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Boundary properties of fractional objects: Flexibility of linear equations and rigidity of minimal graphs

  • Serena Dipierro EMAIL logo , Ovidiu Savin and Enrico Valdinoci

Abstract

The main goal of this article is to understand the trace properties of nonlocal minimal graphs in 3, i.e. nonlocal minimal surfaces with a graphical structure.

We establish that at any boundary points at which the trace from inside happens to coincide with the exterior datum, also the tangent planes of the traces necessarily coincide with those of the exterior datum.

This very rigid geometric constraint is in sharp contrast with the case of the solutions of the linear equations driven by the fractional Laplacian, since we also show that, in this case, the fractional normal derivative can be prescribed arbitrarily, up to a small error.

We remark that, at a formal level, the linearization of the trace of a nonlocal minimal graph is given by the fractional normal derivative of a fractional Laplace problem, therefore the two problems are formally related. Nevertheless, the nonlinear equations of fractional mean curvature type present very specific properties which are strikingly different from those of other problems of fractional type which are apparently similar, but diverse in structure, and the nonlinear case given by the nonlocal minimal graphs turns out to be significantly more rigid than its linear counterpart.

Funding statement: The first and third authors are member of INdAM and are supported by the Australian Research Council Discovery Project DP170104880 NEW “Nonlocal Equations at Work”. The first author’s visit to Columbia has been partially funded by the Fulbright Foundation and the Australian Research Council DECRA DE180100957 “PDEs, free boundaries and applications”. The second author is supported by the National Science Foundation grant DMS-1500438. The third author’s visit to Columbia has been partially funded by the Italian Piano di Sostegno alla Ricerca “Equazioni nonlocali e applicazioni”.

A A classical counterpart of Theorem 1.1

In this appendix, we show that Theorem 1.1 possesses a classical counterpart for the Laplace equation, namely:

Theorem A.1.

Let n2, kN and fCk(B1¯). Then, for every ε>0 there exist fεCk(Rn-1) and uεC(Rn) such that

{Δuε=0in B1{xn>0},uε=0in B1{xn=0},
limxn0uε(x,xn)xn=fε(x)for all xB1,

and

fε-fCk(B1¯)ε.

We observe that formally Theorem A.1 corresponds to Theorem 1.1 with σ:=1. The arguments that we proposed for σ(0,1) can be carried out also when σ=1, and thus prove Theorem A.1. Indeed, for the classical Laplace equation, one can obtain Lemma 2.1 by taking, for instance, the real part of the holomorphic function zzk, with k, and then use this homogeneous solution in the proof of Lemma 2.2. Then, once Lemma 2.2 is proved with σ:=1, one can exploit the proof of Theorem 1.1 presented in Section 2 with σ:=1 and obtain Theorem A.1.

However, in the classical case there is also an explicit polynomial expansion which recovers Lemma 2.2 with σ:=1, that is when (2.3) is replaced by

{Δu=0in Br{xn>0},u=0in Br{xn=0}.

Hence, to establish Theorem A.1, we focus on the following argument of classical flavor (which is not reproducible for σ(0,1) and thus requires new strategies in the case of Theorem 1.1).

Proof of Lemma 2.2 with σ:=1.

Given

κ=(κγ)γn-1|γ|k

in Nk, where Nk is defined in (2.5), we aim at finding a function uC(B1¯) such that

(A.1){Δu=0in B1{xn>0},u=0in B1{xn=0},

and for which there exists ϕCk(B1¯) such that

(A.2)limxn0u(x,xn)xn=ϕ(x)for all xB1,

with, in the notation of (2.4),

(A.3)𝒟kϕ(0)=κ.

To this end, we define ϕ to be the polynomial

ϕ(x):=γn-1|γ|kκγγ!(x)γ.

In this way, (A.3) is automatically satisfied. Furthermore, we observe that Δxjϕ vanishes identically when 2j>k, and therefore we can set

u(x,xn):=jjk/2(-1)jΔxjϕ(x)(2j+1)!xn2j+1=j=0+(-1)jΔxjϕ(x)(2j+1)!xn2j+1.

We observe that

limxn0u(x,xn)xn=limxn0jjk/2(-1)jΔxjϕ(x)(2j+1)!xn2j=ϕ(x),

which establishes (A.2).

Also, we see that u(x,0)=0, and moreover, using the notation J:=j-1,

Δu(x)=Δxu(x)+xn2u(x)
=j=0+(-1)jΔxj+1ϕ(x)(2j+1)!xn2j+1+j=1+(-1)jΔxjϕ(x)(2j-1)!xn2j-1
=j=0+(-1)jΔxj+1ϕ(x)(2j+1)!xn2j+1+J=0+(-1)J+1ΔxJ+1ϕ(x)(2J+1)!xn2J+1
=0,

which gives (A.1). ∎

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Received: 2019-07-03
Revised: 2019-12-15
Published Online: 2020-04-16
Published in Print: 2020-12-01

© 2020 Walter de Gruyter GmbH, Berlin/Boston

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