An isoperimetric result for an energy related to the p -capacity

In this paper, we generalize the notion of relative p -capacity of K with respect to Ω , by replacing the Dirichlet boundary condition with a Robin one. We show that, under volume constraints, our notion of p -capacity is minimal when K and Ω are concentric balls. We use the H -function (see [4, 8]) and a derearrangement technique. MSC 2020: 35J66, 35J92, 35R35.


Introduction
Let p > 1, β > 0 be real numbers.For every open bounded sets Ω ⊂ R n with Lipschitz boundary, and every compact set K ⊆ Ω with Lipschitz boundary, we define We notice that it is sufficient to minimize among all functions v ∈ H 1 (Ω) with v = 1 in K and 0 ≤ v ≤ 1 a.e., moreover if K, Ω are sufficiently smooth, a minimizer u satisfies where ∆ p u = div |∇u| p−2 ∇u is the p-Laplacian of u and ν is the outer unit normal to ∂Ω.If K = Ω, equation (1.2) has to be intended as u = 1 in Ω, and the energy is In general, equation (1.2) has to be intended in the weak sense, that is: for every ϕ ∈ W 1,p (Ω) such that ϕ ≡ 0 in K, ˆΩ|∇u| p−2 ∇u∇ϕ dL n + β ˆ∂Ω u p−1 ϕ dH n−1 = 0. (1.3) In particular if u is a minimizer, letting ϕ = u − 1, we have that Moreover from the strict convexity of the functional, the minimizer is the unique solution to (1.3).
This problem is related to the so-called relative p-capacity of K with respect to Ω, defined as In the case p = 2 it represents the electrostatic capacity of an annular condenser consisting of a conducting surface ∂Ω, and a conductor K, where the electrostatic potential is prescribed to be 1 inside K and 0 outside Ω.Let ω n be the measure of the unit sphere in R n , and let M > ω n , then it is well known that there exists some r ≥ 1 such that This is an immediate consequence of the Pólya-Szegő inequality for the Schwarz rearrangement (see for instance [11,9]).We are interested in studying the same problem for the energy defined in (1.1), which corresponds to changing the Dirichlet boundary condition on ∂Ω into a Robin boundary condition, namely, we consider the following problem In this case, the previous symmetrization techniques cannot be employed anymore.
Problem (1.4) has been studied in the linear case p = 2 in [6], with more general boundary conditions on ∂Ω, namely ∂u ∂ν where Θ is a suitable increasing function vanishing at 0. This problem has been addressed to thermal insulation (see for instance [7,1]).Our main result reads as follows.
Theorem 1.1.Let β > 0 such that Then, for every M > ω n the solution to problem (1.4) is given by two concentric balls in particular we have that either r = 1 or M = ω n r n .Moreover, if K 0 ⊆ Ω 0 are such that and u is the minimizer of E β,p (K 0 , Ω 0 ), then the sets { u = 1 } and { u > 0 } coincide with two concentric balls up to a H n−1 -negligible set.
Remark 1.2.In the case adapting the symmetrization techniques used in [6], it can be proved that a solution to problem (1.4) is always given by the pair (B 1 , B 1 ).
We point out that the proof of the theorem relies on the techniques involving the H-function introduced in [4,8].
The case in which Ω is the Minkowski sum Ω = K + B r (0), the energy E β,p (K, Ω), has been studied in [3] under suitable geometrical constraints.

Proof of the theorem
In order to prove Theorem 1.1, we start by studying the function A similar study of the previous function can also be found in [3].Let For every R > 1, consider ) We have that p−1 . (2.2) Therefore, there are three cases: • if See for instance Figure 1, where In the following, we will need Lemma 2.1.Let R > 1,β > 0 and let u * be the solution of the problem on (B 1 , B R ).Then Proof.Recalling the expressions of u * in (2.1), by straightforward computations we have that Definition 2.2.Let Ω ⊆ R n be an open set, and let U ⊆ Ω be another set.We define the internal boundary of U as and the external boundary of U as Let K ⊆ Ω ⊆ R n be open bounded sets, and let u be the minimizer of E β.p (K, Ω).In the following, we denote by We define Notice that this definition is slightly different from the one given in [5].
Remark 2.5.Notice that if K, and Ω are two concentric balls, the minimizer u is the one written in (2.1), for which the statement of the above Lemma holds for every t ∈ (0, 1).
Lemma 2.6.Let ϕ ∈ L ∞ (Ω).Then there exists t ∈ (0, 1) such that then we evaluate where we used the inequality Multiplying by t p−1 and integrating, we get from which we obtain the conclusion of the proof.
Remark 2.7.Notice that the inequality (2.6) holds as equality if and only if a = b.Therefore, if ϕ = |∇u| u on a set of positive measure, then the inequality in (2.7) is strict, since for small enough t.Therefore, there exists S ⊂ (0, 1) such that L 1 (S) > 0 and for every t ∈ S H(t, ϕ) < E β,p (K, Ω).
In the following, we fix a radius (2.8) Then we have that Proof.In the following, if v is a radial function on B R and r ∈ (0, R), we denote with abuse of notation v(r) = v(x), where x is any point on ∂B r .By Lemma 2.4 we know that for every t ∈ (0, 1) while by Lemma 2.6, for every ϕ ∈ L ∞ (Ω) there exists a t ∈ (0, 1) such that (2.10) We aim to find a suitable ϕ such that, for some t, .11)so that combining (2.10), (2.11), and (2.9) we conclude the proof.In order to construct ϕ, for every t ∈ (0, 1) we define .12) then we set, for every x ∈ Ω,

The functions ϕχ
Notice now that, since ω n r(τ Therefore, substituting in (2.14), we get where we have used polar coordinates to get the last equality.Thus, the claim is proved.
Recalling the definition of ϕ, (2.8) reads then using (2.13) and the definition of H (see Definition 2.3), we have where in the last inequality we have used the isoperimetric inequality and the fact that ϕ is constant on ∂U t .
Remark 2.9.By Remark 2.7, we have that if K and Ω are such that so that, by Lemma 2.4, we have equality in (2.16) for a.e.t ∈ (0, 1).Thus, by the rigidity of the isoperimetric inequality, we get that U t coincides with a ball up to a H n−1 -negligible set for a.e.t ∈ (0, 1).In particular, { u > 0 } = t U t and { u = 1 } = t U t coincide with two balls up to a H n−1 -negligible set.
Proof of Theorem 1.1.
We divide the proof of the minimality of balls into two cases, and subsequently, we study the equality case.
Let us assume that β and recall that in this case the function and a solution to (1.4) is given by the pair (B 1 , B 1 ).
For what concerns the equality case, we will follow the outline of the rigidity problem given in [10, Section 3] (see also [2,Section 2]).Let K 0 ⊆ Ω 0 be such that while, from the minimality of (K 0 , Ω 0 ) we have that . Hence, by the rigidity of the isoperimetric inequality we have that K0 = Ω 0 are balls of radius 1.On the other hand, if K0 = Ω 0 , from the first part of the proof, there exists Therefore, by Remark 2.9, we have that for a.e.t ∈ (0, 1), the superlevel sets U t coincide with balls up to H n−1 -negligible sets, and { u = 1 } and { u > 0 } coincide with balls, up to H n−1 -negligible sets, as well.We only have to show that { u = 1 } and { u > 0 } are concentric balls.To this aim, let us denote by x(t) the center of the ball U t and by r(t) the radius of U t , as already done in (2.12).In addition, we also have that for every ν ∈ S n−1 , so that x(t) is constant and U t are concentric balls for a.e.t ∈ (0, 1).In particular, { u = 1 } = t U t and { u > 0 } = t U t share the same center.
1 , B ρ ) is decreasing.Let u * be the minimizer of E β,p (B 1 , B R ), by Lemma 2.1 condition (2.8) holds and, by Proposition 2.8, we have that a solution to (1.4) is given by the concentric balls (B 1 , B R ). → E β,p (B 1 , B ρ ) increases on [1, α β,p ] and decreases on [α β,p , +∞), and there exist a unique R β,p > α β,p such that E β,p (B 1 , B R β,p ) = E β,p (B 1 , B 1 ).If R ≥ R β,p the function u * , minimizer of E β,p (B 1 , B R ), still satisfies condition (2.8) and, as in the previous case, a solution to (1.4) is given by the concentric balls (B 1 , B R ).On the other hand, if R < R β,p , we can consider u * β,p the minimizer of E β,p (B 1 , B R β,p ).By Lemma 2.1 we have that, for the function u * β,p , condition (2.8) holds and, by Proposition 2.8, we have that if K and Ω are open bounded Lipschitz sets with K ⊆ Ω, |K| = ω n , and |Ω| x), so that, if u(x) = t, then |∇u(x)| = C t > 0. This ensures that we can write (x) ds + ˆK x dL n (x) , and we can infer that x(t) is an absolutely continuous function, since |∇u| > 0 implies that |U t | is an absolutely continuous function as well.Moreover, on ∂U t we have that for every ν ∈S n−1 , C t x (t) • ν − C t r (t) = 1.(2.19)Finally, joining (2.19) and (2.15), and the fact that |∇u| = C t on ∂U t , we get x (t) • ν = 0