5.3. The case j k ≠ 0 : uniform estimates and asymptotics [053W]

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5.3. The case $j_{k}\neq 0$ : uniform estimates and asymptotics

In this subsection, we consider the case $j_{k}\neq 0$ of the homogeneous equation

(5.76)

\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=0.\ z\geq 1,

Under the change of variables given by (5.21) and (5.25), the above equation is transformed into the confluent hypergeometric equation,

(5.77)

y\cdot\frac{d^{2}\mathcal{J}(y)}{dy^{2}}+(\fa-y)\cdot\frac{d\mathcal{J}(y)}{dy}-\fb\cdot\mathcal{J}(y)=0,\ y<0,

where

(5.78)

\displaystyle\begin{cases}\fa=1-\frac{1}{n}\\ \fb=\frac{1}{2}(1-\frac{1}{n})-\frac{\lambda_{k}}{j_{k}\cdot n}.\end{cases}

Since we have shown in Section 5.1 that $\lambda_{k}\geq\frac{j_{k}(n-1)}{2}$ , we have that

(5.79)

\beta\leq 0\ \text{and}\ \ \alpha-\beta\geq 1-\frac{1}{n}>0.

According to the discussion in Appendix A, in our case $y<0$ , the confluent hypergeometric equation (5.77) has two linearly independent solutions

(5.80)

\Ku(\beta,\alpha,y)\equiv\sum\limits_{k=0}^{\infty}\frac{(\beta)_{k}}{(\alpha)_{k}}\cdot\frac{y^{k}}{k!}

and

(5.81)

\Tri(\beta,\alpha,y)\equiv\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{yt}t^{\fa-\fb-1}(1+t)^{\fb-1}dt.

By Item (3) of Lemma A.3, as $y\to-\infty$ , $\Tri(y)$ is a decaying solution to (5.77) for every $\alpha>\beta$ , while Lemma A.5 shows that, in the case $\beta<0$ , the solution $\Ku(y)$ is growing of certain polynomial rate as $y\to-\infty$ . These then yield two linearly independent solutions to the homogeneous equation (5.76),

(5.82)

\begin{cases}\mathcal{G}_{k}(z)=e^{\frac{j_{k}z^{n}}{2}}\cdot\Ku(\beta,\alpha,-j_{k}z^{n}),\\ \mathcal{D}_{k}(z)=e^{\frac{j_{k}z^{n}}{2}}\cdot\Tri(\beta,\alpha,-j_{k}z^{n}).\end{cases}

First we can compute the Wronskian

Proposition 5.6.

For every $k\in\mathbb{N}$ , the Wronskian of $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ is a constant given by

(5.83)

\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot j_{k}^{\frac{1}{n}}.

Proof.

Since $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ solve the homogeneous equation

(5.84)

\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=0

which misses the first order term. Immediately, for all $z\geq 0$ ,

(5.85)

\frac{d}{dz}\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=0,

which implies that the Wronskian $\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))$ is a constant. So it suffices to calculate it at $z=0$ . By the definition of the Wronskian,

	$\displaystyle\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=$	$\displaystyle e^{j_{k}z^{n}}\cdot\Big(\Ku(\beta,\alpha,-j_{k}z^{n})\cdot\frac{d}{dz}\Tri(\beta,\alpha,-j_{k}z^{n})$
(5.86)			$\displaystyle-\frac{d}{dz}\Ku(\beta,\alpha,-j_{k}z^{n})\cdot\Tri(\beta,\alpha,-j_{k}z^{n})\Big).$

To calculate $\frac{d}{dz}\Tri(\beta,\alpha,-j_{k}z^{n})$ , we will apply Kummer’s transformation law to relate $\Tri$ and $\Ku$ , that is,

(5.87)		$\displaystyle\Tri(\beta,\alpha,-j_{k}z^{n})$
	$\displaystyle=$	$\displaystyle e^{-{j_{k}}z^{n}}\cdot\mathcal{U}(\alpha-\beta,\alpha,{j_{k}}z^{n})$
	$\displaystyle=$	$\displaystyle e^{-{j_{k}}z^{n}}\cdot\Big(\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\Ku(\alpha-\beta,\alpha,{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot({j_{k}}z^{n})^{1-\alpha}\Ku(1-\beta,2-\alpha,{j_{k}}z^{n})\Big)$
	$\displaystyle=$	$\displaystyle e^{-{j_{k}}z^{n}}\cdot\Big(\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\Ku(\alpha-\beta,\alpha,{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}z\cdot\Ku(1-\beta,2-\alpha,{j_{k}}z^{n})\Big)$
	$\displaystyle=$	$\displaystyle\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\Ku(\beta,\alpha,-{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}z\cdot\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n}).$

So it follows that

(5.88)		$\displaystyle\frac{d}{dz}\Tri(\beta,\alpha,-{j_{k}}z^{n})$
	$\displaystyle=$	$\displaystyle\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\frac{d}{dz}\Ku(\beta,\alpha,-{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}\cdot\Big(\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n})$
	$\displaystyle+$	$\displaystyle z\cdot\frac{d}{dz}\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n})\Big).$

Since $n\geq 2$ , it directly follows from the definition of $\Ku$ that

(5.89)		$\displaystyle\frac{d}{dz}\Big\|_{z=0}\Ku(\beta,\alpha,-{j_{k}}z^{n})=0,$
(5.90)		$\displaystyle\frac{d}{dz}\Big\|_{z=0}\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n})=0.$

Therefore,

(5.91)		$\displaystyle\frac{d}{dz}\Big\|_{z=0}\Tri(\beta,\alpha,-{j_{k}}z^{n})$	$\displaystyle=$	$\displaystyle\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}\cdot\Ku(1-\alpha+\beta,2-\alpha,0)$
(5.91)			$\displaystyle=$	$\displaystyle\frac{\Gamma(\alpha-1)\cdot{j_{k}}^{\frac{1}{n}}}{\Gamma(\alpha-\beta)}.$

Now evaluate (5.86) at $z=0$ , we have

(5.92)

\mathcal{W}(\mathcal{G}_{k},\mathcal{D}_{k})(z)=\mathcal{W}(\mathcal{G}_{k},\mathcal{D}_{k})(0)=\frac{d}{dz}\Big|_{z=0}\Tri(\beta,\alpha,-{j_{k}}z^{n})=\frac{\Gamma(\alpha-1)\cdot{j_{k}}^{\frac{1}{n}}}{\Gamma(\alpha-\beta)}.

∎

Applying Lemma A.3 and Lemma A.5, immediately we have the following asymptotics for the solutions $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ for fixed $k$ .

Lemma 5.7.

For each fixed $k$ , as $z\to+\infty$ , we have

(5.93)		$\displaystyle\mathcal{G}_{k}(z)$	$\displaystyle\sim\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}\cdot(j_{k}z^{n})^{-\beta}\cdot e^{\frac{j_{k}z^{n}}{2}},$
(5.94)		$\displaystyle\mathcal{D}_{k}(z)$	$\displaystyle\sim(j_{k}z^{n})^{\beta-\alpha}\cdot e^{-\frac{j_{k}z^{n}}{2}}.$

Again we need to derive uniform estimates and asymptotic behavior for $\Ku$ and $\Tri$ . The idea is to first estimate them in terms of certain integrals and then apply Laplace’s method. To start with, we need some preliminary calculations for $\Ku$ and $\Tri$ .

By definition,

(5.95)		$\displaystyle\Tri(\fb,\fa,y)$	$\displaystyle=$	$\displaystyle\frac{e^{y}}{\Gamma(\alpha-\beta)}\int_{0}^{\infty}e^{yt+(\alpha-\beta-1)\log t+(\beta-1)\log(t+1)}dt$
(5.95)			$\displaystyle=$	$\displaystyle\frac{e^{y}}{\Gamma(\alpha-\beta)}\int_{0}^{\infty}e^{yt+(\alpha-\beta-1)\log\frac{t}{t+1}}\cdot\frac{1}{(1+t)^{1+\frac{1}{n}}}dt.$

For simplicity, we denote

(5.96)

F(t)\equiv yt+(\fa-\fb-1)\log\frac{t}{t+1},

then

(5.97)

\Tri(\fb,\fa,y)=\frac{e^{y}}{\Gamma(\alpha-\beta)}\int_{0}^{\infty}e^{F(t)}\cdot\frac{1}{(1+t)^{1+\frac{1}{n}}}dt.

Now we give both upper and lower bounds for $\Ku(\fb,\fa,y)$ by simpler exponential integrals.

Lemma 5.8.

Let $y\leq-1$ , then following holds,

(5.98)

C_{n}^{-1}\cdot\frac{e^{y}(-y)^{\frac{1-2\alpha}{4}}}{\Gamma(\alpha-\beta)}\cdot\int_{\frac{1}{\sqrt{-y}}}^{\infty}e^{G(u)}du\leq\Ku(\fb,\fa,y)\leq C_{n}\cdot\frac{e^{y}(-y)^{\frac{1-2\alpha}{4}}}{\Gamma(\alpha-\beta)}\cdot\int_{0}^{\infty}e^{G(u)}du,

where

(5.99)

G(u)\equiv-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u.

Proof.

To prove this estimate, we need the following integral representation formula for $\Ku(\fb,\fa,y)$ ,

(5.100)

\Ku(\fb,\fa,y)=\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot e^{y}(-y)^{\frac{1-\fa}{2}}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt.

The proof is included in Lemma A.2 of Appendix A.

The key point in the proof of (5.98) is to apply the estimate of $I_{\alpha-1}$ in Proposition 5.5. By definition, $\alpha=1-\frac{1}{n}$ and hence $\alpha-1=-\frac{1}{n}\geq-\frac{1}{2}$ . Applying the upper bound estimate of $I_{\alpha-1}$ in (5.48) of Proposition 5.5,

(5.101)		$\displaystyle I_{\alpha-1}(2\sqrt{-yt})=I_{-\frac{1}{n}}(2\sqrt{-yt})$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot\max\Big\{(2\sqrt{-yt})^{-\frac{1}{n}},(2\sqrt{-yt})^{-\frac{1}{2}}\cdot e^{2\sqrt{-yt}}\Big\}$
(5.101)			$\displaystyle\leq$	$\displaystyle C_{n}\cdot(-yt)^{-\frac{1}{4}}\cdot e^{2\sqrt{-yt}}.$

Substituting the above in (5.100),

(5.102)	$\displaystyle\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot\int_{0}^{\infty}e^{-t+2\sqrt{-yt}}\cdot t^{\frac{2\alpha-3}{4}-\beta}dt$
		$\displaystyle=$	$\displaystyle C_{n}\cdot\int_{0}^{\infty}e^{-t+2\sqrt{-yt}+(\frac{2\alpha-3}{4}-\beta)\log t}dt$
		$\displaystyle=$	$\displaystyle C_{n}\cdot\int_{0}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.$

Therefore,

(5.103)

\displaystyle\Ku(\fb,\fa,y)

\displaystyle\leq

\displaystyle C_{n}\cdot\frac{e^{y}(-y)^{\frac{1-2\alpha}{4}}}{\Gamma(\alpha-\beta)}\cdot\int_{0}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.

Next, $\Ku$ can be also bounded below in a similar way. In fact, we consider the integral domain $t\geq\frac{1}{-y}$ with $y\leq-1$ , then

(5.104)

I_{\alpha-1}(2\sqrt{-yt})\geq C_{n}^{-1}\cdot\frac{e^{2\sqrt{-yt}}}{(-yt)^{\frac{1}{4}}},

and hence

	$\displaystyle\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt$	$\displaystyle\geq\int_{\frac{1}{-y}}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt$
		$\displaystyle\geq C_{n}^{-1}\cdot\int_{\frac{1}{-y}}^{\infty}e^{-t+2\sqrt{-yt}+(\frac{2\alpha-3}{4}-\beta)\log t}dt$
(5.105)			$\displaystyle=C_{n}^{-1}\cdot\int_{\frac{1}{\sqrt{-y}}}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.$

Therefore,

(5.106)

\Ku(\fb,\fa,y)\geq C_{n}^{-1}\cdot\frac{e^{y}(-y)^{\frac{1-2\alpha}{4}}}{\Gamma(\alpha-\beta)}\cdot\int_{\frac{1}{\sqrt{-y}}}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.

∎

Now we set up a few notations for convenience. Let

(5.107)

Q\equiv\alpha-\beta-1\geq-\frac{1}{n},\ \gamma_{n}\equiv\frac{1}{2}+\frac{1}{n},

and recall the notations (5.96) and (5.99),

(5.108)		$\displaystyle F(t)$	$\displaystyle=yt+Q\log\frac{t}{t+1},$
(5.109)		$\displaystyle G(u)$	$\displaystyle=-u^{2}+2(-y)^{\frac{1}{2}}\cdot u+(2Q+\gamma_{n})\cdot\log u.$

By direct calculation

(5.110)		$\displaystyle F^{\prime\prime}(t)$	$\displaystyle=Q(-\frac{1}{t^{2}}+\frac{1}{(t+1)^{2}}),$
(5.111)		$\displaystyle G^{\prime\prime}(u)$	$\displaystyle=-2-\frac{2Q+\gamma_{n}}{u}.$

Notice that $2Q+\gamma_{n}\geq\frac{1}{2}-\frac{1}{n}\geq 0$ . Therefore, $G(u)$ is strictly concave in $\mathbb{R}_{+}$ , and $F$ is strictly concave in $\mathbb{R}$ if $Q>0$ .

We will split our analysis in two different cases:

Case (A): $Q\geq 1$ .

Case (B): $Q\leq 1$ .

Our main focus is Case (A) which is more difficult. The upper bound estimates in Case (B) follows from elementary integral calculations (see Lemma 5.12).

Case (A)

Let $t_{0}>0$ be the unique critical point of $F(t)$ and let $u_{0}>0$ be the unique critical point of $G(u)$ , then $t_{0}$ and $u_{0}$ satisfy the equations

(5.112)		$\displaystyle t_{0}^{2}+t_{0}+\frac{Q}{y}=0,$
(5.113)		$\displaystyle u_{0}^{2}-(-y)^{\frac{1}{2}}\cdot u_{0}-\frac{2Q+\gamma_{n}}{2}=0.$

Immediately we have

(5.114)		$\displaystyle t_{0}$	$\displaystyle=\frac{-1+\sqrt{1+\frac{4Q}{-y}}}{2},$
(5.115)		$\displaystyle u_{0}$	$\displaystyle=\frac{(-y)^{\frac{1}{2}}}{2}\cdot\Big(1+\sqrt{1+\frac{4Q}{-y}+\frac{2\gamma_{n}}{-y}}\Big).$

Now prove the following effective estimates on $\Ku$ and $\Tri$ . The difference from Lemma 5.7 is here the estimates holds uniformly for all $\beta\leq 0$ (recall $\alpha$ is the fixed number $1-\frac{1}{n}$ ).

Proposition 5.9.

There exists some dimensional constant $C_{n}>0$ such that for every $y\leq-1$ , the following estimates hold:

(5.116)		$\displaystyle C_{n}^{-1}\cdot Q^{-\frac{1}{4}-\frac{1}{2n}}\cdot\frac{(-y)^{-1}\cdot e^{y+F(t_{0})}}{\Gamma(Q+1)}$	$\displaystyle\leq\Tri(\beta,\alpha,y)\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot\frac{e^{y+F(t_{0})}}{\Gamma(Q+1)},$
(5.117)		$\displaystyle C_{n}^{-1}\cdot Q^{-\frac{1}{4}}\cdot\frac{(-y)^{\frac{1-2\alpha}{4}}\cdot e^{y+G(u_{0})}}{\Gamma(Q+1)}$	$\displaystyle\leq\Ku(\beta,\alpha,y)\leq C_{n}\cdot\frac{(-y)^{\frac{1-2\alpha}{4}}\cdot e^{y+G(u_{0})}}{\Gamma(Q+1)}.$

Proof.

Our main strategy is to apply Laplace’s method. The basic idea is that the above exponential integrals are concentrated at the critical values $t_{0}$ and $u_{0}$ .

First, we prove the uniform estimate for $\Tri(\fb,\fa,y)$ . By (5.97),

(5.118)

\Tri(\fb,\fa,y)\leq\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{F(t)}dt.

Clearly, the upper bound of $\Tri(\fb,\fa,y)$ follows from the upper bound estimate of $\int_{0}^{\infty}e^{F(t)}dt$ . Write

(5.119)

\int_{0}^{\infty}e^{F(t)}dt=\int_{0}^{2t_{0}}e^{F(t)}dt+\int_{2t_{0}}^{\infty}e^{F(t)}dt.

We will estimate the two terms separately.

To estimate the first term in (5.119), we make a change of variable

(5.120)

t=t_{0}\cdot(1+\xi),\ \xi\in(-1,1),

then Taylor’s theorem gives that

(5.121)	$\displaystyle F(t)-F(t_{0})$	$\displaystyle=$	$\displaystyle F(t_{0}(1+\xi))-F(t_{0})$
		$\displaystyle=$	$\displaystyle F^{\prime}(t_{0})\cdot t_{0}\cdot\xi+\frac{F^{\prime\prime}(\theta)}{2}\cdot t_{0}^{2}\cdot\xi^{2}$
		$\displaystyle=$	$\displaystyle\frac{F^{\prime\prime}(\theta)}{2}\cdot t_{0}^{2}\cdot\xi^{2},$

where $\theta$ is between $t$ and $t_{0}$ . Now we need to estimate the quadratic error term. It is straightforward calculation that

(5.122)

\displaystyle F^{\prime\prime\prime}(t)

\displaystyle=2(\frac{Q}{t^{3}}-\frac{Q}{(t+1)^{3}})>0,

then $F^{\prime\prime}(t)$ is increasing in $t$ . Since $\theta$ is between $t_{0}$ and $t\in[0,2t_{0}]$ , the above monotonicity of $F^{\prime\prime}$ implies $F^{\prime\prime}(\theta)\leq F^{\prime\prime}(2t_{0})<0$ . So the first term of (5.119) becomes

(5.123)		$\displaystyle\int_{0}^{2t_{0}}e^{F(t)}dt$	$\displaystyle=$	$\displaystyle e^{F(t_{0})}\int_{0}^{2t_{0}}e^{F(t)-F(t_{0})}dt$
(5.123)			$\displaystyle\leq$	$\displaystyle e^{F(t_{0})}\cdot t_{0}\cdot\int_{-1}^{1}e^{\frac{F^{\prime\prime}(2t_{0})}{2}\cdot t_{0}^{2}\cdot\xi^{2}}d\xi$

By direct computations, $F^{\prime\prime}(2t_{0})=-\frac{(4t_{0}+1)}{4t_{0}^{2}(2t_{0}+1)^{2}}\cdot Q$ . So we have,

(5.124)	$\displaystyle\int_{0}^{2t_{0}}e^{F(t)}dt$	$\displaystyle\leq$	$\displaystyle e^{F(t_{0})}\cdot t_{0}\cdot\int_{-1}^{1}e^{-\frac{4t_{0}+1}{8(2t_{0}+1)^{2}}\cdot Q\cdot\xi^{2}}d\xi$
		$\displaystyle\leq$	$\displaystyle C_{n}\cdot\frac{t_{0}(2t_{0}+1)}{\sqrt{4t_{0}+1}\cdot\sqrt{Q}}\cdot e^{F(t_{0})}$
		$\displaystyle\leq$	$\displaystyle C_{n}\cdot Q^{\frac{1}{4}}\cdot e^{F(t_{0})},$

where we used that $t_{0}\leq C_{n}\cdot Q^{1/2}$ (since $y\leq-1$ and $Q\geq 1$ ). Immediately, we have

(5.125)

\Tri(\beta,\alpha,y)\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot\frac{e^{y+F(t_{0})}}{\Gamma(\alpha-\beta)}.

Next, we estimate the second term in (5.119). Since we have proved $F^{\prime\prime}(t)<0$ , so this implies that $F^{\prime}(t)$ is decreasing and hence $F^{\prime}(t)\leq F^{\prime}(2t_{0})$ for any $t\geq 2t_{0}$ . Now Taylor’s theorem gives that

(5.126)

F(t)\leq F(2t_{0})+F^{\prime}(2t_{0})\cdot(t-2t_{0}),

which implies that

(5.127)

\int_{2t_{0}}^{\infty}e^{F(t)}dt\leq e^{F(2t_{0})}\int_{2t_{0}}^{\infty}e^{F^{\prime}(2t_{0})\cdot(t-2t_{0})}dt=\frac{e^{F(2t_{0})}}{-F^{\prime}(2t_{0})}.

One can check that $F^{\prime}(2t_{0})=\frac{y(3t_{0}+1)}{2(2t_{0}+1)}<0$ with $0<t_{0}<+\infty$ . Since $F^{\prime}(t)<0$ for all $t>t_{0}$ , so $F(2t_{0})\leq F(t_{0})$ and hence for $y\leq-1$ we have

(5.128)

\int_{2t_{0}}^{\infty}e^{F(t)}dt\leq C_{n}e^{F(t_{0})}.

Combining the above, we have

(5.129)

\int_{0}^{\infty}e^{F(t)}dt\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot e^{F(t_{0})}.

Therefore,

(5.130)

\displaystyle\Tri(\beta,\alpha,y)

\displaystyle\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot\frac{e^{y+F(t_{0})}}{\Gamma(\alpha-\beta)}.

The lower bound estimate for $\Tri(\fb,\fa,y)$ also follows from Laplace’s method and we just sketch the computations.

	$\displaystyle\Tri(\fb,\fa,y)$	$\displaystyle=\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{F(t)}\cdot\frac{1}{(t+1)^{1+\frac{1}{n}}}dt$
		$\displaystyle\geq\frac{e^{y}}{\Gamma(\alpha-\beta)}\int_{t_{0}(y)}^{2t_{0}(y)}e^{F(t)}\cdot\frac{1}{(t+1)^{1+\frac{1}{n}}}dt$
(5.131)			$\displaystyle\geq\frac{e^{y}}{\Gamma(\alpha-\beta)\cdot(1+2t_{0})^{1+\frac{1}{n}}}\int_{t_{0}(y)}^{2t_{0}(y)}e^{F(t)}dt.$

By the concavity of $F(t)$ and the monotonicity of $F^{\prime\prime}(t)$ in the domain $t_{0}\leq t\leq 2t_{0}$ , we have

(5.132)

\int_{t_{0}(y)}^{2t_{0}(y)}e^{F(t)}dt\geq e^{F(t_{0})}\int_{t_{0}(y)}^{2t_{0}(y)}e^{\frac{F^{\prime\prime}(t_{0})}{2}(t-t_{0})^{2}}dt\geq C_{n}\cdot e^{F(t_{0})}\frac{t_{0}(t_{0}+1)}{\sqrt{2t_{0}+1}\cdot\sqrt{Q}}

It is elementary to see that

(5.133)

C_{n}Q^{\frac{1}{2}}(-y)^{-1}\leq t_{0}\leq C_{n}\cdot{Q^{\frac{1}{2}}}

Therefore,

(5.134)

\Tri(\fb,\fa,y)\geq C_{n}\cdot Q^{-\frac{1}{4}-\frac{1}{2n}}\cdot\frac{e^{y}\cdot(-y)^{-1}}{\Gamma(\alpha-\beta)}\cdot e^{F(t_{0})}.

The uniform estimate for $\Ku(\fb,\fa,y)$ stated in (5.117) can be proved in the same way. One just needs to apply Laplace’s method to the integral estimate formula in Lemma 5.8. We can eventually obtain

(5.135)

C_{n}^{-1}\cdot Q^{-\frac{1}{4}}\cdot e^{G(u_{0})}\leq\int_{0}^{\infty}e^{G(u)}du\leq C_{n}\cdot e^{G(u_{0})}.

We omit the computations here.

∎

Converting into the variables $z$ , we obtain

Corollary 5.9.1.

There exists $C_{n}>0$ such that for all $z\geq 1$ , we have

(5.136)		$\displaystyle C_{n}^{-1}\cdot\frac{Q^{-\frac{1}{4}-\frac{1}{2n}}}{\Gamma(Q+1)}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}+F(t_{0}(z))}\cdot(j_{k}z^{n})^{-1}$	$\displaystyle\leq\mathcal{D}_{k}(z)\leq C_{n}\cdot\frac{Q^{\frac{1}{4}}}{\Gamma(Q+1)}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}+F(t_{0}(z))},$
(5.137)		$\displaystyle C_{n}^{-1}\cdot Q^{-\frac{1}{4}}\cdot\frac{(j_{k}\cdot z^{n})^{\frac{1-2\alpha}{4}}}{\Gamma(Q+1)}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}+G(u_{0}(z))}$	$\displaystyle\leq\mathcal{G}_{k}(z)\leq C_{n}\cdot\frac{(j_{k}\cdot z^{n})^{\frac{1-2\alpha}{4}}}{\Gamma(Q+1)}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}+G(u_{0}(z))},$

where $Q\equiv\alpha-\beta-1\geq 1$ .

The next Proposition essentially gives an estimate of the product of $\Ku$ and $\Tri$ .

Proposition 5.10.

There exists some dimensional constant $C_{n}>0$ such that for any $y\leq-1$ , we have

(5.138)

e^{F(t_{0})+G(u_{0})}\leq C_{n}(-y)^{\frac{\gamma_{n}}{2}}e^{-y}e^{-Q}Q^{Q+\frac{\gamma_{n}}{2}}.

In particular we have

(5.139)

\Tri\cdot\Ku\leq C_{n}\cdot\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)^{2}}(-y)^{\frac{1}{n}}Q^{Q}e^{-Q}e^{y}.

Proof.

The calculation in the proof is purely elementary. The order estimate involving the parameter $Q$ will be used at crucial places for our later estimates, so we include the detailed proof. Plugging the critical points formulae (5.114) and (5.115) into the expression of $F$ and $G$ ,

(5.140)

F(t_{0})+G(u_{0})=yt_{0}+(-y)^{\frac{1}{2}}u_{0}-\frac{2Q+\gamma_{n}}{2}+Q\log\frac{t_{0}}{t_{0}+1}+(2Q+\gamma_{n})\log u_{0},

where $Q\equiv\alpha-\beta-1$ and $\gamma_{n}\equiv\frac{1}{2}+\frac{1}{n}$ as before.

First, it is straightforward that

(5.141)

yt_{0}+(-y)^{\frac{1}{2}}u_{0}\leq(-y)+\frac{\gamma_{n}}{2}.

So this implies that

(5.142)	$\displaystyle e^{F(t_{0})+G(u_{0})}$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot e^{-y}\cdot e^{-Q}\cdot\Big(\frac{t_{0}}{1+t_{0}}\Big)^{Q}\cdot u_{0}^{2Q+\gamma_{n}}$
		$\displaystyle=$	$\displaystyle C_{n}\cdot e^{-y}\cdot e^{-Q}\cdot\Big(\frac{t_{0}^{2}}{t_{0}(1+t_{0})}\Big)^{Q}\cdot u_{0}^{2Q+\gamma_{n}}$
		$\displaystyle=$	$\displaystyle C_{n}\cdot e^{-y}\cdot u_{0}^{\gamma_{n}}\cdot e^{-Q}\cdot\frac{(u_{0}t_{0})^{2Q}}{(\frac{Q}{-y})^{Q}},$

where the last equality follows from (5.112).

Now we claim

(5.143)

u_{0}t_{0}\leq(-y)^{-\frac{1}{2}}\cdot(Q+\frac{\gamma_{n}}{2}).

To prove this, we denote $\tau\equiv\frac{2\gamma_{n}}{-y}>0$ and $\widehat{Q}\equiv\frac{4Q}{-y}>0$ . Then using the critical point formulae of $u_{0}$ and $t_{0}$ given by (5.114) and (5.115), we obtain

(5.144)		$\displaystyle u_{0}t_{0}$
	$\displaystyle=$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot\Big(1+\sqrt{1+\widehat{Q}}\Big)\cdot\Big(-1+\sqrt{1+\widehat{Q}+\tau}\Big)$
	$\displaystyle=$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot\Big(-1+\sqrt{1+\widehat{Q}}\cdot\sqrt{1+\widehat{Q}+\tau}+\sqrt{1+\widehat{Q}}-\sqrt{1+\widehat{Q}+\tau}\Big)$
	$\displaystyle\leq$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot\Big(-1+\sqrt{1+\widehat{Q}+\tau}\cdot\sqrt{1+\widehat{Q}+\tau}+\sqrt{1+\widehat{Q}+\tau}-\sqrt{1+\widehat{Q}+\tau}\Big)$
	$\displaystyle=$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot(\widehat{Q}+\tau)$
	$\displaystyle=$	$\displaystyle(-y)^{-\frac{1}{2}}\cdot(Q+\frac{\gamma_{n}}{2}).$

Then it follows that

(5.145)

\frac{(u_{0}t_{0})^{2Q}}{(\frac{Q}{-y})^{Q}}\leq\frac{(Q+\frac{\gamma_{n}}{2})^{2Q}}{Q^{Q}}=Q^{Q}\cdot(1+\frac{\gamma_{n}}{2Q})^{2Q}\leq e^{\gamma_{n}}\cdot Q^{Q}.

Moreover, we notice that

(5.146)

u_{0}^{\gamma_{n}}\leq C_{n}\cdot Q^{\frac{\gamma_{n}}{2}}\cdot(-y)^{\frac{\gamma_{n}}{2}}.

Therefore, combining all the above, we have

(5.147)		$\displaystyle e^{F(t_{0})+G(u_{0})}$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot e^{-y}\cdot u_{0}^{\gamma_{n}}\cdot e^{-Q}\cdot\frac{(u_{0}t_{0})^{2Q}}{(\frac{Q}{-y})^{Q}}$
(5.147)			$\displaystyle\leq$	$\displaystyle C_{n}\cdot(-y)^{\frac{\gamma_{n}}{2}}\cdot e^{-y}\cdot e^{-Q}\cdot Q^{Q+\frac{\gamma_{n}}{2}}.$

∎

In the next subsections, we will also need the following monotonicity formula to study the integral estimates for the above fundamental solutions $\mathcal{G}_{k}$ and $\mathcal{D}_{k}$ .

Lemma 5.11.

Let

(5.148)		$\displaystyle\widehat{F}(z)$	$\displaystyle\equiv-\frac{jz^{n}}{2}+F(t_{0}(z)),$
(5.149)		$\displaystyle\widehat{G}(z)$	$\displaystyle\equiv-\frac{jz^{n}}{2}+G(u_{0}(z)),$

then for all $\eta\geq 0$ , when $z\geq\eta^{\frac{2}{n}}$ , $\widehat{F}(z)+\eta\cdot z^{\frac{n}{2}}$ is decreasing and $\widehat{G}(z)-\eta\cdot z^{\frac{n}{2}}$ is increasing.

Proof.

Let $y=-jz^{n}$ , then it is straightforward that

(5.150)

\displaystyle\frac{d\widehat{F}(y)}{dy}=\frac{1}{2}+t_{0}(y)+F^{\prime}(t_{0}(y))\cdot\frac{dt_{0}(y)}{dy}=\frac{1}{2}+t_{0}(y)=\frac{1}{2}\sqrt{1+\frac{4Q}{-y}}\geq\frac{1}{2}.

This implies that, as $z\geq\eta^{\frac{2}{n}}$ ,

(5.151)

\displaystyle\frac{d(\widehat{F}(z)+\eta z^{\frac{n}{2}})}{dz}=\frac{d\widehat{F}(y)}{dy}\cdot(-nj\cdot z^{n-1})+\frac{n\cdot\eta}{2}\cdot z^{\frac{n}{2}-1}\leq-\frac{n}{2}\cdot z^{\frac{n}{2}-1}(j\cdot z^{\frac{n}{2}}-\eta)\leq 0.

By similar calculations, one can also obtain that $\widehat{G}(z)-\eta\cdot z^{\frac{n}{2}}$ is increasing as $z\geq\eta^{\frac{2}{n}}$ .

∎

Case (B): Now we consider the case when $Q\leq 1$ . As mentioned in the above, this case is easier.

Lemma 5.12.

Let $Q\leq 1$ , then there is some dimensional constant $C_{n}>0$ such that

(5.152)		$\displaystyle C_{n}^{-1}\cdot e^{y}\cdot(-y)^{\beta-\alpha}$	$\displaystyle\leq\Tri(\fb,\fa,y)\leq e^{y}\cdot(-y)^{\beta-\alpha},$
(5.153)		$\displaystyle C_{n}^{-1}\cdot(-y)^{-\beta}$	$\displaystyle\leq\Ku(\fb,\fa,y)\leq C_{n}\cdot(-y)^{-\beta}.$

for all $y\leq-1$ .

Remark 5.12.1.

In the case $Q\leq 1$ , the estimate is optimal in the sense that it coincides with the asymptotic behavior of $\Tri$ and $\Ku$ for fixed $\alpha$ and $\beta$ , as given in Lemma A.3 and Lemma A.5.

Proof.

First, we prove (5.152). Both the upper bound and lower bound estimates can be proved in the similar way:

	$\displaystyle\Tri(\beta,\alpha,y)$	$\displaystyle=\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{yt}t^{\fa-\fb-1}(1+t)^{\fb-1}dt$
		$\displaystyle\leq\frac{e^{y}}{\Gamma(\fa-\fb)}\cdot\int_{0}^{\infty}e^{yt}t^{\fa-\fb-1}dt$
		$\displaystyle=\frac{e^{y}\cdot(-y)^{\beta-\alpha}}{\Gamma(\fa-\fb)}\cdot\int_{0}^{\infty}e^{-u}u^{\fa-\fb-1}du$
(5.154)			$\displaystyle=e^{y}\cdot(-y)^{\beta-\alpha}.$

Similarly,

	$\displaystyle\Tri(\fb,\fa,y)$	$\displaystyle\geq\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{1}e^{yt}t^{\fa-\fb-1}(1+t)^{\fb-1}dt$
		$\displaystyle\geq C_{n}\cdot e^{y}\int_{0}^{1}e^{yt}t^{\fa-\fb-1}dt$
(5.155)			$\displaystyle\geq C_{n}\cdot e^{y}\cdot(-y)^{\fb-\fa}.$

Next, we prove the upper bound estimate for $\Ku$ . Notice in the proof of Lemma 5.9 we do not need the condition $Q\leq 1$ for the upper bound on $\Ku$ . So we have

(5.156)

\Ku(\beta,\alpha,y)\leq C_{n}\cdot\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}\cdot(-y)^{\frac{1-2\alpha}{4}}\cdot e^{y+G(u_{0})}.

To prove (5.153), we need an upper bound estimate for $e^{y+G(u_{0})}$ . This follows from elementary computations. In fact,

e^{y+G(u_{0})}=e^{y-u_{0}^{2}+2\sqrt{-y}u_{0}}\cdot(u_{0})^{2Q+\gamma_{n}}\\ \leq C_{n}\cdot e^{y-u_{0}^{2}+2\sqrt{-y}u_{0}}\cdot(-y)^{Q+\frac{\gamma_{n}}{2}}.

Notice that $u_{0}$ satisfies $G^{\prime}(u_{0})=0$ , i.e.,

(5.157)

u_{0}^{2}-\sqrt{-y}\cdot u_{0}-\frac{2Q+\gamma_{n}}{2}=0,

so we have

(5.158)

e^{y+G(u_{0})}\leq C_{n}\cdot e^{y+\sqrt{-y}u_{0}}\cdot(-y)^{Q+\frac{\gamma_{n}}{2}}.

By (5.115), it is straightforward that

(5.159)

\displaystyle y+\sqrt{-y}u_{0}=\frac{y}{2}\Big(1-\sqrt{1+\frac{4Q+2\gamma_{n}}{-y}}\Big)=\frac{2Q+\gamma_{n}}{1+\sqrt{1+\frac{4Q+2\gamma_{n}}{-y}}}\in[C_{n}^{-1},C_{n}],

for some dimensional constant $C_{n}>0$ . Therefore,

(5.160)

\displaystyle e^{y+G(u_{0})}\leq C_{n}(-y)^{Q+\frac{1}{4}+\frac{1}{2n}},

and hence

(5.161)

\displaystyle\Ku(\fb,\fa,y)\leq C_{n}(-y)^{Q+\frac{1}{n}}=C_{n}(-y)^{-\beta}.

This completes the proof. ∎

Converting into the variables $z$ we obtain

Corollary 5.12.1.

There exists $C_{n}>0$ such that for all $z\geq 1$ , we have

(5.162)		$\displaystyle C_{n}^{-1}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}}\cdot(j_{k}z^{n})^{\beta-\alpha}$	$\displaystyle\leq\mathcal{D}_{k}(z)\leq C_{n}\cdot e^{-\frac{j_{k}\cdot z^{n}}{2}}\cdot(j_{k}z^{n})^{\beta-\alpha},$
(5.163)		$\displaystyle C_{n}^{-1}\cdot e^{\frac{j_{k}\cdot z^{n}}{2}}\cdot(j_{k}z^{n})^{-\beta}$	$\displaystyle\leq\mathcal{G}_{k}(z)\leq C_{n}\cdot e^{\frac{j_{k}\cdot z^{n}}{2}}\cdot(j_{k}z^{n})^{-\beta}.$

We end this subsection by making some remarks regarding the above estimates on $\Ku$ and $\Tri$ . Notice that in the case $Q\equiv\fa-\fb-1\leq 1$ we applied Laplace’s method to turn the problem into estimates on exponential integrals. One may wonder how far the uniform estimates in Lemma 5.9 is from optimal comparing to the non-uniform estimate with the optimal order in Lemma 5.12. We can consider two extreme cases depending on the size of $Q$ compared with $-y$ .

First we assume $\frac{Q^{2}}{-y}\ll 1$ , which obviously includes the case when we fix $Q$ and let $y\rightarrow-\infty$ . Then by definition we see that

(5.164)

t_{0}=\frac{Q}{-y}+O\Big((\frac{Q}{-y})^{2}\Big),

and we get

(5.165)

F(t_{0})=yt_{0}+Q\log\frac{t_{0}}{t_{0}+1}=-Q+Q\log Q-Q\log(-y)+O(\frac{Q}{-y}).

So by Lemma 5.9 we get

(5.166)

C_{n}^{-1}Q^{-\frac{1}{4}-\frac{1}{2n}}e^{y}(-y)^{-Q-1}Q^{Q}e^{-Q}\leq\Tri\leq C_{n}\frac{1}{\Gamma(\alpha-\beta)}e^{y}(-y)^{-Q}Q^{Q}e^{-Q}Q^{\frac{1}{4}}.

Notice by Stirling’s formula for $Q$ large $\Gamma(\alpha-\beta)=Q\Gamma(Q)$ is comparable to $C_{n}Q^{\frac{3}{2}}Q^{Q}e^{-Q}$ . So up to polynomial errors in $Q$ this estimate is optimal comparing with (A.27). Similarly, we have

(5.167)

u_{0}=(-y)^{\frac{1}{2}}\Big(1+\frac{Q+\frac{1}{2}\gamma_{n}}{-y}+O((\frac{Q}{-y})^{2})\Big),

and

(5.168)

G(u_{0})=-u_{0}^{2}+2(-y)^{\frac{1}{2}}u_{0}+(2Q+\gamma_{n})\log u_{0}\leq C_{n}e^{-y}(-y)^{Q+\frac{\gamma_{n}}{2}}.

(5.169)

\Ku\leq C_{n}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}(-y)^{\frac{1-2\alpha}{4}}(-y)^{Q+\frac{\gamma_{n}}{2}}=C_{n}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}(-y)^{-\beta},

which is again optimal comparing with (A.34).

Secondly we assume the other extreme $\frac{Q}{(-y)^{3}}\gg 1$ . In this case we have

(5.170)

t_{0}=\sqrt{\frac{Q}{-y}}-\frac{1}{2}+O(\sqrt{\frac{-y}{Q}}).

Then we get

(5.171)

F(t_{0})=-2\sqrt{-Qy}-\frac{1}{2}y+O(1),

and

(5.172)

C_{n}^{-1}Q^{-\frac{1}{4}-\frac{1}{2n}}e^{\frac{1}{2}y-\sqrt{-Qy}}(-y)^{-Q-1}\leq\Tri(y)\leq C_{n}\frac{1}{\Gamma(\alpha-\beta)}Q^{\frac{1}{4}}(-y)^{-Q}e^{\frac{1}{2}y-2\sqrt{-Qy}}.

Similarly, we get

(5.173)

G(u_{0})=2\sqrt{-Qy}-\frac{y}{2}+(Q+\frac{1}{2}\gamma_{n})\log Q-Q.

(5.174)

\Ku(y)\leq C_{n}\cdot\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}(-y)^{\frac{1-2\alpha}{4}}e^{\frac{1}{2}y+2\sqrt{-Qy}}e^{-Q}Q^{Q+\frac{1}{2}\gamma_{n}}.

In this case even though in the produce $\Tri\cdot\Ku$ there is a good cancellation each of them does behave quite differently from the previous case. This also gives a reason why we do get an optimal estimate (up to polynomial errors in $y$ and $Q$ ) for the product $\Tri\cdot\Ku$ , comparing with (A.27) and (A.34).

5.3. The case j k ≠ 0 : uniform estimates and asymptotics [053W]

5.3. The case jk≠0j_{k}\neq 0: uniform estimates and asymptotics

Proposition 5.6.

Proof.

Lemma 5.7.

Lemma 5.8.

Proof.

Proposition 5.9.

Proof.

Corollary 5.9.1.

Proposition 5.10.

Proof.

Lemma 5.11.

Proof.

Lemma 5.12.

Remark 5.12.1.

Proof.

Corollary 5.12.1.

5.3. The case $j_{k}\neq 0$ : uniform estimates and asymptotics