Uniqueness and nondegeneracy of positive radial solutions of $\mathbf {div\,{\varvec{(}}{\varvec{\rho }} \nabla u{\varvec{)}} +{\varvec{\rho }}{\varvec{(}}-gu+hu^p{\varvec{)}}=0}$

Abstract

We study the uniqueness and nondegeneracy of positive solutions of $\mathrm {div}\,(\rho \nabla u) +\rho (-g u+h u^p)=0$ in a ball, the entire space, an annulus, or an exterior domain under the Dirichlet boundary condition.

Appendices

Appendix 1: The functions a(r), b(r), c(r), G(r) and D(r)

In this appendix, we give detailed expressions of a(r), b(r), c(r), G(r) and D(r) for some specified f(r), g(r) and h(r). In the case $f(r)=r^{n-1}$ (g and h are any functions), we have

$$\begin{aligned} a(r)&=r^\frac{2(n-1)(p+1)}{p+3}h(r)^\frac{-2}{p+3},\qquad b(r) =\frac{r^{\frac{2(n-1)(p+1)}{p+3}-1} }{(p+3)h(r)^{\frac{p+5}{p+3}}} \bigl (2(n-1)h(r)+r h_r(r)\bigr ),\\ c(r)&=\frac{r^{\frac{2(n-1)(p+1)}{p+3}-2}}{(p+3)^2h(r)^{\frac{2(p+4)}{p+3}}} \Bigl ( 2 (n-1) \bigl [n+2-(n-2)p\bigr ] h(r)^2 +(p+5)r^2 h_r(r)^2\\&\qquad -(n-1)(p-5)r h(r) h_r(r) -(p+3)r^2h(r) h_{rr}(r) \Bigr ),\\ G(r)&=\frac{r^{\frac{2(n-1)(p+1)}{p+3}-3}}{2(p+3)^3h(r)^{\frac{2}{p+3}+3}} \biggl ( 4(n-1) \bigl [n+2-(n-2)p\bigr ] \bigl [n-4+(n-2)p\bigr ]h(r)^3\\&\qquad -\Bigl [2(n-1)(p-1)(p+3)^2 r^2 h(r)^3 -4(p+3)^2 r^3 h(r)^2h_r(r)\Bigr ]g(r)\\&\qquad -(p+3)^3 r^3 g_r(r)h(r)^3+(n-1) \bigl [ (2n-3)p(6-p)+6n-33\bigr ]rh(r)^2h_r(r)\\&\qquad +3(n-1)(p-1)(p+5) r^2 h(r) h_r(r)^2 -2(p+4)(p+5) r^3 h_r(r)^3\\&\qquad -3(n-1)(p-1)(p+3) r^2 h(r)^2 h_{rr}(r)\\&\qquad +3(p+3)(p+5) r^3 h(r) h_r(r) h_{rr}(r) -(p+3)^2 r^3 h(r)^2 h_{rrr}(r) \biggr ),\\ D(r)&=\frac{r^{\frac{4 (n-1) (p+1)}{p+3}-2}}{(p+3)^2h(r)^{\frac{2(p+5)}{p+3}} } \Bigl ( 2 (n-1) [(n-2) p+n-4]h(r)^2 +(p+3)^2 r^2 g(r)h(r)^2\\&\qquad -(p+4) r^2 h_r(r)^2+ (p+3) r^2h(r) h_{rr}(r)+(n-1) (p-1) rh(r)h_r(r)\Bigr ). \end{aligned}$$

In the case $f(r)=r^{n-1}$ and $h(r)=1$, we have

$$\begin{aligned} a(r)&=r^{\frac{2 (n-1) (p+1)}{p+3}},\qquad b(r)=\frac{2 (n-1)}{p+3} r^{\frac{2 (n-1) (p+1)}{p+3}-1},\\ c(r)&= \frac{2 (n-1) [n+2-(n-2) p]}{(p+3)^2} r^{\frac{2 (n-1) (p+1)}{p+3}-2},\\ G(r)&= \frac{r^{\frac{2 (n-1) (p+1)}{p+3}-3}}{2 (p+3)^3} \Bigl ( 4 (n-1) [(n-2) p+n-4][n+2-(n-2) p]\\&\quad -2 (n-1) (p-1) (p+3)^2r^2 g(r)-(p+3)^3 r^3 g_r(r) \Bigr ),\\ D(r)&=\frac{r^{\frac{4 (n-1) (p+1)}{p+3}-2}}{(p+3)^2} \Bigl (2 (n-1) [(n-2) p+n-4]+(p+3)^2 r^2 g(r)\Bigr ). \end{aligned}$$

In the case $f(r)=r^{n-1}\exp (r^2/4)$ and $h(r)=1$, we have

$$\begin{aligned} a(r)&=\left( e^{\frac{r^2}{4}} r^{n-1}\right) ^{\frac{2 (p+1)}{p+3}},\qquad b(r)=\frac{\left( 2 n+r^2-2\right) \left( e^{\frac{r^2}{4}} r^{n-1}\right) ^{\frac{2 (p+1)}{p+3}}}{(p+3) r},\\ c(r)&= \frac{\left( e^{\frac{r^2}{ 4}} r^{n-1}\right) ^{\frac{2 (p+1)}{p+3}} }{2 (p+3)^2 r^2} \Bigl ( 4 (n-1) [n+2-(n-2) p]\\&\qquad -2 (2n (p-1)-p+5)r^2-(p-1) r^4\Bigr ),\\ G(r)&= \frac{\left( e^{\frac{r^2}{4}} r^{n-1}\right) ^{\frac{2 (p+1)}{p+3}} }{4 (p+3)^3 r^3} \biggl ( 8 (n-1) [(n-2) p+n-4][n+2-(n-2) p] \\&\qquad -12 (n-1)^2 (p^2-1) r^2 -4(n-1)(p-1)(p+3)^2 r^2g(r) \\&\qquad -2(p+3)^3 r^3 g_r(r) -2(p-1)(p+3)^2 r^4g(r) \\&\qquad -2 r^4 \left( 3 n \left( p^2-1 \right) -(p-6)p+3 \right) -(p^2-1) r^6 \biggr ),\\ D(r)&=\frac{\left( e^{\frac{r^2 }{4}} r^{n-1}\right) ^{\frac{4 (p+1)}{p+3}} }{2 (p+3)^2 r^2} \Bigl ( 4 (n-1) [(n-2) p+n-4] +2 (2 n (p+1)-p+1)r^2\\&\qquad +2 (p+3)^2 r^2 g(r) +(p+1) r^4\Bigr ). \end{aligned}$$

Next, we study the case $G(r)\equiv 0$. Letting $f(r)=r^{n-1}$, $h(r)=r^q$ with $q \in \mathbb R$ and

$$\begin{aligned} g(r)= \frac{r^{-2 (n+1)}}{(p+3)^2} \left( C_1 (p+3)^2 r^{\frac{4 (2 n+p+q+1)}{p+3}}-(2 n+q-2) r^{2 n} [(n-2) p+n-4-q]\right) \end{aligned}$$

with $C_1 \in \mathbb R$, we have

$$\begin{aligned} a(r)&=r^{ \frac{2 ((n-1)(p+1)-q)}{p+3}},\qquad b(r)=\frac{2 n+q-2}{p+3} r^{\frac{(2 n-3) (p+1)-2 (q+1)}{p+3}},\\ c(r)&= \frac{(2 n+q-2) [n+2-(n-2)p+2q]}{(p+3)^2} r^{\frac{2 [(n-2)p+n-4-q]}{p+3}},\\ G(r)&= 0,\qquad D(r)=C_1 r^{2 n-2}. \end{aligned}$$

Letting $f(r)=r^{n-1}\exp (r^2/4)$, $h(r)=r^q$ with $q \in \mathbb R$ and

$$\begin{aligned} g(r)= & {} C_1 \exp \biggl (-\frac{(p-1) r^2}{2 (p+3)}\biggr ) r^{\frac{2 (-(n-1)(p-1)+2 q)}{p+3}} -\frac{p+1}{2(p+3)^2} r^2\\&+\,\frac{(-4 n(p+1)-(p-1)(q-2))}{2(p+3)^2} -\frac{2(2n+q-2)[(n-2)p+n-4-q]}{2(p+3)^2r^2}, \end{aligned}$$

we have

$$\begin{aligned} a(r)&=\left( \exp (r^2/4) r^{n-1}\right) ^{\frac{2 (p+1)}{p+3}} r^{-\frac{2 q}{p+3}},\\ b(r)&=\frac{ \left( \exp (r^2/4) r^{n-1}\right) ^{\frac{2 (p+1)}{p+3}} r^{-\frac{2 q}{p+3}-1} }{p+3}(r^2+2 n+q-2 ),\\ c(r)&= \frac{\left( \exp (r^2/4) r^{n-1}\right) ^{\frac{2 (p+1)}{p+3}} r^{-\frac{2 (p+q+3)}{p+3}} }{2 (p+3)^2} \Bigl (2 (2 n+q-2) [n+2-(n-2)p+2q],\\&\quad - (4 n (p-1)+(p-5) (q-2))r^2 -(p-1) r^4\Bigr ),\\ G(r)&= 0, \qquad D(r)=C_1 \exp (r^2/2) r^{2 n-2}. \end{aligned}$$

Appendix 2: Some properties of $\mathcal X$

In this appendix, we assume $p>1$, (B1) and (B6), and we understand that $\mathcal D$, $\mathcal X$ and $\mathcal L$ are the spaces defined in Sect. 3.

Lemma 9

The space $\mathcal X$ coincides with the completion of $C_0^\infty ((R',R))$ with respect to the norm $\Vert \cdot \Vert _\mathcal X$.

Proof

Since $C_0^\infty ((R',R))\subset \mathcal D$, it is enough to show that each element in $\mathcal D$ is approximated by an element in $C_0^\infty ((R',R))$. Let $u \in \mathcal D$ and let $\varepsilon >0$. We set $C=\max _{R'\le r< R}|u(r)|$. We define $\eta \in W^{1,\infty }(R',R)$ by $\eta (r)= 0$ for $R'< r\le \hat{\delta }$, $\eta (r)= 1$ for $\delta \le r<R$, and

$$\begin{aligned} \eta (r)=\frac{\int _{\hat{\delta }}^{r}\frac{ds}{f(s)}}{\int ^{\delta }_{\hat{\delta }}\frac{ds}{f(s)}} \quad {\mathrm{for}\,\, \hat{\delta }\le r\le \delta ,} \end{aligned}$$

where $\hat{\delta },\delta \in (R',R)$ with $\hat{\delta }< \delta$ are chosen to be

$$\begin{aligned} 2\int _{R'}^{\delta }\bigl (|u_r|^2+|g||u|^2\bigr )f\,dr<\varepsilon \quad \text {and}\quad \int _{\hat{\delta }}^{\delta }\frac{dr}{f(r)}>\frac{C^2}{\varepsilon }. \end{aligned}$$

Then we obtain

$$\begin{aligned} \int _{R'}^R&\bigl (|(\eta u)_r-u_r|^2+|g| |\eta u-u|^2\bigr )f\,dr\\&\le 2\int _{R'}^R|\eta (r)-1|^2 (|u_r|^2+ |g||u|^2\bigr )f(r)\,dr +2 \int _{R'}^R|\eta _r(r)|^2|u|^2f(r)\,dr\\&\le \varepsilon + C^2\biggl ( \int _{\hat{\delta }}^\delta \frac{dr}{f(r)} \biggr )^{-2} \int _{\hat{\delta }}^\delta \frac{dr}{f(r)} \le 2\varepsilon . \end{aligned}$$

By the standard argument with mollifiers, we can infer that our assertion holds.$\square$

Lemma 10

Let $u\in \mathcal X$. Then $u^+,\, u^-,\, |u|\in \mathcal X$.

Proof

Since $u\in H^1_{\mathrm {loc}}(R',R)$, we have

$$\begin{aligned} (u^{+})_r=u_r\cdot 1_{\{ u>0\}} \quad \text {and}\quad (u^{-})_r=u_r\cdot 1_{\{ u<0\}}, \end{aligned}$$

where 1 is the characteristic function. Let $u\in \mathcal X$. By the previous lemma, there exists $\{u_k\}\subset C_0^\infty ((R',R))$ such that $u_k\rightarrow u$ in $\mathcal X$. We choose $K\in C^\infty (\mathbb R)$ such that

$$\begin{aligned} K(t)= {\left\{ \begin{array}{ll} t&{} {\mathrm{for}\, t\ge 1,}\\ 0&{} {\mathrm{for}\, t\le 0,} \end{array}\right. } \quad 0\le K(t)\le t \quad \text {and}\quad 0\le K_t(t)\le 2\quad {\mathrm{for}\, t \in \mathbb R.} \end{aligned}$$

Let $k,m \in \mathbb N$. We claim

$$\begin{aligned} \int _{R'}^{R}&\biggl |\biggl (\frac{1}{m}K(mu_k(r))\biggr )_r-(u^+)_r \biggr |^2f(r)\,dr + \int _{R'}^{R}|g(r)|\biggl |\frac{1}{m}K(mu_k)-u^+\biggr |^2f(r)\,dr\\&\le 2\int _{\{ r\,:\,0<u_k(r)<1/m\}}(u^{+}_k)_r^2f(r)\,dr +4\int _{R'}^{R}|u_{k,r}-u_r|^2f(r)\,dr \\&\quad +4\int _{\{u>0\}}|1_{\{u_k>0\}}-1|^2u_r(r)^2f(r)\,dr +4\int _{\{u<0\}}1_{\{u_k>0\}}u_r(r)^2f(r)\,dr\\&\quad +2\int _{\{ r\,:\,0<u_k(r)<1/m\}}|g(r)|u_k(r)^2f(r)\,dr +2\int _{R'}^{R}|g(r)|\,|u_k-u|^2f(r)\,dr. \end{aligned}$$

Once the claim was shown, choosing a suitable sequence $\{m_k\}$, we can obtain $\{ K(m_k u_k(r))/m_k \}\subset C_0^\infty ((R',R))$ which converges to u in $\mathcal X$. We will show the claim. We have

$$\begin{aligned} \int _{R'}^{R}&\biggl |\biggl (\frac{1}{m}K(mu_k)\biggr )_r-(u^{+})_r \biggr |^2f(r)\,dr\\&\le 2\int _{R'}^{R}|K_r(mu_k(r))(u_k)_r-(u_k^{+})_r|^2f(r)\,dr +2\int _{R'}^{R}|(u_k^{+})_r-(u^{+})_r|^2f(r)\,dr\\&\le 2\int _{\{ r\,:\,0<u_k(r)<1/m\}}(u^{+}_k)_r^2f(r)\,dr\\&\quad +2\int _{R'}^{R}| (u_{k,r}-u_r)1_{\{u_k>0\}} +u_r( 1_{\{u_k>0\}}-1_{\{u>0\}}) |^2f(r)\,dr\\&\le 2\int _{\{ r\,:\,0<u_k(r)<1/m\}}(u^{+}_k)_r^2f(r)\,dr +4\int _{R'}^{R}|u_{k,r}-u_r|^2f(r)\,dr\\&\quad +4\int _{\{u>0\}}|1_{\{u_k>0\}}-1|^2u_r(r)^2f(r)\,dr +4\int _{\{u<0\}}1_{\{u_k>0\}}u_r(r)^2f(r)\,dr. \end{aligned}$$

Here, we used $\int _{\{u=0\}}u_r(r)^2f(r)\,dr=0$. On the other hand, we have

$$\begin{aligned} \int _{R'}^{R}&|g(r)|\biggl |\frac{1}{m}K(mu_k)-u^+\biggr |^2f(r)\,dr\\&\le 2\int _{R'}^{R}|g(r)|\biggl |\frac{1}{m}K(mu_k)-u_k^{+}\biggr |^2f(r)\,dr +2\int _{R'}^{R}|g(r)|\,|u_k^{+}-u^{+}|^2f(r)\,dr\\&\le 2\int _{\{ r\,:\,0<u_k(r)<1/m\}}|g(r)|u_k(r)^2f(r)\,dr +2\int _{R'}^{R}|g(r)|\,|u_k-u|^2f(r)\,dr. \end{aligned}$$

Hence we have shown the claim, and we finish our proof. $\square$

Now, we also assume (B8). We give the following subsolution estimate.

Proposition 6

Let $u\in \mathcal X$ satisfy

$$\begin{aligned} \int _{R'}^{\bar{R}}(u_r\varphi _r+gu\varphi )f\, dr \le \int _{R'}^{\bar{R}}|u|^{p-1}u\varphi hf\, dr \quad {\mathrm{for each}\, \varphi \in \mathcal {X} \mathrm{with}\, \varphi \ge 0.} \end{aligned}$$

Then $\sup _{r\in (R',(R'+\bar{R})/2]}u^+(r)<\infty$.

Lemma 11

Let $u\in \mathcal X$ and let z be a measurable function such that

$$\begin{aligned} \int _{R'}^{\bar{R}}|z^+|^{\frac{\beta +1}{\beta -1}}hf\,dr<\infty \end{aligned}$$

with some $\beta \in (1,\bar{p})$. Assume

$$\begin{aligned} \int _{R'}^{\bar{R}}(u_r\varphi _r+gu\varphi )f\, dr \le \int _{R'}^{\bar{R}}zu\varphi hf\, dr \quad {\mathrm{for\, each}\,\, \varphi \in \mathcal {X}\quad \mathrm{with}\quad \varphi \ge 0.} \end{aligned}$$

Then $\sup _{r\in (R',(R'+\bar{R})/2]}u^+(r)<\infty$.

Proof

We follow the argument in [46, Theorem 2.26]. Let $1/\sqrt{C_1}$ be the infimum value in (3.3), and set

$$\begin{aligned} \gamma =\max \{\beta ,q\}, \quad \bar{z}=z^+ +\frac{|g^-|}{h} \quad \text {and}\quad C_2= \biggl (\int _{R'}^{R'+2t} |\bar{z}|^{\frac{\gamma +1}{\gamma -1}}hf\,dr \biggr )^{\frac{\gamma -1}{\gamma +1}}. \end{aligned}$$

Set also

$$\begin{aligned} \sigma =\frac{(\bar{p}+1)(\gamma -1)}{2(\bar{p}-\gamma )} \quad \text {and}\quad C_3=\max \biggl \{\max _{R'+t\le r\le R'+2t}\frac{144C_1}{h(r)}, \; (12C_1C_2)^{\sigma +1}\biggr \}. \end{aligned}$$

We claim that for each $l>0$, $s>0$ and $r_1, r_2$ with $(R'+\bar{R})/2 \le r_1< r_2\le \bar{R}$, there holds

$$\begin{aligned} \biggl (\int _{R'}^{r_1}\bigl |u| u_{l}|^{s}\bigr |^{\bar{p}+1}hf\,dr \biggr )^{\frac{2}{\bar{p}+1}}\le C_3\biggl (\frac{s+1}{(r_2-r_1)^2}+(s+1)^{\sigma +1}\biggr ) \int _{R'}^{r_2}\bigl |u| u_{l}|^{s}\bigr |^2hf\,dr,\nonumber \\ \end{aligned}$$

(7.16)

where

$$\begin{aligned} u_{l}(r)=\max \{0,\,\min \{u(r),l\}\}. \end{aligned}$$

We will show the claim. Let $l>0$, $s>0$ and $(R'+\bar{R})/2 \le r_1< r_2\le \bar{R}$. Let $\eta \in \mathcal D$. Since we can show $\eta ^2u| u_{l}|^{2s}\in \mathcal X$ by a similar proof of the previous lemma, we have

$$\begin{aligned}&\int _{R'}^{\bar{R}}z\eta ^2u^2| u_{l}|^{2s}hf\,dr\\&\quad \ge \int _{R'}^{\bar{R}} \bigl (u_r(2\eta \eta _ru| u_{l}|^{2s}+\eta ^2u_r| u_{l}|^{2s} +2s\eta ^2u| u_{l}|^{2s-2}u_{l}u_{l,r}) +g\eta ^2u^2| u_{l}|^{2s}\bigr )f\, dr \\&\quad \ge \int _{R'}^{\bar{R}} \biggl (\frac{1}{2}\eta ^2|u_r|^2| u_{l}|^{2s} -2\eta _r^2u^2| u_{l}|^{2s} +2s\eta ^2| u_{l,r}|^2| u_{l}|^{2s}+g\eta ^2u^2| u_{l}|^{2s}\biggr )f\,dr, \end{aligned}$$

which yields

$$\begin{aligned}&\int _{R'}^{\bar{R}}\biggl ( \frac{1}{2}\eta ^2|u_r|^2| u_{l}|^{2s} +2s\eta ^2|u_{l,r}|^2| u_{l}|^{2s}+g^+\eta ^2u^2| u_{l}|^{2s} \biggr )f\,dr\\&\qquad \le \int _{R'}^{\bar{R}} \bigl (2|\eta _r|^2u^2|u_{l}|^{2s}f+\bar{z}\eta ^2u^2| u_{l}|^{2s}hf\bigr ) \,dr. \end{aligned}$$

Noting

$$\begin{aligned} \bigl | \bigl (\eta u| u_{l}|^s\bigr )_r\bigr |^2 \le 3|\eta _r|^2u^2| u_{l}|^{2s}+3\eta ^2| u_r|^2| u_{l}|^{2s} +3s^2\eta ^2| u_{l,r}|^2| u_{l}|^{2s} \end{aligned}$$

and $\eta u| u_{l}|^{s} \in \mathcal {X}$, we have

$$\begin{aligned} \biggl (\int _{R'}^{\bar{R}} \bigl |\eta u| u_{l}|^{s}\bigr |^{\bar{p}+1}hf\,dr\biggr )^{\frac{2}{\bar{p}+1}}&\le C_1\int _{R'}^{\bar{R}} \bigl (\bigl |\bigl (\eta u| u_{l}|^{s}\bigr )_r\bigr |^2 +g^{+}\bigl |\eta u|u_{l}|^s\bigr |^2\bigr )f\,dr\\&\le 6(s+1)C_1\int _{R'}^{\bar{R}} \bigl (3|\eta _r|^2u^2|u_{l}|^{2s}f +\bar{z}\bigl |\eta u|u_{l}|^{s}\bigr |^2hf\bigr )\,dr. \end{aligned}$$

For each $\varepsilon >0$, we have

$$\begin{aligned}&\int _{R'}^{\bar{R}} \bigl (3|\eta _r|^2u^2|u_{l}|^{2s}f +\bar{z}\bigl |\eta u|u_{l}|^{s}\bigr |^2hf\bigr )\,dr\le 3 \int _{R'}^{\bar{R}}| \eta _r|^2u^2| u_{l}|^{2s}f\,dr\\&\quad +C_2 \varepsilon ^2\biggl (\int _{R'}^{\bar{R}} \bigl |\eta u| u_{l}|^{s}\bigr |^{\bar{p}+1}hf\,dr \biggr )^{\frac{2}{\bar{p}+1}} +C_2 \varepsilon ^{-2\sigma } \int _{R'}^{\bar{R}}\bigl |\eta u| u_{l}|^{s}\bigr |^2hf\,dr. \end{aligned}$$

Choosing $\varepsilon ^{-2}=12(s+1)C_1C_2$ and using two inequalities above, we obtain

$$\begin{aligned} \biggl (\int _{R'}^{\bar{R}} \bigl |\eta u| u_{l}|^{s}\bigr |^{\bar{p}+1}hf\,dr\biggr )^{\frac{2}{\bar{p}+1}}&\le 36(s+1)C_1\int _{R'}^{\bar{R}}|\eta _r|^2 u^2| u_{l}|^{2s}f\,dr\\&\quad +(12(s+1)C_1C_2)^{\sigma +1} \int _{R'}^{\bar{R}}\bigl |\eta u| u_{l}|^{s}\bigr |^2hf\,dr. \end{aligned}$$

Now, letting $\eta$ satisfy

$$\begin{aligned} \eta (r)= {\left\{ \begin{array}{ll} 1&{} \text {for}\, R'\le r\le r_1,\\ 0&{} \text {for}\, r\ge r_2, \end{array}\right. } \quad \text {and}\quad |\eta _r(r)|\le \frac{2}{r_2-r_1}, \end{aligned}$$

we can infer that claim (7.16) holds. We set

$$\begin{aligned} \chi =\frac{\bar{p}+1}{2} \quad \text {and}\quad t=\frac{\bar{R}-R'}{2}. \end{aligned}$$

Applying (7.16) with $m\in \mathbb N$, $s=\chi ^m-1$, $r_1=R'+(1+2^{-m})t$ and $r_2=R'+(1+2^{-m+1})t$, and using the Lebesgue convergence theorem, we can infer

$$\begin{aligned}&\biggl (\int _{R'}^{R'+(1+2^{-m})t} |u^+|^{2\chi ^{m+1}}hf\,dr\biggr )^{\frac{1}{2\chi ^{m+1}}}\\&\quad \le \biggl [C_3\biggl ( \frac{(4\chi )^{m}}{t^2} + \chi ^{m(\sigma +1)}\biggr )\biggr ]^{\frac{1}{2\chi ^{m}}} \biggl (\int _{R'}^{R'+(1+2^{-m+1})t}| u^+|^{2\chi ^{m}}hf\,dr \biggr )^{\frac{1}{2\chi ^{m}}}\\&\quad \le \biggl [ C_3\Bigl (\frac{1}{t^2}+1\Bigr )C_4^{m} \biggr ]^{\frac{1}{2\chi ^{m}}} \biggl (\int _{R'}^{R'+(1+2^{-m+1})t}| u^+|^{2\chi ^{m}}hf\,dr\biggr )^{\frac{1}{2\chi ^{m}}}, \end{aligned}$$

where $C_4=\max \{4\chi ,\chi ^{\sigma +1}\}$. So we have

$$\begin{aligned}&\biggl (\int _{R'}^{R'+(1+2^{-m})t} |u^+|^{2\chi ^{m+1}}hf\,dr\biggr )^{\frac{1}{2\chi ^{m+1}}}\\&\quad \le \biggl (C_3\Bigl (\frac{1}{t^2}+1\Bigr )\biggr )^{\frac{1}{2\chi } \sum _{i=1}^{m}\frac{1}{\chi ^{i-1}}}C_4^{\frac{1}{2\chi } \sum _{i=1}^{m}\frac{i}{\chi ^{i-1}}} \biggl (\int _{R'}^{R'+2t}| u^+|^{2\chi }hf\,dr\biggr )^{\frac{1}{2\chi }}\\&\quad \le \biggl (C_3\Bigl (\frac{1}{t^2}+1\Bigr )\biggr )^{\frac{1}{2(\chi -1)} }C_4^{\frac{1}{2(\chi -1)^2}} \biggl (\int _{R'}^{\bar{R}} |u^+|^{\bar{p}+1}hf\,dr\biggr )^{\frac{1}{\bar{p}+1}}. \end{aligned}$$

Letting $m\rightarrow \infty$, we can find that our assertion holds.$\square$

Proof of Proposition 6

First, we note that

$$\begin{aligned} \int _{R'}^{\bar{R}}|u|^{p-1}u\varphi hf\, dr \le \int _{R'}^{\bar{R}}(u^+)^{p-1}u\varphi hf\, dr \quad \text {for each}\,\, \varphi \in \mathcal {X} \,\,\mathrm{with}\,\, \varphi \ge 0. \end{aligned}$$

In the case of $p<\bar{p}$, applying Lemma 11 with $z=(u^+)^{p-1}$ and $\beta =p$, we can see that our assertion holds. So we consider the case $p=\bar{p}$. We set $s=(\bar{p}-1)/2$. Let $l>0$. Using the notations is Lemma 11 and noting $\int _{R'}^{\bar{R}}|u|u_l|^s|^{\bar{p}+1}hf\,dr<\infty$, we have

$$\begin{aligned}&\biggl (\int _{R'}^{\bar{R}} \bigl |\eta u| u_{l}|^{s}\bigr |^{\bar{p}+1}\biggr )^{\frac{2}{\bar{p}+1}}\le C_1\int _{R'}^{\bar{R}} \bigl (\bigl |\bigl (\eta u| u_{l}|^{s}\bigr )_r\bigr |^2 +g^{+}\eta ^2u^2| u_{l}|^{2s}\bigr )f\,dr\\&\quad \le 18(s+1)C_1\int _{R'}^{\bar{R}}|\eta _r|^2u^2| u_{l}|^{2s}f\,dr +6(s+1)C_1\int _{R'}^{\bar{R}}\bar{z}\eta ^2u^2| u_{l}|^{2s}hf\,dr\\&\quad \le 18(s+1)C_1\int _{R'}^{\bar{R}}|\eta _r|^2u^2| u_{l}|^{2s}f\,dr\\&\qquad +6(s+1)C_1\biggl (\int _{R'}^{\bar{R}}|\bar{z}|^{\frac{\bar{p}+1}{\bar{p}-1}}hf\,dr\biggr )^{\frac{\bar{p}-1}{\bar{p}+1}} \biggl (\int _{R'}^{\bar{R}}\bigl |\eta u| u_{l}|^{s}\bigr |^{\bar{p}+1}hf\,dr\biggr )^{\frac{2}{\bar{p}+1}}. \end{aligned}$$

We choose $\delta >0$ satisfying

$$\begin{aligned} 6(s+1)C_1 \biggl (\int _{R'}^{R'+2\delta }| \bar{z}|^{\frac{\bar{p}+1}{\bar{p}-1}}hf\,dr \biggr )^{\frac{\bar{p}-1}{\bar{p}+1}} <\frac{1}{2}. \end{aligned}$$

Then we have

$$\begin{aligned}&\biggl (\int _{R'}^{R'+2\delta }\bigl |\eta u| u_{l}|^{s}\bigr |^{\bar{p}+1}\biggr )^{\frac{2}{\bar{p}+1}} \le 36(s+1)C_1\int _{R'}^{R'+2\delta }|\eta _r|^2u^2| u_{l}|^{2s}f\,dr\\&\qquad \qquad \le \max _{r_1\le r\le r_2} \frac{72(\bar{p}+1)C_1}{\delta ^2 h(r)}\int _{R'}^{R'+2\delta } |u^+|^{\bar{p}+1}hf\,dr. \end{aligned}$$

Letting $l\rightarrow \infty$, we obtain

$$\begin{aligned} \int _{R'}^{R'+\delta }| u^+|^{\frac{(\bar{p}+1)^2}{2}}hf\,dr<\infty . \end{aligned}$$

Since h, f and u are continuous in $(R',R)$, we have $\int _{R'}^{\bar{R}}| u^+|^{\frac{(\bar{p}+1)^2}{2}}hf\,dr<\infty$. Choosing $\beta \in (1,\bar{p})$ such that

$$\begin{aligned} (\bar{p}-1)\frac{\beta +1}{\beta -1}\le \frac{(\bar{p}+1)^2}{2}, \end{aligned}$$

recalling $\bar{p}=p$, and applying Lemma 11 with $z=(u^+)^{\bar{p}-1}$, we can infer that our assertion holds. $\square$

Appendix 3: Proof of Theorem 8

It is enough to show that the unique positive solution ${\bar{u}}$ is a nondegenerate critical point of I in the case $R<\infty$ and $G\equiv 0$ in $(R',R)$; see Remark 19. For each $\delta >0$, we define $g_\delta$, $h_\delta$, $a_\delta$, $b_\delta$, $c_\delta$ and $J_\delta$ by (4.6), (4.7) and (5.2) with $\gamma \equiv 1$; see also (5.9). We also define $S_\delta$ as the set of all positive solutions of

$$\begin{aligned} \left\{ \begin{array}{ll} u_{rr}(r)&{}+\frac{f_r(r)}{f(r)}u_r+g_\delta (r)u+h_\delta (r)u^p=0,\quad R'<r<R, \\ u(R')&{}=0, \quad u(R)=0. \end{array}\right. \end{aligned}$$

(7.17)

We can see that ${\bar{u}}$ is a positive solution of (7.17) for each $\delta >0$.

Since we can prove the next lemma as in Lemma 5, we omit its proof.

Lemma 12

It holds that

$$\begin{aligned} \inf _{0< \delta <1}\inf _{u\in S_\delta }\Vert u\Vert _\mathcal X>0. \end{aligned}$$

Lemma 13

There exist $\delta _{0}\in (0,1)$ such that

$$\begin{aligned} \sup _{0<\delta <\delta _{0}} \sup _{u\in S_\delta } \mathop {\max \phantom {p}}_{R'\le r\le R}u(r)<\infty . \end{aligned}$$

(7.18)

Proof

Suppose that the conclusion does not hold. Then there exist $\{\delta _m\}\subset (0,1)$ with $\delta _m\rightarrow 0$ and $\{u_m\}\subset C([R',R])\cap C^2((R',R))$ such that $u_m \in S_{\delta _m}$ for each $m\in \mathbb N$ and $\theta _m\equiv \max _{R'\le r\le R}u_m(r)\rightarrow \infty$ as $m\rightarrow \infty$. For each $m\in \mathbb N$, we choose $r_m\in (R',R)$ with $\theta _m=u_m(r_m)$ and we define $\{v_m\}$, $\{L_m\}$ and $\{\beta _m\}$ as in the proof of Lemma 6. Without loss of generality, we may assume that $r_m\rightarrow r_*\in [R',R]$, $\lim _{m\rightarrow \infty }\theta _m^{(p-1)/2}(R'-r_m)$ exists in $[-\infty ,0]$ and $\lim _{m\rightarrow \infty }\theta _m^{(p-1)/2}(R-r_m)$ exists in $[0,\infty ]$. Let $L(\subset \mathbb R)$ be the limit closed interval of $\{L_m\}$. We can see that L is unbounded and $0 \in L$. For each $m \in \mathbb N$, we have $v_m(0)=1$, $v_{m,t}(0)=0$ and

$$\begin{aligned} v_{m,tt}(t)&+\frac{(f(\beta _m(t)))_t}{f(\beta _m(t))} v_{m,t}(t) +(1+\delta _m)h(\beta _m(t))v_m(t)^p\\&-\theta _m^{-p+1}\bigl [ g(\beta _m(t))+\delta _m h(\beta _m(t))\bar{u}(\beta _m(t))^{p-1}\bigr ] v_m(t)=0 \end{aligned}$$

for each $t \in L_m$, and hence we have

$$\begin{aligned} v_{m,t}(t)f(\beta _m(t))&=\int _{0}^{t}f(\beta _m(s)) \Bigl [ -(1+\delta _m)h(\beta _m(s))v_m(s)^p\\&\quad +\bigl [ g(\beta _m(s))+\delta _m h(\beta _m(s)) \bar{u}(\beta _m(s))^{p-1}\bigr ]\theta _m^{1-p}v_m(s)\Bigr ] ds \end{aligned}$$

for each $t \in L_m$. We recall $R<\infty$, $f,h\in C^2([R',R])$, $g\in C([R',R])$ and $f>0$ on $[R',R]$. From

$$\begin{aligned} \sup _{t\in L_m} \biggl | \frac{(f(\beta _m(t)))_t}{f(\beta _m(t))} \biggr | \le \frac{\max _{r \in [R',R]}|f_r(r)|}{ \min _{r\in [R',R]}f(r)}\cdot \theta _m^\frac{-(p-1)}{2}\rightarrow 0\quad {\mathrm{as}\,\, m\rightarrow \infty ,} \end{aligned}$$

and the two equalities above, we can see

$$\begin{aligned} \mathop {\varlimsup \phantom {p}}_{m\rightarrow \infty } \sup _{t\in K}|v_{m,t}(t)|<\infty \quad \text {and}\quad \mathop {\varlimsup \phantom {p}}_{m\rightarrow \infty } \sup _{t\in K}|v_{m,tt}(t)|<\infty \end{aligned}$$

for each bounded closed interval $K\subset L$ with $0 \in K$. Taking a subsequence $\{v_{m_i}\}$ of $\{v_m\}$, we can infer that there exists $v\in C^2(L)$ such that $\Vert v_{m_i}-v\Vert _{C^1(K)}\rightarrow 0$ for each bounded closed interval $K\subset L$ with $0 \in K$, v is nonnegative on L, and

$$\begin{aligned}\left\{ \begin{array}{ll} &{}v_{tt}(t)+h(r_*)|v(t)|^{p-1}v(t)=0 \quad \text {for each}\,\, t\in L,\\ &{}v(0)=1, \quad v_{t}(0)=0. \end{array}\right. \end{aligned}$$

However, since L is unbounded, v must be negative somewhere, which is a contradiction. Thus we have shown our assertion. $\square$

Lemma 14

It holds that

$$\begin{aligned} \mathop {\lim \phantom {p}}_{\delta \rightarrow 0} \sup _{u\in S_\delta }\Vert u-\bar{u}\Vert _{C^{1}([R',R])}=0. \end{aligned}$$

Proof

For each $\delta \in (0,\delta _0)$, $u\in S_\delta$ and $r_u \in (R',R)$ with $u_r(r_u)=0$, we have

$$\begin{aligned} f(r)|u_r(r)|&\le \biggl |\int _{r_u}^{r}\bigl ( h_\delta (s)u(s)^p+|g_\delta (s)| \bar{u} (s)^{p-1}u(s)\bigr )f(s)\,ds\biggr | \quad \text {for each}\,\, r\in (R',R), \end{aligned}$$

So we have

$$\begin{aligned} \sup _{\delta \in (0,\delta _0)}\sup _{u \in S_\delta } \Vert u\Vert _{C^1([R',R])}<\infty , \end{aligned}$$

and hence

$$\begin{aligned} \sup _{\delta \in (0,\delta _0)}\sup _{u \in S_\delta }\Vert u\Vert _{C^2([R',R])} <\infty . \end{aligned}$$

Hence by similar arguments as in the proof of Lemma 7, we can infer that our assertion holds.$\square$

Proof of Theorem 8

As already said, it is enough to show that in the case of $R<\infty$ and $G\equiv 0$, ${\bar{u}}$ is a nondegenerate critical point of I. From $(d/dr)J(r;\bar{u})=G(r)\bar{u}(r)^2=0$ for each $r \in (R',R)$, we have

$$\begin{aligned} J(r;\bar{u})=\frac{1}{2}a(R)\bar{u}_r(R)^2>0 \quad {\mathrm{for\, each}\,\, r\in [R',R].} \end{aligned}$$

Letting $\delta \in (0,\delta _0)$ be sufficiently small, we have

$$\begin{aligned} \inf _{u\in S_\delta } \inf _{r \in [R',R]}J_\delta (r;u)>0 \end{aligned}$$

by the previous lemma. By a similar proof of that of Theorem 1, we can see that ${\bar{u}}$ is the unique positive solution of (7.17). Hence, by a similar proof of that of Theorem 3, we can find that ${\bar{u}}$ is a nondegenerate critical point of I. $\square$