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

Authors: Jerico B. Bacani; Gunther Peichl
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The exterior Bernoulli free boundary problem is being considered. The solution to the problem is studied via shape optimization techniques. The goal is to determine a domain having a specific regularity that gives a minimum value for the Kohn-Vogelius-type cost functional while simultaneously solving two PDE constraints: a pure Dirichlet boundary value problem and a Neumann boundary value problem. This paper focuses on the rigorous computation of the first-order shape derivative of the cost functional using the Hölder continuity of the state variables and not the usual approach which uses the shape derivatives of states.


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## Body

## 1. 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 [1–6]. 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, 7–10].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. [11–13]). Alternatively, one can compute on a given domain two auxiliary states: uD which satisfies the Dirichlet condition and uN which satisfies the Neumann condition on the free boundary. The underlying domain is selected such that the difference J(Ω)=|uD-uN|H1(Ω)2 is as small as possible. In fact, if J(Ω)=0 for a domain Ω then uD=uN and (uD,Ω) is a solution of the Bernoulli problem. Sometimes J 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 J require some gradient information. So in this paper, the first-order sensitivity analysis is carried out for the functional J 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. Section2 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 J. Section 4 presents an exhaustive discussion on the first-order shape derivative of J. Finally, Section 5 draws conclusion and observation.


## 2. The Bernoulli Problem

Theexterior Bernoulli free boundary problem is formulated as follows. Given a bounded and connected domain A⊂ℝ2 with a fixed boundary Γ:=∂A and a constant λ<0, one needs to find a bounded connected domain B⊂ℝ2 with a free boundary Σ, containing the closure of A, and an associated state function u:Ω→ℝ, where Ω=B∖A-, such that the overdetermined conditions are satisfied:
(1)-Δu=0inΩ,u=1onΓ,u=0onΣ,∂u∂n=λonΣ.On the other hand, theinterior Bernoulli free boundary problem has the following formulation. Given a bounded and connected domain B⊂ℝ2 with a fixed boundary Γ:=∂B and a constant λ>0, one determines a bounded connected domain A⊂B- with a free boundary Σ and an associated state function u:Ω→ℝ, where Ω=B∖A-, subject to the following constraints:
(2)-Δu=0inΩ,u=0onΓ,u=1onΣ,∂u∂n=λonΣ.
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.The domainΩ for the interior Bernoulli problem (a) and exterior problem (b).
(a)
(b)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
(3)minΣJ1(Σ)≡minΣ12∫Σ‍(∂uD∂n-λ)2ds,
where the state function uD is the solution to the Dirichlet problem
(4)-ΔuD=0inΩ,uD=1onΓ,uD=0onΣ.(2)
Tracking Dirichlet data [11, 13] as
(5)minΣJ2(Σ)≡minΣ12∫Σ‍uN2ds,
where the state function uN is the solution to the Neumann problem
(6)-ΔuN=0inΩ,uN=1onΓ,∂uN∂n=λonΣ.(3)
Minimizing the Kohn-Vogelius type cost functional [12, 15] as
(7)minΩJ(Ω)≡minΩ12∫Ω‍|∇(uD-uN)|2dx,
where state functions uD and uN satisfy (4) and (6), respectively.In this paper, we are just interested in the study of minimizing the Kohn-Vogelius functionalJ.


## 3. Tools in Shape Optimization

### 3.1. Feasible DomainΩ

In this work, we are interested inCk,1-domains, where k≥0. Aside from being Ck,1 we also assume that these are bounded and connected subsets of a bigger set U which is also a bounded connected Ck,1 domain. This U is called the universal or the hold-all domain. The smoothness of these domains can be defined in the following sense (cf. [16]).Consider the standard unit orthonormal basis{e1,e2,…,en} in ℝn. For a point x=(x1,x2,…,xn)∈ℝn, let x′=(x1,x2,…,xn-1)∈ℝn-1 so as to write x=(x′,xn). Consider the unit ball B(0,1) and introduce the subsets
(8)B+(0,1):={x∈B(0,1):xn>0},B-(0,1):={x∈B(0,1):xn<0},B0(0,1):={x∈B(0,1):xn=0}.Definition 1.
A domainΩ⊂ℝn with a nonempty boundary ∂Ω is called a Ck,l-domain, where 0≤k,  0<l≤1, if for every y∈∂Ω there exists a neighborhood Ny of y and a Ck,l diffeomorphism fy:Ny→B(0,1) such that (a)fy(Ny∩Ω)=B+(0,1), (b)fy(Ny∩∂Ω)=B0(0,1), and (c)fy(Ny∩Ω-c)=B-(0,1).To illustrate this forn=2 and k=l=1, see Figure 2.Figure 2
ACk,1-domain Ω, where fy is a diffeomorphism from the neighborhood Ny to the ball B(0,1).Note that ifΩ is a bounded, open, connected set with a C0,1 boundary, then intΩ-=Ω. This was given in [17] and we prove it as follows.Theorem 2 (see [17]).
IfΩ is a bounded open connected subset of ℝn with Lipschitz continuous boundary, then intΩ-=Ω.Proof.
The interior ofΩ- is the largest open set contained in the set Ω-. Moreover, Ω⊆Ω-. It follows that Ω⊆intΩ-. Next, we show that intΩ-⊆Ω. Clearly, intΩ-⊆Ω-. We now show that if x∈intΩ-, then x∉∂Ω.
Supposex∈∂Ω and x∈intΩ-. We need to show that any open set N containing x contains an element not in Ω-. We first note that by definition of C0,1 domain, there exists a neighborhood Nx of x∈∂Ω and a diffeomorphism fx:Nx→B(0,1). Let N be an open set containing x with N⊂Nx. It follows that fx(N) is an open set containing 0 and this set is contained in B(0,1). Hence, there exists z∈fx(N) such that z∈B-(0,1). This implies that fx-1(z)∈N∩Ω-c. Thus, N contains an element not in Ω-, which is a contradiction. Therefore, x∉intΩ-. We have proven that if x∈∂Ω, then x∉intΩ-. Taking the contrapositive of this statement we get that if x∈intΩ-, then x∉∂Ω. Since x∈Ω- but x∉∂Ω, we conclude x∈Ω. Thus, intΩ-⊆Ω. We have shown that Ω⊆intΩ- and intΩ-⊆Ω. Therefore, intΩ-=Ω.


### 3.2. The Perturbation of Identity Technique

Given bounded connected domainsΩ and U of ℝ2, where Ω⊆U, and a linear space Θ of vector fields V, one can deform Ω via the perturbation of identity operator
(9)Tt:U-⟶ℝ2,Tt(x)=x+tV(x),x∈U,-
where V∈Θ. For a given t we denote the deformed domain to be Ωt, which is the image of Ω under Tt.Throughout the paper, we use the usual infinity norms in the spacesC(X;ℝ), C(X;ℝ2), and C(X;ℝ2×2), where X is a compact subset of ℝ2. In addition to this, we also denote the Frobenius norm of M(x) to be
(10)|M(x)|F=(∑i,j=12‍|Mij(x)|2)1/2.
This norm and the infinity norm of the matrix M can be related as
(11)|M(x)|F≤2|M|∞,x∈X.
This can be shown easily. One can also show that if M∈C(X;ℝ2×2) and y∈L2(X;ℝ2), then the vector My is bounded in L2(X;ℝ2). In fact,
(12)|My|L2(X)≤2|M|∞|y|L2(X)
and the proof is trivial. Finally, the symbols |·| or |·|2 will refer to the usual Euclidean norm.The Perturbed Domain
Ω
t. The domains Ωt that are considered in this work are of annulus type with boundary ∂Ωt, which is the union of two disjoint sets Γt and Σt, 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
(13)Θ={V∈C1,1(U-,ℝ2):V|Γ∪∂U=0}.
For t=0, we obtain the reference domain Ω:=Ω0, with a fixed boundary Γ:=Γ0 and a free boundary Σ:=Σ0.The main objective in this subsection is to show thatTt is a diffeomorphism from Ω to Ωt for sufficiently small t. To verify this, we need the following results, which are given and proven in [17].Theorem 3.
IfΩ is a bounded, open, connected set in ℝn such that intΩ-=Ω and φ is a continuous injective mapping from Ω- to ℝn, then
(14)φ(Ω-)=φ(Ω)¯,φ(Ω)=intφ(Ω-),φ(∂Ω)=∂φ(Ω)=∂φ(Ω-).Theorem 4.
Suppose(1)
Ω is a bounded, open, connected set in ℝn such that intΩ-=Ω,(2)
φ0∈C(Ω-,ℝ2) where φ0 is injective,(3)
φ∈C(Ω-,ℝ2)∩C1(Ω,ℝ2) such that
(15)detDφ(x)>0∀x∈Ω,φ(x)=φ0(x)∀x∈∂Ω.Then (i)
φ:Ω-→φ(Ω-) 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), φ(Ω-)=φ0(Ω-).We also consider the following property of a domain, which is also found in [17, page 52].Lemma 5.
IfΩ 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 x,y∈Ω-, one can find a finite sequence of points yk, k=1,…,l+1, satisfying the following properties: (a)
y1=x, yk∈Ω for 2≤k≤l, yl+1=y,(b)
(yk,yk+1)⊂Ω for 1≤k≤l,(c)
∑k=1l‍|yk+1-yk|≤c(Ω)|x-y|.We also recall the useful property of the determinant of the Jacobian ofTt which is given in the next lemma. Here we use the notation
(16)It(x)=detDTt(x),x∈U-.Lemma 6 (see [9, 13]).
Consider the operatorTt defined by (9), where V∈Θ, which is described by (13). Then (i)
It=1+tdivV+t2detDV,(ii)
there existtV,α1,α2>0 such that 0<α1≤It(x)≤α2, for |t|≤tV, x∈U.Proof.
In general, forn-dimensional case, the Jacobian of Tt is given by DTt=(aij), where aij=t(∂Vi/∂xj) if, i≠j, and aii=1+t(∂Vi/∂xi). By definition of the determinant, we can write detDTt as
(17)detDTt=∑σ∈Sn‍sgn(σ)a1,σ(1)a2,σ(2)⋯an,σ(n)=∏i=1n‍(1+t∂Vi∂xi)hhhhh+∑σ∈Fn∖{i}‍sgn(σ)a1,σ(1)a2,σ(2)⋯an,σ(n)hhhhh+∑σ∈Sn∖Fn‍sgn(σ)a1,σ(1)a2,σ(2)⋯an,σ(n)=:I+II+III,
where Sn refers to the set of all permutations of {1,2,…,n}, i is the identity permutation, Fn={σ∈Sn:σ(k)=kforsomepositiveintegerk≤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+tdivV+t2R1(t,V), where R1∈C(ℝ,C0,1(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(∂Vi/∂xj), i≠j. Hence we can write II=t2R2(t,V), where R2 is in C(ℝ,C0,1(U)). All terms of III have factors of the form t(∂Vi/∂xj),  i≠j, and thus we have III=tnR-3(V), which can be written as t2R3(t,V), where R3∈C(ℝ,C0,1(U)). Combining I, II, and III, we get
(18)detDTt=1+tdivV+t2R(t,V)
with R∈C(ℝ,C0,1(U)). In particular, for n=2, the determinant is computed as follows:
(19)detDTt=1+tdivV+t2detDV.
This verifies(i). To show (ii) we first get the lower bound for It(x). Take
(20)tV=min{12(|divV|C(U-)+|detDV|C(U-)),1}.
For |t|≤tV, we obtain
(21)It(x)≥1-|t||divV|C(U-)-|t|2|detDV|C(U-)≥1-|t||divV|C(U-)-|t||detDV|C(U-)=1-|t|(|divV|C(U-)+|detDV|C(U-))≥12.
On the other hand, by triangle inequality we have
(22)It(x)≤1+|t||divV|C(U-)+|t|2|detDV|C(U-)≤1+tV|divV|C(U-)+tV2|detDV|C(U-)hhhhhhhhhhhhhhhhhhhhhhh∀x∈U-.
Hence, we have shown that there are positive constants α1=1/2 and α2=1+tV|divV|C(U-)+tV2|detDV|C(U-) such that α1≤It(x)≤α2 for x∈U-.Considering the theorems and lemmas presented beforehand, we are now ready to prove the following theorem.Theorem 7.
LetΩ and U be nonempty bounded open connected subsets of ℝ2 with Lipschitz continuous boundaries, such that Ω-⊆U, and ∂Ω is the union of two disjoint boundaries Γ and Σ. Let Tt be defined as in (9) where V belongs to Θ, defined as (13).
Then for sufficiently smallt, (1)
Tt:U-→U- is a homeomorphism,(2)
Tt:U→U is a C1,1 diffeomorphism, and in particular, Tt:Ω→Ωt is a C1,1 diffeomorphism,(3)
Γt=Tt(Γ)=Γ,(4)
∂Ωt=Γ∪Tt(Σ).Proof.
First, becauseU is a C0,1 domain, it follows that intU-=U by Theorem 2. Second, T0=I∈C(U-,ℝ2), and it is injective. Third, it is evident that Tt is C1,1 because V is C1,1. For x∈∂U, Tt(x)=x because V vanishes on ∂U. For x∈U, the determinant of the Jacobian of the perturbation of identity operator Tt is given by (19). By Lemma 6, there exists a tV>0, given by (20), such that detDTt(x)>0 for all x∈U and for |t|≤tV. Hence, by applying Theorem 4, we conclude that Tt(U)=U and Tt(U-)=U- for all |t|<tV, and Tt:U-→U- is a homeomorphism. Furthermore, by Theorem 4, we find that Tt:U→U is a C1 diffeomorphism. To show that Tt is a C1,1 diffeomorphism, we are left to show that DTt-1 is Lipschitz continuous. To verify this we use Lemma 5.
Given any two pointsu,v∈U we choose {yk} such that properties (a)–(c) of Lemma 5 are satisfied. For fixed |t|<tV, differentiating the identities Tt∘Tt-1=I and Tt-1∘Tt=I will lead to
(23)DTt(Tt-1(z))DTt-1(z)=I,DTt-1(z)DTt(Tt-1(z))=I,
for all z∈U. Thus,
(24)DTt(Tt-1(yk+1))DTt-1(yk+1)=I,DTt-1(yk)DTt(Tt-1(yk))=I.
This implies(25)|DTt-1(u)-DTt-1(v)|≤∑k=1l‍|DTt-1(yk)-DTt-1(yk+1)|=∑k=1l‍|DTt-1(yk)=∑k=1l‍|×[DTt(Tt-1(yk+1))-DTt(Tt-1(yk))]=∑k=1l‍|×DTt-1(yk+1)|.
Applying the infinity norm in the spaceC(X;ℝ2) we have
(26)|DTt-1(u)-DTt-1(v)|≤∑k=1l‍|DTt-1(yk)||DTt-1(yk+1)|hh×|DTt(Tt-1(yk+1))-DTt(Tt-1(yk))|≤maxy∈U-|DTt-1(y)|2×∑k=1l‍|DTt(Tt-1(yk+1))-DTt(Tt-1(yk))|.
SinceDTt is Lipschitz continuous, we have
(27)|DTt-1(u)-DTt-1(v)|=maxy∈U-|DTt-1(y)|2L1∑k=1l‍|Tt-1(yk+1)-Tt-1(yk)|,
where L1 is the maximum of all Lipschitz constants of DTt for all |t|<tV. Then finally, using the mean value theorem and property (c) in Lemma 5, we obtain
(28)|DTt-1(u)-DTt-1(v)|≤maxy∈U-|DTt-1(y)|3L1∑k=1l‍|yk+1-yk|≤maxy∈U-|DTt-1(y)|3L1c(U)|u-v|.
Hence DTt-1 is Lipschitz continuous which shows that Tt:U→U is a C1,1 diffeomorphism for sufficiently small |t|. Restricting to Ω, this proves that Tt:Ω→Ωt is a C1,1 diffeomorphism. (2) is clear because the fixed boundary is invariant under Tt; that is, Γt:=Tt(Γ)=Γ since V vanishes on Γ. Lastly, using Theorem 3, definition of Tt, (1), and (2), we obtain (3).Corollary 8.
LetΩ and U be two domains of ℝ2 with C1,1 boundary. Then for |t|<tV, where tV is given by (20), the perturbed domain Ωt:=Tt(Ω) is also of class C1,1.Proof.
Giveny∈∂Ωt, we let x=Tt-1(y)∈∂Ω. Then there exists a neighborhood Nx of x and a diffeomorphism gx∈C1,1(Nx,B(0,1)) such that gx(Nx∩Ω)=B+(0,1),  gx(Nx∩∂Ω)=B0(0,1), and gx(Nx∩Ωc)=B-(0,1). We have also shown that Tt defined in Theorem 7 is a C1,1 diffeomorphism. Since Tt-1 is continuous, Tt(Nx)=(Tt-1)-1(Nx)=Ny is a neighborhood of y in U. Define gy:=gx∘Tt-1. This is bijective because gx and Tt-1 are bijective. gy∈C1,1(Ny,B(0,1)) because Tt-1∈C1,1(U-,U-) (hence Tt-1∈C1,1(Ny,Nx)) and gx∈C1,1(Nx,B(0,1)). Also, gy-1=Tt∘gx-1∈C1,1(B(0,1),Ny).
Next, we note that∂Ωt∩Ny=Tt(∂Ω)∩Ny=Tt(∂Ω)∩Tt(Nx). Since Tt is injective, we have ∂Ωt∩Ny=Tt(∂Ω∩Nx). Thus by definition of gy we get
(29)gy(∂Ωt∩Ny)=gy(Tt(∂Ω∩Nx))=gx(∂Ω∩Nx)=B0(0,1).
We also observe the following:
(30)gy(Ωt∩Ny)=gy(Tt(Ω∩Nx))=gx(Ω∩Nx)=B+(0,1),gy(Ω-tc∩Ny)=gy(Tt(Ω-c∩Nx))=gx(Ω-c∩Nx)=B-(0,1).
This shows thatΩt is indeed of class C1,1.Remark 9.
Theorem7 and Corollary 8 tell us that the reference Ω and the perturbed domain Ωt have the same topological structure and regularity under the perturbation of identity operator Tt for sufficiently small t. See Figure 3 for illustration.Figure 3
The action ofTt on a C1,1-domain.Properties of
T
t. In addition to (16) we also use the following notations throughout the work:
(31)Mt(x)=(DTt(x))-T,x∈U-,At(x)=It(x)MtT(x)Mt(x),x∈U-,wt(x)=It(x)|(DTt(x))-Tn(x)|,x∈Σ.Remark 10.
We note the following observations for fixed, sufficiently smallt. (1)
It∈C0,1(U-).(2)
Mt,MtT∈C(U-;ℝ2×2).(3)
At∈C(U-;ℝ2×2).(4)
wt∈C(Σ;ℝ).(5)
V∈C1,1(U-;ℝ2) implies that |V|∞ and |DV|∞ are both finite.
We now provide several properties ofTt.Lemma 11 (see [9, 13, 16, 18]).
Consider the transformationTt, where the fixed vector field V belongs to Θ, defined in (13). Then there exists tV>0 such that Tt and the functions in (16) and (31) restricted to the interval IV=(-tV,tV) have the following regularity and properties.(1)
t↦Tt∈C1(IV,C1,1(U-,ℝ2)).(2)
t↦It∈C1(IV,C0,1(U-)).(3)
t↦Tt-1∈C(IV,C1(U-,ℝ2)).(4)
t↦wt∈C1(IV,C(Σ)).(5)
t↦At∈C(IV,C(U-,ℝ2×2)).(6)
There isβ>0 such that At(x)≥βI for x∈U.(7)
(d/dt)Tt|t=0=V.(8)
(d/dt)Tt-1|t=0=-V.(9)
(d/dt)DTt|t=0=DV.(10)
(d/dt)(DTt)-1|t=0=-DV.(11)
(d/dt)It|t=0=divV.(12)
(d/dt)At|t=0=(divV)I-(DV+(DV)T)≡A.(13)
limt→0wt=1.(14)
(d/dt)wt|t=0=divΣV,where the surface divergence divΣ is defined by
(32)divΣV=divV|Σ-(DVn)·n.We provide proofs for properties (3) and (8). The rest can be seen in [19].Proof.
(3) Supposex∈U-, t,h∈IV, and y=Th-1(x). Then
(33)Tt-1(x)-Th-1(x)=Tt-1(Th(y))-Th-1(Th(y))=Tt-1(Th(y))-Tt-1(Tt(y)).
Using Lemma 5, we connect Th(y) and Tt(y) by a chain yk, k=1,…,l+1, satisfying (i)
y1=Tt(y),   yk∈U for 2≤k≤l,   yl+1=Th(y),(ii)
(yk,yk+1)⊂U for 1≤k≤l,(iii)
∑k=1l‍|yk+1-yk|≤c(U)|Th(y)-Tt(y)|,and then we get
(34)|Tt-1(x)-Th-1(x)|≤∑k=1l‍|Tt-1(yk)-Tt-1(yk+1)|≤maxu∈U-|DTt-1(u)|∑k=1l‍|yk-yk+1|≤c(U)maxu∈U-|DTt-1(u)||Th(y)-Tt(y)|,≤c(U)maxu∈U-|DTt-1(u)||V(y)||t-h|.
Thus,
(35)maxx∈U-|Tt-1(x)-Th-1(x)|≤c(U)maxu∈U-|DTt-1(u)||V|∞|t-h|.
By reducing tV if necessary, we can assume without loss of generality that |tDV|∞<1 for t∈(-tV,tV). This allows us to represent (DTt)-1 as a Neumann series:
(36)DTt-1(x)=(DTt)-1∘Tt-1(x)=∑k=0∞‍(-t)k(DV)k(Tt-1(x)),
and its norm is estimated as follows:
(37)|DTt-1(x)|∞≤∑k=0∞‍|(-t)k(DV)k(Tt-1(x))|∞=11-|t||DV|∞<11-tV|DV|∞=:M.
This shows uniform convergence in t∈IV and x∈U-. Hence, for every ϵ>0 one can choose a δ:=ϵ/c(U)M|V|∞>0 which implies that, for every t,h∈IV, |Tt-1-Th-1|∞<ϵ whenever |t-h|<δ. In other words, t↦Tt-1∈C(IV,C(U-,ℝ2)).
To show thatt↦Tt-1 is continuous from IV to C1(U-,ℝ2), we only need to show that for every t∈IV, |DTt-1-DTh-1|∞<ϵ whenever |t-h|<δ and h∈IV. Let z∈U-. Using (37), estimate |DTt-1(z)-DTh-1(z)| as follows:
(38)|DTt-1(z)-DTh-1(z)|=|(DTt)-1(Tt-1(z))|hh-(DTh)-1(Th-1(z))|=|(DTt)-1(Tt-1(z))hhhh×((DTh)(Th-1(z))hhhhhhhhh-(DTt)(Tt-1(z)))hhhh×(DTh)-1(Th-1(z))|≤M2|(DTh)(Th-1(z))≤M2hh|-(DTt)(Tt-1(z))(DTh)(Th-1(z))|.
Using the definition of Jacobian of a transformation and the regularity of V, we further simplify (38) as follows:
(39)|DTt-1(z)-DTh-1(z)|≤M2|hDV(Th-1(z))-tDV(Tt-1(z))|≤M2|h-t||DV(Ts-1(z))|+M2|t||DV(Th-1(z))-DV(Tt-1(z))|≤M2K|h-t|+M2tVL|Th-1(z)-Tt-1(z)|,
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 z∈U- and using (34) we get
(40)|DTt-1-DTh-1|∞≤α|h-t|,
where α=M2K+M2tVLc(U)M|V|∞. Thus, for any ϵ>0, we choose δ=ϵ/α, so that if 0<|h-t|<δ, then |DTt-1-DTh-1|∞<ϵ. Therefore, t↦Tt-1∈C(IV,C1(U-,ℝ2)).
Proof of property(8) in Lemma 11 is as follows. Given x∈U-, we have Tt(Tt-1(x))=x. This implies that
(41)Tt+h(Tt+h-1(x))-Tt(Tt-1(x))=0.
Manipulating the left hand side of (41), we get
(42)(Tt+h(Tt+h-1(x))-Tt(Tt+h-1(x)))+(Tt(Tt+h-1(x))-Tt(Tt-1(x)))=:A+B.
We first work on A. Applying the definition of Tt, we get
(43)A=Tt+h-1(x)+(t+h)V(Tt+h-1(x))-Tt+h-1(x)-tV(Tt+h-1(x))=hV(Tt+h-1(x)).
Thus,
(44)limh→01hA=V(Tt-1(x)).
Similarly, we can write B as follows:
(45)B=Tt+h-1(x)-Tt-1(x)+tV(Tt+h-1(x))-tV(Tt-1(x)).
Hence, we have
(46)1hB=1h(Tt+h-1(x)-Tt-1(x))+t1h(V(Tt+h-1(x))-V(Tt-1(x))).
Suppose v is a coordinate function of V. By the mean value theorem, we observe that
(47)1h(v(Tt+h-1(x))-v(Tt-1(x)))=Dv(c)1h(Tt+h-1(x)-Tt-1(x)),
where c is a point on the segment joining Tt-1(x) and Tt+h-1(x), and as h tends to infinity, (47) tends to (Dv)(d/dt)Tt-1(x). Thus,
(48)limh→01hB=ddtTt-1(x)+tDV(Tt-1(x))ddtTt-1(x).
Combining (44) and (48), we get
(49)V(Tt-1(x))+(I+tDV(Tt-1(x)))ddtTt-1(x)=0,
which implies that
(50)ddtTt-1(x)=-(I+tDV(Tt-1(x)))-1V(Tt-1(x)).
Evaluating (50) at t=0, we get (d/dt)Tt-1|t=0=-V.


### 3.3. The Method of Mapping

Ifu is defined in Ω and ut is defined in Ωt, then the direct comparison of ut with u is generally not possible since the functions are defined on different domains. To overcome this difficulty, one maps ut back to Ω by composing it with Tt; that is, one defines ut∘Tt: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].Definition 12.
Letu be defined in [0,tV]×U. An element u˙∈Hk(Ω), called the material derivative of u, is defined as
(51)u˙(x):=u˙(0,x):=limt→0+u(t,Tt(x))-u(0,x)t=ddtu(t,x+tV(x))|t=0if the limit exists in (Hk(Ω)).Remark 13.
The material derivative can be written as(52)u˙(x)=limt→0+ut∘Tt(x)-u(x)t=ddt(ut∘Tt(x))|t=0.
It characterizes the behavior of the function u at x∈Ω⊂U in the direction V(x).Definition 14.
Letu be defined in [0,tV]×U. An element u′∈Hk(Ω) is called the shape derivative of u at Ω in the direction V, if the following limit exists in Hk(Ω):
(53)u′(x):=u′(0,x):=limt→0+u(t,x)-u(0,x)t.Remark 15.
The shape derivative ofu is also defined as follows:
(54)u′(x):=∂∂tut(x)|t=0.
We note that if u˙ and ∇u·V exist in Hk(Ω), then the shape derivative can be written as
(55)u′(x)=u˙(x)-(∇u·V)(x).
In general, if u˙(x) and ∇u·V(x) both exist in Wm,p(Ω), then u′(x) also exists in that space.
Domain and Boundary TransformationsLemma 16 (see [18]).
(
1
) Let φt∈L1(Ωt). Then φt∘Tt∈L1(Ω) and
(56)∫Ωt‍φtdxt=∫Ω‍φt∘TtItdx.
(
2
) Let φt∈L1(∂Ωt). Then φt∘Tt∈L1(∂Ω) and
(57)∫∂Ωt‍φtdst=∫∂Ω‍φt∘Ttwtds,
where It and wt are defined in (31).Proofs can be found in [13, 18].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 C0,1 domains while the second theorem requires C1,1 domains. For proofs, see [18].Theorem 17 (domain differentiation formula).
Letu∈C(IV,W1,1(U)) and suppose u˙(0,·):=(d/dt)u(t,Tt(·))|t=0 exists in L1(U). Then
(58)ddt∫Ωt‍u(t,x)dx|t=0=∫Ω‍u′(0,x)dx+∫Σ‍u(0,s)V·nds.Theorem 18 (boundary differentiation formula).
Letu be defined in a neighborhood of Γ. If u∈C(IV,W2,1(U)) and u˙(0,·)∈W1,1(U), then
(59)ddt∫Σt‍u(t,s)ds|t=0=∫Σ‍u′(0,s)ds+∫Σ‍(∂u∂n+u(0,s)κ)V·nds,
where κ is the mean curvature of the free boundary Σ.
The First-Order Eulerian DerivativeDefinition 19.
The Eulerian derivative of the shape functionalJ:Ω→ℝ defined in (7) at the domain Ω in the direction of the deformation field V∈Θ is given by
(60)dJ(Ω;V):=limt→0+J(Ωt)-J(Ω)t,
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.


## 3.1. Feasible DomainΩ

In this work, we are interested inCk,1-domains, where k≥0. Aside from being Ck,1 we also assume that these are bounded and connected subsets of a bigger set U which is also a bounded connected Ck,1 domain. This U is called the universal or the hold-all domain. The smoothness of these domains can be defined in the following sense (cf. [16]).Consider the standard unit orthonormal basis{e1,e2,…,en} in ℝn. For a point x=(x1,x2,…,xn)∈ℝn, let x′=(x1,x2,…,xn-1)∈ℝn-1 so as to write x=(x′,xn). Consider the unit ball B(0,1) and introduce the subsets
(8)B+(0,1):={x∈B(0,1):xn>0},B-(0,1):={x∈B(0,1):xn<0},B0(0,1):={x∈B(0,1):xn=0}.Definition 1.
A domainΩ⊂ℝn with a nonempty boundary ∂Ω is called a Ck,l-domain, where 0≤k,  0<l≤1, if for every y∈∂Ω there exists a neighborhood Ny of y and a Ck,l diffeomorphism fy:Ny→B(0,1) such that (a)fy(Ny∩Ω)=B+(0,1), (b)fy(Ny∩∂Ω)=B0(0,1), and (c)fy(Ny∩Ω-c)=B-(0,1).To illustrate this forn=2 and k=l=1, see Figure 2.Figure 2
ACk,1-domain Ω, where fy is a diffeomorphism from the neighborhood Ny to the ball B(0,1).Note that ifΩ is a bounded, open, connected set with a C0,1 boundary, then intΩ-=Ω. This was given in [17] and we prove it as follows.Theorem 2 (see [17]).
IfΩ is a bounded open connected subset of ℝn with Lipschitz continuous boundary, then intΩ-=Ω.Proof.
The interior ofΩ- is the largest open set contained in the set Ω-. Moreover, Ω⊆Ω-. It follows that Ω⊆intΩ-. Next, we show that intΩ-⊆Ω. Clearly, intΩ-⊆Ω-. We now show that if x∈intΩ-, then x∉∂Ω.
Supposex∈∂Ω and x∈intΩ-. We need to show that any open set N containing x contains an element not in Ω-. We first note that by definition of C0,1 domain, there exists a neighborhood Nx of x∈∂Ω and a diffeomorphism fx:Nx→B(0,1). Let N be an open set containing x with N⊂Nx. It follows that fx(N) is an open set containing 0 and this set is contained in B(0,1). Hence, there exists z∈fx(N) such that z∈B-(0,1). This implies that fx-1(z)∈N∩Ω-c. Thus, N contains an element not in Ω-, which is a contradiction. Therefore, x∉intΩ-. We have proven that if x∈∂Ω, then x∉intΩ-. Taking the contrapositive of this statement we get that if x∈intΩ-, then x∉∂Ω. Since x∈Ω- but x∉∂Ω, we conclude x∈Ω. Thus, intΩ-⊆Ω. We have shown that Ω⊆intΩ- and intΩ-⊆Ω. Therefore, intΩ-=Ω.


## 3.2. The Perturbation of Identity Technique

Given bounded connected domainsΩ and U of ℝ2, where Ω⊆U, and a linear space Θ of vector fields V, one can deform Ω via the perturbation of identity operator
(9)Tt:U-⟶ℝ2,Tt(x)=x+tV(x),x∈U,-
where V∈Θ. For a given t we denote the deformed domain to be Ωt, which is the image of Ω under Tt.Throughout the paper, we use the usual infinity norms in the spacesC(X;ℝ), C(X;ℝ2), and C(X;ℝ2×2), where X is a compact subset of ℝ2. In addition to this, we also denote the Frobenius norm of M(x) to be
(10)|M(x)|F=(∑i,j=12‍|Mij(x)|2)1/2.
This norm and the infinity norm of the matrix M can be related as
(11)|M(x)|F≤2|M|∞,x∈X.
This can be shown easily. One can also show that if M∈C(X;ℝ2×2) and y∈L2(X;ℝ2), then the vector My is bounded in L2(X;ℝ2). In fact,
(12)|My|L2(X)≤2|M|∞|y|L2(X)
and the proof is trivial. Finally, the symbols |·| or |·|2 will refer to the usual Euclidean norm.The Perturbed Domain
Ω
t. The domains Ωt that are considered in this work are of annulus type with boundary ∂Ωt, which is the union of two disjoint sets Γt and Σt, 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
(13)Θ={V∈C1,1(U-,ℝ2):V|Γ∪∂U=0}.
For t=0, we obtain the reference domain Ω:=Ω0, with a fixed boundary Γ:=Γ0 and a free boundary Σ:=Σ0.The main objective in this subsection is to show thatTt is a diffeomorphism from Ω to Ωt for sufficiently small t. To verify this, we need the following results, which are given and proven in [17].Theorem 3.
IfΩ is a bounded, open, connected set in ℝn such that intΩ-=Ω and φ is a continuous injective mapping from Ω- to ℝn, then
(14)φ(Ω-)=φ(Ω)¯,φ(Ω)=intφ(Ω-),φ(∂Ω)=∂φ(Ω)=∂φ(Ω-).Theorem 4.
Suppose(1)
Ω is a bounded, open, connected set in ℝn such that intΩ-=Ω,(2)
φ0∈C(Ω-,ℝ2) where φ0 is injective,(3)
φ∈C(Ω-,ℝ2)∩C1(Ω,ℝ2) such that
(15)detDφ(x)>0∀x∈Ω,φ(x)=φ0(x)∀x∈∂Ω.Then (i)
φ:Ω-→φ(Ω-) 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), φ(Ω-)=φ0(Ω-).We also consider the following property of a domain, which is also found in [17, page 52].Lemma 5.
IfΩ 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 x,y∈Ω-, one can find a finite sequence of points yk, k=1,…,l+1, satisfying the following properties: (a)
y1=x, yk∈Ω for 2≤k≤l, yl+1=y,(b)
(yk,yk+1)⊂Ω for 1≤k≤l,(c)
∑k=1l‍|yk+1-yk|≤c(Ω)|x-y|.We also recall the useful property of the determinant of the Jacobian ofTt which is given in the next lemma. Here we use the notation
(16)It(x)=detDTt(x),x∈U-.Lemma 6 (see [9, 13]).
Consider the operatorTt defined by (9), where V∈Θ, which is described by (13). Then (i)
It=1+tdivV+t2detDV,(ii)
there existtV,α1,α2>0 such that 0<α1≤It(x)≤α2, for |t|≤tV, x∈U.Proof.
In general, forn-dimensional case, the Jacobian of Tt is given by DTt=(aij), where aij=t(∂Vi/∂xj) if, i≠j, and aii=1+t(∂Vi/∂xi). By definition of the determinant, we can write detDTt as
(17)detDTt=∑σ∈Sn‍sgn(σ)a1,σ(1)a2,σ(2)⋯an,σ(n)=∏i=1n‍(1+t∂Vi∂xi)hhhhh+∑σ∈Fn∖{i}‍sgn(σ)a1,σ(1)a2,σ(2)⋯an,σ(n)hhhhh+∑σ∈Sn∖Fn‍sgn(σ)a1,σ(1)a2,σ(2)⋯an,σ(n)=:I+II+III,
where Sn refers to the set of all permutations of {1,2,…,n}, i is the identity permutation, Fn={σ∈Sn:σ(k)=kforsomepositiveintegerk≤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+tdivV+t2R1(t,V), where R1∈C(ℝ,C0,1(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(∂Vi/∂xj), i≠j. Hence we can write II=t2R2(t,V), where R2 is in C(ℝ,C0,1(U)). All terms of III have factors of the form t(∂Vi/∂xj),  i≠j, and thus we have III=tnR-3(V), which can be written as t2R3(t,V), where R3∈C(ℝ,C0,1(U)). Combining I, II, and III, we get
(18)detDTt=1+tdivV+t2R(t,V)
with R∈C(ℝ,C0,1(U)). In particular, for n=2, the determinant is computed as follows:
(19)detDTt=1+tdivV+t2detDV.
This verifies(i). To show (ii) we first get the lower bound for It(x). Take
(20)tV=min{12(|divV|C(U-)+|detDV|C(U-)),1}.
For |t|≤tV, we obtain
(21)It(x)≥1-|t||divV|C(U-)-|t|2|detDV|C(U-)≥1-|t||divV|C(U-)-|t||detDV|C(U-)=1-|t|(|divV|C(U-)+|detDV|C(U-))≥12.
On the other hand, by triangle inequality we have
(22)It(x)≤1+|t||divV|C(U-)+|t|2|detDV|C(U-)≤1+tV|divV|C(U-)+tV2|detDV|C(U-)hhhhhhhhhhhhhhhhhhhhhhh∀x∈U-.
Hence, we have shown that there are positive constants α1=1/2 and α2=1+tV|divV|C(U-)+tV2|detDV|C(U-) such that α1≤It(x)≤α2 for x∈U-.Considering the theorems and lemmas presented beforehand, we are now ready to prove the following theorem.Theorem 7.
LetΩ and U be nonempty bounded open connected subsets of ℝ2 with Lipschitz continuous boundaries, such that Ω-⊆U, and ∂Ω is the union of two disjoint boundaries Γ and Σ. Let Tt be defined as in (9) where V belongs to Θ, defined as (13).
Then for sufficiently smallt, (1)
Tt:U-→U- is a homeomorphism,(2)
Tt:U→U is a C1,1 diffeomorphism, and in particular, Tt:Ω→Ωt is a C1,1 diffeomorphism,(3)
Γt=Tt(Γ)=Γ,(4)
∂Ωt=Γ∪Tt(Σ).Proof.
First, becauseU is a C0,1 domain, it follows that intU-=U by Theorem 2. Second, T0=I∈C(U-,ℝ2), and it is injective. Third, it is evident that Tt is C1,1 because V is C1,1. For x∈∂U, Tt(x)=x because V vanishes on ∂U. For x∈U, the determinant of the Jacobian of the perturbation of identity operator Tt is given by (19). By Lemma 6, there exists a tV>0, given by (20), such that detDTt(x)>0 for all x∈U and for |t|≤tV. Hence, by applying Theorem 4, we conclude that Tt(U)=U and Tt(U-)=U- for all |t|<tV, and Tt:U-→U- is a homeomorphism. Furthermore, by Theorem 4, we find that Tt:U→U is a C1 diffeomorphism. To show that Tt is a C1,1 diffeomorphism, we are left to show that DTt-1 is Lipschitz continuous. To verify this we use Lemma 5.
Given any two pointsu,v∈U we choose {yk} such that properties (a)–(c) of Lemma 5 are satisfied. For fixed |t|<tV, differentiating the identities Tt∘Tt-1=I and Tt-1∘Tt=I will lead to
(23)DTt(Tt-1(z))DTt-1(z)=I,DTt-1(z)DTt(Tt-1(z))=I,
for all z∈U. Thus,
(24)DTt(Tt-1(yk+1))DTt-1(yk+1)=I,DTt-1(yk)DTt(Tt-1(yk))=I.
This implies(25)|DTt-1(u)-DTt-1(v)|≤∑k=1l‍|DTt-1(yk)-DTt-1(yk+1)|=∑k=1l‍|DTt-1(yk)=∑k=1l‍|×[DTt(Tt-1(yk+1))-DTt(Tt-1(yk))]=∑k=1l‍|×DTt-1(yk+1)|.
Applying the infinity norm in the spaceC(X;ℝ2) we have
(26)|DTt-1(u)-DTt-1(v)|≤∑k=1l‍|DTt-1(yk)||DTt-1(yk+1)|hh×|DTt(Tt-1(yk+1))-DTt(Tt-1(yk))|≤maxy∈U-|DTt-1(y)|2×∑k=1l‍|DTt(Tt-1(yk+1))-DTt(Tt-1(yk))|.
SinceDTt is Lipschitz continuous, we have
(27)|DTt-1(u)-DTt-1(v)|=maxy∈U-|DTt-1(y)|2L1∑k=1l‍|Tt-1(yk+1)-Tt-1(yk)|,
where L1 is the maximum of all Lipschitz constants of DTt for all |t|<tV. Then finally, using the mean value theorem and property (c) in Lemma 5, we obtain
(28)|DTt-1(u)-DTt-1(v)|≤maxy∈U-|DTt-1(y)|3L1∑k=1l‍|yk+1-yk|≤maxy∈U-|DTt-1(y)|3L1c(U)|u-v|.
Hence DTt-1 is Lipschitz continuous which shows that Tt:U→U is a C1,1 diffeomorphism for sufficiently small |t|. Restricting to Ω, this proves that Tt:Ω→Ωt is a C1,1 diffeomorphism. (2) is clear because the fixed boundary is invariant under Tt; that is, Γt:=Tt(Γ)=Γ since V vanishes on Γ. Lastly, using Theorem 3, definition of Tt, (1), and (2), we obtain (3).Corollary 8.
LetΩ and U be two domains of ℝ2 with C1,1 boundary. Then for |t|<tV, where tV is given by (20), the perturbed domain Ωt:=Tt(Ω) is also of class C1,1.Proof.
Giveny∈∂Ωt, we let x=Tt-1(y)∈∂Ω. Then there exists a neighborhood Nx of x and a diffeomorphism gx∈C1,1(Nx,B(0,1)) such that gx(Nx∩Ω)=B+(0,1),  gx(Nx∩∂Ω)=B0(0,1), and gx(Nx∩Ωc)=B-(0,1). We have also shown that Tt defined in Theorem 7 is a C1,1 diffeomorphism. Since Tt-1 is continuous, Tt(Nx)=(Tt-1)-1(Nx)=Ny is a neighborhood of y in U. Define gy:=gx∘Tt-1. This is bijective because gx and Tt-1 are bijective. gy∈C1,1(Ny,B(0,1)) because Tt-1∈C1,1(U-,U-) (hence Tt-1∈C1,1(Ny,Nx)) and gx∈C1,1(Nx,B(0,1)). Also, gy-1=Tt∘gx-1∈C1,1(B(0,1),Ny).
Next, we note that∂Ωt∩Ny=Tt(∂Ω)∩Ny=Tt(∂Ω)∩Tt(Nx). Since Tt is injective, we have ∂Ωt∩Ny=Tt(∂Ω∩Nx). Thus by definition of gy we get
(29)gy(∂Ωt∩Ny)=gy(Tt(∂Ω∩Nx))=gx(∂Ω∩Nx)=B0(0,1).
We also observe the following:
(30)gy(Ωt∩Ny)=gy(Tt(Ω∩Nx))=gx(Ω∩Nx)=B+(0,1),gy(Ω-tc∩Ny)=gy(Tt(Ω-c∩Nx))=gx(Ω-c∩Nx)=B-(0,1).
This shows thatΩt is indeed of class C1,1.Remark 9.
Theorem7 and Corollary 8 tell us that the reference Ω and the perturbed domain Ωt have the same topological structure and regularity under the perturbation of identity operator Tt for sufficiently small t. See Figure 3 for illustration.Figure 3
The action ofTt on a C1,1-domain.Properties of
T
t. In addition to (16) we also use the following notations throughout the work:
(31)Mt(x)=(DTt(x))-T,x∈U-,At(x)=It(x)MtT(x)Mt(x),x∈U-,wt(x)=It(x)|(DTt(x))-Tn(x)|,x∈Σ.Remark 10.
We note the following observations for fixed, sufficiently smallt. (1)
It∈C0,1(U-).(2)
Mt,MtT∈C(U-;ℝ2×2).(3)
At∈C(U-;ℝ2×2).(4)
wt∈C(Σ;ℝ).(5)
V∈C1,1(U-;ℝ2) implies that |V|∞ and |DV|∞ are both finite.
We now provide several properties ofTt.Lemma 11 (see [9, 13, 16, 18]).
Consider the transformationTt, where the fixed vector field V belongs to Θ, defined in (13). Then there exists tV>0 such that Tt and the functions in (16) and (31) restricted to the interval IV=(-tV,tV) have the following regularity and properties.(1)
t↦Tt∈C1(IV,C1,1(U-,ℝ2)).(2)
t↦It∈C1(IV,C0,1(U-)).(3)
t↦Tt-1∈C(IV,C1(U-,ℝ2)).(4)
t↦wt∈C1(IV,C(Σ)).(5)
t↦At∈C(IV,C(U-,ℝ2×2)).(6)
There isβ>0 such that At(x)≥βI for x∈U.(7)
(d/dt)Tt|t=0=V.(8)
(d/dt)Tt-1|t=0=-V.(9)
(d/dt)DTt|t=0=DV.(10)
(d/dt)(DTt)-1|t=0=-DV.(11)
(d/dt)It|t=0=divV.(12)
(d/dt)At|t=0=(divV)I-(DV+(DV)T)≡A.(13)
limt→0wt=1.(14)
(d/dt)wt|t=0=divΣV,where the surface divergence divΣ is defined by
(32)divΣV=divV|Σ-(DVn)·n.We provide proofs for properties (3) and (8). The rest can be seen in [19].Proof.
(3) Supposex∈U-, t,h∈IV, and y=Th-1(x). Then
(33)Tt-1(x)-Th-1(x)=Tt-1(Th(y))-Th-1(Th(y))=Tt-1(Th(y))-Tt-1(Tt(y)).
Using Lemma 5, we connect Th(y) and Tt(y) by a chain yk, k=1,…,l+1, satisfying (i)
y1=Tt(y),   yk∈U for 2≤k≤l,   yl+1=Th(y),(ii)
(yk,yk+1)⊂U for 1≤k≤l,(iii)
∑k=1l‍|yk+1-yk|≤c(U)|Th(y)-Tt(y)|,and then we get
(34)|Tt-1(x)-Th-1(x)|≤∑k=1l‍|Tt-1(yk)-Tt-1(yk+1)|≤maxu∈U-|DTt-1(u)|∑k=1l‍|yk-yk+1|≤c(U)maxu∈U-|DTt-1(u)||Th(y)-Tt(y)|,≤c(U)maxu∈U-|DTt-1(u)||V(y)||t-h|.
Thus,
(35)maxx∈U-|Tt-1(x)-Th-1(x)|≤c(U)maxu∈U-|DTt-1(u)||V|∞|t-h|.
By reducing tV if necessary, we can assume without loss of generality that |tDV|∞<1 for t∈(-tV,tV). This allows us to represent (DTt)-1 as a Neumann series:
(36)DTt-1(x)=(DTt)-1∘Tt-1(x)=∑k=0∞‍(-t)k(DV)k(Tt-1(x)),
and its norm is estimated as follows:
(37)|DTt-1(x)|∞≤∑k=0∞‍|(-t)k(DV)k(Tt-1(x))|∞=11-|t||DV|∞<11-tV|DV|∞=:M.
This shows uniform convergence in t∈IV and x∈U-. Hence, for every ϵ>0 one can choose a δ:=ϵ/c(U)M|V|∞>0 which implies that, for every t,h∈IV, |Tt-1-Th-1|∞<ϵ whenever |t-h|<δ. In other words, t↦Tt-1∈C(IV,C(U-,ℝ2)).
To show thatt↦Tt-1 is continuous from IV to C1(U-,ℝ2), we only need to show that for every t∈IV, |DTt-1-DTh-1|∞<ϵ whenever |t-h|<δ and h∈IV. Let z∈U-. Using (37), estimate |DTt-1(z)-DTh-1(z)| as follows:
(38)|DTt-1(z)-DTh-1(z)|=|(DTt)-1(Tt-1(z))|hh-(DTh)-1(Th-1(z))|=|(DTt)-1(Tt-1(z))hhhh×((DTh)(Th-1(z))hhhhhhhhh-(DTt)(Tt-1(z)))hhhh×(DTh)-1(Th-1(z))|≤M2|(DTh)(Th-1(z))≤M2hh|-(DTt)(Tt-1(z))(DTh)(Th-1(z))|.
Using the definition of Jacobian of a transformation and the regularity of V, we further simplify (38) as follows:
(39)|DTt-1(z)-DTh-1(z)|≤M2|hDV(Th-1(z))-tDV(Tt-1(z))|≤M2|h-t||DV(Ts-1(z))|+M2|t||DV(Th-1(z))-DV(Tt-1(z))|≤M2K|h-t|+M2tVL|Th-1(z)-Tt-1(z)|,
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 z∈U- and using (34) we get
(40)|DTt-1-DTh-1|∞≤α|h-t|,
where α=M2K+M2tVLc(U)M|V|∞. Thus, for any ϵ>0, we choose δ=ϵ/α, so that if 0<|h-t|<δ, then |DTt-1-DTh-1|∞<ϵ. Therefore, t↦Tt-1∈C(IV,C1(U-,ℝ2)).
Proof of property(8) in Lemma 11 is as follows. Given x∈U-, we have Tt(Tt-1(x))=x. This implies that
(41)Tt+h(Tt+h-1(x))-Tt(Tt-1(x))=0.
Manipulating the left hand side of (41), we get
(42)(Tt+h(Tt+h-1(x))-Tt(Tt+h-1(x)))+(Tt(Tt+h-1(x))-Tt(Tt-1(x)))=:A+B.
We first work on A. Applying the definition of Tt, we get
(43)A=Tt+h-1(x)+(t+h)V(Tt+h-1(x))-Tt+h-1(x)-tV(Tt+h-1(x))=hV(Tt+h-1(x)).
Thus,
(44)limh→01hA=V(Tt-1(x)).
Similarly, we can write B as follows:
(45)B=Tt+h-1(x)-Tt-1(x)+tV(Tt+h-1(x))-tV(Tt-1(x)).
Hence, we have
(46)1hB=1h(Tt+h-1(x)-Tt-1(x))+t1h(V(Tt+h-1(x))-V(Tt-1(x))).
Suppose v is a coordinate function of V. By the mean value theorem, we observe that
(47)1h(v(Tt+h-1(x))-v(Tt-1(x)))=Dv(c)1h(Tt+h-1(x)-Tt-1(x)),
where c is a point on the segment joining Tt-1(x) and Tt+h-1(x), and as h tends to infinity, (47) tends to (Dv)(d/dt)Tt-1(x). Thus,
(48)limh→01hB=ddtTt-1(x)+tDV(Tt-1(x))ddtTt-1(x).
Combining (44) and (48), we get
(49)V(Tt-1(x))+(I+tDV(Tt-1(x)))ddtTt-1(x)=0,
which implies that
(50)ddtTt-1(x)=-(I+tDV(Tt-1(x)))-1V(Tt-1(x)).
Evaluating (50) at t=0, we get (d/dt)Tt-1|t=0=-V.


## 3.3. The Method of Mapping

Ifu is defined in Ω and ut is defined in Ωt, then the direct comparison of ut with u is generally not possible since the functions are defined on different domains. To overcome this difficulty, one maps ut back to Ω by composing it with Tt; that is, one defines ut∘Tt: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].Definition 12.
Letu be defined in [0,tV]×U. An element u˙∈Hk(Ω), called the material derivative of u, is defined as
(51)u˙(x):=u˙(0,x):=limt→0+u(t,Tt(x))-u(0,x)t=ddtu(t,x+tV(x))|t=0if the limit exists in (Hk(Ω)).Remark 13.
The material derivative can be written as(52)u˙(x)=limt→0+ut∘Tt(x)-u(x)t=ddt(ut∘Tt(x))|t=0.
It characterizes the behavior of the function u at x∈Ω⊂U in the direction V(x).Definition 14.
Letu be defined in [0,tV]×U. An element u′∈Hk(Ω) is called the shape derivative of u at Ω in the direction V, if the following limit exists in Hk(Ω):
(53)u′(x):=u′(0,x):=limt→0+u(t,x)-u(0,x)t.Remark 15.
The shape derivative ofu is also defined as follows:
(54)u′(x):=∂∂tut(x)|t=0.
We note that if u˙ and ∇u·V exist in Hk(Ω), then the shape derivative can be written as
(55)u′(x)=u˙(x)-(∇u·V)(x).
In general, if u˙(x) and ∇u·V(x) both exist in Wm,p(Ω), then u′(x) also exists in that space.
Domain and Boundary TransformationsLemma 16 (see [18]).
(
1
) Let φt∈L1(Ωt). Then φt∘Tt∈L1(Ω) and
(56)∫Ωt‍φtdxt=∫Ω‍φt∘TtItdx.
(
2
) Let φt∈L1(∂Ωt). Then φt∘Tt∈L1(∂Ω) and
(57)∫∂Ωt‍φtdst=∫∂Ω‍φt∘Ttwtds,
where It and wt are defined in (31).Proofs can be found in [13, 18].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 C0,1 domains while the second theorem requires C1,1 domains. For proofs, see [18].Theorem 17 (domain differentiation formula).
Letu∈C(IV,W1,1(U)) and suppose u˙(0,·):=(d/dt)u(t,Tt(·))|t=0 exists in L1(U). Then
(58)ddt∫Ωt‍u(t,x)dx|t=0=∫Ω‍u′(0,x)dx+∫Σ‍u(0,s)V·nds.Theorem 18 (boundary differentiation formula).
Letu be defined in a neighborhood of Γ. If u∈C(IV,W2,1(U)) and u˙(0,·)∈W1,1(U), then
(59)ddt∫Σt‍u(t,s)ds|t=0=∫Σ‍u′(0,s)ds+∫Σ‍(∂u∂n+u(0,s)κ)V·nds,
where κ is the mean curvature of the free boundary Σ.
The First-Order Eulerian DerivativeDefinition 19.
The Eulerian derivative of the shape functionalJ:Ω→ℝ defined in (7) at the domain Ω in the direction of the deformation field V∈Θ is given by
(60)dJ(Ω;V):=limt→0+J(Ωt)-J(Ω)t,
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.


## 4. Main Result

In this section we derive in a rigorous manner the first-order shape derivative of the Kohn-Vogelius functionalJ, 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.


### 4.1. Variational Forms of the Dirichlet and Neumann Problems

We recall that we are considering the shape optimization problem (7) where uD solves the pure Dirichlet problem (4) and uN solves the Neumann problem (6). As in [13], we consider the Hilbert space
(61)HΓ,01={φ∈H1(Ω):φ|Γ=0},
which is endowed with the norm
(62)|φ|12=∫Ω‍|∇φ|2dx,
and a linear manifold defined by
(63)HΓ,v1={φ∈H1(Ω):φ|Γ=v}
for v∈H1/2(Γ).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.FinduD∈H1(Ω) such that
(64)(∇uD,∇ψ)Ω=0∀ψ∈H01(Ω)uD=1onΓ,uD=0onΣ.
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.FinduN∈H1(Ω) such that
(65)(∇uN,∇φ)Ω-(λ,φ)Σ=0∀φ∈HΓ,01(Ω)uN=1onΓ.
It is also well known that (65) has a unique solution.


### 4.2. Analysis of State Variables in Deformed Domains

We now consider the class of perturbed problems:(66)minΩtJ(Ωt)=12∫Ωt‍|∇(uD,t-uN,t)|2dx,
where uD,t solves the pure Dirichlet problem
(67)-ΔuD,t=0inΩt,uD,t=1onΓ,uD,t=0onΣt,
and uN,t solves the Neumann problem
(68)-ΔuN,t=0inΩt,uN,t=1onΓ,∂uN,t∂nt=λonΣt.
Here, nt is the outward unit normal to the deformed free boundary Σt. The variational form of (67) is formulated as follows.FinduD,t∈H1(Ωt) such that
(69)(∇uD,t,∇ψt)Ωt=0∀ψt∈H01(Ωt)uD,t=1onΓ,uD,t=0onΣt.
It is known that (69) has a unique solution.Remark 21.
The functionut:Ωt→ℝ can be referred to as the reference domain by composing ut with Tt; that is,
(70)ut=ut∘Tt:Ω⟶ℝ
and by chain rule of differentiation, we get
(71)(∇ut)∘Tt=(DTt)-T∇ut=:Mt∇ut.LetuD,t be the solution of (69). Applying Lemma 16 for all ψt∈H01(Ωt) we have
(72)(∇uD,t,∇ψt)Ωt=∫Ωt‍∇uD,t·∇ψt=∫Ω‍It(∇uD,t∘Tt)·(∇ψt∘Tt).
Applying (71) and noting that ψt∈H01(Ω) because ψt∈H01(Ωt), we obtain
(73)(∇uD,t,∇ψt)Ωt=∫Ω‍ItMt∇uDt·Mt∇ψt=∫Ω‍ItMtTMt∇uDt·∇ψt=(At∇uDt,∇v)Ω,
where v=ψt∈H01(Ω). Hence, if uD,t solves the variational equation (69), then uDt=uD,t∘Tt satisfies the variational equation,
(74)(At∇uDt,∇v)Ω=0
for all v∈H01(Ω),  uDt=1 on Γ, and uDt=0 on Σ.Now we show thatuDt is the unique solution of (74) in H1(Ω). First, we show that yt=uDt-uD∈H01(Ω) is the unique solution to
(75)(At∇yt,∇v)Ω=-(At∇uD,∇v)Ω
for all v∈H01(Ω). The bilinear form bt(·,·):H01(Ω)×H01(Ω)→ℝ defined by
(76)bt(yt,v)=∫Ω‍At∇yt·∇v,∀yt,v∈H01(Ω)
is continuous, because
(77)|bt(yt,v)|=|∫Ω‍At∇yt·∇v|≤|At∇yt|L2(Ω)|∇v|L2(Ω)≤2|At|∞|∇yt|L2(Ω)|∇v|L2(Ω)=β|yt|H1(Ω)|v|H01(Ω).
The bilinear form is also coercive. To show this we recall that limt→0At=I uniformly on Ω-. This is equivalent to the statement
(78)∀ϵ>0∃δ>0suchthat|t|<δ⟹|At-I|∞<ϵ.
Let ϵ=1/2. So for sufficiently small t, |At-I|∞<1/2, and
(79)bt(yt,yt)=∫Ω‍At∇yt·∇yt=|∫Ω‍(At-I)∇yt·∇yt+|∇yt|2|≥|∇yt|L2(Ω)2-|At-I|∞|∇yt|L2(Ω)2≥12|yt|H01(Ω)2.
So bt(yt,v) is coercive.Next, we show that the functionalf:H01(Ω)→ℝ is bounded:
(80)|〈f,v〉H-1×H01|=|-∫Ω‍At∇uD·∇v|≤2|At|∞|∇uD|L2(Ω)|∇v|L2(Ω)=2|At|∞|uD|H01(Ω)|v|H01(Ω)≤C|v|H01(Ω).
Therefore, by the Lax-Milgram lemma, yt=uDt-uD is the unique solution to the variational equation
(81)(At∇yt,∇v)Ω=-(At∇uD,∇v)Ω,∀v∈H01(Ω).
This implies the existence of a unique solution uDt of (74) as verified below.LetuDt=yt+uD. Using (81) we obtain
(82)(At∇uDt,∇v)Ω=(At∇yt,∇v)Ω+(At∇uD,∇v)Ωhhhhhhhhhhhhhh∀v∈H01(Ω)=-(At∇uD,∇v)Ω+(At∇uD,∇v)Ω=0.
Thus (74) is satisfied. The boundary conditions are also satisfied because on Γ, yt=0 and uD=1 and on Σ, both yt and uD are zero. To show uniqueness, we let uDt and u~Dt be solutions of (74). This implies that there exist yt and y~t such that uDt=yt+uD and u~Dt=y~t+uD, where yt and y~t are solutions to (81). Taking the difference of uDt and u~Dt and considering that solution to (81) is unique, we get uDt=u~Dt.Next, we consider (68) whose variational form is formulated as follows.FinduN,t∈H1(Ωt) such that
(83)(∇uN,t,∇φt)Ωt-(λ,φt)Σt=0∀φt∈HΓ,01(Ωt),uN,t=1onΓ.
Similarly, if uN,t solves the variational problem (83), then uNt solves the variational equation
(84)(At∇uNt,∇ϕ)Ω-(wtλ,ϕ)Σ=0,∀ϕ∈HΓ,01(Ω),
where uNt=1 on Γ.As shown before, the bilinear formbt(·,·):HΓ,01(Ω)×HΓ,01(Ω)→ℝ defined by
(85)bt(zt,v)=∫Ω‍At∇zt·∇v,∀zt,v∈HΓ,01(Ω),
is coercive and continuous. The linear functional f:HΓ,01(Ω)→ℝ defined by 〈f,ϕ〉=∫Σ‍λwtϕ is continuous on HΓ,01(Ω) because
(86)|〈f,ϕ〉H-1×H01|=|∫Σ‍λwtϕ|≤|λ||wt|∞∫Σ‍|ϕ|=|λ||wt|∞|Σ|1/2|γϕ|L2(Σ)≤|λ||wt|∞|Σ|1/2∥γ∥|ϕ|H1(Ω).
By the Lax-Milgram lemma, zt=uNt-1∈HΓ,01(Ω) is the unique solution in HΓ,01(Ω) of
(87)∫Ω‍At∇zt·∇ϕ=λ∫Σ‍wtϕ,∀ϕ∈HΓ,01(Ω).
Let uNt=zt+1. Then, by (87), we get
(88)(At∇uNt,∇ϕ)Ω-(wtλ,ϕ)Σ=(At∇zt,∇ϕ)Ω-(wtλ,ϕ)Σ=0ϕ∈HΓ,01(Ω).
Since zt∈HΓ,01(Ω), uNt=zt+1=1 on Γ. Uniqueness of uNt follows from the uniqueness of zt. Therefore, uNt is the unique solution of the variational problem (84) in H1(Ω).


### 4.3. Hölder Continuity of the States

We show thatuDt and uNt are Hölder continuous on t.Theorem 22 (see [13]).
The solutionsuDt of (74) are uniformly bounded in H1(Ω) for t∈(-tV,tV) and
(89)limt→01t|uDt-uD|H1(Ω)2=0,
where uD is the weak solution of (4).Proof.
We first prove the uniform boundedness ofuDt in H1(Ω) for t∈(-tV,tV). Since uDt-uD∈H01(Ω), by using coercivity of At we get
(90)(At∇(uDt-uD),∇(uDt-uD))Ω≥12|∇(uDt-uD)|H01(Ω)2.
Also, by applying (81), we have
(91)(At∇(uDt-uD),∇(uDt-uD))Ω=-(At∇uD,∇(uDt-uD))Ω≤|At|∞|uD|H1(Ω)|uDt-uD|H01(Ω).
Therefore,
(92)|uDt|H1(Ω)≤|uDt-uD|H01(Ω)+|uD|H1(Ω)≤tV+|uD|H1(Ω).
Now we take the difference between the weak form of (4) and the variational equation (74), to get
(93)(At∇uDt,∇v)Ω-(∇uD,∇v)Ω=0,∀v∈H01(Ω)(At∇1t(uDt-uD),∇v)Ω=-((At-I)t∇uD,∇v)Ω.
Note that uD,t∈HΓ,11(Ωt). This implies that uDt=uD,t∘Tt is in HΓ,11(Ω). Note also that uD∈HΓ,11(Ω). Let yt=(1/t)(uDt-uD). So for sufficiently small t, yt∈H01(Ω). Now choosing v=yt as a test function and by the uniform coercivity of At one obtains
(94)α|∇yt|L2(Ω)2≤(At∇yt,∇yt)Ω=-(1t(At-I)∇uD,∇yt)Ω≤|1t(At-I)|∞|∇uD|L2(Ω)|∇yt|L2(Ω).
If∇yt=0, then the inequality above holds. For ∇yt≠0 we have
(95)α|∇yt|L2(Ω)2≤|1t(At-I)∇uD|L2(Ω)|∇yt|L2(Ω)≤|1t(At-I)|∞|∇uD|L2(Ω)|∇yt|L2(Ω).
Hence,
(96)|1t∇(uDt-uD)|L2(Ω)≤1α|1t(At-I)|∞|∇uD|L2(Ω).
Squaring and multiplying t on both sides of the inequality give us
(97)1t|∇(uDt-uD)|L2(Ω)2≤t1α2|1t(At-I)|∞2|∇uD|L2(Ω)2.
Since |(1/t)(At-I)|∞2 is uniformly bounded in t by Lemma 11,
(98)limt→01t|uDt-uD|H1(Ω)2=limt→01t|∇(uDt-uD)|L2(Ω)2=0
follows.Theorem 23.
The solutionsuNt of (84) are uniformly bounded in H1(Ω) for t∈(-tV,tV) and
(99)limt→01t|uNt-uN|H1(Ω)2=0,
where uN is the solution of (65).Proof.
Subtracting (65) from (84) for all φ∈HΓ,01(Ω) we get
(100)0=(At∇uNt,∇φ)Ω-(wtλ,φ)Σ-(∇uN,∇φ)Ω+(λ,φ)Σ=(At∇uNt-∇uNt+∇uNt-∇uN,∇φ)Ω-(wtλ-λ,φ)Σ.
Hence
(101)(∇(uNt-uN),∇φ)Ω=-((At-I)∇uNt,∇φ)Ω+(wtλ-λ,φ)Σ.
Note that uNt-uN belongs to HΓ,01(Ω). Hence
(102)|uNt-uN|H1(Ω)2=-((At-I)∇uNt,∇(uNt-uN))Ω+λ(wt-1,uNt-uN)Σ.
By Cauchy-Schwarz inequality, we obtain
(103)|uNt-uN|H1(Ω)2≤|(At-I)∇uNt|L2(Ω)|∇(uNt-uN)|L2(Ω)+|λ||wt-1|∞|uNt-uN|L2(Σ).
Furthermore, by trace theorem we have |uNt-uN|L2(Σ)≤c|uNt-uN|H1(Ω). Therefore,
(104)|uNt-uN|H1(Ω)2≤C(|(At-I)∇uNt|L2(Ω)+|λ||wt-1|∞)|uNt-uN|H1(Ω)
holds, where C=max{1,c}. This implies
(105)|uNt-uN|H1(Ω)≤C(|(At-I)∇uNt|L2(Ω)+|wtλ-λ|L2(Σ)),
which entails
(106)1t|uNt-uN|H1(Ω)≤C(|1t(At-I)|L2(Ω)|∇uNt|L2(Ω)≤C(+|λ||Σ|1/2|1t(wt-1)|∞|1t(At-I)|L2(Ω))=tC(|1t(At-I)|L2(Ω)|∇uNt|L2(Ω)tCh(+|λ||Σ|1/2|1t(wt-1)|∞|1t(At-I)|L2(Ω)).
We now show that∇uNt is bounded uniformly in t in L2(Ω). Since uNt-1∈HΓ,01(Ω), we have
(107)α|∇uNt|L2(Ω)2=α|∇(uNt-1)|L2(Ω)2≤∫Ω‍At∇(uNt-1)·∇(uNt-1)=∫Ω‍At∇uNt·∇(uNt-1)=|λ|∫Σ‍wt(uNt-1)≤c|λ||wt|L∞|Σ|1/2|γ||∇uNt|L2(Ω).
Consequently,
(108)|∇uNt|L2(Ω)≤|λ|αc|wt|L∞|Σ|1/2|γ|,
and this shows that ∇uNt is uniformly bounded in L2(Ω) because |wt|∞ is bounded. In addition, At and wt are differentiable at t=0 by Lemma 11. Therefore,
(109)limt→01t|uNt-uN|H1(Ω)2=0.


### 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 uD,uN∈H1(Ω). For C1,1 domains, we show that these solutions also exist in H2(Ω) and more generally in Hk+2(Ω) if domains are of class Ck+1,1, k≥0.To prove higher regularity ofuD, we require the following two theorems, which are proven in [22].Theorem 24 (see [22, page 124]).
LetΩ be a bounded open subset of ℝn with a C1,1 boundary. Consider the Dirichlet boundary value problem:
(110)Au=finΩ,u=gon∂Ω,
where
(111)A=∑i,j=1n‍Di(aijDju)+∑i=1n‍aiDiu+a0u.
Let ai,j be uniformly Lipschitz functions and let ai be bounded measurable functions such that aj,i=ai,j, 1≤i, j≤n and that there exists α>0 with
(112)∑i,j=1n‍ai,j(x)ζiζj≤-α|ζ|2
for all ζ∈ℝn and for almost every x∈Ω-. Assume in addition that either (i)
ai=0,   1≤i≤n and a0≥0 a.e. or(ii)
a0≥β>0 a.e.Then for every f∈Lp(Ω) and every g∈W2-1/p,p(∂Ω), there exists a unique u∈W2,p(Ω) that solves (110).Theorem 25 (see [22, page 128]).
LetΩ be a bounded open subset of ℝn with a Ck+1,1 boundary. Consider the operator A defined by (110) with aij=aji∈Ck,1(Ω-), and assume that there exists α>0 such that (112) holds for all ζ∈ℝn and for every x∈Ω-. Also, consider a real boundary operator B which is either the identity operator (orderd=0) or Bu=∑i=1n‍biDiu with bi∈Ck,1(Ω-), 1≤i≤n(orderd=1) and ∑i=1n‍bini≠0(niistheithcomponentofn) everywhere on Γ:=∂Ω. Furthermore, assume that u∈W2,p(Ω) satisfies Au=f∈Wk,p(Ω) and γBu=g∈W2+k-d-1/p,p(Γ). Then u∈Wk+2,p(Ω).We will also justify the higher regularity ofuN. We use the following results whose proofs are given in the corresponding texts.Theorem 26 (see [23, page 316]).
LetΩ be a bounded open subset of ℝn. Suppose u∈H1(Ω) is a weak solution of the PDE
(113)Lu=finΩ,
where
(114)Lu=-∑i,j=1n‍(aij(x)uxi)xj+∑i=1n‍bi(x)uxi+c(x)u.
Assume furthermore that aij,bi,c∈C∞(Ω) for i,j=1,2,…,n, and f∈C∞(Ω). Then u∈C∞(Ω).Theorem 27 (see [24, page 12]).
LetΩ⊂ℝn be a bounded domain with Ck+1,1 boundary ∂Ω for some nonnegative integer k. Suppose the data f and g of the problem
(115)-Δu=finΩ,u=gon∂Ω
are in Wk,p(Ω) and Wk+2-1/p,p(∂Ω), respectively, for some real number p with 1<p<∞. Then u∈Wk+2,p(Ω).For proof, see [22].Theorem 28 (see [22, page 84]).
LetΩ⊂ℝn be a bounded domain with C1,1 boundary Γ and 1<p<∞. Consider the Neumann problem
(116)-Δu+a0u=finΩ,γ∂u∂n=gonΓ.
If 0<a0∈L∞(Ω), f∈Lp(Ω), and g∈W1-1/p,p(∂Ω), then the weak solution u to (116) exists in W2,p(Ω).For proof, see [22].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.Theorem 29.
LetΩ be a bounded domain with boundary of class C1,1. Let uD,uN∈H1(Ω) be weak solutions of the BVPs (4) and (6), respectively. Then uD and uN also belong to H2(Ω). More generally, if Ω is of class Ck+1,1, where k is a nonnegative integer, then uD and uN are elements of Hk+2(Ω).Proof.
We first consider the solutionuD∈H1(Ω) to the Dirichlet problem (4). We use Theorem 25 to show that uD is an element of H2(Ω). Here, (110) is applied with the following settings.
We considern=2. The domain Ω is of class C1,1. L=-Δ, and hence aij=aji=-1 for i=j and aij=aji=0 for i≠j, with i,j=1,2. We also observe that ∑i,j=12‍aij(x)ζiζj=-(ζ1)2-(ζ2)2=-1|ζ|2, for all ζ=(ζ1,ζ2)T. Thus α=1. Furthermore, we have the following data: f=0∈L2(Ω),  g=1∈H3/2(Γ), g=0∈H3/2(Σ). Therefore, by using Theorem 24, there exists a unique u=uD∈H2(Ω), which is a solution to (4).
For higher regularity ofuD we apply Theorem 25. At first we consider C2,1-domains. In this case, k=1. We have aij=aji=-1 for i=j and aij=aji=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∈H5/2(Γ), g=0∈H5/2(Σ). Therefore, by applying Theorem 25, we have u=uD∈H3(Ω). In general, for smoother domains with Ck+1,1 boundaries, solutions to (4) are elements of Hk+2.
Next, we recall that, forC1,1 domain, there is a weak solution uN∈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 Ck+1,1, then uN is in Hk+2(Ω), where k is a nonnegative integer. For this purpose we need Theorem 26 which implies uN∈C∞(Ω).
Choose a bounded connected domainG with C∞ boundary Γ1 such that A-⊂G and G-⊂B, 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:
(117)-Δw=0inΩ1,w=1onΓ,w=uNonΓ1.
Since Ω1 is bounded with compact boundaries, we have C∞(Γ∪Γ1)⊂Hk+3/2(Γ∪Γ1). So by applying Theorem 27, we get w∈Hk+2(Ω1). Since uN|Ω1 also solves (117), then by uniqueness we have w=uN|Ω1. If A is a domain with C1,1 boundary Γ, then by Theorem 27 we have uN|Ω1∈H2(Ω1). Moreover, if A is a Ck+1,1-domain, then uN|Ω1 is in Hk+2(Ω1).
Second, we consider the following boundary value problem:(118)-Δv+v=uNinΩ2,∂v∂n=∂uN∂nonΓ1,∂v∂n=λonΣ.
Because uN|Ω2∈H1(Ω2), it follows that uN|Ω2∈L2(Ω2). We have also shown that uN∈C∞(Ω). This implies that uN∈C∞(Γ1). Hence, (∂uN/∂n)∈H1/2(Γ1). Also, λ∈H1/2(Σ). Since B is a domain of class C1,1, then by Theorem 28, we infer that (118) has a unique solution v∈H2(Ω2). Note, however, that uN|Ω2 also solves (118). So by uniqueness of the solution, we get uN|Ω2=v. Therefore, uN|Ω2∈H2(Ω2). Now, if domain B is of class Ck+1,1 and uN|Ω2∈H2(Ω2)⊂H1(Ω2), we get v∈H3(Ω2) by applying Theorem 25 and so is uN|Ω2. Doing this recursively, we end up with uN|Ω2∈Hk+2(Ω).
Hence, forCk+1,1-domains Ω(:=B∖A-), if we combine uN|Ω1∈Hk+2(Ω1) and uN|Ω2∈Hk+2(Ω2), we get uN∈Hk+2(Ω1∪Ω2). Moreover, uN is C∞ in a neighborhood of Γ1 because uN∈C∞(Ω). Therefore, uN∈Hk+2(Ω).Remark 30.
In the computation of the first-order shape derivative, since we are dealing withC1,1-domains, we may consider H2-regularity for the solutions uD and uN, as justified by Theorem 29.


### 4.5. The Shape Derivative ofJ

First, we state and prove the following lemma.Lemma 31.
LetΩ⊂ℝn be a bounded Lipschitz domain. Then the following equation:
(119)-∫Ω‍(divF)gdx=∫Ω‍F·∇gdx-∫∂Ω‍g(F·n)ds
is valid for vector field F and scalar function g having the following regularity: (i)
F∈H1(Ω;ℝn) and g∈H1(Ω);(ii)
F∈C1(Ω-;ℝn) and g∈W1,1(Ω).Proof.
First we recall the Gauss’ divergence theorem inℝn saying that if a domain Ω⊂ℝn is a bounded Lipschitz domain, then we have
(120)∫Ω‍divFdx=∫∂Ω‍F·nds
for a vector field F∈C1(Ω-,ℝn). Second we take the divergence of the product of a scalar function g and the vector field F to get
(121)div(gF)=gdivF+F·∇g.
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 (divF)g∈L1(Ω),  F·∇g∈L1(Ω), and the integral ∫∂Ω‍g(F·n) is bounded. Hence, (119) is well defined. Note that the formula
(122)-∫Ω‍(divF)gdx=∫Ω‍F·∇gdx-∫∂Ω‍g(F·n)ds
holds for F∈C1(Ω-;ℝn) and g∈C1(Ω-). We write
(123)-∫Ω‍(divF)gdx=∫Ω‍F·∇gdx-∫∂Ω‍γ0g(γ0F·n)ds,
where γ0:H1(Ω)→L2(∂Ω) is a trace operator. Let F∈H1(Ω;ℝn) and g∈H1(Ω). By density, we pick {Fk}⊂C1(Ω- and {gk}⊂C1(Ω-) such that Fk→F in H1(Ω-;ℝn) and gk→g in H1(Ω). By (122),
(124)-∫Ω‍(divFk)gkdx=∫Ω‍Fk·∇gkdx-∫∂Ω‍gk(Fk·n)ds.
Note that γ0Fk=Fk|∂Ω and γ0gk=gk|∂Ω. Also Fk→F in H1(Ω-;ℝn) implies that divFk→divF in L2(Ω). Moreover, gk→g in H1(Ω) implies ∇gk→∇g in L2(Ω). Furthermore, since γ0∈ℒ(H1(Ω),L2(∂Ω)), γ0Fk→γ0F and γ0gk→γ0g in L2(∂Ω). Therefore, (119) holds for F∈H1(Ω;ℝn) and g∈H1(Ω).
(
ii
) If F∈C1(Ω-;ℝn) and g∈W1,1(Ω), then (divF)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 C1(Ω-) in W1,1(Ω) we can show that (119) is also valid for this case.Now we apply Lemma31 to prove the next lemma.Lemma 32 (see [1]).
LetuD and uN belonging to H2(Ω) satisfy the Dirichlet problem (4) and the Neumann problem (6), respectively. Then
(125)∫Ω‍A∇uD·∇uD=-∫Σ‍(∇uD·n)2V·n,(126)∫Ω‍A∇uN·∇uN=∫Σ‍|∇uN|2V·n-2λ∫Σ‍V·∇uN,
where A is given by property (12) of Lemma 11.Proof.
From Lemma11, we recall the expression for A, which is given by A=(divV)I-((DV)+(DV)T). Our first goal is to derive an expression for ∫Ω‍A∇u·∇v for u,v∈H2(Ω). We begin by writing ∫Ω‍A∇u·∇v as follows:
(127)∫Ω‍A∇u·∇v=∫Ω‍(divV)∇u·∇v-∫Ω‍(DV)∇u·∇v-∫Ω‍(DV)T∇u·∇v.
We manipulate each term on the right-hand side of (127). First, because u,v∈H2(Ω), we have ∇u·∇v∈W1,1(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
(128)-∫Ω‍(divV)∇u·∇v=∫Ω‍∇(∇u·∇v)·V-∫Ω‍(divV)∇u·∇v-∫Σ‍(∇u·∇v)V·n.
The other two terms on the right hand side of (127) are manipulated as follows. The term ∇(∇u·∇v)·V is written as
(129)∇(∇u·∇v)·V=(∇v)T((∇2u)V)+VT((∇2v)∇u),
where ∇2u represents the Hessian of u. Because Hessian is symmetric, we obtain
(130)∇(∇u·∇v)·V=(∇2u∇v+∇2v∇u)·V.
Substituting (130) into (128) we get
(131)∫Ω‍(divV)∇u·∇v=-∫Ω‍((∇2u)∇v+(∇2v)∇u)·V+∫Σ‍(∇u·∇v)V·n.
Next, we expand the expression div((V·∇u)∇v) using (121) as
(132)div((V·∇u)∇v)=(V·∇u)Δv+∇v·(V·∇u)=(V·∇u)Δv+∇v·[(DV)T∇u+∇2uV]=(DV)∇v·∇u+(∇v)T(∇2u)V+(V·∇u)Δv.
Integrating both sides of (132) over Ω, applying Stoke’s theorem, and considering V=0 on Γ we end up with
(133)∫Ω‍(DV)∇v·∇u+∫Ω‍(∇v)T(∇2u)V+∫Ω‍(V·∇u)Δv=∫Σ‍(V·∇u)∇v·n,
or equivalently
(134)∫Ω‍(DV)∇v·∇u=-∫Ω‍(Δv)(V·∇u)-∫Ω‍(∇2u)∇v·V+∫Σ‍∂v∂n(V·∇u).
Interchanging v and u we get
(135)∫Ω‍(DV)∇u·∇v=-∫Ω‍(Δu)V·∇v-∫Ω‍(∇2v)∇u·V+∫Σ‍∂u∂n(V·∇v).
Also, because (DV)T∇v·∇u=(DV)∇u·∇v, we obtain
(136)∫Ω‍(DV)T∇v·∇u=-∫Ω‍(Δu)V·∇v-∫Ω‍(∇2v)∇u·V+∫Σ‍∂u∂n(V·∇v).
Thus,
(137)∫Ω‍(DV)T∇u·∇v=-∫Ω‍(Δv)V·∇u-∫Ω‍(∇2u)∇v·V+∫Σ‍∂v∂n(V·∇u).
Adding (131), (135), and (137) altogether, we express (127) as
(138)∫Ω‍A∇u·∇v=∫Ω‍Δu(V·∇v)+∫Ω‍Δv(V·∇u)-∫Σ‍∂v∂n(V·∇u)-∫Σ‍∂u∂n(V·∇v)+∫Σ‍(∇u·∇v)V·n.
Set u=v=uD in (138). The first two integrals on the right hand side of (138) vanish because -ΔuD=0 in Ω. Moreover, since uD=0 on Σ we have ∇uD=(∂uD/∂n)n. Thus, we can write (138) as follows:
(139)∫Ω‍A∇uD·∇uD=-2∫Σ‍(∇uD·n)(V·∇uD)+∫Σ‍(∇uD·∇uD)V·n=-2∫Σ‍(∂uD∂n)(∂uD∂nV·n)+∫Σ‍(∂uD∂n)2V·n=-∫Σ‍(∇uD·n)2V·n.
Therefore, (125) is satisfied.
On the other hand, by replacing bothu and v by uN and by considering that -ΔuN=0 in Ω and ∂uN/∂n=λ on Σ, we derive (126) as
(140)∫Ω‍A∇uN·∇uN=-2∫Σ‍∂uN∂n(V·∇uN)+∫Σ‍(∇uN·∇uN)V·n=∫Σ‍|∇uN|2V·n-2λ∫Σ‍V·∇uN.Now, we derive the explicit form of the first-order shape derivative ofJ.Theorem 33.
ForC1,1 bounded domain Ω, the first-order shape derivative of the Kohn-Vogelius cost functional
(141)J(Ω)=12∫Ω‍|∇(uD-uN)|2dx
in the direction of a perturbation field V∈Θ, where Θ is defined by (13) and the state functions uD and uN satisfy the Dirichlet problem (4) and the Neumann problem (6), respectively, is given by
(142)dJ(Ω;V)=12∫Σ‍(λ2-(∇uD·n)2=12∫Σ‍(+2λκuN-(∇uN·τ)2)V·nds,
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(143)J(Ω)=12∫Ω‍|∇(uD-uN)|2dx,J(Ωt)=12∫Ωt‍|∇(uD,t-uN,t)|2dxt.
Let zt=uD,t-uN,t and z=uD-uN. Note that if zt=zt∘Tt, then we have Dzt=(Dzt∘Tt)(DTt). Hence
(144)Mt∇zt=(∇zt)∘Tt.
By this together with Lemma 16, we can write J(Ωt) as follows:
(145)J(Ωt)=12∫Ωt‍|∇zt|2dxt=12∫Ω‍It|(∇zt)∘Tt|2dx=12∫Ω‍It|Mt∇zt|2dx=12∫Ω‍〈∇zt,At∇zt〉.
Then we write J(Ωt)-J(Ω) as
(146)J(Ωt)-J(Ω)=12∫Ω‍(〈∇zt,At∇zt〉-〈∇z,∇z〉)dx=12∫Ω‍((|∇zt|2-|∇z|2)〈∇zt,(At-I)∇zt〉+(|∇zt|2-|∇z|2)=12∫Ω‍(+(|∇zt|2-|∇z|2))dx=:J1(t)+J2(t).
Using Lemma 11, the symmetry of A, and noting that zt→z as t→0 we have
(147)J1(t)=12∫Ω‍〈∇zt,(At-I)∇zt〉dx,J1(0)=0.
Hence,
(148)J˙1(0)=limt→0J1(t)-J1(0)t=12limt→0∫Ω‍〈∇zt,At-It∇zt〉dx=12∫Ω‍〈∇z,A∇z〉dx=12∫Ω‍〈∇(uD-uN),A∇(uD-uN)〉dx=12∫Ω‍A∇uD·∇uDdx+12∫Ω‍A∇uN·∇uNdx-∫Ω‍A∇uD·∇uNdx.
To manipulate J2(t) we use the identity a2-b2=(a-b)2+2b(a-b) as
(149)J2(t)=12∫Ω‍(|∇zt|2-|∇z|2)dx=12∫Ω‍|∇(zt-z)|2dx+∫Ω‍〈∇z,∇(zt-z)〉dx=:J21(t)+J22(t).J21(t) is manipulated as follows:
(150)J21(t)=12∫Ω‍|∇(zt-z)|2=12∫Ω‍|∇((uDt-uNt)-(uD-uN))|2≤∫Ω‍|∇(uDt-uD)|2+∫Ω‍|∇(uNt-uN)|2.
Applying Theorems 22 and 23 yields J˙21=0. J22(t) is treated as follows:
(151)J22(t)=∫Ω‍〈∇z,∇(zt-z)〉dx=∫Ω‍〈∇(uD-uN),∇((uDt-uD)-(uNt-uN))〉=∫Ω‍〈∇uD,∇(uDt-uD)〉-∫Ω‍〈∇uN,∇(uDt-uD)〉-∫Ω‍〈∇(uD-uN),∇(uNt-uN)〉dx.
Since uDt-uD∈H01(Ω), the variational equation for uD and uN implies
(152)J22(t)=-∫Ω‍〈∇(uD-uN),∇(uNt-uN)〉dx=∫Ω‍〈∇uN,∇(uD-uN)〉-∫Ω‍〈∇uNt,∇(uD-uN)〉dx.
Choosing uD-uN∈HΓ,01(Ω) as test function and applying (65) and (84), J22(t) can be written as
(153)J22(t)=λ∫Σ‍(uD-uN)-∫Ω‍〈At∇uNt,∇(uD-uN)〉+∫Ω‍〈(At-I)∇uNt,∇(uD-uN)〉=λ∫Σ‍(1-wt)(uD-uN)ds+∫Ω‍〈(At-I)∇uNt,∇(uD-uN)〉.
From this we obtain
(154)J˙22(0)=λ∫Σ‍uNdivΣV+∫Ω‍〈A∇uN,∇(uD-uN)〉,
and therefore,
(155)J˙2(0)=λ∫Σ‍uNdivΣV+∫Ω‍〈A∇uN,∇(uD-uN)〉.
Combining J˙1(0) and J˙2(0) we get
(156)dJ(Ω;V):=limt→0+J(Ωt)-J(Ω)t=J˙1(0)+J˙2(0)=12∫Ω‍A∇uD·∇uD+12∫Ω‍A∇uN·∇uN-∫Ω‍A∇uD·∇uN+λ∫Σ‍(divΣV)uN+∫Ω‍〈A∇uN,∇(uD-uN)〉=12∫Ω‍A∇uD·∇uD-12∫Ω‍A∇uN·∇uN+λ∫Σ‍(divΣV)uN.
We know from the previous section thatuD and uN exist in H2(Ω) since Ω is of class C1,1. Using this smoothness we can now apply Lemma 32 and write (156) as follows:
(157)dJ(Ω;V)=-12∫Σ‍(∇uD·n)2V·n-12[∫Σ‍|∇uN|2V·n-2λ∫Σ‍V·∇uN]+λ∫Σ‍divΣVuN=-12∫Σ‍(∇uD·n)2V·n+λ∫Σ‍(V·∇uN+uNdivΣV)-12∫Σ‍|∇uN|2V·n.
Since
(158)∫Σ‍((∇uN)·V+uNdivΣV)=∫Σ‍(λ+uNdivΣn)V·n
and |∇uN|2=λ2+(∇uN·τ)2, we obtain the first-order shape derivative of J:
(159)dJ(Ω;V)=12∫Σ‍(-λ2-(∇uN·τ)2=12∫Σ‍(+2λ2+2λuNκ-(∇uD·n)2)V·n.


## 4.1. Variational Forms of the Dirichlet and Neumann Problems

We recall that we are considering the shape optimization problem (7) where uD solves the pure Dirichlet problem (4) and uN solves the Neumann problem (6). As in [13], we consider the Hilbert space
(61)HΓ,01={φ∈H1(Ω):φ|Γ=0},
which is endowed with the norm
(62)|φ|12=∫Ω‍|∇φ|2dx,
and a linear manifold defined by
(63)HΓ,v1={φ∈H1(Ω):φ|Γ=v}
for v∈H1/2(Γ).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.FinduD∈H1(Ω) such that
(64)(∇uD,∇ψ)Ω=0∀ψ∈H01(Ω)uD=1onΓ,uD=0onΣ.
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.FinduN∈H1(Ω) such that
(65)(∇uN,∇φ)Ω-(λ,φ)Σ=0∀φ∈HΓ,01(Ω)uN=1onΓ.
It is also well known that (65) has a unique solution.


## 4.2. Analysis of State Variables in Deformed Domains

We now consider the class of perturbed problems:(66)minΩtJ(Ωt)=12∫Ωt‍|∇(uD,t-uN,t)|2dx,
where uD,t solves the pure Dirichlet problem
(67)-ΔuD,t=0inΩt,uD,t=1onΓ,uD,t=0onΣt,
and uN,t solves the Neumann problem
(68)-ΔuN,t=0inΩt,uN,t=1onΓ,∂uN,t∂nt=λonΣt.
Here, nt is the outward unit normal to the deformed free boundary Σt. The variational form of (67) is formulated as follows.FinduD,t∈H1(Ωt) such that
(69)(∇uD,t,∇ψt)Ωt=0∀ψt∈H01(Ωt)uD,t=1onΓ,uD,t=0onΣt.
It is known that (69) has a unique solution.Remark 21.
The functionut:Ωt→ℝ can be referred to as the reference domain by composing ut with Tt; that is,
(70)ut=ut∘Tt:Ω⟶ℝ
and by chain rule of differentiation, we get
(71)(∇ut)∘Tt=(DTt)-T∇ut=:Mt∇ut.LetuD,t be the solution of (69). Applying Lemma 16 for all ψt∈H01(Ωt) we have
(72)(∇uD,t,∇ψt)Ωt=∫Ωt‍∇uD,t·∇ψt=∫Ω‍It(∇uD,t∘Tt)·(∇ψt∘Tt).
Applying (71) and noting that ψt∈H01(Ω) because ψt∈H01(Ωt), we obtain
(73)(∇uD,t,∇ψt)Ωt=∫Ω‍ItMt∇uDt·Mt∇ψt=∫Ω‍ItMtTMt∇uDt·∇ψt=(At∇uDt,∇v)Ω,
where v=ψt∈H01(Ω). Hence, if uD,t solves the variational equation (69), then uDt=uD,t∘Tt satisfies the variational equation,
(74)(At∇uDt,∇v)Ω=0
for all v∈H01(Ω),  uDt=1 on Γ, and uDt=0 on Σ.Now we show thatuDt is the unique solution of (74) in H1(Ω). First, we show that yt=uDt-uD∈H01(Ω) is the unique solution to
(75)(At∇yt,∇v)Ω=-(At∇uD,∇v)Ω
for all v∈H01(Ω). The bilinear form bt(·,·):H01(Ω)×H01(Ω)→ℝ defined by
(76)bt(yt,v)=∫Ω‍At∇yt·∇v,∀yt,v∈H01(Ω)
is continuous, because
(77)|bt(yt,v)|=|∫Ω‍At∇yt·∇v|≤|At∇yt|L2(Ω)|∇v|L2(Ω)≤2|At|∞|∇yt|L2(Ω)|∇v|L2(Ω)=β|yt|H1(Ω)|v|H01(Ω).
The bilinear form is also coercive. To show this we recall that limt→0At=I uniformly on Ω-. This is equivalent to the statement
(78)∀ϵ>0∃δ>0suchthat|t|<δ⟹|At-I|∞<ϵ.
Let ϵ=1/2. So for sufficiently small t, |At-I|∞<1/2, and
(79)bt(yt,yt)=∫Ω‍At∇yt·∇yt=|∫Ω‍(At-I)∇yt·∇yt+|∇yt|2|≥|∇yt|L2(Ω)2-|At-I|∞|∇yt|L2(Ω)2≥12|yt|H01(Ω)2.
So bt(yt,v) is coercive.Next, we show that the functionalf:H01(Ω)→ℝ is bounded:
(80)|〈f,v〉H-1×H01|=|-∫Ω‍At∇uD·∇v|≤2|At|∞|∇uD|L2(Ω)|∇v|L2(Ω)=2|At|∞|uD|H01(Ω)|v|H01(Ω)≤C|v|H01(Ω).
Therefore, by the Lax-Milgram lemma, yt=uDt-uD is the unique solution to the variational equation
(81)(At∇yt,∇v)Ω=-(At∇uD,∇v)Ω,∀v∈H01(Ω).
This implies the existence of a unique solution uDt of (74) as verified below.LetuDt=yt+uD. Using (81) we obtain
(82)(At∇uDt,∇v)Ω=(At∇yt,∇v)Ω+(At∇uD,∇v)Ωhhhhhhhhhhhhhh∀v∈H01(Ω)=-(At∇uD,∇v)Ω+(At∇uD,∇v)Ω=0.
Thus (74) is satisfied. The boundary conditions are also satisfied because on Γ, yt=0 and uD=1 and on Σ, both yt and uD are zero. To show uniqueness, we let uDt and u~Dt be solutions of (74). This implies that there exist yt and y~t such that uDt=yt+uD and u~Dt=y~t+uD, where yt and y~t are solutions to (81). Taking the difference of uDt and u~Dt and considering that solution to (81) is unique, we get uDt=u~Dt.Next, we consider (68) whose variational form is formulated as follows.FinduN,t∈H1(Ωt) such that
(83)(∇uN,t,∇φt)Ωt-(λ,φt)Σt=0∀φt∈HΓ,01(Ωt),uN,t=1onΓ.
Similarly, if uN,t solves the variational problem (83), then uNt solves the variational equation
(84)(At∇uNt,∇ϕ)Ω-(wtλ,ϕ)Σ=0,∀ϕ∈HΓ,01(Ω),
where uNt=1 on Γ.As shown before, the bilinear formbt(·,·):HΓ,01(Ω)×HΓ,01(Ω)→ℝ defined by
(85)bt(zt,v)=∫Ω‍At∇zt·∇v,∀zt,v∈HΓ,01(Ω),
is coercive and continuous. The linear functional f:HΓ,01(Ω)→ℝ defined by 〈f,ϕ〉=∫Σ‍λwtϕ is continuous on HΓ,01(Ω) because
(86)|〈f,ϕ〉H-1×H01|=|∫Σ‍λwtϕ|≤|λ||wt|∞∫Σ‍|ϕ|=|λ||wt|∞|Σ|1/2|γϕ|L2(Σ)≤|λ||wt|∞|Σ|1/2∥γ∥|ϕ|H1(Ω).
By the Lax-Milgram lemma, zt=uNt-1∈HΓ,01(Ω) is the unique solution in HΓ,01(Ω) of
(87)∫Ω‍At∇zt·∇ϕ=λ∫Σ‍wtϕ,∀ϕ∈HΓ,01(Ω).
Let uNt=zt+1. Then, by (87), we get
(88)(At∇uNt,∇ϕ)Ω-(wtλ,ϕ)Σ=(At∇zt,∇ϕ)Ω-(wtλ,ϕ)Σ=0ϕ∈HΓ,01(Ω).
Since zt∈HΓ,01(Ω), uNt=zt+1=1 on Γ. Uniqueness of uNt follows from the uniqueness of zt. Therefore, uNt is the unique solution of the variational problem (84) in H1(Ω).


## 4.3. Hölder Continuity of the States

We show thatuDt and uNt are Hölder continuous on t.Theorem 22 (see [13]).
The solutionsuDt of (74) are uniformly bounded in H1(Ω) for t∈(-tV,tV) and
(89)limt→01t|uDt-uD|H1(Ω)2=0,
where uD is the weak solution of (4).Proof.
We first prove the uniform boundedness ofuDt in H1(Ω) for t∈(-tV,tV). Since uDt-uD∈H01(Ω), by using coercivity of At we get
(90)(At∇(uDt-uD),∇(uDt-uD))Ω≥12|∇(uDt-uD)|H01(Ω)2.
Also, by applying (81), we have
(91)(At∇(uDt-uD),∇(uDt-uD))Ω=-(At∇uD,∇(uDt-uD))Ω≤|At|∞|uD|H1(Ω)|uDt-uD|H01(Ω).
Therefore,
(92)|uDt|H1(Ω)≤|uDt-uD|H01(Ω)+|uD|H1(Ω)≤tV+|uD|H1(Ω).
Now we take the difference between the weak form of (4) and the variational equation (74), to get
(93)(At∇uDt,∇v)Ω-(∇uD,∇v)Ω=0,∀v∈H01(Ω)(At∇1t(uDt-uD),∇v)Ω=-((At-I)t∇uD,∇v)Ω.
Note that uD,t∈HΓ,11(Ωt). This implies that uDt=uD,t∘Tt is in HΓ,11(Ω). Note also that uD∈HΓ,11(Ω). Let yt=(1/t)(uDt-uD). So for sufficiently small t, yt∈H01(Ω). Now choosing v=yt as a test function and by the uniform coercivity of At one obtains
(94)α|∇yt|L2(Ω)2≤(At∇yt,∇yt)Ω=-(1t(At-I)∇uD,∇yt)Ω≤|1t(At-I)|∞|∇uD|L2(Ω)|∇yt|L2(Ω).
If∇yt=0, then the inequality above holds. For ∇yt≠0 we have
(95)α|∇yt|L2(Ω)2≤|1t(At-I)∇uD|L2(Ω)|∇yt|L2(Ω)≤|1t(At-I)|∞|∇uD|L2(Ω)|∇yt|L2(Ω).
Hence,
(96)|1t∇(uDt-uD)|L2(Ω)≤1α|1t(At-I)|∞|∇uD|L2(Ω).
Squaring and multiplying t on both sides of the inequality give us
(97)1t|∇(uDt-uD)|L2(Ω)2≤t1α2|1t(At-I)|∞2|∇uD|L2(Ω)2.
Since |(1/t)(At-I)|∞2 is uniformly bounded in t by Lemma 11,
(98)limt→01t|uDt-uD|H1(Ω)2=limt→01t|∇(uDt-uD)|L2(Ω)2=0
follows.Theorem 23.
The solutionsuNt of (84) are uniformly bounded in H1(Ω) for t∈(-tV,tV) and
(99)limt→01t|uNt-uN|H1(Ω)2=0,
where uN is the solution of (65).Proof.
Subtracting (65) from (84) for all φ∈HΓ,01(Ω) we get
(100)0=(At∇uNt,∇φ)Ω-(wtλ,φ)Σ-(∇uN,∇φ)Ω+(λ,φ)Σ=(At∇uNt-∇uNt+∇uNt-∇uN,∇φ)Ω-(wtλ-λ,φ)Σ.
Hence
(101)(∇(uNt-uN),∇φ)Ω=-((At-I)∇uNt,∇φ)Ω+(wtλ-λ,φ)Σ.
Note that uNt-uN belongs to HΓ,01(Ω). Hence
(102)|uNt-uN|H1(Ω)2=-((At-I)∇uNt,∇(uNt-uN))Ω+λ(wt-1,uNt-uN)Σ.
By Cauchy-Schwarz inequality, we obtain
(103)|uNt-uN|H1(Ω)2≤|(At-I)∇uNt|L2(Ω)|∇(uNt-uN)|L2(Ω)+|λ||wt-1|∞|uNt-uN|L2(Σ).
Furthermore, by trace theorem we have |uNt-uN|L2(Σ)≤c|uNt-uN|H1(Ω). Therefore,
(104)|uNt-uN|H1(Ω)2≤C(|(At-I)∇uNt|L2(Ω)+|λ||wt-1|∞)|uNt-uN|H1(Ω)
holds, where C=max{1,c}. This implies
(105)|uNt-uN|H1(Ω)≤C(|(At-I)∇uNt|L2(Ω)+|wtλ-λ|L2(Σ)),
which entails
(106)1t|uNt-uN|H1(Ω)≤C(|1t(At-I)|L2(Ω)|∇uNt|L2(Ω)≤C(+|λ||Σ|1/2|1t(wt-1)|∞|1t(At-I)|L2(Ω))=tC(|1t(At-I)|L2(Ω)|∇uNt|L2(Ω)tCh(+|λ||Σ|1/2|1t(wt-1)|∞|1t(At-I)|L2(Ω)).
We now show that∇uNt is bounded uniformly in t in L2(Ω). Since uNt-1∈HΓ,01(Ω), we have
(107)α|∇uNt|L2(Ω)2=α|∇(uNt-1)|L2(Ω)2≤∫Ω‍At∇(uNt-1)·∇(uNt-1)=∫Ω‍At∇uNt·∇(uNt-1)=|λ|∫Σ‍wt(uNt-1)≤c|λ||wt|L∞|Σ|1/2|γ||∇uNt|L2(Ω).
Consequently,
(108)|∇uNt|L2(Ω)≤|λ|αc|wt|L∞|Σ|1/2|γ|,
and this shows that ∇uNt is uniformly bounded in L2(Ω) because |wt|∞ is bounded. In addition, At and wt are differentiable at t=0 by Lemma 11. Therefore,
(109)limt→01t|uNt-uN|H1(Ω)2=0.


## 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 uD,uN∈H1(Ω). For C1,1 domains, we show that these solutions also exist in H2(Ω) and more generally in Hk+2(Ω) if domains are of class Ck+1,1, k≥0.To prove higher regularity ofuD, we require the following two theorems, which are proven in [22].Theorem 24 (see [22, page 124]).
LetΩ be a bounded open subset of ℝn with a C1,1 boundary. Consider the Dirichlet boundary value problem:
(110)Au=finΩ,u=gon∂Ω,
where
(111)A=∑i,j=1n‍Di(aijDju)+∑i=1n‍aiDiu+a0u.
Let ai,j be uniformly Lipschitz functions and let ai be bounded measurable functions such that aj,i=ai,j, 1≤i, j≤n and that there exists α>0 with
(112)∑i,j=1n‍ai,j(x)ζiζj≤-α|ζ|2
for all ζ∈ℝn and for almost every x∈Ω-. Assume in addition that either (i)
ai=0,   1≤i≤n and a0≥0 a.e. or(ii)
a0≥β>0 a.e.Then for every f∈Lp(Ω) and every g∈W2-1/p,p(∂Ω), there exists a unique u∈W2,p(Ω) that solves (110).Theorem 25 (see [22, page 128]).
LetΩ be a bounded open subset of ℝn with a Ck+1,1 boundary. Consider the operator A defined by (110) with aij=aji∈Ck,1(Ω-), and assume that there exists α>0 such that (112) holds for all ζ∈ℝn and for every x∈Ω-. Also, consider a real boundary operator B which is either the identity operator (orderd=0) or Bu=∑i=1n‍biDiu with bi∈Ck,1(Ω-), 1≤i≤n(orderd=1) and ∑i=1n‍bini≠0(niistheithcomponentofn) everywhere on Γ:=∂Ω. Furthermore, assume that u∈W2,p(Ω) satisfies Au=f∈Wk,p(Ω) and γBu=g∈W2+k-d-1/p,p(Γ). Then u∈Wk+2,p(Ω).We will also justify the higher regularity ofuN. We use the following results whose proofs are given in the corresponding texts.Theorem 26 (see [23, page 316]).
LetΩ be a bounded open subset of ℝn. Suppose u∈H1(Ω) is a weak solution of the PDE
(113)Lu=finΩ,
where
(114)Lu=-∑i,j=1n‍(aij(x)uxi)xj+∑i=1n‍bi(x)uxi+c(x)u.
Assume furthermore that aij,bi,c∈C∞(Ω) for i,j=1,2,…,n, and f∈C∞(Ω). Then u∈C∞(Ω).Theorem 27 (see [24, page 12]).
LetΩ⊂ℝn be a bounded domain with Ck+1,1 boundary ∂Ω for some nonnegative integer k. Suppose the data f and g of the problem
(115)-Δu=finΩ,u=gon∂Ω
are in Wk,p(Ω) and Wk+2-1/p,p(∂Ω), respectively, for some real number p with 1<p<∞. Then u∈Wk+2,p(Ω).For proof, see [22].Theorem 28 (see [22, page 84]).
LetΩ⊂ℝn be a bounded domain with C1,1 boundary Γ and 1<p<∞. Consider the Neumann problem
(116)-Δu+a0u=finΩ,γ∂u∂n=gonΓ.
If 0<a0∈L∞(Ω), f∈Lp(Ω), and g∈W1-1/p,p(∂Ω), then the weak solution u to (116) exists in W2,p(Ω).For proof, see [22].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.Theorem 29.
LetΩ be a bounded domain with boundary of class C1,1. Let uD,uN∈H1(Ω) be weak solutions of the BVPs (4) and (6), respectively. Then uD and uN also belong to H2(Ω). More generally, if Ω is of class Ck+1,1, where k is a nonnegative integer, then uD and uN are elements of Hk+2(Ω).Proof.
We first consider the solutionuD∈H1(Ω) to the Dirichlet problem (4). We use Theorem 25 to show that uD is an element of H2(Ω). Here, (110) is applied with the following settings.
We considern=2. The domain Ω is of class C1,1. L=-Δ, and hence aij=aji=-1 for i=j and aij=aji=0 for i≠j, with i,j=1,2. We also observe that ∑i,j=12‍aij(x)ζiζj=-(ζ1)2-(ζ2)2=-1|ζ|2, for all ζ=(ζ1,ζ2)T. Thus α=1. Furthermore, we have the following data: f=0∈L2(Ω),  g=1∈H3/2(Γ), g=0∈H3/2(Σ). Therefore, by using Theorem 24, there exists a unique u=uD∈H2(Ω), which is a solution to (4).
For higher regularity ofuD we apply Theorem 25. At first we consider C2,1-domains. In this case, k=1. We have aij=aji=-1 for i=j and aij=aji=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∈H5/2(Γ), g=0∈H5/2(Σ). Therefore, by applying Theorem 25, we have u=uD∈H3(Ω). In general, for smoother domains with Ck+1,1 boundaries, solutions to (4) are elements of Hk+2.
Next, we recall that, forC1,1 domain, there is a weak solution uN∈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 Ck+1,1, then uN is in Hk+2(Ω), where k is a nonnegative integer. For this purpose we need Theorem 26 which implies uN∈C∞(Ω).
Choose a bounded connected domainG with C∞ boundary Γ1 such that A-⊂G and G-⊂B, 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:
(117)-Δw=0inΩ1,w=1onΓ,w=uNonΓ1.
Since Ω1 is bounded with compact boundaries, we have C∞(Γ∪Γ1)⊂Hk+3/2(Γ∪Γ1). So by applying Theorem 27, we get w∈Hk+2(Ω1). Since uN|Ω1 also solves (117), then by uniqueness we have w=uN|Ω1. If A is a domain with C1,1 boundary Γ, then by Theorem 27 we have uN|Ω1∈H2(Ω1). Moreover, if A is a Ck+1,1-domain, then uN|Ω1 is in Hk+2(Ω1).
Second, we consider the following boundary value problem:(118)-Δv+v=uNinΩ2,∂v∂n=∂uN∂nonΓ1,∂v∂n=λonΣ.
Because uN|Ω2∈H1(Ω2), it follows that uN|Ω2∈L2(Ω2). We have also shown that uN∈C∞(Ω). This implies that uN∈C∞(Γ1). Hence, (∂uN/∂n)∈H1/2(Γ1). Also, λ∈H1/2(Σ). Since B is a domain of class C1,1, then by Theorem 28, we infer that (118) has a unique solution v∈H2(Ω2). Note, however, that uN|Ω2 also solves (118). So by uniqueness of the solution, we get uN|Ω2=v. Therefore, uN|Ω2∈H2(Ω2). Now, if domain B is of class Ck+1,1 and uN|Ω2∈H2(Ω2)⊂H1(Ω2), we get v∈H3(Ω2) by applying Theorem 25 and so is uN|Ω2. Doing this recursively, we end up with uN|Ω2∈Hk+2(Ω).
Hence, forCk+1,1-domains Ω(:=B∖A-), if we combine uN|Ω1∈Hk+2(Ω1) and uN|Ω2∈Hk+2(Ω2), we get uN∈Hk+2(Ω1∪Ω2). Moreover, uN is C∞ in a neighborhood of Γ1 because uN∈C∞(Ω). Therefore, uN∈Hk+2(Ω).Remark 30.
In the computation of the first-order shape derivative, since we are dealing withC1,1-domains, we may consider H2-regularity for the solutions uD and uN, as justified by Theorem 29.


## 4.5. The Shape Derivative ofJ

First, we state and prove the following lemma.Lemma 31.
LetΩ⊂ℝn be a bounded Lipschitz domain. Then the following equation:
(119)-∫Ω‍(divF)gdx=∫Ω‍F·∇gdx-∫∂Ω‍g(F·n)ds
is valid for vector field F and scalar function g having the following regularity: (i)
F∈H1(Ω;ℝn) and g∈H1(Ω);(ii)
F∈C1(Ω-;ℝn) and g∈W1,1(Ω).Proof.
First we recall the Gauss’ divergence theorem inℝn saying that if a domain Ω⊂ℝn is a bounded Lipschitz domain, then we have
(120)∫Ω‍divFdx=∫∂Ω‍F·nds
for a vector field F∈C1(Ω-,ℝn). Second we take the divergence of the product of a scalar function g and the vector field F to get
(121)div(gF)=gdivF+F·∇g.
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 (divF)g∈L1(Ω),  F·∇g∈L1(Ω), and the integral ∫∂Ω‍g(F·n) is bounded. Hence, (119) is well defined. Note that the formula
(122)-∫Ω‍(divF)gdx=∫Ω‍F·∇gdx-∫∂Ω‍g(F·n)ds
holds for F∈C1(Ω-;ℝn) and g∈C1(Ω-). We write
(123)-∫Ω‍(divF)gdx=∫Ω‍F·∇gdx-∫∂Ω‍γ0g(γ0F·n)ds,
where γ0:H1(Ω)→L2(∂Ω) is a trace operator. Let F∈H1(Ω;ℝn) and g∈H1(Ω). By density, we pick {Fk}⊂C1(Ω- and {gk}⊂C1(Ω-) such that Fk→F in H1(Ω-;ℝn) and gk→g in H1(Ω). By (122),
(124)-∫Ω‍(divFk)gkdx=∫Ω‍Fk·∇gkdx-∫∂Ω‍gk(Fk·n)ds.
Note that γ0Fk=Fk|∂Ω and γ0gk=gk|∂Ω. Also Fk→F in H1(Ω-;ℝn) implies that divFk→divF in L2(Ω). Moreover, gk→g in H1(Ω) implies ∇gk→∇g in L2(Ω). Furthermore, since γ0∈ℒ(H1(Ω),L2(∂Ω)), γ0Fk→γ0F and γ0gk→γ0g in L2(∂Ω). Therefore, (119) holds for F∈H1(Ω;ℝn) and g∈H1(Ω).
(
ii
) If F∈C1(Ω-;ℝn) and g∈W1,1(Ω), then (divF)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 C1(Ω-) in W1,1(Ω) we can show that (119) is also valid for this case.Now we apply Lemma31 to prove the next lemma.Lemma 32 (see [1]).
LetuD and uN belonging to H2(Ω) satisfy the Dirichlet problem (4) and the Neumann problem (6), respectively. Then
(125)∫Ω‍A∇uD·∇uD=-∫Σ‍(∇uD·n)2V·n,(126)∫Ω‍A∇uN·∇uN=∫Σ‍|∇uN|2V·n-2λ∫Σ‍V·∇uN,
where A is given by property (12) of Lemma 11.Proof.
From Lemma11, we recall the expression for A, which is given by A=(divV)I-((DV)+(DV)T). Our first goal is to derive an expression for ∫Ω‍A∇u·∇v for u,v∈H2(Ω). We begin by writing ∫Ω‍A∇u·∇v as follows:
(127)∫Ω‍A∇u·∇v=∫Ω‍(divV)∇u·∇v-∫Ω‍(DV)∇u·∇v-∫Ω‍(DV)T∇u·∇v.
We manipulate each term on the right-hand side of (127). First, because u,v∈H2(Ω), we have ∇u·∇v∈W1,1(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
(128)-∫Ω‍(divV)∇u·∇v=∫Ω‍∇(∇u·∇v)·V-∫Ω‍(divV)∇u·∇v-∫Σ‍(∇u·∇v)V·n.
The other two terms on the right hand side of (127) are manipulated as follows. The term ∇(∇u·∇v)·V is written as
(129)∇(∇u·∇v)·V=(∇v)T((∇2u)V)+VT((∇2v)∇u),
where ∇2u represents the Hessian of u. Because Hessian is symmetric, we obtain
(130)∇(∇u·∇v)·V=(∇2u∇v+∇2v∇u)·V.
Substituting (130) into (128) we get
(131)∫Ω‍(divV)∇u·∇v=-∫Ω‍((∇2u)∇v+(∇2v)∇u)·V+∫Σ‍(∇u·∇v)V·n.
Next, we expand the expression div((V·∇u)∇v) using (121) as
(132)div((V·∇u)∇v)=(V·∇u)Δv+∇v·(V·∇u)=(V·∇u)Δv+∇v·[(DV)T∇u+∇2uV]=(DV)∇v·∇u+(∇v)T(∇2u)V+(V·∇u)Δv.
Integrating both sides of (132) over Ω, applying Stoke’s theorem, and considering V=0 on Γ we end up with
(133)∫Ω‍(DV)∇v·∇u+∫Ω‍(∇v)T(∇2u)V+∫Ω‍(V·∇u)Δv=∫Σ‍(V·∇u)∇v·n,
or equivalently
(134)∫Ω‍(DV)∇v·∇u=-∫Ω‍(Δv)(V·∇u)-∫Ω‍(∇2u)∇v·V+∫Σ‍∂v∂n(V·∇u).
Interchanging v and u we get
(135)∫Ω‍(DV)∇u·∇v=-∫Ω‍(Δu)V·∇v-∫Ω‍(∇2v)∇u·V+∫Σ‍∂u∂n(V·∇v).
Also, because (DV)T∇v·∇u=(DV)∇u·∇v, we obtain
(136)∫Ω‍(DV)T∇v·∇u=-∫Ω‍(Δu)V·∇v-∫Ω‍(∇2v)∇u·V+∫Σ‍∂u∂n(V·∇v).
Thus,
(137)∫Ω‍(DV)T∇u·∇v=-∫Ω‍(Δv)V·∇u-∫Ω‍(∇2u)∇v·V+∫Σ‍∂v∂n(V·∇u).
Adding (131), (135), and (137) altogether, we express (127) as
(138)∫Ω‍A∇u·∇v=∫Ω‍Δu(V·∇v)+∫Ω‍Δv(V·∇u)-∫Σ‍∂v∂n(V·∇u)-∫Σ‍∂u∂n(V·∇v)+∫Σ‍(∇u·∇v)V·n.
Set u=v=uD in (138). The first two integrals on the right hand side of (138) vanish because -ΔuD=0 in Ω. Moreover, since uD=0 on Σ we have ∇uD=(∂uD/∂n)n. Thus, we can write (138) as follows:
(139)∫Ω‍A∇uD·∇uD=-2∫Σ‍(∇uD·n)(V·∇uD)+∫Σ‍(∇uD·∇uD)V·n=-2∫Σ‍(∂uD∂n)(∂uD∂nV·n)+∫Σ‍(∂uD∂n)2V·n=-∫Σ‍(∇uD·n)2V·n.
Therefore, (125) is satisfied.
On the other hand, by replacing bothu and v by uN and by considering that -ΔuN=0 in Ω and ∂uN/∂n=λ on Σ, we derive (126) as
(140)∫Ω‍A∇uN·∇uN=-2∫Σ‍∂uN∂n(V·∇uN)+∫Σ‍(∇uN·∇uN)V·n=∫Σ‍|∇uN|2V·n-2λ∫Σ‍V·∇uN.Now, we derive the explicit form of the first-order shape derivative ofJ.Theorem 33.
ForC1,1 bounded domain Ω, the first-order shape derivative of the Kohn-Vogelius cost functional
(141)J(Ω)=12∫Ω‍|∇(uD-uN)|2dx
in the direction of a perturbation field V∈Θ, where Θ is defined by (13) and the state functions uD and uN satisfy the Dirichlet problem (4) and the Neumann problem (6), respectively, is given by
(142)dJ(Ω;V)=12∫Σ‍(λ2-(∇uD·n)2=12∫Σ‍(+2λκuN-(∇uN·τ)2)V·nds,
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(143)J(Ω)=12∫Ω‍|∇(uD-uN)|2dx,J(Ωt)=12∫Ωt‍|∇(uD,t-uN,t)|2dxt.
Let zt=uD,t-uN,t and z=uD-uN. Note that if zt=zt∘Tt, then we have Dzt=(Dzt∘Tt)(DTt). Hence
(144)Mt∇zt=(∇zt)∘Tt.
By this together with Lemma 16, we can write J(Ωt) as follows:
(145)J(Ωt)=12∫Ωt‍|∇zt|2dxt=12∫Ω‍It|(∇zt)∘Tt|2dx=12∫Ω‍It|Mt∇zt|2dx=12∫Ω‍〈∇zt,At∇zt〉.
Then we write J(Ωt)-J(Ω) as
(146)J(Ωt)-J(Ω)=12∫Ω‍(〈∇zt,At∇zt〉-〈∇z,∇z〉)dx=12∫Ω‍((|∇zt|2-|∇z|2)〈∇zt,(At-I)∇zt〉+(|∇zt|2-|∇z|2)=12∫Ω‍(+(|∇zt|2-|∇z|2))dx=:J1(t)+J2(t).
Using Lemma 11, the symmetry of A, and noting that zt→z as t→0 we have
(147)J1(t)=12∫Ω‍〈∇zt,(At-I)∇zt〉dx,J1(0)=0.
Hence,
(148)J˙1(0)=limt→0J1(t)-J1(0)t=12limt→0∫Ω‍〈∇zt,At-It∇zt〉dx=12∫Ω‍〈∇z,A∇z〉dx=12∫Ω‍〈∇(uD-uN),A∇(uD-uN)〉dx=12∫Ω‍A∇uD·∇uDdx+12∫Ω‍A∇uN·∇uNdx-∫Ω‍A∇uD·∇uNdx.
To manipulate J2(t) we use the identity a2-b2=(a-b)2+2b(a-b) as
(149)J2(t)=12∫Ω‍(|∇zt|2-|∇z|2)dx=12∫Ω‍|∇(zt-z)|2dx+∫Ω‍〈∇z,∇(zt-z)〉dx=:J21(t)+J22(t).J21(t) is manipulated as follows:
(150)J21(t)=12∫Ω‍|∇(zt-z)|2=12∫Ω‍|∇((uDt-uNt)-(uD-uN))|2≤∫Ω‍|∇(uDt-uD)|2+∫Ω‍|∇(uNt-uN)|2.
Applying Theorems 22 and 23 yields J˙21=0. J22(t) is treated as follows:
(151)J22(t)=∫Ω‍〈∇z,∇(zt-z)〉dx=∫Ω‍〈∇(uD-uN),∇((uDt-uD)-(uNt-uN))〉=∫Ω‍〈∇uD,∇(uDt-uD)〉-∫Ω‍〈∇uN,∇(uDt-uD)〉-∫Ω‍〈∇(uD-uN),∇(uNt-uN)〉dx.
Since uDt-uD∈H01(Ω), the variational equation for uD and uN implies
(152)J22(t)=-∫Ω‍〈∇(uD-uN),∇(uNt-uN)〉dx=∫Ω‍〈∇uN,∇(uD-uN)〉-∫Ω‍〈∇uNt,∇(uD-uN)〉dx.
Choosing uD-uN∈HΓ,01(Ω) as test function and applying (65) and (84), J22(t) can be written as
(153)J22(t)=λ∫Σ‍(uD-uN)-∫Ω‍〈At∇uNt,∇(uD-uN)〉+∫Ω‍〈(At-I)∇uNt,∇(uD-uN)〉=λ∫Σ‍(1-wt)(uD-uN)ds+∫Ω‍〈(At-I)∇uNt,∇(uD-uN)〉.
From this we obtain
(154)J˙22(0)=λ∫Σ‍uNdivΣV+∫Ω‍〈A∇uN,∇(uD-uN)〉,
and therefore,
(155)J˙2(0)=λ∫Σ‍uNdivΣV+∫Ω‍〈A∇uN,∇(uD-uN)〉.
Combining J˙1(0) and J˙2(0) we get
(156)dJ(Ω;V):=limt→0+J(Ωt)-J(Ω)t=J˙1(0)+J˙2(0)=12∫Ω‍A∇uD·∇uD+12∫Ω‍A∇uN·∇uN-∫Ω‍A∇uD·∇uN+λ∫Σ‍(divΣV)uN+∫Ω‍〈A∇uN,∇(uD-uN)〉=12∫Ω‍A∇uD·∇uD-12∫Ω‍A∇uN·∇uN+λ∫Σ‍(divΣV)uN.
We know from the previous section thatuD and uN exist in H2(Ω) since Ω is of class C1,1. Using this smoothness we can now apply Lemma 32 and write (156) as follows:
(157)dJ(Ω;V)=-12∫Σ‍(∇uD·n)2V·n-12[∫Σ‍|∇uN|2V·n-2λ∫Σ‍V·∇uN]+λ∫Σ‍divΣVuN=-12∫Σ‍(∇uD·n)2V·n+λ∫Σ‍(V·∇uN+uNdivΣV)-12∫Σ‍|∇uN|2V·n.
Since
(158)∫Σ‍((∇uN)·V+uNdivΣV)=∫Σ‍(λ+uNdivΣn)V·n
and |∇uN|2=λ2+(∇uN·τ)2, we obtain the first-order shape derivative of J:
(159)dJ(Ω;V)=12∫Σ‍(-λ2-(∇uN·τ)2=12∫Σ‍(+2λ2+2λuNκ-(∇uD·n)2)V·n.


## 5. Conclusion

In this paper we derived the explicit form of the first-order 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 C1,1 boundaries and we need H2 regularity for the state variables.Rewriting the first-order shape derivative asdJ(Ω;V)=∫Σ‍FV·n, where
(160)F=12(-(∇uN·τ)2-(∇uD·n)2+λ2+2λuNκ),
we conclude that J is shape differentiable at Ω. This is because dJ(Ω;V) exists for all V∈Θ and the mapping V↦dJ(Ω;V) is linear and continuous with respect to V∈Θ since
(161)|dJ(Ω;V)|≤|F|L1(Σ)|V|C(Σ)≤|F|L1(Σ)|V|C1,1(U-).We also observe that the shape derivative ofJ which is formulated from the Bernoulli free boundary problem depends on the normal component of the deformation field V at the free boundary Σ; that is, there exists a function gΩ defined on the free boundary Σ such that
(162)dJ(Ω;V)=∫Σ‍gΩV·nds.
This agrees with the Hadamard structure theorem [26, 27].Theorem 34 (see [26, page 318]).
LetΩ be a domain with Ck+1 boundary Γ for some integer k≥0. Assuming that at Ω a shape gradient of J(Ω) exists. Then there exists a scalar distribution g in 𝒟k(Γ)′ such that
(163)dJ(Ω;V)=〈g(Ω),V·n〉𝒟k(Γ)′×𝒟k(Γ),
where 𝒟k(Γ) is the space of Ck functions from Γ to ℝn, and V·n is the normal component of V on Γ.Proof.
See [26].



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