GUANGHAO HONG AND YU YUAN - University of …yuan/papers/2019a.pdf2 GUANGHAO HONG AND YU YUAN at in nit.y And we prove that the exterior Dirichlet problem is uniquely solvable. Throughout

MAXIMAL HYPERSURFACES OVER EXTERIOR

DOMAINS

GUANGHAO HONG AND YU YUAN

Abstract. Exterior problems for the maximal surface equation arestudied. We obtain the precise asymptotic behavior of the exterior so-lution at in�nity. And we prove that the exterior Dirichlet problem isuniquely solvable for admissible boundary data and prescribed asymp-totic behavior at in�nity.

1. Introduction

The maximal surface equation is

div(Dup

1� jDuj2 ) = 0; (1.1)

or equivalently in the non-divergence form

4u+ (Du)0

D2uDu

1� jDuj2 = 0: (1.2)

This equation arises as the Euler equation of the variational problem thatmaximize the area functional

R p1� jDuj2 among the spacelike hypersur-

faces in the Lorentz-Minkowski space Ln+1 (see the de�nitions in Section 2).The graph of a solution to (1.1) is called a maximal hypersurface and thegraph of a solution to the variational problem is called an area maximizinghypersurface.Calabi [Ca68] (n � 4) and Cheng-Yau [CY76] (all dimensions) proved

that every entire maximal hypersurface in Ln+1 or every global solution uto the maximal surface equation (1.1) with jDu (x)j < 1 on Rn must belinear.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 \complementary" one for elliptic equations; see forexample [Be51][Si87] for minimal hypersurfaces, [CL03] for Monge-Ampereequation, [LLY17] for special Lagrangian equation and [HZ18] for in�nityharmonic functions, besides the classic works such as [GS56] for linear ones.We study the exterior problem for the maximal surface equation in thispaper. We obtain the precise asymptotic behavior of the exterior solution

Date: March 13, 2019.

1

2 GUANGHAO HONG AND YU YUAN

at in�nity. And we prove that the exterior Dirichlet problem is uniquelysolvable.Throughout the paper, we assume A � R

n be a bounded closed set. Wesay u is an exterior solution in RnnA if u 2 C2(RnnA) with jDu(x)j < 1 solvethe equation (1.1) in RnnA. Given an exterior solution u; for any bounded

C1 domain U � A; the integral Res[u] :=R@U

@u=@~np1�jDuj2d� is independent of

the choices of U because of the divergence structure of the equation. Thenumber Res[u] can be regarded as the residue of the exterior solution u:

Theorem 1.1. Let u be a smooth exterior solution in RnnA with A being

bounded. Then there exist a vector a 2 �B1 and a constant c 2 R such thatfor n = 2

u(x) = a � x+ (1� jaj2)Res[u] lnpjxj2 � (a � x)2 + c

+Res[u]jajp1� jaj2 jxj(a � x)

jxj2 � (a � x)2 �ln jxjjxj +Ok(jxj�1) (1.3)

and for n � 3

u(x) = a � x+ c� (1� jaj2)Res[u](pjxj2 � (a � x)2)2�n +Ok(jxj1�n) (1.4)

as jxj ! 1 for all k = 0; 1; 2; � � � . The notation '(x) = Ok(jxjm) meansthat jDk'(x)j = O(jxjm�k).On the other hand, for any bounded closed set A, given an admissible

boundary value function g : @A ! R and prescribed asymptotic behaviorat in�nity, the exterior Dirichlet problem for maximal surface equation isuniquely solvable. We say g is admissible if g is bounded and there exists aspacelike function in RnnA such that = g on @A in the sense of (1.1) in[BS82] (see Remark 2.1 in Section 2).

Theorem 1.2. Let A � Rn be a bounded closed set and g : @A ! R be an

admissible boundary value function. Then

(1) n = 2, given any a 2 B1 and d 2 R, there exist a unique smoothsolution u of maximal surface equation on R2nA such that u = g on@A and

u(x) = a � x+ d lnpjxj2 � (a � x)2 +O(1) as x!1;

(2) n � 3, given any a 2 B1 and c 2 R, there exist a unique smoothsolution u of maximal surface equation on RnnA such that u = g on@A and

u(x) = a � x+ c+ o(1) as x!1:

Of course u enjoys �ner asymptotic properties and the relation d = (1�jaj2)Res[u] holds by Theorem 1.1.

The article is organized as follows. In Section 2, we set up some notationsand de�nitions, 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

MAXIMAL HYPERSURFACES OVER EXTERIOR DOMAINS 3

that a spacelike function over an exterior domain can be spacelikely extendedto the whole Rn. This is the starting point of our work. Interestinglythere is a striking similarity between our argument and the argument in[CL03, p.571-572] where Ca�arelli and Li prove the locally convex solution ofdetD2u = 1 over an exterior domain can be extended (after �nitely enlargingthe complementary domain A) to a global convex function. In Section 4, weprove a growth control theorem for the exterior solution u at in�nity. Thisis the key content of this paper. Inspired by Ecker's proof in [Ec86], andrelying on his results there, our argument involves compactness, blowdownanalysis and comparison principle. In Section 5, we prove gradient estimatefor u based on the growth control theorem and Cheng-Yau's estimate onthe second fundamental form. In Section 6, we prove Theorem 1.1. Sincethe equation (1.1) becomes uniformly elliptic by the gradient estimate of theprevious section, the standard tools such as Harnack inequality and Schauderestimate apply. The known radially symmetric solutions play a key role inthe proof. In Section 7, we prove Theorem 1.2. We solve the equation ina series of bigger and bigger ring-shaped domains and use the compactnessmethod to get an exterior solution. We use the Lorentz transformations ofradially symmetric solutions as barrier functions to locate the position ofthe exterior solution. The uniqueness of solutions follows from comparisonprinciple.

2. Notations and preliminary results

We denote the Lorentz-Minkowski space by Ln+1 = fX = (x; t) : x 2Rn; t 2 Rg, with the at metric

Pni=1 dx

2i � dt2. And h�; �i denotes the inner

product in Ln+1 with the signature (+; � � � ;+;�).The light cone at X0 = (x0; t0) 2 Ln+1 is de�ned by

CX0 = fX 2 Ln+1 : hX �X0; X �X0i = 0g:The upper and lower light cones will be denoted by C+

X0and C�

X0respec-

tively.The Lorentz-balls are de�ned by

LR(X0) = fX 2 Ln+1 : hX �X0; X �X0i < R2g:LetM be an n-dimensional hypersurface in Ln+1 which can be represented

as the graph of u 2 C0;1(), where is a open set in Rn. We say that M(or u) is

weakly spacelike if jDuj � 1 a.e. in ,spacelike if ju(x)� u(y)j < jx� yj whenever x; y 2 , x 6= y and the line

segment xy � , andstrictly spacelike if u 2 C1() and jDuj < 1 in .IfM (or u) is strictly spacelike and u 2 C2(), the Lorentz metric on Ln+1

induces a Riemannian metric g onM . Under the coordinates (x1; � � � ; xn) 2, gij = h @X@xi ; @X@xj i = �ij � uiuj , where X = (x; u(x)) is the position vector

on the graph of u, and uk = uxk = @u@xk

for k = 1; � � � ; n. So g = I �


Du(Du)0

, det g = 1� jDuj2, g�1 = I + Du(Du)0

1�jDuj2 and gij = �ij +uiuj

1�jDuj2 . The

second fundamental form is IIij =uijpdet g

and so jIIj2 = gijgkluikujldet g (see (2.3)

in [BS82]) where uij =@2u

@xi@xjand the summation convention on repeated

indices is used. Note that jD2uj � jIIj: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 � R

n be a bounded domain and let ' :@! R be a bounded function. Then the variational problem

supv2K

Z

p1� jDvj2 (2.1)

where K = fv 2 C0;1() : jDvj � 1 a.e. in , v = ' on @g has a uniquesolution u if and only if the set K is nonempty.

Remark 2.1. In above theorem, v = ' on @ means that, for every x0 2 @and every open straight line segment l contained in and with endpoint x0,

limx!x0;x2l

v(x) = '(x0):

Regarding this de�nition and the existence of weakly spacelike extension of', we refer the readers to the discussion in [BS82, p.133, p.148{149].

De�nition 2.1 (Area maximizing hypersurface). A weakly spacelike func-tion u 2 C() ( � R

n is not necessarily bounded) is called area maximizingif it solves the variational problem (2.1) with respect to its own boundaryvalues for every bounded subdomain in . The graph of u is called an areamaximizing hypersurface.

Lemma 2.1 (Closeness of variational solutions [BS82, Lemma 1.3]). If fukgis a sequence of area maximizing functions in and uk ! u in locallyuniformly, then u is also an area maximizing function.

One key result in [BS82, Theorem 3.2] is that if an area maximizinghypersurface contains a segment of light ray, then it contains the whole ofthe ray extended all the way to the boundary or to in�nity. This impliesthe following conclusion.

Theorem 2.2 (The relationship between the variational solutions and thesolutions of maximal surface equation). The solution u of (2.1) is smoothand solves equation (1.1) in

reg u := nsing uwhere

sing u := fxy : x; y 2 @; x 6= y; xy � and j'(x)� '(y)j = jx� yjg:


Furthermore

u(tx+ (1� t)y) = t'(x) + (1� t)'(y); 0 < t < 1

where x; y 2 @ are such that xy � and j'(x)� '(y)j = jx� yj.Remark 2.2 (Solvability of maximal surface equation on bounded do-mains). If the boundary data ' admits a weakly spacelike extension andsatis�es that j'(x) � '(y)j < jx � yj for all x; y 2 @ with xy � andx 6= y, then sing u = ; and hence smooth u solves the equation (1.1) in .

Bartnik proved the following

Theorem 2.3 (Bernstein theorem for variational solutions [Ec86, Theo-rem F]). Entire area maximizing hypersurfaces in Ln+1 are weakly spacelikehyperplanes.

De�nition 2.2 (Isolated singularity [Ec86, p.382]). A weakly spacelike hy-persurface M in Ln+1 containing 0 is called an area maximizing hypersur-face having an isolated singularity at 0 if Mnf0g is area maximizing but Mcannot be extended as an area maximizing hypersurface into 0.

For a weakly spacelike entire or exterior hypersurfaceM (i.e., u is de�nedon R

n or an exterior domain RnnA with A bounded), we de�ne Mr =

r�1M with r > 0 is the graph of ur(x) = r�1(rx). If for some rj ! +1,urj (x) converge locally uniformly to a function u1(x) on R

n or Rnnf0g,then u1 (its graph M1) is called a blowdown of u (M). Note that byweakly spacelikeness, Arzela-Ascoli theorem always ensures the existenceof blowdowns. By Lemma 2.1, u1(x) (M1) is area maximizing on Rn orRnnf0g and u1(0) = 0.Ecker proved that the isolated singularities of area maximizing hypersur-

face are light cone like ([Ec86, Theorem 1.5]). The following lemma will alsobe used in our proof of Theorem 1.1.

Lemma 2.2 ([Ec86, Lemma 1.10]). Let M be an entire area maximizinghypersurface having an isolated sigularity at 0 and assume that some blow-down of M also has an isolated singularity at 0. Then M has to be eitherC+0 or C�

0 .

We also need the following radial, catenoid like solutions to the maximalsurface equation of (1.1) in Rnnf0g; used as barriers in [BS82] and [Ec86].For � 2 R; set

w�(x) :=

Z jxj

0

�pt2(n�1) + �2

dt: (2.2)

For n � 2, the integralR +10

�pt2(n�1)+�2

dt is bounded and we denote this

value as M(�; n). More precisely, by computationZ r

0

�pt2(n�1) + �2

dt =M(�; n)� �

n� 2r2�n +O(r4�3n) (2.3)


for large r. It is obvious that M(�; n) = sign(�)j�j 1n�1M(1; n) ! �1 as

�! �1 and M(�; n)! 0 as �! 0. For n = 2, the integralR +10

�pt2+�2

dt

is in�nite and by computationZ r

0

�pt2 + �2

dt = m(�) + � ln r +O(r�2) (2.4)

for large r, where m(�) =R 10

�pt2+�2

dt+R +11 ( �p

t2+�2� �

t )dt.

De�nition 2.3 (Lorentz transformations, the speed of light is normalizedto 1). For a parameter � 2 (�1; 1), the Lorentz transformation L� : L

n+1 !Ln+1 is de�ned as

L� : (x0; xn; t)! (x0;

xn + �tp1� �2

;�xn + tp1� �2

)

where x0 = (x1; � � � ; xn�1).The Lorentz transformations are isometries of Ln+1. L� maps space-

like (weakly spacelike) surfaces to spacelike (weakly spacelike) surfaces andit maps maximal surfaces (area maximizing surfaces) to maximal surfaces(area maximizing surfaces). Geometrically L� can be seen as a hyperbolicrotation. It maps the light cone f(x; t) 2 L

n+1 : t = jxjg to itself and itmaps the horizontal hyperplanes to the hyperplanes with slope �:

L�(f(x; t) 2 Ln+1 : t = Tg) = f(x; t) 2 Ln+1 : t =p1� �2T + �xng;

for T 2 (�1;+1).More generally, for any vector a 2 B1 we de�ne La := TaLjajT�1a where

Ta is a rotation that keeps t-axis �xed and transforms en to ajaj in R

n (in

case of a = 0 we just de�ne T0 := id).

3. Extension of spacelike hypersurface with hole

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

Theorem 3.1. Let u be a spacelike function in RnnA with A being bounded.Then there exists R� > 0 such that ju (x)� u (y)j < jx � yj for all x; y 2RnnBR� :

Proof. Step 1. We �rst show that there exists a ball BR0(x0) � A suchthat on the boundary oscx2@BR0

(x0)u(x) < 2R0: Without loss of generality

we assume A � B1. We suppose osc@B100u(x) � 200 with max@B100 u(x) =u(100e1) and we will show that osc@B200(100e2)u(x) < 400.Suppose max@B100 u(x) = �min@B100 u(x) because otherwise we can con-

sider u� (max@B100 u+min@B100 u)=2 in place of u. Firstly one can see thatosc@B100u(x) � 202 from the Lipschitz condition on u and the geometry of�B100nB1. So 100 � u(100e1) � 101 and min@B100 u(x) 2 (�101;�100). Sup-pose u(x1) = min@B100 u(x) for some x1 2 @B100. Then jx1 � (�100e1)j � 3


Figure 1. Shift the ball to hide the shadow

because u(x) > u(100e1) � j100e1 � xj > 100 � 200 = �100 for any x 2@B100nB3(�100e1). Thus u(�100e1) 2 (�104;�97). Therefore u(100e2) >u(100e1)�j100e1�100e2j � 100�100p2 > �42 and u(100e2) < u(�100e1)+j � 100e1 � 100e2j < �97 + 100

p2 < 45. In the same way, u(�100e2) 2

(�42; 45). Denote u(100e2) = M . Then u(x) 2 (M � 90;M + 90) for allx 2 B3(�100e2). Let maxx2@B200(100e2)nB3(�100e2) ju(x) �M j := Q < 200.Therefore osc@B200(100e2)u(x) � 2max(Q; 90) < 400. (See Figure 1.)

Step 2. We show that there exists R1 > R0 such that for all R � R1

we have ju(x) � u(y)j < jx � yj for all x; y 2 @BR(x0) with x 6= y: Bymaking a suitable transformation, we may assume x0 = 0, R0 = 1 andmax@B1 u = �min@B1 u = 1 � �0 for some 0 < �0 < 1. Then for R > 1,max@BR

juj � R � �0. For x; y 2 @BR with x 6= y, if the line segmentxy � �BRnB1 then ju(x) � u(y)j < jx � yj. Otherwise, dist(0; xy) < 1

and jx � yj > 2pR2 � 1. If R � 1+�20

2�0then ju(x) � u(y)j � 2(R � �0) �

2pR2 � 1 < jx� yj:Step 3. Set R� := jx0j + R1: Suppose the line segment xy \ @BR1(x0) =

fp; qg and p is closer to x than q. Then ju(x) � u(y)j � ju(x) � u(p)j +


ju(p)�u(q)j+ ju(q)�u(y)j < jx�pj+ jp�qj+ jq�yj = jx�yj. If p = q, theconclusion is also true. If xy\@BR1(x0) = ;; we have ju(x)�u(y)j < jx�yjdirectly. �

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

Theorem 3.2. Let u be a spacelike function in RnnA with A being bounded.Then there exists R� > 0 such that ju (x)� u (y)j < jx � yj for all x; y 2RnnBR� : Moreover, there exists a spacelike function ~u in Rn such that ~u = u

in RnnBR� :

Proof. We only need to prove the second part of the theorem. By Remark2.2, there exists a spacelike function w in BR� such that w = u on @BR� :De�ne ~u := w in BR� and ~u := u in RnnBR� . For x; y 2 Rn with x 6= y; ifboth x and y are in �BR� or R

nnBR� then ju(x)�u(y)j < jx�yj. Otherwise,let fzg = xy \ @BR� , then ju(x) � u(y)j � ju(x) � u(z)j + ju(z) � u(y)j <jx� zj+ jz � yj = jx� yj:If we assume the spacelike function u is also strictly spacelike jDu (x)j < 1

(to exclude spacelike functions such as arctan�), we can get a spacelikeextension inside BR� directly, without relying on the singularity analysis ofvariational solutions to the maximal surface equations of [BS82] containedin Remark 2.2.In fact (cf. [LY, p.61]), for x 2 �BR� set

w (x) = infb2@BR�

fu (b) +m jx� bjg

with m = kDukL1(@BR� )< 1: Then w (x) = u (x) for x 2 @BR� : And for

x; y 2 �BR�

w (y) = infb2@BR�

fu (b) +m jy � bjg� inf

b2@BR�

fu (b) +m jx� bj+m jy � xjg� w (x) +m jy � xj :

Symmetrically w (x) � w (y) +m jx� yj : Hence w is spacelike inside BR� ;ju (x)� u (y)j < m jx� yj < jx� yj :There is another di�erential way to do this extension inside BR� :Without

loss of generality, we assume R� = 1; then oscjxj=1u(x) < 2: For x 2 �B1 set

w (x) = jxj [u (x= jxj)�m] +m

with m = 12

�maxjxj=1 u (x) + minjxj=1 u (x)

�: Then w (x) = u (x) on @B1:

And for x 2 B1n f0g ;jDw (x)j = jDu (x= jxj)j < 1:

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

ju (x= jxj)�mj � maxjxj=1 u (x)�minjxj=1 u (x)2

< 1:


We also reach the same spacelike conclusion of w inside B1: �

4. Growth control of u at infinity

In this section, we show that the linear growth rate of an exterior solutionu at in�nity is uniformly less than one, that is to say, u is controlled notonly by the light cone but by a cone with slop less than one. Meanwhilewe prove that the blowdown of u is unique and is a linear function withslope less than one. We also proved that the graph of u is supported by ahyperplane either from below or from above.

Theorem 4.1. Let u be an exterior solution in RnnA with A being bounded.Then there exist BR � A, 0 < � < 1 and c0 2 R such that

�(1� �)jxj � u(x)� c0 � (1� �)jxjin RnnBR. Moreover, there exists a vector a 2 �B1�� such that

limr!1

u(rx)

r= a � x locally uniformly in Rn:

The function u also enjoys the property that either for some c 2 R, u(x) �a � x+ c in RnnBR and u(y) = a � y + c at some point y 2 @BR or for somec 2 R, u(x) � a �x+c in RnnBR and u(y) = a �y+c at some point y 2 @BR.

Proof. We apply Theorem 3.1. For simplicity of notation, we assume R� = 1.So we have ju(x)�u(y)j < jx�yj for any x; y 2 RnnB1 with x 6= y. We alsoassume max@B1 u = �min@B1 u = 1� �1 for some 0 < �1 < 1. We will showthat �(1� �)jxj � u(x) � (1� �)jxj in RnnB1 for some 0 < � < 1.It is easy to see that �jxj+ �1 � u(x) � jxj � �1 in R

nnB1. So there arefour possibilities for u:

(a) There is 0 < � < 1 such that u(x) � �(1 � �)jxj in RnnB1 and

there is a sequence of points fxjg with 1 < jxj j := Rj ! +1 such thatu(xj) > (1� 1

j )jxj j;(b) The function �u satis�es (a);(c) There are two sequences of points fx�j g with 1 < jx�j j := R�j ! +1

such that u(x+j ) > (1� 1j )jx+j j and u(x�j ) < �(1� 1

j )jx�j j;(d) There is 0 < � < 1 such that �(1� �)jxj � u(x) � (1� �)jxj in RnnB1.We will show that the cases (a)(b)(c) can not happen.

Suppose that u satis�es (a). Let x := limk!1xjkRjk

2 @B1 for some

subsequence fjkg. We assume x = en and consider fjkg as fjg. De�ne

vj(x) :=u(Rjx)Rj

. A subsequence of vj(x) (still denoted as vj(x)) converge

locally uniformly to a function V (x) in Rnnf0g. By Lemma 2.1, V (x) isarea maximizing in Rnnf0g. It is obvious that V (0) = 0, V (en) = 1 andV (x) � �(1 � �)jxj. Thus V (ten) = t for t 2 (0;+1) by weakly space-likeness and Theorem 2.1. If 0 is a removable singularity for V , then Vis a plane by Theorem 2.3 and V has to be V (x) = xn that contradictsV (x) � �(1 � �)jxj. So 0 is an isolated singularity for V . Let V1 be a


blowdown of V , then V1(ten) = t for t 2 (0;+1) and V1(x) � �(1� �)jxj.So 0 is an isolated singularity for V1: By Lemma 2.2, we have V (x) = jxj.Let z 2 @B1 be such that u(z) = min@B1 u := �. For small � > 0,

consider w(x) := �� 1+ �+(1� �)jxj. Since limj!1u(Rjx)Rj

= jxj uniformlyon @B1, for su�ciently large j, u(x) � w(x) on @BRj

. But u(x) � � = w(x)

on @B1 and w(x) is a subsolution to (1.1) in BRjn �B1, so u(x) � w(x) in

BRjn �B1. Let � ! 0, we get u(x) � � � 1 + jxj in BRj

n �B1. Especially,u(2z) � � � 1 + j2zj = � + 1 and hence u(2z) � u(z) � 1 = j2z � zj. Thiscontradicts the fact that u is spacelike \in RnnB1" proved in Theorem 3.1.The case (b) can not happen for the same reason.Now we suppose u satis�es (c). For each j, let wj be the solution of (1.1)

in Bj with wj = u on @Bj . The existence of wj is due to Remark 2.2.For each j, either max@B1 (wj � u) � 0 or min@B1 (wj � u) � 0 (or both).Thus max@B1 (wj � u) � 0 or min@B1 (wj � u) � 0 happens for in�nitelymany j. We assume max@B1 (wj � u) := �j � 0 happens for in�nitelymany j. Let zj 2 @B1 be such that wj(zj) � u(zj) = �j and consider~wj = wj(zj)��j for these j. So ~wj � u in BjnB1 and ~wj(zj) = u(zj). Notethat j ~wj(0)j � j ~wj(zj)j + 1 = juj(zj)j + 1 � 2 for all these j. Therefore,by Arzela-Ascoli a subsequence ~wjk converge locally uniformly to a functionW in R

n. By Lemma 2.1, W is an area maximizing surface. So it is aplane with slope less than or equal to one by Theorem 2.3. Furthermore,we know W � u in RnnB1 and W (z) = u(z) by continuity, where z is anaccumulating point of fzjkg.By assumption of (c), there are fx�j g with jx�j j ! +1 such thatW (x�j ) �

u(x�j ) < �(1� 1j )jx�j j. Thus W has to be a plane with slope 1. We assume

DW (x) = en, so W (x) = xn + u(z) � zn. If zn < 0, then denote ~z =(z0;�zn) 2 @B1 and we have u(~z) � W (~z) = �2zn + u(z) = u(z) + j~z � zj.This contradicts the fact that ju(x) � u(y)j < jx � yj for any x; y 2 @B1

with x 6= y. If zn � 0, then consider the point z + en 2 Rnn �B1 and we haveu(z + en) �W (z + en) = u(z) + 1 = u(z) + j(z + en)� zj. This contradictsthe fact that u is spacelike \in RnnB1" proved in Theorem 3.1.If it is the case that min@B1 (wj � u) � 0 happens for in�nitely many j, we

move up wj by �min@B1 (wj � u) and get a plane W above u by the sameprocess. This time by the assumption that there are fx+j g with jx+j j ! +1such that W (x+j ) � u(x+j ) > (1 � 1

j )jx+j j, we also know the slope of W is

one. Furthermore, W also touches u at some point of @B1. Again, thiscontradicts the fact that u is spacelike \in RnnB1" proved in Theorem 3.1.Therefore only the case (d) can (and must) happens and we have proved

the �rst part of the theorem. In this case we can also construct the planeWin the same way just as we did in the �rst paragraph when we proved theimpossibility of case (c). That is to say, we can place a plane (with slopeless than or equal to 1 � �) either below or above the graph of u in RnnB1

and the plane touches u at some point of @B1. This property implies that


the blowdown of u must be unique and equal to the blowdown of W . Weshow this as follows. Assume that W (x) = c + a � x � u in RnnB1 wherejaj � 1� �. Let V be any blowdown of u, then a � x � V (x) � (1� �)jxj inRn, which implies that 0 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 onlypossible situation is V (x) = a � x. �

5. Gradient estimate

With the strong growth control achieved in the previous section and theknown curvature estimate, we can establish the gradient estimate and ascer-tain Du(1) 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 = (x; u (x)) be a maximal hypersurface, x0 2M andassume that for some � > 0, L2�(x0) \M �� M . Then we have for allx 2 L�(x0)

jIIj2(x) � c(n)�2

(�2 � l2x0(x))2

(5.1)

where c(n) is a constant depending only on the dimension n and lx0(x) =

(jx� x0j2 � ju(x)� u(x0)j2) 12 :If M is an entire maximal hypersurface, then � in (5.1) can be chosen to

be arbitrarily large, so jIIj � 0 and hence the Bernstein Theorem follows.But the following corollary is what we need.

Corollary 5.1. For any 0 < � < 1, there exists a positive constant C(�; n)such that if u solves the equation (1.1) in Rnn �B1 and satis�es �(1� �)jxj �u(x) � (1� �)jxj in RnnB1 then

jIIj(x) � C(�; n)

jxj for jxj � 8

�:

Proof. Fix a point x 2 Rn with jxj � 8� , for any y 2 @B1

ju(x)� u(y)j � ju(x)j+ ju(y)j � (1� �)jxj+ (1� �)

andjx� yj � jxj � 1:

So

lx(y) = (jy � xj2 � ju(y)� u(x)j2) 12 >r�

2jxj:

This means thatLp �

2jxj(x) ��M:

So by letting x = x0 and � =p

�2 jxj in (5.1) we get

jIIj(x) �pc(n)p�2 jxj

=C(�; n)

jxj :


�

Theorem 5.2. Let u be an exterior solution in RnnA. Then for any openset U � A there is � > 0 such that jDuj � 1 � � in R

nnU . Moreover,limx!1Du(x) = a where a is given by Theorem 4.1.

Proof. Assume A � B1 and �(1 � �)jxj � u(x) � (1 � �)jxj in RnnB1.

Denote R := 10� . Since jDu(x)j < 1 for x 2 R

nnU , if jDuj � 1 � � is not

true then there is a sequence of points fxjg such that jDu(xj)j > 1� 1j and

jxj j ! +1. De�ne Rj := R�1jxj j (assume Rj > 1) and vj(x) :=u(Rjx)Rj

.

Then by Theorem 4.1, we have vj(x)! V (x) = a � x.On the other hand, by Corollary 5.1, the curvature jIIj is uniformly

bounded for all vj(x) on the compact set �BR+1nBR�1, so is jD2vj(x)j. Thismeans Dvj(x) ! DV (x) = a in �BR+1nBR�1. Denote limk!1

Rxjkjxjk j

= x 2@BR for some subsequence jk. Then DV (x) = limk!1Dvjk(

Rxjkjxjk j

). But

jDvjk(Rxjkjxjk j

)j = jDu(xj)j > 1 � 1jk

and it implies jDV (x)j = 1. This is a

contradiction.The conclusion limx!1Du(x) = a can be proved in the same compactness

way as above.There is another Harnack way to show the existence of Du (1) ; once jDuj

is uniformly bounded away from one, jDuj � 1 � �: Indeed, each boundedcomponent uk of Du satis�es a uniformly elliptic equation

@xi�Fpipj (Du) @xjuk

�= 0 in RnnA

with F (p) =q1� jpj2: By Moser's Harnack, in fact [Mo61, Theorem 5],

limx!1 uk(x) exists. �

Because we will use Moser's results again in next section, we state themhere in the needed form for convenience.

Theorem 5.3 (Harnack inequality [Mo61, Theorem 1]). Let w be a non-negative solution of

(aij(x)wj)i = 0 (5.2)

in RnnBR0 ; where ��1I � (aij(x)) � �I for for a constant � 2 [1;1): Then

for any R � 10R0

sup@BR

w � � inf@BR

w (5.3)

for � = � (n;�) :

Theorem 5.4 (Behavior at1 [Mo61, Theorem 5]). Let w be a bounded so-lution to the uniformly elliptic equation (5.2) in RnnB1. Then limjxj!1w(x)exists.


6. Asymptotic behavior: proof of Theorem 1.1

Now we are ready to prove Theorem 1.1. We present the proof in thefollowing four subsections. We �rst treat the special case Du (1) = a =0. The general case can be transformed to this special case by a suitablehyperbolic rotation (Lorentz transformation).

6.1. Case a = 0, n = 2.

Step 1. (ju(x)j � c+ d ln jxj 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 limr!1u(rx)r = 0 and limx!1 jDu(x)j = 0. Moreover,

we have either u(x) � c for some c 2 R in RnnB1 and u(y) = c at some pointy 2 @B1 or u(x) � c for some c 2 R in RnnB1 and u(y) = c at some pointy 2 @B1. We assume the former case happens and c = 0, y = e1. That isu(x) � 0 in RnnB1 and u(e1) = 0. Recall the radial barrier w� in (2.2). Set��(x) := w�(x)� w�(e1) and �(x) := ��(x) + max@B1 u. As the �rst stepof the proof, we want to show that u(x) � �(x) in R

nnB1 for su�cientlylarge �.We observe that as long as � is large enough, ��(2e1) can be arbitrarily

close to 1. Since u(2e1) < 1, we can choose �0 such that ��0(2e1) > u(2e1).Now we claim that u(x) � �1(x) in R

nnB1, where �1 := (�+ 1)�0 and theconstant � is from Theorem 5.3 for u. It is easy to see that �1(x) > ��(x)in RnnBR for some R = R (�0;�) large enough. If u(x) � �1(x) in R

nnBR

then u(x) � �1(x) in RnnB1 by comparison principle since u � max@B1 u =

�1(x) on @B1. Suppose u(z) > �1(z) at some point z 2 RnnBR, then

u > ��0 on @Bjzj by Theorem 5.3. Since u � 0 = ��0 on @B1, we haveu � ��0 in BjzjnB1 by comparison principle, especially u(2e1) � ��0(2e1).This is a contradiction. So we proved that u(x) � �1(x) in R

nnB1.

Step 2. (u(x) = c+ d ln jxj+ o(1) for some c and d.)Denote

�� := inff� � 0 : u � � in RnnB1g:By continuity, u � �� in RnnB1. If �� = 0, then 0 � u � max@B1 u inRnnB1. By Theorem 5.4, u has a limit at in�nity. Now we assume �� > 0

and our aim is to show that also u � �� in RnnB1.

For all positive integers k > maxf10; 2�� g, there exist yk such that jykj �

ek2, jyk+1j > jykj and u(yk) > �� 1

k(yk). By (2.4), there exists k such that

for all k � k, we have ��(yk) � u(yk) < ��(y

k) � �� 1k(yk) < 2

k ln jykj.The function w(x) := ��(x)� u(x) satis�es equation (5.2) with

aij(x) =

Z 1

0

�ijp1� jDwtj2 +

wtiw

tj

(p1� jDwtj2)3dt (6.1)

where wt := (1 � t)u + t �� . By Theorem 5.3, we have ��(x) � u(x) <2��k ln jxj on @Bjykj for all k � k. Fix any small � > 0. Note that ��(x) ��(x) > �

2 ln jxj outside some ball. So there exist ~k such that u(x) >


��(x) on @Bjykj for all k � ~k. Thus u � �� in RnnB1 by comparison

principle. By continuity, we have u � �� in RnnB1.

Now we have established that �� u � ��(x) in RnnB1. That is

0 � ��(x) � u � max@B1 u. So by Theorem 5.4, ��(x) � u has a limit atin�nity. Denote this �� = d, then we have

u(x) = c+ d ln jxj+ o(1)

as jxj ! 1 for some constant c. Since we assumed u is bounded below, theconstant d � 0. If u is bounded above, then we have u(x) = c+d ln jxj+o(1)with d � 0.

Step 3. (Improve o(1) to O(jxj�1).)We still assume u � 0 as above. Suppose d > 0. Choose R0 > 10 such

that jDu(x)j < 110 and u(x) < 2d ln jxj when jxj � R0. For any point x

with jxj := 2R � 2R0, de�ne v(y) :=u(Ry+x)

R . Since u satis�es the non-divergence form equation (1.2), v(y) satis�es the equation aij(y)vij(y) =0 for y 2 B1 with aij(y) = �ij +

vivj1�jDvj2 . By Morrey-Nirenberg's C1;�

estimate for 2 dimensional uniformly elliptic non-divergence form equation[GT98,Theorem 12.4], for some � > 0 we have

kvkC1;�(B 12) � CkvkL1(B1) �

C ln jxjjxj (6.2)

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

jDu(x)j = jDv(0)j � C ln jxjjxj ; for jxj � 2R0: (6.3)

Let e be any unit vector, then ve satis�es the equation (aij(y)(ve)j)i = 0

in B1, with aij =�ijp

1�jDvj2 +vivj

(p1�jDvj2)3 . By (6.2), kaijkC�(B 1

2) is bounded

by a universal constant. By Schauder estimate [GT98, Theorem 8.32],

jDve(0)j � CkvekL1(B 12) �

C ln jxjjxj :

Note that Ruee(x) = vee(0), so we have

jD2u(x)j � C ln jxjjxj2 ; for jxj � 2R0: (6.4)

In fact, using bootstrap argument, we have

jDku(x)j � C ln jxjjxjk ; for jxj � 2R0; (6.5)

for all k = 1; 2; � � � .We write equation (1.2) as

4u = �(Du)0D2uDu

1� jDuj2 := f(x); in RnnB2R0 :


Then jf(x)j � C(ln jxj)3jxj4 by (6.3) and (6.4). De�ne K[u](x) := u( x

jxj2 ) forx 2 B 1

2R0

nf0g. Then

4K[u] = jxj�4f( x

jxj2 ) := g(x); in B 12R0

nf0g

with jg(x)j � C(� ln jxj)3. LetN [g] be the Newtonian potential of g in B 12R0

.

Since g is in Lp(B 12R0

) for any p > 0, N [g] is inW 2;p for any p and hence is in

C1;� for any 0 < � < 1. Now K[u]�N [g] is harmonic in B 12R0

nf0g. Noticethat jK[u](x)j � �2d ln jxj+C in B 1

2R0

nf0g, so jK[u]�N [g]j � �2d ln jxj+Cin B 1

2R0

nf0g. Therefore K[u]�N [g] is the sum of c1 ln jxj (for some constantc1) and a harmonic function in B 1

2R0

. So K[u](x) is the sum of c1 ln jxj anda C1;� function in B 1

2R0

. Fix an � 2 (0; 1), for some a�ne function c2+ b �x,we have jK[u](x)� (c1 ln jxj+ c2 + b � x)j � Cjxj1+� in B 1

2R0

nf0g. Go back

to u and we have ju(x)� (�c1 ln jxj+ c2+ b � xjxj2 )j � Cjxj�1�� for jxj � 2R0.

From the result of Step 2, we must have �c1 = d and c2 = c. Thus

u(x) = c+ d ln jxj+O(jxj�1):Step 4. (Improve O(jxj�1) to Ok(jxj�1).)Since d(x) = ~c + d ln jxj + Ok(jxj�1) for some ~c, we consider w(x) :=

d(x) � u(x) � ~c + c = O(jxj�1). The function w satis�es the equation(aijwj)i = 0 with aij given by (6.1). In view of (6.5) and jDk d(x)j � C

jxjk ,

we have jDkwt(x)j � C ln jxjjxjk and hence jDkaij(x)j � C(k)(ln jxj)k

jxjk+2 . For any

point x with jxj := 4R � 4R0, de�ne v(y) :=w(Ry+x)

R . The v satis�es the

equation (~aijvj)i = 0 with ~aij(y) = aij(x + Ry). We have kDk~aijkC0(B1) =

RkkDkaijkC0(BR(x)) � C(k) and so k~aijkCk(B1) � C(k) for all k. Then bySchauder estimate,

Rk�1jDkw(x)j = jDkv(0)j � C(k)kvkL1(B1) �C(k)

jxj2 ;

and hence jDkw(x)j � C(k)jxjk+1 for jxj � 4R0. This means w(x) = Ok(jxj�1)

and hence

u(x) = c+ d ln jxj+Ok(jxj�1):Step 5. (Ascertain the value of d.)

Res[u] =1

2�

Z@Br

@u=@~np1� jDuj2ds

=1

2�

Z 2�

0(d

r+O(r�2))rd� = d+O(r�1):


Letting r !1, we have d = Res[u].

6.2. Case a = 0, n � 3.

Step 1. (ju(x)j � c for large c.)We still assume u � 0 and de�ne �� and � as above. Using the same

method, we can prove u(x) � �(x) for some large � in RnnB1. But in thedimensions n � 3, � is bounded.

Step 2. (u(x) = u1 +O(jxj2�n).)Since u is bounded, applying Theorem 5.4 directly to u, we have u(x) =

u1 + o(1) where u1 := limx!1 u(x). De�ne ��(x) := w�(x) � w�(e1) +min@B1 u and �(x) := w�(x)�w�1(e1) +max@B1 u for � 2 (�1;+1). Wecan choose �1 and �2 such that

limx!1��1(x) = lim

x!1 �2(x) = u1:

By comparison principle,

��1(x) � u(x) � �2(x);

and this means that

u(x) = u1 +O(jxj2�n):Step 3. (u(x) = u1 � djxj2�n +O(jxj1�n) for some d.)We adopt the same strategy as in the step 3 of above subsection: establish

the decay rate of jDu(x)j and jD2u(x)j, make Kelvin transform to u(x)�u1and estimate the Newtonian potential of right hand side. The only di�erenceis that: when we estimate the decay rate of jDu(x)j we can not use Morrey'sC1;� estimate which is only true for 2 dimension, alternatively the �rst

order derivatives of u (so is v(y) := u(x+Ry)�u1R ) satisfy a uniformly elliptic

divergence form equation and thus we can apply De Giorgi-Nash's Theorem[GT98, Chapter 8] to Dv:

kDvkC�(B 12) � CkDvkL2(B 3

4)

� CkvkL2(B1) (Caccioppoli)

� CkvkL1(B1) � Cjxj1�n:This treatment also �ts two dimensional case certainly. We leave the re-maining details to the readers.

Step 4. (Improve O(jxj1�n) to Ok(jxj1�n).)Do the same thing to u(x)� u1 as in the step 4 of above subsection.

Step 5. (Ascertain the value of d.)

Res[u] =1

(n� 2)j@B1jZ@Br

@u=@~np1� jDuj2d�

=1

(n� 2)j@B1jZ@B1

((n� 2)d

rn�1+O(r�n))rn�1dSn�1 = d+O(r�1):


Letting r !1, we have d = Res[u].

6.3. Case jaj > 0, n = 2.

By a rotation, we can assume a = (0; �) with � 2 (0; 1). Make the Lorentztransformation L�� : L2+1 ! L

2+1,

L�� : (x1; x2; t)! (x1;x2 � �tp1� �2

;��x2 + tp1� �2

) := (~x1; ~x2; ~t):

Then the plane ft = �x2g was transformed to the plane f~t = 0g and thegraph of u over R2nA was transformed to another maximal hypersurface

which is the graph of some function (say ~u) de�ned on R2n ~A for some

bounded closed set ~A. The blowdown of ~u is the 0 function. So ~u hasthe asymptotic expansion:

~u(~x) = ~c+ ~d ln j~xj+O(j~xj�1): (6.6)

Transform back and make some direct computations, we can establish theasymptotic expansion of u. The details are as follows.The Lorentz transformation

L� : (~x1; ~x2; ~u(~x1; ~x2))! (~x1;~x2 + �~up1� �2

;�~x2 + ~up1� �2

) = (x1; x2; u(x1; x2)):

(6.7)Use the polar coordinates x1 = r cos �, x2 = r sin � and substitute (6.6) to(6.7), we get

~x2+�~c+� ~d

2ln(r2 cos2 �+~x22)+O(

1pr2 cos2 � + ~x22

) = r sin �p1� �2: (6.8)

We want to solve ~x from (6.8) and substitute it to (6.6) and the third equalityof (6.7), then we will get the expansion of u. We need to solve ~x three timesiteratively.Firstly, we assume sin � 6= 0. From (6.8) we can see

~x2 = r sin �p1� �2(1 +O(

ln r

r)) as r ! +1:

Then

r2 cos2 � + ~x22 = r2(1� �2 sin2 �)(1 +O(ln r

r));

and hence

ln(r2 cos2 � + ~x22) = 2 ln(r

q1� �2 sin2 �) +O(

ln r

r) (6.9)

where O(ln r=r) is independent of small sin �. Substitute (6.9) to (6.8) andsolve ~x2 again,

~x2 = r sin �p1� �2 � �~c� � ~d ln(r

q1� �2 sin2 �) +O(

ln r

r): (6.10)


Now we have

r2 cos2 � + ~x22 = r2(1� �2 sin2 �)(1� 2�p1� �2 ~d sin � ln r

(1� �2 sin2 �)r+O(

1

r));

and

ln(r2 cos2 � + ~x22) = 2 ln(r

q1� �2 sin2 �)� �

p1� �2 ~d sin �

(1� �2 sin2 �)� ln rr

+O(1

r):

(6.11)Substitute (6.11) to (6.8) and solve ~x2 again,

~x2 = r sin �p1� �2 � �~c� � ~d ln(r

q1� �2 sin2 �)

+�p1� �2 ~d sin �

(1� �2 sin2 �)� ln rr

+O(1

r): (6.12)

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

u(r; �) = �r sin � +p1� �2~c+

p1� �2 ~d ln(r

q1� �2 sin2 �)

+�2 ~d sin �

(1� �2 sin2 �)� ln rr

+O(1

r): (6.13)

Notice that we get (6.13) with the assumption sin � 6= 0. If sin � = 0, then(6.8) becomes

~x2 + �~c+� ~d

2ln(r2 + ~x22) +O(

1pr2 + ~x22

) = 0:

Then we have

~x2 = �� ~d ln r(1 + o(1));

r2 + ~x22 = r2(1 +O(1

r));

ln(r2 + ~x22) = 2 ln r +O(1

r);

~x2 = �� ~d ln r � �~c+O(1

r);

and hence

u(r; �) =p1� �2~c+

p1� �2 ~d ln r +O(

1

r):

This means (6.13) is also true for sin � = 0.


Letp1� �2~c := c and

p1� �2 ~d := d. In x coordinates, we have

u(x1; x2) = �x2 + c+ d lnqx21 + (1� �2)x22

+�2djxjx2p

1� �2(x21 + (1� �2)x22)� ln jxjjxj +O(jxj�1):

Getting rid of the assumption a = (0; �), it is not hard to see that

u(x) = a � x+ c+ d lnpjxj2 � (a � x)2

+djajjxj(a � x)p

1� jaj2(jxj2 � (a � x)2) �ln jxjjxj +O(jxj�1):

By the method in Step 4 of Section 6.1, we can improveO(jxj�1) toOk(jxj�1).We omit the details.The remaining task is to compute d in terms of Res[u] and jaj. For

simplicity, we still assume a = (0; �). Consider the ellipse

E� := fx21 + (1� �2)x22 = �2g:Use the polar coordinates, but this time we set x1 = r cos �,

p1� �2x2 =

r sin �. So E� = f(r; �) : r = �; 0 � � < 2�g. On E�:

Du(�) = (d cos �

�+ o(��1); � +

dp1� �2 sin �

�+ o(��1));

the unit outward normal vector

~n(�) = (cos �p

1� �2 sin2 �;

p1� �2 sin �p1� �2 sin2 �

)

and the length element

ds =

p1� �2 sin2 �p

1� �2�d�:

So

@u=@~np1� jDuj2 =

� sin �p1� �2 sin2 �

+d

�p1� �2

p1� �2 sin2 �

+ o(��1)

and hence

Res[u] =1

2�

ZE�

@u=@~np1� jDuj2ds

=1

2�

Z 2�

0

�� sin �p1� �2

d� +1

2�

Z 2�

0

d

1� �2d� + o(1)

=d

1� �2+ o(1):

Letting �! +1, we have

d = (1� �2)Res[u] = (1� jaj2)Res[u]:


6.4. Case jaj > 0, n � 3.

We do the same things as above. Assuming a = (0; �), make the Lorentztransformation L��: graph of u! graph of ~u, then

~u(~x) = ~c� ~dj~xj2�n +O(j~xj1�n)and

L� : (~x0; ~xn; ~u(~x0; ~xn))! (~x0;

~xn + �~up1� �2

;�~xn + ~up1� �2

) = (x0; xn; u(x0; xn)):

Use the polar coordinates x0 = r cos ��, xn = r sin � with ��2 � � � �

2 and

� 2 Sn�2 the unit sphere in Rn�1. Then we are going to solve ~xn from

~xn + �~c� � ~d(r2 cos2 � + ~x2n)2�n2 +O((r2 cos2 � + ~x2n)

1�n2 ) = r sin �

p1� �2:

Suppose sin � 6= 0. We have

~xn = r sin �p1� �2(1 +O(

1

r));

r2 cos2 � + ~x2n = r2(1� �2 sin2 �)(1 +O(1

r));

(r2 cos2 � + ~x2n)2�n2 = r2�n(1� �2 sin2 �)

2�n2 +O(r1�n)

where O(r1�n) is independent of small sin �. So

~u = ~c� ~dr2�n(1� �2 sin2 �)2�n2 +O(r1�n);

and

~xn = r sin �p1� �2 � �~c+ � ~dr2�n(1� �2 sin2 �)

2�n2 +O(r1�n):

Therefore, denotingp1� �2~c := c and

p1� �2 ~d := d,

u(x) = �xn + c� d(jxj2 � �2x2n)2�n2 +O(jxj1�n)

= a � x+ c� d(jxj2 � (a � x)2) 2�n2 +O(jxj1�n):One can verify that the above expansion is also true in the case of sin � = 0.Also O(jxj1�n) can be improved to Ok(jxj1�n). We omit the details.Now we compute d. Assume a = (0; �) and

E� := fjx02 + (1� �2)x2n = �2g:Use the coordinates: x0 = r cos ��,

p1� �2xn = r sin �. So

E� = f(r; �; �) : r = �;��2� � � �

2; � 2 Sn�2g:

On E�:

ui =(n� 2)dxi

rn+O(r�n) for i = 1; � � � ; n� 1


and

un = � +(n� 2)d(1� �2)xn

rn+O(r�n):

The unit outward normal vector

~n = (x1

rp1� �2 sin2 �

; � � � ; xn�1rp1� �2 sin2 �

;(1� �2)xn

rp1� �2 sin2 �

):

The surface element

d� =

p1� �2 sin2 �p

1� �2�n�1 cosn�2 �d�dSn�2:

So

@u=@~np1� jDuj2 =

� sin �p1� �2 sin2 �

+(n� 2)d�1�np

1� �2p1� �2 sin2 �

+O(��n);

and hence

Res[u] =1

(n� 2)j@B1jZE�

@u=@~np1� jDuj2d�

=jSn�2j

(n� 2)j@B1jZ �

2

��2

��n�1 cosn�2 � sin �p1� �2

+(n� 2)d cosn�2 �

1� �2+O(��1)d�

=d

1� �2+O(��1):

We used the fact thatZ �2

��2

jSn�2j cosn�2 �d� = jSn�1j = j@B1j:

Letting �! +1, we have

d = (1� �2)Res[u] = (1� jaj2)Res[u]:

7. Exterior Dirichlet problem: proof of Theorem 1.2

Recall w� is the radially solution de�ned by (2.2). Let a 2 B1, we usewa�(x) to denote the representation function of the hypersurface La(graph

of w�), where the Lorentz transformation La = TaLjajT�1a is de�ned in theend of Section 2. Then the function wa

�(x) has the following properties:wa�(0) = 0, wa

�(x) solves equation (1) in Rnnf0g and (from the argument inthe previous section or by direct calculation) for n = 2

wa�(x) = a � x+

p1� jaj2m(�) +

p1� jaj2� ln

pjxj2 � (a � x)2 + o(1)

and for n � 3

wa�(x) = a �x+

p1� jaj2M(�; n)�

p1� jaj2�n� 2

(jxj2�(a �x)2) 2�n2 +o(jxj2�n)as x!1. The numbers m(�) and M(�; n) are from (2.4) and (2.3).


Now we prove Theorem 1.2. We do this in the following two subsectionscorresponding to the cases n = 2 and n � 3 respectively.

7.1. Case n = 2.

Let A, g, a and d be given as in Theorem 1.2 andp1� jaj2� = d. Choose

constants c� � 0 � c+ such that wa�(x) + c� � g(x) � wa

�(x) + c+ on @A.

We claim that there exists �R(A; g; a; d) > 0 such that for any R � �R thereexists a solution uR of maximal surface equation in BRnA satisfying uR = gon @A and uR = wa

� on @BR.In fact, let be an spacelike extension of g into R2nA. By Theorem 3.1,

there exists R� such that j (x) � (y)j < jx � yj for any x; y 2 @BR� and

x 6= y. Assume j j � G on @BR� . Let R � �R > R�, for any x 2 @BR� andy 2 @BR,

jwa�(y)� (x)j � jwa

�(y)j+G <jaj+ 1

2(R�R�) � jaj+ 1

2jx� yj

provided �R is chosen to be su�ciently large. So we can �nd a spacelikefunction vR on @BRnBR� such that vR = on @BR� and vR = wa

� on @BR.De�ne R by R = in BR�nA and R = vR in BRnBR� . It is not di�cultto see that R is a spacelike function de�ned on BRnA possessing boundaryvalues g and wa

� on @A and @BR respectively. Hence by Remark 2.2, we canget uR by solving the Dirichlet problem. The above claim is proved.By comparison principle, wa

�(x) + c� � uR(x) � wa�(x) + c+ in BRnA.

Choose any sequence of �R < Rj ! 1, by compactness, there exists asubsequence of fuRj

g converging to a function u locally uniformly in RnnA.By Lemma 2.1, u is area maximizing. If u is not maximal, then graph ucontains a segment of light ray and hence the whole of the ray in (RnnA)�Rcontradicting the fact wa

�(x) + c� � u(x) � wa�(x) + c+. Therefore u solves

equation (1.1) in RnnA. Moreover, u = g on @A and

u(x) = a � x+ d lnpjxj2 � (a � x)2 +O(1)

as x!1.Finally, we prove the uniqueness of u. Suppose there is another such

solution v also satisfying v = g on @A and

v(x) = a � x+ d lnpjxj2 � (a � x)2 +O(1)

Then w := u�v satis�es a divergence form elliptic equation in RnnA, w = 0on @A and w is bounded. By [GS56, Theorem 7], w � 0 in RnnA.

7.2. Case n � 3.

Given A, g, a and c as in Theorem 1.2, choose ~R and G such that A 2 B ~R

and jgj � G on @A. Choose �� > 0 such thatp1� jaj2M(��; n) � jcj+ ~R+


G. Denote

�(x) := wa��(x)�

p1� jaj2M(��; n) + c;

+(x) := wa��(x) +

p1� jaj2M(��; n) + c:

One can verify that

�(x) � a � x+ c � +(x) in Rn;

�(x) = a � x+ c+ o(1) as x!1;

�(x) � g � +(x) on @A:

For the same reason as in the two dimensional case in the previous sub-section, there exists �R such that for any R � �R there exists a solution uRin BRnA satisfying uR = g on @A and uR = a � x + c on @BR. And hence�(x) � uR � +(x) in BRnA. In the same way, we can construct asolution u in RnnA satisfying u = g on @A and

u(x) = a � x+ c+ o(1)

as x!1.The uniqueness of u follows from the comparison principle directly.

Acknowledgements

Part of this paper was completed during the �rst author's visit to Uni-versity of Washington (Seattle). His visit was funded by China ScholarshipCouncil. The second author is partially supported by an NSF grant.

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School of Mathematics and Statistics, Xi'an Jiaotong University, Xi'an,

P.R.China 710049.

E-mail address: [email protected]

Department of Mathematics, University of Washington, Seattle, WA 98195,

USA.

E-mail address: [email protected]

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GUANGHAO HONG AND YU YUAN - University of …yuan/papers/2019a.pdf2 GUANGHAO HONG AND YU YUAN at in nit.y And we prove that the exterior Dirichlet problem is uniquely solvable. Throughout