# Optimal Form for Compliance of Membrane Boundary Shift in Nonlinear Case

Authors: Y. Yousfi; I. Hadi; A. Benbrik
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

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## Abstract

In this work, we search the existence shifting compliance optimal form of some boundary membrane, which is not elastic and not isotropic, generating nonlinear PDE. An optimal form of the elastic membrane described by thep-Laplacian is investigated. The boundary perturbation method due to Hadamard is applied in Sobolev spaces.


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## Body

## 1. Introduction and Preliminaries

In this work we will study the geometric shape optimization of forms, where the main idea is to vary the edge position of a form, without changing its topology which remains the same. We use a membrane model as shown in Figure1. At rest the membrane occupies a reference domain Ω whose edge is divided into three disjoint parts:(1)∂Ω=Γ∪ΓN∪ΓDwhere Γ is the free variable part, ΓD is the fixed part of the boundary (Dirichlet boundary conditions), and ΓN is also the free part of the boundary on which we apply the efforts g∈Lp(ΓN)N (Neumann boundary condition). The three of parts of the boundary are supposed to be nonzero surface measurements, as we suppose that the free boundary variable Γ responds to homogenous Neumann condition. So the vertical displacement u is the solution of the following membrane model:(2)-Δu=0inΩu=0onΓD∂u∂n=gonΓN∂u∂n=0onΓWe want to minimize the compliance defined by J(Ω)=∫ΓN‍gudx whenever x∈Ω.Figure 1
Membrane.The shape optimization problem isinfΩ∈Uad⁡JΩ where it remains to define the set of admissible forms.


### 1.1. Existence under a Condition of Regularity

The main idea of this section is to apply a regularity constraint on all the admissible formsUad, to demonstrate a result of existence of optimal forms. The results and demonstrations are mainly due to F. Murat and J. Simons [1, 2]. It rests on a very significant restriction of Uad; in other words, Ω is obtained by applying a regular diffeomorphism T to the reference domain Ω0. We first define a diffeomorphism set:(3)τ=Tsuch  thatT-Id∈W1,∞RN,RN;T-1-Id∈W1,∞RN,RNThen we define a set of the admissible forms obtained by deformation of Ω:(4)CΩ0=Ωsuch  that∃T∈τ;Ω=TΩ0Finally we introduce a pseudo-distance on C(Ω0):(5)dΩ1,Ω2T∈τ/TΩ1=Ω0=inf⁡T-IdW1,∞RN,RN+T-1-IdW1,∞RN,RNThus, we introduce a condition of uniform regularity of the permissible forms, that is to say, Ω open sets close to Ω0 in the sense of pseudo-distance; for each R>0 we pose Uad = {Ω∈C(Ω0) such that d(Ω,Ω0)≤R;ΓD∪ΓN⊂∂Ω and ∫Ω‍dx=V0} where V0 is an imposed volume. The result is the following theorem.Theorem 1.
For all objective functions, the shape optimization probleminfΩ∈Uad⁡J(Ω) admits at least a minimum point.


### 1.2. Derivation from the Domain

The boundary variation method that we study is a classical idea well known and used before by Hadamard [3] in 1907 and many others as [4–12]. We will adopt the same representation as F. Murat and J. Simons [1]. In fact, let Ω0 be an open regular bounded referential domain of RN and the admissible form class C(Ω0) composed of the open sets such as Ω=T(Ω0) where T is a Lipschitz diffeomorphism and(6)T=Id+θwithθ∈W1,∞RN,RNwhere Id is the identity application, and we note Ω=(Id+θ)(Ω0) defined by(7)Ω=x+θxsuchthatx∈Ω0Lemma 2 (See [13]).
For allθ∈W1,∞(RN,RN) satisfying θW1,∞(RN,RN)<1, the application T=Id+θ is a bijection of RN and T∈τ.Definition 3.
LetJ be application from C(Ω0) to R. One says that it is differentiable with respect to the domain at Ω0 if the function(8)Θ→JId+θΩ0is Frechet differentiable at 0 in the Banach space W1,∞(RN,RN). i.e., ∃L, a linear continuous form on W1,∞(RN,RN), such that(9)JId+θΩ0=JΩ0+J′Ω0+∘θ,withlimθ→O⁡∘θθW1,∞=0.The linear form J′(Ω0) depends only on the normal component of θ on the boundary of Ω0.Proposition 4.
LetΩ0 be a regular bounded open set of RN. Let J be a differentiable application on Ω0. If θ1,θ2∈W1,∞(RN,RN) are such that θ1-θ2∈C1(RN,RN) such that θ1.n=θ2.n on ∂Ω0, then the derivative J′(Ω0) is verifying:(10)J′Ω0θ1=J′Ω0θ2


### 1.3. Derivation of Integrals

Since the complianceJ is defined by surface or volume integrals then its differentiation devotes the following tools.Lemma 5 (see [1, 7]).
LetΩ0 be an open set of RN. Let T∈τ and 1≤p≤+∞.Then f∈Lp(T(Ω0)) iff foT∈Lp(Ω0) and one has(11)∫TΩ0fdx=∫Ω0foTdet⁡∇Tdxand∫TΩ0fdet⁡∇T-1dx=∫Ω0foTdxOn the other hand f∈W1,p(T(Ω0)) iff foT∈W1,p(T(Ω0)) and one has(12)∇foT=∇T-1t∇foTProposition 6 (See [13]).
LetΩ0 be a regular bounded open set of RN. Let f∈W1,1(RN) and J be an application from C(Ω0) to R defined by J(Ω)=∫Ω‍f(x)dx. Then J is differentiable in Ω0 and(13)J′Ω0θ=∫Ω0divθxfxdx=∫∂Ω0θxnxfxdSwithθ∈W1,∞RN,RNNow we move to a lemma on the change of variables in surfaces integrals.Lemma 7 (See [1, 7]).
LetΩ0 be an open set of RN. Let T∈τ be a class C1 diffeomorphism of RN, and 1≤p≤+∞.Then f∈Lp(T(Ω0)) iff foT∈Lp(Ω0) and one has(14)∫∂TΩ0fdS=∫∂Ω0f∘Tdet⁡∇T-1∇T-1tnRNdSwhere n is the external normal to ∂Ω0.The surface integral derivative of a function with respect to the domain is given by the following proposition.Proposition 8 (See [13]).
LetΩ0 be a regular bounded open set of RN. Let f∈W2,1(RN) and J be an application from C(Ω0) to R defined by J(Ω)=∫Ω‍f(x)dx. Then J is differentiable in Ω0 and(15)J′Ω0θ=∫∂Ω0∇f.θ+fdivθ-∇θn.n=∫∂Ω0θ.n∂f∂n+HfdSwithθ∈C1RN,RNwhere H is the average curvature of ∂Ω0 defined by H=divn.


### 1.4. Derivation of a Domain Dependent Function

In this section we try to derive a function depending on the domain; for this we use the EulerianU or Lagrangian L derivative. The second is a more reliable concept than the first. Let u(x,Ω) be a function defined for all x∈Ω and depending on Ω. It represents a solution of an PDE posed in Ω. In a point x belonging to both Ω0 and Ω=(Id+θ)(Ω0), we can calculate the differential u(Ω,x): (16)uId+θΩ0,x=uΩ0,x+Uθ,x+oθsuch  thatlimθ↦0⁡oθθ=0U is a linear continuous form in θ; it represents a directional derivative in the direction θ. This definition makes sense in the case where x∈Ω, but it poses a problem if x∈Ω0. Then in this case we use the Lagrangian derivative; for this we build the transported u¯(θ) on Ω0.By changing variables we obtainu¯(θ,x)=u((Id+θ)(Ω0)o(Id+θ))=u((Id+θ)(Ω0),x+θ(x)).To arrive at the derivative Lagrangian by driftingu¯(θ,x) with respect to θ(17)u¯θ,x=uθ,x+Lθ,x+oθsuch  thatlimθ↦0⁡oθθ=0.L is a linear continuous form in θ; it represents a directional derivative in the direction θ.There is a relation between these two derivativesL(θ,x)=U(θ,x)+θ.∇u(Ω0,x).Proposition 9.
LetΩ0 be a regular bounded open set of RN. Let f(Ω) be an application from C(Ω0) to L1(R); one defines its transpose from W1,1(RN) to L1(R)f¯(θ,x)=f((Id+θ)(Ω0)o(Id+θ)) which we suppose to be derivable in 0 and L is considered as its derivative. So the application J1 from C(Ω0) to R defined by J1(Ω)=∫Ω‍f(Ω)dx is differentiable in Ω0 and for all θ∈W1,∞(RN,RN) one has(18)J1′Ω0θ=∫∂ΩfΩ0divθ+LθdSIn the same way, if f¯(θ) is derivable as an application from C1(RN,RN) to L1(∂Ω0), so the application J2 from C(Ω0) to R defined by J2(Ω)=∫∂Ω‍f(Ω)dx is differentiable in Ω0 and for all θ∈C1(RN,RN) one has(19)J2′Ω0θ=∫∂ΩfΩ0divθ-∇θn.n+YθdS


## 1.1. Existence under a Condition of Regularity

The main idea of this section is to apply a regularity constraint on all the admissible formsUad, to demonstrate a result of existence of optimal forms. The results and demonstrations are mainly due to F. Murat and J. Simons [1, 2]. It rests on a very significant restriction of Uad; in other words, Ω is obtained by applying a regular diffeomorphism T to the reference domain Ω0. We first define a diffeomorphism set:(3)τ=Tsuch  thatT-Id∈W1,∞RN,RN;T-1-Id∈W1,∞RN,RNThen we define a set of the admissible forms obtained by deformation of Ω:(4)CΩ0=Ωsuch  that∃T∈τ;Ω=TΩ0Finally we introduce a pseudo-distance on C(Ω0):(5)dΩ1,Ω2T∈τ/TΩ1=Ω0=inf⁡T-IdW1,∞RN,RN+T-1-IdW1,∞RN,RNThus, we introduce a condition of uniform regularity of the permissible forms, that is to say, Ω open sets close to Ω0 in the sense of pseudo-distance; for each R>0 we pose Uad = {Ω∈C(Ω0) such that d(Ω,Ω0)≤R;ΓD∪ΓN⊂∂Ω and ∫Ω‍dx=V0} where V0 is an imposed volume. The result is the following theorem.Theorem 1.
For all objective functions, the shape optimization probleminfΩ∈Uad⁡J(Ω) admits at least a minimum point.


## 1.2. Derivation from the Domain

The boundary variation method that we study is a classical idea well known and used before by Hadamard [3] in 1907 and many others as [4–12]. We will adopt the same representation as F. Murat and J. Simons [1]. In fact, let Ω0 be an open regular bounded referential domain of RN and the admissible form class C(Ω0) composed of the open sets such as Ω=T(Ω0) where T is a Lipschitz diffeomorphism and(6)T=Id+θwithθ∈W1,∞RN,RNwhere Id is the identity application, and we note Ω=(Id+θ)(Ω0) defined by(7)Ω=x+θxsuchthatx∈Ω0Lemma 2 (See [13]).
For allθ∈W1,∞(RN,RN) satisfying θW1,∞(RN,RN)<1, the application T=Id+θ is a bijection of RN and T∈τ.Definition 3.
LetJ be application from C(Ω0) to R. One says that it is differentiable with respect to the domain at Ω0 if the function(8)Θ→JId+θΩ0is Frechet differentiable at 0 in the Banach space W1,∞(RN,RN). i.e., ∃L, a linear continuous form on W1,∞(RN,RN), such that(9)JId+θΩ0=JΩ0+J′Ω0+∘θ,withlimθ→O⁡∘θθW1,∞=0.The linear form J′(Ω0) depends only on the normal component of θ on the boundary of Ω0.Proposition 4.
LetΩ0 be a regular bounded open set of RN. Let J be a differentiable application on Ω0. If θ1,θ2∈W1,∞(RN,RN) are such that θ1-θ2∈C1(RN,RN) such that θ1.n=θ2.n on ∂Ω0, then the derivative J′(Ω0) is verifying:(10)J′Ω0θ1=J′Ω0θ2


## 1.3. Derivation of Integrals

Since the complianceJ is defined by surface or volume integrals then its differentiation devotes the following tools.Lemma 5 (see [1, 7]).
LetΩ0 be an open set of RN. Let T∈τ and 1≤p≤+∞.Then f∈Lp(T(Ω0)) iff foT∈Lp(Ω0) and one has(11)∫TΩ0fdx=∫Ω0foTdet⁡∇Tdxand∫TΩ0fdet⁡∇T-1dx=∫Ω0foTdxOn the other hand f∈W1,p(T(Ω0)) iff foT∈W1,p(T(Ω0)) and one has(12)∇foT=∇T-1t∇foTProposition 6 (See [13]).
LetΩ0 be a regular bounded open set of RN. Let f∈W1,1(RN) and J be an application from C(Ω0) to R defined by J(Ω)=∫Ω‍f(x)dx. Then J is differentiable in Ω0 and(13)J′Ω0θ=∫Ω0divθxfxdx=∫∂Ω0θxnxfxdSwithθ∈W1,∞RN,RNNow we move to a lemma on the change of variables in surfaces integrals.Lemma 7 (See [1, 7]).
LetΩ0 be an open set of RN. Let T∈τ be a class C1 diffeomorphism of RN, and 1≤p≤+∞.Then f∈Lp(T(Ω0)) iff foT∈Lp(Ω0) and one has(14)∫∂TΩ0fdS=∫∂Ω0f∘Tdet⁡∇T-1∇T-1tnRNdSwhere n is the external normal to ∂Ω0.The surface integral derivative of a function with respect to the domain is given by the following proposition.Proposition 8 (See [13]).
LetΩ0 be a regular bounded open set of RN. Let f∈W2,1(RN) and J be an application from C(Ω0) to R defined by J(Ω)=∫Ω‍f(x)dx. Then J is differentiable in Ω0 and(15)J′Ω0θ=∫∂Ω0∇f.θ+fdivθ-∇θn.n=∫∂Ω0θ.n∂f∂n+HfdSwithθ∈C1RN,RNwhere H is the average curvature of ∂Ω0 defined by H=divn.


## 1.4. Derivation of a Domain Dependent Function

In this section we try to derive a function depending on the domain; for this we use the EulerianU or Lagrangian L derivative. The second is a more reliable concept than the first. Let u(x,Ω) be a function defined for all x∈Ω and depending on Ω. It represents a solution of an PDE posed in Ω. In a point x belonging to both Ω0 and Ω=(Id+θ)(Ω0), we can calculate the differential u(Ω,x): (16)uId+θΩ0,x=uΩ0,x+Uθ,x+oθsuch  thatlimθ↦0⁡oθθ=0U is a linear continuous form in θ; it represents a directional derivative in the direction θ. This definition makes sense in the case where x∈Ω, but it poses a problem if x∈Ω0. Then in this case we use the Lagrangian derivative; for this we build the transported u¯(θ) on Ω0.By changing variables we obtainu¯(θ,x)=u((Id+θ)(Ω0)o(Id+θ))=u((Id+θ)(Ω0),x+θ(x)).To arrive at the derivative Lagrangian by driftingu¯(θ,x) with respect to θ(17)u¯θ,x=uθ,x+Lθ,x+oθsuch  thatlimθ↦0⁡oθθ=0.L is a linear continuous form in θ; it represents a directional derivative in the direction θ.There is a relation between these two derivativesL(θ,x)=U(θ,x)+θ.∇u(Ω0,x).Proposition 9.
LetΩ0 be a regular bounded open set of RN. Let f(Ω) be an application from C(Ω0) to L1(R); one defines its transpose from W1,1(RN) to L1(R)f¯(θ,x)=f((Id+θ)(Ω0)o(Id+θ)) which we suppose to be derivable in 0 and L is considered as its derivative. So the application J1 from C(Ω0) to R defined by J1(Ω)=∫Ω‍f(Ω)dx is differentiable in Ω0 and for all θ∈W1,∞(RN,RN) one has(18)J1′Ω0θ=∫∂ΩfΩ0divθ+LθdSIn the same way, if f¯(θ) is derivable as an application from C1(RN,RN) to L1(∂Ω0), so the application J2 from C(Ω0) to R defined by J2(Ω)=∫∂Ω‍f(Ω)dx is differentiable in Ω0 and for all θ∈C1(RN,RN) one has(19)J2′Ω0θ=∫∂ΩfΩ0divθ-∇θn.n+YθdS


## 2. Deriving an Equation with respect to the Domain

### 2.1. Dirichlet Conditions

We consider the following equation with Dirichlet boundary conditions:(20)-Δpu=finΩu=0on∂ΩWith Ω0 a regular bounded open set in RN, f∈W1,p(RN) and Δp(u)=div(|∇u|p-2∇u) with 1<p<∞.Equation (20) admits a unique solution in W01,p(Ω).Remark. For p=2 we obtain the linear operator “Laplacian”.The variational formulation of problem (20) is as follows: ∀v∈W01,p(Ω) we have ∫Ω‍-Δpu.v=∫Ω‍f.v or divf.F=∇f.F+fdiv(F); it implies that(21)fdiv⁡F=divf.F-∇f.FSo∫Ω-Δpu.v=∫Ω-div∇up-2∇u.v=∫Ω-divv.∇up-2∇u+∫Ω∇v∇up-2∇uUsing the Green formula we obtain(22)=∫∂Ωv.∇up-2∇u.ndS+∫Ω∇up-2∇u∇vbut v∈W01,p(Ω); it implies that v=0 on ∂Ω. So the first term equals zero. Then(23)=∫Ω∇up-2∇u∇v=∫Ωf.v∀v∈W01,pΩProposition 10.
LetΩ=(Id+θ)(Ω0); u(Ω) is the solution of the problem (20). We define its transported on Ω0 by(24)u¯θ=uId+θΩ0∘Id+θ∈W01,pΩ0Then the application θ→u¯(θ) from W1,∞(RN,RN) to W1,p(Ω0) is differentiable in 0 and its directional derivative calledLagrangian derivativeL=<u′¯(0), θ> is the unique solution of(25)-ΔpL-divαx∇Lθ,x-div∇Lθ,xp-2∇u¯0,x=g′+divf.θinΩL=0on∂Ωwith g′=-div[(λ(x,θ)+αxρθ)∇u¯(0,x)].Proof.
We consider a test functionw=v∘(Id+θ)∈W01,p(Ω0)⇒v=w∘(Id+θ)-1∈W01,p(Ω). Let U∈W01,p(Ω0) such that u is a solution of problem (20) satisfying u=U∘(Id+θ)-1. We remark that w and U are independent of θ. By a change of variable x=T(y) and the Lemma 5, (23) becomes(26)23⇔∫TΩ0∇up-2∇u∇vdx=∫TΩ0fvdx⇔∫Ω0∇u∘Tp-2∇u∘T∇v∘Tdet⁡∇Tdy=∫Ω0f∘T.v∘Tdet⁡∇Tdy⇔∫Ω0∇T-1t∘∇u∘Tp-2∇T-1t∇u∘T∇T-1t∇v∘Tdet⁡∇Tdy=∫Ω0f∘T.v∘Tdet⁡∇Tdy23⇔∫Ω0∇T-1t∇u¯p-2∇T-1∇T-1t∇u¯∇wdet⁡∇Tdy=∫Ω0f∘T.wdet⁡∇Tdy⇔∫Ω0det⁡∇T∇T-1t∇u¯p-2∇T-1∇T-1t∇u¯∇wdy=∫Ω0f∘T.wdet⁡∇TdyWe  poseAθ,u¯=det⁡∇T∇T-1t∇u¯p-2∇T-1∇T-1t∇u¯and since T=Id+θ we have ∇T=I+∇θ; it implies that(27)Aθ,u¯=det⁡I+∇θI+∇θ-1t∇u¯p-2I+∇θ-1I+∇θ-1t∇u¯Then (23)⇔∫Ω0‍A(θ,u¯)∇wdy=∫Ω0‍|f∘T|.w|det⁡∇T|dy; then we drift with respect to θ in 0.
On the other hand the applicationθ→A(θ,u¯) from W1,∞(RN,RN) to L∞(RN,RN) is differentiable in 0.
In factdet⁡(I+∇θ)=1+div(θ)+∘(θ) with limθ→0∘(θ)L∞(RN)/θW1,∞(RN,RN)=0. Therefore |det⁡(I+∇θ)|=1+div(θ)+∘(θ) because θ is small enough.(28)We  haveI+∇θ-1=I-∇θ+∘θandI+∇θ-1t=I-∇θt+∘θAndu¯θ,x=u¯0,x+Lθ,x+∘θ⇒∇u¯θ,x=∇u¯0,x+∇Lθ,x+∘θBy using [14] we find(29)1-∇θ-1t∇u¯p-2=I-∇θt+∘θ∇u¯0,x+∇Lθ,x+∘θp-2=∇u¯0,x-∇θt∇u¯0,x+∇Lθ,x+∘θp-2=Bθ,u¯=∇u¯0,xp-2-∇θt∇u¯0,xp-2+∇Lθ,xp-2+∘θOn the other hand |det⁡(I+∇θ)|(I+∇θ)-1((I+∇θ)-1)t=(1+div(θ))I-∇θ-(∇θ)t+∘(θ).
Thus(30)Aθ,u¯=Bθ,u¯1+divθI-∇θ-∇θt+∘θ∇u¯0,x+∇Lθ,x+∘θ=Bθ,u¯∇u¯0,x+∇Lθ,x+divθI-∇θ-∇θt∇u¯0,x+∇Lθ,x+∘θ=Bθ,u¯∇u¯0,x+∇Lθ,x+divθI-∇θ-∇θt∇u¯0,x+∘θ=∇u¯0,xp-2∇u¯0,x+∇u¯0,xp-2∇Lθ,x+divθI-∇θ-∇θt∇u¯0,xp-2∇u¯0,x-∇θt∇u-0,xp-2∇u¯0,x-∇θt∇u¯0,xp-2∇Lθ,x-∇θt∇u¯0,xp-2divθI-∇θ-∇θt·∇u¯0,x+∇Lθ,xp-2∇u¯0,x+∇Lθ,xp-2∇Lθ,x+∇Lθ,xp-2divθI-∇θ-∇θt∇u¯0,x+∘θ=∇u¯0,xp-2·∇u¯0,x+∇Lθ,xp-2∇Lθ,x-∇θt∇u¯0,xp-2∇u¯0,x+∇Lθ,xp-2·∇u¯0,x-∇θt∇u¯0,xp-2divθI-∇θ-∇θt∇u¯0,x+∇Lθ,xp-2divθI-∇θ-∇θt∇u¯0,x+∇u¯0,xp-2∇Lθ,x+divθI-∇θ-∇θt∇u¯0,x-∇θt∇u¯0,xp-2∇Lθ,x+∘θ
Therefore, we have(31)
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wThen afterwards we put(32)ρθ=divθ-∇θ-∇θtλx,θ=∇θt∇u¯0,xp-2αx=∇u¯0,xp-2Andg=-divλx,θI+ρθ+αxρθ∇u¯0,xThus∫Ω0Aθ,u¯∇w-∫Ω0∇u¯0,xp-2∇u¯0,x.∇w=∫Ω0∇Lθ,xp-2∇Lθ,x∇w+∫Ω0αx-λx,θ∇Lθ,x∇w+∫Ω0∇Lθ,xp-2I+ρθ∇u¯0,x∇w-∫Ω0λx,θI+ρθ+αxρθ∇u¯0,x∇w+∘θ=∫Ω0divf.θw+∘θ⇔∫Ω0-ΔpL.w+∫Ω0-divαx-λx,θ∇Lθ,xw+∫Ω0-div∇Lθ,xp-2I+ρθ∇u¯0,xw+∫Ω0g.w=∫Ω0divf.θwThus ∀w∈W01,pΩL the Lagrangian of u¯ is a solution of the following differential equation:(33)-ΔpL-divαx-λx,θ∇Lθ,x-div∇Lθ,xp-2I+ρθ∇u¯0,x=g+divf.θinΩL=0on∂ΩRemark 11.
Whenp>2 we will have:(i)
λ(x,θ).∇L(θ,x)=∘(θ)(ii)
|L(θ,x)|p-2ρθ=∘(θ)(iii)
λ(x,θ).ρθ=∘(θ)
ThenL, the Lagrangian of u¯, will be solution of the following reduced differential equation:(34)-ΔpL-divαx∇Lθ,x-div∇Lθ,xp-2∇u¯0,x=g′+divf.θinΩL=0on∂Ωwith g′=-div[(λ(x,θ)+αxρθ)∇u¯(0,x)]


### 2.2. Neumann Conditions

We consider the following equation with Neumann boundary conditions (see [15]):(35)-Δpu+up-2u=finΩ∇up-2∂u∂n=gon∂Ωwhere Ω0 a regular bounded open set in RN, f∈W1,p(RN), g∈W2,p(RN), and Δp(u)=div(|∇u|p-2∇u) with 1<p<∞.The variational formulation of problem (35) is to find u∈W1,p(Ω) such that ∀v∈W1,p(Ω) (v represents a test function) ∫Ω‍-Δp(u).vdx+∫Ω‍|u|p-2u.vdx=∫Ω‍f.vdx or div(f.F)=∇f.F+fdiv(F).It implies thatfdiv(F)=div(f.F)-∇f.F:(36)Then∫Ω-div∇up-2∇u.vdx+∫Ωup-2u.vdx=∫Ωf.vdx⇔∫Ω∇up-2∇u.∇v-∫Ωdivv∇up-2∇u+∫Ωup-2u.vdx=∫Ωf.vdx.By using the Green formula we find(37)∫Ω∇up-2∇u.∇vdx-∫∂Ωv∇up-2∇u.ndS+∫Ωup-2u.vdx=∫Ωf.vdx⇔∫Ω∇up-2∇u.∇vdx+∫Ωup-2u.vdx=∫Ωf.vdx+∫∂Ωg.vdSand∀v∈W1,pΩProposition 12.
LetΩ=(Id+θ)(Ω0); u(Ω) is the solution of problem (35). We define its transport on Ω0 by u¯(θ)=u((Id+θ)(Ω0)∘(Id+θ)∈W01,p(Ω0).
Then the applicationθ→u¯(θ) from C1(RN,RN) to W1,∞(Ω0) is differentiable in 0 on the direction θ and its directional derivative calledLagrangian derivative ofu is L=<u′¯(0),θ> where L∈W1,p(Ω0) is the unique solution of(38)-ΔpL+∇Lθ,xp-2∇Lθ,x=F+divαx-λx,θ∇Lθ,x+div∇Lθ,xp-2I+ρθ∇u¯0,x-Lθ,xp-2u¯0,x-u¯0,xp-2Lθ,x-u¯0,xp-2+∇Lθ,xp-2u¯0,xp-2+Lθ,xdivθinΩGθ-αx-λx,θ∂Lθ,x∂n-∇Lθ,xp-2I+ρθ∂u¯0,x∂n-Lθ,xp-2∂Lθ,x∂n=0on∂Ωwhere F=-div((λ(x,θ)(I+ρθ)+αxρθ)∇u¯(0,x)-f.θ).Proof.
We make a change of variablex=y+θ(y) where y∈Ω0 and x∈Ω in the variational formulation (37). We pose w(y)=v(x)⇔w(y)=v(Id+θ)(y)⇔v=w∘(Id+θ)-1 noticing that w does not depend on θ. Thus by drifting the variational formulation, we obtain by using Lemmas 5 and 7(39)37⇔∫Ω0Aθ,u¯∇wdy+∫Ω0u¯θ,xp-2u¯θ,x.wdet⁡∇Tdy=∫Ω0f∘T.wdet⁡∇Tdy+∫∂Ωg∘Twdet⁡∇T∇T-1t.nRNdSOr(40)∫Ω0Aθ,u¯∇wdy=∫Ω0∇u¯0,xp-2∇u¯0,x.∇w+∫Ω0∇Lθ,xp-2∇Lθ,x.∇w+∫Ω0αx-λx,θ∇Lθ,x∇w+∫Ω0∇Lθ,xp-2I+ρθ∇u¯0,x∇w-∫Ω0λx,θI+ρθ+αxρθ∇u¯0,x∇w+∘θWe have by using [14](41)∫Ω0u¯θ,xp-2u¯θ,x.wdet⁡∇Tdy=∫Ω0u¯0,x+Lθ,xp-2u¯0,x+Lθ,x.w.1+divθdy+∘θ=∫Ω0u¯0,xp-2+Lθ,xp-2u¯0,x+Lθ,x.w.1+divθwdy+∘θ=∫Ω0u¯0,xp-2u¯0,xwdy+∫Ω0Lθ,xp-2Lθ,xwdy+∫Ω0Lθ,xp-2u¯0,xwdy+∫Ω0u¯0,xp-2Lθ,xwdy+∫Ω0u¯0,xp-2+Lθ,xp-2u¯0,x+Lθ,x.w.divθwdy+∘θ(42)And  also∫Ω0f∘T.wdet⁡∇Tdy-∫Ω0fwdy=∫Ω0divf.θwdy+∘θThen∫∂Ωg∘Twdet⁡∇T∇T-1t.nRNdS-∫∂Ω0gwdS=∫∂Ω0∇gθ+gdivθ-∇θn.nwdS+∘θAnd(43)∫Ω0Aθ,u¯∇wdy+∫Ω0u¯θ,xp-2u¯θ,x.wdet⁡∇Tdy-∫Ω0u¯θ,xp-2∇u¯θ,x.∇w-∫Ω0u¯θ,xp-2u¯θ,x.wdy=∫Ω0∇Lθ,xp-2∇Lθ,x.∇w-∫Ω0Lθ,xp-2Lθ,x.wdy+∫Ω0αx-λx,θ∇Lθ,x∇w+∫Ω0∇Lθ,xp-2I+ρθ∇u¯0,x∇w-∫Ω0λx,θI+ρθ+αxρθ∇u¯0,x∇w+∫Ω0Lθ,xp-2u¯0,xwdy+∫Ω0u¯θ,xp-2Lθ,xwdy+∫Ω0u¯0,xp-2+Lθ,xp-2u¯0,x+Lθ,x.w.divθwdy=∫Ω0divf.θwdy+∫∂Ω0∇gθ+gdivθ-∇θn.nwdS+∘θFor(44)
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(45)
We  note
G
θ
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∇
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+
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d
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-
∇
θ
n
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n
+
λ
x
,
θ
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α
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ρ
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∂
u
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x
∂
nSo the transported derivative of u¯(θ,x) at 0 in the direction θ is the Lagrangian L=<u′¯(0),θ> which is the solution of the following equation:(46)-ΔpL+∇Lθ,xp-2∇Lθ,x=F+divαx-λx,θ∇Lθ,x+div∇Lθ,xp-2I+ρθ∇u¯0,x-Lθ,xp-2u¯0,x-u¯0,xp-2Lθ,x-u¯0,xp-2+∇Lθ,xp-2u¯0,xp-2+Lθ,xdivθinΩGθ-αx-λx,θ∂Lθ,x∂n-∇Lθ,xp-2I+ρθ∂u¯0,x∂n-Lθ,xp-2∂Lθ,x∂n=0on∂Ωwhere F=-div((λ(x,θ)(I+ρθ)+αxρθ)∇u¯(0,x)-f.θ).


## 2.1. Dirichlet Conditions

We consider the following equation with Dirichlet boundary conditions:(20)-Δpu=finΩu=0on∂ΩWith Ω0 a regular bounded open set in RN, f∈W1,p(RN) and Δp(u)=div(|∇u|p-2∇u) with 1<p<∞.Equation (20) admits a unique solution in W01,p(Ω).Remark. For p=2 we obtain the linear operator “Laplacian”.The variational formulation of problem (20) is as follows: ∀v∈W01,p(Ω) we have ∫Ω‍-Δpu.v=∫Ω‍f.v or divf.F=∇f.F+fdiv(F); it implies that(21)fdiv⁡F=divf.F-∇f.FSo∫Ω-Δpu.v=∫Ω-div∇up-2∇u.v=∫Ω-divv.∇up-2∇u+∫Ω∇v∇up-2∇uUsing the Green formula we obtain(22)=∫∂Ωv.∇up-2∇u.ndS+∫Ω∇up-2∇u∇vbut v∈W01,p(Ω); it implies that v=0 on ∂Ω. So the first term equals zero. Then(23)=∫Ω∇up-2∇u∇v=∫Ωf.v∀v∈W01,pΩProposition 10.
LetΩ=(Id+θ)(Ω0); u(Ω) is the solution of the problem (20). We define its transported on Ω0 by(24)u¯θ=uId+θΩ0∘Id+θ∈W01,pΩ0Then the application θ→u¯(θ) from W1,∞(RN,RN) to W1,p(Ω0) is differentiable in 0 and its directional derivative calledLagrangian derivativeL=<u′¯(0), θ> is the unique solution of(25)-ΔpL-divαx∇Lθ,x-div∇Lθ,xp-2∇u¯0,x=g′+divf.θinΩL=0on∂Ωwith g′=-div[(λ(x,θ)+αxρθ)∇u¯(0,x)].Proof.
We consider a test functionw=v∘(Id+θ)∈W01,p(Ω0)⇒v=w∘(Id+θ)-1∈W01,p(Ω). Let U∈W01,p(Ω0) such that u is a solution of problem (20) satisfying u=U∘(Id+θ)-1. We remark that w and U are independent of θ. By a change of variable x=T(y) and the Lemma 5, (23) becomes(26)23⇔∫TΩ0∇up-2∇u∇vdx=∫TΩ0fvdx⇔∫Ω0∇u∘Tp-2∇u∘T∇v∘Tdet⁡∇Tdy=∫Ω0f∘T.v∘Tdet⁡∇Tdy⇔∫Ω0∇T-1t∘∇u∘Tp-2∇T-1t∇u∘T∇T-1t∇v∘Tdet⁡∇Tdy=∫Ω0f∘T.v∘Tdet⁡∇Tdy23⇔∫Ω0∇T-1t∇u¯p-2∇T-1∇T-1t∇u¯∇wdet⁡∇Tdy=∫Ω0f∘T.wdet⁡∇Tdy⇔∫Ω0det⁡∇T∇T-1t∇u¯p-2∇T-1∇T-1t∇u¯∇wdy=∫Ω0f∘T.wdet⁡∇TdyWe  poseAθ,u¯=det⁡∇T∇T-1t∇u¯p-2∇T-1∇T-1t∇u¯and since T=Id+θ we have ∇T=I+∇θ; it implies that(27)Aθ,u¯=det⁡I+∇θI+∇θ-1t∇u¯p-2I+∇θ-1I+∇θ-1t∇u¯Then (23)⇔∫Ω0‍A(θ,u¯)∇wdy=∫Ω0‍|f∘T|.w|det⁡∇T|dy; then we drift with respect to θ in 0.
On the other hand the applicationθ→A(θ,u¯) from W1,∞(RN,RN) to L∞(RN,RN) is differentiable in 0.
In factdet⁡(I+∇θ)=1+div(θ)+∘(θ) with limθ→0∘(θ)L∞(RN)/θW1,∞(RN,RN)=0. Therefore |det⁡(I+∇θ)|=1+div(θ)+∘(θ) because θ is small enough.(28)We  haveI+∇θ-1=I-∇θ+∘θandI+∇θ-1t=I-∇θt+∘θAndu¯θ,x=u¯0,x+Lθ,x+∘θ⇒∇u¯θ,x=∇u¯0,x+∇Lθ,x+∘θBy using [14] we find(29)1-∇θ-1t∇u¯p-2=I-∇θt+∘θ∇u¯0,x+∇Lθ,x+∘θp-2=∇u¯0,x-∇θt∇u¯0,x+∇Lθ,x+∘θp-2=Bθ,u¯=∇u¯0,xp-2-∇θt∇u¯0,xp-2+∇Lθ,xp-2+∘θOn the other hand |det⁡(I+∇θ)|(I+∇θ)-1((I+∇θ)-1)t=(1+div(θ))I-∇θ-(∇θ)t+∘(θ).
Thus(30)Aθ,u¯=Bθ,u¯1+divθI-∇θ-∇θt+∘θ∇u¯0,x+∇Lθ,x+∘θ=Bθ,u¯∇u¯0,x+∇Lθ,x+divθI-∇θ-∇θt∇u¯0,x+∇Lθ,x+∘θ=Bθ,u¯∇u¯0,x+∇Lθ,x+divθI-∇θ-∇θt∇u¯0,x+∘θ=∇u¯0,xp-2∇u¯0,x+∇u¯0,xp-2∇Lθ,x+divθI-∇θ-∇θt∇u¯0,xp-2∇u¯0,x-∇θt∇u-0,xp-2∇u¯0,x-∇θt∇u¯0,xp-2∇Lθ,x-∇θt∇u¯0,xp-2divθI-∇θ-∇θt·∇u¯0,x+∇Lθ,xp-2∇u¯0,x+∇Lθ,xp-2∇Lθ,x+∇Lθ,xp-2divθI-∇θ-∇θt∇u¯0,x+∘θ=∇u¯0,xp-2·∇u¯0,x+∇Lθ,xp-2∇Lθ,x-∇θt∇u¯0,xp-2∇u¯0,x+∇Lθ,xp-2·∇u¯0,x-∇θt∇u¯0,xp-2divθI-∇θ-∇θt∇u¯0,x+∇Lθ,xp-2divθI-∇θ-∇θt∇u¯0,x+∇u¯0,xp-2∇Lθ,x+divθI-∇θ-∇θt∇u¯0,x-∇θt∇u¯0,xp-2∇Lθ,x+∘θ
Therefore, we have(31)
∫
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0
A
θ
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∫
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0
∇
u
¯
0
,
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p
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2
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0
,
x
.
∇
w
=
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L
θ
,
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p
-
2
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L
θ
,
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∇
w
+
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,
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θ
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+
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+
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θ
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θ
t
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x
p
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+
d
i
v
θ
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-
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-
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+
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p
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i
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t
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x
=
∫
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0
d
i
v
f
.
θ
.
wThen afterwards we put(32)ρθ=divθ-∇θ-∇θtλx,θ=∇θt∇u¯0,xp-2αx=∇u¯0,xp-2Andg=-divλx,θI+ρθ+αxρθ∇u¯0,xThus∫Ω0Aθ,u¯∇w-∫Ω0∇u¯0,xp-2∇u¯0,x.∇w=∫Ω0∇Lθ,xp-2∇Lθ,x∇w+∫Ω0αx-λx,θ∇Lθ,x∇w+∫Ω0∇Lθ,xp-2I+ρθ∇u¯0,x∇w-∫Ω0λx,θI+ρθ+αxρθ∇u¯0,x∇w+∘θ=∫Ω0divf.θw+∘θ⇔∫Ω0-ΔpL.w+∫Ω0-divαx-λx,θ∇Lθ,xw+∫Ω0-div∇Lθ,xp-2I+ρθ∇u¯0,xw+∫Ω0g.w=∫Ω0divf.θwThus ∀w∈W01,pΩL the Lagrangian of u¯ is a solution of the following differential equation:(33)-ΔpL-divαx-λx,θ∇Lθ,x-div∇Lθ,xp-2I+ρθ∇u¯0,x=g+divf.θinΩL=0on∂ΩRemark 11.
Whenp>2 we will have:(i)
λ(x,θ).∇L(θ,x)=∘(θ)(ii)
|L(θ,x)|p-2ρθ=∘(θ)(iii)
λ(x,θ).ρθ=∘(θ)
ThenL, the Lagrangian of u¯, will be solution of the following reduced differential equation:(34)-ΔpL-divαx∇Lθ,x-div∇Lθ,xp-2∇u¯0,x=g′+divf.θinΩL=0on∂Ωwith g′=-div[(λ(x,θ)+αxρθ)∇u¯(0,x)]


## 2.2. Neumann Conditions

We consider the following equation with Neumann boundary conditions (see [15]):(35)-Δpu+up-2u=finΩ∇up-2∂u∂n=gon∂Ωwhere Ω0 a regular bounded open set in RN, f∈W1,p(RN), g∈W2,p(RN), and Δp(u)=div(|∇u|p-2∇u) with 1<p<∞.The variational formulation of problem (35) is to find u∈W1,p(Ω) such that ∀v∈W1,p(Ω) (v represents a test function) ∫Ω‍-Δp(u).vdx+∫Ω‍|u|p-2u.vdx=∫Ω‍f.vdx or div(f.F)=∇f.F+fdiv(F).It implies thatfdiv(F)=div(f.F)-∇f.F:(36)Then∫Ω-div∇up-2∇u.vdx+∫Ωup-2u.vdx=∫Ωf.vdx⇔∫Ω∇up-2∇u.∇v-∫Ωdivv∇up-2∇u+∫Ωup-2u.vdx=∫Ωf.vdx.By using the Green formula we find(37)∫Ω∇up-2∇u.∇vdx-∫∂Ωv∇up-2∇u.ndS+∫Ωup-2u.vdx=∫Ωf.vdx⇔∫Ω∇up-2∇u.∇vdx+∫Ωup-2u.vdx=∫Ωf.vdx+∫∂Ωg.vdSand∀v∈W1,pΩProposition 12.
LetΩ=(Id+θ)(Ω0); u(Ω) is the solution of problem (35). We define its transport on Ω0 by u¯(θ)=u((Id+θ)(Ω0)∘(Id+θ)∈W01,p(Ω0).
Then the applicationθ→u¯(θ) from C1(RN,RN) to W1,∞(Ω0) is differentiable in 0 on the direction θ and its directional derivative calledLagrangian derivative ofu is L=<u′¯(0),θ> where L∈W1,p(Ω0) is the unique solution of(38)-ΔpL+∇Lθ,xp-2∇Lθ,x=F+divαx-λx,θ∇Lθ,x+div∇Lθ,xp-2I+ρθ∇u¯0,x-Lθ,xp-2u¯0,x-u¯0,xp-2Lθ,x-u¯0,xp-2+∇Lθ,xp-2u¯0,xp-2+Lθ,xdivθinΩGθ-αx-λx,θ∂Lθ,x∂n-∇Lθ,xp-2I+ρθ∂u¯0,x∂n-Lθ,xp-2∂Lθ,x∂n=0on∂Ωwhere F=-div((λ(x,θ)(I+ρθ)+αxρθ)∇u¯(0,x)-f.θ).Proof.
We make a change of variablex=y+θ(y) where y∈Ω0 and x∈Ω in the variational formulation (37). We pose w(y)=v(x)⇔w(y)=v(Id+θ)(y)⇔v=w∘(Id+θ)-1 noticing that w does not depend on θ. Thus by drifting the variational formulation, we obtain by using Lemmas 5 and 7(39)37⇔∫Ω0Aθ,u¯∇wdy+∫Ω0u¯θ,xp-2u¯θ,x.wdet⁡∇Tdy=∫Ω0f∘T.wdet⁡∇Tdy+∫∂Ωg∘Twdet⁡∇T∇T-1t.nRNdSOr(40)∫Ω0Aθ,u¯∇wdy=∫Ω0∇u¯0,xp-2∇u¯0,x.∇w+∫Ω0∇Lθ,xp-2∇Lθ,x.∇w+∫Ω0αx-λx,θ∇Lθ,x∇w+∫Ω0∇Lθ,xp-2I+ρθ∇u¯0,x∇w-∫Ω0λx,θI+ρθ+αxρθ∇u¯0,x∇w+∘θWe have by using [14](41)∫Ω0u¯θ,xp-2u¯θ,x.wdet⁡∇Tdy=∫Ω0u¯0,x+Lθ,xp-2u¯0,x+Lθ,x.w.1+divθdy+∘θ=∫Ω0u¯0,xp-2+Lθ,xp-2u¯0,x+Lθ,x.w.1+divθwdy+∘θ=∫Ω0u¯0,xp-2u¯0,xwdy+∫Ω0Lθ,xp-2Lθ,xwdy+∫Ω0Lθ,xp-2u¯0,xwdy+∫Ω0u¯0,xp-2Lθ,xwdy+∫Ω0u¯0,xp-2+Lθ,xp-2u¯0,x+Lθ,x.w.divθwdy+∘θ(42)And  also∫Ω0f∘T.wdet⁡∇Tdy-∫Ω0fwdy=∫Ω0divf.θwdy+∘θThen∫∂Ωg∘Twdet⁡∇T∇T-1t.nRNdS-∫∂Ω0gwdS=∫∂Ω0∇gθ+gdivθ-∇θn.nwdS+∘θAnd(43)∫Ω0Aθ,u¯∇wdy+∫Ω0u¯θ,xp-2u¯θ,x.wdet⁡∇Tdy-∫Ω0u¯θ,xp-2∇u¯θ,x.∇w-∫Ω0u¯θ,xp-2u¯θ,x.wdy=∫Ω0∇Lθ,xp-2∇Lθ,x.∇w-∫Ω0Lθ,xp-2Lθ,x.wdy+∫Ω0αx-λx,θ∇Lθ,x∇w+∫Ω0∇Lθ,xp-2I+ρθ∇u¯0,x∇w-∫Ω0λx,θI+ρθ+αxρθ∇u¯0,x∇w+∫Ω0Lθ,xp-2u¯0,xwdy+∫Ω0u¯θ,xp-2Lθ,xwdy+∫Ω0u¯0,xp-2+Lθ,xp-2u¯0,x+Lθ,x.w.divθwdy=∫Ω0divf.θwdy+∫∂Ω0∇gθ+gdivθ-∇θn.nwdS+∘θFor(44)
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nSo the transported derivative of u¯(θ,x) at 0 in the direction θ is the Lagrangian L=<u′¯(0),θ> which is the solution of the following equation:(46)-ΔpL+∇Lθ,xp-2∇Lθ,x=F+divαx-λx,θ∇Lθ,x+div∇Lθ,xp-2I+ρθ∇u¯0,x-Lθ,xp-2u¯0,x-u¯0,xp-2Lθ,x-u¯0,xp-2+∇Lθ,xp-2u¯0,xp-2+Lθ,xdivθinΩGθ-αx-λx,θ∂Lθ,x∂n-∇Lθ,xp-2I+ρθ∂u¯0,x∂n-Lθ,xp-2∂Lθ,x∂n=0on∂Ωwhere F=-div((λ(x,θ)(I+ρθ)+αxρθ)∇u¯(0,x)-f.θ).


## 3. Optimality Condition

To calculate the optimality conditions of the following probleminfΩ∈UadJΩ with Uad={Ω∈C(Ω0)and∫x‍dx=V0}, where C(Ω0) is the set of admissible forms obtained by diffeomorphism, the cost function J(Ω) is the compliance defined by(47)JΩ=∫ΩuΩ-uΩ0pdxto reach a target displacement u(Ω0)∈Lp(RN) where the function u(Ω) is solution of the boundary problem posed (resp., Dirichlet or Neumann boundary conditions).We consider the following boundary value problems.Dirichlet Boundary Condition (48) - Δ p u = f  i n  Ω u = 0  o n   ∂ Ωwhere f∈W1,p(RN).Neumann Boundary Condition (49) - Δ p u + u p - 2 u = f  i n   Ω ∇ u p - 2 ∂ u ∂ n = g  o n  ∂ Ωwhere f∈W1,p(RN) and g∈W2,p(RN).The problems admit a unique solutionu(Ω).Theorem 13.
LetΩ0 be a regular bounded open set. The cost function J(Ω) is differentiable and its derivative is(50)J′Ω0θ=∫Ω0uΩ0-u0p-2divθ+puΩ0-u0p-2uΩ0-u0L-θ.∇u0dxL is the Lagrangian derivative and also solution of(51)-ΔpL-divαx∇Lθ,x-div∇Lθ,xp-2∇u¯0,x=g′+divf.θinΩL=0on∂Ωwith g′=-div[(λ(x,θ)+αxρθ)∇u¯(0,x)].Proof.
By applying Proposition4 for the compliance we obtain(52)J′Ω0θ=∫Ω0uΩ0-u0p-2divθ+∫Ω0uΩ0-u0pθJ′Ω0θ=∫Ω0uΩ0-u0p-2divθ+p∫Ω0uΩ0-u0p-2uΩ0-u0∇u-∇u0dxJ′Ω0θ=∫Ω0uΩ0-u0p-2divθ+puΩ0-u0p-2uΩ0-u0L-θ.∇u0dxwhere L is the Lagrangian derivative, solution of the PDE.Remark 14.
From extensive literature which deals with optimum condition calculus for problems asinfΩ∈Uad⁡JΩ we can cite(i)
J1(Ω)=∫∂Ω0‍|u(Ω0)-u0|pdS(ii)
J2(Ω)=∫∂Ω0‍|∇u(Ω0)-∇u0|pdS+∫∂Ω0‍|u(Ω0)-u0|pdSSo, to calculate the gradient of each compliance we use the same argument by the propositions [13].



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*Source: 1689269-2018-09-02.xml*