Asymptotic and exterior Dirichlet problems for the minimal surface equation in the Heisenberg group with a balanced metric

TL;DR

This paper investigates minimal surfaces in the Heisenberg group with a balanced metric, proving the existence of such surfaces with prescribed asymptotic and boundary conditions by solving Dirichlet problems.

Contribution

It establishes the splitting of the Heisenberg group under a balanced metric and constructs minimal surfaces satisfying asymptotic and boundary conditions via Dirichlet problem solutions.

Findings

Heisenberg group with balanced metric splits as a Riemannian product.

Existence of complete properly embedded minimal surfaces in the group.

Solutions to Dirichlet problems yield minimal surfaces with prescribed boundary behavior.

Abstract

It is proved that the Heisenberg group $\operatorname*{Nil}\nolimits_{3}$ with a balanced metric, the sum of the left and right invariant metrics, splits as a Riemannian product $\mathbb{T\times Z}$, where $\mathbb{T}$ is a totally geodesic surface and $\mathbb{Z}$ the center of $\operatorname*{Nil}% \nolimits_{3}.$ It is then proved the existence of complete properly embedded minimal surfaces in $\operatorname*{Nil}\nolimits_{3}$ by solving the asymptotic Dirichlet problem for the minimal surface equation on $\mathbb{T}$. It is also proved the existence of complete properly embedded minimal surfaces foliating an open set of $\operatorname*{Nil}\nolimits_{3}$ having as boundary a given curve $\Gamma$ in $\mathbb{T},$ satisfying the exterior circle condition, by solving the exterior Dirichlet problem for the minimal surface equation in the unbounded connected component of $\mathbb{T}\backslash\Gamma$.

Equations

$$\operatorname*{Nil}\nolimits_{3}=\left\{\left(x,y,z\right):=\left(\begin{array}[c]{ccc}1&x&z\\ 0&1&y\\ 0&0&1\end{array}\right),\text{ }x,y,z\in\mathbb{R}\right\}$$

$${⟨u,v⟩}_g={⟨d{\left(L_g\right)}_g^{-1}u,d{\left(L_g\right)}_g^{-1}v⟩}_e+{⟨d{\left(R_g\right)}_g^{-1}u,d{\left(R_g\right)}_g^{-1}v⟩}_e,$$

$$\left\langle\left(\begin{array}[c]{ccc}0&a_{1}&c_{1}\\ 0&0&b_{1}\\ 0&0&0\end{array}\right),\left(\begin{array}[c]{ccc}0&a_{2}&c_{2}\\ 0&0&b_{2}\\ 0&0&0\end{array}\right)\right\rangle_{e}=a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2},$$

$$\mathbb{T}=\left\{\left(x,y,\frac{xy}{2}\right)\in \underset{3}{\mathrm{Nil}},x,y\in \mathbb{R}\right\}$$

$$e^{i\theta }\left(x,y,z\right)=$$

$$\left(x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta ,\frac{\left(x\sin \theta +y\cos \theta \right)\left(x\cos \theta -y\sin \theta \right)}{2}+z\right).$$

$$\gamma_{\theta}(t)=\left(\begin{array}[c]{ccc}1&\frac{t}{2}\left(\cos\theta-\sin\theta\right)&\frac{\left(\cos\theta-\sin\theta\right)\left(\sin\theta+\cos\theta\right)t^{2}}{8}\\ 0&1&\frac{t}{2}\left(\sin\theta+\cos\theta\right)\\ 0&0&1\end{array}\right),\text{ }t\in\mathbb{R},\text{ }\theta\in\left[0,2\pi\right).$$

$$r\left(x,y,z\right)=\sqrt{2}\sqrt{x^2+y^2}.$$

$$K_p=-4\frac{r^2+12}{{\left(r^2+8\right)}^2},p\in \mathbb{T},$$

$$\left\{\begin{array}[c]{l}\mathcal{M}\left[u\right]:=\operatorname{div}\frac{\nabla u}{\sqrt{1+\left\|\nabla u\right\|^{2}}}=0\text{ in }\mathbb{T}\\ u|_{\partial_{\infty}\mathbb{T}}=\varphi\end{array}\right.$$

$$G(u)=\left\{\left(\begin{array}[c]{ccc}1&x&\frac{xy}{2}+u(x,y)\\ 0&1&y\\ 0&0&1\end{array}\right),\text{ \ \ }x,y\in\mathbb{R}\right\}.$$

$$\left\{\begin{array}[c]{l}\mathcal{M}\left[u\right]=0\text{ in }\Lambda\\ u|_{\partial\Lambda}=0\end{array}\right.$$

$$\underset{r(p)\to \infty }{\mathrm{limsup}}\left|u_{s_1}(p)-u_{s_2}(p)\right|>0$$

$$\phi (x,y,z)=\left(x,y,\frac{xy}{2}+z\right)$$

$$X=\frac{\partial \phi }{\partial x},Y=\frac{\partial \phi }{\partial y},Z=\frac{\partial \phi }{\partial z}.$$

$$⟨X,X⟩$$

$$⟨X,Y⟩$$

$$⟨Y,Z⟩$$

$${\nabla }_{X}X$$

$${\nabla }_{Y}X$$

$${\nabla }_{X}Y$$

$${\nabla }_{Y}Y$$

$${\nabla }_{Y}Z={\nabla }_{Z}Y={\nabla }_{Z}X={\nabla }_{X}Z={\nabla }_{Z}Z=0.$$

$${\gamma }^{\prime }(t)=\frac{1}{2}\left(X+Y\right).$$

$$4{\nabla }_{{\gamma }^{\prime }}{\gamma }^{\prime }(t)$$

$$=\left(-\frac{x{y}^2}{2x^2+2y^2+8}X-\frac{4y+y^3}{2x^2+2y^2+8}Y\right)$$

$$+2\left(\frac{2y+x^{2}y}{2x^2+2y^2+8}X+\frac{2x+x{y}^2}{2x^2+2y^2+8}Y\right)$$

$$+\left(-\frac{4x+x^3}{2x^2+2y^2+8}X-\frac{x^{2}y}{2x^2+2y^2+8}Y\right)=0$$

$${\Vert {\gamma }^{\prime }\Vert }^2$$

$$=\frac{1}{4}\left(\left(2+\frac{y^2}{2}\right)+\left(2+\frac{x^2}{2}\right)-xy\right)=1.$$

$$\Psi\left(\left(\left(\begin{array}[c]{ccc}1&x&\frac{1}{2}xy\\ 0&1&y\\ 0&0&1\end{array}\right),\left(\begin{array}[c]{ccc}1&0&t\\ 0&1&0\\ 0&0&1\end{array}\right)\right)\right)=\left(\begin{array}[c]{ccc}1&x&\frac{1}{2}xy+t\\ 0&1&y\\ 0&0&1\end{array}\right).$$

$$x=\frac{t}{2}\left(\cos \theta -\sin \theta \right),y=\frac{t}{2}\left(\sin \theta +\cos \theta \right)$$

$$x^2+y^2={\left(\frac{t}{2}\left(\cos \theta -\sin \theta \right)\right)}^2+{\left(\frac{t}{2}\left(\sin \theta +\cos \theta \right)\right)}^2=\frac{1}{2}{t}^2.$$

$$t\to \left(\frac{t}{2}\left(\cos \theta -\sin \theta \right),y=\frac{t}{2}\left(\sin \theta +\cos \theta \right),\frac{\left(\cos \theta -\sin \theta \right)\left(\sin \theta +\cos \theta \right)t^2}{8}\right)$$

$$\sqrt{{\Vert X\Vert }^2{\Vert Y\Vert }^2-{⟨X,Y⟩}^2}K=⟨R(X,Y)X,Y⟩=⟨{\nabla }_Y{\nabla }_{X}X-{\nabla }_X{\nabla }_{Y}X,Y⟩.$$

Full text

1 Introduction

The study of minimal surfaces in $3-$dimensional Lie groups with a left invariant metric has recently been attracting the attention of many mathematicians ([10]). In special, in the Heisenberg group

[TABLE]

(see [1], [2], [19], [4], [18], [6]).

Since a left invariant metric on a Lie group is completely determined by its value on the Lie algebra of the group and the group operation, the Riemannian geometry of the group is closely related to the Lie group structure. To bring up this connection, between the group and the Riemannian structures, is one of the motivations for studying minimal surfaces in a Lie group with a left invariant metric. In this paper we consider a balanced metric on $\operatorname*{Nil}\nolimits_{3}$, namely:

[TABLE]

where $e$ is the neutral element of $\operatorname*{Nil}\nolimits_{3}$ (the identity matrix), the inner product of $T_{e}\operatorname*{Nil}\nolimits_{3}$ given by

[TABLE]

$g\in\operatorname*{Nil}\nolimits_{3},$ $u,v\in T_{g}\operatorname*{Nil}\nolimits_{3}$ and $L_{g},R_{g}:\operatorname*{Nil}\nolimits_{3}\rightarrow\operatorname*{Nil}\nolimits_{3}$ are the left and right translations, $L_{g}(x)=gx,$ $R_{g}(x)=xg.$

Clearly, the metric (1) is completely determined by the its value on the Lie algebra of the group and by the group operation and, thus, the geometry of $\operatorname*{Nil}\nolimits_{3}$ with the balanced metric is expected to be closely related to the group structure of $\operatorname*{Nil}\nolimits_{3}$. Indeed, we prove that with the balanced metric $\operatorname*{Nil}\nolimits_{3}$ splits as a Riemannian product of its center and a totally geodesic surface. Actually, the geometry of $\operatorname*{Nil}\nolimits_{3}$ with the balanced metric presents surprising properties, as shown in our first result:

Theorem 1

Setting

[TABLE]

we have:

• (a)

$\mathbb{T}$* is a totally geodesic surface of $\operatorname*{Nil}\nolimits_{3}.$*

• (b)

$\left(\operatorname*{Nil}\nolimits_{3},\left\langle\text{ , }\right\rangle\right)$* is isometric to the Riemannian product $\mathbb{T}\times\mathbb{Z},$ where *\mathbb{Z}=\left\{\left(0,0,z\right)\text{ | }z\in\mathbb{R}\right\}\is the center of $\operatorname*{Nil}\nolimits_{3}$

• (c)

$\mathbb{S}^{1}=\left\{e^{i\theta}\right\}_{\theta\in\mathbb{R}}$* acts isometrically on $\operatorname*{Nil}\nolimits_{3}$ by*

[TABLE]

This action leaves $\mathbb{T}$ invariant and pointwise fixed the center of $\operatorname*{Nil}\nolimits_{3}.$

• (d)

The arc length geodesics of $\mathbb{T}$ passing through the identity $e$ are

[TABLE]

Being $r(p)=d(p,e),$ $d$ the Riemannian distance in $\mathbb{T},$ we have

[TABLE]

• (e)

The sectional curvature $K$ of $\mathbb{T}$ is

[TABLE]

where $r=r(p).$

Notice that from Theorem 1, item (e), it is clear that with the balanced metric $\operatorname*{Nil}\nolimits_{3}$ is not a homogeneous manifold.

The Riemannian splitting $\operatorname*{Nil}\nolimits_{3}\simeq\mathbb{T}\times\mathbb{Z}$ allows the construction of complete properly embedded minimal surfaces of $\operatorname*{Nil}\nolimits_{3}$ with the balanced metric by solving the asymptotic Dirichlet problem in $\mathbb{T}$ (see [16] for details on the this problem). Indeed, it is an immediate consequence of Corollary 1.2 of [15]:

Theorem 2

The asymptotic Dirichlet problem

[TABLE]

has one and only one solution $u\in C^{\infty}\left(\mathbb{T}\right)\cap C^{0}\left(\overline{\mathbb{T}}\right)$ for any $\varphi\in C^{0}\left(\partial_{\infty}\mathbb{T}\right).$

Notice that the graph G(u)=\left\{\left(p,u(p)\right)\text{ | }p\in\mathbb{T}\right\} is a complete minimal surface in $\operatorname*{Nil}_{3}$ with respect to the balanced metric (1). In matricial terms, using the identification $u(x,y)=u(x,y,(xy)/2),$ the graph of $u$ is

[TABLE]

In particular, it follows from Theorem 2 the existence of an infinite number of non congruent foliations of $\operatorname*{Nil}\nolimits_{3}$ by complete properly embedded minimal surfaces transversal to the center of $\operatorname*{Nil}\nolimits_{3}$. We should remark that Theorem 2 is a special case of a family of asymptotic Dirichlet problems which is being investigated on a big class of PDE’s and in general Hadamard manifolds. This is an active area of research which produced a vast numbers of papers in recent years (see [16]).

We also study the exterior Dirichlet problem (EDP) for the minimal surface equation in $\mathbb{T}$ namely, the existence and uniqueness of solutions to the PDE $\mathcal{M}\left[u\right]=0$ on a domain $\Lambda\subset\mathbb{T}$, with a prescribed data at $\partial\Lambda,$ when $\mathbb{T}\backslash\overline{\Lambda}$ is a bounded domain. This problem can of course be considered in any complete non compact Riemannian manifold, and has an old history about which it would be interesting to say some words. It was first studied in the Euclidean space by J. C. C. Nitsche who proved that possible solutions have at most linear growth ([12]). R. Osserman proved the existence of a boundary data at $\partial\Lambda$, even when $\mathbb{R}^{2}\backslash\Lambda$ is a disk, for which the EDP has no solution ([13]). R. Krust proved that the solutions of the EDP having the same Gauss map at infinity form a foliation if there are at least two solutions ([7]). Krust’s result was improved by J. Ripoll and F. Tomi in [17] where they proved the existence of a minimal and of a maximal solutions and also the existence of a boundary data admitting exactly one solution. The results of Krust were extended to arbitrary dimensions by E. Kuwert in [9]. Still in the Euclidean space the EDP was also studied in [8], [14] and, in the Riemannian setting, in [3]. Our result on the EDP in $\operatorname*{Nil}\nolimits_{3}$ is based on [14] and [3]:

Theorem 3

Let $\Omega$ be a $C^{2,\alpha}$ domain of $\mathbb{T}$ satisfying the exterior geodesic circle condition, namely: given $p\in\Omega$, there exists a geodesic circle passing through $p$ which is the boundary of a geodesic disk containing $\Omega.$ Set $\Lambda=\mathbb{T}\backslash\Omega$. Then, for each $s\geq 0$ the exterior Dirichlet problem in $\Lambda$:

[TABLE]

has a solution $u_{s}\in C^{2,\alpha}\left(\overline{\Lambda}\right)$ such that $\sup_{\partial\Lambda}\left\|\nabla u\right\|=s.$ Moreover, the graphs of $u_{s}$ form a foliation of an open subset of $\operatorname*{Nil}\nolimits_{3}$ and

[TABLE]

if $s_{1}\neq s_{2}.$

2 Preliminaries

(i) We shall use the following parametrization $\varphi:\mathbb{R}^{3}\rightarrow\operatorname*{Nil}\nolimits_{3}$ of $\operatorname*{Nil}\nolimits_{3},$

[TABLE]

and the corresponding coordinate vector fields

[TABLE]

The coefficients at $\varphi(x,y,z)$ on the basis $\{X,Y,Z\}$ of the metric (1) are

[TABLE]

and the Riemannian connection $\nabla$ is given, at $\ \varphi(x,y,z),$ by

[TABLE]

and

[TABLE]

(ii) One may see that the curve $\gamma:\mathbb{R\rightarrow T},$ $\gamma(t)=\varphi\left(t/2,t/2,0\right)$ is an arc length geodesic of $\operatorname*{Nil}\nolimits_{3}$. Indeed, note that

[TABLE]

and hence

[TABLE]

since $x=x(t)=t/2=y(t)=y.$ Moreover

[TABLE]

3 Proof of Theorem 1

(a) $\mathbb{T}$ is a totally geodesic surface of $\operatorname*{Nil}\nolimits_{3}$ since one may see that $Z$ is a Killing field orthogonal to $\mathbb{T}$.

(b) An isometry $\Psi:\mathbb{T}\times\mathbb{Z\rightarrow}\operatorname*{Nil}\nolimits_{3}$ is given by

[TABLE]

(c) Follows from direct computations

(d) The first part follows from Section 2 (i) and (c). For the remaining part, given $p\in\mathbb{T}$, $p=\varphi(x,y,0)$, taking $t=\sqrt{2\left(x^{2}+y^{2}\right)}$ there is $\theta\in\mathbb{R}$ such that

[TABLE]

and hence

[TABLE]

Since $t$ is the arc length of the geodesic

[TABLE]

we obtain (2).

(e) Note that the vector fields $X$ and $Y$ are tangent to $\mathbb{T}$ and $\left[X,Y\right]=0.$ Hence

[TABLE]

Using the computations done in the preliminary section, after somewhat long but straightforward calculations one arrives to

[TABLE]

concluding the proof of Theorem 1.

4 The exterior Dirichlet problem

The catenoids, besides being interesting on their own, pieces of them provide explicit examples of solutions of the exterior Dirichlet problem foliating open subsets of $\mathbb{T\times R}$. From straightforward calculations, one has:

Proposition 4

Let $c>0,$ $t_{0}\geq\sqrt{\sqrt{c^{2}+16}-4}$ be given. Then the rotation of the curve

[TABLE]

around around the $z-$ axis is a catenoid that is, a complete rotationally invariant minimal surface in $\operatorname*{Nil}\nolimits_{3}$.

4.1 Proof of Theorem 3

We begin by introducing some notation and remarking some facts to be used in the course of the proof.

Let $r:\Lambda\rightarrow\left[0,\infty\right)$ be the distance function to $\partial\Omega,$ $r(x)=d(x,\partial\Omega),$ where $d$ is the Riemannian distance in $\mathbb{T}$. Given $r,$ set

[TABLE]

One may see that $\Omega_{r}$ satisfies the exterior geodesic circle condition. Then, given $p\in\partial\Omega_{r}\backslash\partial\Omega,$ there is a geodesic circle enclosing a geodesic disk $D^{p}\subset\mathbb{T}$ passing through $p$ and containing $\Omega_{r}.$ By the tangency principle, the curvature of this geodesic circle is smaller than or equal to the curvature of $\partial\Omega_{r}\backslash\partial\Omega$ at $p.$ Take $R(r)>0$ such that

[TABLE]

where $D_{R(r)}$ is the geodesic disk of $\mathbb{T}$ centered at $e$ with radius $R(r).$ By Theorem 1 (e), the maximum of the sectional curvature $K_{r}$ of $D_{R(r)}$ is

[TABLE]

By the Hessian comparison theorem the curvature of the geodesic circles of $\mathbb{T}$ contained in $D_{R(r)},$ oriented inwards, are greater than or equal to the curvature of the geodesic circles of the hyperbolic plane $\mathbb{H}^{2}\left(K_{r}\right)$ of sectional curvature $K_{r}.$ In turn, the curvature $k_{r}$ of any geodesic circles of $\mathbb{H}^{2}\left(K_{r}\right)$ has the lower bound

[TABLE]

Since $\partial\Omega$ is compact it follows from the triangle inequality that

[TABLE]

and thus, there is $\alpha>0$ such that $r-\alpha\leq R(r)\leq r+\alpha$ for all $r\geq 0.$ It follows that

[TABLE]

For the given $s\geq 0$ we now find a subsolution $v_{s}\in C^{2}\left(\overline{\Lambda}\right)$ of

[TABLE]

of the form $v_{s}(x)=f(r(x)),$ $x\in\Lambda,$ for some $f\in C^{2}\left(\left[0,\infty\right)\right),$ such that

[TABLE]

We have

[TABLE]

where $\Delta$ is the Laplacian operator. Since $\Delta v_{s}$ is the curvature the $\partial\Omega_{r}\backslash\partial\Omega,$ assuming that $f^{\prime}(r)\geq 0,$ it follows from (4) that $\mathcal{M}\left[u\right]\geq 0$ if

[TABLE]

Solving the last inequality to have an equality, we obtain

[TABLE]

where $c$ is any constant. Clearly $f(0)=0$ and

[TABLE]

so that we can choose $c$ such that $f^{\prime}(0)=s.$ Since

[TABLE]

it follows that $f$ is bounded. Hence, $v_{s}(x)=f(r(x)),$ $x\in\overline{\Lambda},$ is a subsolution of $\mathcal{M}$ satisfying conditions (5).

Given $m\in\mathbb{N}$, set

[TABLE]

We have $T_{m}\neq\emptyset,$ since $0\in T_{m}$. Give $\varepsilon>0,$ we prove that $t<v_{s+\varepsilon}|_{\partial\Omega_{m}\backslash\partial\Omega}.$ By contradiction, assume that $t\geq v_{s+\varepsilon}|_{\partial\Omega_{m}\backslash\partial\Omega}.$ Since

[TABLE]

there is a neighborhood $U$ of $\partial\Omega$ in $\Omega_{m}$ such that $u_{t}(x)<v_{s+\varepsilon}(x)$ for all $x\in U\backslash\partial\Omega_{m}.$ Since $t<v_{s+\varepsilon}|_{\partial\Omega_{m}\backslash\partial\Omega}$ there exists a domain $V\subset\Omega_{m}$ containing $U$ such that $u_{t}|_{\partial V}=v_{s+\varepsilon}|_{\partial V},$ what is an absurd since $v_{s+\varepsilon}|_{V}$ is a subsolution of \mathcal{M}\in $V$ coinciding with $u_{t}$ at $\partial V$ (this follows from the comparison principle, Proposition 3.1 of [16]). Letting $\varepsilon\rightarrow 0$ we have $t\leq v_{s}|_{\partial\Omega_{m}\backslash\partial\Omega}.$ It follows that $T_{m}$ is bounded and we may set

[TABLE]

We prove that $t_{m}\in T_{m},$ by first proving that there is a constant $C$, not depending on $m,$ such that if $t\in T_{m},$ then $\sup_{\partial\Omega_{m}}\left\|\nabla u_{t}\right\|_{\partial\Omega}\leq C.$ Let $t\in T_{m}$ be given. By definition of $T_{m}$ we have $\sup_{\partial\Omega}\left\|\nabla u_{t}\right\|_{\partial\Omega}\leq s.$

Setting $z_{m}=v_{s}|_{\partial\Omega_{m}\backslash\partial\Omega}$ we have, as proved above, $z_{m}\geq t$. Moreover, the function $w_{s}:=v_{s}-\left(z_{m}-t\right)$ is a subsolution of $\mathcal{M}$ in $\Omega_{m}$ and

[TABLE]

By the comparison principle we obtain

[TABLE]

from what we conclude that

[TABLE]

where

[TABLE]

Since

[TABLE]

for any $t\in T_{m}$ we may then conclude the $C^{1}$ norm of any solution $u_{t}$ for $t\in T_{m}$ has an a-priori bound that depends only on $s$ and $\alpha.$ Hence, from elliptic PDE linear theory there is $C=C(s,\alpha)$ such that $\left\|u_{t}\right\|_{C^{2,\alpha}\left(\overline{\Omega}\right)}\leq C$ for any $t\in T_{m}$ (see Ch. 2.1 of [16]).

Consider now a sequence $\left\{z_{j}\right\}\subset T_{m}$ converging to $t_{m}$ as $j$ goes to infinity. For each $j,$ there is a function $u_{j}\in C^{2,\alpha}(\overline{\Omega}_{m})$ such that $\mathcal{M}\left(u_{j}\right)=0,$ $u_{j}|_{\partial\Omega}=0$ and $u_{j}|_{\Gamma_{m}}=z_{j}.$ Since $\left\|u_{j}\right\|_{C^{2,\alpha}\left(\overline{\Omega}\right)}\leq C$ there is a subsequence of $\left\{u_{j}\right\}$ that converges uniformly in $\overline{\Omega}_{m}$ on the $C^{2}$ norm to a solution $w_{m}\in C^{2}(\overline{\Omega_{m}})$ of $\mathcal{M}=0$ in $\Omega_{m}$. From PDE regularity $w_{m}\in C^{2,\alpha}(\overline{\Omega}_{m})$ ([5]).

The function $w_{m}$ is a solution of $\mathcal{M}=0$ in $\Omega_{m}$ that satisfies $w_{m}|_{\partial\Omega}=0$, $w_{m}|_{\Gamma_{m}}=t_{m}$ and $\sup_{\partial\Omega_{m}}\left|\operatorname{grad}w_{m}\right|\leq s.$ It follows that $t_{m}\in T_{m},$ that is, $w_{m}=u_{t_{m}}$. Moreover, it follows from the implicit function theorem $\sup_{\partial\Omega_{m}}\left|\operatorname{grad}w_{m}\right|<s$ leads to a contradiction so that $\sup_{\partial\Omega}\left|\operatorname{grad}w_{m}\right|=s.$

Since the bound $\left\|w_{m}\right\|_{C^{2,\alpha}\left(\overline{\Lambda}\right)}\leq C$ does not depend on $m$ there is a subsequence of $\{w_{m}\}$ converging uniformly $C^{2}$ on compact subsets of $\overline{\Lambda}$ to a solution $u_{s}\in C^{2,\alpha}(\overline{\Lambda})$ of $\mathcal{M}=0$ in $\Omega$ satisfying $u_{s}|_{\partial\Omega}=0$ and $\sup_{\partial\Omega}\left|\operatorname{grad}u_{s}\right|=s.$

We prove that given if 0\leq s_{1}<s_{2}\then $u_{s_{1}}<u_{s_{2}}$ on $\Lambda.$ Suppose that $u_{s_{i}}=\lim_{m}w_{m}^{i},$ where $w_{m}^{i}\in C^{2,\alpha}\left(\overline{\Omega}_{m}\right)$ is a solution as above $i=1,2$. Given $p\in\Lambda,$ there is $m\in\mathbb{N}$ such that $p\in\Omega_{m}.$ It is clear that $w_{n}^{1}|_{\partial\Omega_{n}\backslash\partial\Omega}<w_{n}^{2}|_{\partial\Omega_{n}\backslash\partial\Omega}$ and then, from the comparison principle, $w_{n}^{1}(p)<w_{n}^{2}(p)$ for all $n\geq m.$ It follows that $u_{s_{1}}(p)\leq u_{s_{2}}(p)$ for all $p\in\Lambda.$ From the maximum principle, $u_{s_{1}}(p)<u_{s_{2}}(p)$ for all $p\in\Lambda.$ This proves that the graphs of $u_{s}$ form a foliation of an open set of $\mathbb{T\times R}$. We also have

[TABLE]

Assume that $\limsup_{r(p)\rightarrow\infty}\left(u_{s_{2}}(p)-u_{s_{1}}(p)\right)=0.$ We claim that for any $\varepsilon>0$, $u_{s_{2}}-\varepsilon<u_{s_{1}}.$ Indeed, if

[TABLE]

is nonempty for some $\varepsilon>0$ then

[TABLE]

is also nonempty and open. Since $u_{s_{2}}-\varepsilon/2$ is also a solution of $\mathcal{M}\left[u\right]=0,$ the comparison principle implies that $U$ is not bounded. Then there is a divergence sequence $p_{n}$ in $\mathbb{T}$ such that $u_{s_{2}}(p_{n})-\varepsilon/2>u_{s_{1}}(p_{n})$ so that $\limsup_{n\rightarrow\infty}\left(u_{s_{2}}(p_{n})-u_{s_{1}}(p)\right)\geq\varepsilon/2,$ contradiction. Letting $\varepsilon$ go to zero in $u_{s_{2}}-\varepsilon<u_{s_{1}}$ we obtain a contradiction with $u_{s_{1}}<u_{s_{2}}$ on $\Lambda.$ This concludes the proof of the theorem.