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\def\rightheadline{EJDE--2000/45\hfil 
Some constancy results for harmonic maps 
\hfil\folio}
\def\leftheadline{\folio\hfil Kewei Zhang
 \hfil EJDE--2000/45}

\def\pretitle{\vbox{\eightrm\noindent\baselineskip 9pt %
 Electronic Journal of Differential Equations,
Vol.~{\eightbf 2000}(2000), No.~45, pp.~1--13.\hfil\break
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\hfill\break 
ftp ejde.math.swt.edu (login: ftp)\bigskip} }

\topmatter
\title
 Some constancy results for harmonic maps \\
from non-contractable domains into spheres 
\endtitle

\thanks 
{\it 1991 Mathematics Subject Classifications:} 58E20 35J65 35J60 \hfil\break\indent
{\it Key words:} harmonic maps, uniqueness, Pohozaev identity, closed geodesics, 
\hfil\break\indent tubular neighbourhoods.\hfil\break\indent
\copyright 2000 Southwest Texas State University  and
University of North Texas.\hfil\break\indent
Submitted March 31, 2000. Published June 14, 2000.
\endthanks

\author  Kewei Zhang  \endauthor
\address 
Kewei Zhang \hfill\break\indent  
Department of Mathematics, Macquarie University, Sydney Australia
\endaddress
\email  kewei\@ics.mq.edu.au
\endemail

\abstract 
  We use the Pohozaev identity on sub-domains of a 
  Euclidean $r$-neighbourhood for a closed or broken curve to show 
  that harmonic maps from such domains into spheres with  constant 
  boundary value remain constant. 
 \endabstract
\endtopmatter
\document

\head \S 1. Introduction\endhead 
In this paper we generalize 
a constancy result for harmonic maps from a non-star shaped domain 
in $\Bbb R^3$ to the sphere $S^2$  obtained by Chou and Zhu 
\cite{CZ}. In \cite{CZ} a special class of non-star shaped domains 
was constructed by rotating a curve which is carefully designed by 
using inversions in Euclidean spaces. The first result of the 
present paper is to generalize this result to domains including 
all smooth rotational ones (Theorem 1). For domains in $\Bbb R^m$ 
with $m\geq 3$, 
 we can show that
the same result holds on a tubular neighbourhood (see e.g. 
\cite{S, I. Cha.9}) 
 of a closed planar curve under a nondegeneracy condition
for closed geodesic in planar domains (Theorem 3). One such 
example is the tubular neighbourhood 
 of a  closed convex curve such as the solid torus. When $m\geq 4$,
we can show that the same claim is true for a thin tubular 
neighbourhood of any smooth embedded curve with an orthogonal 
moving frame. We state the results only for $u:\Omega\subset \Bbb 
R^3\to S^2$ although they can be easily proved for higher 
 dimensional cases. The only exception is Theorem 4 where we can only prove 
the result for domains at least in $\Bbb R^4$.

It is well known that if either $\Omega\subset \Bbb R^2$ is 
contractable \cite{L} or 
 $\Omega\subset \Bbb R^m$ is star-shaped with $m\geq 3$ \cite{W}, 
the constancy result holds. If
one perturb a star-shaped  domain in a $C^2$ manner, one expect to 
have the so-called `nearly star-shaped' domains and the constancy 
result is still true \cite{DZ}. It is also known that if the 
boundary of the domain $\partial\Omega$ is disconnected, the 
constancy result fails \cite{BBC}. 

The method we use is the Pohozaev identity (see \cite{P,PS,CZ}). 
We carefully divide the original domain into sub-domains which are 
thin slices of the original domain 
 such that each sub-domain is star-shaped 
with respect to some specific point on a curve.  We apply Pohozaev 
identity on each of these sub-domains and use the the constacy 
condition $u=u_0$  only on  part of its boundary. We then obtain  
an inequality on each sub-domains. We sum up  the resulting terms 
and use the definition of Riemann integral. In the limit, we 
obtain an inequality connecting two volume integrals.  We reach 
our conclusions by comparing quatities on both sides of the 
inequality. Some results on shortest path in an Euclidean domain 
in $\Bbb R^2$ (or geodesics in such a domain) are used. 

A smooth mapping $u: \Omega\subset\Bbb R^m\to S^n$ is harmonic if
$$-\Delta u=u|Du|^2\quad\text{in}\quad \Omega\tag 1$$ 
and $u$ is a critical point of the total energy 
$E(u)=\int_\Omega |Du|^2dx$. 

Let $\Omega\subset \Bbb R^m$ be piecewise smooth and $u\in 
C^2(\Omega, S^n)\cap C^1(\bar\Omega,S^n)$ be a smooth solution of 
(1). Let  $\nu(x)=(\nu_1(x),\cdots,\nu_m(x))$ be the outward 
normal vector at $x\in\partial\Omega$, and let $h=(h_1,h_2,\cdots, 
h_m)$ be a smooth vector field on $\bar\Omega$. Then (see 
\cite{CZ}) 
$$\frac{\partial}{\partial x_\alpha} \left( 
h_\alpha|Du|^2-2h_\beta\frac{\partial u_k}{\partial x_\alpha} 
\frac{\partial u_k}{\partial x_\beta}\right)=\frac{\partial 
h_\alpha}{\partial x_\alpha}|Du|^2 -2\frac{\partial 
h_\beta}{\partial x_\alpha}\frac{\partial u_k}{\partial x_\alpha} 
\frac{\partial u_k}{\partial x_\beta},\tag 2
$$ 
where the summation 
convention is assumed with $1\leq \alpha,\,\beta\leq m$ and $1\leq 
k\leq n+1$. Recall that a domain $\Omega$ is star-shaped if there 
is a point $x_0\in \Omega$ such that the line segment 
$\overline{x_0x}$ is contained in $\Omega$. For convenience, we 
call $x_0$ the {\it central point} of $\Omega$ if $\Omega$ is 
star-shaped with respect to $x_0$.  

\head \S 2. Main Results\endhead

Theorem 3 below covers Theorems 1 and 2. However, since the 
proofs of both  theorems are needed for establishing Theorem 3, we 
prove them separately. 

\proclaim{Theorem 1} Suppose $\Omega\subset R^3$ is a smooth 
domain  and the orthogonal projection of the domain to the first 
component is an interval $[a,b]$. We assume that there is a 
$\delta>0$, such that for all $a\leq t_1<t_2\leq b$, 
$|t_2-t_1|\leq \delta$, the set $$\Omega_{t_1,t_2}=\{ 
x=(x_1,x_2,x_3)\in\Omega, \; t_1\leq x_1\leq t_2\}$$ 
 is star shaped 
and there is some $t_0\in [t_1,t_2]$ such that    
$x_0=(t_0,0,\cdots,0)$ is a central point. Let $u:\bar\Omega\to 
S^2\subset R^3$ be a smooth harmonic map such that $u(x)=u_0\in 
S^2$ on $\partial\Omega$ for some constant $u_0$. Then $u\equiv 
u_0$ in $\bar\Omega$. 
\endproclaim

\remark{Remark 1} The rotational domains are a special case of 
those defined  in Theorem 1.  More precisely, suppose 
$x_2=f(x_1)>0$ is a smooth function defined in $[a,b]$, then the 
rotation of the the two dimensional region bounded by $f$ and 
$x_1$ axis around $\Bbb R^{m-2}$ defines the domain. In 
particular, the domain we defined is much more general that the 
one given by \cite{CZ}. 
\endremark

Let $\gamma:[0,l]\to \Bbb R^m$ be an simple, smooth and convex 
curve with bounded curvatures. Then it is easy to see that the 
$r$-neighbourhood $$\Omega_r=\{ x\in \Bbb R^m,\; 
\operatorname{dist}(x,\gamma)<r\}$$ is a tubular neighbourhood of 
$\gamma$ with $(m-1)$-dimensional open balls of radius $r$ as its 
fibres. If $\gamma$ is a broken curve, $\Omega_r$ is the union of 
a tubular neighbourhood $\cup_{0<s<l}B_s$ and two half-balls at 
each end of the curve, where $B_s$ is an $m-1$-dimensional open 
ball lying in the normal hyperplane of $\gamma(s)$ and is centred 
at $\gamma(s)$.  

\proclaim{Theorem 2} Let $\gamma\subset R^2$ be a smooth, closed 
convex  curve with maximal curvature $k_0>0$. Let  $\Omega_r$ be 
the $r$-neighbourhood of $\gamma$ in $\Bbb R^3$ with $0<rk_0<1$. 
Suppose $u:\bar\Omega_r\to S^2$ is a smooth harmonic map such that 
$u=u_0$ on the boundary. Then $u\equiv u_0$  in $\bar\Omega_r$. 
\endproclaim

\remark{Remark 2} Let $\Bbb R^3=\Bbb R^2\times \Bbb R$, and 
$\Omega_r$ be the tubular domain of $\gamma$ defined in Theorem 2. 
We see that the boundary of the two-dimensional domain 
$\Omega_r\cap \Bbb R^2\times\{ 0\}$ consists of two disconnected 
convex curves. The inner curve is a closed geodesic of 
$\Omega_r\cap \Bbb R^2\times\{ 0\}$ with curvature $1-rk(s)$. The 
intersection of the normal plane in $\Bbb R^3$ of the inner curve  
and $\Omega_r$ is a disc with radius $r$, so it is a convex 
 two dimensional
set. In fact, use these two properties of $\Omega_r$ to prove 
Theorem 1. In the proof we shall see that if the intersection is a 
copy of the same convex set up to a rigid motion, the proof can 
still go through. We have 
\endremark

\proclaim{Corollary 1} Suppose $\gamma$ is defined as in Theorem 
2. Let $[a,b]\times \gamma$ be a closed cylinder and $G_r$ be the 
$r$-neighbourhood of $ \gamma \times [a,b]$ with $rk_0<1. $ Then 
any smooth harmonic map from $\bar G_r$ to $S^2$ with constant 
boundary value remains 
 a constant
in $\bar G_r$.
\endproclaim

The following result is more general than Theorem 1 and 2. We need  
some technical assumptions on the  geodesic (locally shortest 
path) in the tubular neigbourhood $\Omega_r$ of the curve. It is 
known that for a smooth, compact and connected  manifold $M$ with 
boundary, there is a geodesic of class $C^{1,1}$ connecting any 
two points in $M$. If the fundamental group of $M$ is nontrivial, 
that is $\pi_1(M)\neq \{0\}$, then there is a nontrivial geodesic  
of  class $C^{1,1}$ (\cite{A,S,ABB,ABL,AB,C1,C2} and the 
references therein). In a closed Jordan region on the  plane, 
 the geodesics are  known as the locally
shortest paths \cite{BR, BH}. 

We are interested in tubular neighbourhoods of an embedded
 planar curve $\gamma$. 
(closed or broken). If  $\gamma$  is closed, it is not hard to 
show that there is a closed 
 geodesic (or locally shortest curve) which is homotopic to $\gamma$, 
following the arguments in \cite{BR} or \cite{S}. If $\gamma: 
[0,l]\to\Bbb R^2$ is broken, the $r$-neighbourhood $\Omega_r\cap 
\Bbb R^2\times \{0\}$ of $\gamma$ is a Jordan region for small 
$r>0$ and the shortest path connecting $\gamma(0)$ and $\gamma(l)$ 
is unique \cite{BR}. Furthermore, in both cases, the geodesics are 
of class $C^{1,1}$ (see, for example \cite{C1,C2}). 

The geometric descriptions for geodesics in domains of $\Bbb R^n$ 
can be found, for example, in \cite{AB}: A geodesic contacting the 
boundary in a segment is a geodesic of the boundary (in $\Bbb 
R^2$, it is part of the boundary); a geodesic segment not touching 
the boundary is a 
 straight line segment. A segment on the boundary joins a segment in the ambient space in a
differentiable join. An endpoint on the boundary of a segment not 
touching the boundary is called a {\it switching point}. The 
accumulation points of switching points are called {\it 
intermittent points}. 

We need some technical conditions on the tubular domains which 
exclude the intermittent points. The reason for such assumptions 
is purely for avoiding technical complications. 

\proclaim{ Hypothesis (H1)} If $\gamma\subset \Bbb R^2\times 
\{0\}$ is closed and  $\gamma_0$ is a closed geodesic 
 in $\bar\Omega_r\cap \Bbb R^2\times \{0\}$
 which is  homotopic to $\gamma$, where $\Omega_r\subset \Bbb R^3$ 
is a tubular neighbourhood of $\gamma$. Then 
\roster 
\item"(i)"  $\gamma_0$ has finite number of switching points, 
hence it does not have intermittent points;
\item"(ii)" there is a $\delta>0$, such that for every
straight line segment $\mu\subset \gamma_0$  lying inside 
$\Omega_r\cap\Bbb R^2\times\{0\}$ and any $p,\, q\in\mu$ with 
$|p-q|\leq \delta$,  the sub-domain of $\Omega_r$ bounded by 
 normal planes of $\mu$ passing through $p$ and $q$ respectively is a star-shaped domain
with any point on $\mu$ between $p$ and $q$ a central point.
\endroster
\endproclaim

\proclaim{Hypothesis (H2)} If $\gamma:[0,l]\to \Bbb R^2$ is broken 
with $\gamma(0)=p\neq q=\gamma(l)$. Let $p^\prime = p-\dot 
\gamma(0) r$, $q^\prime=q+\dot \gamma(l)r$. Then for $r>0$ 
sufficiently small, $p^\prime,\, 
q^\prime\in\partial(\Omega_r\cap\Bbb R^2\times\{0\})$. Let 
$\gamma_0$ be the geodesic in $\overline{\Omega_r\cap\Bbb 
R^2\times\{0\}}$ connecting $p^\prime$ and $q^\prime$. Then 
\roster
\item"(i)"  $\gamma_0$ has finite number of switching points, hence it does not have
 intermittent points;
\item"(ii)" there is a $\delta>0$, such that for every
straight line segment $\mu\subset \gamma_0$  lying inside 
$\Omega_r\cap\Bbb R^2\times\{0\}$ and any $a,\, b\in\mu$ with 
$|a-b|\leq \delta$,  the sub-domain of $\Omega_r$ bounded by 
 normal planes of $\mu$ passing through $p$ and $q$ respectively is a star-shaped domain
with any point on $\mu$ between $p$ and $q$ a central point.
\endroster
\endproclaim

\proclaim{Theorem 3} $\gamma\subset R^2$ be a smooth  closed or 
broken curve with maximal curvature $k_0>0$. Let  $\Omega_r$ be 
the $r$-neighbourhood of $\gamma$ in $\Bbb R^3$ with $0<rk_0<1$ 
satisfying (H1) (or (H2) respectively). Then, if 
$u:\bar\Omega_r\to S^2$ is a smooth harmonic map such that $u$ is 
a constant $u_0$ on the boundary, Then $u\equiv u_0$  in 
$\bar\Omega_r$. 
\endproclaim

Notice that if $\gamma$ is not a closed curve,
 $\Omega_r$ is not a $C^2$ domain. However, we can perturb it slightly
near both ends of $\Omega_r$ to make it $C^2$. For simplicity, we 
just prove the result on $\Omega_r$. 
 Theorem 4 below
deals with the constancy problem in general tubular neighbourhoods 
of embedded curves under 
 a technical condition. We assume
that there is a smooth  orthogonal moving frame along the curve 
\cite{S, Ch 1}. Suppose that $\gamma:[0,l]\to \Bbb R^m$ is a 
smooth curve parameterized by its arc-length $s\in [0,l]$. We also 
assume  that there is a smooth orthogonal basis $e_2(s),\cdots, 
e_n(s)$ on the normal hyperplane of $\gamma(s)$. Let $\dot\gamma 
(s)=e_1(s)$. Then $$\aligned &\dot e_1(s)=-k_1(s)e_2,\\ &\dot 
e_j(s)=k_{j-1}(s)e_{j-1}-k_j(s)e_{j+1},\quad 2\leq j\leq m-1,\\ 
&\dot e_m(s)=k_{m-1}e_{m-1}.
\endaligned
$$ We call $k_1(s)$ the first curvature of $\gamma$ and 
 $E(s)$  a moving orthogonal frame along $\gamma$.
We have

\proclaim{Theorem 4} Suppose $n\geq 2$ and $m\geq 4$. Suppose that 
$\gamma$ is an embedded smooth curve (closed or not closed)  in 
$\Bbb R^m$ with   a smooth  orthogonal moving frame. Let $k_1(s)$ 
be the first curvature of $\gamma$ and 
 $\Omega_r$ be its $r$-neighbourhood in $\Bbb R^m$.  Then for 
sufficiently small $r>0$, the only smooth harmonic map $u$ from 
$\bar\Omega_r$ to $S^n$ with  constant boundary value $u_0$ is 
$u\equiv u_0$. 
\endproclaim

\head \S 3. Proofs of the main results \endhead

\demo{Proof of Theorem 1} We divide $[a,b]$ evenly as 
$a=t_0<t_1<\cdots <t_N=b$, with $t_{k+1}-t_k=(b-a)/N$, 
$i=0,1,2,\dots, N$ such that $(b-a)/N<\delta$. Let 
$$\Omega_i=\{ x\in\Omega, \; t_i\leq x_1\leq t_{i+1}\}$$ 
for $i=0,1,\cdots, 
N-1$. From the property of $\Omega$, we see that $\Omega_i$ is 
star-shaped and there is some $t_i^\prime\in [t_k,t_{k+1}]$ such 
that $x^i=(t^\prime_i,0,\cdots,0)$ is a central point of 
$\Omega_i$. We divide the boundary of $\Omega_i$ into three parts: 
$$\partial\Omega_i=\Gamma_i\cup\Gamma_{i+1}\cup S_i,$$ where 
$\Gamma_i=\{ x\in\bar\Omega,\, x_1=t_i\}$, and 
$S_i=\partial\Omega\cup\bar\Omega_i$. Notice that both $\Gamma_0$ 
and $\Gamma_N$ are contained in $\partial\Omega$. Now, we apply 
(2) with $m=3$ to $u$ over the domain $\Omega_i$ for each fixed 
$i$ with $h=x-x^i$. Since $\frac{\partial h_\alpha}{\partial 
x_{\beta}}=\delta_{\alpha,\beta}$ and $\frac{\partial 
h_\alpha}{\partial x_{\alpha}}=3,$, we have, 
$$\frac{\partial}{\partial x_\alpha} \left( 
h_\alpha|Du|^2-2h_\beta\frac{\partial u_k}{\partial x_\alpha} 
\frac{\partial u_k}{\partial x_\beta}\right)=|Du|^2 \tag 3$$ for 
$x\in \Omega_i$. Integrating both sides of (3) and applying the 
divergence theorem, we obtain $$ \int_{\partial\Omega_i}\left( ( 
x_\alpha-x^i_\alpha)|Du|^2-2(x_\beta-x^i_\beta) \frac{\partial 
u_k}{\partial x_\beta}\frac{\partial u_k}{\partial 
x_\alpha}\right)\nu_\alpha dS= \int_{\Omega_i}|Du|^2dx.\tag 4$$ 
Since on $S_k$, $\Gamma_0$ and $\Gamma_N$, we have assumed that 
$u=u_0$, we have $$(x_\beta-x^i_{\beta})\frac{\partial 
u_k}{\partial x_\beta}\frac{\partial u_k}{\partial x_\alpha} 
\nu_\alpha=|Du|^2\langle (x-x^i), \nu\rangle,$$ where 
$\langle\cdot,\cdot\rangle$ denotes the inner product in $\Bbb 
R^3$. Therefore (4) can be rewritten as 
$$\int_{\partial\Omega_i}\left(|Du|^2\langle x-x^i,\nu\rangle
-2\langle Du_k,x-x^i\rangle\langle 
Du_k,\nu\rangle\right)=\int_{\Omega_i}|Du|^2dx.\tag 4'$$ Now, if 
$0<i<N-1$, we have 
$$\aligned \int_{\partial\Omega_i}& 
\left(|Du|^2\langle x-x^i,\nu\rangle -2\langle 
Du_k,x-x^i\rangle\langle Du_k,\nu\rangle\right) dS\\ 
=&-\int_{S_i}|Du|^2\langle x-x^i,\nu \rangle dS\\
&+\int_{\Gamma_{i+1}}\left(|Du|^2\langle x-x^i,\nu\rangle
-2\langle Du_k,x-x^i\rangle\langle Du_k,\nu\rangle\right) dS\\
&-\int_{\Gamma_{i}}\left(|Du|^2\langle x-x^i,\nu\rangle -2\langle 
Du_k,x-x^i\rangle\langle Du_k,\nu\rangle\right) dS\\ 
\leq& 
\int_{\Gamma_{i+1}}\left(|Du|^2\langle x-x^i,\nu\rangle -2\langle 
Du_k,x-x^i\rangle\langle Du_k,\nu\rangle\right) dS\\ 
&-\int_{\Gamma_{i}}\left(|Du|^2\langle x-x^i,\nu\rangle -2\langle 
Du_k,x-x^i\rangle\langle Du_k,\nu\rangle\right) dS, 
\endaligned
\tag 5$$ 
where we choose the normal direction of $\Gamma_i$ as 
towards the positive side of the $x_1$-axis. If $i=0$, we have 
$$\aligned \int_{\partial\Omega_0}&\left(|Du|^2\langle 
x-x^0,\nu\rangle -2\langle Du_k,x-x^0\rangle\langle 
Du_k,\nu\rangle\right) dS\\ 
=&-\int_{S_0}|Du|^2\langle x-x^0,\nu 
\rangle dS\\ &+\int_{\Gamma_{1}}\left(|Du|^2\langle 
x-x^0,\nu\rangle -2\langle Du_k,x-x^0\rangle\langle 
Du_k,\nu\rangle\right) dS\\ &-\int_{\Gamma_{0}} |Du|^2\langle 
x-x^0,\nu\rangle dS\\ 
\leq& \int_{\Gamma_{1}}\left(|Du|^2\langle 
x-x^0,\nu\rangle -2\langle Du_k,x-x^0\rangle\langle 
Du_k,\nu\rangle\right) dS, 
\endaligned
\tag 6$$ where the normal direction of $\Gamma_0$ is the outward 
normal direction of $\Omega_0$. Similarly, When $i=N-1$, we have, 
$$\aligned &\int_{\partial\Omega_{N-1}} \left(|Du|^2\langle 
x-x^{N-1},\nu\rangle -2\langle Du_k,x-x^{N-1}\rangle\langle 
Du_k,\nu\rangle\right) dS\\ &\leq 
-\int_{\Gamma_{N-1}}\left(|Du|^2\langle x-x^{N-1},\nu\rangle 
-2\langle Du_k,x-x^{N-1}\rangle\langle Du_k,\nu\rangle\right) dS.
\endaligned
\tag 7$$ Now we sum up (5), (6) and (7) for $i=0,1,\cdots, N-1$ 
and compare the resulting inequality with (4'), we have $$\aligned 
&\sum^{N-1}_{i=0}\int_{\Omega_i} |Du|^2dx=\int_{\Omega} |Du|^2dx\\
&\leq \sum^{N-1}_{i=0}\int_{\Gamma_{i+1}}\left(|Du|^2\langle 
x^{i+1}-x^i,\nu\rangle -2\langle Du_k,x^{i+1}-x^i\rangle\langle 
Du_k,\nu\rangle\right) dS. 
\endaligned
\tag 8 $$ Since $x^{i+1}-x^i=(t^\prime_{i+1}-t^\prime_i,0,0)$ and 
the normal vector $\nu$ on $\Gamma_{i+1}$ is $(1,0,0)$, we have, 
from (8) that 
$$\int_{\Omega} |Du|^2dx\leq 
\sum^{N-1}_{i=0}\left[\int_{\Gamma_{i+1}}\left(|Du|^2- 
2\left|\frac{\partial u}{\partial 
x_1}\right|^2\right)dS\right](t^\prime_{i+1}-t^\prime_i). 
$$ 
Now we let $N\to\infty$ and use the definition of Riemann integral 
to obtain $$\int_{\Omega} Du|^2dx\leq \int_{\Omega}|Du|^2dx 
-2\int_\Omega \left|\frac{\partial u}{\partial x_1}\right|^2 dx.$$ 
Hence $$\int_\Omega \left|\frac{\partial u}{\partial x_1}\right|^2 
dx=0,\qquad\text{so that}\qquad \frac{\partial u}{\partial 
x_1}=0$$ in $\Omega$. Thus $u\equiv u_0$. \hfill $\boxed{\,}$ 
\enddemo

\demo{Proof of Theorem 2}  Let $\gamma :[0,l]\to \Bbb R^2$ be a 
$C^2$ closed embedded convex curve 
 parameterized by its arc-length, $\gamma(0)=\gamma(l)$. We take  the curve as along
the counter-clockwise direction so that when travelling along the 
curve, the region bounded by it is on the left side. We assume 
that the curvature bound $0\leq k(s)\leq k_0$. We write 
$$\gamma=(x_1(s),x_2(s))\qquad \text{and}\qquad \beta(s)=(-\dot 
x_2(s), \dot x_1(s)),$$ where $\beta(s)$ is the unit normal vector 
pointing towards the interior of the region. Let $\bar\Omega_r$ be 
the closed $r$-neighbourhood in $\Bbb R^3=\Bbb R^2\times \Bbb R$, 
 where $0<rk_0<1$ so that the mapping
$$F: (s, t, z)\to \gamma(s)+t\beta(s)+ze_3^2$$ is periodic in $s$ 
with period $l$. $F$ is smooth and is both one-to-one and onto 
from $[0,l)\times \bar B_r(0)$ to $\bar\Omega_r$, with 
$F(0,\cdot)=F(l,\cdot)$ where $e_3$ is the unit vector  normal to 
the plane and  $$\bar B_r(0)=\{ (t,z)\in \Bbb R^2, \, t^2+z^2\leq 
r^2\}$$ is the closed disc in $\Bbb R^{2}$. The Jacobian of this 
mapping is $1-tk(s)$, where $k(s)$ is the curvature of $\gamma$. 
Now we take the inner curve of the tubular domain defined as
$\gamma_r(s)=\gamma(s)+r\beta(s)$ and divide $[0,l]$ evenly as 
$$0=s_0<s_1<\cdots s_{N-1}<s_N=l,\qquad s_{k+1}-s_k=\frac{l}{N},\; 
k=0,1,\cdots N-1$$ and let $s^\prime_k$ be the mid-point of 
$[s_k,s_{k+1}]$. We let $\Gamma_i$ be the intersection of the 
normal plane of $\gamma_r$ at $s=s_i$ and $\Omega_r$.  We then 
define $\bar \Omega_i$ be the closed sub-domain of $\Omega_r$ 
bounded by $\Gamma_i$ and $\Gamma_{i+1}$. Notice that $\gamma_r$ 
is a closed convex curve so that $\Gamma_N=\Gamma_0$ and 
$\Omega_N=\Omega_0$. Similar to the proof of Theorem 1, we apply 
(4') to each $\Omega_i$ with $h^i(x)=x-\gamma_r(s^\prime_i)$. We 
have 
$$\int_{\partial\omega_i}\left(|Du|^2\langle 
h^i(x),\nu\rangle -2 \langle Du_k,h^i(x)\rangle\langle 
Du_k,\nu\rangle \right) dS=\int_{\omega_i}|Du|^2dx.
\tag 9
$$ 
As in the proof of Theorem 1, we let $I_i$ and $J_i$ be 
the left and right hand sides of (9) and let 
$\partial\Omega_i=\Gamma_i\cup\Gamma_{i+1}\cup S_i$, where 
$S_i=\partial\Omega_i\cap \partial \Omega_r$. Let us first 
consider the surface integral over $S_i\subset \partial\Omega_r$ 
and notice that $u=u_0$ which is a constant on $S_i$, so (5) gives 
$$\int_{S_i}\left(|Du|^2\langle h^i(x),\nu\rangle -2 
\langle Du_k,h^i(x)\rangle\langle Du_k,\nu\rangle \right)dS 
 =-\int_{S_i}\left|\frac{\partial u}{\partial \nu}\right|^2\langle 
h^i,\nu\rangle dS. 
\tag 10$$ 
Since $\langle h^i,\nu\rangle$ is not necessarily 
positive on $S_i$ from our choice of the centre 
$\gamma_r(s^\prime_i)$, we need to find a bound. A general point 
$x\in S_i$ can be written as $$x=\gamma(s)+t\beta(s)+ze_3$$ with 
$t^2+z^2=r^2$, for some $s\in [s_i,s_{i+1}]$ and the outward 
normal vector at $x$ is $$\nu=t\beta(s)+ze_3.$$ Recall that 
$\gamma_r(s^\prime_i)=\gamma(s^\prime_i)+r\beta(s^\prime_i)$, we 
then have 
$$\aligned \langle h^i,\nu\rangle &=\langle 
x-\gamma_r(s^\prime_i),\nu\rangle\\ &=\langle 
\gamma(s)+t\beta(s)+ze_3 
-[\gamma(s^\prime_i)+r\beta(s^\prime_i)],\; 
t\beta(s)+ze_3\rangle\\ &=\langle 
\gamma(s)-[\gamma(s^\prime_i)+r\beta(s^\prime_i)], 
t\beta(s)\rangle+t^2+z^2\\ &=\langle 
\gamma(s)-\gamma(s^\prime_i),t\beta(s)\rangle +r^2-rt\langle 
\beta(s),\beta(s^\prime_i) \rangle\\ &\geq t\langle 
\gamma(s)-\gamma(s^\prime_i),\beta(s)\rangle. 
\endaligned
$$ Here we have used the fact that $|t|\leq r$ and $|\langle 
\beta(s),\beta(s^\prime_i) \rangle|\leq 1$ because $|\beta|=1$. 
Now, by using Taylor's expansion we have, 
$$\aligned t\langle& 
\gamma(s)-\gamma(s^\prime_i),\beta(s)\rangle \\
&=t\langle 
\gamma(s)-\gamma(s^\prime_i),\beta(s)-\beta(s^\prime_i)\rangle + 
t\langle \gamma(s)-\gamma(s^\prime_i),\beta(s^\prime_i)\rangle\\
&=t\langle 
\int^s_{s^\prime_i}\dot\gamma(\tau)d\tau,\int^s_{s^\prime_i}\dot\beta(\eta) 
d\eta\rangle +t\langle \dot 
\gamma(s^\prime_i)(s-s^\prime_i)+\frac{1}{2}\ddot \gamma(\xi_i) 
(s-s^\prime_i)^2, \beta(s^\prime_i)\rangle\\ &\geq 
-tk_0(s-s^\prime_i)^2-\frac{k_0}{2}(s-s^\prime_i)^2 
\geq-\frac{3rk_0}{2}\left(\frac{l}{2N}\right)^2,
\endaligned
\tag11 $$ where we have used the facts that 
$\dot\gamma(s^\prime_i)\perp \beta(s^\prime_i)$,  
$|\dot\beta(\eta)|=k(\eta)\leq k_0$, and 
$|\ddot\gamma(\xi_i)|=|k(x_i)|\leq k_0$, with $x_i$ a 
 point
between $s$ and $s^\prime_i$ given by the Taylor expansion. Thus 
the right hand side of  (10) has the following upper bound 
$$-\int_{S_i}\left|\frac{\partial u}{\partial \nu}\right|^2\langle 
h^i,\nu\rangle dS \leq \frac 
{3rk_0l^2}{16N^2}\int_{S_i}\left|\frac{\partial u}{\partial 
\nu}\right|^2 dS.$$ Now we sum up $I_i$'s as in the proof of 
Theorem 1 to obtain $$\aligned &\sum^{N-1}_{i=0}I_i\leq \frac 
{3rk_0l^2}{8N^2}\sum^{N-1}_{i=0} \int_{S_i}\left|\frac{\partial 
u}{\partial \nu}\right|^2 dS\\ &+\sum^{N-1}_{i=0} 
\int_{\Gamma_{i+1}}\left(|Du|^2 \langle 
\gamma_r(s^\prime_{i+1})-\gamma_r(s^\prime_{i}),\,\nu\rangle 
-2\langle Du_k, 
\gamma_r(s^\prime_{i+1})-\gamma_r(s^\prime_{i})\rangle \langle 
Du_k,\nu\rangle \right) dS\\ &=A_1+A_2, 
\endaligned
\tag 12$$ where 
$$ A_1= \frac {3rk_0l^2}{8N^2}\sum^{N-1}_{i=0} 
\int_{S_i}\left|\frac{\partial u}{\partial \nu}\right|^2 dS =\frac 
{3rk_0l^2}{16N^2}\int_{\partial\Omega}\left|\frac{\partial 
u}{\partial \nu}\right|^2 dS \to 0\tag 13
$$ 
as $N\to \infty$; 
$$A_2=\sum^{N-1}_{i=0} \int_{\Gamma_{i+1}}\Big(|Du|^2 \langle 
\gamma_r(s^\prime_{i+1})-\gamma_r(s^\prime_{i}),\,\nu\rangle 
-2\langle Du_k, 
\gamma_r(s^\prime_{i+1})-\gamma_r(s^\prime_{i})\rangle \langle 
Du_k,\nu\rangle \Big) dS.
$$
 Notice that $\Gamma_N=\Gamma_0$, 
$\nu =\dot\gamma(s_{i+1})$, 
$$\aligned \langle 
\gamma_r(s^\prime_{i+1})-\gamma_r(s^\prime_{i}),\nu\rangle 
=&\langle\dot \gamma_r(s_{i+1})(s^\prime_{i+1}-s^\prime_{i}),\dot 
\gamma(s_{i+1})\rangle+\langle \frac{1}{2}\ddot 
\gamma_r(x_{i+1})(s^\prime_{i+1}-s_{i+1})^2\\
& -\frac{1}{2}\ddot 
\gamma_r(\eta_{i+1})(s_{i+1}-s^\prime_{i})^2,\dot 
\gamma(s_{i+1})\rangle, 
\endaligned
$$ where $\xi_{i+1}$ and $\eta_{i+1}$ are two points in $(s_{i+1}, 
s^\prime_{i+1})$ and $(s^\prime_{i}, s_{i+1})$ respectively. Now 
we have 
$$\aligned \langle\dot 
\gamma_r(s_{i+1})(s^\prime_{i+1}-s^\prime_{i}),\dot 
\gamma(s_{i+1})\rangle &=\langle \dot \gamma(s_{i+1})+r\dot 
\beta(s_{i+1}),\dot \gamma(s_{i+1})\rangle 
(s^\prime_{i+1}-s^\prime_{i})\\ &=[1-rk(s_{i+1})] 
(s^\prime_{i+1}-s^\prime_{i}). 
\endaligned
\tag 14$$ Since $\gamma$ is of $C^2$, there is a constant $C_0>0$ 
such that $|\ddot\gamma(s)|\leq C_0$ for all $s\in [0,l]$. 
Therefore we  also have $$\aligned &\left|\langle \frac{1}{2}\ddot 
\gamma_r(x_{i+1})(s^\prime_{i+1}-s_{i+1})^2 -\frac{1}{2}\ddot 
\gamma_r(\eta_{i+1})(s_{i+1}-s^\prime_{i})^2,\rangle\right|\\ 
&\leq \frac{1}{2}C_0\left[ 
(s^\prime_{i+1}-s_{i+1})^2+(s_{i+1}-s^\prime_{i})^2\right]\\ &\leq 
C_0(s^\prime_{i+1}-s^\prime_{i})^2. 
\endaligned
\tag 15$$ Similarly, we have $$\aligned &\langle 
\gamma_r(s^\prime_{i+1})-\gamma_r(s^\prime_{i}),Du_k\rangle\\ 
&=\langle \dot \gamma (s_{i+1}),Du_k\rangle 
(1-rk(s_{i+1})(s^\prime_{i+1}-s^\prime_{i})\\ &+\langle 
\frac{1}{2}\ddot 
\gamma_r(x^\prime_{i+1})(s^\prime_{i+1}-s_{i+1})^2 
-\frac{1}{2}\ddot 
\gamma_r(\eta^\prime_{i+1})(s_{i+1}-s^\prime_{i})^2,Du_k\rangle, 
\endaligned
\tag 16$$ with $$\aligned &\left| \langle \frac{1}{2}\ddot 
\gamma_r(x_{i+1})(s^\prime_{i+1}-s_{i+1})^2 -\frac{1}{2}\ddot 
\gamma_r(\eta_{i+1})(s_{i+1}-s^\prime_{i})^2,Du_k\rangle\right|\\ 
&\leq C_0|Du_k|(s^\prime_{i+1}-s^\prime_{i})^2.
\endaligned\tag 17
$$ Now we can estimate the second sum in (12)  as follows 
$$\aligned &A_2=\\ &\sum^{N-1}_{i=0} 
\int_{\Gamma_{i+1}}\left(|Du|^2 \langle 
\gamma_r(s^\prime_{i+1})-\gamma_r(s^\prime_{i}),\,\nu\rangle 
-2\langle Du_k, 
\gamma_r(s^\prime_{i+1})-\gamma_r(s^\prime_{i})\rangle \langle 
Du_k,\nu\rangle \right) dS\\ &\leq \int_{\Gamma_{i+1}}\left(|Du|^2 
- 2\langle Du_k,\dot\gamma(s_{i+1})\rangle^2\right)dS
[1-rk(s_{i+1})](s^\prime_{i+1}-s^\prime_{i})\\
&+C_0(s^\prime_{i+1}-s^\prime_{i})\frac{l}{N}\int_{\Gamma_{i+1}}3|Du|^2dS,
\endaligned
\tag 18$$ where we have used the fact that $|\langle 
Du_k,\nu\rangle|\leq |Du_k|$. Now we can estimate the two sums 
$A_1$ and $A_2$ in (12). 
$$\aligned \sum^{N-1}_{i=0}I_i\leq &
A_1+A_2\\ \leq& \frac 
{3rk_0l^2}{16N^2}\int_{\partial\Omega}\left|\frac{\partial 
u}{\partial \nu}\right|^2 dS \\
&+\int_{\Gamma_{i+1}}\left(|Du|^2 - 2\sum^3_{k=1}\langle 
Du_k,\dot\gamma(s_{i+1})\rangle^2\right)dS 
[1-rk(s_{i+1})](s^\prime_{i+1}-s^\prime_{i})\\
&+C_0(s^\prime_{i+1}-s^\prime_{i})\frac{l}{N}\int_{\Gamma_{i+1}}3|Du|^2dS.
\endaligned
\tag 19
$$ 
Passing to the limit $N\to\infty$ in (19) and noticing 
that $$\lim_{N\to\infty}A_1\to 0,\quad 
\lim_{N\to\infty}C_0\sum^{N-1}_{i=0}(s^\prime_{i+1}-s^\prime_{i})\frac{l}{N}\int_{\Gamma_{i+1}}3|Du|^2dS=0$$
because 
$$\sum^{N-1}_{i=0}(s^\prime_{i+1}-s^\prime_{i})\int_{\Gamma_{i+1}}3|Du|^2dS$$
converges to an integral while $l/N\to 0$, we have 
$$\aligned 
&\limsup_{N\to\infty}\sum^{N-1}_{i=0}I_i\\ &\leq 
\int_0^l\int_{\Gamma_s}|Du|^2(1-rk(s))dS\,ds-
2\int_0^l\int_{\Gamma_s}\sum^{3}_{k=1}\langle 
Du_k,\dot\gamma\rangle^2(1-rk(s))dS\,ds, 
\endaligned
\tag 20
$$ 
where $$\Gamma_s=\{ 
\gamma(s)+t\beta(s)+ze_3,\;t^2+z^2\leq r^2\}.$$ Now we sum up the 
right hand side of (10): $$ \sum^{N-1}_{i=0}J_i= 
\int_{\omega_i}|Du|^2 dx =\int_{\Omega_r} |Du|^2 dx.\tag 21$$ We 
now change variables $$x=\gamma(s)+t\beta(s)+ze_3,$$ to obtain $$ 
\int_{\Omega_r}|Du|^2 dx 
=\int^l_0\int_{\Gamma_s}|Du|^2(1-tk(s))dS\,ds. \tag 22$$ Finally we 
obtain, from (20) and (22), $$\aligned 
&\int^l_0\int_{\Gamma_s}|Du|^2(1-tk(s))dS\,ds\\ 
&\leq\int_0^l\int_{\Gamma_s}|Du|^2(1-rk(s))dS\,ds- 
2\int_0^l\int_{\Gamma_s}\sum^{3}_{k=1}\langle 
Du_k,\dot\gamma\rangle^2(1-rk(s))dS\,ds, 
\endaligned
$$ so that $$\int^l_0\int_{\Gamma_s}|Du|^2(r-t)k(s)dS\,ds\leq 
 -2\int_0^l\int_{\Gamma_s}\sum^{3}_{k=1}\langle Du_k,\dot\gamma\rangle^2(1-rk(s))dS\,ds.
\tag 23$$ The first consequence of (23) is 
$$\int_0^l\int_{\Gamma_s}\sum^{3}_{k=1}\langle 
Du_k,\dot\gamma\rangle^2(1-rk(s))dS\,ds=0,$$ so that $\langle 
Du_k,\dot\gamma\rangle=0$ hence for each fixed $(t,z)$, 
$u(\gamma(s)+t\beta(s)+ze_3)$ is independent of $s$.  Now, at 
least in an interval $[a,b]\subset [0,l]$ with $a<b$, $k(s)>0$ 
hence the left hand side of (23) gives $|Du|^2=0$ for $s\in 
[a,b]$. Therefore in $$\{\gamma(s)+t\beta(s)+ze_3,\, s\in [a,b], 
t^2+z^2\leq r\},$$ we have $u=u_0$. Since $u$ is independent of 
$s$, we see that $u\equiv u_0$. \hfill $\boxed{\,}$ 
\enddemo

\demo{Proof of Theorem 3} If the curve $\gamma$ is closed and 
$\Omega_r\subset \Bbb R^3$ is its open $r$-neighbourhood with 
$rk_0<1$, we let $\gamma_0$ be a closed geodesic homotopic to 
$\gamma$. If $\gamma_0$ does not have switching point, $\gamma_0$ 
must be the inner curve $\gamma_r$ of $\Omega_r$ defined in the 
proof of Theorem 2 and it must be convex. Otherwise it is not the 
locally shortest geodesic. Therefore, from Theorem 2, $u\equiv 
u_0$. It is also obvious that if $\gamma_0$ has switching points, 
it  must have at least two such points. Since we assumed that 
$\gamma_0$ has finitely many switching points, we denote them by 
$p_1,\, p_2,\cdots p_m, p_{m+1}$ with $p_1=p_{m+1}$ such that  
$p_k$ and $p_{k+1}$ are two consequent switching points along 
$\gamma_0$. We parameterize $\gamma_0$ by its arc-length 
$\gamma_0:[0,l]\to \bar\Omega_r$ with $\gamma_0(0)=p_1$. let 
$0=s_1\leq s_2<\cdots <s_m<s_{m+1}=l$ be such that 
$\gamma_0(s_k)=p_k$, $k=1,2,\cdots m+1$.  Let $A_k$ be the 
sub-domain of $\Omega_r$ between the normal plane of $\gamma_0$ at 
$p_k$ and $p_{k+1}$. The part of $\gamma_0$ in $A_k$ is either a 
line segment or the shorter part of the boundary of $\Bbb 
R^2\cap\bar\Omega_r\cap\bar A_k$ and is convex. We let 
$\gamma_0\cap\bar D_k=\gamma_k$ with $k=1,2,\cdots,m$. If 
$\gamma_k$ is a line segment, we set $B_k=A_k$. Otherwise we 
simply set $B_k=\emptyset$. On $B_k$ 
 we may use the proof of Theorem 1. 
If it is part of the boundary of $\Omega_r\cap \Bbb R^2$ in the 
plane, we let $C_k=A_k$. Otherwise, we let $C_k=0$. We can use the 
method for proving Theorem 2. If we further divide each $A_k$ 
along $\gamma_0$ and pass to the limit as in the proof of Theorem 
1 or Theorem 2, we obtain 
$$\aligned \sum^m_{k=1}&\mu(k)\int_{B_k} 
|Du|^2dx+ 
\sum^m_{k=1}(1-\mu(k)\int^{s_{k+1}}_{s_k}\int_{\Gamma_s}|Du|^2(1-tk(s))dS\,ds\\
\leq& \sum^m_{k=1}\mu(k)\int_{B_k} |Du|^2dx 
+\sum^m_{k=1}(1-\mu(k))\int^{s_{k+1}}_{s_k}\int_{\Gamma_s}|Du|^2(1-rk(s))dS\,ds 
 \\ &-2(1-rk_0)\sum^m_{k=1}\int^{s_{k+1}}_{s_k}\int_{\Gamma_s} 
\sum^3_{j=1}\langle Du_j,\dot\gamma\rangle^2dS\,ds,
\endaligned
\tag 24$$ where $\mu(k)=1$ if $A_k=B_k$, $\mu(k)=0$ otherwise. We 
have also used the estimate $1-tk(s)\geq 1-rk_0$ and $1\geq 
1-rk_0$ for the last sum on the right hand side of (24). A similar 
argument as in the proof of Theorems 1 and 2 can conclude the 
proof. If $\gamma$ is a broken curve, the proof is similar. Let 
$\gamma(0)=p$, $\gamma(l)=q$. We define two half-balls as 
$$\gathered 
B_r^{-}=\{ -s\dot\gamma(0)+t\beta(0)+ze_3,\, s\geq 
0,\, s^2+t^2+z^2<r^2\}, \\ B_r^{+}=\{ 
s\dot\gamma(l)+t\beta(0)+ze_3,\, s\geq 0,\, s^2+t^2+z^2<r^2\} 
\endgathered
$$
 and let 
$$\Omega=\{ \gamma(s)+t\beta(s)+ze_3,\; t^2+z^2\leq r^2\},$$ 
we have $\bar\Omega_r=\bar B_r^{-}\cup \bar\Omega\cup\bar B_r^{+}$, 
$B_r^{\pm}\cap\Omega=\emptyset$, and $B_r^{-}\cap 
B_r^{+}=\emptyset$ at least when $r>0$ is sufficiently small. We 
now extend $\gamma$ to $\partial\Omega$ smoothly ($C^1$) by 
defining 
$$
\gathered \gamma_-(s)=p-s\dot\gamma(0),\quad  0\leq s<\leq r,\\
 \gamma_+(s)=q+s\dot\gamma(l),\quad 0\leq s\leq r. 
\endgathered
$$ 
Then $\gamma_-\subset \bar B_r^-$, and $\gamma_+\subset \bar 
B_r^+$. We see that $\gamma_-(r),\, 
\gamma_+(r)\in\partial\Omega_r$. If we divide $ B_r^-$ along 
$\gamma_-$ by using normal planes of $\gamma_-$, each sub-domain 
between two planes is star-shaped with respect to points on 
$\gamma_-$ in the sub-domain. Similarly, we can do the same for 
$B_r^+$. Let $\gamma_0=\gamma_-\cup\gamma\cup\gamma_+$. Then if we 
divide $\Omega_r$ along $\gamma_0$ by using normal planes of 
$\gamma_0$ and use the Pohozaev identity on each sub-domain and 
follow the argument for the case of  closed curves, we may
conclude the proof. \hfill $\boxed{\,}$  
\enddemo

\demo{Proof of Theorem 4} We use a similar idea as that in the 
proof of theorem 2 and theorem 3. If $\gamma$ is closed, we divide 
$\Omega_r$ along $\gamma$ itself instead of $\gamma_r$. if 
$\Omega\subset \Bbb R^m$ with $m\geq 4$, formula (4') should be 
changed into $$\int_{\partial\Omega_i}\left(|Du|^2\langle 
x-x^i,\nu\rangle -2\langle Du_k,x-x^i\rangle\langle 
Du_k,\nu\rangle\right)=(m-2)\int_{\Omega_i}|Du|^2dx.$$ Recall that 
the Jacobian of the mapping $$(s,x_2,x_3,\cdots, x_m)\to 
\gamma(s)+x_2e_2(s)+\cdots, x_me_m(s)$$ is $1-x_2k_1(s)$, we may 
follow the arguments similar to the proof of Theorem 2 by using 
$\gamma$ as the central curve of $\Omega_r$ to obtain 
$$\aligned 
(m-2)\int_{\Omega_r}|Du|^2dx=&(m-2)\int^l_0\int_{\Gamma_s}|Du|^2(1-x_2k_1(s))dS\,ds\\
\leq&\int_0^l\int_{\Gamma_s}|Du|^2dS\,ds-
2\int_0^l\int_{\Gamma_s}\sum^{n+1}_{k=1}\langle 
Du_k,\dot\gamma\rangle^2dS\,ds, 
\endaligned
\tag 25$$
 where 
$$\Gamma_s=\{ 
\gamma_s+x_2e_2(s)+\cdots,x_me_m(s),\; x^2_2+\cdots+x_m^2\leq 
r\}.$$
 Since $1-rk_0\leq 1-x_2k_1(s)\leq 1$, we estimate the right 
hand side of (25) as follows: $$\aligned 
&\int_0^l\int_{\Gamma_s}|Du|^2dS\,ds-
2\int_0^l\int_{\Gamma_s}\sum^{n+1}_{k=1}\langle 
Du_k,\dot\gamma\rangle^2dS\,ds\\ &\leq 
\frac{1}{1-rk_0}\int_0^l\int_{\Gamma_s}|Du|^2(1-x_2k_1(s))dS\,ds
-2\int_0^l\int_{\Gamma_s}\sum^{n+1}_{k=1}\langle 
Du_k,\dot\gamma\rangle^2dS\,ds\\ &= 
\frac{1}{1-rk_0}\int_{\Omega_r}|Du|^2dx-2\int_0^l 
\int_{\Gamma_s}\sum^{n+1}_{k=1}\langle 
Du_k,\dot\gamma\rangle^2dS\,ds, 
\endaligned
\tag 26$$
Combining (25) and (26) we obtain $$\left( 
m-2-\frac{1}{1-rk_0}\right)\int_{\Omega_r}|Du|^2dx \leq 
-2\int_0^l\int_{\Gamma_s}\sum^{n+1}_{k=1}\langle 
Du_k,\dot\gamma\rangle^2dS\,ds.\tag 27$$ 
Now, since $m\geq 4$, 
$m-2\geq 2$, we may have 
$$ m-2-\frac{1}{1-rk_0}>0,\quad\text{if }\; 
0<r<\frac{m-3}{k_0(m-2)}.$$ This is possible if $r>0$ is 
sufficiently small, so that $|Du|^2=0$ in $\Omega_r$ hence $u=u_0$ 
in $\bar\Omega_r$. \hfill $\boxed{\,}$ 
\enddemo

\remark{Remark 3} The methods for proving Theorem 4 can be used to 
establish similar uniqueness results for the Dirichlet problem 
$-\Delta u+|u|^{p-1}u=0$ with $u=0$ on $\partial \Omega$ \cite{Z} 
at least for $p>(n+1)/(n-3)$ in a tubular neighbourhood of a 
closed or broken curve in $\Bbb R^n$ with $n\geq 4$. One can only 
divide the domain along the 
 central
curve because the corresponding energy density is 
$F(u,Du)=|Du|^2/2-|u|^{p+1}/(p+1)$ which is not necessarily 
positive. Therefore the approach in Theorem 2 and Theorem 3 of 
using the  shortest path does not improve the result. 
\endremark

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\enddocument 

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