数理解析研究所講究録
1197 巻 2001 年 207-226207

球対称なポテンシャルを含む半線形楕円型方程式の正細細についてギ

On Positive Solutions for Semilinear Elliptic Equations

with Radially Symmetric Potential
東京大学大学院数理科学研究科

廣瀬宗光 (Munemitsu$\mathrm{H}\mathrm{i}\mathrm{r}\mathrm{o}\mathrm{s}\mathrm{e}^{*}$)

Graduate School of Mathematical Sciences, University of Tokyo. This is a joint work with Masahito Ohta (Faculty of Engineering, Shizuoka University).

$\cross$

1. INTRODUCTION AND MAIN RESULTS
In this paper we consider positive solutions of the following nonlinear elliptic equation
with a harmonic potential term$|x|^{2}u$$-\Delta u+(\lambda+|x|^{2})u-|u|^{P^{-1}}u=0$

where$\lambda\in \mathbb{R}$,$x\in \mathbb{R}^{n}$

,

(1.1)

and $p>1$ . This problem arises in the study of standing wave solutions$\psi(t, x)=\exp(i\lambda t)u(x)$for the nonlinear Schr\"odinger equation with a harmonic potential$i \frac{\partial\psi}{\partial t}=-\Delta\psi+|x|^{2}\psi-|\psi|^{p-1}\psi$,$(t, x)\in \mathbb{R}^{1+n}$

,

(1.2)

which is a model equation to describe the Bose-Einstein condensate with attractive interparticle interactions under a magnetic trap (see, e.g., [16]). We see that $\psi(t, x)=\exp(i\lambda t)u(x)$*Supported in part by JSPS Research Felowships for Japanese Young Scientists and Grant-in-Aid for

JSPS Research Fellowships (11-8942).

208
is a solution of (1.2) if $u(x)$ satisfies (1.1). Since (1.2) has two conserved quantities, theenergy and the particle number,$E( \psi):=\int_{\mathrm{R}^{n}}(\frac{1}{2}|\nabla\psi|^{2}+\frac{1}{2}|X|2|\psi|2-\frac{1}{p+1}|\psi|p+1)d_{X}$$\cdot$$N( \psi):=\int_{\mathrm{J}\mathrm{R}^{n}}|\psi|^{2}d_{X}$,

it is natural to study the solutions of (1.1) and (1.2) in the energy space$\Sigma:=\{u\in L^{2}(\mathbb{R}^{n})$Since the embedding$\Sigma\mapsto L^{q}(\mathbb{R}^{n})$:$\int_{\mathbb{R}^{n}}(|u|^{2}+|\nabla u|^{2}+|xu|^{2})dx<\infty\}$.

is compact for $2\leq q<2n/(n-2)l+$ , by the standard

variational method, we can prove that there exists at least one solution of$x\in \mathbb{R}^{n}$,$\{u\in\Sigma,u(-\Delta u+(\lambda+|x|^{2})u-|u|^{p-}1=X)>0\mathrm{f}_{0}\mathrm{r}\mathrm{a}11xu0\in \mathbb{R}^{n}’,$(1.3)

if $n\geq 1,$ $\lambda>-n$ and $1<p<(n+2)/(n-2)^{+}$ (see [5] for details). Here, we note that thecondition $\lambda>-n$ appears naturally to show the existence of solutions for (1.3), because
the first eigenvalue$\mathrm{o}\mathrm{f}-\Delta+|x|^{2}$on$\Sigma$

is equal to$n$and the corresponding eigenfunctions

are $C\exp(-|X|^{2}/2)$ . Recently, the stability of standing waves for (1.2) has been studied by$[5, 21]$. For studying the stability of standing waves, it is important and fundamental toinvestigate the structure of solutions to (1.3) (see, e.g., [2, 3, 6, 13, 14, 17]).

In [8], we have proved the uniqueness of solution for (1.3) when $n\geq 3$ as follows.

Theorem 1.1. (See [8].) Assume $n\geq 3,$ $\lambda>-n$ and $1<p<(n+2)/(n-2)$ . Then (1.3)
has a unique solution.

In this paper, we will show the uniqueness of solution for (1.3) in case $n=2$ . For related, we refer the reader to [9, 10, 18] and the references cited therein.

$\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{S}\mathrm{s}..\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{u}\mathrm{l}\mathrm{t}_{\mathrm{S}}$By a bootstrap argument using the fact that$(-\Delta+|_{X|)^{-}}2+11Lq(\mathbb{R}^{n})=\{v\in W2,q(\mathbb{R}^{n}) :|x|^{2}v\in L^{q}(\mathbb{R}^{n})\}$

209
if $n\geq 1$ and $1<q<\infty$ (see, e.g., [12, Theorem 2.5]), it is shown in a similar way as in [2,Theorem 8.1.1] that all solutions of (1.3) belong to$C^{2}(\mathbb{R}^{n})$and satisfy $\lim_{|x|arrow\infty^{u}}(x)=0$(see [5] for details). Moreover, by [11, Theorem 2], we see that all solutions of (1.3) areradially symmetric about the origin. Therefore, the problem for solutions of (1.3) is reduced
to that for radial solutions of (1.3). Since we are interested in radial solutions $(u=u(r)$with $r=|x|$ ) of (1.3), we study the initial value problem$u”+ \frac{n-1}{r}u’-(\lambda+r^{2})u+|u|^{P^{-1}}u=0$,$r>0$ ,

(1.4)

$\{$

$u(\mathrm{O})=\alpha>0$,$u’(0)=0$ ,where the prime denotes the differentiation with respect to . In Section 2, it will be shown$r$that (1.4) has a unique global solution $u(r)\in C^{2}([0, \infty))$ , which is denoted by $u(r;\alpha)$ . We

classify $u(r;\alpha)$ as follows:(i) $u(r;\alpha)$ is a crossing solution if $u(r;\alpha)$ has a zero in$z\in(\mathrm{O}, \infty)$$(0, \infty)$, i.e., there exists somesuch that $u(z;\alpha)=0$ .

(ii) $u(r;\alpha)$ is an entirely positive solution if $u(r;\alpha)>0$ for all $r\in[0, \infty)$ .

Moreover, we define$\Sigma_{rad}:=\{u\in C^{1}([\mathrm{o}, \infty))$: $\int_{0}^{\infty}(|u|^{2}+|u’|^{2}+|ru|^{2})r-1dnr<\infty\}$ .

Then our main result is the following.

Theorem 1.2. If $n=2,$ $\lambda>-n$ and $1<p<\infty_{f}$ then there exists a unique positive
number$\alpha_{0}$such that the structure of positive solutions to (1.4) is as follows.

(a) For every(b) If$\alpha=\alpha_{0;}$$\alpha\in(\alpha_{0}, \infty),$$u(r;\alpha)$

is a crossing solution.

then $u(r;\alpha)$ is an entirely positive solution with $u(r;\alpha)\in\Sigma_{tad}$ and satisfies$\lim_{farrow\infty}r^{(\lambda}\mathrm{e}\mathrm{x}n+)/2\mathrm{p}(r^{2}/2)u(r;\alpha)\in(0, \infty)$(c) For every$\alpha\in(0, \alpha_{0})_{f}u(r;\alpha)$.

is an entirely positive solution with

(1.5)
$u(r;\alpha)\not\in\Sigma_{\mathrm{r}ad}$.

210
Since, as stated above, all solutions of (1.3) belong to$C^{2}(\mathbb{R}^{n})$and are radially symmetric

about the origin, as a corollary of Theorem 1.2, we have
Theorem 1.3. If $n=2,$ $\lambda>-n$ and $1<p<\infty$ , then (1.3) has a unique solution.
Remark 1.1. The uniqueness question for solutions of (1.3) seems to be open for $n=1$ .

In order to prove Theorem 1.2, we apply the classification theorem by Yanagida andYotsutani $[19, 20]$ . Let$\varphi(r)$be a solution of$\varphi’’+(\frac{n-1}{r}+2r)\varphi’+(n-\lambda)\varphi=0$,$r>0$ ,

(1.6)

$\{$

$\varphi(0)=1$,$\varphi’(0)=0$.

For a solution $u(r)$ of (1.4), if we put$u(r)=\exp(r^{2}/2)\varphi(r)v(r)$

,

(1.7)

then we see that $v(r)$ satisfies$(g(r)v’)’+g(r)K(r)|v|^{\mathrm{P}^{-1}}v=0$,$r>0$ ,

(1.8)

$\{$

$v(\mathrm{O})=\alpha>0$,$v’(\mathrm{O})=0$

,

where
$g(r):=r^{n-1}\exp(r^{2})\varphi(r)^{2}$

We should note that $\varphi(r)>0$ on,$[0, \infty)$$K(r):= \exp(\frac{p-1}{2}r^{2})|\varphi(r)|p-1$

.

(1.9)

if $\lambda>-n$ by (i) of Proposition 2.2 in Section 2.

To see whether $u(r)$ has a zero or not, we have only to check this property for $v(r)$ . For
this purpose, we employ the classification theorem by Yanagida and Yotsutani [20], whichis stated as follows. Let $g(r)$ and $K(r)$ satisfy$g(r)\in C^{\mathit{2}}([0, \infty))$$g(r)>0$on;$(0, \infty)$

;
$(g)$

$\{$

$1/g(r)\not\in L^{1}(0,1)$;$1/g(r)\in L^{1}(1, \infty)$

,

211
and
$K(r)\in C(\mathrm{o}, \infty)$$K(r)\geq 0$

;

and

on

$K(r)\not\equiv \mathrm{O}$$(0, \infty)$

;
$(K)$

$\{$

$h(r)K(r)\in L^{1}(0,1)$ ;$g(r)(h(r)/g(r))^{p}K(r)\in L^{1}(1, \infty)$ ,

where$h(r):=g(r) \int_{f}^{\infty}g(s)-1d_{S}$.

Moreover, define$G(r):= \frac{2}{p+1}g(r)h(r)K(r)-\int_{0}’g(s)K(s)ds$ ,(1.10)$H(r):= \frac{2}{p+1}h(r)^{2}(\frac{h(r)}{g(r)})^{p}K(r)-I,.\infty h(S)(\frac{h(s)}{g(s)})^{p}K(S)ds$(1.11),

and$r_{G}:= \inf\{r\in(0, \infty) :G(r)<0\}$,$r_{H}:= \sup\{r\in(0, \infty) :H(r)<0\}$.

Theorem A (Yanagida and Yotsutani [19, 20]). Assume that $g(r)$ and $K(r)$ satisfy theconditions $(g)$ and $(K)$ . Let $v(r;\alpha)$ be a solution of$(g(r)v)\prime\prime+g(r)K(r)(v)^{\mathrm{p}}+=0$,$r>0$ ,

(1.12)$\{$$v(\mathrm{O})=\alpha>0$,$v’(\mathrm{O})=0$

where $v^{+}:= \max\{v, 0\}_{f}$ and suppose that

$G(r)\not\equiv \mathrm{O}$,on$(0, \infty)$.

(i) If$0<r_{H}\leq r_{G}<\infty$

then there exists a unique positive number$\alpha_{0}$,(1.13)

such that the structure of solutions to(1.12) is as follows.
(a) For every$\alpha\in(\alpha_{0}, \infty)_{\mathrm{z}}v(r;\alpha)$

has a zero in$(0, \infty)$.

212(b) If$\alpha=\alpha_{0}$, then $v(r;\alpha)>0$ on$[0, \infty)$

and$0< \lim_{farrow\infty}(\int_{T}^{\infty}g(s)-1ds)^{-}1rv(;\alpha)<\infty$(c) For every$\alpha\in(0, \alpha 0),$$v(r;\alpha)>0$

on$[0, \infty)$.

(1.14)

and$\lim_{farrow\infty}(\int^{\infty},g(S)-1ds)-1rv(;\alpha)=\infty$.

(1.15)(ii) If $r_{G}<\infty$ and $r_{H}=0(i.e., H(r)\geq 0$ on $[0, \infty))$ , then $v(r;\alpha)$ is positive on$[0, \infty)$

and satisfies (1.15) for every $\alpha>0$ .
(iii) $Ifr_{G}=\infty(i.e., G(r)\geq 0$ on $[0, \infty))$ , then $v(r;\alpha)$ has a zero in

Remark 1.2. Note that if $v(r;\alpha)$ is positive onor (1.15), because$( \int_{f}^{\infty}g(S)^{-}1ds)-1v(r;\alpha)$$[0, \infty)$$(0, \infty)$

for every$\alpha>0$.

, then $v(r;\alpha)$ satisfies either (1.14)

is non-decreasing on$(0, \infty)$.

In Section 3, noting that $g(r)$ and $K(r)$ given by (1.9) satisP the assumptions $(g)$ and$(K)$, we will prove the following proposition.

Proposition 1.1. If $n=2_{f}\lambda>-n$ and $1<p<\infty_{f}$ then condition (1.13) holds for (1.8)
with (1.9).
By the proposition above, we see that $u(r;\alpha)$ is an entirely positive solution forand a crossing solution for$\alpha\in(\alpha_{0}, \infty)$$\alpha\in(0, \alpha_{0}]$. Moreover, for the asymptotic behaviour of entirely

positive solutions, we will prove the following proposition in Section 5.

Proposition 1.2. Let $u(r;\alpha)$ andwith (1.9), respectively. Then, the(i)$u(r;\alpha)\in\Sigma_{\mathrm{r}ad}$

,

(ii)

$v(r;\alpha)$be entirely positive solutions of (1.4) and (1.8)$fol\iota_{\mathit{0}}u\dot{n}ng$$u(r;\alpha)$

three conditions are equivdent:

satisfies (1.5),(iii)$v(r;\alpha)$satisfies (1.14).

Theorem 1.2 follows from Theorem A and Propositions 1.1 and 1.2.

213
2. PRELIMINARIESIn this section, we prepare some results which have been shown in [8]. First, we give thefollowing proposition.
Proposition 2.1. The initial value problem (1.4) has a unique global solution $u(r;\alpha)$ in$C^{2}([\mathrm{o}, \infty))$for every$\alpha>0$.

Next, we study the properties of solutions to the initial value problem (1.6). It is shownthat there exists a unique solution $\varphi(r)\in C^{2}([0, \infty))$ of (1.6) by a standard way. Moreover,

we obtain the following proposition, which plays an important role in what follows.
Proposition 2.2. Assume $\lambda>-n$ and let$\varphi(r)\in C^{2}([\mathrm{o}, \infty))$be the unique solution of(1.6). Then we have(i) $\varphi(r)>0$ on$[0, \infty)$

(ii)
(iii)

if

$[0, \infty)$$\lambda>n$. Moreover, $\varphi’(r)<0$ on. Especially,$m:= \sup_{\mathrm{r}\geq 0}|\frac{r\varphi’(r)}{\varphi(r)}|$is finite, and$\frac{r\varphi’(r)}{\varphi(r)}=\frac{\lambda-n}{2}+O(\frac{1}{r^{2}})$(iv) The limit$\varphi(r)\equiv 1$as$r^{m}\varphi(r)$$rarrow\infty$$L:= \lim_{farrow\infty}r^{(n-}\varphi(\lambda)/2r)$

on$[0, \infty)$$[0, \infty)$for$if- n<\lambda<n$ and $\varphi’(r)>0$ on$\lambda=n$.

is non-decreasing on$[0, \infty)$.

.

exists in$(0, \infty)$.

(v) There exist positive constants $C$ and $R$ such that$|r^{n} \exp(r^{2})\varphi(r)^{2}\int_{f}^{\infty}s^{1}-n\exp(-s^{2})\varphi(S)^{-2}ds-(\frac{1}{2}-\frac{\lambda}{4}r^{-2})|\leq Cr^{-4}$holds for all $r\geq R$ .
(vi)$\frac{r\varphi’(r)}{\varphi(r)}=\frac{\lambda-n}{n}r^{2}+o(r^{2})$as$rarrow \mathrm{O}$.

3. PROOF OF PROPOSITION 1. 1
In this section, we give the proof of Proposition 1.1 by applying Theorem A. In order toapply Theorem$\mathrm{A}$, we first check the conditions imposed on the coefficients of (1.12).

214
Lemma 3.1. If $n=2$ and $\lambda>-n$ , then $g(r)$ and $K(r)$ given by (1.9) satisfy the assumptions $(g)$ and $(K)$ .
Therefore, $g(r)$ and $K(r)$ given by (1.9) are admissible. Inserting their definition (1.9)

into (1.10) and (1.11), we obtain$G(r)= \frac{2}{p+1}r^{2n-}\exp 2(\frac{p+3}{2}r^{2})\varphi(r)\mathrm{P}+3\int_{f}^{\infty}s^{1}-n\exp(-s^{2})\varphi(S)^{-2}d_{S}$(3.1)$- \int_{0}^{r}s^{n}-1\exp(\frac{p+1}{2}S^{2)d}\varphi(s)^{\mathrm{P}}+1s$,$H(r)= \frac{2}{p+1}r^{2n-2}\exp(\frac{p+3}{2}r^{2})\varphi(r)p+3(\int_{f}^{\infty}S^{1}-n)\exp(-s^{2})\varphi(S-2ds)^{p2}+$$- \int_{r}^{\infty}s^{n-1}\exp(\frac{p+1}{2}S^{2})\varphi(S)p+1(\int_{s}^{\infty}t^{1-}n\exp(-t)2\varphi(t)^{-2}dt)^{\mathrm{p}+}1d_{S}$.

In order to show (1.13), we investigate the profiles of $G(r)$ and $H(r)$ . First, we study theincrease and decrease. Differentiating (1.10) and (1.11), we obtain$c’(r)=( \int_{f}^{\infty}g(S)^{-}1d_{S))}-\mathrm{P}-1H’(r=\frac{2}{p+1}g(r)K(r)(\Phi(r)-\frac{p+3}{2}\mathrm{I}$

,

(3.2)

where
$\Phi(r):=(2g’(r)+\frac{g(r)K’(r)}{K(r)})\int_{f}^{\infty}g(_{S})^{-1}ds$$=r^{n-2} \exp(r^{2})\varphi(r)^{2}\{(p+3)(r^{2}+\frac{r\varphi’(r)}{\varphi(r)})+2(n-1)\}$$\cross\int^{\infty}s^{1-n}\exp(-s^{2})\varphi(S)^{-}2ds$

(3.3)

.

In view of (3.2), $G(r)$ and $H(r)$ have the same extremal points, namely those $r>0$ which
satisfy $\Phi(r)=(p+3)/2$ . So, to know the sign of $G’(r)$ and $H’(r)$ , we need to study the
relation between$r=\infty$$\Phi(r)$and $(p+3)/2$ . We first study the behaviour ofby using Proposition 2.2.

Lemma 3.2. If $n=2$ and $\lambda>-n$ , then$\lim_{farrow 0^{\Phi}()}r=+\infty$.

$\Phi(r)$near $r=0$ and215
Lemma 3.3. If $\lambda>-n$ , then$\Phi(r)=\frac{p+3}{2}-\frac{n(p-1)+4}{4}r^{-2}+O(r^{-4})$Noting that$\Phi(r)$is continuous in$[0, \infty)$as$rarrow\infty$.

, we see that there exists at least one crossing

point of $y=\Phi(r)$ and $y=(p+3)/2$ in $(r, y)$ -plane by Lemmas 3.2 and 3.3. The following

lemma will be proved in Section 4.
Lemma 3.4. If $n=2,$ $\lambda>-n$ and 1 $<p<\infty$ , then there exists a unique number$r_{*}\in(0, \infty)$

satisfying $\Phi(r_{*})=(p+3)/2$ such that

Therefore, from (3.2) and Lemma 3.4, we have

Lemma 3.5. If $n=2,$ $\lambda>-n$ and $1<p<\infty$ , then there exists a unique number$r_{*}\in(0, \infty)$

such that $G(r)$ and $H(r)$ are increasing on

Moreover, in order to locate$r_{G}$

and$r_{H}$$[0, r_{*})$and decreasing on$(r_{*}, \infty)$.

, we need to investigate the behaviour of $G(r)$and $H(r)$ near $r=0$ and $r=\infty$ by using Proposition 2.2.

Lemma 3.6. $\wedge Assumen=2_{f}\lambda>-n$ and $1<p<\infty$ . Then we have
(i)
(iii)

$\lim_{farrow\infty}G(r)=-\infty$$\lim_{farrow\infty}H(r)=0$

,

,

,

(ii)

$\lim_{farrow 0}c(r)=0$(iv)$\lim_{farrow 0}H(r)\in[-\infty, 0)$.

Now we prove Proposition 1.1.

Proof of Proposition 1.1. As is already seen in Lemma 3.5, both $G(r)$ and $H(r)$ haveexactly one local$\mathrm{m}\partial_{}\dot{\mathrm{n}}\mathrm{m}\mathrm{u}\mathrm{m}$at $r_{*}\in(0, \infty)$ . Moreover, in view of Lemma 3.6, $H(r)$ isnegative near $r=0$ and positive for large . Thus $H(r_{*})>0$ and $0<r_{H}<r_{*}$ . Fhrthermore,
$r$

216
we obtain $G(r_{*})>0$ from$r_{G}<\infty$$G(\mathrm{O})=0$, and the negativity of $G(r)$ for large$r$yields $0<r_{*}<$; so we conclude that condition (1.13) holds.

$\blacksquare$

4. PROOF OF LEMMA 3.4
For the case$\lambda\geq 0$, we can show Lemma 3.4 in this paper by using same method as theproof of Lemma 3.4 in [8]. So we will consider the-First, we .efine the$\mathrm{d}$$V(r):=r^{2}+\lambda$$\mathrm{f}\mathrm{o}\mathrm{l}1_{0}\mathrm{W}\mathrm{i}.\mathrm{n}\mathrm{g}$$\mathrm{c}\mathrm{a}\mathrm{s}\mathrm{e}-n<\lambda<0$in the following.

functions which we need in this section:,$X(r):=r^{2}+ \frac{r\varphi’(r)}{\varphi(r)}$,$F(r):=(p+3)^{\mathit{2}}r^{2}V(r)-4$,$J(r):=(p+3)x(r)+2$ ,$L(r):=(p+3)X(r)^{2}+4X(r)+(p+3)r^{2}V(r) \equiv\frac{J(r)^{2}+F(r)}{p+3}$ ,$P(r):= \frac{J(r)^{2}}{L(r)}$,$Q(r):=4F(r)X(r)+rF’(r)$ ,$R(r):=16q^{2}r^{4}V(r)3-4q(q+32)r^{2}V(r)2-64(q-4)V(r)$$+4q^{2}r^{4}V(r)V’’(r)-8\mathrm{o}qrV’(r)-16qr^{2}V’’(r)-5q^{2}rV4’(r)^{2}$

where $q:=(p+3)^{2}$ . By these definitions, we can rewrite$\Phi(r)$

as
.

(4.1)

ds–J $(r)$ .

(4.2)$\Phi(r)=J(r)\exp(r^{2})\varphi(r)^{2}\int_{f}^{\infty}s-1\exp(-S^{2})\varphi(s)-2ds$

Differentiating$\Phi(r)$,, we have$r \Phi’(r)=L(r)\exp(r^{2})\varphi(r)^{2}\int_{f}^{\infty}s^{-1}\exp(-s^{2})\varphi(S)-2$(Here, we use the equality $rX’(r)=r^{2}V(r)-x(r)^{2}.$ ) In order to evaluate the sign of

we will investigate the behaviour of $J(r)$ and $L(r)$ for $r\in[0, \infty)$ .

First, we study the profile of $J(r)$ . Note that the following lemma holds.

$\Phi’(r)$,217
Lemma 4.1. There exists a unique positive number$F’(\beta)>0$

such that $F(r)<0$ on$\beta$satisfying

and $F(r)>0$ on$[0, \beta)$$(\beta, \infty)$$\beta>\sqrt{-\lambda},$$F(\beta)=0$

and.

Proof. bivial.

$\blacksquare$

It follows from $\varphi’(0)=0$ and (iii) of Proposition 2.2 that

and$J(0)=2$$\lim_{farrow\infty}J(r)=+\infty$.

Moreover, we obtain the following lemma.

Lemma 4.2. Function $J(r)$ satisfies one of the following conditions:(J1) $J(r)>0$ for all $r\geq 0$ .
(J2) There exist two positive numbers$0$such that $J(r)>0$ on$r_{1}$

and$r_{2}$$(0, r_{1})\cup(r_{2}, \infty)$(J3) $J(\beta)=J’(\beta)=0$ and $J(r)>0$ on

satisfying $r_{1}<\beta<r_{2}$ and $J(r_{1})=J(r_{2})=$and $J(r)<0$ on$(0, \infty)\backslash \beta$$(r_{1}, r_{2})$.

.

Proof. We have the following two equalities:$J’(r)|J(T)=0$$=$$\frac{F(r)}{(p+3)r’}$$J”(r)|J(’)=.J’(t)=0$$=$$\frac{2\{(p+3)^{2}r^{4}+4\}}{(p+3)r2}(>0)$.

Therefore, noting Lemma 4.1, we can see that if $J(r)$ has a zero, then $J(r)$ satisfies (J2) or(J3). Thus we conclude this lemma.

$\blacksquare$

Next, we study the profile of $L(r)$ . Using (iii) and (vi) of Proposition 2.2, we have$L(r)=\lambda(p+5)r^{2}+o(r^{2})$as$rarrow \mathrm{O}$and$\lim_{farrow\infty}L(r)=+\infty$.

Therefore, $L(r)$ is negative for sufficiently small $r>0$ . Moreover, we can show the following

lemma.

218
Lemma 4.3. hnction $L(r)$ satisfies one of the following conditions:(L1) There exists a unique positive number$(0, r_{3})$and $L(r)>0$ on$(r_{3}, \infty)$$(0, r_{5})\backslash r_{4}$satisfying $L(r_{3})=0$ such that $L(r)<0$ on.

(L2) There exist two positive numbersthat $L(r)<0$ on$r_{3}$$r_{4}$

and$r_{5}$

satisfying $L(r_{4})=L’(r_{4})=L(r_{5})=0$ such

and $L(r)>0$ on$(r_{5}, \infty)$.

Proof. Noting the equality $(p+3)L(r)\equiv J(r)^{2}+F(r)$ , we have $L(\beta)>0$ if $J(r)$ satisfies
the condition (J1) or (J2), and $L(\beta)=0$ if $J(r)$ satisfies the condition (J3) from Lemmas

4.1 and 4.2. Moreover, we can see $L(\beta)>0$ for all $r>\beta$ from Lemma 4.1. So it is sufficientto evaluate $L(r)$ for $r<\beta$ . Note that the following equality holds:$L’(r)|_{L}(f)=0= \frac{Q(r)}{(p+3)r}$.

Concerning the profile of $Q(r)$ , we have the following lemma whose proof will be given
below.

Lemma 4.4. There exists a unique number$Q(r)<0$on$(0, r_{6})$$r_{6}\in(0, \beta)$

satisfying $Q(r_{6})=0$ such thatand $Q(r)>0$

on$(r_{6}, \beta]$.

Therefore, similarly to the proof of Lemma 4.2, we can decide the location of zeros for$L(r)$.

$\blacksquare$

Remark 4.1. Set $\hat{r}:=\sup\{r\in(0, \infty) : L(r)<0\}$ . Then we can put$L(r)$

satisfies the condition (L1) or (L2), respectively. Moreover,$\hat{r}\equiv\beta$$\hat{r}=r_{3}$or$\hat{r}=r_{5}$if

if $J(r)$ satisfies thecondition (J3).

Proof of Lemma 4.4. Using (vi) of Proposition 2.2, we obtain$Q(r)=2\lambda(q-4)r2+o(r^{2})$as$rarrow \mathrm{O}$and$Q(\beta)=\beta F’(\beta)>0$

.

(4.3)

219
Therefore, there exists at least one zero in$(0, \beta)$by noting $2\lambda(q-4)<0$ . So we will

evaluate the sign of $Q’(r)|_{Q}(r)=0$ . Since the equality$Q’(r)|Q(f)=0= \frac{rR(r)}{4F(r)}$holds and $F(r)<0$ on$[0, \beta)$, it is sufficient to investigate the profile of $R(r)$ on

Noting that $V(r),$ $V’(r)$ and $V”(r)$ are positive for$R(r)$$=$$r\in(\sqrt{-\lambda}, \beta)$$[0, \beta)$.

, we have$16qrV2(r)2F(r)-4q(q+16)r^{2}V(r)2-64(q-4)V(r)$$+4qr^{2\prime\prime}V(r)F(r)-80_{qr}V’(r)-5q^{2}rV4’(r)^{2}$

$<$

$0$

for

$r\in[\sqrt{-\lambda},$ $\beta)$

Therefore, there exists at least one zero on.

$(0\sqrt{-\lambda}\})$

in view of $R(\mathrm{O})=-64\lambda(q-4)>0$.

Differentiating $R(r)$ , we obtain$R’(r)$$=$$64q^{2}r^{3}V(r)^{\mathrm{a}}+48q^{2}r^{4}V(r)2V’(r)-8q(q+32)rV(r)2-8q(q+32)r^{2}V(r)V’(r)$$-16(9q-16)V’(r)+16q^{2}r^{3}V(r)V\prime\prime(r)-6q^{\mathit{2}}r^{4}V’(r)V’’(r)$$-112qrV\prime\prime(r)+4q^{2}r^{4}V(r)V\prime\prime\prime(r)-16qr^{2}V’’’(r)-20qr^{3}V2’(r)^{2}$.

Then it follows from $R(r)=0$ , which implies$-8\cdot 32qrV2(r)V’(r)$$=$$16q^{3}r^{6}V(r)3V’(r)-4q^{2}(q+32)r^{4}V(r)^{2\prime}V(r)$$-64q^{2}r^{2}V(r)V’(r)+4q^{3}r^{6}V(r)V’(r)V’’(r)$$-80q^{2}r^{3\prime}V(r)^{2}-16qr^{4}V2’(r)V’i(r)-5qrV’(36r)3$ ,220
and $V(r)<0,$ $V’(r)>0$ and $V”(r)>0$ for $r\in(0, \sqrt{-\lambda})$ that
$R(r)|_{R}(f)=0$$=$$64q^{2}r^{3}V(r)3-80_{q^{2}r}4V(r)2V’(r)-256qrV(r)2+16q^{3}r^{6}V(r)^{3\prime}V(r)$$+4q^{3}r^{4}V(r)V’(r)V^{\prime\iota}(r)-22qrV’(24r)V\prime r(r)-5q^{3}rV6’(r)^{3}$$+16(16-q)V’(r)+16q^{2}r^{3}V(r)V’’(r)-112qrV’’(r)$$+4q^{2}r^{4}V(r)V\prime\prime\prime(r)-16qrV2\prime\prime\prime(r)-4qV’(r)\{(qr^{2}V(r)+4)^{2}+16\}$$-8q^{2}r \{(V(r)+\frac{5}{2}rV’(r))^{2}+\frac{25}{4}r^{2}V’(r)^{2}\}$

$<$

for

$0$

$r\in(0,$ $\sqrt{-\lambda})$.

(Note that $V”’(r)=0.$ ) Thus there exists a unique positive number$\{$$R(r)>0$on$R(r_{7})=0$

and$R(r)<0$on$[0, r_{7})$

i.e.,

,

$R’(r_{7})<0$ ,$(r_{7}, \beta)$

i.e.,

,

if

$Q’(r)<0$i.e.,$Q(r)$$Q’(r)=0$$Q’(r)>0$

if

if$r_{7}\in(0, \beta)$

has a zero in$Q(r)$$Q(r)$$[0, r_{7})$such that,has a zero athas a zero in$(r_{7}, \beta)$$r=r_{7}$.

Define $\xi:=\inf\{r\in(0, \infty) : Q(r)=0\}$ . Then $Q’(\xi)\geq 0$ from (4.3), which implies
Moreover, we have$\xi\neq r_{7}$. In fact, if$\xi=r_{7}$,$\xi\geq r_{7}$.

, then $Q(r_{7})=Q’(r_{7})=0$ and $Q”(r_{7})$ must be

non-positive from (4.3). However, it is impossible by the following equality$Q”(r_{7})= \frac{2R(r_{7})+r_{7}R\prime(r_{\tau})}{4F(r_{7})}(>0)$

Therefore, there exists a unique number

on$(0, r_{6})$and $Q(r)>0$ on$(r_{6}, \beta]$$r_{6}\in(r_{7}, \beta)$.

satisfying $Q(r_{6})=0$ such that $Q(r)<0$.

$\blacksquare$

Now we obtain the following lemma about the profile of$\Phi(r)$in view of (4.1), (4.2) andLemmas 4.2 and 4.3.

Lemma 4.5.

$f\mathrm{h}$nction$\Phi(r)$satisfies the following conditions:(i) If $J(r)$ satisfies the condition (J1), then$\Phi(r)$is decreasing on$(0,\hat{r}]$.

221(ii) If $J(r)$ satisfies the condition (J2), thendecreasing on $(0, r_{1})$ and increasing on$\Phi(r)<0$on$(r_{1}, r_{2})$$\hat{r}$satisfies$(\hat{r}, r_{2})$$r_{1}<\hat{r}<r_{2}$. Moreover,$\Phi(r)$is. Especially, $\Phi(r_{1})=\Phi(r_{2})=0$ and.

(iii) If $J(r)$ satisfies the condition (J3), then$\Phi(r)$is decreasing on

and $\Phi(\beta)=0$$(0, \beta)$with $\Phi’(\beta)=0$ .

Proof. If $J(r)$ satisfies the condition (J2), then we have $L(r_{1})=J(r_{1})^{2}+F(r_{1})<0$ and$L(r_{2})=J(r_{2})^{2}+F(r_{2})>0$ because $J(r_{1})=J(r_{2})=0$ and $r_{1}<\beta<r_{\mathit{2}}$$r_{1}<\hat{r}<r_{2}$. Thus we getin view of Lemma 4.3. Moreover, noting Remark 4.1, we can show the increase

and decrease of$\Phi(r)$for each cases automatically fiiom Lemmas 4.2 and 4.3.

It remains to consider the behaviour of$r>\hat{r}$,$\Phi(r)$for$r>r_{2}$or$\blacksquare$(4.4)$r>\beta$

when $J(r)$ satisfies the condition (J1), (J2) or (J3), respectively. Our strategy is to evaluate
the critical values of$\Phi’(r^{*})=0$$\Phi(r)$.

, we will investigate the value of$\mathrm{N}\mathrm{a}m\mathrm{e}\mathrm{l}\mathrm{y}$$\Phi(r^{*})$

for$r^{*}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{i}_{\mathrm{S}}\mathrm{f}\mathrm{y}\mathrm{i}\mathrm{n}\mathrm{g}$. Combining (4.1) and (4.2) with $r=r^{*}$ , we obtain$\Phi(r^{*})=P(r^{*})\equiv\frac{J(r^{*})^{2}}{L(r^{*})}$

.

(4.5)

Now we study the profile of $P(r)$ for (4.4). First, it is easily seen that
$P(r)$

converges to$\frac{p+3}{2}$as$rarrow\infty$with increasing.

(4.6)

Moreover, we can show the following lemma.

Lemma 4.6. Function $P(r)$ satisfies the following conditions:(i) If $J(r)$ satisfies the condition (J1), then there exists a unique positive number

fying $P(\overline{r})=(p+3)/2$ such that
$P(r)> \frac{p+3}{2}$on$(\hat{r},\overline{r})$and$0<P(r)< \frac{p+3}{2}$on$(\overline{r}, \infty)$.

$\overline{r}$satis-

222(ii) If $J(r)$ satisfies the condition (J2), then$0<P(r)< \frac{p+3}{2}$on$(r_{2}, \infty)$.

on$(\beta, \infty)$.

(iii) If $J(r)$ satisfies the condition (J3), then$0<P(r)< \frac{p+3}{2}$Proof. Assume $J(r)$ satisfies the condition (J1). It follows from Lemmas 4.2 and 4.3 that$f arrow\hat{f}\lim_{0+}P(r)=+\infty$

Moreover, for$r>\hat{r}$(4.7)

there exists at least one crossing point of $y=P(r)$ and $y=(p+3)/2$

in $(r, y)$ -plane from (4.6) and (4.7). For$J(\overline{r})^{2}>0$.

; so it must be that$\overline{r}>\beta$, we have$\overline{r}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{i}_{\mathrm{S}}\mathrm{f}\mathrm{y}\mathrm{i}\mathrm{n}\mathrm{g}P(\overline{r})=(p+3)/2$$F(\overline{r})=$holds true. Moreover, we have the following equality,$P’( \overline{r})=\frac{F(\overline{r})f(\overline{r})}{\overline{r}(J(\overline{r})^{2}+F(\overline{r}))2}$

where$f(r)$$:=$$8F(r)-(p+3)rF’(r)$$=$$-2(p-1)(p+3)^{2}r^{\mathit{2}}V(r)-(p+3)^{3}r^{3}V’(r)-32$ .

Therefore, we can see $f(r)<0$ for all $r>\beta$ ; so we obtain $P’(\overline{r})<0$ which implies theuniqueness of crossing point from (4.6) and (4.7).

If $J(r)$ satisfies the condition (J2), then $P(r_{2})=0$ holds. Therefore, from (4.6) and$P’(r)|_{p_{(f}})=(\mathrm{P}+3)/2<0$on$(\beta, \infty),$$y=P(r)$ cannot cross $y=(p+3)/2$ on$(r_{2}, \infty)$.

Moreover, if $J(r)$ satisfies the condition (J3), then we have $\lim_{tarrow\beta}P(r)=0$ by l’Hospital’s
theorem. So it is impossible that$y=\backslash P(r)$crosses $y=(p+3)/2$ on

reason stated above. Thus we finish the proof.

$(\beta, \infty)$by the same$\blacksquare$

223
Now we will prove Lemma 3.4.

Proof of Lemma 3.4. If $J(r)$ satisfies the condition (J1), then $\Phi(r)>0$ for all $r\geq 0$
in view of (4.1). As already seen,$\Phi(r)$is decreasing on$(0,\hat{r}]$$y=(p+3)/2$ at $r=r_{*}\in(0,\overline{r}]$ , then $\Phi(r)$ is decreasing onon$(\hat{r},\overline{r})$, and$\Phi(r)$has a local minimum at some point in$P(r)<(p+3)/2$ on$(\overline{r}, \infty)$$(p+3)/2$ at some point in

at$r=r_{*}\in(\overline{r}, \infty)$. If $y–\Phi(r)$ first crosses$(r_{*},\overline{r}]$because $P(r)>(p+3)/2$$(\overline{r}, \infty)$. Moreover, noting thatand Lemma 3.3, we can see that it is impossible that $\Phi(r)\geq$$(\overline{r}, \infty)$. On the other hand, if $y=\Phi(r)$ first crosses $y=(p+3)/2$, then $\Phi(r)<(p+3)/2$ holds in$(r_{*}, \infty)$by the same reason stated above.

Thus we can see that $y=\Phi(r)$ crosses $y=(p+3)/2$ only once on$(0, \infty)$

If $J(r)$ satisfies the condition (J2), then $0<\Phi(r)<(p+3)/2$ on

noting $\Phi(r_{2})=0,$ $P(r)<(p+3)/2$ and $J(r)>0$ on

crosses $y=(p+3)/2$ once at some point in$(0, r_{1})$$(r_{2}, \infty)$.

$(r_{2}, \infty)$

holds true by

and Lemma 3.3. Thus $y=\Phi(r)$from Lemma 4.5.

Finally, if $J(r)$ satisfies the condition (J3), then $0<\Phi(r)<(p+3)/2$ on$(\beta, \infty)$by thesame reason as the case (J2). Therefore, there exists only one crossing point of $y=\Phi(r)$and $y=(p+3)/2$ in$(0, \beta)$

from Lemma 4.5. Thus we can conclude this lemma.

$\blacksquare$

5. PROOF OF PROPOSITION 1.2
In this section, we give the proof of Proposition 1.2.
First, we show the equivalence between (ii) and (iii). For functions

denote $f_{1}(r)\sim f_{2}(r)$ if the limit$\lim_{\tauarrow\infty^{f_{1}}}(r)/f_{2}(r)$exists in$(0, \infty)$$f_{1}(r)$and$f_{2}(r)$, we. From (1.9) and (iv)and (v) of Proposition 2.2, we have$\int_{f}^{\infty}g(_{S})-1ds\sim r-n\mathrm{p}\mathrm{e}\mathrm{x}(-r^{2})\varphi(r)^{-2}\sim r^{-}\exp(\lambda-r)2$,

from which together with (1.7) and (iv) of Proposition 2.2, the equivalence between (ii) and(iii) follows.

224
Next, we show that (ii) implies (i). Rom (1.4), we see that $u’(r)$ satisfies$(r^{n-1\prime}u(r))’=r^{n-1}\{(\lambda+r^{2})u(r)-|u(r)|^{\mathrm{P}}-1u(r)\}$Since $u(r)$ decays exponentially as$rarrow\infty$

we see that $r^{n-1}u’(r)$ has a limit asintegrating (5.1) on$(r, \infty)$, integrating (5.1) on$rarrow\infty$$(r_{1}, r)$

.

(5.1)

and letting $rarrow\infty$ ,and this limit must be zero by (1.5). Thus,, we see that $u’(r)$ decays exponentially as $rarrow\infty$ , from which

together with (1.5), we obtain that$u(r)\in\Sigma_{\tau ad}$.

Finally, we show that (i) implies (iii). We first note that $u(r)$ and $u’(r)$ decay exponentially as $rarrow\infty$ , because $u(r)$ is a solution of (1.4) in$U(r):=(1+r)^{(n+}\lambda)/2u(r)$$\Sigma_{rad}$(see [1, Lemma 2] and [15]). Set

and$V(r):=\eta(r)(U’(r)+rU(r))$ ,$\eta(r):=r^{n-1}(1+r)^{-()}n+\lambda \mathrm{x}\mathrm{e}\mathrm{p}(-r^{2}/2)$,

then $V(r)$ satisfies$V’(r)= \eta(r)U(r)\{\frac{n+\lambda}{1+r}+\frac{(n-1)(n+\lambda)}{2r(1+r)}-\frac{(n+\lambda)(n+\lambda+2)}{4(1+r)^{2}}-|u(r)|p-1\}$.

Rom the assumption $\lambda>-n$ and the exponential decay of $u(r)$ at infinity, there exists$r_{2}\in(1, \infty)$

such that$\frac{n+\lambda}{1+r}+\frac{(n-1)(n+\lambda)}{2r(1+r)}-\frac{(n+\lambda)(n+\lambda+2)}{4(1+r)^{2}}-|u(r)|^{p-1}>0$

Thus, noting that $U(r)>0$ , we see that $V(r)$ is non-decreasing on$r_{3}>r_{2}$such that $V(r_{3})>0$ , then $V(r)\geq V(r_{3})>0$ for all$U’(r)+rU(r)\geq V(r_{3})/\eta(r)$ ,$r\geq r_{3}$$r\geq r_{3}$for$[r_{2}, \infty)$$r\geq r_{2}$.

. If there exists. This implies that.

However, this is a contradiction, because the left hand side converges to from the definition$0$of $U(r)$ and the exponential decay of $u(r)$ and $u’(r)$ at infinity, while the right hand side

goes to infinity as$rarrow\infty$from the definition of$\eta(r)$. Therefore, we have $V(r)\leq 0$ for all225$r\geq r_{2}$. This shows that$(\exp(r^{2}/2)U(r))’$$=\exp(r^{2}/2)(U^{\iota}(r)+rU(r))$$=r^{1-n}(1+r)^{n+\lambda}\exp(r^{2})V(r)\leq 0$

which implies that $U(r)\leq C_{1}\exp(-r^{2}/2)$ for$r\geq r_{2}$,$r\geq r_{2}$,, where $C_{1}=\exp(r_{2^{2}}/2)U(r_{2})$ . From

the definition of $U(r)$ , we obtain that$u(r)\leq c_{2}r^{-()/2}\exp n+\lambda(-r^{2}/2)$

for some$C_{\mathit{2}}>0$

,

(5.2)

$r\geq r_{4}$and $r_{4}>0$ . As in the proof for the equivalence between (ii) and (iii) above,from (5.2), (1.7), (1.9) and (iv) and (v) of Proposition 2.2, we see that $v(r)$ satisfies$v(r) \leq C_{3}\int_{f}^{\infty}g(s)-1d_{S}$,for some $C_{3}>0$ and $r_{5}>0$ . Since $v(r)$ is an entirely positivefrom (5.3) and Theorem$\mathrm{A}$(5.3)$r\geq r_{5}$$\mathrm{s}\mathrm{o}\mathrm{I}\mathrm{u}\mathrm{t}\mathrm{i}_{\mathrm{o}\mathrm{n}}$of (1.8) with (1.9),, we see that $v(r)$ satisfies (1.14). This completes the proof.

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$\mathrm{S}\mathrm{c}\mathrm{h}_{\Gamma\ddot{\mathrm{o}}}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{r}$equations with unbounded potentials,