Exterior Dirichlet Problem for Translating Solutions of Gauss Curvature Flow in Minkowski Space

and Applied Analysis 3 We conclude that uR are monotone in R and converge locally uniformly to a continuous function u according to Dini’s theorem. To simplify the notation, we will omit the index R and from now on assume that u is a solution of (10) with R fixed sufficiently large.The estimate for the first derivatives is stated in the following lemma. Lemma 5. For R/2 ≤ |x| ≤ R, there is a constant c independent of R such that 󵄨󵄨󵄨∇] (u − u) (x) 󵄨󵄨󵄨 ≤ c R , (14) 󵄨󵄨󵄨∇τu (x) 󵄨󵄨󵄨 ≤ c √R , (15) √1 − c R ≤ |Du (x)| < 1, (16) where ] = x/|x| and τ are unit vectors parallel and orthogonal to x, respectively. Proof. From the convexity of u and Lemma 4, we can prove (14) and (15) by using the similar proof techniques of (2.2) and (2.3) in [10]. Then, we need only to prove (16). Since u is strictly convex, for R/2 ≤ |x| ≤ R, |Du| attains its maximum at ∂B R . In view of (8), we may take 󵄨󵄨󵄨Du 󵄨󵄨󵄨 2 = O(1 − c R ) . (17) Hence for x ∈ ∂B R , by (14) and (17) we have |Du(x)| 2 = 󵄨󵄨󵄨∇τu(x) 󵄨󵄨󵄨 2 + 󵄨󵄨󵄨∇]u(x) 󵄨󵄨󵄨 2 = 󵄨󵄨󵄨∇τu(x) 󵄨󵄨󵄨 2 + 󵄨󵄨󵄨∇]u(x) + ∇](u − u)(x) 󵄨󵄨󵄨 2 ≥ 󵄨󵄨󵄨Du(x) 󵄨󵄨󵄨 2 − 2 󵄨󵄨󵄨∇] (u − u) (x) 󵄨󵄨󵄨 ≥ 1 − c R . (18) On the other hand, by the proof of Theorem 4.1 in [12], max ∂B R |Du| ≤ max ∂B R 󵄨󵄨󵄨Du 󵄨󵄨󵄨 < 1. (19) The lemma is completed. 3. C2 A Priori Estimates In this section, we prove theC a priori estimates for solutions of (10) under the assumption of Theorem 2. As in [12], one obtains that the second derivatives on ∂D are bounded uniformly in R. Furthermore, by considering the function w = a 2 |Du| 2 + log u ξξ (20) for some constant a > 0 and assuming its maximum over (x, ξ) ∈ B R \ D × Sn−1 is attained at an interior, one can prove that max B R \D |Du| 2 ≤ c + max ∂B R ∪∂D 󵄨󵄨󵄨󵄨 D 2 u 󵄨󵄨󵄨󵄨 . (21) Therefore, it suffices to bound |D2u| on the outer boundary ∂B R . Next, we will give estimates for the tangential second derivatives, the mixed second derivatives, and the normal second derivatives on the outer boundary ∂B R , respectively. Theorem6 (tangential second derivatives at the outer boundary). Let x 0 ∈ ∂B R and τ 1 , τ 2 be tangential directions at x 0 . Then we have at x 0 , 󵄨󵄨󵄨󵄨 u τ 1 τ 2 − |x| τ 1 τ 2 󵄨󵄨󵄨󵄨 ≤ c R . (22) Proof. Wemay assume that x 0 = R⋅e n ≡ R⋅(0, . . . , 0, 1).Then ∂B R is represented locally as graph of ω, where ω (x) = √R2 − |x| 2 , x = (x 1 , x 2 , . . . , x n−1 ) ∈ R n−1 . (23) Note that the Dirichlet boundary condition implies (u − u) (x, ω (x)) = 0. (24) We differentiate twice with respect to x i , x j , 1 ≤ i, j ≤ n − 1 to obtain that, at x 0 , (u − u) ij + (u − u) n ω ij + 2(u − u) nj ω i = 0. (25) According to the decay conditions at infinity (8), we have |u ij − |x| ij | = O(1/R). Observing that w i (x 0 ) = 0, ω ij (x 0 ) = − δ ij R . (26) Then, by Lemma 5 we have 󵄨󵄨󵄨󵄨 u ij − |x| ij 󵄨󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨󵄨 −(u − u) n ω ij + (u − |x|) ij 󵄨󵄨󵄨󵄨󵄨 ≤ c R . (27) Theorem7 (mixed second derivatives at the outer boundary). For x 0 ∈ ∂B R , let τ, ] be unit vectors in tangential and normal directions, respectively. Then 󵄨󵄨󵄨uτ] (x0) 󵄨󵄨󵄨 ≤ c √R . (28) The proof is going to be put in three lemmas and will be finished below Lemma 10. Similar to Theorem 6, we may assume that x 0 = R ⋅ e n and represent ∂B R locally as graph of ω with ω(x) = √R − |x|2. We take the logarithm of (3), log det u ij − n + 2 − β 2 log (1 − |Du|2) = 0, (29) and differentiate with respect to x k , u ij u ijk + n + 2 − β 1 − |Du| 2 u i u ik = 0, (30) 4 Abstract and Applied Analysis where (uij) = (u ij ) −1. We introduce the linear differential operator L by Lw := u ij w ij + n + 2 − β 1 − |Du| 2 u i w i (31) and define the linear operator for t < n:


Introduction and Main Results
The Euclidean space R +1 endowed with the Lorentz metric  2 = ∑  =1  2  −  2 +1 is called Minkowski space.We denote it by R ,1 .A space-like hypersurface in R ,1 is a Riemanian -manifold, having an everywhere lightlike normal field ] which we assume to be future directed and thus satisfy the condition ⟨], ]⟩ = −1.Locally, such surfaces can be expressed as graphs of functions  +1 = ( 1 , . . .,   ) : R  → R satisfying the space-like condition |()| < 1 for all  ∈ R  .
If a family of space-like hypersurfaces   = (⋅,) : R  → R ,1 satisfies the evolution equation on some time interval, we say that the surfaces   :=   () are evolved by   -flow, where (⋅, ) is the Gauss curvature of   and  ̸ = 0 is a constant.When the initial surface is a graph over a domain Ω ⊂ R  , (1) is equivalent, up to a diffeomorphism in R  , to with |(⋅, )| < 1, where  is a function defined in Ω×[0, ).The flow (2) was studied in [1] for the special case  = 1.In fact, the authors in [1] used the flow (2) to prove existence and stability of smooth entire strictly convex spacelike hypersurfaces of prescribed Gauss curvature and give a new proof of Theorem 3.5 in [2].
In this paper, we consider strictly convex space-like hypersurfaces of translating solutions to   -flow as graphs over R  \ , where  ⊂ R  is an open domain whose boundary  is a smooth submanifold of R  .We want to look for a function  ∈  ∞ (R  \ ), which solves the problem (3)-(4) with the boundary condition where  ∈  ∞ () is a given function.
There are similar problems for the equation of translating solution of Gauss curvature flow in Eucliden space [9], the equation of prescribed Gauss curvature in Eucliden space [10], and the equation of prescribed Gauss curvature in Minkowski space [11], respectively.It was shown that there are convex solutions to the Dirichlet problems for the three equations on exterior domains, and the solution is close to the rotationally symmetric one at infinity for the first equation and close to a cone for the second and third equation under the assumption that there exists a strictly convex subsolution which is close to a cone up to the third order (see (7) and ( 8)).
In this paper, we will show that the same results as in [10,11] hold for the problem (3)- (5).We would like to point out that (3) is essentially different from the equations in [9][10][11].For example, the equation of prescribed Gauss curvature in Minkowski space, det( 2 ) = (1 − || 2 ) (+2)/2 , has an explicit solution  = √ 1 + || 2 , from which one can easily construct subsolution or supersolution for given Dirichlet problems.However, it is unknown if there is such a solution to (3).In particular, it has no solution in the form of Here and below, we set  = 1/.
The main result of this paper is the following theorem.Then there exists a smooth, strictly convex hypersurface of the exterior Dirichlet problem (3)-( 5) and the solution  is close to a cone in the sense that sup Although the above theorem has an obvious disadvantage that it assumes the existence of a locally strictly convex subsolution, this assumption is reasonable and necessary in some case for the Dirichlet problems on nonconvex domains; see [12] for the details.However, in the special case when  =   0 (0) is a ball and the boundary values are zero, we can construct an explicit subsolution.Theorem 3. Let  =   0 (0) with  0 > 0 and  ≡ 0. If  ≤ 0, then there is a strictly convex subsolution  of (3)-( 5) such that (7) where  > 4 0 and  ⊂   0 for some constant  0 > 1.It is well known from the standard continuity method as in [13] that the Dirichlet problem (10) 3)-( 5) on exterior domain R  \ .
The paper is organized as follows.In Section 2, we prove the  0 and  1 a priori estimates for   .The  2 -estimates are given in Section 3. Finally, we prove Theorem 3 in the last section.

𝐶 0 and 𝐶 1 A Priori Estimates
From now on, we assume  and  as in Theorem 2 and   as in (10); lower indices denote partial derivatives in R  , for example,   = /  .The inverse of the Hessian of  is denoted by (  ) = (  ) −1 .We use the Einstein summation convention.The letter  denotes a constant independent of  which may change its value from line to line throughout the text.
Owing to the maximum principle, we can obtain the following lemma as Lemma 2.2 in [10].

Lemma 4. The functions 𝑢 𝑅 converge locally uniformly to a continuous function
Proof.From the maximum principle we obtain that for any  > 4 0 and for 4 0 <  1 <  2 .Again by the maximum principle, we have We conclude that   are monotone in  and converge locally uniformly to a continuous function  according to Dini's theorem.
To simplify the notation, we will omit the index  and from now on assume that  is a solution of (10) with  fixed sufficiently large.The estimate for the first derivatives is stated in the following lemma.
where ] = /|| and  are unit vectors parallel and orthogonal to , respectively.
Proof.From the convexity of  and Lemma 4, we can prove ( 14) and (15) by using the similar proof techniques of (2.2) and (2.3) in [10].Then, we need only to prove (16).Since  is strictly convex, for /2 ≤ || ≤ , || attains its maximum at   .In view of (8), we may take Hence for  ∈   , by ( 14) and (17) we have On the other hand, by the proof of Theorem 4.1 in [12], max The lemma is completed.

𝐶 2 A Priori Estimates
In this section, we prove the  2 a priori estimates for solutions of (10) under the assumption of Theorem 2. As in [12], one obtains that the second derivatives on  are bounded uniformly in .Furthermore, by considering the function for some constant  > 0 and assuming its maximum over (, ) ∈   \  ×  −1 is attained at an interior, one can prove that max Therefore, it suffices to bound | 2 | on the outer boundary   .
Next, we will give estimates for the tangential second derivatives, the mixed second derivatives, and the normal second derivatives on the outer boundary   , respectively.Theorem 6 (tangential second derivatives at the outer boundary).Let  0 ∈   and  1 ,  2 be tangential directions at  0 .Then we have at  0 , Proof.We may assume that  0 = ⋅  ≡ ⋅(0, . . ., 0, 1).Then   is represented locally as graph of , where Note that the Dirichlet boundary condition implies We differentiate twice with respect to x , x , 1 ≤ ,  ≤  − 1 to obtain that, at  0 , According to the decay conditions at infinity (8), we have Then, by Lemma 5 we have Theorem 7 (mixed second derivatives at the outer boundary).
For  0 ∈   , let , ] be unit vectors in tangential and normal directions, respectively.Then The proof is going to be put in three lemmas and will be finished below Lemma 10.Similar to Theorem 6, we may assume that  0 =  ⋅   and represent   locally as graph of  with ( x) = √  2 − |x| 2 .We take the logarithm of (3), and differentiate with respect to   , where (  ) = (  ) −1 .We introduce the linear differential operator  by and define the linear operator for  < : In the following we restrict attention to the domain where ] = /|| and  is unit vector orthogonal to .
To prove (35), by Lemma 5, we may take 1/(1 − || 2 ) = ().In view of ( 8 In the next lemma, we introduce a function V, which will be the main part of a barrier function to prove Theorem 7.

Lemma 9.
There exists a positive constant  independent of  such that fulfills the estimates provided that  =  3/4 and  is sufficiently large.Here  =  − || is the distance from   .
Proof.In view of  =  3/4 , for  ∈ Ω  ,  = −|| ≤  =  3/4 , and  ≥ , we have We fix  ∈ Ω  and set ] = /||.Let ,   belong to an orthonormal basis for the orthogonal complement of ] which we choose such that the submatrix (   ) is diagonal.Assume that ] and ,   correspond to the indices  and 1, . . .,  − 1, respectively.We use the Einstein summation convention for ,   .The matrix   is positive, and thus testing with the vectors ] ±  gives In view of Direct computations using (17) give By ( 8), (16), and (42) we have Then, Thus, for  large enough, we have Expanding the determinant and using that (   ) is a diagonal matrix give det (  ) = det ( ( By the inequality for arithmetic and geometric means, Hence for large , (50) Note that we have used the fact (3 − 2 − 2)/4 > 1/2 in the last inequality, which is from the assumption  < (3/2) − 2.
Lemma 10.There exists a positive constant  independent of  such that where  =  3/4 and V is as in Lemma 9.
Proof.According to Lemma 9, the fact Θ ≥ 0 on Ω  follows from which holds for  large enough.

Proof of Theorem 3
In this section, we prove Theorem 3, which gives a simple example of a barrier construction.
has a locally strict convex solution in  ∞ (  \ ).Our main task is to show that the  2 -norms of   are uniformly bounded in .
Theorem 11 (double normal  2 -estimates at the outer boundary).Under the assumption of Theorem 2 and the notation of Theorem 7, we have      ]] ( 0 )     ≤ .(59) Proof.As the proof of Lemma 9, we fix  0 ∈   and set ] =  0 /| 0 |.Let ,   belong to an orthonormal basis for the orthogonal complement of ] which we choose such that the submatrix (