\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 32, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2005/32\hfil Domain geometry]
{Domain geometry and the Pohozaev identity}
\author[J. McGough, J. Mortensen, C. Rickett, G. Stubbendieck \hfil EJDE-2005/32\hfilneg]
{Jeff McGough, Jeff Mortensen,\\
Chris Rickett, Gregg Stubbendieck} % in alphabetical order
\address{Jeff McGough \hfill\break
Department of Mathematics and Computer Science,
South Dakota School of Mines and Technology, 501 E St. Joseph St,
Rapid City, SD, 57701 USA}
\email{Jeff.Mcgough@sdsmt.edu}
\address{Jeff Mortensen \hfill\break
Department of Mathematics,
601 AB, University of Nevada-Reno,
Reno, NV 89557 USA}
\email{jm@unr.edu}
\address{Chris Rickett \hfill\break
South Dakota School of Mines and Technology
501 E St. Joseph St, Rapid City, SD, 57701 USA}
\email{cdrickett@hotmail.com}
%Cluster Research Lab Advanced Computing Lab, Los
%Alamos National Lab, Los Alamos NM 87545 USA}
%\email{crickett@lanl.gov}
\address{Gregg Stubbendieck \hfill\break
Department of Mathematics and Computer Science,
South Dakota School of Mines and Technology,
501 E St. Joseph St,
Rapid City, SD, 57701 USA}
\email{Gregg.Stubbendieck@sdsmt.edu}
\date{}
\thanks{Submitted August 26, 2004. Published March 22, 2005.}
\subjclass[2000]{35J20, 35J65}
\keywords{Partial differential equations; variational identities; \hfill\break\indent
Pohozaev identities; numerical methods}
\begin{abstract}
In this paper, we investigate the boundary between existence and
nonexistence for positive solutions of Dirichlet problem
$\Delta u + f(u) = 0$, where $f$ has supercritical growth.
Pohozaev showed that for convex or polar domains, no positive
solutions may be found. Ding and others showed that for domains with
non-trivial topology, there are examples of existence of positive
solutions. The goal of this paper is to illuminate the transition
from non-existence to existence of solutions for the nonlinear
eigenvalue problem as the domain moves from simple (convex) to complex
(non-trivial topology).
To this end, we present the construction of several domains in $R^3$
which are not starlike (polar) but still admit a Pohozaev nonexistence
argument for a general class of nonlinearities. One such domain is a
long thin tubular domain which is curved and twisted in space. It
presents complicated geometry, but simple topology. The construction
(and the lemmas leading to it) are new and combined with established
theorems narrow the gap between non-existence and existence
strengthening the notion that trivial domain topology is the
ingredient for non-existence.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{cor}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\section{Introduction and background}
A fundamental question in differential equations is whether or not a
solution to the differential equation can be found. In the subject of
nonlinear elliptic equations, Pohozaev provided a very useful
tool in addressing this question. The Pohozaev variational identity has been
very successful answering questions of solvability with respect to the
nonlinearity and the domain. The authors have recently focused on the
relation between domain geometry and problem solvability.
In this paper, we continue the thread of investigation by presenting
some of the relations between the geometry of the domain and
solvability of nonlinear eigenvalue problems. The essence of these
problems may be captured into the following problem. Let $\Omega$ be
an open bounded set in $R^N$, $N\geq 3$, with a smooth boundary
$\partial\Omega$ (which means $C^2$ here). We seek $u:\Omega \to \mathbb{R}$
a positive solution to
\begin{equation}\label{semilinear}
\begin{gathered}
\Delta u + f(u) = 0, \quad x\in \Omega , \\
u = 0, \quad x \in \partial \Omega .
\end{gathered}
\end{equation}
where $f$ has critical or supercritical growth,
meaning, $f(u) \geq k \, u^{(N+2)/(N-2)}$ for some positive
constant $k$.
We ask the question ``for a prescribed domain $\Omega$ and a
nonlinearity $f$, can we find a positive solution $u$?''
We restrict our focus to positive solutions and dimension $N \geq 3$;
the latter ensuring the previous growth rate
is defined.
The case of $N=2$ is well studied and existence
of solutions has been demonstrated for general
domains \cite{brezis:nee87,brezis:psn83,defigueiredo:aee82,gidas:abp81,lin:epr89}.
Pohozaev proved that there is no
solution for polar or starlike domains~\cite{pohozaev:ee65}. A
starlike domain is one that there is at least one point in the domain
for which you can see the entire boundary (see
Figure~\ref{punctsphere}). On the other hand, Bahri and Coron,
Ding~\cite{ding:ps89,bahri:nee88}, have shown that a solution exists
when $f(u) = u^{(N+2)/(N-2)}$ and the domain has nontrivial
topology. Figure~\ref{punctsphere} gives an example of a domain with
nontrivial topology. Between these two theorems is a vast complicated
landscape of dimension, topology and growth rates.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.4\textwidth]{starlike1}
\includegraphics[width=0.35\textwidth]{punctsphere}
\caption{Starlike domain and a simple domain with nontrivial topology
\label{punctsphere}}
\end{center}
\end{figure}
The goal of this paper is to narrow the gap between Pohozaev's
nonexistence result and Ding's existence result. It appears that the
dominant factor is domain topology, not domain geometry. No proof of
this assertion is offered here, only mounting evidence of examples for
which domains with negative boundary curvature still present a
nonexistence result. A rather interesting example is found in certain
tubular domains constructed in Section~\ref{tubular}. Our main result
to this end is the construction of the required elements for a
Pohozaev non-existence proof for curved tubular domains. In this
direction, we prove three properties (Lemma's 4, 5, and 6) about the
kernel of Pohozaev's variational identity. The lemmas and the
resulting examples provides a base for building example domains for
further exploration of the solvability question. We feel that the
lemmas and examples are our main contribution in this paper; that
they provide sufficient empirical evidence that the existence or
non-existence of solutions depends on the domain being topologically
trivial. For domain construction, dimension $N=3$ is the difficult
case and our examples will address this case. Before moving into the
construction process, some background is useful.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.75\textwidth]{separation}
\caption{Question: What is the boundary between solution existence
and non-existence?
\label{separation}}
\end{center}
\end{figure}
For Pohozaev's nonexistence result to work, one needs to construct a
vector field, $h: \Omega \to \mathbb{R}^N$, which locally is close to a
radial vector field. Higher dimensions offer plenty of room and
result in admitting more domains. In three dimensions, we must
balance the restrictions arising from domain curvature and the space
requirements of the radial nature of the vector field. The results
that are available are constructed from bending or modifying the base
vector field defined by $h(x) = [x_1,x_2,x_3]=x$.
As mentioned above, the Pohozaev Identity~\cite{pohozaev:ee65} is the
principle tool used here to investigate the relation between domain
geometry and solvability. Published in 1965, it has been a widely
used variational identity in the study of divergence form elliptic
equations. The original identity has been generalized extensively,
and we will focus on the form published by Pucci and
Serrin~\cite{pucci:gvi86}. This paper examines elliptic problems in
divergence form and provides a direct route to the Pohozaev Identity.
Their result is reproduced below. The classical results of
Pohozaev and Pucci-Serrin do
not require that $N=3$ and so are presented in the more general form.
However, our constructions which follow do and so in Section 3 we
restrict ourselves to $N=3$.
\begin{theorem}[Pucci-Serrin]
Let $u$ be a $C^2$ solution to
\begin{equation}\label{el-sep}
\begin{gathered}
\mathop{\rm div}\{ g (x,\nabla u)\} + f(x,u) = 0, \quad x\in\Omega,\\
u = 0,\quad x\in\partial\Omega.
\end{gathered}
\end{equation}
Further, let $h: \Omega \to \mathbb{R}^N$, where $h_i\in C^2(\Omega)\cap
C^1(\overline{\Omega})$,
and $a\in \mathbb{R}$.
Then $u$ satisfies
\begin{equation}\label{pucci-sep}
\begin{aligned}
&\int_{\Omega}\big\{ \mathop{\rm div}(h) [F(x,u)-G(x,\nabla u)] + h \cdot
[F_{x}(x,u)-G_{x}(x,\nabla u)]\\
&- a u f(x,u) + a\nabla u \cdot g(x,\nabla u)
+ \nabla u\cdot Dh \, g(x,\nabla u) \big\}\, dx\\
&= \int_{\partial\Omega} \big[ \nabla u \cdot g(x,\nabla u)
- G(x,\nabla u) \big]
(h\cdot\nu)\, dS
\end{aligned}
\end{equation}
where $g(x,s) = \partial G/ \partial s$, $f(x,s)=\partial F / \partial s$.
\end{theorem}
In our case, application to Equation~\ref{semilinear}, Identity
(\ref{pucci-sep}) becomes:
\begin{equation} \label{identity}
\begin{aligned}
&\int_{\Omega}\left\{ \mbox{\rm div}(h) F(u) - a u f(u) \right\}\, dx\\
&= \frac{1}{2} \int_{\partial\Omega}|\nabla u|^2 (h\cdot\nu)\,dS
+\int_{\Omega}\big\{ \big[\frac{1}{2}\, \mathop{\rm div}(h) - a
\big]|\nabla u|^2
- \nabla u\cdot Dh \, \nabla u \big\}\, dx.
\end{aligned}
\end{equation}
For additional information on the development of variational
identities leading to Pucci-Serrin's result and some applications see
the references contained
in~\cite{mcgough:pon03,mcgough:avi00,schaaf:use92}.
\section{Beyond convexity}
To proceed with the analysis, we recall two geometric definitions. A
domain is said to be convex if for any two arbitrary points in the
domain, a line connecting the two points lies entirely in the domain.
A domain is said to be starlike if there exists some point $x_0$ in
the domain for which $(x-x_0)\cdot \nu >0$, for all $x\in \partial
\Omega$ and $\nu=\nu(x)$ is the boundary normal vector at the point
$x$. Polar domains are often viewed as spheres or ellipsoids, but can
be quite complicated and interesting in their own right (for example
consider the spherical harmonic solutions to the Laplacian). Figure
\ref{starlike} presents an example of a geometrically complex polar
domain.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.4\textwidth]{starlike}
\caption{Non-convex starlike domain.
\label{starlike}}
\end{center}
\end{figure}
We can restate the starlike definition in the language of convex
domains. A domain is said to be starlike if there exists a reference
point in the domain such that for any point in the domain the line
between the reference point and the arbitrary point lies in the
domain.
For those not familiar with Pohozaev's result~\cite{pohozaev:ee65},
it is presented here
for the case where $f(u) = u^{(N+2)/(N-2)}$.
\begin{theorem}[Pohozaev]\label{CriticalExpThm}
Assume that $\Omega$ is a smooth starlike domain, then there are no
positive solutions to
\begin{gather*}
\Delta u + u^{(N+2)/(N-2)} = 0, \quad x\in \Omega ,\\
u=0, \quad x \in \partial\Omega .
\end{gather*}
\end{theorem}
The proof uses a form of the Pucci-Serrin identity and follows a proof by
contradiction argument. We assume a positive solution $u$ exists and select $f(u) =
u^{(N+2)/(N-2)}$, $a=(N-2)/2$, $h(x) = x-x_0$ and plug into
(\ref{identity}). We obtain
$$
\frac{N}{\frac{N+2}{N-2} + 1} - \frac{N-2}{2} >0,
$$
which leads to $0 > 0$, a contradiction.
Many results have followed which generalize the nonlinearity used in
the Pohozaev results. An observation about the vector function $h$ points the
way to generalizing the domain. One notes that
the vector field need not be $x-x_0$ but just to exhibit the
essential properties of a starlike field in a polar domain. This leads
to the definition of $h$-starlike domains~\cite{mcgough:avi00}.
\begin{definition}\label{quasistarlike} \rm
The domain $\Omega$ is said to be {\it $h$-starlike} if there exists
a function $h: \Omega \to \mathbb{R}^N$, $h_i\in
C^1(\overline{\Omega})$, and a positive number $c$ with
\begin{equation}\label{divh}
\begin{gathered}
\mathop{\rm div}(h) |y|^2 - 2 y\cdot Dh y \geq c |y|^2,
\quad x \in \overline{\Omega} , \quad y\in \mathbb{R}^N ,\\[8pt]
h\cdot \nu \geq 0, \quad x\in\partial\Omega ,
\end{gathered}
\end{equation}
where $Dh$ is the derivative map of $h$ and $\nu$ is the outward
unit normal to $\partial\Omega$.
\end{definition}
For computational reasons, we will find it convenient to
reformulate condition (\ref{divh}).
\begin{definition}\label{PTraceDef} \rm
For a vector field $h$ the Pohozaev trace is
$$
P(h)=\mathop{\rm Trace}(Dh)-2|\lambda_1|
$$
where $\lambda_1$ is the largest eigenvalue (in magnitude)
of the symmetrized $Dh$, namely $(Dh+Dh^T)/2$.
\end{definition}
If $\inf_{x\in\Omega}P(h)\ge c>0$, then the condition on $h$ in the first
inequality in (\ref{divh}) is satisfied since
$$
\mathop{\rm div}(h) |y|^2 - 2 y\cdot Dh y \geq P(h) |y|^2,\quad
x \in \overline{\Omega}.
$$
The next theorem can be established by letting $a=c/2$ in (\ref{identity})
and deleting the boundary integral---details can be
found in~\cite{mcgough:pon03}.
\begin{theorem}[Pohozaev]
Assume that there exists an $h$-starlike function for the domain
$\Omega$ with $\inf_{x\in\Omega}P(h)\ge c>0$ and suppose that $f$ satisfies
\begin{equation}\label{fcondition}
\mathop{\rm div}h(x)F(t)-\frac{c}{2}tf(t)\le0
\end{equation}
for $t\ge0$ and $x\in\Omega$. Then there are no positive solutions to
\eqref{semilinear}.
\end{theorem}
If $f(t)=t^{p}$ then $F(t)=t^{p+1}/(p+1)$ and (\ref{fcondition})
is satisfied if $b/(p+1)-c/2\le0$ or $p\ge 2b/c-1$
where $b=\sup_{x\in\Omega}\mathop{\rm div}h(x)$.
In particular, we have that if $\Omega$ is $h$-starlike for a
vector field $h$, then (\ref{up}) below has no positive solutions
for $p\ge 2b/\inf_{x\in\Omega}P(h)-1$.
\begin{equation} \label{up}
\begin{gathered}
\Delta u + u^p = 0, \quad x\in \Omega , \\
u = 0, \quad x \in \partial \Omega .
\end{gathered}
\end{equation}
\begin{definition} \rm
For a given $h$-starlike domain $\Omega$, let the set $Q$ be the
collection of vector fields for which (\ref{divh}) holds.
We define the
Pohozaev critical exponent as
\begin{equation}
P_c = \min_{Q}\frac{2\sup_{x\in\Omega}\mathop{\rm div}h(x)}{\inf_{x\in\Omega}P(h)}-1
\end{equation}
\end{definition}
\begin{remark} \rm
In the subsequent sections we construct example domains
and show that there exists at least one vector field on the domain
for which $\inf_{x\in\Omega}P(h)>0$. In light of the preceding discussion,
it is automatic that
there are no positive solutions for (\ref{up}) on that domain when
$$
p\ge\frac{2\sup_{x\in\Omega}\mathop{\rm div}h(x)}{\inf_{x\in\Omega}P(h)}-1.
$$
We make no attempt to compute the smallest such $p$; i.e., $P_c$.
\end{remark}
\begin{remark} \rm
It is clear that starlike domains are $h$-starlike.
Consider a starlike domain in $\mathbb{R}^N$ with $h(x)=x$. Then
\begin{equation}\label{PCE}
P_c\le\frac{2N}{N-2}-1=\frac{N+2}{N-2}.
\end{equation}
A result reminiscent of Theorem~\ref{CriticalExpThm}.
Indeed, it is not difficult to show that
we actually have equality in (\ref{PCE}). One only
needs to consider the possible eigenvalues of the
symmetrized $Dh$.
\end{remark}
\section{Sectionally starlike domains}
Several $h$-starlike domains are given in \cite{mcgough:avi00}. One
example of $h$-starlike vectors fields which generate $h$-starlike
domains is provided by the re-scaled radial field
$$
h(x_1,x_2,\cdots ,x_N)=(\epsilon x_1 ,x_2,x_3,\cdots,x_N).
$$
Figure~\ref{three_wheel} gives two such sample domains for $N=3$.
For the remainder of the paper, we restrict ourselves to three
dimensional domains: $N=3$.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.4\textwidth]{three_wheel}
\includegraphics[width=0.4\textwidth]{starlike2}
\caption{Non-starlike $h$-starlike domains (three and five disks).
\label{three_wheel}}
\end{center}
\end{figure}
We will refer to a domain as sectionally starlike (convex) if there
exists a curve contained in the domain such that the domain
intersected with the hyperplane normal to the tangent to the curve is
starlike (convex). The previous domain is an example of a sectionally
convex domain. Two questions arise here. Does Pohozaev's
result extend directly to sectionally starlike domains? Is there some
relation between $h$-starlike and sectionally starlike? The answer to
the former question is no. A counter-example is a torus in 3D;
problem~(\ref{semilinear}) is solvable on a torus~\cite{mcgough:pon03}
for the critical exponent problem in $R^3$: $f(u) =
u^{(N+2)/(N-2)} = u^5$. The latter question is addressed below.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.4\textwidth]{torus}
\caption{The torus\label{torus}}
\end{center}
\end{figure}
The solid torus may be generated by moving the center of a disk along
a circle where the disk is orthogonal to the circular path. The curve
that generates the torus, a circle, is a closed curve. The torus has
nontrivial topology (there exists closed curves contained in the torus
which cannot be shrunk to point and remain in the torus). Ding, Bahri
and Coron, Lin and others have presented domains with nontrivial
topology which demonstrate solvability of the critical exponent problem.
The natural place to investigate is sectionally starlike domains with
trivial topology.
The example domains shown in Figure~\ref{three_wheel} are volumes of
revolution. Which means that there is a line segment which may act
as the central curve for the sectional starlike definition. It is not
surprising that a starlike region crossed with an interval produces an
$h$-starlike domain. The more interesting question is whether the
central axis can be curved. For this, we may give an affirmative
answer.
We begin with the vector field defined by $h(x) = [\epsilon x, y,z]$.
Note that this will shrink the $x$ component and allow us to ``bend''
the domain. For clarity in this example, we select $\epsilon = 1/5$,
but noting that this is only for illustration. Next, we may compute
the Jacobian of $h$,
$$
Dh = \begin{bmatrix}
1/5&0&0\\ 0&1&0 \\ 0&0&1 \end{bmatrix} \,.
$$
The eigenvalues are $\{1/5,1,1\}$ and thus by Definition~\ref{PTraceDef}
the Pohozaev trace of $h$ is
$$
P(h)=\mathop{\rm Trace}(Dh) - 2|\lambda_1 | = 1/5.
$$
The vector field has positive Pohozaev trace. To simplify the construction,
we look at the $x$-$y$ cross-section. By following the flow lines,
we may trace out curves
for which the vector field is tangential. This satisfies the boundary
condition for the $h$-starlike definition in the cross-section. We may
construct the flow lines by solving the differential equations
$$
\frac{dx}{dt}= \frac{1}{5}x, \quad \frac{dy}{dt}=y.
$$
The orbits $y = kx^5$ are given in Figure~\ref{field}.
Figure~\ref{boundary} shows the boundary in the $x$-$y$ plane and
a sample domain constructed from this boundary.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.8\textwidth]{field}
\caption{Vector field h with sample orbits. \label{field}}
\end{center}
\end{figure}
The $z$ aspect of the domain turns out not to be very restricted (it
is required to have a starlike cross-section). The essential
requirement is to maintain the non-negative dot product between the
boundary normal and the vector field. A surface with $z$ values
sufficiently large will produce the required dot product. Example
\ref{BasicConst} provides details on how to do this construction.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.4\textwidth]{boundary}
\includegraphics[width=0.4\textwidth]{boundary3d}
\caption{Boundary: (a) XY section, (b) 3D \label{boundary}}
\end{center}
\end{figure}
Following a similar approach we
may create other domains. Using $sin$ and $cos$ we can create vector
fields which do not generate constant eigenvalues (over the domain).
Starting with $h = [x,y ( y-f(x) ) ,z]$, $f(x)=\sin (
x )$, we have
$$
Dh = \begin{bmatrix}
1&0&0\\ -y\cos(x) & 2y-\sin (x) &0\\ 0&0&1
\end {bmatrix} ,
$$
which yields the eigenvalues $\{1,1,2\,y-\sin (x) \} $
and the Pohozaev trace
$$
P(h) = 2+2\,y-\sin ( x ) -2\,\max\{ 1,2\,y-\sin ( x ) \}.
$$
Following the vector
field, orbits can be traced out by integration. Flow lines (in yellow) and vector
fields (in red) are given in Figure~\ref{fields_2and3}. The bounding curve (in
blue) gives the region for which $p>.05$.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{field2}
\includegraphics[width=0.6\textwidth]{field3}
\caption{Flow lines for $h = [x,y ( y-f(x)) ,z]$ (a)
$f(x)=\sin(x)$, (b) $f(x) = \cos (x)$ \label{fields_2and3}}
\end{center}
\end{figure}
To construct sample domains for Figure~\ref{fields_2and3}, we generate
a tubular neighborhood of a curve bounded between the flow lines.
This construction is done for both domains and presented in
Figure~\ref{tubes}. Endcaps for the domain are not generated but
could be placed on in a smooth manner producing a $C^2$ boundary.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.4\textwidth]{sintube}
\includegraphics[width=0.4\textwidth]{costube}
\caption{3D domain for (a) $f(x)=\sin(x)$, (b) $f(x) = \cos (x)$ \label{tubes}}
\end{center}
\end{figure}
We have managed to construct three domains which are $h$-starlike but
not starlike. Is it possible to construct domains like building with
Lego's? In other words, can we start connecting these domains to
construct complicated twisted geometries? We address these questions in the next section.
\section{Tubular neighborhoods\label{tubular}}
The sample domains constructed previously were tubular neighborhoods
of given space curves. The radius of the neighborhood was selected to
ensure that the Pohozaev trace was strictly positive on the closure of
the domain. We now embark on connecting two tubular domains in $R^3$ to
construct domains which curve in the plane and rise out of the plane
of curvature (torsion).
We assume that the two domains can be connected in a smooth fashion.
This means we can cut the end of the domains to be connected so that
the end-section is a planar section, has the same cross-sectional
shape, and is orthogonal to the generating space curve. To join two
domains (see Figure~\ref{join}),
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.25\textwidth]{joina}
\includegraphics[width=0.25\textwidth]{joinb}
\raisebox{8ex}{\large $\Longrightarrow$}
\includegraphics[width=0.35\textwidth]{join}
\caption{Join two domains \label{join}}
\end{center}
\end{figure}
we need to ensure that the Pohozaev trace on the interface (or the
transition region) is defined and positive. First, we need a result
about the Pohozaev trace.
\begin{lemma}\label{superadd}
The Pohozaev trace is superadditive,
$$
P(h_1+h_2) \ge P(h_1)+P(h_2)
$$
\end{lemma}
\begin{proof} For real symmetric
(more generally normal) matrices $A$ and $B$, the spectral radius $r(A)$ is
the same as the
spectral norm. Consequently, we have subadditivity $r(A+B)\le r(A)+r(B)$
and we have that $r(A) = |\lambda_1|$, where $\lambda_1$ is the largest
eigenvalue (in magnitude).
\begin{align*}
&\mathop{\rm div}(h_1+h_2)-2r([Dh_1/2+Dh_1^t/2]+[Dh_2/2+Dh_2^t/2])\\
&\ge \mathop{\rm div}((h_1)+\mathop{\rm div}(h_2)-2r(Dh_1/2+Dh_1^t/2)
-2r(Dh_2/2+Dh_2^t/2) ,
\end{align*}
and thus
$P(h_1+h_2) \ge P(h_1)+P(h_2)$.
\end{proof}
\begin{lemma}\label{rotation}
The Pohozaev trace is invariant under rigid rotations of the vector field.
\end{lemma}
\begin{proof}
Let $h_r$ be a rotation of $h$, and set $h_r = M\circ h\circ M^{-1}$,
where $M$ is an orthogonal matrix. Next, compute
\[
P(h_r) = \mathop{\rm Tr}(M\circ Dh\circ M^{-1})-|\lambda_1 (M\circ Dh\circ M^{-1})|
=\mathop{\rm Tr}(Dh)-|\lambda_1 (Dh)|
= P(h) ,
\]
where as before $\lambda_1(Dh)$ is the largest eigenvalue of $Dh$ in
magnitude.
\end{proof}
The following example shows how one can use a transition function to move
from the zero field to $[x/5,y,z]$.
\begin{example}\label{BasicConst} \rm
Define
$$
p(x) = \begin{cases}
0 &x\le 0 \\
-2x^3+3x^2 &00$ a parameter.
For $s\ge1$, we use the normal vector field
$$n:=r_s\times r_t=[-5\sin(t)a(s-1)^4, \cos(t), \sin(t)] .$$
With this normal,
\begin{equation}\label{BentNormal}
h\cdot n=1+\sin(t)-a\sin(t)(s-1)^4 .
\end{equation}
Analyzing formula (\ref{BentNormal})
we discover that the additional restriction $s\le 1+(2/a)^{1/4}$
must be imposed in order to ensure that $h\cdot n\ge 0$.
At $s=1+(2/a)^{1/4}$ the tube attains
a height of $2+2^{5/4}/a^{1/4}$. Figure \ref{ParamTube}
illustrates the case where $a=2/3^4$. Note: for $0\le s\le1$,
the cylindrical portion of the tube, $h\cdot n=p(s)[1+\sin(t)]\ge0$.
\begin{figure}[ht]
\includegraphics[width=0.7\textwidth]{paramtubealt}
\caption{Parameterized Tube \label{ParamTube}}
\end{figure}
\end{example}
Using the method of example~\ref{BasicConst}, we can
construct two tubes as illustrated in figure~\ref{ParamTubeb}
and glue them together along their cylindrical portions ($0\le s\le1$ in example~\ref{BasicConst}).
It is of course permissible by Lemma~\ref{rotation} to first rotate either of the
tubes about the axis of the cylindrical portion prior to gluing.
In either case, the Pohozaev trace on the union of the tubes will be
positive and bounded away from zero by Lemma~\ref{superadd}.
\begin{figure}[ht]
\includegraphics[width=0.7\textwidth]{paramtubetwoalt}
\caption{Gluing tubes \label{ParamTubeb}}
\end{figure}
The parametrization given in example \ref{BasicConst}
has the advantage that the computations
for $h\cdot n$ are straight-forward. However, it has the disadvantage
that the cross-sections of the tube normal to the generating curve
are not circular on both ends. As a consequence, it is difficult to attach other
tubes to the far end ($s=1+(2/a)^{1/4}$).
\begin{example}\label{CollaredTube} \rm
For $0\le s\le 1$ define
$$ r(s,t):=[s,R\sin(t),R+R\cos(t)]
$$
and for $1\le s\le s_0$ define
$$r(s,t):=[s,0,R+a(s-1)^5]+
R\sin(t)[0,1,0]+
\frac{R\cos(t)}{\sqrt{1+25a^2(s-1)^8}}[5a(s-1)^4,0,-1].
$$
The largest $s$ such that $h\cdot n\ge0$ for all $t$ is
$$
s_{\rm max}:=1+\frac{1}{45}\sqrt [4]{91125\,{\frac {\sqrt [3]{5300+81\,
\sqrt {2451}}}{a}}+
{\frac {20867625}{a\sqrt [3]{5300+81\,\sqrt {2451}}}}+\frac{1822500}{a}}.
$$
Let $v(t):=r(s_0,t)$, we attach
a short right-cylindrical tube to the far end of the tube by the parametrization
\begin{equation}
r(s,t):=v(t)+[s,0,5a(s_0-1)^4s]\,,\quad s_0