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ThesisDefense,2015.12.23 Yi-HsuanLin Thedevelopmentoftheenclosuremethodinananisotropicbackgroundandthestronguniquecontinuationfortheelasticitywithresidualstress

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The development of the enclosure method in an anisotropic background and the strong unique continuation for the elasticity with residual stress

Yi-Hsuan Lin

Department of Mathematics, National Taiwan University

Thesis Defense, 2015.12.23

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Outline

1. The Enclosure Method (introduced by Ikehata) aaaa- Applied to Anisotropic Maxwell system a2. The Strong Unique Continuation Property aaaa- For the Residual Stress System with Gevrey Coecients

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Part 1: The enclosure method for the anisotropic Maxwell system

a

What is the enclosure method ?

(4)

Inverse obstacle problems

Discontinuity in a medium aects propagation of various physical quantity (signal) in a medium.

aCavity, Inclusion, Crack, Obstacle...etc a

Inverse Obstacle Problem is: extract information about the geometry of unknown discontinuity embedded in a known background medium from observed signal propagating inside the medium.

Applications: Nondestructive/Noninvasive testing, Sonar, Radar, EIT, etc.

(5)

Inverse obstacle problems

Discontinuity in a medium aects propagation of various physical quantity (signal) in a medium.

aCavity, Inclusion, Crack, Obstacle...etc a

Inverse Obstacle Problem is: extract information about the geometry of unknown discontinuity embedded in a known background medium from observed signal propagating inside the medium.

Applications: Nondestructive/Noninvasive testing, Sonar, Radar, EIT, etc.

(6)

Inverse obstacle problems

Discontinuity in a medium aects propagation of various physical quantity (signal) in a medium.

aCavity, Inclusion, Crack, Obstacle...etc a

Inverse Obstacle Problem is: extract information about the geometry of unknown discontinuity embedded in a given

background medium from observed signal propagating inside the medium.

Applications: Nondestructive/Noninvasive testing, Sonar, Radar, EIT, etc.

(7)

Inverse obstacle problems

Discontinuity in a medium aects propagation of various physical quantity (signal) in a medium.

aCavity, Inclusion, Crack, Obstacle...etc a

Inverse Obstacle Problem is: extract information about the geometry of unknown discontinuity embedded in a given

background medium from observed signal propagating inside the medium.

Applications: Nondestructive/Noninvasive testing, Sonar, Radar, EIT, etc.

(8)

Inverse obstacle problems

Discontinuity in a medium aects propagation of various physical quantity (signal) in a medium.

aCavity,Inclusion, Crack, Obstacle...etc

(the interface of jump discontinuity of the medium)

Inverse Obstacle Problem is: extract information about the geometry of unknown discontinuity embedded in a given

background medium from observed signal propagating inside the medium.

Applications: Nondestructive/Noninvasive testing, Sonar, Radar, EIT, etc.

(9)

Problem description

We consider unknown obstacles in electromagnetic elds with anisotropic medium lies in Ω.

Let Ω be a bounded domain in R3. D is an inclusion in Ω.

Problem: How to nd D?

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Various methods

There are several methods to retrieve the information of obstacles D inside Ω.

1. Probe Method (Ikehata) 2. Enclosure Method (Ikehata)

3. Linear Sampling Method (Colton-Kirsch) 4. Factorization Method (Kirsch)

5. Singular Source Method (Potthast) a

We focus on theenclosure methodto nd the unknown inclusions.

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The enclosure method

In order to understand ideas, here we consider the simplest case:

∆u0=0 in Ω.

Goal: Reconstruct unknown D by the boundary measurements.

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Consider the function u satisfying

∇ · ((1 + γDχD)∇u) =0 in Ω.

The correspondingDirichlet-to-Neumann(DN) map is given by ΛD(u|∂ Ω) = ∂ u

∂ ν

|∂ Ω, where ν is the unit outer normal on ∂Ω.

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Ideas

Key ideas in the enclosure type method:

(A) Let Λ/0 be the DN map when D = /0 and Λ/0(u0|∂ Ω) =∂ u0

∂ ν

|∂ Ω. By energy method, one can show the following energy integral

Z

∂ Ω

D− Λ/0)f · ¯f dS ≈ Z

D

|∇u0|2dx (1.1) where u0 is the solution of the unperturbed equations (without D,

∆u0=0 in Ω).

(1.1) is true due to the positivity of the equation.

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Special solution

(B) One can nd solutions of the Laplace equation in the following form

u0,d,h= e1h[ω·x−d +i ω·x].

This is thecomplex geometrical optics(CGO) solution of the Laplace equation.

Behavior of u0,h:

(u0,d,h↓0 as h → 0+ for ω · x < d, uo,d ,h↑ ∞ as h → 0+ for ω · x > d.

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Indicator function

From the energy integral, we can dene theindicator functionI: for any f ∈ H1/2(∂ Ω),

I (f ) = Z

∂ Ω

D− Λ/0)f · ¯f dS ,

I (f )is completely determined by the boundary measurements.

We can take f := fd ,h= u0,d,h|∂ Ω into the indicator functional I = I (d , h).

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If D ∩ {ω · x > d} = /0, then I (d,h) → 0 as h → 0+.

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If D ∩ {ω · x > d} 6= /0, then I (d,h) → ∞ as h → 0+.

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Change the direction ω and move the level set {x · ω = d}, we can enclosethe unknown obstacle D.

a

The enclosure reconstruction method consists of: CGO solutions andEnergy integral.

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Anisotropic Maxwell system

This is a joint work with Rulin Kuan and Mourad Sini.

aLet D be an unknown obstacle and let k > 0 be the wave number.

We consider the anisotropic Maxwell system





∇ × E − ik µ H =0 in Ω,

∇ × H + ik ε E =0 in Ω,

ν × E = f on ∂Ω.

(2.1)

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Description of the problem

For the inclusion case, the coecients ε(x) = ε0(x ) − εD(x )χD, where ε0(x )to be a C positive denite matrix-valued function, εD(x )is a matrix-valued function which is regarded as a

perturbation in the unknown obstacle D. µ(x) > 0 is a C scalar function, and ν is the unit outer normal on ∂Ω

As before, D is an inclusion.

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Assume k is not an eigenvalue for the spectral problem to (2.1), then (2.1) is well posed.

Impedance Map: We dene the impedance map ΛD: TH12(∂ Ω) → TH12(∂ Ω)by

ΛD(ν × H|∂ Ω) = (ν × E |∂ Ω),

where TH12(∂ Ω) := {f ∈ H12(∂ Ω)|ν · f =0} and × is the standard cross product in R3.

We denote by Λ/0 the impedance map for the domain without an obstacle.

Inverse Problem: Find D from ΛD.

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Diculties

For the anisotropic Maxwell's system, until now, there areNO CGO solutions. Similar to anisotropic elliptic case, we will construct new special solutions: Theoscillating-decaying(OD) solutions.

aThe OD solutions were rst constructed by G. Nakamura, G.

Uhlmann, J. N. Wang, 2005.

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Construction OD solutions

Consider the anisotropic Maxwell system (

∇ × E − ik µ H =0 in Ω,

∇ × H + ik ε E =0 in Ω, (2.2) where ε is a matrix and µ is a scalar and we want to construct OD solutions.

The rst step of constructing OD solutions is to reduce the anisotropic Maxwell systemto a strongly elliptic system.

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From Maxwell to Elliptic

If E and H are of the following forms





E = −i

1∇ × (µ1(∇ × B)) − ε1(∇ × A) H = i

1∇ × (ε1(∇ × A)) − µ1(∇ × B) with A,B satisfying thestrongly elliptic systems

(LA(A) = µ∇tr (MA∇A) − ∇ × (ε1(∇ × A)) + k2µ A =0 LB(B) = ε∇tr (MB∇B ) − ∇ × (µ1(∇ × B)) + k2ε B =0 , where MA= mµ1I3 and MB = mµ1ε for arbitrary positive constant m. Then (E,H) satises (2.2).

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Oscillating decaying solutions for Elliptic

Then we represent A and B to be two oscillating-decaying solution in the following form: Let ω be a unit vector, then









A = wχA

t,b,t,N,ω+ rχA

t,b,t,N,ω inΩt(ω) := Ω ∩ {x · ω > t}, wχA

t,b,t,N,ω= χt(x0)Qtei τx·ξe−τ(x·ω−t)AAt(x0)b + γχA

t,b,t,N,ω(x , τ), B = wχB

t,b,t,N,ω+ rχB

t,b,t,N,ω in Ωt(ω), wχB

t,b,t,N,ω= χt(x0)Qtei τx·ξe−τ(x·ω−t)ABt(x0)b + γχB

t,b,t,N,ω(x , τ) with suitable decay in τ for γA, γB, rA and rB.

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Oscillating decaying solutions for Maxwell

Via the relations between E,H,A,B, for the inclusion obstacle case, we have









Et= FA1(x )ei τx·ξe−τ(x·ω−t)AAt(x0)b +ΓA,χ1

t,b,t,N,ω+ rχA,1

t,b,t,N,ω

Ht= FA2(x )ei τx·ξe−τ(x·ω−t)AAt(x0)b +ΓA,χ2

t,b,t,N,ω+ rχA,2

t,b,t,N,ω

inΩt(ω),

where FA1(x ) = O(τ), FA2(x ) = O(τ2) are some smooth functions and for |α| = j, j = 1,2, we have

(kΓA,j

χt,b,t,N,ω(x , τ)kL2(Ωt(ω))≤ cτ|α|−3/2e−τ(s−t)aA, krA,j

χt,b,t,N,ω(x , τ)kL2(Ωt(ω))≤ cτj −N+1/2,

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Energy integral

The impedance maps ΛD : ν × H|∂ Ω→ ν × E |∂ Ω and

Λ0: ν × H0|∂ Ω→ ν × E0|∂ Ω. By energy method, one can show that Z

∂ Ω

(ν × H0) · ((ΛD− Λ/0)(ν × H0) × ν)dS ≈ Z

D|∇ × H0|2dx . However, the OD solutions are not well-dene on the whole domain Ω, we cannot obtain the full boundary information from the OD solutions.

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Runge approximation property

In fact, we can nd a sequence of solutions solving the anisotropic Maxwell's system and they will approximate to the OD solution.

This property is called theRunge approximation property. If we set u =

 H E

 and

L := i

 ε1 0 0 µ1I3

  0 ∇×

−∇× 0

 + kI6, then we have

Lu =0,

where Ij means j × j identity matrix for j = 3,6.

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Theorem (Runge approximation property)

Let D and Ω be two open bounded domains with C boundary in R3 withD b Ω. If u ∈ (H(curl,D))2 satises

Lu =0 in D.

Given any compact subset K ⊂ D and any ε > 0, there exists U ∈ (H(curl , Ω))2 such that

LU =0 in Ω, and kU − ukH(curl ,K )< ε, where

kf kH(curl ,K )= kf kL2(K )+ kcurlf kL2(K )

.

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Technical part

Support function: For ω ∈ S2, we dene the support function of D by hD(ω) =infx ∈Dx · ω.

When t = hD(ρ),Ωt(ω) ∩ ∂ D 6= /0. We cannot use the Runge approximation property directly.

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Let η be any positive real number.

This will implyD b Ωt−η(ω).

We denote (Et−η, Ht−η) to be the OD solution dened on Ωt−η(ω).

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By the Runge approximation property, we can nd a sequence of functions {(Eη ,`, Hη ,`)}`=1 satisfying the Maxwell system in Ω such that (Eη ,`, Hη ,`) approximates to (Et−η, Ht−η) as ` → ∞ in

L2(Ωt−η(ω))and in H(curl,D) by interior estimates since D b Ωt−η(ω).

aIn addition, we can show that (Et−η, Ht−η)converges to (Et, Ht) in H(curl , D)as η → 0.

aFrom the energy integral, we can dene the indicator function as follows.

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Indicator function

Indicator function: For ω ∈ S2, τ > 0 and t > 0 we dene the indicator function

Iω(τ, t) := lim

η →0lim

`→∞Iωη ,`(τ, t), where

Iωη ,`(τ, t) := ikτ Z

∂ Ω

(ν × Hη ,`) · ((ΛD− Λ/0)(ν × Hη ,`) × ν)dS.

Goal: We want to characterize the convex hull of the obstacle D from the impedance map ΛD.

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Integral Inequalities

1Iωη ,`| ≤ c Z

D

|∇ × Hη ,`|2dx + k2 Z

µ | gHη ,`|2dx ,

1Iωη ,`| ≥c Z

D

|∇ × Hη ,`|2dx−k2 Z

µ | gHη ,`|2dx ,

where gHη ,`= H − Hη ,` be the reected solution, then gHη ,` satises (

∇ × (ε1∇ × gHη ,`) − k2µ gHη ,`= −∇ × ((ε1(x ) − ε01(x ))∇ × Hη ,`) in Ω, ν × gHη ,`=0 on ∂Ω.

Then we have some estimate for gHη ,`.

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Meyers L

p

estimate

Proposition (Estimate for gHη ,`)

Assume Ω is a smooth domain and D b Ω. Then there exist a positive constant C and δ > 0 such that

k gHη ,`kL2(Ω)≤ C k∇ × Hη ,`kLp(D)

for every p ∈ (4 3,2].

The proof is by using a global Lp estimate for the curl of the solutions of the anisotropic Maxwell system.

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Theorem (Main Theorem)

Let ω ∈ S2, we have the following characterization of hD(ω). (limτ →∞|Iω(τ, t)| =0 when t < hD(ω),

liminfτ →∞|Iω(τ, hD(ω))| >0.

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Conclusion for part 1

•We develop an enclosure type method for identifying inclusion obstacles in anisotropic Maxwell system. Our main tool is the OD solutions for the anisotropic Maxwell system.

a

•Our theory shows that we are able to determine theconvex hull of inclusions by the impedance map.

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Part 2: SUCP for residual stress system with Gevrey coecients

Unique Continuation Property

We call u has the unique continuation property If u ∈ Hloc1 (Ω) satises Lu = 0 in Ω and u vanishes on an open subset of Ω, then u must vanish identically in Ω.

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Strong Unique Continuation Property (SUCP)

We call u has theSUCP if u ∈ Hloc1 (Ω)satises Lu = 0 in Ω and vanishes to innity order at a point x0∈ Ω,i.e., for all N > 0

Z

R≤|x −x0|≤2R|u|2dx = O(RN), R →0, (3.1) then u must vanish identically in Ω. If u is smooth, the condition (3.1) is equivalent to all partial derivatives of u vanishing at x0, which means ∂βf (x0) =0 for all multiindices β.

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Residual Stress System

Let Ω be a open connected domain in R3 and consider the time-harmonic elasticity system

∇ · σ + κ2ρ u =0 in Ω, (3.2) where σ = (σij)3i ,j =1 is the stress tensor eld, κ ∈ R is the

frequency and ρ = ρ(x) > 0 denotes the density of the medium.

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The vector eld u(x) = (ui(x ))3i =1 is the displacement vector.

Suppose that the stress tensor is given by

σ (x ) = T (x ) + (∇u)T (x ) + λ (x )(trE )I +2µ(x)E, where E(x) =∇u + ∇ut

2 is the innitesimal strain and λ(x), µ(x) are the Lam´e parameters.

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The second-rank tensor T (x) = (tij(x ))3i ,j =1 is the residual stress and satises

tij(x ) = tji(x ), ∀i , j =1,2,3 and x ∈ Ω and

∇ · T =

j

jtij =0 in Ω, ∀i = 1,2,3.

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If we dene the elastic tensor C = (Cijkl)3i ,j ,k,l =1 with Cijkl= λ δijδkl+ µ(δjkδjl+ δjkδil) + tjlδik, then (3.2) is equivalent to

∇ · (C ∇u) + κ2ρ u =0 in Ω.

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Brief History

Results on (strong) unique continuation property for theresidual stress systemhas been proved by:

•G. Nakamura and J.N. Wang (2003) proved the unique

continuation property for (3.2) under the condition maxi ,jktijkis small and T (x),λ(x), µ(x) ∈ W2,∞ and ρ(x) ∈ W1,∞.

•C.L. Lin (2004) proved the SUCP for (3.2) under the assumptions that T (0) = 0, maxi ,jktijkis small, λ(x), µ(x) and ρ(x) are in C2.

•G. Uhlmann and J.N. Wang proved unique continuation principle for (3.2) under the conditions T (x),λ(x), µ(x) ∈ W2,∞,

ρ (x ) ∈ W1,∞ and general residual stress.

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Brief History

Results on (strong) unique continuation property for theresidual stress systemhas been proved by:

•G. Nakamura and J.N. Wang (2003) proved the unique

continuation property for (3.2) under the condition maxi ,jktijkis small and T (x),λ(x), µ(x) ∈ W2,∞ and ρ(x) ∈ W1,∞.

•C.L. Lin (2004) proved the SUCP for (3.2) under the assumptions that T (0) = 0, maxi ,jktijkis small, λ(x), µ(x) and ρ(x) are in C2.

•G. Uhlmann and J.N. Wang proved unique continuation principle for (3.2) under the conditions T (x),λ(x), µ(x) ∈ W2,∞,

ρ (x ) ∈ W1,∞ and general residual stress.

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Brief History

Results on (strong) unique continuation property for theresidual stress systemhas been proved by:

•G. Nakamura and J.N. Wang (2003) proved the unique

continuation property for (3.2) under the condition maxi ,jktijkis small and T (x),λ(x), µ(x) ∈ W2,∞ and ρ(x) ∈ W1,∞.

•C.L. Lin (2004) proved the SUCP for (3.2) under the assumptions that T (0) = 0, maxi ,jktijkis small, λ(x), µ(x) and ρ(x) are in C2.

•G. Uhlmann and J.N. Wang (2009) proved unique continuation principle for (3.2) under the conditions T (x),λ(x), µ(x) ∈ W2,∞, ρ (x ) ∈ W1,∞ and general residual stress.

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Reduction - First step

We want to prove the SUCP for

∇ · (C ∇u) + κ2ρ u =0 in Ω.

We dene

Ru = ∇ · (∇uT ) (3.3)

with Ru = ((Ru)1, (Ru)2, (Ru)3), where (Ru)i = ∑jktjkjk2ui, i =1,2,3.

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Set U = (u,v,w)t, where v = ∇ · u, w = ∇ × u and u satises

∇ · (C ∇u) + κ2ρ u =0 in Ω. (3.4) Take divergence on (3.4) and take curl on (3.4), we can nd two equations for v,w. We write dierential equations for u,v,w in the following form.

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(u, v , w )satises

fP1(x , D)u := Ru

µ + ∆u =

1 m=

0

A1,m(u, v )

fP2(x , D)v := Rv

(λ +2µ)+ ∆v = − 1 λ +2µ

jk

∇(tjk) · ∂jk2u +

1

m=0

A2,m(u, v , w ),

fP1(x , D)w := Rw

µ + ∆w = −1 µ

jk

∇(tjk) × ∂jk2u +

1 m=0

A3,m(u, v , w ), where A`,m are m-th order dierential operator, ` = 1,2,3, m = 0,1.

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Diculties

We want to prove the SUCP for general residual stress T . The main diculty is:

a1. The system of (u,v,w) is not decoupled.

a2. We cannot nd a change of coordinates such that the

dierential operators fP1(x , D) and fP2(x , D) are Laplacian at x = 0 simultaneously (If T (0) = 0, then fP1(0,D) = fP2(0,D) = ∆).

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Main Trick

Our main tool is to reduce (3.2) into a specialfourth order elliptic system.

We need to derive suitableCarleman estimates in order to get the SUCP for this elliptic system.

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In general, the SUCP doe not hold even the coecients are smooth,Alinhac (1980) gave a counterexample. Thus, we consider the coecients of (3.2) lie in theGevrey class.

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Gevrey Class

We say that f ∈ C(Ω)belongs to the Gevrey class of order s, denote it as Gs(Ω)(or Gs), if there exist constants c, A and multiindices β such that

|∂βf | ≤ cA|β |(|β |!)s in Ω.

We give several useful properties for Gevrey class Gs. a

The Gevery class collects functions between smooth and analytic.

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Properties for Gevrey functions

1. f ∈ Gs and vanishes to innity order at 0. If s − 1 < ρ, then

|f (x)| ≤ e−|x|1/ρ near x = 0.

2. e−|x|1/ρ∈ Gs provided 1 + ρ = s.

3. (Gevrey regularity) Let P(x,D)u = f in Ω be an elliptic dierential system with coecients and f are in the Gevrey class Gs. Then u ∈ Gs(O)for all bounded O b Ω.

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Properties for Gevrey functions

1. f ∈ Gs and vanishes to innity order at 0. If s − 1 < ρ, then

|f (x)| ≤ e−|x|1/ρ near x = 0.

2. e−|x|1/ρ∈ Gs provided 1 + ρ = s.

3. (Gevrey regularity) Let P(x,D)u = f in Ω be an elliptic dierential system with coecients and f are in the Gevrey class Gs. Then u ∈ Gs(O)for all bounded O b Ω.

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Properties for Gevrey functions

1. f ∈ Gs and vanishes to innity order at 0. If s − 1 < ρ, then

|f (x)| ≤ e−|x|1/ρ near x = 0.

2. e−|x|1/ρ∈ Gs provided 1 + ρ = s.

3. (Gevrey regularity) Let P(x,D)u = f in Ω be an elliptic dierential system with coecients and f are in the Gevrey class Gs. Then u ∈ Gs(O)for all bounded O b Ω.

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We assume P1 and P2 are two strongly elliptic operators, where P1(x , D) :=

jk

a1jk(x )∂x2jxk:=

jk

(µδjk+ tjk)∂x2jxk, (4.1) P2(x , D) :=

jk

a2jk(x )∂x2jxk:=

jk

((λ +2µ)δjk+ tjk)∂x2jxk(4.2) with a1jk(x ) = µ(x)δjk+ tjk(x )and

ajk2(x ) = (λ (x) +2µ(x))δjk+ tjk(x ).

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The elliptic condition means that there exists c0>0 such that for any ξ = (ξi)3i =1∈ R3

jk

a1jk(x )ξjξk =

jk

tjkξjξk+ µ|ξ |2≥ c0|ξ |2 (4.3)

jk

a2jk(x )ξjξk =

jk

tjkξjξk+ (λ +2µ)|ξ|2≥ c0|ξ |2 (4.4) for all x ∈ Ω.

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Colombini, Grammatico and Tataru (2006) derive the following condition for the appropriate index s for Gs. ∃α > 0 such that the eigenvalues λ1`≤ λ2`≤ λ3` of (a`jk(0)) satisfying

α > λ3`− λ1` λ1` and

s <1 + 1 α uniformly in x and for ` = 1,2.

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Main result

Theorem (Main Theorem)

Let the residual stress (tij(x ))3i ,j =1, the Lam´e parameters λ(x), µ (x )and the density of the medium ρ(x) be in the Gevrey class Gs(Ω)with s satisfying s < 1 + 1

α. Then for all u ∈ Hloc2 (Ω; R3) solving (3.2) and for all N > 0

Z

R≤|x |≤2R|u|2dx = O(RN) as R → 0, then u is identically zero in Ω.

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However, it is not easy to prove the main theorem directly for the general residual stress, we will introduce a reduction method to transform (3.2) into a new fourth order elliptic system with principally diagonal leading terms. Moreover, we need to derive suitable Carleman estimates in order to obtain the SUCP.

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Reduction - Second step

Recall that in the rst step, we have reduce the residual stress system into

P1(x , D)u = Ru + µ∆u =

1 m=

0

B1,m(u, v )

P2(x , D)v = Ru + (λ +2µ)∆v = −

jk

∇(tjk) · ∂jk2u +

1 m=

0

B2,m(u, v , w ),

P1(x , D)w = Rw + µ∆w = −

jk

∇(tjk) × ∂jk2u +

1 m=

0

B3,m(u, v , w ), where B`,m are m-th order dierential operators, m = 0,1 and

` =1,2,3.

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Key observation 1

If we act P2 on P1(x , D)u-equation, P1(x , D)w-equation and act P1 on P2(x , D)v equation, we will obtain fourth order elliptic equations P2P1 for u,w and P1P2 for v.

P2(P1(x , D)u) = P2

1

m=0

B1,m(u, v )

! ,

P1(P2(x , D)v ) = P1

jk

∇(tjk) · ∂jk2u +

1

m=0

B2,m(u, v , w )

! ,

P2(P1(x , D)w ) = P2

jk

∇(tjk) × ∂jk2u +

1

m=0

B3,m(u, v , w )

! ,

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Key observation 2

We also observe that P1(x , D)[

jk

∇(tjk) · ∂jk2u] =

m≤3

Dm1(u, v , w ) and

P2(x , D)[

jk

∇(tjk) × ∂jk2u] =

m≤3

Dm2(u, v , w ), where Dm1, Dm2 are m-th order dierential operators.

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Let U = (u,v,w)t, we transform (3.2) into a fourth order principally diagonal elliptic system

P2P1U =

3 m=

0

Bm(U), (4.5)

where Bm(U) is an m-th order dierential operator. Note that the leading coecients of U is principally diagonal and U ∈ Gs since the coecients of (4.5) are in Gs.

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Main result for fourth order elliptic system

Theorem (Main Theorem)

Let P = P2P1 be a fourth order elliptic operator with coecients in Gs. Then the SUCP holds for the elliptic system

PU =

|β |≤3

Bm(U) provided that all the coecients are in Gs. If U has the SUCP, then u has the SUCP.

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The main theorem was rst proved by Colombini and Koch in 2010, they use iterative method to derive the Carleman estimate for higher order elliptic equations. They proved the Carleman estimates in the following type:

Let P1, P2, · · · , PM be second order elliptic operators with Gevrey coecients. Let P = PMPM−1· · · P1 be an 2M order elliptic operator, then the SUCP holds for

Pu =

|β |≤[3M2 ]

aββu

provided aβ∈ Gs, ∀|β| ≤ [3M2 ].

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The proof of the theorem is based on the iteration from the Carleman estimates for the 2nd order elliptic operator. In our case, when M = 2, 2M = 4 and [3M2 ] =3, then we can apply the

Colombini-Koch's result directly.

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Carleman Estimates

We are going to derive theCarleman estimatesfor the weight eτ |x |−α for the fourth order elliptic operator P = P2P1. The main point is that U ∈ Gs and vanishes to innity order, then we have

|U| ≤ e−|x|−α near 0 provided that s < 1 +1 α.

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We have the following Carleman estimates for P:

4

j =0

τ6−2j Z

|x|8−6α|x|2j(1+α)e2τ|x|−α|DjV |2dx

.

2 j =9

τ3−2j Z

|x|4−3α|x|j (1+α)e2τ|x|−α|Dj(P1V )|2dx .

Z

e2τ|x|−α|(P2P1V )|2dx = Z

e2τ|x|−α|PV |2dx .

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Use the vanishing to innity order of U at 0 and the Carleman estimate we can obtain U ≡ 0 in a small neighborhood of 0, then u ≡0 in a small neighborhood of 0.

aFurthermore, by using the unique continuation principal was proved by Uhlmann-Wang's result in 2009, therefore, we can obtain u ≡ 0 in Ω, then we are done.

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Conclusion 2

Under the assumptions for the Gevrey coecients with appropriate indices s, we can prove if u vanishes to innity order at x0∈ Ω, then u vanishes identically in Ω provided that Ω is a simply connected domain.

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e ff f ff f f

f

Thank you for your attention !

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Proof of the Main Theorem

The operator P = P2P1 is strongly elliptic in the Gevrey class Gs, then U is also in the Gevrey class Gs. Therefore, we have the vanishing of innite order implies that

|u| . e−|x|−γ for some γ > α.

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Let χ ∈ C0(R3) be such that χ ≡ 1 for |x| ≤ R and χ ≡ 0 for

|x| ≥2R (R > 0 is small enough). Then we can apply the Carleman estimates to the function χU, which means

C

4

|β |=0

τ6−2|β|

Z

|x|<R

|x|(2|β|−6)(1+α)−2e2τ|x|−α|DβU|2dx (5.1)

Z

e2τ|x|−α|PU|2dx

Z

|x|<R

e2τ|x|−α|PU|2dx + Z

|x|>R

e2τ|x|−α|P(χU)|2

Z

|x|<Re2τ|x|−α|

3

m=0

Ecm(U)|2dx + Z

|x|>Re2τ|x|−α|P(χU)|2,

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If τ is large and R is suciently small, then we have

C

4

|β |=0

τ6−2|β|

Z

|x|<R

|x|(2|β|−6)(1+α)−2e2τ|x|−α|DβU|2dx (5.2)

Z

|x|>R

e2τ|x|−α|P(χU)|2, for some constant C > 0.

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Notice that eτ |x |−α ≥ eτ R−α for |x| < R and eτ |x |−α ≤ eτ R−α for

|x| > R. Therefore, we can use (5.2) to obtain

C

4

|β |=0

τ6−2|β|

Z

|x|<R

|x|(2|β|−6)(1+α)−2|DβU|2dx

Z

|x|>R|P(χU)|2.

Let τ → ∞, we get U = 0 in {|x| < R} for R small, which implies u =0 in {|x| < R}.

參考文獻

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