Mixed formulations for the direct approximation of L 2 -weighted nul. equation

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1 Mixed formulations for the direct approximation of L 2 -weighted null controls for the linear heat equation ARNAUD MÜNCH Université Blaise Pascal - Clermont-Ferrand - France Hammamet, PICOF, 2014 joint work with DIEGO A. DE SOUZA (Sevilla)

2 Introduction Ω R N ; Q T = Ω (0, T ); q T = ω (0, T ) 8 < y t (c(x) y) + d(x, t)y = v 1 ω, in Q T, y = 0, in Σ : T, y(x, 0) = y 0 (x), in Ω. (1) c := (c i,j ) C 1 (Ω; M N (R)); (c(x)ξ, ξ) c 0 ξ 2 in Ω (c 0 > 0), d L (Q T ), y 0 L 2 (Ω); v = v(x, t) is the control y = y(x, t) is the associated state. RESULTS - For any ω Ω, T > 0, y 0 L 2 (Ω), v L 2 (q T ) s.t. y(, T ) = 0 a.e. Ω [FURSIKOV-IMANUVILOV 95, LEBEAU-ROBBIANO 95] GOAL - Approximate numerically such v s. NOTATIONS - Ly := y t (c(x) y) + d(x, t)y; L ϕ := ϕ t (c(x) ϕ) + d(x, t)ϕ

3 Introduction Ω R N ; Q T = Ω (0, T ); q T = ω (0, T ) 8 < y t (c(x) y) + d(x, t)y = v 1 ω, in Q T, y = 0, in Σ : T, y(x, 0) = y 0 (x), in Ω. (1) c := (c i,j ) C 1 (Ω; M N (R)); (c(x)ξ, ξ) c 0 ξ 2 in Ω (c 0 > 0), d L (Q T ), y 0 L 2 (Ω); v = v(x, t) is the control y = y(x, t) is the associated state. RESULTS - For any ω Ω, T > 0, y 0 L 2 (Ω), v L 2 (q T ) s.t. y(, T ) = 0 a.e. Ω [FURSIKOV-IMANUVILOV 95, LEBEAU-ROBBIANO 95] GOAL - Approximate numerically such v s. NOTATIONS - Ly := y t (c(x) y) + d(x, t)y; L ϕ := ϕ t (c(x) ϕ) + d(x, t)ϕ

4 Control of minimal L 2 -norm - Difficulties PRIMAL PROBLEM Minimize J 1 (y, v) := 1 2 q T v(x, t) 2 dx dt (2) over (y, v) C(y 0 ; T ) := { (y, v) : v L 2 (q T ), y solves (1) and satisfies y(, T ) = 0 }. DUAL PROBLEM min ϕ T H J 1 (ϕ T ) := 1 Z ϕ(x, t) 2 dxdt + y 0 (x)ϕ(x, 0)dx 2 q T Ω where ϕ solves the backward heat equation : L ϕ = 0 in Q T, ϕ = 0 on Σ T ; ϕ(, T ) = ϕ T in Ω, (3) H - Hilbert space defined as the completion of L 2 (Ω) w.r.t. ϕ T H := ϕ L 2 (q T ). Coercivity of J1 over H is a consequence of the so-called observability inequality ϕ(, 0) 2 L 2 (Ω) C(ω, T ) q T ϕ(x, t) 2 dx dt ϕ T H. (Numerical) Ill-posedness of the minimization of J1 is due to the hugeness of H.

5 L 2 (0, 1)-norm of the HUM control with respect to time Figure: y 0 (x) = sin(πx) - T = 1 - ω = (0.2, 0.8) - t v(, t) L 2 (0,1) in [0, T ] t

6 Approximation - Difficulties THE MINIMIZATION OF J1 REQUIRES TO FIND A FINITE DIMENSIONAL AND CONFORMAL APPROXIMATION OF H SUCH THAT THE CORRESPONDING DISCRETE ADJOINT SOLUTION SATISFIES (3), WHICH IS IN GENERAL IMPOSSIBLE FOR POLYNOMIAL PIECEWISE APPROXIMATION. IN PRACTICE, THE TRICK INITIALLY INTRODUCED BY GLOWINKSI-LIONS-CARTHEL 94, CONSISTS FIRST TO INTRODUCE A DISCRETE AND CONSISTENT APPROXIMATION OF (1) AND THEN TO MINIMIZE THE CORRESPONDING DISCRETE CONJUGATE FUNCTIONAL. THIS REQUIRES TO PROVE UNIFORM DISCRETE OBSERVABILITY INEQUALITIES (OPEN!) THIS PROPERTY AND THE HUGENESS OF H HAS LEAD MANY AUTHORS TO RELAX THE CONTROLLABILITY PROBLEM AND MINIMIZE OVER L 2 (Ω) THE FUNCTIONAL J 1 (ϕ T ) + ε 2 ϕ T 2 L 2 (Ω) [CARTHEL94,BOYER13, ZHENG10, MUNCH-ZUAZUA10, LABBE-TRELAT06, FERNANDEZCARA-MUNCH13,...] FOR ε = 0 AND WITHIN DUAL TYPE METHODS, STRONG CONVERGENCE OF SOME APPROXIMATIONS IS STILL OPEN!

7 Strong convergent results through a "Primal approach" In [FERNANDEZCARA-MUNCH 12,13], we consider Minimize J(y, v) := 1 2 Q T ρ 2 y 2 dx dt q T ρ 2 0 v 2 dx dt. ρ, ρ 0 C(Q T, R + ) L (Q T δ ) for any δ > 0. Let P the completed space of P 0 = {q C (Q T ) : q = 0 on Σ T } w.r.t. The optimal pair (y, v) is (p, q) P := ρ 2 L p L q dx dt + ρ 2 0 p q dx dt, Q T q T y = ρ 2 L p in Q T, v = ρ 2 0 p 1ω in Q T where p P solves the variational problem (of order 2 in t and 4 in x): (p, q) P = (y 0, q(, 0)) L 2 (Ω) q P. Carleman type weights : ρ(x, t) := exp( β(x) (T t) ), ρ 0(x, t) := (T t) 3/2 ρ(x, t)

8 Conformal space-time finite element approximation For any dimensional space P h P, we can introduce the following approximate problem: (p h, q h ) P = (y 0, q h ) L 2 (Ω), q h P h ; p h P h. (4) P h = { z h C 1,0 x,t (Q T ) : z h K (P 3,x P 1,t )(K ) K Q h, z h = 0 on Σ T }. (5) Theorem (Fernandez-Cara, M, 13) Let p h P h be the unique solution to (4), where P h is given by (5). Let us set y h := ρ 2 L A p h, v h := ρ 2 0 p h 1 qt. Then one has y y h L 2 (Q T ) 0 and v v h L 2 (q T ) 0, as h 0 where (y, v) is the minimizer of J. = NO NEED TO PROVE UNIFORM DISCRETE PROPERTY!!!!!!!

9 Mixed formulation I: the ε > 0 case Let ρ R + and let ρ 0 R defined by R := {w : w C(Q T ); w ρ > 0 in Q T ; w L (Q T δ ) δ > 0} We consider the approximate controllability case (for any ε > 0, the problem reads as follows: 8 >< >: Minimize J ε(y, v) := 1 2 q T ρ 2 0 v 2 dt + 1 2ε y(, T ) 2 L 2 (Ω) (y, v) A(y 0 ; T ) := { (y, v) : v L 2 (q T ), y solves (1) } The corresponding conjugate and well-posed problem is 8 >< >: Minimize J ε (ϕ T ) := 1 2 Subject to ϕ T L 2 (Ω). where ϕ solves q T ρ 2 0 ϕ(x, t) 2 dx dt + ε 2 ϕ T 2 L 2 (Ω) + (y 0, ϕ(, 0)) L 2 (Ω) L ϕ = 0 in Q T, ϕ = 0 on Σ T ; ϕ(, T ) = ϕ T in Ω,

10 Mixed formulation I: the ɛ > 0 case Since ϕ is completely and uniquely determined by the data ϕ T, the main idea of the reformulation is to keep ϕ as main variable. Let Φ 0 := {ϕ C 2 (Q T ), ϕ = 0 on Σ T }. For any η > 0, we define the bilinear form (ϕ, ϕ) Φ0 := ρ 2 0 ϕ ϕ dx dt + ε(ϕ(, T ), ϕ(, T )) L 2 (Ω) + η L q T Q ϕ L ϕ dx dt, T For any ε > 0, let Φ ε be the completion of Φ 0 for this scalar product. We denote the norm over Φ ε by ϕ 2 Φ ε := ρ 1 0 ϕ 2 L 2 (q T ) + ε ϕ(, T ) 2 L 2 (Ω) + η L ϕ 2 L 2, ϕ Φε. (Q T ) Finally, we defined the closed subset W ε of Φ ε by W ɛ = {ϕ Φ ε : L ϕ = 0 in L 2 (Q T )}. Then, we define the following extremal problem : min Ĵε (ϕ) := 1 ρ 2 ϕ W 0 ε 2 ϕ(x, t) 2 dx dt + ε q T 2 ϕ(, T ) 2 L 2 (Ω) + (y 0, ϕ(, 0)) L 2 (Ω). ϕ W ε ϕ(, 0) L 2 (Ω) so Ĵ ε is well-defined over Wε ϕ W ε ϕ(, T ) L 2 (Ω), problems are equivalent.

11 Mixed formulation I: the ɛ > 0 case Since ϕ is completely and uniquely determined by the data ϕ T, the main idea of the reformulation is to keep ϕ as main variable. Let Φ 0 := {ϕ C 2 (Q T ), ϕ = 0 on Σ T }. For any η > 0, we define the bilinear form (ϕ, ϕ) Φ0 := ρ 2 0 ϕ ϕ dx dt + ε(ϕ(, T ), ϕ(, T )) L 2 (Ω) + η L q T Q ϕ L ϕ dx dt, T For any ε > 0, let Φ ε be the completion of Φ 0 for this scalar product. We denote the norm over Φ ε by ϕ 2 Φ ε := ρ 1 0 ϕ 2 L 2 (q T ) + ε ϕ(, T ) 2 L 2 (Ω) + η L ϕ 2 L 2, ϕ Φε. (Q T ) Finally, we defined the closed subset W ε of Φ ε by W ɛ = {ϕ Φ ε : L ϕ = 0 in L 2 (Q T )}. Then, we define the following extremal problem : min Ĵε (ϕ) := 1 ρ 2 ϕ W 0 ε 2 ϕ(x, t) 2 dx dt + ε q T 2 ϕ(, T ) 2 L 2 (Ω) + (y 0, ϕ(, 0)) L 2 (Ω). ϕ W ε ϕ(, 0) L 2 (Ω) so Ĵ ε is well-defined over Wε ϕ W ε ϕ(, T ) L 2 (Ω), problems are equivalent.

12 Mixed formulation I: the ɛ > 0 case We consider the following mixed formulation : find (ϕ ε, λ ε) Φ ε L 2 (Q T ) solution of ( a ε(ϕ ε, ϕ) + b(ϕ, λ ε) = l(ϕ), ϕ Φ ε (6) b(ϕ ε, λ) = 0, λ L 2 (Q T ), where a ε : Φ ε Φ ε R, a ε(ϕ, ϕ) := ρ 2 0 ϕ ϕ dx dt + ε(ϕ(, T ), ϕ(, T )) L 2 (Ω) q T b : Φ ε L 2 (Q T ) R, b(ϕ, λ) := L ϕ λ dx dt Q T l : Φ ε R, l(ϕ) := (y 0, ϕ(, 0)) L 2 (Ω). We have the following result : Theorem (De souza, M 14) 1 The mixed formulation (6) is well-posed. 2 The unique solution (ϕ ε, λ ε) Φ ε L 2 (Q T ) is the unique saddle-point of the Lagrangian L ε : Φ ε L 2 (Q T ) R defined by L ε(ϕ, λ) := 1 aε(ϕ, ϕ) + b(ϕ, λ) l(ϕ). 2 3 The optimal function ϕ ε is the minimizer of Ĵ ε over W ε while the optimal multiplier λ ε L 2 (Q T ) is the state of the heat equation (1) in the weak sense.

13 Mixed formulation I: the ɛ > 0 case We consider the following mixed formulation : find (ϕ ε, λ ε) Φ ε L 2 (Q T ) solution of ( a ε(ϕ ε, ϕ) + b(ϕ, λ ε) = l(ϕ), ϕ Φ ε (6) b(ϕ ε, λ) = 0, λ L 2 (Q T ), where a ε : Φ ε Φ ε R, a ε(ϕ, ϕ) := ρ 2 0 ϕ ϕ dx dt + ε(ϕ(, T ), ϕ(, T )) L 2 (Ω) q T b : Φ ε L 2 (Q T ) R, b(ϕ, λ) := L ϕ λ dx dt Q T l : Φ ε R, l(ϕ) := (y 0, ϕ(, 0)) L 2 (Ω). We have the following result : Theorem (De souza, M 14) 1 The mixed formulation (6) is well-posed. 2 The unique solution (ϕ ε, λ ε) Φ ε L 2 (Q T ) is the unique saddle-point of the Lagrangian L ε : Φ ε L 2 (Q T ) R defined by L ε(ϕ, λ) := 1 aε(ϕ, ϕ) + b(ϕ, λ) l(ϕ). 2 3 The optimal function ϕ ε is the minimizer of Ĵ ε over W ε while the optimal multiplier λ ε L 2 (Q T ) is the state of the heat equation (1) in the weak sense.

14 Mixed formulation I: the ɛ > 0 case We consider the following mixed formulation : find (ϕ ε, λ ε) Φ ε L 2 (Q T ) solution of ( a ε(ϕ ε, ϕ) + b(ϕ, λ ε) = l(ϕ), ϕ Φ ε (6) b(ϕ ε, λ) = 0, λ L 2 (Q T ), where a ε : Φ ε Φ ε R, a ε(ϕ, ϕ) := ρ 2 0 ϕ ϕ dx dt + ε(ϕ(, T ), ϕ(, T )) L 2 (Ω) q T b : Φ ε L 2 (Q T ) R, b(ϕ, λ) := L ϕ λ dx dt Q T l : Φ ε R, l(ϕ) := (y 0, ϕ(, 0)) L 2 (Ω). We have the following result : Theorem (De souza, M 14) 1 The mixed formulation (6) is well-posed. 2 The unique solution (ϕ ε, λ ε) Φ ε L 2 (Q T ) is the unique saddle-point of the Lagrangian L ε : Φ ε L 2 (Q T ) R defined by L ε(ϕ, λ) := 1 aε(ϕ, ϕ) + b(ϕ, λ) l(ϕ). 2 3 The optimal function ϕ ε is the minimizer of Ĵ ε over W ε while the optimal multiplier λ ε L 2 (Q T ) is the state of the heat equation (1) in the weak sense.

15 Mixed formulation I: the ɛ > 0 case We consider the following mixed formulation : find (ϕ ε, λ ε) Φ ε L 2 (Q T ) solution of ( a ε(ϕ ε, ϕ) + b(ϕ, λ ε) = l(ϕ), ϕ Φ ε (6) b(ϕ ε, λ) = 0, λ L 2 (Q T ), where a ε : Φ ε Φ ε R, a ε(ϕ, ϕ) := ρ 2 0 ϕ ϕ dx dt + ε(ϕ(, T ), ϕ(, T )) L 2 (Ω) q T b : Φ ε L 2 (Q T ) R, b(ϕ, λ) := L ϕ λ dx dt Q T l : Φ ε R, l(ϕ) := (y 0, ϕ(, 0)) L 2 (Ω). We have the following result : Theorem (De souza, M 14) 1 The mixed formulation (6) is well-posed. 2 The unique solution (ϕ ε, λ ε) Φ ε L 2 (Q T ) is the unique saddle-point of the Lagrangian L ε : Φ ε L 2 (Q T ) R defined by L ε(ϕ, λ) := 1 aε(ϕ, ϕ) + b(ϕ, λ) l(ϕ). 2 3 The optimal function ϕ ε is the minimizer of Ĵ ε over W ε while the optimal multiplier λ ε L 2 (Q T ) is the state of the heat equation (1) in the weak sense.

16 Mixed formulation I: the ɛ > 0 case The bilinear form a ε is continuous over Φ ε Φ ε, symmetric and positive. The bilinear form b ε is continuous over Φ ε L 2 (Q T ). Furthermore, for any fixed ε, the continuity of the linear form l over Φ ε can be viewed from the energy estimate : ϕ(, 0) 2 L 2 (Ω) C Q T L ϕ 2 dx dt+ ϕ(, T ) 2 L 2 (Ω) max(cη 1, ε 1 ) ϕ 2 Φ ε, ϕ Φ ε. Therefore, the well-posedness is a consequence : a ε is coercive on N (b), where N (b) denotes the kernel of b : N (b) := {ϕ Φ ε : b(ϕ, λ) = 0 for every λ L 2 (Q T )}; b satisfies the usual inf-sup" condition over Φ ε L 2 (Q T ): there exists δ > 0 such that b(ϕ, λ) inf sup δ. (7) λ L 2 (Q T ) ϕ Φ ε ϕ Φε λ L 2 (Q T ) From the definition of a ε, for all ϕ N (b) = W ε, a ε(ϕ, ϕ) = ϕ 2 Φ ε.

17 Mixed formulation I: the ɛ > 0 case The bilinear form a ε is continuous over Φ ε Φ ε, symmetric and positive. The bilinear form b ε is continuous over Φ ε L 2 (Q T ). Furthermore, for any fixed ε, the continuity of the linear form l over Φ ε can be viewed from the energy estimate : ϕ(, 0) 2 L 2 (Ω) C Q T L ϕ 2 dx dt+ ϕ(, T ) 2 L 2 (Ω) max(cη 1, ε 1 ) ϕ 2 Φ ε, ϕ Φ ε. Therefore, the well-posedness is a consequence : a ε is coercive on N (b), where N (b) denotes the kernel of b : N (b) := {ϕ Φ ε : b(ϕ, λ) = 0 for every λ L 2 (Q T )}; b satisfies the usual inf-sup" condition over Φ ε L 2 (Q T ): there exists δ > 0 such that b(ϕ, λ) inf sup δ. (7) λ L 2 (Q T ) ϕ Φ ε ϕ Φε λ L 2 (Q T ) From the definition of a ε, for all ϕ N (b) = W ε, a ε(ϕ, ϕ) = ϕ 2 Φ ε.

18 Mixed formulation I: the ɛ > 0 case For any fixed λ 0 L 2 (Q T ), we define the (unique) element ϕ 0 such that L ϕ 0 = λ 0 in Q T, ϕ = 0 on Σ T, ϕ 0 (, T ) = 0 in Ω. From energy estimates, ρ 2 0 ϕ0 2 dx dt ρ 2 q T Consequently, ϕ 0 Φ ε and b(ϕ 0, λ 0 ) = λ 0 2 L 2 (Q T ) and q T ϕ 0 2 dx dt ρ 2 C Ω,T λ 0 2 L 2 (Q T ). b(ϕ, λ 0 ) sup ϕ Φ ε ϕ Φε λ 0 L 2 (Q T ) b(ϕ 0, λ 0 ) ϕ 0 Φε λ 0 L 2 (Q T ) λ 0 2 L = 2 (Q T ) ρ ϕ0 2 L 2 (q T ) + η λ L 2 (Q T ) λ 0 L 2 (Q T ) Combining the above two inequalities, we obtain b(ϕ 0, λ 0 ) sup ϕ 0 Φ ε ϕ 0 Φε λ 0 L 2 (Q T ) 1 q ρ 2 C Ω,T + η := δ

19 Mixed formulation I: the ɛ > 0 case For any fixed λ 0 L 2 (Q T ), we define the (unique) element ϕ 0 such that L ϕ 0 = λ 0 in Q T, ϕ = 0 on Σ T, ϕ 0 (, T ) = 0 in Ω. From energy estimates, ρ 2 0 ϕ0 2 dx dt ρ 2 q T Consequently, ϕ 0 Φ ε and b(ϕ 0, λ 0 ) = λ 0 2 L 2 (Q T ) and q T ϕ 0 2 dx dt ρ 2 C Ω,T λ 0 2 L 2 (Q T ). b(ϕ, λ 0 ) sup ϕ Φ ε ϕ Φε λ 0 L 2 (Q T ) b(ϕ 0, λ 0 ) ϕ 0 Φε λ 0 L 2 (Q T ) λ 0 2 L = 2 (Q T ) ρ ϕ0 2 L 2 (q T ) + η λ L 2 (Q T ) λ 0 L 2 (Q T ) Combining the above two inequalities, we obtain b(ϕ 0, λ 0 ) sup ϕ 0 Φ ε ϕ 0 Φε λ 0 L 2 (Q T ) 1 q ρ 2 C Ω,T + η := δ

20 Mixed formulation I: the ɛ > 0 case The point (ii) is due to the symmetry and to the positivity of the bilinear form a ε. (iii) Concerning the third point, the first equation of the mixed formulation reads as follows: q T ρ 2 0 ϕε ϕ dx dt+ε(ϕε(, T ), ϕ(, T )) Q T L ϕ(x, t) λ ε(x, t) dx dt = l(ϕ), ϕ Φ ε, or equivalently, since the control is given by v ε := ρ 2 0 ϕε 1ω, v ε ϕ dx dt+(εϕ ε(, T ), ϕ(, T )) L ϕ(x, t) λ ε(x, t) dx dt = l(ϕ), ϕ Φ ε. q T Q T But this means that λ ε L 2 (Q T ) is solution of the heat equation in the transposition sense. Since y 0 L 2 (Ω) and v ε L 2 (q T ), λ ε must coincide with the unique weak solution to (1) (y ε = λ ε) such that λ ε(, T ) = εϕ ε(, T ).

21 Remark 1: Augmentation of the Lagrangian Theorem 2 reduces the search of the approximated control to the resolution of the mixed formulation (6), or equivalently the search of the saddle point for L ε. In general, it is convenient to augment" the Lagrangian, and consider instead the Lagrangian L ε,r defined for any r > 0 by 8 >< L ε,r (ϕ, λ) := 1 aε,r (ϕ, ϕ) + b(ϕ, λ) l(ϕ), 2 >: a ε,r (ϕ, ϕ) := a ε(ϕ, ϕ) + r L ϕ 2 dx dt. Q T Since a ε(ϕ, ϕ) = a ε,r (ϕ, ϕ) on W ε and since the function ϕ ε such that (ϕ ε, λ ε) is the saddle point of L ε verifies ϕ ε W ε, the lagrangian L ε and L ε,r share the same saddle-point.

22 Remark 2: Dual of the dual Proposition Assume r > 0. The following equality holds : sup λ L 2 (Q T ) inf L ε,r (ϕ, λ) = ϕ Φ ε inf λ L 2 (Q T ) J ε,r (λ) + L ε,r (ϕ 0, 0). J ε,r : L 2 (Q T ) L 2 (Q T ) is defined by Jε,r (λ) := 1 (A ε,r λ) λ dx dt b(ϕ 0, λ). 2 Q T ϕ 0 Φ ε solves a ε,r (ϕ 0, ϕ) = l(ϕ), ϕ Φ ε. The linear operator A ε,r from L 2 (Q T ) into L 2 (Q T ) by A ε,r λ := L ϕ, λ L 2 (Q T ) where ϕ Φ ε solves a ε,r (ϕ, ϕ) = b(ϕ, λ), ϕ Φ ε. For any r > 0, A ε,r is a strongly elliptic, symmetric isomorphism from L 2 (Q T ) into L 2 (Q T ).

23 Remark 3: Relaxation of the constraint Similar results with L ϕ ε = 0 in L 2 (0, T ; H 1 (Ω)) = λ ε L 2 (0, T ; H 1 0 (Ω)) b : Φ ε L 2 (0, T ; H0 1 (Ω)) R, b(ϕ, Z T λ) := < L ϕ, λ > H 1,H dt

24 Mixed formulation II: the limit case ε = 0. Let ρ R. Let e Φ 0 = {ϕ C 2 (Q T ) : ϕ = 0 on Σ T } and, for any η > 0, (ϕ, ϕ) eφ0 := ρ 2 0 ϕ ϕ dx dt + η ρ q T Q 2 L ϕ L ϕ dx dt, ϕ, ϕ Φ e 0. T Let e Φ ρ0,ρ be the completion of e Φ 0 for this scalar product. We note the norm over e Φ ρ0,ρ ϕ 2 e Φρ0,ρ := ρ 1 0 ϕ 2 L 2 (q T ) + η ρ 1 L ϕ 2 L 2 (Q T ), ϕ e Φ ρ0,ρ. Finally, we defined the closed subset f W ρ0,ρ of e Φ ρ0,ρ by fw ρ0,ρ = {ϕ e Φ ρ0,ρ : ρ 1 L ϕ = 0 in L 2 (Q T )} We then define the following extremal problem : min Ĵ (ϕ) = 1 ρ 2 ϕ W e 0 ρ0,ρ 2 ϕ(x, t) 2 dx dt + (y 0, ϕ(, 0)) L 2 (Ω). (8) q T For any ϕ W f ρ0,ρ, L ϕ = 0 a.e. in Q T and ϕ = ewρ0 ρ 1,ρ 0 ϕ L 2 (q T ) so that ϕ(, T ) H.

25 Mixed formulation II: the limit case ε = 0. 8 < : ã r (ϕ, ϕ) + b(ϕ, λ) = l(ϕ), ϕ Φ e ρ0,ρ b(ϕ, λ) = 0, λ L 2 (Q T ), (9) where ã r : Φ e ρ0,ρ Φ e ρ0,ρ R, ã r (ϕ, ϕ) = ρ 2 0 ϕ ϕ dx dt + r ρ q T Q 1 L ϕ 2 dx dt T b : e Φ ρ0,ρ L 2 (Q T ) R, b(ϕ, λ) = Q T ρ 1 L ϕ λ dx dt l : e Φρ0,ρ R, l(ϕ) = (y0, ϕ(, 0)) L 2 (Ω). Theorem Let ρ 0 R and ρ R L (Q T ) and assume that there exists a positive constant K such that ρ 0 K ρ c 0, ρ K ρc in Q T. (10) 1 The mixed formulation (9) defined over e Φ ρ0,ρ L 2 (Q T ) is well-posed. 2 The optimal function ϕ is the minimizer of Ĵ over e Φ ρ0,ρ while ρ 1 λ L 2 (Q T ) is the state of the heat equation (1) in the weak sense.

26 Mixed formulation II: the limit case ε = 0. 8 < : ã r (ϕ, ϕ) + b(ϕ, λ) = l(ϕ), ϕ Φ e ρ0,ρ b(ϕ, λ) = 0, λ L 2 (Q T ), (9) where ã r : Φ e ρ0,ρ Φ e ρ0,ρ R, ã r (ϕ, ϕ) = ρ 2 0 ϕ ϕ dx dt + r ρ q T Q 1 L ϕ 2 dx dt T b : e Φ ρ0,ρ L 2 (Q T ) R, b(ϕ, λ) = Q T ρ 1 L ϕ λ dx dt l : e Φρ0,ρ R, l(ϕ) = (y0, ϕ(, 0)) L 2 (Ω). Theorem Let ρ 0 R and ρ R L (Q T ) and assume that there exists a positive constant K such that ρ 0 K ρ c 0, ρ K ρc in Q T. (10) 1 The mixed formulation (9) defined over e Φ ρ0,ρ L 2 (Q T ) is well-posed. 2 The optimal function ϕ is the minimizer of Ĵ over e Φ ρ0,ρ while ρ 1 λ L 2 (Q T ) is the state of the heat equation (1) in the weak sense.

27 Mixed formulation II: the limit case ε = 0. 8 < : ã r (ϕ, ϕ) + b(ϕ, λ) = l(ϕ), ϕ Φ e ρ0,ρ b(ϕ, λ) = 0, λ L 2 (Q T ), (9) where ã r : Φ e ρ0,ρ Φ e ρ0,ρ R, ã r (ϕ, ϕ) = ρ 2 0 ϕ ϕ dx dt + r ρ q T Q 1 L ϕ 2 dx dt T b : e Φ ρ0,ρ L 2 (Q T ) R, b(ϕ, λ) = Q T ρ 1 L ϕ λ dx dt l : e Φρ0,ρ R, l(ϕ) = (y0, ϕ(, 0)) L 2 (Ω). Theorem Let ρ 0 R and ρ R L (Q T ) and assume that there exists a positive constant K such that ρ 0 K ρ c 0, ρ K ρc in Q T. (10) 1 The mixed formulation (9) defined over e Φ ρ0,ρ L 2 (Q T ) is well-posed. 2 The optimal function ϕ is the minimizer of Ĵ over e Φ ρ0,ρ while ρ 1 λ L 2 (Q T ) is the state of the heat equation (1) in the weak sense.

28 Numerical approximations of the mixed formulations Let then Φ ε,h and M ε,h be two finite dimensional spaces parametrized by the variable h such that, for any ε > 0, Φ ε,h Φ ε, M ε,h L 2 (Q T ), h > 0. Then, we can introduce the following approximated problems : find (ϕ h, λ h ) Φ ε,h M ε,h solution of ( a ε,r (ϕ h, ϕ h ) + b(ϕ h, λ h ) = l(ϕ h ), ϕ h Φ ε,h b(ϕ h, λ h ) = 0, λ h M ε,h. (11) r > 0 a ε,r (ϕ, ϕ) r η ϕ 2 Φ ε, ϕ Φ ε a r,ε is coercive on N h (b) = {ϕ h Φ ε,h ; b(ϕ h, λ h ) = 0 λ h M ε,h } Φ ε,h Φ ε. (r L ϕ 2 acts as a (numerical) stabilization term ) L 2 (Q T ) b(ϕ h, λ h ) h > 0, δ h := inf sup > 0. (12) λ h M ε,h ϕ h Φ ε,h ϕ h Φε,h λ h Mε,h

29 Numerical approximations of the mixed formulations Φ ε,h chosen such that L ϕ h L 2 (Q T ) ϕ h Φ ε,h. Φ ε,h = {ϕ h C 1 (Q T ) : ϕ h K P(K ) K T h, ϕ h = 0 on Σ T } P(K ) - Bogner-Fox-Schmit (BFS for short) C 1 -element defined for rectangles. M ε,h = {λ h C 0 (Q T ) : λ h K Q(K ) K T h }, Q(K )- space of affine functions both in x and t on the element K. Let n h = dim Φ h, m h = dim M h. Problem (11) reads as follows : find {ϕ h } R n h and {λ h } R m h such that Aε,r,h Bh T ««{ϕh } Lh =. (13) B h 0 {λ h } 0 «R n h +m h R n h +m h,n h +m h R n h +m h

30 The discrete inf-sup test : inf h δ ε,r,h > 0? b(ϕ h, λ h ) δ ε,r,h := inf sup λ h M ε,h ϕ h Φ ε,h ϕ h Φε,h λ h Mε,h j ff δ := inf : Bh A 1 ε,r,h BT h {λ h} = δ J h {λ h }, {λ h } R m h \ {0}. (14) We obtain that δ ε,r,h C ε,r,h r 1 ; C ε,r,h = O(1) (r large) (15) = APPROXIMATIONS Φ ε,h, M ε,h DO PASS THE TEST!

31 Inf-sup test: δ ε,r,h w.r.t. ε and h. Ω = (0, 1), ω = (0.2, 0.5) T = 1/2, ρ 0 (x, t) := (T t) 3/2 exp 3 4(T t) «, (16) r = 10 2 : h ε = ε = ε = r = 10 2 h ε = ε = ε = δ ε,r,h C ε,r,h r 1 ; C ε,r,h = O(1) (r large) (17) = SPACE (Φ ε,h, M ε,h ) DO PASS THE TEST!

32 Mixed formulation I: the ε > 0 case ε = 10 2 y 0 (x) = sin(πx), Ω = (0, 1) ω = (0.2, 0.5), T = 1/2, C := 0.1, d = h h Figure: ρ 0(v ε v ε,h ) L 2 (qt ) ρ 0 v ε (Left) and yε λ ε,h L 2 (QT ) L 2 (qt y ) ε L 2 (QT ) r = 10 2 ( ), r = 1. ( ) and r = 10 2 ( ). (Right) vs. h for

33 Mixed formulation I: the ɛ > 0 case ε = 10 8 y 0 (x) = sin(πx), Ω = (0, 1) ω = (0.2, 0.5), T = 1/2, C := 0.1, d = h h Figure: ρ 0(v ε v ε,h ) L 2 (qt ) ρ 0 v ε (Left) and yε λ ε,h L 2 (QT ) L 2 (qt y ) ε L 2 (QT ) r = 10 2 ( ), r = 1. ( ) and r = 10 2 ( ). (Right) vs. h for

34 Minimization of J ε,r by a conjugate gradient A ε,r (λ) L 2 (Q T ) r 1 λ L 2 (Q T ) ν(a ε,r ) = A ε,r A 1 ε,r = ν(b h A 1 ε,r,h BT h ) r 1 δ 2 ε,r,h p C ε,r,h h ε = ε = ε = Table: r 1 δ 2 ε,h w.r.t. ε and h ; r = 10 2 ; Ω = (0, 1), ω = (0.2, 0.5), T = 1/2. h iterates - ε = iterates - ε = iterates - ε = κ(a ε,r,h ) - ε = Table: Mixed formulation (6) - r = ω = (0.2, 0.5) ; Conjugate gradient algorithm.

35 Mixed formulation II: the ɛ = 0 case Same features for the limit case, up to a crucial point : make the change of variable ϕ := ρ 0 ψ and solve the mixed formulation w.r.t. (ψ, λ) over ρ 1 0 eφ ρ0,ρ L 2 (Q T ). â : ρ 1 0 eφ ρ0,ρ ρ 1 0 eφ ρ0,ρ R, â(ψ, ψ) = ψ ψ dx dt q T ˆb : ρ 1 0 eφ ρ0,ρ L 2 (Q T ) R, ˆb(ψ, λ) = ρ Q 1 L (ρ 0 ψ) λ dx dt T ˆl : ρ 1 0 eφ ρ0,ρ R, ˆl(ϕ) = (y0, ρ 0 (, 0)ψ(, 0)) L 2 (Ω). 3 If ρ(t) = exp( 4(T t) ) and ρ 0(t) = (T t) 3/2 ρ(t) : ρ 1 L (ρ 0 ψ) = ρ 1 ρ 0 L ψ ρ 1 ρ 0t ψ = (T t) 3/2 L ψ + 32 «(T t)1/2 + K 1 (T t) 1/2 ψ. (18)

36 Mixed formulation II: the ɛ = 0 case h ρ 1 L (ρ 0 ψ h ) L 2 (QT ) ρ 0 (v v h ) L 2 (qt ) ρ 0 v L 2 (qt ) ρ 0 v h L 2 (qt ) ρ 1 λ h L 2 (QT ) y ρ 1 λ h L 2 (QT ) y L 2 (QT ) CG iterates r 1 δ 2 r,h κ(a r,h ) n h =size(a r,h ) y h (, T ) L 2 (0,1) Table: Mixed formulation (9) - r = 10 2 and ε = 0 with ω = (0.2, 0.5).

37 Mixed formulation II: the ɛ = 0 case ω = (0.2, 0.5); y 0 (x) = sin(πx), ε = h h Figure: ρ 0 (v v h ) L 2 (q T ) ρ 0 v L 2 (qt ) r = 10 2 ( ). y ρ 1 λ h L 2 (QT ) (Left) and (Right) vs. h for r = 10 2 ( ), r = 1. ( ) and y L 2 (QT )

38 Mixed formulation II: the ɛ = 0 case ω = (0.2, 0.5); y 0 (x) = sin(πx), ε = Figure: ω = (0.2, 0.5); Approximation v h = ρ 1 0 ψ h of the null control v over Q T - r = 1 and h =

39 Mixed formulation II: the ɛ = 0 case ω = (0.2, 0.5); y 0 (x) = sin(πx), ε = Figure: ω = (0.2, 0.5); Approximation ρ 1 λ h of the controlled state y over Q T - r = 1 and h =

40 Concluding remarks MIXED FORMULATION ALLOWS TO APPROXIMATE DIRECTLY WEIGHTED L 2 CONTROLS FOR THE HEAT EQ. THE MINIMISATION OF Jr (λ) IS VERY ROBUST AND FAST CONTRARY TO THE MINIMISATION OF J (ϕ T ) (INVERSION OF SYMMETRIC DEFINITE POSITIVE AND VERY SPARSE MATRICE WITH DIRECT CHOLESKY SOLVERS) SPACE-TIME FINITE ELEMENT FORMULATION IS VERY WELL-ADAPTED TO MESH ADAPTATION AND TO NON-CYLINDRICAL SITUATION DIRECT APPROACH CAN BE USED OF MANY OTHER CONTROLLABLE SYSTEMS FOR WHICH APPROPRIATE CARLEMAN ESTIMATES ARE AVAILABLE. [CINDEA,MUNCH 2014 CALCOLO] FOR THE WAVE EQ. THE PRICE TO PAY IS TO USED C 1 FINITE ELEMENTS (AT LEAST IN SPACE) UNLESS L ϕ = 0 IS SEEN IN A WEAKER SPACE THAN L 2 (Q T ). A NICE OPEN QUESTION IF THE DISCRETE INF-SUP PROPERTY!?? A SIMPLE STRATEGY IS TO ADD THE LAGRANGIAN THE TERM Lλ h + ρ 2 0 ϕ h 1 ω 2 L 2 (Q T )

41 Concluding remarks THIS APPROACH MAY BE APPLIED FOR INVERSE PROBLEMS, OBSERVATION PROBLEMS, RECONSTRUCTION OF DATA,... Given the observation z L 2 (q T ), find y such that 8 >< Ly = 0 in Q T, y = z in q T, >: y = 0 on Σ T Solve the Least-Squares problem : 1 inf y Y 2 q T (y z) 2 dx dt with Y = {y L 2 (q T ), Ly = 0 in L 2 (Q T ), y = 0 on Σ T }, through a mixed formulation... In progress! THANK YOU FOR YOUR ATTENTION

42 Concluding remarks THIS APPROACH MAY BE APPLIED FOR INVERSE PROBLEMS, OBSERVATION PROBLEMS, RECONSTRUCTION OF DATA,... Given the observation z L 2 (q T ), find y such that 8 >< Ly = 0 in Q T, y = z in q T, >: y = 0 on Σ T Solve the Least-Squares problem : 1 inf y Y 2 q T (y z) 2 dx dt with Y = {y L 2 (q T ), Ly = 0 in L 2 (Q T ), y = 0 on Σ T }, through a mixed formulation... In progress! THANK YOU FOR YOUR ATTENTION