Rapid Stabilization of a Linearized Bilinear $1-D$ Schr\"odinger Equation

Abstract

We consider the one dimensional Schr\"odinger equation with a bilinear control and prove the rapid stabilization of the linearized equation around the ground state. The feedback law ensuring the rapid stabilization is obtained using a transformation mapping the solution to the linearized equation on the solution to an exponentially stable target linear equation. A suitable condition is imposed on the transformation in order to cancel the non-local terms arising in the kernel system. This conditions also insures the uniqueness of the transformation. The continuity and invertibility of the transformation follows from exact controllability of the linearized system.

Full text

arXiv:1611.03738v1 [math.OC] 11 Nov 2016
Rapid Stabilization of a Linearized Bilinear 1-D Schr¨
odinger
Equation
Jean-Michel CORON∗, Ludovick GAGNON†, Morgan MORANCEY‡
Abstract
We consider the one dimensional Schr¨odinger equation with a bilinear control and prove the rapid
stabilization of the linearized equation around the ground state. The feedback law ensuring the rapid
stabilization is obtained using a transformation mapping the solution to the linearized equation on the
solution to an exponentially stable target linear equation. A suitable condition is imposed on the trans-
formation in order to cancel the non-local terms arising in the kernel system. This conditions also insures
the uniqueness of the transformation. The continuity and invertibility of the transformation follows from
exact controllability of the linearized system.
Keywords: Schr¨odinger equation, Rapid stabilization, Integral transform.
2010 Mathematics Subject Classification: 93D15, 93B52.
1 Introduction
1.1 Main result
Let T > 0. Consider the Schr¨odinger equation
(i∂tψ=−∆ψ−u(t)µ(x)ψ, (t, x)∈(0, T )×(0,1),
ψ(t, 0) = ψ(t, 1) = 0, t ∈(0, T ).(1.1)
In (1.1), ψis the complex-valued wave function, of L2-norm 1, of a particle confined in a 1−Din-
finite square potential well. The particle is subjected to an electric field inside of the domain, where
u∈L2((0, T ); R)is the amplitude of the electric field and µ∈H3((0,1); R)is the dipolar moment of the
particle.
Before stating our main result, we set some notations. Let A:D(A)⊂L2((0,1); C)→L2((0,1); C)be
defined by
Aφ:= −∆φ, D(A) := H2((0,1); C)∩H1
0((0,1); C).(1.2)
The eigenvalues and eigenfunctions of Aare given by
λk:= (kπ)2, ϕk(x) := √2 sin(kπx), k ∈N∗.
The eigenstates of (1.1) (u= 0) are given by Φk(t, x) := e−iλktϕk(x). The eigenstate Φ1associated to
the smallest eigenvalue is called the ground state.
Define the space Hs
(0)((0,1); C) := D(As/2)equipped with the inner product
hφ, ψiHs
(0) :=
+∞
X
k=1
λs
khφ, ϕkihψ, ϕki,
where h·,·iis the L2((0,1); C)-inner product. The space Hs
(0) is endowed with the k.kHs
(0) -norm associated
to the Hs
(0)-inner product. We underline that the spaces used in this article can also be explicitly described
∗Universit´e Pierre et Marie Curie, LJLL UMR 7598, 4 place Jussieu, 75005, Paris, France.
†Universit´e de Nice Sophia-Antipolis, CNRS UMR 7351, Laboratoire J.-A. Dieudonn´e, Parc Valrose, 06108, Nice, France.
‡Aix Marseille Universit´e, CNRS, Centrale Marseille, I2M UMR 7373 13453, Marseille, France.
1

by H2
(0)((0,1); C) = H2∩H1
0((0,1); C)and
H3
(0)((0,1); C) = φ∈H3∩H1
0((0,1); C) ; φ′′ (0) = φ′′ (1) = 0,
H5
(0)((0,1); C) = nφ∈H5∩H3
(0)((0,1); C) ; φ(4)(0) = φ(4)(1) = 0o.
We denote by Sthe radius 1sphere of L2((0,1); C).
Throughout this article, we assume that µsatisfies the following assumption.
Hypothesis 1.1 The function µbelongs to H3((0,1); R)and there exists c > 0satisfying
|hµϕ1, ϕki| ≥ c
k3,∀k∈N∗.(1.3)
Remark 1.2 A direct computation shows that, for every function µ∈H3((0,1); R), we have
hµϕ1, ϕki=4
(kπ)3(−1)k+1 µ′(1) −µ′(0)−√2
(kπ)3Z1
0
(µϕ1)′′′(x) cos(kπx)dx (1.4)
and therefore there exists C > 0such that
|hµϕ1, ϕki| ≤ C
k3,∀k∈N∗.(1.5)
Moreover, since
lim
k→+∞Z1
0
(µϕ1)′′′(x) cos(kπx)dx = 0,(1.6)
it follows from (1.4) that (1.3) implies that
µ′(1) 6=µ′(0) and µ′(1) 6=−µ′(0).(1.7)
As proved in [8], Hypothesis 1.1 is not necessary to get local exact controllability of (1.1). However
(see [5]) it is a necessary and sufficient condition to get exact controllability of the following linearized
equation around the ground state





i∂tΨ = −∆Ψ −v(t)µ(x)Φ1(t, x),(t, x)∈(0, T )×(0,1),
Ψ(t, 0) = Ψ(t, 1) = 0, t ∈(0, T ),
Ψ(0, x) = Ψ0(x), x ∈(0,1).
(1.8)
Theorem 1.3 ( [5] ) Let T > 0and assume that µsatisfies Hypothesis 1.1. Then, for every
Ψ0∈nφ∈H3
(0) ;ℜhφ, ϕ1i= 0o=: H0,ΨT∈nφ∈H3
(0) ;ℜhφ, Φ1(T)i= 0o=: HT,(1.9)
there exists v∈L2((0, T ); R)such that the solution Ψof (1.8) with the initial condition Ψ(0, .) = Ψ0
satisfies, Ψ(T , .) = ΨT.
Condition (1.9) means that Ψ0and ΨTlie in the tangent vector space of Sin ϕ1and Φ1(T), respectively.
Due to the linearization of the preservation of the norm for the bilinear problem, the solution of (1.8)
satisfies ℜhΨ(t),Φ1(t)i= 0 for every t≥0.
The main result of this paper is the construction of feedback laws leading to rapid stabilization of the linear
control system (1.8).
Theorem 1.4 Let T > 0. Assume that µsatisfies Hypothesis 1.1. Then, for every λ > 0, there exists
C > 0and a feedback law v(t) = K(Ψ(t, ·)) such that for every Ψ0∈ H0the associated solution of (1.8)
satisfies
kΨ(t, ·)kH3
(0) ≤Ce−λt kΨ0kH3
(0) .
2

For the sake of simplicity we will focus, for the rest of this article, on the rapid stabilization of the
following linear Schr¨odinger equation





i∂tΨ = −∆Ψ −v(t)µ(x)ϕ1(x),(t, x)∈(0, T )×(0,1),
Ψ(t, 0) = Ψ(t, 1) = 0, t ∈(0, T ),
Ψ(0, x) = Ψ0(x), x ∈(0,1).
(1.10)
The only difference between (1.8) and (1.10) is that the control term “v(t)µ(x)Φ1(t, x)” has been replaced
by “v(t)µ(x)ϕ1”. Using again [5, Proposition 4], we get the analogous of Theorem 1.3 that is exact
controllability with L2((0, T ); R)controls of system (1.10) but now in the state space H3
(0)((0,1); C).
We prove the following rapid stabilization result.
Theorem 1.5 Let T > 0. Assume that µsatisfies Hypothesis 1.1. Then, for every λ > 0, there exists
C > 0and a real-valued feedback law v(t) = K(ψ(t, ·)) such that, for every Ψ0∈H3
(0)((0,1); C), the
associated solution of (1.10) satisfies
kΨ(t, ·)kH3
(0) ≤Ce−λt kΨ0kH3
(0) .
Remark 1.6 The final goal would be to achieve local rapid stabilization of the bilinear problem (1.1)
toward the ground state Φ1. To avoid dealing with a moving target, notice that
kψ(t, ·)−Φ1(t, ·)kH3
(0) =keiλ1tψ(t, ·)−ϕ1kH3
(0) .
Thus it is simpler to look at the system satisfied by eiλ1tψ(t, ·). In the same spirit, we will develop the proof
of Theorem 1.5 in this article and detail in Appendix C how we can modify the proof to obtain Theorem 1.4.
The obtained feedback law does not allow us, for now, to obtain rapid stabilization of (1.1).
1.2 A finite dimensional example
Let us explain the general idea of the proof of Theorem 1.5 in a finite dimensional setting. Let A∈Rn×n
and B∈Rn. Consider the control system
x′(t) = Ax(t) + Bu(t)(1.11)
where, at time t, the state is x(t)∈Rnand the control is u(t)∈R. We assume that the system is
controllable, which is equivalent to the Kalman rank condition
rank(B , AB, . . . , An−1B) = n(1.12)
(see e.g. [18, Theorem 1.16]). It is well-known that the controllability allows one to use the pole-shifting
theorem [18, Theorem 10.1]) to design a feedback law u(t) = Kx(t)to obtain the exponential stability
with an arbitrary exponential decay rate of (1.11). Let us present a different approach, more suitable for
PDEs, to this result. Let λ∈Rand denote the identity matrix of size nby I. Consider the target system
y′(t) = (A−λI)y(t) + Bv(t)(1.13)
where, at time t,y(t)∈Rnis the state and v(t)∈Ris the control. A straightforward computation shows
that, for v≡0, the solutions to (1.13) satisfy
ky(t)k ≤ e−(λ−kAk)tky(0)k.
Let us assume, for the moment, that we can design a transformation (T, K )∈Rn×n×R1×nsuch that
if x(t)is the solution of (1.11) with
u(t) := Kx(t) + v(t),(1.14)
then y(t) := T x(t)is the solution of (1.13). Notice that if moreover Tis invertible, then
kx(t)k=kT−1y(t)k ≤ kT−1ke−(λ−kAk)tky(0)k
≤ kT−1ke−(λ−kAk)tkT x(0)k ≤ kT−1kkTke−(λ−kAk)tkx(0)k.
3

Therefore, the exponential stability of (1.11) with an arbitrary exponential decay rate is reduced to find
such a transformation (T, K)with Tinvertible. The transformation (T , K)maps (1.11) into
y′(t) = T x′(t) = T(Ax(t) + BK x(t) + Bv(t)) = (T A +T B K)x(t) + T Bv (t),
Hence this transformation maps (1.11) into (1.13) if and only if
T A +BK =AT −λT, (1.15)
T B =B. (1.16)
One has the following theorem.
Theorem 1.7 ([19]) There exists one and only one (T , K)∈GLn(R)×R1×nsatisfying (1.15)-(1.16).
The proof of Theorem 1.7 provided in [19] relies on the phase variable canonical form (also called
controller form) of (1.11). We present here a different proof (in the case where the eigenvalues of Aare
simple) more suitable to deal with the infinite dimensional setting, with the additional assumption
λ > 0is such that ((λi+λ)I−A)is invertible ∀1≤i≤n. (1.17)
Proof: We first prove that the result holds for (T , K)∈GLn(C)×C1×n. The fact that (T , K)are
real-valued follows from the uniqueness of the transformations and the fact that Aand Bare real-valued.
Denote by {λi, ei}1≤i≤nthe eigenvalues and eigenvectors of A. Then, (1.15)-(1.16) become
((λi+λ)I−A)T ei=−BKei,(1.18)
T Bei=Bei.(1.19)
The proof is then divided in four steps.
Step 1: Existence of a basis of the state space.
Assumption (1.17) implies that that there exists nvectors fi,1≤i≤nsatisfying
((λi+λ)I−A)fi=−B, 1≤i≤n. (1.20)
We begin by proving that the set {fi}forms a basis of Cn. Notice that if Kis known, then one recovers
T eifrom the relation T ei=fiK ei.
Suppose there exists {ai}1≤i≤n⊂C,{ai}1≤i≤n6= 0 such that
n
X
i=1
aifi= 0.(1.21)
Applying Ato this equation and using (1.20), we obtain
n
X
i=1
ai(λi+λ)fi=− n
X
i=1
ai!B.
Applying successively A, we end up with
n
X
i=1
ai(λi+λ)pfi=−
p
X
k=1 n
X
i=1
ai(λi+λ)p−k!Ak−1B, ∀p∈N∗.(1.22)
Note that, for all j∈N∗, each coefficient
n
X
i=1
ai(λi+λ)j,(1.23)
appears in (1.22) for all p≥j+ 1 in front of Ap−j−1B. We distinguish two cases. If there exists j∈N∗
such that there is a coefficient (1.23) that is not equal to zero, then it implies that {Ap−j−1B}p≥j+1 ⊂
4

span{fi}. From the controllability assumption it comes that span{Ap−j−1B}p≥j+1 =Cn. Therefore, in
this case, the set of nvectors {fi}generates the whole space and consequently a basis of Cn.
The remaining case is the situation where every coefficient (1.23) vanishes i.e.
n
X
i=1
ai(λi+λ)j= 0,∀j∈N∗.(1.24)
In this case, consider the entire function
G:z∈C7→
n
X
i=1
aie(λi+λ)z.
From (1.24), we obtain for all j∈N,
G(j)(0) =
n
X
i=1
ai(λi+λ)j= 0.
Therefore G≡0. Let C:= Conv{λi+λ; 1 ≤i≤nsuch that ai6= 0}where, for a nonempty subset R
of C, ConvRis the closed convex-hull of R. The set Chas at least one nonzero extremal, that is a point of
Csuch that there exists at least one hyperplane that meets Conly on this point. One such point must be of
the form λk0+λfor 1≤k0≤n. Therefore, there exists θ∈[0,2π]such that
ℜ(eiθ(λk+λ)) <ℜ(eiθ (λk0+λ)),∀1≤k≤n, k 6=k0.(1.25)
Let z=seiθ where s∈R. We have
e−s(λk0+λ)eiθ G(seiθ) = ak0+
n
X
i=1,i6=k0
aies(λi−λk0)eiθ ≡0.(1.26)
From (1.25) and by letting s→ ∞ in (1.26), we obtain that ak0= 0. It is in contradiction with the fact
that the set Ccontains only nonzero ai. Therefore ai= 0,1≤i≤nso the set {fi}is independant and
consequently a basis of Cn. The two cases were covered which implies that {fi}is a basis of Cn.
Remark 1.8 The latter part of the proof of the existence of a basis could have been done using the Vander-
monde matrix. The proof presented here has the advantage that it may be applied in the infinite dimensional
setting.
Step 2: Existence of the transformation (T, K ).
The transformation is obtained using (1.19). Indeed, let
B=
n
X
i=1
biei.
Notice that, by the controllability assumption, bi6= 0,1≤i≤n. Then,
T B =B⇐⇒ B=
n
X
i=1
biT ei=
n
X
i=1
biKeifi.(1.27)
Since {fi}1≤i≤nis a basis of Cn, there exists {K ei}1≤i≤n⊂Csuch that the last equation is verified,
allowing to define T∈Cn,n and K∈C1,n such that (1.18) and (1.19) hold.
Step 3: Uniqueness of (T , K).
To prove the uniqueness of the transformations (T , K), consider (T1, K1)and (T2, K2)solutions of
(1.18)-(1.19). Therefore (T1−T2, K1−K2)satisfies (1.18) and
(T1−T2)B= 0.(1.28)
5

Since (T1−T2, K1−K2)satisfies (1.18), we use the basis constructed previously and (1.28) to prove that
K1=K2and T1=T2. With the uniqueness of the transformation and the fact that Aand Bare real-
valued, one ensures that the transformations are real-valued since (T , K )is also a solution of (1.18)-(1.19).
Step 4: Invertibility of T.
Let T∈Cn,n and K∈C1,n be such that (1.18) and (1.19) hold. We prove that Tis invertible by
showing that Ker T∗={0}. Let x∈Ker T∗. From (1.15)-(1.16), we obtain
T∗A∗x= (A∗T∗+K∗B∗T∗+λT ∗)x= 0.
Therefore Ker T∗is stable by A∗. From (1.16) it comes that
B∗x=B∗T∗x= 0.
Thus there exists ˜xeigenvector of A∗in Ker T∗⊂Ker B∗. From the controllability assumption and the
Hautus test (see for instance [48, Prop. 1.5.5]) it comes that Ker T∗={0}.

If the functional setting in the infinite dimensional case makes the proof more tricky, the strategy we
use remains the same. Riesz basis results will be used to prove the existence of a basis of the state space and
the invertibility of the transformation will be proved using the approximate controllability of the studied
system.
The main technical difficulty of this paper lies in the decomposition of B(1.27) in the basis of the state
space. Indeed, the control operator Bis admissible but not bounded in the state space. A careful analysis of
the Fourier components of the control operator Ballow us to define a transformation Twhich is bounded
from the state space into itself but the feedback transformation won’t be bounded from the state space into
R. Even so, the transformation Twill be proved to be invertible and the closed-loop linear equation will
be proved to be well-posed in the state space. It is important to note that this technical difficulty is in fact
essential for the invertibility of T(see Remark 3.6). Indeed, in our case, if Bwere to be bounded, then the
transformation Twould be compact and thus not continuously invertible. However, the unboundedness of
Kfrom the state space into Rprevents us to prove directly the well-posedness of the closed-loop nonlinear
equation.
Let us underline that the uniqueness condition T B =B, which was used implicitly in similar previous
works, will be crucial not only to obtain the existence and uniqueness of the transformation, but also to
define rigorously (1.15) in the infinite dimensional setting.
1.3 The linear Schr¨
odinger equation
As presented in the previous paragraph, the strategy to prove the rapid stabilization of the linear equation
(1.10) is inspired by the backstepping method. Recast the equations for Ψ1+iΨ2= Ψ as









∂t Ψ1
Ψ2!= 0−∆
∆ 0 ! Ψ1
Ψ2!+v(t) 0
(µϕ1)(x)!,(t, x)∈(0, T )×(0,1),
Ψ1(t, 0) = Ψ1(t, 1) = 0,Ψ2(t, 0) = Ψ2(t, 1) = 0, t ∈(0, T ),
Ψ1(0, x) = Ψ1
0(x),Ψ2(0, x) = Ψ2
0(x), x ∈(0,1).
(1.29)
where Ψ1
0and Ψ2
0are the real and imaginary part of Ψ0respectively. From now on, all the functional
spaces are real-valued, except when specified. Moreover to deal with those real and imaginary parts, we
denote, for simplicity,
Xs
(0)((0,1); R) := Hs
(0)((0,1); C)∩L2((0,1); R)2,(1.30)
with the product topology. We will use the following operators
A:D(A)−→ X3
(0) B:R−→ C∞×(H3∩H1
0)((0,1); R)T
Ψ1
Ψ27−→ −∆Ψ2
∆Ψ1, a 7−→ a0
(µϕ1)(x),
6

with D(A) := X5
(0). Based on the previous work of the first author and Q. L ¨u ([22, 23]), instead of Volterra
transformations of the second kind usually used for the backstepping approach, we seek for transformations
(T, K)of the form
T:X3
(0) −→ X3
(0)
Ψ1
Ψ27−→ Z1
0k11(x, y)k12(x, y)
k21(x, y)k22(x, y)Ψ1(y)
Ψ2(y)dy, (1.31)
K:D(K)⊂X3
(0) −→ R
Ψ1
Ψ27−→ Z1
0
α1(y)Ψ1(y) + α2(y)Ψ2(y)dy, (1.32)
such that if (Ψ1,Ψ2)Tis solution of (1.29) with
v(t) := KΨ1(t, .)
Ψ2(t, .),(1.33)
then (ξ1(t, .), ξ2(t, .))T:= T(Ψ1(t, .),Ψ2(t, .))Tis solution to









∂t ξ1
ξ2!= 0−∆
∆ 0 ! ξ1
ξ2!−λ ξ1
ξ2!,(t, x)∈(0, T )×(0,1),
ξ1(t, 0) = ξ1(t, 1) = 0, ξ2(t, 0) = ξ2(t, 1) = 0, t ∈ ×(0, T ),
ξ1(0, x) = ξ1
0(x), ξ2(0, x) = ξ2
0(x), x ∈ ×(0,1),
(1.34)
with (ξ1
0, ξ2
0)T=T(Ψ1
0,Ψ2
0)Tand Tis invertible in the state space. The decomposition in real and imagi-
nary part of the solution of (1.10) is made in order to ensure that the feedback v(t) = K(Ψ1(t, .),Ψ2(t, .))T
is real-valued. Note that



ξ1(t)
ξ2(t)


X3
(0) ≤e−λt 


ξ1
0
ξ2
0


X3
(0)
, t ∈[0,+∞).(1.35)
The kernels are defined through the equations they must satisfy for (T , K)to map solutions of (1.29) to
solutions of (1.34). This is done formally in Section 2 together with a more detailed presentation of this
strategy.
1.4 A brief review of previous results
The controllability properties for the Schr ¨odinger equation were mostly studied in the usual (in opposition
to the bilinear model presented here) linear setting. For the control of the linear Schr¨odinger equation with
internal control (localized on a subdomain), we refer to the survey [33] and the references therein. In this
more classical setting we also mention [35, 32, 26] concerning stabilization.
Exact controllability of the bilinear Schr¨
odinger equation.
The first local controllability results on the bilinear Schr¨odinger equation appear in [2, 3, 5]. These local
controllability results have been extended with weaker assumptions in [8], in a more general setting in
infinite time [42] and also in the case of simultaneous controllability of a finite number of particles in
[38, 39]. Note that, despite the infinite speed of propagation, it was proved in [17, 4, 8, 38] that a minimal
amount of time is required for the controllability of some bilinear Schr¨odinger equations. More recently,
local exact controllability has been established in [6] for a Schr ¨odinger-Poisson model in 2D (see also [36]
for approximate controllability) and for the analogue of (1.1) with a control depending on time and space
in dimension less or equal than 3[45].
Approximate controllability and stabilization of the bilinear Schr¨
odinger equation.
The above mentioned results of exact controllability are mostly limited to the one dimensional case. In a
more general setting, the available results deal with approximate controllability. Using geometric control
7

techniques on appropriate Galerkin approximations, approximate controllability in different settings has
been proved [13, 10, 9, 16]. For a detailed presentation, see the survey [11].
However, most of these results are not suitable to prove approximate controllability in higher norms
(typically H3
(0)) and thus approximate controllability for bilinear Schr¨odinger equations has also been
studied from the Lyapunovfunctional point of view [37, 7, 40, 41]. Though it enabled global controllability
results, this strategy usually gives no indication on the convergencerate.
Rapid stabilization.
The strategy used in this article is inspired from backstepping techniques. Initially developed to design,
in a recursive way, more effective feedback laws for globally asymptotic stable finite dimensional system
for which a feedback law and a Lyapunov function are already known (we refer to [30, 46, 18] for a
comprehensive introduction in finite dimension and [20, 34] for an application of this method to partial
differential equations), the backstepping approach was later used in the infinite dimensional setting to
design feedback laws by mapping the system to stabilize to a target stable system. To our knowledge,
this strategy was first introduced in the context of partial differential equations to design a feedback law
for heat equation [1] and, later on, for a class of parabolic PDE [12]. The backstepping-like change of
coordinates of the semi-discretized equation maps the discrete solution to stabilize to a stable solution.
The corresponding continuous mapping obtained is a Volterra transformation of the second kind, seen
as an infinite dimensional backstepping transformation from the triangular domain of definition of the
transformation. This backstepping strategy in infinite dimensions led to a series of work (see [31] for
a global presentation and [29] for the use of the backstepping approach for the rapid stabilization of a
Schr¨odinger equation with a boundary control). The backstepping approach can be used to get stabilization
in finite time as shown in [24, 25]. Moreover it is not limited to linear equations, as shown in [14, 25].
Though using Volterra transformations of the second kind provides easily invertible transformations it is
also limited. Fredholm transformations, from which this paper is inspired, has already been used [22, 23,
21] for rapid stabilization using boundary feedback laws.
Abstract methods have been developed to obtain the rapid stabilization of linear partial differential
equations. Among them, we cite the works [28], [50] and [49], based on the Gramian approach and the
Riccati equations, which could be applied to obtain the rapid stabilization of the linearized equation (1.29)
as (0, µϕ1)∈D(A∗)′. However, it seems difficult to obtain, for various physical systems, the local
rapid stabilization of the nonlinear equation using those methods. For example, at least for the moment,
one does not know how to deduce from [15], where the rapid stabilization of a linearized Korteweg-de
Vries equation is obtained by using the method developed in [49], the rapid stabilization of the associated
nonlinear Korteweg-de Vries equation. This is in contrast with the method used here (linear transformations
to suitable target systems) applied to the same Korteweg-de Vries equation. Indeed, as shown in [22], the
feedback laws obtained by means of this method allows to get the rapid stabilization for the nonlinear
Korteweg-de Vries equation. One may therefore hope that, as in [22], our feedback law Kbeing quite
explicit it might allow to obtain the rapid stabilization of the nonlinear equation. Note however that in
[22] the feedback law Kis continuous, which is not the case in our situation. It makes the application to
the nonlinear system more complicated to study and requires suitable nonlinear modifications of the linear
feedback law K.
1.5 Structure of the article
In Sec. 2, we give a detailed presentation of the strategy used to construct the transformation (T, K )and
give a formal expression of this transformation. In Sec. 3 we prove that this formal transformation Tis well
defined and is continuous in the state space X3
(0). Then, we prove in Sec. 4 that the previous transformation
is indeed invertible in the state space. These properties of Twill follow using Hypothesis 1.1 i.e. exact
controllability of the linearized system. We end the proof of Theorem 1.5 in Sec. 5 by proving that the
constructed feedback Kleads to a well-posed closed-loop system (i.e. the equation (1.29) with vdefined
by (1.33)) and that Tactually maps the closed-loop system to the exponentially stable solutions of (1.34).
In Appendix A we study in a similar way a simplified Saint-Venant equation which exhibits the same
phenomenon but on which we explicitly compute the transformation (T, K ).
8

2 Heuristic construction of the transformations
We recall that we look for a transformation (T, K )of the form (1.31)- (1.32). Let us derive the set of
equations for (T , K)to map solutions of (1.29) to solutions of (1.34). First, to ensure that the boundary
conditions of (1.34) are satisfied, we assume that
kij (0, y) = kij (1, y) = 0,∀y∈(0,1),∀i, j ∈ {1,2}.(2.1)
Using the fact that (Ψ1,Ψ2)T∈X3
(0), and imposing the conditions
kij (x, 0) = kij (x, 1) = 0,∀x∈(0,1),∀i, j ∈ {1,2},(2.2)
formal computations, denoting ∆xand ∆ythe Dirichlet Laplacian with respect to xand yvariables re-
spectively, lead to
∂tξ1(t, x) + ∆ξ2(t, x) + λξ1(t, x)
=Z1
0
k11(x, y)−∆yΨ2(t, y)+k12 (x, y)∆yΨ1(t, y) + v(t)(µϕ1)(y)dy
+Z1
0
∆xk21(x, y)Ψ1(t, y) + ∆xk22 (x, y)Ψ2(t, y )dy
+Z1
0
λk11(x, y)Ψ1(t, y) + λk12 (x, y)Ψ2(t, y )dy
=Z1
0
(∆xk21 + ∆yk12 +λk11) (x, y)Ψ1(t, y)dy
+Z1
0
(∆xk22 −∆yk11 +λk12) (x, y)Ψ2(t, y)dy
+v(t)Z1
0
(µϕ1) (z)k12(x, z)dz.
The boundary conditions (2.2) were imposed to avoid boundary terms in the integrations by parts. Using
the expression (1.32) of the feedback leads to
∂tξ1(t, x) + ∆ξ2(t, x) + λξ1(t, x)
=Z1
0(∆yk12 + ∆xk21 +λk11) (x, y) + α1(y)Z1
0
(µϕ1) (z)k12(x, z)dzΨ1(t, y)dy
−Z1
0(∆yk11 −∆xk22 −λk12) (x, y)−α2(y)Z1
0
(µϕ1) (z)k12(x, z )dzΨ2(t, y)dy. (2.3)
In the same way one gets
∂tξ2(t, x)−∆ξ1(t, x) + λξ2(t, x)
=Z1
0(∆yk22 −∆xk11 +λk21) (x, y) + α1(y)Z1
0
(µϕ1) (z)k22(x, z)dzΨ1(t, y)dy
−Z1
0(∆yk21 + ∆xk12 −λk22) (x, y)−α2(y)Z1
0
(µϕ1) (z)k22(x, z )dzΨ2(t, y)dy. (2.4)
9

If we want (ξ1, ξ2)to satisfy (1.34) then we need to find the functions kij and αjsatisfying







































(∆yk11 −∆xk22 −λk12) (x, y)−α2(y)Z1
0
(µϕ1) (z)k12(x, z)dz= 0,(x, y )∈(0,1)2,
(∆yk12 + ∆xk21 +λk11) (x, y) + α1(y)Z1
0
(µϕ1) (z)k12(x, z)dz= 0,(x, y )∈(0,1)2,
(∆yk21 + ∆xk12 −λk22) (x, y)−α2(y)Z1
0
(µϕ1) (z)k22(x, z)dz= 0,(x, y )∈(0,1)2,
(∆yk22 −∆xk11 +λk21) (x, y) + α1(y)Z1
0
(µϕ1) (z)k22(x, z)dz= 0,(x, y )∈(0,1)2,
kij (x, 0) = kij (x, 1) = 0, x ∈(0,1),
kij (0, y) = kij (1, y) = 0, y ∈(0,1).
(2.5)
A fundamental extra condition. One could try to solve rightaway (2.5) and prove the invertibility of the
transformation Tbut the non-local terms yield a tedious task. To overcome this difficulty, one introduces,
as in the finite dimensional setting, what will be referred throughout this article as the T B =Bcondition,
T0
(µϕ1)(x)=Z1
0k12(x, y)(µϕ1)(y)
k22(x, y)(µϕ1)(y)dy =0
(µϕ1)(x).
Plugging this into (2.5) we obtain that we now seek for a solution to



















(∆yk11 −∆xk22 −λk12) (x, y) = 0,(x, y)∈(0,1)2,
(∆yk12 + ∆xk21 +λk11) (x, y) = 0,(x, y)∈(0,1)2,
(∆yk21 + ∆xk12 −λk22) (x, y)−α2(y)(µϕ1)(x) = 0,(x, y)∈(0,1)2,
(∆yk22 −∆xk11 +λk21) (x, y) + α1(y)(µϕ1)(x) = 0,(x, y)∈(0,1)2,
kij (x, 0) = kij (x, 1) = 0, x ∈(0,1),
kij (0, y) = kij (1, y) = 0, y ∈(0,1),
(2.6)
together to the T B =Bcondition







Z1
0
k12(x, y)(µϕ1)(y)dy = 0, x ∈(0,1),
Z1
0
k22(x, y)(µϕ1)(y)dy = (µϕ1)(x), x ∈(0,1).
(2.7)
Remark 2.1 In [22, 23], the authors were dealing with a boundary control for the Korteweg-de Vries
equation and for the Kuramoto-Sivashinsky equation. In their case, the T B =Bcondition writes
ky(x, L) = 0, x ∈(0, L),
for the former and
kyy (x, L) = 0, x ∈(0, L),
for the latter, where kis the kernel of the Fredholm transformation in each case. Contrary to our frame-
work, this boundary condition appeared naturally from the integration by parts performed in order to
obtain the equation on the kernel. It was not seen as a particular boundary condition although, a careful
analysis of their work shows that the T B =Bcondition was used to prove the uniqueness of the transfor-
mation (T , K). The common ground between their work and this article is the additional regularity that
the kernel needs in order to satisfy the T B =Bcondition. Notice that in what we present in this article,
the relation between the kernels kij and αjis more intricate and considerably modify the analysis.
10

Formal decomposition The global strategy to construct a solution of (2.6)-(2.7) is the following. First
assume that α1and α2are known. This enables us to compute the kernels kij satisfying (2.6) as functions
of α1and α2. Then we prove that we find α1and α2such that (2.7) is satisfied.
We decompose the functions in the following form
kij (x, y) =
+∞
X
n=1
fij
n(x)ϕn(y), αj(y) =
+∞
X
n=1
αj
nϕn(y).(2.8)
This leads to 
















f11
n′′(x) + λnf22
n(x)−λf21
n(x) = α1
n(µϕ1)(x),
f12
n′′(x)−λnf21
n(x)−λf22
n(x) = α2
n(µϕ1)(x),
f21
n′′(x)−λnf12
n(x) + λf11
n(x) = 0,
f22
n′′(x) + λnf11
n(x) + λf12
n(x) = 0,
fij
n(0) = fij
n(1) = 0.
(2.9)
The T B =Bcondition (2.7) becomes













+∞
X
n=1hµϕ1, ϕnif12
n(x) = 0,
+∞
X
n=1hµϕ1, ϕnif22
n(x) = (µϕ1)(x).
(2.10)
As mentioned, we begin by assuming that the feedback law is known. We considertwo sequences (β1
n)n∈N∗
and (β2
n)n∈N∗to be precised later on.
Let (g11
n, g12
n, g21
n, g22
n)Tbe the solution of system (2.9) with right-hand side β1
n(µϕ1),0,0,0Tand let
(h11
n, h12
n, h21
n, h22
n)Tbe the solution of system (2.9) with right-hand side 0, β2
n(µϕ1),0,0T. System (2.9)
being linear, it comes that
fij
n=α1
n
β1
n
gij
n+α2
n
β2
n
hij
n.(2.11)
Decomposing gij
nin the L2-orthonormal basis (ϕk)k∈N∗, if we denote by Ank the following matrix
Ank =



−λk0−λ λn
0−λk−λn−λ
λ−λn−λk0
λnλ0−λk



,(2.12)
system (2.9) leads to
Ank 



hg11
n, ϕki
hg12
n, ϕki
hg21
n, ϕki
hg22
n, ϕki



=



β1
nhµϕ1, ϕki
0
0
0



.
Then 


































g11
n(x) =
+∞
X
k=1
λkλ2
n−λ2−λ2
k
δnk(λ)β1
nhµϕ1, ϕkiϕk(x),
g12
n(x) =
+∞
X
k=1
2λλkλn
δnk(λ)β1
nhµϕ1, ϕkiϕk(x),
g21
n(x) =
+∞
X
k=1
−λλ2+λ2
k+λ2
n
δnk(λ)β1
nhµϕ1, ϕkiϕk(x),
g22
n(x) =
+∞
X
k=1
λnλ2−λ2
k+λ2
n
δnk(λ)β1
nhµϕ1, ϕkiϕk(x),
(2.13)
11

where
δnk(λ) = det(Ank) = λ2+ (λk−λn)2λ2+ (λk+λn)2.(2.14)
The same computations can be carried out for hij
n, leading to the following relations
h11
n=−β2
n
β1
n
g12
n, h12
n=β2
n
β1
n
g11
n,
h21
n=−β2
n
β1
n
g22
n, h22
n=β2
n
β1
n
g21
n.
(2.15)
For the sake of simplicity, we denote by cij
nk and dij
nk the coefficients such that
gij
n(x) =
+∞
X
k=1
cij
nkβ1
nhµϕ1, ϕkiϕk(x), hij
n(x) =
+∞
X
k=1
dij
nkβ2
nhµϕ1, ϕkiϕk(x).(2.16)
Summary of the construction. Finally, using the definition of the transformation (1.31), the decomposi-
tions (2.8), (2.11) and the relations (2.15) it comes that the transformation Twe are looking for is defined
by
TΨ1
Ψ2=





+∞
X
n=1 α1
n
β1
n
g11
n+α2
n
β2
n
h11
nhΨ1, ϕni+
+∞
X
n=1 α1
n
β1
n
g12
n+α2
n
β2
n
h12
nhΨ2, ϕni
+∞
X
n=1 α1
n
β1
n
g21
n+α2
n
β2
n
h21
nhΨ1, ϕni+
+∞
X
n=1 α1
n
β1
n
g22
n+α2
n
β2
n
h22
nhΨ2, ϕni






=
+∞
X
n=1 α1
nhΨ2, ϕni
β1
n−α2
nhΨ1, ϕni
β1
ng12
n
g22
n+α1
nhΨ1, ϕni
β2
n
+α2
nhΨ2, ϕni
β2
nh12
n
h22
n.
(2.17)
In the same spirit, the T B =Bcondition (2.10) becomes
0
µϕ1=
+∞
X
n=1
α1
nhµϕ1, ϕni
β1
ng12
n
g22
n+α2
nhµϕ1, ϕni
β2
nh12
n
h22
n.(2.18)
This ends the heuristic of the construction of the transformation and the feedback law. Indeed, in the
following section we prove that for a suitable choice of the sequences (β1
n)n∈N∗and (β2
n)n∈N∗then
B:= g12
n
g22
n,h12
n
h22
n;n∈N∗,(2.19)
is a Riesz basis of X2
(0) (see Proposition 3.4). Then from (2.18) we get the feedback laws from the expan-
sion of (0, µϕ1)Tin the basis B. Finally, we study the behaviour of the coefficients α1
nand α2
nas ngoes
to infinity to prove that the transformation Tgiven by (2.17) is indeed continuous from X3
(0) to X3
(0).
Remark 2.2 From (2.18) it already appears that the behaviour of the coefficients hµϕ1, ϕni, and thus
Hypothesis 1.1, will play a crucial role.
3 Definition and properties of the transformation
In this section, we make rigorous the heuristic developed in the previous section. In subsection 3.1, we
prove that for a suitable choice of β1
nand β2
nthen Bdefined in (2.19) is a Riesz basis of X2
(0) where the
functions gij
nand hij
nare defined by (2.13) and (2.15). This enables us to define the feedback law and the
transformation Tusing the relations (2.18) and (2.17). However, this does not give enough regularity to
prove that T:X3
(0) →X3
(0). We prove the extra regularity we need on the feedback laws in subsection 3.2.
This leads, in subsection 3.3, to the expected regularity for the transformation T.
12

3.1 Riesz basis property
Let us recall some results on Riesz basis.
Definition 3.1 Let Hbe an Hilbert space and {gn}n∈N∗⊂H. We say that {gn}n∈N∗is ω-independent if
X
n∈N∗
angn= 0,with {an}n∈N∗⊂R=⇒an= 0,∀n∈N∗.
Theorem 3.2 [51, Theorem 15] Let Hbe a separable Hilbert space and let {en}n∈N∗be an orthonormal
basis for H. If {gn}n∈N∗is an ω-independent sequence quadratically close to {en}n∈N∗, that is
X
n∈N∗ken−gnk2
H<+∞,
then {gn}n∈N∗is a Riesz basis for H.
Theorem 3.3 Let Hbe a separable Hilbert space and let {en}n∈N∗be an orthonormal basis for H. If
{gn}n∈N∗is dense in Hand is quadratically close to {en}n∈N∗, i.e.
X
n∈N∗ken−gnk2
H<+∞,
then {gn}n∈N∗is a Riesz basis for H.
Let us provide a proof of Theorem 3.3, stated as a remark in [27, Remark 2.1, p. 318].
Proof:
Let us prove Theorem 3.3 by contradiction. Suppose that the {gn}n∈N∗is dense in Hand is quadrat-
ically close to {en}n∈N∗but that {gn}n∈N∗is not a Riesz basis. Therefore, by Theorem 3.2, there must
exist a non-zero sequence {an}n∈N∗⊂Rsuch that
X
n∈N∗
angn= 0.
Since {gn}n∈N∗is quadratically close to {en}n∈N∗, there exists N∈N∗such that
X
n≥N+1 ken−gnk2
H<1.
Therefore, from [51, Theorem 13], the {en}1≤n≤N∪ {gn}n≥N+1 is a Riesz basis of H. This implies that
there exists k≤Nsuch that ak6= 0. Hence, with the density assumption, we have
H=span{gn|n∈N∗}= span{gn|n∈N∗\ {k}}.
Thus, codim(span{gn|n≥N+ 1})≤N−1.
However
H/span{gn|n≥N+ 1},
is isomorphic to span{en|n≤N}, which is of dimension N, leading to a contradiction.

We will use the previous criteria to prove the following proposition.
Proposition 3.4 Let βi
nbeing defined as (3.1). Then,
B:= g12
n
g22
n,h12
n
h22
n;n∈N∗
is a Riesz basis of X2
(0) and
˜
B:= ( g12
n/λ1/2
n
g22
n/λ1/2
n!, h12
n/λ1/2
n
h22
n/λ1/2
n!;n∈N∗)
is a Riesz basis of X3
(0).
13

To apply the previous criterion for the Riesz basis, we prove that B(resp. ˜
B) is quadratically close to
the orthonormal basis of X2
(0) (resp. X3
(0)) given by
ϕn/λs/2
n
0,0
ϕn/λs/2
n;n∈N∗,with s= 2 (resp. s= 3).
Thus, we choose β1
nand β2
nsuch that g12
nand h22
nare close to ϕn/λni.e. hg12
n, ϕni=hh22
n, ϕni= 1/λn,
that is,
β1
n:= λ(λ2+ 4λ2
n)
2λ3
nhµϕ1, ϕni,
β2
n:= −λ(λ2+ 4λ2
n)
(λ2+ 2λ2
n)λnhµϕ1, ϕni.
(3.1)
Notice that, from Hypothesis 1.1 and Remark 1.2, there exists C > 0such that
n
C≤β1
n≤C n, n
C≤β2
n≤C n, ∀n∈N∗.(3.2)
During the proof of Proposition 3.4, we will use the following lemma. Its proof is purely technical and
postponed to Appendix B.
Lemma 3.5 With the above definition, for s= 2 and s= 3, one gets
X
n∈N∗



ϕn/λs/2
n
0− g12
n/λ(s−2)/2
n
g22
n/λ(s−2)/2
n!




2
Xs
(0)
<+∞,
X
n∈N∗



0
ϕn/λs/2
n− h12
n/λ(s−2)/2
n
h22
n/λ(s−2)/2
n!




2
Xs
(0)
<+∞.
We are now ready to prove Proposition 3.4.
Proof: Let s= 2 or s= 3. In view of Theorem 3.2 and Lemma 3.5, assume that there exists a sequence
{an, bn}n∈N∗such that
X
n∈N∗
ang12
n
g22
n+bnh12
n
h22
n=0
0in Xs
(0).(3.3)
First step: we apply negative powers of the Laplacian to characterize elements of S:= span B.
Recall that Ais defined in (1.2). Let us write (2.9) for gij
nas
Agn=


λn



0 0 0 1
0 0 −1 0
0−1 0 0
1 0 0 0



+λ



0 0 −1 0
0 0 0 −1
1 0 0 0
0 1 0 0







|{z }
=:Jn
gn−



β1
nµϕ1
0
0
0




|{z }
=:Fn
,
where gn= (g11
n, g12
n, g21
n, g22
n)Tand where Agn= (Ag11
n,Ag12
n,Ag21
n,Ag22
n)T.
Since det(Jn) = (λ2+λ2
n)26= 0, we have
A−1gn=J−1
ngn+A−1J−1
nFn,
where
J−1
n=1
λ2+λ2
n




0 0 λ λn
0 0 −λnλ
−λ−λn0 0
λn−λ0 0



.
14

Therefore, using (2.15), we obtain
A−1g12
n=1
λ2
n+λ2λg22
n−λn
β1
n
β2
n
h22
n,
A−1g22
n=1
λ2
n+λ2λn
β1
n
β2
n
h12
n−λg12
n+λnβ1
nA−1(µϕ1),
A−1h12
n=1
λ2
n+λ2λn
β2
n
β1
n
g22
n+λh22
n,
A−1h22
n=−1
λ2
n+λ2λn
β2
n
β1
n
g12
n+λh12
n+λβ2
nA−1(µϕ1).
(3.4)
Thus, applying A−1to (3.3) leads to
c0A−1(µϕ1)
0=X
n∈N∗
λan+λnβ2
n
β1
nbn
λ2
n+λ2g12
n
g22
n+−λnβ1
n
β2
nan+λbn
λ2
n+λ2h12
n
h22
n,(3.5)
where
c0:= X
n∈N∗
λnβ1
nan−λβ2
nbn
λ2
n+λ2.(3.6)
Applying A−1to (3.5) we obtain
c1A−1(µϕ1)
c0A−2(µϕ1)=X
n∈N∗
(λ2−λ2
n)an+ 2λλnβ2
n
β1
nbn
(λ2
n+λ2)2g12
n
g22
n+−2λλnβ1
n
β2
nan+ (λ2−λ2
n)bn
(λ2
n+λ2)2h12
n
h22
n,
(3.7)
where
c1:= X
n∈N∗
2λλnβ1
nan+ (λ2
n−λ2)β2
nbn
(λ2
n+λ2)2.(3.8)
In order to iterate (3.5) and (3.7), notice that applying A−1to the relations (3.4) leads to
A−2g12
n=1
(λ2
n+λ2)2h(λ2
n−λ2)g12
n+ 2λλn
β1
n
β2
n
h12
n+ 2λλnβ1
nA−1(µϕ1)i,
A−2g22
n=1
(λ2
n+λ2)2h(λ2
n−λ2)g22
n+ 2λλn
β1
n
β2
n
h22
ni+λnβ1
n
λ2
n+λ2A−2(µϕ1),
A−2h12
n=1
(λ2
n+λ2)2h(λ2
n−λ2)h12
n−2λλn
β2
n
β1
n
g12
n+ (λ2
n−λ2)β2
nA−1(µϕ1)i,
A−2h22
n=1
(λ2
n+λ2)2h(λ2
n−λ2)h22
n−2λλn
β2
n
β1
n
g22
ni−λβ2
n
λ2
n+λ2A−2(µϕ1).
(3.9)
Applying successively A−2to (3.5) we obtain, for every p∈N(with the convention that the sum from 1
to 0is equal to 0),

c0A−(2p+1)(µϕ1) + Pp
j=1 Pn∈N∗
2λλnβ1
nkj
n+(λ2
n−λ2)β2
nlj
n
(λ2
n+λ2)2j+1 A−2(p−j)−1(µϕ1)
Pp
j=1 Pn∈N∗
λnβ1
nkj
n−λβ2
nlj
n
(λ2
n+λ2)2jA−2(p−j+1)(µϕ1)

=X
n∈N∗
1
(λ2
n+λ2)2p+1 kp+1
ng12
n
g22
n+lp+1
nh12
n
h22
n,
(3.10)
where the coefficients (kj
n, lj
n)are defined by
kj+1
n
lj+1
n=
(λ2−λ2
n) 2λλnβ2
n
β1
n
−2λλnβ1
n
β2
n(λ2−λ2
n)
kj
n
lj
n,with k1
n
l1
n:= 
λan+λnβ2
n
β1
nbn
−λnβ1
n
β2
nan+λbn
.(3.11)
15

Applying successively A−2to (3.7) we obtain, for every p∈N∗,

c1A−(2p+1)(µϕ1) + Pp
j=1 Pn∈N∗
2λλnβ1
n˜
kj
n+(λ2
n−λ2)β2
n˜
lj
n
(λ2
n+λ2)2(j+1) A−2(p−j)−1(µϕ1)
c0A−2(p+1) Pp
j=1 Pn∈N∗
λnβ1
n˜
kj
n−λβ2
n˜
lj
n
(λ2
n+λ2)2j+1 A−2(p−j+1)(µϕ1)

=X
n∈N∗
1
(λ2
n+λ2)2(p+1) ˜
kp+1
ng12
n
g22
n+˜
lp+1
nh12
n
h22
n,
(3.12)
where the coefficients (˜
kj
n,˜
lj
n)are defined by
˜
kj+1
n
˜
lj+1
n=
(λ2−λ2
n) 2λλnβ2
n
β1
n
−2λλnβ1
n
β2
n(λ2−λ2
n)
˜
kj
n
˜
lj
n,with ˜
k0
n
˜
l0
n=an
bn.(3.13)
Assuming that c06= 0, equality (3.5) implies that
A−1(µϕ1)
0∈span B.
Then using (3.7) we obtain
0
c0A−2(µϕ1)+c1A−1(µϕ1)
0∈span B=⇒0
A−2(µϕ1)∈span B.
Iterating with the relations (3.10) and (3.12) it comes that, for every p∈N∗,
A−(2p−1)(µϕ1)
0∈span Band 0
A−2p(µϕ1)∈span B.(3.14)
Notice that if c0= 0 and c16= 0, the same argument can be repeated to obtain (3.14). Actually, one
gets (3.14) as soon as there exists a non-zero coefficient in the left-hand side of (3.10) or in the left-hand
side of (3.12). Thus it comes that either (3.14) holds or, for every j∈N∗,
0 = X
n∈N∗
2λλnβ1
nkj
n+ (λ2
n−λ2)β2
nlj
n
(λ2
n+λ2)2j+1 ,
0 = X
n∈N∗
λnβ1
nkj
n−λβ2
nlj
n
(λ2
n+λ2)2j,
0 = X
n∈N∗
2λλnβ1
n˜
kj
n+ (λ2
n−λ2)β2
n˜
lj
n
(λ2
n+λ2)2(j+1) ,
0 = X
n∈N∗
λnβ1
n˜
kj
n−λβ2
n˜
lj
n
(λ2
n+λ2)2j+1 .
(3.15)
Second step: we prove that if (3.14) holds, then Bis a Riesz basis.
Let (d1, d2)T∈Xs
(0) such that
d1
d2,φ1
φ2= 0,∀φ1
φ2∈span B.
Using (3.14), we obtain that, for every p∈N∗,
0 = d1
d2,A−(2p−1)(µϕ1)
0Xs
(0)
=hd1,A−(2p−1)(µϕ1)iHs
(0)
=X
k∈N∗
λs
khd1, ϕkihϕk,A−(2p−1)(µϕ1)i
=X
k∈N∗
λs+1
khµϕ1, ϕkihd1, ϕki
λ2p
k
.
16

Let d1
k:= hd1, ϕkiand define
G:z∈C7→ X
k∈N∗
λs−1
khµϕ1, ϕkid1
kez/λ2
k∈C.
From uniform convergence on compact sets, it comes that Gis an entire function and the previous relation
imply that, for every p∈N∗,
G(p−1)(0) = X
k∈N∗
λs+1
khµϕ1, ϕkid1
k
λ2p
k
= 0.
Thus, G≡0. If d16= 0, let
n0:= min{n∈N∗;d1
n6= 0}.
It comes that
0 = λs−1
n0hµϕ1, ϕn0id1
n0+X
k>n0
λs−1
khµϕ1, ϕkid1
kexp z1
λ2
k−1
λ2
n0.(3.16)
As hµϕ1, ϕn0i 6= 0, considering zreal and letting zgo to +∞in (3.16) imply d1
n0= 0. Therefore, from
the definition of n0,d1= 0.
Using the exact same strategy for
0
A−2p(µϕ1)∈span B,∀p∈N∗,
implies d2= 0. Thus, it comes that d1=d2= 0 i.e. span B=Xs
(0). From Theorem 3.3, we obtain that
Bis a Riesz basis of X2
(0) (resp. ˜
Bis a Riesz basis of X3
(0)).
Third step: we prove that in the remaining case (3.15), one has an=bn= 0 for any n∈N∗.
Let us define
˜
G:z∈C7→ X
n∈N∗
1
(λ2
n+λ2)2 2λλnβ1
n
(λ2
n−λ2)β2
n,exp (zM )an
bn,(3.17)
with
M:= 1
(λ2
n+λ2)2
(λ2−λ2
n) 2λλnβ2
n
β1
n
−2λλnβ1
n
β2
n(λ2−λ2
n)
.(3.18)
Notice that the matrix appearing in this definition is the one used in the recurrence relations (3.11) and (3.13).
This matrix is diagonalized as follows
M=1
(λ2
n+λ2)2

−iβ2
n
β1
niβ2
n
β1
n
1 1


(λ2−λ2
n) + 2iλλn0
0 (λ2−λ2
n)−2iλλn


i
2
β1
n
β2
n
1
2
−i
2
β1
n
β2
n
1
2


.
From this diagonalization we deduce that
˜
G(z) = X
n∈N∗
exp (λ+iλn)2
(λ2+λ2
n)2z"−iβ1
nan
2
(λ−iλn)2
(λ2
n+λ2)2−β2
nbn
2
(λ+iλn)2
(λ2
n+λ2)2#
+ exp (λ−iλn)2
(λ2+λ2
n)2z"iβ1
nan
2
(λ+iλn)2
(λ2
n+λ2)2+β2
nbn
2
(λ−iλn)2
(λ2
n+λ2)2#.(3.19)
Again, ˜
Gis an entire function from the uniform convergence on compact sets. The recurrence rela-
tion (3.13) implies that, for every p∈N,
˜
G(p)(0) = X
n∈N∗
1
(λ2
n+λ2)2 2λλnβ1
n
(λ2
n−λ2)β2
n, M pan
bn
=X
n∈N∗
2λλnβ1
n˜
kp
n+ (λ2
n−λ2)β2
n˜
lp
n
(λ2
n+λ2)2(p+1)
= 0.
17

Thus we obtain ˜
G≡0.(3.20)
We claim that (λ−iλn)2
(λ2
n+λ2)26=(λ+iλm)2
(λ2
m+λ2)2,(n, m)∈(N∗)2.(3.21)
Indeed, let (n, m)∈(N∗)2be such that
(λ−iλn)2
(λ2
n+λ2)26=(λ+iλm)2
(λ2
m+λ2)2.(3.22)
Taking the modulus of both sides of (3.22), one gets
1
λ2
n+λ2=1
λ2
m+λ2,
which implies that λn=λmand therefore n=m. Hence
(λ−iλn)2
(λ2
n+λ2)2=(λ+iλn)2
(λ2
n+λ2)2,(3.23)
Taking the imaginary part of (3.23) yields a contradiction and proves that (3.21) holds.
For simplicity, we rewrite Gas
G(z) = X
n∈N∗
Cneµnz,
with µnequal to (λ−iλk)2/(λ2
k+λ2)2or (λ+iλk)2/(λ2
k+λ2)2for some k∈N∗and Cnis the
corresponding coefficient in (3.19). Notice that µnare all different, µn→0as n→ ∞ and Cn∈
ℓ2(N∗;C).
We repeat the same argument as in the finite dimensional case. Let
C:= Conv{µn|n∈N∗such that Cn6= 0}.
Consider a nonzero extremal point of C, which is therefore of the form µn0for some n0∈N∗. Hence,
there exists θ∈[0,2π]such that
ℜeiθµn<ℜeiθ µn0,∀n∈N∗\ {n0}(3.24)
By letting z=ρeiθ with ρ∈Rand using (3.20), we obtain
0 = Gρeiθ e−ρeiθµn0=Cn0+X
n∈N∗\{n0}
Cneρ(eiθµn−eiθ µn0).(3.25)
We then let ρ→+∞in (3.25) to obtain, using (3.24), that Cn0= 0 which is a contradiction with the
construction C. It implies that Cn= 0,∀n∈N. The expression of the Cnimplies for all n∈N∗
−iβ1
nan
2
(λ−iλn)2
(λ2
n+λ2)2−β2
nbn
2
(λ+iλn)2
(λ2
n+λ2)2= 0,
iβ1
nan
2
(λ+iλn)2
(λ2
n+λ2)2+β2
nbn
2
(λ−iλn)2
(λ2
n+λ2)2= 0.
One then easily concludes that an=bn= 0 by using (3.22) and the fact that β1
n6= 0 and β2
n6= 0 for all
n∈N∗. Theorem 3.2 thus implies the Riesz basis property.

18

3.2 Definition and regularity of the feedback law
So far, we have obtained from (2.9) the expression of the kernels kij with respect to the Fourier coefficients
αi
nof the feedback
KΨ1
Ψ2:=
+∞
X
n=1
α1
nhΨ1, ϕni+α2
nhΨ2, ϕni.(3.26)
The regularity of the kernels and, consequently, the regularity of T, will be directly related to the decay
rate of those coefficients as n→+∞. It remains to use the T B =Bcondition (2.18) to construct Kand
T.As µϕ1∈H2
(0) and Bis a Riesz basis of X2
(0) (see Proposition 3.4), it comes that there exist sequences
(a1
n)n∈N∗,(a2
n)n∈N∗∈ℓ2(N∗,R)such that
0
µϕ1=
+∞
X
n=1
a1
ng12
n
g22
n+a2
nh12
n
h22
n,in X2
(0).(3.27)
Getting back to the T B =Bcondition (2.18), we define
α1
n:= β1
n
hµϕ1, ϕnia1
n, α2
n:= β2
n
hµϕ1, ϕnia2
n.(3.28)
Then, following the heuristic of Sec. 2, the transformation Tis defined by (2.17) and the feedback law K
is defined by (3.26).
Unfortunately, the regularity of the coefficients is only (aj
n)n∈N∗∈ℓ2(N∗,R)for the X2
(0) Riesz basis
and will not be sufficient to prove that Tis continuous in X3
(0).
Remark 3.6 Recall that we assumed controllability of the linearized system i.e. Hypothesis 1.1 . From [5,
Remark 2], it follows that µ′(0) ±µ′(1) 6= 0 and then that (µϕ1)6∈ H3
(0). Thus, (0, µϕ1)Tcannot be
decomposed in the Riesz basis of X3
(0). This would have led to more regularity for the sequences (αj
n)n.
It is fortunate since, as it will be noticed in Remark 4.2, if (µϕ1)∈H3
(0), the obtained transformation
Twould have been compact in X3
(0) and thus not invertible in X3
(0).
Performing a suitable decomposition of the function µϕ1, we provethat the coefficients of the feedback
law satisfy the following regularity.
Proposition 3.7 We define the sequence (hn)n∈N∗by
hn:= 4
n3π2(−1)n+1µ′(1) −µ′(0).
Then the sequences (α1
n)n∈N∗and (α2
n)n∈N∗defined in (3.28) satisfy
α1
n
n3∈ℓ2(N∗,R),1
n3α2
n−λnβ2
n
hn
hµϕ1, ϕni∈ℓ2(N∗,R).
Remark 3.8 As a corollary, it comes that, for every j∈ {1,2},
αj
n
n3n∈N∗∈ℓ∞(N∗,R).
Proof:
First step: splitting of the problem. We start using the ideas developed in [44] to extend the regulariza-
tion result [5, Lemma 1] to higher dimensions. Let
h:x∈(0,1) 7→ −π√2
3(µ′(0) + µ′(1))x3−3µ′(0)x2+ (2µ′(0) −µ′(1))x.
It comes that
g:= µϕ1−h∈H3
(0).(3.29)
19

The Fourier coefficients of hare given by
hh, ϕki=hk=4
k3π2(−1)k+1µ′(1) −µ′(0),∀k∈N∗,(3.30)
which is the leading term in the asymptotic expansion of hµϕ1, ϕkigiven in [5, Remark 2]. Let us split the
left-hand side of (3.27) in two parts
0
µϕ1=0
g+0
h.
As g∈H3
(0), using the Riesz basis of X3
(0) we get the existence of sequences (δ1
n)n∈N∗and (δ2
n)n∈N∗
such that 0
g=
+∞
X
n=1hµϕ1, ϕni"λ1/2
nδ1
n
β1
n g12
n/λ1/2
n
g22
n/λ1/2
n!+λ1/2
nδ2
n
β2
n h12
n/λ1/2
n
h22
n/λ1/2
n!#.(3.31)
The coefficients of the decomposition in a Riesz basis being ℓ2sequences, Hypothesis 1.1 and the behaviour
of coefficients βj
nin (3.2) imply that, for every j∈ {1,2},
δj
n
n3n∈N∗∈ℓ2(N∗,R).(3.32)
Second step: decomposition of h.Using the Riesz basis Bof X2
(0), we get coefficients (ρ1
n)n∈N∗and
(ρ2
n)n∈N∗such that 0
h=
+∞
X
n=1hµϕ1, ϕniρ1
n
β1
ng12
n
g22
n+ρ2
n
β2
nh12
n
h22
n.(3.33)
Recall that the basis Bis obtained as a perturbation of the L2-orthonormal basis. To highlight this, we
define the sequences (γj
n)n∈N∗,j∈ {1,2}, such that
ρ1
n=λnβ1
nγ1
n,
ρ2
n=λnβ2
nhh, ϕni
hµϕ1, ϕni+λnβ2
nγ2
n.(3.34)
From (2.16), (3.33), and (3.34), we get, for every k∈N∗,
0
hh, ϕki=
+∞
X
n=1hµϕ1, ϕnihµϕ1, ϕkiρ1
nc12
nk
c22
nk+ρ2
nd12
nk
d22
nk
=
+∞
X
n=1hµϕ1, ϕnihµϕ1, ϕkihh, ϕni
hµϕ1, ϕniλnβ2
nd12
nk
d22
nk
+
+∞
X
n=1hµϕ1, ϕnihµϕ1, ϕkiλnβ1
nγ1
nc12
nk
c22
nk+λnβ2
nγ2
nd12
nk
d22
nk.
which, using (2.13), (2.14), (2.15), and (3.1), is equivalent to
− P+∞
n=1hµϕ1, ϕkihh, ϕniλnβ2
nd12
nk
P+∞
n=1,n6=khµϕ1, ϕkihh, ϕniλnβ2
nd22
nk!
=
+∞
X
n=1hµϕ1, ϕnihµϕ1, ϕkiλnβ1
nγ1
nc12
nk
c22
nk+λnβ2
nγ2
nd12
nk
d22
nk,
for every k∈N∗. Hence, if we define
˜
h1
˜
h2:= −
+∞
X
k=1 
P+∞
n=1hh, ϕnihµϕ1, ϕkiλnβ2
nd12
nkϕk
P+∞
n=1,n6=khh, ϕnihµϕ1, ϕkiλnβ2
nd22
nkϕk
,(3.35)
20

we get, using also (2.16),
˜
h1
˜
h2=
+∞
X
n=1
+∞
X
k=1hµϕ1, ϕniλnγ1
nhµϕ1, ϕkiβ1
nc12
nk
hµϕ1, ϕkiβ1
nc22
nk+λnγ2
nhµϕ1, ϕkiβ2
nd12
nk
hµϕ1, ϕkiβ2
nd22
nkϕk
=
+∞
X
n=1
λ3/2
nhµϕ1, ϕni"γ1
n g12
n/λ1/2
n
g22
n/λ1/2
n!+γ2
n h12
n/λ1/2
n
h22
n/λ1/2
n!#.(3.36)
Finally, if (˜
h1,˜
h2)T∈X3
(0) (which will be proved in the next two steps), it comes that, for every j∈ {1,2},
λ3/2
nhµϕ1, ϕniγj
nn∈ℓ2(N∗,R),
which, thanks to Hypothesis 1.1, implies that
(γj
n)n∈ℓ2(N∗,R).(3.37)
Using (3.27), (3.28), (3.29), (3.31), (3.33), and (3.34), we obtain
α1
n
n3=ρ1
n
n3+δ1
n
n3=λnβ1
n
n3γ1
n+δ1
n
n3,
α2
n
n3=ρ2
n
n3+δ2
n
n3=λnβ2
n
n3γ2
n+λnβ2
n
n3hh, ϕni
hµϕ1, ϕni+δ2
n
n3,
which, with (1.3), (3.2), (3.30), (3.32), and (3.37) will end the proof of Proposition 3.7.
Third step: H3
(0) regularity of ˜
h1.We start by proving that ˜
h1∈H3
(0) i.e.
+∞
X
k=1 k3h˜
h1, ϕki
2=
+∞
X
k=1 k3
+∞
X
n=1hµϕ1, ϕkihh, ϕniλnβ2
nd12
nk
2
<+∞.(3.38)
From (3.2) and (3.30) it comes that there exists C > 0such that
|hh, ϕniλnβ2
n| ≤ C, ∀n∈N∗.
Using (1.5) together with (2.13), (2.14), (2.15) and (2.16), it suffices to prove that
M12 :=
+∞
X
k=1 
+∞
X
n=1
d12
nk
2
=
+∞
X
k=1 
+∞
X
n=1
λk(λ2+λ2
k−λ2
n)
(λ2+ (λk−λn)2) (λ2+ (λk+λn)2)
2
<+∞.(3.39)
Notice that
M12 ≤2
+∞
X
k=1


λ2λk
λ2(λ2+ 4λ2
k)2
+


+∞
X
n=1
n6=k
d12
nk


2

.
As +∞
X
k=1 λ2λk
λ2(λ2+ 4λ2
k)2
<+∞,
we only deal with the second term. Straightforward computations lead to
+∞
X
n=1
n6=k
d12
nk =1
2
+∞
X
n=1
n6=kλk−λn
λ2+ (λk−λn)2+λk+λn
λ2+ (λk+λn)2.(3.40)
21

The second term of this sum is easily dealt with. As λ2+ (λk+λn)2≥(λk+λn)2it comes that
+∞
X
n=1
n6=k
λk+λn
λ2+ (λk+λn)2≤
+∞
X
n=1
n6=k
1
λk+λn
≤
+∞
X
n=1 Zn
n−1
1
λk+π2x2dx
=1
2kπ .
Thus,
+∞
X
k=1



+∞
X
n=1
n6=k
λk+λn
λ2+ (λk+λn)2


2
<+∞.(3.41)
We now turn back to the first term of the right-hand side of (3.40). We have

+∞
X
n=1
n6=k
λk−λn
λ2+ (λk−λn)2≤
+∞
X
n=1
n6=k
1
|λk−λn|=
k−1
X
n=1
1
λk−λn
+
+∞
X
n=k+1
1
λn−λk
.
Straightforward computations lead to
k−1
X
n=1
1
λk−λn
=1
2kπ2
k−1
X
n=1 1
k−n+1
k+n
=1
2kπ2

k−1
X
j=1
1
j+
2k−1
X
j=k+1
1
j

=1
2kπ2

2k−1
X
j=1
1
j−1
k
.
In the same spirit, for N≥k+ 1,
0≥
N
X
n=k+1
1
λk−λn
=1
2kπ2
N+k
X
j=2k+1
1
j−1
2kπ2
N−k
X
j=1
1
j
and thus,
0≤
+∞
X
n=k+1
1
λn−λk≤1
2kπ2
2k
X
j=1
1
j−1
2kπ2
N+k
X
j=n−k+1
1
j≤1
2kπ2
2k
X
j=1
1
j.
Finally, 
+∞
X
n=1
n6=k
λk−λn
λ2+ (λk−λn)2≤
+∞
X
n=1
n6=k
1
|λk−λn|≤1
2kπ2
2
2k−1
X
j=1
1
j−1
2k
,(3.42)
which belongs to ℓ2(N∗)as P2k−1
j=1 1
j∼
k→+∞ln(2k−1). From (3.40), this proves that ˜
h1∈H3
(0).
Fourth step: H3
(0) regularity of ˜
h2.We end the proof of Proposition 3.7 by proving that ˜
h2∈H3
(0) i.e.,
by (3.35),
+∞
X
k=1 k3h˜
h2, ϕki
2=
+∞
X
k=1

k3
+∞
X
n=1
n6=k
hµϕ1, ϕkihh, ϕniλnβ2
nd22
nk
2
<+∞.(3.43)
22

From (3.2) and (3.30) it comes that there exists C > 0such that
|hh, ϕniλnβ2
n| ≤ C, ∀n∈N∗.
Using (1.5) it suffices to prove that
+∞
X
k=1

+∞
X
n=1
n6=k
d22
nk
2
=
+∞
X
k=1

+∞
X
n=1
n6=k
λ(λ2+λ2
k+λ2
n)
(λ2+ (λk−λn)2) (λ2+ (λk+λn)2)
2
<+∞.(3.44)
Notice that
λ(λ2+λ2
k+λ2
n)
(λ2+ (λk−λn)2) (λ2+ (λk+λn)2)=λ
21
λ2+ (λk+λn)2+1
λ2+ (λk−λn)2
≤λ
2λk+λn
λ2+ (λk+λn)2+|λk−λn|
λ2+ (λk−λn)2
≤λ
2λk+λn
λ2+ (λk+λn)2+1
|λk−λn|.
From (3.41) and (3.42), it comes that
+∞
X
k=1

+∞
X
n=1
n6=k
λ(λ2+λ2
k+λ2
n)
(λ2+ (λk−λn)2) (λ2+ (λk+λn)2)
2
<+∞.
This ends the proof of Proposition 3.7.

3.3 Domain of definition and continuity of the transformation
The regularity of the coefficients obtained in the previous section is sufficient to define a continuous oper-
ator Tin the state space.
Proposition 3.9 The transformation Tdefined on X3
(0) by (2.17) and (3.28) is linear continuous in X3
(0).
Proof: Let (Ψ1,Ψ2)T∈X3
(0). From (2.17), using the Riesz basis property of Proposition 3.4, it comes
that TΨ1
Ψ2belongs to X3
(0) if the following sequences,
λ1/2
nα1
nhΨ2, ϕni
β1
n−λ1/2
nα2
nhΨ1, ϕni
β1
n!n∈N∗
,
λ1/2
nα1
nhΨ1, ϕni
β2
n
+λ1/2
nα2
nhΨ2, ϕni
β2
n!n∈N∗
,
belong to ℓ2(N∗,R). From (3.2) and Remark 3.8, it comes that, for every i, j, k ∈ {1,2},
+∞
X
n=1 
λ1/2
nαj
nhΨi, ϕni
βk
n
2
≤C
+∞
X
n=1 
αj
n
n3n3hΨi, ϕni
2
≤C
+∞
X
n=1 n3hΨi, ϕni2
=CkΨik2
H3
(0) .
Finally, we obtain 


TΨ1
Ψ2


X3
(0) ≤C


Ψ1
Ψ2


X3
(0)
.

23

4 Invertibility of the transformation
This section aims at proving the invertibility of T. As a first step, we prove in subsection 4.1 that Tis a
Fredholm operator. In subsection 4.1, we prove that the analogous of (1.15) in finite dimension holds on a
certain functional space. This will be used in subsection (4.3) to obtain the invertibility of T.
4.1 Fredholm form
The goal of this subsection is the proof of the following result.
Proposition 4.1 There exists e
T:X3
(0) →X3
(0) invertible such that T−e
Tis a compact operator.
The proof of this proposition rely on the study of the feedback law done in Proposition 3.7.
Proof: Let e
Tbe the transformation defined by (2.17) where the coefficients α1
nand α2
nare respectively
replaced by
˜α1
n= 0,˜α2
n=hh, ϕni
hµϕ1, ϕniλnβ2
n.
From Proposition 3.7 (recall that hn=hh, ϕniis given in (3.30)), it follows that defining ˆαj
n=αj
n−˜αj
n,
we get that (ˆαj
n/n3)n∈ℓ2(N∗,R).
The computations done in the proof of Proposition 3.9 show that e
Tis a linear continuous operator of
X3
(0). We prove that e
Tis invertible. For any (Ψ1,Ψ2)T∈X3
(0),
e
TΨ1
Ψ2=
+∞
X
n=1 −˜α2
nλ1/2
n
β1
nhΨ1, ϕni g12
n/λ1/2
n
g22
n/λ1/2
n!+˜α2
nλ1/2
n
β2
nhΨ2, ϕni h12
n/λ1/2
n
h22
n/λ1/2
n!.
From (1.5), (1.7), (3.2), and (3.30), we have that, for every j∈ {1,2},(βj
nλ3/2
n/(˜α2
nλ1/2
n))n∈ℓ∞(N∗,R).
Then, the Riesz basis property of Proposition 3.4 ends the proof of the invertibility of e
T.
Finally, we prove that T−e
Tis compact using the Hilbert-Schmidt criterion, i.e., we prove that
+∞
X
n=1 


(T−e
T)ϕn/λ3/2
n
0



2
X3
(0)
+


(T−e
T)0
ϕn/λ3/2
n



2
X3
(0)
<+∞.(4.1)
From (2.17) and the definition of e
Tit comes that
T−e
TΨ1
Ψ2=
+∞
X
n=1 ˆα1
nhΨ2, ϕni
β1
n−ˆα2
nhΨ1, ϕni
β1
ng12
n
g22
n
+
+∞
X
n=1 ˆα1
nhΨ1, ϕni
β2
n
+ˆα2
nhΨ2, ϕni
β2
nh12
n
h22
n.
Thus,



(T−e
T)ϕn/λ3/2
n
0



2
X3
(0)
=



−ˆα2
n
λ3/2
nβ1
ng12
n
g22
n+ˆα1
n
λ3/2
nβ2
nh12
n
h22
n




2
X3
(0)
=



−ˆα2
n
λnβ1
n g12
n/λ1/2
n
g22
n/λ1/2
n!+ˆα1
n
λnβ2
n h12
n/λ1/2
n
h22
n/λ1/2
n!




2
X3
(0)
.
24

The first term is estimated in the following way





ˆα2
n
λnβ1
n g12
n/λ1/2
n
g22
n/λ1/2
n!




2
X3
(0)
≤2




ˆα2
n
λnβ1
n g12
n/λ1/2
n−ϕn/λ3/2
n
g22
n/λ1/2
n!




2
X3
(0)
+ 2 



ˆα2
n
λnβ1
nϕn/λ3/2
n
0



2
X3
(0)
≤2ˆα2
n
λnβ1
n2



 g12
n/λ1/2
n−ϕn/λ3/2
n
g22
n/λ1/2
n!




2
X3
(0)
+ 2 ˆα2
n
λnβ1
n2
.
We proved in Lemma 3.5 that
+∞
X
n=1 



 g12
n/λ1/2
n−ϕn/λ3/2
n
g22
n/λ1/2
n!




2
X3
(0)
<+∞.
As (ˆα2
n/n3)n∈ℓ2(N∗,R), this gives, using (3.2), that
+∞
X
n=1 




ˆα2
n
λnβ1
n g12
n/λ1/2
n
g22
n/λ1/2
n!




2
X3
(0)
<+∞.
The term
+∞
X
n=1 




ˆα1
n
λnβ2
n h12
n/λ1/2
n
h22
n/λ1/2
n!




2
X3
(0)
,
is dealt with in the exact same way, ending the proof of Proposition 4.1.

Remark 4.2 Notice that the key point in proving the invertibility of e
Tis that for any n∈N∗,hh, ϕni 6= 0.
We also underline that the compactness of T−e
Tcomes from the fact that (ˆαj
n/n3)n∈N∗∈ℓ2(N∗,R).
If we had µϕ1∈H3
(0) (this is not possible due to Hypothesis 1.1, see [5, Remark 2]), then from (2.18)
we would have obtained that (αj
n/n3)n∈N∗∈ℓ2(N∗,R). This would have led to the compactness of the
transformation Tpreventing its invertibility.
4.2 Operator equality
We prove that the formal operator equality
T(A+BK) = AT −λT (4.2)
holds true on an appropriate functional space. Recall that Kis defined by (3.26).
Remark 4.3 Notice that due to the regularity of the coefficients αj
nobtained in Proposition 3.7, the oper-
ator Kis not defined on X3
(0). Otherwise, taking any (0, ψ)T∈X3
(0) with ψ /∈Hs
(0) for any s > 3would
imply K0
ψ=X
n∈N∗
α2
nhψ, ϕni=X
n∈N∗
α2
n
n3(n3hψ, ϕni)<+∞,
and therefore (α2
n/n3)n∈N∗∈l2(N∗,R)which is in contradiction with Proposition 3.7.
25

Due to the previous remark, the functional setting for (4.2) to hold needs to be specified. Let us first deal
with K. Proposition 3.7 implies that, for every (Ψ1,Ψ2)T∈H3
(0) ×H4
(0),
X
n∈N∗|α1
nhΨ1, ϕni+α2
nhΨ2, ϕni| ≤ X
n∈N∗
α1
n
n3
2!1/2
kΨ1kH3
(0) +CX
n∈N∗
λ3/2
nhΨ2, ϕni
≤ X
n∈N∗
α1
n
n3
2!1/2
kΨ1kH3
(0) +C X
n∈N∗
1
λn!1/2 X
n∈N∗
λ4
nhΨ2, ϕni2!1/2
≤CkΨ1kH3
(0) +CkΨ2kH4
(0) .
This shows that Kis well defined and continuous on H3
(0) ×H4
(0).
We now turn to A+BK. Recall that we expect
(A+BK)Ψ1
Ψ2=0−∆
∆ 0 Ψ1
Ψ2+KΨ1
Ψ2 0
µϕ1.(4.3)
We define
D(A+BK) := Ψ1
Ψ2∈(H5∩H3
(0))×H5
(0) ; ∆Ψ1+KΨ1
Ψ2µϕ1∈H3
(0).
and then define A+BK on D(A+BK )by (4.3). Note that (Ψ1,Ψ2)T∈D(A+BK )if and only if
(Ψ1,Ψ2)∈(H5∩H3
(0))×H5
(0) and
∆2Ψ1(x) + 2KΨ1
Ψ2µ′(x)ϕ′
1(x) = 0, x ∈ {0,1}.(4.4)
We now prove the density of D(A+BK).
Lemma 4.4 The domain D(A+BK )is dense in X3
(0).
Proof: Let us prove that D(A+BK)⊥={0}in X3
(0). Let (Ψ1,Ψ2)T∈D(A+BK )⊥.
First step: we prove that Ψ1= 0.
Let k∈N∗. Consider φ1:= ϕk. From the asymptotic behaviour of α2
n(see Proposition 3.7), there
exists N∈N∗such that ∀n≥N,α2
n6= 0. Therefore, setting φ2=−α1
kϕn/α2
n∈H5
(0) implies that
Kφ1
φ2=α1
k+α2
n−α1
k
α2
n= 0
which means that (4.4) is satisfied and (φ1, φ2)T∈D(A+BK). Thus,
0 = *Ψ1
Ψ2, ϕk
−α1
k
α2
nϕn!+X3
(0)
=Ψ1, ϕkH3
(0) −α1
k
α2
nΨ2, ϕnH3
(0)
=Ψ1, ϕkH3
(0) −λ3/2
nα1
k
α2
nλ3/2
nΨ2, ϕn
−→
n→+∞Ψ1, ϕkH3
(0)
.
Indeed, since Ψ2∈H3
(0), it implies that n3hΨ2, ϕnin≥N∈ℓ2(N∗;R)and Remark 1.2, (3.2) and (3.34)
imply that λ3/2
nα1
k/α2
nn∈ℓ∞(N∗\ {1, ..., N −1};R). We conclude that
Ψ1, ϕkH3
(0)
= 0,∀k∈N∗,
and thus Ψ1= 0.
26

Second step : we prove that Ψ2= 0.
For all k∈N∗, let φ2=ϕk. If there exists n∈N∗such that α1
n6= 0, then setting φ1=−α2
kϕn/α1
n
implies that Kφ1
φ2= 0 and therefore (φ1, φ2)T∈D(A+BK). Thus,
0 = *Ψ1
Ψ2, −α2
k
α1
nϕn
ϕk!+X3
(0)
=Ψ2, ϕkH3
(0)
.
Otherwise, for every φ1∈H3
(0),
Kφ1
φ2=α2
k.
Then we consider φ1solution to (−∆φ1=α2
k(µϕ1)
φ1(0) = φ1(1) = 0.
Since µϕ1∈H3∩H1
0, then φ1∈H5∩H3
(0) and, since (4.4) is satisfied, (φ1, φ2)T∈D(A+BK).
Moreover, as Ψ1= 0,
0 = Ψ1
Ψ2,φ1
φ2X3
(0)
=Ψ2, ϕkH3
(0)
.
This proves that (Ψ1,Ψ2)T= 0.

Let us now turn our attention to the kernel system. More precisely, we prove the following.
Proposition 4.5 For (Ψ1,Ψ2)T∈D(A+BK),
T(A+BK)Ψ1
Ψ2= (A−λI)TΨ1
Ψ2,in X3
(0).
Proof: Let (Ψ1,Ψ2)T∈D(A+BK ). Then
(A+BK)Ψ1
Ψ2=
−∆Ψ2
∆Ψ1+KΨ1
Ψ2µϕ1
.
Therefore
T(A+BK)Ψ1
Ψ2=X
n∈N∗α1
n
β1
n∆Ψ1+KΨ1
Ψ2µϕ1, ϕn+α2
n
β1
n∆Ψ2, ϕng12
n
g22
n
+−α1
n
β2
n∆Ψ2, ϕn+α2
n
β2
n∆Ψ1+KΨ1
Ψ2µϕ1, ϕnh12
n
h22
n
=KΨ1
Ψ2X
n∈N∗α1
n
β1
ng12
n
g22
n+α2
n
β2
nh12
n
h22
nhµϕ1, ϕni
−X
n∈N∗
λnα1
n
β1
nΨ1, ϕn+α2
n
β1
nΨ2, ϕng12
n
g22
n
+λnα1
n
β2
nΨ2, ϕn−α2
n
β2
nΨ1, ϕnh12
n
h22
n
= X
n∈N∗
α1
nhΨ1, ϕni+α2
nhΨ2, ϕni!0
µϕ1
+X
n∈N∗−λnα1
n
β1
nΨ1, ϕn+α2
n
β1
nΨ2, ϕng12
n
g22
n
+λnα1
n
β2
nΨ2, ϕn−α2
n
β2
nΨ1, ϕnh12
n
h22
n,in X2
(0),
27

using the expression of Tin (2.17), the T B =Bcondition (2.18) and the expression of Kgiven in (3.26).
Moreover, by the definition of gij
n, hij
ngiven in (2.13) and by the relations (2.15), we have, on one hand,
using (2.17) once more,
(A−λI)TΨ1
Ψ2,ϕk
0(L2)2
=TΨ1
Ψ2, A∗ϕk
0(L2)2−λTΨ1
Ψ2,ϕk
0(L2)2
=TΨ1
Ψ2,−λϕk
λkϕk(L2)2
=X
n∈N∗α1
n
β1
nhΨ2, ϕni − α2
n
β1
nhΨ1, ϕniλkhg22
n, ϕki − λhg12
n, ϕki
+α1
n
β2
nhΨ1, ϕni+α2
n
β2
nhΨ2, ϕniλkhh22
n, ϕki − λhh12
n, ϕki
=X
n∈N∗α1
n
β1
nhΨ2, ϕni − α2
n
β1
nhΨ1, ϕniλnhg11
n, ϕki
+α1
n
β2
nhΨ1, ϕni+α2
n
β2
nhΨ2, ϕniλnhh11
n, ϕki
=X
n∈N∗α1
n
β2
nhΨ2, ϕni − α2
n
β2
nhΨ1, ϕniλnhh12
n, ϕki
−α1
n
β2
nhΨ1, ϕni+α2
n
β2
nhΨ2, ϕniλnhg12
n, ϕki
=T(A+BK)Ψ1
Ψ2,ϕk
0(L2)2
.
On the other hand, using again (2.13), (2.15) and (2.17)
(A−λI)TΨ1
Ψ2,0
ϕk(L2)2
=TΨ1
Ψ2,−λkϕk
−λϕk(L2)2
=X
n∈N∗α1
n
β1
nhΨ2, ϕni − α2
n
β1
nhΨ1, ϕni−λhg22
n, ϕki − λkhg12
n, ϕki
+α1
n
β2
nhΨ1, ϕni+α2
n
β2
nhΨ2, ϕni−λhh22
n, ϕki − λkhh12
n, ϕki
=X
n∈N∗α1
n
β2
nhΨ2, ϕni − α2
n
β2
nhΨ1, ϕniλnhh22
n, ϕki
−α1
n
β1
nhΨ1, ϕni+α2
n
β1
nhΨ2, ϕniλnhg22
n, ϕki
+X
n∈N∗α1
nhΨ1, ϕni+α2
nhΨ2, ϕnihµϕ1, ϕki
=T(A+BK)Ψ1
Ψ2,0
ϕk(L2)2
.
Indeed, (2.13) and (2.15) imply
−λkhh12
n, ϕki − λhh22
n, ϕki=λnhh21
n, ϕki+β2
nhµϕ1, ϕki
=−λn
β2
n
β1
nhg22
n, ϕki+β2
nhµϕ1, ϕki.
Consequently, by the definition of D(A+BK)and by the continuity of Tfrom X3
(0) into itself, T(A+
BK ) = (A−λI )Tholds in X3
(0). Notice that all the previous infinite sums are converging due to the
regularity assumptions on the functions of D(A+BK).
28


We conclude this section by noting that, following Proposition 4.5, if (Ψ1
Ψ2)T∈D(A+BK), then TΨ1
Ψ2∈X5
(0). Indeed,
AT Ψ1
Ψ2=T(A+BK)Ψ1
Ψ2−λT Ψ1
Ψ2,
where the right-hand side of the last equation is in X3
(0).
4.3 Invertibility
Let us now turn to the invertibility of Tand prove the following result.
Proposition 4.6 The operator Tis invertible from X3
(0) into itself.
Proof: In the previous section, we have proven that e
Tis invertible and T−e
Tis a compact operator.
Consequently, the index of Tis equal to zero. Thus, it is sufficient to prove that Ker T∗={0}to show the
invertibility of T.
Let us rewrite the operator equality (4.2) under the form
T(A+BK +λI +ρI) = AT +ρT = (A+ρI )T , (4.5)
where ρ∈Cwill be chosen later. Assume for the moment that ρ∈Cis such that
(A+ρI +BK +λI)is an invertible operator from D(A+BK )to X3
(0),(4.6)
(A+ρI)is an invertible operator from D(A)to X3
(0).(4.7)
Note that here, the vector spaces D(A+BK), D(A)and X3
(0) are complexified. Then, from (4.5), one has
(A+ρI)−1T=T(A+BK +λI +ρI)−1.(4.8)
Consider (χ1, χ2)T∈Ker T∗. Then, for all (φ1, φ2)T∈X3
(0), we have
0 = h(A+ρI)−1Tφ1
φ2−T(A+BK +λI +ρI)−1φ1
φ2,χ1
χ2iX3
(0)
=hφ1
φ2, T ∗((A+ρI)−1)∗χ1
χ2iX3
(0) − h(A+BK +λI +ρI)−1φ1
φ2, T ∗χ1
χ2iX3
(0)
=hφ1
φ2, T ∗((A+ρI)−1)∗χ1
χ2iX3
(0) .
Thus the space Ker T∗, which is of finite dimension, is stable by ((A+ρI)∗)−1. Hence, if
Ker T∗6={0},(4.9)
((A+ρI)∗)−1has an eigenfunction in Ker T∗. This eigenfunction is also an eigenfunction of (A∗)−1.
Thus, if (4.9) holds, there exists ν∈Cand (χ1, χ2)T∈Ker T∗\ {0}such that
(A∗)−1χ1
χ2=−∆−1χ2
∆−1χ1=νχ1
χ2.(4.10)
Hence, for every j∈N∗,
ν2hχ1, ϕji=νh−∆−1χ2, ϕji=1
λj
νhχ2, ϕji=λjh∆−1χ1, ϕji=−1
λ2
jhχ1, ϕji.(4.11)
Therefore,
(ν2+1
λ2
j
)hχ1, ϕji= 0.(4.12)
29

Note that χ1= 0 together with (4.11) implies χ2= 0. Hence, since (χ1, χ2)T6= 0,χ16= 0, which with
(4.12) implies that there exists one and only one k∈N∗such that
ν=±i1
λk
, χ1=ckϕk, ck∈C\ {0}.
Furthermore, from (4.10), we obtain χ2=∓ickϕk. Finally, we have, by the T B =Bcondition (2.18),
±ickλ3
khµϕ1, ϕki=hµϕ1, χ2iH1
0,H5
(0)
=h0
µϕ1,χ1
χ2i(H1
0)2,(H5
(0))2
=hT0
µϕ1,χ1
χ2i(H1
0)2,(H5
(0))2
=h0
µϕ1, T ∗χ1
χ2i(H1
0)2,(H5
(0))2
=0.
Since hµϕ1, ϕki 6= 0, we conclude that ckmust be zero, which implies that (4.9) does not hold and
therefore Ker T∗={0}.
It remains to prove the existence of ρ∈Csuch that (4.6) and (4.7) hold. Let κ:= ρ+λ. Applying
A−1to A+BK +κI yields the operator
I+A−1BK +κA−1:D(A+BK)→D(A),(4.13)
where A−1B= (∆−1(µϕ1),0)T. Let us prove that the set of κ∈Csuch that I+A−1BK +κA−1is
invertible from D(A+BK )to D(A)is non-empty.
First, if K(A−1B)6=−1, then the operator I+A−1BK :D(A+BK)→D(A)is invertible and the
proof is over. Indeed, to solve
(I+A−1BK)ψ=f, (4.14)
for any f∈D(A), one applies Kto (4.14) (K(ψ),K(A−1B)and K(f)are well-defined in this case)
leading to
K(ψ)(1 + K(A−1B)) = K(f).
Since K(A−1B)6=−1, we use the expression of K(ψ)in (4.14) to obtain
ψ=f−A−1BK(f)
1 + K(A−1B).
Suppose then that K(A−1B) = −1. It corresponds to the case where A−1B∈D(A+BK). Notice that
0is an eigenvalue of I+A−1BK of algebraic multiplicity 1. Then, from [47], there exists an open set
Ω⊂Cof 0∈Csuch that there exist an holomorphic function κ∈Ω7→ λ(κ)∈Cand an holomorphic
function κ∈Ω7→ x(κ)∈D(A+BK )such that
x(0) = A−1B=∆−1(µϕ1)
0,
(I+A−1BK +κA−1)x(κ) = λ(κ)x(κ).(4.15)
If λ(κ)6= 0 in a small neighborhood of 0, then I+A−1BK +κA−1is invertible for κclose to 0and the
proof is over. Suppose then that λ(κ) = 0 in a small neighborhood of 0. In this case, consider the power
serie expansion of xaround 0
x=∆−1(µϕ1)
0+
∞
X
k=1
κkx1
k
x2
k.
Notice that since x∈D(A+BK )and A−1B∈D(A+BK), we obtain that (x1
k, x2
k)T∈D(A). At the
zeroth order, (4.15) writes
∆−1(µϕ1)
0+∆−1(µϕ1)
0K∆−1(µϕ1)
0=0
0,
30

from the hypothesis K(A−1B) = −1. At the higher order, we have
x1
k
x2
k+∆−1(µϕ1)
0Kx1
k
x2
k+κA−1x1
k−1
x2
k−1=0
0,(4.16)
where (x1
0, x2
0)T:= (∆−1(µϕ1),0)T. Taking Kof (4.16) yields
KA−1x1
k
x2
k= 0,∀k≥0.
By successively taking A−1and Kof (4.16), we obtain
KA−nx1
k
x2
k= 0,∀k≥0,∀n≥1.(4.17)
Therefore from (4.16) and (4.17)
KA−1−2n0
(µϕ1)=X
j∈N
α1
jh(−1)n∆−1−2n(µϕ1), ϕni
=X
j∈N
(−1)nα1
jhµϕ1, ϕni
λ1+2n
j
=0.(4.18)
Consider the entire function
H(z) := X
j∈N
α1
jhµϕ1, ϕnie−z/λ2
j
λ3
j
.
From (4.18), we obtain that H(p)(0) = 0 and therefore H≡0. By letting z→ −∞ and by Hypothesis
1.1, we deduce that
α1
j= 0,∀j≥1.
In the same fashion,
KA−2n0
(µϕ1)=X
j∈N
α2
jh(−1)n∆−2n(µϕ1), ϕni
=X
j∈N
(−1)nα1
jhµϕ1, ϕni
λ2n
j
=0.(4.19)
Consider the entire function
ˆ
H(z) := X
j∈N
α2
jhµϕ1, ϕnie−z/λ2
j
λ2
j
.
From (4.19), we obtain that ˆ
H(p)(0) = 0 and therefore H≡0. By letting z→ −∞ and by Hypothesis
1.1, we deduce that
α2
j= 0,∀j≥1.
From Proposition 3.7, we know that α2
j6= 0,∀j≥1. Hence a contradiction either with K(A−1B) = −1,
which implies the invertibility of I+A−1BK +κA−1for all κ∈C, or with the fact that λ(κ) = 0 in a
small neighborhood of 0, which implies that I+A−1BK +κA−1is invertible in a small neighborhood of
0. Since (A+ρI)has discrete eigenvalues, it is possible in those two cases to choose ρsuch that (4.6)-(4.7)
are satisfied.
31

5 Well-posedness of the closed-loop linear system and rapid stabi-
lization
This section ends the proof of Theorem 1.5. Due to Remark 4.3, the feedback Kis not well defined for
functions in X3
(0). In subsection 5.1, we give a meaning to the solution (Ψ1,Ψ2)Tof the closed-loop
system









∂t Ψ1
Ψ2!= 0−∆
∆ 0 ! Ψ1
Ψ2!+K Ψ1
Ψ2! 0
(µϕ1)(x)!,(t, x)∈(0, T )×(0,1),
Ψ1(t, 0) = Ψ1(t, 1) = 0,Ψ2(t, 0) = Ψ2(t, 1) = 0, t ∈(0, T ),
Ψ1(0, x) = Ψ1
0(x),Ψ2(0, x) = Ψ2
0(x), x ∈(0,1)
(5.1)
by proving that A+BK generates a C0−semigroup. Finally we conclude to the exponential stability using
the operator equality of Proposition 4.5 and the invertibility of the transformation T.
5.1 Well-posedness of the closed-loop linear system
Let us first show,
Proposition 5.1 The operator (A+BK)defined on D(A+BK)generates a C0−semigroup on X3
(0).
Thus, there exists a unique solution C([0, T ]; X3
(0))of (1.29) with v(t) = Kψ1(t, .)
ψ2(t, .).
Proof: We prove that (A+BK)is the infinitesimal generator of a C0−semigroup on X3
(0).
First step. The density of D(A+BK )in X3
(0) was proven in Lemma 4.4.
Second step. Let us prove that (A+BK)is closed. Let (ψ1
n, ψ2
n)T∈D(A+BK)such that
ψ1
n
ψ2
n−→
n→+∞ψ1
ψ2,in X3
(0),
(A+BK)ψ1
n
ψ2
n−→
n→+∞φ1
φ2,in X3
(0).
We have
(A+BK)ψ1
n
ψ2
n=
−∆ψ2
n
∆ψ1
n+Kψ1
n
ψ2
nµϕ1
.
Hence
−∆ψ2
n−→
n→+∞−∆ψ2,in H1
0
−∆ψ2
n−→
n→+∞φ1,in H3
(0),
and, consequently, −∆ψ2=φ1,ψ2∈H5
(0) and ψ2
n−→
n→+∞ψ2in H5
(0). Therefore, Kψ1
ψ2is well-
defined and
Kψ1
n
ψ2
n−Kψ1
ψ2=X
k∈N∗
α1
khψ1
n−ψ1, ϕki+α2
khψ2
n−ψ2, ϕki
≤ X
k∈N∗α1
k
k32!1/2
kψ1
n−ψ1kH3
(0) + X
k∈N∗α2
k
k42!1/2
kψ2
n−ψ2kH4
(0) −→
n→+∞0.
32

We obtain
∆ψ1
n+Kψ1
n
ψ2
nµϕ1−→
n→+∞φ2in H3
(0),
∆ψ1
n+Kψ1
n
ψ2
nµϕ1−→
n→+∞∆ψ1+Kψ1
ψ2µϕ1in H1
0.
We conclude that ∆ψ1+Kψ1
ψ2µϕ1=φ2and therefore (ψ1, ψ2)T∈D(A+BK ).
Third step. Let us now prove the dissipativity of (A+BK). Since Tis invertible from X3
(0) into itself,
we define the norm k · kT:= kT· kX3
(0) , which is equivalent to the X3
(0) norm. We denote h·,·iTthe
associated inner product.
Consider (ψ1, ψ2)T∈D(A+BK). From Proposition 4.5, we have, since Tψ1
ψ2∈X5
(0),
h(A+BK)ψ1
ψ2,ψ1
ψ2iT=hT(A+BK)ψ1
ψ2, T ψ1
ψ2iX3
(0)
=hAT ψ1
ψ2, T ψ1
ψ2iX3
(0) −λ


ψ1
ψ2



2
T
=−λ


ψ1
ψ2



2
T≤0.
Consider now the dissipativity of (A+BK )∗. First, let (ψ1, ψ2)T∈D(A+BK )and (φ1, φ2)T∈
X3
(0). We have
h(A+BK)ψ1
ψ2,φ1
φ2iT=hT(A+BK)ψ1
ψ2, T φ1
φ2iX3
(0)
=hψ1
ψ2,(T−1A∗T−λI)φ1
φ2iT.
which implies that D((A+BK)∗) = n(φ1, φ2)T∈X3
(0) Tφ1
φ2∈X5
(0)o. Then, for (φ1, φ2)T∈
D((A+BK)∗), we have
h(A+BK)∗φ1
φ2,φ1
φ2iT=h(T−1A∗T−λI)φ1
φ2,φ1
φ2iT
=hA∗Tφ1
φ2, T φ1
φ2iX3
(0) −λ


Tφ1
φ2



2
X3
(0)
≤0.
Thus from the Lumer-Philipps theorem (see e.g. [43, Corollary 4.4]) we obtain that A+BK generates
aC0−semigroup on X3
(0).

5.2 Proof of the rapid stabilization
This section is dedicated to the proof of the main result, the rapid stabilization stated in Theorem 1.5.
Proof: To begin, let us assume that (Ψ1
0,Ψ2
0)T∈D(A+BK). Then, from Proposition 5.1, we get that
the associated solution of (1.29) with the feedback law v(t) = K(Ψ1(t, .),Ψ2(t, .))Tis given by
Ψ1(t, .)
Ψ2(t, .)=et(A+BK)Ψ1
0
Ψ2
0.(5.2)
33

Let ξ1(t, .)
ξ2(t, .):= TΨ1(t, .)
Ψ2(t, .). Using the equality of operators given in Proposition 4.5, it comes that
d
dtξ1(t, .)
ξ2(t, .)=Td
dtΨ1(t, .)
Ψ2(t, .)
=T(A+BK)Ψ1(t, .)
Ψ2(t, .)
= (AT −λT )Ψ1(t, .)
Ψ2(t, .)
=Aξ1(t, .)
ξ2(t, .)−λξ1(t, .)
ξ2(t, .).
Thus, we obtain the following exponential stability of (ξ1(t, .), ξ2(t, .))T



ξ1(t, .)
ξ2(t, .)


X3
(0) ≤e−λt 


ξ1
0
ξ2
0


X3
(0)
=e−λt 


TΨ1
0
Ψ2
0


X3
(0)
.
From the continuity and invertibility of Tin X3
(0) (see Proposition 4.6) it comes that



Ψ1(t, .)
Ψ2(t, .)


X3
(0) ≤
T−1



ξ1(t, .)
ξ2(t, .)


X3
(0)
≤
T−1
kTke−λt 


Ψ1
0
Ψ2
0


X3
(0)
.(5.3)
Finally, let Ψ0∈H3
(0). Then, (ℜ(Ψ0)
ℑ(Ψ0))T∈X3
(0). The stability estimate (5.3) and the density proved in Lemma 4.4 ends the proof of
Theorem 1.5.

Acknowledgements. This work has been partially carried out thanks to the support of ARCHIMEDE
LabEx (ANR-11-LABX- 0033), the A*MIDEX project (ANR-11-IDEX-0001-02) funded by the “In-
vestissements d’Avenir” French government program managed by the ANR, the ANR grant EMAQS
No.ANR-2011-BS01-017-01, the ERC avanced grant 266907 (CPDENL) of the 7th Research Framework
Programme (FP7) and the FQRNT.
A Simplified Saint-Venant Equation Example
Let us provide an explicit transformation (T , K)which allows to stabilize exponentially rapidly the sim-
plified Saint-Venant equation









ht+vx= 0,(t, x)∈(0, T )×(0,1),
vt+hx=−u(t),(x, t)∈(0, T )×(0,1),
h(t, 0) = v(t, 1) = 0, t ∈(0, T ),
h(0, x) = h0(x), v(0, x) = v0(x), x ∈(0,1),
(A.1)
which is controllable in time T > 2.
Let H:= hxand V:= vx. Then, the equation on (H, V )writes















Ht+Vx= 0,(t, x)∈(0, T )×(0,1),
Vt+Hx= 0,(t, x)∈(0, T )×(0,1),
H(t, 1) = −u(t), t ∈(0, T ),
V(t, 0) = 0, t ∈(0, T ),
H(0, x) = H0(x), V (0, x) = V0(x), x ∈(0,1),
34

with H0= (h0)xand V0= (v0)x. Consider now R1:= H+Vand R2:= H−V. Then















R1
t+R1
x= 0,(x, t)∈(0, T )×(0,1),
R2
t−R2
x= 0,(x, t)∈(0, T )×(0,1),
(R1+R2)(t, 1) = −2u(t), t ∈(0, T ),
(R1−R2)(t, 0) = 0, t ∈(0, T ),
R1(0, x) = R1
0(x), R2(0, x) = R2
0(x), x ∈(0,1),
with R1
0:= H0+V0and R2
0:= H0−V0. Let us consider a transformation which maps (R1, R2)to a
solution of a target stable system, that is, e
R1:= e−λxR1/cosh(λ)and e
R2:= eλxR2/cosh(λ), for λ > 0.
A straightforward computation leads to















e
R1
t+e
R1
x+λe
R1= 0,(x, t)∈(0, T )×(0,1),
e
R2
t−e
R2
x+λe
R2= 0,(x, t)∈(0, T )×(0,1),
(e
R1+e
R2)(t, 1) = −2e−λ(u(t)/cosh(λ)) + 2 tanh(λ)R2(t, 1), t ∈(0, T )
(e
R1−e
R2)(t, 0) = 0, t ∈(0, T ),
e
R1(0, x) = e
R1
0(x),e
R2(0, x) = e
R2
0(x), x ∈(0,1).
(A.2)
Hence, the exponential stability of (A.2) is obtained if −2e−λu(t)/cosh(λ) + 2 tanh(λ)R2(t, 1) = 0. In
terms of the original variables, it implies that
0 = −2e−λu(t)/cosh(λ) + 2 tanh(λ)R2(t, 1)
=−2e−λu(t)/cosh(λ) + 2 tanh(λ)(hx(t, 1) −vx(t, 1))
=−2e−λu(t)/cosh(λ)−2 tanh(λ)(vx(t, 1) + u(t)),
that is, u(t) = −tanh(λ)vx(t, 1).
The target system of (A.1) is given by









e
ht+evx+λe
h= 0,(x, t)∈(0, T )×(0,1),
evt+e
hx+λev= 0,(x, t)∈(0, T )×(0,1),
e
h(t, 0) = ev(t, 1) = 0, t ∈(0, T ),
e
h(0, x) = e
h0(x),ev(0, x) = ev0(x), x ∈(0,1),
(A.3)
One can recover an explicit expression of the transformation (T , K)leading to this target system using
e
hx+evx=e−λx(hx+vx)/cosh(λ),
e
hx−evx=eλx(hx−vx)/cosh(λ),(A.4)
which boils down to
e
hx= (cosh(λx)hx−sinh(λx)vx)/cosh(λ)
evx= (−sinh(λx)hx+ cosh(λx)vx)/cosh(λ).
Using the boundary conditions, one obtains the explicit transformation T
e
h(x) = 1
cosh(λ)cosh(λx)h(x)−λZx
0
sinh(λy)h(y)dy −sinh(λx)v(x) + λZx
0
cosh(λy)v(y)dy
=1
cosh(λ)Z1
0δx=ycosh(λy)−λ1(0,x)(y) sinh(λy)h(y)dy
+Z1
0λ1(0,x)(y) cosh(λy)−δx=ysinh(λy)v(y)dy,
ev(x) = 1
cosh(λ)sinh(λ)h(1) −sinh(λx)h(x)−Z1
x
λcosh(λy)h(y)dy
35

+ cosh(λx)v(x) + λZ1
x
sinh(λy)v(y)dy
=1
cosh(λ)Z1
0δy=1 sinh(λ)−δx=ysinh(λy)−λ1(x,1)(y) cosh(λy)h(y)dy
+Z1
0λ1(x,1)(y) sinh(λy) + δx=ycosh(λy)v(y)dy.
Moreover, the explicit transformation Kwrites
u(t) = tanh(λ)Z1
0
δ′
y=1v(t, y)dy.
If one writes, in the same spirit as (2.6), the kernels equation for (A.1), then one obtains that the kernels of
the transformations (T , K)exhibited here are the solution of this system. One also verifies that, thanks to
the factor 1/cosh(λ), the T B =Bcondition is verified by the transformation T. Getting back to (A.4)
one sees that the inverse of Tcan be computed explicitly performing similar computations.
Moreover, the Fourier coefficients of the kernels system associated to (A.1) have the same expression
as (2.13), where the eigenvalues/eigenfunctions are replaced by those associated with (A.1) and the Fourier
coefficients of the control operator µϕ1are replaced by the one of (A.1), that is 1.
It is also noticeable that the Fourier coefficients of the kernel α2are
α2
n= (−1)n(πn) tanh(λ),
which is adequate with the perturbation argument used in Proposition 3.7 to obtain the Fourier coefficients
from the T B =Bcondition. One notice that, for (A.1), α1≡0. Whether α1≡0or not in the case of the
linearized Schr¨odinger equation cannot be verified with our analysis.
B Quadratically close families
This section is devoted to the proof of Lemma 3.5.
Proof: Let s= 2 or 3. To simplify the notations, let,
˜cij
nk := cij
nkβ1
nhµϕ1, ϕki,˜
dij
nk := dij
nkβ2
nhµϕ1, ϕki,
where cij
nk and dij
nk are defined by (2.16) and βj
nare defined by (3.1).
First step: let us prove that
X
n∈N∗



ϕn/λs/2
n
0− g12
n/λ(s−2)/2
n
g22
n/λ(s−2)/2
n!




2
Xs
(0)
<+∞.
First, by denoting k=m+n, we have using in particular (1.3), (1.5), (2.13) and (2.15)
X
n∈N∗



ϕn
λs/2
n−g12
n
λ(s−2)/2
n



2
Hs
(0)
=X
n∈N∗X
k∈N∗\{n}
λs/2
k˜c12
nk
λ(s−2)/2
n
2
=X
n∈N∗X
0<|m|<n
m∈Z
+X
n<m
m∈N∗λs
n+m
λs
nλn˜c12
nn+m2
=λ4X
n∈N∗X
0<|m|<n
m∈Z
+X
n<m
m∈N∗λs
n+m
λs
n
(λ2+ 4λ2
n)λn+m
δnn+m(λ)λn
2hµϕ1, ϕn+mi
hµϕ1, ϕni
2
≤Cλ4X
n∈N∗X
0<|m|<n
m∈Z
+X
n<m
m∈N∗λn+m
λns−1
λ2+ 4λ2
n
δnn+m(λ)
2
.(B.1)
36

The two sums of (B.1) are dealt with separately.
Consider first the case where 0<|m|< n and m∈Z. We have, using in particular (2.14)
(λ2+ 4λ2
n)
δnn+m(λ)=1
m2n2
λ2
n4+ 4π4
λ2
n4+π2+π21 + m
n2 λ2
(mn)2+m
nπ2+ 2π22
≤1
m2n2
λ4+ 4π4
π8,(B.2)
and
λn+m=1 + m
n2λn≤4λn.(B.3)
Thus, for the first term of the right-hand side of (B.1)
X
n∈N∗X
0<|m|<n
m∈Zλn+m
λns−1
(λ2+ 4λ2
n)
δnn+m(λ)
2
≤CX
n∈N∗X
0<|m|<n
m∈Z
1
m4n4<+∞.(B.4)
Consider now the case where m > n. We have
(λ2+ 4λ2
n)
δnn+m(λ)=n4
m8
λ2
n4+ 4π4
λ2
m4+1 + n
m2π2+n
m2π22λ2
m4+π2+ 2 n
mπ22,
≤n4
m8
λ2+ 4π4
π8(B.5)
and λn+m
λn
=(n+m)2
n2=m2
n21 + n
m2≤4m2
n2.(B.6)
Thus, for the second term of the right-hand side of (B.1) yields,
X
n∈N∗X
n<m
m∈N∗λn+m
λns−1
(λ2+ 4λ2
n)
δnn+m(λ)
2
≤CX
n∈N∗X
n<m
m∈N∗n
m8−2(s−1) 1
m8
≤CX
n∈N∗X
n<m
m∈N∗
1
m4
1
n4<+∞.(B.7)
Then inequalities (B.1), (B.4) and (B.7) imply that
X
n∈N∗



ϕn
λs/2
n−g12
n
λ(s−2)/2
n



2
Hs
(0)
<+∞.(B.8)
The other sum is treated in a similar way. Indeed,
X
n∈N∗



g22
n
λ(s−2)/2
n



2
Hs
(0)
=X
n∈N∗X
k∈N∗
λs/2
k˜c22
nk
λ(s−2)/2
n
2
=X
n∈N∗X
0<|m|<n
m∈Z
+X
n<m
m∈N∗λn+m
λnsλn˜c22
nn+m2+X
n∈N∗λn˜c22
nn2
=λ2
4X
n∈N∗X
0<|m|<n
m∈Z
+X
n<m
m∈N∗λn+m
λns
λ2+ 4λ2
n
δnn+m(λ)
2
λ2−λ2
n+m+λ2
n
λn
2hµϕ1, ϕn+mi
hµϕ1, ϕni
2
+X
n∈N∗λ
2λn2
≤C+CX
n∈N∗X
0<|m|<n
m∈Z
+X
n<m
m∈N∗λn+m
λns−3
λ2+ 4λ2
n
δnn+m(λ)
2
λ2−λ2
n+m+λ2
n
λn
2
.(B.9)
37

Consider first the case where 0<|m|< n. We have

λ2−λ2
n+m+λ2
n
λn≤Cn3|m|
n2
λ2
n3m−π44 + 6 m
n+ 4 m
n2+m
n3≤Cn|m|λ2+ 15π4.
Using (B.2), it comes that
X
n∈N∗X
0<|m|<n
m∈Zλn+m
λns−3
λ2+ 4λ2
n
δnn+m(λ)
2
λ2−λ2
n+m+λ2
n
λn
2
≤CX
n∈N∗X
0<|m|<n
m∈Zλn+m
λns−31
n2m2.
This sum is clearly finite for s= 3. For s= 2, we notice that the previous series is the general term of a
Cauchy product. Indeed,
X
0<|m|<n
m∈Zλn+m
λns−31
n2m2=X
0<|m|<n
m∈Z
1
m2
1
(n+m)2
≤2
n−1
X
m=1
1
m2
1
(n−m)2+2
n2X
m∈N∗
1
m2.
Thus, for s∈ {2,3},
X
n∈N∗X
0<|m|<n
m∈Zλn+m
λns−3
λ2+ 4λ2
n
δnn+m(λ)
2
λ2−λ2
n+m+λ2
n
λn
2
<+∞.(B.10)
Consider now the case where m > n. We have

λ2−λ2
n+m+λ2
n
λn=
m4λ2
m4−1 + n
m4π4+n
m4
n2π2≤Cm4
n2,
and λn+m
λns−3
≤1.
Using (B.5), it then comes that,
X
n∈N∗X
n<m
m∈N∗λn+m
λns−3
λ2+ 4λ2
n
δnn+m(λ)
2
λ2−λ2
n+m+λ2
n
λn
2
≤CX
n∈N∗X
n<m
m∈N∗m4
n2
n4
m82
≤CX
n∈N∗X
n<m
m∈N∗
1
m2
1
n2<+∞.(B.11)
Then, inequalities (B.9), (B.10) and (B.11) imply that
X
n∈N∗



g22
n
λ(s−2)/2
n



2
Hs
(0)
<+∞.(B.12)
Together with (B.8) it ends the first step.
Second step: let us prove that
X
n∈N∗



0
ϕn/λs/2
n− h12
n/λ(s−2)/2
n
h22
n/λ(s−2)/2
n!




2
Xs
(0)
<+∞.
38

This proof is very similar to the first step. Thus we give the expressions of the different sums but we
do not detail every computation:
X
n∈N∗



h12
n
λ(s−2)/2
n



2
Hs
(0)
=X
n∈N∗X
k∈N∗
λs/2
k˜
d12
nk
λ(s−2)/2
n
2
=X
n∈N∗X
0<|m|<n
m∈Z
+X
n<m
m∈N∗λn+m
λnsλn˜
d12
nn+m
2+X
n∈N∗λn˜
d12
nn
2
=λ2X
n∈N∗X
0<|m|<n
m∈Z
+X
n<m
m∈N∗λs
n+m
λs
n
λ2+ 4λ2
n
δnn+m(λ)
2
λn+mλ2+λ2
n+m−λ2
n
λ2+ 2λ2
n
2hµϕ1, ϕn+mi
hµϕ1, ϕni
2
+X
n∈N∗
λ2λn
λ2+λ2
n
2
<+∞,
and
X
n∈N∗



ϕn
λs/2
n−h22
n
λ(s−2)/2
n



2
Hs
(0)
=X
n∈N∗X
k∈N∗\{n}
λs/2
k˜
d22
nk
λ(s−2)/2
n
2
=X
n∈N∗X
0<|m|<n
m∈Z
+X
n<m
m∈N∗λs
n+m
λs
nλn˜
d22
nn+m
2
=λ4X
n∈N∗X
0<|m|<n
m∈Z
+X
n<m
m∈N∗λs
n+m
λs
n
λ2+ 4λ2
n
δnn+m(λ)
2
λ2+λ2
n+m+λ2
n
λ2+ 2λ2
n
2hµϕ1, ϕn+mi
hµϕ1, ϕni
2
<+∞.
This completes the proof of Lemma 3.5.

C Rapid stabilization of the linearized system
In this section we detail how Theorem 1.4 can be obtained from the results developed in this article.
Let Ψ0∈ H0and Ψthe associated solution for a control u. Then, if we define, e
Ψ(t, ·) := Ψ(t, ·)eiλ1t,
it comes that e
Ψsatisfies





i∂te
Ψ = −∆e
Ψ−λ1e
Ψ−u(t)µϕ1,(t, x)∈(0, T )×(0,1)
e
Ψ(t, 0) = e
Ψ(t, 1) = 0, t ∈(0, T )
e
Ψ(0,·) = Ψ0, x ∈(0,1)
(C.1)
and ke
Ψ(t, .)kH3
(0) =kΨ(t, .)kH3
(0) .
Rapid stabilization of (C.1). Notice that system (C.1) is almost identical to (1.10) except that the spec-
trum of the underlying operator is shifted by λ1. This modifies the state space. Indeed, for every t≥0,
0 = ℜhΨ(t, .),Φ1(t, .)i=ℜhe
Ψ(t, .), ϕ1i.
Thus, e
Ψ∈ H0. Notice that due to Theorem 1.3, one gets that system (C.1) is exactly controllable in H0.
The rest of the analysis is barely modified. For instance, the Riesz bases have instead the form
B=g12
n
g22
n;n≥2∪h12
n
h22
n;n≥1,
39

where, for example,
g12
n(x) =
+∞
X
k=1
2λ(λk−λ1)(λn−λ1)
(λ2+ (λk−λn)2)(λ2+ (λk+λn−2λ1)2)β1
nhµϕ1, ϕkiϕk(x),
with β1
nchosen such that hg12
n, ϕni= 1/λn.
This leads to the existence of a feedback law ˜
K ℜ(e
Ψ)
ℑ(e
Ψ)!such that, for the closed-loop system




 ℜ(e
Ψ(t, .))
ℑ(e
Ψ(t, .))!



X3
(0)
≤Ce−λt 


ℜ(Ψ0)
ℑ(Ψ0)


X3
(0)
.
Rapid stabilization of (1.8). Due to the previous relation, e
Ψ(t, .) = Ψ(t, .)eiλ1tit comes that if Ψis the
solution of (1.8) with the feedback law
v(t) = ˜
Kcos(λ1t)ℜ(Ψ(t, .)) −sin(λ1t)ℑ(Ψ(t, .))
sin(λ1t)ℜ(Ψ(t, .)) + cos(λ1t)ℑ(Ψ(t, .)),
then
kΨ(t, .)kH3
(0) ≤Ce−λt kΨ0kH3
(0) .
References
[1] A. Balogh and M. Krsti´c. Infinite dimensional backsteppingstyle feedback transformations for a heat
equation with an arbitrary level of instability. Eur. J. Control, 8(3):165–175, 2002.
[2] K. Beauchard. Local controllability of a 1-D Schr¨odinger equation. J. Math. Pures Appl. (9),
84(7):851–956, 2005.
[3] K. Beauchard and J.-M. Coron. Controllability of a quantum particle in a moving potential well. J.
Funct. Anal., 232(2):328–389, 2006.
[4] K. Beauchard, J.-M. Coron, and H. Teismann. Minimal time for the bilinear control of Schr¨odinger
equations. Systems Control Lett., 71:1–6, 2014.
[5] K. Beauchard and C. Laurent. Local controllability of 1D linear and nonlinear Schr ¨odinger equations
with bilinear control. J. Math. Pures Appl. (9), 94(5):520–554, 2010.
[6] K. Beauchard and C. Laurent. Exact controllability of the 2D Schr¨odinger-Poisson system. Preprint,
2016.
[7] K. Beauchard and M. Mirrahimi. Practical stabilization of a quantum particle in a one-dimensional
infinite square potential well. SIAM J. Control Optim., 48(2):1179–1205, 2009.
[8] K. Beauchard and M. Morancey. Local controllability of 1D Schr¨odinger equations with bilinear
control and minimal time. Math. Control Relat. Fields, 4(2):125–160, 2014.
[9] U. Boscain, M. Caponigro, T. Chambrion, and M. Sigalotti. A weak spectral condition for the con-
trollability of the bilinear Schr¨odinger equation with application to the control of a rotating planar
molecule. Comm. Math. Phys., 311(2):423–455, 2012.
[10] U. Boscain, M. Caponigro, and M. Sigalotti. Multi-input Schr ¨odinger equation: controllability, track-
ing, and application to the quantum angular momentum. preprint, arXiv:1302.4173, 2013.
[11] U. Boscain, T. Chambrion, and M. Sigalotti. On some open questions in bilinear quantum control.
preprint, arXiv:1304.7181, 2013.
40

[12] D. M. Boˇskovi´c, A. Balogh, and M. Krsti´c. Backstepping in infinite dimension for a class of parabolic
distributed parameter systems. Math. Control Signals Systems, 16(1):44–75, 2003.
[13] N. Boussaid, M. Caponigro, and T. Chambrion. Regular propagators of bilinear quantum systems.
preprint, June 2014.
[14] E. Cerpa and J.-M. Coron. Rapid stabilization for a Korteweg-de Vries equation from the left Dirichlet
boundary condition. IEEE Trans. Automat. Control, 58(7):1688–1695, 2013.
[15] E. Cerpa and E. Cr´epeau. Rapid exponential stabilization for a linear Korteweg-de Vries equation.
Discrete Contin. Dyn. Syst. Ser. B, 11(3):655–668, 2009.
[16] T. Chambrion, P. Mason, M. Sigalotti, and U. Boscain. Controllability of the discrete-spectrum
Schr¨odinger equation driven by an external field. Ann. Inst. H. Poincar´
e Anal. Non Lin´
eaire,
26(1):329–349, 2009.
[17] J.-M. Coron. On the small-time local controllability of a quantum particle in a moving one-
dimensional infinite square potential well. C. R. Math. Acad. Sci. Paris, 342(2):103–108,2006.
[18] J.-M. Coron. Control and nonlinearity, volume 136 of Mathematical Surveys and Monographs.
American Mathematical Society, Providence, RI, 2007.
[19] J.-M. Coron. Stabilization of control systems and nonlinearities. In Proceedings of the 8th Inter-
national Congress on Industrial and Applied Mathematics, pages 17–40. Higher Ed. Press, Beijing,
2015.
[20] J.-M. Coron and B. d’Andr´ea Novel. Stabilization of a rotating body beam without damping. IEEE
Trans. Automat. Control, 43(5):608–618, 1998.
[21] J.-M. Coron, L. Hu, and G. Olive. Stabilization and controllability of first-order integro-differential
hyperbolic equations. Preprint, arXiv:1511.01078, 2015.
[22] J.-M. Coron and Q. L¨u. Local rapid stabilization for a Korteweg-de Vries equation with a Neumann
boundary control on the right. J. Math. Pures Appl. (9), 102(6):1080–1120, 2014.
[23] J.-M. Coron and Q. L ¨u. Fredholm transform and local rapid stabilization for a Kuramoto–Sivashinsky
equation. J. Differential Equations, 259(8):3683–3729, 2015.
[24] J.-M. Coron and H.-M. Nguyen. Null controllability and finite time stabilization for the heat equations
with variable coefficients in space in one dimension via backstepping approach. 2015. Preprint, hal-
01228895, version 1.
[25] J.-M. Coron, R. Vazquez, M. Krsti´c, and G. Bastin. Local exponential H2stabilization of a 2×2
quasilinear hyperbolic system using backstepping. SIAM J. Control Optim., 51(3):2005–2035,2013.
[26] B. Dehman, P. G´erard, and G. Lebeau. Stabilization and control for the nonlinear Schr¨odinger equa-
tion on a compact surface. Math. Z., 254(4):729–749, 2006.
[27] I. C. Gohberg and M. G. Kre˘ın. Introduction to the theory of linear nonselfadjoint operators. Trans-
lated from the Russian by A. Feinstein. Translations of Mathematical Monographs, Vol. 18. American
Mathematical Society, Providence, R.I., 1969.
[28] V. Komornik. Rapid boundary stabilization of linear distributed systems. SIAM J. Control Optim.,
35(5):1591–1613, 1997.
[29] M. Krsti´c, B.-Z. Guo, and A. Smyshlyaev. Boundary controllers and observers for the linearized
Schr¨odinger equation. SIAM J. Control Optim., 49(4):1479–1497, 2011.
[30] M. Krsti´c, I. Kanellakopoulos, and P.V. Kokotovic. Nonlinear and Adaptive Control Design. Adap-
tative and Learning Systems for Signal Processing, Communications, and Control. John Wiley and
Sons, July 1995. ISBN: 978-0-471-12732-1.
41

[31] M. Krsti´c and A. Smyshlyaev. Boundary control of PDEs, volume 16 of Advances in Design and
Control. Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 2008. A course
on backstepping designs.
[32] I. Lasiecka, R. Triggiani, and X. Zhang. Global uniqueness, observability and stabilization of non-
conservative Schr¨odinger equations via pointwise Carleman estimates. II. L2(Ω)-estimates. J. Inverse
Ill-Posed Probl., 12(2):183–231, 2004.
[33] C. Laurent. Internal control of the Schr¨odinger equation. Math. Control Relat. Fields, 4(2):161–186,
2014.
[34] W.-J. Liu and M. Krsti ´c. Backstepping boundary control of Burgers’ equation with actuator dynamics.
Systems Control Lett., 41(4):291–303, 2000.
[35] E. Machtyngier and E. Zuazua. Stabilization of the Schr¨odinger equation. Portugal. Math.,
51(2):243–256, 1994.
[36] F. M´ehats, Y. Privat, and M. Sigalotti. On the controllability of quantum transport in an electronic
nanostructure. SIAM J. Appl. Math., 74(6):1870–1894, 2014.
[37] M. Mirrahimi. Lyapunov control of a quantum particle in a decaying potential. Ann. Inst. H. Poincar´
e
Anal. Non Lin´
eaire, 26(5):1743–1765, 2009.
[38] M. Morancey. Simultaneous local exact controllability of 1D bilinear Schr¨odinger equations. Ann.
Inst. H. Poincar´
e Anal. Non Lin´
eaire, 31(3):501–529, 2014.
[39] M. Morancey and V. Nersesyan. Simultaneous global exact controllability of an arbitrary number of
1d bilinear Schr ¨odinger equations. J. Math. Pures Appl. (9), 103(1):228–254, 2015.
[40] V. Nersesyan. Growth of Sobolev norms and controllability of the Schr¨odinger equation. Comm.
Math. Phys., 290(1):371–387, 2009.
[41] V. Nersesyan. Global approximate controllability for Schr¨odinger equation in higher Sobolev norms
and applications. Ann. Inst. H. Poincar´
e Anal. Non Lin´
eaire, 27(3):901–915, 2010.
[42] V. Nersesyan and H. Nersisyan. Global exact controllability in infinite time of Schr¨odinger equation.
J. Math. Pures Appl. (9), 97(4):295–317, 2012.
[43] A. Pazy. Semigroups of linear operators and applications to partial differential equations, volume 44
of Applied Mathematical Sciences. Springer-Verlag, New York, 1983.
[44] J.-P. Puel. A regularity property for Schr ¨odinger equations on bounded domains. Rev. Mat. Complut.,
26(1):183–192, 2013.
[45] J.-P. Puel. Local exact bilinear control of the Schr¨odinger equation. Preprint, 2016.
[46] E.D. Sontag. Mathematical control theory, volume 6 of Texts in Applied Mathematics. Springer-
Verlag, New York, second edition, 1998. Deterministic finite-dimensional systems.
[47] B. Sz.-Nagy. Perturbations des transformations lin´eaires ferm´ees. Acta Sci. Math. Szeged, 14:125–
137, 1951.
[48] M. Tucsnak and G. Weiss. Observation and control for operator semigroups. Birkh¨auser Advanced
Texts: Basler Lehrb¨ucher. Birkh¨auser Verlag, Basel, 2009.
[49] J. M. Urquiza. Rapid exponential feedback stabilization with unbounded control operators. SIAM J.
Control Optim., 43(6):2233–2244 (electronic), 2005.
[50] A. Vest. Rapid stabilization in a semigroup framework. SIAM J. Control Optim., 51(5):4169–4188,
2013.
[51] R.M. Young. An introduction to nonharmonic Fourier series, volume 93 of Pure and Applied Math-
ematics. Academic Press Inc. [Harcourt Brace Jovanovich Publishers], New York, 1980.
42