Symmetry and asymmetry in a multi-phase overdetermined problem

A celebrated theorem of Serrin asserts that one overdetermined condition on the boundary is enough to obtain radial symmetry in the so-called one-phase overdetermined torsion problem. It is also known that imposing just one overdetermined condition on the boundary is not enough to obtain radial symmetry in the corresponding multi-phase overdetermined problem. In this paper, we show that, in order to obtain radial symmetry in the two-phase overdetermined torsion problem, two overdetermined conditions are needed. Moreover, it is noteworthy that this pattern does not extend to multi-phase problems with three or more layers, for which we show the existence of non-radial configurations satisfying countably infinitely many overdetermined conditions on the outer boundary.


Introduction
Let N denote the set of positive integers.For some fixed number m ∈ N, let us introduce the problem setting and notation related to multi-phase (m-layered) elliptic overdetermined problems.
Finally, let n denote the outward unit normal vector to Ω k (k = 1, . . ., m) and ∂ n be the corresponding normal derivative.Similarly, let (∂ n ) j denote the so-called, jth order normal derivative, which is given by the following expression: where the summation is taken over all multi-indices β = (β 1 , . . ., β N ) of length k and In this paper, we will consider the following boundary value problem, which will be referred to as the multi-phase torsion problem: − div(σ∇u) = 1 in Ω, u = 0 on ∂Ω. (1.2) We recall that even when no additional smoothness assumptions are imposed on Ω k , the weak solution to (1.2) is defined as the unique function u ∈ H 1 0 (Ω) satisfying It is a well-known fact (see [AS]) that, under suitable smoothness assumptions, the boundary value problem (1.2) can be rewritten as the following transmission problem (the name diffraction problem is also used, see [LU,Chapter 16]): where the quantity • , the jump through the interface ∂Ω k , is defined as follows: for any where f j := f D j (j = 1, . . ., m).
If the domains Ω k are concentric balls then, by unique solvability, there exist real The aim of this paper is to investigate to what extent the reverse implication holds.It may not be surprising to know that the answer depends on the number of layers m.
The case m = 1 was solved by Serrin.By adapting the famous reflection principle of Alexandrov (see [Al1] and Theorem C of page 5), he showed the following symmetry result: We refer the interested reader to the survey paper [Ma] for an overview of some qualitative and quantitative results related to Theorems A and C.
As shown by the author and Yachimura by making use of a perturbation method relying on the implicit function theorem, one overdetermined condition is not enough to obtain symmetry when m = 2.

Theorem B ([CY1]
).Let m = 2.Then, there exist infinitely many pairs of domains Ω 1 ⊂⊂ Ω 2 that are not concentric balls but such that the solution u of (1.2) satisfies Regarding the study of nontrivial solutions to the above two-phase overdetermined problem, further analysis has been carried out concerning local bifurcation ( [CY2]) and stability ( [CPY]) respectively.
The difference in behaviors between one-phase and two-phase elliptic overdetermined problems presented by Theorems A and B admits the following heuristic interpretation.
The one-phase overdetermined problem of Theorem A has one constraint (the overdetermined condition) and one degree of freedom (the shape of Ω).On the other hand, the two-phase overdetermined problem of Theorem B also has one constraint (the overdetermined condition) but two degrees of freedom (the shapes of both Ω 1 and Ω 2 ).In other words, Theorem B shows that, when the number of degrees of freedom exceeds that of constraints, the overdetermined problem admits nontrivial solutions.
By combining the soap bubble theorem of Alexandrov [Al1] and a symmetry result by Sakaguchi [Sak], we obtain the following result: Theorem I. Let m = 2 and let Ω 1 ⊂⊂ Ω 2 be bounded domains of class C 2 such that In light of this result, together with Theorem A, one might be tempted to formulate the following (false!)conjecture.
Then, problem (1.2) admits a solution u of class C m is a neighborhood of ∂Ω m satisfying for some constants c k ∈ R if and only if the sets Ω k are concentric balls.
We will show that Conjecture 1.1 does not hold for m ≥ 3.As a matter of fact, we are able to exhibit a counterexample as follows.
This paper is organized as follows.In Section 2, we show Theorem I while the subsequent sections are devoted to the proof of Theorem II.In Section 3, we introduce some preliminary results concerning the shape differentiability of state functions in two-phase problems.In Section 4, we study the invertibility properties of a linearized operator (Dirichlet-to-Neumann map).Then, in Section 5 we combine the results of the two preceding sections to give a proof of Theorem II by means of the implicit function theorem.
Finally, Section 6 is devoted to some comments on the proof of Theorem II and how it relates to the existing literature.
2 Proof of Theorem I In this section, we will give a simple proof of Theorem I. To this end, we will need the following symmetry results.

Theorem C ([Al1]
).A compact hypersurface, embedded in R N , that has constant mean curvature must be a sphere.

Theorem D ([Sak]
).Let m = 2. Let Ω 2 be an open ball and let Ω 1 ⊂⊂ Ω 1 be a bounded open set of class C 2 with finitely many connected components such that We remark that both theorems above were originally stated in a more general setting (see also [Al2]) but the formulations above are enough for our purposes.
Theorem I now follows by combining the two theorems above.
Proof of Theorem I. Let m = 2 and let the sets Ω 1 and Ω 2 satisfy the hypotheses of the theorem.In what follows, we will assume that problem (1.2) admits a solution u of class for some real constants c 1 , c 2 and then show that (Ω 1 , Ω 2 ) must be concentric balls.The reverse implication is trivial and therefore omitted.
Since the solution u is of class C 2 in a neighborhood of ∂Ω 2 , the decomposition formula for the Laplace operator ( [HP,Proposition 5.4.12])combined with (2.1) yields: where ∆ τ = div τ •∇ τ denotes the tangential Laplacian (that is, the tangential divergence of the tangential gradient, also known as "Laplace-Beltrami operator", see [HP,Definitions 5.4.5,5.4.6 and 5.4.11]) on ∂Ω 2 and H is the (additive) mean curvature given by the tangential divergence of the outward unit normal n (notice that, under this definition, the mean curvature of a ball of radius R is N −1 R ).The terms in (2.2) can be rearranged to show that the mean curvature H is constant on the entire ∂Ω 2 .Thus, by applying Theorem C to each connected component of ∂Ω 2 we obtain that ∂Ω 2 is the disjoint union of a finite number of spheres with the same radius and orientation.This leaves us with just one possibility, that is, ∂Ω 2 is a sphere and Ω 2 is a ball.The conclusion readily follows from Theorem D.
Remark 2.1.Notice that in the proof above we did not use the connectedness of Ω 1 (nor In [BS, Ba], the authors showed the radial symmetry of the solutions to a similar multiphase overdetermined problem in R N (in the elliptic and parabolic settings respectively), where the overdetermined condition considered requires the solution u to be constant of each interface.In our setting, the following analogous result holds: Corollary 2.2.Let m ∈ N and let Ω k , k ∈ {1, . . ., m}, be as in the introduction.If the solution u to (1.4) satisfies for some real constants α 1 , . . ., α m−1 and c, then the sets Ω k are concentric balls and the function u is radial.
Proof.We will show the claim by induction on the number of layers m.The base case m = 1 is exactly Serrin's result, Theorem A. In what follows, let us assume that the claim holds when the number of layers is strictly less than m and then show that the claim holds for m as well.Let u be the solution to (1.4) with m layers Ω 1 ⊂⊂ • • • ⊂⊂ Ω m and assume that u satisfies the overdetermined conditions (2.3).Let By construction, u solves the transmission problem (1.4) with m − 1 layers Ω 2 ⊂⊂ • • • ⊂⊂ Ω m .Moreover, u also satisfies the overdetermined conditions (2.3) (starting from the "first" interface ∂Ω 2 ).Thus, by the inductive hypothesis, Ω 2 , . . ., Ω m are concentric balls and u is radial.As a consequence, u is also radial and, since ∂Ω 1 is a level set of u, the remaining set Ω 1 is also a ball concentric with Ω m .This concludes the proof.

Preliminaries on shape derivatives
In this section, we are going to introduce the main definitions and known results concerning shape calculus for two-phase problems that are going to be useful in the proof of Theorem II.The experienced reader might therefore skip this section.
The machinery of shape derivatives is the right tool to give a quantitative description of how v ξ,η depends on the perturbations (ξ, η).The main technical difficulties lie in the following two points: firstly, the functions v ξ,η depend on two parameters, and secondly, each v ξ,η lies in a different function space depending on the pair (ξ, η).To overcome these difficulties, let Θ := C 2,α (R N , R N ) and consider the following construction.For small θ ∈ Θ, set and let v θ be the unique solution to (3.2) with respect to the pair (D θ , Ω θ ).Moreover, set Then, the (first-order) shape derivative of v θ at θ = 0 is defined as where V (0)[θ] denotes the Fréchet derivative of V at θ = 0 in the direction θ ∈ Θ.Notice that the definition (3.4) is given in such a way as to be compatible with a formal application of partial differentiation with respect to θ in (3.3).
Lemma 3.1.We have the following: (ii) Let U be a neighborhood of ∂Ω 2 that does not intersect Ω 1 and set .
(iv) Following (3.4), let v [ξ] := v [E(ξ, 0)] denote the shape derivative of v θ with respect to the outer perturbation ξ only.Then, v [ξ] is independent of the extension operator E and can be characterized as the unique solution to the following boundary value problem: (3.5) Sketch of the proof.In the case of the Laplace operator (σ ≡ 1) with homogeneous Dirichlet boundary conditions (f ≡ 0), the claims (i) − (iv) are well-known results that can be obtained by a standard procedure that combines the implicit function theorem (Theorem E, page 12) and the Schauder regularity theory [GT, Chapter 6]: see, for instance, [HP, Section 5.3 and the final remark therein].Also, the case of two-phase boundary value problems with homogeneous Dirichlet boundary conditions has been dealt with in [Ca, Appendix], while the case of general boundary conditions for the Laplace operator has been briefly covered in [HP,Section 5.6] and the references therein.Finally, as far as the extension operator E is concerned, we refer the interested reader to [GT,Section 6.9].
Here, we will just limit to showing a simple trick that allows us to reduce to the cases mentioned above.For small θ ∈ Θ, let v θ be the solution to (3.2) with respect to the pair (D θ , Ω θ ) and let where K θ := (Id + θ)(K).Notice that, by construction, w θ is a weak solution to where the function ) is given by F θ := 1+div(σ θ ∇f ).Also, notice that, by construction, F θ admits an extension to R N that is independent of ξ.Therefore, all results (i) − (iv) hold for w θ .Since v θ = w θ + f by definition, it is clear that (i) − (iii) hold for v θ as well.To conclude, we just need to check that v [ξ] solves (3.5).To this end, notice that, by definition, we have satisfies the equation in (3.5).On the other hand, notice that W (0) = V (0) − f holds by construction, while, by (iv), w [ξ] satisfies the boundary condition The claim (iv) for v θ follows by combining the identities above.

The two-phase Dirichlet-to-Neumann map
As in the previous section, let Ω 1 and Ω 2 be concentric balls of radii R ∈ (0, 1) and 1 respectively and let σ 2 := 1.Let us introduce the following two-phase Dirichlet-to- where w[ξ] is the unique solution to the following boundary value problem be a maximal family of linearly independent solutions to the eigenvalue problem By the method of separation of variables, it can be shown that the spherical harmonics form an orthonormal basis of eigenfunctions of N in L 2 (∂Ω 2 ).The eigenvalues of N have been computed in [CY1].
Proof.Since the eigenvalues of (4.2) have been computed in [CY1], in what follows we only need to check that the right-hand side in (4.2) vanishes if and only if k = 0. Furthermore, since N (Y 0,i ) = 0 by construction, it will suffice to show that N (Y k,i ) = 0 for k ∈ N. To this end, let ϕ denote the numerator in the right-hand side of (4.2), that is We will show that ϕ(R) > 0 for all R ∈ (0, 1] and k ∈ N, proving the claim.First, notice that ϕ is a decreasing function of R. Thus, for all R ∈ (0, 1], which is what we wanted to show. Let C i,α * (∂Ω 2 ) denote the set of all functions in C i,α (∂Ω 2 ) with zero average over ∂Ω 2 (i ∈ {1, 2}).Notice that, by Lemma 4.1, N fixes the eigenspaces of the Laplace-Beltrami operator, whence N is a well-defined operator from First of all, let us consider the Sobolev space H 1 (Ω 2 ) endowed with the (equivalent) norm and consider the bilinear form B : Notice that, by construction, B is bilinear, continuous, and coercive.Fix now an element . By the Lax-Milgram theorem, there exists a unique function Now, if we restrict the identity above to φ in H 1 0 (Ω 2 ), then we realize that w must satisfy In other words, w satisfies the equation and the transmission conditions: Moreover, integration by parts in (4.3) and the arbitrariness of the trace of φ ∈ H 1 (Ω 2 ) on ∂Ω 2 yield Now, since w is the solution to the transmission problem (4.5), (4.6) and (4.7), we can inductively bootstrap its regularity in a classical way by means of the standard elliptic regularity estimates [GT,Chapter 8] and the Schauder boundary estimates [GT,Chapter 6] (see for example the argument in the proof of [KS,Proposition 5.2] after (5.7)).We obtain that w is of class C 2,α in an open neighborhood of ∂Ω 2 (whose closure does not intersect Ω 1 ).In particular, the function is a well defined element of C 2,α (∂Ω 2 ).This, together with (4.5), (4.6) and (4.7), implies that w is the solution to (4.1).In particular, by (4.8) and (4.7), , where p = (ξ(λ), λ) and λ ∈ Λ .
Before giving the proof of Theorem II, let us first define some auxiliary functions.Set R 0 := 0 and let u 0 denote the following radial function: where the value of the sum m−1 j=k+1 is set to be zero if m − 1 < k + 1 (empty sum).It is easy to check that u 0 solves the transmission problem (1.4).The following auxiliary function will also play a crucial role in our construction: (5.3)This is nothing but the solution to problem (1.4) in the case σ 1 = σ 2 = σ 3 .
Let α ∈ (0, 1).For sufficiently small η ∈ C 2,α (∂Ω 1 ) and ξ ∈ C 2,α (∂Ω 2 ), let D η and Ω ξ be the bounded domains whose boundaries are given by (3.1).Moreover, let v ξ,η be the solution to the following boundary value problem associated to the pair (D η , Ω ξ ): (5.4) where (5.5) Proof of Theorem II.We will construct nontrivial domains D η and Ω ξ with η ∈ C 2,α (∂Ω 1 ) and ξ ∈ C 2,α (∂Ω 2 ) such that the solution v ξ,η to (5.4) satisfies (5.6)where ∂ n ξ is the normal derivative in the outward direction on ∂Ω ξ .In other words, if (5.6) holds, then the function solves the transmission problem (1.4) with respect to (5.1).Moreover, since the function v defined above is radial in Ω m \ Ω ξ , in particular, it satisfies (1.5).We are therefore left with the problem of finding a nontrivial pair of functions (ξ, η) such that (5.6) holds.To this end, consider the following mapping: (5.8) Let us first clarify the notation employed in the definition of Ψ.Here, n stands for the outward unit normal vector to the unperturbed boundary ∂Ω 2 (we recall that the outer normal to ∂Ω ξ is denoted by n ξ ) and Id + ξn is the natural pullback mapping from ∂Ω ξ to ∂Ω 2 .Moreover, J τ (ξ) is the tangential Jacobian associated with the mapping Id + ξn (that is, it is the multiplicative term that appears in the integrand of a surface integral after the corresponding change of variables, see [HP,(5.67)-(5.68)]).It is known that both n ξ and J τ (ξ) are differentiable with respect to ξ at ξ = 0 (see [HP,Proposition 5.4.14 and Lemma 5.4.15].We remark that in [HP,Proposition 5.4.14]only Gâteaux differentiability is shown.This notwithstanding, the Fréchet differentiability of the normal can be shown analogously or by noticing that a smooth extension of n can be written as the normalized gradient of some subharmonic function vanishing on the boundary).These facts together with Lemma 3.1 imply that Ψ is a well-defined and Fréchet differentiable mapping from a neighborhood of (0, 0) ∈ C 2,α * (∂Ω 2 ) × C 2,α (∂Ω 1 ) into C 1,α (∂Ω 2 ).We just need to check that, for small (ξ, η) ∈ C 2,α * (∂Ω 2 ) × C 2,α (∂Ω 1 ), the image Ψ(ξ, η) is indeed a function of zero average over ∂Ω 2 .To this end, notice that div(σ ξ,η ∇v ξ,η ) = −1 = σ 3 ∆v 0 in Ω ξ and thus the divergence theorem and a change of variables yield the desired identity: Moreover, since the tangential Jacobian J τ never vanishes, it is clear that, by definition, Ψ(ξ, η) = 0 if and only if v ξ,η satisfies (5.6).
In what follows, we will apply the implicit function theorem (Theorem E) to the mapping Ψ : To this end, it will be sufficient to show that the partial Fréchet derivative of Ψ with respect to ξ at ξ = 0 is a bijection between C 2,α * (∂Ω 2 ) and C 1,α * (∂Ω 2 ).A direct computation with (5.2), (5.3), (5.5) and (4.1) at hand yields: We remark that the computation above is also dramatically simplified because the Fréchet derivative of n ξ • (Id + ξn) at ξ = 0 is tangent to ∂Ω 2 , that is, orthogonal to n (as a matter of fact, it is equal to −∇ τ ξ by [HP,Proposition 5.4.14]).Finally, since σ 3 = σ 2 = 1, the conclusion follows from Lemma 4.3.

Some final comments
In this section, we give some comments on the various topological and regularity assumptions used in this paper.

On the topological assumptions in Theorem D
Theorem D of page 5 ensures spherical symmetry in a two-phase setting under the topological assumption that Ω 1 has finitely many connected components and D 1 := Ω 2 \ Ω 1 is connected.Notice that, by Theorem II, we know that the connectedness of D 1 is necessary.Indeed, since Theorem II holds in the "two-phase-three-layer" case (that is, m = 3 and σ 1 = σ 3 ), there exists a non-trivial triplet of domains D η ⊂⊂ Ω ξ ⊂⊂ Ω 3 such that (1.5) holds.Renaming the domains as gives the desired counterexample to Theorem D for disconnected D 2 = Ω 2 \ Ω 1 .To the best of my knowledge, it is still an open question whether there exists a counterexample to Theorem where Ω 1 has infinitely many connected components without developing a microstructure.
in the appropriate Banach spaces becomes an ill-conditioned problem.As a result, Fredholmness is lost and the proofs of the analogs of Lemmas 4.2-4.3fail.As a rule of thumb, we can say that this sort of ill-conditioning usually happens when the "free boundary" (in this case ∂Ω ξ ) and the "overdetermined boundary" (that is, the boundary where the tracking takes place, in this case, ∂Ω m ) do not coincide.