Étude de quelques problèmes inverses pour le système de Stokes. Application aux poumons.

Transcription

1 Étude de quelques problèmes inverses pour le système de Stokes. Application aux poumons. Anne-Claire Egloffe Directrices de thèse : Céline Grandmont et Muriel Boulakia Le 19 novembre 2012

2 Plan 1 Introduction 2 State of the art 3 Stability estimates 4 Back to the initial problem 5 Conclusion

3 Introduction Plan 1 Introduction 2 State of the art 3 Stability estimates 4 Back to the initial problem 5 Conclusion

4 E tude de quelques proble mes inverses pour le syste me de Stokes. Application aux poumons. Introduction Motivations Molding of human lung realized by E. R. Weibel. Reconstructed bronchial tree, [Baffico, Grandmont, Maury 10]. airflow in the lungs, [Baffico, Grandmont, Maury 10], blood flow in the cardiovascular system, [Quarteroni, Veneziani 03], [Vignon-Clementel, Figueroa, Jansen, Taylor 06].

5 Introduction Modeling of the respiratory tract [Baffico, Grandmont, Maury 10] Decomposition of the respiratory tract into three stages: the upper part (up to the sixth generation), the distal part (from the seventh to the 17 th generation), the acini.

6 Introduction Modeling of the respiratory tract ρ tu + ρ(u )u µ u + p = 0, in (0, T ) Ω, div u = 0, in (0, T ) Ω, u = 0, on (0, T ) Γ l, µ n pn = P 0n, on (0, T ) Γ 0, µ n pn = Pan R i( Γ u n)n, on (0, T ) Γ i i, mẍ + kx = f ext + SP a, on (0, T ), Sẋ = N i=1 Γ u n = i Γ u n, on (0, T ). 0 (1) u: fluid velocity, p: fluid pressure, x: position of the diaphragm, R i : airflow resistance, k: stiffness of the diaphragm.

7 Introduction Modeling of the respiratory tract ρ tu + ρ(u )u µ u + p = 0, in (0, T ) Ω, div u = 0, in (0, T ) Ω, u = 0, on (0, T ) Γ l, µ n pn = P 0n, on (0, T ) Γ 0, µ n pn = Pan R i( Γ u n)n, on (0, T ) Γ i i, mẍ + kx = f ext + SP a, on (0, T ), Sẋ = N i=1 Γ u n = i Γ u n, on (0, T ). 0 (1) u: fluid velocity, p: fluid pressure, x: position of the diaphragm, R i : airflow resistance, k: stiffness of the diaphragm. Physiological interpretation: R i in asthma, k in emphysema. To identify R i et k from measurements available at mouth.

8 Introduction Spirometry Figure: Evaluation of lung function with a spirometer Figure: Normal profil Figure: Restrictive lung disease Figure: Obstructive lung disease

9 E tude de quelques proble mes inverses pour le syste me de Stokes. Application aux poumons. Introduction Direct problem: existence of solution Ω = Γl S N i=0 Γi, Γi Γj =, Γi Γl. Difficulty: to estimate the nonlinear term Z N Z ρx ρ (u ) u u = u 2 u n. 2 i=0 Γi Ω 3 Existence of a solution u L2 (0, t ; D(A)), with D(A) H 2 + (Ω) By replacing the boundary conditions Z pn = Pi n Ri u n n, on Γi, n Γi by Z pn n = Pi Ri u n, on Γi, n Γi u τk = 0, on Γi, for k = 1,..., d 1, we obtain more regularity in space: u L2 (0, t ; D(A)), with D(A) H 2 (Ω)

10 Introduction Inverse problem no coupling with the spring, no nonlinear convective term, the dissipative boundary conditions on Γ i : ( ) n pn + R i u n n = 0, Γ i are replaced by Robin boundary conditions simplified geometry. pn + qu = 0, n

11 Introduction Our problem Let T > 0 and Ω R d be a bounded connected open set. tu u + p = 0, in (0, T ) Ω, div u = 0, in (0, T ) Ω, u = 0, on (0, T ) Γ l, n pn = g, on (0, T ) Γ 0, pn + qu = 0, on (0, T ) Γout, n u(0) = u 0, on Ω. Let Γ Γ 0 and (u j, p j ) be a solution of (P qj ) for j = 1, 2. (P q)

12 Introduction Our problem Let T > 0 and Ω R d be a bounded connected open set. tu u + p = 0, in (0, T ) Ω, div u = 0, in (0, T ) Ω, u = 0, on (0, T ) Γ l, n pn = g, on (0, T ) Γ 0, pn + qu = 0, on (0, T ) Γout, n u(0) = u 0, on Ω. Let Γ Γ 0 and (u j, p j ) be a solution of (P qj ) for j = 1, 2. (P q)

13 Introduction Our problem Let T > 0 and Ω R d be a bounded connected open set. tu u + p = 0, in (0, T ) Ω, div u = 0, in (0, T ) Ω, u = 0, on (0, T ) Γ l, n pn = g, on (0, T ) Γ 0, pn + qu = 0, on (0, T ) Γout, n u(0) = u 0, on Ω. Let Γ Γ 0 and (u j, p j ) be a solution of (P qj ) for j = 1, 2. (P q)

14 Introduction Our problem Let T > 0 and Ω R d be a bounded connected open set. tu u + p = 0, in (0, T ) Ω, div u = 0, in (0, T ) Ω, u = 0, on (0, T ) Γ l, n pn = g, on (0, T ) Γ 0, pn + qu = 0, on (0, T ) Γout, n u(0) = u 0, on Ω. Let Γ Γ 0 and (u j, p j ) be a solution of (P qj ) for j = 1, 2. (P q) Identifiability: Does M (0,T ) Γ (u 1, p 1) = M (0,T ) Γ (u 2, p 2) imply q 1 = q 2? Stability: Is it possible to obtain stability estimate like ( ) (q 1 q 2) (0,T ) Γout f (u 1 u 2) (0,T ) Γ + (p 1 p 2) (0,T ) Γ, where f : R + R + is an increasing function such that lim x 0 f(x) = 0?

15 Introduction Our problem Let T > 0 and Ω R d be a bounded connected open set. tu u + p = 0, in (0, T ) Ω, div u = 0, in (0, T ) Ω, u = 0, on (0, T ) Γ l, n pn = g, on (0, T ) Γ 0, pn + qu = 0, on (0, T ) Γout, n u(0) = u 0, on Ω. Let Γ Γ 0 and (u j, p j ) be a solution of (P qj ) for j = 1, 2. (P q) Thanks to boundary conditions on Γ out: To do: estimate ( 1 u 1 (q 2 q 1 ) = n ) 2 (p 1 p 2 )n + q 2 (u 1 u 2 ), n ( 1 n ) 2 n (0,T ) Γ out + (p 1 p 2 ) (0,T ) Γout + (u 1 u 2 ) (0,T ) Γout by boundary terms on (0, T ) Γ.

16 State of the art Plan 1 Introduction 2 State of the art 3 Stability estimates 4 Back to the initial problem 5 Conclusion

17 State of the art State of the art Unique continuation properties: [Fabre, Lebeau 96], [Regbaoui 99], [Lin, Uhlmann,Wang 10],... Inverse problems: [Imanuvilov, Yamamoto 00], [Alvarez, Conca, Friz, Kavian, Ortega 05], [Ballerini 10], [Conca, Schwindt, Takahashi 12],... Related field: [Fernández-Cara, Guerrero, Imanuvilov, Puel 04], [Imanuvilov, Puel, Yamamoto 09],...

18 State of the art State of the art for the Laplace equation Stationary case: using analytic functions theory, [Chaabane, Jaoua 99], [Alessandrini, Del Piero, Rondi 03], [Chaabane, Fellah, Jaoua, Leblond 04], [Sincich 07],... using Carleman inequalities, [Bellassoued, Cheng, Choulli 08], [Cheng, Choulli, Lin 08],... Nonstationary case: Up to our knowledge, largely open question.

19 Stability estimates Plan 1 Introduction 2 State of the art 3 Stability estimates 4 Back to the initial problem 5 Conclusion

20 Stability estimates General setting Let d N and Ω R d be a connected bounded open set. We consider: u + p = 0, in Ω, div u = 0, in Ω, u = 0, on Γ l, (P q) n pn = g, on Γ 0, pn + qu = 0, on Γout. n

21 Stability estimates General setting Let d N and Ω R d be a connected bounded open set. We consider: u + p = 0, in Ω, div u = 0, in Ω, u = 0, on Γ l, (P q) n pn = g, on Γ 0, pn + qu = 0, on Γout. n Identifiability: Let x 0 Γ 0, r > 0. Assume that g is non identically zero on Γ 0. Let (u j, p j ) H 1 (Ω) L 2 (Ω) be the solution of (P qj ) for j = 1, 2. If u 1 = u 2 on Γ 0 B(x 0, r), then q 1 = q 2 on Γ out. Remark It is a corollary of Fabre-Lebeau unique continuation result.

22 Stability estimates General setting Let d N and Ω R d be a connected bounded open set. We consider: u + p = 0, in Ω, div u = 0, in Ω, u = 0, on Γ l, (P q) n pn = g, on Γ 0, pn + qu = 0, on Γout. n Stability estimates: Let Γ Γ 0. We obtained two kind of results: - two logarithmic stability estimates of type q 1 q 2 L 2 (K) ( with 0 < λ < 1 and K {x Γ out/u 1 (x) 0}, - a Lipschitz stability estimate of type ln ( C C 1 u 1 u 2 L 2 (Γ) + p 1 p 2 L 2 (Γ) + p 1 n p 2 n L 2 (Γ) ( q 1 q 2 L (Γ out ) C u 1 u 2 L 2 (Γ) + p 1 p 2 L 2 (Γ) + p 1 n p 2 n )) λ, L 2 (Γ) ), under the a priori assumption that the Robin coefficient is piecewise constant on Γ out.

23 Stability estimates A logarithmic stability estimate Let d N and Ω R d be a connected bounded open set. We consider: u + p = 0, in Ω, div u = 0, in Ω, u = 0, on Γ l, (P q) n pn = g, on Γ 0, pn + qu = 0, on Γout. n Γl The boundary of the domain is composed of three parts Γ0 Γout Γ 0 Γ out Γ l = Ω, pairwise disjoint: Figure: Example of such an open set Ω in dimension 2. - Γ 0 Γ out =, - Γ 0 Γ l =, - Γ l Γ out =.

24 Stability estimates Logarithmic stability estimate Theorem (M. Boulakia, A.-C. E., C. Grandmont) Let α > 0, M 1 > 0, M 2 > 0, k N such that k + 2 > d. Assume that: 2 - Γ Γ 0 and Γ out are of class C, - g H 1 2 +k (Γ 0 ) is non identically zero and g H 1 2 +k (Γ0 ) M 1, - q j H s (Γ out), with s > d 1 2 and s k, is such that q j α a. e. on Γ out and q j H 1 2 +k (Γout ) M 2. Let K be a compact subset of {x Γ out/u 1 0} and m > 0 be such that u 1 m on K. Then, β (0, 1), C(α, M 1, M 2 ) > 0 and C 1 (α, M 1, M 2 ) > 0 such that q 1 q 2 L 2 (K) 1 m ( ( ln C(α, M 1, M 2 ) C 1(α,M 1,M 2) u 1 u 2 L 2 (Γ) + p 1 p 2 L 2 (Γ) + p 1 n p 2 n L 2 (Γ) )) 3β 4. Note that [Bellassoued, Cheng, Choulli 08] has a similar result for the laplacian.

25 Stability estimates Sketch of the proof Let us denote by u = u 1 u 2 and p = p 1 p 2. (q 2 q 1 )u 1 = q 2 u + pn, on Γout. ( n q 1 q 2 L 2 (K) 1 ) m C(M 2) u L 2 (K) + n + p L 2 (K). L 2 (K) Using the continuity of the trace mapping: H 3 2 +ɛ (ω) L ( 2 (K) L 2 ) (K) v v K, v, n K and the following interpolation inequality: we obtain v H 3 2 +ɛ (ω) C v θ H 1 (ω) v 1 θ H 3 (ω), q 1 q 2 L 2 (K) 1 ( ) m C(α, M 1, M 2 ) u θ H 1 (ω) + p θ L 2. (ω)

26 Stability estimates Sketch of the proof Let us denote by u = u 1 u 2 and p = p 1 p 2. (q 2 q 1 )u 1 = q 2 u + pn, on Γout. ( n q 1 q 2 L 2 (K) 1 ) m C(M 2) u L 2 (K) + n + p L 2 (K). L 2 (K) Using the continuity of the trace mapping: H 3 2 +ɛ (ω) L ( 2 (K) L 2 ) (K) v v K, v, n K and the following interpolation inequality: we obtain v H 3 2 +ɛ (ω) C v θ H 1 (ω) v 1 θ H 3 (ω), q 1 q 2 L 2 (K) 1 ( ) m C(α, M 1, M 2 ) u θ H 1 (ω) + p θ L 2. (ω)

27 Stability estimates Sketch of the proof Let us denote by u = u 1 u 2 and p = p 1 p 2. (q 2 q 1 )u 1 = q 2 u + pn, on Γout. ( n q 1 q 2 L 2 (K) 1 ) m C(M 2) u L 2 (K) + n + p L 2 (K). L 2 (K) Using the continuity of the trace mapping: H 3 2 +ɛ (ω) L ( 2 (K) L 2 ) (K) v v K, v, n K and the following interpolation inequality: we obtain v H 3 2 +ɛ (ω) C v θ H 1 (ω) v 1 θ H 3 (ω), q 1 q 2 L 2 (K) 1 ( ) m C(α, M 1, M 2 ) u θ H 1 (ω) + p θ L 2. (ω) To do ( u H 1 (ω) + p L 2 (ω) f u L 2 (Γ) + p L 2 (Γ) + p n increasing function. L 2 (Γ) ), where f : R + R + is an

28 Stability estimates Sketch of the proof Theorem Assume that Ω is of class C. Let 0 < ν 1 and let Γ be a nonempty open subset of the 2 boundary of Ω. There exists d 0 > 0 such that for all γ ( 0, ν), for all d > d 0, there exists c > 0, such that we have u 3 +ν + p 3 +ν u H 1 (Ω) + p H 1 (Ω) c H 2 (Ω) H 2 (Ω) ( ( )) u 3 H 2 ln d +ν + p γ, 3 (Ω) H 2 +ν (Ω) u L 2 (Γ) + p L 2 (Γ) + n L 2 (Γ) + p n L 2 (Γ) { for all couples (u, p) H 3 2 +ν (Ω) H 3 2 +ν u + p = 0, in Ω, (Ω) solution of div u = 0, in Ω.

29 Stability estimates Sketch of the proof Theorem Assume that Ω is of class C. Let 0 < ν 1 and let Γ be a nonempty open subset of the 2 boundary of Ω. There exists d 0 > 0 such that for all γ ( 0, ν), for all d > d 0, there exists c > 0, such that we have u 3 +ν + p 3 +ν u H 1 (Ω) + p H 1 (Ω) c H 2 (Ω) H 2 (Ω) ( ( )) u 3 H 2 ln d +ν + p γ, 3 (Ω) H 2 +ν (Ω) u L 2 (Γ) + p L 2 (Γ) + n L 2 (Γ) + p n L 2 (Γ) { for all couples (u, p) H 3 2 +ν (Ω) H 3 2 +ν u + p = 0, in Ω, (Ω) solution of div u = 0, in Ω. End of the proof of the logarithmic stability estimate: q 1 q 2 L2 (K) 1 ( ) m C(α, M 1, M 2 ) u θ H 1 (ω) + p θ L 2 (ω). We apply the previous theorem in ω with ν = 1 2 and with γ suitably chosen.

30 Stability estimates Remark (u, p) H 3 (ω) H 2 (ω), C > 0, θ = 3 ( 1 2ɛ ), u H +ɛ 2 (ω) C u θ H 1 (ω) u 1 θ H 3 (ω), q 1 q 2 L 2 (K) 1 m ( ( ln C(α, M 1, M 2 ) dc 1(α,M 1,M 2) u 1 u 2 L 2 (Γ) + p 1 p 2 L 2 (Γ) + p 1 n p 2 n L 2 (Γ) (u, p) H k+2 (ω) H k+1 (ω), k N such that k + 2 > d 2, )) 3β 4 C > 0, θ = 1/2 + k 1 + k ɛ 1 + k, u H 3 2 +ɛ (ω) C u θ H 1 (ω) u 1 θ H k+2 (ω), q 1 q 2 L2 (K) 1 m ( ( ln C(α, M 1, M 2 ) dc 1(α,M 1,M 2) u 1 u 2 L 2 (Γ) + p 1 p 2 L 2 (Γ) + p 1 n p 2 n L 2 (Γ) )) 1/2+k 1+k

31 Stability estimates A quantitative estimate of a unique continuation result Assume that Ω is of class C. Let 0 < ν 1 2 and let Γ Ω be a nonempty open subset. Theorem There exists d 0 > 0 such that for all γ ( 0, ν), for all d > d 0, there exists c > 0, such that we have u 3 H 2 +ν + p 3 (Ω) H 2 +ν (Ω) u H 1 (Ω) + p H 1 (Ω) c u 3 H 2 ln d +ν + p γ, 3 (Ω) H 2 +ν (Ω) u L 2 + p (Γ) L 2 + (Γ) n L 2 + p (Γ) n L 2 (Γ) for all β ( 0, ν), there exists c > 0, such that for all ɛ > 0, we have ( ) u H 1 (Ω) + p H 1 (Ω) e c ɛ u L 2 (Γ) + p L 2 (Γ) + n + p L 2 (Γ) n L 2 (Γ) for all couples (u, p) H 3 2 +ν (Ω) H 3 2 +ν (Ω) solution of + ɛ β ( u 3 + p +ν 3 H 2 (Ω) H 2 +ν ), (Ω) { u + p = 0, in Ω, div u = 0, in Ω.

32 Stability estimates A quantitative estimate of a unique continuation result We adapt to the Stokes system a quantitative estimate of unique continuation results for the Laplace equation: Remarques sur l observabilité pour l équation de Laplace, Kim-Dang Phung, The proof is based on local Carleman estimates: - inside the domain [Hörmander], - near the boundary [Lebeau-Robbiano 95].

33 Stability estimates For all (u, p) H 2 3 +ν (Ω) H 3 2 +ν (Ω) solution of { u + p = 0, in Ω, div u = 0, in Ω, (2) u H 1 ( ω Ω) + p H 1 ( ω Ω) e c ɛ ( u H 1 (ω) + p H 1 (ω) ) + ɛβ ( u H 3 2 +ν (Ω) + p H 3 2 +ν (Ω) ) u H 1 (ω) + p H 1 (ω) ( c u ɛ H 1 (Γ) + p H 1 (Γ) + n + p L 2 (Γ) n L 2 (Γ) ) + ɛ s ( u H 1 (Ω) + p H 1 (Ω) ) and for all (u, p) H 1 (Ω) L 2 (Ω) solution of (2) u H 1 ( ω) + p L 2 ( ω) c ɛ ( ) ( ) u H 1 (ω) + p L 2 (ω) + ɛ s u H 1 (Ω) + p L 2 (Ω)

34 Stability estimates Carleman estimate Proposition (Lebeau-Robbiano) Let K = {x R n + / x R0}. Let P be a second-order differential operator whose coefficients are C in a neighborhood of K defined by P (x, x) = 2 xn + R(x, 1 i x ). Let us denote by r(x, ξ ) the principal symbol of R and assume that r(x, ξ ) R and that there exists a constant c > 0 such that (x, ξ ) K R n 1, we have r(x, ξ ) c ξ 2. Let φ= φ(x) C be a function defined in a neighborhood of K. We assume that the function φ satisfies the Hörmander hypoellipicity property on K and xn φ(x) 0, x K. Then, there exists c > 0 and h 1 > 0 such that for all h (0, h 1) we have: y(x) 2 e 2φ(x)/h dx + h 2 y(x) 2 e 2φ(x)/h dx ch 3 P (x, x)y(x) 2 e 2φ(x)/h dx R n R + n R + n + + c ( y(x, 0) 2 + h x R n 1 y(x, 0) 2 + h xn y(x, 0) 2 )e 2φ(x,0)/h dx, for all function y C (R n + ) with support in K.

35 E tude de quelques proble mes inverses pour le syste me de Stokes. Application aux poumons. Stability estimates Estimates near the boundary Key - points of the proof: go back to the half plane by passing in geodesic normal coordinates, suitably choose the weight function φ, apply the Carleman inequality twice: one time to the velocity and another time to the pressure. More precisely, for the first one: φ(x) = eλxn + Hardy inequality For the second one: φ(x) = e λ(xn + x )

36 Stability estimates Estimate inside the domain - To prove the third one, we use the classical local Carleman inequality for the Laplace equation inside the domain. - Caccioppoli inequality allows to discard of the gradient of p in the right hand side. - In particular, we prove a three balls inequality (see below). Lemma (Three balls inequality) Let 0 < r 1 < r 2 < r 3 and q R d. There exist C > 0, α > 0 such that for all function (u, p) H 1 (B(q, r 3 )) H 1 (B(q, r 3 )) solution of { u + p = 0, div u = 0, in B(q, r 3 ) there exists α > 0 such that: u H 1 (B(q,r 2 )) + p L 2 (B(q,r 2 )) ( ) α ( 1 α C u H 1 (B(q,r 1 )) + p L 2 (B(q,r 1 )) u H 1 (B(q,r 3 )) + p L 2 (B(q,r 3 ))). Useful to prove the Lipschitz stability estimate when the Robin coefficient is piecewise constant.

37 Stability estimates Extension We also proved another logarithmic stability estimate valid in dimension d = 2. Main differences: Regularity on Ω Regularity needed on (u, p) Valid in dimension C 3,1 (u, p) H 4 (Ω) H 3 (Ω) 2 locally C (u, p) H k+2 (Ω) H k+1 (Ω) in any dimension d for k N be such that k + 2 > d 2 We can extend previous stability estimates to the nonstationary problem by using inequalities coming from analytic semigroup properties. It leads to measurements in infinite time: u 1 u 2 L (0,+ ;L 2 (Γ)) + p 1 p 2 L (0,+ ;L 2 (Γ)) + p 1 n p 2 n L (0,+ ;L 2 (Γ)) [M. Boulakia, A.-C. E., C. Grandmont, Accepted for publication in Mathematical control and related field]..

38 Stability estimates A Lipschitz stability estimate Let d = 2, 3 and Ω R d be a connected bounded open set. We consider: u + p = 0, in Ω, div u = 0, in Ω, u = 0, on Γ l, n pn = g, on Γ 0, n pn + qu = 0, on Γ out. We assume that q is piecewise constant on Γ out = N i=1 Γ i: q Γi = q i, with q i R + for i = 1,..., N. Remark Due to the mixed boundary conditions, we do no have global regularity on the solution.

39 Stability estimates A Lipschitz stability estimate Theorem (A.-C. E.) Let m > 0, M 1 > 0, R M > 0. Assume that: - Γ Γ 0 is of class C and is such that ( Γ Γ l ) ( Γ Γout ) =, - Γ out is of class C 2,1, - g H 3 2 (Γ 0 ) is non identically zero and g H 3 2 (Γ0 ) M 1, - q j Γi = qj i with qi j R+ and qj i R M for i = 1,..., N and j = 1, 2. { ) } We assume that there exits x i x Γ i /d (x, Ω\Γ i > 0 such that u 2 (x i ) > m for all i = 1,..., N. Then, C(m, R M, M 1, N) > 0 such that q 1 q 2 L (Γ out ) ( C(m, R M, M 1, N) u 1 u 2 L 2 (Γ) + p 1 p 2 L 2 (Γ) + p 1 n p 2 n L 2 (Γ) ). Note that [Sincich 07] has a similar result for the Laplace equation.

40 Stability estimates Sketch of the proof We consider ( ) u 1 u 2 p 1 p 2 (w, π) = N j=1 q1 j q2 j, N j=1 q1 j q2 j. Since for j = 1, 2, q j is piecewise constant, (w, π) is solution of: w + π = 0, in Ω, div w = 0, in Ω, w = 0, on Γ l, w n πn = 0, on Γ 0, w n πn + q (q 2 q 1 ) 1w = N j=1 q1 j q2 j u 2, on Γ out. We prove that there exists C > 0 such that: w L 2 (Γ) + π L 2 (Γ) + π n C. L 2 (Γ)

41 Stability estimates Sketch of the proof To do so, we use: - previous unique continuation estimates for the Stokes system, - a sequence of balls which approachs the boundary, - Hölder regularity of the solution in a neighborhood of the boundary. Figure: Scheme of how information is transmitted.

42 Back to the initial problem Plan 1 Introduction 2 State of the art 3 Stability estimates 4 Back to the initial problem 5 Conclusion

43 E tude de quelques proble mes inverses pour le syste me de Stokes. Application aux poumons. Back to the initial problem Back to the dissipative boundary conditions Let (uk, pk ) be solution of u + p div u u pn n R pn + Ri u n n n Γi with Ri = Rik for i = 1,..., N and k = 1, 2. = = = = 0, 0, 0, P0 n, = 0, in Ω, in Ω, on Γl, on Γ0, on Γi, for i = 1,..., N,

44 E tude de quelques proble mes inverses pour le syste me de Stokes. Application aux poumons. Back to the initial problem Back to the dissipative boundary conditions Let (uk, pk ) be solution of u + p div u u pn n R pn + Ri u n n n Γi = = = = 0, 0, 0, P0 n, = 0, in Ω, in Ω, on Γl, on Γ0, on Γi, for i = 1,..., N, with Ri = Rik for i = 1,..., N and k = 1, 2. In a domain like Ω = Γl S Γi Γj =, Γi Γl. N i=0 Γi,

45 Back to the initial problem Back to the dissipative boundary conditions Let (u k, p k ) be solution of u + p = 0, in Ω, div u = 0, in Ω, u = 0, on Γ l, n ( pn = P 0n, on Γ 0, ) n pn + R i u n n = 0, on Γ Γi i, for i = 1,..., N, with R i = R k i for i = 1,..., N and k = 1, 2. We denote by (v, π) = (u 1 u 2, p 1 p 2 ). Thanks to the boundary conditions on Γ i, we have: ) ( ) (Ri 2 R1 i ( Γ ) u 1 n = Ri 2 v n + v i Γ i n πn. Ri 1 R2 i Γ u 1 n R2 i v n i Γ + 1 v i K n where K Γ i is a non empty set. L 2 (K) + 1 K π L 2 (K),

46 Back to the initial problem Back to the dissipative boundary conditions Let (u k, p k ) be solution of u + p = 0, in Ω, div u = 0, in Ω, u = 0, on Γ l, n ( pn = P 0n, on Γ 0, ) n pn + R i u n n = 0, on Γ Γi i, for i = 1,..., N, with R i = R k i for i = 1,..., N and k = 1, 2. We denote by (v, π) = (u 1 u 2, p 1 p 2 ). Thanks to the boundary conditions on Γ i, we have: ) ( ) (Ri 2 R1 i ( Γ ) u 1 n = Ri 2 v n + v i Γ i n πn. Ri 1 R2 i Γ u 1 n R2 i v n i Γ + 1 v i K n where K Γ i is a non empty set. Non local term! L 2 (K) + 1 K π L 2 (K),

47 Back to the initial problem A Lipschitz stability estimate in a particular case We assume that N = 1 (only one outlet). Assume that: - R m R j 1 R M, for j = 1, 2 - P 0 be a nonzero constant. Then, there exists C(R M, R m, P 0 ) > 0 such that ( R1 1 R2 1 C(R M, R m, P 0 ) u 1 u 2 L 2 (Γ 0 ) + p 1 p 2 L 2 (Γ 0 ) + p 1 n p 2 n if u 1 u 2 L 2 (Γ 0 ) + p 1 p 2 L 2 (Γ 0 ) + p 1 n p 2 is small enough. n L 2 (Γ 0 ) L 2 (Γ 0 ) ) Key point: Since u is divergence-free, u n = u n. Γ 0 Γ 1 Remark The constant involved in the stability estimate depends only on the data.

48 Back to the initial problem Back to the initial model: numerical point of view ρ tu + ρ(u )u µ u + p = 0, in (0, T ) Ω, div u = 0, in (0, T ) Ω, u = 0, on (0, T ) Γ l, µ n pn = P 0n, on (0, T ) Γ 0, µ n pn = Pan R i( Γ u n)n, on (0, T ) Γ i i, mẍ + kx = f ext + SP a, on (0, T ), Sẋ = N i=1 Γ u n = i Γ u n, on (0, T ). 0 Volume curve Flow-volume curve

49 Back to the initial problem Back to the initial model: numerical point of view ρ tu + ρ(u )u µ u + p = 0, in (0, T ) Ω, div u = 0, in (0, T ) Ω, u = 0, on (0, T ) Γ l, µ n pn = P 0n, on (0, T ) Γ 0, µ n pn = Pan R i( Γ u n)n, on (0, T ) Γ i i, mẍ + kx = f ext + SP a, on (0, T ), Sẋ = N i=1 Γ u n = i Γ u n, on (0, T ). 0 Volume curve Flow-volume curve

50 Back to the initial problem - Resolution of the direct problem with FreeFem++ [Devys, Grandmont, Grec, Maury 10] - Resolution of the inverse problem: we use Genetic algorithm [Dumas] Physiological data: - m = 0, 3 kg, total mass of the lung, - S = 0, 011 m 2, surface of the moving box, - E = 3, N m 5, the lung elastance, - k 0 = E S 2 = 40, 172 N m 1, the stiffness of the spring, - R i = 1, P a s m 3, the resistance at the outlet Γ i. Figure: Domain Ω.

51 Back to the initial problem Presentation of results Estimation of the stiffness constant Volume curve Flow-volume loop parameter reference obtained parameter Estimation of two resistances Volume curve reference parameters parameters obtained Flow-volume loop reference parameters parameters obtained

52 Conclusion Plan 1 Introduction 2 State of the art 3 Stability estimates 4 Back to the initial problem 5 Conclusion

53 Conclusion Conclusion Can we relax the regularity assumption needed on the boundary? [Bourgeois 10] [Bourgeois, Dardé 10]

54 Conclusion Conclusion Can we relax the regularity assumption needed on the boundary? [Bourgeois 10] [Bourgeois, Dardé 10] Can we obtain logarithmic stability estimates on the whole Γ out or on any compact κ Γ out?

55 Conclusion Conclusion Can we relax the regularity assumption needed on the boundary? [Bourgeois 10] [Bourgeois, Dardé 10] Can we obtain logarithmic stability estimates on the whole Γ out or on any compact κ Γ out? Can we obtain stability estimate for the unsteady problem in finite time?

56 Conclusion Conclusion Can we relax the regularity assumption needed on the boundary? [Bourgeois 10] [Bourgeois, Dardé 10] Can we obtain logarithmic stability estimates on the whole Γ out or on any compact κ Γ out? Can we obtain stability estimate for the unsteady problem in finite time? Can we obtain stability estimate with less measurement?

57 Conclusion Merci pour votre attention!