Proof. [03HT]

Original official author HTML, exact retained edition. Historical TeX conversion verdicts remain unchanged. Cited-edition alignment and mathematical self-containment are not assessed.

Complete original source context · Original author HTML

Proof.

We will prove that there exists some constant $\eta_{0}<\eta<\eta_{0}+\underline{\delta}/10$ such that

(4.136)

\frac{\mathcal{D}_{k}(z)}{\mathcal{W}_{k}(z)}\leq C_{0}\cdot Q_{k}\cdot e^{\eta z}

and

(4.137)

\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\leq C_{0}\cdot Q_{k}\cdot e^{\eta z},

where the positive constant $C_{0}>0$ is independent of the index $k$ .

We prove (4.136) and (4.137) in two different cases.

In the first case, $k\in\mathbb{Z}_{+}$ satisfies $j_{k}=0$ . The fundamental solutions have an explicit form

(4.138)

\mathcal{F}_{k}(z)\equiv e^{\sqrt{\lambda_{k}}\cdot z}

and

(4.139)

\mathcal{U}_{k}(z)\equiv e^{-\sqrt{\lambda_{k}}\cdot z}.

Immediately,

(4.140)

\mathcal{W}_{k}(z)=\mathcal{W}(\mathcal{F}_{k}(z),\mathcal{U}_{k}(z))=2\sqrt{\lambda_{k}}

and hence for $\eta>\eta_{0}$ ,

(4.141)

\displaystyle\frac{|\mathcal{D}_{k}(z)|}{|\mathcal{W}_{k}(z)|}=\frac{\mathcal{F}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{U}_{k}(r)|\xi_{k}(r)\cdot r|dr\leq\frac{Q_{k}e^{\sqrt{\lambda_{k}}\cdot z}}{\sqrt{\lambda_{k}}}\int_{z}^{\infty}e^{(-\sqrt{\lambda_{k}}+\eta)\cdot r}dr\leq C_{0}\cdot Q_{k}e^{\eta z}.

Similarly,

(4.142)

\displaystyle\frac{|\mathcal{G}_{k}(z)|}{|\mathcal{W}_{k}(z)|}=\frac{\mathcal{U}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z_{0}}^{z}\mathcal{F}_{k}(r)|\xi_{k}(r)\cdot r|dr\leq\frac{Q_{k}e^{-\sqrt{\lambda_{k}}\cdot z}}{\sqrt{\lambda_{k}}}\int_{z_{0}}^{z}e^{(\sqrt{\lambda_{k}}+\eta)\cdot r}dr\leq C_{0}\cdot Q_{k}e^{\eta z}.

In the latter case $j_{k}\in\mathbb{Z}_{+}$ and $k\in\mathbb{Z}_{+}$ , we will prove the uniform estimates. A crucial point is to apply the monotonicity in Lemma 4.9. In fact,

	$\displaystyle\frac{\mathcal{D}_{k}(z)}{\mathcal{W}_{k}(z)}$	$\displaystyle=\frac{\mathcal{F}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{U}_{k}(r)\xi_{k}(r)\cdot rdr$
(4.143)			$\displaystyle\leq\frac{C_{0}e^{\widehat{F}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{U}_{k}(r)}\xi_{k}(r)\cdot rdr\leq\frac{C_{0}e^{\widehat{F}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{U}_{k}(r)+\eta^{\prime}r}dr,$

where $\eta^{\prime}>\eta_{0}$ . We choose $\epsilon\in(\underline{\delta}/100,\underline{\delta}/10)$ and denote $\eta\equiv\eta^{\prime}+\epsilon$ , then by Lemma 4.9

	$\displaystyle\frac{e^{\widehat{F}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{U}_{k}(r)+\eta^{\prime}r}dr$	$\displaystyle=\frac{e^{\widehat{F}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{U}_{k}(r)+\eta r}\cdot e^{-\epsilon r}dr$
		$\displaystyle\leq\frac{C_{0}\cdot Q_{k}\cdot e^{\widehat{F}_{k}(z)+\widehat{U}_{k}(z)+\eta z}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{-\epsilon r}dr$
(4.144)			$\displaystyle\leq C_{0}\cdot Q_{k}\cdot\frac{e^{\widehat{F}_{k}(z)+\widehat{U}_{k}(z)+\eta z}}{\mathcal{W}_{k}(z)}\leq C_{0}\cdot Q_{k}\cdot e^{\eta z}.$

The proof of (4.136) is done.

Next, for the estimate (4.137),

	$\displaystyle\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}$	$\displaystyle=\frac{\mathcal{U}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{1}^{z}\mathcal{F}_{k}(r)\xi_{k}(r)\cdot rdr$
(4.145)			$\displaystyle\leq\frac{C_{0}e^{\widehat{U}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{1}^{z}e^{\widehat{F}_{k}(r)+\eta^{\prime}r}dr\leq\frac{C_{0}\cdot Q_{k}z\cdot e^{\widehat{U}_{k}(z)+\widehat{F}_{k}(z)+\eta^{\prime}z}}{\mathcal{W}_{k}(z)}\leq C_{0}\cdot Q_{k}\cdot e^{\eta z}.$

This completes the proof of the proposition.

∎