Appendix: Proof of Proposition 4.1, the concentration-compactness analysis
The Proof of Proposition 4.1 consists of a sequence of lemmas.
Assume \(\lambda _n\rightarrow 0,\) \(u_n\in X,\) satisfies the equation
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}\left( a(u_n)\nabla u_n\nabla \varphi +\frac{1}{2}a'(u_n)|\nabla u_n|^2\varphi \right) \text{ d }x+\int _{{\mathbb {R}}^3}V(x) u_n\varphi \text{ d }x\\&\quad +\int _{{\mathbb {R}}^3}b(u_n)|\nabla u_n|^2\text{ d }x\int _{{\mathbb {R}}^3} \left( b(u_n)\nabla u_n\nabla \varphi +\frac{1}{2}b'(u_n)|\nabla u_n|^2\varphi \right) \text{ d }x=\int _{{\mathbb {R}}^3}f_{\lambda _n}(u_n)\varphi \text{ d }x,\ \\&\quad \text{ for } \varphi \in C_0^{\infty }({\mathbb {R}}^3), \end{aligned}\nonumber \\ \end{aligned}$$
(a1)
and there exists a constant L, independent of n, such that
$$\begin{aligned} \int _{{\mathbb {R}}^3}(1+u_n^2)|\nabla u_n|^2\text{ d }x+\int _{{\mathbb {R}}^3}V(x)u_n^2\text{ d }x\le L. \end{aligned}$$
(a2)
Lemma A1
Assume \(u_n\rightharpoonup u\) in \(H_{V},\) \(u_n\nabla u_n\rightharpoonup u\nabla u\) in \(L^2({\mathbb {R}}^3)\) and \(K_n=\int _{{\mathbb {R}}^3}b(u_n)|\nabla u_n|^2\text{ d }x\rightarrow K_{\infty }.\) u satisfies the equation
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}\left( a(u)\nabla u\nabla \varphi +\frac{1}{2}a'(u)|\nabla u|^2\varphi \right) \text{ d }x+\int _{{\mathbb {R}}^3}V(x) u\varphi \text{ d }x\\&\quad +K_{\infty }\left( \int _{{\mathbb {R}}^3}(b(u)\nabla u\nabla \varphi +\frac{1}{2}b'(u)|\nabla u|^2\varphi )\text{ d }x\right) =\int _{{\mathbb {R}}^3}(u^{11}+\mu |u|^{q-2}u)\varphi \text{ d }x,\ \\&\quad \text{ for } \varphi \in C_0^{\infty }({\mathbb {R}}^3). \end{aligned} \nonumber \\ \end{aligned}$$
(a3)
Proof
Notice that \(u_n(x){\mathop {\rightarrow }\limits ^{a.e.}} u(x),\) and
$$\begin{aligned} & f_{\lambda _n}(u_n(x))=|u_n(x)|^{q-2}u_n(x)|m_{\lambda _n}(u_n(x))|^{12-q}\\ & \quad +\mu |u_n(x)|^{q-2}u_n(x) {\mathop {\rightarrow }\limits ^{a.e.}}u^{11}(x)+\mu |u(x)|^{q-2}u(x). \end{aligned}$$
Lemma \(A_1\) can be proved as Lemma 2.2 of Guo et al. (2016). \(\square \)
Lemma A2
Let \(T>0,\) let \(u^T\) be the truncated function: \(u^T(x)=u(x)\) if \(u(x)\le T,\) \(u^T(x)=\pm T\) if \(\pm u(x)\ge T.\) Then \(u_n^T\rightarrow u^T\) in \(H_V.\)
Proof
Multiply by \(u_n^T\) and \(u^T\) the equations \((a_1),(a_3)\) respectively, we obtain
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}a(u_n^T)|\nabla u_n^T|^2\text{ d }x+\frac{1}{2}\int _{{\mathbb {R}}^3}a'(u_n)|\nabla u_n|^2u_n^T\text{ d }x+\int _{{\mathbb {R}}^3}V(x)u_nu_n^T\text{ d }x\\&\quad +K_n\left( \int _{{\mathbb {R}}^3}b(u_n^T)|\nabla u_n^T|^2\text{ d }x+\frac{1}{2}\int _{{\mathbb {R}}^3}b'(u_n)|\nabla u_n|^2u_n^T\text{ d }x\right) =\int _{{\mathbb {R}}^3}f_{\lambda _n}(u_n)u_n^T\text{ d }x.\\ \end{aligned}\nonumber \\ \end{aligned}$$
(a4)
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}a(u^T)|\nabla u^T|^2\text{ d }x+\frac{1}{2}\int _{{\mathbb {R}}^3}a'(u)|\nabla u|^2u^T\text{ d }x+\int _{{\mathbb {R}}^3}V(x)uu^T\text{ d }x\\&\qquad +K_{\infty }\left( \int _{{\mathbb {R}}^3}b(u^T)|\nabla u^T|^2\text{ d }x+\frac{1}{2}\int _{{\mathbb {R}}^3}b'(u)|\nabla u|^2u^T\text{ d }x\right) \\&\quad =\int _{{\mathbb {R}}^3}(u^{11}+\mu |u|^{q-2}u)u^T\text{ d }x. \end{aligned}\nonumber \\ \end{aligned}$$
(a5)
Since \(u_n\in X\) is radial function, X is compactly imbedding to \(L^q({\mathbb {R}}^3),2<q<12,\) that is a subsequence of \(u_n\) strongly converge to u in \(L^q({\mathbb {R}}^3)\), provided \(u_n\rightharpoonup u\) in \(H_V,\) \(u_n\nabla u_n\rightharpoonup u\nabla u\) in \(L^2({\mathbb {R}}^3).\) We have
$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^3}f_{\lambda _n}(u_n)u_n^T\text{ d }x\rightarrow \int _{{\mathbb {R}}^3}(u^{11}+\mu |u|^{q-2}u)u^T\text{ d }x, \ \ \text{ as } \ \ n\rightarrow \infty . \end{aligned} \end{aligned}$$
(a6)
By lower semi-continuity, the limits of terms on the left hand side of \((a_4)\) are larger than or equal to the corresponding terms on the left side of \((a_5)\), hence taking \((a_6)\) into account, we obtain
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}a(u_n^T)|\nabla u_n^T|^{2}\text{ d }x\rightarrow \int _{{\mathbb {R}}^3}a(u^T)|\nabla u^T|^2\text{ d }x,\\&\int _{{\mathbb {R}}^3}V(x)u_nu_n^T\text{ d }x\rightarrow \int _{{\mathbb {R}}^3}V(x)uu^T\text{ d }x, \end{aligned}\nonumber \\ \end{aligned}$$
(a7)
and \(u_n^T\rightarrow u^T\) in \(H_V.\) \(\square \)
Corollary A3
Assume the profile decomposition (4.11) holds. Then \(v=k(u)\) satisfies the equation
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}A(g(v))\nabla v\nabla \varphi \text{ d }x +\frac{1}{2}\int _{{\mathbb {R}}^3}A'(g(v))g'(v)|\nabla v|^2\varphi \text{ d }x +\int _{{\mathbb {R}}^3}V(x)g(v)g'(v)\text{ d }x\\&\quad +K_{\infty }\left( \int _{{\mathbb {R}}^3}B(g(v))\nabla v\nabla \varphi \text{ d }x +\frac{1}{2}\int _{{\mathbb {R}}^3}B'(g(v))g'(v)|\nabla v|^2\varphi \text{ d }x\right) \\&\quad =\int _{{\mathbb {R}}^3}f_{\lambda }(g(v))g'(v)\varphi \text{ d }x, \ \\&\quad \text{ for } \ \varphi \in C_0^{\infty }({\mathbb {R}}^3) \end{aligned}\nonumber \\ \end{aligned}$$
(a8)
and \(v_n^T=(k(u_n))^T\rightarrow v^T\) in \(H_V\).
Lemma A4
It holds that
$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^3}A(g(v_n))\nabla v_n\nabla \varphi \text{ d }x&=\int _{{\mathbb {R}}^3}A_{\infty }\nabla v_n\nabla \varphi \text{ d }x\\&\quad +\int _{{\mathbb {R}}^3}(A(g(v))-A_{\infty })\nabla v\nabla \varphi \text{ d }x +o(1)\Vert \varphi \Vert _{{\mathfrak {D}}}, \end{aligned} \end{aligned}$$
(a9)
where \(A_{\infty }=\lim _{|t|\rightarrow \infty }A(t).\)
Proof
Using the fact \(v_n^T\rightarrow v^T\) in \({\mathfrak {D}},\) we can prove Lemma \(A_4\). For the details, see the proof of Lemma 2.2 (Guo et al. 2016). \(\square \)
Lemma A5
Assume \(\sigma _{n,k}^{-\frac{1}{2}}v_n(\sigma _{n,k}^{-1}\cdot )\rightharpoonup U_k\) in \({\mathfrak {D}},\) \(w=|U_k|.\) Then w satisfies the inequality
$$\begin{aligned} \begin{aligned}&\alpha \int _{{\mathbb {R}}^3}\nabla w\nabla \varphi \text{ d }x \le 2^{5}\int _{{\mathbb {R}}^3} w^5\varphi \text{ d }x,\ \ \ \ \text{ for } \ \ \ \varphi \in {\mathfrak {D}},\varphi \ge 0, \end{aligned} \end{aligned}$$
(a10)
where \(\alpha =A_{\infty }+K_{\infty }B_{\infty },\) \(B_{\infty }=\lim _{|t|\rightarrow \infty }B(t).\)
Proof
\(v_n\) satisfies the equation
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}A(g(v_n))\nabla v_n\nabla \varphi \text{ d }x +\frac{1}{2}\int _{{\mathbb {R}}^3}A'(g(v_n))g'(v_n)|\nabla v_n|^2\varphi \text{ d }x +\int _{{\mathbb {R}}^3}V(x_n)g(v_n)g'(v_n)\varphi \text{ d }x\\&K_{n}\left( \int _{{\mathbb {R}}^3}B(g(v_n))\nabla v_n\nabla \varphi \text{ d }x +\frac{1}{2}\int _{{\mathbb {R}}^3}B'(g(v_n))g'(v_n)|\nabla v_n|^2\varphi \text{ d }x\right) \\&\quad =\int _{{\mathbb {R}}^3}f_{\lambda }(g(v_n))g'(v_n)\varphi \text{ d }x. \end{aligned} \nonumber \\ \end{aligned}$$
(a11)
Denote \(w_n=|v_n|,\) \(w_n\rightharpoonup |v|=w,\) \(\sigma _n^{-\frac{1}{2}}w_n(\sigma _n^{-1}\cdot )\rightharpoonup w\) in \({\mathfrak {D}}.\) For \(\varepsilon >0,\varphi \in C_0^{\infty }({\mathbb {R}}^3),\varphi \ge 0,\) we have
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}A(g(v_n))\nabla v_n\nabla (\frac{v_n}{\sqrt{\varepsilon ^2+v_n^2}}\varphi )\text{ d }x +\frac{1}{2}\int _{{\mathbb {R}}^3}A'(g(v_n))g'(v_n)\left| v_{n} \right| ^{2} \frac{v_n}{\sqrt{\varepsilon ^2+v_n^2}}\varphi \text{ d }x\\&\quad =\int _{{\mathbb {R}}^3}A(g(v_n))\nabla v_n\frac{v_n}{\sqrt{\varepsilon ^2+v_n^2}}\nabla \varphi \text{ d }x +\int _{{\mathbb {R}}^3}A(g(v_n))\frac{\varepsilon ^2|\nabla v_n|^2}{(\varepsilon ^2+v_n^2)^{3/2}}\varphi \text{ d }x\\&\qquad +\frac{1}{2}\int _{{\mathbb {R}}^3}A'(g(v_n))g'(v_n)\left| v_{n} \right| ^{2}\frac{v_n}{\sqrt{\varepsilon ^2+v_n^2}}\varphi \text{ d }x\\&\quad \ge \int _{{\mathbb {R}}^3}A(g(v_n))\nabla v_n\frac{v_n}{\sqrt{\varepsilon ^2+v_n^2}}\nabla \varphi \text{ d }x\rightarrow \int _{{\mathbb {R}}^3}A(g(v_n))\nabla w_n\nabla \varphi \text{ d }x, \text{ as } \varepsilon \rightarrow 0. \end{aligned} \nonumber \\ \end{aligned}$$
(a12)
In a similar way we treat the other terms in (a11) and obtain
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}A(g(v_n))\nabla w_n\nabla \varphi \text{ d }x +K_{n}\int _{{\mathbb {R}}^3}B(g(v_n))\nabla w_n\nabla \varphi \text{ d }x\\&\quad \le \int _{{\mathbb {R}}^3}|f_{\lambda _n}(g(v_n))|g'(v_n)\varphi \text{ d }x,\ \ \text{ for } \ \ \varphi \in C_0^{\infty }({\mathbb {R}}^3),\varphi \ge 0. \end{aligned} \nonumber \\ \end{aligned}$$
(a13)
By Lemma \(A_4\) (which is valid for \(w_n=|v_n|\) as well as \(v_n\).)
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}A(g(v_n))\nabla w_n\nabla \varphi \text{ d }x =\int _{{\mathbb {R}}^3}A_{\infty }\nabla w_n\nabla \varphi \text{ d }x\\&\quad +\int _{{\mathbb {R}}^3}(A(g(v))-A_{\infty })\nabla w\nabla \varphi \text{ d }x+o_{n}(1)\Vert \varphi \Vert _{{\mathfrak {D}}}. \end{aligned} \nonumber \\ \end{aligned}$$
(a14)
For \(\varphi \in C_0^{\infty }({\mathbb {R}}^3),\) \(\varphi \ge 0,\) choose \(\varphi _n=\sigma _n^{\frac{1}{2}}\varphi (\sigma _n\cdot )\) as test function in (a14) (for simplicity we use \(\sigma _n\) instead of \(\sigma _{n,k}\)), then
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}A(g(v_n))\nabla w_n\nabla \varphi _n\text{ d }x\\&\quad =\int _{{\mathbb {R}}^3}A_{\infty }\nabla w\nabla \varphi _n\text{ d }x +\int _{{\mathbb {R}}^3}(A(g(v))-A_{\infty })\nabla w\nabla \varphi _n\text{ d }x+o(1)\Vert \varphi _n\Vert \\&\quad =\int _{{\mathbb {R}}^3}A_{\infty }\nabla w(\sigma _n^{-\frac{1}{2}}\cdot )\nabla \varphi \text{ d }x +\int _{{\mathbb {R}}^3}(A(g(v))-A_{\infty })\nabla w\nabla \varphi _n\text{ d }x+o(1)\Vert \varphi \Vert \\&\quad \rightarrow A_{\infty }\int _{{\mathbb {R}}^3}\nabla w\nabla \varphi \text{ d }x,\ \ \ \text{ as } \ \ \ n\rightarrow \infty . \end{aligned} \nonumber \\ \end{aligned}$$
(a15)
Similarly, \(K_{n}\int _{{\mathbb {R}}^3}B(g(v_n))\nabla w_n\nabla \varphi _n\text{ d }x\rightarrow K_{\infty }B_{\infty }\int _{{\mathbb {R}}^3}\nabla w\nabla \varphi \text{ d }x\) as \(n\rightarrow \infty .\) Finally,
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}f_{\lambda _n}(g(v_n))g'(v_n)\varphi _n\text{ d }x\\&\quad =\int _{{\mathbb {R}}^3}\frac{|g(v_n)|^{11}+\mu |g(v_n)|^{q-1}}{\sqrt{M^2+g^2(v_n)}}\varphi _n\text{ d }x\\&\quad \le \int _{{\mathbb {R}}^3}(g^{10}(v_n)+\mu |g(v_n)|^{q-2})\varphi _n\text{ d }x\\&\quad \le \int _{{\mathbb {R}}^3}(2^{5}w_n^5+cw_n^{\frac{q}{2}-1})\varphi _n\text{ d }x \rightarrow 2^{5}\int _{{\mathbb {R}}^3}w^5\varphi \text{ d }x. \end{aligned} \nonumber \\ \end{aligned}$$
(a16)
The inequality (a10) follows. \(\square \)
Corollary A6
The index set \(\Lambda \) is a finite set.
Proof
By Lemma \(A_5,\)
$$\begin{aligned} \begin{aligned} \alpha \int _{{\mathbb {R}}^3}|\nabla U_k|^2\text{ d }x \le 2^{5}\int _{{\mathbb {R}}^3}U_k^6\text{ d }x \le 2^{5}S^3\int _{{\mathbb {R}}^3}|\nabla U_k|^2\text{ d }x, \end{aligned} \end{aligned}$$
where S is the best constant for the imbedding \( {\mathfrak {D}}\) into \(L^{6}({\mathbb {R}}^3).\) Hence there exists \(m>0\) such that \(\int _{{\mathbb {R}}^3}|\nabla U_k|^2\text{ d }x\ge m.\) By the profile decomposition (4.11), \(\Lambda \) is a finite set. \(\square \)
Remark A7
By Lemma \(A_5,\) there exists \(c,\alpha >0\) such that
$$\begin{aligned} \left| v(x) \right| \le c e^{-\alpha |x|},\ \ |U_k(x)|\le c(1+|x|^2)^{-\frac{1}{2}},\ \ x\in {\mathbb {R}}^3. \end{aligned}$$
By using the profile decomposition, the following Lemma \(A_8\) and Lemma \(A_9\) can be proved in a similar way as in Devillanova and Solimini (2002), Cao et al. (2012), Guo et al. (2016).
Lemma A8
Let \(3<p_2<6<p_1,\) then
$$\begin{aligned} \Vert v_n\Vert _{p_1,p_2,\sigma _n}\le c, \end{aligned}$$
where \(\sigma _n=\min \{\sigma _{n,k},k\in \Lambda \}\) and the norm \(\Vert \cdot \Vert _{p_1,p_2,\sigma _n}\) is defined as follows
$$\begin{aligned} \begin{aligned} \Vert v\Vert _{p_1,p_2,\sigma _{n}}&=\inf \{\alpha | \text{ there } \text{ exists } v_1,v_2 \text{ such } \text{ that } |v(x)|\le v_1(x)+v_2(x),\ x\in {\mathbb {R}}^3\\&\Vert v_1\Vert _{L^{p_1}({\mathbb {R}}^3)}\le \alpha ,\ \Vert v_2\Vert _{L^{p_2}({\mathbb {R}}^3)}\le \alpha \sigma _n^{\frac{1}{2}-\frac{3}{p_2}}\}; \end{aligned} \end{aligned}$$
Lemma A9
There exists a constant c, independent of n, such that
$$\begin{aligned} |v_n(x)|\le c \text{ for } x\in T_n, \ \ \ \int _{T_n}|\nabla v_n|^2\text{ d }x\le c\sigma _n^{-\frac{1}{2}}, \end{aligned}$$
where
$$\begin{aligned} T_n=\{x|x\in {\mathbb {R}}^3,\ \sigma _n^{-\frac{1}{2}}\le |x|\le 2\sigma _n^{-\frac{1}{2}}\}. \end{aligned}$$
Remark A10
The definition of \(\Vert \cdot \Vert _{{p_1,p_2,\sigma _{n}}}\) was introduced in Devillanova and Solimini (2002). By Lemma \(A_7\), we have \(\Vert v\Vert _{p_1,p_2,\sigma _{n}}\le c\) and
$$\begin{aligned} \begin{aligned}&\Vert \Sigma _{k\in \Lambda }\sigma _{n,k}^{\frac{1}{2}}U_k(\sigma _{n,k}\cdot )\Vert _{L^{p_2}({\mathbb {R}}^3)} \le \Sigma _{k\in \Lambda }\left( \int _{{\mathbb {R}}^3}|\sigma _{n,k}^{\frac{1}{2}}U_k(\sigma _{n,k}\cdot )|^{p_2}\text{ d }x \right) ^{\frac{1}{p_2}}\\&=\Sigma _{k\in \Lambda }\sigma _{n,k}^{\frac{1}{2}-\frac{3}{p_2}}\left( \int _{{\mathbb {R}}^3}|U_k(x)|^{p_2}\text{ d }x \right) ^{\frac{1}{p_2}}\le c\Sigma _{k\in \Lambda }\sigma _{n,k}^{\frac{1}{2}-\frac{3}{p_2}} \left( \int _{{\mathbb {R}}^3}(1+x^2)^{-\frac{p_2}{2}}\text{ d }x \right) ^{\frac{1}{p_2}}\\&\le c\Sigma _{k\in \Lambda }\sigma _{n,k}^{\frac{1}{2}-\frac{3}{p_2}}\le c\sigma _{n}^{\frac{1}{2}-\frac{3}{p_2}}. \end{aligned} \end{aligned}$$
Since \(v_n=v+\sum _{k\in \Lambda }\sigma _{n,k}^{\frac{1}{2}}U_k(\sigma _{n,k}\cdot )+r_n\) and \(r_n\) is small in \(L^6({\mathbb {R}}^3),\) we may expect \(\Vert v_n\Vert _{p_1,p_2,\sigma _n}\) to be bounded.
Due to the estimates in \(T_n,\) the region \(T_n\) is called safe region.
We need the following local Pohožaev identity.
Lemma A11
Assume v satisfies the following equation
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}(A(g(v))\nabla v\nabla \varphi +\frac{1}{2}A'(g(v))g'(v)|\nabla v|^2\varphi )\text{ d }x+\int _{{\mathbb {R}}^3}V(x)g(v)g'(v)\varphi \text{ d }x\\&\qquad +\int _{{\mathbb {R}}^3}B(g(v))|\nabla v|^2\text{ d }x\cdot \int _{{\mathbb {R}}^3}(B(g(v))\nabla v\nabla \varphi +\frac{1}{2}B'(g(v))g'(v)|\nabla v|^2\varphi )\text{ d }x\\&\quad =\int _{{\mathbb {R}}^3}f_{\lambda }(g(v))g'(v)\varphi \text{ d }x, \ \ \ \text{ for } \ \ \ \varphi \in C_0^{\infty }({\mathbb {R}}^3). \end{aligned} \nonumber \\ \end{aligned}$$
(a17)
Then
$$\begin{aligned} & \int _{{\mathbb {R}}^3}(3F_{\lambda }(g(v))-\frac{1}{2}f_{\lambda }(g(v)g'(v)))\psi \text{ d }x -\frac{3}{2}\int _{{\mathbb {R}}^3}V(x)g^2(v)\psi \text{ d }x\nonumber \\ & \qquad +\frac{1}{2}\int _{{\mathbb {R}}^3}V(x)g(v)g'(v)v\psi \text{ d }x\nonumber \\ & \qquad -\frac{1}{2}\int _{{\mathbb {R}}^3}(x,\nabla V)g^{2}(v)\text{ d }x\ \ +\frac{1}{4}\int _{{\mathbb {R}}^3}A'(g(v))g'(v)|\nabla v|^2v\psi \text{ d }x\nonumber \\ & \qquad +\frac{1}{4}\int _{{\mathbb {R}}^3}B'(g(v))|\nabla v|^2v\psi \text{ d }x\nonumber \\ & \quad =-\frac{1}{2}\int _{{\mathbb {R}}^3}A(g(v))v(\nabla v,\nabla \psi )\text{ d }x -\int _{{\mathbb {R}}^3}A(g(v))(x,\nabla v)(\nabla v,\nabla \psi )\text{ d }x\nonumber \\ & \qquad +\frac{1}{2}\int _{{\mathbb {R}}^3}A(g(v))|\nabla v|^2(x,\nabla \psi )\text{ d }x+\frac{1}{2}\int _{{\mathbb {R}}^3}V(x)g^2(v)(x,\nabla \psi )\text{ d }x\nonumber \\ & \qquad +\int _{{\mathbb {R}}^3}B(g(v))|\nabla v|^2\text{ d }x \{-\frac{1}{2}\int _{{\mathbb {R}}^3}B(g(v))v(\nabla v,\nabla \psi )\text{ d }x\nonumber \\ & \qquad -\int _{{\mathbb {R}}^3}B(g(v))(x,\nabla v)(\nabla v,\nabla \psi )\text{ d }x \nonumber \\ & \qquad +\frac{1}{2}\int _{{\mathbb {R}}^3}B(g(v))|\nabla v|^2(x,\nabla \psi )\text{ d }x \}- \int _{{\mathbb {R}}^3}F_{\lambda }(g(v))(x,\nabla \psi )\text{ d }x,\ \text{ for } x\in C_0^{\infty }(B_{\delta }(0)). \nonumber \\ \end{aligned}$$
(a18)
Proof
Multiplying (a17) by \(\varphi =(x,\nabla v)\psi \) and integrating by parts we obtain
$$\begin{aligned} \begin{aligned}&-\frac{1}{2}\int _{{\mathbb {R}}^3}A(g(v))|\nabla v|^2\psi \text{ d }x +\int _{{\mathbb {R}}^3}A(g(v))(x,\nabla v)(\nabla v,\nabla \psi )\text{ d }x\\&\qquad -\frac{1}{2}\int _{{\mathbb {R}}^3}A(g(v))|\nabla v|^2(x,\nabla \psi )\text{ d }x -\frac{3}{2}\int _{{\mathbb {R}}^3}V(x)g^2(v)\psi \text{ d }x\\&\qquad -\frac{1}{2}\int _{{\mathbb {R}}^3}(x, \nabla V)g^2(v)\psi \text{ d }x-\frac{1}{2}\int _{{\mathbb {R}}^3}V(x)g^2(v)(x,\nabla \psi )\text{ d }x\\&\qquad +\int _{{\mathbb {R}}^3}B(g(v))|\nabla v|^2\text{ d }x\{-\frac{1}{2}\int _{{\mathbb {R}}^3}B(g(v))|\nabla v|^2\psi \text{ d }x\\&\qquad +\int _{{\mathbb {R}}^3}B(g(v))(x,\nabla v)(\nabla v,\nabla \psi )\text{ d }x -\frac{1}{2}\int _{{\mathbb {R}}^3}B(g(v))|\nabla v|^2(x,\nabla \psi )\text{ d }x\}\\&\quad =-3\int _{{\mathbb {R}}^3}F_{\lambda }(g(v))\psi \text{ d }x -\int _{{\mathbb {R}}^3}F_{\lambda }(g(v))(x,\nabla \psi )\text{ d }x. \end{aligned} \nonumber \\ \end{aligned}$$
(a19)
Multiplying (a17) by \(\varphi =v\psi ,\) we obtain
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^3}A(g(v))|\nabla v|^2\psi \text{ d }x +\int _{{\mathbb {R}}^3}A(g(v))\nabla v v\nabla \psi \text{ d }x+\frac{1}{2}\int _{{\mathbb {R}}^3}A'(g(v))g'(v) v\psi \text{ d }x\\&\qquad +\int _{{\mathbb {R}}^3}V(x)g(v)g'(v)\left| \nabla v\right| ^{2} v\psi \text{ d }x+\int _{{\mathbb {R}}^3}B(g(v))|\nabla v|^2\text{ d }x\left\{ \int _{{\mathbb {R}}^3}B(g(v))|\nabla v|^2\psi \text{ d }x \right. \\&\qquad \left. +\int _{{\mathbb {R}}^3}B(g(v))\nabla v v\nabla \psi \text{ d }x +\frac{1}{2}\int _{{\mathbb {R}}^3}B'(g(v))g'(v)|\nabla v|^2v\psi \text{ d }x\right\} \\&\quad =\int _{{\mathbb {R}}^3}f_{\lambda }(g(v))g'(v)v\psi \text{ d }x. \end{aligned}\nonumber \\ \end{aligned}$$
(a20)
The local Pohožaev identity (a18) follows from (a19), (a20) by eliminating the term
$$\begin{aligned} \int _{{\mathbb {R}}^3}A(g(v))|\nabla v|^2\psi \text{ d }x+ \int _{{\mathbb {R}}^3}B(g(v))|\nabla v|^2\text{ d }x\cdot \int _{{\mathbb {R}}^3}B(g(v))|\nabla v|^2\psi \text{ d }x. \end{aligned}$$
This completes the proof. \(\square \)
Proof of Proposition 4.1. Apply the local Pohožaev identity to the solution \(v_n\) and choose a function \(\psi \in C_0^{\infty }(B_{ \delta }(0))\) such that \(\psi (x)=1\) for \(|x|\le \sigma _n^{-\frac{1}{2}};\) \(\psi (x)=0\) for \(|x|\ge 2\sigma _n^{-\frac{1}{2}}\) and \(|\nabla \psi |\le 2\sigma _n^{\frac{1}{2}}.\) In the region \(T_n,\) \(|x|\le 2\sigma _n^{-\frac{1}{2}}.\) The right hand side of (a18), by Lemma \(A_9\)
$$\begin{aligned} RHS\le c\int _{T_n}(|\nabla v_n|^2+|\nabla v_n||v_n|\sigma _n^{-\frac{1}{2}}+|v_n|^6+|v_n|)\text{ d }x \le c\sigma _n^{-\frac{1}{2}}. \end{aligned}$$
(a21)
We estimate the left hand side of (a18). By Lemma 4.1,
$$\begin{aligned} \begin{aligned}&3F_{\lambda }(g(v))-\frac{1}{2}f_{\lambda }(g(v))g'(v)v\\&\quad =3H_{\lambda }(g(v))-\frac{1}{2}h_{\lambda }(g(v))g'(v)v+\frac{3}{q}\mu |g(v)|^q-\frac{1}{2}\mu |g(v)|^{q-2} g(v)g'(v)v\\&\quad \ge \frac{1}{4}h_{\lambda }(g(v))(g(v)-2g'(v)v)+ \left( \frac{3}{q}-\frac{1}{4}\right) \mu |g(v)|^q\\&\qquad +\frac{1}{4}\mu |g(v)|^{q-2} g(v)(g(v)-2g'(v)v)\\&\quad \ge c\mu |g(v)|^q-c|g(v)|^{11}\frac{\ln g(v)}{\sqrt{M^2+g^2(v)}}-c|g(v)|^{q-1}\frac{\ln g(v)}{\sqrt{M^2+g^2(v)}}\\&\quad \ge c\mu |g(v)|^{q}-c, \end{aligned}\nonumber \\ \end{aligned}$$
(a22)
In the above we have used the assumption \(q\in (10,12).\)
Also we have
$$\begin{aligned} \left| \frac{3}{2}V(x)g^2(v)-\frac{1}{2}V(x)g(v)g'(v)v+\frac{1}{2}(x,\nabla v)g^2(v) \right| \le \varepsilon |g(v)|^q+c. \end{aligned}$$
(a23)
Notice that \(A'(t)t\ge 0,\) \(B'(t)t\ge 0\) for \(t\in {\mathbb {R}}.\) The left side side of (a18)
$$\begin{aligned} LHS\ge c\int _{B_n}|g(v_n)|^{q}\text{ d }x-\int _{{\mathbb {R}}^3}\psi \text{ d }x\ge c\int _{B_n}|g(v_n)|^{q}\text{ d }x -c\sigma _n^{-\frac{3}{2}}, \end{aligned}$$
(a24)
where
$$\begin{aligned} B_n=\{x|x\in {\mathbb {R}}^3,|x|\le \sigma _n^{-\frac{1}{2}} \}. \end{aligned}$$
Choose L large enough such that \(\int _{B_1(0)}|U_1|^{\frac{q}{2}}\text{ d }x=m>0.\) For n large enough \(B_{L\sigma _n^{-1}}(0)\subseteq B_{\sigma _n^{-\frac{1}{2}}}(0)=B_n.\) We have
$$\begin{aligned} \begin{aligned} \int _{B_n}|g(v_n)|^{q}\text{ d }x&\ge \int _{B_{L\sigma _n}(0)}|g(v_n)|^{q}\text{ d }x =\int _{B_{L}(0)}\left| \frac{g(v_n(\sigma _n^{-1}x))}{\sigma _n^{\frac{1}{4}}} \right| ^{q}\text{ d }x \cdot \sigma _n^{\frac{q}{4}-3}\\&\sim \int _{B_{L}(0)}(2|U|)^{\frac{q}{2}}\text{ d }x\sigma _n^{\frac{q}{4}-3}, \end{aligned} \nonumber \\ \end{aligned}$$
(a25)
since
$$\begin{aligned} \left| \frac{g(v_n(\sigma _n^{-1}x))}{\sigma _n^{\frac{1}{4}}}\right| =\frac{|g(v_n(\sigma _n^{-1}x))|}{|v_n(\sigma _n^{-1}x)|^{\frac{1}{2}}} \cdot |\sigma _n^{-\frac{1}{2}}v_n(\sigma _n^{-1}x)|^{\frac{1}{2}}{\mathop {\rightarrow }\limits ^{a.e.}} \sqrt{2|U_1(x)|} \text{ as } |x|\rightarrow \infty . \end{aligned}$$
We arrive at a contradiction for \(\sigma _n\) sufficiently large that
$$\begin{aligned} \sigma _n^{\frac{q}{4}-3}\le c\sigma _n^{-\frac{1}{2}},\ \ \ \text{ and } \ \ q>10. \end{aligned}$$
