$\displaystyle\frac{F(V,\sqrt{-\frac{3}{2}V};\gamma,z)}{G(V,\sqrt{-\frac{3}{2}V};\gamma,z)}+\frac{1}{2}\sqrt{\frac{-3}{2V}}<0.$
(6.18)

We observe that this inequality is (4.10) with $k(\gamma,z)$ replaced by
$\frac{3}{2}$. As $G(V,\sqrt{-\frac{3}{2}V};\gamma,z)<0$ for any
$V\in[V_{1},-\frac{2}{3}C_{8}^{2}(z))$ by Lemma 3.10, it suffices to show that
for any $\gamma\in(\gamma_{1},3]$, $z\in(\hat{z}_{m}(\gamma),z_{2}(\gamma)]$,
$V\in[V_{1},-\frac{2}{3}C_{8}^{2}(z))$, and $m=1,2$, the following function
(cf. (4.12)) is positive.

$\displaystyle\mathfrak{B}_{\frac{3}{2}}(V,z,m):=(m-1-m\gamma)V^{2}+(-4+2m+\frac{m+1}{2}-m\gamma+m(\gamma-2)\gamma
z)V-m\gamma z-1+\frac{3mz}{1+V}.$ (6.19)

The strategy is the following:

  1. 1.

For $m=1,2$, we will show that it is enough to check the sign of
$(1-\frac{2}{3}C^{2}_{8}(z))\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z),z,m)$.

  2. 2.

For $m=1,2$, we will show that
$\frac{d^{2}}{dz^{2}}[(1-\frac{2}{3}C^{2}_{8}(z))\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z),z,m)]<0$
so that
$(1-\frac{2}{3}C^{2}_{8}(z))\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z),z,m)$
is a concave function. Since
$\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z_{2}),z_{2},m)=0$ by the
definition of $z_{2}$, it is sufficient to check the sign of
$(1-\frac{2}{3}C^{2}_{8}(\hat{z}_{m}(\gamma)))\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(\hat{z}_{m}(\gamma)),\gamma,\hat{z}_{m}(\gamma),m)>0$.

$\underline{\text{Step 1}}:$ First of all, by Lemma 2.1,

$\displaystyle V_{1}(\gamma)\geq V_{1}(\gamma_{1})=\sqrt{2}-2.$ (6.20)

When $m=1$, by using (6.20) and $0<1+V<1$, we have

$\displaystyle\frac{\partial\mathfrak{B}_{\frac{3}{2}}(V,z,1)}{\partial V}$
$\displaystyle<-2\gamma
V_{1}-1-\gamma+[(\gamma-2)\gamma-3]z<(3-2\sqrt{2})\gamma-1+[(\gamma-2)\gamma-3]z<0.$
(6.21)

Therefore, to establish the positivity of $\mathfrak{B}_{\frac{3}{2}}(V,z,1)$,
it suffices to show

$\displaystyle(1-\frac{2}{3}C^{2}_{8}(z))\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z),z,1)\geq
0.$

As for $m=2$, we compute

$\displaystyle(1+V)\mathfrak{B}_{\frac{3}{2}}(V,z,2)=(1-2\gamma)V^{3}+(\frac{5}{2}-4\gamma+2(\gamma-2)\gamma
z)V^{2}+(\frac{1}{2}-2\gamma+2(\gamma-3)\gamma z)V+2(3-\gamma)z-1.$

For any $\gamma\in(\gamma_{1},3]$, $z\in(\hat{z}_{m}(\gamma),z_{2}(\gamma)]$
and $V\in[V_{1},-\frac{2}{3}C_{8}^{2}(z))$, by using (6.16),

$\displaystyle\frac{\partial^{2}}{\partial
V^{2}}[(1+V)\mathfrak{B}_{\frac{3}{2}}(V,z,2)]$
$\displaystyle=6(1-2\gamma)V+5-8\gamma+4(\gamma-2)\gamma z$
$\displaystyle<\frac{12(2\gamma-1)}{\gamma+1}+5-8\gamma+\frac{(\gamma-2)\gamma}{2}=\frac{\gamma^{3}-17\gamma^{2}+40\gamma-14}{2(\gamma+1)}<0$

where the negative sign of the cubic polynomial is shown in Proposition E.9.
Thus, $(1+V)\mathfrak{B}_{\frac{3}{2}}(V,z,2)$ is a concave function in $V$.
It is then enough to check the signs of
$\mathfrak{B}_{\frac{3}{2}}(V_{1},z,2)$ and
$\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z),z,2)$. We now compute
$\mathfrak{B}_{\frac{3}{2}}(V_{1},z,2)$,

$\displaystyle\mathfrak{B}_{\frac{3}{2}}(V_{1},z,2)$
$\displaystyle=\frac{3\gamma(\gamma-3)}{(\gamma+1)^{2}}+\frac{6(\gamma^{2}(3-\gamma)+\gamma+1)}{\gamma^{2}-1}z$
$\displaystyle>\frac{3\gamma(\gamma-3)}{(\gamma+1)^{2}}+\frac{6(\gamma^{2}(3-\gamma)+\gamma+1)}{20(\gamma^{2}-1)}=\frac{3(-\gamma^{4}+12\gamma^{3}-36\gamma^{2}+32\gamma+1)}{10(\gamma-1)(\gamma+1)^{2}}>0$

where we have used (6.16) in the second line, while the positive sign of the
last inequality is shown in Proposition E.10. Thus, in order to show
$(1+V)\mathfrak{B}_{\frac{3}{2}}(V,z,2)>0$ for any $\gamma\in(\gamma_{1},3]$,
$z\in(\hat{z}_{m}(\gamma),z_{2}(\gamma)]$ and
$V\in[V_{1},-\frac{2}{3}C_{8}^{2}(z))$, it is sufficient to show that

$\displaystyle(1-\frac{2}{3}C^{2}_{8}(z))\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z),z,2)\geq
0.$

$\underline{\text{Step 2}}:$ Our goal is to show
$(1-\frac{2}{3}C^{2}_{8}(z))\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z),\gamma,z,m)\geq
0$ for any $\gamma\in(\gamma_{1},3]$,
$z\in(\hat{z}_{m}(\gamma),z_{2}(\gamma)]$,
$V\in[V_{1},-\frac{2}{3}C_{8}^{2}(z))$, and $m=1,2$. We will first show
$(1-\frac{2}{3}C^{2}_{8}(z))\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z),\gamma,z,m)$
is a concave function in $z$. For notational convenience, we will denote

$\displaystyle\Im(z):=-\frac{2}{3}C^{2}_{8}(z).$ (6.22)

By using Lemma 2.6, (6.14) and (6.15), we obtain

$\displaystyle-\frac{1}{2}=\Im(\hat{z}_{m}(\gamma))<\Im(z)\leq\Im(z_{2})=-\frac{7-\sqrt{33}}{4}.$
(6.23)

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

$\displaystyle\Im^{\prime}(z)$
$\displaystyle=-\frac{4}{3}C_{8}C^{\prime}_{8}(z)>0,$ (6.24)
$\displaystyle\Im^{\prime\prime}(z)$
$\displaystyle=-\frac{4}{3}[C_{8}C^{\prime\prime}_{8}(z)+(C^{\prime}_{8})^{2}]>0.$
(6.25)

We rewrite
$(1-\frac{2}{3}C^{2}_{8}(z))\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z),\gamma,z,m)$
as

$\displaystyle(1+\Im(z))\mathfrak{B}_{\frac{3}{2}}(\Im(z),z,m)$
$\displaystyle=(m-1-m\gamma)\Im^{3}(z)+(3m+\frac{m+1}{2}-5-2m\gamma+m(\gamma-2)\gamma
z)\Im^{2}(z)$ $\displaystyle\ \ +(2m+\frac{m+1}{2}-5-m\gamma+m(\gamma-3)\gamma
z)\Im(z)+m(3-\gamma)z-1,$

and compute the second $z$ derivative to obtain

$\displaystyle\frac{d^{2}}{dz^{2}}[(1+\Im(z))\mathfrak{B}_{\frac{3}{2}}(\Im(z),z,m)]:=\Im^{\prime\prime}(z)A(\gamma,z,m)+\Im^{\prime}(z)B(\gamma,z,m),$

where

$\displaystyle A(\gamma,z,m)=$
$\displaystyle\,3(m-1-m\gamma)\Im^{2}(z)+2[3m+\frac{m+1}{2}-5-2m\gamma+m(\gamma-2)\gamma
z]\Im(z)$ $\displaystyle+2m+\frac{m+1}{2}-5-m\gamma+m(\gamma-3)\gamma z,$
$\displaystyle B(\gamma,z,m)=$
$\displaystyle\,6(m-1-m\gamma)\Im(z)\Im^{\prime}(z)+2[3m+\frac{m+1}{2}-5-2m\gamma+m(\gamma-2)\gamma
z]\Im^{\prime}(z)$
$\displaystyle+4m(\gamma-2)\gamma\Im(z)+2m(\gamma-3)\gamma.$

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

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

where

$\displaystyle B_{1}(\gamma,z,m)$
$\displaystyle=6(m-1-m\gamma)\Im(z)+7m-9-4m\gamma+2m(\gamma-2)\gamma z,$
$\displaystyle B_{2}(\gamma,z,m)$
$\displaystyle=4m(\gamma-2)\gamma\Im(z)+2m(\gamma-3)\gamma.$

Clearly, $B_{2}(\gamma,z,m)<0$. Regarding $B_{1}(\gamma,z,m)$, by using (6.16)
and (6.23), we obtain

$\displaystyle B_{1}(\gamma,z,m)$
$\displaystyle<6(m-1-m\gamma)\Im(\hat{z}_{m}(\gamma))+7m-9-4m\gamma+2m\gamma
z_{M}=4m-6-\frac{3m\gamma}{4}<0$

for any $m=1,2$ and $\gamma\in(\gamma_{1},3]$.

For $A(\gamma,z,m)$, we notice that

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

Thus, to show $A(\gamma,z,m)<0$, it is enough to verify
$A(\gamma,\hat{z}_{m}(\gamma),m)<0$. Using (6.16) and (6.23),

$\displaystyle
A(\gamma,\hat{z}_{m}(\gamma),m)=\frac{m\gamma}{4}-\frac{m+3}{4}-m\gamma\hat{z}_{m}(\gamma)<\frac{m\gamma}{5}-\frac{m+3}{4}<0$

for $m=1,2$ and $\gamma\in(\gamma_{1},3]$. Hence,
$(1+\Im(z))\mathfrak{B}_{\frac{3}{2}}(\Im(z),z,m)$ is a concave function in
$z$. It is enough to check the sign at two ends points of $z$. By the
definition (6.13) of $z_{2}$,
$(1+\Im(z_{2}))\mathfrak{B}_{\frac{3}{2}}(\Im(z_{2}),\gamma,z_{2},m)=0$. For
$(1+\Im(\hat{z}_{m}(\gamma)))\mathfrak{B}_{\frac{3}{2}}(\Im(\hat{z}_{m}(\gamma)),\hat{z}_{m}(\gamma),m)$,
we evaluate

$\displaystyle(1+\Im(\hat{z}_{m}(\gamma)))\mathfrak{B}_{\frac{3}{2}}(\Im(\hat{z}_{m}(\gamma))),\hat{z}_{m}(\gamma)),m)=\begin{cases}&\frac{\gamma-2}{8}+\frac{(12-\gamma^{2})\hat{z}_{m}(\gamma)}{4}\text{
when }m=1,\\\
&\frac{\gamma-3}{4}+\frac{(12-\gamma^{2})\hat{z}_{m}(\gamma)}{2}\text{ when
}m=2.\end{cases}$

Obviously,
$(1+\Im(\hat{z}_{m}(\gamma)))\mathfrak{B}_{\frac{3}{2}}(\Im(\hat{z}_{m}(\gamma))),\hat{z}_{m}(\gamma)),1)>0$.
For
$(1+\Im(\hat{z}_{m}(\gamma)))\mathfrak{B}_{\frac{3}{2}}(\Im(\hat{z}_{m}(\gamma))),\hat{z}_{m}(\gamma)),2)$,
by using (6.16), we have

$\displaystyle(1+\Im(\hat{z}_{m}(\gamma)))\mathfrak{B}_{\frac{3}{2}}(\Im(\hat{z}_{m}(\gamma))),\hat{z}_{m}(\gamma)),2)>\frac{\gamma-3}{4}+\frac{12-\gamma^{2}}{40}=\frac{-\gamma^{2}+10\gamma-18}{40}>0$

for any $\gamma\in(\gamma_{1},3]$. This completes the proof. ∎

##  7 Solving to right

In the previous sections, for each $\gamma\in(1,3]$, we established the
existence of $z_{std}(\gamma;P_{*})$ and a range of $z$ which
$z_{std}(\gamma;P_{*})$ must belong to. This $z_{std}$ allows the solution
$C(V;\gamma,z_{std}(\gamma;P_{*}),P_{*})$ of (2.1) to pass smoothly from
$P_{1}$ through the triple point $P_{*}$, where $P_{*}$ is either $P_{6}$ or
$P_{8}$. The remaining goal is to extend this smooth solution from the triple
point $P_{*}$ to the origin $P_{0}$ while ensuring that it remains within the
second quadrant in the phase plane (that is, that we retain both $V<0$ and
$C>0$ up to time the flow meets the origin). To prove this property for the
solution associated to $z_{std}$, we will in fact prove the stronger property
that the local solution to the right of the sonic $P_{*}$ always extends to
$P_{0}$ within the second quadrant for all $z$ within the range containing
$z_{std}(\gamma;P_{*})$.

For notational convenience, we define unified notation for the various
possible ranges of $z$ containing $z_{std}$ from the results of Sections 4–6.

$\displaystyle\mathring{\mathcal{Z}}(\gamma;P_{*})=\begin{cases}&(z_{g}(\gamma),z_{M}(\gamma)]\quad\text{for
}\gamma\in(1,2]\text{ at }P_{6},\\\
&(z_{1}(\gamma),z_{M}(\gamma)]\quad\text{for
}\gamma\in(\gamma_{\star},\gamma_{1}]\text{ at }P_{8},\\\
&(z_{2}(\gamma),z_{M}(\gamma)]\quad\text{for }\gamma\in(\gamma_{1},3]\text{ at
}P_{8}.\end{cases}$ (7.1)

In this section, we will show that for any $\gamma\in(1,3]$ and
$z\in\mathring{\mathcal{Z}}(\gamma;P_{*})$, the local analytic solutions
around $P_{*}$ constructed in Theorem 3.8 continue to the origin $P_{0}$ in
the second quadrant.

From the phase portrait analysis, three possibilities arise for the extension
of the local analytic solution.

  1. 1.

The trajectory intersects the negative $V$-axis before reaching $V=0$.

  2. 2.

The trajectory intersects the positive $C$-axis when $V=0$.

  3. 3.

The trajectory converges to $P_{0}$ within the second quadrant.

To rule out the first two possibilities, we will use suitable barrier
functions to establish an invariant region for the solutions ensuring
convergence to $P_{0}$ within the second quadrant. We begin with the extension
for the local analytic solution to the right of the triple point.

######  Lemma 7.1.

Let $\gamma\in(1,3]$ and $z\in\mathring{\mathcal{Z}}$ be given and let
$P_{*}=(V_{*},C_{*})$ be either $P_{6}$ or $P_{8}$. Consider the local,
analytic solution $C:[V_{*},V_{*}+\epsilon]\to\mathbb{R}_{+}$, guaranteed by
Theorem 3.8. This solution extends smoothly to the right within the second
quadrant onto the domain $C:[V_{*},V_{0})\to\mathbb{R}_{+}$, where
$V_{0}=\min\\{C^{-1}(0),0\\}$. Furthermore, except at the triple point, the
solution enjoys $\frac{dC}{dV}<0$, $F<0$, $G>0$, $D<0$.

###### Proof.

The result follows from our choice of $c_{1}$ (cf. (3.18)),
$F_{C}(V_{*},C_{*})>0$ (cf. (3.7)), $G_{C}(V_{*},C_{*})<0$ (cf. (3.5)). ∎

Next, we will eliminate the first possibility.

######  Lemma 7.2.

For any $\gamma\in(1,3]$ and $z\in\mathring{\mathcal{Z}}$, the solution
constructed in Lemma 7.1 does not intersect the negative $V$-axis (i.e., in
the notation of that Lemma, $V_{0}=0$).

###### Proof.

We argue by contradiction. Suppose the solution intersects the negative
$V$-axis before reaching $V=0$, and let $(\overline{V},0)$ denote the point of
intersection of the solution trajectory with the $V$-axis. Consider the
initial value problem:

$\begin{cases}&\cfrac{dC(V)}{dV}=\cfrac{F(V,C;\gamma,z)}{G(V,C;\gamma,z)},\\\
&C(\overline{V})=0.\end{cases}$

In a small rectangular region around $(\overline{V},0)$, it is evident that
$\frac{F(V,C;\gamma,z)}{G(V,C;\gamma,z)}$ is continuously differentiable. By
the standard theorem for existence and uniqueness of solutions to ODEs with
locally Lipschitz right hand side (e.g. [44]), this initial value problem
possesses a unique solution on the interval
$(\overline{V}-\epsilon,\overline{V}+\epsilon)$ for sufficiently small
$\epsilon$. However, we see trivially that $C(\overline{V})\equiv 0$ solves
this problem, and so must be the unique solution, leading to a contradiction.
∎

We remark that Lemma 7.2 yields $V_{0}=0$ in Lemma 7.1. The rest of this
section is devoted to ruling out the second possibility by the barrier
argument.

###  7.1 Connecting $P_{6}$ to the origin

In this subsection, we prove that the local analytic solutions around $P_{6}$
constructed in Theorem 3.8 for any $\gamma\in(1,2]$ and
$z\in(z_{g}(\gamma),z_{M}(\gamma)]$ continue to the origin and stay below the
barrier curve $B_{1}(V)=\sqrt{-V}$. In particular, this implies that the
solution trajectories to the right of the sonic point will stay between $C=0$
and $B_{1}(V)$ and must therefore strictly decrease to the origin.

######  Lemma 7.3.

For any $\gamma\in(1,2]$ and $z\in(z_{g}(\gamma),z_{M}(\gamma)]$, the solution
constructed in Lemma 7.1 with $P_{*}=P_{6}$, always lies below the curve
$B_{1}(V)=\sqrt{-V}$ for $V\in[V_{6},0)$.

###### Proof.

According to (2.12), $P_{6}$ consistently remains below the curve $C=B_{1}(V)$
for any $\gamma\in(1,2]$. In other words, $C_{6}<B_{1}(V_{6})$. Additionally,
as per Lemma 7.1, $\frac{dC}{dV}<0$ holds for $V\in[V_{6},0)$. Observing that
$-C_{6}^{2}>V_{6}$, we therefore see that the inequality $C(V)<B_{1}(V)$ holds
trivially for $V\in[V_{6},-C_{6}^{2}]$. Consequently, our analysis can be
confined to $V\in(-C_{6}^{2},0)$. Thus, employing the barrier argument (2.20),
we verify the validity of the following inequality for any $\gamma\in(1,2]$,
$z\in(z_{g}(\gamma),z_{M}(\gamma)]$, and $V\in(-C_{6}^{2},0)$:

$\displaystyle\frac{F(V,\sqrt{-V};\gamma,z)}{G(V,\sqrt{-V};\gamma,z)}+\frac{1}{2\sqrt{-V}}<0.$
(7.2)

Observe that the inequality is (4.10) with $k(\gamma,z)$ replaced by $1$. So,
by Lemma 7.1 which assures that $G(V,\sqrt{-V};\gamma,z)>0$ for any
$V\in(-C_{6}^{2},0)$, our task reduces to proving

$\mathfrak{B}_{1}(V,z,m)=\frac{2F(V,\sqrt{-V};\gamma,z)}{\sqrt{-V}}+\frac{G(V,\sqrt{-V};\gamma,z)}{-V}<0,$

for $m=1,2$, $\gamma\in(1,2]$, $z\in(z_{g}(\gamma),z_{M}(\gamma)]$,
$V\in(-C_{6}^{2},0)$, where we recall from (6.8) that

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

We note first that

$\displaystyle\frac{\partial\mathfrak{B}_{1}(V,z,m)}{\partial V}$
$\displaystyle=(m-1-m\gamma)(1+2V)+m-2+m(\gamma-2)\gamma
z-\frac{2mz}{(1+V)^{2}}$ $\displaystyle<(m-1-m\gamma)(1-2C_{6}^{2})<0$ (7.3)

for $V\in[-C_{6}^{2},0)$, since
$C_{6}\leq\frac{\sqrt{\gamma}}{\sqrt{\gamma}+\sqrt{2}}\leq\frac{1}{2}$ by
using (2.12). Thus, it is sufficient to check the negativity of
$\mathfrak{B}_{1}(-C^{2}_{6}(\gamma,z),z,m)$.

For any fixed $\gamma\in(1,2]$, we write $C_{6}=C_{6}(z)$ and
$C_{6}^{\prime}(z)=\frac{dC_{6}}{dz}$. By (7.3), Lemma 2.6 and the chain rule,
we check the sign of the $z$-derivative of
$\mathfrak{B}_{1}(-C^{2}_{6}(z),z,m)$:

$\displaystyle\frac{d\mathfrak{B}_{1}(-C^{2}_{6}(z),z,m)}{dz}=\frac{\partial\mathfrak{B}_{1}}{\partial
V}(-C^{2}_{6}(z),z,m)(-2C_{6}(z)C_{6}^{\prime}(z))+\frac{\partial\mathfrak{B}_{1}}{\partial
z}(-C^{2}_{6}(z),z,m)>0,$

where we also used $\frac{\partial\mathfrak{B}_{1}}{\partial
z}(-C^{2}_{6}(z),z,m)=-m(\gamma-2)\gamma
C^{2}_{6}(z)+m(\frac{2}{1-2C^{2}_{6}(z)}-\gamma)>0$ for any $\gamma\in(1,2]$.
As a consequence, it is now enough to show that
$\mathfrak{B}_{1}(-C^{2}_{6}(\gamma,z_{M}),z_{M})<0$ to accomplish (7.2). By
(2.11) and (2.9), we have $C_{6}(\gamma,z_{M})=\sqrt{\gamma z_{M}}$ and hence,
we obtain

$\displaystyle\mathfrak{B}_{1}(-C^{2}_{6}(\gamma,z_{M}),z_{M},m)$
$\displaystyle=(m-1-m\gamma)\gamma^{2}z^{2}_{M}-(-3+2m-m\gamma+m(\gamma-2)\gamma
z_{M})\gamma z_{M}-m\gamma z_{M}-1+\frac{2mz_{M}}{1-\gamma z_{M}}$
$\displaystyle=(3m-1)\gamma^{2}z^{2}_{M}-2m\gamma^{3}z^{2}_{M}+3(1-m)\gamma
z_{M}+m\gamma^{2}z_{M}-1+\frac{2mz_{M}}{1-\gamma z_{M}}.$

When $m=1$, we have

$\displaystyle\mathfrak{B}_{1}(-C^{2}_{6}(\gamma,z_{M}),z_{M},1)$
$\displaystyle=2(1-\gamma)\gamma^{2}z^{2}_{M}+\frac{\gamma^{2}z_{M}-\gamma^{3}z^{2}_{M}-1+(\gamma+2)z_{M}}{1-\gamma
z_{M}}.$ (7.4)

With $\gamma>1$, the first term in the equation above is negative. To find an
explicit expression for the second term, first, by (2.9), we see that

$\displaystyle 1-\gamma z_{M}=1-\frac{\gamma}{\gamma+2+2\sqrt{2\gamma}}>0$

and

$\displaystyle\gamma^{2}z_{M}-\gamma^{3}z^{2}_{M}-1+(\gamma+2)z_{M}$
$\displaystyle=\frac{\gamma^{2}(\sqrt{\gamma}+\sqrt{2})^{2}-\gamma^{3}-(\sqrt{\gamma}+\sqrt{2})^{4}+(\gamma+2)(\sqrt{\gamma}+\sqrt{2})^{2}}{(\sqrt{\gamma}+\sqrt{2})^{4}}$
$\displaystyle=\frac{(\gamma^{2}-2\sqrt{2\gamma})(\sqrt{\gamma}+\sqrt{2})^{2}-\gamma^{3}}{(\sqrt{\gamma}+\sqrt{2})^{4}}<0,$

where we have used $\gamma\leq 2$. Therefore, from (7.4), we deduce that
$\mathfrak{B}_{1}(-C^{2}_{6}(\gamma,z_{M}),z_{M},1)<0$ and hence
$\mathfrak{B}_{1}(V,z,1)<0$ for all $V\in(-C_{6}^{2},0)$,
$z\in(z_{g}(\gamma),z_{M}(\gamma)]$.

When $m=2$, we have

$\displaystyle\mathfrak{B}_{1}(-C^{2}_{6}(\gamma,z_{M}),z_{M},2)$
$\displaystyle=4(1-\gamma)\gamma^{2}z^{2}_{M}+\frac{4\gamma^{2}z^{2}_{M}-2\gamma
z_{M}+2\gamma^{2}z_{M}-\gamma^{3}z^{3}_{M}-2\gamma^{3}z^{2}_{M}-1+4z_{M}}{1-\gamma
z_{M}}.$ (7.5)

Since $\gamma>1$, the first term is again negative. For the second term, we
again apply (2.9) to rearrange the numerator as

$\displaystyle 4\gamma^{2}z^{2}_{M}-2\gamma
z_{M}+2\gamma^{2}z_{M}-\gamma^{3}z^{3}_{M}-2\gamma^{3}z^{2}_{M}-1+4z_{M}$
$\displaystyle=(4\gamma z_{M}-1)\gamma
z_{M}-\gamma^{3}z^{3}_{M}-2\gamma^{3}z_{M}^{2}+2(\gamma^{2}+1-\sqrt{2\gamma}-\gamma)z_{M}$
$\displaystyle=\frac{3\gamma-2-2\sqrt{2\gamma}}{\gamma+2+2\sqrt{2\gamma}}\gamma
z_{M}-\gamma^{3}z^{3}_{M}+2(\gamma^{2}+1-\sqrt{2\gamma}-\gamma-\gamma^{3}z_{M})z_{M}.$
(7.6)

The first two terms are negative because $\gamma\in(1,2]$. As for the last
term, when $\gamma\in(1,\sqrt[3]{2}]$, we have

$\displaystyle\gamma^{2}-\sqrt{2\gamma}<0.$

Thus, $\mathfrak{B}_{1}(-C^{2}_{6}(\gamma,z_{M}),z_{M},2)<0$ when
$\gamma\in(1,\sqrt[3]{2}]$. When $\gamma\in(\sqrt[3]{2},2]$, by using
$z_{M}\geq\frac{1}{8}$, we obtain

$\displaystyle\gamma^{2}+1-\sqrt{2\gamma}-\gamma-\gamma^{3}z_{M}$
$\displaystyle<\gamma^{2}+1-\sqrt{2\gamma}-\gamma-\frac{\gamma^{3}}{8}.$ (7.7)

This upper bound is increasing as a function of $\gamma$ as

$\displaystyle\frac{d}{d\gamma}\left(\gamma^{2}+1-\sqrt{2\gamma}-\gamma-\frac{\gamma^{3}}{8}\right)=-\frac{3\gamma^{2}}{8}+2\gamma-1-\frac{1}{\sqrt{2\gamma}}\geq-\frac{3\gamma^{2}}{8}+2\gamma-1-\frac{1}{\sqrt{2\sqrt[3]{2}}}>0.$

Hence, for any $\gamma\in(\sqrt[3]{2},2]$, we have

$\displaystyle\gamma^{2}+1-\sqrt{2\gamma}-\gamma-\frac{\gamma^{3}}{8}\leq 0,$

which, combined with (7.5), (7.6) and (7.7), leads to
$\mathfrak{B}_{1}(-C^{2}_{6}(\gamma,z_{M}),z_{M},2)<0$ in the remaining range
$\gamma\in[\sqrt[3]{2},2]$ and hence we have established
$\mathfrak{B}_{1}(V,z,2)<0$ for any $V\in[-C^{2}_{6},0)$,
$z\in(z_{g}(\gamma),z_{M}(\gamma)]$. This concludes the proof. ∎

###  7.2 Connecting $P_{8}$ to the origin

In this subsection we prove analogous results around $P_{8}$ for
$\gamma\in(\gamma_{\star},3]$ to those of the previous subsection, that is, we
show that the local analytic solutions around $P_{8}$ for
$z\in\mathring{\mathcal{Z}}(\gamma;P_{8})$ converge to the origin $P_{0}$
within the second quadrant by employing a barrier argument. We split into two
cases: $\gamma\in(\gamma_{\star},\gamma_{1}]$ and $\gamma\in(\gamma_{1},3]$.

####  7.2.1 $P_{8}$ for $\gamma\in(\gamma_{\star},\gamma_{1}]$

######  Lemma 7.4.

For any $\gamma\in(\gamma_{\star},\gamma_{1}]$ and
$z\in(z_{1}(\gamma),z_{M}(\gamma)]$, the solution constructed in Lemma 7.1
with $P_{*}=P_{8}$, always lies below the curve $B_{1}(V)=\sqrt{-V}$ for
$V\in(V_{8},0)$.

###### Proof.

By the definition of $z_{1}(\gamma)$ (cf. (6.2)) and Lemma 2.6, $C_{8}\leq
B_{1}(V_{8})$ for any $\gamma\in(\gamma_{\star},\gamma_{1}]$ and
$z\in(z_{1}(\gamma),z_{M}(\gamma)]$. Since $\frac{dC}{dV}<0$ by Lemma 7.1, the
solution stays below the curve $B_{1}(V)$ for $V\in(V_{8},-C_{8}^{2}]$. Thus,
it suffices to show the claim for $V\in(-C_{8}^{2},0)$. By employing the
barrier argument (2.20), we will establish the following inequality for any
$\gamma\in(\gamma_{\star},\gamma_{1}]$, $z\in(z_{1}(\gamma),z_{M}(\gamma)]$,
and $V\in(-C_{8}^{2},0)$:

$\displaystyle\frac{F(V,\sqrt{-V};\gamma,z)}{G(V,\sqrt{-V};\gamma,z)}+\frac{1}{2\sqrt{-V}}<0.$
(7.8)

By Lemma 7.1, $G(V,\sqrt{-V};\gamma,z)>0$ for all $V\in(-C_{8}^{2},0)$ and
hence, as in Lemma 6.2 (cf. (6.8)), we again see that it is sufficient to
prove that

$\displaystyle\mathfrak{B}_{1}(V,z,m)=(m-1-m\gamma)V^{2}+(-3+2m-m\gamma+m(\gamma-2)\gamma
z)V-m\gamma z-1+\frac{2mz}{1+V}<0.$

We first derive an upper bound of $\frac{\partial\mathfrak{B}_{1}}{\partial
V}$:

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

because $(\gamma-2)\gamma-2<0$ for $\gamma\in(\gamma_{\star},\gamma_{1}]$. By
(2.12), we observe that for any $\gamma\in(\gamma_{\star},\gamma_{1}]$,

$\displaystyle\frac{9}{20}<\frac{\sqrt{\gamma_{\star}}}{\sqrt{\gamma_{\star}}+\sqrt{2}}<\frac{\sqrt{\gamma}}{\sqrt{\gamma}+\sqrt{2}}\leq
C_{8}\leq C_{8}(z_{1})=\frac{\sqrt{5}-1}{2}.$ (7.9)

Thus, $2V+1>-2C_{8}^{2}+1>0$, which implies

$\displaystyle\frac{\partial\mathfrak{B}_{1}(V,z,m)}{\partial V}<0.$ (7.10)

It is therefore sufficient to show that
$\mathfrak{B}_{1}(-C^{2}_{8}(\gamma,z),z,m)<0$. We use the same notation as in
Lemma 6.2, denoting $\mathfrak{J}(z)=-C^{2}_{8}(\gamma,z)$. For any fixed
$\gamma\in(\gamma_{\star},\gamma_{1}]$, we write
$\mathfrak{J}^{\prime}(z)=\frac{d\mathfrak{J}(z)}{dz}$ and
$\mathfrak{J}^{\prime\prime}(z)=\frac{d^{2}\mathfrak{J}(z)}{dz^{2}}$. We then
derive the $z$ derivative of $\mathfrak{B}_{1}(\mathfrak{J}(z),z,m)$ as

$\displaystyle\frac{\partial\mathfrak{B}_{1}(\mathfrak{J}(z),z,m)}{\partial
z}=\frac{\partial\mathfrak{B}_{1}}{\partial
V}(\mathfrak{J}(z),z,m)\mathfrak{J}^{\prime}(z)+\frac{\partial\mathfrak{B}_{1}}{\partial
z}(\mathfrak{J}(z),z,m)=:A(\gamma,z,m)\mathfrak{J}^{\prime}(z)+mB(\gamma,z),$

where

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

We claim that both $A(\gamma,z,m)$ and $B(\gamma,z)$ are negative. For
$A(\gamma,z,m)$, in the case $m=1$, we have

$\displaystyle A(\gamma,z,1)=-2\gamma\mathfrak{J}(z)-1-\gamma+(\gamma-2)\gamma
z<-\gamma(1+2\mathfrak{J}(z))-1+|\gamma-2|\gamma z_{M}<0,$

where we have used $|\gamma-2|\gamma z_{M}<1$ in the last inequality. When
$m=2$, by (7.9) and $z<z_{M}(\gamma)<z_{M}(\gamma_{\star})<\frac{1}{5}$ for
any $\gamma\in(\gamma_{\star},\gamma_{1}]$ and
$z\in(z_{1}(\gamma),z_{M}(\gamma)]$, we see

$\displaystyle A(\gamma,z,2)=(1-2\gamma)(1+2\mathfrak{J}(z))+2(\gamma-2)\gamma
z<\frac{1-2\gamma+2|\gamma-2|\gamma}{5}<0$

for any $\gamma\in(\gamma_{\star},\gamma_{1}]$. Hence, $A(\gamma,z,m)<0$.

For $B(\gamma,z)$, we first observe that when
$\gamma\in(\gamma_{\star},\gamma_{1}]$, by (7.9),

$\displaystyle-\gamma+(\gamma-2)\gamma\mathfrak{J}(z)=-\gamma[1-(\gamma-2)\mathfrak{J}(z)]<0.$
(7.11)

For the remaining two terms, by Lemma 2.6 and Lemma 6.1, we have
$\mathfrak{J}^{\prime}(z)>0$ and $\mathfrak{J}^{\prime\prime}(z)>0$.
Therefore, for any $z\in(z_{1}(\gamma),z_{M}(\gamma)]$,

$\displaystyle\frac{d}{dz}[1+\mathfrak{J}(z)-z\mathfrak{J}^{\prime}(z)]=-z\mathfrak{J}^{\prime\prime}(z)<0.$

Hence, using also (7.11),

$\displaystyle
B(\gamma,z)<\frac{2}{1+\mathfrak{J}(z)}-\frac{2z\mathfrak{J}^{\prime}(z)}{(1+\mathfrak{J}(z))^{2}}=\frac{2}{(1+\mathfrak{J}(z))^{2}}(1+\mathfrak{J}(z)-z\mathfrak{J}^{\prime}(z))<\frac{2}{(1+\mathfrak{J}(z))^{2}}(1+\mathfrak{J}(z_{1})-z_{1}\mathfrak{J}^{\prime}(z_{1})).$

Recalling (2.6) and (6.3), we have

$\displaystyle C_{8}(z_{1})=\frac{\sqrt{5}-1}{2}\text{ and
}w(z_{1})=\sqrt{5}-2-(\gamma-2)z_{1}>0,$

where the positivity of $w(z_{1})$ is due to Remark 2.5. By direct
computation, using (2.6), (2.8), and (6.3), we obtain

$\displaystyle 1+\mathfrak{J}(z_{1})-z_{1}\mathfrak{J}^{\prime}(z_{1})$
$\displaystyle=1-C_{8}^{2}(z_{1})+2z_{1}C_{8}(z_{1})C_{8}^{\prime}(z_{1})$
$\displaystyle=\frac{C_{8}(z_{1})}{w(z_{1})}[\sqrt{5}-2+[(\sqrt{5}-4)\gamma-2(\sqrt{5}-2)]z_{1}]$
$\displaystyle=\frac{C_{8}(z_{1})}{w(z_{1})}\frac{(5-3\sqrt{5})\gamma+4\sqrt{5}-8}{2(\sqrt{5}+1+\gamma)}<0$

for any $\gamma\in(\gamma_{\star},\gamma_{1}]$, and therefore we obtain
$B(\gamma,z)<0$.

Therefore, we have obtained $\frac{\partial}{\partial
z}\mathfrak{B}_{1}(\mathfrak{J}(z),z,m)<0$. Hence, we conclude that for any
fixed $\gamma\in(\gamma_{\star},\gamma_{1}]$, $m=1,2$,
$z\in(z_{1}(\gamma),z_{M}(\gamma)]$, and $V\in(-C^{2}_{8},0)$, applying also
(7.10), we have
$\mathfrak{B}_{1}(V,z,m)<\mathfrak{B}_{1}(\mathfrak{J}(z),z,m)<\mathfrak{B}_{1}(\mathfrak{J}(z_{1}),z_{1},m)=0$
by the definition of $z_{1}$ (cf. (6.2)). ∎

####  7.2.2 $P_{8}$ for $\gamma\in(\gamma_{1},3]$

######  Lemma 7.5.

For any $\gamma\in(\gamma_{1},3]$ and $z\in(z_{2}(\gamma),z_{M}(\gamma)]$, the
solution constructed in Lemma 7.1 with $P_{*}=P_{8}$, always lies below the
curve $B_{\frac{3}{2}}(V)=\sqrt{-\frac{3}{2}V}$ for $V\in(V_{8},0)$.

###### Proof.

Using a similar argument as in Lemma 7.4, it suffices to verify the following
inequality:

$\displaystyle\frac{F(V,\sqrt{-\frac{3}{2}V};\gamma,z)}{G(V,\sqrt{-\frac{3}{2}V};\gamma,z)}+\frac{1}{2}\sqrt{\frac{-3}{2V}}<0$

for any $m=1,2$, $\gamma\in(\gamma_{1},3]$,
$z\in(z_{2}(\gamma),z_{M}(\gamma)]$ and $V\in(-\frac{2}{3}C_{8}^{2},0)$. Given
that $G(V,\sqrt{-\frac{3}{2}V},\gamma,z)>0$ for any
$V\in(-\frac{2}{3}C_{8}^{2},0)$ as shown in Lemma 7.1, and by the same
calculations as in Lemma 6.3 (cf. (6.19)), it is sufficient to demonstrate
that

$\displaystyle\mathfrak{B}_{{\scriptstyle\frac{3}{2}}}(V,z,m)=(m-1-m\gamma)V^{2}+(-4+2m+\frac{m+1}{2}-m\gamma+m(\gamma-2)\gamma
z)V-mz\gamma-1+\frac{3mz}{1+V}<0$

for any $m=1,2$, $\gamma\in(\gamma_{1},3]$,
$z\in(z_{2}(\gamma),z_{M}(\gamma)]$ and $V\in(-\frac{2}{3}C^{2}_{8},0)$. The
$V$ derivative of $\mathfrak{B}_{\frac{3}{2}}$ is given by

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

When $m=1$, the same argument as in (6.21) implies
$\frac{\partial\mathfrak{B}_{{\scriptstyle\frac{3}{2}}}(V,z,1)}{\partial
V}<0$. When $m=2$, we have

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

Note that $(\gamma-2)\gamma-\frac{3}{(1+V)^{2}}<0$ for
$\gamma\in(\gamma_{1},3]$ and $V\in(-\frac{2}{3}C_{8}^{2},0)$. Moreover,
$2(1-2\gamma)V+\frac{3}{2}-2\gamma<0$, since for any fixed
$\gamma\in(\gamma_{1},3]$,

$\displaystyle
2(1-2\gamma)V+\frac{3}{2}-2\gamma<-2(1-2\gamma)\frac{2}{3}C_{8}^{2}+\frac{3}{2}-2\gamma=\frac{\sqrt{33}-4}{2}+(5-\sqrt{33})\gamma<0.$

Hence,

$\displaystyle\frac{\partial\mathfrak{B}_{\frac{3}{2}}(V,z,m)}{\partial V}<0.$

It is therefore sufficient to show that
$\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C^{2}_{8}(z),z)<0$. Let
$\Im(z)=-\frac{2}{3}C^{2}_{8}(z)$ and, for any fixed $\gamma$, we write
$\Im^{\prime}(z)=\frac{d\Im(z)}{dz}$ and
$\Im^{\prime\prime}(z)=\frac{d^{2}\Im(z)}{dz^{2}}$. By Lemma 2.6 and Lemma
6.1, $\Im^{\prime}(z)>0$ and $\Im^{\prime\prime}(z)>0$. The $z$ derivative of
$\mathfrak{B}_{\frac{3}{2}}(\mathfrak{J}(z),z,m)$ can therefore be written as

$\displaystyle\frac{d\mathfrak{B}_{\frac{3}{2}}(\Im(z),z,m)}{dz}=\frac{\partial\mathfrak{B}_{\frac{3}{2}}}{\partial
V}(\Im(z),z,m)\Im^{\prime}(z)+\frac{\partial\mathfrak{B}_{\frac{3}{2}}}{\partial
z}(\Im(z),z,m)=:A(\gamma,z,m)\Im^{\prime}(z)+mB(\gamma,z),$

where

$\displaystyle
A(\gamma,z,m):=2(m-1-m\gamma)\Im(z)-4+2m+\frac{m+1}{2}-m\gamma+m\gamma(\gamma-2)z,$
$\displaystyle
B(\gamma,z):=(\gamma-2)\gamma\Im(z)-\gamma+\frac{3(1+\Im(z))-3z\Im^{\prime}(z)}{(1+\Im(z))^{2}}.$

We claim that both $A(\gamma,z,m)$ and $B(\gamma,z)$ are negative. We first
check $A(\gamma,z,m)$. When $m=1$,

$\displaystyle A(\gamma,z,1)=-2\gamma\Im(z)-1-\gamma+(\gamma-2)\gamma
z=-\gamma(1+2\Im(z))-1+(\gamma-2)\gamma z<0,$ (7.12)

where we have used $(\gamma-2)\gamma z\leq(\gamma-2)\gamma
z_{M}\leq\frac{\gamma}{\gamma+2+2\sqrt{2\gamma}}\leq\frac{3}{5+2\sqrt{6}}$ and
$1+2\Im(z)>1+2\Im(z_{2})=\frac{\sqrt{33}-5}{2}$ in the last inequality. When
$m=2$, using
$1+2\Im(z)>1+2\Im(z_{2})=\frac{\sqrt{33}-5}{2}>\frac{3}{5+2\sqrt{6}}\geq\gamma
z_{M}\geq\gamma z$,

$\displaystyle A(\gamma,z,2)$
$\displaystyle=2(1-2\gamma)\Im(z)+\frac{3}{2}-2\gamma+2(\gamma-2)\gamma
z=2(2-\gamma)[1+2\Im(z)-\gamma z]-3(1+2\Im(z))+\frac{1}{2}$
$\displaystyle\leq-3(1+2\Im(z))+\frac{1}{2}\leq-3(\frac{\sqrt{33}-5}{2})+\frac{1}{2}=\frac{16-3\sqrt{33}}{2}<0.$
(7.13)

For $B(\gamma,z)$, the first term is trivially negative for any
$\gamma\in(\gamma_{1},3]$ since $\Im(z)<0$. We will show that the remaining
term is negative as well. By using (2.11), we obtain that for any
$z\in(z_{2}(\gamma),z_{M}(\gamma)]$,

$\displaystyle 0<C_{8}(z_{2})=1-\frac{2}{3}C_{8}^{2}(z_{2})<1+\Im(z)\leq
1-\frac{2}{3}C_{8}^{2}(z_{M})=1-\frac{2\gamma z_{M}(\gamma)}{3}.$

Since $\frac{d}{dz}(-z\Im^{\prime}(z)+1+\Im(z))=-z\Im^{\prime\prime}(z)<0$,

$\displaystyle\frac{3(-z\Im^{\prime}(z)+1+\Im(z))}{(1+\Im(z))^{2}}<\frac{3|-z_{2}\Im^{\prime}(z_{2})+1+\Im(z_{2})|}{(1+\Im(z_{2}))^{2}}.$
(7.14)

Recalling (2.6), Remark 2.5 and (6.14), we have

$\displaystyle C_{8}(z_{2})=\frac{\sqrt{33}-3}{4}\text{ and
}w(z_{2})=\frac{\sqrt{33}-5}{2}-(\gamma-2)z_{2}>0.$

Thus, by a direct computation, we obtain

$\displaystyle\frac{3|-z_{2}\Im^{\prime}(z_{2})+1+\Im(z_{2})|}{(1+\Im(z_{2}))^{2}}$
$\displaystyle=\frac{3}{C_{8}^{2}(z_{2})}|1-\frac{2}{3}C^{2}_{8}(z_{2})+\frac{4}{3}z_{2}C_{8}(z_{2})C_{8}^{\prime}(z_{2})|$
$\displaystyle=\frac{1}{C_{8}(z_{2})}\frac{|(33-7\sqrt{33})\gamma+12\sqrt{33}-48|}{(\sqrt{33}-7)\gamma+12}$
$\displaystyle=\frac{4}{\sqrt{33}-3}|\sqrt{33}-\frac{48}{(\sqrt{33}-7)\gamma+12}|$
$\displaystyle\leq\frac{4}{\sqrt{33}-3}(\sqrt{33}-\frac{48}{(\sqrt{33}-7)\gamma_{1}+12})<\gamma$
(7.15)

for any $\gamma\in(\gamma_{1},3]$. Therefore, combining (7.12)–(7.15), we have
found

$\displaystyle\frac{d\mathfrak{B}_{\frac{3}{2}}(\Im(z),z,m)}{dz}=$
$\displaystyle\,A(\gamma,z,m)\Im^{\prime}(z)+mB(\gamma,z)$ $\displaystyle<$
$\displaystyle\,m\Big{(}(\gamma-2)\gamma\Im(z)-\gamma+\frac{3(1+\Im(z))-3z\Im^{\prime}(z)}{(1+\Im(z))^{2}}\Big{)}$
$\displaystyle<$
$\displaystyle\,m\Big{(}-\gamma+\frac{3|-z_{2}\Im^{\prime}(z_{2})+1+\Im(z_{2})|}{(1+\Im(z_{2}))^{2}}\Big{)}<0.$

Hence, for any fixed $\gamma\in(\gamma_{1},3]$, $m=1,2$,
$z\in(z_{2}(\gamma),z_{M}(\gamma)]$, and $V\in(-\frac{2}{3}C^{2}_{8},0)$, we
have
$\mathfrak{B}_{\frac{3}{2}}(V,z,m)<\mathfrak{B}_{\frac{3}{2}}(\Im(z),z,m)<\mathfrak{B}_{\frac{3}{2}}(-\frac{2}{3}C_{8}^{2}(z_{2}),z_{2},m)=0$
by the definition of $z_{2}$ (cf. (6.13)). ∎

##  8 Proof of the main theorem

We now prove Theorem 2.9.

$(i)$ follows by $(ii)$, $(iii)$, and $(iv)$.

By Theorem 4.4, there exists a $z_{std}(\gamma)$ such that the local real
analytic solution $C(V;\gamma,z,P_{*})$ around $P_{*}$ for either
$P_{*}=P_{6}$ or $P_{*}=P_{8}$ given by Theorem 3.8 extends on the left to
$P_{1}$.

$(ii)$. Let $\gamma\in(1,\gamma_{\star}]$ be fixed. By Lemma 4.7, any such
$z_{std}(\gamma)$ must connect $P_{1}$ to $P_{6}$ and, by Proposition 5.3,
$z_{std}(\gamma)\in(z_{g}(\gamma),z_{M}(\gamma)]$, that is,
$z_{std}(\gamma)\in\mathring{\mathcal{Z}}(\gamma;P_{6})$. Then, Lemma 5.4
gives that in fact $z_{std}(\gamma)$ is unique. By Lemma 7.3, this unique
solution extends to $P_{0}$ in the second quadrant to give a unique connection
from $P_{1}$ to $P_{0}$ which passes through $P_{6}$ and is monotone.

$(iii)$. Let $\gamma\in(\gamma_{\star},2)$ be fixed. Given such a
$z_{std}(\gamma)$, if $P_{1}$ is connected analytically to $P_{6}$, by
Proposition 5.3, we must have
$z_{std}(\gamma)\in(z_{g}(\gamma),z_{M}(\gamma)]$ and so, by Lemma 7.3, the
solution extends to the right of $P_{6}$ to connect to $P_{0}$ within the
second quadrant. On the other hand, if the solution connects $P_{1}$ to
$P_{8}$ analytically, then by Lemma 6.2,
$z_{std}(\gamma)\in(z_{1}(\gamma),z_{M}(\gamma)]$, and, applying Lemma 7.4,
the solution again extends inside the second quadrant to connect to $P_{0}$.
Thus, in either case, we have
$z_{std}(\gamma)\in\mathring{\mathcal{Z}}(\gamma;P_{*})$ and have obtained a
monotone analytic solution connecting $P_{1}$ to $P_{0}$ through a single
triple point.

$(iv)$. Let $\gamma\in[2,3]$ be fixed. By Proposition 4.9, the solution for
$z=z_{std}(\gamma)$ must connect $P_{1}$ to $P_{8}$ analytically and, by
Lemmas 6.2 and 6.3, $z_{std}(\gamma)\in\mathring{\mathcal{Z}}(\gamma;P_{8})$.
Therefore, by Lemma 7.4 and Lemma 7.5, in each case, the solution extends to
the right in the second quadrant to connect $P_{8}$ to $P_{0}$ and we again
have obtained an analytic, monotone solution connecting $P_{1}$ to $P_{0}$.

Acknowledgements. JJ and JL are supported in part by the NSF grants
DMS-2009458 and DMS-2306910. MS is supported by the EPSRC Post-doctoral
Research Fellowship EP/W001888/1.

##  Appendix A Calculation for Taylor Expansion

The purpose of this Appendix is to establish the proof of Lemma 3.4. To this
end, we begin from (2.1) in the form

$(1+V)F(C,V)-(1+V)C^{\prime}(V)G(C,V)=0.$

We write the left hand side of this equation as a power series in $v=V-V_{*}$
as

$(1+V)F(C,V)-(1+V)C^{\prime}(V)G(C,V)=\sum_{\ell=0}^{\infty}\mathcal{C}_{\ell}v^{\ell}.$

Substituting in (3.20), (3.21), (3.22), and (3.23), the first term of this
identity expands as

$\displaystyle(1+v+V_{*})F(V,C)$ $\displaystyle=$
$\displaystyle\,(1+v+V_{*})C(V)\big{\\{}C^{2}(V)[1+\frac{mz}{(1+v+V_{*})}]-a_{1}(1+v+V_{*})^{2}+a_{2}(1+v+V_{*})-a_{3}\big{\\}}$
$\displaystyle=$
$\displaystyle\,vC^{3}(V)+(1+V_{*}+mz)C^{3}(V)+\Big{[}-a_{1}v^{3}+[-3a_{1}(1+V_{*})+a_{2}]v^{2}+[-3a_{1}(1+V_{*})^{2}+2a_{2}(1+V_{*})-a_{3}]v$
$\displaystyle+[-a_{1}(1+V_{*})^{3}+a_{2}(1+V_{*})^{2}-a_{3}(1+V_{*})]\Big{]}C(V)$
$\displaystyle=$
$\displaystyle\,\sum_{\ell=0}^{\infty}(c^{3})_{\ell}v^{\ell+1}+(1+V_{*}+mz)\sum_{\ell=0}^{\infty}(c^{3})_{\ell}v^{\ell}-a_{1}\sum_{\ell=0}^{\infty}c_{\ell}v^{\ell+3}+\Big{[}-3a_{1}(1+V_{*})+a_{2}\Big{]}\sum_{\ell=0}^{\infty}c_{\ell}v^{\ell+2}$
(A.1)
$\displaystyle+\Big{[}-3a_{1}(1+V_{*})^{2}+2a_{2}(1+V_{*})-a_{3}\Big{]}\sum_{\ell=0}^{\infty}c_{\ell}v^{\ell+1}+\Big{[}-a_{1}(1+V_{*})^{3}+a_{2}(1+V_{*})^{2}-a_{3}(1+V_{*})\Big{]}\sum_{\ell=0}^{\infty}c_{\ell}v^{\ell}.$

Expanding also the second term, we obtain

$\displaystyle(1+v+V_{*})C^{\prime}(V)G(V,C)$ $\displaystyle=$
$\displaystyle\,(1+v+V_{*})C^{\prime}(V)\bigg{\\{}C^{2}(V)\Big{[}(m+1)(v+V_{*})+2mz\Big{]}-(v+V_{*})(1+v+V_{*})(\lambda+v+V_{*})\bigg{\\}}$
$\displaystyle=$
$\displaystyle\,(m+1)v^{2}C^{\prime}(V)C^{2}(V)+\Big{[}(m+1)(1+2V_{*})+2mz\Big{]}vC^{\prime}(V)C^{2}(V)+(1+V_{*})\Big{[}(m+1)V_{*}+2mz\Big{]}C^{\prime}(V)C^{2}(V)$
$\displaystyle-\Big{[}v^{4}+(\lambda+2+4V_{*})v^{3}+[6V_{*}^{2}+(3\lambda+6)V_{*}+2\lambda+1]v^{2}+[4V_{*}^{3}+(3\lambda+6)V_{*}^{2}+(4\lambda+2)V_{*}+\lambda]v$
$\displaystyle+V_{*}(1+V_{*})^{2}(\lambda+V_{*})\bigg{]}C^{\prime}(V)$
$\displaystyle=$
$\displaystyle\,\frac{m+1}{3}\sum_{\ell=1}^{\infty}\ell(c^{3})_{\ell}v^{\ell+1}+\frac{(m+1)(1+2V_{*})+2mz}{3}\sum_{\ell=1}^{\infty}\ell(c^{3})_{\ell}v^{\ell}+\frac{(1+V_{*})\Big{[}(m+1)V_{*}+2mz\Big{]}}{3}\sum_{\ell=1}^{\infty}\ell(c^{3})_{\ell}v^{\ell-1}$
$\displaystyle-\sum_{\ell=1}^{\infty}\ell
c_{\ell}v^{\ell+3}-(\lambda+2+4V_{*})\sum_{1}^{\infty}\ell
c_{\ell}v^{\ell+2}-\Big{[}6V_{*}^{2}+(3\lambda+6)V_{*}+2\lambda+1\Big{]}\sum_{\ell=1}^{\infty}\ell
c_{\ell}v^{\ell+1}$ (A.2)
$\displaystyle-\Big{[}4V_{*}^{3}+(3\lambda+6)V_{*}^{2}+(4\lambda+2)V_{*}+\lambda\Big{]}\sum_{\ell=1}^{\infty}\ell
c_{\ell}v^{\ell}-V_{*}(1+V_{*})^{2}(\lambda+V_{*})\sum_{\ell=1}^{\infty}\ell
c_{\ell}v^{\ell-1}.$

We now proceed to study the difference of (A.1) and (A.2) and to group terms
at each order in $v$ to simplify the resulting identity for
$\mathcal{C}_{\ell}$. First, at order zero, we have

$\displaystyle\mathcal{C}_{0}=$
$\displaystyle\,(1+V_{*}+mz)c_{0}^{3}+\Big{[}-a_{1}(1+V_{*})^{3}+a_{2}(1+V_{*})^{2}-a_{3}(1+V_{*})\Big{]}c_{0}-\frac{(1+V_{*})\Big{[}(m+1)V_{*}+2mz\Big{]}}{3}(3c_{0}^{2}c_{1})$
$\displaystyle+V_{*}(1+V_{*})^{2}(\lambda+V_{*})c_{1}$ $\displaystyle=$
$\displaystyle\,\Big{[}c_{0}^{2}(1+V_{*}+mz)-a_{1}(1+V_{*})^{3}+a_{2}(1+V_{*})^{2}-a_{3}(1+V_{*})\Big{]}c_{0}$
$\displaystyle-\Big{[}c_{0}^{2}[(m+1)V_{*}+2mz]-V_{*}(1+V_{*})(\lambda+V_{*})\Big{]}\Big{]}(1+V_{*})c_{1}$
$\displaystyle=$
$\displaystyle\,F(V_{*},C_{*})(1+V_{*})c_{0}-G(V_{*},C_{*})(1+V_{*})c_{1}=0.$
(A.3)

Next, the first order coefficient in $v$ simplifies as

$\displaystyle\mathcal{C}_{1}=$
$\displaystyle\,c_{0}^{3}+(1+V_{*}+mz)3c_{0}^{2}c_{1}+\Big{[}-3a_{1}(1+V_{*})^{2}+2a_{2}(1+V_{*})-a_{3}\Big{]}c_{0}+\Big{[}-a_{1}(1+V_{*})^{3}+a_{2}(1+V_{*})^{2}-a_{3}(1+V_{*})\Big{]}c_{1}$
$\displaystyle-\frac{(m+1)(1+2V_{*})+2mz}{3}3c_{0}^{2}c_{1}-\frac{(1+V_{*})\Big{[}(m+1)V_{*}+2mz\Big{]}}{3}2(3c_{0}^{2}c_{2}+3c_{0}c_{1}^{2})$
(A.4)
$\displaystyle+\Big{[}4V_{*}^{3}+(3\lambda+6)V_{*}^{2}+(4\lambda+2)V_{*}+\lambda\Big{]}c_{1}+V_{*}(1+V_{*})^{2}(\lambda+V_{*})2c_{2}.$

In order to simplify this identity, we recall that as $G(V_{*},C_{*})=0$, we
have $c_{0}^{2}((m+1)+2mz)=V_{*}(1+V_{*})(\lambda+V_{*})$ and thus, recalling
also (3.2), we have the auxiliary identity

$-[(m+1)(1+2V_{*})+2mz]c_{0}^{2}+\Big{[}4V_{*}^{3}+(3\lambda+6)V_{*}^{2}+(4\lambda+2)V_{*}+\lambda\Big{]}=-G_{V}(V_{*},C_{*})(1+V_{*}).$
(A.5)

Substituting this along with the other identities in (3.2) into (A.4) and
grouping terms, we find $\mathcal{C}_{1}$ simplifies to

$\displaystyle\mathcal{C}_{1}=$
$\displaystyle\,2\Big{\\{}V_{*}(1+V_{*})^{2}(\lambda+V_{*})-(1+V_{*})\Big{[}(m+1)V_{*}+2mz\Big{]}c_{0}^{2}\Big{\\}}c_{2}+\Big{\\{}-2(1+V_{*})\Big{[}(m+1)V_{*}+2mz\Big{]}c_{0}\Big{\\}}c_{1}^{2}$
$\displaystyle+\Big{\\{}3(1+V_{*}+mz)c_{0}^{2}+\Big{[}-a_{1}(1+V_{*})^{3}+a_{2}(1+V_{*})^{2}-a_{3}(1+V_{*})\Big{]}-[(m+1)(1+2V_{*})+2mz]c_{0}^{2}$
$\displaystyle\quad+\Big{[}4V_{*}^{3}+(3\lambda+6)V_{*}^{2}+(4\lambda+2)V_{*}+\lambda\Big{]}\Big{\\}}c_{1}+c_{0}^{3}+\Big{[}-3a_{1}(1+V_{*})^{2}+2a_{2}(1+V_{*})-a_{3}\Big{]}c_{0}$
$\displaystyle=$
$\displaystyle-2G(V_{*},C_{*})(1+V_{*})c_{2}-G_{C}(V_{*},C_{*})(1+V_{*})c_{1}^{2}+[F_{C}(V_{*},C_{*})-G_{V}(V_{*},C_{*})](1+V_{*})c_{1}+F_{V}(V_{*},C_{*})c_{0}$
$\displaystyle=$ $\displaystyle-
G_{C}(V_{*},C_{*})(1+V_{*})c_{1}^{2}+[F_{C}(V_{*},C_{*})-G_{V}(V_{*},C_{*})](1+V_{*})c_{1}+F_{V}(V_{*},C_{*})c_{0}.$
(A.6)

Finally, we group coefficients at order $\ell\geq 2$, recalling the convention
that $c_{k}=0$ if $k<0$, to obtain the coefficient

$\displaystyle\mathcal{C}_{\ell}=$
$\displaystyle\,(c^{3})_{\ell-1}+\big{(}1+V_{*}+mz)(c^{3})_{\ell}-a_{1}c_{\ell-3}+\big{(}-3a_{1}(1+V_{*})+a_{2}\big{)}c_{\ell-2}$
$\displaystyle+\big{(}-3a_{1}(1+V_{*})^{2}+2a_{2}(1+V_{*})-a_{3}\big{)}c_{\ell-1}+\big{(}-a_{1}(1+V_{*})^{3}+a_{2}(1+V_{*})^{2}-a_{3}(1+V_{*})\big{)}c_{\ell}$
$\displaystyle-\frac{m+1}{3}(\ell-1)(c^{3})_{\ell-1}-\frac{(m+1)(1+2V_{*})+2mz}{3}\ell(c^{3})_{\ell}-\frac{(1+V_{*})\big{(}(m+1)V_{*}+2mz\big{)}}{3}(\ell+1)(c^{3})_{\ell+1}$
$\displaystyle+(\ell-3)c_{\ell-3}+(\lambda+2+4V_{*})(\ell-2)c_{\ell-2}+\big{(}6V_{*}^{2}+(3\lambda+6)V_{*}+2\lambda+1\big{)}(\ell-1)c_{\ell-1}$
(A.7)
$\displaystyle+\big{(}4V_{*}^{3}+(3\lambda+6)V_{*}^{2}+(4\lambda+2)V_{*}+\lambda\big{)}\ell
c_{\ell}+V_{*}(1+V_{*})^{2}(\lambda+V_{*})(\ell+1)c_{\ell+1}.$

Recalling that

$\displaystyle(c^{3})_{\ell}=$
$\displaystyle\,\sum_{i+j+k=\ell}c_{i}c_{j}c_{k}=3c_{0}^{2}c_{\ell}+\sum_{\begin{subarray}{c}i+j+k=\ell\\\
i,j,k\leq\ell-1\end{subarray}}c_{i}c_{j}c_{k}$
$\displaystyle(c^{3})_{\ell+1}=$
$\displaystyle\,\sum_{i+j+k=\ell+1}c_{i}c_{j}c_{k}=3c_{0}^{2}c_{\ell+1}+6c_{0}c_{1}c_{\ell}+\sum_{\begin{subarray}{c}i+j+k=\ell+1\\\
i,j,k\leq\ell-1\end{subarray}}c_{i}c_{j}c_{k},$

we isolate the highest order terms in $c_{\ell}$ and $c_{\ell+1}$ from (A.7)
as

$\displaystyle
c_{\ell+1}(\ell+1)\Big{(}3c_{0}^{2}\big{(}-\frac{(1+V_{*})\big{(}(m+1)V_{*}+2mz\big{)}}{3}+V_{*}(1+V_{*})^{2}(\lambda+V_{*})\Big{)}=-c_{\ell+1}(\ell+1)(V_{*}+1)G(V_{*},C_{*})=0,$
$\displaystyle
c_{\ell}\Big{(}3(1+V_{*}+mz)c_{0}^{2}-a_{1}(1+V_{*})^{3}+a_{2}(1+V_{*})^{2}-a_{3}(1+V_{*})-\big{(}(m+1)(1+2V_{*})+2mz)\ell
c_{0}^{2}$
$\displaystyle\quad-2(1+V_{*})\big{(}(m+1)V_{*}+2mz\big{)}(\ell+1)c_{0}c_{1}+\big{(}4V_{*}^{3}+(3\lambda+6)V_{*}^{2}+(4\lambda+2)V_{*}+\lambda\big{)}\ell\Big{)}$
$\displaystyle=c_{\ell}(1+V_{*})\Big{(}F_{C}-\ell
G_{V}-(\ell+1)c_{1}G_{C}\Big{)}$ $\displaystyle=A_{\ell}c_{\ell},$

where we have again applied (3.2) and (A.5) and where $A_{\ell}$ is as defined
in Lemma 3.4. Thus, substituting these into (A.7) and grouping terms by order
of $c_{k}$, we have obtained that

$\displaystyle\mathcal{C}_{\ell}=$
$\displaystyle\,A_{\ell}c_{\ell}-\frac{(1+V_{*})\Big{[}(m+1)V_{*}+2mz\Big{]}}{3}(\ell+1)\sum_{\begin{subarray}{c}i+j+k=\ell+1\\\
i,j,k\leq\ell-1\end{subarray}}c_{i}c_{j}c_{k}$
$\displaystyle+\Big{[}(1+V_{*}+mz)-\frac{(m+1)(1+2V_{*})+2mz}{3}\ell\Big{]}\sum_{\begin{subarray}{c}i+j+k=\ell\\\
i,j,k\leq\ell-1\end{subarray}}c_{i}c_{j}c_{k}$
$\displaystyle+\Big{[}1-\frac{m+1}{3}(\ell-1)\Big{]}\sum_{\begin{subarray}{c}i+j+k=\ell-1\\\
i,j,k\geq 0\end{subarray}}c_{i}c_{j}c_{k}$
$\displaystyle+\Big{[}\big{[}6V_{*}^{2}+(3\lambda+6)V_{*}+2\lambda+1\big{]}(\ell-1)-3a_{1}(1+V_{*})^{2}+2a_{2}(1+V_{*})-a_{3}\Big{]}c_{\ell-1}$
(A.8)
$\displaystyle+\Big{[}(\lambda+2+4V_{*})(\ell-2)-3a_{1}(1+V_{*})+a_{2}\Big{]}c_{\ell-2}+\Big{[}\ell-3-a_{1}\Big{]}c_{\ell-3},$

which, recalling the definition of $B_{\ell}$ from Lemma 3.4, concludes the
proof.∎

##  Appendix B $z_{m}<z_{g}$ for $\gamma\in(1,2]$

As the expression for $z_{g}$ depends on $m$, we will prove the inequality
first in the case $m=1$ and then for $m=2$. Recall first that $z_{m}$ is
defined by (3.16),

$\displaystyle
z_{m}=\cfrac{(\gamma-1)}{(2\gamma-1)(\gamma+1)}=\cfrac{(\gamma-1)}{2\gamma^{2}+\gamma-1}.$

On the other hand, for $z_{g}$, when $m=1$, we have

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

Therefore, it is sufficient to check
$\gamma^{2}+\gamma-1>\gamma\sqrt{\gamma^{2}+(\gamma-1)^{2}}$ for
$\gamma\in(1,2]$. Since

$\displaystyle(\gamma^{2}+\gamma-1)^{2}-\gamma^{2}(\gamma^{2}+(\gamma-1)^{2})=-(\gamma-1)(\gamma^{2}(\gamma-3)+1-\gamma)>0$

for any $\gamma\in(1,2]$, we conclude

$\displaystyle z_{m}<z_{g}$

for any $\gamma\in(1,2]$ as desired.

When $m=2$, we have

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

Therefore, in order to show $z_{g}>z_{m}$, it is enough to check
$2\gamma^{2}+3\gamma-3>\sqrt{(2\gamma^{2}-\gamma+1)^{2}+2\gamma(\gamma-1)[4\gamma(\gamma-1)+\frac{8}{3}]}$.
Since

$\displaystyle(2\gamma^{2}+3\gamma-3)^{2}-(2\gamma^{2}-\gamma+1)^{2}+2\gamma(\gamma-1)[4\gamma(\gamma-1)+\frac{8}{3}]=\frac{8}{3}(\gamma-1)(3-\gamma)(3\gamma^{2}-1)>0$

for any $\gamma\in(1,2]$, we again conclude the claimed inequality.

##  Appendix C Proof of (4.9)

Let $\gamma\in(1,2]$ and $z\in[z_{g},z_{M}]$, where we recall that $z_{g}$ is
defined as in (5.8). In this section, for notational convenience, we will use
$G_{C}$, $G_{V}$, $F_{C}$, $F_{V}$, $R$ to represent their evaluations at
$P_{6}=(V_{6},C_{6})$ where $G_{C}$, $G_{V}$, $F_{C}$, $F_{V}$, $R$ are given
in (3.5)–(3.8) and (3.4) respectively.

Recall from (3.18) that the slope of the curve $C=C(V)$ at
$P_{6}=(V_{6},C_{6})$ is given by

$\displaystyle\frac{dC}{dV}|_{V=V_{6},C=C_{6}}=c_{1}=\frac{F_{C}-G_{V}+R}{2G_{C}}.$

Thus (4.9) holds if and only if

$\displaystyle\frac{F_{C}-G_{V}+R}{2G_{C}}-\frac{1+V_{6}}{2V_{6}}$
$\displaystyle=\frac{V_{6}F_{C}-V_{6}G_{V}+V_{6}R-(1+V_{6})G_{C}}{2V_{6}G_{C}}<0.$

Note that $G_{C}<0$ by (3.5) and hence $V_{6}G_{C}>0$. Therefore, showing
(4.9) is equivalent to proving that the numerator is negative for each
$\gamma\in(1,2]$ and $z\in[z_{g},z_{M}]$. Using (3.4), it is then sufficient
to show

$\displaystyle
V_{6}^{2}[(F_{C}-G_{V})^{2}+4F_{V}G_{C}]-[V_{6}F_{C}-V_{6}G_{V}-(1+V_{6})G_{C}]^{2}>0,$
(C.1)

which is equivalent to

$\displaystyle\mathcal{Q}:=4V_{6}^{2}F_{V}G_{C}-(1+V_{6})^{2}G_{C}^{2}+2(1+V_{6})V_{6}(F_{C}-G_{V})G_{C}>0.$
(C.2)

The rest of this section is devoted to the proof of (C.2). We first rewrite
$\mathcal{Q}$ by using various identities satisfied by
$V_{6},C_{6},F_{C},F_{V},G_{C},G_{V}$:

$\displaystyle\mathcal{Q}=$
$\displaystyle\,G_{C}\Big{[}2V_{6}^{2}(m(\gamma-1)wC_{6}-2F_{C})-C_{6}^{2}G_{C}+2C_{6}V_{6}(F_{C}+mwC_{6}+G_{C})\Big{]}$
$\displaystyle=$ $\displaystyle\,G_{C}V_{6}C_{6}\Big{[}-2V_{6}^{2}+2m[\gamma
w+(\gamma-2)z]V_{6}+2m([2-\gamma]z+w)+2\Big{]}$

where we have used $2F_{V}=m(\gamma-1)wC_{6}-2F_{C}$ and
$-G_{V}=mwC_{6}+G_{C}$ from (3.11) and (3.12) as well as $C_{6}=1+V_{6}$ in
the first line and used (3.7) and (3.5) in the second line. Recalling the
formula for $V_{6}$ in (2.4) and using $V_{6}^{2}=((\gamma-2)z-1)V_{6}-2z$, we
next rearrange the bracket as a linear function in $w$ whose coefficients are
polynomials in $\gamma,z$ so that

$\displaystyle\mathcal{Q}=$
$\displaystyle\,G_{C}V_{6}C_{6}\Big{[}A(\gamma,z,m)w+B(\gamma,z,m)\Big{]}$

where

$\displaystyle A(\gamma,z,m)$
$\displaystyle=2m-1-m\gamma+(\gamma-2+2m-3m\gamma+m\gamma^{2})z,$
$\displaystyle B(\gamma,z,m)$
$\displaystyle=1-m\gamma+(6m+(2+m)\gamma+2m\gamma^{2})z+(4m-4+4(1-2m)\gamma+(5m-1)\gamma^{2}-m\gamma^{3})z^{2}.$

Since $G_{C}V_{6}C_{6}>0$, our aim is to show $A(\gamma,z,m)w+B(\gamma,z,m)>0$
for each $\gamma\in(1,2]$, $z\in[z_{g},z_{M}]$ and $m=1,2$. We will treat
$m=1$ and $m=2$ separately.

Case 1: $m=1$. When $m=1$, we have

$\displaystyle A(\gamma,z,1)$ $\displaystyle=1-\gamma+\gamma(\gamma-2)z<0,$
$\displaystyle B(\gamma,z,1)$
$\displaystyle=1-\gamma+(2\gamma^{2}+3\gamma+6)z-\gamma(\gamma-2)^{2}z^{2}$

for all $\gamma\in(1,2]$. On the other hand, for any fixed $\gamma\in(1,2]$,

$\displaystyle\frac{d}{dz}(A(\gamma,z,1)w+B(\gamma,z,1))$
$\displaystyle=2\gamma^{2}+3\gamma+6-2\gamma(\gamma-2)^{2}z+A(\gamma,z,1)\frac{dw}{dz}+\gamma(\gamma-2)w$
$\displaystyle=\gamma(2\gamma+(\gamma-2)w)+2\gamma(1-(\gamma-2)^{2}z)+6+\gamma-A(\gamma,z,1)\frac{(\gamma+2)-(\gamma-2)^{2}z}{w}$
$\displaystyle>0,$

where the last inequality follows from $-1<\gamma-2\leq 0$, $0\leq w<1$ and
$z<1$ for any $\gamma\in(1,2]$ and $z\in[z_{m},z_{M}]$. Here we extend the
domain of $z$ into $[z_{m},z_{M}]$ to facilitate computations. Since
$A(\gamma,z,1)w+B(\gamma,z,1)$ is increasing in $z$, it is then enough to
check that $A(\gamma,z_{m},1)w(z_{m})+B(\gamma,z_{m},1)>0$. From (3.16), we
obtain the following relation between $w(z_{m})$ and $z_{m}$:

$\displaystyle V_{6}(\gamma,z_{m})=\frac{-2}{\gamma+1}\ \Longleftrightarrow\
\frac{-1+(\gamma-2)z_{m}-w(z_{m})}{2}=\frac{-2}{\gamma+1}\
\Longleftrightarrow\ w(z_{m})=\frac{3-\gamma}{\gamma+1}+(\gamma-2)z_{m}.$
(C.3)

Hence, using (3.16) again we get for any $\gamma\in(1,2]$,

$\displaystyle A(\gamma,z_{m},1)w(z_{m})+B(\gamma,z_{m},1)$
$\displaystyle=(1-\gamma+\gamma(\gamma-2)z_{m})(\frac{3-\gamma}{\gamma+1}+(\gamma-2)z_{m})+1-\gamma+(2\gamma^{2}+3\gamma+6)z_{m}-\gamma(\gamma-2)^{2}z_{m}^{2}$
$\displaystyle=\frac{4(\gamma-1)(\gamma^{2}+2)}{(\gamma+1)^{2}(2\gamma-1)}$
$\displaystyle>0.$

Therefore, we deduce that $A(\gamma,z,1)w+B(\gamma,z,1)>0$ for any
$\gamma\in(1,2]$ and $z\in[z_{m},z_{M}]$, which in turn implies (C.2).

Case 2: $m=2$. When $m=2$, the sign of
$A(\gamma,z,2)=3-2\gamma+(2-5\gamma+2\gamma^{2})z$ changes for
$\gamma\in(1,2]$ and $z\in[z_{g},z_{M}]$ and the argument for $m=1$ is not
applicable. We will employ another approach. We first decompose $\mathcal{Q}$
in (C.2) into two parts

$\mathcal{Q}=:I+2\,II$

where

$\displaystyle I$
$\displaystyle:=2V_{6}^{2}F_{V}G_{C}-(1+V_{6})^{2}G_{C}^{2},$ $\displaystyle
II$ $\displaystyle:=V_{6}^{2}F_{V}G_{C}+(1+V_{6})V_{6}(F_{C}-G_{V})G_{C}.$

We claim that $I$ and $II$ are both positive. For $I$, we first rewrite it by
again using the identities (3.7), (3.5), (3.11), and (3.12), as well as
$C_{6}=1+V_{6}$ to get

$\displaystyle I$ $\displaystyle=G_{C}[2V_{6}^{2}F_{V}-(1+V_{6})^{2}G_{C}]$
$\displaystyle=G_{C}\Big{[}V_{6}^{2}(2(\gamma-1)wC_{6}-4C_{6}(C_{6}+2z))-2V_{6}C_{6}^{2}(C_{6}+2\gamma
z)\Big{]}$
$\displaystyle=G_{C}\Big{[}2(\gamma-1)wV_{6}^{2}C_{6}-4V_{6}^{2}C_{6}^{2}-8zV_{6}^{2}C_{6}-2V_{6}C_{6}^{3}-4\gamma
zV_{6}C_{6}^{2}\Big{]}$
$\displaystyle=G_{C}V_{6}C_{6}\Big{[}2(-V_{6}-C_{6})C_{6}+4z(-2V_{6}-\gamma
C_{6})+(2(\gamma-1)w-2C_{6})V_{6}\Big{]}.$

The goal is to show the bracket is positive. For any $\gamma\in(1,2]$ and
$z\in[z_{g},z_{M}]$, by Lemma 2.6, we have $V_{6}\leq
V_{6}(z_{M}(2))=-\frac{1}{2}$. Thus, $-V_{6}-C_{6}\geq 0$ since
$1+V_{6}=C_{6}$. Next, we show that the remainder of the bracket is also
positive. By using $C_{6}=1+V_{6}$, we have

$\displaystyle 4z(-2V_{6}-\gamma
C_{6})+(2(\gamma-1)w-2C_{6})V_{6}=-2V_{6}^{2}-(2+(8+4\gamma)z-2(\gamma-1)w)V_{6}-4\gamma
z=:p_{1}(V_{6}).$

We observe that $p_{1}(V_{6})$ is a quadratic polynomial of $V_{6}$. Since the
coefficient of $V_{6}^{2}$ is negative and $-1<V_{6}\leq-\frac{1}{2}$ for
$\gamma\in(1,2]$ and $z\in[z_{g},z_{M}]$, to show $p_{1}(V_{6})>0$, it is
sufficient to check that $p_{1}(-1)>0$ and $p_{1}(-\frac{1}{2})>0$. Note that

$\displaystyle p_{1}(-1)$
$\displaystyle=-2+2+(8+4\gamma)z-2(\gamma-1)w-4\gamma z=8z-2(\gamma-1)w$
$\displaystyle>8z_{m}-2(\gamma-1)w(z_{m})$

for any $\gamma\in(1,2]$ and $z\in[z_{g},z_{M}]$ because $\frac{dw(z)}{dz}<0$.
By (3.16) and (C.3), we deduce that

$\displaystyle p_{1}(-1)>8z_{m}-2(\gamma-1)w(z_{m})$
$\displaystyle=\frac{8(\gamma-1)}{(2\gamma-1)(\gamma+1)}-2(\gamma-1)[\frac{3-\gamma}{\gamma+1}+\frac{(\gamma-2)(\gamma-1)}{(2\gamma-1)(\gamma+1)}]$
$\displaystyle=\frac{2(\gamma-1)(\gamma^{2}-4\gamma+5)}{(2\gamma-1)(\gamma+1)}>0.$

For $p_{1}(-\frac{1}{2})$, observe that

$\displaystyle p_{1}(-\frac{1}{2})$
$\displaystyle=-\frac{1}{2}+1+(4+2\gamma)z-(\gamma-1)w-4\gamma
z=\frac{1}{2}+(4-2\gamma)z-(\gamma-1)w$
$\displaystyle>\frac{1}{2}+(4-2\gamma)z_{m}-(\gamma-1)w(z_{m})$

for any $\gamma\in(1,2]$ and $z\in[z_{g},z_{M}]$ because $\frac{dw(z)}{dz}<0$.
By (3.16) and (C.3), we have

$\displaystyle
p_{1}(-\frac{1}{2})>\frac{1}{2}+(4-2\gamma)z_{m}-(\gamma-1)w(z_{m})$
$\displaystyle=\frac{1}{2}+\frac{(4-2\gamma)(\gamma-1)}{(2\gamma-1)(\gamma+1)}-(\gamma-1)[\frac{3-\gamma}{\gamma+1}+\frac{(\gamma-2)(\gamma-1)}{(2\gamma-1)(\gamma+1)}]$
$\displaystyle=\frac{2\gamma^{3}-12\gamma^{2}+23\gamma-11}{2(2\gamma-1)(\gamma+1)}>0$

where the positive sign is shown in Proposition E.11. Therefore, we conclude
that $I>0$ since $G_{C}V_{6}C_{6}>0$.

For $II$, by using (3.5), (3.7), (3.11) and (3.12), we have the following:

$\displaystyle II$
$\displaystyle=G_{C}V_{6}\Big{[}V_{6}(F_{V}+F_{C})+F_{C}+C_{6}(2wC_{6}+G_{C})\Big{]}$
$\displaystyle=G_{C}V_{6}\Big{[}\frac{2(\gamma-1)w}{2}V_{6}C_{6}+2C_{6}(C_{6}+2z)+C_{6}(2wC_{6}+2V_{6}(C_{6}+2\gamma
z))\Big{]}$
$\displaystyle=G_{C}V_{6}C_{6}\Big{[}2w(\frac{\gamma-1}{2}V_{6}+C_{6})+4z(1+\gamma
V_{6})+2(1+V_{6})^{2}\Big{]}$
$\displaystyle=G_{C}V_{6}C_{6}\Big{[}\tilde{A}(\gamma,z)w+\tilde{B}(\gamma,z)\Big{]},$

where

$\displaystyle\tilde{A}(\gamma,z)$
$\displaystyle=\frac{1-\gamma}{2}+\frac{\gamma^{2}-7\gamma+2}{2}z<0,$
$\displaystyle\tilde{B}(\gamma,z)$
$\displaystyle=-\frac{\gamma-1}{2}+(\gamma^{2}+\gamma+2)z+\frac{(2-\gamma)(\gamma^{2}-7\gamma+2)}{2}z^{2}$

for $\gamma\in(1,2]$. We will first show $\tilde{B}(\gamma,z)>0$ for any
$\gamma\in(1,2]$ and $z\in[z_{m},z_{M}]$. When $\gamma=2$,
$\tilde{B}(2,z)=-\frac{1}{2}+8z\geq-\frac{1}{2}+8z_{m}=-\frac{1}{2}+\frac{8}{15}>0$.
For any fixed $\gamma\in(1,2)$, $\tilde{B}(\gamma,z)$ is a quadratic
polynomial of $z$. Since $\frac{(2-\gamma)(\gamma^{2}-7\gamma+2)}{2}<0$,
$\tilde{B}(\gamma,z)$ has global maximum at
$z=\frac{(\gamma^{2}+\gamma+2)}{(2-\gamma)(\gamma^{2}-7\gamma+2)}>\frac{1}{2}$
since $2(\gamma^{2}+\gamma+2)>\gamma^{2}-7\gamma+2$ and $z\leq
z_{M}=\frac{1}{\gamma+2+2\sqrt{2\gamma}}<\frac{1}{2}$ for any
$\gamma\in(1,2)$. Hence, to show $\tilde{B}(\gamma,z)>0$, it is sufficient to
check the sign of $\tilde{B}(\gamma,z_{m})$ for any $\gamma\in(1,2)$. Direct
computations show that

$\displaystyle\tilde{B}(\gamma,z_{m})$
$\displaystyle=-\frac{\gamma-1}{2}+\frac{(\gamma^{2}+\gamma+2)(\gamma-1)}{(2\gamma-1)(\gamma+1)}+\frac{(2-\gamma)(\gamma^{2}-7\gamma+2)(\gamma-1)^{2}}{2(2\gamma-1)^{2}(\gamma+1)^{2}}$
$\displaystyle=\frac{(\gamma-1)(-\gamma^{4}+12\gamma^{3}-14\gamma^{2}+24\gamma-9)}{2(2\gamma-1)(\gamma+1)}>0,$

where the positive sign is verified in Proposition E.12. Now since
$\tilde{B}(\gamma,z)>0$, to show $\tilde{A}(\gamma,z)w+\tilde{B}(\gamma,z)>0$,
it is sufficient to check the sign of
$\tilde{B}^{2}(\gamma,z)-\tilde{A}^{2}(\gamma,z)w^{2}>0$. By direct
computations,

$\displaystyle\tilde{B}^{2}(\gamma,z)-\tilde{A}^{2}(\gamma,z)w^{2}$
$\displaystyle=2z\Big{[}-2\gamma^{2}+\gamma+1+(4\gamma^{3}+4\gamma^{2}-6\gamma+4)z+(-2\gamma^{4}+13\gamma^{3}+5\gamma^{2}-16\gamma+4)z^{2}\Big{]}$
$\displaystyle=2z\Big{[}-2\gamma^{2}+\gamma+1+2(\gamma+2)(2\gamma^{2}-2\gamma+1)z-(\gamma^{2}-7\gamma+2)(2\gamma^{2}+\gamma-2)z^{2}\Big{]}=:2zp_{2}(z).$

Clearly $p_{2}(z)$ is a quadratic polynomial in $z$. We notice that
$-(\gamma^{2}-7\gamma+2)(2\gamma^{2}+\gamma-2)>0$, and $\tilde{B}(\gamma,z)$
has global minimum at
$z=\frac{(\gamma+2)(2\gamma^{2}-2\gamma+1)}{(\gamma^{2}-7\gamma+2)(2\gamma^{2}+\gamma-2)}<0$
for any $\gamma\in(1,2]$. Hence, in order to prove $p_{2}(z)>0$ on
$[z_{g},z_{M}]$, it is enough to check the sign of $p_{2}(z_{g})$ for each
$\gamma\in(1,2]$. From (5.20), we write $z_{g}$ as

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

We then compute $p_{2}(z_{g})[q(\gamma)+(2\gamma^{2}-\gamma+1)]^{2}$ to obtain

$\displaystyle p_{2}(z_{g})[q(\gamma)+(2\gamma^{2}-\gamma+1)]^{2}$
$\displaystyle=$
$\displaystyle(-2\gamma^{2}+\gamma+1)[q(\gamma)+(2\gamma^{2}-\gamma+1)]^{2}+4(\gamma+2)(2\gamma^{2}-2\gamma+1)(\gamma-1)[q(\gamma)+(2\gamma^{2}-\gamma+1)]$
$\displaystyle-4(\gamma-1)^{2}(\gamma^{2}-7\gamma+2)(2\gamma^{2}+\gamma-2)$
$\displaystyle=$
$\displaystyle(-2\gamma^{2}+\gamma+1)[2(2\gamma^{2}-\gamma+1)^{2}+2\gamma(\gamma-1)[4\gamma(\gamma-1)+\frac{8}{3}]]+4(\gamma+2)(2\gamma^{2}-2\gamma+1)(\gamma-1)(2\gamma^{2}-\gamma+1)$
$\displaystyle-4(\gamma-1)^{2}(\gamma^{2}-7\gamma+2)(2\gamma^{2}+\gamma-2)+2q(\gamma)[(-2\gamma^{2}+\gamma+1)(2\gamma^{2}-\gamma+1)+2(\gamma+2)(2\gamma^{2}-2\gamma+1)(\gamma-1)]$
$\displaystyle=$
$\displaystyle\frac{2(\gamma-1)^{2}(-36\gamma^{4}+114\gamma^{3}-4\gamma^{2}-83\gamma+15)}{3}+2(\gamma-1)^{2}(4\gamma-3)q(\gamma)>0,$

where the positive sign of the quartic polynomial is shown in Proposition
E.13. Therefore, we deduce that $\tilde{A}(\gamma,z)w+\tilde{B}(\gamma,z)>0$
and hence $II>0$. This completes the proof of (C.2) for $m=2$.

##  Appendix D Proof of (6.6) and (6.17)

In this section, we consider specific values of the parameters:
$z=z_{1}(\gamma)$ or $z_{2}(\gamma)$, and $\kappa=1$ or $\frac{3}{2}$. For
notational convenience, we use $G_{C}$, $G_{V}$, $F_{C}$, $F_{V}$, $R$ to
represent their evaluations at $P_{8}=(V_{8},C_{8})$ where $G_{C}$, $G_{V}$,
$F_{C}$, $F_{V}$, $R$ are given in (3.5)–(3.8) and (3.4) respectively.
Recalling (3.18) , the two inequalities (6.6) and (6.17) can be written as

$\displaystyle\frac{F_{C}-G_{V}+R}{2G_{C}}<-\frac{1}{2}\sqrt{\frac{\kappa}{-V_{8}}}=-\frac{\kappa}{2C_{8}},$

where $\kappa=1$ corresponds to (6.6) and $\kappa=\frac{3}{2}$ is equivalent
to (6.17). From (3.5), we see that $G_{C}<0$. Moreover, $R>0$, and hence it is
equivalent to prove that

$R^{2}>(-G_{C}\sqrt{\frac{\kappa}{-V_{8}}}-F_{C}+G_{V})^{2}.$

Expanding $R$ using the definition in (3.4), we find that this is equivalent
to

$\displaystyle-\frac{\kappa
G_{C}}{V_{8}}+\frac{2\kappa}{C_{8}}(F_{C}-G_{V})-4F_{V}>0.$ (D.1)

By using (3.7), (3.11), (3.13) and $w=2C_{8}-1-(\gamma-2)z$ which is given by
(2.6), we compute

$\displaystyle 4F_{V}$
$\displaystyle=4(-\frac{\gamma-1}{2}mwC_{8}-2C_{8}(C_{8}+mz))=-2m(\gamma-1)wC_{8}-8C_{8}^{2}-8mzC_{8}$
$\displaystyle=-[4m(\gamma-1)+8]C_{8}^{2}+2m(\gamma-1)C_{8}+2m(\gamma^{2}-2\gamma-3)zC_{8}.$
(D.2)

Thanks to (3.5), (3.7), (3.13) , $C_{8}^{2}=-\kappa V_{8}$,
$w=2C_{8}-1-(\gamma-2)z$ and $C_{8}=1+V_{8}$, we obtain

$\displaystyle-\frac{\kappa G_{C}}{V_{8}}+\frac{2\kappa}{C_{8}}(F_{C}-G_{V})$
$\displaystyle=2\kappa\left(-m\gamma z-1+V_{8}+2C_{8}+2mz-mw\right)-4m\gamma
zC_{8}$ $\displaystyle=2\kappa\left[m-2+(3-2m)C_{8}\right]-4m\gamma zC_{8}.$
(D.3)

Now, we are ready to show (6.6) and (6.17) hold.

######  Lemma D.1.

For any $\gamma\in(\gamma_{\star},\gamma_{1}]$, $z=z_{1}$ and $\kappa=1$,
(D.1) holds, and therefore so does (6.6).

###### Proof.

To show that (6.6) holds, by using (D.1), it is enough to check the positivity
of $-\frac{\kappa G_{C}}{V_{8}}+\frac{2\kappa}{C_{8}}(F_{C}-G_{V})-4F_{V}$.
When $m=1$, by using (D.2), (D.3), $V_{8}(z_{1})=-C_{8}^{2}(z_{1})$, we obtain
that, for any $\gamma\in(\gamma_{\star},\gamma_{1}]$,

$\displaystyle-\frac{G_{C}}{V_{8}}+\frac{2}{C_{8}}(F_{C}-G_{V})-4F_{V}$
$\displaystyle=2C_{8}-2+4(\gamma+1)C_{8}^{2}-2(\gamma-1)C_{8}+2(3-\gamma^{2})z_{1}C_{8}$
$\displaystyle=-2\gamma(2V_{8}+C_{8})+2[(3-\gamma^{2})z_{1}C_{8}+1]$
$\displaystyle>(7-3\sqrt{5})\gamma_{\star}+2[(3-\gamma_{1}^{2})z_{1}(\gamma_{\star})C_{8}+1]>0,$

where we have used (6.2) and (6.3) in the last two inequalities.

When $m=2$, by using (D.2), (D.3) and $V_{8}(z_{1})=-C_{8}^{2}(z_{1})$, we
have, for any $\gamma\in(\gamma_{\star},\gamma_{1}]$,

$\displaystyle-\frac{G_{C}}{V_{8}}+\frac{2}{C_{8}}(F_{C}-G_{V})-4F_{V}$
$\displaystyle=8\gamma C_{8}^{2}-(4\gamma-2)C_{8}+4(3-\gamma^{2})z_{1}C_{8}$
$\displaystyle=-4\gamma(2V_{8}+C_{8})+2C_{8}+4(3-\gamma^{2})z_{1}C_{8}$
$\displaystyle>2(7-3\sqrt{5})\gamma+2[2(3-\gamma_{1}^{2})z_{1}(\gamma_{\star})+1]C_{8}>0,$

where we have used (6.2) and (6.3) in the last two inequalities. ∎

######  Lemma D.2.

For any $\gamma\in(\gamma_{1},3]$, $z=z_{2}$ and $\kappa=\frac{3}{2}$, (D.1)
holds and therefore so does (6.17).

###### Proof.

To show that (6.17) holds, by using (D.1), it is enough to check the
positivity of $-\frac{\kappa
G_{C}}{V_{8}}+\frac{2\kappa}{C_{8}}(F_{C}-G_{V})-4F_{V}$. When $m=1$, by using
(D.2), (D.3) and $C_{8}^{2}=-\frac{3}{2}V_{8}$, we have, for any
$\gamma\in(\gamma_{1},3]$,

$\displaystyle-\frac{3}{2}\frac{G_{C}}{V_{8}}+\frac{3}{C_{8}}(F_{C}-G_{V})-4F_{V}$
$\displaystyle=3C_{8}-3+4(\gamma+1)C_{8}^{2}-2(\gamma-1)C_{8}+2(3-\gamma^{2})z_{2}C_{8}$
$\displaystyle=-2\gamma(3V_{8}+C_{8})-V_{8}+2[(3-\gamma^{2})z_{2}C_{8}+1]$
$\displaystyle>2(6-\sqrt{3}3)\gamma-V_{8}+2(-6z_{2}(\gamma_{1})C_{8}+1)>0,$

where we have used (6.13) and (6.14) in the last two equalities.

When $m=2$, by using (D.2), (D.3) and $C_{8}^{2}=-\frac{3}{2}V_{8}$, we
obtain, for any $\gamma\in(\gamma_{1},3]$,

$\displaystyle-\frac{3}{2}\frac{G_{C}}{V_{8}}+\frac{3}{C_{8}}(F_{C}-G_{V})-4F_{V}$
$\displaystyle=8\gamma C_{8}^{2}-(4\gamma-1)C_{8}+4(3-\gamma^{2})z_{2}C_{8}$
$\displaystyle=-4\gamma(3V_{8}+C_{8})+[4(3-\gamma^{2})z_{2}+1]C_{8}$
$\displaystyle>-4\gamma(3V_{8}+C_{8})+[-24z_{2}(2)+1]C_{8}$
$\displaystyle=4(6-\sqrt{3}3)\gamma+16\sqrt{33}-93>0,$

where we have used (6.13) and (6.14) in the last two inequalities. ∎

##  Appendix E Calculation of polynomials

######  Lemma E.1 ([24], Exercise 10.14).

Consider the cubic polynomial

$\displaystyle p(x)=ax^{3}+bx^{2}+cx+d$

where $a\neq 0$, $b$, $c$ and $d$ are all real numbers. Define the
discriminant $\Delta$ of $p(x)$ as

$\displaystyle\Delta=b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd.$ (E.1)

Then,

  1. 1.

If $\Delta=0$, then $p(x)$ has a multiple root and all its roots are real;

  2. 2.

If $\Delta>0$, then $p(x)$ has three distinct real roots;

  3. 3.

If $\Delta<0$, then $p(x)$ has one real root and two complex conjugate roots.

######  Lemma E.2 ([24], Exercise 10.30 and Exercise 10.36).

Consider the quartic polynomial

$\displaystyle p(x)=ax^{4}+bx^{3}+cx^{2}+dx+e$

where $a\neq 0$, $b$, $c$, $d$ and $e$ are all real numbers. Define the
discriminant $\Delta$ of $p(x)$ as

$\displaystyle\Delta=$ $\displaystyle
18abcd^{3}+18b^{3}cde-80abc^{2}de-6ab^{2}d^{2}e+144a^{2}cd^{2}e+144ab^{2}ce^{2}-128a^{2}c^{2}e^{2}-192a^{2}bde^{2}+b^{2}c^{2}d^{2}$
$\displaystyle-4b^{3}d^{3}-4c^{3}d^{3}-4b^{2}c^{3}e+16ac^{4}e-27a^{2}d^{4}-27b^{4}e^{2}+256a^{3}e^{3}.$
(E.2)

Then,

  1. 1.

If $\Delta=0$, then $p(x)$ has a multiple root;

  2. 2.

If $\Delta>0$, then the roots of $p(x)$ are either all real or all complex;

  3. 3.

If $\Delta<0$, then $p(x)$ has two real roots and two complex conjugate roots.

######  Lemma E.3 ([43], Sturm’s Theorem).

Take any polynomial $p(x)$, and let $p_{0}(x),\ldots p_{m}(x)$ denote the
Sturm chain corresponding to $p(x)$. Take any interval $(a,b)$ such that
$p_{i}(a),p_{i}(b)\neq$ 0 , for any $i\in\\{0,1,\dots,m\\}$. For any constant
$c$, let $\sigma(c)$ denote the number of changes in sign in the sequence
$p_{0}(c),\ldots p_{m}(c)$. Then $p(x)$ has $\sigma(a)-\sigma(b)$ distinct
roots in the interval $(a,b)$.

######  Proposition E.4.

For any $\gamma\in(1,2]$, the cubic polynomial

$\displaystyle p(\gamma)=-\gamma^{3}-7\gamma^{2}+\frac{106}{3}\gamma-36<0.$

###### Proof.

Since the discriminant of $p(x)$ is

$\displaystyle\Delta=\frac{-189956}{27}<0,$

$p(x)$ has only one real root by Lemma E.1. Moreover, $p(-\infty)>0$ and
$p(0)=-36<0$, which implies that the real root must be negative. Thus,
$p(\gamma)<0$ for $\gamma\in(1,2]$. ∎

######  Proposition E.5.

For any $\gamma\in(1,2]$, the quintic polynomial

$\displaystyle p(\gamma)=81(\gamma-1)(\gamma+1)^{4}-8\gamma(3\gamma-1)^{4}<0.$

###### Proof.

By direct computations, the Sturm chain of $p(\gamma)$ is given by

$\displaystyle p_{0}(\gamma)$
$\displaystyle=-81-251\gamma-66\gamma^{2}-270\gamma^{3}+1107\gamma^{4}-567\gamma^{5},$
$\displaystyle p_{1}(\gamma)$
$\displaystyle=-251-132\gamma-810\gamma^{2}+4428\gamma^{3}-2835\gamma^{4},$
$\displaystyle p_{2}(\gamma)$
$\displaystyle=\frac{52816}{525}+\frac{36944\gamma}{175}+\frac{720\gamma^{2}}{7}-\frac{41616\gamma^{3}}{175},$
$\displaystyle p_{3}(\gamma)$
$\displaystyle=-\frac{92164800}{83521}-\frac{126201600\gamma}{83521}+\frac{162187200\gamma^{2}}{83521},$
$\displaystyle p_{4}(\gamma)$
$\displaystyle=-\frac{14651587904}{271832505}-\frac{5446905536\gamma}{453054175},$
$\displaystyle p_{5}(\gamma)$
$\displaystyle=-\frac{3876577982771200}{86724949081}.$

Therefore, $\sigma(-\infty)=3$ and $\sigma(\infty)=2$. Thus, $p(\gamma)$ only
has one real root. Since $p(-\infty)>0$ and $p(1)<0$, $p(\gamma)<0$ for any
$\gamma\in(1,2]$. ∎

######  Proposition E.6.

For any $\gamma\in(\gamma_{\star},\gamma_{1}]$, the quartic polynomial

$\displaystyle
p(\gamma)=\frac{8(\gamma-2)^{2}\gamma^{2}}{625}+\frac{2(1-2\gamma)\gamma(\gamma+1)}{25}+(1-2\gamma)(2-\gamma)<0.$

###### Proof.

By (E.2), the discriminant of $p(\gamma)$ is

$\displaystyle\Delta=-\frac{1586137650624}{3814697265625}<0.$

By Lemma E.2, $p(x)$ has two real roots. Since $p(-\infty)>0$,
$p(1)=-\frac{717}{625}<0$, $p(\frac{5}{2})=-\frac{39}{50}<0$, and
$p(+\infty)>0$, $p(\gamma)<0$ for any $\gamma\in(\gamma_{\star},\gamma_{1}]$.
∎

######  Proposition E.7.

For any $\gamma\in(\gamma_{\star},\gamma_{1}]$, the quadratic polynomial

$\displaystyle
p(\gamma)=(3\sqrt{2}-4)\gamma+1-\sqrt{2}+\frac{2}{25}[-(2-\sqrt{2})\gamma^{2}+(3-2\sqrt{2})\gamma+2\sqrt{2}+2]>0.$
(E.3)

###### Proof.

Since $p(-\infty)<0$, $p(1)=\frac{58\sqrt{2}-77}{25}>0$,
$p(3)=\frac{264\sqrt{2}-361}{25}>0$, and $p(\infty)<0$. So, $p(\gamma)>0$ for
any $\gamma\in(\gamma_{\star},\gamma_{1}]$. ∎

######  Proposition E.8.

For any $\gamma\in(\gamma_{\star},\gamma_{1}]$, the quadratic polynomial

$\displaystyle
p(\gamma)=2(3\sqrt{2}-4)\gamma+3-3\sqrt{2}+\frac{4}{25}[-(2-\sqrt{2})\gamma^{2}+(3-2\sqrt{2})\gamma+2\sqrt{2}+2]>0.$
(E.4)

###### Proof.

Since $p(-\infty)<0$, $p(\frac{3}{2})=\frac{182\sqrt{2}-253}{25}>0$,
$p(3)=\frac{503\sqrt{2}-697}{25}>0$, and $p(\infty)<0$. So, $p(\gamma)>0$ for
any $\gamma\in(\gamma_{\star},\gamma_{1}]$. ∎

######  Proposition E.9.

For any $\gamma\in(\gamma_{1},3]$, the cubic polynomial

$\displaystyle p(\gamma)=\gamma^{3}-17\gamma^{2}+40\gamma-14<0.$

###### Proof.

Since $p(-\infty)<0$, $p(\infty)>0$, $p(1)=10>0$,
$p(\gamma_{1})=11\sqrt{2}-18<0$, $p(3)=-20<0$. Thus, $p(x)<0$ for any
$\gamma\in(\gamma_{1},3]$. ∎

######  Proposition E.10.

For any $\gamma\in(\gamma_{1},3]$, the quartic polynomial

$\displaystyle p(\gamma)=-\gamma^{4}+12\gamma^{3}-36\gamma^{2}+32\gamma+1>0.$

###### Proof.

Since the discriminant of $p(\gamma)$ is

$\displaystyle\Delta=-495616<0,$

by Lemma E.2, $p(x)$ has two real roots. Thus, $p(-\infty)<0$, $p(1)=8>0$,
$p(3)=16$, and $p(\infty)<0$ implies that $p(x)>0$ for any
$\gamma\in(\gamma_{1},3]$. ∎

######  Proposition E.11.

For any $\gamma\in(1,2]$, the cubic polynomial

$\displaystyle p(\gamma)=2\gamma^{3}-12\gamma^{2}+23\gamma-11>0.$

###### Proof.

Since the discriminant of $p(\gamma)$ is

$\displaystyle\Delta=-964<0,$

by Lemma E.1, $p(x)$ has only one real root. Thus, as $p(-\infty)<0$ and
$p(1)=2$, we must have that $p(x)>0$ for any $\gamma\in(1,2]$. ∎

######  Proposition E.12.

For any $\gamma\in(1,2]$, the quartic polynomial

$\displaystyle p(\gamma)=-\gamma^{4}+12\gamma^{3}-14\gamma^{2}+24\gamma-9>0.$

###### Proof.

Since the discriminant of $p(\gamma)$ is

$\displaystyle\Delta=-30235392<0,$

by Lemma E.2, $p(x)$ has two real roots. Thus, as $p(-\infty)<0$,
$p(\infty)<0$, $p(1)=12>0$ and $p(2)=63>0$, we have that $p(x)>0$ for any
$\gamma\in(1,2]$. ∎

######  Proposition E.13.

For any $\gamma\in(1,2]$, the quartic polynomial

$\displaystyle
p(\gamma)=-36\gamma^{4}+114\gamma^{3}-4\gamma^{2}-83\gamma+15>0.$

###### Proof.

Note that the first derivative of $p(\gamma)$ is

$\displaystyle\frac{dp(\gamma)}{d\gamma}=-144\gamma^{3}+342\gamma^{2}-8\gamma-83.$

Since $\frac{dp}{d\gamma}(-\infty)>0$, $\frac{dp}{d\gamma}(\infty)<0$,
$\frac{dp}{d\gamma}(0)=-83<0$, $\frac{dp}{d\gamma}(1)=107>0$, and
$\frac{dp}{d\gamma}(2)=117>0$, $\frac{dp(\gamma)}{d\gamma}>0$ for any
$\gamma\in(1,2]$. Thus, $p(1)=6$ implies that $p(x)>0$ for any
$\gamma\in(1,2]$. ∎

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