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

and Applied Analysis 3 smoothness of these domains can be defined in the following sense (cf. [16]). Consider the standard unit orthonormal basis {e 1 , e 2 , . . . , e n } in R. For a point x = (x 1 , x 2 , . . . , x n ) ∈ R, let x = (x 1 , x 2 , . . . , x n−1 ) ∈ R so as to write x = (x, x n ). Consider the unit ball B(0, 1) and introduce the subsets B + (0, 1) := {x ∈ B (0, 1) : n > 0} , B − (0, 1) := {x ∈ B (0, 1) : n < 0} , B 0 (0, 1) := {x ∈ B (0, 1) : n = 0} . (8) Definition 1. A domain Ω ⊂ R with a nonempty boundary ∂Ω is called a C-domain, where 0 ≤ k, 0 < l ≤ 1, if for every y ∈ ∂Ω there exists a neighborhood N y of y and a C 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) = B 0 (0, 1), and (c) f y (N y ∩ Ω c ) = B − (0, 1). To illustrate this for n = 2 and k = l = 1, see Figure 2. Note that if Ω is a bounded, open, connected set with a C 0,1 boundary, then int Ω = Ω. This was given in [17] and we prove it as follows. Theorem 2 (see [17]). IfΩ is a bounded open connected subset of R with Lipschitz continuous boundary, then int Ω = Ω. Proof. The interior of Ω is the largest open set contained in the set Ω. Moreover, Ω ⊆ Ω. It follows that Ω ⊆ int Ω. Next, we show that int Ω ⊆ Ω. Clearly, int Ω ⊆ Ω. We now show that if x ∈ int Ω, then x ∉ ∂Ω. Suppose x ∈ ∂Ω and x ∈ int Ω. We need to show that any open set N containing x contains an element not in Ω. We first note that by definition of C 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 f x (z) ∈ N ∩ Ω c. Thus, N contains an element not in Ω, which is a contradiction. Therefore, x ∉ int Ω. We have proven that if x ∈ ∂Ω, then x ∉ int Ω. Taking the contrapositive of this statement we get that if x ∈ int Ω, then x ∉ ∂Ω. Since x ∈ Ω but x ∉ ∂Ω, we conclude x ∈ Ω. Thus, int Ω ⊆ Ω. We have shown that Ω ⊆ int Ω and int Ω ⊆ Ω. Therefore, int Ω = Ω. 3.2. The Perturbation of Identity Technique. Given bounded connected domains Ω and U of R, where Ω ⊆ U, and a linear space Θ of vector fields V, one can deform Ω via the perturbation of identity operator T t : U 󳨀→ R 2 , T t ( x) = x + tV (x) , x ∈ U, (9) where V ∈ Θ. For a given t we denote the deformed domain to be Ω t , which is the image of Ω under T t . Throughout the paper, we use the usual infinity norms in the spaces C(X;R), C(X;R), and C(X;R × ), where X is a y


Introduction
The Bernoulli problem is the prototype of a stationary free boundary problem.It arises in various applications such as electrochemical machining, potential flow in fluid mechanics, tumor growth, optimal insulation, molecular diffusion, and steel and glass production [1][2][3][4][5][6].A characteristic feature of free boundary problems is that not only the state variable is unknown but also the domain on which the state equation is posed.This represents a significant theoretical as well as numerical challenge.One can characterize the Bernoulli problem, at least along general lines, by finding a connected domain as well as a function which is harmonic on this domain.One component on the boundary is known.The other one is determined by a set of overdetermined boundary conditions (a Dirichlet condition and a Neumann condition) for the state.If the free boundary component is strictly exterior to the fixed part of the boundary, the problem is called exterior Bernoulli problem and interior Bernoulli problem otherwise.For more discussions related to interior and exterior Bernoulli problems, we refer the reader to [1,4,[7][8][9][10].
Recent strategies to compute a numerical solution are based on reformulating the Bernoulli problem as a shape optimization problem.This can be achieved in several ways.For a given domain, one can choose one of the boundary conditions on the free boundary to obtain a well-posed state equation.The domain is determined by the requirement that the other condition on the free boundary is satisfied in a least squares sense (cf.[11][12][13]).Alternatively, one can compute on a given domain two auxiliary states:   which satisfies the Dirichlet condition and   which satisfies the Neumann condition on the free boundary.The underlying domain is selected such that the difference (Ω) = |  −   | 2  1 (Ω) is as small as possible.In fact, if (Ω) = 0 for a domain Ω then   =   and (  , Ω) is a solution of the Bernoulli problem.Sometimes  is called Kohn-Vogelius functional since Kohn and Vogelius were among the first who used such a functional in the context of inverse problems [14].Standard algorithms to minimize  require some gradient information.So in this paper, the first-order sensitivity analysis is carried out for the functional  for the exterior Bernoulli problem.The main contribution in this paper is the application of a shape optimization technique that leads to the explicit expression for the shape derivative of the cost functional.This is done through variational means similar to the techniques developed in [9,10,13], wherein we use the Hölder continuity of the state variables satisfying the Dirichlet and Neumann problems but we do not introduce any adjoint variables.In our approach, we also bypass the use of the material derivatives of the states (which was done in [1]) and the use of states' shape derivatives.
The rest of the paper is structured as follows.Section 2 presents the Bernoulli free boundary problem and its shape optimization formulations.Section 3 provides a list of shape optimization tools that are needed in the analysis for the shape derivatives of the Kohn-Vogelius cost functional .Section 4 presents an exhaustive discussion on the first-order shape derivative of .Finally, Section 5 draws conclusion and observation.

The Bernoulli Problem
The exterior Bernoulli free boundary problem is formulated as follows.Given a bounded and connected domain  ⊂ R 2 with a fixed boundary Γ :=  and a constant  < 0, one needs to find a bounded connected domain  ⊂ R 2 with a free boundary Σ, containing the closure of , and an associated state function  : Ω → R, where Ω =  \ , such that the overdetermined conditions are satisfied: On the other hand, the interior Bernoulli free boundary problem has the following formulation.Given a bounded and connected domain  ⊂ R 2 with a fixed boundary Γ :=  and a constant  > 0, one determines a bounded connected domain  ⊂  with a free boundary Σ and an associated state function  : Ω → R, where Ω =  \ , subject to the following constraints: In both problems n is the outward unit normal vector to Σ.
The difference in the domains of these two types of Bernoulli problems is depicted in Figure 1.Methods of shape optimization can be employed in solving the exterior Bernoulli free boundary problem (1).As we observe, this boundary problem is ill-posed due to the fact that we have overdetermined conditions on the free boundary Σ.So to overcome the difficulty of solving it, one can reformulate it as one of the following shape optimization problems which involves now a well-posed state equation.(1) Tracking Neumann data [11,12] as where the state function   is the solution to the Dirichlet problem (2) Tracking Dirichlet data [11,13] as where the state function   is the solution to the Neumann problem (3) Minimizing the Kohn-Vogelius type cost functional [12,15] as where state functions   and   satisfy ( 4) and ( 6), respectively.
In this paper, we are just interested in the study of minimizing the Kohn-Vogelius functional .

Tools in Shape Optimization
3.1.Feasible Domain Ω.In this work, we are interested in  ,1 -domains, where  ≥ 0. Aside from being  ,1 we also assume that these are bounded and connected subsets of a bigger set  which is also a bounded connected  ,1 domain.This  is called the universal or the hold-all domain.The smoothness of these domains can be defined in the following sense (cf.[16]).
To illustrate this for  = 2 and  =  = 1, see Figure 2. Note that if Ω is a bounded, open, connected set with a  0,1 boundary, then int Ω = Ω.This was given in [17] and we prove it as follows.
Proof.The interior of Ω is the largest open set contained in the set Ω.Moreover, Ω ⊆ Ω.It follows that Ω ⊆ int Ω. Next, we show that int Ω ⊆ Ω.Clearly, int Ω ⊆ Ω.We now show that if  ∈ int Ω, then  ∉ Ω.
Suppose  ∈ Ω and  ∈ int Ω.We need to show that any open set  containing  contains an element not in Ω.We first note that by definition of  0,1 domain, there exists a neighborhood   of  ∈ Ω and a diffeomorphism   :   → (0, 1).Let  be an open set containing  with  ⊂   .It follows that   () is an open set containing 0 and this set is contained in (0, 1).Hence, there exists  ∈   () such that  ∈  − (0, 1).This implies that  −1  () ∈  ∩ Ω  .Thus,  contains an element not in Ω, which is a contradiction.Therefore,  ∉ int Ω.We have proven that if  ∈ Ω, then  ∉ int Ω. Taking the contrapositive of this statement we get that if  ∈ int Ω, then  ∉ Ω.Since  ∈ Ω but  ∉ Ω, we conclude  ∈ Ω.Thus, int Ω ⊆ Ω.We have shown that Ω ⊆ int Ω and int Ω ⊆ Ω.Therefore, int Ω = Ω.

The Perturbation of Identity Technique.
Given bounded connected domains Ω and  of R 2 , where Ω ⊆ , and a linear space Θ of vector fields V, one can deform Ω via the perturbation of identity operator where V ∈ Θ.For a given  we denote the deformed domain to be Ω  , which is the image of Ω under   .
Throughout the paper, we use the usual infinity norms in the spaces (; R), (; R 2 ), and (; R 2 × 2 ), where  is a compact subset of R 2 .In addition to this, we also denote the Frobenius norm of () to be This norm and the infinity norm of the matrix  can be related as This can be shown easily.One can also show that if  ∈ (; R 2 × 2 ) and  ∈  2 (; R 2 ), then the vector  is bounded in  2 (; R 2 ).In fact,          2 () ≤ 2|| ∞          2 () (12) and the proof is trivial.Finally, the symbols | ⋅ | or | ⋅ | 2 will refer to the usual Euclidean norm.
The Perturbed Domain Ω  .The domains Ω  that are considered in this work are of annulus type with boundary Ω  , which is the union of two disjoint sets Γ  and Σ  , referred to as the fixed and free boundaries, respectively.These domains are obtained through the operator defined in (9), where V belongs to Θ, which is defined as For  = 0, we obtain the reference domain Ω := Ω 0 , with a fixed boundary Γ := Γ 0 and a free boundary Σ := Σ 0 .The main objective in this subsection is to show that   is a diffeomorphism from Ω to Ω  for sufficiently small .To verify this, we need the following results, which are given and proven in [17].
where  0 is injective, ,  is a bijection,  is continuous, and  −1 is continuous), We also consider the following property of a domain, which is also found in [17,   We also recall the useful property of the determinant of the Jacobian of   which is given in the next lemma.Here we use the notation Lemma 6 (see [9,13]).Consider the operator   defined by (9), where V ∈ Θ, which is described by (13).
Considering the theorems and lemmas presented beforehand, we are now ready to prove the following theorem.

Theorem 7.
Let Ω and  be nonempty bounded open connected subsets of R 2 with Lipschitz continuous boundaries, such that Ω ⊆ , and Ω is the union of two disjoint boundaries Γ and Σ.Let   be defined as in (9) where V belongs to Θ, defined as (13).
Then for sufficiently small , (1)   :  →  is a homeomorphism, (2)   :  →  is a  1,1 diffeomorphism, and in particular, Proof.First, because  is a  0,1 domain, it follows that int  =  by Theorem 2. Second,  0 =  ∈ (, R 2 ), and it is injective.Third, it is evident that For  ∈ ,   () =  because V vanishes on .For  ∈ , the determinant of the Jacobian of the perturbation of identity operator   is given by (19).By Lemma 6, there exists a   > 0, given by (20), such that det   () > 0 for all  ∈  and for || ≤   .Hence, by applying Theorem 4, we conclude that   () =  and   () =  for all || <   , and   :  →  is a homeomorphism.Furthermore, by Theorem 4, we find that   :  →  is a  1 diffeomorphism.To show that   is a  1,1 diffeomorphism, we are left to show that  −1  is Lipschitz continuous.To verify this we use Lemma 5.
Given any two points , V ∈  we choose {  } such that properties (a)-(c) of Lemma 5 are satisfied.For fixed || <   , differentiating the identities   ∘  −1  =  and  −1  ∘   =  will lead to for all  ∈ .Thus, This implies Applying the infinity norm in the space (; R 2 ) we have Since   is Lipschitz continuous, we have where  1 is the maximum of all Lipschitz constants of   for all || <   .Then finally, using the mean value theorem and property (c) in Lemma 5, we obtain Hence  −1  is Lipschitz continuous which shows that   :  →  is a  1,1 diffeomorphism for sufficiently small ||.2) is clear because the fixed boundary is invariant under   ; that is, Γ  :=   (Γ) = Γ since V vanishes on Γ. Lastly, using Theorem 3, definition of   , (1), and (2), we obtain (3).
Remark 9. Theorem 7 and Corollary 8 tell us that the reference Ω and the perturbed domain Ω  have the same topological structure and regularity under the perturbation of identity operator   for sufficiently small .See Figure 3 for illustration.
Properties of   .In addition to (16) we also use the following notations throughout the work: Remark 10.We note the following observations for fixed, sufficiently small .
Proof.(3) Suppose  ∈ , , ℎ ∈   , and Using Lemma 5, we connect  ℎ () and   () by a chain   ,  = 1, . . .,  + 1, satisfying and its norm is estimated as follows: This shows uniform convergence in  ∈   and  ∈ .Hence, for every  > 0 one can choose a  := /()|V| ∞ > 0 which implies that, for every , ℎ ∈   , To show that   →  −1  is continuous from   to  1 (, R 2 ), we only need to show that for every  ∈   , Using the definition of Jacobian of a transformation and the regularity of V, we further simplify (38) as follows: where  is the Lipshitz constant for V and  is upper bound for |V| ∞ .Taking the maximum of both sides of the inequality for all  ∈  and using (34) we get where  =  2  +  2   ()|V| ∞ .Thus, for any  > 0, we choose  = /, so that if 0 < |ℎ − | < , then Proof of property (8) in Lemma 11 is as follows.Given  ∈ , we have   ( −1  ()) = .This implies that Manipulating the left hand side of (41), we get We first work on .Applying the definition of   , we get Thus, Similarly, we can write  as follows: Hence, we have Suppose V is a coordinate function of V.By the mean value theorem, we observe that where  is a point on the segment joining  −1  () and  −1 +ℎ (), and as ℎ tends to infinity, (47) tends to (V)(/) −1  ().Thus, Combining ( 44) and (48), we get which implies that Evaluating (50) at  = 0, we get (/) −1  | =0 = −V.

The Method of Mapping.
If  is defined in Ω and   is defined in Ω  , then the direct comparison of   with  is generally not possible since the functions are defined on different domains.To overcome this difficulty, one maps   back to Ω by composing it with   ; that is, one defines   ∘   : Ω → R. With this new mapping one can define the material and the shape derivatives of states, the domain and boundary integral transformations, and derivatives of integrals, as well as the Eulerian derivative of the shape functional.This technique is called the method of mapping.
Material and Shape Derivatives.The material and shape derivatives of state variables are defined as follows [20,21].
Remark 13.The material derivative can be written as It characterizes the behavior of the function  at  ∈ Ω ⊂  in the direction V().
Definition 14.Let  be defined in [0,   ] × .An element   ∈   (Ω) is called the shape derivative of  at Ω in the direction V, if the following limit exists in   (Ω): () :=   (0, ) Remark 15.The shape derivative of  is also defined as follows: We note that if u and ∇ ⋅  exist in   (Ω), then the shape derivative can be written as In general, if u () and ∇ ⋅ V() both exist in  , (Ω), then   () also exists in that space.
Domain and Boundary Differentiation.We recall some results concerning the derivative of integrals with respect to the domain of integration.For the first theorem, it is sufficient to have  0,1 domains while the second theorem requires  1,1 domains.For proofs, see [18].
The First-Order Eulerian Derivative Definition 19.The Eulerian derivative of the shape functional  : 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.  is said to be shape differentiable at Ω if (Ω; V) exists for all V ∈ Θ and is linear and continuous with respect to V.

Main Result
In this section we derive in a rigorous manner the first-order shape derivative of the Kohn-Vogelius functional , defined by (7), subject to the Dirichlet and Neumann boundary value problems (BVPs) ( 4) and ( 6), respectively.Our strategy bypasses the material or shape derivatives of states.In the derivation, we have employed techniques used in [9,10,13] but there is no need to use adjoint variables.This section discusses the variational forms of the PDEs, the state variables in the perturbed domains, the Hölder continuity of the state variables, and the higher regularity of the solutions to the BVPs.The rest of the proof is presented in the last part of this section.

Variational Forms of the Dirichlet and Neumann Problems.
We recall that we are considering the shape optimization problem (7) where   solves the pure Dirichlet problem (4) and   solves the Neumann problem (6).As in [13], we consider the Hilbert space which is endowed with the norm and a linear manifold defined by First, we determine the variational equations for the Dirichlet and the Neumann problems.The variational form of the Dirichlet problem ( 4) is given by the following.
Find   ∈  1 (Ω) such that Equation (64) can be shown to have a unique solution using Theorem 2.4.2.5 of [22].Similarly, the variational form of the Neumann problem ( 6) is formulated as follows.
Find   ∈  1 (Ω) such that It is also well known that (65) has a unique solution.

Analysis of State Variables in Deformed Domains
. We now consider the class of perturbed problems: min where  , solves the pure Dirichlet problem (68) Here, n  is the outward unit normal to the deformed free boundary Σ  .The variational form of (67) is formulated as follows.
Find  , ∈  1 (Ω  ) such that It is known that (69) has a unique solution.
Let    =   +   .Using (81) we obtain Thus ( 74) is satisfied.The boundary conditions are also satisfied because on Γ,   = 0 and   = 1 and on Σ, both   and   are zero.To show uniqueness, we let    and ũ  be solutions of (74).This implies that there exist   and ỹ such that    =   +   and ũ  = ỹ +   , where   and ỹ are solutions to (81).Taking the difference of    and ũ  and considering that solution to (81) is unique, we get    = ũ  .Next, we consider (68) whose variational form is formulated as follows.

Hölder Continuity of the States. We show that 𝑢 𝑡
and    are Hölder continuous on .
Proof.Subtracting (65) from ( 84) for all  ∈  1 Γ,0 (Ω) we get Hence Note that    −   belongs to Furthermore, by trace theorem we have holds, where  = max{1, }.This implies We now show that Consequently, and this shows that ∇   is uniformly bounded in  2 (Ω) because |  | ∞ is bounded.In addition,   and   are differentiable at  = 0 by Lemma 11.Therefore, 4.4.Higher Regularity of the Solutions.In this section we will show that the solutions to the PDEs ( 4) and ( 6) have higher regularity.We begin by considering the state variable   ,   ∈  1 (Ω).For  1,1 domains, we show that these solutions also exist in  2 (Ω) and more generally in  +2 (Ω) if domains are of class  +1,1 ,  ≥ 0.
To prove higher regularity of   , we require the following two theorems, which are proven in [22].
We will also justify the higher regularity of   .We use the following results whose proofs are given in the corresponding texts.
Using the theorems presented above, we will now prove our claim that the solutions to the PDEs ( 4) and ( 6) have indeed higher regularity.This result is given in the following theorem.
Proof.We first consider the solution   ∈  1 (Ω) to the Dirichlet problem (4).We use Theorem 25 to show that   is an element of  2 (Ω).Here, (110) is applied with the following settings.
Next, we recall that, for  1,1 domain, there is a weak solution   ∈  1 (Ω) to the boundary value problem (6).We also show that the solution actually lies in  2 (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  +1,1 , then   is in  +2 (Ω), where  is a nonnegative integer.For this purpose we need Theorem 26 which implies   ∈  ∞ (Ω).
Proof.First we recall the Gauss' divergence theorem in R  saying that if a domain Ω ⊂ R  is a bounded Lipschitz domain, then we have for a vector field F ∈  1 (Ω, R  ).Second we take the divergence of the product of a scalar function  and the vector field Then, integrating both sides over Ω and applying the divergence theorem to the vector field F we obtain (119).
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 ∇(∇ ⋅ ∇V) ⋅ V is written as where ∇ 2  represents the Hessian of .Because Hessian is symmetric, we obtain Substituting (130) into (128) we get Next, we expand the expression div ((V ⋅ ∇)∇V) using (121) as div Integrating both sides of (132) over Ω, applying Stoke's theorem, and considering V = 0 on Γ we end up with Interchanging V and  we get Also, because (V)  ∇V ⋅ ∇ = (V)∇ ⋅ ∇V, we obtain Thus, Adding (131), (135), and (137) altogether, we express (127) as Set  = V =   in (138).The first two integrals on the right hand side of (138) vanish because −Δ  = 0 in Ω.Moreover, since   = 0 on Σ we have ∇  = (  /n)n.Thus, we can write (138) as follows: Therefore, (125) is satisfied.
On the other hand, by replacing both  and V by   and by considering that −Δ  = 0 in Ω and   /n =  on Σ, we derive (126) as Now, we derive the explicit form of the first-order shape derivative of .
Theorem 33.For  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   and   satisfy the Dirichlet problem (4) and the Neumann problem (6), respectively, is given by where n is the unit exterior normal vector to Σ,  is a unit tangent vector to Σ, and  is the mean curvature of Σ.
Proof.First we consider the functionals defined on the reference domain and perturbed domains   (156) We know from the previous section that   and   exist in  2 (Ω) since Ω is of class  1,1 .Using this smoothness we can now apply Lemma 32 and write (156) as follows: (159)

Conclusion
In this paper we derived the explicit form of the firstorder Eulerian shape derivative of the Kohn-Vogelius cost functional given by (7) in a rigorous manner.As seen in the presentation, we can avoid working on the shape derivatives of the states and apply their Hölder continuity instead.We employed techniques similar to [9,13] but it was not necessary to introduce adjoint variables.For the shape derivative of the cost functional to be well defined we observe that we can consider domains with

Figure 1 :
Figure 1: The domain Ω for the interior Bernoulli problem (a) and exterior problem (b).