Applications of global Carleman inequalities
to controllability and inverse problems
Jean-Pierre Puel
1
May 2008
1
J.-P. Puel: ([email protected]) Laboratoire de Mathématiques de Versailes, Université de Versailles Saint-Quentin, 45 avenue des Etats Unis, 78035 Versailles
Chapter 1
Introduction
These notes contain three main parts corresponding to three applications of global
Carleman inequalities.
The first part will describe what we call exact controllability to trajectories. These
terms will be defined precisely later on but for linear systems, it corresponds to null
(or zero) controllability and roughly speaking it means that we look for an exact
controllability when the targets to be reached are points on trajectories of the same
operator. These notions are relevant for dissipative systems such as heat equations
or Navier Stokes equations or related systems. Indeed, for these equations, one
cannot expect to obtain exact controllability in the usual sense (see for example [?]
or [?]) as for wave-type equations. The notion of approximate controllability, which
has been extensively studied (see for example [?]) and which will be quite useful in
a first step here, is not really satisfactory for real purposes. Exact controllability to
trajectories corresponds to real objectives and, when applicable, is a powerful tool.
As already mentionned, the main mathematical tool to obtain exact controlability
to trajectories is the use of global Carleman inequalities. The method along these
lines has been given in the works of Fursikov-Imanuvilov (e.g. [?]) and Imanuvilov
([?], [?]) and we will use essential ideas coming from these works, even if we revisited
the question in order to clarify the presentation. In particular we will completely
prove the basic Carleman inequalities for diffusion-convection equations.
The second part will be devoted to a data assimilation problem. This question is of
great importance in meteorology (wheather prediction), climatology, study of ocean
movements for example. At the moment, the wheather prediction models make use
of what is called variational data assimilation (see [?] for an interesting presentation), based on optimal control methods. To give an idea of the importance of this
question one has to know that it requires more than a half of the computational time
effectively used for weather prediction. In the usual setting the problem is ill-posed
1
which of course generates instabilities and it requires a Tychonov regularization.
We will present here a new approach of the problem, for which we will prove (on
basic examples) well-posedness using the same type of Carleman inequalities as before (cf [?] and [?]). This method leads to new algorithms which can be effectively
implemented.
The third part deals with inverse problems for wave equations and Schrödinger
equations. The question is to retrieve an unknown (time independent) potential
in a wave (or Schrödinger) equation from boundary additional measurements on
a part of the boundary. For stationary equations, it is usual to assume that we
know the whole Dirichlet to Neumann map and the results concern uniqueness and
not stability (see [?] for example). Here we have a single, but time dependent,
measurement and we will show uniqueness and stability for the inverse problem.
For these questions we will present results obtained by [?] for the wave equation and
by [?] for Schrödinger equation.
These notes correspond to a series of lectures given for a summer school at the
Universidade Federal de Rio de Janeiro in February 2003, and an important part
was already contained in a D.E.A. course taught at the Université Pierre et Marie
Curie in Paris in 2001.
2
Chapter 2
Exact controllability to
trajectories
We will essentially treat here the case of a convection-diffusion equation for sake of
simplicity. But the problem can be explained (at least formally) for a very general
evolution system.
Suppose we have a (natural) Hilbert space H. By natural we mean that H is “usual”
and can be described easily like an L2 space for example. We assume that we have
an unbounded linear operator L on H and a nonlinear operator N such that the
following problem is well posed for every initial data Y 0 ∈ H and every right hand
side F in a suitable space
(2.0.1)
(2.0.2)
dY
+ LY + N (Y ) = F, t ∈ (0, T ),
dt
Y (0) = Y 0 .
We now take
F = F 0 + Bv
where F 0 is a fixed function and B is a linear operator from the space of “controls”v
to the space of right hand sides. As an important example, if the problem is set on
spatial domain which is an open subset Ω of IRN and ω is an open subset of Ω, we
may consider the case of Bv = v.χω where χω is the characteristic function of ω.
The solution of (??) for each control v is then denoted by Y (v).
We now consider another (uncontrolled) trajectory of the same operator, with the
same fixed right hand side F 0 (this could be somewhat relaxed) and a different
0
initial data Y , which can be viewed as an “ideal” trajectory
(2.0.3)
dY
+ LY + N (Y ) = F 0 , t ∈ (0, T ),
dt
3
(2.0.4)
0
Y (0) = Y .
Notice that an important example is the case when Y is a stationary solution of (??).
This stationary solution may be unstable (in a usual sense to be defined precisely).
The question of exact controllability to trajectories can be now formulated as follows:
Does there exist v (in the space of controls which has to be prescribed) such that
at time T we have
Y (v)(T ) = Y (T ) ?
In fact we may have two different questions : Can this be done for any initial value
0
Y 0 ? In this case we have a global problem. Or can this be done for Y 0 − Y small
enough ? In that case we have a local problem.
If we can solve this question, after time T , we can take the control to be v = 0 and,
because of the well-posedness, the controlled trajectory will fit to the ideal one.
Notice that in the case when Y is an unstable stationary solution of (??), this
corresponds to stabilizing (exactly !) an unstable state !
Usually problems to be considered are nonlinear but the only strategy we know for
studying these questions is to consider first a linear version of the problem, to study
exact controllability to trajectories for it, then to use a fixed point argument to
obtain the result for the original nonlinear problem. For the linear case (take for
example N = 0) it is evident to see that writing Z(v) = Y (v) − Y , then Z(v) is
solution of
(2.0.5)
(2.0.6)
dZ(v)
+ LZ(v) = Bv, t ∈ (0, T ),
dt
0
Z(v)(0) = Z 0 = Y 0 − Y .
The question is then : Does there exist v (in an appropriate space) such that
Z(v)(T ) = 0? This notion is known as the null-controllability for Z or the zerocontrollability. Of course the two notions are equivalent for linear systems but they
are not for nonlinear systems !
Many important systems can be put into the framework of (??), for example the
Navier-Stokes system, or Navier-Stokes equations coupled with one or several diffusionconvection equations, or the primitive system of atmosphere modelling the motions
of the atmosphere and therefore the basic model in meteorology or climatology, etc.
Of course, the more complex the system is, the more difficult and technical the
mathematics involved are. In the sequel we will treat completely the case of a linear
diffusion-convection equation with a distributed control and afterwards we will give
some ideas and results for extensions.
4
2.1
Case of a linear diffusion-convection equation
Let T > 0 and let Ω be a bounded open regular subset of IRN . We denote by Γ the
boundary of Ω which is supposed to be of class C 2+α . We consider an operator L
depending on time t which is elliptic for each time t and which is defined by
(2.1.7)
Lz = −
N
X
N
X
∂z
∂
∂
(ai,j
)+
(bi z) + a0 z,
∂x
∂x
∂x
i
j
i
i,j=1
i=1
where
(2.1.8)
∀i, j = 1, · · · , N, ai,j ∈ W 1,∞ (Ω × (0, T )), ai,j = aj,i
(2.1.9)
∀i = 1, · · · , N, bi , a0 ∈ L∞ (Ω × (0, T ))
and the coefficients ai,j satisfy an ellipticity condition uniformly in t
N
(2.1.10) ∃β > 0, ∀(x, t) ∈ Ω × (0, T ), ∀ξ ∈ IR ,
N
X
ai,j (x, t)ξj ξi ≥ β|ξ|2
i,j=1
Let ω be an open subset of Ω. For each control v ∈ L2 (0, T ; L2 (ω)) we consider the
following controlled diffusion-convection equation
(2.1.11)
(2.1.12)
(2.1.13)
∂y
+ Ly = f 0 + v.χω in Ω × (0, T ),
∂t
y = 0 on Γ × (0, T ),
y(x, 0) = y 0 (x) in Ω,
where y 0 ∈ L2 (Ω) and f 0 ∈ L2 (0, T ; H −1 (Ω)) for example.
It is well known (cf for example [?]) that for every v ∈ L2 (0, T ; L2 (ω)) there
exists a unique solution y = y(v) to equation (??) with y ∈ C([0, T ]; L2 (Ω)) ∩
L2 (0, T ; H01 (Ω)).
Let us now consider an “ideal” uncontrolled trajectory starting at time t = 0 from
the initial data y 0 ∈ L2 (Ω)
(2.1.14)
(2.1.15)
(2.1.16)
∂y
+ Ly = f 0 in Ω × (0, T ),
∂t
y = 0 on Γ × (0, T ),
y(x, 0) = y 0 (x) in Ω,
Again we have a unique solution for (??) y ∈ C([0, T ]; L2 (Ω))∩L2 (0, T ; L2 (Ω)). The
question of exact controllability to trajectories is then, for every y 0 ∈ L2 (Ω) and
y 0 ∈ L2 (Ω), to find v ∈ L2 (0, T ; L2 (ω)) such that y(T ) = y(T ).
5
Remark 2.1.1 We have taken here the same right hand side f 0 in both equations
(??) and (??) for sake of simplicity. We could have taken in (??) a right hand side
0
0
f 6= f 0 but the conditions we would have to impose on f 0 − f are not easy to state
0
correctly. Nevertheless the proof will be given with g 0 = f 0 − f 6= 0.
Of course, as already mentionned above, as we deal here with a linear equation, the
problem is equivalent to the following null-controllability one. Let z be the solution
to
(2.1.17)
(2.1.18)
(2.1.19)
∂z
+ Lz = v.χω in Ω × (0, T ),
∂t
z = 0 on Γ × (0, T ),
z(x, 0) = z 0 (x) in Ω,
where z 0 ∈ L2 (Ω). We then look for v ∈ L2 (0, T ; L2 (ω)) such that
(2.1.20)
z(T ) = 0.
The main result of this chapter is the following
Theorem 2.1.2 Under the previous hypotheses (??), (??), (??), for every open
subset ω of Ω, for every time T > 0 and for every z 0 ∈ L2 (Ω), there exists a control
v ∈ L2 (0, T ; L2 (ω)) such that (??) holds. Moreover we can obtain a control v of
minimal norm in L2 (0, T ; L2 (ω)) among the admissible controls (such that (??) is
satisfied).
Proof of Theorem ?? will require several steps which will be developped in the next
sections. There are various strategies which lead to the result, all based on the
same Carleman inequalities. First of all, we will develop a method starting from
approximate controllability because it is quite natural, adaptable to many other
situations and also easier to understand. We will then give a second approach which
turns out to be useful for extension to nonlinear problems.
Remark 2.1.3 For some further extensions such as treating the case of nonlinear
diffusion-convection equations, we sometimes need to obtain a control v in a smaller
space like Lr (O, T ; Lq (Ω)) with r and q larger than 2. We can obtain the previous
theorem with this class of controls but this requires some minor modifications in the
proof for example a careful use of regularity results for the heat equation.
6
2.2
Approximate null-controllability
We now take g 0 ∈ L2 (0, T ; L2 (Ω)) for the moment (further conditions will be needed
later on) and we consider the following variant of (??)
∂z
+ Lz = g 0 + v.χω in Ω × (0, T ),
∂t
z = 0 on Γ × (0, T ),
(2.2.21)
(2.2.22)
z(x, 0) = z 0 (x) in Ω,
(2.2.23)
Let be a stricly positive number. In a first step towards the proof of Theorem ??,
we want to find a control v such that the solution of (??)-(??) satisfies
|z(T )|L2 (Ω) ≤ .
(2.2.24)
We also want to select such a control in order to have a unique determination of it
and we will take a control which minimizes the L2 -norm among admissible controls.
For sake of completeness, we will give here a complete result on approximate controllability following [?] and [?]. We will show that the set of reachable states for
equation (??)-(??) is dense in L2 (Ω) or, equvalently, that given any z 1 ∈ L2 (Ω) and
any > 0, we can find a control v such that the solution of (??)-(??) reaches at
time T the ball of radius centered at z 1 .
In order to state this result precisely, we need to introduce the adjoint problem of
equation (??)-(??).
Let us consider the operator L∗ as the formal adjoint of L, defined by
(2.2.25)
L∗ ξ = −
N
X
N
X
∂
∂ξ
∂ξ
(ai,j
)−
bi
+ a0 ξ.
∂xj
∂xi
∂xi
i,j=1
i=1
We then consider for every ξ 0 ∈ L2 (Ω) the solution ξ of the (backward) adjoint
equation
∂ξ
(2.2.26)
−
+ L∗ ξ = 0 in Ω × (0, T ),
∂t
(2.2.27)
ξ = 0 on Γ × (0, T ),
ξ(T ) = ξ 0 in Ω.
(2.2.28)
We know that for every ξ 0 ∈ L2 (Ω), there exists a unique solution ξ to (??)-(??)
with ξ ∈ C([0, T ]; L2 (Ω)) ∩ L2 (0, T ; H01 (Ω)). We can now define a functional J (ξ 0 )
by
(2.2.29)
J (ξ 0 ) =
1
2
Z TZ
0
ω
|ξ|2 dxdt + |ξ 0 |L2 (Ω) +
Z
Z
z 0 ξ(0)dx −
Ω
1 0
Z TZ
z ξ dx +
Ω
7
0
Ω
g 0 ξdxdt.
Theorem 2.2.1 Under the above hypotheses (??), (??), (??) on coefficients of the
operator L, given a time T > 0 and a non empty open subset ω of Ω, for every
g 0 ∈ L2 (0, T ; L2 (Ω)), for every z 0 ∈ L2 (Ω), z 1 ∈ L2 (Ω) and for every > 0, there
exists a control v ∈ L2 (0, T ; L2 (ω)) such that the corresponding solution z of
(??)-(??) satisfies
(2.2.30)
|z (T ) − z 1 |L2 (Ω) ≤ .
Moreover we take take v of minimum norm in L2 (0, T ; L2 (ω)) by choosing v = ξ .χω
where ξ is the solution of (??)-(??) corresponding to the initial value ξ0 which is
such that
(2.2.31)
J (ξ0 ) = min J (ξ 0 ).
ξ 0 ∈L2 (Ω)
Sketch of the proof.
Let us define first the set of reachable states for equation (??)-(??) by
(2.2.32)
R(T ) = {z(T ), z is solution of (??)-(??), v ∈ L2 (0, T ; L2 (ω))}.
We call z̄ the solution of
∂ z̄
+ Lz̄ = g 0 in Ω × (0, T ),
∂t
z̄ = 0 on Γ × (0, T ),
z̄(x, 0) = z 0 (x) in Ω,
and we define z̃ as the solution of
∂ z̃
+ Lz̃ = v.χω in Ω × (0, T ),
∂t
z̃ = 0 on Γ × (0, T ),
(2.2.33)
(2.2.34)
(2.2.35)
z̃(x, 0) = 0 in Ω.
Therefore we have
R(T ) = z̄(T ) + R0 (T )
where R0 (T ) is the vector space such that
(2.2.36)
R0 (T ) = {z̃(T ), z̃ is solution of (??)-(??), v ∈ L2 (0, T ; L2 (ω))}.
The density of R(T ) in L2 (Ω) is equivalent to the density of R0 (T ) in L2 (Ω) and
from Hahn-Banach Theorem, as R0 (T ) is a vector space, this is also equivalent to
prove that the orthogonal space R0⊥ (T ) in L2 (Ω) is reduced to {0}.
8
Let ξ 0 ∈ R0⊥ (T ) and let ξ be the corresponding solution of (??)-(??). Multiplying
(??) by ξ we obtain
∀v ∈ L2 (0, T ; L2 (ω)),
Z TZ
Z
z̃(T ).ξ 0 dx = 0.
v.ξdxdt =
0
ω
Ω
Then ξ = 0 in ω × (0, T ).
Using the unique continuation property for (??)-(??), we see that ξ = 0 in the whole
cylinder Ω × (0, T ) and therefore ξ0 = 0.
The unique continuation property in this context has first been obtained by [?] when
the coefficients are regular, but it will also be a consequence of further results we
shall obtain in the next section.
In [?] it is also proved that the unique continuation property implies that the functional J defined in (??) is strictly convex and coercive, so that it has a unique
minimum in L2 (Ω) and then problem (??) has a unique solution ξ0 .
If we define the set of admissible controls V as the set of controls v ∈ L2 (0, T ; L2 (ω))
such that (??) holds, then this set V is non empty and it is easy to see that it is
closed and convex. Therefore, we can look for the (unique) element of this set of
minimum norm in L2 (0, T ; L2 (ω)) which we call v . Using the Fenchel-Rockafellar
duality result (see for example [?]), we obtain that the dual of this problem is the
minimization of J and that v = ξ .χω where ξ is the solution of (??)-(??)
corresponding to the initial value ξ0 .
This proves Theorem ?? but for sake of completeness let us show that when ξ0 6= 0
(the other case being trivial), if we take v = ξ .χω , then the corresponding solution
z of (??)-(??) satisfies (??). We use unambiguous usual notations.
As ξ0 is the solution of (??) we have
< J0 (ξ0 ), ζ 0 >= 0, ∀ζ 0 ∈ L2 (Ω),
i.e., if ζ denotes the solution of (??)-(??) corresponding to ζ 0 ,
Z TZ
Z
ξ ζdxdt + 0
ω
Ω
ξ0
ζ 0 dx +
|ξ0 |L2 (Ω)
Z
0
z ζ(0)dx −
Ω
Z
Z TZ
1 0
z ζ dx +
Ω
0
g 0 ζdxdt = 0.
Ω
Now taking v = ξ .χω in (??) and taking the scalar product of this equation with
ζ we obtain
Z
z (T )ζ(T )dx −
Ω
Z TZ
Z
z (0)ζ(0)dx =
Ω
0
so that
Z
Ω
(z (T ) − z 1 )ζ 0 dx = −
Z
Ω
Ω
g 0 ζdxdt +
Z TZ
ξ ζdxdt,
0
ω
ξ0
ζ 0 dx, ∀ζ 0 ∈ L2 (Ω),
|ξ0 |L2 (Ω)
9
which shows (??).
We have to notice that the control v we have obtained obviously verifies
|v |2L2 (0,T ;L2 (ω)) =
(2.2.37)
Z TZ
0
|ξ |2 dxdt.
ω
Remark 2.2.2 If we need to obtain controls in smaller spaces than L2 (0, T ; L2 (ω)),
we have to change the functional J as it is done in [?]. Notice that in some cases,
we have to be careful with the use of duality results. It happens to be better to take
the minimization of J as a primal problem and the problem of minimal norm for
controls v as the dual problem. This is the case in particular when we look for
controls in L∞ (ω × (0, T )). This case is treated in [?].
An immediate consequence of Theorem ?? which will be used for the proof of Theorem ?? is the following corollary which only considers the case where z 1 = 0.
Corollary 2.2.3 For every > 0, there exists a control v ∈ L2 (0, T ; L2 (ω)) such
that the corresponding solution z of (??)-(??) satisfies (??). Moreover, we can take
v of minimal norm in L2 (0, T ; L2 (ω)) among the admissible controls by choosing
v = ξ .χω where ξ is the solution of (??)-(??) corresponding to the initial value
ξ0 which is such that
(2.2.38)
J0 (ξ0 ) = min J0 (ξ 0 ),
ξ 0 ∈L2 (Ω)
where
(2.2.39)
2.3
J0 (ξ 0 )
1
=
2
Z TZ
0
2
ω
Z
0
|ξ| dxdt + |ξ |L2 (Ω) +
0
Z TZ
g 0 ξdxdt.
z ξ(0)dx +
0
Ω
Ω
Null-controllability modulo observability inequality
We take from now on ξ0 minimizing J0 . If we take ξ 0 = 0 the corresponding solution
of (??)-(??) is 0 and therefore J0 (0) = 0. Then we have for every > 0
J0 (ξ0 ) ≤ 0,
(2.3.40)
which says, from the definition of J0 that
(2.3.41) |v |2L2 (0,T ;L2 (ω)) =
Z TZ
0
Z
|ξ |2 dxdt ≤ −2(
ω
Ω
z 0 ξ (0)dx +
Z TZ
0
g 0 ξ dxdt).
Ω
Let us introduce a weight ρ, which will be precisely defined later on, but which, for
the moment, only satisfies ρ > 0 in Ω × (0, T ). Using Hölder inequality we obtain
provided that
ρg 0 ∈ L2 (0, T ; L2 (Ω))
10
Z TZ
(2.3.42)
0
|ξ |2 dxdt ≤
ω
1
1
1
2(|z 0 |2L2 (Ω) + |ρg 0 |2L2 (0,T ;L2 (Ω)) ) 2 (|ξ (0)|2L2 (Ω) + | ξ |2L2 (0,T ;L2 (Ω)) ) 2 .
ρ
We would like to deduce from this last inequality a bound for |v |2L2 (0,T ;L2 (ω)) =
RT R
|ξ |2 dxdt. This will be the case if for some suitable weight ρ there exists a
constant C (independent of ) such that we have the following inequality
0
ω
(2.3.43)
1
|ξ (0)|2L2 (Ω) + | ξ |2L2 (0,T ;L2 (Ω)) ≤ C
ρ
Z TZ
0
|ξ |2 dxdt.
ω
Such an inequality will be called an observability inequality and its proof will be the
subject of the next sections. Let us for the moment suppose that inequality (??)
holds true. Then we easily obtain that for every > 0
(2.3.44)
|v |2L2 (0,T ;L2 (ω)) =
Z TZ
0
ω
|ξ |2 dxdt ≤ 4C(|z 0 |2L2 (Ω) + |ρg 0 |2L2 (0,T ;L2 (Ω)) ).
Therefore v is bounded in L2 (0, T ; L2 (ω)) independently of and for a subsequence
(still denoted by ) we have
(2.3.45)
v * v in L2 (0, T ; L2 (ω)) weakly.
Let us now go back to equation (??)-(??) where we take v = v , the solution being
then denoted by z . As v * v in L2 (0, T ; L2 (ω)) weakly, we have, from classical
results for parabolic equations,
(2.3.46)
z → z in L2 (0, T ; H01 (Ω)) ∩ C([0, T ]; L2 (Ω)),
where z is the solution to (??)-(??) corresponding to v. We know that z satisfies
(??). Therefore passing to the limit in we immediately obtain that z satisfies
(2.3.47)
z(T ) = 0,
which is exactly (??) and this will prove the first part of Theorem ??.
Remark 2.3.1 In fact, assuming that the observability inequality (??) is valid we
have proved a stronger result as we have taken in (??) g 0 6= 0 but we have to assume
that ρg 0 ∈ L2 (0, T ; L2 (Ω)).
Now the set of functions v such that (??) holds (admissible controls) is not empty and
it is immediate to see that it is a closed convex set of L2 (0, T ; L2 (ω)). Therefore there
exists a unique admissible control v0 of minimal norm in L2 (0, T ; L2 (ω)). But, as for
11
the corresponding solution of (??)-(??) denoted by z(v0 ) we have z(v0 )(T ) = 0, v0 is
a fortiori admissible for approximate null-controllability when > 0, and therefore
we have for every > 0, |v |L2 (0,T ;L2 (ω)) ≤ |v0 |L2 (0,T ;L2 (ω)) . Then the (weak) limit
v of (a subsequence of) v satisfies |v|L2 (0,T ;L2 (ω)) ≤ |v0 |L2 (0,T ;L2 (ω)) and z(T ) = 0
which implies v = v0 . Then by a standard argument we see that the whole sequence
v converges to v0 and also that the convergence is strong in L2 (0, T ; L2 (ω)).
Now, in order to finish the proof of Theorem ??, it remains to show the observability
inequality (??) and to choose of course a suitable weight ρ. Such an inequality says
that the knowledge of the solution of (??)-(??) on a small cylinder ω ×(0, T ) governs
the (weighted) behaviour of the solution on the whole domain Ω×(0, T ) and its final
value (notice that (??) is a backward equation so that the final value is z(0)). This
must require a way to propagate the information inside the spacial domain and will
be a consequence of a global Carleman inequality, the proof of which will be the
object of the next section.
2.4
Globlal Carleman inequality
We first have to define the weight function that we will use. There are several
possible choices and we follow here the 2-parameters choice of [?] with some slight
modification.
2.4.1
Weight functions
We have to choose a weight which we called ρ in the last section but this choice
will require long arguments and we have to begin with a basic choice of weight
depending only on the space variables. This weight is fundamental in the sense
that, roughly speaking, information will propagate in space along the gradient lines
of this function.
Lemma 2.4.1 Let ω0 be an open set such that ω0 ⊂ ω (for example ω0 can be a
small open ball). Then there exists ψ ∈ C 2 (Ω) such that
ψ(x) > 0, ∀x ∈ Ω,
ψ(x) = 0, ∀x ∈ Γ,
|∇ψ(x)| 6= 0, ∀x ∈ Ω − ω0 .
Proof. This proof is very technical and can be omitted in a first step.
As Ω is regular, we can first choose a function θ ∈ C 2 (IRN ) such that
Ω = {x ∈ IRN , θ(x) > 0} and |∇θ(x)| =
6 0, ∀x ∈ Γ. This can be done locally,
and then extended globally using a partition of unity. From Morse’s density theorem ([?]), there exists a sequence of Morse functions (θk ) (i.e. such that their
12
gradients vanishes only at a finite number of points) such that θk → θ in C 2 (Ω)
when k → +∞ (θk does not necessary vanish on the boundary). Moreover we can
take θk > 0 as θ > 0 on Ω. Let us define C = {x ∈ IRN , ∇θ(x) = 0} as the set of
critical points of θ. As |∇θ(x)| =
6 0, ∀x ∈ Γ, there exists an open neighborhood V
of Γ in IRN and δ > 0 such that
∀x ∈ V , |∇θ(x)| ≥ δ.
Let ϕ ∈ D(V ) such that ϕ(x) = 1, ∀x ∈ Γ and 0 ≤ ϕ ≤ 1. We set
µk (x) = θk (x) + ϕ(x)(θ(x) − θk (x)).
Then µk (x) = 0, ∀x ∈ Γ, µk (x) > 0, ∀x ∈ Ω and moreover
∀x ∈ Ω − V , ∇µk (x) = ∇θk (x).
Now if x ∈ Ω ∩ V , we have
∇µk (x) = ∇θk (x) + ϕ(x)(∇θ(x) − ∇θk (x)) + ∇ϕ(x)(θ(x) − θk (x))
so that for k ≥ k0 , k0 large enough we have
|∇µk (x)| ≥ |∇θk (x)| − 2||ϕ||C 1 (Ω) ||θ − θk ||C 1 (Ω)
≥ δ − 2||ϕ||C 1 (Ω) ||θ − θk ||C 1 (Ω)
≥
δ
.
2
Let us choose k ≥ k0 and set µ(x) = µk (x). Then µ is a Morse function because
the points where its gradient vanishes are among the points where ∇θk vanishes.
Moreover we have µ(x) = 0, ∀x ∈ Γ.
Let now x1 , x2 , · · · , xr be the critical points of µ. Then for i = 1, · · · , r we have
xi ∈ Ω − V . We can find r disjoint regular paths l1 , · · · , lr such that for i = 1, · · · , r,
li ∈ C ∞ ([0, 1]; IRN ),
li (t) ∈ Ω − V , ∀t ∈ [0, 1],
li (t1 ) 6= li (t2 ), ∀t1 , t2 ∈ [0, 1], t1 6= t2 ,
li (1) = xi , and li (0) ∈ ω0 ,
∀s, t ∈ [0, 1], li (s) 6= lj (t), when i 6= j,
and we can find r functions f1 , · · · , fr such that for i = 1, · · · , r
fi ∈ C ∞ (IRN , IRN ), and
dli
(t) = fi (li (t)), ∀t ∈ (0, 1).
dt
13
Now, for i = 1, · · · , r we can find open neighborhoods Wi of the sets {li (t), t ∈ [0, 1]}
such that
Wi ⊂ Ω − V , and Wi ∩ Wj = ∅ if i 6= i.
Then we take functions ei ∈ D(Wi ) such that ei (li (t)) = 1, ∀t ∈ [0, 1] and we set
gi (x) = ei (x)fi (x).
Let us consider the differential equation
dx
(t) = gi (x(t)), ∀t ∈ (0, 1),
dt
x(0) = x.
We denote by Sti : IRN → IRN the operator which maps x to x(t). We then have
S1i (li (0)) = xi , i = 1, · · · , r.
We now define
S(x) = S11 ◦ S12 ◦ · · · ◦ S1r (x).
We can se that if x ∈ Ω − (
Sr
i=1 Wi ),
then S(x) = x and therefore
∀x ∈ V, S(x) = x.
On the other hand, each S1i is a diffeomorphism from Ω into itself, so is S and ∇S
is invertible.
Let us now set
ψ(x) = µ(S(x)).
Then we have ψ(x) = 0, ∀x ∈ Γ. Moreover, as ∇S is invertible, if ∇ψ(x) = 0, this
means that S(x) ∈ {x1 , · · · , xr }. But we know that S1j = Id on Ω − Wj so that
S(li (0)) = S1i (li (0)) = xi .
As S is a diffeomorphism, we see that
S(x) ∈ {x1 , · · · , xr } ⇒ x ∈ {l1 (0), · · · , lr (O)} ⇒ x ∈ ω0 .
Therefore
∇ψ(x) = 0 ⇒ x ∈ ω0 ,
and ψ satifies all conditions of the lemma. This finishes the proof of Lemma ??.
14
We will now use the function ψ given by Lemma ?? to build new weights. Let us
define for λ > 0 and for an integer k ≥ 1
(2.4.48)
ϕ(x, t) =
eλ(ψ(x)+m1 )
,
tk (T − t)k
(2.4.49)
η(x, t) =
eλ(|ψ|L∞ (Ω) +m2 ) − eλ(ψ(x)+m1 )
,
tk (T − t)k
where the positive constants m1 and m2 will be chosen below.
Remark 2.4.2 For our purpose here, we only need to take k = 1 but for further
extensions it happens that we sometimes need to take k > 1 and this does not make
any change in the sequel.
We now want to choose the constants m1 and m2 so that the numerator of η is
positive which implies that m2 > m1 and also that we can bound (modulo some
∂2η
2
3
constants) ∂η
∂t and ∂t2 respectively by ϕ and ϕ . A simple calculation shows that
for example, a possible choice of these constants is
(2.4.50)
m1 = |ψ|L∞ (Ω) + 2 ; m2 = |ψ|L∞ (Ω) + 3.
We now have for every λ > 0 the following properties which will be helpful for our
calculations
(2.4.51)
(2.4.52)
(2.4.53)
(2.4.54)
∇ϕ = λϕ∇ψ ; ∇η = −λϕ∇ψ,
T
T
T
1 ≤ ( )2k ϕ ; ϕ ≤ ( )2k ϕ2 ; ϕ ≤ ( )4k ϕ3
2
2
2
∂ϕ
T 2(k−1) 2
∂2ϕ
T
| | ≤ kT ( )
ϕ ; | 2 | ≤ k(k + 1)T 2 ( )4(k−1) ϕ3 ,
∂t
2
∂t
2
∂η
T 2(k−1) 2
∂2η
T
| | ≤ kT ( )
ϕ ; | 2 | ≤ k(k + 1)T 2 ( )4(k−1) ϕ3 .
∂t
2
∂t
2
We can notice that η tends rapidly to +∞ when t → T or t → 0 but that η is
uniformly bounded in Ω × [δ, T − δ] if δ > 0.
Our final weight will depend on a second positive parameter s and will be of the
form e−sη(x,t) . We can see that, for fixed s, this function tends very rapidly to 0
when t → T or t → 0.
2.4.2
Proof of a global Carleman inequality
We want to prove a Carleman inequality for the solution of equation (??)-(??) but
we will take the general case of a parabolic equation. We still consider a backward
15
equation because (??) is backward but of course there is no change for a forward
equation. As we will see we only need to consider the principal part of L∗ and as
the coefficients ai,j are symmetric it is equivalent to take the principal part of L
that we call L0 . So we define
L0 u = −
(2.4.55)
N
X
∂
∂u
(ai,j
),
∂xi
∂xj
i,j=1
and for g ∈ L2 (0, T ; L2 (Ω)) we consider the solution u of the following backward
parabolic equation
−
(2.4.56)
(2.4.57)
∂u
+ L0 u = g, in Ω × (0, T ),
∂t
u = 0, on Γ × (0, T ),
(2.4.58)
u(T ) = u0 , in Ω.
We can now state the global Carleman inequality
Theorem 2.4.3 There exist parameters s0 > 0 and λ0 > 0 and there exists a
constant C > 0 depending only on Ω, ω0 , ψ, on β defined in (??) and on the
coefficients ai,j such that for every s > T 2k s0 , for every λ > λ0 and for every
solution of (??) we have
(2.4.59)
1
s
Z T Z −2sη
e
sλ
0
2
ϕ
Ω
Z TZ
(|
N
X
∂2u 2
∂u 2
|
| +
| )dxdt +
∂t
∂xi ∂xj
i,j=1
−2sη
ϕe
Z
0
T
3 4
|∇u| dxdt + s λ
Z TZ
Ω
Z
C(
0
2
0
e−2sη |g|2 dxdt + s3 λ4
Z TZ
Ω
0
ϕ3 e−2sη |u|2 dxdt ≤
Ω
ϕ3 e−2sη |u|2 dxdt).
ω
Proof. For s > 0 and λ > 0 we define
(2.4.60)
w(x, t) = e−sη(x,t) u(x, t).
We can see that
(2.4.61)
w(x, 0) = w(x, T ) = 0.
We now compute in terms of w the operator
(2.4.62)
e−sη (−
∂(esη w)
+ L0 (esη w)) = e−sη g.
∂t
16
We have
∂(esη w)
∂w
∂η
= esη (
+ s w)
∂t
∂t
∂t
and because of (??)
∂w
∂ψ
∂(esη w)
= esη (
− sλϕ
w)
∂xj
∂xj
∂xj
so that
(2.4.63)
∂
∂(esη w)
(ai,j
)=
∂xi
∂xj
∂
∂w
∂ψ ∂w
∂ψ ∂w
esη (
(ai,j
) − sλϕai,j (
+
)−
∂xi
∂xj
∂xj ∂xi ∂xi ∂xj
∂
∂ψ
∂ψ ∂ψ
∂ψ ∂ψ
sλϕ
(ai,j
)w − sλ2 ϕai,j
w + s2 λ2 ϕ2 ai,j
w).
∂xi
∂xj
∂xj ∂xi
∂xj ∂xi
After minor modifications and using the symmetry of coefficients ai,j we then obtain
(2.4.64)
P w = P1 w + P2 w = gs
where
(2.4.65)P1 w = −
N
N
X
X
∂w
∂ψ ∂w
∂ψ ∂ψ
ϕai,j
+ 2sλ
ϕai,j
+ 2sλ2
w,
∂t
∂xj ∂xi
∂xj ∂xi
i,j=1
i,j=1
N
X
∂w
∂ψ ∂ψ
∂η
∂
2 2
(2.4.66)P2 w = −
(ai,j
)−s λ
ϕ2 ai,j
w − s w,
∂x
∂x
∂x
∂x
∂t
i
j
j
i
i,j=1
i,j=1
N
X
(2.4.67) gs = e−sη g + sλ2
N
X
ϕai,j
i,j=1
N
X
∂
∂ψ
∂ψ ∂ψ
w − sλ
ϕ
(ai,j
)w.
∂xj ∂xi
∂xi
∂xj
i,j=1
We now take the L2 -norm of each term in (??) and we obtain
Z TZ
2
|P1 w| dxdt +
(2.4.68)
0
Ω
Z TZ
0
Z TZ
2
|P2 w|2 dxdt +
Ω
Z T
Z
P1 wP2 wdxdt =
0
Ω
0
We shall now compute the term
Z TZ
P1 wP2 wdxdt;
0
Ω
17
Ω
|gs |2 dxdt.
using (??) and (??). This computation will give 9 terms Ik,l . In the sequel, by C
we mean various constants independent of s, λ and T as we want to keep track of
the powers of s, λ and T involved. In order to organize the calculations we will give
particular importance to terms of the order of
sλ2
Z TZ
0
and
s3 λ4
ϕ|∇w|2 dxdt
Ω
Z TZ
0
ϕ3 |w|2 dxdt.
Ω
We take s ≥ 1 and λ ≥ 1 and we denote by
sp λq A
all terms which can be bounded by
Csp λq
Z TZ
0
ϕ|∇w|2 , p ≤ 1, q ≤ 2, p + q ≤ 2
Ω
and by
sp λq B
the terms which can be bounded by
Csp λq
Z TZ
0
ϕ3 |w|2 dxdt, p ≤ 3, q ≤ 4, p + q ≤ 6.
Ω
These terms will be neglectible as we will see later on. We have the following
successive results.
I1,1 =
Z TZ
N
∂w X
0
=
Ω
∂
∂w
(ai,j
)dxdt
∂t i,j=1 ∂xi
∂xj
N Z T Z
X
i,j=1 0
=
Z Z
N
X
1 T
i,j=1
= −
2
0
Ω
N
X
∂
∂w ∂w
1
(ai,j
)dxdt −
∂t
∂xj ∂xi
2
i,j=1
Z Z
N
X
1 T
i,j=1
Then
(2.4.69)
Ω
∂ ∂w
∂w
(
)ai,j
dxdt
∂t ∂xi
∂xj
2
0
Ω
(
Z TZ
(
0
Ω
∂ai,j ∂w ∂w
)
dxdt
∂t ∂xj ∂xi
∂ai,j ∂w ∂w
)
dxdt , because of (??).
∂t ∂xj ∂xi
I1,1 = T 2k A.
18
2 2
I1,2 = s λ
=
Z TZ
N
∂w X
s2 λ2
2
∂t
0
Ω
N
X
Z TZ
i,j=1 0
ϕ2 ai,j
i,j=1
ϕ2 ai,j
Ω
∂ψ ∂ψ
wdxdt
∂xj ∂xi
∂ψ ∂ψ ∂
|w|2 dxdt
∂xj ∂xi ∂t
N Z T Z
∂ai,j 2
∂ψ ∂ψ
s2 λ2 X
(
|w|2 dxdt
= −
)ϕ ai,j
2 i,j=1 0 Ω ∂t
∂xj ∂xi
N Z T Z
X
− s2 λ2
i,j=1 0
ϕ
Ω
∂ϕ
∂ψ ∂ψ
|w|2 dxdt.
ai,j
∂t
∂xj ∂xi
Because of (??) we have
I1,2 = s2 λ2 (T 2k + T 2k−1 )B.
(2.4.70)
I1,3 = s
Z TZ
∂w ∂η
w=
s
2
Z TZ
∂η ∂
0
Ω ∂t ∂t
0
Z TZ
s
∂2η 2
= −
|w| dxdt,
2 0 Ω ∂t2
and using (??)
(2.4.71)
Ω
∂t ∂t
(|w|2 )dxdt
I1,3 = sT 4k−2 B.
before procceeding our calculations we have to notice that, if ν is the unit exterior
normal on the boundary Γ, as ψ and w vanish on Γ we have for x ∈ Γ
(2.4.72)
∇ψ(x, t) = (∇ψ.ν)ν and ∇w(x, t) = (∇w.ν)ν.
I2,1 = −2sλ
N
X
Z TZ
ϕai,j
Ω
i,j,k,l=1 0
= −2sλ
N
X
Z TZ
i,j,k,l=1 0
+ 2sλ2
N
X
Γ
+ 2sλ
N
X
ϕ(∇ψ.ν)|∇w.ν|2 ai,j νi νj ak,l νk νl dγdt
Z TZ
ϕ(ai,j
i,j,k,l=1 0
Ω
Z TZ
i,j,k,l=1 0
ϕ
Ω
∂ψ ∂w ∂
∂w
(ak,l
)dxdt
∂xj ∂xi ∂xk
∂xl
∂ψ ∂w
∂ψ ∂w
)(ak,l
)dxdt
∂xj ∂xi
∂xl ∂xk
∂w ∂w
∂
∂ψ
(ai,j
)ak,l
dxdt
∂xk
∂xj
∂xi ∂xl
19
N
X
+ 2sλ
ϕai,j
i,j,k,l=1 0
N
X
= −sλ
Ω
Z TZ
i,j,k,l=1 0
+ 2sλ
N
X
2
Z TZ
ϕ(ai,j
N
X
Ω
Z TZ
N
X
Ω
Z TZ
ϕai,j
i,j,k,l=1 0
N
X
− sλ
Ω
Z TZ
N
X
− sλ
Ω
Z TZ
ϕai,j
i,j,k,l=1 0
∂ψ ∂ψ
∂w ∂w
ak,l
dxdt
∂xj ∂xi
∂xl ∂xk
∂
∂ψ
∂w ∂w
dxdt
(ai,j
)ak,l
∂xi
∂xj
∂xl ∂xk
ϕ
i,j,k,l=1 0
∂ψ ∂w
∂ψ ∂w
)(ak,l
)dxdt
∂xj ∂xi
∂xl ∂xk
∂
∂w ∂w
∂ψ
)ak,l
(ai,j
dxdt
∂xk
∂xj
∂xi ∂xl
ϕ
i,j,k,l=1 0
− sλ2
ϕ(∇ψ.ν)|∇w.ν|2 ai,j νi νj ak,l νk νl dγdt
Γ
i,j,k,l=1 0
+ 2sλ
∂ψ
∂w ∂ 2 w
ak,l
dxdt
∂xj
∂xl ∂xk ∂xi
Z TZ
Ω
∂ψ ∂ak,l ∂w ∂w
dxdt.
∂xj ∂xi ∂xl ∂xk
Therefore we have
(2.4.73)
N
X
I2,1 = −sλ
Z TZ
i,j,k,l=1 0
N
X
+ 2sλ2
Γ
Z TZ
ϕ(ai,j
i,j,k,l=1 0
− sλ
N
X
2
ϕ(∇ψ.ν)|∇w.ν|2 ai,j νi νj ak,l νk νl dγdt
Ω
Z TZ
ϕai,j
i,j,k,l=1 0
Ω
∂ψ ∂w
∂ψ ∂w
)(ak,l
)dxdt
∂xj ∂xi
∂xl ∂xk
∂ψ ∂ψ
∂w ∂w
ak,l
dxdt
∂xj ∂xi
∂xl ∂xk
+ sλA.
I2,2 = −2s3 λ3
N
X
Z TZ
i,j,k,l=1 0
= −s3 λ3
N
X
Ω
Z TZ
i,j,k,l=1 0
= 3s3 λ4
N
X
ϕ3 ai,j
Ω
Z TZ
i,j,k,l=1 0
ϕ3 ai,j
ϕ3 ai,j
Ω
20
∂ψ ∂w
∂ψ ∂ψ
ak,l
wdxdt
∂xj ∂xi
∂xl ∂xk
∂ψ ∂
∂ψ ∂ψ
(|w|2 )ak,l
dxdt
∂xj ∂xi
∂xl ∂xk
∂ψ ∂ψ
∂ψ ∂ψ
ak,l
|w|2 dxdt
∂xj ∂xi
∂xl ∂xk
N
X
3 3
+ s λ
Z TZ
i,j,k,l=1 0
N
X
+ s3 λ3
ϕ3
Ω
Z TZ
i,j,k,l=1 0
∂
∂ψ
∂ψ ∂ψ
(ai,j
)ak,l
|w|2 dxdt
∂xi
∂xj
∂xl ∂xk
ϕ3 ai,j
Ω
∂ψ ∂
∂ψ ∂ψ
(ak,l
)|w|2 dxdt,
∂xj ∂xi
∂xl ∂xk
so that
(2.4.74)
N
X
I2,2 = 3s3 λ4
Z TZ
i,j,k,l=1 0
ϕ3 ai,j
Ω
∂ψ ∂ψ
∂ψ ∂ψ
ak,l
|w|2 dxdt
∂xj ∂xi
∂xl ∂xk
+s3 λ3 B.
2
I2,3 = −2s λ
N Z T Z
X
i,j=1 0
= −s2 λ
Ω
N Z T Z
X
i,j=1 0
= s λ
i,j=1 0
i,j=1 0
2
+ s λ
i,j=1 0
so that
(2.4.75)
Ω
∂η
∂ψ ∂
ai,j
|w|2 dxdt
∂t
∂xj ∂xi
∂η
∂ψ ∂ψ
ai,j
|w|2 dxdt
∂t
∂xj ∂xi
ϕ
∂ψ ∂ψ
∂ϕ
ai,j
|w|2 dxdt
∂t
∂xj ∂xi
Ω
N Z T Z
X
∂ψ ∂w ∂η
wdxdt
∂xj ∂xi ∂t
ϕ
Ω
N Z T Z
X
− s2 λ2
ϕ
Ω
N Z T Z
X
2 2
ϕai,j
ϕ
∂η ∂
∂ψ
(ai,j
)|w|2 dxdt,
∂t ∂xi
∂xj
I2,3 = s2 λ2 T 2k−1 B.
I3,1 = −2sλ
2
N
X
Z TZ
ϕai,j
i,j,k,l=1 0
= 2sλ3
N
X
Z TZ
ϕai,j
i,j,k,l=1 0
+ 2sλ2
N
X
N
X
Ω
Z TZ
ϕ
i,j,k,l=1 0
+ 2sλ2
Ω
Ω
∂ψ ∂ψ ∂ψ
∂w
ak,l
wdxdt
∂xj ∂xi ∂xk
∂xl
∂
∂w
∂ψ ∂ψ
(ai,j
)ak,l
wdxdt
∂xk
∂xj ∂xi
∂xl
Z TZ
i,j,k,l=1 0
∂ψ ∂ψ
∂
∂w
w
(ak,l
)dxdt
∂xj ∂xi ∂xk
∂xl
ϕai,j
Ω
21
∂ψ ∂ψ
∂w ∂w
ak,l
dxdt
∂xj ∂xi
∂xl ∂xk
and we have
I3,1 = 2sλ
(2.4.76)
N
X
2
Z TZ
ϕai,j
i,j,k,l=1 0
√
2k
+sλ3 T
Ω
∂ψ ∂ψ
∂w ∂w
ak,l
dxdt
∂xj ∂xi
∂xl ∂xk
AB.
Now we directly get
N
X
(2.4.77) I3,2 = −2s3 λ4
Z TZ
i,j,k,l=1 0
and
(2.4.78)
2 2
I3,3 = −2s λ
Ω
Z TZ
ϕ
0
Ω
ϕ3 (ai,j
∂ψ ∂ψ
∂ψ ∂ψ
)(ak,l
)|w|2 dxdt,
∂xj ∂xi
∂xl ∂xk
∂η
∂ψ ∂ψ
(ai,j
)|w|2 dxdt = s2 λ2 T 2k−1 B.
∂t
∂xj ∂xi
Grouping all different terms Ik,l we obtain
Z TZ
2
0
P1 wP2 wdxdt = (sλ + λ2 T 2k )A + (s3 λ3 + s2 λ4 T 2k + s2 λ2 T 2k−1 + sT 4k−2 )B
Ω
+2sλ
N
X
2
Z TZ
ϕai,j
i,j,k,l=1 0
N
X
+2s3 λ4
Z TZ
i,j,k,l=1 0
−2sλ
N
X
+4sλ2
ϕ3 (ai,j
Ω
Z TZ
i,j,k,l=1 0
N
X
Ω
Γ
∂ψ ∂ψ
∂ψ ∂ψ
)(ak,l
)|w|2 dxdt
∂xj ∂xi
∂xl ∂xk
ϕ(∇ψ.ν)|∇w.ν|2 ai,j νi νj ak,l νk νl dγdt
Z TZ
i,j,k,l=1 0
∂ψ ∂ψ
∂w ∂w
dxdt
ak,l
∂xj ∂xi
∂xl ∂xk
ϕ(ai,j
Ω
∂ψ ∂w
∂ψ ∂w
)(ak,l
)dxdt.
∂xj ∂xi
∂xl ∂xk
We know that ∇ψ.ν < 0 on Γ so that the boundary integral (with the − sign) is
positive. Using (??) we then have, writing
A(s, λ, T ) = (sλ + λ2 T 2k )A
and
B(s, λ, T ) = (s3 λ3 + s2 λ4 T 2k + s2 λ2 T 2k−1 + sT 4k−2 )B,
Z TZ
(2.4.79)
P1 wP2 wdxdt ≥ A(s, λ, T ) + B(s, λ, T )
2
0
+2sλ2 β 2
Ω
Z TZ
0
ϕ|∇ψ|2 |∇w|2 dxdt + 2s3 λ4 β 4
Ω
Z TZ
0
22
Ω
ϕ3 |∇ψ|4 |w|2 dxdt.
Because |∇ψ| =
6 0 on Ω − ω0 (see Lemma ??), there exists δ > 0 such that
β|∇ψ| ≥ δ on Ω − ω0 ,
so that
Z TZ
P1 wP2 wdxdt + 2sλ2 δ 2
2
0
Z TZ
0
Ω
2 2
≥ 2sλ δ
Z TZ
2
ϕ|∇w|2 dxdt + 2s3 λ4 δ 4
0
ω0
3 4 4
Z TZ
ϕ|∇w| dxdt + 2s λ δ
0
Z TZ
0
Ω
ϕ3 |w|2 dxdt
ω0
ϕ3 |w|2 dxdt + A(s, λ, T ) + B(s, λ, T ).
Ω
We also have
Z TZ
0
Z TZ
2
|gs | dxdt ≤
Ω
0
e−2sη |g|2 dxdt + B(s, λ, T ),
Ω
so that
Z TZ
0
Z TZ
2
|P1 w| dxdt +
0
Ω
|P2 w|2 dxdt
Ω
Z TZ
2 2
2
0
Z TZ
0
Z TZ
ϕ|∇w| dxdt + 2s λ δ
+ 2sλ δ
≤
3 4 4
0
Ω
e−2sη |g|2 dxdt + 2sλ2 δ 2
Z TZ
0
Ω
3 4 4
Z TZ
ϕ|∇w|2 dxdt
ω0
ϕ3 |w|2 dxdt + A(s, λ, T ) + B(s, λ, T ).
+ 2s λ δ
0
ϕ3 |w|2 dxdt
Ω
ω0
Because of the form of A(s, λ, T ) and B(s, λ, T ), we can eliminate them by choosing
s and λ sufficiently large, say s ≥ (T 2k + T 2k−1 )s0 and λ ≥ λ0 where s0 and λ0 are
independent of T . Therefore we have for s ≥ (T 2k + T 2k−1 )s0 and λ ≥ λ0
Z TZ
|P1 w|2 dxdt +
(2.4.80)
0
2 2
Z TZ
+sλ δ
Z TZ
0
Ω
0
2
3 4 4
|P2 w|2 dxdt
Ω
Z TZ
ϕ3 |w|2 dxdt
ϕ|∇w| dxdt + s λ δ
0
≤
Z TZ
Ω
0
e−2sη |g|2 dxdt + 2sλ2 δ 2
Ω
0
+2s3 λ4 δ 4
Ω
Z TZ
ϕ|∇w|2 dxdt
ω0
Z TZ
0
ϕ3 |w|2 dxdt.
ω0
R R
Now we want to get rid of the term 2sλ2 δ 2 0T ω0 ϕ|∇w|2 dxdt in the right hand side
of (??). To this aim let us introduce a cut-off function θ such that
(2.4.81)
θ ∈ D(ω), 0 ≤ θ ≤ 1, θ(x) = 1 ∀x ∈ ω0 .
23
We multiply P2 w by ϕθ2 w which gives
Z TZ
0
P2 wϕθ2 wdxdt = −s
0
Ω
N
X
Z TZ
2
+
0
Ω
ai,j
i,j=1
N
X
Z TZ
Z TZ
∂η 2 2
w ϕθ dxdt +
∂t
Ω
0
∂w ∂θ
wϕθdxdt + λ
∂xj ∂xi
−s2 λ2
Ω i,j=1
N
X
Z TZ
0
0
ai,j
∂w ∂ψ
wϕθ2 dxdt
∂xj ∂xi
ϕ3 ai,j
∂ψ ∂ψ 2 2
w θ dxdt.
∂xj ∂xi
Ω i,j=1
N
X
Z TZ
∂w ∂w 2
ϕθ dxdt
∂xj ∂xi
ai,j
Ω i,j=1
Therefore, using (??) we have
Z TZ
0
Z TZ
0
ω
Z TZ
P2 wϕθ2 wdxdt + λ
ϕθ2 |∇w|2 dxdt ≤ C(
Ω
2 2
+ (s λ + sT
2k−1
0
Z TZ
3
1
1
ϕ 2 θ|∇w|ϕ 2 wdxdt
ω
2
ϕ w dxdt)
)
0
ω
We then obtain for s ≥ (T 2k + T 2k−1 )s0 and λ ≥ λ0 ,
2sλ2 δ 2
Z TZ
0
ϕ|∇w|2 dxdt ≤
ω0
1
2
Z TZ
0
|P2 w|2 dxdt + Cs3 λ4
Z TZ
0
Ω
ϕ3 w2 dxdt.
ω
From (??) this gives
Z TZ
|P1 w|2 dxdt +
(2.4.82)
0
Z TZ
0
Ω
+sλ
Z TZ
2
2
3 4
Z TZ
ϕ3 |w|2 dxdt
ϕ|∇w| dxdt + s λ
0
≤ C(
|P2 w|2 dxdt
Ω
0
0
Ω
Z TZ
e−2sη |g|2 dxdt + s3 λ4
Ω
Z TZ
0
Ω
ϕ3 |w|2 dxdt).
ω
We want to give this inequality in terms of u instead of w. We know that w = e−sη u.
Therefore we have
Z TZ
0
3
2
ϕ |w| dxdt =
Ω
Z TZ
0
ϕ3 e−2sη |u|2 dxdt
Ω
and because of
∇u = esη (∇w − sλϕ∇ψw),
Z TZ
−2sη
ϕe
0
Ω
2
|∇u| dxdt ≤ C(
Z TZ
2
2
2
Z TZ
ϕ|∇w| dxdt + s λ
0
Ω
0
24
Ω
ϕ3 |w|2 dxdt).
This immediately gives one part of (??). In order to obtain the complete inequality
(??), we use the explicit form of P1 w and P2 w which from (??) give for s ≥ (T 2k +
T 2k−1 )s0 and λ ≥ λ0
1
s
Z TZ
0
Ω
1 ∂w 2
|
| ≤ C(
ϕ ∂t
Z TZ
0
e−2sη |g|2 dxdt + s3 λ4
Z TZ
0
Ω
ϕ3 |w|2 dxdt),
ω
and
1
s
Z TZ
0
Ω
N
1 X
∂w 2
∂
(ai,j
)| dxdt ≤ C(
|
ϕ i,j=1 ∂xi
∂xj
Z TZ
Developping the expression
−2sη
e
0
2
3 4
|g| dxdt + s λ
Z TZ
0
Ω
ϕ3 |w|2 dxdt).
ω
N
X
∂
∂
w
(ai,j
( √ )) and using again estimate (??),
∂x
∂x
ϕ
i
j
i,j=1
s ≥ (T 2k + T 2k−1 )s0 and λ ≥ λ0 , we obtain
1
s
Z TZ
0
Ω
|
N
X
∂
∂
w
(ai,j
( √ ))|2 dxdt ≤ C(
∂xi
∂xj
ϕ
i,j=1
Z TZ
0
e−2sη |g|2 dxdt + s3 λ4
Z TZ
0
Ω
ϕ3 |w|2 dxdt).
ω
Using elliptic regularity results in Ω, we obtain from this last inequality
1
s
0
N
X
∂2
w
|
( √ )|2 dxdt ≤ C(
∂x
∂x
ϕ
Ω i,j=1
i
j
Z TZ
Using the development of
Z TZ
e
0
−2sη
2
3 4
|g| dxdt + s λ
Z TZ
0
Ω
ϕ3 |w|2 dxdt).
ω
w
∂2
( √ ) together with (??), s ≥ (T 2k + T 2k−1 )s0 and
∂xi ∂xj
ϕ
λ ≥ λ0 gives now
1
s
Z TZ
0
Ω
N
∂2w 2
1 X
|
(
| )dxdt ≤ C(
ϕ i,j=1 ∂xi ∂xj
Z TZ
e
0
−2sη
2
3 4
|g| dxdt + s λ
Ω
Z TZ
0
We know that
∂u
∂(esη w)
∂w
∂η
=
= esη (
+ s w)
∂t
∂t
∂t
∂t
and
∂2u
∂ 2 (esη w)
=
=
∂xi ∂xj
∂xi ∂xj
∂2w
∂ψ ∂w
∂ψ ∂w
esη (
− sλϕ(
+
)−
∂xi ∂xj
∂xj ∂xi ∂xi ∂xj
∂2ψ
∂ψ ∂ψ
∂ψ ∂ψ
sλϕ
w − sλ2 ϕ
w + s2 λ2 ϕ2
w).
∂xi ∂xj
∂xj ∂xi
∂xj ∂xi
25
ω
ϕ3 |w|2 dxdt).
We then obtain
Z T Z −2sη
e
1
s
0
ϕ
Ω
Z TZ
N
X
∂u 2
∂2u 2
| )dxdt
| +
|
∂t
∂x
∂x
i
j
i,j=1
e−2sη |g|2 dxdt + s3 λ4
≤ C(
0
(|
Z TZ
0
Ω
ϕ3 e−2sη |u|2 dxdt),
ω
with C independent of s, λ and T and this finishes the proof of Theorem ??.
2.4.3
Case of a general diffusion-convection operator
For the moment we have proved a Carleman inequality for a solution of (??) where
only operator L0 is considered. If we take the general case of operator L∗ like in
(??) we can obtain a similar inequality. Let us consider the general problem
−
(2.4.83)
(2.4.84)
∂u
+ L∗ u = h, in Ω × (0, T ),
∂t
u = 0, on Γ × (0, T ),
(2.4.85)
u(T ) = u0 , in Ω.
We write
(2.4.86)
|a0 |∞ = |a0 |L∞ (Ω) and |b|∞ = max |bi |L∞ (Ω) ,
i=1,···,N
and we have the following result corresponding to Theorem ??.
Theorem 2.4.4 Let s0 , λ0 be defined as in Theorem ??. There exists a constant
C > 0 depending on Ω, ω0 , ψ, on β defined in (??) and on coefficients ai,j such
that, if
2
3
s1 = C(1 + |a0 |∞
+ |b|2∞ ),
(2.4.87)
for every s > T 2k s1 + T 2k−1 s0 , for every λ > λ0 and for every solution of (??) we
have
(2.4.88)
1
s
Z T Z −2sη
e
0
sλ2
ϕ
Ω
Z TZ
Z
0
T
N
X
∂u 2
∂2u 2
| +
|
| )dxdt +
∂t
∂x
∂x
i
j
i,j=1
ϕe−2sη |∇u|2 dxdt + s3 λ4
Z TZ
Ω
Z
C(
0
(|
0
e−2sη |h|2 dxdt + s3 λ4
Ω
Z TZ
0
26
ω
ϕ3 e−2sη |u|2 dxdt ≤
Ω
ϕ3 e−2sη |u|2 dxdt).
Proof.
It is a simple consequence of the previous result. In fact let us define
(2.4.89)
g =h+
N
X
bi
i=1
∂u
− a0 u.
∂xi
Then the solution of (??) is solution of (??) with g defined by (??). We can then
apply Theorem ?? and obtain (??) for s ≥ (T 2k + T 2k−1 )s0 and λ ≥ λ0 .
On the other hand the value of g gives immediately
Z TZ
0
Z TZ
e−2sη |g|2 dxdt ≤ C(
Ω
0
|b|2∞ T 2k
Z TZ
0
Ω
e−2sη |h|2 dxdt +
Ω
ϕe−2sη |∇u|2 dxdt + |a0 |2∞ T 6k
Z TZ
0
ϕ3 e−2sη |u|2 dxdt),
Ω
and therefore
1
s
Z T Z −2sη
e
sλ
0
2
ϕ
Ω
Z TZ
(|
−2sη
ϕe
Z
0
T
2
3 4
|∇u| dxdt + s λ
Z TZ
Ω
Z
C(
0
N
X
∂2u 2
∂u 2
|
| +
| )dxdt +
∂t
∂xi ∂xj
i,j=1
0
e−2sη |h|2 dxdt + s3 λ4
Ω
C|b|2∞ T 2k
Z TZ
0
Z TZ
0
Ω
ϕ3 e−2sη |u|2 dxdt ≤
Ω
ϕ3 e−2sη |u|2 dxdt) +
ω
ϕe−2sη |∇u|2 dxdt + C|a0 |2∞ T 6k
Z TZ
0
ϕ3 e−2sη |u|2 dxdt.
Ω
Therefore by choosing
(2.4.90)
1
2
3
s1 = s0 + C 3 |a0 |∞
+ C|b|2∞ ,
taking s ≥ T 2k s1 + T 2k−1 s0 and λ ≥ λ0 , we can absorb both of the two last terms
in the right hand side of the previous inequality by the left hand side. This gives
immediately the result of Theorem ??.
2.5
Observability inequality
In order to complete the proof of Theorem ?? which gives the result of exact controllability to trajectories, we need to prove the observability inequality (??) with a
suitable weight function ρ.
27
Directly from (??) we have for s ≥ T 2k s1 + T 2k−1 s0 and λ ≥ λ0 ,
(2.5.91)
s3 λ4
Z TZ
0
ϕ3 e−2sη |u|2 dxdt ≤ C(
Z TZ
0
Ω
3 4
e−2sη |h|2 dxdt
Ω
Z TZ
ϕ3 e−2sη |u|2 dxdt).
+ s λ
0
ω
Now we can fix parameters s and λ in the admissible range and for example we take,
in order to simplify notations, for C large enough,
2
3
s = CT 2k ((1 + |a0 |∞
+ |b|2∞ +
1
), λ = λ0 .
T
We shall now only keep track of the dependence of the constants on T , |a0|∞ and
|b|∞ .
On a time interval ( T4 , 3T
4 ), we have
η≤
2
C
3 +|b|2 + 1 )
−2sη
−C(1+|a0 |∞
∞ T
,
so
that
e
≥
e
2k
T
and
ϕ≥
C
.
T 2k
From the last inequality (??), we have
Z
(2.5.92)
3T
4
T
4
Z
2
|u| dxdt ≤ C(T, |a0 |∞ , |b|∞ )
Z TZ
0
Ω
+C(T, |a0 |∞ , |b|∞ )T 6k
Z TZ
0
Ω
ϕ3 e−2sη |u|2 dxdt,
ω
where
(2.5.93)
e−2sη |h|2 dxdt
2
3
2
1
C(T, |a0 |∞ , |b|∞ ) = CeC(1+|a0 |∞ +|b|∞ + T )
Let us consider a cut-off function of time t θ such that
dθ
C
(t)| ≤ , ∀t ∈ [0, T ],
dt
T
T
3T
θ(t) = 1 ∀t ∈ [0, ], θ(t) = 0 ∀t ∈ [ , T ].
2
4
θ ∈ C ∞ ([0, T ]), 0 ≤ θ(t) ≤ 1, |
We define
ũ(x, t) = θ(t)u(x, t).
28
Then ũ satisfies the following equation
−
(2.5.94)
(2.5.95)
∂ ũ
dθ
+ L∗ ũ = θh − u, in Ω × (0, T ),
∂t
dt
ũ = 0, on Γ × (0, T ),
(2.5.96)
ũ(T ) = 0, in Ω.
We now use a classical energy estimate for ũ. We have to notice that ũ(x, t) = u(x, t)
T
3T
on [0, T2 ] and that dθ
dt = 0 on [0, 2 ] ∪ [ 4 , T ]. Multiplying the previous equation by
ũ and integrating over (t, T ) for t ∈ (0, T ), we easily find that
Z TZ
|ũ(t)|2L2 (Ω) + β
+C(|a0 |∞ +
t
|∇ũ|2 dxdt ≤ C
Ω
C
|ũ| dxdt + 2
T
Ω
t
Z
t
Z TZ
|b|2∞ )
3T
4
Z
2
3T
4
Z
|h|2 dxdt
Ω
Z
T
2
|u|2 dxdt.
Ω
From Gronwall’s Lemma, we therefore have
|ũ(t)|2L2 (Ω)
C(|a0 |∞ +|b|2∞ )T
≤ Ce
Z
(
3T
4
0
1
|h| dxdt + 2
T
Ω
Z
2
Z
3T
4
T
2
Z
|u|2 dxdt).
Ω
Using (??) we obtain in particular
Proposition 2.5.1 For every solution u of (??)-(??), we have the following observability inequality.
(2.5.97)
|u(0)|2L2 (Ω)
C(|a0 |∞ +|b|2∞ )T
≤ Ce
Z
(
3T
4
Z
0
Z TZ
+C(T, |a0 |∞ , |b|∞ )(
−2sη
e
0
|h|2 dxdt
Ω
2
|h| dxdt + T
6k−2
Ω
Z TZ
0
ϕ3 e−2sη |u|2 dxdt)),
ω
where C(T, |a0 |∞ , |b|∞ ) is defined by (??).
Applying (??) to ξ which is solution to (??) or to (??) with h = 0, we easily obtain
one part of the desired observability inequality (??), namely
(2.5.98)
|ξ (0)|2L2 (Ω)
≤ CT
6k−2 K(T,|a0 |∞ ,|b|∞ )
Z TZ
e
0
ϕ3 e−2sη |ξ |2 dxdt).
ω
where
2
3
(2.5.99) K(T, |a0 |∞ , |b|∞ ) = C(1 + |a0 |∞
+ |b|2∞ + T |a0 |∞ + T |b|2∞ +
29
1
).
T
This completes the proof of Theorem ?? in the case g 0 = 0 as it is announced. Here
we have been keeping track of the dependence of the constants on T , |a0 |∞ and |b|∞
in order to treat nonlinear problemes later on.
Let us now for simplicity forget about this dependence and define
(
η(x, t) if t ∈ [ T2 , T ],
η(x, T2 ) if t ∈ [0, T2 ]
(
ϕ(x, t) if t ∈ [ T2 , T ],
ϕ(x, T2 ) if t ∈ [0, T2 ].
η̃(x, t) =
(2.5.100)
and
(2.5.101)
ϕ̃(x, t) =
We obtain from the above inequalities and from Carleman estimate
|u(0)|2L2 (Ω)
(2.5.102)
≤C
Z TZ
0
Z TZ
ϕ̃3 e−2sη̃ |u|2 dxdt
+
e−2sη̃ |h|2 dxdt + C
Z
0
T
0
Ω
Ω
Z
ϕ̃3 e−2sη̃ |u|2 dxdt.
ω
This gives the desired observability inequality (??) with the weight
ρ=
esη̃
3
.
ϕ̃ 2
This completes the proof of Theorem ?? in the case we have a right hand side g 0
satisfying
esη̃ 0
2
2
3 g ∈ L (0, T ; L (Ω)).
ϕ̃ 2
2.6
Another strategy. Complementary results
In this section we will give another strategy to prove exact controllability to trajectories. This strategy is again based on the global Carleman estimate obtained
above, and it will enable us to improve the controllability results. Roughly speaking
we will show that we can obtain exact controllability to trajectories with a control
which is exponentially decreasing near time T , the corresponding state being also
exponentially decreasing when t → T .
From now on in this section we fix the parameters s and λ such that
s > T 2k s1 + T 2k−1 s0 and λ > λ0
30
where s0 , s1 and λ0 are defined in Theorem ?? and Theorem ?? so that the Carleman
estimate (??) is valid.
Moreover we have previously defined η̃ and ϕ̃ in (??) and (??) and we have shown
above that inequality (??) is valid. We consider the general linear problem
∂z
+ Lz = g 0 + v.χω , in Ω × (0, T ),
∂t
z = 0, on Γ × (0, T ),
(2.6.103)
(2.6.104)
z(x, 0) = z 0 (x), in Ω,
(2.6.105)
We look for v such that
(2.6.106)
z(T ) = 0.
We will assume that
Z T Z 2sη̃
e
(2.6.107)
0
Ω
ϕ̃3
|g 0 |2 dxdt < +∞
We will prove the following result.
Theorem 2.6.1 Under the previous hypotheses (??), (??), (??), for every open
subset ω of Ω, for every time T > 0 and for every z 0 ∈ L2 (Ω), if g 0 satisfies (??),
there exists a control v and the corresponding solution z of (??)-(??) such that (??)
holds and
Z TZ
(2.6.108)
e2sη̃ |z|2 dxdt +
H(z, v) =
0
Z T Z 2sη̃
e
Ω
0
ω
ϕ̃3
|v|2 dxdt < +∞.
Moreover, v (and therefore z) can be chosen such that H(z, v) is minimum (among
admissible functions).
Notice that the weight e2sη̃ tends exponentially to +∞ when t → T , so that z and
v must tend exponentially to zero when t → T .
Proof. For > 0 let us consider the functional
1
H (z, v) = |z(T )|2L2 (Ω) +
Z TZ
e
0
2sη̃
2
|z| dxdt +
Ω
Z T Z 2sη̃
e
0
ω
ϕ̃3
|v|2 dxdt.
We first study the optimal control problem : find (z , v ) ∈ W such that
(2.6.109)
H (z , v ) = min H (z, v)
(z,v)∈W
where W is the set of functions (z, v) such that H(z, v) < ∞ and z is solution of
(??)-(??).
31
Notice that when z is solution of (??)-(??), then z(T ) ∈ L2 (Ω). It is not a priori
clear that W is not empty. But the set {(z, v), H(z, v) < +∞} is a Hilbert space
and if W is not empty, it is a closed convex set in this space, so that problem (??)
has a unique solution.
Take a small parameter δ > 0 and define
(
(2.6.110)
δ
g (x, t) =
g 0 (x, t) if t ∈ [0, T − 2δ],
0 if t ∈ [T − 2δ, T ]
We define the problem
(2.6.111)
(2.6.112)
∂z
+ Lz = g δ + v.χω , in Ω × (0, T ),
∂t
z = 0, on Γ × (0, T ),
z(x, 0) = z 0 (x), in Ω,
(2.6.113)
First of all we consider the null controllability problem for (??)-(??) on the time
interval (0, T − δ). As g δ = 0 on the interval (T − 2δ, T − δ) we know from the
results of the previous sections that there exists a control v δ ∈ L2 (0, T − δ; L2 (ω))
such that the corresponding solution z δ of (??)-(??) satisfies z δ (T − δ) = 0. Let
us extend v δ by 0 on the interval (T − δ, T ). Then z δ is also extended by 0 on the
same interval and is solution of (??)-(??) on the time interval (0, T ) with of course
z δ (T ) = 0. Then it is clear that H(z δ , v δ ) < +∞. Let us define W δ as the set of
functions (z, v) such that H(z, v) < +∞ and z is solution of (??)-(??) on (0, T ).
We now study the optimal control problem : find (zδ , vδ ) ∈ W δ such that
H (zδ , vδ ) =
(2.6.114)
min
(z,v)∈W δ
H (z, v)
As (z δ , v δ ) ∈ W δ , W δ is not empty and it is closed and convex in {(z, v), H(z, v) <
+∞}. Therefore, from standard arguments, there exists a unique solution (zδ , vδ ) of
problem (??). We can write the optimality system for this problem corresponding
to the first order optimality conditions, using an adjoint state pδ . We obtain
(2.6.115)
(2.6.116)
∂zδ
+ Lzδ = g δ + vδ .χω , in Ω × (0, T ),
∂t
zδ = 0, on Γ × (0, T ),
zδ (x, 0) = z 0 (x), in Ω,
(2.6.117)
(2.6.118)
(2.6.119)
(2.6.120)
−
∂pδ
+ L ∗ pδ = e2sη̃ zδ , in Ω × (0, T ),
∂t
pδ = 0, on Γ × (0, T ),
1 δ
pδ (x, T ) =
z (x, T ), in Ω,
32
pδ +
(2.6.121)
e2sη̃ δ
v = 0 in ω × (0, T ).
ϕ̃3 Multiplying equation (??) by zδ and using (??) we obtain
Z TZ
1 δ
|z (T )|2L2 (Ω) +
0
Ω
e2sη̃ |zδ |2 dxdt =
Z TZ
0
Z TZ
Z
Ω
g δ pδ dxdt+
pδ (0)z 0 dx+
Ω
0
ω
vδ pδ dxdt.
Because of (??) this gives
H (zδ , vδ )
1
= |zδ (T )|2L2 (Ω) +
Z TZ
2sη̃
e
0
Ω
|zδ |2 dxdt
Z TZ
(2.6.122)
=
0
Z T Z 2sη̃
e
≤(
0
ϕ̃3
Ω
1
Z TZ
|g δ |2 dxdt + |z 0 |2L2 (Ω) ) 2 (
0
Ω
Ω
+
Z T Z 2sη̃
e
0
ω
g δ pδ dxdt +
ϕ̃3
Z
Ω
|vδ |2 dxdt
pδ (0)z 0 dx
1
ϕ̃3 e−2sη̃ |pδ |2 dxdt + |pδ (0)|2L2 (Ω) ) 2 .
First of all we notice that
∀δ > 0,
Z T Z 2sη̃
e
0
ϕ̃3
Ω
δ 2
|g | dxdt ≤
Z T Z 2sη̃
e
0
Ω
ϕ̃3
|g 0 |2 dxdt.
Moreover from the observability inequality (??), (??) and (??) we have
Z TZ
3 −2sη̃
ϕ̃ e
0
Ω
Z TZ
≤ C(
0
Ω
|pδ |2 dxdt
+
|pδ (0)|2L2 (Ω)
e2sη̃ |zδ |2 dxdt +
ω
ϕ̃3
2sη̃
≤ C(
Z T Z 2sη̃
e
0
Z TZ
e
0
Ω
|zδ |2 dxdt
Z TZ
+
0
ω
ϕ̃3 e−2sη̃ |pδ |2 dxdt)
|vδ |2 dxdt)
≤ CH (zδ , vδ ).
Therefore we obtain
(2.6.123) ∀ > 0, ∀δ > 0, H (zδ , vδ ) ≤ C(
Z T Z 2sη̃
e
0
Ω
ϕ̃3
|g 0 |2 dxdt + |z 0 |2L2 (Ω) ).
In particular this implies that, independently of δ (and ) the quantities
Z T Z 2sη̃
e
0
ω
|v δ |2 dxdt
ϕ̃3 Z TZ
and
0
Ω
e2sη̃ |zδ |2 dxdt
are bounded. Let us fix for the moment and let δ tend to 0. We can extract
subsequences such that
vδ * v and zδ * z
33
such that
Z T Z 2sη̃
e
0
gδ
ω
g0
ϕ̃3
2
|v | dxdt < +∞ and
Z TZ
0
e2sη̃ |z |2 dxdt < +∞.
Ω
L2 (0, T ; L2 (Ω))
As
→
in
it is clear from (??) that z is solution of (??)-(??)
with control v . Therefore using the lower semicontinuity for the weak topology of
the functional H we have
(2.6.124)
Z T Z 2sη̃
e
H (z , v ) ≤ C(
0
Ω
ϕ̃3
|g 0 |2 dxdt + |z 0 |2L2 (Ω) ) < +∞
and (z , v ) ∈ W. Then problem (??) has a unique solution and we call again this
solution (z , v ) which of course satisfies (??).
Now we let tend to 0. Again we can extract subsequences such that
v * v , z * z and z (T ) → 0 in L2 (Ω)
with
Z T Z 2sη̃
e
Z TZ
e2sη̃ |z|2 dxdt < +∞
ϕ̃3
0
Ω
and z is solution of (??)-(??) with control v. Therefore z verifies
0
|v|2 dxdt < +∞ and
ω
z(T ) = 0
which is (??). We have
H (z , v ) ≥ H(z , v )
and because of the lower semicontinuity for the weak topology of H
lim inf H(z , v ) ≥ H(z, v).
→0
On the other hand, because of (??) and (??), we have
H (z , v ) ≤ H (z, v) = H(z, v).
This shows that
H (z , v ) → H(z, v)
and that (z, v) realizes the minimum of H among functions (z̃, ṽ) satisfying (??)(??), (??) such that H(z̃, ṽ) < +∞. The proof of Theorem ?? is then complete.
2.7
Some semilinear operators
2.8
The case of Navier-Stokes equations
2.9
Some comments
34
Chapter 3
Data assimilation problems
35
Chapter 4
Some inverse problems for the
wave equation
36
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