# Classification for positive singular solutions to critical sixth order
equations

João Henrique Andrade  and  Juncheng Wei  Department of Mathematics,
University of British Columbia
V6T 1Z2, Vancouver-BC, Canada
and
Institute of Mathematics and Statistics, University of São Paulo
05508-090, São Paulo-SP, Brazil<EMAIL_ADDRESS><EMAIL_ADDRESS>Department of Mathematics, University of British Columbia
V6T 1Z2, Vancouver-BC, Canada<EMAIL_ADDRESS>
###### Abstract.

We classify entire positive singular solutions to a family of critical sixth
order equations in the punctured space with a non-removable singularity at the
origin. More precisely, we show that when the origin is a non-removable
singularity, solutions are given by a singular radial factor times a periodic
solution to a sixth order IVP with constant coefficients. On the technical
level, we combine integral sliding methods and qualitative analysis of ODEs,
based on a conservation of energy result, to perform a topological two-
parameter shooting technique. We first use the integral representation of our
equation to run a moving spheres technique, which proves that solutions are
radially symmetric with respect to the origin. Thus, in Emden–Fowler
coordinates, we can reduce our problem to the study of an sixth order
autonomous ODE with constant coefficients. The main heuristics behind our
arguments is that since all the indicial roots of the ODE operator are
positive, it can be decomposed into the composition of three second order
operators satisfying a comparison principle. This allows us to define a
Hamiltonian energy which is conserved along solutions, from which we extract
their qualitative properties, such as uniqueness, boundedness, asymptotic
behavior, and classification.

###### Key words and phrases:

Tri-Laplacian, Critical exponent, Sixth order equation, Liouville-type
theorem, Emden–Fowler solutions

###### 2020 Mathematics Subject Classification:

35J60, 35B09, 35J30, 35B40

This research is partially supported by São Paulo Research Foundation (FAPESP)
#2020/07566-3 and #2021/15139-0 and Natural Sciences and Engineering Research
Council of Canada (NSERC)

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Integral representation
    2. 2.2 Kelvin transform
    3. 2.3 Radial Emden–Fowler coordinates
    4. 2.4 Decomposition
    5. 2.5 Conservation of energy
  3. 3 Radial symmetry
    1. 3.1 Integrability
    2. 3.2 Superharmonicity
    3. 3.3 Integral moving spheres method
  4. 4 ODE analysis
    1. 4.1 Boundedness
    2. 4.2 Comparison
    3. 4.3 Asymptotics
    4. 4.4 Ordering
    5. 4.5 Uniqueness
    6. 4.6 Classification
    7. 4.7 Shooting technique
  5. 5 Proof of the main theorem

##  1\. Introduction

We are interested in classifying (classical) positive singular solutions $u\in
C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ with $n\geqslant 7$ (which will always
be assumed so forth) to the following family of critical sixth order PDEs on
the punctured space

$\displaystyle(-\Delta)^{3}u=c_{n}u^{\frac{n+6}{n-6}}\quad{\rm
in}\quad\mathbb{R}^{n}\setminus\\{0\\}.$ ($\mathcal{P}_{6,\infty}$)

Here $\Delta^{3}=\Delta\circ\Delta\circ\Delta$ is the tri-Laplacian and
$f(u):=c_{n}|u|^{p-2}u$ with $p=\frac{2n}{n-6}:=2^{\\#}$ is critical in the
sense of the compact Sobolev embedding of $H^{3}(\mathbb{R}^{n})$ and $c_{n}$
is a normalizing geometric constant given by

$c_{n}:=\frac{n(n-6)(n^{4}-20n^{2}+64)}{64}.$

We say that a positive solution $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$
has a removable singularity at the origin if $\lim_{x\rightarrow
0}u(x)<\infty$, that is, it can be continuously extended across the origin.
Otherwise, we say that the origin is a non-removable singularity. Let us call
these solutions non-singular and singular, respectively.

Let us mention that the second-named author and X. Xu [38] studied non-
singular solutions to ($\mathcal{P}_{6,\infty}$). On this subject, they are
based on a sliding technique to prove that all its positive non-singular
solutions are radially symmetric with respect to some point and have a closed
expression. This result confirmed a conjecture by E. Lieb [27] about the
classification of extremal function for the Sobolev inequality. These results
can be stated as follows

######  Theorem A.

Let $u\in C^{6}(\mathbb{R}^{n})$ be a positive non-singular solution to
($\mathcal{P}_{6,\infty}$). Then, there exist $x_{0}\in\mathbb{R}^{n}$ and
$\mu\in\mathbb{R}$ such that $u$ is radially symmetric with respect to $x_{0}$
and monotonically decreasing. Moreover, it holds

$u(x)=\left(\frac{2\mu}{\mu^{2}+|x-x_{0}|^{2}}\right)^{\frac{6-n}{2}}.$

The elements in this family are called spherical solutions and they are
denoted by $u_{x_{0},\mu}$.

Our main result in this manuscript extends Theorem A for the case when the
origin is a non-removable singularity and can be stated as

######  Theorem 1.

Let $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ be a positive singular
solution to ($\mathcal{P}_{6,\infty}$). Then, $u$ is radially symmetric with
respect to the origin and monotonically decreasing. Moreover, there exist
$a_{0}\in\mathbb{(}0,a^{*}_{n})$ and $T\in[0,T_{a_{0}}]$ such that

$u(x)=|x|^{\frac{6-n}{2}}v_{a}(-\ln|x|+T).$

Here $v_{a}$ is the unique periodic bounded positive solution to the sixth
order IVP $($or Cauchy problem$)$ below

$\displaystyle\begin{cases}v^{(6)}-K_{4}v^{(4)}+K_{2}v^{(2)}-K_{0}v+c_{n}v^{\frac{n+6}{n-6}}=0\quad{\rm
in}\quad\mathbb{R}\\\ v(0)=a_{0},\quad v^{(2)}(0)=a_{2},\quad
v^{(4)}(0)=a_{4},\quad{\rm and}\quad
v^{(1)}(0)=v^{(3)}(0)=v^{(5)}(0)=0.\end{cases}$ ($\mathcal{O}_{6,\infty}$)

Here $K_{0},K_{2},K_{4}$ are dimensional constants $($see (2.3)$)$,
$a_{2},a_{4}$ depend on $a_{0}$, and $a_{*}=K_{0}^{(n-6)/12}$. The elements in
this family are called $($sixth order$)$ Emden–Fowler solutions and are
denoted by $u_{a,T}$.

######  Remark 2.

Notice that the assumption that our solutions are classical is not
restrictive. In fact, by a standard regularity lifting technique from [6], one
can show that any weak solution $u\in H^{3}(\mathbb{R}^{n}\setminus\\{0\\})$
to ($\mathcal{P}_{6,\infty}$) also salves ($\mathcal{P}_{6,\infty}$) in
classical sense, that is, $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ or even
smooth $u\in C^{\infty}(\mathbb{R}^{n}\setminus\\{0\\})$ and
($\mathcal{P}_{6,\infty}$) holds pointwise.

Recently, T. Jin and J. Xiong [23] used variational techniques to prove
existence of solutions to ($\mathcal{P}_{6,\infty}$). In fact, they considered
a integral equation generalizing ($\mathcal{P}_{6,\infty}$), which includes
non-local versions of the nonlinearity (see also [8]). The full classification
result for the critical fractional case is still unknown. For other ranges of
the power-nonlinearity, S. Luo et al. [29, 30] proved a monotonicity formula
for solutions to related higher order poly-harmonic equations. We refer the
reader to [19, 20, 35, 11, 10, 41, 40, 18, 32] for more results on this
subject.

Eq. ($\mathcal{P}_{6,\infty}$) arises naturally in the study of the lack of
compactness for the critical Sobolev embedding. Besides, its applicability in
PDEs, our result has also a connection with differential geometry. In this
language, ($\mathcal{P}_{6,\infty}$) is the locally conformally flat version
of a more general geometric equation, the so-called sixth order GJMS equation
[12, 16], which is driven by the $P^{6}$ the sixth order GJMS operator. In
this geometrical language, our main results classify the metrics with constant
(sixth order) $Q^{6}$-curvature on
$(\mathbb{S}^{n}\setminus\\{p,-p\\},g_{0})$, where $g_{0}$ is the standard
round metric on the punctured unit sphere and $Q^{6}_{g_{0}}\equiv
2^{-5}{n(n^{4}-20n^{2}+64)}$. Indeed, the metric $g=u^{6/{(n-6)}}g_{0}$ has
constant $Q^{6}$-curvature if, and only, if $u\in
C^{\infty}(\mathbb{S}^{n}\setminus\\{0\\})$ is a (smooth) positive solution to
the geometric PDE

$P^{6}_{g_{0}}u=c_{n}u^{\frac{n+6}{n-6}}\quad{\rm
in}\quad\mathbb{S}^{n}\setminus\\{p,-p\\},$

where

$P_{g_{0}}^{6}=\left(-\Delta_{g_{0}}+\frac{(n-6)(n+4)}{4}\right)\left(-\Delta_{g_{0}}+\frac{(n-4)(n+2)}{4}\right)\left(-\Delta_{g_{0}}+\frac{n(n-2)}{4}\right).$

For more details on this subject, we refer the interested reader to [24, 9, 5,
13] and the references therein. It is also worth mentioning the relations of
($\mathcal{P}_{6,\infty}$) with tri-harmonic maps [3].

######  Remark 3.

In [23], the asymptotic behavior of solutions to a local equation on the
punctured ball of radius $R>0$ is studied. More precisely, it is proved that
solutions to

$\displaystyle(-\Delta)^{3}u=c_{n}u^{\frac{n+6}{n-6}}\quad{\rm in}\quad
B_{R}\setminus\\{0\\},$ ($\mathcal{P}_{6,R}$)

satisfy

$u(x)=(1+\mathrm{o}(1))u_{0}(x)\quad{\rm as}\quad x\rightarrow 0,$

where $u_{0}\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ solves
($\mathcal{P}_{6,\infty}$). Our main theorem may be applied to improve the
asymptotic behavior of solutions to ($\mathcal{P}_{6,R}$) near the origin.
Based on [25, 1, 33], we conjecture that assuming $-\Delta u>0$ and
$\Delta^{2}u>0$, one can find $\beta_{0}>0$ such that

$u(x)=(1+\mathcal{O}(|x|^{\beta_{0}}))u_{0}(x)\quad{\rm as}\quad x\rightarrow
0.$

A result like this have direct implications on the study of the $($sixth
order$)$ singular GJMS.

Now let us compare our results to the ones in the fourth and second order
cases. First, we consider positive solutions $u\in
C^{4}(\mathbb{R}^{n}\setminus\\{0\\})$ the fourth order critical equation

$\displaystyle(-\Delta)^{2}u=\frac{n(n-4)^{2}(n^{2}-4)}{16}u^{\frac{n+4}{n-4}}\quad{\rm
in}\quad\mathbb{R}^{n}\setminus\\{0\\},$ ($\mathcal{P}_{4,\infty}$)

where $n\geqslant 5$, $\Delta^{2}=\Delta\circ\Delta$ is the bi-Laplacian.
Notice that ($\mathcal{P}_{4,\infty}$) is critical in the sense of the compact
Sobolev embedding $H^{2}(\mathbb{R}^{n})$. On this subject, we should mention
that when the origin is a non-removable singularity, C. S. Lin [28] (see also
[39]) obtained radial symmetry for positive solutions to
($\mathcal{P}_{4,\infty}$) using the asymptotic moving planes technique.
Recently, Z. Guo, et al. [17] proved the existence of periodic solutions by
applying a mountain pass theorem and conjectured that all ODE solutions should
be periodic. Later on, R. L. Frank and T. König [15] answered this conjecture,
obtaining more accurate results concerning the classification for entire
positive singular solutions to ($\mathcal{P}_{4,\infty}$).

######  Theorem B.

Let $u\in C^{4}(\mathbb{R}^{n}\setminus\\{0\\})$ be a positive solution to
($\mathcal{P}_{4,\infty}$). Then, $u$ is radially symmetric with respect to
the origin and monotonically decreasing. Moreover,

  * (i)

Assume that $u$ is non-singular. Then, there exist $x_{0}\in\mathbb{R}^{n}$
and $\mu\in\mathbb{R}$ such that

$u(x)=\left(\frac{2\mu}{\mu^{2}+|x-x_{0}|^{2}}\right)^{\frac{4-n}{2}}.$

  * (ii)

Assume that $u$ is singular. Then, there exist
$a_{0}\in\mathbb{(}0,[n(n-4)/(n^{2}-4)]^{n-4/8})$ and $T\in[0,T_{a_{0}}]$ such
that

$u(x)=|x|^{\frac{4-n}{2}}v_{a}(-\ln|x|+T),$

where $v_{a}$ is the unique periodic bounded solution to the fourth order IVP
below

$\displaystyle\begin{cases}v^{(4)}-\frac{n(n-4)+8}{2}v^{(2)}+\frac{n^{2}(n-4)^{2}}{16}v-\frac{n(n-4)^{2}(n^{2}-4)}{16}v^{\frac{n+4}{n-4}}=0\quad{\rm
in}\quad\mathbb{R}\\\ v(0)=a_{0},\quad v^{(2)}(0)=a_{2},\quad{\rm and}\quad
v^{(1)}(0)=v^{(3)}(0)=0.\end{cases}$ ($\mathcal{C}_{4,\infty}$)

Second, we consider positive singular solutions $u\in
C^{2}(\mathbb{R}^{n}\setminus\\{0\\})$ the second order critical equation
below

$\displaystyle-\Delta u=\frac{n(n-4)}{8}u^{\frac{n+2}{n-2}}\quad{\rm
in}\quad\mathbb{R}^{n}\setminus\\{0\\},$ ($\mathcal{P}_{2,\infty}$)

where $n\geqslant 3$, $\Delta$ is the Laplacian. Let us notice that
($\mathcal{P}_{2,\infty}$) is critical in the sense of the compact Sobolev
embedding of $H^{1}(\mathbb{R}^{n})$. All the aforementioned classification
results were inspired by the classical theorems of R. H. Fowler [14] (see also
[4]) and T. Aubin [2] and G. Talenti [36] on the study of conformally
equivalent metrics with constant scalar curvature, the so-called Yamabe
problem.

######  Theorem C.

Let $u\in C^{2}(\mathbb{R}^{n}\setminus\\{0\\})$ be a positive solution to
($\mathcal{P}_{2,\infty}$). Then, $u$ is radially symmetric with respect to
the origin and monotonically decreasing. Moreover,

  * (i)

if $u$ is non-singular, then there exist $x_{0}\in\mathbb{R}^{n}$ and
$\mu\in\mathbb{R}$ such that

$u(x)=\left(\frac{2\mu}{\mu^{2}+|x-x_{0}|^{2}}\right)^{\frac{2-n}{2}},$

which are called spherical solutions and they are denoted by $u_{x_{0},\mu}$.

  * (ii)

if $u$ is singular, then there exist
$a_{0}\in\mathbb{(}0,[(n-2)/n]^{(n-2)/4})$ and $T\in[0,T_{a_{0}}]$ such that

$u(x)=|x|^{\frac{2-n}{2}}v_{a}(-\ln|x|+T),$

where $v_{a}$ is the unique periodic bounded solution to the second order IVP
below

$\displaystyle\begin{cases}v^{(2)}-\frac{(n-2)^{2}}{4}v+\frac{n(n-2)}{4}v^{\frac{n+2}{n-2}}=0\quad{\rm
in}\quad\mathbb{R}\\\ v(0)=a_{0}\quad{\rm and}\quad v^{(1)}(0)=0.\end{cases}$
($\mathcal{C}_{2,\infty}$)

Let us describe our strategy to prove Theorem 1. First, we prove integrability
and superharmonicity properties for solutions to ($\mathcal{P}_{6,\infty}$),
which we use to perform an integral sliding technique and prove that positive
singular solutions to ($\mathcal{P}_{6,\infty}$) are radially symmetric.
Second, we perform a change of variables to transform
($\mathcal{P}_{6,\infty}$) into ($\mathcal{O}_{6,\infty}$) a sixth order IVP
with constant coefficients, which can be decomposed into three second order
problems satisfying a maximum principle. This allows us to define a
Hamiltonian energy that is conserved along solutions, which, we use to prove
some qualitative properties for positive solutions to
($\mathcal{O}_{6,\infty}$). Third, we use a two-parameter shooting technique
to finish the proof of our classification result. By undoing the Emden–Fowler
change of variables, this can be easily translated into a result about
solutions to ($\mathcal{P}_{6,\infty}$).

The main difficulties in our approach are the lack of maximum principle for
higher order operators and the possibly chaotic behavior enjoyed by solutions
to ODEs driven by these operators. The former can be overcome either by an
integral representation formula or a decomposition into second order
operators. To deal with the later issue, we use the conservation of the energy
and qualitative properties previously proved. These properties assure that
this sixth order ODE behaves as a second order one.

We remark that ($\mathcal{P}_{6,\infty}$), ($\mathcal{P}_{4,\infty}$), and
($\mathcal{P}_{2,\infty}$) are particular cases of a more general class of
equations, which we describe as follows. For any $m\in\mathbb{N}^{+}$, we are
interested in classifying (smooth) positive singular solutions $u\in
C^{2m}(\mathbb{R}^{n}\setminus\\{0\\})$ with $n>2m:=N$ (which will always be
assumed so forth) to the following family of critical even order poly-harmonic
PDEs

$\displaystyle(-\Delta)^{m}u=c_{n,N}u^{\frac{n+N}{n-N}}\quad{\rm
in}\quad\mathbb{R}^{n}\setminus\\{0\\}.$ ($\mathcal{P}_{N,\infty}$)

Here $(-\Delta)^{m}$ is the $m$-poly-Laplacian and $c_{n,N}$ is a normalizing
constant given by

$c_{n,N}=2^{N}{\Gamma\left(\frac{n+N}{4}\right)^{2}}{\Gamma\left(\frac{n-N}{4}\right)^{-2}},$
(1)

where $\Gamma(s)=\int_{0}^{\infty}\tau^{s-1}e^{-\tau}\mathrm{d}\tau$ is the
standard Gamma function. The power $\frac{n+N}{n-N}:=2_{m}^{*}-1$ is critical
in the sense of the compact Sobolev embedding $H^{m}(\mathbb{R}^{n})$.

We remark that almost all of our technical results in this manuscript extend
to this class of even-order poly-harmonic operators. In this regard, a
classification result for entire singular solutions to
($\mathcal{P}_{N,\infty}$) in the sense of Theorem 1 is believed to be true.
Let us state this conjecture as follows

######  Conjecture 4.

Let $u\in C^{N}(\mathbb{R}^{n}\setminus\\{0\\})$ be a positive singular
solution to ($\mathcal{P}_{6,\infty}$). Then, $u$ is radially symmetric with
respect to the origin and monotonically decreasing. Moreover, there exist
$a_{0}\in\mathbb{(}0,a^{*}_{n,N})$ and $T\in[0,T_{a_{0}}]$ such that

$u(x)=|x|^{\frac{N-n}{2}}v_{a}(\ln|x|+T).$

Here $v_{a}$ is the unique periodic bounded positive solution to the following
higher order IVP

$\displaystyle\begin{cases}\sum_{j=0}^{m}(-1)^{j+1}K_{2j,N}(n)v^{(2j)}=c_{n,N}v^{\frac{n+N}{n-N}}\quad{\rm
in}\quad\mathbb{R}\\\ v^{(2j)}(0)=a_{j}\quad{\rm and}\quad
v^{(2j-1)}(0)=0\quad{\rm for}\quad j=1,\dots,m.\end{cases}$
($\mathcal{C}_{N,\infty}$)

Here $K_{2j,N}(n),a^{*}_{n,N}>0$ are dimensional constants for all
$j=1,\dots,m$, namely

$a^{*}_{n,N}=K_{0}^{\frac{n-N}{2N}}\quad{\rm and}\quad
K_{2j,N}(n)=\sigma_{j}\left(\pm\sqrt{\lambda_{1,N}(n)},\dots,\pm\sqrt{\lambda_{m,N}(n)}\right),$

where $\sigma_{j}$ are the $j$-th harmonic polynomial and

$\lambda_{j,N}(n)=\frac{n-N+4j}{2}$

are the indicial roots of the constant coefficients ODE operator.

We explain the plan for the rest of the manuscript. In Section 2, we introduce
some definitions and prove some preliminary results that will be used
subsequently, such as integral representation formulas, Kelvin transform,
radial Emden–Fowler coordinates, decomposition result, and conservation of
energy. In Section 3, we apply the integral moving spheres method to prove
that solutions to ($\mathcal{P}_{6,\infty}$) are radially symmetric. In
Section 4, we perform some ODE analysis to prove qualitative properties for
solutions to ($\mathcal{O}_{6,\infty}$), which we use to perform a topological
shooting method and classify these solutions. In Section 5, we combine the
energy conservation, the radial symmetry, and the ODE analysis to prove
Theorem 1.

##  2\. Preliminaries

This section aims to introduce some necessary background definitions and
results for developing the sliding methods and the asymptotic analysis that
will be later used in this manuscript.

###  2.1. Integral representation

Now we use a Green identity to transform the differential equation
($\mathcal{P}_{6,\infty}$) into an integral equation. In this way, we can
avoid using the classical form of the maximum principle, and a sliding method
is available [34, 26], which will be used to classify solutions.

The next result uses the Green identity to convert ($\mathcal{P}_{6,\infty}$)
into an integral system.

######  Proposition 2.1.

Let $u\in C^{6}(\mathbb{R}^{n})$ be a positive solution to
($\mathcal{P}_{6,\infty}$). Then $($up to constant$)$, it follows

$u(x)=\int_{\mathbb{R}^{n}}|x-y|^{6-n}{f}(u)\mathrm{d}y\quad{\rm
in}\quad\mathbb{R}^{n}\setminus\\{0\\}.$ ($\mathcal{I}_{6}$)

###### Proof.

It directly follows from [8, Theorem 4.3] (see also [7]). ∎

###  2.2. Kelvin transform

Later we will employ the moving spheres technique, which is based on the
$2m$-order Kelvin transform. For this, given $x_{0}\in\mathbb{R}^{n}$ and
$\mu>0$, we need to establish the concept of inversion about a sphere
$\partial B_{\mu}(x_{0})$, which is given by
$\mathcal{I}_{x_{0},\mu}(x)=x_{0}+\mathcal{K}_{x_{0},\mu}(x)^{2}(x-x_{0})$,
where $\mathcal{K}_{x_{0},\mu}(x)=\mu/|x-x_{0}|$.

######  Definition 2.2.

For any $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$, let us consider the
sixth order Kelvin transform about the sphere with center at
$x_{0}\in\mathbb{R}^{n}$ and radius $\mu>0$ defined by

$u_{x_{0},\mu}(x)=\mathcal{K}_{x_{0},\mu}(x)^{n-6}u\left(\mathcal{I}_{x_{0},\mu}(x)\right).$

The next proposition states that solutions to ($\mathcal{P}_{6,\infty}$) are
invariant under the Kelvin transform. This holds because of the conformal
invariance enjoyed by this family of critical equations.

######  Proposition 2.3.

If $u$ is a solution to ($\mathcal{P}_{6,\infty}$), then $u_{x_{0},\mu}$ is a
solution to

$(-\Delta)^{3}u_{x_{0},\mu}=f(u_{x_{0},\mu})\quad{\rm
in}\quad\mathbb{R}^{n}\setminus\\{0,x_{0}\\}.$

###### Proof.

It directly follows by using [38]. ∎

###  2.3. Radial Emden–Fowler coordinates

This is section is devoted to constructing a change of variables that
transforms the singular PDE ($\mathcal{P}_{6,\infty}$) problem into the nice
ODE problem with constant coefficients. Here we only consider functions that
are radially symmetric, that is, $u=u(r)$ with $r=|x|$. Later on, we will
prove that this in fact holds for positive singular solutions to
($\mathcal{P}_{6,\infty}$) (see Proposition 3.1).

######  Definition 2.4.

For any $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ positive $($radial$)$
solution to ($\mathcal{P}_{6,\infty}$), we define the $($sixth order$)$
Emden–Fowler $($or logarithmic-cylindrical$)$ change of variables given by

$v(t)=|x|^{\frac{6-n}{2}}u(r)\quad{\rm with}\quad t=\ln r.$

The power $\gamma_{n}:=\frac{6-n}{2}$ is chosen by conformal invariance.

This change of variables is used to transform the singular PDE
($\mathcal{P}_{6,\infty}$) problem into the nice ODE problem with constant
coefficients

$v^{(6)}-K_{4}v^{(4)}+K_{2}v^{(2)}+g(v)=0\quad{\rm in}\quad\mathbb{R}.$
($\mathcal{O}_{6}$)

The nonlinear term is $g:C^{6}(\mathbb{R})\rightarrow C^{6}(\mathbb{R})$ is
defined as

$g(v):=f(v)-K_{0}v\quad{\rm where}\quad f(v):={c}_{n}|v|^{\frac{12}{n-6}}v.$

The operator $P^{3}_{\rm rad}:C^{6}(\mathbb{R})\rightarrow C(\mathbb{R})$ is
the so-called (sixth order) logarithmic cylindrical Paneitz operator
(restricted to radial functions), and is given by

$P^{3}_{\rm
rad}=\partial_{t}^{(6)}-K_{4}\partial_{t}^{(4)}+K_{2}\partial_{t}^{(2)}-K_{0},$
(2.1)

where

$\displaystyle K_{0}$ $\displaystyle=\frac{1}{64}(n-6)^{2}(n-2)^{2}(n+2)^{2},$
$\displaystyle K_{2}$
$\displaystyle=\frac{1}{16}(3n^{4}-24n^{3}+72n^{2}-96n+304),$ (2.2)
$\displaystyle K_{4}$ $\displaystyle=\frac{1}{4}(3n^{2}-12n+44).$

###  2.4. Decomposition

We prove a decomposition result for the tri-Laplacian operator written in
Emden–Fowler coordinates. Namely, we decompose $P^{3}_{\rm rad}$ into a
composition of three second order operators satisfying a comparison principle.
In what follows, given $\lambda>0$, we denote
$L_{\lambda}:=\partial_{t}^{(2)}-\lambda$.

We start with the definition of indicial roots for a linear operator.

######  Definition 2.5.

Let ${L}\in(C^{6}(\mathbb{R}))^{\prime}$ be a linear operator. The indicial
roots of $L$ at $+\infty$ $($resp. $-\infty$$)$ are $\lambda\in\mathbb{R}$ for
which there is a non-zero function $v\in{C}^{6}(\mathbb{R})$ and
$\lambda^{\prime}<\lambda$ $($resp. $\lambda^{\prime}>\lambda$$)$ such that
$e^{-\lambda^{\prime}t}L(e^{\lambda t}v(t))\rightarrow 0$ as
$t\rightarrow+\infty$ $($resp. $t\rightarrow-\infty$$)$. We denote by
$\mathcal{I}(L)\subset\mathbb{R}$ its set of indicial roots.

######  Remark 2.6.

Notice that when ${L}\in(C^{6}(\mathbb{R}))^{\prime}$ has constant
coefficients, that is, $L=\sum_{j=0}^{6}k_{j}\partial_{t}^{(j)}$, it follows
that $\mathcal{I}(L)=\\{p_{L}(0)\\}^{-1}$, where
$p_{L}(\lambda)=\sum_{j=0}^{6}k_{j}\lambda^{j}$ is the indicial polynomial. In
particular, a direct computation shows that the indicial equation associated
to (2.1) are given by

$p_{\lambda}(P^{3}_{\rm
rad})=\lambda^{6}-K_{4}\lambda^{4}+K_{2}\lambda^{2}-K_{0}=0,$ (2.3)

which have a strictly positive discriminant, that is,

${\rm
disc}_{\lambda}(p):=-27K_{0}^{2}-4K_{2}^{2}-4K_{4}^{2}+K_{4}^{2}K_{2}^{2}+18K_{4}K_{2}K_{0}>0.$
(2.4)

This in turns guarantees the existence of distinct real roots
$\pm\sqrt{\lambda_{1}},\pm\sqrt{\lambda_{2}},\pm\sqrt{\lambda_{3}}\in\mathbb{R}$
solving (2.3).

Next, we prove a decomposition of $P^{3}_{\rm rad}$ into three second order
operators.

######  Proposition 2.7.

The following decomposition holds

$P^{3}_{\rm rad}=L_{\lambda_{1}}\circ L_{\lambda_{2}}\circ L_{\lambda_{3}},$
(2.5)

where

$\lambda_{1}=\frac{n-6}{2},\quad\lambda_{2}=\frac{n-2}{2},\quad{\rm
and}\quad\lambda_{3}=\frac{n+2}{2}.$ (2.6)

###### Proof.

Initially, recall that for $q>6-n$, the following formula holds in the sense
of distributions,

$(-\Delta)^{3}(|x|^{q})=\prod_{j=0}^{3}(q-2j)\prod_{j=1}^{3}(q-2j+n)|x|^{q-6}\quad{\rm
in}\quad\mathbb{R}^{n}\setminus\\{0\\},$

which implies that $\\{u_{j}\\}_{j\in\\{1,\dots,6\\}}\subset
C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ is a basis for the space of radial
solutions to the homogeneous tri-harmonic equation, where
$u_{j}(|x|)=|x|^{q_{j}}$ with $q_{j}=2(j-1)$ if $j=1,2,3$ and $2(j-m)-n$ if
$j=4,3,5$. In other terms, $(-\Delta)^{3}u_{j}=0$ in
$\mathbb{R}^{n}\setminus\\{0\\}$ for any $j=1,2,3,4,5,6$. Hence, under the
Emden–Fowler change of variables, we have that
$\\{v^{\pm}_{1},v^{\pm}_{2},v^{\pm}_{3}\\}\subset C^{6}(\mathbb{R})$, where
$v^{\pm}_{j}(t)=e^{\pm\sqrt{\lambda_{j}}t}$ with
$\lambda_{j}=q_{j}-\gamma_{n}$ for $j=1,2,3$, forms a basis for the space of
solutions to ($\mathcal{O}_{6}$), that is,
$\\{v^{\pm}_{1},v^{\pm}_{2},v^{\pm}_{3}\\}\subset C^{6}(\mathbb{R})$ is
linearly independent and $P^{3}_{\rm rad}v^{\pm}_{1}=P^{3}_{\rm
rad}v^{\pm}_{2}=P^{3}_{\rm rad}v^{\pm}_{3}=0$ in $\mathbb{R}$. Then, the set
of indicial roots of $\mathcal{I}(P^{3}_{\rm
rad})=\\{\pm\sqrt{\lambda_{1}},\pm\sqrt{\lambda_{2}},\pm\sqrt{\lambda_{3}}\\}\subset\mathbb{R}$.
From this, it is straightforward to check that (2.5) holds. ∎

###  2.5. Conservation of energy

In this section, we find a quantity that is conserved along solutions to
($\mathcal{O}_{6}$). These will be called the Hamiltonian energy associated
with the sixth order ODE ($\mathcal{O}_{6}$).

Let us start with the classical definition of Hamiltonian energy for a
solution $v\in C^{6}(\mathbb{R})$ to ($\mathcal{O}_{6}$), which is obtained by
multiplying ($\mathcal{O}_{6}$) by $v^{(1)}$ and integrating by parts.

######  Definition 2.8.

For any real function $v\in C^{6}(\mathbb{R})$, let us define its Hamiltonian
energy $($with respect to ($\mathcal{O}_{6}$)$)$ $\mathcal{H}:\mathbb{R}\times
C^{6}(\mathbb{R})\rightarrow\mathbb{R}$ by

$\mathcal{H}(t,v):=\left(v^{(5)}v^{(1)}-v^{(4)}v^{(2)}+\frac{1}{2}{v^{(3)}}^{2}\right)-{K_{4}}\left(v^{(3)}v^{(1)}-\frac{1}{2}{v^{(2)}}^{2}\right)+\frac{K_{2}}{2}{v^{(1)}}^{2}-\frac{K_{0}}{2}v^{2}+F(v),$
(2.7)

where

$G(v):=F(v)-\frac{K_{0}}{2}v^{2}\quad{\rm and}\quad
F(v):=\hat{c}_{n}|v|^{\frac{2n}{n-6}}$

and

$\hat{c}_{n}:=\frac{(n-6)^{2}(n^{4}-20n^{2}+64)}{128}.$

######  Remark 2.9.

Notice that the real roots of $G$ are the equilibrium solutions to
($\mathcal{O}_{6}$), namely

$v\equiv 0\quad{\rm and}\quad v\equiv\pm a^{*}_{n},$ (2.8)

where

$a^{*}_{n}:=K_{0}^{\frac{n-6}{12}}=\left(\frac{(n-6)(n-2)(n+2)}{8}\right)^{\frac{n-6}{6}}.$
(2.9)

Next, it is direct to prove that this energy is conserved along with solutions
to ($\mathcal{O}_{6}$). We emphasize that this conservation is a local
property and valid on the maximal interval of existence and does not require
any a priori boundedness assumption.

######  Proposition 2.10.

If $v\in C^{6}(\mathbb{R})$ is a positive solution to ($\mathcal{O}_{6}$),
then

$\dfrac{\partial}{\partial t}\mathcal{H}(t,v)\equiv 0.$

In other terms, there exists $\mathcal{H}_{v}\in\mathbb{R}$ such that

$\mathcal{H}(v)(t)=\mathcal{H}(t,v)\equiv\mathcal{H}(v):=\mathcal{H}_{v}.$
(2.10)

###### Proof.

It is a direct computation. ∎

######  Remark 2.11.

It is convenient to write the Hamiltonian energy as

$\displaystyle\mathcal{H}(v)=\frac{1}{2}{v^{(3)}}^{2}+\mathcal{E}_{2}(v)v^{(2)}+\mathcal{E}_{1}(v)v^{(1)}+G(v),$
(2.11)

where

$\displaystyle\mathcal{E}_{1}(v):=v^{(5)}-{K_{4}}v^{(3)}+\frac{K_{2}}{2}v^{(1)}\quad{\rm
and}\quad\mathcal{E}_{2}(v):=-v^{(4)}+\frac{K_{4}}{2}v^{(2)}$ (2.12)

are the so-called auxiliary energy summand functions.

##  3\. Radial symmetry

In this section, using the integral moving spheres technique, we prove that
singular solutions to ($\mathcal{P}_{6,\infty}$) are radially symmetric. The
main ingredient in the proof is the integral version of the moving spheres
technique contained in [21, 22, 26, 23]. For this, we are based on the
conformal invariance enjoyed by solutions to ($\mathcal{P}_{6,\infty}$) in
Proposition 2.3.

######  Proposition 3.1.

Let $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ be a positive singular
solution to equation ($\mathcal{P}_{6,\infty}$). Then, $u$ is radially
symmetric about the origin and monotonically decreasing.

Before we give the proof of the rotational symmetry for singular solutions to
($\mathcal{P}_{6,\infty}$), we need to establish a set of preliminary results.

###  3.1. Integrability

We show that any positive singular solution to ($\mathcal{P}_{6,\infty}$) is
distributional.

######  Lemma 3.2.

Let $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ be a positive singular
solution to ($\mathcal{P}_{6,\infty}$). Then, it holds $u\in
L^{\frac{n+6}{n-6}}(\mathbb{R}^{n})$. In particular, $u$ is a distribution
solution to ($\mathcal{P}_{6,\infty}$).

###### Proof.

For any $0<\varepsilon\ll 1$, let us consider $\eta_{\varepsilon}\in
C^{\infty}(\mathbb{R}^{n})$ with $0\leqslant\eta_{\varepsilon}\leqslant 1$
satisfying

$\eta_{\varepsilon}(x)=\begin{cases}0,&\mbox{if}\ |x|\leqslant\varepsilon\\\
1,&\mbox{if}\ |x|\geqslant 2\varepsilon,\end{cases}$ (3.1)

and $|D^{(j)}\eta_{\varepsilon}|\leqslant C_{0}\varepsilon^{-j}$ in
$\mathbb{R}^{n}$ for $j=1,2,3,4,5,6$ and some $C_{0}>0$. Define
$\xi_{\varepsilon}=\left(\eta_{\varepsilon}\right)^{\frac{n+6}{2}}$.
Multiplying ($\mathcal{P}_{6,\infty}$) by $\xi_{\varepsilon}$, and integrating
by parts in $B_{r}$ with $r\in(1/2,1)$, we obtain

$\int_{\mathbb{R}^{n}}f(u)\xi_{\varepsilon}\mathrm{d}x=\int_{\mathbb{R}^{n}}u(-\Delta)^{3}\xi_{\varepsilon}\mathrm{d}x.$

One can verify that there exists $C_{1}>0$ such that

$\left|(-\Delta)^{3}\xi_{\varepsilon}\right|\leqslant
C_{1}\varepsilon^{-6}\eta_{\varepsilon}^{\frac{n-6}{2}}\chi_{\\{\varepsilon\leqslant|x|\leqslant
2\varepsilon\\}}=C_{1}\varepsilon^{-6}\xi_{\varepsilon}^{\frac{n-6}{n+6}}\chi_{\\{\varepsilon\leqslant|x|\leqslant
2\varepsilon\\}},$

which, by Hölder’s inequality, gives us

$\displaystyle\left|\int_{\mathbb{R}^{n}}u(-\Delta)^{3}\xi_{\varepsilon}\mathrm{d}x\right|$
$\displaystyle\leqslant
C_{1}\varepsilon^{-6}\int_{\\{\varepsilon\leqslant|x|\leqslant
2\varepsilon\\}}u\xi_{\varepsilon}^{\frac{n-6}{n+6}}\mathrm{d}x$
$\displaystyle\leqslant\tilde{C}_{1}\varepsilon^{-6}\varepsilon^{\frac{12n}{n+6}}\left(\int_{\\{\varepsilon\leqslant|x|\leqslant
2\varepsilon\\}}f(u)\xi_{\varepsilon}\mathrm{d}x\right)^{\frac{n-6}{n+6}}$
$\displaystyle\leqslant\hat{C}_{1}\left(\int_{\\{\varepsilon\leqslant|x|\leqslant
2\varepsilon\\}}f(u)\xi_{\varepsilon}\mathrm{d}x\right)^{\frac{n-6}{n+6}}.$

Thus, it follows

$\int_{\mathbb{R}^{n}}f(u)\xi_{\varepsilon}\mathrm{d}x\leqslant\hat{C}_{1}\left(\int_{\\{\varepsilon\leqslant|x|\leqslant
2\varepsilon\\}}f(u)\xi_{\varepsilon}\mathrm{d}x\right)^{\frac{n-6}{n+6}},$

from which one can find a constant $C_{2}>0$ (independent of $\varepsilon$)
such that

$\int_{\mathbb{R}^{n}}f(u)\xi_{\varepsilon}\mathrm{d}x\leqslant C_{2}.$

Now letting $\varepsilon\rightarrow 0$, we conclude that $u\in
L^{\frac{n+6}{n-6}}(\mathbb{R}^{n})$ and the integrability follows.

For any nonnegative $\phi\in C^{\infty}_{c}(\mathbb{R}^{n})$, we multiply
($\mathcal{P}_{6,\infty}$) by $\widetilde{\phi}=\eta_{\varepsilon}\phi$, where
$\eta_{\varepsilon}$ is given by (3.1). Then, using that $u\in
L^{\frac{n+6}{n-6}}(\mathbb{R}^{n})$ and integrating by parts, we get

$\int_{\mathbb{R}^{n}}u(-\Delta)^{3}\left(\eta_{\varepsilon}\phi\right)\mathrm{d}x=\int_{\mathbb{R}^{n}}f(u)\eta_{\varepsilon}\phi\mathrm{d}x.$
(3.2)

By a direct computation, we find that
$\Delta^{3}(\eta_{\varepsilon}\phi)=\eta_{\varepsilon}\Delta^{3}\phi+\psi_{\varepsilon}$,
where

$\psi_{\varepsilon}=6\nabla\eta_{\varepsilon}\nabla\Delta^{2}\phi-15\Delta\eta_{\varepsilon}\Delta^{2}\phi+20\nabla\Delta\eta_{\varepsilon}\nabla\Delta\phi-15\Delta^{2}\eta_{\varepsilon}\Delta\phi+6\nabla\Delta^{2}\eta_{\varepsilon}\nabla\phi-\phi\Delta^{3}\eta_{\varepsilon}.$

Furthermore, using Hölder’s inequality again, one can find $C_{3}>0$ such that

$\displaystyle\left|\int_{\mathbb{R}^{n}}u\psi_{\varepsilon}\mathrm{d}x\right|\leqslant
C_{3}\left(\int_{\\{\varepsilon\leqslant|x|\leqslant
2\varepsilon\\}}f(u)\mathrm{d}x\right)^{\frac{n-6}{n+6}}\rightarrow
0\quad\mbox{as}\quad\varepsilon\rightarrow 0.$

Finally, letting $\varepsilon\rightarrow 0$ in (3.2), and applying the
dominated convergence theorem the proof follows. ∎

###  3.2. Superharmonicity

We show that positive singular solutions to ($\mathcal{P}_{6,\infty}$) satisfy
a superharmonicity property for lower orders powers of the Laplacian.

######  Lemma 3.3.

Let $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ be a positive singular
solution to ($\mathcal{P}_{6,\infty}$). Then, $u$ is a superharmonic in the
distributional sense, that is, $\phi\in C^{\infty}_{c}(\mathbb{R}^{n})$, one
has

$\int_{\mathbb{R}^{n}}\Delta u\Delta^{2}\phi\mathrm{d}x\leqslant 0\quad{\rm
for\ all}\quad\phi\in C^{\infty}_{c}(\mathbb{R}^{n}).$

Moreover, in the classical sense

$-\Delta u\geqslant 0\quad{\rm in}\quad\mathbb{R}^{n}\setminus\\{0\\}.$ (3.3)

###### Proof.

Proceeding similarly to Lemma 3.2, one can prove that $u\in L_{\rm
loc}^{\frac{n+6}{n-6}}(\mathbb{R}^{n})$. Let $\eta_{\varepsilon}\in
C^{\infty}(\mathbb{R}^{n})$ be the cut-off function given by (3.1) and
$\phi\in C_{c}^{\infty}(\mathbb{R}^{n})$ be a nonnegative test function. Then,
by multiplying ($\mathcal{P}_{6,\infty}$) by $\eta_{\varepsilon}\phi$, and
integrating by parts, we get

$\displaystyle 0\geqslant-\int_{\mathbb{R}^{n}}\eta_{\varepsilon}\phi
f(u)\mathrm{d}x=\int_{\mathbb{R}^{n}}\Delta^{2}\left(\eta_{\varepsilon}\phi\right)\Delta
u\mathrm{d}x=\int_{\mathbb{R}^{n}}\Delta
u\left(\Delta^{2}\phi\eta_{\varepsilon}+\varsigma^{\varepsilon}\right)\mathrm{d}x,$

where
$\varsigma^{\varepsilon}:=4\langle\nabla\Delta\phi,\nabla\eta_{\varepsilon}\rangle+6\Delta\phi\Delta\eta_{\varepsilon}+4\langle\nabla\phi,\nabla\Delta\eta_{\varepsilon}\rangle+\phi\Delta^{2}\eta_{\varepsilon}$.
Notice that $\varsigma^{\varepsilon}(x)\equiv 0$ when
$|x|\leqslant\varepsilon$ or $|x|\geqslant 2\varepsilon$, and
$|\Delta\varsigma^{\varepsilon}(x)|\leqslant C_{1}\varepsilon^{-6}$, for some
$C_{1}>0$. In addition, since $n-6-\frac{n(n-6)}{n+6}>0$, the following
estimate holds

$\displaystyle\left|\int_{\mathbb{R}^{n}}\Delta
u\varsigma^{\varepsilon}\mathrm{d}x\right|$
$\displaystyle\leqslant\int_{\mathbb{R}^{n}}u|\Delta\varsigma^{\varepsilon}|\mathrm{d}x$
$\displaystyle\leqslant
C_{1}\varepsilon^{-6}\left(\int_{\\{\varepsilon\leqslant|x|\leqslant
2\varepsilon\\}}f(u)\mathrm{d}x\right)^{\frac{n-6}{n+6}}\varepsilon^{n\left(1-\frac{n-6}{n+6}\right)}$
$\displaystyle\leqslant C_{1}\varepsilon^{n-6-\frac{n(n-6)}{n+6}}\rightarrow
0\quad\mbox{as}\quad\varepsilon\rightarrow 0,$

which implies

$\displaystyle\int_{\mathbb{R}^{n}}\Delta
u\Delta\phi\mathrm{d}x=\lim_{\varepsilon\rightarrow
0}\int_{\mathbb{R}^{n}}\left(\Delta
u\Delta^{2}\varsigma^{\varepsilon}\mathrm{d}x+\eta_{\varepsilon}\Delta\phi\Delta
u\right)\mathrm{d}x=\int_{\mathbb{R}^{n}}\phi f(u)\mathrm{d}x\leqslant 0.$

Thus, $-\Delta u$ is superharmonic in the whole space $\mathbb{R}^{n}$ in the
distributional sense, which gives the first part of the proof.

To prove the second statement, given $0<\varepsilon\ll 1$, let us consider
$\widetilde{u}^{\varepsilon}:=-\Delta u+\varepsilon$. Using Lemma 3.2, there
exists a constant $C_{2}>0$, depending only on $n$ and $s$, such that for all
$|x|\geqslant 4$, it holds

$\sum_{j=0}^{5}|x|^{\gamma_{n}+j}|D^{(j)}u(x)|\leqslant C_{2},$

which yields that $\lim_{|x|\rightarrow\infty}|\Delta u(x)|=0$. Whence, for
any $0<\varepsilon\ll 1$, there exists $R_{\varepsilon}\gg 1$ such that
$\widetilde{u}^{\varepsilon}>{\varepsilon}/{2}$ for $|x|\geqslant
R_{\varepsilon}$. Finally, using that $\widetilde{u}^{\varepsilon}$ is
superharmonic in $\mathbb{R}^{n}$ in the distributional sense, we have that
$\widetilde{u}^{\varepsilon}\geqslant 0$ in $\mathbb{R}^{n}\setminus\\{0\\}$,
which, by passing to the limit as $\varepsilon\rightarrow 0$, provides
$-\Delta u\geqslant 0$ in $\mathbb{R}^{n}\setminus\\{0\\}$. The last
inequality concludes the proof of the lemma. ∎

######  Lemma 3.4.

Let $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ be a positive singular
solution to ($\mathcal{P}_{6,\infty}$). Then, $-\Delta u$ is a superharmonic
in the distributional sense, that is, $\phi\in
C^{\infty}_{c}(\mathbb{R}^{n})$, one has

$\int_{\mathbb{R}^{n}}\Delta^{2}u\Delta\phi\mathrm{d}x\geqslant 0\quad{\rm
for\ all}\quad\phi\in C^{\infty}_{c}(\mathbb{R}^{n}).$

Moreover, in the classical sense

$\Delta^{2}u\geqslant 0\quad{\rm in}\quad\mathbb{R}^{n}\setminus\\{0\\}.$
(3.4)

###### Proof.

The proof is similar to the one of Lemma 3.3, so we omit it. ∎

######  Remark 3.5.

Notice that these superharmonicity identities in (3.3) and (3.3) can be
respectively rewritten in radial Emden–Fowler coordinates $($see Subsection
2.3$)$ as follows

$v^{(2)}+(n-5)v^{(1)}+\frac{(n-6)(n-4)}{4}v<0$

and

$v^{(4)}+2(n-3)v^{(3)}-{(3n^{2}-18n+22)}v^{(2)}+{(n^{3}-9n^{2}+22n-12)}v^{(1)}+\frac{(n-6)(n-4)(n-2)}{16}v>0.$

In addition, we have the decomposition

$v^{(4)}-{(\nu^{+}_{n}+\nu^{-}_{n})}v^{(2)}+(\nu^{+}_{n}\nu^{-}_{n})v>0,$

where

$\nu^{\pm}:=\sqrt{5\pm\sqrt{2n^{4}-24n^{3}+88n^{2}-96n+16}}.$

###  3.3. Integral moving spheres method

We prove the main result of this subsection, namely, the radial symmetry of
the blow-up limit solutions.

######  Lemma 3.6.

Let $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ be a positive singular
solution to ($\mathcal{P}_{6,\infty}$). For any $x\in\mathbb{R}^{n}$,
$z\in\mathbb{R}^{n}\setminus\left(\\{0\\}\cup B_{\mu}(x)\right)$ and
$\mu\in(0,1)$, it holds that $u(z)-(u)_{x,\mu}(z)>0$.

###### Proof.

Let $u$ be a positive singular solution to ($\mathcal{P}_{6,\infty}$). Using
the identities in [26, page 162], one has

$\left(\frac{\mu}{|z-x|}\right)^{n-6}\int_{|y-x|\geqslant\mu}{\left|\mathcal{I}_{x,\mu}(z)-y\right|^{n-6}}{f(u(y))}\mathrm{d}y=\int_{|y-x|\leqslant\mu}\left|z-y\right|^{n-6}{f(u(y))}\mathrm{d}y$

and

$\left(\frac{\mu}{|z-x|}\right)^{n-6}\int_{|y-x|\leqslant\mu}{\left|\mathcal{I}_{x,\mu}(z)-y\right|^{n-6}}{f(u(y))}\mathrm{d}y=\int_{|y-x|\geqslant\mu}\left|z-y\right|^{n-6}{f(u(y))}\mathrm{d}y,$

which yields

${(u)}_{x,\mu}(z)=\int_{\mathbb{R}^{n}}\left|z-y\right|^{n-6}{f(u(y))}\mathrm{d}y\quad\mbox{for}\quad
z\in\mathcal{I}_{x,\mu}(\mathbb{R}^{n}).$ (3.5)

Consequently, for any $x\in\mathbb{R}^{n}$ and $\mu<1$, we have that for
$z\in\mathbb{R}^{n}\setminus\\{0\\}\cup B_{\mu}(x)$,

$u(z)-(u)_{x,\mu}(z)=\int_{|y-x|\geqslant\mu}E(x,y,\mu,z)\left[f(u(y))-f(u_{x,\mu}(y))\right]\mathrm{d}y,$

where

$E(x,y,z,\mu):={|z-y|^{6-n}}-\left(\frac{|z-x|}{\mu}\right)^{6-n}{\left|\mathcal{I}_{x,\mu}(z)-y\right|^{6-n}}$
(3.6)

is used to estimate the difference between $u$ and its Kelvin transform
$u_{x,\mu}$. Finally, using its decay properties, it is straightforward to
check that $E(x,y,z,\mu)>0$ for all $|z-x|>\mu>0$, which concludes the proof.
∎

Next, let us introduce the critical sliding parameter as the supremum for
which an inequality relating a component function and its Kelvin transform is
satisfied.

######  Definition 3.7.

Given $x\in\mathbb{R}^{n}$, let us define

$\mu_{*}(x)=\sup\left\\{\mu>0:(u)_{x,r}\leqslant u\ {\rm in}\
\mathbb{R}^{n}\setminus B_{r}(x)\ {\rm for\ any}\ 0<r<\mu\right\\}.$ (3.7)

Since $u$ is such that $-\Delta u>0$ and $\Delta^{2}u>0$, we get
$\mu^{*}(x)>0$.

The next lemma is essentially the moving spheres technique in its integral
form. This method provides the exact form for any blow-up limit solution to
($\mathcal{P}_{6,\infty}$), which depends on whether the critical sliding
parameter $\mu^{*}(x)$ is finite or infinite.

######  Lemma 3.8.

Let $u\in C^{6}(\mathbb{R}^{n}\setminus\\{0\\})$ be a positive singular
solution to ($\mathcal{P}_{6,\infty}$), $z\in\mathbb{R}^{n}$ and
$\mu^{*}(z)>0$ given by (3.7). The following holds:

  * (i)

if $\mu^{*}(x)<\infty$ is finite, then $u_{x,\mu^{*}(x)}=u$ in
$\mathbb{R}^{n}\setminus\\{0,x\\}$.

  * (ii)

if $\mu^{*}(x_{0})=\infty$, for some $x_{0}\in\mathbb{R}^{n}$, then
$\mu^{*}(x)=\infty$ for all $x\in\mathbb{R}^{n}\setminus\\{0\\}$.

###### Proof.

Without loss of generality, we may assume $x_{0}=0$. Let us fix
$\mu^{*}=\mu^{*}(0)$ and $(u)_{\mu}=(u)_{0,\mu}$. By the definition of
$\mu^{*}(x)>0$, we have $(u)_{\mu^{*}}(x)\leqslant u(x)$ for all
$|x|\geqslant\mu^{*}$. Thus, by (3.5), with $x=0$ and
$\mu=|x|\geqslant\mu^{*}$, and the positivity of the kernel $E(0,y,z,\mu)$
given by (3.6), either $u_{\mu^{*}}(y)=u(y)$ for all $|x|\geqslant\mu^{*}$ or
$(u)_{\mu^{*}}(y)<(u)(y)$ for all $|x|>\mu^{*}$. In the former case, the
conclusion easily follows. In the sequel, we assume that the last condition
holds. Hence, the integral representation in Proposition 2.1 yields

$\displaystyle\liminf_{|z|\rightarrow\infty}|z|^{n-6}\left[u(z)-(u)_{\mu^{*}}(z)\right]$
$\displaystyle=\liminf_{|z|\rightarrow\infty}\int_{|y|\geqslant\mu^{*}}|z|^{n-6}E(0,y,z,\mu^{*})\left[{f}(u(y))-{f}(u_{\mu^{*}}(y))\right]\mathrm{d}y$
$\displaystyle\geqslant\int_{|y|\geqslant\mu^{*}}\left(1-\left(\frac{\mu^{*}}{|y|}\right)^{n-6}\right)\left[{f}(u(y))-{f}(u_{\mu^{*}}(y))\right]\mathrm{d}y>0,$

which implies that there exists $\varepsilon_{1}\in(0,1)$ satisfying
$u(z)-(u)_{\mu^{*}}(z)\geqslant{\varepsilon_{1}}{|z|^{6-n}}$ for all
$|z|\geqslant\mu^{*}+1$. Moreover, there exists
$\varepsilon_{2}\in(0,\varepsilon_{1})$ such that, for $|z|\geqslant\mu^{*}+1\
\mbox{and}\ \mu^{*}\leqslant\mu\leqslant\mu^{*}+\varepsilon_{2}$, we find

$\left(u-(u)_{\mu^{*}}\right)(z)\geqslant{\varepsilon_{1}}{|z|^{6-n}}+\left((u)_{\mu^{*}}-(u)_{\mu}\right)(z)\geqslant\frac{\varepsilon_{1}}{2}|z|^{6-n}.$
(3.8)

Whence, for any $\varepsilon\in\left(0,\varepsilon_{2}\right)$ (to be chosen
later), $\mu^{*}\leqslant\mu\leqslant\mu^{*}+\varepsilon$, and
$\mu\leqslant|y|\leqslant\mu^{*}+1$, we have

$\displaystyle\left(u-(u)_{\mu^{*}}\right)(z)$
$\displaystyle=\int_{|y|\geqslant\mu}E(0,y,z,\mu)\left[{f}(u(y))-{f}(u_{\mu}(y))\right]\mathrm{d}y$
$\displaystyle\geqslant\int_{\mu^{*}\leqslant|y|\leqslant\mu^{*}+1}E(0,y,z,\mu)\left[{f}(u_{\mu^{*}}(y))-{f}(u_{\mu}(y))\right]\mathrm{d}y$
$\displaystyle+\int_{\mu^{*}+2\leqslant|y|\leqslant\mu^{*}+3}E(0,y,z,\mu)\left[{f}(u(y))-{f}(u_{\mu}(y))\right]\mathrm{d}y.$

Now using (3.8), there exists $\delta_{1}>0$ such that
${f}(u(y))-{f}(u_{\mu}(y))\geqslant\delta_{1}$ for
$\mu^{*}+2\leqslant|y|\leqslant\mu^{*}+3$. Since $E(0,y,z,\mu)=0$ for all
$|z|=\mu$ and

$\nabla_{z}E(0,y,z,\mu)\cdot
z\big{|}_{|z|=\mu}=(n-4)|z-y|^{8-n}\left(|z|^{2}-|y|^{2}\right)>0\quad\mbox{for
all}\quad\mu^{*}+2\leqslant|y|\leqslant\mu^{*}+3,$

where $\delta_{2}>0$ is a constant independent of $\varepsilon$. Then, there
exists $C_{1}>0$ such that, for
$\mu^{*}\leqslant\mu\leqslant\mu^{*}+\varepsilon$, we get

$\left|{f}(u(y))-{f}(u_{\mu^{*}}(y))\right|\leqslant
C_{1}(\mu-\mu^{*})\leqslant C_{1}\varepsilon\quad\mbox{for
all}\quad\mu^{*}\leqslant\mu\leqslant|y|\leqslant\mu^{*}+1.$

Furthermore, recalling that $\mu\leqslant|z|\leqslant\mu^{*}+1$, one can find
$C_{2}>0$ such that

$\displaystyle\int_{\mu\leqslant|y|\leqslant\mu^{*}}E(0,y,z,\mu)\mathrm{d}y$
$\displaystyle\leqslant\left|\int_{\mu\leqslant|y|\leqslant\mu^{*}+1}\left[{|y-z|^{6-n}}-{\left|\mathcal{I}_{\mu}(z)-y\right|^{8-n}}+\left(\frac{\mu}{|z|}-1\right)^{n-6}\left|\mathcal{I}_{\mu}(z)-y\right|^{n-4}\right]\mathrm{d}y\right|$
$\displaystyle\leqslant
C_{2}\left|\mathcal{I}_{\mu}(z)-z\right|+C_{2}(|z|-\mu)\leqslant
C_{2}(|z|-\mu),$

which, provides that for small $0<\varepsilon\ll 1$,
$\mu^{*}\leqslant\mu\leqslant\mu^{*}+\varepsilon$, and
$\mu\leqslant|z|\leqslant\mu^{*}+1$, it follows

$\displaystyle\left(u-(u)_{\mu^{*}}\right)(z)$ $\displaystyle\geqslant-
C_{2}\varepsilon\int_{\mu\leqslant|y|\leqslant\mu^{*}+1}E(0,y,z,\mu)\mathrm{d}y+\delta_{1}\delta_{2}(|z|-\mu)\int_{\mu^{*}+2\leqslant|z|\leqslant\mu^{*}+3}\mathrm{d}y$
$\displaystyle\geqslant\left(\delta_{1}\delta_{2}\int_{\mu^{*}+2\leqslant|y|\leqslant\mu^{*}+3}\mathrm{d}z-C_{2}\varepsilon\right)(|z|-\mu)$
$\displaystyle\geqslant 0.$

This is a contradiction to the definition of $\mu^{*}(x)>0$. Therefore, the
first part of the lemma is established.

Next, by the definition of $\mu^{*}(x)>0$, we know that $u_{x,\mu}(z)\leqslant
u(z)$ for all $0<\mu<\mu^{*}(x)$, with $|z-x|\geqslant\mu$; thus, multiplying
it by $|z|^{n-6}$, and taking the limit as $|z|\rightarrow\infty$, yields

$\ell=\liminf_{|z|\rightarrow\infty}|z|^{n-6}|\mathcal{U}(z)|\geqslant\mu^{n-6}|\mathcal{U}(z)|\quad\mbox{for
all}\quad 0<\mu<\mu^{*}(x).$ (3.9)

On the other hand, if $\mu^{*}(x_{0})<\infty$, multiplying the identity
obtained in (i) by $|z|^{n-6}$ and passing to the limit when
$|z|\rightarrow\infty$, we obtain

$\ell=\lim_{|z|\rightarrow\infty}|z|^{n-6}f(u(z))=\mu^{*}(x_{0})^{n-6}f(x_{0})<\infty.$
(3.10)

Finally, by (3.9) and (3.10), if there exists $x_{0}\in\mathbb{R}^{n}$ such
that $\mu^{*}(x_{0})<\infty$, then $\mu^{*}(x)<\infty$ for all
$x\in\mathbb{R}^{n}$. ∎

###### Proof of Proposition 3.1.

Using Lemma 3.8, we can apply [21, Proposition 2.1] to conclude that positive
singular solutions to ($\mathcal{P}_{6,\infty}$) are radially symmetric with
respect to the origin. The proof that they are radially monotonically
decreasing will be given in the next section (See Proposition 4.1 (iii)). ∎

##  4\. ODE analysis

This section is devoted to the classification of (positive) solutions to the
problem ($\mathcal{O}_{6}$). The positivity will not play a role here. Our
strategy here is inspired by the methods in [37, 15, 17] and relies on the
decomposition in Proposition 2.7. It is outstanding to recover such properties
for solutions to sixth order ODE with possibly chaotic dynamical behavior near
equilibrium points.

The next classification result will be the key part of the proof of our main
theorem

######  Proposition 4.1.

Let $v\in C^{6}(\mathbb{R})$ be a solution to ($\mathcal{O}_{6}$).

  1. (i)

Then,

$\inf_{\mathbb{R}}|v|\leqslant a^{*}_{n}$ (4.1)

with equality holding if, and only if, $v$ is a nonzero constant function.

  2. (ii)

Conversely, if $a_{0}\in(0,a^{*}_{n})$, then there exists a unique $($up to
translations$)$ bounded solution $v\in C^{6}(\mathbb{R})$ to
($\mathcal{O}_{6}$) such that $\inf_{\mathbb{R}}|v|=a_{0}$. This solution is
periodic, has a unique local maximum and minimum per period, and is symmetric
with respect to its local extrema.

  3. (iii)

Moreover, if $v$ is positive, then $v^{(1)}-\gamma_{n}v<0$ in $\mathbb{R}$.

We begin with some preliminary remarks.

######  Remark 4.2.

Notice that $v\equiv 0$ and $v\equiv a^{*}_{n}$, where $a^{*}_{n}>0$ is given
by (2.9) are exactly the only three constant solutions to ($\mathcal{O}_{6}$).
Moreover, if $v(\cdot)$ is a solution to ($\mathcal{O}_{6}$),

  * •

then $v(-\cdot)$ also is a solution to ($\mathcal{O}_{6}$) since it contains
only even order derivatives;

  * •

then $-v(\cdot)$ also is a solution to ($\mathcal{O}_{6}$) since $G$ is odd;

  * •

then $v(\cdot+T)$ for any $T\in\mathbb{R}$ is also a solution to
($\mathcal{O}_{6}$) since it is autonomous.

To prove our main proposition in this section, we will need some auxiliary
results quantifying the intuition that the set of bounded solutions to the
higher order equation ($\mathcal{O}_{6}$) behaves in some respects similar to
the set of solutions to a second order equation. As we pointed out before, for
this it is crucial that the relation (2.4) holds.

###  4.1. Boundedness

The following lemma states that solutions to ($\mathcal{O}_{6}$) are bounded,
which is one of the key new results in this manuscript. The proof is based
solely on the conservation of energy, as so is independent of the rest of the
argument.

######  Lemma 4.3.

Let $v\in C^{6}(\mathbb{R})$ be a solution to ($\mathcal{O}_{6}$). Then $v$ is
bounded.

###### Proof.

Initially, by transforming $\tau=-t$, without loss of generality it is enough
to prove that $v$ is bounded on $[0,\infty)$.

Indeed, for any solution $v\in C^{6}(\mathbb{R})$, we set

$\mathcal{Z}_{+}(v)=Z_{+}^{(1)}(v)\cap Z_{+}^{(2)}(v),$ (4.2)

where

$Z_{+}^{(1)}(v)=\\{t\geqslant 0:v^{(1)}(t)=0\\}\quad{\rm and}\quad
Z_{+}^{(2)}(v)=\\{t\geqslant 0:v^{(2)}(t)=0\\}.$

In what follows, we have two cases to study:

Case 1. $\mathcal{Z}_{+}(v)$ is bounded.

In this case, using Lemma 4.8, we get that $v$ is monotone for $t\gg 1$ large
enough, and thus admits a finite limit
$a_{+}:=\lim_{t\rightarrow\infty}v(t)<\infty$. Therefore, $v$ is bounded on
$[0,\infty)$.

Next, we show that the other possibility cannot happen.

Case 2. $\mathcal{Z}_{+}(v)$ is unbounded.

Now, since $F(v)\rightarrow\infty$ as $|v|\rightarrow\infty$, there exists an
$R>|v(0)|$ such that $F(v)>H_{v}$ for all $|v|\geqslant R$. Also, notice that
$|v|<R$ in $[0,\infty)$. Indeed, by contradiction assume that
$M_{R}:=\\{t\geqslant 0:|v(t)|\geqslant R\\}\neq\varnothing$. We set
$t^{*}:=\inf_{\mathbb{R}_{+}}M_{R}$.

In addition, because $|v(0)|<R$, we must have $t^{*}>0$ and $|v(t^{*})|=R$.
Replacing $v(t)$ by $-v(t)$ if necessary and observing that it does not change
the set $\mathcal{Z}_{+}(v)$, we may assume that $v(t^{*})=R$. Hence, we also
find $v^{(1)}(t^{*})\geqslant 0$ and $v^{(2)}(t^{*})\geqslant 0$. Using that
$\mathcal{Z}_{+}(v)$ is unbounded, we have that
$-\infty<T:=\inf(\mathcal{Z}_{+}(v)\cap(t^{*},\infty))$ is well-defined,
which, by continuity of $v^{(1)}$ and $v^{(2)}$, yields
$v^{(1)}(T)=v^{(2)}(T)=0$ and $\min\\{v^{(1)},v^{(2)}\\}\geqslant 0$ in
$[t^{*},T]$. Consequently, we get $v(T)\geqslant v(t^{*})=R$, from which we
deduce

$\mathcal{H}(v)(T)=\frac{1}{2}v^{(3)}(T)^{2}+F(v(T))\geqslant
F(v(T))>\mathcal{H}_{v},$

which is a contradiction with the energy conservation (2.10) in Proposition
2.10.

The proof is finished. ∎

###  4.2. Comparison

The first result states that any bounded entire solution to
($\mathcal{O}_{6}$) is uniquely determined by only two (instead of $6$)
initial values.

######  Lemma 4.4.

Let $v_{1},v_{2}\in C^{6}(\mathbb{R})$ be bounded solutions to
($\mathcal{O}_{6}$). If $v_{1}(0)=v_{2}(0)$ and
$v_{1}^{(1)}(0)=v_{2}^{(1)}(0)$. Then $v_{1}\equiv v_{2}$.

###### Proof.

Let $v_{1},v_{2}\in C^{6}(\mathbb{R})$ be bounded solutions to
($\mathcal{O}_{6}$) which satisfy the initial conditions $v_{1}(0)=v_{2}(0)$
and $v_{1}^{(1)}(0)=v_{2}^{(1)}(0)$. By interchanging $v_{1}$ and $v_{2}$ or
replacing $v_{1}(t)$ and $v_{2}(t)$ by $v_{1}(-t)$ and $v_{2}(-t))$, we may
assume without loss of generality that $v_{1}^{(j)}(0)\geqslant
v_{2}^{(j)}(0)$ for $j=2,3,4,5$.

Suppose, by contradiction, that $v_{1}\not\equiv v_{2}$. Then, by standard
uniqueness result for ODEs, we have that $v_{1}^{(j)}(0)\neq v_{2}^{(j)}(0)$
for $j=2,3,4,5$. In both cases, we deduce from our hypotheses on the initial
conditions $v(t)>w(t)$ on $(0,\tau)$ for some sufficiently small $0<\tau\ll
1$.

Let $\\{\lambda_{1},\lambda_{2},\lambda_{3}\\}\subset\mathbb{R}$ be given by
(2.6). Using the decomposition in Proposition 2.7, let us set the auxiliary
functions $\\{\Phi_{1},\Phi_{2},\Phi_{3}\\}\subset C^{6}(\mathbb{R})$ defined
as

$\Phi_{1}:=L_{\lambda_{1}},\quad\Phi_{2}:=L_{\lambda_{1}}\circ
L_{\lambda_{2}},\quad\mbox{and}\quad\Phi_{3}:=L_{\lambda_{1}}\circ
L_{\lambda_{2}}\circ L_{\lambda_{3}}.$ (4.3)

By the contradiction assumption, we have

$(\Phi_{2}(v_{1})-\Phi_{2}(v_{2}))(0)\geqslant 0\quad\text{ and
}\quad(\Phi_{2}(v_{1})-\Phi_{2}(v_{2}))^{(1)}(0)\geqslant 0$ (4.4)

and

$(\Phi_{1}(v_{1})-\Phi_{1}(v_{2}))(0)\geqslant 0\quad\text{ and
}\quad(\Phi_{1}(v_{1})-\Phi_{1}(v_{2}))^{(1)}(0)\geqslant 0.$ (4.5)

Also, using that $v_{1}$ and $v_{2}$ satisfy ($\mathcal{O}_{6}$), we find

$-L_{\lambda_{3}}(\Phi_{2}(v_{1})-\Phi_{2}(v_{2}))=f(v_{2}(t))-f(v_{1}(t))\quad\text{for
all}\quad t\in\mathbb{R}.$

Next, using that $v_{1}(t)>v_{2}(t)$ on $(0,\tau)$ and the strictly
monotonicity of since the function $u\mapsto f(u)$, we get

$-L_{\lambda_{3}}(\Phi_{2}(v_{1})-\Phi_{2}(v_{2}))(t)>0\quad\text{ for all
}\quad t\in(0,\tau).$ (4.6)

Hence, from (4.4) and (4.6), we easily find that
$(\Phi_{2}(v_{1})-\Phi_{2}(v_{2}))(t)\geqslant 0$ for $t\in(0,\tau)$, or
equivalently,

$-L_{\lambda_{2}}(\Phi_{1}(v_{1})-\Phi_{1}(v_{2}))(t)=\Phi_{2}(v_{1})-\Phi_{2}(v_{2})(t)\geqslant
0\quad\text{ for all }\quad t\in(0,\tau).$ (4.7)

Whence, from (4.5) and (4.7), we easily find that
$(\Phi_{1}(v_{1})-\Phi_{1}(v_{2}))(t)\geqslant 0$ for $t\in(0,\tau)$, or
equivalently,

$(v_{1}-v_{2})^{(2)}(t)\geqslant\lambda_{1}(v_{1}-v_{2})(t)>0\quad\text{ for
all }\quad t\in(0,\tau).$ (4.8)

By the hypotheses, we have $(v_{1}-v_{2})^{(1)}(0)\geqslant 0$, which combined
with $\lambda_{3}>0$ and (4.8) implies that $(v_{1}-v_{2})^{(1)}(t)>0$ for all
$t\in(0,\tau)$. Thus, $v_{1}-v_{2}$ is strictly increasing on $(0,\tau)$, and
since $0<\tau\ll 1$ was arbitrarily small such that $v_{1}-v_{2}>0$ on
$(0,\tau)$, we conclude that $v_{1}-v_{2}>0$ for all $t\in\mathbb{R}$.

Repeating the above arguments for the interval $(0,\infty)$ instead of
$(0,\tau)$, we get from (4.8) that $(v_{1}-v_{2})^{(1)}$ is positive and
strictly increasing on $(0,\infty)$, which clearly contradicts the boundedness
of $v_{1}-v_{2}$. This shows that $v_{1}\equiv v_{2}$, and concludes the proof
of this lemma. ∎

As a consequence, we have the following symmetry result

######  Corollary 4.5.

Let $v\in C^{6}(\mathbb{R})$ be a bounded solution to ($\mathcal{O}_{6}$).

  * (i)

If $v^{(1)}(t_{0})=0$ for some $t_{0}\in\mathbb{R}$, then $v$ is symmetric
with respect to $t_{0}$, that is $v(t_{0}+t)=v(t_{0}-t)$ for all
$t\in\mathbb{R}$;

  * (ii)

If $v(t_{0})=0$ for some $t_{0}\in\mathbb{R}$, then $v$ is anti-symmetric with
respect to $t_{0}$, that is, $v(t_{0}-t)=-v(t_{0}+t)$ for all
$t\in\mathbb{R}$.

###### Proof.

To prove (i), notice that since ($\mathcal{O}_{6}$) is autonomous, we may
assume $t_{0}=0$. Moreover, if $v$ is a solution to ($\mathcal{O}_{6}$), then
$\hat{v}(t):=v(-t)$ also is. Hence $v(0)=\hat{v}(0)$ and, by assumption,
$v^{(1)}(0)=\hat{v}^{(1)}(0)=0$, which by Lemma 4.4 gives $v\equiv\hat{v}$.

To prove (ii), we use the same strategy. ∎

With a similar proof, we can also prove the following comparison

######  Lemma 4.6.

Let $v_{1},v_{2}\in C^{6}(\mathbb{R})$ be bounded solutions to
($\mathcal{O}_{6}$). Suppose that

$\displaystyle v_{1}(0)$ $\displaystyle\geqslant v_{2}(0)$ $\displaystyle
v_{1}^{(1)}(0)$ $\displaystyle\geqslant v_{2}^{(1)}(0)$ $\displaystyle
L_{\lambda_{1}}(v_{1}(0))$ $\displaystyle\geqslant L_{\lambda_{1}}(v_{2}(0))$
$\displaystyle L_{\lambda_{1}}^{(1)}(v_{1}(0))$ $\displaystyle\geqslant
L_{\lambda_{1}}^{(1)}(v_{2}(0))$ $\displaystyle(L_{\lambda_{1}}\circ
L_{\lambda_{2}})(v_{1}(0))$ $\displaystyle\geqslant(L_{\lambda_{1}}\circ
L_{\lambda_{2}})(v_{2}(0))$ $\displaystyle(L_{\lambda_{1}}\circ
L_{\lambda_{2}})^{(1)}(v_{1}(0))$ $\displaystyle\geqslant(L_{\lambda_{1}}\circ
L_{\lambda_{2}})^{(1)}(v_{2}(0)).$

Then $v_{1}\equiv v_{2}$.

###### Proof.

Follows the same strategy as in Lemma 4.4 ∎

###  4.3. Asymptotics

This subsection is devoted to the study of some asymptotic properties for
solutions to ($\mathcal{O}_{6}$).

First, it is convenient to introduce the following definition

######  Definition 4.7.

For any $v\in C^{6}(\mathbb{R})$ solution to ($\mathcal{O}_{6}$), let us
define its asymptotic set by

$\mathcal{A}(v):=\left\\{\ell\in\mathbb{R}\cup\\{\pm\infty\\}:\lim_{t\rightarrow\pm\infty}v(t)=\ell\right\\}.$

In other words, $\mathcal{A}(v)$ is the set of all possible limits at infinity
of $v$.

The content of the following lemma shows that ($\mathcal{O}_{6}$) does not
possess a solution that tends to either plus or minus infinity at infinity,
that is, a solution that blows up does it in finite time.

######  Lemma 4.8.

Let $v\in C^{6}(\mathbb{R})$ be a solution to ($\mathcal{O}_{6}$). If
$\ell_{\pm}:=\lim_{t\rightarrow\pm\infty}v(t)\in\mathbb{R}\cup\\{\pm\infty\\}$
exists, then $\ell_{\pm}\in\mathbb{R}$.

###### Proof.

Initially, notice that by interchanging $v(t)$ by $v(-t)$, it is only
necessary to consider the case $t\rightarrow+\infty$.

Suppose by contradiction that the lemma does not hold. Thus, one of the
following two possibilities shall happen: either the asymptotic limit of $v$
is a finite constant $\ell_{+}>0$, which does not belong to the asymptotic set
$\mathcal{A}$, or the limit blows up, that is, $\ell_{+}=+\infty$.

Let us consider these two cases separately:

Case 1: $\ell_{+}\in[0,\infty)\setminus\\{0,a^{*}_{n}\\}$.

By assumption, we have

$\displaystyle\lim_{t\rightarrow\infty}\left(K_{0}v(t)-f(v(t))\right)=\kappa,\quad\hbox{where}\quad\kappa:=K_{0}\ell_{+}-f(\ell_{+})\neq
0,$ (4.9)

which implies

$K_{0}v(t)-f(v(t))=v^{(6)}(t)-K_{4}v^{(4)}(t)+K_{2}v^{(2)}(t).$ (4.10)

Hence, combining (4.9) with (4.10) implies that for any $\varepsilon>0$ there
exists $T\gg 1$ sufficiently large satisfying

$\kappa-\varepsilon<v^{(6)}(t)-K_{4}v^{(4)}(t)+K_{2}v^{(2)}(t)<\kappa+\varepsilon.$
(4.11)

Now, integrating (4.11), we obtain

$\int_{T}^{t}(\kappa-\varepsilon)\mathrm{d}\tau<\int_{T}^{t}\left[v^{(6)}(\tau)-K_{4}v^{(4)}(\tau)+K_{2}v^{(2)}(\tau)\right]\mathrm{d}\tau<\int_{T}^{t}(\kappa+\varepsilon)\mathrm{d}\tau,$

which yields

$(\kappa-\varepsilon)(t-T)+C_{1}(T)<v^{(5)}(t)-K_{4}v^{(3)}(t)+K_{2}v^{(1)}(t)<(\kappa+\varepsilon)(t-T)+C_{1}(T),$
(4.12)

for some $C_{1}(T)>0$. Defining $\delta:=\sup_{t\geqslant
T}|v(t)-v(T)|<\infty$, we obtain

$\left|\int_{T}^{t}K_{4}v^{(3)}(\tau)\mathrm{d}\tau\right|\leqslant|K_{4}|\delta\quad{\rm
and}\quad\left|\int_{T}^{t}K_{2}v^{(1)}(\tau)\mathrm{d}\tau\right|\leqslant|K_{2}|\delta,$

which, by integrating $\eqref{bound5}$, gives us

$\frac{(\kappa-\varepsilon)}{2}(t-T)^{2}+L(t)<v^{(4)}(t)<\frac{(\kappa+\varepsilon)}{2}(t-T)^{2}+R(t),$
(4.13)

where $L(t),R(t)\in\mathcal{O}(t^{2})$, namely

$L(t)=C_{1}(T)(T-t)-|K_{4}|\delta-|K_{2}|\delta+C_{2}(T)$

and

$R(t)=C_{1}(T)(T-t)+|K_{4}|\delta+|K_{2}|\delta+C_{2}(T),$

for some $C_{2}(T)>0$.

Then, repeating the same integration procedure in (4.13), we find

$\frac{(\kappa-\varepsilon)}{6!}(t-T)^{6}+\mathcal{O}(t^{6})<v(t)<\frac{(\kappa+\varepsilon)}{6!}(t-T)^{6}+\mathcal{O}(t^{6})\quad\mbox{as}\quad
t\rightarrow\infty.$ (4.14)

Therefore, since $\kappa\neq 0$ we can choose $0<\varepsilon\ll 1$
sufficiently small such that $\kappa-\varepsilon$ and $\kappa+\varepsilon$
have the same sign. Finally, by passing to the limit as $t\rightarrow\infty$
on inequality (4.14), we obtain that $v$ blows-up and $\ell_{+}=\infty$, which
is contradiction. This concludes the proof of the first case.

The second case requires a suitable choice of test functions, which is
inspired in [31].

Case 2: $\ell_{+}=+\infty$.

Let $\phi_{0}\in C^{\infty}({[0,\infty]})$ be a nonnegative function
satisfying $\phi_{0}>0$ in $[0,2)$,

$\phi_{0}(z)=\begin{cases}1,&\ {\rm for}\ 0\leqslant z\leqslant 1,\\\ 0,&\
{\rm for}\ z\geqslant 2,\end{cases}$

and, let us fix the positive constants

$M_{j}:=\int_{0}^{2}\frac{|\phi_{0}^{(j)}(z)|}{|\phi_{0}(z)|}\,\mathrm{d}z\quad{\rm
for}\quad j=1,2,3,4,5,6.$ (4.15)

Using the contradiction assumption, we may assume that there exists $T>0$ such
that for $t>T$, it follows

$v^{(6)}(t)-K_{4}v^{(4)}(t)+K_{2}v^{(2)}(t)=K_{0}v(t)-f(v(t))\geqslant\frac{1}{2}f(v(t))$
(4.16)

and

$v^{(5)}(t)-K_{4}v^{(3)}(t)+K_{2}v^{(1)}(t)=\frac{1}{2}\int_{T}^{t}f(v(\tau))\mathrm{d}\tau+C_{1}(T).$
(4.17)

Besides, as a consequence of (4.17), we can find $T^{*}>T$ satisfying

$v^{(5)}(T^{*})-K_{4}v^{(3)}(T^{*})+K_{2}v^{(1)}(T^{*}):=\upsilon>0.$

Furthermore, since ($\mathcal{O}_{6}$) is autonomous, we may suppose without
loss of generality that $T^{*}=0$. Then, multiplying inequality (4.16) by
$\phi(t)=\phi_{0}(\tau/t)$, and by integrating, we find

$\int_{0}^{T^{\prime}}\left[v^{(6)}(\tau)\phi(\tau)-K_{4}v^{(4)}(\tau)\phi(\tau)+K_{2}v^{(2)}(\tau)\phi(\tau)\right]\mathrm{d}\tau\geqslant\frac{1}{2}\int_{0}^{T^{\prime}}v(\tau)\mathrm{d}\tau,$

where $T^{\prime}=2T$. Moreover, integration by parts combined with
$\phi^{(j)}(T^{\prime})=0$ for $j=0,1,2,3,4,5$ implies

$\int_{0}^{T^{\prime}}\left[v(\tau)\phi^{(6)}(\tau)-K_{4}v(\tau)\phi^{(4)}(\tau)+K_{2}v(\tau)\phi^{(2)}(\tau)\right]\mathrm{d}\tau\geqslant\frac{1}{2}\int_{0}^{T^{\prime}}f(v(\tau))\mathrm{d}\tau+\upsilon.$
(4.18)

On the other hand, applying the Young inequality on the right-hand side of
(4.18), it follows

$v(\tau)|\phi^{(j)}(\tau)|=\varepsilon
v^{\frac{n+6}{n-6}}(\tau)\phi(\tau)+C_{\varepsilon}\frac{|\phi^{(j)}(\tau)|^{\frac{n+6}{12}}}{\phi(\tau)^{\frac{n-6}{12}}}\quad{\rm
for}\quad j=0,1,2,3,4,5.$ (4.19)

Hence, combining (4.19) and (4.18), we have that for $0<\varepsilon\ll 1$
sufficiently small, it follows that there exists $\widetilde{C}_{1}>0$
satisfying

$\widetilde{C}_{1}\int_{0}^{T^{\prime}}\left[\frac{|\phi^{(6)}(\tau)|^{\frac{n+6}{12}}}{\phi(\tau)^{\frac{n-6}{12}}}+\frac{|\phi^{(4)}(\tau)|^{\frac{n+6}{12}}}{\phi(\tau)^{\frac{n-6}{12}}}+\frac{|\phi^{(2)}(\tau)|^{\frac{n+6}{12}}}{\phi(\tau)^{\frac{n-6}{12}}}\right]\mathrm{d}\tau\geqslant\frac{1}{6}\int_{0}^{T^{\prime}}f(v(\tau))\mathrm{d}\tau+\upsilon.$

Now, by (4.15), one can find $\widetilde{C}_{2}>0$ such that

$\widetilde{C}_{2}\left(M_{6}T^{-\frac{n+4}{2}}-M_{4}T^{-\frac{n+3}{3}}+M_{2}T^{-\frac{n}{6}}\right)\geqslant\frac{1}{6}\int_{0}^{T}f(v(\tau))\mathrm{d}\tau.$
(4.20)

Therefore, passing to the limit in (4.20) the left-hand side converges,
whereas the right-hand side blows-up; this is a contradiction.

For proving the second part, let us notice that

$\lim_{t\rightarrow\infty}\mathcal{H}(t,v)=\left(F(\ell_{+})-\frac{K_{0}}{2}\ell_{+}^{2}\right)\geqslant
0,$

which implies $\ell^{*}=0$ and $\mathcal{H}(v)=0$. ∎

######  Lemma 4.9.

Let $v\in C^{6}(\mathbb{R})$ be a positive solution to ($\mathcal{O}_{6}$).
Assume that $v$ is a non-constant function, and denote by
$\\{s_{k}\\}_{k\in\mathbb{N}}$ and $\\{S_{k}\\}_{k\in\mathbb{N}}$ its sets of
local minimum and maximum points of $v$, respectively. Then, it holds that
$v(S_{k})>a^{*}_{n}$ and $v(s_{k})<a^{*}_{n}$ for any $k\in\mathbb{N}$.

###### Proof.

Let us assume by contradiction that there exists a maximum point $t=t_{0}$ of
$v$ such that $v(t_{0})\leqslant a^{*}_{n}$ and $v^{(1)}(t_{0})=0$,
$v^{(2)}(t_{0})\leqslant 0$. Notice that from ($\mathcal{O}_{6}$), we have two
possibilities which we describe as follows

Case 1. $v(t_{0})=a^{*}_{n}$, $v^{(2)}(t_{0})=0$, $v^{(4)}(t_{0})=0$, and
$v^{(6)}(t_{0})=0$.

Initially, we observe that $v^{(3)}(t_{0})=0$ and $v^{(5)}(t_{0})=0$.
Otherwise, without loss of generality, we may suppose that $v^{(3)}(t_{0})>0$
or $v^{(5)}(t_{0})>0$. If, $v^{(3)}(t_{0})>0$, by an elementary analysis, we
deduce that there exists $\delta>0$ satisfying $v^{(2)}(t)>0$ and
$v^{(1)}(t)>0$ in $(t_{0},t_{0}+\delta)$. This contradicts the fact the point
$t=t_{0}$ is a local maximum of $v(t)$. If, $v^{(5)}(t_{0})>0$, we can apply
the same strategy. Then, by the uniqueness, we get that $v(t)\equiv
a^{*}_{n}$, which is impossible.

Case 2.
$v^{(6)}(t_{0})=K_{4}v^{(4)}(t_{0})-K_{2}v^{(2)}(t_{0})-g(v(t_{0}))<0$.

Now, notice that for some small $\varepsilon>0$ it follows $v^{(6)}(t)<0$ for
$t\in(t_{0},t_{0}+\varepsilon)$ . In this case, we assume
$v^{(3)}(t_{0})\leqslant 0$ and $v^{(5)}(t_{0})\leqslant 0$, which gives us
$v^{(1)}(t)<0$, $v^{(3)}(t)<0$ and $v^{(5)}(t)<0$ for
$t\in(t_{0},t_{0}+\varepsilon)$. Now, setting

$t_{1}=\sup\left\\{\tilde{t}>t_{0}:v^{(1)}(t)<0,\ v^{(3)}(t)<0,\ {\rm and}\
t\in(t_{0},\tilde{t})\right\\}.$

we have that since $v$ oscillates, it follows that $t_{1}<\infty$ and either
$v^{(1)}(t_{1})=0$, $v^{(3)}(t_{1})=0$, or $v^{(5)}(t_{1})=0$ because of
continuity. This implies that $\mathcal{E}_{1}(v(t_{1}))v^{(1)}(t_{1})=0$.
Moreover, since $v^{(2)}$ is decreasing in $(t_{0},t_{1})$, combined with
$v^{(2)}(t_{0})\leqslant 0$, we obtain that $H_{2}(v)v^{(2)}$ is increasing in
$(t_{0},t_{1})$. Putting these together and remembering that
$v^{(1)}(t_{0})=0$, it follows from Proposition 2.10 that the equality holds

$\frac{1}{2}{v^{(3)}}(t_{1})^{2}-\frac{K_{2}}{2}v^{(1)}(t_{1})^{2}+G(v(t_{1}))=\frac{1}{2}{v^{(3)}}(t_{0})^{2}+G(v(t_{0})),$

which yields

$G(v(t_{0}))-G(v(t_{1}))=\frac{1}{2}({v^{(3)}}(t_{1})-{v^{(3)}}(t_{0}))^{2}-\frac{K_{2}}{2}v^{(1)}(t_{1})^{2}>0.$

On the other hand, since $v(t_{1})<v(t_{0})\leqslant a^{*}_{n}$, it is
straightforward to see that $G(v(t_{1}))>G(v(t_{0}))$; this contradicts the
last identity. The proof of the first case is finished.

Similarly, we can prove the second part for local minimum points of $v$, and
this concludes the proof of the lemma. ∎

The following lemma shows that when a solution tends to a limit monotonically,
then it has to be an equilibrium point.

######  Lemma 4.10.

Let $v\in C^{6}(\mathbb{R})$ be a bounded solution to
($\mathcal{P}_{6,\infty}$). If $v$ is eventually monotone at infinity, that
is, $\lim_{t\rightarrow\pm\infty}{\rm sign}(v^{(1)}(t))\neq 0$, then

$\lim_{t\rightarrow\pm\infty}v(t)\in\\{0,\pm a^{*}_{n}\\}\quad{\rm
and}\quad\lim_{t\rightarrow\pm\infty}v^{(j)}(t)=0\quad{\rm for}\quad
j=1,\dots,5.$

In other terms,

$\mathcal{A}(v)=\\{0,\pm a^{*}_{n}\\}\quad{\rm
and}\quad\mathcal{A}(v^{(j)})=\\{0\\}\quad{\rm for}\quad j=1,\dots,5.$

###### Proof.

By interchanging $v(t)$ by $v(-t)$, we may assume that
$\lim_{t\rightarrow+\infty}{\rm sign}(v^{(1)}(t))\neq 0$. Then, there exists
the limit

$\lim_{t\rightarrow+\infty}v(t):=\ell_{0}$

and $v$ increases towards $\ell_{0}$ as $t\rightarrow+\infty$. Since $v$ is
bounded, $\ell_{0}<+\infty$.

Claim 1: $\lim_{t\rightarrow}v^{(4)}(t)=0$ and
$\lim_{t\rightarrow}v^{(2)}(t)=0$ as $t\rightarrow+\infty$.

Let us define the function
$\Upsilon_{1}(v)=v^{(5)}-K_{4}v^{(3)}+K_{2}v^{(1)}$, which satisfies
$\Upsilon_{1}^{(2)}(v)=v^{(1)}g(v)$. Now we have to consider two cases:

Case 1: $g^{(1)}(\ell_{0})\neq 0$

In this case, $g^{(1)}(v)$ has a sign for $T\gg 1$ large enough, that is,
either $g^{(1)}(v)\geqslant 0$ or $g^{(1)}(v)\leqslant 0$ for large $T\gg 1$.
In addition, since $v^{(}1)\geqslant 0$, it follows
$\lim_{t\rightarrow\infty}{\rm sign}(\Upsilon_{1}^{(2)}(v(t)))\neq 0$, which
implies that $\Upsilon_{1}^{(2)}(v(t))$ has a sign for $T\gg 1$ large enough;
thus by a comparison principle $\Upsilon_{1}(v(t))$ also does. Furthermore,
because $\Upsilon_{1}(v(t))=v^{5}(t)-K_{4}v^{(3)}(t)+K_{2}v^{(1)}(t)$ has a
sign for $T\gg 1$ large enough, we find the the following limit exists:

$\lim_{t\rightarrow+\infty}\left(v^{(4)}(t)-K_{4}v^{(2)}(t)+K_{2}v(t)\right):=\ell_{1}.$

Consequently, we get

$\lim_{t\rightarrow+\infty}\left(v^{(4)}(t)-K_{4}v^{(2)}(t)\right):=\ell_{1}-K_{2}\ell_{0}.$
(4.21)

Defining $\Upsilon_{2}(v):=v^{(2)}-K_{4}v$, we get
$\lim_{t\rightarrow+\infty}\Upsilon_{2}^{(2)}(v)\neq 0$, that is,
$\Upsilon_{2}(v(t))$ has a sign for $T\gg 1$ large enough. Hence, by a
comparison principle $\Upsilon_{2}(v(t))$ also does. Moreover, since
$\Upsilon_{2}(v(t))=v^{(2)}(t)-K_{4}v(t)$ has a sign for $T\gg 1$ large
enough, we get that the the following limit exists

$\lim_{t\rightarrow+\infty}\left(v^{(2)}(t)-K_{4}v(t)\right):=\ell_{2}.$

Therefore, we conclude $v^{(2)}(t)\rightarrow\ell_{2}+K_{4}\ell_{0}$ as
$t\rightarrow\infty$, which by the boundedness of $v$, proves that
$v^{(2)}(t)\rightarrow 0$ as $t\rightarrow+\infty$. Finally, going back to
(4.21), we find $v^{(4)}(t)\rightarrow 0$ as $t\rightarrow+\infty$.

Case 2: $g^{(1)}(\ell_{0})=0$

In this case, we can define
$\widetilde{\Upsilon}_{1}(v)=v^{(5)}-K_{4}v^{(3)}+\frac{K_{2}}{2}v^{(1)}$ and
$\widetilde{\Upsilon}_{2}(v)=v^{(4)}-\frac{K_{4}}{2}v^{(2)}$ and proceed as
before to conclude the proof.

We finished the proof of the first claim

At last, using that $v^{(2)}(t)\rightarrow 0$ and $v^{(4)}(t)\rightarrow 0$ as
$t\rightarrow+\infty$, one can prove that $v^{(1)}(t)\rightarrow 0$ and
$v^{(3)}(t)\rightarrow 0$ as $t\rightarrow+\infty$. Since, we can write
($\mathcal{O}_{6}$) as

$v^{(6)}=K_{4}v^{(4)}-K_{2}v^{(2)}-g(v)\quad{\rm in}\quad\mathbb{R},$

it holds

$\lim_{x\rightarrow+\infty}v^{(6)}(t):=\ell_{3}=g(\ell_{0}).$

Therefore, $v$ is bounded, we find $\ell_{3}=0$, and thus
$\ell_{0}\in\mathcal{A}(v)$. By this discussion, it is easy now to see that
$\mathcal{A}(v^{(j)})=0$ for $j=1,2,3,4,5$, which concludes the proof of the
lemma. ∎

###  4.4. Ordering

Now we prove that, as in the lower order cases, the Hamiltonian energy is a
parameter that orders bounded solutions in the $(v,v^{(1)})$-phase plane.

Our first technical lemma is a sign property for some terms of the Hamiltonian
energy associated to ($\mathcal{O}_{6}$), which will be important in our
upcoming arguments. Although, our results ares similar in spirit to the one in
[37, Lemma 5], we provide a shorter proof, which also works in our sixth order
situation.

######  Lemma 4.11.

Let $v\in C^{6}(\mathbb{R})$ be a bounded solution to ($\mathcal{O}_{6}$). The
following sign identities hold

  * (i)

${\rm sign}(v^{(1)})={\rm sign}(\mathcal{E}_{1}(v))$;

  * (ii)

${\rm sign}(v^{(2)})={\rm sign}(\mathcal{E}_{2}(v))$.

where $\mathcal{E}_{1}(v),\mathcal{E}_{2}(v)$ the auxiliary functions given by
(2.12).

###### Proof.

The proof will be divided into three steps.

Step 1. There exist $\alpha_{1},\alpha_{2}>0$ (depending only on
$K_{2},K_{4}$) such that

$0<\Psi_{1,\alpha_{1}}(v)v^{(1)}<\mathcal{E}_{1}(v)v^{(1)}\quad{\rm and}\quad
0<\Psi_{2,\alpha_{2}}(v)v^{(2)}<\mathcal{E}_{2}(v)v^{(2)},$

where

$\Psi_{1,\alpha_{1}}(v)=v^{(5)}-{K_{4}}v^{(3)}+\alpha_{1}v^{(1)}\quad{\rm
and}\quad\Psi_{2,\alpha_{2}}(v)=-v^{(4)}+\alpha_{2}v^{(2)}.$

Notice that since

$\mathcal{E}_{1}(v)v^{(1)}=\Psi_{1,\alpha_{1}}(v)v^{(1)}-\sqrt{\left(\frac{K_{2}}{2}-\alpha_{1}\right)}{v^{(1)}}^{2},$

and

$\mathcal{E}_{2}(v)v^{(2)}=\Psi_{2,\alpha_{2}}(v)v^{(2)}-\sqrt{\left(\frac{K_{4}}{2}-\alpha_{2}\right)}{v^{(2)}}^{2},$

there exists $C_{1},C_{2}\geqslant 0$ (depending only on $K_{0},K_{2}$) such
that

$\mathcal{E}_{1}(v)v^{(1)}=\Psi_{1,\alpha_{1}}(v)v^{(1)}+\sqrt{C_{1}}{v^{(1)}}^{2}>\Psi_{1,\alpha_{1}}(v)v^{(1)},$
(4.22)

and

$\mathcal{E}_{2}(v)v^{(2)}=\Psi_{2,\alpha_{2}}(v)v^{(2)}+\sqrt{C_{2}}{v^{(2)}}^{2}>\Psi_{2,\alpha_{2}}(v)v^{(2)},$
(4.23)

which proves the first claim.

Now, we are left to study the sign of these auxiliary terms.

Let us start with the term related to the first derivative. For this, we
observe that $\Psi_{1,\alpha_{1}}(v)$ can be written as a positive fourth
order operator on $v^{(1)}$.

Step 2. ${\rm sign}(v^{(1)})={\rm sign}(\Psi_{1,\alpha_{1}}(v))$.

We divide the remaining part of the proof into will be divided into two cases.

Case 1. If $v^{(1)}\geqslant 0$, then $\Psi_{1,\alpha_{1}}(v)\geqslant 0$.

Let $t_{0}\in\mathbb{R}$ be arbitrary. We may assume that
$v^{(1)}(t_{0})\geqslant 0$. We see from Corollary 4.5 that Claim 2 holds if
$v^{(1)}(t_{0})=0$. We thus assume that $v^{(1)}(t_{0})>0$. Since $v$ is
bounded there exist $-\infty\leqslant t_{1}<t_{0}<t_{2}\leqslant\infty$ such
that $v^{(1)}(t_{1})=v^{(1)}(t_{2})=0$ and $v^{(1)}<0$ in $(t_{1},t_{2})$. By
Corollary 4.5 and Lemma 4.10, we get that $v^{(3)}(t_{1})=v^{(3)}(t_{2})=0$,
which implies $\Psi_{1,\alpha_{1}}(v(t_{1}))=\Psi_{1,\alpha_{1}}(v(t_{2}))=0$.
Thus, we arrive at the following system

$\begin{cases}(L_{\alpha_{1}^{+}}\circ
L_{\alpha_{1}^{-}})(v^{(1)})=\Psi_{1,\alpha_{1}}(v)\quad{\rm
in}\quad(t_{1},t_{2}).\\\
L_{\alpha^{-}_{1}}(v^{(1)}(t_{1}))=L_{\alpha^{-}_{1}}(v^{(1)}(t_{2}))=v^{(1)}(t_{1})=v^{(1)}(t_{2})=0,\end{cases}$
(4.24)

where $\alpha_{1}^{+}+\alpha_{1}^{-}=K_{4}$ and
$\alpha_{1}^{+}\alpha_{1}^{-}=\alpha_{1}$.

Next, suppose by contradiction that $\Psi_{1,\alpha_{1}}(v(t_{0}))\leqslant
0$. Hence, using the maximum principle in (4.24), it follows that
$v^{(1)}(t_{0})<0$, which is a contradiction. The first case is proved.

For the second case, the idea is similar.

Case 2. If $v^{(1)}<0$, then $\Psi_{1,\alpha_{1}}(v)<0$.

When $v^{(1)}(t_{0})<0$, as before one can find $-\infty\leqslant
t_{1}<t_{0}<t_{2}\leqslant\infty$ such that $v^{(1)}<0$ in $(t_{1},t_{2})$ and
(4.24) holds. Using the same contradiction argument, suppose that there exists
$t_{0}\in(t_{1},t_{2})$ such that $\Psi_{1,\alpha_{1}}(v(t_{0}))\geqslant 0$.
By the minimum principle applied to (4.24), we find $v^{(1)}(t_{0})\geqslant
0$, which is also a contradiction. This finishes the proof of the second
claim.

The study of the remaining term is less complicated and direct application of
a comparison principle. Indeed, notice that $\Psi_{2,\alpha_{2}}(v)$ can be
written as a positive second order operator on $v^{(2)}$.

Step 3. ${\rm sign}(v^{(2)})={\rm sign}(\Psi_{2,{\alpha_{2}}}(v))$.

As before, we must study two cases as follows.

Case 1. If $v^{(2)}\geqslant 0$, then $\Psi_{2,{\alpha_{2}}}(v)\geqslant 0$.

In fact, we may assume that $v^{(2)}(t_{0})\geqslant 0$. We know from
Corollary 4.5 that Claim 2 holds if $v^{(2)}(t_{0})=0$. We thus assume that
$v^{(2)}(t_{0})>0$. Since $v$ is bounded there exist $-\infty\leqslant
t_{1}<t_{0}<t_{2}\leqslant\infty$ such that $v^{(2)}(t_{1})=v^{(2)}(t_{2})=0$
and $v^{(2)}>0$ in $(t_{1},t_{2})$. By Corollary 4.5 and Lemma 4.10, we get
that $v^{(4)}(t_{1})=v^{(4)}(t_{2})=0$, which implies
$L_{\alpha_{2}}(v^{(2)}(t_{1}))=L_{\alpha_{2}}(v^{(2)}(t_{2}))=0$.

Now, it is convenient to write

$\begin{cases}-L_{\alpha_{2}}(v^{(2)})=\Psi_{2,{\alpha_{2}}}(v)\quad{\rm
in}\quad(t_{1},t_{2})\\\
L_{\alpha_{2}}(v^{(2)}(t_{1}))=L_{\alpha_{2}}(v^{(2)}(t_{2}))=0.\end{cases}$
(4.25)

It is straightforward to check that $\Psi_{2,{\alpha_{2}}}(v(t_{0})>0$.
Otherwise, we would have $\Psi_{2,{\alpha_{2}}}(v(t_{0})<0$. Thus, by the
maximum principle, $v^{(2)}(t_{0})<0$, which contradicts the assumption.

The second case has a similar proof.

Case 2. If $v^{(2)}<0$, then $\Psi_{2,{\alpha_{2}}}(v)<0$.

When $v^{(2)}(t_{0})<0$, as before one can find $-\infty\leqslant
t_{1}<t_{0}<t_{2}\leqslant\infty$ such that $v^{(2)}<0$ in $(t_{1},t_{2})$ and
(4.25) holds. Again, the same argument by contradiction based on the minimum
principle yields $\Psi_{2,\alpha_{2}}(v(t_{0}))<0$.

The conclusion is a direct consequence of Steps 1, 2 and 3. ∎

######  Remark 4.12.

Using the same ideas one can also verify the sign identity below

${\rm sign}(v^{(1)})={\rm sign}(\mathcal{E}_{1}(v)-v^{(5)})={\rm
sign}(\Psi_{1,\alpha_{1}}(v)-v^{(5)})={\rm
sign}(-K_{4}v^{(3)}+\alpha_{1}v^{(1)}),$ (4.26)

where $\alpha_{1}\in(0,\frac{K_{2}}{2}]$ is such that (4.22) holds.

As a consequence of the last lemma, we find two energy inequalities for
solutions to ($\mathcal{O}_{6}$).

######  Corollary 4.13.

Let $v\in C^{6}(\mathbb{R})$ be a bounded solution to ($\mathcal{O}_{6}$). The
following energy identity holds:

$-\mathcal{R}(v):=-\mathcal{H}(v)+G(v)+\frac{1}{2}{v^{(3)}}^{2}\leqslant 0.$
(4.27)

###### Proof.

Initially, notice that using (2.11), we get

$-\mathcal{R}(v):=-[\mathcal{E}_{1}(v)v^{(1)}+\mathcal{E}_{2}(v)v^{(2)}],$
(4.28)

which combined with Lemma 4.11 implies (4.27), and so the proof is concluded.
∎

The next comparison lemma is a preparation for proving the energy ordering.
Subsequently, we will focus on the situation
$v_{1}^{(1)}(0)>v_{2}^{(1)}(0)\geqslant 0$. By symmetry, the same strategy
applies otherwise. This result is a weaker version of Lemma 4.4, and strongly
relies on the energy identity in Corollary 4.13. It shows that to prove
uniqueness of solutions to ($\mathcal{O}_{6}$), one only needs three initial
conditions instead of six as usual.

######  Lemma 4.14.

Let $v_{1},v_{2}\in C^{6}(\mathbb{R})$ be bounded solutions to
($\mathcal{O}_{6}$). Suppose that
$\mathcal{H}(v_{1})\leqslant\mathcal{H}(v_{2})$, and

$\displaystyle v_{1}(0)=v_{2}$ $\displaystyle(0),\quad
v_{1}^{(1)}(0)>v_{2}^{(1)}(0)>0,\quad v_{1}^{(2)}(0)=v_{2}^{(2)}(0),\quad{\rm
and}\quad v_{1}^{(3)}(0)=v_{2}^{(3)}(0).$ (4.29)

Then, $v_{1}\equiv v_{2}$.

###### Proof.

Initially, using (4.29), it is not hard to check that

$L_{\lambda_{1}}(v_{1}(0))\geqslant L_{\lambda_{1}}(v_{2}(0))\quad{\rm
and}\quad L_{\lambda_{1}}^{(1)}(v_{1}(0))\geqslant
L_{\lambda_{1}}^{(1)}(v_{2}(0)).$ (4.30)

Next, from (4.22) and (4.28) at $t=0$, we have

$-\mathcal{R}(v_{1}(0))={-\mathcal{H}(v_{1}(0))+G(v_{1}(0))+\frac{1}{2}{v_{1}^{(3)}}(0)^{2}}$

and

$-\mathcal{R}(v_{2}(0))=-\mathcal{H}(v_{2}(0))+G(v_{2}(0))+\frac{1}{2}{v_{2}^{(3)}(0)}^{2}.$

Also, using (4.29), it follows

$-\mathcal{H}(v_{1}(0))+G(v_{1}(0))+\frac{1}{2}{v_{1}^{(3)}(0)}^{2}\leqslant-\mathcal{H}(v_{2}(0))+G(v_{2}(0))+\frac{1}{2}{v_{2}^{(3)}(0)}^{2}\leqslant
0.$

This combined with Corollary 4.13 gives us

$-\mathcal{R}(v_{1}(0))\leqslant-\mathcal{R}(v_{2}(0))\leqslant 0.$ (4.31)

From this, we obtain two identities as follows

$0\leqslant\left[\mathcal{E}_{1}(v_{2})v^{(1)}_{2}-\mathcal{E}_{1}(v_{1})v_{1}^{(1)}\right](0)\leqslant\left[\mathcal{E}_{1}(v_{2}(0))-\mathcal{E}_{1}(v_{1}(0))\right]v_{2}^{(1)}(0)$

and

$0\leqslant\left[\mathcal{E}_{2}(v_{2})v_{2}^{(2)}-\mathcal{E}_{2}(v_{1})v_{1}^{(2)}\right](0)\leqslant\left[\mathcal{E}_{2}(v_{2}(0))-\mathcal{E}_{2}(v_{1}(0))\right]v_{2}^{(2)}(0),$

which yields

$\mathcal{E}_{1}(v_{2}(0))\geqslant\mathcal{E}_{1}(v_{1}(0))\quad{\rm
and}\quad\mathcal{E}_{2}(v_{2}(0))\geqslant\mathcal{E}_{2}(v_{1}(0)).$ (4.32)

Now, by combining (4.32) and (4.29), we get respectively

$v_{1}^{(5)}(0)\geqslant v_{2}^{(5)}(0)\quad{\rm and}\quad
v_{1}^{(4)}(0)\leqslant v_{2}^{(4)}(0).$

Therefore, we conclude

$(L_{\lambda_{1}}\circ
L_{\lambda_{2}})(v_{1}(0))\geqslant(L_{\lambda_{1}}\circ
L_{\lambda_{2}})(v_{2}(0)).$ (4.33)

and

$(L_{\lambda_{1}}\circ
L_{\lambda_{2}})^{(1)}(v_{1}(0))\geqslant(L_{\lambda_{1}}\circ
L_{\lambda_{2}})^{(1)}(v_{2}(0)).$ (4.34)

Finally, using (4.29), (4.30), (4.33) and (4.34), we can apply Lemma 4.6 to
finish the proof. ∎

In what follows, let $[t_{1},t_{2}]\subset\mathbb{R}$ denote intervals where
the functions $v^{(1)},v^{(2)}$ are strictly monotone on $(t_{1},t_{2})$. In
this fashion, denoting the inverse of $v(t)$ by $t(v)$, we define the change
of variables

$s=v\quad\text{ and }\quad\zeta(s)=\left[v^{(1)}(t(s))\right]^{2}.$ (4.35)

Now for $s\in[s_{1},s_{2}]=[v(t_{1}),v(t_{2})]$, we have

$\zeta^{(1)}(s)=2v^{(2)}(t(s))\quad{\rm
and}\quad\zeta^{(2)}(s)=2v^{(3)}(t(s)).$

If $t_{1}=-\infty$, then we write $\zeta^{(1)}(s_{1})=\lim_{s\rightarrow
s_{1}}\zeta^{(1)}(t)$ and $\zeta^{(2)}(s_{1})=\lim_{s\rightarrow
s_{1}}\zeta^{(2)}(t)$. In the same way, if $t_{2}=+\infty$, then we write
$\zeta^{(1)}(s_{2})=\lim_{s\rightarrow s_{2}}\zeta^{(1)}(t)$ and
$\zeta^{(2)}(s_{2})=\lim_{s\rightarrow s_{2}}\zeta^{(2)}(t)$. Notice that
these two limits exist as a consequence of Lemma 4.10.

We are ready prove the energy ordering lemma, which is another crucial result
in this manuscript.

######  Lemma 4.15.

Let $v_{1},v_{2}\in C^{6}(\mathbb{R})$ be bounded solutions to
($\mathcal{O}_{6}$). Suppose that $v_{1}(0)=v_{2}(0)$ and either
$v_{1}^{(1)}(0)>v_{2}^{(1)}(0)\geqslant 0$ or
$v_{1}^{(1)}(0)<v_{2}^{(1)}(0)\leqslant 0$, then
$\mathcal{H}(v_{1})>\mathcal{H}(v_{2})$.

###### Proof.

We divide the proof into three steps as follows.

The first step states that there are at most two bounded solutions on the
unstable manifold of each equilibrium point.

Step 1. Suppose that $v_{1},v_{2},v^{(1)}_{1},v^{(1)}_{2}$ are non-constant
functions, and that there exists
$\ell_{0}\in\mathcal{A}(v_{1})\cap\mathcal{A}(v_{2})$ such that

$\lim_{t\rightarrow\pm\infty}v_{1}(t)=\lim_{t\rightarrow\pm\infty}v_{2}(t)=\ell_{0}\quad{\rm
and}\quad\lim_{t\rightarrow\pm\infty}v^{(1)}_{1}(t)=\lim_{t\rightarrow\pm\infty}v^{(1)}_{2}(t)=0.$

Then, it follows that $v_{1}(t)$ decreases to $\ell_{0}$ and $v_{2}(t)$
increases to $\ell_{0}$ as $t\rightarrow\pm\infty$, or vice versa. In
addition, $v^{(1)}_{1}(t)$ increases to $0$ and $v^{(1)}_{2}(t)$ decreases to
$0$ as $t\rightarrow\pm\infty$, or vice versa.

Initially, by Corollary 4.5, $v_{1},v_{2}$ and $v^{(1)}_{1},v^{(1)}_{2}$ can
only tend monotonically to their respective limits. Also, using symmetry, we
may suppose by contradiction that $v_{1}$ and $v_{2}$ both decrease towards
$\ell_{0}$ as $t\rightarrow-\infty$ and $v^{(1)}_{1}$ and $v^{(1)}_{2}$ both
increase towards $0$ as $t\rightarrow-\infty$. In other words, there exists
$T\gg 1$ such that $v_{1}(t)>0$, $v_{2}(t)>0$, $v_{1}^{(1)}(t)<0$, and
$v_{2}^{(1)}(t)<0$ for $t\in(-\infty,T)$.

We finish the contradiction argument with the following claim:

Claim 1. $v_{1}\equiv v_{2}$.

Indeed, for $s\in(\ell_{0},\ell_{0}+\varepsilon_{0})$, where
$0<\varepsilon_{0}\ll 1$ is small enough, let $\zeta_{1}$ and $\zeta_{2}$
correspond to $v_{1}$ and $v_{2}$ respectively by the change of variables
defined in (4.35). Notice that one must have
$\zeta^{(1)}_{1}\neq\zeta^{(1)}_{2}$ and $\zeta^{(2)}_{1}\neq\zeta^{(2)}_{2}$
on $(\ell_{0},\ell_{0}+\varepsilon_{0})$ . Otherwise, by Lemma 4.4, we would
get $v_{1}\equiv v_{2}$. Without loss of generality, we may assume
$\zeta^{(1)}_{1}>\zeta^{(1)}_{2}$ and $\zeta^{(2)}_{1}>\zeta^{(2)}_{2}$ on
$(\ell_{0},\ell_{0}+\varepsilon_{0})$. Since both $\zeta_{1},\zeta^{(1)}_{1}$
and $\zeta_{2},\zeta^{(1)}_{2}$ are differentiable on
$(\ell_{0},\ell_{0}+\varepsilon_{0})$ and, by Lemma 4.10,
$\zeta_{1}(\ell_{0})=\zeta_{2}(\ell_{0})=0$ and
$\zeta_{1}^{(1)}(\ell_{0})=\zeta_{2}^{(1)}(\ell_{0})=0$. Therefore, there
exists $s_{0}\in(\ell_{0},\ell_{0}+\varepsilon_{0})$ such that
$\zeta^{(1)}_{1}(s_{0})\geqslant\zeta^{(1)}_{2}(s_{0})$ and
$\zeta^{(2)}_{1}(s_{0})\geqslant\zeta^{(2)}_{2}(s_{0})$.

We divide the analysis into two cases:

Case 1. Both $\zeta_{2}^{(1)}(s_{0})\geqslant 0$ and
$\zeta_{2}^{(2)}(s_{0})\geqslant 0$.

There exist $t_{1},t_{2}\in\mathbb{R}$ such that

$\displaystyle v_{1}(t_{1})=v_{2}(t_{2})$ $\displaystyle=s_{0},\quad
v_{1}^{(1)}(t_{1})>v_{2}^{(1)}(t_{2})>0,\quad v_{1}^{(2)}(t_{1})\geqslant
v_{2}^{(2)}(t_{2}),\quad\text{and}\quad v_{1}^{(3)}(t_{1})\geqslant
v_{2}^{(3)}(t_{2}).$

By translating, $v_{1}$ and $v_{2}$ by $t_{1}$ and $t_{2}$ respectively, we
find

$\displaystyle v_{1}(0)=$ $\displaystyle v_{2}(0),\quad
v_{1}^{(1)}(0)>v_{2}^{(1)}(0)>0,\quad v_{1}^{(2)}(0)\geqslant
v_{2}^{(2)}(0),\quad\text{and}\quad v_{1}^{(3)}(0)\geqslant v_{2}^{(3)}(0).$
(4.36)

Since both $v_{1}$ and $v_{2}$ tend to $\ell_{0}$ monotonically as
$t\rightarrow-\infty$, we conclude from Lemma 4.10 that
$\mathcal{A}^{-}(v_{1})=\mathcal{A}^{-}(v_{2})=\ell_{0}$ and
$\mathcal{A}^{-}(v^{(j)}_{1})=\mathcal{A}^{-}(v^{(j)}_{2})=0$ for
$j=1,2,3,4,5$. Therefore, $\mathcal{H}(v_{1})=\mathcal{H}(v_{2})$. In the
light of (4.26), it is straightforward to check that (4.36) instead of (4.29)
is sufficient for the proof in Lemma 4.14 to go on unchanged. Hence
$v_{1}\equiv v_{2}$. This finishes the proof of the first case.

Case 2. Either $\zeta_{2}^{(1)}(s_{0})<0$ or $\zeta_{2}^{(2)}(s_{0})<0$.

Without loss of generality, we assume $\zeta_{2}^{(1)}(s_{0})<0$; the same
idea applies in the other situation. Notice that $\zeta^{(1)}_{2}>0$ on
$(\ell_{0},s_{0}]$ and $\zeta^{(1)}_{2}(\ell_{0})=0$. Then, there exists a
$\bar{s}_{0}\in(\ell_{0},s_{0})$ such that $\zeta_{2}^{(1)}(\bar{s}_{0})=0$.
If $\zeta_{1}^{(1)}(\bar{s}_{0})\geqslant 0$, then
$\zeta_{1}^{(1)}(\bar{s}_{0})\geqslant\zeta_{2}^{(1)}(\bar{s}_{0})=0$, which
is covered by the last case. If $\zeta_{1}^{(1)}(\bar{s}_{0})<0$, then
$\zeta_{1}^{(1)}(\bar{s}_{0})<\zeta_{2}^{(1)}(\bar{s}_{0})$ and
$\zeta_{1}^{(1)}(s_{0})\geqslant\zeta_{2}^{(1)}(s_{0})$, and by continuity
there exists $\tilde{s}_{0}\in(\bar{s}_{0},s_{0}]$ such that
$\zeta_{1}^{(1)}(\tilde{s}_{0})=\zeta_{2}^{(1)}(\tilde{s}_{0})$. Hence, one
can find $t_{1},t_{2}\in\mathbb{R}$ such that

$\displaystyle v_{1}(t_{1})=v_{2}($ $\displaystyle t_{2})=\tilde{s}_{0},\quad
v_{1}^{(1)}(t_{1})>v_{2}^{(1)}(t_{2})>0,\quad
v_{1}^{(2)}(t_{1})=v_{2}^{(2)}(t_{2}),\quad\text{and}\quad
v_{1}^{(3)}(t_{1})\geqslant v_{2}^{(3)}(t_{2}).$

Since $\mathcal{H}(v_{1})=\mathcal{H}(v_{2})$ as above, we may apply Lemma
4.14 to obtain $v_{1}\equiv v_{2}$, which finishes the proof of the second
case.

The proof of Step 1 is concluded.

The next step states that whenever the energy inequality is violated one can
find points on which the second and the third derivatives of these two bounded
solutions coincide.

Step 2. Suppose that $v_{1}(0)=v_{2}(0)$ and
$v_{1}^{(1)}(0)>v_{2}^{(1)}(0)\geqslant 0$, and
$\mathcal{H}(v_{1})\leqslant\mathcal{H}(v_{2})$. Then, there exist
$t_{1},t_{2}\in\mathbb{R}$ such that

$\displaystyle v_{1}(t_{1})=$ $\displaystyle v_{2}(t_{2}),\quad
v_{1}^{(1)}(t_{1})>v_{2}^{(1)}(t_{2})\geqslant 0,\quad
v_{1}^{(2)}(t_{1})=v_{2}^{(2)}(t_{2}),\quad\text{and}\quad
v_{1}^{(3)}(t_{1})=v_{2}^{(3)}(t_{2}).$ (4.37)

Let $[\hat{t}_{1},\hat{t}_{2}]$ be the largest interval containing $t=0$ on
which $v_{1}^{(1)},v_{1}^{(2)},v_{1}^{(3)}$ are positive, and let
$[t_{1},t_{2}]$ be the largest interval containing $t=0$ on which
$v_{2}^{(1)},v_{2}^{(2)},v_{2}^{(3)}$ are positive. We can now apply the
change variables (4.35) again. Let $\zeta_{1}$ correspond to $v_{1}$ on
$[\hat{s}_{1},\hat{s}_{2}]=[v_{1}(\hat{t}_{1}),v_{1}(\hat{t}_{2})]$, and let
$\zeta_{2}$ correspond to $v_{2}$ on
$[s_{1},s_{2}]=[v_{2}(t_{1}),v_{2}(t_{2})]$. Clearly, as a consequence of
Lemma 4.4, we have $\zeta_{1}>\zeta_{2}$ on $(s_{1},s_{2})$. Finally, if
$-\infty<t_{1}$, then it follows from Lemma 4.10 that $\hat{s}_{1}<s_{1}$,
whereas if $t_{1}=-\infty$, then this follows from Step 1. The same happens to
$\hat{s}_{2}>s_{2}$.

We fix the notation $\boldsymbol{\zeta}=(\zeta^{(1)},\zeta^{(2)})$ and
$|\boldsymbol{\zeta}|^{2}={\zeta^{(1)}}^{2}+{\zeta^{(2)}}^{2}$. We will prove
that the norm function has opposite sign on the edges of $[s_{1},s_{2}]$.

Claim 2.
$|\boldsymbol{\zeta}_{1}(s_{1})|^{2}<|\boldsymbol{\zeta}_{2}(s_{1})|^{2}$.

One can see that $\zeta_{2}(s_{1})=v_{2}^{(1)}(t_{1})=0$,
$\zeta_{2}^{(1)}(s_{1})=2v_{2}^{(2)}(t_{1})=0$ and
$\zeta_{2}^{(2)}(s_{1})=2v_{2}^{(3)}(t_{1})\geqslant 0$. Indeed, if
$t_{1}=-\infty$, this follows from Lemma 4.10, whereas if $-\infty<t_{1}$,
then it holds because $v(t_{1})$ and $v^{(1)}(t_{1})$ are minimums. Notice
that $v_{1}(\hat{t}_{1})=\hat{s}_{1}<s_{1}$ and
$v_{1}(\hat{t}_{2})=\hat{s}_{2}>s_{2}>s_{1}$, thus let
$t_{*}\in(\hat{t}_{1},\hat{t}_{2})$ be such that
$v_{1}(t_{*})=v_{2}(t_{1})=s_{1}$. By (2.11) and Lemma 4.11, we obtain

$\displaystyle\frac{|\boldsymbol{\zeta}_{2}(s_{1})|^{2}-|\boldsymbol{\zeta}_{1}(s_{1})|^{2}}{8}$
$\displaystyle=\left(\frac{1}{2}v_{2}^{(2)}(t_{1})^{2}-\frac{1}{2}v_{1}^{(2)}(t_{*})^{2}\right)+\left(\frac{1}{2}v_{2}^{(3)}(t_{1})^{2}-\frac{1}{2}v_{1}^{(3)}(t_{*})^{2}\right)$
$\displaystyle\geqslant\left[\mathcal{H}(v_{2})-G(s_{1})\right]-\left[\mathcal{H}(v_{1})-G(s_{1})-\mathcal{R}(v_{1}(t_{*}))\right]$
$\displaystyle=[\mathcal{H}(v_{2})-\mathcal{H}(v_{1})]+\mathcal{R}(v_{1}(t_{*})).$

Next, using that
$v_{1}^{(2)}(t_{*})=\sqrt{\zeta^{(1)}_{1}(s_{1})}>\sqrt{\zeta^{(1)}_{2}(s_{1})}=0$,
combined with Corollary 4.13, we conclude $\mathcal{R}(v_{1}(t_{*}))>0$.
Hence, since $\mathcal{H}(v_{2})-\mathcal{H}(v_{1})\geqslant 0$, the desired
conclusion holds.

Claim 3.
$|\boldsymbol{\zeta}_{1}(s_{2})|^{2}>|\boldsymbol{\zeta}_{2}(s_{2})|^{2}$.

Indeed, one can see that $\zeta_{2}(s_{2})=0$ and
$\zeta_{2}^{(1)}(s_{2})=2v_{2}^{(2)}(t_{1})\leqslant 0$. Indeed, if
$t_{2}=\infty$, this follows from Lemma 4.10, whereas if $t_{2}<+\infty$, then
it holds because $v(t_{2})$ is a maximum. Notice that
$v_{2}({t}_{1})={s}_{1}>\hat{s}_{1}$ and
$v_{2}({t}_{2})={s}_{2}<\hat{s}_{2}<\hat{s}_{1}$, thus let
$t^{*}\in({t}_{1},{t}_{2})$ be such that
$v_{2}(t^{*})=v_{1}(\hat{t}_{2})=\hat{s}_{2}$. By (2.11) and Lemma 4.11, we
obtain

$\displaystyle\frac{|\boldsymbol{\zeta}_{2}(s_{2})|^{2}-|\boldsymbol{\zeta}_{1}(s_{2})|^{2}}{8}=[\mathcal{H}(v_{2})-\mathcal{H}(v_{1})]+\mathcal{R}(v_{1}(t_{2})).$

As before, we conclude the proof of this claim.

By continuity, we find $s_{*}\in(s_{1},s_{2})$ such that
$|\boldsymbol{\zeta}_{1}(s_{*})|=|\boldsymbol{\zeta}_{2}(s_{*})|$, which
implies ${\zeta}^{(1)}_{1}(s_{*})={\zeta}^{(1)}_{2}(s_{*})$ and
${\zeta}^{(2)}_{1}(s_{*})={\zeta}^{(2)}_{2}(s_{*})$. The proof is finished by
undoing the transformation (4.35)

Finally, we are based on the last step to conclude the proof of the energy
ordering.

Step 3. $\mathcal{H}(v_{1})>\mathcal{H}(v_{2})$.

By symmetry, we may assume $v_{1}^{(1)}(0)>v_{2}^{(1)}(0)\geqslant 0$. Suppose
by contradiction that $\mathcal{H}(v_{1})\leqslant\mathcal{H}(v_{2})$, then,
by Step 2, there exist $t_{1},t_{2}\in\mathbb{R}$ such that (4.37) holds.
Thus, by translation invariance, we may take $t_{1}=t_{2}=0$, which gives us
that (4.29) are satisfied. Therefore, using Lemma 4.14 we obtain that
$v_{1}\equiv v_{2}$, which contradicts our assumptions, and so Step 3 is
proved.

The proof of the lemma is finished. ∎

###  4.5. Uniqueness

We prove uniqueness, up to translations, of homoclinic solutions to
($\mathcal{O}_{6}$).

######  Lemma 4.16.

Let $v_{1},v_{2}\in C^{6}(\mathbb{R})$ be bounded positive solutions to
($\mathcal{O}_{6}$). Suppose that
$\lim_{|t|\rightarrow\infty}v_{1}(t)=\lim_{|t|\rightarrow\infty}v_{2}(t)=0$
and $v_{1}^{(1)}(0)=v_{2}^{(1)}(0)=0$. Then, it follows $v_{1}\equiv v_{2}$.

###### Proof.

We divide the proof into claims

Claim 1. $t_{0}=0$ is the only zero of $v_{1}^{(1)}$ and $v_{2}^{(1)}$.

Indeed, if $v_{1}^{(1)}$ had another zero at, say, $t_{1}>0$, then by
repeatedly applying Corollary 4.5, we get that $v$ must be periodic of period
$2t_{1}>0$. In particular $0<v(0)=v(2kt_{1})$ for all $k\in\mathbb{N}$, which
contradicts the fact $v(t)\rightarrow 0$ as $t\rightarrow+\infty$. The
argument for $v_{2}^{(1)}$ is analogous. Hence, we get

$v_{1}^{(1)}(t)<0\quad\text{and}\quad v_{2}^{(1)}(t)<0\quad\text{for all}\quad
t>0;$ (4.38)

this proves the first claim.

Next, by Lemma 4.10 and Proposition 2.10, we find

$\mathcal{H}_{v_{1}}=\lim_{t\rightarrow+\infty}\mathcal{H}_{v_{1}}(t)=G(0)=0\quad\text{and}\quad\mathcal{H}_{v_{2}}=\lim_{t\rightarrow+\infty}\mathcal{H}_{v_{2}}(t)=G(0)=0.$
(4.39)

From which, we see that if $v_{1}(0)=v_{2}(0)$, then, by Lemma 4.4, the proof
is concluded.

Claim 2. $v_{1}(0)=v_{2}(0)$.

As a matter of fact, let us suppose by contradiction that $v_{1}(0)>v_{2}(0)$.
We claim that this implies $v_{1}>v_{2}$ everywhere. Indeed, otherwise there
would exist $t_{1}>0$ such that $v_{1}>v_{2}$ on $[0,t_{1})$ and
$v_{1}(t_{1})=v_{2}(t_{1})$. Then, by (4.38), we infer that
$v_{1}^{(1)}(t_{1})\leqslant v_{2}^{(1)}(t_{1})<0$. If
$v_{1}^{(1)}(t_{1})=v^{(1)}_{2}(t_{0})$, then Lemma 4.4 implies $v_{1}\equiv
v_{2}$, contradicting $v_{1}(0)>v_{2}(0)$. If
$v_{1}^{(1)}(t_{1})<v_{2}^{(1)}(t_{1})<0$, then, by Lemma 4.15, we obtain
$\mathcal{H}_{v_{1}}>\mathcal{H}_{v_{2}}$, which contradicts (4.39).

For any $R>0$, since $v_{1},v_{2}\in C^{6}(\mathbb{R})$ satisfy
($\mathcal{O}_{6}$), using integration by parts, we get

$\displaystyle 0$
$\displaystyle=\int_{-T}^{T}v_{2}\left(v_{1}^{(6)}-K_{4}v_{1}^{(4)}+K_{2}v_{1}^{(2)}+g(v_{1})\right)$
$\displaystyle=B(T)+\int_{-T}^{T}v_{1}\left(v_{2}^{(6)}-K_{4}v_{2}^{(4)}+K_{2}v_{2}^{(2)}+g(v_{2})\right)+\int_{-T}^{T}v_{2}v_{1}\left(g(v_{2})-g(v_{1})\right)$
$\displaystyle=B(T)+\int_{-T}^{T}v_{2}v_{1}\left(g(v_{2})-g(v_{1})\right),$

where the term $B(T)$ denotes all the boundary terms coming from the
integrations by parts. By Lemma 4.10, we have $B(T)\rightarrow 0$ as
$R\rightarrow+\infty$. However, since
$\int_{-T}^{T}v_{2}v_{1}(f(v_{2})-f(v_{1}))$ is a negative and strictly
decreasing function of $T$, we obtain a contradiction by choosing $T\gg 1$
large enough.

The lemma is proved. ∎

A direct consequence of this uniqueness result is that for the concrete values
of $K_{0},K_{2},K_{4}$ given by (2.3) and $p=\frac{2n}{n-6}$, one can compute
the homoclinic solution explicitly. This is precisely the only place in the
proof where the closed form of these parameters enters.

######  Corollary 4.17.

Let $v\in C^{6}(\mathbb{R})$ be a bounded solution to ($\mathcal{O}_{6}$). If
$\lim_{|t|\rightarrow\infty}v(t)=0$, then there exists $T\in\mathbb{R}$ such
that

$v(t)=c_{n}(2\cosh(t-T))^{-\gamma_{n}}.$

###### Proof.

A straightforward calculation shows that
$v_{0}(t)=c_{n}(2\cosh(t))^{-\gamma_{n}}$ solves ($\mathcal{O}_{6}$). Using
the conditions on $v$, we have that $v$ has a global maximum at some
$T\in\mathbb{R}$. Since $v^{(1)}(T)=0$, we can apply Lemma 4.16 to conclude
that $v(\cdot+T)=v_{0}$, and this finishes the proof. ∎

###  4.6. Classification

In what follows, we state the most important auxiliary result in the proof of
our main proposition.

######  Lemma 4.18.

Let $v\in C^{6}(\mathbb{R})$ be a solution to ($\mathcal{O}_{6}$). One of the
following three alternatives holds:

  * (a)

$v\equiv\pm a^{*}_{n}$ or $v\equiv 0$;

  * (b)

$v(t)=\pm c_{n}(2\cosh(t-T))^{-\gamma_{n}}$ for some $T\in\mathbb{R}$;

  * (c)

$v$ is periodic, has a unique local maximum and minimum per period, and is
symmetric with respect to its local extrema.

###### Proof.

Let $v\in C^{6}(\mathbb{R})$ be a solution to ($\mathcal{O}_{6}$) and let
$\\#Z^{(1)}({v})$ be its critical set as defined previously in (4.2). We now
distinguish several cases with respect to the cardinality of the critical set.

Case 1. $\\#Z^{(1)}({v})=0$.

In this case, $v$ is strictly monotone. In fact, up to replacing $v(t)$ by
$v(-t)$, we may assume that $v$ is strictly increasing, and so, by Lemma 4.8,
it follows $\ell_{\pm}=\lim_{t\rightarrow\pm\infty}v(t)\in\mathbb{R}$. By
Lemma 4.10, we are reduced to studying three situations as follows.

If $\ell_{-}=0$ and $\ell_{+}=a^{*}_{n}$, then using Lemma 4.10, we get
$\lim_{t\rightarrow-\infty}\mathcal{H}_{v}(t)=G(0)=0$, whereas
$\lim_{t\rightarrow+\infty}\mathcal{H}_{v}(t)=G(a^{*}_{n})<0$, a contradiction
to energy conservation. In the same way, a contradiction is obtained if
$\ell_{-}=-a^{*}_{n}$ and $\ell_{+}=0$.

It remains to consider the case $\ell_{-}=-a^{*}_{n}$ and
$\ell_{+}=a^{*}_{n}$. As before, we can apply Lemma 4.10

$\lim_{|t|\rightarrow\infty}\mathcal{H}_{v}(t)=G(a^{*}_{n})<0.$ (4.40)

On the other hand, by Corollary 4.13, we have that the following sign identity
holds

$\mathcal{H}_{v}(t)\geqslant\frac{1}{2}v^{(3)}(t)^{2}+\mathcal{R}(v(t))+G(v(t)).$
(4.41)

However, by evaluating the energy at some $t=t_{0}$ such that $v(t_{0})=0$ and
using (4.41), we obtain
$\mathcal{H}_{v}(t_{0})\geqslant\frac{1}{2}v^{(3)}(t_{0})^{2}+\mathcal{R}(0)+G(0)\geqslant
0$, which contradicts (4.40) and the energy conservation. The conclusion is
that $Z^{(1)}({v})=\varnothing$ cannot occur.

Case 2. $\\#Z^{(1)}({v})=1$.

Up to a translation, we may assume $Z^{(1)}({v})=\\{0\\}$. Thus, $v$ is
strictly monotone on $(-\infty,0)$ and $(0,\infty)$, which yields that
$\ell_{\pm}=\lim_{t\rightarrow\pm\infty}v(t)\in\mathbb{R}\cup\\{\pm\infty\\}$.
By Lemma 4.8, we have $\ell_{\pm}\in\mathbb{R}$, and so $v$ is bounded, and,
by Corollary 4.5, also even. Then, we find $\ell_{+}=\ell_{-}$. By Lemma 4.10,
only three cases can occur: $\ell_{+}=\ell_{-}=0$ or $\ell_{+}=\ell_{-}=\pm
a^{*}_{n}$. In this case, monotonicity implies that either $v>0$ or $v<0$, and
we conclude $v(t)=\pm c_{n}(2\cosh(t))^{-\gamma_{n}}$ by Corollary 4.17.

For the remaining cases, by otherwise replacing $v(t)$ by $-v(t)$, let us
assume without loss of generality that $\ell_{+}=\ell_{-}=a_{*}$. Using that
$v$ is strictly monotone on $[0,\infty)$, $v(0)\neq a^{*}_{n}$, and
$\mathcal{H}_{v}(0)\geqslant\frac{1}{2}v^{(3)}(0)^{2}+G(0)\geqslant 0$, one
can see that since $F$ attains its global minimal value only at $v=\pm
a^{*}_{n}$ it holds $v(0)=-a^{*}_{n}$. Hence $v$ changes sign; i.e., there
exists $t_{0}\in\mathbb{R}$ such that $v\left(t_{0}\right)=0$. By Corollary
4.5, we get that $v$ is anti-symmetric with respect to $t_{0}$, which is a
contradiction with $\ell_{\pm}>0$. Therefore, we obtained that if
$\\#Z^{(1)}({v})=1$, then $v(t)=\pm c_{n}(2\cosh(t))^{-\gamma_{n}}$.

Case 3. $\\#Z^{(1)}(v)\geqslant 2$.

By continuity of $v^{(1)}$, we find that unless $v$ is constant (and hence
$v\equiv\pm a^{*}_{n}$ or $v\equiv 0)$, the closed set $Z^{(1)}(v)$ cannot be
dense; that is, there exist real numbers $t_{1}<t_{2}$ such that
$v^{(1)}(t_{1})=v^{(1)}(t_{2})=0$ and $v^{(1)}\neq 0$ on $(t_{1},t_{2})$.
Then, by Lemma 4.3, we get that $v$ is bounded and therefore, we can use
Corollary 4.5 as in the proof of Lemma 4.16 to find that $v$ must be periodic
of period $2(t_{2}-t_{1})$. In addition, since $v$ is strictly monotone on
$(t_{1},t_{2})$, there is (strictly) only one maximum and minimum per period
interval. The symmetry with respect to the extrema follows from Corollary 4.5.
∎

###  4.7. Shooting technique

Finally, we are ready to prove the main result in this section.

###### Proof of Proposition 4.1.

The proof will be divided into three steps as follows:

Step 1. Proof of (i).

By Lemma 4.18, the only possible case on which the estimate (4.1) may fail to
is when $v$ is periodic. In this case, $v$ possesses a local minimum at, say,
$t_{0}\in\mathbb{R}$. Note that if $v$ has a zero then (4.1) is automatically
fulfilled, so we may assume that $v$ has a fixed sign and, up to replacing $v$
by $-v$, we may assume that $v>0$. However, using Lemma 4.9, either $v$ is
constant, and so $v\equiv a^{*}_{n}$, or $v(t_{0})<a^{*}_{n}$, which yields
that (4.1) holds with strict inequality, which in turn proves (i).

Step 2. Proof of (ii).

The value $a_{0}\in(0,a^{*}_{n})$ will be considered to be fixed throughout
the following argument. For any vector
$\boldsymbol{b}=(b_{1},b_{2})\in\mathbb{R}^{2}_{+}:=\\{\boldsymbol{b}:b_{j}\geqslant
0\;{\rm for}\;j=1,2\\}$, where $b_{j}=a_{2j}$ for $j=1,2$, let us denote by
$v_{\boldsymbol{b}}\in C^{6}(\mathbb{R})$ the unique solution to
($\mathcal{O}_{6}$) with the initial values

$v(0)=a_{0},\quad v^{(2)}(0)=b_{1},\quad v^{(4)}(0)=b_{2},\quad{\rm and}\quad
v^{(1)}(0)=v^{(3)}(0)=v^{(5)}(0)=0,$ (4.42)

and by $T_{\boldsymbol{b}}\in(0,\infty]$ its maximal forward time of
existence.

We now proceed to our shooting technique. Since $G(v)\rightarrow+\infty$ as
$v\rightarrow+\infty$, for any $C(a_{0})>0$, which will be chosen later, there
exists $R(a_{0})>0$ such that $G(v)>C(a_{0})$ for all $v>R(a_{0})$. In this
fashion, let us define the shooting decomposition sets

$S_{1}:=\left\\{\boldsymbol{b}\in\mathbb{R}^{2}_{+}:v_{\boldsymbol{b}}(t)<0\
{\rm for\ some}\ t\in(0,T_{\boldsymbol{b}})\right\\},$

and

$S_{2}:=\left\\{\boldsymbol{b}\in\mathbb{R}^{2}_{+}:v_{\boldsymbol{b}}(t)>R(a_{0})\
{\rm for\ some}\ t\in(0,T_{\boldsymbol{b}})\ {\rm and}\ v_{\boldsymbol{b}}>0\
{\rm in}\ [0,t]\right\\}.$

Clearly, $S_{1},S_{2}\subset\mathbb{R}^{m-1}_{+}$ are open because of the
standard continuous dependence of solutions on the initial conditions. Thus,
we are left to show that they are both non-empty and disjoint. This is the
content of the following claims.

Claim 1: $(b^{*},+\infty)^{2}\subset S_{2}\neq\varnothing$, where
$b^{*}=\max\\{\beta_{1},\beta_{2}\\}$ are the greatest solutions to the
algebraic equation $-K_{4}\beta^{4}+K_{2}\beta^{2}-\widehat{K}_{0}=0$ in
$\mathbb{R}$.

In fact, suppose that $b_{j}>b^{*}$ for any $j=1,2$. Hence, writing
($\mathcal{O}_{6}$) as

$v_{\boldsymbol{b}}^{(6)}=K_{4}v_{\boldsymbol{b}}^{(4)}-K_{2}v_{\boldsymbol{b}}^{(2)}-g(v_{\boldsymbol{b}})\quad{\rm
in}\quad\mathbb{R}$ (4.43)

and combining with (4.42), we find that $v_{\boldsymbol{b}}^{(6)}$ increases
initially. Thus, $v_{\boldsymbol{b}}^{(2)}(0)$ and
$v_{\boldsymbol{b}}^{(4)}(0)$ increase initially, and since the right-hand
side of (4.43) is positive initially, it is easy to check that they stay
positive on $[0,T_{\boldsymbol{b}})$. Whence, $v_{\boldsymbol{b}}^{(6)}>0$ in
$[0,T_{\boldsymbol{b}})$, which implies that $v^{(j)}_{\boldsymbol{b}}$ for
$j=0,\dots,5$ all keep increasing on $[0,T_{\boldsymbol{b}})$. Consequently,
if $T_{\boldsymbol{b}}=+\infty$ then $v_{\boldsymbol{b}}$ is unbounded. On the
other hand, since $g$ is locally Lipschitz, if $T_{\boldsymbol{b}}<+\infty$,
then $v_{\boldsymbol{b}}(t)\rightarrow+\infty$ as $t\rightarrow
T_{\boldsymbol{b}}$ . To summarize, $v_{\boldsymbol{b}}^{(1)}>0$ in
$[0,T_{\boldsymbol{b}})$ and $\lim_{t\rightarrow
T_{\boldsymbol{b}}}v_{\boldsymbol{b}}(t)=+\infty$ when $b_{j}\geqslant b^{*}$
for any $j=1,2$. Based on this, we can restrict our search to the so-called
shooting parameter set $\boldsymbol{b}\in[0,b^{*}]^{2}=:S$.

Claim 2: $(S_{1}\cap S_{2})\cap S=\varnothing$.

In fact, notice that for all $\boldsymbol{b}\in S$, we have the uniform energy
bound

$\mathcal{H}({v_{\boldsymbol{b}}})(0)\leqslant-
b_{2}b_{1}+b_{1}^{2}+G(a_{0})\leqslant(b_{1}-b_{2})b_{1}+G(a_{0}):=C(a_{0}).$
(4.44)

This implies that whenever $\boldsymbol{b}\in[0,b^{*}]^{2}$ and
$v_{\boldsymbol{b}}(t_{0})>R(a_{0})$, we must have
$v_{\boldsymbol{b}}^{(1)}(t_{0})\neq 0$, otherwise, by using Corollary 4.13,
the following identity holds

$\displaystyle\mathcal{H}({v_{\boldsymbol{b}}}(t_{0}))=\frac{1}{2}{{{v}_{\boldsymbol{b}}^{(3)}(t_{0})}}^{2}+\mathcal{R}({{v}_{\boldsymbol{b}}}(t_{0}))+G({{v}_{\boldsymbol{b}}}(t_{0}))\geqslant
C(a_{0}),$

where
$\mathcal{R}({{v}_{\boldsymbol{b}}}(t_{0}))=\mathcal{E}_{2}({{v}_{\boldsymbol{b}}}(t_{0})){{v}_{\boldsymbol{b}}}^{(2)}(t_{0})>0$
is given by (4.27). The last inequality contradicts (4.44) and the
conservation of energy in Proposition 2.10. In particular, if
$v_{\boldsymbol{b}}$ enters the interval $(R(a_{0}),+\infty)$, then
$v_{\boldsymbol{b}}$ cannot leave it again, and hence is certainly not a
periodic solution.

Claim 3: $0\in S_{1}\neq\varnothing$.

On the other hand, if $\boldsymbol{b}=\boldsymbol{0}$, we see from (4.43) that
$v_{\boldsymbol{0}}^{(6)}(0)=g(a_{0})<0$, and hence $v_{\boldsymbol{0}}$,
$v^{(2)}_{\boldsymbol{0}}$, and $v^{(4)}_{\boldsymbol{0}}$ are strictly
decreasing on some small interval $t\in(0,\tau)$ for some $\tau>0$. Since
$g(v)<0$ for $v\in(0,a_{0})$, we deduce from (4.43) that
$v_{\boldsymbol{0}}^{(j)}$ for $j=1,\dots,5$ stay strictly negative until
$v_{\boldsymbol{0}}$ reaches a negative value. Therefore, if
$\boldsymbol{b}=\boldsymbol{0}$, there must be $t_{0}$ such that
$v_{\boldsymbol{0}}(t_{0})<0$.

Since our shooting parameter interval $\mathbb{R}^{2}_{+}$ is connected, we
deduce that $S_{1}\cup S_{2}\neq\mathbb{R}^{2}_{+}$. Hence there exists
$\boldsymbol{b}^{*}\in\mathbb{R}^{2}_{+}\setminus\\{0\\}$ and a corresponding
solution $v^{*}:=v_{\boldsymbol{b}^{*}}$ such that $0\leqslant v^{*}\leqslant
R(a_{0})$, which means that $v^{*}$ is bounded. Combining this with the fact
that $g$ is locally Lipschitz, we get that $T_{\boldsymbol{b}^{*}}=\infty$. By
even reflection, we find a solution defined on $\mathbb{R}$, which we still
refer to as $v^{*}$. Since
$\boldsymbol{b}^{*}\in\mathbb{R}^{2}_{+}\setminus\\{0\\}$, we know $v^{*}$ has
a strict local minimum at the origin. Therefore, using the classification from
Lemma 4.18, $v^{*}$ must be periodic. Moreover, it has a unique local maximum
and minimum per period and is symmetric with respect to its extrema. The
uniqueness of $v^{*}$ up to translations follows from Lemma 4.15.

Step 3. Proof of (iii).

Using that $\lambda_{1}=\gamma_{n}$, we need to prove that $w:=v^{(1)}/v$
satisfies $w(t)<\lambda_{1}$ for all $t\geqslant 0$. Using Lemma 4.18, one can
find $t_{0}\in\mathbb{R}$ such that $v^{(1)}(t_{0})=0$, and so $w(t_{0})=0$.
Then, we reduced to prove the following claim

Claim 4. $M:=\\{t>t_{0}:w(t)\geqslant\lambda_{1}\\}=\varnothing$.

Indeed, notice that $w\in C^{5}(\mathbb{R})$ satisfies

$w^{(1)}=-w^{2}+\lambda_{1}+\frac{\Phi_{1}}{v},$ (4.45)

where $\Phi_{1}=\partial_{t}^{(2)}v-\lambda_{1}v$. Now, suppose by
contradiction that $M\neq\varnothing$ and let $t_{1}:=\inf M$. It is easy to
see that $t_{1}>t_{0}$, which yields $w^{(1)}(t_{1})\geqslant 0$. On the other
hand, since $w(t_{1})=\lambda_{1}$, (4.45) implies

$w^{(1)}(t_{1})=\frac{\Phi_{1}(t_{1})}{v(t_{1})},$

which combined with $v(t_{1})>0$ yields

$\Phi_{1}(t_{1})>0.$ (4.46)

Next, using the notation in the proof of Lemma 4.4, we can rewrite
($\mathcal{O}_{6}$) as

$-L_{\lambda_{1}\lambda_{2}}(\Phi_{2}):=-\Phi_{2}^{(2)}+\lambda_{1}\lambda_{2}\Phi_{2}=-f(v)\quad{\rm
in}\quad\mathbb{R},$

where $\Phi_{1},\Phi_{2},\Phi_{3}$ are defined in (4.3). According to Lemma
4.18, $v$ attains its maximum in $\mathbb{R}$. Since $v>0$, the maximum
principle implies that $\Phi_{2}<0$ in $\mathbb{R}$. Thus, since

$-L_{\lambda_{1}}(\Phi_{1}):=-\Phi_{1}^{(2)}+\lambda_{1}\Phi_{1}=\Phi_{2}\quad{\rm
in}\quad\mathbb{R},$

the maximum principle again implies that $\Phi_{1}<0$ in $\mathbb{R}$. This is
a contradiction with (4.46), which finishes the proof. ∎

##  5\. Proof of the main theorem

In this section, we are based on Proposition 3.1 and Proposition 4.1 to prove
Theorem 1.

###### Proof of Theorem 1.

First, using Proposition 3.1, it follows that $u$ is radially symmetric.
Notice that $v(t)=e^{\gamma_{n}t}u(e^{t})$ satisfies ($\mathcal{O}_{6}$), we
are in position to apply the classification result from Proposition 4.1. Hence
either $v\equiv a^{*}_{n}$ is constant or $v$ is periodic. Indeed, the only
case that remains to be excluded is that
$v(t)=c_{n}(2\cosh(t-T))^{\gamma_{n}}$ for some $T\in\mathbb{R}$. However, if
this holds, we would have that $v(t)\sim c_{n}e^{\gamma_{n}t}$ as
$t\rightarrow-\infty$ and hence the singularity of $u$ would be removable,
which is a contradiction since $0$ is a non-removable singularity. Thus,
either $v$ is constant or periodic.

Second, fixing $a_{0}:=\inf v$ and using Proposition 4.1 (i), it follows that
$a_{0}\in(0,a^{*}_{n}]$, and $a_{0}=a^{*}_{n}$ if and only if $v\equiv
a^{*}_{n}$. Moreover, when $a_{0}\in(0,a^{*}_{n}$), the function $v$ is
periodic with minimal value $a_{0}$. Therefore, by the Proposition 4.1 (ii),
we get that $v(t)=v_{a}(t+T)$ for some $T\in\mathbb{R}$.

Finally, a simple computation shows in Emden–Fowler coordinates that
$\partial^{(1)}_{r}u<0$ in $(0,+\infty)$ is equivalent to
$v^{(1)}<\gamma_{n}v$ in $\mathbb{R}$, which is true because of Proposition
4.1 (iii). This concludes the proof of our main theorem. ∎

Acknowledgments. This paper was finished when the first-named author held a
Post-doctoral position at the University of British Columbia, whose
hospitality he would like to acknowledge. He also wishes to express gratitude
to Professor João Marcos do Ó for his constant support and several valuable
conversations.

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