On the solutions to p -Poisson equation with Robin boundary conditions when p goes to + ∞

We study the behaviour, when p → + ∞ , of the ﬁrst p -Laplacian eigenvalues with Robin boundary conditions and the limit of the associated eigenfunctions. We prove that the limit of the eigenfunctions is a viscosity solution to an eigenvalue problem for the so-called ∞ -Laplacian. Moreover, in the second part of the paper, we focus our attention on the p -Poisson equation when the datum f belongs to L ∞ (Ω) and we study the behaviour of solutions when p → ∞ . MSC 2020: 35J92, 35J94, 35P15.


Introduction
Let β be a positive parameter and let Ω be a bounded and open set of R n , n ≥ 2, with Lipschitz boundary.
We study the ∞-Laplacian eigenvalue problem with Robin boundary conditions where ∆ ∞ , the so-called ∞-Laplacian, is defined by We refer to problem (1) as the ∞-Laplacian eigenvalue problem because it can be seen as limit, in some sense, of the p-Laplacian eigenvalues problem (3) By classical arguments, one can show that the infimum in (3) is achieved and in what follows we will denote by u p ∈ W 1,p (Ω) the eigenfunction corresponding to the first eigenvalue Λ p .
In this paper, we firstly prove that lim p→+∞ and we give a geometric characterization of this quantity, precisely: where R Ω denotes the inradius of Ω, i.e. the radius of the l argest ball contained in Ω.Thereafter, we prove that Λ ∞ is the first eigenvalue of the infinite Laplacian, in the sense that equation (1) admits non-trivial solutions only if Λ ≥ Λ ∞ .Similar results, in the case of Dirichlet and Neumann boundary conditions were obtained in [JLM99, JL05, BDM89, EKNT15, RS16].
More specifically, in [JLM99, JL05], Juutinen, Lindqvist and Manfredi have studied the Dirichlet case as p → +∞.They provided a complete characterization of the limiting solutions in terms of geometric quantities.Indeed, the first eigenvalue of the p-Laplace operator {λ D p } happens to satisfy The related eigenfunctions v D p also converge (up to a subsequence) to some Lipschitz function v D ∞ .Most important, the Authors show that there exists a natural viscosity formulation of the eigenvalue problem for the ∞-Laplacian, for which λ D ∞ and v D ∞ turn out to be the first eigenvalue and first eigenfunction, respectively.
The Neumann case seems to be more subtle.It was investigated in [EKNT15,RS16] and similarly to the Dirichlet case, the Authors established that the first non-trivial eigenvalues of the p-Laplacian {λ N p } satisfy lim where diam(Ω) is the intrinsic diameter of Ω, i.e. the supremum of the geodetic distance between two points of Ω.However, while both first eigenvalues and first eigenfunctions converge (as p → ∞) and are solutions to some appropriate eigenvalue problem for the ∞-Laplacian, in [EKNT15], the Authors are able to prove that they actually converge to the first eigenvalue e first eigenfunction only if the domain Ω is convex.Whether or not the same holds true in the general case, it is still an open problem.
In the second part of the paper, we focus our attention on the study of the limit of the p-Poisson equation with Robin boundary conditions: We prove that there exists (up to a subsequence) a limiting solution v ∞ as p → ∞ and we establish conditions on f which are equivalent to the uniqueness of v ∞ .
The ∞-Poisson problem for Dirichlet boundary conditions was already studied in [BDM89] by Bhattacharya, DiBenedetto and Manfredi, while, to the best of our knowledge, similar results have not been addressed in the case of Neumann boundary conditions.

Notations and Preliminaries
Throughout this article, | • | will denote the Euclidean norm in R n , and We denote by d(x, ∂Ω) the distance function from the boundary, defined as for an exhaustive discussion about this function and its properties see [GT01].Moreover, we recall that the inradius R Ω of Ω is The following lemma makes us understand why (4) can be seen as a limit problem of (3).
Proof.The proof of this lemma can be found in [RS15].

Viscosity solutions
Before going on, we recall the definition of viscosity solutions to a boundary value problem, see [CIL92] for more details.
Definition 2.1.We consider the following boundary value problem where Viscosity supersolution A lower semi-continuous function u is a viscosity supersolution to (9) if, whenever we fix x 0 ∈ Ω, for every φ ∈ C 2 (Ω) such that u(x 0 ) = φ(x 0 ) and x 0 is a strict minimum in Ω for u − φ, then • if x 0 ∈ Ω, the following holds Viscosity subsolution An upper semi-continuous function u is a viscosity subsolution to (9) if, whenever we fix x 0 ∈ Ω, for every φ ∈ C 2 (Ω) such that u(x 0 ) = φ(x 0 ) and x 0 is a strict maximum in Ω for u − φ, then • if x 0 ∈ Ω, the following holds Viscosity solution A continuous function u is a viscosity solution to (9) if it is both a super and subsolution.
Remark 2.1.The condition u − φ has a strict maximum or minimum can be relaxed: it is sufficient to ask that u − φ has a local maximum or minimum in a ball B R (x 0 ) for some positive R.

The ∞-eigenvalue problem
Let us start this section observing that Lemma 2.1 brings to the following estimate We can say something more Lemma 3.1.Let { Λ p } p>1 be the sequence of the first eigenvalues of the p-Laplacian operator with Robin boundary condition.Then, where Λ ∞ is defined in (4).Moreover, if { u p } p>1 is the sequence of eigenfunctions associated to {Λ p } p>1 , then there exists a function u ∞ ∈ W 1,∞ (Ω) such that, up to a subsequence, weakly in L q (Ω), ∀q.
Proof.As a consequence of (10), the sequence { u p } p>1 of eigenfunctions associated to Λ p is uniformly bounded in W 1,q (Ω): indeed, if q < p, by Hölder inequality, where the constant C is independent of p.By a classical argument of diagonalization, see for instance [BDM89], we can extract a subsequence u p j such that Moreover, from (12) and ( 13), the following inequality holds Now we want to show that the limit u ∞ solves (1) in viscosity sense, but before we need the following proposition Proposition 3.2.A continuous weak solution u to (2) is a viscosity solution to (2).
Proof.The proof is similar to the one in [JLM99,EKNT15] for the p-Laplacian with other boundary conditions.
We only write explicitly the proof that u ∞ satisfies the boundary conditions in the viscosity sense.
Let u be a continuous weak solution to (2), let x 0 ∈ ∂Ω and let us consider a function φ such that φ(x 0 ) = u(x 0 ) and such that u − φ has a strict minimum at x 0 .Then Assume by contradiction that both terms are negative.If we choose r sufficiently small, in Ω ∩ B r (x 0 ), we have Then using the definition of weak solution, we have which gives a contradiction.

Now we can prove the following
Theorem 3.3.Let u ∞ be the function given in Theorem 3.1.Then u ∞ is a viscosity solution to Proof.We divide the proof in two steps.
Step 1 u ∞ is a viscosity supersolution.Let x 0 ∈ Ω and let φ ∈ C 2 (Ω) be such that u ∞ − φ has a strict minimum in x 0 .We want to show min Notice that u p − φ has a minimum in x p and x p → x 0 .If we set φ p (x) = φ(x) + c p with c p = u p (x p ) − φ(x p ) → 0 when p goes to infinity, we have that u p (x p ) = φ p (x p ) and u p − φ p has a minimum in x p , so Proposition 3.2 implies This gives us |∇φ(x 0 )| − Λ ∞ φ(x 0 ) ≥ 0 since, otherwise, the right-hand side of (17) would go to infinity, in contradiction with the fact that φ ∈ C 2 (Ω).Moreover −∆ ∞ φ(x 0 ) ≥ 0, just taking the limit.
Then, min If for infinitely many x p ∈ Ω (16) holds true, then we get If for infinitely many p, x p ∈ ∂Ω the following holds true Only two cases can occur: • − ∂φ ∂ν (x 0 ) > 0, then letting p to infinity in the following That is Step 2 u ∞ is a viscosity subsolution.
Let us fix so it is enough to prove that only one of the two terms in the bracket is non positive.For instance, assume that −∆ ∞ φ(x 0 ) > 0, we can argue as in ( 16), but now, all the inequality involving the second order differential operator are reversed and we get As −∆ ∞ φ(x 0 ) > 0, the term in the big parenthesis is non-negative, we can erase everything to the power 1/p, obtaining which shows that u ∞ is a viscosity subsolution to (15).Similar arguments to step 1 give us the boundary conditions for viscosity subsolution.
We are also able to give a geometric characterization of Λ ∞ .
Lemma 3.4.Let Λ ∞ be the quantity defined in (4) and let R Ω be the inradius of Ω.Then .
In order to prove the reverse inequality, we take The following facts can occur We choose x ∈ Ω and y equal to the point on the boundary such that |x − y| = d(x, ∂Ω).So, we have and then With the same choice of x and y, we have .
and then the desired inequality Theorem 3.5.Let Λ ∞ be the quantity defined in (4).Then where Ω ♯ is the ball centered at the origin with the same measure of Ω.
Proof.The Faber-Krahn inequality for the first eigenvalue of the p-Laplacian with Robin boundary condition (for instance see [BD10]) states that Letting p go to infinity, we have This can follow also from the geometric characterization in Lemma 3.4 R Ω as the ball maximizes the inradius among sets of given volume.
Remark 3.1.One can easily prove that the function 1 β + d(x, ∂Ω) is an eigenfunction if the domain Ω = B R (x 0 ).This is not true if Ω is a square: see for instance [JLM99].

The first Robin ∞-eigenvalue
Now we want to show that Λ ∞ is the first eigenvalue of (1), that is the smallest Λ such that Proof.Let Λ be an eigenvalue to (1), let u be a corresponding eigenfunction.We normalize it in this way Then u is viscosity subsolution to For every ε > 0 and γ > 0, let us consider the function It is well known (see [GT01]) that in a small tubular neighbourhood Γ µ of ∂Ω, the boundary ∂Ω and the distance function d(x, ∂Ω) share the same regularity: so both d(x, ∂Ω) and g ε,γ are C 2 (Γ µ ).
Moreover, by a direct computation, one can check that if Hence, Theorem 2.1 in [Jen93] ensures that Assume by contradiction that m ε < − ε β , and set v = g ε,γ − m ε .We observe that v ≥ u in Ω and v(x 0 ) = u(x 0 ), where x 0 is the point which achieves the minimum on the boundary, so we can use it as test function in the definition of viscosity subsolution for u.
Assuming γ < ε 2R Ω , we obtain So we have letting ε and γ go to zero, it follows which concludes the proof.

The p-Poisson equation
Let f be a function in L ∞ (Ω) and let β > 0. We consider the following p-Poisson equation with Robin boundary conditions A function v p is a weak solution to (18) if it satisfies It is well known that the solution to this equation is the unique minimum of the functional Indeed, it is possible to prove the existence and the uniqueness of the minimum of the functional thanks to the so-called direct method of calculus of variation, see for instance [Dac08, Giu94, Lin06, AGM21].
If we let formally p go to ∞ in (20), we obtain the functional The limit procedure imposes two extra constraints to (21), namely The following proposition holds true.
Proposition 4.1.Let v p be the solution to (18).Then there exists a subsequence v p j j such that Proof.Choosing ϕ = v p in (19), we have and Young inequality gives Taking into account (3), we get where the constant C is independent of p, thanks to Lemma 2.1.Hence If p > m, Hölder inequality gives Then there exists v p j such that We are also able to link the so obtained function v ∞ with the functional (21), indeed Theorem 4.2.The functional Proof.Let v ∞ a limit of a subsequence of {v p } and let us assume it is not a minimum of J ∞ .This implies that ∃ϕ ∈ W 1,∞ (Ω) such that We want to show that there exists a function φ and an exponent p, such that J p (φ) < J p (v p ), which contradicts the minimality of v p .
First of all, we recall that exists a sequence and so there exists i for which we still have Two cases can occur: Then which is a contradiction.
Considering φ = αϕ with α ∈ (0, 1): We now choose p i : Proposition 4.3.Let v p be the solution to (18) and let v ∞ be any limit of a subsequence of as we have proven that ∇v ∞ ∞ ≤ 1.This holds true for every x, y in Ω.In particular, we can choose y equal to the point on the boundary which realizes |x − y| = d(x, ∂Ω).So, we have Remark 4.1.We explicitly observe that the estimate ϕ(x) ≤ β + d(x, ∂Ω) holds for every admissible function ϕ.Proposition 4.4.Assume f > 0 in Ω.Then the sequence of solution to (18 The function 1 β + d(x, ∂Ω) is a competitor and then Then (28) gives v ∞ (x) = 1 β + d(x, ∂Ω).Since every subsequence of { v p } has a subsequence converging to 1 β + d(x, ∂Ω) weakly in W 1,m (Ω), the whole sequence { v p } converges to 1 β + d(x, ∂Ω) weakly in W 1,m (Ω), and in particular, in C α (Ω) and its gradient weakly in L m (Ω).
It remains to prove the strong convergence in W 1,m (Ω).Clarkson's inequality implies for p, q > m From (25) we deduce and then lim sup Remark 4.2.If Suppf ⊂ Ω, then we can deduce that v ∞ (x) = 1 β + d(x, ∂Ω) for all x ∈ Suppf , while in Ω \ Suppf inequality (28) can be strict.Proof.Suppose that R ⊂ Suppf and let w be a minimum of (26)in the class {ϕ ∈ W 1,∞ (Ω) : and arguing as in Remark 4.2 we have Assume by contradiction that there exists x ∈ (Suppf ) c such that setting η = ∇d(x, ∂Ω), we choose the smallest t such that y = x + tη belongs to ∂(Suppf ) (Lemma 4.6 will justify this choice), thus where the last equality follows from Lemma 4.6.So we have ∇w L ∞ (Ω) > 1 that is a contradiction.
Assume now that w(x) = 1 β + d(x, ∂Ω) is the unique extremal of (26), and assume by contradiction that R ⊂ Supp f .Then, thanks to the Aronsson theorem (see [Aro67]), we can construct a function ϕ different from w which coincide with w in Supp f and such that it is admissible for (21).Then ϕ is a minimum too.This contradicts the fact that w is the unique minimum of (21), so R ⊂ Suppf .
We have to prove Lemma 4.6 to complete the proof.
Lemma 4.6.Let x ∈ Ω \ R and set η = ∇(d(x, ∂Ω)).Let us consider y t = x + tη, then there exists T such that y T ∈ R and y t / ∈ R for all t < T .Moreover, Proof.Consider the following Cauchy problem in a maximal interval [0, T ).We have that These considerations give us the following: • T < ∞, otherwise d is unbounded, and this is a contradiction as Ω is bounded; • γ(T ) ∈ R, otherwise one can extend the solution for t > T , in contradiction with the fact that [0, T ) is the maximal interval.
In the end, if y = γ(T ), we have This conclude the proof.

The limit PDE
We have proved that any limit v ∞ of subsequence { v p } is a minimum of the functional (26) defined in W 1,∞ (Ω).Now we want to understand if such limits are solutions to a certain PDE, which, in some sense, is the Euler-Lagrange equation of the functional (26).If we consider the following sequence So f q (x) ≤ −3C for x ∈ ∂B R (x 0 ), and f q (x 0 ) ≥ −2C.Then f q (x) attains its maximum at some point x q in the interior of B R (x 0 ).and it holds Moreover, the sequence { x q } must converge to x 0 .Assume by contradiction that |∇ϕ(x 0 )| > 1, so there exists δ ∈ (0, 1) such that |∇ϕ(x 0 )| ≥ 1 + δ.Choosing q large enough, we have By Lemma 1.1 of Part III in [BDM89], where γ is a constant independent of q.
Proof.The proof follows the same techniques of Proposition 3.2.We stress that, if we let p → ∞ in (36), we get −∆ ∞ ϕ(x 0 ) ≥ 0, so v ∞ is a supersolution to −∆ ∞ u = 0. Now, we only have to prove that v ∞ is a viscosity solution to ∆ ∞ ϕ = 0 in ({ f > 0 }) c .If we fix x 0 ∈ ({ f > 0 }) c , and ϕ ∈ C 2 (Ω) such that v ∞ − ϕ has a strict maximum in x 0 , we can choose R such that B R (x 0 ) ⊂ ({ f > 0 }) c .The function v p − ϕ has a maximum x p → x 0 , and x p ∈ B R (x 0 ) for p large enough.

Definition 4. 1 .
We denote by R the set of discontinuity of the function ∇d(x, ∂Ω).This set consists of points x ∈ Ω for which d(x, ∂Ω) is achieved by more than one point y on the boundary.Then it holds true the following Theorem 4.5.Let f be a non-negative function inΩ, then function v ∞ (x) = 1 β + d(x,∂Ω) is the unique extremal function of (26) if and only if R ⊂ Suppf .