The wave equation in elastodynamic Wave propagation in a non-homogeneous anisotropic elastic medium occupying a bounded domain R d, d = 2, 3, with boundary Γ, is described by the linear wave equation: ρ 2 v τ 2 = f, in (, T ), (1) τ = C ɛ, (2) v = v, v =, (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, t is the time variable, T is a final time, and f(x, t) R d, is a given source function. 1
Further, ɛ is the strain tensor with components coupled to τ by Hooke s law ɛ ij = ɛ ij (v) = 1 2 ( v i x j + v j x i ) (4) τ ij = d k=1 d C ijkl ɛ kl, (5) l=1 where C is a cyclic symmetric tensor, satisfying C ijkl = C klij = C jkli. (6) If the constants C ijkl (x) do not depend on x, the material of the body is said to be homogeneous. If the constants C ijkl (x) do not depend on the choice of the coordinate system, the material of the body is said to be isotropic at the point x. Otherwise, the material is anisotropic at the point x. 2
In the isotropic case C can be written as C ijkl = λδ ij δ kl + µ(δ ij δ kl + δ il δ jk ), (7) where δ ij is Kronecker symbol, in which case (5) takes the form of Hooke s law d τ ij = λδ ij ɛ kk + 2µɛ ij, (8) k=1 where λ and µ are the Lame s coefficients, depending on x, given by µ = E 2(1 + ν), λ = Eν (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 >, µ > E >, < ν < 1/2. (1) 3
4
ρ 2 v 1 2 ρ 2 v 2 2 ρ 2 v 3 2 (( ) v 1 λ + 2µ + λ v 2 + λ v 3 ) x 1 x 1 x 2 x 3 ( ( v 1 µ + v 2 )) x 2 x 2 x 1 ( ( v 1 µ + v 3 )) = f1, x 3 x 3 x 1 ( v 2 (λ + 2µ) + λ v 1 + λ v 3 ) x 2 x 2 x 1 x 3 ( ( v 1 µ + v 2 )) x 1 x 2 x 1 ( ( v 2 µ + v 3 )) = f2, x 3 x 3 x 2 ( v 3 (λ + 2µ) + λ v 2 + λ v 1 ) x 3 x 3 x 2 x 1 ( ( v 3 µ + v 2 )) x 2 x 2 x 3 ( ( v 1 µ + v 3 ) ) = f 3, x 1 x 3 5 x 1
or in more compact form ρ 2 v (µ v) ((λ + µ) v) = f. (11) 2 6
The inverse scattering problem We consider the elastics system in a non-homogeneous isotropic medium in a bounded domain R d, d = 2, 3, with boundary Γ: ρ 2 v (µ v) ((λ + µ) v) = f in (, T ), (12) 2 v(x, t) v(x, ) = v (t), =, v Γ =. (13) where v = (v 1, v 2, v 3 ) T, v = ( v 1, v 2, v 3 ) 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 L 2 (); < ρ min < ρ(x) < ρ max, and minimizes the quantity E(v, ρ, λ, µ) = 1 T 2 < λ min < λ(x) < λ max, < µ min < µ(x) < µ max } (14) (v ṽ) 2 δ obs dxdt + 1 2 γ 1 + 1 2 γ 2 µ 2 dx + 1 2 γ 3 ρ 2 dx λ 2 dx, (15) where ṽ is observed data at a finite set of observation points x obs, v satisfies (12) and thus depends on ρ, µ, λ, δ obs = δ(x obs ) is a sum of delta-functions corresponding to the observation points, and γ i, 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 ( α v ρ + 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 (ᾱ, v, ρ, λ, µ) L (α, v, ρ, µ, λ)(ᾱ, v, ρ, µ, λ) =, (16) where L is the gradient of L and we assume that α(, T ) = ᾱ(, T ) =, α = and v(, ) = v(, ) =. 9
= L (α, v, ρ, µ, λ)(ᾱ) = α = L v T ( ρ ᾱ v + µ ᾱ v (17) + (λ + µ) v ᾱ fᾱ ) dxdt, T (α, v, ρ, µ, λ)( v) = (v ṽ) v δ obs dxdt (18) + T ρ α v + µ α v + (λ + µ) v α dxdt, 1
= L (α, v, ρ, µ, λ)( ρ) = ρ = L (α, v, ρ, µ, λ)( µ) = µ = L λ (α, v, ρ, µ, λ)( λ) = T + γ 1 T + γ 2 T + γ 3 α(x, t) ρ ρ dx, x. v(x, t) ρ dxdt (19) ( α v + v α) µ dxdt µ µ dx, x. v α λ dxdt (2) λ λ dx, x. 11
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 2 α 2 (µ α) ((λ + µ) α) = (v ṽ)δ obs, x, < t < T, α(t ) = α(t ) =, (21) α = on Γ (, T ), and (19) - (2) 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 initial guess ρ, µ, λ and computing a sequence ρ n, µ n, λ n in the following steps: 1. Compute the solution v n = (v n 1, v n 2, v n 3 ) of the forward problem (12-13) with ρ = ρ n, µ = µ n, λ = λ n. 2. Compute the solution α n = (α n 1, α n 2, α n 3 ) of the adjoint problem (22). 3. Update the ρ, µ, λ according to ρ n+1 (x) = ρ n (x) β n( T µ n+1 (x) = µ n (x) β n( T λ n+1 (x) = λ n (x) β n( T α n (x, t) v n (x, t) dt + γ 1 ρ n (x) ), α n v n + v n α n + γ 2 µ n (x) ), v n α n + γ 3 λ n (x) ), 13
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 (, T ) in the usual way denoting by K h = {K} a partition of the domain into elements K (triangles in R 2 and tetrahedra in R 3 with h = h(x) being a mesh function representing the local diameter of the elements), and we let J k = {J} be a partition of the time interval I = (, T ) into time intervals J = (t k 1, t k ] of uniform length τ = t k t k 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 V h, W v h and W α h defined by : V h := {v L 2 () : v P (K), K K h }, W v := {v [ H 1 ( I) ] 3 : v(, ) =, v Γ = }, W α := {α [ H 1 ( I) ] 3 : α(, T ) = α Γ = }, W v h := {v W v : v K J [P 1 (K) P 1 (J)] 3, K K h, J J k }, W α h := {v W α : v K J [P 1 (K) P 1 (J)] 3, K K h, J J k }, where P 1 (K) and P 1 (J) are the set of linear functions on K and J, respectively. 15
The finite element method now reads: Find ρ h V h, µ h V h, λ h V h, α h Wh α, v h Wh v, such that L (α h, v h, ρ h, µ h, λ h )(ᾱ, v, ρ, µ, λ) = ρ V h, µ V h, λ V h, ᾱ W α h, v W v h. 16
Fully discrete scheme Expanding v, α in terms of the standard continuous piecewise linear functions ϕ i (x) in space and ψ i (t) in time and substituting this into (17-18), the following system of linear equations is obtained: M(v k+1 2v k + v k 1 ) = τ 2 τ 2 M(α k+1 2α k + α k 1 ) = τ 2 τ 2 ρ F k τ 2 ρ µk(1 6 vk 1 + 2 3 vk + 1 6 vk+1 ) ρ (λ + µ)dvk, (22) ρ Sk τ 2 ρ µk(1 6 αk 1 + 2 3 αk + 1 6 αk+1 ) ρ (λ + µ)dαk, (23) 17
with initial conditions : v() =, v(), (24) α(t ) =, α(t ). (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, S k 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. 18
The explicit formulas for the entries in system (22-23) at the element level can be given as: M e i,j = (ϕ i, ϕ j ) e, (26) K e i,j = ( ϕ i, ϕ j ) e, (27) D e i,j = ( ϕ i, ϕ j ) e, (28) F e j,m = (f, ϕ j ψ m ) e J, (29) S e j,m = (v ṽ, ϕ j ψ m ) e J, (3) where (.,.) e denotes the L 2 (e) scalar product. The matrix M e is the contribution from element e to the global assembled matrix in space M, K e is the contribution from element e to the global assembled matrix K,D e is the contribution from element e to the global assembled matrix D, F e and S e 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 M L, the diagonal approximation obtained by taking the row sum of M. By multiplying (22) - (23) with (M L ) 1 and replacing the terms 1 6 vk 1 + 2 3 vk + 1 6 vk+1 and 1 6 αk 1 + 2 3 αk + 1 6 αk+1 by v k and α k, respectively, we obtain an efficient explicit formulation: v k+1 = τ 2 τ 2 α k 1 = τ 2 τ 2 ρ (M L ) 1 F k + 2v k τ 2 ρ (λ + µ)(m L ) 1 Dv k v k 1, ρ (M L ) 1 S k + 2α k τ 2 ρ (λ + µ)(m L ) 1 Dα k α k+1. ρ µ(m L ) 1 Kv k (31) ρ µ(m L ) 1 Kα k (32) 2
The discrete version of gradients takes the form: = L (α, v, ρ(x), µ(x), λ(x))( ρ) (33) ρ T α h v h = ρ dxdt + γ 1 ρ h ρ dx, ρ V h. = L (α, v, ρ(x), µ(x), λ(x))( µ) (34) µ T = ( α h v h + v h α h ) µ dxdt + γ 2 µ h µ dx, x. = L λ (α, v, ρ(x), µ(x), λ(x))( λ) (35) T = v h α h λ dxdt + γ3 λ h λ dx, x. 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(α h, v h, ρ h, µ h, λ h ) = 1 2 L (α h, v h, ρ h, µ h, λ h )((α, v, ρ, µ, λ) (α h, v h, ρ h, µ h, λ h )) + R = 1 2 L (α h, v h, ρ h, µ h, λ h ))(α α h, v v h, ρ ρ h, µ µ h, λ λ h ) + R, where R denotes (a small) second order term. Using the Galerkin orthogonality and the splitting α α h = (α αh I ) + (αi h α h), v v h = (v vh I ) + (vi h v h), ρ ρ h = (ρ ρ I h ) + (ρi h ρ h), µ µ h = (µ µ I h ) + (µi h µ h), λ λ h = (λ λ I h ) + (λi h λ h), where (αh I, vi h, ρi h, µi h, λi h ) denotes an interpolant of (α, v, ρ, µ, λ) Wh α W h v V h V h V h, and neglecting the term R, 22
we get: e 1 2 L (α h, v h, ρ h, µ h, λ h ))(α α h, v v h, ρ ρ h, µ µ h, λ λ h ) = 1 2 (I 1 + I 2 + I 3 + I 4 + I 5 ), (36) 23
where I 1 = T ρ h (α α I h ) v h + µ h (α α I h) v h + (λ h + µ h )( v h (α αh)) I f(α αh)) I dxdt, T I 2 = (v h ṽ)(v vh) I δ obs dxdt + T T ρ h α h (v v I h ) + (λ h + µ h ) (v vh) I α h dxdt, T α h (x, t) I 3 = I 4 = I 5 = T + µ h α h (v v I h) v h (x, t) (ρ ρ I h) dxdt + ( α h v h + v h α h ) (µ µ I h) dxdt + ( v h α h )(λ λ I h) dxdt + ρ h (ρ ρ I h) dx, λ h (λ λ I h) dx. µ h (µ µ I h) dx, 24
Defining the residuals R v1 = f, R v2 = µ h max 2 S K h 1 [ ] k s v h, R v3 = ρ h 2 τ 1 [ ] v ht, R α1 = v h ṽ, R α2 = µ h max 2 S K h 1 [ ] k s α h, R α3 = ρ h 2 τ 1 [ ] α ht, R ρ1 = α h v h, R ρ2 = ρ h, R µ1 = α h v h + α h v h, R µ2 = µ h, R λ1 = α h v h, R λ2 = λ h, (37) 25
and interpolation errors in the form [ ] [ ] σ α = Cτ αh + Ch αh, (38) n [ ] [ ] σ v = Cτ vh + Ch vh, (39) n σ ρ = C [ρ h ], σ µ = C [µ h ], σ λ = C [λ h ] (4) 26
we obtain the following a posteriori estimate e 1 ( T T R v1 σ α dxdt + R v2 σ α dxdt + 2 T T + R α1 σ v dxdt + R α2 σ v dxdt + + + T R ρ1 σ ρ dxdt + R µ2 σ µ dx + T R ρ2 σ ρ dx + R λ1 σ λ dxdt + T T T R µ1 σ µ R λ2 σ λ dx ). R v3 σ α R α3 σ v dxdt dxdt dxdt 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 K h and an initial time partition J k of the time interval (, T ). 2. Compute the solution v n = (v n 1, v n 2, v n 3 ) on K h and J k of the forward problem (12-13) with ρ = ρ n, µ = µ n, λ = λ n. 3. Compute the solution α n = (α n 1, α n 2, α n 3 ) of the adjoint problem ρ 2 α 2 µ α (λ+µ) ( α) = (v ṽ)δ obs, x, < t < T on K h and J k. 28
4. Update the ρ, µ, λ according to ρ n+1 (x) = ρ n (x) β n( T µ n+1 (x) = µ n (x) β n( T λ n+1 (x) = λ n (x) β n( T α n (x, t) v n (x, t) dt + γ 1 ρ n (x) ), α n v n + v n α n + γ 2 µ n (x) ), v n α n + γ 3 λ n (x) ). Make steps 1 4 as long the gradient quickly decreases. 5. Refine all elements, where (R ρ1 + R ρ2 )σ ρ + (R µ1 + R µ2 )σ µ + (R λ1 + R λ2 )σ λ > tol and construct a new mesh K h and a new time partition J k. Here tol is a tolerance chosen by the user. Return to 1. 29