Radial solutions for fully nonlinear elliptic equations of Monge–Ampère type

with D2u being the Hessian matrice of u, M(x, u, Du) being a given symmetric matrix function, and g being a positive function. In (1.1), the operator F is elliptic for u if D2u + M(x, u, Du) > 0, and at the same time, a C2-solution u is called an elliptic solution. Equations of the form (1.1) come from geometric optics [1], optimal transportation [2], conformal geometry [3], isometric embedding [4], and reflector–refractor problems [5– 8]. In optimal transportation, the function u is called the potential function and it satisfies the optimal transportation equation


Introduction
We firstly consider the Monge-Ampère-type equations with D 2 u being the Hessian matrice of u, M(x, u, Du) being a given symmetric matrix function, and g being a positive function. In (1.1), the operator F is elliptic for u if D 2 u + M(x, u, Du) > 0, and at the same time, a C 2 -solution u is called an elliptic solution.
Equations of the form (1.1) come from geometric optics [1], optimal transportation [2], conformal geometry [3], isometric embedding [4], and reflector-refractor problems [5][6][7][8]. In optimal transportation, the function u is called the potential function and it satisfies the optimal transportation equation with the vector field Y : D × R × R n → R n , Y = Y (x, z, p) being independent of z, D being a domain in R n , det Y p = 0, Y coming from a cost functionθ : R n × R n → R,θ =θ (x, y) being determined by the equations andη being a known nonnegative functionη : D × R × R n → R. The optimal transportation equation is of the form (1.1).
Monge-Ampère-type equations have got a lot of interest [9,10]. For M ≡ 0, the equation (1.1) becomes the standard Monge-Ampère equation det D 2 u = g(x, u, Du) which was investigated in [11][12][13], etc. For instance, the a priori estimates and existence of solutions [11], as well as the symmetry, existence, and nonexistence of solutions [13] were studied. The blow-up solutions to the Monge-Ampère equation and fully nonlinear equations and convex solutions of the Monge-Ampère systems can be found in [14][15][16]. If M = 0 and M is linear in, or independent of, the gradient Du [4], one has very similar a priori estimates as those of the standard Monge-Ampère equations. But if M is nonlinear in Du, the situation is very different. The interior a priori estimates of (1.1) were got in [8] for dimension two, and in [1] for all dimensions. The interior C 2 -estimates and the interior regularity of (1.1) were obtained by Ma-Trudinger-Wang [17] under some structure conditions on the matrix M and under a generalized target convexity condition. Under weaker conditions, Trudinger-Wang [18] have proved the global C 2 -estimates and regularity to the second boundary value problem of (1.1). However, the C 2,α -estimates are few. When M = M(x, Du), g = g(x), Liu-Trudinger-Wang [19] have proved the interior C 2,α -estimates of (1.1) under appropriate assumptions and Huang-Jiang-Liu [20] have got the boundary C 2,α -estimates of solutions. Jiang-Trudinger-Yang [21] have proved that the Dirichlet problem of (1.1) has a unique classical solution. Dai and Li [22] obtained the necessary and sufficient condition for the existence of subsolutions to (1.1) with M = κI, κ ≥ 0, and g = g(u).
When studying partial differential equations, we are interested in knowing whether the solutions are symmetric about some plane, or monotone in some direction. Gidas-Ni-Nirenberg [23] obtained the first symmetry results. Utilizing the moving plane method, they showed that if the solutions to the Dirichlet problem ofu = g(u), g ∈ C 0,1 (R) in a ball are C 2 and positive, then they are symmetric. Alexandrov [24] first introduced the distinguished the moving plane method and then Serrin [25] developed it. Later, many authors generalized the symmetry results. Especially, Li [26] extended the symmetry results to fully nonlinear elliptic equations G(x, u, Du, D 2 u) = 0, with G being uniformly elliptic on smooth domains, and the author of [27] studied the symmetry of viscosity solutions to fully nonlinear parabolic equations -u t + G(x, t, u, Du, D 2 u) = 0 in a singular domain. Zhang-Wang [13] have obtained that the classical solutions to the standard Monge-Ampère equations det D 2 u = e -u are symmetric.
Let D ⊂ R n be a domain. Consider the symmetry of solutions for the Dirichlet problem (1.2) We suppose that 0 ∈ D and the following hypotheses hold: (F) g ≥ g 0 > 0, g 0 is a constant, g is Lipschitz continuous in u.
We assume that m ij (x), i, j ∈ {1, 2, . . . , n} satisfy the following hypotheses: The symmetry results are the following. Now we discuss the existence and nonexistence of classical solutions for the equations with exponential terms on the right-hand side which usually appear in geometry. Consider where κ > 0 is a constant, I is the identity matrix, and B H = B H (0) is the ball centered at the origin and having radius H. Let D ⊂ R n be a bounded domain and 0 ∈ D. In the following, we will consider the Hessian equation where λ(D 2 u) = (λ 1 , . . . , λ n ) is the eigenvalues of D 2 u, is the kth elementary symmetric function of λ(D 2 u), k = 1, . . . , n and g is a known positive function of D.
Notice that if k = 1, equation (1.4) corresponds to u = g(u), which is semilinear, and if k = n, then we get det D 2 u = g(u), which is the standard Monge-Ampère equation which is fully nonlinear. For 1 < k < n, equation (1.4) is fully nonlinear and elliptic, and it is important in conformal geometry, etc., see [28]. We will restrict the class of functions to guarantee that (1.4) is elliptic. Let u ∈ C 2 (D) and k = {λ ∈ R n |σ l (λ) > 0, l = 1, . . . , k} be a convex cone, and let its vertex be the origin. If λ(D 2 u) ∈ k , then u is said to be uniformly k-convex. If λ(D 2 u) ∈ k , then u is said to be k-convex. According to [28] For any y ∈ ∂D, let ν(y) = (ν 1 , . . . , ν n-1 ), where ν j , j = 1, . . . , n -1, represent the principal curvatures of y ∈ ∂D. If ν(y) ∈ k-1 , then the domain D is said to be uniformly (k -1)convex. We always assume that the domain D in (1.4) is uniformly (k -1)-convex.
Equation (1.4) has seen a lot of investigation. If g = g(x), independent of u, the existence of solutions is considered, see [28][29][30], among other sources. If g = g(x, u), the multiplicity of (1.4) for k = 1 and k = n is investigated in [13,31,32]. For instance, Zhang-Wang [13] obtained the multiplicity of solutions for the Dirichlet problem of standard Monge-Ampère equation det D 2 u = e -u and det D 2 u = e τ u . For 1 < k < n, Jacobsen [33] considered the multiplicity of solutions of The author of [34] studied the multiplicity of solutions of Li [35] investigated the existence of classical solutions for the Dirichlet problem of the [36,37] investigated the symmetry of solutions to σ k (λ(D 2 u)) = g(x, u, Du).
In this paper, using the radial symmetry results for Hessian equations, we consider the following problem: (1.5) Lettingᾱ be a parameter, we also consider the problem (1.6) Theorem 1.4 There exists some constantᾱ * > 0 such that for anyᾱ >ᾱ * , there does not exist any solution of (1.6).
We arrange the paper as follows: Some maximum principles of linear elliptic equations are proved in Sect. 2. We get the symmetry of solutions for problem (1.2) in Sect. 3. Theorems 1.1 and 1.2 will be demonstrated in Sect. 4, and Theorems 1.3 and 1.4 will be demonstrated in Sect. 5.

Maximum principles for "narrow regions"
Let D ⊂ R n be a domain. In this section, a few maximum principles for the linear elliptic equation will be given. These maximum principles may be found in other references, but here for convenience we list them and give their proofs. We always assume in this section that a ij (x), θ (x) are continuous and bounded in D, and (a ij (x)) is uniformly elliptic, that is, there are some constants 0 < λ ≤ such that Proof We use contradiction arguments to prove the lemma. Letx ∈ D be some point satisfying u(x) < 0. Define Then from the fact that h(x) > 0 in D, we know that q(x) < 0. Let x 0 ∈ D be one minimum point of q(x), then q(x 0 ) < 0. By a direct computation we get that On the one hand, since q(x 0 ) is a minimum of q(x), we have On the other hand, by (2.1)-(2.3), and considering u(x 0 ) < 0, we get, at point x 0 , By virtue of (2.4) and (2.5), we then have This contradicts (2.6), and the proof of Lemma 2.1 is completed.
Using the same idea of the above arguments, we can extend Lemma 2.1 to unbounded domains provided that u is "nonnegative" at infinity.

Lemma 2.2 Let D be unbounded. If, in addition to the conditions in Lemma
Proof We consider the same q(x) as above in the proof of Lemma 2.1. Obviously, assumption (2.7) means that we cannot obtain the minimum of q(x) at infinity. So the minimum point of q is in the interior of D. Similar to the proof of Lemma 2.1, we finish the proof of Lemma 2.2.
In the following, we give some particular h(x) such that θ (x) satisfies (2.3) in the narrow regions, thus the maximum principle can be applied to those narrow regions. Let be a narrow region with width l -2ε, then we can choose For the constants and λ, if (a ij (x)) is uniformly elliptic, let ρ(x) = -λ/l 2 , so ρ(x) can be very negative when l is sufficiently small and Now applying Lemma 2.1, we have This contradicts the equation in (1.2). Next we prove that u = 0 in D. On the contrary, let some point x 0 ∈ D satisfy u(x 0 ) = 0. From the above proof, we know that u(x 0 ) is the maximum of u in D. Then at x 0 , D 2 u ≤ 0. We get a contradiction by the similar arguments as the above.
Proof of Theorem 1.1 Let Then we have (-1, 1, . . . , 1) and Q T being the transpose of Q. By the hypothesis (H 1 ), we know that From the Lipschitz continuity of g, for some constantθ 0 > 0, |θ (x, t)| ≤θ 0 . Since M is independent of Du, from [4,10], we know that the equation in (1.2) has the second derivative estimate for the solutions. So F ij is uniformly elliptic. Obviously, we have u = u t on ∂D t ∩ {x 1 = t}. In addition, we know that the reflection of ∂D ∩ ∂D t is in the interior of D, then, by Lemma 3.1, we get for t < 0, u t < 0 on ∂D ∩ ∂D t . Thus on ∂D t , h t ≥ 0. Thanks to Corollary 2.3, we know that h t ≥ 0 in D t if the width of D t is small enough.
Towards the right, the plane can be moved. Let We will prove that On the contrary, if t 0 < 0, we will demonstrate that the plane can still be moved a little to the right, and this contradicts the definition of t 0 . Indeed, if t 0 < 0, then the image of ∂D t 0 ∩ ∂D through the {x 1 = t 0 } reflection is inside D. But u < 0 in D by Lemma 3.1. It follows that on ∂D t 0 ∩ ∂D, h t 0 > 0. By the strong maximum principle for the linear elliptic equations, Assume that d 0 is the maximum width of such narrow regions in which we can still utilize the "narrow region principle. " Suppose that is a small positive constant satisfying < min{-t 0 , d 0 /2}. Let the narrow region be Now we take into account the function h t 0 + (x) on U t 0 + . Similarly to equation (3.2), . To see the boundary condition, we first note that it is satisfied on ∂U t 0 + ∩ ∂D, and on ∂U t 0 + ∩ {x 1 = t 0 + } the same holds, by virtue of the definition of h t 0 + . To illustrate that it is also true on ∂U t 0 + ∩{x 1 = t 0 -d 0 2 }, we use the continuity argument. If (t 0 -d 0 2 , x 2 , . . . , x n ) ∈ ∂D, it is clear that If (t 0 -d 0 2 , x 2 , . . . , x n ) ∈ D t 0 , by (3.3), we can get that there is a positive constant δ 0 such that Since for sufficiently small, h t 0 is continuous in t 0 , then we have Therefore the boundary condition in (3.4) holds for such a small . Now applying Corollary 2.3, we get Consequently, h t 0 + (x) ≥ 0 in D t 0 + which contradicts the definition of t 0 . Thus t 0 = 0, particularly h 0 ≥ 0 in D 0 . This implies for x 1 < 0 that If the plane is moved from right to left, then similarly It follows that This implies that u is symmetric about the plane {x : x 1 = 0}. For the sake of completing the proof of Theorem 1.1, it is needed to prove that u is decreasing in x 1 < 0. In fact, h t = 0 on the plane {x 1 = t, t < 0}, and h t > 0 on ∂D ∩ ∂D t . Due to the strong maximum principle, h t (x) > 0 in D t . We then apply the Hopf lemma and conclude that 2u x 1 = ∂h t ∂x 1 < 0.
Hence Theorem 1.1 is proved.
Since u is increasing in [0, H], we get κr + u (r) n ≥ e -u(r) r 0 ns n-1 ds = r n e -u(r) .
Set r = H, then Therefore, So if H is large enough, (1.3) has no solution.
Then for any H > 0, u(x) is one subsolution to (1.3). As a result, problem (1.3) has a unique elliptic solution.  (3), if a subsolution is found, then the existence of solution can be obtained from the global estimates in [35] and an argument similar to the following Lemma 5.3.

Existence and nonexistence results of Hessian equations
Let u ∈ C 3 (D) be a solution to (1.5). By Theorem 4.2 (see [38]), the equation in (1.5) is uniformly elliptic. Thanks to the Theorem 2.1 in [23], we have Theorem 5.1 Suppose that D is symmetric about a hyperplane. Any solution u ∈ C 3 (D) to (1.5) is also symmetric about the hyperplane.
Due to Theorem 5.1, if we let the x 1 axis be any direction, then we obtain Corollary 5.2 Let D be a ball. All solutions of (1.5) are radially symmetric with respect to the origin and monotone increasing along the radii.
Using the above symmetry results, we are able to prove Theorem 1.3. We first apply the upper and lower solution method to build a solution of (1.5) through iteration. We can also refer the reader to [39]. But we will give the proof for the sake of readers' convenience. Proof Let u 0 = u and x ∈ ∂D.
In addition, we have Then we have u 1 ≥ u 0 in D due to the comparison principle. So e -u 1 ≤ e -u 0 = S k (D 2 (u 1 )) in D. So u k ≤ u k+1 by induction, and then for any k, e -u k+1 ≤ S k (D 2 (u k+1 )). Due to Theorem 3.1 (see [30]), the C 1,1 (D) norm of u can control the C 1,1 (D) norm of u k . Thus the C 1,1 (D) norm of u can control the C 3,α (D) norm of u k+1 and so the sequence {u k } converges to u in C 2 (D). Therefore u is a solution to (1.5) when we take the limit in (5.1).
Proof of Theorem 1.3 According to Corollary 5.2, any solution to (1.5) is radially symmetric. Hence set u(x) = u(r), r = |x|. In addition, u (r) > 0 and u(0) is the minimum of u.
Then we have (4.1) and (4.2). A direct calculation gives that here C k-1 n-1 = (n-1)! (k-1)!(n-k)! . From (5.2), we deduce that By the integration for r from 0 to r ∈ [0, H] and using the fact that u (0) = 0, we have Since u is increasing, we know that e -u is then decreasing, and therefore get Assume that for some constantC > 0, u(0) = -C. By integrating with respect to r, we derive Particularly, we obtain from u(H) = 0 that So H ≤ (2k) Proof We only need to choose a subsolution to confine u in D 1 and utilize Lemma 5.3, and then we get the results.
Proof From Lemma 5.4, we know that if (1.5) has a solution in μD, then for any μ ∈ (0, μ), (1.5) has a solution in μ D; and if (1.5) has no solution in μD, then for any μ > μ, (1.5) has no solution in μ D. Define This completes the proof of Theorem 5.5.
Then we deduce that u -φ satisfies Consequently u -φ cannot achieve its maximum in D. This is a contradiction to the fact that x 0 is its maximum point in D.
From Lemma 5.6, we know that Theorem 1.4 holds.
Proof Assume that the sequence {(ᾱ n , u n )} satisfies (1.6) and u n → ∞,ᾱ n →ᾱ 0 > 0, n → ∞. Let v n = u n / u n . Then v n satisfies the equation S k D 2 v n = 1 u n k e -ᾱ n u n = 1 u n k e -ᾱ n u n v n . (5.5) For any D 1 ⊂⊂ D, from Lemma 5.7, we know that u n → ∞ uniformly in D 1 . In particular, v n → 1 uniformly in D 1 , which implies that S k (D 2 v n ) → 0 for each x ∈ D 1 . But, from (5.5) and the fact thatᾱ n and v n < 0 are bounded away from zero in D 1 , we derive that in D 1 , S k (D 2 v n ) → ∞. This is a contradiction. Henceᾱ → 0 as u → ∞. Thus we complete the proof.
The results of [13] and our results motivate the following conjecture.