Digital Object Identifier (DOI) 10.1007/s00205-009-0243-8
What is the Optimal Shape of a Pipe?
Antoine Henrot & Yannick Privat
Communicated by G. Dal Maso
Abstract
We consider an incompressible fluid in a three-dimensional pipe, following the Navier–Stokes system with classical boundary conditions. We are interested in the following question: is there any optimal shape for the criterion “energy dissipated by the fluid”? Moreover, is the cylinder the optimal shape? We prove that there exists an optimal shape in a reasonable class of admissible domains, but the cylin- der is not optimal. For that purpose, we define the first order optimality condition, thanks to the adjoint state and we prove that it is impossible that the adjoint state be a solution of this over-determined system when the domain is the cylinder. At last, we show some numerical simulations for that problem.
1. Introduction
The shape optimization problems in fluid mechanics are very important and gave rise to many works. Most often, these works have a numerical character due to the intrinsic difficulty of the Navier–Stokes equations. For a first bibliography on the topic, we refer, for example, to [7,9,11,14,16].
In this work, we are interested in one of the simplest problem: what shape must have a pipe in order to minimize the energy dissipated by a fluid? For us, a pipe (of “length” L) will be a three dimensional domainΩ contained in the strip {(x1,x2,x3) ,0 < x3 <L}. We will assume that the inlet E :=∂Ω∩ {x3 =0}
(where∂Ω denotes the boundary ofΩ) and the outlet S :=∂Ω ∩ {x3 = L}are two fixed identical discs and that the volume ofΩ is imposed. The unknown (or free) part of the boundary ofΩwill be denoted byΓ (so∂Ω=E∪Γ ∪S).
In the pipeΩ, we consider the flow of a viscous, incompressible fluid with a velocity u and a pressure p satisfying the Navier–Stokes system. We assume that the velocity profile u0at the inlet E is of parabolic type; on the lateral boundary Γ, we assume no-slip condition u = 0 and we control the outlet by imposing
an “outlet-pressure” condition on S. We will assume that the viscosityµis large enough in order that the solution of the system is unique (see [19]). The criterion that we want to minimize, with respect to the shapeΩ, is the energy dissipated by the fluid (or viscosity energy) defined by J(Ω):=2µ
Ω|ε(u)|2dx whereεis the stretching tensor.
We will first prove an existence theorem. To obtain this result, we work in the class of admissible domains which satisfy anε-cone property (see [4,9]). Then we are interested in symmetry properties of the optimal domain. For the Stokes model, we are only able to prove that the optimum has one plane of symmetry.
It is not completely clear to see whether the optimum should be axially symmet- ric. In a series of papers Arumugam and Pironneau [2], and Pironneau and Arumugam [15], proved for a similar, but much simpler problem that one has to build riblets on the lateral boundary to reduce the drag. Nevertheless, it is a natural question to ask whether the cylinder should be the optimum for our problem. We will show that it is not the case. For that purpose, we define the first order optimality condition. This condition can be easily expressed in terms of the adjoint state and gives an over-determined condition on the lateral boundaryΓ. Then we prove that it is impossible that the adjoint state be a solution of this over-determined system when the domain is the cylinder.
This paper is organized as follows. At Section2, we state the shape optimiza- tion problem, we prove existence and symmetry. Section3is devoted to the proof of the main theorem. We give in Section4some numerical results and concluding remarks.
These results have been announced in the Note [10].
2. The shape optimization problem
Let us give the notations used in this paper. We consider a generic three- dimensional domainΩcontained in a compact set
D:=
(x1,x2,x3) ,x12+x22R02,0x3L
where R0and L are two positive constants. We will denote by∂Ωthe boundary of Ω. In the sequel, we will assume that the inlet E ofΩdefined by E :=∂Ω∩{x3= 0}and the outlet S defined by S:=∂Ω∩ {x3=L}are two fixed identical discs of radius R<R0centered on the x3axis. We will also assume that the volume of all the domainsΩ is imposed, say|Ω| =V =πR2L. We decompose the boundary ofΩas the disjoint union∂Ω =E∪Γ ∪S andΓ, the lateral boundary is the main unknown or the shape we want to design.
Let us now make precise the state equation. We consider the flow of a viscous incompressible fluid intoΩ. We denote by u=(u1,u2,u3)(letters in bold will cor- respond to vectors) its velocity and by p its pressure. As usual in fluid mechanics, we introduceεthe stretching tensor defined by:
ε(u)= 1
2 ∂ui
∂xj +∂uj
∂xi
1i,j3
.
We will consider the Navier–Stokes system (except for Theorem3where the Stokes system will be considered). As boundary conditions, we assume that the velocity profile u0at the inlet E = {x3 =0}is of parabolic type; on the lateral boundary Γ, we assume adherence or no-slip condition u=0 and we control the outlet by imposing an “outlet-pressure” condition on S = {x3 = L}. Therefore, the PDE system satisfied by the velocity and the pressure is:
⎧⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎩
−µu+ ∇p+ ∇u·u=0 x∈Ω,
div u=0 x∈Ω
u=u0:=
0,0,c(x12+x22−R2)
x∈E
u=0 x∈Γ
−pn+2µε(u)·n=h:=(2µcx1,2µcx2,−p1) x∈S.
(1)
whereµ > 0 denotes the viscosity of the fluid, n the exterior unit normal vector (on S we have n=(0,0,1)). At last, the constant c which appears in the bound- ary condition on E and S is assumed to be negative. The sign of c can physically be explained. Indeed, in the case whereΩ is a cylinder, the flow is driven by a Poiseuille law (simplified physical law derived from the Navier–Stokes system which describes a slow viscous incompressible flow through a constant circular section). Then, this constant c can be written c = p41−µLp0, where p1denotes the constant value of the pressure at the outlet S while p0is the constant value of the pressure at the inlet E.
This choice of the boundary condition ensures that the solution of (1) will be given by a parabolic profile when Ω is a cylinder. More precisely, if Ω is the cylinder of radius R and height L, the solution of (1) is explicitly given by:
u(x1,x2,x3)=
0,0,c(x12+x22−R2)
p(x1,x2,x3)=4µc(x3−L)+p1. (2) More generally, ifΩis a regular domain, we have a classical existence and unique- ness result for such systems, see for example [3,19].
Theorem 1. Let us assume that u0belongs to the Sobolev space(H3/2(E))3and h ∈ (H1/2(S))3. If the viscosityµis large enough, the problem (1) has a unique solution(u,p)∈ H1(Ω)×L2(Ω).
The criterion we want to minimize is the energy dissipated by the fluid (or viscosity energy) defined by:
J(Ω):=2µ
Ω|ε(u)|2dx, (3) whereεis the stretching tensor:
ε(u)= 1
2 ∂ui
∂xj +∂uj
∂xi
1i,j3
.
To make the statement precise, we also need to define the class of admissible domains or shapes. We will consider a first general class:
OV def=
Ωbounded and simply connected domain inR3:
|Ω| =V, Π0∩Ω =E, ΠL∩Ω =S, (4) whereΠ0andΠL denote respectively the planes{x3=0}and{x3=L}.
To prove an existence result, we need to restrict the class of admissible domains.
It is a very classical feature in shape optimization, since these problems are often ill-posed, see [1,9]. We adopt here the choice made by Chenais [4] which consists of assuming some kind of uniform regularity. More precisely, we will consider domains which satisfy an uniform cone condition, we say that these domains have theε-cone property, we refer to [4,5or9] for the precise definition. So, we define the class
OεV := {Ω ∈OV :Ωhas theε-cone property} (5) Lemma 1. The classOεV is closed for the Hausdorff distance.
Proof. We recall that the class of open sets with theε-cone property is closed for the Hausdorff convergence (see Theorem 2.4.10 in [9]). Moreover, the convergence also holds for characteristic functions, so the volume constraint is preserved. So it remains just to prove that the properties defining the inlet E and the outlet S are preserved. Let (Ωn)n∈Nbe a sequence of domains inOVε which converges, for the Hausdorff distance, to a domainΩ. We want to prove thatΠ0∩Ω = E andΠL ∩Ω = S. The first inclusionΠ0∩Ω ⊂ E is just a consequence of the stability of inclusion for the Hausdorff convergence of compact sets. Let us prove the reverse inclusion: let x0 ∈ E and n ∈ N. SinceΩn has theε-cone property, there exists a unit vectorξnsuch that the cone C(ε,x0, ξn)be contained inΩn. Up to a subsequence, one can assume that(ξn)converges to some unit vectorξ and that the sequence of cones C(ε,x0, ξn)converges (for the Hausdorff distance) to the cone C(ε,x0, ξ). By stability with respect to inclusion, one has
∀n∈N, C(ε,x0, ξn)⊂Ωn
C(ε,x0, ξn)−−−−→H
n→+∞ C(ε,x0, ξ) Ωn
−−−−→H n→+∞ Ω
⎫⎪
⎪⎬
⎪⎪
⎭⇒C(ε,x0, ξ)⊂Ω.
Therefore x0∈Ω, and since x0∈ E ⊂Π0, the reverse inclusion is proved.
We are now in position to give our existence result.
Theorem 2. The problem
min J(Ω)
Ω∈OεV, (6)
where J is defined in (3) with u the velocity, solution of the Navier–Stokes problem (1), andOVε is defined in (5), has a solution.
Proof. Let (Ωn)n∈N, be a minimizing sequence inOεV. Since the open sets Ωn
are contained in a fixed compact set D, there exists a subsequence, still denoted byΩnwhich converges (for the Hausdorff distance, but also for the other usual topologies) to some set Ω. Moreover, according to Lemma1,Ω belongs to the classOεV.
To prove the existence result, it remains to prove continuity (or lower semi- continuity) of the criterion J . For any n∈N, we denote by unand pnthe solution of the Navier–Stokes system (1) onΩn. Due to the homogeneous Dirichlet bound- ary condition on the lateral boundaryΓ, we can extend by zero unand pnoutside Ωn. So we can consider that the functions are all defined on the box D and the integrals overΩnand over D will be the same. Let us first remark that(un)is uni- formly bounded in H1(D). Indeed, the sequence
Ωn|ε(un)|2dx=
D|ε(un)|2dx is bounded by definition and the result follows using Korn’s inequality on the set
D together with a Poincaré’s inequality (see below proof of Proposition1).
Therefore, according to reflexivity of H1and the Rellich–Kondrachov’s theo- rem, there exists a vector u∈ [H1(D)]3and a subsequence, still denoted unsuch that:
un H1
u and un Lq
−→u, ∀q ∈ [1,6[.
It remains to prove that u is the velocity solution of the Navier–Stokes system on Ω. Let us write the variational formulation of (1). For any function w satisfying
w∈ [H1(D)]3:w=0 on E∪Γ and divw=0 in D, and for all n∈N, the function unverifies:
D(2µε(un):ε(w)+ ∇un·un·w)dx=
S
h.un·w ds (7) Since we have weak convergence of un, this results in:
Dε(un):ε(w)dx−−−−→
n→+∞
Dε(u):ε(w)dx. Let us now have a look to the trilinear term. We already know that∇un
L2(D)
∇u.
Moreover, from Cauchy–Schwarz’s inequality and Sobolev’s embedding theorem, we have:
(un−u)·w2[L2(D)]3 3 i=1
Ω(un,i −ui)4dx
Ωw4i dx 3un−u2[L4(D)]3w2[L4(D)]3. Then(un·w)n∈Nconverges strongly in L2(D)to u·w. Therefore,
D∇un·un·w dx−−−−→
n→+∞
D∇u·u·w dx.
Finally, weak convergence of unin[H1(D)]3implies weak convergence of the trace in L2(S)and the boundary term
Sh.un·w ds in (7) converges to
Sh.u·w ds.
Therefore, u satisfies the variational formulation (7) (and also the boundary condi- tion u=u0on E because every unsatisfies it). To conclude, it remains to prove that u is zero on the lateral boundaryΓ. It is actually a consequence of the convergence in the sense of compacts ofΩntoΩ, and the fact thatΩis Lipschitz and then stable in the sense of Keldys. We refer to Theorems 2.4.10 and 3.4.7 in [9].
We are now concerned with symmetry properties of the minimizer. When the state system is Stokes instead of Navier–Stokes the following result can be proved:
Theorem 3. There exists a minimizer of the problem (6) (with the Stokes system as state equation) which has a plane of symmetry containing the vertical axis.
Moreover, any minimizer of class C2has such a plane symmetry.
Proof. LetΩdenotes (one of) the minimizer(s) of problem (6) and D the vertical axis x1=x2=0. Among every plane containing D, at least one, say P0, cutsΩ in two sub-domainsΩ1andΩ2of same volume (volume equals to V/2).
Let us now introduce the two quantities J1and J2defined by:
J1:=2µ
Ω1
|ε(u)|2dx and J2:=2µ
Ω2
|ε(u|2dx,
so J(Ω) = J1+ J2. Without loss of generality, one can assume J1 J2. Let us now consider the new domainΩ =Ω1∪σ(Ω1), whereσ denotes the plane symmetry with respect to P0. We also introduce the functions(u,p)defined by
u(x)=
u(x) if x∈Ω1
u(σ (x)) if x∈σ(Ω1) and p(x)=
p(x) if x∈Ω1
p(σ(x)) if x∈σ (Ω1) It is clear thatu∈ [H1(Ω)] 3,p∈ L2(Ω) and divu=0. Moreover
2µ
Ω|ε(u)|2dx=4µ
Ω1
|ε(u)|2dx=2 J1J(Ω).
Now, it is well known that the solution of our Stokes problem can also be defined as the unique minimizer of the functional
ψΩ(v)):=2µ
Ω|ε(v)|2dx on the space
V(Ω):= {v∈ H1(Ω):div v=0, v|E =u0 and v|Γ =0}.
Therefore, we have:
J(Ω) = min
v∈V(Ω)
2µ
Ω|ε(v)|2dx
2µ
Ω|ε(u)|2dx J(Ω), (8)
this proves thatΩ, which has the same volume asΩand is symmetric with respect to P0, is also a minimizer of J .
Now, let us prove that ifΩis regular enough (actually C2but one can weaken as shown by the proof below), it must coincide withΩ, and therefore is symmetric.
Necessarily, we must have the equality in the chain of inequalities (8). It proves, in particular, thatu is the solution of the Stokes problem onΩ. But sinceu coincides with u onΩ1by definition, one can use the analyticity of the solution of the Stokes problem (see for example [12]) to claim thatu=u onΩ∩Ω. Now, if Ωwould not coincide withΩ, we would have a part of the boundary ofΩ, sayγincluded inΩ.
By assumption,Ωbeing C2, the solution of the Stokes problem is continuous up to the boundary (see [8]) and thereforeu should vanish onγ. By analyticity, it would imply that it vanishes identically: a contradiction with the boundary condition on
E.
As explained in the introduction, one can wonder whether the minimizer has more symmetry. In particular, could the cylinder be the minimizer? The following theorem proves that it is not the case. It is the main result of this paper. The proof is absolutely not obvious and will be given at the next section. Let us remark that the following result also holds for the Stokes equation. The proof in the Stokes case follows the same lines and is a little bit simpler, see [17] for details.
Theorem 4. The cylinder is not the solution of the shape optimization problem min J(Ω)
Ω ∈OV, (9)
where J is defined in (3) with u the velocity, solution of the Navier–Stokes problem (1), andOV is defined in (4).
3. Proof of the main theorem
In all this section,Ωwill now denote the cylinder{x12+x22<R2,0<x3<L}.
3.1. Computation of the shape derivative
Let us consider a regular vector field V: R3 →R3with compact support in the strip 0<x3<L. For small t, we defineΩt =(I +t V)Ω, the image ofΩby a perturbation of identity and f(t):= J(Ωt). We recall that the shape derivative of J atΩ with respect to V is f(0). We will denote it by d J(Ω;V). To compute it, we first need to compute the derivative of the state equation. We use here the classical results of the shape derivative as in [9,13,18]. The derivative of(u,p)is the solution of the following linear system:
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
−µu+ ∇u·u+ ∇u·u+ ∇p=0 x∈Ω
div u=0 x∈Ω
u=0 x∈ E
u= −∂u
∂n(V·n) x∈Γ
−pn+2µε(u)·n=0 x∈ S.
(10)
Now, we have (see [9,18]) d J(Ω,V)=4µ
Ωε(u):ε(u)dx+2µ
Γ |ε(u)|2(V·n)ds. (11) It is more convenient to work with another expression of the shape derivative.
For that purpose, we need to introduce an adjoint state.
Proposition 1. Let us consider(v,q), solution of the following adjoint problem:
⎧⎪
⎪⎨
⎪⎪
⎩
−µv+ ∇u·v− ∇v·u+ ∇q = −2µu x∈Ω
div v=0 x∈Ω
v=0 x∈ E∪Γ
−qn+2µε(v)·n+(u·n)v−4µε(u)·n=0 x∈S.
(12)
If the viscosityµ is large enough, then the problem (12) has a unique solution (v,q). Moreover, this solution belongs to C1(Ω)×C0(Ω).
Proof. The existence and uniqueness of the solution is a standard application of Lax–Milgram’s lemma. We introduce the Hilbert space
V(Ω):= {u∈H1(Ω):div u=0}.
the bilinear formαand the linear formdefined by α(v,w):=
Ω(2µε(v):ε(w)+ ∇w·u·v+ ∇u·w·v)dx ,w :=4µ
Ωε(u):ε(w)dx.
To prove ellipticity of the bilinear formαwe use Korn’s inequality:
∇v[L2(Ω)]3 C1(v[L2(Ω)]3+ ε(v)[L2(Ω)]3).
and a Poincaré inequality:
v[L2(Ω)]3 C2
Ω|ε(v)|2dx. (13) These two inequalities yield (we also use the explicit expression of u given in (2) to estimate the integrals containing u):
α(v,v)
µmin(1,C2)
C1+1 − |c|(R2+2R)
v2[H1(Ω)]3.
andαis elliptic as soon asµ > |c|(Rmin2+2R(1,)(CC2)1+1). Now, existence and uniqueness of the solution follow from a standard application of Lax–Milgram’s lemma together with De Rham’s lemma to recover the pressure.
It remains to prove the regularity of the solution. The C∞regularity inΩon the one-hand and on the smooth surfaces E, S and the interior of the lateral boundary Γ on the other hand is standard (refer to [8]). The only point which is not clear is the C1regularity on the circles E∩Γ and S∩Γ. To prove it, one can use the
cylindrical symmetry which is proved later (without any regularity assumptions) in Theorem2. This symmetry allows us to consider a two-dimensional problem in the rectangle(0,R)×(0,L)into the variables r =(x21+x22)1/2and x3. For that problem, one needs to prove regularity at the corners(R,0)and(R,L). For that purpose, one extends the solution by reflection around the line r = R, this leads to a partial differential equation in the rectangle(0,2R)×(0,L)whose solution coincides with our solution in the first half of the rectangle. The C1regularity, up to the boundary, of the solution of this elliptic PDE is standard and the result follows.
Let us come back to the computation of the shape derivative. We prove Proposition 2. With the previous notations, the shape derivative of the criterion J is given by
d J(Ω,V)=2µ
Γ
ε(u):ε(v)− |ε(u)|2
(V.n)ds. (14) Proof. Using Green’s formula in (11), one gets
d J(Ω,V)=4µ
Ωε(u):ε(u)dx+2µ
Γ |ε(u)|2(V.n)ds
= −2µ
Ω((u+ ∇divu)·u)dx+4µ
∂Ωε(u)·n·uds +2µ
∂Ω|ε(u)|2(V·n)ds
Now, let us multiply the first equation of the adjoint problem (12) by uand integrate overΩ, one obtains
−µ
Ωv·udx+
Ω∇q·udx+
Ω(∇u)T ·v·udx
−
Ω∇v·u·udx= −2µ
Ωu·udx.
Using one integration by parts and the boundary conditions satisfied by uand v, we get
Ω
2µε(u)·ε(v)− ∇v·u·u+ ∇u·u·v dx−
S
σ(v,q)·n·uds +
S
(u·v)(u·n)−(u·n)(u·v) ds
−
Γσ (v,q)·n)·uds = −2µ
Ωu·udx.
In the same way, if we multiply the first equation of the problem (10) by v and integrate overΩ, we obtain
−µ
Ωu·v dx+
Ω∇p·v dx+
Ω∇u·u·v dx+
Ω∇u·u·v dx=0
and
Ω
2µε(u)·ε(v)+ ∇u·u·v− ∇v·u·u dx +
S
−σ(u,p)·n·v+(u·v)(u·n) ds=0.
Coming back to the shape derivative expression d J(Ω,V)= −2µ
Ω((u+ ∇div u)·u)dx+4µ
∂Ωε(u)·n·uds +2µ
∂Ω|ε(u)|2(V·n)ds
= A+4µ
∂Ωε(u)·n·uds+2µ
∂Ω|ε(u)|2(V·n)ds, where we set A:= −2µ
Ω((u+ ∇divu)·u)dx. Using the previous identities, we get for A
A=
Γ∪S(qn−2µε(v)·n)·uds−
S
(u·n)(v·u)ds.
Therefore, according to (12) d J(Ω,V)=
Γ∪S
(qn−2µε(v)·n)·uds−
S
(u·n)(v·u)ds +4µ
S∪Γε(u)·n·uds+2µ
Γ|ε(u)|2(V.n)ds
=
Γ (qn−2µε(v)·n+4µε(u)·n)·uds+2µ
Γ|ε(u)|2(V.n)ds
= −
Γ
(qn−2µε(v)·n+4µε(u)·n)· ∂u
∂n −2µ|ε(u)|2
(V·n)ds To get the (more symmetric) expression given in (14), one can use the following elementary properties. Since u (and v) is divergence-free and vanishes onΓ, we have on this boundary:
– n· ∂∂un =0.
– ε(u)·n·∂∂un = |ε(u)|2. – (ε(v)·n)· ∂∂un =ε(u):ε(v).
Proposition2follows.
3.2. Analysis of the PDE (12)
We will prove the following symmetry result for the solution of the adjoint system. It shows that the solution has the same symmetry as the cylinder.
Lemma 2. With the same assumptions onµas in Proposition1, there exist(w, w3)∈ [H1((0,R)×(0,L))]2andq˜ ∈L2((0,R)×(0,L))such that, for any(x1,x2,x3) inΩ
(i) vi(x1,x2,x3)=xiw(r,x3), for i∈ {1,2};
(ii) v3(x1,x2,x3)=w3(r,x3); (iii) q(x1,x2,x3)= ˜q(r,x3).
where r =(x12+x22)1/2.
Proof. Let us introduce the differential operatorLθ defined by Lθ =x1 ∂
∂x2 −x2 ∂
∂x1.
Lθ corresponds actually to the differentiation with respect to the polar angleθ. Let us set
vi =Lθ(vi), ∀i ∈ {1,2,3} and q =Lθ(q). (15) By applying the operator Lθ to the equation (12) we get the following system (where we have used the explicit expression of the solution u given in (2))
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎩
−µv1+2cx1v3−2cx2v3−c(x12+x22−R2)∂v1
∂x3 + ∂q
∂x1− ∂q
∂x2 =0 x∈Ω
−µv2+2cx2v3+2cx1v3−c(x12+x22−R2)∂v2
∂x3 + ∂q
∂x2+ ∂q
∂x1 =0 x∈Ω
−µv3−c(x21+x22−R2)∂v3
∂x3+ ∂q
∂x3 =0 x∈Ω
∂v1
∂x1+∂v2
∂x2+ ∂v3
∂x3−∂v1
∂x2+ ∂v2
∂x1 =0 x∈Ω
v1=v2=v3=0 x∈E∪Γ
µ ∂v1
∂x3+ ∂v3
∂x1
−µ∂v3
∂x2 +c(x12+x22−R2)v1= −4µcx2 x∈S, µ
∂v2
∂x3+ ∂v3
∂x2
+µ∂v3
∂x1+c(x12+x22−R2)v2=4µcx1 x∈S, 2µ∂v3
∂x3+c(x12+x22−R2)v3=q x∈S, (16) Let us now introduce the following new functions
– z1=v1+v2; – z2=v2−v1; – z3=v3.
According to system (12), the system (16) is rewritten in terms of z1, z2, z3
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
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⎨
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⎪⎩
−µz1+2cx1z3−c(x12+x22−R2)∂z1
∂x3+ ∂q
∂x1 =0 x∈Ω
−µz2+2cx2z3−c(x12+x22−R2)∂z2
∂x3+ ∂q
∂x2 =0 x∈Ω
−µz3−c(x12+x22−R2)∂z3
∂x3+ ∂q
∂x3 =0 x∈Ω
∂z1
∂x1+ ∂z2
∂x2 +∂z3
∂x3 =0 x∈Ω
z1=z2=z3=0 x∈E∪Γ
µ ∂z1
∂x3 +∂z3
∂x1
+z1c(x12+x22−R2)=0 x∈S, µ
∂z2
∂x3 +∂z3
∂x2
+z2c(x12+x22−R2)=0 x∈S, 2µ∂z3
∂x3+c(x21+x22−R2)z3=q x∈S,
(17)
This adjoint problem has a unique solution ifµis large enough (see proposition1), therefore
z1=z2=v3=q ≡0.
The fact thatv3 =Lθ(v3)andq = Lθ(q)vanish proves points ii and iii of the lemma. Now let us make precise the properties of functions v1, v2. It has been proved thatLθ(v1)= −v2andLθ(v2)=v1. Therefore, applying once more the operatorLθyieldsLθ◦Lθ(v1)+v1=0. This implies that there exist two functions αandβ in the space H1((0,R)×(0,L)), such that
v1=x1α(r,x3)+x2β(r,x3).
Moreover, sinceLθ(v1)= −v2, we get
v2= −x1β(r,x3)+x2α(r,x3).
To finish the proof, it remains to check that the functionβ is identically zero. For that purpose, let us write down the partial differential equation satisfied byβ. From the two first equations of system (12) and the boundary condition, we can prove thatβ satisfies the following system
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⎩
−µ ∂2β
∂r2 +3 r
∂β
∂r +∂2β
∂x32
−c(r2−R2)∂β
∂x3
=0 (r,x3)∈(0,R)×(0,L) β(r,0)=β(R,x3)= ∂β
∂r(0,x3)=0 (r,x3)∈(0,R)×(0,L) µ∂β
∂n +c(r2−R2)β=0 (r,x3)∈(0,R)× {L}
(18)
It remains to prove that the zero function is the unique solution of the previous system. Multiplying the equation byβ and integrating on the rectangle in polar coordinates gives, using the boundary conditions
0=µ
Ω
∂β
∂r 2
+ ∂β
∂x3
2
r dr dx3+µ L
0
β2(0,x3)dx3
+c 2
R
0 (r2−R2)β2(r,L)r dr.
Since c <0 and r < R, we get ∂β∂r ≡0 in(0,R)×(0,L)andβ2(0,x3)=0 for any x3∈(0,L). Thenβ ≡0 which gives the desired result.
3.3. The optimality condition
We argue by contradiction. Let us assume that the cylinderΩ is optimal for the criterion J . We first write down the first order optimality condition. From the explicit expression (2) of u, we have
ε(u)=
⎛
⎝ 0 0 cx1
0 0 cx2
cx1 cx2 0
⎞
⎠.
Therefore
|ε(u)|2=2c2(x12+x22), and|ε(u)|2=2c2R2is constant onΓ.
Now the first order optimality condition ensures the existence of a Lagrange multiplierλ∈R, such that d J(Ω,V)=λdVol(Ω,V)for any vector field V. Due to the expression of the shape derivatives of J and the volume, it is written as
2µ
Γ
ε(u):ε(v)− |ε(u)|2
(V.n)ds=λ
Γ(V·n)ds.
This implies thatε(u): ε(v)is constant onΓ. Now, from the expression ofε(u) onΓ, we deduce
ε(u):ε(v)|Γ = c 2
x1∂v3
∂x1 +x2∂v3
∂x2+x1∂v1
∂x3+x2∂v2
∂x3
= c 2
x1∂v3
∂x1 +x2∂v3
∂x2
= c R 2
∂v3
∂n |Γ,
becausev1|Γ =v2|Γ =0. Therefore the optimality condition is written as
∃ξ ∈R: ∂v3
∂n =ξ onΓ. (19)
Now, we give another useful lemma
Lemma 3. If the cylinderΩ is optimal and using the notations of Lemma2, we have
∂q
∂n|Γ =∂q˜
∂r|{r=R} =0.
Proof. Let us write the adjoint problem (12) in term of the functionsw,w3etq.˜ We get
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⎩
−µ ∂2w
∂r2 +1 r
∂w
∂r +∂2w
∂x23
+1 r
∂q˜
∂r +2cw3−c(r2−R2)∂w
∂x3
=0 inΩ
−µ ∂2w3
∂r2 +1 r
∂w3
∂r +∂2w3
∂x23
+1 r
∂q˜
∂x3 −c(r2−R2)∂w3
∂x3 = −8µc inΩ 2w+r∂w
∂r +∂w3
∂x3 =0 inΩ
w(r,0)=w3(r,0)=w(R,x3)=w3(R,x3)=0 µ
∂w
∂x3+1 r
∂w3
∂r
+c(r2−R2)w=4µc on S 2µ∂w3
∂x3 +c(r2−R2)w3= ˜q on S.
(20) Sincew|{r=R} =w3|{r=R} =0, we have∂∂wx
3|{r=R} =∂w∂x33|
{r=R} =0 and∂∂2w
x32|{r=R}
=0.
In particular, from the divergence-free condition, we obtain ∂w∂r|
{r=R} =0.
Now, let us differentiate the divergence-free condition with respect to r , we get
∀(r,x3)∈(0,R)×(0,L), 3∂w
∂r +r∂2w
∂r2 + ∂2w3
∂r∂x3 =0.
Now, ∂w∂r3|
{r=R} = ξ (it is the optimality condition (19)) ; therefore, we have
∂2w3
∂x3∂r|{r=R} =0. Combining this last result with ∂w∂r|
{r=R} =0, results in
∂2w
∂r2|{r=R} =0.
We let r going to R in the first equation of problem (20) and we use the previous identities to get
∂q˜
∂r|{r=R} =0.
3.4. An auxiliary function
Using notation of Lemma2, we introduce now two new functions – w0: [0,R] × [0,L] −→R
(r,x3)−→
x3
0
w(r,z)dz .
– ψ: [0,R] × [0,L] −→R x3−→
R
0
2π
0
q˜(r,x3)−2cr2w0(r,x3) dθr dr
.
We will also denote by Tz the horizontal section of the cylinder{x∈Ω :x3=z}.
The following lemma is the key point of the proof.
Lemma 4. The functionψis affine.
Proof. The couple(v,q)satisfies the following PDE
−µv+ ∇q+ ∇u·v− ∇v·u= −2µu.
Let us compute the divergence of both sides of the previous equality. Using the expression of u in the cylinderΩ, we obtain that(v,q)verifies
q+4cv3+2c
x1∂v3
∂x1+x2∂v3
∂x2
−2c
x1∂v1
∂x3 +x2∂v2
∂x3
=0. (21) Let us integrate this equation on a slide
ω:= {(x1,x2,x3)∈Ω;z−x3z+} (we will denote by e the inlet ofωand s its outlet). We get
ωq+4cv3dx+2c
ω
x1∂v3
∂x1+x2∂v3
∂x2
−2c
x1∂v1
∂x3 +x2∂v2
∂x3
dx=0.
Now, from Green’s formula, we have
ωx1∂v3
∂x1
dx=
∂ωx1v3n1ds−
ωv3dx=
∂ω∩Γ x1v3n1ds−
ωv3dx
= −
ωv3dx in the same way
ωx2∂v3
∂x2
dx= −
ωv3dx. Therefore
4c
ωv3dx+2c
ω
x1∂v3
∂x1 +x2∂v3
∂x2
dx=0,
so
ωq dx=2c
ω
x1∂v1
∂x3
+x2∂v2
∂x3
dx. (22)
Let us consider the left-hand side of (22). From Lemma3is
ωq dx=
s∪e
∂q
∂n ds. (23)
Now, let us consider the right-hand side of (22). Integrating by parts yields
–
ωx1∂v1
∂x3
dx=
∂ωx1v1n3ds=
e∪s
x1v1n3ds.
–
ωx2∂v2
∂x3
dx=
∂ωx2v2n3ds=
e∪s
x2v2n3ds.
Combining this result with (23) gives
s
∂q
∂x3−2c(x1v1+x2v2)
ds=
e
∂q
∂x3−2c(x1v1+x2v2)
ds, (24) what can also be rewritten for any(z−,z+)∈(0,L)2:
R
0
∂q˜
∂x3(r,z−)−2cr2w(r,z−)
r dr= R
0
∂q˜
∂x3(r,z+)−2cr2w(r,z+)
r dr. (25) Now, sinceψ(z)=2πR
0 (˜q(r,z)−2cr2w0(r,z))r dr , we have by differentiating, for all z in[0,L],
ψ(z)=2π R
0
∂q˜
∂x3 −2cr2∂w0
∂x3
r dr =2π R
0
∂q˜
∂x3−2cr2w
r dr.
Now, identity (25) proves thatψis a constant function which gives the desired result.
We are now in position to make precise the value of the constantξ appearing in the first order optimality condition (19). For that purpose, we use the symmetry result given in Lemma2together with Equation (20). In this equation, let us inte- grate between x3 =0 and x3 = z ∈ (0,L). Sincew3(r,0)=0, we get for any (r,z)∈ [0,R] × [0,L]:
2w0(r,z)+r∂w0
∂r (r,z)+w3(r,z)=0.
Let us differentiate this last relation with respect to r . This yields 3∂w0
∂r +∂2w0
∂r2 +∂w3
∂r =0. (26)
Now, in (20), we differentiate the divergence equation with respect to r , and we make r → R. We obtain
∂w
∂r|Γ = ∂2w
∂r2|Γ =0.
Letting r go to R in (26) and interverting the limit and integral gives, using the previous equality
∂v3
∂n |Γ =0.
So we conclude thatξ =0 and the optimality condition is rewritten as
∂v3
∂n |Γ =0. (27)
