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Carleman estimates for elliptic PDE and applications Sylvain - - PowerPoint PPT Presentation

Cauchy Linear Weight General Weights Carleman estimates for elliptic PDE and applications Sylvain Ervedoza Institut de Mathmatiques de Toulouse & CNRS Monastir - Mai 2017 Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE


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Cauchy Linear Weight General Weights

Carleman estimates for elliptic PDE and applications

Sylvain Ervedoza

Institut de Mathématiques de Toulouse & CNRS

Monastir - Mai 2017

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights

Goal

Carleman estimates for elliptic PDE and applications Cauchy Problem for the Laplace operator. Carleman estimates with a linear weight. Applications to the Calderón problem. Carleman estimates with a general weight : the role of strict pseudo-convexity. Applications to the unique continuation property for the Laplace operator.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights Full Info Partial Info Intermediate

Outline

1

The Cauchy problem for the Laplace operator The case of full informations The case of partial data Intermediate case

2

Carleman estimates with a linear weight Goal Proof of the Carleman estimate : Fourier techniques Proof of the Carleman estimate : Multiplier techniques More general geometric settings Application to the Calderón Problem

3

More general Carleman Weights The case of a strip The case of a strip with a multiplier technique The general case More on unique continuation

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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The Cauchy problem

The Cauchy Problem for the Laplace operator We look for u solution of    ∆u = f , in Ω, u = gD,

  • n Γ,

∂nu = gN,

  • n Γ.

Known data of the problem : The geometric setting, the domain Ω and a part Γ of ∂Ω where the Dirichlet and Neumann conditions are known. f source term in Ω. gD Dirichlet data on Γ. gN Neumann data on Γ. Physically (Electronic) : u is a potential, ∇u is a current.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Case Γ = ∂Ω, Cauchy problem with full information

The Cauchy problem, Case Γ = ∂Ω We look for u solution of    ∆u = f , in Ω, u = gD,

  • n ∂Ω,

∂nu = gN,

  • n ∂Ω.

When Γ = ∂Ω, the information on the boundary is redundant : Using classical elliptic theory, one can solve ∆u = f , in Ω, u = gD,

  • n ∂Ω,

OR ∆u = f , in Ω, ∂nu = gN,

  • n ∂Ω.

yielding u ∈ H2(Ω) for f ∈ L2(Ω) and gD ∈ H3/2(∂Ω) f ∈ L2(Ω) and gN ∈ H1/2(∂Ω), in the case

  • Ω f =
  • ∂Ω gN.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights Full Info Partial Info Intermediate

Case Γ = ∂Ω, Cauchy problem with full information

The Cauchy problem, Case Γ = ∂Ω We look for u solution of    ∆u = f , in Ω, u = gD,

  • n ∂Ω,

∂nu = gN,

  • n ∂Ω.

Consequences (When Γ = ∂Ω) : (f , gD) completely determines u, thus gN. (f , gN) completely determines u, thus gD. A correct functional setting is given by f ∈ L2(Ω), gD ∈ H3/2(∂Ω), gN ∈ H1/2(∂Ω), u ∈ H2(Ω).

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy problem with partial data Γ = ∂Ω

The Cauchy Problem for the Laplace operator We look for u solution of    ∆u = f , in Ω, u = gD,

  • n Γ,

∂nu = gN,

  • n Γ.

Known data of the problem : The geometric setting, the domain Ω and a part Γ of ∂Ω where the Dirichlet and Neumann conditions are known. f source term in Ω. gD Dirichlet data on Γ. gN Neumann data on Γ.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights Full Info Partial Info Intermediate

Cauchy problem with partial data

The Cauchy Problem for the Laplace operator We look for u solution of    ∆u = f , in Ω, u = gD,

  • n Γ,

∂nu = gN,

  • n Γ.

To simplify the study, we will restrict ourselves to the following simplified geometric setting : Ω = (0, 1) × Rd−1 is a vertical strip (d ≥ 1). Γ = {0} × Rd−1 is one side of the strip.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy problem with Ω = (0, 1) × Rd−1, Γ = {0} × Rd−1

We look for u solution of    ∆u = f , in Ω, u = gD,

  • n Γ,

∂nu = gN,

  • n Γ,

i.e.    ∂11u + ∆′u = f , for (x1, x′) ∈ (0, 1) × Rd−1, u(0, x′) = gD(x′), for x′ ∈ Rd−1, ∂1u(0, x′) = gN(x′), for x′ ∈ Rd−1, where ∆′ = ∂22 + · · · + ∂dd is the Laplace operator in the variable x′ = (x2, · · · , xd) ∈ Rd−1.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Partial Fourier transform

We introduce ˆ u(x1, ξ′) = Fx′→ξ′u(x1, ·), ˆ f (x1, ξ′) = Fx′→ξ′f (x1, ·), ˆ gD(ξ′) = Fx′→ξ′gD(·), ˆ gN(ξ′) = Fx′→ξ′gN(·), where Fx′→ξ′ is the Fourier transform in the variable x′, defined for function v ∈ S (Rd−1) of x′ by ∀ξ′ ∈ Rd−1, (Fx′→ξ′v)(ξ′) =

  • Rd−1 v(x′) e−ix′·ξ′ dx′.

Fx′→ξ′ enjoys the same properties as the usual Fourier transform : It is an isometry of L2(Rd−1). Its action on derivatives is easily understood : Fx′→ξ′(∆x′v)(ξ′) = −|ξ′|2(Fx′→ξ′v)(ξ′).

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Therefore, the Cauchy problem    ∂11u + ∆′u = f , for (x1, x′) ∈ (0, 1) × Rd−1, u(0, x′) = gD(x′), for x′ ∈ Rd−1, ∂1u(0, x′) = gN(x′), for x′ ∈ Rd−1, now reads    ∂11 ˆ u − |ξ′|2 ˆ u = ˆ f , for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ u(0, ξ′) = ˆ gD(ξ′), for ξ′ ∈ Rd−1, ∂1 ˆ u(0, ξ′) = ˆ gN(ξ′), for ξ′ ∈ Rd−1, Advantages : We are back to analyze a family of ODE indexed by the frequency parameter ξ′ ∈ Rd−1.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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At ξ′ ∈ Rd−1 fixed, one can solve    ∂11 ˆ u − |ξ′|2 ˆ u = ˆ f , for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ u(0, ξ′) = ˆ gD(ξ′), for ξ′ ∈ Rd−1, ∂1 ˆ u(0, ξ′) = ˆ gN(ξ′), for ξ′ ∈ Rd−1, using Duhamel formula : ˆ u(x1, ξ′) = cosh(|ξ′|x1)ˆ gD(ξ′) + sinh(|ξ′|x1) |ξ′| ˆ gN(ξ′) + x1 sinh(|ξ′|(x1 − x)) |ξ′| ˆ f (x, ξ′) dx. As the Fourier transform is invertible and this identity holds ∀ξ′, the Cauchy problem is solved. We even have a fomula to give the solution !

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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At ξ′ ∈ Rd−1 fixed, one can solve    ∂11 ˆ u − |ξ′|2 ˆ u = ˆ f , for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ u(0, ξ′) = ˆ gD(ξ′), for ξ′ ∈ Rd−1, ∂1 ˆ u(0, ξ′) = ˆ gN(ξ′), for ξ′ ∈ Rd−1, using Duhamel formula : ˆ u(x1, ξ′) = cosh(|ξ′|x1)ˆ gD(ξ′) + sinh(|ξ′|x1) |ξ′| ˆ gN(ξ′) + x1 sinh(|ξ′|(x1 − x)) |ξ′| ˆ f (x, ξ′) dx. As the Fourier transform is invertible and this identity holds ∀ξ′, the Cauchy problem is solved. We even have a fomula to give the solution ! This last sentence is partially wrong !

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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What we have proved : If we know that there exists u solving    ∆u = f , in Ω, u = gD,

  • n Γ,

∂nu = gN,

  • n Γ,

then it is unique, and it is given by the formula ˆ u(x1, ξ′) = cosh(|ξ′|x1)ˆ gD(ξ′) + sinh(|ξ′|x1) |ξ′| ˆ gN(ξ′) + x1 sinh(|ξ′|(x1 − x)) |ξ′| ˆ f (x, ξ′) dx. However, we never said that such u exist !

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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In particular, the formula ˆ u(x1, ξ′) = cosh(|ξ′|x1)ˆ gD(ξ′) + sinh(|ξ′|x1) |ξ′| ˆ gN(ξ′) + x1 sinh(|ξ′|(x1 − x)) |ξ′| ˆ f (x, ξ′) dx. may not belong to S′(Rd−1) and therefore, the inverse Fourier transform may not exist. We only have : ∃C > 0, s.t. ∀x1 ∈ (0, 1), |ˆ u(x1, ξ′)| ≤ C exp(|ξ′|x1)

gD(ξ′)| + |ˆ gN(ξ′)| + x1 |ˆ f (x, ξ′)| dx

  • .

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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We thus obtain 1

  • Rd−1 |ˆ

u(x1, ξ′)|2 exp(−2|ξ′|x1) dξ′dx1 ≤ C

  • gD2

L2(Rd−1) + gN2 L2(Rd−1) + f 2 L2((0,1)×Rd−1)

  • .

The left hand-side defines a norm on u but weaker than any norm

  • f the form L2(0, 1; H−k(Rd−1)).

This is a prototype setting in which the linear mapping u ∈ H2(Ω) → (f , gD, gN) ∈ L2(Ω) × H3/2(Γ) × H1/2(Γ) is well-defined, injective, but its inverse is not continuous for reasonable topologies. Therefore, the Cauchy problem is ill-posed in the Hadamard sense.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Another question

We briefly presented the case of Full informations Γ = ∂Ω. Partial informations Γ ⊂ ∂Ω. What is in between ?        ∂11u + ∆′u = f , for (x1, x′) ∈ (0, 1) × Rd−1, u(0, x′) = gD(x′), for x′ ∈ Rd−1, ∂1u(0, x′) = gN(x′), for x′ ∈ Rd−1, u(1, x′) = g1(x′), for x′ ∈ Rd−1. In this case, one can solve ∆u = f in Ω with full Dirichlet data, hence gN is fully determined by gD, g1. Question Can we give a better formula than the one in the case of partial informations ?

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Same strategy as before : After Fourier transform,        ∂11 ˆ u − |ξ′|2 ˆ u = ˆ f , for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ u(0, ξ′) = ˆ gD(ξ′), for ξ′ ∈ Rd−1, ∂1 ˆ u(0, ξ′) = ˆ gN(ξ′), for ξ′ ∈ Rd−1, ˆ u(1, ξ′) = ˆ g1(ξ′), for ξ′ ∈ Rd−1. Again, family of ODE of order 2 with 3 boundary conditions !.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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We decompose the operator ∂11 − |ξ′|2 = (∂1 − |ξ′|)(∂1 + |ξ′|) = (∂1 + |ξ′|)(∂1 − |ξ′|). Therefore, one can rewrite ∂11 ˆ u − |ξ′|2 ˆ u = ˆ f , for x1 ∈ (0, 1), as (∂1 − |ξ′|)ˆ u = ˆ v, for x1 ∈ (0, 1), (∂1 + |ξ′|)ˆ v = ˆ f , for x1 ∈ (0, 1), OR (∂1 + |ξ′|)ˆ u = ˆ w, for x1 ∈ (0, 1), (∂1 − |ξ′|) ˆ w = ˆ f , for x1 ∈ (0, 1). Is there one decomposition better than the other ? Which one ?

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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How to chose the correct decomposition of the operator ? With the boundary conditions : We have more informations (2 boundary conditions) on the left point x1 = {0} than on the right end point (1 boundary condition). On the other hand, if one considers the ODE (∂1 + a)z = 0 in (0, 1), then we can get the two following formulas, z(x1) = z(0) exp(−ax1), OR z(x1) = z(1) exp(a(1 − x1)). In particular, If a > 0, the first formula is better (more stable) than the second one. Information comes from the left. If a < 0, the first formula is worst (less stable) than the second

  • ne. Information comes from the right.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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We therefore choose to write            (∂1 − |ξ′|)ˆ u = ˆ v, for x1 ∈ (0, 1), (∂1 + |ξ′|)ˆ v = ˆ f , for x1 ∈ (0, 1), ˆ u(0, ξ′) = ˆ gD(ξ′), , ∂1 ˆ u(0, ξ′) = ˆ gN(ξ′), ˆ u(1, ξ′) = ˆ g1(ξ′), that we solve in two steps :

  • ∂1ˆ

v + |ξ′|ˆ v = ˆ f for x1 ∈ (0, 1), ˆ v(0, ξ′) = ˆ gN(ξ′) − |ξ′|ˆ gD(ξ′), then    ∂1 ˆ u − |ξ′|ˆ u = ˆ v for x1 ∈ (0, 1), ˆ u(0, ξ′) = ˆ gD(ξ′), ˆ u(1, ξ′) = ˆ g1(ξ′).

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Solving

  • ∂1ˆ

v + |ξ′|ˆ v = ˆ f for x1 ∈ (0, 1), ˆ v(0, ξ′) = ˆ gN(ξ′) − |ξ′|ˆ gD(ξ′), we find ˆ v(x1, ξ′) = exp(−|ξ′|x1)ˆ v(0, ξ′)+ x1 exp

  • −|ξ′|(x1 − x)

ˆ f (x, ξ′) dx, Solving    ∂1 ˆ u − |ξ′|ˆ u = ˆ v for x1 ∈ (0, 1), ˆ u(0, ξ′) = ˆ gD(ξ′), ˆ u(1, ξ′) = ˆ g1(ξ′). we find ˆ u(x1, ξ′) = e−|ξ′|(1−x)ˆ g1(ξ′) − 1

x1

exp

  • −|ξ′|(x − x1)
  • ˆ

v(x, ξ′) dx.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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This leads ˆ u(x1, ξ′) =e−|ξ′|(1−x)ˆ g1(ξ′) − ˆ v(0, ξ′) 1

x1

exp

  • −|ξ′|(2x − x1)
  • dx

− 1 ˆ f (˜ x, ξ′) 1

max{x1,˜ x}

exp(−|ξ′|(2x − x1 − ˜ x)) dx

  • d ˜

x, with ˆ v(0, ξ′) = ˆ gN(ξ′) − |ξ′|ˆ gD(ξ′). Explicit formula with only decaying exponentials. We can provide estimates from the above formula in reasonable norms.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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For some C independent of ξ′,

  • ˆ

v(0, ξ′) 1

x1

exp

  • −|ξ′|(2x − x1)
  • dx
  • L2(0,1)

≤ C 1 + |ξ′|3/2 |ˆ v(0, ξ′)|, and

  • 1

ˆ f (˜ x, ξ′) 1

max{x1,˜ x}

exp(−|ξ′|(2x − x1 − ˜ x)) dx

  • d ˜

x

  • L2(0,1)

≤ C 1 + |ξ′|

  • 1

|ˆ f (˜ x, ξ′)| exp(−|ξ′||x1 − ˜ x|) d ˜ x

  • L2(0,1)

≤ C 1 + |ξ′|

  • ˆ

f (x1, ξ′)1x1∈(0,1) ∗x1 exp(−|ξ′||x1|)1x1∈(−2,2)

  • L2(0,1)

≤ C 1 + |ξ′|

  • ˆ

f (·, ξ′)

  • L2(0,1)
  • exp(−|ξ′||x1|)1x1∈(−2,2)
  • L1(−2,2)

≤ C 1 + |ξ′|2

  • ˆ

f (·, ξ′)

  • L2(0,1) ,

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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For C independent of ξ′, (1+|ξ′|4)

  • ˆ

u(·, ξ′)

  • 2

L2(0,1) ≤ C(1+|ξ′|)3/2|ˆ

g1(ξ′)|+C(1+|ξ′|)|ˆ v(0, ξ′)|2 + C

  • ˆ

f (·, ξ′)

  • 2

L2(0,1) .

Integrating in ξ′ ∈ Rd−1 and using Parseval’s identity u2

L2(0,1;H2(Rd−1)) ≤ C f 2 L2((0,1)×Rd−1) + C g12 H3/2(Rd−1)

+ C

  • Rd−1(1 + |ξ′|)|ˆ

v(0, ξ′)|2 dξ′. Using ∂11u = f − ∆′u ∈ L2((0, 1) × Rd−1), u2

H2((0,1)×Rd−1)) ≤ C f 2 L2((0,1)×Rd−1) + C g12 H3/2(Rd−1)

+ C

  • Rd−1(1 + |ξ′|)|ˆ

v(0, ξ′)|2 dξ′.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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From the explicit form of ˆ v(0, ξ′) = ˆ gN(ξ′) − |ξ′|ˆ gD(ξ′)

  • Rd−1(1+|ξ′|)|ˆ

v(0, ξ′)|2 dξ′ ≤ C gD2

H3/2(Rd−1)+C gN2 H1/2(Rd−1) .

We thus conclude uH2((0,1)×Rd−1) ≤ C f L2((0,1)×Rd−1) + C g1H3/2(Rd−1) + C gDH3/2(Rd−1) + C gNH1/2(Rd−1) . (1)

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Note that we also have, from the well-posedness of the elliptic problem with Dirichlet boundary conditions that uH2((0,1)×Rd−1) ≤ C f L2((0,1)×Rd−1) + C gDH3/2(Rd−1) , which seems stronger than what we proved. In fact, this estimate can also be proved along the same lines of the one above. But more importantly, the strategy developed to prove the estimate (1) can be adapted to prove Carleman estimates.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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The formula ˆ u(x1, ξ′) =ˆ g1(ξ′)e−|ξ′|(1−x1) − (ˆ gN(ξ′) − |ξ′|ˆ gD(ξ′)) 1

x1

exp

  • −|ξ′|(2x − x1)
  • dx

− 1 ˆ f (˜ x, ξ′) 1

max{x1,˜ x}

exp(−|ξ′|(2x − x1 − ˜ x)) dx

  • d ˜

x, is thus “stable”. However, it does not provide easily the property    f = 0 in Ω gD = 0 on Γ, gN = 0 on Γ ⇒ u = 0 in Ω. Carleman estimates.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Outline

1

The Cauchy problem for the Laplace operator The case of full informations The case of partial data Intermediate case

2

Carleman estimates with a linear weight Goal Proof of the Carleman estimate : Fourier techniques Proof of the Carleman estimate : Multiplier techniques More general geometric settings Application to the Calderón Problem

3

More general Carleman Weights The case of a strip The case of a strip with a multiplier technique The general case More on unique continuation

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Goal

The goal of this section is to get a stable estimate for solutions of    ∆u = f , in Ω, u = gD,

  • n ∂Ω,

∂nu = gN,

  • n Γ.

allowing to prove that    f = 0 in Ω gD = 0 on Γ, gN = 0 on Γ, ⇒ u = 0 in Ω, even when Γ = ∂Ω. The strategy then consists in mixing the previous computations, by considering norms which allow a linear exponential growth.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

We will first focus on the case of a strip Ω = (0, 1) × Rd−1, Γ = {0} × Rd−1, with homogeneous Dirichlet boundary conditions to simplify the presentation :    ∂11u + ∆′u = f , for (x1, x′) ∈ (0, 1) × Rd−1, u(0, x′) = u(1, x′) = 0, for x′ ∈ Rd−1, ∂1u(0, x′) = gN(x′), for x′ ∈ Rd−1. (2) A Carleman estimate ∃C > 0, s.t. for all k ≥ 1, any solution u of (2) with f ∈ L2(Ω) and gN ∈ H1/2(Γ) satisfies k

  • ue−kx1
  • L2((0,1)×Rd−1) +
  • ∇ue−kx1
  • L2((0,1)×Rd−1)

≤ C

  • fe−kx1
  • L2((0,1)×Rd−1) + Ck1/2 gNL2(Rd−1) .

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Theorem ∃C > 0, s.t. ∀k ≥ 1, any solution u of (2) with f ∈ L2(Ω) and gN ∈ H1/2(Γ) satisfies k

  • ue−kx1
  • L2((0,1)×Rd−1) +
  • ∇ue−kx1
  • L2((0,1)×Rd−1)

≤ C

  • fe−kx1
  • L2((0,1)×Rd−1) + Ck1/2 gNL2(Rd−1) .

(3) Estimate (3) is a Carleman estimate with linear weight : Weighted norms appear containing exponential terms. Free parameter k ≥ 1, which can be made arbitrarily large. C does not depend on k. The weight e−kx1 is larger on the side on which the measurements are done.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Some consequences

Carleman estimate for a Laplace operator with potential Let q ∈ L∞((0, 1) × Rd−1). There exists a constant C > 0 such that for all k ≥ 1, any solution u ∈ L2((0, 1) × Rd−1) of    ∂11u + ∆′u + qu = f , for (x1, x′) ∈ (0, 1) × Rd−1, u(0, x′) = u(1, x′) = 0, for x′ ∈ Rd−1, ∂1u(0, x′) = gN(x′), for x′ ∈ Rd−1, with source term f ∈ L2((0, 1) × Rd−1) and Neumann data gN ∈ L2(Rd−1) satisfies (3).

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Proof

Apply the Carleman estimate (3) to u with f replaced by f − qu : ∃C > 0, ∀k ≥ 1, k

  • ue−kx1
  • L2((0,1)×Rd−1) +
  • ∇ue−kx1
  • L2((0,1)×Rd−1)

≤ C

  • (f − qu)e−kx1
  • L2((0,1)×Rd−1) + Ck1/2 gNL2(Rd−1) .

Using

  • (f − qu)e−kx1
  • L2((0,1)×Rd−1) ≤
  • fe−kx1
  • L2((0,1)×Rd−1)

+ qL∞((0,1)×Rd−1)

  • ue−kx1
  • L2((0,1)×Rd−1) .

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Hence, taking k ≥ kq = 2C qL∞((0,1)×Rd−1), k

  • ue−kx1
  • L2((0,1)×Rd−1) +
  • ∇ue−kx1
  • L2((0,1)×Rd−1)

≤ 2C

  • fe−kx1
  • L2((0,1)×Rd−1) + 2Ck1/2 gNL2(Rd−1) .

Now, ∃C > 0 s.t. ∀k ≥ 1, k

  • ue−kx1
  • L2((0,1)×Rd−1) +
  • ∇ue−kx1
  • L2((0,1)×Rd−1)

≤ 2C

  • fe−kx1
  • L2((0,1)×Rd−1) + 2Ck1/2 gNL2(Rd−1) ,

as the case k ∈ [1, kq] can be handled by immediate bounds on the estimate for k = kq.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Corollary Let q ∈ L∞((0, 1) × Rd−1), and u ∈ L2((0, 1) × Rd−1) be the solution of    ∂11u + ∆′u + qu = f , for (x1, x′) ∈ (0, 1) × Rd−1, u(0, x′) = 0, for x′ ∈ Rd−1, ∂1u(0, x′) = gN(x′), for x′ ∈ Rd−1, with u(1, x′) ∈ H3/2(Rd−1) and Neumann data gN = 0 and source term f ∈ L2((0, 1) × Rd−1) satisfying, for some a ∈ (0, 1), f (x1, x′) = 0 for x1 ∈ (0, a), x′ ∈ Rd−1. Then u vanishes in (0, a) × Rd−1.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Proof

If u(1, x′) = 0, apply the Carleman estimate to u : for all k ≥ 1, k

  • ue−kx1
  • L2((0,1)×Rd−1) ≤ C
  • fe−kx1
  • L2((0,1)×Rd−1) .

But on one hand, k

  • ue−kx1
  • L2((0,1)×Rd−1) ≥ k
  • ue−kx1
  • L2((0,a)×Rd−1) ≥ ke−ka uL2((0,a)×Rd

On the other hand, as f vanishes in (0, a) × Rd−1,

  • fe−kx1
  • L2((0,1)×Rd−1) ≤ e−ka f L2((0,1)×Rd−1) .

Using the Carleman estimate then yields, for all k ≥ 1, k uL2((0,a)×Rd−1) ≤ C f L2((0,1)×Rd−1) . Taking the limit k → ∞, u necessarily vanishes in (0, a) × Rd−1.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

If u(1, x′) ∈ H3/2(Rd−1), u ∈ H2(Rd−1). We therefore set v(x) = η(x1)u(x), in Ω, where η = η(x1) is a smooth cut-off function taking value 1 on (0, a), vanishing in x1 = 1. Then v solves    ∂11v + ∆′v + qv = ηf − [∂11, η]u, for (x1, x′) ∈ (0, 1) × Rd−1, v(0, x′) = v(1, x′) = 0, for x′ ∈ Rd−1, ∂1v(0, x′) = 0, for x′ ∈ Rd−1, Thus, by the previous case, v(x1, x′) = 0 for x1 ∈ (0, a), i.e. u(x1, x′) = 0 for (x1, x′) ∈ (0, a) × Rd−1.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Recall the Carleman estimate

Theorem ∃C > 0, s.t. ∀k ≥ 1, any solution u of    ∂11u + ∆′u = f , for (x1, x′) ∈ (0, 1) × Rd−1, u(0, x′) = u(1, x′) = 0, for x′ ∈ Rd−1, ∂1u(0, x′) = gN(x′), for x′ ∈ Rd−1, with f ∈ L2(Ω) and gN ∈ H1/2(Rd−1) satisfies k

  • ue−kx1
  • L2((0,1)×Rd−1) +
  • ∇ue−kx1
  • L2((0,1)×Rd−1)

≤ C

  • fe−kx1
  • L2((0,1)×Rd−1) + Ck1/2 gNL2(Rd−1) .

Two proofs : By Fourier techniques. By multiplier techniques.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Proof by Fourier techniques

Since we have to prove estimates on ue−kx1 in terms of fe−kx1, it is convenient to work on the conjugate variables : U(x1, x′) = u(x1, x′)e−kx1, F(x1, x′) = f (x1, x′)e−kx1, for (x1, x′) ∈ (0, 1) × Rd−1. The equation on u then rewrites in terms of U, as U(x1, x′) = ekx1u(x1, x′) :    ∂11U + 2k∂1U + k2U + ∆′U = F, for (x1, x′) ∈ (0, 1) × Rd−1, U(0, x′) = U(1, x′) = 0, for x′ ∈ Rd−1, ∂1U(0, x′) = gN(x′), for x′ ∈ Rd−1,

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Similarly as before, we take the partial Fourier transform in the x′-variable : ˆ U(x1, ξ′) = Fx′→ξ′U(x1, ·), ˆ F(x1, ξ′) = Fx′→ξ′F(x1, ·), ˆ gN(ξ′) = Fx′→ξ′gN(·). We obtain a family of ODE indexed by ξ′ :    ∂11 ˆ U + 2k∂1 ˆ U + k2 ˆ U − |ξ′|2 ˆ U = ˆ F, for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ U(0, ξ′) = ˆ U(0, ξ′) = 0, for ξ′ ∈ Rd−1, ∂1 ˆ U(0, x′) = ˆ gN(ξ′), for ξ′ ∈ Rd−1.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

The operator factorized as follows : ∂11 + 2k∂1 + k2 − |ξ′|2 = (∂1 + k + |ξ′|)(∂1 + k − |ξ′). Therefore, we introduce the function ˆ V = (∂1 + k − |ξ|) ˆ U, (x1, ξ′) ∈ (0, 1) × Rd−1. Back to          ∂1 ˆ U + (k − |ξ′|) ˆ U = ˆ V for (x1, ξ′) ∈ (0, 1) × Rd−1, ∂1 ˆ V + (k + |ξ′|) ˆ V = ˆ F for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0 for ξ′ ∈ Rd−1, ˆ V (0, ξ′) = ˆ gN(ξ′) for ξ′ ∈ Rd−1.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

         ∂1 ˆ U + (k − |ξ′|) ˆ U = ˆ V for (x1, ξ′) ∈ (0, 1) × Rd−1, ∂1 ˆ V + (k + |ξ′|) ˆ V = ˆ F for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0 for ξ′ ∈ Rd−1, ˆ V (0, ξ′) = ˆ gN(ξ′) for ξ′ ∈ Rd−1. For fixed ξ′ ∈ Rd−1, we can then solve this system in two steps :

1 Compute ˆ

V in terms of ˆ F and of ˆ gN : ∂1 ˆ V + (k + |ξ′|) ˆ V = ˆ F for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ V (0, ξ′) = ˆ gN(ξ′) for ξ′ ∈ Rd−1.

2 Compute ˆ

U in terms of ˆ V : ∂1 ˆ U + (k − |ξ′|) ˆ U = ˆ V for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0 for ξ′ ∈ Rd−1,

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

The computation of ˆ V is straightforward : ∂1 ˆ V + (k + |ξ′|) ˆ V = ˆ F for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ V (0, ξ′) = ˆ gN(ξ′) for ξ′ ∈ Rd−1. yields : ˆ V (x1, ξ′) = e−(k+|ξ′|)x1 ˆ gN(ξ′)+ x1 exp(−(k+|ξ′|)(x1−x)) ˆ F(x, ξ′) dx. The computation of ˆ U also is straightforward, as it solves : ∂1 ˆ U + (k − |ξ′|) ˆ U = ˆ V for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0 for ξ′ ∈ Rd−1, But now we have two possible formulae, depending whether we use the boundary condition at x1 = 0 or at x1 = 1.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

∂1 ˆ U + (k − |ξ′|) ˆ U = ˆ V for (x1, ξ′) ∈ (0, 1) × Rd−1, ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0 for ξ′ ∈ Rd−1, Two formulae : ˆ U(x1, ξ′) = x1 exp(−(k − |ξ′|)(x1 − x)) ˆ V (x, ξ′) dx, ˆ U(x1, ξ′) = − 1

x1

exp(−(k − |ξ′|)(x1 − x)) ˆ V (x, ξ′) dx. We have to do a choice : The first formula for |ξ′| ≤ k, Low frequency case. The second formula for |ξ′| ≥ k, High frequency case.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

The low frequency case |ξ′| ≤ k

Here, we use the first formula : ˆ U(x1, ξ′) = ˆ gN(ξ′) exp(−(k − |ξ′|)x1) x1 exp(−2|ξ′|x) dx + x1 ˆ F(˜ x, ξ′)e−k(x1−˜

x)

x1

˜ x

e|ξ′|(˜

x+x1−2x) dx

  • d ˜

x. Direct estimates :

  • ˆ

gN(ξ′) exp(−(k − |ξ′|)x1) x1 exp(−2|ξ′|x) dx

  • L2(0,1)

≤ C|ˆ gN(ξ′)|

  • 1

1 + |ξ′| 1 1 + (k − |ξ′|)1/2

C k1/2 |ˆ gN(ξ′)|.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

The low frequency case |ξ′| ≤ k

On the other hand, writing x1 ˆ F(˜ x, ξ′)e−k(x1−˜

x)

x1

˜ x

e|ξ′|(˜

x+x1−2x) dx

  • d ˜

x = 1 ˆ F(˜ x, ξ′)H(x1 − ˜ x, ξ′) d ˜ x with H(X, ξ′) = 1X>0e−(k−|ξ′|)X X e−2|ξ′|x dx, we obtain

  • x1

ˆ F(˜ x, ξ′)e−k(x1−˜

x)

x1

˜ x

e|ξ′|(˜

x+x1−2x) dx

  • d ˜

x

  • L2(0,1)

≤ C

  • ˆ

F(·, ξ′)

  • L2(0,1)
  • H(·, ξ′)
  • L1(0,2)

≤ C

  • ˆ

F(·, ξ′)

  • L2(0,1)
  • 1

1 + k − |ξ′| 1 1 + |ξ′|

  • ≤ C

k

  • ˆ

F(·, ξ′)

  • L2(0,1) .

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

The low frequency case |ξ′| ≤ k

We thus get, for ξ′ ∈ Rd−1 with |ξ′| ≤ k,

  • ˆ

U(x1, ξ′)

  • L2(0,1) ≤

C k1/2 |ˆ gN(ξ′)| + C k

  • ˆ

F(·, ξ′)

  • L2(0,1) ,

i.e. ∀ξ′ ∈ Rd−1 with |ξ′| ≤ k k2

  • ˆ

U(x1, ξ′)

  • 2

L2(0,1) ≤ Ck|ˆ

gN(ξ′)|2 + C

  • ˆ

F(·, ξ′)

  • 2

L2(0,1) ,

with C independent of ξ′ and k.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

The high frequency case |ξ′| ≥ k

Here, we use the other formula : ˆ U(x1, ξ′) = − ˆ gN(ξ′)e−(k−|ξ′|)x1 1

x1

e−2|ξ′|x dx − 1 ˆ F(˜ x, ξ′)e−k(x1−˜

x)e−|ξ′||x1−˜ x|

1−max{˜

x,x1}

e−2|ξ′|x dx

  • For the first term :
  • −ˆ

gN(ξ′)e−(k−|ξ′|)x1 1

x1

e−2|ξ′|x dx

  • L2(0,1)

≤ C 1 + |ξ′|

  • −ˆ

gN(ξ′)e−(k+|ξ′|)x1

  • L2(0,1)

≤ C 1 + |ξ′||ˆ gN(ξ′)| 1 (k + |ξ′|)1/2 ≤ C k3/2 |ˆ gN(ξ′)|.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

The high frequency case |ξ′| ≥ k

The second term can be bounded as follows :

1 ˆ F(˜ x, ξ′)e−k(x1−˜

x)e−|ξ′||x1−˜ x|

1−max{˜

x,x1}

e−2|ξ′|x dx

  • d ˜

x

  • L2(0,1)

  • 1

| ˆ F(˜ x, ξ′)|e−k(x1−˜

x)e−|ξ′||x1−˜ x|

1−max{˜

x,x1}

e−2|ξ′|x dx

  • d ˜

x

  • L2(0

  • 1

| ˆ F(˜ x, ξ′)|e(k−|ξ′|)|x1−˜

x|

∞ e−2|ξ′|x dx

  • d ˜

x

  • L2(0,1)

≤ C |ξ′|

  • | ˆ

F|(x1, ξ′)1x1∈(0,1) ∗x1 e(k−|ξ′|)|x1|

  • L2(0,1)

≤ C |ξ′|

  • ˆ

F(·, ξ′)

  • L2(0,1)
  • e(k−|ξ′|)|x1|
  • L1(−2,2)

≤ C |ξ′|(1 − k + |ξ′|)

  • F(·, ξ′)
  • L2(0,1) ≤ C

k

  • ˆ

F(·, ξ′)

  • L2(0,1) .

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

The high frequency case |ξ′| ≥ k

We thus get, for ξ′ ∈ Rd−1 with |ξ′| ≥ k,

  • ˆ

U(x1, ξ′)

  • L2(0,1) ≤

C k1/2 |ˆ gN(ξ′)| + C k

  • ˆ

F(·, ξ′)

  • L2(0,1) ,

i.e. ∀ξ′ ∈ Rd−1 with |ξ′| ≥ k k2

  • ˆ

U(x1, ξ′)

  • 2

L2(0,1) ≤ Ck|ˆ

gN(ξ′)|2 + C

  • ˆ

F(·, ξ′)

  • 2

L2(0,1) ,

with C independent of ξ′ and k. This is the same estimate as the one obtained for ξ′ ∈ Rd−1 with |ξ′| ≤ k.

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Low and High-frequency together

As the estimates obtained for |ξ′| ≤ k and for |ξ′| ≥ k are the same, ∃C > 0, such that ∀ξ′ ∈ Rd−1, k2

  • ˆ

U(x1, ξ′)

  • 2

L2(0,1) ≤ Ck|ˆ

gN(ξ′)|2 + C

  • ˆ

F(·, ξ′)

  • 2

L2(0,1) .

Integrating this estimate with respect to ξ′ ∈ Rd−1 and using Parseval’s identity : k2 U2

L2((0,1)×Rd−1) ≤ Ck gN2 L2(Rd−1) + C F2 L2((0,1)×Rd−1) .

i.e. k2

  • ue−kx1
  • 2

L2((0,1)×Rd−1) ≤ Ck gN2 L2(Rd−1)+C

  • fe−kx1
  • 2

L2((0,1)×Rd−1) .

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

In order to estimate ∇ue−kx1 in L2(Ω), we multiply the equation ∆u = f , for (x1, x′) ∈ (0, 1) × Rd−1, u(0, x′) = u(1, x′) = 0, for x′ ∈ Rd−1, by ue−2kx1 : −

  • (0,1)×Rd−1 |∇u|2e−2kx1 dx1dx′+2 k2
  • (0,1)×Rd−1 |u|2e−2kx1 dx1dx′

=

  • (0,1)×Rd−1 fue−2kx1 dx1dx′,

so that

  • ∇ue−kx1
  • 2

L2((0,1)×Rd−1) ≤ k2

  • ue−kx1
  • 2

L2((0,1)×Rd−1)

+

  • fe−kx1
  • L2((0,1)×Rd−1)
  • ue−kx1
  • L2((0,1)×Rd−1) .

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

We can thus use the proved estimate k2

  • ue−kx1
  • 2

L2((0,1)×Rd−1) ≤ Ck gN2 L2(Rd−1)+C

  • fe−kx1
  • 2

L2((0,1)×Rd−1) .

in

  • ∇ue−kx1
  • 2

L2((0,1)×Rd−1) ≤ k2

  • ue−kx1
  • 2

L2((0,1)×Rd−1)

+

  • fe−kx1
  • L2((0,1)×Rd−1)
  • ue−kx1
  • L2((0,1)×Rd−1) .

We therefore get

  • ∇ue−kx1
  • 2

L2((0,1)×Rd−1) ≤ Ck gN2 L2(Rd−1)+C

  • fe−kx1
  • 2

L2((0,1)×Rd−1) .

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Comments on this proof

To sum up : This proof yields a formula to get u in terms of the known quantities. For low frequency |ξ′| ≤ k, the information comes from the left. For high frequency |ξ′| ≥ k, the information comes from the left and from the right. Once ue−kx1 is estimated, a weighted type regularity estimate give an estimate on ∇ue−kx1.

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Proof by multiplier techniques

U = ue−kx1 solves the equation    ∂11U + 2k∂1U + k2U + ∆′U = F, for (x1, x′) ∈ (0, 1) × Rd−1, U(0, x′) = U(1, x′) = 0, for x′ ∈ Rd−1, ∂1U(0, x′) = gN(x′), for x′ ∈ Rd−1, Multiplying the equation by ∂1U we easily get 2k ∂1U2

L2((0,1)×Rd−1) =

  • (0,1)×Rd−1 F∂1U dx1dx′

+ 1 2

  • Rd−1 |∂1U(0, x′)|2 dx′ − 1

2

  • Rd−1 |∂1U(1, x′)|2 dx′,

so that k ∂1U2

L2((0,1)×Rd−1) ≤ C

k F2

L2((0,1)×Rd−1) + C gN2 L2(Rd−1) .

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But U satisfies homogeneous boundary conditions at x1 = 0 and x1 = 1 Poincaré’s inequality applies : U2

L2((0,1)×Rd−1) ≤ C ∂1U2 L2((0,1)×Rd−1) .

Thus, k2 U2

L2((0,1)×Rd−1) ≤ C F2 L2((0,1)×Rd−1) + Ck gN2 L2(Rd−1) .

Remains to estimate ∇U.

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Cauchy Linear Weight General Weights Goal Proof 1 Proof 2 Geometry Calderon

Multiplying    ∂11U + 2k∂1U + k2U + ∆′U = F, for (x1, x′) ∈ (0, 1) × Rd−1, U(0, x′) = U(1, x′) = 0, for x′ ∈ Rd−1, ∂1U(0, x′) = gN(x′), for x′ ∈ Rd−1, by U, we obtain

  • ∇U2

L2((0,1)×Rd−1) − k2 U2 L2((0,1)×Rd−1)

  • ≤ FL2((0,1)×Rd−1) UL2((0,1)×Rd−1) ,

As U is already suitably estimated, ∇U2

L2((0,1)×Rd−1) ≤ Ck2 U2 L2((0,1)×Rd−1) + C F2 L2((0,1)×Rd−1) .

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We thus have k2 U2

L2((0,1)×Rd−1) + ∇U2 L2((0,1)×Rd−1)

≤ C F2

L2((0,1)×Rd−1) + Ck gN2 L2(Rd−1) .

Using then that u = Uekx1, |u|e−kx1 ≤ |U|, |∇u|e−kx1 ≤ |∇U| + k|U|. Therefore, k2

  • ue−kx1
  • 2

L2((0,1)×Rd−1) +

  • ∇ue−kx1
  • 2

L2((0,1)×Rd−1)

≤ C

  • fe−kx1
  • 2

L2((0,1)×Rd−1) + Ck gN2 L2(Rd−1) .

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Comments on the second proof

This second proof is much shorter than the first one. However, It is not very precise on the boundary terms. It seems to strongly use the Hilbertian property of L2. In particular, the first strategy is more precise and can handled : Non-homogeneous Dirichlet boundary conditions in H3/2(∂Ω) and Neumann boundary data in H1/2(Γ). Lp settings. Still, the second strategy generalizes easily in more general geometric setting....

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More general geometric settings

Theorem Let Ω be a smooth (C 2) bounded domain of Rd. ∃C > 0 s.t. ∀k ∈ Rd with |k| ≥ 1, any solution u ∈ L2(Ω) of ∆u = f , for x ∈ Ω u(x) = 0, for x ∈ ∂Ω, with source term f ∈ L2(Ω) satisfies |k|2

  • ue−k·x
  • 2

L2(Ω) +

  • ∇ue−k·x
  • 2

L2(Ω) +

1 |k|2

  • ue−k·x
  • 2

H2(Ω)

≤ C|k|

  • ∂nue−k·x
  • 2

L2(Γk) + C

  • fe−k·x
  • 2

L2(Ω) ,

where Γk = {x ∈ ∂Ω, | k · nx < 0}.

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Proof by multiplier techniques

Setting U = ue−k·x and F = fe−k·x, U satisfies ∆U + 2k · ∇U + |k|2U = F for x ∈ Ω, U(x) = 0, for x ∈ ∂Ω. Multiplying the equation by U, we first derive ∇U2

L2(Ω) ≤ C|k|2 U2 L2(Ω) + C F2 L2(Ω) .

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Multiplying ∆U + 2k · ∇U + |k|2U = F for x ∈ Ω, U(x) = 0, for x ∈ ∂Ω. by k · ∇U, we derive

Fk · ∇U dx = 2

|k · ∇U|2 dx +

  • ∂Ω

∂nUk · ∇U dσ − 1 2

  • ∂Ω

k · nx|∇U|2 dσ = 2

|k · ∇U|2 dx + 1 2

  • ∂Ω

k · nx|∂nU|2 dσ, where we used that, as U = 0 on ∂Ω, ∇U = (∂nU)nx on ∂Ω. Therefore, ∃C independent of k ∈ Rd s.t.

|k · ∇U|2 dx ≤ C

|F|2 dx + C |k|

  • Γk

|∂nU|2 dσ.

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But Ω is bounded, so by Poincaré’s inequality |k|2 U2

L2(Ω) ≤ C

|k · ∇U|2 dx. Thus, ∀k ∈ Rd, |k|2 U2

L2(Ω) ≤ C F2 L2(Ω) + C |k| ∂nU2 L2(Γk) .

Based on the estimate of ∇U in terms of U, we thus derive |k|2 U2

L2(Ω) + ∇U2 L2(Ω) ≤ C F2 L2(Ω) + C |k| ∂nU2 L2(Γk) .

Recalling that ue−k·x = U and ∇u e−k·x = ∇U + kU, we immediately conclude the Carleman estimate. The H2-norm of ue−k·x comes from the equation satisfied by U.

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Comments on the proof

Proof based on the multiplier technique developed in the case of the strip. Mainly the same proof as in the case of the strip. A similar proof based on a Fourier decomposition technique would be much more intricate. Note that the observation set Γk = {x ∈ ∂Ω, | k · nx < 0} depends

  • n the direction of k.

Geometry of the domain and the observation part Γ of the boundary and of the wave functions are linked.

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Application to the Calderón Problem

Calderón problem (or Electrical Impedance Tomography) This corresponds to a medical imaging technique which consists in the recovery of the conductivity of a tissue (or a material) by applying currents on the surface on the body and measuring the electrical potentials on the surface of the body. Let Ω be a bounded domain of Rd, and consider the elliptic problem div (σ∇u) = 0, for x ∈ Ω, u(x) = gd(x), for x ∈ ∂Ω. Here, σ = σ(x) is a scalar function modeling the conductivity of the material, σ is unknown.

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Mathematical Formulation Recover σ knowing the map Λσ : gd → σ∂nu, where u denotes the solution div (σ∇u) = 0, for x ∈ Ω, u(x) = gd(x), for x ∈ ∂Ω. Physically : Λσ is the so-called Voltage-to-current map. gd is a voltage imposed on the boundary of the object. We can measure the current (σ∇u) · nx on the boundary ∂Ω. Mathematically : Λσ is the Dirichlet to Neumann map.

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Mathematical framework

Calderón Problem Can we determine the conductivity of the material σ from the knowledge on the Voltage-to-Current map ? div (σ∇u) = 0, for x ∈ Ω, u(x) = gd(x), for x ∈ ∂Ω. Assumption 1 : σ ∈ C 0(Ω), ∃C∗ > 0, s.t. ∀x ∈ Ω, 1 C∗ ≤ σ(x) ≤ C∗. If gd ∈ H1/2(∂Ω) yields u ∈ H1(Ω) and (σ∇u) · nx = σ∂nu ∈ H−1/2(∂Ω). Λσ : H1/2(∂Ω) → H−1/2(∂Ω).

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The Calderón problem is an inverse problem. Subproblems : Uniqueness : If Λσ1 = Λσ2, can we deduce σ1 = σ2 ? Stability : If Λσ1 − Λσ2 is small (in suitable norms), can we deduce that σ1 − σ2 is small (in suitable norms) ? Reconstruction : Given Λσ, can we compute σ ? In the following, we will focus on the uniqueness problem, i.e. on the injectivity of the map Λ : σ → Λσ.

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Injectivity of Λ : σ → Λσ

Main difficulty The map Λ is non-linear. In here, we shall further assume the following : σ and ∂nσ are known on the boundary. √σ belongs to C 2(Ω). Under these assumptions, we can transform the Calderón problem in the recovery of a potential.

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Liouville’s transform

If u solves div (σ∇u) = 0, for x ∈ Ω, u(x) = gd(x), for x ∈ ∂Ω, Then v = σ1/2u in Ω, solves ∆v + qv = 0, for x ∈ Ω, v(x) = hd(x), for x ∈ ∂Ω. with q= −∆(σ1/2) σ1/2 , in Ω hd = σ1/2gd, ∂nv = σ−1/2(σ∂nu) + ∂n(σ1/2)gd, on ∂Ω.

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We therefore introduce the map ˜ Λ : q ∈ L∞(Ω) → ˜ Λq, where ˜ Λq : hd ∈ H1/2(∂Ω) → ∂nv ∈ H−1/2(∂Ω), where ∂nv is given by ∆v + qv = 0, for x ∈ Ω, v(x) = hd(x), for x ∈ ∂Ω. Claim Under the previous assumptions on σ, Λσ1 = Λσ2 is equivalent to ˜ Λq1 = ˜ Λq2, where q1 = −∆(σ1/2

1

) σ1/2

1

, q2 = −∆(σ1/2

2

) σ1/2

2

.

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Goal

Λ : q ∈ L∞(Ω) → Λq, where Λq : hd ∈ H1/2(∂Ω) → ∂nv ∈ H−1/2(∂Ω), where ∂nv is given by ∆v + qv = 0, for x ∈ Ω, v(x) = hd(x), for x ∈ ∂Ω. Theorem Assume d ≥ 3. q1, q2 ∈ L∞(Ω), Λq1 = Λq2 ⇒ q1 = q2.

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Preliminary properties of Λq

Proposition Given q ∈ L∞(Ω), the map Λq is self-adjoint on H1/2(∂Ω). i.e. ∀h1, h2 ∈ H1/2(∂Ω), Λqh1, h2H−1/2(∂Ω),H1/2(∂Ω) = h1, Λqh2H1/2(∂Ω),H−1/2(∂Ω),

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Proof

Let q ∈ L∞(Ω) and, for h1 and h2 in H1/2(∂Ω), set ∆v1 + qv1 = 0, for x ∈ Ω, v1(x) = h1(x), for x ∈ ∂Ω, ∆v2 + qv2 = 0, for x ∈ Ω, v2(x) = h2(x), for x ∈ ∂Ω. Thus, Λqh1, h2H−1/2(∂Ω),H1/2(∂Ω) = ∂nv1, v2H−1/2(∂Ω),H1/2(∂Ω) =

∆v1v2 dx +

∇v1 · ∇v2 dx = −

qv1v2 dx +

∇v1 · ∇v2 dx = h1, Λqh2H1/2(∂Ω),H−1/2(∂Ω).

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A polarization formula

Proposition Let q1 and q2 in L∞(Ω). Then for all h1, h2 in H1/2(∂Ω), the solutions v1, v2 of ∆v1 + q1v1 = 0, in Ω, v1(x) = h1(x),

  • n ∂Ω,

∆v2 + q2v2 = 0, in Ω, v2(x) = h2(x),

  • n ∂Ω,

satisfy (Λq1 − Λq2)h1, h2H−1/2(∂Ω),H1/2(∂Ω) =

(q2 − q1)v1v2 dx. In particular, if Λq1 = Λq2, for all v1 and v2 as above,

(q2 − q1)v1v2 dx = 0.

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Proof

We compute Λq1h1, h2H−1/2(∂Ω),H1/2(∂Ω) = ∂nv1, v2H−1/2(∂Ω),H1/2(∂Ω) =

∆v1v2 dx +

∇v1 · ∇v2 dx = −

q1v1v2 dx +

∇v1 · ∇v2 dx. Similar computations yield h1, Λq2h2H1/2(∂Ω),H−1/2(∂Ω) = −

q2v1v2 dx +

∇v1 · ∇v2 dx. As Λq2 is self-adjoint, h1, Λq2h2H1/2(∂Ω),H−1/2(∂Ω) = Λq2h1, h2H−1/2(∂Ω),H1/2(∂Ω). Subtraction of the two above identities gives the result.

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Idea

In particular, if Λq1 = Λq2, for all v1 and v2 as before,

(q2 − q1)v1v2 dx = 0. Therefore, our next goal is to generate a dense set of functions of the form v1v2, where ∆v1 + q1v1 = 0, in Ω, v1(x) = h1(x),

  • n ∂Ω,

∆v2 + q2v2 = 0, in Ω, v2(x) = h2(x),

  • n ∂Ω,

Therefore, our goal is to find function v1v2 which approximates the Fourier basis x → exp(−iξx).

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Looking for specific solutions

Remark If ρ ∈ Cd satisfies ρ · ρ = 0, then ∆(eρ·x) = 0 in Rd. There are many such ρ ∈ Cd ! If ρ = a + ib, with a, b ∈ Rd, ρ · ρ = 0 ⇔ |a| = |b|, a · b = 0. NB : If ρ · ρ = 0 and ρ ∈ Rd, then ρ = 0 of course !

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Complex Geometric Optics solutions

Theorem Let q ∈ L∞(Ω). For all ρ = a + ib ∈ Cd with ρ · ρ = 0 and |a| ≥ 1, there exists a solution vρ ∈ L2(Ω) of ∆vρ + qvρ = 0 in Ω, that can be written as vρ(x) = eρ·x + ea·xrρ(x), with rρ satisfying the estimate rρL2(Ω) ≤ C |a|. In other words, vρ(x) ≃ eρ·x when |a| → ∞.

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Proof

Remark that vρ(x) = eρ·x + ea·xrρ(x) solves ∆vρ + qvρ = 0 iff e−a·x(∆ + q)(ea·xrρ) = −qeib·x. We set ˜ q = −qeib·x. To be proved ∃rρ ∈ L2(Ω) solution of e−a·x(∆ + q)(ea·xrρ) = ˜ q in Ω, with rρL2(Ω) ≤ C |a|. Here, the difficult part is the estimate on rρ.

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A duality approach

The function rρ solves e−a·x(∆ + q)(ea·xrρ) = ˜ q iff for all w ∈ D(Ω),

  • ea·x(∆ + q)(e−a·xw)
  • dx =

˜ qw dx,

  • r, by density, for all w ∈ H2

0(Ω).

Consequence The set {rρ solving e−a·x(∆ + q)(ea·xrρ) = ˜ q} is an affine space of direction {ea·x(∆ + q)(e−a·xw), w ∈ H2

0(Ω)}⊥L2(Ω).

The function rρ of minimal L2(Ω) norm : rρ ∈ {ea·x(∆ + q)(e−a·xw), w ∈ H2

0(Ω)}.

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We claim that {ea·x(∆ + q)(e−a·xw), w ∈ H2

0(Ω)}

= {ea·x(∆ + q)(e−a·xw), w ∈ H2

0(Ω)}.

Indeed, ea·x(∆ + q)(e−a·xw) = ∆w − 2a · ∇w + |a|2w + qw, so the Carleman estimate applied to e−axw yields |a|2 w2

L2(Ω) +

1 |a|2 w2

H2(Ω) ≤ C

  • ea·x(∆ + q)(e−a·xw)
  • 2

L2(Ω) .

In particular, for all |a| ≥ 1,

  • ea·x(∆ + q)(e−a·xw)
  • L2(Ω) is equivalent to wH2

0(Ω) .

hence the closedness of the above vector space.

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Therefore, the solution rρ of minimal L2 norm writes, for some W ∈ H2

0(Ω),

rρ = ea·x(∆ + q)(e−a·xW ), and satisfies, for all w ∈ H2

0(Ω), for all w ∈ D(Ω),

  • ea·x(∆ + q)(e−a·xw)
  • dx =

˜ qw dx, We take w = W :

|ea·x(∆ + q)(e−a·xW )|2 dx =

˜ qW dx≤ ˜ qL2 W L2.

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We got

|ea·x(∆ + q)(e−a·xW )|2 dx ≤ ˜ qL2 W L2 , while the Carleman estimate implies, for all |a| ≥ 1 and w ∈ H2

0(Ω)

|a|2 w2

L2(Ω) ≤ C

  • ea·x(∆ + q)(e−a·xw)
  • 2

L2(Ω) .

Consequently, rρL2(Ω) =

|ea·x(∆ + q)(e−a·xW )|2 dx 1/2 ≤ C ˜ qL∞ |a| ≤ C qL∞ |a| .

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Back to the Calderón problem

Recall if Λq1 = Λq2, for all v1 and v2 with ∆v1 + q1v1 = 0, in Ω, ∆v2 + q2v2 = 0, in Ω, we have

(q2 − q1)v1v2 dx = 0. For each ξ ∈ Rd, we will chose (v1,n)n∈N ≃ (eρ1,n·x)n∈N, (v2,n)n∈N ≃ (eρ2,n·x)n∈N, with    ρ1,n · ρ1,n = 0, limn→∞ |ρ1,n| = ∞, ρ2,n · ρ2,n = 0, limn→∞ |ρ2,n| = ∞, ρ1,n + ρ2,n = −iξ.

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Choosing ρ1,n, ρ2,n

We want :    ρ1,n · ρ1,n = 0, limn→∞ |ρ1,n| = ∞, ρ2,n · ρ2,n = 0, limn→∞ |ρ2,n| = ∞, ρ1,n + ρ2,n = −iξ. As d ≥ 3, we start by choosing α, β ∈ Rd such that |α| = |β| = 1 and α · β = β · ξ = α · ξ = 0. We then set, for n large enough, ρ1,n = nα + i

  • γnβ − ξ

2

  • ,

ρ2,n = −nα − i

  • γnβ + ξ

2

  • ,

where γ2

n = n2 − |ξ|2

4 .

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End of the proof of the uniqueness

With the constructed ρ1,n, ρ2,n, we obtain v1,n(x) = eρ1,n·x + r1,nenα·x, v2,n(x) = eρ2,n·x + r2,ne−nα·x. with r1,nL2(Ω) ≤ C q1L∞ |n| , r2,nL2(Ω) ≤ C q2L∞ |n| . Therefore, as Λq1 = Λq2,

(q2 − q1)e−iξx dx = −

(q2−q1)

  • e−i(γnβ+ξ/2)xrn,1(x) + ei(γnβ−ξ/2)xrn,2(x) + rn,1rn,2
  • dx.

Taking the limit n → ∞,

(q2 − q1)e−iξx dx = 0.

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Therefore, ∀ξ ∈ Rd,

(q2 − q1)e−iξx dx = 0. ⇒ q1 = q2 in Ω. Our result is proved : Λq1 = Λq2 ⇒ q1 = q2.

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Comments

Here, we studied the Calderón problem for a scalar conductivity σ with the knowledge of the full Dirichlet to Neumann map.

  • With partial Dirichlet to Neumann map, the Calderón problem

can be solved provided there exists suitably “Limiting Carleman Weights”. [Kenig Sjöstrand Uhlmann 2007]

  • For anisotropic σ, the Calderón problem is still relevant,

corresponding to anisotropic materials. But there are counterexamples in this case ! ! [Tartar].

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In the case of anisotropic conductivities, Tartar proposed a simple construction to show that uniqueness for the Calderón problem cannot hold. Indeed, given any C 2 diffeomorphism Ψ on Ω with Ψ = Id on the boundary ∂Ω, then Λ˜

σ = Λσ,

where ˜ σ(y) = DΨT(Ψ−1(y)) × σ(Ψ−1(y)) × DΨ(Ψ−1(y)) |det (DΨ(Ψ−1(y)))|

  • .

This shows in particular that one cannot distinguish between σ and ˜ σ when allowing anisotropic conductivities.

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Cloaking ?

Actually, this counterexample is the basis of several recent works on invisibility (cloaking), see [Uhlmann 2009]. The idea is to construct diffeomorphisms that approximate the singular transformation between B(0, 2) \ {0} and B(0, 2) \ B(0, 1) given by Ψ(x) = x |x|

  • 1 + |x|

2

  • .

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Outline

1

The Cauchy problem for the Laplace operator The case of full informations The case of partial data Intermediate case

2

Carleman estimates with a linear weight Goal Proof of the Carleman estimate : Fourier techniques Proof of the Carleman estimate : Multiplier techniques More general geometric settings Application to the Calderón Problem

3

More general Carleman Weights The case of a strip The case of a strip with a multiplier technique The general case More on unique continuation

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General Weight functions

The goal of this section is to make precise what weight functions can be used in Carleman estimates. In a strip : what weight functions ϕ(x1) ? ∃C > 0, s.t. ∀s ≥ 1, ∀u ∈ H2 ∩ H1

0(Ω),

s# uesϕL2((0,1)×Rd−1) ≤ C ∆uesϕL2((0,1)×Rd−1) + Cs∗

  • ∂nu(0, ·)esϕ(0)
  • L2(Rd−1) .

In general geometries : Specific issue : Can we made the observation set arbitrarily small ?

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Back to the case of a strip

Ω = (0, 1) × Rd−1, Γ = {0} × Rd−1. Our goal is to understand for which functions ϕ = ϕ(x1) one can get an estimate of the form ∃C > 0, s.t. ∀s ≥ 1, ∀u ∈ H2 ∩ H1

0(Ω),

s# uesϕL2((0,1)×Rd−1) ≤ C ∆uesϕL2((0,1)×Rd−1) + Cs∗

  • ∂nu(0, ·)esϕ(0)
  • L2(Rd−1) .

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

We start from    ∂11u + ∆′u = f , for (x1, x′) ∈ (0, 1) × Rd−1, u(0, x′) = u(1, x′) = 0, for x′ ∈ Rd−1, ∂1u(0, x′) = gN(x′), for x′ ∈ Rd−1, and we set U(x1, x′) = u(x1, x′)esϕ(x1), F(x1, x′) = f (x1, x′)esϕ(x1), Gn(x′) = gn(x′)esϕ(0). The equations now read    ∂11U − 2s∂1ϕ∂1U + (s2|∂1ϕ|2 − s∂11ϕ)U + ∆′U = F, in Ω, U(0, x′) = U(1, x′) = 0, in Rd−1, ∂1U(0, x′) = GN(x′), in Rd−1.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Factorization of the operator

After partial Fourier transform,    ∂11 ˆ U − 2s∂1ϕ∂1 ˆ U + (s2|∂1ϕ|2 − s∂11ϕ)U − |ξ′|2 ˆ U = ˆ F, in (0, 1) × ˆ U(0, ξ′) = ˆ U(0, ξ′) = 0, in Rd−1, ∂1 ˆ U(0, ξ′) = ˆ GN(ξ′), in Rd−1, We then recognize that ∂11 − 2s∂1ϕ∂1 + (s2|∂1ϕ|2 − s∂11ϕ) − |ξ′|2 = (∂1 − s∂1ϕ)2 − |ξ′|2 = (∂1 − s∂1ϕ − |ξ′|)(∂1 − s∂1ϕ + |ξ′|) = (∂1 − s∂1ϕ + |ξ′|)(∂1 − s∂1ϕ − |ξ′|).

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Which conditions on ϕ ?

To see what are the needed conditions on ϕ, note that we have to consider the two ODE with operators (∂1−s∂1ϕ − |ξ′|), (∂1−s∂1ϕ + |ξ′|). Claim The sign of −s∂1ϕ − |ξ′| and of −s∂1ϕ + |ξ′| indicates how the information propagates.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Conditions on ϕ ?

We are considering operators of the form : (∂1−s∂1ϕ − |ξ′|), (∂1−s∂1ϕ + |ξ′|). Therefore, for all ξ′ ∈ Rd−1, we want to be able to drive the information from the left to the right. ⇒ To handle the case ξ′ = 0, we need inf{−s∂1ϕ} > 0, i.e. max

x1∈[0,1] ∂1ϕ(x1) < 0.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Thus, to solve    ∂11 ˆ U − 2s∂1ϕ∂1 ˆ U + (s2|∂1ϕ|2 − s∂11ϕ)U − |ξ′|2 ˆ U = ˆ F, in (0, 1), ˆ U(0, ξ′) = ˆ U(0, ξ′) = 0, ∂1 ˆ U(0, ξ′) = ˆ GN(ξ′), we introduce ˆ V (x1, ξ′) = (∂1−s∂1ϕ − |ξ′|) ˆ U(x1, ξ′) :          ∂1 ˆ U + (−s∂1ϕ − |ξ′|) ˆ U = ˆ V in (0, 1) ∂1 ˆ V + (−s∂1ϕ + |ξ′|) ˆ V = ˆ F in (0, 1) ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0, ˆ V (0, ξ′) = ˆ GN(ξ′). No problem to solve ˆ V . What happens for ˆ U ? ∂1 ˆ U + (−s∂1ϕ − |ξ′|) ˆ U = ˆ V in (0, 1) ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0,

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Back to the ODE case

(∂1+a)z = f in (0, 1), z(0) = z(1) = 0, Then, using a primitive A of a, i.e. A′ = a, we have two formulas : z(x1) = x1 eA(˜

x)−A(x1)f (˜

x)d ˜ x, = − 1

x1

eA(˜

x)−A(x1)f (˜

x)d ˜ x. Recall the goal : Have bounds without exponential factors. If inf[0,1] a > 0, information propagates from the left. If sup[0,1] a < 0, information propagates from the right. What happens otherwise, i.e. if a cancels in [0, 1] ?

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

The case a cancels in [0, 1]

We focus on the strict cases, in which a changes sign. We have A′ = a and for each x1 ∈ (0, 1), either ∀x < x1, A(x) < A(x1) or ∀x > x1, A(x) < A(x1). We assumed that a vanishes in some point x∗ ∈ [0, 1] and changes

  • sign. Thus

A′(x∗) = 0 and A attains a strict local maximum in x∗. There cannot exist other strict local maximizers ˜ x of A. Otherwise, ∃xm between ˜ x and x∗ which is a local minimizer, and the above condition is violated at x1 = xm. The condition is equivalent to ∀x < x1 < x∗, A(x) < A(x1)(< A(x∗)), ∀x > x1 > x∗, A(x) < A(x1)(< A(x∗)),

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Recall A′ = a and A′(x∗) = 0. The condition reads ∀x < x1 < x∗, A(x) < A(x1)(< A(x∗)), ∀x > x1 > x∗, A(x) < A(x1)(< A(x∗)), Therefore, one should have ∀x1 < x∗, a(x1) > a(x∗) = 0, ∀x1 > x∗, a(x1) < a(x∗) = 0.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

We consider ∂1 ˆ U + (−s∂1ϕ − |ξ′|) ˆ U = ˆ V in (0, 1) ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0, corresponding to a(x1, ξ′) = −s∂1ϕ(x1) − |ξ′|. For being suitable for Carleman estimates, we require, for all ξ′ ∈ Rd−1, that if ∃x∗,ξ such that a(x∗,ξ, ξ′) = 0, then ∀x1 < x∗,ξ, a(x1, ξ′) > a(x∗,ξ, ξ′) = 0, ∀x1 > x∗,ξ, a(x1, ξ′) < a(x∗,ξ, ξ′) = 0. Claim This latter condition is equivalent to a convexity condition on ϕ.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

A convexity condition

a(x1, ξ′) = −s∂1ϕ(x1) − |ξ′|. Condition ∀(x∗, ξ′) ∈ Rd−1 with a(x∗, ξ′) = 0, ∀x1 < x∗, a(x1, ξ′) > a(x∗, ξ′) = 0, ∀x1 > x∗, a(x1, ξ′) < a(x∗, ξ′) = 0. Obviously, this condition is satisfied if −s∂1ϕ is strictly decreasing, i.e. if inf[0,1] ∂11ϕ > 0. Reciprocally, this condition implies inf[0,1] ∂11ϕ > 0 : ∀x∗ ∈ (0, 1), ∃ξ′ ∈ Rd−1 s.t. a(x∗, ξ′) = 0. Thus, ∀x1 < x∗, a(x1, ξ′) > 0, i.e. ∂1ϕ(x1) < ∂1ϕ(x∗).

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Conditions on ϕ = ϕ(x1)

Ω = (0, 1) × Rd−1, Γ = {0} × Rd−1. Conditions on ϕ = ϕ(x1) ϕ = ϕ(x1) is smooth, max

[0,1]{∂1ϕ} < 0 and min [0,1]{∂11ϕ} > 0.

Theorem ∃C > 0, s.t. ∀s ≥ 1, ∀u ∈ H2 ∩ H1

0(Ω),

s3/2 uesϕL2((0,1)×Rd−1) + s1/2 ∇uesϕL2((0,1)×Rd−1) ≤ C ∆uesϕL2((0,1)×Rd−1) + Cs1/2

  • ∂nu(0, ·)esϕ(0)
  • L2(Rd−1) .

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Comparison with the linear weight

With a linear weight, we got : ∀k ≥ 1, k

  • ue−kx1
  • L2((0,1)×Rd−1) +
  • ∇ue−kx1
  • L2((0,1)×Rd−1)

≤ C

  • ∆ue−kx1
  • L2((0,1)×Rd−1) + Ck1/2 ∂nuL2(Rd−1) .

With a strictly convex weight ϕ, ∀s ≥ 1, s3/2 uesϕL2((0,1)×Rd−1) + s1/2 ∇uesϕL2((0,1)×Rd−1) ≤ C ∆uesϕL2((0,1)×Rd−1) + Cs1/2

  • ∂nu(0, ·)esϕ(0)
  • L2(Rd−1) .

Not the same powers of the parameter ! Note that a linear weight is not strictly convex.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Explanation (1)

In both cases, we first solve ∂1 ˆ V + (−s∂1ϕ + |ξ′|) ˆ V = ˆ F in (0, 1) ˆ V (0, ξ′) = ˆ GN(ξ′). so that there exists a C > 0 independent of s and ξ′ such that s

  • ˆ

V (·, ξ′)

  • L2(0,1) ≤ C| ˆ

GN(ξ′)| + C

  • ˆ

F(·, ξ′)

  • L2(0,1)

Only uses max{∂1ϕ} < 0. So it applies also when sϕ(x1) = −sx1.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Explanation (2)

We then have to solve ∂1 ˆ U + (−s∂1ϕ − |ξ′|) ˆ U = ˆ V in (0, 1) ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0, In the context of linear weights, this equation is ∂1 ˆ U + (k − |ξ′|) ˆ U = ˆ V in (0, 1) ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0. In particular, when |ξ′| = k, we cannot have better than

  • ˆ

U(·, ξ′)

  • L2(0,1) ≤ C
  • ˆ

V (·, ξ′)

  • L2(0,1) .

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Explanation (3)

For strictly convex weights, estimating ˆ U solving ∂1 ˆ U + (−s∂1ϕ − |ξ′|) ˆ U = ˆ V in (0, 1) ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0, yields : ∃C > 0 independent of s and ξ′ s.t. s1/2

  • ˆ

U(·, ξ′)

  • L2(0,1) ≤ C
  • ˆ

V (·, ξ′)

  • L2(0,1) .

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Explanation (4)

To prove the estimate on ˆ U, multiply ∂1 ˆ U + (−s∂1ϕ − |ξ′|) ˆ U = ˆ V in (0, 1) ˆ U(0, ξ′) = ˆ U(1, ξ′) = 0, by (−s∂1ϕ − |ξ′|) ˆ U, integrate and take the real part : 1 | ˆ U(x1, ξ′)|2(s∂11ϕ + (−s∂1ϕ − |ξ′|)2) dx1 ≤ ˆ V (·, ξ′)L2(0,1)(−s∂1ϕ − |ξ′|) ˆ UL2(0,1). Consequently, s1/2

  • ˆ

U(·, ξ′)

  • L2(0,1) ≤ C
  • ˆ

V (·, ξ′)

  • L2(0,1) .

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

A proof of the Carleman estimate by a multiplier technique

Ω = (0, 1) × Rd−1, Γ = {0} × Rd−1. Conditions on ϕ = ϕ(x1) ϕ = ϕ(x1) is smooth, max

[0,1]{∂1ϕ} < 0 and min [0,1]{∂11ϕ} > 0.

Theorem ∃C > 0, s.t. ∀s ≥ 1, ∀u ∈ H2 ∩ H1

0(Ω),

s3/2 uesϕL2((0,1)×Rd−1) + s1/2 ∇uesϕL2((0,1)×Rd−1) ≤ C ∆uesϕL2((0,1)×Rd−1) + Cs1/2

  • ∂nu(0, ·)esϕ(0)
  • L2(Rd−1) .

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Proof

U(x1, x′) = u(x1, x′)esϕ(x1), F(x1, x′) = ∆u(x1, x′)esϕ(x1), Gn(x′) = gn(x′)esϕ(0). The equations now read    ∂11U − 2s∂1ϕ∂1U + (s2|∂1ϕ|2 − s∂11ϕ)U + ∆′U = F, in Ω, U(0, x′) = U(1, x′) = 0, in Rd−1, ∂1U(0, x′) = GN(x′), in Rd−1.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Multiply the equation by s∂1ϕ∂1U :

Fs∂1ϕ∂1U =

  • ∂11U − 2s∂1ϕ∂1U + (s2|∂1ϕ|2 − s∂11ϕ)U + ∆′U
  • s∂1ϕ∂1U

= s 2

  • Rd−1 ∂1ϕ|∇U|2
  • x1=1

x1=0

+ s 2

  • (0,1)×Rd−1 ∂11ϕ|∇′U|2

s 2∂11ϕ|∂1U|2 + 2s2(∂1ϕ)2|∂1U|2 + 3s3 2 ∂11ϕ(∂1ϕ)2|U|2

  • +

s2 2 ((∂11ϕ)2 + ∂1ϕ∂111ϕ)|U|2. We recall that max

[0,1]{∂1ϕ} < 0 and min [0,1]{∂11ϕ} > 0.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Therefore,

s 2∂11ϕ|∂1U|2 + 2s2(∂1ϕ)2|∂1U|2 + 3s3 2 ∂11ϕ(∂1ϕ)2|U|2

  • ≤ C

|Fs∂1U| + Cs

  • Γ

|∂1U|2 + s 2

∂11ϕ|∇′U|2 + Cs2

|U|2. We need to estimate s

  • Ω ∂11ϕ|∇′U|2.

Multiply the equation by s∂11ϕU :

Fs∂11ϕU =

  • ∂11U − 2s∂1ϕ∂1U + (s2|∂1ϕ|2 − s∂11ϕ)U + ∆′U
  • s∂11ϕU

= −s

∂11ϕ|∇U|2 + s 2

∂(4)

1 ϕ|U|2 + s2

∂1(∂1ϕ∂11ϕ)|U|2 + s3

∂11ϕ(∂1ϕ)2|U|2 − s2

(∂11ϕ)2|U|2.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Combining the two estimates, we obtain

s 2∂11ϕ|∂1U|2 + s2(∂1ϕ)2|∂1U|2 + s3∂11ϕ(∂1ϕ)2|U|2 ≤ C

|F|2 + Cs

  • Γ

|∂1U|2 + Cs2

|U|2. Taking s large enough,

  • s|∂1U|2 + s2|∂1U|2 + s3|U|2

≤ C

|F|2 + Cs

  • Γ

|∂1U|2. We have an estimate on U, hence on ∇U from the previous estimates.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Comments on the proof

A direct approach consists in multiplying the equation of U by 2s∂1ϕ∂1U + s∂11ϕU. But the two above terms do not play the same role : s∂1ϕ∂1U is a multiplier chosen for analyzing the propagation in the x1 variable. The other term, s∂11ϕU, rather corresponds to an energy method to compensate the “bad” term

  • |∇′U|2 coming from

the multiplier s∂1ϕ∂1U. In particular, s∂11ϕU is a weaker order term in U than s∂1ϕ∂1U.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

The general case Ω

Carleman estimate Let Ω be a bounded domain. Assume that ϕ ∈ C 4(Ω), ∃α, β > 0, s.t. inf

x∈Ω

|∇ϕ(x)| ≥ α, and      −|∇ϕ|2∆ϕ + 2D2ϕ(∇ϕ, ∇ϕ) ≥ β|∇ϕ|2. inf

x∈Ω ξ∈Rd, |ξ|=1

  • ∆ϕ + 2D2ϕ(ξ, ξ)
  • ≥ β.

∃C > 0 s.t. for all s ≥ 1 and u ∈ H2 ∩ H1

0(Ω),

s3 esϕu2

L2(Ω) + s esϕ∇u2 L2(Ω)

≤ C

  • esϕf 2

L2(Ω) + s esϕ∂nu2 L2(Γϕ)

  • ,

where Γϕ = {x ∈ ∂Ω, ∂nϕ(x) > 0}.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Remarks

On the conditions on the weight functions : If ϕ = ϕ(x1), the two conditions are satisfied iff min[0,1]{|∂1ϕ|} > 0 and min[0,1]{∂11ϕ} > 0. If we further impose Γϕ to correspond to x1 = 0 in the case of a strip, one should have ∂1ϕ < 0. The regularity of ϕ can be lowered to C 2(Ω). The conditions on the weight functions used here are slightly too strong, see later. We have the following result : Lemma Let Γ ⊂ ∂Ω with Γ = ∅. ∃ϕ satisfying the conditions of the Theorem s.t. Γϕ ⊂ Γ.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Proof

Set U = u exp(sϕ) and F = ∆uesϕ. Then U solves ∆U − 2s∇ϕ · ∇U + s2|∇ϕ|2U − s∆ϕ U = F in Ω, U = 0

  • n ∂Ω.

We thus write P1U = ∆U + s2|∇ϕ|2U, P2U = −2s∆ϕ U − 2s∇ϕ · ∇U, RU = −s∆ϕ U. so that P1U + P2U = F + RU. P1, P2 roughly correspond to the self-adjoint and skew-adjoint parts of the operator.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Taking the L2 norm of both sides of the identity P1U + P2U = F + RU,

|P1U|2 +

|P2U|2 + 2

P1U P2U ≤ 2

|F|2 + 2

|RU|2. Main idea Compute

  • Ω P1U P2U and bound it from below.

After tedious computations, one gets :

P1U P2U = s3

  • −|∇ϕ|2∆ϕ + 2D2ϕ(∇ϕ, ∇ϕ)
  • |U|2

−s

∆2ϕ |U|2+s

∆ϕ|∇U|2 + 2s

D2ϕ(∇U, ∇U)−s

  • ∂Ω

∂nϕ|∂nU|2.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Denote by Ii,j the cross product between the i-th term of P1 and the j-th term of P2. I11 = −2s

∆ϕ U ∆U = 2s

∆ϕ|∇U|2 − s

∆2ϕ |U|2 I12 = −2s

∆U ∇ϕ · ∇U = 2s

∇U · ∇(∇ϕ · ∇U) − 2s

  • ∂Ω

∂nϕ|∂nU|2 = 2s

D2ϕ(∇U, ∇U) − s

∆ϕ |∇U|2 − s

  • ∂Ω

∂nϕ|∂nU|2 I21 = −2s3

|∇ϕ|2∆ϕ|U|2 I22 = −2s3

|∇ϕ|2∇ϕ · ∇U U = −s3

|∇ϕ|2∇ϕ · ∇(|U|2) = s3

  • |∇ϕ|2∆ϕ + 2D2ϕ(∇ϕ, ∇ϕ)
  • |U|2.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

We got

P1U P2U = s3

  • −|∇ϕ|2∆ϕ + 2D2ϕ(∇ϕ, ∇ϕ)
  • |U|2

−s

∆2ϕ |U|2+s

∆ϕ|∇U|2 + 2s

D2ϕ(∇U, ∇U)−s

  • ∂Ω

∂nϕ|∂nU|2. From our assumptions on ϕ, we obtain, for s ≥ s0 large enough, that s3

|U|2 + s

|∇U|2 ≤ C

P1U P2U + Cs

  • Γϕ

|∂nU|2. Thus, s3

|U|2 + s

|∇U|2 ≤ C

|F|2 + Cs2

|U|2 + Cs

  • Γϕ

|∂nU|2. Hence taking s0 larger if necessary, for all s ≥ s0, s3

|U|2 + s

|∇U|2 ≤ C

|F|2 + Cs

  • Γϕ

|∂nU|2.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Comments on the Lemma

Lemma Let Γ ⊂ ∂Ω with Γ = ∅. ∃ϕ satisfying the ∃α, β > 0, s.t. inf

x∈Ω

|∇ϕ(x)| ≥ α, and      −|∇ϕ|2∆ϕ + 2D2ϕ(∇ϕ, ∇ϕ) ≥ β|∇ϕ|2. inf

x∈Ω ξ∈Rd, |ξ|=1

  • ∆ϕ + 2D2ϕ(ξ, ξ)
  • ≥ β.

s.t. Γϕ ⊂ Γ. In particular, Γ can be arbitrarily small.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Proof of the Lemma

We first construct a smooth function ψ in C 4(Ω) such that inf

x∈Ω

{|∇ψ(x)|} > 0, and ∀x ∈ ∂Ω \ Γ, ∂nψ(x) < 0. Theorem

([Fursikov Imanuvilov ’96], see also [Tucsnak Weiss, App. III])

There exists such ψ. We then look for ϕ of the form ϕ(x) = eλψ(x) for some λ > 1 chosen later large enough.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Explicit computations yield ∇ϕ = λ∇ψϕ ∆ϕ = λ2|∇ψ|2ϕ + λ∆ψϕ D2ϕ = λ2∇ψT∇ψϕ + λD2ψϕ. By construction, Γϕ ⊂ Γ. Taking λ large enough − |∇ϕ|2∆ϕ + 2D2ϕ(∇ϕ, ∇ϕ) = λ4|∇ψ|4ϕ3 − λ3D2ψ(∇, ∇ψ)ϕ3 + λ3∆ψ|∇ψ|2ϕ3> 0. ∆ϕ|ξ|2 + 2D2ϕ(ξ, ξ) = λ2|∇ψ|2|ξ|2ϕ+λ∆ψ|ξ|2ϕ+2λ2(∇ψ · ξ)2ϕ+λD2ψ(ξ, ξ)ϕ> 0.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Remarks on this construction

The difficult point is Theorem

([Fursikov Imanuvilov ’96], see also [Tucsnak Weiss, App. III])

There exists ψ such that inf

x∈Ω

{|∇ψ(x)|} > 0, and ∀x ∈ ∂Ω \ Γ, ∂nψ(x) < 0. Once such ψ is constructed, we can chose ϕ = f (ψ) for any function f sufficiently convex, for instance of the form f (y) = eλy for sufficiently large λ. In other words, f increases the convexity in the direction of ∇ψ compared to the curvature of the level sets {ψ = c}. This process is called a “convexification” process.

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Comments on the conditions on ϕ

We imposed : inf

x∈Ω

|∇ϕ(x)| ≥ α, and      −|∇ϕ|2∆ϕ + 2D2ϕ(∇ϕ, ∇ϕ) ≥ β|∇ϕ|2. inf

x∈Ω ξ∈Rd, |ξ|=1

  • ∆ϕ + 2D2ϕ(ξ, ξ)
  • ≥ β.

No critical point in the domain. A kind of convexity condition. One can weaken this convexity condition into a weaker convexity condition.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Sharp convexity conditions

Carleman estimates for the Laplace operator holds for weight functions ϕ such that inf

x∈Ω

|∇ϕ(x)| ≥ α, and ∃β > 0, s.t ∀x ∈ Ω, ∀ξ ∈ Rd with ∇ϕ(x) · ξ = 0 and |∇ϕ(x)| = |ξ|, D2ϕx(∇ϕ(x), ∇ϕ(x)) + D2ϕx(ξ, ξ) ≥ β|∇ϕ(x)|2. This is the strict pseudo-convexity condition, see [Hörmander]. NB : A linear weight is degenerate, and belongs to the family of “Limiting Carleman Weights”.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Origin of the strict pseudo convexity condition

Write the conjugated operator Ps = esϕ∆(e−sϕ ·) as Ps = As + Bs, with A∗

s = As, B∗ s = −Bs

Then if PsU = F, we have F2 = AsU2 + BsU2 +

  • AsUBsU +
  • BsUAsU

=

  • U(A2

s + B∗ s Bs + [As, Bs])U + Boundary Terms.

In terms of pseudo-differential operator and taking the symbols of the operators, we will be able to derive estimates on U if the symbols of A2

s + B∗ s Bs + [As, Bs] is positive. In terms of symbols,

this means as(x, ξ) = 0 and bs(x, ξ) = 0 ⇒ {a, b}(x, ξ) > 0.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

In our case, As = −∆ − s2|∇ϕ|2, Bs = 2s∇ϕ · ∇, so the symbols

  • f the operators read

a(x, ξ) = |ξ|2 − |∇ϕ(x)|2 b(x, ξ) = 2iξ · ∇ϕ(x) {a, b}(x, ξ) = D2ϕx(∇ϕ(x), ∇ϕ(x)) + D2ϕx(ξ, ξ). Therefore, as(x, ξ) = 0 and bs(x, ξ) = 0 ⇒ {a, b}(x, ξ) > 0 is equivalent to ∃β > 0, s.t ∀x ∈ Ω, ∀ξ ∈ Rd with ∇ϕ(x) · ξ = 0 and |∇ϕ(x)| = |ξ|, D2ϕx(∇ϕ(x), ∇ϕ(x)) + D2ϕx(ξ, ξ) ≥ β|∇ϕ(x)|2. See [Le Rousseau Lebeau 2010] for more details.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Unique continuation

Theorem Let q ∈ L∞(Ω) and W ∈ (L∞(Ω))d, and let Γ be a non-empty

  • pen subset of the boundary ∂Ω. Then any solution u ∈ H1

0(Ω) of

∆u + qu + W · ∇u = 0 for x ∈ Ω, u = 0 for x ∈ ∂Ω, which further satisfies ∂nu = 0 on Γ vanishes identically on Ω : u = 0 in Ω. This property is called the unique continuation property through Γ. When q is constant, this shows that any eigenvector of −∆ + W · ∇ cannot identically vanish on Γ. NB : Here, we handled the zero order term q and the first order term W · ∇.

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Cauchy Linear Weight General Weights Strip Multiplier(1) General Unique Cont.

Proof

Choose a suitable Carleman weight function ϕ so that Γϕ ⊂ Γ, and then apply the Carleman estimate to u : s3 esϕu2

L2(Ω) + s esϕ∇u2 L2(Ω) ≤ C esϕ∆u2 L2(Ω)

≤ C

  • q2

L∞ esϕu2 L2(Ω) + W 2 L∞ esϕ∇u2 L2(Ω)

  • ,

Taking s large enough, we easily obtain u = 0 in Ω.

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More singular potentials ?

We can prove that for all u ∈ H1

0(Ω) with source term in L2(Ω) and

ϕ an admissible Carleman weight, s3 esϕu2

L2(Ω) + 1

s esϕu2

H2(Ω)

≤ C

  • esϕ∆u2

L2(Ω) + s esϕ∂nu2 L2(Γϕ)

  • .

Following, by interpolation, for all θ ∈ (0, 1) we get s3−4θ esϕu2

H2θ(Ω) ≤ C

  • esϕ∆u2

L2(Ω) + s esϕ∂nu2 L2(Γϕ)

  • .

Arguing as above allow then to derive unique continuation properties if q ∈ L2d/3+(Ω). Using Lp Carleman inequalities, [Jerison Kenig ’85] improves the result to q ∈ Ld/2+(Ω), which is sharp, see e.g. [Koch Tataru ’02].

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Thank you for your attention !

Sylvain Ervedoza Mai 2017 Carleman estimates for elliptic PDE and applications