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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 930868, 11 pages
http://dx.doi.org/10.1155/2012/930868
Research Article

A Problem Concerning Yamabe-Type Operators of Negative Admissible Metrics

1Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China
2Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China

Received 14 February 2012; Accepted 29 February 2012

Academic Editor: Yonghong Yao

Copyright © 2012 Jin Liang and Huan Zhu. 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.

Abstract

This paper is about a problem concerning nonlinear Yamabe-type operators of negative admissible metrics. We first give a result on 𝜎𝑘 Yamabe problem of negative admissible metrics by virtue of the degree theory in nonlinear functional analysis and the maximum principle and then establish an existence and uniqueness theorem for the solutions to the problem.

1. Introduction

Let (𝑀,𝑔) be a compact closed, connected Riemannian manifold of dimension 𝑛3. In 2003, Gursky-Viaclovsky [1] introduced a modified Schouten tensor as follows:𝐴𝑡𝑔=1𝑛2Ric𝑔𝑡𝑅𝑔𝑔2(𝑛1),𝑡1,(1.1) where Ric𝑔 and 𝑅𝑔 are the Ricci tensor and the scalar curvature of 𝑔, respectively.

Define𝜎𝑘(𝜆)=1𝑖1𝑖𝑘𝑛𝜆𝑖1𝜆𝑖𝑘𝜆for𝜆=1,,𝜆𝑛𝑛,Ω+𝑘=𝜆𝜆=1,,𝜆𝑛𝑛;𝜎𝑗.(𝜆)>0,1𝑗𝑘(1.2) The 𝜎𝑘 Yamabe problem is to find a metric ̃𝑔 conformal to 𝑔, such that𝜎𝑘𝜆̃𝑔𝐴̃𝑔=1,𝜆̃𝑔𝐴̃𝑔Ω+𝑘on𝑀,(1.3) where 𝜆̃𝑔(𝐴̃𝑔) denotes the eigenvalue of 𝐴̃𝑔 with respect to the metric ̃𝑔. This problem has attracted great interest since the work of Viaclovsky in [2] (cf., e.g., [27] and references therein).

Assume Ω𝑘=Ω+𝑘. Then the 𝜎𝑘 Yamabe problem in negative cone𝜎𝑘𝜆̃𝑔𝐴̃𝑔=1,𝜆̃𝑔𝐴̃𝑔Ω𝑘on𝑀,(1.4) is still elliptic (see [1]).

Definition 1.1. A metric ̃𝑔 conformal to 𝑔 is called negative admissible if 𝜆̃𝑔𝐴𝑡̃𝑔Ω𝑘on𝑀.(1.5)
Under the conformal relation ̃𝑔=𝑒2𝑧𝑔, the transformation law for the modified Schouten tensor above is as follows:𝐴𝜏̃𝑔=𝐴𝜏𝑔2𝑧1𝜏𝑛2(Δ𝑧)𝑔2𝜏2|𝑧|2𝑔+𝑑𝑧𝑑𝑧.(1.6) We consider the following nonlinear equation: 𝑃𝜆(𝑍)=𝛽𝑔(𝑍)=𝜑(𝑥,z),𝜆𝑔(𝑍)Ωon𝑀,(1.7) where 𝑍=2𝑧+1𝑡𝑛2(Δ𝑧)𝑔+2𝑡2|𝑧|2𝑔𝑑𝑧𝑑𝑧𝐴𝑡𝑔,(1.8)𝛽𝐶(Ω+)𝐶0(Ω+) is a symmetric function and is homogeneous of degree one normalized, and 𝜑 is a positive 𝐶 function satisfying the monotone condition: thereexiststwoconstants𝛾<0<𝜑𝛾with𝑥,𝛾<𝛽𝜆𝑔𝐴𝑡𝑔<𝜑𝑥,𝛾,𝑥𝑀.(1.9) For this equation, we have the following.

Theorem 1.2. Let (𝑀,𝑔) be a compact, closed, connected Riemannian manifold of dimension 𝑛3 and 𝐴𝑡𝑔Ω,for𝑡<1.(1.10) Suppose that Ω+,Ω𝑅𝑛 are open convex symmetric cones with vertex at the origin, satisfying Ω𝑛ΩΩ1,Ω=Ω+,(1.11) where Ω1𝜆=𝜆=1,,𝜆𝑛;𝑛𝑖=1𝜆𝑖,Ω>0𝑛𝜆=𝜆=1,,𝜆𝑛;𝜆𝑖.>0for1𝑖𝑛(1.12) Let 𝛽 satisfy(i)𝛽>0 in Ω+, 𝛽𝑖=𝜕𝛽/𝜕𝜆𝑖>0 on Ω+, and 𝛽(𝑒)=1 on Ω+, where 𝑒=(1,,1).(1.13)(ii)𝛽 is concave on Ω+, and 𝛽(𝜆)𝜚𝜎1(𝜆),𝜆Ω+,(1.14) where 𝜚 is a positive constant. Moreover, assume that 𝜑(𝑥,𝑧) is a positive 𝐶 satisfying condition (1.9). Then there exists a solution to (1.7).

Theorem 1.3. Let (𝑀,𝑔) be a compact, closed, connected Riemannian manifold of dimension 𝑛3 and 𝐴𝑡𝑔Ω,for𝑡<1.(1.15) Let (𝛽,Ω+) be those as in Theorem 1.2. Then there exist a function 𝜙 and a positive number 𝜆, such that 𝜙 is a solution to the eigenvalue problem 𝑃𝜆(𝑈)=𝛽𝑔(𝑈)=Λ,(1.16) where 𝑈=𝐴𝑡̃𝑔=2𝜙+1𝑡𝑛2(Δ𝜙)𝑔+2𝑡2||||𝜙2𝑔𝑑𝜙𝑑𝜙𝐴𝑡𝑔(1.17) for conformal metric ̃𝑔=𝑒2𝜙 and 𝜆𝑔(𝑈) denotes the eigenvalue of 𝑈 with respect to metric 𝑔.

Remark 1.4. (1) (𝜙,Λ) is unique in Theorem 1.3 under the sense that, if there is another solution (𝜙,Λ) satisfying (1.16), then Λ=Λ,𝜙=𝜙+𝑐(1.18) for some constant 𝑐.
(2) Λ is called the eigenvalue related to fully nonlinear Yamabe-type operators of negative admissible metrics, and 𝜙 is called an eigenfunction with respect to Λ.

2. Proof of Theorem 1.2

To prove Theorem 1.2, firstly, let us give the following proposition.

Proposition 2.1. Suppose all the conditions in Theorem 1.2 are satisfied. Then every 𝐶2 solution 𝑧 to (1.7) with 𝛾𝑧𝛾(2.1) satisfies 𝛾<𝑧<𝛾.(2.2)

Proof. Assume 𝑧 is a solution to (1.7) with 𝛾𝑧. Denote ̃𝑧=𝑧𝛾,𝑧𝑠𝑍=𝑠𝑧+(1𝑠)𝛾,𝑠=2𝑧𝑠+1𝑡𝑛2Δ𝑧𝑠𝑔+2𝑡2|𝑧𝑠|2𝑔𝑑𝑧𝑠𝑑𝑧𝑠𝐴𝑡𝑔.(2.3) It is easy to verify that 𝑍𝑠Ω+.
Write𝑄[𝑧]=𝑃(𝑍)𝜑(𝑥,𝑧).(2.4) Then 𝑄[𝑧]𝛾𝑄=0𝑃𝐴𝑡𝑔+𝜑𝑥,𝛾.(2.5) On the other hand, 𝑄[𝑧]𝛾𝑄=10𝑑𝑄𝑧𝑑𝑠𝑠=𝑑𝑠10𝑇𝑖𝑗𝑍𝑠𝑑𝑠𝐷𝑖𝑗̃𝑧+𝑏𝑖𝐷𝑖̃𝑧+𝑐̃𝑧=𝐿(̃𝑧)(2.6) for some bound 𝑏𝑖 and constant 𝑐, where 𝑇𝑖𝑗=𝑃𝑖𝑗+1𝑡𝑛2𝑙𝑃𝑙𝑙𝛾𝑖𝑗𝑃0,𝑖𝑗=𝜕𝑃𝜕𝑍𝑖𝑗0(2.7) by condition (ii).
Therefore, we know that 𝐿 is an elliptic operator, and𝐿(̃𝑧)<0with̃𝑧0.(2.8) By the maximum principle, we get ̃𝑧>0. That is, 𝑧>𝛾.(2.9) Similarly, we can derive 𝑧<𝛾,(2.10) for solution 𝑧 with 𝑧𝛾.

Thus, we have the following Gradient and Hessian estimates for solutions to (1.7).

Lemma 2.2. Let 𝑧 be a 𝐶3 solution to (1.7) for some 𝑡<1 satisfying 𝛾<𝑧<𝛾. Then 𝑧𝐿<𝐶1,(2.11) where 𝐶1 depends only upon 𝛾,𝛾,𝑔,𝑡,𝜑.
Moreover,2𝑧𝐿<𝐶2,(2.12) where 𝐶2 depends only upon 𝛾,𝛾,𝑔,𝑡,𝜑,𝐶1.

Proof of Theorem 1.2. We now prove Theorem 1.2 using a priori estimates in Lemma 2.2, the maximum principle in Proposition 2.1, and the degree theory in nonlinear functional analysis (cf., e.g., [8]).
For each 0𝜏1, let𝛽𝜏(𝜆)=𝛽𝜏𝜆+(1𝜏)𝜎1(𝜆)𝑒,(2.13) (here 𝑒=(1,,1) as in Section 1) which is defined on Ω+𝜏=𝜆𝑛;𝜏𝜆+(1𝜏)𝜎1(𝜆)𝑒Ω+.(2.14) We consider the problem 𝑃𝜏𝑍+(1𝜏)𝜎1(𝑍)𝑒=𝜏𝜑(𝑥,𝑧)+(1𝜏)𝜎1𝐴𝑡𝑔𝑒2𝑧(2.15) on 𝑀, where 𝑍=2𝑧+1𝑡𝑛2(Δ𝑧)𝑔+2𝑡2||||𝑧2𝑔𝑑𝑧𝑑𝑧𝐴𝑡𝑔.(2.16) Since 𝐴𝑡𝑔Ω, we have 𝜎1𝐴𝑡𝑔>0(2.17) by condition (ii). Hence for 𝜏=0, it follows from the maximum principle that 𝑧=0 is the unique solution.
In view of Proposition 2.1, we see that, for each 𝜏[0,1], every 𝐶2 solution 𝑧𝜏 to (2.15) with 𝛾𝑧𝜏𝛾 satisfies𝛾<𝑧𝜏<𝛾.(2.18) This, together with Lemma 2.2, shows that for each 𝜏[0,1] and solution 𝑧𝜏 to (2.15) with 𝛾𝑧𝜏𝛾, the following estimate holds 𝑧𝜏𝐶2<𝐶,(2.19) for some constant 𝐶 independent of 𝜏.
This estimate yields uniform ellipticity, and by virtue of the concavity condition (ii), the well-known theory of Evans-Krylov, and the standard Schauder estimate (cf. [9]), we know that there exists a constant 𝐾 independent of 𝜏 such that𝑧𝜏𝐶4,𝛼<𝐾,(2.20) where 𝑧𝜏 is a 𝐶2 solution to (2.15) with 𝛾𝑧𝜏𝛾.
Set𝑆𝜏𝛾=<𝑧𝜏<𝛾𝑧𝜏𝐶4,𝛼<𝐾𝑍Ω+𝜏,(2.21) and define 𝑇𝜏𝐶4,𝛼𝐶2,𝛼 by 𝑇𝜏(𝑧)=𝑃𝜏𝑍+(1𝜏)𝜎1(𝑍)𝑒𝜏𝜑(𝑥,𝑧)(1𝜏)𝜎1𝐴𝑡𝑔𝑒2𝑧.(2.22) Then, by (2.19), we see that there is no solution to the equation 𝑇𝜏(𝑧)=0on𝜕𝑆𝜏.(2.23) So the degree of 𝑇𝜏 is well defined and independent of 𝜏. As mentioned above, there is a unique solution at 𝜏=0. Therefore 𝑇deg0,𝑆0,00.(2.24) Since the degree is homotopy invariant, we have 𝑇deg1,𝑆1,00.(2.25) Thus, we conclude that (1.7) has a solution in 𝑆1.
The proof of Theorem 1.2 is completed.

3. Proof of Theorem 1.3

Proof of Theorem 1.3. Take a look at the following equation: 𝑃(𝑢)=𝑃2𝑢+1𝑡𝑛2(Δ𝑢)𝑔+2𝑡2|𝑢|2𝑔𝑑𝑢𝑑𝑢𝐴𝑡𝑔𝑒𝑢=𝜆.(3.1) We will prove that, for small 𝜆>0, (3.1) has a unique smooth solution.
Since 𝜕𝑃/𝜕𝑢<0, the uniqueness of the solution to (3.1) follows from the maximum principle.
Next, we show the existence of the solution to (3.1) by using Theorem 1.2.
It follows from𝐴𝑡𝑔Ω(3.2) that, for 𝜆>0 small enough, we can find two constants 𝛾<0<𝛾, such that 𝑒𝛾+𝜆<𝑃𝐴𝑡𝑔<𝑒𝛾+𝜆.(3.3) That is, condition (1.9) for 𝜑(𝑥,𝑧) in Theorem 1.2 is satisfied. Therefore, by the result in Theorem 1.2, the existence of unique solution to (3.1) is established for small 𝜆>0.
Set𝐸={𝜆>0;(3.1)hasasolution}.(3.4) Since 𝐸, we can define Λ=sup𝜆𝐸𝜆.(3.5) We claim Λ is finite. Actually, 𝜆<𝑃2𝑢+1𝑡𝑛2(Δ𝑢)𝑔+2𝑡2|𝑢|2𝑔𝑑𝑢𝑑𝑢𝐴𝑡𝑔.(3.6) If we assume that at 𝑥0, 𝑢 achieves its maximum, then 2𝑢0, and so 𝜆<𝑃2𝑢+1𝑡𝑛2(Δ𝑢)𝑔𝐴𝑡𝑔𝑃𝐴𝑡𝑔.(3.7) This means that Λ𝑃𝐴𝑡𝑔.(3.8)
For any sequence 𝜆𝑖𝐸 with 𝜆𝑖Λ, let 𝑢𝜆𝑖 be the corresponding solution to (3.1) with 𝜆=𝜆𝑖.
First, we claim thatinf𝑀𝑢𝜆𝑖as𝑖.(3.9) Suppose this is not true, that is, inf𝑀𝑢𝜆𝑖𝐶0(3.10) for a positive constant 𝐶0. Then, by (3.1), at any maximum point 𝑥0 of 𝑢𝜆𝑖, max𝑀𝑢𝜆𝑖𝐶(3.11) for some constant 𝐶 depending only on 𝑃(𝐴𝑡𝑔). Then the apriori estimates imply that 𝑢𝜆𝑖 (by taking a subsequence) converges to a smooth function 𝑢0 in 𝐶, such that 𝑢0 satisfies (3.1) for 𝜆=𝜆0. Since the linearized operator of (3.1) is invertible, by the standard implicit function theorem, we have a solution to (3.1) for 𝜆=𝜆0+𝛿with𝛿>0smallenough.(3.12) This is a contradiction. Hence (3.9) holds.
Next, we prove thatmax𝑀𝑢𝜆𝑖as𝑖.(3.13)
We divided our proof into two steps.
Step 1. Let Λ=𝑃𝐴𝑡𝑔.(3.14) Then, following the above argument, 𝑢𝜆𝑖𝜙0inC,(3.15) and (Λ,𝑢0) is a solution to (3.1). Assume 𝑢0 attains its maximum at 𝑦0. Then at 𝑦0, 2𝑢00,𝑢0=0.(3.16) Therefore, 𝑒𝑢0(𝑦0)𝑃𝐴𝑡𝑔Λ=0.(3.17) So 𝑢0𝑦0=.(3.18) That means that (3.13) holds.Step 2. Let 𝑃𝐴𝑡𝑔Λ=𝜛>0.(3.19) Then, if (3.13) is not true, that is, max𝑀𝑢𝜆𝑖𝐶0(3.20) for a positive constant 𝐶0, write 𝑧𝜆𝑖=𝑢𝜆𝑖max𝑀𝑢𝜆𝑖.(3.21) Then we have max𝑀𝑧𝜆𝑖0,inf𝑀𝑧𝜆𝑖,(3.22) as 𝑖.
On the other hand, 𝑧𝜆𝑖 satisfies𝑃2𝑧𝜆𝑖+1𝑡𝑛2Δ𝑧𝜆𝑖𝑔+2𝑡2|𝑧𝜆𝑖|2𝑑𝑧𝜆𝑖𝑑𝑧𝜆𝑖𝐴𝑡𝑔=𝑒max𝑀𝑢𝜆𝑖𝑒𝑧𝜆𝑖+𝜆𝑖.(3.23) Since at any minimum point 𝑧0 of 𝑧𝜆𝑖, 2𝑧𝜆𝑖0,𝑧𝜆𝑖=0.(3.24) Consequently, at 𝑧0, we obtain 𝑒max𝑀𝑢𝜆𝑖𝑒𝑧𝜆𝑖𝑃𝐴𝑡𝑔Λ>0.(3.25) Thus, it is easy to verify that 𝑧𝜆𝑖 is bounded from below as 𝑖. This is a contradiction. So we see that (3.13) is true.
By a priori estimates results again, we deduce that 𝑧𝜆𝑖 converges to a smooth function 𝑧 in 𝐶 and 𝑧 satisfies (1.16) with 𝜆=Λ.
Finally, let us prove the uniqueness.
Denote𝑍=2𝑧+1𝑡𝑛2(Δ𝑧)𝑔+2𝑡2|𝑧|2𝑔𝑑𝑧𝑑𝑧𝐴𝑡𝑔,(3.26) and for any smooth functions 𝑧0 and 𝑧1, set 𝑣=𝑧1𝑧0,𝑧𝑠=𝑠𝑧1+(1𝑠)𝑧0,𝑍𝑠=2𝑧𝑠+1𝑡𝑛2Δ𝑧𝑠𝑔+2𝑡2|𝑧𝑠|2𝑔𝑑𝑧𝑠𝑑𝑧𝑠𝐴𝑡𝑔.(3.27) Then we get 𝑃𝑍1𝑍𝑃0=10𝑑𝑃𝑍𝑑𝑠𝑠=10𝑃𝑖𝑗+1𝑡𝑛2𝑙𝑃𝑙𝑙𝛾𝑖𝑗𝑍𝑠𝑑𝑠𝑣𝑖𝑗+𝑏𝑙𝑣𝑙(3.28) for some bounded 𝑏𝑙. Thus, if 𝑧0=𝜙,𝑧1=𝜙(3.29) are two solutions to (1.16) for some 𝜆 and 𝜆, respectively, then 𝑎𝑖𝑗 is positive definite. Therefore, 𝜙=𝜙+𝑐(3.30) for some constant 𝑐 by the maximum principle.

Acknowledgment

The authors acknowledge support from the NSF of China (11171210) and the Chinese Academy of Sciences.

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