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On control strategies for avoiding loss of information in Non Destructive Testing
Philippe Destuynder, Caroline Fabre
To cite this version:
Philippe Destuynder, Caroline Fabre. On control strategies for avoiding loss of information in Non
Destructive Testing. 2015. �hal-01204660�
ON CONTROL STRATEGIES FOR AVOIDING LOSS OF INFORMATION IN NON DESTRUCTIVE TESTING
PHILIPPEDESTUYNDER
Département d’ingénierie mathématique, laboratoire M2N Conservatoire National des Arts et Métiers
292, rue saint Martin, 75003 Paris France
CAROLINEFABRE
Laboratoire de mathématiques d’Orsay, UMR 8628 Univ Paris-Sud, CNRS, Université Paris-Saclay
Orsay 91405 France
ABSTRACT. This study takes place in the framework of non destructive testing. Based on energy invariants, one can derive boundary quantities which could help to detect damages in a structure. Since only a small part of a structure is equipped with sensors, a lot of information may be lost through the boundary where there is no sensor. In this paper we study several control strategies in order to drive most of the information contained in the energy invariant used, to the sensors.
1. Introduction. Let us consider an open, nonempty, bounded and connected setΩ⊂R2 with a boundaryΓ =∂Ωas shown on figure1. There are at least two parts on the bound- aryΓ. One denoted byΓm, equipped with strain sensors and the complementary -sayΓf- which is unequipped. On all the boundary, homogeneous Neumann boundary conditions are prescribed. The unit normal toΓoutwardΩis denoted byν. We assume that there is a hole insideΩoccupying the open setT with boundary∂T. The goal of the method that we describe in this paper, consists in detecting this hole from boundary measurements.
The difficulty is that only a part of the external boundary ofΩis equipped with sensors.
Therefore, a part of the information is lost. One can find such situation in [1] and [8]. One suggestion is to use an optimal control strategy in order to avoid if possible this loss of in- formation through the unequipped boundary onΓ\∂T. A control is then set on a sub-open set ofΩsayO1- denoted byf and is computed in order to minimize the loss. In fact, the cost of the optimal control is introduced with a small parameter -sayε- and when it tends to zero, we discuss the exact controllability of the quantity onΓ\∂Twhich is the loss of information that we try to cancel. The paper is organized as follows.
1. Introduction;
2. The mathematical model;
3. The invariant used;
4. The criterion used and the optimality relations;
5. Robustness of the optimal control;
Date: September 24, 2015 13:4.
2010Mathematics Subject Classification. Primary: 35C07, 65M15; Secondary: 35M12, 65T60.
Key words and phrases.Non destructive testing, Wave equation, Fracture mechanics, Inverse problems.
1
6. Computation of approximate or exact control(s);
7. Conclusion.
Finally, we refer to [5], [7], [12], [13] for a state of the art on the NDT subject.
2. The mathematical model. Let us consider the open setΩas shown on figure1and two open and nonempty setsO1andO2inΩwithO1∪O2⊂ΩandO1∩O2=∅.
We introduce two functionsf andgwith compact supports respectively inO1andO2. Notice that in what followsgis a given data whereasf is a control.
We writeΓ = Γf ∪Γm∪∂T.It is assumed that in neighborhoods of the connection points betweenΓfandΓmthe unit normalνsatisfiesν.e1= 0onΓf andν.e2= 0onΓm. This implies that all the angles at the corners betweenΓfandΓmare±π
2.
Furthermore, in all the paper,χOdenotes the charactheristic function of a setO.
FIGURE1. The open setΩ
2.1. The Helmoltz equation used. Let us introduce the following mathematical model set in the open setΩ.
Findu∈H1(Ω) such that:
−ω2u−div(c2∇u) =f +ginΩ,
∂u
∂ν = 0onΓ.
(1)
As soon asω2is not solution of the following eigenvalue problem (concerning the values ofλ, one can refer to [6], [17]):
findw∈H1(Ω), λ∈R+, ∀v∈H1(Ω), Z
Ω
c2∇w.∇v=λ Z
Ω
wv, (2)
Fredholm’s theorem (see [2], [9]) enables one to ensure that there is a unique solution to (1) for any couple(f, g)∈ L2(O1)×L2(O2). The set of eigenvaluesλnsolution of the previous model (2) is denoted byΛ = {λn, n ∈N}. Let us point out thatλ0 = 0.The wave velocityccan be discontinuous for instance in case of a heterogeneous material. For
sake of simplicity, it is assumed that it is constant in a neighbourhood of the boundary of Ω.
Remark 1. From classical results, one can prove some regularity results for ubut the functional framework (spaceH1(Ω)) only enables one to defineuon the boundary ofΩ as a function of the spaceH1/2(Γ). Thus, the derivative ofualongΓbelongs to the space H−1/2(Γ)(see [2], [17]). Nevertheless, using the domain derivative tool introduced by J.
Hadamard [10], it is possible to derive estimates on this term inL2norm. Let us explain how in the next subsection.
2.2. Energy estimates onu. Let us introduce a vector fieldθof R2 whose components are denoted byθαand are assumed to be at least in the spaceW1,∞(Ω). In all this section, we assume that the support ofθis contained in a close neighbourhood of the boundaryΓf where there is no sensor (see figure2). Furthermore, we assume that this neighbourhood doesn’t meet the closure of the setsO1,O2andT.
FIGURE2. The support of the vector fieldθ
Let us introduce the bilinear continuous symmetrical form (Dsθis the symmetrical part of the jacobian matrixDθ):
(u, v)∈H1(Ω)2 →d(u, v) =ω2[ Z
Γ
uv θ.ν− Z
Ω
uv div(θ)]
+ Z
Ω
c2[(∇u.∇v)div(θ)−2(Dsθ∇u.∇v)].
(3)
We prove the following lemma:
Lemma 2.1. (i) For everyθ ∈ W1,∞(Ω), every (f, g) ∈ L2(Ω)2and usolution of the Helmoltz’s equation (1), one has
d(u, u) = Z
Γ
c2|∂u
∂s|2θ.ν. (4)
(ii) There existsθ∈ W1,∞(Ω), such that for every(f, g)∈ L2(Ω)2andusolution of the Helmoltz equation (1), one has
d(u, u) = Z
Γf
c2|∂u
∂s|2ν12. (5)
(iii) There exists a positive constant c0 such that for every (f, g) ∈ L2(Ω)2 and u solution of the Helmoltz equation (1), one has
Z
Γ
c2|∂u
∂s|2≤c0||u||21,2,Ω. (6) Proof -(i) Assuming in a first step, a regularity ofu(u ∈ H2(Ω) is sufficient), one obtains from (1) the following identities whereDθis the Jacobian matrix associated to the vector fieldθ:
0 =−ω2 Z
Ω
u∇u.θ− Z
Ω
div(c2∇u)∇u.θ=
−ω2 Z
Γ
u2
2 θ.ν+ω2 2
Z
Ω
u2div(θ)− Z
Γ
c2∂u
∂ν∇u.θ+ Z
Γ
c2
2|∇u|2θ.ν
− Z
Ω
c2
2|∇u|2div(θ) + Z
Ω
c2(Dθ∇u.∇u).
From the boundary conditions satisfied byu, the previous expression leads to the next one:
Z
Γ
c2|∂u
∂s|2θ.ν=ω2[ Z
Γ
u2θ.ν− Z
Ω
u2div(θ)] + Z
Ω
c2[|∇u|2div(θ)−2(Dθ∇u.∇u)]. (7) Equation (7) ends the proof of assertion (i).
Let us now turn to the second point (ii). Since the term on the lefthanside is an integral over all the boundary of the open setΩ, it could include a contribution from the boundary
∂T of the holeT. Butθ is zero far from the neighbourhood ofΓf which doesn’t meet the holeT. Therefore this boundary integral is restricted toΓf ∪Γm. Let us chooseθas follows, using the coordinates shown on figure2, denoting byOthe neighbourhoodΓfand byx= (x1, x2)a generic point inΩ:
η∈W1,∞(O) with η(x) = 0ifx /∈O, η(x) = 1ifx∈Γf andν1(x)6= 0,
η(x) = 0ifx∈Γm,
and θ=
ν1η(x) 0
.
(8)
OnΓf, the functionν1= (e1, ν)is assumed to be smooth enough (for instanceC1(Γf)) and zero in a close vicinity of the junctions betweenΓf andΓm. The functionθ is then vanishing at the extremities ofΓf and onΓm.Therefore one has from (7):
Z
Γf
c2|∂u
∂s|2ν12=ω2[ Z
Γf
u2θ.ν− Z
Ω
u2div(θ)] + Z
Ω
c2[|∇u|2div(θ)−2(Dθ∇u.∇u)], (9) which is the point (ii) (5). Let us notice that one can easily obtain that there exists a positive constant sayc0such that:
Z
Γf
c2|∂u
∂s|2ν12≤c0||u||21,2,Ω. (10)
The general case of solution with finite energy can be easily obtained with a density argument.
Let us now turn to the proof of the third point (iii). Let us notice thatuis not globally H2(Ω)since the velocity is not constant. Its proof is based on suitable choices of vector fieldsθin assertion (7). LetTbe an open set withT¯⊂T.Withθ∈C0∞(T)withθ=νon
∂T, we get (5) on∂T.
LetK⊂Γfbe a compact set withK∩∂Γf =∅.Withθ=ηνonΓf,η= 1onKand η= 0onΓm, (7) leads to
∂u
∂s ∈L2loc(Γf) andu∈H1(Ω) → ∂u
∂s ∈ L2loc(Γf)is linear and continuous. It is easy to obtain an analogous result onΓmthus ∂u
∂s ∈L2loc(Γm). The last point concerns the corners. Thanks to our hypothesis on the corners betweenΓf andΓm(the angles are±π/2) and du to the fact thatcis constant there, it is known that the functionuis aH2-one near the corner.
This easily ends Lemma’s proof.
Remark 2. Ifuis not solution of the Helmoltz equation (1), the previous equality is not necessarily true.
Remark 3. Lemma2.1(ii) proves thatdis a positive bilinear form (with the suitable choice of θ). One can ask if it is definite. We give a positive answer under some geometric conditions. Ifuis regular (say C2(Ω)) and ifd(u, u) = 0then ∂u
∂s = 0onΓf where ν16= 0. Butuis locally (in the neighbourhoodOfofΓf) solution of:
−ω2u−div(c2∇u) = 0inOf and∂u
∂ν = 0onΓf.
The solution of this model can be computed using local coordinates(s, ξ)(see figure3) assuming that the boundaryΓfis smooth enough.
Let us set:
q= ∂u
∂s inOf.
It is solution of a partial differential equation onOfand satisfies both:
q= ∂q
∂ν = 0onΓf.
Then, one can conclude from Holmgrem theorem thatq= 0inOf and thusuis a function of the only coordinateξ(see figure3) inOf. Becauseuis therefore constant onΓf, one has (Ris the radius of curvature ofΓfand the velocitycis assumed to be constant for sake of simplicity) along the the line normal toΓf
−c2∂2u
∂ξ2 − c2 R+ξ
∂u
∂ξ −ω2u= 0.
Let us consider for instance a point M0 of Γf whereR = ∞ (inflection point). The equation satisfied byubecomes along the normal toΓfat this point:
−c2∂2u
∂ξ2 −ω2u= 0,
FIGURE3. Local coordinates nearΓf
which implies (∂u
∂ξ(0) = 0):
u(ξ) =Acos(ωξ c ).
Assuming thatOf can be prolongated up to this next boundary ofΩ, the line considered crosses necessarily another point -say M1- of the boundary of the open set Ωand one should have at this new point (∂u
∂s = 0along the line):
∂u
∂ν(M1) = ∂u
∂ξ(M1)ν(M0).ν(M1) = 0.
Assuming thatν(M0).ν(M1)6= 0, and that (Dis the lenght betweenM0andM1):
ω6∈ πc DZ,
if the line doesn’t meet the open setO1∪O2, and one can conclude thatu= 0along this line and therefore, from Holmgrem’s theorem again,u= 0onΩ−(O1∪O2). Of course, any regular functions with compact supports inO1and/orO2satisfiesd(u, u) = 0thus our remark seems to be optimal.
3. The invariant used. Let us come back to the analogous formula (7) but derived with nowθ=e1(there is no more truncationη). Therefore one should add the terms onO1and O2which contain respectively the support of the two functionsf andg.This leads to the
following expression:
Z
Γ
c2|∂u
∂s|2ν1−ω2 Z
Γ
u2ν1− Z
O1
f ∂u
∂x1 − Z
O2
g∂u
∂x1 = 0. (11) But the integral overΓf includes two contributions: one whereν1 = 0, and a second one denoted byΓ1f, whereν16= 0. The contributions onΓmand on the boundary of the defect (the hole)ie∂T, are taken into account in this expression. For anyω, f andggiven, we define the functionalGby:
G(ω, f, g) = Z
Γm
c2|∂u
∂s|2ν1−ω2 Z
Γm
u2ν1− Z
O1
f ∂u
∂x1
− Z
O2
g ∂u
∂x1
. (12) This quantity can be mesured as far asuis measured onΓmbut also on O1andO2. In fact from (11) one can state that the term representing the loss of information in the global expression ofGis:
∆(ω, f, g) =− Z
Γf∪Γ∂T
c2|∂u
∂s|2ν1+ω2 Z
Γf∪Γ∂T
u2ν1. (13) If one can ensure by a judicious choice offthat the integral overΓfare zero, the functional Gbecomes an indicator of the presence of the hole. Indeed, let us notice that if there is no hole one can claim thatG= 0.
Conversely, if for everyω∈[ω1, ω2],(ω1< ω2) and for everyg∈L2(O2),one hasG= 0 then it is possible to discuss the non existence of a defect as in [3] and [4]. For instance, if Γf is flat, there would’nt be any contribution from the boundaryΓfin the expression ofG.
The terms onΓfcan be viewed as a loss of information and more precisely as a perturbator with respect to the meaning ofG. This is the reason why it is suggested in this paper to use the functionf as a control variable in order to cancel this parasite term onΓf. This control should be chosen such that the information measured onΓmis still meaningful. In fact, one can observe that the term is vanishing if
[−ωu±c∂u
∂s]ν1= 0 on Γf. Hence, we suggest to define a controlf such that -for instance-
[−ωu+c∂u
∂s]ν1= 0 on Γf.
4. The criterion used and the optimality relations. For anyf ∈L2(O1)we define the criterion:
f ∈L2(O1)→Jε(f) =1 2
Z
Γf
(−ωu+c∂u
∂s)2ν12ds+ε 2
Z
O1
f2dx. (14) It is a positive and strictly convex function off. Furthermore, it is continuous from Lemma 2.1. In fact, for the numerical implementation and from (12) it is more accurate to use the following expression forJεeven if for the continous formulation it is strictly the same. Let
us make it explicit (τ1= (e2, ν)andRis the radius of curvature ofΓf):
Jε(f) = 1 2 Z
Γf
(ω2u2ν12−2ωcu∂u
∂sν12+c2(∂u
∂s)2ν12)ds+ε 2
Z
O1
f2dx
= Z
Γf
ων1(ων1+cτ1
R )u2+1 2 Z
Ω
(c2|∇u|2−ω2u2)div(θ)− Z
Ω
c2(Dθ∇u.∇u)
+ε 2 Z
O1
f2dx.
(15)
The advantage of this last expression is that it is stable in the spaceH1(Ω) which is the right one in a numerical approximation of the initial model (1). Let us now introduce the model which can be used in order to increase the efficiency of the strategy for detecting defects.
min
f∈L2(O1)
Jε(f). (16)
The classical optimization results (see [11], [16]) can be applied to the previous model and the solution is denoted byfε. The corresponding solution isuε.This leads to the result that we summarize hereafter:
Theorem 4.1. Letg∈L2(O2)andω2∈/ Λ. For sake of simplicity, it is also assumed that the wave velocityc, is constant on the boundaryΓf. For anyε >0there is a unique so- lution to the problem (16). Furthermore, we introduce the adjoint statepwhich is solution of:
−ω2pε−div(c2∇pε) = 0inΩ,
∂pε
∂ν = 0onΓm∪Γ∂T, c2∂pε
∂ν = (ω2ν12+ 2τ1ν1ωc
R )uε−2τ1ν1c2 R
∂uε
∂s −c2ν12∂2uε
∂s2 onΓf,
(17)
whereτ1 = e2.ν and Ris the radius of curvature of the boundaryΓf. The optimality condition for the solutionf of (16) is:
pε+εfε= 0inO1. (18) Thus the solution of (16) is equivalent to solve (1)-(17) and (18).
Remark 4. In practical implementation it is easier and more stable to use the expression ofJεgiven at (15). But this expression is equivalent to (14) for the continuous model. The solution method which can be used can be a direct solving of the linear system (1)-(17) and (18). Another possibility is to use the preconditioned (by the diagonal of the quadratic form with respect tofrepresentingJε) conjugate gradient algorithm which is a very popular and efficient method.
5. Robustness of the optimal control. The question that we discuss in this section con- cerns the asymptotic behaviour of the solutionf whenε → 0. Let us assume that there exists at least one control functionfe ∈L2(O1)such that ifueis the solution of (1) with this choice forf, one has:
(−ωue+c∂ue
∂s )ν1= 0onΓf.
We say thatfeis an exact control for the criterion chosen. In this case, one has the follow- ing result.
Theorem 5.1. Letfεbe the unique solution of problem (16). The corresponding solution to equation (1) is denoted byuε. There exists a unique function-sayf0∈L2(O1)- and the corresponding solutionu0of (1) which satisfy the following requirements:
1.(−ωu0+c∂u0
∂s)ν1= 0onΓf; 2. lim
ε→0||fε−f0||0,O1 = 0 and lim
ε→0||uε−u0||1,Ω= 0;
3. f0is the minimal exact control inL2(O1).
(19)
Proof Let us first observe that:
Jε(fε)≤Jε(fe).
Hence:
||fε||0,O1 ≤ ||fe||0,O1, Z
Γf
|(−ωuε+c∂uε
∂s )ν1|2ds≤ε||fe||20,O
1.
This enables one to deduce that there is a subsequencefε0 and an elementf0 ∈L2(O1) such that:
fε0 * f0inL2(O1)-weak.
Classically, uε0 is weakly convergent tou0solution of (1) associated tof0, in the space H1(Ω). From the semi-continuity of convex functions, one can deduce that:
(−ωu0+c∂u0
∂s )ν1= 0.
This proves thatf0is an exact control.
From the property thatf0is an exact control one deduces that on the one hand:
Z
O1
|fε0|2≤ Z
O1
|f0|2, and on the other hand:
Z
O1
|fε0−f0|2= Z
O1
Z
O1
|fε0|2−2 Z
O1
fε0f0+ Z
O1
|f0|2
≤ Z
O1
|f0|2−2 Z
O1
fε0f0+ Z
O1
|f0|2= 2 Z
O1
f0(f0−fε0).
Therefore:
lim
ε0→0||fε0−f0||0,O1 = 0.
Let us imagine that there are two accumulation points to the sequencefε-sayf0andf1- corresponding respectively to two subsequences offεdenoted byfε0 andfε1. Both are exact control and this leads to:
||f0||0,O1 ≤lim inf
ε0→0 ||fε0||0,O1≤ ||f1||0,O1 ≤lim inf
ε1→0 ||fε1||0,O1 ≤ ||f0||0,O1. Or else:
||f0||0,O1 =||f1||0,O1.
In other words, because the functionsf0andf1are both exact controls, one has:
Jε(f0) =Jε(f1).
ButJεis stricly convex, thusf0=f1. Finally we proved the uniqueness of the accumu- lation point and all the sequencefεconverges to this single pointf0. From the inequality:
∀fe∈L2(O1), (exact control), ||f0||0,O1≤lim
ε→0inf||fε||0,O1 ≤ ||fe||0,O1 we also deduce thatf0is the exact control with the minimalL2(O1)-norm. 2
6. Computation of approximate or exact control(s). A first possibility consists in min- imizing the functionalJε. But the result can’t be stable if there is no exact control, at least as far as one is concerned with the continuous model. Let us discuss two direct methods for computing an approximation of an exact control (if it exists).
6.1. The asymptotic strategy. The first idea is to look for an asymptotic expansion ofuε, pε,fεforεsmall enough. Let us set:
uε=u0+εu1+. . . , pε=p0+εp1+. . . , fε=f0+εf1+. . . .
(20)
Remark 5. At this step, nothing guaranties thatf0exists but if it is the case, it is natural to use the same notation as the one used for denoting the limit offεdiscussed in the previous section. In fact, in the following of this section, we proceed by necessary conditions which should be satisfied byf0.
By introducing the assumed expressions (20) into the equations satisfied by(uε, pε, fε) and by equating the term of same power in the resulting expressions, one obtains:
1. Order 0:
−ω2u0−div(c2∇u0) =f0χO1+gχO2inΩ
∂u0
∂ν = 0on∂Ω,
−ω2p0−div(c2∇p0) = 0inΩ
∂p0
∂ν = 0onΓm∪∂T;
c2∂p0
∂ν = (ω2ν12+ 2τ1ν1ωc
R )u0−2τ1ν1c2 R
∂u0
∂s −c2ν12∂2u0
∂s2 onΓf, p0= 0inO1.
(21)
2. Order 1:
−ω2u1−div(c2∇u1) =f1χO1inΩ
∂u1
∂ν = 0on∂Ω,
−ω2p1−div(c2∇p1) = 0inΩ
∂p1
∂ν = 0onΓm∪∂T, c2∂p1
∂ν = (ω2ν12+ 2τ1ν1ωc
R )u1−2τ1ν1c2 R
∂u1
∂s −c2ν12∂2u1
∂s2 onΓf, p1+f0= 0inO1.
(22)
3. and so on . . .
6.1.1. Treatment of the order 0. Assuming that the boundary Γf is smooth enough, the first point is that from Holmgrem theorem one can state thatp0is zero. This implies that:
(ω2ν12+ 2τ1ν1ωc
R )u0−2τ1ν1c2 R
∂u0
∂s −c2ν12∂2u0
∂s2 = 0onΓf. Or else, using the variational formulation forp0:
(−ωu0+c∂u0
∂s )ν1= 0onΓf.
6.1.2. Treatment of the order 1. Fromf0 = −p1inO1,one deduces thatu0 should be solution of:
−ω2u0−div(c2∇u0) =−p1χO1+gχO2 inΩ,
∂u0
∂ν = 0on∂Ω.
Let us define the sub-open set ofΓf denoted byΓ1f and such that one has on this subset ν16= 0(see figure (4)).
Let us point out thatu1is unknown. The functionp1will be defined as soon as∂p1
∂ν on Γ1f will be known. Let us construct a relation concerning this term which will ensure that the condition onΓ1f concerningu0will be satisfied.
From the equation satisfied byu0one should have for any smooth enough functionq:
−ω2 Z
Ω
u0q− Z
Ω
div(c2∇u0)q=− Z
O1
p1q+ Z
O2
gq. (23)
FIGURE4. Definition ofΓ1f
From two integrations by parts and assuming that for a given smooth enough function µ, (for instance in the spaceH1(Γ1f)), the functionqis solution of:
−ω2q−div(c2∇q) = 0inΩ, ∂q
∂ν =−(c∂µ
∂s+ωµ)onΓ1f, ∂q
∂ν = 0on∂Ω\Γ1f, one obtains the following identity which should be satisfied by any functionµsmooth enough in order to make sense to the expressions used:
∀µ∈H1(Γ1f), − Z
Γ1f
c2u0(c∂µ
∂s +ωµ) + Z
O1
p1q= Z
O2
gq. (24)
But, at this step, nothing is done in order to prescribe the condition which should be satisfied byu0onΓ1f(whereν16= 0).
Assuming for instance thatµ∈H01(Γ1f), one can rewrite the relation (24) as follows:
∀µ∈H01(Γ1f), Z
Γ1f
c2(c∂u0
∂s −ωu0)µ+ Z
O1
p1q= Z
O2
gq. (25)
Now we apply the condition that we are wishing foru0. This implies that:
∀µ∈H01(Γ1f), Z
O1
p1q= Z
O2
gq. (26)
6.1.3. Characterization of an exact control. We are now able to characterize the exact control withL2(O1)−minimun norm. Let us introduce the linear forml(.)and the bilinear
and continuous formβ(., .)defined on the spaceH01(Γ1f)by:
µ∈H01(Γ1f), l(µ) = Z
O2
gq(µ),
(ξ, µ)∈H01(Γ1f)→β(ξ, µ) = Z
O1
p1(ξ)q(µ), wherep1(ξ)andq(µ)are respectively solutions of:
−ω2p1(ξ)−div(c2∇p1(ξ)) = 0, −ω2q(µ)−div(c2∇q(µ)) = 0inΩ,
∂p1(ξ)
∂ν = 0, ∂q(µ)
∂ν = 0onΓm∪∂T∪ {Γf\Γ1f},
∂p1
∂ν =−(c∂ξ
∂s+ωξ), ∂q
∂ν =−(c∂µ
∂s +ωµ)onΓ1f.
(27)
We then consider the following problem:
findξ∈H01(Γ1f)such that:
∀µ∈H01(Γ1f), β(ξ, µ) =l(µ).
(28) Let us assume that there is a solution -sayξ- to equation (28). By comparing (28) and (25), one deduces that:
∀µ∈H01(Γ1f), Z
Γ1f
(c∂u0
∂s −ωu0)µ= 0, or elsec∂u0
∂s −ωu0= 0. (29) This proves that−p1χO1 in an exact control foru0.In fact,u0 would satisfy the re- quired condition on Γ1f. The main question is now to study the variational model (28).
Unfortunately, this is a difficult problem that we are still not able to solve completely. On the one hand, the bilinear formβ(., .)is symmetrical, positive and clearly continuous on H01(Γ1f). On the other hand,β(ξ, ξ) = 0implies thatp1(ξ) = 0on the open setO1. But, p1 is solution of the equation (22). Thus -assuming that all the necessary hypothe- ses are satisfied, one can claim from Holmgrem’s theorem thatp1 = 0onΩand finally that:
c∂ξ
∂s+ωξ= 0 onΓ1f.
By solving this ordinary differential equation on each connected component of Γ1f, one obtains:
ξ=Ce−ωcs,
and from the boundary condition satisfied byξ, one finally gets thatξ = 0. Hence, the kernel ofβ(., .)is reduced to{0}. Because the mapping:
ξ∈H01(Γ1f),→p β(ξ, ξ)
is a continous norm on the spaceH01(Γ1f), one can define (see [14]) the completed spaceVβ ofH01(Γ1f)with respect to this norm. As soon as the functiongbelongs to the dual space Vβ0, the Lax-Milgram theorem enables to claim that (28) has a unique solution inVβ. The point is to characterize this completed space (Vβand its dualVβ0). This could be derived
from an inverse inequality but it is not established excepted in very particular geometry (and will be the study of forthcoming paper). Nevertheless in practical applications one can use the fact that the solution of the acoustic model is close to an eigenmode as soon as the frequency used (ω
2π) is close to an eigenfrequency of the structure. This is the discussion suggested in the next subsection.
6.1.4. What happens whenω2 is close to an eigenvalue of the model. Let us consider a particular case which is similar to a finite dimensional situation. This is the case when the pulsationω is very close to the square root of an eigenvalue of the acoustic operator (see equation2). This is the resonance case. The solution would be close to an eigenfunction of the Helmotz’s operator and this is the property used on order to construct an approximate control. Let us assume for sake of brevity, that the eigenvalueλi0is single and let us set:
ω2=λi0−υ. (30)
For anyξ∈H01(Γ1f), (in fact one could extend the expression toξ∈L2(Γ1f)), the solution p1is :
p1=X
i≥1
− Z
Γ1f
(c∂ξ
∂s+ωξ)wi λi−ω2 wi
=− Z
Γ1f
(c∂ξ
∂s+ωξ)wi0
υ wi0− X
i≥1, i6=i0
Z
Γ1f
(c∂ξ
∂s+ωξ)wi
υ+λi−λi0 wi.
(31)
Hence assuming that|υ|<< δ= min
i≥1, i6=i0
|λi−λi0|, one has:
||p1+ Z
Γ1f
(c∂ξ
∂s+ωξ)wi0
υ wi0||0,O1 ≤ ||p1+ Z
Γ1f
(c∂ξ
∂s+ωξ)wi0
υ wi0||0,Ω
= [X
i6=i0
| Z
Γ1f
(c∂ξ
∂s+ωξ)wi
υ+λi−λi0 |2]1/2≤ 1
δ− |υ|||p1c||0,Ω,
(32)
wherep10is defined by:
p1c =p1−p1i
0, withp1i
0 =− Z
Γ1f
(c∂ξ
∂s+ωξ)wi0
υ wi0. (33)
Let us choose a function ξ = αwi0ν1 ∈ H01(Γ1f)whereαis constant. The coefficient ν1 ensures that this choice forξis effectively zero on the boundary ofΓ1f as soon as the boundaryΓf is at leastC1in order to avoid discontinuity ofν1. Then one has (Ris the radius of curvature of the boundaryΓ1f; its is assumed to be zero at all the extremities of connected components ofΓ1f):
Z
Γ1f
(c∂ξ
∂s+ωξ)wi0 =α Z
Γ1f
(ων1+cτ1
2R)w2i0.
Hence, assuming thatων1+cτ1
2R 6= 0(which is compatible with the fact that we choose ωfor a given structure), one can claim that||p1i
0||0,1 '1/υ. Finally, there is a constantc1 such that with this choice ofξ, one has:
||p1−p1i0||0,O1≤c1|υ| ||p1i0||0,O1 (34) The coefficientαis defined from (28) by:
α= Z
O2
gp1i0 Z
O1
(p1i0)2
, (35)
and the approximation of the controlf0is given by:
fi0
0 =−αp1i0χO1, (36)
wherep1i0 is defined by solving (22), and introducingξ=ξ0=wi0ν1in the expression of p1i0given at (33).
Remark 6. Clearly the choice, made forξ is restrictive. But if the frequencyω/2πis close to an eigenfrequency of the structure, it is justified as far as the the solution of the Helmoltz’s equation of our model is also close to the corresponding eigenmode. It is possi- ble, without any new difficulty, to expand the method with a finite number of eigenmodes, instead of only one has we did. But in such a case the choice ofξ is done in the finite dimensional space spanned by a choice of vectorswiν1restricted to the boundaryΓ1f (see [15]). In this case, the coefficientαis replaced by a vector and defined by solving a linear system deduced from (28) as for (35).
Remark 7. The control can be one of the two methods described in this paper. The first one consists in finding the optimal control defined in section (4) at equation (16) with a particular choice of the parammeter ε. In fact, the Helmoltz’s equation will be replaced in a numerical approximation by a finite element model or by a modal model using the eigenmodes computed from a finite element method. In fact, a splitting of the spectrum will be used in order to focus on the eigenmodes which will be concerned by the non destructive strategy suggested. From the analysis performed in section5, the solution of the optimal control model (in the reduced basis spanned by the selected eigenvectors), one can claim that the solution converges asε→0to the one of the limit model that has been studied in this subsection.
6.2. The simplist method. The goal of the control method that has been introduced in section4, is to reduce the loss of information on the boundaryΓ1f. The criterionJεwhich has been introduced at (15) contains a regulation term on the control which represents the cost of the control. We studied in section6the asymptotic behaviour of the solution (optimal control) when the marginal costεtends to zero. In this section, we discuss another strategy which is not yet mathematically founded but which can be used as a simple -and even simplist- method. In the following we shall use the notationu0=ug+zwhereugis solution of:
ug∈H1(Ω), such that:
∀v∈H1(Ω), −ω2 Z
Ω
ugv+ Z
Ω
c2∇ug.∇v= Z
O2
gv,
(37)
and our goal is still to definezsuch that(−ωu0+c∂u0
∂s )ν1would be as smallest as possible onΓ1f. We try in this section to avoid the difficulty connected to the small parameterε.
The basic idea is to try to solve the terms of order one forp1in the assumed asymptotic expansion (27).
For anyξ∈L2(Γf), let us setp1∈H1(Ω)solution of:
−ω2p1−div(c2∇p1) = 0inΩ,
∂p1
∂ν = 0 onΓm∪∂T∪ {Γf\Γ1f},
∂p1
∂ν =ξonΓ1f.
(38)
Then, let us introducez∈H1(Ω)solution of:
−ω2z−div(c2∇z) =−p1χO1 inΩ,
∂z
∂ν = 0on∂Ω.
We introduce the mappingLfromL2(Γf)into itself by (see5for the regularity property of ∂z
∂s onΓf):
ξ∈L2(Γ1f)→L(ξ) =ν1(−ωz+c∂z
∂s).
For another functionµ∈L2(Γ1f), we define the functionydefined similarly tozwithξ.
Let us now consider two functions(ξ, µ) ∈ [L2(Γ1f)]2 and let us define the symmetrical and positive bilinear form:
a(ξ, µ) = Z
Γ1f
(−ωz+c∂z
∂s)(−ωy+c∂y
∂s)ν12, and the linear forml(.):
µ∈L2(Γ1f)→l(µ) =− Z
Γ1f
(−ωug+c∂ug
∂s )(−ωy+c∂y
∂s)ν12. There is a kernelKtoa(., .)which is such that:
K={µ∈L2(Γ1f), (−ωy+c∂y
∂s)ν1= 0onΓ1f}. (39) From the linearity and continuity of the mapping:
µ∈L2(Γ1f)→(−ωy+c∂y
∂s)ν1∈L2(Γ1f),
one can claim thatK is a closed sub-set ofL2(Γ1f). More precisely,Kis a finite dimen- sional sub-space ofL2(Γ1f)because it corresponds to functions such that: y = Akeωcs on each connected component ofΓ1f (and we assume that there is a finite number of such components).
Let us denote by:
V=L2(Γ1f)/Kc (40)
the completed space of L2(Γ1f)/K equipped with the norm (which is a semi-norm on L2(Γ1f)) defined by:
µ∈L2(Γ1f)→p a(µ, µ).
One can observe that:
∀µ∈K, l(µ) = 0.
Therefore, it makes sense to try to solve the following variational problem:
findξ∈Vsuch that:
∀µ∈V, a(ξ, µ) =l(µ).
(41) It is equivalent to minimize over the spaceL2(Γ1f)the functional:
G(ξ) =1
2a(ξ, ξ)−l(ξ), or else, becausezis a function ofξ:
min
ξ∈L2(Γ1f)
1 2
Z
Γ1f
| −ω(z+ug) +c∂(z+ug)
∂s |2ν12.
If the minimum is zero, then the controlf =−p(ξ)χO1 is exact. Let us explain why. In this case, onceξdefined, we observe that the functionh=ug+zsatisfies the condition:
(−ωh+c∂h
∂s)ν1= 0 onΓf.
Hence the functionp1obtained by solving (38) with the value ofξfound by solving (41) is such thatfe = −p1χO1 is an exact control. One could object that it is necessary to characterize the completed of the quotient spaceV /Kcin order to ensure that we are able to compute practicallyfe as an element of the spaceL2(O1). Obviously the question is simplified as far as one restricts the spaceL2(Γ1f)to a finite dimensional subspace which is the case of a finite element approximation of the solution of (41).
The goal of this section is to start a discussion and to underline the difference with the asymptotic method described in subsection6.1
7. Conclusion. In this paper we have discussed the possibility to extend the strategy in- troduced in [3] for detecting defects in a rectangular structure. In the present, case the geometry can be more general but a control function has been introduced in order to re- strict the loss of information through the lateral boundary of the open setΩwhich is not equipped with sensors and which is not parallel to the axisx1. This control function is solution of an optimal control model and can be computed for any given excitation (named gin this paper) and for any -also given- frequency ω
2π.
The main result is that the boundary criterion which permits this reduction, is stable be- cause of an hidden regularity of the Helmoltz’s equation. Furthermore, it has been proved that the control method is robust (stable) when the cost of the control (the parameterε) tends to zero as far as the frequency is chosen close to an eigenfrequency of the structure.
Another step would be to prove an inverse inequality in order to characterize the func- tional spaceVβ(see section6) in which the functionξof section5should be. But let us underline that this choice of frequency -quite close to an eigenfrequency of the structure- enables one to get an almost finite dimensional control model for which the question is solved.
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E-mail address:philippe.destuynder@cnam.fr E-mail address:caroline.fabre@u-psud.fr
