New Type of Positive Bubble Solutions for a Critical Schrödinger Equation

Abstract

In this paper, we investigate the following critical elliptic equation $-\Delta u+V(y)u=u^{\frac{N+2}{N-2}},\;u>0,\;\text {in}\,{\mathbb {R}}^{N},\;u\in H^{1}({\mathbb {R}}^{N}),$ where V(y) is a bounded non-negative function in ${\mathbb {R}}^{N}.$ Assuming that $V(y)=V(|\hat{y}|,y^{*}),y=(\hat{y},y^{*})\in {\mathbb {R}}^{4}\times {\mathbb {R}}^{N-4}$ and gluing together bubbles with different concentration rates, we obtain new solutions provided that $N\ge 7,$ whose concentrating points are close to the point $(r_{0},y^{*}_{0})$ which is a stable critical point of the function $r^{2}V(r,y^{*})$ satisfying $r_{0}>0$ and $V(r_{0},y^{*}_{0})>0.$ In order to construct such new bubble solutions for the above problem, we first prove a non-degenerate result for the positive multi-bubbling solutions constructed in Peng et al. (J Funct Anal 274:2606–2633, 2018) by some local Pohozaev identities, which is of great interest independently. Moreover, we give an example which satisfies the assumptions we impose.

Appendices

Appendix A: Some Pohozaev Identities

Set

$$\begin{aligned} -\Delta u+V(|y'|,y'')u=u^{2^*-1}, \end{aligned}$$

(A.1)

and

$$\begin{aligned} -\Delta \eta +V(|y'|,y'')\eta =(2^*-1)u^{2^*-2}\eta . \end{aligned}$$

(A.2)

Then standard arguments give $u,\, \eta \in L^\infty ({\mathbb {R}}^N)$ and

$$\begin{aligned} |u(y)|,\;\; |\eta (y)|\le \frac{C}{(1+ |y|)^{N-2}}. \end{aligned}$$

(A.3)

Suppose that $\Omega$ is a smooth domain in ${\mathbb {R}}^N$.

We have the following identities which are used in Sect. 2 by proving the non-degeneracy of the multi-bubbling solutions obtained in [20].

Lemma A.1

There holds

$$\begin{aligned}&-\int _{\partial \Omega }\frac{\partial u}{\partial \nu }\frac{\partial \eta }{\partial y_i}-\int _{\partial \Omega }\frac{\partial \eta }{\partial \nu }\frac{\partial u}{\partial y_i} +\int _{\partial \Omega }\langle \nabla u,\nabla \eta \rangle \nu _i+\int _{\partial \Omega }Vu\eta \nu _i -\int _{\partial \Omega }u^{2^*-1}\eta \nu _i\\&\quad =\int _{\Omega }\frac{\partial V}{\partial y_i}u\eta , \end{aligned}$$

(A.4)

and

$$\begin{aligned}&\int _{\Omega }u\eta \langle \nabla V,y-x_0\rangle +2\int _{\Omega }V\eta u =-\int _{\partial \Omega }u^{2^*-1}\eta \langle \nu ,y-x_0\rangle -\int _{\partial \Omega }\frac{\partial u}{\partial \nu }\langle \nabla \eta ,y-x_0\rangle \\&\quad -\int _{\partial \Omega }\frac{\partial \eta }{\partial \nu }\langle \nabla u,y-x_0\rangle +\int _{\partial \Omega }\langle \nabla u,\nabla \eta \rangle \langle \nu ,y-x_0\rangle +\int _{\partial \Omega }Vu\eta \langle \nu ,y-x_0\rangle \\&\quad +\frac{2-N}{2}\int _{\partial \Omega }\eta \frac{\partial u}{\partial \nu }+\frac{2-N}{2}\int _{\partial \Omega }u\frac{\partial \eta }{\partial \nu }. \end{aligned}$$

(A.5)

Proof of (A.4)

First we have

$$\begin{aligned} \int _{\Omega }(-\Delta u+Vu)\frac{\partial \eta }{\partial y_i}=\int _{\Omega }u^{2^*-1}\frac{\partial \eta }{\partial y_i}, \end{aligned}$$

and

$$\begin{aligned} \int _{\Omega }(-\Delta \eta +V\eta )\frac{\partial u}{\partial y_i}=\int _{\Omega }(2^*-1)u^{2^*-2}\eta \frac{\partial u}{\partial y_i}, \end{aligned}$$

which implies that

$$\begin{aligned}&\int _{\Omega }\Big (-\Delta u\frac{\partial \eta }{\partial y_i}+(-\Delta \eta )\frac{\partial u}{\partial y_i}+V u\frac{\partial \eta }{\partial y_i}+V\eta \frac{\partial u}{\partial y_i}\Big ) \nonumber \\&\quad =\int _{\Omega }\Big (u^{2^*-1}\frac{\partial \eta }{\partial y_i}+(2^*-1)u^{2^*-2}\eta \frac{\partial u}{\partial y_i}\Big ). \end{aligned}$$

(A.6)

It is easy to check that

$$\begin{aligned} \int _{\Omega }\Big (u^{2^*-1}\frac{\partial \eta }{\partial y_i}+(2^*-1)u^{2^*-2}\eta \frac{\partial u}{\partial y_i}\Big )=\int _{\Omega }\frac{\partial (u^{2^*-1}\eta )}{\partial y_i} =\int _{\partial \Omega }u^{2^*-1}\eta \nu _i. \end{aligned}$$

(A.7)

Moreover, similar to (2.7) in [15], we have

$$\begin{aligned} \int _{\Omega }\Big (-\Delta u\frac{\partial \eta }{\partial y_i}+(-\Delta \eta )\frac{\partial u}{\partial y_i} \Big )=-\int _{\partial \Omega }\frac{\partial u}{\partial \nu }\frac{\partial \eta }{\partial y_i}-\int _{\partial \Omega }\frac{\partial \eta }{\partial \nu }\frac{\partial u}{\partial y_i}+\int _{\partial \Omega }\langle \nabla u,\nabla \eta \rangle \nu _i, \end{aligned}$$

(A.8)

and

$$\begin{aligned} \int _{\Omega }\Big (Vu\frac{\partial \eta }{\partial y_i}+V\eta \frac{\partial u}{\partial y_i}\Big )=\int _{\Omega }V \frac{\partial }{\partial y_i}(u\eta ) =\int _{\partial \Omega }Vu\eta \nu _i-\int _{\Omega }u\eta \frac{\partial V}{\partial y_i}. \end{aligned}$$

(A.9)

It follows from (A.6) to (A.9) that (A.4) holds. $\square$

Proof of (A.5)

It is easy to check that

$$\begin{aligned}&\int _{\Omega }\big ((-\Delta u+V u)\langle \nabla \eta ,y-x_0\rangle +(-\Delta \eta +V\eta )\langle \nabla u,y-x_0\rangle \big )\\&\quad =\int _{\Omega }\big (u^{2^*-1}\langle \nabla \eta ,y-x_0\rangle +(2^*-1)u^{2^*-2}\eta \langle \nabla u,y-x_0\rangle \big ). \end{aligned}$$

(A.10)

We find that

$$\begin{aligned}&\int _{\Omega }\big (u^{2^*-1}\langle \nabla \eta ,y-x_0\rangle +(2^*-1)u^{2^*-2}\eta \langle \nabla u,y-x_0\rangle \big )\\&\quad =\int _{\Omega }\langle \nabla (u^{2^*-1}\eta ),y-x_0\rangle =\int _{\partial \Omega }u^{2^*-1}\eta \langle \nu ,y-x_0\rangle -N\int _{\Omega }u^{2^*-1}\eta . \nonumber \\ \end{aligned}$$

(A.11)

Also similar to (2.10) in [15], we have

$$\begin{aligned}&\int _{\Omega }\big (-\Delta u\langle \nabla \eta ,y-x_0\rangle +(-\Delta \eta )\langle \nabla u,y-x_0\rangle \big )\\&\quad =-\int _{\partial \Omega }\frac{\partial u}{\partial \nu }\langle \nabla \eta ,y-x_0\rangle -\int _{\partial \Omega }\frac{\partial \eta }{\partial \nu }\langle \nabla u,y-x_0\rangle \\&\qquad +\int _{\partial \Omega }\langle \nabla u,\nabla \eta \rangle \langle \nu ,y-x_0\rangle +(2-N)\int _{\Omega }\langle \nabla u,\nabla \eta \rangle . \end{aligned}$$

(A.12)

On the other hand, there holds

$$\begin{aligned} 2^*\int _{\Omega }u^{2^*-1}\eta&=\int _{\Omega }\big ((-\Delta u \eta +u(-\Delta \eta )+V u\eta +V\eta u\big )\\&=2\int _{\Omega }\langle \nabla u,\nabla \eta \rangle -\int _{\partial \Omega }\eta \frac{\partial u}{\partial \nu }-\int _{\partial \Omega }u\frac{\partial \eta }{\partial \nu }+2\int _{\Omega }Vu\eta , \end{aligned}$$

(A.13)

which yields

$$\begin{aligned} \int _{\Omega }\langle \nabla u,\nabla \eta \rangle =\frac{2^*}{2}\int _{\Omega }u^{2^*-1}\eta +\frac{1}{2}\int _{\partial \Omega }\eta \frac{\partial u}{\partial \nu }+\frac{1}{2}\int _{\partial \Omega }u\frac{\partial \eta }{\partial \nu }-\int _{\Omega }Vu\eta . \end{aligned}$$

(A.14)

Moreover, we obtain

$$\begin{aligned}&\int _{\Omega }\big (V u\langle \nabla \eta ,y-x_0\rangle +V\eta \langle \nabla u,y-x_0\rangle \big ) =\int _{\Omega }V\langle \nabla (u\eta ),y-x_0\rangle \\&=\int _{\partial \Omega }Vu\eta \langle \nu ,y-x_0\rangle -\int _{\Omega }u\eta \langle \nabla V,y-x_0\rangle -N\int _{\Omega }Vu\eta . \end{aligned}$$

(A.15)

Therefore, from (A.10) to (A.15) we know that (A.5) holds. $\square$

Appendix B: The Green’s Functions

In this section, we mainly study the Green’s function of $L_m$ (see the definition of (2.4)).

For any function g defined in ${\mathbb {R}}^N$, we define its corresponding function $g^\star \in H_s$ as follows.

First we define ${{\textbf {A}}}_j$ as

$$\begin{aligned} {{\textbf {A}}}_j z= \Bigl ( r \cos \big (\theta +\frac{2j \pi }{m}\big ), r \sin \big (\theta +\frac{2j \pi }{m}\big ), z''\Big ),\quad j=1, \ldots , m, \end{aligned}$$

where $z= (z', z'')\in {\mathbb {R}}^N$, $z'= (r\cos \theta , r\sin \theta )\in {\mathbb {R}}^2$, $z''\in {\mathbb {R}}^{N-2}$, while

$$\begin{aligned} {{\textbf {B}}}_i z ={\left\{ \begin{array}{ll} \bigl ( z_1,\ldots , z_{i-1}, -z_i, z_{i+1},\ldots , z_N),\quad &{} i=2, \\ \bigl ( z_1,\ldots , z_{i-1}, z_i, z_{i+1},\ldots , z_N),\quad &{}i=1,3,4,\cdot \cdot \cdot ,N. \end{array}\right. } \end{aligned}$$

Let

$$\begin{aligned} {\hat{g}}(y)=\frac{1}{m}\sum _{j=1}^m g({{\textbf {A}}}_j y), \end{aligned}$$

and

$$\begin{aligned} g^\star (y) = \frac{1}{(N-1)}\sum _{i=2}^N \frac{1}{2} \bigl ( {\hat{g}} (y)+ {\hat{g}} ({{\textbf {B}}}_i y)\bigr ). \end{aligned}$$

Then $g^\star \in H_s.$

Noting that $\delta _x$ is not in $H_s,$ we consider

$$\begin{aligned} L_m u = \delta _x^\star , \quad u\in H_s. \end{aligned}$$

(B.1)

The solution of (B.1) is denoted as $G_m(y, x)$. We want to point out that

$$\begin{aligned} \delta _x^\star =\frac{1}{N-1}\sum _{i=2}^N \frac{1}{2} \Bigl ( \frac{1}{m}\sum _{j=1}^m \delta _{{{\textbf {A}}}_j x}+ \frac{1}{m}\sum _{j=1}^m \delta _{{{\textbf {B}}}_i {{\textbf {A}}}_j x}\Bigr ). \end{aligned}$$

Proposition B.1

Assume that $V(y)\ge 0$ is bounded in ${\mathbb {R}}^{N}.$ The solution $G_m(y, x)$ of (B.1) satisfies

$$\begin{aligned} |G_m(y, x)|\le \frac{1}{N-1}\sum _{i=2}^N \frac{1}{2} \Bigg (\frac{1}{m}\sum _{j=1}^m \frac{C}{|y- {{\textbf {A}}}_j x|^{N-2}}+ \frac{1}{m}\sum _{j=1}^m \frac{C}{|y- {{\textbf {B}}}_i {{\textbf {A}}}_j x|^{N-2}}\Bigg ) \end{aligned}$$

for all $x\in B_R(0)$, where $R>0$ is any fixed large constant.

Proof

Let $\omega _1= \frac{C_N}{|y-x|^{N-2}}$, which satisfies $-\Delta \omega _1= \delta _x$ in ${\mathbb {R}}^N$. Let $\omega _2$ be the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta \omega +V(y)\omega = (2^*-1) u_m^{2^*-2} \omega _1,&{} \text {in}\; B_{2R}(0),\\ \omega =0,&{} \text {on}\; \partial B_{2R}(0). \end{array}\right. } \end{aligned}$$

Then $\omega _2\ge 0$ and

$$\begin{aligned} \omega _2(y)= & {} \int _{B_{2R}(0)}G(z, y) (2^*-1) u_m^{2^*-2} \omega _1\le C \int _{B_R(0)}\frac{1}{|y-z|^{N-2}}\frac{1}{|z-x|^{N-2}}\,\text {d}z\\\le & {} \frac{C}{|y-x|^{N-4}}, \end{aligned}$$

where G(z, y) is the Green’s function of the positive operator $-\Delta +V(y)$ in $B_{2R}(0)$ with zero boundary condition. We can continue this process to find $\omega _i$, which is the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta \omega +V(y)\omega = (2^*-1) u_m^{2^*-2} \omega _{i-1},&{} \text {in}\; B_{2R}(0),\\ \omega =0,&{} \text {on}\; \partial B_{2R}(0), \end{array}\right. } \end{aligned}$$

and satisfies

$$\begin{aligned} \begin{aligned} 0\le \omega _i (y)&= \int _{B_{2R}(0)}G(z, y) (2^*-1) u_m^{2^*-2} \omega _{i-1}\\&\le C \int _{B_{2R}(0)}\frac{1}{|y-z|^{N-2}}\frac{1}{|z-x|^{N-2(i-1)}}\,\text {d}z \le \frac{C}{|y-x|^{N-2i}}. \end{aligned} \end{aligned}$$

Let i be large satisfying $\omega _i\in L^\infty (B_{2R}(0))$. Define

$$\begin{aligned} \omega =\sum _{l=1}^i \omega _l, \end{aligned}$$

and $\upsilon = G_m(y, x)- \xi \omega ^\star$, where $\xi (y)=\xi (|y|)\in C^\infty _0(B_{2R}(0))$, $\xi =1$ in $B_{\frac{3}{2} R}(0)$ and $0\le \xi \le 1$. Then we have

$$\begin{aligned} L_m \upsilon = g, \end{aligned}$$

(B.2)

where $g\in L^\infty \cap H_s$ and $g=0$ in ${\mathbb {R}}^N\setminus B_{2R}(0)$. Applying Proposition 2.1, (B.2) has a solution $\upsilon \in H_s.$

We still need to prove that $|\upsilon (y)|\le \frac{C}{|y|^{N-2}}$ as $|y|\rightarrow +\infty$.

First, we claim that $|\upsilon |\le C |g|_{L^\infty ({\mathbb {R}}^N)}$. Indeed, assume that there are $g_n\in L^\infty \cap H_s$, $\upsilon _n$ satisfying (B.2), with $|g_{n}|_{L^\infty ({\mathbb {R}}^N)}\rightarrow 0$ and $|\upsilon _n|_{L^\infty ({\mathbb {R}}^N)}=1$. Then, $\upsilon _n\rightarrow \upsilon$ in $C^1_{loc}({\mathbb {R}}^N)$, which satisfies $L_m \upsilon =0$. Therefore $\upsilon =0$. On the other hand, we have

$$\begin{aligned} |\upsilon _n(y)| \le C_N \int _{{\mathbb {R}}^N}\frac{1}{|z-y|^{N-2}}\big |(2^*-1) u_m^{2^*-2} \upsilon _n\,\big |\text {d}z+ C_N \nonumber \\ \int _{{\mathbb {R}}^N}\frac{1}{|z-y|^{N-2} } |g_n| \,\text {d}z. \end{aligned}$$

(B.3)

Hence we obtain that $|\upsilon _n(y)|\le \frac{C}{|y|^2}$ as $|y|\rightarrow +\infty ,$ which contradicts to $|\upsilon _n|_{L^\infty ({\mathbb {R}}^N)}=1$.

So $\upsilon$ is bounded. Then it follows from (B.3) that $|\upsilon (y)|\le \frac{C}{(1+|y|)^2}$. Also, we have

$$\begin{aligned} |\upsilon (y)|\le C\int _{{\mathbb {R}}^N}\frac{1}{|z-y|^{N-2} } u_m^{2^*-2}\frac{C}{(1+|y|)^2} +\frac{C}{(1+|y|)^{N-2}}\le \frac{C}{(1+|y|)^4}. \end{aligned}$$

Repeating this process, we can prove $|\upsilon (y)|\le \frac{C}{|y|^{N-2}}$ as $|y|\rightarrow +\infty .$ $\square$

Appendix C: Basic Estimates

For each fixed k and j, $k\ne j$, we consider the following function

$$\begin{aligned} g_{k,j}(y)=\frac{1}{(1+|y-x_{j}|)^{\alpha }}\frac{1}{(1+|y-x_{k}|)^{\beta }}, \end{aligned}$$

(C.1)

where $\alpha \ge 1$ and $\beta \ge 1$ are two constants.

Lemma C.1

([25, Lemma B.1]) For any constants $0<\delta \le \min \{\alpha ,\beta \}$, there is a constant $C>0$, such that

$$\begin{aligned} g_{k,j}(y)\le \frac{C}{|x_{k}-x_{j}|^{\delta }}\Big (\frac{1}{(1+|y-x_{k}|)^{\alpha +\beta -\delta }}+\frac{1}{(1+|y-x_{j}|)^{\alpha +\beta -\delta }}\Big ). \end{aligned}$$

Lemma C.2

([25, Lemma B.2]) For any constant $0<\delta <N-2$, there is a constant $C>0$, such that

$$\begin{aligned} \int \frac{1}{|y-z|^{N-2}}\frac{1}{(1+|z|)^{2+\delta }}\text {d}z\le \frac{C}{(1+|y|)^{\delta }}. \end{aligned}$$

Let us recall that

$$\begin{aligned} Z_{t,{\tilde{y}}^{*},\mu }(y)=\sum _{j=1}^{n}{\hat{\zeta }}(y) U_{p_{j},\mu }=[N(N-2)]^{\frac{N-2}{4}}\sum _{j=1}^{n} {\hat{\zeta }}(y)\Bigg (\frac{\mu }{1+\mu ^{2}|y-p_{j}|^{2}}\Bigg )^{\frac{N-2}{2}}. \end{aligned}$$

Just by the same argument as that of Lemma B.3 in [20], we can prove

Lemma C.3

Suppose that $N\ge 5.$ Then there is a small constant $\iota >0$, such that

$$\begin{aligned} \int \frac{1}{|y-z|^{N-2}}Z^{\frac{4}{N-2}}_{t,{\tilde{y}}^{*},\mu }(z)\sum _{j=1}^{n}\frac{1}{(1+\mu |z-p_{j}|)^{\frac{N-2}{2}+\iota }}\text {d}z \!\le \! \sum _{j=1}^{n}\frac{C}{(1+\mu |y-p_{j}|)^{\frac{N-2}{2}\!+\!\tau +\iota }}. \end{aligned}$$

Appendix D: An Example of the Potential $V(r,y^{*})$

Here we give an example of $V({\hat{y}},y^{*})$ which satisfies the assumptions (V) and $({\tilde{V}}).$ We define

$$\begin{aligned} V(r,y^*)= \left\{ \begin{array}{ll} r^2-4r\Big (\sum \limits _{j=5}^Ny_j\Big )+\Big (\sum \limits _{j=5}^Ny_j^2\Big )+1,&{}B_\rho (r_0,y_0^*),\\ \ge 0, &{}{\mathbb {R}}^N\backslash B_\rho (r_0,y_0^*), \end{array} \right. \end{aligned}$$

where $\rho$ is the same as that of [20] and $(r_0,y_0^*)$ is defined below. By some direct computations, we can check that

$$\begin{aligned} f(r,y^*):=r^2V(r,y^*)=r^4-4r^3\Big (\sum \limits _{j=5}^Ny_j\Big )+r^{2}\Big (\sum \limits _{j=5}^Ny_j^2\Big )+r^2. \end{aligned}$$

We have

$$\begin{aligned} \frac{\partial f}{\partial r}=4r^3-12r^2\Big (\sum \limits _{j=5}^Ny_j\Big )+2r\Big (\sum \limits _{j=5}^Ny_j^2\Big )+2r, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial f}{\partial y_i}=-4r^3+2r^2y_i,\quad i=5,\ldots ,N. \end{aligned}$$

Suppose that $\frac{\partial f}{\partial r}=0,\,\frac{\partial f}{\partial y_i}=0$, we obtain

$$\begin{aligned} y_i=2r,\quad \hbox {for}\; i=5,\ldots ,N, r_0=\sqrt{\frac{1}{8N-34}},\quad y_{0,i}=2\sqrt{\frac{1}{8N-34}}(i=5,...,N). \end{aligned}$$

Therefore, $(r_0,y^{*}_{0})$ is a critical point of the function $f(r,y^*)$ and $V(r_0,y^{*}_{0})=\frac{1}{2}>0$. Also

$$\begin{aligned}&\frac{\partial ^2f}{\partial r^2}=12r^2-24r\Big (\sum \limits _{j=5}^Ny_j\Big )+2\Big (\sum \limits _{j=5}^Ny_j^2\Big )+2, \\&\frac{\partial ^2f}{\partial r\partial y_i}=-12r^2+4ry_i,\quad i=5,\ldots ,N, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial ^2f}{\partial y_i^2}=2r^2(i=5,\ldots ,N),\quad \frac{\partial ^2f}{\partial y_i\partial y_j}=0(i,j=5,...,N,i\ne j). \end{aligned}$$

By direct computation, we obtain

$$\begin{aligned} B= \left( \begin{array}{cccc} \frac{\partial ^2f}{\partial r^2}&{}\frac{\partial ^2f}{\partial r\partial y_5}&{}\cdots &{}\frac{\partial ^2f}{\partial r\partial y_N}\\ \frac{\partial ^2f}{\partial r\partial y_5}&{}\frac{\partial ^2f}{\partial y_5\partial y_5}&{}\cdots &{}0\\ \cdots &{}\cdots &{}\cdots &{}\cdots \\ \frac{\partial ^2f}{\partial r\partial y_N}&{}0&{}\cdots &{}\frac{\partial ^2f}{\partial y_N\partial y_N} \end{array} \right) _{(N-3)\times (N-3)}. \end{aligned}$$

Also by some tedious computation, we can obtain the eigenvalues of the matrix B are as follows:

$$\begin{aligned} |\lambda I-B|_{(r_0,y_0^*)}&=\!\Big [\big (\lambda -\frac{\partial ^2f}{\partial r^2}\big )(\lambda -2r^2)\!-\!\sum \limits _{j=5}^N(-12r^2\!+\!4ry_{j})^2\Big ] (\lambda \!-\!2r^2)^{N-5}\Big |_{(r_0,y_0^*)}\nonumber \\&=0, \end{aligned}$$

which implies that

$$\begin{aligned} \Big (\lambda -\frac{\partial ^2f}{\partial r^2}\mid _{(r_0,y_0^*)}\Big )(\lambda -2r_0^2)&=\sum \limits _{j=5}^N(-12r_0^2+4r_0y_{0j})^2,\quad \text {or}\quad (\lambda -2r^2)^{N-5}\Big |_{(r_0,y_0^*)}\nonumber \\&=0. \end{aligned}$$

By further computation, we can check that $\min \{\lambda _1,\lambda _2\}<0$ and

$$\begin{aligned} \lambda _3=\cdots =\lambda _{N-3}=2r_0^2>0. \end{aligned}$$

Hence the assumption (V) holds.

On the other hand, we recall

$$\begin{aligned} V(r,y)=r^2-4r\Bigg (\sum \limits _{j=5}^Ny_j\Bigg )+\Bigg (\sum \limits _{j=5}^Ny_j^2\Bigg )+1,\quad \text {in}\;B_\rho (r_0,y_0^*). \end{aligned}$$

We obtain in $B_\rho (r_0,y_0^*)$

$$\begin{aligned}&\frac{\partial V}{\partial r}=2r-4\sum \limits _{j=5}^Ny_j,\quad \frac{\partial V}{\partial y_i}=2y_i-\frac{4y_i}{r}\Bigg (\sum \limits _{j=5}^Ny_j\Bigg ),i=1,2,3,4, \nonumber \\&\frac{\partial V}{\partial y_k}=-4r+2y_k,\quad k=5,\ldots ,N, \nonumber \\&\frac{\partial ^2V}{\partial r^2}=2,\quad \frac{\partial ^2V}{\partial r\partial y_i}=\frac{2y_{0,i}}{r_{0}}(i=1,2,3,4),\quad \frac{\partial ^2V}{\partial r\partial y_k}=-4(k=5,\ldots ,N), \end{aligned}$$

(D.1)

$$\begin{aligned}&\frac{\partial ^2V}{\partial y_i\partial y_i}=2-\frac{4r^2-4y_i^2}{r^3}\Big (\sum \limits _{j=5}^Ny_j\Big )(i=1,2,3,4), \quad \frac{\partial V}{\partial y_k^2}=2(k=5,\ldots ,N) \end{aligned}$$

(D.2)

and

$$\begin{aligned} \frac{\partial ^2V}{\partial y_k\partial y_j}=0(j\ne k, k,j=5,\ldots ,N). \end{aligned}$$

(D.3)

Then from (D.1) to (D.3), we obtain in $B_\rho (r_0,y_0^*)$

$$\begin{aligned} \Delta V&=\sum \limits _{j=1}^N\frac{\partial ^2V}{\partial y_i\partial y_i}=\sum \limits _{i=1}^4\Big (2-\frac{4r^2-4y_i^2}{r^3}(y_5+\cdots +y_N)\Big )+2(N-4)\\&=2N-\frac{12}{r}(y_5+\cdots +y_N). \end{aligned}$$

Hence

$$\begin{aligned} \Delta V\Big |_{(r_{0},y^{*}_{0})}=96-22N. \end{aligned}$$

For $y_0^*=(y_{0,5},y_{0,6}\ldots ,y_{0,N}),$ one has

$$\begin{aligned}&\frac{\partial \Delta V}{\partial y_1}\Big |_{(r_0,y_0^*)}=\frac{12y_1}{r^3}(y_5+\cdots +y_N)\Big |_{(r_0,y_0^*)}=\frac{24y_{0,1}}{r_0^2}(N-4), \\&\frac{\partial (\Delta V)}{\partial y_k}\Big |_{(r_0,y_0^*)}=-\frac{12}{r_0},k=5,\ldots ,N. \end{aligned}$$

Since $y_{0,i}=2r_0(i=5,\ldots ,N),$ we obtain

$$\begin{aligned} A_{i,l}= \left\{ \begin{array}{ll} \Big [ 2-(\frac{24y_{0,i}(N-4)}{2r_0^2(96-22N)}+\frac{\nu _1}{\langle \nu ,x_1\rangle })(34-8N)r_0], \text {when}\;i=l=1; \\ -4, \text {when}\;i=1, l=2,3,...,N-3; \\ \cos \frac{2i\pi }{m}(-4-(\frac{-6}{r_0(96-22N)}+\frac{\nu _{i+1}}{\langle \nu ,x_{1}\rangle })(34-8N)r_0),\\ \text {when}\;i=2,3,...,N-3,l=1; \\ 2, \text {when}\;i=l,i,l=2,3,...,N-3, \\ 0, \text {when}\;i\ne l,\,i,l=2,3,...,N-3. \end{array} \right. \end{aligned}$$

(D.4)

Therefore, from (D.4) we have

$$\begin{aligned}&\det (A_{i,l})_{(N-3)\times (N-3)}\\&\quad =\prod _{i=2}^{N-3}A_{i,i}\Bigg (A_{1,1}-\sum _{k=2}^{N-3}\frac{A_{k,1}A_{1,k}}{A_{k,k}}\Bigg )\\&\quad =\prod _{i=2}^{N-3}2 \Bigg ( 2-(\frac{24y_{0,i}(N-4)}{2r_0^2(8-22(N-4))}+\frac{\nu _1}{\langle \nu ,x_1\rangle })(34-8N)r_0\Bigg ) \\&\qquad +\sum _{k=2}^{N-3}2\cos \frac{2i\pi }{m}(-4-(\frac{-6}{r_0(8-22(N-4))}+\frac{\nu _{i+3}}{\langle \nu ,x_{1}\rangle })(34-8N)r_0) \Bigg )\\&\quad \ne 0 . \end{aligned}$$

Hence, the assumption $({\tilde{V}})$ also holds.