Existence of hypercylinder expanders of the inverse mean curvature flow

TL;DR

This paper provides a new proof for the existence of hypercylinder expanders in the inverse mean curvature flow, demonstrating their properties and asymptotic behavior in a radially symmetric setting.

Contribution

It introduces a novel proof of the existence and properties of hypercylinder expanders as homothetic solitons in inverse mean curvature flow, including their asymptotic and convexity features.

Findings

Existence of a unique even solution r(y) with specified boundary conditions.

The solution r(y) tends to infinity as y approaches infinity.

The derivative r'(y) approaches a finite limit a_1 with 0 ≤ a_1 < ∞.

Abstract

We will give a new proof of the existence of hypercylinder expander of the inverse mean curvature flow which is a radially symmetric homothetic soliton of the inverse mean curvature flow in $\mathbb{R}^n\times \mathbb{R}$, $n\ge 2$, of the form $(r,y(r))$ or $(r(y),y)$ where $r=|x|$, $x\in\mathbb{R}^n$, is the radially symmetric coordinate and $y\in \mathbb{R}$. More precisely for any $\lambda>\frac{1}{n-1}$ and $\mu>0$, we will give a new proof of the existence of a unique even solution $r(y)$ of the equation $\frac{r''(y)}{1+r'(y)^2}=\frac{n-1}{r(y)}-\frac{1+r'(y)^2}{\lambda(r(y)-yr'(y))}$ in $\mathbb{R}$ which satisfies $r(0)=\mu$, $r'(0)=0$ and $r(y)>yr'(y)>0$ for any $y\in\mathbb{R}$. We will prove that $\lim_{y\to\infty}r(y)=\infty$ and $a_1:=\lim_{y\to\infty}r'(y)$ exists with $0\le a_1<\infty$. We will also give a new proof of the existence of a constant $y_1>0$ such that $r''(y_1)=0$, $r''(y)>0$ for any $0<y<y_1$ and $r''(y)<0$ for any $y>y_1$.

Equations

$$\frac{\partial }{\partial t}F(x,t)=-\frac{\nu }{H}\forall x\in M^n,0<t<T$$

$$-\frac{\nu (p)}{H(p)}=\lambda X(p{)}^{\perp }\forall p\in \Sigma$$

$$-<H\nu ,X>=\frac{1}{\lambda }\iff -<{\Delta }_{g}X,X>=\frac{1}{\lambda }\forall X\in \Sigma$$

$$\frac{r^{\prime \prime }(y)}{1+r^{\prime }(y{)}^2}=\frac{n-1}{r(y)}-\frac{1+r^{\prime }(y{)}^2}{\lambda (r(y)-y{r}^{\prime }(y))},r(y)>0$$

$$y_{rr}+\frac{n-1}{r}\cdot (1+y_r^2)y_r-\frac{(1+y_r^2{)}^2}{\lambda (r{y}_r-y)}=0$$

$$\{\begin{array}{rl}& \frac{r^{\prime \prime }(y)}{1+r^{\prime }(y{)}^2}=\frac{n-1}{r(y)}-\frac{1+r^{\prime }(y{)}^2}{\lambda (r(y)-y{r}^{\prime }(y))},r(y)>0,\forall y\in \mathbb{R}\\ & r(0)=\mu ,r^{\prime }(0)=0\end{array}$$

$$r(y)>y{r}^{\prime }(y)\forall y\in \mathbb{R}$$

$$r^{\prime \prime }(0)=\left(n-1-\frac{1}{\lambda }\right)\frac{1}{\mu }.$$

$$r^{\prime }(y)>0\forall y>0,$$

$$a_1:=\underset{y\to \infty }{\lim }{r}^{\prime }(y)\text{mbox}existsand0\le a_1<\infty ,$$

$$\underset{y\to \pm \infty }{\lim }r(y)=\infty .$$

$$\{\begin{array}{rl}& r^{\prime \prime }(y)>0\forall 0<y<y_1\\ & r^{\prime \prime }(y)<0\forall y>y_1\\ & r^{\prime \prime }(y_1)=0.\end{array}$$

$$\{\begin{array}{rl}& \frac{r^{\prime \prime }}{1+r^{\prime }{}^2}=\frac{n-1}{r}-\frac{1+r^{\prime }{}^2}{\lambda (r-y{r}^{\prime })},r(y)>0,\forall y>0\\ & r(0)=\mu ,r^{\prime }(0)=0\end{array}$$

$$r(y)>y{r}^{\prime }(y)\forall y>0$$

$$\{\begin{array}{rl}& \frac{r^{\prime \prime }}{1+r^{\prime }{}^2}=\frac{n-1}{r}-\frac{1+r^{\prime }{}^2}{\lambda (r-y{r}^{\prime })},r(y)>0,\text{mbox}in[0,y_0)\\ & r(0)=\mu ,r^{\prime }(0)=0\end{array}$$

$$r(y)>y{r}^{\prime }(y)\text{mbox}in[0,y_0)$$

$${\mathcal{X}}_{\epsilon }:=\left\{(g,h):g,h\in C\left([0,\epsilon ];\mathbb{R}\right)\right\}$$

$$||(g,h)|{|}_{{\mathcal{X}}_{\epsilon }}=\max \left\{\Vert g{\Vert }_{L^{\infty }(0,\epsilon )},\Vert h(s){\Vert }_{L^{\infty }(0,\epsilon )}\right\}.$$

$$\Phi (g,h):=\left({\Phi }_1(g,h),{\Phi }_2(g,h)\right),$$

$$\{\begin{array}{rl}& {\Phi }_1(g,h)(y):=\mu +{\int }_0^{y}h(s)ds\\ & {\Phi }_2(g,h)(y):={\int }_0^y(1+h(s{)}^2)\left(\frac{n-1}{g(s)}-\frac{1+h(s{)}^2}{\lambda (g(s)-sh(s))}\right)ds.\end{array}$$

$${\mathcal{D}}_{\epsilon ,\eta }:=\left\{(g,h)\in {\mathcal{X}}_{\epsilon }:||(g,h)-(\mu ,0)|{|}_{{\mathcal{X}}_{\epsilon }}\le \eta \right\}.$$

$$|h(s)|\le \eta \le \mu /4\text{mbox}and\frac{3\mu }{4}\le g(s)\le \frac{5\mu }{4}\forall 0\le s\le \epsilon$$

$$\Rightarrow$$

$$|{\Phi }_1(g,h)(y)-\mu |\le {\int }_0^y|h(s)|ds\le \eta \epsilon \le \eta \forall 0\le y\le \epsilon .$$

$$|{\Phi }_2(g,h)(y)|\le (1+{\eta }^2)\left(\frac{4(n-1)}{3\mu }+\frac{2(1+{\eta }^2)}{|\lambda |\mu }\right)\epsilon \le \eta \forall 0\le y\le \epsilon$$

$${\epsilon }_1=\min \left(\frac{1}{2},\eta (1+{\eta }^2{)}^{-1}{\left(\frac{4(n-1)}{3\mu }+\frac{2(1+{\eta }^2)}{|\lambda |\mu }\right)}^{-1}\right).$$

$$|h_i(s)|\le \eta \le \mu /4,\frac{3\mu }{4}\le g_i(s)\le \frac{5\mu }{4}\text{mbox}and{g}_i(s)-s{h}_i(s)\ge \frac{\mu }{2}>0\forall 0\le s\le \epsilon ,i=1,2.$$

$$|{\Phi }_1(g_1,h_1)(y)-{\Phi }_1(g_2,h_2)(y)|\le {\int }_0^y|h_1(s)-h_2(s)|ds\le \delta \epsilon \le \frac{\delta }{2}\forall 0\le y\le \epsilon$$

$$|{\Phi }_2(g_1,h_1)(y)-{\Phi }_2(g_2,h_2)(y)|$$

$$\le$$

$$+(n-1){\int }_0^y(1+h_2(s{)}^2)\left|\frac{g_2(s)-g_1(s)}{g_1(s)g_2(s)}\right|ds$$

$$+{\int }_0^y(1+h_2(s{)}^2)\left|\frac{1+h_1(s{)}^2}{\lambda (g_1(s)-s{h}_1(s))}-\frac{1+h_2(s{)}^2}{\lambda (g_2(s)-s{h}_2(s))}\right|ds\forall 0\le y\le \epsilon$$

Full text

Key words: inverse mean curvature flow, hypercylinder expander solution, existence, asymptotic behaviour

AMS 2010 Mathematics Subject Classification: Primary 35K67, 35J75 Secondary 53C44

1 Introduction

Consider a family of immersions $F:M^{n}\times[0,T)\to{\mathbb{R}}^{n+1}$ of $n$-dimensional hypersurfaces in ${\mathbb{R}}^{n+1}$. We say that $M_{t}=F_{t}(M^{n})$, $F_{t}(x)=F(x,t)$, moves by the inverse mean curvature flow if

[TABLE]

where $H(x,t)>0$ and $\nu$ are the mean curvature and unit interior normal of the surface $F_{t}$ at the point $F(x,t)$. Recently there are a lot of study on the inverse mean curvature flow by P. Daskalopoulos, C. Gerhardt, K.M. Hui [H], G. Huisken, T. Ilmanen, K. Smoczyk, J. Urbas and others [DH], [G], [HI1], [HI2], [HI3], [S], [U]. Although there are a lot of study on the inverse mean curvature flow on the compact case, there are not many results for the non-compact case.

Recall that [DLW] a $n$-dimensional submanifold $\Sigma$ of ${\mathbb{R}}^{n+1}$ with immersion $X:\Sigma\to{\mathbb{R}}^{n+1}$ and non-vanishing mean curvature $H$ is called a homothetic soliton for the inverse mean curvature flow if there exists a constant $\lambda\neq 0$ such that

[TABLE]

where $X(p)^{\perp}$ is the component of $X(p)$ that is normal to the tangent space $T_{X(p)}(X(\Sigma))$ at $X(p)$. As proved by G. Drugan, H. Lee and G. Wheeler in [DLW] (1.1) is equivalent to

[TABLE]

where $g$ is the induced metric of the immersion $X:\Sigma\to{\mathbb{R}}^{n+1}$. If the homothetic soliton of the inverse mean curvature flow is a radially symmetric solution in ${\mathbb{R}}^{n}\times{\mathbb{R}}$, $n\geq 2$, of the form $(r,y(r))$ or $(r(y),y)$ where $r=|x|$, $x\in{\mathbb{R}}^{n}$, is the radially symmetric coordinate, $y\in{\mathbb{R}}$, then by (1.2) and a direct computation $r(y)$ satisfies the equation

[TABLE]

or equivalently $y(r)$ satisfies the equation,

[TABLE]

where $r^{\prime}(y)=\frac{dr}{dy}$, $r^{\prime\prime}(y)=\frac{d^{2}r}{dy^{2}}$ and $y_{r}(r)=\frac{dy}{dr}$, $y_{rr}(r)=\frac{d^{2}y}{dr^{2}}$ etc. In the paper [DLW] G. Drugan, H. Lee and G. Wheeler stated the existence and asymptotic behaviour of hypercylinder expanders which are homothetic soliton for the inverse mean curvature flow with $\lambda>1/n$. However there are no proof of the existence result in that paper except for the case $\lambda=\frac{1}{n-1}$ and the proof of the asymptotic behaviour of hypercylinder expanders there are very sketchy. In this paper I will give a new proof of the existence of hypercylinder expanders for the inverse mean curvature flow with $\lambda>\frac{1}{n-1}$. We will also give a new proof of the asymptotic behaviour of these hypercylinder expanders.

More precisely I will prove the following main results.

Theorem 1.1**.**

For any $n\geq 2$, $\lambda>\frac{1}{n-1}$ and $\mu>0$, there exists a unique even solution $r(y)\in C^{2}({\mathbb{R}})$ of the equation

[TABLE]

which satisfies

[TABLE]

and

[TABLE]

Theorem 1.2**.**

(cf. Theorem 20 of [DLW]) Let $n\geq 2$, $\lambda>\frac{1}{n-1}$, $\mu>0$, and $r(y)\in C^{2}({\mathbb{R}})$ be the unique solution of (1.4). Then

[TABLE]

[TABLE]

and

[TABLE]

Moreover there exists a constant $y_{1}>0$ such that

[TABLE]

Since (1.4) is invariant under reflection $y\to-y$, by uniqueness of solution of ODE the solution of (1.4) is an even function and Theorem 1.1 is equivalent to the following theorem.

Theorem 1.3**.**

For any $n\geq 2$, $\lambda>\frac{1}{n-1}$ and $\mu>0$, there exists a unique solution $r(y)\in C^{2}([0,\infty))$ of the equation

[TABLE]

which satisfies

[TABLE]

and (1.6).

2 Existence and asymptotic behaviour of solution

In this section we willl prove Theorem 1.2 and Theorem 1.3. We first start with a lemma.

Lemma 2.1**.**

For any $n\geq 2$, $\lambda\neq 0$ and $\mu>0$, there exists a constant $y_{0}>0$ such that the equation

[TABLE]

has a unique solution $r(y)\in C^{2}([0,y_{0}))$ which satisfies

[TABLE]

Moreover (1.6) holds.

Proof: Uniqueness of solution of (2.1) follows from standard ODE theory. Hence we only need to prove existence of solution of (2.1). We will use a modification of the fixed point argument of the proof of Lemma 2.1 of [H] to prove the existence of solution of (2.1). Let $0<\varepsilon<1$. We now define the Banach space

[TABLE]

with a norm given by

[TABLE]

For any $(g,h)\in{\mathcal{X}}_{\varepsilon},$ we define

[TABLE]

where for any $0<y\leq\varepsilon,$

[TABLE]

For any $0<\eta\leq\mu/4$, let

[TABLE]

Note that ${\mathcal{D}}_{\varepsilon,\eta}$ is a closed subspace of ${\mathcal{X}}_{\varepsilon}$. We will show that if $\varepsilon\in(0,1)$ is sufficiently small, the map $(g,h)\mapsto\Phi(g,h)$ will have a unique fixed point in ${\mathcal{D}}_{\varepsilon,\eta}$.

We first prove that $\Phi({\mathcal{D}}_{\varepsilon,\eta})\subset{\mathcal{D}}_{\varepsilon,\eta}$ if $\varepsilon\in(0,1)$ is sufficiently small. Let $(g,h)\in{\mathcal{D}}_{\varepsilon,\eta}$. Then

[TABLE]

and

[TABLE]

Hence by (2.3), (2.4) and (2.5),

[TABLE]

if $0<\varepsilon\leq\varepsilon_{1}$ where

[TABLE]

Thus by (2.6) and (2.7), $\Phi({\mathcal{D}}_{\varepsilon,\eta})\subset{\mathcal{D}}_{\varepsilon,\eta}$ for any $0<\varepsilon\leq\varepsilon_{1}$.

We now let $0<\varepsilon\leq\varepsilon_{1}$. Let $(g_{1},h_{1}),(g_{2},h_{2})\in{\mathcal{D}}_{\varepsilon,\eta}$ and $\delta:=||(g_{1},h_{1})-(g_{2},h_{2})||_{{\mathcal{X}}_{\varepsilon}}$. Then by (2.4) and (2.5),

[TABLE]

Hence by (2.8), we have

[TABLE]

and

[TABLE]

where

[TABLE]

Now by (2.8),

[TABLE]

and

[TABLE]

[TABLE]

By (2), (2) and (2),

[TABLE]

We now let

[TABLE]

and $0<\varepsilon\leq\varepsilon_{2}$. Then by (2.9) and (2.13),

[TABLE]

Hence $\Phi$ is a contraction map on ${\mathcal{D}}_{\varepsilon,\eta}$. Then by the Banach fix point theorem the map $\Phi$ has a unique fix point. Let $(g,h)\in{\mathcal{D}}_{\varepsilon,\eta}$ be the unique fixed point of the map $\Phi$. Then

[TABLE]

Hence

[TABLE]

and

[TABLE]

By (2.5), (2.14) and (2), $g\in C^{1}([0,\varepsilon))\cap C^{2}(0,\varepsilon)$ satisfies (2.1) and (2.2) with $y_{0}=\varepsilon$. Letting $y\to 0$ in (2.1) we get (1.6) and hence $g\in C^{2}([0,\varepsilon))$ and the lemma follows. $\square$

By an argument similar to the proof of Lemma 2.1 we have the following lemma.

Lemma 2.2**.**

For any $n\geq 2$, $\lambda\neq 0$, $\mu>0$, $M_{1}>0$, $\delta_{0}>0$, $r_{0},r_{1}\in{\mathbb{R}}$, satisfying

[TABLE]

there exists a constant $\delta_{1}\in(0,y_{0}/2)$ depending on $\lambda$, $\delta_{0}$, $y_{0}$ and $M_{1}$ such that for any $y_{0}/2<y_{1}<y_{0}$ the equation

[TABLE]

has a unique solution $r(y)\in C^{2}([y_{1},y_{1}+\delta_{1}))$ which satisfies

[TABLE]

Lemma 2.3**.**

Let $n\geq 2$, $0<\lambda\neq\frac{1}{n-1}$, $\mu>0$ and $y_{0}>0$. Suppose $r(y)\in C^{2}([0,y_{0}))$ is the solution of (2.1) which satisfies (2.2). Then the following holds.

1. (i)

If $\lambda>\frac{1}{n-1}$, then

[TABLE] 2. (ii)

If $0<\lambda<\frac{1}{n-1}$, then

[TABLE]

Proof: By Lemma 2.1, (1.6) holds. We divide the proof into two cases:

Case 1: $\lambda>\frac{1}{n-1}$

By (1.6), $r^{\prime\prime}(0)>0$. Hence there exists a constant $\delta>0$ such that $r^{\prime}(s)>0$ for any $0<s<\delta$. Let $(0,a_{1})$, $\delta\leq a_{1}\leq y_{0}$, be the maximal interval such that

[TABLE]

Suppose $a_{1}<y_{0}$. Then $r^{\prime}(a_{1})=0$ and hence $r^{\prime\prime}(a_{1})\leq 0$. On the other hand by (2.1),

[TABLE]

and contradiction arises. Hence $a_{1}=y_{0}$ and (i) follows.

Case 2: $0<\lambda<\frac{1}{n-1}$

By (1.6), $r^{\prime\prime}(0)<0$. Hence there exists a constant $\delta>0$ such that $r^{\prime}(s)<0$ for any $0<s<\delta$. Let $(0,a_{1})$, $\delta\leq a_{1}\leq y_{0}$, be the maximal interval such that

[TABLE]

Suppose $a_{1}<y_{0}$. Then $r^{\prime}(a_{1})=0$ and hence $r^{\prime\prime}(a_{1})\geq 0$. On the other hand by (2.1),

[TABLE]

and contradiction arises. Hence $a_{1}=y_{0}$ and (ii) follows. $\square$

Lemma 2.4**.**

Let $n\geq 2$, $\lambda>\frac{1}{n-1}$, $\mu>0$ and $y_{0}>0$. Suppose $r(y)\in C^{2}([0,y_{0}))$ is the solution of (2.1) which satisfies (2.2). Then there exist a constant $\delta_{1}>0$ such that

[TABLE]

Proof: Let $w(y)=r(y)-yr^{\prime}(y)$, $a_{1}=\min_{0\leq y\leq y_{0}/2}w(y)$, $a_{2}=\frac{\mu}{\lambda(n-1)}$ and $a_{3}=\frac{1}{2}\min(a_{1},a_{2})$. Then $a_{1}>0$ and $a_{3}>0$. By Lemma 2.3,

[TABLE]

Suppose there exists $y_{1}\in(y_{0}/2,y_{0})$ such that $w(y_{1})<a_{3}$. Let $(a,b)$ be the maximal interval containing $y_{1}$ such that $w(y)<a_{3}$ for any $y\in(a,b)$. Then $a>y_{0}/2$, $w(a)=a_{3}$ and

[TABLE]

By (2.1), (2.19), (2.20), and a direct computation,

[TABLE]

Hence

[TABLE]

and contradiction arises. Hence no such $y_{1}$ exists and $w(y)\geq a_{3}$ for any $y\in(0,y_{0})$. Thus (2.18) holds with $\delta_{1}=a_{3}$.

$\square$

Lemma 2.5**.**

Let $n\geq 2$, $\lambda>\frac{1}{n-1}$, $\mu>0$ and $y_{0}>0$. Suppose $r(y)\in C^{2}([0,y_{0}))$ is the solution of (2.1) which satisfies (2.2). Then there exists a constant $M_{1}>0$ such that

[TABLE]

and

[TABLE]

Proof: By (2.1), (2.2) and Lemma 2.3,

[TABLE]

Integrating (2.23) over $(0,y_{0})$,

[TABLE]

By Lemma 2.3 and (2.24), (2.21) holds with

[TABLE]

By (2.21) we get (2.22) and the lemma follows. $\square$

Lemma 2.6**.**

Let $n\geq 2$, $\lambda>\frac{1}{n-1}$, $\mu>0$ and $y_{0}>0$. Suppose $r(y)\in C^{2}([0,y_{0}))$ is the solution of (2.1) which satisfies (2.2). Then either

[TABLE]

or there exists a constant $y_{1}\in(0,y_{0})$ such that $r^{\prime\prime}(y_{1})=0$ and

[TABLE]

Proof: We will use a modification of the proof of Lemma 15 of [DLW] to prove this lemma. By (1.6), $r^{\prime\prime}(0)>0$. Hence there exists a constant $\delta>0$ such that $r^{\prime\prime}(s)>0$ for any $0<s<\delta$. Let $(0,y_{1})$, $\delta\leq y_{1}\leq y_{0}$, be the maximal interval such that

[TABLE]

If $y_{1}=y_{0}$, then (2.25) holds. If $y_{1}<y_{0}$, then $r^{\prime\prime}(y_{1})=0$. By Lemma 2.3 and (2.1),

[TABLE]

Hence there exist a constant $0<\delta^{\prime}<y_{0}-y_{1}$ such that $r^{\prime\prime}(y)<0$ for any $y_{1}<y<y_{1}+\delta^{\prime}$. Let $(y_{1},z_{0})$ be the maximal interval such that

[TABLE]

If $z_{0}<y_{0}$, then $r^{\prime\prime}(z_{0})=0$ and $r^{\prime\prime}(z_{0})\geq 0$. On the other hand by Lemma 2.3 and (2.27),

[TABLE]

and contradiction arises. Hence $z_{0}=y_{0}$ and (2.26) follows. $\square$

We are now ready to prove Theorem 1.3.

Proof of Theorem 1.3: By Lemma 2.1 there exists a constant $y_{0}^{\prime}>0$ such that (2.1) has a unique solution $r(y)\in C^{2}([0,y_{0}^{\prime}))$ which satisfies (1.6) and (2.2) in $(0,y_{0}^{\prime})$. Let $(0,y_{0})$ be the maximal interval of existence of solution $r(y)\in C^{2}([0,y_{0}))$ of (2.1) which satisfies (2.2) and (1.6). Suppose $y_{0}<\infty$. By Lemma 2.2, Lemma 2.4 and Lemma 2.5, there exists a constant $\delta_{1}\in(0,y_{0})$ such that for any $y_{0}/2<y_{1}<y_{0}$ there exists a unique solution $r_{1}(y)\in C^{2}([y_{1},y_{1}+\delta_{1}))$ of (2.16) which satisfies (2.17) with $r_{0}=r(y_{1})$ and $r_{1}=r^{\prime}(y_{1})$. Let $y_{1}\in\left(y_{0}-\frac{\delta_{1}}{2},y_{0}\right)$ and let $r_{1}(y)\in C^{2}([y_{1},y_{1}+\delta_{1}))$ be the unique solution of (2.16) given by Lemma 2.2 which satisfies (2.17) with $r_{0}=r(y_{1})$ and $r_{1}=r^{\prime}(y_{1})$. We then extend $r(y)$ to a solution of (1.11) in $(0,y_{1}+\delta_{1})$ by setting $r(y)=r_{1}(y)$ for any $y_{0}\leq y<y_{1}+\delta_{1}$. Since $y_{1}+\delta_{1}>y_{0}$, this contradicts the maximality of the interval $(0,y_{0})$. Hence $y_{0}=\infty$ and there exists a unique solution $r(y)\in C^{2}([0,\infty))$ of the equation (1.11) which satisfies (1.12) and (1.6) and the theorem follows. $\square$

Proof of Theorem 1.2: We will give a simple proof different from the sketchy proof of this result in [DLW] here. By (i) of Lemma 2.3 (1.7) holds. By Lemma 2.6 either

[TABLE]

or there exists $y_{1}>0$ such that (1.10) holds. Suppose (2.28) holds. Then

[TABLE]

and $a_{1}>0$. We now divide the proof into two cases.

Case 1: $a_{1}=\infty$

Then there exists $y_{2}>0$ such that

[TABLE]

By (1.11) and (2.30),

[TABLE]

which contradicts (2.28). Hence $a_{1}\neq\infty$.

Case 2: $a_{1}<\infty$

By (1.12),

[TABLE]

Now by (2.29) and the l’Hosiptal rule,

[TABLE]

By (1.11), (2.29), (2.31), (2.32) and the l’Hosiptal rule,

[TABLE]

which contradicts (2.28). Hence $a_{1}<\infty$ does not hold. Thus by Case 1 and Case 2 (2.28) cannot hold. Hence there exists $y_{1}>0$ such that (1.10) holds.

By (1.10) and Lemma 2.3, (1.8) holds. By (1.7),

[TABLE]

Since by (1.10) $(r(y)-yr^{\prime}(y))^{\prime}=-yr^{\prime\prime}(y)>0$ for any $y>y_{1}$,

[TABLE]

Suppose

[TABLE]

Then

[TABLE]

By (1.8), (2.33) and (2.34),

[TABLE]

Suppose $a_{4}>0$. Then there exists $y_{2}>y_{1}$ such that

[TABLE]

which contradicts (2.34). Hence

[TABLE]

Letting $y\to\infty$ in (1.11), by (2.35) and (2.36),

[TABLE]

which contradicts (1.10). Hence (2.34) does not hold and $a_{2}=\infty$. Thus (1.9) holds and the theorem follows.

$\square$