# On self-similar converging shock waves

Juhi Jang, Jiaqi Liu and Matthew Schrecker Department of Mathematics,
University of Southern California, Los Angeles, CA 90089, USA, and Korea
Institute for Advanced Study, Seoul, Korea. Email<EMAIL_ADDRESS>of Mathematics, University of Southern California, Los Angeles, CA 90089, USA.
Email<EMAIL_ADDRESS>of Mathematics, University of Bath, Claverton
Down, Bath, UK. Email<EMAIL_ADDRESS>
###### Abstract

In this paper, we rigorously prove the existence of self-similar converging
shock wave solutions for the non-isentropic Euler equations for
$\gamma\in(1,3]$. These solutions are analytic away from the shock interface
before collapse, and the shock wave reaches the origin at the time of
collapse. The region behind the shock undergoes a sonic degeneracy, which
causes numerous difficulties for smoothness of the flow and the analytic
construction of the solution. The proof is based on continuity arguments,
nonlinear invariances, and barrier functions.

###### Contents

  1. 1 Introduction
  2. 2 Basic structure of phase portrait and main result
    1. 2.1 Roots of $F$, $G$ and $D$
    2. 2.2 Main result and methodology
  3. 3 Local smooth solutions around sonic points
    1. 3.1 Choice of branch at $P_{6}$ and $P_{8}$
    2. 3.2 Analyticity at $P_{6}$ and $P_{8}$
  4. 4 Solving to left: basic setup and constraints on connections
    1. 4.1 Connection from the sonic point to the initial point $P_{1}$
    2. 4.2 No connection to $P_{8}$ for $\gamma\in(1,\gamma_{\star}]$
    3. 4.3 No connection to $P_{6}$ for $\gamma\in[2,3]$
  5. 5 Solving to left: $P_{6}$
    1. 5.1 Existence for $\gamma\in(1,2]$
    2. 5.2 Uniqueness of $z_{std}$ for $P_{6}$ when $\gamma\in(1,2]$
  6. 6 Solving to left: $P_{8}$
    1. 6.1 Conditional existence for $\gamma\in(\gamma_{\star},\gamma_{1}]$
    2. 6.2 Conditional existence for $\gamma\in(\gamma_{1},3]$
  7. 7 Solving to right
    1. 7.1 Connecting $P_{6}$ to the origin
    2. 7.2 Connecting $P_{8}$ to the origin
      1. 7.2.1 $P_{8}$ for $\gamma\in(\gamma_{\star},\gamma_{1}]$
      2. 7.2.2 $P_{8}$ for $\gamma\in(\gamma_{1},3]$
  8. 8 Proof of the main theorem
  9. A Calculation for Taylor Expansion
  10. B $z_{m}<z_{g}$ for $\gamma\in(1,2]$
  11. C Proof of (4.9)
  12. D Proof of (6.6) and (6.17)
  13. E Calculation of polynomials

##  1 Introduction

The converging shock wave problem is a classical hydrodynamical problem in gas
dynamics, where a spherical shock originates from infinity or a large radius
(for example, by a spherical piston) in a spherically symmetric medium and
propagates towards the center of symmetry, becoming stronger as it approaches
the origin. In finite time, the spherical shock collapses at the center. The
problem was first discussed by Guderley in his seminal work [20] (see also
Landau [27] and Stanyukovich [42]). Due to a wide range of applications such
as detonation, laser fusion and chemical reactions, the theory of converging
shocks has attracted a lot of attention in the mathematics and physics
communities over several decades [1, 17, 20, 25, 27, 28, 35, 38, 45, 47], and
is still an active area of research [19, 26, 36]. In addition, imploding shock
waves are frequently used as a test problem in scientific computing and
algorithms for compressible flows [19, 36]. A rigorous analysis, therefore, is
not only of mathematical interest but also of practical importance as it lays
out foundational evidence in support of these applications.

It has been long known since Guderley that for an inviscid perfect gas, only a
particular choice of similarity exponent would lead to a converging self-
similar radially symmetric shock wave. Despite many works [17, 20, 27, 28]
regarding the numerical value of such a similarity exponent and the
corresponding self-similar solutions based on phase portraits and numerics, a
rigorous construction of self-similar converging shock wave solutions that are
smooth away from the shock interface has remained elusive. In this paper, we
give a rigorous construction of self-similar converging shock wave solutions
described by the non-isentropic compressible Euler equations for an ideal
perfect gas.

The Euler system for compressible gas flows in radial symmetry is given by the
system of PDEs

$\displaystyle\rho_{t}+\frac{1}{r^{m}}(r^{m}\rho u)_{r}$ $\displaystyle=0,$
(1.1) $\displaystyle(\rho u)_{t}+\frac{1}{r^{m}}\big{(}r^{m}(\rho
u^{2})\big{)}_{r}+p_{r}$ $\displaystyle=0,$
$\displaystyle\Big{[}\rho\big{(}e+\cfrac{u^{2}}{2}\big{)}\Big{]}_{t}+\frac{1}{r^{m}}\Big{[}r^{m}\rho
u\big{(}e+\cfrac{u^{2}}{2}+\cfrac{p}{\rho}\big{)}\Big{]}_{r}$
$\displaystyle=0,$

where $\rho=\rho(t,r)\geq 0$ is the density, $u=u(r,t)$ is the radial fluid
velocity, $p(t,r)\geq 0$ is the pressure, and $e(t,r)$ is the specific
internal energy. Here $(t,r)\in\mathbb{R}\times\mathbb{R}_{+}$ and $m=1,2$
distinguishes flows with cylindrical or spherical symmetry. The equations in
(1.1) stand for the conservation of mass, momentum, and energy respectively.
We consider an ideal perfect gas whose equation of state is given by

$\displaystyle p=(\gamma-1)\rho e=(\gamma-1)c_{v}\rho\theta,$ (1.2)

where $\gamma>1$ and $c_{v}$ are positive constants. The specific entropy $S$
is related to

$\displaystyle
p\rho^{-\gamma}=\text{Constant}\cdot\text{exp}(\frac{S}{c_{v}}).$ (1.3)

By the conservation laws (1.1), the entropy $S$ remains constant along
particle trajectories in smooth regions of the flow:

$\displaystyle S_{t}+uS_{r}=0.$ (1.4)

The sound speed is given by

$\displaystyle c=\sqrt{\cfrac{\gamma p}{\rho}}.$ (1.5)

By taking $u$, $\rho$ and $c$ to be the main unknowns, the system (1.1) takes
the form away from vacuum

$\displaystyle\rho_{t}+(\rho u)_{r}+\cfrac{m\rho u}{r}$ $\displaystyle=0,$
(1.6) $\displaystyle u_{t}+uu_{r}+\frac{1}{\gamma\rho}(\rho c^{2})_{r}$
$\displaystyle=0,$ (1.7) $\displaystyle
c_{t}+uc_{r}+\frac{\gamma-1}{2}c(u_{r}+\frac{mu}{r})$ $\displaystyle=0.$ (1.8)

The system (1.6)–(1.8) admits a three-parameter family of invariant scalings:
the scaling transformation

$\displaystyle\rho(t,r)\to\nu^{\kappa}\rho(\frac{t}{\nu^{\lambda}},\frac{r}{\nu}),\quad
u(t,r)\to\nu^{1-\lambda}u(\frac{t}{\nu^{\lambda}},\frac{r}{\nu}),\quad
c(t,r)\to\nu^{1-\lambda}c(\frac{t}{\nu^{\lambda}},\frac{r}{\nu}),$ (1.9)

for $\nu>0,\ \lambda>0,\ \kappa\in\mathbb{R}$, leaves the system invariant.
This scaling symmetry is intimately connected to the existence of self-similar
solutions. Self-similarity is an important concept in hydrodynamics due to its
universal nature and the possibility that self-similar solutions are
attractors for different physical phenomena in fluid and gas dynamics [39,
47]. In the physics literature [47], two kinds of self-similar solutions have
been discussed: Type I if all self-similar parameters are completely
determined from a dimensional analysis and Type II otherwise. Converging self-
similar shock waves emerge as Type II solutions as the speed of collapse,
which is a free parameter, is determined only a posteriori through the
regularity requirement of solutions. To analyze the converging shock wave
problem, inspired by the scaling symmetry (1.9), we introduce the similarity
variable111This is consistent with some of the literature, for instance by
Morawetz [34] and Lazarus [28], while other authors use the equivalent
similarity variable $y={r}/{|t|^{\frac{1}{\lambda}}}$ (see [3, 4, 5, 13, 32]).

$\displaystyle x=\frac{t}{r^{\lambda}},$ (1.10)

and the ansatz

$\displaystyle u(t,r)$ $\displaystyle=-\frac{r}{\lambda
t}V(x)=-\frac{r^{1-\lambda}}{\lambda}\frac{V(x)}{x},$ (1.11) $\displaystyle
c(t,r)$ $\displaystyle=-\frac{r}{\lambda
t}C(x)=-\frac{r^{1-\lambda}}{\lambda}\frac{C(x)}{x},$ (1.12)
$\displaystyle\rho(t,r)$ $\displaystyle=r^{\kappa}R(x)$ (1.13)

where $\lambda>1$ and $\kappa$ are free parameters. This self-similar ansatz
applied to (1.4) in any region where the flow is smooth leads to an algebraic
relation between $V$, $C$ and $R$:

$\displaystyle
R(x)^{q+1-\gamma}(\cfrac{C(x)}{x})^{2}|1+V(x)|^{q}\equiv\text{constant},$
(1.14)

where $q=\tfrac{2(\lambda-1)}{m+1}$. Therefore by plugging (1.11)–(1.13) to
the Euler system (1.6)–(1.8) and using (1.14), we obtain the system of ODEs
for two unknowns $V(x)$, $C(x)$:

$\displaystyle\cfrac{dV}{dx}$ $\displaystyle=-\frac{1}{\lambda
x}\frac{G(V(x),C(x);\gamma,z)}{D(V(x),C(x))}\ \ \text{ and }\ \
\cfrac{dC}{dx}=-\frac{1}{\lambda
x}\frac{F(V(x),C(x);\gamma,z)}{D(V(x),C(x))},$ (1.15)

where

$\displaystyle D(V,C)$ $\displaystyle=(1+V)^{2}-C^{2},$ (1.16) $\displaystyle
G(V,C;\gamma,z)$ $\displaystyle=C^{2}[(m+1)V+2mz]-V(1+V)(\lambda+V),$ (1.17)
$\displaystyle F(V,C;\gamma,z)$
$\displaystyle=C\big{\\{}C^{2}[1+\frac{mz}{(1+V)}]-a_{1}(1+V)^{2}+a_{2}(1+V)-a_{3}\big{\\}},$
(1.18)

and

$\displaystyle z=\cfrac{\lambda-1}{m\gamma},\ \ \
a_{1}=1+\cfrac{m(\gamma-1)}{2},\ \ \
a_{2}=\cfrac{m(\gamma-1)+mz\gamma(\gamma-3)}{2},\ \ \
a_{3}=\cfrac{mz\gamma(\gamma-1)}{2}.$ (1.19)

The derivation of the ODE system is standard and we have adopted the notation
used by Lazarus [28].

We seek a solution for which the shock converges towards the origin for $t<0$
along a self-similar path which is described by a constant value of the
similarity variable $x$,

$\displaystyle x\equiv-1\quad\text{so that}\quad
r_{shock}=(-t)^{\frac{1}{\lambda}},\ \ t<0,$ (1.20)

and the shock reaches the origin at $t=0$. Moreover, the flows on either side
of the shock are assumed to be similarity flows with the same values of
$\gamma$, $\lambda$, and $\kappa$ in (1.11)-(1.13). Under this assumption, we
still require that the jump in the similarity variables is consistent with the
standard Rankine-Hugoniot jump conditions across the shock. Let the subscript
0 and 1 denote evaluation immediately ahead of and behind the shock. The
Rankine-Hugoniot conditions and Lax entropy condition, reformulated in the
self-similar variables, are

$\displaystyle 1+V_{1}$
$\displaystyle=\frac{\gamma-1}{\gamma+1}(1+V_{0})+\frac{2C_{0}^{2}}{(\gamma+1)(1+V_{0})},$
$\displaystyle C_{1}^{2}$
$\displaystyle=C_{0}^{2}+\frac{\gamma-1}{2}[(1+V_{0}^{2})-(1+V_{1})^{2}],$
(1.21) $\displaystyle R_{1}(1+V_{1})$ $\displaystyle=R_{0}(1+V_{0}),$
$\displaystyle C_{0}^{2}$ $\displaystyle<(1+V_{0})^{2}.$ (1.22)

We assume that the fluid ahead of the shock is at rest and at a constant
density and pressure. Then, by (1.13), we have $\kappa=0$ and $R(x)$ is a
constant. For convenience, we let

$\displaystyle R(x)\equiv 1\quad\text{for}\quad-\infty<x<-1.$ (1.23)

Also, by (1.5), the sound speed $c$ is also constant ahead of the shock. As we
assume $\lambda>1$, it implies that $C$ must vanish identically there. By the
assumption that the fluid is at rest before the shock, (1.11) implies that $V$
also must vanish identically there. Therefore, we have

$\displaystyle V(x)=C(x)\equiv 0\qquad\text{for}\quad-\infty<x<-1$ (1.24)

so that $(V_{0},C_{0},R_{0})=(0,0,1)$. Obviously, (1.22) is satisfied. Then,
applying (1.21), we get

$\displaystyle V_{1}$ $\displaystyle=-\frac{2}{\gamma+1},$ (1.25)
$\displaystyle C_{1}$
$\displaystyle=\frac{\sqrt{2\gamma(\gamma-1)}}{\gamma+1},$ (1.26)
$\displaystyle R_{1}$ $\displaystyle=\frac{\gamma+1}{\gamma-1}.$ (1.27)

As we are interested in solutions such that $u$, $c$ and $\rho$ are well-
behaved at any location $r>0$ at $t=0$, we seek solutions such that

$\displaystyle u(0,r)=-\frac{r^{1-\lambda}}{\lambda}\lim_{x\rightarrow
0}\frac{V(x)}{x}<\infty,\quad
c(0,r)=-\frac{r^{1-\lambda}}{\lambda}\lim_{x\rightarrow
0}\frac{C(x)}{x}<\infty,\quad\rho(0,r)=R(0).$ (1.28)

In particular, we require

$\displaystyle V(0)=C(0)=0.$ (1.29)

The converging shock wave problem is, for given adiabatic index $\gamma$, to
find a smooth solution to (1.15) for $-1<x<0$ connecting the shock interface
represented by $(V_{1},C_{1})$ at $x=-1$ to the ultimate collapsed state
$(0,0)$ at $x=0$. Together with the pre-shock state (1.24), such a piecewise
smooth solution to (1.15) gives rise to a collapsing shock solution to the
Euler system (1.6)–(1.8).

A key difficulty in solving the collapsing shock wave problem is that
singularities of the dynamical system (1.15) may occur when $D=0$ or $x=0$.
The moveable singularity $D=0$ is associated with the so-called sonic
singularity (the condition $D=0$ means exactly that the fluid speed and sound
speed coincide), while the singularity at $x=0$ is a removable singularity
which is due to the symmetry assumption. For smooth solutions, if $D=0$ at
some point $x=x_{sonic}$, $G$ and $F$ must vanish at $x=x_{sonic}$. For our
problem, $D(V_{1},C_{1})<0$ (cf. (2.17)) and $D(0,0)=1>0$ and hence any smooth
solution must pass through a sonic point ($D=0$) at which $G=F=0$. This triple
vanishing property is not satisfied by generic values of $\lambda$, but it is
expected that there exists a particular value of $\lambda$ allowing smooth
passage through the sonic point. The main result of this paper is the
existence (and, for a certain range of $\gamma$, uniqueness) of this $\lambda$
which yields a converging shock wave solution.

######  Theorem 1.1 (Informal statement).

(i) Let $\gamma\in(1,3]$. Then there exists a collapsing shock solution to the
non-isentropic Euler equations (1.6)-(1.8).

(ii) Moreover, suppose $\gamma\in(1,\frac{5}{3}]$. Then there is a unique
blow-up speed $\lambda$ such that the aforementioned solution exists.

The precise statement of Theorem 1.1 will be given in Theorem 2.9 after we
discuss the basic structure of the phase portrait plane associated with the
ODE system (1.15) and introduce the set of important parameters appearing in
our analysis in Section 2. We remark that self-similar collapsing shock waves
are solutions of unbounded amplitude (cf. (1.28)) and their continuation to
expanding shock solutions are genuine weak solutions to the Euler system
(1.6)–(1.8), as shown by Jenssen-Tsikkou [25]. Before moving forward, we
mention some works on compressible Euler flows with a focus on weak solutions
and singularities.

The study of the compressible Euler equations has a long history and a
correspondingly vast literature, much of it focused on the one-dimensional
problem. As is well known, a fundamental difficulty in the analysis of the
compressible Euler equations stems from the expected formation of
singularities in the solutions, a phenomenon known since the time of Riemann
and Stokes. For a survey of the literature on the 1D Euler equations,
including existence of weak solutions and formation of singularities, we refer
to [17, 18, 27] and the references therein.

Although there is no general theory for the existence of weak solutions for
the multi-dimensional problem, in recent years, the existence of weak entropy
solutions for the isentropic system under the assumption of spherical symmetry
has been established in [10, 11, 37] using the vanishing viscosity method from
artificial viscosity solutions of certain auxiliary problems. This has been
extended to cover more physical, density-dependent viscosities in [12]. The
weak solutions constructed in these works are based on a finite energy method
that allows for discontinuous and unbounded solutions to arise, especially at
the origin. Earlier results, [9, 30, 31], gave existence results on gases in
an exterior region surrounding a solid ball, and relied on boundedness of
solutions.

The formation of singularities in the multi-dimensional compressible Euler
equations was first rigorously established in [41]. To better understand the
structure of the singularities, there has been much interest in the study of
shock formation in solutions of the multi-dimensional compressible Euler
equations. The first rigorous results are those in spherical symmetry of [46],
which studies the formation and development of shocks in spherical symmetry
for perturbations of constant data for the non-isentropic system. The work
[14] on shock formation for irrotational, isentropic, relativistic gases gives
a truly multi-dimensional result and sharp understanding of the geometry of
the solution at the blow-up time (see also [15, 16]). In recent years, there
have been further exciting developments on shock formation to allow for non-
trivial vorticity and entropy and to remove symmetry assumptions [2, 7, 29]
while still showing the finite time formation of a singularity with sharp
asymptotic behavior on approach to the blowup. Moreover, in [6] the authors
have established the local-in-time continuation of a shock solution from the
first blow-up time for the full, non-isentropic Euler equations; see also a
recent work [40] for the maximal development problem.

As well as these shock solutions, other kinds of strong singularity have also
been areas of active interest, especially the implosion solutions of
Merle–Raphaël–Rodnianski–Szeftel, constructed in [32] and whose finite-
codimension stability is established in [33]. These solutions of the
isentropic Euler equations with $\gamma$-law pressure (excluding a countable
set of $\gamma\in(1,\infty)$) have been constructed using a self-similar ODE
analysis, and the authors must also handle the presence of triple points in
the phase plane (the sonic points), through which the solutions must pass
smoothly. The existence of these solutions has been extended to cover a wider
range of $\gamma$ in [5] and to allow for non-radial perturbations in [8].
Following these works, the construction of continuous (but not necessarily
smooth) implosion solutions to the non-isentropic Euler equations has been
achieved in [26] using a combination of analytic and numerical techniques.
This result also discusses the continuation of the blowup solution past the
first blowup time with an expanding shock wave solution.

##  2 Basic structure of phase portrait and main result

In this section, we discuss the basic structure of the phase portrait of the
ODE system (1.15) and the main result of the paper along with the methodology.
In our analysis, we will primarily make use of the following ODE associated
with the system (1.15)

$\displaystyle\cfrac{dC}{dV}=\cfrac{F(V,C;\gamma,z)}{G(V,C;\gamma,z)},$ (2.1)

which makes the phase portrait analysis more accessible in the $(V,C)$ plane.

We denote the initial data point by

$\displaystyle P_{1}=(V_{1},C_{1})$ (2.2)

in the $(V,C)$ plane.

######  Lemma 2.1.

For $\gamma\in(1,3]$, the initial data points $V_{1}(\gamma)$ and
$C_{1}(\gamma)$ given in (1.25) and (1.26) are monotone increasing with
respect to $\gamma$.

###### Proof.

The result follows from direct computation:

$\displaystyle V_{1}^{\prime}(\gamma)=\cfrac{2}{(\gamma+1)^{2}}>0,\quad
C_{1}^{\prime}(\gamma)=\cfrac{\gamma}{(\gamma+1)^{2}\sqrt{2\gamma(\gamma-1)}}>0.$

∎

###  2.1 Roots of $F$, $G$ and $D$

In this subsection, we summarize the critical points of the dynamical system
(1.15) and some fundamental monotonicity properties with respect to the
parameters $z$ and $\gamma$.

Triple points $F=G=D=0$. The triple points at which $F=G=D=0$ are crucial to
understanding the dynamics of solutions to the ODE system (1.15). On the one
hand, at these points, generic trajectories will suffer a loss of regularity.
On the other hand, at least one such point must be passed through for a
trajectory to reach from the initial data $P_{1}$ to the origin.

######  Lemma 2.2 ([28]).

The solutions to $F=G=D=0$ are

$\displaystyle P_{2}$ $\displaystyle=(-1,0),$ (2.3) $\displaystyle P_{6}$
$\displaystyle=(V_{6},C_{6})=(\frac{-1+(\gamma-2)z-w}{2},1+V_{6}),$ (2.4)
$\displaystyle P_{7}$
$\displaystyle=(V_{7},C_{7})=(\frac{-1+(\gamma-2)z-w}{2},-1-V_{7}),$ (2.5)
$\displaystyle P_{8}$
$\displaystyle=(V_{8},C_{8})=(\frac{-1+(\gamma-2)z+w}{2},1+V_{8}),$ (2.6)
$\displaystyle P_{9}$
$\displaystyle=(V_{9},C_{9})=(\frac{-1+(\gamma-2)z+w}{2},-1-V_{9}),$ (2.7)

where

$\displaystyle w(z)=+\sqrt{1-2(\gamma+2)z+(\gamma-2)^{2}z^{2}}.$ (2.8)

######  Remark 2.3.

Since $w\geq 0$, we will always have $V_{8}\geq V_{6}$ and $C_{8}\geq C_{6}$.

######  Remark 2.4.

(1.25) and (1.26) imply $C_{1}>1+V_{1}$ immediately behind the shock, while
the condition (1.29) implies $C(0)<1+V(0)$. Since we require that $u$ and $c$
are all well behaved at any location away from the origin, the trajectory must
at least continuously pass through the line $D(V,C)=0$ at some
$x_{0}\in(-1,0)$. Comparing this with the ODE system (1.15), we see that we
must have $F(x_{0})=G(x_{0})=0$ to ensure continuity. Thus, the trajectory can
only pass through the sonic line $D=0$ at $P_{6}$ or $P_{8}$. As a
consequence, $w(z)$ given by (2.8) must be a real number, which gives us the
constraint

$\displaystyle z\leq z_{M}(\gamma)=(\sqrt{\gamma}+\sqrt{2})^{-2}.$ (2.9)

Recall that $z=\frac{\lambda-1}{m\gamma}$ from (1.19). That is, equivalently,
we must have

$\displaystyle\lambda\leq\lambda_{M}=m\gamma
z_{M}+1=\frac{m\gamma}{(\sqrt{\gamma}+\sqrt{2})^{2}}+1.$ (2.10)

We will henceforth restrict the range of parameters $\lambda$ and $z$ to
$(1,\lambda_{M}]$ and $(0,z_{M}]$, respectively, and will use both $z$ and
$\lambda$ as convenient.

######  Remark 2.5.

For $z\in(0,z_{M}]$, the function $w(z)$ defined by (2.8) is a decreasing
function in $z$ and satisfies

$\displaystyle 0\leq w(z)<1,$

where $w(z_{M})=0$ and $w(z)\rightarrow 1$ when $z\rightarrow 0$.

The following lemma establishes the monotonicity properties of the locations
of $P_{6}$ and $P_{8}$ with respect to $z$.

######  Lemma 2.6.

For any $\gamma\in(1,3]$ , $V_{6}$ and $C_{6}$ are strictly increasing, and
$V_{8}$ and $C_{8}$ are strictly decreasing with respect to $z\nearrow z_{M}$.

###### Proof.

For any fixed $\gamma\in(1,3]$, we write $V_{6}=V_{6}(z)$ and so
$V_{6}^{\prime}(z)=\frac{dV_{6}}{dz}$. We compute

$\displaystyle
2V_{6}^{\prime}(z)=\gamma-2+\cfrac{(\gamma+2)-(\gamma-2)^{2}z}{\sqrt{1-2(\gamma+2)z+(\gamma-2)^{2}z^{2}}}.$

For any $z\in(0,z_{M}]$, we have
$w(z)=\sqrt{1-2(\gamma+2)z+(\gamma-2)^{2}z^{2}}<1$ and
$(\gamma-2)^{2}z<1<\gamma+2$. Hence

$2V_{6}^{\prime}(z)\geq\gamma-2+\big{(}\gamma+2-(\gamma-2)^{2}z\big{)}=2\gamma-(\gamma-2)^{2}z>0.$

Arguing similarly for $V_{8}$, we have

$\displaystyle 2V_{8}^{\prime}(z)$
$\displaystyle=\gamma-2-\cfrac{(\gamma+2)-(\gamma-2)^{2}z}{\sqrt{1-2(\gamma+2)z+(\gamma-2)^{2}z^{2}}}$
$\displaystyle<\gamma-2-\big{(}\gamma+2-(\gamma-2)^{2}z\big{)}<0.$

Since $C_{6}=1+V_{6}$ and $C_{8}=1+V_{8}$, the desired results follow. ∎

From the definitions of $P_{6}$ and $P_{8}$ in (2.4) and (2.6) and $z_{M}$ in
(2.9), we have

$\displaystyle
V_{6}(z_{M})=V_{8}(z_{M})=\frac{-\sqrt{2}}{\sqrt{\gamma}+\sqrt{2}},\quad
C_{6}(z_{M})=C_{8}(z_{M})=\frac{\sqrt{\gamma}}{\sqrt{\gamma}+\sqrt{2}}.$
(2.11)

Therefore, by Lemma 2.6, we have that

$\displaystyle-1\leq V_{6}$
$\displaystyle\leq\frac{-\sqrt{2}}{\sqrt{\gamma}+\sqrt{2}}\leq V_{8}\leq
0,\quad 0\leq C_{6}\leq\frac{\sqrt{\gamma}}{\sqrt{\gamma}+\sqrt{2}}\leq
C_{8}\leq 1.$ (2.12)

Figure 1: $m=1$, $\gamma=1.5$, $z=0.14$, green line: $D=0$, orange line:
$G=0$, blue line: $F=0$, red line: solution trajectory

Double roots $F=G=0$. In addition to the triple points, there are also a
number of stationary points of the ODE system (1.15) at which $F=G=0$ but
$D\neq 0$. To simplify notation, we define

$\displaystyle H(V)=\sqrt{\cfrac{V(1+V)(\lambda+V)}{(m+1)V+2mz}}.$

Then, the double points of the system may be directly computed as in the
following lemma (cf. [28]).

######  Lemma 2.7.

The solutions to $F=G=0$ and $D\neq 0$ are

$\displaystyle P_{0}$ $\displaystyle=(0,0),$ (2.13) $\displaystyle P_{3}$
$\displaystyle=(V_{3},C_{3})=(-\lambda,0),$ (2.14) $\displaystyle P_{4}$
$\displaystyle=(V_{4},C_{4})=(\frac{-2\lambda}{\gamma+1+m(\gamma-1)},H(V_{4})),$
(2.15) $\displaystyle P_{5}$
$\displaystyle=(V_{5},C_{5})=(\frac{-2\lambda}{\gamma+1+m(\gamma-1)},-H(V_{5})).$
(2.16)

######  Remark 2.8.

Since the solution $C$ of (1.15) must remain positive before the collapse
$t<0$ in order to be physically meaningful, the points $P_{2}$, $P_{3}$,
$P_{5}$, $P_{7}$ and $P_{9}$ do not play a role in the construction of the
solution before the collapse.

We observe that $D$, $F$, and $G$ at $P_{1}$ satisfy the following sign
conditions:

$\displaystyle D(V_{1},C_{1})$ $\displaystyle<0,$ (2.17) $\displaystyle
F(V_{1},C_{1})$ $\displaystyle>0,$ (2.18) $\displaystyle G(V_{1},C_{1})$
$\displaystyle<0.$ (2.19)

Further details on the signs of $D$, $F$, and $G$ can be found in [28]. By
(2.17), $P_{1}$ will be always located above the sonic line $D(V,C)=0$ as in
Figure 1.

###  2.2 Main result and methodology

Many authors have claimed that, for each $\gamma\in(1,3]$, there exists a
$\lambda_{std}\text{ or }z_{std}$ such that the corresponding trajectory
exists from $P_{1}$ to the origin $P_{0}$, analytically passes through the
triple point $P_{6}$ or $P_{8}$ and is monotone decreasing to the origin,
therefore describing a collapsing shock solution of the compressible Euler
equations (see, for example, [17, 20, 25, 27, 28]). The goal of this paper is
to prove rigorously the existence of such a $z_{std}$ and the corresponding
analytic solution to (2.1).

The self-similar solutions that we construct are built by concatenating two
trajectories in the phase-plane in such a way that we obtain an analytic
solution of the ODE (2.1).

  * •

The first trajectory connects $P_{1}$ to either $P_{6}$ or $P_{8}$ in the 2nd
quadrant of the $(V,C)$-plane. To ensure the trajectory passes through $P_{6}$
or $P_{8}$ analytically, we need the trajectory to enter the triple point
$P_{6}$ or $P_{8}$ with a specific slope.

  * •

The second trajectory connects either $P_{6}$ or $P_{8}$ to the origin
$P_{0}=(0,0)$, which is a stable node for (1.15). Since the first trajectory
passes through $P_{6}$ or $P_{8}$ analytically, this second one is uniquely
determined by the slope at $P_{6}$ or $P_{8}$.

Directly solving the initial value problem for (2.1) poses complexity due to
the non-linearity of $F(V,C;\gamma,z)$ and $G(V,C;\gamma,z)$ and the two
parameters $\gamma$ and $z$. One significant challenge in this problem lies in
the non-trivial nature of solutions around the triple points $P_{6}$ and
$P_{8}$, which can be entered either along a primary or a secondary direction
by solutions of (2.1). Along the dominant, primary direction, the solutions
will be only of finite regularity, and so we require the solutions to connect
along the secondary direction to ensure analyticity. This property of analytic
connection fails for generic choices of the parameter $z$, and so the isolated
value, $z_{std}$, that enables this analytic connection must be carefully
constructed. Moreover, it emerges that, for some ranges of $\gamma\in(1,3]$,
the solution emanating from the initial condition $P_{1}$ will converge to
$P_{6}$, while for other $\gamma$, it will converge to $P_{8}$. We must
therefore understand which of the triple points the solution from $P_{1}$
should connect to in order to identify $z_{std}$ and establish an analytic
connection.

To address these challenges effectively, we employ barrier functions for a
number of purposes (cf. Definition 2.1). First, to exclude connection from
$P_{1}$ to $P_{8}$ (respectively $P_{6}$) for small (respectively large)
values of $\gamma$. Second, to establish an appropriate interval of candidate
values of $z$ containing $z_{std}$. Employing the barrier function
$B_{k_{M}}(V)=\sqrt{\frac{C^{2}_{6}(z_{M})}{V_{6}(z_{M})}V}$, we exclude
connections to $P_{8}$ for small $\gamma$, and we will also exclude connection
to $P_{6}$ for $\gamma\geq 2$. In addition, the function $B_{1}(V)=\sqrt{-V}$
is essential for establishing connection from $P_{1}$ to $P_{8}$ for
intermediate values of $\gamma$. This motivates the following definitions.

  * •

$\gamma_{\star}$ is defined to be the value such that $P_{1}$ lies on the
curve $C=B_{k_{M}}(V)$ (see (4.16)). $\gamma_{\star}\approx 1.7$.

  * •

$\gamma_{1}$ is defined to be the value such that $P_{1}$ lies on the curve
$C=B_{1}(V)$ (see (6.1)). $\gamma_{1}=1+\sqrt{2}$.

As mentioned above, we also need to limit the window of possible $z$ values
for which we may have an analytic connection from $P_{1}$ to either $P_{6}$ or
$P_{8}$. This leads us to the following definitions of key values of $z$.

  * •

For any $\gamma\in(1,3]$, $z_{M}(\gamma)$ is defined to be the value such that
$V_{6}(z_{M})=V_{8}(z_{M})$, which means $P_{6}$ and $P_{8}$ are coincident
(see (2.9)).

  * •

For any $\gamma\in(1,3]$, $z_{m}(\gamma)$ is defined to be the value such that
$V_{6}(z_{m})=V_{1}$, so that $P_{6}$ lies on the vertical line through
$P_{1}$ (see (3.16)).

  * •

For any $\gamma\in(1,2]$, $z_{g}(\gamma)$ is defined to be the value such that
$V_{4}=V_{6}$, which means $P_{4}$ and $P_{6}$ are coincident (see (5.8)).

  * •

For any $\gamma\in(\gamma_{\star},\gamma_{1}]$, $z_{1}(\gamma)$ is defined to
be the value such that the curve $C=B_{1}(V)$ intersects the sonic line at
$P_{8}$ (see (6.2)).

  * •

For any $\gamma\in(\gamma_{1},3]$, $z_{2}(\gamma)$ is defined to be the value
such that the curve $C=\sqrt{-\frac{3}{2}V}$ intersects the sonic line at
$P_{8}$ (see (6.13)).

We now state the main result of the paper.

######  Theorem 2.9.

(i) For all $\gamma\in(1,3]$, there is a monotone decreasing analytic solution
to (2.1) connecting $P_{1}$ to the origin.

(ii) For $\gamma\in(1,\gamma_{\star}]$, the solution is unique (in the sense
of unique $z$) and it connects $P_{1}$ to the origin via $P_{6}$. The value of
$z_{std}$ lies in $(z_{g},z_{M}]$.

(iii) For $\gamma\in(\gamma_{\star},2)$, if such a solution connects through
$P_{6}$, then $z_{std}\in(z_{g},z_{M}]$ and $z_{std}$ gives the only such
connection through $P_{6}$. If a solution connects through $P_{8}$, then
$z_{std}\in(z_{1},z_{M}]$.

(iv) For $\gamma\in[2,3]$, any such solution must connect through $P_{8}$ with
a $z_{std}$ value $z_{std}\in(z_{1},z_{M}]$ if $\gamma\in[2,\gamma_{1}]$ or
$z_{std}\in(z_{2},z_{M}]$ if $\gamma\in(\gamma_{1},3]$.

######  Remark 2.10.

To see that these solutions from Theorem 2.9 do indeed give solutions of the
original self-similar problem (in the $x$ variable) is straightforward. Note
that, given $C(V)$, one can solve for $V(x)$ (and hence $C(x)$) via the ODE
$V^{\prime}(x)=-\frac{1}{\lambda x}\frac{G(V(x),C(V(x)))}{D(V(x),C(V(x)))}$
simply by integrating, away from critical points. As $C(V)$ is an analytic
function in $V$ and $G$ and $D$ are both analytic in $(V,C)$ with simple zeros
at the triple points $P_{6}$ and $P_{8}$, repeated application of the chain
rule establishes that the solution $V(x)$ remains smooth (indeed, analytic),
as it passes through a unique sonic point $x_{\textup{sonic}}$ where
$(V(x_{\textup{sonic}}),C(x_{\textup{sonic}}))$ is either $P_{6}$ or $P_{8}$.
As $G(0,0)=0$, it is clear that if, for some $x_{0}\in(x_{\textup{sonic}},0)$,
we hit $V(x_{0})=C(x_{0})=0$, then the ODE has a local, unique solution, which
is the identically zero solution. But this extends backwards for all $x$,
contradicting the initial data and the sonic time. So the solution cannot hit
zero except at $x=0$.

Our strategy for proving the existence of the solutions constructed in Theorem
2.9 proceeds in three key stages, inspired by recent mathematical
constructions of self-similar gravitational collapse [21, 22, 23], where the
authors developed the shooting methods for self-similar non-autonomous ODE
systems to connect smoothly two behaviors at the center and at the far field
through the sonic point. First, in Section 3, we construct local, analytic
solutions around each of the triple points. That is, for all
$z\in[z_{m},z_{M}]$, we construct a local solution around $P_{6}$ and we
construct a local solution around $P_{8}$ for all $z\in(0,z_{M}]$. In order to
show the local existence of such solutions, we first choose a local branch at
the triple points along the secondary direction of (2.1) with a negative slope
$c_{1}<0$ (cf. Section 3.1) and derive a formal recurrence relation for the
Taylor coefficients of a power series
$C(V;\gamma,z)=\sum_{k=0}^{\infty}c_{k}(\gamma,z)(V-V_{*}(\gamma,z))^{k}$.
Once we have found the recurrence relation for the higher order coefficients,
a series of combinatorial estimates and an inductive argument allow us to
bound coefficients to all orders and establish the convergence of the series
in Theorem 3.8.

The second main step of the proof is to show the existence, for each
$\gamma\in(1,3]$, of a $z_{std}$ such that the local analytic solution from
either $P_{6}$ or $P_{8}$, extended backwards in $V$, connects to $P_{1}$.
This is achieved in Section 4.1 via a continuity argument. We show first that
the solution from $P_{6}$ for $z=z_{m}$ always passes below $P_{1}$ in the
phase space, while there always exists a $z\in(0,z_{M})$ such that the
solution from $P_{8}$ passes above $P_{1}$. Then, depending on whether the
solution for $z=z_{M}$ passes above or below $P_{1}$, we may apply a
continuity argument to either $P_{6}$ or $P_{8}$ to establish the connection.

The third main step in the construction is to prove that the solution
connecting $P_{1}$ smoothly to either $P_{6}$ or $P_{8}$ then continues to
connect to the origin. In fact, the behavior of connecting to the origin is
not limited only to the solution that connects to $P_{1}$, but holds for a
non-trivial interval of $z$ around $z_{std}$, as the origin is an attractive
point in the phase plane. A key difficulty is that the solution must connect
from inside the second quadrant, else the velocity changes sign before
collapse. We cannot, a priori, exclude the possibility that, for some range of
$z$, the solution passes through the $C$-axis for some positive value of $C$
before converging to the origin from the first quadrant. To show that this
does not occur, we apply careful barrier arguments to gain an upper bound on
the solution which traps it into a region in the second quadrant in which it
must converge to the origin. This notion is made precise in the following
definition.

######  Definition 2.1 (Lower barrier function and upper barrier function).

We say that a differentiable function $B(V)$ is a lower barrier for $C(V)$ on
$(V_{a},V_{b})$ if $B(V)<C(V)$ on $(V_{a},V_{b})$, and a upper barrier if
$B(V)>C(V)$ on $(V_{a},V_{b})$.

In practice, $C(V)$ will be the solution of (2.1) and $B(V)$ is a specific
differentiable function where we design $B$ such that at one end point, $C(V)$
is greater or less than $B(V)$, and show that the solution $C(V)$ stays above
or below $B(V)$ as $V$ moves to the other end point. The latter part will be
achieved by nonlinear invariances of (2.1). Suppose we intend to show that $B$
is a lower barrier for $C$ and that $C(V_{a})>B(V_{a})$ (respectively
$C(V_{b})>B(V_{b})$). We assume for a contradiction that there exists
$\overline{V}\in(V_{a},V_{b})$ such that $C(\overline{V})=B(\overline{V})$. By
simple continuity and compactness arguments, there exists a minimal
(respectively maximal) such $\overline{V}$, from which we deduce that, at
$\overline{V}$, we must have

$\frac{d}{dV}(C-B)\big{|}_{\overline{V}}\leq 0,\ \text{ respectively }\
\frac{d}{dV}(C-B)\big{|}_{\overline{V}}\geq 0.$

To derive a contradiction, we therefore prove that, whenever
$C(\overline{V})=B(\overline{V})$, then we must have

$\frac{d}{dV}(C-B)\big{|}_{\overline{V}}>0,\ \text{ respectively }\
\frac{d}{dV}(C-B)\big{|}_{\overline{V}}<0.$ (2.20)

As the self-similar blowup speed $z$ varies, the associated solutions from the
triple points $P_{6}$ and $P_{8}$ efficiently explore a large portion of the
phase space, with the solutions from $P_{8}$ in particular moving far up in
the phase plane. In order, therefore, to apply the precise barrier arguments
that will force the solution to the right of the triple point to converge to
the origin, we in fact require better control on the range of $z$ (depending
on $\gamma$) for which the solution to the left connects to $P_{1}$, else we
lose effective control on the trajectory to the right and cannot exclude the
possibility that the trajectory passes through $V=0$ away from the origin.
This improved control on $z$ also allows us to make more quantitative and
qualitative statements concerning the behavior of the imploding shock
solution, especially for $\gamma\in(1,\gamma_{\star}]$.

To this end, we first limit the range of $\gamma$ for which the connecting
solution may come from $P_{6}$ or $P_{8}$. This is achieved in Sections
4.2–4.3, in which we employ our first barrier arguments to the left in order
to show that for $\gamma\in(1,\gamma_{\star}]$, the solution must connect to
$P_{6}$, and for $\gamma\in[2,3]$, it must connect to $P_{8}$.

Following this, in Section 5, we improve the range of $z$ for which the
solution from $P_{6}$ (given $\gamma\in(1,2]$) may connect to $P_{1}$,
tightening the range $z\in[z_{m},z_{M}]$ to the much sharper
$z\in(z_{g},z_{M}]$ by showing that the trajectory is bounded from above, for
this range of $z$, by the solution to a simpler ODE that allows for explicit
integration and estimation. This improvement ensuring $z_{std}>z_{g}$ is
essential, as the structure of the phase portrait changes fundamentally as
$P_{4}$ crosses $P_{6}$ at $z=z_{g}$. We are then able also to show in Lemma
5.4 that, for $\gamma\in(1,2]$, there is at most one value of
$z\in(z_{g},z_{M}]$ for which the solution from $P_{6}$ may connect to $P_{1}$
by studying the derivative $\frac{\partial}{\partial z}(\frac{dC}{dV})$.

The next section, Section 6, contains the analogous sharpening of the possible
range of $z$ for solutions from $P_{8}$. In it, we show that, for
$\gamma\in(\gamma_{\star},\gamma_{1}]$, solutions with $z\in(0,z_{1}]$ cannot
connect to $P_{1}$ by employing the barrier $B_{1}(V)=\sqrt{-V}$, while for
$\gamma\in(\gamma_{1},3]$, solutions with $z\in(0,z_{2}]$ cannot connect to
$P_{1}$ by employing the barrier $B_{\frac{3}{2}}(V)=\sqrt{-\frac{3}{2}V}$
(cf. the definitions of $z_{1}$ and $z_{2}$ above).

Having established these tighter ranges of $z$, depending on $\gamma$, for the
existence of the imploding shock solution, in Section 7 we are then able to
prove that the solution must connect to the origin within the second quadrant.
A simple proof in Lemma 7.2 shows that the trajectories can never hit the
$V$-axis, and so it suffices to find upper barriers connecting to the origin.
Indeed, for $\gamma\in(1,2]$, we show that the solutions from $P_{6}$ for all
$z\in[z_{g},z_{M}]$ admit $B_{1}(V)=\sqrt{-V}$ as an upper barrier, and the
solutions from $P_{8}$ for any $\gamma\in(\gamma_{\star},\gamma_{1}]$ and
$z\in(z_{1},z_{M}]$ admit the same upper barrier. Finally, for the remaining
range, $\gamma\in(\gamma_{1},3]$ and $z\in(z_{2},z_{M}]$, the barrier
$B_{\frac{3}{2}}(V)=\sqrt{-\frac{3}{2}V}$ is an upper barrier for the solution
to the right.

Finally, in Section 8, we put together the earlier results in order to
establish the proof of the main theorem.

##  3 Local smooth solutions around sonic points

In this section, we show the existence of local analytic solutions around the
triple point $P_{*}=P_{6}\text{ or }P_{8}$:

$C(V)=C_{*}+\sum_{\ell=1}^{\infty}c_{\ell}(V-V_{*})^{\ell},$

where the Taylor coefficients $c_{\ell}=c_{\ell}(\gamma,z)$ and with a choice
of branch having a negative slope $c_{1}<0$. The first step is to show that it
is always possible to choose a branch with $c_{1}<0$ for the admissible range
$z\in(0,z_{M}]$ at $P_{8}$ and $z\in[z_{m},z_{M}]$ at $P_{6}$ (see Section
3.1). The second step is to derive a recursive formula to define $c_{\ell}$
for $\ell\geq 2$ and prove the convergence of the Taylor series with positive
radius of convergence (see Section 3.2).

###  3.1 Choice of branch at $P_{6}$ and $P_{8}$

Throughout this section, for ease of notation, we will denote by $P_{*}$
either $P_{6}$ or $P_{8}$. From (2.1), we have $\frac{dC}{dV}=\frac{0}{0}$ at
$P_{*}$. Therefore, for smooth solutions, by using L’Hôpital’s rule, we see
that the slope $c_{1}$ at $P_{*}$ must solve the quadratic equation

$\displaystyle-
G_{C}(V_{*},C_{*})c_{1}^{2}+(F_{C}(V_{*},C_{*})-G_{V}(V_{*},C_{*}))c_{1}+F_{V}(V_{*},C_{*})=0,$
(3.1)

where

$\displaystyle G_{C}(V_{*},C_{*})$ $\displaystyle=\frac{\partial G}{\partial
C}\Big{|}_{(V_{*},C_{*})}=2C_{*}[(m+1)V_{*}+2mz],$ (3.2) $\displaystyle
G_{V}(V_{*},C_{*})$ $\displaystyle=\frac{\partial G}{\partial
V}\Big{|}_{(V_{*},C_{*})}=(m+1)C_{*}^{2}-3V_{*}^{2}-2(\lambda+1)V_{*}-\lambda,$
$\displaystyle F_{C}(V_{*},C_{*})$ $\displaystyle=\frac{\partial F}{\partial
C}\Big{|}_{(V_{*},C_{*})}=3C_{*}^{2}[1+\frac{mz}{(1+V_{*})}]-a_{1}(1+V_{*})^{2}+a_{2}(1+V_{*})-a_{3},$
$\displaystyle F_{V}(V_{*},C_{*})$ $\displaystyle=\frac{\partial F}{\partial
V}\Big{|}_{(V_{*},C_{*})}\\!\begin{aligned}
&=C_{*}\Big{\\{}-mz-2a_{1}(1+V_{*})+a_{2}\Big{\\}}\\\
&=-mzC_{*}-2a_{1}(1+V_{*})^{2}+a_{2}(1+V_{*})\\\
&=C_{*}^{2}-3a_{1}(1+V_{*})^{2}+2a_{2}(1+V_{*})-a_{3}.\end{aligned}$

Solving the quadratic equation (3.1), we get

$\displaystyle c_{1}=\frac{F_{C}(V_{*},C_{*})-G_{V}(V_{*},C_{*})\pm
R(V_{*},C_{*})}{2G_{C}(V_{*},C_{*})},$ (3.3)

where

$R(V_{*},C_{*})=\sqrt{(F_{C}(V_{*},C_{*})-G_{V}(V_{*},C_{*}))^{2}+4F_{V}(V_{*},C_{*})G_{C}(V_{*},C_{*})}.$
(3.4)

Since the first trajectory should be monotone decreasing from $P_{1}$ to
$P_{*}$, we demand the slope $c_{1}$ at $P_{*}$ to be negative. In particular,
solutions for (3.1) must be real, which requires the expression under the
square root of (3.4) to be non-negative.

In order to establish the necessary conditions for $R$ to be real and to
understand the possible solutions of (3.1), we analyze the properties of the
four partial derivatives in (3.2). Using $G(V_{*},C_{*})=0$,
$F(V_{*},C_{*})=0$ and $C_{*}=1+V_{*}$, we see that

$\displaystyle G_{C}(V_{*},C_{*})$
$\displaystyle=\frac{G_{C}(V_{*},C_{*})C_{*}}{1+V_{*}}=2V_{*}(\lambda+V_{*})<0\
\text{ since }-1<V_{*}<0,\ \ \lambda>1,$ (3.5) $\displaystyle
G_{V}(V_{*},C_{*})$
$\displaystyle=(m+1)C_{*}^{2}-(\lambda+V_{*})-2V_{*}(\lambda+V_{*})-V_{*}(1+V_{*}),$
(3.6) $\displaystyle F_{C}(V_{*},C_{*})$
$\displaystyle=2C_{*}^{2}[1+\frac{(\lambda-1)}{\gamma(1+V_{*})}]=2C_{*}[C_{*}+\frac{(\lambda-1)}{\gamma}]>0,$
(3.7) $\displaystyle F_{V}(V_{*},C_{*})$
$\displaystyle=-\frac{(\lambda-1)}{\gamma}C_{*}-a_{1}C_{*}^{2}+a_{3}-[1+\frac{(\lambda-1)}{\gamma
C_{*}}]C_{*}^{2}.$ (3.8)

Summing (3.7) with (3.8) and summing (3.5) with (3.6) and applying the
definitions of $a_{1}$ and $a_{3}$ from (1.19), we find the simpler identities

$\displaystyle F_{C}(V_{*},C_{*})+F_{V}(V_{*},C_{*})$
$\displaystyle=-\frac{m(\gamma-1)}{2}C_{*}^{2}+\frac{(\gamma-1)(\lambda-1)}{2},$
(3.9) $\displaystyle G_{C}(V_{*},C_{*})+G_{V}(V_{*},C_{*})$
$\displaystyle=mC_{*}^{2}-(\lambda-1).$ (3.10)

In turn, these identities imply, recalling (2.4) and (2.6),

$\displaystyle F_{C}(V_{*},C_{*})+F_{V}(V_{*},C_{*})$
$\displaystyle=-\frac{\gamma-1}{2}(G_{C}(V_{*},C_{*})+G_{V}(V_{*},C_{*})),$
(3.11) $\displaystyle G_{C}(V_{6},C_{6})+G_{V}(V_{6},C_{6})$
$\displaystyle=-mwC_{6}<0,$ (3.12) $\displaystyle
G_{C}(V_{8},C_{8})+G_{V}(V_{8},C_{8})$ $\displaystyle=mwC_{8}>0.$ (3.13)

As a direct consequence, we first obtain the following.

######  Lemma 3.1.

At $P_{8}$, for any $\gamma\in(1,3]$ and $\lambda\in(1,\lambda_{M}]$,
$R(V_{8},C_{8})$ is real and strictly positive, and the two solutions of (3.1)
must have different signs.

###### Proof.

We first obtain the sign of $F_{V}(V_{8},C_{8})$ from (3.11), (3.13) and
(3.7):

$\displaystyle
F_{V}(V_{8},C_{8})=-\frac{\gamma-1}{2}(G_{C}(V_{8},C_{8})+G_{V}(V_{8},C_{8}))-F_{C}(V_{8},C_{8})<0.$

Thus, as we also have $G_{C}(V_{8},C_{8})<0$ from (3.5), it is clear that $R$
is real and positive.
Applying (3.3), the product of the two solutions of (3.1) is given by
$-2\frac{F_{V}(V_{8},C_{8})}{G_{C}(V_{8},C_{8})}$. As we have just shown that
$F_{V}(V_{8},C_{8})$ and $G_{C}(V_{8},C_{8})$ are both negative, we conclude
the proof. ∎

######  Remark 3.2.

The situation at $P_{6}$ is different. For $\lambda$ sufficiently close to 1,
$R(V_{6},C_{6})\not\in\mathbb{R}$, and so we require an appropriate range of
$\lambda$ (equivalently of $z$) which guarantees the above properties at
$P_{6}$. As the first trajectory connecting $P_{1}$ and $P_{6}$ is supposed to
be monotone decreasing, it is sufficient to consider only those $V_{6}\geq
V_{1}$. We therefore denote by $\lambda_{m}$ (equivalently $z_{m}$) the value
such that

$\displaystyle V_{6}(\lambda_{m})=V_{1}.$ (3.14)

By a straightforward calculation, we have

$\displaystyle\lambda_{m}$
$\displaystyle=\cfrac{m\gamma(\gamma-1)}{(2\gamma-1)(\gamma+1)}+1,$ (3.15)
$\displaystyle z_{m}$
$\displaystyle=\cfrac{(\gamma-1)}{(2\gamma-1)(\gamma+1)}.$ (3.16)

It is straightforward to check that $\lambda_{m}<\lambda_{M}$ for any
$\gamma\in(1,3]$. By Lemma 2.6, we have $V_{6}(\lambda)\geq V_{1}$ for any
$\lambda\in[\lambda_{m},\lambda_{M}]$.

Moreover, by (2.17),we have

$C_{6}(\lambda_{m})=1+V_{6}(\lambda_{m})=1+V_{1}<C_{1}.$ (3.17)

We now show that within the new sonic window the quadratic equation (3.1) at
$P_{6}$ has two real solutions with different signs.

######  Lemma 3.3.

At $P_{6}$, for any $\gamma\in(1,3]$ and $\lambda\in[\lambda_{m},\lambda_{M}]$
where $\lambda_{m}$ is given by (3.15) and $\lambda_{M}$ is given by (2.10),
the two solutions of (3.1) are both real and have different signs.

###### Proof.

If $F_{V}(V_{6},C_{6})<0$, then by the same argument as in Lemma 3.1, $R$ must
be real and one of the solutions must be negative. Suppose
$F_{V}(V_{6},C_{6})\geq 0$. By (3.2), as $C_{6}>0$, we then have

$-mz-2a_{1}(1+V_{6})+a_{2}\geq 0.$

As $a_{1}=1+\frac{m(\gamma-1)}{2}>0,$ this is is equivalent to

$V_{6}\leq\frac{a_{2}-mz}{2a_{1}}-1.$

We will now show that, in fact, for all $\lambda\in[\lambda_{m},\lambda_{M}]$,
the reverse inequality holds. Given $\gamma\in(1,3]$, we always have

$\displaystyle\frac{a_{2}-mz}{2a_{1}}-1-V_{6}(\lambda_{m})$
$\displaystyle=\frac{a_{2}-mz}{2a_{1}}-1+\frac{2}{\gamma+1}$
$\displaystyle=\frac{1}{2a_{1}}\Big{\\{}\frac{m(\gamma-1)+(\gamma-3)(\lambda-1)}{2}-\frac{\lambda-1}{\gamma}-\frac{m(\gamma-1)^{2}+2(\gamma-1)}{\gamma+1}\Big{\\}}$
$\displaystyle=\frac{1}{2a_{1}}\frac{\gamma(\gamma-1)(-m\gamma+3m-4)+(\gamma+1)(\gamma^{2}-3\gamma-2)(\lambda-1)}{2\gamma(\gamma+1)}$
$\displaystyle=\begin{cases}&\frac{1}{2a_{1}}\frac{(\gamma^{2}-3\gamma-2)(\lambda-1)-\gamma(\gamma-1)}{2\gamma},\quad\quad\text{when
}m=1\\\
&\frac{1}{2a_{1}}\frac{(\gamma+1)(\gamma^{2}-3\gamma-2)(\lambda-1)-2\gamma(\gamma-1)^{2}}{2\gamma(\gamma+1)},\quad\quad\text{when
}m=2.\end{cases}$

In each case, we see that $\frac{a_{2}-mz}{2a_{1}}-1-V_{6}(\lambda_{m})<0$,
and so which means $V_{6}(\lambda_{m})>\frac{a_{2}-mz}{2a_{1}}-1$. By Lemma
2.6, $V_{6}$ is strictly increasing in $\lambda\in(1,\lambda_{M})$. We
conclude that for $\gamma\in(1,3]$ and any
$\lambda\in[\lambda_{m},\lambda_{M}]$, we always have $F_{V}<0$. This means
that two slopes at $P_{6}$ have different signs and
$R(V_{6},C_{6})\in\mathbb{R}$. ∎

In conclusion, for each $\gamma\in(1,3]$ and for the appropriate range of $z$
at $P_{*}$, there exists exactly one negative slope

$\displaystyle
c_{1}=\frac{F_{C}(V_{*},C_{*})-G_{V}(V_{*},C_{*})+R(V_{*},C_{*})}{2G_{C}(V_{*},C_{*})}<0$
(3.18)

which will be our choice of branch. Here we have used $G_{C}<0$ by (3.5).

###  3.2 Analyticity at $P_{6}$ and $P_{8}$

As shown in the previous section, to have the first trajectory with negative
slope, the ranges of $\lambda$ at $P_{6}$ and $P_{8}$ are taken differently.
For notational convenience, we define

$\displaystyle\Lambda=\begin{cases}[\lambda_{m},\lambda_{M}]&\text{if
}P_{*}=P_{6},\\\ (1,\lambda_{M}]&\text{if }P_{*}=P_{8}.\end{cases}$ (3.19)

We write the formal Taylor series around the point $P_{*}$ as

$\displaystyle C(V)=\sum_{\ell=0}^{\infty}c_{\ell}(V-V_{*})^{\ell},$ (3.20)

where $c_{0}=C_{*}$. In a neighborhood of $(V_{*},C_{*})$, we formally have

$\displaystyle\frac{dC}{dV}=\sum_{\ell=1}^{\infty}\ell
c_{\ell}(V-V_{*})^{\ell-1}.$ (3.21)

Now, to simplify notation, we set

$\left\\{\begin{aligned} &v=V-V_{*}\\\
&(c^{2})_{\ell}=\sum_{\begin{subarray}{c}i+j=\ell\\\ i,j\geq
0\end{subarray}}c_{i}c_{j},\\\
&(c^{3})_{\ell}=\sum_{\begin{subarray}{c}i+j+k=\ell\\\ i,j,k\geq
0\end{subarray}}c_{i}c_{j}c_{k}.\end{aligned}\right.$ (3.22)

With this notation, the following quantities have a simple expression:

$\displaystyle C^{2}$
$\displaystyle=(\sum_{\ell=0}^{\infty}c_{\ell}v^{\ell})^{2}=\sum_{\ell=0}^{\infty}(c^{2})_{\ell}v^{\ell},$
(3.23) $\displaystyle C^{3}$
$\displaystyle=(\sum_{1}^{\infty}c_{\ell}v^{\ell})^{3}=\sum_{\ell=0}^{\infty}(c^{3})_{\ell}v^{\ell},$
$\displaystyle C^{\prime}C^{2}$
$\displaystyle=\frac{1}{3}(C^{3})^{\prime}=\frac{1}{3}\sum_{\ell=1}^{\infty}\ell(c^{3})_{\ell}v^{\ell-1}.$

######  Lemma 3.4.

Suppose $C(V)$ defined by (3.20) is an analytic solution of (2.1). Then the
following identity holds:

$\displaystyle\sum_{\ell\geq
2}(A_{\ell}c_{\ell}-B_{\ell})v^{\ell}+[-G_{C}(V_{*},C_{*})c_{1}^{2}+[F_{C}(V_{*},C_{*})-G_{V}(V_{*},C_{*})]c_{1}+F_{V}(V_{*},C_{*})]c_{0}v=0,$
(3.24)

where for each $\ell\geq 2$,

$\displaystyle A_{\ell}$
$\displaystyle=C_{*}\big{[}F_{C}(V_{*},C_{*})-G_{C}(V_{*},C_{*})c_{1}-\ell[G_{V}(V_{*},C_{*})+G_{C}(V_{*},C_{*})c_{1}]\big{]},$
(3.25) $\displaystyle B_{\ell}$
$\displaystyle=\frac{(1+V_{*})\Big{[}(m+1)V_{*}+2mz\Big{]}}{3}(\ell+1)\sum_{\begin{subarray}{c}i+j+k=\ell+1\\\
i,j,k\leq\ell-1\end{subarray}}c_{i}c_{j}c_{k}$
$\displaystyle-\Big{[}(1+V_{*}+mz)-\frac{(m+1)(1+2V_{*})+2mz}{3}\ell\Big{]}\sum_{\begin{subarray}{c}i+j+k=\ell\\\
i,j,k\leq\ell-1\end{subarray}}c_{i}c_{j}c_{k}$
$\displaystyle-\Big{[}1-\frac{m+1}{3}(\ell-1)\Big{]}\sum_{\begin{subarray}{c}i+j+k=\ell-1\end{subarray}}c_{i}c_{j}c_{k}$
(3.26)
$\displaystyle-\Big{[}\big{[}6V_{*}^{2}+(3\lambda+6)V_{*}+2\lambda+1\big{]}(\ell-1)-3a_{1}(1+V_{*})^{2}+2a_{2}(1+V_{*})-a_{3}\Big{]}c_{\ell-1}$
$\displaystyle-\Big{[}(\lambda+2+4V_{*})(\ell-2)-3a_{1}(1+V_{*})+a_{2}\Big{]}c_{\ell-2}-\Big{[}\ell-3-a_{1}\Big{]}c_{\ell-3},$

where we use the convention $c_{n}=0$ for $n<0$.

###### Proof.

The identity follows by substituting (3.20), (3.21), (3.22), and (3.23) into

$(1+V)F(V,C)=(1+V)\frac{dC}{dV}G(V,C)$

and grouping the coefficients of $v^{\ell}$. For the details, we refer to
Appendix A. ∎

Since we are seeking an analytic solution around the sonic point $P_{*}$, we
demand that (3.24) holds for all $|v|<\epsilon$ where $\epsilon>0$ is
sufficiently small. We therefore require that the coefficient of $v^{\ell}$
should be zero at every order $\ell\in\mathbb{N}$. In Section 3.1, we have
already shown the existence of $c_{1}<0$ satisfying

$\displaystyle-
G_{C}(V_{*},C_{*})c_{1}^{2}+[F_{C}(V_{*},C_{*})-G_{V}(V_{*},C_{*})]c_{1}+F_{V}(V_{*},C_{*})=0.$

For $\ell\geq 2$, we directly obtain the recursive relation for $c_{\ell}$,

$\displaystyle A_{\ell}c_{\ell}=B_{\ell},$ (3.27)

where we note from (3.26) that $B_{\ell}$ involves only coefficients $c_{i}$
for $0\leq i\leq\ell-1$. To ensure the solvability of $c_{\ell}$ for all
$\ell\geq 2$, it is obvious that we require $A_{\ell}\neq 0$, and so we need
the following non-vanishing condition:

$\displaystyle
F_{C}(V_{*},C_{*})-G_{C}(V_{*},C_{*})c_{1}-\ell[G_{V}(V_{*},C_{*})+G_{C}(V_{*},C_{*})c_{1}]\neq
0\text{ for any }\ell\geq 2.$ (NVC)

In the following lemma, we show that (NVC) holds for any $\lambda\in\Lambda$.

######  Lemma 3.5.

Let $\gamma\in(1,3]$, $P_{*}\in\\{P_{6},P_{8}\\}$. Then, for any
$\lambda\in\Lambda$ and $\ell\geq 2$, (NVC) is satisfied.

###### Proof.

Recalling (3.6) and $C_{*}=1+V_{*}$, we first see

$\displaystyle G_{V}(V_{*},C_{*})$
$\displaystyle=(m+1)C_{*}^{2}-(\lambda+V_{*})-2V_{*}(\lambda+V_{*})-V_{*}(1+V_{*})$
$\displaystyle=(m-2)V_{*}^{2}+2(m-\lambda)V_{*}+(m+1-\lambda).$

When $m=1$, this gives that

$\displaystyle G_{V}(V_{*},C_{*})=-V_{*}^{2}+2(1-\lambda)V_{*}+2-\lambda$

is quadratic with respect to $V_{*}$. Furthermore, $G_{V}(-1,0)=\lambda-1>0$
and $G_{V}(0,1)=2-\lambda\geq
2-\lambda_{M}=1-\frac{\gamma}{\sqrt{\gamma}+\sqrt{2}}>0$ for any
$\gamma\in(1,3]$. Thus, as $V_{*}\in(-1,0)$ for all $\lambda\in\Lambda$, we
obtain $G_{V}(V_{*},C_{*})>0$.

When $m=2$, we have

$\displaystyle
G_{V}(V_{*},C_{*})=2(2-\lambda)V_{*}+3-\lambda=2(2-\lambda)(1+V_{*})+\lambda-1>0.$

Therefore, for any $\gamma\in(1,3]$ and $\lambda\in\Lambda$,
$G_{V}(V_{*},C_{*})>0$. Denote

$\displaystyle
f(\ell)=:F_{C}(V_{*},C_{*})-G_{C}(V_{*},C_{*})c_{1}-\ell[G_{V}(V_{*},C_{*})+G_{C}(V_{*},C_{*})c_{1}].$

Notice that $f(\ell)$ is a linear equation with respect to $\ell$. When
$\ell=1$, using (3.18), we see
$f(1)=F_{C}(V_{*},C_{*})-G_{V}(V_{*},C_{*})-2G_{C}(V_{*},C_{*})c_{1}=-R(V_{*},C_{*})<0$.
Thus, $f(\ell)<0$ for any $\ell\geq 2$ since we have just shown
$G_{V}(V_{*},C_{*})>0$ and $G_{C}(V_{*},C_{*})c_{1}>0$ by (3.5) and (3.18).

In conclusion,(NVC) is satisfied for any $\lambda\in\Lambda$ at $P_{*}$. ∎

Since $A_{\ell}\neq 0$ by Lemma 3.5, we can rewrite

$\displaystyle c_{\ell}=\cfrac{B_{\ell}}{A_{\ell}}.$ (3.28)

In the following, we estimate the growth of $B_{L}$ under the inductive growth
assumption on $c_{\ell}$ for $2\leq\ell\leq L-1$.

######  Lemma 3.6.

For any fixed $\gamma\in(1,3]$ and $\lambda\in\Lambda$, let $\alpha\in(1,2)$
be given. Then, there exists a constant $K_{*}=K_{*}(\gamma)>1$ such that if
$K\geq K_{*}$ and $L\geq 5$, then if also the following inductive assumption
holds,

$\displaystyle|c_{\ell}|\leq\cfrac{K^{\ell-\alpha}}{\ell^{3}},\quad
2\leq\ell\leq L-1,$ (3.29)

then we have

$\displaystyle|B_{L}|\leq\beta\cfrac{K^{L-\alpha}}{L^{2}}\Big{(}\cfrac{1}{K^{\alpha-1}}+\frac{1}{K}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\Big{)}$
(3.30)

for some constant $\beta=\beta(\gamma,\lambda)$.

For the proof, we will require the following result from [23] to estimate
certain combinations of coefficients.

######  Lemma B.1 ([23, Lemma B.1]).

There exists a universal constant $a>0$ such that for all $L\in\mathbb{N}$,
the following inequalities hold

$\displaystyle\sum_{\begin{subarray}{c}i+j+k=L\\\ i,j,k\geq
1\end{subarray}}\cfrac{1}{i^{3}j^{3}k^{3}}$
$\displaystyle\leq\cfrac{a}{L^{3}},$
$\displaystyle\sum_{\begin{subarray}{c}i+j=L\\\ i,j\geq
1\end{subarray}}\cfrac{1}{i^{3}j^{3}}$ $\displaystyle\leq\cfrac{a}{L^{3}}.$

###### Proof of Lemma 3.6.

First, by using the induction assumption (3.29) and Lemma B.1, we have

$\displaystyle\Big{|}\sum_{\begin{subarray}{c}i+j+k=L\\\ i,j,k\leq
L-1\end{subarray}}c_{i}c_{j}c_{k}\Big{|}$
$\displaystyle=\Big{|}6c_{0}c_{1}c_{L-1}+3c_{0}\sum_{j=2}^{L-2}c_{j}c_{L-j}+3c_{1}^{2}c_{L-2}+3c_{1}\sum_{j=2}^{L-3}c_{j}c_{L-1-j}+\sum_{\begin{subarray}{c}i+j+k=L\\\
i,j,k\geq 2\end{subarray}}c_{i}c_{j}c_{k}\Big{|}$ $\displaystyle\leq
6\big{|}c_{0}c_{1}\big{|}\cfrac{K^{L-1-\alpha}}{(L-1)^{3}}+3|c_{0}|K^{L-2\alpha}\sum_{j=2}^{L-2}\frac{1}{j^{3}(L-j)^{3}}$
$\displaystyle\qquad+3c_{1}^{2}\cfrac{K^{L-2-\alpha}}{(L-2)^{3}}+3|c_{1}|K^{L-1-2\alpha}\sum_{j=2}^{L-3}\frac{1}{j^{3}(L-1-j)^{3}}+K^{L-3\alpha}\sum_{\begin{subarray}{c}i+j+k=L\\\
i,j,k\geq 2\end{subarray}}\frac{1}{i^{3}j^{3}k^{3}}$
$\displaystyle\lesssim\cfrac{K^{L-1-\alpha}}{(L-1)^{3}}+\cfrac{K^{L-2\alpha}}{L^{3}}+\cfrac{K^{L-2-\alpha}}{(L-2)^{3}}+\cfrac{K^{L-1-2\alpha}}{(L-1)^{3}}+\cfrac{K^{L-3\alpha}}{L^{3}}$
$\displaystyle\lesssim\cfrac{K^{L-1-\alpha}}{L^{3}},$

where we have used that $c_{0}$ and $c_{1}$ are bounded by a constant
depending on $\gamma$ and $\lambda$ as well as the assumptions
$\alpha\in(1,2)$ and $K\geq 1$ and, moreover, $L-2>2$, so that the inductive
assumption applies still to $c_{L-2}$. Note that as $L\geq 5$, there exists a
universal constant $C>0$ such that
$\frac{L}{L-1},\frac{L}{L-2},\frac{L}{L-3}\leq C$. Next, a similar argument
yields

$\displaystyle\Big{|}\sum_{\begin{subarray}{c}i+j+k=L-1\end{subarray}}c_{i}c_{j}c_{k}\Big{|}$
$\displaystyle=\Big{|}3c_{0}^{2}c_{L-1}+6c_{0}c_{1}c_{L-2}+3c_{0}\sum_{j=2}^{L-3}c_{j}c_{L-1-j}+3c_{1}^{2}c_{L-3}+3c_{1}\sum_{j=2}^{L-4}c_{j}c_{L-2-j}+\sum_{\begin{subarray}{c}i+j+k=L-1\\\
i,j,k\geq 2\end{subarray}}c_{i}c_{j}c_{k}\Big{|}$
$\displaystyle\lesssim\cfrac{K^{L-1-\alpha}}{(L-1)^{3}}+\cfrac{K^{L-2-\alpha}}{(L-2)^{3}}+\cfrac{K^{L-1-2\alpha}}{(L-1)^{3}}+\cfrac{K^{L-3-\alpha}}{(L-3)^{3}}+\cfrac{K^{L-2-2\alpha}}{(L-2)^{3}}+\cfrac{K^{L-1-3\alpha}}{(L-1)^{3}}$
$\displaystyle\lesssim\cfrac{K^{L-1-\alpha}}{L^{3}},$

where we again note that as $L\geq 5$, $L-3\geq 2$, so that the inductive
assumption applies to $c_{L-3}$. Again using similar arguments, we bound

$\displaystyle\Big{|}\sum_{\begin{subarray}{c}i+j+k=L+1\\\ i,j,k\leq
L-1\end{subarray}}c_{i}c_{j}c_{k}\Big{|}$
$\displaystyle=\Big{|}3c_{0}\sum_{\begin{subarray}{c}j+k=L+1\\\ j,k\leq
L-1\end{subarray}}c_{j}c_{k}+3c_{1}^{2}c_{L-1}+3c_{1}\sum_{\begin{subarray}{c}j+k=L\\\
j,k\leq L-2\end{subarray}}c_{j}c_{k}+\sum_{\begin{subarray}{c}i+j+k=L+1\\\
i,j,k\geq 2\end{subarray}}c_{i}c_{j}c_{k}\Big{|}$
$\displaystyle\lesssim\cfrac{K^{L+1-2\alpha}}{(L+1)^{3}}+\cfrac{K^{L-1-\alpha}}{(L-1)^{3}}+\cfrac{K^{L-2\alpha}}{L^{3}}+\cfrac{K^{L+1-3\alpha}}{(L+1)^{3}}$
$\displaystyle\lesssim\cfrac{K^{L+1-2\alpha}}{L^{3}}.$

Now we estimate $B_{L}$, recalling the definition in (3.26), by employing
these three combinatorial estimates as

$\displaystyle|B_{L}|$
$\displaystyle\lesssim(L+1)\bigg{(}\Big{|}\sum_{\begin{subarray}{c}i+j+k=L+1\\\
i,j,k\leq
L-1\end{subarray}}c_{i}c_{j}c_{k}\Big{|}+\Big{|}\sum_{\begin{subarray}{c}i+j+k=L\\\
i,j,k\leq
L-1\end{subarray}}c_{i}c_{j}c_{k}\Big{|}+\Big{|}\sum_{\begin{subarray}{c}i+j+k=L-1\end{subarray}}c_{i}c_{j}c_{k}\Big{|}+|c_{\ell-1}|+|c_{\ell-2}|+|c_{\ell-3}|\bigg{)}$
$\displaystyle\lesssim(L+1)\bigg{(}\cfrac{K^{L+1-2\alpha}}{L^{3}}+\cfrac{K^{L-1-\alpha}}{L^{3}}+\cfrac{K^{L-1-\alpha}}{L^{3}}+\cfrac{K^{L-1-\alpha}}{L^{3}}+\cfrac{K^{L-2-\alpha}}{L^{3}}+\cfrac{K^{L-3-\alpha}}{L^{3}}\bigg{)}$
$\displaystyle\lesssim\cfrac{K^{L-\alpha}}{L^{2}}\Big{(}\cfrac{1}{K^{\alpha-1}}+\frac{1}{K}\Big{)},$

where have used that there exists a universal constant $C>0$ such that
$\frac{L+1}{L}\leq C$ for all $L\geq 5$. ∎

We next justify the inductive growth assumption on $c_{\ell}$.

######  Lemma 3.7.

For any fixed $\gamma\in(1,3]$ and $\lambda\in\Lambda$, let $\alpha\in(1,2)$
be given. Let $c_{\ell}$ be the coefficients in the formal Taylor expansion of
$C(V)$ around $(V_{*},C_{*})$ solving the recursive relation of Lemma 3.4.
Then there exists a constant $K=K(\gamma,\lambda)>1$ such that $c_{\ell}$
satisfies the bound

$\displaystyle|c_{\ell}|\leq\cfrac{K^{\ell-\alpha}}{\ell^{3}}.$ (3.31)

###### Proof.

We argue by induction on $\ell$. When $\ell=2,3,4$, it is clear from (3.31),
the forms of $A_{\ell}$ and $B_{\ell}$ defined by (3.25)–(3.26), and the non-
vanishing condition (NVC) that there exists a constant $K(\gamma,\lambda)$
such that $c_{2}$, $c_{3}$, and $c_{4}$ satisfy the bounds. Suppose for some
$L\geq 5$, (3.31) holds for all $2\leq\ell\leq L-1$. Then we may apply Lemma
3.6 and with the recursive relation (3.28), we obtain

$\displaystyle|c_{L}|\leq\cfrac{\beta}{|A_{L}|}\cfrac{K^{L-\alpha}}{L^{2}}\Big{(}\cfrac{1}{K^{\alpha-1}}+\frac{1}{K}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\Big{)},$

where
$A_{L}=C_{*}\big{[}F_{C}(V_{*},C_{*})-G_{C}(V_{*},C_{*})c_{1}-L[G_{V}(V_{*},C_{*})+G_{C}(V_{*},C_{*})c_{1}]\big{]}$.
As $A_{L}$ is linear in $L$ and non-zero for all $L$, there exists constants
$\eta_{1}=\eta_{1}(\gamma,\lambda)$ and $\eta_{2}=\eta_{2}(\gamma,\lambda)$
such that $\eta_{1}L\leq|A_{L}|\leq\eta_{2}L$ for all $L\geq 5$. Therefore,

$\displaystyle|c_{L}|\leq\beta\frac{1}{\eta_{1}L}\cfrac{K^{\ell-\alpha}}{L^{2}}\Big{(}\cfrac{1}{K^{\alpha-1}}+\frac{1}{K}\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\Big{)}.$

Choosing $K$ sufficiently large, as $\alpha>1$, it is clear that the estimate
(3.31) holds for $\ell=L$, thus concluding the proof. ∎

We are now ready to prove the main result of this section.

######  Theorem 3.8.

For any fixed $\gamma\in(1,3]$ and $\lambda\in\Lambda$, there exists
$\epsilon=\epsilon(\gamma,\lambda)>0$ such that the Taylor series

$\displaystyle C(V)=\sum_{\ell=0}^{\infty}c_{\ell}(V-V_{*})^{\ell}$ (3.32)

converges absolutely on the interval $(V_{*}-\epsilon,V_{*}+\epsilon)$.
Moreover, $C(V)$ is the unique analytic solution to (2.1).

###### Proof.

Let $\alpha\in(1,2)$ be fixed and suppose $|V-V_{*}|<\epsilon$, where
$\epsilon>0$ is to be chosen later. By (3.31) in Lemma 3.7, there exists a
constant $K(\alpha,\gamma,\lambda)$ such that

$\displaystyle|\sum_{\ell=2}^{\infty}c_{\ell}(V-V_{*})^{\ell}|\leq\sum_{\ell=2}^{\infty}\cfrac{K^{\ell-\alpha}}{\ell^{3}}\epsilon^{\ell}\leq\sum_{\ell=2}^{\infty}(K\epsilon)^{\ell}<\infty,$

provided $\epsilon<\frac{1}{K}$. Thus,
$\sum_{\ell=0}^{\infty}c_{\ell}(V-V_{*})^{\ell}$ converges absolutely for
$V\in(V_{*}-\epsilon,V_{*}+\epsilon)$ with $0<\epsilon<\frac{1}{K}$. ∎

######  Remark 3.9.

Notice that the local analytic solution $C(V)$ obtained in Theorem 3.8 depends
on $\gamma$ and $z$. Since all the coefficients $c_{\ell}$ for any $\ell\geq
1$ are continuous functions of $(\gamma,\lambda)$ for $\gamma\in(1,3]$ and
$\lambda\in\Lambda$ by (NVC) and (3.5), by standard compactness and uniform
convergence, we deduce that $C=C(V;\gamma,\lambda)$ is a continuous function
of $(\gamma,\lambda)$ (equivalently continuous in $(\gamma,z)$) on its domain.

######  Lemma 3.10.

The local analytic solution (3.32), propagated by (2.1) to the left of the
sonic point $P_{*}$, except at the sonic point itself, remains strictly away
from the zeros of $F$, $G$, and $D$. In addition, the solution to the left of
the sonic point $P_{*}$ satisfies $F>0$, $G<0$, $D<0$ and $\frac{dC}{dV}<0$.

###### Proof.

The result is owing to our choice of the negative branch $c_{1}$ (see (3.18)).
The sign conditions then follow from $F_{C}(V_{*},C_{*})>0$ and
$G_{C}(V_{*},C_{*})<0$ from (3.7) and (3.5). ∎

##  4 Solving to left: basic setup and constraints on connections

In this section, we introduce the basic setup for the continuity argument for
the first trajectory to the left of the sonic point and also show that the
solutions of (1.15) starting from the initial point $P_{1}$ can’t connect to
$P_{8}$ for $\gamma\in(1,\gamma_{\star}]$ and can’t connect to $P_{6}$ for
$\gamma\in[2,3]$. We will use $z$ only (instead of $\lambda$) in the following
sections. The corresponding range of $z$ for $P_{*}$ is given by

$\displaystyle\mathcal{Z}(\gamma;P_{*})=\begin{cases}&[z_{m}(\gamma),z_{M}(\gamma)]\quad\text{
when }P_{*}=P_{6},\\\ &(0,z_{M}(\gamma)]\quad\text{ when
}P_{*}=P_{8}.\end{cases}$ (4.1)

###  4.1 Connection from the sonic point to the initial point $P_{1}$

By Theorem 3.8, the following problem

$\displaystyle\begin{cases}&\cfrac{dC}{dV}=\cfrac{F(V,C;\gamma,z)}{G(V,C;\gamma,z)},\quad
V\in[V_{1},V_{*}],\\\ &C(V_{*})=C_{*},\\\
&\frac{dC}{dV}(V_{*})=c_{1},\end{cases}$ (4.2)

where $c_{1}$ is as defined in (3.18), has a local analytic solution. To prove
the existence of a trajectory connecting $P_{1}$ and $P_{*}$, we will first
show that the local analytic solution obtained from Theorem 3.8 extends
smoothly as a strictly monotone decreasing solution to the left to
$C:[V_{1},V_{*}]\to\mathbb{R}_{+}$. Secondly, we will show that for any
$\gamma\in(1,3]$, there exists a $z_{std}\in\mathcal{Z}(\gamma;P_{*})$ such
that the local analytic solution from either $P_{6}$ or $P_{8}$ can be
extended smoothly to $P_{1}$ by using a continuity argument.

######  Lemma 4.1.

Let $P_{*}=(V_{*},C_{*})$ be either $P_{6}$ or $P_{8}$ and suppose
$C:[V_{*}-\epsilon,V_{*}]\to\mathbb{R}_{+}$ is the local, analytic solution to
(4.2) guaranteed by Theorem 3.8. Then the solution extends smoothly to the
left to $C:[V_{1},V_{*}]\to\mathbb{R}_{+}$.

###### Proof.

We argue by contradiction. Suppose that the maximal time of existence of the
solution is $(V_{0},V_{*}]$ for some $V_{0}\geq V_{1}$. By Remark 3.10,
$C(V)>C(V_{*})$ on ($V_{0},V_{*})$. Moreover, as the right hand side of the
ODE (4.2) is locally Lipschitz away from zeros of $G$, we see that the only
obstruction to continuation past $V_{0}$ is blow-up of $C$, i.e.,
$\limsup_{V\to V_{0}^{+}}C(V)=\infty$.

Now, from the explicit forms of $F$ and $G$, we observe that there exists
$M=M(\gamma,z)>0$ such that if $C\geq M$, for all
$V\in[V_{1},V_{*}-\epsilon]$, we have

$\displaystyle G(V,C;\gamma,z)=$
$\displaystyle\,C^{2}[(m+1)V+2mz]-V(1+V)(\lambda+V)\leq\frac{1}{2}C^{2}[(m+1)V+2mz]<0,$
$\displaystyle F(V,C;\gamma,z)=$
$\displaystyle\,C\big{\\{}C^{2}[1+\frac{mz}{(1+V)}]-a_{1}(1+V)^{2}+a_{2}(1+V)-a_{3}\big{\\}}\leq\frac{3}{2}C^{3}[1+\frac{mz}{(1+V)}]$

and $F>0$. Thus,

$\displaystyle\frac{dC}{dV}=\frac{F(V,C;\gamma,z)}{G(V,C;\gamma,z)}\geq
3\frac{C[1+\frac{mz}{(1+V)}]}{[(m+1)V+2mz]}.$

Thus, as $V$ is contained in a bounded set, we see that there exists a
constant $A>0$ such that whenever $C(V)\geq M$, we have

$\frac{d\log C}{dV}\geq-A,$

and hence $C$ is necessarily bounded on $(V_{0},V_{*})$, contradicting the
assumption. ∎

By Theorem 3.8 and Lemma 4.1, (4.2) has a smooth solution on $[V_{1},V_{*}]$.
We use $C(V;\gamma,z,P_{*})$ to denote the solution of (4.2) at
$V\in[V_{1},V_{*}]$. By the fundamental theorem of calculus,

$\displaystyle
C(V;\gamma,z,P_{*})=\int_{V_{*}(\gamma,z)}^{V}\frac{dC(V;\gamma,z)}{dV}dV+C_{*}(\gamma,z)=\int_{V_{*}(\gamma,z)}^{V}\cfrac{F(V,C;\gamma,z)}{G(V,C;\gamma,z)}dV+C_{*}(\gamma,z).$
(4.3)

It is clear from the expressions for $(V_{6},C_{6})$ and $(V_{8},C_{8})$ in
(2.4), (2.6) as well as the continuous dependence of the local, analytic
solution on $\gamma$, $z$ (cf. Remark 3.9), and the continuity properties of
$F$ and $G$ that this is a continuous function with respect to $\gamma$ and
$z$. Moreover, the initial value $V_{1}$ only depends on $\gamma$. Hence, for
any fixed $\gamma$ and a sonic point $P_{*}\in\\{P_{6},P_{8}\\}$, if we can
show that there exists a $\underline{z}\in\mathcal{Z}(\gamma;P_{*})$ such that
$C(V_{1};\gamma,z,P_{*})\leq C_{1}$ and a
$\overline{z}\in\mathcal{Z}(\gamma;P_{*})$ such that
$C(V_{1};\gamma,z,P_{*})\geq C_{1}$, then we conclude that there exists a
$z_{std}$ such that $\underline{z}\leq z_{std}\leq\overline{z}$ and
$C(V_{1};\gamma,z_{std},P_{*})=C_{1}$. This motivates the introduction of
upper and lower solutions:

######  Definition 4.1.

(Upper and lower solution). Let $\gamma\in(1,3]$ and
$z\in\mathcal{Z}(\gamma;P_{*})$. Let
$C(\cdot;\gamma,z,P_{*}):[V_{1},V_{*}]\to\mathbb{R}_{+}$ be the analytic
solution obtained from Theorem 3.8 and Lemma 4.1. We say that
$\overline{z}(\gamma;P_{*})$ gives an upper solution for $P_{*}$ if

$\displaystyle C(V_{1};\gamma,\overline{z}(\gamma;P_{*}),P_{*})>C_{1}.$ (4.4)

We say that $\underline{z}(\gamma;P_{*})$ gives a lower solution for $P_{*}$
if

$\displaystyle C(V_{1};\gamma,\underline{z}(\gamma;P_{*}),P_{*})<C_{1}.$ (4.5)

The proof of the existence of an analytic solution connecting $P_{1}$ and
either $P_{6}$ or $P_{8}$ proceeds as follows. We will first show that $P_{6}$
always admits a lower solution and $P_{8}$ always admits an upper solution. It
will then follow that, depending on whether
$C(V;\gamma,z_{M},P_{6})=C(V;\gamma,z_{M},P_{8})$ gives a lower or an upper
solution (or connects to $P_{1}$), at least one of $P_{6}$ and $P_{8}$ has
both an upper and a lower solution, thus concluding the proof.

First we show the existence of a lower solution for $P_{6}$.

######  Lemma 4.2.

Let $\gamma\in(1,3]$. Then there exists
$\underline{z}(\gamma;P_{6})\in\mathcal{Z}(\gamma;P_{6})$ such that
$C(V;\gamma,\underline{z}(\gamma),P_{6})$ is a lower solution for $P_{6}$.

###### Proof.

This follows simply from Remark 3.2 and the monotonicity of
$C(V;\gamma,{z},P_{6})$. When $z=z_{m}(\gamma)$, we have
$V_{6}(z_{m}(\gamma))=V_{1}(\gamma)$ and
$C(V_{1};\gamma,z_{m}(\gamma),P_{6})=C_{6}(z_{m}(\gamma))<C_{1}(\gamma)$ by
(3.17). Thus, $C(V;\gamma,z_{m}(\gamma),P_{6})$ gives a lower solution for
$P_{6}$. ∎

Next we show the existence of an upper solution for $P_{8}$.

######  Lemma 4.3.

Let $\gamma\in(1,3]$. Then there exists
$\overline{z}(\gamma;P_{8})\in\mathcal{Z}(\gamma;P_{8})$ such that
$C(V;\gamma,\overline{z}(\gamma;P_{8}),P_{8})$ is an upper solution for
$P_{8}$.

###### Proof.

By Lemma 2.1, $C_{1}(\gamma)\leq C_{1}(3)=\frac{\sqrt{3}}{2}<1$ for each
$\gamma\in(1,3]$. By Lemma 2.6, $C_{8}(\gamma,z)$ is monotone decreasing with
respect to $z$ for $0<z\leq z_{M}$. Since $C_{8}(\gamma,0)=1$, there exists a
sufficiently small $\overline{z}(\gamma;P_{8})\in\mathcal{Z}(\gamma;P_{8})$
such that $C_{8}(\overline{z}(\gamma;P_{8}))>C_{1}$. By the monotonicity of
$C(V;\gamma,\overline{z}(\gamma;P_{8}),P_{8})$ with respect to $V$, we
conclude that $C(V;\gamma,\overline{z}(\gamma;P_{8}),P_{8})>C_{1}(\gamma)$.
Thus, $C(V;\gamma,\overline{z}(\gamma;P_{8}),P_{8})$ is an upper solution for
$P_{8}$. ∎

We now prove the main result of this section.

######  Theorem 4.4.

Let $\gamma\in(1,3]$. Then there exists a $P_{*}\in\\{P_{6},P_{8}\\}$ and a
corresponding $z_{std}(\gamma;P_{*})\in\mathcal{Z}(\gamma;P_{*})$ such that
the local analytic solution obtained from Theorem 3.8 extends smoothly from
$P_{*}$ to $P_{1}$.

###### Proof.

By Lemma 4.1, the domain of the local analytic solution extends smoothly
(analytically) to $V=V_{1}$. It remains to show that for each $\gamma\in(1,3]$
there exist $P_{*}$ and $z_{std}(\gamma;P_{*})\in\mathcal{Z}(\gamma;P_{*})$
such that $C(V_{1};\gamma,z_{std}(\gamma),P_{*})=C_{1}$. Recall that when
$z=z_{M}$, $P_{6}$ coincides with $P_{8}$. Therefore, for $P_{*}=P_{6}=P_{8}$
with $z=z_{M}$, there are three possibilities:

  1. 1.

If $C(V_{1};\gamma,z_{M}(\gamma),P_{*})<C_{1}(\gamma)$, then $z=z_{M}(\gamma)$
gives a lower solution for $P_{8}$. Then, by using the continuity argument and
Lemma 4.3, there exists a
$z_{std}\in(\overline{z}(\gamma;P_{8}),z_{M}(\gamma))$ such that
$C(V_{1};\gamma,z_{std},P_{8})=C_{1}$.

  2. 2.

If $C(V_{1};\gamma,z_{M}(\gamma),P_{*})=C_{1}(\gamma)$, then $z=z_{M}(\gamma)$
gives the solution.

  3. 3.

If $C(V_{1};\gamma,z_{M}(\gamma),P_{*})>C_{1}(\gamma)$, then $z=z_{M}(\gamma)$
gives an upper solution for $P_{6}$. Then, by using the continuity argument
and Lemma 4.2, there exists a
$z_{std}\in(\underline{z}(\gamma;P_{6}),z_{M}(\gamma))$ such that
$C(V_{1};\gamma,z_{std},P_{6})=C_{1}$.

This concludes the proof. ∎

###  4.2 No connection to $P_{8}$ for $\gamma\in(1,\gamma_{\star}]$

Now that we have established the existence of an analytic solution to (2.1)
connecting $P_{1}$ to either $P_{6}$ or $P_{8}$ for each $\gamma\in(1,3]$, we
seek to understand better the nature of the solutions in order to connect the
solution through the triple point to the origin. The first step in showing
this is to prove that, for $\gamma\in(1,\gamma_{\star}]$, for $\gamma_{\star}$
defined below in (4.16), the connection must be to $P_{6}$.

For notational convenience, we define a constant

$\displaystyle k(\gamma,z)=-\frac{(1+V_{6}(\gamma,z))^{2}}{V_{6}(\gamma,z)},$
(4.6)

and, for each $\gamma\in(1,2]$, $z\in[z_{m},z_{M}]$, we define a barrier
function

$B_{k}(V)=\sqrt{-k(\gamma,z)V}.$ (4.7)

######  Lemma 4.5.

For any $\gamma\in(1,2]$ and $z\in[z_{m},z_{M}]$, the curve $C=B_{k}(V)$ is a
lower barrier (in the sense of Definition 2.1) for the solution of

$\displaystyle\begin{cases}&\cfrac{dC}{dV}=\cfrac{F(V,C;\gamma,z)}{G(V,C;\gamma,z)},\quad
V\in[V_{1},\overline{V}],\\\ &C(\overline{V})=\overline{C},\end{cases}$ (4.8)

where $\overline{V}\in(V_{1},V_{6})$ and $\overline{C}>B_{k}(\overline{V})$.
In particular, for any $z\in[z_{g},z_{M}]$ the curve $C=B_{k}(V)$ is a lower
barrier for the solution of the problem (4.2) with $P_{*}=P_{6}$. Recall that
$z_{g}\in\mathcal{Z}(\gamma;P_{6})$ is the value of $z$ such that
$P_{4}=P_{6}$, see (5.8).

###### Proof.

We begin by showing that the second claim follows from the first one. Observe
that, assuming the first part of the lemma is proved, it is sufficient to
verify that there exists some interval $[\overline{V},V_{6})$ such that the
solution $C(V)$ to the problem (4.2) satisfies $C(V)>B_{k}(V)$ for
$V\in(\overline{V},V_{6})$. This claim follows one we verify that the
derivative at $V_{6}$ satisfies the inequality

$\displaystyle\frac{dC}{dV}(V_{6})=c_{1}<-\frac{1}{2}\sqrt{\cfrac{k(\gamma,z)}{-V_{6}}}=\frac{1+V_{6}}{2V_{6}}.$
(4.9)

The proof of this inequality for $\gamma\in(1,2]$ and $z\in[z_{g},z_{M}]$ is
given in Appendix C.

We therefore focus on proving the first claim. Suppose that $C(V)$ is a
solution to the problem (4.8). We will apply the barrier argument (cf. 2.20)
to show that as the initial point $C(\overline{V})>B_{k}(\overline{V})$, this
inequality is propagated by the ODE. As the solution to the ODE (4.8) remains
monotone by Lemma 3.10, it is clear that it cannot meet a sonic point. Our
goal is to show that for any $\gamma\in(1,2]$, $z\in[z_{g},z_{M}]$ and
$V\in[V_{1},V_{6})$,

$\displaystyle\cfrac{F(V,\sqrt{-k(\gamma,z)V};\gamma,z)}{G(V,\sqrt{-k(\gamma,z)V};\gamma,z)}+\frac{1}{2}\sqrt{\cfrac{k(\gamma,z)}{-V}}<0.$
(4.10)

Since $G(V,\sqrt{-k(\gamma,z)V};\gamma,z)<0$ for any $\gamma\in(1,2]$,
$z\in[z_{m},z_{M}]$ and $V\in[V_{1},V_{6})$ by Lemma 3.10, it is sufficient to
show that

$\displaystyle
F(V,\sqrt{-k(\gamma,z)V};\gamma,z)+\frac{1}{2}\sqrt{\cfrac{k(\gamma,z)}{-V}}G(V,\sqrt{-k(\gamma,z)V};\gamma,z)>0.$
(4.11)

By direct computations, we obtain

$\displaystyle\cfrac{2}{\sqrt{-k(\gamma,z)V}}\left(F(V,\sqrt{-k(\gamma,z)V};\gamma,z)+\frac{1}{2}\sqrt{\cfrac{k(\gamma,z)}{-V}}G(V,\sqrt{-k(\gamma,z)V};\gamma,z)\right)$
$\displaystyle=(m-1-m\gamma)V^{2}+[-2-m(\gamma-1)+m\gamma
z(\gamma-2)+(m-1)k(\gamma,z)]V+\cfrac{2mk(\gamma,z)z}{1+V}-1-m\gamma z$
$\displaystyle=:\mathfrak{B}_{k}(V,z,m).$ (4.12)

Since $\cfrac{\sqrt{-k(\gamma,z)V}}{2}>0$, (4.11) is equivalent to the
positivity of $\mathfrak{B}_{k}(V,z,m)$:

$\displaystyle\mathfrak{B}_{k}(V,z,m)>0$ (4.13)

for any $\gamma\in(1,2]$, $z\in[z_{m},z_{M}]$ and $V\in[V_{1},V_{6})$.

As $\mathfrak{B}_{k}(V_{6},z,m)=0$ for any $\gamma\in(1,2]$ and
$z\in[z_{m},z_{M}]$ (due to $B_{k}(V_{6})=C_{6}$ and the vanishing of $F$ and
$G$ at $(V_{6},C_{6})$), we will conclude that $\mathfrak{B}_{k}(V,z,m)>0$ by
demonstrating that for $V\in[V_{1},V_{6})$,

$\frac{\partial\mathfrak{B}_{k}(V,z,m)}{\partial V}<0.$ (4.14)

The $V$ derivative of $\mathfrak{B}_{k}$ is given by

$\displaystyle\frac{\partial\mathfrak{B}_{k}(V,z,m)}{\partial
V}=2(m-1-m\gamma)V-2-m(\gamma-1)+m\gamma
z(\gamma-2)+(m-1)k(\gamma,z)-\cfrac{2mk(\gamma,z)z}{(1+V)^{2}}.$

When $m=1$, for any $V\in[V_{1},V_{6})$,

$\displaystyle\frac{\partial\mathfrak{B}_{k}(V,z,1)}{\partial V}=-2\gamma
V-1-\gamma+(\gamma-2)\gamma z+\cfrac{2(1+V_{6})^{2}z}{V_{6}(1+V)^{2}}<-2\gamma
V_{1}-1-\gamma+(\gamma-2)\gamma z+\cfrac{2z}{V_{6}}=:I,$

where we have used $-2\gamma V<-2\gamma V_{1}$ and
$\cfrac{2(1+V_{6})^{2}z}{V_{6}(1+V)^{2}}<\cfrac{2z}{V_{6}}$ for any
$V\in(V_{1},V_{6})$. Recalling $\eqref{V after shock}$ and $\eqref{P6}$, we
deduce

$\displaystyle I$
$\displaystyle=-\frac{(\gamma-1)^{2}}{\gamma+1}+[(\gamma-2)\gamma+\frac{2}{V_{6}}]z<-\frac{(\gamma-1)^{2}}{\gamma+1}+[(\gamma-2)\gamma+\frac{2}{V_{1}}]z=-\frac{(\gamma-1)^{2}}{\gamma+1}+[\gamma(\gamma-3)-1]z<0$

for any $\gamma\in(1,2]$ and $z\in[z_{m},z_{M}]$, which in turn leads to
(4.14) for $m=1$.

When $m=2$, for any $V\in[V_{1},V_{6})$,

$\displaystyle\frac{\partial\mathfrak{B}_{k}(V,z,2)}{\partial V}$
$\displaystyle=2(1-2\gamma)V-2\gamma+2(\gamma-2)\gamma
z-\frac{(1+V_{6})^{2}}{V_{6}}+\frac{4(1+V_{6})^{2}z}{V_{6}(1+V)^{2}}$
$\displaystyle<2(1-2\gamma)V_{1}-2\gamma+2(\gamma-2)\gamma
z-\frac{(1+V_{6})^{2}}{V_{6}}+\frac{4z}{V_{6}}=:II+III$

since $(1-2\gamma)V<(1-2\gamma)V_{1}$ and
$\cfrac{2(1+V_{6})^{2}z}{V_{6}(1+V)^{2}}<\cfrac{2z}{V_{6}}$ for any
$V\in(V_{1},V_{6})$, where $II$ and $III$ denote

$\displaystyle II$ $\displaystyle:=2(1-2\gamma)V_{1}-2\gamma+2(\gamma-2)\gamma
z\ \ \text{ and }\ \ III:=-\frac{(1+V_{6})^{2}}{V_{6}}+\frac{4z}{V_{6}}.$

By (1.25),

$\displaystyle II=-\frac{2(\gamma-1)(\gamma-2)}{\gamma+1}+2(\gamma-2)\gamma
z.$

Using (2.4) and (2.8), we rewrite $III$ as

$\displaystyle III$
$\displaystyle=\frac{-1-(\gamma-2)^{2}z^{2}-w^{2}-2(\gamma-2)z+2w+2(\gamma-2)zw+16z}{4V_{6}}$
$\displaystyle=\frac{-(1+(\gamma-2)^{2}z^{2}-2(\gamma+2)z)-w^{2}+2w+2(\gamma-2)zw+(12-4\gamma)z}{4V_{6}}$
$\displaystyle=\frac{2(1-w)w+4(3-\gamma)z+2[(\gamma-2)w+2]z}{4V_{6}}.$

By Remark 2.5 and the fact that $-1<V_{6}<0$ for any $\gamma\in(1,2]$ and
$z\in[z_{m},z_{M}]$, we have

$\displaystyle III$
$\displaystyle<-\frac{2(1-w)w+4(3-\gamma)z+2[(\gamma-2)w+2]z}{4}<-\frac{4(3-\gamma)+2[(\gamma-2)w+2]}{4}z<(\gamma-\frac{7}{2})z.$

Then, $II+III$ is bounded by

$\displaystyle
II+III<-\frac{2(\gamma-1)(\gamma-2)}{\gamma+1}+2(\gamma-2)\gamma
z+(\gamma-\frac{7}{2})z<-\frac{2(\gamma-1)(\gamma-2)}{\gamma+1}+4(\gamma-2)\gamma
z$

where we have used $\gamma-\frac{7}{2}<2(\gamma-2)\gamma$ for any
$\gamma\in(1,2]$. Since $z_{m}=\frac{\gamma-1}{(2\gamma-1)(\gamma+1)}$ by
(3.16),

$\displaystyle
II+III<-\frac{2(\gamma-1)(\gamma-2)}{\gamma+1}+4(\gamma-2)\gamma z$
$\displaystyle\leq-\frac{2(\gamma-1)(\gamma-2)}{\gamma+1}+\frac{4\gamma(\gamma-1)(\gamma-2)}{(2\gamma-1)(\gamma+1)}$
$\displaystyle=\frac{2(\gamma-1)(\gamma-2)}{\gamma+1}(\frac{2\gamma}{2\gamma-1}-1)\leq
0.$

Therefore, (4.14) holds for $m=2$, $\gamma\in(1,2]$, $z\in[z_{m},z_{M}]$ and
$V\in[V_{1},V_{6})$, thereby completing the proof. ∎

Next, we establish a uniform upper barrier for the forward solution trajectory
of (2.1) with the initial value $P_{1}$ for a particular range of $\gamma$ and
$z\in[z_{m}(\gamma),z_{M}(\gamma)]$ to demonstrate there is no connection from
$P_{1}$ to $P_{8}$.

By (4.6) and (2.11),

$\displaystyle
k(\gamma,z_{M})=-\frac{(1+V_{6}(z_{M}))^{2}}{V_{6}(z_{M})}=\frac{\gamma}{2+\sqrt{2\gamma}}.$
(4.15)

We define $\gamma_{\star}$ to be the value such that

$\displaystyle
C_{1}(\gamma_{\star})=\sqrt{-k(\gamma_{\star},z_{M})V_{1}(\gamma_{\star})}.$
(4.16)

######  Lemma 4.6.

For any $\gamma\in(1,\gamma_{\star}]$ and $z\in[z_{m},z_{M}]$, the curve
$B_{k_{M}}(V)=\sqrt{-k(\gamma,z_{M})V}$ is an upper barrier for the solution
of

$\displaystyle\begin{cases}&\cfrac{dC}{dV}=\cfrac{F(V,C;\gamma,z)}{G(V,C;\gamma,z)},\quad
V\in[V_{1},V_{6}),\\\ &C(V_{1})=C_{1}.\end{cases}$ (4.17)

###### Proof.

We begin by verifying that $P_{1}=(V_{1},C_{1})$ lies on or below the curve
defined by $B_{k_{M}}(V)$. Note that

$\displaystyle
C_{1}-\sqrt{-k(\gamma,z_{M})V_{1}}=\frac{\sqrt{2\gamma(\gamma-1)}}{\gamma+1}-\sqrt{\frac{\sqrt{2}\gamma}{(\sqrt{\gamma}+\sqrt{2})(\gamma+1)}}=\cfrac{\frac{2\gamma(\gamma-1)}{\gamma+1}-\frac{\sqrt{2}\gamma}{\sqrt{\gamma}+\sqrt{2}}}{\sqrt{2\gamma(\gamma-1)}+\sqrt{\frac{\sqrt{2}\gamma(\gamma+1)}{\sqrt{\gamma}+\sqrt{2}}}}\,.$

Since

$\displaystyle\frac{d}{d\gamma}\left[\frac{2\gamma(\gamma-1)}{\gamma+1}-\frac{\sqrt{2}\gamma}{\sqrt{\gamma}+\sqrt{2}}\right]$
$\displaystyle=\frac{2(\gamma^{2}+2\gamma-1)}{(\gamma+1)^{2}}-\frac{\sqrt{2\gamma}+4}{2(\sqrt{\gamma}+\sqrt{2})^{2}}$
$\displaystyle=-\frac{4}{(\gamma+1)^{2}}-\frac{1}{\sqrt{2}(\sqrt{\gamma}+\sqrt{2})}-\frac{1}{(\sqrt{\gamma}+\sqrt{2})^{2}}+2>0$

because $-\frac{4}{(\gamma+1)^{2}}>-1$,
$-\frac{1}{\sqrt{2}(\sqrt{\gamma}+\sqrt{2})}>-\frac{1}{3}$, and
$-\frac{1}{(\sqrt{\gamma}+\sqrt{2})^{2}}>-\frac{1}{2}$ for any
$\gamma\in(1,\gamma_{\star}]$, and since
$C_{1}(\gamma_{\star})=\sqrt{-k(\gamma_{\star},z_{M})V_{1}}$ by the definition
of $\gamma_{\star}$, we deduce that $C_{1}-\sqrt{-k(\gamma,z_{M})V_{1}}\leq 0$
when $\gamma\leq\gamma_{\star}$ with equality only when
$\gamma=\gamma_{\star}$. Hence, $P_{1}$ is located below the curve
$B_{k_{M}}(V)$ for $\gamma<\gamma_{\star}$, and $P_{1}$ lies on the curve
$B_{k_{M}}(V)$ when $\gamma=\gamma_{\star}$.

We will now employ a barrier argument (cf.(2.20)) to establish that the curve
$B_{k_{M}}(V)$ serves as an upper barrier for the solution of (4.17).
Specifically, we will show that for all $\gamma\in(1,\gamma_{\star}]$,
$z\in[z_{m},z_{M}]$, and $V\in[V_{1},V_{6})$,

$\displaystyle\cfrac{F(V,\sqrt{-k(\gamma,z_{M})V};\gamma,z)}{G(V,\sqrt{-k(\gamma,z_{M})V};\gamma,z)}+\frac{1}{2}\sqrt{\cfrac{k(\gamma,z_{M})}{-V}}<0.$

By Lemma 3.10, $G(V,\sqrt{-k(\gamma,z_{M})V};\gamma,z)<0$ for any
$\gamma\in(1,\gamma_{\star}]$, $z\in[z_{m},z_{M}]$ and $V\in[V_{1},V_{6})$.
Hence, using the same procedure as outlined in Lemma 4.5 and recalling (4.12),
it is enough to show that

$\displaystyle\mathfrak{B}_{k(z_{M})}(V,z,m)$
$\displaystyle:=(m-1-m\gamma)V^{2}+[-2-m(\gamma-1)+m(\gamma-2)\gamma
z+(m-1)k(\gamma,z_{M})]V$
$\displaystyle\quad+\cfrac{2mk(\gamma,z_{M})z}{1+V}-1-m\gamma z>0.$

As $-1<V_{6}<0$ for any $z\in[z_{m},z_{M}]$, by Lemma 2.6, we have

$\displaystyle\frac{\partial k(\gamma,z)}{\partial
z}=(\frac{1}{V_{6}^{2}}-1)\frac{\partial V_{6}}{\partial z}>0.$ (4.18)

Thus, we have

$\displaystyle k(\gamma,z_{M})\geq k(\gamma,z)\quad\text{for any
}z\in[z_{m},z_{M}].$ (4.19)

When $m=1$, recalling (4.12) and using (4.19) and (4.13), we deduce that

$\displaystyle\mathfrak{B}_{k(z_{M})}(V,z,1)\geq\mathfrak{B}_{k}(V,z,1)>0$

for any $z\in[z_{m},z_{M}]$ and $V\in[V_{1},V_{6})$.

When $m=2$, by (2.12), we have
$-1<V_{6}\leq\frac{-\sqrt{2}}{\sqrt{\gamma}+\sqrt{2}}\leq-\frac{1}{2}$ for any
$\gamma\in(1,2]$ and $z\in[z_{m},z_{M}]$. Thus,

$\displaystyle
k(\gamma,z_{M})V+\frac{4zk(\gamma,z_{M})}{1+V}=\frac{k(\gamma,z_{M})}{1+V}(V^{2}+V+4z)>\frac{k(\gamma,z_{M})}{1+V}(V_{6}^{2}+V_{6}+4z)$

for any $V\in[V_{1},V_{6})$. By direct computations, we obtain

$\displaystyle V_{6}^{2}+V_{6}+4z$
$\displaystyle=\frac{2(\gamma-2)^{2}z^{2}+2(2-\gamma)zw+2(6-\gamma)z}{4}>0.$

Therefore, by (4.19) and and (4.13), we obtain
$\mathfrak{B}_{k(z_{M})}(V,z,2)\geq\mathfrak{B}_{k}(V,z,2)>0$, thereby
completing the proof. ∎

######  Proposition 4.7.

For any $\gamma\in(1,\gamma_{\star}]$, the analytic solution to (2.1) which
connects $P_{1}$ to either $P_{6}$ or $P_{8}$, guaranteed by Theorem 4.4 with
the initial condition $C(V_{1})=C_{1}$, can only connect to $P_{6}$.

###### Proof.

When $z=z_{M}$, we have $P_{6}=P_{8}$, thus obviating the need for further
discussion. If $z\in(0,z_{M})$, by Theorem 4.4, it is equivalent to
demonstrating that the solution trajectory can not connect to $P_{8}$. We will
discuss $z\in(0,z_{m}]$ and $z\in[z_{m},z_{M})$ separately.

Let $z\in(0,z_{m}]$ be given. We observe that when $C_{8}(z)\geq
C_{1}|_{\gamma=2}$, the solution trajectory cannot connect to $P_{8}$, since
the solution of (2.1) with $C(V_{1})=C_{1}$ is decreasing by Lemma 3.10. We
further note that $z=\frac{2}{3(\gamma+4)}$ leads to $C_{8}(z)=C_{1}(2)$. By
Lemma 2.1 and Lemma 2.6, $C_{8}(z)\geq C_{1}|_{\gamma=2}>C_{1}(\gamma)$ for
any $\gamma\in(1,\gamma_{\star}]$ and $z\leq\frac{2}{3(\gamma+4)}$. On the
other hand, it is easy to check $z_{m}<\frac{2}{3(\gamma+4)}$:

$\displaystyle
z_{m}-\frac{2}{3(\gamma+4)}=\frac{\gamma-1}{(2\gamma-1)(\gamma+1)}-\frac{2}{3(\gamma+4)}=\frac{(2-\gamma)(\gamma-5)}{(\gamma-\frac{1}{2})(\gamma+1)(\gamma+4)}<0,$

and hence, the conclusion follows for $z\in(0,z_{m}]$.

When $z\in[z_{m},z_{M})$, we have
$C(V_{6};\gamma,z)<\sqrt{-k(\gamma,z_{M})V_{6}(z)}$ by Lemma 4.6. Therefore,
in order to show that this solution can not connect to $P_{8}$, it is
sufficient to show that

$\displaystyle\sqrt{-k(\gamma,z_{M})V_{6}(z)}<C_{8}(z),\ \text{ equvalently }\
-k(\gamma,z_{M})V_{6}(z)-C_{8}^{2}(z)<0.$

Since
$-k(\gamma,z_{M})V_{6}(z_{M})-C_{8}^{2}(z_{M})=C_{6}(z_{M})^{2}-C_{8}(z_{M})^{2}=0$
by (2.11) and (4.15), the proof will be complete upon showing that
$-k(\gamma,z_{M})V_{6}(z)-C_{8}^{2}(z)$ is monotone increasing in $z$. Now,
differentiating with respect to $z$ (for any fixed
$\gamma\in(1,\gamma_{\star})$), and recalling $\frac{dC_{8}}{dz}<0$,

$\displaystyle\frac{d}{dz}\left(-k(\gamma,z_{M})V_{6}(z)-C_{8}^{2}(z)\right)=-\sqrt{\frac{\gamma}{2}}C_{8}(z_{M})\frac{dV_{6}}{dz}-2C_{8}(z)\frac{dC_{8}}{dz}>C_{8}(z_{M})\left(-\sqrt{\frac{\gamma}{2}}\frac{dV_{6}}{dz}-2\frac{dC_{8}}{dz}\right).$

The inner bracket is

$\displaystyle-\sqrt{\frac{\gamma}{2}}\frac{dV_{6}}{dz}-2\frac{dC_{8}}{dz}$
$\displaystyle=-\sqrt{\frac{\gamma}{2}}(\frac{\gamma-2}{2}+\frac{1}{2}\frac{(\gamma+2)-(\gamma-2)^{2}z}{w})-(\gamma-2-\frac{(\gamma+2)-(\gamma-2)^{2}z}{w})$
$\displaystyle=(2-\gamma)(\frac{1}{2}\sqrt{\frac{\gamma}{2}}+1)+\frac{(\gamma+2)-(\gamma-2)^{2}z}{w}(1-\frac{1}{2}\sqrt{\frac{\gamma}{2}})>0$

where we have used $z<z_{M}=\frac{1}{\gamma+2+2\sqrt{2\gamma}}<\frac{1}{4}$
and $\gamma_{\star}<2$ to conclude the positivity. ∎

###  4.3 No connection to $P_{6}$ for $\gamma\in[2,3]$

In this subsection, we shall employ another barrier function $B_{s}(V)$ to
demonstrate that for $\gamma\in[2,3]$, the solution trajectory originating at
$P_{1}$ and propagated by (2.1) can only establish a connection with $P_{8}$.

We define

$\displaystyle B_{s}(V)=-\sqrt{\frac{\gamma}{2}}V.$

From (2.11), we observe that

$\displaystyle\frac{C_{8}(z_{M})}{V_{8}(z_{M})}=-\sqrt{\frac{\gamma}{2}}.$
(4.20)

First we will show that the solution trajectory of (2.1) starting from $P_{1}$
remains above the curve $B_{s}(V)$ for
$V\in[V_{1},-\sqrt{\frac{2}{\gamma}}C_{8}(z))$.

######  Lemma 4.8.

For any $\gamma\in[2,3]$ and $z\in(0,z_{M}]$, the curve
$B_{s}(V)=-\sqrt{\frac{\gamma}{2}}V$ is a lower barrier for the solution of

$\displaystyle\begin{cases}&\cfrac{dC}{dV}=\cfrac{F(V,C;\gamma,z)}{G(V,C;\gamma,z)},\quad
V\in[V_{1},-\sqrt{\frac{2}{\gamma}}C_{8}(z)),\\\ &C(V_{1})=C_{1}.\end{cases}$
(4.21)

###### Proof.

To show that $B_{s}(V)$ is a lower barrier of the solution of (4.21), we first
verify that the initial point $P_{1}=(V_{1},C_{1})$ lies on or above the curve
$B_{s}(V)$. This follows from

$\displaystyle
C_{1}+\sqrt{\frac{\gamma}{2}}V_{1}=\frac{\sqrt{2\gamma(\gamma-1)}}{\gamma+1}-\sqrt{\frac{\gamma}{2}}\frac{2}{\gamma+1}=\frac{\sqrt{2\gamma}}{\gamma+1}(\sqrt{\gamma-1}-1)\geq
0$

for any $\gamma\in[2,3]$ where the equality holds when $\gamma=2$.

Next, we employ a barrier argument (cf. (2.20)) to show that $B_{s}(V)$ is a
lower barrier for the solution trajectory of (4.21). Specifically, we aim to
prove that for any $\gamma\in[2,3]$, $z\in(0,z_{M}]$, and
$V\in[V_{1},-\sqrt{\frac{2}{\gamma}}C_{8}(z))$,

$\displaystyle\cfrac{F(V,-\sqrt{\frac{\gamma}{2}}V;\gamma,z)}{G(V,-\sqrt{\frac{\gamma}{2}}V;\gamma,z)}+\sqrt{\frac{\gamma}{2}}>0.$
(4.22)

By Lemma 3.10, $G(V,-\sqrt{\frac{\gamma}{2}}V;\gamma,z)<0$ for any
$\gamma\in[2,3]$, $z\in(0,z_{M}]$ and
$V\in[V_{1},-\sqrt{\frac{2}{\gamma}}C_{8}(z))$, and so it is sufficient to
prove that

$\displaystyle
F(V,-\sqrt{\frac{\gamma}{2}}V;\gamma,z)+\sqrt{\frac{\gamma}{2}}G(V,-\sqrt{\frac{\gamma}{2}}V;\gamma,z)<0.$
(4.23)

By (1.18) and (1.17), we have

$\displaystyle
F(V,-\sqrt{\frac{\gamma}{2}}V;\gamma,z)+\sqrt{\frac{\gamma}{2}}G(V,-\sqrt{\frac{\gamma}{2}}V;\gamma,z)=-m\sqrt{\frac{\gamma}{2}}V^{2}\Big{[}(-\gamma+\frac{1}{2})V+\frac{z\gamma
V}{2(1+V)}+\frac{(\gamma z-1)(\gamma-1)}{2}-z\gamma\Big{]}.$

Since $-m\sqrt{\frac{\gamma}{2}}V^{2}<0$, it is sufficient to show that

$\displaystyle\mathfrak{B}(V,z):=(-\gamma+\frac{1}{2})V+\frac{z\gamma
V}{2(1+V)}+\frac{(\gamma z-1)(\gamma-1)}{2}-z\gamma>0$ (4.24)

for any $\gamma\in[2,3]$, $z\in(0,z_{M}]$ and
$V\in[V_{1},-\sqrt{\frac{2}{\gamma}}C_{8}(z))$.

By Lemma 2.6, $C_{8}(z)>C_{8}(z_{M})$ for any $z\in(0,z_{M})$. Thus,

$\displaystyle-\sqrt{\frac{2}{\gamma}}C_{8}(z)<-\sqrt{\frac{2}{\gamma}}C_{8}(z_{M})=V_{8}(z_{M}).$

Therefore, if we can establish the validity of (4.24) for all
$V\in[V_{1},V_{8}(z_{M}))$, it trivially holds for all
$V\in[V_{1},-\sqrt{\frac{2}{\gamma}}C_{8}(z))$. Notice that for any
$z\in(0,z_{M}]$ and $V\in[V_{1},V_{8}(z_{M}))$, we have

$\displaystyle\frac{\partial\mathfrak{B}(V,z)}{\partial
V}=-\gamma+\frac{1}{2}+\frac{z\gamma}{2(1+V)^{2}}\leq-\gamma+\frac{1}{2}+\frac{\gamma
z_{M}}{2(1+V_{1})^{2}}=\left(\frac{(\gamma+1)^{2}}{2(\gamma-1)^{2}}z_{M}-1\right)\gamma+\frac{1}{2}.$

Given that $z_{M}(\gamma)=\frac{1}{\gamma+2+2\sqrt{2\gamma}}$ and
$\frac{\gamma+1}{\gamma-1}=1+\frac{2}{\gamma-1}$ are both positive and
monotonically decreasing functions in $\gamma$, it follows that
$\frac{(\gamma+1)^{2}}{(\gamma-1)^{2}}z_{M}(\gamma)-1$ is also monotone
decreasing in $\gamma$. Hence for all $\gamma\in[2,3]$, we have

$\displaystyle\frac{(\gamma+1)^{2}}{2(\gamma-1)^{2}}z_{M}-1\leq-\frac{7}{16}\
\ \text{ and }\ \
\left(\frac{(\gamma+1)^{2}}{2(\gamma-1)^{2}}z_{M}-1\right)\gamma+\frac{1}{2}\leq-\frac{7}{8}+\frac{1}{2}=-\frac{3}{8}<0,$

which implies that for any $\gamma\in[2,3]$, $z\in(0,z_{M}]$ and
$V\in[V_{1},V_{8}(z_{M}))$,

$\displaystyle\mathfrak{B}(V,z)>\mathfrak{B}(V_{8}(z_{M}),z).$

To finish the proof of (4.22), it is now sufficient to show that
$\mathfrak{B}(V_{8}(z_{M}),z)\geq 0$. By (2.11), $V_{8}(z_{M})$ is independent
of $z$ so that

$\displaystyle\frac{\partial\mathfrak{B}(V_{8}(z_{M}),z)}{\partial
z}=\frac{\gamma
V_{8}(z_{M})}{2(1+V_{8}(z_{M}))}+\frac{\gamma(\gamma-3)}{2}<0.$

Hence, we obtain

$\displaystyle\mathfrak{B}(V_{8}(z_{M}),z)\geq\mathfrak{B}(V_{8}(z_{M}),z_{M}),$

where the equality holds when $z=z_{M}$. By (4.20) and Lemma 2.2,

$\displaystyle\mathfrak{B}(V_{8}(z_{M}),\gamma,z_{M})=\cfrac{F(V_{8}(z_{M}),C_{8}(z_{M}),\gamma,z_{M})+\sqrt{\frac{\gamma}{2}}G(V_{8}(z_{M}),C_{8}(z_{M}),\gamma,z_{M})}{-m\sqrt{\frac{\gamma}{2}}V_{8}(z_{M})^{2}}=0.$

In conclusion, we have shown that for $\gamma\in[2,3]$, $z\in(0,z_{M}]$ and
$V\in[V_{1},V_{8}(z_{M}))$, $\mathfrak{B}(V,z)>0$, thereby completing the
proof. ∎

######  Proposition 4.9.

For any $\gamma\in[2,3]$, the analytic solution to (2.1) connecting $P_{1}$ to
either $P_{6}$ or $P_{8}$, guaranteed by Theorem 4.4, can only connect to
$P_{8}$.

###### Proof.

When $z=z_{M}$, the points $P_{6}$ and $P_{8}$ coincide, rendering any further
discussion unnecessary. For $z\in(0,z_{M})$, by Theorem 4.4, it is equivalent
to showing that the solution trajectory cannot connect to $P_{6}$.

By (2.12) and Lemma 2.6, for any $\gamma\in[2,3]$ and $z\in(0,z_{M})$, it
holds

$\displaystyle C_{6}(z)<C_{8}(z_{M}).$

Moreover, from (2.4) and (2.6),

$\displaystyle V_{6}(z)+\sqrt{\frac{2}{\gamma}}C_{8}(z)$
$\displaystyle=\frac{\sqrt{\frac{2}{\gamma}}-1+(\sqrt{\frac{2}{\gamma}}+1)(\gamma-2)z+(\sqrt{\frac{2}{\gamma}}-1)w}{2}<\frac{\sqrt{\frac{2}{\gamma}}-1+(\sqrt{\frac{2}{\gamma}}+1)(\gamma-2)z_{M}}{2}=0$

since $z_{M}=\frac{1}{(\sqrt{\gamma}+\sqrt{2})^{2}}$, which shows that
$(V_{1},V_{6}(z))\subset(V_{1},-\sqrt{\frac{2}{\gamma}}C_{8}(z))$. Thus,
$P_{6}$ always lies below the curve
$\\{B_{s}(V)\,|\,V\in(V_{1},-\sqrt{\frac{2}{\gamma}}C_{8}(z))\\}$ for
$z\in(0,z_{M})$. On the other hand, by Lemma 4.8, the solution trajectory of
(4.21) is always above $B_{s}(V)$ on
$(V_{1},-\sqrt{\frac{2}{\gamma}}C_{8}(z))$. Therefore, we conclude that the
solution cannot connect to $P_{6}$. ∎

##  5 Solving to left: $P_{6}$

In this section, we will refine our analysis and the existence result around
$P_{6}$ for $\gamma\in(1,2]$ by deriving an appropriate upper bound for the
backwards solution of (2.1) starting from $P_{6}$ to determine a more precise
sonic window for $z_{std}(\gamma;P_{6})$. In addition, we will prove that, for
$\gamma\in(1,2]$, the value $z_{std}(\gamma;P_{6})$ is unique when it exists.

###  5.1 Existence for $\gamma\in(1,2]$

Recall from Section 2.2 that, for $\gamma\in(1,2]$, $z_{g}$ is defined to be
the value of $z$ such that $V_{4}(z_{g})=V_{6}(z_{g})$. In this subsection, we
rigorously demonstrate that for $\gamma\in(1,2]$ and $z\in[z_{m},z_{g}]$, the
analytic solution to (2.1) backwards from $P_{6}$ guaranteed by Theorem 3.8
and defined on the domain $[V_{1},V_{6}]$ by Lemma 4.1 is indeed a lower
solution for $P_{6}$, which yields an improvement of the range of
$z_{std}(\gamma;P_{6})$ to $[z_{g},z_{M}]$. We remark that $z_{m}<z_{g}$ for
any $\gamma\in(1,2]$ and $m=1,2$. A proof of this simple fact may be found in
Appendix B.

In what follows, recalling the definitions (1.17)–(1.19), we use the notation

$\displaystyle G(V,C)$ $\displaystyle=C^{2}g_{1}(V)-g_{2}(V),$ (5.1)
$\displaystyle F(V,C)$ $\displaystyle=C(C^{2}f_{1}(V)-f_{2}(V))$ (5.2)

where

$\displaystyle g_{1}(V)$ $\displaystyle=(m+1)V+2mz,$ (5.3) $\displaystyle
g_{2}(V)$ $\displaystyle=V(1+V)(m\gamma z+1+V),$ (5.4) $\displaystyle
f_{1}(V)$ $\displaystyle=1+\frac{mz}{(1+V)},$ (5.5) $\displaystyle f_{2}(V)$
$\displaystyle=a_{1}(1+V)^{2}-a_{2}(1+V)+a_{3}.$ (5.6)

We rewrite $\frac{dC}{dV}=\frac{F(V,C)}{G(V,C)}$ as

$\displaystyle\cfrac{d\log{C}}{dV}=\frac{1}{C}\frac{dC}{dV}=\cfrac{C^{2}f_{1}(V)-f_{2}(V)}{C^{2}g_{1}(V)-g_{2}(V)}.$
(5.7)

For $\gamma\in(1,2]$, $z_{g}=z_{g}(\gamma)$ is defined to be the value such
that

$V_{4}(\gamma,z_{g})=V_{6}(\gamma,z_{g}).$

In fact, there exists $\gamma_{g}\in(2,3)$ such that $z_{g}$ defined in this
way is well-defined for $\gamma\in(1,\gamma_{g}]$, while for
$\gamma\in(\gamma_{g},3]$, $V_{4}$ meets $V_{8}$ at $z_{g}$ (defined in an
equivalent manner). A detailed discussion of $\gamma_{g}$ and $z_{g}$ is given
in [28]. However, for our analysis, we require an understanding of $z_{g}$
only in the range $\gamma\in(1,2]$. The value $z_{g}$ admits an explicit
representation as

$\displaystyle
z_{g}=\begin{cases}\cfrac{\sqrt{\gamma^{2}+(\gamma-1)^{2}}-\gamma}{\gamma(\gamma-1)}\quad&\text{
when }m=1,\\\
\cfrac{\sqrt{(2\gamma^{2}-\gamma+1)^{2}+2\gamma(\gamma-1)[4\gamma(\gamma-1)+\frac{8}{3}]}-(2\gamma^{2}-\gamma+1)}{\gamma[4\gamma(\gamma-1)+\frac{8}{3}]}\quad&\text{
when }m=2.\end{cases}$ (5.8)

We claim that for any $\gamma\in(1,2]$, $z\in[z_{m},z_{g}]$ gives a lower
solution for $P_{6}$. Recalling the definition of a lower solution, (4.5), it
is enough to show that

$\displaystyle\log
C(V_{1};\gamma,z,P_{6}):=-\int_{V_{1}}^{V_{6}(z)}\frac{d\log C}{dV}dV+\log
C_{6}(z)<\log C_{1}.$ (5.9)

Solving this inequality directly is not a trivial task, since the integral is
implicit as the integrand involves not only $V$ but also $C$ (cf. (5.7)). To
simplify our approach and avoid the complications associated with this
implicit integral, we will derive an explicit lower bound for $\frac{d\log
C}{dV}$ for any $\gamma\in(1,2]$, $z\in[z_{m},z_{g}]$ and
$V\in[V_{1},V_{6}(z))$.

######  Lemma 5.1.

For any $\gamma\in(1,2]$ and $z\in[z_{m},z_{g}]$, the solution obtained from
Theorem 3.8 and Lemma 4.1 satisfies

$\displaystyle-\int_{V_{1}}^{V_{6}(z)}\frac{d\log
C}{dV}dV<-\int_{V_{1}}^{V_{6}(z)}\frac{f_{1}(V)}{g_{1}(V)}dV.$ (5.10)

###### Proof.

By direct computations, we have

$\displaystyle\cfrac{d\log{C}}{dV}-\frac{f_{1}(V)}{g_{1}(V)}=\cfrac{C^{2}f_{1}(V)-f_{2}(V)}{C^{2}g_{1}(V)-g_{2}(V)}-\frac{f_{1}(V)}{g_{1}(V)}=-\frac{g_{1}(V)f_{2}(V)-f_{1}(V)g_{2}(V)}{[C^{2}g_{1}(V)-g_{2}(V)]g_{1}(V)}.$

We will show this function is positive for any $\gamma\in(1,2]$,
$z\in[z_{m},z_{g}]$, and $V\in[V_{1},V_{6}(z))$.

By (2.12) and the fact that for $\gamma\in(1,2]$, $z<z_{M}<\frac{1}{5}$, we
have $g_{1}(V)<0$ for $V\in[V_{1},V_{6})$. On the other hand, by Lemma 3.10,
$G(V,C)=C^{2}g_{1}(V)-g_{2}(V)<0$ for any $V\in[V_{1},V_{6})$. Therefore, it
is sufficient to show that

$\displaystyle q(V):=g_{1}(V)f_{2}(V)-f_{1}(V)g_{2}(V)<0.$

Note that $q(V)$ is a cubic polynomial in $V$. Also, $F(V,C)=G(V,C)=0$ at
$P_{4}$, $P_{6}$ and $P_{8}$ which implies
$C_{k}^{2}g_{1}(V_{k})=g_{2}(V_{k})$ and $C_{k}^{2}f_{1}(V_{k})=f_{2}(V_{k})$
for $k=4,6,8$. Consequently, $V_{4}$, $V_{6}$ and $V_{8}$ are three roots of
$g_{1}f_{2}-g_{2}f_{1}=0$. Thus,

$\displaystyle
q(V)=\frac{m[(m+1)(\gamma-1)+2]}{2}(V-V_{4})(V-V_{6})(V-V_{8}).$

According to (2.12), we have $V_{6}\leq V_{8}$ with the equality when
$z=z_{M}$. Therefore, the sign of $q(V)$ depends on the location of $V_{4}$.
If we can show that $V_{4}\geq V_{6}$ for $z\in[z_{m},z_{g}]$, then $q(V)<0$
for any $V\in[V_{1},V_{6})$. We claim that

$\displaystyle V_{4}(z)\geq V_{6}(z)\ \text{ for }\ z\in[z_{m},z_{g}]$ (5.11)

where the equality holds when $z=z_{g}$. By using (2.15), we have

$\displaystyle\frac{dV_{4}(z)}{dz}=\frac{d}{dz}\left(\frac{-2m\gamma
z-2}{(m+1)\gamma+1-m}\right)=\frac{-2m\gamma}{(m+1)\gamma+1-m}<0,$

which implies $V_{4}$ is a decreasing function in $z$. By Lemma 2.6,
$V_{6}(z)$ is an increasing function in $z$. From the definition of $z_{g}$
(5.8), $V_{4}(z_{g})=V_{6}(z_{g})$ for any $\gamma\in(1,2]$. We have shown
(5.11), which leads to

$\displaystyle\cfrac{d\log{C}}{dV}-\frac{f_{1}(V)}{g_{1}(V)}>0.$ (5.12)

This completes the proof of (5.10). ∎

Motivated by Lemma 5.1, we define

$\displaystyle\delta(V_{1};z):=-\int_{V_{1}}^{V_{6}(z)}\frac{f_{1}(V;z)}{g_{1}(V;z)}dV+\log
C_{6}(z)$ (5.13)

where we have used the notations $f_{1}(V;z)$ and $g_{1}(V;z)$ for $f_{1}(V)$
and $g_{1}(V)$ to emphasize the dependence of $f_{1}$ and $g_{1}$ on $z$. By
Lemma 5.1, we have for any $\gamma\in(1,2]$ and $z\in(z_{m},z_{g}]$,

$\displaystyle\log C(V_{1};\gamma,z,P_{6})<\delta(V_{1};z).$

Our next step is to show that for any $\gamma\in(1,2]$, $z=z_{g}$ gives a
lower solution for $P_{6}$.

######  Lemma 5.2.

For any $\gamma\in(1,2]$ and $z=z_{g}$,

$\displaystyle\delta(V_{1};z_{g})<\log C_{1}.$ (5.14)

###### Proof.

We first evaluate $\delta(V_{1};z)$ in (5.13) by using (5.5) and (5.3) to
calculate the integral explicitly as

$\displaystyle\delta(V_{1};z)$
$\displaystyle=\cfrac{(m^{2}-m)z+(m+1)}{(2mz-m-1)(m+1)}\log\frac{(m+1)V_{6}+2mz}{(m+1)V_{1}+2mz}+\cfrac{mz}{2mz-m-1}\log\frac{1+V_{1}}{1+V_{6}}+\log(1+V_{6}).$
(5.15)

By (1.25) and (1.26),

$\displaystyle 1+V_{1}$ $\displaystyle=\frac{\gamma-1}{\gamma+1}\text{ and
}\log C_{1}=\frac{1}{2}\log(1+V_{1})+\frac{1}{2}\log\frac{2\gamma}{\gamma+1}.$
(5.16)

To study the remainder of the expression for $\delta(V_{1};z)$, we treat $m=1$
and $m=2$ separately.

When $m=1$, by (2.15) and (5.8),

$\displaystyle V_{6}(z_{g})$ $\displaystyle=V_{4}(z_{g})=-\frac{\gamma
z_{g}+1}{\gamma}=-z_{g}-\frac{1}{\gamma},$ (5.17) $\displaystyle z_{g}$
$\displaystyle=\frac{\sqrt{\gamma^{2}+(\gamma-1)^{2}}-\gamma}{\gamma(\gamma-1)}=\frac{\gamma-1}{\gamma(\sqrt{\gamma^{2}+(\gamma-1)^{2}}+\gamma)}<\frac{1}{2}.$
(5.18)

Together with (5.15) and (5.16), we then have

$\displaystyle\delta(V_{1};z_{g})-\log
C_{1}=\frac{1}{2}\log\frac{(1+V_{4})(\gamma+1)}{2\gamma}+\frac{1}{2(z_{g}-1)}\log[\frac{(V_{4}+z_{g})(1+V_{1})}{(V_{1}+z_{g})(1+V_{4})}]=:\frac{1}{2}(I+II).$

For $I$, we have $1+V_{4}<1$ and $\frac{\gamma+1}{2\gamma}<1$ for any
$\gamma\in(1,2]$. Thus, $I<0$. From (5.18), $\frac{1}{z_{g}-1}>-2$. Moreover,

$\displaystyle I+II$
$\displaystyle<\log\frac{(1+V_{4})(\gamma+1)}{2\gamma}-2\log[\frac{(V_{4}+z_{g})(1+V_{1})}{(V_{1}+z_{g})(1+V_{4})}]$
$\displaystyle=\log[\frac{(\gamma+1)(\gamma-1)}{2\gamma^{2}}(2-z_{g}(\gamma+1))^{2}(1-\frac{1}{(\sqrt{\gamma^{2}+(\gamma-1)^{2}}+\gamma)})^{3}]$
$\displaystyle<\log\frac{(\gamma+1)(\gamma-1)}{\gamma^{2}}+\log\frac{2-z_{g}(\gamma+1)}{2}+\log[(2-z_{g}(\gamma+1))(1-\frac{1}{\sqrt{5}+2})^{3}]$
$\displaystyle<\log\frac{2(\sqrt{5}+1)^{3}}{(\sqrt{5}+2)^{3}}<0$

where we have used that $2-z_{g}(\gamma+1)<2$ and
$\cfrac{1}{(\sqrt{\gamma^{2}+(\gamma-1)^{2}}+\gamma)}$ is a decreasing
function. This concludes the proof in the case $m=1$.

When $m=2$, for any $\gamma\in(1,2]$, by (1.25), (2.15), (2.4) and (5.8),

$\displaystyle V_{4}$ $\displaystyle=-\frac{4\gamma z_{g}+2}{3\gamma-1},$
(5.19) $\displaystyle z_{g}$
$\displaystyle=\frac{2(\gamma-1)}{\sqrt{(2\gamma^{2}-\gamma+1)^{2}+2\gamma(\gamma-1)[4\gamma(\gamma-1)+\frac{8}{3}]}+(2\gamma^{2}-\gamma+1)}.$
(5.20)

We first claim that $z_{g}<\frac{1}{8}$. By direct computations, for any
$\gamma\in(1,2]$,

$\displaystyle(16(\gamma-1)-(2\gamma^{2}-\gamma+1))^{2}-(2\gamma^{2}-\gamma+1)^{2}-2\gamma(\gamma-1)[4\gamma(\gamma-1)+\frac{8}{3}]$
$\displaystyle=8(\gamma-1)(-\gamma^{3}-7\gamma^{2}+\frac{106}{3}\gamma-36)<0$

where the cubic polynomial is negative for any $\gamma\in(1,2]$ as shown in
Proposition E.4. This then implies

$16(\gamma-1)-(2\gamma^{2}-\gamma+1)<\sqrt{(2\gamma^{2}-\gamma+1)^{2}+2\gamma(\gamma-1)[4\gamma(\gamma-1)+\frac{8}{3}]},$

and hence

$\frac{2(\gamma-1)}{\sqrt{(2\gamma^{2}-\gamma+1)^{2}+2\gamma(\gamma-1)[4\gamma(\gamma-1)+\frac{8}{3}]}+(2\gamma^{2}-\gamma+1)}<\frac{1}{8},$

that is,

$\displaystyle z_{g}<\frac{1}{8},$ (5.21)

as claimed. By (5.15), (5.16), (5.19) and (5.20), we have

$\displaystyle\delta(V_{1};z_{g})-\log C_{1}$
$\displaystyle=\frac{2z_{g}+3}{2(12z_{g}-9)}\log[\frac{(3V_{4}+4z_{g})^{2}}{(3V_{1}+4z_{g})^{2}}\frac{2\gamma}{\gamma-1}]+\frac{6z_{g}-9}{4(12z_{g}-9)}\log\frac{(1+V_{4})^{4}}{(1+V_{1})^{4}}+\frac{7z_{g}-3}{12z_{g}-9}\log\frac{\gamma-1}{2\gamma}$
$\displaystyle<\frac{2z_{g}+3}{2(12z_{g}-9)}\log[\frac{(3V_{4}+4z_{g})^{2}}{(3V_{1}+4z_{g})^{2}}\frac{2\gamma}{\gamma-1}]+\frac{6z_{g}-9}{4(12z_{g}-9)}\log\frac{(\gamma-1)(1+V_{4})^{4}}{2\gamma(1+V_{1})^{4}}$

where we have used (5.21) in the inequality. Now, we show that both terms are
negative. We compute

$\displaystyle\frac{(3V_{4}+4z_{g})^{2}}{(3V_{1}+4z_{g})^{2}}\frac{2\gamma}{\gamma-1}=(\cfrac{\frac{-12\gamma
z_{g}-6}{3\gamma-1}+4z_{g}}{\frac{-6}{\gamma+1}+4z_{g}})^{2}\frac{2\gamma}{\gamma-1}=\frac{(-6-4z_{g})^{2}}{(-6+4z_{g}(\gamma+1))^{2}}\frac{2\gamma(\gamma+1)^{2}}{(3\gamma-1)^{2}(\gamma-1)}.$

Note that $\frac{2\gamma(\gamma+1)^{2}}{(3\gamma-1)^{2}(\gamma-1)}>1$ for
$\gamma\in(1,2]$. Also, $z_{g}>0$ implies

$\displaystyle\frac{(-6-4z_{g})^{2}(\gamma+1)^{2}}{(-6+4z_{g}(\gamma+1))^{2}}>1.$

Hence,

$\displaystyle\frac{2z_{g}+3}{2(12z_{g}-9)}\log[\frac{(3V_{4}+4z_{g})^{2}}{(3V_{1}+4z_{g})^{2}}\frac{2\gamma}{\gamma-1}]<0$
(5.22)

because $12z_{g}-9<0$. As for the second term, we first note that

$\displaystyle\frac{(\gamma-1)(1+V_{4})^{4}}{2\gamma(1+V_{1})^{4}}$
$\displaystyle=\frac{(\gamma+1)^{4}}{2\gamma}\cfrac{(3(\gamma-1)-4\gamma
z_{g})^{4}}{(\gamma-1)^{3}(3\gamma-1)^{4}}$
$\displaystyle=\frac{81(\gamma+1)^{4}}{(3\gamma-1)^{4}}\frac{\gamma-1}{2\gamma}(1-\frac{8}{3(\sqrt{(\frac{2\gamma^{2}-\gamma+1}{\gamma})^{2}+8[(\gamma-1)^{2}+\frac{2(\gamma-1)}{3\gamma}]}+\frac{2\gamma^{2}-\gamma+1}{\gamma}})^{4}.$

Moreover, for any $\gamma\in(1,2]$,

$\displaystyle(\frac{2\gamma^{2}-\gamma+1}{\gamma})^{\prime}$
$\displaystyle=2-\frac{1}{\gamma^{2}}>0\ \ \text{ and }\ \
((\gamma-1)^{2}+\frac{2(\gamma-1)}{3\gamma})^{\prime}=\frac{2}{3\gamma^{2}}+2\gamma-2>0,$

which implies

$\displaystyle(1-\frac{8}{3(\sqrt{(\frac{2\gamma^{2}-\gamma+1}{\gamma})^{2}+8[(\gamma-1)^{2}+\frac{2(\gamma-1)}{3\gamma}]}+\frac{2\gamma^{2}-\gamma+1}{\gamma}})^{4}<(1-\frac{8}{3(\sqrt{(\frac{7}{2})^{2}+8[1+\frac{1}{3}]}+\frac{7}{2}})^{4}<\frac{1}{4}.$

Therefore, for any $\gamma\in(1,2]$, we deduce that

$\displaystyle\frac{(\gamma-1)(1+V_{4})^{4}}{2\gamma(1+V_{1})^{4}}<\frac{81(\gamma-1)(\gamma+1)^{4}}{8\gamma(3\gamma-1)^{4}}<1$

where the last inequality is shown in Proposition E.5. We then have

$\displaystyle\frac{6z_{g}-9}{4(12z_{g}-9)}\log\frac{(\gamma-1)(1+V_{4})^{4}}{2\gamma(1+V_{1})^{4}}<0,$
(5.23)

thereby completing the proof. ∎

######  Proposition 5.3.

For any $\gamma\in(1,2]$, $z\in[z_{m},z_{g}]$ gives a lower solution for
$P_{6}$.

###### Proof.

We first show $\delta(V_{1};z)$ is an increasing function. From (5.13) and
(5.15)

$\displaystyle\frac{d\delta(V_{1};z)}{dz}=$
$\displaystyle-\int_{V_{1}}^{V_{6}(z)}\frac{-2m+m(m-1)V}{(1+V)((m+1)V+2mz)^{2}}dV-\frac{dV_{6}(z)}{dz}\frac{f_{1}(V_{6}(z);z)}{g_{1}(V_{6}(z);z)}+\frac{1}{C_{6}(z)}\frac{dC_{6}(z)}{dz}>0$

where we have used $\frac{dV_{6}(z)}{dz}=\frac{dC_{6}(z)}{dz}>0$ by Lemma 2.6
and $\frac{f_{1}(V_{6}(z);z)}{g_{1}(V_{6}(z);z)}<0$ by (5.12). By Lemma 5.1
and Lemma 5.2, we then have

$\displaystyle\log
C(V_{1};\gamma,z,P_{6})<\delta(V_{1};z)\leq\delta(V_{1};z_{g})<\log C_{1}$

for any $\gamma\in(1,2]$ and $z\in[z_{m},z_{g}]$. This finishes the proof. ∎

###  5.2 Uniqueness of $z_{std}$ for $P_{6}$ when $\gamma\in(1,2]$

Recall $z_{std}(\gamma;P_{6})$ is the value of $z$ such that the solution
$C(V;\gamma,z_{std}(\gamma;P_{6}),P_{6})$ (cf. (4.3)) satisfies

$\displaystyle\begin{cases}&\cfrac{dC}{dV}=\cfrac{F(V,C;\gamma,z_{std})}{G(V,C;\gamma,z_{std})},\\\
&C(V_{1})=C_{1},\quad C(V_{6})=C_{6},\\\
&\frac{dC}{dV}(V_{6},C_{6})=c_{1}.\end{cases}$ (5.24)

By Proposition 4.9, we know that for $\gamma\in[2,3]$ the solution can only
connect to $P_{8}$ and therefore, we focus on $\gamma\in(1,2]$ for further
analysis of $z_{std}(\gamma;P_{6})$. Within this range of $\gamma$, we
demonstrate the uniqueness of $z_{std}$ for $P_{6}$. This is achieved by
showing that for any fixed $\gamma\in(1,2]$, the solution trajectories
$C(V;\gamma,z,P_{6})$ of (4.2) starting from $P_{6}$ do not intersect for
different values of $z\in[z_{g},z_{M}]$. In particular, at most one such
trajectory can connect to $P_{1}$.

######  Lemma 5.4.

For any $\gamma\in(1,2]$ fixed, the solution trajectories
$C(V;\gamma,z,P_{6})$ do not intersect for different $z\in[z_{g},z_{M}]$ in
the interval $[V_{1},V_{6})$.

###### Proof.

We argue by contradiction. For any fixed $\gamma\in(1,2]$, we write
$C^{\prime}(V,C,z)=\frac{dC(V,C,z)}{dV}$. We suppose that there exist
$z_{s},z_{t}\in[z_{g},z_{M}]$ and $z_{s}<z_{t}$ such that
$C(V;\gamma,z_{s},P_{6})$ and $C(V;\gamma,z_{t},P_{6})$ intersect at a point
$(V_{0},C_{0})$ where $V_{1}\leq V_{0}<V_{6}(z_{s})$. By the continuity of the
solution curves with respect to both $V$ and $z$ (see Remark 3.9), we may
assume without loss of generality that $(V_{0},C_{0})$ is the first such
intersection point to the left of $P_{6}(z_{s})$ and $P_{6}(z_{t})$. In
particular, there are no other intersection points within the triangular
region enclosed by the curves
$\\{(V,C(V;\gamma,z_{s},P_{6}))\,|\,V\in[V_{0},V_{6}(z_{s})]\\}$,
$\\{(V,C(V;\gamma,z_{t},P_{6}))\,|\,V\in[V_{0},V_{6}(z_{t})]\\}$ and
$\\{(V_{6}(z),C_{6}(z))\,|\,z\in[z_{s},z_{t}]\\}$. Then, we have
$C(V_{0},z)=C_{0}$ for all $z\in[z_{s},z_{t}]$ so that for all
$z\in(z_{s},z_{t})$,

$\displaystyle\frac{\partial C}{\partial z}(V_{0},C_{0},z)=0$ (5.25)

and $C^{\prime}(V_{0},C_{0},z_{s})\leq C^{\prime}(V_{0},C_{0},z_{t})$. By the
Mean Value Theorem, there exists a $\tilde{z}\in(z_{s},z_{t})$ such that
$\frac{\partial C^{\prime}}{\partial z}(V_{0},C_{0},\tilde{z})\geq 0$. We will
show $\frac{\partial C^{\prime}}{\partial z}(C_{0},V_{0},z)<0$ for all
$z\in(z_{s},z_{t})$ to reach the contradiction.

By direct computations from the explicit forms of $F$ and $G$ from
(1.17)–(1.18) and using (5.25), we have, for any $z\in(z_{s},z_{t})$

$\displaystyle\frac{\partial C^{\prime}}{\partial z}(C_{0},V_{0},z)$
$\displaystyle=\cfrac{C_{0}(\frac{mC_{0}^{2}}{1+V_{0}}+\frac{m\gamma(\gamma-3)}{2}(1+V_{0})-\frac{m\gamma(\gamma-1)}{2})G(V_{0},C_{0},z)-(2mC_{0}^{2}-m\gamma
V_{0}(1+V_{0}))F(V_{0},C_{0},z)}{G^{2}(V_{0},C_{0},z)}.$

Since $z_{s},z_{t}\in[z_{g},z_{M}]$, it is sufficient to show that

$\displaystyle\frac{\partial C^{\prime}}{\partial z}(C_{0},V_{0},z)<0$ (5.26)

for any $z\in(z_{g},z_{M})$. Substituting (1.18) and (1.17) into the above
formula and simplifying the expression, we arrive at

$\displaystyle\frac{\partial C^{\prime}}{\partial
z}(C_{0},V_{0},z)=\cfrac{mC_{0}(C_{0}^{2}-(1+V_{0})^{2})[((m-1)V_{0}-2)C_{0}^{2}+(m+1)\frac{\gamma-1}{2}\gamma
V_{0}^{2}(1+V_{0})]}{G^{2}(V_{0},C_{0},z)(1+V_{0})}.$

Notice that, for any $V_{0}\in[V_{1},V_{6})$ and $z\in(z_{g},z_{M})$

$\displaystyle C_{0}>0,\quad C_{0}^{2}-(1+V_{0})^{2}>0,\quad
G^{2}(V_{0},C_{0},z)(1+V_{0})>0.$

Thus in order to show $\frac{\partial C^{\prime}}{\partial
z}(C_{0},V_{0},z)<0$, it is enough to show

$\displaystyle
h(V_{0},C_{0},z):=((m-1)V_{0}-2)C_{0}^{2}+(m+1)\frac{\gamma-1}{2}\gamma
V_{0}^{2}(1+V_{0})<0.$

By Lemma 4.5, $C=\sqrt{\frac{(1+V_{6})^{2}}{V_{6}}V}$ is a lower barrier of
$C(V;\gamma,z,P_{6})$ with $z\in(z_{g},z_{M})$. Hence,

$\displaystyle h(V_{0},C_{0},z)$
$\displaystyle\leq\frac{(1+V_{6})^{2}}{V_{6}}((m-1)V_{0}-2)V_{0}+(m+1)\frac{\gamma-1}{2}\gamma
V_{0}^{2}(1+V_{0})$
$\displaystyle=V_{0}\Big{[}\frac{(1+V_{6})^{2}}{V_{6}}((m-1)V_{6}-2)+(m+1)\frac{\gamma-1}{2}\gamma
V_{0}(1+V_{0})\Big{]}$
$\displaystyle<\frac{V_{0}(1+V_{6})}{V_{6}}\Big{[}(1+V_{6})((m-1)V_{6}-2)+(m+1)\frac{\gamma-1}{2}\gamma
V_{6}^{2}\Big{]}$

because $V_{0}<V_{6}<V_{6}(z_{M}|_{\gamma=2})=-\frac{1}{2}$ by Lemma 2.6 and
$m=1,2$. Denote

$\displaystyle q(V_{6}):=(1+V_{6})((m-1)V_{6}-2)+(m+1)\frac{\gamma-1}{2}\gamma
V_{6}^{2}.$

Our goal is to show $q(V_{6})<0$. Again, by Lemma 2.6, we have
$-\frac{1}{2}=V_{6}(z_{M}|_{\gamma=2})>V_{6}>V_{6}(z_{g})=V_{4}(z_{g})$ for
any $z\in(z_{g},z_{M})$. Thus, if $q(V_{4}(z_{g}))\leq 0$ and
$q(-\frac{1}{2})\leq 0$, then $q(V_{6})<0$ for all $z\in(z_{g},z_{M})$ since
the coefficient of $V_{6}^{2}$ is positive. We first note that

$\displaystyle
q(-\frac{1}{2})=-1-\frac{m-1}{4}+\frac{(m+1)(\gamma-1)\gamma}{8}<0$

for any $\gamma\in(1,2]$ and $m=1,2$. For $q(V_{4}(z_{g}))$, we claim that
$V_{4}(z_{g})$ is a negative zero of $q$. To this end, by using
$V_{4}(z_{g})=V_{6}(z_{g})$ and $C_{4}(z_{g})=C_{6}(z_{g})$, we rewrite
$q(V_{4}(z_{g}))$ as

$\displaystyle q(V_{4}(z_{g}))$
$\displaystyle=\frac{1}{1+V_{4}(z_{g})}\left[(1+V_{6}(z_{g}))^{2}((m-1)V_{4}(z_{g})-2)+(m+1)\tfrac{\gamma-1}{2}\gamma(1+V_{4}(z_{g}))V_{4}(z_{g})^{2}\right]$
$\displaystyle=\frac{1}{1+V_{4}(z_{g})}\left[C_{4}(z_{g})^{2}((m-1)V_{4}(z_{g})-2)+(m+1)\tfrac{\gamma-1}{2}\gamma(1+V_{4}(z_{g}))V_{4}(z_{g})^{2}\right].$

Using (2.15), we replace $C_{4}(z_{g})$ by $H(V_{4}(z_{g}))$ to obtain

$\displaystyle
q(V_{4}(z_{g}))=\frac{V_{4}(z_{g})}{(m+1)V_{4}(z_{g})+2mz}\tilde{q}(V_{4}(z_{g})),$

where

$\displaystyle\tilde{q}(V_{4}(z_{g}))$ $\displaystyle=(V_{4}(z_{g})+1+m\gamma
z_{g})((m-1)V_{4}(z_{g})-2)+(m+1)\tfrac{\gamma-1}{2}\gamma
V_{4}(z_{g})((m+1)V_{4}(z_{g})+2mz_{g})$
$\displaystyle=\begin{cases}2\gamma(\gamma-1)V_{4}(z_{g})^{2}+2(\gamma(\gamma-1)z_{g}-1)V_{4}(z_{g})-2(1+\gamma
z_{g}),&m=1,\\\
(\tfrac{9}{2}\gamma(\gamma-1)+1)V_{4}(z_{g})^{2}+(2\gamma(3\gamma-2)z_{g}-1)V_{4}(z_{g})-2(1+2\gamma
z_{g}),&m=2.\end{cases}$

It is routine to check that $\tilde{q}$ has two roots
$-z_{g}-\frac{1}{\gamma}$, $\frac{1}{\gamma-1}$ when $m=1$ and $-\frac{4\gamma
z_{g}+2}{3\gamma-1}$ and $\frac{2}{3\gamma-2}$ when $m=2$. Therefore by (5.17)
and (5.19), $\tilde{q}(V_{4}(z_{g}))=0$ and $q(V_{4}(z_{g}))=0$. This finishes
the proof of $q(V_{6})<0$ and $\frac{\partial C^{\prime}}{\partial
z}(C_{0},V_{0},z)<0$ for all $z\in(z_{g},z_{M})$, which contradicts our
assumption. Therefore we conclude that $C(V;\gamma,z_{s},P_{6})$ can not
intersect $C(V;\gamma,z_{t},P_{6})$ if $z_{s}\neq z_{t}$. ∎

##  6 Solving to left: $P_{8}$

In this section, we again employ suitable barrier functions to delineate a
more precise range for $z$ in which $z_{std}(\gamma;P_{8})$ resides. By
Proposition 4.7, we may focus on $\gamma\in(\gamma_{\star},3]$ for further
analysis of $z_{std}(\gamma;P_{8})$.

###  6.1 Conditional existence for $\gamma\in(\gamma_{\star},\gamma_{1}]$

As described in Section 2.2, we define $\gamma_{1}$ to be the value such that
$C_{1}=\sqrt{-V_{1}}$. A simple calculation then establishes that

$\displaystyle\gamma_{1}=1+\sqrt{2}.$ (6.1)

By Lemma 2.1, $P_{1}$ is below $B_{1}(V)=\sqrt{-V}$ when
$\gamma\leq\gamma_{1}$. We will show that the solution trajectory, originating
at $P_{1}$ and propagated by (2.1), remains below $C=B_{1}(V)$ within a
specific range of $V$, the exact bounds of which will be established
subsequently, when $\gamma\leq\gamma_{1}$.

For each $\gamma$, we define $z_{1}$ to be the value such that
$C_{8}(z_{1})=\sqrt{-V_{8}(z_{1})}$. Then

$\displaystyle z_{1}=\frac{\sqrt{5}-1}{2(1+\sqrt{5}+\gamma)}.$ (6.2)

Since $V_{8}(z_{1})=C_{8}(z_{1})-1$ and $1-C_{8}^{2}(z_{1})=C_{8}(z_{1})$,

$\displaystyle V_{8}(z_{1})=\frac{\sqrt{5}-3}{2},\quad
C_{8}(z_{1})=\frac{\sqrt{5}-1}{2}.$ (6.3)

For any fixed $\gamma\in(1,3]$, we write $C_{8}=C_{8}(z)$,
$C^{\prime}_{8}(z)=\frac{dC_{8}(z)}{dz}$ and
$C^{\prime\prime}_{8}(z)=\frac{d^{2}C_{8}(z)}{dz^{2}}$. We first show the
concavity of $C_{8}^{2}$ with respect to $z$, which will be crucial for
subsequent arguments.

######  Lemma 6.1.

For any fixed $\gamma\in(1,3]$, and any $z\in(0,z_{M})$,

$\displaystyle C_{8}C^{\prime\prime}_{8}+(C^{\prime}_{8})^{2}<0.$

###### Proof.

By direct computations, we have

$\displaystyle C^{\prime\prime}_{8}=-\frac{4\gamma}{w^{3}}<0.$

For any $\gamma\in(1,3]$ and $z\in(0,z_{M})$, recalling (2.4), we have

$\displaystyle 1+(\gamma-2)z+w=2C_{6}+2w>2w.$

Hence,

$\displaystyle 4C_{8}C_{8}^{\prime\prime}+4(C_{8}^{\prime})^{2}$
$\displaystyle=(1+(\gamma-2)z+w)\frac{-8\gamma}{w^{3}}+4(C_{8}^{\prime})^{2}\leq(2C^{\prime}_{8})^{2}-\frac{16\gamma}{w^{2}}=(2C^{\prime}_{8}-\frac{4\sqrt{\gamma}}{w})(2C^{\prime}_{8}+\frac{4\sqrt{\gamma}}{w}).$

By Lemma 2.6, $C_{8}^{\prime}<0$ and thus
$2C^{\prime}_{8}-\frac{4\sqrt{\gamma}}{w}<0$. We will show
$2C^{\prime}_{8}+\frac{4\sqrt{\gamma}}{w}>0$. By using $0<w<1$ (cf. Remark
2.5), we see that

$\displaystyle(2C^{\prime}_{8}+\frac{4\sqrt{\gamma}}{w})w$
$\displaystyle=4\sqrt{\gamma}-\gamma-2+(\gamma-2)w+(\gamma-2)^{2}z>4\sqrt{\gamma}-\gamma-3=(3-\sqrt{\gamma})(\sqrt{\gamma}-1)>0$

for any $\gamma\in(1,3]$, thereby completing the proof. ∎

######  Lemma 6.2.

For any $\gamma\in(\gamma_{\star},\gamma_{1}]$, $z\in(0,z_{1}(\gamma)]$ gives
a upper solution for $P_{8}$.

###### Proof.

In view of Lemma 4.3, we may determine the existence of
$\overline{z}(\gamma;P_{8})$ such that for all
$z\leq\overline{z}(\gamma;P_{8})$, it holds that $C_{8}(\gamma,z)\geq
C_{1}(\gamma)$ and each $z\in(0,\overline{z}(\gamma;P_{8})]$ gives rise to an
upper solution for $P_{8}$. However, the expression for such
$\overline{z}(\gamma;P_{8})$, derived from the equation
$C_{8}(\gamma,\overline{z}(\gamma;P_{8}))=C_{1}(\gamma)$, is intricate and
inconvenient. For that reason, we use another value of $z$, which has a
closed-form representation. Specifically, we define $\tilde{z}_{m}(\gamma)$ to
be the value satisfying

$\displaystyle
C_{8}(\gamma,\tilde{z}_{m}(\gamma))=C_{1}(\gamma_{1})=\sqrt{2-\sqrt{2}}$ (6.4)

so that

$\displaystyle\tilde{z}_{m}(\gamma)=\frac{-2^{\frac{1}{4}}\sqrt{1+\sqrt{2}}+2^{\frac{3}{4}}\sqrt{1+\sqrt{2}}}{2\sqrt{2}+2^{\frac{5}{4}}\sqrt{1+\sqrt{2}}+\gamma}.$

It is easy to check that for any $\gamma\in(\gamma_{\star},\gamma_{1}]$,

$\displaystyle\tilde{z}_{m}(\gamma)\geq\tilde{z}_{m}(\gamma_{1})>\frac{2}{25}.$
(6.5)

By Lemma 2.6 and Lemma 2.1, for any $\gamma\in(\gamma_{\star},\gamma_{1}]$ and
$z\leq\tilde{z}_{m}(\gamma)$, we have

$\displaystyle C_{8}(\gamma,\tilde{z}_{m}(\gamma))=C_{1}(\gamma_{1})\geq
C_{1}(\gamma)=C_{8}(\gamma,\overline{z}(\gamma;P_{8})).$

It then follows that $\tilde{z}_{m}(\gamma)\leq\overline{z}(\gamma;P_{8})$ and
hence, each $z\in(0,\tilde{z}_{m}(\gamma)]$ serves as an upper solution for
$P_{8}$.

To complete the proof, we must demonstrate that each
$z\in(\tilde{z}_{m}(\gamma),z_{1}(\gamma)]$ serves as an upper solution for
$P_{8}$. To achieve this, we employ the barrier function $B_{1}(V)=\sqrt{-V}$.
It suffices to establish that, for any $\gamma\in(\gamma_{\star},\gamma_{1}]$
and $z\in(\tilde{z}_{m}(\gamma),z_{1}(\gamma)]$, $B_{1}(V)=\sqrt{-V}$ is a
lower barrier for the solution $C(V)$ of (4.2) with $P_{*}=P_{8}$. By Lemma
3.10, $\frac{dC}{dV}<0$ to the left of the triple point $P_{8}$. Therefore,

$C(V)>C_{8}=B_{1}(-C_{8}^{2})\geq B_{1}(V)\quad\text{ for
}V\in[-C_{8}^{2},V_{8})$

and $C(V_{8})>B_{1}(V_{8})$ for $z\in(\tilde{z}_{m}(\gamma),z_{1}(\gamma))$,
$C(V_{8})=B_{1}(V_{8})$ for $z=z_{1}(\gamma)$, and

$\displaystyle\frac{dC}{dV}|_{V=V_{8}(z_{1}),C=C_{8}(z_{1}),z=z_{1}}<-\frac{1}{2\sqrt{-V_{8}(z_{1})}},$
(6.6)

which guarantees the existence of $\overline{V}<V_{8}$ sufficiently close to
$V_{8}$ so that also for $z=z_{1}$, the solution $C(V)$ to (4.2) enjoys
$C(V)>B_{1}(V)$ for $V\in[\overline{V},V_{8})$. The proof of (6.6) is given in
Lemma D.1. Hence, by the barrier argument (2.20), we want to show

$\displaystyle\frac{F(V,\sqrt{-V};\gamma,z)}{G(V,\sqrt{-V};\gamma,z)}+\frac{1}{2\sqrt{-V}}<0.$
(6.7)

Since this inequality is nothing but (4.10) with $k(\gamma,z)$ replaced by
$1$, and $G(V,\sqrt{-V};\gamma,z)<0$ for any $V\in[V_{1},-C_{8}^{2}(z))$ by
Lemma 3.10, our goal is to show the positivity of the following function (cf.
(4.12)):

$\displaystyle\mathfrak{B}_{1}(V,z,m):=(m-1-m\gamma)V^{2}+(-3+2m-m\gamma+m(\gamma-2)\gamma
z)V-m\gamma z-1+\frac{2mz}{1+V}$ (6.8)

for each $\gamma\in(\gamma_{\star},\gamma_{1}]$,
$z\in(\tilde{z}_{m}(\gamma),z_{1}(\gamma)]$, $V\in[V_{1},-C_{8}^{2}(z))$ and
$m=1,2$. Our strategy is the following:

  1. 1.

For $m=1,2$, we will show that it is enough to check the sign of
$(1-C^{2}_{8}(z))\mathfrak{B}_{1}(-C^{2}_{8}(z),z,m)$.

  2. 2.

For $m=1,2$, we will show that
$\frac{d^{2}}{dz^{2}}[(1-C^{2}_{8}(z))\mathfrak{B}_{1}(-C^{2}_{8}(z),z,m)]<0$
so that $(1-C^{2}_{8}(z))\mathfrak{B}_{1}(-C^{2}_{8}(z),z,m)$ is a concave
function. Since $\mathfrak{B}_{1}(-C^{2}_{8}(z_{1}),z_{1},m)=0$ by the
definition of $z_{1}$, it is sufficient to check the sign of
$(1-C^{2}_{8}(\tilde{z}_{m}(\gamma)))\mathfrak{B}_{1}(-C^{2}_{8}(\tilde{z}_{m}(\gamma),\tilde{z}_{m}(\gamma),m)>0$.

For any fixed $\gamma\in(\gamma_{\star},\gamma_{1}]$, we write
$C_{8}=C_{8}(z)$, $C^{\prime}_{8}(z)=\frac{dC_{8}(z)}{dz}$ and
$C^{\prime\prime}_{8}(z)=\frac{d^{2}C_{8}(z)}{dz^{2}}$.

$\underline{\text{Step 1}}:$ When $m=1$, for each
$z\in(\tilde{z}_{m}(\gamma),z_{1}(\gamma)]$ and $V\in[V_{1},-C_{8}^{2}(z))$,
using $-\gamma(2V+1)\leq-\gamma(2V_{1}+1)$ and $-\frac{2z}{(1+V)^{2}}<-2z$, we
derive

$\displaystyle\frac{\partial\mathfrak{B}_{1}(V,z,1)}{\partial V}$
$\displaystyle<-\gamma(2V_{1}+1)-1+[(\gamma-2)\gamma-2]z=\frac{-(\gamma-1)^{2}}{\gamma+1}+[(\gamma-2)\gamma-2]z<0$

where we have used $(\gamma-2)\gamma-2<0$ for any
$\gamma\in(\gamma_{\star},\gamma_{1}]$. Since $\mathfrak{B}_{1}(V,z,1)$ is a
decreasing function in $V$ and $1-C^{2}_{8}(\gamma,z)>0$, it is sufficient to
check the sign of $(1-C^{2}_{8}(z))\mathfrak{B}_{1}(-C^{2}_{8}(z),z,1)$.

When $m=2$, we compute $(1+V)\mathfrak{B}_{1}(V,z,2)$ to obtain

$\displaystyle(1+V)\mathfrak{B}_{1}(V,z,2)=(1-2\gamma)V^{3}+2(1-2\gamma+(\gamma-2)\gamma
z)V^{2}+(-2\gamma+2(\gamma-3)\gamma z)V-2\gamma z-1+4z.$

We next show that $\frac{\partial}{\partial
V}[(1+V)\mathfrak{B}_{1}(V,z,2)]<0$. Note that

$\displaystyle\frac{\partial}{\partial
V}[(1+V)\mathfrak{B}_{1}(V,z,2)]=3(1-2\gamma)V^{2}+4(1-2\gamma+(\gamma-2)\gamma
z)V+(-2\gamma+2(\gamma-3)\gamma z),$

which is a quadratic polynomial of $V$. If the discriminant of the polynomial
is negative for any $\gamma\in(\gamma_{\star},\gamma_{1}]$,
$z\in(\tilde{z}_{m}(\gamma),z_{1}(\gamma)]$, then
$\frac{d}{dV}[(1+V)\mathfrak{B}_{1}(V,z)]$ will always be negative because
$3(1-2\gamma)<0$. The discriminant of the polynomial is given by

$\displaystyle\Delta$ $\displaystyle=16(1-2\gamma+(\gamma-2)\gamma
z)^{2}-12(1-2\gamma)(-2\gamma+2(\gamma-3)\gamma z)=:8p(z)$

where

$\displaystyle
p(z):=2(\gamma-2)^{2}\gamma^{2}z^{2}+(1-2\gamma)\gamma(\gamma+1)z+(1-2\gamma)(2-\gamma).$

When $\gamma=2$, it is clear that $p(z)<0$. When $\gamma\neq 2$, $p(z)$ is a
quadratic polynomial in $z$. It has a local minimum at
$z=\frac{(2\gamma-1)(\gamma+1)}{4(\gamma-2)^{2}\gamma}>\frac{1}{4}>z_{1}$.
Thus, by (6.5), to verify the negativity of $\Delta$, it is sufficient to
check the negativity of $p(\frac{2}{25})$. This condition is checked in
Proposition E.6. Therefore, we have shown

$\displaystyle\frac{\partial}{\partial V}[(1+V)\mathfrak{B}_{1}(V,z,2)]<0,$

and hence, to show that $\mathfrak{B}_{1}(V,z,2)>0$, it is enough to check the
sign of $(1-C^{2}_{8}(z))\mathfrak{B}_{1}(-C^{2}_{8}(z),z,2)$.

$\underline{\text{Step 2}}:$ Our next goal is to show
$(1-C^{2}_{8}(z))\mathfrak{B}_{1}(-C^{2}_{8}(z),z,m)>0$ for any
$z\in(\tilde{z}_{m}(\gamma),z_{1}(\gamma)]$, $V\in[V_{1},-C_{8}^{2}(z))$, and
$m=1,2$. We will first show
$(1-C^{2}_{8}(z))\mathfrak{B}_{1}(-C^{2}_{8}(z),z,m)$ is a concave function in
$z$.
For notational convenience, we will write

$\displaystyle\mathfrak{J}(z):=-C^{2}_{8}(z).$ (6.9)

By using Lemma 2.6, (6.3) and (6.4), we obtain

$\displaystyle\sqrt{2}-2=\mathfrak{J}(\tilde{z}_{m}(\gamma))<\mathfrak{J}(z)\leq\mathfrak{J}(z_{1})=-\frac{3-\sqrt{5}}{2}.$
(6.10)

Note that by using Lemma 2.6 and Lemma 6.1, we obtain

$\displaystyle\mathfrak{J}^{\prime}(z)$
$\displaystyle=-C_{8}C^{\prime}_{8}(z)>0,$ (6.11)
$\displaystyle\mathfrak{J}^{\prime\prime}(z)$
$\displaystyle=-[C_{8}C^{\prime\prime}_{8}(z)+(C^{\prime}_{8})^{2}]>0.$ (6.12)

We rewrite $(1-C^{2}_{8}(z))\mathfrak{B}_{1}(-C^{2}_{8}(z),z,m)$ as

$\displaystyle(1+\mathfrak{J}(z))\mathfrak{B}_{1}(\mathfrak{J}(z),z,m)=$
$\displaystyle\,(m-1-m\gamma)\mathfrak{J}^{3}(z)+(3m-4-2m\gamma+m(\gamma-2)\gamma
z)\mathfrak{J}^{2}(z)$ $\displaystyle+(2m-4-m\gamma+m(\gamma-3)\gamma
z)\mathfrak{J}(z)+m(2-\gamma)z-1,$

and compute the second $z$ derivative to obtain

$\displaystyle\frac{d^{2}}{dz^{2}}[(1+\mathfrak{J}(z))\mathfrak{B}_{1}(\mathfrak{J}(z),z,m)]:=\mathfrak{J}^{\prime\prime}(z)A(\gamma,z,m)+\mathfrak{J}^{\prime}(z)B(\gamma,z,m),$

where

$\displaystyle A(\gamma,z,m)=$
$\displaystyle\,3(m-1-m\gamma)\mathfrak{J}^{2}(z)+2[3m-4-2m\gamma+m(\gamma-2)\gamma
z]\mathfrak{J}(z)+2m-4-m\gamma+m(\gamma-3)\gamma z,$ $\displaystyle
B(\gamma,z,m)=$
$\displaystyle\,6(m-1-m\gamma)\mathfrak{J}(z)\mathfrak{J}^{\prime}(z)+2[3m-4-2m\gamma+m(\gamma-2)\gamma
z]\mathfrak{J}^{\prime}(z)$
$\displaystyle+4m(\gamma-2)\gamma\mathfrak{J}(z)+2m(\gamma-3)\gamma.$

We claim $A(\gamma,z,m)$ and $B(\gamma,z,m)$ are negative. We first check
$B(\gamma,z,m)<0$. We decompose $B(\gamma,z,m)$ into two parts

$\displaystyle
B(\gamma,z,m)=\mathfrak{J}^{\prime}(z)B_{1}(\gamma,z,m)+B_{2}(\gamma,z,m),$

where

$\displaystyle B_{1}(\gamma,z,m)$
$\displaystyle=6(m-1-m\gamma)\mathfrak{J}(z)+6m-8-4m\gamma+2m(\gamma-2)\gamma
z,$ $\displaystyle B_{2}(\gamma,z,m)$
$\displaystyle=2m\gamma[(2(\gamma-2)\mathfrak{J}(z)+(\gamma-3)].$

For $B_{1}(\gamma,z,m)$, by using $m-1-m\gamma<0$, (6.10),
$|(\gamma-2)\gamma|\leq 1$, and $z<z_{M}<\frac{1}{5}$, we obtain

$\displaystyle B_{1}(\gamma,z,m)$
$\displaystyle<6(m-1-m\gamma)\mathfrak{J}(\tilde{z}_{m}(\gamma))+6m-8-4m\gamma+\frac{2m}{5}$
$\displaystyle=4-6\sqrt{2}-\frac{m(28-30\sqrt{2})}{5}+m(8-6\sqrt{2})\gamma<0$

for any $m=1,2$ and $\gamma\in(\gamma_{\star},\gamma_{1}]$.

For $B_{2}(\gamma,z,m)$, when $\gamma\geq 2$, it is clear that
$B_{2}(\gamma,z,m)<0$. When $\gamma\in(\gamma_{\star},2)$, by using (6.10),

$\displaystyle
B_{2}(\gamma,z,m)<2m\gamma(2(\gamma-2)\mathfrak{J}(\tilde{z}_{m}(\gamma))+\gamma-3)=2m\gamma[(2\sqrt{2}-3)\gamma+5-4\sqrt{2}]<0.$

Hence, we have shown that $B_{2}(\gamma,z,m)<0$ for any $m=1,2$,
$\gamma\in(\gamma_{\star},\gamma_{1}]$ and
$z\in(\tilde{z}_{m}(\gamma),z_{1}(\gamma)]$.

Regarding $A(\gamma,z,m)$, we observe that

$\displaystyle\frac{dA(\gamma,z,m)}{dz}=\mathfrak{J}^{\prime}(z)B_{1}(\gamma,z,m)+\frac{1}{2}B_{2}(\gamma,z,m)<0.$

Therefore, in order to show $A(\gamma,z,m)<0$, it is enough to verify
$A(\gamma,\tilde{z}_{m}(\gamma),m)<0$. When $m=1$, by using (6.10) and
$B_{2}(\gamma,z,1)<0$, we have

$\displaystyle A(\gamma,\tilde{z}_{m}(\gamma),1)$
$\displaystyle=-3\gamma\mathfrak{J}^{2}(\tilde{z}_{m}(\gamma))+2(-1-2\gamma)\mathfrak{J}(\tilde{z}_{m}(\gamma))-2-\gamma+\frac{\tilde{z}_{m}(\gamma)}{2}B_{2}(\gamma,\tilde{z}_{m}(\gamma),1)$
$\displaystyle<(8\sqrt{2}-11)\gamma+2(1-\sqrt{2})<0.$

When $m=2$, by using (6.10), (6.5) and $(2\sqrt{2}-3)\gamma+5-4\sqrt{2}<0$, we
have

$\displaystyle A(\gamma,\tilde{z}_{m}(\gamma),2)$
$\displaystyle=3(1-2\gamma)\mathfrak{J}^{2}(\tilde{z}_{m}(\gamma))+2(2-4\gamma)\mathfrak{J}(\tilde{z}_{m}(\gamma))-2\gamma+2\gamma[2(\gamma-2)\mathfrak{J}(\tilde{z}_{m}(\gamma))+\gamma-3]\tilde{z}_{m}(\gamma)$
$\displaystyle<(16\sqrt{2}-22)\gamma+10-8\sqrt{2}+\frac{4\gamma[(2\sqrt{2}-3)\gamma+5-4\sqrt{2}]}{25}$
$\displaystyle=\frac{2}{25}\Big{[}(4\sqrt{2}-6)\gamma^{2}+(192\sqrt{2}-265)\gamma+125-100\sqrt{2}\Big{]}=:p(\gamma).$

Since $p(\gamma)$ has a global maximum at
$\gamma=\frac{192\sqrt{2}-265}{12-8\sqrt{2}}>3>\gamma_{1}$ and
$p(\gamma_{1})=\frac{484-346\sqrt{2}}{25}<0$, $p(\gamma)<0$ for any
$\gamma\in(\gamma_{\star},\gamma_{1}]$. We conclude that $A(\gamma,z,m)<0$.

Hence, $(1+\mathfrak{J}(z))\mathfrak{B}_{1}(\mathfrak{J}(z),z,m)$ is a concave
function in $z$. It is then enough to check the sign for the function at two
ends points of $z$. By definition of $z_{1}$ (cf. (6.2)),
$(1+\mathfrak{J}(z_{1}))\mathfrak{B}_{1}(\mathfrak{J}(z_{1}),z_{1},m)=0$. At
$z=\tilde{z}_{m}(\gamma)$, since
$1+\mathfrak{J}(\tilde{z}_{m}(\gamma)=\sqrt{2}-1>0$, we only need to show
$\mathfrak{B}_{1}(\mathfrak{J}(\tilde{z}_{m}(\gamma)),\tilde{z}_{m}(\gamma),m)>0$.
By direct computations,

$\displaystyle\mathfrak{B}_{1}(\mathfrak{J}(\tilde{z}_{m}(\gamma)),\tilde{z}_{m}(\gamma),m)$
$\displaystyle=m(3\sqrt{2}-4)\gamma+(1-\sqrt{2})(2m-1)+m[(\sqrt{2}-2)\gamma^{2}+(3-2\sqrt{2})\gamma+2\sqrt{2}+2]\tilde{z}_{m}(\gamma)$
$\displaystyle>m(3\sqrt{2}-4)\gamma+(1-\sqrt{2})(2m-1)+\frac{2m}{25}[-(2-\sqrt{2})\gamma^{2}+(3-2\sqrt{2})\gamma+2\sqrt{2}+2]>0,$

where we have used $(\sqrt{2}-2)\gamma^{2}+(3-2\sqrt{2})\gamma+2\sqrt{2}+2>0$
for each $\gamma\in(\gamma_{\star},\gamma_{1}]$ and (6.5) in the second line,
while the positive sign of the last inequality is shown in Proposition E.7 and
Proposition E.8 for $m=1$ and $m=2$ respectively. ∎

###  6.2 Conditional existence for $\gamma\in(\gamma_{1},3]$

In this subsection, we will employ the barrier function
$B_{\frac{3}{2}}(V)=\sqrt{-\frac{3}{2}V}$ to delineate a narrower and more
precise range for the potential location of $z_{std}(\gamma;P_{8})$. With the
choice of the barrier function, we define $z_{2}(\gamma)$ to be the value such
that $P_{8}$ lies on the curve $C=B_{\frac{3}{2}}(V)$:

$\displaystyle z_{2}=\frac{\sqrt{33}-3}{6+2\sqrt{33}+4\gamma}.$ (6.13)

Since $V_{8}(z_{2})=C_{8}(z_{2})-1$ and
$1-\frac{2}{3}C_{8}^{2}(z_{2})=C_{8}(z_{2})$,

$\displaystyle V_{8}(z_{2})=\frac{\sqrt{33}-7}{4},\quad
C_{8}(z_{2})=\frac{\sqrt{33}-3}{4}.$ (6.14)

######  Lemma 6.3.

For any $\gamma\in(\gamma_{1},3]$, any $z\in(0,z_{2}(\gamma)]$ gives an upper
solution for $P_{8}$.

###### Proof.

Similar to our approach in Lemma 6.2, we define $\overline{z}(\gamma;P_{8})$
as the solution to $C_{8}(\gamma,\overline{z}(\gamma;P_{8}))=C_{1}(\gamma)$
and we further introduce $\hat{z}_{m}(\gamma)$, which satisfies the equation:

$\displaystyle C_{8}(\gamma,\hat{z}_{m}(\gamma))$
$\displaystyle=C_{1}(3)=\frac{\sqrt{3}}{2}$ (6.15)

so that

$\displaystyle\hat{z}_{m}(\gamma)=\frac{\sqrt{3}}{12+8\sqrt{3}+2\gamma}.$

It is easy to check that for any $\gamma\in(\gamma_{1},3]$,

$\displaystyle\frac{1}{8}>z_{M}(2)>z_{M}(\gamma)>\hat{z}_{m}(\gamma)\geq\hat{z}_{m}(3)>\frac{1}{20}.$
(6.16)

By Lemma 2.6 and Lemma 2.1, for any $\gamma\in(\gamma_{1},3]$ and
$z\leq\hat{z}_{m}(\gamma)$, we have

$\displaystyle C_{8}(\gamma,\hat{z}_{m}(\gamma))=C_{1}(3)\geq
C_{1}(\gamma)=C_{8}(\gamma,\overline{z}(\gamma;P_{8})),$

and it follows that $\hat{z}_{m}(\gamma)\leq\overline{z}(\gamma;P_{8})$.
Consequently, each $z\in(0,\hat{z}_{m}(\gamma)]$ serves as an upper solution
for $P_{8}$.

To conclude the proof, we need to establish that each
$z\in(\hat{z}_{m}(\gamma),z_{2}(\gamma)]$ gives an upper solution for $P_{8}$.
To this end, we will employ the barrier function $B_{\frac{3}{2}}(V)$. Hence,
it suffices to establish that, for any $\gamma\in(\gamma_{1},3]$ and
$z\in(\hat{z}_{m}(\gamma),z_{2}(\gamma)]$, $B_{\frac{3}{2}}(V)$ is a lower
barrier for the solution $C(V)$ of (4.2) with $P_{*}=P_{8}$. By Lemma 3.10,
$\frac{dC}{dV}<0$ to the left of the triple point $P_{8}$. Therefore,
$C(V)>B_{\frac{3}{2}}(V)$ for $V\in[-\frac{2}{3}C_{8}^{2},V_{8}]$ and
$z\in(\hat{z}_{m}(\gamma),z_{2}(\gamma))$, while
$C(V_{8})=B_{\frac{3}{2}}(V_{8})$ for $z=z_{2}(\gamma)$ and it satisfies

$\displaystyle\frac{dC}{dV}|_{V=V_{8}(z_{2}),C=C_{8}(z_{2}),z=z_{2}}<-\frac{1}{2\sqrt{-V_{8}(z_{2})}},$
(6.17)

so that $C(V)>B_{\frac{3}{2}}(V)$ for $V\in[\overline{V},V_{8})$ for some
$\overline{V}<V_{8}$. The proof of (6.17) is given in Lemma D.2. Now, by using
the barrier argument (cf. (2.20)), it is sufficient to show that