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Stabilization methods for the Korteweg-de Vries equation∗
Eduardo Cerpa†
Abstract
These notes are concerned with the stabilization of the Korteweg-de Vries equation,which is an important partial differential equation describing approximately long waves inwater of relatively shallow depth. Internal and boundary feedback control laws exponentiallystabilizing the system to the origin are designed. Three different tools are used: a dampingmethod, a Gramian-based approach and the Backstepping method.
Key words. Korteweg-de Vries equation, stabilization, internal control, boundary control
Contents
1 Introduction 21.1 Internal control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Boundary control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2.1 Controlling from the right . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Controlling from the left . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Preliminaries 52.1 Linear system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Nonlinear system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3 Internal control 93.1 Linear system on a noncritical interval . . . . . . . . . . . . . . . . . . . . . . . . 93.2 Linear system on a critical interval . . . . . . . . . . . . . . . . . . . . . . . . . . 103.3 Nonlinear system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
4 Boundary control from the right 124.1 Control design in the finite-dimensional case . . . . . . . . . . . . . . . . . . . . . 134.2 Control design in the infinite-dimensional case . . . . . . . . . . . . . . . . . . . . 154.3 Spectral analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.4 Ingham’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.5 Control design for our system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.6 Numerical simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
5 Boundary control from the left 235.1 Control design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245.2 Stability of the linear system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.3 Stability of the nonlinear system . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
6 Discussion 296.1 Internal control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.2 Boundary control from the right . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.3 Boundary control from the left . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
∗Lecture notes for a 3-hour course given by the author at Summer School of Automatic Control held in Grenoblein September 2012.†Departamento de Matematica, Universidad Tecnica Federico Santa Marıa, Casilla 110-V, Valparaıso, Chile
1
1 Introduction
In 1895 D. J. Korteweg and G. de Vries published the article [17] deriving the equation (up torescaling)
ut + uxxx + uux = 0, x ∈ R, t ≥ 0,
where u = u(t, x) models for a time t the amplitude of the water wave at position x. This nonlin-ear dispersive partial differential equation, named Korteweg-de Vries equation and abbreviatedas KdV, describes approximately long waves in water of relatively shallow depth.
In these notes, we are interested in control properties of the KdV equation posed on a boundedinterval. In this case, as suggested in [1], the extra term ux should be incorporated in the equationin order to obtain an appropriate model for water waves in a uniform channel when coordinatesx is taken with respect to a fixed frame. Thus, the equation considered here is
ut + ux + uxxx + uux = 0, x ∈ [0, L], t ≥ 0,
for some L > 0. From the nature of this equation (third-order in space variable), one boundarycondition at the left end-point and two boundary conditions at the right end-point have to beimposed. Here, the boundary conditions act on
u(t, 0), u(t, L), and ux(t, L).
We consider a control system where the state, at each time, is given by the solution of theKdV equation and where the control is some term in the equation. If the control is a termdistributed in a region of the domain (for instance, a source term), one calls that an internalcontrol. On the other hand, if the control is a term on a region of the boundary (for instance, aboundary condition), one calls that a boundary control.
A system is said to be asymptotically stable if the solution of the system without controlconverges as the time goes to infinity to a stationary solution of the partial differential equation.If this convergence holds with a control depending at any time on the state, one says that thesystem is stabilizable by means of a feedback control law. The input (control) depends on theoutput (a full or partial measure of the state) in a closed form. Some good references concerningthe control of partial differential equations are the review by Russell [33], and the books by Lions[22] and by Coron [8].
These notes are concerned first with the stabilization of KdV by considering an internalcontrol and homogeneous boundary data. We prove that this feedback control introduces adamping phenomena and forces the solutions of KdV to decay exponentially to zero in L2-norm.This part is based on the results obtained in the article [29]. Secondly, we consider the boundarystabilization problem. By applying two different methods, a Gramian-based approach and theBackstepping method, we design some exponentially stabilizing feedback laws acting at the end-points of the domain. In this part, we focus on the results in [6, 4].
For a complete revision on control results for the KdV equation, including controllability andstabilizability, we recommend the survey [32] and the tutorial [3].
1.1 Internal control
For some domains (0, L), there are solutions of the linear KdV equation whose energy does notdecay to zero as the time goes to infinity. Thus, it is not clear whenever the solutions of thenonlinear KdV equation go to zero. In order to avoid this phenomena, we add to the equationan internal control term F localized on a small subdomain of [0, L]
ut + ux + uxxx + uux = F.
The goal is to design a control which dissipates enough energy to force the decay of thesolutions in L2-norm. We look for a control feedback law in the form F = F (u). We considercontrols in the form F (u) = −au, where a ∈ L∞(0, L) satisfies{
a(x) ≥ a0 > 0, ∀x ∈ O,where O is nonempty open subset of (0, L).
(1)
2
In this way, we have to study the equation ut + ux + uxxx + au+ uux = 0,u(t, 0) = u(t, L) = ux(t, L) = 0,u(0, ·) = u0(·).
(2)
A natural strategy is to consider first the linearized equation around the origin ut + ux + uxxx + au = 0,u(t, 0) = u(t, L) = ux(t, L) = 0,u(0, ·) = u0(·),
(3)
and prove the exponential decay of its solutions.
Theorem 1 [29] Let L > 0 and a = a(x) satisfying (1). There exist C > 0 and ω > 0 such that
‖u(t, ·)‖L2(0,L) ≤ Ce−ωt‖u0‖L2(0,L), ∀t ≥ 0
for any solution of (3) with u0 ∈ L2(0, L).
Using a perturbative argument, a local version of this theorem for equation (2) is also givenin [29] by adding a smallness condition on the initial data.
Theorem 2 [29] Let L > 0 and a = a(x) satisfying (1). There exist C, r > 0 and ω > 0 suchthat
‖u(t, ·)‖L2(0,L) ≤ Ce−ωt‖u0‖L2(0,L), ∀t ≥ 0
for any solution of (2) with ‖u0‖L2(0,L) ≤ r.
The goal of Section 3 is to explain the proofs of Theorems 1 and 2.
Remark 3 Other alternative approach is dealing directly with the nonlinear system (2) withoutpassing by the linear one (3). Thus, the following semi-global stabilization result can be proven[29, 27]. Let L > 0, a = a(x) satisfying (1), and R > 0. There exist C = C(R) > 0 andω = ω(R) > 0 such that
‖u(t, ·)‖L2(0,L) ≤ Ce−ωt‖u0‖L2(0,L), ∀t ≥ 0
for any solution of (2) with ‖u0‖L2(0,L) ≤ R.The semi-global character comes from the fact that even if we are able to chose any radius R
for the initial data, the decay rate ω depends on R. This result has been first proved in [29] byassuming the extra condition
∃δ > 0, (0, δ) ∪ (L− δ, L) ⊂ O(4)
which has been removed by Pazoto in [27].
Remark 4 These results still hold for nonlinearities other than uux. For instance, in [31] theyconsider the nonlinearity upux with p < 4 (generalized KdV equation) and in [21] the nonlinearityu4ux (critical generalized KdV equation). Other feedback laws can also be considered. In [26]
they introduce the feedback control F (u) = (− d2
dx2u)−11O instead of F (u) = −au. Notice that wehave denoted by 1O the characteristic function of the subset O.
1.2 Boundary control
We also address the boundary stabilizability problem for the KdV equation on a bounded domain.The boundary here has two points, x = 0 and x = L, and surprisingly the control propertiesof the system are different depending on the location of the control inputs. For this reason, weapply different methods to control from the left and from the right. In both cases the goal is thesame: to design some feedback control laws exponentially stabilizing the system to the origin. Inaddition, our design allows to deal with the rapid stabilization problem: to build, for any ω > 0,a feedback law such that the closed-loop system has an exponential decay rate ω at least.
3
1.2.1 Controlling from the right
We consider a system with homogeneous Dirichlet boundary conditions where the control actson the Neumann boundary condition at the right endpoint. In [43] Urquiza has generalized toinfinite-dimensional control systems what Russell called the Bass method (see [34, pages 117-118]). We apply this method, well-adapted to time-reversible systems, to build and operatorFω : H1
0 (0, L) −→ R such that the following result holds.
Theorem 5 [6] Let ω > 0. The closed-loop system{ut + ux + uxxx = 0, u(0, ·) = u0,u(t, 0) = u(t, L) = 0, ux(t, L) = ux(t, 0) + Fω(u(t)),
is globally well posed in H10 (0, L). Moreover, the solutions decay to zero with an exponential rate
of 2ω, i.e.,∃C > 0,∀u0 ∈ H1
0 (0, L), ‖u(t, ·)‖H10≤ Ce−2ωt‖u0‖H1
0.
The goal of Section 4 is to prove Theorem 5.
Remark 6 Slemrod in [38] (internal control) and Komornik in [15] (boundary control) alsouse a Gramian-based approach. They were inspired by the method introduced independently byKleinman in [14] and by Luke in [25] in a finite-dimensional framework.
1.2.2 Controlling from the left
In this part, we consider a control acting on the left Dirichlet boundary condition. Given L > 0,the linear control system is{
ut + ux + uxxx = 0, u(0, ·) = u0,u(t, 0) = Kω, u(t, L) = 0, ux(t, L) = 0,
(5)
and the nonlinear one is{ut + ux + uxxx + uux = 0, u(0, ·) = u0,u(t, 0) = Kω, u(t, L) = 0, ux(t, L) = 0.
(6)
By using the backstepping method, we are able to find a feedback control lawKω : L2(0, L) −→R exponentially stabilizing the origin. The design is based on the linear system around the origin.The linear closed-loop system is exponentially stable.
Theorem 7 [4] For any ω > 0, there exist a feedback control law Kω = Kω(u(t, ·)) and D > 0such that
‖u(t, ·)‖L2(0,L) ≤ De−ωt‖u0‖L2(0,L), ∀t ≥ 0,(7)
for any solution of (5).
The same result is obtained, by using a perturbative argument, for the nonlinear KdV equa-tion. A smallness condition on the initial data appears.
Theorem 8 [4] For any ω > 0, there exist a feedback control law Kω = Kω(u(t, ·)), r > 0 andD > 0 such that
‖u(t, ·)‖L2(0,L) ≤ De−ωt‖u0‖L2(0,L), ∀t ≥ 0,(8)
for any solution of (6) satisfiying ‖u0‖L2(0,L) ≤ r.
In both cases the feedback law Kω is explicitly defined as follows
Kω(u(t, ·)) =
∫ L
0
k(0, y)u(t, y)dy,(9)
where the function k = k(x, y) will be characterized as the solution of a given partial differentialequation depending on ω. The proofs of Theorems 7 and 8 will be given in Section 5.
4
Remark 9 The backstepping method is very known as a ODE control method (see [18] and [8,Section 12.5]). The first extensions to PDE have appeared in [10] and [24]. Later on, Krstic andhis collaborators introduced a modification of the method by means of an integral transformationof the PDE. This invertible transformation map the original PDE into an asymptotically stableone. In this context, the first continuous backstepping designs were proposed for the heat equation[23, 40]. The applications to wave equation appeared later in [19, 42, 39]. An excellent startingpoint to get inside this method is the book [20] by Krstic and Smyshlyaev.
2 Preliminaries
In this section we state the well-posedness framework for equations considered in these notes.Some inequalities useful before are proven here.
2.1 Linear system
Looking at the linear control system ut + ux + uxxx = 0,u(t, 0) = u(t, L) = ux(t, L) = 0,u(0, ·) = u0,
(10)
we find the underlying spatial operator defined by
D(A) = {w ∈ H3(0, L)/w(0) = w(L) = w′(L) = 0},
A : w ∈ D(A) ⊂ L2(0, L) 7−→ (−w′ − w′′′) ∈ L2(0, L).
It can be proven that operator A is dissipative, which means that∫ L
0
wA(w)dx ≤ 0, ∀w ∈ D(A).
Its adjoint operator A∗, defined by
D(A∗) = {w ∈ H3(0, L)/w(0) = w(L) = w′(0) = 0},
A∗ : w ∈ D(A∗) ⊂ L2(0, L) 7−→ (w′ + w′′′) ∈ L2(0, L),
is also dissipative and therefore A generates a strongly continuous semigroup of contractions andthe following result follows (see [28] and [30]).
Proposition 10 For any u0 ∈ D(A), there exists a unique solution u ∈ C([0, T ];D(A)) ∩C1([0, T ];L2(0, L)) to (10).
Next, we obtain some useful inequalities in order to state the well-posedness framework withless regular data. Pick up a regular function q = q(t, x). By multiplying (10) by qu and integratingby parts we get after some computations
(11) −∫ s
0
∫ L
0
(qt + qx + qxxx)|u|2 dxdt+
∫ L
0
q(s, x)|u(s, x)|2 dx−∫ L
0
q(0, x)|u(0, x)|2 dx
+ 3
∫ s
0
∫ L
0
qx|ux|2 dxdt+
∫ s
0
q(t, 0)|ux(t, 0)|2 dt = 0.
By choosing q = 1 in (11), we get
(12)
∫ L
0
|u(s, x)|2 dx+
∫ s
0
|ux(t, 0)|2 dt =
∫ L
0
|u0(x)|2 dx.
5
From that we deduce
(13) maxs∈[0,T ]
∫ L
0
|u(s, x)|2 dx ≤∫ L
0
|u0(x)|2 dx,
which implies that the solution belongs to C([0, T ];L2(0, L)) provided that u0 ∈ L2(0, L). More-over, from (13) we get
‖u‖2L2(0,T ;L2(0,L)) ≤ T‖u0‖2L2(0,L),(14)
and from (12) we deduce that
(15)
∫ T
0
|ux(t, 0)|2 dt ≤∫ L
0
|u0(x)|2 dx,
which implies that ux(t, 0) ∈ L2(0, T ) provided that u0 ∈ L2(0, L). This is a hidden regularityproperty for this equation.
By choosing q = x and s = T in (11), we get
(16)
∫ T
0
∫ L
0
|u|2 dx dt+
∫ L
0
x|u0(x)|2 dx =
∫ L
0
x|u(T, x)|2 dx+ 3
∫ T
0
∫ L
0
|ux|2 dxdt
From (16) we obtain
(17) 3
∫ T
0
∫ L
0
|ux|2 dx dt ≤∫ T
0
∫ L
0
|u|2 + L
∫ L
0
|u0(x)|2 dx
and by using (14), we get
(18)
∫ T
0
∫ L
0
|ux|2 dx dt ≤(L+ T
3
)(∫ L
0
|u0(x)|2 dx
)
which implies that the solution belongs to L2(0, T ;H1(0, L)) provided that u0 ∈ L2(0, L).Furthermore, by choosing q = (T − t) in (11), we get
(19) T
∫ L
0
|u0(x)|2 dx ≤ T∫ T
0
|ux(t, 0)|2 dt+
∫ T
0
∫ L
0
|u|2 dt.
which will be useful later.By using the density of D(A) in L2(0, L) and inequalities (13), (15) and (18), we can extend
the notion of solution for less regular data to obtain the following.
Proposition 11 Let u0 ∈ L2(0, L). Then there exists a unique solution u ∈ C([0, T ];L2(0, L))∩L2(0, T ;H1(0, L)) of (10). Moreover, there exists C > 0 such that the solutions of (10) satisfy
‖u‖C([0,T ];L2(0,L)) + ‖u‖L2(0,T ;H1(0,L)) ≤ C‖u0‖L2(0,L)
and the extra trace regularity
‖ux(·, 0)‖L2(0,T ) ≤ ‖u0‖L2(0,L).
In order to be able to consider the nonlinear KdV equation we need a well-posedness resultwith a nontrivial right hand side.
Proposition 12 [30, Proposition 4.1] Let u0 ∈ L2(0, L) and f ∈ L1(0, T ;L2(0, L)). Then thereexists a unique solution u ∈ C([0, T ];L2(0, L)) ∩ L2(0, T ;H1(0, L)) of (10). Moreover, thereexists C > 0 such that the solutions of (10) satisfy
‖u‖C([0,T ];L2(0,L)) + ‖u‖L2(0,T ;H1(0,L)) ≤ C(‖u0‖2L2(0,L) + ‖f‖2L1(0,T ;L2(0,L))
)1/2.
6
Proof. Since previous results, we only have to consider the case u0 = 0. In addition, thesemigroup theory gives that the (unique) solution u belongs to C([0, T ];L2(0, L)) if the right-hand side belongs to L1(0, T ;L2(0, L)) and there exists a constant C > 0 such that
‖u‖C([0,T ];L2(0,L)) ≤ C‖f‖L1(0,T ;L2(0,L)).
The only thing we have to prove is that u ∈ L2(0, T ;H1(0, L)). In fact, there exists a constantC > 0 such that
‖u‖L2(0,T ;H1(0,L)) ≤ C‖f‖L1(0,T ;L2(0,L)).
To prove that, as in (16) we can write
2
∫ T
0
∫ L
0
xfu dxdt = −∫ T
0
∫ L
0
|u|2 dx dt+
∫ L
0
x|u(T, x)|2 dx+ 3
∫ T
0
∫ L
0
|ux|2 dxdt
and therefore
(20) 3
∫ T
0
∫ L
0
|ux|2 dxdt ≤∫ T
0
∫ L
0
|u|2 dx dt+ 2L
∫ T
0
∫ L
0
fu dxdt
≤ T‖f‖2L1(0,T ;L2(0,L)) + 2L
∫ T
0
‖f‖L2(0,L)‖y‖L2(0,L) ≤ (T + 2L)‖f‖2L1(0,T ;L2(0,L)),
which ends the proof.�As we have seen, the following space is very important in this regularity framework. We
define the space
B := C([0, T ], L2(0, L)) ∩ L2(0, T,H1(0, L))(21)
endowed with the norm
‖u‖B = maxt∈[0,T ]
‖u(t)‖L2(0,L) +
(∫ T
0
‖u(t)‖2H1(0,L)dt
)1/2
.
2.2 Nonlinear system
We want to consider the KdV equation. For that, the first step is to show that the nonlinearityuux can be considered as a source term of the linear equation.
Proposition 13 [30, Proposition 4.1] Let u ∈ L2(0, T ;H1(0, L)). Then uux ∈ L1(0, T ;L2(0, L))and the map u ∈ L2(0, T ;H1(0, L)) 7−→ uux ∈ L1(0, T ;L2(0, L)) is continuous.
Proof. Let u, z ∈ L2(0, T ;H1(0, L)). Denoting with the constant K the norm of the embeddingH1(0, L) ↪→ L∞(0, L), we have
(22) ‖uux − zzx‖L1(0,T ;L2(0,L)) ≤∫ T
0
‖(u− z)ux‖L2(0,L)dt+
∫ T
0
‖z(ux − zx)‖L2(0,L)dt
≤∫ T
0
‖u− z‖L∞(0,L)‖ux‖L2(0,L)dt+
∫ T
0
‖z‖L∞(0,L)‖ux − zx‖L2(0,L)dt
≤ K∫ T
0
‖u− z‖H1(0,L)‖u‖H1(0,L)dt+K
∫ T
0
‖z‖H1(0,L)‖u− z‖H1(0,L)dt
≤ K(‖u‖L2(0,T ;H1(0,L)) + ‖z‖L2(0,T ;H1(0,L)))‖u− z‖L2(0,T ;H1(0,L))
By taking z = 0 we see that uux ∈ L1(0, T ;L2(0, L)) provided that u ∈ L2(0, T ;H1(0, L)).From (22), we also get that the map u ∈ L2(0, T ;H1(0, L)) 7−→ uux ∈ L1(0, T ;L2(0, L)) iscontinuous.
�
7
Let us state the well posedness property proved by Coron and Crepeau in [9] for the followingnonlinear KdV equation ut + ux + ux + uux = f,
u(t, 0) = u(t, L) = ux(t, L) = 0,u(0, ·) = u0.
(23)
Proposition 14 [9, Proposition 14] Let L > 0 and T > 0. There exist ε > 0 and C > 0 suchthat, for every f ∈ L1(0, T, L2(0, L)) and u0 ∈ L2(0, L) such that
‖f‖L1(0,T,L2(0,L)) + ‖u0‖L2(0,L) ≤ ε,
there exists a unique solution of (23) which satisfies
‖u‖B ≤ C(‖f‖L1(0,T,L2(0,L)) + ‖u0‖L2(0,L)).(24)
Proof.Let f ∈ L1(0, T, L2(0, L)) and u0 ∈ L2(0, L) as in the theorem with ε to be chosen later.
Given z ∈ B, we consider the map M : B → B defined by M(z) = u where u is the solution of ut + ux + ux = f − zzx,u(t, 0) = u(t, L) = ux(t, L) = 0,u(0, ·) = u0.
Clearly u ∈ B is a solution of (23) if and only if u is a fixed point of the map M . FromProposition 12 and equation (22) we get the existence of a constant D such that
‖M(z)‖B ≤ D{‖f‖L1(0,T ;L2(0,L)) + ‖u0‖L2(0,L) + ‖z‖2B
}and
‖M(z1)−M(z2)‖B ≤ D(‖z1‖B + ‖z2‖B)(‖z1 − z2‖B)
We consider M restricted to the closed ball {z ∈ B/‖z‖B ≤ R} with R to be chosen later.
We can write‖M(z)‖B ≤ D
{ε+ ‖z‖2B
}and
‖M(z1)−M(z2)‖B ≤ 2DR(‖z1 − z2‖B)
If R, ε are small enough so that
R <1
2D, ε <
R
2D,
we can apply the Banach fixed point theorem and prove that a unique fixed point of M exists.�
Remark 15 The smallness condition given by ε in Proposition 14 can be removed in order toget a global well-posedness result. See [11, 7].
In addition, we have the following result whose proof we omit.
Proposition 16 [9, Proposition 15] Let T > 0 and let L > 0. There exists C > 0 such thatfor every (u01, u02) ∈ L2(0, L)2 and (f1, f2) ∈ L1(0, T, L2(0, L))2 for which there exist solutionsu1 = u1(t, x) and u2 = u2(t, x) of (23), one has the following estimates:∫ T
0
∫ L
0
|∂xu1(t, x)− ∂xu2(t, x)|2dxdt ≤ eC(1+‖u1‖2L2(0,T,H1(0,L))+‖u2‖2L2(0,T,H1(0,L))
)
·(‖f1 − f2‖2L1(0,T,L2(0,L)) + ‖u01 − u02‖2L2(0,L)
),
∫ L
0
|u1(t, x)− u2(t, x)|2dx ≤ eC(1+‖u1‖2L2(0,T,H1(0,L))+‖u2‖2L2(0,T,H1(0,L))
)
·(‖f1 − f2‖2L1(0,T,L2(0,L)) + ‖u01 − u02‖2L2(0,L)
),
for every t ∈ [0, T ].
8
3 Internal control
This section is devoted to the proof of Theorems 1, and 2. We define for each time the energy ofthe solution of our KdV equation as its L2-norm
E(t) =
∫ L
0
|u(t, x)|2 dx.
We are interested in the long-time behavior of the energy E(t). More precisely we want toprove the exponential decay of E(t) as t goes to infinity. Let us start considering the linearequation {
ut + ux + uxxx = 0, u(0, ·) = u0,u(t, 0) = u(t, L) = ux(t, L) = 0,
(25)
By performing integration by parts in the equation∫ L
0
(ut + ux + uxxx)u dx = 0
we get
d
dt
∫ L
0
|u(t, x)|2 dx = −|ux(t, 0)|2 ≤ 0.(26)
From here we see that the energy is not increasing but we can not be sure that it is decreasing.For instance, if L = 2π and u0 = (1− cos(x)), the solution is u(t, x) = (1− cos(x)) which satisfiesux(t, 0) = 0 for any t ≥ 0. Thus, in this case the energy is constant. The value L = 2π is not theonly length with this phenomena. In [30], Rosier found out all these numbers of critical values.More precisely, given T > 0, there exists a constant C > 0 such that the solutions of (25) satisfythe observability inequality
∀u0 ∈ L2(0, L), ‖ux(·, 0)‖L2(0,T ) ≥ C‖u0‖L2(0,L)(27)
if and only if L /∈ N where
(28) N :=
{2π
√k2 + k`+ `2
3; k, ` ∈ N∗
}.
First, we will see that the stability holds for the linear system on a noncritical domain (L /∈ N ).Second, by introducing some damping into the system we deal with the linear case on a criticaldomain (L ∈ N ). Finally, for the nonlinear system we will get a local result by applying aperturbative method.
3.1 Linear system on a noncritical interval
We consider system (3) in the no damping case a(x) = 0 and for a noncritical domain L /∈ N .By integrating (26) in time on (0, 1) and using (27) with T = 1 we obtain the existence of Csuch that∫ L
0
|u(1, x)|2 dx−∫ L
0
|u0(x)|2 dx = −∫ 1
0
|ux(s, 0)|2 ds ≤ − 1
C2
∫ L
0
|u0(x)|2 dx,
that implies ∫ L
0
|u(1, x)|2 dx ≤ C2 − 1
C2
∫ L
0
|u0(x)|2 dx.
Of course we also have∫ L
0
|u(t+ 1, x)|2 dx ≤ C2 − 1
C2
∫ L
0
|u(t, x)|2 dx,(29)
9
which gives the exponential decay to the origin of the solutions. Indeed, let k ≤ t ≤ k+ 1. From
(26), (29) and denoting γ := C2−1C2 < 1, we have
(30) E(t) ≤ E(k) ≤ γE(k − 1) ≤ γ2E(k − 2) ≤ · · · ≤ γkE(0)
=γk+1
γE(0) =
1
γe(k+1) ln(γ)E(0) ≤ 1
γe−t| ln(γ)|E(0)
which ends the proof of Thereom 1 with a = 0 in a noncritical domain.
3.2 Linear system on a critical interval
We consider here a critical domain (L ∈ N ) and therefore if there is no damping there are so-lutions of the linear system which do not decay to zero. In this way a dissipative mechanismis needed in this case. We consider an internal damping given by the term a(x)u in (3) witha = a(x) possibly localized in a small subdomain of (0, L).
Proof of Theorem 1.By performing integration by parts in the equation∫ L
0
(ut + ux + uxxx + au)u dx = 0
we get
d
ds
∫ L
0
|u(s, x)|2 dx = −|ux(s, 0)|2 −∫ L
0
a(x)|u(s, x)|2 dx ≤ 0(31)
and then by integrating on (0, t) we obtain∫ L
0
|u(t, x)|2 dx−∫ L
0
|u0(x)|2 dx = −∫ t
0
|ux(s, 0)|2 ds−∫ t
0
∫ L
0
a(x)|u(s, x)|2 dxds.
The same proof as before runs if we are able to prove that ∀T > 0,∃C > 0 such that
∀u0 ∈ L2(0, L), ‖ux(·, 0)‖2L2(0,T ) +
∫ T
0
∫ L
0
a(x)|u(s, x)|2 dxdt ≥ C2‖u0‖2L2(0,L).(32)
From direct computations (multiplying the equation by (T − t)u and integrating by parts aswe did to prove (19)), we obtain
‖u0‖2L2(0,L) ≤1
T‖u‖2L2(0,T ;L2(0,L)) + ‖ux(·, 0)‖2L2(0,T ) + 2
∫ T
0
∫ L
0
a(x)|u(t, x)|2 dxdt(33)
and therefore we will be done if we prove that there exists a constant K > 0 such that
K‖u‖2L2(0,T ;L2(0,L)) ≤ ‖ux(·, 0)‖2L2(0,T ) +
∫ T
0
∫ L
0
a(x)|u(t, x)|2 dxdt(34)
We proceed by contradiction by assuming that (34) does not hold, i.e. we assume that ∀K > 0there exists a solution u such that
K‖u‖2L2(0,T ;L2(0,L)) > ‖ux(·, 0)‖2L2(0,T ) +
∫ T
0
∫ L
0
a(x)|u(t, x)|2 dxdt(35)
By using this successively with K = 1/n, we obtain a sequence {un}n∈N of solutions such that‖un‖L2(0,T ;L2(0,L)) = 1 (if not, we could normalize. This is due to the linearity of the equation)and
1
n> ‖unx(·, 0)‖2L2(0,T ) +
∫ T
0
∫ L
0
a(x)|un(t, x)|2 dxdt(36)
10
From here we have that
‖unx(·, 0)‖L2(0,L) → 0,
∫ T
0
∫ L
0
a(x)|un(t, x)|2 dxdt→ 0, as n→∞
Moreover, from (33) we see that the corresponding initial conditions {un0}n∈N are bounded inL2(0, L).
The goal now is to pass to the limit and get a contradiction. By using different multipliers(as in the proof of (15)) we get that {un}n∈N are bounded in L2(0, T ;H1(0, L)) which impliesthat {unxxx}n∈N are bounded in L2(0, T ;H−2(0, L)). From the equation we see that unt = (−unx−unxxx − aun) is bounded in L2(0, T ;H−2(0, L)) and we can apply the Aubin-Lions lemma belowto conclude that {un}n∈N is relatively compact in L2(0, T ;L2(0, L)). Hence, we can assume that{un}n∈N converges in L2(0, T ;L2(0, L)) to some u with ‖u‖L2(0,T ;L2(0,L)) = 1 and such thatau = 0.
Lemma 17 (Aubin-Lions, see [37, Corollary 4]) Let X0 ⊂ X ⊂ X1 be three Banach spaces withX0, X1 reflexive spaces. Suppose that X0 is compactly embedded in X and that X is continuouslyembedded in X1. Then {h ∈ Lp(0, T ;X0) | h ∈ Lq(0, T ;X1)} embeds compactly in Lp(0, T ;X) forany 1 < p, q <∞.
Moreover, the initial data {un0}n∈N also converge due to (33). This is important to see thatthe u-limit satisfies
ut + ux + uxxx = 0.(37)
In order to get a contradiction, we have to use the property (1) of a = a(x). Recall that O isa nonempty open subset of (0, L) as small as we want. From (1) we get that the limit solution iszero in O, i.e.
u(t, x) = 0, ∀x ∈ O, ∀t ∈ (0, T ).
As the equation (37) is linear, latter condition implies the solution u is zero everywhere becauseof the Holmgrem’s Uniqueness Theorem (unique continuation principle). This is a contradictionto the fact that ‖u‖L2(0,T ;L2(0,L)) = 1 and consequently we have proved Theorem 1.�
Remark 18 In the proof of Remark 3, some nonlinear unique continuation principle are needed[36, 44] in order to get the contradiction.
3.3 Nonlinear system
We prove Theorem 2, i.e. the exponential decay of small amplitude solutions. This is basicallya linear result deduced from applying Theorem 1 to system 2.
Proof of Theorem 2.Consider ‖u0‖L2(0,L) ≤ r with r to be chosen later. The solution u of (2) can be written as
u = u1 + u2 where u1 is the solution of u1t + u1x + u1xxx + au1 = 0,u1(t, 0) = u1(t, L) = u1x(t, L) = 0,u1(0, x) = u0
(38)
and u2 is the solution of u2t + u2x + u2xxx + au2 = −uux,u2(t, 0) = u2(t, L) = u2x(t, L) = 0,u2(0, x) = 0
(39)
Thus, from (29) and the energy estimates for linear systems (see Proposition 12 and (22) inSection 2), we have
(40) ‖u(t, ·)‖L2(0,L) ≤ ‖u1(t, ·)‖L2(0,L) + ‖u2(t, ·)‖L2(0,L)
≤ γ‖u0‖L2(0,L) + C‖uux‖L1(0,T ;L2(0,L)) ≤ γ‖u0‖L2(0,L) + C‖u‖2L2(0,T ;H1(0,L))
11
with γ < 1. Of course we need somewhere a nonlinear estimate and it is here in order to dealwith the last term in previous inequality. Let us multiply equation (2) by xu and integrate toobtain
(41) 3
∫ T
0
∫ L
0
|ux|2dxdt+
∫ L
0
x|u(T, ·)|2dx+ 2
∫ T
0
∫ L
0
xa|u|2dxdt
=
∫ T
0
∫ L
0
|u|2dxdt+
∫ L
0
x|u0|2dx− 2
∫ T
0
∫ L
0
xux|u|2dxdt
Using
3
∫ T
0
∫ L
0
xux|u|2dxdt = −∫ T
0
∫ L
0
|u|3dxdt
into (41) we get
‖u‖2L2(0,T ;H1(0,L)) ≤(3T + L)
3‖u0‖L2(0,L) +
2
9
∫ T
0
∫ L
0
|u|3dxdt(42)
As u ∈ L2(0, T ;H1(0, L)) and H1(0, L) embeds into C([0, L]), we have
(43)
∫ T
0
∫ L
0
|u|3dxdt ≤∫ T
0
‖u‖L∞(0,L)
∫ L
0
|u|2dxdt ≤ C∫ T
0
‖u‖H1(0,L)
∫ L
0
|u|2dxdt
≤ C‖u‖2L∞(0,T ;L2(0,L))
∫ T
0
‖u‖H1(0,L)dt ≤ CT 1/2‖u0‖2L2(0,L)‖u‖L2(0,T ;H1(0,L))
Thanks to this and (42) we obtain
‖u‖2L2(0,T ;H1(0,L)) ≤(8T + 2L)
3‖u0‖2L2(0,L) +
TC
27‖u0‖4L2(0,L)(44)
which combined with (40) gives the existence of C > 0 such that
‖u(t, ·)‖L2(0,L) ≤ ‖u0‖L2(0,L)
{γ + C‖u0‖L2(0,L) + C‖u0‖3L2(0,L)
}(45)
Given ε > 0 small enough such that (γ+ε) < 1, we can take r small enough so that r+r3 < εC ,
in order to have‖u(t, ·)‖L2(0,L) ≤ (γ + ε)‖u0‖L2(0,L)
The rest of the proof runs as before thanks to the fact that (γ + ε) < 1. Thus, we end theproof of Theorem 2.
�
Remark 19 In previous result, constants C and µ can be chosen as close as we want to thecorresponding constants for the linear result given by Theorem 1. Of course, smaller ε is chosen,smaller is the radius r defining the set of initial data for which the exponential decay rate is valid.
4 Boundary control from the right
In this section, we will prove Theorem 5. Let L > 0 be fixed. Let us consider the following linearcontrol system for the KdV equation with homogeneous Dirichlet boundary conditions ut + ux + uxxx = 0,
u(t, 0) = u(t, L) = 0,ux(t, L) = F (t),
(46)
12
where the state is u(t, ·) : [0, L] → R and the control is F (t) ∈ R. In [45], Zhang considers thefollowing feedback law with α ∈ (0, 1) and L = 1
(47) F (t) = αux(t, 0).
One obtains, for the closed-loop system, that the energy satisfies
d
dt
∫ 1
0
|u(t, x)|2dx = −(1− α)|ux(t, 0)|2.
Thus, a decay to zero of the solutions is naturally expected. In fact, Zhang proves that theclosed-loop system is well posed in L2(0, 1) and that there exist ω > 0 and C > 0 such that
(48) ∀u0 ∈ L2(0, 1),∀t > 0, ‖u(t, ·)‖L2(0,1) ≤ Ce−ωt‖u0‖L2(0,1),
where u is the solution of (46)-(47) with initial data u0 and L = 1. That means, the feedback law(47) stabilizes the control system (46) to the origin. Then, Rosier proves in [30] that for somevalues of L, called critical values (see the definition of the set N in (28)), there exist some initialconditions u0 such that the corresponding solution of (46) with F = 0, conserves its L2-norm.These solutions also satisfy
ux(t, 0) = 0,
and therefore a feedback law as (47) does not stabilize the system. (Note that 1 /∈ N). Later, in[29], Perla, Vasconcellos and Zuazua prove that (48) actually holds for (46) with F = 0, providedthat the length L of the interval is not critical (see Section 3).
Here, we are interested in the design of some feedback laws
(49) F (t) = Fω(u(t, ·)),
such that the closed-loop system (46) and (49) has an exponential decay rate ω in (48) as largeas desired. In order to get this stabilizability property we use a method due to Urquiza [43].
4.1 Control design in the finite-dimensional case
We follow [8, Section 10.3] in order to explain the method on a finite dimensional control system
x = Ax+Bu, x(0) = x0,(50)
where for n,m ∈ N, A ∈ Mn×n(R), B ∈ Mn×m(R) and for each time t, the state is x(t) ∈ Rnand the control is u(t) ∈ Rm. The state x0 is the initial data. The solution of (50) is given by
x(t) = eAtx0 +
∫ t
0
e(t−s)ABu(s)ds
It is well known that the system is controllable in time T if and only if the symmetric andnonnegative Gramian matrix
C =
∫ T
0
e(T−t)ABB∗e(T−t)A∗dt
is invertible. (Notation M∗ stands for the transpose matrix of M). For instance, if C is invertible,then the system is driven from x0 to x1 in time T (for any x0, x1 ∈ Rn) by applying the control
u(s) = B∗e(T−s)A∗C−1(x1 − eTAx0), ∀s ∈ [0, T ].
Let us see how the Gramian matrix can also be used to stabilize the system. Let us supposethe system is controlable. Thus, the matrix
CT = e−TACe−TA∗
=
∫ T
0
e−tABB∗e−tA∗dt
is invertible and we can define the feedback control
u(t) = −B∗C−1T x(t).(51)
By applying a Lyapunov method, it can be easily proven the following.
13
Theorem 20 [8, Theorem 10.16] There exist M,µ > 0 such that every solution of (50)-(51)satisfies
|x(t)| ≤Me−µt|x(0)|, ∀t ≥ 0.
Proof. Because of the linearity of the system, it suffices to prove that x(t)→ 0 as t→∞.We define V : z ∈ Rn 7→ z∗C−1T z ∈ R which satisfies
V (z) > V (0), ∀z ∈ Rn \ {0}, and V (z)→∞ as |z| → ∞
Take a look at V on a trajectory by defining
p(t) = V (x(t)), ∀t ∈ R
We getp = −|B∗C−1T x|2 + x∗(A∗C−1T + C−1T A)x− x∗C−1T BB∗C−1T x
and using the identity
(A∗C−1T + C−1T A) = −C−1T e−TABB∗e−TA∗C−1T + C−1T BB∗C−1T
we can writep = −|B∗C−1T x|2 − |B∗e−TA
∗C−1T x|2
In order to apply LaSalle invariance principle, we have to check that
p(t) = 0,∀t ∈ R =⇒ x(t) = 0,∀t ∈ R.
Let us suppose p(t) = 0, for any t. This implies
B∗C−1T x(t) = 0(52)
and
B∗e−TA∗C−1T x(t) = 0(53)
From the way the control is chosen, we see that x(t) = Ax(t). Differentiating (52) with respectto time we obtain
B∗C−1T Ax(t) = 0, ∀t ∈ R
but C−1T Ax(t) = −A∗C−1T x(t) and therefore
B∗A∗C−1T x(t) = 0, ∀t ∈ R
By doing the same again k times, we get
B∗(A∗)kC−1T x(t) = 0, ∀t ∈ R
and therefore x(t)∗C−1T is always orthogonal to the columns of the Kalman matrix defined by
[B|AB|A2B| . . . |An−1B]
From the controllability of (A,B) and the invertibility of C−1T , we conclude that x(t) = 0 for anytime t. �
Now, as we want to impose an exponential decay rate equals to ω, we make the changey = eωtx. The system becomes
y = (A+ ωId)y +Bv
where Id is the identity matrix and the control is given by v = eωtu. The controllability of thissystem is equivalent to the controllability of (50). Then, the feedback control given by
v(t) = −B∗(∫ T
0
e−t(A+ωId)BB∗e−t(A∗+ωId)dt
)−1y(t).
14
gives the exponential decay of y but we do not know exactly the rate µ. By coming back to x,we get
|x(t)| ≤Me−ωt|x(0)|, ∀t ≥ 0,
which is what we were looking for. An improvement of this method can be realized if we considerthe matrix
Cω,∞ =
∫ ∞0
e−t(A+ωId)BB∗e−t(A∗+ωId)dt
In fact, we obtain(A+ ωId)Cω,∞ + Cω,∞(A+ ωId)
∗ = BB∗
and then if we use the control
u(t) = −B∗C−1ω,∞x(t)(54)
in (50), then we obtain
(A−BB∗C−1ω,∞) = Cω,∞(−A∗ − 2ωId)C−1ω,∞
In particular, if the operator A satisfies A∗ = −A, then the eigenvalues of the closed-loopsystem x = (A−BB∗C−1ω,∞)x are exactly the eigenvalues of A shifted 2ω units to the left in thecomplex plane. That means, the solutions of (50)-(54) satisfy
|x(t)| ≤Me−2ωt|x(0)|, ∀t ≥ 0.
4.2 Control design in the infinite-dimensional case
In [43] Urquiza adapted previous method and delivered a rigorous proof in an infinite-dimensionalframework. Let us state his result on the following abstract control system{
y(t) = Ay(t) +Bκ(t),y(0) = y0,
(55)
with state y(t) in a Hilbert space Y and control κ(t) in a Hilbert space U . Here, the initialcondition y0 ∈ Y , A is a skew-adjoint operator (i.e. A∗ = −A) in Y whose domain is dense in Y ,and B is an unbounded operator from U to Y . Let us assume that these operators satisfy thefollowing hypothesis.
(H1) The skew-adjoint operator A is the infinitesimal generator of a strongly continuous groupon Y .
(H2) The operator B : U → D(A)′ is linear continuous.
(H3) Regularity property. For every 0 < T <∞ there exists CT > 0 such that∫ T
0
‖B∗e−tA∗y‖2Udt ≤ CT ‖y‖2Y , y ∈ D(A∗).
(H4) Controllability property. There exist T > 0 and cT > 0 such that∫ T
0
‖B∗e−tA∗y‖2Udt ≥ cT ‖y‖2Y , y ∈ D(A∗).
Then, one has the following result whose proof mainly relies on general results about the algebraicRiccati equation associated with the linear quadratic regulator problem (see [12]).
Theorem 21 (see [43, Theorem 2.1]) Consider operators A and B under assumptions (H1)-(H4). For any ω > 0, we have
15
(i) The symmetric positive operator Λω defined by
(Λωx, z)Y =
∫ ∞0
(B∗e−τ(A+ωI)∗x,B∗e−τ(A+ωI)∗z
)Udτ, ∀x, z ∈ Y,
is coercive and is an isomorphism on Y .
(ii) Let Fω := −B∗Λ−1ω . The operator A + BFω with D(A + BFω) = Λω(D(A∗)) is the in-finitesimal generator of a strongly continuous semigroup on Y.
(iii) The closed-loop system (system (55) with the feedback law κ = Fω(y)) is exponentially stablewith a decay rate equals to 2ω, that is,
∃C > 0,∀y0 ∈ Y, ‖et(A+BFω)y0‖Y ≤ Ce−2ωt‖y0‖Y .
As one can see, the feedback operator is built in an explicit way. This fact and the free choiceof the parameter ω are the main advantages of this method. The first point to check in order tobe able to apply this theorem to our linear KdV control system (46) is (H1). As we easily see,hypothesis (H1) holds if we take as control, the function v defined by
v(t) = F (t)− yx(t, 0).
Hence our system becomes symmetric with respect to the space variable ut + ux + uxxx = 0,u(t, 0) = u(t, L) = 0,ux(t, L)− ux(t, 0) = v(t).
(56)
We can rewrite (56) in the abstract form (55) by defining the operators A and B as follows
D(A) :={w ∈ H3(0, L);w(0) = w(L) = 0, w′(0) = w′(L)
},
Aw := −w′ − w′′′,B : s ∈ R 7→ Ls ∈ D(A∗)′,
Ls : z ∈ D(A∗) 7→ szx(L) ∈ R.It is not difficult to see that A∗ = −A and that
(Aw,w)L2(0,L) = 0, ∀w ∈ D(A).
Hence, from classical semigroup results, one sees that the operator A satisfies (H1). We also seethat (H2) holds for the operator B. Hypothesis (H3) and (H4) are more delicate to show. Asour operator B stands for a boundary control, we will see that assumption (H3) is a sharp traceregularity. Concerning (H4), it is an observability inequality.
In order to prove (H3) and (H4) we first study the asymptotic behavior of the eigenvalues ofthe operator A. Then, we apply a classical Ingham’s inequality to prove that (H3) and (H4) holdfor our control system.
4.3 Spectral analysis
It is not difficult to see that the skew-adjoint operator A has a compact resolvent. Hence thespectrum σ(A) of A consists only of eigenvalues. Furthermore the spectrum of A is a discretesubset of iR and the eigenfunctions form an orthonormal basis of L2(0, L). For the work to bedone here, we require very detailed informations about the asymptotic behavior of the eigen-elements of the operator A. Let us denote by (iλk)k∈Z the eigenvalues of A and by (φk)k∈Z itseigenfunctions.
Proposition 22 The real numbers (λk)k∈Z have the asymptotic form
λk =8π3k3
L3+O(k2) as k → ±∞.
16
Proof. The eigenvalue problem to be considered is −φ′ − φ′′′ = iλφ,
φ(0) = φ(L) = 0,φ′(0) = φ′(L).
To each λ corresponds at least a real a such that λ = 2a(4a2 − 1). Thus, the three solutions of
z3 + z + iλ = 0
read asz0 =
√|3a2 − 1| − ai, z1 = −
√|3a2 − 1| − ai, z2 = 2ai.
We distinguish 3 cases.1. If 3a2 − 1 < 0.In this case, it is easy to see that the eigenfunction φ of A associated to the eigenvalue λ =2a(4a2 − 1) may be written
φ(x) = e−iax(α cos(√
1− 3a2x) + β sin(√
1− 3a2x)) + γe2iax,
where α, β and γ are some constants such that φ(0) = φ(L) = 0 and φ′(0) = φ′(L). That means,such that
α+ γ = 0,
(57) e−iaL(α cos(√
1− 3a2L) + β sin(√
1− 3a2L))− αe2iaL = 0,
(58) − iaα+ β√
1− 3a2 + 2iaγ = −iae−iaL(α cos(√
1− 3a2L) + β sin(√
1− 3a2L))
2iaγe2iaL + e−iaL√
1− 3a2(−α sin(√
1− 3a2L) + β cos(√
1− 3a2L)).
From (57), one obtains
β = αe3iaL − cos(
√1− 3a2L)
sin(√
1− 3a2L).
Taking the real part of equation (58), one obtains that a must satisfy
(59)√
1− 3a2 cos(2aL) = 3a sin(aL) sin(√
1− 3a2L)
+√
1− 3a2 cos(aL) cos(√
1− 3a2L).
The number of parameters a satisfying (59) is finite and depends on L. As if a satisfies (59),then (−a) so, we find in this case 2NL eigenvalues
{λ−NL, ..., λ−1, λ1, ..., λNL
}
2. If 3a2 − 1 = 0.We don’t find any eigenfunction in this case. In fact, here
z0 = z1 =i√
3
3, z2 =
−2i√
3
3or z0 = z1 =
−i√
3
3, z2 =
2i√
3
3
and the candidate function to be an eigenfunction cannot satisfy the boundary conditions.3. If 3a2 − 1 > 0.In this case, it is easy to see that the eigenfunction φ of A associated to the eigenvalue λ =2a(4a2 − 1) may be written as
φ(x) = e−iax(α cosh(√
3a2 − 1x) + β sinh(√
3a2 − 1x)) + γe2iax
17
where α, β and γ are some constants such that φ(0) = φ(L) = 0 and φ′(0) = φ′(L). That means,such that
α+ γ = 0,
e−iaL(α cosh(√
3a2 − 1L) + β sinh(√
3a2 − 1L))− αe2iaL = 0,(60)
(61) − iaα+ 2iaγ + β√
3a2 − 1 =
− iae−iaL(α cosh(√
3a2 − 1L) + β sinh(√
3a2 − 1L)) + 2iaγe2iaL
+ e−iaL√
3a2 − 1(α sinh(√
3a2 − 1L) + β cosh(√
3a2 − 1L))
We deduce from (60)-(61) that
β = αe3iaL − cosh(
√3a2 − 1L)
sinh(√
3a2 − 1L),
and
− 3a+ =(β)√
3a2 − 1 = −3a cos(2aL)
+√
3a2 − 1 sin(−aL)(sinh(√
3a2 − 1L) + <(β) cosh(√
3a2 − 1L))
+√
3a2 − 1 cos(aL)=(β) cosh(√
3a2 − 1L).
From these equations, one obtains that a satisfies the following one
(62)√
3a2 − 1 cos(2aL)− 3a sin(aL) sinh(√
3a2 − 1L)
−√
3a2 − 1 cos(aL) cosh(√
3a2 − 1L) = 0.
If one neglects the terms e−L√3a2−1 as a→ ±∞, one gets
eL√3a2−1 =
cos(2aL)
2 cos(aL− π/3)
and hence there exists, for k ∈ N large enough, a unique solution ak+NL(respectively a−k−NL
)in the interval [
kπ
L, (k + 1)
π
L
],(respectively
[−(k + 1)
π
L,−k π
L
] )defined by equation (62) and given asymptotically by
(63) ak =5π
6L+kπ
L+O(
1
k)(respectively a−k = − 5π
6L− kπ
L+O(
1
k)).
The associated eigenfunction, φk is
φk(x) = αk
[e−iakx
(cosh(
√3a2k − 1x)
+e3iakL − cosh(
√3a2k − 1L)
sinh(√
3a2k − 1L)sinh(
√3a2k − 1x)
)− e2iakx
],
where αk is chosen in such a way that ‖φk‖L2(0,L) = 1. Asymptotically, one sees that (αk)
converge to 1/√L as k goes to ∞.
Thus, from (63), one deduces the asymptotic behavior of the eigenvalues and therefore the proofof this proposition is complete. �
18
Remark 23 We easily deduce from equations (59) and (62) that
∀k ∈ Z, ak = −a−k and λk = −λ−k.
Remark 24 Similar asymptotical behaviors have been found out in [35] and [45].
From the proof of the last proposition, we deduce the following lemma.
Lemma 25 There exists a constant C > 0 such that
limk→±∞
|φ′k(L)||k|
= C
Proof. By using the formulae for the eigenfunctions φk, we get
φ′k(L) = αk
[−3iake
2iakL +e−iakL
√3a2k − 1
sinh(L√
3a2k − 1)
(e3iakL cosh(L
√3a2k − 1)− 1
)]This fact together with (63) allows us to find that
limk→±∞
|φ′k(L)||k|
=2π√
3
L3/2> 0 .
�
4.4 Ingham’s inequality
Given the asymptotic behavior of the eigenvalues of A, we have to modify the choice of the statespace L2(0, L) in order to prove (H3) and (H4). From the previous section, we know that {φk}k∈Zis a basis of L2(0, L). Thus, for any f ∈ L2(0, L) there exists a unique sequence {fk}k∈Z with∑k∈Z |fk|2 <∞ such that
f =∑k∈Z
fkφk and ‖f‖L2(0,L) =(∑k∈Z|fk|2
)1/2.
Let us now define some useful spaces.
Definition 1 Let us denote by Z the linear hull of the basis functions {φk}k∈Z. Then Z is adense subspace of L2(0, L). For any s ∈ R we define the space Hs as the completion of Z withrespect to the norm defined by∥∥∥∑
k∈Zckφk
∥∥∥s
:=(∑k∈Z
(1 + |λk|)23 s|ck|2
)1/2.
In each space Hs, one has the orthonormal basis {(1 + |λk|)−s3φk}k∈Z.
With this definition we can state the following well-posedness result whose proof is directfrom the previous analysis.
Proposition 26 For any z0 =∑k∈Z z
k0φk ∈ Hs, there exists a unique solution of the homoge-
neous problem {zt + zx + zxxx = 0, z(0, ·) = z0,z(t, 0) = z(t, L) = 0, zx(t, L)− zx(t, 0) = 0,
which belongs to C(R, Hs) and is given by
z(t, x) =∑k∈Z
eiλktzk0φk(x).
Moreover, as {λk}k∈Z ⊂ R, one has that
∀t ∈ R, ‖z(t, ·)‖s = ‖z0‖s.
19
Now, we are interested in the regularity needed to obtain zx(·, L) ∈ L2(0, T ) for any T > 0.As one has at least formally,
zx(t, L) =∑k∈Z
eiλktzk0φ′k(L),
one sees the importance of Lemma 25 which gives us the asymptotic behavior of φ′k(L) as k tendsto ±∞. In order to find the regularity needed, we use the following classical result mainly dueto Ingham (see [13] and [16]).
Lemma 27 (Ingham’s inequality) Let T > 0. Let {βk}k∈Z ⊂ R be a sequence of pairwisedistinct real numbers such that
lim|k|→+∞
βk+1 − βk = +∞.
Then there exist two strictly positive constants C1 and C2 such that for any sequence {γk}k∈Zsatisfying
∑k∈Z γ
2k <∞, the series f(t) =
∑k∈Z γke
iβkt converges in L2(0, T ) and satisfies
C1
∑k∈Z
γ2k ≤∫ T
0
|f(t)|2dt ≤ C2
∑k∈Z
γ2k.
Let us apply this lemma. Let z0 =∑k∈Z z
k0φk ∈ Hs for some s ≥ 0. We take βk = λk and
γk = zk0φ′k(L). From the asymptotic behavior, we have that if s ≥ 1, then∑
k∈Z|zk0 |2|φ′k(L)|2 <∞.
This together with Proposition 22 allow us to apply the Ingham’s inequality and get the existenceof two constants C1, C2 > 0 such that for any z0 ∈ Hs, with s ≥ 1,
C1
∑k∈Z|zk0 |2|φ′k(L)|2 ≤
∫ T
0
|zx(t, L)|2dt ≤ C2
∑k∈Z|zk0 |2|φ′k(L)|2 .(64)
We can estimate by above the right-hand side in terms of the H1-norm of z0, and consequently,in terms of any Hs-norm with s ≥ 1. To get inequalities (H3) and (H4) in the space H1, we needto estimate by below the left-hand side in terms of the H1-norm of z0. In order to do that wecan not lose any coefficient zk0 . Thus, the condition
∀k ∈ Z, φ′k(L) 6= 0,
is required. From the work of Rosier in [30], we know that if L satisfies
L /∈ N :=
{2π
√k2 + kl + l2
3; k, l ∈ N∗
},(65)
then, there exist no µ ∈ C, ϕ ∈ H3(0, L)\{0} satisfying{µϕ+ ϕ′ + ϕ′′′ = 0,ϕ(0) = ϕ(L) = ϕ′(0) = ϕ′(L) = 0.
In particular, this implies that φ′k(L) 6= 0 for any k ∈ Z and therefore from (64), we get theexistence of positive constants cT and CT such that
cT ‖z0‖2H1≤∫ T
0
|zx(t, L)|2dt ≤ CT ‖z0‖2H1, ∀z0 ∈ H1.(66)
The left-hand inequality in (66) is called an observability inequality and it is equivalent to anexact controllability result. This ends the proof of (H3) and (H4).
20
4.5 Control design for our system
In this section, we apply Urquiza’s method to our linear control system (56). Let us design thefeedback laws allowing us to get the rapid stabilization result. We first define, for any q0 andψ0 ∈ H1, the bilinear form
aω(q0, ψ0) :=
∫ ∞0
e−2ωτqx(τ, L)ψx(τ, L)dτ,
where q and ψ are the respective solutions of{qτ + qx + qxxx = 0, q(0, .) = q0,q(τ, 0) = q(τ, L) = 0, qx(τ, L)− qx(τ, 0) = 0
and {ψτ + ψx + ψxxx = 0, ψ(0, .) = ψ0,ψ(τ, 0) = ψ(τ, L) = 0, ψx(τ, L)− ψx(τ, 0) = 0.
We then define the operator Λω : H1 −→ H−1 assumed to satisfy
< Λωq0, ψ0 >H−1,H1= aω(q0, ψ0), ∀q0, ψ0 ∈ H1.
Finally, we define the following operator
Fω : H1 −→ Rz −→ Fω(z) := −q′0(L),
where q0 is the solution of the following Lax-Milgram problem
aω(q0, ψ0) =< z, ψ0 >H−1,H1, ∀ψ0 ∈ H1 .
As hypothesis (H1)-(H4) are satisfied, Theorem 21 can be applied and one easily gets Theorem5: The closed-loop system{
ut + ux + uxxx = 0, u(0, .) = y0,u(t, 0) = u(t, L) = 0, ux(t, L)− ux(t, 0) = Fω(u(t)),
is globally well posed in H1. Moreover, the solutions decay to zero with an exponential rate of2ω, i.e.,
∃C > 0,∀u0 ∈ H1, ‖u(t, ·)‖H1 ≤ Ce−2ωt‖u0‖H1 .
4.6 Numerical simulations
Let ω > 0 be fixed. We use the Galerkin method and an approximation by modal superpositionto decompose our solutions. Let (iλk, φk) be the eigenmodes of −φ
′k − φ′′′k = iλkφk,
φk(0) = φk(L) = 0,φk(L)− φk(0) = 0.
Let us defineVN = Span{φ−N , ..., φ−1, φ1, ..., φN}, ∀N ∈ N∗.
For any z0 ∈ H1 let q0N ∈ VN be the unique solution of the variational equation
aω(q0N , ψ0N ) =
∫ ∞0
e−2ωτqN x(τ, L)ψN x(τ, L)dτ =
∫ L
0
z0(x)ψ0N (x)dx, ∀ψ0
N ∈ VN ,
where qN and ψN are the respective solutions of
(67)
qN τ + qN x + qN xxx = 0,qN (τ, 0) = qN (τ, L) = 0,qN x(τ, L)− qN x(τ, 0) = 0,qN (0, .) = q0N
21
and
(68)
ψN τ + ψN x + ψN xxx = 0,ψN (τ, 0) = ψN (τ, L) = 0,ψN x(τ, L)− ψN x(τ, 0) = 0,ψN (0, .) = ψ0
N .
We define the discrete operator,
LN : z0 ∈ H1 7−→ q0N ∈ VN .
As (q0N , ψ0N ) ∈ VN × VN we can write
q0N =
N∑k=−N
γ0kφk(x), and ψ0N =
N∑k=−N
γ0kφk(x)
and we easily deduce from (67) and (68) that
qN (τ, x) =
N∑k=−N
eiλkτγ0kφk(x), and ψN (τ, x) =
N∑k=−N
eiλkτ γ0kφk(x).
Let us define mk = φ′k(L) for k ∈ N∗. Then
aω(q0N , ψ0N ) =
N∑k,j=−N
γ0k γ0jmkmj
2ω − iλk − iλj.
In order to solve the stabilization problem we write the problem in a weak form where theboundary term appears. We multiply (56) by w ∈ D(A) and get by integration by parts∫ L
0
utwdx−∫ L
0
u(wx + wxxx)dx = v(t)wx(L).
00.2
0.40.6
0.81
00.2
0.40.6
0.81!3
!2
!1
0
1
2
3
xt
y
00.2
0.40.6
0.81
0
0.2
0.4
0.6
0.8
1!3
!2
!1
0
1
2
3
xt
y
Figure 1: Evolution of the solution u when ω = 2 (left) and ω = 3 (right).
We take as an approximation of the controlled solution, uN : [0,∞]→ VN solution of
(69)
∫ L
0
uNt (t, x)wdx−∫ L
0
uN (t, x)(wx + wxxx)dx = vN (t)wx(L), ∀w ∈ D(A),
where the stabilizing control is chosen as
vN (t) =∂
∂xLN (uN (t, .))
22
with initial data uN (0) = PN (u0) (PN is the orthogonal projection on VN ). Let q0N (t) =LN (uN (t, .)). It satisfies
N∑k,j=−N
γ0k γ0jmkmj
2ω − iλk − iλj=
∫ L
0
N∑−N
yk(t)φk(x)
N∑−N
γ0l φl(x)
=
N∑k,l=−N
yk(t)γ0l
∫ L
0
φk(x)φl(x) =
N∑k=−N
yk(t)γ0k
Figure 2: Time-evolution of the norm ‖u‖H1compared with e−ωt‖u0‖H1
for ω = 2 and ω = 3.
Let Aω be the matrix with coefficients
(Aω)l,j =mlmj
2ω − iλl − iλj.
Let J be the one anti-diagonal matrix, Γ0N the vector of coefficients γ0k and YN (t) the vector of
coefficients yk(t). Then we getΓ0N = A−1ω JYN (t)
and the control is
vN (t) =
N∑k=−N
[A−1ω JYN (t)]kmk.
So with (69), choosing w = φ−k for k = −N, ...,−1, 1, ..., N , we get
(70) Y ′N (t)− iD(λ−N , . . . , λN )YN (t) = KYN (t)
where D(λ−N , . . . , λN ) is a diagonal matrix with coefficients λ−N , . . . , λN and K is the matrix
(K)l,j = m−l
N∑k=−N
[A−1ω J ]kjmk.
The matrix equation (70) can easily be solved. Results are drawn on some figures for N = 10and the initial condition given by Y 0
N (k) = 1, for k = −N, ...,−1, 1, ...N . On Figure 1 we showthe evolution of the solution for ω = 2 and ω = 3. Note particularly, on Figure 2, the excellentagreement between theoretical and numerical results for the time-evolution of the H1-norm.
5 Boundary control from the left
This section is devoted to prove Theorems 7 and 8.
23
5.1 Control design
The backstepping method applied here is based on the linear part of the equation. In this way,we consider the control system linearized around the origin{
ut + ux + uxxx = 0,u(t, 0) = κ(t), u(t, L) = 0, ux(t, L) = 0.
(71)
Given a positive parameter ω, we look for a transformation Π : L2(0, L) → L2(0, L) definedby
v(x) = Π(u(x)) := u(x)−∫ L
x
k(x, y)u(y)dy,(72)
such that the trajectory u = u(t, x), solution of (71) with
κ(t) =
∫ L
0
k(0, y)u(t, y)dy,(73)
is map into the trajectory v = v(t, x), solution of the linear system{vt + vx + vxxx + ωv = 0,v(t, 0) = 0, v(t, L) = 0, vx(t, L) = 0.
(74)
For system (74), called the target system, we have for any t ≥ 0
d
dt
∫ L
0
|v(t, x)|2dx = −|vx(t, 0)|2 − 2ω
∫ L
0
|v(t, x)|2dx ≤ −2ω
∫ L
0
|v(t, x)|2dx(75)
and therefore we easily obtain for v = v(t, x) the exponential decay at rate ω
‖v(t, ·)‖L2(0,L) ≤ e−ωt‖v(0, ·)‖L2(0,L), ∀t ≥ 0.(76)
In Section 5.2, we prove that, thanks to the invertibility of the map Π, the exponential decay(76) also holds for system (71).
Naturally, we can wonder if this decay rate is sharp. Let us notice that the eigenvalues ofsystem (74) are the eigenvalues of{
vt + vx + vxxx = 0,v(t, 0) = 0, v(t, L) = 0, vx(t, L) = 0,
(77)
shifted to the left ω units. Thus, we are lead to study the eigenvalues σ of{−φ′(x)− φ′′′(x) = σφ(x),φ(0) = 0, φ(L) = 0, φ′(L) = 0.
(78)
Surprisingly, the location of the eigenvalues of (78) depends on the length of the interval.From [30], we know that there exist some eigenvalues located on the imaginary axis if and only
if L ∈ {2π√
k2+k`+`2
3
/(k, `) ∈ N2}, which is called the set of critical lengths for this problem. In
Figure 3, the first five eigenvalues of system (77) are plot for different values of L. In case (a),L = 1 (non-critical) and the first eigenvalue σ1 is approximately −72. The system behaves like adissipative one. In (b), L = 2π (critical) and we have σ1 = 0. The system has one conservativecomponent given by the eigenfunction φ(x) = 1− cos(x). In (c), L = 2π
√7/3 and the first two
eigenvalues are imaginary numbers σ1 = 0.2i and σ2 = −0.2i. This examples show the differentbehaviors system (74) can have and the important role played by the parameter ω in our design.In conclusion, the decay in (76) is optimal for some values of L.
24
-6000 -5000 -4000 -3000 -2000 -1000 0-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
HaL-20 -15 -10 -5 0
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
HbL-5 -4 -3 -2 -1 0
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
HcL
Figure 3: The first five eigenvalues of system (77) for (a) L = 1, (b) L = 2π, and (c) L = 2π√
7/3,respectively.
Let us focus in the key step, which is finding the kernel k = k(x, y) such that v(t, x) =Π(u(t, x)) satisfies (74). For that, we perform the following computations
vt(t, x) = ut(t, x)−∫ Lxut(t, y)k(x, y)dy
= ut(t, x) +∫ Lx
(uy(t, y) + uyyy(t, y)) k(x, y)dy
= ut(t, x)−∫ Lxu(t, y) (ky(x, y) + kyyy(x, y)) dy
− k(x, x)(u(t, x) + uxx(t, x))+ ky(x, x)ux(t, x)− kyy(x, x)u(t, x) + k(x, L)u(t, L)+ k(x, L)uxx(t, L)− ky(x, L)ux(t, L) + kyy(x, L)u(t, L),
(79)
vx(t, x) = ux(t, x) + k(x, x)u(t, x)−∫ Lxkx(x, y)u(t, y)dy,(80)
vxx(t, x) = uxx(t, x) + u(t, x) ddxk(x, x) + k(x, x)ux(t, x)
+ kx(x, x)u(t, x)−∫ Lxkxx(x, y)u(t, y)dy,
(81)
and
vxxx(t, x) = uxxx(t, x) + u(t, x) d2
dx2 k(x, x) + 2ux(t, x) ddxk(x, x)
+ k(x, x)uxx(t, x) + u(t, x) ddxkx(x, x) + kx(x, x)ux(t, x)
+ kxx(x, x)u(t, x)−∫ Lxkxxx(x, y)u(t, y)dy.
(82)
Thus, given ω ∈ R and using (71), we have
(83) vt(t, x) + vx(t, x) + vxxx(t, x) + ωv(t, x) =
−∫ L
x
u(t, y)(kxxx(x, y) + kx(x, y) + kyyy(x, y) + ky(x, y) + ωk(x, y)
)dy
+ k(x, L)uxx(t, L) + ux(t, x)(ky(x, x) + kx(x, x) + 2
d
dxk(x, x)
)+ u(t, x)
(ω + kxx(x, x)− kyy(x, x) +
d
dxkx(x, x) +
d2
dx2k(x, x)
).
After the above computations and since ddxk(x, x) = kx(x, y) + ky(x, y), we obtain that one
has (74) for every u solution of (71) with (73) if the kernel k = k(x, y) defined in the triangleT := {(x, y)
/x ∈ [0, L], y ∈ [x, L]} satisfies
kxxx(x, y) + kyyy(x, y) + kx(x, y) + ky(x, y) = −ωk(x, y), in T ,k(x, L) = 0, in [0, L],k(x, x) = 0, in [0, L],
kx(x, x) =ω
3(L− x), in [0, L].
(84)
25
Let us make the following change of variable
t = y − x, s = x+ y,(85)
and define G(s, t) := k(x, y). We have k(x, y) = G(x+ y, y − x) and therefore
kx = Gs −Gt, ky = Gs +Gt,(86)
kxx = Gss − 2Gst +Gtt, kyy = Gss + 2Gst +Gtt,(87)
kxxx = Gsss − 3Gsst + 3Gstt −Gttt,(88)
kyyy = Gsss + 3Gsst + 3Gstt +Gttt.(89)
Now, the function G = G(s, t), defined in T0 := {(s, t)/t ∈ [0, L], s ∈ [t, 2L− t]}, satisfies
6Gtts(s, t) + 2Gsss(s, t) + 2Gs(s, t) = −ωG(s, t), in T0,G(s, 2L− s) = 0, in [L, 2L],
G(s, 0) = 0, in [0, 2L],
Gt(s, 0) =ω
6(s− 2L), in [0, 2L].
(90)
Let us transform this system into an integral one. We write the equation in variables (η, ξ),integrate ξ in (0, τ) and use that 6Gts(η, 0) = ω. Next, we integrate τ in (0, t) and use thatGs(η, 0) = 0. Finally, we integrate η in (s, 2L − t) and use that G(2L − t, t) = 0. Thus, we canwrite the following integral form for G = G(s, t)
(91) G(s, t) = −ωt6
(2L− t− s)
+1
6
∫ 2L−t
s
∫ t
0
∫ τ
0
(2Gsss(η, ξ) + 2Gs(η, ξ) + ωG(η, ξ)
)dξdτdη.
To prove that such a function G = G(s, t) exists, we use the method of successive approxi-mations. We take as an initial guess
G1(s, t) = −ωt6
(2L− t− s)(92)
and define the recursive formula as follows,
Gn+1(s, t) =1
6
∫ 2L−t
s
∫ t
0
∫ τ
0
(2Gnsss(η, ξ) + 2Gns (η, ξ) + ωGn(η, ξ)
)dξdτdη.(93)
Performing some computations, we get for instance
G2(s, t) =1
108
{t3(ω − ω2L+
ω2t
4
)(2L− t− s
)+t3ω2
4
[(2L− t)2 − s2
]},(94)
and more generally the following formula
Gk(s, t) =
k∑i=1
(aikt
2k−1 + bikt2k)[
(2L− t)i − si],(95)
where the coefficients satisfy bkk = 0 and more importantly, there exist positive constants M,Bsuch that, for any k ≥ 1 and any (s, t) ∈ T0∣∣Gk(s, t)
∣∣ ≤M Bk
(2k)!(t2k−1 + t2k).(96)
This implies that the series∑∞n=1G
n(s, t) is uniformly convergent in T0. Therefore the seriesdefines a continuous function G : T0 → R
G(s, t) =
∞∑n=1
Gn(s, t)(97)
26
and we get a solution of our integral equation. Indeed, we can write
(98) G = G1 +
∞∑n=1
Gn+1
= G1 +1
6
∞∑n=1
∫ 2L−t
s
∫ t
0
∫ τ
0
(2Gnsss(η, ξ) + 2Gns (η, ξ) + ωGn(η, ξ)
)dξdτdη
= G1 +1
6
∫ 2L−t
s
∫ t
0
∫ τ
0
(2
∞∑n=1
Gnsss(η, ξ) + 2
∞∑n=1
Gns (η, ξ) + ω
∞∑n=1
Gn(η, ξ))dξdτdη
= G1 +1
6
∫ 2L−t
s
∫ t
0
∫ τ
0
(2Gsss(η, ξ) + 2Gs(η, ξ) + ωG(η, ξ)
)dξdτdη.
where we have used that the corresponding series∑n≥1G
ns and
∑n≥1G
nsss are also uniformly
convergent.Once we have found the function G = G(s, t), we get the existence of the kernel k = k(x, y).
It is easy to see that the map Π : L2(0, L) → L2(0, L), defined by (72), is continuous andconsequently we have the existence of a positive constant Dκ such that
‖Π(f)‖L2(0,L) ≤ Dκ‖f‖L2(0,L), ∀f ∈ L2(0, L).(99)
In Figure 4, we plot the gain kernel k(0, y) (see (72)) as a function of y ∈ [0, L] for differentlengths (a) L = 1 (non-critical), (b) L = 2π (critical) and (c) L = 2π
√7/3 (critical). The kernel
functions are defined with ω = 1. This illustrate the fact that case (a) is easier to stabilizesthan case (b), which is easier to stabilizes than case (c). This is due to the location of thecorresponding open-loop eigenvalues as shown in Figure 3.
0.0 0.2 0.4 0.6 0.8 1.0-0.08
-0.06
-0.04
-0.02
0.00
HaL0 1 2 3 4 5 6
-6
-5
-4
-3
-2
-1
0
HbL0 2 4 6 8
-50
-40
-30
-20
-10
0
HcL
Figure 4: Gain kernel k(0, y) corresponding to ω = 1 for (a) L = 1, (b) L = 2π, and (c)L = 2π
√7/3, respectively.
5.2 Stability of the linear system
We know that the target system (74) is exponentially stable. In order to get the same conclusionfor the linear system (71) and get Theorem 7, the method we are applying uses the inversetransformation Π−1. For that, we introduce a kernel function `(x, y) which satisfies
`xxx(x, y) + `yyy(x, y) + `x(x, y) + `y(x, y) = ω`(x, y), in T ,`(x, L) = 0, in [0, L],`(x, x) = 0, in [0, L],
`x(x, x) =ω
3(L− x), in [0, L],
(100)
The existence and uniqueness of such a kernel ` = `(x, y) are proven in the same way thanfor the kernel k = k(x, y) previously. Once we have defined ` = `(x, y), it is easy to see that thetransformation Π−1 is characterized by
u(x) = Π−1(v(x)) := v(x) +
∫ L
x
`(x, y)v(y)dy.(101)
27
Let us see that k(x, y) and `(x, y) are related by the formula
`(x, y)− k(x, y) =
∫ y
x
k(x, η)`(η, y)dη,(102)
which in fact proves that Π−1 maps a trajectory of (74) into a trajectory of (71) with controlκ(t) defined by (73). Indeed, by plugging (101) into (72)and using Fubini’s theorem we get, forany v ∈ L2(0, L), that for any x ∈ [0, L]∫ L
x
(`(x, y)− k(x, y))v(y)dy =
∫ L
x
∫ y
x
k(x, η)`(η, y)v(y)dηdy,(103)
which proves (102) for any (x, y) ∈ T .The map Π−1 : L2(0, L) → L2(0, L) is continuous and therefore we get the existence of a
positive constant D` such that
‖Π−1(f)‖L2(0,L) ≤ D`‖f‖L2(0,L), ∀f ∈ L2(0, L).(104)
Let us prove that system (71)-(73) is exponentially stable. In fact, given u0 ∈ L2(0, L), wedefine
v0(x) = Π(u0(x)) := u0(x)−∫ L
x
k(x, y)u0(y)dy.(105)
The solution of (74) with initial condition v(0, x) = v0(x) satisfies (75), i.e.
‖v(t, ·)‖L2(0,L) ≤ e−ωt‖v0(·)‖L2(0,L), ∀t ≥ 0.(106)
Moreover, the solution of (71) is given by u(t, x) = Π−1(v(t, x)). Thus, from (99),(104) and(106) we have for any t ≥ 0
(107) ‖u(t, ·)‖L2(0,L) ≤ D`‖v(t, ·)‖L2(0,L) ≤ D`e−ωt‖v0(·)‖L2(0,L) ≤ D`Dke
−ωt‖u0(·)‖L2(0,L),
which proves the exponential decay at rate ω for system (71) with feedback law (73). This endsthe proof of Theorem 7.
5.3 Stability of the nonlinear system
Let us prove Theorem 8. Let u = u(t, x) be a solution of the nonlinear equation (46) with thecontrol κ given by (73). Then, v = Π(u(t, x)) satisfies
(108) vt(t, x) + vx(t, x) + vxxx(t, x) + ωv(t, x)
= −
(v(t, x) +
∫ L
x
`(x, y)v(t, y)dy
)(vx(t, x) +
∫ L
x
`x(x, y)v(t, y)dy
)with homogeneous boundary conditions
v(t, 0) = 0, v(t, L) = 0, vx(t, L) = 0.(109)
We multiply (108) by v and integrate in (0, L) to obtain
d
dt
∫ L
0
|v(t, x)|2dx = −|vx(t, 0)|2 − 2ω
∫ L
0
|v(t, x)|2dx− 2
∫ L
0
v(t, x)F (t, x)dx
where the term F = F (t, x) is given by
(110) F (t, x) = v(t, x)
∫ L
x
`x(x, y)v(t, y)dy + vx(t, x)
∫ L
x
`(x, y)v(t, y)dy
+
(∫ L
x
`(x, y)v(t, y)dy
)(∫ L
x
`x(x, y)v(t, y)dy
).
28
We can prove that there exists a positive constant C = C(‖`‖C1(T )) such that
∣∣∣2 ∫ L
0
v(t, x)F (t, x)dx∣∣∣ ≤ C (∫ L
0
|v(t, x)|2)3/2
(111)
and therefore, if there exists t0 ≥ 0 such that ‖v(t0, ·)‖L2(0,L) ≤ ωC , then we obtain
d
dt
∫ L
0
|v(t, x)|2dx ≤ −ω∫ L
0
|v(t, x)|2dx, ∀t ≥ t0.(112)
Thus, we get
‖v(t, ·)‖L2(0,L) ≤ e−ω2 t‖v(0, ·)‖L2(0,L), ∀t ≥ 0,(113)
provided that
‖v(0, ·)‖L2(0,L) ≤ω
C.(114)
As we did for the linear system, by using the continuity of the transformations Π and Π−1 (see(99) and (104)) and (113), we obtain the exponential decay of the nonlinear equation (46). From(114), we have to add a smallness condition on the initial data of system (46). This concludesthe proof of Theorem 8.
6 Discussion
6.1 Internal control
In these notes we have introduced an internal damping mechanism in order to be sure the energyof the system decreases to zero in an exponential way. We have proved a local result for the KdVequation.
The following problem remains open. Let u0 ∈ L2(0, L) and u the solution of ut + ux + uxxx + uux = 0,u(t, 0) = u(t, L) = ux(t, L) = 0,u(0, ·) = u0.
Does the solution u decay to zero as t goes to infinity?We know that in general the solutions of the linear equation decay to zero if and only if L
is not critical (remember definition of set N in (28)). Thus, we wonder if the nonlinearity givesus the stability in the critical cases? That could seem strange, but a similar phenomena appearswhen studying the controllability of the system from the right Neumann boundary condition.The linear system is controllable if and only if L is not critical but in despite of that the nolinearsystem is always controllable [30, 9, 2, 5].
In order to answer the question, a really nonlinear method is needed because with a first-orderapproximation one obtains the linear system which has some solutions conserving its L2-norm.
6.2 Boundary control from the right
By using a finite-dimensional method based on the Gramian matrix we have design some feedbackcontrols which make the linear KdV equation stable with an exponential decay rate as large asdesired. This method can not be applied if the underlying spatial operator is not skew-adjoint.For that we consider a first order boundary condition on (ux(t, L)− ux(t, 0) instead of ux(t, L).
A major difficulty in order to consider the nonlinear KdV equation is to deal with the technicalpoint of well-posedness of the equation with the convenient boundary conditions. Is the system ut + ux + uxxx + uux = 0,
u(t, 0) = u(t, L) = 0,ux(t, L)− ux(t, 0) = 0,
well-posed in L2(0, L) or H1(0, L)? With this boundary conditions there is no Kato smoothingeffect allowing us to deal with the nonlinearity uux in the well-posedness framework.
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6.3 Boundary control from the left
The backstepping method has been applied to build some boundary feedback laws, which locallystabilize the Korteweg-de Vries equation posed on a finite interval. Our control acts on theDirichlet boundary condition at the left hand side of the interval where the system evolves. Theclosed-loop system is proven to be locally exponentially stable with a decay rate that can bechosen to be as large as we want.
The situation where we act on the right end-point is different. If we consider homogeneousDirichlet condition on the left and one or two control inputs at the right hand side of theinterval, then we are not able to prove the backstepping method works with the transformation(72). Indeed, when imposing (vt+vx+vxxx+ωv = 0) on the target system, we get the followingexpression at x = L to vanish
k(x, L)uxx(t, L) + k(x, L)u(t, L) + kyy(x, L)u(t, L)− ky(x, L)ux(t, L).(115)
As we do not have to our disposal uxx(t, L), the first term above arises the condition k(x, L) = 0.Even if we do not care about the two last terms in (115), in order to keep w(t, 0) = u(t, 0) = 0,we have to impose k(0, y) = 0 for any y ∈ (0, L). With these four boundary restrictions (theother two are on k(x, x)), the third order kernel equation satisfied by k = k(x, y) becomesoverdetermined. Therefore, it is not clear if such a function k = k(x, y) exists. A natural idea todeal with controls at x = L is to use the transformation
v(t, x) = u(t, x)−∫ x
0
k(x, y)u(t, y)dy,(116)
instead of (72). However, it is not clear if that approach works. In fact, if we do that, we haveto deal now with the extra condition ky(x, 0) = 0 for any x ∈ (0, L). This is due to the fact thatwhen imposing vt+vx+vxxx+ωv = 0 on the target system, we get the extra term ux(t, 0)ky(x, 0)to be cancelled. As previously, this fourth restriction gives an overdetermined kernel equationfor k = k(x, y). Moreover, the existence of critical lengths when only one control is consideredat the right end-point suggests that either the existence of the kernel or the inversibility of thecorresponding map Π should fail for some spatial domains.
A still open problem is the output feedback control problem. We believe it could be solvedby applying the backstepping approach in order to built some observers as done in [41, 19] forthe heat and the wave equations respectively.
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