\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small {\em Electronic Journal of
Differential Equations}, Vol. 2003(2003), No. 37, pp. 1--13.
\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu
or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/37\hfil Radial solutions]
{Radial solutions of singular nonlinear biharmonic equations and
applications to conformal geometry}

\author[P. J. McKenna \& W. Reichel \hfil EJDE--2003/37\hfilneg]
{P. J. McKenna \& Wolfgang Reichel}

\address{P. J. McKenna\hfill\break
Department of Mathematics, University of Connecticut,
Storrs, CT 06269, USA}
\email{mckenna@math.uconn.edu}

\address{Wolfgang Reichel \hfill\break
Mathematisches Institut, Universit\"at Basel,
Rheinsprung 21, CH-4051 Basel, Switzerland}
\email{reichel@math.unibas.ch}


\date{}
\thanks{Submitted February 12, 2003. Published April 10, 2003.}
\subjclass[2000]{35J60}
\keywords{Singular biharmonic equation, conformal invariance}


\begin{abstract}
 Positive entire solutions of the singular biharmonic equation
 $\Delta^2 u + u^{-q}=0$ in $\mathbb{R}^n$ with $q>1$ and $n\geq 3$
 are considered. We prove that there are infinitely many radial
 entire solutions with different growth rates close to quadratic.
 If $u(0)$ is kept fixed we show that a unique minimal entire
 solution exists, which separates the entire solutions from those
 with compact support. For the special case $n=3$ and $q=7$
 the function $U(r) = \sqrt{1/\sqrt{15}+r^2}$ is the minimal entire
 solution if $u(0)=15^{-1/4}$ is kept fixed.
\end{abstract}

\maketitle

\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

We consider positive $C^4$-solutions of the equation
\begin{equation}
\Delta^2 u +u^{-q}=0 \quad\mbox{in } D\subset\mathbb{R}^n.
\label{basic}
\end{equation}
A solution is called \emph{entire} if it exists in all of $\mathbb{R}^n$.
In a recent paper \cite{choi} Choi and Xu studied (\ref{basic}) in 
$\mathbb{R}^3$.
They proved that under the restriction of exact linear growth at infinity,
i.e., $\lim_{x\to\infty} u(x)/|x|=\alpha>0$ and $q=7$, problem (\ref{basic})
admits (up to translation) only one kind of entire solution given by
\begin{equation}
U(x) = \alpha\sqrt{1/\sqrt{15\alpha^8}+|x|^2}.
\label{ast}
\end{equation}
Moreover Choi and Xu prove that (\ref{basic}) has no linear growth solution
if $4<q<7$. Thus, (\ref{basic}) in dimension $n=3$ and with $q=7$ is a very
distinguished case. Indeed, notice that for this choice of parameters
$(\ref{basic})$ amounts to
$$
\Delta^2 u = (n-4)u^\frac{n+4}{n-4}.
$$
This equation has explicit geometric relevance as explained in
Section \ref{geometry}.


The remarkable result of Choi and Xu led to our present investigation, where
we analyze the global structure of the set of all radial solutions of
(\ref{basic}). In our main result of Theorem \ref{separatrix}, Section
\ref{main_result}, we prove
that (\ref{basic}) possesses infinitely many entire solutions with almost
quadratic, superlinear
growth rates. Moreover, one entire solution is distinguished from the others:
the minimal entire solution, which coincides with the solution
$U(x) = \alpha\sqrt{1/\sqrt{15\alpha^8}+|x|^2}$ found by Choi and Xu.

The existence of infinitely many entire solutions with different growth rates
for the conformally invariant equation $\Delta^2 u + u^{-7}=0$ in 
$\mathbb{R}^3$ is
in striking contrast to the conformally invariant equation $\Delta^2 u =
u^\frac{n+4}{n-4}$ in $\mathbb{R}^n$ with $n\geq 5$ and the second order
equation $-\Delta u = u^\frac{n+2}{n-2}$ in $\mathbb{R}^n$ with $n\geq 3$. In 
both cases there exists a unique one-parameter family of positive entire 
solutions, cf. Juncheng Wei and Xingwang Xu \cite{wei_xu}, and Wenxiong Chen 
and Congming Li~\cite{chen_li}.

\section{Geometric relevance} \label{geometry}

\subsection{Yamabe's problem}\label{yamabe_section}

Let $g=(g_{ij})$ be the standard Euclidean metric on $\mathbb{R}^n$, $n\geq 3$ 
with $g_{ij}=\delta_{ij}$. Let $\bar g = u^{4/(n-2)}g$ be a second metric 
derived from $g$ by the positive conformal factor $u:\mathbb{R}^n\to 
\mathbb{R}$. Then $u$ satisfies Yamabe's equation
\begin{equation}
-\Delta u = \frac{n-2}{4(n-1)}R_{\bar g} u^{(n+2)/(n-2)},
\label{yamabe}
\end{equation}
where $R_{\bar g}$ is the scalar curvature of $\bar g$, cf. Aubin \cite{AU}.
If one looks for constant scalar curvature $R_{\bar g}=\pm n(n-1)$ then
(\ref{yamabe}) has the following explicit solutions
$$
U(r) = \big(\frac{2a}{a^2\pm r^2}\big)^{(n-2)/2} \quad\mbox{with }
r = |x-x_0|, \; a>0.
$$
The corresponding metric is
$$
\bar g_{ij} = \big(\frac{2a}{a^2\pm r^2}\big)^2\delta_{ij}.
$$
In case of ``$+$'' one finds that $(\mathbb{R}^n,\bar g)$ is isometrically 
isomorphic
to a sphere $\mathbb{S}^n_a$ of radius $a$ equipped with standard Euclidian 
metric scaled by $1/a^2$. Moreover, Wenxiong Chen and Congming Li showed in
\cite{chen_li} that (\ref{yamabe}) has no other positive solutions.
In case of ``$-$'', the solution $U$ blows up on
$\partial B_a(x_0)$ and one finds that $(B_a(x_0),\bar g)$ is
isometrically isomorphic to the hyperbolic space
$$
\mathbb{H}^n_a=\{(y_1,\ldots,y_{n+1})\in\mathbb{R}^{n+1}: y_1^2+\ldots+y_n^2-
y_{n+1}^2=-a^2\}
$$
with standard Lorentz-Minkowski metric
$g(v,w)=\frac{1}{a^2}(v_1w_1+\ldots v_nw_n-v_{n+1}w_{n+1})$. The explicit form
of these solutions and their uniqueness on balls $B_a(0)$ was proved by
Loewner, Nirenberg \cite{LoeNi}.

\subsection{A fourth order analog of Yamabe's equation}

For $n\neq 4$ let $\bar g$ be given as $\bar g_{ij} =
u^{4/(n-4)}\delta_{ij}$. Then the conformal factor $u:\mathbb{R}^n\to 
\mathbb{R}$ satisfies
\begin{equation}
\Delta^2 u = \frac{n-4}{2} Q_{\bar g} u^\frac{n+4}{n-4},
\label{higher}
\end{equation}
where
$$
Q_{\bar g} = \frac{-1}{2n-2}\Delta R_{\bar g} + \frac{n^3-4n^2+16n-16}
{8(n-1)^2(n-2)^2}R_{\bar g}^2 - \frac{2}{(n-2)^2}|Ric_{\bar g}|^2
$$
and $R_{\bar g}$, $Ric_{\bar g}$ are scalar curvature and  Ricci curvature of
$\bar g$, respectively. Generalizations of (\ref{higher}) to the case
where $g$, $\bar g$ are conformally related Riemannian metrics on a
Riemannian manifold lead to more complicated fourth order equations involving
the Paneitz operator instead of $\Delta^2$, cf. Chang \cite{CHA} and
Chang, Yang \cite{CHAYA}.

The quantity $Q_{\bar g}$ is a curvature term with $Q_{\bar g}\equiv 0$ in
dimension $n=2$. If we assume
$Q_{\bar g}\equiv\frac{1}{8}n(n^2-4)$ then via a scaling (\ref{basic}) and
(\ref{higher}) are equivalent. In this case (\ref{higher}) has the following
explicit solutions
$$
U(r) = \big(\frac{2a}{a^2\pm r^2}\big)^{(n-4)/2} \quad\mbox{ with }
r = |x-x_0|, \; a>0
$$
producing the same metrics as before, i.e, the metrics representing 
$\mathbb{S}^n$ and $\mathbb{H}^n$. For the case of ``$+$'' and $n\geq 5$, 
uniqueness of the above family
was shown by Juncheng Wei and Xingwang Xu \cite{wei_xu}. In the case $n=3$
uniqueness fails, as it follows from our main result Theorem \ref{separatrix}.
For the case of ``$-$'' uniqueness for (\ref{higher}) on a ball is open to
the best of our knowledge.

\subsection{The uniqueness result of Choi and Xu}

The uniqueness result of Choi and Xu in dimension $n=3$ has the following
geometric meaning:
\emph{if $u$ is asymptotically linear as $|x|\to \infty$ then 
$(\mathbb{R}^3,\bar g)$ is isometrically isomorphic to a standard sphere 
$\mathbb{S}^3$}. The requirement of $u$ being asymptotically linear means 
that the metric $\bar g = u^{-4}\delta_{ij}$ on $\mathbb{R}^3$ can be pulled 
back via inverse stereographic projection to a metric on $\mathbb{S}^3$. 
However our main result shows
that many other radial solutions $u$ of (\ref{higher}) exist -- in
striking contrast to (\ref{yamabe}) for $n\geq 3$ and (\ref{higher}) for
$n\geq 5$. These metrics cannot be realized as metrics on $\mathbb{S}^3$ but 
only on $\mathbb{S}^3\setminus\{P\}$, i.e., on the sphere with one point 
removed.

A second Theorem of Choi and Xu states the following: if $u$ is an
arbitrary entire solution of (\ref{basic}) such that the scalar curvature
$R_{\bar g}$ of the metric $\bar g_{ij} = u^{4/(n-4)}\delta_{ij}$ is
everywhere non-negative then $u$ must be of the form \eqref{ast}.
Geometrically this means: \emph{if $u$ induces a metric $\bar g$ with
everywhere non-negative scalar curvature then $(\mathbb{R}^3,\bar g)$ is
isometrically isomorphic to a standard sphere $\mathbb{S}^3$.}

This shows also, that the special solution $U(r)$ of type \eqref{ast} is
distinguished from the infinitely many other solutions $u(r)$ found in
Theorem \ref{separatrix}, since they induce metrics on $\mathbb{R}^3$ with
sign-changing scalar curvature.

\section{Radial solutions of $\Delta^2u = -u^{-q}$ for $n\geq 3$}
\label{main_result}

We restrict our analysis to radial solutions of (\ref{basic}). For this class
of solutions the biharmonic operator simplifies to
$(\frac{d^2}{dr^2}+\frac{n-1}{r}\frac{d}{dr})^2$. Therefore we investigate the
initial value problem
\begin{gather}
\Big(\frac{d^4}{dr^4}+\frac{2(n-1)}{r}\frac{d^3}{dr^3}
+\frac{(n-1)(n-3)}{r^2}(\frac{d^2}{dr^2}-\frac{1}{r}\frac{d}{dr})
\Big) u +u^{-q}=0,\label{rad1}\\
u(0)=1, u'(0)=0, u''(0)=\delta, u'''(0)=0\label{rad2}.
\end{gather}
In contrast to \emph{entire} solutions, which exist on $(0,\infty)$,
we say that a solution $u$ has \emph{compact support} if $u$ is positive on
some interval $(0,R)$ and $u(R)=0$, since then the solution stops to exist.
In the following we will say that a real-valued function 
$f(s), s\in\mathbb{R}$ is \emph{increasing} if $s_1<s_2$ implies 
$f(s_1)\leq f(s_2)$. It is \emph{strictly increasing} if $s_1<s_2$ implies 
$f(s_1)<f(s_2)$. Similarly we use the word \emph{decreasing} and 
\emph{strictly decreasing}. We have the following results:

\begin{theorem} \label{thm1}
All solutions of \eqref{rad1}-\eqref{rad2} are strictly ordered with
respect to $\delta$.
For $n\geq 3$ and $q>1$ the following types of solutions are known:
there exists a value $\delta_0>0$ such that
\begin{itemize}
\item[(a)] for $-\infty<\delta<\delta_0$ every solution has compact support,
\item[(b)] for $\delta\geq \delta_0$ every solution is entire,
\item[(c)] the entire solution $u_0$ with $u_0''(0)=\delta_0$ is a
separatrix, i.e.,
\begin{align*}
 u_0 & =  \sup\{u: \mbox{ $u$ is a compact support solution}\}\\
 & =  \inf\{u: \mbox{ $u$ is an entire solution}\},
\end{align*}
\item[(d)] if $R(\delta)$ is the first zero of the solution $u$ with
$u''(0)=\delta$, $-\infty<\delta<\delta_0$ then $R(\delta)$ is a
continuous, strictly monotone function with $R(\delta)\to\infty$ as
$\delta\to\delta_0$ and $R(\delta)\to 0$ as $\delta\to -\infty$,
\item[(e)] for $\epsilon>0$ sufficiently small there exist
solutions which grow faster than $r^{2-\epsilon}$,
\item[(f)] no solution grows faster than $r^2$.
\end{itemize}
\label{separatrix}
\end{theorem}

Our proof depends on the construction of suitable sub-, supersolutions
and the use of the comparison principle. This technique depends on the fact
that (\ref{rad1}) can be rewritten as a second-order system
\begin{equation}
(r^{n-1}u')'= r^{n-1}U,\quad (r^{n-1}U')'+r^{n-1}u^{-q}=0.
\label{system}
\end{equation}
Notice that $U(0)=nu''(0)$, $U'(0)=\frac{n+1}{2}u'''(0)$. The next lemma
is a comparison result between upper and lower solutions of (\ref{system}).
In spirit it follows from corresponding comparison results of Walter
\cite{walter} for quasimonotone systems.

\begin{lemma}[Comparison Principle] \label{lm2}
Let $(v,V)$ and $(w,W)$ be two pairs of $C^2$-functions on the interval
$[0,R)$ with $v,w>0$ on $[0,R)$ and with
\begin{gather*}
(r^{n-1}v')'=r^{n-1}V,\quad (r^{n-1}w')'= r^{n-1}W,\vspace{\jot}\\
(r^{n-1}V')'+r^{n-1}v^{-q}\leq 0,\quad
(r^{n-1}W')'+r^{n-1}w^{-q}\geq 0
\end{gather*}
on $(0,R)$. Then the following holds:
\begin{itemize}
\item[(a)] (Weak comparison)
If $v(0)\leq w(0)$, $v'(0)=w'(0)=0$ and $V(0)\leq W(0)$,
$V'(0)=W'(0)=0$ then $v\leq w$, $v'\leq w'$, $V\leq W$ and $V'\leq W'$ on
$[0,R)$.
\item[(b)] (Strong comparison)
If for some $\rho>0$ we have $v<w$ on the interval $(0,\rho)$
then $v<w$, $v'<w'$, $V<W$, $V'<W'$ on $(0,R)$. A simple way to achieve
$v<w$ initially is to have {\bf one} strict inequality in the initial
conditions.
\end{itemize}
\label{comparison}
\end{lemma}

\begin{proof} We begin with proving part (b). Suppose $v<w$ initially on a
small interval $(0,\rho)$. By the second differential inequality we find
$(r^{n-1}V')'<(r^{n-1}W')'$ on $(0,\rho)$. By a double integration we get
$V<W$ on $(0,\rho)$. Now let $(0,c)$ be the largest interval on which
$V<W$, and suppose for contradiction that $c<R$. Using $V<W$ and the first
differential inequality we get $(r^{n-1}v')'< (r^{n-1}w')'$ on $(0,c)$ and
by a double integration $v'<w'$ and $v<w$ on $(0,c)$. Inserting this into
the second differential inequality and integrating twice we get
$V'<W'$ on $(0,c)$ and hence $V<W$ on the semi-closed interval $(0,c]$.
This contradicts the assumption that $(0,c)$ is the largest interval for which
$V<W$. Therefore we have strict inequalities
between $v,v',V,V'$ and $w,w',W,W'$ on the entire interval $(0,R)$.


Part (a) follows from (b) in the following way. Let $f:= (r^{n-1}V')'+v^{-q}$
be the defect in the differential inequality for $V$, and let
$(v_\epsilon,V_\epsilon)$ be the solution of the initial value problem
\begin{gather*}
(r^{n-1}v_\epsilon')' = r^{n-1}V_\epsilon,
\quad v_\epsilon(0)=v(0)-\epsilon, v_\epsilon'(0)=0,\\
(r^{n-1}V_\epsilon')' + r^{n-1}v_\epsilon^{-q}=f\leq 0,
\quad V_\epsilon(0)=V(0), V_\epsilon'(0)=0.
\end{gather*}
Note that for $\epsilon=0$ the uniqueness of the initial value problem
gives $(v_0,V_0)=(v,V)$. Moreover, for $\epsilon>0$ the pairs
$(v_\epsilon,V_\epsilon)$ and $(w,W)$ satisfy the hypotheses of part (b),
and we can deduce that $(v_\epsilon,v_\epsilon',V_\epsilon,V_\epsilon')<
(w,w',W,W')$ on $(0,R)$. Letting $\epsilon$ tend to $0$, the strict inequality
becomes a weak inequality. This proves the lemma.\end{proof}

Lemma \ref{comparison} will be applied to problem (\ref{rad1})-(\ref{rad2}) in
the following way. Suppose two positive $C^4$-functions $v(r),w(r)$ are given
with
\begin{gather*}
\Delta^2 v+v^{-q}\leq 0,  \quad v(0)\leq 1, \quad v'(0)=0, \quad
v''(0)\leq\delta, \quad  v'''(0)=0\\
\Delta^2 w+w^{-q}\geq 0,  \quad w(0)\geq 1, \quad w'(0)=0, \quad
w''(0)\geq\delta, \quad w'''(0)=0
\end{gather*}
then $v,w$ are called a sub,- supersolutions relative to the
initial value problem
$$
\Delta^2 u+u^{-q}= 0, \quad u(0)= 1, u'(0)=0, \quad u''(0)=\delta,\quad
u'''(0)=0.
$$
Lemma \ref{comparison} applied to $(u,u''+\frac{n-1}{r}u')$ with either
$(v,v''+\frac{n-1}{r}v')$ or $(w,w''+\frac{n-1}{r}w')$ yields the
conclusion that $v\leq u\leq w$, $v'\leq u'\leq w'$ on their common interval
of existence. Moreover, strict inequality holds as soon as one strict 
inequality holds in the initial conditions for the function or its second 
derivative.

\begin{lemma} \label{lm3}
There exists a value $\tilde\delta>0$ such that for all $\delta\leq
\tilde\delta$ the solution of {\rm (\ref{rad1})-(\ref{rad2})} has compact
support.
\label{compact}
\end{lemma}

\begin{proof} Consider the function $w(r)=\epsilon r^2(A-r^2)+1$ for 
$\epsilon, A>0$, which is positive on $(0,\sqrt{A})$.
Then $\Delta^2 w = -8\epsilon n(n+2)$. In order to have $w$ as a
compact-support supersolution we need $\Delta^2 w+w^{-q}\geq 0$, i.e.
$$
-8\epsilon n(n+2)+\big(\epsilon r^2(A-r^2)+1\big)^{-q}\geq 0 \quad
\mbox{ on } (0,\sqrt{A}).
$$
The maximum of $w$ over $(0,\sqrt{A})$ is obtained at $\sqrt{A/2}$. Therefore
the above equation is satisfied provided
$$
\epsilon \frac{A^2}{4}+1\leq \big(8\epsilon n(n+2)\big)^{-1/q}.
$$
For a given value of $\epsilon$ such that  $0<\epsilon < 1/(8n(n-2))$ the
largest possible value of of $A$ is given by
$$
A(\epsilon) := \frac{2}{\sqrt{\epsilon}}\sqrt{(8\epsilon n(n+2))^{-1/q}-1}
$$
and clearly $A(\epsilon)\to \infty$ for $\epsilon\to 0$ and
$A(\epsilon)\to 0$ as $\epsilon\to 1/(8n(n+2))$. Furthermore
$w(0)=1, w'(0)=w'''(0)=0$, and  $w''(0)=2\epsilon A(\epsilon)$ has the
properties that $w''(0)\to 0$ as $\epsilon\to 0$ and as
$\epsilon\to 1/(8n(n+2))$. The second derivative of $w$ at $0$ is maximal
for $\epsilon=(1-1/q)^q\,1/(8n(n+2))$ and the value of $w''(0)$ is
$\tilde\delta := 4\sqrt{(1-1/q)^q/(8n(n+2)(q-1))}$. As a result we have
that every solution of (\ref{rad1})-(\ref{rad2}) with
$\delta\leq \tilde\delta$ stays below $w(r)$, and hence it has compact
support. \end{proof}


\begin{lemma} \label{lm4}
No entire solution of {\rm (\ref{rad1})-(\ref{rad2})} grows
faster than $r^2$.
\label{fastest}
\end{lemma}

\begin{proof} Let $w(r)=Ar^2+1$. Then $\Delta^2 w=0$ and hence $w$ is a
supersolution. Given an arbitrary entire solution $u$ of
(\ref{rad1})-(\ref{rad2}) we can choose $2A>u''(0)$. Hence $u<w$ by the
comparison principle of Lemma \ref{comparison}. \end{proof}

\begin{lemma}
There exists $\epsilon_0=\epsilon_0(n)>0$ such that for every $\epsilon\in
(0,\epsilon_0)$ there exists a value $\bar\delta=\bar\delta(\epsilon)$ with
the property that every solution of
{\rm (\ref{rad1})-(\ref{rad2})} with $\delta\geq \bar\delta$ is entire and
grows faster than $r^{2-\epsilon}$.
\label{entire}
\end{lemma}

\begin{proof} For the
construction of an entire subsolution we consider
$v(r) = (1+b^2 r^2)^{1-\frac{\epsilon}{2}}$. Let $p(r)$ be the function such
that
$$
\Delta^2 v + p(r) v^{-q}=0 \quad \mbox {on } (0,\infty).
$$
With the help of MAPLE we compute
$$
p(r) = 2 b^4\epsilon(1-\epsilon/2)(1+b^2r^2)^{-3-\frac{\epsilon}{2}(1+q)+q}
\rho(r)
$$
where
$$
\rho(r) = \big( b^4r^4(\epsilon^2-2n\epsilon+2\epsilon-2n+n^2)
+b^2r^2(-4\epsilon-2n\epsilon+2n^2-8)+n^2+2n\big)
$$
For small $\epsilon<\epsilon_0(n)$ and for $n\geq 3$ we see that
$$
c(1+b^2r^2)^2 \leq \rho(r) \leq C(1+b^2r^2)^2
\quad \mbox{on } (0,\infty)
$$
for two constants $c,C$ independent of $b$ and $\epsilon$. Hence
$$
p(r)\geq 2c b^4\epsilon(1-\epsilon/2)(1+b^2r^2)^{-1-\frac{\epsilon}{2}(1+q)+q}
\geq 1
$$
if we ensure that $\epsilon<2(q-1)/(1+q)$ and if we choose $b$ sufficiently 
large. For this choice of the parameters $\epsilon$ and $b$ we find
$$
\Delta^2 v + v^{-q}\leq 0 \quad \mbox{on } (0,\infty),
$$
i.e., $v$ is indeed a subsolution with $v(0)=1$, $v'(0)=v'''(0)=0$ and
$v''(0)=b^2(2-\epsilon)$. Together
with the supersolution $w=Ar^2+1$ for large $A$ from Lemma \ref{fastest},
they give rise to an entire solution with growth rate larger than
$r^{2-\epsilon}$.\end{proof}

\begin{proof}[Proof of Theorem \ref{separatrix}]
Part (a): By Lemma \ref{compact}
there is a $\tilde\delta>0$ such that the solution of
(\ref{rad1})-(\ref{rad2}) with $u''(0)=\tilde\delta$ has compact support.
Via the comparison principle we see that for $-\infty<\delta<\tilde\delta$ the
solutions also have compact support. Therfore, we may define
$$
\delta_0 := \sup\big\{\delta: \mbox{ the solution with $u''(0)=\delta$ has
compact support.}\big\}
$$
By the entire solution found in Lemma \ref{entire} the value $\delta_0$ is
finite and positive, and any solution with $u''(0)>\delta$ must be entire.
As we will see in the proof of Part (d), the first zero $R(\delta)$
tends to $\infty$ as $\delta\to\delta_0$. Therefore the separatrix-solution
$u$ with $u''(\delta_0)$ must be entire, too. Hence Part (b) and (c) are
established. Part (e) follows from Lemma \ref{entire}, Part (f) from Lemma
\ref{fastest}.

Part (d): Let $R(\delta)$ be the first zero of the solution $u$
with $u''(0)=\delta$. Notice that $u(r) = -\int_r^{R(\delta)} u'(t)\,dt$ and
that $u$ is absolutely continuous on $[0,R(\delta)]$. We want to show that
$R$ is a continuous function of
$\delta$ with $R\to 0$ as $\delta\to -\infty$ and $R\to\infty$ as
$\delta\to\delta_0$. By the comparison principle the function $R(\delta)$
is monotone in $\delta$. Moreover, for two solution $u_1$, $u_2$ with
$u_1''(0)=\delta_1<\delta_2=u_2''(0)$ we find by the comparison principle
that $(u_2-u_1)'>0$, i.e. the gap between the solutions is increasing.
Therefore $R(\delta)$ is a strictly monotone function of $\delta$, and hence
continuity can only fail if $R(\delta)$ has jump-discontinuities, which
are excluded by the continuous dependence of the solution on initial values.
Next we assume for contradiction that $R(\delta)$ tends to a finite
limit as $\delta\to\delta_0$. Since the solutions $u$ with
$u''(0)>\delta_0$ must be entire by the definition of $\delta_0$
we get again a contradiction to the continuous dependence principle.
Similarly, $R(\delta)$ cannot approach a positive limit as
$\delta\to -\infty$. \end{proof}

\section{Entire solutions of $\Delta^2u = -u^{-q}$ for $n=3$}

Since radially symmetric functions in $\mathbb{R}^3$ satisfy 
$\Delta^2 u(r) = u^{(iv)}+4u'''/r$ we can prove the following variant of the 
comparison principle of Lemma \ref{comparison}:

\begin{lemma}[Comparison principle for $n=3$] \label{lm6}
Let $(v,w)$ be a pair of $C^4$-functions on the interval
$[0,R)$ with $v,w>0$ on $[0,R)$ and with
$$
v^{(iv)}+\frac{4}{r}v''' +v^{-q}\leq 0,\quad w^{(iv)}+\frac{4}{r}w'''\geq 0
\quad \mbox{on } (0,R).
$$
Then the following holds:
\begin{itemize}
\item[(a)] (Weak comparison)
If $v(0)\leq w(0)$, $v'(0)=w'(0)=0$, $v''(0)\leq w''(0)$, $v'''(0)=w'''(0)=0$
then $v\leq w$, $v'\leq w'$, $v''\leq w''$ and $v'''\leq w'''$ on
$[0,R)$.
\item[(b)] (Strong comparison)
If for some $\rho>0$ we have $v<w$ on the interval $(0,\rho)$
then $v<w$, $v'<w'$, $v''<w''$, $v'''<w'''$ on $(0,R)$. A simple way to
achieve $v<w$ initially is to have {\bf one} strict inequality in the initial
conditions.
\end{itemize}
\label{comparison_3}
\end{lemma}

\begin{proof} First we notice that the differential inequality for $v$ is
equivalent to
\begin{equation}
\begin{gathered}
v_1'  =  v_2, \quad v_1(0)=1,\\
v_2'  =  v_3, \quad v_2(0)=0, \\
v_3'  =  v_4, \quad v_3(0)=\delta, \\
v_4' \leq  -v_4/r - v_1^{-q}, \quad v_4(0)=0,
\end{gathered}
\label{true}
\end{equation}
with a similar system for $w$. System (\ref{true}) is of quasimonotone type,
i.e., the $j$-th right-hand side, $j=1,\ldots,4$ is increasing in the
variables $v_k$ for all $k\in \{1,\ldots,4\}\setminus\{j\}$. Thus the
comparison theorem from Walter \cite{walter} for quasimonotone systems of
ordinary differential equations applies.
\end{proof}

Let $u: [0,\infty)\to \mathbb{R}$ and let $a,b,A,B$ be positive constants
(depending on $u$). We say that $u$ has

\begin{tabular}{ll}
\emph{linear growth} & if $ar+b\leq u(r)\leq Ar+B$,\\
\emph{exactly linear growth} & if $\lim_{r\to\infty} u(r)/r>0$,\\
\emph{at least linear growth} & if $u(r)\geq ar+b$,\\
\emph{superlinear growth} & if $u(r)\geq ar+b$, but $u$ does
 not have linear growth.
\end{tabular} \smallskip

The following theorem gives a more detailed
picture of the entire radial solutions in dimension $n=3$. A stronger version
of part (b) including non-radial solutions has been obtained by Choi and
Xu \cite{choi} as part of their main result. Here we give a different proof.
To the best of our knowledge, part (a) is new for $q>7$.

\begin{theorem}
For $n=3$ entire solutions of {\rm (\ref{rad1})-(\ref{rad2})} have the
following properties:
\begin{itemize}
\item[(a)] For $q\geq 7$ a unique solution with linear growth exist. It
coincides with the sepa\-ratrix and has exactly linear growth. For $q=7$ it
is given by $\sqrt{1+r^2/\sqrt{15}}$.
\item[(b)] For $4<q<7$ there is no solution with linear growth.
\item[(c)] For $4<q<7$ the separatrix has superlinear growth.
\item[(d)] For $1<q\leq 4$ the separatrix has at least linear growth.
\end{itemize}
\label{special_3}
\end{theorem}

It is unknown whether in (b) and (c) the condition $4<q<7$ can be
weakened to $0<q<7$.

\noindent
{\bf Interpretation.} To understand Part (b) of Theorem \ref{special_3}
let us say that $u^\frac{2n}{n-4}=u^{-6}$ has ``critical growth''
and set $F(u)=u^{1-q}/(1-q)$. Then $F(u)/u^{-6}$ is decreasing
for $1<q<7$. In other words $u^{-q}$ for $1<q<7$ has ``subcritical growth''.
For ``subcritical growth'' it is well known that (\ref{yamabe}) and
(\ref{higher}) with positive exponents have no entire positive solution.
Part (b) of Theorem \ref{special_3} is such a subcritical non-existence
result for certain solution classes.

It was observed by Choi and Xu that the uniqueness of the explicit solution
$\sqrt{1+r^2/\sqrt{15}}$ can also be obtained by a geometric condition 
different to the linear growth condition:

\begin{corollary}[Choi and Xu] \label{coro8}
 For $n=3$ and $q=7$ the uniqueness result of
Part (a) of Theorem \ref{special_3} holds in the class of functions $u$ such
that the scalar curvature of $\bar g = u^\frac{4}{n-4}\delta_{ij}$ is
everywhere positive.
\end{corollary}

\begin{proof} The scalar curvature of $\bar g$ is given as
(see Section \ref{yamabe_section})
\begin{equation}
R_{\bar g} = -4\frac{n-1}{n-2}u^\frac{n+2}{4-n}\Delta u^\frac{n-2}{n-4}
\label{scalar_curvature}
\end{equation}
which reduces in dimension $n=3$ and for a radial function $u$ to
\begin{equation}
R_{\bar g}= \frac{8}{r^2}u^5(r^2 u'u^{-2})'.
\label{scalar_curvature_3}
\end{equation}
Hence, positivity of $R_{\bar g}$ implies that $r^2u'u^{-2}$ is increasing.
Integration of the inequality
$
r^2u'u^{-2} \geq c_0 \mbox{ for } r \geq r_0
$
implies $1/u(r)-1/u(s)\geq c_0/r -c_0/s$ for $r_0<r<s$. Letting $s\to \infty$ 
we obtain $u(r)\leq r/c_0$ for $r\geq r_0$. Since $u$ is also convex, 
cf. Lemma \ref{convex}, this
implies that $u$ has linear growth, and Theorem \ref{special_3} applies.
\end{proof}

For radially symmetric functions in $\mathbb{R}^3$ we find
$$\Delta^2 u(r) = u^{(iv)}+
4u'''/r = r^{-4}(r^4 u''')'.$$

\begin{lemma}\label{lm9}
For $n=3$ a solution $u$ of {\rm (\ref{rad1})-(\ref{rad2})} is
entire if and only if it is convex.
\label{convex}
\end{lemma}

\noindent
{\bf Remark.}\; It is unknown whether this classification of entire solutions
also holds for $n\geq 4$.


\begin{proof}[Proof of Lemma \ref{lm9}]
If $u$ is convex then $u$ is entire. Now suppose there exists
$r_0$ such that $u''(r_0)<0$. Since $(r^4 u''')'\leq 0$ and $u'''(0)=0$ we
find that $u'''\leq 0$, i.e. $u''$ is decreasing. Therefore $u''$ stays
negative once it is negative at $r_0$. Thus $u$ cannot be entire.
\end{proof}

\begin{lemma} \label{estimates_3}
Suppose $n=3$. Let $u(r)$ be an entire solution of
{\rm (\ref{rad1})-(\ref{rad2})} with linear growth. Then there exists a
constant $C>0$ such that $0\leq u'(r)\leq C$ if $q>3$ and moreover:
\begin{gather*}
0  \leq  u''  \leq  \begin{cases}
C(1+r)^{-3} & \mbox{if } q>5,\\
C(1+r)^{-3}\log(2+r) & \mbox{if } q=5,\\
C(1+r)^{-q+2} & \mbox{if } 2<q<5,
\end{cases}\\
0 \geq   u''' \geq \begin{cases}
-C(1+r)^{-4} & \mbox{if } q>5,\\
-C(1+r)^{-4}\log(2+r) & \mbox{if } q=5,\\
-C(1+r)^{-q+1} & \mbox{if } 1<q<5.
\end{cases}
\end{gather*}
\end{lemma}

\begin{proof} We begin with the estimate for $u'''$. Since
$(r^4 u ''')'\leq 0$ and $u'''(0)=0$ we see that $u'''\leq 0$. Hence
$$
0\geq u'''(r) = \frac{-1}{r^4}\int_0^r s^4 u(s)^{-q}\,ds \geq
\frac{-C}{r^4}\int_0^r (as+b)^{4-q}\,ds.
$$
The estimate follows by performing the integration. Likewise, since
$$
u''(r)-u''(s)=\int_s^r u'''(t)\,dt
$$
and since $\int^\infty u'''(t)\,dt$ converges for $q>2$ we have that
$\lim_{r\to\infty} u''(r)$ exists and vanishes due the assumption of linear
growth. Moreover $u''>0$ by Lemma \ref{convex}. Thus
$$
0 \leq u''(s) = \int_s^\infty -u'''(t)\,dt,
$$
and the estimate for $u''$ follows by integrating the one for $u'''$. A final
integration leads to $0\leq u'(r)=\int_0^r u''(t)\,dt<\infty$ provided $q>3$.
\end{proof}

\begin{lemma}
Suppose $n=3$. Let $q>4$ and suppose $u$ is a solution of
{\rm (\ref{rad1})-(\ref{rad2})} with linear growth. Then
$\lim_{r\to \infty}u-ru'$ exists.
\label{bounded}
\end{lemma}

\begin{proof} Integration by parts yields
\begin{align*}
\frac{1}{2}\int_0^r u^{-q}s^3\,ds & =  \frac{-1}{2}\int_0^r (s^4 u''')'
\frac{1}{s}\,ds\\
& =  -\frac{r^3}{2}u'''(r)-\frac{r^2}{2}u''(r)+ru'(r)-u(r) + u(0).
\end{align*}
By the assumption $q>4$ we find that the left-hand side converges as
$r\to \infty$. Moreover, by Lemma \ref{estimates_3}, $r^3 u''', r^2 u''\to 0$
as $r\to\infty$. Hence
$$
\lim_{r\to \infty} u-ru' = u(0)-\frac{1}{2}\int_0^\infty u^{-q}r^3\,dr,
$$
as claimed. \end{proof}

\begin{lemma}
Suppose $n=3$. Let $q>4$ and suppose $u$ is a solution of 
{\rm (\ref{rad1})-(\ref{rad2})} with linear growth. Then 
$u-ru'\geq 0$ on $[0,\infty)$.
\label{positive}
\end{lemma}

\begin{proof} 
Let $h=u-ru'$. We derive a differential inequality for $h$. By direct 
computation 
$$\Delta^2 h = -7u^{(iv)}-ru^{(v)}-\frac{8}{r}u'''.
$$ 
Differentiation of (\ref{rad1}) and multiplication by $r$ yields
$$
ru^{(v)} +4u^{(iv)}-\frac{4}{r}u''' = rqu^{-1-q}u'.
$$
Substituting this into the expression for $\Delta^2h$ and using 
(\ref{rad1}) again we get
\begin{eqnarray}
\Delta^2 h & = & -rqu^{-1-q}u'+ 3u^{-q} \nonumber\\
 & = & 3u^{-1-q} h + (3-q)ru^{-1-q} u' \label{new_diq} \\
& \leq & 3u^{-1-q}h \nonumber 
\end{eqnarray}
since $q>4$ by assumption. Notice that $h$ is decreasing since 
$h'=-ru''<0$. Suppose for contradiction that $h(r)<0$ for $r\geq r_0$. 
Then (\ref{new_diq}) implies $\Delta^2 h<0 $ on $(r_0,\infty)$, i.e., 
$h'''r^4$ is decreasing on $(r_0,\infty)$. Now we distinguish two cases:

\medskip

Case (a): $h'''$ is negative somewhere in $(r_0,\infty)$. Then $h'''$ stays 
negative, say, on $(r_1,\infty)$, i.e. $h'$ is a concave, negative function on 
$(r_1,\infty)$. This implies that $h$ is unbounded below, which is impossible 
by Lemma \ref{bounded}. 

\medskip

Case (b): $h'''\geq 0$ in $(r_0,\infty)$. Then $h'''$ is decreasing on 
$(r_0,\infty)$, i.e., $h''$ is concave on $(r_0,\infty)$. For large enough 
$r_1$ we have either case (b1) $h''<0$ on $(r_1,\infty)$ or case (b2) $h''>0$ 
on $(r_1,\infty)$. In case (b1) $h$ is a concave decreasing 
function contradicting Lemma \ref{bounded}. Hence we can assume case (b2), 
i.e., $h''>0$ on $(r_1,\infty)$. Thus $h''$ is a positive concave function on 
$(r_1,\infty)$ and hence increasing at $\infty$. However, by 
Lemma \ref{estimates_3}, we have $r^2 h''(r)\to 0 $ as $r\to\infty$. This is 
incompatible with the fact that $h''$ is positive increasing at $\infty$, and 
finishes the discussion of case (b). 
\end{proof}

\begin{lemma}[Poho\v{z}aev's identity] \label{lm13}
 Suppose $n\geq 3$. Let $u$ be an entire
solution of {\rm (\ref{rad1})-(\ref{rad2})}. Then the following identity holds
\begin{align*}
&\int_0^\rho u^{1-q}\Big(\frac{n}{1-q}-
\frac{n-4}{2}\Big)r^{n-1}\,dr \\
 &=  -\frac{\rho^n}{2} (u'')^2+\frac{\rho^n}{1-q}u^{1-q}
+\frac{n}{2}\rho^{n-1}u'u''+\frac{(n-1)(n-4)}{2}\rho^{n-2}uu''\\
 &\quad +\frac{n-4}{2}\rho^{n-1}uu'''
+\rho^nu'u'''
-\frac{n-1}{2}\rho^{n-2}(u')^2
-\frac{(n-1)(n-4)}{2}\rho^{n-3}uu'
\end{align*}
for every $\rho>0$.
\end{lemma}

\begin{proof} The result follows from a general identity of Pucci, Serrin
\cite{PS}, Proposition~4, for the one-dimensional Lagrangian
$\mathcal{F}=(\frac{1}{2}(u''+\frac{n-1}{r}u')^2+\frac{1}{1-q}u^{1-q})r^{n-1}$.
\end{proof}


\begin{proof}[Proof of Theorem \ref{special_3}]
 Part (a): For $q\geq 7$ the function $U(r)=\sqrt{1+r^2/\sqrt{15}}$ is a 
subsolution. Therefore every compact support
solution stays below $U(r)$, and thus the separatrix must stay below $U(r)$.
Hence, the separatrix $S(r)$ has linear growth. Let $t(r)$ be the slope of the
tangent of $S(r)$. By convexity, $t(r)$ is increasing, and by the
upper bound $U(r)$ we see that $t(r)$ is bounded, i.e. convergent
with $t=\lim_{r\to\infty} t(r)$. Hence the separatrix $S(r)$ has exactly
linear growth with slope $t$.

Let $u$ be an entire, linear growth solution with $u(0)=1$. If
$u''(0)<S''(0)$ then by Lemma~\ref{estimates_3} we have $u'''(r)< S'''(r)$,
and hence $c := (S-u)''(0)< (S-v)''(r)$, i.e. $u''(r) \leq S''(r)-c$. However,
since $S''(r)\to 0$ as $r\to \infty$ by Lemma~\ref{estimates_3}, we get that
$u''(r)<0$ for large $r$. Hence $u$ becomes concave eventually, stays concave,
and hence cannot be entire. Now suppose $u''(0)>S''(0)$. Then $u''(r)\geq
S''(r)+u''(0)-S''(0)>0$, i.e. $\liminf_{r\to\infty} u''(r)>0$. This
contradicts Lemma \ref{estimates_3}. Hence, among all solutions with linear
growth, $S(r)$ is unique. This uniqueness shows that in case $q=7$ the 
separatrix $S(r)$ must have the explicit form $\sqrt{1+r^2/\sqrt{15}}$.

Part (b): Suppose $4<q<7$. Then
in the right-hand side of Poho\v{z}aev's identity all
terms except the last two converge individually to $0$ as $\rho\to\infty$. The
last two terms are
$$
uu'-\rho (u')^2=u'(u-\rho u')\geq 0
$$
by Lemma \ref{positive}. Hence the $\liminf$ of the entire right
hand side is $\geq 0$. In contrast, the left-hand side is negative. Hence no
linear growth solution exists.

Part (c): For $4<q<7$ the separatrix cannot have linear growth. By
convexity it has a least linear growth, i.e, it has superlinear growth.

Part (d): The reasoning of Part (c) shows that the separatrix
has at least linear growth. However, linear growth itself can no longer be
excluded.
\end{proof}


\section{Compact support solutions of $\Delta^2u = -u^{-q}$ for $n=3$}

Let $u$ be a function with support $[0,R]$ and let $a,A$ be a positive
constants (depending on $u$). We say that $u$ has




\begin{tabular}{ll}
\emph{square root growth} & if $a\sqrt{R-r}\leq u(r)\leq A\sqrt{R-r}$\\
\emph{exactly square root growth} & if $\lim_{r\to R} u(r)/\sqrt{R-r}>0$,\\
\emph{at least square root growth} & if $a\sqrt{R-r}\leq u(r)$.
\end{tabular}

\begin{theorem} \label{thm14}
For $n=3$ compact support solutions of {\rm (\ref{rad1})-(\ref{rad2})} have
the following properties:
\begin{itemize}
\item[(a)] for $6<q<7$ there is no compact support solution with square-root
growth at its zero,
\item[(b)] for $1<q\leq 6$ there is no compact support solution with a least
square-root growth at its zero.
\item[(c)] For $q=7$ there are compact support solutions with exactly square
root growth given by $\alpha\sqrt{1/\sqrt{15\alpha^8}-|x|^2}$, $\alpha>0$ 
(here we have dropped the requirement $u(0)=1$).
\end{itemize}
\label{special_compact}
\end{theorem}

Unlike  the entire solution situation, we do not know how to uniquely select the
special solutions of Theorem \ref{special_compact}(c). In fact,
the requirement of negative scalar curvature is not enough:

\begin{corollary} \label{cor_compact}
For $q=7$ there are infinitely many solutions
with support $[0,R]$ which generate a metric with negative scalar curvature.
Each of them can be pulled back by stereographic projection to a metric
on hyperbolic space $\mathbb{H}^3_R$.
\end{corollary}

One might conjecture that exact square root growth near its zero uniquely
selects the explicit solution of Theorem \ref{special_compact}(c).
So far, we do not know whether this holds true. Geometrically the square
root condition means that $\bar g = u^{-4}\delta_{ij}$ can be pulled back
to $\mathbb{H}^3$ via inverse stereographic projection and the resulting metric
$\hat g$ on $\mathbb{H}^3$ has the property that 
$\hat g/g_{\mathbb{H}^3}\to {\rm const.}$ at
$\infty$.

\begin{lemma} \label{lm16}
Suppose $n=3$. Let $u(r)$ be a compact support solution of
{\rm (\ref{rad1})-(\ref{rad2})} with at least square-root growth.
Then there exists a constant $C>0$ such that
\begin{gather*}
|u'|  \leq  \begin{cases}
C(R-r)^{3-\frac{q}{2}} & \mbox{if } q>6,\\
C|\log(R-r)| & \mbox{if } q =6,\\
C  & \mbox{if } 1<q<6,
\end{cases}\\
|u''|  \leq   \begin{cases}
C(R-r)^{2-\frac{q}{2}} & \mbox{if } q>4,\\
C|\log(R-r)| & \mbox{if } q=4,\\
C & \mbox{if } 1<q<4,
\end{cases}\\
|u'''|  \leq  \begin{cases}
C(R-r)^{1-\frac{q}{2}} & \mbox{if } q > 2,\\
C|\log(R-r)| & \mbox{if } q=2,\\
C & \mbox{if } 1<q<2.
\end{cases}
\end{gather*}
\end{lemma}

\begin{proof} We begin with the estimate for $u'''$. As in
Lemma \ref{estimates_3} we see that $u'''\leq 0$. Hence
$$
0\geq u'''(r) = \frac{-1}{r^4}\int_0^r s^4 u(s)^{-q}\,ds \geq
-C\int_0^r (R-r)^{-q/2}\,ds.
$$
The estimate follows by performing the integration. Likewise,
$$
|u''(r)|\leq C+\int_0^r |u'''(t)|\,dt
$$
leads to the estimate on $u''$, and a further integration leads to
$$
|u'(r)|\leq \int_0^r |u''(t)|\,dt<\infty
$$ and its subsequent estimate.
\end{proof}



\begin{proof}[Proof of Theorem \ref{special_compact}]
 Part (b): If $1<q<6$ then
$u'$ is bounded, which is incompatible with the assumption that $u$ has at
least square root growth. The same holds if $q=6$ since integration of the
$u'$ estimates yields $|u|\leq C(R-r)|\log(R-r)|$, which is again incompatible
with the square root lower bound.

Part (a): Assume $q>6$. We use again Poho\v{z}aev's identity for
$\rho\in (0,R)$. For the terms in the right-hand side we find
$(u'')^2\leq C(R-\rho)^{4-q}$, $|u''u'|\leq C(R-\rho)^{5-q}$,
$|u'u'''|\leq C(R-\rho)^{4-q}$, $|uu''|\leq C(R-\rho)^{(5-q)/2}$,
$|uu'''|\leq C(R-\rho)^{(3-q)/2}$. It turns out that the most singular term
is $(R-\rho)^{4-q}$. However, the remaining term $u^{1-q}\approx
(R-\rho)^{(1-q)/2}$ is more singular provided $q<7$. Here we use that $u$ is
bounded above by multiples of $\sqrt{R-r}$. Hence, the
right-hand side converges to $-\infty$ with rate $-(R-\rho)^{(1-q)/2}$ as
$\rho\to R$. The left-hand side is also negative for $q<7$. If we use that
$u$ is bounded below by multiples of $\sqrt{R-r}$ then we see that the left
hand side converges to $-\infty$ with the less singular rate
$-(R-\rho)^{(3-q)/2}$. Hence there is no solution with square root growth for
$6<q<7$. \end{proof}



\begin{proof}[Proof of Corollary \ref{cor_compact}]
If we prescribe $u(0)=\alpha>0$
then by Theorem \ref{separatrix} there exists $\delta_0>0$ such that every
solution with $u''(0)=\delta<\delta_0$ has compact support. Moreover, the
zero $R(\delta)$ ranges continuously between $0$ and $+\infty$ if $\delta$
ranges between $-\infty$ and $\delta_0$. Hence for $\alpha$ and $R>0$
there exists $\delta^\ast$ such that the solution with
$u''(0)=\delta^\ast(\alpha)$ has exactly support $[0,R]$. If
$\alpha\to \infty$ then necessarily $\delta^\ast(\alpha)\to-\infty$, i.e.
the solutions are concave and decreasing. Hence $u'r^2/u^2$ is decreasing and
thus $u$ generates a metric with negative scalar curvature, cf.
(\ref{scalar_curvature_3}). \end{proof}

\section{Open questions}

We finish with a selection of questions which remain open.
\begin{itemize}
\item [(1)] For $n\geq 4$ the equation $\Delta^2 u = -u^{-q}$ in 
$\mathbb{R}^n$ has non-radial 
positive entire solutions given by $u(x',x_n)=v(|x'|)$,  where $v(r)$ is a 
radial positive entire solution satisfying $\Delta^2 v = -v^{-q}$ in 
$\mathbb{R}^{n-1}$. This leaves the question whether in $\mathbb{R}^3$ 
non-radial positive entire solution exist. The above construction does not 
work, since Theorem \ref{separatrix} requires $n-1\geq 3$ for the existence of 
a solution $v$.
\item[(2)] Do there exist positive entire solutions of $\Delta^2 u = -u^{-q}$ 
in $\mathbb{R}^2$? A positive answer would resolve question (1).
\item [(3)] Can one find an explicit formula for the
 growth rate of the separatrix in terms of $q$?
\item [(4)] For $n=3$,  can one drop the assumption $q>4$ in
Theorem \ref{special_3}?
\item [(5)] Suppose $q=7$ and $n=3$. Are the explicit compact support
solutions $\alpha\sqrt{1/\sqrt{15\alpha^8}-|x|^2}$ unique in the class of
solutions having square root growth near their zero?
\item [(6)] In $\mathbb{R}^3$ the equation $\Delta^2u = u^{-7}$ arises from
(\ref{higher}) by assuming $Q_{\bar g}={\rm const.}<0$. The function
$U(r)=\sqrt[4]{4/3}\sqrt{r}$ is a solution, and the generated metric
$\bar g=U^{-4}\delta_{ij}$ has constant scalar curvature $8/3$. In what class
of solutions is $U(r)$ unique?
\item [(7)] In what class of solutions are $u(r)=
\left(\frac{2a}{a^2-r^2}\right)^\frac{n-4}{2}$ unique boundary-blow up
solutions of $\Delta^2 u = \frac{n}{16}(n-4)(n^2-4)u^\frac{n+4}{n-4}$ in
balls $B_a(0)\subset\mathbb{R}^n$ for $n\geq 5$?
\end{itemize}


\begin{thebibliography}{99}

\bibitem{AU} Th. Aubin, Some nonlinear problems in Riemannian Geometry.
Springer Monographs in Mathematics, Springer Verlag, 1998.

\bibitem{CHA} Sun-Yung A. Chang, {\sl On a fourth-order partial differential
equation in conformal geometry.} M. Christ (ed.) et al., Harmonic analysis and
partial differential equations. Essays in honor of A.P. Calder\'{o}n's 75th
birthday. The University of Chicago Press, Chicago Lectures in Mathematics,
127--150 (1999).

\bibitem{CHAYA} Sun-Yung A. Chang and Paul C. Yang, {\sl On  fourth order
curvature invariant.} Th. Branson (ed.), Spectral problems in geometry and
arithmetic. Providence, Amer. Math. Soc. Contemp. Math. {\bf 237}, 9--28
(1999).

\bibitem{chen_li} Wenxiong Chen and Congming Li, {\sl Classification of
solutions of some nonlinear elliptic equations.} Duke Math. J. {\bf 63}
(1991), 615--622.

\bibitem{choi} Y.S. Choi and Xingwang Xu, {\sl Nonlinear biharmonic equations
with negative exponents.} Preprint (2000).

\bibitem{LoeNi} C. Loewner and L. Nirenberg, {\sl Partial differential
equations invariant under conformal or projective transformations.}
L. Ahlfors (ed.), Contributions to analysis. Academic Press, New York,
245--272 (1974).

\bibitem{PS} P. Pucci and J. Serrin, {\sl A general variational identity.}
Indiana Univ. Math. J. {\bf 35} (1986), 681--703.

\bibitem{walter} W. Walter, Ordinary differential equations, Graduate texts
in mathematics {\bf 182}, Springer Verlag, 1998.

\bibitem{wei_xu} Juncheng Wei and Xingwang Xu, {\sl Classification of
solutions of higher order conformally invariant equations.} Math. Ann.
{\bf 313} (1999), 207--228.

\end{thebibliography}

\end{document}


\begin{proof} The function $v(s)=su(1/s)$ satisfies
$$
\Delta^2 v = v^{(iv)} +\frac{4}{s} v'''= -s^{q-7} v^{-q} \quad\mbox{in }
(0,\infty).
$$
Moreover,
\begin{gather}
v'(s)  =  u(\frac{1}{s})-\frac{1}{s}u'(\frac{1}{s}), \quad
v''(s) = \frac{1}{s^3}u''(\frac{1}{s}), \label{kt1}\\
v'''(s)  =  \frac{-3}{s^4}u''(\frac{1}{s})-\frac{1}{s^5}u'''(\frac{1}{s}),
\label{kt2} \\
v^{(iv)}(s)  =
\frac{12}{s^5}u''(\frac{1}{s})+\frac{8}{s^6}u'''(\frac{1}{s})
+\frac{1}{s^7}u^{(iv)}(\frac{1}{s}). \label{kt3}
\end{gather}
Let $w =\Delta v= v''+\frac{2}{s} v'$. For $q<4$ we find
$$
\int_0^s -t^{q-7}v^{-q}t^2\,dt = \int_0^s (w''+2\frac{w'}{t})t^2\,dt
= \int_0^s (t^2 w')'\,ds = s^2w'(s),
$$
since $\lim_{s\to 0} s^2 w'(s)=0$ by (\ref{kt1}), (\ref{kt2}) and
Lemma \ref{estimates_3}. Hence $w'\leq 0$, and since $w(s)\to 0$ as
$s\to \infty$ this shows that $w\geq 0$. Hence $s^2w(s) = (v's^2)'\geq 0$.
Notice that $v'$ stays bounded as $s\to 0$, cf. Lemma \ref{bounded}.
Hence $v's^2\geq 0$ for all $s\in (0,\infty)$. Together with (\ref{kt1}),
this shows that $u(1/s)-u'(1/s)/s\geq 0$, which is equivalent to our claim.
\end{proof}