Inverse problems for linear hyperbolic equations using mixed formulations

2.2 MB·English

by  Nicolae Cindea

Inverse problems for linear hyperbolic equations
using mixed formulations
Nicolae Cˆındea∗ Arnaud Mu¨nch†
5
1 February 3, 2015
0
2

n
a
Abstract
J
1 Weintroduceinthisdocumentadirectmethodallowingtosolvenumericallyinversetype
3 problems for linear hyperbolic equations. We first consider the reconstruction of the full
solutionofthewaveequationposedinΩ×(0,T)-ΩaboundedsubsetofRN -fromapartial

C] distributed observation. We employ a least-squares technique and minimize the L2-norm of
the distance from the observation to any solution. Taking the hyperbolic equation as the
O
mainconstraintoftheproblem,theoptimalityconditionsarereducedtoamixedformulation
.
h involving both the state to reconstruct and a Lagrange multiplier. Under usual geometric
at optic conditions, we show the well-posedness of this mixed formulation (in particular the
m inf-sup condition) and then introduce a numerical approximation based on space-time finite
elements discretization. We prove the strong convergence of the approximation and then
[
discussed several examples for N =1 and N =2. The problem of the reconstruction of both
1
the state and the source term is also addressed.
v
4 keywords :Linear wave equation, Inverse problem, Finite elements methods; Mixed formula-
1
tion
1
0 AMS number: 35L10, 65M12, 93B40.
0
.
2
1 Introduction - Inverse problems for hyperbolic equations
0
5
1 Let Ω be a bounded domain of RN (N ≥ 1) whose boundary ∂Ω is Lipschitz and let T > 0. We
:
v note QT := Ω×(0,T) and ΣT := ∂Ω×(0,T). We are concerned in this work with inverse type
i problems for linear hyperbolic equation of the following type
X
r 
a  ytt−∇·(c(x)∇y)+d(x,t)y =f, (x,t)∈QT
y =0, (x,t)∈Γ (1)
T
 (y(·,0),y (·,0))=(y ,y ), x∈Ω.
t 0 1
We assume that c ∈ C1(Ω,R) with c(x) ≥ c > 0 in Ω, d ∈ L∞(Q ), (y ,y ) ∈ H := L2(Ω)×
0 T 0 1
H−1(Ω) and f ∈X :=L2(0,T;H−1(Ω)).
For any (y ,y ) ∈ H and any f ∈ X, there exists exactly one solution y to (1), with y ∈
0 1
C0([0,T];L2(Ω))∩C1([0,T];H−1(Ω)) (see [20]).
In the sequel, for simplicity, we shall use the following notation:
Ly :=y −∇·(c(x)∇y)+d(x,t)y. (2)
tt
∗LaboratoiredeMath´ematiques,Universit´eBlaisePascal(Clermont-Ferrand2),UMRCNRS6620,Campusde
C´ezeaux,63177,Aubi`ere,France. E-mails: [email protected].
†LaboratoiredeMath´ematiques,Universit´eBlaisePascal(Clermont-Ferrand2),UMRCNRS6620,Campusde
C´ezeaux,63177,Aubi`ere,France. E-mail: [email protected].
1
1 INTRODUCTION - INVERSE PROBLEMS FOR HYPERBOLIC EQUATIONS 2
and X(cid:48) :=L2(0,T;H1(Ω)).
0
Letnowω beanynonemptyopensubsetofΩandletq :=ω×(0,T)⊂Q . Atypicalinverse
T T
problem for (1) is the following one : from an observation or measurement y in L2(q ) on the
obs T
sub-domainq ,wewanttorecoverasolutiony oftheboundaryvalueproblem(1)whichcoincides
T
with the observation on q .
T
Introducing the operator P :L2(Q )→X ×L2(q ) defined by P y :=(Ly,y ), the problem
T T |qT
is reformulated as :
find y ∈L2(Q ) solution of P y =(f,y ). (IP)
T obs
Fromtheuniquecontinuationpropertyfor(1), ifthesetq satisfiessomegeometricconditions
T
and if y is a restriction to q of a solution of (1), then the problem is well-posed in the sense
obs T
that the state y corresponding to the pair (y ,f) is unique.
obs
Inviewoftheunavoidableuncertaintiesonthedatay (comingfrommeasurements,numerical
obs
approximations,etc),theproblemneedstoberelaxed. Inthisrespect,themostnatural(andwidely
used in practice) approach consists to introduce the following extremal problem (of least-squares
type)
 1
minimize over H J(y ,y ):= (cid:107)y−y (cid:107)2
0 1 2 obs L2(qT) (LS)
where y solves (1),
sinceyisuniquelyandfullydeterminedfromf andthedata(y ,y ). Heretheconstrainty−y =
0 1 obs
0inL2(q )isrelaxed;however,ify isarestrictiontoq ofasolutionof(1),thenproblems(LS)
T obs T
and (IP) obviously coincide. A minimizing sequence for J in H is easily defined in term of the
solution of an auxiliary adjoint problem. Apart from a possible low decrease of the sequence near
extrema,themaindrawback,whenonewantstoprovetheconvergenceofadiscreteapproximation
isthat, itisingeneralnotpossibletominimizeoveradiscretesubspaceof{y;Ly−f =0}subject
to the equality (in X) Ly − f = 0. Therefore, the minimization procedure first requires the
discretization of the functional J and of the system (1); this raised the issue of uniform coercivity
property (typically here some uniform discrete observability inequality for the adjoint solution)
of the discrete functional with respect to the approximation parameter. As far as we know, this
delicate issue has received answers only for specific and somehow academic situations (uniform
Cartesian approximation of Ω, constant coefficients in (1)). We refer to [12, 17, 19, 22] and the
references therein.
More recently, a different method to solve inverse type problems like (IP) has emerged and
use so called Luenberger type observers: this consists in defining, from the observation on q ,
T
an auxiliary boundary value problem whose solution possesses the same asymptotic behavior in
time than the solution of (1): the use of the reversibility of the hyperbolic equation then allows
to reconstruct the initial data (y ,y ). We refer to [8, 24] and the references therein. But, for
0 1
the same reasons, on a numerically point of view, these method require to prove uniform discrete
observability properties.
Inaseriesofworks,Klibanovandco-workersusedifferentapproachestosolveinverseproblems
(we refer to [18] and the references therein): they advocate in particular the quasi-reversibility
method which reads as follows : for any ε>0, find y ∈A the solution of
ε
(cid:104)Py ,Py(cid:105) +ε(cid:104)y ,y(cid:105) =(cid:104)(f,y ),Py(cid:105) , (QR)
ε X×L2(qT) ε A obs X(cid:48)×L2(qT),X×L2(qT)
for all y ∈ A, where A denotes a Hilbert space subset of L2(Q ) so that Py ∈ X ×L2(q ) for
T T
all y ∈ A and ε > 0 a Tikhonov like parameter which ensures the well-posedness. We refer for
instanceto[13]wherethelateralCauchyproblemforthewaveequationwithnonconstantdiffusion
isaddressedwithinthismethod. Remarkthat(QR)canbeviewedasaleast-squaresproblemsince
thesolutiony minimizesoverAthefunctionaly →(cid:107)Py−(f,y )(cid:107)2 +ε(cid:107)y(cid:107)2. Eventually,
ε obs X×L2(qT) A
if y is a restriction to q of a solution of (1), the corresponding y converges in L2(Q ) toward
obs T ε T
2 RECOVERING THE SOLUTION FROM A PARTIAL OBSERVATION 3
tothesolutionof (IP)asε→0. There, unlikeinProblem(LS), theunknownisthestatevariable
y itself (as it is natural for elliptic equations) so that any standard numerical methods based on a
conformal approximation of the space A together with appropriate observability inequalities allow
to obtain a convergent approximation of the solution. In particular, there is no need to prove
discreteobservabilityinequalities. Werefertothebook[2]. Wealsomention[6,5]whereasimilar
technique has been used recently to solve the inverse obstacle problem associated to the Laplace
equation, which consists in finding an interior obstacle from boundary Cauchy data.
Inthespiritoftheworks[18,6,13],weexplorethedirectresolutionoftheoptimalityconditions
associated to the extremal problem (LS), without Tikhonov parameter while keeping y as the
unknown of the problem. This strategy, which avoids any iterative process, has been successfully
applied in the closed context of the exact controllability of (1) in [12] and [7, 10]. The idea is to
takeintoaccountthestateconstraintLy−f =0withaLagrangemultiplier. Thisallowstoderive
explicitly the optimality systems associated to (LS) in term of an elliptic mixed formulation and
therefore reformulate the original problem. Well-posedness of such new formulation is related to
an observability inequality for the homogeneous solution of the hyperbolic equation.
The outline of this paper is as follow. In Section 2, we consider the least-squares problem (P)
and reconstruct the solution of the wave equation from a partial observation localized on a subset
q ofQ . Forthat, inSection2.1, weassociateto(P)theequivalentmixedformulation(7)which
T T
reliesontheoptimalityconditionsoftheproblem. Assumingthatq satisfiestheclassicalgeometric
T
opticcondition(Hypothesis1,see(H)),wethenshowthewell-posednessofthismixedformulation,
in particular, we check the Babuska-Brezzi inf-sup condition (see Theorem 2.1). Interestingly, in
Section 2.2, we also derive a equivalent dual extremal problem, which reduces the determination
ofthestatey totheminimizationofanellipticfunctionalwithrespecttotheLagrangemultiplier.
In Section 3, we apply the same procedure to recover from a partial observation both the state
and the source term. Section 4 is devoted to the numerical approximation, through a conformal
space-time finite element discretization. The strong convergence of the approximation (y ,f ) is
h h
shownasthediscretizationparameterhtendstozero. Inparticular,wediscussthediscreteinf-sup
property of the mixed formulation. We present numerical experiments in Section 5 for Ω = (0,1)
and Ω ⊂ R2, in agreement with the theoretical part. We consider in particular time dependent
observation zones. Section 6 concludes with some perspectives.
2 Recovering the solution from a partial observation: a
mixed re-formulation of the problem
Inthissection,assumingthattheinitial(y ,y )∈H areunknown,weaddresstheinverseproblem
0 1
(IP). Without loss of generality, in view of the linearity of the system (1), we assume that the
source term f ≡0.
We consider the non empty vectorial space Z defined by
Z :={y :y ∈C([0,T],L2(Ω))∩C1([0,T],H−1(Ω)),Ly ∈X}. (3)
and then introduce the following hypothesis :
Hypothesis 1 There exists a constant C =C(ω,T,(cid:107)c(cid:107) ,(cid:107)d(cid:107) ) such that the following
obs C1(Ω) L∞(Ω)
estimate holds :
(cid:18) (cid:19)
(cid:107)y(·,0),y (·,0)(cid:107)2 ≤C (cid:107)y(cid:107)2 +(cid:107)Ly(cid:107)2 , ∀y ∈Z. (H)
t H obs L2(qT) X
Condition (H) is a generalized observability inequality for the solution of the hyperbolic equa-
tion: for constant coefficients, this estimate is known to hold if the triplet (ω,T,Ω) satisfies a
2 RECOVERING THE SOLUTION FROM A PARTIAL OBSERVATION 4
geometric optic condition. We refer to [1]. In particular, T should be large enough. Upon the
same condition, (H) also holds in the non-cylindrical situation where the domain ω varies with
respecttothetimevariable: wereferto[7]fortheonedimensionalcase. Fornonconstantvelocity
c and potential d, we refer to [10] and the references therein.
Then, within this hypothesis, for any η >0, we define on Z the bilinear form
(cid:90)(cid:90) (cid:90) T
(cid:104)y,y(cid:105) := yydxdt+η (cid:104)Ly, Ly(cid:105) dt ∀y,y ∈Z. (4)
Z H−1(Ω)
qT 0
In view of (H), this bilinear form defines a scalar product over Z. Moreover, endowed to this
scalar product, we easily obtain that Z is a Hilbert space (see [7], Corollary 2.4). We note the
(cid:112)
corresponding norm by (cid:107)y(cid:107) := (cid:104)y,y(cid:105) .
Z Z
Then, we consider the following extremal problem :
 1
 infJ(y):= (cid:107)y−y (cid:107)2 ,
2 obs L2(qT) (P)
subject to y ∈W
where W is the closed subspace of Z defined by
W :={y ∈Z; Ly =0 in X}
and endowed with the norm of Z.
The extremal problem (P) is well posed : the functional J is continuous over W, is strictly
convex and is such that J(y)→+∞ as (cid:107)y(cid:107) →∞. Note also that the solution of (P) in W does
W
not depend on η.
RemindthatfromthedefinitionofZ,LybelongstoX. Similarly,theuniquenessofthesolution
is lost if the hypothesis (H) is not fulfilled, for instance if T is not large enough. Eventually, from
(H), the solution y in Z of (P) satisfies (y(·,0),y (·,0))∈H, so that problem (P) is equivalent to
t
the minimization of J with respect to (y ,y )∈H as in problem (IP), Section 1.
0 1
We also recall that for any z ∈Z there exists a positive constant C such that
Ω,T
(cid:18) (cid:19)
(cid:107)z(cid:107)2 ≤C (cid:107)z(·,0),z (·,0)(cid:107)2 +(cid:107)Lz(cid:107)2 . (5)
L2(QT) Ω,T t H X
This equality and (H) imply that
(cid:18) (cid:19)
(cid:107)z(cid:107)2 ≤C C (cid:107)z(cid:107)2 +(1+C )(cid:107)Lz(cid:107)2 , ∀z ∈Z. (6)
L2(QT) Ω,T obs L2(qT) obs X
2.1 Direct approach
In order to solve (P), we have to deal with the constraint equality which appears in the space
W. Proceeding as in [12], we introduce a Lagrangian multiplier λ ∈ X(cid:48) and the following mixed
formulation: find (y,λ)∈Z×X(cid:48) solution of
(cid:40)
a(y,y)+b(y,λ) = l(y), ∀y ∈Z
(7)
b(y,λ) = 0, ∀λ∈X(cid:48),
where
(cid:90)(cid:90)
a:Z×Z →R, a(y,y):= yydxdt, (8)
qT
(cid:90) T
b:Z×X(cid:48) →R, b(y,λ):= (cid:104)λ, Ly(cid:105) dt, (9)
H1(Ω),H−1(Ω)
0
0
(cid:90)(cid:90)
l:Z →R, l(y):= y ydxdt. (10)
obs
qT
2 RECOVERING THE SOLUTION FROM A PARTIAL OBSERVATION 5
System (7) is nothing else than the optimality system corresponding to the extremal problem (P).
Precisely, the following result holds :
Theorem 2.1 Under the hypothesis (H),
(i) The mixed formulation (7) is well-posed.
(ii) The unique solution (y,λ) ∈ Z × X(cid:48) to (7) is the unique saddle-point of the Lagrangian
L:Z×X(cid:48) →R defined by
1
L(y,λ):= a(y,y)+b(y,λ)−l(y).
2
(iii) We have the estimate
(cid:112)
(cid:107)y(cid:107) =(cid:107)y(cid:107) ≤(cid:107)y (cid:107) , (cid:107)λ(cid:107) ≤2 C +η(cid:107)y (cid:107) . (11)
Z L2(qT) obs L2(qT) X(cid:48) Ω,T obs L2(qT)
Proof- We use classical results for saddle point problems (see [4], chapter 4).
We easily obtain the continuity of the bilinear form a over Z ×Z, the continuity of bilinear b
over Z×X(cid:48) and the continuity of the linear form l over Z. In particular, we get
(cid:107)l(cid:107) =(cid:107)y (cid:107) , (cid:107)a(cid:107) =1, (cid:107)b(cid:107) =η−1/2. (12)
Z(cid:48) obs L2(qT) (Z×Z)(cid:48) (Z×X(cid:48))(cid:48)
Moreover, the kernel N(b)={y ∈Z; b(y,λ)=0 ∀λ∈X(cid:48)} coincides with W: we easily get
a(y,y)=(cid:107)y(cid:107)2, ∀y ∈N(b)=W.
Z
Therefore,inviewof[4,Theorem4.2.2],itremainstochecktheinf-supconstantproperty: ∃δ >0
such that
b(y,λ)
inf sup ≥δ. (13)
λ∈X(cid:48)y∈Z (cid:107)y(cid:107)Z(cid:107)λ(cid:107)X(cid:48)
We proceed as follows. For any fixed λ∈X(cid:48), we define y as the unique solution of
Ly =−∆λ in Q , (y(·,0),y (·,0))=(0,0) on Ω, y =0 on Σ . (14)
T t T
We get b(y,λ)=(cid:107)λ(cid:107)2 and
X(cid:48)
(cid:107)y(cid:107)2 =(cid:107)y(cid:107)2 +η(cid:107)λ(cid:107)2 .
Z L2(qT) X(cid:48)
(cid:112)
Using (5), the estimate (cid:107)y(cid:107) ≤ C (cid:107)λ(cid:107) implies that y ∈Z and that
L2(qT) Ω,T X(cid:48)
b(y,λ) 1
sup ≥ >0
(cid:112)
y∈Z (cid:107)y(cid:107)Z(cid:107)λ(cid:107)X(cid:48) CΩ,T +η
leading to the result with δ =(C +η)−1/2.
Ω,T
The third point is the consequence of classical estimates (see [4], Theorem 4.2.3.) :
(cid:18) (cid:19)
1 1 (cid:107)a(cid:107)
(cid:107)y(cid:107) ≤ (cid:107)l(cid:107) , (cid:107)λ(cid:107) ≤ 1+ (cid:107)l(cid:107)
Z α Z(cid:48) X(cid:48) δ α Z(cid:48)
0 0
where
a(y,y)
α := inf . (15)
0 y∈N(b) (cid:107)y(cid:107)2Z
Estimates (12) and the equality α =1 lead to the results. Eventually, from (12), we obtain that
0
2
(cid:107)λ(cid:107) ≤ (cid:107)y (cid:107)
X(cid:48) δ obs L2(qT)
2 RECOVERING THE SOLUTION FROM A PARTIAL OBSERVATION 6
and that δ ≥(C +η)−1/2 to get (11). (cid:50)
Ω,T
In practice, it is very convenient to ”augment” the Lagrangian (see [16]) and consider instead
the Lagrangian L defined for any r >0 by
r
1
L (y,λ):= a (y,y)+b(y,λ)−l(y),
r 2 r
a (y,y):=a(y,y)+r(cid:107)Ly(cid:107)2 .
r X
Sincea (y,y)=a(y,y)onW, theLagrangianLandL sharethesamesaddle-point. Thepositive
r r
number r is an augmentation parameter.
Remark 1 Assuming additional hypotheses on the regularity of the solution λ, precisely Lλ ∈
L2(Q ) and (λ,λ ) ∈ H1(Ω)×L2(Ω), we easily prove, writing the optimality condition for
T t |t=0,T 0
L, that the multiplier λ satisfies the following relations :
(cid:40)
Lλ=−(y−y )1 in Q , λ=0 in Σ ,
obs ω T T
(16)
λ=λ =0 on Ω×{0,T}.
t
Therefore,λ(definedintheweaksense)isanexactcontrolledsolutionofthewaveequationthrough
the control −(y−y )1 ∈L2(q ).
obs ω T
• If y is the restriction to q of a solution of (1), then the unique multiplier λ must van-
obs T
ish almost everywhere. In that case, we have sup inf L (y,λ) = inf L (y,0) =
λ∈Λ y∈Y r y∈Y r
inf J (y) with
y∈Y r
1 r
J (y):= (cid:107)y−y (cid:107)2 + (cid:107)Ly(cid:107)2 . (17)
r 2 obs L2(QT) 2 X
The corresponding variational formulation is then : find y ∈Z such that
(cid:90)(cid:90) (cid:90) T
a (y,y)= yydxdt+r (cid:104)Ly, Ly(cid:105) dt=l(y), ∀y ∈Z.
r H−1(Ω)
qT 0
• In the general case, the mixed formulation can be rewritten as follows: find (z,λ) ∈ Z ×X(cid:48)
solution of
(cid:40)
(cid:104)P y,P y(cid:105) +(cid:104)Ly,λ(cid:105) =(cid:104)(0,y ),P y(cid:105) , ∀y ∈Z,
r r X×L2(qT) X,X(cid:48) obs r X×L2(qT) (18)
(cid:104)Ly,λ(cid:105) =0, ∀λ∈X(cid:48)
X,X(cid:48)
√
with P y :=( rLy,y ). This approach may be seen as generalization of the (QR) problem
r |qT
(see (QR)), where the variable λ is adjusted automatically (while the choice of the parameter
ε in (QR) is in general a delicate issue).
System (16) can be used to define a equivalent saddle-point formulation, very suitable at the
numerical level. Precisely, we introduce - in view of (16) - the space Λ by
Λ:={λ:λ∈C([0,T];H1(Ω))∩C1([0,T];L2(Ω)),
0
Lλ∈L2(Q ),λ(·,0)=λ (·,0)=0}.
T t
(cid:90)(cid:90)
Endowed with the scalar product (cid:104)λ,λ(cid:105) := (λλ+LλLλ)dxdt, we check that Λ is a Hilbert
Λ
QT
space. Then, for any parameter α ∈ (0,1), we consider the following mixed formulation : find
(y,λ)∈Z×Λ such that
2 RECOVERING THE SOLUTION FROM A PARTIAL OBSERVATION 7
(cid:40)
a (y,y)+b (y,λ) = l (y), ∀y ∈Z
r,α α 1,α (19)
b (y,λ)−c (λ,λ) = l (λ), ∀λ∈Λ,
α α 2,α
where
(cid:90)(cid:90) (cid:90) T
a :Z×Z →R, a (y,y):=(1−α) yydxdt+r (Ly,Ly) dt,
r,α r,α H−1(Ω)
qT 0
(cid:90) T (cid:90)(cid:90)
b :Z×Λ→R, b (y,λ):= (cid:104)λ,Ly(cid:105) dt−α yLλdxdt,
α α H1(Ω),H−1(Ω)
0
0 qT
(cid:90)(cid:90)
c :Λ×Λ→R, c (λ,λ):=α LλLλ, dxdt
α α
QT
(cid:90)(cid:90)
l :Z →R, l (y):=(1−α) y ydxdt,
1,α 1,α obs
qT
(cid:90)(cid:90)
l :Λ→R, l (λ):=−α y Lλdxdt.
2,α 2,α obs
qT
From the symmetry of a and c , we easily check that this formulation corresponds to the
r,α α
saddle point problem :

sup inf L (y,λ),
 λ∈Λy∈Z r,α
α
Lr,α(y,λ):=Lr(y,λ)− 2(cid:107)Lλ+(y−yobs)1ω(cid:107)2L2(QT).
Proposition 2.1 Underthehypothesis (H),foranyα∈(0,1),theformulation(19)iswell-posed.
Moreover, the unique pair (y,λ) in Z×Λ satisfies
(cid:18)(1−α)2 α2(cid:19)
θ (cid:107)y(cid:107)2 +θ (cid:107)λ(cid:107)2 ≤ + (cid:107)y (cid:107)2 . (20)
1 Z 2 Λ θ θ obs L2(qT)
1 2
with
(cid:18) (cid:19) (cid:18) (cid:19)
r 1 1
θ :=min 1−α, , θ := min α, .
1 η 2 2 C
Ω,T
Proof- We easily get the continuity of the bilinear forms a , b and c :
r,α α α
r
|a (y,y)|≤max(1−α, )(cid:107)y(cid:107) (cid:107)y(cid:107) , ∀y,y ∈Z,
r,α η Z Z
1
|b (y,λ)|≤max(α,√ )(cid:107)y(cid:107) (cid:107)λ(cid:107) , ∀y ∈Z,∀λ∈Λ,
α η Z Λ
|c (λ,λ)≤α(cid:107)λ(cid:107) (cid:107)λ(cid:107) , ∀λ,λ∈Λ
α Λ Λ
and of the linear form l and l : (cid:107)l (cid:107) =(1−α)(cid:107)y (cid:107) and (cid:107)l (cid:107) =α(cid:107)y (cid:107) .
1 2 1 Z(cid:48) obs L2(qT) 2 Λ(cid:48) obs L2(qT)
Moreover, since α∈(0,1), we also obtain the coercivity of a and of c : precisely,
r,α α
(cid:18) (cid:19)
r
a (y,y)≥min 1−α, (cid:107)y(cid:107)2, ∀y ∈Z,
r,α η Z
(cid:18) (cid:19)
1−m
c (λ,λ)≥min αm, (cid:107)λ(cid:107)2 ∀λ∈Λ, ∀m∈(0,1).
α C Λ
Ω,T
The result [4, Prop 4.3.1] implies the well-posedness and the estimate (20) taking m=1/2. (cid:50)
Theα-terminL isastabilizationterm: itensuresacoercivitypropertyofL withrespect
r,α r,α
to the variable λ and automatically the well-posedness. In particular, there is no need to prove
any inf-sup property for the application b .
α
2 RECOVERING THE SOLUTION FROM A PARTIAL OBSERVATION 8
Proposition 2.2 If the solution (y,λ) ∈ Z × X(cid:48) of (7) enjoys the property λ ∈ Λ, then the
solutions of (7) and (19) coincide.
Proof- The hypothesis of regularity and the relation (16) imply that the solution (y,λ)∈Z×X(cid:48)
of(7)isalsoasolutionof(19). Theresultthenfollowsfromtheuniquenessofthetwoformulations.
(cid:50)
2.2 Dual formulation of the extremal problem (7)
As discussed at length in [12], we may also associate to the extremal problem (P) an equivalent
probleminvolvingonlythevariableλ. Again,thisisparticularlyinterestingatthenumericallevel.
This requires a strictly positive augmentation parameter r.
For any r >0, let us define the linear operator P from X(cid:48) into X(cid:48) by
r
P λ:=−∆−1(Ly), ∀λ∈X(cid:48)
r
where y ∈Z is the unique solution to
a (y,y)=b(y,λ), ∀y ∈Z. (21)
r
Theassumptionr >0isnecessaryhereinordertoguaranteethewell-posednessof(21). Precisely,
for any r >0, the form a defines a norm equivalent to the norm on Z.
r
The following important lemma holds:
Lemma 2.1 For any r >0, the operator P is a strongly elliptic, symmetric isomorphism from X(cid:48)
r
into X(cid:48).
Proof- From the definition of a , we easily get that (cid:107)P λ(cid:107) ≤ r−1(cid:107)λ(cid:107) and the continuity of
r r X(cid:48) X(cid:48)
P . Next, consideranyλ(cid:48) ∈X(cid:48) anddenoteby y(cid:48) thecorrespondinguniquesolutionof(21)so that
r
P λ(cid:48) :=−∆−1(Ly(cid:48)). Relation (21) with y =y(cid:48) then implies that
r
(cid:90) T
(cid:104)P λ(cid:48),λ(cid:105) dt=a (y,y(cid:48)) (22)
r H1(Ω) r
0
0
andthereforethesymmetryandpositivityofP . Thelastrelationwithλ(cid:48) =λandtheobservability
r
estimate (H) imply that P is also positive definite.
r
Finally, let us check the strong ellipticity of P , equivalently that the bilinear functional
r
(λ,λ(cid:48))→(cid:82)T(cid:104)P λ,λ(cid:48)(cid:105) dt is X(cid:48)-elliptic. Thus we want to show that
0 r H01(Ω),H01(Ω)
(cid:90) T
(cid:104)P λ,λ(cid:105) dt≥C(cid:107)λ(cid:107)2 , ∀λ∈X(cid:48) (23)
r H1(Ω) X(cid:48)
0
0
for some positive constant C. Suppose that (23) does not hold; there exists then a sequence
{λ } of X(cid:48) such that
n n≥0
(cid:90) T
(cid:107)λ (cid:107) =1, ∀n≥0, lim (cid:104)P λ ,λ (cid:105) dt=0.
n X(cid:48) r n n H1(Ω)
n→∞ 0
0
Let us denote by y the solution of (21) corresponding to λ . From (22), we then obtain that
n n
lim r(cid:107)Ly (cid:107)2 +(cid:107)y (cid:107)2 =0. (24)
n→∞ n X n L2(qT)
From (21) with y =y and λ=λ , we have
n n
(cid:90) T (cid:90)(cid:90)
(cid:10)r(−∆−1)Ly −λ ,(−∆−1)Ly(cid:11) dt+ y ydxdt=0, ∀y ∈Z. (25)
n n H1(Ω) n
0 0 qT
2 RECOVERING THE SOLUTION FROM A PARTIAL OBSERVATION 9
We define the sequence {y } as follows :
n n≥0
Ly =rLy +∆−1λ , in Q ,
 n n n T
y =0, in Σ ,
n T
y (·,0)=y (·,0)=0, in Ω,
n n,t
so that, for all n, y is the solution of the wave equation with zero initial data and source term
n
(cid:112)
rLy +∆λ in X. Using again (5), we get (cid:107)y (cid:107) ≤ C (cid:107)rLy +∆λ (cid:107) , so that y ∈Z.
n n n L2(qT) Ω,T n n X n
Then, using (25) with y =y we get
n
(cid:107)r(−∆−1)Ly −λ (cid:107) ≤(cid:112)C (cid:107)y (cid:107) .
n n X(cid:48) Ω,T n L2(qT)
Then, from (24), we conclude that lim (cid:107)λ (cid:107) = 0 leading to a contradiction and to the
n→+∞ n X(cid:48)
strong ellipticity of the operator P . (cid:50)
r
The introduction of the operator P is motivated by the following proposition :
r
Proposition 2.3 For any r >0, let y ∈Z be the unique solution of
0
a (y ,y)=l(y), ∀y ∈Z
r 0
and let J(cid:63)(cid:63) :X(cid:48) →X(cid:48) be the functional defined by
r
1(cid:90) T
J(cid:63)(cid:63)(λ)= (cid:104)P λ,λ(cid:105) dt−b(y ,λ).
r 2 r H01(Ω) 0
0
The following equality holds :
sup inf L (y,λ)=− inf J(cid:63)(cid:63)(λ) +L (y ,0).
r r r 0
λ∈X(cid:48)y∈Z λ∈X(cid:48)
The proof is classical and we refer for instance to [12] in a similar context. This proposition
reduces the search of y, solution of problem (P), to the minimization of J(cid:63)(cid:63). The well-posedness
r
is a consequence of the ellipticity of the operator P .
r
Remark 2 The results of this section apply if the distributed observation on q is replaced by a
T
NeumannboundaryobservationonasufficientlylargesubsetΣ of∂Ω×(0,T)(i.e. assuming ∂y =
T ∂ν
y ∈ L2(Σ ) is known on Σ ). This is due to the following generalized observability inequality:
obs T T
there exists a positive constant C =C(ω,T,(cid:107)c(cid:107) ,(cid:107)d(cid:107) ) such that the following estimate
obs C1(Ω) L∞(Ω)
holds :
(cid:107)y(·,0),yt(·,0)(cid:107)2H01(Ω)×L2(Ω) ≤Cobs(cid:18)(cid:13)(cid:13)(cid:13)(cid:13)∂∂νy(cid:13)(cid:13)(cid:13)(cid:13)2L2(ΣT)+(cid:107)Ly(cid:107)2L2(QT)(cid:19), ∀y ∈Z (26)
which holds if the triplet (Q ,Σ ,T) satisfies the geometric condition as before (we refer to [10]
T T
and the references therein). Actually, it suffices to re-define the form a in (8) by a(y,y) :=
(cid:82)(cid:82) ∂y∂y dσdx and the form l by l(y):=(cid:82)(cid:82) ∂yy dσdx for all y,y ∈Z.
ΣT ∂ν∂ν ΣT ∂ν obs
Remark 3 We emphasize that the mixed formulation (7) has a structure very closed to the one
we get when we address - using the same approach - the null controllability of (1): more precisely,
the control of minimal L2(q )-norm which drives to rest the initial data (y ,y )∈H1(Ω)×L2(Ω)
T 0 1 0
is given by v =ϕ1 where (ϕ,λ)∈Φ×L2(0,T;H1(Ω)) solves the mixed formulation
qT 0
(cid:40)
a(ϕ,ϕ)+b(ϕ,λ) = l(ϕ), ∀ϕ∈Φ
(27)
b(ϕ,λ) = 0, ∀λ∈L2(0,T;H1(Ω)),
0
3 RECOVERINGTHESOURCEANDTHESOLUTIONFROMAPARTIALOBSERVATION10
where
(cid:90)(cid:90)
a:Φ×Φ→R, a(ϕ,ϕ)= ϕ(x,t)ϕ(x,t)dxdt
qT
(cid:90) T
b:Φ×L2(0,T;H1(Ω))→R, b(ϕ,λ)= (cid:104)Lϕ,λ(cid:105) dt
0 H−1,H1
0
0
(cid:90) 1
l:Φ→R, l(ϕ)=−(cid:104)ϕ (·,0),y (cid:105) + ϕ(·,0)y dx.
t 0 H−1(Ω),H1(Ω) 1
0
0
with Φ=(cid:8)ϕ∈L2(Q ), ϕ=0 on Σ such that Lϕ∈L2(0,T;H−1(Ω))(cid:9). We refer to [12].
T T
Remark 4 Reversing the order of priority between the constraint y − y = 0 in L2(q ) and
obs T
Ly−f = 0 in X, a possibility could be to minimize the functional y → (cid:107)Ly−f(cid:107) over y ∈ Z
X
subject to the constraint y−y = 0 in L2(q ) via the introduction of a Lagrange multiplier in
obs T
L2(q ). The proof of the following inf-sup property : there exists δ >0 such that
T
(cid:82)(cid:82)
λydxdt
inf sup qT ≥δ
λ∈L2(qT)y∈Z (cid:107)λ(cid:107)L2(qT)(cid:107)y(cid:107)Y
associated to the corresponding mixed-formulation is however unclear. If a ε-term is added as in
(QR), this difficulty disappears (we refer again to the book [18]).
3 Recovering the source and the solution from a partial ob-
servation: a mixed re-formulation of the problem
Given a partial observation y of the solution on the subset q ⊂ Q , we now consider the
obs T T
reconstruction of the full solution as well as the source term f assumed in X. We assume that the
initial data (y ,y )∈H are unknown.
0 1
Thesituationisdifferentwithrespecttotheprevioussection,sincewithoutadditionalassump-
tiononf, thecouple(y,f)isnotunique. Considerthecaseofasourcef supportedinasetwhich
is near ∂Ω×(0,T) and disjoint from q : from the finite propagation of the solution, the source
T
f will not affect the solution y in q . On the other hand, the determination of a couple (y,f)
T
which solves (1) such that y coincides with y is straightforward : it suffices to ”extend” y on
obs
Q \q appropriately to preserve the boundary conditions, then compute Ly and recover a source
T T
term. However, we emphasize that, on a practical viewpoint, the extension of y out of q is not
obs T
obvious. Moreover, this strategy does not offer any control on the object f.
We briefly show that we can apply the method developed in Section 2 which allows a robust
reconstruction and then consider the case of uniqueness via additional condition on f.
We assume again that (H) holds. We note Y :=Z×X and define on Y the bilinear form, for
any ε,η >0
(cid:90)(cid:90) (cid:90) T
(cid:104)(y,f),(y,f)(cid:105) := yydxdt+η (cid:104)Ly−f,Ly−f(cid:105) dt
Y H−1(Ω)
qT 0 (28)
(cid:90) T
+ε (cid:104)f,f(cid:105) dt, ∀(y,f),(y,f)∈Y.
H−1(Ω)
0
Inviewof(H),thisbilinearformdefinesascalarproductoverY. Moreover,endowedtothisscalar
product, we easily obtain that Y is a Hilbert space (we refer to [7]). We note the corresponding
(cid:112)
norm by (cid:107)(y,f)(cid:107) := ((y,f),(y,f)) .
Y Y