数理解析研究所講究録
第 1838 巻 2013 年 64-7064

UNIQUENESS OF POSITIVE RADIAL SOLUTIONS OF$\Delta u+g(r)u+h(r)u^{p}=0$ AND ITS APPLICATIONS
横浜国立大学大学院工学研究院

塩路直樹 (NAOKI SHIOJI)

FACULTY OF ENGINEERING, YOKOHAMA NATIONAL UNIVERSITY

防衛大学校情報工学科

渡辺宏太郎 (KOHTARO WATANABE)

DEPARTMENT OF COMPUTER SCIENCE, NATIONAL DEFENSE ACADEMY

1. INTRODUCTION AND MAIN RESULTS
We consider the problem$[Matrix]$(1.1)

are appropriate functions.
:where $n\geq 2,$ $R\in(0, \infty], p\in(1, \infty)$ andHere, $u(R)=0$ in the case $R=\infty$ means $u(x)arrow 0$ as $|x|arrow\infty$ . Such a problem has
been studied by many researchers; see [1,3, 5,8,9, 12-18, 20-27,30, 32-36] and others.
In this note, we introduce a result obtained in [28].
$g,$$h$$(0, R)arrow \mathbb{R}$with $n\geq 2$ and $p\in(1, \infty)$ . Let $g\in C([O, R))\cap$
Theorem 1. Let $0<R\leq\infty,$$C^{1}((0, R))$ and $h\in C^{2}([0, R))\cap C^{3}((0, R))$ such thatis positive on $[0, R)$ . with $R’=0.$
Assume the following.
$n\in \mathbb{R}$$h$(i) In the case of $R<\infty,$ $g\in C([O, R]),$ $h\in C^{2}([0, R])$ and $h(R)>0$ are alsosatisfied.
(ii) There exists $\kappa\in[0, R]$ such that
$G(r)\geq 0on(O, \kappa)$

and$G(r)\leq 0on(\kappa, R)$,

where$G(r)= \frac{r^{p+3^{-3}}2n-1\ovalbox{\tt\small REJECT}+1}{2(p+3)^{3}h(r)^{\frac{2}{p+3}+3}}(4(n-1)[n+2-(n-2)p][n-4+(n-2)p]h(r)^{3}$$+[2(n-1)(p-1)(p+3)^{2}r^{2}h(r)^{3}-4(p+3)^{2}r^{3}h(r)^{2}h_{r}(r)]g(r)$$+(p+3)^{3}r^{3}g_{r}(r)h(r)^{3}$$+(n-1)[(2n-3)p(6-p)+6n-33]rh(r)^{2}h_{r}(r)$$+3(n-1)(p-1)(p+5)r^{2}h(r)h_{r}(r)^{2}-2(p+4)(p+5)r^{3}h_{r}(r)^{3}$

65$-3(n-1)(p-1)(p+3)r^{2}h(r)^{2}h_{rr}(r)$$+3(p+3)(p+5)r^{3}h(r)h_{r}(r)h_{rr}(r)-(p+3)^{2}r^{3}h(r)^{2}h_{rrr}(r))$(iii) In the case of $R=\infty,$$G^{-}\not\equiv 0$.

is satisfied.

Then in the case of $R<\infty$ , problem (1.1) has at most one positive solution, and in
the case of $R=\infty$ , problem (1.1) has at most one positive solution which satisfies$J(r;u)arrow 0$ as, where$u$$rarrow\infty$$a(r)=r^{2n-1}\simeq_{p+3}\ovalbox{\tt\small REJECT}+旦_{}h(r)^{\frac{-2}{p+3}},$,$b(r)= \frac{r^{2n-}p+3^{-1}\ovalbox{\tt\small REJECT}^{1+1}}{(p+3)h(r)p^{\frac{+5}{+3}}R}(2(n-1)h(r)+rh_{r}(r))$$\ovalbox{\tt\small REJECT}_{-2}2n-1+1$$c(r)= \frac{rp+3}{(p+3)^{2}h(r)^{2}p+3\Delta e+A^{4}}(2(n-1)[n+2-(n-2)p]h(r)^{2}+(p+5)r^{2}h_{r}(r)^{2}$$-(n-1)(p-5)rh(r)h_{r}(r)-(p+3)r^{2}h(r)h_{rr}(r))$ ,$J(r;u)= \frac{1}{2}a(r)u_{r}(r)^{2}+b(r)u_{r}(r)u(r)+\frac{1}{2}c(r)u(r)^{2}$$+ \frac{1}{2}a(r)g(r)u(r)^{2}+\frac{1}{p+1}a(r)h(r)u(r)^{p+1}.$

Remark 1. In [32, Theorems 2.1 and 2.2], Yanagida obtained a closely related result.
By the theorem above, we can obtain the following; see [13, Theorem 0.1].

Corollary 1 (Kabeya-Tanaka). Letwith $n\geq 2$ . Let $p>1$ and $g\in C^{2}([0, \infty))$such $that- \infty<\inf_{r\in[0,\infty)}g(r)\leq\sup_{r\in[0,\infty)}g(r)<0$ , and set$n\in \mathbb{N}$$L= \frac{2(n-1)[(n-2)p+n-4]}{(p+3)^{2}}$and$\beta=\frac{2(n-1)(p-1)}{p+3}.$

Assume that$g_{r}(r)r^{3}+\beta g(r)r^{2}-(\beta-2)L<0$for each$r\geq 0$

in the case of $n=2$ , and that $p<(n+2)/(n-2)$ and$\sup_{r>0}(g_{rr}(r)r^{2}+(3+\beta)g_{r}(r)r+2\beta g(r))<0$

in the case of $n\geq 3$ . Then the problem(1.2)$u\in H^{1}(\mathbb{R}^{n}) , \triangle u(x)+g(|x|)u(x)+u(x)^{p}=0 in\mathbb{R}^{n}$has a unique positive radial solution.

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Next, we consider the problem(1.3)$\{\begin{array}{ll}u_{rr}(r)+\frac{n-1}{r}u_{r}+g(r)u(r)+h(r)u(r)^{p}=0, R’<r<R,u(R’)=0, u(R)=0.

\end{array}$

The uniqueness of a positive solution of such a problem was studied in [4,6, 7, 10, 11, 19,24, 29-31].
The following is also obtained in [28].
$p\in(1, \infty),$ $g\in C([R’, R))\cap C^{1}((R’, R))$ ,Theorem 2. Let $0<R’<R\leq\infty,$
and be$h\in C^{2}([R’, R))\cap C^{3}((R’, R))$ such thatis positive on $[R’, R)$ . Let
the functions given in Theorem 1. Assume the following.
$n\in \mathbb{R},$$h$$a,$$b,$$c,$$G$$J$(i) In the case of $R<\infty,$ $g\in C([R’, R]),$ $h\in C^{2}([R’, R])$ and $h(R)>0$ are alsosatisfied.
(ii) There exists $\kappa\in[R’, R]$ such that
$G(r)\geq 0$on$(R’, \kappa)$and$G(r)\leq 0$on$(\kappa, R)$.

Then in the case of $R<\infty$ , problem (1.3) has at most one positive solution, and in
the case of $R=\infty$ , problem (1.3) has at most one positive solution which satisfies$u$$J(r;u)arrow 0$

as$rarrow\infty.$

Remark 2. For the case $h(r)\equiv 1$ , a similar result is obtained by Felmer-Mart\’inez-Tanaka;
see [10, Theorem 1.1].
2. APPLICATIONSIn this section, we give examples of Theorem 1. First, we give a comment on the scalar
field equation$\Delta u(x)-u(x)+u(x)^{p}=0$in$\mathbb{R}^{n},$$u(x)arrow 0$

as$|x|arrow\infty.$

The unique existence of its positive solution was established by Kwong [18]. Since theuniqueness of its positive solution can be derived from Corollary 1, of course, it can bealso done by Theorem 1.
Next, we consider the following Brezis-Nirenberg problem.
(2.1)$\{\begin{array}{l}\Delta_{S^{n}}u+\lambda u+u^{p}=0 in D,u=0on\partial D.\end{array}$is the unit sphere inHere,is a natural number with $n\geq 3,$$D=\{X\in S^{n} : X_{n+1}>\cos\theta_{1}\}$ with
Laplace-Beltrami operator on$S^{n}$$n$$S^{n},$$\mathbb{R}^{n+1},$$\Delta_{S^{n}}$is the,$\theta_{1}\in(0, \pi)$67$1<p\leq(n+2)/(n-2)$ and$\lambda<\lambda_{1}$, wherethe Dirichlet boundary condition.
Let$P:S^{n}\backslash \{(0, \ldots, 0, -1)\}arrow \mathbb{R}^{n}$$\lambda_{1}$is the first eigenvalue$of-\triangle_{S^{n}}$

on

$D$

with

be the stereographic projection defined by$P(X_{1}, \ldots, X_{n}, X_{n+1})=\frac{1}{X_{n+1}+1}(X_{1}, \ldots, X_{n})$for $X\in S^{n}\backslash \{(0, \ldots, 0, -1)\}.$

Then we can see $P(D)=B_{R}$ , where $B_{R}=\{x\in \mathbb{R}^{n}:|x|<R\}$ with
$R= \frac{\sin\theta_{1}}{1+\cos\theta_{1}}.$Let
for$u$

be a positive solution of (2.1) and define
by $u(P^{-1}x)=(1+|x|^{2})^{\frac{n-2}{2}}v(x)$. Then we see that is a positive solution of$v:\overline{B_{R}}arrow \mathbb{R}$$x\in\overline{B_{R}}$$v$$\Delta v+\frac{n(n-2)+4\lambda}{(1+|x|^{2})^{2}}v+4(1+|x|^{2})^{(n-2)p-(n+2)}$$\overline{2}v^{p}=0$

in

$v=0$

on

$B_{R},$$\{$$\partial B_{R}.$We setand$g(r)= \frac{n(n-2)+4\lambda}{(1+r^{2})^{2}}$We can see that$G$$h(r)=4(1+r^{2})^{\frac{(n-2)p-(n+2)}{2}}$for $r\geq 0.$

in Theorem 1 is given by$G(r)= \frac{2^{L_{\frac{1}{3}}^{-}}p+(n-1)}{(p+3)^{3}}r^{2n-1+1}p(1\ovalbox{\tt\small REJECT}_{+3^{-3}}+r^{2})^{\frac{n+2-(n-2)p}{p+3}3}(1-r^{2})(Ar^{4}+Br^{2}+A)$

,

where
$A=(n-2)^{2}( \frac{n+2}{n-2}-p)(p+\frac{n-4}{n-2})$$=(p+3)[3n^{2}-6n-(n^{2}-4n+4)p]-8(n-1)^{2},$$B=(p+3)[-6n^{2}+12n+(2n^{2}+4\lambda-4)p+2\lambda p^{2}-6\lambda-12]+16(n-1)^{2}.$Then we can infer the following. For the details, see [28].
Theorem 3. Letwith $n\geq 3,1<p\leq(n+2)/(n-2)$ andthat one of the following conditions:$n\in \mathbb{N}$(i)(ii)$\theta_{1}\in(0, \pi/2]$$\theta_{1}\in(\pi/2, \pi)$

and
and

$\theta_{1}\in(0, \pi)$. Assume$\lambda<\lambda_{1},$$\frac{6+(6-4n)p}{(p+3)(p-1)}\leq\lambda<\lambda_{1}.$Then (2.1) has at most one positive radial solution. Moreover, if $\lambda\geq-n(n-2)/4$ is alsosatisfied, then (2.1) has at most one positive solution.

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Remark 3. It holds that$\frac{6+(6-4n)p}{(p+3)(p-1)}\leq-\frac{n(n-2)}{4},$and if $p=(n+2)/(n-2)$ then the constants in the both sides in the inequality above
coincide.
Remark 4. In the case of $n=3$ , Bandle-Benguria obtained a sharper result. For thedetails, see [2].

Remark 5. In the case of $R>1$ , we cannot apply Yanagida’s uniqueness theorem [32,Theorem 2.1]. Indeed, by his notation, we have$G(r;n-2)= \frac{2(4\lambda+n(n-2))r^{n-1}(1-r^{2})}{(r^{2}+1)^{3}}.$

So one of his assumptions $G(r;n-2)\leq 0$ on $(0, R)$ is not satisfied even if $\lambda>-n(n-2)/4.$
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