Nonlinear Hyperbolic Waves in Relativistic Gases of Massive Particles with Synge Energy

Abstract

In this article, we study some fundamental properties of nonlinear waves and the Riemann problem of Euler’s relativistic system when the constitutive equation for energy is that of Synge for a monatomic rarefied gas or its generalization for diatomic gas. These constitutive equations are the only ones compatible with the relativistic kinetic theory for massive particles in the whole range from the classical to the ultra-relativistic regime. They involve modified Bessel functions of the second kind and this makes Euler’s relativistic system rather complex. Based on delicate estimates of the Bessel functions, we prove: (i) a limit on the speed of sound of $1{/}\sqrt{3}$ times the speed of light (which a fortiori implies subluminality, that is causality), (ii) the genuine non-linearity of the acoustic waves, (iii) the compatibility of Rankine–Hugoniot relations with the second law of thermodynamics (entropy growth through all Lax shocks), and (iv) the unique resolvability of the initial value problem of Riemann (if we include the possibility of vacuum as in the non-relativistic context).

Appendices

A Modified Bessel Functions and Properties

In this Appendix, we recall expressions of the modified Bessel functions and their basic properties. Moreover, with our observation, a simple corollary is also presented. Now we give the modified Bessel functions and collect their basic properties.

Lemma 1

[3, 54, 55] Let $K_j(\gamma )$ be the Bessel functions defined by

$$\begin{aligned} K_j(\gamma )=\frac{(2^j)j!}{(2j)!}\frac{1}{\gamma ^j}\int _{\lambda =\gamma }^{\lambda =\infty }e^{-\lambda }(\lambda ^2-\gamma ^2)^{j-1/2}\mathrm{{d}}\lambda ,\quad (j\ge 0). \end{aligned}$$

(1)

Then the following identities hold:

$$\begin{aligned} K_j(\gamma )= & {} \frac{2^{j-1}(j-1)!}{(2j-2)!}\frac{1}{\gamma ^j}\int _{\lambda =\gamma }^{\lambda =\infty }e^{-\lambda }(\lambda ^2-\gamma ^2)^{j-3/2}\mathrm{{d}}\lambda ,\quad (j>0),\nonumber \\ K_{j+1}(\gamma )= & {} 2j\frac{K_j(\gamma )}{\gamma }+K_{j-1}(\gamma ),\quad (j\ge 1), \nonumber \\ K_j(\gamma )< & {} K_{j+1}(\gamma ),\quad (j\ge 0), \end{aligned}$$

(2)

and

$$\begin{aligned}&\frac{{\text {d}}}{\mathrm{{d}}\gamma }\left( \frac{K_j(\gamma )}{\gamma ^j}\right) =-\frac{K_{j+1}(\gamma )}{\gamma ^j},\quad (j\ge 0), \end{aligned}$$

(3)

$$\begin{aligned}&\displaystyle K_{j}(\gamma )=\sqrt{\frac{\pi }{2\gamma }}e^{-\gamma }\left( \gamma _{j,n}(\gamma )\gamma ^{-n}+\sum _{m=0}^{n-1}A_{j,m}\gamma ^{-m}\right) ,\quad (j\ge 0,~n\ge 1), \end{aligned}$$

(4)

where expressions of the coefficients in (4) are

$$\begin{aligned}&A_{j,0}=1\nonumber \\&A_{j,m}=\frac{(4j^2-1)(4j^2-3^2)\cdots (4j^2-(2m-1)^2)}{m!8^m},\quad (j\ge 0,~m\ge 1), \nonumber \\&|\gamma _{j,n}(\gamma )|\le 2e^{[j^2-1/4]\gamma ^{-1}}|A_{j,n}|,\quad (j\ge 0,~n\ge 1). \end{aligned}$$

(5)

On the other hand, according to [55] (on page 80), the Bessel functions defined in (1) can also be written in the following form:

$$\begin{aligned} K_0(\gamma )=&-\sum ^{\infty }_{m=0}\frac{(\frac{1}{2}\gamma )^{2m}}{m!m!}\Big [\ln \Big (\frac{\gamma }{2}\Big )-\psi (m+1)\Big ],\nonumber \\ K_n(\gamma )=&\frac{1}{2}\sum ^{n-1}_{m=0}(-1)^m\frac{(n-m-1)!}{m!}\Big (\frac{1}{2}\gamma \Big )^{-n+2m}\nonumber \\&+(-1)^{n+1}\sum ^{\infty }_{m=0}\frac{(\frac{1}{2}\gamma )^{n+2m}}{m!(m+n)!}\nonumber \\&\quad \times \Big [\ln \Big (\frac{\gamma }{2}\Big ) -\frac{1}{2}\psi (n+m)-\frac{1}{2}\psi (n+m+1)\Big ], \nonumber \\ \psi (1)=&-C_E,\quad \psi (m+1)=-C_E+\sum ^{m}_{k=1}\frac{1}{k},\quad m\ge 1,\nonumber \\ K_1(\gamma )=&\frac{1}{\gamma }+\sum ^{\infty }_{m=0}\frac{(\frac{1}{2}\gamma )^{2m+1}}{m!(m+1)!} \Big [\ln \Big (\frac{\gamma }{2}\Big )-\frac{1}{2}\psi (m+1)-\frac{1}{2}\psi (m+2)\Big ], \end{aligned}$$

(6)

where $C_E=0.5772157\ldots$ is the Euler’s constant.

From Lemma 1, we immediately have the following corollary:

Corollary 2

For $K_j(j\ge 0)$ defined in Lemma 1, it holds that

$$\begin{aligned}&K_1^2\le 3K_0K_2, \end{aligned}$$

(7)

$$\begin{aligned}&3\left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2+\frac{6}{\gamma }\frac{K_0(\gamma )}{K_1(\gamma )}-1\ge 0. \end{aligned}$$

(8)

Proof

From (1), we have

$$\begin{aligned} K_0(\gamma )&=\int _{\lambda =\gamma }^{\lambda =\infty }e^{-\lambda }(\lambda ^2-\gamma ^2)^{-1/2}\mathrm{{d}}\lambda ,\\ K_1(\gamma )&=\frac{1}{\gamma }\int _{\lambda =\gamma }^{\lambda =\infty }e^{-\lambda }(\lambda ^2-\gamma ^2)^{1/2}\mathrm{{d}}\lambda ,\\ K_2(\gamma )&=\frac{1}{3\gamma ^2}\int _{\lambda =\gamma }^{\lambda =\infty }e^{-\lambda }(\lambda ^2-\gamma ^2)^{3/2}\mathrm{{d}}\lambda . \end{aligned}$$

These equations imply (7) by H$\ddot{o}$lder’s inequality.

Taking $j=1$ in (2) and inserting it to (7) yield

$$\begin{aligned} K_1^2(\gamma )\le 3K_0(\gamma )\left( \frac{2}{\gamma }K_1(\gamma )+K_0(\gamma )\right) . \end{aligned}$$

Then (8) follows. $\square$

B Estimates of the Ratio $\pmb {\frac{K_0(\gamma )}{K_1(\gamma )}}$

In this subsection, we concentrate on estimates of $\frac{K_0(\gamma )}{K_1(\gamma )}$. Our estimates are divided into two cases according to different expressions of $K_m (m\ge 1)$ given in Lemma 1: the case $\gamma \in (0, \sqrt{2}]$ by (6) and the case $\gamma \in [1.1, \infty )$ by (1). Here $\gamma _0=1.1229189\ldots$ is a constant satisfying

$$\begin{aligned} \ln \Big (\frac{\gamma }{2}\Big )+C_E=0. \end{aligned}$$

We first estimate $\frac{K_0(\gamma )}{K_1(\gamma )}$ for the first case $\gamma \in (0, \sqrt{2}]$ by using the expressions (6).

Proposition 8

For $\gamma \in [\gamma _0, \sqrt{2}]$, it holds that

$$\begin{aligned} \frac{K_0(\gamma )}{K_1(\gamma )}\le 1-\frac{\gamma _0-1}{\gamma }, \end{aligned}$$

(9)

and for $\gamma \in (0, \gamma _0]$, we have $\Big (\frac{K_0(\gamma )}{K_1(\gamma )}\Big )^2+\frac{2}{\gamma }\frac{K_0(\gamma )}{K_1(\gamma )}-1>0$. Moreover, we have

$$\begin{aligned} \frac{\gamma }{\sqrt{\gamma ^2+1}+1}\le \frac{K_0(\gamma )}{K_1(\gamma )}\le \gamma \left[ \frac{11}{16}-\left( \ln \left( \frac{\gamma }{2}\right) +C_E\right) \right] . \end{aligned}$$

(10)

Proof

We first prove (9). From (6), we get for $\gamma \in [\gamma _0, \sqrt{2}]$ that

$$\begin{aligned} \gamma \frac{K_0(\gamma )}{K_1(\gamma )}\le \frac{-[\ln \left( \frac{\gamma }{2}\right) +C_E]\Big [\gamma ^2+\frac{\gamma ^4}{4}+\frac{\gamma ^6e^{\frac{\gamma ^2}{28}}}{64}\Big ]+\frac{\gamma ^4}{4}+ -\frac{3\gamma ^6 e^{\frac{\gamma ^2}{28}}}{128}}{[\ln \left( \frac{\gamma }{2}\right) +C_E]\Big [\frac{\gamma ^2}{2}+\frac{\gamma ^4}{16}+\frac{\gamma ^6e^{\frac{\gamma ^2}{28}}}{32\times 12}\Big ]+1-\frac{\gamma ^2}{4}- \frac{5\gamma ^4}{64}+ -\frac{5\gamma ^6 e^{\frac{\gamma ^2}{40}}}{32\times 36}}. \end{aligned}$$

Here we point out that the coefficient $e^{\frac{\gamma ^2}{28}}$ and the estimate $\gamma ^2+\frac{\gamma ^4}{4}+\frac{\gamma ^6e^{\frac{\gamma ^2}{28}}}{64}$ are chosen proper upper bounds which are convenient for computation and can guarantee the derivation of (9). The choice is not unique. Then, (9) holds if we can show

$$\begin{aligned}&[\ln \left( \frac{\gamma }{2}\right) +C_E]\Big [\gamma ^2+\frac{\gamma ^4}{4}+\frac{\gamma ^6e^{\frac{\gamma ^2}{28}}}{64}\Big ]+\frac{\gamma ^4}{4}+ -\frac{3\gamma ^6 e^{\frac{\gamma ^2}{28}}}{128}\\&\quad \le \Big \{[\ln \left( \frac{\gamma }{2}\right) +C_E]\Big [\frac{\gamma ^2}{2}+\frac{\gamma ^4}{16}+\frac{\gamma ^6e^{\frac{\gamma ^2}{28}}}{32\times 12}\Big ]+1-\frac{\gamma ^2}{4}- \frac{5\gamma ^4}{64}+ -\frac{5\gamma ^6 e^{\frac{\gamma ^2}{40}}}{32\times 36}\Big \}\\&\qquad \times (1+\gamma -\gamma _0); \end{aligned}$$

namely,

$$\begin{aligned} f(\gamma ):=&1+\gamma -\gamma _0-\frac{1}{4}(1+\gamma -\gamma _0)\gamma ^2-\frac{\gamma ^4}{64}[5(\gamma -\gamma _0)+21]\nonumber \\&-\Big [5(1+\gamma -\gamma _0)e^{\frac{\gamma ^2}{40}}+27e^{\frac{\gamma ^2}{28}}\Big ]\frac{\gamma ^6}{32\times 36}\nonumber \\&+\Big [\ln \left( \frac{\gamma }{2}\right) +C_E\Big ] \Big \{\frac{\gamma ^2}{2}(3+\gamma -\gamma _0)+\frac{\gamma ^4}{16}(5+\gamma -\gamma _0)\nonumber \\&+\frac{\gamma ^6}{32\times 12}\Big [(1+\gamma -\gamma _0)e^{\frac{\gamma ^2}{40}}+6e^{\frac{\gamma ^2}{28}}\Big ]\Big \}\ge 0 . \end{aligned}$$

(11)

Note that for $\gamma \in [\gamma _0, \sqrt{2}]$,

$$\begin{aligned} f'(\gamma )=&1-\frac{1}{4}\gamma ^2-\frac{\gamma }{2}(1+\gamma -\gamma _0)-\frac{\gamma ^3}{16}[5(\gamma -\gamma _0)+21] -\frac{5\gamma ^4}{64}\\&-\frac{\gamma ^5}{32\times 6}\Big [5(1+\gamma -\gamma _0)e^{\frac{\gamma ^2}{40}}+27e^{\frac{\gamma ^2}{28}}\Big ]\\&+\frac{\gamma }{2}(3+\gamma -\gamma _0)+\frac{\gamma ^3}{16}(5+\gamma -\gamma _0)\\&+\frac{\gamma ^5}{32\times 12}\Big [(1+\gamma -\gamma _0)e^{\frac{\gamma ^2}{40}}+6e^{\frac{\gamma ^2}{28}}\Big ]\\&-\frac{\gamma ^6}{32\times 36}\Big \{5\Big [1+\frac{\gamma (1+\gamma -\gamma _0)}{20}\Big ]e^{\frac{\gamma ^2}{40}}+\frac{27\gamma }{14}e^{\frac{\gamma ^2}{28}}\Big ]\Big \}\\&+\Big [\ln \left( \frac{\gamma }{2}\right) +C_E\Big ] \Big \{(3+\gamma -\gamma _0)\gamma +\frac{\gamma ^2}{2}+\frac{\gamma ^3}{4}(5+\gamma -\gamma _0)+\frac{\gamma ^4}{16}\\&+\frac{\gamma ^5}{64}\Big [(1+\gamma -\gamma _0)e^{\frac{\gamma ^2}{40}}+6e^{\frac{\gamma ^2}{28}}\Big ]\\&+\frac{\gamma ^6}{32\times 12}\Big [\Big (1+\frac{\gamma (1+\gamma -\gamma _0)}{20}\Big )e^{\frac{\gamma ^2}{40}} +\frac{3\gamma }{7}e^{\frac{\gamma ^2}{28}}\Big ]\Big \}. \end{aligned}$$

We can further obtain that for $\gamma \in [\gamma _0, \sqrt{2}]$,

$$\begin{aligned} f''(\gamma )=&4-\gamma _0+\gamma +\frac{(2\gamma _0-7)\gamma ^2}{4}-\gamma ^3\\&-\frac{\gamma ^4}{128}\Big [13(1+\gamma -\gamma _0)e^{\frac{\gamma ^2}{40}} +68e^{\frac{\gamma ^2}{28}}\Big ]\\&-\frac{\gamma ^6}{64}\Big \{\Big [3+\frac{3\gamma (1+\gamma -\gamma _0)}{20}\Big ]e^{\frac{\gamma ^2}{40}} +\frac{27\gamma }{7}e^{\frac{\gamma ^2}{28}}\Big \}\\&-\frac{\gamma ^6}{32\times 36}\Big \{ \Big [\frac{1-\gamma _0+3\gamma }{4}+\frac{\gamma ^2(1+\gamma -\gamma _0)}{80}\Big ]e^{\frac{\gamma ^2}{40}}\\&+\Big (\frac{27}{14}+\frac{27\gamma ^2}{256}\Big )e^{\frac{\gamma ^2}{28}}\Big \}+\Big [\ln \left( \frac{\gamma }{2}\right) +C_E\Big ] \Big \{3-\gamma _0+3\gamma +\\&+\frac{(15-3\gamma _0)\gamma ^2}{4}+\gamma ^3 \frac{5\gamma ^4}{64}\Big [(1+\gamma -\gamma _0)e^{\frac{\gamma ^2}{40}}+6e^{\frac{\gamma ^2}{28}}\Big ]\\&+\frac{\gamma ^5}{32}\Big [\Big (1+\frac{\gamma (1+\gamma -\gamma _0)}{20}\Big )e^{\frac{\gamma ^2}{40}} +\frac{3\gamma }{7}e^{\frac{\gamma ^2}{28}}\Big ]+\frac{\gamma ^6}{32\times 12}\\&\quad \times \Big \{ \Big [\frac{1-\gamma _0+3\gamma }{20}+\frac{\gamma ^2(1+\gamma -\gamma _0)}{400}\Big ]e^{\frac{\gamma ^2}{40}} \\&+\Big (\frac{3}{7}+\frac{3\gamma ^2}{98}\Big )e^{\frac{\gamma ^2}{28}}\Big \}\Big \} < 0. \end{aligned}$$

Then we have

$$\begin{aligned} f(\gamma )\ge \min \{f(\gamma _0), f(\sqrt{2})\} \end{aligned}$$

for $\gamma \in [\gamma _0, \sqrt{2}]$. On the other hand, we have

$$\begin{aligned} f(\gamma _0)=&1-\frac{\gamma _0^2}{4}-\frac{21\gamma _0^4}{64}-\frac{\gamma _0^6}{32\times 36}\Big [5e^{\frac{\gamma _0^2}{40}}+27e^{\frac{\gamma _0^2}{28}}\Big ]>0,\quad \text{ and }\\ f(\sqrt{2})=&1+\sqrt{2}-\gamma _0-\frac{1+\sqrt{2}-\gamma _0}{2}-\frac{5(\sqrt{2}-\gamma _0)+21}{16}\\&-\Big [\dfrac{5(1+\sqrt{2}-\gamma _0)}{144}e^{\frac{1}{20}}+\dfrac{27}{144}e^{\frac{1}{14}}\Big ]+\Big [\ln (\frac{\sqrt{2}}{2})+C_E\Big ] \\&\quad \times \Big \{3+\sqrt{2}-\gamma _0+\frac{5+\sqrt{2}-\gamma _0}{4}\\&+\frac{1}{48}\Big [(1+\sqrt{2}-\gamma _0)e^{\frac{1}{20}}+6e^{\frac{1}{14}}\Big ]\Big \}>0. \end{aligned}$$

Then (11) holds.

Now we turn to the proof of (10). We first verify the left inequality of (10). For $\gamma \in (0,\gamma _0],$ we use (6) to get

$$\begin{aligned} \frac{K_0(\gamma )}{K_1(\gamma )}\ge \frac{-\left[ \ln \left( \frac{\gamma }{2}\right) +C_E\right] \gamma -\frac{1}{4}\left[ \ln \left( \frac{\gamma }{2}\right) +C_E-1\right] \gamma ^3 }{1+\frac{1}{2}\left[ \ln \left( \frac{\gamma }{2}\right) +C_E-\frac{1}{2}\right] \gamma ^2+\frac{1}{16}\left[ \ln \left( \frac{\gamma }{2}\right) +C_E-\frac{5}{4}\right] \gamma ^4 }. \end{aligned}$$

(10) holds if we have

$$\begin{aligned} \overline{f}(\gamma )=&-\Big [\ln \left( \frac{\gamma }{2}\right) +C_E\Big ]\Big [1+\sqrt{\gamma ^2+1}+\frac{\gamma ^2(3+\sqrt{\gamma ^2+1})}{4}+\frac{\gamma ^4}{16}\Big ]\\&+\frac{\gamma ^2(2+\sqrt{\gamma ^2+1})}{4}+\frac{5\gamma ^4}{64}-1 >0. \end{aligned}$$

In fact, for $\gamma \in (0,\gamma _0)$, we have

$$\begin{aligned} \overline{f}'(\gamma )=&-\frac{1}{\gamma }\Big [1+\sqrt{\gamma ^2+1}+\frac{\gamma ^2(3+\sqrt{\gamma ^2+1})}{4}+\frac{\gamma ^4}{16}\Big ]\\&+\frac{\gamma ^3}{4\sqrt{\gamma ^2+1}} +\frac{\gamma (2+\sqrt{\gamma ^2+1})}{2} +\frac{5\gamma ^3}{16}-\Big [\ln \left( \frac{\gamma }{2}\right) +C_E\Big ]\\&\quad \times \Big [\frac{\gamma }{\sqrt{\gamma ^2+1}}+\frac{\gamma (3+\sqrt{\gamma ^2+1})}{2} +\frac{\gamma ^3}{4\sqrt{\gamma ^2+1}}+\frac{\gamma ^3}{4}\Big ]\\<&(1+\sqrt{\gamma ^2+1})\Big [-\frac{1}{\gamma }+\frac{\gamma }{4}+\frac{\gamma ^3}{4(\gamma ^2+1)}\Big ]\\&-\gamma \left[ \ln \left( \frac{\gamma }{2}\right) +C_E\right] (3+\sqrt{\gamma ^2+1})\\<&-\frac{3(1+\sqrt{\gamma ^2+1})}{5\gamma }-\gamma \left[ \ln \left( \frac{\gamma }{2}\right) +C_E\right] (3+\sqrt{\gamma ^2+1})\\ <&0. \end{aligned}$$

Then $\overline{f}(\gamma )\ge \overline{f}(\gamma _0)>0$. The left inequality of (10) holds.

We finally treat the right inequality of (10). For $\gamma \in (0, \gamma _0]$, we use (6) to get

$$\begin{aligned} \gamma \frac{K_0(\gamma )}{K_1(\gamma )}\le \frac{-\left[ \ln \left( \frac{\gamma }{2}\right) +C_E\right] \gamma ^2-\frac{1}{4}\left[ \ln \left( \frac{\gamma }{2}\right) +C_E-1\right] \gamma ^4e^{\frac{\gamma ^2}{10}}}{1+\frac{1}{2}\left[ \ln \left( \frac{\gamma }{2}\right) +C_E-\frac{1}{2}\right] \gamma ^2e^{\frac{5\gamma ^2}{16}}}. \end{aligned}$$

To prove the right inequality of (10), we only need to derive the following inequality:

$$\begin{aligned}&-\left[ \ln \left( \frac{\gamma }{2}\right) +C_E\right] \gamma ^2-\frac{1}{4}\left[ \ln \left( \frac{\gamma }{2}\right) +C_E-1\right] \gamma ^4e^{\frac{\gamma ^2}{10}}\\&\quad \le \Big [1+\frac{1}{2}\Big (\ln \left( \frac{\gamma }{2}\right) +C_E-\frac{1}{2}\Big ) \gamma ^2 e^{\frac{5\gamma ^2}{16}}\Big ]\gamma ^2\left[ \frac{11}{16}-\left( \ln \left( \frac{\gamma }{2}\right) +C_E\right) \right] ; \end{aligned}$$

that is,

$$\begin{aligned} \tilde{f}(\gamma )=&\gamma ^2\Big \{-\Big [\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )-1\Big ]e^{\frac{\gamma ^2}{10}} +\Big [2\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )^2\nonumber \\&-\frac{19}{8}\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )+\frac{11}{16}\Big ]e^{\frac{5\gamma ^2}{16}}\Big \}-\frac{11}{4}\le 0 \end{aligned}$$

(12)

for $\gamma \in (0, \gamma _0]$. Note the fact that

$$\begin{aligned} \tilde{f}'(\gamma )=&2\gamma \Big \{-\Big [\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )-1\Big ]e^{\frac{\gamma ^2}{10}} +\Big [2\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )^2\\&-\frac{19}{8}\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )+\frac{11}{16}\Big ]e^{\frac{5\gamma ^2}{16}}\Big \}-\gamma e^{\frac{\gamma ^2}{10}}-\frac{19\gamma }{8}e^{\frac{5\gamma ^2}{16}}\\&+4\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )\gamma e^{\frac{5\gamma ^2}{16}}+\frac{\gamma ^3}{5}\Big [-\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )+1\Big ]e^{\frac{\gamma ^2}{10}}\\&+\frac{5\gamma ^3}{8}\Big [2\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )^2-\frac{19}{8}\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big ) +\frac{11}{16}\Big ]e^{\frac{5\gamma ^2}{16}}\\ \ge&\gamma \Big \{\Big (\frac{\gamma ^2}{5}+1\Big )e^{\frac{\gamma ^2}{10}} +\Big [-\frac{3}{4}\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )+\frac{55\gamma ^2}{128}-1\Big ]e^{\frac{5\gamma ^2}{16}}\Big \}\\ >&0. \end{aligned}$$

Here we used the estimate $-\frac{3}{4}\Big (\ln \left( \frac{\gamma }{2}\right) +C_E\Big )+\frac{55\gamma ^2}{128}\ge \frac{1}{2}$ for $\gamma \in (0, \gamma _0]$. Then we have

$$\begin{aligned} \tilde{f}(\gamma )\le \gamma ^2\Big (e^{\frac{\gamma ^2}{10}} +\frac{11}{16}e^{\frac{5\gamma ^2}{16}}\Big )-\frac{11}{4}<0 \end{aligned}$$

for $\gamma \in (0, \gamma _0]$. (12) is verified. $\square$

For later use, we also need two different estimates:

Proposition 9

Let $\gamma \in (\sqrt{2}, \infty )$. Then $\frac{K_0(\gamma )}{K_1(\gamma )}$ satisfies:

$$\begin{aligned} 1-\frac{1}{2\gamma }\le \frac{K_0(\gamma )}{K_1(\gamma )}\le 1-\frac{1}{2\gamma }+\frac{3}{8\gamma ^2}+\frac{3}{16\gamma ^3}. \end{aligned}$$

(13)

Moreover, for $\gamma \in (2, \infty )$, it holds that

$$\begin{aligned}&\frac{K_0(\gamma )}{K_1(\gamma )}\ge 1-\frac{1}{2\gamma }+\frac{3}{8\gamma ^2}-\frac{3}{8\gamma ^3}+\frac{63}{128\gamma ^4}-\frac{31}{20\gamma ^5}, \nonumber \\&\frac{K_0(\gamma )}{K_1(\gamma )}\le 1-\frac{1}{2\gamma }+\frac{3}{8\gamma ^2}-\frac{3}{8\gamma ^3}+\frac{63}{128\gamma ^4}+\frac{7}{8\gamma ^5}. \end{aligned}$$

(14)

Proof

Compared to the proof of (13), the proof of (14) is more tedious but simpler. For brevity, we only prove (13). From (5), one has

$$\begin{aligned} A_{0,1}&=-\frac{1}{8}, \quad A_{0,2}=\frac{9}{2\times 8^2}, \quad A_{0,3}=-\frac{75}{2\times 8^3}, \nonumber \\ A_{0,4}&=\frac{3\times 25\times 49}{8^5}, \quad A_{0,5}=-\frac{15\times 49\times 81}{8^6},\quad \text{ and } \end{aligned}$$

(15)

$$\begin{aligned} A_{1,1}&=\frac{3}{8}, \quad A_{1,2}=-\frac{15}{2\times 8^2}, \quad A_{1,3}=\frac{105}{2\times 8^3}, \nonumber \\ A_{1,4}&=-\frac{105\times 45}{8^5}, \quad A_{1,5}=\frac{21\times 45\times 77}{8^6}. \end{aligned}$$

(16)

Moreover, for $\gamma >0$,

$$\begin{aligned}&r_{0,3}\le 2e^{-\frac{1}{4\gamma }}|A_{0,3}|= \frac{75e^{-\frac{1}{4\gamma }}}{8^3},\quad r_{1,3}\le 2e^{\frac{3}{4\gamma }}|A_{1,3}|= \frac{105e^{\frac{3}{4\gamma }}}{8^3}, \end{aligned}$$

(17)

$$\begin{aligned}&r_{0,4}\le 2e^{-\frac{1}{4\gamma }}|A_{0,4}|= \frac{75\times 49e^{-\frac{1}{4\gamma }}}{4\times 8^5},\nonumber \\&r_{1,4}\le 2e^{\frac{3}{4\gamma }}|A_{1,4}|= \frac{105\times 45e^{\frac{3}{4\gamma }}}{4\times 8^4}, \nonumber \\&r_{0,5}=2e^{-\frac{1}{4\gamma }}|A_{0,5}|\le \frac{15\times 49\times 81}{4\times 8^5}e^{-\frac{1}{4\gamma }},\nonumber \\&r_{1,5}=2e^{\frac{3}{4\gamma }}|A_{1,5}|\le \frac{21\times 45\times 77}{4\times 8^5}e^{\frac{3}{4\gamma }}. \end{aligned}$$

(18)

Firstly, we show that the inequality on the left side of (13) is true. We use (5), (15), (16) and (17) to get

$$\begin{aligned} \frac{K_0(\gamma )}{K_1(\gamma )}\ge \frac{ 1-\frac{1}{8\gamma }+\frac{9}{128\gamma ^2}-\frac{75}{8^3\gamma ^3}e^{-\frac{1}{4\gamma }}}{1+\frac{3}{8\gamma }-\frac{15}{128\gamma ^2}+\frac{105}{8^3\gamma ^3}e^{\frac{3}{4\gamma }}}. \end{aligned}$$

Then, it suffices to show that

$$\begin{aligned}&1-\frac{1}{8\gamma }+\frac{9}{128\gamma ^2}-\frac{75}{8^3\gamma ^3}e^{-\frac{1}{4\gamma }}\\&\quad \ge \left( 1+\frac{3}{8\gamma }-\frac{15}{128\gamma ^2}+\frac{105}{8^3\gamma ^3}e^{\frac{3}{4\gamma }}\right) \left( 1-\frac{1}{2\gamma }\right) \\&\quad =1-\frac{1}{8\gamma }-\left( \frac{3}{16}+\frac{15}{128}\right) \frac{1}{\gamma ^2}\\&\qquad +\left( \frac{105}{8^3}e^{\frac{3}{4\gamma }}+\frac{15}{256}\right) \frac{1}{\gamma ^3}-\frac{105}{2\times 8^3\gamma ^4}e^{\frac{3}{4\gamma }}; \end{aligned}$$

namely,

$$\begin{aligned}&\frac{3}{8\gamma ^2}+\frac{105}{2\times 8^3\gamma ^4}e^{\frac{3}{4\gamma }}\ge \left( \frac{75}{8^3}e^{-\frac{1}{4\gamma }}+\frac{105}{8^3}e^{\frac{3}{4\gamma }}+\frac{15}{256}\right) \frac{1}{\gamma ^3},\nonumber \\&\gamma ^2-\left( \frac{25}{64}e^{-\frac{1}{4\gamma }}+\frac{35}{64}e^{\frac{3}{4\gamma }}+\frac{5}{32}\right) \gamma +\frac{35}{128}e^{\frac{3}{4\gamma }}\ge 0. \end{aligned}$$

(19)

Denote $f_3(\gamma )=:\gamma ^2-\left( \frac{25}{64}e^{-\frac{1}{4\gamma }}+\frac{35}{64}e^{\frac{3}{4\gamma }}+\frac{5}{32}\right) \gamma +\frac{35}{128}e^{\frac{3}{4\gamma }}$. It is easy to check that

$$\begin{aligned} f_3(1.1)>0,\qquad f_3'(\gamma )>0 ~~\text{ for } ~~\gamma \ge 1. \end{aligned}$$

Then (19) holds for $\gamma \in [1.1,\infty )\subset (\sqrt{2}, \infty )$.

We now continue to verify the inequality on the right side of (13). Similarly, from (5), (15), (16) and (18), we get

$$\begin{aligned} \frac{K_0(\gamma )}{K_1(\gamma )}\le \frac{ 1-\frac{1}{8\gamma }+\frac{9}{128\gamma ^2}-\frac{75}{2\times 8^3\gamma ^3}+\frac{75\times 49e^{-\frac{1}{4\gamma }}}{4\times 8^4\gamma ^4}}{1+\frac{3}{8\gamma }-\frac{15}{128\gamma ^2}+\frac{105}{2\times 8^3\gamma ^3}-\frac{105\times 45e^{\frac{3}{4\gamma }}}{4\times 8^4\gamma ^4}}. \end{aligned}$$

The proof can be completed if we can show the following inequality:

$$\begin{aligned}&1-\frac{1}{8\gamma }+\frac{9}{128\gamma ^2}-\frac{75}{2\times 8^3\gamma ^3}+\frac{75\times 49e^{-\frac{1}{4\gamma }}}{4\times 8^4\gamma ^4}\\&\quad \le \Big (1+\frac{3}{8\gamma }-\frac{15}{128\gamma ^2}+\frac{105}{2\times 8^3\gamma ^3}-\frac{105\times 45e^{\frac{3}{4\gamma }}}{4\times 8^4\gamma ^4}\Big )\\&\quad \qquad \times \Big (1-\frac{1}{2\gamma }+\frac{3}{8\gamma ^2}+\frac{3}{16\gamma ^3}\Big )\\&\quad =1-\frac{1}{8\gamma }+\frac{9}{128\gamma ^2}+\left( \frac{3}{16}+\frac{9}{64}+\frac{15}{256}+\frac{105}{1024}\right) \frac{1}{\gamma ^3}\\&\qquad +\left( \frac{9}{128}-\frac{45}{1024}-\frac{105}{4\times 8^3}-\frac{105\times 45}{4\times 8^4}e^{\frac{3}{4\gamma }}\right) \frac{1}{\gamma ^4}\\&\qquad +\left( \frac{-45\times 4+315}{2\times 8^4}+\frac{105\times 45}{8^5}e^{\frac{3}{4\gamma }}\right) \frac{1}{\gamma ^5}\\&\qquad +\left( \frac{315}{4\times 8^4}-\frac{315\times 45}{4\times 8^5}e^{\frac{3}{4\gamma }}\right) \frac{1}{\gamma ^6}-\frac{315\times 45}{8^6}\frac{e^{\frac{3}{4\gamma }}}{\gamma ^7}. \end{aligned}$$

This inequality can be simplified as

$$\begin{aligned}&9\gamma ^4+\left( \frac{9}{8}-\frac{195}{128}-\frac{105\times 45e^{\frac{3}{4\gamma }}+75\times 49e^{-\frac{1}{4\gamma }}}{2\times 8^3}\right) \gamma ^3\nonumber \\&\quad + \Big (-\frac{45}{128}+\frac{315}{8^3}+\frac{105\times 45e^{\frac{3}{4\gamma }}}{4\times 8^3}\Big )\gamma ^2 \nonumber \\&\quad + \Big (\frac{315}{2\times 8^3}-\frac{315\times 45e^{\frac{3}{4\gamma }}}{2\times 8^4}\Big )\gamma -\frac{315\times 45e^{\frac{3}{4\gamma }}}{2\times 8^4}\ge 0. \end{aligned}$$

(20)

Denote the function on the left side of (20) as $f_4(\gamma )$. It can be verified that

$$\begin{aligned} f_4(\sqrt{2})>0,\qquad f'_4(\gamma )>0 ~~\text{ for }~~ \gamma \in [1,\infty ) . \end{aligned}$$

Therefore, (20) holds for $\gamma \in [\frac{5}{4}, \infty )$. $\square$

C Essential Estimates

In this part, we present estimates essential to the analysis in our paper.

Based on Lemma 1 and Corollary 2, the following two estimates hold:

Proposition 10

Let $\gamma \in (0,\infty )$ and $K_j(\gamma ) (j\ge 0)$ be the functions defined in Lemma 1. Then it holds that

$$\begin{aligned}&\gamma ^2\left( \frac{K_1(\gamma )}{K_2(\gamma )}\right) ^2+3\gamma \frac{K_1(\gamma )}{K_2(\gamma )}- \gamma ^2-3<0, \end{aligned}$$

(21)

$$\begin{aligned}&\gamma \left( \frac{K_1(\gamma )}{K_2(\gamma )}\right) ^3+4\left( \frac{K_1(\gamma )}{K_2(\gamma )}\right) ^2-\gamma \frac{K_1(\gamma )}{K_2(\gamma )}-1<0. \end{aligned}$$

(22)

Proof

We first prove (21). Its proof is divided into two cases: $\gamma \in (0,\sqrt{2}]$ and $\gamma \in (\sqrt{2},\infty )$. Firstly, by (2), we can rewrite (21) as

$$\begin{aligned} (\gamma ^2+3)\left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2+\left( \gamma +\frac{12}{\gamma }\right) \frac{K_0(\gamma )}{K_1(\gamma )}+\frac{12}{\gamma ^2}-\gamma ^2-2>0. \end{aligned}$$

(23)

Noting that $K_0(\gamma ), K_1(\gamma )>0$ for $\gamma \in (0,\infty )$ and

$$\begin{aligned} \frac{12}{\gamma ^2}-\gamma ^2-2>0,\quad \gamma \in (0,\sqrt{2}], \end{aligned}$$

(21) holds when $\gamma \in (0,\sqrt{2}]$. For the case $\gamma \in (\sqrt{2},\infty )$, we use (13) to get

$$\begin{aligned}&(\gamma ^2+3)\left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2+\left( \gamma +\frac{12}{\gamma }\right) \frac{K_0(\gamma )}{K_1(\gamma )}+\frac{12}{\gamma ^2}-\gamma ^2-2 \\&\quad \ge (\gamma ^2+3)\left( 1-\frac{1}{2\gamma }\right) ^2+\left( \gamma +\frac{12}{\gamma }\right) \left( 1-\frac{1}{2\gamma }\right) +\frac{12}{\gamma ^2}-\gamma ^2-2\\&\quad =\gamma ^2-\gamma +\frac{13}{4}-\frac{3}{\gamma }+\frac{3}{4\gamma ^2}+\gamma -\frac{1}{2}+\frac{12}{\gamma }-\frac{6}{\gamma ^2}-\gamma ^2-2 \\&\quad =\frac{3}{4}+\frac{9}{\gamma }-\frac{21}{4\gamma ^2}>0. \end{aligned}$$

This yields (23). Then (21) follows.

Now we turn to prove (22). The proof is also done in two cases, $\gamma \in (0,\sqrt{2}]$ and $\gamma \in (\sqrt{2},\infty )$, separately. We use (2) again to rewrite (22) as

$$\begin{aligned} \left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^3+\left( \gamma +\frac{6}{\gamma }\right) \left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2+\frac{12}{\gamma ^2}\frac{K_0(\gamma )}{K_1(\gamma )}-\gamma -\frac{4}{\gamma }+\frac{8}{\gamma ^3}>0. \end{aligned}$$

(24)

We first show that (24) is true when $\gamma \in (0,\sqrt{2})$. For this purpose, we use (8) to get

$$\begin{aligned}&\left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^3+\left( \gamma +\frac{6}{\gamma }\right) \left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2+\frac{12}{\gamma ^2}\frac{K_0(\gamma )}{K_1(\gamma )}-\gamma -\frac{4}{\gamma }+\frac{8}{\gamma ^3} \\&\quad =\frac{K_0(\gamma )}{K_1(\gamma )}\left[ \left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2+\frac{2}{\gamma }\frac{K_0(\gamma )}{K_1(\gamma )}-\frac{1}{3}\right] \\&\qquad +\left( \gamma +\frac{4}{\gamma }\right) \left[ \left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2 +\frac{2}{\gamma }\frac{K_0(\gamma )}{K_1(\gamma )}-\frac{1}{3}\right] \\&\qquad +\frac{1}{3}\left( \gamma +\frac{4}{\gamma }\right) +\Big (\frac{1}{3}-2+\frac{4}{\gamma ^2}\Big ) \frac{K_0(\gamma )}{K_1(\gamma )}-\gamma -\frac{4}{\gamma }+\frac{8}{\gamma ^3} \\&\quad>\left( \frac{4}{\gamma ^2}-\frac{5}{3}\right) \frac{K_0(\gamma )}{K_1(\gamma )}-\frac{2\gamma }{3}-\frac{8}{3\gamma }+\frac{8}{\gamma ^3}>0, \end{aligned}$$

when $\gamma \in (0,\sqrt{2}]$. Here we have used the simple estimates: for $\gamma \in (0,\sqrt{2}]$,

$$\begin{aligned} \frac{4}{\gamma ^2}-\frac{5}{3}>0,\quad -\frac{2\gamma }{3}-\frac{8}{3\gamma }+\frac{8}{\gamma ^3}\ge 0. \end{aligned}$$

When $\gamma \in (\sqrt{2},\infty )$, in a fashion similar to the proof of (21), we use (13) to obtain

$$\begin{aligned}&\Big (\frac{K_0(\gamma )}{K_1(\gamma )}\Big )^3+\left( \gamma +\frac{6}{\gamma }\right) \left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2 +\frac{12}{\gamma ^2}\frac{K_0(\gamma )}{K_1(\gamma )}-\gamma -\frac{4}{\gamma }+\frac{8}{\gamma ^3}\nonumber \\&\quad \ge \left( \gamma +1+\frac{11}{2\gamma }\right) \left( 1-\frac{1}{\gamma }+\frac{1}{4\gamma ^2}\right) + \frac{12}{\gamma ^2}+\frac{2}{\gamma ^3}-\gamma -\frac{4}{\gamma }\nonumber \\&\quad =\gamma +\frac{19}{4\gamma }-\frac{21}{4\gamma ^2}+\frac{11}{8\gamma ^3}+ \frac{12}{\gamma ^2}+\frac{2}{\gamma ^3}-\gamma -\frac{4}{\gamma } \nonumber \\&\quad =\frac{3}{4\gamma }+\frac{27}{4\gamma ^2}+\frac{27}{8\gamma ^3}>0. \end{aligned}$$

$\square$

Remark 3

In the proof of Proposition 10, instead of working on the ratio $\frac{K_1(\gamma )}{K_2(\gamma )}$ directly, we transformed the ratio $\frac{K_1(\gamma )}{K_2(\gamma )}$ into the ratio $\frac{K_0(\gamma )}{K_1(\gamma )}$ and divided our proof of (21) and (22) into two cases, $\gamma \in (0,\sqrt{2}]$ and $\gamma \in (\sqrt{2},\infty )$. The motivations for this are as follows: from the expansion of $K_j(\gamma )$ in (4) and (5), we can see that it works well for $\gamma$ which is a little larger than 1, and vice versa; the estimate of the remaining term $|r_{j,n}(\gamma )|$ seems more accurate when j is smaller due to the coefficient $e^{[j^2-1/4]\gamma ^{-1}}$ in the estimate, which increase more rapidly than the normal exponential function; when $\gamma$ is small, we can make use of the simple inequality (8) from the observation (7).

Remark 4

Estimates (21) and (22) imply the two Conjectures in [42]. Moreover, these two estimates are of essential importance in this paper.

Proposition 11

Let $\gamma \in (0,\infty )$ and $K_j(\gamma ) (j\ge 0)$ be the functions defined in Lemma 1. Then it holds that

$$\begin{aligned} \gamma ^2\left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^3+2\gamma \left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2-(\gamma ^2 +2)\frac{K_0(\gamma )}{K_1(\gamma )}- \gamma <0. \end{aligned}$$

(25)

Proof

For $\gamma \le 2,$ it is straightforward to get (25) by the fact $\frac{K_0(\gamma )}{K_1(\gamma )}<1$. For the case $\gamma >2$, we use (13) to get

$$\begin{aligned} \frac{K_0(\gamma )}{K_1(\gamma )}\le 1-\frac{1}{2\gamma }+\frac{1}{2\gamma ^2}, \end{aligned}$$

and

$$\begin{aligned}&\gamma ^2\left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2+2\gamma \left( \frac{K_0(\gamma )}{K_1(\gamma )}\right) ^2-(\gamma ^2 +2)\frac{K_0(\gamma )}{K_1(\gamma )}- \gamma \\&\quad \le \gamma \frac{K_0(\gamma )}{K_1(\gamma )}\Big [\gamma \Big (1-\frac{1}{2\gamma } +\frac{1}{2\gamma ^2}\Big )^2+2\Big (1-\frac{1}{2\gamma }+\frac{1}{2\gamma ^2}\Big )\Big ]\\&\qquad -(\gamma ^2 +2)\frac{K_1(\gamma )}{K_2(\gamma )}- \gamma \\&\quad \le \gamma \frac{K_0(\gamma )}{K_1(\gamma )}\Big [\gamma \Big (1-\frac{1}{\gamma } +\frac{5}{4\gamma ^2}-\frac{1}{2\gamma ^3}+\frac{1}{4\gamma ^4}\Big )+2-\frac{1}{\gamma }+\frac{1}{\gamma ^2}\Big )\Big ]\\&\qquad -(\gamma ^2 +2)\frac{K_0(\gamma )}{K_1(\gamma )}- \gamma \\&\quad \le \frac{K_0(\gamma )}{K_1(\gamma )}\Big (\gamma -\frac{3}{4}+\frac{1}{2\gamma }+\frac{1}{4\gamma ^2}\Big )-\gamma <0. \end{aligned}$$

$\square$

D Proof of Proposition 2 for the Genuine Nonlinearity

Proof of Proposition 2

We only need to prove (46). If this is done, (47) follows immediately from (45) and (46). From (42), we can further obtain that

$$\begin{aligned} e_{pp}=&\frac{e_{p}}{p}-\frac{e}{p^2}+\frac{1}{\partial _\gamma p}\partial _\gamma \left( \frac{p}{\partial _\gamma p}\frac{\hbox {d}}{\mathrm{{d}}\gamma }\left( \gamma \frac{K_1(\gamma )}{K_2(\gamma )}\right) \right) \nonumber \\ =&\frac{1}{p}\frac{\gamma \left( \frac{K_1(\gamma )}{K_2(\gamma )}\right) ^2 +4\frac{K_1(\gamma )}{K_2(\gamma )}-\gamma }{\gamma \left( \frac{K_1(\gamma )}{K_2(\gamma )}\right) ^2+3\frac{K_1(\gamma )}{K_2(\gamma )} -\gamma -\frac{4}{\gamma }} \nonumber \\&+\frac{1}{p}\frac{1}{\gamma \left( \frac{K_1(\gamma )}{K_2(\gamma )}\right) ^2+3\frac{K_1(\gamma )}{K_2(\gamma )}-\gamma -\frac{4}{\gamma }}\nonumber \\&\quad \times \frac{\hbox {d}}{\mathrm{{d}}\gamma }\Big (\frac{\gamma \Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^2 +4\frac{K_1(\gamma )}{K_2(\gamma )}-\gamma }{\gamma \Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^2+3\frac{K_1(\gamma )}{K_2(\gamma )} -\gamma -\frac{4}{\gamma }}\Big ). \end{aligned}$$

(26)

Then we use (42) and (26) to get

$$\begin{aligned}&(e+p)e_{pp}-2e_{p}(e_{p}-1)\nonumber \\&\quad =e_{p}(-2e_{p}+3)+\frac{e}{p}\Big (e_{p}-\frac{e}{p}-1\Big ) +\frac{e+p}{\partial _\gamma p}\partial _\gamma \Big (\frac{p}{\partial _\gamma p}\frac{\hbox {d}}{\mathrm{{d}}\gamma }\Big (\gamma \frac{K_1(\gamma )}{K_2(\gamma )}\Big )\Big ) \nonumber \\&< -9+\Big (\gamma \frac{K_1(\gamma )}{K_2(\gamma )}+3\Big )\frac{\frac{K_1(\gamma )}{K_2(\gamma )}+\frac{4}{\gamma }}{\gamma \Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^2+3\frac{K_1(\gamma )}{K_2(\gamma )}-\gamma -\frac{4}{\gamma }}\nonumber \\&\qquad +\Big (\gamma \frac{K_1(\gamma )}{K_2(\gamma )}+4\Big )\Bigg [\frac{\frac{1}{\gamma }}{\gamma \Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^2 +3\frac{K_1(\gamma )}{K_2(\gamma )}-\gamma -\frac{4}{\gamma }}-\Big (\frac{K_1(\gamma )}{K_2(\gamma )}+\frac{4}{\gamma }\Big )\nonumber \\&\qquad \quad \times \frac{\Big [2\gamma \Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^3+10\Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^2 +\Big (\frac{9}{\gamma }-2\gamma \Big )\frac{K_1(\gamma )}{K_2(\gamma )}-4+\frac{4}{\gamma ^2}\Big ]}{\Big (\gamma \Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^2+3\frac{K_1(\gamma )}{K_2(\gamma )}-\gamma -\frac{4}{\gamma }\Big )^3}\Bigg ]\nonumber \\&= -9+\frac{\left( \gamma \frac{K_1(\gamma )}{K_2(\gamma )}+4\right) \left( \frac{K_1(\gamma )}{K_2(\gamma )}+\frac{4}{\gamma }\right) }{\left( \gamma \left( \frac{K_1(\gamma )}{K_2(\gamma )}\right) ^2+3\frac{K_1(\gamma )}{K_2(\gamma )}-\gamma -\frac{4}{\gamma }\right) ^3}\times \mathcal {I}_1(\gamma ) \Big ], \end{aligned}$$

(27)

where

$$\begin{aligned} \mathcal {I}_1(\gamma )=&\gamma ^2\Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^4+4\gamma \Big (\frac{K_1(\gamma )}{K_2(\gamma )}\Big )^3 -(2\gamma ^2+9)\left( \frac{K_1(\gamma )}{K_2(\gamma )}\right) ^2\nonumber \\&-\Big (4\gamma +\frac{33}{\gamma }\Big )\frac{K_1(\gamma )}{K_2(\gamma )}+\gamma ^2+12+\frac{12}{\gamma ^2}. \end{aligned}$$

Noting that

$$\begin{aligned} \gamma \left( \frac{K_1(\gamma )}{K_2(\gamma )}\right) ^2+3\frac{K_1(\gamma )}{K_2(\gamma )}-\gamma -\frac{4}{\gamma }<0, \end{aligned}$$

in order to show (46), it suffices to prove

$$\begin{aligned} \mathcal {I}_1(\gamma )>0 \end{aligned}$$

(28)

in (27). By using $K_2(\gamma )=\frac{2}{\gamma }K_1(\gamma )+K_0(\gamma )$, we can rewrite (28) as

$$\begin{aligned} \mathcal {I}_2(\gamma )=&\Big (\gamma ^2+12+\frac{12}{\gamma ^2}\Big )\Big (\frac{K_0(\gamma )}{K_1(\gamma )}\Big )^4 +\Big (4\gamma +\frac{63}{\gamma }+\frac{96}{\gamma ^3}\Big )\Big (\frac{K_0(\gamma )}{K_1(\gamma )}\Big )^3 \nonumber \\&+\Big (-2\gamma ^2-9+\frac{90}{\gamma ^2}+\frac{288}{\gamma ^4}\Big )\Big (\frac{K_0(\gamma )}{K_1(\gamma )}\Big )^2\nonumber \\&+\Big (-4\gamma -\frac{52}{\gamma }-\frac{12}{\gamma ^3}+\frac{384}{\gamma ^5}\Big )\frac{K_0(\gamma )}{K_1(\gamma )}\nonumber \\&+\gamma ^2-\frac{52}{\gamma ^2}-\frac{72}{\gamma ^4}+\frac{192}{\gamma ^6}>0. \end{aligned}$$

(29)

Now we come to prove (29). It is easy to find that (29) holds for $\gamma \in (0, r_0]$.

Now we turn to show that (29) holds for $\gamma \in (r_0, \infty )$. Rewrite $\mathcal {I}_2(\gamma )$ as

$$\begin{aligned} \mathcal {I}_2(\gamma )&=\Big (\frac{K_0(\gamma )}{K_1(\gamma )}-1+\frac{1}{2\gamma }\Big )\Big [\Big (\gamma ^2+12+\frac{12}{\gamma ^2}\Big )\Big (\frac{K_0(\gamma )}{K_1(\gamma )}\Big )^3\\&\quad +\Big (\gamma ^2+\frac{7\gamma }{2}+ 12+\frac{57}{\gamma }+\frac{12}{\gamma ^2}+\frac{90}{\gamma ^3}\Big )\Big (\frac{K_0(\gamma )}{K_1(\gamma )}\Big )^2\\&\quad +\Big (-\gamma ^2+3\gamma +\frac{5}{4}+\frac{51}{\gamma }+\frac{147}{2\gamma ^2}+{84}{\gamma ^3}+\frac{243}{\gamma ^4}\Big )\frac{K_0(\gamma )}{K_1(\gamma )}\\&\quad -\gamma ^2-\frac{\gamma }{2}-\frac{1}{4}-\frac{13}{8\gamma }+\frac{48}{\gamma ^2}+\frac{141}{4\gamma ^3}+\frac{201}{\gamma ^4}+\frac{525}{2\gamma ^5}\Big ]\\&\quad -\frac{3}{2\gamma }-\frac{51}{16\gamma ^2}+\frac{45}{4\gamma ^3}+\frac{891}{8\gamma ^4}+\frac{162}{\gamma ^5}+\frac{243}{4\gamma ^6}\\&=\Big (\frac{K_0(\gamma )}{K_1(\gamma )}-1+\frac{1}{2\gamma }\Big )\Big \{\Big (\frac{K_0(\gamma )}{K_1(\gamma )}-1+\frac{1}{2\gamma }\Big )\\&\times \Big [\Big (\gamma ^2+12+\frac{12}{\gamma ^2}\Big ) \Big (\frac{K_0(\gamma )}{K_1(\gamma )}\Big )^2\\&\quad +\Big (2\gamma ^2+3\gamma +24+\frac{51}{\gamma }+\frac{24}{\gamma ^2}+\frac{84}{\gamma ^3}\Big )\frac{K_0(\gamma )}{K_1(\gamma )}\\&\quad +\gamma ^2+5\gamma +\frac{95}{4}+\frac{90}{\gamma }+\frac{72}{\gamma ^2}+\frac{156}{\gamma ^3}+\frac{201}{\gamma ^4}\Big ]\\&\quad +4\gamma +21+\frac{153}{2\gamma }+\frac{75}{\gamma ^2}+\frac{621}{4\gamma ^3}+\frac{324}{\gamma ^4}+\frac{162}{\gamma ^5}\Big \}\\&\quad -\frac{3}{2\gamma }-\frac{51}{16\gamma ^2}+\frac{45}{4\gamma ^3}+\frac{891}{8\gamma ^4}+\frac{162}{\gamma ^5}+\frac{243}{4\gamma ^6}. \end{aligned}$$

Note that

$$\begin{aligned} -\frac{3}{2\gamma }-\frac{51}{16\gamma ^2}+\frac{45}{4\gamma ^3}+\frac{881}{8\gamma ^4}+\frac{162}{\gamma ^5}+\frac{243}{4\gamma ^6}>0, \quad \text{ for }~~\gamma \le 4. \end{aligned}$$

Then (29) holds for $\gamma \in (0, 4]$.

Finally we show (29) for $\gamma >4$. For the case $\gamma >4$, we use (14) to get

$$\begin{aligned} \frac{K_0(\gamma )}{K_1(\gamma )}\ge 1-\frac{1}{2\gamma }+\frac{3}{8\gamma ^2}-\frac{3}{8\gamma ^3}. \end{aligned}$$

Moreover, we have

$$\begin{aligned}&\Big (4\gamma +21+\frac{153}{2\gamma }\Big )\Big (\frac{3}{8\gamma ^2}-\frac{3}{8\gamma ^3}\Big ) -\frac{3}{2\gamma }-\frac{51}{16\gamma ^2}\\&\qquad +\frac{45}{4\gamma ^3}+\frac{891}{8\gamma ^4}+\frac{162}{\gamma ^5}+\frac{243}{4\gamma ^6}\\&\quad =\Big (\frac{513}{16}+\frac{45}{4}\Big )\frac{1}{\gamma ^3} +\frac{1323}{16\gamma ^4}+\frac{162}{\gamma ^5}+\frac{243}{4\gamma ^6}>0. \end{aligned}$$

This proves (29) for $\gamma >4$. $\square$