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non-smooth domains
Konstantinos Gkikas, Laurent Véron
To cite this version:
Konstantinos Gkikas, Laurent Véron. The spherical p-harmonic eigenvalue problem in non-smooth
domains. Journal of Functional Analysis, Elsevier, 2018, 274, pp.1155-1176. �hal-01501604�
in non-smooth domains
Konstantinos Gkikas
∗Laurent Véron
†Abstract
We prove the existence of p-harmonic functions under the formu(r, σ) =r−βω(σ)in any coneCS generated by a spherical domainSand vanishing on∂CS. We prove the uniqueness of the exponentβ and of the normalized functionωunder a Lipschitz condition onS.
2010 Mathematics Subject Classification.35J72; 35J92.
Key words.p-Laplacian operator; polar sets; Harnack inequality; boundary Harnack inequality;p-Martin boundary.
Contents
1 Introduction 1
2 Existence 2
2.1 Estimates . . . 2
2.2 Approximations from inside . . . 5
2.3 Approximations from outside . . . 9
3 Uniqueness 10 3.1 Uniqueness of exponentβ . . . 10
3.2 Uniqueness of eigenfunction . . . 11
3.2.1 The convex case . . . 11
3.2.2 Proof of Theorem E . . . 12
1 Introduction
Letp >1,Sa domain of the unit sphereSN−1ofRN andCS :={(r, σ) :r > 0, σ∈S}the positive cone generated byS. If one looks forp-harmonic functions inCS under the formu(x) = u(r, σ) = r−βω(σ)vanishing on∂CS\ {0}, thenωsatisfies thesphericalp-harmonic eigenvalue problemonS
−div0
β2ω2+|∇0ω|2p−22
∇0ω
= (p−1)β(β−β0)
β2ω2+|∇0ω|2p−22
ω inS
ω= 0 in∂S
(1.1)
1
withβ0 = Np−1−p and werediv0 and∇0 denote the divergence operator and the covariant gradient on SN−1endowed with the metric induced by its isometric inbedding intoRN. Separable solutions play a key role for describing the boundary behaviour and the singularities of solutions of a large variety of quasilinear equations. WhenN = 2the equation is completely integrable and has been solved by Kroll in the regular caseβ <0and Kichenassamy and Véron in the the singular caseβ >0. In higher dimension, Tolksdorff [15] proved the following:
Theorem AIfS is a smooth spherical domain, there exist two couples(βS, ωS)and (βS0, ωS0)where βS >0andβS0 <0,ωS andωS0 are positiveC2(S)-functions vanishing on∂Swhich solve(1.1)with (β, ω) = (βS, ωS) or(β, ω) = (βS0, ω0S). FurthermoreβS andβS0 are unique, andωS and ωS0 are unique up to an homothety.
A more general and transparent proof has been obtained by Porretta and Véron [13], but always in the case of a smooth spherical domain. The aim of this article is to extend Theorem A to a general spherical domain. If we consider an increasing sequence of smooth domains{Sk}such thatSk ⊂Sk ⊂Sk+1
and∪kSk=Swe prove the following:
Theorem BAssume thatScis not polar. Then the sequence of theβSk>0from Theorem A is decreasing and converges toβS >0. There existsωS ∈W01,p(S)∩L∞(S)weak solution of(1.1)withβ =βS. FurthermoreβS>0is the largest exponentβsuch that(1.1)admits a positive solutionωS ∈W01,p(S).
Under a mild assumption onSit is possible to approximate it by a decreasing sequence of smooth domainsSk0 such thatSk0 ⊂S0k ⊂Sk−10 and∩kSk0 =S
Theorem CAssume thatS =
o
S. Then the sequenceβS0
k > 0is increasing and converges toβˆS >0 and there existsωˆS ∈W01,p(S)∩L∞(S)weak solution of(1.1)withβ = ˆβS. FurthermoreβˆS is the smallest exponentβsuch that(1.1)admits a positive solutionωS∈W01,p(S).
We prove the uniqueness of the exponentβ, under a Lipschitz assumption onS.
Theorem D Assume thatS is a Lipschitz domain, then βS = ˆβS and if ω and ω0 are two positive solutions of(1.1)inW01,p(S), there exists a constantc >0such thatc−1ω0 ≤ω≤cω0.
The proof of Theorem C is based upon a sharp form of boundary Harnack inequality proved in [10],
lnωω(σ0(σ11))−lnωω(σ0(σ22))
≤c1|σ1−σ2|α ∀σ1, σ2∈S, (1.2) for somec1 =c1(N, p, S)>0andα∈(0,1). Actually we have a stronger result, much more delicate to obtain.
Theorem ELetSbe a Lipschitz subdomain ofSN−1. Then two positive solutions of(1.1)inW01,p(S) are proportional.
The proof is based upon a non trivial adaptation of a series of deep results of Lewis and Nyström [10] concerning thep-Martin boundary of domains.
AcknowledgementsThis article has been prepared with the support of the collaboration programs ECOS C14E08.
2 Existence
2.1 Estimates
Through this article we assume thatScis not polar, or equivalently that it has positivecS1,pN−1-capacity.
Lemma 2.1. Assumep >1. Then any solutionω∈W01,p(S)of(1.1)satisfies
kωkCγ(S)≤c1kωkLp(S), (2.1)
ifp > N−1whereγ= 1−Np−1 ifp > N−1and
kωkL∞(S)≤c1kωkLp(S), (2.2)
if1< p≤N−1, wherec1>0depends onp,N,β.
Proof. Multiplying the equation byωand using Hölder’s inequality, we derive (i)
Z
S
β2ω2+|∇0ω|2p2
dS≤(β(pβ−(p−1)β0))p2 Z
S
|ω|pdS if p≥2, (ii)
Z
S
β2ω2+|∇0ω|2p2
dS≤βp−1(pβ−(p−1)β0) Z
S
|ω|pdS if 1< p <2.
(2.3)
Notice that these inequalities hold for allp >1. Ifp > N−1 (2.1)follows by Morrey’inequality. Here after we assume1< p≤N−1. Letα≥1andk >0. Thenζ = min{|ω|, k}α−1ωis an admissible test function, hence
1- Ifp≥2, Z
S
β2ω2+|∇0ω|2p−22
h∇0ω.∇0ζidS= (p−1)β(β−β0) Z
S
β2ω2+|∇0ω|2p−22 ωζdS
≤c2
Z
S
|∇0ω|p−2ω2min{|ω|, k}α−1dS+c2βp Z
S
|ω|pmin{|ω|, k}α−1dS
≤c2 Z
S
|ω|pmin{|ω|, k}α−1dS
p−2p Z
S
|∇0ω|pmin{|ω|, k}α−1dS p2
+c2βp Z
S
|ω|pmin{|ω|, k}α−1dS,
(2.4)
wherec2=c2(N, p, β)>0. Since Z
S
β2ω2+|∇0ω|2p−22
h∇0ω.∇0ζidS≥c3(p) Z
S
|∇0ω|pmin{|ω|, k}α−1dS, it implies that there existsc4=c4(N, p, β)such that
Z
S
|∇0ω|pmin{|ω|, k}α−1dS≤c4 Z
S
|ω|pmin{|ω|, k}α−1dS, (2.5) which yields
Z
S
|∇0j(ω)|pdS≤c4
Z
S
|j(ω)|pdS, (2.6)
wherej(ω) = min{|ω|, k}α−1p ω.
2- If1< p <2, then Z
S
β2ω2+|∇0ω|2p−22
h∇0ω.∇0ζidS= Z
S
β2ω2+|∇0ω|2p−22
|∇0ω|2min{|ω|, k}α−1dS
+ (α−1) Z
S∩{|ω|<k}
β2ω2+|∇0ω|2p−22
|∇0ω|2|ω|α−1dS.
(2.7)
Since Z
S
β2ω2+|∇0ω|2p−22
|∇0ω|2min{|ω|, k}α−1dS= Z
S
β2ω2+|∇0ω|2p2
min{|ω|, k}α−1dS
−β2 Z
S
β2ω2+|∇0ω|2p−22
min{|ω|, k}α−1ω2dS
≥ Z
S
|∇0ω|pmin{|ω|, k}α−1dS−β2 Z
S
β2ω2+|∇0ω|2p−22
min{|ω|, k}α−1ω2dS, we derive
Z
S
|∇0ω|pmin{|ω|, k}α−1dS≤βp−1(pβ−(p−1)β0) Z
S
|ω|pmin{|ω|, k}α−1dS, (2.8) which leads to(2.6). Lettingk→ ∞we infer by Fatou’s lemma,
Z
S
∇0|ω|α−1p +1
p
dS≤c4 Z
S
|ω|α−1+pdS. (2.9)
Ifp < N−1we derive from Sobolev inequality and puttingq=α−1 +pands= N−1−pN−1 >1 Z
S
|ω|sqdS 1s
≤c5 Z
S
|ω|qdS, (2.10)
andc5>0depends onN,pandβ. Iterating this estimate by Moser’s method we derive(2.10).
Ifp=N−1we have for1≤m < p−1andm∗=m(N−1)N−1−m c6
Z
S
|ω|(α−1p +1)m∗dS mpm∗
≤ Z
S
∇0|ω|α−1p +1
m
dS mp
≤ |S|mp−1c4
Z
S
|ω|α−1+pdS, andc6=c6(N, p), hence
Z
S
|ω|tqdS 1t
≤c5
Z
S
|ω|qdS, (2.11)
witht=p(Nm(N−1−m)−1) =N−1−mm . The proof follows again by Moser’s iterative scheme.
Proposition 2.2. LetS1andS2be two subdomains ofSN−1such thatS1⊂S1⊂S2andS2not polar.
Letβj >0, j=1,2, such that there exist positive solutionsωj∈W01,p(Sj)solutions of
−div0
βj2ω2j+|∇0ωj|2p−22
∇0ωj
= (p−1)βj(βj−β0)
βj2ωj2+|∇0ωj|2p−22
ωj inSj
ωj= 0 in∂Sj.
(2.12) Thenβ1≥β2.
Proof. Setuj(r, σ) =r−βjωj(σ),CSj = (0,∞)×SJ and assumeβ1 < β2. By Harnack inequality ω2≥c >0onS1, thus
u2(r, σ)≥cr−β2 a.e. inCS1. For >0there existr>0such that
u2(x)≥u1(x) ∀x∈CS1∩Br.
Letδ >0, there existsRδ >0such that
u1(x)≤δ ∀x∈CS1∩BRcδ. Henceζ = (u1−u2−δ)+ ∈W01,p(QrS,Rδ
1 ), whereQrS,Rδ
1 = {x ∈ CS1 : r <|x| < Rδ}. This implies
0 = Z
Qr,RδS
1
|∇u1|p−2∇u1− |∇(u1)|p−2∇(u1).∇ζ dx
= Z
Qr,RδS
1 ∩{u1−u2≥δ}
|∇u1|p−2∇u1− |∇(u1)|p−2∇(u1).∇(u1−u2) dx.
Therefore∇(u1−u2−δ)+= 0a.e. inQrS,Rδ
1 , which leads tou1−u2≤δin the same set. Letting δ →0 yieldsRδ → ∞, thus we obtainu1 ≤u2inCS1 ∩Bcr henceu1 ≤ 0inCS1, contradiction.
2.2 Approximations from inside
Proof of Theorem B. Let{Sk}be an increasing sequence of smooth domains such thatSk⊂Sk ⊂Sk+1. We denote by{(βSk, ωk)}the corresponding sequence of solutions of(1.1)withβ=βSkandω=ωk. The sequence{βSk}is uniquely determined by [15], it admits a limitβ:=βS, and theωkare the unique positive solutions such that
Z
Sk
|ωk|dS= 1.
Ifp≥2, we have Z
Sk
|∇0ωk|pdS≤ Z
Sk
β2Skω2k+|∇0ωk|2p−22
|∇0ωk|2dS
= (p−1)βSk(βSk−β0) Z
Sk
β2S
kω2k+|∇0ωk|2p−22 ωk2dS
≤2
(p−4)+
2 (p−1)βSk(βSk−β0) Z
Sk
βSp−2
k ωkp+|∇0ωk|p−2ωk2 dS
≤c7(N, p, βSk) Z
Sk
ωpkdS+1 2 Z
Sk
|∇0ωk|pdS.
SinceβSk≤β1, we derive
Z
Sk
|∇0ωk|pdS≤c8, (2.13)
from the normalization assumption withc8= 2c7(N, p, β1).
If1< p <2, we have Z
Sk
|∇0ωk|pdS≤ Z
Sk
βS2
kωk2+|∇0ωk|2p2 dS
≤βSk(pβSk+ (p−1)β0) Z
Sk
β2S
kω2k+|∇0ωk|2p−22 ωk2dS
≤βkp−1(pβSk+ (p−1)β0) Z
Sk
ωkpdS,
and we obtain(2.13)withc8=β1p−1(pβ1+ (p−1)β0).
Next we extend ωk by 0 inSkc. Then there existsω ∈ W01,p(S) such that ωk * ω weakly in W01,p(S), up to subsequence that we still denote{ωk}, andωk→ωinLp(S).
Step 1: We claim that∇0ωkconverges to∇0ωlocally inLp(S).
Leta∈Sandr >0such thatB4r(a)⊂S. Then fork≥k0,B2r(a)⊂Sk. Letζ∈C0∞(B2r(a))such that0≤ζ≤1,ζ= 1inBr(a). For test function we chooseηk =ζ(ω−ωk), then
Z
Sk
β2Skω2k+|∇0ωk|2p−22
h∇0ωk.∇0ηkidS= (p−1)βSk(βSk−β0) Z
Sk
βS2kω2k+|∇0ωk|2p−22
ωkηkdS.
By the above inequality, we have Z
B2r(a)
β2ω2+|∇0ω|2p−22
∇0ω− βS2
kω2k+|∇0ωk|2p−22
∇0ωk.∇0ηk
dS
= Z
B2r(a)
β2ω2+|∇0ω|2p−22
h∇0ω.∇0ηkidS
−(p−1)βSk(βSk−β0) Z
Sk
βS2
kωk2+|∇0ωk|2p−22
ωkηkdS.
Using the weak convergence of the gradient, we have
k→∞lim Z
B2r(a)
β2ω2+|∇0ω|2p−22
h∇0ω.∇0ηkidS= 0.
Sinceωkis uniformly bounded inW01,p(S)andωk→ωinLp(S), we have
k→∞lim Z
B2r(a)
βS2
kωk2+|∇0ωk|2p−22
ωkηkdS= 0, and
k→∞lim Z
B2r(a)
(ω−ωk)
β2ω2+|∇0ω|2p−22
∇0ω−
βS2kωk2+|∇0ωk|2p−22
∇0ωk.∇0ζ
dS= 0.
Combining the above relations we infer
k→∞lim Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22
∇0ω−
β2Skω2k+|∇0ωk|2p−22
∇0ωk.∇0(ω−ωk)
dS= 0.
(2.14) Next we write
Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22
∇0ω− βS2
kωk2+|∇0ωk|2p−22
∇0ωk.∇0(ω−ωk)
dS
=1 2 Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22 +
βS2
kωk2+|∇0ωk|2p−22
|∇0(ω−ωk)|2dS
+1 2 Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22
−
βS2kωk2+|∇0ωk|2p−22
×
|∇0ω|2+β2ω2−β2S
kω2k− |∇0ωk|2 dS
−1 2 Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22
− βS2
kωk2+|∇0ωk|2p−22
β2ω2−β2S
kωk2 dS.
(2.15)
Ifp≥2, we have from(2.4), Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22
∇0ω− βS2
kωk2+|∇0ωk|2p−22
∇0ωk.∇0(ω−ωk)
dS
≥1 2 Z
B2r(a)
ζ
|∇0ω|p−2+|∇0ωk|p−2
|∇0(ω−ωk)|2dS
−1 2 Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22
− βS2
kωk2+|∇0ωk|2p−22
β2ω2−β2S
kωk2 dS
≥min{2−1,22−p} Z
B2r(a)
ζ|∇0(ω−ωk)|pdS
−1 2 Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22
− βS2
kωk2+|∇0ωk|2p−22
β2ω2−β2S
kωk2 dS.
(2.16) Sinceωk →ωinLp(S),βSk→βandωk, ωare uniformly bounded inW01,p(S), we derive
Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22
− βS2
kω2k+|∇0ωk|2p−22
β2ω2−βS2
kωk2
dS→0 ask→ ∞. Jointly with(2.14)we infer that
k→∞lim Z
Br(a)
|∇0(ω−ωk)|pdS= 0. (2.17)
If1< p <2, then Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22
∇0ω− β2S
kω2k+|∇0ωk|2p−22
∇0ωk.∇0(ω−ωk)
dS
= Z
B2r(a)
ζ
β2S
kω2k+|∇0ω|2p−22
∇0ω− βS2
kωk2+|∇0ωk|2p−22
∇0ωk.∇0(ω−ωk)
dS
+ Z
B2r(a)
ζ
β2ω2+|∇0ω|2p−22
− βS2
kωk2+|∇0ω|2p−22
∇0ω.∇0(ω−ωk)
dS.
(2.18) Up to extracting a subsequence, we have thatωk →ω a.e. inS and that there existsΦ∈L1(S)such that
|ωk|p+|ω|p≤Φ a.e. inS and ∀k≥1. (2.19) Since
β2Skω2k+|∇0ω|2p−22
|∇ω| ≤
βS2kω2k+|∇0ω|2p−12
≤βSp−1
k ωp−1k +|∇0ω|p−1, and
β2ω2+|∇0ω|2p−22
|∇ω| ≤βp−1ωp−1+|∇0ω|p−1, we derive that
β2ω2+|∇0ω|2p−22
−
βS2kω2k+|∇0ω|2p−22
|∇0ω| ≤2
βp−1Φp−1+|∇0ω|p−1 , which implies that
ζ
β2ω2+|∇0ω|2p−22
−
βS2kωk2+|∇0ω|2p−22
∇0ω→0 inLp0(S)
wherep0is the conjugate ofp, and finally Z
B2r(a)
ζh
β2ω2+|∇0ω|2p−22
− β2S
kω2k+|∇0ω|2p−22
∇0ω.∇0(ω−ωk)idS→0 ask→ ∞.
(2.20) For the last term on the right-hand side of(2.18), we have, forγ∈R+andA,B∈RN,
γ+|B|2p−22 B−
γ+|A|2p−22 A=
Z 1 0
d dt
γ+|tB+ (1−t)A|2p−22
(tB+ (1−t)A)
dt
= Z 1
0
γ+|tB+ (1−t)A|2p−22 dt
(B−A)
+ (p−2) Z 1
0
γ+|tB+ (1−t)A|2p−42
htB+ (1−t)A.B−Ai(tB+ (1−t)A)dt.
This implies h
γ+|B|2p−22 B−
γ+|A|2p−22
A.B−Ai= Z 1
0
γ+|tB+ (1−t)A|2p−22 dt
|B−A|2
+ (p−2) Z 1
0
γ+|tB+ (1−t)A|2p−42
htB+ (1−t)A.B−Ai2dt.
We observe that Z 1
0
γ+|tB+ (1−t)A|2p−42
htB+ (1−t)A.B−Ai2dt
≤ |B−A|2 Z 1
0
γ+|tB+ (1−t)A|2p−22 dt, and since1< p <2, we finally obtain
h
γ+|B|2p−22 B−
γ+|A|2p−22
A.B−Ai
≥(p−1) Z 1
0
γ+|tB+ (1−t)A|2p−22 dt
|B−A|2
≥(p−1)|B−A|2
γ+ 1 +|B|2+|A|2p−22 .
(2.21)
We plug this estimate into(2.18)withγ=β2kωk2,A=∇0ωandB=∇0ωk, then Z
B2r(a)
ζh β2S
kω2k+|∇0ω|2p−22
∇0ω− βS2
kωk2+|∇0ωk|2p−22
∇0ωk.∇0(ω−ωk)idS
≥ Z
B2r(a)
ζ|∇0(ω−ωk)|2
βk2ω2k+ 1 +|∇0ωk|2+|∇0ω|2p−22 dS.
(2.22)
Setφ(.) =βk2ω2k+ 1 +|∇0ωk|2+|∇0ω|2, then Z
Br(a)
|∇0ω− ∇0ωk|pdS= Z
Br(a)
|∇0ω− ∇0ωk|pφp(p−2)4 φ−p(p−2)4 dS
≤ Z
Br(a)
|∇0ω− ∇0ωk|2φp−22 dS
!p2 Z
Br(a)
φp2dS
!2−p2 .
(2.23)
Jointly with(2.14)and(2.22)we conclude that(2.17). Step 1 follows by a standard covering argument.
Step 2: We claim thatωkconverges toωinW01,p(S).
Up to a subsequence that we denote again by{k}, we can assume thatωk →ωand∇0ωk → ∇0ωa.e.
inS. Letζ ∈C0∞(S), then there existsk∈Nsuch that the supportKofζis a compact subset ofSk
for allk≥k. If1< p <2,
βS2kωk2+|∇0ωk|2p−22
|∇0ωk| ≤ |∇0ωk|p−1,
which bounded inLp0(K), then uniformly integrable inKand by Vitali’s convergence theorem βS2
kω2k+|∇0ωk|2p−22
∇0ωk →
β2ω2+|∇0ω|2p−22
∇0ω, inL1loc(S). Similarly
β2Skωk2+|∇0ωk|2p−22
ωk →
β2ω2+|∇0ω|2p−22 ω, inL1loc(S). Ifp≥2
βS2kωk2+|∇0ωk|2p−22
|∇0ωk| ≤c
|ωk|p−1+|∇0ωk|p−1 ,
and we conclude again by Vitali’s convergence theorem that the previous convergences hold. Since Z
Sk
β2S
kω2k+|∇0ωk|2p−22
h∇0ωk.∇0ζidS= (p−1)βSk(βSk−β0) Z
Sk
β2S
kω2k+|∇0ωk|2p−22 ωkζdS
we conclude thatωis a weak solution of(1.1)withβ=βS.
2.3 Approximations from outside
Proof of Theorem C. SinceSchas a non-empty interior, the existence of a sequence{ω0k}corresponding to solutions of(1.1)inSk0 withβ =βSk0 is the consequence of [13]. The fact that{βSk0}is increasing follows from Proposition 2.2. We denote byβˆ:= ˆβSits limit, and it is smaller or equal toβS. Estimates (2.4)are valid withS0k,ωk0 andβSk0 instead ofS,ωandβ. If we extendω0kby0inSk0cthese estimates are valid with SN−1 instead ofS0k. Then up to a subsequence the existsω ∈ W1,p(SN−1) and a subsequence stil denoted by{k}such thatωk0 * ωweakly inW1,p(SN−1), strongly inLp(SN−1)and a.e. inSN−1. Furthermore, as in the proof of Theorem A, for any compact setK ⊂S,∇0ωk0 → ∇0ω0 inLp(K). This is sufficient to assert thatωis a weak solution of
−div0
βˆ2ω02+|∇0ω0|2p−22
∇0ω0
= (p−1) ˆβ( ˆβ−β0)
βˆ2ω2+|∇0ω0|2p−22
ω0 inS.
Moreoverω0bS0
k belongs toW01,p(Sk0)for allk. Sinceωk0 = 0inSkc and converges a.e. toω, this last function vanishes a.e. in∪kSkc = (∩kSk)c =Sc. Thereforeωvanishes a.e. inScand since it is quasi continuous, it vanishes,(1−p)- quasi everywhere inSc. From Netrusov’s theorem (see [1, Th 10.1.1]- (iii)) there exists a sequence{ηn} ⊂ C0∞(S)which converges toω inW1,p(S), thusω ∈ W01,p(S).
3 Uniqueness
3.1 Uniqueness of exponent β
Proof of Theorem D. IfS is Lipschitz,CS is also Lipschitz. We fixz ∈ S ≈ SN−1∩∂CS and we apply [10, Th 2] inGz =CS ∩B1
2(z)to two separablep-harmonic functionsu(r, σ) = r−βω(σ)and u0(r, σ) =r−β0ω0(σ). There existγ∈(0,12),c10>0andα∈(0,1)such that
ln u(y1)
u0(y1)−ln u(y2) u0(y2)
≤c10|y1−y2|α ∀y1, y2∈CS∩Bγ(z). (3.24) Assume|y1|=|y2|= 1, then
ln ω(y1)
ω0(y1)−ln ω(y2) ω0(y2)
≤c10|y1−y2|α ∀y1, y2∈S∩Bγ(z). (3.25) We denote by`(x, y)the geodesic distance onSN−1and by`(x, K)the geodesic distance from a point x∈SN−1to a subsetK. Since the setSγ ={σ∈S :`(σ, ∂S)≤γ2}can be covered by a finite number of balls with center on∂S, we infer that
ln ω(y1)
ω0(y1)−ln ω(y2) ω0(y2)
≤c11 ∀y1, y2∈Sγ. (3.26) InS\Sγ
2
we can use Harnack inequality to obtain
−c12≤lnω(y1)
ω(y2) ≤c12 ∀y1, y2∈S\Sγ
2
s.t.`(y1, y2)≤ γ2. (3.27) Hence there exists a constantc13>0such that(3.27)holds for anyy1, y2∈S\Sγ
2
, withc12replaced byc13. Furthermore ω0 satisfies the same inequality in S\Sγ
2
. Combining the two inequalities we obtain
−2c13≤lnω(y1)
ω(y2)−lnω0(y1)
ω0(y2)≤2c13 ∀y1, y2∈S\Sγ
2
. (3.28)
Combining this estimate with(3.25)we derive that it holds for ally1, y2∈S. This implies e−2c13 ω(y2)
ω0(y2) ≤ ω(y1)
ω0(y1) ≤e2c13ω(y2)
ω0(y2) ∀y1, y2∈S. (3.29) Assume now that there exist two exponentsβ > β0 > 0 such thatr−βω(.)andr−β0ω0(.)arep- harmonic and positive in the coneCSand vanishes on∂CS. Putθ=ββ0,η=ω0θand
T(η) =−div0
β2η2+|∇0η|2p−22
∇0η
−(p−1)β(β−β0)
β2η2+|∇0η|2p−22 η, then
T(η) =−θp−2
β02ω02+|∇0ω0|2p−22
(β−β0)ω02+ (p−1)θ(θ−1)|∇0ω0|2
≤0.
Up to multiplyingω0 byλ, we can assume thatη ≤ ωand that the graphs of η andω are tangent in S. Sinceω0 ≤ cω,η = o(ω)near∂S. Hence there existsσ0 ∈ S such thatω(σ0) = η(σ0)and the
coincidence set ofηandωis a compact subset ofS. We putw=ω−η, since∇ω(σ0) =∇η(σ0)we proceed as in [14, Th 4.1] (see also [4] in the flat case) and derive thatwsatisfies, in a system of local coordinates(σ1, ..., σN−1)nearσ0,
Lw:=−X
`,j
∂
∂σ`
Aj,`
∂w
∂σj
+X
j
Cj
∂w
∂σ`+Cw≥0,
where the matrix(Aj,`)is smooth, symmetric and positive nearσ0and theCjandCare bounded. Hence wis locally zero. By a standard argument of connectedness, this implies that the zero set ofwmust be
empty, contradiction. Henceβ=β0.
3.2 Uniqueness of eigenfunction
The proof is based upon a delicate adaptation of the characterisation of thep-Martin boundary obtained in [10], but we first give a proof in the convex case.
3.2.1 The convex case
Theorem 3.1. AssumeS is a convex spherical subdomain. Then two positive solutions of(1.1)are proportional.
Proof. We recall that a domainSis (geodesically) convex if a minimal geodesic joining two points of S is contained in S. IfS ⊂ SN−1 is convex, the cone CS is convex too. SinceS is convex, it is Lipschitz and by Theorem D,βS = ˆβS :=β. Letωandω0be two positive solutions of(1.1)satisfying supSω= supSω0 = 1. We denote byuω(x) =|x|−βω(.)anduω0(x) =|x|−βω0(.)the corresponding separablep-harmonic functions defined inCS. If0< a < b, we setCSa,b =CS∩(Bb\Ba). Then for 0< <1we denote byuthe unique function which satisfies
−∆pu= 0 inCS,1 u=−βω inCS∩∂B
u= 0 in (CS∩∂B1)∪ ∂CS∩(B1\B) .
(3.30)
Then
(uω−1)+≤u≤uω inCS,1. (3.31)
Furthermore7→ uis increasing. When ↓0,u ↑u0whereu0is positive andp-harmonic inCS1,0, vanishes on∂CS1,0\ {0}and satisfies(3.30)with= 0. In particular
r→0limrβu0(r, σ) =ω(σ) locally uniformly inS. (3.32) We construct the same approximationu0inCS,1withω0instead ofω. Mutadis mutandis(3.31)holds andu0↑u00which is positive andp-harmonic inCS1, satisfies
(uω0−1)+≤u00≤uω0 inCS1,0, and thus
r→0limrβu00(r, σ) =ω0(σ) locally uniformly inS. (3.33) However, by [10, Th 4]u0andu00are proportional. Combined with(3.32),(3.33)it implies the claim.
3.2.2 Proof of Theorem E
In what follows we borrow most of our construction from [10] that we adapt to the case of an infinite cone a make explicit for the sake of completeness. The nextnondegeneracy propertyof positivep-harmonic functions is proved in [10, Lemma 4.28].
Proposition 3.2. Let Ω ⊂ RN be a bounded Lipschitz domain and 1 < p < ∞. Then there exist constantsρ >0,c14, c15>0depending respectively onΩ(forρ), andp,N and the Lipschitz normM of∂Ω(forc14andc15) with the property that for anyw∈∂Ωand any positivep-harmonic functionu inΩ, continuous inΩ∩B2ρ(w)and vanishing on∂Ω∩Bρ(w), one can findξ∈SN−1, independent of u, such that
c−114 u(y)
dist(y, ∂Ω)≤ h∇u(y), ξi ≤ |∇u(y)| ≤c14
u(y)
dist(y, ∂Ω), (3.34) for ally∈CS∩Bρ|w|
c15
(w).
IfΩis replaced by a coneCS, the nondegeneracy property still holds uniformly on∂CS\ {0}.
Corollary 3.3. Let1< p <∞,S⊂SN−1is a Lipschitz domain andCSthe cone generated byS.
(i) Then there exist constantsρ < 12,c14, c15>0depending respectively onS(forρ), andp,Nand the Lipschitz normMof∂Sanddiam(S)(forc14andc15) with the property that for anyw∈∂CSand any positivep-harmonic functionuinCS, continuous inCS∩B2ρ|w|(w)and vanishing on∂CS∩Bρ|w|(w) continuous, one can findξ∈SN−1, independent ofu, such that
c−114 u(y)
dist(y, ∂CS)≤ h∇u(y), ξi ≤ |∇u(y)| ≤c14 u(y)
dist(y, ∂CS), (3.35) for ally∈B ρ
c15
(w)∩CS.
(ii) Then there exist positive constantsκandc16, c17depending onS(forκ), andp,Nand the Lipschitz normM of∂Sanddiam(S)(forc16, c17such that for anya >0and any positivep-harmonic function uinCSa vanishing on∂CS∩Bac, there holds
c−116 u(y)
dist(y, ∂CS) ≤ |∇u(y)| ≤c16
u(y)
dist(y, ∂CS) ∀y∈CSc17a s.t.dist(y, ∂CS)≤κ|y|. (3.36) Letω, ω0 ∈W01,p(S)∩C(S)be positive solutions(1.1). Sinceωω0 is bounded from above and from below inS by positive constants, we can assume, as in the proof of Theorem D, thatω ≥ω0 inSand that the graphs ofωandω0are tangent. hence, ifω6=ω0, thenω > ω0 inSand there exists a sequence {σn}converging toσ0∈∂Sasn→ ∞such that
n→∞lim ω0(σn)
ω(σn) = 1.
We defineδ1= sup{δ >0 :δω < ω0}. Fort∈(δ1,1), we set φt= sup{ω0, tω} and ψt= inf
t δ1
ω0, ω
(3.37) We also set
vφt(r, σ) =r−βφt(σ) and vψt(r, σ) =r−βψt(σ) ∀(r, σ)∈(0,∞)×S. (3.38)