The wave equation in elastodynamic - Chalmerslarisa/www/Aposter_elastic.pdf · The wave equation in...

29
The wave equation in elastodynamic Wave propagation in a non-homogeneous anisotropic elastic medium occupying a bounded domain Ω R d ,d =2, 3, with boundary Γ, is described by the linear wave equation: ρ 2 v ∂t 2 -5· τ = f, in Ω × (0,T ), (1) τ = C , (2) v = v 0 , ∂v ∂t =0, (3) where v (x, t) R d , is the displacement, τ is the stress tensor, ρ(x) is the density of the material depending on x Ω, t is the time variable, T is a final time, and f (x, t) R d , is a given source function. 1

Transcript of The wave equation in elastodynamic - Chalmerslarisa/www/Aposter_elastic.pdf · The wave equation in...

The wave equation in elastodynamic

Wave propagation in a non-homogeneous anisotropic elastic medium

occupying a bounded domain Ω ⊂ Rd, d = 2, 3, with boundary Γ, is

described by the linear wave equation:

ρ∂2v

∂t2−5 · τ = f, in Ω × (0, T ), (1)

τ = C ε, (2)

v = v0,∂v

∂t= 0, (3)

where v(x, t) ⊂ Rd, is the displacement, τ is the stress tensor, ρ(x) is the

density of the material depending on x ∈ Ω, t is the time variable, T is a

final time, and f(x, t) ⊂ Rd, is a given source function.

1

Further, ε is the strain tensor with components

εij = εij(v) =1

2

( ∂vi

∂xj

+∂vj

∂xi

)

(4)

coupled to τ by Hooke’s law

τij =d

k=1

d∑

l=1

Cijkl εkl, (5)

where C is a cyclic symmetric tensor, satisfying

Cijkl = Cklij = Cjkli. (6)

If the constants Cijkl(x) do not depend on x, the material of the body issaid to be homogeneous. If the constants Cijkl(x) do not depend on thechoice of the coordinate system, the material of the body is said to beisotropic at the point x. Otherwise, the material is anisotropic at the pointx.

2

In the isotropic case C can be written as

Cijkl = λδijδkl + µ(δijδkl + δilδjk), (7)

where δij is Kronecker symbol, in which case (5) takes the form of

Hooke’s law

τij = λδij

d∑

k=1

εkk + 2µεij , (8)

where λ and µ are the Lame’s coefficients, depending on x, given by

µ =E

2(1 + ν), λ =

(1 + ν)(1 − 2ν), (9)

where E is the modulus of elasticity (Young modulus) and ν is the

Poisson’s ratio of the elastic material. We have that

λ > 0, µ > 0 ⇐⇒ E > 0, 0 < ν < 1/2. (10)

3

4

ρ∂2v1∂t2

−∂

∂x1

((

λ+ 2µ) ∂v1∂x1

+ λ∂v2∂x2

+ λ∂v3∂x3

)

−∂

∂x2

(

µ(∂v1∂x2

+∂v2∂x1

))

−∂

∂x3

(

µ(∂v1∂x3

+∂v3∂x1

))

= f1,

ρ∂2v2∂t2

−∂

∂x2

(

(λ+ 2µ)∂v2∂x2

+ λ∂v1∂x1

+ λ∂v3∂x3

)

−∂

∂x1

(

µ(∂v1∂x2

+∂v2∂x1

))

−∂

∂x3

(

µ(∂v2∂x3

+∂v3∂x2

))

= f2,

ρ∂2v3∂t2

−∂

∂x3

(

(λ+ 2µ)∂v3∂x3

+ λ∂v2∂x2

+ λ∂v1∂x1

)

−∂

∂x2

(

µ(∂v3∂x2

+∂v2∂x3

))

−∂

∂x1

(

µ(∂v1∂x3

+∂v3∂x1

))

= f3,5

or in more compact form

ρ∂2v

∂t2−∇ · (µ∇v) −∇((λ+ µ)∇ · v) = f. (11)

6

The inverse scattering problem

We consider the elastics system in a non-homogeneous isotropic medium

in a bounded domain Ω ⊂ Rd, d = 2, 3, with boundary Γ:

ρ∂2v

∂t2−∇ · (µ∇v) −∇((λ+ µ)∇ · v) = f in Ω × (0, T ), (12)

v(x, 0) = v0(t),∂v(x, t)

∂t= 0, v|Γ = 0. (13)

where v = (v1, v2, v3)T ,∇v = (∇v1,∇v2,∇v3)

T , and ∇ · v is the

divergence of the vector field v.

7

The inverse problem for (12 - 13) can be formulated as follows: find a

controls ρ, µ, λ which belongs to the set of admissible controls

U = ρ, µ, λ ∈ L2(Ω); 0 < ρmin < ρ(x) < ρmax,

0 < λmin < λ(x) < λmax,

0 < µmin < µ(x) < µmax (14)

and minimizes the quantity

E(v, ρ, λ, µ) =1

2

∫ T

0

Ω

(v − v)2δobs dxdt+1

2γ1

Ω

ρ2 dx

+1

2γ2

Ω

µ2 dx+1

2γ3

Ω

λ2 dx, (15)

where v is observed data at a finite set of observation points xobs, vsatisfies (12) and thus depends on ρ, µ, λ, δobs =

δ(xobs) is a sum of

delta-functions corresponding to the observation points, and

γi, i = 1, 2, 3 are a regularization parameters.

8

To approach this minimization problem, we introduce the Lagrangian

L(α, v, ρ, µ, λ) = E(v, ρ, λ, µ) +∫ T

0

Ω

(

− ρ∂α

∂t

∂v

∂t+ µ∇α∇v + (λ+ µ)∇ · v ∇ · α− fα

)

dxdt,

and search for a stationary point with respect to (α, v, ρ, µ, λ) satisfying

for all (α, v, ρ, λ, µ)

L′(α, v, ρ, µ, λ)(α, v, ρ, µ, λ) = 0, (16)

where L′ is the gradient of L and we assume that

α(·, T ) = α(·, T ) = 0, α = 0 and v(·, 0) = v(·, 0) = 0.

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0 =∂L

∂α(α, v, ρ, µ, λ)(α) =

∫ T

0

Ω

(

− ρ∂α

∂t

∂v

∂t+ µ∇α∇v (17)

+ (λ+ µ)∇ · v ∇ · α− fα)

dxdt,

0 =∂L

∂v(α, v, ρ, µ, λ)(v) =

∫ T

0

Ω

(v − v) v δobs dxdt (18)

+

∫ T

0

Ω

−ρ∂α

∂t

∂v

∂t+ µ∇α∇v

+ (λ+ µ)∇ · v ∇ · α dxdt,

10

0 =∂L

∂ρ(α, v, ρ, µ, λ)(ρ) =

∫ T

0

Ω

∂α(x, t)

∂t

∂v(x, t)

∂tρ dxdt (19)

+ γ1

Ω

ρρ dx, x ∈ Ω.

0 =∂L

∂µ(α, v, ρ, µ, λ)(µ) =

∫ T

0

Ω

(∇α∇v + ∇ · v ∇ · α)µ dxdt

+ γ2

Ω

µµ dx, x ∈ Ω.

0 =∂L

∂λ(α, v, ρ, µ, λ)(λ) =

∫ T

0

Ω

∇ · v ∇ · α λ dxdt (20)

+ γ3

Ω

λλ dx, x ∈ Ω.

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The equation (17) is a weak form of the state equation (12 - 13), the

equation (18) is a weak form of the adjoint state equation

ρ∂2α

∂t2−∇ · (µ∇α) −∇((λ+ µ)∇ · α) = −(v − v)δobs, x ∈ Ω, 0 < t < T,

α(T ) =∂α(T )

∂t= 0, (21)

α = 0 on Γ × (0, T ),

and (19) - (20) expresses stationarity with respect to ρ(x), µ(x), λ(x).

12

To solve the minimization problem we shall use a discrete form of the

following steepest descent or gradient method starting from an initialguess ρ0, µ0, λ0 and computing a sequence ρn, µn, λn in the followingsteps:

1. Compute the solution vn = (vn1 , v

n2 , v

n3 ) of the forward problem (12 -

13) with ρ = ρn, µ = µn, λ = λn .

2. Compute the solution αn = (αn1 , α

n2 , α

n3 ) of the adjoint problem (22).

3. Update the ρ, µ, λ according to

ρn+1(x) = ρn(x) − βn(

∫ T

0

∂αn(x, t)

∂t

∂vn(x, t)

∂tdt+ γ1ρ

n(x))

,

µn+1(x) = µn(x) − βn(

∫ T

0

∇αn∇vn + ∇ · vn∇ · αn + γ2µn(x)

)

,

λn+1(x) = λn(x) − βn(

∫ T

0

∇ · vn∇ · αn + γ3λn(x)

)

,

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Finite element discretization.

We now formulate a finite element method for (16) based on using

continuous piecewise linear functions in space and time. We discretize

Ω × (0, T ) in the usual way denoting by Kh = K a partition of the

domain Ω into elements K (triangles in R2 and tetrahedra in R3 with

h = h(x) being a mesh function representing the local diameter of the

elements), and we let Jk = J be a partition of the time interval

I = (0, T ) into time intervals J = (tk−1, tk] of uniform length

τ = tk − tk−1. In fully discrete form the resulting method corresponds to

a centered finite difference approximation for the second order time

derivative and a usual finite element approximation of the Laplacian.

To formulate the finite element method for (16) we introduce the finite

14

element spaces Vh, W vh and Wα

h defined by :

Vh := v ∈ L2(Ω) : v ∈ P0(K), ∀K ∈ Kh,

W v := v ∈[

H1(Ω × I)]3

: v(·, 0) = 0, v|Γ = 0,

Wα := α ∈[

H1(Ω × I)]3

: α(·, T ) = α|Γ = 0,

W vh := v ∈W v : v|K×J ∈ [P1(K) × P1(J)]3, ∀K ∈ Kh, ∀J ∈ Jk,

Wαh := v ∈Wα : v|K×J ∈ [P1(K) × P1(J)]3, ∀K ∈ Kh, ∀J ∈ Jk,

where P1(K) and P1(J) are the set of linear functions on K and J,

respectively.

15

The finite element method now reads: Find

ρh ∈ Vh, µh ∈ Vh, λh ∈ Vh, αh ∈Wαh , vh ∈W v

h , such that

L′(αh, vh, ρh, µh, λh)(α, v, ρ, µ, λ) = 0

∀ρ ∈ Vh, µ ∈ Vh, λ ∈ Vh, α ∈Wαh , v ∈W v

h .

16

Fully discrete scheme

Expanding v, α in terms of the standard continuous piecewise linear

functions ϕi(x) in space and ψi(t) in time and substituting this into (17 -

18), the following system of linear equations is obtained:

M(vk+1 − 2vk + vk−1) =

τ2

ρF k −

τ2

ρµK(

1

6v

k−1 +2

3v

k +1

6v

k+1)

−τ2

ρ(λ+ µ)Dv

k, (22)

M(αk+1 − 2αk + αk−1) = −

τ2

ρSk −

τ2

ρµK(

1

k−1 +2

k +1

k+1)

−τ2

ρ(λ+ µ)Dα

k, (23)

17

with initial conditions :

v(0) = 0, v(0) ≈ 0, (24)

α(T ) = 0, α(T ) ≈ 0. (25)

Here, M is the mass matrix in space, K is the stiffness matrix, D is the

divergence matrice, k = 1, 2, 3 . . . denotes the time level, F k, Sk are the

load vectors, v is the unknown discrete field values of v, α is the

unknown discrete field values of α and τ is the time step.

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The explicit formulas for the entries in system (22 - 23) at the element

level can be given as:

Mei,j = (ϕi, ϕj)e, (26)

Kei,j = (5ϕi,5ϕj)e, (27)

Dei,j = (5 · ϕi,5 · ϕj)e, (28)

F ej,m = (f, ϕjψm)e×J , (29)

Sej,m = (v − v, ϕjψm)e×J , (30)

where (., .)e denotes the L2(e) scalar product. The matrix Me is the

contribution from element e to the global assembled matrix in space M ,

Ke is the contribution from element e to the global assembled matrix

K,De is the contribution from element e to the global assembled matrix

D, F e and Se are the contributions from element e to the assembled

source vectors F and vector of the right hand side of (22),

correspondingly .

19

To obtain an explicit scheme we approximate M with the lumped mass

matrix ML, the diagonal approximation obtained by taking the row sum

of M . By multiplying (22) - (23) with (ML)−1 and replacing the terms1

6v

k−1 + 2

3v

k + 1

6v

k+1 and 1

k−1 + 2

k + 1

k+1 by vk and α

k,

respectively, we obtain an efficient explicit formulation:

vk+1 =

τ2

ρ(ML)−1F k + 2vk −

τ2

ρµ(ML)−1Kv

k (31)

−τ2

ρ(λ+ µ)(ML)−1Dv

k − vk−1,

αk−1 = −

τ2

ρ(ML)−1Sk + 2αk −

τ2

ρµ(ML)−1Kα

k (32)

−τ2

ρ(λ+ µ)(ML)−1Dα

k − αk+1.

20

The discrete version of gradients takes the form:

0 =∂L

∂ρ(α, v, ρ(x), µ(x), λ(x))(ρ) (33)

=

∫ T

0

Ω

∂αh

∂t

∂vh

∂tρ dxdt+ γ1

Ω

ρhρ dx, ∀ρ ∈ Vh.

0 =∂L

∂µ(α, v, ρ(x), µ(x), λ(x))(µ) (34)

=

∫ T

0

Ω

(∇αh∇vh + ∇ · vh ∇ · αh)µ dxdt

+ γ2

Ω

µhµ dx, x ∈ Ω.

0 =∂L

∂λ(α, v, ρ(x), µ(x), λ(x))(λ) (35)

=

∫ T

0

Ω

∇ · vh ∇ · αh λ dxdt+ γ3

Ω

λhλ dx, x ∈ Ω.

21

An a posteriori error estimate for the Lagrangian

We start by writing an equation for the error e in the Lagrangian as

e = L(α, v, ρ, µ, λ) − L(αh, vh, ρh, µh, λh)

=1

2L′(αh, vh, ρh, µh, λh)((α, v, ρ, µ, λ) − (αh, vh, ρh, µh, λh)) +R

=1

2L′(αh, vh, ρh, µh, λh))(α− αh, v − vh, ρ− ρh, µ− µh, λ− λh) +R,

where R denotes (a small) second order term. Using the Galerkin

orthogonality and the splitting

α− αh = (α− αIh) + (αI

h − αh), v − vh =

(v − vIh) + (vI

h − vh), ρ− ρh = (ρ− ρIh) + (ρI

h − ρh), µ− µh =

(µ− µIh) + (µI

h − µh), λ− λh = (λ− λIh) + (λI

h − λh), where

(αIh, v

Ih, ρ

Ih, µ

Ih, λ

Ih) denotes an interpolant of

(α, v, ρ, µ, λ) ∈Wαh ×W v

h × Vh × Vh × Vh, and neglecting the term R,

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we get:

e ≈1

2L′(αh, vh, ρh, µh, λh))(α− αh, v − vh, ρ− ρh, µ− µh, λ− λh)

=1

2(I1 + I2 + I3 + I4 + I5), (36)

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where

I1 =

∫ T

0

Ω

−ρh

∂(α− αIh)

∂t

∂vh

∂t+ µh∇(α− αI

h)∇vh

+ (λh + µh)(∇ · vh∇ · (α− αIh)) − f(α− αI

h)) dxdt,

I2 =

∫ T

0

Ω

(vh − v)(v − vIh) δobs dxdt

+

∫ T

0

Ω

−ρh

∂αh

∂t

∂(v − vIh)

∂t+ µh∇αh∇(v − vI

h)

+ (λh + µh)∇ · (v − vIh)∇ · αh dxdt,

I3 = −

∫ T

0

Ω

∂αh(x, t)

∂t

∂vh(x, t)

∂t(ρ− ρI

h) dxdt+

Ω

ρh (ρ− ρIh) dx,

I4 =

∫ T

0

Ω

(∇αh∇vh + ∇ · vh∇ · αh) (µ− µIh) dxdt+

Ω

µh (µ− µIh) dx,

I5 =

∫ T

0

Ω

(∇ · vh∇ · αh)(λ− λIh) dxdt+

Ω

λh (λ− λIh) dx.

24

Defining the residuals

Rv1=

∣f∣

∣, Rv2=µh

2max

S⊂∂Kh−1

k

[

∂svh

]∣

∣, Rv3=ρh

2τ−1

[

∂vht

]∣

∣,

Rα1=

∣vh − v∣

∣, Rα2=µh

2max

S⊂∂Kh−1

k

[

∂sαh

]∣

∣, Rα3=ρh

2τ−1

[

∂αht

]∣

∣,

Rρ1=

∂αh

∂t

∂vh

∂t

∣, Rρ2= |ρh|,

Rµ1= |∇αh∇vh + ∇ · αh∇ · vh

∣, Rµ2= |µh|,

Rλ1= |∇ · αh∇ · vh|, Rλ2

= |λh|,

(37)

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and interpolation errors in the form

σα = Cτ

[

∂αh

∂t

]∣

+ Ch

[

∂αh

∂n

]∣

, (38)

σv = Cτ

[

∂vh

∂t

]∣

+ Ch

[

∂vh

∂n

]∣

, (39)

σρ = C∣

∣[ρh]∣

∣, σµ = C∣

∣[µh]∣

∣, σλ = C∣

∣[λh]∣

∣ (40)

26

we obtain the following a posteriori estimate

∣e∣

∣ ≤1

2

(

∫ T

0

Ω

Rv1σα dxdt+

∫ T

0

Ω

Rv2σα dxdt+

∫ T

0

Ω

Rv3σα dxdt

+

∫ T

0

Ω

Rα1σv dxdt+

∫ T

0

Ω

Rα2σv dxdt+

∫ T

0

Ω

Rα3σv dxdt

+

∫ T

0

Ω

Rρ1σρ dxdt+

Ω

Rρ2σρ dx+

∫ T

0

Ω

Rµ1σµ dxdt

+

Ω

Rµ2σµ dx+

∫ T

0

Ω

Rλ1σλ dxdt+

Ω

Rλ2σλ dx

)

.

27

Adaptive algorithm

In the computations below we use the following variant of the gradient

method with adaptive mesh selection:

1. Choose an initial mesh Kh and an initial time partition Jk of the time

interval (0, T ).

2. Compute the solution vn = (vn1 , v

n2 , v

n3 ) on Kh and Jk of the

forward problem (12 - 13) with ρ = ρn, µ = µn, λ = λn .

3. Compute the solution αn = (αn1 , α

n2 , α

n3 ) of the adjoint problem

ρ∂2α

∂t2−µ4α−(λ+µ)∇(∇·α) = −(v− v)δobs, x ∈ Ω, 0 < t < T

on Kh and Jk .

28

4. Update the ρ, µ, λ according to

ρn+1(x) = ρn(x) − βn(

∫ T

0

∂αn(x, t)

∂t

∂vn(x, t)

∂tdt+ γ1ρ

n(x))

,

µn+1(x) = µn(x) − βn(

∫ T

0

∇αn∇vn + ∇ · vn∇ · αn + γ2µn(x)

)

,

λn+1(x) = λn(x) − βn(

∫ T

0

∇ · vn∇ · αn + γ3λn(x)

)

.

Make steps 1 − 4 as long the gradient quickly decreases.

5. Refine all elements, where

(Rρ1+Rρ2

)σρ + (Rµ1+Rµ2

)σµ + (Rλ1+Rλ2

)σλ > tol and

construct a new mesh Kh and a new time partition Jk. Here tol is a

tolerance chosen by the user. Return to 1.

29