#  On the basin of attraction of a critical three-cycle of a model for the
secant map

Ernest Fontich  Departament de Matemàtiques i Informàtica, Universitat de
Barcelona (UB), Gran Via de les Corts Catalanes 585, 08007 Barcelona,
Catalonia, Spain and Centre de Recerca Matemàtica (CRM), 08193 Bellaterra,
Barcelona, Catalonia, Spain<EMAIL_ADDRESS>,  Antonio Garijo  Departament
d’Enginyeria Informàtica i Matemàtiques, Universitat Rovira i Virgili, 43007
Tarragona, Catalonia, Spain<EMAIL_ADDRESS>and  Xavier Jarque
Departament de Matemàtiques i Informàtica, Universitat de Barcelona, Gran Via,
585, 08007 Barcelona, Catalonia and Centre de Recerca Matemàtica, Edifici C,
Campus Bellaterra, 08193 Bellaterra, Catalonia<EMAIL_ADDRESS>
###### Abstract.

We consider the secant method $S_{p}$ applied to a real polynomial $p$ of
degree $d+1$ as a discrete dynamical system on $\mathbb{R}^{2}$. If the
polynomial $p$ has a local extremum at a point $\alpha$ then the discrete
dynamical system generated by the iterates of the secant map exhibits a
critical periodic orbit of period 3 or three-cycle at the point
$(\alpha,\alpha)$. We propose a simple model map $T_{a,d}$ having a unique
fixed point at the origin which encodes the dynamical behaviour of $S_{p}^{3}$
at the critical three-cycle. The main goal of the paper is to describe the
geometry and topology of the basin of attraction of the origin of $T_{a,d}$ as
well as its boundary. Our results concern global, rather than local, dynamical
behaviour. They include that the boundary of the basin of attraction is the
stable manifold of a fixed point or contains the stable manifold of a two-
cycle, depending on the values of the parameters of $d$ (even or odd) and
$a\in\mathbb{R}$ (positive or negative).
Keywords: Root finding algorithms, secant map, stable manifold, center
manifold, basin of attraction.

The first author is supported by the grant PID2021-125535NB-I00. The second
and third authors were supported by MICIU/AEI grant PID2020-118281GB-C32(33)
Both grant are funded by MICIU/AEI/10.13039/501100011033/FEDER,UE. The second
author is supported by Generalitat de Catalunya 2021SGR-633. We want to thank
the Thematic Research Programme Modern holomorphic dynamics and related
fields, Excellence Initiative – Research University programme at the
University of Warsaw. Finally, this work has also been funded through the
María de Maeztu Program for Centers and Units of Excellence in R&D, grant
CEX2020-001084-M funded by MICIU/AEI/10.13039/501100011033/FEDER,UE

##  1\.  Introduction

A major goal in applied and theoretical mathematical modelling is to find
stable equilibria which determines the expected behaviour of the phenomenon we
are analyzing. Those equilibria are given by real (or complex) numbers, real
(or complex) finite dimensional vectors, or functions belonging to an infinite
dimensional space, depending on the nature of the model under consideration.

In the majority of cases, the stable equilibria determining the evolutionary
steady states of any model turn out to be solutions of non-linear equations.
In general, we cannot solve these equations explicitly. Accordingly, there is
a long history of research studying different algorithms which efficiently
find their solutions.

Among these algorithms the ones given by the special kind of discrete
dynamical systems, known as root-finding algorithms, have been shown to be the
most useful. Roughly speaking, a root-finding algorithm is a system such that
for most of the initial conditions the asymptotic behaviour of the
corresponding iterative process tends to one of the solutions of the non-
linear equation determining the equilibria of the model. We observe that the
condition of convergence for most of the initial conditions is a global
phenomenon rather than (only) a local one. Indeed, this is the reason why the
global dynamics of root-finding algorithms has been an important subject of
study for dynamicists.

Moreover, when the model has more than one steady state the phase portrait of
the root-finding algorithm splits into regions where the iterates of the seeds
converge to different equilibria. Consequently, two natural questions arise.
On the one hand, about the boundaries of these regions: do they have easy
geometry and topology? what about the restricted dynamics over these
boundaries? do they have positive measure? On the other hand, about the stable
steady states: are there other stable behaviour of the algorithm unrelated to
the steady states of the model? If they exist, there would be open sets of the
dynamical plane where the root-finding algorithm is full of bad initial
conditions. The answers to all these questions has had a great influence on
the study of theoretical as well as applied discrete dynamical systems.

There is no discussion that the most famous root-finding algorithm is the
well-known Newton’s method. More concretely, assume the equation we want to
solve is $p(x)=0$, where, to simplify the exposition, we assume $p(x)$ to be a
polynomial with $x\in\mathbb{R}$ or $x\in\mathbb{C}$ but the method extends to
higher dimensional problems. Newton’s method is the study of the dynamical
system

(1.1)
$x_{n+1}=N_{p}\left(x_{n}\right):=x_{n}-\frac{p(x_{n})}{p^{\prime}(x_{n})},\qquad
x_{0}\in\mathbb{R}\quad\text{or}\quad x_{0}\in\mathbb{C}.$

It is easy to see that $N_{p}(\alpha)=\alpha$ if and only if $p(\alpha)=0$ (if
$p^{(\ell)}(\alpha)=0,\ 1\leq\ell\leq k$, one can still use (1.1) after some
modifications) and moreover if $x_{0}\approx\alpha$ then
$\\{x_{n}:=N^{n}_{p}(x_{0})\\}\to\alpha$ as $n\to\infty$. Consequently,
Newton’s method is a (one dimensional) dynamical systems whose fixed points
correspond to the roots of $p$ and they are local attractors. In fact, it is
somehow the canonical roof-finding algorithm. The natural dynamical space for
one dimensional Newton’s method applied to degree $d$ polynomials is
$\mathbb{C}$ (rather than $\mathbb{R}$) due to the fundamental theorem of
algebra, and it defines one of the most studied family of rational
(holomorphic) maps on the Riemann sphere [Bla94, Bla84, Shi09, HSS01]. See
also [Ber93, BFJK18] for Newton’s method applied to transcendental entire
maps.

Despite its fundamental role, Newton’s method has some limitations and the
literature has explored other root finding algorithms trying to avoid these
weakness (for instance avoiding to compute the derivatives if their
computational cost is too high) or to improve the efficiency of the method
under certain hypothesis (for instance improving the local speed of
convergence of the method to the root(s) of $p$).

Another basic root-finding algorithm is the secant method given by the
dynamical system generated by the iterates of the 2-dimensional map

(1.2) $S(x,y)=S_{p}(x,y):=\left(y,y-p(y)\frac{x-y}{p(x)-p(y)}\right).$

However, in contrast to Newton’s method, secant’s method is a two dimensional
system and it does not required to compute any derivative of $p$. Nonetheless,
as before, we have that $S(\alpha,\alpha)=(\alpha,\alpha)$ if and only if
$p(\alpha)=0$. Moreover if $(x_{0},y_{0})\approx(\alpha,\alpha)$ then

$\\{\left(x_{n},y_{n}\right):=S^{n}(x_{0},y_{0})\\}\to(\alpha,\alpha)$

at least for simple roots of $p$ (see [GJ20] for multiple roots). We refer to
[GJ19, BF18], and references therein, for a detailed discussion of the phase
plane ($\mathbb{R}^{2}$ and $\mathbb{C}^{2}$) of the secant method.

One (unexpected) fact of the real secant map is that there are no finite
periodic points of period two or three in $\mathbb{R}^{2}$. However, it has
finite periodic points of period four and some of them determine the geometry
and topology of the boundaries of the immediate basin of attraction of its
fixed points (see [GGJ21]). Moreover, if we extend the domain of the secant
method to infinity (that is, if we extend the secant map to $\mathbb{RP}^{2}$
or $\mathbb{CP}^{2}$) a new three-cycle phenomenon arises. Indeed, in [BF18]
(see also [GJ19]) the authors showed that if $c\in\mathbb{R}$ satisfies that
$p^{\prime}(c)=0$ (critical point) and $p(c)p^{\prime\prime}(c)\neq 0$ the
secant method exhibits a critical three-cycle at $(c,c)$ given by

$(c,c)\xmapsto{\text{S}}(c,\infty)\xmapsto{\text{S}}(\infty,c)\xmapsto{\text{S}}(c,c).$

Moreover, the three-cycle has a basin of attraction whose geometry varies
depending on the degree of the polynomial. However, its geometry and topology
is quite similar among polynomials of the same degree. These basins, and their
disparate geometry can be visualized in red in Figure 1 for concrete
polynomials of degree of different parity.

The main goal of this work is to go deeper into the understanding of the
geometry of the basin of the critical three-cycle by means of a model which
captures the relevant information and allows us to give an accurate
description.

$-1$$0$$1$$-1$$0$$1$

$-0.2$$-0.1$$0$$0.1$$0.2$$-0.2$$-0.1$$0$$0.1$$0.2$

$-1$$0$$1$$-1$$0$$1$

$-0.4$$-0.2$$0$$0.2$$0.4$$-0.4$$-0.2$$0$$0.2$$0.4$

Figure 1. Phase planes of the secant map applied to the polynomials
$p(x)=1-2x^{2}+x^{3}$ (first row) and $p(x)=1-8x^{2}+8x^{4}$ (second row). In
all the pictures we show in red the set of points converging towards the
critical three-cycle $\\{(0,0),(0,\infty),(\infty,0)\\}$. The second column is
a zoom near the origin of the first one.

Following the approach in [BF18] we assume, without lost of generality, that
$c=0$ and $p(0)=1$. Thus, assuming also that $\mbox{deg}(p)=d+1$, the
polynomial $p$ writes as

(1.3) $p(x)=1+a_{2}x^{2}+\ldots+a_{d+1}x^{d+1},$

where $d\geq 2$ and $a_{2}a_{d+1}\neq 0$. Using the natural extension of $S$
at infinity, via the charts $\varphi_{1}(x,y)=(1/x,y)$ and
$\varphi_{2}(x,y)=(x,1/y)$, and some computations (explicit in [BF18]) the
expression of $S^{3}$ near the origin is given by

(1.4) $S^{3}\left(\begin{array}[]{l}x\\\
y\end{array}\right)=\left(\begin{array}[]{l}y-\frac{(-a_{2})^{d}}{a_{d+1}}(x+y)^{d}\\\
y-2\frac{(-a_{2})^{d}}{a_{d+1}}(x+y)^{d}\end{array}\right)+\mathcal{O}_{d+1},$

where $\mathcal{O}_{d+1}$ indicates terms bounded by $(|x|+|y|)^{d+1}$. The
expression (1.4) motivates the introduction of the following model map
$T_{a,d}$ which encodes the dominant terms of $S^{3}$ near the origin.
Concretely,

(1.5) $T_{a,d}\left(\begin{array}[]{l}x\\\
y\end{array}\right)=\left(\begin{array}[]{l}y-a(x+y)^{d}\\\
y-2a(x+y)^{d}\end{array}\right),$

where $a\neq 0$ is a parameter. We now are ready to state the main results of
this paper about the basin of attraction of the origin of (1.5), defined as

(1.6) $\mathcal{A}_{a,d}(0)=\\{(x,y)\in\mathbb{R}^{2}\,|\
T_{a,d}^{n}(x,y)\to(0,0)\,\hbox{ as }\,n\to\infty\\},$

depending on the parameters $a$ and $d$. Obviously, the origin is the unique
fixed point of $T_{a,d}$ and $DT_{d}(0,0)$ has eigenvalues $0$ and $1$. The
$0$ eigenvalue guarantees that $\mathcal{A}_{a,d}(0)\neq\emptyset$ but a
complete topological and geometric description depends on the motion over the
center manifold. The main theorem describes $\mathcal{A}_{a,d}(0)$ as well as
its boundary $\partial\mathcal{A}_{a,d}(0)$ depending on $a$ and $d$.

######  Theorem A.

Let $\mathcal{A}_{a,d}(0)$ be the basin of attraction of the origin for the
map $T_{a,d}$.

  1. (a)

If $d$ is even and $a\neq 0$ then $\mathcal{A}_{a,d}(0)$ is a compact set
which is homeomorphic to a closed topological disk and the boundary of
$\mathcal{A}_{a,d}(0)$ is the stable manifold of the origin. See Figure 2(a).

  2. (b)

If $d$ is odd and $a>0$ then $\mathcal{A}_{a,d}(0)$ is an open, simply
connected, unbounded set. Moreover, $\partial\mathcal{A}_{a,d}(0)$ contains
the stable manifold of a hyperbolic two-cycle $\\{p_{0},p_{1}\\}$ lying on
$\partial\mathcal{A}_{a,d}(0)$. See Figure 2(b).

  3. (c)

If $d$ is odd and $a<0$ then $\mathcal{A}_{a,d}(0)$ is the stable manifold of
the origin. Moreover, $\mathcal{A}_{a,d}(0)$ is unbounded. See Figure 2(c).

(a) $d=2$ and $a=1$.  (b) $d=3$ and $a=1$. (c) $d=3$ and $a=-1$.

Figure 2.  Phase plane of the map $T_{a,d}$ for different values of $a$ and
$d$. The basin of attraction $\mathcal{A}_{a,d}$ is shown in red.

$y=1$$y=-1$$p_{0}$$p_{1}$

Figure 3. The homoclinic intersection between the stable (blue) and the
unstable (yelow) manifolds of the hyperbolic two-cycle $\\{p_{0},p_{1}\\}$.
The picture also illustrates that the stable manifold of the cycle is related
to the boundary of $\mathcal{A}_{d}(0)$ (in red) for $d\geq 3$ odd (the
picture is done for $d=3$).

We finish with an important remark, somehow complementary, on the previous
result to calibrate their value. On the one hand, from construction, system
(1.5) encodes the information of system (1.4) as long as $(x,y)\approx(0,0)$.
But if one reads carefully Theorem A, we see that it does not refer to the
dynamics in a given small neighbourhood of the origin, as for instance Theorem
A(b) is showing that $\mathcal{A}_{a,d},\ a>0,$ is unbounded. Hence, a priory,
there is no reason to argue that Theorem A can be transported to explain the
red regions in Figure 1. However, comparing the top right picture in Figure 1
with Figure 2(a) ($d$ even), or comparing the bottom right picture in Figure 1
with Figure 2(b) ($d$ odd) one can immediately see that (1.5) and (1.4) share
more than expected. A better explanation for this connection, somehow global,
will require future work.

In a companion paper [FGJ24] we study in more depth the boundary of
$\mathcal{A}_{a,d}(0)$ when $d$ is odd and $a$ is positive and we show there
is a (topologically transversally) homoclinic intersection between the stable
and the unstable manifolds of the hyperbolic two-cycle $\\{p_{0},p_{1}\\}$ and
there are infinitely many periodic points (somehow chaotic dynamics) in
$\partial\mathcal{A}_{a,d}(0)$. See Figure 3.

The paper is organized as follows. In Section 2 we show that $T_{a,d}$ reduces
to three cases: $d$ even with $a=1$, and $d$ odd with $a=\pm 1$. In Section 3
we study the series expansions of the stable and center invariant manifolds of
the origin. Theorems A(a) and A(b) are proven in Sections 4 and 5,
respectively. Finally, in Section 6 we prove Theorem A(c).

##  2\. Preliminaries and local dynamics near the origin

A preliminary simple step is to show that, given $d\geq 2$ fixed, for most
values of $a\neq 0$ the maps of the family $T_{a,d}$ in (1.5) are conjugate to
each other, so that we only need to deal with one or two particular values of
$a$. See Corollary 2.2 below.

######  Lemma 2.1.

We have the following statements.

  1. (a)

If $d$ is even and $a_{1}$ and $a_{2}$ are such that $a_{1}a_{2}\neq 0$ then
$T_{a_{1},d}$ is conjugate to $T_{a_{2},d}$.

  2. (b)

If $d$ is odd and $a_{1}$ and $a_{2}$ are such that $a_{1}a_{2}>0$ then
$T_{a_{1},d}$ is conjugate to $T_{a_{2},d}$.

###### Proof.

The conjugation will be a rescaling. Given any $\mu\in\mathbb{R}$ we have that

$T_{a,d}(\mu x,\mu
y)=\mu\left(y-a\mu^{d-1}(x+y)^{d},y-2a\mu^{d-1}(x+y)^{d}\right)=\mu
T_{a\mu^{d-1},d}(x,y).$

Given $a_{1},a_{2}\neq 0$ we take

$\mu:=(a_{2}/a_{1})^{1/(d-1)}.$

If $d$ is even and $a_{1}$ and $a_{2}$ are two parameters with $a_{1}a_{2}\neq
0$ we immediately have

$T_{a_{1},d}(\mu x,\mu y)=\mu T_{a_{2},d}(x,y).$

If $d$ is odd the same is true but the existence of the $(d-1)$-root requires
the condition $a_{1}a_{2}>0$. ∎

######  Corollary 2.2.

To study the dynamics of the family of maps given by (1.5) it is enough to
consider the cases $\\{a=1,\ d\geq 2\\}$ and $\\{a=-1,\ d\geq 3,\ d\
\mbox{odd}\\}$.

To avoid heavy notation (depending on the parameter $a=\pm 1$) in what follows
we assume $a=1$. We will deal with the case $a=-1$ (for $d$ odd) in Section 3,
Remark 3.2, and in Section 6. In particular, when $a=1$, we will use the
simplified notation

(2.1) $T_{d}(x,y):=T_{1,d}(x,y)=\left(\begin{array}[]{l}y-(x+y)^{d}\\\
y-2(x+y)^{d}\end{array}\right).$

######  Lemma 2.3.

We have

  * (a)

If $d$ is even, $T_{d}$ sends $\mathbb{R}^{2}$ onto $\\{x\geq y\\}$. The map
$T_{d}$ has two inverses

(2.2) $T^{-1}_{\pm,d}(x,y)=\left(-2x+y\pm\left(x-y\right)^{1/d},\,2x-y\right)$

which determine two one to one maps $T^{-1}_{+,d}:\\{x\geq
y\\}\to\\{x\geq-y\\}$ and $T^{-1}_{-,d}:\\{x\geq y\\}\to\\{x\leq-y\\}$.

Moreover, for any $x_{0}\in\mathbb{R}$, $T_{d}$ maps the line $y=-x+x_{0}$
onto the line $y=x-x_{0}^{d}$ in a one-to-one way.

  * (b)

If $d$ is odd, the map $T_{d}:\mathbb{R}^{2}\to\mathbb{R}^{2}$ is a
homeomorphism onto $\mathbb{R}^{2}$ and its inverse map is real analytic in
$\mathbb{R}^{2}\setminus\\{x=y\\}$, but not differentiable on $\\{x=y\\}$. Its
inverse is given by

(2.3) $T^{-1}_{d}(x,y)=\left(-2x+y+\left(x-y\right)^{1/d},\,2x-y\right).$

Moreover, for any $x_{0}\in\mathbb{R}$, the map $T_{d}$ maps bijectively the
line $y=-x+x_{0}$ onto the line $y=x-x_{0}^{d}$.

###### Proof.

All statements come from direct computations. ∎

##  3\. Local dynamics around the origin: The stable and the center manifolds

The origin is the only fixed point of the map $T_{d}$ in (2.1). In this
section we obtain information on the local dynamics near the origin from the
analytic expressions (in series expansion) of the (local) invariant manifolds.
The derivative of $T_{d}$ at $(x,y)$ is given by

(3.1) $DT_{d}(x,y)=\left(\begin{array}[]{cc }-d(x+y)^{d-1}&\ \
1-d(x+y)^{d-1}\\\ -2d(x+y)^{d-1}&\ \ 1-2d(x+y)^{d-1}\end{array}\right),$

and therefore

$DT_{d}(0,0)=\left(\begin{array}[]{cc }0&1\\\ 0&1\end{array}\right).$

The matrix $DT_{d}(0,0)$ is independent of the parameter $d$. Its eigenvalues
are $0$ and $1$ with associated eigenvectors $v_{1}=(1,0)$ and $v_{2}=(1,1)$,
respectively. In other words the direction $v_{1}$ is super-attracting while
the direction $v_{2}$ is neutral. It follows from the general theory of
invariant manifolds of fixed points of maps that there is a stable invariant
manifold of $(0,0)$ being tangent to $v_{1}$ and a (non-unique) center
invariant manifold of $(0,0)$ being tangent to $v_{2}$. We denote them
$W^{s}_{d}(0)$ and $W^{c}_{d}(0)$, respectively. According to the general
theory, $W^{s}_{d}(0)$ is analytic and $W^{c}_{d}(0)$ is $C^{k}$ for all
$k\geq 1$. Even if $W^{c}_{d}(0)$ may not be unique, all its Taylor
coefficients are uniquely determined.

More concretely, the local invariant manifolds can be parametrized as graphs

(3.2)
$W^{s}_{d,\mathrm{loc}\,}(0)=\\{(x,\varphi^{s}_{d}(x))\mid\,|x|<\varepsilon_{0}\\}\qquad\text{
and }\qquad
W^{c}_{d,\mathrm{loc}\,}(0)=\\{(x,\varphi^{c}_{d}(x))\mid\,|x|<\varepsilon_{0}\\},$

for some $\varepsilon_{0}>0$, where

(3.3)
$\varphi^{s}_{d}(x)=\sum_{n=2}^{\infty}\alpha_{n}(d)x^{n}\qquad\mbox{and}\qquad\varphi^{c}_{d}(x)=x+\sum_{n=2}^{\infty}\beta_{n}(d)x^{n}.$

We also denote by $R^{s}_{d}$ and $R^{c}_{d}$ the maps which encode the
induced dynamics on the invariant manifolds. Thus, locally, we have

(3.4)
$T_{d}\left(x,\varphi_{d}^{s}(x)\right)=\left(R^{s}_{d}(x),\varphi_{d}^{s}(R^{s}_{d}(x))\right)\qquad\mbox{and}\qquad
T_{d}\left(x,\varphi_{d}^{c}(x)\right)=\left(R^{c}_{d}(x),\varphi_{d}^{c}(R^{c}_{d}(x))\right),$

respectively.

See [HPS77, CFdlL03, CFdlL05] for a general discussion on the theory of local
invariant manifolds. In the next lemma we provide the structure of the Taylor
expansion of $\varphi^{s}_{d}$ and $\varphi^{c}_{d}$. The lower order terms
will determine the local dynamics near the origin.

######  Lemma 3.1.

Let $d\geq 2$. The Taylor series of $\varphi^{s}_{d}$ and $\varphi^{c}_{d}$
have the following structure

(3.5)
$\varphi^{s}_{d}(x)=x^{d}\displaystyle\sum_{k=0}^{\infty}\alpha_{d+k(d-1)}(d)x^{k(d-1)}\qquad\mbox{and}\qquad\varphi^{c}_{d}(x)=x+x^{d}\displaystyle\sum_{k=0}^{\infty}\beta_{d+k(d-1)}(d)x^{k(d-1)}.$

Moreover, $\alpha_{d}(d)=2$, $\alpha_{2d-1}(d)=4d$, $\beta_{d}(d)=-2^{d}$ and
$\beta_{2d-1}(d)=-3d2^{2d-1}$ and thus we have that

$\varphi_{d}^{s}(x)=2x^{d}+\mathcal{O}(x^{2d-1})\qquad\mbox{and}\qquad\varphi_{d}^{c}(x)=x-2^{d}x^{d}+\mathcal{O}(x^{2d-1})$

and the one-dimensional dynamics induced by $T_{d}$ on the stable and center
manifolds are governed by

(3.6) $R^{s}_{d}:x\mapsto x^{d}+\mathcal{O}(x^{2d-1})\qquad\mbox{and}\qquad
R^{c}_{d}:x\mapsto x-4^{4}x^{d}+\mathcal{O}(x^{2d-1}),$

respectively.

See Figure 4(a) and (b) for the induced dynamics of the map $T_{d}$ on the
invariant manifolds.

###### Proof.

To simplify the notation below we introduce the symbol $\\{\cdot\\}_{n}$ so
that if $\Phi$ is a formal series around the origin, we write

$\Phi(x)=\sum_{n\geq 0}\\{\Phi\\}_{n}\ x^{n}.$

We prove (3.5) for the case of the stable manifold $W^{s}_{d}(0)$ (see (3.3)).
Using that the stable manifold is an invariant graph for $T_{d}$ we obtain
that if $W^{s}_{d}(0)=\mathrm{graph}\,\varphi^{s}_{d}$ then

(3.7)
$\varphi^{s}_{d}(x)-2\left[x+\varphi^{s}_{d}(x)\right]^{d}=\varphi^{s}_{d}\left(\varphi^{s}_{d}(x)-\left[x+\varphi^{s}_{d}(x)\right]^{d}\right).$

From the above equation, some computations show that, on the one hand
$\alpha_{2}(2)=2$ and $\alpha_{2}(d)=0$ for all $d\geq 3$, and on the other
hand, for all $n\geq 3$ we have that $\alpha_{n}(d)$ in (3.3) can be written
recursively as

(3.8)
$\alpha_{n}(d)=2\left\\{\left(x+\displaystyle\sum_{j=2}^{n-1}\alpha_{j}(d)x^{j}\right)^{d}\right\\}_{n}+\displaystyle\sum_{i=2}^{n-1}\alpha_{i}(d)\left\\{\left(\displaystyle\sum_{j=2}^{n-1}\alpha_{j}(d)x^{j}-\left(x+\displaystyle\sum_{j=2}^{n-1}\alpha_{j}(d)x^{j}\right)^{d}\right)^{i}\right\\}_{n}.$

Proving (3.5) for the stable manifold is equivalent to see that in (3.3) the
coefficient $\alpha_{n}(d)=0$ for all $n\geq 2$ such that $n-d$ is not a
multiple of $d-1$, or equivalently, not of the form $n=d+k(d-1)$ for $k\geq
0$. We argue by induction. We claim that for any $N\geq 1$, up to order

$n=d+(N-1)(d-1)$

the stable manifold writes as

(3.9)
$x^{d}\displaystyle\sum_{k=0}^{N-1}\alpha_{d+k(d-1)}(d)x^{k(d-1)}=:x^{d}\Psi(x^{d-1}).$

When $N=1$ the result is true since $\alpha_{n}(d)=0$ for $2\leq n\leq d-1$
and $\alpha_{d}(d)=2$. Indeed, from (3.8), $a_{j}(d)=0$ implies
$\alpha_{j+1}(d)=0$ for all $j=1,\ldots,d-2$. Also, $\alpha_{d}(d)=2$ since we
have a unique term of degree $d$ with coefficient $2$ associated to the first
$\\{\Phi\\}_{n}$ term in the right hand side of (3.8).

Assuming the claim is true for $N$, we are going to prove that in the right
hand side of (3.8) the coefficients $\alpha_{n+j}(d)x^{n+j}$, $j\geq 1$, are
involved in terms of order $n+d$ or higher. This is easy to check for $j=1$.
For $j>1$ the coefficients appear in terms of order bigger or equal than
$n+d+1$. In the right hand side of (3.8) the first term is

$2(x+x^{d}\Psi(x^{d-1})+\alpha_{n+1}(d)x^{n+1}+\dots)^{d}$

and the lower term in which $\alpha_{n+1}(d)$ appears is
$2dx^{d-1}\alpha_{n+1}(d)x^{n+1}=2d\alpha_{n+1}(d)x^{n+d}$. The second term of
the right hand side of (3.8) can be written as

$\alpha_{d}(d)\left(x^{d}\Psi(x^{d-1})+\alpha_{n+1}(d)x^{n+1}+\dots-(x+x^{d}\Psi(x^{d-1})+\alpha_{n+1}(d)x^{n+1}+\dots)^{d}\right)^{d}+\dots$

and the lower term in which $\alpha_{n+1}(d)$ appears is
$2d(2x^{d})^{d-1}\alpha_{n+1}(d)x^{n+1}=\mathcal{O}(x^{n+d(d-1)+1})$. This
finishes the induction.

Once the expression of $\varphi^{s}_{d}$ given in (3.5) is proved and the
first terms of the expansion have been calculated we only need to justify the
expression in (3.6). For this we compute the image of a point on the stable
invariant manifold only using the lowest term of the series expansion

$\displaystyle T_{d}\left(x,\varphi^{s}_{d}(x)\right)$
$\displaystyle=\left(2x^{d}-\left(x+2x^{d}\right)^{d},\,2x^{d}-2\left(x+2x^{d}\right)^{d}\right)$
$\displaystyle=\left(x^{d}+\mathcal{O}(x^{2d-1}),-4dx^{2d-1}+\mathcal{O}(x^{3d-2)})\right).$

Therefore, the one-dimensional dynamics is given by

$x\mapsto x^{d}+\mathcal{O}(x^{2d-1}).$

Similar computations provide the result for $\varphi^{c}_{d}$. ∎

######  Remark 3.2.

Using the same arguments as the ones in Lemma 3.1 one can get similar results
for the case $d$ odd and $a=-1$. The difference is the sign of some leading
coefficients. More precisely if $d$ is odd and $a=-1$ in the definition of
$T_{d}$ we have that $\alpha_{d}(d)=-2$, $\alpha_{2d-1}(d)=4d$,
$\beta_{d}(d)=2^{d}$ and $\beta_{2d-1}(d)=-3d2^{2d-1}$ and hence we have that

$\varphi_{d}^{s}(x)=-2x^{d}+\mathcal{O}(x^{2d-1})\qquad\text{and}\qquad\varphi_{d}^{c}(x)=x+2^{d}x^{d}+\mathcal{O}(x^{2d-1})$

and the one-dimensional dynamics induced by $T_{d}$ over the stable and center
manifold are governed by

$x\mapsto-x^{d}+\mathcal{O}(x^{2d-1})\qquad\text{and}\qquad x\mapsto
x+4^{4}x^{d}+\mathcal{O}(x^{2d-1}),$

respectively. See Figure 4(c) for the induced dynamics of the map $T_{d}$ over
the invariant manifolds in this case.

(a) $d$ even and $a=1$$W^{s}(0)$$W^{c}(0)$(b) $d$ odd and
$a=1$$W^{c}(0)$$W^{s}(0)$(c) $d$ odd and $a=-1$$W^{c}(0)$$W^{s}(0)$

Figure 4. Local dynamics of $T_{a,d}$ near the origin.

We close this section by completing the discussion above, for the case $d$
even. We have shown in Lemma 3.1 that many coefficients of the series
expansion of the stable and center manifolds are zero (no mater the parity of
$d$). Next we prove that, for $d$ even, all non-zero coefficients of
$\varphi^{s}_{d}$ are positive.

######  Lemma 3.3.

Let $d$ be even. Then, $\alpha_{\ell}(d)\geq 0$ for all $\ell\geq 0$.

###### Proof.

In Lemma 3.1 we proved that the coefficients of the series expansion of the
analytic expression of the local stable manifold at the origin satisfy certain
properties. In particular we proved that all coefficients $\alpha_{\ell}(d),\
\ell\geq 0,$ of the monomials $x^{\ell}$ with $\ell\neq d+k(d-1)$ for some
$k\geq 0$ are zero. Moreover we also proved that $\alpha_{d}(d)=2$ for all
$d\geq 2$.

We fix $d\geq 2$ even. To simplify the notation we remove the dependence of
the coefficients with respect to $d$; that is, we write
$\alpha_{k}:=\alpha_{k}(d)$. Let $\gamma(x)$ be the auxiliary analytic
function given by the series expansion

$\gamma(x)=\sum_{k=d}^{\infty}\gamma_{k}x^{k}:=\left(x+\displaystyle\sum_{k=d}^{\infty}\alpha_{k}x^{k}\right)^{d}.$

Note that $\gamma_{k+1}$ depends on $\alpha_{j}$, $d\leq j\leq k$. The lemma
follows from the following claim.

Claim: If $n\geq d$ then, for all $d\leq k\leq n$, we have $\alpha_{k}\geq
2\gamma_{k}\geq 0$.

We prove the claim by induction. For $n=d$ it is obviously true because
$\alpha_{d}=2$ and $\gamma_{d}=1$. Assuming the claim is true for $n$, from
(3.8) we can write

(3.10)
$\alpha_{n+1}=2\left\\{\left(x+\displaystyle\sum_{k=d}^{n}\alpha_{k}x^{k}\right)^{d}\right\\}_{n+1}+\left\\{\displaystyle\sum_{i=d}^{n}\alpha_{i}\left[\displaystyle\sum_{k=d}^{n}\alpha_{k}x^{k}-\left(x+\displaystyle\sum_{k=d}^{n}\alpha_{k}x^{k}\right)^{d}\right]^{i}\right\\}_{n+1}.$

The induction assumption implies

$\alpha_{k}\geq
2\left\\{\left(x+\displaystyle\sum_{j=d}^{k}\alpha_{j}x^{j}\right)^{d}\right\\}_{k}\geq\left\\{\left(x+\displaystyle\sum_{j=d}^{k}\alpha_{j}x^{j}\right)^{d}\right\\}_{k},\qquad
d\leq k\leq n.$

This implies that all coefficients of the terms of order $n+1$ of the second
term of the right hand side of (3.10) are non-negative (because $i\geq d\geq
2$). Then, we conclude from (3.10) that $\alpha_{n+1}\geq 2\gamma_{n+1}\geq
0$, and the claim follows.

∎

##  4\. Proof of Theorem A(${\rm a}$): The case $d$ even and $a=1$

Let $d\geq 2$ be even. From Corollary 2.2 we can take $a=1$ to cover all cases
$(a\neq 0$). We simplify the notation writing $T_{d}:=T_{1,d}$ and
$\mathcal{A}_{d}(0):=\mathcal{A}_{1,d}(0)$. We will show that the origin
belongs to the boundary of the basin, that the basin is contained in the upper
half plane and that its boundary is the stable manifold of the origin.

Let us introduce some notation. Given $(x_{0},y_{0})\in\mathbb{R}^{2}$ we will
write $(x_{k},y_{k})=T_{d}^{k}(x_{0},y_{0})$ for $k\geq 0$. Set

$R_{d}:=\big{(}1-\frac{1}{d}\big{)}(2d)^{\frac{-1}{d-1}}.$

Note that $R_{2}=1/8$ and, in general, $R_{d}<1$. Finally let

$\mathcal{T}=\\{(x,y)\mid\ y\leq
x\\}\qquad\text{and}\qquad\mathcal{T}_{R_{d}}=\\{(x,y)\in\mathcal{T}\mid\
y\geq 0,\ 0\leq x\leq R_{d}\\}.$

Since the proof of Theorem A(a) is quite long we split the arguments into
several lemmas. The first one is just an observation.

######  Lemma 4.1.

We have that $(x_{k},y_{k})\in\mathcal{T}$ for $k\geq 1$ and then the
sequences $\\{x_{k}\\}_{k\geq 1}$ and $\\{y_{k}\\}_{k\geq 1}$ are
monotonically decreasing. Moreover, while $x_{k}>-y_{k}$ the sequences are
strictly decreasing.

###### Proof.

The first assertion follows from Lemma 2.3(a). The second one follows directly
from the inequalities:

(4.1) $\displaystyle x_{k+1}$ $\displaystyle=y_{k}-(x_{k}+y_{k})^{d}\leq
y_{k}\leq x_{k},\quad k\geq 1\quad(k\geq 0\ \ \text{if}\ \
(x_{0},y_{0})\in\mathcal{T}),$ (4.2) $\displaystyle y_{k+1}$
$\displaystyle=y_{k}-2(x_{k}+y_{k})^{d}\leq y_{k},\quad\qquad k\geq 0.$

∎

$R_{d}$$r_{0}$$(p,p)$$(-p,p)$$\mathcal{A}_{d}(0)$${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\Gamma}$$q$$m=2$$\Gamma_{+}$$\Gamma_{+}$$q^{+}$$q^{-}$$\Gamma_{-}$$y=x$$y=-x$$\rho_{2}$$\Omega_{0}^{+}$$\Omega_{0}^{-}$$\Omega_{r}$$J_{\ell}$$J_{r}$

Figure 5. Sketch of the construction of $\mathcal{A}_{d}(0)$.

Next lemma shows that $\mathcal{A}_{d}(0)$ is a bounded set.

######  Lemma 4.2.

$\mathcal{A}_{d}(0)$ is a bounded set. More concretely,

$\mathcal{A}_{d}(0)\subset\mathcal{K}:=(-5R_{d},R_{d})\times(0,2R_{d})\cup\\{(0,0)\\}.$

###### Proof.

We decompose

$\mathbb{R}^{2}\setminus\mathcal{K}=\mathcal{K}_{1}\cup\mathcal{K}_{2}\cup\mathcal{K}_{3}\cup\mathcal{K}_{4},$

where

$\displaystyle\mathcal{K}_{1}$ $\displaystyle=\\{(x,y)\mid\ y\leq
0\\}\setminus\\{(0,0)\\},$ $\displaystyle\mathcal{K}_{2}$
$\displaystyle=\\{(x,y)\mid\ x\geq R_{d}\\},$ $\displaystyle\mathcal{K}_{3}$
$\displaystyle=\\{(x,y)\mid\ y\geq 2R_{d}\\},$ $\displaystyle\mathcal{K}_{4}$
$\displaystyle=\\{(x,y)\mid\ x\leq-5R_{d},\ 0<y<2R_{d}\\}$

and we argue that $\mathcal{A}_{d}(0)\cap\mathcal{K}_{j}=\emptyset$, $1\leq
j\leq 4$, so that $\mathcal{A}_{d}(0)\subset\mathcal{K}$. We will use the
following property: by the invariance of $\mathcal{A}_{d}(0)$ by $T_{d}$ and
its inverses we have that $T_{\pm,d}^{-1}$,
$(x_{0},y_{0})\in\mathcal{A}_{d}(0)$ if and only if
$(x_{k},y_{k})\in\mathcal{A}_{d}(0)$ for some $k\geq 0$.

Let $(x_{0},y_{0})\in\mathcal{K}_{1}$. Since $(x_{0},y_{0})\neq(0,0)$, if
$y_{0}=0$ then $y_{1}=-2x_{0}^{d}<0$ and if $y_{0}<0$ then $y_{1}\leq
y_{0}<0$. Hence, in both cases $y_{1}<0$ and by (4.2) the sequence of iterates
cannot converge to $(0,0)$.

Next, we claim that if $(x_{0},y_{0})\in\mathcal{K}_{2}$ then
$(x_{1},y_{1})\in\mathcal{K}_{1}$. Indeed, we consider the line
$\\{x=x_{0}\\}$ with $x_{0}\geq R_{d}$ and we look at the second component of
its image $\Psi_{1}(y):=\pi_{y}T_{d}(x_{0},y)=y-2(x_{0}+y)^{d}$. Since $d\geq
2$ is even, $\lim_{t\to\pm\infty}\Psi(y)=-\infty$ and therefore $\Psi_{1}$ has
a global maximum. Actually, it has a unique maximum whose location is obtained
from the condition $\Psi_{1}^{\prime}(y)=0$ and is
$y^{(m)}:=1/(2d)^{1/(d-1)}-x_{0}$. Then
$\Psi^{(m)}_{1}:=\Psi_{1}(y^{(m)})=1/(2d)^{1/(d-1)}-2/(2d)^{d/(d-1)}-x_{0}=R_{d}-x_{0}$
and therefore $y_{1}\leq 0$. Moreover, $(x_{1},y_{1})\neq(0,0)$ because the
only preimage of $(0,0)$ is $(0,0)\notin\mathcal{K}_{2}$.

Next we take $(x_{0},y_{0})\in\mathcal{K}_{3}$ and we claim that
$(x_{1},y_{1})\in\mathcal{K}_{1}\cup\mathcal{K}_{2}$. Indeed, consider the
line $\\{y=y_{0}\\}$ with $y_{0}\geq 2R$. Its image is contained in the line
$\\{(u,v)\mid\ v=2u-y_{0}\\}$ and it is contained in
$\mathcal{K}_{1}\cup\mathcal{K}_{2}$ because we have that either $u\geq R_{d}$
and the claim is true or $u<R_{d}$ and then $v=2u-y_{0}<2R_{d}-y_{0}\leq 0$
and $(x_{1},y_{1})\neq(0,0)$.

Finally, if $(x_{0},y_{0})\in\mathcal{K}_{4}$ we claim that
$(x_{1},y_{1})\in\mathcal{K}_{1}\cup\mathcal{K}_{2}$. Indeed, notice that
$x_{0}+y_{0}<-5R_{d}+2R_{d}=-3R_{d}$ and then
$(x_{0}+y_{0})^{d}>(3R_{d})^{d}$. If $x_{1}\geq R_{d}$ the claim is true. If
$x_{1}<R_{d}$ then
$y_{1}=y_{0}-2(x_{0}+y_{0})^{d}=x_{1}-(x_{0}+y_{0})^{d}<R_{d}-(3R_{d})^{d}=(1-\frac{3^{d}}{2d}(1-\frac{1}{d})^{d-1})R_{d}<0$.
∎

######  Lemma 4.3.

We have that $(x_{0},y_{0})\in\mathcal{A}_{d}(0)$ if and only if
$(x_{k},y_{k})\in\mathcal{T}_{R_{d}}$ for all $k\geq 1$.

###### Proof.

Assume $(x_{0},y_{0})\in\mathcal{A}_{d}(0)$. By Lemma 2.3(a), $x_{k}\geq
y_{k}$ for all $k\geq 1$ and by Lemma 4.2, $x_{k}<R_{d}$ for all $k\geq 0$.
Since the sequences $\\{x_{k}\\}$ and $\\{y_{k}\\}$ are decreasing for $k\geq
1$, if there exists $m>0$ such that $y_{m}<0$, then $y_{k}\leq y_{m}<0$ for
all $k\geq m$ and $(x_{k},y_{k})$ cannot converge to $(0,0)$. Then, $y_{k}\geq
0$ for all $k$ and the limit $y^{\star}=\lim_{k\to\infty}y_{k}$ exists,
$y^{\star}\geq 0$ and then $x_{k}\geq y_{k}\geq 0$. As a consequence
$(x_{k},y_{k})\in\mathcal{T}_{R_{d}}$ for all $k\geq 1$.

Conversely, let $(x_{0},y_{0})\in\mathbb{R}^{2}$ and assume that
$(x_{k},y_{k})\in\mathcal{T}_{R_{d}}$ for all $k\geq 1$. Since the sequence
$\\{y_{k}\\}_{k\geq 0}$ is strictly decreasing and bounded from below by 0
there exists the limit $y^{\star}=\lim_{k\to\infty}y_{k}\geq 0$. From the
recurrence

$y_{k+1}=y_{k}-2(x_{k}+y_{k})^{d}$

we obtain that $\lim_{k\to\infty}(x_{k}+y_{k})$ exists and it is 0. This
implies that $-y^{\star}=\lim_{k\to\infty}x_{k}\geq 0$. Then $y^{\star}=0$ and
$(x_{0},y_{0})\in\mathcal{A}_{d}(0)$. ∎

We now turn the attention to $\partial\mathcal{A}_{d}(0)$. Our goal is to
prove that $\partial\mathcal{A}_{d}(0)$ coincides with the global stable
manifold of the origin, $W^{s}_{d}(0)$.

######  Lemma 4.4.

$W^{s}_{d}(0)$ cuts the line $\\{x=y\\}$ at some point $(p,p)$ with
$0<p<R_{d}$.

###### Proof.

In Lemma 3.3 it is proven that the local expression of $W^{s}_{d}(0)$ is given
by the graph of an analytic function $\varphi_{d}^{s}(x)=2x^{d}+\dots$ whose
series expansion in the $x$-variable has all its coefficients non-negative,
and therefore there exists $\rho_{1}>0$ such that
$\gamma=\\{(x,\varphi_{d}^{s}(x))\mid\ x\in(0,\rho_{1})\\}$ is contained in
$\mathcal{T}_{R_{d}}$. We claim that the extension of this local piece
$\gamma$ of $W^{s}_{d}(0)$ eventually leaves $\mathcal{T}_{R_{d}}$. Indeed,
assume the contrary. We globalize $\gamma$ iterating with $T^{-1}_{+,d}$ (see
(2.2)). Let $(x_{0},y_{0})\in\gamma$ and denote

$\left(x_{-k},y_{-k}\right)=T^{-k}_{+,d}(x_{0},y_{0}),\qquad k\geq 0.$

We have

(4.3) $y_{-k-1}=y_{-k}+2(x_{-k}-y_{-k})>y_{-k}.$

If all $\left(x_{-k},y_{-k}\right)\in\mathcal{T}_{R_{d}}$ we have that the
sequence $\\{y_{-k}\\}_{k\geq 0}$ is strictly increasing and bounded, and we
conclude that there exists $y^{\star}>0$ such that

(4.4) $\displaystyle y^{\star}=\lim_{k\to\infty}y_{-k}>0.$

Moreover, from (4.3) we have that $x_{-k}=(y_{-k}+y_{-k-1})/2\to y^{\star}$.
Now, using the recurrence

$x_{-k-1}=-2x_{-k}+y_{-k}+(x_{-k}-y_{-k})^{1/d},$

we get that $y^{\star}=0$, which provides a contradiction with (4.4). Finally,
since we have seen that $\mathcal{A}_{d}(0)$ does not meet
$\\{y=0\\}\setminus\\{(0,0)\\}$ nor $\\{x=R_{d}\\}$ the globalization of
$\gamma$ has to cross $\\{x=y\\}$. ∎

We denote by $\Gamma$ the piece of the stable manifold $W^{s}_{d}(0)$ from
$(0,0)$ to $(p,p)$ contained in $\mathcal{T}_{R_{d}}$. We plot $\Gamma$ in red
colour in Figure 5. Let $\varphi^{s}_{d}$ be given in (3.2).

######  Lemma 4.5.

The following properties for $\varphi_{d}^{s}$ hold.

  * (a)

There exists a unique point
$\bar{q}=\left(\bar{q}_{x},\bar{q}_{y}\right)\in\Gamma$ whose tangent vector
has slope $m=1/2$.

  * (b)

If we denote by $r_{0}>0$ the radius of convergence of $\varphi_{d}^{s}$ (as a
function of a complex variable) then $0<r_{0}<R_{d}$, $\varphi^{s}_{d}$ is
increasing and convex in the interval $(0,r_{0})$ and decreasing and convex
$(-r_{0}-2\varphi^{s}_{d}(r_{0}),0)$.

###### Proof.

We observe that since all coefficients of the series expansion of
$\varphi_{d}^{s}$ are non-negative (see Lemma 3.3) we conclude from Vivanti-
Pringsheim’s Theorem [Evg78] that $\varphi^{s}_{d}$ as a function of a complex
variable has a singularity at $x=r_{0}>0$ and

$\varphi_{0}:=\varphi^{s}_{d}(r_{0})=\displaystyle\sum_{k\geq
d}\alpha_{k}r_{0}^{k}.$

In fact we have that $r_{0}<R_{d}<\infty$ and $\varphi_{0}<2R_{d}$ since
$\mathrm{graph}\,\varphi_{d}^{s}\subset W_{d}^{s}(0)\subset\mathcal{A}_{d}(0)$
and by Lemma 4.2, $\mathcal{A}_{d}(0)\subset\mathcal{K}$. In particular
$\varphi^{s}_{d}|_{(0,r_{0})}$ is an increasing and convex function and

$\displaystyle\lim_{x\to
r_{0}^{-}}\left(\varphi^{s}_{d}\right)^{\prime}(x)=+\infty.$

Indeed, if $\left(\varphi^{s}_{d}\right)^{\prime}(r_{0})<\infty$, then
$\left(\varphi^{s}_{d}\right)^{\prime}$ could be extended in a differentiable
way for $x>r_{0}$. The graph close to $x=r_{0}$ will be the image by
$T^{-1}_{+,d}$ of a piece of the graph of $\varphi^{s}_{d}$, say $\gamma_{2}$,
closer to the origin. The piece $\gamma_{2}$ does not contain the point
$(p,p)$ since its image is the point $(-p,p)$ outside $\mathcal{T}_{R_{d}}$.
Therefore $T^{-1}_{+,d}$ is analytic on $\gamma_{2}$ and then
$\varphi^{s}_{d}$ would be analytic in a neighborhood of $r_{0}$ which
provides a contradiction. This proves statement (a) and provides the existence
of $r_{0}$.

We have the symmetry ($d$ is even)

(4.5) $T_{d}\left(x,y\right)=T_{d}\left(-2y-x,y\right).$

Hence if $(x,y)\in W_{d}^{s}(0)$ then also $(-2y-x,y)\in W_{d}^{s}(0)$. More
concretely, since $(x,\varphi^{s}_{d}(x))\in W^{s}_{d}(0)$,
$(-x-2\varphi^{s}_{d}(x),\varphi^{s}_{d}(x))\in W^{s}_{d}(0)$ and then

(4.6) $\varphi^{s}_{d}(-x-2\varphi^{s}_{d}(x))=\varphi^{s}_{d}(x).$

This means that $\varphi^{s}_{d}$ is defined for
$x\in(-r_{0}-2\varphi^{s}_{d}(r_{0}),0)$. Moreover, taking derivatives in
(4.6) we get

(4.7)
$\displaystyle(\varphi^{s}_{d})^{\prime}(-x-2\varphi^{s}_{d}(x))(-1-2(\varphi^{s}_{d})^{\prime}(x))=(\varphi^{s}_{d})^{\prime}(x),$
(4.8)
$\displaystyle(\varphi^{s}_{d})^{\prime\prime}(-x-2\varphi^{s}_{d}(x))(-1-2(\varphi^{s}_{d})^{\prime}(x))^{2}+(\varphi^{s}_{d})^{\prime}(-x-2\varphi^{s}_{d}(x))(-2(\varphi^{s}_{d})^{\prime\prime}(x))=(\varphi^{s}_{d})^{\prime\prime}(x),$

and hence we can conclude that $\varphi^{s}_{d}$ is decreasing and convex in
$(-r_{0}-2\varphi^{s}_{d}(r_{0}),0)$. Indeed, substituting
$(\varphi^{s}_{d})^{\prime}(-x-2\varphi^{s}_{d}(x))$ from (4.7) into (4.8) we
obtain

$(\varphi^{s}_{d})^{\prime\prime}(-x-2\varphi^{s}_{d}(x))(-1-2(\varphi^{s}_{d})^{\prime}(x))^{2}=(\varphi^{s}_{d})^{\prime\prime}(x)\left(1-\frac{2(\varphi^{s}_{d})^{\prime}(x)}{1+2(\varphi^{s}_{d})^{\prime}(x)}\right)>0.$

From the previous properties there exists a unique point
$\bar{q}=\left(\bar{q}_{x},\bar{q}_{y}\right)\in\Gamma$ whose tangent vector
has slope $m=1/2$. ∎

Let

$\begin{split}\Omega_{0}^{+}&=\\{(x,y)\in\mathcal{T}_{R_{d}}\mid\,y\geq\varphi_{d}^{s}(x),\,0\leq
x\leq\rho_{2}\\},\\\
\Omega_{0}^{-}&=\\{(x,y)\in\mathcal{T}_{R_{d}}\mid\,0<y<\varphi_{d}^{s}(x),\,0\leq
x\leq\rho_{2}\\},\end{split}$

with
$\rho_{2}<\min\left\\{\frac{1}{2}\frac{1}{(4d)^{1/(d-1)}},\,\bar{q}_{x}\right\\}$.
See Figure 5.

######  Lemma 4.6.

The domain $\Omega_{0}^{+}$ is invariant by $T_{d}$ and
$\Omega_{0}^{+}\subset\mathcal{A}_{d}(0)$. Moreover,
$\Omega_{0}^{-}\cap\mathcal{A}_{d}(0)=\emptyset$.

###### Proof.

Let $(x,y):=(x_{0},y_{0})\in\Omega_{0}^{+}$. The sequence $\\{x_{k}\\}_{k\geq
0}$ is decreasing by Lemma 4.1 . So, it is enough to show that
$y_{1}-\varphi^{s}_{d}(x_{1})\geq 0$. Indeed

(4.9) $\displaystyle y_{1}-\varphi^{s}_{d}(x_{1})$
$\displaystyle=y-2(x+y)^{d}-\varphi^{s}_{d}(y-(x+y)^{d})$
$\displaystyle=y-\varphi^{s}_{d}(x)+\varphi^{s}_{d}(x)-\varphi^{s}_{d}(y-(x+y)^{d})-2(x+y)^{d}$
$\displaystyle=y-\varphi^{s}_{d}(x)+H(x,y),$

where

(4.10) $H(x,y)=\varphi^{s}_{d}(x)-\varphi^{s}_{d}(y-(x+y)^{d})-2(x+y)^{d}.$

Taking into account that $\varphi^{s}_{d}$ satisfies the invariance equation

(4.11)
$\varphi^{s}_{d}(x)-2(x+\varphi^{s}_{d}(x))^{d}=\varphi^{s}_{d}(\varphi^{s}_{d}(x)-(x+\varphi^{s}_{d}(x))^{d}),$

$H$ can be rewritten as

(4.12) $\displaystyle H(x,y)$
$\displaystyle=\varphi^{s}_{d}(\varphi^{s}_{d}(x)-(x+\varphi^{s}_{d}(x))^{d})-\varphi^{s}_{d}(y-(x+y)^{d})+2(x+\varphi^{s}_{d}(x))^{d}-2(x+y)^{d}$
$\displaystyle=\int_{0}^{1}\Big{[}\frac{d}{dt}\varphi^{s}_{d}(\xi_{t}-(x+\xi_{t})^{d})(1-d(x+\xi_{t})^{d-1})+2d(x+\xi_{t})^{d-1}\Big{]}(\varphi_{d}^{s}(x)-y)\,dt$
$\displaystyle=:(\varphi^{s}_{d}(x)-y)\widehat{H}(x,y),$

where $\xi_{t}=y+t(\varphi^{s}_{d}(x)-y)$ and hence since
$(x,y)\in\Omega_{0}^{+}$ we have $0<\varphi_{d}^{s}(x)\leq\xi_{t}\leq y\leq
x\leq\rho_{2}$.

From (4.9) and (4.12) we have

$y_{1}-\varphi^{s}_{d}(x_{1})=\left(y-\varphi^{s}_{d}(x)\right)\left(1-\widehat{H}(x,y)\right),$

so that it is enough to see that $\widehat{H}(x,y)<1$.

Given $x\in(0,\rho_{2})$, we introduce $\Psi_{2}(\xi)=\xi-(x+\xi)^{d}$ for
$\xi\in(0,x)$. The function $\Psi_{2}(x)$ is concave and we have that
$\Psi_{2}(0)=-x^{d}<0$, with $-x^{d}>-\rho_{2}^{d}$ and
$\Psi_{2}(x)=x-(2x)^{d}=x(1-2^{d}x^{d-1})>x(1-1/(2d))>0$ by one of the
conditions in the definition of $\rho_{2}$. Hence,
$\xi_{t}-(x+\xi_{t})^{d}\geq-\rho_{2}^{d}$ for all $t\in[0,1]$. Note that, by
the fact that the coefficients of the expansion of $\varphi^{s}_{d}$ are non-
negative (Lemma 3.3), at the symmetric point the absolute value of the
derivative is smaller, i.e. for $x\in[0,r_{0})$,
$|(\varphi^{s}_{d})^{\prime}(-x)|\leq(\varphi^{s}_{d})^{\prime}(x)$ so that,
for $-x^{d}<\zeta<0$,
$|(\varphi^{s}_{d})^{\prime}(\zeta)|\leq(\varphi^{s}_{d})^{\prime}(-\zeta)\leq(\varphi^{s}_{d})^{\prime}(x^{d})\leq(\varphi^{s}_{d})^{\prime}(\rho_{2})\leq
1/2$. Then

(4.13) $|(\varphi^{s}_{d})^{\prime}(\xi_{t}-(x+\xi_{t})^{d})|<1/2.$

Moreover,

(4.14) $2d(x+\xi_{t})^{d-1}<2d(2x)^{d-1}\leq 1/2.$

By (4.13) and (4.14) we obtain $\widehat{H}(x,y)<1$ and therefore we obtain
that the iterates stay in the same side of $\mathrm{graph}\,\varphi^{s}_{d}$.

Now we deal with $\Omega^{-}_{0}$. To prove that
$\Omega_{0}^{-}\cap\mathcal{A}_{d}(0)=\emptyset$ we will see that if
$(x_{0},y_{0})\in\Omega_{0}^{-}$ then not all its iterates can remain in
$\Omega_{0}^{-}$. Assume the contrary. To simplify the estimates we do a
($x$-depending) translation to put the (local) stable manifold at $\\{y=0\\}$.
Actually, we make the change $C(x,y)=(x,y+\varphi^{s}_{d}(x))$. The
transformed map is

$\widehat{T}_{d}\left(\begin{array}[]{l}x\\\
y\end{array}\right)=\left(\begin{array}[]{l}F(x,y)\\\
G(x,y)\end{array}\right):=\left(\begin{array}[]{l}y+\varphi^{s}_{d}(x)-(x+y+\varphi^{s}_{d}(x))^{d}\\\
y+\varphi^{s}_{d}(x)-2(x+y+\varphi^{s}_{d}(x))^{d}-\varphi^{s}_{d}\big{(}F(x,y)\big{)}\end{array}\right).$

The domain $\Omega^{-}_{0}$ is transformed into

$\widehat{\Omega}^{-}_{0}=\\{(x,y)\mid\
0<x<\rho_{2},\,-\varphi^{s}_{d}(x)<y<0\\}.$

Let $(x_{0},y_{0})\in\widehat{\Omega}^{-}_{0}$. We use again the notation
$(x_{k},y_{k})=\widehat{T}_{d}^{k}(x_{0},y_{0})$ for $k\geq 0$.

Let $\rho_{3}\in(0,\rho_{2}]$ be such that

$0<\varphi^{s}_{d}(x)<3x^{d}\qquad\text{ for }\quad x\in(0,\rho_{3}).$

Assume that $(x_{k},y_{k})\in\widehat{\Omega}^{-}_{0}$ for all $k\geq 0$.
Since $d$ is even, we also have $0<x_{k+1}\leq
y_{k}+\varphi^{s}_{d}(x_{k})<\varphi^{s}_{d}(x_{k})\leq x_{k}$. Then
$\\{x_{k}\\}$ is also decreasing and

$x_{k}=y_{k-1}+\varphi_{d}^{s}(x_{k-1})-\left(x_{k-1}+y_{k-1}+\varphi_{d}^{s}(x_{k-1})\right)^{d}\leq\varphi^{s}_{d}(x_{k-1})-x^{d}_{k-1}<2x^{d}_{k-1},$

and inductively we get

(4.15)
$x_{k}<2^{\left(\frac{d^{k}-1}{d-1}\right)}x_{0}^{d^{k}}<(2^{1/(d-1)}x_{0})^{d^{k}}.$

Note that since $2^{1/(d-1)}x_{0}<2^{1/(d-1)}\rho_{2}<1/2$ then $x_{k}\to 0$.
We have

$G(x,0)=\varphi^{s}_{d}(x)-2(x+\varphi^{s}_{d}(x))^{d}-\varphi^{s}_{d}(\varphi^{s}_{d}(x)-(x+\varphi^{s}_{d}(x))^{d})=0$

by the invariance equation (4.11), and

$G_{1}(x):=\frac{\partial G}{\partial
y}(x,0)=1-2d(x+\varphi^{s}_{d}(x))^{d-1}-(\varphi^{s}_{d})^{\prime}(F(x,0))(1-d(x+\varphi^{s}_{d}(x))^{d-1})=1-2dx^{d-1}+\dots$

so that

$G(x,y)=G_{1}(x)y+G_{2}(x,y)\qquad\text{with}\quad
G_{2}(x,y)=\mathcal{O}(y^{2}).$

There exists $\rho_{4}\in(0,\rho_{3}]$ such that

$G_{1}(x)>1-\nu x^{d-1}\qquad\text{and}\qquad|G_{2}(x,y)|<M|y|^{2},\quad
x\in(0,\rho_{4}),\quad(x,y)\in\widehat{\Omega}^{-}_{0},$

for some $\nu>2d$ and $M>0$. Then, taking an iterate $(x_{k},y_{k})$ such that
$x_{k}<\rho_{4}$ and relabeling it by $(x_{0},y_{0})$ and, starting again the
iteration, we have

$\displaystyle y_{k+1}=G(x_{k},y_{k})\leq$ $\displaystyle(1-\nu
x_{k}^{d-1})y_{k}+My_{k}^{2}$ (4.16) $\displaystyle<$ $\displaystyle(1-\nu
x_{k}^{d-1}-M\varphi^{s}_{d}(x_{k}))y_{k}\leq(1-bx_{k}^{d-1})y_{k},$

where $b=\nu+3\rho_{4}M$. Iterating (4.16) we obtain

$\displaystyle y_{k}$
$\displaystyle<\prod_{j=0}^{k-1}(1-bx_{j}^{d-1})y_{0}=y_{0}\exp\sum_{j=0}^{k-1}\log(1-bx_{j}^{d-1}).$

The series $\sum\log(1-bx_{j}^{d-1})$ is convergent since $bx_{j}^{d-1}$ tends
to zero and $\log(1+x)>(2\log 2)x$ if $x\in(-1/2,0)$. Then,
$y_{k}<y_{0}\exp(S_{0})$ where
$S_{0}=\sum_{j=0}^{\infty}\log(1-bx_{j}^{d-1})$. This means that $y_{k}$ is
less than some negative number so that $y_{k}$ cannot converge to 0 and
therefore $(x_{0},y_{0})\notin\mathcal{A}_{d}(0)$.

If $(x_{0},y_{0})\in\Omega_{0}^{-}$, assume that all its iterates stay in
$\mathcal{T}_{R_{d}}$. Then the sequences $\\{x_{k}\\}$ and $\\{y_{k}\\}$ are
decreasing and there exists $m\geq 0$ such that $x_{m}<\rho_{4}$ and, by the
previous estimates, $(x_{m},y_{m})\notin\mathcal{A}_{d}(0)$. ∎

###### Proof of Theorem A(a).

Let $\Omega_{r}$ be the closure of the bounded domain whose boundary is the
simple closed curve formed by the concatenation of $\Gamma$ and
$J_{r}:=\\{(x,y)\mid\ x=y,\,0<x<p\\}$ (the meaning of $r$ is right, in
contrast of the later notation $J_{\ell}$ for left). See Figure 5. The domain
$\Omega_{r}$ is invariant by $T_{d}$ since the iterates cannot jump across the
boundary. Moreover, there exists $m\geq 1$ such that
$T^{m}_{d}(\Omega_{r})\subset\Omega^{+}_{0}$. Then
$\Omega_{r}\subset\mathcal{A}_{d}(0)$. By Lemma 4.3, to obtain
$\mathcal{A}_{d}(0)$ we only need to take one preimage of $\Omega_{r}$ by
$T_{d}$.

In the light of Lemma 2.3(a) we write

$\Gamma_{\pm}:=T^{-1}_{\pm,d}\left(\Gamma\right),\qquad\Omega_{\pm}:=T^{-1}_{\pm,d}\left(\Omega_{r}\right).$

Clearly, the sets $\Omega_{\pm}$ are contained in $\mathcal{A}_{d}(0)$. The
boundaries of $\Omega_{\pm}$ are the images of the boundaries of $\Omega_{r}$
by $T_{\pm,d}^{-1}$. Consequently, we have

$\partial\Omega_{+}=\Gamma_{+}\cup
J_{\ell}\qquad\text{and}\qquad\partial\Omega_{-}=\Gamma_{-}\cup J_{\ell},$

where $\Gamma_{+}:=T_{+,d}^{-1}(\Gamma)$ is a curve contained in
$\\{y\geq-x\\}$ which joints $(0,0)$ with $(-p,p)$,
$\Gamma_{-}:=T_{-,d}^{-1}(\Gamma)$ is a curve contained in $\\{y\leq-x\\}$
which joints $(0,0)$ with $(-p,p)$, and
$J_{\ell}:=T_{+,d}^{-1}(J_{r})=\\{(x,y)\mid\ y=-x,\,-p<x<0\\}$. Notice that
every point in $\Gamma\setminus\\{(0,0)\cup(p,p)\\}$ has two preimages while

$T^{-1}_{+,d}\left(0,0\right)=T^{-1}_{-,d}\left(0,0\right)=(0,0)\qquad\mbox{and}\qquad
T^{-1}_{+,d}\left(p,p\right)=T^{-1}_{-,d}\left(p,p\right)=\left(-p,p\right).$

See Figure 5. Accordingly, the curves $\Gamma_{\pm}$ joint the points
$\left(0,0\right)$ and $\left(-p,p\right)$, they are mapped bijectively onto
$\Gamma$ by $T_{d}$ and determine the boundary of the basin of attraction of
the origin. That is,

$\mathcal{A}_{d}(0)=\Omega_{+}\cup\Omega_{-}\cup\Gamma_{+}\cup\Gamma_{-}.$

In Figure 5 we draw $\Gamma_{-}$ in green and $\Gamma_{+}$ in blue. This
finishes the proof of Theorem A(a). ∎

We can add some extra information about the geometry of $W_{d}^{s}(0)$. See
Figure 5. On the one hand, direct computations from (2.2) imply that if
$u=\left(u_{1},u_{2}\right)$ is the tangent vector of $W_{d}^{s}(0)$ at the
point $(p,p)$ then

$DT^{-1}_{\pm,d}(p,p)(u)=\left(\begin{array}[]{c}\infty\\\
2u_{1}-u_{2}\end{array}\right)\approx\left(\begin{array}[]{l}1\\\
0\end{array}\right)\quad\mbox{since}\quad
DT^{-1}_{\pm,d}(p,p)=\left(\begin{array}[]{ll}\infty&\infty\\\
2&-1\end{array}\right),$

where here $\infty$ has to be understood as a limit. Concretely, the tangent
vector of $W_{d}^{s}(0)$ at the point $(-p,p)$ is horizontal.

Moreover, taking into account the symmetry (4.5), the points with highest
value of $y$ in $W_{d}^{s}(0)$ should be symmetric. Actually, they coincide
with the two points $q^{\pm}=\left(q_{x}^{\pm},q_{y}^{\pm}\right)$ which are
mapped by $T_{d}$ to a point $q=\left(q_{x},q_{y}\right)\in\Gamma$ whose
tangent vector has slope $m=2$.

##  5\. Proof of Theorem A(${\rm b}$): The case $d$ odd and $a=1$

For the whole section we assume that $d\geq 3$ is odd and $a=1$. The proof of
Theorem A(b) is quite long and therefore we will split it into several lemmas
and propositions. Roughly speaking the strategy is as follows. First we will
see that $\mathcal{A}_{d}(0)$ is open, simply connected and that
$[-1/2,0]\times\\{0\\}\subset\mathcal{A}_{d}(0)$ (Proposition 5.1). Second we
will show that there exists a hyperbolic two-cycle of saddle type whose
unstable manifold intersects $[-1/2,0]\times\\{0\\}$. From this we will show
that the two-cycle as well as its stable manifold belong to
$\partial\mathcal{A}_{d}(0)$ (Proposition 5.6). And finally we will see that
$\partial\mathcal{A}_{d}(0)$ is unbounded (Proposition 5.7).

Let $b\in(0,1/2]$. We denote by $Q_{b}\subset\mathbb{R}^{2}$ be the compact
convex polygon bounded by the straight segments
$A_{b}:=\\{(2b^{d},y)\in\mathbb{R}^{2}\mid\ y\in[0,2b^{d}]\\}$,
$B_{b}:=\\{(x,2b^{d})\in\mathbb{R}^{2}\mid\ x\in[0,2b^{d}]\\}$,
$C_{b}:=\\{(x,2x+2b^{d})\in\mathbb{R}^{2}\mid\ x\in[-b^{d},0]\\}$,
$D_{b}:=\\{(-b^{d},y)\in\mathbb{R}^{2}\mid\ y\in[-b^{d},0]\\}$,
$E_{b}:=\\{(x,-b^{d})\in\mathbb{R}^{2}\mid\ x\in[-b^{d},0]$},
$F_{b}:=\\{(x,\frac{1}{2}x-b^{d})\in\mathbb{R}^{2}\mid\ x\in[0,2b^{d}]\\}$. We
denote $Q^{\star}:=Q_{1/2}$.

######  Proposition 5.1.

We have that $Q^{\star}\subset\mathcal{A}_{d}(0)$. In particular,
$[-1/2,0]\times\\{0\\}\subset\mathcal{A}_{d}(0)$. Moreover,
$\mathcal{A}_{d}(0)$ is open and simply connected.

###### Proof.

From Lemma 3.1 the origin is asymptotically stable and therefore
$\mathcal{A}_{d}(0)$ is open (see also Figure 4). The family
$\\{Q_{b}\\}_{b\in(0,1/2]}$ is a neighbourhood basis of the origin.

We claim that $T_{d}(Q_{b})\subset{\mathrm{int}\,}(Q_{b}),\ b\in(0,1/2]$. See
Figure 6 for a sketch of $Q_{b}$ and its image.

Assume the claim is true. This implies that $Q^{\star}\subset\mathcal{A}(0)$.
Since the image of $[-1/2,0]\times\\{0\\}$ by $T_{d}$ is the segment
$\\{(x,2x)\in\mathbb{R}^{2}\mid\ x\in[0,(1/2)^{d}]\\}\subset Q^{\star}$ we
conclude that $[-1/2,0]\times\\{0\\}\subset\mathcal{A}_{d}(0)$ as desired.
Moreover, since there exists an open simply connected neighborhood $Q^{\star}$
containing $(0,0)$ and contained in $\mathcal{A}_{d}(0)$, the origin is
asymptotically stable (our proof demonstrates again that the origin is
asymptotically stable). Finally, $\mathcal{A}_{d}(0)=\bigcup_{k\geq
0}T_{d}^{-k}(Q^{\star})$. Since $T_{d}$ is one to one, $T_{d}^{-k}(Q^{\star})$
is also open and simply connected, for all $k$. Furthermore, since
$T_{d}^{-k-1}(Q^{\star})\supset T_{d}^{-k}(Q^{\star})$, we conclude that
$\mathcal{A}_{d}(0)$ is open and simply connected as well.

The rest of the proof is devoted to prove the claim. Hereafter we remove from
the notation the dependence of the whole construction with respect to the
parameter $b$, unless strictly necessary. The proof consists in studying the
image of each side of the boundary of $Q$ by $T_{d}$. We will get that the
image of the boundary of $Q$ is contained in $\mathrm{int}\,(Q)$ and therefore
$T_{d}(Q)\subset\mathrm{int}\,(Q)$.

We denote by $\Gamma_{A}$ the image of the segment $A$ under $T_{d}$ and
similarly for the other pieces of the boundary. Next, we prove that each image
is contained in $\mathrm{int}\,(Q)$. See Figure 6.

The image $\Gamma_{A}=T_{d}(A)$. We parametrize $\Gamma_{A}$ as follows

$\Gamma_{A}=T_{d}(A)=\left\\{\left(\Psi_{1}(y):=y-(2b^{d}+y)^{d},\
\Psi_{2}(y):=y-2(2b^{d}+y)^{d}\right),\,y\in[0,2b^{d}]\right\\}.$

We check that $\Gamma_{A}\subset\mathrm{int}\,Q\cap\\{y<x\\}$. The condition
$y<x$ is equivalent to $\Psi_{2}(y)<\Psi_{1}(y)$ for $y\in[0,2b^{d}]$ which is
clearly true. The condition $y>\frac{1}{2}x-b^{d}$ is equivalent to
$\Psi_{2}(y)>\frac{1}{2}\Psi_{1}(y)-b^{d}$ for $y\in[0,2b^{d}]$ which can be
written as $\chi_{1}(y):=\frac{1}{2}y+b^{d}-\frac{3}{2}(2b^{d}+y)^{d}>0$ for
$y\in[0,2b^{d}]$. We have that
$\chi_{1}(0)=b^{d}(1-\frac{3}{2}2^{d}b^{d^{2}-d})>0$ and
$\chi_{1}(2b^{d})=2b^{d}(1-\frac{3}{4}4^{d}b^{d^{2}-d})>0$ since $b\in(0,1/2]$
and $d\geq 3$. Also
$\chi_{1}^{\prime\prime}(y)=-d(d-1)\frac{3}{2}(2b^{d}+y)^{d-2}<0$, therefore
$\chi_{1}(y)>0$. Finally, $\Psi_{1}(y)<y\leq 2b^{d}$ and
$\Psi_{2}(y)\geq-b^{d}$. The first claim is immediate. For the second we
consider the auxiliary function $\chi_{2}(y):=y+b^{d}-2(2b^{d}+y)^{d}$. We
have $\chi_{2}(0)=b^{d}(1-2^{d+1}b^{d^{2}-d})>0$ and
$\chi_{2}(2b^{d})=3b^{d}(1-\frac{1}{3}2^{2d+1}b^{d^{2}-d})>0$ since $d\geq 3$.
Moreover, $\chi_{2}^{\prime\prime}(y)=-2d(d-1)(2b^{d}+y)^{d-2}<0$ and hence
$\chi_{2}(y)>0$.

$\hat{a}=(2b^{d},0)$$\bullet$$Q_{b}$$F$$\Gamma_{A}$$A$$\Gamma_{B}$$B$$\Gamma_{C}$$C$$\Gamma_{D}$$D$$\Gamma_{E}$$E$$\Gamma_{F}$$\hat{b}=(2b^{d},2b^{d})$$\bullet$$\hat{c}=(0,2b^{d})$$\bullet$$\hat{d}=(-b^{d},0)$$\bullet$$\hat{e}=(-b^{d},-b^{d})$$\bullet$$\hat{f}=(0,-b^{d})$$\bullet$$y=x$$y=2x$$T_{d}(\hat{a})$$T_{d}(\hat{b})$$T_{d}(\hat{c})$$T_{d}(\hat{d})$$T_{d}(\hat{e})$$T_{d}(\hat{f})$

Figure 6. Sketch of the closed set $Q_{b}$ and its image $T_{d}(Q_{b})$ for
$d\geq 5$.

The image $\Gamma_{B}=T_{d}(B)$. We parametrize $\Gamma_{B}$ by $x$:

$\Gamma_{B}=T_{d}(B)=\left\\{\left(\Psi_{1}(x):=2b^{d}-(2b^{d}+x)^{d},\
\Psi_{2}(x):=2b^{d}-2(2b^{d}+x)^{d}\right),\,x\in[0,2b^{d}]\right\\}.$

It is immediate to see that

$\Psi_{1}^{\prime}(x)=-d(2b^{d}+x)^{d-1}<0\qquad\text{and
}\qquad\Psi_{2}^{\prime}(x)=-2d(2b^{d}+x)^{d-1}<0.$

Therefore,

$\Gamma_{B}\subset[\Psi_{1}(2b^{d}),\Psi_{1}(0)]\times[\Psi_{2}(2b^{d}),\Psi_{2}(0)]\subset(0,2b^{d})\times[0,2b^{d})\subset\mathrm{int}\,(Q).$

The image $\Gamma_{C}=T_{d}(C)$. We parametrize $\Gamma_{C}$ by $x$:

$\Gamma_{C}=\left\\{\left(\Psi_{1}(x):=2x+2b^{d}-(3x+2b^{d})^{d},\
\Psi_{2}(x):=2x+2b^{d}-2(3x+2b^{d})^{d}\right),\,x\in[-b^{d},0]\right\\}.$

Similarly as before

$\Psi_{1}^{\prime}(x)=2-3d(3x+2b^{d})^{d-1}>2-3d2^{d-1}b^{d^{2}-d}>0\\\ $

and

$\Psi_{2}^{\prime}(x)=2-6d(3x+2b^{d})^{d-1}>2-6d2^{d-1}b^{d^{2}-d}>0.$

Then,

$\Gamma_{C}\subset[\Psi_{1}(-b^{d}),\Psi_{1}(0)]\times[\Psi_{2}(-b^{d}),\Psi_{2}(0)]\subset(0,2b^{d})\times(0,2b^{d})\subset\mathrm{int}\,(Q).$

The image $\Gamma_{D}=T_{d}(D)$. We parametrize $\Gamma_{D}$ by $y$: Clearly,

$\Gamma_{D}=\left\\{\left(\Psi_{1}(y):=y-(-b^{d}+y)^{d},\
\Psi_{2}(y):=y-2(-b^{d}+y)^{d}\right),\,y\in[-b^{d},0]\right\\}.$

First, we check that $\Gamma_{D}\subset[-b^{d},2b^{d}]\times[-b^{d},2b^{d}]$.
Indeed,

$\Psi^{\prime}_{1}(y)=1-d(-b^{d}+y)^{d-1}>0,\qquad\text{and}\qquad\Psi^{\prime}_{2}(y)=1-2d(-b^{d}+y)^{d-1}>0.$

Then

$\displaystyle-b^{d}<\Psi_{1}(-b^{d})\leq\Psi_{1}(y)\leq\Psi_{1}(0)<2b^{d},$
$\displaystyle-b^{d}<\Psi_{2}(-b^{d})\leq\Psi_{2}(y)\leq\Psi_{2}(0)<2b^{d}.$

The condition $\psi_{2}(y)<2\psi_{1}(y)+2b^{d}$ reads
$y-2(-b^{d}+y)^{d}<2(y-(-b^{d}+y)^{d})+2b^{d}$ which is satisfied if
$y+2b^{d}>0$ which is the case.

The condition $\psi_{2}(y)>\frac{1}{2}\psi_{1}(y)-b^{d}$ reads
$y-2(-b^{d}+y)^{d}>\frac{1}{2}(y-(-b^{d}+y)^{d})-b^{d}$ which is satisfied if
$\chi_{3}(y):=\frac{1}{2}y-\frac{3}{2}(-b^{d}+y)^{d}+b^{d}>0$. This is indeed
true because $\chi_{3}(-b^{d})=\frac{1}{2}b^{d}+\frac{3}{2}(2b^{d})^{d}>0$ and
$\chi_{3}^{\prime}(y)=\frac{1}{2}-\frac{3}{2}d(-b^{d}+y)^{d-1}>0$ (since
$d\geq 3$ and $b\in(0,1/2]$).

The image $\Gamma_{E}=T_{d}(E)$. We parametrize $\Gamma_{E}$ by $x$:

$\Gamma_{E}=\left\\{\left(\Psi_{1}(x):=-b^{d}-(-b^{d}+x)^{d},\
\Psi_{2}(x):=-b^{d}-2(-b^{d}+x)^{d}\right),\,x\in[-b^{d},0]\right\\}.$

In this case we will check that
$\Gamma_{E}\subset(-b^{d},0)\times(-b^{d},0)\subset\mathrm{int}\,(Q)$. Indeed,
directly from the expression of the parametrization we have,
$\Psi_{1}(x)>-b^{d}$, $\Psi_{2}(x)>-b^{d}$,
$\Psi_{1}(x)<-b^{d}-(-2b^{d})^{d}=-b^{d}(1-2^{d}b^{d^{2}-d})<0$ and
$\Psi_{2}(x)<-b^{d}(1-2^{d+1}b^{d^{2}-d})<0$.

The image $\Gamma_{F}=T_{d}(F)$. We parametrize $\Gamma_{F}$ by $x$:

$\Gamma_{F}=\left\\{\left(\Psi_{1}(x):=\frac{1}{2}x-b^{d}-\left(\frac{3}{2}x-b^{d}\right)^{d},\
\Psi_{2}(x):=\frac{1}{2}x-b^{d}-2\left(\frac{3}{2}x-b^{d}\right)^{d}\right),\,x\in[0,2b^{d}]\right\\}.$

In this case we will also check that
$\Gamma_{F}\subset(-b^{d},0)\times(-b^{d},0)\subset\mathrm{int}\,(Q)$. We have
that

$\Psi_{1}(0)=-b^{d}-(-b^{d})^{d}>-b^{d},\qquad\Psi_{1}(2b^{d})=-(2b^{d})^{d}<0.$

Moreover, since
$\Psi^{\prime}_{1}(x)=\frac{1}{2}-\frac{3}{2}d\left(\frac{3}{2}x-b^{d}\right)^{d-1}$
which is positive because
$\frac{3}{2}d\left(\frac{3}{2}x-b^{d}\right)^{d-1}\leq\frac{3}{2}d2^{d-1}b^{d(d-1)}\leq\frac{3}{2}d2^{-d^{2}+2d-1)}\leq
9/32<1/2$ we get $\Psi_{1}(x)\in(-b^{d},0)$.

Concerning $\Psi_{2}$, $\Psi_{2}(0)=-b^{d}+2\left(b^{d}\right)^{d}>-b^{d}$ and
$\Psi_{2}(2b^{d})=-2(2b^{d})^{d}<0$. However, it is not (always) monotone.
Depending on $b$ and $d$ it may have a maximum at some $x_{c}\in(0,b^{d})$.
The value of $x_{c}$ is obtained from the condition
$\Psi^{\prime}_{2}(x_{c})=0$. It is the positive solution of

$\left(\frac{3}{2}x_{c}-b^{d}\right)^{d-1}=\frac{1}{6d}.$

In case $x_{c}$ belongs to the interval $(0,2b^{d})$ we have that
$\Psi_{2}(x_{c})=\frac{1}{2}x_{c}-b^{d}-\left(\frac{3}{2}x_{c}-b^{d}\right)^{d}=\frac{1}{2}x_{c}-b^{d}-\left(\frac{1}{6d}\right)^{d/(d-1)}<-\left(\frac{1}{6d}\right)^{d/(d-1)}<0$.
Thus, $\Psi_{2}(x)<0$ for $x\in[0,2b^{d}]$. This finishes the proof. ∎

Following the steps of the strategy of the proof of Theorem A(b) described at
the beginning of the section. We start checking that
$\\{p_{0}=(0,1),p_{1}=(0,-1)\\}$ forms a hyperbolic two-cycle.

Since $DT_{d}(p_{0})=DT_{d}(p_{1})$, the chain rule implies that

(5.1)
$DT_{d}^{2}(p_{0})=DT_{d}^{2}(p_{1})=DT_{d}(p_{0})DT_{d}(p_{1})=\left(\begin{array}[]{ll}3d^{2}-2d&\
\ 3d^{2}-4d+1\\\ 6d^{2}-2d&\ \ 6d^{2}-6d+1\end{array}\right).$

A direct computation shows that the eigenvalues and eigenvectors of
$DT_{d}^{2}(p_{j}),\ j=0,1$, are given by

$\lambda^{\pm}_{d}=\frac{1}{2}\left(9d^{2}-8d+1\pm(3d-1)\sqrt{9d^{2}-10d+1}\right)$

and

(5.2) $(1,m^{\pm}_{d})=\left(1,\ \frac{4d}{1-d\pm\sqrt{9d^{2}-10d+1}}\right),$

respectively. Finally, it is straightforward to check that both eigenvalues
are strictly positive. Moreover, $\lambda^{-}_{d}$ is strictly decreasing and
$\lambda^{+}_{d}$ is strictly increasing, with respect to the parameter $d$.
We also have

$\lim\limits_{d\to\infty}\lambda^{-}_{d}=1/9$ | and | $1/9<\lambda^{-}_{d}\leq\lambda^{-}_{3}=29-8\sqrt{13}\approx 0.1556$
---|---|---
$\lim\limits_{d\to\infty}\lambda^{+}_{d}=\infty$ | and | $\lambda^{+}_{d}\geq\lambda^{+}_{3}=29+8\sqrt{13}\approx 57.8444.$

On the other hand, $m^{-}_{d}$ is negative and strictly increasing while
$m^{+}_{d}$ is positive and strictly decreasing (both with respect to the
parameter $d$). We also have

$\lim\limits_{d\to\infty}m^{-}_{d}=-1$ | and | $-1.3028\approx\frac{-6}{1+\sqrt{13}}=m^{-}_{3}\leq m^{-}_{d}<-1$
---|---|---
$\lim\limits_{d\to\infty}m^{+}_{d}=2$ | and | $2<m^{+}_{d}\leq m^{+}_{3}=\frac{6}{\sqrt{13}-1}\approx 2.3028$

Therefore, the two cycle $\\{p_{0},p_{1}\\}$ is a hyperbolic saddle point. In
what follows we will denote by $W^{s}:=W^{s}_{\\{p_{0},p_{1}\\}}$ and
$W^{u}:=W^{u}_{\\{p_{0},p_{1}\\}}$ the (global) stable and unstable manifolds
of the periodic orbit $\\{p_{0},p_{1}\\}$, respectively. We split
$W^{u}=W^{u}_{p_{0}}\cup W^{u}_{p_{1}}$ where $W^{u}_{p_{j}}$ is the (global)
unstable manifold of the fixed point $p_{j}$ for the map $T_{d}^{2}$, $j=1,2$.
Similarly, $W^{s}=W^{s}_{p_{0}}\cup W^{s}_{p_{1}}$ for the stable manifold.
Consequently, we remark that $W^{s}$ and $W^{u}$ refer to the manifolds
associated to the hyperbolic periodic orbit $\\{p_{0},p_{1}\\}$, and hence
they are not the manifolds associated to the origin (with a similar notation)
studied and considered in Sections 3 and 4.

To simplify the notation, unless strictly necessary, we drop the dependence of
$\lambda_{d}^{\pm}$ and $m_{d}^{\pm}$ with respect to the parameter $d$. Thus,
we will write

$\lambda^{\pm}:=\lambda^{\pm}_{d}\quad\quad\mbox{and}\quad\quad
m^{\pm}:=m^{\pm}_{d}.$

We introduce $m^{\star}=\frac{7}{2}$.

Next lemma gives a precise description of the geometry of $W^{u}_{p_{0}}$ that
we will use to prove that $\\{p_{0},p_{1}\\}\subset\partial\mathcal{A}_{d}(0)$
and finally to prove that $W^{s}\subset\partial\mathcal{A}_{d}(0)$.

######  Lemma 5.2.

Let $\mathcal{D}$ be the closed triangle determined by the vertices

$p_{1}=(0,-1),\qquad\left(\frac{1}{m^{+}+1},\frac{-1}{m^{+}+1}\right)\qquad\text{and}\qquad\left(\frac{1}{m^{\star}+1},\frac{-1}{m^{\star}+1}\right).$

Then, there is a local piece of $W^{u}_{p_{0}}$ (attached to $p_{0}$) tangent
to the line $y=1+m^{+}x$ contained in $T_{d}(\mathcal{D})$. Moreover, if we
parametrize $W^{u}_{p_{0}}\cap\\{y\leq 1\\}$ as
$W^{u}_{p_{0}}:=\\{\varphi(t)\mid\ t\geq 0\\}$, with $\varphi(0)=(0,1)$ and
$\varphi(t)\subset\mathrm{int}\,(T_{d}(\mathcal{D}))$ for $t\in(0,t_{0})$ and
$\varphi(t_{0})\in\partial T_{d}(\mathcal{D})$ then

$\varphi(t_{0})\subset\partial
T_{d}(\mathcal{D})\cap\\{y=x\\}=\left\\{(s,s)\mid\ \frac{1}{m^{+}+1}\leq
s\leq\frac{1}{m^{\star}+1}\right\\}.$

See Figure 7 (right).

###### Proof.

The triangle $\mathcal{D}$ can also be represented as

$\mathcal{D}=\\{(t,-1+mt)\,|\ t\in[0,1/(m+1)],\ m\in[m^{+},m^{\star}]\\}.$

The proof of this lemma will follow from an accurate description of the sets
$T_{d}(\mathcal{D})$ and $T_{d}^{-1}(\mathcal{D})$, their relative position
and geometry in the plane, and the behaviour of the map
$T_{d}^{-2}:T_{d}(\mathcal{D})\to T_{d}^{-1}(\mathcal{D})$. The shape of
$T_{d}(\mathcal{D})$. We consider the decomposition of $\mathcal{D}$ into the
segments

(5.3) $\ell_{m}=\\{(t,-1+mt),\ t\in[0,1/(m+1)]\\},\qquad\text{with}\qquad
m\in[m^{+},m^{\star}].$

If we write
$\gamma(t):=\gamma_{m}(t)=T_{d}\left(\ell_{m}\right):=(x_{m}(t),y_{m}(t))=:\left(x(t),y(t)\right)$
we have

(5.4) $x(t)=mt-1-((m+1)t-1)^{d}\qquad\text{and}\qquad
y(t)=mt-1-2((m+1)t-1)^{d}.$

Thus, the first derivatives of $x(t)$ and $y(t)$ are given by

$x^{\prime}(t)=m-d(m+1)((m+1)t-1)^{d-1}\quad\text{and}\quad
y^{\prime}(t)=m-2d(m+1)((m+1)t-1)^{d-1}.$

Easy computations show that $x^{\prime}(t)$ and $y^{\prime}(t)$ vanish at the
points

$r^{\pm}=\frac{1}{m+1}\left[1\pm\left({\frac{m}{d(m+1)}}\right)^{1/(d-1)}\right]\quad\text{and}\quad
s^{\pm}=\frac{1}{m+1}\left[1\pm\left({\frac{m}{2d(m+1)}}\right)^{1/(d-1)}\right],$

respectively. Moreover,

$0<r^{-}<s^{-}<\frac{1}{m+1}<s^{+}<r^{+}\qquad\text{and}\qquad
x\left(\frac{1}{m+1}\right)=y\left(\frac{1}{m+1}\right)=\frac{-1}{m+1},$

where $t=1/(m+1)$ corresponds to the common maximum of $x^{\prime}(t)$ and
$y^{\prime}(t)$. See Figure 7 (left). In summary, the components $x(t)$ and
$y(t)$ of the curve $\gamma(t)$ are polynomial functions in $t$, having a
unique minimum in the interval $[0,1/(m+1)]$ located at $t=r^{-}$ and
$t=s^{-}$, respectively, and sharing the same negative value, $-1/(m+1)$, at
$t=1/(m+1)$. See the middle picture in Figure 7.

$(t,x^{\prime}(t))$$(t,y^{\prime}(t))$$r^{-}$$s^{-}$$\frac{1}{m+1}$$\frac{-1}{m+1}$$1$$m$$\frac{1}{m+1}$$r^{-}$$s^{-}$$(t,y(t))$$(t,x(t))$$T_{d}(\mathcal{D})$$\gamma_{m^{+}}(r^{-})$$\gamma_{m^{+}}(s^{+})$$(0,1)$$y=x$$y=-x$${W}^{u}_{p_{0}}$$\mathcal{D}$$(0,-1)$$\bullet$$\bullet$

Figure 7. Left figure shows the graphs of the curves $x^{\prime}(t)$ and
$y^{\prime}(t)$. The center picture shows the graphs of the curves $x(t)$ and
$y(t)$. Finally, the right picture displays a sketch of the triangle
$\mathcal{D}$ and its image $T_{d}(\mathcal{D})$.

To conclude the description of the shape of $\gamma$, see Figure 7 (right).
Let us prove that its image can be represented as the union of two graphs with
respect to the variable $x$ (i.e., it admits a piecewise parametrization of
graphs with respect to $x$). We write

$\gamma(x)=(x,\gamma^{(2)}(x)):=\left(x,y\left(t(x)\right)\right),$

where $t(x):=x^{-1}(t)$ is one of the two branches of the inverse of $x(t)$.
Some direct computations show that

(5.5)
$\begin{split}\frac{d\gamma^{(2)}}{dx}&=\frac{dy}{dt}\left(\frac{dx}{dt}\right)^{-1},\\\
\frac{d^{2}\gamma^{(2)}}{dx^{2}}&=\left(\frac{dx}{dt}\right)^{-3}\left(\frac{d^{2}y}{dt^{2}}\frac{dx}{dt}-\frac{dy}{dt}\frac{d^{2}x}{dt^{2}}\right)=\left(\frac{dx}{dt}\right)^{-3}\left(d(d-1)m(m+1)^{2}\left((m+1)t-1\right)\right)^{d-2},\end{split}$

everything evaluated at the corresponding branch of $t=t(x)$. We denote
$\gamma^{(2)}_{\rm u}:=\gamma^{(2)}_{{\rm u},m}$ and
$\gamma^{(2)}_{\ell}:=\gamma^{(2)}_{{\ell},m}$ the functions corresponding to
the upper (concave) and lower (convex) graphs, respectively.

From the previous discussion, $dx/dt$ has a unique zero at $t=r^{-}$ and it is
monotone in the whole interval $[0,1/(m+1)]$, see Figure 7 (left). For the
upper branch, corresponding to $0\leq t<r^{-}$, we have $dx/dt<0$ and
therefore $\gamma^{(2)}_{\rm u}(x)$ is increasing and concave (see (5.5)) and
for the lower branch, $r^{-}<t\leq 1/(m+1)$, we have $dx/dt>0$ and therefore
$\gamma^{(2)}_{\ell}(x)$ is convex having a minimum at $x(s^{-})$ (see again
(5.5)). See Figure 7 (right), we have drawn (qualitatively) the curve $\gamma$
for the values $m=m^{+}$ and $m=m^{\star}$. We remark that $\gamma^{(2)}_{\rm
u,m^{+}}(x)$ is tangent at $p_{0}$ to the line $y=m^{+}x+1$ since it is the
image of the side of $\mathcal{D}$ tangent to $W^{u}_{p_{1}}$. Since $T_{d}$
sends the line $\\{y=-x\\}$ to $\\{y=x\\}$ the images of all the curves
$\gamma_{m}$ end up at $\\{y=x\\}$. All together determines the shape of
$T_{d}(\mathcal{D})$. Moreover, in the light of the above arguments we have
that

$\partial T_{d}(\mathcal{D})=\gamma^{(2)}_{{\rm
u},m^{+}}\cup\gamma^{(2)}_{{\ell},m^{+}}\cup\gamma^{(2)}_{{\rm
u},m^{\star}}\cup\gamma^{(2)}_{{\ell},m^{\star}}\cup\\{(x,x)\mid\
\frac{-1}{m^{+}+1}<x<\frac{-1}{m^{\star}+1}\\}.$

Hereafter we will refer to

$\gamma^{(2)}_{{\rm u},m^{+}}\cup\gamma^{(2)}_{{\rm
l},m^{+}}\qquad\text{and}\qquad\gamma^{(2)}_{{\rm
u},m^{\star}}\cup\gamma^{(2)}_{{\rm l},m^{\star}}$

as the left and right the boundaries of $T_{d}(\mathcal{D})$, respectively.
See Figure 7 (right) and Figure 8.

The shape of $T_{d}^{-1}(\mathcal{D})$. We consider the same decomposition of
$\mathcal{D}$ into the segments $\ell_{m}$ as in (5.3). We denote
$\Gamma(t):=\Gamma_{m}(t)=T_{d}^{-1}\left(\ell_{m}\right):=(\alpha_{m}(t),\beta_{m}(t))=:(\alpha(t),\beta(t))$.
We have

(5.6) $\alpha(t)=(m-2)t-1+((1-m)t+1)^{1/d},\quad\beta(t)=(2-m)t+1,\quad
t\in[0,1/(m+1)].$

Therefore, the first and second derivatives of $\alpha(t)$ and $\beta(t)$ are
given by

$\begin{split}&\alpha^{\prime}(t)=m-2+\frac{1-m}{d}((1-m)t+1)^{(1-d)/d},\ \ \
\qquad\beta^{\prime}(t)=2-m,\\\
&\alpha^{\prime\prime}(t)=\frac{1-d}{d^{2}}(1-m)^{2}((1-m)t+1)^{(1-2d)/d}<0,\quad\beta^{\prime\prime}(t)=0.\end{split}$

Clearly $\beta^{\prime}(t)<0$ since $m\geq m^{+}>2$. Next, we focus the
attention on $\alpha^{\prime}(t)$. The line $t=1/(m-1)$ is a vertical
asymptote (outside the domain $(0,1/(m+1))$) and simple computations show that
$\alpha^{\prime}(t)=0$ if and only if $t=t^{\pm}$ where

$t^{\pm}:=t^{\pm}_{m}=\frac{1}{m-1}\pm\left(\frac{m-1}{d^{d}(m-2)^{d}}\right)^{1/(d-1)}.$

Some further computations show that

(5.7) $t^{-}_{m^{+}}<0,\qquad
t^{-}_{m^{\star}}>\frac{1}{m^{\star}+1}\qquad\text{and}\qquad
t_{m}^{+}>\frac{1}{m-1},\qquad\forall m\in[m^{+},m^{\star}].$

Since
$\alpha^{\prime}_{m^{+}}(0)=4(d-1)\left(1-d+\sqrt{1-10d+9d^{2}}\right)^{-1}+1/d-2<0$
and $\alpha^{\prime}_{m^{\star}}(0)>\frac{1}{2}(3-5/d)>0$, it follows from the
previous arguments and (5.7) that $\alpha_{m^{+}}(t)$ is monotonically
decreasing and $\alpha_{m\star}(t)$ is monotonically increasing in the
considered domain. Consequently, $\Gamma_{m^{+}}$ and $\Gamma_{m^{\star}}$ can
be expressed as graphs of monotone functions of the form

$\Gamma(x)=(x,\Gamma^{(2)}(x)):=\left(x,\beta\left(t(x)\right)\right),$

where $t(x):=\alpha^{-1}(t)$ for $m=m^{+}$ and $m=m^{\star}$, respectively. We
have

(5.8)
$\begin{split}\frac{d\Gamma^{(2)}}{dx}&=\frac{d\beta}{dt}\left(\frac{d\alpha}{dt}\right)^{-1},\\\
\frac{d^{2}\Gamma^{(2)}}{dx^{2}}&=-\left(\frac{d\alpha}{dt}\right)^{-3}\frac{d\beta}{dt}\frac{d^{2}\alpha}{dt^{2}}=-\left(\frac{d\alpha}{dt}\right)^{-3}\left(\frac{d-1}{d^{2}}\frac{(m-2)(m-1)^{2}}{\left((1-m)t+1\right)^{(2d-1)/d}}\right),\end{split}$

everything evaluated at $t=t(x)$. Indeed, taking into account (5.8), when
$m=m^{+}$, $\Gamma^{(2)}$ is increasing and convex, while when $m=m^{\star}$,
$\Gamma^{(2)}$ is decreasing and concave. From (5.6) we conclude that

$\beta(t)>\beta(1/(m+1))=3/(m+1)>3/(m^{\star}+1)=2/3,$

and then the preimage $T^{-1}(\mathcal{D})$ is above the line $\\{y=2/3\\}$.
Finally we notice that the image by $T^{-1}_{d}$ of the segment
$\mathcal{D}\cap\\{y=-x\\}$ is contained in (the graph of)

$x=\phi(y):=-y+\left(\frac{2}{3}y\right)^{1/d}.$

Then

$\phi^{\prime}(y)=-1+\frac{2}{3d}\left(\frac{2}{3}y\right)^{(1-d)/d}<-1+\frac{3}{2d}<-\frac{1}{2},$

and the function $y=\phi^{-1}(x)$ is decreasing in the corresponding domain.

In particular the curve $\Gamma_{m^{+}}(x)$ belongs to the second quadrant,
while $\Gamma_{m^{\star}}(x)$ belongs to the first one. In Figure 8 we display
$T_{d}(\mathcal{D})$ and $T_{d}^{-1}(\mathcal{D})$ and we can see the relative
position of these two sets and the initial triangle $\mathcal{D}$. We
emphasize that, from arguments above, $\gamma_{m^{+}}(x)$ and
$\Gamma_{m^{+}}(x)$ are both tangent to the line $y=m^{+}x+1$, but, using the
convexity properties, $\gamma_{m^{+}}(x)$ is below this line while
$\Gamma_{m^{+}}(x)$ is above it, hence their relative position illustrated in
Figure 8 is the right one.

$T_{d}(\mathcal{D})$$T_{d}^{-1}(\mathcal{D})$$y=m^{+}x-1$$y=m^{+}x+1$$(0,1)$$y=-x$$y=x$$\mathcal{D}$$(0,-1)$

Figure 8. A qualitative representation of the triangle $\mathcal{D}$ and the
sets $T_{d}(\mathcal{D})$ and $T_{d}^{-1}(\mathcal{D})$.

Consider now $T_{d}^{-2}:T_{d}(\mathcal{D})\to T_{d}^{-1}(\mathcal{D})$. From
the stable/unstable manifold theorem and the relative position and geometry of
the sets $T_{d}(\mathcal{D})$ and $T_{d}^{-1}(\mathcal{D})$, we can conclude
that, locally, $W^{u}_{p_{0}}$ exists, it is tangent to $y=m^{+}x+1$, and it
is contained in $T_{d}(\mathcal{D})$. In particular we also conclude that
there is a local piece of $W^{u}_{p_{1}}$ attached to $p_{1}$ belonging to
$\mathcal{D}$.

We claim that ${W}^{u}_{p_{0}}$ may only leave $T_{d}(\mathcal{D})$ through
the piece of the boundary given by $\partial T_{d}(\mathcal{D})\cap\\{y=x\\}$
(later we will see that $W^{u}_{p_{0}}$ does leave $T_{d}(\mathcal{D})$
through this boundary). To check the claim we first observe that
$W^{u}_{p_{0}}\cap\\{y\leq 1\\}$ can be parametrized
$W^{u}_{p_{0}}:=\\{\varphi(t)\mid\ t\geq 0\\}$, with $\varphi(0)=(0,1)$ (see
[CFdlL03]). Second, we suppose it leaves $T_{d}(\mathcal{D})$ either for the
left or the right boundaries of $T_{d}(\mathcal{D})$, and we get a
contradiction.

Let $p=\varphi(t_{0})$ for some $t_{0}>0$ such that $\\{\varphi(t),\
t\in(0,t_{0})\\}\subset\mathrm{int}\,(T_{d}(\mathcal{D}))$ for all
$t\in(0,t_{0})$ and $p\in\partial T_{d}(\mathcal{D})\setminus\\{x=y\\}$ (that
is, it leaves $T_{d}(\mathcal{D})$ through the left or right boundaries of
it). Consider $q:=T_{d}^{-2}(p)$. Since $W^{u}_{p_{0}}$ is invariant by
$T_{d}^{-2}$ we have that $q=\varphi(t_{q})$ for some $t_{q}\in(0,t_{p})$.
However, we also have

$q\in\partial\left(T_{d}^{-1}(\mathcal{D})\right)\setminus
T_{d}^{-1}\left(\partial\mathcal{D}\cap\\{y=-x\\}\right),$

which provides a contradiction (see Figure 8). ∎

Let $\mathcal{E}$ be the closed triangle determined by the vertices
$\tau_{0}=(0,1)$, $\tau_{1}=(-1/2,0)$ and $\tau_{2}=(-1/3,0)$.

Next two lemmas refer to the set $T_{d}(\mathcal{D})\cap\\{y\geq 0\\}$. First
we show that this set belongs to the triangular region $\mathcal{E}$ and,
second, we will give relevant information of the dynamics of
$T^{-2}|_{\mathcal{E}}$ (and therefore in $T_{d}(\mathcal{D})\cap\\{y\geq
0\\}$). All together implies two main properties for $W^{u}_{p_{0}}$. On the
one hand, we will prove that
$W^{u}_{p_{0}}\cap[-1/2,0]\times\\{0\\}\neq\emptyset$ and, on the other hand,
we will prove that $W^{u}_{p_{0}}\cap\\{y=x\\}\neq\emptyset$ (such
intersection happens on the fourth quadrant). See Figure 9.

######  Lemma 5.3.

We have

$T_{d}(\mathcal{D})\cap\\{y\geq 0\\}\subset\mathcal{E}.$

###### Proof.

We will check that the left and the right boundaries of
$T_{d}(\mathcal{D})\cap\\{y\geq 0\\}$, given by pieces of the curves
$\gamma_{m^{+}}$ and $\gamma_{m^{\star}}$, respectively, are contained in
$\mathcal{E}$. By the discussion we did when analyzing the shape of
$T_{d}(\mathcal{D})$ we know that both curves can be written as graphs of
concave functions that only intersect at the point $p_{0}=(0,1)$ (see Lemma
5.2). Moreover the slope $s_{m^{+}}$ of $\gamma_{m^{+}}$ at $t=0$ is $m^{+}>2$
(the slope of the left side of $\mathcal{E}$). Indeed,

$2<s_{m^{+}}\leq 6/(\sqrt{13}-1)\lessapprox 2.3028.$

All together implies that the statement of the lemma follows from proving that
the (only) intersection of $\gamma_{m^{\star}}$ with $y=0$ happens at a point
$x<-1/3$.

Consider the second component $y(t)$ of $\gamma_{m^{\star}}$ defined for
$t\in[0,1/(m^{\star}+1)]$. Since $y(0)=1$,
$y(1/(m^{\star}+1))=-1/(m^{\star}+1)<0$ and $y^{\prime\prime}(t)>0$ there
exists a unique $t_{1}\in[0,1/(m^{\star}+1)]$ such that $y(t_{1})=0$. See
Figure 7 (center).

When $d=3$ we can localize $t_{1}$ with some precision. Let $t_{1}^{-}=1/18$
and $t_{1}^{+}=1/16$. Both values belong to $[0,r^{-}]$ where the functions
$x(t)$ and $y(t)$ are decreasing. See (5.4). We have

$y(t_{1}^{-})=-\frac{29}{36}+2\Big{(}\frac{27}{36}\Big{)}^{3}>\frac{1}{30}\qquad\text{and}\qquad
y(t_{1}^{+})=-\frac{25}{32}+2\Big{(}\frac{23}{32}\Big{)}^{3}<\frac{-1}{30}.$

This means that $t_{1}\in(1/18,1/16)$ and that, since $\gamma_{m^{\star}}$ is
the graph of a concave function,
$x(t_{1})<x(t_{1}^{-})=-\frac{29}{36}+\Big{(}\frac{27}{36}\Big{)}^{3}<-\frac{1}{3}$.
By concavity we get that $\gamma_{m^{\star}}\cap\\{y\geq
0\\}\subset\mathcal{E}$.

To deal with the general value of $d$ odd, $d\geq 3$, we will see that if we
consider the intersection point $(x(t_{1}),0)$ as a function of $d$, it is
decreasing so that for $d\geq 3$, $x(t_{1})<-\frac{1}{3}$. Indeed, we compute
the derivative of $x(t_{1})$ with respect to $d$.

We write $y=y(t,d)$. Let $t_{1}(d)$ be the parameter such that
$y(t_{1}(d),d)=0$. We want to compute $(x(t_{1}(d),d))^{\prime}$, where prime
stands for the derivative with respect to $d$.

Derivating impliticly (and simplifying notation) we have

$(t_{1}(d))^{\prime}=-\frac{\partial y}{\partial d}(t_{1}(d),d)/\frac{\partial
y}{\partial t}(t_{1}(d),d)=:-\left(\frac{\partial y}{\partial
d}/\frac{\partial y}{\partial t}\right)|_{(t_{1}(d),d)}=:-\left(\frac{\partial
y}{\partial d}/\frac{\partial y}{\partial t}\right).$

Then,

$\displaystyle(x(t_{1}(d),d))^{\prime}$ $\displaystyle=\frac{\partial
x}{\partial t}\times(t_{1}(d))^{\prime}+\frac{\partial x}{\partial d}$
$\displaystyle=\frac{\partial x}{\partial t}\times\Big{(}-\frac{\partial
y}{\partial d}/\frac{\partial y}{\partial t}\Big{)}+\frac{\partial x}{\partial
d}$ $\displaystyle=\Big{(}1/\frac{\partial y}{\partial
t}\Big{)}\times\Big{(}\frac{\partial x}{\partial d}\times\frac{\partial
y}{\partial t}-\frac{\partial x}{\partial t}\times\frac{\partial y}{\partial
d}\Big{)}.$

Taking the corresponding derivatives from equation (5.4) and simplifying we
get

$(x(t_{1}(d),d))^{\prime}=-\Big{(}1/\frac{\partial y}{\partial
t}\Big{)}m^{\star}(1-(m^{\star}+1)t)^{d}\log(1-(m^{\star}+1)t)<0.$

Indeed, we are evaluating the right side of the above equation at the point
$(t_{1}(d),d)$ with $0<t_{1}(d)<r^{-}(d)$. Thus, we have

$\left(1/\frac{\partial y}{\partial
t}\right)|_{(t_{1}(d),d)}<0\qquad\text{and}\qquad 0<1-(m^{\star}+1)t<1.$

∎

Next lemma tell us that, while the iterates by $T_{d}^{-2}$ remain in
$\mathcal{E}$, the sequence of their second coordinates of them is strictly
increasing. See Figure 9.

######  Lemma 5.4.

Let $(f(x,y),g(x,y)):=T_{d}^{-2}(x,y)$. If $(x,y)\in\mathcal{E}$ then
$g(x,y)\geq y$ and the equality only holds when $(x,y)=(0,1)$.

###### Proof.

From (2.3) we have that $g(x,y)=3y-6x+2(x-y)^{1/d}$, and then $g(x,y)\geq y$
in $\mathcal{E}$ if and only if

$G(x,y):=2y-6x+2(x-y)^{1/d}>0,\qquad\forall(x,y)\in\mathcal{E}\setminus\\{(0,1)\\}.$

To prove this inequality we will show that $G$ restricted to $\mathcal{E}$ has
a global minimum $G=0$ at $(0,0)$ which is only attained at $(0,0)$. A direct
computation shows that the partial derivatives of $G$ cannot vanish
simultaneously, therefore the minimum has to be attained at the boundary of
$\mathcal{E}$. It is clear that the restriction of the function $G$ on each of
the three segments of $\partial\mathcal{E}$ is given by

$\begin{split}\chi_{1}(x):=&G(x,0)=2\left(-3x+x^{1/d}\right),\quad\quad\quad\quad\quad\quad\quad\
x\in[-1/2,-1/3],\\\
\chi_{2}(x):=&G(x,2x+1)=2\left(1-x-(1+x)^{1/d}\right),\quad\quad
x\in[-1/2,0],\\\
\chi_{3}(x):=&G(x,3x+1)=2\left(1-(1+2x)^{1/d}\right),\quad\quad\quad\
x\in[-1/3,0].\end{split}$

Using elementary methods we can check that indeed $\chi_{1}(x)>0$,
$\chi_{2}(x)\geq 0$ and $\chi_{3}(x)\geq 0$ in the indicated intervals and
that $\chi_{2}(x)=0$, $\chi_{3}(x)=0$ only hold when $x=0$. ∎

######  Lemma 5.5.

The unstable manifold $W^{u}_{p_{0}}$ crosses the interval
$I_{0}:=T_{d}(\mathcal{D})\cap\\{y=x\\}$ at some point
$(\widehat{p},\widehat{p})$ such that
$\frac{1}{m^{+}+1}<\widehat{p}<\frac{1}{m^{\star}+1}$. Moreover, the piece of
$W^{u}_{p_{0}}$ from $(0,0)$ to $(\widehat{p},\widehat{p})$ is contained in
$T_{d}(\mathcal{D})$. We also have that this piece of $W^{u}_{p_{0}}$ cuts the
segment $(-1/2,-1/3)\times\\{0\\}\subset\mathbb{R}^{2}$.

###### Proof.

We first prove the existence of the point $(\widehat{p},\widehat{p})$. A
completely analogous procedure will be used in the proof of Proposition 5.7
and in Section 6. Let

$I_{0}:=T_{d}(\mathcal{D})\cap\\{y=x\\}=\left\\{(s,s)\mid\
\frac{1}{m^{+}v+1}\leq s\leq\frac{1}{m^{\star}+1}\right\\}.$

The image $T^{-2}_{d}(I_{0})$ is a curve, which is a piece of the boundary of
$T_{d}^{-1}(\mathcal{D})$ that, be previous arguments, has to cross the left
and right boundaries of $T_{d}(\mathcal{D})$. Actually, it can be parametrized
as $s\mapsto(-3s+(2s)^{1/d},3s)$. In the study of the shape of
$T_{d}^{-1}(\mathcal{D})$ we have seen that
$T_{d}^{-1}(\mathcal{D})\subset\\{y>\frac{1}{m^{\star}+1}=\frac{2}{3}\\}$. We
define

$I_{1}=T_{d}^{2}(T_{d}^{-2}\left(I_{0}\right)\cap T_{d}(\mathcal{D}))\subset
I_{0}$

and, in general,

$I_{n}=T_{d}^{2n}(T_{d}^{-2n}\left(I_{n-1}\right)\cap
T_{d}(\mathcal{D})),\qquad n\geq 1.$

It is clear that $I_{n}\subset I_{n-1}$ for all $n\geq 1$.

Then, $I_{\infty}=\cap_{n\geq 0}I_{n}$ is compact and contains the points in
$T_{d}(\mathcal{D})$ such that all their negative iterates by $T_{d}^{2}$ are
in $T_{d}(\mathcal{D})$. Moreover, by Lemma 5.4, the sequence of the second
components of these iterates is increasing and has to converge to 1. Then,
those points must belong to $W^{u}_{p_{0}}$ and therefore there exists
$(\widehat{p},\widehat{p})\in I_{0}$ such that

$(\widehat{p},\widehat{p})\subset W^{u}_{p_{0}}\cap\\{y=x\\}\subset
T_{d}(\mathcal{D})\cap\\{y=x\\}\neq\emptyset.$

From Lemma 5.2 the piece of $W^{u}_{p_{0}}$ from $(0,0)$ to
$(\widehat{p},\widehat{p})$ must be contained in $T_{d}(\mathcal{D})$. Hence
Lemma 5.3 implies that $W^{u}_{p_{0}}$ cuts the segment
$(-1/2,-1/3)\times\\{0\\}\subset\mathbb{R}^{2}$. ∎

$y=2x+1$$y=3x+1$$x=-1/3$$x=-1/2$$p_{0}$$T_{d}^{-2}$$T_{d}^{-2}$$T_{d}^{-2}(T_{d}(\mathcal{D}))$$T_{d}(\mathcal{D})$$\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\mathcal{E}$

Figure 9. $T_{d}(\mathcal{D})\cap\\{y\geq 0\\}\subset\mathcal{E}$. Points in
$\mathcal{E}$ are mapped “up” by $T_{d}^{-2}$.

Next propositions are devoted to show the two further properties of
$\mathcal{A}_{d}(0)$ claimed in Theorem A(b). First we show that the stable
manifold of the periodic orbit belongs to $\partial\mathcal{A}_{d}(0)$ and
second we show that $\mathcal{A}_{d}(0)$ is unbounded, which follows because
the stable manifold is unbounded.

######  Proposition 5.6.

Let $d\geq 3$ odd. Then, $W^{s}\subset\partial\mathcal{A}_{d}(0)$.

###### Proof.

From Proposition 5.1 we know that $W^{u}_{p_{0}}$ crosses the interval
$(-1/2,-1/3)\times\\{0\\}\subset\mathbb{R}^{2}$. Let $(x_{0},0)\in
W^{u}_{p_{0}}\cap(-1/2,-1/3)\times\\{0\\}$ be the first intersection point of
$W^{u}_{p_{0}}$ with the segment. From Proposition 5.1 we also have that
$[x_{0},0]\times\\{0\\}\subset\mathcal{A}_{d}(0)$. We have (recall that
$T^{-1}_{d}(p_{0})=p_{1}$ and that when $d$ is odd $T_{d}$ is one-to-one)

$\displaystyle\bigcup_{n=0}^{\infty}T_{d}^{-n}\left([x_{0},0]\times\\{0\\}\right)\subset\mathcal{A}_{d}(0)\qquad\text{and}\qquad
p_{j}\in{\rm Acc}\left(\\{T_{d}^{-n}(x_{0})\\}_{n\geq 0}\right),\quad j=0,1,$

where ${\rm Acc}(X)$ denotes the set of accumulation points of $X$. Since, of
course, $\\{p_{0},p_{1}\\}\not\in\mathcal{A}_{d}(0)$ we conclude that
$p_{j}\in\partial\mathcal{A}_{d}(0),\ j=0,1$. Now, let $q$ be any point in
$W^{s}_{p_{0}}$ and $U$ a small disc centered at $q$ and let $\Sigma\subset U$
be a transversal segment to $W^{s}_{p_{0}}$ through $q$. On the one hand,
$W^{s}_{p_{0}}\cap U$ is not contained in $\mathcal{A}_{d}(0)$. On the other
hand, by the $\lambda$-Lemma [Pal69], the iterates by $T_{d}^{2}$ of the
points in $\Sigma$ (close enough to $W^{s}_{p_{0}}$) accumulate to
$W^{u}_{p_{0}}$. Therefore, we would have

$(x_{0},0)\in{\rm Acc}\left(\\{T_{d}^{n}(\Sigma)\\}_{n\geq 0}\right).$

Since $(x_{0},0)\in\mathcal{A}_{d}(0)$ and $\mathcal{A}_{d}(0)$ is open we
conclude that $U$ contains points of $\mathcal{A}_{d}(0)$. If $q\in
W^{s}_{p_{1}}$ then $T_{d}(q)\in W^{s}_{p_{0}}$ and the conclusion is the
same. All together implies that $q\in\partial\mathcal{A}_{d}(0)$, as desired.
∎

######  Proposition 5.7.

$\mathcal{A}_{d}(0)$ is unbounded.

###### Proof.

From the previous lemma it is enough to see that the stable manifold $W^{s}$
of the hyperbolic two-cycle $\\{p_{0},p_{1}\\}$ is unbounded. We start
introducing some notation. See Figure 10. Let $Q_{2}^{\star}$ and
$Q_{4}^{\star}$ be the closed unbounded subsets of the second and fourth
quadrant defined as follows.

$Q_{2}^{\star}:=\\{(x,y)\mid\,x\leq 0,\,y\geq 1\\}\quad\mbox{and}\quad
Q_{4}^{\star}:=\\{(x,y)\mid\,x\geq 0,\,y\leq-1\\}.$

Next we split the above sets into three pieces. Concretely,

$Q_{2}^{\star}=\bigcup_{j=1}^{3}E_{j}\qquad\mbox{and}\qquad
Q_{4}^{\star}=\bigcup_{j=1}^{3}D_{j},$

where

$E_{1}=\\{x\leq\left(\frac{y+1}{2}\right)^{\frac{1}{d}}-y\mid\ y\geq 1\\}$, |  |  $D_{1}=\\{0\leq x\leq y^{\frac{1}{d}}-y\mid\ y\leq-1\\}$,
---|---|---
$E_{2}=\\{\left(\frac{y+1}{2}\right)^{\frac{1}{d}}-y\leq x\leq y^{\frac{1}{d}}-y\mid\ y\geq 1\\}$, |  |  $D_{2}=\\{y^{\frac{1}{d}}-y\leq x\leq\left(\frac{y-1}{2}\right)^{\frac{1}{d}}-y\mid\ y\leq-1\\}$,
$E_{3}=\\{y^{\frac{1}{d}}-y\leq x\leq 0\mid\ y\geq 1\\}$, |  |  $D_{3}=\\{x\geq\left(\frac{y-1}{2}\right)^{\frac{1}{d}}-y\mid\ y\leq-1\\}$.

We denote by $\\{\mathcal{J}_{\ell},\mathcal{I}_{\ell}\\}$ with $\ell=1,2$ the
straight boundaries of the above sets. That is,

$\mathcal{J}_{1}=\\{(x,1)\mid\ x\leq 0\\},$ |  |  $\mathcal{I}_{1}=\\{(x,-1)\mid\ x\geq 0\\}$,
---|---|---
$\mathcal{J}_{2}=\\{(0,y)\mid\ y\geq 1\\},$ |  |  $\quad\mathcal{I}_{2}=\\{(0,y)\mid\ y\leq-1\\}$.

Finally, we denote by $\\{\gamma^{\pm},\sigma^{\pm}\\}$ the other boundaries
of the sets $E_{j}$ and $D_{j}$. That is,

(5.9) $\begin{split}\gamma^{+}&=E_{2}\cap
E_{3}=\left\\{\left(y^{\frac{1}{d}}-y,y\right)\mid\ y\geq
1\right\\}=\\{y=(x+y)^{d}\mid\ x\leq 0,\ y\geq 1\\},\\\ \gamma^{-}&=E_{1}\cap
E_{2}=\left\\{\Big{(}\Big{(}\frac{y+1}{2}\Big{)}^{\frac{1}{d}}-y,y\Big{)}\mid\
y\geq 1\right\\}=\\{y=-1+2(x+y)^{d}\mid\ x\leq 0,\ y\geq 1\\},\\\
\sigma^{-}&=D_{1}\cap D_{2}=\Big{\\{}\Big{(}y^{\frac{1}{d}}-y,y\Big{)}\mid\
y\leq-1\Big{\\}}=\\{y=(x+y)^{d}\mid\ x\geq 0,\ y\leq-1\\},\\\
\sigma^{+}&=D_{2}\cap
D_{3}=\Big{\\{}\Big{(}\Big{(}\frac{y-1}{2}\Big{)}^{\frac{1}{d}}-y,y\Big{)}\mid\
y\leq-1\Big{\\}}=\\{y=1+2(x+y)^{d}\mid\ x\geq 0,\ y\leq-1\\}.\end{split}$

$\sigma^{+}$$\sigma^{-}$$\mathcal{J}_{1}$$\mathcal{J}_{2}$$\mathcal{I}_{2}$$\mathcal{I}_{1}$$\gamma^{+}$$\gamma^{-}$$L$$x^{\star}$$D_{2}$$L_{2}$$E_{2}$$T_{d}(L)$$T_{d}(L_{2})$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$x^{-}$$(x^{-},s_{1})$$\beta(t_{1})=-t_{1}$$x_{0}$$(x_{0},t_{2})$$\alpha(t_{2})=\frac{t_{2}+1}{2}$$\alpha_{2}(s_{1})=\frac{s_{1}-1}{2}$$y=\frac{2d}{1-2d}x_{0}-1$$y=\frac{2d}{1-2d}x^{-}+1$$\gamma^{-}$$\gamma^{-}$

Figure 10. The sets and curves used in the proof of Proposition 5.7.

See Figure 10 for a qualitative picture and the relative position of all
curves and sets. One can check that by construction we have
$T_{d}(\gamma^{-})=\mathcal{I}_{1}$, $T_{d}(\gamma^{+})=\mathcal{I}_{2}$,
$T_{d}(\sigma^{-})=\mathcal{J}_{2}$ and $T_{d}(\sigma^{+})=\mathcal{J}_{1}$.
Consequently,

$T_{d}(D_{2})=\bigcup_{j=1}^{3}E_{j}\qquad\text{and}\qquad
T_{d}(E_{2})=\bigcup_{j=1}^{3}D_{j}.$

We also notice that the curves $\gamma^{\pm}(y)$ and $\sigma^{\pm}(y)$ are
graphs of monotonically decreasing functions of $y$. Indeed, for instance, if
we write $\gamma^{\pm}=\\{\left(\gamma_{1}^{\pm}(y),y\right)\mid\ y\geq 1\\}$
then

$\frac{d\gamma_{1}^{+}}{dy}(y)=\frac{1}{d}\left(\frac{1}{y}\right)^{\frac{d-1}{d}}-1\leq\frac{1}{d}-1<0\qquad\mbox{and}\qquad\frac{d\gamma_{1}^{-}}{dy}(y)=\frac{1}{2d}\left(\frac{2}{y+1}\right)^{\frac{d-1}{d}}-1\leq\frac{1}{2d}-1<0.$

Let

$\Omega:=T_{d}^{-1}(E_{2}).$

According to the previous discussion it is clear that $\Omega\subset D_{2}$
(remember that $\partial E_{2}=\gamma^{+}\cup\gamma^{-}$). We also claim that
$\partial\Omega$ is given by two curves contained in $D_{2}$ which can be
written as graphs of monotone functions (of $y$ as well as of $x$). Of course
$\partial\Omega=T_{d}^{-1}\left(\gamma^{-}\right)\cup
T_{d}^{-1}\left(\gamma^{+}\right)$. Using (2.3) and (5.9) we have

$T_{d}^{-1}\left(\gamma^{-}(y)\right)=\left(\begin{array}[]{c}\xi_{1}(y)\\\
\xi_{2}(y)\end{array}\right):=\left(\begin{array}[]{c}3y-2^{\frac{d-1}{d}}\left(y+1\right)^{\frac{1}{d}}+\left[2^{-\frac{1}{d}}\left(y+1\right)^{\frac{1}{d}}-2y\right]^{\frac{1}{d}}\\\
-3y+2^{\frac{d-1}{d}}(y+1)^{\frac{1}{d}}\end{array}\right),\qquad y\geq 1.$

Thus, we have

$\begin{split}\frac{d\xi_{1}}{dy}(y)&=3-\frac{1}{d}\left(\frac{y+1}{2}\right)^{\frac{1-d}{d}}+\frac{1}{d}\left[\left(\frac{y+1}{2}\right)^{\frac{1}{d}}-2y\right]^{\frac{1-d}{d}}\left(\frac{1}{2d}\left(\frac{y+1}{2}\right)^{\frac{1-d}{d}}-2\right)\\\
&\geq 3-\frac{1}{d}>0\end{split}$

and

$\frac{d\xi_{2}}{dy}(y)=-3+\frac{1}{d}\left(\frac{y+1}{2}\right)^{\frac{1-d}{d}}\leq-3+\frac{1}{d}<0.$

Therefore, using the same formulas as the ones in (5.8),
$T_{d}^{-1}\left(\gamma^{-}\right)$ can be written as a graph of a
monotonically decreasing function (with respect to $y$ as well as $x$).
Similar computations lead to the same conclusion for
$T_{d}^{-1}\left(\gamma^{+}\right)$.

Claim 1. Let $\lambda:=d^{2}/(1-2d)^{2}$. Let $x_{0}>0$ and
$(x_{0},y_{0})\in\Omega$. We denote $(x_{2},y_{2})=T_{d}^{2}(x_{0},y_{0})$.
Then,

$0\leq x_{2}<\lambda\,x_{0}.$

Given $x_{0}>0$, let $L:=\\{x=x_{0}\\}\cap D_{2}=\\{(x_{0},t)\mid\
t_{1}=t_{1}(x_{0})\leq t\leq t_{2}(x_{0})=t_{2}\\},$ where
$(x_{0},t_{1})\in\sigma^{-}$ and $(x_{0},t_{2})\in\sigma^{+}$ and hence

$t_{1}=(x_{0}+t_{1})^{d}\qquad\text{and}\qquad t_{2}=2(x_{0}+t_{2})^{d}+1.$

The image of $L$ by $T_{d}$ can be represented by

$\Gamma_{1}(t)=\left(\begin{array}[]{c}\alpha(t)\\\
\beta(t)\end{array}\right):=T_{d}\left(\begin{array}[]{c}x_{0}\\\
t\end{array}\right)=\left(\begin{array}[]{c}t-(x_{0}+t)^{d}\\\
t-2(x_{0}+t)^{d}\end{array}\right),\qquad t\in[t_{1},t_{2}].$

Since $d-1$ is even and the fact that if $(x,y)\in D_{2}$ we have
$y>2(x+y)^{d}+1$ and $x+y<0$,

(5.10) $\alpha^{\prime}(t)=1-d\left(x_{0}+t\right)^{d-1}\leq
1-d\left(\frac{t-1}{2}\right)^{(d-1)/d}\leq 1-d<0.$

This means that $\alpha(t)$ is strictly decreasing in $t$ with

(5.11) $\alpha(t_{2})=(t_{2}+1)/2\leq\alpha(t)\leq\alpha(t_{1})=0.$

Similarly we have $\beta^{\prime}(t)=1-2d\left(x_{0}+t\right)^{d-1}\leq
1-2d<0$ and $\beta(t_{2})=1\leq\beta(t)\leq\beta(t_{1})=-t_{1}.$

This implies that $\Gamma_{1}(t)$ can be seen as the graph of an increasing
function joining ${\cal J}_{1}$ with ${\cal J}_{2}$. Therefore, it crosses
transversally the boundary of $E_{2}$. Let
$\left(x^{+},y^{+}\right)\in\gamma^{+}$ and
$\left(x^{-},y^{-}\right)\in\gamma^{-}$ be the corresponding intersections.
From (5.11) and (5.10) we have

(5.12) $(t_{2}+1)/2\leq x^{-}\leq x\qquad\text{for
all}\quad(x,y)\in\Gamma_{1}\cap E_{2}.$

Now, given $\xi\in[x^{-},0]$ we consider the new vertical segment in $E_{2}$,

$L_{2}:=\\{x=\xi\\}\cap E_{2}.$

By its definition $T_{d}(L_{2})$ is a curve joining $\mathcal{I}_{1}$ and
$\mathcal{I}_{2}$, parametrized by

$\Gamma_{2}(s)=\left(\begin{array}[]{c}\alpha_{2}(s)\\\
\beta_{2}(s)\end{array}\right):=T_{d}\left(\begin{array}[]{c}\xi\\\
s\end{array}\right)=\left(\begin{array}[]{c}s-(\xi+s)^{d}\\\
s-2(\xi+s)^{d}\end{array}\right),\qquad s\in[s_{1},s_{2}],$

where $s_{1}=s_{1}(\xi)$, $s_{2}=s_{2}(\xi)$ and

$s_{1}=2(\xi+s_{1})^{d}-1>1\qquad\text{and}\qquad s_{2}=(\xi+s_{2})^{d}.$

A similar computation to the one in (5.10) gives that
$\beta_{2}^{\prime}(s)<\alpha_{2}^{\prime}(s)\leq 1-d<0$ and then

$\alpha_{2}(s_{2})\leq\alpha_{2}(s)\leq\alpha_{2}(s_{1})=\frac{s_{1}-1}{2},\qquad
s\in[s_{1},s_{2}].$

The claim will follow from
$\alpha_{2}(s_{1}(\xi))=\frac{s_{1}(\xi)-1}{2}\leq\lambda x_{0}$ for all
$\xi\in[x^{-},0]$.

Clearly $\sigma^{+}$ is above its tangent line at the point $(0,-1)$ which is
given by $y=\frac{2d}{1-2d}x-1$ (to get the slope of the line we can use
implicit derivation to $y=1+2(x+y)^{d}$ at $(x,y)=(0,-1)$). As a consequence,
since $(x_{0},t_{2})\in\sigma^{+}$,

(5.13) $t_{2}+1>\frac{2d}{1-2d}x_{0}.$

Similarly, $\gamma^{-}$ is below its tangent line at the point $(0,1)$ given
by $y=\frac{2d}{1-2d}x+1$. Consequently, since
$(\xi,s_{1}(\xi))\in\gamma^{-}$,

(5.14) $s_{1}(\xi)-1<\frac{2d}{1-2d}\xi.$

Using (5.12), (5.13) and (5.14) we have that

$\alpha_{2}(s_{1}(\xi))=\frac{s_{1}(\xi)-1}{2}<\frac{d}{1-2d}\xi\leq\frac{d}{1-2d}x^{-}\leq\frac{d}{1-2d}\frac{t_{2}+1}{2}\leq\left(\frac{d}{1-2d}\right)^{2}x_{0}=\lambda\,x_{0}.$

Claim 2. Let $x=x_{0}>0$. Then, there exists a point $(x_{0},y)\in\Omega$ such
that $(x_{0},y)\in W^{s}$. In particular, from Proposition 5.6, we conclude
that $\partial\mathcal{A}(0)$ is unbounded.

Since $(0,-1)$ is hyperbolic we already know that $W^{s}_{p_{1}}$ exists and
consists of the points such that their $\omega$-limit with respect to
$T_{d}^{2}$ is $(0,-1)$.

We recall that $\Omega\subset T_{d}^{2}(\Omega)=Q^{\star}_{4}$. Thus,

(5.15) $T_{d}^{-2}(\Omega)\subset\Omega.$

Let $K_{0}:=\\{x=x_{0}\\}\cap\Omega$. Clearly, $T_{d}(K_{0})$ is a curve
connecting $\gamma^{-}$ and $\gamma^{+}$ and so, by construction,
$T_{d}^{2}\left(K_{0}\right)$ is a curve connecting $\mathcal{I}_{1}$ and
$\mathcal{I}_{2}$ and crossing $\partial\Omega$ at exactly two points
(remember that $T_{d}$ is one-to-one): one in
$T_{d}^{-1}\left(\gamma^{-}\right)$ and the other in
$T_{d}^{-1}\left(\gamma^{+}\right)$. We write
$K_{1}=T_{d}^{-2}(T_{d}^{2}\left(K_{0}\right)\cap\Omega)\subset
K_{0}\subset\Omega$.

Repeating this procedure we can define recursively

$K_{j}=T_{d}^{-2j}(T_{d}^{2j}(K_{j-1})\cap\Omega)\subset K_{j-1},\qquad j\geq
1.$

Therefore, $\\{K_{j}\\}_{j\geq 0}$ is a sequence of nested compact sets and
therefore

$\bigcap_{j\geq 0}K_{j}\neq\emptyset.$

Now, we check that if $(x_{0},y_{0})\in\bigcap_{j\geq 0}K_{j}$, then
$(x_{0},y_{0})\in W^{s}$. Indeed, let $(x_{0},y_{0})\in\bigcap_{j\geq
0}K_{j}$. By the definition of $K_{j}$,

$(x_{2j},y_{2j})=T_{d}^{2j}(x_{0},y_{0})\in
T_{d}^{2j}(K_{j-1})\cap\Omega\subset T_{d}^{2j}(K_{0})\cap\Omega,\qquad j\geq
0.$

We can prove by induction that $x_{2j}<\lambda^{j}x_{0}$ for all $j\geq 1$.
Since $(x_{0},y_{0})\in K_{0}\cap\Omega$, by Claim 1, $x_{2}<\lambda x_{0}$.
Assuming the statement is true for $j-1$, since $(x_{2j-2},y_{2j-2})\in
T_{d}^{2j-2}(K_{0})\cap\Omega$, then
$(x_{2j},y_{2j})=T_{d}^{2}(x_{2j-2},y_{2j-2})$ satisfies $x_{2j}<\lambda
x_{2j-2}$. We conclude that $x_{2j}\to 0$. Since $(x_{2j},y_{2j})\in\Omega$ we
also have $y_{2j}\to-1$. Since $x_{0}$ is arbitrarily large we obtain that the
invariant manifold is unbounded. ∎

##  6\. Proof of Theorem A${\text{(}c)}$: The case $d$ odd and $a=-1$

According to Remark 3.2, under the parameter values $d$ odd and $a=-1$, the
dynamics on the center manifold of the origin is repelling and therefore the
only points tending to the origin under iteration are the ones of the stable
manifold of (0,0). Hence, it remains to show that the stable manifold is
unbounded.

It follows from (1.5) that for $d$ odd and $a=-1$ the map $T_{d}$ is a
homeomorphism and we have

(6.1) $T_{d}\left(\begin{array}[]{l}x\\\
y\end{array}\right)=\left(\begin{array}[]{l}y+(x+y)^{d}\\\
y+2(x+y)^{d}\end{array}\right)\quad\text{and}\quad
T_{d}^{-1}\left(\begin{array}[]{l}x\\\
y\end{array}\right)=\left(\begin{array}[]{l}-2x+y-(x-y)^{1/d}\\\
2x-y\end{array}\right).$

In a similar way as in the proof of Proposition 5.7 we introduce a domain,
which we expect to contain $W^{s}$, and we prove that contains points,
arbitrarily far away, such that all their iterates are in the domain and
moreover tend to the origin so that indeed it contains $W^{s}$.

We will take this domain in the fourth quadrant $Q_{4}:=\\{x\geq 0,\,y\leq
0\\}$. We define $D^{0}\subset Q_{4}$ by the condition
$T_{d}^{2}(D^{0})=Q_{4}$. Since $T_{d}^{2}$ is a homeomorphism the boundary of
$D^{0}$ is obtained by taking the preimage of the boundary of $Q_{4}$ with
respect to $T_{d}^{2}$.

Consequently, the boundary of $D^{0}$ is the union of the images of the curves

(6.2) $\displaystyle\sigma^{+}_{0}(t)$
$\displaystyle:=(\alpha^{+}_{0}(t),\beta^{+}_{0}(t))=T_{d}^{-2}(t,0)=\Big{(}6t+2t^{1/d}+(4t+t^{1/d})^{1/d},-6t-2t^{1/d}\Big{)},\
\ \ \ t\geq 0,$ $\displaystyle\sigma^{-}_{0}(t)$
$\displaystyle:=(\alpha^{-}_{0}(t),\beta^{-}_{0}(t))=T_{d}^{-2}(0,-t)=\Big{(}3t+2t^{1/d}+(2t+t^{1/d})^{1/d},-3t-2t^{1/d}\Big{)},\
\ t\geq 0.$

We have that $(\alpha^{\pm}_{0})^{\prime}(t)>0$ and
$(\beta^{\pm}_{0})^{\prime}(t)<0$. Therefore, the curves $\sigma^{\pm}_{0}$
are graphs of well defined decreasing functions
$h^{\pm}_{0}=\beta^{\pm}_{0}\circ(\alpha^{\pm}_{0})^{-1}$ from $[0,\infty)$
onto $(-\infty,0]$.

By construction, the set $D^{1}:=T_{d}(D^{0})=T_{d}^{-1}(Q_{0})$ is the domain
limited by the curves $\sigma^{\pm}_{1}:=T_{d}(\sigma^{\pm}_{0}(t)),\ t\geq
0$. Concretely, these curves are

(6.3) $\displaystyle\sigma^{+}_{1}(t)$
$\displaystyle=(\alpha^{+}_{1}(t),\beta^{+}_{1}(t))=T_{d}(\sigma^{+}_{0}(t))=\big{(}-2t-t^{1/d},2t\big{)},\quad
t\geq 0,$ $\displaystyle\sigma^{-}_{1}(t)$
$\displaystyle=(\alpha^{-}_{1}(t),\beta^{-}_{1}(t))=T_{d}(\sigma^{-}_{0}(t))=\big{(}-t-t^{1/d},t\big{)},\
\ \ \quad t\geq 0.$

Similarly, since $(\alpha^{\pm}_{1})^{\prime}(t)<0$ and
$(\beta^{\pm}_{1})^{\prime}(t)>0$, we have that $\sigma^{\pm}_{1}$ are graphs
of decreasing functions
$h^{\pm}_{1}=\beta^{\pm}_{1}\circ(\alpha^{\pm}_{1})^{-1}$ from $(-\infty,0]$
onto $[0,\infty)$.

Finally, $D^{2}:=T_{d}(D^{1})$ is the full closed fourth quadrant $Q_{4}$. For
notational convenience we also define

$\displaystyle\sigma^{+}_{2}(t)$
$\displaystyle:=T_{d}(\sigma^{+}_{1}(t))=\big{(}t,0\big{)},\qquad t\geq 0,$
$\displaystyle\sigma^{-}_{2}(t)$
$\displaystyle:=T_{d}(\sigma^{-}_{1}(t))=\big{(}0,-t\big{)},\ \quad t\geq 0.$

Since $T_{d}$ is a homeomorphism, the curves $\sigma^{+}_{1}$ and
$\sigma^{-}_{1}$ are the only preimages of the curves $\sigma^{+}_{2}$ and
$\sigma^{-}_{2}$, respectively. They only intersect at the origin. The same
happens with $\sigma^{+}_{0}$ and $\sigma^{-}_{0}$. Moreover, $\sigma^{+}_{1}$
is above $\sigma^{-}_{1}$ and $\sigma^{+}_{0}$ is above $\sigma^{-}_{0}$.
Also, we will use that $\sigma^{+}_{1}$ is below $\\{y=-x\\}$ and
$\sigma^{-}_{0}$ is above $\\{y=-x\\}$. Indeed, these claims can be checked
from (6.2) and (6.3) after some computations.

######  Lemma 6.1.

If $(x_{0},y_{0})\in D^{0}$ then $(x_{2},y_{2}):=T_{d}^{2}(x_{0},y_{0})\in
Q_{4}$ and $x_{2}\leq x_{0}/2$.

###### Proof.

The first part of the statement follows from the previous construction. We
have to prove the inequality. We define

$D^{0}_{\rho}=\\{(x,y)\in D^{0}\mid\ x\leq\rho\\},\quad
D^{1}_{\rho}=\\{(x,y)\in D^{1}\mid\ x\geq-\rho\\},\quad
D^{2}_{\rho}=\\{(x,y)\in D^{2}\mid\ x\leq\rho\\}.$

We will prove that, for any $\rho>0$,

$T_{d}(D^{0}_{\rho})\subset D^{1}_{\rho/2}\qquad\text{and}\qquad
T_{d}(D^{1}_{\rho/2})\subset D^{2}_{\rho/2}.$

For the first inclusion we consider the segments $\\{x=r\\}\cap D^{0}_{\rho}$
with $0\leq r\leq\rho$, parametrized by $s\in[s_{-},s_{+}]\subset[-r,0]$, with
$s_{-}$ and $s_{+}$ such that
$(r,s_{-})\in\\{\mathrm{image}\,\sigma_{0}^{-}\\}$ and
$(r,s_{+})\in\\{\mathrm{image}\,\sigma_{0}^{+}\\}$. In particular, we have
that there exists $t_{1}\geq 0$ such that

$\sigma^{-}_{0}(t_{1})=\Big{(}3t_{1}+2t_{1}^{1/d}+(2t_{1}+t_{1}^{1/d})^{1/d},-3t_{1}-2t_{1}^{1/d}\Big{)}=(r,s_{-}).$

The image of the segment can be represented by

$\tau_{1}(s)=\big{(}\tau_{1}^{x}(s),\tau_{1}^{y}(s)\big{)}:=T_{d}(r,s)=\big{(}s+(r+s)^{d},s+2(r+s)^{d}\big{)},\qquad
s\in[s_{-},s_{+}].$

Since $d-1$ is even, $(\tau_{1}^{x})^{\prime}(s)$ and
$(\tau_{1}^{y})^{\prime}(s)$ are positive. This implies that the minimum of
$\tau_{1}^{x}(s)$ is attained at the value $s=s_{-}$. This point is sent by
$T_{d}$ to $(-t_{1}-t_{1}^{1/d},\,t_{1})$ and we have

$-t_{1}-t_{1}^{1/d}=\frac{1}{2}(-3t_{1}-2t_{1}^{1/d})+\frac{1}{2}t_{1}\geq\frac{1}{2}s_{-}\geq-\frac{1}{2}r\geq-\frac{1}{2}\rho.$

Now let $\widetilde{r}$ be such that $-\rho/2\leq\widetilde{r}\leq 0$ and we
consider the image of the segment
$\\{x=\widetilde{r}\\}\cap\mathcal{D}^{1}_{\rho}$, parametrized by
$\widetilde{s}\in[\widetilde{s}_{-},\widetilde{s}_{+}]\subset[0,-\widetilde{r}]$.
We write

$\tau_{2}(\widetilde{s})=\big{(}\tau_{2}^{x}(\widetilde{s}),\tau_{2}^{y}(\widetilde{s})\big{)}:=T_{d}(\widetilde{r},\widetilde{s})=\big{(}\widetilde{s}+(\widetilde{r}+\widetilde{s})^{d},\widetilde{s}+2(\widetilde{r}+\widetilde{s})^{d}\big{)}.$

Since $\sigma^{+}_{1}$ is below $\\{y=-x\\}$, $\widetilde{r}+\widetilde{s}<0$.
In this case we also have that $(\tau_{2}^{x})^{\prime}(\widetilde{s})$ and
$(\tau_{2}^{y})^{\prime}(\widetilde{s})$ are positive. Then, a bound of the
maximum of $\tau_{2}^{x}(\widetilde{s})$ is obtained from

$\tau_{2}^{x}(\widetilde{s})\leq\tau_{2}^{x}(-\widetilde{r})=-\widetilde{r}\leq\rho/2.$

This implies $T_{d}(D^{1}_{\rho/2})\subset D^{0}_{\rho/2}$. ∎

Now we take $\rho>0$ arbitrary and define

$I^{0}=D^{0}\cap\\{x=\rho\\}.$

Its image by $T_{d}^{2}$ is a curve in $Q_{4}$ that joints a point in
$\\{\mathrm{image}\,\sigma_{2}^{-}\\}$ and a point in
$\\{\mathrm{image}\,\sigma_{2}^{+}\\}$. Then this curve has to cross
$\\{\mathrm{image}\,\sigma_{0}^{-}\\}$ and
$\\{\mathrm{image}\,\sigma_{0}^{+}\\}$. The set
$I^{1}=T_{d}^{-2}(T_{d}^{2}(I^{0})\cap D^{0})\subset I^{0}$ contains points
such that they, together with their second iterates, belong to $I^{0}$.
Repeating this procedure we define, as in the final part of the proof of
Proposition 5.7,

$I^{k}=T_{d}^{-2k}(T_{d}^{2k}(I^{k-1})\cap D^{0}).$

Clearly, $I^{k}\subset I^{k-1}$ so that $I^{k}$ is a sequence of nested
compact sets as well.

Then $I^{\infty}=\cap_{n\geq 0}I^{n}\neq\emptyset$. By this construction, if
$(x_{0},y_{0})\in I^{\infty}$, $(x_{2k},y_{2k})=T_{d}^{2k}(x_{0},y_{0})\in
T_{d}^{2k}(I^{k-1})\cap D^{0}\subset D^{0}$ for all $k\geq 0$. Then, by Lemma
6.1,

$0<x_{2k}<\left(\frac{1}{2}\right)^{k}x_{0}$

and as $(x_{2k},y_{2k})\in D^{0}$, the iterates $(x_{2k},y_{2k})$ converge to
$(0,0)$ which implies that $(x_{0},y_{0})\in W^{s}_{0}$. Since $\rho$ is
arbitrary, $W^{s}_{0}$ is unbounded.

######  Remark 6.2.

Since the curves that determine th e boundary of $D^{0}$ are very close they
provide a very good approximation for the stable manifold, even far away from
the origin.

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