Two- and Multi-phase Quadrature Surfaces

In this paper we shall initiate the study of the two- and multi-phase quadrature surfaces (QS), which amounts to a two/multi-phase free boundary problems of Bernoulli type. The problem is studied mostly from a potential theoretic point of view that (for two-phase case) relates to integral representation $$ \int_{\partial \Omega^+} g h (x) \ d\sigma_x - \int_{\partial \Omega^-} g h (x) \ d\sigma_x= \int h d\mu \ , $$ where $d\sigma_x$ is the surface measure, $\mu= \mu^+ - \mu^-$ is given measure with support in (a priori unknown domain) $\Omega$, $g$ is a given smooth positive function, and the integral holds for all functions $h$, which are harmonic on $\overline \Omega$. Our approach is based on minimization of the corresponding two- and multi-phase functional and the use of its one-phase version as a barrier. We prove several results concerning existence, qualitative behavior, and regularity theory for solutions. A central result in our study states that three or more junction points do not appear.


Introduction
The current paper concerns the so-called quadrature identities for surface integrals, for the harmonic class of functions, and for given measures. Our primary purpose is to generalize the concept of quadrature surface (henceforth QS) to the two-and multi-phase counterpart.
The free boundary problem studied here has some "new" components that might be interesting to free boundary and potential theory community. From potential theory point of view, we consider here a completely new problem dealing with the two-phase version of the problem of gravi-equivalent bodies 1 , in particular, the existence of surfaces that "surround" the body is essential hearth of matter. On the other hand, the free boundary communities, specially those working with regularity theory, would find an interesting extension of the concept of two-phase Bernoulli problem, with the zero set having non-void interior. This obviously makes the problem a three phase problem with the third phase being free of fluid.
1.1. One-phase QS. Let Ω ⊂ IR N (N ≥ 2) be a bounded domain with reasonably smooth boundary, and µ be a measure with support contained in Ω. Then we say that ∂Ω is a quadrature surface with respect to µ if the overdetermined Cauchy problem (1.1) ∆u = −µ in Ω u = 0, ∂u ∂ν = −1 on ∂Ω, has a solution. Here ν is the outward normal to the boundary ∂Ω.
For a better understanding, we recall the definition of one phase quadrature domains from [13]: Given density functions 0 ≤ g, h ∈ L ∞ (IR N ) and a Radon measure µ, we say that Ω is a quadrature domain for µ, for the given densities g and h if Ω is a bounded open set in IR N such that supp µ ⊂ Ω, (1.2) Quadrature domains can be obtained as supports of local minimizers for the one phase functional (1.9) where f , g ∈ L ∞ (IR N ) are suitably chosen and satisfy suitable conditions to allow a minimum for the functional. It was shown in [13] that a local minimum of the functional J 1 f,g satisfies (1.10) ∆u = −f in Ω = {u > 0}, u = 0, |∇u| = g on ∂Ω.
For general measures, e.g. Dirac masses, the functional may not have lower bound, and hence the minimization may not work. However, there is an easy way out of this problem, by smoothing out the measure and solving the approximate problem, and then considering the limit problem. Indeed, for a given measure µ, one uses radial mollifiers,μ for approximating µ. For f =μ − h letũ ≥ 0 denote local minimum for J {f,g} so that it satisfies the equation (1.10). See [13] for details.
Equation (1.10) can be rewritten in the sense of distributions as In terms of the measure ν defined in (1.4), the above identity can be written as µ + ∆u = ν, so that u is the difference of Newtonian potentials for the measures µ and ν. The set Ω is a quadrature domain for µ if and only if supp(µ) ⊂ Ω. Now if we let h = 0 then ν is the surface measure, and solution to this problem corresponds to (one phase) quadrature surfaces.
1.2. Two-phase model. The two-phase counterpart of the functional (1.9) is for given functions f 1 , f 2 , g, where u + (x) := max{u(x), 0} and u − (x) := max{−u(x), 0}. The two phase functional with g = 0, i.e., J {f,g=0} was studied in the paper [11]. In this paper we are interested in showing existence of "two-phase" quadrature surfaces corresponding to a measure µ = µ + − µ − . Thus, assume that supp(µ + ) = ∅, supp(µ − ) = ∅ and that supp(g 2 ) has positive measure. We look for minimizer of the functional (1.12) where the functions f 1 correspond to mollification of µ + and f 2 is mollification of µ − . Here we expect that the (local) minimizer of (1.12) will satisfy Our approach in proving existence of minimizers to the two-phase functional, shall follow that of [11]. By relating the two phase functional to the one phase functionals, one can efficiently generate solutions to the two phase problems by using suitable conditions ensuring existence of one phase solutions.
As mentioned earlier, our problem produces three different phases, rather than two. More exactly, and contrary to the classical Bernoulli-type free boundaries, the interior of the set {u = 0} is non-void in our case. In particular, one has a triple junction free boundary points, where all three phases meet. This type of Bernoulli-free boundary is subject of current study by the third author, and his collaborators, see [3].
Remark 1.1. The general case where one replaces g 2 χ {u =0} with g 2 1 χ {u>0} + g 2 2 χ {u<0} is not treated in this paper, but can be handled in much the same way as our situation. The functional, in this general case leads to the Bernoulli condition on the two-phase boundary, and the standard one-phase boundary condition holds on onephase boundary points.

Notation
Here, for the reader convenience, we present some notations, which will be used during the exposition of the paper:

Existence of minimizers
In this section, we give some conditions for existence of minimizers of the variational functional (1.12). We begin by proving the following comparison lemma for J {f 1 ,f 2 ,g} , similar to Lemma 1.1 in [12,13].
In particular, this shows that the minimizer of the functional J {f 1 ,f 2 ,g} , if it exists, can be assumed to have support inside the union of supports of the minimizers for the corresponding one phase functionals, J 1 {f 1 ,g} and J 1 {f 2 ,g} .
is the one phase functional. Then, we conclude that In fact, for any ϕ ∈ H 1 (IR N ), we can write Thus, as long as there are conditions on f 1 , f 2 and g which guarantee existence of one phase minimizers with compact support, we can minimize the functional J {f 1 ,f 2 ,g} on H 1 (IR N ). In particular, we recall Theorem 1.4 of [13] which give conditions for existence of minimizer for the one-phase functional J 1 {f,g} , viz., at least one of f ≤ −c 1 < 0 or g ≥ c 0 > 0 hold outside a compact set for some positive constants c 0 , c 1 .
Proof. Since (f 1 , g + ) and (−f 2 , −g − ) both satisfy the conditions (A1) − (A4), then we get existence of minimizers for the one phase functionals J 1 f 1 ,g and J 1 f 2 ,g . Thus, minimizing J {f 1 ,f 2 ,g} over the set W := {u ∈ H 1 (IR N ) : u 1 ≤ u ≤ U 1 } and repeating the proof of Proposition 2.1 of [11], we obtain a minimizer for the two phase functional J {f 1 ,f 2 ,g} . Here we note that J {f 1 ,f 2 ,g} ≥ J {f 1 ,f 2 ,g=0} . Theorem 3.1 will be used to prove the existence of "two phase quadrature surface" in Section 7. This is the case when each f i in (1.12) is replaced by µ i * ψ, where µ i * ψ is a mollified version of a positive Radon measure µ i with compact support. We will restrict ourselves to the case when measures µ i are "sufficiently concentrated" as defined in [13], which we refer to as measures satisfying Sakai's concentration condition defined as follows.
Definition 3.1 (Sakai's concentration condition). We say that the Radon measure µ satisfies Sakai's concentration condition if for every x ∈ supp(µ) where c > 0 is a fixed constant such that 0 ≤ g(x) ≤ c. Here g is the function given in (1.12).

Free boundary condition for the Minimizer
The free boundary of u, denoted Γ u = Γ, is defined as A point x ∈ Γ u is said to be a one phase free boundary point if there exists r > 0 such that and it is said to be a two phase free boundary point if for all r > 0, The set of one phase free boundary points of u is defined as while the set of two phase points, denoted Γ ′′ u = Γ ′′ is Finally, the set of branch points Γ * = Γ * u is the intersection of Γ ′ and Γ ′′ ; The free boundary Γ u of a solution u to a two-phase problem can thus be decomposed as Here we show that under suitable conditions, a local minimizer u of the functional J := J {f,g} satisfies (4.2) ∆u + f L | Ω = gH N −1 | ∂Ω in Ω := supp(u).
Observe that, for any ϕ ∈ C 2 0 (Ω), for the functional and after integration by parts we have that u satisfies One may now show in a standard way that the minimizer satisfies the Bernoulli boundary condition in a weak sense, and in the strong sense H n−1 -almost everywhere on the free boundary. Here weak sense refers to , and for all vector fields Θ ∈ C 0 (B r (z 1 ), IR N ). Here ν denotes the outward normal vector to the boundary of the sets. Analysis of the free boundary in neighborhood of the branch points is very technical and relies on the results proved in the paper [3]. The following theorem summarizes the regularity properties of free boundary: i) The weak free boundary condition (4.4) holds for minimizers. ii) For any point z ∈ Γ ′′ \ Γ ⋆ (two-phase and non-branch points) we have The free boundary has finite (N − 1)-dimensional Hausdorff measure. iv) Close to the branch points, the free boundary consists of two C 1,α graphs in a universal neighborhood of the branch point.
The proof of (i) is straightforward and similar to that of Theorem 2.4 in [2], by use of domain variation. Note that our model, as formulated, does not require the condition meas{u = 0} = 0. This is because in our situation, the function λ(u) of [2] is and hence λ(0) = 0. From Theorem 7.1 and Remark 7.1 in [2], it follows that the set of one phase boundary points Γ ′ u has finite (N − 1)-Hausdorff measure. Furthermore, due to the choice g 1 = g 2 we have made here, the two phase free boundaries are level surfaces of harmonic functions.
Proof of iii) also uses non-degeneracy of both phases. 2 For interested reader we refer to [3] Section 3, for further local measure theoretic properties of the free boundary.
The proof of iv) is a deep result, using chains of technical arguments. The core idea is that due to non-degeneracy of both phases, a blow up (scaling of the type u r (x) := u(rx + z)/r at any branch point z) leads to a global two-phase solution, which is classified and shown to be a two-plane solutions (i.e. a broken linear function L(x) := a + x + − a − x − ). One may then reiterate the blow-up argument, but this time by linearization technique, i.e. considering the limits of (u r − L(rx))/r. One proves that these limits exists and will solve a so-called thinobstacle problem, which in turn is well-studied. Enough information and knowledge about the regularity theory of their free boundary is available in literature. From here on, one may then show that our model is a (close enough) perturbation of the limiting problem and hence we can derive regularity of the free boundary for our original problem.

Qualitative properties
In this section we discuss some qualitative properties of minimizers of (1.12), that has already been established for the one-phase case. The complications, with the two-phase case makes similar properties much more hard to show. Here we apply the moving plane method to obtain convexity or monotonicity of the level sets of minimizers. To this aim, for a fixed unit vector n ∈ IR N , and for t ∈ IR we set T t = {x · n = t}, T − t = {x · n < t}, and T + t = {x · n > t}. For x ∈ IR N let x t denote the reflection of x with respect to T t . We also set ϕ t (x) ≡ ϕ(x t ), for a function ϕ and if Ω ⊂ IR N we define , and assume that for some unit vector n ∈ IR N and some t 0 ∈ IR we have According to Lemma 3.1 we obtain Since u is a smallest minimizer, then we conclude u ≤ v t , which completes the proof.
Corollary 5.2. Let u be a solution of Suppose that µ = c + δ z + − c − δ z − is Dirac measure and g 2 ≡ constant > 0. Then the solution of (5.7) is symmetric with respect to the line joining the points z + and z − .
Proof. The Corollary 5.2 follows from Theorem 5.1 by choosing f 1 = c + ρ and f 2 = c − ρ where ρ is a radially symmetric mollification of Dirac measure. It follows that Ω is has rotational symmetry with respect to the axis L : Remark 5.1. Observe that symmetry of Ω + and/or Ω − will depend on the weights c + and c − . In particular, if say µ ± is sufficiently concentrated around the point z ± so that Ω + ∩ Ω − = ∅, then Ω ± is a ball and u ± is bounded, radially symmetric.
6. Quadrature Identities 6.1. Two-phase Quadrature surfaces. In this section we discuss the concept of two-phase Quadrature surfaces. The one phase problem has been well studied in the literature and we refer the reader to the following works [13,15,19,21,22].
As for two-phase quadrature domains case (see [11]), here again the key point is that the measures have to be concentrated enough and also in balance. Indeed, if the measures µ 1 , say, has a very high density on its support, but not the other, then the support of the corresponding u 1 will have the possibility of covering the support of µ 2 . This naturally makes it impossible to find a two-phase QS for our measures. Finding right conditions for this balance is a question to be answered in the future. Here we will illustrate this for measures that satisfy Sakai's concentration condition. Let Then the system (1.13), for f i = µ i − λ i , can be rewritten as follows: where (6.10) For quadrature surfaces we need to take λ 1 = λ 2 = 0. Then for arbitrary h ∈ HL 1 (Ω + ∪Ω − ), we write Greens second identity: Apparently equation (6.11) leads to Thus, one can formally write (leaving the verification to the reader) which finally gives It is easy to see that the standard mollifier technique (see [13]) will also work in this case, and we may replace the measures with smooth functions, with support close to the support of measures.
If we reduce the test class h to be subharmonic in Ω 1 and super-harmonic in Ω 2 (due to negative sign in front of the integral) then the equality in (6.12) is replaced with an inequality (≥).
Observe also if we take µ 2 = 0, then Ω 2 = ∅, and we get the definition of a one-phase quadrature surface Theorem 6.1. Let µ 1 and µ 2 be given Radon measures with compact supports, that satisfy Sakai's concentration condition as in Definition 3.1. Suppose that for each µ i the corresponding one-phase quadrature surface ∂Q i (see (6.13)) is such that (6.14) Then, we have a solution to our two-phase free boundary problem (6.9) along with supp(µ 1 ) ⊂ supp({u > 0}), and supp(µ 2 ) ⊂ supp({u < 0}).
Let the constant M > 0 is taken such that max sup i=1,2 and consider the following measure, which satisfies Sakai's condition According to Lemma 1.  [13]). The proof of this result relies on the so-called symmetric decreasing rearrangement technique, and we refer for its background to the book [18].
Let v 1 be a minimizer to the functional over the set {ϕ ∈ H 1 (IR N ) : ϕ ≥ 0}. Then using the same arguments as in the proof of Theorem 1.4 in [13], one can easily conclude that supp(v 1 ) ⊂ B R (z 0 ), and therefore supp(v 1 )∩ Ω = ∅. Apparently supp(ν 1 ) ⊂ {v 1 > 0}, which implies that J 1 ν 1 ,g (v 1 ) < 0. Now, simple computation gives that . This contradicts the minimality of u. Thus supp(µ 1 ) \ supp({u > 0}) = ∅, and this implies supp(µ 1 ) ⊂ supp({u > 0}). This completes the proof of Theorem 6.2. Examples of two-phase QS. Due to Theorem 3.1, minimizers for J {f 1 ,f 2 ,g} always exist in the following special cases: a) f 1 ≡ 0 and both g + > 0, f 2 ≤ c 2 < 0 outside a compact set or f 2 ≡ 0 and both g − > 0, f 1 ≤ c 1 < 0 outside a compact set. b) A simpler two phase functional where it is assumed that either g ≥ 0 or g ≤ 0. These conditions, however, do not say anything about whether the quadrature identity (1.8) is admitted by the domain and the measure. Here we discuss simple examples of two phase QS, some of which are generated using one phase QS and symmetry arguments.
Example 1 (Plane Symmetric QS): Let (u, Ω) be a one phase QS for a measure f , and g as appearing in the functional (1.9), see also equation (1.1). Consider further a hyperplane T , not intersecting Ω, and an odd reflection of the solution u with respect to a plane T . This will trivially give a two phase quadrature surface whereΩ,f denotes the reflection of Ω, respectively f , in the plane T . A less trivial example can be constructed as follows: Let g, and f be as before for the one phase QS. Let further T + := {x : x 1 > 0} and minimize the one-phase functional in the set T + , with zero Dirichlet boundary values on ∂T + . Suppose further that the support of u reaches all the way to the plane ∂T + . This will formally solve Then odd reflection of u with respect to the plane T gives a quadrature surface symmetric about the plane T . Example 2 (Spherically symmetric QS): A different example would be an annular two phase quadrature surface. That is, a quadrature surface Γ = ∂Ω = ∂Ω 1 ∪ ∂Ω 2 such that Ω = Ω 2 \ Ω 1 is an annular domain with inner boundary ∂Ω 1 and outer boundary ∂Ω 2 with g + = c 1 > 0 on ∂Ω 1 and g − = c 2 < 0 on ∂Ω 2 (or vice versa). For a construction of a spherical annular two phase quadrature surface, we proceed as follows. Consider a uniformly distributed (and sufficiently large) 3 measure µ on the sphere S 2 : |x| = 2 (or defined in a εtubular neighborhood of S 2 ), and solve the one phase free boundary problem It is not hard to calculate explicitly what R > 1 is, but we surely know that there is at least one such R. Now a two phase solution can be obtained by extending u by an odd Kelvin inversion of u with respect to the sphere |x| = 1.
Then the extended function, which for simplicity is labeled u, satisfies Here ν = µ −μ, whereμ is the even Kelvin reflection of µ in |x| = 1. Then the boundary of the new domain is the required quadrature surface. Example 3 (Non-trivial two-phase QS): The above examples illustrates that it is not trivial to give explicit examples of QS that do not carry information from the one-phase problem. Here we shall give one slightly more complicated example, which again is constructed by reflection of one-phase QS. Nevertheless, the reflection is more elaborate than standard ones, and it is defined through the so-called Schwarz function [10], which is defined as follows: Let Γ be any analytic curve dividing the complex plane C into at least two components. Denote by Γ + one of these components. The Schwarz function, which we denote by S(z) for z = x + iy, is a function which is analytic in a neighborhood, say N , of Γ satisfying S(z) =z on Γ. Next fix a point z 0 ∈ Γ, and suppose that B r (z) ⊂ N so that S(z) is analytic in B r (z). We assume without loss of generality that r = 1, otherwise we just scale the curve with r. Choosing g = 1 and µ a "smooth" measure with support in the open set Γ + ∩ B 1 (z), let (u, Ω) be a minimizer of the functional J {µ,g} with zero Dirichlet data on Γ ∩ B 1 (z). We may further assume that µ satisfies conditions so that there exists a solution to the one-phase equation (1.1) in Γ + ∩ B 1 (z), cf. equation (6.2). Next, we invoke the anti-conformal reflection R Γ associated to Γ, and defined as R Γ (z) = S(z) (see [10,Chapter 6]).
For z ∈ Γ − we define u(z) = −u(R Γ z), and in this way we extend u across Γ as a solution to our problem with negative measure −µ(R Γ z). This creates an example of a two-phase free boundary for our problem.
6.3. Solutions with unbounded support. There are not many trivial examples of twophase QS with unbounded support, however, there exist a few. The most simple example is naturally when we take µ 1 to be Dirac mass at origin and µ 2 ≡ 0. Then the appropriate sphere is both one-and two-phase QS, that can easily be verified, using integration by parts. Continuing on this path, if we assume both the measures to be identically zero, then for g 1 = g 2 = constant one can show that an appropriate linear function is a solution to our problem. Quadrature surfaces, with unbounded supports (for both phases) can be constructed from bounded ones, by a simple procedure. Indeed, if we already have a QS, for a measure µ, we may consider minimizing the corresponding functional in B R , where we now put Dirichlet data on |x| = R, that corresponds to h(x) = (x − x 0 ) · a for some x 0 and vector a, such that supp(µ) ⊂ {h > 0}. Any (global) minimizer u R to this problem will have the property that its support stretches all the way up to sphere |x| = R, due to the boundary values. Such domains give rise to (partial) QS, which amounts to being QS for the class of harmonic functions on supp(u R ), that vanishes on the sphere. By letting R tend to infinity, along with using barrier arguments for control of the (linear growth) one can show that there is a limit (at least for a subsequence of R) which satisfies a quadrature identity. The heuristic argument here can be made easily rigorous by some footwork, and is left to interested reader. For quadrature domains there are at least two references the authors are aware of [5,20]. Similar methods can be applied to a two-phase QS, without much efforts.
It is interesting to mention that unbounded two-phase QS may behave much more differently than their one-phase counterpart. Indeed, we expect that two-phase quadrature surfaces, with both phases being unbounded, have to behave like plane solutions at infinity. This can be seen easily if the QS is smooth, by shrinking the solution through any sequence u j = u(R j x)/R j and obtaining a new unbounded solution u ∞ , without any measure (these are called Null QS). One can then classify Null QS, which are solutions to ∆u ∞ = 0 outside the zero set of u ∞ , and have the property that |∇u + | 2 − |∇u − | 2 = constant, where the constant is the limit value of g 2 1 − g 2 2 at infinity (see Remark 1.1). Since in our case we have taken g 1 = g 2 , this implies that we actually obtain a limiting function that is harmonic with linear growth, and hence a plane. This proves our claim. For more general values of g 1 , g 2 , one may still prove a similar result, but that would require using strong tools, such as monotonicity formulas, which is outside the scope of this paper.
We close this paragraph by remarking that bounded two-phase QS are uniformly bound. This follows from the fact that two-phase QS are smaller than the union of the corresponding two one-phases, which in turn are uniformly bounded. Again, the details are left to the readers.
6.4. Uniqueness. In [22], it was shown that ∂Ω is a quadrature surface with respect to the measure µ if and only if there is a solution to the Cauchy problem (6.5) where ν is an outward normal. Furthermore, it was proved that if µ = cδ x , c > 0 and δ x Dirac measure then ∂Ω is a sphere centered at x. Uniqueness for QS in general fails, unless one has some geometric restriction. This is already known for the one-phase problem. Since the functional representing the problem is not convex, one expects that local minima as well as stationary points may give rise to solutions to our free boundary problems. 4 For the one phase case there are indeed examples of non-uniqueness for QS worked out by A. Henrot [15]. Therefore, a uniqueness question is even more complicated in the two-phase case and it seems that the only way to achieve partial results is by imposing strong geometric or other type of restrictions on the solutions, and the data involved.
In one phase problem it is well-known that a QS (u, Ω) for a single (multiple of) Dirac mass c 0 δ z , at the point z, is the appropriate sphere ∂Ω = ∂B r (z), with r = r(c 0 ), provided ∂Ω is smooth enough (usually C 1 suffices). The same question for the two-phase problem, for µ = c + δ z + − c − δ z + seems to be much harder to find an answer to. 6.5. Null Quadrature surfaces. In this section we shall let g ≡ 1, and discuss the so-called unbounded QS, with zero measures, or so-called null-quadrature surfaces.
A null QS, is a quadrature surface with zero measure (see [16] for the quadrature domain counterpart). The one phase null-QS refers to a domain Ω such that one can find a harmonic function u in Ω, with zero Dirichlet data and |∇u| = g on ∂Ω.
Obviously Ω cannot be bounded (due to maximum principle). So one may then wonder about the behavior of u at infinity.
In order to understand the concept of null-QS better, we shall consider it from a potential theoretic point of view, which is more instructive. We define, in analogy with null quadrature domains, a null quadrature surface to be the boundary of a domain Ω such that ∂Ω h(x) dσ x = 0 for all functions h, harmonic in Ω, and integrable over ∂Ω. This is the one-phase case of a null-QS, for g ≡ 1.
Let us give a few examples of one-phase null-QS. The simplest example is the half-space, Ω = {a · x > 0}, (|a| = 1) with the corresponding function u = a · x. A second example is A third example is the exterior of any ball Ω = IR n \ B r (x 0 ), where the function u = b|x − x 0 | 2−n + c (for appropriate b, c) solves the free boundary problem. Naturally cylinders can be built, with exterior of balls as base.
More complicated examples can be given using the construction of H. Alt and L. Caffarelli [1], which is a cone , 0 using polar coordinates x(r, ϕ, θ) = r(cos ϕ sin θ, sin ϕ sin θ, cos θ) in IR 3 . The function f (θ) = 2 + cos θ log 1 − cos θ 1 + cos θ is a solution of (sin θf ′ ) ′ + 2 sin θf = 0, f ′ ( π 2 ) = 0, and θ 0 ≈ 33.534 • is the unique zero of f between 0 and π 2 . The function u is harmonic in {u > 0} and ∂ ν u = 1 on ∂{u > 0}\{0}, i.e., the free boundary condition is satisfied everywhere on the surface of the cone, but at the origin. At the origin one has lim inf x→0 |∇u|(x) < 1. However, since the free boundary is satisfied at every other free boundary point, and that the solution function u(x) behaves linearly at infinity, one obtains (by simple drill of integrations by parts) that ∂{u > 0} is a quadrature surface.
Other less regular, and very complicated examples, are the so-called pseudospheres of John Lewis [17]. 5 These objects are much more complicated that fail to be smooth at some points, with gradient of the potential functions being unbounded at some boundary points. Nevertheless, they admit QS identities, and hence are Quadrature surfaces.
It is worth mentioning that a recent example of [14] (cf. also [23]) solving our PDE with unbounded support, has growth that is exponential and that does not qualify as a QS, in our sense.
The two-phase null-QS corresponds to a similar integral identity as before, but without any measure where Ω + ∩ Ω − = ∅.
As shown in [3] the two-plane solutions (as they called it) are given by All these are global minimizers; for the last example to be a minimizer one needs a ≥ 1 (see [3], Lemma 4). It is, however, not clear whether these are the only two-phase null QS. Indeed, a (null)-QS does not need to be a minimizer of our functional, and the only requirement is that it satisfies a quadrature identity.

Multi phase QS
7.1. The model equation. It is apparent that once the seed of the idea of two-phase QS (or any free boundary problem) has taken root, one may think of more complex situations where multi-phases are involved. In this section we shall rely on the above results for two-phase QS case, and provide setting of a multi-phase problem, as done previously in segregation problems [8], or quadrature domain theory [4]. Recently, the same approaches have been applied in shape optimization problems as well [6].
The exact formulation of the multi-phase problem is done using the two phase version of it as follows: Given m positive measures µ i , we want to find functions u i ≥ 0, (i = 1, . . . , m), with mutually 5 John Lewis constructed such objects for Dirac masses, but the same can be done for any measure µ with high enough concentration, such that the Greens potential G µ D of µ with respect to some domain D ⊂ supp(µ), has the property that |∇G µ D | > 1 on ∂D.
In other words, for each pair (i, j) with i = j, the function u i − u j solves a two-phase versions of our problem outside the union of the supports of the other functions.
A natural question that arises is: does the proposed model cover the two phase case? The answer is yes, because as for Multi-phase Quadrature domains (Theorem 2 in [4]), one can show the similar equivalence result. 7.2. Existence of minimizers for multi-phase case. In this section we will adapt the existence analysis, which has been done for multi-phase quadrature domains [4]. We start with the definition of the minimization sets K and S. Define K = {(u 1 , u 2 , . . . , u m ) ∈ (H 1 (IR N )) m : u i ≥ 0, for all i = 1, · · · , m}, and S = {(u 1 , u 2 , . . . , u m ) ∈ (H 1 (IR N )) m : u i ≥ 0, and u i · u j = 0, for all i = j}.
Obviously we have S ⊂ K. Next we define where each function f i and g are satisfying conditions (Ã1) − (Ã4). 6 (7.8) is compact for all i = 1, 2, ..., m (Ã3) g ≥ 0 (Ã4) either for all i we have f i ≤ −c i < 0 or g ≥ c 0 > 0 hold outside a compact set for some positive constants c 0 , c i .
In light of Lemma 1 in [4], one can show that for every minimizer (u 1 , . . . , u m ) of G over K, each component u i is going to minimize corresponding one-phase functional Hence, following [13] we say that the vector (u 1 , . . . , u m ) is a largest (smallest) minimizer of G over K, if for every i, each component u i is accordingly the largest (smallest) minimizer (in the sense considered in [13]) of J 1 {f i ,g} (ϕ) over the set {ϕ ∈ H 1 (IR N ) : ϕ ≥ 0}.
Theorem 7.1. Let f i (x), g(x) satisfy the conditions (Ã1) − (Ã4). Then G(u 1 , . . . , u m ) has at least one minimizer (ū 1 ,ū 2 , . . . ,ū m ) in S, and also all minimizers have compact support. Moreover, the following inclusion of supports holds: For any minimizer (ū 1 ,ū 2 , . . . ,ū m ) of G over S, and the largest minimizer (v 1 , v 2 , . . . , v m ) of G over K, we have Proof. The functional G(u 1 , u 2 , . . . , u m ) is lower semi-continuous, coercive and convex. Since the set S is closed, then the existence of a minimizer follows from standard arguments of calculus of variations. Note that the minimizer is not necessarily unique. For simplicity, we make the following notations: To see the ordering of the supports (equation (7.10)) one can proceed as in Lemma 3.1, which clearly will imply the following inquality SinceŪ ∈ S and V ∈ K, then min(Ū , V ) ∈ S. Therefore Observe that max(Ū , V ) ∈ K and V = (v 1 , v 2 , . . . , v m ) is the largest minimizer to G(u 1 , u 2 , . . . , u m ) in K. Hence, for all i = 1, · · · , m. We recall that each component v i is the largest minimizer to the functional J 1 {f i ,g} (ϕ) over the set {ϕ ∈ H 1 (IR N ) : ϕ ≥ 0}. For these functionals and under more general setting it has been proved (see [13,Theorem 1.4] ) that all minimizers have support in a fixed compact set. Thus, supp(v i ) is compact, which in turn yields the compactness of supp(ū i ), for all i = 1, · · · , m. This completes the proof of Theorem.
Following the proof of Proposition 1 in [4], we can prove a similar result for the functional (7.7). Proposition 7.1. If (ū 1 ,ū 2 , . . . ,ū m ) is a minimizer to the functional (7.7) subject to the set S, then the following holds in the sense of distributions: (7.11) where Ω i = {ū i > 0}.
In the light of Two-phase QS and Proposition 7.1, we give a definition of Multi-phase version as follows Definition 7.1 (Multi-phase Quadrature surface). Suppose we are given m bounded positive measures µ i and disjoint domains Ω i such that supp(µ i ) ⊂ Ω i . If for every harmonic functions h ∈ HL 1 (Ω i ∪ Ω j ), such that h is continuous across ∂Ω i ∩ ∂Ω j , and h = 0 on ∪ k =i,j ∂Ω k , the following QI holds , and a given smooth positive function g.
If we extend the test class h to the subharmonics in Ω i and super-harmonics in Ω j (due to negative sign in front of the integral) then the equality in (7.12) is replaced with an inequality (≥).
The analogue of the Theorem 6.1 for multi-phase case is the following result below.
Theorem 7.2. Let µ i be given Radon measures with compact supports, that satisfy Sakai's condition as in Definition 3.1. Suppose that for each µ i the corresponding one-phase quadrature surface ∂Q i (see (6.13)) is such that (7.13) Q i ∩ supp(µ j ) = ∅, for every i = j.
The proof of this result repeats the same lines as in Theorem 6.1, and therefore is omitted.
7.3. Analysis of junction points. In this section our goal is to show the absence of triple junction points in IR N , away from the support of the measures µ i . More exactly we shall show that for multi-phase QS, there is at most two phases that can meet at each point. In the case of multi-phase quadrature domains it was shown (see [4]) that a triple junction may actually appear. For the proof of the main result, in this section, we will need the multi-phase counterpart (see [24]) of a celebrated Caffarelli-Jerison-Kenig monotonicity formula [7]. Theorem 7.3. ( [24])[Three-phase monotonicity formula] Let B 1 ⊂ IR N be the unit ball in IR N and let u i ∈ H 1 (B 1 ), i = 1, 2, 3, be three non-negative Sobolev functions such that ∆u i + 1 ≥ 0, ∀i = 1, 2, 3, and u i · u j = 0 a.e. in B 1 , ∀i = j.
Then there are dimensional constants ε > 0 and C N > 0 such that for each r ∈ (0, 1) we have Lemma 7.1 (Non-degeneracy). Let (u 1 , u 2 , . . . u m ) ∈ S be a minimizer to (7.7). Then there exist a constant D N,f i ,g > 0, depending on dimension N, and functions f i , g, such that for every Here, we set Ω i = {u i > 0}, and i = 1, 2, . . . , m.
Proof. To see this for some fixed i, we set then following the proof of Lemma 2.8 in [13], we conclude that the largest minimizer v β,i , ofG r,i over K β,i , vanishes in the ball B r/4 , provided r and β are small enough. The upper thresholds for the constants r and β can be taken as follows: with force term f i (see the proof of Lemma 2.8 in [13]). On the other hand due to Harnack inequality for the component u i we have where the constants C 1 and C 2 are depending only on dimension N. Using the following rescaling property β 0 (r, l, M i ) = rβ 0 (1, l, rM i ), one can easily achieve (7.14) u i (x) < β 0 (r, l, M i ) on ∂B r/2 , by letting Since β 0 (1, l, 0) > 0, then taking r small enough we will have (7.14), provided 1 r ∂Br where C > 0 is some constant. Our aim is to prove that u i = 0 in B r/4 . We take v β,i to be the largest minimizer ofG r,i over the set K β,i for β = β 0 (r, l, M i ). We define a new function w to be min(u i , v β,i ) in B r/2 and equal u i in IR N \B r/2 . The inequality (7.14) implies (u 1 , . . . , u i−1 , w, u i+1 , . . . , u m ) ∈ S, and therefore G(u 1 , . . . , u m ) ≤ G(u 1 , . . . , u i−1 , w, u i+1 , . . . , u m ).
This leads to Using the same arguments as in the proof of Lemma 3.1, we can obtain the following inequality Thus, in the light of (7.15) and (7.16) we getG r,i (v β,i ) ≥G r,i (max(u i , v β,i )), which in turn implies max(u i , v β,i ) ≤ v β,i . The latter inequality follows from the fact that v β,i is a largest minimizer toG r,i over the set K β,i , and max(u i , v β,i ) ∈ K β,i . Hence, u i ≤ v β,i in B r/2 and this gives that u i = 0 in B r/4 . Thus, we have proved that for every component u i there exists a dimensional constant C N > 0, depending also on l and M i such that for every sufficiently small r > 0 the following statement is true: This basically gives the desired non-degeneracy property. Similarly, it can be shown that the statement (7.17) remains true with B kr in place of B r/4 for any 0 < k < 1. In this case the constant C N will also depend on k. Proof. Invoking Theorem 7.1 in [9], by choosing F (W ) as in equation (1.6) in the same paper, we may conclude that u i is locally C α . From here one may apply Lemma 5.2 in [2] to conclude (7.18) ∂Br(z) with C universal constant, depending only on the distance between z and the support of measures (in our case). It should be remarked that in Lemma 5.2 of [2] we have to take only two functions at a time, so as to apply the monotonicity function. The latter can be found in more advanced forms in [7]. Further the importance of initial Holder regularity is needed in Lemma 5.2 (equation (5.12)) in [2]. From (7.18) we may now infer Lipschitz regularity as done in the proof of Theorem 5.3 in [2], where one also needs that |u i | is a sub solution, which is fulfilled by our solutions.
Remark 7.1. It is noteworthy that several recent papers prove Lipschitz regularity of solutions for two and multi-phase problems with heuristic arguments, without stressing the need for initial partial regularity. It needs to be stressed that the conditions on the functions in the monotonicity formula of [2], and the succeeding ones, have been relaxed considerably, and in general one can avoid continuity of solutions. Nevertheless, for applying the formula to prove regularity of solutions in free boundary problems, one does need to begin with some initial partial regularity. This part of the problem is too often neglected and not taken seriously. This has been highlighted in our proof of Lipschitz regularity of solutions in [9] where we begin with solutions satisfying Hölder regularity. To the author's best knowledge, the C α -regularity for multi-phase problems is by no means an easy problem, and cannot be done as that of Theorem 2.1 in [2]. Theorem 7.4. Let (u 1 , u 2 , . . . u m ) ∈ S be a minimizer to (7.7). Then there is a universal constant R 0 > 0 (depending only on the norms) such that for any point z i,j ∈ ∂{u i > 0} ∩ ∂{u j > 0} we have |z i 1 i 2 − z i 3 i 4 | > R 0 , provided (i 1 , i 2 ) = (i 3 , i 4 ). Here i k ∈ {1, · · · , m}.
In particular triple junction points cannot appear, and that two different class of two-phase points stay uniformly away from each other.
This theorem can be set in relation to segregation problems that have been in focus lately, see [6]. A particular application of this theorem is that in segregation problems, where multiphase Bernoulli type free boundaries appear in the limit, one can claim that more than two phases cannot meet at the same time.
Proof. We first notice that by compactness, and non-degeneracy, if there is a sequence z k i 1 ,i 2 , z k i 3 ,i 4 , (k = 1, 2, · · · ) of two-phase points of different classes such that |z k i 1 ,i 2 − z k i 3 ,i 4 | → 0, then the limit point w = lim k z k i 1 ,i 2 = lim k z k i 3 ,i 4 is a triple junction point. Hence to prove the theorem, it suffices to show that triple junction points do not exist.
By non-degeneracy (see Lemma 7.1) for each z i ∈ ∂{u i > 0} there exists a point y i ∈ ∂B r/2 with u i (y i ) ≥ D N,f i ,g · r.
From Lemma 7.2, u i is Lipschitz regular, and therefore u i (x) > 0 in B εr (y i ), for some small enough ε > 0. Thus there exists a positive constant c 0 > 0 such that for any small enough r > 0 the following property holds: Since all sets {u i > 0} are disjoint, then there exists a dimensional constant α 0 > 0 for every i = 1, 2, . . . , m.
We will need the following version of Poincare inequality: For every function v ∈ H 1 (B r ) we have (7.19) |{v The proof of this inequality can be found implicitly in [1,Lemma 3.2]. Another reference is Lemma 4.5 in [6]. In view of non-degeneracy property and inequality (7.19) we arrive at: Thus, there exists a universal constant L i > 0 depending only on N, f i , g, such that (7.20) Br(z i ) |∇u i | 2 ≥ L i · r N . Now, let the origin be a possible triple junction point for components u i 1 , u i 2 and u i 3 , away from the measures µ i . Our aim is to apply the multi-phase version of Caffarelli-Jerison-Kenig monotonicity formula around the origin and come to a contradiction. Since the triple junction point is away from the measures µ i , the constants L i do not depend on f i in a small neighborhood of the origin and therefore are the same.
First we recall the following inequality obtained in [7, Remark 1.5] (see also [24]): Suppose that u ∈ H 1 (B 2 ) is a nonnegative Sobolev function such that ∆u + 1 ≥ 0 on B 2 ⊂ IR N . Then, there is a dimensional constant Q N > 0, such that (7.21)