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_11$ 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 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} $$ and clearly $A(\epsilon)\to \infty$ for $\epsilon\to 0$ and $A(\epsilon)\to 0$ as $\epsilon\to 1/(8n(n+2))$. Furthermore $w(0)=1, w'(0)=w'''(0)=0$, and $w''(0)=2\epsilon A(\epsilon)$ has the properties that $w''(0)\to 0$ as $\epsilon\to 0$ and as $\epsilon\to 1/(8n(n+2))$. The second derivative of $w$ at $0$ is maximal for $\epsilon=(1-1/q)^q\,1/(8n(n+2))$ and the value of $w''(0)$ is $\tilde\delta := 4\sqrt{(1-1/q)^q/(8n(n+2)(q-1))}$. As a result we have that every solution of (\ref{rad1})-(\ref{rad2}) with $\delta\leq \tilde\delta$ stays below $w(r)$, and hence it has compact support. \end{proof} \begin{lemma} \label{lm4} No entire solution of {\rm (\ref{rad1})-(\ref{rad2})} grows faster than $r^2$. \label{fastest} \end{lemma} \begin{proof} Let $w(r)=Ar^2+1$. Then $\Delta^2 w=0$ and hence $w$ is a supersolution. Given an arbitrary entire solution $u$ of (\ref{rad1})-(\ref{rad2}) we can choose $2A>u''(0)$. Hence $u 0$ such that for every $\epsilon\in (0,\epsilon_0)$ there exists a value $\bar\delta=\bar\delta(\epsilon)$ with the property that every solution of {\rm (\ref{rad1})-(\ref{rad2})} with $\delta\geq \bar\delta$ is entire and grows faster than $r^{2-\epsilon}$. \label{entire} \end{lemma} \begin{proof} For the construction of an entire subsolution we consider $v(r) = (1+b^2 r^2)^{1-\frac{\epsilon}{2}}$. Let $p(r)$ be the function such that $$ \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). $$ Then the following holds: \begin{itemize} \item[(a)] (Weak comparison) If $v(0)\leq w(0)$, $v'(0)=w'(0)=0$, $v''(0)\leq w''(0)$, $v'''(0)=w'''(0)=0$ then $v\leq w$, $v'\leq w'$, $v''\leq w''$ and $v'''\leq w'''$ on $[0,R)$. \item[(b)] (Strong comparison) If for some $\rho>0$ we have $v 0$,\\ \emph{at least linear growth} & if $u(r)\geq ar+b$,\\ \emph{superlinear growth} & if $u(r)\geq ar+b$, but $u$ does not have linear growth. \end{tabular} \smallskip The following theorem gives a more detailed picture of the entire radial solutions in dimension $n=3$. A stronger version of part (b) including non-radial solutions has been obtained by Choi and Xu \cite{choi} as part of their main result. Here we give a different proof. To the best of our knowledge, part (a) is new for $q>7$. \begin{theorem} For $n=3$ entire solutions of {\rm (\ref{rad1})-(\ref{rad2})} have the following properties: \begin{itemize} \item[(a)] For $q\geq 7$ a unique solution with linear growth exist. It coincides with the sepa\-ratrix and has exactly linear growth. For $q=7$ it is given by $\sqrt{1+r^2/\sqrt{15}}$. \item[(b)] For $4 0$ such that $0\leq u'(r)\leq C$ if $q>3$ and moreover: \begin{gather*} 0 \leq u'' \leq \begin{cases} C(1+r)^{-3} & \mbox{if } q>5,\\ C(1+r)^{-3}\log(2+r) & \mbox{if } q=5,\\ C(1+r)^{-q+2} & \mbox{if } 25,\\ -C(1+r)^{-4}\log(2+r) & \mbox{if } q=5,\\ -C(1+r)^{-q+1} & \mbox{if } 12$ we have that $\lim_{r\to\infty} u''(r)$ exists and vanishes due the assumption of linear growth. Moreover $u''>0$ by Lemma \ref{convex}. Thus $$ 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 $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~~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 $60$ (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 } 14,\\ C|\log(R-r)| & \mbox{if } q=4,\\ C & \mbox{if } 12,\\ C|\log(R-r)| & \mbox{if } q=2,\\ C & \mbox{if } 16$. 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 $60$ 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. 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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}