Linearized internal functionals for anisotropic conductivities - - PowerPoint PPT Presentation

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Linearized internal functionals for anisotropic conductivities

Chenxi Guo

Chenxi Guo Chenxi Guo Chenxi Guo

joint work with Guillaume Bal Guillaume Bal Guillaume Bal Guillaume Bal and Fran Fran Fran Franç ç ç çois Monard

• is Monard
• is Monard
• is Monard
• Dept. of Applied Physics and Applied Mathematics, Columbia University.

June 20th, 2012 UC Irvine Conference in honor of Gunther Uhlmann

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2

1.Background

, 1

( ) ( )

n ij i j i j

u u γ γ

=

−∇⋅ ∇ ≡ − ∂ ∂ =

∑

is a real-valued symmetric positive definite tensor with bounded coefficients, satisfying a uniform ellipticity condition for some

γ

2 1

( ) x κ ξ ξ γ ξ κ ξ

−

≤ ⋅ ≤

1 κ ≥

n

R ξ∈ x∈Ω

| u g

∂ Ω=

( ) Ω

• Conductivity equation: rules the equilibrium distribution of the electrostatic

potential u inside the domain , in response to a prescribed boundary voltage g. Electrical Impedance Tomography(EIT). Calderon’s problem

| u v γ

∂Ω

∇ ⋅

→

• Internal measurement

power density of a solution u

[ ]( ): ( ) ( ) ( ) H g x u x x u x

γ

γ = ∇ ⋅ ∇

Application: Hybrid imaging How to construct power densities: Ammari et al. (2008), Kuchment-Kunyansky (2010)

Ω

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History (non-linear case):

( ) u γ −∇ ⋅ ∇ =

| u g

∂Ω=

( ) Ω

• Isotropic case

n

I σ γ =

from only one measurement

2

H u σ = ∇

2

( ) H u u ∇⋅ ∇ = ∇

( ) Ω | u g

∂ Ω=

• Newton-based method see (Gebauer and Scherzer (2009))
• Theoretically, by Bal (2012).

[ ]( ) : ( ) ( ) ( ) H g x u x x u x

γ

γ = ∇ ⋅ ∇

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4

The conductivity equation

• )

' ( = ∇ ⋅ ∇ u

unknown

γ

), (Ω

• known

g

u

=

Ω ∂

|

Fixing a few boundary condtions

) 1 (

with

' i

u solving (1) with

Measurement operator :

1, , m

g g ⋯

' '

' ) ' ( ) ' ( ' :

j i ij

u u H H H ∇ ⋅ ∇ = = → γ γ γ γ

i

g g ≡

m j i ≤ ≤ , 1

Problem: recover ' γ from

• Explicit reconstruction using a large number of functionals in an isotropic

case in dimension 3. see (Bal,Monard, Bonnetier and Triki)

2 2

u ∇

α

σ

where α not necessary 2

1

• Generalized to dimension n, isotropic tensor with more general type of

measurements

• reconstruction formulas for the anisotropic two-dimensional problem

see Monard and Bal (2012). see Monard and Bal (2012).

• Isotropic, 2 dimensional setting, see (Capdeboscq et al. (2009))

H

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Linearization of the problem The partial differential equation

• )

' ( = ∇ ⋅ ∇ u

unknown

γ

), (Ω

• known

g

u

=

Ω ∂

|

Fix a few boundary conditions

) 1 (

with

' i

u solving (1) with

Measurement operator :

1, , m

g g ⋯

' '

' ) ' ( ) ' ( ' :

j i ij

u u H H H ∇ ⋅ ∇ = = → γ γ γ γ

i

g g ≡ m j i ≤ ≤ , 1 Fréchet derivative:

• )

( '

2

ε ο εγ γ γ + + =

known

) (

2 '

ε ο ε + + =

i i i

v u u PDE (1) of order and

• )

( = ∇ ⋅ ∇ −

known i

u γ

• known

i i

g u =

Ω ∂

|

• )

( ) (

i unknown i

u v ∇ ⋅ ∇ = ∇ ⋅ ∇ − γ γ ), (Ω

| =

Ω ∂ i

v Non-linear problem: recover

' γ

from

H

) (ε

• )

1 (

• ),

(Ω

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6

• isotropic case in dimension 2 and 3 with numerical implementation

see (Kuchment and Kunyansky (2011))

• isotropic case, studied using pseudo-differential calculus, inversion modulo a

compact operator. see (Kuchment and Steinhauer (2011)) References on the linearized problem The measurements look like

) ( ) (

2

ε ο γ γ γ ε γ + ∇ ⋅ ∇ + ∇ ⋅ ∇ + ∇ ⋅ ∇ + ∇ ⋅ ∇ =

i j j i j i j i ij

v u v u u u u u H

Linearized measurements

• i

j j i j i known ij

v u v u u u dH ∇ ⋅ ∇ + ∇ ⋅ ∇ + ∇ ⋅ ∇ = γ γ γ

↓ Linearized problem: recover γ from

ij

dH

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7

Study of the principal symbol

) , ( ξ x M ij

• f

ij

dH

Recall the equations and measurements:

• )

( ) (

i unknown i

u v ∇ ⋅ ∇ = ∇ ⋅ ∇ − γ γ ), (Ω | =

Ω ∂ i

v

• )

( = ∇ ⋅ ∇ −

known i

u γ

), (Ω

• known

i i

g u =

Ω ∂

|

• i

j j i j i known ij

v u v u u u dH ∇ ⋅ ∇ + ∇ ⋅ ∇ + ∇ ⋅ ∇ = γ γ γ

Denote ) ( : ∇ ⋅ −∇ = γ L

)) ( (

1 i i

u L v ∇ ⋅ ∇ =

−

γ

suppose γ compactly supported inside Ω

Ω

γ

supp

Insert

)) ( (

1 i i

u L v ∇ ⋅ ∇ =

−

γ

into

ij

dH

and express as a pseudo-DO

y d d y x M x M e x dH

n n R

R ij ij y x i n ij

∫∫

× − − ⋅ −

+ = ξ γ ξ ξ π γ

ξ

) ( : )) , (

|

) , ( ( ) 2 ( ) , (

1 ) (

) ( ) , ( ξ ο ξ = x M ij ) ( ) , (

|

1 1 − −

= ξ ο ξ x M ij

↓ ↓

pricipal symbol the symbol of order -1

2.Microlocal inversion

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8 y d d y x M x M e x dH

n n R

R ij ij y x i n ij

∫∫

× − − ⋅ −

+ = ξ γ ξ ξ π γ

ξ

) ( : )) , ( | ) , ( ( ) 2 ( ) , (

1 ) (

) ( ) , ( ξ ο ξ = x M ij ) ( ) , ( |

1 1 − −

= ξ ο ξ x M ij

↓ ↓

pricipal symbol the symbol of order -1

Goal: determine under which conditions the operator

m j i ij

dH dH

≤ ≤

=

, 1

} {

is an elliptic pseudo-differential operator Conclusion: with only principal symbols ,

) , ( ξ x M ij

m j i ij

dH dH

≤ ≤

=

, 1

} {

will never be elliptic, no matter how large m Define

2 1

) ( γ = A

∧

= ξ ξ

0 :

A

and

i i

u A V ∇ = :

rewrite the principal symbol

) , ( ξ x M ij

) , (

~

ξ x M ij

) , ( A x M A

ij

ξ =

( ) ( )

i j i j j i

V V V V V V ξ ξ ξ ξ = − ⋅ − ⋅ ⊙ ⊙ ⊙

1( ) 2 U V U V V U = ⊗ + ⊗ ⊙

Notation:

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Lemma: For any

j i, and vector fields

j i V

V ,

defined as above

i i

u A V ∇ = :

the symbol

) , (

~

ξ x M ij

satisfies

~

( , ):

ij

M x ξ ξ η = ⊙

for all

1 −

∈

n

S η

and

η ξ ⊥

Conclusion: the

ij

M can never control a subspace of

) (R Sn

• f dimension

1 − n

Basic Hypothesis: the gradients

n i i

u

1

} {

=

∇

form a frame in

n

R

Lemma: Suppose that the vector fields

n i i

V

1

} {

= form a basis of

If a matrix

) (R S P

n

∈

~

( , ):

ij

M x P ξ = n j i ≤ ≤ ≤ 1

Then P is of the form

P ξ η = ⊙ for some vector η

satisfying

η ξ ⊥

Conclusion: the only

• nly
• nly
• nly directions that are not controlled

by the principal symbols are

P ξ η = ⊙

n

R

η ξ ⊥

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The second term of

ij

dH

Goal: invert

m j i ij

dH dH

≤ ≤

=

, 1

} {

microlocally from

) , ( | ) , (

1

ξ ξ x M x M

ij ij −

+

• case

γ is constant

) , ( | 1

~

ξ x M

ij −

y d d y x M x M e x dH

n n R

R ij ij y x i n ij

∫∫

× − − ⋅ −

+ = ξ γ ξ ξ π γ

ξ

) ( : )) , ( | ) , ( ( ) 2 ( ) , (

1 ) (

1

1[ (( )( 2 ) )

i j n j

A H V I V ξ ξ ξ ξ ξ

−

= − ⋅ − ⊗ + ⊗

1

) , ( | A x M A

ij

ξ

−

= (( )( 2 ) ]sym

j i n i

H V I V ξ ξ ξ ξ + ⋅ − ⊗ + ⊗

2 i i

H A u A = ∇ where Lemma Suppose

γ is constant, pick

i i

x u = n i ≤ ≤ 1

and add an additional solution denoted by

1 + n

u

with full-rank Hessian

1 2 +

∇

n

u

If there exist

, ( )

n

P Q S R ∈

such that for all

1 , i j n ≤ ≤

~ ~ 1

( | )( , ) :( 1 )

ij ij

M M x P Q ξ

−

+ + − =

~ ~ , 1 1 , 1

( | )( , ):( 1 )

i n i n

M M x P Q ξ

+ − + +

+ − = then

P Q = =

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• case 0

γ is not constant

Basic Hypotheses

• 1

( , )

n

u u ∇ ∇ ⋯

form a frame

• )

( = ∇ ⋅ ∇ −

known i

u γ

• known

i i

g u =

Ω ∂

|

• Add one addition such that

1 n

u +

1 n j j n j

u u µ

+

∇ +∇ =

∑

we construct the matrix

1

[ , , ]

n

Z µ µ = ∇ ∇ ⋯

assume Z to be invertible Remark: in the case

γ

constant, the above hypothesis is automatically satisfied, by choosing

i i

u x = 1 i n ≤ ≤

1

1 2

t n

u x Qx

+ =

Q

invertible

Goal: under the above hypotheses, can control all bad directions

ξ η ⊙

~ 1

|

ij

M

−

Recall

y d d y x M x M e x dH

n n R

R ij ij y x i n ij

∫∫

× − − ⋅ −

+ = ξ γ ξ ξ π γ

ξ

) ( : )) , ( | ) , ( ( ) 2 ( ) , (

1 ) (

) ( = Q tr

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12

Lemma Assume that

1

[ , , ]

n

Z µ µ = ∇ ∇ ⋯

is invertible, if

~ 1

| :

nc ij

M ξ η

−

= ⊙

and

~ , 1 1

| :

nc i n

M ξ η

+ −

= ⊙

for any

ξ η ⊥

1 , i j n ≤ ≤

and

y d d y x M x M e x dH

n n R

R ij ij y x i n ij

∫∫

× − − ⋅ −

+ = ξ γ ξ ξ π γ

ξ

) ( : )) , ( | ) , ( ( ) 2 ( ) , (

1 ) (

• i

j j i j i known ij

v u v u u u dH ∇ ⋅ ∇ + ∇ ⋅ ∇ + ∇ ⋅ ∇ = γ γ γ )) ( (

1 i i

u L v ∇ ⋅ ∇ =

−

γ

Recall:

= η

then Theorem Suppose that we have measurements

1 ,

{ }

ij i j n

dH

≤ ≤ and , 1 1

{ }

i n i n

dH

+ ≤ ≤

such that

1

[ , , ]

n

Z µ µ = ∇ ∇ ⋯

is invertible and

1

( , )

n

u u ∇ ∇ ⋯

form a frame. we denote the operator as follows

dH

, 1 1 ,

: ( , , )

ij i n i j n

dH dH dH

+ ≤ ≤

= ⋯

Then the operator

( 1) ( 1) 2 2 2 2

:{ ( ')} { ( ')}

n n n n n

G dH L L

+ + +

Ω ⇒ Ω

• is semi-Fredholm, where

G

is the restriction operator to

2(

') L Ω

Proof: construct a parametrix Q

Q dH Id −

• such that is of order -1

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3.Explicit reconstruction

• )

( = ∇ ⋅ ∇ −

known i

u γ

), (Ω

• known

i i

g u =

Ω ∂

|

), (Ω | =

Ω ∂ i

v

• )

( ) (

i unknown i

u v ∇ ⋅ ∇ = ∇ ⋅ ∇ − γ γ

• i

j j i j i known ij

v u v u u u dH ∇ ⋅ ∇ + ∇ ⋅ ∇ + ∇ ⋅ ∇ = γ γ γ

{

Let us now add another solution of (＊) (＊) u with corresponding v at order ε

) ( '

2

ε ο ε + + = v u u

u ∇ may be expressed in the basis

1, n

u u ∇ ∇ ⋯

n j j j

u u µ ∇ +∇ =

∑

Suppose

1

[ ] :[ | | ]

n

U u u ∇ = ∇ ∇ ⋯

and

1

: [ | | ]

n

Z Z Z = ⋯

be invertible, where

i i

Z µ = ∇

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In the end, we obtain the strongly coupled elliptic system where With 0

: ( ) L γ =−∇⋅ ∇ define

1 2 1

: ( ) L f L u H

−

∈ ∈ Ω ↦

where u solves

( ) u f γ −∇⋅ ∇ = | u ∂Ω=

By Lax-Milgram theorem, and by Rellich imbedding,

1 2 1

: ( ) ( ) L L H

−

Ω Ω ↦

Applying

1

L−

to the elliptic system yields

1

( ) ,

n i ij j i j

v W v f γ

=

−∇ ⋅ ∇ + ⋅∇ =

∑

|

i

v

∂Ω=

1 i n ≤ ≤

ij

W depends on γ

and

) , , ( 1

n

u u ⋯

i

f

depends on

pq

dH

and their derivatives is compact.

1 1 1

( ) :

n i ij j i i j

v L W v h L f

− − =

+ ⋅∇ = =

∑

1 1 1

: ( ) : ( ) ( )

ij ij ij

P H v Pv L W v H

−

Ω ∋ → = ⋅∇ ∈ Ω compact if

ij

W

bounded.

1

( , , ),

n

v v v = ⋯

1

( , , ),

n

h h h = ⋯

( ) I P v h + =

where

P

compact (＊) (＊) satisfies a Fredholm alternative, if -1 is not an eigenvalue of

1 1

( ) ( ) i H H

v C dH

Ω Ω

≤

P

we obtain and and we obtain the stability:

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2 1

( ) ( ) L H

C dH γ

Ω Ω

≤

• i

j j i j i known ij

v u v u u u dH ∇ ⋅ ∇ + ∇ ⋅ ∇ + ∇ ⋅ ∇ = γ γ γ

( ( ) 2 )

pq ij ij ip q j i j

dH H u H u H v u γ γ γ = ∇ ⊗ ∇ − ∇ ∇ ⊙

• nce

are reconstructed

) , ( 1

n

v v ⋯

the relations can be inverted for γ and we get the stability

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16

• case

γ is constant

• )

( ) (

i unknown i

u v ∇ ⋅ ∇ = ∇ ⋅ ∇ − γ γ ), (Ω

|

=

Ω ∂ i

v

• )

( = ∇ ⋅ ∇ −

known i

u γ

), (Ω

• known

i i

g u =

Ω ∂

|

• i

j j i j i known ij

v u v u u u dH ∇ ⋅ ∇ + ∇ ⋅ ∇ + ∇ ⋅ ∇ = γ γ γ

Taking

Id = γ

using harmonic polynomials as solutions of (＊) (＊)

i i

u x = 1 i n ≤ ≤

' 2 1

1 2

n p p p

u a x

=

= ∑

p

a ≠

{

Then we get n additional measurements

'

: ' ' '

i i i i

dH u u u v v u γ = ∇ ∇ +∇ ∇ +∇ ⋅∇ ⊙ ⊙

Together with

i j j i j i ij

v u v u u u dH ∇ ⋅ ∇ + ∇ ⋅ ∇ + ∇ ⋅ ∇ = γ 1 i n ≤ ≤ 1 , i j n ≤ ≤

=

∑

p p

a

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For this choice of , the elliptic system is decoupled, hence solvable.

'

: ' ' '

i i i i

dH u u u v v u γ = ∇ ∇ +∇ ∇ +∇ ⋅∇ ⊙ ⊙

i j j i j i ij

v u v u u u dH ∇ ⋅ ∇ + ∇ ⋅ ∇ + ∇ ⋅ ∇ = γ

1 i n ≤ ≤

1 , i j n ≤ ≤

{

n n n + + ) 1 ( 2 1

' 1 1 1

1 1 ( )

n n n i j j p p ip j i i p pp j p p i i

v a x dH dH a dH a a

= = =

∆ = ∂ −∂ +∂ + ∂

∑ ∑ ∑

ij ij j i i j

dH v v γ = −∂ −∂

we obtain a standard estimate for

i

v

1 1

( ) ( ) i H H

v C dH

Ω Ω

≤

notice that the RHS is a linear combination

'

,

ij i

dH dH and their first derivatives

1 2

( ) ( ) ij H L

C dH γ

Ω Ω ≤

⇒

we lose one derivative

• )

( '

2

ε ο εγ γ γ + + =

known

) (

2 '

ε ο ε + + =

i i i

v u u

• f

stability: then

) , , ( 1

n

u u ⋯