# Constant $Q$-curvature metrics with Delaunay ends: The nondegenerate case

João Henrique Andrade  ,  Rayssa Caju  ,  João Marcos do Ó  ,  Jesse Ratzkin
and  Almir Silva Santos
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, Federal University of Paraíba
58051-900, João Pessoa-PB, Brazil<EMAIL_ADDRESS>Department of
Mathematics, Federal University of Paraíba
58051-900, João Pessoa-PB, Brazil<EMAIL_ADDRESS>Department of Mathematics,
Universität Würzburg
97070, Würzburg-BA, Germany<EMAIL_ADDRESS>Department of Mathematics, Federal University of Sergipe
49100-000, Sao Cristovão-SE, Brazil<EMAIL_ADDRESS>
###### Abstract.

We construct a one-parameter family of solutions to the positive singular
$Q$-curvature problem on compact nondegenerate manifolds of dimension bigger
than four with finitely many punctures. If the dimension is at least eight we
assume that the Weyl tensor vanishes to sufficiently high order at the
singular points. On a technical level, we use perturbation methods and gluing
techniques based on the mapping properties of the linearized operator both in
a small ball around each singular point and in its exterior. Main difficulties
in our construction include controlling the convergence rate of the Paneitz
operator to the flat bi-Laplacian in conformal normal coordinates and matching
the Cauchy data of the interior and exterior solutions; the latter difficulty
arises from the lack of geometric Jacobi fields in the kernel of the
linearized operator. We overcome both these difficulties by constructing
suitable auxiliary functions.

###### Key words and phrases:

Paneitz operator, $Q$-curvature equation, Gluing construction, Prescribed
asymptotics, Isolated singularities, Geometric equation

###### 2020 Mathematics Subject Classification:

35J60, 35B09, 35J30, 35B40

This work was partially supported by São Paulo Research Foundation (FAPESP)
#2020/07566-3 and #2021/15139-0, Paraíba State Research Foundation (FAPESQ)
#3034/2021, and National Council for Scientific and Technological Development
(CNPq) #312340/2021-4, #403349/2021-4, and #429285/2016-7

###### Contents

  1. 1 Introduction
  2. 2 Delaunay-type solutions
    1. 2.1 Classification results and some estimates
    2. 2.2 Variations of Delaunay-type solutions
  3. 3 Function spaces
  4. 4 Model operators
    1. 4.1 Interior Poisson operator
    2. 4.2 Exterior Poisson operator
    3. 4.3 Navier-to-Neumann operator
  5. 5 Interior problem
    1. 5.1 Conformal normal coordinates
    2. 5.2 Linear analysis
      1. 5.2.1 Case $j>n$
      2. 5.2.2 Case $j=0$
      3. 5.2.3 Case $j=1,\ldots,n$
    3. 5.3 Nonlinear analysis
      1. 5.3.1 Fixed point argument
  6. 6 Exterior problem
    1. 6.1 Linear analysis
    2. 6.2 Nonlinear analysis
  7. 7 One-point gluing procedure
    1. 7.1 High Fourier modes
    2. 7.2 Constant functions
    3. 7.3 Coordinate functions
  8. 8 Proof of the main result

##  1\. Introduction

In the last several decades, much of the geometric analysis community has
explored curvature problems similar to and extending the classical Yamabe
problem regarding scalar curvature. The present paper constructs several new
metrics on a punctured manifold with constant (fourth order) $Q$-curvature.
Let $(M^{n},g)$ be a closed (compact and without boundary) Riemannian manifold
of dimension $n\geq 5$ and

$Q_{g}=\displaystyle-\frac{1}{2(n-1)}\Delta_{g}R_{g}+\frac{n^{3}-4n^{2}+16n-16}{8(n-1)^{2}(n-2)^{2}}R_{g}^{2}-\frac{2}{(n-2)^{2}}|{\rm
Ric}_{g}|^{2},$ (1.1)

where $R_{g}$ is the scalar curvature, ${\rm Ric}_{g}$ is the Ricci curvature,
and $\Delta_{g}$ is the Laplace-Beltrami operator. It is well-known that
$Q_{g}$ transforms under a conformal change of metric, denoted by
$\widetilde{g}=u^{{4}/{(n-4)}}g$ with $u\in C^{\infty}(M)$ and $u>0$,
according to the rule

$Q_{\widetilde{g}}=\frac{2}{n-4}u^{-\frac{n+4}{n-4}}P_{g}u,$ (1.2)

where the Paneitz operator $P_{g}$ is given by

$P_{g}u=\Delta_{g}^{2}u+\operatorname{div}_{g}\left(\frac{4}{n-2}\operatorname{Ric}_{g}(\nabla
u,\cdot)-\frac{(n-2)^{2}+4}{2(n-1)(n-2)}R_{g}\langle\nabla
u,\cdot\rangle\right)+\frac{n-4}{2}Q_{g}u.$ (1.3)

We refer the interested reader to [8, 9, 22, 10, 16, 45] for a thorough
background on the $Q$-curvature and on the Paneitz operator.

We are interested in the problem of prescribing the $Q$-curvature of a
conformal metric. For this, we adopt the normalization
$Q_{\widetilde{g}}={n(n^{2}-4)}/{8}$, which makes this curvature equals that
of the sphere ($\mathbb{S}^{n},g_{0})$ with its standard round metric. With
this normalization, we define the $Q$-curvature operator
$H_{g}:C^{4,\alpha}(M)\rightarrow C^{0,\alpha}(M)$, for some $\alpha\in(0,1)$,
by

$H_{g}(u)=P_{g}u-\frac{n(n-4)(n^{2}-4)}{16}|u|^{\frac{8}{n-4}}u,$ (1.4)

so that (1.2) can be reformulated as $H_{g}(u)=0$. The operator $H_{g}$ is
conformally covariant, in that

$H_{\widetilde{g}}(\phi)=u^{-\frac{n+4}{n-4}}H_{g}(u\phi)\quad{\rm for\
all}\quad\phi\in C^{\infty}(M).$ (1.5)

For a given finite set of points $\Lambda$, our main goal in this paper is to
find positive smooth solutions to the singular problem

$\begin{cases}H_{g}(u)=0\quad{\rm on}\quad M\backslash\Lambda\\\
\displaystyle\liminf_{x\rightarrow p}u(x)=\infty\quad\mbox{for each
}p\in\Lambda.\end{cases}$ ($\mathcal{Q}$)

When $(M^{n},g)$ is conformally equivalent to the round sphere
$(\mathbb{S}^{n},g_{0})$ and $\Lambda$ is a single point there exists no
solution to ($\mathcal{Q}$), as it is was proved by C. S. Lin [36] and J. Wei
and X. Xu [52]. The first result proving existence of solutions to the
positive singular $Q$-curvature problem is due to S. Baraket and S. Rebhi [4].
They provided a partial answer to the existence problem as follows

######  Theorem A ([4]).

For each natural number $k$ there exists a finite set $\Lambda$ with
cardinality $2k$ such that $\mathbb{S}^{n}\backslash\Lambda$ carries a
complete, constant $Q$-curvature metric conformal to the round metric.

Using variational bifurcation theory and topological methods R. G. Bettiol, P.
Piccione and Y. Sire [6] obtained the existence of infinitely many complete
metrics with constant positive $Q$-curvature on
$\mathbb{S}^{n}\backslash\mathbb{S}^{k}$ with $0\leq k\leq{(n-4)}/{2}$,
conformal to the round metric. Recently, for arbitrary closed Riemannian
manifolds, A. Hyder and Y. Sire [27] studied the positive singular
$Q$-curvature problem when $\Lambda$ is a smooth submanifold of $M$ with
Hausdorff dimension ${\rm dim}_{\mathcal{H}}(\Lambda)$ strictly between $0$
and $(n-4)/2$. Motivated by the results in [40], they established the
following

######  Theorem B ([27]).

Let $(M^{n},g)$ be a closed Riemannian manifold with $n\geq 5$ with semi-
positive $Q$-curvature and nonnegative scalar curvature. Let $\Lambda$ be a
connected smooth closed submanifold of $M$ with $0<{\rm
dim}_{\mathcal{H}}(\Lambda)<{(n-4)}/{2}$. Then, there exists an infinite-
dimensional family of complete metrics on $M\backslash\Lambda$ with positive
constant $Q$-curvature.

Let $(M^{n},g)$ be a closed Riemannian manifold with dimension $n\geq 5$ and
let $\Lambda$ be a finite set of points. Under two natural conditions on the
metric, namely, nondegeneracy and vanishing of the Weyl tensor $W_{g}$ up to
some suitable order, we prove existence of a one-parameter family of solutions
to ($\mathcal{Q}$). We say that $g$ is nondegenerate if the linearized
operator $L_{g}:C^{4,\alpha}(M)\rightarrow C^{0,\alpha}(M)$ is invertible for
some $\alpha\in(0,1)$. Here

$L_{g}u=P_{g}u-\frac{n(n^{2}-4)(n+4)}{16}u,$ (1.6)

that is, $L_{g}$ is the linearization of $H_{g}$ at the function $1$. Let
$\mathcal{M}$ be the space of smooth metrics in $M$. For $5\leq n\leq 7$, we
define $\mathcal{W}_{\Lambda}^{d}=\left\\{g\in\mathcal{M}:g\mbox{ is
nondegenerate}\right\\}$. For $n\geq 8$, and $d\in\mathbb{N}$, we set

$\mathcal{W}^{d}_{\Lambda}=\left\\{g\in\mathcal{M}:\begin{aligned} \
&\mbox{$\nabla_{g}^{j}W_{g}(p)=0$ for all $p\in\Lambda$ and $j=0,\dots,d$},\\\
&\mbox{\quad\quad\quad\quad\quad and $g$ is
nondegenerate.}\end{aligned}\right\\}.$

Our main theorem extends the results in [50, 13] to the context of constant
$Q$-curvature metrics

######  Theorem 1.1.

Let $\left(M^{n},g\right)$ be a closed Riemannian manifold with dimension
$n\geq 5$ and let $\Lambda$ be a finite set of points. Assume that
$Q_{g}={n(n^{2}-4)}/{8}$ and $g\in\mathcal{W}^{d}_{\Lambda}$ for
$d=\left[{(n-8)}/{2}\right]$. Then, there exist a constant $\varepsilon_{0}>0$
and a one-parameter family of complete metrics
$\\{\tilde{g}_{\varepsilon}\\}_{\varepsilon\in\left(0,\varepsilon_{0}\right)}$
on $M\backslash\Lambda$, such that each $\tilde{g}_{\varepsilon}$ is conformal
to $g$ with $Q$-curvature equal to $n(n^{2}-4)/8$. Near each singular point
$p\in\Lambda$ the metric $\tilde{g}_{\varepsilon}$ is asymptotic to one of the
Delaunay metrics described in Section 2. Moreover,
$\tilde{g}_{\varepsilon}\rightarrow g$ uniformly on compact sets of
$M\backslash\Lambda$ as $\varepsilon\rightarrow 0$.

In contrast with Theorem B, the assumption that the scalar curvature is
nonnegative is not necessary. This set of hypotheses allows one to recover a
maximum principle for the Paneitz operator, and it was first introduced by M.
Gursky and A. Malchiodi [21]. Our techniques are point-fixed based and do not
rely on such a principle. This condition was also used in [3] to prove
compactness of solutions to ($\mathcal{Q}$) in the spherical setting.

The vanishing condition on the Weyl tensor also appears in the scalar
curvature setting. Indeed, in [50], a similar hypothesis was used to prove
existence of solutions to the Yamabe problem with finitely many singularities,
which generalizes the results due to A. Byde [13] in the conformally flat
setting, at least around the singularities (see also [48]). Both results were
extended for the case of fully nonlinear equations and strongly coupled
systems [51, 14]. The vanishing condition is also one of the essential pieces
in the program proposed by R. M. Schoen to establish compactness results for
the Yamabe equation [49, 30], and it comes naturally from the Weyl vanishing
conjecture [49]; this was proved to be true for $3\leq n\leq 24$ [38, 34, 35,
30], and disproved otherwise [39].

The study of compactness for solutions to the $Q$-curvature equation is still
under development. Recently, G. Li [32] obtained compactness of solutions for
$5\leq n\leq 7$ assuming the conditions in [21] and without any hypothesis on
the Weyl tensor. Independently, this was also proved by Y. Y. Li and J. Xiong
[33] for $5\leq n\leq 9$ assuming the Weyl tensor to vanish at singular points
of a sequence of blowing-up solutions and the weaker hypothesis in [23]. In
addition, since the Weyl tensor and its covariant derivatives appear in the
expansion of the Green function for Paneitz operator, they observed that for
$n\geq 8$ its vanishing at the singular points is required. In our situation,
a similar phenomenon happens. When $5\leq n\leq 7$, we do not need any
assumption on the Weyl tensor in Theorem 1.1. Nevertheless, for $n\geq 8$ we
assume that it vanishes up to order $\left[{(n-8)}/{2}\right]$ at singular
points. This order comes up naturally in our method but is not known to be
optimal (see Remark 5.17).

We expect that the set of nondegenerate metrics is Baire generic with respect
to the Gromov-Hausdorff topology, similarly to what Beig, Chruściel and Schoen
[5] proved in the scalar curvature setting.

We construct the metrics in Theorem 1.1 by gluing together known examples of
constant $Q$-curvature metrics, a technique that usually requires a
nondegeneracy assumption. For example, Y.-J. Lin [37] used a similar
definition of nondegeneracy to perform a connected sum construction of
manifolds with constant $Q$-curvature. The most important example of a
degenerate manifold is the standard round sphere $(\mathbb{S}^{n},g_{0})$. To
see this, notice that the linearized operator (1.6) is given by

$L_{g_{0}}=\left(\Delta_{g_{0}}-\frac{n^{2}-4}{2}\right)\left(\Delta_{g_{0}}+n\right),$

which annihilates the restrictions of linear functions on $\mathbb{R}^{n+1}$
to $\mathbb{S}^{n}$.

As mentioned above, there exists no solution to the singular $Q$-curvature
problem in the round sphere with a one-point singularity. In contrast with the
conformally flat case, for a generic choice of the background metric, we can
find solutions to ($\mathcal{Q}$) in a closed manifold with one point removed.

To provide an example of a nondegenerate metric, consider the round sphere
$(\mathbb{S}^{m}(k),g_{k})$ with sectional curvature $k={(n-1)}{(m-1)}$ and
$n:=2m\geq 5$. This implies that the spectrum of the Laplacian is given by
$\operatorname{Spec}(\Delta_{g_{k}})=\\{i(m+i-1)k:\;i=0,1,\ldots\\}$. With
this normalization, the product metric $g=g_{k}+g_{k}$ in
$\mathbb{S}^{m}(k)\times\mathbb{S}^{m}(k)$ satisfies ${\rm Ric}_{g}=(n-1)g$,
$R_{g}=n(n-1)$ and $Q_{g}={n(n^{2}-4)}/{8}$, and
$L_{g}=(\Delta_{g}-\frac{n^{2}-4}{2})(\Delta_{g}+n)$, where
$\Delta_{g}=\Delta_{g_{k}}+\Delta_{g_{k}}$. Hence, it is not difficult to show

$\operatorname{Spec}(\Delta_{g})=\left\\{\frac{n-1}{m-1}\left(i(i+m-1)+j(j+m-1)\right):\;i,j=0,1,\ldots\right\\}.$

This implies that $n=2m\not\in\mbox{Spec}(\Delta_{g})$. Therefore,
$(\mathbb{S}^{3}(k)\times\mathbb{S}^{3}(k),g)$ is nondegenerate, and our main
theorem applies. Notice that in this dimension we do not require any
conditions on the Weyl tensor, and that this manifold is not locally
conformally flat.

We employ a standard gluing strategy based on mapping properties of the
linearized operator around an approximate solution in proving the existence of
solutions to ($\mathcal{Q}$). This strategy typically has three parts:
interior analysis, exterior analysis, and gluing procedure. First, we fix a
ball of sufficiently small radius around each isolated singularity. The
interior analysis then consists of constructing a solution inside this ball
being uniformly bounded and satisfying proper estimates on its norm. To this
end, the hypothesis on vanishing the Weyl tensor up to some orders is
necessary. Second, in the exterior analysis, we use the nondegeneracy
condition to study ($\mathcal{Q}$) on the complement of this ball, proving
that the solution operator is also uniformly bounded. Finally, we prove the
existence of an isomorphism, which we call the Navier-to-Neumann operator, and
use it to match the interior and exterior solutions up to third order on the
boundary. This is accomplished in parts, projecting the equation on its low
and high Fourier modes.

We encounter several difficulties in carrying out the strategy outlined in the
previous paragraph. First, one needs to control the convergence rate of the
Paneitz operator to the flat bi-Laplacian in conformal normal coordinates. In
dimension $n=5,6,7,$ this convergence comes naturally from local expansions in
conformal normal coordinates, but in dimension $n\geq 8,$ we only obtain
reasonable convergence rates after constructing an auxiliary function. Gluing
the interior and exterior solutions on the boundary up to third order is also
problematic because of the lack of geometric Jacobi fields on the kernel of
the linearized operator in the low-frequency mode, making the system involving
the gluing variables under-determined. To circumvent this problem, we
construct auxiliary functions and insert the missing variables into the
problem, making it solvable.

For the sake of clarity, we provide an intuitive picture of our method. Figure
1 shows summands for our gluing construction before any modifications, namely
the manifold minus a small geodesic ball and a punctured ball with the flat
metric.

$M\setminus B_{r}(p)$$B_{r}(p)\setminus\\{p\\}$ Figure 1. This figure shows
the summands in their original states.

Figure 2 shows the summands after we have conformally deformed each of them.
We use the Green’s function of the flat bi-Laplacian on the left and the
Delaunay metric on the right.

Figure 2. This figure shows the summands after conformal modification.

Now we explain the summands in our gluing procedure more precisely. The first
summand is the underlying nondegenerate manifold with a point removed
$M\backslash B_{r}(p)$, where $p\in\Lambda$ satisfies the Weyl vanishing
hypotheses $\mathcal{W}_{\Lambda}^{d}$. The second summand is one end of a
Delaunay metric. The Delaunay metric are all possible constant $Q$-curvature
metrics on the cylinder $\mathbb{R}\times\mathbb{S}^{n-1}$ (or, equivalently,
on $\mathbb{S}^{n}\backslash\\{p_{1},p_{2}\\}$) and we describe them in detail
in Section 2. Most importantly, one can characterize a Delaunay metric by
choosing a necksize parameter $\varepsilon$ and when this necksize is small
the conformal factor of the Delaunay metric is close to the Green’s function
for the flat bi-Laplacian. Thus, after conformally modifying the given metric
$g$ on $M$ by a multiple of the Green’s function whose pole is the gluing base
point $p$, we see that the two summands are geometrically sufficiently close,
making it possible for us to glue them together. After describing the Delaunay
metrics, we introduce some appropriate function spaces in Section 3. Next, we
construct and study our model operators in Section 4. Then, we construct our
solution operators for the geometric problem as perturbations of these model
operators. In this same section, we also prove that the associated Navier-to-
Neumann operator is an isomorphism. In Section 5 we complete our interior
analysis, using a fixed-point argument and the hypothesis on the Weyl tensor
to obtain a solution to ($\mathcal{Q}$) in the interior of a ball of small
radius around a singular point with fine estimates. In Section 6 we use the
nondegeneracy hypothesis to find a solution to ($\mathcal{Q}$) in the exterior
of this ball. In Section 7 we use the Navier-to-Neumann operator to match the
interior and exterior solutions on the boundary of the small ball around a
unique singular point. At the beginning of this section we explain why
matching the interior and exterior solutions to third order across their
common boundaries suffices to give us a smooth, global solution. In Section 8,
we discuss the necessary modifications to extend the gluing construction of
the last section for the case of multiple points; this yields the proof of
Theorem 1.1.

Acknowledgements: We would like to thank the referees for a close read of and
valuable comments regarding an earlier version of this manuscript.
Additionally the first four authors would like to thank the last author for
performing above and beyond the call of duty.

##  2\. Delaunay-type solutions

We introduce the (logarithmic) cylindrical coordinates. We present the
Delaunay-type solutions together with a classification result and some
estimates. Finally, we consider deformations of the Delaunay-type solutions.
These solutions will be used as approximating solutions to the interior
problem. In what follows we will use dots to denote derivatives with respect
to $t$.

If $g=u^{\frac{4}{n-4}}\delta$ is a complete metric in
$(\mathbb{R}^{n}\backslash\\{0\\},\delta)$ with constant $Q$-curvature
$Q_{g}=n(n^{2}-4)/8$, where $\delta$ is the Euclidean metric in
$\mathbb{R}^{n}$, then the function $u$ satisfies the following equation

$\Delta^{2}u=\frac{n(n^{2}-4)(n-4)}{16}u^{\frac{n+4}{n-4}}.$ (2.1)

It was proved by C.-S. Lin [36] that solutions to (2.1) in
$\mathbb{R}^{n}\backslash\\{0\\}$ with a nonremovable singularity at the
origin are radially symmetric. If we consider the conformal diffeomorphism

$\Phi:(\mathbb{R}\times\mathbb{S}^{n-1},\,g_{\rm
cyl}:=dt^{2}+d\theta^{2})\rightarrow(\mathbb{R}^{n}\backslash\\{0\\},\delta),$
(2.2)

given by $\Phi(t,\theta)=e^{-t}\theta$, we find $\Phi^{*}\delta=e^{-2t}g_{\rm
cyl}$. Also, if $g=u^{\frac{4}{n-4}}\delta$, then
$\Phi^{*}g=v^{\frac{4}{n-4}}g_{\rm cyl}$, where

$v(t)=e^{\frac{4-n}{2}t}u(e^{-t}\theta)=|x|^{\frac{n-4}{2}}u(x).$

In this logarithm cylindrical coordinates (also known as Endem-Fowler
coordinates), (2.1) is equivalent to

$P_{\rm cyl}v=\frac{n(n-4)(n^{2}-4)}{16}v^{\frac{n+4}{n-4}},$

where

$\displaystyle P_{\rm cyl}$ $\displaystyle=$
$\displaystyle\partial_{t}^{4}+\Delta_{\mathbb{S}^{n-1}}^{2}+2\Delta_{\mathbb{S}^{n-1}}\partial_{t}^{2}-\frac{n^{2}-4n+8}{2}\partial_{t}^{2}-\frac{n(n-4)}{2}\Delta_{\mathbb{S}^{n-1}}+\frac{n^{2}(n-4)^{2}}{16}.$

Restricting to radial functions reduces the last PDE to the ODE below

$\ddddot{v}-\frac{n^{2}-4n+8}{2}\ddot{v}+\frac{n^{2}(n-4)^{2}}{16}v-\frac{n(n-4)(n^{2}-4)}{16}v^{\frac{n+4}{n-4}}=0.$
(2.3)

Following the notation of [18], we write this ODE as

$\ddddot{v}-A\ddot{v}=f(v),$

with $A^{2}=4B+4(n-2)^{2}>4B$ and

$f(v)=Cv^{\frac{n+4}{n-4}}-Bv=\frac{n(n-4)}{16}v\left((n^{2}-4)v^{\frac{8}{n-4}}-n(n-4)\right),$
(2.4)

where

$A=\frac{n(n-4)+8}{2},\quad B=\frac{n^{2}(n-4)^{2}}{16}\quad\mbox{ and }\quad
C=\frac{n(n-4)(n^{2}-4)}{16}.$ (2.5)

###  2.1. Classification results and some estimates

Now we will present some recent results concerning the classification of
solutions to (2.3). Notice that one can find a first integral for this ODE
defined as

$H(x,y,z,w)=-xz+\frac{1}{2}y^{2}+\frac{n^{2}-4n+8}{4}z^{2}-\frac{n^{2}(n-4)^{2}}{32}w^{2}+\frac{(n-4)^{2}(n^{2}-4)}{32}w^{\frac{2n}{n-4}}.$
(2.6)

This Hamiltonian energy is constant along solutions to (2.3),
$H_{\varepsilon}=H(\dddot{v}_{\varepsilon},\ddot{v}_{\varepsilon},\dot{v}_{\varepsilon},v_{\varepsilon})=$
constant. Let

$v_{\rm sph}(t)=(\cosh t)^{\frac{4-n}{2}}\;\;\;{\rm and}\;\;\;v_{\rm
cyl}=\left(\frac{n(n-4)}{n^{2}-4}\right)^{\frac{n-4}{8}}$

be the spherical and cylindrical solutions, whose energies are

$H_{\rm sph}=0\quad{\rm and}\quad H_{\rm
cyl}=-\frac{(n-4)(n^{2}-4)}{8}\left(\frac{n(n-4)}{n^{2}-4}\right)^{\frac{n}{4}}<0,$

respectively. It was observed in [47] (see also [46]) that $H_{\varepsilon}<0$
and it is strictly decreasing function of $\varepsilon<0$.

In a recent paper, R. Frank and T. König [18] provided a full characterization
to the solutions to the ODE (2.3), as stated below

######  Theorem C ([18]).

Let $v\in C^{4}(\mathbb{R})$ be a solution to (2.3). Then $\inf|v|\leq v_{\rm
cyl},$ with equality if and only if $v$ is a non-zero constant. Moreover, one
of the following three alternatives holds:

  1. (a)

$v\equiv\pm v_{\rm cyl}$ or $v\equiv 0$.

  2. (b)

$v(t)=\cosh^{\frac{4-n}{2}}(t-T)$ for some $T\in\mathbb{R}$.

  3. (c)

Let $\varepsilon\in(0,v_{\rm cyl})$. Then there is a unique (up to
translations) bounded solution $v_{\varepsilon}\in C^{4}(\mathbb{R})$ of (2.3)
with minimal value $\varepsilon$. This solution is periodic, has a unique
local maximum and minimum per period and is symmetric with respect to its
local extrema.

We call $v_{\varepsilon}$ the _Delaunay-type solution_ with necksize
$\varepsilon$, and denote its minimal period by $T_{\varepsilon}$. As
$\varepsilon\rightarrow 0$ one has $T_{\varepsilon}\rightarrow\infty$ and
$v_{\varepsilon}(t+T_{\varepsilon}/2)\rightarrow\cosh^{\frac{4-n}{2}}(t)$
uniformly on compact sets. Also, for all $\varepsilon\in(0,v_{\rm cyl})$, we
have that $0<v_{\varepsilon}<1$, see [47] and [2]. We observe that
$\varepsilon=v_{\varepsilon}(0)=\min v_{\varepsilon}$,
$\dot{v}_{\varepsilon}(0)=0$ and $\ddot{v}_{\varepsilon}(0)\geq 0$. We write
the corresponding solution to (2.1) as
$u_{\varepsilon}(x)=|x|^{\frac{4-n}{2}}v_{\varepsilon}(-\log|x|)$.

Let then $\lambda$, $\mu$ and $\omega$ be real numbers such that
$\lambda+\mu=A$, $\lambda\mu=\omega$ and $A^{2}\geq 4\omega$. We see that
$\lambda$ and $\mu$ satisfy $\lambda^{2}-A\lambda+\omega=0$. Thus

$\lambda=\frac{A}{2}-\frac{1}{2}\sqrt{A^{2}-4\omega}\;\;\;\;\;\;{\rm
and}\;\;\;\;\;\;\mu=\frac{A}{2}+\frac{1}{2}\sqrt{A^{2}-4\omega},$ (2.7)

implying that the equation (2.3) is equivalent to

$\left\\{\begin{array}[]{l}\ddot{v}-\lambda v=w\\\ \ddot{w}-\mu w=f(v)+\omega
v,\end{array}\right.$ (2.8)

where $f$ is defined in (2.4). We can choose $\omega$ such that

$B-\frac{n(n+4)(n^{2}-4)}{16}\varepsilon^{\frac{8}{n-4}}<\omega<\frac{A^{2}}{4}$

and this implies that $f^{\prime}(v)+\omega>0$.

The proof of the next lemma is similar of [7, Lemma 5] and we include its
proof for completeness.

######  Lemma 2.1.

Let $\lambda\leq\gamma\leq\mu$. If $v$ is a positive solution to (2.3), then

$sign(\dddot{v}(t)-\gamma\dot{v}(t))=-sign(\dot{v}(t))$

for all $t\in\mathbb{R}$, where $sign(0)=0$.

###### Proof.

Let $t_{0}\in\mathbb{R}$. If $\dot{v}(t_{0})=0$, it follows by [18, Corollary
5] that $v(t_{0}+t)=v(t_{0}-t)$ for all $t\in\mathbb{R}$. Thus
$\dddot{v}(t_{0})=0$ the result follows. Then assume that $\dot{v}(t_{0})>0$.
In this case, since $v$ is bounded there exist real numbers $t_{1}$ and
$t_{2}$ such that $\dot{v}(t_{1})=\dot{v}(t_{2})=0$ and $\dot{v}(t)>0$ for all
$t\in(t_{1},t_{2})$. Again we have $\dddot{v}(t_{1})=\dddot{v}(t_{2})=0$.
Define $y=\dddot{v}-\gamma\dot{v}$. Thus, using (2.8) we get

$\begin{array}[]{rcl}\dot{y}&=&\ddddot{v}-\gamma\ddot{v}=\ddddot{v}-\lambda\ddot{v}+(\lambda-\gamma)\ddot{v}=\ddot{w}+(\lambda-\gamma)\ddot{v}\\\
&=&\mu w+f(v)+\omega v+(\lambda-\gamma)\ddot{v}\end{array}$

and

$\begin{array}[]{rcl}\ddot{y}&=&\mu\dot{w}+\dot{v}(f^{\prime}(v)+\omega)+(\lambda-\gamma)\dddot{v}\\\
&=&\mu(\dddot{v}-\lambda\dot{v})+\dot{v}(f^{\prime}(v)+\omega)+(\lambda-\gamma)\dddot{v}\\\
&=&\mu(y+\gamma\dot{v}-\lambda\dot{v})+\dot{v}(f^{\prime}(v)+\omega)+(\lambda-\gamma)(y+\gamma\dot{v})\\\
&=&(A-\gamma)y+\dot{v}(f^{\prime}(v)+\omega+(\gamma-\lambda)(\mu-\gamma)).\end{array}$

This implies that

$\left\\{\begin{array}[]{l}\ddot{y}-(A-\gamma)y=\dot{v}(f^{\prime}(v)+\omega+(\gamma-\lambda)(\mu-\gamma))>0\\\
y(t_{1})=y(t_{2})=0.\end{array}\right.$

Therefore, by the strong maximum principle, $y<0$ in $(t_{1},t_{2})$. In
particular, $y(t_{0})<0$.

If $\dot{v}(t_{0})<0$, the proof is similar. ∎

Using the Lemma 2.1 and that the energy (2.6) is negative along the Delaunay
solution we get that

$\left(\frac{A}{2}-\lambda\right)\dot{v}_{\varepsilon}^{2}+\frac{1}{2}\ddot{v}_{\varepsilon}^{2}\leq\dot{v}_{\varepsilon}(\dddot{v}_{\varepsilon}-\lambda\dot{v}_{\varepsilon})+\frac{B}{2}v_{\varepsilon}^{2}-\frac{n-4}{2n}Cv_{\varepsilon}^{\frac{2n}{n-4}}<\frac{B}{2}v_{\varepsilon}^{2},$
(2.9)

with $A/2-\lambda>0$, see (2.7).

######  Proposition 2.2.

For any $\varepsilon\in(0,v_{\rm cyl})$ and for all $t\geq 0$ the Delaunay-
type solution $v_{\varepsilon}$ satisfies the estimates

$\displaystyle\left|v_{\varepsilon}(t)-\alpha_{\varepsilon}\cosh\left(\frac{n-4}{2}t\right)-\beta_{\varepsilon}\cosh\left(\frac{n}{2}t\right)\right|\leq$
$\displaystyle\leavevmode\nobreak\
c_{n}\varepsilon^{\frac{n+4}{n-4}}e^{\frac{n+4}{2}t},$
$\displaystyle\left|\dot{v}_{\varepsilon}(t)-\frac{n-4}{2}\alpha_{\varepsilon}\sinh\left(\frac{n-4}{2}t\right)-\frac{n}{2}\beta_{\varepsilon}\sinh\left(\frac{n}{2}t\right)\right|\leq$
$\displaystyle\leavevmode\nobreak\
c_{n}\varepsilon^{\frac{n+4}{n-4}}e^{\frac{n+4}{2}t},$
$\displaystyle\left|\ddot{v}_{\varepsilon}(t)-\left(\frac{n-4}{2}\right)^{2}\alpha_{\varepsilon}\cosh\left(\frac{n-4}{2}t\right)-\left(\frac{n}{2}\right)^{2}\beta_{\varepsilon}\cosh\left(\frac{n}{2}t\right)\right|\leq$
$\displaystyle\leavevmode\nobreak\
c_{n}\varepsilon^{\frac{n+4}{n-4}}e^{\frac{n+4}{2}t},$
$\displaystyle\left|\dddot{v}_{\varepsilon}(t)-\left(\frac{n-4}{2}\right)^{3}\alpha_{\varepsilon}\sinh\left(\frac{n-4}{2}t\right)-\left(\frac{n}{2}\right)^{3}\beta_{\varepsilon}\sinh\left(\frac{n}{2}t\right)\right|\leq$
$\displaystyle\leavevmode\nobreak\
c_{n}\varepsilon^{\frac{n+4}{n-4}}e^{\frac{n+4}{2}t},$
$\displaystyle\left|\ddddot{v}_{\varepsilon}(t)-\left(\frac{n-4}{2}\right)^{4}\alpha_{\varepsilon}\cosh\left(\frac{n-4}{2}t\right)-\left(\frac{n}{2}\right)^{4}\beta_{\varepsilon}\cosh\left(\frac{n}{2}t\right)\right|\leq$
$\displaystyle\leavevmode\nobreak\
c_{n}\varepsilon^{\frac{n+4}{n-4}}e^{\frac{n+4}{2}t},$

for all $t\geq 0$, where

$\displaystyle\frac{n}{2(n-2)}\varepsilon<\alpha_{\varepsilon}$
$\displaystyle\displaystyle:=\frac{1}{2(n-2)}\left(\frac{n^{2}}{4}\varepsilon-\ddot{v}_{\varepsilon}(0)\right)<\frac{n}{4}\varepsilon$
(2.10) $\displaystyle-\frac{n-4}{4}\varepsilon<\beta_{\varepsilon}$
$\displaystyle\displaystyle:=\frac{1}{2(n-2)}\left(\ddot{v}_{\varepsilon}(0)-\frac{(n-4)^{2}}{4}\varepsilon\right)<-\frac{(n-4)(n+2)}{8n}\varepsilon^{\frac{n+4}{n-4}},$
(2.11)

for some positive constants $C_{n}$ and $c_{n}$ which depends only on $n$.

###### Proof.

First, note that by (2.9) we have

$|\ddot{v}_{\varepsilon}(0)|<\frac{n(n-4)}{4}\varepsilon$ (2.12)

and then we obtain (2.10). First, note that if $v_{\varepsilon}$ is any
solution to (2.3), then

$e^{\frac{n}{2}t}\left(e^{-nt}\left(e^{2t}\left(e^{(n-4)t}\left(e^{\frac{4-n}{2}t}v_{\varepsilon}\right)^{\prime}\right)^{\prime}\right)^{\prime}\right)^{\prime}=Cv_{\varepsilon}^{\frac{n+4}{n-4}}.$

Thus

$\displaystyle v_{\varepsilon}(t)=$
$\displaystyle\alpha_{\varepsilon}\cosh\frac{n-4}{2}t+\beta_{\varepsilon}\cosh\frac{n}{2}t$
(2.13)
$\displaystyle+\frac{n(n-4)(n^{2}-4)}{16}e^{\frac{n-4}{2}t}\int_{0}^{t}e^{(4-n)s}\int_{0}^{s}e^{-2x}\int_{0}^{x}e^{ny}\int_{0}^{y}e^{-\frac{n}{2}z}v_{\varepsilon}^{\frac{n+4}{n-4}}dzdydxds.$

Using that $v_{\varepsilon}\geq\varepsilon$ we obtain

$\displaystyle v_{\varepsilon}(t)$
$\displaystyle\geq\alpha_{\varepsilon}\cosh\frac{n-4}{2}t+\beta_{\varepsilon}\cosh\frac{n}{2}t$
$\displaystyle+\frac{n(n-4)(n^{2}-4)}{16}\varepsilon^{\frac{n+4}{n-4}}\left(\frac{2}{n^{2}(n-2)}\cosh\frac{n}{2}t-\frac{2}{(n-2)(n-4)^{2}}\cosh\frac{n-4}{2}t+\frac{16}{n^{2}(n-4)^{2}}\right),$

for all $t\geq 0$. Since $\alpha_{\varepsilon}>0$ we conclude the upper bound
in (2.10) and (2.11). Using (2.12) we obtain the lower bound of (2.11). Also,
using (2.13) we get

$0<v_{\varepsilon}(t)<\varepsilon\cosh\left(\frac{n-4}{2}t\right)\leq\varepsilon
e^{\frac{n-4}{2}|t|}$ (2.14)

This and (2.13) implies the estimates. ∎

For the next result we will write $f=O^{(l)}(K|x|^{k})$ to mean
$|\partial_{r}^{i}f|\leq c_{n}K|x|^{k-i}$, for all $i\in\\{1,\ldots,l\\}$ and
for some positive constant $c_{n}$ which depends only on $n$. Here $K>0$ is a
constant.

######  Corollary 2.3.

For any $\varepsilon\in(0,v_{\rm cyl})$ and any
$x\in\mathbb{R}^{n}\backslash\\{0\\}$ with $|x|\leq 1$, the Delaunay-type
solution $u_{\varepsilon}(x)$ satisfies the estimates

$u_{\varepsilon}(x)=\frac{\alpha_{\varepsilon}}{2}(1+|x|^{4-n})+\frac{\beta_{\varepsilon}}{2}(|x|^{2}+|x|^{2-n})+O^{(4)}(\varepsilon^{\frac{n+4}{n-4}}|x|^{-n}),$

where $\alpha_{\varepsilon}$ and $\beta_{\varepsilon}$ are given by (2.10).

###### Proof.

Remember that
$u_{\varepsilon}(x)=|x|^{\frac{4-n}{2}}v_{\varepsilon}(-\log|x|)$. Besides,
since $t=-\log|x|$, we get

$\displaystyle|x|^{\frac{4-n}{2}}\cosh\left(\frac{n-4}{2}t\right)$
$\displaystyle=\frac{1}{2}(1+|x|^{4-n}),$
$\displaystyle|x|^{\frac{4-n}{2}}\cosh\left(\frac{n}{2}t\right)$
$\displaystyle=\frac{1}{2}(|x|^{2}+|x|^{2-n}).$

From this we get the expansion for $u_{\varepsilon}(x)$, since $0<|x|\leq 1$
implies that $t=-\log|x|>0$. Also we have

$\displaystyle|x|^{\frac{4-n}{2}}\sinh\left(\frac{n-4}{2}t\right)$
$\displaystyle=\frac{1}{2}(|x|^{4-n}-1)$

and

$|x|\partial_{r}u_{\varepsilon}(x)=\frac{4-n}{2}u_{\varepsilon}(x)-|x|^{\frac{4-n}{2}}\dot{v}_{\varepsilon}(-\log|x|).$
(2.15)

Using the estimates of the derivative of $v_{\varepsilon}$ given by
Proposition 2.2 we obtain the estimates for the radial derivatives of
$u_{\varepsilon}$. ∎

###  2.2. Variations of Delaunay-type solutions

Applying rigid motions to the solutions to the PDE (2.1), we can obtain some
important families of solutions. We focus in two types of variations:
translations along the Delaunay axis and translations at infinity.

We can obtain the first family of solutions by simply noting that if $u$ is a
solution to (2.1), then $R^{\frac{4-n}{2}}u(R^{-1}x)$, for $R>0$, also solves
this equation. Applying this transformation to a Delaunay-type solution yields
the family

$\mathbb{R}^{+}\ni R\mapsto|x|^{\frac{4-n}{2}}v_{\varepsilon}(-\log|x|+\log
R).$

To construct the second family, define the standard inversion
$I:\mathbb{R}^{n}\backslash\\{0\\}\rightarrow\mathbb{R}^{n}\backslash\\{0\\}$
given by $I(x)=x|x|^{-2}$. It is well known that $I^{*}\delta=|x|^{-4}\delta$.
Hence, if $g=u^{\frac{4}{n-4}}\delta$, then $I^{*}g=(|x|^{4-n}u\circ
I)^{\frac{4}{n-4}}\delta.$ Recall that the Kelvin transform $\mathcal{K}$ is
given by

$\mathcal{K}(u)(x)=|x|^{4-n}u(x|x|^{-2}).$

By (1.2), we have

$\mathcal{K}(u)^{-\frac{n+4}{n-4}}P_{\delta}(\mathcal{K}(u))=(u\circ
I)^{-\frac{n+4}{n-4}}P_{\delta}u\circ I,$
$\Delta^{2}(\mathcal{K}(u))=\mathcal{K}(|x|^{8}\Delta^{2}u).$

Now suppose that $u$ is a solution to (2.1), then

$\begin{array}[]{rcl}\displaystyle\Delta^{2}\mathcal{K}(u)(x)&=&\displaystyle\mathcal{K}(|x|^{8}\Delta^{2}u)(x)=\mathcal{\mathcal{missing}}K(c(n)|x|^{8}u^{\frac{n+4}{n-4}})\\\
&=&\displaystyle c(n)|x|^{-4-n}(u\circ
I(x))^{\frac{n+4}{n-4}}=c(n)(\mathcal{K}(u)(x))^{\frac{n+4}{n-4}}.\end{array}$

Therefore, $\mathcal{K}(u)$ still is a solution to (2.1). The function
$u(x-a)$ also solves (2.1), but with a singularity in $a\in\mathbb{R}^{n}$
instead of the origin.

For the purpose of this paper, we consider the family of solutions

$u_{\varepsilon,R,a}(x)=\mathcal{K}(\mathcal{K}(u_{\varepsilon})(\cdot-a))(x)=|x-a|x|^{2}|^{\frac{4-n}{2}}v_{\varepsilon}\left(-\log|x|+\log\left|\frac{x}{|x|}-a|x|\right|+\log
R\right).$ (2.16)

This function has a singular point when $x=0$ and $x=a/|a|$. In the particular
case when $a=0$, as a direct consequence of Corollary 2.3 the following
asymptotic expansion

$\displaystyle u_{\varepsilon,R}(x)$
$\displaystyle=\frac{\alpha_{\varepsilon}}{2}\left(R^{\frac{4-n}{2}}+R^{\frac{n-4}{2}}|x|^{4-n}\right)+\frac{\beta_{\varepsilon}}{2}(R^{-\frac{n}{2}}|x|^{2}+R^{\frac{n}{2}}|x|^{2-n})+O^{(4)}(R^{\frac{n+4}{2}}\varepsilon^{\frac{n+4}{n-4}}|x|^{-n}),$
(2.17)

where $\alpha_{\varepsilon}$ and $\beta_{\varepsilon}$ are given by (2.10).

######  Proposition 2.4.

There exists a constant $r_{0}\in(0,1)$, such that for any $x$ and $a$ in
$\mathbb{R}^{n}$ with $|x|\leq 1$, $|a||x|<r_{0}$ , $R>0$ and
$\varepsilon\in(0,v_{\varepsilon})$ the solution $u_{\varepsilon,R,a}$
satisfies the estimate

$u_{\varepsilon,R,a}(x)=u_{\varepsilon,R}(x)+\left((n-4)u_{\varepsilon,R}(x)+|x|\partial_{r}u_{\varepsilon}(x)\right)\langle
a,x\rangle+O(|a|^{2}|x|^{\frac{8-n}{2}}),$ (2.18)

and if $R\leq|x|$ the estimate

$u_{\varepsilon,R,a}(x)=u_{\varepsilon,R}(x)+\left((n-4)u_{\varepsilon,R}(x)+|x|\partial_{r}u_{\varepsilon}(x)\right)\langle
a,x\rangle+O(|a|^{2}\varepsilon R^{\frac{4-n}{2}}|x|^{2}).$ (2.19)

###### Proof.

First, using the Taylor expansion we get

$|x-a|x|^{2}|^{\frac{4-n}{2}}=|x|^{\frac{4-n}{2}}+\frac{n-4}{2}\langle
a,x\rangle|x|^{\frac{4-n}{2}}+O(|a|^{2}|x|^{\frac{8-n}{2}})$ (2.20)

and

$\log\left|\frac{x}{|x|}-a|x|\right|=-\langle a,x\rangle+O(|a|^{2}|x|^{2}),$
(2.21)

for $|a||x|<r_{0}$ and some $r_{0}\in(0,1)$. Also

$\displaystyle v_{\varepsilon}$
$\displaystyle\left(-\log|x|+\log\left|\frac{x}{|x|}-a|x|\right|+\log
R\right)=v_{\varepsilon}\left(-\log|x|+\log R\right)$
$\displaystyle+\dot{v}_{\varepsilon}(-\log|x|+\log
R)\log\left|\frac{x}{|x|}-a|x|\right|+\ddot{v}_{\varepsilon}(-\log|x|+\log
R+t_{a,x})\left(\log\left|\frac{x}{|x|}-a|x|\right|\right)^{2}$
$\displaystyle=v_{\varepsilon}\left(-\log|x|+\log
R\right)-\dot{v}_{\varepsilon}(-\log|x|+\log R)\langle
a,x\rangle+\dot{v}_{\varepsilon}(-\log|x|+\log R)O(|a|^{2}|x|^{2})$
$\displaystyle+\ddot{v}_{\varepsilon}(-\log|x|+\log
R+t_{a,x})O(|a|^{2}|x|^{2}),$

for some $t_{a,x}\in\mathbb{R}$ with
$0<|t_{a,x}|<\left|\log\left|\frac{x}{|x|}-a|x|\right|\right|$. Note that
$t_{a,x}\rightarrow 0$ when $|a||x|\rightarrow 0$.

Note that, using (2.9) and (2.12) it follows that

$|\dot{v}_{\varepsilon}|\leq c_{n}v_{\varepsilon}\;\;\;\;{\rm
and}\;\;\;\;|\ddot{v}_{\varepsilon}|\leq c_{n}v_{\varepsilon}$ (2.22)

for some positive constant $c_{n}$ that depends only on $n$. Thus, by (2.16)
and (2.20) we obtain

$u_{\varepsilon,R,a}(x)=u_{\varepsilon,R}(x)+\left(\frac{n-4}{2}u_{\varepsilon,R}(x)-|x|^{\frac{4-n}{2}}\dot{v}_{\varepsilon}(-\log|x|+\log
R)\right)\langle a,x\rangle+O(|a^{2}||x|^{\frac{8-n}{2}}).$

By (2.15) this implies (2.18). Moreover, by (2.9) and (2.14), since
$-\log|x|+\log R\leq 0$ for $R\leq|x|$, then we get that
$v_{\varepsilon}(-\log|x|+\log R)$ and $v_{\varepsilon}(-\log|x|+\log
R+t_{a,x})$ are bounded by $c_{n}\varepsilon
R^{\frac{4-n}{2}}|x|^{\frac{n-4}{2}}$, for some positive constant $c_{n}$
which depends only on $n$. Therefore, we get (2.19). ∎

##  3\. Function spaces

In this section, we define some function spaces that will be useful in this
work. The first one is the weighted Hölder spaces in the punctured ball. It is
well-established in the literature that these spaces are the most convenient
spaces to define the linearized operator. The second one appears so naturally
in our results that it is more helpful to put its definition here. Finally,
the third one is the weighted Hölder spaces in which the exterior analysis
will be carried out. These are the same weighted spaces as in [41, 29].

######  Definition 3.1.

For each $k\in\mathbb{N}$, $r>0$, $0<\alpha<1$ and $\sigma\in(0,r/2)$, let
$u\in C^{k}(B_{r}(0)\backslash\\{0\\})$, set

$\|u\|_{(k,\alpha),[\sigma,2\sigma]}=\sup_{|x|\in[\sigma,2\sigma]}\left(\sum_{j=0}^{k}\sigma^{j}|\nabla^{j}u(x)|\right)+\sigma^{k+\alpha}\sup_{|x|,|y|\in[\sigma,2\sigma]}\frac{|\nabla^{k}u(x)-\nabla^{k}u(y)|}{|x-y|^{\alpha}}.$

Then, for any $\mu\in\mathbb{R}$, the space
$C^{k,\alpha}_{\mu}(B_{r}(0)\backslash\\{0\\})$ is the collection of functions
$u$ that are locally in $C^{k,\alpha}(B_{r}(0)\backslash\\{0\\})$ and for
which the norm

$\|u\|_{(k,\alpha),\mu,r}=\sup_{0<\sigma\leq\frac{r}{2}}\sigma^{-\mu}\|u\|_{(k,\alpha),[\sigma,2\sigma]}$

is finite.

The one result about these that we shall use frequently, and without comment,
is that to check if a function $u$ is an element of some $C_{\mu}^{0,\alpha}$,
say, it is sufficient to check that $|u(x)|\leq C|x|^{\mu}$ and $|\nabla
u(x)|\leq C|x|^{\mu-1}$. In particular, the function $|x|^{\mu}$ is in
$C_{\mu}^{k,\alpha}$ for any $k,$ $\alpha$, or $\mu$.

Notice that $C_{\mu}^{k,\alpha}\subseteq C_{\delta}^{l,\alpha}$ if
$\mu\geq\delta$ and $k\geq l$, and $\|u\|_{(l,\alpha),\delta}\leq
C\|u\|_{(k,\alpha),\mu}$ for all $u\in C_{\mu}^{k,\alpha}$.

######  Definition 3.2.

For each $k\in\mathbb{N}$, $0<\alpha<1$ and $r>0$. Let $\phi\in
C^{k}(\mathbb{S}_{r}^{n-1})$, set

$\|\phi\|_{(k,\alpha),r}:=\|\phi(r\cdot)\|_{C^{k,\alpha}(\mathbb{S}^{n-1})}.$

Then, the space $C^{k,\alpha}(\mathbb{S}^{n-1}_{r})$ is the collection of
functions $\phi\in C^{k}(\mathbb{S}^{n-1}_{r})$ for which the norm
$\|\phi\|_{(k,\alpha),r}$ is finite.

Next, consider an $n$-dimensional compact Riemannian manifold $(M^{n},g)$ and
a coordinate system $\Psi:B_{r_{1}}(0)\rightarrow M$ on $M$ centered at a
point $p=\Psi(0)\in M$, where $B_{r_{1}}(0)\subset\mathbb{R}^{n}$ is the ball
of radius $r_{1}$. For $0<r<s\leq r_{1}$ define
$M_{r}:=M\backslash\Psi(B_{r}(0))$ and $\Omega_{r,s}:=\Psi(A_{r,s}),$ where
$A_{r,s}:=\\{x\in\mathbb{R}^{n};r\leq|x|\leq s\\}$.

######  Definition 3.3.

For all $k\in\mathbb{N}$, $\alpha\in(0,1)$ and $\nu\in\mathbb{R}$, the space
$C_{\nu}^{k,\alpha}(M\backslash\\{p\\})$ is the space of functions $v\in
C_{\rm loc}^{k,\alpha}(M\backslash\\{p\\})$ for which the following norm is
finite

$\|v\|_{C_{\nu}^{k,\alpha}(M\backslash\\{p\\})}:=\|v\|_{C^{k,\alpha}(M_{r_{1}/2})}+\|v\circ\Psi\|_{(k,\alpha),\nu,r_{1}},$

where the norm $\|\cdot\|_{(k,\alpha),\nu,r_{1}}$ is the one defined in
Definition 3.1.

For all $0<r<s\leq r_{1}$, we can also define the spaces
$C_{\mu}^{k,\alpha}(\Omega_{r,s})$ and $C_{\mu}^{k,\alpha}(M_{r})$ to be the
space of restriction of elements of $C_{\mu}^{k,\alpha}(M\backslash\\{p\\})$
to $M_{r}$ and $\Omega_{r,s}$, respectively. These spaces is endowed with the
following norm

$\|f\|_{C^{k,\alpha}_{\mu}(\Omega_{r,s})}:=\sup_{r\leq\sigma\leq\frac{s}{2}}\sigma^{-\mu}\|f\circ\Psi\|_{(k,\alpha),[\sigma,2\sigma]}$

and

$\|h\|_{C^{k,\alpha}_{\mu}(M_{r})}:=\|h\|_{C^{k,\alpha}(M_{r_{1}/2})}+\|h\|_{C^{k,\alpha}_{\mu}(\Omega_{r,r_{1}})}.$

Notice that these norms are independent of the extension of the functions $f$
and $h$ to $M_{r}$.

##  4\. Model operators

In this section, we follow [28, Sections 11 and 13] to construct a Poisson
operator for the bi-Laplacian in the Euclidean space, which will be crucial to
perform the gluing construction in Section 7.

Let us set some notation. We introduce $(e_{j},\lambda_{j})$, the eigendata of
the Laplacian on $\mathbb{S}^{n-1}$. That is,
$\Delta_{\mathbb{S}^{n-1}}e_{j}+\lambda_{j}e_{j}=0$. We assume that the
eigenvalues are counted with multiplicity and that the eigenfunctions are
normalized so that their $L^{2}$-norm is equal to 1. It is well known that
$(e_{j},-\lambda_{j}^{2})$ are the eigendata of the bi-Laplacian on
$\mathbb{S}^{n-1}$, that is,
$\Delta_{\mathbb{S}^{n-1}}^{2}e_{j}=\lambda_{j}^{2}e_{j}$. Thus, for $\phi\in
L^{2}(\mathbb{S}^{n-1})$ we use the following decomposition

$\phi(\theta)=\sum_{j=0}^{\infty}\phi_{j}e_{j}(\theta).$

Define the orthogonal projections

$\pi^{\prime},\pi^{\prime\prime}:L^{2}(\mathbb{S}_{r}^{n-1})\rightarrow
L^{2}(\mathbb{S}_{r}^{n-1})$

by

$\pi^{\prime}\left(\sum_{j=0}^{\infty}\phi_{j}e_{j}\right)=\sum_{j=0}^{n}\phi_{j}e_{j},\quad\pi^{\prime\prime}(\phi)=\phi-\pi^{\prime}(\phi).$
(4.1)

It is appropriate to consider the product of Hölder spaces
$C_{\delta}^{4,\alpha}(\mathbb{S}_{r}^{n-1},\mathbb{R}^{2}):=C_{\delta}^{4,\alpha}(\mathbb{S}_{r}^{n-1})\times
C_{\delta}^{4,\alpha}(\mathbb{S}_{r}^{n-1})$, where $r>0$, $\alpha\in(0,1)$
and $\delta\in\mathbb{R}$. We also write the projection onto the high Fourier
modes as

$\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}_{r}^{n-1},\mathbb{R}^{2})):=\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}_{r}^{n-1}))\times\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}_{r}^{n-1})).$

###  4.1. Interior Poisson operator

First, we can prove the existence of the Poisson operator for the interior of
the punctured unit ball.

######  Proposition 4.1.

Let $0<\alpha<1$ be a fixed constant. There exists a bounded linear operator
$\mathcal{P}_{1}:\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}^{n-1}))\times
C^{4,\alpha}(\mathbb{S}^{n-1})\rightarrow
C^{4,\alpha}_{2}(B_{1}(0)\backslash\\{0\\}),$ such that for all
$\phi_{0}\in\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}^{n-1}))$ and
$\phi_{2}\in C^{4,\alpha}(\mathbb{S}^{n-1})$ it holds

$\left\\{\begin{array}[]{rclcc}\Delta^{2}\mathcal{P}_{1}(\phi_{0},\phi_{2})&=&0&{\rm
in}&B_{1}(0)\backslash\\{0\\}\\\
\pi^{\prime\prime}(\mathcal{P}_{1}(\phi_{0},\phi_{2}))&=&\displaystyle\phi_{0}&{\rm
on}&\partial B_{1}\\\ \Delta\mathcal{P}_{1}(\phi_{0},\phi_{2})&=&\phi_{2}&{\rm
on}&\partial B_{1}.\end{array}\right.$ (4.2)

Moreover, if $\pi^{\prime\prime}(\phi_{2})=0$, then

$\mathcal{P}_{1}(0,\phi_{2})=|x|^{2}\left(\frac{(\phi_{2})_{0}}{2n}+\frac{1}{6n-4}\sum_{j=1}^{n}(\phi_{2})_{j}x_{j}\right).$
(4.3)

###### Proof.

By linearity, we reduce the proof to two cases.

Case 1: $\phi_{2}\equiv 0$.

By [28, Proposition 11.25] and [44, Lemma 6.2], there exists a smooth function
$v_{\phi_{0}}\in C_{2}^{2,\alpha}(B_{1}(0)\backslash\\{0\\})$ satisfying the
following

$\begin{cases}\Delta v_{\phi_{0}}=0&\quad\mbox{ in
}B_{1}(0)\backslash\\{0\\}\\\
\pi^{\prime\prime}(v_{\phi_{0}})=\phi_{0}&\quad\mbox{ on }\partial
B_{1}.\end{cases}$

Thus, we define $\mathcal{P}_{1}(\phi_{0},0):=v_{\phi_{0}}$.

Case 2: $\phi_{0}\equiv 0$. This case is divided into two steps.

Step 1: $\pi^{\prime\prime}(\phi_{2})=0$.

By a simple calculation we see that the function in (4.3) satisfies (4.2) and

$\sup_{B_{1}\backslash\\{0\\}}|x|^{-2}\mathcal{P}_{1}(0,\phi_{2})\leq
c\|\phi_{1}\|_{C^{4,\alpha}(\mathbb{S}^{n-1})}$

for some constant $c>0$ independent of $\phi_{2}$.

Step 2: $\pi^{\prime}(\phi_{2})=0$.

Initially, by linearity, we may assume that
$\|{\phi}_{2}\|_{{C}^{4,\alpha}\left(\mathbb{S}^{n-1}\right)}=1$. Hence the
solution to the problem (4.2), denoted by $v_{0}$, can be obtained as the
limit $\rho\rightarrow 0$ of solutions $v_{\rho}$ to

$\begin{cases}\Delta^{2}v_{\rho}=0&\quad{\rm in}\quad B_{1}\setminus
B_{\rho}\\\ v_{\rho}=0&\quad{\rm on}\quad\partial B_{1}\\\ \Delta
v_{\rho}=\phi_{2}&\quad{\rm on}\quad\partial B_{1}\\\ \Delta
v_{\rho}=v_{\rho}=0&\quad{\rm on}\quad\partial B_{\rho}.\\\ \end{cases}$ (4.4)

Claim 1: There exists a constant $c>0$, independent of $\phi_{2}$ and $\rho$,
such that

$\sup_{B_{1}\setminus B_{\rho}}|x|^{-2}\left|v_{\rho}\right|\leq c.$

Arguing by contradiction, there would exist a sequence of numbers
$\\{\rho_{i}\\}$ and functions $v_{\rho_{i}}$ solving (4.4) for $\rho_{i}$
with $\sup_{B_{1}\setminus
B_{\rho_{i}}}|x|^{-2}\left|v_{\rho_{i}}\right|\rightarrow\infty$ as
$i\rightarrow\infty$. For each $r\in(\rho_{i},1)$ the function
$v_{\rho_{i}}(r\cdot)$ is $L^{2}$-orthogonal to $e_{0},\ldots,e_{n}$ on
$\mathbb{S}^{n-1}$. Now choose
$(r_{i},\theta_{i})\in(\rho_{i},1)\times\mathbb{S}^{n-1}$ such that

$A_{i}:=\sup_{\left(\rho_{i},1\right)\times\mathbb{S}^{n-1}}r^{-2}\left|v_{\rho_{i}}(r\theta)\right|=r_{i}^{-2}\left|v_{\rho_{i}}(r_{i}\theta_{i})\right|\rightarrow\infty\quad\mbox{as}\quad
i\rightarrow\infty.$

For $|x|\in[\rho_{i}r_{i}^{-1},r_{i}^{-1}]$, define

$\widetilde{v}_{\rho_{i}}(x)=r_{i}^{-2}{A_{i}^{-1}}v_{\rho_{i}}\left(r_{i}x\right)$

which satisfies

$\sup_{\left(\rho_{i}r_{i}^{-1},r_{i}^{-1}\right)\times\mathbb{S}^{n-1}}r^{-2}\left|\widetilde{v}_{\rho_{i}}(r\theta)\right|=|\widetilde{v}_{\rho_{i}}(\theta_{i})|=1.$
(4.5)

Arguing as in the proof of Lemma 5.8, in page 5.2.1, we conclude that the
sequences $\rho_{i}r_{i}^{-1}$ and $r_{i}^{-1}$ remain bounded away from 1. Up
to a subsequence, we can assume that the sequence
$\rho_{i}r_{i}^{-1}\rightarrow\tau_{1}\in[0,1)$ and
$r_{i}^{-1}\rightarrow\tau_{2}\in(1,+\infty]$. Using (4.5) and Schauder
estimates like in [19], by Arzelà–Ascoli Theorem we can suppose that
$\widetilde{v}_{\rho_{i}}$ converges to some biharmonic function
$v_{\infty}\in C^{4}(B_{\tau_{2}}\backslash B_{\tau_{1}})$ satisfying
$v_{\infty}(\theta_{\infty})=1$, for some $\theta_{\infty}$. If $\tau_{1}=0$
we consider $B_{0}=\\{0\\}$. Moreover, $v_{\infty}=\Delta v_{\infty}=0$ on the
boundary and by (4.5) we conclude that $v_{\infty}$ is not identically zero
and $|v_{\infty}(x)|\leq|x|^{2}$. Using an analogous argument as in the end of
the proof of the Lemma 5.8, we obtain a contradiction.

The estimates for the derivatives follow from Schauder’s regularity (see [19,
43]). This finishes the proof of the proposition. ∎

Using a simple scaling argument, we can also define the Poisson operator in a
more general context.

######  Corollary 4.2.

Let $0<\alpha<1$, $\mu\leq 2$ and $r>0$ be a fixed constants. There exists a
bounded linear operator
$\mathcal{P}_{r}:\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}_{r}^{n-1}))\times
C^{4,\alpha}(\mathbb{S}_{r}^{n-1}))\rightarrow
C^{4,\alpha}_{2}(B_{r}(0)\backslash\\{0\\})$ satisfying that for each
$\phi_{0}\in\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}_{r}^{n-1}))$ and
$\phi_{2}\in C^{4,\alpha}(\mathbb{S}_{r}^{n-1})$

$\displaystyle\left\\{\begin{array}[]{rclcc}\Delta^{2}\mathcal{P}_{r}(\phi_{0},\phi_{2})&=&0&{\rm
in}&B_{r}(0)\backslash\\{0\\}\\\
\pi^{\prime\prime}(\mathcal{P}_{r}(\phi_{0},\phi_{2}))&=&\displaystyle\phi_{0}&{\rm
on}&\partial B_{r}\\\
\Delta\mathcal{P}_{r}(\phi_{0},\phi_{2})&=&r^{-2}\phi_{2}&{\rm on}&\partial
B_{r}.\\\ \end{array}\right.$

Moreover, there exists a constant $C>0$ independently of $r$ such that

$\left\|\mathcal{P}_{r}\left(\phi_{0},\phi_{2}\right)\right\|_{(4,\alpha),\mu,r}\leq
Cr^{-2}\left\|(\phi_{0},\phi_{2})\right\|_{(4,\alpha),r},$

and if $\pi^{\prime\prime}(\phi_{2})=0$, then

$\mathcal{P}_{r}(0,\phi_{2})=r^{-2}|x|^{2}\left(\frac{(\phi_{2})_{0}}{2n}+\frac{1}{6n-4}r^{-1}\sum_{j=1}^{n}(\phi_{0})_{j}x_{j}\right).$

###### Proof.

Let us consider

$\mathcal{P}_{r}\left(\phi^{r}_{0},\phi^{r}_{2}\right)(x)=\mathcal{P}_{1}(\phi_{0},\phi_{2})\left(r^{-1}x\right),$

where $\phi^{r}_{i}(\theta):=\phi_{i}(r\theta)$ for $i=0,2$. This operator is
obviously bounded and satisfies

$\left\|\mathcal{P}_{r}\left(\phi^{r}_{0},\phi^{r}_{2}\right)\right\|_{(4,\alpha),\mu,r}=r^{-\mu}\left\|\mathcal{P}_{1}(\phi^{r}_{0},\phi^{r}_{2})\right\|_{(4,\alpha),\mu,1},$

which concludes the proof. ∎

###  4.2. Exterior Poisson operator

Similar to the last proposition, we prove the existence of a Poisson operator
for the exterior problem. This time we denote our boundary data as
$\psi_{0},\psi_{2}\in C^{4,\alpha}(\mathbb{S}^{n-1})$.

######  Proposition 4.3.

For each $0<\alpha<1$ there exists a bounded linear operator
$\mathcal{Q}_{1}:C^{4,\alpha}(\mathbb{S}^{n-1},\mathbb{R}^{2})\rightarrow
C^{4,\alpha}_{4-n}(\mathbb{R}^{n}\backslash B_{1})$ such that for all
$(\psi_{0},\psi_{2})\in C^{4,\alpha}(\mathbb{S}^{n-1},\mathbb{R}^{2})$

$\displaystyle\left\\{\begin{array}[]{rclcc}\Delta^{2}\mathcal{Q}_{1}(\psi_{0},\psi_{2})&=&0&{\rm
in}&\mathbb{R}^{n}\backslash B_{1}\\\
\mathcal{Q}_{1}(\psi_{0},\psi_{2})&=&\psi_{0}&{\rm on}&\partial B_{1}\\\
\Delta\mathcal{Q}_{1}(\psi_{0},\psi_{2})&=&\psi_{2}&{\rm on}&\partial
B_{1}.\\\ \end{array}\right.$ (4.9)

More specifically, we have that

$\mathcal{Q}_{1}(\psi_{0},\psi_{2})=|x|^{2-n}\sum_{j=0}^{\infty}|x|^{-j}\left((\psi_{0})_{j}+\frac{(\psi_{2})_{j}}{c_{2}(n,j)}(|x|^{2}-1)\right),$
(4.10)

where

$c_{2}(n,j)=\left(4-n+\frac{n-\sqrt{n^{2}+8+4\lambda_{j}}}{2}\right)\left(2+\frac{n-\sqrt{n^{2}+8+4\lambda_{j}}}{2}\right)$

and $\lambda_{j}$ is the $j$th eigenvalue of $\Delta_{\mathbb{S}^{n-1}}$.

######  Remark 4.4.

Recall that the eigenvalues of $\Delta_{\mathbb{S}^{n-1}}$ all have the form
$\lambda_{j}=l(n-2+l)$ for some $l=0,1,2,3,\dots$ In this form, the expression
for $c_{2}(n,j)$ simplifies to $(4-n-l)(2-l)$.

###### Proof.

Since the expression (4.10) satisfies (4.9), we need only to prove that

$\|\mathcal{Q}_{1}(\psi_{0},\psi_{2})\|_{(4,\alpha),4-n}\leq
c\|(\psi_{0},\psi_{2})\|_{(4,\alpha)}.$

To prove this, we consider two cases as follows.

Case 1: $\psi_{2}\equiv 0$.

This case is similar to the Case 1 of Proposition 4.1. See [28, Proposition
13.25].

Case 2: $\psi_{0}\equiv 0$.

We verify directly that there exists $c>0$ such that

$\sup_{\mathbb{R}^{n}\backslash
B_{1}}|x|^{n-4}|(|x|^{4-n}-|x|^{2-n})(\psi_{2})_{0}|\leq
c\|\psi_{2}\|_{L^{\infty}\left(\mathbb{S}^{n-1}\right)}.$

Notice that since $\mathbb{S}^{n-1}$ is compact, using Morrey’s inequality, we
have $H^{s}(\mathbb{S}^{n-1})\hookrightarrow
C^{0,\alpha}(\mathbb{S}^{n-1})\hookrightarrow L^{\infty}(\mathbb{S}^{n-1}),$
for $s>n/2$ and $\alpha=1-\frac{n}{2s}$. Considering $(\psi_{2})_{0}=0$, by
the definition of Sobolev fractional norm on the sphere (see, for instance,
[12, page 406]), there exists $c>0$ independently of $r>0$, such that

$\displaystyle\sup_{\mathbb{S}^{n-1}}r^{n-4}|\mathcal{Q}_{1}(0,\psi_{2})(r\;\cdot)|\leq
cr^{n-4}\|\mathcal{Q}_{1}(0,\psi_{2})(r\;\cdot)\|_{H^{s}\left(\mathbb{S}^{n-1}\right)}\leq
c\left[\sum_{j=1}^{\infty}\left(1+\lambda_{j}^{s}\right)\left|(\psi_{2})_{j}\right|^{2}r^{-j}\right]^{1/2}.$

Also, since $j>0$, observe that

$\sup_{r\geq 2}\sup_{j\geq 1}\left(1+\lambda_{j}^{s}\right)r^{-j}<\infty.$

Thus, we have

$\sup_{\mathbb{R}^{n}\backslash
B_{2}}|x|^{n-4}|\mathcal{Q}_{1}(0,\psi_{2})|\leq
c\|\psi_{2}\|_{L^{2}\left(\mathbb{S}^{n-1}\right)}\leq
c\|\psi_{2}\|_{L^{\infty}\left(\mathbb{S}^{n-1}\right)}.$

The maximum principle in [17], then implies that

$\sup_{\mathbb{R}^{n}\backslash
B_{1}}|x|^{n-4}|\mathcal{Q}_{1}(0,\psi_{2})|\leq
c\|\psi_{2}\|_{L^{\infty}\left(\mathbb{S}^{n-1}\right)}.$

Finally, using the last estimate, we can use standard (rescaled) elliptic
estimates to prove the estimates for the derivatives, which completes the
proof of the proposition. ∎

######  Corollary 4.5.

Let $0<\alpha<1$ and $r>0$. There exists a bounded linear operator

$\mathcal{Q}_{r}:C^{4,\alpha}(\mathbb{S}^{n-1}_{r},\mathbb{R}^{2})\rightarrow
C^{4,\alpha}_{4-n}(\mathbb{R}^{n}\backslash B_{r})$

satisfying that for each $(\psi_{0},\psi_{2})\in
C^{4,\alpha}(\mathbb{S}_{r}^{n-1},\mathbb{R}^{2})$

$\displaystyle\left\\{\begin{array}[]{rclcc}\Delta^{2}\mathcal{Q}_{r}(\psi_{0},\psi_{2})&=&0&{\rm
in}&\mathbb{R}^{n}\backslash B_{r}\\\
\mathcal{Q}_{r}(\psi_{0},\psi_{2})&=&\psi_{0}&{\rm on}&\partial B_{r}\\\
\Delta\mathcal{Q}_{r}(\psi_{0},\psi_{2})&=&r^{-2}\psi_{2}&{\rm on}&\partial
B_{r}.\\\ \end{array}\right.$

Moreover, there exists a constant $C>0$ such that

$\|\mathcal{Q}_{r}(\psi_{0},\psi_{2})\|_{C^{4,\alpha}_{4-n}(\mathbb{R}^{n}\backslash
B_{1})}\leq
Cr^{n-4}\|(\psi_{0},\psi_{2})\|_{C^{4,\alpha}(\mathbb{S}_{r}^{n-1},\mathbb{R}^{2})}.$

###### Proof.

As before, let us consider

$\mathcal{Q}_{r}\left(\psi^{r}_{0},\psi^{r}_{2}\right)(x)=\mathcal{Q}_{1}(\psi_{0},\psi_{2})\left(r^{-1}x\right),$

where $\psi^{r}_{i}(\theta):=\psi_{i}(r\theta)$ for $i=0,2$. The proof now
follows the same lines as in the interior case. ∎

###  4.3. Navier-to-Neumann operator

Finally we prove the main result of this section, which is the construction of
an isomorphism we call the Navier-to-Neumann operator. To this end, we combine
a Liouville-type result for bounded entire biharmonic functions and a standard
result from the theory of pseudodifferential operators.

######  Proposition 4.6.

The operator
$\mathcal{Z}_{1}:\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}^{n-1},\mathbb{R}^{2}))\rightarrow\pi^{\prime\prime}(C^{1,\alpha}(\mathbb{S}^{n-1},\mathbb{R}^{2}))$
given by

$\mathcal{Z}_{1}(\phi_{0},\phi_{2})=\left(\partial_{r}(\mathcal{P}_{1}(\phi_{0},\phi_{2})-\mathcal{Q}_{1}(\phi_{0},\phi_{2})),\partial_{r}(\Delta\mathcal{P}_{1}(\phi_{0},\phi_{2})-\Delta\mathcal{Q}_{1}(\phi_{0},\phi_{2}))\right)$

is an isomorphism.

###### Proof.

Notice that $\mathcal{Z}_{1}$ is a linear third order elliptic self-adjoint
pseudodifferential operator with principal symbol
$\sigma_{\mathcal{Z}_{1}}(\xi)=-2(|\xi|,|\xi|^{3})$. Hence, since
$\sigma_{\mathcal{Z}_{1}}(\xi)\neq 0$ whenever $\xi\neq 0$, $\mathcal{Z}_{1}$
is surjective by a classical result from the theory of linear operators.

Now it is sufficient to prove that this operator is also injective. Let us
take
$(\phi_{0},\phi_{2})\in{C}^{4,\alpha}\left(\mathbb{S}^{n-1},\mathbb{R}^{2}\right)$
satisfying $\mathcal{Z}_{1}(\phi_{0},\phi_{2})=0$ and define
$w_{(\phi_{0},\phi_{2})}\in C^{1,\alpha}(\mathbb{R}^{n})$ as

$w_{(\phi_{0},\phi_{2})}(x)=\begin{cases}\overline{w}_{(\phi_{0},\phi_{2})},\;{\rm
if}\;x\in B_{1}\\\ \widetilde{w}_{(\phi_{0},\phi_{2})}\;{\rm
if}\;x\in\mathbb{R}^{n}\setminus B_{1},\par\end{cases}$

where

$\overline{w}_{(\phi_{0},\phi_{2})}:=r^{1-n}\sum_{j=2}^{\infty}\left((\phi_{0})_{j}e_{j}+\frac{(\phi_{2})_{j}}{c_{2}(n,j)}(r^{2}-1)\right).$

and $\widetilde{w}_{(\phi_{0},\phi_{2})}$ is the bounded entire biharmonic
function defined in the exterior region $\mathbb{R}^{n}\setminus B_{1}$. Using
the Lioville theorem in [26], it follows that $w\equiv 0$ in
$C^{1,\alpha}(\mathbb{R}^{n})$, and so $(\phi_{0},\phi_{2})\equiv 0$ in
${C}^{4,\alpha}\left(\mathbb{S}^{n-1},\mathbb{R}^{2}\right)$,which proves the
proposition. ∎

As in the proofs of Corollaries 4.2 and 4.5 we obtain

$r\partial_{r}(\mathcal{P}_{r}(\phi^{r}_{0},\phi^{r}_{2})-\mathcal{Q}_{r}(\phi^{r}_{0},\phi^{r}_{2}))(r\theta)=\partial_{r}(\mathcal{P}_{1}(\phi_{0},\phi_{2})-\mathcal{Q}_{1}(\phi_{0},\phi_{2}))(\theta)$

and

$r^{3}\partial_{r}\Delta(\mathcal{P}_{r}(\phi^{r}_{0},\phi^{r}_{2})-\mathcal{Q}_{r}(\phi^{r}_{0},\phi^{r}_{2}))(r\theta)=\partial_{r}(\Delta\mathcal{P}_{1}(\phi_{0},\phi_{2})-\Delta\mathcal{Q}_{1}(\phi_{0},\phi_{2}))(\theta).$

Therefore, we obtain

######  Corollary 4.7.

The operator
$\mathcal{Z}_{r}:\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}_{r}^{n-1},\mathbb{R}^{2}))\rightarrow\pi^{\prime\prime}(C^{1,\alpha}(\mathbb{S}_{r}^{n-1},\mathbb{R}^{2}))$
given by

$\mathcal{Z}_{r}(\phi_{0},\phi_{2})=\left(r\partial_{r}(\mathcal{P}_{r}(\phi_{0},\phi_{2})-\mathcal{Q}_{r}(\phi_{0},\phi_{2})),r^{3}\partial_{r}(\Delta\mathcal{P}_{r}(\phi_{0},\phi_{2})-\Delta\mathcal{Q}_{r}(\phi_{0},\phi_{2}))\right)$

is an isomorphism, with norm bounded independent of $r$.

##  5\. Interior problem

Following the strategy described in the introduction, our goal in this section
is to construct a solution in a small punctured ball centered at the singular
point. Due to the conformal invariance of the $Q$-curvature equation, we will
use conformal normal coordinates centered at the singular point. In these
coordinates, we show the metric is sufficiently close to the Euclidean metric
using the vanishing assumption on the Weyl tensor. We use these coordinates to
show that the first terms in the expansion of $P_{g}(u_{\varepsilon,R})$ are
$L^{2}$-orthogonal to the low Fourier modes on the sphere, see Lemma 5.6. This
allows us to use an auxiliary function which will be important to define the
fixed point problem. By the expansion of the metric, it should be possible to
perturb a Delaunay-type solution to an exact solution to equation
($\mathcal{Q}$).

###  5.1. Conformal normal coordinates

Let $(M,g_{0})$ be a smooth Riemannian manifold of dimension $n\geq 5$. Given
a point $p\in M$, there exist $r>0$ and a smooth function $\mathcal{F}$ on $M$
such that the conformal metric $g=\mathcal{F}^{\frac{4}{n-4}}g_{0}$ satisfies

$\det g=1,\quad{\rm in}\quad B_{r}(p)$ (5.1)

in $g$-normal coordinates. In these coordinates $\mathcal{F}=1+O(|x|^{2})$ and
$g=\delta+O(|x|^{2})$. Additionally,

$R_{ij}(0)=0,\;\;R_{g}=O(|x|^{2})\;\;{\rm
and}\;\;\Delta_{g}R_{g}(0)=-\frac{1}{6}|W_{g}(0)|^{2},$ (5.2)

where $W_{g}$ is the Weyl tensor of the metric $g$. J. M. Lee and T. H. Parker
[31] first proved the existence of such coordinates, and later J. G. Cao [15]
and M. Günther [20] derived refined expansions of various curvature terms in
conformal normal coordinates.

######  Remark 5.1.

For the remainder of the paper we let $d=\left[\dfrac{n-8}{2}\right]$ for
$n\geq 8$ and $d=-1$ for $5\leq n\leq 7$.

Using the results in [24] and [25] we obtain improved estimates on curvature
terms if we assume that the Weyl tensor vanishes to sufficiently high order.
If the Weyl tensor satisfies

$\nabla^{l}W_{g}(p)=0,\;\;\mbox{for }l=0,1,\ldots,d,$ (5.3)

then

$g=\delta+O(|x|^{d+3}).$ (5.4)

This implies that $R_{g}=O(|x|^{\max\\{2,d+1\\}})$ and ${\rm
Ric}_{g}=O(|x|^{\max\\{1,d+1\\}})$. Therefore, by (1.1), we have
$Q_{g}=O(|x|^{\max\\{0,d-1\\}})$

Note that the Weyl tensor is conformally invariant, which implies that the
condition (5.3) is satisfied for any metric in the conformal class of $g$.
Also, for dimensions $5\leq n\leq 7$, the condition on $W_{g}$ is vacuous.

In conformal normal coordinates, we will always write

$g=\exp(h),$

where $h$ is a two-tensor on $\mathbb{R}^{n}$ satisfying $h_{ij}(x)x_{j}=0$
and $tr(h(x))=0$ (See [30]). Note that by $\eqref{eq035}$, we have
$h=O(|x|^{d+3})$.

Motivated by results in [1, 11] we prove the following lemma.

######  Lemma 5.2.

We have

$R_{jk}=\frac{1}{2}(\partial_{i}\partial_{j}h_{ki}+\partial_{i}\partial_{k}h_{ij}-\partial_{i}\partial_{i}h_{jk})+O(|\partial
h|^{2})+O(|h||\partial^{2}h|)$ (5.5)

and

$R_{g}=\partial_{i}\partial_{j}h_{ij}+O(|\partial
h|^{2})+O(|h||\partial^{2}h|)$ (5.6)

###### Proof.

First we observe that $g^{-1}=\exp(-h)$. The Ricci tensor is given by

$R_{jk}=\partial_{i}\Gamma_{jk}^{i}-\partial_{j}\Gamma_{ik}^{i}+\Gamma_{jk}^{p}\Gamma_{ip}^{i}-\Gamma_{ik}^{p}\Gamma_{jp}^{i}.$

Note that
$\Gamma_{ik}^{i}=\frac{1}{2}g^{il}\partial_{k}g_{il}=\frac{1}{2}\partial_{k}\log\det
g=0$, by (5.1). Therefore,
$R_{jk}=\partial_{i}\Gamma_{jk}^{i}-\Gamma_{ik}^{p}\Gamma_{jp}^{i},$ where the
Christoffel symbols are given by
$\Gamma_{jk}^{i}=\frac{1}{2}g^{il}(\partial_{j}g_{kl}+\partial_{k}g_{lj}-\partial_{l}g_{jk}).$
Thus, $\Gamma_{ij}^{k}=O(|\partial h|)$ and

$\partial_{i}\Gamma_{jk}^{i}=\frac{1}{2}(\partial_{i}\partial_{j}h_{ki}+\partial_{i}\partial_{k}h_{ij}-\partial_{i}\partial_{i}h_{jk})+O(|\partial
h|^{2})+O(|h||\partial^{2}h|).$

This implies (5.5). Since $R_{g}=g^{jk}R_{jk}$ and $tr(h(x))=0$ we obtain the
result. ∎

######  Lemma 5.3.

Let $\\{x_{i}\\}$ be conformal normal coordinates centered at $p$. If $u$ is
radial with respect to these coordinates then in a small geodesic ball
centered at $p$ it holds

$\nabla_{i}\nabla_{j}u=\frac{x_{i}x_{j}}{r^{2}}u^{\prime\prime}-\frac{x_{i}x_{j}}{r^{3}}u^{\prime}+\frac{\delta_{ij}}{r}u^{\prime}+O(|\partial
h|)|u^{\prime}|,$ (5.7)

and

$\Delta_{g}u=\Delta u=u^{\prime\prime}+\frac{n-1}{r}u^{\prime},$

where $\Delta$ is the Euclidean laplacian and we use ′ to denote
differentiation with respect to $r$.

The equation (5.7) follows from (5.1) and [21, Lemma 2.6].

######  Lemma 5.4.

We have $x_{j}x_{k}\partial_{i}\partial_{j}h_{ki}=x_{j}x_{k}\Delta
h_{jk}=x_{i}x_{j}\partial_{i}\partial_{k}h_{jk}=0.$

###### Proof.

Recall that $h_{ki}x_{k}=0$ and $trh=0$. Taking a derivative, we get
$\partial_{j}h_{ki}x_{k}+h_{ij}=0$ and

$x_{k}\partial_{i}h_{ki}=0,$ (5.8)

which implies that $\partial_{j}h_{ki}x_{k}x_{j}=0$. Differentiating again we
see
$\partial_{i}\partial_{j}h_{ki}x_{k}x_{j}+\partial_{j}h_{ii}x_{j}+\partial_{i}h_{ki}x_{k}=0$,
and so $x_{j}x_{k}\partial_{i}\partial_{j}h_{ki}=0$. Thus
$\partial_{i}h_{jk}x_{j}+h_{ik}=0$, which implies that
$\partial_{i}h_{jk}x_{j}x_{k}=0$. Finally,
$\partial_{i}\partial_{i}h_{jk}x_{j}x_{k}+\partial_{i}h_{ik}x_{k}+\partial_{i}h_{ji}x_{j}=0$.
Therefore $x_{j}x_{k}\Delta h_{jk}=0$. ∎

######  Lemma 5.5.

The functions $\partial_{i}\partial_{j}h_{ij}$,
$\Delta\partial_{i}\partial_{j}h_{ij}$ and
$x_{k}\partial_{k}\partial_{i}\partial_{j}h_{ij}$ are $L^{2}$-orthogonal to
the functions $\\{1,x_{1},\ldots,x_{n}\\}$ in $\mathbb{S}^{n-1}_{r}$.

###### Proof.

First observe that
$\partial_{i}\partial_{j}h_{ij}=\operatorname{div}(\partial_{j}h_{ij}e_{i})$
and $\partial_{j}h_{kj}=\operatorname{div}(h_{kj}e_{j})$. Using integration by
parts and (5.8), we get

$\int_{B_{r}}\partial_{i}\partial_{j}h_{ij}=\frac{1}{r}\int_{\mathbb{S}^{n-1}_{r}}x_{i}\partial_{j}h_{ij}=0$
(5.9)

and

$\int_{B_{r}}x_{k}\partial_{i}\partial_{j}h_{ij}=-\int_{B_{r}}\partial_{j}h_{kj}+\frac{1}{r}\int_{\mathbb{S}^{n-1}_{r}}x_{i}\partial_{j}h_{ij}=-\frac{1}{r}\int_{\mathbb{S}^{n-1}_{r}}h_{kj}x_{j}=0.$

Since
$\partial_{k}\partial_{i}\partial_{j}h_{ij}=\operatorname{div}(\partial_{k}\partial_{j}h_{ij}e_{i})$
and
$\partial_{k}\partial_{j}h_{lj}=\operatorname{div}(\partial_{k}h_{lj}e_{j})$,
using Lemma 5.4, (5.8) and (5.9), we obtain

$\int_{B_{r}}x_{k}\partial_{k}\partial_{i}\partial_{j}h_{ij}=-\int_{B_{r}}\partial_{k}\partial_{j}h_{kj}+\frac{1}{r}\int_{\mathbb{S}^{n-1}_{r}}x_{k}x_{i}\partial_{k}\partial_{j}h_{ij}=0$

and

$\begin{array}[]{rcl}\displaystyle\int_{B_{r}}x_{l}x_{k}\partial_{k}\partial_{i}\partial_{j}h_{ij}&=&\displaystyle-\int_{B_{r}}(x_{l}\partial_{k}\partial_{j}h_{kj}+x_{k}\partial_{k}\partial_{j}h_{lj})+\frac{1}{r}\int_{\mathbb{S}^{n-1}_{r}}x_{l}x_{k}x_{i}\partial_{k}\partial_{j}h_{ij}\\\
\\\
&=&\displaystyle-\int_{B_{r}}x_{k}\partial_{k}\partial_{j}h_{lj}=\int_{B_{r}}\partial_{k}h_{lk}-\frac{1}{2}\int_{\mathbb{S}^{n-1}_{r}}x_{k}x_{j}\partial_{k}h_{lj}=0.\end{array}$

By the previous results we get

$\int_{B_{r}}\Delta\partial_{i}\partial_{j}h_{ij}=\frac{1}{r}\int_{S_{r}}x_{k}\partial_{k}\partial_{i}\partial_{j}h_{ij}=0$

and

$\int_{B_{r}}x_{l}\Delta\partial_{i}\partial_{j}h_{ij}=-\int_{B_{r}}\partial_{l}\partial_{i}\partial_{j}h_{ij}=-\frac{1}{r}\int_{\mathbb{S}^{n-1}_{r}}x_{j}\partial_{l}\partial_{i}h_{ij}=\frac{1}{r}\int_{\mathbb{S}^{n-1}_{r}}\partial_{i}h_{il}=0.$

∎

######  Lemma 5.6.

In conformal normal coordinates at $p$, for sufficiently small $r=|x|$, one
has

$P_{g}u_{\varepsilon,R}=\Delta^{2}u_{\varepsilon,R}+\mathscr{P}(u_{\varepsilon,R})+\mathscr{R}$

where $\Delta$ is the Euclidean laplacian and

$\begin{array}[]{rcl}\mathscr{P}(u_{\varepsilon,R})&=&\displaystyle\frac{4}{n-2}\left(\frac{u^{\prime}_{\varepsilon,R}}{r}-\frac{(n-2)^{2}+4}{8(n-1)}\Delta
u_{\varepsilon,R}\right)\partial_{i}\partial_{j}h_{ij}+\frac{6-n}{2(n-1)}x_{i}\partial_{i}\partial_{j}\partial_{k}h_{jk}\frac{u^{\prime}_{\varepsilon,R}}{r}\\\
\\\
&&\displaystyle-\frac{n-4}{4(n-1)}u_{\varepsilon,R}\Delta\partial_{i}\partial_{j}h_{ij},\end{array}$

is $L^{2}$-orthogonal to the functions $\\{1,x_{1},\ldots,x_{n}\\}$ and
$\mathscr{R}=O(r^{d+4-\frac{n}{2}+\max\\{d,0\\}})$.

###### Proof.

By (1.3)

$\begin{array}[]{rcl}P_{g}u_{\varepsilon,R}&=&\displaystyle\Delta^{2}_{g}u_{\varepsilon,R}+\frac{4}{n-2}\langle{\rm
Ric}_{g},\nabla^{2}u_{\varepsilon,R}\rangle-\frac{(n-2)^{2}+4}{2(n-1)(n-2)}R_{g}\Delta_{g}u_{\varepsilon,R}\\\
&&\displaystyle+\frac{6-n}{2(n-1)}\langle\nabla R_{g},\nabla
u_{\varepsilon,R}\rangle+\frac{n-4}{2}Q_{g}u_{\varepsilon,R}.\end{array}$
(5.10)

Now by (5.5) and (5.7), we obtain

$\displaystyle\langle{\rm Ric}_{g},\nabla^{2}u_{\varepsilon,R}\rangle=$
$\displaystyle\leavevmode\nobreak\
g^{ij}g^{kl}R_{jk}\nabla_{i}\nabla_{l}u_{\varepsilon,R}$ $\displaystyle=$
$\displaystyle\leavevmode\nobreak\ \displaystyle
g^{ij}g^{kl}R_{jk}\left(\frac{x_{i}x_{l}}{r^{2}}u_{\varepsilon,R}^{\prime\prime}-\frac{x_{i}x_{l}}{r^{3}}u_{\varepsilon,R}^{\prime}+\frac{\delta_{il}}{r}u_{\varepsilon,R}^{\prime}+O(|\partial
h|)|u_{\varepsilon,R}^{\prime}|\right)$ $\displaystyle=$
$\displaystyle\leavevmode\nobreak\ \displaystyle
R_{g}\frac{u_{\varepsilon,R}^{\prime}}{r}+R_{jk}x_{j}x_{k}\left(\frac{u_{\varepsilon,R}^{\prime\prime}}{r^{2}}-\frac{u_{\varepsilon,R}^{\prime}}{r^{3}}\right)+O(|h||Ric|)\left(\frac{|u_{\varepsilon,R}^{\prime}|}{r}+|u_{\varepsilon,R}^{\prime\prime}|\right)$
$\displaystyle+O(|Ric||\partial h|)|u_{\varepsilon,R}^{\prime}|.$

Using that $h=O(r^{d+3})$, $Ric=O(r^{\max\\{1,d+1\\}})$ and
$u_{\varepsilon,R}=O(r^{\frac{4-n}{2}})$, as well (5.5) and (5.6) we obtain

$\begin{array}[]{rcl}\langle{\rm
Ric}_{g},\nabla^{2}u_{\varepsilon,R}\rangle&=&\displaystyle\partial_{i}\partial_{j}h_{ij}\frac{u_{\varepsilon,R}^{\prime}}{r}+\left(\partial_{i}\partial_{j}h_{ki}x_{j}x_{k}-\frac{1}{2}\partial_{i}\partial_{i}h_{jk}x_{j}x_{k}\right)\left(\frac{u_{\varepsilon,R}^{\prime\prime}}{r^{2}}-\frac{u_{\varepsilon,R}^{\prime}}{r^{3}}\right)\\\
&&+O(r^{d+4-\frac{n}{2}+\max\\{d,0\\}})\\\
&=&\displaystyle\partial_{i}\partial_{j}h_{ij}\frac{u_{\varepsilon,R}^{\prime}}{r}+O(r^{d+4-\frac{n}{2}+\max\\{d,0\\}}),\end{array}$
(5.11)

where in the last equality we used (5.8) and (5.9). Now observe that

$\langle\nabla R_{g},\nabla
u_{\varepsilon,R}\rangle=g^{ij}\nabla_{i}R_{g}\nabla_{j}u_{\varepsilon,R}=\partial_{i}\partial_{j}\partial_{k}h_{jk}x_{i}\frac{u^{\prime}_{\varepsilon,R}}{r}+O(r^{2d+4-\frac{n}{2}}).$

Using (1.1) and (5.6) we get

$R_{g}\Delta_{g}u_{\varepsilon,R}=\partial_{i}\partial_{j}h_{ij}\Delta
u_{\varepsilon,R}+O(r^{2d+4-\frac{n}{2}})$

and

$Q_{g}=-\frac{1}{2(n-1)}\Delta\partial_{i}\partial_{j}h_{ij}+O(r^{2d+2}).$

This finishes the proof.

∎

######  Lemma 5.7.

In conformal normal coordinates one has

$(\Delta^{2}-\Delta_{g}^{2})u_{\varepsilon,R,a}=O(|x|^{d+2-\frac{n}{2}}),$

where $\Delta$ is the Euclidean Laplacian.

###### Proof.

First we write

$(\Delta^{2}-\Delta_{g}^{2})u_{\varepsilon,R,a}=(\Delta^{2}-\Delta_{g}^{2})(u_{\varepsilon,R,a}-u_{\varepsilon,R})+(\Delta^{2}-\Delta_{g}^{2})u_{\varepsilon,R}.$

Since $u_{\varepsilon,R}$ is a radial function, using (5.1) and Lemma 2.6 in
[21] we have $\Delta_{g}^{2}u_{\varepsilon,R}=\Delta^{2}u_{\varepsilon,R}$.

Now, note that by Lemma 2.4 we have
$v=u_{\varepsilon,R,a}(x)-u_{\varepsilon,R}(x)=O(|x|^{\frac{6-n}{2}})$. Using
the Laplacian in coordinates and (5.4) we obtain

$\Delta_{g}v=\Delta
v+\partial_{i}g^{ij}\partial_{j}v+(g^{ij}-\delta^{ij})\partial_{i}\partial_{j}v,$
(5.12)

which implies

$\Delta_{g}^{2}v=\Delta^{2}v+\partial_{i}g^{ij}\partial_{j}\Delta
v+(g^{ij}-\delta^{ij})\partial_{i}\partial_{j}\Delta
v+\Delta_{g}(\partial_{i}g^{ij}\partial_{j}v+(g^{ij}-\delta^{ij})\partial_{i}\partial_{j}v).$
(5.13)

This implies the result. ∎

###  5.2. Linear analysis

In order to solve the singular Q-curvature problem we seek functions $u$ that
are close to a function $u_{0}$ such that $(u+u_{0})(x)\rightarrow+\infty$ as
$x\rightarrow p$, $u+u_{0}$ is a positive function, and $H_{g}(u_{0}+u)=0$.
This is done by considering the linearization about $u_{0}$ of the operator
$H_{g}$, defined in (1.4), which is given by

$L_{g}^{u_{0}}(u)=\frac{\partial}{\partial
t}\bigg{|}_{t=0}H_{g}(u_{0}+tu)=P_{g}u-\frac{n(n^{2}-4)(n+4)}{16}u_{0}^{\frac{8}{n-4}}u.$
(5.14)

Thus we can write

$H_{g}(u_{0}+u)=H_{g}(u_{0})+L_{g}^{u_{0}}(u)+\mathcal{R}^{u_{0}}(u),$

where the nonlinear term $\mathcal{R}^{u_{0}}(u)$ is independent of the metric
and given by

$\begin{array}[]{rcl}\mathcal{R}^{u_{0}}(u)&=&\displaystyle-\frac{n(n^{2}-4)(n-4)}{16}\left[|u_{0}+u|^{\frac{8}{n-4}}(u_{0}+u)-u_{0}^{\frac{n+4}{n-4}}-\frac{n+4}{n-4}u_{0}^{\frac{8}{n-4}}u\right].\end{array}$
(5.15)

This section is devoted to functional properties of the linearized operator
around a Delaunay solution in $\mathbb{R}^{n}$. In this case the $Q$-curvature
operator (1.4) is given by

$H_{\delta}(u)=\Delta^{2}u-\frac{n(n^{2}-4)(n-4)}{16}|u|^{\frac{8}{n-4}}u,$

and its linearization (5.14) around a solution $u_{\varepsilon,R,a}$ (see
(2.16)), denoted by $L_{\varepsilon,R,a}$, is given by

$L_{\varepsilon,R,a}(v)=\Delta^{2}v-\frac{n(n^{2}-4)(n+4)}{16}u_{\varepsilon,R,a}^{\frac{8}{n-4}}v.$
(5.16)

We abbreviate $L_{\varepsilon,R,0}=L_{\varepsilon,R}$ and
$L_{\varepsilon,1,0}=L_{\varepsilon}$.

Remember that if $g=u^{\frac{4}{n-4}}\delta$, then
$\Phi^{*}g=v^{\frac{4}{n-4}}g_{\rm cyl}$, where
$v(t)=e^{\frac{4-n}{2}t}u(e^{-t}\theta)=|x|^{\frac{n-4}{2}}u(x)$, i.e.
$u(x)=|x|^{\frac{4-n}{2}}v(-\log|x|)$, where $\Phi$ is defined in (2.2).
Hence, using (1.5) we see

$L_{\varepsilon,R}(u)=|x|^{-\frac{n+4}{2}}\mathcal{L}_{\varepsilon,R}(e^{\frac{n-4}{2}t}u\circ\Phi)\circ\Phi^{-1},$

where

$\mathcal{L}_{\varepsilon,R}(v)=P_{g_{\rm
cyl}}(v)-\displaystyle\frac{n(n+4)(n^{2}-4)}{16}v_{\varepsilon,R}^{\frac{8}{n-4}}v$

and

$P_{g_{\rm cyl}}(v)=\Delta_{g_{\rm cyl}}^{2}v-\frac{n(n-4)}{2}\Delta_{g_{\rm
cyl}}v-4\partial_{t}^{2}v+\frac{n^{2}(n-4)^{2}}{16}v.$

Using $\Delta_{g_{\rm cyl}}v=\partial_{t}^{2}v+\Delta_{\mathbb{S}^{n-1}}v$ we
obtain

$\begin{array}[]{rcl}\displaystyle\mathcal{L}_{\varepsilon,R}(v)&=&\displaystyle\partial_{t}^{4}v+\Delta_{\mathbb{S}^{n-1}}^{2}v+2\Delta_{\mathbb{S}^{n-1}}\partial_{t}^{2}v-\frac{n(n-4)}{2}\Delta_{\mathbb{S}^{n-1}}v-\frac{n(n-4)+8}{2}\partial_{t}^{2}v\\\
&&\displaystyle+\frac{n^{2}(n-4)^{2}}{16}v-\frac{n(n+4)(n^{2}-4)}{16}v_{\varepsilon,R}^{\frac{8}{n-4}}v.\end{array}$
(5.17)

Now we focus on solving the following linear problem with Navier boundary
conditions

$\left\\{\begin{array}[]{lcl}\mathcal{L}_{\varepsilon}(w)=f&{\rm in}&D_{R}\\\
\Delta w=w=0&{\rm on}&\partial D_{R},\end{array}\right.$ (5.18)

where $D_{R}:=(-\log R,\infty)\times\mathbb{S}^{n-1}$ and
$\mathcal{L}_{\varepsilon}$ is given by (5.17) with $R=1$.

We will follow closely the program due to R. Mazzeo and F. Pacard [41]. Due to
the geometry of the domain, it is natural to approach our problem by means of
a classical separation of variables, decomposing both $w$ and $f$ in Fourier
series and using the projections (4.1).

Projecting the linear problem (5.18) along (the space generated by) each
eigenfunction $e_{j}$ we obtain an infinite dimensional system of ODE’s

$\left\\{\begin{array}[]{lcl}\mathcal{L}^{j}_{\varepsilon}(w_{j})=f_{j}&{\rm
in}&(-\log R,\infty)\\\ \Delta w_{j}(-\log R)=w_{j}(-\log
R)=0,\end{array}\right.$

where the ordinary differential operator $\mathcal{L}^{j}_{\varepsilon}$ is
given by

$\begin{array}[]{rcl}\mathcal{L}_{\varepsilon}^{j}(w_{j})&=&\displaystyle\ddddot{w}_{j}-\left(2\lambda_{j}+\frac{n(n-4)+8}{2}\right)\ddot{w}_{j}\\\
&&\displaystyle+\left(\lambda_{j}^{2}+\frac{n(n-4)}{2}\lambda_{j}+\frac{n^{2}(n-4)^{2}}{16}-\frac{n(n+4)(n^{2}-4)}{16}v_{\varepsilon}^{\frac{8}{n-4}}\right)w_{j}.\end{array}$
(5.19)

The problem will be divided in three cases: $j>n$, $j=0$ and $j=1,\ldots,n$.

####  5.2.1. Case $j>n$

We consider the projections $w^{\prime\prime}=\pi^{\prime\prime}(w)$ and
$f^{\prime\prime}=\pi^{\prime\prime}(f)$ of $w$ and $f$ to the high Fourier
modes

$\left\\{\begin{array}[]{lcl}\mathcal{L}_{\varepsilon}(w^{\prime\prime})=f^{\prime\prime}&{\rm
in}&D_{R}\\\ \Delta w^{\prime\prime}=w^{{}^{\prime\prime}}=0&{\rm on}&\partial
D_{R}.\end{array}\right.$ (5.20)

First we solve the problem

$\left\\{\begin{array}[]{rcl}\mathcal{L}_{\varepsilon}(w_{T})(t,\theta)=\overline{f}&{\rm
in}&D_{R}^{T}\\\ \Delta w_{T}=w_{T}=0&{\rm on}&\partial
D_{R}^{T},\end{array}\right.$ (5.21)

where $D_{R}^{T}:=(-\log R,T)\times\mathbb{S}^{n-1}$. This solution can be
obtained variationally. Consider the energy function $\mathcal{E}_{T}$ given
by

$\mathcal{E}_{T}(w)=\langle P_{g_{\rm
cyl}}w,w\rangle-\int_{D_{R}^{T}}\left(\frac{n(n+4)(n^{2}-4)}{16}v_{\varepsilon}^{\frac{8}{n-4}}w^{2}+fw\right)d\theta
dt,$

where

$\begin{array}[]{rcl}\langle P_{g_{\rm
cyl}}w,w\rangle&=&\displaystyle\int_{D_{R}^{T}}\left(\Delta_{g_{\rm
cyl}}w\Delta_{g_{\rm cyl}}u-4A_{g_{\rm cyl}}(\nabla w,\nabla u)\right.\\\
&=&\displaystyle\int_{D_{R}^{T}}\left((\Delta_{g_{\rm
cyl}}w)^{2}+\frac{n(n-4)}{2}|\nabla_{g_{\rm
cyl}}w|^{2}+4\dot{w}^{2}+\frac{n^{2}(n-4)^{2}}{16}w^{2}\right)d\theta
dt.\end{array}$

It is easy to see that critical points of the functional $\mathcal{E}_{T}$ are
weak solutions to (5.21). Observe that

$\displaystyle\mathcal{E}_{T}(w)\geq$ $\displaystyle\leavevmode\nobreak\
\displaystyle\int_{D_{R}^{T}}\left(\lambda^{2}w^{2}+\ddot{w}^{2}-2\ddot{w}w\lambda+\frac{n(n-4)}{2}\lambda
w^{2}+\frac{n(n-4)+8}{2}\dot{w}^{2}\right.$ $\displaystyle\leavevmode\nobreak\
\displaystyle\left.+\frac{n^{2}(n-4)^{2}}{16}w^{2}-\frac{n(n+4)(n^{2}-4)}{16}v_{\varepsilon}^{\frac{8}{n-4}}w^{2}-fw\right)d\theta
dt$ $\displaystyle\geq$ $\displaystyle\leavevmode\nobreak\
\displaystyle\int_{D_{R}^{T}}\left(\ddot{w}^{2}+\left(2\lambda+\frac{n(n-4)+8}{2}\right)\dot{w}^{2}\right.$
$\displaystyle\leavevmode\nobreak\
\displaystyle\left.+\left(\lambda^{2}+\frac{n(n-4)}{2}\lambda+\frac{n^{2}(n-4)^{2}}{16}-\frac{n(n+4)(n^{2}-4)}{16}v_{\varepsilon}^{\frac{8}{n-4}}\right)w^{2}-fw\right)d\theta
dt.$

When $\lambda\geq 2n$ we have

$\lambda^{2}+\frac{n(n-4)}{2}\lambda+\frac{n^{2}(n-4)^{2}}{16}-\frac{n(n+4)(n^{2}-4)}{16}>0.$
(5.22)

This implies that $\mathcal{E}_{T}$ is coercive and convex in the space

$\displaystyle H^{2}_{0}(D_{R}^{T})^{\perp}:=\left\\{u\in
H^{2}(D_{R}^{T});\Delta u=u=0\mbox{ in }\partial D_{R}^{T}\ {\rm and}\
\int_{\mathbb{S}^{n-1}}u(\cdot,\theta)e_{j}(\theta)d\theta=0,\;\;j=0,\ldots,n\right\\}.$

The existence of a unique minimizer for $\mathcal{E}_{T}$ is immediate. The
standard elliptic theory yields the expected regularity issues for $w_{T}$ in
terms of the regularity of $f$.

######  Lemma 5.8.

For $2-n<\mu<2$, let $\delta=\frac{n-4}{2}+\mu$. Then there exist constants
$\varepsilon_{0}>0$ and $C>0$ independent of $T$ and $R$ such that for all
$\varepsilon\in(0,\varepsilon_{0})$, we have

$\sup_{(t,\theta)\in D_{R}^{T}}e^{\delta t}|w_{T}(t,\theta)|\leq
C\sup_{(t,\theta)\in D_{R}^{T}}e^{\delta t}|\overline{f}(t,\theta)|$ (5.23)

###### Proof.

The proof is by contradiction. Suppose that the estimate (5.23) does not hold.
This implies that we can find a sequence
$(\varepsilon_{i},T_{i},R_{i},w_{T_{i}},f_{i})$ such that

  1. (1)

$\varepsilon_{i}\rightarrow 0$ as $i\rightarrow\infty$.

  2. (2)

$\left\\{\begin{array}[]{lcl}\mathcal{L}_{\varepsilon}(w_{T_{i}})=f_{i}&{\rm
in}&D_{R_{i}}^{T_{i}}\\\ \Delta w_{T_{i}}=w_{T_{i}}=0&{\rm on}&\partial
D_{R_{i}}^{T_{i}}\end{array}\right.,$ for every $i\in\mathbb{N}$.

  3. (3)

$\displaystyle\sup_{(t,\theta)\in D_{R_{i}}^{T_{i}}}|e^{\delta
t}f_{i}^{\prime\prime}(t,\theta)|=1$, for every $i\in\mathbb{N}$.

  4. (4)

$\displaystyle\sup_{(t,\theta)\in D_{R_{i}}^{T_{i}}}e^{\delta
t}|w_{T_{i}}(t,\theta)|\rightarrow\infty$ as $i\rightarrow\infty$.

Now choose $t_{i}\in[-\log R_{i},T_{i}]$ such that

$A_{i}:=\sup_{(t,\theta)\in D_{R_{i}}^{T_{i}}}e^{\delta
t}|w_{T_{i}}(t,\theta)|=\sup_{\theta\in\mathbb{S}^{n-1}}e^{\delta
t_{i}}|w_{T_{i}}(t_{i},\theta)|.$

Define $w_{i}(t,\theta):=A_{i}^{-1}e^{\delta t_{i}}w_{T_{i}}(t+t_{i},\theta)$
and $f_{i}(t,\theta):=A_{i}^{-1}e^{\delta
t_{i}}f^{\prime\prime}(t+t_{i},\theta),$ for all $i\in\mathbb{N}$ and all
$(t,\theta)\in D_{i}:=(-\log R_{i}-t_{i},T_{i}-t_{i})\times\mathbb{S}^{n-1}$.
Then

$\sup_{(t,\theta)\in D_{i}}e^{\delta
t}|w_{i}(t,\theta)|=\sup_{\theta\in\mathbb{S}^{n-1}}|w_{i}(0,\theta)|=1$
(5.24)

and

$\sup_{(t,\theta)\in D_{i}}e^{\delta t}f_{i}(t,\theta)\rightarrow 0\;\;\mbox{
as }i\rightarrow\infty.$ (5.25)

Additionally

$\left\\{\begin{array}[]{lcl}\mathcal{L}_{i}(w_{i})=f_{i}&{\rm in}&D_{i}\\\
\Delta w_{i}=w_{i}=0&{\rm on}&\partial D_{i},\end{array}\right.$

where

$\mathcal{L}_{i}(w)=\displaystyle\Delta_{g_{\rm
cyl}}^{2}w-\displaystyle\frac{n(n-4)}{2}\Delta_{g_{\rm
cyl}}w-4\partial^{2}_{t}w+\displaystyle\frac{n^{2}(n-4)^{2}}{16}w-\frac{n(n+4)(n^{2}-4)}{16}v_{\varepsilon_{i}}(t+t_{i})^{\frac{8}{n-4}}w.$

Up to a subsequence, we can assume that $-\log
R_{i}-t_{i}\rightarrow\tau_{1}\in[-\infty,0]$ and
$T_{i}-t_{i}\rightarrow\tau_{2}\in[0,+\infty]$, as $i\rightarrow+\infty$.
Suppose $\tau_{1}=0$. Since $w_{i}(-\log R_{i}-t_{i},\theta)=0$ for every
$\theta\in\mathbb{S}^{n-1}$ and (5.24) holds, then $\\{|\nabla w_{i}|\\}$
blows up a region of the type $[-\log R_{i}-t_{i},-\log
R_{i}-t_{i}+1]\times\mathbb{S}^{n-1}$ as $i\rightarrow\infty$. On the other
hand, using (5.24) and (5.25) we see that $|w_{i}|$ and $|f_{i}|$ are bounded
by a positive constant times $e^{\delta(\log R_{i}+t_{i})}$ in the region
$[-\log R_{i}-t_{i},-\log R_{i}-t_{i}+1]\times\mathbb{S}^{n-1}$. Using
estimates as in [19, 43], we see that $|\nabla w_{i}|$ is also bounded by the
same quantities in $[-\log R_{i}-t_{i},-\log
R_{i}-t_{i}+1]\times\mathbb{S}^{n-1}$, which implies that $|\nabla w_{i}|$ is
uniformly bounded as $i\rightarrow+\infty$, contradicting the blow-up of
$|\nabla w_{i}|$. Therefore $\tau_{1}\not=0$. In the same way we conclude that
$\tau_{2}\not=0$.

Set $v_{i}(t):=v_{\varepsilon_{i}}(t+t_{i})$. By Proposition 2.2 we have that
over every compact set of $\mathbb{R}$ the $C^{4}$-norm of the functions
$v_{i}$ are uniformly bounded. Using the Arzelà–Ascoli Theorem, we deduce that
there is a subsequence, still denoted by $v_{i}$, and a function
$v_{\infty}\in C_{\rm loc}^{3}(\mathbb{R})$ such that $v_{i}\rightarrow
v_{\infty}$ in $C_{\rm loc}^{3}(\mathbb{R})$.

Claim 1:  Either $v_{\infty}\equiv 0$ or $0<v_{\infty}<1$. In the second case
we have $\displaystyle\lim_{t\rightarrow\pm\infty}v_{\infty}(t)=0$.

Since for each $i\in\mathbb{N}$ the function $v_{i}$ satisfies the ODE

$\ddddot{v}_{i}=\frac{n^{2}-4n+8}{2}\ddot{v}_{i}-\frac{n^{2}(n-4)^{2}}{16}v_{i}+\frac{n(n-4)(n^{2}-4)}{16}v_{i}^{\frac{n+4}{n-4}}$

and the right hand side converges to
$\frac{n^{2}-4n+8}{2}\ddot{v}_{\infty}-\frac{n^{2}(n-4)^{2}}{16}v_{\infty}+\frac{n(n-4)(n^{2}-4)}{16}v_{\infty}^{\frac{n+4}{n-4}}$
in $C^{0}_{\rm loc}(\mathbb{R})$, we conclude that $v_{i}\rightarrow
v_{\infty}$ in $C^{4}_{\rm loc}(\mathbb{R})$. Therefore, $v_{\infty}$
satisfies the ODE (2.3). Now, the claim follows by Theorem C, since
$v_{i}\rightarrow v_{\infty}$ in $C^{4}_{\rm loc}(\mathbb{R})$ as
$\varepsilon_{i}:=\min v_{i}\rightarrow 0$.

Combining (5.24), (5.25), and Schauder estimates (see [19]) the Arzela-Ascoli
theorem gives us a subsequence, which we still denote by $w_{i}$, that
converges to a function $w_{\infty}\in C^{2,\alpha}_{\rm
loc}((\tau_{1},\tau_{2})\times\mathbb{S}^{n-1})$ satisfying the equation

$\mathcal{L}_{\infty}(w_{\infty})=0\;\mbox{ in
}(\tau_{1},\tau_{2})\times\mathbb{S}^{n-1}$ (5.26)

in the sense of distributions, where

$\displaystyle\displaystyle\mathcal{L}_{\infty}=$
$\displaystyle\leavevmode\nobreak\ \displaystyle\Delta_{g_{\rm
cyl}}^{2}-\displaystyle\frac{n(n-4)}{2}\Delta_{g_{\rm
cyl}}-4\partial_{t}^{2}+\displaystyle\frac{n^{2}(n-4)^{2}}{16}-\displaystyle\frac{n(n+4)(n^{2}-4)}{16}v_{\infty}^{\frac{8}{n-4}}$
$\displaystyle=$ $\displaystyle\leavevmode\nobreak\
\displaystyle\partial_{t}^{4}+\Delta_{\mathbb{S}^{n-1}}^{2}+2\Delta_{\mathbb{S}^{n-1}}\partial_{t}^{2}-\frac{n(n-4)}{2}\Delta_{\mathbb{S}^{n-1}}-\frac{n(n-4)+8}{2}\partial_{t}^{2}$
$\displaystyle\leavevmode\nobreak\
\displaystyle+\frac{n^{2}(n-4)^{2}}{16}-\frac{n(n+4)(n^{2}-4)}{16}v_{\infty}^{\frac{8}{n-4}}.$

If $\tau_{i}$ is a real number, then the boundary condition becomes $\Delta
w_{\infty}(\tau_{i})=w_{\infty}(\tau_{i})=0$. It is important to point out
that by (5.24) we have that
$\displaystyle\sup_{\theta\in\mathbb{S}^{n-1}}|w_{\infty}(0,\theta)|=1.$ Thus,
$w_{\infty}$ is nontrivial. Decompose $w_{\infty}$ as

$w_{\infty}(t,\theta)=\sum_{j\geq n+1}w_{\infty}^{j}(t)e_{j}(\theta).$

Projecting (5.26) along the eigenfuctions $e_{j}$, for $j\geq n+1$, we obtain

$\displaystyle\ddddot{w}_{\infty}^{j}-$
$\displaystyle\left(2\lambda_{j}+\frac{n(n-4)+8}{2}\right)\ddot{w}_{\infty}^{j}$
$\displaystyle+\left(\lambda_{j}^{2}+\frac{n(n-4)}{2}\lambda_{j}+\frac{n^{2}(n-4)^{2}}{16}-\frac{n(n+4)(n^{2}-4)}{16}v_{\infty}^{\frac{8}{n-4}}\right)w_{\infty}^{j}=0.$

After multiplying this equation by $w_{\infty}^{j}$, we use integration by
parts and (5.22) to deduce that $w_{\infty}^{j}\equiv 0$, which is a
contradiction. To justify the integration by parts, note that if both
$\tau_{1}$ and $\tau_{2}$ are finite, we use that
$w_{\infty}^{j}(\tau_{1})=0$. Otherwise, since
$\displaystyle\lim_{r\rightarrow\pm\infty}v_{\infty}=0$ in case
$0<v_{\infty}<1$, we deduce that

$w_{\infty}^{j}(t)\sim e^{\pm\mu_{j}^{\pm}t},\;\;\;\mbox{ as
}|t|\rightarrow+\infty,$

where

$\mu_{j}^{\pm}:=\frac{\sqrt{n(n-4)+8+4\lambda_{j}\pm
4\sqrt{(n-2)^{2}+4\lambda_{j}}}}{2},$

and $\lambda_{j}\geq 2n$. We remark that $\mu_{j}^{\pm}\geq\frac{n}{2}$. On
the other hand, the condition $|w_{\infty}|\leq e^{-\delta t}$, together with
the fact that $|\delta|<\frac{n}{2}$, implies that $|w_{\infty}^{j}(t)|\leq
Ce^{-|t|\frac{n}{2}}$, as $|t|\rightarrow+\infty$. Therefore, in this case we
can also integrate by parts. This finishes the proof of the lemma ∎

Using a scaling argument as in [42, Proposition 6.2] we can see that if
$f^{\prime\prime}\in C^{0,\alpha}_{\delta}(D_{R})$, then $w_{T}\in
C^{4,\alpha}_{\delta}(D_{R}^{T})$. If $\gamma\in(-n/2,n/2)$ then there exist
$\varepsilon_{0}>0$ and a positive constant $C$, which does not depend on
$\varepsilon$, $R$ and $T$, such that for all
$\varepsilon\in(0,\varepsilon_{0}]$ we have

$\|w_{T}\|_{C^{2,\alpha}_{\delta}(D_{R}^{T})}\leq
C\|\overline{f}\|_{C^{0,\alpha}_{\delta}(D_{R}^{T})}.$

Therefore, letting $T\rightarrow+\infty$, we obtain the existence of a
solution $\ddot{w}$ to (5.20) fulfilling the estimate

$\|w^{{}^{\prime\prime}}\|_{C^{4,\alpha}_{\delta}(D_{R})}\leq
C\|f^{\prime\prime}\|_{C^{0,\alpha}_{\delta}(D_{R})},$

with $C>0$ independent of $R$ and $\varepsilon$.

####  5.2.2.  Case $j=0$

We start by considering the projection of the problem (5.18) along the
eigenfunction $\varphi_{0}$. We get

$\left\\{\begin{array}[]{lcl}\mathcal{L}^{0}_{\varepsilon}(w_{0})=f_{0}&{\rm
in}&(-\log R,\infty)\\\ \ddot{w}_{0}(-\log R)=0\\\ w_{0}(-\log
R)=0,\end{array}\right.$

where

$\begin{array}[]{rcl}\mathcal{L}_{\varepsilon}^{0}(w_{0})&=&\displaystyle\ddddot{w}_{0}-\frac{n(n-4)+8}{2}\ddot{w}_{0}+\left(\frac{n^{2}(n-4)^{2}}{16}-\frac{n(n+4)(n^{2}-4)}{16}v_{\varepsilon}^{\frac{8}{n-4}}\right)w_{0}.\end{array}$

Since we no longer have (5.22), it is necessary to use a different approach.
Choose an extension of $f_{0}$ to $\mathbb{R}$, still denoted by $f_{0}$. For
$T>-\log R$ consider the unique solution $w_{T}$ of the problem

$\mathcal{L}^{0}_{\varepsilon}(w_{T})=f_{0}\mbox{ in }(-\infty,T)$

with $\dddot{w}_{T}(T)=\ddot{w}_{T}(T)=\dot{w}_{T}(T)=w_{T}(T)=0.$

######  Lemma 5.9.

For $\mu>0$, let $\delta=\frac{n-4}{2}+\mu$. Then there are constants
$\varepsilon_{0}>0$ and $C>0$ independent of $T$ and $R>0$, such that, for
$\varepsilon\in(0,\varepsilon_{0}]$, we have

$\sup_{t\in(-\log R,T)}e^{\delta t}|w_{T}(t)|\leq C\sup_{t\in(-\log
R,T)}e^{\delta t}|f_{0}(t)|.$ (5.27)

###### Proof.

The proof is similar to that of Lemma 5.8. Suppose that (5.27) does not hold.
Then there is a sequence $(\varepsilon_{i},T_{i},R_{i},w_{T_{i}},f_{0,i})$
such that

  1. (1)

$\varepsilon_{i}\rightarrow 0$ as $i\rightarrow\infty$;

  2. (2)

$\mathcal{L}^{0}_{\varepsilon_{i}}(w_{T_{i}})=f_{0,i}$ in $(-\infty,T_{i})$,
with
$\dddot{w}_{T_{i}}(T_{i})=\ddot{w}_{T_{i}}(T_{i})=\dot{w}_{T_{i}}(T_{i})=w_{T_{i}}(T_{i})=0,$
for all $i\in\mathbb{N}$.

  3. (3)

$\displaystyle\sup_{t\in(-\log R_{i},T_{i})}e^{\delta t}|f_{0}(t,\theta)|=1$,
for every $i\in\mathbb{N}$;

  4. (4)

$\displaystyle\sup_{t\in(-\log R_{i},T_{i})}e^{\delta
t}|w_{T_{i}}(t,\theta)|\rightarrow\infty$ as $i\rightarrow\infty$.

There exists $t_{i}\in[-\log R_{i},T_{i}]$ such that

$A_{i}:=\sup_{t\in[-\log R_{i},T_{i}]}e^{\delta t}|w_{T_{i}}(t)|=e^{\delta
t_{i}}|w_{T_{i}}(t_{i})|.$

Now set $w_{i}(t)=A_{i}^{-1}e^{\delta t_{i}}w_{T_{i}}(t+t_{i})$ and
$f_{i}(t)=A_{i}^{-1}e^{\delta t_{i}}f_{0,i}(t+t_{i}),$ for $t\in[-\log
R_{i}-t_{i},T_{i}-t_{i}]$ and every $i\in\mathbb{N}$. Note that $|w_{i}(t)|$
and $|f_{i}(t)|$ are bounded by a positive constant times $e^{-\delta t}$.
Besides

$\mathcal{L}_{i}^{0}(w_{i})=f_{i}\textrm{ in }(-\log R_{i}-t_{i},T_{i}-t_{i})$
(5.28)

with
$\dddot{w}_{i}(T_{i}-t_{i})=\ddot{w}_{i}(T_{i}-t_{i})=\dot{w}_{i}(T_{i}-t_{i})=w_{i}(T_{i}-t_{i})=0$,
as well as

$\sup_{t\in[-\log R_{i}-t_{i},T_{i}-t_{i}]}e^{\delta
t}|w_{i}(t)|=|w_{i}(0)|=1$

and

$\sup_{t\in[-\log R_{i}-t_{i},T_{i}-t_{i}]}e^{\delta t}|f_{i}(t)|\rightarrow
0,\;\;\mbox{ as }i\rightarrow\infty,$

where

$\mathcal{L}_{i}^{0}(w_{i})=\ddddot{w}_{i}-\frac{n(n-4)+8}{2}\ddot{w}_{i}+\left(\frac{n^{2}(n-4)^{2}}{16}-\frac{n(n+4)(n^{2}-4)}{16}v_{i}^{\frac{8}{n-4}}\right)w_{i}$

and $v_{i}(t)=v_{\varepsilon_{i}}(t+t_{i})$. Up to a subsequence, $-\log
R_{i}-t_{i}\rightarrow\tau_{1}\in[-\infty,0]$ and
$T_{i}-t_{i}\rightarrow\tau_{2}\in[0,\infty]$.

Claim: $\tau_{2}>0$.

If $\tau_{2}=0$, since $|w_{i}(0)|=1$ and $w_{i}(T_{i}-t_{i})=0$, then
$\dot{w}_{i}$ must blow up in $[T_{i}-t_{i}-1,T_{i}-t_{i}]$, as
$i\rightarrow\infty$. Using that $|w_{i}(t)|,|f_{i}(t)|\leq
Ce^{-\delta(T_{i}-t_{i})}$ in this interval and the boundary condition in
(5.28), we integrate the ODE to obtain that

$|\dot{w}_{i}(t)|\leq Ce^{-\delta(T_{i}-t_{i})},$

in the same interval. Thus, since $T_{i}-t_{i}\rightarrow 0$, we obtain a
contradiction. The claim follows.

The claim implies that the interval $(\tau_{1},\tau_{2})$ is not empty.

As in the proof of Lemma 5.8 we can suppose that $v_{i}\rightarrow v_{\infty}$
in $C^{4}_{\rm loc}(\mathbb{R})$, where $v_{\infty}$ satisfies (2.3) and is
such that either $v_{\infty}\equiv 0$ or $0<v_{\infty}<1$ with
$\displaystyle\lim_{t\rightarrow\pm\infty}v_{\infty}(t)=0$. Passing to a
subsequence if necessary, there exists a function $w_{\infty}$ such that
$w_{i}\rightarrow w_{\infty}$ in $C^{3}_{\rm loc}(\tau_{1},\tau_{2})$ and
satisfies the equation

$\mathcal{L}_{\infty}^{0}(w_{\infty})=\ddddot{w}_{\infty}-\frac{n(n-4)+8}{2}\ddot{w}_{\infty}+\left(\frac{n^{2}(n-4)^{2}}{16}-\frac{n(n+4)(n^{2}-4)}{16}v_{\infty}^{\frac{8}{n-4}}\right)w_{\infty}=0$

in $(\tau_{1},\tau_{2})$. By the normalization $|w_{i}(0)|=1$ we have
$|w_{\infty}(0)|=1$, and so $w_{\infty}$ is not trivial.

If $\tau_{2}<\infty$, then
$\dddot{w}_{\infty}(\tau_{2})=\ddot{w}_{\infty}(\tau_{2})=\dot{w}_{\infty}(\tau_{2})=w_{\infty}(\tau_{2})=0$
implies that $w_{\infty}\equiv 0$, which is a contradiction. If
$\tau_{2}=\infty$, we use the fact that $v_{\infty}\rightarrow 0$ as
$|t|\rightarrow\infty$ to conclude that

$w_{\infty}(t)\sim e^{\pm\mu t}\quad\mbox{ as }\quad t\rightarrow+\infty,$

where $\mu=\frac{n}{2}$ or $\frac{n-4}{2}$. Using that $|w_{\infty}(t)|\leq
e^{-\delta t}$ with $\delta>\frac{n-4}{2}$, we get that $w_{\infty}\equiv 0$,
which again is a contradiction. ∎

Since the estimate (5.27) is independent of the parameters $T$, $R$ and
$\varepsilon$, we let $T\rightarrow+\infty$ to obtain a function $w_{0}$ which
satisfies

$\mathcal{L}_{\varepsilon}^{0}(w_{0})=f_{0}$

in $\mathbb{R}$ with the estimate

$\sup_{t\in(-\log R,+\infty)}e^{\delta t}|w_{0}(t)|\leq C\sup_{t\in(-\log
R,+\infty)}e^{\delta t}|f_{0}(t)|,$

where $\delta>\frac{n-4}{2}$ and the positive constant $C$ does not depend on
$\varepsilon\in(0,\varepsilon_{0})$ and $R$. If $f_{0}$ belongs to
$C^{0,\alpha}_{\delta}(-\log R,+\infty)$, then $w_{0}$ belongs to
$C^{4,\alpha}_{\delta}(-\log R,+\infty)$ and there exists a positive constant
$C$ which does not depend on $\varepsilon$ and $R$ such that

$\|w_{0}\|_{C^{4,\alpha}_{\delta}(-\log R,+\infty)}\leq
C\|f_{0}\|_{C^{0,\alpha}_{\delta}(-\log R,+\infty)}.$

There is no reason why the function $w_{0}$ satisfies the boundary condition
$\ddot{w}_{0}=w_{0}=0$ in $t=-\log R$. Although it would be nice to find a
right inverse with these conditions, we can not control the boundary condition
in the right function space.

####  5.2.3. Case $j=1,\ldots,n$

Projecting (5.18) along the eigenfunction $\varphi_{j}$, with $j=1,\ldots,n$,
we get

$\left\\{\begin{array}[]{lcl}\mathcal{L}^{j}_{\varepsilon}(w_{j})=f_{j}&{\rm
in}&(-\log R,\infty)\\\ \ddot{w}_{j}(-\log R)=0\\\ w_{j}(-\log
R)=0,\end{array}\right.$

where $\mathcal{L}_{\varepsilon}^{j}$ is given by (5.19). The proof is similar
to the case $j=0$. First we consider an extension of $f_{j}$, still denoted by
$f_{j}$, to $\mathbb{R}$ and find the unique solution $w_{T}$ of

$\mathcal{L}^{j}_{\varepsilon}(w_{T})=f_{j}\textrm{ in }(-\infty,T),$

with $\dddot{w}_{T}(T)=\ddot{w}_{T}(T)=\dot{w}_{T}(T)=w_{T}(T)=0.$ In a
similar manner of the case $j=0$ we can prove the following lemma.

######  Lemma 5.10.

For $\mu>1$, let $\delta=\frac{n-4}{2}+\mu$. Then there are constants
$\varepsilon_{0}>0$ and $C>0$ independent of $T$ and $R>0$, such that, for all
$\varepsilon\in(0,\varepsilon_{0}]$, we have

$\sup_{t\in(-\log R,T)}e^{\delta t}|w_{T}(t,\theta)|\leq C\sup_{t\in(-\log
R,T)}e^{\delta t}|f_{0}(t,\theta)|.$

The proof of this lemma is similar to that of the Lemma 5.9, so we will not
include it here. The fact that $\lambda_{j}=n-1$ for each $j=1,2,\dots,n$
forces us to choose a weight $\delta>\frac{n-2}{2}$. As in the case of $j=0$,
as $T\rightarrow+\infty$ we get, for every $j=1,\ldots,n$ a solution $w_{j}$
to the equation

$\mathcal{L}_{\varepsilon}^{j}(w_{j})=f_{j}$

in the whole $\mathbb{R}$. Moreover, if $f_{j}\in C^{0,\alpha}_{\delta}(-\log
R,+\infty)$ with $\delta>\frac{n-2}{2}$, then $w_{j}\in
C^{4,\alpha}_{\delta}(-\log R,+\infty)$ and there exists a positive constant
$C$, which does not depend on $R$ and $\varepsilon$, such that

$\|w_{j}\|_{C^{4,\alpha}_{\delta}(-\log R,+\infty)}\leq
C\|f_{j}\|_{C^{0,\alpha}_{\delta}(-\log R,+\infty)}.$

Combining the previous three lemmas we obtain the following proposition,

######  Proposition 5.11.

Let $R>0$, $\alpha\in(0,1)$ and $\mu\in(1,2)$. Then there exists
$\varepsilon_{0}>0$ such that, for all $\varepsilon\in(0,\varepsilon_{0})$,
there is an operator

$G_{\varepsilon,R}:C^{0,\alpha}_{\mu-4}(B_{1}(0)\backslash\\{0\\})\rightarrow
C^{4,\alpha}_{\mu}(B_{1}(0)\backslash\\{0\\})$

with the norm bounded independently of $\varepsilon$ and $R$, such that for
all $f\in C^{0,\alpha}_{\mu-4}(B_{1}(0)\backslash\\{0\\})$, the function
$w:=G_{\varepsilon,R}(f)$ solves the equation

$\left\\{\begin{array}[]{rcl}L_{\varepsilon,R}(w)=f&{\rm
in}&B_{1}(0)\backslash\\{0\\}\\\
\pi^{\prime\prime}(w|_{\mathbb{S}^{n-1}})=0&{\rm on}&\partial B_{1}(0)\\\
\pi^{\prime\prime}(\Delta w|_{\mathbb{S}^{n-1}})=0&{\rm on}&\partial
B_{1}(0).\end{array}\right.$ (5.29)

Moreover, if
$f\in\pi^{\prime\prime}(C^{0,\alpha}_{\mu-4}(B_{1}(0)\backslash\\{0\\}))$,
then $w\in\pi^{\prime\prime}(C^{4,\alpha}_{\mu}(B_{1}(0)\backslash\\{0\\}))$
and we may take $\mu\in(2-n,2)$.

We will work with the solution $u_{\varepsilon,R,a}$ given in (2.16) and with
functions defined in $B_{r}(0)\backslash\\{0\\}$ with $0<r<1$. Then it is
convenient to study the operator $L_{\varepsilon,R,a}$ in weighted function
spaces defined in $B_{r}(0)\backslash\\{0\\}$.

Our construction requires a right-inverse for the linearized operator in a
ball of radius $r>0$, which will be very small. We obtain the solution
operator in $B_{r}(0)\backslash\\{0\\}$ from the solution operator in the
punctured unit ball by rescaling.

First, we observe that if $f\in
C^{0,\alpha}_{\mu-4}(B_{1}(0)\backslash\\{0\\})$ and $w\in
C^{4,\alpha}_{\mu}(B_{1}(0)\backslash\\{0\\})$ satisfy (5.29), then the
function $f_{r}(x)=r^{-4}f(r^{-1}x)$ and $w_{r}(x)=w(r^{-1}x)$ satisfy

$\left\\{\begin{array}[]{rcl}L_{\varepsilon,rR}(w_{r})=f_{r}&{\rm
in}&B_{r}(0)\backslash\\{0\\}\\\
\pi^{\prime\prime}(w_{r}|_{\mathbb{S}_{r}^{n-1}})=0&{\rm on}&\partial
B_{r}(0)\\\ \pi^{\prime\prime}(\Delta w_{r}|_{\mathbb{S}_{r}^{n-1}})=0&{\rm
on}&\partial B_{r}(0).\end{array}\right.$

Also, it is not difficulty to see that

$\|w_{r}\|_{(4,\alpha),\mu,r}\leq c\|f_{r}\|_{(0,\alpha),\mu-4,r},$

where $c>0$ is a constant that does not depend on $\varepsilon$, $R$, $a$ and
$r$.

Finally using a perturbation argument we obtain the following two
propositions.

######  Proposition 5.12.

Let $R>0$, $\alpha\in(0,1)$ and $\mu\in(1,2)$. Then there exists
$\varepsilon_{0}>0$ such that, for all $\varepsilon\in(0,\varepsilon_{0})$,
$a\in\mathbb{R}^{n}$ and $0<r\leq 1$ with $|a|r\leq r_{0}$ for some
$r_{0}\in(0,1)$, there is an operator

$G_{\varepsilon,R,a,r}:C^{0,\alpha}_{\mu-4}(B_{r}(0)\backslash\\{0\\})\rightarrow
C^{4,\alpha}_{\mu}(B_{r}(0)\backslash\\{0\\})$

with the norm bounded independently of $\varepsilon$, $R$, $a$ and $r$, such
that for all $f\in C^{0,\alpha}_{\mu-4}(B_{r}(0)\backslash\\{0\\})$, the
function $w:=G_{\varepsilon,R,a,r}(f)$ solves the equation

$\left\\{\begin{array}[]{rcl}L_{\varepsilon,R,a}(w)=f&{\rm
in}&B_{r}(0)\backslash\\{0\\}\\\
\pi^{\prime\prime}(w|_{\mathbb{S}_{r}^{n-1}})=0&{\rm on}&\partial B_{r}(0)\\\
\pi^{\prime\prime}(\Delta w|_{\mathbb{S}_{r}^{n-1}})=0&{\rm on}&\partial
B_{r}(0).\end{array}\right.$

Moreover, if
$f\in\pi^{\prime\prime}(C^{0,\alpha}_{\mu-4}(B_{r}(0)\backslash\\{0\\}))$,
then $w\in\pi^{\prime\prime}(C^{4,\alpha}_{\mu}(B_{r}(0)\backslash\\{0\\}))$
and we may take $\mu\in(2-n,2)$.

###### Proof.

Observe that

$(L_{\varepsilon,R,a}-L_{\varepsilon,R})v=c(n)(u_{\varepsilon,R,a}^{\frac{8}{n-4}}-u_{\varepsilon,R}^{\frac{8}{n-4}})v.$

By (2.16) we have

$u_{\varepsilon,R,a}^{\frac{8}{n-4}}(x)=|x-a|x|^{2}|^{-4}v_{\varepsilon}^{\frac{8}{n-4}}\left(-\log|x|+\log\left|\frac{x}{|x|}-a|x|\right|+\log
R\right).$

However, $|x-a|x|^{2}|^{-4}=|x|^{-4}+O(|a||x|^{-3}),$
$\log\left|x/|x|-a|x|\right|=O(|a||x|)$ and

$\displaystyle
v_{\varepsilon}^{\frac{8}{n-4}}\left(-\log|x|+\log\left|\frac{x}{|x|}-a|x|\right|+\log
R\right)=v_{\varepsilon}^{\frac{8}{n-4}}\left(-\log|x|+\log R\right)$
$\displaystyle+\frac{8}{n-2}\int_{0}^{\log\left|\frac{x}{|x|}-a|x|\right|}\left(v_{\varepsilon}^{\frac{12-n}{n-4}}\dot{v}_{\varepsilon}\right)\left(-\log|x|+\log
R+t\right)dt.$

Therefore, using (2.22) and fact that $0<v_{\varepsilon}<1$ we get

$\begin{array}[]{rcl}|(u_{\varepsilon,R,a}^{\frac{8}{n-4}}-u_{\varepsilon,R}^{\frac{8}{n-4}})(x)|&\leq&\displaystyle
c_{n}|x|^{-4}\int_{0}^{O(|a||x|)}v_{\varepsilon}^{\frac{8}{n-4}}(-\log|x|+\log
R+t)dt+O(|a||x|^{-3})\\\ &\leq&c_{n}|a||x|^{-3}.\end{array}$ (5.30)

This implies that

$\|u_{\varepsilon,R,a}^{\frac{8}{n-4}}(x)-u_{\varepsilon,R}^{\frac{8}{n-4}}(x)\|_{(0,\alpha),[\sigma,2\sigma]}\leq
c_{n}|a|\sigma^{-3},$

which implies that

$\|(L_{\varepsilon,R,a}-L_{\varepsilon,R})v\|_{(0,\alpha),\mu-4,r}\leq
c_{n}|a|r\|v\|_{(4,\alpha),\mu,r}.$

Therefore, by an perturbation argument we obtain the result. ∎

###  5.3. Nonlinear analysis

The main goal of this section is to construct constant $Q$-curvature metrics
in the punctured ball $B_{r}(p)$ that are close to the deformed Delaunay
metrics $g_{\varepsilon,R,a}=u_{\varepsilon,R,a}^{\frac{4}{n-4}}\delta$ with
partially prescribed boundary Navier boundary data. More precisely, given
$\phi_{0},\phi_{2}\in C^{4,\alpha}(\mathbb{S}^{n-1}_{r})$ we want to find a
complete metric $g_{\rm int}$ in the punctured ball
$B_{r}(p)\backslash\\{p\\}$ with $Q$-curvature equal to $\frac{n(n^{2}-4)}{8}$
where

$g_{\rm
int}=(u_{\varepsilon,R,a}{+\Upsilon}+v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v)^{\frac{4}{n-4}}g.$

Here

$\Upsilon:=-\frac{\beta_{\varepsilon}}{2}R^{-\frac{n}{2}}|x|^{2},$ (5.31)

where $\beta_{\varepsilon}$ is defined in (2.11),
$v_{\phi_{0},\phi_{2}}=\mathcal{P}_{r}(\phi_{0},\phi_{2})$, where
$\mathcal{P}_{r}$ is the Poisson operator we constructed in Corollary 4.1, and

$w_{\varepsilon,R}=\left\\{\begin{array}[]{rl}0,&5\leq n\leq 9\\\
-G_{\varepsilon,R,a}(\mathscr{P}(u_{\varepsilon,R})),&n\geq
10,\end{array}\right.$

where $G_{\varepsilon,R,a}$ is given in Proposition 5.12 and $\mathscr{P}$ is
given in Lemma 5.6. Note that $\Upsilon$ is biharmonic.

Following (1.4) we arrive at the boundary value problem

$H_{g}(u_{\varepsilon,R,a}{+\Upsilon}+v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v)=0,\quad\pi^{\prime\prime}(\left.v\right|_{\partial
B_{r}})=0,\quad\pi^{\prime\prime}(\left.(\Delta v)\right|_{\partial
B_{r}})=0.$ (5.32)

for some $0<r\leq r_{1}$. We can expand (5.32) as

$\begin{array}[]{r}L_{\varepsilon,R,a}(v)=\displaystyle(\Delta^{2}-P_{g})(u_{\varepsilon,R,a}{+\Upsilon}+v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v)-\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v)\\\
{-L_{\varepsilon,R,a}(w_{\varepsilon,R})}+\dfrac{n(n^{2}-4)(n+4)}{16}u_{\varepsilon,R,a}^{\frac{8}{n-4}}({\Upsilon+}v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R})=:\mathscr{M}_{\varepsilon,R,a}(v),\end{array}$
(5.33)

where $L_{\varepsilon,R,a}$ is defined in (5.16), $P_{g}$ is the Paneitz-
Branson operator (1.3),
$\mathcal{R}_{\varepsilon,R,a}:=\mathcal{R}^{u_{\varepsilon,R,a}}$ is defined
in (5.15).

We first show that for a suitable choice of parameters the right hand side
$\mathscr{M}_{\varepsilon,R,a}(v)$ lies in the space
$C^{0,\alpha}_{\mu-4}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$, which is the
domain of the right inverse $G_{\varepsilon,R,a}$ constructed in Proposition
5.12. This will allow us to consider the map $\mathcal{N}_{\varepsilon}$,
defined as

$\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},\cdot):\mathscr{B}\rightarrow
C^{4,\alpha}_{\mu}(B_{r_{\varepsilon}}(0)\backslash\\{0\\}),\quad\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},v)=G_{\varepsilon,R,a,r}\left(\mathscr{M}_{\varepsilon,R,a}(v)\right),$
(5.34)

where $\mathscr{B}$ is a small ball in
$C^{0,\alpha}_{\mu-4}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$. Therefore, to
solve the equation (5.32), it is sufficient to find a fixed point of the map
$\mathcal{N}_{\varepsilon}$. We use the two auxiliary functions
$w_{\varepsilon,R}$ and $\Upsilon$ to account for different phenomena. Without
$w_{\varepsilon,R}$ the right hand side of (5.33) has terms of order
$O(|x|^{d+1-\frac{n}{2}})$, which is not enough to assure that
$\mathscr{M}_{\varepsilon,R,a}(v)\in
C_{\mu-4}^{0,\alpha}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$ for
$\mu\in(1,2)$. We choose $\Upsilon$ precisely to cancel the third term in the
expansion of the Delaunay-type solution given in (2.17).

######  Remark 5.13.

Throughout this work let $\delta_{0}>0$ and $m>\delta_{2}>\delta_{1}>0$ be
sufficiently small, but fixed, constants. We fix $s=\frac{2}{n-4}-\delta_{0}$
and $r_{\varepsilon}=\alpha_{\varepsilon}^{s}$ and choose $a\in\mathbb{R}^{n}$
such that $|a|r_{\varepsilon}\leq 1$. Finally we let $b\in\mathbb{R}$ satisfy
$|b|\leq 1/2$ and choose
$R=\left(\frac{\alpha_{\varepsilon}}{2+2b}\right)^{\frac{2}{n-4}}$.

From the parameter choices in Remark 5.13 and (2.16) it follows that there are
positive constants $c_{1}$ and $c_{2}$ independent of $R$ and $a$ such that

$c_{1}\varepsilon|x|^{\frac{4-n}{2}}\leq u_{\varepsilon,R,a}(x)\leq
c_{2}|x|^{\frac{4-n}{2}},$

for each $x$ in $B_{r_{\varepsilon}}(0)\backslash\\{0\\}$. Furthermore, since
$s>8/(n(n-4))$ when $\delta_{0}$ is sufficiently small, there exists a
constant $c_{3}$ such that

$c_{1}\varepsilon|x|^{\frac{4-n}{2}}\leq u_{\varepsilon,R,a}+\Upsilon\leq
c_{3}|x|^{\frac{4-n}{2}}.$ (5.35)

######  Lemma 5.14.

Let $\mu\in(1,3/2)$. There exist $\varepsilon_{0}>0$ and $m>\delta_{1}>0$,
such that for each $\varepsilon\in(0,\varepsilon_{0})$, $a\in\mathbb{R}^{n}$
with $|a|r_{\varepsilon}\leq 1$, and for all $v_{i}\in
C^{4,\alpha}_{\mu}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$, $i=0,1$, and
$w\in C^{4,\alpha}_{5+d-\frac{n}{2}}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$
with $\|w\|_{(4,\alpha),5+d-\frac{n}{2},r_{\varepsilon}}\leq c$ and
$\|v_{i}\|_{(4,\alpha),\mu,r_{\varepsilon}}\leq
cr_{\varepsilon}^{m-\mu-\delta_{1}}$ there exists a constant $C>0$ indepenent
of $\varepsilon$, $R$ and $a$ such that $\mathcal{R}_{\varepsilon,R,a}$
satisfies

$\|\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}w+v_{1})-\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}w+v_{0})\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\leq
C\varepsilon^{\delta^{\prime}}\|v_{1}-v_{0}\|_{(4,\alpha),\mu,r_{\varepsilon}}$

and

$\|\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{1}+w)\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\leq{C\varepsilon^{\delta^{\prime}}r_{\varepsilon}^{m-\mu}},$

for some $\delta^{\prime}>0$.

###### Proof.

First, by (5.15), we can write

$\displaystyle\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}w+v_{1})-\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}w+v_{0})=-\frac{n(n+4)^{2}}{2}(v_{1}-v_{0})\int_{0}^{1}\int_{0}^{1}(u_{\varepsilon,R,a}+sz_{t})^{\frac{12-n}{n-4}}z_{t}dtds$

where $z_{t}={\Upsilon+}w+tv_{1}+(1-t)v_{0}$. Combining this last displayed
equation with (5.35) implies

$c\varepsilon|x|^{\frac{4-n}{2}}\leq
u_{\varepsilon,R,a}(x){+\Upsilon(x)}+v_{i}(x)+w(x)\leq C|x|^{\frac{4-n}{2}}$

for sufficiently small $\varepsilon>0$. Thus

$\max_{0\leq s,t\leq
1}\|(u_{\varepsilon,R,a}+sz_{t})^{\frac{12-n}{n-4}}\|_{(0,\alpha),[\sigma,2\sigma]}\leq
C\varepsilon^{\lambda_{n}}\sigma^{\frac{n-12}{2}},$

where $\lambda_{n}=0$ for $5\leq n\leq 12$ and $\lambda_{n}=\frac{12-n}{n-4}$
for $n\geq 13$, which in turn yields

$\displaystyle\sigma^{4-\mu}\|\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}w+v_{1})-\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}w+v_{0})\|_{(0,\alpha),[\sigma,2\sigma]}\leq$
$\displaystyle\leq
C\varepsilon^{\lambda_{n}}\sigma^{\frac{n-4}{2}}\|v_{1}-v_{0}\|_{(4,\alpha),\mu,r_{\varepsilon}}\left(\|w\|_{(0,\alpha),[\sigma,2\sigma]}+\|v_{1}\|_{(0,\alpha),[\sigma,2\sigma]}+\|v_{2}\|_{(0,\alpha),[\sigma,2\sigma]}{+\|\Upsilon\|_{(0,\alpha),[\sigma,2\sigma]}}\right)$
$\displaystyle\leq
C\varepsilon^{\lambda_{n}-\frac{4}{n-4}+\frac{n}{2}s}\|v_{1}-v_{0}\|_{(4,\alpha),\mu,r_{\varepsilon}}$

and

$\displaystyle\sigma^{4-\mu}\|\mathcal{R}_{\varepsilon,R,a}(\Upsilon+v_{1}+w)\|_{(0,\alpha),[\sigma,2\sigma]}\leq$
$\displaystyle\leavevmode\nobreak\
C\varepsilon^{\lambda_{n}}\sigma^{\frac{n-4}{2}-\mu}(\|v_{1}\|^{2}_{(0,\alpha),[\sigma,2\sigma]}+\|w\|^{2}_{(0,\alpha),[\sigma,2\sigma]}+\|\Upsilon\|^{2}_{(0,\alpha),[\sigma,2\sigma]})$
$\displaystyle\leq$ $\displaystyle\leavevmode\nobreak\
C\varepsilon^{\lambda_{n}-\frac{8}{n-4}+s\frac{n+4}{2}}r_{\varepsilon}^{-\mu},$

with $\lambda_{n}-\frac{4}{n-4}+\frac{n}{2}s>0$ and
$\lambda_{n}-\frac{8}{n-4}+s\frac{n+4}{2}>0$. From these inequalities we
obtain the result. ∎

######  Lemma 5.15.

Let $5\leq n\leq 9$, let $\mu\in(1,3/2)$, and let $c$ and $\kappa$ be fixed,
positive constants. There exists $\varepsilon_{0}>0$ such that for each
$\varepsilon\in(0,\varepsilon_{0})$, for all $v\in
C^{4,\alpha}_{\mu}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$, $\phi_{0}\in
C^{4,\alpha}(\mathbb{S}_{r_{\varepsilon}}^{n-1})$ and
$\phi_{2}\in\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}_{r_{\varepsilon}}^{n-1}))$,
with $\|v\|_{(4,\alpha),\mu,r_{\varepsilon}}\leq
cr_{\varepsilon}^{{m}-\mu-\delta_{1}}$ and
$\|(\phi_{0},\phi_{2})\|_{(4,\alpha),r_{\varepsilon}}\leq\kappa
r_{\varepsilon}^{{m}-\delta_{1}}$, the right-hand side of $\eqref{p2}$ belongs
to $C^{0,\alpha}_{\mu-4}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$.

###### Proof.

By Corollary 4.2, the norm of $v_{\phi_{0},\phi_{2}}+v$ is bounded by a
positive constant times $r_{\varepsilon}^{{m}-\delta_{1}}$, so Lemma 5.14
implies the term
$\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+v)$ belongs to
$C^{0,\alpha}_{\mu-4}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$. We also have
$u_{\varepsilon,R,a}^{\frac{8}{n-4}}v_{\phi_{0},\phi_{2}}=O(|x|^{-2})=O(|x|^{\mu-4})$.

To prove the remaining term of (5.33) has the right decay, we write

$(\Delta^{2}-P_{g})(u_{\varepsilon,R,a}+v_{\phi_{0},\phi_{2}}+v)=(\Delta^{2}-\Delta^{2}_{g})u_{\varepsilon,R,a}+(\Delta^{2}-P_{g})(v_{\phi_{0},\phi_{2}}+v)+(\Delta_{g}^{2}-P_{g})u_{\varepsilon,R,a}$

and estimate each of these terms separately. Observe that
$v_{\phi_{0},\phi_{2}}=O(|x|^{2})$ and $v=O(|x|^{\mu})$, and so the second
terms decays like $O(|x|^{\mu-4})$. By Lemma 5.7 we also have the estimate
$(\Delta^{2}-\Delta^{2}_{g})u_{\varepsilon,R,a}=O(|x|^{\mu-4})$, and so it
only remains to estimate $(\Delta_{g}-P_{g})u_{\varepsilon,R,a}$.

Observe that (5.2) gives us $R_{g}=O(|x|^{2})$, ${\rm Ric}_{g}=O(|x|)$ and
$Q_{g}=O(1)$, and so from (5.10) we obtain directly that
$(\Delta_{g}^{2}-P_{g})u_{\varepsilon,R,a}=O(|x|^{1-\frac{n}{2}})$ for $5\leq
n\leq 7$. For dimension $8$ and $9$, (5.11) yields

$\langle{\rm
Ric}_{g},\nabla^{2}u_{\varepsilon,R}\rangle=R_{g}\frac{u^{\prime}}{r}+O(|x|^{4-\frac{n}{2}}).$

Thus, using that $R_{g}=O(|x|^{2})$, $Q_{g}=O(1)$ and (5.10) we obtain
$(\Delta_{g}^{2}-P_{g})u_{\varepsilon,R}=O(|x|^{2-\frac{n}{2}})$. Now, using
(2.18) and (5.10) we get
$(\Delta_{g}^{2}-P_{g})(u_{\varepsilon,R,a}-u_{\varepsilon,R})=O(|x|^{2-\frac{n}{2}})$.
Therefore, $(\Delta_{g}^{2}-P_{g})u_{\varepsilon,R,a}=O(|x|^{2-\frac{n}{2}})$
for $n=8$ or $9$. In either case we have
$(\Delta_{g}^{2}-P_{g})u_{\varepsilon,R,a}=O(|x|^{\mu-4})$. This completes the
proof, since $\Upsilon=O(|x|^{2})$. ∎

Let us now consider the case $n\geq 10$. Since
$(P_{g}-\Delta_{g}^{2})u_{\varepsilon,R,a}=O(|x|^{d+1-\frac{n}{2}})$,
unfortunately $(P_{g}-\Delta_{g}^{2})u_{\varepsilon,R,a}\not\in
C^{0,\alpha}_{\mu-4}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$ for any
$\mu>1$, and so we cannot use the right inverse of $L_{\varepsilon,R,a}$
directly. To overcome this difficulty we will consider the expansion (2.18)
and the fact that, by Lemma 5.6, $\mathscr{P}(u_{\varepsilon,R})$ is
orthogonal to $\\{1,x_{1},\ldots,x_{n}\\}$. If follows from this fact and
Proposition 5.12 that there exists
$w_{\varepsilon,R}\in\pi^{\prime\prime}(C^{4,\alpha}_{5+d-\frac{n}{2}}(B_{r_{\varepsilon}}(0)\backslash\\{0\\}))$
such that

$L_{\varepsilon,R,a}(w_{\varepsilon,R})=-\mathscr{P}(u_{\varepsilon,R}).$
(5.36)

This auxiliary function $w_{\varepsilon,R}$ will eliminate the terms that were
preventing us from applying the right inverse to the right hand-side of
(5.33). From this, the expansion (2.18) and (5.33) we obtain

$\begin{array}[]{rcl}L_{\varepsilon,R,a}(v)&=&\displaystyle(\Delta^{2}-P_{g})(u_{\varepsilon,R,a}-u_{\varepsilon,R}{+\Upsilon}+v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v)\\\
&&+\dfrac{n(n^{2}-4)(n+4)}{16}u_{\varepsilon,R,a}^{\frac{8}{n-4}}({\Upsilon+}v_{\phi_{0},\phi_{2}})-\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v)-\mathscr{R},\end{array}$
(5.37)

where Lemma 5.6 implies $\mathscr{R}=O(|x|^{2d+4-\frac{n}{2}})$.

######  Lemma 5.16.

Suppose $n\geq 10$, $\mu\in(1,3/2)$, $\kappa>0$ and $c>0$ be fixed constants.
Under the same assumption of Lemma 5.15 the right-hand side of $\eqref{eq051}$
belongs to $C^{0,\alpha}_{\mu-4}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$.

###### Proof.

The terms $(\Delta^{2}-P_{g})(v_{\phi_{0},\phi_{2}}+v)$,
$u_{\varepsilon,R,a}^{\frac{8}{n-4}}v_{\phi_{0},\phi_{2}}$ and
$\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v)$
are similar to the proof in Lemma 5.15, while the term $\mathscr{R}\in
C^{0,\alpha}_{\mu-4}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$ because
$2d+4-\frac{n}{2}>\mu-4$.

Since $w_{\varepsilon,R}=O(|x|^{5+d-\frac{n}{2}})$, (5.4) and (5.13) imply
$(\Delta^{2}-\Delta_{g}^{2})w_{\varepsilon,R}=O(|x|^{2+2d-\frac{n}{2}})$, with
$2+2d-n/2>\mu-4$. Observe that $\Delta_{g}-P_{g}^{2}$ is a second order
operator and $3+d-n/2>\mu-4$. Thus
$(\Delta^{2}-P_{g})w_{\varepsilon,R}=O(|x|^{\mu-4})$. In an analogous way, by
(2.18), we obtain
$(\Delta^{2}-P_{g})(u_{\varepsilon,R,a}-u_{\varepsilon,R})=O(|x|^{2+d-\frac{n}{2}})$
with $2+d-\frac{n}{2}>\mu-4$. This completes the proof, since
$\Upsilon=O(|x|^{2})$. ∎

######  Remark 5.17.

The vanishing of the Weyl tensor up to order $d=\left[\frac{n-8}{2}\right]$ is
sharp in the following sense: If the Weyl tensor vanishes up to order $d-1$,
then for $n\geq 10$, $g_{ij}=\delta_{ij}+O(|x|^{d+2})$ and

$(\Delta^{2}-\Delta^{2}_{g})(u_{\varepsilon,R,a}-u_{\varepsilon,R})=O(|x|^{1+d-\frac{n}{2}}),$

which implies that
$(\Delta^{2}-\Delta^{2}_{g})(u_{\varepsilon,R,a}-u_{\varepsilon,R})\not\in
C^{0,\alpha}_{\mu-4}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$.

####  5.3.1. Fixed point argument

In this section we prove the map

$\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},v)=G_{\varepsilon,R,a,r}(\mathscr{M}_{\varepsilon,R,a}(v)),$

defined in (5.34) has a fixed point, where $\mathscr{M}_{\varepsilon,R,a}(v)$
is the right hand side of (5.33) for $5\leq n\leq 9$, and the right hand side
of (5.37) for $n\geq 10$.

######  Lemma 5.18.

For all $\mu\in\mathbb{R}$ and $v\in
C^{4,\alpha}_{\mu}(B_{r}(0)\backslash\\{0\\})$ there exists a constant $c>0$
that does not depend on $r$ and $\mu$ such that

$\displaystyle\|(P_{g}-\Delta^{2})v\|_{(0,\alpha),\mu-4,r}\leq
cr^{d+3}\|v\|_{(4,\alpha),\mu,r}$

###### Proof.

First we use (5.4) and (5.13) to obtain

$\sigma^{4-\mu}\|(\Delta_{g}^{2}-\Delta^{2})v\|_{(0,\alpha),[\sigma,2\sigma]}\leq
C\sigma^{d+3-\mu}\|v\|_{(4,\alpha),[\sigma,2\sigma]}.$

Also, by (1.3) we get

$\sigma^{4-\mu}\|(P_{g}-\Delta_{g}^{2})v\|_{(0,\alpha),[\sigma,2\sigma]}\leq
C\sigma^{\max\\{3,d+3\\}-\mu}\|v\|_{(4,\alpha),[\sigma,2\sigma]}.$

Combining these two inequalities implies the result. ∎

######  Lemma 5.19.

Given $a\in\mathbb{R}^{n}$ with $|a|r_{\varepsilon}^{1-\delta_{2}}\leq 1$ and
$\|(\phi_{0},\phi_{2})\|_{(4,\alpha),r_{\varepsilon}}\leq\kappa
r_{\varepsilon}^{m-\delta_{1}}$, if $\varepsilon>0$ is sufficiently small
there exists a constant $\beta>0$ such that

$\|u_{\varepsilon,R,a}^{\frac{8}{n-4}}({\Upsilon+}v_{\phi_{0},\phi_{2}})\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\leq
c\varepsilon^{\beta}r_{\varepsilon}^{m-\mu}$

where $m$ is defined in Remark 5.13.

###### Proof.

First we note that by Remark 5.13 we obtain $\log
R=\frac{2}{n-4}\log\alpha_{\varepsilon}+\frac{2}{4-n}\log(2+2b)$. By (2.20)
and (2.21) we have

$|x|^{\frac{4-n}{2}}\left|\frac{x}{|x|}-a|x|\right|^{\frac{4-n}{2}}=|x|^{\frac{4-n}{2}}(1+O(r_{\varepsilon}^{\delta_{2}}))\quad\mbox{
and }\quad\log\left|\frac{x}{|x|}-a|x|\right|=O(r_{\varepsilon}^{\delta_{2}})$
(5.38)

respectively. This implies that if $r_{\varepsilon}^{1+\ell}\leq|x|\leq
r_{\varepsilon}$ and $\ell\geq 0$ is small enough, we have

$\left(\frac{2}{n-4}-s\right)\log\alpha_{\varepsilon}-C\leq-\log|x|+\log\left|\frac{x}{|x|}-a|x|\right|+\log
R\leq\left(\frac{2}{n-4}-s(1+\ell)\right)\log\alpha_{\varepsilon}+C.$

Observe $2/(n-4)-s>0$ by Remark 5.13. Thus, for $0<\ell<\frac{2}{s(n-4)}-1$
and $\varepsilon>0$ small enough (2.14) implies

$v_{\varepsilon}\left(-\log|x|+\log\left|\frac{x}{|x|}-a|x|\right|+\log
R\right)\leq c_{n}r_{\varepsilon}^{\frac{n-4}{2}}.$

By (2.16) and (5.38) we have $u_{\varepsilon,R,a}(x)\leq
C|x|^{\frac{4-n}{2}}r_{\varepsilon}^{\frac{n-4}{2}},$ and so

$u_{\varepsilon,R,a}(x)^{\frac{8}{n-4}}\leq c_{n}|x|^{-4}r_{\varepsilon}^{4}$

for all $r_{\varepsilon}^{1+\ell}\leq|x|\leq r_{\varepsilon}$. Then, for
$\frac{1}{4}r_{\varepsilon}^{1+\ell}\leq\sigma\leq\frac{1}{2}r_{\varepsilon}$,
we have

$\displaystyle\sigma^{4-\mu}\|u_{\varepsilon,R}^{\frac{8}{n-4}}(\Upsilon+v_{\phi_{0},\phi_{2}})\|_{(0,\alpha),[\sigma,2\sigma]}\leq$
$\displaystyle\leavevmode\nobreak\
c\sigma^{-\mu}r_{\varepsilon}^{4}(\|v_{\phi_{0},\phi_{2}}\|_{(0,\alpha),[\sigma,2\sigma]}+\|\Upsilon\|_{(0,\alpha),[\sigma,2\sigma]})$
$\displaystyle\leq$ $\displaystyle\leavevmode\nobreak\
c(r_{\varepsilon}^{m-\mu+4-\delta_{1}}+r_{\varepsilon}^{6-\mu}\alpha_{\varepsilon}^{-\frac{4}{n-4}})\leq\leavevmode\nobreak\
cr_{\varepsilon}^{\beta}r_{\varepsilon}^{m-\mu},$

with $6s-4/(n-4)>0$ by Remark 5.13. For $0\leq\sigma\leq
4^{-1}r_{\varepsilon}^{1+\ell}$, we use $(2-\mu)\ell-\delta_{1}>0$ and
$(2+\mu(2-\mu))s-\frac{4}{n-4}>0$ to see

$\displaystyle\sigma^{4-\mu}\|u_{\varepsilon,R}^{\frac{8}{n-4}}(\Upsilon+v_{\phi_{0},\phi_{2}})\|_{(0,\alpha),[\sigma,2\sigma]}\leq$
$\displaystyle\leavevmode\nobreak\
c\sigma^{-\mu}(\|v_{\phi_{0},\phi_{2}}\|_{(0,\alpha),[\sigma,2\sigma]}+\|\Upsilon\|_{(0,\alpha),[\sigma,2\sigma]})$
$\displaystyle\leq$ $\displaystyle\leavevmode\nobreak\
cr_{\varepsilon}^{(2-\mu)\ell+\delta_{1}}r_{\varepsilon}^{m-\mu}+r_{\varepsilon}^{2-\mu+(2-\mu)\ell}\alpha_{\varepsilon}^{-\frac{4}{n-4}}\leq
r_{\varepsilon}^{\beta}r_{\varepsilon}^{m-\mu},$

which completes our proof. ∎

######  Proposition 5.20.

Let $\mu\in(1,3/2)$, $\tau>0$ and $\kappa>0$ be fixed constants. There exists
$\varepsilon_{0}>0$ such that for each $\varepsilon\in(0,\varepsilon_{0})$,
$a\in\mathbb{R}^{n}$ with $|a|r_{\varepsilon}^{1-\delta_{2}}\leq 1$,
$\phi_{0}\in C^{4,\alpha}(\mathbb{S}_{r_{\varepsilon}}^{n-1})$ and
$\phi_{2}\in\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}_{r_{\varepsilon}}^{n-1}))$
with $\|(\phi_{0},\phi_{2})\|_{(4,\alpha),r_{\varepsilon}}\leq\kappa
r_{\varepsilon}^{{m}-\delta_{1}}$, there exists a fixed point of the map
$\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},\cdot)$ in the ball of radius
$\tau r_{\varepsilon}^{{m}-\mu}$.

###### Proof.

We will prove that the map
$\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},\cdot)$ is a contraction for
$\varepsilon>0$ small enough by proving the following two inequalities

$\|\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},0)\|_{(4,\alpha),\mu,r_{\varepsilon}}<\frac{1}{2}\tau
r_{\varepsilon}^{{m}-\mu}$ (5.39)

and

$\|\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},v_{1})-\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},v_{2})\|_{(4,\alpha),\mu,r_{\varepsilon}}<\frac{1}{2}\|v_{1}-v_{2}\|_{(4,\alpha),\mu,r_{\varepsilon}}$
(5.40)

for $v_{i}\in C^{4,\alpha}_{\mu}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$,
$i=1,2$ with $\|v_{i}\|_{(4,\alpha),\mu}\leq\tau r_{\varepsilon}^{{m}-\mu}$.

Case $\mathbf{5\leq n\leq 9:}$ Since the right inverse
$G_{\varepsilon,R,a,r_{\varepsilon}}$ given by Proposition 5.12 is bounded in-
dependently of $\varepsilon$, $R$, $a$ and $r$, the inequality (5.39) follows
once we estimate the $C^{0,\alpha}_{\mu-4}$-norm of the right hand side of
(5.33) when $v=0$. Using (2.18), Lemma 5.3 and the estimates in the proof of
Lemma 5.7 we get

$\displaystyle\|(\Delta^{2}-\Delta^{2}_{g})u_{\varepsilon,R,a}\|_{(0,\alpha),\mu-4,r_{\varepsilon}}=$
$\displaystyle\leavevmode\nobreak\ \
\|(\Delta^{2}-\Delta^{2}_{g})(u_{\varepsilon,R,a}-u_{\varepsilon,R})\|_{(0,\alpha),\mu-4,r_{\varepsilon}}$
$\displaystyle\leq$ $\displaystyle\leavevmode\nobreak\
c|a|r_{\varepsilon}^{6+d-\frac{n}{2}-\mu}\leq
cr_{\varepsilon}^{\delta_{2}}r_{\varepsilon}^{{m}-\mu}.$

Here we have used $|a|r_{\varepsilon}^{1-\delta_{2}}\leq 1$, with
$\delta_{2}>0$. By the proof of Lemma 5.15 we have
$(\Delta_{g}^{2}-P_{g})u_{\varepsilon,R,a}=O(|x|^{2+d-\frac{n}{2}})$, which
implies that

$\|(\Delta_{g}^{2}-P_{g})u_{\varepsilon,R,a}\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\leq
cr^{\delta_{2}}r_{\varepsilon}^{{m}-\mu}.$ (5.41)

By Corollary 4.2 and Lemma 5.18 we have

$\|(\Delta^{2}-P_{g})v_{\phi_{0},\phi_{2}}\|_{(0,\alpha),\mu-4,r}\leq
cr^{d+3}\|v_{\phi_{0},\phi_{2}}\|_{(4,\alpha),\mu,r}\leq
cr_{\varepsilon}^{d+3-\mu}\|(\phi_{0},\phi_{2})\|_{(4,\alpha),r}$ (5.42)

and

$\|(\Delta^{2}-P_{g})\Upsilon\|_{(0,\alpha),\mu-4,r}\leq
c\varepsilon^{-\frac{4}{n-4}+s(d+5-m)}r_{\varepsilon}^{m-\mu},$ (5.43)

since $\Upsilon=O(\varepsilon^{-\frac{4}{n-4}}|x|^{2})$. We also have

$\|Q_{g}v_{\phi_{0},\phi_{2}}\|_{(0,\alpha),\mu-4,r}\leq
cr_{\varepsilon}^{4-\mu}\|(\phi_{0},\phi_{2})\|_{(4,\alpha),r}.$ (5.44)

Using Lemma 5.14, one has

$\|\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}})\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\leq{c\varepsilon^{\delta^{\prime}}r^{m-\mu}},$
(5.45)

where $\delta^{\prime}>0$. Therefore, by (5.41), (5.42), (5.43), (5.44),
(5.45) and Lemma 5.19 we obtain (5.39) for $\varepsilon>0$ small enough. Now,
using (5.34) with $\mathscr{M}_{\varepsilon,R,a}(v)$ as the right hand side of
(5.33) we obtain

$\displaystyle\|\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},v_{1})-\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},v_{2})\|_{(4,\alpha),\mu,r_{\varepsilon}}\leq
C\left(\|(\Delta^{2}-P_{g})(v_{1}-v_{2})\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\right.$
$\displaystyle\left.+\|\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+v_{1})-\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+v_{2})\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\right).$

By Lemmas 5.18 and 5.14 we obtain directly (5.40) for $\varepsilon>0$ small
enough.

Case $\mathbf{n\geq 10}$: Now we use (5.34) with
$\mathscr{M}_{\varepsilon,R,a}$ as the right hand side of (5.37). As in the
previous case, to prove (5.39) it is enough to estimate the
$C^{0,\alpha}_{\mu-4}$-norm of the right hand side of (5.37) when $v=0$.
Similar to the proof of Lemma 5.18, we obtain

$\|(\Delta^{2}-P_{g})w_{\varepsilon,R}\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\leq
c\sigma^{8+2d-\frac{n}{2}-\mu}\|w_{\varepsilon,R}\|_{(4,\alpha),5+d-\frac{n}{2},r_{\varepsilon}}.$
(5.46)

By Lemma 5.14 we know that

$\|\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R})\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\leq\varepsilon^{\delta^{\prime}}r_{\varepsilon}^{m-\mu},$
(5.47)

where $\delta^{\prime}>0$. Therefore, using
$\mathscr{R}=O(r_{\varepsilon}^{2d+4-\frac{n}{2}})$, Lemma 5.19, (5.42),
(5.46), (5.47) and a similar calculation to that of (5.41) we conclude that
(5.39) holds. Moreover,

$\displaystyle\|\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},v_{1})-\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},v_{2})\|_{(4,\alpha),\mu,r_{\varepsilon}}\leq
C\left(\|(\Delta^{2}-P_{g})(v_{1}-v_{2})\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\right.$
$\displaystyle\left.+\|\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v_{1})-\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v_{2})\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\right).$

By Lemma 5.14

$\displaystyle\|\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v_{1})-\mathcal{R}_{\varepsilon,R,a}({\Upsilon+}v_{\phi_{0},\phi_{2}}+w_{\varepsilon,R}+v_{2})\|_{(0,\alpha),\mu-4,r_{\varepsilon}}\leq
c\varepsilon^{\delta^{\prime}}\|v_{1}-v_{2}\|_{(4,\alpha),\mu,r_{\varepsilon}},$

with $\delta^{\prime}>0$. From this and Lemma 5.18, we obtain (5.40) for
$\varepsilon$ sufficiently small. ∎

######  Theorem 5.21.

Let $\mu\in(1,3/2)$, $\tau>0$, $\kappa>0$ and $\delta_{2}>\delta_{1}>0$ be
fixed constants. There exists $\varepsilon_{0}>0$, such that for each
$\varepsilon\in(0,\varepsilon_{0}]$, $|b|\leq 1/2$, $a\in\mathbb{R}^{n}$,
$\phi_{2}\in C^{4,\alpha}(\mathbb{S}_{r_{\varepsilon}}^{n-1})$ and
$\phi_{0}\in\pi^{\prime\prime}(C^{4,\alpha}(\mathbb{S}_{r_{\varepsilon}}^{n-1}))$,
with $|a|r_{\varepsilon}^{1-\delta_{2}}\leq 1$,
$\|(\phi_{0},\phi_{2})\|_{(4,\alpha),r_{\varepsilon}}\leq\kappa
r_{\varepsilon}^{{m}-\delta_{1}}$, there exists
$U_{\varepsilon,R,a,\phi_{0},\phi_{2}}\in
C^{4,\alpha}_{\mu}(B_{r_{\varepsilon}}(0)\backslash\\{0\\})$ satisfying

$\displaystyle\left\\{\begin{array}[]{lcl}H_{g}(u_{\varepsilon,R,a}{+\Upsilon}+w_{\varepsilon,R}+v_{\phi_{0},\phi_{2}}+U_{\varepsilon,R,a,\phi_{0},\phi_{2}})=0&{\rm
in}&B_{r_{\varepsilon}}(0)\backslash\\{0\\}\\\ \pi^{\prime\prime}(\Delta
U_{\varepsilon,R,a,\phi_{0},\phi_{2}})=\pi^{\prime\prime}(U_{\varepsilon,R,a,\phi_{0},\phi_{2}})=0&{\rm
on}&\partial B_{r_{\varepsilon}}(0)\\\ \end{array}\right.$

where $w_{\varepsilon,R}\equiv 0$ when $5\leq n\leq 9$,
$w_{\varepsilon,R}\in\pi^{\prime\prime}(C^{4,\alpha}_{5+d-\frac{n}{2}}(B_{r_{\varepsilon}}(0)\backslash\\{0\\}))$
is the solution of $\eqref{eq068}$ when $n\geq 10$. Moreover,

$\|U_{\varepsilon,R,a,\phi_{0},\phi_{2}}\|_{(4,\alpha),\mu,r_{\varepsilon}}\leq\tau
r_{\varepsilon}^{{m}-\mu}$ (5.48)

and

$\|U_{\varepsilon,R,a,\phi_{1},\phi_{2}}-U_{\varepsilon,R,a,\tilde{\phi}_{0},\tilde{\phi}_{2}}\|_{(4,\alpha),\mu,r_{\varepsilon}}\leq
cr^{\delta_{3}-\mu}\|(\phi_{0},\phi_{2})-(\tilde{\phi}_{0},\tilde{\phi}_{2})\|_{(4,\alpha),r_{\varepsilon}},$
(5.49)

for some constants $c>0$ and $\delta_{3}>0$ that do not depend on
$\varepsilon$, $R$, $a$ and $(\phi_{0},\phi_{2})$,
$(\tilde{\phi}_{0},\tilde{\phi}_{2})$.

###### Proof.

The solution $U_{\varepsilon,R,a,\phi_{0},\phi_{2}}$ is the fixed point of the
map $\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},\cdot)$ given by
Proposition 5.20 with the estimate (5.48). Using this fact and (5.40), we
obtain that

$\displaystyle\|U_{\varepsilon,R,a,\phi_{0},\phi_{2}}-U_{\varepsilon,R,a,\tilde{\phi}_{0},\tilde{\phi}_{2}}\|_{(4,\alpha),\mu,r_{\varepsilon}}$
$\displaystyle\leq$ $\displaystyle\leavevmode\nobreak\
2\|\mathcal{N}_{\varepsilon}(R,a,\phi_{0},\phi_{2},U_{\varepsilon,R,a,\phi_{0},\phi_{2}})-\mathcal{N}_{\varepsilon}(R,a,\tilde{\phi}_{0},\tilde{\phi}_{2},U_{\varepsilon,R,a,\tilde{\phi}_{0},\tilde{\phi}_{2}})\|_{(4,\alpha),\mu,r_{\varepsilon}}.$

The estimate (5.49) now follows from the definition of
$\mathcal{N}_{\varepsilon}$ together with the estimates obtained in
Proposition 5.20. ∎

##  6\. Exterior problem

In the previous section we exploited the conformal structure of the equation
to perturb a Delaunay-type solution to a exact solution to the Q-curvature
problem in a small neighborhood of the singularity. To find a solution on the
complement of this neighborhood whose boundary data matches the solution
constructed in the previous section, we need to choose the approximate
solution that will be perturbed on the exterior domain carefully. Such
approximate solution must be close to the Delaunay-type solution
$u_{\varepsilon,R,a}$ near to the boundary. Our asymptotic expansion in
Corollary 2.3 shows that as $\varepsilon\rightarrow 0$ the function
$u_{\varepsilon,R,a}$ is close to the Green’s function of the flat bi-
Laplacian with a pole on the singularity, at least in a sufficiently small
annulus. Consequently, since our manifold has constant $Q$-curvature (the
constant function $1$ is a solution to $H_{g_{0}}(1)=0$), the natural
approximate solution on the exterior domain will be the constant 1 plus the
Green function with a pole at $p$.

Let $r_{1}\in(0,1)$ and $\Psi:B_{r_{1}}(0)\rightarrow M$ be a normal
coordinate system with respect to $g=\mathcal{F}^{\frac{4}{n-4}}g_{0}$ on M
centered in $p$, where $\mathcal{F}$ is defined in Section 5.1. In these
coordinates we have $g_{ij}=\delta_{ij}+O(|x|^{2})$ and
$\mathcal{F}=1+O(|x|^{2})$, which implies that
$(g_{0})_{ij}=\delta_{ij}+O(|x|^{2})$. Remember that $g_{0}$ has constant
$Q$-curvature equal to $n(n^{2}-8)/8$. Denote by $G_{p}(x)$ the Green’s
function with pole at $p$ for
$L_{g_{0}}=P_{g_{0}}-\frac{n(n^{2}-4)(n+4)}{16}$, the linearization of
$H_{g_{0}}$ at the constant function 1. The Green’s function $G_{p}$ exists by
our hypothesis that $g_{0}$ is nondegenerate. By a similar argument to the one
in Proposition 2.7 of [23] we can normalize $G_{p}$ such that
$\displaystyle\lim_{x\rightarrow 0}|x|^{n-4}G_{p}(x)=1$. This implies that
$|G_{p}(x)|\leq C|x|^{4-n}$, for all $x\in B_{r_{1}}(0)$. Our main goal in
this section is to solve the following problem

$H_{g_{0}}(1+\lambda G_{p}+u)=0\quad{\rm on}\quad M\backslash B_{r}(p),$ (6.1)

with $\lambda\in\mathbb{R}$, $r\in(0,r_{1})$ and prescribed boundary data.
Similarly to the strategy of the previous section, expanding this equation we
obtain

$H_{g_{0}}(1+\lambda G_{p}+u)=L_{g_{0}}(u)+\mathcal{R}(\lambda G_{p}+u)=0,$

since $H_{g_{0}}(1)=0$ and $L_{g_{0}}(G_{p})=0$. Here $\mathcal{R}$ is defined
in (5.15) with $u_{0}=1$. In order to apply a fixed point theorem in this
case, it is necessary to show that the linearized operator in $M\backslash
B_{r}(p)$ has a right inverse. At this point, the nondegeneracy of the metric
will be necessary.

###  6.1. Linear analysis

The goal of this subsection is to find a right inverse of the operator
$L_{g_{0}}$. To this aim, given $0<r<s$ consider in $\mathbb{R}^{n}$ the
annulus $\Omega_{r,s}:=B_{s}(0)\backslash B_{r}(0)$. We start proving the
following result.

######  Proposition 6.1.

Let $\nu\in(4-n,5-n)$ fixed and $0<2r<s$. There exists an operator
$\mathcal{G}_{r,s}:C^{0,\alpha}_{\nu-4}(\Omega_{r,s})\rightarrow
C^{4,\alpha}_{\nu}(\Omega_{r,s})$ such that, for all $f\in
C^{0,\alpha}_{\nu-4}(\Omega_{r,s})$, the function $w=\mathcal{G}_{r,s}(f)$
satisfies

$\left\\{\begin{array}[]{lcl}\Delta^{2}w=f&{\rm in}&\Omega_{r,s}\\\ \Delta
w=w=0&{\rm on}&\partial B_{s}.\end{array}\right.$ (6.2)

Moreover, there exists a positive constant $C$ that does not depend on $s$ and
$r$ such that

$\|\mathcal{G}_{r,s}(f)\|_{C^{4,\alpha}_{\nu}(\Omega_{r,s})}\leq
C\|f\|_{C^{0,\alpha}_{\nu-4}(\Omega_{r,s})}.$

###### Proof.

The existence of $w$ satisfying (6.2) follows by standard elliptic theory. Now
we need to prove the existence of a constant $c>0$ which does not depend on
$r$ and $s$ such that

$\sup_{\Omega_{r,s}}r^{-\nu}|w|\leq c\sup_{\Omega_{r,s}}r^{4-\nu}|f|,$ (6.3)

Suppose by contradiction that the estimate does not hold. Then there exist a
sequence of numbers $0<r_{i}<s_{i}$, a sequence of functions $f_{i}$
satisfying $\sup_{\Omega_{r_{i},s_{i}}}r^{4-\nu}|f_{i}|=1$, and a sequence of
solutions $w_{i}$ to the problem

$\left\\{\begin{array}[]{rcl}\Delta^{2}w_{i}=f_{i}&{\rm
in}&\Omega_{r_{i},s_{i}}\\\ \Delta w_{i}=w_{i}=0&{\rm
on}&\partial\Omega_{r_{i},s_{i}}\end{array}\right.$

such that
$A_{i}:=\sup_{\Omega_{r_{i},s_{i}}}r^{-\nu}\left|{w}_{i}\right|\rightarrow+\infty.$
Consider $(t_{i},\theta_{i})\in(r_{i},s_{i})\times\mathbb{S}^{n-1}$, a point
where this supremum $A_{i}$ is attained. Up to a subsequence, we can assume
that $r_{i}/t_{i}\rightarrow\tau_{1}\in[0,1]$ and
$s_{i}/t_{i}\rightarrow\tau_{2}\in[1,+\infty]$. With a similar argument
applied in page 5.2.1 we can prove that $\tau_{1}$ and $\tau_{2}$ is not equal
to $1$.

Define the sequence of rescaled functions

$\hat{w}_{i}(r\theta):=\frac{t_{i}^{-\nu}}{A_{i}}{w}_{i}\left(t_{i}r\theta\right)\quad\text{
and
}\quad\hat{f}_{i}(r\theta):=\frac{t_{i}^{2-\nu}}{A_{i}}{f}_{i}\left(t_{i}r\theta\right).$

Note that $1\in(\tau_{1},\tau_{2})$, $\hat{w}_{i}(\theta_{i})=1$ and
$|\hat{w}_{i}(r\theta)|\leq r^{\nu}$. By Arzelá-Ascoli’s Theorem we can
assume, up to a subsequence, the sequence of functions $\hat{w}_{i}$ converges
uniformly on compact subsets of $(\tau_{1},\tau_{2})\times\mathbb{S}^{n-1}$ to
a biharmonic function $\hat{w}_{\infty}$ which is not identically zero.
Moreover, $\hat{w}_{\infty}$ is bounded by $r^{\nu}$ on
$(\tau_{1},\tau_{2})\times\mathbb{S}^{n-1}$, and
$\Delta\hat{w}_{\infty}=\hat{w}_{\infty}=0$ on the boundary if $\tau_{1}>0$ or
if $\tau_{2}<\infty$. Writing

$\hat{w}_{\infty}(r,\theta)=\sum_{j=0}^{\infty}\hat{w}_{j}(r)e_{j}(\theta),$

where $\hat{w}_{j}$ is given explicitly by

$\hat{w}_{j}:=a_{j}^{-}r^{j}+a_{j}^{+}r^{2-n-j}+b_{j}^{-}r^{2+j}+b_{j}^{+}r^{4-n-j}.$