On the First-Order Shape Derivative of the Kohn-Vogelius Cost Functional of the Bernoulli Problem

Definition 1. A domain Ω ⊂ ℝⁿ with a nonempty boundary ∂Ω is called a C^k,l-domain, where 0 ≤ k,  0 < l ≤ 1, if for every y ∈ ∂Ω there exists a neighborhood N_y of y and a C^k,l diffeomorphism f_y : N_y → B(0,1) such that (a)  f_y(N_y∩Ω) = B⁺(0,1), (b)  f_y(N_y∩∂Ω) = B⁰(0,1), and $$.

Theorem 2 (see [17].)If Ω is a bounded open connected subset of ℝⁿ with Lipschitz continuous boundary, then $$.

Proof. The interior of $$ is the largest open set contained in the set $$. Moreover, $$. It follows that $$. Next, we show that $$. Clearly, $$. We now show that if $$, then x ∉ ∂Ω.

Suppose x ∈ ∂Ω and $$. We need to show that any open set N containing x contains an element not in $$. We first note that by definition of C^0,1 domain, there exists a neighborhood N_x of x ∈ ∂Ω and a diffeomorphism f_x : N_x → B(0,1). Let N be an open set containing x with N ⊂ N_x. It follows that f_x(N) is an open set containing 0 and this set is contained in B(0,1). Hence, there exists z ∈ f_x(N) such that z ∈ B⁻(0,1). This implies that $$. Thus, N contains an element not in $$, which is a contradiction. Therefore, $$. We have proven that if x ∈ ∂Ω, then $$. Taking the contrapositive of this statement we get that if $$, then x ∉ ∂Ω. Since $$ but x ∉ ∂Ω, we conclude x ∈ Ω. Thus, $$. We have shown that $$ and $$. Therefore, $$.

Theorem 3. If Ω is a bounded, open, connected set in ℝⁿ such that $$ and φ is a continuous injective mapping from $$ to ℝⁿ, then

Theorem 4. Suppose

• (1)

  Ω is a bounded, open, connected set in ℝⁿ such that $$,

• (2)

  $$ where φ₀ is injective,

• (3)

  $$ such that

  $$ ()

Then

• (i)

  $$ is a homeomorphism (i.e., φ is a bijection, φ is continuous, and φ⁻¹ is continuous),

• (ii)

  φ : Ω → φ(Ω) is a C¹-diffeomorphism (i.e., φ is a bijection, φ ∈ C¹(Ω), φ⁻¹ ∈ C¹(φ(Ω))),

• (iii)

  φ(Ω) = φ₀(Ω), $$.

Lemma 5. If Ω is a bounded, open, connected subset of ℝⁿ having a Lipschitz continuous boundary, then there is a number c(Ω) such that, for any given points $$, one can find a finite sequence of points y_k, k = 1, …, l + 1, satisfying the following properties:

• (a)

  y₁ = x, y_k ∈ Ω for 2 ≤ k ≤ l, y_l+1 = y,

• (b)

  (y_k, y_k+1) ⊂ Ω for 1 ≤ k ≤ l,

• (c)

  $$.

Lemma 6 (see [9], [13].)Consider the operator T_t defined by (9), where V ∈ Θ, which is described by (13). Then

• (i)

  I_t = 1 + t div   V + t²det DV,

• (ii)

  there exist t_V, α₁, α₂ > 0 such that 0 < α₁ ≤ I_t(x) ≤ α₂, for |t | ≤ t_V, x ∈ U.

Proof. In general, for n-dimensional case, the Jacobian of T_t is given by DT_t = (a_ij), where a_ij = t(∂V_i/∂x_j) if, i ≠ j, and a_ii = 1 + t(∂V_i/∂x_i). By definition of the determinant, we can write det DT_t as

where S_n refers to the set of all permutations of {1,2, …, n}, i is the identity permutation, F_n = {σ ∈ S_n : σ(k) = k    for  some  positive  integer  k   ≤   n}, and sgn (σ) is either 1 (if the number of inversions is even) or −1 (if the number of inversions is odd). We observe that the expression I can be written as I = 1 + tdiv V + t²R₁(t, V), where R₁ ∈ C(ℝ, C^0,1(U)). We also observe that, for n ≥ 2, each term of the expression II has at least 2 factors that are of the form t(∂V_i/∂x_j), i ≠ j. Hence we can write II = t²R₂(t, V), where R₂ is in C(ℝ, C^0,1(U)). All terms of III have factors of the form t(∂V_i/∂x_j),  i ≠ j, and thus we have $$, which can be written as t²R₃(t, V), where R₃ ∈ C(ℝ, C^0,1(U)). Combining I, II, and III, we get with R ∈ C(ℝ, C^0,1(U)). In particular, for n = 2, the determinant is computed as follows:

This verifies (i). To show (ii) we first get the lower bound for I_t(x). Take

For |t | ≤ t_V, we obtain On the other hand, by triangle inequality we have Hence, we have shown that there are positive constants α₁ = 1/2 and $$ such that α₁ ≤ I_t(x) ≤ α₂ for $$.

Theorem 7. Let Ω and U be nonempty bounded open connected subsets of ℝ² with Lipschitz continuous boundaries, such that $$, and ∂Ω is the union of two disjoint boundaries Γ and Σ. Let T_t be defined as in (9) where V belongs to Θ, defined as (13).

Then for sufficiently small t,

• (1)

  $$ is a homeomorphism,

• (2)

  T_t : U → U is a C^1,1 diffeomorphism, and in particular, T_t : Ω → Ω_t is a C^1,1 diffeomorphism,

• (3)

  Γ_t = T_t(Γ) = Γ,

• (4)

  ∂Ω_t = Γ ∪ T_t(Σ).

Proof. First, because U is a C^0,1 domain, it follows that $$ by Theorem 2. Second, $$, and it is injective. Third, it is evident that T_t is C^1,1 because V is C^1,1. For x ∈ ∂U, T_t(x) = x because V vanishes on ∂U. For x ∈ U, the determinant of the Jacobian of the perturbation of identity operator T_t is given by (19). By Lemma 6, there exists a t_V > 0, given by (20), such that det DT_t(x) > 0 for all x ∈ U and for |t | ≤ t_V. Hence, by applying Theorem 4, we conclude that T_t(U) = U and $$ for all |t | < t_V, and $$ is a homeomorphism. Furthermore, by Theorem 4, we find that T_t : U → U is a C¹ diffeomorphism. To show that T_t is a C^1,1 diffeomorphism, we are left to show that $$ is Lipschitz continuous. To verify this we use Lemma 5.

Given any two points u, v ∈ U we choose {y_k} such that properties (a)–(c) of Lemma 5 are satisfied. For fixed |t| < t_V, differentiating the identities $$ and $$ will lead to

for all z ∈ U. Thus,

This implies

Applying the infinity norm in the space C(X; ℝ²) we have

Since DT_t is Lipschitz continuous, we have

where L₁ is the maximum of all Lipschitz constants of DT_t for all |t | < t_V. Then finally, using the mean value theorem and property (c) in Lemma 5, we obtain Hence $$ is Lipschitz continuous which shows that T_t : U → U is a C^1,1 diffeomorphism for sufficiently small |t|. Restricting to Ω, this proves that T_t : Ω → Ω_t is a C^1,1 diffeomorphism. (2) is clear because the fixed boundary is invariant under T_t; that is, Γ_t : = T_t(Γ) = Γ since V vanishes on Γ. Lastly, using Theorem 3, definition of T_t, (1), and (2), we obtain (3).

Corollary 8. Let Ω and U be two domains of ℝ² with C^1,1 boundary. Then for |t | < t_V, where t_V is given by (20), the perturbed domain Ω_t : = T_t(Ω) is also of class C^1,1.

Proof. Given y ∈ ∂Ω_t, we let $$. Then there exists a neighborhood N_x of x and a diffeomorphism g_x ∈ C^1,1(N_x, B(0,1)) such that g_x(N_x∩Ω) = B⁺(0,1),  g_x(N_x∩∂Ω) = B⁰(0,1), and g_x(N_x∩Ω^c) = B⁻(0,1). We have also shown that T_t defined in Theorem 7 is a C^1,1 diffeomorphism. Since $$ is continuous, $$ is a neighborhood of y in U. Define $$. This is bijective because g_x and $$ are bijective. g_y ∈ C^1,1(N_y, B(0,1)) because $$ (hence $$) and g_x ∈ C^1,1(N_x, B(0,1)). Also, $$.

Next, we note that ∂Ω_t∩N_y = T_t(∂Ω)∩N_y = T_t(∂Ω)∩T_t(N_x). Since T_t is injective, we have ∂Ω_t∩N_y = T_t(∂Ω∩N_x). Thus by definition of g_y we get

We also observe the following:

This shows that Ω_t is indeed of class C^1,1.

Remark 9. Theorem 7 and Corollary 8 tell us that the reference Ω and the perturbed domain Ω_t have the same topological structure and regularity under the perturbation of identity operator T_t for sufficiently small t. See Figure 3 for illustration.

Remark 10. We note the following observations for fixed, sufficiently small t.

• (1)

  $$.

• (2)

  $$.

• (3)

  $$.

• (4)

  w_t ∈ C(Σ; ℝ).

• (5)

  $$ implies that |V|_∞ and |DV|_∞ are both finite.

We now provide several properties of T_t.

Lemma 11 (see [9], [13], [16], [18].)Consider the transformation T_t, where the fixed vector field V belongs to Θ, defined in (13). Then there exists t_V > 0 such that T_t and the functions in (16) and (31) restricted to the interval I_V = (−t_V, t_V) have the following regularity and properties.

• (1)

  $$.

• (2)

  $$.

• (3)

  $$.

• (4)

  t ↦ w_t ∈ C¹(I_V, C(Σ)).

• (5)

  $$.

• (6)

  There is β > 0 such that A_t(x) ≥ βI for x ∈ U.

• (7)

  (d/dt)T_t|_t=0 = V.

• (8)

  $$.

• (9)

  (d/dt)DT_t|_t=0 = DV.

• (10)

  (d/dt)(DT_t) ⁻¹|_t=0 = −DV.

• (11)

  (d/dt)I_t|_t=0 = div V.

• (12)

  (d/dt)A_t|_t=0 = (div V)I − (DV + (DV) ^T) ≡ A.

• (13)

  lim   _t→0w_t = 1.

• (14)

  (d/dt)w_t|_t=0 = div _ΣV,

where the surface divergence div _Σ is defined by

Proof. (3) Suppose $$, t, h ∈ I_V, and $$. Then

Using Lemma 5, we connect T_h(y) and T_t(y) by a chain y_k, k = 1, …, l + 1, satisfying

• (i)

  y₁ = T_t(y), y_k ∈ U for 2 ≤ k ≤ l, y_l+1 = T_h(y),

• (ii)

  (y_k, y_k+1) ⊂ U for 1 ≤ k ≤ l,

• (iii)

  $$,

and then we get Thus, By reducing t_V if necessary, we can assume without loss of generality that |tDV|_∞ < 1 for t ∈ (−t_V, t_V). This allows us to represent $$ as a Neumann series: and its norm is estimated as follows: This shows uniform convergence in t ∈ I_V and $$. Hence, for every ϵ > 0 one can choose a δ : = ϵ/c(U)M|V|_∞ > 0 which implies that, for every t, h ∈ I_V, $$ whenever |t − h | < δ. In other words, $$.

To show that $$ is continuous from I_V to $$, we only need to show that for every t ∈ I_V, $$ whenever |t − h| < δ and h ∈ I_V. Let $$. Using (37), estimate $$ as follows:

Using the definition of Jacobian of a transformation and the regularity of V, we further simplify (38) as follows: where L is the Lipshitz constant for DV and K is upper bound for |DV|_∞. Taking the maximum of both sides of the inequality for all $$ and using (34) we get where α = M²K + M²t_VLc(U)M|V|_∞. Thus, for any ϵ > 0, we choose δ = ϵ/α, so that if 0 < |h − t| < δ, then $$. Therefore, $$.

Proof of property (8) in Lemma 11 is as follows. Given $$, we have $$. This implies that

Manipulating the left hand side of (41), we get We first work on A. Applying the definition of T_t, we get Thus, Similarly, we can write B as follows: Hence, we have Suppose v is a coordinate function of V. By the mean value theorem, we observe that where c is a point on the segment joining $$ and $$, and as h tends to infinity, (47) tends to $$. Thus, Combining (44) and (48), we get which implies that Evaluating (50) at t = 0, we get $$.

Definition 12. Let u be defined in [0, t_V] × U. An element $$, called the material derivative of u, is defined as

Remark 13. The material derivative can be written as

It characterizes the behavior of the function u at x ∈ Ω ⊂ U in the direction V(x).

Definition 14. Let u be defined in [0, t_V] × U. An element u^′ ∈ H^k(Ω) is called the shape derivative of u at Ω in the direction V, if the following limit exists in H^k(Ω):

Remark 15. The shape derivative of u is also defined as follows:

We note that if $$ and ∇u · V exist in H^k(Ω), then the shape derivative can be written as In general, if $$ and ∇u · V(x) both exist in W^m,p(Ω), then u^′(x) also exists in that space.

Domain and Boundary Transformations

Lemma 16 (see [18].)(1) Let φ_t ∈ L¹(Ω_t). Then φ_t∘T_t ∈ L¹(Ω) and

(2) Let φ_t ∈ L¹(∂Ω_t). Then φ_t∘T_t ∈ L¹(∂Ω) and

where I_t and w_t are defined in (31).

Theorem 17 (domain differentiation formula). Let u ∈ C(I_V, W^1,1(U)) and suppose $$ exists in L¹(U). Then

Theorem 18 (boundary differentiation formula). Let u be defined in a neighborhood of Γ. If u ∈ C(I_V, W^2,1(U)) and $$, then

where κ is the mean curvature of the free boundary Σ.

The First-Order Eulerian Derivative

Definition 19. The Eulerian derivative of the shape functional J : Ω → ℝ defined in (7) at the domain Ω in the direction of the deformation field V ∈ Θ is given by

if the limit exists.

Remark 20. J is said to be shape differentiable at Ω if dJ(Ω; V) exists for all V ∈ Θ and is linear and continuous with respect to V.

Remark 21. The function u_t : Ω_t → ℝ can be referred to as the reference domain by composing u_t with T_t; that is,

and by chain rule of differentiation, we get

Theorem 22 (see [13].)The solutions $$ of (74) are uniformly bounded in H¹(Ω) for t ∈ (−t_V, t_V) and

where u_D is the weak solution of (4).

Proof. We first prove the uniform boundedness of $$ in H¹(Ω) for t ∈ (−t_V, t_V). Since $$, by using coercivity of A_t we get

Also, by applying (81), we have Therefore,

Now we take the difference between the weak form of (4) and the variational equation (74), to get

Note that $$. This implies that $$ is in $$. Note also that $$. Let $$. So for sufficiently small t, $$. Now choosing v = y^t as a test function and by the uniform coercivity of A_t one obtains

If ∇y^t = 0, then the inequality above holds. For ∇y^t ≠ 0 we have

Hence, Squaring and multiplying t on both sides of the inequality give us Since $$ is uniformly bounded in t by Lemma 11, follows.

Theorem 23. The solutions $$ of (84) are uniformly bounded in H¹(Ω) for t ∈ (−t_V, t_V) and

where u_N is the solution of (65).

Proof. Subtracting (65) from (84) for all $$ we get

Hence Note that $$ belongs to $$. Hence By Cauchy-Schwarz inequality, we obtain Furthermore, by trace theorem we have $$. Therefore, holds, where C = max {1, c}. This implies which entails

We now show that $$ is bounded uniformly in t in L²(Ω). Since $$, we have

Consequently, and this shows that $$ is uniformly bounded in L²(Ω) because |w_t|_∞ is bounded. In addition, A_t and w_t are differentiable at t = 0 by Lemma 11. Therefore,

Theorem 24 (see [22], page 124.)Let Ω be a bounded open subset of ℝⁿ with a C^1,1 boundary. Consider the Dirichlet boundary value problem:

where Let a_i,j be uniformly Lipschitz functions and let a_i be bounded measurable functions such that a_j,i = a_i,j,   1 ≤ i,  j ≤ n and that there exists α > 0 with for all ζ ∈ ℝⁿ and for almost every $$. Assume in addition that either

• (i)

  a_i = 0,   1 ≤ i ≤ n and a₀ ≥ 0 a.e. or

• (ii)

  a₀ ≥ β > 0 a.e.

Then for every f ∈ L^p(Ω) and every g ∈ W^2−1/p,p(∂Ω), there exists a unique u ∈ W^2,p(Ω) that solves (110).

Theorem 25 (see [22], page 128.)Let Ω be a bounded open subset of ℝⁿ with a C^k+1,1 boundary. Consider the operator A defined by (110) with $$, and assume that there exists α > 0 such that (112) holds for all ζ ∈ ℝⁿ and for every $$. Also, consider a real boundary operator B which is either the identity operator (order  d = 0) or $$ with $$, 1 ≤ i ≤ n  (order d = 1) and $$ everywhere on Γ : = ∂Ω. Furthermore, assume that u ∈ W^2,p(Ω) satisfies Au = f ∈ W^k,p(Ω) and γBu = g ∈ W^{2+k−d−1/p,p}(Γ). Then u ∈ W^k+2,p(Ω).

Theorem 26 (see [23], page 316.)Let Ω be a bounded open subset of ℝⁿ. Suppose u ∈ H¹(Ω) is a weak solution of the PDE

where Assume furthermore that a^ij, bⁱ, c ∈ C^∞(Ω) for i, j = 1,2, …, n, and f ∈ C^∞(Ω). Then u ∈ C^∞(Ω).

Theorem 27 (see [24], page 12.)Let Ω ⊂ ℝⁿ be a bounded domain with C^k+1,1 boundary ∂Ω for some nonnegative integer k. Suppose the data f and g of the problem

are in W^k,p(Ω) and W^k+2−1/p,p(∂Ω), respectively, for some real number p with 1 < p < ∞. Then u ∈ W^k+2,p(Ω).

Theorem 28 (see [22], page 84.)Let Ω ⊂ ℝⁿ be a bounded domain with C^1,1 boundary Γ and 1 < p < ∞. Consider the Neumann problem

If 0 < a₀ ∈ L^∞(Ω), f ∈ L^p(Ω), and g ∈ W^1−1/p,p(∂Ω), then the weak solution u to (116) exists in W^2,p(Ω).

Theorem 29. Let Ω be a bounded domain with boundary of class C^1,1. Let u_D, u_N ∈ H¹(Ω) be weak solutions of the BVPs (4) and (6), respectively. Then u_D and u_N also belong to H²(Ω). More generally, if Ω is of class C^k+1,1, where k is a nonnegative integer, then u_D and u_N are elements of H^k+2(Ω).

Proof. We first consider the solution u_D ∈ H¹(Ω) to the Dirichlet problem (4). We use Theorem 25 to show that u_D is an element of H²(Ω). Here, (110) is applied with the following settings.

We consider n = 2. The domain Ω is of class C^1,1. L = −Δ, and hence a_ij = a_ji = −1 for i = j and a_ij = a_ji = 0 for i ≠ j, with i, j = 1,2. We also observe that $$, for all ζ = (ζ₁, ζ₂) ^T. Thus α = 1. Furthermore, we have the following data: f = 0 ∈ L²(Ω),  g = 1 ∈ H^3/2(Γ), g = 0 ∈ H^3/2(Σ). Therefore, by using Theorem 24, there exists a unique u = u_D ∈ H²(Ω), which is a solution to (4).

For higher regularity of u_D we apply Theorem 25. At first we consider C^2,1-domains. In this case, k = 1. We have a_ij = a_ji = −1 for i = j and a_ij = a_ji = 0 for i ≠ j, i, j = 1,2. The operator B is the identity operator, thus of order d = 0. From the first consequence, it is known that u ∈ H²(Ω) satisfies −Δu = 0 and u = g on ∂Ω, where g = 1 ∈ H^5/2(Γ), g = 0 ∈ H^5/2(Σ). Therefore, by applying Theorem 25, we have u = u_D ∈ H³(Ω). In general, for smoother domains with C^k+1,1 boundaries, solutions to (4) are elements of H^k+2.

Next, we recall that, for C^1,1 domain, there is a weak solution u_N ∈ H¹(Ω) to the boundary value problem (6). We also show that the solution actually lies in H²(Ω) and if the domain is more regular, then so is the solution. More precisely, we want to show that if Ω is a domain whose boundary is of class C^k+1,1, then u_N is in H^k+2(Ω), where k is a nonnegative integer. For this purpose we need Theorem 26 which implies u_N ∈ C^∞(Ω).

Choose a bounded connected domain G with C^∞ boundary Γ₁ such that $$ and $$, where A and B are the domains described in Section 2. Let Ω₁ be the annulus having boundaries Γ and Γ₁, and let Ω₂ be the other annulus with boundaries Γ₁ and Σ. First, we consider the following elliptic problem on Ω₁:

Since Ω₁ is bounded with compact boundaries, we have C^∞(Γ ∪ Γ₁) ⊂ H^k+3/2(Γ ∪ Γ₁). So by applying Theorem 27, we get w ∈ H^k+2(Ω₁). Since $$ also solves (117), then by uniqueness we have $$. If A is a domain with C^1,1 boundary Γ, then by Theorem 27 we have $$. Moreover, if A is a C^k+1,1-domain, then $$ is in H^k+2(Ω₁).

Second, we consider the following boundary value problem:

Because $$, it follows that $$. We have also shown that u_N ∈ C^∞(Ω). This implies that u_N ∈ C^∞(Γ₁). Hence, (∂u_N/∂n) ∈ H^1/2(Γ₁). Also, λ ∈ H^1/2(Σ). Since B is a domain of class C^1,1, then by Theorem 28, we infer that (118) has a unique solution v ∈ H²(Ω₂). Note, however, that $$ also solves (118). So by uniqueness of the solution, we get $$. Therefore, $$. Now, if domain B is of class C^k+1,1 and $$, we get v ∈ H³(Ω₂) by applying Theorem 25 and so is $$. Doing this recursively, we end up with $$.

Hence, for C^k+1,1-domains $$, if we combine $$ and $$, we get u_N ∈ H^k+2(Ω₁ ∪ Ω₂). Moreover, u_N is C^∞ in a neighborhood of Γ₁ because u_N ∈ C^∞(Ω). Therefore, u_N ∈ H^k+2(Ω).

Remark 30. In the computation of the first-order shape derivative, since we are dealing with C^1,1-domains, we may consider H²-regularity for the solutions u_D and u_N, as justified by Theorem 29.

Lemma 31. Let Ω ⊂ ℝⁿ be a bounded Lipschitz domain. Then the following equation:

is valid for vector field F and scalar function g having the following regularity:

• (i)

  F ∈ H¹(Ω; ℝⁿ) and g ∈ H¹(Ω);

• (ii)

  $$ and g ∈ W^1,1(Ω).

Proof. First we recall the Gauss’ divergence theorem in ℝⁿ saying that if a domain Ω ⊂ ℝⁿ is a bounded Lipschitz domain, then we have

for a vector field $$. Second we take the divergence of the product of a scalar function g and the vector field F to get Then, integrating both sides over Ω and applying the divergence theorem to the vector field gF we obtain (119).

(i) If F ∈ H¹(Ω; ℝⁿ) and g ∈ H¹(Ω), then (div F)g ∈ L¹(Ω),  F · ∇g ∈ L¹(Ω), and the integral ∫_∂Ω g(F · n) is bounded. Hence, (119) is well defined. Note that the formula

holds for $$ and $$. We write where γ₀ : H¹(Ω) → L²(∂Ω) is a trace operator. Let F ∈ H¹(Ω; ℝⁿ) and g ∈ H¹(Ω). By density, we pick $$ and $$ such that F_k → F in $$ and g_k → g in H¹(Ω). By (122), Note that γ₀F_k = F_k|_∂Ω and γ₀g_k = g_k|_∂Ω. Also F_k → F in $$ implies that div F_k → div F in L²(Ω). Moreover, g_k → g in H¹(Ω) implies ∇g_k → ∇g in L²(Ω). Furthermore, since γ₀ ∈ ℒ(H¹(Ω), L²(∂Ω)), γ₀F_k → γ₀F and γ₀g_k → γ₀g in L²(∂Ω). Therefore, (119) holds for F ∈ H¹(Ω; ℝⁿ) and g ∈ H¹(Ω).

(ii) If $$ and g ∈ W^1,1(Ω), then (div F)g ∈ L¹(Ω) and F · ∇g ∈ L¹(Ω). Note that γg ∈ L¹(∂Ω) (cf. [25, page 316]), where γg refers to the trace of g on ∂Ω, hence g(F · n) ∈ L¹(∂Ω). Therefore, (119) is also well defined for this case. Using similar arguments as above and using the density of $$ in W^1,1(Ω) we can show that (119) is also valid for this case.

Lemma 32 (see [1].)Let u_D and u_N belonging to H²(Ω) satisfy the Dirichlet problem (4) and the Neumann problem (6), respectively. Then

where A is given by property (12) of Lemma 11.

Proof. From Lemma 11, we recall the expression for A, which is given by A = (div V)I − ((DV)+(DV) ^T). Our first goal is to derive an expression for ∫_Ω A∇u · ∇v for u, v ∈ H²(Ω). We begin by writing ∫_Ω A∇u · ∇v as follows:

We manipulate each term on the right-hand side of (127). First, because u, v ∈ H²(Ω), we have ∇u · ∇v ∈ W^1,1(Ω). Hence we can use (119) by taking g = ∇u · ∇v and by choosing F = V. In addition, we take into account that V vanishes on the fixed boundary Γ. This leads to The other two terms on the right hand side of (127) are manipulated as follows. The term ∇(∇u · ∇v) · V is written as where ∇²u represents the Hessian of u. Because Hessian is symmetric, we obtain Substituting (130) into (128) we get Next, we expand the expression div  ((V · ∇u)∇v) using (121) as

Integrating both sides of (132) over Ω, applying Stoke’s theorem, and considering V = 0 on Γ we end up with

or equivalently Interchanging v and u we get Also, because (DV) ^T∇v · ∇u = (DV)∇u · ∇v, we obtain Thus, Adding (131), (135), and (137) altogether, we express (127) as Set u = v = u_D in (138). The first two integrals on the right hand side of (138) vanish because −Δu_D = 0 in Ω. Moreover, since u_D = 0 on Σ we have ∇u_D = (∂u_D/∂n)n. Thus, we can write (138) as follows: Therefore, (125) is satisfied.

On the other hand, by replacing both u and v by u_N and by considering that −Δu_N = 0 in Ω and ∂u_N/∂n = λ on Σ, we derive (126) as

Theorem 33. For C^1,1 bounded domain Ω, the first-order shape derivative of the Kohn-Vogelius cost functional

in the direction of a perturbation field V ∈ Θ, where Θ is defined by (13) and the state functions u_D and u_N satisfy the Dirichlet problem (4) and the Neumann problem (6), respectively, is given by where n is the unit exterior normal vector to Σ, τ is a unit tangent vector to Σ, and κ is the mean curvature of Σ.

Proof. First we consider the functionals defined on the reference domain and perturbed domains

Let z_t = u_D,t − u_N,t and z = u_D − u_N. Note that if z^t = z_t∘T_t, then we have Dz^t = (Dz_t∘T_t)(DT_t). Hence By this together with Lemma 16, we can write J(Ω_t) as follows: Then we write J(Ω_t) − J(Ω) as Using Lemma 11, the symmetry of A, and noting that z^t → z as t → 0 we have Hence, To manipulate J₂(t) we use the identity a² − b² = (a − b) ² + 2b(a − b) as J₂₁(t) is manipulated as follows: Applying Theorems 22 and 23 yields $$. J₂₂(t) is treated as follows: Since $$, the variational equation for u_D and u_N implies Choosing $$ as test function and applying (65) and (84), J₂₂(t) can be written as From this we obtain and therefore, Combining $$ and $$ we get

We know from the previous section that u_D and u_N exist in H²(Ω) since Ω is of class C^1,1. Using this smoothness we can now apply Lemma 32 and write (156) as follows:

Since and |∇u_N|² = λ² + (∇u_N · τ) ², we obtain the first-order shape derivative of J: