Topological Derivatives in Shape Optimization

Transcription

1 Topological Derivatives in Shape Optimization Jan Sokołowski Institut Élie Cartan 28 mai 2012

2 Shape optimization Well posedness of state equation - nonlinear problems, compressible Navier-Stokes equations Stability of solutions to state equation with respect to the shape - nonlinear problems, compressible Navier-Stokes equations Formula for the shape gradient of shape functional - nonlinear problems, compressible Navier-Stokes equations Asymptotic analysis and the formula for the topological derivative of shape functional - anisotropic elasticity, compressible Navier-Stokes equations Numerical methods for the shape optimization problems - anisotropic elasticity, compressible Navier-Stokes equations

3 Compressible Navier-Stokes Equations weak renormalized global solutions of compressible Navier-Stokes equations : shape optimization problems are well posed strong local solutions of compressible Navier-Stokes equations : work and drag shape functionals are differentiable P.I. Plotnikov and J.S., Compressible Navier-Stokes Equations. Theory and Shape Optimization. Monografie Matematyczne, Vol. 73, Springer, Birkhauser Basel, 470 pages, 2013, to appear in 2012.

4 Anisotropic Elasticity asymptotic analysis of solutions to PDE s in singularly perturbed domains, eg., inclusions and cavities in anisotropic elasticity : topological derivatives of energy functionals combine the descriptions from applied mathematics and mechanics closed formulae in continuous case, easy to implement numerically

5 Asymptotic Analysis vs Shape Sensitivity asymptotic analysis is complementary tool for the shape sensitivity analysis : the way to evaluate the polarization tensors classical asymptotic analysis : the influence of boundary singularities on the energy functionals (Russian school) the regularity required for the asymptotic analysis shape sensitivity analysis leads to the same result on the boundary under weaker regularity assumptions, since the shape gradient is a distribution topological derivatives in the interior are obtained as the singular limits of shape derivatives conclusion : the topological derivative on the boundary of the energy functionals for elliptic system can be recovered from the shape derivative

6 REFERENCES J.S. and A. Zochowski : SICON, T. Lewinski and J.S. : IJSS, 2003 (mechanics) S.A. Nazarov and J.S. : JMPA, 2003 (mathematics) monograph : A.A. Novotny and J.S. to be published by Springer (mathematics and mechanics)

7 Topological derivative singular and regular domain perturbations : asymptotic analysis elementary examples relation between shape gradient and the topological derivative on the boundary energy for scalar equation energy for elasticity system multiscale elasticity

8 Topological derivative Let us consider a bounded domain Ω R d, with d 2, which is subject to a nonsmooth perturbation confined in a small region ω ε ( x) = x + εω of size ε, as shown in fig. 1. Here, x is an arbitrary point of Ω and ω is a fixed domain of R d. We introduce a characteristic function x χ(x), x R d, associated to the unperturbed domain, namely χ = 1 Ω, such that (1) Ω = χ, R d where Ω is the Lebesgue measure of Ω. Then, we define a characteristic function associated to the topologically perturbed domain of the form x χ ε ( x;x), x R d. In the case of a perforation, for instance, χ ε ( x) = 1 Ω 1 ωε ( x) and the perforated domain is obtained as Ω ε = Ω \ ω ε.

9 Topological derivative We assume that a given shape functional ψ(χ ε ( x)), associated to the topologically perturbed domain, admits the following topological asymptotic expansion (2) ψ(χ ε ( x)) = ψ(χ) + f (ε)t ( x) + o(f (ε)), where ψ(χ) is the shape functional associated to the original (unperturbed) domain, f (ε) is a positive function such that f (ε) 0, when ε 0. The function x T ( x) is called the topological derivative of ψ at x. Therefore, this derivative can be seen as a first order correction of ψ(χ) to approximate ψ(χ ε ( x)). In fact, after rearranging (2) we have (3) ψ(χ ε ( x)) ψ(χ) f (ε) = T ( x) + o(f (ε)) f (ε). The limit passage ε 0 in the above expression leads to ψ(χ ε ( x)) ψ(χ) (4) T ( x) = lim. ε 0 f (ε)

10 Topological derivative Since we are dealing with singular perturbations associated to nucleation of holes, the shape functionals ψ(χ ε ( x)) and ψ(χ) are associated to topologically different domains. Therefore, the above limit is not trivial to be calculated. In particular, we need to perform an asymptotic analysis of the shape functional ψ(χ ε ( x)) with respect to the small parameter ε. The notation for the shape functional we are going to introduce is the following : (5) ψ(χ ε ) := J Ωε (u ε ) and ψ(χ ε ) := J χε (u ε ) for singular and regular perturbations, respectively, where u ε is solution to the perturbed boundary values problems. FIGURE: The topological derivative concept.

11 Topological derivative In order to fix these ideas, we present simple examples. The first one concerns singular domain perturbation. The second example is associated to regular perturbation of the integrand.

12 Exemple 1 Example 1 Let us consider a shape functional of the form (6) ψ(χ ε ( x)) := g(x), Ω ε where the function g : R d R is assumed to be continuous, χ ε ( x) = 1 Ω 1 ωε ( x) and Ω ε = Ω \ ω ε. Since ω ε 0 as ε 0, we have (7) ψ(χ) := g(x). We are looking for an asymptotic expansion of the form (2), namely ψ(χ ε ( x)) = g(x) + g(x) g(x) Ω ε ω ε ω ε = g(x) g(x) Ω ω ε (8) = ψ(χ) ω ε g( x) + o( ω ε ). For the simple example, the term g( x) is called the topological derivative of the shape functional ψ, that is, T ( x) = g( x), and we recognize f (ε) = ω ε ε d. Ω

13 Example 2 Example 2 Let us now consider a shape functional of the form (9) ψ(χ ε ( x)) := g ε (x), Ω where χ ε ( x) = 1 Ω (1 γ)1 ωε ( x) and g ε = χ ε g is defined as { g(x) if x Ω \ ωε ( x) (10) g ε (x) := γg(x) if x ω ε ( x), with g standing by a continuous function of the form g : R d R. It means that the set Ω includes the interface ω ε. This is the case of shape functionals depending on the characteristic function of small sets. Since ω ε 0 as ε 0, we have (11) ψ(χ) := g(x). Ω

14 Example 2 We are looking for an asymptotic expansion of the form (2), namely ψ(χ ε ( x)) = g(x) + γg(x) Ω\ω ε ω ε = g(x) + γg(x) + g(x) g(x) Ω\ω ε ω ε ω ε ω ε = g(x) (1 γ) g(x) Ω ω ε (12) = ψ(χ) ω ε (1 γ)g( x) + o( ω ε ). For the simple example, the term (1 γ)g( x) is called the topological derivative of the shape functional ψ, that is, T ( x) = (1 γ)g( x), and we recognize f (ε) = ω ε ε d. In addition, we note that for the limit case γ 0, we have T ( x) = g( x). It means that the first example can be seen as the singular limit of this last one.

15 Relation Between Shape and Topological Changes The topological derivative can be considered as the singular limit of the shape derivative, so it is a generalization of the classical tool in shape optimization. In order to fix this idea, let us present an example concerning the relation between shape and topological derivatives.

16 Relation Between Shape and Topological Changes Example 3 Let as consider a bar under torsion effects. The cross section of the shaft is represented by an open bounded domain Ω R 2, as shown in fig. 2. Following the Prandtl s approach, we consider the complementary dissipation energy shape functional of the form (13) ψ(χ) = 1 u 2 bu, 2 Ω where the scalar field u is solution to the following boundary value problem Find u, such that (14) u = b in Ω, u = 0 on Ω. with the constant b used to denote a rigid twist of the cross-section of the shaft. The shape derivative of ψ(χ) is given by (15) dψ(χ;v) = 1 ( n u) 2 n V. 2 Ω where V is the shape change velocity field Ω

17 Shape and topological derivatives The topological asymptotic expansion for the above problem is given by (16) ψ(χ ε ( x)) = ψ(χ) + πε 2 u( x) 2 + o(ε 2 ) x Ω. where we recognize the topological derivative as T ( x) = u( x) 2. This last result can be extend to the boundary by using the odd extension of the problem, as shown in fig. 2. In addition, since u = 0 on Ω, then its tangential derivative vanishes on the boundary, namely τ u = 0 on Ω. Therefore, the topological asymptotic expansion on the boundary leads to (17) ψ(χ ε ( x)) = ψ(χ) πε2 ( n u( x)) 2 + o(ε 2 ) x Ω.

18 Shape and topological derivatives It means that in this simple example, the topological derivative T ( x) = 1 2 ( nu( x)) 2 coincides with the density of the shape derivative at the point x Ω. FIGURE: Cross section of a shaft under torsion effects : original and singulary perturbed domains.

19 Energy Change-Poisson problem It is sufficient to define V (18) { V = 0 on Ω V = n on B ε, where n = (x x)/ε, with x B ε, is the normal unit vector field pointing toward the center of the circular hole B ε. FIGURE: Topologically perturbed domain by the nucleation of a small circular hole.

20 Energy Change The shape functional in the unperturbed domain Ω is given by (19) ψ(χ) := J Ω (u) = 1 2 Ω u 2 bu + qu, Ω Γ N where the scalar function u is the solution to the variational problem : Find u U, such that (20) u η = bη qη η V, Γ N Ω In the above equation, b is a source-term assumed to be constant everywhere. The set U and the space V are respectively defined as Ω (21) (22) U := {ϕ H 1 (Ω) : ϕ ΓD = u}, V := {ϕ H 1 (Ω) : ϕ ΓD = 0}.

21 Energy Change In addition, Ω = Γ D Γ N with Γ D Γ N =, where Γ D and Γ N are Dirichlet and Neumann boundaries, respectively. Thus u is a Dirichlet data on Γ D and q is a Neumann data on Γ N, both functions are assumed to be smooth enough. FIGURE: The Poisson equation defined in the unperturbed domain Ω.

22 Energy Change The strong formulation associated to the variational problem (20) leads to the so-called Poisson equation, namely : Find u, such that u = b in Ω, (23) u = u on Γ D, n u = q on Γ N,

23 Energy Change Now, let us state the same problem in the perturbed domain Ω ε. In this case, the total potential energy reads (24) ψ(χ ε ) := J Ωε (u ε ) = 1 u ε 2 bu ε + qu ε, 2 Ω ε Ω ε Γ N where the scalar function u ε solves the variational problem : Find u ε U ε, such that (25) u ε η = bη qη η V ε. Ω ε Ω ε Γ N

24 Energy Change The set U ε and the space V ε must be defined according to the boundary condition on B ε. In particular, we have (26) (27) U ε := {ϕ H 1 (Ω ε ) : ϕ ΓD = u, βϕ Bε = 0}, V ε := {ϕ H 1 (Ω ε ) : ϕ ΓD = 0, βϕ Bε = 0}, with β {0,1}. This notation have to be interpreted as follows : When β = 1, we have homogeneous Dirichlet boundary condition on B ε, since u ε = 0 and η = 0 on B ε. When β = 0, u ε and η are free on the boundary of the hole B ε, then we have homogeneous Neumann boundary condition on B ε.

25 Energy Change FIGURE: The Poisson equation defined in the perturbed domain Ω ε.

26 Energy Change The strong formulation associated to the variational problem (25) reads Find u ε, such that u ε = b in Ω ε, (28) u ε = u on Γ D, n u ε = q on Γ N, βu ε + (1 β) n u ε = 0 on B ε.

27 Shape sensitivity analysis. Proposition Let J Ωε (u ε ) be the shape functional defined by (24). Then, the derivative of this functional with respect to the small parameter ε is given by (29) J Ωε (u ε ) = Σ ε V, Ω ε where V is the shape change velocity field defined through (18) and Σ ε is a generalization of the classical Eshelby energy-momentum tensor given by (30) Σ ε = 1 2 ( u ε 2 2bu ε )I u ε u ε.

28 Shape sensitivity analysis. Proof By making use of Reynolds transport theorem, the shape derivative (the material or total derivative with respect to the parameter ε) of the functional (24) is given by (31) J Ωε (u ε ) = 1 (( u ε 2 ) + u ε 2 divv) 2 Ω ε b( u ε + u ε divv) + q u ε, Ω ε Γ N since Γ N remains fixed according to (18). Next, by using the concept of material derivative of spatial fields we find that the first term of the above right-hand side integral can be written as ( u ε 2 ) = ( u ε u ε ) (32) = 2 u ε u ε 2 V u ε u ε = 2 u ε u ε 2( u ε u ε ) V,

29 Proof From these last results we obtain J Ωε (u ε ) = 1 (( u ε 2 2bu ε )I 2 u ε u ε ) V 2 Ω ε (33) + u ε u ε b u ε + q u ε. Ω ε Ω ε Γ N where we have made use of the identity divv = I V. Since u ε is a variation of u ε in the direction of the velocity field V, then u ε V ε. Finally, by taking u ε as test function in the variational problem (25), we have that the last three terms of the above equation vanish. Now, we can prove that the shape derivative of the functional J Ωε (u ε ) can be written in terms of quantities concentrated on the boundary Ω ε. In fact, the following result holds :

30 Boundary shape gradient Let J Ωε (u ε ) be the shape functional defined by (24). Then, the derivative of this functional with respect to the small parameter ε is given by (34) J Ωε (u ε ) = Σ ε n V, Ω ε where V is the shape change velocity field defined through (18) and tensor Σ ε is given by (30). PROOF We have (35) J Ωε (u ε ) = (36) Σ ε n V divσ ε V. Ω ε Ω ε divσ ε = 1 2 ( u ε u ε 2bu ε ) div( u ε u ε ) = ( u ε ) u ε b u ε u ε div u ε ( u ε ) u ε = ( u ε ) u ε ( u ε ) u ε ( u ε + b) u ε. Thus, since u ε = ( u ε ) and u ε = b, we observe that (37) divσ ε V = 0 V, Ω ε

31 Corollary We have (38) J Ωε (u ε ) = = Σ ε n V Ω ε Σ ε n V + B ε Ω Σ ε n V. Since we are dealing with an uniform expansion of the circular hole, then by taking into account the associated velocity field defined through (18), namely, V = n on B ε and V = 0 on Ω, we finally obtain (39) d dε ψ(χ ε) = J Ωε (u ε ) = Σ ε n n. B ε

32 How to evaluate topological derivative? Let ψ(χ ε ( x)) be the shape functional associated to the topologically perturbed domain, which admits, for ε small enough, the topological asymptotic expansion of the form (40) ψ(χ ε ( x)) = ψ(χ) + f (ε)t ( x) + R(f (ε)), where ψ(χ) is the shape functional associated to the original (unperturbed) domain, the positive function f (ε) is such that f (ε) 0, with ε 0, and the function T ( x) is the topological derivative of the shape functional ψ. We assume that the remainder R(f (ε)) = o(f (ε)) has the following additional property R (f (ε)) 0, when ε 0. Then, the topological derivative can be written as 1 (41) T ( x) = lim ε 0 f (ε) d dε ψ(χ ε( x)), where d dε ψ(χ ε( x)) is the derivative of ψ(χ ε ( x)) with respect to the small parameter ε.

33 Proof Let us calculate the total derivative of the expansion (40) with respect to the real parameter ε, that is (42) After division by f (ε) we have d dε ψ(χ ε( x)) = f (ε)t ( x) + R (f (ε))f (ε). (43) T ( x) = 1 d f (ε) dε ψ(χ ε( x)) + R (f (ε)). Finally, by taking the limit ε 0 in the above expression, the result holds since R (f (ε)) 0, when ε 0

34 Asymptotics of solutions. Neumann case For the Neumann case β = 0, let us start with the ansatz for an expansion of u ε in the form u ε (x) = u(x) + εw(ε 1 x) + ũ ε (x) = u( x) + u( x) (x x) + 1 u(y)(x x) (x x) 2 (44) + εw(ε 1 x) + ũ ε (x), where y is an intermediate point between x and x. On the boundary of the hole B ε we have n u ε Bε = 0. Thus, the normal derivative of the above expansion, evaluated on B ε, leads to (45) u( x) n ε u(y)n n + ε n w(ε 1 x) + n ũ ε (x) = 0.

35 Thus, we can choose w such that (46) ε n w(ε 1 x) = u( x) n on B ε. In the new variable ξ = ε 1 x, which implies ξ w(ξ ) = ε w(ε 1 x), the following exterior problem is considered, and formally obtained with ε 0 : Find w, such that (47) ξ w = 0 in R 2 \ B 1, w 0 at, ξ w n = u( x) n on B 1. The above boundary value problem admits an explicit solution, namely (48) w(ε 1 x) = ε u( x) (x x). x x 2

36 Now we can construct ũ ε in such a way that it compensates the discrepancies introduced by the higher-order terms in ε as well as by the boundary-layer w on the exterior boundary Ω. It means that the remainder ũ ε must be the solution to the following boundary value problem : (49) Find ũ ε, such that ũ ε = 0 in Ω ε, ũ ε = εw on Γ D, n ũ ε = ε n w on Γ N, n ũ ε = ε u(y)n n on B ε, where, under appropriate regularity assumption, clearly ũ ε = O(ε). However, this estimate can be improved, namely ũ ε = O(ε 2 ), and the expansion for u ε reads (50) u ε (x) = u(x) + ε2 x x 2 u( x) (x x) + O(ε2 ).

37 FIGURE: Polar coordinate system (r,θ) centered at the point x Ω.

38 Topological derivative evaluation From an orthonormal curvilinear coordinate system n and τ on the boundary B ε, the gradient u ε can be decomposed into its normal and tangential components, that is (51) u ε Bε = ( n u ε )n + ( τ u ε )τ. Therefore, on the boundary of the hole B ε we observe that Σ ε n n Bε = 1 2 ( u ε 2 2bu ε )In n ( u ε u ε )n n = 1 2 ( u ε 2 2bu ε ) ( u ε n) 2 = 1 2 (( nu ε ) 2 + ( τ u ε ) 2 2bu ε ) ( n u ε ) 2 (52) = 1 2 (( τu ε ) 2 ( n u ε ) 2 2bu ε ). Now, we shall analyze the behavior of the total potential energy with respect to the parameter ε by taking into account each type of boundary condition on B ε, namely, Neumann.

39 PROOF The shape derivative of the cost functional (39) reads (53) d dε ψ(χ ε) = J Ωε (u ε ) = 1 (( τ u ε ) 2 2bu ε ), 2 B ε where we have used (52). In addition, we have that the following expansion for u ε holds in the neighborhood of the hole (50) u ε (x) = u(x) + ε2 u( x) (x x) + o(ε) x x 2 (54) = u( x) + u( x) (x x) + ε2 u( x) (x x) + o(ε), x x 2 which can be written in a polar coordinate system (r,θ) centered at the point x (see fig. 6), namely (55) u ε (r,θ) = ϕ 0 + r2 + ε 2 (ϕ 1 cosθ + ϕ 2 sinθ) + o(ε), r where we have introduced the notation u( x) = ϕ 0 and u( x) = (ϕ 1,ϕ 2 ).

40 The tangential derivative of u ε is obtained in the following way τ u ε (x) = 1 r θ u ε (r,θ) (56) = r2 + ε 2 r 2 (ϕ 1 sinθ ϕ 2 cosθ) + o(ε), τ u ε (x) Bε = 2(ϕ 1 sinθ ϕ 2 cosθ) + o(ε), where the last expression was obtained by taking r = ε in the previous one. In addition, considering the above expansion in (53) and after explicitly evaluating the integral on the boundary of the hole B ε, we obtain d dε ψ(χ ε) = 1 2π (4(ϕ 1 sinθ ϕ 2 cosθ) 2 2bϕ 0 )εdθ + o(ε 2 ) 2 0 (57) = 2πε((ϕ1 2 + ϕ2 2 ) bϕ 0) + o(ε 2 ), Therefore, the above result together with the relation between shape and topological derivative given by (41) leads to 1 ( ) (58) T = lim ε 0 f 2πε((ϕ1 2 (ε) + ϕ2 2 ) bϕ 0) + o(ε 2 ).

41 Now, in order to identify the main term of the above expansion, we choose (59) f (ε) = πε 2, which leads to (60) T = (ϕ ϕ2 2 ) + bϕ 0. Recalling that u( x) = ϕ 0 and u( x) = (ϕ 1,ϕ 2 ), the final formula for the topological derivative becomes (61) T ( x) = u( x) 2 + bu( x). This leads to the topological asymptotic expansion of the energy shape functional, namely (62) ψ(χ ε ( x)) = ψ(χ) πε 2 ( u( x) 2 bu( x)) + o(ε 2 ).

42 The Navier Problem The shape functional in the unperturbed domain Ω is given by (63) ψ(χ) := J Ω (u) = 1 2 Ω σ(u) u s b u q u, Ω Γ N where the vector function u is the solution to the variational problem : (64) Find u U, such that σ(u) η s = b η + Ω Ω q η Γ N η V, with σ(u) = C u s.

43 In the above equation, C is the constitutive tensor given by (65) C = E ((1 ν)i + νi I), 1 ν2 where I and I are the second and fourth order identity tensors, respectively, E is the Young modulus and ν the Poisson ratio, both considered constants everywhere. For the sake of simplicity, we also assume that the thickness of the elastic body is constant and equal to one. The set U and the space V are respectively defined as (66) (67) U := {ϕ H 1 (Ω;R 2 ) : ϕ ΓD = u}, V := {ϕ H 1 (Ω;R 2 ) : ϕ ΓD = 0}. In addition, Ω = Γ D Γ N with Γ D Γ N =, where Γ D and Γ N are Dirichlet and Neumann boundaries, respectively. Thus u is a Dirichlet data on Γ D and q is a Neumann data on Γ N, both assumed to be smooth enough.

44 FIGURE: The Navier system defined in the unperturbed domain Ω.

45 The strong system associated to the variational problem (64) reads : Find u, such that divσ(u) = b in Ω, (68) σ(u) = C u s u = u on Γ D, σ(u)n = q on Γ N. Since the Young modulus E and the Poisson ratio ν are considered constants, the above boundary value problem reduces itself to the well-known Navier equations, namely (69) µ u (λ + µ) (divu) = b in Ω, with the Lame s coefficients µ and λ respectively given by (70) µ = E 2(1 + ν) and λ = νe 1 ν 2.

46 Now, let us state the same problem in the perturbed domain Ω ε. In this case, the total potential energy reads (71) ψ(χ ε ) := J Ωε (u ε ) = 1 σ(u ε ) u s ε b u ε q u ε, 2 Ω ε Ω ε Γ N where the vector function u ε solves the variational problem : (72) Find u ε U ε, such that σ(u ε ) η s = b η + q η η V ε, Ω ε Ω ε Γ N with σ(u ε ) = C u s ε.

47 The set U ε and the space V ε are defined as (73) (74) U ε := {ϕ H 1 (Ω ε ;R 2 ) : ϕ ΓD = u}, V ε := {ϕ H 1 (Ω ε ;R 2 ) : ϕ ΓD = 0}. We have homogeneous Neumann condition on the boundary of the hole. FIGURE: The Navier system defined in the perturbed domain Ω ε.

48 The strong system associated to the variational problem (72) reads : Find u ε, such that divσ(u ε ) = b in Ω ε, σ(u (75) ε ) = C u s ε u ε = u on Γ D, σ(u ε )n = q on Γ N, σ(u ε )n = 0 on B ε.

49 Shape Sensitivity Analysis Let J Ωε (u ε ) be the shape functional defined by (71). Then, the derivative of this functional with respect to the small parameter ε is given by (76) J Ωε (u ε ) = Σ ε V, Ω ε where V is the shape change velocity field defined through (18) and Σ ε is a generalization of the classical Eshelby energy-momentum tensor given by (77) Σ ε = 1 2 (σ(u ε) u s ε 2b u ε )I u ε σ(u ε ).

50 PROOF By Reynolds transport theorem the shape derivative (the material or total derivative with respect to the parameter ε) of the functional (71) is given by J Ωε (u ε ) = 1 ((σ(u ε ) u s ε) + (σ(u ε ) u s 2 ε)divv) Ω ε (78) b ( u ε + u ε divv) q u ε. Ω ε Γ N Next, by using the concept of material derivative of spatial fields we find that the first term of the above right-hand side integral can be written as (79) (σ(u ε ) u s ε) = 2σ(u ε ) u s ε 2σ(u ε ) ( u ε V) s = 2σ(u ε ) u s ε 2 u ε σ(u ε ) V, since σ(u ε ) = C u s ε. From this last result we obtain J Ωε (u ε ) = 1 ((σ(u ε ) u s ε 2b u ε )I 2 u ε σ(u ε )) V 2 Ω ε (80) + σ(u ε ) u s ε b u ε q u ε. Ω ε Ω ε Γ N

51 Boundary shape gradient Let J Ωε (u ε ) be the shape functional defined by (71). Then, the derivative of this functional with respect to the small parameter ε is given by (81) J Ωε (u ε ) = Σ ε n V, Ω ε with V standing by the shape change velocity field defined through (18) and tensor Σ ε is given by (77).

52 PROOF By making use of the other version of the Reynolds transport theorem the shape derivative of the functional (71) is (82) J Ωε (u ε ) = 1 (σ(u ε ) u s 2 ε) + 1 (σ(u ε ) u s Ω ε 2 ε)n V Ω ε b u ε (b u ε )n V q u ε. Ω ε Ω ε Γ N Next, by using the concept of shape derivatives of spatial fields, we find that the first term of the above right-hand side integral can be written as (83) (σ(u ε ) u s ε) = 2σ(u ε ) ( u ε) s, since σ(u ε ) = σ (u ε ) and ( ) is the partial derivative of ( ) with respect to ε. Now, let us use the relation between material and spatial derivatives of vector fields, namely ϕ = ϕ ( ϕ)v, which leads to (84) (σ(u ε ) u s ε) = 2σ(u ε ) ( u ε ( u ε )V) s = 2σ(u ε ) u s ε 2σ(u ε ) (( u ε )V) s.

53 From this last result we obtain J Ωε (u ε ) = 1 (σ(u ε ) u s ε 2b u ε )n V 2 Ω ε (σ(u ε ) (( u ε )V) s b ( u ε )V) Ω ε (85) + σ(u ε ) u s ε b u ε q u ε. Ω ε Ω ε Γ N Since u ε is a variation of u ε in the direction of the velocity field V, then u ε V ε. Now, by taking into account that u ε is the solution to the variational problem (72), we have that the last three terms of the above equation vanish. By using a tensor relation and after applying the divergence theorem, we have J Ωε (u ε ) = 1 (σ(u ε ) u s ε 2b u ε )n V ( u ε )V σ (u ε )n 2 Ω ε Ω ε (86) + (divσ(u ε ) + b) ( u ε )V. Ω ε

54 Then, since σ (u ε ) = σ(u ε ), we can rewrite the above equation in the compact form (87) J Ωε (u ε ) = Σ ε n V + Ω ε (divσ(u ε ) + b) ( u ε )V. Ω ε Considering that u ε is also solution to the strong system, namely divσ(u ε ) = b, the last term of the above equation vanishes, which leads to the required result. According to the obtained result we have (88) J Ωε (u ε ) = = Σ ε n V Ω ε Σ ε n V + B ε Ω Σ ε n V. Since we are dealing with an uniform expansion of the circular hole B ε, then in view of the associated velocity field (18), namely, V = n on B ε and V = 0 on Ω, we finally obtain d (89) dε ψ(χ ε) = J Ωε (u ε ) = Σ ε n n. B ε It means that the shape derivative of the cost functional can be represented by a boundary integral or distributions supported on B ε

55 Asymptotic Analysis FIGURE: Polar coordinate system (r,θ) centered at the point x Ω.

56 Let us start with the ansatz for an expansion of u ε in the form (90) u ε (x) = u(x) + w ε (x) + ũ ε (x). After applying the operator σ we have σ(u ε (x)) = σ(u(x)) + σ(w ε (x)) + σ(ũ ε (x)) (91) = σ(u( x)) + σ(u(y))(x x) + σ(w ε (x)) + σ(ũ ε (x)), where y is an intermediate point between x and x. On the boundary of the hole B ε we have σ(u ε )n Bε = 0. Thus, the normal projection of the above expansion, evaluated on B ε, leads to (92) σ(u( x))n ε( σ(u(y))n)n + σ(w ε (x))n + σ(ũ ε (x))n = 0.

57 Thus, we can choose σ(w ε ) such that (93) σ(w ε (x))n = σ(u( x))n on B ε. Now, the following exterior boundary value problem is considered and formally obtained with ε 0 : Find σ(w ε ), such that divσ(w (94) ε ) = 0 in R 2 \ B ε, σ(w ε ) 0 at, σ(w ε )n = σ(u( x))n on B ε.

58 The above boundary value problem admits an explicit solution (see, for instance, the book by Little 1973 ), which can be written in a polar coordinate system (r,θ) centered at the point x (see fig. 9) as σ rr ε 2 ) (95) (w ε (r,θ)) = ϕ 1 r 2 ϕ 2 (4 ε2 r 2 3ε4 r 4 cos2θ, (96) (97) σ θθ (w ε (r,θ)) = ϕ 1 ε 2 ε 4 r 2 3ϕ 2 r 4 cos2θ, ( σ rθ (w ε (r,θ)) = ϕ 2 2 ε2 r 2 3ε4 r 4 The coefficients ϕ 1 and ϕ 2 are given by ) sin2θ. (98) ϕ 1 = 1 2 (σ 1(u( x)) + σ 2 (u( x))), ϕ 2 = 1 2 (σ 1(u( x)) σ 2 (u( x))), where σ 1 (u( x)) and σ 2 (u( x)) are the eigenvalues of tensor σ(u( x)), which can be expressed as (99) σ 1,2 (u( x)) = 1 ( ) trσ(u( x)) ± 2σ 2 D (u( x)) σ D (u( x)), with σ D (u( x)) standing by the deviatoric part of the stress tensor σ(u( x)),

59 namely (100) σ D (u( x)) = σ(u( x)) 1 2 trσ(u( x))i. In addition, σ rr (ϕ), σ θθ (ϕ) and σ rθ (ϕ) are the components of tensor σ(ϕ) in the polar coordinate system, namely, σ rr (ϕ) = e r σ(ϕ)e r, σ θθ (ϕ) = e θ σ(ϕ)e θ and σ rθ (ϕ) = σ θr (ϕ) = e r σ(ϕ)e θ, with e r and e θ used to denote the canonical basis associated to the polar coordinate system (r,θ), such that, e r = e θ = 1 and e r e θ = 0. See fig. 9. Now we can construct ũ ε in such a way that it compensates the discrepancies introduced by the higher-order terms in ε as well as by the boundary-layer w ε on the exterior boundary Ω.

60 It means that the remainder ũ ε must be the solution to the following boundary value problem : Find ũ ε, such that divσ(ũ ε ) = 0 in Ω ε, σ(ũ (101) ε ) = C ũ s ε ũ ε = w ε on Γ D, σ(ũ ε )n = σ(w ε )n on Γ N, σ(ũ ε )n = εh on B ε. where ũ ε = O(ε 2 ). Finally, the expansion for σ(u ε ) in the polar coordinate system (r,θ) reads ( ) ) σ rr (u ε (r,θ)) = ϕ 1 1 ε2 r 2 + ϕ 2 (1 4 ε2 r 2 + 3ε4 r 4 cos2θ + O(ε 2 ), ( ) ) σ θθ (u ε (r,θ)) = ϕ ε2 r 2 ϕ 2 (1 + 3 ε4 r 4 cos2θ + O(ε 2 ), ( σ rθ (u ε (r,θ)) = ϕ ε2 r 2 3ε4 r 4 ) sin2θ + O(ε 2 ), where we have used the fact that σ rr (u( x)) = ϕ 1 + ϕ 2 cos2θ, σ θθ (u( x)) = ϕ 1 ϕ 2 cos2θ and σ rθ (u( x)) = ϕ 2 sin2θ.

61 Topological Derivative Evaluation From an orthonormal curvilinear coordinate system n and τ defined on the boundary B ε (see fig. 9), the tensors σ(u ε ) and u ε can be decomposed as (102) (103) σ(u ε ) Bε = σ nn (u ε )(n n) + σ nτ (u ε )(n τ) + σ τn (u ε )(τ n) + σ ττ (u ε )(τ τ), u ε Bε = n u n ε(n n) + τ u n ε(n τ) + n u τ ε(τ n) + τ u τ ε(τ τ). Therefore, we observe that (104) σ(u ε )n Bε = σ nn (u ε )n + σ τn (u ε )τ = 0, which implies in (105) σ nn (u ε ) = σ τn (u ε ) = σ nτ (u ε ) = 0 on B ε. since σ(u ε ) = σ(u ε ).

62 In addition, Σ ε n n Bε = 1 2 (σ(u ε) u s ε 2b u ε )In n u ε σ(u ε )n n = 1 2 (σ(u ε) u s ε 2b u ε )n n σ(u ε )n ( u ε )n = 1 2 (σ(u ε) u s ε 2b u ε ) (106) = 1 2 (σ ττ (u ε ) τ u τ ε 2b u ε ). On the other hand, the constitutive tensor C is invertible, namely, (107) C 1 = 1 ((1 + ν)i νi I)), E which implies in u s ε = C 1 σ(u ε ) τ u τ ε = 1 E (σ ττ (u ε ) νσ nn (u ε )) = 1 E σ ττ (u ε ) on B ε.

63 Therefore, we have (108) Σ ε n n Bε = 1 2 ( ) 1 E (σ ττ (u ε )) 2 2b u ε. The shape derivative of the cost functional (89) reads d (109) dε ψ(χ ε) = J Ωε (u ε ) = 1 ( ) 1 2 E (σ ττ (u ε )) 2 2b u ε. From formula (102), we have that the following expansion for σ ττ (u ε ) holds in the neighborhood of the hole (110) σ ττ (u ε (x)) = σ θθ (u ε (r,θ)) ( = ϕ ε2 r 2 B ε ) ϕ 2 (1 + 3 ε4 r 4 ) cos2θ + O(ε), σ ττ (u ε (x)) Bε = 2ϕ 1 4ϕ 2 cos2θ + O(ε).

64 The following expansion for u ε holds on the boundary of the hole (111) u ε (x) Bε = ϕ 0 + O(ε), where ϕ 0 = u( x). We obtain d 2π ( ) 1 dε ψ(ε) = E (2ϕ 1 4ϕ 2 cos2θ) 2 2b ϕ 0 εdθ + O(ε 2 ) ( ) 2 (112) = 2πε E (ϕ ϕ2 2 ) b ϕ 0 + O(ε 2 ), This leads to 1 (113) T = lim ε 0 f (ε) We choose f (ε) = πε 2, which results in [ ( ) ] 2 2πε E (ϕ ϕ2 2 ) b ϕ 0 + O(ε 2 ). (114) T = 2 E (ϕ ϕ2 2 ) + b ϕ 0. Recalling that ϕ 1 = (σ 1 (u( x)) + σ 2 (u( x)))/2, ϕ 2 = (σ 1 (u( x)) σ 2 (u( x)))/2 and ϕ 0 = u( x), the final formula for the topological derivative evaluated at x Ω becomes :

65 In terms of the principal stresses σ 1 (u( x)) and σ 2 (u( x)) (115) T ( x) = 1 [ (σ 1 (u( x)) + σ 2 (u( x))) 2 + 2E In terms of the stress tensor σ(u( x)) 2(σ 1 (u( x)) σ 2 (u( x))) 2] + b u( x). (116) T ( x) = 1 [ ] 4σ(u( x)) σ(u( x)) tr 2 σ(u( x)) +b u( x). 2E In terms of the stress tensor σ(u( x)) and the strain tensor u s ( x) (117) T ( x) = Pσ(u( x)) u s ( x) + b u( x), where P is the Pólya-Szegö polarization tensor, given in this particular case by the following isotropic fourth order tensor (118) P = ν I 1 3ν 2(1 ν 2 ) I I.

66 Multiscale Elasticity Problem The homogenization-based multi-scale constitutive framework presented, among others, in Sanchez-Palencia monograph, 1980, is adopted here in the estimation of the macroscopic elastic response from the knowledge of the underlying microstructure. The main idea behind this well-established family of constitutive theories is the assumption that any point x of the macroscopic continuum is associated to a local Representative Volume Element (RVE) whose domain is denoted by Ω µ, with boundary Ω µ, as shown in fig. 10. x FIGURE: Macroscopic continuum with a locally attached microstructure.

67 The Homogenized Elasticity Tensor Crucial to the developments presented here is the closed form of the homogenized elasticity tensor C for the multi-scale model defined in the above. The components of the homogenized elasticity tensor C, in the orthonormal basis {e i }, for i = 1,2,3, of the Euclidean space (refer to fig. 10), can be written as (119) (C) ijkl = 1 (σ µ (u µkl )) ij. V µ Ω µ where V µ denotes the total volume of the RVE, e.g. the volume of the cub in the sketch shown in fig. 10. The canonical microscopic displacement field u µkl is solution to the equilibrium equation of the form Sanchez-Palencia, 1980 (120) σ µ (u µkl ) η s = 0 η V µ, Ω µ

68 The Homogenized Elasticity Tensor We assume that the microscopic stress tensor field σ µ (u µkl ) satisfies (121) σ µ (u µkl ) = C µ u s µ kl, where C µ is the microscopic constitutive tensor given by (122) C µ = E ( µ I + ν ) µ I I, 1 + ν µ 1 2ν µ with I and I the second and fourth order identity tensors, respectively, E µ the Young modulus and ν µ Poisson ratio of the RVE. Without loss of generality, u µkl (y), with y Ω µ, may be decomposed into a sum (123) u µkl (y) := u + (e k e l )y + ũ µkl (y), of a constant (rigid) RVE displacement coinciding with the macroscopic displacement field u at the point x Ω, a linear field (e k e l )y, and a canonical microscopic displacement fluctuation field ũ µkl (y).

69 The Homogenized Elasticity Tensor The microscopic displacement fluctuation field ũ µkl is solution to the following canonical set of variational problems Sanchez-Palencia, 1980 Findũ µkl V µ, such that (124) σ µ (ũ µkl ) η s + C µ (e k s e l ) η s = 0 η V µ, Ω µ Ω µ with σ µ (u µkl ) = C µ u µkl. The complete characterization of the multi-scale constitutive model is obtained by defining the subspace V µ U µ of kinematically admissible displacement fluctuations. In general, different choices produce different macroscopic responses for the same RVE. The analysis can be focussed eg, on media with periodic microstructure. In this case, the geometry of the RVE cannot be arbitrary and must represent a cell whose periodic repetition generates the macroscopic continuum. In addition, the displacement fluctuations must satisfy periodicity on the boundary of the RVE. Accordingly, we have (125) V µ := { ϕ U µ : ϕ(y + ) = ϕ(y ) (y +,y ) P }, where P is the set of pairs of points, defined by a one-to-one periodicity correspondence, lying on opposing sides of the RVE boundary.

70 The Homogenized Elasticity Tensor Finally, the minimally constrained space of kinematically admissible displacements U µ is defined as { } (126) U µ := ϕ H 1 (Ω µ ;R 3 ) : ϕ = 0, ϕ s n = 0. Ω µ Ω µ where n is the outward unit normal to the boundary Ω µ and s denotes the symmetric tensor product between vectors. From the definition of the homogenized elasticity tensor (119), we have (C) ijkl = 1 e i σ µ (u µkl )e j V µ Ω µ = 1 (127) σ µ (u µkl ) (e i e j ). V µ Ω µ On the other hand, the additive decomposition (123) allows us to write (128) e i e j = ((e i e j )y) s since (e i e j )y = u µij (y) ũ µij (y) u. = (u µij (y) ũ µij (y) u) s = (u µij (y) ũ µij (y)) s.

71 The Homogenized Elasticity Tensor Therefore, by combining these two results, we obtain (129) (C) ijkl = 1 σ µ (u µkl ) (u µij ũ µij ) s V µ Ω µ = 1 σ µ (u µkl ) u s µ V ij, µ Ω µ since u µkl satisfies the equilibrium equation (120) and ũ µij V µ.