# Stationary solutions for dyadic mixed model of the Euler equation. A
complete spectrum.

Carlo Metta<EMAIL_ADDRESS>University of Florence. Department of
Mathematics, Informatics and Statistics, Italy.  ISTI-CNR, Via G. Moruzzi,
Pisa, Italy.

###### Abstract

Dyadic models of the Euler equations were introduced as toy models to study
the behaviour of an inviscid fluid in turbulence theory. In 1974 Novikov
proposed a generalized mixed dyadic model that extends both Katz-Pavlovic and
Obukhov models giving birth to a more complex structure: no results were found
in literature until 2015 where blow up in finite time for smooth solutions and
existence of self-similar solution for particular values of the model
parameters were shown by Jeong I.J. We extend such partial results by giving a
complete spectrum of existence and uniqueness results for two cardinal classes
of finite energy stationary solutions, namely constant and self-similar
solutions.

###### keywords:

Euler Equation, Dyadic Model, Turbulence Theory, Partial Differential
Equations

##  1 Introduction

In this work we prove existence and uniqueness results for special classes of
stationary solutions of the following system of ordinary differential
equations:

$\begin{split}\frac{dY_{n}(t)}{dt}=\delta_{1}&[k_{n}Y^{2}_{n-1}(t)-k_{n+1}Y_{n}(t)Y_{n+1}(t)]\\\
-\delta_{2}&[k_{n}Y^{2}_{n+1}(t)-k_{n-1}Y_{n}(t)Y_{n-1}(t)]\end{split}$ (1)

for $n\geq 1$, where $k_{n}=2^{\beta n}$ for some $\beta>0$,
$\delta_{1},\delta_{2}\geq 0$ non negative parameters, and boundary conditions
$Y_{-1}\equiv 0$, $Y_{n}(0)=y_{n}$.

Model (1) belongs to the family of dyadic shell models developed in turbulence
theory in order to deepen the study of turbulent fluid dynamics [7]. These
models are consistent with but simpler than the cardinal Euler equation. They
are often used as toy model to experiment and apply novel and useful
techniques related to Euler dynamics (for some effective results in this
direction see for example [5], [21]).

In the Fourier representation of three-dimensional Euler equation, the
transfer of energy from large to small scales is described as a flux of energy
from small wave numbers to large wave numbers. The idea behind shell models is
to divide the space into concentric spheres with exponentially growing radius
$k_{n}=\lambda^{n}$, $\lambda>1$. We then call $n$-th shell the set of wave
numbers contained in the $n$-th sphere and not contained in the $(n-1)$-th
sphere. The square of the scalar variable $Y_{n}(t)$ is linked with the energy
of a fluid velocity vector field restricted to $n$-th shell frequency. The
quantity

$E(t)=\sum_{n\geq 0}Y_{n}^{2}(t)$

is then called the _energy_ of the solution $Y_{n}(t)$. One can easily check
from (1) that energy is _formally_ conserved in time. It is natural to
restrict the study to non-negative and finite energy solutions.

In order to state regularity results for dyadic models it is common to define
counterpart of the Sobolev norms in the space of sequences. The $H^{s}$ norm
of a solution $Y=(Y_{0},Y_{1},\ldots)$ at a certain time $t$ is given by the
formula

$||Y_{n}(t)||_{s}^{2}=\sum_{n\geq 0}2^{2sn}Y_{n}^{2}(t).$

In particular, the energy of a solution is the square of its $H^{0}$ norm,
usually called the $\ell^{2}$ norm.

Shell models investigate the energy cascade flow with a system of coupled
nonlinear ordinary differential equation of the form:

$\frac{d}{dt}Y_{n}=k_{n}G_{n}[Y,Y]-\nu_{n}Y_{n}+f_{n},$

where the variable $Y_{n}$ represents the evolution over time of the velocity
over a wavelength of scale $k_{n}$. The nonlinear function
$G_{n}[\cdot,\cdot]$ is chosen to preserve some suitable properties inherited
from the original nonlinear terms of Euler equation. Moreover, it is common
for $G_{n}[\cdot,\cdot]$ to couple only scales that are close to each other
(for instance nearest and next-to-nearest shells). The model is said to be
either inviscid or viscous depending on whether the friction coefficient
$\nu=\\{\nu_{n}\\}_{n}$ is equal to 0. Similarly, the model is called unforced
if there is no forcing term in the equation ($f_{n}=0$), otherwise forced, if
we add a positive forcing (constant) term to the first node in order to
simulate a force sustaining turbulent fluids.

The constraints to have local interaction, quadratic non-linearity, preserving
total energy (or total helicity), and phase-space evolution do not fix
uniquely the form $G_{n}[\cdot,\cdot]$. Consequently, many models have been
developed even in recent years in order to study different aspects of
turbulent fluids.

It is worth noting that model (1) reduces to what in the literature is called
Katz-Pavlovic dyadic model [15] (even though it had already been introduced by
Novikov [10]) and Obukhov dyadic model [19] by setting respectively
$\delta_{2}=0$ and $\delta_{1}=0$, thus we expect it to carry both Katz-
Pavlovic and Obukhov dynamics giving birth to a more complex structure: even
simple uniqueness and positiveness properties proved for example in [2], [3],
[8], [15], [16] do not hold anymore.

As observed many times in literature, for example by Waleffe [22], Katz
Pavlovic and Obukhov models constitute the two basic blocks of all models
satisfying four characteristic features derived from the spectral form of
Euler equation: (i) quadratic non-linearity, (ii) appropriate scaling property
of dimensionless coefficients, (iii) energy conservation, (iv) and nearest
neighbor coupling. All of these except the last one descend from the Euler
equation: the last condition is a simplification to make the problem more
tractable. Thus, it is natural to investigate the dynamics of the generalized
model (1), which from now on we refer to as the _mixed (linear) dyadic model_
, where we use the _linear_ attribute to distinguish from the more complex
family of _tree models_ (see [1], [4], [6]).

In Section 2 we start by introducing main properties of model (1), focusing on
existence and uniqueness results for _constant_ and _self-similar_ solutions.
These results show a transition from a region, in the parameter space, where
uniqueness is known to another one where it is false or open. They are
consistent with and extend previous result found in literature. A K41 scaling
property is also proved. Section 3 is devoted to present a different proof of
Theorem 10 in [3]. The original proof was based on complex analysis and
numerical simulations. Since it was not clear whether it can be extended to
the case $\delta_{1},\delta_{2}>0$, we prefer to adopt another strategy by
means of a novel pull-back technique. In Section 4 and 5 we then extend such
technique proving the main results of Section 2, namely existence and
uniqueness statements respectively for constant and self-similar solutions.

##  2 Mixed dyadic model

In [2] positiveness property for weak solutions of the Katz-Pavlovic linear
model was proved. This property plays a crucial role in many cardinal results,
like the exponential global attraction of finite energy solutions to the
unique constant solution (see [9]).
Unfortunately, the positiveness property does not hold anymore in the mixed
dyadic model. Indeed, by the variation of constants formula

$\begin{split}Y_{n}(t)&=Y_{n}(t_{0})\cdot
e^{-\int_{t_{0}}^{t}[\delta_{1}k_{n+1}Y_{n+1}(s)-\delta_{2}k_{n-1}Y_{n-1}(s)]ds}+\\\
&+\int_{t_{0}}^{t}k_{n}[\delta_{1}Y_{n-1}^{2}(s)-\delta_{2}Y_{n+1}^{2}(s)]\cdot
e^{-\int_{s}^{t}[\delta_{1}k_{n+1}Y_{n+1}(z)-\delta_{2}k_{n-1}Y_{n-1}(z)]dz}ds.\end{split}$
(2)

In the Katz-Pavlovic model, i.e. $\delta_{2}=0$, one can easily derive that
positiveness holds for positive initial condition. However, in the mixed model
(1) the following condition

$\delta_{1}Y_{n-1}^{2}(t)-\delta_{2}Y_{n+1}^{2}(t)\geq 0$ (3)

does not hold in general.

Because of its complex dynamics, there were no results in literature until
2015 [13], where the authors show that smooth solutions blows up in finite
time when $\delta_{2}/\delta_{1}$ is small enough, extending previous results
which proved blow-up in the KP model [8], [12], [15], [16], [22]. Only later
in 2019 [14] the existence of self-similar solution for particular value of
parameters $(\delta_{1},\delta_{2})$ was proved, within a local uniqueness
theorem. Moreover, in [18], the author proved the following theorem about the
existence of weak solution of the mixed dyadic model for every initial
condition $y\in\ell^{2}$, in the case $\delta_{1}=\delta_{2}=1$.

######  Theorem 2.1 (Montagnani [18]).

Consider the infinite dimensional shell model

$\frac{d}{dt}Y_{n}(t)=k_{n}Y_{n-1}^{2}(t)-k_{n+1}Y_{n}(t)Y_{n+1}(t)-k_{n}Y_{n+1}^{2}(t)+k_{n-1}Y_{n}(t)Y_{n-1}(t),$
$Y(0)=y.$

Then, for any initial condition $y\in\ell^{2}$ there exists at least one
solution $Y(t)$ on $[0,T]$.

Extending Theorem 2.1 to general parameters $\delta_{1},\delta_{2}$ is
straightforward.

In the next sections we prove existence and uniqueness results for both
constant and self-similar solutions, for every couple of parameter
$(\delta_{1},\delta_{2})$ in the positive quadrant. A K41 scaling property is
also proved. These findings are consistent with the partial results found in
[14], although we prefer to consider model (1) as originally expressed in [11]
and [16].

###  2.1 Constant solutions

We recall that a _constant_ solution $Y=\\{Y_{n}(t)\\}$ of (1) is a solution
that is time independent, i.e. $Y_{n}(t)=a_{n}$ for all $t\geq 0$ and some
$a_{n}\in\mathbb{R}_{\geq 0}$. From the point of view of fluid dynamics theory
it is common to restrict ourselves to study positive constant solutions with
the additional condition of being finite energy, i.e.

$\sum_{n=1}^{\infty}a_{n}^{2}<\infty.$

In order to consider (1) as a model of turbulence dynamics, it is natural to
add a constant forcing term ($F>0$) to sustain turbulent regime:

$\frac{dY_{0}(t)}{dt}=-\delta_{1}k_{1}Y_{0}(t)Y_{1}(t)-\delta_{2}Y^{2}_{1}(t)+F.$
(4)

In particular, if $Y_{n}(t)=a_{n}$ is a constant solution, then

$\delta_{1}[k_{n}a_{n-1}^{2}-k_{n+1}a_{n}a_{n+1}]=\delta_{2}[k_{n}a_{n+1}^{2}-k_{n-1}a_{n}a_{n-1}],$
(5)

and the first terms $a_{0},a_{1}$ are related to $F$ by the following relation

$\delta_{1}k_{1}a_{0}a_{1}+\delta_{2}a_{1}^{2}=F.$ (6)

Despite their simplicity, constant solutions are one of the most important
class of solutions. In the case $\delta_{2}=0$ there exists a unique constant
solution of the form

$a_{n}=C_{F}\cdot k_{n}^{-1/3},\,\,\,\,\,n\geq 0,$

where $C_{F}$ is a constant depending on the starting force $F$ and the
scaling $k_{n}^{-1/3}$ is reminiscent of the famous Kolmogorov K41 [17] law as
well. Moreover, the existence of a constant solution in a forced system is an
example of anomalous dissipation in turbulent fluids. In [9], the authors
showed that, given a forcing term $F$, the only constant solution is an
exponential global attractor for every finite energy solution.
The existence of a global attractor for an inviscid system is, perhaps,
surprising. However, it is perfectly consistent with the concept of anomalous
or turbulent dissipation conjectured by Onsager [20]. It is then natural to
investigate whether constant solutions continue to exist in the mixed model
and how the competition of Katz-Pavlovic and Obukhov dynamics affects the
behaviour of such solutions.

Later we prove the following result:

######  Theorem 2.2.

The forced mixed model $(\ref{mixedmodeliniziale})$ admits positive constant
solutions for every choice of coefficient $\delta_{1},\delta_{2}>0$.

In particular:

  * 1.

if $\frac{\delta_{1}}{\delta_{2}}<k_{1}^{-4/3}$, then for every $F>0$ there
exist infinitely many positive finite energy constant solution
$\\{a_{n}\\}_{n\geq 0}$, one for every choice of $a_{0}>0$;

  * 2.

if $\frac{\delta_{1}}{\delta_{2}}>k_{1}^{-4/3}$, then for every $F>0$ there
exists just one value $a_{0}>0$ from which stems a positive finite energy
constant solution $\\{a_{n}\\}_{n\geq 0}$.

Moreover, any such solution satisfies Kolmogorov’s scaling law

$\lim_{n\rightarrow\infty}\frac{a_{n}}{k_{n}^{-1/3}}=C$

for some positive constant $C>0$.

Theorem 2.2 offers a complete picture of how the two opposite dynamics compete
against each other to shape the behaviour of fixed points (constant
solutions). If the total impulse of Obukhov dynamics we assign to the model
does not exceed $k_{1}^{4/3}$ times the corresponding Katz-Pavlovic dynamics,
then existence and uniqueness of fixed point is preserved. It is an
interesting open question whether this unique fixed point still attracts all
the other finite energy solutions like in [9]. On the other hand, if
$\frac{\delta_{1}}{\delta_{2}}<k_{1}^{-4/3}$ it is not surprising that all
constant solutions become finite energy since the regularization phenomenon
related to Obukhov dynamics ([16]) prevails over Katz-Pavlovic.

###  2.2 Self-similar solutions

When the forcing term is absent no non-trivial fixed point are yet known in
literature for Katz-Pavlovic and Obukhov model. However, there exists another
special class of stationary solution closely related to constant solutions.

######  Definition 2.1.

A self-similar solution is a finite energy solution $Y$ such that there exists
a differentiable function $\phi(t)$ and a sequence of real numbers
$a=(a_{n})_{n\geq 1}$ such that $Y_{n}(t)=a_{n}\cdot\phi(t)$ for all $n\geq 1$
and all $t\geq 0$.

######  Observation.

It is easy to observe that positive self-similar solutions satisfying Katz-
Pavlovic model have the form

$Y_{n}(t)=\frac{a_{n}}{t-t_{0}},$ (7)

for some $t>t_{0}$ and $t_{0}<0$.
Indeed, if a positive solution is of the form (7), then

$\begin{split}-\frac{a_{n}}{(t-t_{0})^{2}}=\frac{dY_{n}(t)}{dt}=&k_{n-1}Y_{n-1}^{2}(t)-k_{n}Y_{n}(t)Y_{n+1}(t)\\\
=&k_{n-1}\frac{a_{n-1}^{2}}{(t-t_{0})^{2}}-k_{n}\frac{a_{n}a_{n+1}}{(t-t_{0})^{2}},\end{split}$

that leads us to the sequence $\\{a_{n}\\}_{n}$ satisfying

$a_{n}a_{n+1}=\frac{a^{2}_{n-1}}{2^{\beta}}+2^{-\beta n}a_{n}.$

Although it is possible for the first terms $a_{1},a_{2},\ldots,a_{n_{0}}$ to
be zero, if $a_{n_{0}+1}>0$ then all the subsequent coefficients must be
positive too:

$a_{n+1}=2^{-\beta n}+\frac{a_{n-1}^{2}}{2^{\beta}a_{n}}>0,\,\,\,\,\,\,\forall
n\geq n_{0}+1.$ (8)

Thus, without loss of generality, one can look for solutions with $a_{0}=0$
and $a_{n}>0$. Theoretically, every choice of $a_{1}>0$ would give a self-
similar solution following recursion (8), but we are only interested in finite
energy solution. Unexpectedly, in [3] it was proved the existence of just one
special $a_{1}>0$ that satisfies this condition in the Katz-Pavlovic setting.

######  Theorem 2.3.

(Barbato, Flandoli, Morandin [3]) Given $t_{0}<0$, there exists a unique
positive self-similar solution with $a_{1}\neq 0$. Moreover, given $t_{0}<0$
and $n_{0}\geq 0$, there exists a unique positive self-similar solution with

$a_{1}=a_{2}=\cdots a_{n_{0}}=0,a_{n_{0}+1}>0.$

In addition, the coefficients $a_{n}$ have the property

$\lim_{n\rightarrow\infty}\frac{a_{n}}{k_{n}^{-1/3}}=C_{n_{0}},$

for some constant $C_{n_{0}}$.

Theorem 2.3 was originally proved by using complex analysis argument and with
the help of numerical computation, and it was not clear whether such argument
can be extended to the mixed dyadic model. In the next section we present a
different proof based on a pullback technique that will be useful in other
proofs.
Theorem 2.3 shows also that the Kolmogorov scaling law appears in these
special solutions, phenomenologically associated to decaying turbulence. But
it is an open problem to understand whether all other solutions approach the
self-similar ones and in which sense.
The existence of finite energy self-similar solutions is of theoretical
interest in itself, in comparison with analogous investigations for Euler and
Navier-Stokes equations, moreover the existence of such solutions has a number
of implications. For instance, they realize perfectly the decay rate $t^{-1}$,
coherently with Theorem 8 and 9 in [3]. It has been conjectured that the set
of all self-similar solutions (set depending on $t_{0}\in\mathbb{R}$ and
$n_{0}\geq 0$) attracts all other finite energy solutions. If this is the
case, the decay rate $t^{-1}$ would be the true one for all solutions.
Moreover, self-similar solutions offer an easy example of lack of uniqueness
as shown by next observation.

######  Observation.

It is possible to prove that for some initial conditions in $H=\ell^{2}$ with
all negative components there exist infinitely many finite energy solutions.
Indeed, by Theorem 2.3, there exists a self-similar solution $Y$ whose total
energy is strictly decreasing. Let $T>0$, then $X(t)=-Y(T-t)$ is a local
solution on $[0,T]$ by a simple time inversion property. For any time
$s\in[0,T]$, let’s consider the solution $X^{s}$ obtained by attaching $X$ on
$[0,s]$ to a Leray-Hopf solution on $[s,\infty)$ with initial condition
$X(s)=-Y(T-s)\in H$. The energy of this solution strictly increases on $[0,s]$
and then is non-increasing on $[s,\infty)$. Thus, to different values of $s$
correspond finite energy solutions which are really different, but all with
the same negative initial condition $-Y(T)$.

Once more, it is natural to address the question of existence and possibly
uniqueness of self-smilar solutions also for the mixed dyadic model (1). As
observed previously, it is still easy to prove that self-similar solutions in
the unforced ($F=0$) mixed model have the form

$Y_{n}(t)=\frac{a_{n}}{t-t_{0}},\,\,\,\,\,a_{0}=0,$ (9)

with $t>t_{0}$ and $t_{0}<0$.
If a positive solution is of the form (9), then

$\begin{split}-\frac{a_{n}}{(t-t_{0})^{2}}=\frac{dY_{n}(t)}{dt}=&\frac{\delta_{1}}{(t-t_{0})^{2}}(k_{n}a_{n-1}^{2}-k_{n+1}a_{n}a_{n+1})\\\
-&\frac{\delta_{2}}{(t-t_{0})^{2}}(k_{n}a_{n+1}^{2}-k_{n-1}a_{n}a_{n-1}),\end{split}$
(10)

that leads us to the sequence $\\{a_{n}\\}_{n\geq 1}$ satisfying

$-\frac{a_{n}}{k_{n}}=\delta_{1}(a_{n-1}^{2}-k_{1}a_{n}a_{n+1})-\delta_{2}(a_{n+1}^{2}-k_{1}^{-1}a_{n}a_{n-1}).$

It is still possible for the first terms $a_{1},a_{2},\ldots,a_{n_{0}}$ to be
zero, although if $a_{n_{0}+1}>0$ then all the subsequent coefficients must be
not zero: indeed, from the latter relation if $a_{n_{0}}=0$ then

$\begin{split}-\frac{a_{n_{0}+1}}{k_{n_{0}+1}}&=\delta_{1}(-k_{1}a_{n_{0}+1}a_{n_{0}+2})-\delta_{2}(a_{n_{0}+2}^{2})\iff\\\
\frac{a_{n_{0}+1}}{k_{n_{0}+1}}&=\delta_{2}a_{n_{0}+2}^{2}+\delta_{1}k_{1}a_{n_{0}+1}a_{n_{0}+2}\end{split}$

and $a_{n_{0}+1}>0$ implies $a_{n_{0}+2}\neq 0$. Since we are interested in
positive solutions, without loss of generality one can set $a_{0}=0$ and
$a_{n}>0$ for every $n\geq 1$.
In [14] self-similar solutions were found for $\delta_{1}=1$ and small enough
$\delta_{2}$, within a local uniqueness theorem. In the next section we prove
the following result:

######  Theorem 2.4.

Given $t_{0}<0$, and $k_{1}^{-4}\leq\delta_{1}/\delta_{2}\leq 1$, there exist
self-similar solutions of the unforced ($F=0$) model (1). In particular

  * 1.

if $k_{1}^{-4}\leq\delta_{1}/\delta_{2}<k_{1}^{-4/3}$ then for every $a_{1}>0$
there exists a self-similar solution $\\{a_{n}\\}_{n\geq 0}$ of (1);

  * 2.

if $k_{1}^{-4/3}<\delta_{1}/\delta_{2}\leq 1$ there exists just one self-
similar solution $\\{a_{n}\\}_{n\geq 0}$ of (1).

In addition, any such self-similar solution satisfies Kolmogorov’s scaling law

$\lim_{n\rightarrow\infty}\frac{a_{n}}{k_{n}^{-1/3}}=C$

for some positive constant $C>0$.

Theorem 2.4 divides the positive plane in four subregions: above the line
$\delta_{1}/\delta_{2}=k_{1}^{-4}$ and below $\delta_{1}/\delta_{2}=1$ Theorem
2.4 does not give any information about existence of self-similar solutions;
between the lines $\delta_{1}/\delta_{2}=k_{1}^{-4}$ and
$\delta_{1}/\delta_{2}=k_{1}^{-4/3}$ we have existence but not uniqueness;
between the lines $\delta_{1}/\delta_{2}=k_{1}^{-4/3}$ and
$\delta_{1}/\delta_{2}=1$ we have existence and uniqueness of self-similar
solution.
However, as we will see later, upper and lower bounds for the ratio
$\delta_{1}/\delta_{2}$ can be further refined. Numerical simulations suggest
the existence of a _true_ bound $L_{true}<k_{1}^{-4}$ such that Theorem 2.4
holds in the wider domain $\delta_{2}\cdot L_{true}\leq\delta_{1}$. This
result is consistent with and complements Theorem 2.3 and Theorem 1 in [14] by
giving a full spectrum of behaviour of self-similar solution in the mixed
dyadic model. Once again, we stress that when the impulse $\delta_{2}$ given
by Obukhov dynamics exceeds $k_{1}^{4/3}$ times the opposite impulse
$\delta_{1}$, the regularization phenomenon occurs and all self-similar
solutions become finite energy. On the contrary, when the Katz-Pavlovic
dynamics prevails in the sense of second part of Theorem 2.4, then the
dynamics resembles the one described by Theorem 2.3 where just one finite
energy self-similar solution exists.

##  3 Proof of Theorem 2.3

The original proof of Theorem 2.3 was based on complex analysis and numerical
simulations, while in [14] another proof was presented based on the analysis
of certain dynamical systems on the plane. Since it was not clear whether such
result could be fully extended to the whole parameters positive plane
$(\delta_{1},\delta_{2})$, we prefer to adopt another strategy by means of a
novel methodology of pull-back suggested in [6] for studying constant solution
of more structured dyadic models. Pull-back is a key dynamical idea often used
in the literature to construct special solutions of differential equations,
like periodic solutions. However, its utility is usually associated to the
non-autonomy of certain systems: for example when coefficients are time-
dependent and one is looking for infinite-time objects, the natural ways is to
start from minus infinity. It is extremely interesting that this technique is
successful not just in the Katz-Pavlovic framework, completely autonomous, but
also in the generalized mixed dyadic model, as we will prove in last sections.

As already mentioned in (8), we are interested in the following recursion

$a_{n+1}=2^{-\beta n}+\frac{a_{n-1}^{2}}{2^{\beta}a_{n}}>0,\,\,\,\,\,\,\forall
n\geq 1,\beta>0,$ (11)

where, without loss of generality, we set $a_{0}=0$ and $a_{1}>0$. Our main
goal is to prove the existence of one and only one $a_{1}>0$ such that the
self-similar solution that stems from $(a_{0},a_{1})$ following the rule (11)
has finite energy.

In order to lighten the notation, we prove Theorem 2.3 in the case $\beta=1$.
The general case will be a straightforward consequence.

First of all we start by proving the following proposition.

######  Proposition 3.5.

Consider the recursion

$\begin{split}a_{n+1}&=\frac{a^{2}_{n-1}}{2a_{n}}+\epsilon_{n},\,\,\,\epsilon_{n}=2^{-n},\,\,\,n\geq
1\\\ a_{0}&=0.\end{split}$ (12)

There is one and only one $u=\\{u_{n}\\}_{n\in\mathbb{N}}$ positive finite
energy sequence satisfying (12). Moreover, such $\\{u_{n}\\}$ lies in $H^{s}$
for any $s<1/3$.

###### Proof.

We structure our proof in three different steps. At first, we prove the
existence of solution for recursion (12), then we show regularity of such
solution, finally we prove uniqueness among positive solutions with finite
energy.

Step (1): Existence.

We start by considering the following two definition.

######  Definition 3.1.

We call strong self-similar any positive sequence
$\\{a_{n}\\}_{n\in\mathbb{N}}$ satisfying the recurrence:

$\begin{split}a_{n+1}&=\frac{a_{n-1}^{2}}{2a_{n}}+\epsilon_{n},\,\,\,\epsilon_{n}=2^{-n},\,\,\,n\geq
1\\\ a_{0}&=0.\end{split}$ (13)

######  Definition 3.2.

We call weak self-similar any positive sequence
$\\{\tilde{a}_{n}\\}_{n\in\mathbb{N}}$ satisfying the recurrence:

$\begin{split}\tilde{a}_{n+1}&=\frac{\tilde{a}_{n-1}^{2}}{\tilde{a}_{n}}+\zeta_{n},\,\,\,\zeta_{n}=\epsilon_{n}\cdot
2^{\frac{n-2}{3}},\,\,\,n\geq 1\\\ \tilde{a}_{0}&=0.\end{split}$ (14)

######  Remark.

It is easy to verify that if $\\{\tilde{a}_{n}\\}_{n\in\mathbb{N}}$ is a weak
self-similar sequence then
$\\{a_{n}=\frac{\tilde{a}_{n}}{2^{n/3}}\\}_{n\in\mathbb{N}}$ is a strong self-
similar sequence. Conversely, for any strong self-similar sequence it is
possible to recover the corresponding weak sequence from the equality above.

In order to prove existence of strong self-similar sequence we use a pull back
technique: we first consider recursion (14) backwards fixing a large $N>2$ and
two appropriate starting values $\tilde{a}_{N+1}$ and $\tilde{a}_{N}$, then
compute $\tilde{a}_{n}$ for lower coefficients $n<N$; finally we let
$N\rightarrow\infty$ proving convergence by compactness to a weak self-similar
solution and finally recovering a strong self-similar sequence from the remark
above.

Thus, for any fixed $N>2$ we are interested in the following truncated
reversed recursion:

$\begin{split}&(\tilde{a}_{n-1}^{(N)})^{2}=\tilde{a}_{n}^{(N)}(\tilde{a}_{n+1}^{(N)}-\zeta_{n}),\,\,\,n\leq
N,\,\,\,\ \zeta_{n}=\epsilon_{n}\cdot 2^{\frac{n-2}{3}}\\\
&\tilde{a}_{N+1}^{(N)}=\tilde{a}_{N}^{(N)}=L>0,\\\
&\tilde{a}_{n}^{(N)}=0,\,\,\,n>N+1,\end{split}$ (15)

where the initial value L will be chosen later accordingly to our
requirements.

By taking a close look at recursion (15) we observe that the right-hand side
must be non negative for every $n\leq N$. Thus, we first have to assess under
which conditions recursion (15) is well-defined. The following proposition
poses sufficient conditions for existence of weak self-similar solution.

######  Proposition 3.6.

Consider the system of equations (15) and let $M>0$ be such that

$M=\sum_{i=1}^{\infty}\zeta_{i}<\infty.$

Then any initial value $L\geq M$ gives rise to a well-defined sequence
$\\{\tilde{a}_{n}^{(N)}\\}$ for every $N>2$.

###### Proof.

We start by proving that every $\\{\tilde{a}_{n}^{(N)}\\}$ that satisfies
(15), when well-defined, is weakly increasing. We proceed by induction on $n$.

First two base cases are easy to verify:

$\tilde{a}_{N+1}^{(N)}=L=\tilde{a}_{N}^{(N)},$
$\tilde{a}_{N-1}^{(N)}=\sqrt{\tilde{a}_{N}^{(N)}(\tilde{a}_{N+1}^{(N)}-\zeta_{n-1})}=\sqrt{L(L-\zeta_{n-1})}<L=\tilde{a}_{N}^{(N)}.$

For the inductive step, we consider by hypothesis

$\tilde{a}_{n+2}^{(N)}\geq\tilde{a}_{n+1}^{(N)}\,\,\,(i),\,\,\,\tilde{a}_{n+3}^{(N)}\geq\tilde{a}_{n+2}^{(N)}\implies\tilde{a}_{n+3}^{(N)}-\zeta_{n+1}\geq\tilde{a}_{n+2}^{(N)}-\zeta_{n}\,\,\,(ii)$

and multiplying together inequalities $(i)$ and $(ii)$ we get:

$(\tilde{a}_{n+1}^{(N)})^{2}=\tilde{a}_{n+2}^{(N)}(\tilde{a}_{n+3}^{(N)}-\zeta_{n+1})\geq\tilde{a}_{n+1}^{(N)}(\tilde{a}_{n+2}^{(N)}-\zeta_{n})=(\tilde{a}_{n}^{(N)})^{2},$

proving the claim $\tilde{a}_{n+1}^{(N)}\geq\tilde{a}_{n}^{(N)}$.

It is also immediate to verify by induction that
$\\{\tilde{a}_{n}^{(N)}\\}\leq L$ for every $n\leq N$.

Let us consider again the increasing property in the following form:

$(\tilde{a}_{n+1}^{(N)})^{2}\geq(\tilde{a}_{n}^{(N)})^{2}=\tilde{a}_{n+1}^{(N)}(\tilde{a}_{n+2}^{(N)}-\zeta_{n}),$

and dividing both sides for the positive term $\tilde{a}_{n+1}^{(N)}$ we
finally get

$\tilde{a}_{n+2}^{(N)}-\tilde{a}_{n+1}^{(N)}\leq\zeta_{n}.$

Applying a recursive argument to the inequality above it is possible to show

$\tilde{a}_{N}^{(N)}-\tilde{a}_{1}^{(N)}\leq\sum_{i=1}^{N-1}\zeta_{i}\leq\sum_{i=1}^{\infty}\zeta_{i}=M.$

We finally deduce that any initial value L satisfying

$0\leq L-M\leq\tilde{a}_{n}^{(N)}\leq L$

gives rise to a well-defined truncated sequence. In particular, it is
sufficient that $L\geq M$, concluding the proof. ∎

Proposition 3.6 tells that for every $N>2$, $\\{\tilde{a}_{n}^{(N)}\\}_{n}$
lies in the compact set $[L-M,L]$, thus by compactness and a diagonal
extraction argument we can choose a subsequence $(N_{i})_{i}\in\mathbb{N}$
such that $\tilde{a}_{n}^{(N_{i})}$ converges for all $n\in\mathbb{N}$ to some
number $\tilde{a}_{n}$. The sequence $\tilde{a}=\\{\tilde{a}_{n}\\}_{n}$
satisfies recursion (14) by construction. In order to say that
$\\{\tilde{a}_{n}\\}_{n}$ is a weak self-similar sequence it is left to prove
that $\tilde{a}_{0}=0$. The next Proposition shows how to choose the starting
value $L$ in order to achieve this goal.

######  Proposition 3.7.

Let us consider recursion (15) and a subsequence $(N_{i})_{i}\in\mathbb{N}$
such that $\tilde{a}_{n}^{(N_{i})}$ converges to $\tilde{a}_{n}$. There exists
a starting value $L>0$ such that $\tilde{a}_{n}$ is weak self-similar, in
particular

$\tilde{a}_{0}=\lim_{N_{i}\rightarrow\infty}\tilde{a}_{0}^{(N_{i})}=0$ (16)

###### Proof.

Proposition 3.6 proves that, for every $N_{i}>2$, recursion
$\tilde{a}_{n}^{(N)}$ is monotonously increasing. Thus, the idea is to the set
the starting value $L$ as low as possible to produce a suitable recursion that
proves the claim. With this regard, up to extracting one more subsequence of
$(N_{i})$, for every fixed $N_{i}>2$ consider the following

$I^{(N_{i})}=\inf\\{L>0\,\,|\,\,\text{the recursion that starts
at}\,\,\tilde{a}_{N_{i}}^{(N_{i})}=\tilde{a}_{N_{i}-1}^{(N_{i})}=L\,\,\text{is
well-defined}\\}$

Thanks to Proposition 3.6, $I^{(N_{i})}$ exists and $0<I^{(N_{i})}\leq M$.
Moreover, the sequence $I^{N_{i}}$ is non-decreasing in $N_{i}$. Indeed, if
$N_{i}<N_{j}$ then the sequences $\tilde{a}_{n}^{(N_{i})}$,
$\tilde{a}_{n}^{(N_{j})}$ built from the same starting value satisfy
$\tilde{a}_{n}^{(N_{i})}\geq\tilde{a}_{n}^{(N_{j})}$ thanks to the monotonic
property. From the latter follows that the minimum starting value from which
stems a well-defined $(N_{i})$-truncated recursion is less or equal than the
equivalent for an $(N_{j})$-truncated recursion.
We then claim that from the following starting value

$L^{*}=\sup_{N_{i}>2}I^{(N_{i})}$

stems a limit sequence $\tilde{a}_{n}$ that is weak self-similar.
First, we observe that $L^{*}$ is finite since $I^{(N_{i})}\leq M$ for every
$N_{i}>2$. The sequence $\tilde{a}_{n}$ satisfies recursion (15), hence the
only property left to prove is

$\tilde{a}_{0}=\lim_{N_{i}\rightarrow\infty}\tilde{a}_{0}^{(N_{i})}=0.$

On the contrary, suppose $\tilde{a_{0}}>\epsilon>0$ for some $\epsilon>0$ and
consider $f^{(N_{i})}:\mathbb{R}_{>0}\rightarrow\mathbb{R}_{\geq 0}$ defined
by

$f^{(N_{i})}(L)=\tilde{a}_{0}^{(N_{i})}.$

The function $f$ maps a positive number $L$ into the first element of the
truncated $(N_{i})$ recursion with starting values identical to $L$. By
hypothesis, there is an index $N^{*}$ such that
$f^{(N_{i})}(L^{*})>\epsilon>0$ for every $N_{i}>N^{*}$. Hence, by definition
of $L^{*}$ and a continuity argument, there is a small enough value $\zeta>0$
and an index $N^{\zeta}>N^{*}$ such that $f^{(N_{i})}(L^{*}-\zeta)>\epsilon>0$
for every $N_{i}>N^{\zeta}$. Since $I^{N_{i}}$ is monotonous, the latter would
imply that

$L^{*}=\sup_{N_{i}>2}I^{N_{i}}\leq L^{*}-\zeta,$

that is an absurd.

∎

Proposition 3.7 proves that $\\{\tilde{a}_{n}\\}_{n}$ is a weak self-similar
sequence, and $a=\\{a_{n}=\frac{\tilde{a}_{n}}{2^{n/3}}\\}_{n}$ is the
corresponding strong self-similar sequence.

Furthermore, we observe that the condition

$M=\sum_{i=1}^{\infty}\zeta_{i}<\infty$

is equivalent to

$\sum_{n=1}^{\infty}\epsilon_{n}\cdot
2^{n/3}=\sum_{n=1}^{\infty}2^{-2n/3}<\infty,$

as required by Proposition 3.5.

Step (2): Regularity.

We are now ready to prove that any strong self-similar sequence
$a=\\{a_{n}\\}_{n\in\mathbb{N}}$ has finite energy, i.e.

$\sum_{n=1}^{\infty}a_{n}^{2}<\infty.$

Moreover, such $\\{a_{n}\\}$ lies in $H^{s}$ for any $s<1/3$.

######  Proposition 3.8.

For every $s<1/3$ any well-defined strong self-similar sequence has finite
energy and lies in $H^{s}$.

###### Proof.

In Step (1) we have already shown that any weak self-similar sequence built
from $L$ satisfies

$0\leq L-M\leq\tilde{a}_{n}\leq L.$

By recovering the correct expression for the related strong self-similar
sequence, we derive

$\sum_{n=1}^{\infty}2^{2sn}\cdot
a^{2}_{n}=\sum_{n=1}^{\infty}2^{2sn}\cdot\frac{\tilde{a}_{n}}{2^{2n/3}}\leq
L\cdot\sum_{n=1}^{\infty}2^{n(2s-2/3)}.$

Finally, from the latter equation it follows that any strong self-similar
sequence built from $L$ lies in $H^{s}$ for every $s<1/3$. ∎

Step (3): Uniqueness.

We now prove uniqueness among strong self-similar sequence with finite energy.

######  Proposition 3.9.

Let $\\{b_{n}\\}_{n}$ be a solution of recursion (8) different from
$\\{a_{n}\\}_{n}$. Then exists $\alpha>0$ such that for every $n\geq 3$:

$\begin{split}&b_{n}\geq a_{n}\cdot 2^{\alpha
n},\,\,\,\,\,\,\,\,\,\,n\,\,\,odd\\\ &b_{n}\leq a_{n}\cdot 2^{-\alpha
n},\,\,\,\,\,\,n\,\,\,even\end{split}$

or

$\begin{split}&b_{n}\geq a_{n}\cdot 2^{\alpha
n},\,\,\,\,\,\,\,\,\,\,n\,\,\,even\\\ &b_{n}\leq a_{n}\cdot 2^{-\alpha
n},\,\,\,\,\,\,n\,\,\,odd\end{split}$

###### Proof.

We prove only the first case of the proposition, the second being similar. We
start by considering odd values of $n$ and finally even values.

Case (1): $n$ odd. By induction over $n$. By hypothesis $\\{b_{n}\\}_{n}$ is
different from $\\{a_{n}\\}_{n}$, so without loss of generality we can suppose
$b_{3}>a_{3}$ and the existence of a real number $\alpha_{1}>0$ so that

$b_{3}\geq a_{3}\cdot 2^{3\alpha_{1}},$

moreover, by Definition 3.5

$b_{4}=\frac{b_{2}^{2}}{2b_{3}}+\epsilon_{3}<\frac{a_{2}^{2}}{2a^{3}}+\epsilon_{3}=a_{4},$

thus there exists also a real number $\alpha_{2}>0$ so that

$b_{4}\leq a_{4}\cdot 2^{-4\alpha_{2}},$

finally, by setting $\alpha=\min\\{\alpha_{1},\alpha_{2}\\}$ we have proved
base cases of induction.
If $n$ is an odd number, then by hypothesis we have

$b_{n+1}=\frac{b^{2}_{n-1}}{2b_{n}}+\epsilon_{n}\geq\frac{2^{2\alpha(n-1)}a^{2}_{n-1}}{2^{-\alpha
n}(2a_{n})}+\epsilon_{n}.$

In what follows we will show that

$\frac{2^{2\alpha(n-1)}a^{2}_{n-1}}{2^{-\alpha n}(2a_{n})}+\epsilon_{n}\geq
2^{\alpha(n+1)}(\frac{a^{2}_{n-1}}{2a_{n}}+\epsilon_{n})=2^{\alpha(n+1)}a_{n+1},$

concluding the proof.
Let us first rewrite the latter inequality in the more compact form

$(a_{n+1}-\epsilon_{n})\cdot(2^{3\alpha n-2\alpha}-2^{\alpha
n+\alpha})\geq\epsilon_{n}\cdot(2^{\alpha n+\alpha}-1).$ (17)

We structure the proof in two different steps.

Step (1): $a_{n+1}\geq\epsilon_{n-1}$.

Let’s rewrite the claim in terms of the corresponding $\tilde{a}_{n}$ weak
sequence:

$a_{n+1}\geq\epsilon_{n-1}\Longleftrightarrow\frac{\tilde{a}_{n+1}}{2^{(n+1)/3}}\geq\epsilon_{n-1}\Longleftrightarrow\tilde{a}_{n+1}\geq
2^{\frac{-(2n-4)}{3}}.$

In the latter inequality the left side in an increasing function of $n$ while
the right side is decreasing, thus it is enough to verify the inequality for
the smallest meaningful value of odd $n$, i.e. $n=5$:

$\tilde{a}_{6}\geq 2^{-2}.$

In what follows we are going to prove the stronger relation:

$\tilde{a}_{4}\geq 2^{-2}.$

The latter is equivalent to $a_{4}\geq 2^{-\frac{10}{3}}$ for the related
strong self-similar sequence. We recall that without loss of generality we
have chosen $a_{0}=0$ and $a_{1}\in\mathbb{R}$ and, by a direct calculation,
thanks to (13) we derive

$a_{4}=\frac{1}{8}\cdot\frac{a_{1}^{2}+\frac{3}{2}}{a_{1}^{2}+\frac{1}{2}}.$

Finally

$a_{4}\geq
2^{-\frac{10}{3}}\iff\frac{1}{8}\cdot\frac{a_{1}^{2}+\frac{3}{2}}{a_{1}^{2}+\frac{1}{2}}\geq
2^{-\frac{10}{3}}\iff\frac{a_{1}^{2}+\frac{3}{2}}{a_{1}^{2}+\frac{1}{2}}\geq
2^{-\frac{1}{3}},$

with the last inequality been true for every choice $a_{1}\in\mathbb{R}$.

Step (2): $(2^{3\alpha n-2\alpha}-2^{\alpha n+\alpha})\geq(2^{\alpha
n+\alpha}-1)$.

First we rewrite above inequality in the form

$2^{3\alpha n-2\alpha}+1\geq 2^{\alpha n+\alpha+1},$

then observing that both sides are increasing function of $n$ and left sides
grows faster than right sides, it is again enough to prove the claim for the
smallest meaningful value $n=5$.
By letting $n=5$ we obtain:

$\begin{split}2^{13\alpha}-2^{6\alpha+1}+1=(2^{\alpha}-1)\cdot[&(2^{12\alpha}-2^{5\alpha})+(2^{11\alpha}-2^{4\alpha})+(2^{10\alpha}-2^{3\alpha})+\\\
&(2^{9\alpha}-2^{2\alpha})+(2^{8\alpha}-2^{\alpha})+(2^{7\alpha}-1)+2^{6\alpha}]>0\end{split}$

because it is a product of positive numbers. By multiplying together
inequalities in Step (1) and Step (2) one can derive (17).

Case (2): $n$ even. By induction over $n$. In the previous case we have
already shown that exists a real number $\alpha>0$ so that

$b_{3}\geq a_{3}\cdot 2^{3\alpha},\,\,\,b_{4}\leq a^{4}\cdot 2^{-4\alpha}.$

If $n$ is an even number, then by hypothesis we have

$b_{n+1}=\frac{b_{n-1}^{2}}{2b_{n}}+\epsilon_{n}\leq\frac{2^{-2\alpha(n+1)}\cdot
a_{n-1}^{2}}{2^{\alpha n}a_{n}}+\epsilon_{n}.$

We will now show that

$\frac{2^{-2\alpha(n+1)}\cdot a_{n-1}^{2}}{2^{\alpha n}\cdot
a_{n}}+\epsilon_{n}\leq
2^{-\alpha(n+1)}(\frac{a_{n-1}^{2}}{2a_{n}}+\epsilon_{n})=2^{-\alpha(n+1)}a_{n+1}.$

concluding the proof.

First, we rewrite inequality above as follows:

$(a_{n+1}-\epsilon_{n})(2^{-3\alpha n-2\alpha}-2^{-\alpha
n-\alpha})\leq\epsilon_{n}(2^{-\alpha n-\alpha}-1).$ (18)

In the previous case we have already shown that $a_{n+1}\geq\epsilon_{n-1}$,
thus as a fortiori argument we have the following:

$2^{n}a_{n+1}\geq\epsilon_{n-1}.$

Moreover, observing that for every $\alpha>0$

$(2^{-3\alpha n-2\alpha}-2^{-\alpha n-\alpha})<0,\,\,\,\,(2^{-\alpha
n-\alpha}-1)<0,$

in order to prove (18) it is enough to show that

$2^{-n}(2^{-3\alpha n-2\alpha}-2^{-\alpha n-\alpha})\geq(2^{-\alpha
n-\alpha}-1),$

or equivalently

$2^{3\alpha n+2\alpha+n}+1\geq 2^{2\alpha n+\alpha}(2^{n}+1).$

Both sides are increasing functions of $n$ and left side increases faster than
right side, so it is enough to prove the claim for the smallest admissible
even $n$, i.e. $n=6$. Namely, we need to prove

$2^{20\alpha+6}+1\geq 2^{13\alpha+6}+2^{13\alpha}.$

Let’s consider the function

$f(\alpha)=2^{20\alpha+6}+1-2^{13\alpha+6}-2^{13\alpha}.$

It is easy to notice that $f(0)=0$ and $f(x)$ has positive derivative on the
positive x-asis, namely

$\frac{df}{dx}=5\cdot 2^{13}\cdot\log_{2}(2^{7x+8}-169)>0,\,\,\,\,x\geq 0,$

this proves the claim. ∎

Proposition 3.9 tells us that every solution $b_{n}$ different from $a_{n}$
cannot have finite energy. Moreover, any other solution except $a_{n}$ cannot
lie in any space $H^{s}$ even for negative values of $s$. ∎

We now observe that Theorem 2.3 is an immediate consequence of Proposition
3.5.

##  4 Proof of Theorem 2.2

This section is entirely devoted to proving Theorem 2.2.

Let us start by considering the following recursive sequence:

$\begin{split}&b_{0}=C>0,\\\
&b_{n+1}=\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}b_{n}^{-2}+4\delta_{2}^{2}b_{n}^{-1}}}{2\delta_{2}}\end{split}$
(19)

for some positive starting value $C>0$ and some positive coefficient
$\delta_{1},\delta_{2}$ such that $\delta_{1}/\delta_{2}<k_{1}^{-4/3}$.
Lemma 4.10 tells useful information about the sequence $\\{b_{n}\\}_{n}$ and
its asymptotic behaviour.

######  Lemma 4.10.

For every starting value $C>0$, and positive coefficient
$\delta_{1},\delta_{2}$ such that $\delta_{1}/\delta_{2}<k_{1}^{-4/3}$, the
recursive sequence (19) satisfies

$\lim_{n\rightarrow\infty}b_{n}=1.$

###### Proof.

Since recursion (19) admits $1$ as unique fixed point, we first observe that
if $C=1$ then $b_{n}\equiv 1$ for every $n\geq 0$.

Without loss of generality let us suppose $C<1$ (the case $C>1$ being
specular). We will prove the following properties

  1. 1.

$b_{2n+1}>1,b_{2n}<1$, $\forall n\geq 0$;

  2. 2.

$1<b_{2n+1}<b_{2n-1}$ and $0<b_{2n}<b_{2n+2}<1$, $\forall n\geq 0$;

  3. 3.

$\lim_{n\rightarrow\infty}b_{2n+1}=\lim_{n\rightarrow\infty}b_{2n}=1$,

the statement will follow trivially.

We start observing that $b_{1}>1$:

$\begin{split}b_{1}&=\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}C^{-2}+4\delta_{2}^{2}C^{-1}}}{2\delta_{2}}\\\
&>\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}+4\delta_{2}^{2}}}{2\delta_{2}}=1.\end{split}$

Let’s now suppose $b_{2n-1}>1$ for some $n>0$. Then we have

$b_{2n+1}=\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}b_{2n}^{-2}+4\delta_{2}^{2}b_{2n}^{-1}}}{2\delta_{2}},$

moreover, by inductive hypothesis and definition

$\begin{split}b_{2n}=&\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}b_{2n-1}^{-2}+4\delta_{2}^{2}b_{2n-1}^{-1}}}{2\delta_{2}}\\\
<&\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}+4\delta_{2}^{2}}}{2\delta_{2}}=1,\end{split}$

hence $b_{2n}^{-1}>1$, and finally

$\begin{split}b_{2n+1}=&\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}b_{2n}^{-2}+4\delta_{2}^{2}b_{2n}^{-1}}}{2\delta_{2}}\\\
&>\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}+4\delta_{2}^{2}}}{2\delta_{2}}=1,\end{split}$

proving property (1).
We now focus on the first part of property (2) (the second being identical).
By definition we can write

$\begin{split}b_{2n+1}<b_{2n-1}\iff&\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}b_{2n}^{-2}+4\delta_{2}^{2}b_{2n}^{-1}}}{2\delta_{2}}<b_{2n-1}\\\
\iff&4\delta_{1}\delta_{2}k_{1}^{4/3}b_{2n}^{-2}+4\delta_{2}^{2}b_{2n}^{-1}<4b_{2n-1}^{2}\delta_{2}^{2}+4\delta_{1}\delta_{2}b_{2n-1}k_{1}^{4/3}\\\
\iff&\delta_{1}k_{1}^{4/3}b_{2n}^{-2}+\delta_{2}b_{2n}^{-1}<b_{2n-1}^{2}\delta_{1}+\delta_{2}b_{2n-1}k_{1}^{4/3}\\\
\iff&\delta_{1}k_{1}^{4/3}(b_{2n}^{-2}-b_{2n-1})<\delta_{2}(b_{2n-1}^{2}-b_{2n}^{-1}).\end{split}$

By hypothesis, we set $\delta_{1}k_{1}^{4/3}<\delta_{2}$, thus it is enough to
require

$b_{2n}^{-2}-b_{2n-1}<b_{2n-1}^{2}-b_{2n}^{-1}$

within the positive condition on the right side $0<b_{2n-1}^{2}-b_{2n}^{-1}$.
We observe that the above two inequalities are both satisfied if
$b_{2n}^{-1}<b_{2n-1}$, indeed:

$b_{2n}^{-1}<b_{2n-1}\implies b_{2n}^{-1}+b_{2n}^{-2}<b_{2n-1}+b_{2n-1}^{2}$

and

$b_{2n}^{-1}<b_{2n-1}\implies b_{2n}^{-2}<b_{2n-1}^{2}\implies
0<b_{2n}^{-1}<b_{2n}^{-2}<b_{2n-1}^{2},$

the latter being true due to $b_{2n}<1$.

We are now left to prove the sufficient condition $b_{2n}^{-1}<b_{2n-1}$ or
equivalently

$\begin{split}&b_{2n}^{-1}=\frac{2\delta_{2}}{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}b_{2n-1}^{-2}+4\delta_{2}^{2}b_{2n-1}^{-1}}}<b_{2n-1}\\\
&\iff\frac{2\delta_{2}}{b_{2n-1}}+\delta_{1}k_{1}^{4/3}<\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}b_{2n-1}^{-2}+4\delta_{2}^{2}b_{2n-1}^{-1}}\\\
&\iff
4\delta_{1}^{2}b_{2n-1}^{-2}+4\delta_{1}\delta_{2}k_{1}^{4/3}b_{2n-1}^{-1}<4\delta_{1}\delta_{2}k_{1}^{4/3}b_{2n-1}^{-2}+4\delta_{2}^{2}b_{2n-1}^{-1}\\\
&\iff\delta_{2}b_{2n-1}^{-2}+\delta_{1}k_{1}^{4/3}b_{2n-1}^{-1}<\delta_{1}k_{1}^{4/3}b_{2n-1}^{-2}+\delta_{2}b_{2n-1}^{-1}\\\
&\iff\delta_{2}+\delta_{1}k_{1}^{4/3}b_{2n-1}<\delta_{1}k_{1}^{4/3}+\delta_{2}b_{2n-1}\\\
&\iff(\delta_{2}-\delta_{1}k_{1}^{4/3})<(\delta_{2}-\delta_{1}k_{1}^{4/3})\cdot
b_{2n-1},\end{split}$

finally the latter inequality holds because
$\delta_{2}-\delta_{1}k_{1}^{4/3}>0$ and $b_{2n-1}>1$.
We can now say that $b_{2n+1}$ admits limit
$\lim_{n\rightarrow\infty}b_{2n+1}=L\geq 1$. Suppose $L>1$, then

$L=\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}g(L)^{-2}+4\delta_{2}^{2}g(L)^{-1}}}{2\delta_{2}}$
(20)

where

$g(L)=\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}L^{-2}+4\delta_{2}^{2}L^{-1}}}{2\delta_{2}}$

By a direct calculation equation (20) holds if and only if

$\begin{split}&4\delta_{2}^{2}L^{2}+4\delta_{1}\delta_{2}k_{1}^{4/3}L=4\delta_{1}\delta_{2}k_{1}^{4/3}g(L)^{-2}+4\delta_{2}^{2}g(L)^{-1}\\\
&\iff\delta_{2}L^{2}+\delta_{1}k_{1}^{4/3}L=\delta_{1}k_{1}^{4/3}g(L)^{-2}+\delta_{2}g(L)^{-1}\\\
&\iff\delta_{2}(L^{2}-g(L)^{-1})=\delta_{1}k_{1}^{4/3}(g(L)^{-2}-L).\end{split}$

Let’s take a closer look at the last equation. By hypothesis
$\delta_{2}>\delta_{1}k_{1}^{4/3}$, so just one of the following could hold:

  * 1.

$L^{2}-g(L)^{-1}>0$ and $(L^{2}-g(L)^{-1})<(g(L)^{-2}-L)$:

from the second inequality we recover $(L^{2}+L)<(g(L)^{-2}+g(L)^{-1})$.
Thanks to the properties we have already proved, it is not hard from the
latter to deduce $L<g(L)^{-1}$. It is now time to expand the right-hand side
to obtain:

$\begin{split}&L<g(L)^{-1}=\frac{2\delta_{2}}{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}L^{-2}+4\delta_{2}^{2}L^{-1}}}\\\
&\iff
4\delta_{1}\delta_{2}k_{1}^{4/3}+4\delta_{2}^{2}L<4\delta_{2}^{2}+4\delta_{1}k_{1}^{4/2}L\\\
&\iff
L(\delta_{2}-\delta_{1}k_{1}^{4/3})<(\delta_{2}-\delta_{1}k_{1}^{4/3})\iff
L<1\end{split}$

that is absurd.

  * 2.

$L^{2}-g(L)^{-1}<0$ and $(L^{2}-g(L)^{-1})>(g(L)^{-2}-L)$:

again, from our assumptions:

$\begin{split}&L^{2}<g(L)^{-1}\iff
L^{2}<\frac{2\delta_{2}}{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}L^{-2}+4\delta_{2}^{2}L^{-1}}}\\\
&\iff\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}L^{-2}+4\delta_{2}^{2}L^{-1}<(2\delta_{2}/L^{2}+\delta_{1}k_{1}^{4/3})^{2}\\\
&\iff
4\delta_{1}\delta_{2}k_{1}^{4/3}L^{-2}+4\delta_{2}^{2}L^{-1}<4\delta_{2}^{2}L^{-4}+4\delta_{1}\delta_{2}k_{1}^{4/3}L^{-2}\\\
&\iff L^{3}<1\iff L<1\\\ \end{split}$

again against our assumption.

  * 3.

$L^{2}=g(L)^{-1}$ and $L=g(L)^{-2}$:

if $x,y$ are two positive real numbers such that

$x^{2}=y,\,\,\,\,x=y^{2},$

the only solution is $x=y=1$.

We conclude that $L=\lim_{n\rightarrow\infty}{b_{2n+1}}=1$. With same argument
one can show $\lim_{n\rightarrow\infty}{b_{2n}}=1$, concluding the proof. ∎

We now state and prove an equivalent for Lemma 4.10 when
$\delta_{1}/\delta_{2}>k_{1}^{-4/3}$.
First, for every $N>1$ consider the following recursive backward sequence
$\\{(b^{*}_{n})^{(N)}\\}_{n}$:

$\begin{split}&(b^{*}_{N})^{(N)}=C^{*}>0,\\\
&(b^{*}_{n})^{(N)}=\frac{-\delta_{2}k_{1}^{-4/3}+\sqrt{\delta_{2}^{2}k_{1}^{-8/3}+4\delta_{1}\delta_{2}k_{1}^{-4/3}((b^{*}_{n+1})^{(N)})^{-2}+4\delta_{1}^{2}((b^{*}_{n+1})^{(N)})^{-1}}}{2\delta_{1}}\end{split}$
(21)

for any $0\leq n<N$, some positive starting value $C^{*}>0$ and some positive
coefficient $\delta_{1},\delta_{2}$ such that
$\delta_{1}/\delta_{2}>k_{1}^{-4/3}$.
Lemmas 4.11 and 4.12 tell useful information about sequence
$\\{(b^{*}_{n})^{(N)}\\}_{n}$ and its asymptotic behaviour.

######  Lemma 4.11.

For every starting value $C^{*}>0$, any $N>1$ and any positive coefficients
$\delta_{1}$, $\delta_{2}$ such that $\delta_{1}/\delta_{2}>k_{1}^{-4/3}$, the
recursive sequence $\\{(b_{n}^{*})^{(N)}\\}_{n}$ defined above satisfies

  1. 1.

$0<(b_{N}^{*})^{(N)}<(b_{n}^{*})^{(N)}<(b_{N-1}^{*})^{(N)}<\frac{1}{C^{*}}$,
if $C^{*}<1$, $\forall n\geq 1$.

  2. 2.

$0<(b_{N-1}^{*})^{(N)}<(b_{n}^{*})^{(N)}<(b_{N}^{*})^{(N)}=C^{*}$, if
$C^{*}>1$, $\forall n\geq 1$.

  3. 3.

$(b_{n}^{*})^{(N)}\equiv 1$, if $C^{*}=1$, $\forall n\geq 1$.

###### Proof.

This proof is entirely equivalent to the one we proposed for Lemma 4.10 by
swapping $\delta_{1}$ and $\delta_{2}$ coefficients. The only statement left
to prove is

$(b_{N-1}^{*})^{(N)}<\frac{1}{C^{*}},\,\,\,\text{if}\,\,\,C^{*}<1.$

By a direct calculation we have

$\begin{split}(b_{N-1}^{*})^{(N)}<\frac{1}{C^{*}}\iff&\frac{-\delta_{2}k_{1}^{-4/3}+\sqrt{\delta_{2}^{2}k_{1}^{-8/3}+4\delta_{1}\delta_{2}k_{1}^{-4/3}/C^{*2}+4\delta_{1}^{2}/C^{*}}}{2\delta_{1}}<\frac{1}{C^{*}}\\\
\iff&\delta_{2}^{2}k_{1}^{-8/3}+\frac{4\delta_{1}\delta_{2}k_{1}^{-4/3}}{C^{*2}}+\frac{4\delta_{1}^{2}}{C^{*}}<\delta_{2}^{2}k_{1}^{-8/3}+\frac{4\delta_{1}\delta_{2}k_{1}^{-4/3}}{C^{*}}+\frac{4\delta_{1}^{2}}{C^{*2}}\\\
\iff&\frac{\delta_{2}k_{1}^{-4/3}}{C^{*2}}+\frac{\delta_{1}}{C^{*}}<\frac{\delta_{2}k_{1}^{-4/3}}{C^{*}}+\frac{\delta_{1}}{C^{*2}}\\\
\iff&C^{*}(\delta_{1}-\delta_{2}k_{1}^{-4/3})<(\delta_{1}-\delta_{2}k_{1}^{-4/3})\iff
C^{*}<1,\end{split}$

due to the assumption $\delta_{1}/\delta_{2}>k_{1}^{-4/3}$. ∎

Lemma 4.11 tells that for every $N>1$, $\\{(b^{*}_{n})^{(N)}\\}$ lies in the
compact set $[0,C^{*}]$ if $C^{*}\geq 1$ or $[0,1/C^{*}]$ if $C^{*}<1$, thus
by compactness and a diagonal extraction argument we can choose a subsequence
$(N_{i})_{i}\in\mathbb{N}$ such that $(b_{n}^{*})^{(N_{i})}$ converges for all
$n\in\mathbb{N}$ to some number $\tilde{b}_{n}^{*}$. The sequence
$\tilde{b}=\\{\tilde{b}_{n}^{*}\\}_{n}$ satisfies the following equation by
construction

$\tilde{b}^{*}_{n}=\frac{-\delta_{2}k_{1}^{-4/3}+\sqrt{\delta_{2}^{2}k_{1}^{-8/3}+4\delta_{1}\delta_{2}k_{1}^{-4/3}(\tilde{b}^{*}_{n+1})^{-2}+4\delta_{1}^{2}(\tilde{b}^{*}_{n+1})^{-1}}}{2\delta_{1}}.$
(22)

######  Lemma 4.12.

For every starting value $C^{*}>0$, and positive coefficient
$\delta_{1},\delta_{2}$ such that $\delta_{1}/\delta_{2}>k_{1}^{-4/3}$, the
recursive sequence (22) satisfies

$\lim_{n\rightarrow\infty}\tilde{b}_{n}^{*}=1.$

###### Proof.

It is equivalent to the proof of Lemma 4.10 by swapping $\delta_{1}$ and
$\delta_{2}$ coefficients. ∎

We are now ready to prove Theorem 2.2.

Let us start by considering equation (5) written in the following form

$0=\delta_{1}(a_{n-1}^{2}-k_{1}a_{n}a_{n+1})-\delta_{2}(a_{n+1}^{2}-k_{1}^{-1}a_{n}a_{n-1}).$

We already focus our interest into positive solutions with no zero term, so
dividing by $a_{n}$ both sides and changing variable with
$b_{n}=\frac{a_{n}}{a_{n-1}}$ we obtain

$0=\delta_{1}(b_{n}^{-2}-k_{1}b_{n+1})-\delta_{2}(b_{n+1}^{2}-k_{1}^{-1}b_{n}^{-1}).$

We now apply a further change of variable $a_{n}=\tilde{a}_{n}/k_{n}^{1/3}$
and consequently $b_{n}=\tilde{b}_{n}/k_{1}^{1/3}$ to finally get

$0=\delta_{1}(k_{1}^{2/3}\tilde{b}_{n}^{-2}-k_{1}^{2/3}\tilde{b}_{n+1})-\delta_{2}(k_{1}^{-2/3}\tilde{b}_{n+1}^{2}-k_{1}^{-2/3}\tilde{b}_{n}^{-1}).$

We can solve the above equation of degree two restricting ourselves only to
positive solutions

$\begin{split}&\tilde{b}_{0}=a_{1}/F>0,\\\
&\tilde{b}_{n+1}=\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{n}^{-2}+4\delta_{2}^{2}\tilde{b}_{n}^{-1}}}{2\delta_{2}}.\end{split}$
(23)

Lemma 4.10 shows that $\lim_{n\rightarrow\infty}\tilde{b}_{n}=1$ every time
$\delta_{1}/\delta_{2}<k_{1}^{-4/3}$. Thus,
$\lim_{n\rightarrow\infty}b_{n}=k_{1}^{-1/3}$ and
$\lim_{n\rightarrow\infty}a_{n}/a_{n-1}=k_{1}^{-1/3}<1$, proving Theorem 2.2
in the case $\delta_{1}/\delta_{2}<k_{1}^{-4/3}$.
In the same fashion, one can consider a backward change of variable
$b_{n}=\frac{a_{n-1}}{a_{n}}$ and obtain

$0=\delta_{1}(b_{n}^{2}-k_{1}b_{n+1}^{-1})-\delta_{2}(b_{n+1}^{-2}-k_{1}^{-1}b_{n}).$

We now apply a further change of variable $a_{n}=\tilde{a}_{n}/k_{n}^{1/3}$
and consequently $b_{n}=\tilde{b}_{n}/k_{1}^{1/3}$ and finally get

$0=\delta_{1}(k_{1}^{2/3}\tilde{b}_{n}^{2}-k_{1}^{2/3}\tilde{b}_{n+1}^{-1})-\delta_{2}(k_{1}^{-2/3}\tilde{b}_{n+1}^{-2}-k_{1}^{-2/3}\tilde{b}_{n}).$

As before we can solve the above equation of degree two restricting ourselves
only to positive solution

$\begin{split}\tilde{b}_{0}&=a_{0}/a_{1}>0\\\
\tilde{b}_{n}&=\frac{-\delta_{2}k_{1}^{-4/3}+\sqrt{\delta_{2}^{2}k_{1}^{-8/3}+4\delta_{1}\delta_{2}k_{1}^{-4/3}\tilde{b}_{n+1}^{-2}+4\delta_{1}^{2}\tilde{b}_{n+1}^{-1}}}{2\delta_{1}}.\end{split}$
(24)

Lemma 4.11 states $\lim_{n\rightarrow\infty}\tilde{b}_{n}=1$ every time
$\delta_{1}/\delta_{2}>k_{1}^{-4/3}$. Thus
$\lim_{n\rightarrow\infty}b_{n}=k_{1}^{-1/3}$ and
$\lim_{n\rightarrow\infty}a_{n-1}/a_{n}=k_{1}^{1/3}>1$, proving Theorem 2.2
also in the case $\delta_{1}/\delta_{2}>k_{1}^{-4/3}$.

##  5 Proof of Theorem 2.4

Now that we have successfully proved Theorem 2.2, we observe that equation
(10) for self-similar sequences differs from equation (5) for constant
solution only by a perturbation term $\frac{a_{n}}{k_{n}}$. Thus, we adapt our
proof to take care of this extra term.

Indeed, without loss of generality, we can set $a_{0}=0$ and $a_{n}>0$ for
every $n>0$ in equation (10), then dividing by $a_{n}$ both sides and changing
variable with $b_{n}=\frac{a_{n}}{a_{n-1}}$ we obtain

$\delta_{1}(b_{n}^{-2}-k_{1}b_{n+1})-\delta_{2}(b_{n+1}^{2}-k_{1}^{-1}b_{n}^{-1})-\frac{1}{a_{n}k_{n}}=0.$

We now apply a further change of variable $a_{n}=\tilde{a}_{n}/k_{n}^{1/3}$
and consequently $b_{n}=\tilde{b}_{n}/k_{1}^{1/3}$ and finally get

$\delta_{1}(k_{1}^{2/3}\tilde{b}_{n}^{-2}-k_{1}^{2/3}\tilde{b}_{n+1})-\delta_{2}(k_{1}^{-2/3}\tilde{b}_{n+1}^{2}-k_{1}^{-2/3}\tilde{b}_{n}^{-1})-\frac{\tilde{a}_{n}}{k_{n}^{2/3}}=0.$

We can solve the above equation of degree two restricting ourselves only to
positive solution

$\begin{split}&\tilde{b}_{1}=a_{2}/a_{1}>0,\\\
&\tilde{b}_{n+1}=\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{n}^{-2}+4\delta_{2}^{2}\tilde{b}_{n}^{-1}+4\delta_{2}k_{1}^{2/3}\epsilon_{n}}}{2\delta_{2}},\end{split}$
(25)

where $\epsilon_{n}=(\tilde{a}_{n}k_{n}^{2/3})^{-1}=(a_{n}k_{n})^{-1}$.

In the next Lemma we prove that $\lim_{n\rightarrow\infty}\epsilon_{n}=0$.

######  Lemma 5.13.

If $\\{a_{n}\\}_{n}$ is a positive self-similar sequence satisfying equation
(7) with $k_{1}^{-4}\leq\delta_{1}/\delta_{2}\leq 1$, then

$\lim_{n\rightarrow\infty}a_{n}k_{n}=\infty.$

###### Proof.

Let’s consider a change of variable $a_{n}=c_{n}/k_{n}$ in equation (7) to
obtain

$-c_{n}=\delta_{1}(k_{1}^{2}c_{n-1}^{2}-c_{n}c_{n+1})-\delta_{2}(k_{1}^{-2}c_{n+1}^{2}-c_{n}c_{n-1}).$

It is helpful to express $c_{n+1}$ as function of previous terms

$c_{n+1}=\frac{-\delta_{1}c_{n}+\sqrt{\delta_{1}^{2}c_{n}^{2}+4\delta_{2}k_{1}^{-2}c_{n}+4\delta_{1}\delta_{2}c_{n-1}^{2}+4\delta_{2}^{2}k_{1}^{-2}c_{n}c_{n-1}}}{2\delta_{2}k_{1}^{-2}}.$

We first prove that $c_{n+1}>c_{n-1}$ for every $n\geq 1$.

Indeed, we have

$\begin{split}&c_{n+1}=\frac{-\delta_{1}c_{n}+\sqrt{\delta_{1}^{2}c_{n}^{2}+4\delta_{2}k_{1}^{-2}c_{n}+4\delta_{1}\delta_{2}c_{n-1}^{2}+4\delta_{2}^{2}k_{1}^{-2}c_{n}c_{n-1}}}{2\delta_{2}k_{1}^{-2}}>c_{n-1}\\\
&\iff
4\delta_{2}k_{1}^{-2}c_{n}+4\delta_{1}\delta_{2}c_{n-1}^{2}+4\delta_{2}^{2}k_{1}^{-2}c_{n}c_{n-1}>4\delta_{2}^{2}k_{1}^{-4}c_{n-1}^{2}+4\delta_{1}\delta_{2}k_{1}^{-2}c_{n}c_{n-1}\\\
&\iff
k_{1}^{-2}c_{n}+\delta_{1}c_{n-1}^{2}+\delta_{2}k_{1}^{-2}c_{n}c_{n-1}>\delta_{2}k_{1}^{-4}c_{n-1}^{2}+\delta_{1}k_{1}^{-2}c_{n}c_{n-1}\\\
&\iff
k_{1}^{-2}c_{n}+c_{n-1}^{2}(\delta_{1}-\delta_{2}k_{1}^{-4})+k_{1}^{-2}c_{n}c_{n-1}(\delta_{2}-\delta_{1})>0,\end{split}$
(26)

and the last inequality holds thanks to the assumption
$\delta_{2}k_{1}^{-4}\leq\delta_{1}\leq\delta_{2}$.

Thus, there is a positive value $D>0$ so that $c_{n}\geq D$ for every $n\geq
1$.

With a similar argument it is possible to prove the existence of $M>1$ so that
$c_{n+1}>c_{n-1}\cdot M$, which will conclude the proof.
Indeed,

$\begin{split}&c_{n+1}=\frac{-\delta_{1}c_{n}+\sqrt{\delta_{1}^{2}c_{n}^{2}+4\delta_{2}k_{1}^{-2}c_{n}+4\delta_{1}\delta_{2}c_{n-1}^{2}+4\delta_{2}^{2}k_{1}^{-2}c_{n}c_{n-1}}}{2\delta_{2}k_{1}^{-2}}>c_{n-1}\cdot
M\\\ &\iff
k_{1}^{-2}c_{n}+\delta_{1}c_{n-1}^{2}+\delta_{2}k_{1}^{-2}c_{n}c_{n-1}>\delta_{2}k_{1}^{-4}M^{2}c_{n-1}^{2}+\delta_{1}k_{1}^{-2}Mc_{n}c_{n-1}.\end{split}$

Last inequality further simplifies as follows

$\begin{split}&k_{1}^{-2}c_{n}+c_{n-1}^{2}(\delta_{1}-\delta_{2}k_{1}^{-4}M^{2})+k_{1}^{-2}c_{n}c_{n-1}(\delta_{2}-\delta_{1}M)>0\\\
&\iff
k_{1}^{-2}D+D^{2}(\delta_{1}+k_{1}^{-2}\delta_{2})(1-k_{1}^{-2}M)>0.\end{split}$

Finally, by hypothesis $\lambda>1$ and $\beta>0$, hence it is possible to
choose $1<M\leq k_{1}^{2}$ in the latter relation, so that left-hand sides
becomes a sum of positive term. ∎

######  Remark.

We notice that upper and lower bounds on Theorem 2.4 arise from inequality
(26). Condition $\delta_{2}k_{1}^{-4}\leq\delta_{1}\leq\delta_{2}$ is
sufficient in order to satisfy inequality (26), although the first term
$k_{1}^{-2}c_{n}$ gives a small but significant positive contribute.
Consequently, upper and lower bounds for the ratio $\delta_{1}/\delta_{2}$ can
be further refined. Numerical simulation suggests the existence of a _true_
bound $L_{true}<k_{1}^{-4}$ such that theorem 2.4 holds in the wider domain
$\delta_{2}\cdot L_{true}\leq\delta_{1}$. This is consistent with Theorem 2.3
where $\delta_{1}=1$ and $\delta_{2}=0$ but also with Theorem 1 in [14] where
the author proved existence of self-similar solutions for $\delta_{1}=1$ and
$0\leq\delta_{2}<\beta_{0}$ for some small $\beta_{0}>0$.

The following Lemma is an equivalent of Lemma 4.10 for the sequence (25).

######  Lemma 5.14.

Let’s suppose $\tilde{b}_{1}=a_{2}/a_{1}=C>0$, and
$k_{1}^{-4}\leq\delta_{1}/\delta_{2}\leq k_{1}^{-4/3}$. If
$\tilde{b}_{1}\geq\tilde{b}_{3}$ then

  * 1.

$C\geq\tilde{b}_{3}\geq\tilde{b}_{5}\geq\ldots\geq\tilde{b}_{2n+1}\geq\ldots>1$
for all $n\geq 0$;

  * 2.

$0<\tilde{b}_{2}\leq\tilde{b}_{4}\leq\tilde{b}_{6}\leq\ldots\tilde{b}_{2n}\leq\ldots\leq
1+\sqrt{\epsilon_{1}\frac{k_{1}^{2/3}}{\delta_{2}}}$ for all $n\geq 1$.

Otherwise, if $\tilde{b}_{1}<\tilde{b}_{3}$ then

  * 1.

$C\leq\tilde{b}_{3}\leq\tilde{b}_{5}\leq\ldots\leq\tilde{b}_{2n+1}\leq\ldots\leq
1+\sqrt{\epsilon_{2}\frac{k_{1}^{2/3}}{\delta_{2}}}$ for all $n\geq 0$;

  * 2.

$\tilde{b}_{2}\geq\tilde{b}_{4}\geq\tilde{b}_{6}\geq\ldots\tilde{b}_{2n}\geq\ldots>1$
for all $n\geq 1$.

Moreover,
$\lim_{n\rightarrow\infty}\tilde{b}_{2n+1}=\lim_{n\rightarrow\infty}\tilde{b}_{2n}=1$.

###### Proof.

We consider only the case $\tilde{b}_{1}\geq\tilde{b}_{3}$ (the other being
specular).
Let’s first observe that

$\begin{split}\tilde{b}_{4}=&\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{3}^{-2}+4\delta_{2}^{2}\tilde{b}_{3}^{-1}+4\delta_{2}k_{1}^{2/3}\epsilon_{3}}}{2\delta_{2}}\\\
&\leq\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{3}^{-2}+4\delta_{2}^{2}\tilde{b}_{3}^{-1}+4\delta_{2}k_{1}^{2/3}\epsilon_{1}}}{2\delta_{2}}\end{split}$

thus $\tilde{b}_{2}\leq\tilde{b}_{4}$ if and only if

$\begin{split}&\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{1}^{-2}+4\delta_{2}^{2}\tilde{b}_{1}^{-1}+4\delta_{2}k_{1}^{2/3}\epsilon_{1}}\\\
&\leq\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{3}^{-2}+4\delta_{2}^{2}\tilde{b}_{3}^{-1}+4\delta_{2}k_{1}^{2/3}\epsilon_{1}}\\\
&\iff
4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{1}^{-2}+4\delta_{2}^{2}\tilde{b}_{1}^{-1}\leq
4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{3}^{-2}+4\delta_{2}^{2}\tilde{b}_{3}^{-1}\\\
&\iff\delta_{1}k_{1}^{4/3}\tilde{b}_{1}^{-2}+\delta_{2}\tilde{b}_{1}^{-1}\leq\delta_{1}k_{1}^{4/3}\tilde{b}_{3}^{-2}+\delta_{2}\tilde{b}_{3}^{-1}\\\
\end{split}$

and, remembering $\delta_{1}/\delta_{2}\leq k_{1}^{-4/3}$, the latter is
implied by $\tilde{b}_{1}\geq\tilde{b}_{3}$.
The following cascade of implications is then an immediate consequence

$\tilde{b}_{1}\geq\tilde{b}_{3}\implies\tilde{b}_{2}\leq\tilde{b}_{4}\implies\tilde{b}_{3}\geq\tilde{b}_{5}\implies\tilde{b}_{4}\leq\tilde{b}_{6}\implies\ldots$
$\implies\tilde{b}_{2n-1}\geq\tilde{b}_{2n+1}\implies\tilde{b}_{2n}\leq\tilde{b}_{2n+2},\,\,\,\
\forall n\geq 1.$

We now say that $\\{\tilde{b}_{2n+1}\\}_{n}$ admits a finite limit, say
$L_{1}$: with the same argument used in Lemma 4.10, thanks to Lemma 5.13 one
can easily prove $L_{1}=1$.
We now prove the upper bound

$\tilde{b}_{2n}\leq
1+\sqrt{\epsilon_{1}\frac{k_{1}^{2/3}}{\delta_{2}}},\,\,\,\ \forall n\geq 1.$

We first stress that $\tilde{b}_{1}\geq\tilde{b}_{3}$ only if
$C=\tilde{b}_{1}>1$, thus we can write

$\begin{split}\tilde{b}_{2n}=&\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{2n-1}^{-2}+4\delta_{2}^{2}\tilde{b}_{2n-1}^{-1}+4\delta_{2}k_{1}^{2/3}\epsilon_{2n-1}}}{2\delta_{2}}\\\
\leq&\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{2n-1}^{-2}+4\delta_{2}^{2}\tilde{b}_{2n-1}^{-1}}+\sqrt{4\delta_{2}k_{1}^{2/3}\epsilon_{2n-1}}}{2\delta_{2}}\\\
\leq&\frac{-\delta_{1}k_{1}^{4/3}+\sqrt{\delta_{1}^{2}k_{1}^{8/3}+4\delta_{1}\delta_{2}k_{1}^{4/3}\tilde{b}_{2n-1}^{-2}+4\delta_{2}^{2}\tilde{b}_{2n-1}^{-1}}+\sqrt{4\delta_{2}k_{1}^{2/3}\epsilon_{1}}}{2\delta_{2}}\\\
\leq&\,1+\sqrt{\epsilon_{1}\frac{k_{1}^{2/3}}{\delta_{2}}}\end{split}$

where the last inequality is a direct consequence of $C>1$ and
$b_{2n-1}>L_{1}=1$.
We know also that $\\{\tilde{b}_{2n}\\}_{n}$ admit finite limit, say $L_{2}$:
with similar argument used in Lemma 4.10 and thanks to Lemma 5.13 we conclude
$L_{2}=1$. ∎

Lemmas 4.10 and 5.14 show that $\lim_{n\rightarrow\infty}\tilde{b}_{n}=1$
every time $k_{1}^{-4}\leq\delta_{1}/\delta_{2}\leq k_{1}^{-4/3}$. Thus
$\lim_{n\rightarrow\infty}b_{n}=k_{1}^{-1/3}$ and
$\lim_{n\rightarrow\infty}a_{n}/a_{n-1}=k_{1}^{-1/3}<1$, proving Theorem 2.4
in the case $k_{1}^{-4}\leq\delta_{1}/\delta_{2}\leq k_{1}^{-4/3}$.

We now mimic again the proof of Theorem 2.2 also in the case
$k_{1}^{-4/3}<\delta_{1}/\delta_{2}\leq 1$, by considering a backward change
of variable $b_{n}=\frac{a_{n-1}}{a_{n}}$ to obtain

$-\frac{1}{a_{n}k_{n}}=\delta_{1}(b_{n}^{2}-k_{1}b_{n+1}^{-1})-\delta_{2}(b_{n+1}^{-2}-k_{1}^{-1}b_{n}).$

By applying a further change of variable $a_{n}=\tilde{a}_{n}/k_{n}^{1/3}$ and
consequently $b_{n}=\tilde{b}_{n}/k_{1}^{1/3}$ we finally get

$-\frac{1}{\tilde{a}_{n}k_{n}^{2/3}}=\delta_{1}(k_{1}^{2/3}\tilde{b}_{n}^{2}-k_{1}^{2/3}\tilde{b}_{n+1}^{-1})-\delta_{2}(k_{1}^{-2/3}\tilde{b}_{n+1}^{-2}-k_{1}^{-2/3}\tilde{b}_{n}).$

As before we can solve the above backward equation of degree two restricting
ourselves only to positive solutions. For every $N>1$ let be

$\begin{split}&\tilde{b}_{N}^{(N)}=C^{*}>0,\\\
&\tilde{b}_{n}^{(N)}=\frac{-\delta_{2}k_{1}^{-4/3}+\sqrt{\delta_{2}^{2}k_{1}^{-8/3}+4\delta_{1}\delta_{2}k_{1}^{-4/3}(\tilde{b}_{n+1}^{(N)})^{-2}+4\delta_{1}^{2}(\tilde{b}_{n+1}^{(N)})^{-1}-4\delta_{1}\epsilon_{n}^{*}}}{2\delta_{1}},\end{split}$

for any $0\leq n<N$, with
$\epsilon_{n}^{*}=\frac{1}{\tilde{a}_{n}^{(N)}k_{n}^{2/3}}=\frac{1}{a_{n}^{(N)}k_{n}}$.

Lemma 5.15 shows that there is $C^{*}>0$ so that the sequence
$\\{\tilde{b}_{n}^{(N)}\\}_{n}$ is well-defined and lies uniformly in a
compact set for every $N>0$.

######  Lemma 5.15.

For every $k_{1}^{-4/3}<\delta_{1}/\delta_{2}\leq 1$, there is $C^{*}>0$ so
that $\tilde{b}_{n}^{(N)}$ is well-defined for every $0<n\leq N$.
Moreover, there is $M^{*}>1$ and $N^{*}>0$ so that the sequence
$\\{\tilde{b}_{n}^{(N)}\\}_{n}$ satisfies

$0<\frac{1}{M^{*}}\leq\tilde{b}_{n}^{(N)}\leq M^{*}$

for every $N>N^{*}$ and every $N^{*}<n\leq N$.

###### Proof.

By definition $\frac{\tilde{a_{N-1}^{(N)}}}{\tilde{a_{N}^{(N)}}}=C^{*}>0$, and
by Lemma 5.13 we can choose the free parameter $C^{*}$ small enough so that

$4\delta_{1}\epsilon_{n}^{*}\leq
4\delta_{1}\max\\{\epsilon_{1}^{*},\epsilon_{2}^{*}\\}\leq\delta_{2}k_{1}^{-8/3}$

for every $n>0$, this implies that the square root in the expression of
sequence $\\{\tilde{b}_{n}^{(N)}\\}_{n}$ is well-defined.

We now prove the statement by induction over $n\leq N$. Indeed, for every
$C^{*}$ there is $M^{*}>1$ so that

$\frac{1}{M^{*}}\leq C^{*}=\tilde{b}_{N}^{(N)}\leq M^{*}$

Let’s suppose now $\frac{1}{M^{*}}\leq\tilde{b}_{n}^{(N)}\leq M^{*}$ for some
$n\leq N$. By definition

$\begin{split}\tilde{b}_{n-1}^{(N)}\leq
M^{*}&\iff\frac{4\delta_{1}\delta_{2}k_{1}^{-4/3}}{(\tilde{b}_{n}^{(N)})^{2}}+\frac{4\delta_{1}^{2}}{\tilde{b}_{n}^{(N)}}\leq
4\delta_{1}^{2}M^{*2}+4\delta_{1}\delta_{2}k_{1}^{-4/3}M^{*}+4\delta_{1}\epsilon_{n}^{*}\\\
&\iff\frac{\delta_{2}k_{1}^{-4/3}}{(\tilde{b}_{n}^{(N)})^{2}}+\frac{\delta_{1}}{\tilde{b}_{n}^{(N)}}\leq\delta_{1}M^{*2}+\delta_{2}k_{1}^{-4/3}M^{*}+\epsilon_{n}^{*}.\end{split}$

By hypothesis $\frac{1}{M^{*}}\leq\frac{1}{\tilde{b}_{n}^{(N)}}\leq M^{*}$, so
it is enough to prove

$\begin{split}&\delta_{2}k_{1}^{-4/3}M^{*2}+\delta_{1}M^{*2}\leq\delta_{1}M^{*2}+\delta_{2}k_{1}^{-4/3}M^{*}\\\
&\iff\delta_{2}k_{1}^{-4/3}(M^{*}-1)\leq\delta_{1}(M^{*}-1),\end{split}$

the latter being true thanks to $M^{*}>1$ and
$\delta_{1}>\delta_{2}k_{1}^{-4/3}$. This proves the right side of our claim.
Again, by definition

$\begin{split}\tilde{b}_{n-1}^{(N)}\geq\frac{1}{M^{*}}&\iff\frac{4\delta_{1}\delta_{2}k_{1}^{-4/3}}{(\tilde{b}_{n}^{(N)})^{2}}+\frac{4\delta_{1}^{2}}{\tilde{b}_{n}^{(N)}}\geq\frac{4\delta_{1}^{2}}{M^{*2}}+\frac{4\delta_{1}\delta_{2}k_{1}^{-4/3}}{M^{*}}+4\delta_{1}\epsilon_{n}^{*}\\\
&\iff\frac{\delta_{2}k_{1}^{-4/3}}{(\tilde{b}_{n}^{(N)})^{2}}+\frac{\delta_{1}}{\tilde{b}_{n}^{(N)}}\geq\frac{\delta_{1}}{M^{*2}}+\frac{\delta_{2}k_{1}^{-4/3}}{M^{*}}+\epsilon_{n}^{*}.\end{split}$

We observe that it is enough to prove

$\begin{split}&\frac{\delta_{2}k_{1}^{-4/3}}{M^{*2}}+\frac{\delta_{1}}{M^{*}}\geq\frac{\delta_{1}}{M^{*2}}+\frac{\delta_{2}k_{1}^{-4/3}}{M^{*}}+\epsilon_{n}^{*}\\\
&\iff\delta_{2}k_{1}^{-4/3}+\delta_{1}M^{*}\geq\delta_{1}+\delta_{2}k_{1}^{-4/3}M^{*}+\epsilon_{n}^{*}M^{*2}\\\
&\iff(\delta_{1}-\delta_{2}k_{1}^{-4/3})\frac{M^{*}-1}{M^{*2}}\geq\epsilon_{n}^{*}.\end{split}$

By Lemma 5.13, $\lim_{n\rightarrow\infty}\epsilon_{n}^{*}=0$, so there is
$N^{*}>0$ so that

$(\delta_{1}-\delta_{2}k_{1}^{-4/3})\frac{M^{*}-1}{M^{*2}}\geq\epsilon_{n}^{*}$

for every $N^{*}<n\leq N$.

∎

Lemma 5.15 shows that exists $N^{*}$ such that for every $N>0$,
$\\{\tilde{b}_{n}^{(N)}\\}$ lies in a compact set, thus by compactness and a
diagonal extraction argument we can choose a subsequence
$(N_{i})_{i}\in\mathbb{N}$ such that $\tilde{b}_{n}^{(N_{i})}$ converges for
all $n\in\mathbb{N}$ to some number $\tilde{b}^{*}_{n}$. The sequence
$\tilde{b}^{*}=\\{\tilde{b}^{*}_{n}\\}_{n}$ satisfies the following equation
by construction

$\tilde{b}^{*}_{n}=\frac{-\delta_{2}k_{1}^{-4/3}+\sqrt{\delta_{2}^{2}k_{1}^{-8/3}+4\delta_{1}\delta_{2}k_{1}^{-4/3}(\tilde{b}^{*}_{n+1})^{-2}+4\delta_{1}^{2}(\tilde{b}^{*}_{n+1})^{-1}-4\delta_{1}\epsilon_{n}^{*}}}{2\delta_{1}}.$

Finally, by the same argument used in Lemmas 4.11 and 5.13 we deduce
$\lim_{n\rightarrow\infty}\tilde{b}^{*}_{n}=1$ every time
$\delta_{1}/\delta_{2}\geq k_{1}^{-4/3}$. Thus, the related solution
$a_{n}^{*}$ satisfies
$\lim_{n\rightarrow\infty}\tilde{b}^{*}_{n}=a_{n-1}^{*}/a_{n}^{*}=k_{1}^{1/3}>1$.
Moreover, by a continuity argument similar to Proposition 3.7 we can pick a
suitable $C^{*}$ so that $b_{0}=0$ and consequently $a_{0}=0$, proving Theorem
2.4 statements also in the case $k_{1}^{-4/3}<\delta_{1}/\delta_{2}\leq 1$.

## Acknowledgements

The author thanks Prof. F. Morandin, for suggesting the problem, reviewing
drafts of the work and for helpful discussions, Prof. F. Flandoli and Prof.
L.A. Bianchi for their insightful comments and review.

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