\documentclass[reqno]{amsart} \usepackage{graphicx} \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