problem Nonuniqueness of solutions to the L -Minkowski Advances in Mathematics

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Advances in Mathematics

www.elsevier.com/locate/aim

Nonuniqueness of solutions to the L

-Minkowski problem

Huaiyu Jian^a,b, Jian Lu^b,c,∗, Xu-Jia Wang^b

aDepartmentofMathematics,TsinghuaUniversity,Beijing100084,China

bCentreforMathematicsandItsApplications,AustralianNationalUniversity, Canberra,ACT0200,Australia

cDepartmentofAppliedMathematics,ZhejiangUniversityofTechnology, Hangzhou310023,China

a r t i c l e i n f o a b s t r a c t

Articlehistory:

Received8July2014

Receivedinrevisedform19May 2015

Accepted21May2015 Availableonline12June2015 CommunicatedbyErwinLutwak

Keywords:

Monge–Ampèreequation Minkowskiproblem Nonuniqueness

Inthispaper weshow byexamplethat there isno uniform estimatefortheLp-Minkowskiproblem.Asaresultweobtain thenonuniquenessofsolutionstotheproblem.

© 2015ElsevierInc.All rights reserved.

1. Introduction

WeconsiderthefollowingLp-Minkowskiproblem,

det(∇²H+HI) =f H^p−1 onSⁿ, (1.1)

✩ ThefirstauthorwassupportedbytheNaturalScienceFoundationofChina(11131005,11271118);the secondauthorwassupportedbytheNaturalScienceFoundationofChina(11401527);andthethirdauthor wassupportedbytheAustralianResearchCouncil(DP120102718andFL130100118).

* Correspondingauthor.

E-mailaddresses:[email protected](H.Y. Jian),[email protected](J. Lu), [email protected](X.-J. Wang).

http://dx.doi.org/10.1016/j.aim.2015.05.010 0001-8708/© 2015ElsevierInc.All rights reserved.

whereH isthesupportfunctionofaconvexbodyK=KHintheEuclideanspaceRⁿ⁺¹, I is the unit matrix, f is agiven positive function on the unit sphere Sⁿ, and p is a constant.Themain resultofthepaperis

Theorem1.1. Foranyp∈(−n−1,0),thereexistsapositivefunctionf ∈C^∞(Sⁿ)such that equation (1.1)admits twodifferentsolutions.

To prove Theorem 1.1, we will construct a smooth, positive function f, which is radially symmetricinx1,· · ·,xn, andalso symmetricinxn+1, suchthatequation (1.1) has asolutionwith small volume.By[7] there isanother solutionwhose volume hasa positivelowerbound dependingonlyonn,infSⁿf and sup_Snf.Thereforewehavetwo different solutionsforequation (1.1).

The Lp-Minkowski problem was introduced by Lutwak [16] and has been studied by manyauthors [4,5,7,11,14,15,17–21].Oneof themain questionsistheuniquenessof solutions. In particular, a solution to (1.1) is also a self-similar solution to the Gauss curvature flow,ofwhichtheuniquenesshasbeen extensivelystudied.

Denoteq= 1−p.Whenf ≡1,theuniquenesshasbeenobtainedinthecaseq=nfor alln[9]andinthecasen= 2 andq= 1[1].Whenn= 1,q= 1,andf issymmetricwith respect to theorigin, namelyf(θ)=f(θ+π), theuniqueness wasobtained in[10,13].

When q <−n, theuniquenesscanbeobtainedbythemaximum principle[7]. Onecan alsofindsomeuniquenessresultsinthecaseq <0 forsymmetricf in[16].Inthediscrete caseauniquenesswasestablishedfortheL₀-Minkowskiproblem in[20].

Ontheotherhand,itiswellknownthatwhenq=n+ 2,allellipsoidswiththevolume oftheunitballaresolutionsof(1.1)withf ≡1.Theuniquenessismoredelicateinthe caseq > n+ 2,atleastwhenn= 1[3].In[7]itwasshownthatthesolutionmaynotbe uniqueifq∈(1,n+ 2) andisverycloseton+ 2.Incontrasttotheuniquenessin[10,13]

for n= 1,q= 1,and symmetricf,asurprising nonuniqueness resultwasdiscovered in [22] fornonsymmetricf,alsointhecasen= 1, q= 1.

The uniqueness of solutions for the case 0 < q < n+ 2 attracted much attentions as itis relatedtothelimitshape ofGauss curvatureflows ofconvexhypersurfacesand has received considerable investigations. See [1,2,6,8,9] and the above discussion. The uniquenessofsolutionshasbeenconjecturedforanumberofspecialcases,includingin particular the case f ≡ 1 and q = 1 [12]. Our theorem above shows that for general positivefunctionf, theremaybe morethanonesolutionto(1.1), forallq∈(1,n+ 2) and all dimensions n. It implies that the limit shape of anisotropic curvature flows is usually notunique.

We will firstprove Theorem 1.1 for the case p∈ (−n−1,−1) in Sections 2 and 3, andthenforthecasep∈[−1,0) inSection4.InSection2weconstructanf_εsuchthat equation (1.1)hasasolutionHε ofsmallvolume.TheninSection3weshowthatthere is a solutionto (1.1) of whichthe volumehas apositivelower bound. In Section4we extendtheexampletothecasep∈[−1,0).

2. Asolutionwithsmallvolume

Denoteδ=n+ 2−q∈(0,n+ 1).Let

Mε= diag(1,· · ·,1, ε) = I 0

0 ε

(2.1) beamatrix,where0< ε≤1 isasmallconstantandasbeforeIistheunitn×nmatrix.

Considertheequation

det(∇²h+hI)(x) =|Mεx|^−δ, x∈Sⁿ. (2.2) Equation(2.2) is justthe classical Minkowskiproblem.As its right hand sidesatisfies the necessary condition

Sⁿx_k|M_εx|^−δ = 0 for all1 ≤k ≤n+ 1, there is a solution, whichisuniqueupto translation,totheequation.

Leth_ε be theuniquesolution to (2.2)suchthatthe centreofthe associated convex bodyK_hε is locatedat theorigin.Notethath_ε isradiallysymmetricinx₁,· · ·,x_n and symmetricinxn+1.Define

Hε(x) = (detMε)^n+q2 ·M_ε⁻¹x·hε

M_ε⁻¹x Mε⁻¹x

, (2.3)

thenHεisthesupportfunctionofaconvexbodyKHε, andKHε canbeobtainedfrom Khε bymaking thecoordinate transformx→M_ε⁻¹xand then makingadilation x→ (detMε)^n+q2 x.

Lemma2.1. The functionHε satisfiestheequation

det(∇²Hε+HεI) = ˆh^q_ε Hε^q

on Sⁿ, (2.4)

where

ˆhε(x) =hε

M_ε⁻¹x Mε⁻¹x

. (2.5)

Proof. Let

uε(x) =Mε−1

x·hε

Mε−1

M_ε⁻¹xx

. (2.6)

Bytheinvarianceofthequantityhⁿ⁺²_ε det(∇²hε+hεI) underunimodularaffinetrans- formations,seeProposition 7.1in[7]or formula (2.12)in[15],wehave

det(∇²u_ε+u_εI)(x) = det(∇²h_ε+h_εI)

M_ε⁻¹x

|Mε⁻¹x|

· (detM_ε⁻¹)²

|Mε⁻¹x|ⁿ⁺².

Here wenote thatM_ε⁻¹ isnotaunimodulartransformation.Since hε satisfiesequation (2.2),we have

det(∇²uε+uεI)(x) =|M_ε⁻¹x|^δ· (detM_ε⁻¹)²

|Mε⁻¹x|ⁿ⁺²

= 1

(detM_ε)²|Mε⁻¹x|^q. (2.7) Therefore bythedefinitionofH_ε,(2.3),onegets

det(∇²H_ε+H_εI)(x) = (detM_ε)^n+q2n det(∇²u_ε+u_εI)(x)

= 1

(detM_ε)^n+q2q Mε⁻¹x^q

= hˆ^q_ε Hε^q

. 2

ToestimatethevolumeoftheconvexbodyKHε,oneneedstostudytheconvexbody Khε, or equivalently the support function hε. When δ ∈ (0,n), we have the following uniform estimates.

Lemma 2.2. When 2 < q < n+ 2, there exists a positive constant C, independent of ε∈(0,1],such that

C⁻¹≤hε≤C on Sⁿ. (2.8)

Proof. Onecan easily see thatthe area of∂Khε is uniformly bounded from above. In fact,since

|Mεx| ≥

1−x²_n+1 ∀ x∈Sⁿ andε∈(0,1], and notingthatδ=n+ 2−q < n, wehave

area(∂Khε) =

Sⁿ

det(∇²hε+hεI)

=

Sⁿ

|M_εx|^−δ

≤

Sⁿ

(1−x²_n+1)^−δ/2

≤C,

whereC is apositiveconstantdependingonn, δbutindependentof ε.Bytheisoperi- metricinequality,weobtain

vol(K_hε)≤C_narea(∂K_hε)ⁿ⁺¹ⁿ ≤C. (2.9) Assumeh_ε attains its maximum at point x_ε ∈ Sⁿ, namely h_ε(x_ε) = max_Sⁿh_ε. By convexityandrecallingthathεissymmetric, wehave

hε(x)≥maxhε· xε, x ∀x∈Sⁿ. Observingthat

|Mεx| ≤1 ∀x∈Sⁿ and ε∈(0,1], byequation(2.2),wehave

vol(Khε) = 1 n+ 1

Sⁿ

hεdet(∇²hε+hεI)

= 1

n+ 1

Sⁿ

h_ε(x)|M_εx|^−δ

≥ 1 n+ 1

Sⁿ

hε(x)

≥C_nmaxh_ε·

S_εⁿ

x_ε, x

=Cnmaxhε,

whereS_εⁿ ={x∈Sⁿ: xε, x>0}. Thereforeweobtainfrom (2.9)that

maxhε≤Cnvol(Khε)≤C. (2.10) Thesecondinequalityof(2.8)isproved.

Toprovethefirstinequalityof(2.8),wemakeuseoftheconceptofminimumellipsoid ofaconvex body.LetE_hε betheminimumellipsoid ofK_hε.Thenwe have

1

n+ 1Ehε ⊂Khε ⊂Ehε.

Bythesymmetry ofKhε,thecentreofEhε isat theorigin.LetR1,ε≥ · · · ≥Rn+1,ε be thelengthsof thesemi-axisofEhε.Then

R1,ε≤(n+ 1) maxhε, Rn+1,ε≤(n+ 1) minhε.

Hence

vol(Khε)≤vol(Eε)

≤ω_n+1Rⁿ_1,εR_n+1,ε

≤C_n(maxh_ε)ⁿ·minh_ε,

where ωn+1 isthevolumeoftheunitballinRⁿ⁺¹.By(2.10) itfollowsthat 1≤Cn(maxhε)ⁿ⁻¹·minhε.

Thefirstinequalityof(2.8)follows. 2 Now let

fε= ˆh^q_ε. (2.11)

In viewof (2.5)and (2.8), there exist two positiveconstants C₁, C₂, independent ofε, suchthat

C₁≤f_ε≤C₂. (2.12)

From (2.4),Hε isasolutionto

det(∇²Hε+HεI) = fε

Hε^q

onSⁿ. (2.13)

By(2.3),wehavethevolumeestimate

vol(KHε) = (detMε)^2n+2n+q ·detM_ε⁻¹·vol(Khε)

= (detM_ε)^n+qδ vol(K_hε). (2.14) Note thatdetM_ε=ε.Hencewhen0< δ < n,thevolumeofK_Hε canbe assmallaswe want,providedεissufficientlysmall.

Remark. From the transform (2.3) and estimate (2.8), one easily sees thatthere is no uniform upperboundforthesolutionHε, whenε>0 issmall.

3. Avariationalsolution

Inthepaper[7],thevariationalproblem sup

h

yinf∈Kh

J[h(x)−y·x] : vol(Kh) = 1

(3.1)

is studied. Here h is the support function of a non-degenerate convex body Kh with volumevol(Kh).ThefunctionalJ[h] isgivenby

J[h] = 1 q−1

Sⁿ

f h^q−1.

Forconveniencewerecallthefollowingexistenceresultin[7].

Lemma3.1.(See[7].)Let1< q < n+2andf ∈L^∞(Sⁿ)beapositivefunctionsatisfying (2.12). Then the variational problem (3.1) admits a maximiser H which satisfies, for someLagrangemultiplier λ>0,

det(∇²H+HI) = λf

H^q onSⁿ. (3.2)

Whenf(x)=f(−x),onecanconsidertheproblem(3.1)intherestrictedclassh(x)= h(−x).Onecanfollow thesameargumentsasin[7]toshowthatamaximiser H exists anditscorrespondingconvex bodyis centrallysymmetric.

Bythis lemma,weobtainasolutiontoequation (1.1). Let

H˜ =λ^−n+q1 H. (3.3)

ThenH˜ isasolutiontoequation

det(∇²H˜+ ˜HI) = f

H˜^q onSⁿ. (3.4)

Thevolumeofthecorresponding convexbodyKH˜ is

vol(KH˜) =λ^−n+1n+q. (3.5) Weneed to estimate theLagrange constantλ. It is given as follows. MultiplyingH toboth sidesofequation(3.2)andtakingintegration,wehave

1 = vol(K_H) = 1 n+ 1

Sⁿ

Hdet(∇²H+HI)

= λ

n+ 1

Sⁿ

f H^q−1

= (q−1)λ

n+ 1 J[H]. (3.6)

TheBlaschke–Santaloinequalityforcentrallysymmetricconvexbodiesisgivenby V(h)

Sⁿ

1

hⁿ⁺¹dS(x)≤ ω_n² n+ 1,

where h(x) = h(−x) is the support function of any centrally symmetric convex body and ωn isthevolumeoftheunitball inRⁿ.Usingthis inequality,wehave

J(H) = 1 q−1

Sⁿ

f H^q−1

=supf q−1

Sⁿ

1 H^q−1

_n+1^q−1

Sⁿ

1 ^n+2−q_n+1

≤C_nsupf q−1 .

On the other hand, let h0 be the constant function ω_n+1^−1/(n+1) which is the support functionoftheballcentred attheoriginwhosevolumeisequalto 1.Fromthevariational characterisation ofH,wehave

J(H)≥H(h₀) =C_n supf q−1 . From (3.6),wethusobtain

C_n⁻¹inf

Sⁿf ≤λ⁻¹≤Cnsup

Sⁿ

f.

By(3.5),weseethatthereisapositiveconstantCn dependingonlyonn,suchthat C_n⁻¹(inf

Sⁿf)^n+1n+q ≤vol(KH˜)≤Cn(sup

Sⁿ

f)^n+1n+q. (3.7) Now we let f =fε be the function in(2.13). Denote the corresponding variational solutionbyH˜ε,whichisthesupportfunctionofaconvexbodyKH˜ε.Byvirtueof(2.12) and (3.7),onehastheestimates

C3≤vol(KH˜ε)≤C4,

where C3,C4 arepositiveconstantsdependingonlyonn,q,C1 andC2.Butrecallthat forthesolutionH_εgivenin(2.3),thevolumevol(K_Hε)→0 asε→0.HenceH˜_εandH_ε are two different solutions.We have therefore obtainedthe nonuniqueness of solutions to (2.13)forsmallε>0,inthecaseq∈(2,n+ 2).

Remark.Asf isrotationallysymmetricwithrespecttothexn+1-axis,onemayconsider thesupremumin(3.1)inthefamilyofrotationallysymmetricconvexbodiesandobtain a rotationally symmetric solution. Therefore for equation (2.13) the two solutions we obtained are both rotationally symmetric with respect to the x_n+1-axis, and so are smooth.

Wealsoremarkthattheideaofourproofisbasedontheobservationthatvol(KHε) issmall.Butwhenq= 1 andq=n+ 2,onecanprovethatforanysolutionto(1.1),the volume of the associated convex body hasa positive lower bound. Therefore our con- structionofmorethanonesolution doesnotcoverthecaseq= 1.Infact,asmentioned intheintroduction,whenn= 1 andq= 1,thesolutionisuniquewhenf is symmetric [10,13].

4. Thecase1< q ≤2

WehaveprovedTheorem 1.1forthecase2< q < n+ 2.Inthissectionweshowthat the solutions Hε and H˜ε given in (2.3) and (3.3) are different even when 1 < q ≤ 2, providedεaresufficientlysmall.

RecallthatK_hε isradiallysymmetricinx₁,· · ·,x_nandsymmetricinx_n+1.Wedenote R1 =R1,ε =hε(1,0,· · ·,0),Rn+1=Rn+1,ε=hε(0,· · ·,0,1).Namely R1 and Rn+1 are respectivelythevaluesofhεontheequatorandatthenorthpoleofthesphereSⁿ. Lemma4.1. Thereexistsaconstant C >0,independentof ε∈(0,1],suchthat

R_n+1< CR₁. (4.1)

Proof. Denote S_η = {x ∈ Sⁿ : |x_n+1| < η} and S_η^c = Sⁿ\S_η, where η ∈ (0,1) is a constant.LetΓη ={p∈∂Khε :G(p)∈Sη}and Γ^c_η =∂Khε\Γη,where GistheGauss mapof∂Khε, namelyG(p) istheunitouternormalof∂Khε at thepointp.

If(4.1)isnottrue,wehave

area(Γ_1/2)>>area(Γ^c_1/2).

Ontheotherhand,byequation (2.2)we have area(Γ1/2) =

S1/2

det(∇²hε+hεI) =

S1/2

|Mεx|^−δ. (4.2)

Notethat

sup

Sη

|Mεx|^−δ≤inf

S^c_η|Mεx|^−δ ∀η∈(0,1) andε∈(0,1].

Hencetheright handsideof(4.2)

≤C_n

S^c_1/2

|M_εx|^−δ =C_n

S_1/2^c

det(∇²h_ε+h_εI) =C_narea(Γ^c_1/2). (4.3)

Wereachacontradiction. 2

Byequation(2.2),wehaveasε→0 that area(∂K_hε) =

Sⁿ

|M_εx|^−δ

=

(C+o(1))ε^q−2 if 1< q <2,

(C+o(1))|logε| ifq= 2, (4.4) where C is apositiveconstantindependentof ε.Onthe otherhand,byLemma 4.1we have

C⁻¹R₁ⁿ≤area(∂Khε)≤CR1n

.

From (4.4)itfollowsthat

C⁻¹ε^q−2n ≤R₁≤Cε^q−2n if 1< q <2,

C⁻¹|logε|^1/n≤R1≤C|logε|^1/n ifq= 2. (4.5) Observing that

vol(Khε) =CRⁿ₁Rn+1, by(2.14) weobtain

vol(K_Hε) =ε^n+qδ vol(K_hε)

=

Cε(q−1)(n+q−2)

n+q R_n+1 if 1< q <2,

Cε^n+qδ |logε|R_n+1 ifq= 2. (4.6) Now considerthesolutionH˜ε givenin(3.3).From (3.7), wehave

vol(KH˜ε)≥C(inf

Sⁿfε)^n+1n+q. Notingthatfε= ˆh^q_ε andthatinfSⁿˆhε= infSⁿhε,weseethat

vol(KH˜ε)≥C(inf

Sⁿhε)^q(n+1)n+q

≥CR

q(n+1) n+q

n+1 . (4.7)

Inorderto provethatH˜ε andHε aretwodifferentsolutions,weneedthat,

vol(KH˜ε)>vol(KHε). (4.8)

By(4.6)and(4.7),itsufficesto have Rn+1>

Cε^n+q−2n if 1< q <2,

Cε|logε|ⁿ⁺²ⁿ ifq= 2. (4.9) Soitsufficesto givealowerbound forR_n+1 tocomplete Theorem 1.1.Infactwehave thefollowing

Lemma4.2. For1< q≤2,wehave R_n+1≈

ε^{(2−q)(n−1)n} if 1< q <2,

|logε|¹⁻ⁿⁿ if q= 2. (4.10) Herethenotation “≈” means theratio of thetwosides has uniform positiveupper and lowerbounds.

Proof. Letuεbeas (2.6),Kuε be thecorrespondingconvexbody.From(2.7),we have area(∂K_uε) =

Sⁿ

1

ε²|Mε⁻¹x|^q = (C+o(1))ε⁻¹, ∀q >1. (4.11) Denote

r1=r1,ε=uε(1,0,· · ·,0) =R1, r_n+1 =r_n+1,ε=u_ε(0,· · ·,0,1) =ε⁻¹R_n+1. Since

area(∂Kuε)≈r₁ⁿ+rⁿ₁⁻¹rn+1, multiplyingbothsidesbyε,andnoting(4.11), weobtain

1≈εRⁿ₁ +Rⁿ⁻¹₁ Rn+1. Onaccountof(4.5),εRⁿ₁ <<1,whichleadsto

1≈Rⁿ₁⁻¹R_n+1. (4.12)

By(4.5)again,wehave(4.10). 2

Combining (4.9)and (4.10), weobtain (4.8).Therefore thecase 1< q ≤2 ofTheo- rem 1.1isalsoproved.

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