Classification of finite Morse index solutions to the polyharmonic Hénon equation

Abstract

We present Liouville-type results for stable solutions and finite Morse index solutions of the polyharmonic Hénon elliptic equation

$$\begin{aligned} (-\Delta )^mu=|x|^a|u|^{p-1}u,\quad \mathrm {in}\ {\mathbb {R}}^n, \end{aligned}$$

where $m\ge 3$, $p>1$, $a\ge 0$ and n is a large dimension. To construct the classification theorem of homogeneous stable solutions, we exhibit a concise and explicit form for a critical exponent $p_{a}(n,m)$ which is known as the Joseph-Lundgren exponent when taking $a=0$. Based on this, we obtain the desired results by establishing the monotonicity formula, applying some energy estimates and finally using a blow-down analysis.

Appendix A The generalized Joseph-Lundgren exponent

In this section, we consider the inequality

$$\begin{aligned} pJ_{0,m}(k)-J_{0,m}(\tau )>0, \end{aligned}$$

(A.1)

where $k=\frac{2m+a}{p-1}$, $\tau =\frac{n-2m}{2}$. We wish to prove the existence and give an expression of the generalized Joseph-Lundgren exponent for polyharmonic Hénon equation (1.1), and investigate some of its properties. Throughout this section, we assume that $n>2m$, $a\ge 0$ and $m\ge 1$.

1.1 The concise expression of the critical exponent

Since Gamma function holds the recurrence relation

$$\begin{aligned} \Gamma (s+1)=s\Gamma (s), \end{aligned}$$

by the definition of $J_{t,m}$ (see (2.1)), we have

$$\begin{aligned}J_{0,m}(x) = 2^{2m} \frac{\Gamma \left( \frac{n}{2}-\frac{x}{2} \right) \Gamma \left( m+\frac{x}{2} \right) }{ \Gamma \left( \frac{x}{2} \right) \Gamma \left( \frac{n-2m}{2} - \frac{x}{2} \right) } .\end{aligned}$$

Hence (A.1) is equivalent to the following inequality:

$$\begin{aligned} p\frac{\Gamma \left( \frac{n}{2}-\frac{k}{2} \right) \Gamma \left( m+\frac{k}{2}\right) }{\Gamma \left( \frac{k}{2} \right) \Gamma \left( \frac{n-2m}{2}-\frac{k}{2} \right) } - \left( \frac{\Gamma \left( \frac{n+2m}{4} \right) }{\Gamma \left( \frac{n-2m}{4} \right) } \right) ^2 >0. \end{aligned}$$

(A.2)

We have assumed that p is supercritical, i.e. $p>\frac{n+2m+2a}{n-2m}$. The purpose of this section is to determine the range of p such that the inequality (A.1) holds.

We consider the equation with variable p:

$$\begin{aligned} p\frac{\Gamma \left( \frac{n}{2}-\frac{k}{2} \right) \Gamma \left( m+\frac{k}{2}\right) }{\Gamma \left( \frac{k}{2} \right) \Gamma \left( \frac{n-2m}{2}-\frac{k}{2} \right) } - \left( \frac{\Gamma \left( \frac{n+2m}{4} \right) }{\Gamma \left( \frac{n-2m}{4} \right) } \right) ^2 =0. \end{aligned}$$

(A.3)

Since $k = \frac{2m+a}{p-1}$, we have that (A.3) is equivalent to

$$\begin{aligned} \begin{aligned}&\frac{\Gamma \left( \frac{2m+a+k}{2}+1 \right) \Gamma \left( \frac{2m+k}{2}\right) \Gamma \left( \frac{n}{2}-\frac{k}{2} \right) }{ \Gamma \left( \frac{2m+a+k}{2}\right) \Gamma \left( \frac{k}{2} +1 \right) \Gamma \left( \frac{n-2m}{2}-\frac{k}{2} \right) } - \left( \frac{\Gamma \left( \frac{n+2m}{4} \right) }{\Gamma \left( \frac{n-2m}{4} \right) } \right) ^2=0. \end{aligned} \end{aligned}$$

(A.4)

In order to put this equation in a more symmetric and convenient form, we take a key transformation similar as that introduced in [18]:

$$\begin{aligned} k:=\frac{n-(2m+2+a)}{2} + \lambda \sqrt{n}, \end{aligned}$$

(A.5)

transforming (A.4) to the following function w.r.t. $\lambda$:

$$\begin{aligned} \begin{aligned}&\frac{ \Gamma \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) \Gamma \left( \frac{1}{4} n + \frac{1}{2} m +\frac{1}{4}a + \frac{1}{2} - \frac{1}{2} \lambda \sqrt{n} \right) }{ \Gamma \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a - \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) \Gamma \left( \frac{1}{4}n - \frac{1}{2}m - \frac{1}{4}a + \frac{1}{2} + \frac{1}{2} \lambda \sqrt{n} \right) }\\&\quad \quad \times \frac{ \Gamma \left( \frac{1}{4} n + \frac{1}{2} m -\frac{1}{4}a - \frac{1}{2} + \frac{1}{2} \lambda \sqrt{n} \right) }{ \Gamma \left( \frac{1}{4}n - \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} - \frac{1}{2} \lambda \sqrt{n} \right) }=\left( \frac{\Gamma \left( \frac{1}{4} n+ \frac{1}{2} m \right) }{ \Gamma \left( \frac{1}{4} n - \frac{1}{2} m \right) } \right) ^2. \end{aligned} \end{aligned}$$

(A.6)

Notice that

$$\begin{aligned} p> & {} \frac{n+2m+2a}{n-2m} \quad \Leftrightarrow \quad 0<k<\frac{n-2m}{2} \\ \Leftrightarrow \quad \lambda _1:= & {} \frac{1}{2}\left( -\frac{n-2m}{\sqrt{n}}+\frac{2+a}{\sqrt{n}}\right)<\lambda <\frac{2+a}{2\sqrt{n}} =: \lambda _2, \end{aligned}$$

so we only need to consider (A.4) in the interval $\left( 0,\frac{n-2m}{2} \right)$ i.e. find the solution of the equation (A.6) in the interval $\left( \lambda _1,\lambda _2 \right)$.

To obtain further properties of equation (A.6), we introduce the logarithmic derivative of the gamma function, known as the Digamma Function

$$\begin{aligned} \Psi (x)=\frac{{\mathrm d}}{{\mathrm {d}x}} \left( \ln (\Gamma (x)) \right) = \frac{\Gamma '(x)}{\Gamma }, \end{aligned}$$

and recall the following expansion (see[1], p. 260)

$$\begin{aligned} \Psi ^{(i)}(x)=(-1)^{i+1}i!\sum _{j=0}^{\infty } \frac{1}{(x+j)^{i+1}},\quad i=1,2,\cdots ; x\ne 0,-1,-2,\cdots . \end{aligned}$$

(A.7)

We take the logarithm on both sides of equation (A.6), then it becomes

$$\begin{aligned} \begin{aligned}&f(n,m,a,\lambda )\\&\quad :=\ln \Gamma \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) + \ln \Gamma \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} - \frac{1}{2}\lambda \sqrt{n} \right) \\&\quad \quad + \ln \Gamma \left( \frac{1}{4}n + \frac{1}{2}m - \frac{1}{4}a - \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) - \ln \Gamma \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a - \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) \\&\quad \quad - \ln \Gamma \left( \frac{1}{4}n - \frac{1}{2}m - \frac{1}{4}a + \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) - \ln \Gamma \left( \frac{1}{4}n - \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} - \frac{1}{2}\lambda \sqrt{n} \right) \\&\quad \quad - 2\ln \Gamma \left( \frac{1}{4}n + \frac{1}{2}m \right) -2\ln \Gamma \left( \frac{1}{4}n - \frac{1}{2}m \right) =0. \end{aligned} \end{aligned}$$

(A.8)

Taking derivative w.r.t $\lambda$, we have

$$\begin{aligned} \begin{aligned}&g(n,m,a,\lambda )\\&\quad :=\frac{d}{d\lambda } f(n,m,a,\lambda ) \\&\quad = \frac{1}{2}\sqrt{n} \left[ \Psi \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) - \Psi \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} - \frac{1}{2}\lambda \sqrt{n} \right) \right. \\&\quad \quad +\Psi \left( \frac{1}{4}n + \frac{1}{2}m - \frac{1}{4}a - \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) -\Psi \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a - \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) \\&\quad \quad \left. -\Psi \left( \frac{1}{4}n - \frac{1}{2}m - \frac{1}{4}a + \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) + \Psi \left( \frac{1}{4}n - \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} - \frac{1}{2}\lambda \sqrt{n} \right) \right] . \end{aligned} \end{aligned}$$

(A.9)

We first show that $f(n,m,a,\lambda )$ is a concave function w.r.t $\lambda$ in $\big (\lambda _1,\lambda _2\big )$.

Lemma A.1

Assume that $\lambda _1<\lambda < \lambda _2$, then $g(n,m,a,\lambda )$ is decreasing w.r.t $\lambda$.

Proof

This can be obtained directly by calculation.

$$\begin{aligned} \begin{aligned} g'(\lambda ):=&\frac{d}{d\lambda } g(n,m,a,\lambda )\\ =&\frac{1}{4}n \left[ \Psi ' \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) + \Psi ' \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} - \frac{1}{2}\lambda \sqrt{n} \right) \right. \\&+\Psi ' \left( \frac{1}{4}n + \frac{1}{2}m - \frac{1}{4}a - \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) -\Psi ' \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a - \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) \\&\left. -\Psi ' \left( \frac{1}{4}n - \frac{1}{2}m - \frac{1}{4}a + \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) - \Psi ' \left( \frac{1}{4}n - \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} - \frac{1}{2}\lambda \sqrt{n} \right) \right] . \end{aligned} \end{aligned}$$

Taking derivative w.r.t m,

$$\begin{aligned} \begin{aligned} \frac{d}{dm}g'(\lambda ) =&\frac{1}{8}n \left[ \underbrace{\Psi '' \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) }_{T_1} + \underbrace{\Psi '' \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} - \frac{1}{2}\lambda \sqrt{n} \right) }_{T_2}\right. \\&+ \underbrace{\Psi '' \left( \frac{1}{4}n + \frac{1}{2}m - \frac{1}{4}a - \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) }_{T_3} - \underbrace{\Psi '' \left( \frac{1}{4}n + \frac{1}{2}m + \frac{1}{4}a - \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) }_{T_4}\\&\left. + \underbrace{\Psi '' \left( \frac{1}{4}n - \frac{1}{2}m - \frac{1}{4}a + \frac{1}{2} + \frac{1}{2}\lambda \sqrt{n} \right) }_{T_5} + \underbrace{\Psi '' \left( \frac{1}{4}n - \frac{1}{2}m + \frac{1}{4}a + \frac{1}{2} - \frac{1}{2}\lambda \sqrt{n} \right) }_{T_6}\right] . \end{aligned} \end{aligned}$$

By (A.7), $\Psi '' <0$ and $\Psi ''$ is increasing. Therefore, we have

$$\begin{aligned} T_3-T_4<0, \end{aligned}$$

and

$$\begin{aligned} T_1+T_2+T_5+T_6<0. \end{aligned}$$

This implies that $g'(\lambda )$ is decreasing w.r.t m. Then we take $m=1$, $g'(\lambda )$ becomes

$$\begin{aligned} \begin{aligned} g'(\lambda ,m=1) =&\frac{1}{4}n \left[ \Psi '\left( \frac{1}{4}n + \frac{1}{4}a + 1 + \frac{1}{2}\lambda \sqrt{n} \right) - \Psi '\left( \frac{1}{4}n + \frac{1}{4}a + \frac{1}{2}\lambda \sqrt{n} \right) \right. \\&\left. + \Psi '\left( \frac{1}{4}n + \frac{1}{4}a + 1 - \frac{1}{2}\lambda \sqrt{n} \right) - \Psi '\left( \frac{1}{4}n + \frac{1}{4}a - \frac{1}{2}\lambda \sqrt{n} \right) \right] . \end{aligned} \end{aligned}$$

Since $\Psi '$ is decreasing, we obtain that $g'(\lambda ,m=1) <0$. Therefore, for any $m\ge 1$, there holds $g'(\lambda )<0$. $\square$

The following lemmas investigate the behaviors of the function $f(n,m,a,\lambda )$ w.r.t the variable $\lambda$ and n.

Lemma A.2

There holds

$$\begin{aligned} g\left( n,m,a,\lambda \right) |_{\lambda =\lambda _2}<0. \end{aligned}$$

(A.10)

Proof

Since

$$\begin{aligned} \begin{aligned}&g\left( n,m,a,\lambda \right) |_{\lambda =\lambda _2}\\&\quad = \frac{1}{2}\sqrt{n}\left[ \Psi \left( \frac{1}{4}n+\frac{1}{2}m+\frac{1}{2}a+1 \right) -\Psi \left( \frac{1}{4}n+\frac{1}{2}m+\frac{1}{2}a \right) + \Psi \left( \frac{1}{4}n-\frac{1}{2}m \right) \right. \\&\left. \qquad - \Psi \left( \frac{1}{4}n-\frac{1}{2}m+1 \right) \right] \\&\quad =\left. \frac{1}{2}\sqrt{n}\left[ \Psi \left( \frac{1}{4}n+\frac{1}{2}m+\frac{1}{2}a+x \right) - \Psi \left( \frac{1}{4}n-\frac{1}{2}m+x \right) \right] \right| _{x=0}^{x=1}, \end{aligned} \end{aligned}$$

and the fact that $\Psi '$ is decreasing, we have

$$\begin{aligned} \begin{aligned}&\frac{d}{dx}\left[ \Psi \left( \frac{1}{4}n+\frac{1}{2}m+\frac{1}{2}a+x \right) - \Psi \left( \frac{1}{4}n-\frac{1}{2}m+x \right) \right] \\&\quad = \Psi '\left( \frac{1}{4}n+\frac{1}{2}m+\frac{1}{2}a+x \right) - \Psi '\left( \frac{1}{4}n-\frac{1}{2}m+x \right) <0. \end{aligned} \end{aligned}$$

Thus we obtain the inequality (A.10). $\square$

Lemma A.3

There holds

$$\begin{aligned} f(n,m,a,\lambda )|_{\lambda =\lambda _2}>0. \end{aligned}$$

(A.11)

Proof

This can be checked by computation.

$$\begin{aligned} \begin{aligned}&f(n,m,a,\lambda )|_{\lambda = \lambda _2} \\&\quad = \ln \Gamma \left( \frac{1}{4}n+\frac{1}{2}m+\frac{1}{2}a+1 \right) - \ln \Gamma \left( \frac{1}{4}n+\frac{1}{2}m+\frac{1}{2}a \right) - \ln \Gamma \left( \frac{1}{4}n-\frac{1}{2}m+1 \right) \\&\qquad + \ln \Gamma \left( \frac{1}{4}n-\frac{1}{2}m \right) \\&\quad = \left. \left[ \ln \Gamma \left( \frac{1}{4}n+\frac{1}{2}m+\frac{1}{2}a+x \right) -\ln \Gamma \left( \frac{1}{4}n-\frac{1}{2}m+x \right) \right] \right| _0^1 \end{aligned} \end{aligned}$$

Since $\Psi '>0$, we have

$$\begin{aligned} \begin{aligned}&\frac{d}{dx}\left[ \ln \Gamma \left( \frac{1}{4}n+\frac{1}{2}m+\frac{1}{2}a+x \right) -\ln \Gamma \left( \frac{1}{4}n-\frac{1}{2}m+x \right) \right] \\&\quad =\Psi \left( \frac{1}{4}n+\frac{1}{2}m+\frac{1}{2}a+x \right) -\Psi \left( \frac{1}{4}n-\frac{1}{2}m+x \right) >0. \end{aligned} \end{aligned}$$

Then we obtained (A.11). $\square$

Lemma A.4

When taking $\lambda =\lambda _1$, $f(n,m,a,\lambda )$ is decreasing w.r.t. the variable n.

Proof

 

$$\begin{aligned} \begin{aligned} f(n,m,a,\lambda )|_{\lambda =\lambda _1} =&\ln \Gamma \left( m+\frac{1}{2}a+1\right) + \ln \Gamma \left( \frac{1}{2}n\right) +\ln \Gamma \left( m\right) -\ln \Gamma \left( m+\frac{1}{2}a\right) \\&-\ln \Gamma \left( \frac{1}{2}n-m\right) -2\ln \Gamma \left( \frac{1}{4}n+\frac{1}{2}m\right) +2\ln \Gamma \left( \frac{1}{4}n-\frac{1}{2}m\right) . \end{aligned} \end{aligned}$$

Taking derivative w.r.t. n,

$$\begin{aligned}&\frac{d}{dn}\Big [ f(n,m,a,\lambda )|_{\lambda =\lambda _1} \Big ]\\&= \frac{1}{2}\left[ \Psi \left( \frac{1}{2}n\right) - \Psi \left( \frac{1}{2}n-m\right) - \Psi \left( \frac{1}{4}n+\frac{1}{2}m\right) + \Psi \left( \frac{1}{4}n-\frac{1}{2}m\right) \right] \\&= \left. \left[ \Psi \left( \frac{1}{2}n-m+x\right) -\Psi \left( \frac{1}{4}n-\frac{1}{2}m+x\right) \right] \right| ^m_0. \end{aligned}$$

Since $\Psi ''<0$, we have

$$\begin{aligned} \Psi '\left( \frac{1}{2}n-m+x\right) -\Psi '\left( \frac{1}{4}n-\frac{1}{2}m+x\right) <0, \end{aligned}$$

i.e. $\Psi \left( \frac{1}{2}n-m+x\right) -\Psi \left( \frac{1}{4}n-\frac{1}{2}m+x\right)$ is decreasing w.r.t. x. Thus,

$$\begin{aligned} \frac{d}{dn}\Big [ f(n,m,a,\lambda )|_{\lambda =\lambda _1} \Big ]<0. \end{aligned}$$

$\square$

Lemma A.5

There hold that

$$\begin{aligned} f(n,m,a,\lambda )|_{n=2m+4,\lambda =\lambda _1}>0, \end{aligned}$$

(A.12)

and

$$\begin{aligned} f(n,m,a,\lambda )|_{\lambda =\lambda _1}\rightarrow -\infty \text { as } n\rightarrow +\infty . \end{aligned}$$

(A.13)

Proof

(1) Taking $n=2m+4,\lambda =\lambda _1$, then

$$\begin{aligned} f(n,m,a,\lambda )= & {} \ln \Gamma \left( m+\frac{1}{2}a+1\right) +\ln \Gamma (m+2)+\ln \Gamma (m)\\&-\ln \Gamma \left( m+\frac{1}{2}a\right) -2\ln \Gamma (m+1)\\= & {} \left. \left[ \ln \Gamma \left( m+\frac{1}{2}a+x\right) -\ln \Gamma (1+x) + \ln \Gamma \left( m+1+x\right) \right. \right. \\&\left. \left. -\ln \Gamma (m+x) \right] \right| ^1_0. \end{aligned}$$

Since $\Psi '>0$, we have

$$\begin{aligned} \Psi \left( m+\frac{1}{2}a+1\right) - \Psi (1+x)>0 , \quad \Psi (m+1+x) - \Psi (1+x)>0, \end{aligned}$$

and then

$$\begin{aligned} \begin{aligned}&\frac{d}{dx} \left[ \ln \Gamma \left( m+\frac{1}{2}a+x\right) -\ln \Gamma (1+x) + \ln \Gamma \left( m+1+x\right) -\ln \Gamma (m+x) \right] \\&\quad = \Psi \left( m+\frac{1}{2}a+x\right) -\Psi (1+x) + \Psi \left( m+1+x\right) -\Psi (m+x)>0. \end{aligned} \end{aligned}$$

Hence we obtain (A.12).

(2) For n satisfying $n>2m$ and $n=4l$, $l\in {\mathbb {N}}$, by direct computation,

$$\begin{aligned} \begin{aligned}&\frac{\Gamma \left( \frac{1}{2}n+1\right) \Gamma ^2\left( \frac{1}{4}n-\frac{1}{2}m\right) }{\Gamma \left( \frac{1}{2}n-m\right) \Gamma ^2\left( \frac{1}{4}n+\frac{1}{2}m\right) }\\&\quad = \frac{\left( \frac{1}{2}n-m\right) \left( \frac{1}{2}n-m+1\right) \cdots \left( \frac{1}{2}n-1\right) \frac{1}{2}n}{\left( \frac{1}{4}n-\frac{1}{2}m\right) ^2\left( \frac{1}{4}n-\frac{1}{2}m+1\right) ^2\cdots \left( \frac{1}{4}n+\frac{1}{2}m-1\right) ^2}\\&\quad = \frac{O(n^{m+1})}{O(n^{2m})}. \end{aligned} \end{aligned}$$

Thus, with $m\ge 1$ and n large enough, we have

$$\begin{aligned} \ln \frac{\Gamma \left( \frac{1}{2}n+1\right) \Gamma ^2\left( \frac{1}{4}n-\frac{1}{2}m\right) }{\Gamma \left( \frac{1}{2}n-m\right) \Gamma ^2\left( \frac{1}{4}n+\frac{1}{2}m\right) } <0. \end{aligned}$$

that is,

$$\begin{aligned} 2\ln \Gamma \left( \frac{1}{4}n-\frac{1}{2}m \right) - \ln \Gamma \left( \frac{1}{2}n-m \right) -2\ln \Gamma \left( \frac{1}{4}n+\frac{1}{2}m \right) < -\ln \Gamma \left( \frac{1}{2}n+1 \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} f(n,m,a,\lambda )|_{\lambda =\lambda _1} =&\ln \Gamma \left( m+\frac{1}{2}a+1\right) + \ln \Gamma \left( m\right) - \ln \Gamma \left( m+\frac{1}{2}a\right) + \ln \Gamma \left( \frac{1}{2}n\right) \\&-\ln \Gamma \left( \frac{1}{2}n-m\right) -2\ln \Gamma \left( \frac{1}{4}n+\frac{1}{2}m\right) +2\ln \Gamma \left( \frac{1}{4}n-\frac{1}{2}m\right) \\ \le&\ln \Gamma \left( m+\frac{1}{2}a+1\right) + \ln \Gamma \left( m\right) - \ln \Gamma \left( m+\frac{1}{2}a\right) + \ln \Gamma \left( \frac{1}{2}n\right) \\&- \ln \Gamma \left( \frac{1}{2}n+1\right) \\ =&\ln \Gamma \left( m+\frac{1}{2}a+1\right) + \ln \Gamma \left( m\right) - \ln \Gamma \left( m+\frac{1}{2}a\right) -\ln n, \end{aligned} \end{aligned}$$

for $n=4l$, $l\in {\mathbb {N}}$. Furthermore, by Lemma A.4, $f(n,m,a,\lambda )|_{\lambda =\lambda _1}$ is decreasing w.r.t n, hence for $n >2m$, we prove (A.13). $\square$

By Lemma A.4 and Lemma A.5, we can immediately conclude that

Corollary A.1

The equation w.r.t the variable n

$$\begin{aligned} f(n,m,a,\lambda )|_{\lambda =\lambda _1}=0, \end{aligned}$$

i.e.

$$\begin{aligned} \frac{2m+a}{2m}\frac{\Gamma \left( m\right) \Gamma \left( \frac{1}{2}n\right) }{\Gamma \left( \frac{n-2m}{2}\right) } - \frac{\Gamma \left( \frac{n+2m}{4}\right) ^2}{\Gamma \left( \frac{n-2m}{4}\right) ^2} = 0. \end{aligned}$$

(A.14)

has merely one solution for $n>2m$, which we denote by $n_0(m,a)$. Furthermore,

$$\begin{aligned} 2m+4<n_0(m,a).\end{aligned}$$

Corollary A.2

Assume that $n>n_0(m,a)$, then there holds

$$\begin{aligned} g(n,m,a,\lambda )|_{\lambda =\lambda _1}>0. \end{aligned}$$

Proof

Assume that $g(n,m,a,\lambda )|_{\lambda =\lambda _1}\le 0$, then by Lemma A.1 and Lemma A.2,

$$\begin{aligned} g(n,m,a,\lambda )<0 \quad \text { for any } \lambda \in \left( \lambda _1,\lambda _2\right) . \end{aligned}$$

This implies that $f(n,m,a,\lambda )$ is decreasing w.r.t. $\lambda$ for any $\lambda \in \left( \lambda _1,\lambda _2\right)$. Then, combining with (A.11), we have

$$\begin{aligned} f(n,m,a,\lambda )|_{\lambda =\lambda _1}>f(n,m,a,\lambda )|_{\lambda =\lambda _2}>0. \end{aligned}$$

This contradicts the fact that

$$\begin{aligned} f(n,m,a,\lambda )|_{\lambda =\lambda _1}<0 \quad \text { for any } n>n_0(m,a).\end{aligned}$$

Hence the prove of the conclusion is complete. $\square$

With above results, we immediately have the following propositions.

Proposition A.1

Assume that $n>2m$, $m\ge 1$, $a\ge 0$. $n_0(m,a)$ is defined as the unique solution of (A.14).

• (1) If $n> n_0(m,a)$, then for any fixed n, m, a, there exists only one solution $\lambda _{n,m,a} \in \big ( \lambda _1,\lambda _2 \big )$ such that

  $$\begin{aligned} f(n,m,a,\lambda )=0. \end{aligned}$$

  Moreover, there holds

  $$\begin{aligned} f(n,m,a,\lambda ) \left\{ \begin{aligned}< 0, \quad \ \text { if } \lambda _1<\lambda<\lambda _{n,m,a},\\ > 0, \quad \ \text { if } \lambda _{n,m,a}<\lambda <\lambda _2. \end{aligned} \right. \end{aligned}$$

• (2) If $n \le n_0(m,a)$, then the inequality

  $$\begin{aligned} f(n,m,a,\lambda )>0 \end{aligned}$$

  holds for any $\lambda \in (\lambda _1,\lambda _2)$.

Proof

(1) If $n > n_0(m,a)$, then by Lemma A.1, Lemma A.2 and Corollary A.2, there exists $\lambda _0 \in \big ( \lambda _1,\lambda _2 \big )$ such that

$$\begin{aligned} \frac{d}{d\lambda }f(n,m,a,\lambda )|_{\lambda =\lambda _0}=0, \end{aligned}$$

and

$$\begin{aligned}&\frac{d}{d\lambda }f(n,m,a,\lambda )>0 \quad \text { for } \lambda _1<\lambda<\lambda _0; \\&\frac{d}{d\lambda }f(n,m,a,\lambda )<0 \quad \text { for } \lambda _0<\lambda <\lambda _2. \end{aligned}$$

These imply that $f(n,m,a,\lambda )$ is increasing in $\big ( \lambda _1,\lambda _0 \big )$ and decreasing in $\big (\lambda _0,\lambda _2\big )$. By Lemma A.3, Lemma A.4 and Corollary A.1, we have

$$\begin{aligned} f(n,m,a,\lambda )|_{\lambda =\lambda _1} <0, \end{aligned}$$

and

$$\begin{aligned} f(n,m,a,\lambda )|_{\lambda =\lambda _0}> f(n,m,a,\lambda )|_{\lambda =\lambda _2} > 0. \end{aligned}$$

Then there exists a unique $\lambda _{n,m,a} \in \big ( \lambda _1,\lambda _0 \big )$ such that

$$\begin{aligned} f(n,m,a,\lambda )|_{\lambda =\lambda _{n,m,a}} = 0. \end{aligned}$$

Furthermore,

$$\begin{aligned} f(n,m,a,\lambda )< 0 \quad \text { for } \lambda _1<\lambda <\lambda _{n,m,a}, \end{aligned}$$

and

$$\begin{aligned} f(n,m,a,\lambda ) > 0 \quad \text { for } \lambda _{n,m,a}<\lambda <\lambda _2. \end{aligned}$$

(2) If $n < n_0(m,a)$, since $g(n,m,a,\lambda )$ is decreasing for $\lambda _1<\lambda <\lambda _2$ (by Lemma A.2), and

$$\begin{aligned}&\frac{d}{d\lambda }f(n,m,a,\lambda )|_{\lambda =\lambda _2} < 0, \\&f(n,m,a,\lambda )|_{\lambda =\lambda _1}> 0, \\&f(n,m,a,\lambda )|_{\lambda =\lambda _2} > 0, \end{aligned}$$

then whether $g(n,m,a,\lambda )|_{\lambda =\lambda _1}$ is negative or not, there holds

$$\begin{aligned} f(n,m,a,\lambda ) > 0 \quad \text { for any } \lambda _1<\lambda <\lambda _2. \end{aligned}$$

$\square$

Recalling the transform $k=\frac{n-(2m+2+a)}{2}+\lambda \sqrt{n}$, we can immediately return to k and p, and have the following

Proposition A.2

Assume that $n>2m$, $m\ge 1$, $a\ge 0$. $n_0(m,a)$ is defined as the unique solution of (A.14).

• (1) If $n> n_0(m,a)$, then the inequality (A.1) holds if

  $$\begin{aligned} 0< \frac{n-(2m+2+a)}{2}+\lambda _{n,m,a}\sqrt{n}< k < \frac{n-2m}{2}, \end{aligned}$$

  i.e.

  $$\begin{aligned} \frac{n+2m+2a}{n-2m}< p < 1+\frac{4m+a}{n-2m-2-a+2\lambda _{n,m,a}\sqrt{n}}. \end{aligned}$$

• (2) If $n \le n_0(m,a)$, then the inequality (A.1) holds for any k satisfying

  $$\begin{aligned} 0< k < \frac{n-2m}{2}, \end{aligned}$$

  i.e.

  $$\begin{aligned} \frac{n+2m+2a}{n-2m}< p <+\infty . \end{aligned}$$

This proposition directly yields Proposition 1.1.

1.2 On the estimate $\lambda _{n,m,a}>-(1+\frac{a}{2})$

In this part, we prove the estimate that $-\lambda _{n,m,a}<1+\frac{a}{2}$, which is critical in the proof of monotonicity formula in Sect. 3.

Proposition A.3

Assume that $n>2m$, $m\ge 1$, $a\ge 0$ and that $\lambda _{n,m,a} \in \big ( \lambda _1,\lambda _2 \big )$ is defined by Proposition A.1, then there holds

$$\begin{aligned} \lambda _{n,m,a}>-\left( 1+\frac{a}{2}\right) . \end{aligned}$$

Proof

This result is equivalent to the following inequality

$$\begin{aligned} f(n,m,a,\lambda )|_{\lambda =-\left( 1+\frac{a}{2}\right) } <0. \end{aligned}$$

(A.15)

If $\lambda _1=\frac{1}{2}\left( -\frac{n-2m}{\sqrt{n}}+\frac{2+a}{\sqrt{n}} \right) \ge -\left( 1+\frac{a}{2}\right)$, (A.15) holds immediately.

It remains to consider the case $\lambda _1 < -\left( 1+\frac{a}{2}\right)$, that is, $n-\sqrt{n}(2+a)>2+a+2m.$ Let $\frac{1}{2}\sqrt{n}=t$, then t satisfies

$$\begin{aligned} t^2-\left( 1+\frac{a}{2} \right) t > \frac{1}{2}+\frac{1}{4}a+\frac{1}{2}m. \end{aligned}$$

We set

$$\begin{aligned} \begin{aligned}&H(t,m,a)\\&\quad :=f(n,m,a,\lambda )|_{\lambda =-\left( 1+\frac{a}{2}\right) }\\&\quad = \ln \Gamma \left( t^2+\frac{1}{2}m+\frac{1}{4}a+\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) + \ln \Gamma \left( t^2+\frac{1}{2}m+\frac{1}{4}a+\frac{1}{2}+\left( 1+\frac{a}{2}\right) t \right) \\&\quad \quad + \ln \Gamma \left( t^2+\frac{1}{2}m-\frac{1}{4}a-\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) - \ln \Gamma \left( t^2+\frac{1}{2}m+\frac{1}{4}a-\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) \\&\quad \quad - \ln \Gamma \left( t^2-\frac{1}{2}m-\frac{1}{4}a+\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) - \ln \Gamma \left( t^2-\frac{1}{2}m+\frac{1}{4}a+\frac{1}{2}+\left( 1+\frac{a}{2}\right) t \right) \\&\quad \quad - 2\ln \Gamma \left( t^2+\frac{1}{2}m\right) + 2\ln \Gamma \left( t^2-\frac{1}{2}m\right) . \end{aligned} \end{aligned}$$

Now we prove the conclusion by induction.

(1) When $m=1$,

$$\begin{aligned} \begin{aligned} H(t,m,a) =&\ln \Gamma \left( t^2+1+\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) - \ln \Gamma \left( t^2+\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) \\&+ \ln \Gamma \left( t^2+1+\frac{1}{4}a+\left( 1+\frac{a}{2}\right) t \right) - \ln \Gamma \left( t^2+\frac{1}{4}a+\left( 1+\frac{a}{2}\right) t \right) \\&+ 2\ln \Gamma \left( t^2-\frac{1}{2} \right) - 2\ln \Gamma \left( t^2+\frac{1}{2} \right) \\ =&\ln \frac{\Gamma \left( t^2+1+\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) \Gamma \left( t^2+1+\frac{1}{4}a+\left( 1+\frac{a}{2}\right) t \right) \Gamma ^2\left( t^2-\frac{1}{2} \right) }{ \Gamma \left( t^2+\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) \Gamma \left( t^2+\frac{1}{4}a+\left( 1+\frac{a}{2}\right) t \right) \Gamma ^2\left( t^2+\frac{1}{2} \right) }\\ =&\ln \frac{ \left( t^2+\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) \left( t^2+\frac{1}{4}a+\left( 1+\frac{a}{2}\right) t \right) }{ \left( t^2-\frac{1}{2} \right) ^2 } \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} t^2>\frac{1}{2}+\frac{1}{4}a+\frac{1}{2}m+\left( 1+\frac{a}{2}\right) t>1+\frac{1}{4}a, \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned}&\left( t^2+\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) \left( t^2+\frac{1}{4}a+\left( 1+\frac{a}{2}\right) t \right) - \left( t^2-\frac{1}{2} \right) ^2\\&\quad = -\left( 1+\frac{a}{2} \right) \left[ \frac{1}{2}at^2 + \frac{1}{4}\left( 1 - \frac{a}{2} \right) \right] \\&\quad< -\left( 1+\frac{a}{2} \right) \left[ \frac{1}{2}a\left( 1+\frac{1}{4}a \right) + \frac{1}{4}\left( 1 - \frac{a}{2} \right) \right] \\&\quad = -\frac{1}{16}(a+2)^2(a+1)<0. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} H(t,m,a)|_{m=1}<0. \end{aligned}$$

(2) When $m=2$,

$$\begin{aligned} \begin{aligned} H(t,m,a) =&\ln \Gamma \left( t^2 + \frac{3}{2} + \frac{1}{4}a - \left( 1+\frac{a}{2}t \right) \right) + \ln \Gamma \left( t^2 + \frac{3}{2} + \frac{1}{4}a + \left( 1+\frac{a}{2}t \right) \right) \\&+ \ln \Gamma \left( t^2 + \frac{1}{2} - \frac{1}{4}a - \left( 1+\frac{a}{2}t \right) \right) - \ln \Gamma \left( t^2 + \frac{1}{2} + \frac{1}{4}a - \left( 1+\frac{a}{2}t \right) \right) \\&- \ln \Gamma \left( t^2 - \frac{1}{2} - \frac{1}{4}a - \left( 1+\frac{a}{2}t \right) \right) - \ln \Gamma \left( t^2 - \frac{1}{2} + \frac{1}{4}a + \left( 1+\frac{a}{2}t \right) \right) \\&- 2\ln \Gamma (t^2+1) + 2\ln \Gamma (t^2-1)\\ =&\ln \left[ \frac{1}{{ (t^2-1)^2 t^4 }} \left( t^2 + \frac{1}{2} + \frac{1}{4}a - \left( 1+\frac{a}{2}t \right) \right) \left( t^2 - \frac{1}{2} + \frac{1}{4}a + \left( 1+\frac{a}{2}t \right) \right) \right. \\&\times \left. \left( t^2 + \frac{1}{2} + \frac{1}{4}a + \left( 1+\frac{a}{2}t \right) \right) \left( t^2 - \frac{1}{2} - \frac{1}{4}a - \left( 1+\frac{a}{2}t \right) \right) \right] . \end{aligned} \end{aligned}$$

By computation,

$$\begin{aligned} \begin{aligned} G(t,a) :=&\left( t^2 + \frac{1}{2} + \frac{1}{4}a - \left( 1+\frac{a}{2}t \right) \right) \left( t^2 - \frac{1}{2} + \frac{1}{4}a + \left( 1+\frac{a}{2}t \right) \right) \\&\times \left( t^2 + \frac{1}{2} + \frac{1}{4}a + \left( 1+\frac{a}{2}t \right) \right) \left( t^2 - \frac{1}{2} - \frac{1}{4}a - \left( 1+\frac{a}{2}t \right) \right) - t^4(t^2-1)^2\\ =&\underbrace{A_6 t^6 + A_5 t^5 + A_4 t^4 + A_3 t^3 }_{T_1} + \underbrace{A_2 t^2 + A_1 t + A_0}_{T_2}. \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} A_6&:= -\frac{1}{2} a(a+3),&A_5&= -\frac{1}{4} a(a+2),\\ A_4&:= \frac{1}{16} \big ( a^4+6a^3+16a^2+20a-8 \big ), \\ A_3&:= \frac{1}{16}a^2(a+2)^2,&\ A_2&= -\frac{1}{32}(a+2)^2(3a+4), \\ A_1&:= -\frac{1}{64}a(a+2)^3,&A_0&= -\frac{1}{256}(a+2)^3(a-2). \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} t^2>\left( 1+\frac{a}{2}\right) t+\frac{3}{2}+\frac{1}{4}a, \end{aligned}$$

and

$$\begin{aligned} A_6 \le 0, \quad A_2<0, \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned} T_1<&A_6\left[ \left( 1+\frac{a}{2}\right) t+\frac{3}{2}+\frac{1}{4}a \right] t^4 + A_5 t^5 + A_4 t^4 + A_3 t^3 = B_5 t^5 + B_4 t^4 + A_3 t^3\\ <&B_5\left[ \left( 1+\frac{a}{2}\right) t+\frac{3}{2}+\frac{1}{4}a \right] t^3 + B_4 t^4 + A_3 t^3 = C_4 t^4 + C_3 t^3, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} B_5&:= -\frac{1}{4}a(a+2)(a+4)\le 0,\\ B_4&:= \frac{1}{16}(a+2)(a-2)(a^2+4a+2),\\ C_4&:= -\frac{1}{16}(a+2)(a^3+10a^2+22a+4)<0,\\ C_3&:= -\frac{1}{2} a(a+2)(a+3)\le 0, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} T_2 <&A_2\left[ \left( 1+\frac{a}{2}\right) t+\frac{3}{2}+\frac{1}{4}a \right] + A_1 t + A_0 = B_1 t + B_0, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} B_1&:= -\frac{1}{16}(a+2)^3(a+1)<0,\\ B_0&:= -\frac{1}{256}(a+2)^2 \left( 7a^2+44a+44 \right) <0. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} T_1+T_2<0, \end{aligned}$$

that is,

$$\begin{aligned} G(t,a)<0, \end{aligned}$$

which implies that

$$\begin{aligned} H(t,m,a)|_{m=2}<0. \end{aligned}$$

(3) Assume that for any $l\ge 2$, there holds

$$\begin{aligned} H(t,m,a)|_{m=l-1}<0. \end{aligned}$$

Taking $m=l+1$, then

$$\begin{aligned} \begin{aligned} H(t,m,a) =&\ln \Gamma \left( t^2+\frac{1}{2}(l+1)+\frac{1}{4}a+\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) \\&+ \ln \Gamma \left( t^2+\frac{1}{2}(l+1)+\frac{1}{4}a+\frac{1}{2}+\left( 1+\frac{a}{2}\right) t \right) \\&+ \ln \Gamma \left( t^2+\frac{1}{2}(l+1)-\frac{1}{4}a-\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) \\&- \ln \Gamma \left( t^2+\frac{1}{2}(l+1)+\frac{1}{4}a-\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) \\&- \ln \Gamma \left( t^2-\frac{1}{2}(l+1)-\frac{1}{4}a+\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) \\&- \ln \Gamma \left( t^2-\frac{1}{2}(l+1)+\frac{1}{4}a+\frac{1}{2}+\left( 1+\frac{a}{2}\right) t \right) \\&- 2\ln \Gamma \left( t^2+\frac{1}{2}(l+1)\right) + 2\ln \Gamma \left( t^2-\frac{1}{2}(l+1)\right) . \end{aligned} \end{aligned}$$

To prove that $H(t,m,a)|_{m=l+1} <0$ is to show that the following inequality holds:

$$\begin{aligned}&\tilde{H}(l+1)\\&:=\frac{\Gamma \left( t^2+\frac{1}{2}(l+1)+\frac{1}{4}a+\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) \Gamma \left( t^2+\frac{1}{2}(l+1)+\frac{1}{4}a+\frac{1}{2}+\left( 1+\frac{a}{2}\right) t \right) }{\Gamma \left( t^2+\frac{1}{2}(l+1)+\frac{1}{4}a-\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) \Gamma \left( t^2-\frac{1}{2}(l+1)-\frac{1}{4}a+\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) }\\&\times \frac{\Gamma \left( t^2+\frac{1}{2}(l+1)-\frac{1}{4}a-\frac{1}{2}-\left( 1+\frac{a}{2}\right) t \right) \Gamma ^2\left( t^2-\frac{1}{2}(l+1)\right) }{\Gamma \left( t^2-\frac{1}{2}(l+1)+\frac{1}{4}a+\frac{1}{2}+\left( 1+\frac{a}{2}\right) t \right) \Gamma ^2\left( t^2+\frac{1}{2}(l+1)\right) } <1, \end{aligned}$$

which, by the recurrence relation of Gamma function, is equivalent to

$$\begin{aligned} \begin{aligned}&\frac{\left( t^2+\frac{1}{2}l+\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) \left( t^2+\frac{1}{2}l+\frac{1}{4}a+\left( 1+\frac{a}{2}\right) t \right) \left( t^2+\frac{1}{2}l-1-\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) }{\left( t^2+\frac{1}{2}l-1+\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) }\\&\qquad \times \frac{\left( t^2-\frac{1}{2}l-\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) \left( t^2-\frac{1}{2}l+\frac{1}{4}a+\left( 1+\frac{a}{2}\right) t \right) }{ \left( t^2-\frac{1}{2}l-\frac{1}{2} \right) ^2 \left( t^2+\frac{1}{2}l-\frac{1}{2} \right) ^2 }\tilde{H}(l-1)<1. \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} H(t,m,a)|_{m=l-1}<0 \quad \text { i.e. } \tilde{H}(l-1)<1, \end{aligned}$$

and

$$\begin{aligned} \frac{t^2+\frac{1}{2}l-1-\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t}{t^2+\frac{1}{2}l-1+\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t} \le 1, \end{aligned}$$

we only need to show that

$$\begin{aligned} \begin{aligned} G(t,l,a):=&\left( t^2+\frac{1}{2}l+\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) \left( t^2+\frac{1}{2}l+\frac{1}{4}a+\left( 1+\frac{a}{2}\right) t \right) \\&\times \left( t^2-\frac{1}{2}l-\frac{1}{4}a-\left( 1+\frac{a}{2}\right) t \right) \left( t^2-\frac{1}{2}l+\frac{1}{4}a+\left( 1+\frac{a}{2}\right) t \right) \\&- \left( t^2-\frac{1}{2}l-\frac{1}{2} \right) ^2 \left( t^2+\frac{1}{2}l-\frac{1}{2} \right) ^2 <0. \end{aligned} \end{aligned}$$

By computation,

$$\begin{aligned} G(t,l,a) = D_6 t^6 + D_5 t^5 + D_4 t^4 + D_3 t^3 + D_2 t^2 + D_1 t + D_0, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} D_6&:= -\frac{1}{2}a(a+3), \quad \quad \quad \quad \quad \quad D_5 := -\frac{1}{4}a(a+2),\\ D_4&:= \frac{1}{16} \Big ( a^4+6a^3+16a^2-4(l-6)a-8 \Big ),\\ D_3&:= \frac{1}{16} a(a+2)\Big ( a^2+2a+4(1-l) \Big ),\\ D_2&:= -\frac{1}{32} \Big ( \big (2l+1\big )a^3+4l\big (l+3\big )a^2+4l\big (5l+2\big )a+32l^2-16 \Big ),\\ D_1&:= -\frac{1}{64} a(a+2)(a+2l)^2, \quad \quad D_0 := \frac{1}{256} \big ( -a^4 -4la^3 +16l^3a +32l^2 -16 \big ). \end{aligned} \end{aligned}$$

Noticing that

$$\begin{aligned} t^2>\left( 1+\frac{a}{2}\right) t+\frac{1}{2}+\frac{1}{4}a+\frac{1}{2}\big (l+1\big ) = \left( 1+\frac{a}{2}\right) t + 1 + \frac{1}{4}a + \frac{1}{2}l, \end{aligned}$$

as the case $m=2$, we can similarly compute that

$$\begin{aligned} G(t,l,a)< & {} D_6\left[ \left( 1+\frac{a}{2}\right) t+1+\frac{1}{4}a+\frac{1}{2}l\right] t^4 + D_5 t^5 + D_4 t^4 \\&+ D_3 t^3 + D_2 t^2 + D_1 t + D_0 \\= & {} E_5 t^5 + E_4 t^4 + D_3 t^3 + D_2 t^2 + D_1 t + D_0\\< & {} E_5 \left[ \left( 1+\frac{a}{2}\right) t+1+\frac{1}{4}a+\frac{1}{2}l\right] t^3 + E_4 t^4 + D_3 t^3 + D_2 t^2 + D_1 t + D_0\\= & {} F_4 t^4 + F_3 t^3 + D_2 t^2 + D_1 t + D_0\\< & {} F_4 t^4 + F_3 \left[ \left( 1+\frac{a}{2}\right) t+1+\frac{1}{4}a+\frac{1}{2}l\right] t + D_2 t^2 + D_1 t + D_0\\= & {} F_4 t^4 + P_2 t^2 + P_1 t + D_0 \\< & {} F_4 t^4 + P_2 \left[ \left( 1+\frac{a}{2}\right) t+1+\frac{1}{4}a+\frac{1}{2}l\right] + P_1 t + D_0\\= & {} F_4 t^4 + Q_1 t + Q_0, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} E_5 :=&-\frac{1}{4}a(a+2)(a+4) \le 0,\\ E_4 :=&\frac{1}{16}\Big ( a^4+4a^3+\big ( 2-4l \big )a^2 - 16la-8 \Big ),\\ F_4 :=&-\frac{1}{16}\Big ( a^4+12a^3+2\big (2l+19\big )a^2+16(l+2)a+8 \Big ) < 0,\\ F_3 :=&\left( 1+\frac{1}{4}a+\frac{1}{2}l \right) E_5+D_3 = -\frac{1}{8} a\big (a+2\big ) \Big ( \big (l+3\big )a+6(l+1) \Big ) \le 0,\\ P_2 :=&\frac{1}{32}\Big ( 2\big (l+3\big )a^4 + \big (22l+37\big )a^3 + 4\big (l^2+17l+18\big )a^2 \\&+ 4\big (5l^2+14l+12\big )a + 32l^2 - 16 \Big ),\\ P_1 :=&- \frac{1}{64} a(2 + a) \Big ( 2(l+3)a^2 + (4l^2 + 32l + 37)a + 28l^2 + 72l + 48 \Big ),\\ Q_1 :=&-\frac{1}{32}(a + 2)\Big ( (l + 3)a^4 + 2(6l + 11)a^3 + (4l^2 + 52l + 54)a^2 \\&+ 8(3l^2 + 8l + 6)a + 16l^2 -8 \Big )\\ Q_0 :=&- \frac{1}{256}\Big (4(l + 3)a^5 + (8l^2 + 84l + 123)a^4 + 8(12l^2 + 58l + 55)a^3 \\&+ 32(2l^3 + 14l^2 + 20l + 11)a \\&+ 8(2l^3 + 43l^2 + 118l + 84)a^2 + 16(8l^3 + 14l^2 -4l -7)\Big ). \end{aligned} \end{aligned}$$

Since $F_4<0$, $Q_1<0$ and $Q_0<0$, we have

$$\begin{aligned} G(t,l,a) <0, \end{aligned}$$

Hence we have shown that for any $l \ge 2$, if

$$\begin{aligned} H(t,m,a)|_{m=l-1}<0, \end{aligned}$$

then there also holds

$$\begin{aligned} H(t,m,a)|_{m=l+1}<0. \end{aligned}$$

Combining (1)-(3), we have obtained that if $\lambda _1 < -\left( 1+\frac{a}{2}\right)$, the inequality (A.15) holds for any integer $m\ge 1$. This complete the proof. $\square$

Remark A.1

For the case $a=0$, this estimate has been proved to be sharp. Furthermore,

$$\begin{aligned} \lim \limits _{n\rightarrow +\infty }\lambda _{n,m,0} = -1,\quad \quad \text { for any fixed } m>0. \end{aligned}$$

For more details about the location of $\lambda _{n,m,0}$ and further discussion, see [18, 19].