Minimal time issues for the observability of Grushin-type equations

Transcription

1 Intro Proofs Further Minimal time issues for the observability of Grushin-type equations Karine Beauchard (1) Jérémi Dardé (2) (2) (1) ENS Rennes (2) Institut de Mathématiques de Toulouse GT Contrôle LJLL 09/03/2018

2 Outline Intro Proofs Further 1 Introduction Context Results 2 Proofs Strategy Control on the two lateral boundaries Boundary condition at x = 0 Observation at one end 3 Further comments

3 Outline Intro Proofs Further Context Results 1 Introduction Context Results 2 Proofs Strategy Control on the two lateral boundaries Boundary condition at x = 0 Observation at one end 3 Further comments

4 Outline Intro Proofs Further Context Results 1 Introduction Context Results 2 Proofs Strategy Control on the two lateral boundaries Boundary condition at x = 0 Observation at one end 3 Further comments

5 Context Intro Proofs Further Context Results Null-controllability problem at T for the heat equation through Γ Let Ω R d, Γ Ω. The heat equation with control function v L 2 (0, T ; L 2 (Γ)): t u x u = 0, in (0, T ) Ω, u(t, x) = v(t, x)1 Γ (x), in (0, T ) Ω, u(0, x) = u 0 (x) in Ω. Given u 0 L 2 (Ω), can we find a control function v L 2 (0, T ; L 2 (Γ)) s.t. u(t, ) = 0? YES [Fursikov Imanuvilov 96, Lebeau Robbiano 95] For any time T > 0 and any non-empty open subset Γ Ω.

6 Intro Proofs Further Context Results By duality [Fattorini-Russell 71], Null-controllability Observability. Observability of the heat equation through (0, T ) Γ Let Ω R d, Γ Ω. We consider the following heat equation: t z x z = 0, in (0, T ) Ω, z(t, x) = 0, in (0, T ) Ω, z(0, x) = z 0 (x) in Ω. Does there exist a constant C > 0 such that for all z 0 H 1 0 (Ω) z(t, ) L 2 (Ω) C νz L 2 ((0,T ) Γ)? YES Proof by Carleman estimates [Fursikov Imanuvilov 96].

7 Intro Proofs Further Context Results General motivation Understand what happens for degenerate parabolic equations. Typical examples: Degeneracies on the boundary of the domain: { t z x (x 2α x z) = 0, (t, x) (0, T ) (0, L), z(t, 0) = z(t, L) = 0, t (0, T ). See [Cannarsa Martinez Vancostenoble 16] for latest developments. Degeneracies inside the domain Ω = ( L, L) (0, π): { t z xx z x 2α yy z = 0, (t, x, y) (0, T ) Ω, z(t, x, y) = 0, (t, x, y) (0, T ) Ω.

8 Intro Proofs Further Context Results Focus on the case of interior degeneracies In Ω = ( L, L) (0, π), Grushin type operators: { t z xx z x 2α yy z = 0, (t, x, y) (0, T ) Ω, z(t, x, y) = 0, (t, x, y) (0, T ) Ω. Boundary observation through Γ Ω: z(t ) L 2 (Ω) C νz L 2 ((0,T ) Γ). Internal/distributed Observation through ω Ω: z(t ) L 2 (Ω) C z L 2 ((0,T ) ω).

9 Known results Intro Proofs Further Context Results Ω = ( L, L) (0, π), Grushin type operators: { t z xx z x 2α yy z = 0, (t, x, y) (0, T ) Ω, z(t, x, y) = 0, (t, x, y) (0, T ) Ω. [Beauchard Cannarsa Guglielmi 14] α < 1: Observable in any time T > 0 from any open subset ω Ω or Γ Ω. Like the usual heat equation. α > 1: Not observable from ω, whatever T > 0, when ω {x = 0} =. The strong degeneracy prevents from any observability result.

10 Intro Proofs Further Context Results The case α = 1 Ω = ( L, L) (0, π), Grushin type operators: { t z xx z x 2 yy z = 0, (t, x, y) (0, T ) Ω, z(t, x, y) = 0, (t, x, y) (0, T ) Ω. If ω = ω x (0, π), and ω x {0} (0, π) =, there exists a critical time T ω > 0 such that The equation is not observable in any time T < T ω. For any T > T ω, the equation is observable: z(t ) L 2 (Ω) C z L 2 ((0,T ) ω). cf [Beauchard Cannarsa Guglielmi 14]. If there is an horizontal strip which does not meet ω, there is never observability whatever T > 0 is [Koenig 17].

11 Intro Proofs Further Context Results The case α = 1 Ω = ( L, L) (0, π), Grushin type operators: { t z xx z x 2 yy z = 0, (t, x, y) (0, T ) Ω, z(t, x, y) = 0, (t, x, y) (0, T ) Ω. If Γ = Γ x (0, π), with Γ x = {L}, { L} or { L, L}, there exists a critical time TΓ > 0 such that The equations are not observable in any time T < TΓ. For any T > TΓ, the equations are observable: z(t ) L 2 (Ω) C xz L 2 ((0,T ) Γ). cf [Beauchard Cannarsa Guglielmi 14]: T Γ L2 /2.

12 Outline Intro Proofs Further Context Results 1 Introduction Context Results 2 Proofs Strategy Control on the two lateral boundaries Boundary condition at x = 0 Observation at one end 3 Further comments

13 Intro Proofs Further Context Results Observations from two ends Ω = ( L, L) (0, π): t z xx z x 2 yy z = 0 in (0, T ) Ω, z(t, x, y) = 0 in (0, T ) Ω, z(0, x, y) = z 0 (x, y) in Ω, Theorem [K. Beauchard, J. Dardé & S.E. 2018] Let T > L 2 /2 and Γ = { L, L} (0, π). Then there exists C > 0 such that for all smooth solutions z, z(t ) L 2 (Ω) C xz L 2 ((0,T ) Γ). Sharp time: No observability if T < L 2 /2. See [Beauchard Cannarsa Guglielmi 2014]. Same as in [Beauchard Miller Morancey 2015], but proof.

14 Intro Proofs Further Context Results Observation from one end Ω = ( L, L) (0, π): t z xx z x 2 yy z = 0 in (0, T ) Ω, z(t, x, y) = 0 in (0, T ) Ω, z(0, x, y) = z 0 (x, y) in Ω, Theorem [K. Beauchard, J. Dardé & S.E. 2018] Let T > L 2 /2 and Γ = {L} (0, π). Then there exists C > 0 such that for all smooth solutions z, z(t ) L 2 (Ω) C xz L 2 ((0,T ) Γ). Same time as for Γ = { L, L} (0, π). Sharp time: No controllability if T < L 2 /2. See [Beauchard Cannarsa Guglielmi 2014].

15 Intro Proofs Further Context Results Grushin equations in non-symmetric domains Ω = ( L, L + ) (0, π): { t z xx z x 2 yy z = 0 in (0, T ) Ω, z(t, x, y) = 0, (t, x, y) (0, T ) Ω. Theorem [K. Beauchard, J. Dardé & S.E. 2018] Let T > L 2 +/2 and Γ = {L + } (0, π). Then there exists C > 0 such that for all smooth solutions z, z(t ) L 2 (Ω) C xz L 2 ((0,T ) Γ). Sharp time Agmon estimates.

16 Intro Proofs Further Context Results The role of boundary conditions Ω = (0, L) (0, π): { t z xx z x 2 yy z = 0 in (0, T ) Ω, z(t, x, y) = 0, (t, x, y) (0, T ) ( Ω \ {0} (0, π)). + Dirichlet conditions z(t, 0, y) = 0 in (0, T ) (0, π) or Neumann conditions x z(t, 0, y) = 0 in (0, T ) (0, π). Theorem [K. Beauchard, J. Dardé & S.E. 2018] When Γ = {L} (0, 1), the critical time for observability is T = L 2 /6 in the Dirichlet case; T = L 2 /2 in the Neumann case. The BC at x = 0 plays an important role.

17 Comments Intro Proofs Further Context Results Our results also apply to Heisenberg equations in tensorized domains: t 2 x (x y + z ) 2. see [Beauchard Cannarsa 17] for previous results. Some inverse problems similar to the ones of [Beauchard Cannarsa Yamamoto 14]. Slightly more general settings for Grushin equations: Ω = Ω x Ω y, with Ω y of any dimension, Operators of the form t 2 x q(x) 2 2 y with q C 3 ( L, L + ), q(0) = 0, inf x q > 0. Critical time T = 1 L+ q (0) 0 q(x) dx when observed from L +.

18 Outline Intro Proofs Further Strategy 2 ends BC in 0 One end 1 Introduction Context Results 2 Proofs Strategy Control on the two lateral boundaries Boundary condition at x = 0 Observation at one end 3 Further comments

19 Outline Intro Proofs Further Strategy 2 ends BC in 0 One end 1 Introduction Context Results 2 Proofs Strategy Control on the two lateral boundaries Boundary condition at x = 0 Observation at one end 3 Further comments

20 Intro Proofs Further Strategy 2 ends BC in 0 One end Each result concerns observability properties for the equation: ( t x 2 x 2 y 2 )z(t, x, y) = 0, (t, x, y) (0, T ) Ω, z(t, x, y) = 0, (t, x, y) (0, T ) Ω, z(0,.,.) = z 0 L 2 (Ω). We shall take advantage of the tensorized form of the problem: We use Fourier series. z(t, x, y) = n z n (t, x) sin(ny). ( t x 2 + n 2 x 2 )z n (t, x) = 0, (t, x) (0, T ) ( L, L), z n (t, L) = z n (t, L) = 0, t (0, T ), z n (0,.) = z 0,n H0 1 ( L, L),

21 Intro Proofs Further Strategy 2 ends BC in 0 One end Use Fourier series. ( t x 2 + n 2 x 2 )z n (t, x) = 0, (t, x) (0, T ) ( L, L), z n (t, L) = z n (t, L) = 0, t (0, T ), z n (0,.) = z 0,n L 2 ( L, L),? Uniform? observability problems If observed from L and L: L L z n (T, x) 2 dx C If observed from L: L L T 0 z n (T, x) 2 dx C ( x z n (t, L) 2 + x z n (t, L) 2 )dt. T 0 x z n (t, L) 2 dt.

22 Intro Proofs Further Strategy 2 ends BC in 0 One end To prove a uniform observability result in time T, we use A careful analysis of the cost of observability of a family of 1-d heat equations as n, for a given T 0 > 0 (small): z n (T 0 ) L 2 ( L,L) CeAn observation L 2 (0,T 0 ), with C and A independent of n N. The dissipation of each semigroup: z n (T ) L 2 ( L,L) e µn(t T 0) z n (T 0 ) L 2 ( L,L), with µ independent of n. Uniform observability provided T T 0 + A µ. As T 0 is arbitrarily small, this gives T > A/µ.

23 Intro Proofs Further Strategy 2 ends BC in 0 One end Follows the same strategy as in [Beauchard Cannarsa Guglielmi 14]. The dissipation rates of each semi-group are known (µ = 1 in the case x ( L, L)) It mainly remains to analyze the cost of observability of a family of 1-d heat equations, in the asymptotics n and for a given T 0 > 0 (small): z n (T 0 ) L 2 ( L,L) CeAn observation L 2 (0,T 0 ). Remark To get a sharp result, we should obtain sharp estimates on A.

24 Outline Intro Proofs Further Strategy 2 ends BC in 0 One end 1 Introduction Context Results 2 Proofs Strategy Control on the two lateral boundaries Boundary condition at x = 0 Observation at one end 3 Further comments

25 Intro Proofs Further Strategy 2 ends BC in 0 One end Observation from both boundaries ±L ( t x 2 + n 2 x 2 )z n (t, x) = 0, (t, x) (0, T ) ( L, L), z n (t, L) = z n (t, L) = 0, t (0, T ), z n (0,.) = z 0,n H0 1 ( L, L), Then ( w n (t, x) = z n (t, x)exp n coth(2nt) ) (L 2 x 2 ) 2 satisfies ( t x 2 + 2θ n (t)x x + θ n(t) L2 ) 2 + θ n(t) w n (t, x) = 0 w n (t, L) = w n (t, L) = 0, where we set θ n (t) = n coth(2nt). No terms in x 2 anymore!

26 Intro Proofs Further Strategy 2 ends BC in 0 One end (Carleman type) Energy estimates: L L ( x w n (T 0 ) 2 2L n 2 L 2 sinh(2nt 0 ) 2 T0 ) 2 w n(t 0 ) 2 Given T 0 > 0, n 0 N, s.t. for all n n 0, L L w n (T 0 ) 2 = L L 0 dx ( x w n (t, L) 2 + x w n (t, L) 2 )dt. ( ) z n (T 0, x) 2 exp n coth(2nt 0 )(L 2 x 2 ) C T0 0 ( x z n (t, L) 2 + x z n (t, L) 2 )dt Observability cost in exp(nl 2 /2) at time T 0. To be combined with the dissipation in exp( n(t T 0 )).

27 Intro Proofs Further Strategy 2 ends BC in 0 One end Why this choice? see [Dardé Ervedoza, 16, 17]. The fundamental solution of ( t xx + x 2 )K (t, x, y) = δ t=0 δ x=y is given by the Mehler kernel K (t, x, y) = ( 1 coth(2t) (2π sinh(2t)) 1/2 e ) x 2 + y 2 2x y 2 sinh(2t). Scaling (t, x, y) (nt, nx, ny) and y il, gives K n (t, x, il) = ( 1 coth(2nt) (2π sinh(2nt)) 1/2 e n The weight function is the exponential envelop of 1/K n (t, x, il). ) x 2 L 2 2.

28 Outline Intro Proofs Further Strategy 2 ends BC in 0 One end 1 Introduction Context Results 2 Proofs Strategy Control on the two lateral boundaries Boundary condition at x = 0 Observation at one end 3 Further comments

29 Intro Proofs Further Strategy 2 ends BC in 0 One end The role of the boundary condition at x = 0 Corresponding to Dirichlet boundary conditions: { ( t x 2 + n 2 x 2 )z n (t, x) = 0, (t, x) (0, T ) (0, L), z n (t, 0) = z n (t, L) = 0, t (0, T ), Corresponding to Neumann boundary condition { ( t x 2 + n 2 x 2 )z n (t, x) = 0, (t, x) (0, T ) (0, L), x z n (t, 0) = z n (t, L) = 0, t (0, T ), Cost of observability in exp(nl 2 /2) by symmetry arguments. Dissipation in exp( 3nt) for Dirichlet BC, exp( nt) for Neumann BC. T = L 2 /6 for Dirichlet BC, T = L 2 /2 for Neumann BC.

30 Outline Intro Proofs Further Strategy 2 ends BC in 0 One end 1 Introduction Context Results 2 Proofs Strategy Control on the two lateral boundaries Boundary condition at x = 0 Observation at one end 3 Further comments

31 Intro Proofs Further Strategy 2 ends BC in 0 One end Observation at one end ( t x 2 + n 2 x 2 )z n (t, x) = 0, (t, x) (0, T ) ( L, L), z n (t, L) = z n (t, L) = 0, t (0, T ), z n (0,.) = z 0,n H0 1 ( L, L), Goal Get a uniform observability inequality z n (T ) L 2 ( L,L) C exp(a n) xz n (t, L) L 2 (0,T ), with a precise knowledge of A.

32 Intro Proofs Further Strategy 2 ends BC in 0 One end Proof done in several steps: Passing the information from the right x = L to the left of the singularity x = ε, ε > 0 small. Passing the information from the singularity x = 0 to the extreme left of the domain L. A gluing argument. The first two steps are done by Carleman estimates.

33 Intro Proofs Further Strategy 2 ends BC in 0 One end From x = L to x = 0 (or ε) ϕ R,n (t, x) = nθ(t)ψ R (x) + θ(t), (t, x) (0, T ) ( ε, L) with θ and Ψ R as follows 1/t for t < T /4, θ C 1 for t (T /3, 2T /3), (0, T ), θ(t) = 1/(T t) for t > 3T /4, 1 for t (0, T ), Ψ R (x) = L2 x ε(L x), x ( ε, L). Remark ϕ R,n (t, x) n coth(2nt) L2 x 2. 2

34 Intro Proofs Further Strategy 2 ends BC in 0 One end Then n 0 > 0 and C > 0 s.t. for all n n 0, for all u n satisfying ( t x 2 + n 2 x 2 )u n = 0, (t, x) (0, T ) ( ε, L), u n (t, ε) = u n (t, L) = 0, t (0, T ), u n (0,.) = u 0,n H0 1 ( ε, L), we have n 3/2 θ 3/2 u n e ϕ R,n θ L 2 ((0,T ) ( ε,l)) Cn1/2 1/2 x u n (t, L + )e θ(t) L 2 (0,T A typical Carleman estimate: Weight function adapted to the potential x 2, inspired by the Mehler kernel. The Carleman parameter is chosen = n. Sources terms can be handled easily.

35 Intro Proofs Further Strategy 2 ends BC in 0 One end From x = 0 to x = L There, the potential n 2 x 2 improves the observability property. We choose a weight function: ϕ L,n (t, x) = nθ(t)a ( ) x 2 nθ(t) 2 + 2Lx, (t, x) (0, T ) ( L, 0), where A is a suitable positive constant.

36 Intro Proofs Further Strategy 2 ends BC in 0 One end n 0 > 0 and C > 0 s.t. for all n n 0, for all u n satisfying ( t x 2 + n 2 x 2 )u n = 0, (t, x) (0, T ) ( L, 0), u n (t, L) = u n (t, 0) = 0, t (0, T ), u n (0,.) = u 0,n H0 1 ( L, 0). we have n 3/4 θ 3/2 u n e ϕ L,n θ L 2 ((0,T ) ( L,0)) Cn1/4 1/2 x u n (t, 0)e nθ(t)a L 2 (0,T The weight function ϕ L,n (t, x) = nθ(t)a ( ) x 2 nθ(t) 2 + 2Lx, ϕ L,n is essentially constant in space nθ(t)a. Variations (in x) of the weight function are of lower order. A will be chosen to match Ψ R,n (t, 0) for the gluing argument.

37 Outline Intro Proofs Further 1 Introduction Context Results 2 Proofs Strategy Control on the two lateral boundaries Boundary condition at x = 0 Observation at one end 3 Further comments

38 Intro Proofs Further Further comments Our results also apply to Non-symmetric geometric settings. Heisenberg equations in tensorized domains: t 2 x (x y + z ) 2. see [Beauchard Cannarsa 17] for previous results. Some inverse problems similar to the ones of [Beauchard Cannarsa Yamamoto 14]. Slightly more general settings for Grushin equations: Ω = Ω x Ω y, with Ω y of any dimension, Operators of the form t 2 x q(x) 2 2 y with q C 3 ( L, L + ), q(0) = 0, inf x q > 0. Critical time T = 1 L+ q (0) 0 q(x) dx when observed from L +.

39 Intro Proofs Further Open problems Determine the correct geometric condition in non-tensorized settings for which observability holds for Grushin operators. Determine the time required for observability for Kolmogorov like operators t vv + v 2 x, or t vv + v 2 ( x ) 1/2. See [Beauchard Helffer Henry Robbiano 15]. Precisely describe the reachable sets in each of the above situations, and how it evolves in time. See [Dardé Ervedoza 16].

40 Intro Proofs Further Merci pour votre attention! Ref: Minimal time issues for the observability of Grushin-type equations, Karine Beauchard, Jérémi Dardé,, Available on HaL.