SYMMETRY OF SOLUTIONS TO A CLASS OF MONGE-AMP`ERE EQUATIONS

. We study the symmetry of solutions to a class of Monge-Amp`ere type equations from a few geometric problems. We use a new transform to analyze the asymptotic behavior of the solutions near the inﬁnity. By this and a moving plane method, we prove the radially symmetry of the solutions.

1. Introduction. The radial symmetry for the positive solutions of the equations ∆u + f (u) = 0 for x ∈ R n was studied in [14] under the ground state conditions: u(x) → 0 as |x| → ∞. Afterwards, a lot of works have been published along this direction, dealing with various symmetry problems for semi-linear elliptic equations. See, for example [1,2,4,6,11,25] and the references therein. The radial symmetry for quasi-linear elliptic equations and fully nonlinear elliptic equations were also studied in [13,12,29] and in [2,7,24,26] respectively. However, all those symmetric results, up to authors' knowledge, need the a priori ground state condition.
In this paper, we study the radial symmetry for the solutions to the following equations of Monge-Ampère equation type: where n ≥ 2 is a positive integer.
Recall that a C 2 -function h(x) is called strictly convex in Ω if its hessian matrix [D 2 h(x)] is positive definite in Ω.
Since the solutions we considered for equation (1) are strictly convex, we may assume that the origin is the minimum point of u so that u(0) = 0, Du(0) = 0 (2) and D 2 u(0) = I (the unit matrix) (3) by translating, rotating and scaling the coordinates. Note that equation (1) is invariant under these coordinates transforms.
The solutions of (1) do not satisfy the ground state condition. In fact, because of the convexity it tends to infinity as x → ∞. Usually, it satisfies 1248 FAN CUI AND HUAIYU JIAN which will bring us many technique difficulties. Our new idea to overcome the difficulties is to use the transform , v(y) = 1 u(x) (5) introduced in [22], which is similar to the Kelvin transform in conformal geometry. This transform reduces equation (1) to an equation of the same type but with an isolated singular point in R n . However, by the advantages of this transform we will obtain the behavior of v near ∞ from that of u near 0. The latter can be easily understood because of (2), (3) and the convexity. To state our main results, we need the following structure assumptions on F : is strictly decreasing for s ∈ (0, ∞). Theorem 1.1. Suppose that F satisfies A1 and A2. If u ∈ C 4 (R n ) is strictly convex and satisfies (1)-(4) and then u is radially symmetric about the origin.
For the existence, uniqueness and the regularity of radially symmetric solution for (1), see [5] and the references therein.
Observe that the convexity and (2) implies that which will be used in Section 2 and Section 3 for the proof of Theorem 1.1. Our study is motivated by a few geometric equations. The first one is the equation of the form An important special case is p = n + 2, which describes the Minkowski problem in Centroaffine Geometry, and K 1 (x) is the Centroaffine Gauss curvature function of the convex hypersurface as the graph of u. See, for example [9,17,20,21,23]. In particular, when K 1 is a positive constant, (8) is exactly the equation for affine hyperbolic affine spheres [8]. For general p, Equation (8) is reduced from the study of Minkowski problem for complete noncompact convex hypersurfaces [10] and L p -Minkowski problem [18,19,21,27]. Using Legendre transform we see that (8) is reduced to which is the special case of (1) if K 1 (x) = K 1 (|x|).
The second model motivating our study is the equation in the form of where p = n+2 2 − 1 2α for some number α. Equation (11) is the equation of translating solutions to Gauss curvature flow ν = [ K K2(x,v(x),Dv(x)) ] α where ν is the velocity along the normal direction of the moving hyper-surface and K is its Gauss curvature. See [22,30] for the details. When K 2 is a positive constant and α = 1 n+2 , the classical results of Jorgens (n = 2, [16]), Calabi (n ≤ 5, [3]) and Pogorelov (n ≥ 2, [28]) assert that any convex solution to (11) must a quadratic polynomial. Obviously, it is radially symmetric under (2) and (3). When K 2 is a positive constant and α ∈ (0, 1 2 ), there exists a radially symmetric solution to (11) as well as infinitely many smooth, non-rotationally symmetric convex solutions to (11). See Theorem 6 in [30] and Theorem 1.2 in [22] respectively.
The existence for (1), especially for (8) (or (10)) and (11), are better understood. See the references mentioned above. But little is known for the uniqueness or the behavior of the solutions near ∞. Theorem 1.1 can be viewed as the first step in understanding this respect. In fact it can be viewed as a generalization of the classical results of Jorgens, Calabi and Pogorelov as mentioned above.

A transform and asymptotic behavior.
In this section, we use the transform introduced in [21] to reduce the behavior of u near the origin to that of the new function near ∞. The advantage of this transform is to preserve the form of equation (1). We will study the properties of the new functions.
Consider the transform where we assume u ∈ C 2 and u > 0. Then .
Denoting u i = ∂u ∂xi and v i = ∂v ∂yi , we have From this we solve u i as which implies and where v ij := v yiyj and b k = y k y · Dv − v .

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Since Lemma 2.1. Suppose that a strictly convex function u ∈ C 2 (R n ) satisfies (1), (2) and (4). Let y and v be given by (12). Then and v ∈ C 2 (R n \{0}) is a strictly convex and positive function, satisfying equation (2) and (3) where ξ is a point in the segment connecting 0 and x. On the other hand, we have, by (4) and the strictly convexity, that lim x→∞ Since x · Du(x) − u(x) > 0 for all x ∈ R n \{0} by the convexity of u and (2). Hence by (14). Therefor, (15) implies that [v kl ] n×n is positive definite and so v is strictly convex for y ∈ R n \{0}. Finally, (17) follows directly from (1), and (12)- (16). (2) and (3), there is a function h ∈ C 1 (B r (0)) such that h(0) = 0 and for any x near 0, where Proof. By (2), (3) and Taylor expansion, we have, by Taylor expansion again, that |u xixj x k (ξ)| when x near 0. Letting we obtain the desired result. (2) and (3). If y and v(y) are given by (12), then there exists a r 0 > 0 such that for all λ ≥ 3M, where M is the same constant as in Lemma 2.2.

Since Lemma 2.2 implies
), as y → ∞, we have, for any λ > 0, any |y| > r (large enough) such that where ξ is a point in the segment connecting y v(y) and y λ v λ (y) . Observing that and this, together with the fact |y| < |y λ | for all λ > 0 with y 1 < λ, implies lim y→∞ y λ v λ (y) = 0 uniformly for all λ > 0, we can choose large r such that for all λ > 0 and all |y| > r with ), ∀|y| > r with y 1 < λ (22) and

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Hence, we have, by Lemma 2.2, that for all |y| > r with y 1 < λ, Substituting this inequality into (22) and using (20) and (21), we prove the Lemma.
3. Proof of Theorem 1.1. In this section, we gives the proof of Theorem 1.1. First, we state two preliminary results for the sake of convenience.
where Ω is an open set in R n , a ij , b i , C(x) ∈ L ∞ (Ω ) and the matrix [a ij (x)] is positive in Ω for any compact set Ω ⊂Ω. Then either w ≡ 0 in Ω or w(x) > 0 for all x ∈ Ω. Moreover, if w(x 0 ) > 0 for some x 0 ∈ Ω and w(x) = 0 for somex ∈ ∂Ω which is smooth nearx, then ∂w ∂ν (x) < 0 where ν is the unit outer normal of ∂Ω. The proof can be found in [15].
Proof of Theorem 1.1. Assume u ∈ C 4 (R n ) is strictly convex and satisfies (1)-(4) and (6). Let y and v(y) be given by (12). It is enough to prove that v is radially symmetric about the origin in R n \{0}.
It follows Lemma 2.1 that v ∈ C 4 (R n \{0}) is strictly convex, satisfying for all x ∈ R n \{0}, where and We will use the moving planes method to prove that v is radially symmetric with respect to the origin. This needs to show that v is symmetric in any direction with respect to the origin. Since equation (30) is invariant under orthogonal transforms, it is sufficient to do this in one direction. Without loss of generality, we will do it in e 1 -direction. In a word, to show Theorem 1.1, it is enough to prove that . Denote the term on the right hand side of equation (30) by G(v)(x), i.e., where I(v) is defined as (31). Let One key to use the moving planes method is to introduce the following differential operator: By assumption A1 and a mean value theorem, we have the following obvious result. If v ∈ C 2 (B r (0)\{0}) is bounded, positive and strictly convex, then for any compact Ω in B r (0)\{0, 0 λ }, there exist a constant C 1 > 0 independent of λ ∈ (0, ∞) (but depending on Ω) and piecewise continuous functions {a ij λ (x)}, {b i λ (x)}, C λ (x), (all depending on the v and its derivatives up to second order in Ω), such that and We will complete the proof of Theorem 1.1 by four Steps.
Step 1. (i) There exists aλ > 4M (depending only on the function v) such that (ii) Given λ ∈ (0, ∞). If To show (i), we use the fact lim y→∞ v(y) = +∞ in (32) to find aλ > 3M such that for all where M, r 0 are the same as in Lemmas 2.2 and 2.3. This, together with Lemma 2.3 again, implies This proves (i).

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Step 2. If 0 < a < b, then there exists a R = R(a, b) > 0 such that for any λ ∈ (a, b), To show this, we use (6) and Lemma 2.4 to find that for any λ > 0 and any x ∈ R n such that x 1 > λ and |x| is large enough, Since H(x) = O(1) as x → ∞ and lim x→∞ |x| |x λ | = 1 uniformly for λ ∈ (a, b), there is a R 1 > 2b such that for all λ ∈ (a, b) On the other hand, it follows from Lemma 2.4 again that where ξ is a point in the segment connecting x and x λ . Observing that ξ → ∞ as Consequently, (38) is also true for all x ∈ (λ)\(B R (0) {0 λ }). This proves Step 2.
Let Ω 0 be the symmetric set of Ω λ0 0 with respect to the plane x 1 = λ 0 . Then Ω 0 is an open set andx is its interior point, and ∂v ∂x 1 (x λ0 ) < 0 ∀x ∈ Ω 0 .