Proof. Multiplying $\nabla \overline {w}\cdot x$ in both sides of (1.9) and integrating on $\mathbb{R}^N$ , then we have

(2.5) \begin{equation} \begin{aligned} 0 &= \mathrm{Re}\int _{\mathbb{R}^{N}}\Delta ^2w\nabla \overline {w}\cdot xdx+\operatorname {Re}\int _{\mathbb{R}^{N}}w\nabla \overline {w}\cdot xdx-\mathrm{Re}\int _{\mathbb{R}^{N}}w\Delta |w|^2\nabla \overline {w}\cdot xdx\\[3pt] & \quad -\mathrm{Re}\int _{\mathbb{R}^{N}}|w|^{p-1}w\nabla \overline {w}\cdot xdx=I_1+I_2+I_3+I_4. \end{aligned} \end{equation}

For the first term $I_1$ , a direct calculation gives that

\begin{equation*}\begin{aligned}I_1 & =\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{jj}\partial _{ii}w\partial _{k}\overline {w}x_{k}dx=\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{jj}w\partial _{ii}(\partial _{k}\overline {w}x_{k})dx\\[2pt] & =\operatorname {Re}\int _{\mathbb{R}^{N}}\partial _{jj}w\left (2\partial _{ii}\overline {w}+x_{k}\partial _{k}\left (\partial _{ii}\overline {w}\right ))dx\right .\\[2pt] & =2\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{jj}w\partial _{ii}\overline {w}dx+\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{jj}wx_{k}\partial _{k}\left (\partial _{ii}\overline {w}\right )dx.\end{aligned}\end{equation*}

Since

\begin{equation*}\begin{aligned}&\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{jj}wx_{k}\partial _{k}(\partial _{ii}\overline {w}\:)dx \\[2pt]& =-\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{ii}\overline {w}\:\partial _{k}(\partial _{jj}w)x_{k}dx-\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{ii}\overline {w}\:\partial _{jj}w\partial _{k}x_{k}dx\\[2pt] & =-\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{ii}\overline {w}\:\partial _{k}(\partial _{jj}w)x_{k}dx-N\int _{\mathbb{R}^{N}}|\Delta w|^2dx\\[2pt] & =\frac {N}{2}\int _{\mathbb{R}^{N}}|\Delta w|^2dx-N\int _{\mathbb{R}^{N}}|\Delta w|^2dx,\end{aligned}\end{equation*}

we discover that

(2.6) \begin{equation} I_1=2\|\Delta w\|_2^2+\frac {N}{2}\|\Delta w\|_2^2-N\|\Delta w\|_2^2=\bigg (2-\frac {N}{2}\bigg )\|\Delta w\|_2^2. \end{equation}

For the second term $I_2$ , we have

\begin{equation*}\begin{aligned}I_2 & =\mathrm{Re}\int _{\mathbb{R}^{N}}wx_{k}\partial _{k}\overline {w}\:dx=-\mathrm{Re}\int _{\mathbb{R}^{N}}\overline {w}\partial _{k}\left (wx_{k}\right )dx\\[2pt] & =-\mathrm{Re}\int _{\mathbb{R}^{N}}\overline {w}\partial _{k}wx_{k}dx-\mathrm{Re}\int _{\mathbb{R}^{N}}\overline {w}w\partial _{k}x_{k}dx,\end{aligned}\end{equation*}

which implies that

(2.7) \begin{equation} I_2=-\frac {N{}^{}}{2}\|w\|_2^2. \end{equation}

For the third term $I_3$ , it is found that

\begin{align*} I_3 & =-\mathrm{Re}\int _{\mathbb{R}^{N}}w\partial _{jj}|w|^2\partial _{k}\overline {w}x_{k}dx=\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{j}(w\partial _{k}\overline {w}x_{k})\partial _{j}|w|^2dx\\[2pt] & =\operatorname {Re}\int _{\mathbb{R}^{N}}\partial _{j}w\partial _{k}\overline {w}x_{k}\partial _{j}|w|^2dx+\operatorname {Re}\int _{\mathbb{R}^{N}}w\partial _{jk}\overline {w}x_{k}\partial _{j}|w|^2dx\\[2pt] & \,\,\,\,+\operatorname {Re}\int _{\mathbb{R}^{N}}w\partial _{k}\overline {w}\delta _{jk}\partial _{j}|w|^2dx\\[2pt] & =\operatorname {Re}\int _{\mathbb{R}^{N}}\partial _{j}w\partial _{k}\overline {w}x_{k}\partial _{j}|w|^2dx+\operatorname {Re}\int _{\mathbb{R}^{N}}w\partial _{jk}\overline {w}x_{k}\partial _{j}|w|^2dx+\frac 12\|\nabla |w|^2\|_2^2, \end{align*}

where

\begin{align*} \begin{aligned} &\mathrm{Re}\int _{\mathbb{R}^{N}} w\partial _{jk}\overline {w}x_{k}\partial _{j}|w|^2dx=-\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{k}(wx_{k}\partial _{j}|w|^2)\partial _{j}\overline {w}dx & & \\[2pt] & =-\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{k}wx_{k}\partial _{j}|w|^2\partial _{j}\overline {w}dx-\mathrm{Re}\int _{\mathbb{R}^{N}}w\partial _{k}x_{k}\partial _{j}|w|^2\partial _{j}\overline {w}dx & & \\[2pt] & \,\,\,\,-\operatorname {Re}\int _{\mathbb{R}^{N}}wx_{k}\partial _{jk}|w|^2\partial _{j}\overline {w}dx & & \\[2pt] & =-\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{k}wx_{k}\partial _{j}|w|^2\partial _{j}\overline {w}dx-I_3^1-I_3^2. & & \end{aligned} \end{align*}

Note that $\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{k}wx_{k}\partial _{j}|w|^2\partial _{j}\overline {w}=\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{j}w\partial _{k}\overline {w}x_{k}\partial _{j}|w|^2$ , so it yields

\begin{equation*}I_3=-I_3^1-I_3^2+\frac 12\|\nabla |w|^2\|_2^2.\end{equation*}

By a series of calculations, we discover that

\begin{equation*}-I_3^1=-N\mathrm{Re}\int _{\mathbb{R}^{N}}w\partial _{j}|w_{}|^2\partial _{j}\overline {w}dx=-\frac {N}{2}\int _{\mathbb{R}^{N}}\partial _{j}|w|^2\partial _{j}|w|^2dx=-\frac {N}{2}\|\nabla |w|^2\|_2^2,\end{equation*}

and

\begin{equation*}\begin{aligned}-I_3^2 & =-\frac 12\mathrm{Re}\int _{\mathbb{R}^{N}}x_{k}\partial _{jk}|w|^2\partial _{j}|w|^2dx=\frac 12\mathrm{Re}\int _{\mathbb{R}^{N}}\partial _{k}\left (x_{k}\partial _{j}|w|^2\right )\partial _{j}|w|^2dx\\[2pt] & =\frac {N}{2}\|\nabla |w|^2\|_2^2+\frac 12\mathrm{Re}\int _{\mathbb{R}^{N}}x_{k}\partial _{jk}|w|^2\partial _{j}|w|^2dx.\end{aligned}\end{equation*}

Thus, we get that

(2.8) \begin{equation} I_3=\left ({-}\frac {N}{2}+\frac {N}{4}+\frac 12\right )\|\nabla |w|^2\|_2^2=\frac {2-N}{4}\|\nabla |w|^2\|_2^2. \end{equation}

For the fourth term $I_4$ , there appears the relation

(2.9) \begin{equation} I_4 =-\mathrm{Re}\int _{\mathbb{R}^{N}}|w|^{p-1}w\nabla \overline {w}\cdot xdx=-\frac {1}{p+1}\mathrm{Re}\int _{\mathbb{R}^{N}}\nabla (|w|^{p+1})\cdot xdx=\frac {N}{p+1}\|w\|_{p+1}^{p+1}. \end{equation}

Adding (2.6)–(2.9) into (2.5), we conclude that $Z(w)=0$ . Moreover, once $w\in H$ is a weak solution of equation (1.9), then a simple calculation yields that $\langle I^{\prime }(w),w\rangle =0.$

Proof. (1) As mentioned earlier, we know that $h_{w}$ attains its maximum. We further introduce a variable $s=t^{N+p+1}$ such that

\begin{equation*}h_{w}(s)= \frac {s^{\frac {N-2}{N+p+1}}}{2}\|\Delta w\|_2^2+\frac {s^{\frac {N+2}{N+p+1}}}{2}\|w\|_2^2+\frac {s^{\frac {N+2}{N+p+1}}}{4}\|\nabla |w|^2\|_2^2 -\frac {s}{p+1}\|w\|_{p+1}^{p+1}.\end{equation*}

Here, $h_w(s)$ is a concave function and exists a maximum point for $p\gt 1$ . Suppose that $\bar {t}$ is a unique point, at which this maximum is achieved. We declare that $\bar {t}$ is the unique critical point of $h_w$ . In fact, if $h_w$ has another maximum point $\bar {s}\neq \bar {t}$ and let $h_w(\bar {s})=h_w(\bar {t})$ . From the strict convexity of $h_w$ , we know that, for any $\lambda \in (0,1)$ , $h_w(\lambda \bar {t}+(1-\lambda )\bar {s})\gt \lambda h_w(\bar {t})+(1-\lambda )h_w(\bar {s})= h_w(\bar {t})$ , which is in contradiction with $\bar {t}$ being the maximum point of $h_w$ . Hence, $\bar {t}$ is the unique maximum point of $h_w$ , which means that $\bar {t}$ is the unique critical point of $h_w$ . And the function $h_w$ is positive and increasing in $0\lt t\lt \bar {t}$ and decreasing in $t\gt \bar {t}$ . Specifically, for any $w\neq 0$ , $\bar {t}\in \mathbb{R}$ is the unique value such that $w_{\bar {t}}$ belongs to $\mathcal{M}$ , and $I(w_{\bar {t}})$ get a global maximum point at $t=\bar {t}$ .

(2) Since $L(w)\lt 0$ , there appears the relations

\begin{equation*}L(w)=\frac {N-2}{2}\|\Delta w\|_2^2+\frac {N+2}{2}\|w\|_2^2+\frac {N+2}{4}\|\nabla |w|^2\|_2^2-\frac {N+p+1}{p+1}\|w\|_{p+1}^{p+1}\lt 0,\end{equation*}

and

(2.11) \begin{equation} \frac {N-2}{2}\bar {t}^{N-2}\|\Delta w\|_2^2+\frac {N+2}{2}\bar {t}^{N+2}\|w\|_2^2+\frac {N+2}{4}\bar {t}^{N+2}\|\nabla |w|^2\|_2^2 -\frac {N+p+1}{p+1}\bar {t}^{N+p+1}\|w\|_{p+1}^{p+1}=0. \end{equation}

Furthermore, we have

(2.12) \begin{equation} \begin{aligned}\bar {t}^{N+p+1}L(w) & =\frac {N-2}{2}\bar {t}^{N+p+1}\|\Delta w\|_2^2+\frac {N+2}{2}\bar {t}^{N+p+1}\|w\|_2^2\\[3pt] & \,\,\, +\frac {N+2}{4}\bar {t}^{N+p+1}\|\nabla |w|^2\|_2^2-\frac {N+p+1}{p+1}\bar {t}^{N+p+1}\|w\|_{p+1}^{p+1}\lt 0. \end{aligned} \end{equation}

In view of (2.11) and (2.12), we conclude that

\begin{equation*}\frac {N-2}{2}(\bar {t}^{N+p+1}-\bar {t}^{N-2})\|\Delta w\|_2^2+\frac {N+2}{2}(\bar {t}^{N+p+1}-\bar {t}_{}^{N+2})\|w\|_2^2 +\frac {N+2}{4}(\bar {t}^{N+p+1}-\bar {t}^{N+2})\|\nabla |w|^2\|_2^2\lt 0,\end{equation*}

which implies that $\bar {t}\lt 1$ .

Proof. For any $w\in {\mathcal{M}}$ , we have from the definition of $L(w)$ that

\begin{equation*}\frac {N+2}{2}\|w\|_2^2+\frac {N-2}{2}\|\Delta w\|_2^2\leq \frac {N+p+1}{p+1}\|w\|_{p+1}^{p+1}.\end{equation*}

Using interpolation inequality with $0\lt l\lt 1$ , we get that

\begin{equation*}\begin{aligned} & \frac {N+p+1}{p+1}\|w\|_{p+1}^{p+1} \leq \frac {C(N+p+1)}{p+1}\|w\|_2^{l(p+1)}\|\Delta w\|_2^{\left (1-l\right )(p+1)},\end{aligned}\end{equation*}

where $\frac {1}{p+1}=\frac l2+(\frac 12-\frac 2N)(1-l)$ . Using Young inequality $ab\leq \epsilon a^{p_1}+(\frac {q_1}{p_1}\epsilon )^{-\frac {q_1}{p_1}}b^{q_1}$ with $p_1=\frac {2}{l({p}+1)},$ $q_1=\frac {2}{2-l({p}+1)},$ then we have

\begin{equation*}\begin{aligned}\frac {N+p+1}{p+1}\|w\|_{p+1}^{p+1}&\leq \frac {C(N+p+1)}{p+1}\|w\|_2^{l(p+1)}\|\Delta w\|_2^{\left (1-l\right )(p+1)}\\[3pt] &\leq \frac {N+2}{2}\|w\|_2^2+C\|\Delta w\|_{2}^{^{\frac {2N}{N-4}}}.\end{aligned}\end{equation*}

Thus, it follows that

\begin{equation*}\frac {N+2}{2}\|w\|_2^2+\frac {N-2}{2}\|\Delta w\|_2^2\leq \frac {N+2}{2}\|w\|_2^2+C\|\Delta w\|_2^{\frac {2N}{N-4}}.\end{equation*}

Namely, we have

\begin{equation*} \frac 12\|\Delta w\|_2^2\leq C\|\Delta w\|_2^{\frac {2N}{N-4}},\end{equation*}

which implies that $\|\Delta w\|_2^2$ is bounded away from zero on $\mathcal{M}$ . From the definition of $L(w)$ , we know that

\begin{equation*}-\frac {N+p+1}{p+1}\|w\|_{p+1}^{p+1}=-\frac {N-2}{2}\|\Delta w\|_2^2-\frac {N+2}{2}\|w\|_2^2-\frac {N+2}{4}\|\nabla |w|^2\|_2^2.\end{equation*}

So, we have

\begin{equation*}-\frac {1}{p+1}\|w\|_{p+1}^{p+1}=-\frac {N-2}{2(N+p+1)}\|\Delta w\|_2^2-\frac {N+2}{2(N+p+1)}\|w\|_2^2-\frac {N+2}{4(N+p+1)}\|\nabla |w|^2\|_2^2.\end{equation*}

Finally, from the definition of $I(w)$ , we deduce that

\begin{equation*}I(w)=\frac {p+3}{2(N+p+1)}\|\Delta w\|_2^2+\frac {p-1}{2\left (N+p+1\right )}\|w\|_2^2+\frac {p-1}{4\left (N+p+1\right )}\|\nabla |w|^2\|_2^2\gt 0,\end{equation*}

which shows that the number $m\gt 0$ .

Proof. For any $w\in {\mathcal{M}}$ , we have from the definition of $L(w)$ that

\begin{equation*}\frac {N+2}{2}\|w\|_2^2+\frac {N+2}{4}\|\nabla |w|^2\|_2^2\leq \frac {N+p+1}{p+1}\|w\|_{p+1}^{p+1}.\end{equation*}

Using interpolation inequality with $0\lt l\lt 1$ , we get that

\begin{equation*}\begin{aligned} & \frac {N+p+1}{p+1}\|w\|_{p+1}^{p+1} \leq \frac {C(N+p+1)}{p+1}\|w\|_2^{l(p+1)}\|w\|_{\frac {4N}{N-2}}^{\left (1-l\right )(p+1)},\end{aligned}\end{equation*}

where $\frac {1}{p+1}=\frac l2+\frac {N-2}{4N}(1-l)$ . Using Young inequality with $ab\leq \epsilon a^{p_2}+(\frac {q_2}{p_2}\epsilon )^{-\frac {q_2}{p_2}}b^{q_2}$ , we have

\begin{equation*}\begin{aligned}\frac {N+p+1}{p+1}\|w\|_{p+1}^{p+1}&\leq \frac {C(N+p+1)}{p+1}\|w\|_2^{l(p+1)}\|w\|_{\frac {4N}{N-2}}^{\left (1-l\right )(p+1)}\\[3pt]& \leq \frac {N+2}{2}\|w\|_2^2+C\|w\|_{^{\frac {4N}{N-2}}}^{^{\frac {4N}{N-2}}},\end{aligned}\end{equation*}

where $p_2=\frac {2}{l({p}+1)}, q_2=\frac {2}{2-l({p}+1)}$ . Thus, it follows that

\begin{equation*}\frac {N+2}{2}\|w\|_2^2+\frac {N+2}{4}\|\nabla |w|^2\|_2^2\leq \frac {N+2}{2}\|w\|_2^2+C\|w\|_{\frac {4N}{N-2}}^{\frac {4N}{N-2}}.\end{equation*}

Using Sobolev inequality, we have

\begin{equation*}\begin{aligned} & \frac 14\|\nabla |w|^2\|_2^2\leq C\|w\|_{\frac {4N}{N-2}}^{\frac {4N}{N-2}}\leq C\|\nabla |w|^2\|_2^{\frac {2N}{N-2}},\end{aligned}\end{equation*}

which implies that $\|\nabla |w|^2\|_2^2$ is bounded away from zero on $\mathcal{M}$ . Furthermore, using similar arguments as Lemma2.3, we can conclude that the number $m\gt 0$ .

Proof. For any $w\in \mathcal{M}$ and $t\in (0,1)$ , a direct computation gives that

\begin{equation*}\begin{aligned} & I(w_{t})-t^{N+p+1}I(w)\\&\quad = \left (\frac {t^{N-2}}{2}-\frac {t^{N+p+1}}{2}\right )\|\Delta w\|_2^2+\left (\frac {t^{N+2}}{2}-\frac {t^{N+p+1}}{2}\right )\|w\|_2^2+\left (\frac {t^{N+2}}{4}-\frac {t^{N+p+1}}{4}\right )\|\nabla |w|^2\|_2^2. \end{aligned}\end{equation*}

Taking $t$ small enough such that there exists a positive fixed constant $\gamma =\min \Big \{\frac {t^{N-2}}{2}-\frac {t^{N+p+1}}{2},\frac {t^{N+2}}{4}-\frac {t^{N+p+1}}{4}\Big \}\gt 0.$ By Lemmas2.1 and 2.2, we know $I(w_{t})\leq I(w)$ . So it follows that

\begin{equation*}(1-t^{N+p+1})I(w) \geq I\left (w_{t}\right )-t^{N+p+1}I(w) \geq \gamma \left (\|\Delta w\|_2^2+\|w\|_2^2+\|\nabla |w|^2\|_2^2\right ).\end{equation*}

Thus, we complete the proof by choosing $c=\frac {\gamma }{1-t^{N+p+1}}.$

Proof. Let $\{u_n\}$ be bounded in $H^{2}(\mathbb{R}^{N})$ . By Hölder and Sobolev inequalities, we get

\begin{equation*}\begin{aligned}\int _{B_{y}(k)}|u|^{\beta }dx & \leq \left (\int _{B_{y}(k)}|u|^{q}dx\right )^{\frac {(1-\alpha )\beta }{q}}\left (\int _{B_{y}(k)}|u|^{2^{*}}dx\right )^{\frac {\alpha \beta }{^{2^{*}}}}\\ & \leq C\left (\int _{B_{y}(k)}|u|^{q}dx\right )^{\frac {(1-\alpha )\beta }{q}}\left (\int _{B_{y}(k)}(|\Delta u|^2+|\nabla u|^2+u^2)dx\right )^{\frac {\alpha \beta }{^2}},\end{aligned}\end{equation*}

where $0\lt \alpha =\frac {2^{*}(\beta -q)}{2^{*}\beta -q\beta }\lt 1$ and $2^*=\frac {2N}{N-4}$ . If $\beta \geq \frac 2\alpha$ , then $\alpha \beta \geq 2$ and

\begin{equation*}\begin{aligned}\int _{B_{y}(k)}|u|^{\beta }dx\leq C\left (\int _{B_{y}(k)}|u|^{q}dx\right )^{\frac {(1-\alpha )\beta }{q}}\|u\|_{H^2}^{\alpha \beta -2}\int _{B_{y}(k)}(|\Delta u|^2+|\nabla u|^2+u^2)dx.\end{aligned}\end{equation*}

Covering $\mathbb{R}^N$ by a family of balls $\{B_{y_i}(k)\}$ such that each point is contained in at most $l$ such balls. Additionally, summing up these inequalities over this family of balls, it follows that

\begin{equation*}\int _{\mathbb{R}^{N}}|u|^{\beta }dx\leq lC\sup _{y\in \mathbb{R}^{N}}\left (\int _{B_{y}(k)}|u|^{q}dx\right )^{\frac {(1-\alpha ){\beta }}{q}}\|u\|_{H^2}^{\alpha \beta }.\end{equation*}

Substituting $u=u_n$ in above inequalities, we obtain the result for case $\beta \geq \frac 2\alpha$ . If $2\lt \beta \lt \frac 2\alpha$ , we write $\beta =2\tau +(1-\tau )\frac 2\alpha$ for some $\tau \in (0,1)$ . Applying interpolation inequality, we deduce that

\begin{equation*}\|u\|_{\beta }^{\beta }\leq \|u\|_2^{2\tau }\|u\|_{\frac 2\alpha }^{(1-\tau )\frac 2\alpha },\end{equation*}

and the result in this case is established.

Proof. The proof’s process is similar to Lemma 2 of [Reference Poppenberg, Schmitt and Wang24], and the detail is omitted.

Proof. From the definition (2.10) of $m$ , we can choose $w_n\in \mathcal{M}$ such that $I(w_{n})\to m$ . Furthermore, applying Lemma2.5, it follows that $\{w_{n}\}$ are bounded in $H$ and $\{w_{n}^2\}$ are bounded in $H^1$ . So there exist the subsequences of $\{w_{n}\}$ (still denoted by $\{w_{n}\}$ ) satisfying $w_{n}\rightharpoonup w$ in $H$ , $w_{n}^2\rightharpoonup w^2$ in $H^1$ . This can show that $\{w_{n}\}$ are bounded in $L^{p+1}(\mathbb{R}^{N})$ , where $1\lt p\lt +\infty$ when $2\lt N\leq 4$ , $1\lt p\lt \frac {N+4}{N-4}$ when $4\lt N\leq 6$ and $1\lt p\lt \frac {3N+2}{N-2}$ when $N\gt 6$ . The detail of its proof can be divided into three steps.

Step 1. Showing that $\|w_{n}\|_{p+1}^{p+1}\nrightarrow 0$ .

It follows from Lemma2.5 and

\begin{equation*}I(w_n)=\frac {1}{2}\|\Delta w_n\|_2^2+\frac {1}{2}\|w_n\|_2^2+\frac {1}{4}\|\nabla |w_n|^2\|_2^2-\frac {1}{p+1}\|w_n\|_{p+1}^{p+1}\rightarrow m\gt 0\end{equation*}

that $\|w_{n}\|_{H}+\|w_{n}^2\|_{H^1}\not \to 0.$ By using Lemmas2.1 and 2.2, we have for any $t\gt 1$ that

\begin{equation*}\begin{aligned} m&\leftarrow I(w_{n}) \geq I(w_{nt})\\[2pt] & =\frac {t^{N-2}}{2}\|\Delta w_{n}\|_2^2+\frac {t^{N+2}}{2}\|w_{n}\|_2^2+\frac {t^{N+2}}{4}\|\nabla |w_{n}|^2\|_2^2-\frac {t^{N+p+1}}{p+1}\|w_{n}\|_{p+1}^{p+1}\\[2pt] & \geq \frac {t^{N-2}}{2}\big(\|\Delta w_{n}\|_2^2+\|w_{n}\|_2^2+\|\nabla |w_{n}|^2\|_2^2\big)-\frac {t^{N+p+1}}{p+1}\|w_{n}\|_{p+1}^{p+1}\\[2pt] & \geq \frac {t^{N-2}}{2}\delta -\frac {t^{N+p+1}}{p+1}\|w_{n}\|_{p+1}^{p+1},\end{aligned}\end{equation*}

where $\delta$ is a fixed positive constant. Taking $t\gt 1$ large enough such that $\frac {t^{N-2}\delta }{2}\gt 2m$ , which means that there exists a lower bound for $\|w_{n}\|_{p+1}^{p+1}$ such that

(2.14) \begin{equation} \|w_{n}\|_{p+1}^{p+1}\rightarrow D\in (0,\infty ). \end{equation}

$\mathbf{Step\,2}$ . Splitting by concentration-compactness.

Applying (2.14) and Lemma2.6, it follows that there exist $\alpha \gt 0$ and $\{x_{n}\}\subset \mathbb{R}^{N}$ such that

(2.15) \begin{equation} \int _{B_{x_{n}}(1)}\lvert w_{n}\rvert ^{p+1}dy\gt \alpha \gt 0. \end{equation}

Choosing $\eta _R(t)$ to be a smooth function defined on $[0,+\infty )$ and then satisfying

$(\mathrm{1})$ $\eta _{R}(t)=1\,\mathrm{for}\,0\leq t\leq R$ ,

$(\mathrm{2})$ $\eta _{R}(t)=0\,\mathrm{for}\,t\geq 2R$ ,

$(\mathrm{3})$ $\eta _{R}^{\prime }(t)\leq \frac {2}{R}$ .

Let $\theta _{n}(x)=\eta _{R}(|x-x_{n}|)w_{n}(x), \lambda _{n}(x)=(1-\eta _{R}(|x-x_{n}|))w_{n}(x).$ It is easy to verify that $w_{n}=\theta _{n}+\lambda _{n}$ . Note that, in particular,

(2.16) \begin{equation} \liminf _{n\to \infty }\int _{B_{x_n}(R)}\left |\theta _n\right |^{p+1}dy\geq \alpha . \end{equation}

$\mathbf{Step\,3}$ . The minimum of $I|_{\mathcal{M}}$ is achieved.

Note that $w_{n}\rightharpoonup w$ in $H$ , $w_{n}^2\rightharpoonup w^2$ in $H^1$ . We define the functions $w_{n}({\cdot} +x_{n}):=\widetilde {w}_{n}$ , where $x_n$ is given in (2.15). Thus, it is found that $\widetilde {w}_{n}\rightharpoonup \widetilde {w}$ in $H$ , $\widetilde {w}_{n}^2\rightharpoonup \widetilde {w}^2$ in $H^1$ . Under such circumstances, we obtain that $I(w)=I(\widetilde {w})$ . Furthermore, by (2.16), one gets that

\begin{equation*}\alpha \leq \liminf _{n\to \infty }\int _{\mathbb{R}^{N}}\left |\theta _{n}\right |^{p+1}dy\leq \lim _{n\to +\infty }\int _{B_0(2R)}\left |\widetilde {w}_{n}\right |^{p+1}dy=\int _{B_0(2R)}\left |\widetilde {w}\right |^{p+1}dy,\end{equation*}

which imply $\widetilde {w}\neq 0$ and then $w\neq 0$ .

Next, we shall claim that $w\in \mathcal{M}$ . In fact, if $w\notin \mathcal{M}$ , then we will discuss the following two cases.

$\mathbf{Case\,1}$ : If $L(w)\lt 0$ , employing Lemma2.2, we know that there exists a point $t\in (0,1)$ such that $w_t\in \mathcal{M},$ which, together with Fatou’s lemma, yields that

\begin{equation*}\begin{aligned} m & =\liminf _{n\to \infty }\Big [I(w_{n})-\frac {1}{N+p+1}L(w_{n})\Big ] & \\ & =\frac {1}{2(N+p+1)}\liminf _{n\to \infty }\Big \{\left (p+3\right )\|\Delta w_{n}\|_2^2+\left (p-1\right )\|w_{n}\|_2^2+\frac {p-1}{2}\|\nabla |w_{n}|^2\|_2^2\Big \} & \\ & \geq \frac {1}{2\left (N+p+1\right )}\Big \{(p+3)\|\Delta w\|_2^2+\left (p-1\right )\|w\|_2^2+\frac {p-1}{2}\|\nabla |w|^2\|_2^2\Big \} & \\ & \gt \frac {1}{2\left (N+p+1\right )}\Big \{t^{N-2}(p+3)\|\Delta w\|_2^2+t^{N+2}\left (p-1\right )\|w\|_2^2+t^{N+2}\frac {p-1}{2}\|\nabla |w|^2\|_2^2\Big \}\,\,\,\, & \\ \ & =I(w_{t})-\frac {1}{N+p+1}L(w_{t})\geq m. & \end{aligned} \end{equation*}

So, case (1) is impossible.

$\mathbf{Case\,2}$ : When $L(w)\gt 0$ , we set $\xi _{n}\,:\!=\,w_{n}-w$ . In view of Lemma2.7 and Brézis-Lieb lemma in [Reference Brezis and Lieb33], there appears a relation

(2.17) \begin{equation} L(w_n)\geq L(w)+L(\xi _{n})+o_{n}(1), \end{equation}

which, along with $L(w_n)=0$ , gives that $\limsup \limits _{n\to \infty } L(\xi _{n})\lt 0.$ Thus, Lemma2.2 shows that there exists a $t_n\in (0,1)$ such that $(\xi _{n})_{t_{n}}\in \mathcal{M}.$

In addition, it is easy to see that $\limsup \limits _{n\to \infty } t_n\lt 1$ ; in fact, arguments by contradiction, along a subsequence $t_{n}\to 1$ , make that $L(\xi _{n})=L(\xi _{n})_{t_{n}}+o_{n}(1)=o_{n}(1);$ this is also impossible. Since $w_n\in \mathcal{M},$ in view of (2.13) and (2.17), it is found that

\begin{align*} m+o_{n}(1)& =I(w_{n})-\frac {1}{N+p+1}L(w_{n})\\ & =\frac {1}{2(N+p+1)}\Big \{(p+3)\|\Delta w_{n}\|_2^2+\left (p-1\right )\|w_{n}\|_2^2+\frac {\left (p-1\right )}{2}\|\nabla |w_{n}|^2\|_2^2\Big \}\\ & \geq \frac {p+3}{2(N+p+1)}\big \{\|\Delta w\|_2^2+\|\Delta \xi _{n}\|_2^2\big \}+\frac {p-1}{2(N+p+1)}\big \{\|w\|_2^2+\|\xi _{n}\|_2^2\big \}\\ &\,\,\,\,+\frac {p-1}{4\left (N+p+1\right )}\big \{{\|\nabla |w|^2\|_2^2+\|\nabla |\xi _{n}|^2\|_2^2}\big \}\,\,\,\,\\ & =I(w)-\frac {1}{N+p+1}L(w)+I(\xi _{n})-\frac {1}{N+p+1}L(\xi _{n})\\ & \gt I(w)-\frac {1}{N+p+1}L(w)+I(\xi _{n})_{t_{n}}-\frac {1}{N+p+1}L(\xi _{n})_{t_{n}}\\ & =I(\xi _{n})_{t_{n}}+\frac {p+3}{2(N+p+1)}\|\Delta w\|_2^2+\frac {p-1}{2(N+p+1)}\|w\|_2^2\\ & \,\,\,\,+\frac {p-1}{4\left (N+p+1\right )}\|\nabla |w|^2\|_2^2\geq m. \end{align*}

This case is also impossible. Hence, we deduce that $w\in \mathcal{M}$ .

Next, by using Lebesgue dominated convergence theorem, Fatou’s lemma and $w_n\in \mathcal{M}$ , it follows that

\begin{equation*}\begin{aligned}m & \leq I(w)-\frac {1}{N+p+1}L(w)\\[2pt] & =\frac {1}{2(N+p+1)}\left\{(p+3)\|\Delta w\|_2^2+\left (p-1\right )\|w\|_2^2+\frac {p-1}{2}\|\nabla |w|^2\|_2^2\right\}\\[2pt] & \leq \frac {1}{2(N+p+1)}\liminf _{n\to \infty }\left\{(p+3){}\|\Delta w_{n}\|_2^2+\left (p-1\right )\|w_{n}\|_2^2+\frac {p-1}{2}\|\nabla |w_{n}|^2\|_2^2\right\}\\[2pt] & =\liminf _{n\to \infty }\bigg[I(w_{n})-\frac {1}{N+p+1}L(w_{n})\bigg]\\[2pt] & =\liminf _{n\to \infty }I(w_{n})=m.\end{aligned}\end{equation*}

Thus, we get that $\|w_n\|_{H}\to \|w\|_{H}$ and $I(w)=m$ , which means that there exists a minimum $m$ of $I|_{\mathcal{M}}$ . In fact, by arguments similar to those in Lemma2.5 of [Reference Liu, Wang and Wang34], $w$ is indeed a solution of (1.9). We thereby complete the proof of this lemma.

Proof. Suppose that $\tilde {w}\in \mathcal{M}$ is a minimizer of the functional $I|_{\mathcal{M}}$ . Then, applying Lemma2.2, we see that

\begin{equation*}I(\tilde {w})=\inf _{w\in H}\max _{t\gt 0}I(w_{t})=m.\end{equation*}

Proceeding by contradiction, we assume that $\tilde {w}$ is not a weak solution to equation (1.9). Thus, we can find that there exists $\phi \in C_0^{\infty }(\mathbb{R}^{N})$ such that

\begin{equation*}\begin{aligned} \langle I^{\prime }(\tilde {w}),\phi \rangle =\int _{\mathbb{R}^{N}}(\Delta \tilde {w}\Delta \phi +\nabla (\tilde {w}^2)\nabla (\tilde {w}\phi )+\tilde {w}\phi -|\tilde {w}|^{p-1}\tilde {w}\phi )dy\lt -1.\end{aligned}\end{equation*}

Owing to the continuous property of $I^{\prime }(\tilde {w}_t)$ , we can choose a sufficiently small $\varepsilon \gt 0$ such that

\begin{equation*}\left \langle I^{\prime }(\tilde {w}_{t}+\sigma \phi ),\phi \right \rangle \leq -\frac {1}{2},\quad |t-1|\leq \varepsilon ,\,\,|\sigma |\leq \varepsilon .\end{equation*}

Furthermore, we introduce a cut-off function $\zeta (t)$ in form

\begin{equation*}\left .\zeta (t)\,:\!=\,\left \{\begin{array}{ll}\zeta (t)=1,&\quad |t-1|\leq \frac \varepsilon 2, \\[2pt] 0\leq \zeta (t)\leq 1,&\quad \frac \varepsilon 2\lt |t-1|\lt \varepsilon , \\[2pt] \zeta (t)=0,&\quad |t-1|\geq \varepsilon , \end{array}\right .\right .\end{equation*}

and then define the function

\begin{equation*}\gamma (t)\,:\!=\,\left \{\begin{array}{ll}\tilde {w}_t,&\quad \left |t-1\right |\geq \varepsilon ,\\ \tilde {w}_t+\varepsilon \zeta (t)\phi , &\quad \left |t-1\right |\lt \varepsilon .\end{array}\right .\end{equation*}

Thus, note that $\gamma (t)$ is a continuous curve endowed with metric $(H,d_{H})$ . We further choose a smaller $\varepsilon$ such that $d_{H}(\gamma (t),0)\gt 0$ for $|t-1|\lt \varepsilon$ .

First, we shall claim that $\sup \limits _{t\geq 0}I(\gamma (t))\lt m$ . In fact, if $\left |t-1\right |\geq \varepsilon$ , we have

\begin{equation*}I(\gamma (t))=I(\tilde {w}_{t})\lt I(\tilde {w})=m.\end{equation*}

If $\left |t-1\right |\lt \varepsilon$ , we define a $\mathrm{C}^1$ map $\sigma \mapsto I(\tilde {w}_{t}+\sigma \zeta (t)\phi )\in \mathbb{R}$ , where $\sigma \in \left [0,\varepsilon \right ]$ . Using the mean value theorem, it is easy to see that for a suitable $\bar {\sigma }\in (0,\sigma ),$

\begin{equation*}\begin{aligned}I (\tilde {w}_{t}+\sigma \zeta (t)\phi ) &=I(\tilde {w}_{t})+\langle I^{\prime }(\tilde {w}_{t}+\bar {\sigma }\zeta (t)\phi ,\zeta (t)\phi \bar {\sigma }\rangle \\ & \leq I(\tilde {w}_{t})-\frac {1}{2}\lt m.\end{aligned}\end{equation*}

Next, since $I(\tilde {w}_{t})=h_{\tilde {w}}(t)$ and $h_{\tilde {w}}^{^{\prime }}(1)=0=L(\tilde {w})$ , we observe that

\begin{equation*}L(\gamma (1-\varepsilon ))\gt 0,\,\text{for}\,t=1-\varepsilon ,\end{equation*}

and

\begin{equation*}L(\gamma (1+\varepsilon ))\lt 0,\,\text{for}\,t=1+\varepsilon .\end{equation*}

Applying the continuity of $t\mapsto L(\gamma (t)),$ we obtain that there exists a $t_{0}\in (1-\varepsilon ,1+\varepsilon )$ such that $L(\gamma (t_0))=0$ , that is

\begin{equation*} (\gamma (t_0))=(\tilde {w}_{t_0}+\varepsilon \zeta (t_0)\phi )\in \mathcal{M}.\end{equation*}

Thus, by the definition of $m$ , we have that

\begin{equation*}I(\gamma (t_0))\lt m,\end{equation*}

which is a contradiction. We deduce that $\tilde {w}$ is a weak solution of system (1.9). By similar arguments to those in steps 2 and 3 of Lemma2.8, it follows that $\tilde {w}\neq 0$ . Since any solution of (1.9) belongs to $\mathcal{M}$ , its minimizer is a ground state solution. Thereby, we conclude that the Cauchy problem (1.1) has the standing wave solutions in form (1.8). This completed the proof of Theorem1.1.

Proof. First, we choose a sequence of open, unit cubes of $\mathbb{R}^{N}$ denoted by $(Q_{n})_{n\geq 0}$ , which satisfy $Q_{n}\cap Q_l=\emptyset$ ( $n\neq l$ ) and $\overline {\cup _{n\geq 0}(Q_{n})}=\mathbb{R}^{N}$ . By the additivity of integration over domains, we obtain that for $\alpha \in \mathbb{R}$ ,

\begin{equation*}\int \limits _{\mathbb{R}^{N}}\big |u\big |^{\alpha +2}dx=\sum \limits _{n=0}^{\infty }\int \limits _{Q_{n}}\big |u\big |^{\alpha +2}dx,\end{equation*}

and

\begin{equation*}\int \limits _{\mathbb{R}^{N}}|\Delta u|^2dx=\sum \limits _{n=0}^{\infty }\int \limits _{Q_{n}}|\Delta u|^2dx.\end{equation*}

We further elaborate the analysis through the following two steps.

$\mathbf{Step\,1}$ . There exists constant $C$ independent of $n$ such that for all $u\in H(Q_{n})$ ,

(3.3) \begin{equation} \begin{aligned} & \int \limits _{Q_{n}}|u|^{\frac {2N+4}N}dx\leq C\bigg (\int \limits _{Q_{n}}|u|^2dx\bigg )^{\frac 2N}\int \limits _{Q_{n}}|\Delta u|^2dx.\quad \end{aligned} \end{equation}

In fact, for $4\lt N\leq 6$ , using the Gargliardo–Nirenberg theorem, we get that

(3.4) \begin{equation} \begin{aligned}\bigg (\int \limits _{Q_{n}}\big |u\big |^{p+1}dx\bigg )^{\frac {1}{p+1}} & \leq C\bigg (\int \limits _{Q_{n}}\big |\Delta u\big |^2dx\bigg )^{\frac {\theta }{2}}\bigg (\int \limits _{Q_{n}}\big |u^2\big |dx\bigg )^{\frac {1-\theta }{2}}\\ & =C\bigg (\int \limits _{Q_{n}}\big |\Delta u\big |^2dx\bigg )^{{\frac {N(p -1)}{8(p+1)}}}\bigg (\int \limits _{Q_{n}}\big |u^2\big |dx\bigg )^{\frac 12-{\frac {N(p-1)}{8(p+1)}}},\end{aligned} \end{equation}

where $\frac {1}{p+1}=\theta \left (\frac {1}{2}-\frac {2}{N}\right )+\frac {(1-\theta )}2,\,1\lt p\lt \frac {N+4}{N-4}.$ Furthermore, taking $p=\frac {N+8}N$ in (3.4), we deduce that the inequality (3.3) holds for $4\lt N\leq 6$ . Especially, choosing $p=\frac {11}3$ for case $N=3$ and selecting $p=3$ for case $N=4$ , it is easy to see that (3.3) holds. Finally, in the aforementioned calculations, the fact that the constant $C$ is independent of $n$ follows from the translation invariance of $|Q_{n}|$ .

$\mathbf{Step\,2}$ . Summing up $n$ in the above inequality (3.3), it follows that

(3.5) \begin{equation} \begin{aligned}\int \limits _{\mathbb{R}^{N}}|u|^{\frac {2N+8}N}dx\leq C\bigg (\sup \limits _{n\in \mathbb{N}}\int \limits _{Q_{n}}|u|^2dx\bigg )^{\frac 4N}\bigg (\int \limits _{\mathbb{R}^{N}}|\Delta u|^2dx\bigg ).\end{aligned} \end{equation}

Furthermore, changing $u(x)$ to $u(rx)$ in (3.5) gives that

\begin{equation*}\begin{aligned} & \int \limits _{\mathbb{R}^{N}}|u\left (y\right )|^{\frac {2N+8}N}d(r^{-1}y)\\ & \leq C\bigg (\sup \limits _{n\in \mathbb{N}}\int \limits _{Q_{n}}|u\left (y\right )|^2d(r^{-1}y)\bigg )^{\frac 4N}\bigg (\int \limits _{\mathbb{R}^{N}}|\Delta u\left (y\right )|^2d(r^{-1}y)\bigg ).\end{aligned}\end{equation*}

By direct calculations, we have that

\begin{equation*}\begin{aligned} & r^{-N}\int \limits _{\mathbb{R}^{N}}|u(y)|^{\frac {2N+8}{N}}dy\\ & \leq C\bigg (r^{-N}\sup \limits _{n\in \mathbb{N}}\int \limits _{Q_{n}}|u\left (y\right )|^2dy\bigg )^{\frac 4N}\bigg (r^{4-N}\int \limits _{\mathbb{R}^{N}}|\Delta u\left (y\right )|^2dy\bigg )\\ & \leq Cr^{-N}\bigg (\sup \limits _{n\in \mathbb{N}}\int \limits _{Q_{n}}|u\left (y\right )|^2dy\bigg )^{\frac 4N}\bigg (\int \limits _{\mathbb{R}^{N}}|\Delta u\left (y\right )|^2dy\bigg ),\end{aligned}\end{equation*}

which, along with definition (3.1) of $\rho (u,r)$ , yields that (3.2) holds.

Proof. The demonstrations of this lemma employ reasoning analogous to 1.7.5 of [Reference Cazenave35]. The core modification lies in substituting the original condition $\sup \limits _{n\geq 0}\|\nabla u_{n}\|_2^2\lt \infty$ with the condition stated above $\sup \limits _{n\geq 0}\|\Delta u_{n}\|_2^2\lt \infty$ . Given the substantial overlap in technical manoeuvres with prior arguments, we elect to omit the details here.

Proof. Let the functions $\gamma ({\cdot} )$ and $y({\cdot} ,\cdot )$ , and the sequences $(u_{n_{l}})_{l\geq 0}$ and $\left (r_{l}\right )_{l\geq 0}$ , be defined as in Lemmas3.1 and 3.3. Choosing $\widetilde {r}$ sufficiently large so that $\gamma (\widetilde {r})\gt \frac a{2}$ and letting $y_{l}=y(u_{n_{l}},\widetilde {r})$ . By standard subsequence arguments, we may assume that there exists $u\in H^2$ such that

(3.15) \begin{equation} u_{n_l}({\cdot} -y_l)\rightharpoonup u\quad \mathrm{in\,}{H^2}(\mathbb{R}^N)\mathrm{\,as\,}l\to \infty . \end{equation}

Now, the proof proceeds in the following three steps:

$\mathbf{Step\,1}$ . Establishing that part (1) holds.

When $\sigma =a$ , we can assert that if $u$ is given by (3.15), then

(3.16) \begin{equation} \|u\|_{2}^2=a. \end{equation}

Further, by applying (3.16), we can show that (1) holds.

We first prove that the claim (3.16) holds. Since the embedding $H^2(B_{R})\hookrightarrow L^{2}(B_{R})$ is compact, it follows that

(3.17) \begin{equation} \int \limits _{\{|x|\lt R\}}|u(x)|^2dx=\lim \limits _{l\to \infty }\int \limits _{\{|x-y_{l}|\lt R\}}|u_{n_{l}}(x)|^2dx,\quad \text{for every}\,R\gt 0. \end{equation}

On the other hand, by the previous arguments, we have that $\rho (u_{n_{l}},\widetilde {r})\gt \frac a{2}$ for $l$ large enough. Choosing $\varepsilon \lt \frac a{2}$ and taking $\tau$ large so that $\rho (u_{n_{l}},\tau )\gt a-\varepsilon$ . Thus, we have that for $l$ large enough,

\begin{equation*}\int \limits _{\{|x-y_{l}|\lt \widetilde {r}\}}|u_{n_{l}}|^2dx+\int \limits _{\{|x-y\left (u_{n_{l}}\tau )|\lt \tau \}\right .}|u_{n_{l}}|^2dx\gt \frac {a}{2}+a-\varepsilon \gt a,\end{equation*}

where $\{|x-y_{l}|\lt \widetilde {r}\}\cap \{|x-y(u_{n_{l}},\tau )|\lt \tau \}\neq \varnothing$ . Particularly, when $R=\widetilde {r}+2\tau$ , it follows that $\{|x-y\left (u_{n_{l}},\tau \right )|\lt \tau \}\subset \{|x-y_{l}|\lt R\}$ . So for $l$ large enough, there appears the relation

\begin{equation*}\int \limits _{\{|x-y_{l}|\lt R\}}|u_{n_{l}}|^2dx\geq \int \limits _{\{|x-y\left (u_{n_{l}},\tau )|\lt \tau \}\right .}|u_{n_{l}}|^2dx\gt a-\varepsilon ,\end{equation*}

which together with (3.17) yields that

\begin{equation*}\|u\|_2^2\geq \int \limits _{\{|x|\lt R\}}|u\left (x\right )|^2dx\geq a-\varepsilon .\end{equation*}

Letting $\varepsilon \to 0$ , we get that (3.16) holds.

Furthermore, $1\lt p\lt +\infty$ , when $2\lt N\leq 4$ and $1\lt p\lt \frac {N+4}{N-4}$ , when $4\lt N\leq 6$ , similar arguments as in (3.4) give that

(3.18) \begin{equation} \|u_{{n_{l}}}\|_{p+1}\leq C\|\Delta u_{{n_{l}}}\|_2^{\theta }\|u_{{n_{l}}}\|_2^{1-\theta },\,\, \end{equation}

where $\frac {1}{p+1}=\theta \left (\frac {1}{2}-\frac {2}{N}\right )+\frac {(1-\theta )}2$ and $0\lt \theta \lt 1$ . In view of (3.6), (3.16), and (3.18), we deduce that

(3.19) \begin{equation} \begin{aligned}\|u_{{n_{l}}}\|_{p+1}^{p+1}\leq C\|\Delta u_{{n_{l}}}\|_2^{\theta (p+1)}\|u_{{n_{l}}}\|_2^{(1-\theta)(p+1)}\lt +\infty , \end{aligned} \end{equation}

which shows that (1) holds.

$\mathbf{Step\,2}$ . Demonstrating that part (2) holds.

If $\sigma =0$ , it follows from Lemma3.3 that $\rho (u_{n_{l}},r_{l})\to 0$ as $l\to \infty$ , which implies that

(3.20) \begin{equation} \|u\|_2^2=\lim \limits _{l\to \infty }\|u_{{n_{l}}}\|_2^2\leq \lim \limits _{l\to \infty }\sup \limits _{y\in \mathbb{R}^{N}}\int \limits _{\{|x-y^{}|\lt r\}}|u_{{n_{l}}}|^2dx=0. \end{equation}

The same arguments as (3.19), we have from (3.6) and (3.20) that

(3.21) \begin{equation} \begin{aligned}\|u_{{n_{l}}}\|_{p+1}^{p+1} & \leq C\|\Delta u_{{n_{l}}}\|_2^{\theta (p+1)}\|u_{{n_{l}}}\|_2^{(1-\theta)(p+1)}\\[4pt] & \leq C\|u_{{n_{l}}}\|_2^{(1-\theta)(p+1)}\rightarrow 0,\,\,\text{as}\,l\rightarrow \infty .\end{aligned} \end{equation}

So part (2) holds.

$\mathbf{Step\,3}$ . Showing that part (3) holds.

Let us select $\theta ,\varphi \in C^\infty ([0,\infty ))$ with $0\leq \theta ,\varphi \leq 1$ and

\begin{equation*}\begin{aligned}\theta (r)\equiv 1\quad \mathrm{for\,}0\leq r\leq \frac {1}{2},\quad \theta (r)\equiv 0\quad \mathrm{for\,}r\geq \frac {3}{4},\\[3pt] \varphi (r)\equiv 0\quad \mathrm{for\,}0\leq r\leq \frac {3}{4},\quad \varphi (r)\equiv 1\quad \mathrm{for\,}r\geq 1,\end{aligned}\end{equation*}

and further define

\begin{equation*}v_{l}=\theta _{l}u_{n_{l}},\quad w_{l}=\varphi _{l}u_{n_{l}},\end{equation*}

where

\begin{equation*}\theta _{l}(x)=\theta \bigg (\frac {|x-y(u_{n_{l}},\frac {r_{l}}{2})|}{r_{l}}\bigg ),\quad \varphi _{l}(x)=\varphi \bigg (\frac {|x-y(u_{n_{l}},\frac {r_{l}}{2})|}{r_{l}}\bigg ).\end{equation*}

Note that the properties (3.8)–(3.10) are apparent and can be directly deduced. In addition, a series of calculations gives that

\begin{equation*}\begin{aligned} \rho \Big (u_{n_{l}},\frac {r_{l}}{2}\Big )&=\int \limits _{\left \{|x-y(u_{n_{l}},\frac {r_{l}}{2})|\leq \frac {r_{l}}{2}\right \}\big .}|u_{n_{l}}|^2dx\\[3pt] & \leq \int \limits _{\mathbb{R}^{N}}|\theta _{l}u_{n_{l}}|^2dx=\int \limits _{\mathbb{R}^{N}}|v_{{l}}|^2dx\\[3pt] & \leq \int \limits _{\left \{|x-y(u_{n_{l}},\frac {r_{l}}{2})|\leq r_{l}\right \}\big .}|u_{n_{l}}|^2dx\\[3pt] & \leq \int \limits _{\left \{|x-y(u_{n_{l}},r_{l})|\leq r_{l}\right \}\big .}|u_{n_{l}}|^2dx\leq \rho (u_{n_{l}},r_{l}),\end{aligned}\end{equation*}

which together with Lemma3.1 yields that

(3.22) \begin{equation} \|v_{l}\|_{2}^2\longrightarrow \sigma ,\,\,\text{as}\,\,l\rightarrow +\infty . \end{equation}

We denote $z_{l}=u_{n_{l}}-v_{l}-w_{l}$ and then obtain $|z_{l}|\leq |u_{n_{l}}|$ . In addition, it is found that

\begin{equation*}\begin{aligned} \int \limits _{\mathbb{R}^{N}}|z_{l}|^2dx & \leq \int \limits _{{\left \{\frac {r_{l}}{2}\leq |x-y(u_{n_{l}},\frac {r_{l}}{2})|\leq r_{l}\right \}}}|u_{n_{l}}|^2dx\\[3pt] & =\int \limits _{\left \{|x-y(u_{n_{l}},\frac {r_{l}}{2})|\leq r_{l}\right \}}|u_{n_{l}}|^2dx-\int \limits _{{\left \{|x-y(u_{n_{l}},\frac {r_{l}}{2})|\leq \frac {r_{l}}{2}\right \}}}|u_{n_{l}}|^2dx\\[3pt] & \leq \int \limits _{\left \{|x-y(u_{n_{l}},r_{l})|\leq r_{l}\right \}}|u_{n_{l}}|^2dx-\int \limits _{\left \{|x-y(u_{n_{l}},\frac {r_{l}}{2})|\leq \frac {r_{l}}{2}\right \}}|u_{n_{l}}|^2dx\\[3pt] & =\rho (u_{n_{l}},r_{l})-\rho \Big (u_{n_{l}},\frac {r_{l}}{2}\Big ),\end{aligned}\end{equation*}

which along with Lemma3.1 yields that

(3.23) \begin{equation} \|z_{l}\|_{2}^2\longrightarrow 0,\,\,as\,\,{l\to \infty }. \end{equation}

In view of (3.8), (3.22) and (3.23), we deduce that

\begin{equation*}\|w_{l}\|_{2}^2\longrightarrow a-\sigma ,\,\,as\,\,{l\to \infty }.\end{equation*}

Thus, we complete the proof of (3.11).

Furthermore, it is easy to see that

(3.24) \begin{equation} |\nabla |u_{n_{l}}|^2|^2-|\nabla |v_{l}|^2|^2-|\nabla |w_{l}|^2|^2=|\nabla |u_{n_{l}}|^2|^2-|\nabla |\theta _{l}u_{n_{l}}|^2|^2-|\nabla |\varphi _{l}u_{n_{l}}|^2|^2. \end{equation}

Through a series of calculations, we have

(3.25) \begin{equation} \begin{aligned}|\nabla |u_{n_{l}}|^2|^2 & =|\nabla (u_{n_{l}}\overline {u_{n_{l}}})|^2=|u_{n_{l}}\nabla \overline {u_{n_{l}}}+\overline {u_{n_{l}}}\nabla u_{n_{l}}|^2\\[2pt] & =2|u_{n_{l}}|^2|\nabla u_{n_{l}}|^2+(u_{n_{l}})^2(\nabla \overline {u_{n_{l}}})^2+\overline {(u_{n_{l}})^2(\nabla \overline {u_{n_{l}}})^2}\\[2pt] & =2|u_{n_{l}}|^2|\nabla u_{n_{l}}|^2+2\operatorname {Re}\left ((u_{n_{l}})^2(\nabla \overline {u_{n_{l}}})^2\right )\!,\end{aligned} \end{equation}

and

(3.26) \begin{equation} \begin{aligned}|\nabla |\theta _{l}u_{n_{l}}|^2|^2 & =|\nabla ((\theta _{l}u_{n_{l}})(\theta _{l}\overline {u_{n_{l}}}))|^2\\[2pt] & =|2\theta _{l}|u_{n_{l}}|^2\nabla \theta _{l}+\theta _{l}^2\overline {u_{n_{l}}}\nabla u_{n_{l}}+\theta _{l}^2u_{n_{l}}\nabla \overline {u_{n_{l}}}|^2\\[2pt] & =4\theta _{l}^2|u_{n_{l}}|^4|\nabla \theta _{l}|^2+2\theta _{l}^4|u_{n_{l}}|^2|\nabla u_{n_{l}}|^2\\[2pt] & \,\,\,+4\theta _{l}^3|u_{n_{l}}|^2\nabla \theta _{l}(u_{n_{l}}\nabla \overline {u_{n_{l}}}+\overline {u_{n_{l}}}\nabla u_{n_{l}})\\[2pt] & \,\,\,+\theta _{l}^4\left (\overline {u_{n_{l}}}^2(\nabla u_{n_{l}})^2+u_{n_{l}}^2(\nabla \overline {u_{n_{l}}})^2\right )\\[2pt] & =4\theta _{l}^2|u_{n_{l}}|^4|\nabla \theta _{l}|^2+2\theta _{l}^4|u_{n_{l}}|^2|\nabla u_{n_{l}}|^2\\[2pt] & \,\,\,+8\theta _{l}^3|u_{n_{l}}|^2\nabla \theta _{l}\operatorname {Re}(u_{n_{l}}\nabla \overline {u_{n_{l}}})+2\theta _{l}^4\operatorname {Re}\left (u_{n_{l}}^2(\nabla \overline {u_{n_{l}}})^2\right )\!.\end{aligned} \end{equation}

Using the similar calculations, we have

(3.27) \begin{equation} \begin{aligned}|\nabla |\varphi _{l}u_{n_{l}}|^2|^2 & =4\varphi _{l}^2|u_{n_{l}}|^4|\nabla \varphi _{l}|^2+2\varphi _{l}^4|u_{n_{l}}|^2|\nabla u_{n_{l}}|^2 & \\[2pt] & \,\,\,\,+8\varphi _{l}^3|u_{n_{l}}|^2\nabla \varphi _{l}\operatorname {Re}(u_{n_{l}}\nabla \overline {u_{n_{l}}})+2\varphi _{l}^4\operatorname {Re}\left (u_{n_{l}}^2(\nabla \overline {u_{n_{l}}})^2\right )\!. \end{aligned} \end{equation}

Inserting (3.25)–(3.27) into (3.24), it follows that

(3.28) \begin{equation} \begin{aligned} & |\nabla |u_{n_{l}}|^2|^2-|\nabla |v_{l}|^2|^2-|\nabla |w_{l}|^2|^2\\[2pt] & =2\left (1-\theta _{l}^4-\varphi _{l}^4\right )|u_{n_{l}}|^2|\nabla u_{n_{l}}|^2\\[2pt] & \,\,\,-4|u_{n_{l}}|^4\left (\theta _{l}^2|\nabla \theta _{l}|^2+\varphi _{l}^2|\nabla \varphi _{l}|^2\right )\\[2pt] & \,\,\,-8\left (\theta _{l}^3\nabla \theta _{l}+\varphi _{l}^3\nabla \varphi _{l}\right )|u_{n_{l}}|^2\operatorname {Re}(u_{n_{l}}\nabla \overline {u_{n_{l}}})\\[2pt] & \,\,\,+2\left (1-\theta _{l}^4-\varphi _{l}^4\right )\operatorname {Re}\left (u_{n_{l}}^2(\nabla \overline {u_{n_{l}}})^2\right )\\[2pt] & \geq -\frac {C}{r_{l}^4}|u_{n_{l}}|^4-\frac {C}{r_{l}^4}|u_{n_{l}}|^3|\nabla u_{n_{l}}|.\end{aligned} \end{equation}

Taking the limit on both sides of (3.24) as $l\rightarrow +\infty$ , we can get that (3.13) holds. Adapting the same way, we can conclude (3.12).

Next, we denote $s=\frac {|x-y(u_{n_{l}},\frac {r_{l}}{2})|}{r_{l}}$ and discuss through the following three cases.

$\mathbf{Case\,1}$ . If $s\in [0,\frac {1}{2}]$ , we have $\theta _{l}(s)=1$ and $\varphi _{l}(s)=0$ , which implies that $v_{l}=u_{n_{l}}$ and $w_{l}=0$ . In this situation, (3.14) obviously holds.

$\mathbf{Case\,2}$ . When $s\in [\frac {3}{4},+\infty )$ , we have $\theta _{l}(x)=0$ and $\varphi _{l}(x)=1$ , which implies that $w_{l}=u_{n_{l}}$ and $v_{l}=0$ . (3.14) clearly holds.

$\mathbf{Case\,3}$ . For $s\in (\frac {1}{2},\frac {3}{4})$ , we first verify that

\begin{equation*}\left ||u_{n_{l}}|^{p+1}-|v_{l}|^{p+1}-|w_{l}|^{p+1}\right |\leq C|u_{n_{l}}|^{p}|z_{l}|.\end{equation*}

In fact, if $s\in (\frac {1}{2},\frac {3}{4})$ , then $\varphi _{l}(s)=0$ and $0\lt \theta _{l}(s)\lt 1$ . Thus, we have $w_{l}=0$ and

\begin{equation*}|z_{l}|=|u_{n_{l}}-v_{l}|=|u_{n_{l}}-\theta _{l}u_{n_{l}}|=|u_{n_{l}}(1-\theta _{l})|.\end{equation*}

Furthermore, by a series of calculations, we find that

(3.29) \begin{equation} \begin{aligned} & \left ||u_{n_{l}}|^{p+1}-|\theta _{l}u_{n_{l}}|^{p+1}\right |\leq |u_{n_{l}}|^{p+1}\left (1-\theta _{l}^{p+1}\right )\\ & =|u_{n_{l}}|^{p+1}\left (1-\theta _{l}\right )\left (1+\theta _{l}+\theta _{l}^2+\cdots +\theta _{l}^{p}\right )\\ & \leq C|u_{n_{l}}|^{p+1}\left (1-\theta _{l}\right )\leq C|u_{n_{l}}|^{p}(u_{n_{l}}-u_{n_{l}}\theta _{l})\\ & =C|u_{n_{l}}|^{p}|z_{l}|.\end{aligned} \end{equation}

Using Hölder inequality, we can get that

\begin{equation*}\int _{\mathbb{R}^{N}}|u_{n_{l}}|^{p}|z_{l}|dx\leq \|u_{n_{l}}\|_{p+1}^{p}\|z_{l}\|_{p+1}^{p+1}.\end{equation*}

Let $1\lt p\lt +\infty$ , when $2\lt N\leq 4$ and $1\lt p\lt \frac {N+4}{N-4}$ , when $4\lt N\leq 6$ . In the same way as (3.19) and (3.21), we have

\begin{equation*}\|z_{l}\|_{p+1}^{p+1}\leq C\|\Delta z_{l}\|_2^{\theta (p+1)}\|z_{l}\|_2^{(1-\theta)(p+1)},\end{equation*}

where $\frac {1}{p+1}=\theta \left (\frac {1}{2}-\frac {2}{N}\right )+\frac {(1-\theta )}2$ and $0\lt \theta \lt 1$ . Note that $z_l$ is bounded in $H^2$ and converges to $0$ in $L^2$ . Hence, (3.14) holds. We thereby complete the proof of this lemma.

Proof. The proof of this lemma follows the same lines as Lemma3.2. When $N\gt 6$ , by applying Gargliardo–Nirenberg theorem, we have

(3.31) \begin{equation} \begin{aligned}\bigg (\int \limits _{Q_{n}}\big |u\big |^{p+1}dx\bigg )^{\frac {2}{p+1}} & \leq C\bigg (\int \limits _{Q_{n}}\big |\nabla |u|^2\big |^2dx\bigg )^{\frac {\theta }{2}}\bigg (\int \limits _{Q_{n}}\big |u^2\big |dx\bigg )^{1-\theta }\\ & =C\bigg (\int \limits _{Q_{n}}\big |\nabla |u|^2\big |^2dx\bigg )^{{\frac {N(p-1)}{\left (N+2\right )(p+1)}}}\bigg (\int \limits _{Q_{n}}\big |u^2\big |dx\bigg )^{1-{\frac {2N(p-1)}{\left (N+2\right )(p+1)}}},\end{aligned} \end{equation}

where $\frac {2}{p+1}=\theta \left (\frac {1}{2}-\frac {1}{N}\right )+(1-\theta),\,1\lt p\lt \frac {3N+2}{N-2}.$ Especially, taking $p=\frac {3N+4}N$ in (3.31), then we deduce that there exists constant $C$ independent of $n$ such that for all $u\in \bar {H}(Q_{n})$ ,

(3.32) \begin{equation} \begin{aligned} & \int \limits _{Q_{n}}|u|^{\frac {4N+4}N}dx\leq C\bigg (\int \limits _{Q_{n}}|u|^2dx\bigg )^{\frac 2N}\int \limits _{Q_{n}}|\nabla |u|^2|^2dx.\quad \end{aligned} \end{equation}

Summing on $n$ in (3.32), we get that

(3.33) \begin{equation} \begin{aligned}\int \limits _{\mathbb{R}^{N}}|u|^{\frac {4N+4}N}dx\leq C\bigg (\sup \limits _{n\in \mathbb{N}}\int \limits _{Q_{n}}|u|^2dx\bigg )^{\frac 2N}\bigg (\int \limits _{\mathbb{R}^{N}}|\nabla |u|^2|^2dx\bigg ).\end{aligned} \end{equation}

Subsequently, changing $u(x)$ to $u(rx)$ in (3.33) and a direct calculations gives that

\begin{equation*}\begin{aligned} & r^{-N}\int \limits _{\mathbb{R}^{N}}|u(y)|^{\frac {4N+4}{N}}dy\\ & \leq C\bigg (r^{-N}\sup \limits _{n\in \mathbb{N}}\int \limits _{Q_{n}}|u\left (y\right )|^2dy\bigg )^{\frac 2N}\bigg (r^{2-N}\int \limits _{\mathbb{R}^{N}}|u\left (y\right )\nabla u\left (y\right )|^2dy\bigg )\\ & \leq Cr^{-N}\bigg (\sup \limits _{n\in \mathbb{N}}\int \limits _{Q_{n}}|u\left (y\right )|^2dy\bigg )^{\frac 2N}\bigg (\int \limits _{\mathbb{R}^{N}}|\nabla |u\left (y\right )|^2|^2dy\bigg ).\end{aligned}\end{equation*}

Finally, by applying the definition (3.1) of $\rho (u,r)$ , we conclude that (3.30) holds.

Proof. $\mathbf{Step\,1}$ . Proving that $0\lt \nu \lt \infty$ . For $u\in \Gamma$ and $\lambda \gt 0$ , we define the functional in form

\begin{equation*}u_{\lambda }(x)=\lambda ^{d}u(\lambda x).\end{equation*}

By a direct calculations, one can verify that $u_{\lambda }\in \Gamma$ and

\begin{equation*}E(u_{\lambda })=\frac {\lambda ^{2d+4-N}}{2}\|\Delta u\|_2^2+\frac {\lambda ^{2(2d+1)-N}}{4}\|\nabla |u|^2\|_2^2-\frac {\lambda ^{d(p+1)-N}}{p+1}\|u\|_{p+1}^{p+1}.\end{equation*}

Choosing $0\lt d\lt \min \{\frac 4{p-1},\frac 2{p-3}\}$ such that $d(p+1)-N\lt \min \{2d+4-N,4d+2-N\}$ . So it is easy to see that $E(u_{\lambda })\lt 0$ for $\lambda$ small enough. By the definition (1.13), it follows that $\nu \gt 0$ .

From Lemma 3 in [Reference Li and Di31], we know that

\begin{equation*}\begin{aligned}\|u\|_{p+1}^{p+1} & =C\left (p,N\right )\|\nabla |u|^2\|_2^{\frac {pN}{N+2}}\|u\|_2^{^{p+1-\frac {2pN}{N+2}}}\\ & \leq C\left (p,N\right )\left (\|\nabla |u|^2\|_2^2\right )^{\frac {pN}{2(N+2)}}\left (\|u\|_2^2\right )^{\frac {p+1}{2}-\frac {pN}{N+2}}\!.\end{aligned}\end{equation*}

Applying condition $1\lt p\lt \frac {N+2}{N-2}$ , it it found that $\frac {pN}{2(N+2)}\lt 1$ . Taking $\|u\|_2^2=\|u_0\|_2^2=\tau$ and using Young inequality with $\epsilon$ , we get that

(3.43) \begin{equation} E(u)\geq \frac {1}{2}\|\Delta u\|_2^2+\left (\frac 14-\epsilon \right )\|\nabla |u|^2\|_2^2-\frac {1}{p+1}C(\varepsilon ,p,N,\tau )\geq -c_2. \end{equation}

We thereby conclude that $-\nu \geq -c_2\gt -\infty .$

$\mathbf{Step\,2}$ . Showing that the minimizing sequence $u_n$ of (1.14) is bounded in $H$ and $\|u_{n}\|_{p+1}^{p+1}$ is bounded from below.

In fact, by (3.43), one has that $\|\nabla |u_{n}|^2\|_2^2+\|\Delta u_n\|_2^2=\|u_n\|_{\bar H}^2\lt +\infty$ for space dimension $2\lt N\leq 6$ , and $\|u_n\|_{H^2}^2\lt +\infty$ for space dimension $N\gt 6$ is bounded. Furthermore, $u_n\in \Gamma$ implies that $u_n$ of (1.12) is bounded in $L^2$ . Since $\nu \gt 0$ , it is easy to see that $E(u_n)\leq -\frac {\nu }{2}$ for $n$ large enough. And a series of calculation yields that

(3.44) \begin{equation} \|u_{n}\|_{p+1}^{p+1}=-(p+1)E(u)+\frac {p+1}{2}\|\Delta u\|_2^2+\frac {p+1}{4}\|\nabla |u|^2\|_2^2\geq \frac {p+1}{2}\nu . \end{equation}

$\mathbf{Step\,3}$ . Establishing that part (1) holds.

In fact, suppose that $\tilde {u}_{n}$ satisfy $\|\tilde {u}_{n}\|_{2}^{2}\to \sigma$ and $E(\tilde {u}_{n})\to -\nu ,$ where $\sigma$ was introduced in Lemmas3.3 and 3.6. Furthermore, we introduce

\begin{equation*}u_{n}=\sqrt {\frac {\tau }{\|\tilde {u}_{n}\|_2^2}}\tilde {u}_{n},\end{equation*}

and then show that $u_n$ is a minimizing sequence of (1.14). By rescaling, without loss of generality, we may assume that $\tau =1.$ In addition, we shall apply Lemmas3.3 and 3.6 to $u_n$ (minimizing sequence) with $a=1$ and prove that

(3.45) \begin{equation} \sigma =1. \end{equation}

From the proof of above Step 2, we know that $\|u_{n}\|_{p+1}^{p+1}$ is bounded from below, which implies that $\sigma \gt 0$ by Lemmas3.3 and 3.6. Arguing by contradiction, we suppose that

(3.46) \begin{equation} 0\lt \sigma \lt 1. \end{equation}

In view of (3.12)–(3.14) in Lemma3.4 and (3.40)–(3.42) in Lemma3.7 by the definition of $E(u)$ , one has

\begin{equation*}\liminf _{l\to \infty }(E(u_{n_{l}})-E(v_{l})-E(w_{l}))\geq 0,\end{equation*}

which implies that

\begin{equation*}\limsup _{l\to \infty }(E(v_{l})+E(w_{l}))\leq \limsup _{l\to \infty }E(u_{n_{l}})\leq -\nu . \end{equation*}

So we have

(3.47) \begin{equation} \begin{aligned} \limsup _{l\to \infty }(E(v_{l})+E(w_{l}))\leq -\nu . \end{aligned} \end{equation}

On the other hand, for any $\lambda \gt 0$ and $u\in H$ , we discover that

\begin{equation*}E(\lambda u)=\frac {1}{2}\|\Delta \lambda u\|_2^2+\frac 14\|\nabla |\lambda u|^2\|_2^2-\frac {1}{p+1}\|\lambda u\|_{p+1}^{p+1},\end{equation*}

which, together with the presentation of $E(u)$ , gives that

\begin{equation*}\begin{aligned}\frac {1}{\lambda ^2}E(\lambda u) & =\frac {1}{2}\|\Delta u\|_2^2+\frac {\lambda ^2}{4}\|\nabla |u|^2\|_2^2-\frac {\lambda ^{p-1}}{p+1}\|u\|_{p+1}^{p+1}\\[3pt] & =E(u)+\frac {\lambda ^2-1}{4}\|\nabla |u|^2\|_2^2-\frac {\lambda ^{p-1}-1}{p+1}\|u\|_{p+1}^{p+1}.\end{aligned}\end{equation*}

Thus, we get that

(3.48) \begin{equation} E(u)=\frac {1}{\lambda ^2}E(\lambda u)+\frac {1-\lambda ^2}{4}\|\nabla |u|^2\|_2^2+\frac {\lambda ^{p-1}-1}{p+1}\|u\|_{p+1}^{p+1}. \end{equation}

Combining (3.47) and (3.48), we have

\begin{equation*}\begin{gathered}E(v_{l})\geq -\frac {\nu }{a_{l}^{2}}+\frac {1-a_{l}^2}{4}\|\nabla |v_{l}|^2\|_2^2+\frac {a_{l}^{p-1}-1}{p+1}\|v_{l}\|_{p+1}^{p+1}\quad \mathrm{for}\,a_{l}=\frac {1}{\|v_{l}\|_2},\\ E(w_{l})\geq -\frac {\nu }{b_{l}^{2}}+\frac {1-b_{l}^2}{4}\|\nabla |w_{l}|^2\|_2^2+\frac {b_{l}^{p-1}-1}{p+1}\|w_{l}\|_{p+1}^{p+1}\quad \mathrm{for}\,b_{l}=\frac {1}{\|w_{l}\|_2},\end{gathered}\end{equation*}

where $a_{l}(v_{l})\in \Gamma$ and $b_{l}(w_{l})\in \Gamma$ . By the definition of $\|u\|_{p+1}^{p+1}$ , we have

\begin{equation*}\begin{aligned} & E(v_{l})+E(w_{l}) & \\ & =a_{l}^{-2}E(v_{l})+b_{l}^{-2}E(w_{l})+\frac {1-a_{l}^2}{4}\|\nabla |v_{l}|^2\|_2^2+\frac {1-b_{l}^2}{4}\|\nabla |w_{l}|^2\|_2^2 & \\[2pt] & \,\,\,\,+\left (a_{l^{}}^{p-1}-1\right )\bigg [-E(v_{l})+\frac 12\|\Delta v_{l}\|_2^2+\frac 14\|\nabla |v_{l}|^2\|_2^2\bigg ] & \\[2pt] & \,\,\,\,+\left (b_{l^{}}^{p-1}-1\right )\bigg [-E(w_{l})+\frac 12\|\Delta w_{l}\|_2^2+\frac 14\|\nabla |w_{l}|^2\|_2^2\bigg ] & \\[2pt] & =a_{l}^{-2}E(v_{l})+b_{l}^{-2}E(w_{l})-\left (a_{l}^{p-1}-1\right )E(v_{l})-\left (b_{l}^{p-1}-1\right )E(w_{l}) & \\[2pt] & \,\,\,\,+\frac {a_{l}^{p-1}-a_{l}^2}{4}\|\nabla |v_{l}|^2\|_2^2+\frac {b_{l}^{p-1}-b_{l}^2}{4}\|\nabla |w_{l}|^2\|_2^2 & \\[2pt] & \,\,\,\,+\frac {a_{l}^{p-1}-1}{2}\|\Delta v_{l}\|_2^2+\frac {b_{l}^{p-1}-1}{2}\|\Delta w_{l}\|_2^2. & \end{aligned}\end{equation*}

Since $\|v_l\|_2^2\rightarrow \sigma$ and $\|w_l\|_2^2\rightarrow 1-\sigma$ , we have that $a_{l}\gt 1, b_{l}\gt 1$ for $l$ large enough. According to the conditions $3\lt p\lt +\infty$ when $2\lt N\leq 4$ ; $3\lt p\lt 3+\frac {16-2N}{N-4}$ when $4\lt N\leq 6$ ; $3\lt p\lt 3+\frac {8}{N-2}$ when $N\gt 6$ , we also obtain the following relations:

\begin{equation*} -\left (a_{l}^{p-1}-1\right )E(v_{l})-\left (b_{l}^{p-1}-1\right )E(w_{l})\gt 0, \end{equation*}

\begin{equation*} \frac {a_{l}^{p-1}-a_{l}^2}{4}\|\nabla |v_{l}|^2\|_2^2+\frac {b_{l}^{p-1}-b_{l}^2}{4}\|\nabla |w_{l}|^2\|_2^2\gt 0 ,\end{equation*}

\begin{equation*} \frac {a_{l}^{p-1}-1}{2}\|\Delta v_{l}\|_2^2+\frac {b_{l}^{p-1}-1}{2}\|\Delta w_{l}\|_2^2\gt 0, \end{equation*}

Applying $\|v_l\|_2^2\rightarrow \sigma$ and $\|w_l\|_2^2\rightarrow 1-\sigma$ again, we know that $a_{l}^{-2}\to \sigma$ and $ b_{l}^{-2}\to 1-\sigma$ . So we can obtain that

\begin{equation*}\operatorname *{lim}_{l\to \infty }\inf (E(v_{l})+E(w_{l}))\geq a_{l}^{-2}E(v_{l})+b_{l}^{-2}E(w_{l})\gt -\nu ,\end{equation*}

which contradicts (3.47). Hence, $\sigma =1$ .

Since $\sigma =a=1$ , by using Lemmas3.4 and 3.7 again, we have that there exists a sequence $\{y_{l}\}\subset \mathbb{R}^{N}$ and some $u\in H$ such that $u_{n_{l}}({\cdot} -y_{l})\to u$ in $L^2$ with $u\in \Gamma$ . The weak lower semi-continuity of the $H$ norm means that

\begin{equation*}E(u)\leq \lim _{l\to \infty }E(u_{n_{l}})=-\nu ,\end{equation*}

which, along with the definition of $\nu$ , yields that $E(u)=-\nu$ . Thus, we get that $E(u_{n_l})\to E(u)$ . Moreover, using (3.14) and (3.42), we have for $l\to \infty$ ,

\begin{equation*}\|\Delta u_{n_{l}}\|_2^2\to \|\Delta u\|_2^2,\ \ \|\nabla |u_{n_{l}}|^2\|_2^2\to \|\nabla |u|^2\|_2^2,\end{equation*}

which implies that $u_{n_{l}}({\cdot} -y_1)\to u$ strongly in $H.$ Since $\|w\|_2^2=\|u_0\|_2^2$ , we know that the minimizer of (1.14) can be attained at some $w\in H.$ Similar to corollary 8.3.8 of [Reference Cazenave35], one can verify that $w\in H$ is also a ground-state solution of the equation

(3.49) \begin{equation} \Delta ^2w+w-w\Delta |w|^2=|w|^{p-1}w, \quad \text{for}\, x\in \mathbb{R}^N. \end{equation}

Consequently, $u(x,t)=e^{i\omega t}w(x)$ is a standing wave solution of (1.1).

$\mathbf{Step\,4}$ . Demonstrating that part (2) holds.

Let $3\lt p\lt +\infty$ when $2\lt N\leq 4$ and $3\lt p\lt 3+\frac {16-2N}{N-4}$ when $4\lt N\leq 6$ . The stability of standing waves solution will be discussed in $H^2$ -norm. Arguing by contradiction, we suppose that there exists a sequence $\{\psi _{n}\}_{n=1}^{\infty }\subset H$ , a sequence $\{t_{n}\}_{n=1}^{\infty }\subset \mathbb{R}$ and some $\epsilon _{0}\gt 0$ meeting

(3.50) \begin{equation} \|\psi _{n}-w\|_{H^2}\to 0,\quad n\to \infty , \end{equation}

but the corresponding solution $s_{n}(x,t)$ with the initial value $\varphi _{n}$ satisfies

(3.51) \begin{equation} \inf _{\theta \in \mathbb{R}}\inf _{y\in \mathbb{R}^{N}}\left \|s_{n}({\cdot} ,t_{n})-e^{i\theta }w({\cdot} -y)\right \|_{H^2}\geq \epsilon _0. \end{equation}

For convenience, we denote

\begin{equation*}z_{n}(x)=s_{n}(x,t_{n}).\end{equation*}

To get (3.51), we need to prove that

(3.52) \begin{equation} \inf _{w\,\text{is a minimizer of }(1:14)}\|z_{n}-w\|_{H^2}\geq \varepsilon . \end{equation}

From Propositions (1.2) and (3.50), it is found that for $n\to \infty$ ,

(3.53) \begin{equation} \|z_{n}\|_2^2={\psi }\|z_{n}\|_2^2\to \|w\|_2^2, \end{equation}

(3.54) \begin{equation} E(z_{n})=E(\psi _{n})\to E(w)=-\nu . \end{equation}

(3.53) and (3.54) show that $\{z_{n}\}_{n=1}^{\infty }$ is also a minimizing sequence for variational problem (1.14). Based on standard constrained variational methods and convexity arguments, we observe that all ground states of Equation (1.9) are unique modulo translations and phase rotations. Hence, following similar arguments to these in the proof of Step 3 above, we have that there exist a sequence $\{y_{n}\}_{n\in \mathbb{N}}\subset \mathbb{R}^{N}$ and a minimizer $w^{*}$ of (1.14) such that for $n\to \infty$ ,

(3.55) \begin{equation} \|z_{n}-w^{*}(\cdot-y_{n})\|_{H^2}=\|z_{n}-w\|_{H^2}\to0. \end{equation}

Since $w^{*}(\cdot-y_{n})$ is also a minimizer of (1.14), we can note that (3.55) contradicts (3.52).

$\mathbf{Step\,5}$ . Establishing that part (3) holds.

Let $3\lt p\lt 3+\frac {8}{N-2}$ and $N\gt 6$ . The stability of standing wave solutions will be analysed in $\bar {H}$ -norm. The proof’s processes follow exactly along the same line as above Step 4, and the detail is then omitted. We thereby finish the proof of this theorem.