\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 $$\Delta^2 u +u^{-q}=0 \quad\mbox{in } D\subset\mathbb{R}^n. \label{basic}$$ 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 $$U(x) = \alpha\sqrt{1/\sqrt{15\alpha^8}+|x|^2}. \label{ast}$$ Moreover Choi and Xu prove that (\ref{basic}) has no linear growth solution if 40. $$The corresponding metric is$$ \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\} $$with standard Lorentz-Minkowski metric g(v,w)=\frac{1}{a^2}(v_1w_1+\ldots v_nw_n-v_{n+1}w_{n+1}). The explicit form of these solutions and their uniqueness on balls B_a(0) was proved by Loewner, Nirenberg \cite{LoeNi}. \subsection{A fourth order analog of Yamabe's equation} For n\neq 4 let \bar g be given as \bar g_{ij} = u^{4/(n-4)}\delta_{ij}. Then the conformal factor u:\mathbb{R}^n\to \mathbb{R} satisfies $$\Delta^2 u = \frac{n-4}{2} Q_{\bar g} u^\frac{n+4}{n-4}, \label{higher}$$ where$$ 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 $$(r^{n-1}u')'= r^{n-1}U,\quad (r^{n-1}U')'+r^{n-1}u^{-q}=0. \label{system}$$ 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 v0 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} $$and clearly A(\epsilon)\to \infty for \epsilon\to 0 and A(\epsilon)\to 0 as \epsilon\to 1/(8n(n+2)). Furthermore w(0)=1, w'(0)=w'''(0)=0, and w''(0)=2\epsilon A(\epsilon) has the properties that w''(0)\to 0 as \epsilon\to 0 and as \epsilon\to 1/(8n(n+2)). The second derivative of w at 0 is maximal for \epsilon=(1-1/q)^q\,1/(8n(n+2)) and the value of w''(0) is \tilde\delta := 4\sqrt{(1-1/q)^q/(8n(n+2)(q-1))}. As a result we have that every solution of (\ref{rad1})-(\ref{rad2}) with \delta\leq \tilde\delta stays below w(r), and hence it has compact support. \end{proof} \begin{lemma} \label{lm4} No entire solution of {\rm (\ref{rad1})-(\ref{rad2})} grows faster than r^2. \label{fastest} \end{lemma} \begin{proof} Let w(r)=Ar^2+1. Then \Delta^2 w=0 and hence w is a supersolution. Given an arbitrary entire solution u of (\ref{rad1})-(\ref{rad2}) we can choose 2A>u''(0). Hence u0 such that for every \epsilon\in (0,\epsilon_0) there exists a value \bar\delta=\bar\delta(\epsilon) with the property that every solution of {\rm (\ref{rad1})-(\ref{rad2})} with \delta\geq \bar\delta is entire and grows faster than r^{2-\epsilon}. \label{entire} \end{lemma} \begin{proof} For the construction of an entire subsolution we consider v(r) = (1+b^2 r^2)^{1-\frac{\epsilon}{2}}. Let p(r) be the function such that$$ \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 withu''(0)=\deltahas 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). $$Then the following holds: \begin{itemize} \item[(a)] (Weak comparison) If v(0)\leq w(0), v'(0)=w'(0)=0, v''(0)\leq w''(0), v'''(0)=w'''(0)=0 then v\leq w, v'\leq w', v''\leq w'' and v'''\leq w''' on [0,R). \item[(b)] (Strong comparison) If for some \rho>0 we have v0,\\ \emph{at least linear growth} & if u(r)\geq ar+b,\\ \emph{superlinear growth} & if u(r)\geq ar+b, but u does not have linear growth. \end{tabular} \smallskip The following theorem gives a more detailed picture of the entire radial solutions in dimension n=3. A stronger version of part (b) including non-radial solutions has been obtained by Choi and Xu \cite{choi} as part of their main result. Here we give a different proof. To the best of our knowledge, part (a) is new for q>7. \begin{theorem} For n=3 entire solutions of {\rm (\ref{rad1})-(\ref{rad2})} have the following properties: \begin{itemize} \item[(a)] For q\geq 7 a unique solution with linear growth exist. It coincides with the sepa\-ratrix and has exactly linear growth. For q=7 it is given by \sqrt{1+r^2/\sqrt{15}}. \item[(b)] For 40 such that 0\leq u'(r)\leq C if q>3 and moreover: \begin{gather*} 0 \leq u'' \leq \begin{cases} C(1+r)^{-3} & \mbox{if } q>5,\\ C(1+r)^{-3}\log(2+r) & \mbox{if } q=5,\\ C(1+r)^{-q+2} & \mbox{if } 25,\\ -C(1+r)^{-4}\log(2+r) & \mbox{if } q=5,\\ -C(1+r)^{-q+1} & \mbox{if } 12 we have that \lim_{r\to\infty} u''(r) exists and vanishes due the assumption of linear growth. Moreover u''>0 by Lemma \ref{convex}. Thus$$ 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 40,\\ \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 } 1 2,\\ 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. 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. 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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}