数理解析研究所講究録
第 1547 巻 2007 年 18-26

18

Asymptotic forms of weakly increasing
positive solutions of quasilinear ordinary
differential equations(準線型常微分方程式の弱増加正値解の漸近形)
広島大学・理学研究科 宇佐美広介 (Hiroyuki USAMI)
Department of Mathematics, Graduate School of Science,
Hiroshima University

1

Introduction

This talk is a joint work with Professor Ken-ichi Kamo (Sapporo Medical University,
Japan). Let us consider the equations of the form(E)$(|u’|^{\alpha-1}u’)’+p(t)|u|^{\beta-1}u=0$

under the following conditions:and are positive constants satisfying;$(A_{2})p(t)$ is a-function defined nearsatisfying the asymptotic conditionas.
for some$(A_{1})\alpha$$\alpha<\beta$$\beta$$C^{1}$$\sigma\in R$$p(t)\sim t^{-\sigma}$

By condition$+\infty$$(A_{2})$$tarrow\infty$

equation (E) can be rewritten in the form$(|u’|^{\alpha-1}u’)’+t^{-\sigma}(1+\epsilon(t))|u|^{\beta-1}u=0$,(E)where $\epsilon(t)=t^{\sigma}p(t)-1$ satisfies. Of course, here and in what follows the“$tarrow\infty$$f(t)\sim
g(t)$$\lim_{tarrow\infty}f(t)/g(t)=1$
symbolasmeans that
. A function is defined) andto be a solution of equation (E) if) and it satisfies
equation (E) on) for sufficiently large .
easilyIt isseen that all positive solutions $u(t)$ of (E) are classified into the following$\lim_{tarrow\infty}\epsilon(t)=0$$u$$u\in C^{1}[t_{1}, \infty$$[t_{1}, \infty$$|u’|^{\alpha-1}u’\in C^{1}[t_{1},\infty$$t_{1}$

three types according as their asymptotic behavior as(I) (as ymptotically linear solution):$u(t)\sim c_{1}t$$tarrow\infty$

for some constant:$c_{1}>0$;(II) (weakly increasing solution):$u’(t)\downarrow 0$,

and$u(t)\uparrow\infty$;(III) (asymptotically constant solution):$u(t)\sim c_{1}$for some constant$c_{1}>0$.

Concerning qualitative properties of positive solutions, the study of asymptotic behavior of asymptotically linear solutions and asymptotically constant solutions are rather

19

easy, because their first approximations are given by definition. On the other hand, we
can not easily find how the weakly increasing positive solutions behave except for the caseof $\alpha=1([1,4])$ .
In [4, Section 20], equation (E) with $\alpha=1$ has been considered systematically, andasymptotic forms of weakly increasing positive solutions are given by means of the parameters and . When, as far as the authors are aware, there are no works inwhich asymptotic forms of weakly increasing positive solutions are studied systematically.
Motivated by these facts we have been making an attempt to find out asymptotic forms
of weakly increasing positive solutions of (E) for the general case $\alpha>0$ .
To gain an insight into our problem, we consider the typical equation$\beta$$\alpha\neq 1$$\sigma$$(|u’|^{\alpha-1}u’)’+t^{-\sigma}|u|^{\beta-1}u=0$,(E)where $\sigma\in R$ is a constant. Note that equation (E) can be regarded as a perturbed
equation of this equation. Equation
has a weakly increasing positive solution of the$(c>0,0<\rho<1)$ if and only if $\alpha+1<\sigma<\beta+1$ . This solution is uniquely
form
given by,(1)$(E_{0})$$ct^{\rho},$$u_{0}(t)=\hat{C}t^{k}$where$k= \frac{\sigma-\alpha-1}{\beta-\alpha}\in(0,1)$,$\hat{C}=\{\alpha(1-k)k^{\alpha}\}^{\frac{1}{\beta-\alpha}}$.

(2)

From this simple observation we can see that asymptotic forms of weakly increasing
positive solutions of (E) may be strongly affected by that of the coefficient function $p(t)$ .
Furthermore we conjecture that weakly increasing positive solutions of (E) behave likegiven by (1) and (2) if
is suMciently small at . It should be noted that the
number appearing in (2) plays important roles in the sequel.
We have shown in [3] that the above conjecture is true in some case as seen from thefollowing theorem:$u$$u_{0}(t)$$|\epsilon(t)|$$\infty$$k$

Theorem 1 Let
furthermore that either$\alpha\leq 1$

and $1/2<k<1(\Leftrightarrow(\alpha+\beta+2)/2<\sigma<\beta+1)$ . Suppose$\int^{\infty}\frac{\epsilon(t)^{2}}{t}dt<\infty$(3)$\int^{\infty}|\epsilon’(t)|dt<\infty$

(4)

$or$

holds. Then, every weakly increasing positive solution$u(t)\sim u_{0}(t)$where$u_{0}$

as$u$$tarrow\infty$of (E) has the asymptotic form,

is given by (1) and (2).

In today’s talk we report that our conjecture is still valid for other cases; that is, wewill prove the following:

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and $0<k<1/2(\Leftrightarrow\alpha+1<\sigma<(\alpha+\beta+2)/2)$ . Supposethateither(3)or (4) holds. Then, the same conclusion as in Theorem 1
furthermore
holds.

Theorem 2 Let$\alpha\geq 1$

Remark. (i) In Theorems 1 and 2, the differentiability of is unnecessary when (3)is assumed.
$Th\infty rems1$ and 2 have been(ii) When $\alpha=1$ andobtained by [1] and [4,Corollaries 20.2 and 20.3].
$p$$\epsilon(t)\equiv 0,$

We note that existence results of weakly increasing positive solutions to (E) are notwell known. But we can show many concrete examples of those equations that have weakly
increasing positive solutions. Some of existence results of weakly increasing solutions forthe case $\alpha=1$ are found in $[6,7]$ .
The paper is organized as follows. In Section 2 we give preparatory lemmas employedlater. In Section 3 we give the proof of Theorem 2. Other related results are found in[3,5,6,7].

2

Preparatory lemmas

Lemma 3 Let $w\in C^{1}[t_{0}, \infty$ ), $w’(t)=O(1)$ as $tarrow\infty$ , and. Then, $\lim_{tarrow\infty}w(t)=0$ .

$w\in L^{\lambda}[t_{0}, \infty$$\lambda>0$) for some

Proof. We have$|w(t)|^{\lambda}w(t)$$=$$|w(t_{0})|^{\lambda}w(t_{0})+ \int_{l_{0}}^{l}(|w(s)|^{\lambda}w(s))’ds$.

$|w(t_{0})|^{\lambda}w(t_{0})+( \lambda+1)\int_{l_{0}}^{t}|w(s)|^{\lambda}w’(s)ds$

By our assumptions the last integral converges as. Henceexists. Since), the limit must be . The proof is completed.
$tarrow\infty$$w\in L^{\lambda}[t_{0}, \infty$$\lim_{tarrow\infty}|w(t)|^{\lambda}w(t)\in R$$0$

Lemma 4 Let $\sigma\in(\alpha+1,\beta+1)$ . Then every weakly increasing positive solution of (E), whereis the exact solution ofsatisfies $u(t)=O(u_{0}(t))$ and $u’(t)=O(u_{0}’(t))$ as
given by (1) and (2).
$u$$tarrow\infty$$u_{0}$$(E_{0})$

Proof. We may assume that

and

$u$

$u,$

$u’>0$ for $t\geq t_{1}$ . Since$u’(t)^{\alpha}= \int_{t}^{\infty}p(s)u(s)^{\beta}ds$,$u’(t)^{\alpha} \geq u(t)^{\beta}\int^{\infty}p(s)ds$,

is increasing, we have$u$satisfies for large

$t$

(5)

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that is$u’(t)u(t)^{-A} \circ\geq(\int^{\infty}p(s)ds)^{\frac{1}{a}}$An integration of this inequality on the interval$[t, \infty$.

) will give,$u(t) \leq C_{1}\{\int^{\infty}(\int_{s}^{\infty}p(r)dr)^{1/\alpha}ds\}^{-\alpha/(\beta-\alpha)}\equiv C_{2}u_{0}(t)$

where

and

$C_{1}$$C_{2}$

are positive constant. Furthermore, by (5) we find that$u’(t)=( \int^{\infty}p(s)u(s)^{\beta}ds)^{\iota/\alpha}\leq C_{3}\int^{\infty}s^{-\sigma+k\beta}ds=C_{4}t^{k-1}=O(u_{0}’(t))$

where$C_{3}$

and$C_{4}$

as$tarrow\infty$,

are positive constants. This completes the proof.

Lemma 5 Let $\sigma\in(\alpha+1,\beta+1)_{y}$ and a weakly increasing positive solution of equ ation(E). Put $s=\log u_{0}(t)$ and $v=u/u_{0}$ . Then(i) $v,\dot{v}=O(1)$ as, and $v+\dot{v}>0$ nearwhere $\cdot=d/ds$ ;(ii) $v(s)$ satisfies near
the equation$u$$sarrow\infty$$\infty,$$\infty$$\ddot{v}-a\dot{v}-bu+b(\dot{v}+v)^{1-\alpha}v^{\beta}+b\delta(s)(\dot{v}+v)^{1-\alpha}v^{\beta}=0$

,

(6)

where
$a= \frac{1}{k}-2>0$,$b= \frac{1-k}{k}>0$,

and$\delta(s)=\epsilon(t)$.

Proof. We will prove only (i), because (ii) can be proved by direct computations.
We assume that $u’>0$ . Since $u=u_{0}v$ , the boundedness of follows from Lemma 4.
Noting $du/dt=\hat{C}kt^{k-1}(v+\dot{v})$ , we have $v+\dot{v}>0$ . On the other hand, since $dt/ds=t/k$ ,we have$u,$$v$$| \dot{v}|=|\frac{d}{dt}(\frac{u}{u_{0}})\frac{dt}{ds}|=|\frac{u’u_{0}-u_{0}’u}{u_{0}^{2}}|\frac{t}{k}\leq C\frac{t^{k-1}t^{k}t}{t^{2k}}=O(1)$as$sarrow\infty$.

This completes the proof.

Lemma 6 Let the assumptions of Theorem 2 holds, andfor sufficiently large .
$\dot{v}\in L^{2}[s_{0}, \infty)$$v$

be as in Lemma 5. Then$s_{0}$

Proof. We note that conditions (3) and (4), respectively, are equivalent to$\int^{\infty}\delta(s)^{2}ds<$

(7)

科科

and
$\int^{\infty}|\dot{\delta}(s)|ds<\infty$

.

(8)

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We multiply the both sides of (6) by; and so we obtain$\dot{v}$. Since$\alpha\geq 1$, we have$(1+\delta(s))(\dot{v}+v)^{1-\alpha}\dot{v}\leq$$(1+\delta(s))v^{1-\alpha}\dot{v}$$a\dot{v}^{2}\leq\dot{v}\ddot{v}-bv\dot{v}+b(1+\delta(s))v^{1-\alpha+\beta}\dot{v}$

An integration on the interval$[s_{0}, s]$

.

gives

$a \int_{*0}\dot{v}^{2}dr\leq\frac{\dot{v}^{2}}{2}-\frac{bv^{2}}{2}+\frac{bv^{2-\alpha+\beta}}{2-\alpha+\beta}+\int_{so}\delta(r)v^{1-\alpha+\beta}\dot{v}dr+const$that is$a \int_{\epsilon_{0}}^{*}\dot{v}^{2}dr\leq b\int_{so}^{s}\delta(r)v^{1-\alpha+\beta}\dot{v}dr+O(1)$as$sarrow\infty$

;

(9)

.

Here we have employed (i) of Lemma 5. Let the integral condition (3) hold; that is, let(7) hold. By the Schwarz inequality we have$a \int_{*0}^{s}\dot{v}^{2}dr\leq C_{1}(\int_{s_{0}}^{\infty}\backslash \delta(r)^{2}dr)^{1/2}(\int_{t_{0}}^{f}\dot{v}^{2}dr)^{1/2}+O(1)$for some constant $C_{1}>0$ . Therefore
parts, we obtain from (9)$\dot{v}\in L^{2}[s_{0}, \infty$). Next let (4) hold. Using integral by$a \int_{\iota_{0}}^{\epsilon}\dot{v}^{2}dr\leq\frac{\dot{v}^{2}}{2}-\frac{b}{2}v^{2}+\frac{b[1+\delta(r)]v^{2-\alpha+\beta}}{2-\alpha+\beta}-\frac{b}{2-\alpha+\beta}\int\dot{\delta}(r)v^{2-\alpha+\beta}dr+const$

As before by noting (i) of Lemma 5, we find that$\dot{v}\in L^{2}[s_{0}, \infty$.

). This completes the proof.

Proof of Theorem 2

3

We are now in a position to prove the main result Theorem 2:
Proof of Theorem 2. To this end it suffices to show that $\lim_{\iotaarrow\infty}v(s)=1$ , whereis the function introduced in Lemma 5. The proof is divided into three steps.

$v(s)$

Step 1. We claim that $\lim\inf_{*arrow\infty}v(s)>0$ ; namely $\lim\inf_{tarrow\infty}u(t)/u_{0}(t)>0$ . The
proof is done by contradiction.
Suppose to the contrary that $\lim\inf_{sarrow\infty}v(s)=0$ . Firstly, we suppose that $v(s)$decreases to as. This means that $u(t)/u_{0}(t)$ decreases to as. Accordingly
we have$sarrow\infty$$0$$tarrow\infty$$0$$u’(t)^{\alpha}= \int^{\infty}p(r)u(r)^{\beta}dr=\int^{\infty}p(r)u_{0}(r)^{\beta}(\frac{u(r)}{u_{0}(r)})^{\beta}dr$,$\leq(\frac{u(t)}{u_{0}(t)})^{\beta}\int^{\infty}p(r)u_{0}(r)^{\beta}dr=C_{1}t^{1-\sigma}u(t)^{\beta}$

where $C_{1}>0$ is a constant. Consequently we obtain the differential inequality $u’\leq$some constant $C_{2}>0$ near . But this differential inequality implies$C_{2}t^{(1-\sigma)/\alpha}u^{\beta/\alpha}for$$\infty$23

that $u(t)/u_{0}(t)\equiv v(s)\geq C_{3}>0$ for some constant $C_{3}>0$ . This is an bvious contradiction.
Next sppose that $\lim\inf_{sarrow\infty}v(s)=0$ andchanges sign in any neighborhood of. Since $v(s)$ takes local maxima in the region, there are thefollowing sequencesandsatisfying$0$$\dot{v}(s)$$v\geq(1+\delta(s))^{-1/\langle\beta-\alpha)}$$\infty$$\{\underline{s}_{n}\}$$\{\overline{s}_{n}\}$$\underline{s}_{n}<\overline{s}_{n}<\underline{s}_{n+1}$

and$\dot{v}(\underline{s}_{n})=\dot{v}(\overline{s}_{n})=0$

,

,

.

$\lim_{n\prec\infty}s=\lim_{narrow\infty}\overline{s}_{n}=\infty\sim$$\lim_{narrow\infty}v(\underline{s}_{n})=0$,$v(\overline{s}_{n})\geq(1+\delta(\overline{s}_{n}))^{-1/(\beta-\alpha)}$.

Now, we decompose in the form $\alpha=m-\rho$ , where $m\in N$ and $\rho>0$ . Although there
are infinitely many such choices of decomposition for , we fix one choice for a moment.
We rewrite equation (6) as$\alpha$$\alpha$$\ddot{v}-a\dot{v}-bv+b(\dot{v}+v)^{-m+1+\rho}v^{\beta}+b\delta(s)(\dot{v}+v)^{-m+1+\rho}v^{\beta}=0$

We multiply the both sides byinterval
to obtain$(v+\dot{v})^{m}\dot{v}$.

and then integrate the resulting equation on the$[\underline{s}_{\mathfrak{n}},\overline{s}_{n}]$$\int_{s}^{\overline{\iota}_{n}}\ddot{v}\dot{v}(vr+\dot{v})^{m}dr-a\int_{l}^{\overline{*}}n(vm+\dot{v})^{m}\dot{v}^{2}dr-b\int_{h}^{\overline{*}}nv\dot{v}(v+\dot{v})^{m}dr$$+b \int_{n}^{\mathfrak{n}}(v+\dot{v})^{1+\rho}\dot{v}v^{\beta}dr+b\int_{\underline{\epsilon}_{n}}^{\overline{\epsilon}_{\hslash}}\delta(r)(v+\dot{v})^{1+\rho}\dot{v}v^{\beta}dr=0$

.

(10)

The binomial expansion implies that

dr-b$\sum_{k=0}^{m}c_{k}\underline{\int_{\epsilon}^{r_{n}}}\ddot{v}\dot{v}^{k+1}v^{m-k}dr-a\int_{\underline{\epsilon}_{n}}^{n}(v+\dot{v})^{m}\dot{v}^{2}$$\sum_{k=0}^{m}c_{k}\underline{\int_{*}^{\overline{\epsilon}_{n}}}v^{m-k+1}\dot{v}^{k+1}dr$,$+b \int_{l}^{\overline{s}_{\mathfrak{n}}}(varrow+\dot{v})^{1+\rho}\dot{v}v^{\beta}dr+b\int_{s}^{\overline{*}n}\delta(r)(v-+\dot{v})^{1+\rho}\dot{v}v^{\beta}dr=0$

where
are the binomial coefficients. Now, we evaluate each term in the left handside. For $k\in\{0,1, \ldots, m-1\}$ we obtain$c_{k}=(_{k}^{m})$$\int_{f}^{\overline{\iota}_{n}}\ddot{v}\dot{v}^{k+1}v^{m-k}dr=r\int_{-n}^{\overline{s}_{n}}\frac{d}{dr}(\frac{\dot{v}^{k+2}}{k+2})v^{m-k}dr$$=- \frac{m-k}{k+2}\int_{g_{*}}^{\overline{s}_{n}}\dot{v}^{k+3}v^{m-k-1}dr=o(1)$

$0$

as

$narrow\infty$$narrow\infty$.

. Hence the first term of the left hand side
. The second term is dominated by, and hence

For $k=m$ obviously we haveof (10) tends toas$\int_{\underline{\epsilon}_{n}}^{n}\ddot{v}\dot{v}^{k+1}dr=0$$Const\int_{\underline{\epsilon}_{n}}^{\overline{l}_{\mathbb{R}}}\dot{v}^{2}dr$24

it tends to zero as $narrow\infty$ . Next, we compute the third term. For $k\in\{1,2, \ldots, m\}$ wehave. For $k=0$ we have$|\underline{\int_{*}^{\overline{s}_{n}}}v^{m-k+1}\dot{v}^{k+1}dr|\leq const\underline{\int_{s}^{\overline{s}_{n}}}\dot{v}^{2}dr$$\underline{\int_{*}^{\overline{\epsilon}_{n}}}v^{m+1}\dot{v}dr=\frac{1}{m+2}(v(\overline{s}_{\mathfrak{n}})^{m+2}-v(\underline{s}_{n})^{m+2})=\frac{v(\overline{s}_{n})^{m+2}}{m+2}+o(1)$as$narrow\infty$.

Therefore the third term is equal to$o(1)- \frac{bv(\overline{s}_{n})^{m+2}}{m+2}$

as$narrow\infty$.

To evaluate the fourth term we employ the mean value theorem to obtain$(v+\dot{v})^{1+\rho}=v^{1+\rho}+(1+\rho)(v+\theta(r)\dot{v})^{\rho}\dot{v}$where$\theta(r)$is a quantity between$0$,

and 1. Hence we can compute$\int_{b}^{\overline{s}_{\hslash}}(v+\dot{v})^{1+\rho}\dot{v}v^{\beta}dr=\int_{L}^{\overline{*}n}v^{1+\rho+\beta}\dot{v}dr+(1+\rho)\int_{1}^{\overline{*}n}(v+\theta(r)\dot{v})^{\rho}\dot{v}^{2}v^{\beta}dr$as$= \frac{v(\overline{s}_{n})^{2+\rho+\beta}-v(\underline{s}_{n})^{2+\rho+\beta}}{2+\rho+\beta}+(1+\rho)\int_{\underline{s}_{n}}^{\overline{l}n}o(1)\dot{v}^{2}dr=\frac{v(\overline{s}_{n})^{2+\rho+\beta}}{2+\rho+\beta}+o(1)$$narrow\infty$.

Finally by Schwarz’s inequality we find that the last term is dominated by the quantity

const

as

$( \int_{l}^{\overline{s}_{\hslash}}arrow\delta(r)^{2}dr)^{1/2}(\int_{\underline{\epsilon}_{n}}^{\overline{*}}n\dot{v}^{2}dr)^{1/2}=o(1)$$narrow\infty$.

Consequently, from (10) we obtain the formula$o(1)- \frac{b}{m+2}v(\overline{s}_{n})^{m+2}+\frac{b}{2+\rho+\beta}v(\overline{s}_{n})^{2+\rho+\beta}+o(1)=0$

as$narrow\infty$.

This implies that $\lim_{narrow\infty}v(\overline{s}_{n})=[(m+2+\beta-\alpha)/(m+2)]^{1/\beta}$ . Since $m$ can be moved
arbitrarily, this is an obvious contradiction. Therefore $\lim\inf_{*arrow\infty}v>0$ .
Step 2. We claim that $\lim_{*arrow\infty}\dot{v}(s)=0$ . Since $\lim\inf_{arrow\infty}v(s)>0$ by Step 1,
we find that $\lim\inf_{tarrow\infty}u(t)/u_{0}(t)>0$ . Integrating equation (5), we further find that$\lim\inf_{tarrow\infty}u’(t)/u_{0}’(t)>0$ . Since $v+\dot{v}=u’(t)/u_{0}’(t)$ , we obtain $\lim\inf_{*arrow\infty}(v+\dot{v})>0$.
Recalling equation (6), we find that $\ddot{v}(s)=O(1)$ as. Since we have already known
that), Lemma 3 shows that.
Step 3. We claim that. To see this, we integrate (6) multiplied by ab:$sarrow\infty$$\dot{v}\in L^{2}[s_{0}, \infty$$\lim_{arrow\infty}\dot{v}=0$$\lim_{\epsilonarrow\infty}v(s)=1$$\frac{\dot{v}^{2}}{2}-a\int_{\epsilon_{0}}^{*}\dot{v}^{2}dr-\frac{b}{2}v^{2}+b\int_{s0}^{s}(\dot{v}+v)^{1-\alpha}v^{\beta}\dot{v}dr$$+b \int_{l_{0}}^{l}\delta(r)(\dot{v}+v)^{1-\alpha}v^{\beta}\dot{v}dr=const$

.

(11)

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Suppose that condition (3); namely (7) holds. Since), the first, and thethird integrals in the left hand side of (11) converge as $sarrow\infty$ . The mean value theorem
shows that,
(12)$\dot{v}\in L^{2}[s_{0}, \infty$$(v+\dot{v})^{1-\alpha}=v^{1-\alpha}+(1-\alpha)(v+\theta(r)\dot{v})^{-\alpha}\dot{v}$where$\theta(r)$is a quantity satisfying $0<\theta(r)<1$ . Therefore,$/s_{0^{*}}( \dot{v}+v)^{1-\alpha}v^{\beta}\dot{v}dr=\int_{\epsilon_{0}}^{s}\{v^{1-\alpha+\beta}\dot{v}+(1-\alpha)(v+\theta(r)\dot{v})^{-\alpha}v^{\beta}\dot{v}^{2}\}dr$$= \frac{v(s)^{2+\beta-\alpha}}{2+\beta-\alpha}+\int_{*0}^{l}o(1)\dot{v}^{2}dr+const$.

So we find that the $function-v^{2}/2+v^{2+\beta-\alpha}/(2+\beta-\alpha)$ has a finite limit. This fact showsthat $\ell=\lim_{sarrow\infty}v(s)\in(O, \infty)$ exists. Next suppose that (4); namely (8) holds. We haveby (12)$\int_{\epsilon_{0}}^{*}\delta(r)(\dot{v}+v)^{1-\alpha}v^{\beta}\dot{v}dr=\int_{s_{0}}\{\delta(r)v^{1-\alpha+\beta}\dot{v}+\delta(r)(1-\alpha)(v+\theta(r)\dot{v})^{-\alpha}v^{\beta}\dot{v}^{2}\}dr$$= \frac{\delta(s)v^{2+\beta-\alpha}}{2+\beta-\alpha}-\frac{1}{2+\beta-\alpha}\int_{\epsilon 0}^{\epsilon}\dot{\delta}(r)v^{2+\beta-\alpha}dr+const+/0^{*}0(1)\dot{v}^{2}dr$as. Hence, as before we know that the $function-v^{2}/2+v^{2+\beta-\alpha}/(2+\beta-\alpha)$ hasa finite limit. Therefore $\ell=\lim_{*arrow\infty}v(s)\in(O, \infty)$ exists.
Finally, we let $sarrow\infty$ in equation (6). Then, we have.
$\ell=1$$\dot{v}=o(1)$, we must have
, implying
Since. The proof of Theorem 2
is completed.
$sarrow\infty$$\lim_{sarrow\infty}\ddot{v}(s)=b(\ell-\ell^{1+\beta-\alpha})$$\lim_{sarrow\infty}\ddot{v}(s)=0$

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Hiroyuki Usami:
Department of Mathematics, Graduate School of Science,
Hiroshima University,
Higashi-Hiroshima, 739-8521, Japan

usami@mis.hiroshima-u.ac.jp