JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 10.1155/2016/4649150 4649150 Research Article Uniqueness of Solutions to a Nonlinear Elliptic Hessian Equation http://orcid.org/0000-0001-5014-9979 Li Siyuan 1 Conca Carlos School of Mathematics and Applied Statistics Faculty of Engineering and Information Sciences University of Wollongong Wollongong NSW 2522 Australia uow.edu.au 2016 1122016 2016 29 06 2016 06 11 2016 2016 Copyright © 2016 Siyuan Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Through an Alexandrov-Fenchel inequality, we establish the general Brunn-Minkowski inequality. Then we obtain the uniqueness of solutions to a nonlinear elliptic Hessian equation on Sn.

1. Introduction

According to a general Brunn-Minkowski inequality, we obtain a proof of the uniqueness of solutions to the following fully nonlinear elliptic Hessian equation: (1)σkuij+uδij=fup-1on  Sn,where u is the support function of convex bodies, uij are the second-order covariant derivations of u with respect to any orthonormal frame e1,e2,,en on Sn, δij is the standard Kronecker symbol, Sn is the unit sphere of n-dimension, f is a positive function defined on Sn, k{1,2,,n}, p>1, and σk is the kth elementary symmetric function defined as follows: for λ=(λ1,λ2,,λn)Rn, (2)σkλ=1i1<i2<<iknλi1λi2λik.The definition can be extended to any symmetric matrix WRn×n by σk(W)=σk(λ(W)), where λ(W)=(λ1(W),λ2(W),,λn(W)) is the eigenvalue vector of W.

Equation (1) arrives from the geometry of convex bodies. A compact convex subset of Euclidean (n+1)-space Rn+1 with nonempty interiors is called a convex body. An important concept related to a convex body Q is its support function.

Definition 1.

Let M (the boundary of a convex body Q) be a smooth, closed, uniformly convex hypersurface enclosing the origin in Rn+1. Assume that M is parameterized by its inverse Gauss map X:SnMRn+1; the support function u of M (or Q) is defined by (3)ux=x,Xx,xSn,where ·,· denotes the standard inner product in Rn+1.

u is convex after being extended as a function of homogeneous degree 1 in Rn+1. Conversely, any continuous convex function u of homogeneous degree 1 determines a convex body as follows: (4)Q=yRn+1:y·xux,  xSn.From some basic concepts to support function, Minkowski sum [see Definition 4], and mixed volumes [see Definition 5], Minkowski developed a set of theories related to convex bodies. If k=n and p=1, (1) is the Monge-Ampère equation corresponding to the classical Minkowski problem (5)detuij+uδij=fon  Sn,which has been solved by Nirenberg , Pogorelov [2, 3], Cheng and Yau , and many others. When p=1, (1) is the classical Christoffel-Minkowski problem: (6)σkuij+uδij=fon  Sn.A necessary condition  for (6) to have a solution is (7)Snxifxds=0,i=1,2,,n+1,where ds is the standard area form on Sn. Guan et al.  obtained that (7) is sufficient for (6) to have an admissible solution [see Definition 6].

Firey  generalized the Minkowski sum to p-sum [see Definition 4] from p=1 to p1 in 1962. Later, Lutwak  extended the classical surface area measure to the p-sum cases. Also in , Lutwak first introduced the general Minkowski problem, which is called Lp-Minkowski problem thereafter. In the smooth category, Lp-Minkowski problem is equivalent to considering the following Monge-Ampère equation: (8)detuij+uδij=fup-1on  Sn.The uniqueness of Lp-Minkowski problem for p>1 and pn+1 (the uniqueness holds up to a dilation if p=n+1) has been solved in . However, the uniqueness for p<1 is difficult and still open. In , Jian et al. obtained that, for any -n-1<p<0, there exists a positive function fC(Sn) to guarantee that (8) has two different solutions, which means that we need more conditions to consider the uniqueness.

When considering cases 1k<n, attention is paid to the generalized Christoffel-Minkowski problem. In the smooth category, we need to study the k-Hessian equation (1).

For (1), Hu et al.  got the existence and uniqueness of solutions to (1) when 1k<n and p>k+1 under appropriate conditions. However, the uniqueness of (1) when p<1 has not been solved well. In this paper, we study the uniqueness of (1) for p>1.

Our main result is the following.

Theorem 2.

Suppose u is a positive admissible solution of (9)σkuij+uδij=fup0on  Sn,where 1k<n, kZ, p0R+k, and f is a positive function defined on the unit sphere Sn and then the uniqueness holds. If p0=k, the uniqueness holds up to a dilation, which means that if u solves (9), then au:aR+ are the whole solutions of (9).

Remark 3.

Here, we rewrite (1) by (9), where p0=p-1.

The organization of this paper is as follows. In Section 2, we show some basic concepts and lemmas which have been obtained by Guan et al. in . In Section 3, we prove two useful propositions according to the methods in . In the last section, we prove the main theorem.

2. Preliminaries Definition 4.

Given two convex bodies Q1 and Q2 in Rn+1 with respective support functions u1, u2, and λ,μ0 (λ+μ>0), the Minkowski sum λQ1+μQ2Rn+1 is defined by the convex body whose support function is λu1+μu2.

For p1, let Q1 and Q2 be two convex bodies containing the origin in Rn+1 in their interiors, and λ,μ0 (λ+μ>0). The convex body λQ1+pμQ2, whose support function is given by λu1p+μu2p1/p, is called Firey’s p-sum of Q1 and Q2, where “+p” means the p-summation and “” means Firey’s multiplication.

Definition 5.

Let Q1,Q2,,Qr be convex bodies in Rn+1 and the volume of their Minkowski sum (10)Q=λ1Q1+λ2Q2++λrQr,λi0,is an (n+1)th degree homogeneous polynomial of the family λ1,λ2,,λr. Specially, the volume of Q is (11)VolQ=Volλ1Q1+λ2Q2++λrQr=i1,i2,,in+1=1rλi1λi2λin+1VQi1,Qi2,,Qin+1,where the functions V are symmetric. Then V(Q1,Q2,,Qn+1) is called the Minkowski mixed volume of Q1,Q2,,Qn+1.

Definition 6.

For k1,2,,n, let Γk be the convex cone in Rn which is determined by (12)Γk=λRn:σ1λ>0,σ2λ>0,,σkλ>0.A function uC2(Sn) is called k-convex if (13)Wx=uijx+uxδijΓk,xSn,and u is called an admissible solution to (1) if u is k-convex and satisfies (1).

Definition 7.

Let A1,A2,,Am be symmetric real k×k matrices, λ1,λ2,,λmR; the determinant of λ1A1++λmAm is a homogeneous polynomial of degree k in λ1,λ2,,λm. Namely, (14)detλ1A1++λmAm=i1,,ik=1mλi1λikDkAi1,,Aik.In fact, the coefficient λi1λik depends only on Ai1,,Aik; then they are uniquely determined. Dk(A1,,Ak) is called the mixed discriminant of A1,,Ak.

For later applications, we collect some results here which have been proved in .

Lemma 8.

Let u1,u2,,un+1 be the support function of convex bodies Q1,Q2,,Qn+1, respectively. Denoting Minkowski mixed volume V(Q1,Q2,,Qn+1) by V(u1,u2,,un+1) and (15)Wm=umij+umδij,m=1,2,,n+1,then(16)Vu1,u2,,un+1=Snu1DnW2,W3,,Wn+1ds,where Dn(W2,W3,,Wn+1) is the mixed discriminant [see Definition 7] of W2,W3,,Wn+1.

Remark 9.

For all 1kn, setting uk+2==un+1=1, then (17)Vu1,,uk+1,1,,1Vk+1u1,u2,,uk+1=Snu1DkW2,W3,,Wk+1ds,where Dk(W2,W3,,Wk+1) is the mixed discriminant of W2,W3,,Wk+1. Furthermore, if u1=u2==un+1=u, denote V(u1,u2,,un+1)V(u) and Vk+1(u1,u2,,uk+1)Vk+1(u); then (18)Vu=Snudetuij+uδijds,Vk+1u=Snuσkuij+uδijds.

Lemma 10.

V is a symmetric multilinear form on (C2(Sn))n+1.

Lemma 11.

For any function uC2(Sn), W=uij+uδij, 1k<n, we have the Minkowski type integral formula, (19)SnuσkWds=Snσk+1Wds,where ds is the standard area element on Sn.

The following is a form of Alexandrov-Fenchel inequality for positive k-convex functions which comes from .

Lemma 12 (Alexandrov-Fenchel inequality).

If u1,u2,,uk are k-convex, u1 is positive, and there exists l2,3,,k such that ul0 on Sn, then, for any vC2(Sn), (20)Vk+12v,u1,u2,,ukVk+1u1,u1,u2,,ukVk+1v,v,u2,,uk,with equality if and only if v=au1+i=1n+1aixi for some constants a,a1,,an+1.

3. Two Important Propositions

Now we prove two important propositions. The methods we use are from .

Proposition 13.

Suppose u0,u1>0 are k-convex; then (21)Vk+11/k+11-tu0+tu11-tVk+11/k+1u0+tVk+11/k+1u1,t0,1,with equality if and only if u0=au1+i=1n+1aixi for some constants a,a1,,an+1.

Proof.

We only need to prove that (22)Ft=Vk+11/k+11-tu0+tu1is concave on 0,1. Setting ut=(1-t)u0+tu1, t0,1, we have (23)Ft=Vk+11/k+1ut,ut,,utk+1.By the symmetric multilinear property of V, it is obvious that (24)Ft=Vk+11/k+1-1ut,,utk+1Vk+1-u0+u1,ut,,utk,(25)Ft=kVk+11/k+1-2ut,,utk+1·Vk+1ut,,utk+1·Vk+1-u0+u1,-u0+u1,ut,,utk-1-Vk+12-u0+u1,ut,,utk0,where the last inequality uses (20); thus F is a concave function on 0,1. The equality condition is checked easily.

Proposition 14 (general Brunn-Minkowski inequality).

Supposing u0,u1>0 are k-convex, then(26)Snu1σku0ij+u0δijdsVk+11/k+1u1Vk+11-1/k+1u0,with equality if and only if u0=au1+i=1n+1aixi for some constants a,a1,,an+1.

Proof.

Setting(27)Ft=Vk+11/k+11-tu0+tu1-1-tVk+11/k+1u0-tVk+11/k+1u1,then F(0)=F(1)=0. By (21), F(t)0; thus F(0)0; namely, (28)Vk+11/k+1-1u0Vk+1-u0+u1,u0,,u0k+Vk+11/k+1u0-Vk+11/k+1u10.Then(29)Vk+11/k+1-1u0Sn-u0+u1σku0ij+u0δijds+Vk+11/k+1u0Vk+11/k+1u1.By (19), (30)Vk+11/k+1-1u0Snu1σku0ij+u0δijdsVk+11/k+1u1,and then(31)Snu1σku0ij+u0δijdsVk+11/k+1u1Vk+11-1/k+1u0.

4. Proof of Theorem <xref ref-type="statement" rid="thm1">2</xref>

Now we prove Theorem 2. The main methods are from [7, 12].

Proof.

Assuming that (9) has two solutions u and v, then we consider the equation in the following three cases.

Case  1 (p0>k). Supposing x0 is the maximum value point of G=u/v, then at x0, we have (32)0=lnG=uu-vv,02lnG=2uu-u2u2-2vv-v2v2=2uu-2vv;that is, (33)2uu2vv.Hence (34)fup0x0=ukx0σkuiju+δijx0ukx0σkvijv+δijx0=ukx0vkx0fvp0x0;therefore(35)up0-kx0vp0-kx0Gx0=ux0vx01;then(36)uv1.Similarly, we have v/u1. Thus uv.

Case  2 (0<p0<k). We have (37)u-p0σkuij+uδij=v-p0σkvij+vδij;then(38)Vk+1u=Snuσkuij+uδijds=Snuvp0+1vσkvij+vδijdsSnuσkvij+vδijdsp0+1Snvσkvij+vδijds-p0Vk+1p0+1/k+1uVk+1kp0+k/k+1vVk+1-p0v=Vk+1p0+1/k+1uVk+11-p0+1/k+1v,where we have used Hölder inequality in the first inequality and used (26) in the second one. Hence Vk+1(u)=Vk+1(v), which forces both the equalities to hold. By the equality condition, there exists a constant aR such that v=au. By (9), we know a=1. Therefore, uv.

Case  3 (p0=k). According to Case  2, when p0=k, we have (39)Vk+1u=Snuσkuij+uδijds=Snuvk+1vσkvij+vδijdsSnuσkvij+vδijdsk+1Snvσkvij+vδijds-kVk+1uVk+1kvVk+1-kv=Vk+1u;then all the equalities hold. Thus there exists aR, such that v=au. Therefore {au:aR+} are the whole solutions of (9).

Now we complete the proof of Theorem 2.

Competing Interests

The author declares no competing interests.

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