 ∂ x v 1 -v p 2 ∂ x v 2 ∥ L 1 ((0,θ),L 2 (0,L) . Finally by Proposition 2.6 ∥Φ(v 1 ) -Φ(v 2 )∥ B θ ≤ C 3 (θ (2-p)/4 + √ θ) ∥v 1 ∥ B θ ∥v 2 ∥ p-1 B θ + ∥v 1 ∥ p B θ + ∥v 2 ∥ p B θ ∥v 1 -v 2 ∥ B θ ,
for some C 3 > 0. Then, taking θ > 0, small enough such that 3C 3 (θ (2-p)/4 + √ θ)R p < 1, we get that Φ is a contraction and by the Banach fixed point theorem and a density argument, we deduce the local in time well-posedness. To conclude the global well-posedness result, we need some a priori estimates. Let T > 0 and u be a strong solution of (1.6). Multiplying (1.6) by u and integrating over (0, L) × (0, s), we get after some integration by parts

1 2 L 0 u(s, •) 2 dx + 1 2 s 0 ∂ 2 x u(t, 0) 2 -sat(α∂ 2 x u(t, 0)) 2 dt = 1 2 L 0 u 2 0 dx. Using (1.2) we derive 1 2 L 0 u 2 (s, •)dx + C α s 0 ∂ 2 x u(t, 0) 2 dt ≤ 1 2 L 0 u 2 0 dx. (2.21)
Then, taking the supremum for s ∈ [0, T ] in (2.21),

∥u∥ 2 C([0,T ],L 2 (0,L)) ≤ ∥u 0 ∥ 2 L 2 (0,L) . (2.22)
On the other hand, taking s = T in (2.21),

T 0 ∂ 2 x u(t, 0) 2 dt ≤ 1 2 C -1 α ∥u 0 ∥ 2 L 2 (0,L) . (2.23)
Now, we multiply the first line of (1.6) by xu and integrate on (0, L) × (0, T ),

1 2 L 0 xu 2 (T, •)dx + 3 2 T 0 1 0 ∂ x u 2 dxdt + 5 2 T 0 L 0 ∂ 2 x u 2 dxdt = 1 2 L 0 xu 2 0 (x) + 1 p + 2 T 0 L 0 u p+2 dxdt + 1 2 T 0 L 0 u 2 dxdt + L 2 T 0 ∂ 2 x u(t, 0) 2 dt.
For the term involving

T 0 L 0 u 2 dxdt T 0 L 0 u 2 (t, x)dx ≤ T ∥u∥ 2 C([0,T ],L 2 (0,L)) .
For the term involving

T 0 L 0 u p+2 (t, x)dxdt, observe that 1 p + 2 T 0 L 0 u p+2 dxdt ≤ 1 p + 2 T 0 ∥u(t, •)∥ p L ∞ (0,L) ∥u(t, •)∥ 2 L 2 (0,L) dt ≤ 2 p/2 p + 2 T 0 ∥∂ x u(t, •)∥ p/2 L 2 (0,L) ∥u(t, •)∥ 2+p/2 L 2 (0,L) dt.
Therefore using (2.22) we have

1 p + 2 T 0 L 0 u p+2 dxdt ≤ C 2 p/2 p + 2 ∥u 0 ∥ 2+p/2 L 2 (0,L) T 0 ∥∂ x u(t, •)∥ p/2 L 2 (0,L) dt ≤ C 2 p/2 T 1-p/4 p + 2 ∥u 0 ∥ 2+p/2 L 2 (0,L) T 0 ∥∂ x u(t, •)∥ 2 L 2 (0,L) dt p/4 ≤ C p T ∥u 0 ∥ 8+2p 4-p L 2 (0,L) + 1 2 T 0 ∥∂ x u(t, •)∥ 2 L 2 (0,L) dt , (2.24)
for some positive constant C p depending on p and C. Thus, from (2.22), (2.23) and Young's inequality we derive

T 0 1 0 ∂ x u 2 dxdt + T 0 L 0 ∂ 2 x u 2 dxdt ≤ C p (1 + T )∥u 0 ∥ 2 L 2 (0,L) + T ∥u 0 ∥ 8+2p 4-p L 2 (0,L) . (2.25)
Finally, with (2.22) and (2.25) and arguing by density, the proof of Theorem 2.7 is finished. □ 2.2. Well-posedness with internal feedback. In this section, we study the global well-posedness of (1.4), some of the computations are quite similar to those exposed in Section 2.1 for that we focus in the main difference.

2.2.1.

Well-posedness of linear system. Let us first investigate the well-posedness of the linear Kawahara system, i.e when we drop the nonlinear internal terms u p ∂ x u and sat(au):

     ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = ∂ 2 x u(t, L) = 0, t ∈ (0, ∞), u(0, x) = u 0 (x),
x ∈ (0, L).

(2.26)

Thus we write the above system as:

u(t) = Au(t), t > 0, u(0) = u 0 , (2.27)
where the operator A can be written as

Au = -u ′ -u ′′′ + u ′′′′′ with the domain D(A) = u ∈ H 5 (0, L) ∩ H 2 0 (0, L) : u xx (L) = 0 . Remark 2.8.
Observe that contrary to the case of boundary saturation, we firts analyse a system which is linear so that we use linear semigroup theory for the study of (2.26). In what follows, we add the nonlinear terms as source terms. In the case of boundary saturation this strategy it is not to clear, and for that reason we just drop the internal nonlinear term and we used nonlinear semigroup arguments. That made the well-posedness of (2.26) easier that the boundary case.

It can be checked that the operator A and its adjoint operator

A * u = u ′ + u ′′′ -u ′′′′′ with the domain D(A * ) = u ∈ H 5 (0, L) ∩ H 2 0 (0, L) : u xx (0) = 0 ,
are both are dissipative. Therefore by Lumer-Philips theorem, it follows that A generates a strongly continuous semigroup of contraction {S(t)} t≥0 . Hereinafter, we have the following well-posedness result for the linear Kawahara system: Theorem 2.9. Let us take u 0 ∈ D(A). Then the equation (2.26) possesses a unique strong solution u ∈ C([0, T ]; D(A)∩C 1 (0, T ; L 2 (0, L)). If the initial data u 0 ∈ L 2 (0, L), the unique mild solution of (2.26) lies in C([0, T ]; L 2 (0, L)) ∩ L 2 (0, T ; H 2 (0, L)). Moreover, we have a generic positive constant C > 0 such that the solution satisfies the following estimates

• ∥u∥ C([0,T ];L 2 (0,L)) + ∥u∥ L 2 (0,T ;H 2 (0,L)) ≤ C ∥u 0 ∥ L 2 (0,L) , • ∥u xx (•, 0)∥ L 2 (0,T ) ≤ C ∥u 0 ∥ L 2 (0,L) .
Next, we study the well-posedness result for the linear Kawahara equation with a source term:

     ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u = f, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = ∂ 2 x u(t, L) = 0, t ∈ (0, ∞), u(0, x) = u 0 (x),
x ∈ (0, L).

(2.28) Thus, we write the above system in the form of an abstract nonhomogeneous evolution equation:

u = Au + f, t > 0, u(0) = u 0 . (2.29)
By classical semigroup theory result, it follows that for any f ∈ L 1 loc (0, ∞; L 2 (0, L)), the solution of the system (2.29)

u(t) = S(t)u 0 + t 0 S(t -s)f (s)ds belongs to C([0, ∞); L 2 (0, L)).
We also have the following estimates: Theorem 2.10. Let us take f ∈ L 1 (0, T ; L 2 (0, L)). Then, the solution u satisfies the following estimates

∥u∥ B T + ∂ 2 x u(•, 0) L 2 (0,T ) ≤ C(1 + √ T ) ∥y 0 ∥ L 2 (0,L) + ∥f ∥ L 1 ((0,T ),L 2 (0,L)) . (2.30)
Now we are in a position to prove the global well-posedness result for the system (1.4).

Theorem 2.11. Let u 0 ∈ L 2 (0, L). Then there exists a unique mild solution u ∈ C([0, ∞), L 2 (0, L)) ∩ L 2 (0, ∞), H 2 (0, L) of (1.4).Moreover, for any T > 0, there exists C > 0, such that the solution of the system (1.4) with (1.5), satisfies the following:

∥u∥ 2 L 2 (0,T ;H 2 (0,L)) ≤ C ∥u 0 ∥ 2 L 2 (0,L) + T ∥u 0 ∥ 8+2p 4-p L 2 (0,L) , (2.31)
∂ 2 x u(•, 0) L 2 (0,L) ≤ C ∥u 0 ∥ L 2 (0,L) , (2.32)
∥u 0 ∥ 2 L 2 (0,L) ≤ 1 T T 0 L 0 u 2 (t, x)dxdt + T 0 |∂ 2 x u(t, 0)| 2 dt + 2 T 0 L 0 sat(a(x)u(t, x))u(t, x)dxdt. (2.33)
Proof. Let u 0 ∈ D(A) and R, θ > 0 to be chosen later. Consider again the ball

B B θ (0, R) := {v ∈ B θ , ∥v∥ B θ ≤ R}. Let Ψ, the map defined by Ψ : B θ → B θ , Ψ(v) = u, where u solves      ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u = -v p ∂ x v -sat(a(x)v(t, x)), (t, x) ∈ (0, θ) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, ∂ 2 x u(t, L) = 0, t ∈ (0, θ), u(0, x) = u 0 (x),
x ∈ (0, L).

Observe that Φ is well-defined, since by Proposition 2.6 v p ∂ x v ∈ L 1 ((0, θ), L 2 (0, L)) and by [START_REF] Marx | Stabilization of a linear Korteweg-de Vries equation with a saturated internal control[END_REF]Proposition 3.4]

∥sat(av)∥ L 1 ((0,θ);L 2 (0,L) ≤ 3 √ La 1 √ θ∥v∥ B θ .
By Theorem 2.10, Proposition 2.6 and the above inequality we get

∥Φ(v)∥ B θ ≤C(1 + √ θ) ∥u 0 ∥ L 2 (0,L) + ∥v p ∂ x v∥ L 1 ((0,θ),L 2 (0,L)) + ∥sat(a(x)v(t, x))∥ L 1 ((0,θ),L 2 (0,L)) ≤C 1 (1 + √ θ)∥u 0 ∥ L 2 (0,L) + C 2 (θ (2-p)/4 + √ θ)∥v∥ p+1 B T + C 2 √ θ∥v∥ B T ,
for some C 1 , C 2 > 0. Then taking R = 3C 1 ∥u 0 ∥ L 2 (0,L) and θ > 0 small enough such that

   C 2 (θ 2-p 4 R p + √ θR p + √ θ) < 1 3 , (1 + √ θ) < 2, we have that ∥Ψ(v)∥ B θ ≤ R and thus Φ maps B B θ (0, R) into itself. Now take v 1 , v 2 ∈ B B θ (0, R), and consider w = u 1 -u 2 = Ψ(v 1 ) -Ψ(v 2 )
. We observe that w solves

     ∂ t w + ∂ x w + ∂ 3 x w -∂ 5 x w = f + h, (t, x) ∈ (0, θ) × (0, L), w(t, 0) = w(t, L) = ∂ x w(t, 0) = ∂ x w(t, L) = 0, ∂ 2 x u(t, L) = 0, t ∈ (0, θ), w(0, x) = 0,
x ∈ (0, L),

where f = -v p 1 ∂ x v 1 + v p 2 ∂ x v 2 and h = -sat(a(x)v 1 (t, x)) + sat(a(x)v 2 (t, x))
. Similar integration by parts as in Theorem 2.10 a allows to write

∥w∥ B θ ≤ C ∥h∥ L 1 ((0,θ),L 2 (0,L) + ∥f ∥ L 1 ((0,θ);L 2 (0,L) .
(2.34)

Further we have (see [START_REF] Marx | Stabilization of a linear Korteweg-de Vries equation with a saturated internal control[END_REF]Proposition 3.4])

∥h∥ L 1 ((0,θ);L 2 (0,L) = ∥ -sat(av 1 ) + sat(av 2 )∥ L 1 ((0,θ);L 2 (0,L) ≤ 3 √ La 1 √ θ∥v 1 -v 2 ∥ B θ .
Using this estimate along with (2.19) in (2.34) we obtain

∥Ψ(v1) -Ψ(v2)∥B θ =∥w∥B θ ≤ C ∥f ∥ L 1 ((0,θ),L 2 (0,L) + 3 √ La1 √ θ∥v1 -v2∥B θ ≤C (θ (2-p)/4 + √ θ) ∥v1∥B θ ∥v2∥ p-1 B θ + ∥v1∥ p B θ + ∥v2∥ p B θ + 3 √ La1 √ θ ∥u -v∥B θ .
Then, taking θ > 0, small enough such that C (θ

(2-p)/4 + √ θ)R p + 3 √ La 1 √ θ < 1,
we get that Ψ is a contraction and by the Banach fixed point theorem and a density argument, we deduce the local in time well-posedness. To conclude the global well-posedness result, we need some a priori estimates. Let T > 0 and u be a strong solution of (1.4). Multiplying the equation (1.4) by u and integrating on [0, L] we have after integration by parts and boundary data 1 2

d dt L 0 u 2 dx + 1 2 ∂ 2 x u(t, 0) 2 + L 0 sat(a(x)u(t, x))u(t, x)dx = 0. (2.35)
Integrating over [0, T ] and using the fact that sat is an odd function, one can prove the estimate (2.32).

Similarly to (2.22), integrating (2.35) over [0, s] we can show that

∥u∥ 2 C([0,T ],L 2 (0,L)) ≤ ∥u 0 ∥ 2 L 2 (0,L) . (2.36)
Multiplying first equation of (1.4) by xu and taking integration on both sides over [0, T ]×(0, L), applying successive integration by parts and using boundary conditions, we obtain

1 2 L 0 xu 2 (T, •)dx + 3 2 T 0 L 0 u 2 x dx + 5 2 T 0 L 0 u 2 xx dxdt + T 0 L 0 xusat(au)dxdt = 1 2 L 0 xu 2 0 (•)dx + 1 2 T 0 L 0 u 2 dxdt + 1 p + 2 T 0 L 0 u p+2 dxdt.
Noting the fact sat is an odd function, using (2.36) and following the computations of (2.24) we finally have (2.31). Next multiplying first equation of (1.4) by (T -t)u and taking integration over [0, L] on both sides, we write

L 0 (T -t)u∂ t udx + L 0 (T -t)u∂ x udx + L 0 (T -t)u∂ 3 x udx - L 0 (T -t)u∂ 5 x udx + L 0 (T -t)u∂ x udx + L 0 (T -t)sat(a(x)u(t, x))u(t, x)dx = 0. (2.37)
Integrating over [0, T ] we have

1 2 T 0 L 0 u 2 dxdt - T 2 L 0 u 2 0 dx + 1 2 T 0 (T -t)∂ 2 x u(t, 0) 2 dt + T 0 L 0 (T -t)sat(a(x)u(t, x))u(t, x)dxdt = 0.
Thus we get the estimate (2.33). □

Exponential stabilization

In this part we prove our main results with respect to the exponential stability of systems (1.4) and (1.6). Both proofs are based on showing an appropriated observability inequality, using a contradiction argument and compactness ideas. The main difference in the proof of internal and boundary saturation is that for the boundary case we first prove such a observability inequality by analyzing system (2.1) and then we pass to the fully nonlinear system (1.6) by a perturbation approach, due to this argument our exponential stability result for (1.6) it is of local nature. From the other side, in the proof of the observability inequality for (1.4) we work directly with the nonlinear system which allow us to obtain a semiglobal exponential stability result. In what follows, we consider here the classical L 2 -energy,

E(t) = 1 2 L 0 u(t, x) 2 dx. (3.1)
3.1. Boundary saturation. In this part we are going to prove our main result in the case of boundary saturation related to the exponential stability of system (1.6). In the case of boundary saturation, the following unique continuation property is crucial in the proof

Lemma 3.1. ([3, Lemma 4.3]) Let L > 0, if λ, ψ ∈ C × H 5 (0, L) satisfies      λψ(x) + ψ ′ (x) + ψ ′′′ (x) -ψ ′′′′′ (x) = 0, x ∈ (0, L) ψ(0) = ψ(L) = ψ ′ (0) = ψ ′ (L) = 0, ψ ′′ (0) = ψ ′′ (L) = 0, (3.2)
then ψ ≡ 0.

For a strong solution of (1.6) (or (2.1)) performing integration by parts, we get using |α| < 1

Ė(t) = 1 2 sat(α∂ 2 x u(t, 0)) 2 -∂ 2 x u(t, 0) 2 ≤ -C α ∂ 2 x u(t, 0) 2 ≤ 0. (3.3)
Thus, the energy is a non-increasing function of time. In fact, we will prove that the energy decays exponentially to 0. Our approach will be based on an observability inequality for the system (2.1).

Proposition 3.2. Let L > 0, |α| < 1 and u 0 ∈ L 2 (0, L). For any solution u of (2.1), the following inequality holds:

L 0 u 2 0 dx ≤ C obs T 0 ∂ 2 x u(t, 0) 2 dt, (3.4)
for some C obs > 0, that does not depend on u 0 .

Proof. In order to prove the observability inequality, we follow [START_REF] Rosier | Exact boundary controllability for the Korteweg-de Vries equation on a bounded domain[END_REF] and use a contradiction argument. Let us suppose that (3.4) is false, then there exists (u n 0 ) n∈N ⊂ L 2 (0, L) such that L 0 (u n 0 ) 2 dx = 1, and

T 0 ∂ 2,n x u(t, 0) 2 dt -→ 0
when n → ∞ and where u n is the associate solution of (2.1) with initial data u n 0 . Now, using (2.10) with f = 0 we get that (u n ) n∈N is bounded in L 2 ((0, T ), H 2 (0, L)). Then as

∂ t u n = -∂ x u n -∂ 3 x u n + ∂ 5 x u n , we can see that (∂ t u n ) n∈N is bounded in L 2 ((0, T ), H -3 (0, L)), thus by Aubin-Lions lemma (u n ) n∈N is relatively compact in L 2 ((0, T ), L 2 (0, L)), therefore we can assume that u n -→ u in L 2 ((0, T ), L 2 (0, L)). Now, using (2.11) we derive that (u n 0 ) n∈N is a Cauchy sequence in L 2 (0, L) . Let u 0 = lim n→∞ u n 0 , then L 0 u 2 0 dx = 1, and T 0 ∂ 2 x u(t, 0) 2 dt = 0
Thus, as sat loc (0) = 0, we conclude that y solves the following linear equation:

         ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = 0, t ∈ (0, ∞), ∂ x u(t, 0) = ∂ x u(t, L) = 0, t ∈ (0, ∞), ∂ x u(t, 0) = ∂ 2 x u(t, 0) = ∂ 2 x u(t, L) = 0, t ∈ (0, ∞). (3.5)
But by Lemma 3.1 we can show that there is no function satisfying this system and ∥u 0 ∥ L 2 (0,L) = 1, which gives us the contradiction and finishes the proof of Proposition 3.2. □ Now, we prove our main result related with the exponential stability in the boundary saturated case.

Theorem 3.3. Let L > 0, |α| < 1 and u 0 ∈ L 2 (0, L). For any solution u of (2.1), the energy of the system decays exponentially to zero i.e, there exists C, γ > 0 such that E(t) ≤ CE(0)e -γt , for all t ≥ 0. Moreover, if ∥u 0 ∥ L 2 (0,L) ≤ ε for ε > 0 small enough, the same property holds for (1.6).

Proof. Let u 0 ∈ D(A b ), using (3.3) and ( 2.3) we get for a strong solution of (2.1)

Ė(t) ≤ -C α ∂ 2 x u(t, 0) 2 , thus we can see E(T ) -E(0) ≤ -C α T 0 ∂ 2 x u(t, 0) 2 dt.
Using the last expression, (3.3) and (3.4) we get

E(T ) ≤ E(0) ≤ C obs T 0 ∂ 2 x u(t, 0) 2 dt ≤ C obs C α (E(0) -E(T )),
which implies that

E(T ) ≤ γE(0), with γ = C obs C α + C obs < 1. ( 3
E(mT ) ≤ γE((m -1)T ) ≤ • • • ≤ γ m E(0).
Hence, we have E(mT ) ≤ e -µmT E(0) where µ = 1 T ln 1 γ > 0. Let t > 0 then there exists m ∈ N * such that (m -1)T < t ≤ mT , and then using again the non-increasing property of the energy we get

E(t) ≤ E((m -1)T ) ≤ e -µ(m-1)T E(0) ≤ 1 γ e -µt E(0).
Finally, we conclude the exponential stability of (2.1) by a density argument.

We study now the nonlinear case, note that as (1.6) is also invariant by translation in time, thus to prove the stability, the objective is to show that (3.6) holds for u solution of (1.6) if the initial data it is small enough. This idea was used in several works, for instance [START_REF] Parada | Stability analysis of a Korteweg-de Vries equation with saturated boundary feedback[END_REF][START_REF] Parada | Delayed stabilization of the Korteweg-de Vries equation on a star-shaped network[END_REF]. Let ∥u 0 ∥ L 2 (0,L) ≤ ϵ for ϵ > 0 to fix later. Take u the solution of (1.6), then we can split u as u = u + ũ, where

         ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, t ∈ (0, ∞), ∂ 2 x u(t, L) = sat loc (α∂ 2 x u(t, 0)), t ∈ (0, ∞), u(0, x) = u 0 (x),
x ∈ (0, L),

and          ∂ t ũ + ∂ x ũ + ∂ 3 x ũ -∂ 5 x ũ = u p ∂ x u, (t, x) ∈ (0, ∞) × (0, L), ũ(t, 0) = ũ(t, L) = ∂ x ũ(t, 0) = ∂ x ũ(t, L) = 0, t ∈ (0, ∞), ∂ 2 x ũ(t, L) = h(t), t ∈ (0, ∞), ũ(0, x) = 0, x ∈ (0, L).
where

h(t) = sat(α∂ 2 x u(t, 0)) -sat(α∂ 2 x u(t, 0)). Now, we observe that ∥u(T, •)∥ L 2 (0,L) ≤ ∥u(T, •)∥ L 2 (0,L) + ∥ũ(T, •)∥ L 2 (0,L) . (3.7)
From the other side as u satisfies (2.1), we get using (3.6) and u(0, x) = u 0 (x) that there exists γ < 1 such that ∥u(T )∥ L 2 (0,L) ≤ γ∥u 0 ∥ L 2 (0,L) .

(3.8) For ũ, by (2.10) and Proposition 2.6,

∥ũ(T, •)∥ L 2 (0,L) ≤ ∥ũ∥ C([0,T ],L 2 (0,L)) ≤ C∥u p ∂ x u∥ L 1 ((0,T ),L 2 (0,L)) ≤ C∥u∥ p+1 B T , using (2.22)-(2.25), we derive ∥u∥ p+1 B T ≤ C ∥u 0 ∥ p+1 L 2 (0,L) + ∥u 0 ∥ (4+p)(p+1) 4-p L 2 (0,L)
.

Thus, we get for ũ ∥ũ(T,

•)∥ L 2 (0,L) ≤ C∥u 0 ∥ p L 2 (0,L) + C∥u 0 ∥ p(6+p) 4-p L 2 (0,L) ∥u 0 ∥ L 2 (0,L) .
Finally, as ∥u 0 ∥ L 2 (0,L) ≤ ϵ we deduce from (3.7)

∥u(T, •)∥ L 2 (0,L) ≤ ∥u(T, •)∥ L 2 (0,L) + ∥ũ(T, •)∥ L 2 (0,L) ≤ γ∥u 0 ∥ L 2 (0,L) + (Cϵ p + Cϵ p(6+p) 4-p )∥u 0 ∥ L 2 (0,L) = (γ + Cϵ p + Cϵ p(6+p) 4-p )∥u 0 ∥ L 2 (0,L) .
We conclude by choosing ϵ > 0 small enough such that γ = γ + Cϵ p + Cϵ p(6+p) 4-p < 1. □

Internal saturation.

We pass now to study the case of internal saturation. We are interested to prove the following exponential stabilization result:

Theorem 3.4 (Semi global stabilization). Let us consider the system (1.4). Then for any r > 0 there exists a constant µ(r) > 0 such that for any u 0 ∈ L 2 (0, L) satisfying ∥u 0 ∥ L 2 (0,L) ≤ r, the solution of (1.4) satisfy the following:

E(t) ≤ e -µt E(0), ∀t ≥ 0,
where E(•) is defined in (3.1).

Before going to prove Theorem 3.4, first we state some useful results borrowed from the article [START_REF]Global stabilization of a Korteweg-de Vries equation with saturating distributed control[END_REF].

Lemma 3.5 ([20]

). Let r > 0 and a : [0, L] → R be a function which satisfies (1.5). Let us define

k(r) = min M a 1 r , 1 ,
where M and a 1 are defined in (1.2) and (1.5) respectively. The following two hold:

• For s ∈ L 2 (0, L) with ∥s∥ L 2 (0,L) ≤ r, we have Using (2.36), we have ∥u(t, •)∥ L 2 (0,L) ≤ r, where r > 0 be any number such that ∥u 0 ∥ L 2 ≤ r. Applying first estimate (3.9) of Lemma 3.5 we deduce 1 2

(sat 2 (a(x)s(x)) -k(r)a(x)s(x)) s(x) ≥ 0, ∀x ∈ [0, L]. ( 3
d dt L 0 u 2 dx ≤ - L 0 k(r)a 0 u 2 (t, x)dx. (3.12)
Thus the proof is complete with µ = min{a 0 , k(r)}.

3.2.2.

Proof of Theorem 3.4 with ω ⊂ [0, L] and sat = sat 2 . In this case, we will use a contradiction argument as [START_REF] Rosier | Global stabilization of the generalized Korteweg-de Vries equation posed on a finite domain[END_REF]. First, let us state a unique continuation lemma for the Kawahara equation, which plays an integral part in the proof of stabilization result:

Lemma 3.6 (Unique continuation principle, [START_REF] Gao | Global Carleman estimate for the Kawahara equation and its applications[END_REF]). Let u 0 ∈ L 2 (0, L) and u ∈ B T be the solution of the equation

     ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u + u p ∂ x u = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = ∂ 2 x u(t, L) = 0, t ∈ (0, ∞), u(0, x) = u 0 (x),
x ∈ (0, L), (3.13) such that u(t, x) = 0, ∀ x ∈ ω, ∀ t ∈ (0, T ), with ω ⊂ (0, L). Then u(t, x) = 0, ∀ x ∈ (0, L), ∀ t ∈ (0, T ).

For the sake of completeness, we give a sketch of the proof of Lemma 3.6 in the Appendix B. Let us consider ∥u 0 ∥ L 2 (0,L) ≤ r. Using (2.36), we deduce

∥u(t, •)∥ L 2 (0,L) ≤ r, ∀t ≥ 0. (3.14) Integrating (2.35) over [0, T ] we have L 0 u 2 (T, •)dx = L 0 u 2 0 dx - T 0 ∂ 2 x u(t, 0) 2 dt -2 T 0 L 0 sat(a(x)u(t, x))u(t, x)dxdt (3.15)
As (3.14) holds, we apply the first part of Lemma 3.5, and deduce

L 0 u 2 (T, •)dx ≤ L 0 u 2 0 dx - T 0 ∂ 2 x u(t, 0) 2 dt -2 T 0 L 0 k(r)a(x)|u(t, x)| 2 dxdt. (3.16)
Next, for the time being we assume the following result:

Lemma 3.7. Let us take T > 0 and let r be any positive number. Then for any u 0 ∈ L 2 (0, L), with ∥u 0 ∥ L 2 (0,L) ≤ r, the solution of the equation (1.4) satisfies the following

∥u 0 ∥ 2 L 2 (0,L) ≤ C 2 T 0 |∂ 2 x u(t, 0)| 2 dt + 2 T 0 L 0 k(r)a(x)|u(t, x)| 2 dxdt . (3.17)
Therefore combining (3.16) and (3.17), we finally have the existence of γ ∈ (0, 1) such that

∥u(kT, •)∥ 2 L 2 (0,L) ≤ γ k ∥u 0 ∥ 2 L 2 (0,L) , ∀ k ≥ 0. (3.18) Using (2.36), we have ∥u(t, •)∥ 2 L 2 (0,L) ≤ ∥u(kT, •)∥ 2 L 2 (0,L) , where kT ≤ t ≤ (k + 1)T. Finally, we obtain that ∥u(t, •)∥ L 2 (0,L) ≤ 1 γ e -µt ∥u 0 ∥ L 2 (0,L) , ∀ t ≥ 0, (3.19)
where µ = -log γ T . 3.2.3. Proof of Lemma 3.7. Thanks to (2.33), we can say that to prove Lemma 3.7, it is enough to show that the following

∥u∥ 2 L 2 (0,T ;L 2 (0,L) ≤ C 2 T 0 |∂ 2 x u(t, 0)| 2 dt + 2 T 0 L 0 k(r)a(x)|u(t, x)| 2 dxdt , (3.20)
for some C > 0, provided ∥u 0 ∥ L 2 (0,L) ≤ r. We prove this inequality by contradiction approach. Suppose that (3.20) is not true. Hence there exists a sequence of functions u n 0 ∈ L 2 (0, L) with ∥u n 0 ∥ L 2 (0,L) ≤ r and {u n } n∈N ∈ B T be the corresponding solution of the equation

     ∂ t u n + ∂ x u n + ∂ 3 x u n -∂ 5 x u n + u np ∂ x u n + sat(a(x)u n (t, x)) = 0, (t, x) ∈ (0, ∞) × (0, L), u n (t, 0) = u n (t, L) = ∂ x u n (t, 0) = ∂ x u n (t, L) = ∂ 2 x u n (t, L) = 0, t ∈ (0, ∞), u n (0, x) = u n 0 (x), x ∈ (0, L), (3.21)
where p ∈ [1, 2), and

lim n→∞ ∥u n ∥ 2 L 2 (0,T ;L 2 (0,L)) T 0 |∂ 2 x u n (t, 0)| 2 dt + 2 T 0 L 0 k(r)a|u n (t, x)| 2 dxdt = +∞. (3.22) Also (2.36) gives that ∥u n (t, •)∥ L 2 (0,L) ≤ r, ∀ t ≥ 0. (3.23) Let us denote λ n = ∥u n ∥ L 2 (0,T ;L 2 (0,L)) and v n (t, x) = u n (t, x) λ n .
{λ n } n∈N is a bounded sequence by (3.23).

Therefore it has a subsequence, denoted by {λ n } n∈N such that there exists λ ≥ 0 with

λ n → λ as n → ∞. (3.24)
Note that v n satisfies the following equation:

       ∂ t v n + ∂ x v n + ∂ 3 x v n -∂ 5 x v n + λ n v np ∂ x v n + sat(a(x)λ n v n (t, x)) λ n = 0, (t, x) ∈ (0, ∞) × (0, L), v n (t, 0) = v n (t, L) = ∂ x v n (t, 0) = ∂ x v n (t, L) = ∂ 2 x v n (t, L) = 0, t ∈ (0, ∞), ∥v n ∥ L 2 (0,T ;L 2 (0,L) = 1. (3.25)
and also by (3.22) we further have

T 0 |∂ 2 x v n (t, 0)| 2 dt + 2 T 0 L 0 k(r)a|v n (t, x)| 2 dxdt → 0 as n → ∞. (3.26)
In addition, using (2.33), we have that {∥v n (0, •)∥ L 2 (0,L) } n∈N is a bounded sequence and due to the following

   ∥v n ∥ 2 L 2 (0,T ;H 2 (0,L)) ≤ C ∥v n (0, •)∥ 2 L 2 (0,L) + T ∥v n (0, •)∥ 8+2p 4-p L 2 (0,L) ∥v n (t, •)∥ L 2 (0,L) ≤ ∥v n (0, •)∥ L 2 (0,L) , (3.27)
the sequence {v n } n∈N is bounded in the space L 2 (0, T ; H 2 (0, L)) and L ∞ ([0, T ]; L 2 (0, L)). Therefore there exists a subsequence still denoted by

{v n } n∈N and a function v ∈ L ∞ (0, T ; L 2 (0, L)) ∩ L 2 (0, T ; H 2 (0, L)) such that v n * ⇀ v in L ∞ (0, T ; L 2 (0, L)) v n ⇀ v in L 2 (0, T ; H 2 (0, L)) (3.28)
To derive the equation satisfied by v, we need to pass the limit in (3.25) and thus, in the following steps, we need to take care of the convergence of the terms related to the nonlinearity. First of all, we have

∥v np ∂ x v n ∥ L 1 ((0,T ),L 2 (0,L)) ≤ C(T (2-p)/4 + √ T )∥v n ∥ p B T ∥v n ∥ B T ≤ C. (3.29)
In the next step, we prove that

{(v n ) p+1 } n∈N is bounded in L 2 (0, T ; L 2 (0, L)). T 0 L 0 (v n ) 2p+2 dxdt ≤ ∥v n ∥ 2 L 2 (0,T ;H 1 (0,L)) ∥v n ∥ 3+ϵ L ∞ (0,T ;L 2 (0,L) ≤ C, (3.30)
where p = 1 + ϵ, 0 < ϵ < 1. This shows that

{v np ∂ x v n } n∈N is bounded in the space L 2 (0, T ; H -1 (0, L)) ⊂ L 2 (0, T ; H -3 (0, L)). Also, we know that {∂ 5 x v n } n∈N is bounded in L 2 (0, T ; H -3 (0, L))
. Furthermore, it is easy to check the following estimate (see [START_REF]Global stabilization of a Korteweg-de Vries equation with saturating distributed control[END_REF]Lemma 3.2] for more details)

sat(a(x)λ n v n (t, x)) λ n L 2 (0,T ;L 2 (0,L)) ≤ C ∥v n ∥ L 2 (0,T ;H 1 (0,L)) . (3.31)
Thus combining we have that

∂ t v n = -∂ x v n + ∂ 3 x v n -∂ 5 x v n + λ n v np ∂ x v n + sat(a(x)λnv n (t,x))
λn is bounded in L 2 (0, T ; H -3 (0, L)). Thanks to the Lions-Aubin lemma with the embedding H 2 (0, L) ⊂ L 2 (0, L) ⊂ H -3 (0, L), {v n } n∈N is relatively compact in L 2 (0, T ; L 2 (0, L)). It follows that there exists a subsequence, again denoted by {v n } n∈N such that v n → v in L 2 (0, T ; L 2 (0, L)). Further, it implies (v n ) p+1 → v p+1 a.e. (t, x) ∈ (0, T ) × (0, L). Since {(v n ) p+1 } n∈N is bounded in L 2 (0, T ; L 2 (0, L)), therefore there exists some h ∈ L 2 (0, T ; L 2 (0, L)) such that (v n ) p+1 ⇀ h in L 2 (0, T ; L 2 (0, L)). Uniqueness of limit gives h = v p+1 a.e. and we have (v n ) p+1 → v p+1 in D ′ ((0, T ) × (0, L)). Taking the derivative with respect to space, we further have

λ n v np ∂ x v n → λv p ∂ x v in D ′ ((0, T ) × (0, L)).
As {v n } n∈N and {∂ t v n } n∈N are bounded in the spaces L ∞ (0, T ; L 2 (0, L)) and L 2 (0, T ; H -3 (0, L)) respectively. Using the embedding L 2 (0, L) ⊂ H -1 (0, L) ⊂ H -3 (0, L) (see [START_REF] Simon | Compact sets in the space L p (0, T ; B)[END_REF]Corollary 4]), we get by for a subsequence denoted by

{v n } n∈N that v n → v in C([0, T ]; H -1 (0, L)).
Thus in a nutshell, interpolation theory gives us the following

               v n * ⇀ v in L ∞ (0, T ; L 2 (0, L)) v n ⇀ v in L 2 (0, T ; H 2 (0, L)) v n → v in L 2 (0, T ; L 2 (0, L)) v n → v in C([0, T ]; H -1 (0, L)) λ n v np ∂ x v n → λv p ∂ x v in D ′ ((0, T ) × (0, L)).
(3.32) Moreover, from (3.26) we deduce:

T 0 |∂ 2 x v(t, 0)| 2 dt + 2 T 0 L 0 k(r)a(x)|v(t, x)| 2 dxdt ≤ lim inf n→∞ T 0 |∂ 2 x v n (t, 0)| 2 dt + 2 T 0 L 0 k(r)a(x)|v n (t, x)| 2 dxdt = 0.
Therefore a(x)k(x)v = 0, x ∈ (0, L) and ∂ 2 x v(•, 0) = 0, t ∈ (0, T ). This further implies v = 0, x ∈ ω, t ∈ (0, T ). We also obtain the limit v solves the following equation in the sense of distribution

     ∂ t v + ∂ x v + ∂ 3 x v -∂ 5 x v + λv p ∂ x v = 0, (t, x) ∈ (0, ∞) × (0, L), v(t, 0) = v(t, L) = ∂ x v(t, 0) = ∂ x v(t, L) = ∂ 2
x v(t, L) = 0, t ∈ (0, ∞), ∥v∥ L 2 (0,T ;L 2 (0,L)) = 1. (3.33) by unique continuation principle Lemma 3.6, we have v = 0, x ∈ (0, L), t ∈ (0, T ). This contradicts the fact ∥v∥ L 2 (0,T ;L 2 (0,L)) = 1. Therefore Lemma 3.7 holds.

3.2.4. Proof of Theorem 3.4 with ω ⊂ [0, L] and sat = sat loc . We will prove the exponential stabilization result Theorem 3.4 with sat = sat loc as the same strategy of above. It is clear that the stabilizability follows if we can prove the following lemma: Lemma 3.8. Let us take T > 0 and let r be any positive number. Then for any u 0 ∈ L 2 (0, L), with ∥u 0 ∥ L 2 (0,L) ≤ r, the solution of the equation (1.4) satisfies the following

∥u 0 ∥ 2 L 2 (0,L) ≤ C 2 T 0 |∂ 2 x u(t, 0)| 2 dt + 2 T 0 L 0 sat loc (au(t, x))u(t, x)dxdt . (3.34)
Proof. Using (2.33), we can say that to prove Lemma 3.7, it is enough to show that the following

∥u∥ 2 L 2 (0,T ;L 2 (0,L) ≤ C 2 T 0 |∂ 2 x u(t, 0)| 2 dt + 2 T 0 L 0 sat loc (au(t, x))u(t, x)dxdt , (3.35)
for some C > 0, provided ∥u 0 ∥ L 2 (0,L) ≤ r. Using contradiction we will prove the estimate (3.35). If possible, let us assume that there exists a sequence of functions u n 0 ∈ L 2 (0, L) with ∥u n 0 ∥ L 2 (0,L) ≤ r and {u n } n∈N ∈ B T be the corresponding solution of the equation Thanks to (2.36), we obtain ∥u n (t, •)∥ L 2 (0,L) ≤ r, ∀ t ≥ 0.

     ∂ t u n + ∂ x u n + ∂ 3 x u n -∂ 5 x u n + u np ∂ x u n + sat loc (a(x)u n (t, x)) = 0, (t, x) ∈ (0, ∞) × (0, L), u n (t, 0) = u n (t, L) = ∂ x u n (t, 0) = ∂ x u n (t, L) = ∂ 2 x u n (t, L) = 0, t ∈ (0, ∞), u n (0, x) = u n 0 (x), x ∈ (0, L), (3.36
(3.38) Also, using (2.33), we have that {v n (0, •)} n∈N is a bounded sequence in L 2 (0, L) and we further deduce

∥v n ∥ 2 L 2 (0,T ;H 2 (0,L)) ≤ C ∥v n (0, •)∥ 2 L 2 (0,L) + T ∥v n (0, •)∥ 8+2p 4-p L 2 (0,L) ≤ Cr 2 + CT r 8+2p 4-p . (3.39)
Using Poincare inequality, we have

∥u n (t, •)∥ L ∞ (0,L) ≤ C ∥u n ∥ H 2 (0,L) ∀ t ∈ [0, T ].
This further gives

∥u n (•, x)∥ 2 L 2 (0,T ) ≤ C 1 , (3.40)
where C 1 = C 2 (Cr 2 + CT r 8+2p 4-p ). Let us consider the subsets Ω i ⊂ [0, T ] defined as

Ω i = {t ∈ [0, T ] : ∥u n (t, •)∥ L ∞ (0,L) > i}. (3.41)
Its complement is then given by

Ω c i = {t ∈ [0, T ] : ∥u n (t, •)∥ L ∞ (0,L) ≤ i}. (3.42) Therefore we have ∥u n ∥ 2 L 2 (0,T ;L ∞ (0,L)) ≥ ∥u n ∥ 2 L 2 (Ωi;L ∞ (0,L)) ≥ i 2 µ(Ω i ), (3.43)
where µ(Ω i ) is the Lebesgue measure of the set Ω i . Next, using (3.40) we have µ(Ω i ) ≤ C1 i 2 . This implies that max T -

C 1 i 2 , 0 ≤ µ(Ω c i ) ≤ T. (3.44)
Thanks to the estimate (3.10) of Lemma 3.5, we have

T 0 L 0 sat loc (au n )u n dtdx = Ωi L 0 sat loc (au n )u n dtdx + Ω c i L 0 sat loc (au n )u n dtdx ≥ Ω c i L 0 sat loc (au n )u n dtdx ≥ Ω c i L 0 ak(i)(u n ) 2 dtdx. (3.45)
Let us denote λ n = ∥u n ∥ L 2 (0,T ;L 2 (0,L)) and v n (t, x) = u n (t, x) λ n . {λ n } n∈N is a bounded sequence by (3.23).

Therefore it has a subsequence, denoted by {λ n } n∈N such that there exists λ ≥ 0 with

λ n → λ as n → ∞. (3.46)
Note that v n satisfies the following equation:

       ∂ t v n + ∂ x v n + ∂ 3 x v n -∂ 5 x v n + λ n v np ∂ x v n + sat loc (a(x)λ n v n (t, x)) λ n = 0, (t, x) ∈ (0, ∞) × (0, L), v n (t, 0) = v n (t, L) = ∂ x v n (t, 0) = ∂ x v n (t, L) = ∂ 2 x v n (t, L) = 0, t ∈ (0, ∞), ∥v n ∥ L 2 (0,T ;L 2 (0,L) = 1, ( 3
.47) and also by (3.37) and (3.45) and we further have

T 0 |∂ 2 x v n (t, 0)| 2 dt + 2 T 0 L 0 k(i)a(x)|v n (t, x)| 2 dxdt → 0 as n → ∞. (3.48)
Moreover, it is easy to check the following estimate (see [START_REF]Global stabilization of a Korteweg-de Vries equation with saturating distributed control[END_REF]Lemma 3.2])

sat loc (a(x)λ n v n (t, x)) λ n L 2 (0,T ;L 2 (0,L)) ≤ C ∥v n ∥ L 2 (0,T ;H 1 (0,L)) . (3.49)
Like previous section, we have that

∂ t v n = -∂ x v n + ∂ 3 x v n -∂ 5 x v n + λ n v np ∂ x v n + sat(a(x)λnv n (t,x))
λn is bounded in L 2 (0, T ; H -3 (0, L)). Thanks to the Lions-Aubin lemma with the embedding H 2 (0, L) ⊂ L 2 (0, L) ⊂ H -3 (0, L), {v n } n∈N is relatively compact in L 2 (0, T ; L 2 (0, L)). It follows that there exists a subsequence, again denoted by {v n } n∈N such that v n → v in L 2 (0, T ; L 2 (0, L)). Moreover with (3.48) we have ak

(i)v(t, x) = 0, ∀ x ∈ (0, L), ∀ t ∈ Ω c i and ∂ 2 x v(t, 0) = 0, ∀ t ∈ (0, T ). As k(i) is strictly positive, we further have v(t, x) = 0, ∀ x ∈ ω, ∀ t ∈ ∪ i∈N Ω c i and ∂ 2 x v(t, 0) = 0, ∀t ∈ (0, T ). As µ(∪ i∈N Ω c i ) = T, we have v(t, x) = 0, ∀x ∈ ω, ∀ t ∈ ∪ i∈N Ω c i and ∂ 2
x v(t, 0) = 0, ∀ t ∈ (0, T ). This implies that v(t, x) = 0, ∀ x ∈ ω with almost every t ∈ (0, T ). We also obtain that the limit v solves the following equation in the sense of distribution

     ∂ t v + ∂ x v + ∂ 3 x v -∂ 5 x v + λv p ∂ x v = 0, (t, x) ∈ (0, ∞) × (0, L), v(t, 0) = v(t, L) = ∂ x v(t, 0) = ∂ x v(t, L) = ∂ 2 x v(t, L) = 0, t ∈ (0, ∞), ∥v∥ L 2 (0,T ;L 2 (0,L)) = 1. (3.50)
Using the unique continuation principle Lemma 3.6 for Kawahara equation we have v(t, x) = 0, ∀ x ∈ (0, L). Which is a contradiction. Hence the proof follows. □

Conclusion and final comments

In this work the exponential stabilization of the nonlinear Kawahara equation was addressed. The control was considered acting internally or in the boundary. In both cases, a global well-posedness result was obtained. In the case of boundary saturation a local exponential stability result was derived while for internal saturated actuator a semiglobal exponential stability result was shown. The main strategies in the proof of those results were observability inequalities, contradiction arguments and compactness ideas. We finish this work with several comments and remarks, in order: 4.1. Sign of the first-order derivative. One can wonder if our results are still valid if we modify slightly our systems. This, extension is in fact not true in all the cases. To fix ideas, consider the following equation

     ∂ t u -∂ x u + ∂ 3 x u -∂ 5 x u + u p ∂ x u = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, ∂ 2 x u(t, L) = sat loc (α∂ 2 x u(t, 0)), t ∈ (0, ∞), u(0, x) = u 0 (x),
x ∈ (0, L). (4.1) Observe that (4.1) it is almost the same equation than (1.6), the unique difference is the sign of the first order derivative. If we reply the arguments Theorem 3.3 to show exponential stability, we are ask to analyze the solutions of

         ∂ t u -∂ x u + ∂ 3 x u -∂ 5 x u = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = 0, t ∈ (0, ∞), ∂ x u(t, 0) = ∂ x u(t, L) = 0, t ∈ (0, ∞), ∂ x u(t, 0) = ∂ 2 x u(t, 0) = ∂ 2 x u(t, L) = 0, t ∈ (0, ∞). (4.2)
For this system, it was shown in [START_REF] Araruna | Energy decay for the modified Kawahara equation posed in a bounded domain[END_REF], the existence of a critical set N , such that if L ∈ N , there exists a steady-state solution of (4.2). Therefore, the argument does not follow. In the case of the KdV equation similar phenomena appears, but for the KdV equation studied in [START_REF] Rosier | Exact boundary controllability for the Korteweg-de Vries equation on a bounded domain[END_REF] the set of critical lengths is totally know, that allows in [START_REF] Parada | Stability analysis of a Korteweg-de Vries equation with saturated boundary feedback[END_REF] to show the exponential stability with saturated boundary control if the length is not critical. By contrast, for (4.2) it is not known if the set N represents all the critical lengths or just some of them. With this in mind, the exponential stability with boundary saturation holds as long the following unique continuation property is true for give L > 0

Assumption 4.1. If λ, ψ ∈ C × H 5 (0, L) satisfies      λψ(x) -ψ ′ (x) + ψ ′′′ (x) -ψ ′′′′′ (x) = 0, x ∈ (0, L) ψ(0) = ψ(L) = ψ ′ (0) = ψ ′ (L) = 0, ψ ′′ (0) = ψ ′′ (L) = 0, then ψ ≡ 0.
Now we consider the same equation, but we drop the first order term; i.e.

         ∂ t u + ∂ 3 x u -∂ 5 x u = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = 0, t ∈ (0, ∞), ∂ x u(t, 0) = ∂ x u(t, L) = 0, t ∈ (0, ∞), ∂ x u(t, 0) = ∂ 2 x u(t, 0) = ∂ 2 x u(t, L) = 0, t ∈ (0, ∞). (4.3)
In [START_REF] Vasconcellos | Stabilization of the linear Kawahara equation with localized damping[END_REF][START_REF] Vasconcellos | Stabilization of the linear Kawahara equation with localized damping[END_REF], it was shown that the set of critical lengths related to this equation it is empty. Therefore, we expect that our stability result by saturated boundary feedback is still valid in this context. In the case of internal saturation the situation is better. Consider the following system

     ∂ t u + η∂ x u + ∂ 3 x u -∂ 5
x u + u p ∂ x u + sat(a(x)u(t, x)) = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, ∂ 2

x u(t, L) = 0, t ∈ (0, ∞), u(0, x) = u 0 (x),

x ∈ (0, L),

where η = 0, -1 and a ∈ L ∞ (0, L) satisfies (1.5). Straightforward modifications of Theorem 2.11 allow us to show that there exists a unique mild solution u ∈ C([0, ∞), L 2 (0, L)) ∩ L 2 (0, ∞), H 2 (0, L) of (4.4). With respect to the stabilization, following the proof of Theorem 3.4 we are asking to prove similar unique continuation principle as Lemma 3.6, for a solution of

         ∂ t u + η∂ x u + ∂ 3 x u -∂ 5
x u + u p ∂ x u = 0, (t, x) ∈ (0, ∞) × (0, L), u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = 0, ∂ 2

x u(t, L) = 0, t ∈ (0, ∞), u(t, x) = 0, (t, x) ∈ (0, ∞) × ω, u(0, x) = u 0 (x),

x ∈ (0, L).

In particular, we can write

∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u + u p ∂ x u = (1 -η)∂ x u.
Then, we can apply the Carleman estimate (B.1) and absorb the remaining right hand-side terms by the left hand side terms. From where we can conclude the semiglobal exponential stability of (4.4).

4.2. More general nonlinearities. In the introduction of this work, we mention that in [START_REF] Araruna | Energy decay for the modified Kawahara equation posed in a bounded domain[END_REF] the internal stabilization was studied in the case p ∈ [1, 4), but a smallness condition in the initial data was required for p ∈ [START_REF] Boyd | Weakly non-local solitons for capillary-gravity waves: fifth-degree korteweg-de vries equation[END_REF][START_REF] Capistrano-Filho | Massera's theorems for a higher order dispersive system[END_REF]. This is one of the main reasons why we only considered the case p ∈ [1, 2). In fact, if the initial data is too small, then by continuity, the solution remains small, which could imply that sat(au) = au, then the system is not really saturated. One can prove the above result in the same spirit of [START_REF] Rosier | Global stabilization of the generalized Korteweg-de Vries equation posed on a finite domain[END_REF]. Let us consider a function ψ ∈ C ∞ (0, 1) with ψ > 0 in (0, 1), ψ(0) = ψ(1) = 0, ∥ψ∥ L ∞ (0,1) = 1, |ψ x | > 0 in [0, 1] \ ω, ψ x (0) > 0, ψ x (1) < 0. For any ϵ > 0, we consider a(t, x) = e µ(ψ(x)+3) -e 5µ (t -ϵ)(T -t) , ξ = e λa , φ = e µ(ψ(x)+3) (t -ϵ)(T -t) , t ∈ (ϵ, T ). x u 2 + s 3 φ 3 ξ 2 ∂ Finally we have the following:

T ϵ L 0 sφξ 2 ∂ 4 x u 2 + s 3 φ 3 ξ 2 ∂ 3 x u 2 + s 5 φ 5 ξ 2 ∂ 2 x u 2 + s 7 φ 7 ξ 2 ∂ x u 2 + s 9 φ 9 ξ 2 u 2 ≤ C T ϵ ω sφξ 2 ∂ 4 x u 2 + s 9 φ 9 ξ 2 u 2 .
Therefore u = 0 in (0, T ) × ω implies that u = 0 in (ϵ, T ) × (0, L) for all ϵ > 0. As a result we get have u = 0, (t, x) ∈ (0, T ) × (0, L).

. 9 )• 3 . 2 . 1 .

 9321 For s ∈ L ∞ (0, L) with |s(x)| ≤ r, we have(sat loc (a(x)s(x)) -k(r)a(x)s(x)) s(x) ≥ 0, ∀x ∈ [0, L].(3.10) Proof of Theorem 3.4 with ω = [0, L] and sat = sat 2 . In this case, the proof is direct. Indeed, thanks to the estimate (2.35)(x)u(t, x))u(t, x)dx.(3.11)

2 x 2 T 0 L 0 sat

 2200 u n (t, 0)| 2 dt + loc (au(t, x))u(t, x)dxdt = +∞.(3.37)

5 .

 5 Acknowledgment. Subrata Majumdar received partial financial support from the Department of Atomic Energy and National Board for Higher Mathematics Fellowship, Grant No. 0203/16(21)/2018-R&D-II/10708, during the initial stage of this work. Presently, this research is supported by institute post-doctoral fellowship of IIT Bombay. Hugo Parada received partial financial support from the Project IRS CONSPIR and the post-doctoral fellowship of Labex CIMI under grant agreement ANR-11-LABX-0040.

  Now we will use a global Carleman estimate [13, Proposition 1] for Kawahara equation. Let us denote the following spaceU ϵ = u ∈ L 2 (ϵ, T ; H 5 (0, L))| u(t, 0) = u(t, L) = ∂ x u(t, 0) = ∂ x u(t, L) = ∂ 2 x u(t, L) = 0, P u = ∂ t u + ∂ x u + ∂ 3 x u -∂ 5 x u ∈ L 2 (ϵ, T ; L 2 (0, L)) Proposition B.2.There exists positive constants µ 0 > 1, C 0 > 0 and C 1 > 0 such that for µ = µ 0 and for every λ ≥ C 0 (T + T 2 ), for all u ∈ U ϵ , we have

0 sφξ 2 ∂ 4 x u 2 + s 3 φ 3 ξ 2 ∂ 3 x u 2 + s 5 φ 5 ξ 2 ∂ 2 x u 2 + s 7 φ 7 ξ 2 ∂ x u 2 + s 9 φ 9 ξ 2 u 2 ≤ C T ϵ ω ξ 2 s 7 φ 7 ξ 2 ∂ x u 2 .ϵ ω ξ 2 s 7 φ

 0222222222227 H 5 (0, L)) ∩ L 2 (t 3 , T ; H 7 (0, L)). Therefore, we eventually have u ∈ C([ϵ, T ]; H 5 (0, L)) ∩ L 2 (ϵ, T ; H 7 (0, L)). Thus we can use (B.1) for the solution u as u ∈ U ϵ .Plugging P u = u p ∂ x u in the Global Carleman estimate (B.1), we haveT ϵ L |∂ x uu p | 2 + sφξ 2 ∂ 4 x u 2 + s 9 φ 9 ξ 2 u 2 . (B.2)Note that u ∈ L ∞ ((ϵ, T ) × (0, L)). Therefore one can find a large constantC 1 = ∥u∥ p L ∞ ((ϵ,T )×(0,L))and consider s large enough so we can absorb the last term in the right hand sides by the term 1 Indeed chosse s ≥ C 1 (T + T 2 ) we haveC T |u p ∂ x u| 2 ≤ 7 ξ 2 ∂ x u 2 .

  3 x u 2 + s 5 φ 5 ξ 2 ∂ 2 x u 2 + s 7 φ 7 ξ 2 ∂ x u 2 + s 9 φ 9 ξ 2 u 2 |P u| 2 + sφξ 2 ∂ 4 x u 2 + s 9 φ 9 ξ 2 u 2 . (B.1) B.1. Proof of Lemma 3.6. As, u ∈ L 2 (0, T ; H 2 (0, L)), for any ϵ ∈ (0, T ), there exists a candidate t 1 ∈ (0, ϵ) such that u(t 1 , •) ∈ H 2 (0, L). Using Proposition B.1, we get that u ∈ C([t 1 , T ]; H 2 (0, L)) ∩ L 2 (t 1 , T ; H 4 (0, L)). Similar argument gives a time t 2 ∈ (t 1 , ϵ) such that u(t 1 , •) ∈ H 4 (0, L). Again from Proposition B.1, it follows that that u ∈ C([t 2 , T ]; H 4 (0, L))∩L 2 (t 2 , T ; H 6 (0, L)). Proceeding with similar argument we finally have t 3 ∈ (t 2 , ϵ) such that u ∈ C([t 3 , T ];

	T
	≤ C	ξ 2
	ϵ	ω

Appendix A. Proof of Lemma 2.2 In this part, we prove that the operator T is well-defined. Indeed, take ψ defined by

We can easily observe that ψ(0

which is equivalent to (A 0 b -λI)φ = -f . Now, let (λ, y) a pair eigenvalue, eigenfunction of A 0 b , i.e A 0 b y = λy, in particular we have -y ′ -y ′′′ + y ′′′′′ = λy, -y ′ -y ′′′ + y ′′′′′ = λy.

We multiply the first equation by y, the second one by y and we integrate over (0, L), we get after some integration by parts

By adding these two equations, we deduce

Then as λ > 0, we get λ / ∈ σ(A 0 b ), thus A 0 b -λI is invertible and finally, φ = -(A 0 b -λI) -1 f , is well defined and also z.

This section is devoted to the proof of the unique continuation principle for the Kawahara equation with interior observation. Let us first recall the following existence result of the Kawahara equation in a regular set-up.

Proposition B.1. Let us assume that u 0 ∈ H 5 (0, L). Then the equation (3.13) with u(0, •) = u 0 possesses a unique solution u ∈ C([0, T ], H 5 (0, L))∩L 2 (0, T ; H 7 (0, L)). Also by interpolation, if u 0 ∈ H s (0, L), then the equation (3.33) possesses a unique solution u ∈ C([0, T ], H s (0, L)) ∩ L 2 (0, T ; H s+2 (0, L)), 0 ≤ s ≤ 5.


Leila Amgoud
email: amgoud@irit.fr

Mathieu Serrurier
email: serrurier@irit.fr

Arguing and Explaining Classifications

Keywords: I.2.3 [Deduction and Theorem Proving]: Nonmonotonic reasoning and belief revision ; I.2.11 [Distributed Articial Intelligence]: Intelligent agents Human Factors, Theory Argumentation, Classication

Argumentation is a promising approach used by autonomous agents for reasoning about inconsistent knowledge, based on the construction and the comparison of arguments. In this paper, we apply this approach to the classication problem, whose purpose is to construct from a set of training examples a model (or hypothesis) that assigns a class to any new example.

We propose a general formal argumentation-based model that constructs arguments for/against each possible classication of an example, evaluates them, and determines among the conicting arguments the acceptable ones. Finally, a \valid" classication of the example is suggested. Thus, not only the class of the example is given, but also the reasons behind that classication are provided to the user as well in a form that is easy to grasp.

We show that such an argumentation-based approach for classication oers other advantages, like for instance classifying examples even when the set of training examples is inconsistent, and considering more general preference relations between hypotheses. Moreover, we show that in the particular case of concept learning, the results of version space theory are retrieved in an elegant way in our argumentation framework.

INTRODUCTION

Argumentation has become an Articial Intelligence keyword for the last fteen years, especially in sub-elds such as non monotonic reasoning, inconsistency-tolerant reasoning, multiple-source information systems [START_REF] Amgoud | Inferring from inconsistency in preference-based argumentation frameworks[END_REF][START_REF] Prakken | Argument-based extended logic programming with defeasible priorities[END_REF][START_REF] Simari | A mathematical treatment of defeasible reasoning and its implementation[END_REF][START_REF] Chesñevar | Integrating defeasible argumentation with fuzzy art neural networks for pattern classication[END_REF]. Argumentation follows basically three steps: i) to construct arguments and counter-arguments for a statement, ii) to select the \acceptable" ones and, nally, iii) to determine whether the statement can be accepted or not. This paper claims that argumentation can also be used as an alternative approach for the problem of classication. Classication aims at building models that describe a concept from a set of training examples. The models are intended to be suciently general in order to be reused on new examples. When the concept to learn is binary, i.e. examples of that concept can be either true or false, the problem is called concept learning.

In our argumentation-based approach, the classication problem is reformulated as follows: given a set of examples (the training ones, and/or additional examples) and a set of hypotheses, what should be the class of a given example? To answer this question, arguments are constructed in favor of all the possible classications of that example. A classication can come either from an hypothesis, or from a training example. The obtained arguments may be conicting since it may be the case that the same example is aected to dierent classes. Finally, a \valid" classication of the example is suggested. Thus, not only the class of the example is given, but also the reasons behind that classication are provided to the user as well in a form that is easy to grasp.

We show that such an argumentation-based approach for classication oers other advantages, like for instance classifying examples even when the set of training examples is inconsistent, and considering more general preference relations between hypotheses. Moreover, we show that in the particular case of concept learning, the results of the version space theory developed by Mitchell in [START_REF] Mitchell | Generalization as search[END_REF] are retrieved in an elegant way in our argumentation framework. We show that the acceptability semantics dened in [START_REF] Dung | On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games[END_REF] allow us to identify and characterize the version space as well as its lower and upper bounds. In sum, this paper proposes a formal theoretical framework for handling, analysing and explaining the problem of classication. The framework presents the following features that make it original and exible: 2. it allows one to reason directly on the set of hypotheses;

3. examples are classied on the basis of the whole set of hypotheses rather than only one hypothesis as it is the case in standard classication models. Indeed, in the standard approach, a unique hypothesis is chosen, and all the examples are classied on the basis of that hypothesis.

4. it presents several original and intuitive decision criteria for choosing the class of an example.

The paper is organized as follows. We rst present the classication problem, then we introduce the basic argumentation framework of Dung [START_REF] Dung | On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games[END_REF]. The third section introduces our argumentation-based framework for classication as well as its properties. = c2. Otherwise, is said to be inconsistent. Regarding the hypothesis space r, it may be, for instance, decision trees, propositional sets of rules, neural nets, etc.

CLASSIFICATION PROBLEM

An hypothesis h is a mapping from to g (i.e. h: U 3 g). Before dening the output of the framework, let us rst introduce a key notion, that of soundness. Definition 2 (Soundness). Let h P r. An hypothesis h is sound with respect to a training example (x; c) P i h(x) = c. If V(xi; ci) P , h is sound w.r.t (xi; ci), then h is said to be sound with . The general task of classication is to identify an h P r that is sound with respect to the training examples. This hypothesis will be next used for classifying new examples. The most common approach for identifying this hypothesis is to use a greedy exploration of the hypothesis space, guided by a preference relation on hypothesis. This preference relation may be encoded by a utility function or by a syntactic or a semantic relation. Utility functions are generally based on the accuracy of the hypothesis (proportion of well classied examples) weighted by some complexity criteria (number of rules, etc.). Utility functions encode usually a total order. Syntactic relations may be for instance entailment or subsumption in the logical case. In this case it encodes a partial preorder on r.

Example 1 (Learning the concept sunny day). In this example, the feature space is a pair (attribute, value). Three attributes are considered: pressure, temperature, and humidity. The concept to learn is supposed to be binary, thus g = f0; 1g. Four Let us suppose that the hypothesis space r is the space of constraints on the values of each attribute. Indeed, the constraints are conjunctions of accepted values of attributes.

The special constraint Y (resp. ?) means that no (resp. all) values of attributes are accepted. If a vector of values of attributes match all the constraints, then it is considered as a positive example, otherwise it is a negative one. The hypotheses hY; Y; Yi and h?; ?; ?i are respectively the lower and the upper bound of the hypothesis space r.

ABSTRACT ARGUMENTATION FRAME-WORK

Argumentation is a reasoning model that follows the following steps:

1. Constructing arguments and counter-arguments.

2. Dening the strengths of those arguments.

3. Evaluating the acceptability of the dierent arguments.

4. Concluding or dening the justied conclusions.

In [START_REF] Dung | On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games[END_REF], an argumentation system is dened as follows: Definition 3 (Argumentation system). An argumentation system (AS) is a pair he, i. e is a set of arguments and e ¢ e is a defeat relation. We say that an argument A defeats an argument B i (A; B) P (or A B).

Note that to each argumentation system is associated an oriented graph whose nodes are the dierent arguments, and the edges represent the defeasibility relationship between them. Among all the conicting arguments, it is important to know which arguments to keep for inferring conclusions or for making decisions. In [START_REF] Dung | On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games[END_REF], dierent semantics for the notion of acceptability have been proposed. Let us recall them here. Definition 4 (Conflict-free, Defence). Let f e. f is conict-free i there exist no Ai, Aj P f such that Ai Aj . f defends an argument Ai i for each argument Aj P e, if Aj Ai, then there exists A k P f such that A k Aj .

Definition 5

(Acceptability semantics). Let f be a conict-free set of arguments, and let p: 2 A U 3 2 A be a function such that p(f) = fA j f defends Ag.

f is a complete extension i f = p(f). f is a grounded extension i it is the minimal (w.r.t.

set-inclusion) complete extension.

f is a preferred extension i it is a maximal (w.r.t.

set-inclusion) complete extension.

f is a stable extension i it is a preferred extension that defeats all arguments in enf. Let fi1, : : :, ing be the set of all possible extensions under a given semantics.

Note that there is only one grounded extension which may be empty. It contains all the arguments that are not defeated, and also the arguments which are defended directly or indirectly by non-defeated arguments. The last step of an argumentation process consists of determining, among all the conclusions of the dierent arguments, the \good" ones, called justied conclusions.

AN ARGUMENTATION FRAMEWORK FOR CLASSIFICATION

The aim of this section is to propose an instantiation of the general and abstract framework of Dung that allows the classication of examples. Throughout this section, we will consider a features space , a concept space g = fc1; : : : ; cng, a (maybe inconsistent) set of m > 0 training examples, a hypotheses space r that is equipped with an arbitrary preference relation #. Thus, # r ¢ r, and r is supposed to be a partial preorder.

In order to instantiate the abstract framework of Dung, one needs to dene the set e of arguments as well as the defeat relation between those arguments.

In our particular application, one needs to argue about particular classications, thus arguments are constructed in favor of assigning particular classes from g to an example in . Indeed, an argument in favor of the pair (x; c) represents the reason for assigning the class c to the example x. Two reasons can be distinguished: eS T = Y.

Proof. The rst point follows from the above denition, and from the fact that an hypothesis h cannot be empty. The second point follows directly from the rst property, i.e. jj = jeSj, and the assumption that T = Y.

Let us illustrate the notion of argument through example 1.

Example 2. In example 1, there are exactly four arguments with an empty support, and they correspond to the training examples: e ; = fa1 = hY, (pressure, low) (temperature, medium) (humidity,high), 0i, a2 = hY, (pressure, medium) (temperature, medium) (humidity, low), 1i, a3 = hY, (pressure, low) (temperature, medium) (humidity, medium), 0i, a4 = hY, (pressure, medium) (temperature, high) (humidity, medium), 1ig. There are also arguments with a nonempty support such as: ha5 = h ? , medium high, ?i, (pressure, low) (temperature, high) (humidity, high), 1ig, a6 = hh medium high, ?, ?i, (pressure, low) (temperature, high) (humidity, high), 0i, a7 = hh medium, medium high, ?i, (pressure, low) (temperature, high) (humidity, high), 0i.

In [START_REF] Amgoud | Inferring from inconsistency in preference-based argumentation frameworks[END_REF][START_REF] Prakken | Argument-based extended logic programming with defeasible priorities[END_REF][START_REF] Simari | A mathematical treatment of defeasible reasoning and its implementation[END_REF], it has been argued that arguments may have dierent strengths depending on the quality of information used to construct them. In [START_REF] Simari | A mathematical treatment of defeasible reasoning and its implementation[END_REF], for instance, arguments built from specic information are stronger than arguments built from more general ones. In our particular application, it is clear that arguments with an empty support are stronger than arguments with a non-empty one. This reects the fact that classications given by training examples take precedence over ones given by hypotheses in r. It is also natural to consider that arguments using most preferred hypothesis are stronger than arguments with less preferred ones. Definition 7 (Comparing arguments). Let hh; x; ci, hh 0 ; x 0 ; c 0 i be two arguments of e. hh; x; ci is preferred to hh 0 ; x 0 ; c 0 i, denoted by hh; x; ci Pref hh 0 ; x 0 ; c 0 i, i: h = Y and h 0 T = Y, or h # h 0 .

Property 2. The relation Pref is a partial preorder.

Proof. This is due to the fact that the relation # is a partial preorder.

Now that the set of arguments is dened, it is possible to dene the defeasibility relation between arguments in e.

Here again, there are two ways in which an argument A can attack another argument B:

1. by rebutting its conclusion. This situation occurs when the two arguments have contradictory conclusions, i.e. the same example is classied in dierent ways.

2. by undercutting its support. (Rebutting). Let hh; x; ci, hh 0 ; x 0 ; c 0 i be two arguments of e. hh; x; ci rebuts hh 0 ; x 0 ; c 0 i i x x 0 , c T = c 0 . Example 3. In example 2, we have for instance : a5 rebuts a6, a5 rebuts a7, a6 rebuts a5, and a7 rebuts a5. Definition 9 (Undercutting). Let hh; x; ci, hh 0 ; x 0 ; c 0 i be two arguments of e. hh; x; ci undercuts hh 0 ; x 0 ; c 0 i i h = Y and h 0 (x) T = c. Example 4. In example 2, we have for instance : a1 undercuts a5, a2 undercuts a5, a3 undercuts a5, and a4 undercuts a5.

Note that the rebutting and undercutting relations are used in most argumentation systems that handle inconsistency in knowledge bases. Property 3. If is consistent, then @ A, B P eS such that A rebuts B, or A undercuts B.

Proof. Let A = hY; x; ui, B = hY; x 0 ; u 0 i P such that A rebuts B. According to Denition 8, x x 0 and u T = u 0 . This contradicts the fact that is consistent.

The two above conict relations are brought together in a unique relation, called \Defeat". Definition 10 (Defeat). Let A = hh; x; ci, B = hh 0 ; x 0 ; c 0 i be two arguments of e. A defeats B i:

1. A rebuts (resp. undercuts) B, and 2. (A Pref B), or (not(A Pref B) and not(B Pref A))

Example 5. With the argument dened in ex. 2 we have for instance : a1 defeats a5, a2 defeats a5, a3 defeats a5, a4 defeats a5, a5 defeats a6, a5 defeats a7 and a6 defeats a5.

From the above denition, it is easy to check that an argument with an empty-support cannot be defeated by an argument with a non-empty support. Property 4. V A P eS, @ B P eneS s.t B defeats A. Proof. Let A P eS and B P eneS such that B defeats A. This means that B rebuts A (because according to Denition 9, an argument with a non-empty support cannot undercut an argument with an empty one. Moreover, according to Denition 10, we have either B Pref A, or (not(B Pref A) and not(A Pref B)). This is impossible because according to Denition 7, arguments in eS are always preferred to arguments with a non-empty support.

The argumentation system for classication is then the following: Definition 11 (Argumentation system). An argumentation system for classication (ASC) is a pair he; def eati, where e is the set of arguments (see Denition 6) and defeat is the relation dened in Denition 10.

Let us now identify the acceptable arguments of the above ASC. It is clear that the arguments that are not defeated at all will be acceptable. Let denote that set of undefeated arguments.

Proposition 1. If is consistent, then eS . Proof. Let A P eS. Let us assume that WB P e such that B defeats A. According to Property 4, B = P eneS. Thus, B P eS. Moreover, B defeats A means that B rebuts A. This means then that A classies a training example in u, and B classies an equivalent example in u 0 T = u. This contradicts the fact that the set is consistent.

As said in Section 3, one of the acceptability semantics is the so-called `grounded extension'. Such an extension is unique and maybe empty. However, we show that when the set of training examples is consistent, this grounded extension is not empty. Proposition 2 (Grounded extension). If is consistent, then the argumentation system he; def eati has a non empty grounded extension i.

Proof. This is due to the fact that eS T = Y and eS . Note that the system he; def eati is not always nite. By nite we mean that each argument is defeated by a nite number of arguments. This is due to the fact that r and are not always nite. Proposition 3. If r and are nite, then the system he; def eati is nite. When an argumentation system is nite, its characteristic function p is continuous. Consequently, the least xed point of this function can be dened by an iterative application of p to the empty set. Proposition 4. If the argumentation system he; def eati is nite, then the grounded extension i is:

i = [ p i0 (Y) = [ [ i1 p i ()]:
Let us now analyze the other acceptability semantics, namely preferred and stable ones. In general, the ASC has at least one preferred extension that may be empty. However, as for the case of grounded extension, we can show that in the particular case of a consistent set of training examples, the ASC has at least one non-empty preferred extension.

Proposition 5. If is consistent, then the ASC he; def eati has n ! 1 non-empty preferred extensions.

Proof. In [START_REF] Dung | On the acceptability of arguments and its fundamental role in nonmonotonic reasoning, logic programming and n-person games[END_REF], it has been shown that the grounded extension is included in very preferred extension. Since the grounded extension is not empty (according to Proposition 2, then there exists at least one non-empty preferred extension).

In general, the preferred extensions of an argumentation system are not stable. However, we can show that when the set g contains only two possible classes, this means that the concept to learn is binary, these extensions coincide. This result is due to the fact that the oriented graph associated to the above ASC has no odd length circuits in this case. However, it may contain circuits of even length. Proposition 6. If g = fc1; c2g with c1 T = c2, then:

The graph associated with the system he; def eati has no odd length circuits.

The preferred extensions and stable extensions of the system he; def eati coincide. Part 2: This is a consequence of the fact that there is no odd circuits in the system. Moreover, in this case the intersection of all the preferred (stable) extensions coincides with the grounded extension. Proposition 7. Let he; def eati be an ASC. Let i be its grounded extension, and i1; : : : ; in its preferred (stable) extensions. If g = fc1; c2g with c1 T = c2, then i = T i=1;:::;n ii.

The last step of an argumentation process consists of dening the status of the conclusions, in our case, the classi-

cation of examples.
In what follows we present two decision criteria: The rst one, called universal vote, consists of accepting those classications that are in any extension. However, it is clear that this kind of voting may not classify all the examples. Thus, we propose a second criterion, called majority vote, that allows to associate a class with each example. The conclusions here are the ones that are supported by a majority of arguments that appear in the dierent extensions. Formally:

Definition 12. Let he; def eati be a ASC, and i1; : : : ; in its extensions under a given semantics. Let x P and c P g. Universal vote: x is universally classied in c i Vii, W <h; x; c> P ii. U V denotes the set of all (x; c), such that x is universally classied in c.

Majority vote:

x is majoritarily classied in c i jf< h; x; c > jWii; < h; x; c >P iigj ! jf< h; x; c 0 > jc 0 T = c; Wii; < h; x; c 0 >P iigj. M V denotes the set of all (x; c), such that x is majoritarily classied in c.

The universally classied examples are those that are supported by arguments in all the extensions. From a classication point of view, these correspond to examples classied by the most preferred hypotheses. It is easy to check that the set of universally classied examples is included in the set of majoritarily classied ones.

Property 5. Let he; def eati be a ASC, and i1; : : : ; in its extensions under a given semantics :

U V M V
We can show that the above argumentation framework delivers \safe" results, since its sets of conclusions U V , M V are consistent. Formally: Proposition 8. Let he; def eati be a ASC, and U V , M V its sets of conclusions. The sets U V and M V are consistent.

RETRIEVING VERSION SPACE THEORY

As said before, concept learning is a particular case of classication, where the concept to learn is binary. In [START_REF] Mitchell | Generalization as search[END_REF], Mitchell has proposed the famous general and abstract framework, called version space learning, for concept learning. That framework takes as input a consistent set of training examples on the concept to learn. g contains only two classes, denoted respectively by 0 and 1. Thus, g = f0; 1g. The set r is supposed to be equipped with a \particular" partial preorder # that reects the idea that some hypothesis are more general than others in the sense that they classify positively more examples. This preorder denes a lattice on the hypothesis space. Formally:

Definition 13 (Generality order on hypothesis).

Let h1, h2 P r. h1 is more general than h2, denoted by h1 # h2, i fx P jh1(x) = 1g fx P jh2(x) = 1g.

The framework identies the version space, which is the set of all the hypothesis of r that are sound with . The idea is that a \good" hypothesis should at least classify the training examples correctly.

Definition 14 (Version space).

= fh P rj h is sound with g Version space learning aims at identifying the upper and the lower bounds of this version space . The upper bound will contain the most general hypothesis, i.e the ones that classify more examples, whereas the lower bound will contain the most specic ones, i.e the hypothesis that classify less examples.

Definition 15

(General hypotheses). The set of general hypothesis is G = fh P r j h is sound with and @ h 0 P r with h 0 sound with , and h 0 # hg.

Definition 16

(Specific hypotheses). The set of specic hypothesis is S = fh P r j h is sound with and @ h 0 P r with h 0 sound with , and h # h 0 g.

From the above denition, we have the following simple property characterizing the elements of . Property 6. [START_REF] Mitchell | Generalization as search[END_REF] = fh P rjWh1 P S; Wh2 P G; h2 # h # h1g

In [START_REF] Mitchell | Generalization as search[END_REF], an algorithm that computes the version space by identifying its upper and lower bounds S and G has been proposed.

The above framework has some limits. First, nding the version space is not sucient for classifying examples out of the training set. This is due to possible conicts between hypothesis. Second, it has been shown that the complexity of the algorithm that identies S and G is very high. In order to palliate that limit, learning algorithms try in general to reach only one hypothesis in the version space by using heuristical exploration of r (from general to specic exploration, for instance FOIL [START_REF] Quinlan | Learning logical denitions from relations[END_REF], or from specic to general exploration, for instance PROGOL [START_REF] Muggleton | Inverse entailment and Progol[END_REF]). That hypothesis is then used for classifying new objects. Moreover, it is obvious that this framework does not support inconsistent set of examples: Property 7. [START_REF] Mitchell | Generalization as search[END_REF] If the set is inconsistent, then the version space = Y.

A consequence of the above result is that no concept can be learned. This problem may appear in the case of noisy training data set. Let us now show how the above ASC can retrieve the results of the version space learning, namely the version space and its lower and upper bounds. Before doing that, we start rst by introducing some useful notations. Let Hyp be a function that returns for a given set of arguments, their non empty supports. In other words, this function returns all the hypothesis used to build arguments: Definition 17. Let T e.

Hyp(T ) = fh j W hh, x, ui P T and h T = Yg Now we will show that the argumentation-based model for concept learning computes in an elegant way the version space . Proposition 9. Let he; def eati be a ASC. Let i be its grounded extension, and i1; : : : ; in its preferred (stable) extensions. If the set is consistent then: Hyp(i) = Hyp(i1) = : : : = Hyp(in) = where is the version space.

Proof. Let ii be an extension under a given semantics. Hyp(ii) : Let h P Hyp(ii), then W hh; x; ui P ii.

Let us assume that W(xi; ui) P such that h(xi) T = ui. This means hY; xi; uii undercuts hh; x; ui (according to Denition 9). Consequently, hY; xi; uii defeats hh; x; ui. However, according to Property 1, hY; xi; uii P eS, thus hY; xi; uii P ii. Contradiction because ii is an extension, thus by denition it is conict-free.

Hyp(ii): Let h P , and let us assume that h = P Hyp(ii). Since h P , then V(xi; ui) P , h(xi) = ui (1) Let (x; u) P , thus h(x) = u and consequently hh; x; ui P e. Moreover, since h = P Hyp(i), then hh; x; ui = P E. Thus, W hh 0 ; x 0 ; u 0 i that defeats hh; x; ui.

Case 1: h 0 = Y. This means that hY; x 0 ; u 0 i undercuts hh; x; ui and h(x 0 ) T = u 0 Contradiction with (1).

Case 2: h 0 T = Y. This means that hh 0 ; x 0 ; u 0 i rebuts hh; x; ui. Consequently, x x 0 and u T = u 0 . However, since h P , then h is sound with . Thus, hY; x; ui defeats hh 0 ; x 0 ; u 0 i, then hY; x; ui defeats hh; x; ui. Since hY; x; ui P , then hh; x; ui P p(g) and consequently, hh; x; ui P ii. Contradiction.

The above result is of great importance. It shows that to get the version space, one only needs to compute the grounded extension.

We can also show that if a given argument is in an extension ii, then any argument based on an hypothesis from the version space that supports the same conclusion is in that extension. Formally:

Proposition 10. Let he; def eati be a ASC, and i1; : : : ; in its extensions under a given semantics. If < h; x; u >P ii, then Vh 0 P s.t. h 0 T = h if h 0 (x) = u then < h 0 ; x; u >P ii.

Using the grounded extension, one can characterize the upper and the lower bounds of the version space. The upper bound corresponds to the most preferred arguments (w.r.t Pref) of the grounded extension, whereas the lower bound corresponds to the less preferred ones.

Proposition 11. Let he; def eati be a ASC, and i its grounded extension. G = fh j W <h; x; u> P i s.t V <h 0 ; x 0 ; u 0 > P i, not (<h 0 ; x 0 ; u 0 > Pref <h; x; u>)g. S = fh j W <h; x; u> P i s.t V <h 0 ; x 0 ; u 0 > P i, not (<h; x; u> Pref <h 0 ; x 0 ; u 0 >)g.

Proof.

G = fh j W <h; x; u> P i s.t V <h 0 ; x 0 ; u 0 > P i, not (<h 0 ; x 0 ; u 0 > Pref <h; x; u>)g.

Let h P G, thus h P , and Vh 0 P , h # h 0 . Since h P , thus, h P Hyp(i), with i an extension. Then, Whh; x; ui P i. Since h # h 0 for any h 0 P , then h # h 0 for any h 0 P Hyp(i). Thus, hh; x; ui Pref hh 0 ; x 0 ; u 0 i, V hh 0 ; x 0 ; u 0 i P i. Let hh; x; ui P i such that V hh 0 ; x 0 ; u 0 i P i, and not(hh 0 ; x 0 ; u 0 iPrefhh; x; ui). Thus, h P Hyp(i), and V h 0 P Hyp(i), not(h 0 # h), thus h P G. S = fh j W <h; x; u> P i s.t V <h 0 ; x 0 ; u 0 > P i, not (<h; x; u> Pref <h 0 ; x 0 ; u 0 >)g. Let h P S, thus @h 0 P such that h # h 0 . Since h P S, then h P and consequently, h P Hyp(i). This means that Whh; x; ui P i. Let us assume that Whh 0 ; x 0 ; u 0 i P i such that hh; x; ui P ref hh 0 ; x 0 ; u 0 i, thus h # h 0 . Contradiction with the fact that h P S. Let hh; x; ui P i such that V hh 0 ; x 0 ; u 0 i P i, and not(hh; x; uiP ref hh 0 ; x 0 ; u 0 i), thus not(h # h 0 ). Since h P , and Vh 0 P , not(h # h 0 ), then h P S.

CONCLUSION

Recently, some researchers have tried to use argumentation techniques in machine learning [START_REF] Zabkar | Argument based machine learning in a medical domain[END_REF][START_REF] Mozina | Argument based rule learning[END_REF]. The basic idea behind their work is to improve existing algorithms in learning by providing arguments. However, they don't exploit the whole power of argumentation theory. This paper has proposed, to the best of our knowledge, the rst framework for classication that is completely argumentation-based, and that uses Dung's semantics.

This framework considers the classication problem as a process that follows four main steps: it rst constructs arguments in favor of classications of examples from a set of training examples, and a set of hypothesis. Conicts between arguments may appear when two arguments classify the same example in dierent classes. Once the arguments identied, it is possible to compare them on the basis of their strengths. The idea is that arguments coming from the set of training examples are stronger than arguments built from the set of hypothesis. Similarly, arguments based on most preferred hypothesis are stronger than arguments built from less preferred hypothesis. We have shown that acceptability semantics of the ASC retrieves and even characterizes the version space and its upper and lower bounds. Thus, the argumentation-based approach gives another interpretation of the version space as well as its two bounds in terms of arguments. We have also shown that when the set of training examples is inconsistent, it is still possible to classify examples. Indeed, in this particular case, the version space is empty as it is the case in the version space learning framework. A last and not least feature of our framework consists of dening the class of each example on the basis of all the hypothesis and not only one, and also to suggest four intuitive decision criteria for that purpose. A rst extension of this framework would be to explore the proof theories in argumentation that test directly whether a given argument is in the grounded extension without computing this last. This means that one may know the class of an example without exploring the whole hypothesis space.

1 .

 1 it handles i) the case of a consistent set of training examples; ii) the case of an inconsistent set of training examples; and iii) the case of an empty set of training examples;

  (according to Denition 9). Consequently, Example(A) Example(B) Example(C), and Value(A) T = Value(B), Value(B) T = Value(C). Due to the fact that = f0; 1g, we have Value(A) = Value(C). This contradicts the assumption that C rebuts A.

  1. (x; c) is a training example in , 2. there exists a hypothesis h P r that classies x in c. that from the above denition, for any training example (xi; ci) P , WhY; xi; cii P e. Let eS = fhY; x; ci P eg (i.e. the set of arguments coming from the training examples). Since the set of training examples is not empty, then eS is not empty as well. Property 1. Let be a set of training examples. jj = jeSj 1 .

	Definition 6 A = hh; x; ci such that: (Argument). An argument is a triplet
	1. h P r, x P , c P g
	2. If h T = Y, then c = h(x)
	3. If

h = Y, then (x; c) P h is called the support of the argument, and (x; c) its conclusion. Let Example(A) = x, and Class(A) = c. Let e be the set of arguments built from (r; ; g). Note

  This occurs when the support of B classies in a dierent way the example of the conclusion of A. However, this relation is only restricted to training examples. Indeed, only arguments built from training examples are allowed to undercut other arguments. The idea behind this is that training examples are the only, in some sense, certain information one has, and thus cannot be defeated by hypothesis. However, hypothesis have controversial status.

	Definition 8

  Proof Sketch.Part 1: Let A; B; C be three arguments such that A defeats B, B defeats C , and C defeats A. Case 1: Let us suppose that A P eS. According to Property 3, B P eneS. According to Property 4, C should be in eneS. Contradiction because according to Property 4, C cannot defeat A, which is in eS. Case 2: Let us suppose that A; B; C P eneS. This means that A rebuts B, B rebuts C , and C rebuts A