COMMUNICATIONS ON
PURE AND APPLIED ANALYSIS
Volume 8, Number 1, January 2009
doi:10.3934/cpaa.2009.8.311
pp. 311–333
LOCAL EXACT CONTROLLABILITY TO THE TRAJECTORIES
OF THE BOUSSINESQ SYSTEM VIA A FICTITIOUS CONTROL
ON THE DIVERGENCE EQUATION
Manuel González-Burgos
Dpto. E.D.A.N., University of Sevilla, Aptdo. 1160, 41080 Sevilla, Spain
Sergio Guerrero
Université Pierre et Marie Curie-Paris6, UMR 7598 Laboratoire Jacques-Louis Lions,
Paris, F-75005 France
Jean-Pierre Puel
Laboratoire de Mathématiques de Versailles, Université de Versailles St Quentin,
45 avenue des Etats Unis, 78035 Versailles Cedex, France
Abstract. In this paper, we deal with the three-dimensional Boussinesq system. We prove the local exact controllability to the trajectories of this system
when the control is supported in a small set.
The main objective of this paper is to present a new method to control
systems associated to equations of fluid dynamics. This method consists of
controlling the same system with an additional control acting on the divergence
condition in a first step and lifting this condition in a second step. In this paper,
we have chosen to apply this technique to the Boussinesq system.
1. Introduction. Let Ω ⊂ RN (with N ≥ 2) be a bounded connected open set
whose boundary Γ = ∂Ω is of class C 2 . Let ω ⊂ Ω be a nonempty open subset of
Ω and let T > 0. We will use the notation Q = Ω × (0, T ) and Σ = ∂Ω × (0, T ) and
we will denote by n(x) the outward unit normal to Ω at the point x ∈ ∂Ω.
Throughout this article we will denote by C a generic positive constant (usually
depending on Ω and ω) and χω will denote the characteristic function of ω.
We consider the controlled Boussinesq system

yt − µ∆y + ∇ · (y ⊗ y) + ∇p = −θg + v1 χω in Q,





in Q,
 θt − ∆θ + ∇ · (θy) = v2 χω
∇·y = 0
in Q,
(1)



y
=
0,
θ
=
0
on
Σ,



y(0) = y 0 , θ(0) = θ0
in Ω,
where µ is the viscosity, g is a constant vector field representing gravity acceleration,
v1 is a priori a N -components vector control and v2 is a scalar control, both controls
acting in ω.
2000 Mathematics Subject Classification. Primary: 93C20, 76D55; Secondary: 93B05.
Key words and phrases. Controllability, Boussinesq system, control of fluid flows, fictitious
control.
The first author is partially supported by D.G.E.S. (Spain), Grant MTM2006-07932. The third
author is supported by ANR C-QUID 06-BLAN-0052 .
311
312
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
We will use the following vector spaces which are usual in the mathematical
context of incompressible fluids:

N

 V = {ϕ ∈ D(Ω) : ∇ · ϕ = 0 in Ω},
2
H = {w ∈ L (Ω)N : ∇ · w = 0 in Ω and w · n = 0 on ∂Ω} and

 V = {w ∈ H 1 (Ω)N : ∇ · w = 0 in Ω}.
0
For p, q ∈ [1, ∞), we will also use the notation:
X
p,q
= {u : u ∈ Lp (0, T ; W 2,q (Ω) ∩ W01,q (Ω)), ut ∈ Lp (0, T ; Lq (Ω))},
X p = X p,p .
We consider an “ideal” trajectory (y, θ), together with a pressure p, which is a
regular enough solution of the uncontrolled Boussinesq system


 y t − µ∆y + ∇ · (y ⊗ y) + ∇p = −θg in Q,



in Q,

 θ t − ∆θ + y · ∇θ = 0
∇·y =0
in Q,
(2)



y = 0, θ = 0
on Σ,



0

0
y(0) = y , θ(0) = θ
in Ω.
The question of local controllability to this trajectory (y, θ) is to look for controls
(v1 , v2 ) such that, at time T , we have
y(T ) = y(T ), θ(T ) = θ(T ) in Ω,
0
0
0
provided that (y − y ) and (θ0 − θ ) are small enough in suitable norms.
Taking the differences
z = y − y, q = p − p, ρ = θ − θ,
we obtain the new system

zt − µ∆z + ∇ · (z ⊗ z + y ⊗ z + z ⊗ y) + ∇q = −ρg + v1 χω





ρ
 t − ∆ρ + ∇ · (zρ) + ∇ · yρ + zθ = v2 χω
∇·z = 0



z = 0, θ = 0



z(0) = z 0 , ρ(0) = ρ0
in Q,
in Q,
in Q,
on Σ,
in Ω
(3)
and we look for controls (v1 , v2 ) in a space which will be specified later on such that
z(T ) = 0, ρ(T ) = 0 in Ω,
(4)
provided that (z 0 , ρ0 ) are small enough in suitable spaces.
In order to prove this local result, we will use a fixed point argument and it is
therefore natural to introduce the following “linearized” system:

zt − µ∆z + ∇ · (a ⊗ z + z ⊗ b) + ∇q = −ρg + v1 χω in Q,





ρ
in Q,
 t − ∆ρ + ∇ · (cρ) + ∇ · (zd) = v2 χω
∇·z = 0
in Q,
(5)


 z = 0, ρ = 0
on
Σ,



z(0) = z 0 , ρ(0) = ρ0
in Ω,
where the following hypothesis will be imposed on the coefficients appearing in (5)
(
a, b, c ∈ L∞ (Q)N , d ∈ L∞ (Q), ∇ · a = ∇ · b = ∇ · c = 0 in Q,
(6)
at , bt , ct ∈ Lp (0, T ; Lq (Ω)N ), dt ∈ Lp (0, T ; Lq (Ω)), p, q ∈ (1, ∞),
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
b ∈ L∞ (0, T ; W 1,∞ (Ω)N ),
313
d ∈ L∞ (0, T ; W 1,∞ (Ω)).
(7)
One of our goals will be to prove the null controllability for this system (5), i.e.
to show that for every initial data (z 0 , ρ0 ) in a suitable space, there exist controls
(v1 , v2 ) in a space which will be made precise such that (4) holds. These questions
have already been studied by several authors, in particular [8] and [10] using a direct
approach and in particular a global Carleman estimate for the adjoint system which
is difficult to obtain as it requires elimination of a local term on the pressure (this
will be made more explicit in the next section).
The main interest of the present article is to change the strategy for obtaining the
local exact controllability to the trajectories of system (1). This strategy consists
in introducing, in a first step, a fictitious control in the divergence term. In order
to obtain a controllability result with this additional control, we will use again a
global Carleman estimate for the adjoint system, but a simpler one, in which we can
accept local terms in the pressure variable. In a second step, we will remove this
fictitious control in order to obtain the local controllability result for system (1).
Therefore we now introduce a new controllability problem. Let us take a non
empty open set ω1 with ω 1 ⊂ ω and let us consider ζ ∈ C0∞ (RN ) such that
0 ≤ ζ(x) ≤ 1, ∀x ∈ RN , ζ(x) = 1, ∀x ∈ ω1 , Supp ζ ⊂ ω.
We introduce the following linear system

zt − µ∆z + ∇ · (a ⊗ z + z ⊗ b) + ∇q = −ρg + v1 χω





 ρt − ∆ρ + ∇ · (cρ) + ∇ · (zd) = v2 χω
∇ · z = v3 ζ



z = 0, ρ = 0



z(0) = z 0 , ρ(0) = ρ0
in Q,
in Q,
in Q,
on Σ,
in Ω,
(8)
where a, b, c and d satisfy (6), z 0 ∈ H, ρ0 ∈ L2 (Ω) and v1 , v2 , v3 are the controls.
Control v3 , which is our fictitious control, must of course satisfy a compatibility
condition due to the homogeneous Dirichlet boundary condition on the velocity z.
In section 2 we will study null controllability for system (8), i.e. we will look for
controls (v1 , v2 , v3 ) such that
z(T ) = 0, ρ(T ) = 0 in Ω.
In a second step we will be able to get rid of the fictitious control by showing
a regularity result on admissible controls and then by lifting, in a local way, the
control appearing in the divergence term. This will be done in Section 3.
The nonlinear problem will be treated afterwards in Section 4 using a Kakutani
fixed point argument.
The method is presented here for the Boussinesq system only for sake of clarity
but one of its main interests is that it can be easily adapted to a large class of weakly
coupled systems involving Navier-Stokes equations and several diffusion convection
equations.
2. Null controllability for a linearized Boussinesq system with a control
in the divergence term. We will here study null controllability for problem (8)
with controls v1 , v2 and v3 . First of all we have to give a precise sense to this
problem when the controls have weak regularity and also to the null controllability
problem.
314
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
2.1. Definition of weak solutions for problem (8) and existence result.
We begin this section by making formal calculations in order to explain the correct
sense of weak solutions for (8). To this end, let us assume that (z, ρ) is a solution to
system (8) associated to (v1 , v2 , v3 ) ∈ L2 (Q)N +2 and consider “regular” functions
ϕ = (ϕ1 , ..., ϕN ) and r with ∇ · ϕ = 0 in Q and ϕ = 0, r = 0 on Σ. We formally take
the scalar product of the equations in (8) by ϕ and r respectively and integrate by
parts. If (·, ·)H and (·, ·) denote, respectively, the scalar product in H and L2 (Ω),
we obtain
ZZ

0

(z(T
),
ϕ(T
))
−
(z
,
ϕ(0))
+
y · (−ϕt − µ∆ϕ) dx dt

H
H


Q
ZZ
ZZ
N ZZ
X
∂ϕi



(aj zi + zj bi )
dx dt = −
ρg · ϕ dx dt +
v1 · ϕ dx dt
 −
∂xj
Q
Q
ω×(0,T )
i,j=1
and
ZZ

0


(ρ(T
),
r(T
))
−
(ρ
,
r(0))
+
ρ(−rt − ∆r) dx dt


Q
ZZ
N ZZ
N ZZ
X
X

∂r
∂r


ρci
dx dt −
dzi
dx dt =
v2 r dx dt.
 −
∂xi
∂xi
Q
Q
ω×(0,T )
i=1
i=1
Let us suppose that (ϕ, r) satisfies (together with a pressure π) the following
system

−ϕt − µ∆ϕ − (a · ∇) ϕ − ∇ϕ b − d∇r + ∇π = θ in Q,





in Q,
 −rt − ∆r − c · ∇r + g · ϕ = σ
∇·ϕ =0
in Q,
(9)



ϕ = 0, r = 0
on Σ,



ϕ(T ) = ϕ0 , r(T ) = r0
in Ω.
We then obtain
ZZ
ZZ

0
0


z · θ dx dt +
πv3 ζ dx dt + (ρ(T ), r0 )
 (z(T ), ϕ )H − (z , ϕ(0))H +
Q
ω×(0,T )
ZZ
ZZ
ZZ

0

ρσ dx dt =
v1 · ϕ dx dt +
v2 r dx dt.
 − (ρ , r(0)) +
Q
ω×(0,T )
ω×(0,T )
This can be rewritten in the form
ZZ
ZZ

0
0

′ ,V + (ρT , r ) +

hz
,
ϕ
i
z
·
θ
dx
dt
+
ρσ dx dt
T
V

Q
Q
ZZ


(v1 · ϕ + v2 r − v3 ζπ) dx dt + hz 0 , ϕ(0)iV ′ ,V + (ρ0 , r(0))
 =
(10)
ω×(0,T )
with
zT = z(T ) and ρT = ρ(T ) in Ω.
We will now give a correct formulation of problem (8) in a weak form based on
(10). Let us take
(θ, σ, ϕ0 , r0 ) ∈ X ≡ L2 (0, T ; L2 (Ω)N ) × L2 (0, T ; H −1(Ω)) × V × L2 (Ω).
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
315
Then problem (9) has a unique solution (ϕ, π, r) (π being defined up to a constant
for each time t) with
(ϕ, π) ∈ Y ≡ C([0, T ]; V ) ∩ L2 (0, T ; H 2 (Ω)N ∩ V ) × L2 (0, T ; H 1(Ω)/R),
r ∈ C([0, T ]; L2 (Ω)) ∩ L2 (0, T ; H01(Ω)).
Moreover, the mapping
(θ, σ, ϕ0 , r0 ) ∈ X 7→ (ϕ, π, r) ∈ Y × C([0, T ]; L2(Ω)) ∩ L2 (0, T ; H01 (Ω))
is linear and continuous.
We will use the following notation
L2ζ (ω) = {v ∈ L2 (ω) :
Z
ζv dx = 0}.
ω
2
For given v1 ∈ L2 (0, T ; L2(ω)N ), v2 ∈ L2 (0, T ; L (ω)), v3 ∈ L2 (0, T ; L2ζ (ω)), z 0 ∈ V ′
and ρ0 ∈ L2 (Ω), we define
ZZ
L(θ, σ, ϕ0 , r0 ) =
(v1 · ϕ + v2 r − v3 ζπ) dx dt + hz 0 , ϕ(0)iV ′ ,V + (ρ0 , r(0)),
ω×(0,T )
0
0
with (θ, σ, ϕ , r ) ∈ X and where (ϕ, π, r) is the solution to (9) associated to
(θ, σ, ϕ0 , r0 ). Of course L is a continuous linear form on X ≡ L2 (0, T ; L2(Ω)N ) ×
L2 (0, T ; H −1 (Ω)) × V × L2 (Ω) which is a Hilbert space. Therefore, from Riesz’s
theorem, there exist
z ∈ L2 (0, T ; L2 (Ω)N ), ρ ∈ L2 (0, T ; H01 (Ω)), zT ∈ V ′ , ρT ∈ L2 (Ω)
(uniquely defined) such that for every (θ, σ, ϕ0 , r0 ) ∈ X,
ZZ
Z T
L(θ, σ, ϕ0 , r0 ) =
z · θ dx dt +
hσ, ρiH −1 ,H01 dt + hzT , ϕ0 iV ′ ,V + (ρT , r0 ). (11)
Q
0
We can now formulate the definition of weak solutions to problem (8) and the
corresponding existence and uniqueness result.
Definition 2.1. Let (v1 , v2 , v3 ) ∈ L2 (0, T ; L2 (ω)N )×L2 (0, T ; L2 (ω))×L2 (0, T ; L2ζ (ω))
and (z 0 , ρ0 ) ∈ V ′ × L2 (Ω) be given. We say that (z, ρ, zT , ρT ) is a weak solution
(by transposition) of problem (8) if
z ∈ L2 (0, T ; L2 (Ω)N ), ρ ∈ L2 (0, T ; H01 (Ω)), zT ∈ V ′ , ρT ∈ L2 (Ω)
and if they satisfy (11) for every (θ, σ, ϕ0 , r0 ) ∈ L2 (0, T ; L2 (Ω)N )×L2 (0, T ; H −1(Ω))×
V × L2 (Ω).
Theorem 2.2. For every (v1 , v2 , v3 ) ∈ L2 (0, T ; L2(ω)N ) × L2 (0, T ; L2(ω))
×L2 (0, T ; L2ζ (ω)) and (z 0 , ρ0 ) ∈ V ′ × L2 (Ω), there exists a unique weak solution (by
transposition) of problem (8).
Of course the proof of Theorem 2.2 has already been given.
Remark 1. If we have a “strong” solution (z, ρ) of problem (8), it will be of
course the unique weak solution (by transposition) with zT = z(T ) and ρT =
ρ(T ). This will be the case if, for example, we can find F ∈ L2 (0, T ; H01 (Ω)N )
with Ft ∈ L2 (0, T ; H −1 (Ω)N ) such that ∇ · F = v3 ζ in Q. Then by a standard
translation, the problem is reduced to a classical linear Boussinesq system. In fact
as v3 ∈ L2 (0, T ; L2ζ (ω)) we know that there exists F ∈ L2 (0, T ; H01 (Ω)N ) such that
∇ · F = v3 ζ in Q. But without assuming regularity on the time derivative of v3 we
do not have any information on Ft .
316
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
Remark 2. We can give a partial interpretation of the meaning of (11). Let us
take
ϕ ∈ D((0, T ); V), r ∈ D((0, T ); D(Ω)), π = 0
and define
θ = −ϕt − µ∆ϕ − (a · ∇)ϕ − ∇ϕb − d∇r,
σ = −rt − ∆r − c · ∇r + g · ϕ.
We then have

+∇ · (a ⊗ z +z ⊗ b) + ρg, ϕiD′ ,D + hρt −∆ρ+∇ · (cρ) + ∇ · (zd), riD′ ,D
 hzt −µ∆z
ZZ
Z T
ZZ
ZZ
z · θ dxdt+ hσ, ρiH −1 ,H01 dt =
v1 · ϕ dxdt+
v2 r dx, dt.
 =
Q
0
ω×(0,T )
ω×(0,T )
′
Using de Rham’s lemma we deduce that there exists q ∈ D (Q) such that, in the
sense of distributions in Q, the following holds
zt − µ∆z + ∇ · (a ⊗ z + z ⊗ b) + ∇q = −ρg + v1 χω
ρt − ∆ρ + ∇ · (cρ) + ∇ · (zd) = v2 χω .
Let us now take ϕ = 0, r = 0 and π ∈ D(Q) so that θ = ∇π and σ = 0. Reasoning
as above, we obtain
ZZ
ZZ
z · ∇π dx dt =
πv3 ζ dx dt
Q
ω×(0,T )
whence
∇ · z = v3 ζ in D′ (Q).
We know that ρ ∈ L2 (0, T ; H01 (Ω)) and, from the equation satisfied by ρ, this
implies ρt ∈ L2 (0, T ; H −1(Ω)) so that ρ ∈ C([0, T ]; L2 (Ω)) and ρ(0) and ρ(T ) are
well defined in L2 (Ω). It is then classical to show that we have
ρ(0) = ρ0
and ρ(T ) = ρT
in Ω.
2
On the other hand we only know that z ∈ L (0, T ; H∇ (Ω)) where
H∇ (Ω) = {z ∈ L2 (Ω)N : ∇ · z ∈ L2 (Ω)}.
1
Therefore we can give a sense to the normal trace of z with z·n ∈ L2 (0, T ; H − 2 (∂Ω))
and it is then classical to show that z · n = 0. But we cannot give a sense to the
tangential trace of z. In the same way, as we do not have any regularity for the
pressure q, we do not have any information on the time derivative zt and this
prevents us from giving a sense to the values of z at times t = 0 and t = T . We can
only say that “formally” we have z(0) = z 0 and z(T ) = zT in Ω.
2.2. Null controllability for weak solutions of (8). From the results given
above, we know that for every z 0 ∈ V ′ , ρ0 ∈ L2 (Ω), v1 ∈ L2 (0, T ; L2 (ω)N ), v2 ∈
L2 (0, T ; L2 (ω)) and v3 ∈ L2 (0, T ; L2ζ (ω)) there exists a unique weak solution (by
transposition) (z, ρ, zT , ρT ) of (8) with
z ∈ L2 (0, T ; L2(Ω)N ), ρ ∈ L2 (0, T ; H01(Ω))∩C([0, T ]; L2 (Ω)), zT ∈ V ′ , ρT ∈ L2 (Ω).
Definition 2.3. We say that (8) is null controllable at time T > 0 if for every
(z 0 , ρ0 ) ∈ V ′ × L2 (Ω), there exist v1 ∈ L2 (0, T ; L2 (ω)N ), v2 ∈ L2 (0, T ; L2(ω)) and
v3 ∈ L2 (0, T ; L2ζ (ω)) such that the corresponding solution (z, ρ, zt , ρT ) to (8) satisfies
zT = 0 and ρT = 0 in Ω.
The object of this subsection is to show the following result:
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
317
Theorem 2.4. Let us assume that ω is a non empty open subset of Ω, T > 0 and
the coefficients a, b, c, d satisfy
a, b, c ∈ L∞ (Q)N , d ∈ L∞ (Q).
Then problem (8) is null controllable at time T . Moreover, we can obtain controls
v1 , v2 and v3 with supplementary integrability properties.
We will obtain the proof of this result as a consequence of an observability inequality satisfied by the solutions to the corresponding adjoint problem to (8),

−ϕt − µ∆ϕ − (a · ∇) ϕ − ∇ϕ b − d∇r + ∇π = 0 in Q,





in Q,
 −rt − ∆r − c · ∇r + g · ϕ = 0
∇·ϕ =0
in Q,
(12)


 ϕ = 0, r = 0
on
Σ,



ϕ(T ) = ϕ0 , r(T ) = r0
in Ω,
which itself is a consequence of global Carleman estimates for the decoupled system.
First of all we have to introduce some weight functions which are essential to obtain
these Carleman estimates.
Let us fix ω2 ⊂⊂ ω1 , a new open set. We know (cf. [7] Lemma 1.1 or [14]) that
there exists η 0 ∈ C 2 (Ω) such that
η 0 > 0 in Ω, η 0 = 0 on ∂Ω, |∇η 0 | > 0 in Ω \ ω2 .
From this function η 0 and for s, λ ≥ 1 and m > 4, we construct the following weight
functions:

0
0
0
e(5/4)λmkη k∞ − eλ(mkη k∞ +η (x))



α(x,
t)
=
,

t4 (T − t)4
(13)
λ(mkη 0 k∞ +η 0 (x))


 ξ(x, t) = e

.
t4 (T − t)4
These precise weights were also considered in [5] and [10].
Let us introduce a notation corresponding to weighted Sobolev norms
ZZ
ZZ
I(s, λ; ϕ) = s−1
e−2sα ξ −1 |ϕt |2 dx dt + s−1
e−2sα ξ −1 |∆ϕ|2 dx dt
Q
Q
ZZ
ZZ
2
−2sα
2
3 4
+ sλ
e
ξ|∇ϕ| dx dt + s λ
e−2sα ξ 3 |ϕ|2 dx dt.
Q
Q
We now consider separately a backward Stokes equation and a backward heat equation. To be precise, we consider the systems


 −ϕt − µ∆ϕ + ∇π = f1 , ∇ · ϕ = 0 in Q,
ϕ=0
on Σ,
(14)

 ϕ(T ) = ϕ0
in Ω,
and


 −rt − ∆r = f2
r=0

 r(T ) = r0
in Q,
on Σ,
in Ω,
(15)
with f1 ∈ L2 (Q)N , f2 ∈ L2 (Q), ϕ0 ∈ H and r0 ∈ L2 (Ω).
Using the results of [7], [13] and [5] we can obtain the following Carleman estimates.
318
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
Proposition 1. There exist three positive constants C0 , s0 and λ0 depending on Ω
and ω1 such that, for every f1 ∈ L2 (Q)N , f2 ∈ L2 (Q), ϕ0 ∈ H and r0 ∈ L2 (Ω), the
solutions (ϕ, π) to (14) and r to (15) satisfy, respectively,
ZZ
 ZZ
−2sα
2
2 2

e
|∇π| dx dt + s λ
e−2sα ξ 2 |π|2 dx dt + I(s, λ; ϕ)



Q
Q



ZZ
ZZ


≤ C0 s
e−2sα ξ |f1 |2 dx dt + s3 λ4
e−2sα ξ 3 |ϕ|2 dx dt
(16)
Q
ω
×(0,T
)

1

!


ZZ



2 2

+
s
λ
e−2sα ξ 2 |π|2 dx dt

ω1 ×(0,T )
and
I(s, λ; r) ≤ C0
ZZ
e
−2sα
2
3 4
|f2 | dx dt + s λ
Q
ZZ
e
−2sα 3
!
2
ξ |r| dx dt , (17)
ω1 ×(0,T )
for every s ≥ s0 (T 4 + T 8 ) and λ ≥ λ0 .
For the sake of completeness, the proof of this proposition is given in an appendix,
at the end of this paper.
Remark 3. We can notice that the same weight e−2sα is present in both sides of
(16) and (17). We also point out that the right hand side of (16) contains a local
term on the pressure π.
As an immediate consequence of (16), (17) and the previous remark, we obtain
the following Carleman estimate for the linear (backward) Boussinesq system (9):
Proposition 2. Let us assume that a, b, c ∈ L∞ (Q)N and d ∈ L∞ (Q). Then, there
exist three positive constants C1 , s1 and λ1 depending on Ω and ω1 such that, for
every ϕ0 ∈ H and every r0 ∈ L2 (Ω), we have
 ZZ
ZZ

−2sα
2
2 2

e
|∇π|
dx
dt
+
s
λ
e−2sα ξ 2 |π|2 dx dt + I(s, λ; ϕ) + I(s, λ; r)



Q
Q

Z Z

ZZ


−2sα
2
2
3 4
≤ C1
e
(sξ|θ| +|σ| )dxdt +s λ
e−2sα ξ 3 (|ϕ|2 +|r|2 )dxdt
Q
ω1 ×(0,T )


!

ZZ



2
2
−2sα
2
2

e
ξ |π| dxdt ,

 +s λ
ω1 ×(0,T )
(18)
for s ≥ s1 (T 4 + T 8 ) = s2 and λ ≥ λ1 (1 + kak∞ + kbk∞ + kck∞ + kdk∞ ) = λ2 .
−2sα 3
From now on we fix s ≥ s2 and λ ≥ λ2 . On the set Ω×( T4 , 3T
ξ
4 ) the weight e
is bounded from below by a positive constant. It is then easy to obtain, using
standard energy estimates, the following observability inequality for system (12).
Proposition 3. Let us assume that s ≥ s2 and λ ≥ λ2 . Then there exists a positive
constant C2 (depending on Ω, ω, T , s, λ, ||a||∞ , ||b||∞ , ||c||∞ and ||d||∞ ) such that
for every solution (ϕ, π, r) of problem (12) we have
||ϕ(0)||2V + ||r(0)||2L2
ZZ
ZZ
≤ C2
e−2sα ξ 3 (|ϕ|2 + |r|2 ) dx dt +
ω1 ×(0,T )
ω1 ×(0,T )
e
−2sα 2
2
!
ξ |π| dx dt (. 19)
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
319
Now, let us fix ϕ0 ∈ H and r0 ∈ L2 (Ω) and let us consider the corresponding solution (ϕ, π, r) to system (12). From Proposition 3 we readily obtain the
observability inequality
!
ZZ
ZZ
2
2
2
2
2
||ϕ(0)||V + ||r(0)||L2 ≤ C3
|ϕ| + |r| dx dt +
|π| dx dt
ω1 ×(0,T )
ω1 ×(0,T )
where C3 is a new positive constant only depending on Ω, ω, T , ||a||∞ , ||b||∞ ,
||c||∞ and ||d||∞ . The proof of Theorem 2.4 is a standard consequence of this
observability inequality for system (12). In fact, this proof is included in the proof
of Proposition 4, where we will also prove that the controls have further integrability
properties.
3. Regularity of controls and removing the fictitious control. This section
is devoted to prove the null controllability of system (5). To this end, we will
solve again the null controllability problem for system (8) at time T with control
functions v1 , v2 and v3 which are more regular. This regularity property will allow
us to remove the fictitious control v3 and prove the null controllability of system (5).
On the other hand, this new null controllability result for system (8) is a consequence
of a refined Carleman inequality for system (12) (see Lemma 3.3). This Carleman
inequality will be deduced combining inequality (18) and the parabolic regularizing
effect of system (12).
The second result of this paper provides the null controllability of system (5). In
order to correctly state this result, for q ∈ [1, ∞), let us introduce qb given by
 −1

1
2
N


−
if q < ,
q
N
2
qb =
(20)

N

 qb ∈ [2, ∞)
if q ≥ .
2
The controllability result for system (5) is given in the following theorem:
Theorem 3.1. Let a, b, c and d satisfy (6) and (7) with p ≥ 2 and q ≥ max{1,
2N/(N + 4)}. Assume z 0 ∈ H and ρ0 ∈ L2 (Ω). Then, we can find controls
v1 ∈ L2 (Q)N and v2 ∈ L2 (Q) with Supp vi ⊂ ω × [0, T ] (1 ≤ i ≤ 2) such that the
solution of (5) satisfies (4). Moreover, v1 ∈ Lp (0, T ; Lqb(Ω)N ), v2 ∈ Ls (Q) for every
s ∈ [2, ∞) and one has
kv1 kLp (Lqb ) + kv2 kLs ≤ C(kz 0 kH + ||ρ0 ||L2 ),
(21)
for a positive constant C which depends on Ω, ω, T , s, ||a||∞ , ||b||∞ , ||c||∞ , ||d||∞ ,
||at ||Lp (Lq ) , ||bt ||Lp (Lq ) , ||ct ||Lp (Lq ) and ||dt ||Lp (Lq ) .
3.1. An improved Carleman inequality. In order to prove Theorem 3.1, we
will systematically use the following result, which is a consequence of the regularizing effect of the Stokes and heat operators with homogeneous Dirichlet boundary
conditions:
Lemma 3.2. Assume that
a, b, c ∈ L∞ (Q)N ,
d ∈ L∞ (Q),
∇ · a = ∇ · b = ∇ · c = 0 in Q.
320
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
Then, if f ∈ Ll (Q)N and h ∈ Ll (Q) with l ∈ [2, ∞), the solution

−ϕt − µ∆ϕ − (a · ∇) ϕ − ∇ϕ b − d∇r + ∇π = f





 −rt − ∆r − c · ∇r + g · ϕ = h
∇·ϕ =0


 ϕ = 0, r = 0



ϕ(T ) = 0, r(T ) = 0
(ϕ, π, r) to
in Q,
in Q,
in Q,
on Σ,
in Ω,
(22)
satisfies (ϕ, π, r) ∈ (X l )N × Ll (0, T ; W 1,l (Ω)) × X l and
||(ϕ, r)||X l + ||π||Ll (W 1,l ) ≤ M1 (||f ||Ll + ||h||Ll )
with M1 = M1 (Ω, T, ||a||∞ , ||b||∞ , ||c||∞ , ||d||∞ ) a positive function increasing with
respect its last four arguments.
The proof of this lemma can be easily deduced if we combine a boot-strap argument and the results on parabolic regularity stated in [9].
We are now ready to show an improved Carleman inequality for the solutions to
system (12):
Lemma 3.3. Let (ϕ, π) and r be the solution to (12) associated to ϕ0 ∈ H, r0 ∈
L2 (Ω). Let ℓ ∈ (N + 2, ∞) be given. Assume that a, b, c and d satisfy (6). Then, if
s ≥ s2 ≡ s1 (T 4 + T 8 ) and λ ≥ λ2 ≡ λ1 (1 + kak∞ + kbk∞ + kck∞ + kdk∞ ), one has

N
∗

ϕ
e = (sξ ∗ )−(M+7)/4 e−sα ϕ ∈ X ℓ ∩ L∞ (0, T ; W 1,∞(Q)N ∩ V ),




∗ −(M+7)/4 −sα∗

e
r ∈ X ℓ ∩ L∞ (0, T ; W 1,∞ (Q)),
 re = (sξ )
∗
π
e = (sξ ∗ )−(M+7)/4 e−sα π ∈ Lℓ (0, T ; W 1,ℓ(Ω)),




ϕ
et ∈ Lp (0, T ; W 2,q (Ω)N ∩ V ), ret ∈ Lp (0, T ; W 2,q (Ω)),



π
et ∈ Lp (0, T ; W 1,q (Ω)),
where M ∈ N, α∗ and ξ ∗ are given by
(
)


(N + 2)(ℓe − 2)

 M = min m ∈ N : m ≥
, ℓe = max{p, q, ℓ},
4ℓe


 α∗ (t) = max α(x, t) and ξ ∗ (t) = min ξ(x, t),
x∈Ω
x∈Ω
Furthermore, one has

π ||2Lℓ (W 1,ℓ ) + ||e
πt ||2Lp (W 1,q )
e re)||2(X ℓ )∩L∞ (W 1,∞ ) + ||(ϕ
et , ret )||2Lp (W 2,q ) + ||e

 ||(ϕ,
!
ZZ
ZZ
−2sα 3
2
2
2 2
−2sα 2
2
3
4

e
ξ |ϕ| + |r| + s λ
e
ξ |π| ,
 ≤C s λ
ω1 ×(0,T )
ω1 ×(0,T )
(23)
for a positive constant C which depends on Ω, ω, T , ||a||∞ , ||b||∞ , ||c||∞ , ||d||∞ ,
||at ||Lp (Lq ) , ||bt ||Lp (Lq ) , ||ct ||Lp (Lq ) and ||dt ||Lp (Lq ) .
Proof. All along this proof, we will use the following property: given
∗
γ
e(t) = sn e−sα
(t)
l
(ξ ∗ (t))
n, l ∈ Z and γ
b(t) = (sξ ∗ (t))−1/4 ,
then for every s ≥ s1 (T 4 + T 8 ), one has
(
∗
l+5/4
|e
γ ′ (t)| ≤ C(Ω, ω)sn+5/4 e−sα (t) (ξ ∗ (t))
,
|b
γ ′ (t)| ≤ C(Ω, ω),
(24)
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
321
for all t ∈ (0, T ). On the other hand, without loss of generality we will assume that
ℓ ≥ p and ℓ ≥ q and thus, ℓe = ℓ.
Let us define

∗
ϕ
e = (sξ ∗ )−(6+i)/4 e−sα ϕ = (sξ ∗ )−1/4 ϕ
ei−1 ,

 i
∗ −(6+i)/4 −sα∗
∗ −1/4
rei = (sξ )
e
r = (sξ )
rei−1 ,


∗ −(6+i)/4 −sα∗
∗ −1/4
π
ei = (sξ )
e
π = (sξ )
π
ei−1 ,
∗
for 1 ≤ i ≤ M + 1, where we have denoted (ϕ
e0 , re0 , π
e0 ) = (sξ ∗ )−3/2 e−sα (ϕ, r, π).
On the other hand, let us introduce ℓi such that
1
1
2i
= −
,
ℓi
2 N +2
if 1 ≤ i ≤ M − 1, and
 −1

1
2M


−
2 N +2
=


 ℓ
N +2
,
4
ℓM
N +2
if M ≥
.
4
Observe that, thanks to the choice of M , we have ℓi ∈ (2, ℓ), for 1 ≤ i ≤ M − 1,
and ℓM ≥ ℓ.
In order to simplify the notation, from now on, C will stand for a generic positive constant which depends on Ω, ω, T , ||a||∞ , ||b||∞ , ||c||∞ , ||d||∞ , ||at ||Lp (Lq ) ,
||bt ||Lp (Lq ) , ||ct ||Lp (Lq ) and ||dt ||Lp (Lq ) .
if M <
Step 1. We will first prove that for every i, 1 ≤ i ≤ M , one has
N
(ϕ
ei , π
ei , rei ) ∈ X ℓi
× Lℓi (0, T ; W 1,ℓi (Ω)) × X ℓi ,
and

||(ϕ
ei , rei )||2X ℓi + ||e
πi ||2Lℓi (W 1,ℓi )


ZZ
ZZ
2 2
2
2
3 4
−2sα 3

e
ξ |ϕ| + |r| + s λ
 ≤C s λ
ω1 ×(0,T )
ω1 ×(0,T )
e−2sα ξ 2 |π|2
!
.
(25)
It is easy to check that (ϕ
e1 , π
e1 , re1 ) is the solution to (22) with f = f1 = ρ1 (t)ϕ
and h = h1 = ρ1 (t)r, where the function ρ1 is given by
∗
−7/4
ρ1 (t) = −∂t s−7/4 e−sα (t) (ξ ∗ (t))
.
Taking into account the global Carleman inequality (18) and inequality (24), we
get ∆f1 ∈ L2 (Q)N , f1,t ∈ L2 (Q)N , ∆h1 ∈ L2 (Q), h1,t ∈ L2 (Q) and

2
2
2
||∆f1
||2L2 + Z||f

Z1,t ||L2 + ||∆h1 ||L2 + ||h1,t ||L2
ZZ


−1

−2sα∗ ∗ 3/2
2
2
−2sα∗ ∗ −1
2
2

 ≤ C s3/2
e
(ξ )
|ϕ| +|r| +s
e
(ξ )
|ϕt | +|∆ϕ|
Q
Q
!
ZZ
ZZ



−2sα
3
2
2
2
2
−2sα
2
2

3
4

e
ξ |ϕ| + |r| + s λ
e
ξ |π| .
 ≤C s λ
ω1 ×(0,T )
ω1 ×(0,T )
(26)
Then, f1 ∈ (X 2 )N and h1 ∈ X 2 . Owing to the continuous embedding X 2 ֒→
Lℓ1 (Q), one has that f1 ∈ Lℓ1 (Q)N and h1 ∈ Lℓ1 (Q). Therefore, we can apply
322
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
e1 , π
e1 , re1 ) (the solution to (22) corresponding to (f1 , h1 )) and deduce
Lemma 3.2 to (ϕ
N
(ϕ
e1 , π
e1 , re1 ) ∈ X ℓ1
× Lℓ1 (0, T ; W 1,ℓ1 (Ω)) × X ℓ1 .
From (26), inequality (25) holds for i = 1.
Now, let us assume that
N
(ϕ
ek−1 , π
ek−1 , rek−1 ) ∈ X ℓk−1
× Lℓk−1 (0, T ; W 1,ℓk−1 (Ω)) × X ℓk−1
and inequality (25) holds for i = k − 1, with 2 ≤ k ≤ M . We readily obtain that
(ϕ
ek , π
ek , rek ) is the solution to (22) associated to the functions f = fk ≡ γ0 (t)ϕ
ek−1
and h = hk ≡ γ0 (t)e
rk−1 , with γ0 given by
γ0 (t) = −∂t ((sξ ∗ (t))−1/4 ).
Observe that |γ0 (t)| ≤ C, for every t ∈ (0, T ). Thus, the continuous embedding
N
X ℓk−1 ֒→ Lℓk (Q) and (25) written for i = k − 1 give fk ∈ Lℓk (Q) , hk ∈ Lℓk (Q)
and

ZZ


2
2
3 4

||f
||
+
||h
||
≤
C
s
λ
e−2sα ξ 3 |ϕ|2 + |r|2
k L ℓk
k L ℓk


ω1 ×(0,T )
!
ZZ


2 2
−2sα 2
2


+s λ
e
ξ |π| .

ω1 ×(0,T )
As a consequence of this last inequality and Lemma 3.2, we can deduce
N
(ϕ
ek , π
ek , rek ) ∈ X ℓk
× Lℓk (0, T ; W 1,ℓk (Ω)) × X ℓk
and inequality (25) for i = k.
Let us point out that X ℓM ֒→ L∞ (0, T ; W 1,∞ (Q)) since ℓM > N + 2. In particular, ϕ
eM ∈ L∞ (0, T ; W 1,∞ (Q))N , reM ∈ L∞ (0, T ; W 1,∞ (Q)) and

ZZ


3 4
−2sα 3
2
2
2

||(
ϕ
e
,
r
e
)||
≤
C
s
λ
e
ξ
|ϕ|
+
|ψ|
∞ (W 1,∞ )

M
M
L

ω1 ×(0,T )
!
(27)
ZZ


2
2
−2sα
2
2


+s λ
e
ξ |π|

ω1 ×(0,T )
is satisfied. This completes step 1.
Step 2. To conclude the proof of Lemma 3.3, we will prove that ϕ
e= ϕ
eM+1
satisfies the three last statements.
First, we observe that (ϕ
eM+1 π
eM+1 , reM+1 ) fulfill system (22) with f = fM+1 =
−∂t ((sξ ∗ )−1/4 )ϕ
eM and h = hM+1 = −∂t ((sξ ∗ )−1/4 )e
rM . Let us consider (u, π0 , η)
the solution of system (22) for f = feM+1 and h = e
hM+1 where
(
feM+1 = ∂t fM+1 + (at · ∇)ϕ
eM + ∇ϕ
eM and
e
hM+1 = ∂t e
hM+1 + ct · ∇rM .
Our aim is to prove that (u, π0 , η) = ∂t (ϕ
eM+1 π
eM+1 , reM+1 ). Indeed, thanks to the
assumptions (6) and inequality (27), we have that
(at · ∇)ϕ
eM , ∇ϕ
eM bt , dt ∇e
rM ∈ Lp (0, T ; Lq (Ω)N ), ct · ∇e
rM ∈ Lp (0, T ; Lq (Ω))
and their norms are estimated by the right hand side of (27). Furthermore, since
(
2
∂t fM+1 = −∂tt
((sξ ∗ )−1/4 )ϕ
eM − ∂t ((sξ ∗ )−1/4 )∂t ϕ
eM and
2
∗ −1/4
∗ −1/4
∂t hM+1 = −∂tt ((sξ )
)e
rM − ∂t ((sξ )
)∂t reM ,
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
323
are bounded in Lℓ (Q), they are also in Lp (0, T ; Lq (Ω)).
Consequently, (u, π0 , η) ∈ (X p,q )N × Lp (0, T ; W 1,q (Ω)) × X p,q and its norm is
also bounded by the right hand side of (27).
Finally, by a density argument (on the coefficients a, b, c and d) one can identify
(u, π0 , η) = ∂t (ϕ
eM+1 , π
eM+1 , reM+1 ). The proof of Lemma 3.3 is complete.
3.2. Null controllability of system (5): Proof of Theorem 3.1. In this Section, we will prove the null controllability of system (5). As said above, we will first
improve Theorem 2.4, obtaining the null controllability for system (8) with regular
control functions.
We recall here the system under study:


 zt − µ∆z + ∇ · (a ⊗ z + z ⊗ b) + ∇q = −ρg + v1 χω in Q,



in Q,
 ρt − ∆ρ + ∇ · (cρ) + ∇ · (zd) = v2 χω
∇ · z = v3 ζ
in Q,
(28)


 z = 0, ρ = 0
on Σ,



z(0) = z 0 , ρ(0) = ρ0
in Ω,
with z 0 ∈ V ′ , ρ0 ∈ L2 (Ω).
The null controllability of system (28) with regular controls is given in the following proposition:
Proposition 4. Assume that a, b, c and d satisfy (6), z 0 ∈ V ′ and ρ0 ∈ L2 (Ω).
Then, we can find controls (v1 , v2 , v3 ) satisfying

(v1 , v2 ) ∈ L∞ (Q)N +1



 v ∈ L2 (0, T ; L2 (ω)) ∩ Ll (0, T ; W 1,l (Ω)) ∀l ∈ (1, ∞),
3
ζ
(29)

v3,t ∈ Lp (0, T ; W 1,q (Ω)),



v3 (x, 0) = v3 (x, T ) = 0 a.e. in Ω,
and
k(v1 , v2 )k∞ + kv3 kLl (W 1,l ) + kv3,t kLp (W 1,q ) ≤ C kz 0 kV ′ + ||ρ0 ||L2
(30)
(C is a positive constant which depends on Ω, ω, T , ||a||∞ , ||b||∞ , ||c||∞ , ||d||∞ ,
||at ||Lp (Lq ) , ||bt ||Lp (Lq ) , ||ct ||Lp (Lq ) and ||dt ||Lp (Lq ) ) such that the corresponding solution (z, ρ, zT , ρT ) to (28) satisfies zT = 0 and ρT = 0.
Proof. From now on, let us set s = s1 (T 4 + T 8 ) and λ = λ1 (1 + kak∞ + kbk∞ +
kck∞ + kdk∞ ), with s1 and λ1 provided by Proposition 2.
Let us take ǫ > 0 and consider the optimal control problem:

ZZ

1


min
β1−1 (|v1 |2 + |v2 |2 ) + ζβ2−1 |v3 |2 dx dt

 (v )3 ∈U
2
i i=1
ad
ω×(0,T )
(31)


1
2
2


+
kzT kV ′ + ||ρT ||L2 ,

2ǫ
where z and ρ, together with q, is the solution to (28) associated to v1 , v2 and v3
and Uad , the set of admissible controls, is given by
Uad = {(v1 , v2 , v3 ) : (v1Z,Zv2 ) ∈ L2 (0, T ; L2(ω)N ) × L2 (0, T ; L2(ω)),
v3 ∈ L2 (0, T ; L2ζ (ω)),
(β1−1 |v1 |2 + |v2 |2 ) + ζβ2−1 |v3 |2 dx dt < +∞}.
ω×(0,T )
324
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
Of course, Uad is a nonempty set. Recall that the weight functions are given by
β1 = e−2sα ξ 3 , β2 = e−2sα ξ 2 .
As the mapping (v1 , v2 , v3 ) ∈ Uad → (zT , ρT ) ∈ V ′ × L2 (Ω) is linear continuous,
the following result is straightforward (cf. [11]).
Proposition 5. There exists a unique optimal control (v1ǫ , v2ǫ , v3ǫ ) ∈ Uad for problem (31). Moreover, if we denote by S the canonical isomorphism between V and
V ′ (defined by the Stokes system), (v1ǫ , v2ǫ , v3ǫ ) is characterized by the following optimality system:
 ǫ
ǫ
ǫ
ǫ
ǫ
ǫ
ǫ

 zt − µ∆z + ∇ · (a ⊗ z + z ⊗ b) + ∇q = −ρ g + v1 χω in Q,

ǫ
ǫ
ǫ
ǫ
ǫ


in Q,
 ρt − ∆ρ + ∇ · (cρ ) + ∇ · (z d) = v2 χω
ǫ
ǫ
∇ · z = v3 ζ
in Q,
(32)


 z ǫ = 0, ρǫ = 0
on
Σ,


 ǫ
z (0) = z 0 , ρǫ (0) = ρ0
in Ω,

−ϕǫt − µ∆ϕǫ − (a · ∇) ϕǫ − ∇ϕǫ b − d∇rǫ + ∇π ǫ = 0 in Q,




−rtǫ − ∆rǫ − c · ∇rǫ + g · ϕǫ = 0
in Q,



∇ · ϕǫ = 0
in Q,
ǫ
ǫ


ϕ = 0, r = 0
on Σ,




1
1
 ϕǫ (T ) = − S −1 z ǫ , rǫ (T ) = − ρǫ
in Ω,
T
ǫ
ǫ T
 ǫ
v1 = β1 ϕǫ χω ,



 v ǫ = β rǫ χ ,
1
ω
2
R
(33)
 ǫ
β2 π ǫ ζ dx

ǫ
ǫ
ǫ
ω

R
.
 v3 = β2 (−π + c ) with c =
ω β2 ζ dx
From the correct definition of (z ǫ , ρǫ , zTǫ , ρǫT ) by transposition, using as test function (ϕǫ , rǫ ) we obtain
hzTǫ , ϕǫ (T )iV ′ ,V + (ρǫT , rǫ (T ))
=
hz 0 , ϕǫ (0)iV ′ ,V + (ρ0 , rǫ (0)) +
ZZ
ω×(0,T )
and thanks to (33),
 ZZ



ω×(0,T )
β1−1 |v1ǫ |2 + |v2ǫ |2 dx dt +
(v1ǫ · ϕǫ + v2ǫ rǫ − v3ǫ ζπ ǫ ),
ZZ
ζβ2 |π ǫ |2 dx dt
ω×(0,T )

1

 +
kzTǫ k2V ′ + ||ρǫT ||2L2 = −hz 0 , ϕǫ (0)iV ′ ,V −
ǫ
On the other hand, one has
ZZ
ZZ
β1−1 (|v1ǫ |2 +|v2ǫ |2 )+ζβ2−1 |v3ǫ |2 =
ω×(0,T )
ω×(0,T )
Z
(34)
ρ0 rǫ (0) dx.
Ω
β1 (|ϕǫ |2 +|rǫ |2 )+β2 ζ|π̂ ǫ |2 ,
where π̂ ǫ is the realization of π ǫ (which is defined up to a constant) such that
Z
β2 π̂ ǫ ζ dx = 0,
Ω
i.e., π̂ ǫ = π ǫ − cǫ with cǫ given in (33).
(35)
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
325
Coming back to (34), we obtain

1
−1/2 ǫ 2
−1/2 ǫ 2
−1/2 ǫ 2

v1 kL2 + kβ1
v2 kL2 + kζ 1/2 β2
v3 kL2 + kzTǫ k2V ′
 kβ1
ε
1
δ
1


+ ||ρǫT ||2L2 ≤
kϕǫ (0)k2V + ||rǫ (0)||2L2 +
kz 0 k2V ′ + ||ρ0 ||2L2 ,
ǫ
2
2δ
for every ǫ > 0 and δ > 0. Using now inequality (19) for (ϕǫ , π̂ ǫ , rǫ ) (with θ ≡ σ ≡ 0)
with δ sufficiently small and taking into account equality (35), we deduce

 kβ1−1/2 v1ǫ k2L2 + kβ1−1/2 v2ǫ k2L2 + kζ 1/2 β2−1/2 v3ǫ k2L2
(36)
1
1

+ kzTǫ k2V ′ + kρǫT k2L2 ≤ C kz 0 k2V + ||ρ0 ||2L2 ,
ǫ
ε
for every ǫ > 0.
Let us now see that (v1ǫ , v2ǫ ) lies in L∞ (Q)N +1 together with the estimate
||v1ǫ ||∞ + ||v2ǫ ||∞ ≤ C kz 0 k2V ′ + ||ρ0 ||2L2 .
(37)
γ1 ϕ
eǫ χω2 , γ
e2 reǫ χω2 , γ
e3 ζe
π ǫ χω2 ) with
To this end, we rewrite (v1ǫ , v2ǫ , v3ǫ ) = (e
∗
and
ϕ
eǫ = (sξ ∗ )−(7+M)/4 e−sα ϕǫ ,
∗
∗
reǫ = (sξ ∗ )−(7+M)/4 e−sα rǫ ,
∗
π
eǫ = (sξ ∗ )−(7+M)/4 e−sα π̂ ǫ ,
∗
γ
e1 = γ
e2 = e−2sα+sα ξ 3 (sξ ∗ )(7+M)/4 , γ
e3 = e−2sα+sα ξ 2 (sξ ∗ )(7+M)/4 .
Observe that the function γ
ei (i = 1, 2, 3) are uniformly bounded in Q. Thanks to
Lemma 3.3, equality (34) and inequalities (23) and (36), we deduce that (v1ǫ , v2ǫ ) ∈
L∞ (Q)N +1 and that (37) holds.
On the other hand, v3ǫ ∈ L2 (0, T ; L2ζ (ω)) and also
∇v3ǫ = ∇e
γ3 ζ π
eǫ + γ
e3 ∇ζ π
eǫ + γ
e3 ζ ∇e
πǫ ,
ǫ
v3,t
= ζ (e
γ3,t π
eǫ + e
γ3 π
etǫ ) .
The definition of the weight function γ
e3 , together with Lemma 3.3 and (36) gives
v3ǫ ∈ Ll (0, T ; W 1,l (Ω)),
ǫ
v3,t
∈ Lp (0, T ; W 1,q (Ω)),
for every l ∈ (1, ∞), and
ε
||v3ǫ ||Ll (W 1,l ) + ||v3,t
||Lp (W 1,q ) ≤ C kz 0 k2V ′ + ||ρ0 ||2L2 .
We are now ready to pass to the limit with respect to ǫ and finish the proof. To
this end, we denote by L2 (0, T ; β1−1 ; L2 (ω)) the space of square integrable functions with respect to the weight β1−1 on ω × (0, T ) and in an analogous way
L2 (0, T ; β2−1 ; L2ζ (ω)).
We have obtained that

ǫ
ǫ
∞
N +1
∩ L2 (0, T ; β1−1 ; L2 (ω)N +1 ),

 {(v1 , v2 )}ǫ>0 ∈ a bounded set of L (Q)
ǫ
ℓ
1,ℓ
{v3 }ǫ>0 ∈ a bounded set of L (0, T ; W (Ω)) ∩ L2 (0, T ; β2−1 ; L2ζ (ω)),


ǫ
{v3,t
}ǫ>0 ∈ a bounded set of Lp (0, T ; W 1,q (Ω)).
We can then extract subsequences still denoted v1ǫ , v2ǫ , v3ǫ such that
 ǫ ǫ
(v1 , v2 ) ⇀ (v1 , v2 ) in L2 (0, T ; β1−1 ; L2 (ω)N +1 ) weakly,



 (v ǫ , v ǫ ) ⇀ (v , v ) in L∞ (Q)N +1 weakly-∗,
1 2
1 2
ǫ
 v3 ⇀ v3 in Lℓ (0, T ; W 1,ℓ (Ω)) ∩ L2 (0, T ; β2−1 ; L2ζ (ω)) weakly,


 ǫ
v3,t ⇀ v3,t in Lp (0, T ; W 1,q (Ω)) weakly,
326
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
with ℓ ∈ (1, ∞) and (v1 , v2 ) ∈ L∞ (Q)N +1 ∩ L2 (0, T ; β1−1 ; L2 (ω)N +1 ) and v3 ∈
Lℓ (0, T ; W 1,ℓ(Ω)) ∩ L2 (0, T ; β2−1 ; L2ζ (ω)) with v3,t ∈ Lp (0, T ; W 1,q (Ω)).
Because of linearity and continuity of the mapping (v1 , v2 , v3 ) → (z, ρ, zT , ρT )
where (z, ρ, zT , ρT ) is a weak solution (by transposition) of (8), we see from (32)
that
zTǫ ⇀ zT in V ′ weakly, ρǫT ⇀ ρT in L2 (Ω) weakly,
where (z, ρ, zT , ρT ) is solution of (8) associated to (v1 , v2 , v3 ). From (36), we have
zT = 0 and ρT = 0 so that (v1 , v2 , v3 ) realizes the null controllability property for
problem (8) and, evidently, (29) and (30) hold. This finishes the proof of Proposition 4.
Remark 4. We have in fact, as announced in Theorem 2.4, obtained controls with
the additional integrability property :
ZZ
ZZ
J0 (v1 , v2 , v3 ) =
β1−1 (|v1 |2 + |v2 |2 ) dx dt +
ζβ2−1 |v3 |2 dx dt < +∞.
ω×(0,T )
ω×(0,T )
In fact it can be easily shown that the controls we have obtained as limits of
(v1ǫ , v2ǫ , v3ǫ ) realize the minimum of J0 among controls such that null controllability is achieved. The convergence of (v1ǫ , v2ǫ , v3ǫ ) towards (v1 , v2 , v3 ) is then strong.
Proof of Theorem 3.1. As advanced above, we will obtain the proof of this result
eliminating the fictitious control acting on the divergence equation in (28) and, to
this end, we will use Proposition 4 and the following result:
Theorem 3.4. Let O ⊂ RN be a bounded Lipschitz domain and r ∈ (1, ∞) and let
us denote
Z
L10 (O) = {u ∈ L1 (O) :
u dx = 0}.
O
Then there exists R ∈ L(W01,r (O) ∩ L10 (O); W02,r (O)N ) such that ∇ · (Rw) = w in
O, for all w ∈ W01,r (O) ∩ L10 (O). Moreover, if s ∈ (r, ∞), then R|W 1,s (O)∩L1 (O) ∈
0
0
L(W01,s (O) ∩ L10 (O); W02,s (O)N ).
For a proof of this result see for instance [2] (Theorem 2.4, p. 72).
Applying Proposition 4, we infer the existence of controls (b
v1 , b
v2 , vb3 ) such that (29)
and (30) hold and the corresponding solution (b
z , ρb, zbT , ρbT ) to problem (28) satisfies
zbT = 0 and ρbT = 0. Let us consider O ⊂⊂ ω a Lipschitz set such that Supp ζ ⊂ O.
We have that v3 (t)ζ ∈ W01,q (O) ∩ L10 (O) a.e. in (0, T ). Therefore we can apply the
previous theorem for r = q ∈ (1, ∞) and define
Z(t) = R(v3 (t)ζ),
a.e. t ∈ (0, T ).
Let us fix ℓ ∈ [q, ∞). From (29), (30) and the properties of the operator R, it is
not difficult to see that Z ∈ Lℓ (0, T ; W02,ℓ(O)N ), Z is derivable with respect to time
and Zt (t) = R(v3,t ζ). Thus, Zt ∈ Lp (0, T ; W02,q (O)N ), Z ∈ C 0 ([0, T ]; W02,q (O)N ),
Z(0) = Z(T ) = 0 in O, and
||Z||Lℓ (0,T ;W 2,ℓ (O)N ) + ||Zt ||Lp (0,T ;W 2,q (O)N ) + ||Z||L∞ (0,T ;W 2,q (O)N )
0
0
0
≤ C kz 0 kH + ||ρ0 ||L2 ,
(38)
for a positive constant C which depends on Ω, ω, T , ||a||∞ , ||b||∞ , ||c||∞ , ||d||∞ ,
||at ||Lp (Lq ) , ||bt ||Lp (Lq ) , ||ct ||Lp (Lq ) and ||dt ||Lp (Lq ) .
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
327
Now, we extend by 0 the function Z to the whole domain Ω and we denote this
e We define
extension by Z.
(
et − µ∆Ze + ∇ · a ⊗ Z
e+Z
e⊗b ,
v1 = vb1 + Z
e
v2 = vb2 + ∇ · (dZ).
Choosing ℓ ≥ max{p, qb} we deduce that v1 ∈ Lp (0, T ; Lqb(Ω)N ). Also, choosing
ℓ = s we get v2 ∈ Ls (Q). From (38) we readily obtain (21).
Finally, let us check that (v1 , v2 ) fulfills the statement of Theorem 3.1. Firstly,
the hypotheses on p and q imply (v1 , v2 ) ∈ L2 (Q)N +1 and so problem (5) admits a
unique weak solution (z, q, ρ) with
(z, ρ) ∈ L2 (0, T ; V ) ∩ C 0 ([0, T ]; H) × L2 (0, T ; H01 (Ω)) ∩ C 0 ([0, T ]; L2 (Ω)) .
e q = qb and ρ = ρb.
Secondly, a simple calculation provides the identities z = zb − Z,
0
2
N +1
We then deduce that (b
z , ρb) ∈ C ([0, T ]; L (Ω)
) and (b
z , ρb) = (b
zT , ρbT ) ≡ (0, 0).
e
e ) = 0 in Ω, we see that z(0) = z 0 , ρ(0) = ρ0 , z(T ) = 0
Finally, since Z(0)
= Z(T
and ρ(T ) = 0 in Ω. This ends the proof of Theorem 3.1.
4. Local null controllability of system (3). In this Section we will establish a
result of local exact controllability for system (1) to the trajectories of (1). To this
end, we will consider (y, θ) a regular trajectory of (1), i.e., a regular solution to (2),
and we will look for controls (v1 , v2 ) such that the solution (y, θ) to (1) satisfies
(y(T ), θ(T )) = (y(T ), θ(T )) in Ω.
As said above, if we take the new variables z = y − y, q = p − p and ρ = θ − θ, the
controllability to the trajectories of system (1) amounts to a local null controllability
result for system (3) with
0
z 0 = y 0 − y0 and ρ0 = θ0 − θ .
The local null controllability of system (3) has already been studied and established by several authors (see for instance [8], [10], [6], ...) and most of them show
the nonlinear controllability result combining a global Carleman inequality for the
linearized adjoint system (12) and an appropriate version of the inverse mapping
theorem. In the present work we will obtain the null controllability of (3) at time
T as a consequence of Theorem 3.1 using a fixed point argument.
All along this section we will assume the following regularity hypotheses on the
0
initial data (y 0 , θ ) and the trajectory (y, θ) ∈ L2 (0, T ; V ) × L2 (0, T ; H01(Ω)), weak
solution to (2):
 0 0
(y , θ ) ∈ W 2−2/p0 ,p0 (Ω)N ∩ V × W 2−2/p0 ,p0 (Ω) ∩ H01 (Ω)






with p0 > N/2 + 1 and p0 ≥ max{p, q}
(39)
∞
(y,
θ)
∈
L
(0, T ; W 1,∞ (Ω)N +1 ),





 (y t , θt ) ∈ Lp (0, T ; Lq (Ω)N +1 ) with p ≥ 2, q ≥ 1, N + 1 < 2.
2q p
Remark 5. Observe that, since the couple (y, θ) is a solution of (2), the hypothesis
(y, θ) ∈ L∞ (Q)N +1 readily implies (39).
The main result of this section reads as follows:
328
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
Theorem 4.1. Under assumption (39), there exists a positive constant δ = δ(Ω, ω,
T, y, θ) such that if (y 0 , θ0 ) ∈ W 2−2/p0 ,p0 (Ω)N ∩ V × W 2−2/p0 ,p0 (Ω) ∩ H01 (Ω) ,
with p0 > N/2 + 1 and p0 ≥ max{p, q}, and
0
||(y 0 − y0 , θ0 − θ )||W 2−2/p0 ,p0 ≤ δ,
2
N +1
then, there exists a control (v1 , v2 ) ∈ L (Q)
y(T ) = θ(T ) = 0
(40)
such that
in Ω.
Proof. In order to obtain the proof of this result, we will use a fixed point argument.
To be precise, we will use the Kakutani Theorem (see for instance [16]):
Theorem 4.2. Assume that
1. Λ : K 7→ 2K is an upper semi-continuous multi-valued mapping,
2. K 6= ∅, K is a compact, convex set in a locally convex space X,
3. the set Λ(w) is a nonempty, closed and convex set for all w ∈ K.
Then, Λ has a fixed point w in K, i.e., there exists w ∈ K such that w ∈ Λ(w).
In order to apply this result, let us set X = L2 (0, T ; H) and
K = {w ∈ X : w ∈ L2 (0, T ; V ), wt ∈ Lp (0, T ; Lq (Ω)N ), w ∈ L∞ (Q)N ,
||w||L2 (V ) + ||wt ||Lp (Lq ) + ||w||∞ ≤ R},
where R is a fixed positive constant. For each w ∈ K, we consider the linear null
controllability problem for system (5) with
a ≡ c ≡ w + y,
b≡y
and d ≡ θ.
Thanks to the assumption on w, y and θ, we have q ≥ max{1, 2N/(N + 4)} and we
can apply Theorem 3.1 to system (5) (with qb given by (20) and s ≥ max{p, qb}) and
conclude the existence of (v1 , v2 ) ∈ Lp (0, T ; Lqb(Ω)N +1 ) ⊂ L2 (Q)N +1 such that the
weak solution (z, q, ρ) (q is unique up a constant) to

zt − µ∆z + (w + y) · ∇z + z · ∇y + ∇q = −ρg + v1 χω in Q,





in Q,

 ρt − ∆ρ + (w + y) · ∇ρ + z · ∇θ = v2 χω
∇·z =0
in Q,
(41)


 z = 0, ρ = 0
on Σ,



0

z(0) = y 0 − y0 , ρ(0) = θ0 − θ
in Ω,
satisfies (4) and
0
||v1 ||Lp (Lqb ) + ||v2 ||Lp (Lqb ) ≤ C0 ||y 0 − y 0 ||H + ||θ0 − θ ||L2 (Ω)
(42)
for a positive constant C0 = C0 (Ω, ω, T, y, θ, R) which is increasing with respect to
R.
Using again the results in [9] on maximal regularity for the heat and Stokes
operators with homogeneous Dirichlet boundary conditions, we get
(
N
N
z ∈ (X p0 ) + X p,bq , ρ ∈ X p0 + X p,bq ,
q ∈ Lp0 (0, T ; W 1,p0 (Ω)) + Lp (0, T ; W 1,bq (Ω)),
and
0
||(z, ρ)||X p0 +X p,qb ≤ C ||(y 0 − y 0 , θ0 − θ )||W 2−2/p0 ,p0 + ||v1 ||Lp (Lqb ) + ||v2 ||Lp (Lqb )
(43)
for a new positive constant C = C(Ω, y, θ, R).
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
329
Let us remark that when q ≥ N/2, qb can be arbitrarily chosen in [2, ∞). Thus,
we choose it in such a way that
N
1
+ < 1.
2b
q p
When q < N/2, from (20) and (39), one also has the previous inequality. Therefore,
in both cases, we obtain the continuous embedding X p,bq ֒→ L∞ (Q). On the other
hand, since p0 ≥ N/2 + 1, we also deduce X p0 ֒→ L∞ (Q). Also, as a consequence
of the choice of p0 , we get X p0 + X p,bq ⊂ X p,q . We infer that (z, ρ) ∈ L∞ (Q)N +1 ,
zt ∈ Lp (0, T ; Lq (Ω)N ) and, thanks to inequalities (42) and (43),
≤
||z||L2 (V ) + ||ρ||L2 (H01 )) + ||(zt , ρt )||Lp (Lq ) + ||(z, ρ)||∞
0
C1 ||(y 0 − y 0 , θ0 − θ )||W 2−2/p0 ,p0
(44)
for a new positive constant C1 = C1 (Ω, ω, T, y, θ, R) also increasing with respect to
R.
Let us introduce a set-valued mapping on the Hilbert space X = L2 (0, T ; H) and
let us check that it possesses a fixed point. To this end, for each w ∈ K, let us
define the set of admissible controls by
UR (w) = {(v1 , v2 ) ∈ L2 (Q)N +1 : (v1 , v2 ) ∈ Lp (0, T ; Lqb(Ω)N +1 ), (42) holds and
the corresponding solution (z, q, ρ) to (41) satisfies (4)}.
On the other hand, let us set
ΛR (w) = {y ∈ L2 (0, T ; V ) : (y, q, ρ) is solution to (41) associated to
(v1 , v2 ) ∈ UR (w) and satisfies (44)}.
Now we define the set-valued mapping ΛR as
ΛR : w ∈ K ⊂ X 7→ ΛR (w) ⊂ X,
and we will show that, if (40) holds for an appropriate δ > 0, this mapping possesses
a fixed point z ∈ K. Of course, the existence of this fixed point z will prove the
statement of Theorem 4.1.
Let us prove that K and ΛR fulfill the statement of Theorem 4.2. Firstly, we
readily check that K is a nonempty convex compact set of the Hilbert space X.
Secondly, the considerations above prove that ΛR (w) is a nonempty set. Also,
due to the linearity of system (41), we obtain that ΛR (w) is, for every w ∈ K, a
closed convex set of X. Moreover, if we choose δ = R/C1 (Ω, ω, T, y, θ, R) in (40),
from (44), we deduce that ΛR (w) ⊂ K for all w ∈ K.
Finally, let us prove that the mapping ΛR is upper semi-continuous in K. To
this end, let us consider two sequences {wn }n≥1 , {zn }n≥1 ⊂ K and w, z ∈ K such
that
zn ∈ ΛR (wn ) and wn → w
and zn → z
in X ≡ L2 (0, T ; H).
330
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
Our goal is to prove that z ∈ ΛR (w). Observe that zn together with ρn and qn is
solution to

zn,t − µ∆zn + (wn + y) · ∇zn + zn · ∇y + ∇qn = −ρn g + v1,n χω in Q,





in Q,

 ρn,t − ∆ρn + (wn + y) · ∇ρn + zn · ∇θ = v2,n χω
∇ · zn = 0
in Q,



zn = 0, ρn = 0
on Σ,



0

0
0
0
zn (0) = y − y , ρn (0) = θ − θ , zn (T ) = 0, ρn (T ) = 0
in Ω,
(45)
with (v1,n , v2,n ) ∈ UR (wn ). From the definitions of UR (wn ) and ΛR (wn ) we readily
obtain that for a subsequence (still denoted by n), one has

(v1,n , v2,n ) ⇀ (v1 , v2 ) weakly in Lp (0, T ; Lqb(Ω)N +1 ),



2
2
2


 wn → w in L (0, T ; H), zn → z in L (0, T ; H), ρn → ρ in L (Q),
2
2
1
zn ⇀ z weakly in L (0, T ; V ), ρn ⇀ ρ weakly in L (0, T ; H0 (Ω)),



(zn , ρn ) ⇀ (z, ρ) weakly-* in L∞ (Q)N +1 and



(zn,t , ρn,t ) ⇀ (zt , ρt ) weakly in Lp (0, T ; Lq (Ω)N +1 )
with (v1 , v2 ) ∈ Lp (0, T ; Lqb(Ω)N +1 ) and (z, ρ) ∈ L2 (0, T ; V ) ∩ L2 (0, T ; H01 (Ω)) with
(z, ρ) ∈ L∞ (Q)N +1 , (zt , ρt ) ∈ Lp (0, T ; Lq (Ω)N +1 ). Of course, (v1 , v2 ) and (z, ρ)
satisfy respectively inequalities (42) and (44). On the other hand, thanks to the
previous convergences we can pass to the limit in problem (45) and obtain that (z, ρ),
together with a pressure q, is the weak solution to (41) corresponding to w ∈ K and
(v1 , v2 ) and fulfills (4). Summarizing, we have obtained that (v1 , v2 ) ∈ UR (w) and
z ∈ ΛR (w), that is to say, the multi-valued mapping ΛR is upper semi-continuous
in K. We have then completed the proof of Theorem 4.1.
Appendix A. Appendix. Proof of Proposition 1. Let us first apply the Carleman inequality for the heat system with right hand side in L2 (Q) to each component
of ϕ. See [7] for the proof and for the explicit dependence with respect to s, λ and
T see for instance [4]. We get
Z Z
ZZ

−2sα
2


I(s,
λ;
ϕ)
≤
C
e
|∇π|
+
e−2sα |f1 |2 dx dt


Q
Q
!
ZZ
(46)

3
4
−2sα
3
2


+s λ
e
ξ |ϕ| dx dt ,

ω2 ×(0,T )
for s ≥ C(T 7 + T 8 ) and λ ≥ C. We remark here that, thanks to the definition of ξ
(see (13)), we have
ξ −1 ≤ 2−8 T 8 .
Now, we must localize the global term of the pressure in (46). To this end, we
employ similar arguments to those developed in [5] and [10].
Let us then look at the (weak) equation satisfied by the pressure, which can be
found taking the divergence operator in the first equation of (12):
∆π(t) = ∇ · g1 (t) in Ω, a.e. t ∈ (0, T ).
−1
(47)
Watching the right hand side of (47) like a H
term, we apply the main result
in [13], say, a Carleman estimate for the weak solutions to elliptic equations. This
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
331
gives the existence of two constants σ > 1 and λ > 1, such that
 Z
Z
Z

2ση
2
2 2
2ση 2
2

e |∇π(t)| dx + σ λ
e η |π(t)| dx ≤ C σ
e2ση η|f1 (t)|2 dx

Ω
ZΩ
ZΩ

2ση
2
2 2
2ση 2
2
1/2 2σ
2

e |∇π(t)| dx + σ λ
e η |π(t)| dx
 +σ e kπ(t)kH 1/2 (∂Ω) +
ω2
ω2
(48)
for any σ ≥ σ and any λ ≥ λ. Here, we have denoted for each λ > 0,
η(x) = eλη
0
(x)
x ∈ Ω.
The next step will be now to eliminate the local term in ∇π. For this end, we define
χ ∈ C 2 (Ω) such that
supp χ ⊂ ω1 ,
χ ≡ 1 in ω2 ,
0 ≤ χ ≤ 1.
Let us integrate by parts several times:
Z
Z
 Z
2ση
2
2ση
2


e
|∇π(t)|
dx
≤
e
χ|∇π(t)|
dx
=
−
e2ση χ∆π(t)π(t) dx

ω2 Z
ω1
ω1
Z

2ση
2ση

e (∇χ · ∇π(t))π(t) dx − 2σλ
e ηχ(∇η 0 · ∇π(t))π(t) dx
 −
ω1
(49)
ω1
From (47) and new integrations by parts, we get the following for the first term:
 Z
Z

2ση

e
χ∆π(t)π(t)
dx
=
e2ση χ∇ · f1 (t) π(t) dx



ω
ω

1 Z
1
Z

=−
e2ση ∇χ · f1 (t) π(t) dx − 2σλ
e2ση ηχ∇η 0 · f1 (t) π(t) dx

ω1
ω1
Z





e2ση χf1 (t) · ∇π(t) dx.
 −
ω1
Consequently, for σ, λ ≥ C, we have
 Z

 −
e2ση χ∆π(t)π(t) dx
ω1
R
R

2ση 2
2
2ση
2
 ≤ C σ 2 λ2
e
η
|π(t)|
dx
+
e
|f
(t)|
dx
+
1
ω1
ω1
1
2
R
ω1
e2ση χ|∇π(t)|2 dx
(50)
Analogous computations for the second and the third terms in (49), give
Z
Z
−
e2ση (∇χ · ∇π(t))π(t) dx − 2σλ
e2ση ηχ(∇η 0 · ∇π(t))π(t) dx
ω1
ω1
Z
2 2
2ση 2
2
≤ Cσ λ
e η |π(t)| dx
(51)
ω1
for σ, λ ≥ C.
Combining (50) and (51) with (49), we obtain
Z
Z
Z
e2ση |∇π(t)|2 dx ≤ C σ 2 λ2
e2ση η 2 |π(t)|2 dx +
ω2
ω1
ω1
e2ση |f1 (t)|2 dx
for σ, λ ≥ C(Ω, ω), which together with (48) yields
 Z
Z

2ση
2
2 2

e |∇π(t)| dx + σ λ
e2ση η 2 |π(t)|2 dx

Ω Z
Ω
Z

2ση
2
1/2 2σ
2
2 2

e
η|f1 (t)| dx+σ e kπ(t)kH 1/2 (∂Ω) +σ λ
 ≤C σ
Ω
ω1
e
η |π(t)| dx
2ση 2
2
(52)
332
MANUEL GONZÁLEZ-BURGOS, SERGIO GUERRERO AND JEAN-PIERRE PUEL
for σ, λ ≥ C and for almost every t ∈ (0, T ).
The next step will be to integrate in t this inequality. Thus, let us set
0
s
σ= 4
eλmkη k∞ ,
t (T − t)4
let us multiply the expression (52) by
(
0
e(5/4)λmkη k∞
exp −2s 4
t (T − t)4
)
and then integrate it between t = 0 and t = T . This provides
ZZ
ZZ
 ZZ
−2sα
2
2 2
−2sα 2
2


e
|∇π|
dxdt+s
λ
e
ξ
|π|
dxdt
≤
C
s
e−2sα ξ|f1 |2 dxdt


Q
Q
Q
!
Z T
ZZ
∗

1/2

1/2

e−2sα ξ 2 |π|2 dx dt
e−2sα ξ ∗ kπ(t)k2H 1/2 (∂Ω) dt + s2 λ2
 +s
ω1 ×(0,T )
0
0
for any λ ≥ C and any CT 8 e−λmkη k∞ . Plugging this inequality into (46), we find
ZZ
 ZZ
−2sα
2
2 2

e
|∇π|
dx
dt
+
s
λ
e−2sα ξ 2 |π|2 dx dt + I(s, λ; ϕ)



Q
Q



Z T
ZZ


∗
≤ C s1/2
e−2sα ξ ∗ 1/2 (t)kπ(t)k2H 1/2 (∂Ω) dt+s2 λ2
e−2sα ξ 2 |π|2 dxdt
0
ω1 ×(0,T )


!


ZZ
ZZ



3
4
−2sα
3
2
−2sα
2

e
ξ |ϕ| dx dt + s
e
ξ|f1 | dx dt
 +s λ
ω2 ×(0,T )
Q
(53)
for any s ≥ C(T 7 + T 8 ) and any λ ≥ C.
Now, we will eliminate the term of the trace of the pressure. The tool here will be
classical a priori estimates of strong solutions of the Stokes system. More precisely,
let us introduce for
∗
β = s1/4 e−sα ξ ∗ 1/4 ,
the functions
ϕ
b = βϕ,
π
b = βπ
and let us see which system they fulfill:

−ϕ
bt − µ∆ϕ
b + ∇b
π = β f 1 − βt ϕ



 ∇·ϕ
b=0

ϕ
b=0



ϕ(T
b )=0
in Q,
in Q,
on Σ,
in Ω.
A priori estimates for this system (see, for instance, [15]) give
kb
π kL2 (H 1 ) ≤ C kβ f1 kL2 (Q)N + kβt ϕkL2 (Q)N ,
for a positive constant C = C(Ω). From the definition of β, ξ and α (see (13)
above), we deduce
∗ 1
βt = s1/4 e−sα ( ξ ∗ −3/4 (ξ ∗ )t − s(α∗ )t ξ ∗ 1/4 )
4
≤ Cs1/4 T e−sα (ξ 1/2 + sξ 3/2 ) ≤ Cs5/4 T e−sα ξ 3/2
CONTROLLABILITY OF THE BOUSSINESQ SYSTEM
333
for s ≥ CT 8 . Therefore, using the continuity of the trace operator, we have
Z T
ZZ
ZZ
2
1/2
−2sα 1/2
2
5/2 2
−2sα 3
2
kb
π kH 1/2 (∂Ω) dt ≤ C s
e
ξ |f1 | dxdt+s T
e
ξ |ϕ| dxdt
0
Q
Q
8
for s ≥ CT .
Finally, connecting this estimate of the trace of the pressure with (53), we deduce
the desired inequality (16).
This ends the proof of proposition 1.
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Received April 2008; revised August 2008.
E-mail address: [email protected]
E-mail address: [email protected]
E-mail address: [email protected]