Maximal hypersurfaces over exterior domains

TL;DR

This paper investigates the exterior problem for the maximal surface equation, establishing the asymptotic behavior at infinity and proving unique solvability under certain boundary and asymptotic conditions.

Contribution

It provides a detailed analysis of the exterior maximal surface problem, including asymptotics and uniqueness results, which were not previously fully understood.

Findings

Precise asymptotic behavior of solutions at infinity.

Unique solvability of the exterior Dirichlet problem.

Conditions for admissible boundary data and prescribed asymptotics.

Abstract

In this paper, we study the exterior problem for the maximal surface equation. We obtain the precise asymptotic behavior of the exterior solution at infinity. And we prove that the exterior Dirichlet problem is uniquely solvable given admissible boundary data and prescribed asymptotic behavior at infinity.

Equations

$$div(\frac{Du}{\sqrt{1-|Du{|}^2}})=0,$$

$$△u+\frac{(Du{)}^{{}^{\prime }}{D}^{2}uDu}{1-|Du{|}^2}=0.$$

$$u(x)$$

$$+Res[u]|a|\sqrt{1-|a{|}^2}\frac{|x|(a\cdot x)}{|x{|}^2-(a\cdot x{)}^2}\cdot \frac{\ln |x|}{|x|}+O_k(|x{|}^{-1})$$

$$u(x)=a\cdot x+c-(1-|a{|}^2)Res[u](\sqrt{|x{|}^2-(a\cdot x{)}^2}{)}^{2-n}+O_k(|x{|}^{1-n})$$

$$u(x)=a\cdot x+d\ln \sqrt{|x{|}^2-(a\cdot x{)}^2}+O(1)\text{mbox}asx\to \infty ;$$

$$u(x)=a\cdot x+c+o(1)\text{mbox}asx\to \infty .$$

$$C_{X_0}=\{X\in {\mathbb{L}}^{n+1}:⟨X-X_0,X-X_0⟩=0\}.$$

$$L_R(X_0)=\{X\in {\mathbb{L}}^{n+1}:⟨X-X_0,X-X_0⟩<R^2\}.$$

$$\underset{v\in K}{\sup }{\int }_{\Omega }\sqrt{1-|Dv{|}^2}$$

$$\underset{x\to x_0,x\in l}{\lim }v(x)=\phi (x_0).$$

$$regu:=\Omega \backslash singu$$

$$singu:=\{\overline{xy}:x,y\in \partial \Omega ,x\ne y,\overline{xy}\subset \Omega \text{mbox}and|\phi (x)-\phi (y)|=|x-y|\}.$$

$$u(tx+(1-t)y)=t\phi (x)+(1-t)\phi (y),0<t<1$$

$$w_{\lambda }(x):={\int }_0^{|x|}\frac{\lambda }{\sqrt{t^{2(n-1)}+{\lambda }^2}}dt.$$

$${\int }_0^r\frac{\lambda }{\sqrt{t^{2(n-1)}+{\lambda }^2}}dt=M(\lambda ,n)-\frac{\lambda }{n-2}{r}^{2-n}+O(r^{4-3n})$$

$${\int }_0^r\frac{\lambda }{\sqrt{t^2+{\lambda }^2}}dt=m(\lambda )+\lambda \ln r+O(r^{-2})$$

$$L_{\kappa }:(x^{\prime },x_n,t)\to (x^{\prime },\frac{x_n+\kappa t}{\sqrt{1-{\kappa }^2}},\frac{\kappa x_n+t}{\sqrt{1-{\kappa }^2}})$$

$$L_{\kappa }(\{(x,t)\in {\mathbb{L}}^{n+1}:t=T\})=\{(x,t)\in {\mathbb{L}}^{n+1}:t=\sqrt{1-{\kappa }^2}T+\kappa x_n\},$$

$$w\left(x\right)=\underset{b\in \partial B_{R^{\ast }}}{\inf }\left\{u\left(b\right)+m\left|x-b\right|\right\}$$

$$w\left(y\right)$$

$$\le \underset{b\in \partial B_{R^{\ast }}}{\inf }\left\{u\left(b\right)+m\left|x-b\right|+m\left|y-x\right|\right\}$$

$$\le w\left(x\right)+m\left|y-x\right|.$$

$$w\left(x\right)=\left|x\right|\left[u\left(x/\left|x\right|\right)-m\right]+m$$

$$\left|Dw\left(x\right)\right|=\left|Du\left(x/\left|x\right|\right)\right|<1.$$

$$\left|u\left(x/\left|x\right|\right)-m\right|\le \frac{{\max }_{|x|=1}u\left(x\right)-{\min }_{|x|=1}u\left(x\right)}{2}<1.$$

$$-(1-ϵ)|x|\le u(x)-c_0\le (1-ϵ)|x|$$

$$\underset{r\to \infty }{\lim }\frac{u(rx)}{r}=a\cdot x\mathrm{locally}\mathrm{uniformly}\mathrm{in}{\mathbb{R}}^n.$$

$$|II{|}^2(x)\le \frac{c(n){\rho }^2}{({\rho }^2-l_{x_0}^2(x){)}^2}$$

$$|II|(x)\le \frac{C(ϵ,n)}{|x|}\text{mbox}for|x|\ge \frac{8}{ϵ}.$$

$$|u(x)-u(y)|\le |u(x)|+|u(y)|\le (1-ϵ)|x|+(1-ϵ)$$

$$|x-y|\ge |x|-1.$$

Full text

1. Introduction

The maximal surface equation is

[TABLE]

or equivalently in the non-divergence form

[TABLE]

This equation arises as the Euler equation of the variational problem that maximize the area functional $\int\sqrt{1-|Du|^{2}}$ among the spacelike hypersurfaces in the Lorentz-Minkowski space $\mathbb{L}^{n+1}$ (see the definitions in section 2). The graph of a solution to (1.1) is called a maximal hypersurface and the graph of a solution to the variational problem is called an area maximizing hypersurface.

Calabi [Ca68] ($n\leq 4$) and Cheng-Yau [CY76] (all dimensions) proved that every entire maximal hypersurface in $\mathbb{L}^{n+1}$ or every global solution $u$ to the maximal surface equation (1.1) with $\left|Du\left(x\right)\right|<1$ on $\mathbb{R}^{n}$ must be linear.

The Dirichlet problem for bounded domain was studied by Bartnik-Simon [BS82] and the isolated singularity problem was studied by Ecker [Ec86]. The exterior problem is a “complimentary” one for elliptic equations; see for example [Be51][Si87] for minimal hypersurfaces, [CL03] for Monge-Ampere equation, [LLY17] for special Lagrangian equation and [HZ18] for infinity harmonic functions, besides the classic works such as [GS56] for linear ones. We study the exterior problem for the maximal surface equation in this paper. We obtain the precise asymptotic behavior of the exterior solution at infinity. And we prove that the exterior Dirichlet problem is uniquely solvable.

Throughout the paper, we assume $A\subset\mathbb{R}^{n}$ be a bounded closed set. We say $u$ is an exterior solution in $\mathbb{R}^{n}\backslash A$ if $u\in C^{2}(\mathbb{R}^{n}\backslash A)$ with $|Du(x)|<1$ solve the equation (1.1) in $\mathbb{R}^{n}\backslash A$. Given an exterior solution $u,$ for any bounded $C^{1}$ domain $U\supset A,$ the integral $Res[u]:=\int_{\partial U}\frac{\partial u/\partial\vec{n}}{\sqrt{1-|Du|^{2}}}d\sigma$ is independent of the choices of $U$ because of the divergence structure of the equation. The number $Res[u]$ can be regarded as the residue of the exterior solution $u.$

Theorem 1.1**.**

Let $u$ be a smooth exterior solution in $\mathbb{R}^{n}\backslash A$ with A\being bounded. Then there exist a vector $a\in\mathring{B}_{1}$ and a constant $c\in\mathbb{R}$ such that for $n=2$

[TABLE]

and for $n\geq 3$

[TABLE]

as $|x|\rightarrow\infty$ for all $k=0,1,2,\cdots$. The notation $\varphi(x)=O_{k}(|x|^{m})$ means that $|D^{k}\varphi(x)|=O(|x|^{m-k})$.

On the other hand, for any bounded closed set $A$, given an admissible boundary value function $g:\partial A\rightarrow\mathbb{R}$ and prescribed asymptotic behavior at infinity, the exterior Dirichlet problem for maximal surface equation is uniquely solvable. We say $g$ is admissible if $g$ is bounded and there exists a spacelike function $\psi$ in $\mathbb{R}^{n}\backslash A$ such that $\psi=g$ on $\partial A$ in the sense of (1.1) in [BS82] (see Remark 2.1 in Section 2).

Theorem 1.2**.**

Let $A\subset\mathbb{R}^{n}$ be a bounded closed set and $g:\partial A\rightarrow\mathbb{R}$ be an admissible boundary value function. Then

1. (1)

$n=2$, given any $a\in B_{1}$ and $d\in\mathbb{R}$, there exist a unique smooth solution $u$ of maximal surface equation on $\mathbb{R}^{2}\backslash A$ such that $u=g$ on $\partial A$ and

[TABLE] 2. (2)

$n\geq 3$, given any $a\in B_{1}$ and $c\in\mathbb{R}$, there exist a unique smooth solution $u$ of maximal surface equation on $\mathbb{R}^{n}\backslash A$ such that $u=g$ on $\partial A$ and

[TABLE]

Of course $u$ enjoys finer asymptotic properties and the relation $d=(1-|a|^{2})Res[u]$ holds by Theorem 1.1.

The article is organized as follows. In Section 2, we set up some notations and definitions, and we collect some results from [CY76], [BS82], and [Ec86] that are needed in the proofs of the later sections. In Section 3, we prove that a spacelike function over an exterior domain can be spacelikely extended to the whole $\mathbb{R}^{n}$. This is the starting point of our work. Interestingly there is a striking similarity between our argument and the argument in [CL03, p.571-572] where Caffarelli and Li prove the locally convex solution of $\det D^{2}u=1$ over an exterior domain can be extended (after finitely enlarging the complimentary domain $A$) to a global convex function. In Section 4, we prove a growth control theorem for the exterior solution $u$ at infinity. This is the key content of this paper. Inspired by Ecker’s proof in [Ec86], and relying on his results there, our argument involves compactness, blowdown analysis and comparison principle. In Section 5, we prove gradient estimate for $u$ based on the growth control theorem and Cheng-Yau’s estimate on the second fundamental form. In Section 6, we prove Theorem 1.1. Since the equation (1.1) becomes uniformly elliptic by the gradient estimate of the previous section, the standard tools such as Harnack inequality and Schauder estimate apply. The known radially symmetric solutions play a key role in the proof. In Section 7, we prove Theorem 1.2. We solve the equation in a series of bigger and bigger ring-shaped domains and use the compactness method to get an exterior solution. We use the Lorentz transformations of radially symmetric solutions as barrier functions to locate the position of the exterior solution. The uniqueness of solutions follows from comparison principle.

2. Notations and preliminary results

We denote the Lorentz-Minkowski space by $\mathbb{L}^{n+1}=\{X=(x,t):x\in\mathbb{R}^{n},t\in\mathbb{R}\}$, with the flat metric $\sum_{i=1}^{n}dx_{i}^{2}-dt^{2}$. And $\langle\cdot,\cdot\rangle$ denotes the inner product in $\mathbb{L}^{n+1}$ with the signature $(+,\cdots,+,-)$.

The light cone at $X_{0}=(x_{0},t_{0})\in\mathbb{L}^{n+1}$ is defined by

[TABLE]

The upper and lower light cones will be denoted by $C^{+}_{X_{0}}$ and $C^{-}_{X_{0}}$ respectively.

The Lorentz-balls are defined by

[TABLE]

Let $M$ be an $n$-dimensional hypersurface in $\mathbb{L}^{n+1}$ which can be represented as the graph of $u\in C^{0,1}(\Omega)$, where $\Omega$ is a open set in $\mathbb{R}^{n}$. We say that $M$ (or $u$) is

weakly spacelike if $|Du|\leq 1$ a.e. in $\Omega$,

spacelike if $|u(x)-u(y)|<|x-y|$ whenever $x,y\in\Omega$, $x\neq y$ and the line segment $\overline{xy}\subset\Omega$, and

strictly spacelike if $u\in C^{1}(\Omega)$ and $|Du|<1$ in $\Omega$.

If $M$ (or $u$) is strictly spacelike and $u\in C^{2}(\Omega)$, the Lorentz metric on $\mathbb{L}^{n+1}$ induces a Riemannian metric $g$ on $M$. Under the coordinates $(x_{1},\cdots,x_{n})\in\Omega$, $g_{ij}=\langle\frac{\partial X}{\partial x_{i}},\frac{\partial X}{\partial x_{j}}\rangle=\delta_{ij}-u_{i}u_{j}$, where $X=(x,u(x))$ is the position vector on the graph of $u$, and $u_{k}=u_{x_{k}}=\frac{\partial u}{\partial x_{k}}$ for $k=1,\cdots,n$. So $g=I-Du(Du)^{{}^{\prime}}$, $\det g=1-|Du|^{2}$, $g^{-1}=I+\frac{Du(Du)^{{}^{\prime}}}{1-|Du|^{2}}$ and $g^{ij}=\delta_{ij}+\frac{u_{i}u_{j}}{1-|Du|^{2}}$. The second fundamental form is $II_{ij}=\frac{u_{ij}}{\sqrt{\det g}}$ and so $|II|^{2}=\frac{g^{ij}g^{kl}u_{ik}u_{jl}}{\det g}$ (see (2.3) in [BS82]) where $u_{ij}=\frac{\partial^{2}u}{\partial x_{i}\partial x_{j}}$ and the summation convention on repeated indices is used. Note that $|D^{2}u|\leq|II|.$

The following fundamental results were achieved by Bartnik and Simon in [BS82].

Theorem 2.1** **(Solvability of variational problem on bounded domains [BS82, Proposition

1.1]).

Let $\Omega\subset\mathbb{R}^{n}$ be a bounded domain and let $\varphi:\partial\Omega\rightarrow R$ be a bounded function. Then the variational problem

[TABLE]

where $K=\{v\in C^{0,1}(\Omega):|Dv|\leq 1$ a.e. in $\Omega$, $v=\varphi$ on $\partial\Omega\}$ has a unique solution $u$ if and only if the set $K$ is nonempty.

Remark 2.1**.**

In above theorem, $v=\varphi$ on $\partial\Omega$ means that, for every $x_{0}\in\partial\Omega$ and every open straight line segment $l$ contained in $\Omega$ and with endpoint $x_{0}$,

[TABLE]

Regarding this definition and the existence of weakly spacelike extension of $\varphi$, we refer the readers to the discussion in [BS82, p.133, p.148–149].

Definition 2.1** (Area maximizing hypersurface).**

A weakly spacelike function $u\in C(\Omega)$ ($\Omega\subset\mathbb{R}^{n}$ is not necessarily bounded) is called area maximizing if it solves the variational problem (2.1) with respect to its own boundary values for every bounded subdomain in $\Omega$. The graph of $u$ is called an area maximizing hypersurface.

Lemma 2.1** (Closeness of variational solutions [BS82, Lemma 1.3]).**

If $\{u_{k}\}$ is a sequence of area maximizing functions in $\Omega$ and $u_{k}\rightarrow u$ in $\Omega$ locally uniformly, then $u$ is also an area maximizing function.

One key result in [BS82, Theorem 3.2] is that if an area maximizing hypersurface contains a segment of light ray, then it contains the whole of the ray extended all the way to the boundary or to infinity. This implies the following conclusion.

Theorem 2.2** **(The relationship between the variational solutions and the solutions of

maximal surface equation).

The solution $u$ of (2.1) is smooth and solves equation (1.1) in

[TABLE]

where

[TABLE]

Furthermore

[TABLE]

where $x,y\in\partial\Omega$ are such that $\overline{xy}\subset\Omega$ and $|\varphi(x)-\varphi(y)|=|x-y|$.

Remark 2.2** (Solvability of maximal surface equation on bounded domains).**

If the boundary data $\varphi$ admits a weakly spacelike extension and satisfies that $|\varphi(x)-\varphi(y)|<|x-y|$ for all $x,y\in\partial\Omega$ with $\overline{xy}\subset\Omega$ and $x\neq y$, then $sing\ u=\emptyset$ and hence smooth $u$ solves the equation (1.1) in $\Omega$.

Bartnik proved the following

Theorem 2.3** (Bernstein theorem for variational solutions [Ec86, Theorem F]).**

Entire area maximizing hypersurfaces in $L^{n+1}$ are weakly spacelike hyperplanes.

Definition 2.2** (Isolated singularity [Ec86, p.382]).**

A weakly spacelike hypersurface $M$ in $L^{n+1}$ containing [math] is called an area maximizing hypersurface having an isolated singularity at [math] if $M\backslash\{0\}$ is area maximizing but $M$ cannot be extended as an area maximizing hypersurface into [math].

For a weakly spacelike entire or exterior hypersurface $M$ (i.e., $u$ is defined on $\mathbb{R}^{n}$ or an exterior domain $\mathbb{R}^{n}\backslash A$ with $A$ bounded), we define $M_{r}=r^{-1}M$ with $r>0$ is the graph of $u_{r}(x)=r^{-1}(rx)$. If for some $r_{j}\rightarrow+\infty$, $u_{r_{j}}(x)$ converge locally uniformly to a function $u_{\infty}(x)$ on $\mathbb{R}^{n}$ or $\mathbb{R}^{n}\backslash\{0\}$, then $u_{\infty}$ (its graph $M_{\infty}$) is called a blowdown of $u$ ($M$). Note that by weakly spacelikeness, Arzela-Ascoli theorem always ensures the existence of blowdowns. By Lemma 2.1, $u_{\infty}(x)$ ($M_{\infty}$) is area maximizing on $\mathbb{R}^{n}$ or $\mathbb{R}^{n}\backslash\{0\}$ and $u_{\infty}(0)=0$.

Ecker proved that the isolated singularities of area maximizing hypersurface are light cone like ([Ec86, Theorem 1.5]). The following lemma will also be used in our proof of Theorem 1.1.

Lemma 2.2** ([Ec86, Lemma 1.10]).**

Let $M$ be an entire area maximizing hypersurface having an isolated sigularity at [math] and assume that some blowdown of $M$ also has an isolated singularity at [math]. Then $M$ has to be either $C_{0}^{+}$ or $C_{0}^{-}$.

We also need the following radial, catenoid like solutions to the maximal surface equation of (1.1) in $\mathbb{R}^{n}\backslash\{0\},$ used as barriers in [BS82] and [Ec86]. For $\lambda\in\mathbb{R},$ set

[TABLE]

For $n\geq 2$, the integral $\int_{0}^{+\infty}\frac{\lambda}{\sqrt{t^{2(n-1)}+\lambda^{2}}}dt$ is bounded and we denote this value as $M(\lambda,n)$. More precisely, by computation

[TABLE]

for large $r$. It is obvious that $M(\lambda,n)=sign(\lambda)|\lambda|^{\frac{1}{n-1}}M(1,n)\rightarrow\pm\infty$ as $\lambda\rightarrow\pm\infty$ and $M(\lambda,n)\rightarrow 0$ as $\lambda\rightarrow 0$. For $n=2$, the integral $\int_{0}^{+\infty}\frac{\lambda}{\sqrt{t^{2}+\lambda^{2}}}dt$ is infinite and by computation

[TABLE]

for large $r$, where $m(\lambda)=\int_{0}^{1}\frac{\lambda}{\sqrt{t^{2}+\lambda^{2}}}dt+\int_{1}^{+\infty}(\frac{\lambda}{\sqrt{t^{2}+\lambda^{2}}}-\frac{\lambda}{t})dt$.

Definition 2.3** (Lorentz transformations, the speed of light is normalized to 1).**

For a parameter $\kappa\in(-1,1)$, the Lorentz transformation $L_{\kappa}:\mathbb{L}^{n+1}\rightarrow\mathbb{L}^{n+1}$ is defined as

[TABLE]

where $x^{\prime}=(x_{1},\cdots,x_{n-1})$.

The Lorentz transformations are isometries of $\mathbb{L}^{n+1}$. $L_{\kappa}$ maps spacelike (weakly spacelike) surfaces to spacelike (weakly spacelike) surfaces and it maps maximal surfaces (area maximizing surfaces) to maximal surfaces (area maximizing surfaces). Geometrically $L_{\kappa}$ can be seen as a hyperbolic rotation. It maps the light cone $\{(x,t)\in\mathbb{L}^{n+1}:t=|x|\}$ to itself and it maps the horizontal hyperplanes to the hyperplanes with slope $\kappa$:

[TABLE]

for $T\in(-\infty,+\infty)$.

More generally, for any vector $a\in B_{1}$ we define $L_{a}:=T_{a}L_{|a|}T_{a}^{-1}$ where $T_{a}$ is a rotation that keeps $t$-axis fixed and transforms $e_{n}$ to $\frac{a}{|a|}$ in $\mathbb{R}^{n}$ (in case of $a=0$ we just define $T_{0}:=id$).

3. Extension of spacelike hypersurface with hole

We start our proofs for the two main theorems by extending any spacelike function over an exterior domain to a global spacelike function, after finitely enlarging the bounded complimentary domain.

Theorem 3.1**.**

Let $u$ be a spacelike function in $\mathbb{R}^{n}\backslash A$ with $A$ being bounded. Then there exists $R^{\ast}>0$ such that $\left|u\left(x\right)-u\left(y\right)\right|<|x-y|$ for all $x,y\in$ $\mathbb{R}^{n}\backslash B_{R^{\ast}}.$

Proof.

Step 1. We first show that there exists a ball $B_{R_{0}}(x_{0})\supset A$ such that on the boundary $osc_{x\in\partial B_{R_{0}}(x_{0})}u(x)<2R_{0}.$ Without loss of generality we assume $A\subset B_{1}$. We suppose $osc_{\partial B_{100}}u(x)\geq 200$ with $\max_{\partial B_{100}}u(x)=u(100e_{1})$ and we will show that $osc_{\partial B_{200}(100e_{2})}u(x)<400$.

Suppose $\max_{\partial B_{100}}u(x)=-\min_{\partial B_{100}}u(x)$ because otherwise we can consider $u-(\max_{\partial B_{100}}u+\min_{\partial B_{100}}u)/2$ in place of $u$. Firstly one can see that $osc_{\partial B_{100}}u(x)\leq 202$ from the Lipschitz condition on $u$ and the geometry of $\bar{B}_{100}\backslash B_{1}$. So $100\leq u(100e_{1})\leq 101$ and $\min_{\partial B_{100}}u(x)\in(-101,-100)$. Suppose $u(x_{1})=\min_{\partial B_{100}}u(x)$ for some $x_{1}\in\partial B_{100}$. Then $|x_{1}-(-100e_{1})|\leq 3$ because $u(x)>u(100e_{1})-|100e_{1}-x|>100-200=-100$ for any $x\in\partial B_{100}\backslash B_{3}(-100e_{1})$. Thus $u(-100e_{1})\in(-104,-97)$. Therefore $u(100e_{2})>u(100e_{1})-|100e_{1}-100e_{2}|\geq 100-100\sqrt{2}>-42$ and $u(100e_{2})<u(-100e_{1})+|-100e_{1}-100e_{2}|<-97+100\sqrt{2}<45$. In the same way, $u(-100e_{2})\in(-42,45)$. Denote $u(100e_{2})=M$. Then $u(x)\in(M-90,M+90)$ for all $x\in B_{3}(-100e_{2})$. Let $\max_{x\in\partial B_{200}(100e_{2})\backslash B_{3}(-100e_{2})}|u(x)-M|:=Q<200$. Therefore $osc_{\partial B_{200}(100e_{2})}u(x)\leq 2\max(Q,90)<400$. (See Figure 1.)

Step 2. We show that there exists $R_{1}>R_{0}$ such that for all $R\geq R_{1}$ we have $|u(x)-u(y)|<|x-y|$ for all $x,y\in\partial B_{R}(x_{0})$ with $x\neq y.$ By making a suitable transformation, we may assume $x_{0}=0$, $R_{0}=1$ and $\max_{\partial B_{1}}u=-\min_{\partial B_{1}}u=1-\epsilon_{0}$ for some $0<\epsilon_{0}<1$. Then for $R>1$, $\max_{\partial B_{R}}|u|\leq R-\epsilon_{0}$. For $x,y\in\partial B_{R}$ with $x\neq y$, if the line segment $\overline{xy}\subset\bar{B}_{R}\backslash B_{1}$ then $|u(x)-u(y)|<|x-y|$. Otherwise, $dist(0,\overline{xy})<1$ and $|x-y|>2\sqrt{R^{2}-1}$. If $R\geq\frac{1+\epsilon_{0}^{2}}{2\epsilon_{0}}$ then $|u(x)-u(y)|\leq 2(R-\epsilon_{0})\leq 2\sqrt{R^{2}-1}<|x-y|.$

Step 3. Set $R^{\ast}:=|x_{0}|+R_{1}.$ Suppose the line segment $\overline{xy}\cap\partial B_{R_{1}}(x_{0})=\{p,q\}$ and $p$ is closer to $x$ than $q$. Then $|u(x)-u(y)|\leq|u(x)-u(p)|+|u(p)-u(q)|+|u(q)-u(y)|<|x-p|+|p-q|+|q-y|=|x-y|$. If $p=q$, the conclusion is also true. If $\overline{xy}\cap\partial B_{R_{1}}(x_{0})=\emptyset,$ we have $|u(x)-u(y)|<|x-y|$ directly. ∎

For completeness, we include the promised full spacelike extension result here, which is not needed in the proof of our two main theorems.

Theorem 3.2**.**

Let $u$ be a spacelike function in $\mathbb{R}^{n}\backslash A$ with $A$ being bounded. Then there exists $R^{\ast}>0$ such that $\left|u\left(x\right)-u\left(y\right)\right|<|x-y|$ for all $x,y\in$ $\mathbb{R}^{n}\backslash B_{R^{\ast}}.$ Moreover, there exists a spacelike function $\tilde{u}$ in $\mathbb{R}^{n}$ such that $\tilde{u}=u$ in $\mathbb{R}^{n}\backslash B_{R^{\ast}}.$

Proof.

We only need to prove the second part of the theorem. By Remark 2.2, there exists a spacelike function $w$ in $B_{R^{\ast}}$ such that $w=u$ on $\partial B_{R^{\ast}}.$ Define $\tilde{u}:=w$ in $B_{R^{\ast}}$ and $\tilde{u}:=u$ in $\mathbb{R}^{n}\backslash B_{R^{\ast}}$. For $x,y\in\mathbb{R}^{n}$ with $x\neq y,$ if both $x$ and $y$ are in $\bar{B}_{R^{\ast}}$ or $\mathbb{R}^{n}\backslash B_{R^{\ast}}$ then $|u(x)-u(y)|<|x-y|$. Otherwise, let $\{z\}=\overline{xy}\cap\partial B_{R^{\ast}}$, then $|u(x)-u(y)|\leq|u(x)-u(z)|+|u(z)-u(y)|<|x-z|+|z-y|=|x-y|.$

If we assume the spacelike function $u$ is also strictly spacelike $\left|Du\left(x\right)\right|<1$ (to exclude spacelike functions such as $\arctan\lambda$), we can get a spacelike extension inside $B_{R^{\ast}}$ directly, without relying on the singularity analysis of variational solutions to the maximal surface equations of [BS82] contained in Remark 2.2.

In fact (cf. [LY, p.61]), for $x\in\bar{B}_{R^{\ast}}$ set

[TABLE]

with $m=\left\|Du\right\|_{L^{\infty}\left(\partial B_{R^{\ast}}\right)}<1.$ Then $w\left(x\right)=u\left(x\right)$ for $x\in\partial B_{R^{\ast}}.$ And for $x,y\in\bar{B}_{R^{\ast}}$

[TABLE]

Symmetrically $w\left(x\right)\leq w\left(y\right)+m\left|x-y\right|.$ Hence $w$ is spacelike inside $B_{R^{\ast}},$ $\left|u\left(x\right)-u\left(y\right)\right|<m\left|x-y\right|<\left|x-y\right|.$

There is another differential way to do this extension inside $B_{R^{\ast}}.$ Without loss of generality, we assume $R^{\ast}=1,$ then $osc_{\left|x\right|=1}u(x)<2.$ For $x\in\bar{B}_{1}$ set

[TABLE]

with $m=\frac{1}{2}\left[\max_{\left|x\right|=1}u\left(x\right)+\min_{\left|x\right|=1}u\left(x\right)\right].$ Then $w\left(x\right)=u\left(x\right)$ on $\partial B_{1}.$ And for $x\in B_{1}\backslash\left\{0\right\},$

[TABLE]

The Lipschitz norm of $w$ at $x=0$ is also less than one because

[TABLE]

We also reach the same spacelike conclusion of $w$ inside $B_{1}.$ ∎

4. Growth control of $u$ at infinity

In this section, we show that the linear growth rate of an exterior solution $u$ at infinity is uniformly less than one, that is to say, $u$ is controlled not only by the light cone but by a cone with slop less than one. Meanwhile we prove that the blowdown of $u$ is unique and is a linear function with slope less than one. We also proved that the graph of $u$ is supported by a hyperplane either from below or from above.

Theorem 4.1**.**

Let $u$ be an exterior solution in $\mathbb{R}^{n}\backslash A$ with A\being bounded. Then there exist $B_{R}\supset A$, $0<\epsilon<1$ and $c_{0}\in\mathbb{R}$ such that

[TABLE]

in $\mathbb{R}^{n}\backslash B_{R}$. Moreover, there exists a vector $a\in\bar{B}_{1-\epsilon}$ such that

[TABLE]

The function $u$ also enjoys the property that either for some $c\in\mathbb{R}$, $u(x)\geq a\cdot x+c$ in $\mathbb{R}^{n}\backslash B_{R}$ and $u(y)=a\cdot y+c$ at some point $y\in\partial B_{R}$ or for some $c\in\mathbb{R}$, $u(x)\leq a\cdot x+c$ in $\mathbb{R}^{n}\backslash B_{R}$ and $u(y)=a\cdot y+c$ at some point $y\in\partial B_{R}$.

Proof.

We apply Theorem 3.1. For simplicity of notation, we assume $R^{\ast}=1$. So we have $|u(x)-u(y)|<|x-y|$ for any $x,y\in\mathbb{R}^{n}\backslash B_{1}$ with $x\neq y$. We also assume $\max_{\partial B_{1}}u=-\min_{\partial B_{1}}u=1-\epsilon_{1}$ for some $0<\epsilon_{1}<1$. We will show that $-(1-\epsilon)|x|\leq u(x)\leq(1-\epsilon)|x|$ in $\mathbb{R}^{n}\backslash B_{1}$ for some $0<\epsilon<1$.

It is easy to see that $-|x|+\epsilon_{1}\leq u(x)\leq|x|-\epsilon_{1}$ in $\mathbb{R}^{n}\backslash B_{1}$. So there are four possibilities for $u$:

(a) There is $0<\epsilon<1$ such that $u(x)\geq-(1-\epsilon)|x|$ in $\mathbb{R}^{n}\backslash B_{1}$ and there is a sequence of points $\{x_{j}\}$ with $1<|x_{j}|:=R_{j}\rightarrow+\infty$ such that $u(x_{j})>(1-\frac{1}{j})|x_{j}|$;

(b) The function $-u$ satisfies (a);

(c) There are two sequences of points $\{x^{\pm}_{j}\}$ with $1<|x^{\pm}_{j}|:=R^{\pm}_{j}\rightarrow+\infty$ such that $u(x^{+}_{j})>(1-\frac{1}{j})|x^{+}_{j}|$ and $u(x^{-}_{j})<-(1-\frac{1}{j})|x^{-}_{j}|$;

(d) There is $0<\epsilon<1$ such that $-(1-\epsilon)|x|\leq u(x)\leq(1-\epsilon)|x|$ in $\mathbb{R}^{n}\backslash B_{1}$.

We will show that the cases (a)(b)(c) can not happen.

Suppose that $u$ satisfies (a). Let $\hat{x}:=\lim_{k\rightarrow\infty}\frac{x_{j_{k}}}{R_{j_{k}}}\in\partial B_{1}$ for some subsequence $\{j_{k}\}$. We assume $\hat{x}=e_{n}$ and consider $\{j_{k}\}$ as $\{j\}$. Define $v_{j}(x):=\frac{u(R_{j}x)}{R_{j}}$. A subsequence of $v_{j}(x)$ (still denoted as $v_{j}(x)$) converge locally uniformly to a function $V(x)$ in $\mathbb{R}^{n}\backslash\{0\}$. By Lemma 2.1, $V(x)$ is area maximizing in $\mathbb{R}^{n}\backslash\{0\}$. It is obvious that $V(0)=0$, $V(e_{n})=1$ and $V(x)\geq-(1-\epsilon)|x|$. Thus $V(te_{n})=t$ for $t\in(0,+\infty)$ by weakly spacelikeness and Theorem 2.1. If [math] is a removable singularity for $V$, then $V$ is a plane by Theorem 2.3 and $V$ has to be $V(x)=x_{n}$ that contradicts $V(x)\geq-(1-\epsilon)|x|$. So [math] is an isolated singularity for $V$. Let $V_{\infty}$ be a blowdown of $V$, then $V_{\infty}(te_{n})=t$ for $t\in(0,+\infty)$ and $V_{\infty}(x)\geq-(1-\epsilon)|x|$. So [math] is an isolated singularity for $V_{\infty}.$ By Lemma 2.2, we have $V(x)=|x|$.

Let $z\in\partial B_{1}$ be such that $u(z)=\min_{\partial B_{1}}u:=\lambda$. For small $\delta>0$, consider $w(x):=\lambda-1+\delta+(1-\delta)|x|$. Since $\lim_{j\rightarrow\infty}\frac{u(R_{j}x)}{R_{j}}=|x|$ uniformly on $\partial B_{1}$, for sufficiently large $j$, $u(x)\geq w(x)$ on $\partial B_{R_{j}}$. But $u(x)\geq\lambda=w(x)$ on $\partial B_{1}$ and $w(x)$ is a subsolution to (1.1) in $B_{R_{j}}\backslash\bar{B}_{1}$, so $u(x)\geq w(x)$ in $B_{R_{j}}\backslash\bar{B}_{1}$. Let $\delta\rightarrow 0$, we get $u(x)\geq\lambda-1+|x|$ in $B_{R_{j}}\backslash\bar{B}_{1}$. Especially, $u(2z)\geq\lambda-1+|2z|=\lambda+1$ and hence $u(2z)-u(z)\geq 1=|2z-z|$. This contradicts the fact that $u$ is spacelike “in $\mathbb{R}^{n}\backslash B_{1}"$ proved in Theorem 3.1.

The case (b) can not happen for the same reason.

Now we suppose $u$ satisfies (c). For each $j$, let $w_{j}$ be the solution of (1.1) in $B_{j}$ with $w_{j}=u$ on $\partial B_{j}$. The existence of $w_{j}$ is due to Remark 2.2. For each $j$, either $\max_{\partial B_{1}}\left(w_{j}-u\right)\geq 0$ or $\min_{\partial B_{1}}\left(w_{j}-u\right)\leq 0$ (or both). Thus $\max_{\partial B_{1}}\left(w_{j}-u\right)\geq 0$ or $\min_{\partial B_{1}}\left(w_{j}-u\right)\leq 0$ happens for infinitely many $j$. We assume $\max_{\partial B_{1}}\left(w_{j}-u\right):=\lambda_{j}\geq 0$ happens for infinitely many $j$. Let $z_{j}\in\partial B_{1}$ be such that $w_{j}(z_{j})-u(z_{j})=\lambda_{j}$ and consider $\tilde{w}_{j}=w_{j}(z_{j})-\lambda_{j}$ for these $j$. So $\tilde{w}_{j}\leq u$ in $B_{j}\backslash B_{1}$ and $\tilde{w}_{j}(z_{j})=u(z_{j})$. Note that $|\tilde{w}_{j}(0)|\leq|\tilde{w}_{j}(z_{j})|+1=|u_{j}(z_{j})|+1\leq 2$ for all these $j$. Therefore, by Arzela-Ascoli a subsequence $\tilde{w}_{j_{k}}$ converge locally uniformly to a function $W$ in $\mathbb{R}^{n}$. By Lemma 2.1, $W$ is an area maximizing surface. So it is a plane with slope less than or equal to one by Theorem 2.3. Furthermore, we know $W\leq u$ in $\mathbb{R}^{n}\backslash B_{1}$ and $W(z)=u(z)$ by continuity, where $z$ is an accumulating point of $\{z_{j_{k}}\}$.

By assumption of (c), there are $\{x_{j}^{-}\}$ with $|x_{j}^{-}|\rightarrow+\infty$ such that $W(x_{j}^{-})\leq u(x_{j}^{-})<-(1-\frac{1}{j})|x_{j}^{-}|$. Thus $W$ has to be a plane with slope 1. We assume $DW(x)=e_{n}$, so $W(x)=x_{n}+u(z)-z_{n}$. If $z_{n}<0$, then denote $\tilde{z}=(z^{\prime},-z_{n})\in\partial B_{1}$ and we have $u(\tilde{z})\geq W(\tilde{z})=-2z_{n}+u(z)=u(z)+|\tilde{z}-z|$. This contradicts the fact that $|u(x)-u(y)|<|x-y|$ for any $x,y\in\partial B_{1}$ with $x\neq y$. If $z_{n}\geq 0$, then consider the point $z+e_{n}\in\mathbb{R}^{n}\backslash\bar{B}_{1}$ and we have $u(z+e_{n})\geq W(z+e_{n})=u(z)+1=u(z)+|(z+e_{n})-z|$. This contradicts the fact that $u$ is spacelike “in $\mathbb{R}^{n}\backslash B_{1}"$ proved in Theorem 3.1.

If it is the case that $\min_{\partial B_{1}}\left(w_{j}-u\right)\leq 0$ happens for infinitely many $j$, we move up $w_{j}$ by $-\min_{\partial B_{1}}\left(w_{j}-u\right)$ and get a plane $\hat{W}$ above $u$ by the same process. This time by the assumption that there are $\{x_{j}^{+}\}$ with $|x_{j}^{+}|\rightarrow+\infty$ such that $\hat{W}(x_{j}^{+})\geq u(x_{j}^{+})>(1-\frac{1}{j})|x_{j}^{+}|$, we also know the slope of $\hat{W}$ is one. Furthermore, $\hat{W}$ also touches $u$ at some point of $\partial B_{1}$. Again, this contradicts the fact that $u$ is spacelike “in $\mathbb{R}^{n}\backslash B_{1}"$ proved in Theorem 3.1.

Therefore only the case (d) can (and must) happens and we have proved the first part of the theorem. In this case we can also construct the plane $W$ in the same way just as we did in the first paragraph when we proved the impossibility of case (c). That is to say, we can place a plane (with slope less than or equal to $1-\epsilon$) either below or above the graph of $u$ in $\mathbb{R}^{n}\backslash B_{1}$ and the plane touches $u$ at some point of $\partial B_{1}$. This property implies that the blowdown of $u$ must be unique and equal to the blowdown of $W$. We show this as follows. Assume that $W(x)=c+a\cdot x\leq u$ in $\mathbb{R}^{n}\backslash B_{1}$ where $|a|\leq 1-\epsilon$. Let $V$ be any blowdown of $u$, then $a\cdot x\leq V(x)\leq(1-\epsilon)|x|$ in $\mathbb{R}^{n}$, which implies that [math] is a removable singularity of $V$ by Ecker stated in Lemma 2.2. Then $V$ is an entire solution and must be a plane. The only possible situation is $V(x)=a\cdot x$. ∎

5. Gradient estimate

With the strong growth control achieved in the previous section and the known curvature estimate, we can establish the gradient estimate and ascertain $Du(\infty)$ in this section. We state the curvature estimate of Cheng-Yau [CY76] in the following improved extrinsic form carried out by Schoen (see [Ec86, Theorem 2.2]).

Theorem 5.1**.**

Let $M=\left(x,u\left(x\right)\right)$ be a maximal hypersurface, $x_{0}\in M$ and assume that for some $\rho>0$, $L_{2\rho}(x_{0})\cap M\subset\subset M$. Then we have for all $x\in L_{\rho}(x_{0})$

[TABLE]

where $c(n)$ is a constant depending only on the dimension $n$ and $l_{x_{0}}(x)=(|x-x_{0}|^{2}-|u(x)-u(x_{0})|^{2})^{\frac{1}{2}}.$

If $M$ is an entire maximal hypersurface, then $\rho$ in (5.1) can be chosen to be arbitrarily large, so $|II|\equiv 0$ and hence the Bernstein Theorem follows. But the following corollary is what we need.

Corollary 5.1**.**

For any $0<\epsilon<1$, there exists a positive constant $C(\epsilon,n)$ such that if $u$ solves the equation (1.1) in $\mathbb{R}^{n}\backslash\bar{B}_{1}$ and satisfies $-(1-\epsilon)|x|\leq u(x)\leq(1-\epsilon)|x|$ in $\mathbb{R}^{n}\backslash B_{1}$ then

[TABLE]

Proof.

Fix a point $x\in\mathbb{R}^{n}$ with $|x|\geq\frac{8}{\epsilon}$, for any $y\in\partial B_{1}$

[TABLE]

and

[TABLE]

So

[TABLE]

This means that

[TABLE]

So by letting $x=x_{0}$ and $\rho=\sqrt{\frac{\epsilon}{2}}|x|$ in (5.1) we get

[TABLE]

∎

Theorem 5.2**.**

Let $u$ be an exterior solution in $\mathbb{R}^{n}\backslash A$. Then for any open set $U\supset A$ there is $\theta>0$ such that $|Du|\leq 1-\theta$ in $\mathbb{R}^{n}\backslash U$. Moreover, $\lim_{x\rightarrow\infty}Du(x)=a$ where $a$ is given by Theorem 4.1.

Proof.

Assume $A\subset B_{1}$ and $-(1-\epsilon)|x|\leq u(x)\leq(1-\epsilon)|x|$ in $\mathbb{R}^{n}\backslash B_{1}$. Denote $\hat{R}:=\frac{10}{\epsilon}$. Since $|Du(x)|<1$ for $x\in\mathbb{R}^{n}\backslash U$, if $|Du|\leq 1-\theta$ is not true then there is a sequence of points $\{x_{j}\}$ such that $|Du(x_{j})|>1-\frac{1}{j}$ and $|x_{j}|\rightarrow+\infty$. Define $R_{j}:=\hat{R}^{-1}|x_{j}|$ (assume $R_{j}>1$) and $v_{j}(x):=\frac{u(R_{j}x)}{R_{j}}$. Then by Theorem 4.1, we have $v_{j}(x)\rightarrow V(x)=a\cdot x$.

On the other hand, by Corollary 5.1, the curvature $|II|$ is uniformly bounded for all $v_{j}(x)$ on the compact set $\bar{B}_{\hat{R}+1}\backslash B_{\hat{R}-1}$, so is $|D^{2}v_{j}(x)|$. This means $Dv_{j}(x)\rightarrow DV(x)=a$ in $\bar{B}_{\hat{R}+1}\backslash B_{\hat{R}-1}$. Denote $\lim_{k\rightarrow\infty}\frac{\hat{R}x_{j_{k}}}{|x_{j_{k}}|}=\hat{x}\in\partial B_{\hat{R}}$ for some subsequence $j_{k}$. Then $DV(\hat{x})=\lim_{k\rightarrow\infty}Dv_{j_{k}}(\frac{\hat{R}x_{j_{k}}}{|x_{j_{k}}|})$. But $|Dv_{j_{k}}(\frac{\hat{R}x_{j_{k}}}{|x_{j_{k}}|})|=|Du(x_{j})|>1-\frac{1}{j_{k}}$ and it implies $|DV(\hat{x})|=1$. This is a contradiction.

The conclusion $\lim_{x\rightarrow\infty}Du(x)=a$ can be proved in the same compactness way as above.

There is another Harnack way to show the existence of $Du\left(\infty\right),$ once $\left|Du\right|$ is uniformly bounded away from one, $\left|Du\right|\leq 1-\theta.$ Indeed, each bounded component $u_{k}$ of $Du$ satisfies a uniformly elliptic equation

[TABLE]

with $F\left(p\right)=\sqrt{1-\left|p\right|^{2}}.$ By Moser’s Harnack, in fact [Mo61, Theorem 5], $\lim_{x\rightarrow\infty}u_{k}(x)$ exists. ∎

Because we will use Moser’s results again in next section, we state them here in the needed form for convenience.

Theorem 5.3** (Harnack inequality [Mo61, Theorem 1]).**

Let $w$ be a nonnegative solution of

[TABLE]

in $\mathbb{R}^{n}\backslash B_{R_{0}},$ where $\Lambda^{-1}I\leq(a_{ij}(x))\leq\Lambda I$ for for a constant $\Lambda\in[1,\infty).$ Then for any $R\geq 10R_{0}$

[TABLE]

for $\Gamma=\Gamma\left(n,\Lambda\right).$

Theorem 5.4** (Behavior at $\infty$ [Mo61, Theorem 5]).**

Let $w$ be a bounded solution to the uniformly elliptic equation (5.2) in $\mathbb{R}^{n}\backslash B_{1}$. Then $\lim_{|x|\rightarrow\infty}w(x)$ exists.

6. Asymptotic behavior: proof of Theorem 1.1

Now we are ready to prove Theorem 1.1. We present the proof in the following four subsections. We first treat the special case $Du\left(\infty\right)=a=0$. The general case can be transformed to this special case by a suitable hyperbolic rotation (Lorentz transformation).

6.1. Case $a=0$, $n=2$

Step 1. ($|u(x)|\leq c+d\ln|x|$ for large $c$ and $d$.)

We still assume $R=1$ in Theorem 4.1. By Theorem 4.1 and Theorem 5.2, we known that $\lim_{r\rightarrow\infty}\frac{u(rx)}{r}=0$ and $\lim_{x\rightarrow\infty}|Du(x)|=0$. Moreover, we have either $u(x)\geq c$ for some $c\in\mathbb{R}$ in $\mathbb{R}^{n}\backslash B_{1}$ and $u(y)=c$ at some point $y\in\partial B_{1}$ or $u(x)\leq c$ for some $c\in\mathbb{R}$ in $\mathbb{R}^{n}\backslash B_{1}$ and $u(y)=c$ at some point $y\in\partial B_{1}$. We assume the former case happens and $c=0$, $y=e_{1}$. That is $u(x)\geq 0$ in $\mathbb{R}^{n}\backslash B_{1}$ and $u(e_{1})=0$. Recall the radial barrier $w_{\lambda}$ in (2.2). Set $\phi_{\lambda}(x):=w_{\lambda}(x)-w_{\lambda}(e_{1})$ and $\psi_{\lambda}(x):=\phi_{\lambda}(x)+\max_{\partial B_{1}}u$. As the first step of the proof, we want to show that $u(x)\leq\psi_{\lambda}(x)$ in $\mathbb{R}^{n}\backslash B_{1}$ for sufficiently large $\lambda$.

We observe that as long as $\lambda$ is large enough, $\phi_{\lambda}(2e_{1})$ can be arbitrarily close to 1. Since $u(2e_{1})<1$, we can choose $\lambda_{0}$ such that $\phi_{\lambda_{0}}(2e_{1})>u(2e_{1})$. Now we claim that $u(x)\leq\psi_{\lambda_{1}}(x)$ in $\mathbb{R}^{n}\backslash B_{1}$, where $\lambda_{1}:=(\Gamma+1)\lambda_{0}$ and the constant $\Gamma$ is from Theorem 5.3 for $u$. It is easy to see that $\psi_{\lambda_{1}}(x)>\Gamma\phi_{\lambda}(x)$ in $\mathbb{R}^{n}\backslash B_{R}$ for some $R=R\left(\lambda_{0},\Gamma\right)$ large enough. If $u(x)\leq\psi_{\lambda_{1}}(x)$ in $\mathbb{R}^{n}\backslash B_{R}$ then $u(x)\leq\psi_{\lambda_{1}}(x)$ in $\mathbb{R}^{n}\backslash B_{1}$ by comparison principle since $u\leq\max_{\partial B_{1}}u=\psi_{\lambda_{1}}(x)$ on $\partial B_{1}$. Suppose $u(z)>\psi_{\lambda_{1}}(z)$ at some point $z\in\mathbb{R}^{n}\backslash B_{R}$, then $u>\phi_{\lambda_{0}}$ on $\partial B_{|z|}$ by Theorem 5.3. Since $u\geq 0=\phi_{\lambda_{0}}$ on $\partial B_{1}$, we have $u\geq\phi_{\lambda_{0}}$ in $B_{|z|}\backslash B_{1}$ by comparison principle, especially $u(2e_{1})\geq\phi_{\lambda_{0}}(2e_{1})$. This is a contradiction. So we proved that $u(x)\leq\psi_{\lambda_{1}}(x)$ in $\mathbb{R}^{n}\backslash B_{1}$.

Step 2. ($u(x)=c+d\ln|x|+o(1)$ for some $c$ and $d$.)

Denote

[TABLE]

By continuity, $u\leq\psi_{\lambda^{\ast}}$ in $\mathbb{R}^{n}\backslash B_{1}$. If $\lambda^{\ast}=0$, then $0\leq u\leq\max_{\partial B_{1}}u$ in $\mathbb{R}^{n}\backslash B_{1}$. By Theorem 5.4, $u$ has a limit at infinity. Now we assume $\lambda^{\ast}>0$ and our aim is to show that also $u\geq\phi_{\lambda^{\ast}}$ in $\mathbb{R}^{n}\backslash B_{1}$.

For all positive integers $k>\max\{10,\frac{2}{\lambda^{\ast}}\}$, there exist $y^{k}$ such that $|y^{k}|\geq e^{k^{2}}$, $|y^{k+1}|>|y^{k}|$ and $u(y^{k})>\psi_{\lambda^{\ast}-\frac{1}{k}}(y^{k})$. By (2.4), there exists $\hat{k}$ such that for all $k\geq\hat{k}$, we have $\psi_{\lambda^{\ast}}(y^{k})-u(y^{k})<\psi_{\lambda^{\ast}}(y^{k})-\psi_{\lambda^{\ast}-\frac{1}{k}}(y^{k})<\frac{2}{k}\ln|y^{k}|$. The function $w(x):=\psi_{\lambda^{\ast}}(x)-u(x)$ satisfies equation (5.2) with

[TABLE]

where $w^{t}:=(1-t)u+t\psi_{\lambda^{\ast}}$. By Theorem 5.3, we have $\psi_{\lambda^{\ast}}(x)-u(x)<\frac{2\cdot\Gamma}{k}\ln|x|$ on $\partial B_{|y^{k}|}$ for all $k\geq\hat{k}$. Fix any small $\delta>0$. Note that $\psi_{\lambda^{\ast}}(x)-\phi_{\lambda^{\ast}-\delta}(x)>\frac{\delta}{2}\ln|x|$ outside some ball. So there exist $\tilde{k}$ such that $u(x)>\phi_{\lambda^{\ast}-\delta}(x)$ on $\partial B_{|y^{k}|}$ for all $k\geq\tilde{k}$. Thus $u\geq\phi_{\lambda^{\ast}-\delta}$ in $\mathbb{R}^{n}\backslash B_{1}$ by comparison principle. By continuity, we have $u\geq\phi_{\lambda^{\ast}}$ in $\mathbb{R}^{n}\backslash B_{1}$.

Now we have established that $\phi_{\lambda^{\ast}}\leq u\leq\psi_{\lambda^{\ast}}(x)$ in $\mathbb{R}^{n}\backslash B_{1}$. That is $0\leq\psi_{\lambda^{\ast}}(x)-u\leq\max_{\partial B_{1}}u$. So by Theorem 5.4, $\psi_{\lambda^{\ast}}(x)-u$ has a limit at infinity. Denote this $\lambda^{\ast}=d$, then we have

[TABLE]

as $|x|\rightarrow\infty$ for some constant $c$. Since we assumed $u$ is bounded below, the constant $d\geq 0$. If $u$ is bounded above, then we have $u(x)=c+d\ln|x|$+o(1) with $d\leq 0$.

Step 3. (Improve $o(1)$ to $O(|x|^{-1})$.)

We still assume $u\geq 0$ as above. Suppose $d>0$. Choose $R_{0}>10$ such that $|Du(x)|<\frac{1}{10}$ and $u(x)<2d\ln|x|$ when $|x|\geq R_{0}$. For any point $x$ with $|x|:=2R\geq 2R_{0}$, define $v(y):=\frac{u(Ry+x)}{R}$. Since $u$ satisfies the non-divergence form equation (1.2), $v(y)$ satisfies the equation $a_{ij}(y)v_{ij}(y)=0$ for $y\in B_{1}$ with $a_{ij}(y)=\delta_{ij}+\frac{v_{i}v_{j}}{1-|Dv|^{2}}$. By Morrey-Nirenberg’s $C^{1,\alpha}$ estimate for 2 dimensional uniformly elliptic non-divergence form equation [GT98,Theorem 12.4], for some $\alpha>0$ we have

[TABLE]

where $C$ is a universal constant. In particular, it means that

[TABLE]

Let $e$ be any unit vector, then $v_{e}$ satisfies the equation $(a_{ij}(y)(v_{e})_{j})_{i}=0$ in $B_{1}$, with $a_{ij}=\frac{\delta_{ij}}{\sqrt{1-|Dv|^{2}}}+\frac{v_{i}v_{j}}{(\sqrt{1-|Dv|^{2}})^{3}}$. By (6.2), $\|a_{ij}\|_{C^{\alpha}(B_{\frac{1}{2}})}$ is bounded by a universal constant. By Schauder estimate [GT98, Theorem 8.32],

[TABLE]

Note that $Ru_{ee}(x)=v_{ee}(0)$, so we have

[TABLE]

In fact, using bootstrap argument, we have

[TABLE]

for all $k=1,2,\cdots$.

We write equation (1.2) as

[TABLE]

Then $|f(x)|\leq\frac{C(\ln|x|)^{3}}{|x|^{4}}$ by (6.3) and (6.4). Define $K[u](x):=u(\frac{x}{|x|^{2}})$ for $x\in B_{\frac{1}{2R_{0}}}\backslash\{0\}$. Then

[TABLE]

with $|g(x)|\leq C(-\ln|x|)^{3}$. Let $N[g]$ be the Newtonian potential of $g$ in $B_{\frac{1}{2R_{0}}}$. Since $g$ is in $L^{p}(B_{\frac{1}{2R_{0}}})$ for any $p>0$, $N[g]$ is in $W^{2,p}$ for any $p$ and hence is in $C^{1,\alpha}$ for any $0<\alpha<1$. Now $K[u]-N[g]$ is harmonic in $B_{\frac{1}{2R_{0}}}\backslash\{0\}$. Notice that $|K[u](x)|\leq-2d\ln|x|+C$ in $B_{\frac{1}{2R_{0}}}\backslash\{0\}$, so $|K[u]-N[g]|\leq-2d\ln|x|+C$ in $B_{\frac{1}{2R_{0}}}\backslash\{0\}$. Therefore $K[u]-N[g]$ is the sum of $c_{1}\ln|x|$ (for some constant $c_{1}$) and a harmonic function in $B_{\frac{1}{2R_{0}}}$. So $K[u](x)$ is the sum of $c_{1}\ln|x|$ and a $C^{1,\alpha}$ function in $B_{\frac{1}{2R_{0}}}$. Fix an $\alpha\in(0,1)$, for some affine function $c_{2}+b\cdot x$, we have $|K[u](x)-(c_{1}\ln|x|+c_{2}+b\cdot x)|\leq C|x|^{1+\alpha}$ in $B_{\frac{1}{2R_{0}}}\backslash\{0\}$. Go back to $u$ and we have $|u(x)-(-c_{1}\ln|x|+c_{2}+b\cdot\frac{x}{|x|^{2}})|\leq C|x|^{-1-\alpha}$ for $|x|\geq 2R_{0}$. From the result of Step 2, we must have $-c_{1}=d$ and $c_{2}=c$. Thus

[TABLE]

Step 4. (Improve $O(|x|^{-1})$ to $O_{k}(|x|^{-1})$.)

Since $\psi_{d}(x)=\tilde{c}+d\ln|x|+O_{k}(|x|^{-1})$ for some $\tilde{c}$, we consider $w(x):=\psi_{d}(x)-u(x)-\tilde{c}+c=O(|x|^{-1})$. The function $w$ satisfies the equation $(a_{ij}w_{j})_{i}=0$ with $a_{ij}$ given by (6.1). In view of (6.5) and $|D^{k}\psi_{d}(x)|\leq\frac{C}{|x|^{k}}$, we have $|D^{k}w^{t}(x)|\leq\frac{C\ln|x|}{|x|^{k}}$ and hence $|D^{k}a_{ij}(x)|\leq\frac{C(k)(\ln|x|)^{k}}{|x|^{k+2}}$. For any point $x$ with $|x|:=4R\geq 4R_{0}$, define $v(y):=\frac{w(Ry+x)}{R}$. The $v$ satisfies the equation $(\tilde{a}_{ij}v_{j})_{i}=0$ with $\tilde{a}_{ij}(y)=a_{ij}(x+Ry)$. We have $\|D^{k}\tilde{a}_{ij}\|_{C^{0}(B_{1})}=R^{k}\|D^{k}a_{ij}\|_{C^{0}(B_{R}(x))}\leq C(k)$ and so $\|\tilde{a}_{ij}\|_{C^{k}(B_{1})}\leq C(k)$ for all $k$. Then by Schauder estimate,

[TABLE]

and hence $|D^{k}w(x)|\leq\frac{C(k)}{|x|^{k+1}}$ for $|x|\geq 4R_{0}$. This means $w(x)=O_{k}(|x|^{-1})$ and hence

[TABLE]

Step 5. (Ascertain the value of $d$.)

[TABLE]

Letting $r\rightarrow\infty$, we have $d=Res[u]$.

6.2. Case $a=0$, $n\geq 3$

Step 1. ($|u(x)|\leq c$ for large $c$.)

We still assume $u\geq 0$ and define $\phi_{\lambda}$ and $\psi_{\lambda}$ as above. Using the same method, we can prove $u(x)\leq\psi_{\lambda}(x)$ for some large $\lambda$ in $\mathbb{R}^{n}\backslash B_{1}$. But in the dimensions $n\geq 3$, $\psi_{\lambda}$ is bounded.

Step 2. ($u(x)=u_{\infty}+O(|x|^{2-n})$.)

Since $u$ is bounded, applying Theorem 5.4 directly to $u$, we have $u(x)=u_{\infty}+o(1)$ where $u_{\infty}:=\lim_{x\rightarrow\infty}u(x)$. Define $\phi_{\lambda}(x):=w_{\lambda}(x)-w_{\lambda}(e_{1})+\min_{\partial B_{1}}u$ and $\psi_{\lambda}(x):=w_{\lambda}(x)-w_{\lambda_{1}}(e_{1})+\max_{\partial B_{1}}u$ for $\lambda\in(-\infty,+\infty)$. We can choose $\lambda_{1}$ and $\lambda_{2}$ such that

[TABLE]

By comparison principle,

[TABLE]

and this means that

[TABLE]

Step 3. ($u(x)=u_{\infty}-d|x|^{2-n}+O(|x|^{1-n})$ for some $d$.)

We adopt the same strategy as in the step 3 of above subsection: establish the decay rate of $|Du(x)|$ and $|D^{2}u(x)|$, make Kelvin transform to $u(x)-u_{\infty}$ and estimate the Newtonian potential of right hand side. The only difference is that: when we estimate the decay rate of $|Du(x)|$ we can not use Morrey’s $C^{1,\alpha}$ estimate which is only true for 2 dimension, alternatively the first order derivatives of $u$ (so is $v(y):=\frac{u(x+Ry)-u_{\infty}}{R}$) satisfy a uniformly elliptic divergence form equation and thus we can apply De Giorgi-Nash’s Theorem [GT98, Chapter 8] to $Dv.$

[TABLE]

This treatment also fits two dimensional case certainly. We leave the remaining details to the readers.

Step 4. (Improve $O(|x|^{1-n})$ to $O_{k}(|x|^{1-n})$.)

Do the same thing to $u(x)-u_{\infty}$ as in the step 4 of above subsection.

Step 5. (Ascertain the value of $d$.)

[TABLE]

Letting $r\rightarrow\infty$, we have $d=Res[u]$.

6.3. Case $|a|>0$, $n=2$

By a rotation, we can assume $a=(0,\eta)$ with $\eta\in(0,1)$. Make the Lorentz transformation $L_{-\eta}:\mathbb{L}^{2+1}\rightarrow\mathbb{L}^{2+1}$,

[TABLE]

Then the plane $\{t=\eta x_{2}\}$ was transformed to the plane $\{\tilde{t}=0\}$ and the graph of $u$ over $\mathbb{R}^{2}\backslash A$ was transformed to another maximal hypersurface which is the graph of some function (say $\tilde{u}$) defined on $\mathbb{R}^{2}\backslash\tilde{A}$ for some bounded closed set $\tilde{A}$. The blowdown of $\tilde{u}$ is the [math] function. So $\tilde{u}$ has the asymptotic expansion:

[TABLE]

Transform back and make some direct computations, we can establish the asymptotic expansion of $u$. The details are as follows.

The Lorentz transformation

[TABLE]

Use the polar coordinates $x_{1}=r\cos\theta$, $x_{2}=r\sin\theta$ and substitute (6.6) to (6.7), we get

[TABLE]

We want to solve $\tilde{x}$ from (6.8) and substitute it to (6.6) and the third equality of (6.7), then we will get the expansion of $u$. We need to solve $\tilde{x}$ three times iteratively.

Firstly, we assume $\sin\theta\neq 0$. From (6.8) we can see

[TABLE]

Then

[TABLE]

and hence

[TABLE]

where $O(\ln r/r)$ is independent of small $\sin\theta$. Substitute (6.9) to (6.8) and solve $\tilde{x}_{2}$ again,

[TABLE]

Now we have

[TABLE]

and

[TABLE]

Substitute (6.11) to (6.8) and solve $\tilde{x}_{2}$ again,

[TABLE]

Substitute (6.11) to (6.6) and then substitute (6.6) and (6.12) to the third equality of (6.7), we have

[TABLE]

Notice that we get (6.13) with the assumption $\sin\theta\neq 0$. If $\sin\theta=0$, then (6.8) becomes

[TABLE]

Then we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and hence

[TABLE]

This means (6.13) is also true for $\sin\theta=0$.

Let $\sqrt{1-\eta^{2}}\tilde{c}:=c$ and $\sqrt{1-\eta^{2}}\tilde{d}:=d$. In $x$ coordinates, we have

[TABLE]

Getting rid of the assumption $a=(0,\eta)$, it is not hard to see that

[TABLE]

By the method in Step 4 of Section 6.1, we can improve $O(|x|^{-1})$ to $O_{k}(|x|^{-1})$. We omit the details.

The remaining task is to compute $d$ in terms of $Res[u]$ and $|a|$. For simplicity, we still assume $a=(0,\eta)$. Consider the ellipse

[TABLE]

Use the polar coordinates, but this time we set $x_{1}=r\cos\theta$, $\sqrt{1-\eta^{2}}x_{2}=r\sin\theta$. So $E_{\rho}=\{(r,\theta):r=\rho,0\leq\theta<2\pi\}$. On $E_{\rho}$:

[TABLE]

the unit outward normal vector

[TABLE]

and the length element

[TABLE]

So

[TABLE]

and hence

[TABLE]

Letting $\rho\rightarrow+\infty$, we have

[TABLE]

6.4. Case $|a|>0$, $n\geq 3$

We do the same things as above. Assuming $a=(0,\eta)$, make the Lorentz transformation $L_{-\eta}$: graph of $u\rightarrow$ graph of $\tilde{u}$, then

[TABLE]

and

[TABLE]

Use the polar coordinates $x^{\prime}=r\cos\theta\xi$, $x_{n}=r\sin\theta$ with $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$ and $\xi\in S^{n-2}$ the unit sphere in $\mathbb{R}^{n-1}$. Then we are going to solve $\tilde{x}_{n}$ from

[TABLE]

Suppose $\sin\theta\neq 0$. We have

[TABLE]

[TABLE]

[TABLE]

where $O(r^{1-n})$ is independent of small $\sin\theta$. So

[TABLE]

and

[TABLE]

Therefore, denoting $\sqrt{1-\eta^{2}}\tilde{c}:=c$ and $\sqrt{1-\eta^{2}}\tilde{d}:=d$,

[TABLE]

One can verify that the above expansion is also true in the case of $\sin\theta=0$. Also $O(|x|^{1-n})$ can be improved to $O_{k}(|x|^{1-n})$. We omit the details.

Now we compute $d$. Assume $a=(0,\eta)$ and

[TABLE]

Use the coordinates: $x^{\prime}=r\cos\theta\xi$, $\sqrt{1-\eta^{2}}x_{n}=r\sin\theta$. So

[TABLE]

On $E_{\rho}$:

[TABLE]

and

[TABLE]

The unit outward normal vector

[TABLE]

The surface element

[TABLE]

So

[TABLE]

and hence

[TABLE]

We used the fact that

[TABLE]

Letting $\rho\rightarrow+\infty$, we have

[TABLE]

7. Exterior Dirichlet problem: proof of Theorem 1.2

Recall $w_{\lambda}$ is the radially solution defined by (2.2). Let $a\in B_{1}$, we use $w_{\lambda}^{a}(x)$ to denote the representation function of the hypersurface $L_{a}$(graph of $w_{\lambda}$), where the Lorentz transformation $L_{a}=T_{a}L_{\left|a\right|}T_{a}^{-1}$ is defined in the end of Section 2. Then the function $w_{\lambda}^{a}(x)$ has the following properties: $w_{\lambda}^{a}(0)=0$, $w_{\lambda}^{a}(x)$ solves equation (1) in $\mathbb{R}^{n}\backslash\{0\}$ and (from the argument in the previous section or by direct calculation) for $n=2$

[TABLE]

and for $n\geq 3$

[TABLE]

as $x\rightarrow\infty$. The numbers $m(\lambda)$ and $M(\lambda,n)$ are from (2.4) and (2.3).

Now we prove Theorem 1.2. We do this in the following two subsections corresponding to the cases $n=2$ and $n\geq 3$ respectively.

7.1. Case $n=2$

Let $A$, $g$, $a$ and $d$ be given as in Theorem 1.2 and $\sqrt{1-|a|^{2}}\lambda=d$. Choose constants $c^{-}\leq 0\leq c^{+}$ such that $w_{\lambda}^{a}(x)+c^{-}\leq g(x)\leq w_{\lambda}^{a}(x)+c^{+}$ on $\partial A$. We claim that there exists $\breve{R}(A,g,a,d)>0$ such that for any $R\geq\breve{R}$ there exists a solution $u_{R}$ of maximal surface equation in $B_{R}\backslash A$ satisfying $u_{R}=g$ on $\partial A$ and $u_{R}=w_{\lambda}^{a}$ on $\partial B_{R}$.

In fact, let $\psi$ be an spacelike extension of $g$ into $\mathbb{R}^{2}\backslash A$. By Theorem 3.1, there exists $R^{\ast}$ such that $|\psi(x)-\psi(y)|<|x-y|$ for any $x,y\in\partial B_{R^{\ast}}$ and $x\neq y$. Assume $|\psi|\leq G$ on $\partial B_{R^{\ast}}$. Let $R\geq\breve{R}>R^{\ast}$, for any $x\in\partial B_{R^{\ast}}$ and $y\in\partial B_{R}$,

[TABLE]

provided $\breve{R}$ is chosen to be sufficiently large. So we can find a spacelike function $v_{R}$ on $\partial B_{R}\backslash B_{R^{\ast}}$ such that $v_{R}=\psi$ on $\partial B_{R^{\ast}}$ and $v_{R}=w_{\lambda}^{a}$ on $\partial B_{R}$. Define $\Psi_{R}$ by $\Psi_{R}=\psi$ in $B_{R^{\ast}}\backslash A$ and $\Psi_{R}=v_{R}$ in $B_{R}\backslash B_{R^{\ast}}$. It is not difficult to see that $\Psi_{R}$ is a spacelike function defined on $B_{R}\backslash A$ possessing boundary values $g$ and $w_{\lambda}^{a}$ on $\partial A$ and $\partial B_{R}$ respectively. Hence by Remark 2.2, we can get $u_{R}$ by solving the Dirichlet problem. The above claim is proved.

By comparison principle, $w_{\lambda}^{a}(x)+c^{-}\leq u_{R}(x)\leq w_{\lambda}^{a}(x)+c^{+}$ in $B_{R}\backslash A$. Choose any sequence of $\breve{R}<R_{j}\rightarrow\infty$, by compactness, there exists a subsequence of $\{u_{R_{j}}\}$ converging to a function $u$ locally uniformly in $\mathbb{R}^{n}\backslash A$. By Lemma 2.1, $u$ is area maximizing. If $u$ is not maximal, then graph $u$ contains a segment of light ray and hence the whole of the ray in $(\mathbb{R}^{n}\backslash A)\times\mathbb{R}$ contradicting the fact $w_{\lambda}^{a}(x)+c^{-}\leq u(x)\leq w_{\lambda}^{a}(x)+c^{+}$. Therefore $u$ solves equation (1.1) in $\mathbb{R}^{n}\backslash A$. Moreover, $u=g$ on $\partial A$ and

[TABLE]

as $x\rightarrow\infty$.

Finally, we prove the uniqueness of $u$. Suppose there is another such solution $v$ also satisfying $v=g$ on $\partial A$ and

[TABLE]

Then $w:=u-v$ satisfies a divergence form elliptic equation in $\mathbb{R}^{n}\backslash A$, $w=0$ on $\partial A$ and $w$ is bounded. By [GS56, Theorem 7], $w\equiv 0$ in $\mathbb{R}^{n}\backslash A$.

7.2. Case $n\geq 3$

Given $A$, $g$, $a$ and $c$ as in Theorem 1.2, choose $\tilde{R}$ and $G$ such that $A\in B_{\tilde{R}}$ and $|g|\leq G$ on $\partial A$. Choose $\lambda^{\ast}>0$ such that $\sqrt{1-|a|^{2}}M(\lambda^{\ast},n)\geq|c|+\tilde{R}+G$. Denote

[TABLE]

One can verify that

[TABLE]

For the same reason as in the two dimensional case in the previous subsection, there exists $\breve{R}$ such that for any $R\geq\breve{R}$ there exists a solution $u_{R}$ in $B_{R}\backslash A$ satisfying $u_{R}=g$ on $\partial A$ and $u_{R}=a\cdot x+c$ on $\partial B_{R}$. And hence $\Psi^{-}(x)\leq u_{R}\leq\Psi^{+}(x)$ in $B_{R}\backslash A$. In the same way, we can construct a solution $u$ in $\mathbb{R}^{n}\backslash A$ satisfying $u=g$ on $\partial A$ and

[TABLE]

as $x\rightarrow\infty$.

The uniqueness of $u$ follows from the comparison principle directly.

Acknowledgements

Part of this paper was completed during the first author’s visit to University of Washington (Seattle). His visit was funded by China Scholarship Council. The second author is partially supported by an NSF grant.