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

Vincenzo Amato; Alba Lia Masiello; Carlo Nitsch; Cristina Trombetti

doi:10.1515/anona-2022-0258

Open Access Published by De Gruyter July 29, 2022

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

Vincenzo Amato , Alba Lia Masiello , Carlo Nitsch and Cristina Trombetti

From the journal Advances in Nonlinear Analysis

https://doi.org/10.1515/anona-2022-0258

Abstract

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 article, 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 → ∞ .

Keywords: p-Laplacian; Robin boundary conditions; eigenvalue problem; infinity Laplacian

MSC 2010: 35J92; 35J94; 35P15

1 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

(1) min { ∣ ∇ u ∣ − Λ u , − Δ ∞ u } = 0 in Ω , − min ∣ ∇ u ∣ − β u , − ∂ u ∂ ν = 0 on ∂ Ω ,

where Δ ∞ , the so-called ∞ -Laplacian, is defined by

Δ ∞ u = ⟨ D 2 u ⋅ ∇ u , ∇ u ⟩ .

We refer to problem (1) as the ∞ -Laplacian eigenvalue problem because it can be seen as the limit, in some sense, of the p-Laplacian eigenvalue problem

(2) − Δ p u = Λ p ∣ u ∣ p − 2 u in Ω , ∣ ∇ u ∣ p − 2 ∂ u ∂ ν + β p ∣ u ∣ p − 2 u = 0 on ∂ Ω .

A function u p ∈ W 1 , p ( Ω ) is a weak solution to (2) if it satisfies

∫ Ω ∣ ∇ u p ∣ p − 2 ∇ u p ∇ φ d x + β p ∫ ∂ Ω ∣ u p ∣ p − 2 u p φ d ℋ n − 1 = Λ p ∫ Ω ∣ u p ∣ p − 2 u p φ d x , ∀ φ ∈ W 1 , p ( Ω ) ,

where β and Λ p are both positive.

It is well known that the first eigenvalue of the p-Laplacian is the minimum of the following Rayleigh quotient:

(3) Λ p = inf w ∈ W 1 , p ( Ω ) ∥ w ∥ L p ( Ω ) = 1 ∫ Ω ∣ ∇ w ∣ p d x + β p ∫ ∂ Ω ∣ w ∣ p d ℋ n − 1 .

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 article, we first prove that

(4) lim p → + ∞ ( Λ p ) 1 / p = Λ ∞ ≕ inf w ∈ W 1 , ∞ ( Ω ) ∥ w ∥ L ∞ ( Ω ) = 1 max { ∥ ∇ w ∥ L ∞ ( Ω ) , β ∥ w ∥ L ∞ ( ∂ Ω ) } ,

and we give a geometric characterization of this quantity, precisely:

(5) Λ ∞ = 1 1 / β + R Ω ,

where R Ω denotes the inradius of Ω , i.e. the radius of the largest 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 [4,7,11,12,16].

More specifically, in [11,12], 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 p D } happens to satisfy

lim p → ∞ ( λ p D ) 1 / p = λ ∞ D ≔ 1 R Ω .

The related eigenfunctions v p D 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 [7,16] and similarly to the Dirichlet case, the authors established that the first non-trivial eigenvalues of the p-Laplacian { λ p N } satisfy

lim p → ∞ ( λ p N ) 1 ∕ p = λ ∞ N ≔ 2 diam ( Ω ) ,

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 problems for the ∞ -Laplacian, in [7], the authors are able to prove that they actually converge to the first eigenvalue and 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 article, we focus our attention on the study of the limit of the p-Poisson equation with Robin boundary conditions:

(6) − Δ p v = f in Ω , ∣ ∇ v ∣ p − 2 ∂ v ∂ ν + β p ∣ v ∣ p − 2 v = 0 on ∂ Ω ,

when f ∈ L ∞ ( Ω ) is a non-negative function.

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 [4] by Bhattacharya et al., while, to the best of our knowledge, similar results have not been addressed in the case of Neumann boundary conditions.

2 Notations and preliminaries

Throughout this article, ∣ ⋅ ∣ will denote the Euclidean norm in R n , and ℋ k ( ⋅ ) , for k ∈ [ 0 , n ) , will denote the k -dimensional Hausdorff measure in R n .

We denote by d ( x , ∂ Ω ) the distance function from the boundary, defined as

(7) d ( x , ∂ Ω ) = inf y ∈ ∂ Ω ∣ x − y ∣ ,

for an exhaustive discussion about this function and its properties see [9]. Moreover, we recall that the inradius R Ω of Ω is

(8) R Ω = sup x ∈ Ω inf y ∈ ∂ Ω ∣ x − y ∣ = ∥ d ( ⋅ , ∂ Ω ) ∥ L ∞ ( Ω ) .

The following lemma makes us understand why (4) can be seen as a limit problem of (3).

Lemma 2.1

Given f , g ∈ W 1 , ∞ ( Ω ) , then

lim p → ∞ ∫ Ω ∣ f ∣ p + ∫ Ω ∣ g ∣ p 1 / p = max { ∥ f ∥ ∞ , ∥ g ∥ ∞ } .

Proof

The proof of this lemma can be found in [15].□

2.1 Viscosity solutions

Before going on, we recall the definition of viscosity solutions to a boundary value problem, see [5] for more details.

Definition 2.1

We consider the following boundary value problem:

(9) F ( x , u , ∇ u , D 2 u ) = 0 in Ω , B ( x , u , ∇ u ) = 0 on ∂ Ω ,

where F : R n × R × R n × R n × n → R and B : R n × R × R n → R are two continuous functions.

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
F ( x 0 , ϕ ( x 0 ) , ∇ ϕ ( x 0 ) , D 2 ϕ ( x 0 ) ) ≥ 0 ;
if x 0 ∈ ∂ Ω , the following holds
max { F ( x 0 , ϕ ( x 0 ) , ∇ ϕ ( x 0 ) , D 2 ϕ ( x 0 ) ) , B ( x 0 , ϕ ( x 0 ) , ∇ ϕ ( x 0 ) ) } ≥ 0 .

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
F ( x 0 , ϕ ( x 0 ) , ∇ ϕ ( x 0 ) , D 2 ϕ ( x 0 ) ) ≤ 0 ;
if x 0 ∈ ∂ Ω , the following holds
min { F ( x 0 , ϕ ( x 0 ) , ∇ ϕ ( x 0 ) , D 2 ϕ ( x 0 ) ) , B ( x 0 , ϕ ( x 0 ) , ∇ ϕ ( x 0 ) ) } ≤ 0 .

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 .

3 The ∞ -eigenvalue problem

Let us start this section by observing that Lemma 2.1 brings to the following estimate:

(10) limsup p → ∞ Λ p 1 / p ≤ Λ ∞ .

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,

(11) lim p → ∞ ( Λ p ) 1 p = Λ ∞ ,

where Λ ∞ is defined in (4).

Moreover, if { u p } p > 1 is the sequence of eigenfunctions associated with { Λ p } p > 1 , then there exists a function u ∞ ∈ W 1 , ∞ ( Ω ) such that, up to a subsequence,

u p → u ∞ uniformly in Ω ∇ u p → ∇ u ∞ weakly in L q ( Ω ) , ∀ q .

Proof

As a consequence of (10), the sequence { u p } p > 1 of eigenfunctions associated with Λ p is uniformly bounded in W 1 , q ( Ω ) : indeed, if q < p , by Hölder’s inequality,

(12) ∥ ∇ u p ∥ L q ( Ω ) ≤ ∥ ∇ u p ∥ L p ( Ω ) ∣ Ω ∣ 1 q − 1 p ≤ Λ p 1 / p ∣ Ω ∣ 1 q − 1 p ≤ C ,

(13) ∥ u p ∥ L q ( Ω ) ≤ ∥ u p ∥ L p ( Ω ) ∣ Ω ∣ 1 q − 1 p ≤ ∣ Ω ∣ 1 q − 1 p ≤ C ,

where the constant C is independent of p.

By a classical argument of diagonalization, see for instance [4], we can extract a subsequence u p j such that

u p j → u ∞ uniformly ⇒ ‖ u p j ‖ L p j → ∥ u ∞ ∥ L ∞ , ∇ u p j → ∇ u ∞ weakly in L q ( Ω ) , ∀ q > 1 .

Moreover, from (12) and (13), the following inequality holds

∥ ∇ u ∞ ∥ L q ( Ω ) ∥ u ∞ ∥ L q ( Ω ) ≤ liminf p → ∞ ∥ ∇ u p ∥ L q ( Ω ) ∥ u p ∥ L q ( Ω ) ≤ liminf p → ∞ ∥ ∇ u p ∥ L p ( Ω ) ∥ u p ∥ L q ( Ω ) ∣ Ω ∣ 1 q − 1 p ≤ ∣ Ω ∣ 1 q ∥ u ∞ ∥ L q ( Ω ) liminf p → ∞ ( Λ p ) 1 p .

Letting q → ∞ we obtain

∥ ∇ u ∞ ∥ L ∞ ( Ω ) ≤ liminf p → ∞ ( Λ p ) 1 p .

Similarly,

β ∥ u p ∥ L q ( ∂ Ω ) ≤ β ∥ u p ∥ L p ( ∂ Ω ) ∣ ∂ Ω ∣ 1 q − 1 p ≤ Λ p 1 / p ∣ ∂ Ω ∣ 1 q − 1 p ≤ C ,

gives us

β ∥ u ∞ ∥ L ∞ ( ∂ Ω ) ≤ liminf p → ∞ ( Λ p ) 1 p ,

hence

Λ ∞ ≤ liminf p → ∞ Λ p 1 / p . □

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 [7, 12] 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

(14) max − ∣ ∇ ϕ ( x 0 ) ∣ p − 2 Δ ϕ ( x 0 ) − ( p − 2 ) ∣ ∇ ϕ ( x 0 ) ∣ p − 4 Δ ∞ ϕ ( x 0 ) − Λ p ∣ ϕ ( x 0 ) ∣ p − 2 ϕ ( x 0 ) , ∣ ∇ ϕ ( x 0 ) ∣ p − 2 ∂ ϕ ( x 0 ) ∂ ν + β p ∣ ϕ ( x 0 ) ∣ p − 2 ϕ ( x 0 ) ≥ 0 .

Assume by contradiction that both terms are negative. If we choose r sufficiently small, in Ω ¯ ∩ B r ( x 0 ) , we have

− ∣ ∇ ϕ ( x ) ∣ p − 2 Δ ϕ ( x ) − ( p − 2 ) ∣ ∇ ϕ ( x ) ∣ p − 4 Δ ∞ ϕ ( x ) − Λ p ∣ ϕ ( x ) ∣ p − 2 ϕ ( x ) < 0

and, in ∂ Ω ∩ B r ( x 0 ) ,

∣ ∇ ψ ( x ) ∣ p − 2 ∂ ψ ( x ) ∂ ν + β p ∣ ψ ( x ) ∣ p − 2 ψ ( x ) < 0 , where ψ = ϕ + m 2 .

Then

∫ { ψ > u } ∩ B r ( x 0 ) ∣ ∇ ψ ∣ p − 2 ∇ ψ ∇ ( ψ − u ) d x < Λ p ∫ { ψ > u } ∩ B r ( x 0 ) ∣ ϕ ∣ p − 2 ϕ ( ψ − u ) d x − β p ∫ ∂ Ω ∩ B r ( x 0 ) ∩ { ψ > u } ∣ ψ ∣ p − 2 ψ ( ψ − u ) d ℋ n − 1 ,

using the definition of weak solution, we have

C ( N , p ) ∫ { ψ > u } ∩ B r ( x 0 ) ∣ ∇ ψ − ∇ u ∣ p d x ≤ ∫ { ψ > u } ∩ B r ( x 0 ) ⟨ ∣ ∇ ψ ∣ p − 2 ∇ ψ − ∣ ∇ u ∣ p − 2 ∇ u , ∇ ( ψ − u ) ⟩ d x < Λ p ∫ { ψ > u } ∩ B r ( x 0 ) ( ∣ ϕ ∣ p − 2 ϕ − ∣ u ∣ p − 2 u ) ( ψ − u ) d x − β p ∫ ∂ Ω ∩ B r ( x 0 ) ∩ { ψ > u } ( ∣ ψ ∣ p − 2 ψ − ∣ u ∣ p − 2 u ) ( ψ − u ) d ℋ n − 1 < 0 ,

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

(15) min { ∣ ∇ u ∣ − Λ ∞ u , − Δ ∞ u } = 0 in Ω , − min ∣ ∇ u ∣ − β u , − ∂ u ∂ ν = 0 on ∂ Ω .

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 { ∣ ∇ ϕ ( x 0 ) ∣ − Λ ∞ ϕ ( x 0 ) , − Δ ∞ ϕ ( x 0 ) } ≥ 0 .

Note 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

(16) − ∣ ∇ ϕ p ( x p ) ∣ p − 2 Δ ϕ p ( x p ) − ( p − 2 ) ∣ ∇ ϕ p ( x p ) ∣ p − 4 Δ ∞ ϕ ( x p ) − Λ p ∣ ϕ p ( x p ) ∣ p − 2 ϕ p ( x p ) ≥ 0 .

Now dividing by ( p − 2 ) ∣ ∇ ϕ p ( x p ) ∣ p − 4 , we obtain

(17) − Δ ∞ ϕ p ( x p ) − ∣ ∇ ϕ p ( x p ) ∣ 2 Δ ϕ p ( x p ) p − 2 ≥ ∣ ∇ ϕ p ( x p ) ∣ 4 ( p − 2 ) ϕ p ( x p ) Λ p 1 / p ϕ p ( x p ) ∣ ∇ ϕ p ( x p ) ∣ p .

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, − Δ ∞ ϕ ( x 0 ) ≥ 0 , just taking the limit.

Then, min { ∣ ∇ ϕ ( x 0 ) ∣ − Λ ∞ ϕ ( x 0 ) , − Δ ∞ ϕ ( x 0 ) } ≥ 0 and u ∞ is a viscosity supersolution.

Let us fix x 0 ∈ ∂ Ω , ϕ ∈ C 2 ( Ω ¯ ) such that u − ϕ has a strict minimum in x 0 , our aim is to prove that

max min { ∣ ∇ ϕ ( x 0 ) ∣ − Λ ∞ ϕ ( x 0 ) , − Δ ∞ ϕ ( x 0 ) } , − min ∣ ∇ ϕ ( x 0 ) ∣ − β ϕ ( x 0 ) , − ∂ ϕ ∂ ν ( x 0 ) ≥ 0 .

If for infinitely many x p ∈ Ω (16) holds true, then we obtain

min { ∣ ∇ ϕ ( x 0 ) ∣ − Λ ∞ ϕ ( x 0 ) , − Δ ∞ ϕ ( x 0 ) } ≥ 0 .

If for infinitely many p, x p ∈ ∂ Ω the following holds true

∣ ∇ ϕ p ( x p ) ∣ p − 2 ∂ ϕ p ( x p ) ∂ ν + β p ∣ ϕ p ( x p ) ∣ p − 2 ϕ p ( x p ) ≥ 0 ,

then

∣ ∇ ϕ p ( x p ) ∣ p − 2 − ∂ ϕ p ( x p ) ∂ ν ≤ β p ∣ ϕ p ( x p ) ∣ p − 2 ϕ p ( x p ) .

Only two cases can occur:

− ∂ ϕ ∂ ν ( x 0 ) ≤ 0 ;
− ∂ ϕ ∂ ν ( x 0 ) > 0 , then letting p to infinity in the following
∣ ∇ ϕ p ( x p ) ∣ p − 2 − ∂ ϕ p ( x p ) ∂ ν 1 / p ≤ ( β p ∣ ϕ p ( x p ) ∣ p − 2 ϕ p ( x p ) ) 1 / p .
we obtain ∣ ∇ ϕ ( x 0 ) ∣ ≤ β ϕ ( x 0 ) .

That is

− min ∣ ∇ ϕ ( x 0 ) ∣ − β ϕ ( x 0 ) , − ∂ ϕ ∂ ν ( x 0 ) ≥ 0 .

Step 2 u ∞ is a viscosity subsolution.

Let us fix x 0 ∈ Ω , ϕ ∈ C 2 ( Ω ) such that u ∞ − ϕ has a strict maximum. We want to prove that

min { ∣ ∇ ϕ ( x 0 ) ∣ − Λ ∞ ϕ ( x 0 ) , − Δ ∞ ϕ ( x 0 ) } ≤ 0 ,

so it is enough to prove that only one of the two terms in the bracket is non-positive.

For instance, assume that − Δ ∞ ϕ ( x 0 ) > 0 , we can argue as in (16), but now, all the inequalities involving the second-order differential operator are reversed and we obtain

Λ p ϕ p p − 1 ( x p ) ≥ ( p − 2 ) ∣ ∇ ϕ p ( x p ) ∣ p − 4 − ∣ ∇ ϕ p ( x p ) ∣ 2 Δ ϕ p ( x p ) p − 2 − Δ ∞ ϕ p ( x p ) .

As − Δ ∞ ϕ ( x 0 ) > 0 , the term in the big parenthesis is non-negative, we can erase everything to the power 1 / p , obtaining

Λ ∞ ϕ ( x 0 ) ≥ ∣ ∇ ϕ ( x 0 ) ∣ ,

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

Λ ∞ = min x 0 ∈ Ω 1 1 β + d ( x 0 , ∂ Ω ) = 1 1 β + R Ω .

Proof

The function 1 β + d ( x , ∂ Ω ) ∈ W 1 , ∞ ( Ω ) , moreover

∥ ∇ ( 1 / β + d ( x , ∂ Ω ) ) ∥ L ∞ ( Ω ) = 1 and β ∥ 1 / β + d ( x , ∂ Ω ) ∥ L ∞ ( ∂ Ω ) = 1 .

Then

Λ ∞ ≤ min x 0 ∈ Ω 1 1 β + d ( x 0 , ∂ Ω ) .

In order to prove the reverse inequality, we take w ∈ W 1 , ∞ ( Ω ) such that ∥ w ∥ L ∞ ( Ω ) = 1 .

The following facts can occur.

If β ∥ w ∥ L ∞ ( ∂ Ω ) ≤ ∥ ∇ w ∥ L ∞ ( Ω ) , then
max { ∥ ∇ w ∥ L ∞ ( Ω ) , β ∥ w ∥ L ∞ ( ∂ Ω ) } = ∥ ∇ w ∥ L ∞ ( Ω ) .
We choose x ∈ Ω and y equal to the point on the boundary such that ∣ x − y ∣ = d ( x , ∂ Ω ) . So, we have
∣ w ( x ) ∣ ≤ ∣ w ( x ) − w ( y ) ∣ + ∣ w ( y ) ∣ ≤ ∥ ∇ w ∥ L ∞ ( Ω ) ∣ x − y ∣ + ∥ w ∥ L ∞ ( ∂ Ω ) ≤ ∥ ∇ w ∥ L ∞ ( Ω ) d ( x , ∂ Ω ) + 1 β ∥ ∇ w ∥ L ∞ ( Ω ) = ∥ ∇ w ∥ L ∞ ( Ω ) 1 β + d ( x , ∂ Ω ) ≤ ∥ ∇ w ∥ L ∞ ( Ω ) ∥ 1 / β + d ( x , ∂ Ω ) ∥ L ∞ ( Ω ) ,
and then
∥ ∇ w ∥ L ∞ ( Ω ) ∥ w ∥ L ∞ ( Ω ) ≥ 1 ∥ 1 / β + d ( x , ∂ Ω ) ∥ L ∞ ( Ω ) .
If β ∥ w ∥ L ∞ ( ∂ Ω ) > ∥ ∇ w ∥ L ∞ ( Ω ) , then
max { ∥ ∇ w ∥ L ∞ ( Ω ) , β ∥ w ∥ L ∞ ( ∂ Ω ) } = β ∥ w ∥ L ∞ ( ∂ Ω ) .
With the same choice of x and y , we have
∣ w ( x ) ∣ ≤ ∣ w ( x ) − w ( y ) ∣ + ∣ w ( y ) ∣ ≤ ∥ ∇ w ∥ L ∞ ( Ω ) ∣ x − y ∣ + ∥ w ∥ L ∞ ( ∂ Ω ) ≤ β ∥ w ∥ L ∞ ( ∂ Ω ) d ( x , ∂ Ω ) + ∥ w ∥ L ∞ ( ∂ Ω ) = β ∥ w ∥ L ∞ ( ∂ Ω ) d ( x , ∂ Ω ) + 1 β ≤ β ∥ w ∥ L ∞ ( ∂ Ω ) ∥ 1 / β + d ( x , ∂ Ω ) ∥ L ∞ ( Ω ) .
Hence,
β ∥ w ∥ L ∞ ( ∂ Ω ) ∥ w ∥ L ∞ ( Ω ) ≥ 1 ∥ 1 / β + d ( x , ∂ Ω ) ∥ L ∞ ( Ω ) .

Finally, we obtain ∀ w ∈ W 1 , ∞ ( Ω ) : ∥ w ∥ L ∞ ( Ω ) = 1 ,

max { ∥ ∇ w ∥ L ∞ ( Ω ) , β ∥ w ∥ L ∞ ( ∂ Ω ) } ≥ 1 ∥ 1 / β + d ( x , ∂ Ω ) ∥ L ∞ ( Ω ) ,

and then the desired inequality

Λ ∞ ≥ min x 0 ∈ Ω 1 1 β + d ( x 0 , ∂ Ω ) . □

Theorem 3.5

Let Λ ∞ be the quantity defined in (4). Then

Λ ∞ ( Ω ) ≥ Λ ∞ ( Ω ♯ ) ,

where Ω ♯ is the ball centred 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 [3]) states that

Λ p ( Ω ) ≥ Λ p ( Ω ♯ ) .

Letting p go to infinity, we have

Λ ∞ ( Ω ) ≥ Λ ∞ ( Ω ♯ ) .

This can follow also from the geometric characterization in Lemma 3.4

Λ ∞ = 1 1 β + 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 [12].

3.1 The first Robin ∞ -eigenvalue

Now we want to show that Λ ∞ is the first eigenvalue of (1), that is the smallest Λ such that

min { ∣ ∇ u ∣ − Λ u , − Δ ∞ u } = 0 in Ω , − min ∣ ∇ u ∣ − β u , − ∂ u ∂ ν = 0 on ∂ Ω

admits a non-trivial solution.

Theorem 3.6

Let Ω be a bounded and open set of class C 2 in R n . If for some Λ , problem (1) admits a non-trivial eigenfunction u , then Λ ≥ Λ ∞ .

Proof

Let Λ be an eigenvalue to (1), let u be a corresponding eigenfunction. We normalize it in this way

max x ∈ Ω u ( x ) = 1 Λ .

Then u is viscosity subsolution to

min { ∣ ∇ u ∣ − 1 , − Δ ∞ u } = 0 in Ω .

For every ε > 0 and γ > 0 , let us consider the function

g ε , γ = 1 β + ( 1 + ε ) d ( x , ∂ Ω ) − γ d ( x , ∂ Ω ) 2 .

It is well known (see [9]) 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

γ < ε 2 R Ω ,

then g ε , γ is a viscosity supersolution to

min { ∣ ∇ g ε , γ ∣ − 1 , − Δ ∞ g ε , γ } = 0 in Ω .

Hence, Theorem 2.1 in [10] ensures that

m ε = inf x ∈ Ω ( g ε , γ ( x ) − u ( x ) ) = inf x ∈ ∂ Ω ( g ε , γ ( x ) − u ( x ) ) .

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 2 R Ω , we obtain

∇ v ( x ) = [ 1 + ε − 2 γ d ( x , ∂ Ω ) ] ∇ d ( x , ∂ Ω ) , ∣ ∇ v ( x 0 ) ∣ = 1 + ε − 2 γ d ( x 0 , ∂ Ω ) > 1 , − ∂ v ∂ ν ( x 0 ) = − [ 1 + ε − 2 γ d ( x 0 , ∂ Ω ) ] ∇ d ( x 0 , ∂ Ω ) ⋅ ν > 0 , − Δ ∞ v ( x 0 ) = 2 γ [ 1 + ε − 2 γ d ( x 0 , ∂ Ω ) ] 2 ∣ ∇ d ( x 0 , ∂ Ω ) ∣ 4 > 0 .

The fact that m ε < − ε β implies

∣ ∇ v ( x 0 ) ∣ − β v ( x 0 ) = ε + β m ε < 0 .

Therefore,

− min ∣ ∇ v ∣ − β v , − ∂ v ∂ ν > 0 and min { ∣ ∇ v ∣ − 1 , − Δ ∞ v } > 0

against the fact that

min min { ∣ ∇ v ∣ − 1 , − Δ ∞ v } , − min ∣ ∇ v ∣ − β v , − ∂ v ∂ ν ≤ 0 .

So we have

g ε , γ ( x ) − u ( x ) ≥ m ε ≥ − ε β ,

letting ε and γ go to zero, it follows

1 β + d ( x , ∂ Ω ) ≥ u ( x ) ∀ x ∈ Ω .

Hence,

1 Λ ∞ = max x ∈ Ω 1 β + d ( x , ∂ Ω ) ≥ max x ∈ Ω u ( x ) = 1 Λ ,

which concludes the proof.□

4 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

(18) − Δ p v = f in Ω , ∣ ∇ v ∣ p − 2 ∂ v ∂ ν + β p ∣ v ∣ p − 2 v = 0 on ∂ Ω .

A function v p is a weak solution to (18) if it satisfies

(19) ∫ Ω ∣ ∇ v p ∣ p − 2 ∇ v p ∇ φ d x + β p ∫ ∂ Ω ∣ v p ∣ p − 2 v p φ d ℋ n − 1 = ∫ Ω f φ d x , ∀ φ ∈ W 1 , p ( Ω ) .

It is well known that the solution to this equation is the unique minimum of the functional

(20) J p ( φ ) = 1 p ∫ Ω ∣ ∇ φ ∣ p d x + β p p ∫ ∂ Ω ∣ φ ∣ p d ℋ n − 1 ( x ) − ∫ Ω f φ d x .

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 [1,6, 8,13].

If we let formally p go to ∞ in (20), we obtain the functional

(21) φ ⟶ min ∫ Ω − f φ d x φ ∈ W 1 , ∞ ( Ω ) .

The limit procedure imposes two extra constraints to (21), namely

∥ ∇ φ ∥ ∞ ≤ 1 , β ∥ φ ∥ L ∞ ( ∂ Ω ) ≤ 1 .

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

v p j → v ∞ uniformly , ∇ v p j → ∇ v ∞ weakly in L m ( Ω ) , ∀ m > 1 .

Moreover,

∥ ∇ v ∞ ∥ ∞ ≤ 1 β ∥ v ∞ ∥ L ∞ ( ∂ Ω ) ≤ 1 .

Proof

Choosing φ = v p in (19), we have

∫ Ω ∣ ∇ v p ∣ p + β p ∫ ∂ Ω v p p = ∫ Ω f v p ,

and Young’s inequality gives

(22) ∫ Ω ∣ ∇ v p ∣ p + β p ∫ ∂ Ω v p p − 1 ε p p p ∫ Ω v p p ≤ ε p p ′ p ′ ∫ Ω f p ′ .

Taking into account (3), we obtain

1 − 1 p Λ p ε p p ∫ Ω ∣ ∇ v p ∣ p + β p ∫ ∂ Ω v p p ≤ ∫ Ω ∣ ∇ v p ∣ p + β p ∫ ∂ Ω v p p − 1 ε p p p ∫ Ω v p p ≤ ε p p ′ p ′ ∫ Ω f p ′ .

Choosing ε p such that 1 − 1 p Λ p ε p p = 1 2 , we have

(23) ∫ Ω ∣ ∇ v p ∣ p + β p ∫ ∂ Ω v p p ≤ 2 ε p p ′ p ′ ∫ Ω f p ′ ≤ C ∫ Ω f p ′ ,

where the constant C is independent of p, thanks to Lemma 2.1.

Hence,

∫ Ω ∣ ∇ v p ∣ p 1 / p ≤ C ∫ Ω f p ′ 1 / p ≤ ( C ∣ Ω ∣ ∥ f ∥ ∞ p ′ ) 1 / p , β p ∫ ∂ Ω v p p 1 / p ≤ C ∫ Ω f p ′ 1 / p ≤ ( C ∣ Ω ∣ ∥ f ∥ ∞ p ′ ) 1 / p .

Analogously,

(24) ∫ Ω v p p 1 / p ≤ C ∫ Ω f p ′ 1 / p .

If p > m , Hölder’s inequality gives

(25) ∫ Ω ∣ ∇ v p ∣ m 1 / m ≤ ∫ Ω ∣ ∇ v p ∣ p 1 / p ∣ Ω ∣ 1 / m − 1 / p ≤ ( C ∥ f ∥ ∞ p ′ ) 1 / p ∣ Ω ∣ 1 / m , β ∫ ∂ Ω v p m 1 / m ≤ β p ∫ ∂ Ω v p p 1 / p ∣ ∂ Ω ∣ 1 / m − 1 / p ≤ ( C ∥ f ∥ ∞ p ′ ) 1 / p ∣ ∂ Ω ∣ 1 / m , ∫ Ω v p m 1 / m ≤ ∫ Ω v p p 1 / p ∣ Ω ∣ 1 / m − 1 / p ≤ C ( ∥ f ∥ ∞ p ′ ) 1 / p ∣ Ω ∣ 1 / m .

Then there exists v p j such that

v p j → v ∞ , uniformly, ∇ v p j → ∇ v ∞ weakly in L m ( Ω ) , ∀ m > 1 ,

moreover

∥ ∇ v ∞ ∥ m ≤ liminf j → ∞ ∥ ∇ v p j ∥ m ≤ lim j → ∞ ( C ∥ f ∥ ∞ p ′ ) 1 / p j ∣ Ω ∣ 1 / m = ∣ Ω ∣ 1 / m β ∥ v ∞ ∥ L m ( ∂ Ω ) = β lim j → ∞ ∥ v p j ∥ L m ( ∂ Ω ) ≤ lim j → ∞ ( C ∥ f ∥ ∞ p ′ ) 1 / p j ∣ ∂ Ω ∣ 1 / m = ∣ ∂ Ω ∣ 1 / m ,

and then

∥ ∇ v ∞ ∥ L ∞ ( Ω ) ≤ 1 β ∥ v ∞ ∥ L ∞ ( ∂ Ω ) ≤ 1 . □

We are also able to link the so-obtained function v ∞ with the functional (21), indeed.

Theorem 4.2

The functional

(26) J ∞ ( φ ) = − ∫ Ω f φ φ ∈ W 1 , ∞ ( Ω )

admits at least one minimum φ ¯ satisfying ∥ ∇ φ ¯ ∥ L ∞ ( Ω ) ≤ 1 and β ∥ φ ¯ ∥ L ∞ ( ∂ Ω ) ≤ 1 .

Moreover, if v ∞ is a limit of a subsequence of { v p } , then v ∞ is a minimizer of (26).

Proof

Let v ∞ be 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

− ∫ Ω f φ < − ∫ Ω f v ∞ .

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 there exists a sequence v p i ⇀ v ∞ in W 1 , m ( Ω ) ∀ m . Then

∫ Ω f v p i → ∫ Ω f v ∞ ,

and so there exists i ¯ for which we still have

(27) − ∫ Ω f φ < − ∫ Ω f v p i ∀ i > i ¯ .

Two cases can occur:

∃ i > i ¯
∫ Ω ∣ ∇ φ ∣ p i + β p i ∫ ∂ Ω ∣ φ ∣ p i ≤ ∫ Ω ∣ ∇ v p i ∣ p i + β p i ∫ ∂ Ω v p i p i .
Then
J p i ( φ ) = 1 p ∫ Ω ∣ ∇ φ ∣ p i + β p i p ∫ ∂ Ω ∣ φ ∣ p i − ∫ Ω f φ < 1 p ∫ Ω ∣ ∇ v p i ∣ p i + β p i p ∫ ∂ Ω v p i p i − ∫ Ω f v p i = J p i ( v p i ) ,
which is a contradiction.
∀ i > i ¯
∫ Ω ∣ ∇ φ ∣ p i + β p i ∫ ∂ Ω ∣ φ ∣ p i > ∫ Ω ∣ ∇ v p i ∣ p i + β p i ∫ ∂ Ω v p i p i .
Considering ϕ = α φ with α ∈ ( 0 , 1 ) :
− ∫ Ω f ϕ = − α ∫ Ω f φ < − ∫ Ω f v p i < 0 ∀ i > i ¯ ,
we have
∫ Ω ∣ ∇ ϕ ∣ p i + β p i ∫ ∂ Ω ∣ ϕ ∣ p i = α p i ∫ Ω ∣ ∇ φ ∣ p i + β p i ∫ ∂ Ω ∣ φ ∣ p i .
Moreover,
M ≤ ∫ Ω f v p i = ∫ Ω ∣ ∇ v p i ∣ p i + β p i ∫ ∂ Ω v p i p i ∫ Ω ∣ ∇ φ ∣ p i + β p i ∫ ∂ Ω ∣ φ ∣ p i ≤ ∣ Ω ∣ ∥ ∇ φ ∥ L ∞ ( Ω ) p i + β p i ∣ ∂ Ω ∣ ∥ φ ∥ L ∞ ( ∂ Ω ) p i ≤ ∣ Ω ∣ + ∣ ∂ Ω ∣ .
We now choose p i :
0 ← i → ∞ α p i ≤ M ∣ Ω ∣ + ∣ ∂ Ω ∣ ≤ ∫ Ω ∣ ∇ v p i ∣ p i + β p i ∫ ∂ Ω v p i p i ∫ Ω ∣ ∇ φ ∣ p i + β p i ∫ ∂ Ω ∣ φ ∣ p i
obtaining
∫ Ω ∣ ∇ ϕ ∣ p i + β p i ∫ ∂ Ω ∣ ϕ ∣ p i ≤ ∫ Ω ∣ ∇ v p i ∣ p i + β p i ∫ ∂ Ω v p i p i . □

Proposition 4.3

Let v p be the solution to (18) and let v ∞ be any limit of a subsequence of { v p } p > 1 . Then

(28) v ∞ ( x ) ≤ 1 β + d ( x , ∂ Ω ) .

Proof

We note that

∣ v ∞ ( x ) − v ∞ ( y ) ∣ ≤ ∣ x − y ∣ ,

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

v ∞ ( x ) ≤ v ∞ ( y ) + d ( x , ∂ Ω ) ≤ 1 β + d ( x , ∂ Ω ) ,

as v ∞ also satisfies β ∥ v ∞ ∥ L ∞ ( ∂ Ω ) ≤ 1 .□

Remark 4.1

We explicitly observe that the estimate φ ( x ) ≤ 1 β + d ( x , ∂ Ω ) holds for every admissible function φ .

Proposition 4.4

Assume f > 0 in Ω . Then the sequence of solutions to (18) converges strongly in W 1 , m ( Ω ) , for all m > 1 , to

v ¯ ∞ ( x ) = 1 β + d ( x , ∂ Ω ) .

Proof

Let v ∞ be any limit of a subsequence { v p j } ⊂ { v p } . Theorem 4.2 gives that v ∞ is a minimum of the functional J ∞ , in the class { φ ∈ W 1 , ∞ ( Ω ) : ∥ ∇ φ ∥ ∞ ≤ 1 , β ∥ φ ∥ L ∞ ( ∂ Ω ) ≤ 1 } .

The function 1 β + d ( x , ∂ Ω ) is a competitor and then

(29) ∫ Ω f v ∞ − 1 β − d ( x , ∂ Ω ) ≥ 0 .

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

∫ Ω ∣ ∇ v p + ∇ v q ∣ m 2 m + ∫ Ω ∣ ∇ v p − ∇ v q ∣ m 2 m ≤ 1 2 ∫ Ω ∣ ∇ v p ∣ m + 1 2 ∫ Ω ∣ ∇ v q ∣ m .

From (25) we deduce

lim p → ∞ ∫ Ω ∣ ∇ v p ∣ m ≤ ∣ Ω ∣ ,

then semicontinuity of L m -norm gives

limsup p , q ∫ Ω ∣ ∇ v p + ∇ v q ∣ m 2 m ≤ ∣ Ω ∣ = ∫ Ω ∣ ∇ d ( x , ∂ Ω ) ∣ m ≤ liminf p , q ∫ Ω ∣ ∇ v p + ∇ v q ∣ m 2 m ,

and then

limsup p , q ∫ Ω ∣ ∇ v p − ∇ v q ∣ m 2 m = 0 . □

Remark 4.2

If Supp f ⊂ Ω , then we can deduce that v ∞ ( x ) = 1 β + d ( x , ∂ Ω ) for all x ∈ Supp f , while in Ω ⧹ Supp f inequality (28) can be strict.

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 in Ω , then function v ¯ ∞ ( x ) = 1 β + d ( x , ∂ Ω ) is the unique extremal function of (26) if and only if ℛ ⊂ Supp f .

Proof

Suppose that ℛ ⊂ Supp f and let w be a minimum of (26) in the class { φ ∈ W 1 , ∞ ( Ω ) : ∥ ∇ φ ∥ ∞ ≤ 1 , β ∥ φ ∥ L ∞ ( ∂ Ω ) ≤ 1 } . By Remark 4.1, we have

w ( x ) ≤ 1 β + d ( x , ∂ Ω ) ∀ x ∈ Ω ,

and arguing as in Remark 4.2 we have

w ( x ) = 1 β + d ( x , ∂ Ω ) ∀ x ∈ supp f ¯ .

Assume by contradiction that there exists x ∈ ( Supp f ¯ ) c such that

w ( x ) < 1 β + d ( x , ∂ Ω ) ,

setting η = ∇ d ( x , ∂ Ω ) , we choose the smallest t such that y = x + t η belongs to ∂ ( Supp f ) (Lemma 4.6 will justify this choice), thus

∥ ∇ w ∥ L ∞ ( Ω ) ∣ y − x ∣ ≥ w ( y ) − w ( x ) > d ( y , ∂ Ω ) − d ( x , ∂ Ω ) = ∇ d ( ξ , ∂ Ω ) ⋅ ( y − x ) = ∣ y − x ∣ ,

where the last equality follows from Lemma 4.6.

So we have ∥ ∇ w ∥ L ∞ ( Ω ) > 1 , which 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 [2]), we can construct a function φ different from w , which coincides 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 ⊂ Supp f .□

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 + t η , then there exists T such that y T ∈ ℛ and y t ∉ ℛ for all t < T . Moreover,

∇ d ( x + t η , ∂ Ω ) = η ∀ t ∈ [ 0 , T ) .

Proof

Consider the following Cauchy problem:

(30) γ ˙ ( t ) = ∇ d ( γ ( t ) , ∂ Ω ) , γ ( 0 ) = x

in a maximal interval [ 0 , T ) . We have that

L ( γ ) = ∫ 0 T ∣ γ ˙ ( t ) ∣ d s = T ;
d d t d ( γ ( t ) , ∂ Ω ) = ∇ d ( γ ( t ) , ∂ Ω ) γ ˙ ( t ) = 1 , then
T = ∫ 0 T d d t d ( γ ( t ) , ∂ Ω ) d t = d ( γ ( T ) , ∂ Ω ) − d ( x , ∂ Ω ) .

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

d ( y , ∂ Ω ) = d ( x , ∂ Ω ) + T = d ( x , ∂ Ω ) + L ( γ ) ,

L ( γ ) ≥ ∣ y − x ∣ , and they are equal if and only if γ is a segment.

If L ( γ ) > ∣ y − x ∣ , then

d ( y , ∂ Ω ) = d ( x , ∂ Ω ) + L ( γ ) > d ( x , ∂ Ω ) + ∣ y − x ∣ ≥ ∣ y − z ∣

with z ∈ ∂ Ω such that d ( x , ∂ Ω ) = ∣ x − z ∣ , and this is a contradiction, because d ( y , ∂ Ω ) is the infimum. Then L ( γ ) = ∣ y − x ∣ and, remembering the fact γ ˙ ( t ) = ∇ d ( γ ( t ) , ∂ Ω ) , whose norm is 1, then γ is a segment and

∇ d ( γ ( t ) , ∂ Ω ) = η ∀ t ∈ [ 0 , T ) .

This concludes the proof.□

4.1 The limit partial differential equation (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).

Proposition 4.7

Let f ∈ L ∞ ( Ω ) ∩ C ( Ω ¯ ) be a non-negative function. Then any limit v ∞ of a subsequence { v p } satisfies

(31) ∣ ∇ v ∞ ∣ ≤ 1 in the viscosity sense .

Proof

The proof follows the techniques contained in [4,12,17,14].

If x 0 ∈ Ω , let φ ∈ C 2 ( Ω ) be such that v ∞ − φ has a local maximum at x 0

( v ∞ − φ ) ( x 0 ) ≥ ( v ∞ − φ ) ( x ) , ∀ x ∈ B R ( x 0 ) ,

then

(32) ∣ ∇ φ ( x 0 ) ∣ ≤ 1 .

Indeed, let

C = sup q > 1 max { ∥ v q ∥ L ∞ ( B R ( x 0 ) ) ; ∥ φ ∥ L ∞ ( B R ( x 0 ) ) } ,

we consider the following sequence:

f q ( x ) = v q ( x ) − φ ( x ) − k ∣ x − x 0 ∣ a , k = 4 C R a , a > 2 .

So f q ( x ) ≤ − 3 C for x ∈ ∂ B R ( x 0 ) , and f q ( x 0 ) ≥ − 2 C . Then f q ( x ) attains its maximum at some point x q in the interior of B R ( x 0 ) , and it holds

∇ v q ( x q ) = ∇ φ ( x q ) + k a ( x q − x 0 ) ∣ x q − x 0 ∣ a − 2 .

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

∣ ∇ v q ( x q ) ∣ > 1 + δ 2 − k a ∣ x q − x 0 ∣ a − 1 ≥ 1 + δ 4 , ∀ q ≥ q ¯ .

By Lemma 1.1 of Part III in [4],

(33) ∣ ∇ v q ( x ) ∣ ≤ γ R n 1 q ∫ B R 2 ( x 0 ) ( 1 + ∣ ∇ v q ∣ ) q d x 1 q , ∀ x ∈ B R 2 ( x 0 ) ,

where γ is a constant independent of q .

For q sufficiently large, this contradicts ∣ ∇ v q ( x q ) ∣ > 1 + δ 4 .□

Proposition 4.8

Let f ∈ L ∞ ( Ω ) ∩ C ( Ω ) be a non-negative function, then a continuous weak solution to (18) is a viscosity solution.

Proof

The proof follows the same techniques of Proposition 3.2.□

Theorem 4.9

Let f ∈ L ∞ ( Ω ) ∩ C ( Ω ) be a non-negative function. Then any v ∞ satisfies

(34) ∣ ∇ v ∞ ∣ = 1 on { f > 0 } in the viscosity sense

(35) − Δ ∞ v ∞ = 0 on ( { f > 0 } ¯ ) c in the viscosity sense

Proof

We first prove (34). Let x 0 ∈ Ω ∩ { f > 0 } and let φ ∈ C 2 ( Ω ) such that v ∞ − φ has a strict minimum in x 0 . We want to show

∣ ∇ φ ( x 0 ) ∣ ≥ 1 .

Let us denote by x p the minimum of v p − φ , we have that x p → x 0 , then x p ∈ B R ( x 0 ) ⊂ { f > 0 } for p large enough. Setting φ p ( x ) = φ ( x ) + c p , then c p = v p ( x p ) − φ ( x p ) → 0 when p goes to infinity. We note that v p ( x p ) = φ p ( x p ) and v p − φ p has a minimum in x p , so by Proposition 4.8,

− ∣ ∇ φ p ( x p ) ∣ p − 2 Δ φ p ( x p ) − ( p − 2 ) ∣ ∇ φ p ( x p ) ∣ p − 4 Δ ∞ φ ( x p ) ≥ f ( x p ) > 0 .

Dividing by ( p − 2 ) ∣ ∇ φ p ( x p ) ∣ p − 4 , we obtain

(36) − Δ ∞ φ p ( x p ) − ∣ ∇ φ p ( x p ) ∣ 2 Δ φ p ( x p ) p − 2 ≥ f ( x p ) ( p − 2 ) ∣ ∇ φ p ( x p ) ∣ p − 4 .

This gives us ∣ ∇ φ ( x 0 ) ∣ ≥ 1 , otherwise the right-hand side would go to infinity, in contradiction with the fact that φ ∈ C 2 ( Ω ) .

We stress that, if we let p → ∞ in (36), we obtain − Δ ∞ φ ( x 0 ) ≥ 0 , so v ∞ is a supersolution to − Δ ∞ u = 0 .

Now, we only have to prove that v ∞ is a viscosity solution to Δ ∞ 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.

The definition of viscosity subsolution implies

− ∣ ∇ φ p ( x p ) ∣ p − 2 Δ φ p ( x p ) − ( p − 2 ) ∣ ∇ φ p ( x p ) ∣ p − 4 Δ ∞ φ ( x p ) ≤ f ( x p ) = 0 .

Without loss of generality, we may assume ∣ ∇ φ ( x 0 ) ∣ ≠ 0 . Dividing both sides of the last equation by ( p − 2 ) ∣ ∇ φ p ( x p ) ∣ p − 4 , we obtain

− Δ ∞ φ p ( x p ) ≤ ∣ ∇ φ p ( x p ) ∣ 2 Δ φ p ( x p ) p − 2 .

Letting p → ∞ , we obtain

− Δ ∞ φ ( x 0 ) ≤ 0 .

Analogously, if φ ∈ C 2 ( Ω ) is such that v ∞ − φ has a minimum at x 0 , a symmetric argument shows that − Δ ∞ φ ( x 0 ) ≥ 0 .□

Some examples

Example 4.3

We consider the case Ω = B 1 ( 0 ) and f = 1 . The uniqueness of the solution and the invariance under rotation of the p-Laplacian ensure that the solution v is radially symmetric and

r n − 1 Δ p v p = d d r ( r n − 1 ∣ v p ′ ∣ p − 2 v p ′ ) .

Setting α = 1 / ( p − 1 ) , we have

v p ( x ) = − p − 1 n α p ∣ x ∣ p p − 1 + 1 ( n β p ) α + p − 1 n α p .

Then

v ∞ ( x ) = − ∣ x ∣ + 1 β + 1 = 1 β + d ( x , ∂ Ω ) .

Example 4.4

We fix 0 < ε < 1 and we consider

f = 1 if x ∈ B ε ( 0 ) 0 if x ∈ B 1 ( 0 ) ⧹ B ε ( 0 ) .

In this case, v p is radially symmetric, and

v p = p − 1 n α p ε p p − 1 − ∣ x ∣ p p − 1 + ε n α ( p − 1 ) n α ( p − n ) 1 − ε p − n p − 1 + ε n α ( n β p ) α if x ∈ B ε ( 0 ) ε n α ( p − 1 ) n α ( p − n ) 1 − ∣ x ∣ p − n p − 1 + ε n α ( n β p ) α if x ∈ B 1 ( 0 ) ⧹ B ε ( 0 ) .

Letting p go to infinity, we obtain

v ∞ = 1 β + d ( x , ∂ Ω ) .

Conflict of interest: The authors state no conflict of interest.

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Received: 2021-11-19

Revised: 2022-03-21

Accepted: 2022-05-09

Published Online: 2022-07-29

This work is licensed under the Creative Commons Attribution 4.0 International License.

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

Abstract

1 Introduction

2 Notations and preliminaries

Lemma 2.1

Proof

2.1 Viscosity solutions

Definition 2.1

Remark 2.1

3 The ∞ -eigenvalue problem

Lemma 3.1

Proof

Proposition 3.2

Proof

Theorem 3.3

Proof

Lemma 3.4

Proof

Theorem 3.5

Proof

Remark 3.1

3.1 The first Robin ∞ -eigenvalue

Theorem 3.6

Proof

4 The p-Poisson equation

Proposition 4.1

Proof

Theorem 4.2

Proof

Proposition 4.3

Proof

Remark 4.1

Proposition 4.4

Proof

Remark 4.2

Definition 4.1

Theorem 4.5

Proof

Lemma 4.6

Proof

4.1 The limit partial differential equation (PDE)

Proposition 4.7

Proof

Proposition 4.8

Proof

Theorem 4.9

Proof

Some examples

Example 4.3

Example 4.4

References

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