Proof. $(i)$ We borrow the idea from [Reference Cazenave4]. Let $u$ be a solution of problem (1.1). Thus,

\[ -\Delta u+\frac{\omega}{\left(1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)}u=\frac{1}{\left(1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)}|u|^{p-2}u. \]

Let $A=1+b\int _{\mathbb {R}^2}|\nabla u|^2\,{\rm d}x$ ($A<\infty$), $B=\frac {1+b\int _{\mathbb {R}^2}|\nabla u|^2\,{\rm d}x}{\omega ^2}$. Changing $u(x)$ to

\[ u(x)=\left(A/\omega\right)^{\frac{1}{p-2}}v(\sqrt{\omega/A}x), \]

we may assume that $v$ satisfies

(2.3)\begin{equation} {-}\Delta v+v=\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}|v|^{p-2}v=B|v|^{p-2}v. \end{equation}

Note that (2.3) can be written in the form

(2.4)\begin{equation} \mathcal{F}^{{-}1}((1+4\pi^2|\xi|^2)\mathcal{F}v)=B|v|^{p-2}v, \end{equation}

where $\mathcal {F}$ is the Fourier transform. For $K>0$, set

\[ \begin{array}{@{}ll} \displaystyle v_K & =\left\{\begin{array}{@{}ll} \displaystyle \pm K,\ \pm v>K,\\ \displaystyle v,\ |v|\leq K. \end{array}\right. \end{array} \]

Observing that, for $\beta \geq 1$,

\[ \int_{\mathbb{R}^2}|\nabla(|v_K|^{\beta-1}v)|^2\,{\rm d}x=\int_{\mathbb{R}^2}|v_K|^{2\beta-2}|\nabla v|^2\,{\rm d}x+(\beta^2-1)\int_{\mathbb{R}^2}|v_K|^{2\beta-2}|\nabla v_K|^2\,{\rm d}x, \]

we get

(2.5)\begin{align} & \displaystyle\frac{1}{\beta+1}\int_{\mathbb{R}^2}[|\nabla(|v_K|^{\beta-1}v)|^2+(|v_K|^{\beta-1}v)^2]\,{\rm d}x\nonumber\\ & =\displaystyle\frac{1}{\beta+1}\int_{\mathbb{R}^2}|v_K|^{2\beta-2}|\nabla v|^2\,{\rm d}x+(\beta-1)\int_{\mathbb{R}^2}|v_K|^{2\beta-2}|\nabla v_K|^2\,{\rm d}x\nonumber\\ & \quad +\frac{1}{\beta+1}\int_{\mathbb{R}^2}(|v_K|^{\beta-1}v)^2\,{\rm d}x\nonumber\\ & \leq \displaystyle\int_{\mathbb{R}^2}|v_K|^{2\beta-2}|\nabla v|^2\,{\rm d}x+\int_{\mathbb{R}^2}(|v_K|^{\beta-1}v)^2\,{\rm d}x\nonumber\\ & \quad +2(\beta-1)\int_{\mathbb{R}^2}|v_K|^{2\beta-2}|\nabla v_K|^2\,{\rm d}x. \end{align}

Taking the test function $\varphi =|v_K|^{2\beta -2}v$ in (2.3), we have

(2.6)\begin{align} & \int_{\mathbb{R}^2}|v_K|^{2\beta-2}|\nabla v|^2\,{\rm d}x+2(\beta-1)\int_{\mathbb{R}^2}|v_K|^{2\beta-2}|\nabla v_K|^2\,{\rm d}x+\int_{\mathbb{R}^2}(|v_K|^{\beta-1}v)^2\,{\rm d}x\nonumber\\ & \quad = B\int_{\mathbb{R}^2}|v_K|^{2\beta-2}|v|^2|v|^{p-2}\,{\rm d}x. \end{align}

It follows from (2.5) and (2.6) that

(2.7)\begin{equation} \frac{1}{\beta+1}\int_{\mathbb{R}^2}[|\nabla(|v_K|^{\beta-1}v)|^2+(|v_K|^{\beta-1}v)^2]\,{\rm d}x\leq B\int_{\mathbb{R}^2}|v_K|^{2\beta-2}v^2|v|^{p-2}\,{\rm d}x. \end{equation}

On one hand, by the Sobolev inequality, we have

\[ \int_{\mathbb{R}^2}[|\nabla(v_K|^{\beta-1}v)|^2+(|v_K|^{\beta-1}v)^2]\,{\rm d}x\geq S_{2p}\left[\int_{\mathbb{R}^2}(|v_K|^{\beta-1}v)^{2p}\,{\rm d}x\right]^{\frac{1}{p}}, \]

where

\[ S_{2p}=\inf_{u\in H^1(\mathbb{R}^2)\backslash\{0\}}\frac{\int_{\mathbb{R}^2}(|\nabla u|^2+u^2)\,{\rm d}x}{(\int_{\mathbb{R}^2}|u|^{2p}\,{\rm d}x)^{\frac{1}{p}}}. \]

On the other hand, by the Hölder inequality, we deduce that

\begin{align*} \int_{\mathbb{R}^2}|v_K|^{2\beta-2}v^2|v|^{p-2}\,{\rm d}x& =\displaystyle\left[\int_{\mathbb{R}^2}(|v_K|^{2\beta-2}v^2)^{\frac{2p}{p+2}}\,{\rm d}x\right]^{\frac{p+2}{2p}} \left[\int_{\mathbb{R}^2}|v|^{2p}\,{\rm d}x\right]^{\frac{p-2}{2p}}\\ & \leq S_{2p}^{-\frac{p-2}{2}}\|v\|^{p-2}\left[\int_{\mathbb{R}^2}(|v_K|^{2\beta-2}v^2)^{\frac{2p}{p+2}}\,{\rm d}x\right]^{\frac{p+2}{2p}}. \end{align*}

The symbol $\|\cdot \|$ is used only for the norm in $H^1(\mathbb {R}^2)$. Applying (2.7) and letting $K\rightarrow \infty$, we obtain

\[ \frac{S_{2p}}{\beta+1}|v|_{L^{2\beta p}}^{2\beta}\leq S_{2p}^{-\frac{p-2}{2}}\|v\|^{p-2}|v|_{L^{\frac{4\beta p}{p+2}}}^{2\beta}, \]

where $|\cdot |_{L^q}$ denotes the standard norm in $L^q(\mathbb {R}^2)$. It follows from the above inequality that

(2.8)\begin{equation} |v|_{L^{2\beta p}}\leq C(\beta+1)^{\frac{1}{2\beta}}\|v\|^\frac{p-2}{2\beta}|v|_{L^{\frac{4\beta p}{p+2}}}. \end{equation}

Set

\[ 2p\beta_{n-1}=\frac{4p}{p+2}\beta_n, \]

so that

\[ \beta_n=\left(\frac{p+2}{2}\right)^{n-1}, \quad n\in\mathbb{N}. \]

Consequently, $\beta _n\rightarrow \infty$ as $n\rightarrow \infty$. Therefore,

\[ \lim_{n\rightarrow\infty}(\beta_n+1)^{\frac{1}{2\beta_n}} =\displaystyle\lim_{n\rightarrow\infty}e^{\frac{\log(1+\beta_n)}{2\beta_n}} =1. \]

Doing iteration by (2.8), we obtain $v\in L^\infty (\mathbb {R}^2)$. Thus $|v|^{p-2}v\in L^2(\mathbb {R}^2)\cap L^\infty (\mathbb {R}^2)$, so that $v\in W^{2,q}(\mathbb {R}^2)$ and $|v|^{p-2}v\in W^{1,q}(\mathbb {R}^2)$ for all $2\leq q<\infty$. Therefore, it follows from (2.4) that $(-\Delta +I)\partial _jv\in L^q(\mathbb {R}^2)$, i.e., $\mathcal {F}^{-1}((1+4\pi ^2|\xi |^2)\mathcal {F}\partial _jv)\in L^q(\mathbb {R}^2)$. Thus $\partial _jv\in H^{2,q}(\mathbb {R}^2)=W^{2,q}(\mathbb {R}^2)$, and so $v\in W^{3,q}(\mathbb {R}^2)$. By the Sobolev's embedding theorem, $v\in C^{2,\delta }(\mathbb {R}^2)$ for $0<\delta <1$. Therefore, $|D^\beta v(x)|\rightarrow 0$ as $|x|\rightarrow \infty$ for all $|\beta |\leq 2$, so that $|D^\beta u(x)|\rightarrow 0$ as $|x|\rightarrow \infty$.

$(ii)$ Let $\varepsilon >0$ and $\theta _\varepsilon (x)=e^{\frac {|x|}{1+\varepsilon |x|}}$. $\theta _\varepsilon$ is bounded, Lipschitz continuous, and $|\nabla \theta _\varepsilon |\leq \theta _\varepsilon$ a.e. Taking the test function $\theta _\varepsilon v\in H^1(\mathbb {R}^2)$ in (2.3), we get

(2.9)\begin{equation} \int_{\mathbb{R}^2}\nabla v\nabla(\theta_\varepsilon v)\,{\rm d}x+\int_{\mathbb{R}^2}\theta_\varepsilon|v|^2\,{\rm d}x=\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\mathbb{R}^2}\theta_\varepsilon|v|^p\,{\rm d}x. \end{equation}

Noting that

\begin{align*} \int_{\mathbb{R}^2}\nabla v\nabla(\theta_\varepsilon v)\,{\rm d}x& =\displaystyle\int_{\mathbb{R}^2}\theta_\varepsilon|\nabla v|^2\,{\rm d}x+\int_{\mathbb{R}^2}v\nabla\theta_\varepsilon\nabla v\,{\rm d}x\\ & \geq\displaystyle\int_{\mathbb{R}^2}\theta_\varepsilon|\nabla v|^2\,{\rm d}x-\int_{\mathbb{R}^2}|v||\nabla\theta_\varepsilon||\nabla v|\,{\rm d}x\\ & \geq\displaystyle\int_{\mathbb{R}^2}\theta_\varepsilon|\nabla v|^2\,{\rm d}x-\int_{\mathbb{R}^2}\theta_\varepsilon|v||\nabla v|\,{\rm d}x\\ & \geq\displaystyle\frac{1}{2}\int_{\mathbb{R}^2}\theta_\varepsilon|\nabla v|^2\,{\rm d}x-\frac{1}{2}\int_{\mathbb{R}^2}\theta_\varepsilon|v|^2\,{\rm d}x, \end{align*}

we obtain

(2.10)\begin{align} \frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\mathbb{R}^2}\theta_\varepsilon|v|^p\,{\rm d}x& =\displaystyle\int_{\mathbb{R}^2}\nabla v\nabla(\theta_\varepsilon v)\,{\rm d}x+\int_{\mathbb{R}^2}\theta_\varepsilon|v|^2\,{\rm d}x\nonumber\\ & \geq\displaystyle\frac{1}{2}\int_{\mathbb{R}^2}\theta_\varepsilon|v|^2\,{\rm d}x. \end{align}

Since $u\in H^1(\mathbb {R}^2)$, there exists $R>0$ such that $|v(x)|^{p-2}\leq \frac {1+b\int _{\mathbb {R}^2}|\nabla u|^2\,{\rm d}x}{4\omega ^2}$ for $|x|\geq R$. Thus

\begin{align*} & \displaystyle\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\mathbb{R}^2}\theta_\varepsilon|v|^p\,{\rm d}x\\ & \quad =\displaystyle\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\{|x|\leq R\}}\theta_\varepsilon|v|^p\,{\rm d}x+\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\{|x|\geq R\}}\theta_\varepsilon|v|^p\,{\rm d}x\\ & \quad \leq \displaystyle\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\{|x|\leq R\}}e^{|x|}|v|^p\,{\rm d}x+\frac{1}{4}\int_{\{|x|\geq R\}}\theta_\varepsilon|v|^{2}\,{\rm d}x\\ & \quad \leq \displaystyle\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}e^{R}\int_{\{|x|\leq R\}}|v|^p\,{\rm d}x+\frac{1}{4}\int_{\mathbb{R}^2}\theta_\varepsilon|v|^{2}\,{\rm d}x. \end{align*}

It follows from (2.10) that

\[ \int_{\mathbb{R}^2}\theta_\varepsilon|v|^2\,{\rm d}x\leq4\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}e^{R}\int_{\{|x|\leq R\}}|v|^p\,{\rm d}x. \]

Letting $\varepsilon \downarrow 0$, we deduce that

(2.11)\begin{equation} \int_{\mathbb{R}^2}e^{|x|}|v|^2\,{\rm d}x<{+}\infty. \end{equation}

Since $v$ is globally Lipschitz continuous by $(i)$, $|v(x)|^{4}e^{|x|}$ is bounded. Similarly, applying $\partial _j$ to equation (2.3) and multiplying the resulting equation by $\theta _\varepsilon \partial _j u$ for $j=1,\,2$, we have

\[ \int_{\mathbb{R}^2}\nabla(\partial_j v)\nabla(\theta_\varepsilon (\partial_j v))\,{\rm d}x+\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_j v|^2\,{\rm d}x=\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_j v|^p\,{\rm d}x. \]

Applying the fact

\begin{align*} \int_{\mathbb{R}^2}\nabla (\partial_j v)\nabla(\theta_\varepsilon (\partial_j v))\,{\rm d}x& = \displaystyle\int_{\mathbb{R}^2}\theta_\varepsilon|\nabla \partial_j v|^2\,{\rm d}x+\int_{\mathbb{R}^2}(\partial_j v)\nabla\theta_\varepsilon\nabla (\partial_j v)\,{\rm d}x\\ & \geq \displaystyle\int_{\mathbb{R}^2}\theta_\varepsilon|\nabla (\partial_j v)|^2\,{\rm d}x-\int_{\mathbb{R}^2}|\partial_j v||\nabla\theta_\varepsilon||\nabla (\partial_j v)|\,{\rm d}x\\ & \geq \displaystyle\int_{\mathbb{R}^2}\theta_\varepsilon|\nabla (\partial_j v)|^2\,{\rm d}x-\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_j v||\nabla (\partial_j v)|\,{\rm d}x\\ & \geq \displaystyle\frac{1}{2}\int_{\mathbb{R}^2}\theta_\varepsilon|\nabla (\partial_j v)|^2\,{\rm d}x-\frac{1}{2}\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_j v|^2\,{\rm d}x, \end{align*}

we have

\begin{align*} \displaystyle\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_j v|^p\,{\rm d}x& =\displaystyle\int_{\mathbb{R}^2}\nabla (\partial_j v)\nabla(\theta_\varepsilon (\partial_j v))\,{\rm d}x+\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_j v|^2\,{\rm d}x\\ & \geq \displaystyle\frac{1}{2}\int_{\mathbb{R}^2}\theta_\varepsilon|\nabla (\partial_j v)|^2\,{\rm d}x+\frac{1}{2}\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_j v|^2\,{\rm d}x. \end{align*}

Consequently

(2.12)\begin{equation} \frac{1}{2}\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_j v|^2\,{\rm d}x\leq\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_jv|^p\,{\rm d}x. \end{equation}

By $(i)$, there exists $R>0$ such that $|\partial _jv(x)|^{p-2}\leq \frac {1+b\int _{\mathbb {R}^2}|\nabla u|^2\,{\rm d}x}{4\omega ^2}$ for $|x|\geq R$. Thus

\begin{align*} & \displaystyle\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_jv|^p\,{\rm d}x\\ & \quad =\displaystyle\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\{|x|\leq R\}}\theta_\varepsilon|\partial_jv|^p\,{\rm d}x\\ & \quad +\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\{|x|\geq R\}}\theta_\varepsilon|\partial_jv|^p\,{\rm d}x\\ & \quad \leq \displaystyle\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}\int_{\{|x|\leq R\}}e^{|x|}|\partial_jv|^p\,{\rm d}x+\frac{1}{4}\int_{\{|x|\geq R\}}\theta_\varepsilon|\partial_jv|^{2}\,{\rm d}x\\ & \quad \leq \displaystyle\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}e^{R}\int_{\{|x|\leq R\}}|\partial_jv|^p\,{\rm d}x+\frac{1}{4}\int_{\mathbb{R}^2}\theta_\varepsilon|\partial_jv|^{2}\,{\rm d}x. \end{align*}

It follows from (2.12) that

\[ \int_{\mathbb{R}^2}\theta_\varepsilon|\partial_jv|^2\,{\rm d}x\leq4\frac{1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x}{\omega^2}e^{R}\int_{\{|x|\leq R\}}|\partial_jv|^p\,{\rm d}x. \]

Letting $\varepsilon \downarrow 0$, we deduce that

\[ \int_{\mathbb{R}^2}e^{|x|}|\nabla v|^2\,{\rm d}x<{+}\infty. \]

Since $\nabla v$ is globally Lipschitz continuous, similarly, we deduce that $|\nabla v(x)|^{4}e^{|x|}$ is bounded. The proof is now finished.

Proof. A similar argument to the proof of Theorem 8.1.5 in [Reference Cazenave4], we observe that $d_N>0$. Indeed, consider $u\in N$, it implies that $V(u)=0$, namely

\[ \int_{\mathbb{R}^2}|u|^2\,{\rm d}x=\frac{2}{\omega p}\int_{\mathbb{R}^2}|u|^p\,{\rm d}x. \]

It follows from the Gagliardo–Nirenberg inequality that there exists $C$ independent of $u$ such that

\[ \int_{\mathbb{R}^2}|u|^p\,{\rm d}x\leq C\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^{\frac{p-2}{2}}\int_{\mathbb{R}^2}|u|^2\,{\rm d}x. \]

From the above information, there exists $c>0$ such that $\int _{\mathbb {R}^2}|\nabla u|^2\,{\rm d}x\geq c$, and so

\begin{align*} S(u)& =\displaystyle\frac{1}{2}\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-V(u)\\ & =\displaystyle\frac{1}{2}\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & > 0 \end{align*}

for all $u\in N$, which implies that $d_N>0$.

We recall the definition of the Schwarz symmetrization [Reference Berestycki and Lions3]. If $u\in L^2(\mathbb {R}^2)$ is a nonnegative function, we denote by $u^*$ the unique spherically symmetric, nonnegative, non-increasing function such that

\[ |\{x\in\mathbb{R}^2: u^*(x)>\lambda\}|=|\{x\in\mathbb{R}^2: |u(x)|>\lambda\}| \ \mathrm{for~all} \ \lambda>0. \]

In particular,

(2.14)\begin{equation} \begin{cases} \displaystyle\int_{\mathbb{R}^2}|u^*|^q\,{\rm d}x=\int_{\mathbb{R}^2}|u|^q\,{\rm d}x,\ \mathrm{for}\ 1\leq q<{+}\infty,\\ \displaystyle\int_{\mathbb{R}^2}|\nabla u^*|^2\,{\rm d}x\leq\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x. \end{cases} \end{equation}

The proof of lemma 2.2 is divided into three steps.

Step 1. we claim that the minimization problem (2.13) has a solution.

Similar to the proof of [Reference Berestycki, Gallouët and Kavian2], it is clear that $N\neq \emptyset$. Let $\{v_n\}\subset N$ be a minimizing sequence of $S$, that is, $V(v_n)=0$, and $S(v_n)\rightarrow d_N$. Let $w_n=|v_n|^*$. It follows from (2.14) that $V(w_n)=V(v_n)$, and hence $\{w_n\}$ is also a minimizing sequence of $S$. Define now $u_n$ by $u_n(x)=w_n(\lambda ^{1/2}_nx)$, where

\[ \lambda_n=\frac{\int_{\mathbb{R}^2}|w_n|^2\,{\rm d}x}{\gamma}, \]

where

\[ \gamma=\frac{2(4bd_N+1)-2\sqrt{4bd_N+1}}{\omega b(p-2)}. \]

We deduce that

\begin{align*} \displaystyle\int_{\mathbb{R}^2}|u_n|^2\,{\rm d}x& = \displaystyle\int_{\mathbb{R}^2}|w_n|^2\frac{1}{\lambda_n}\,{\rm d}x\\ & = \displaystyle\gamma,\\ V(u_n)& =\displaystyle\frac{1}{p}\int_{\mathbb{R}^2}|w_n|^{p}\frac{1}{\lambda_n}\,{\rm d}x-\frac{\omega}{2}\int_{\mathbb{R}^2}|w_n|^2\frac{1}{\lambda_n}\,{\rm d}x\\ & =\displaystyle\frac{1}{\lambda_n}\left(\frac{1}{p}\int_{\mathbb{R}^2}|w_n|^{p}\,{\rm d}x-\frac{\omega}{2}\int_{\mathbb{R}^2}|w_n|^2\,{\rm d}x\right)\\ & =\displaystyle\frac{1}{\lambda_n}\left(\frac{1}{p}\int_{\mathbb{R}^2}(|v_n|^*)^{p}\,{\rm d}x-\frac{\omega}{2}\int_{\mathbb{R}^2}(|v_n|^*)^2\,{\rm d}x\right)\\ & =\displaystyle\frac{1}{\lambda_n}\left(\frac{1}{p}\int_{\mathbb{R}^2}|v_n|^{p}\,{\rm d}x-\frac{\omega}{2}\int_{\mathbb{R}^2}|v_n|^2\,{\rm d}x\right)\\ & = 0, \end{align*}

and

\begin{align*} \displaystyle T(u_n)& =\displaystyle\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x+b\left(\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x\right)^2\\ & =\displaystyle\int_{\mathbb{R}^2}|\nabla w_n|^2\lambda_n\cdot\frac{1}{\lambda_n}\,{\rm d}x+b\left(\int_{\mathbb{R}^2}|\nabla w_n|^2\lambda_n\cdot\frac{1}{\lambda_n}\,{\rm d}x\right)^2\\ & =\displaystyle\int_{\mathbb{R}^2}|\nabla w_n|^2\,{\rm d}x+b\left(\int_{\mathbb{R}^2}|\nabla w_n|^2\,{\rm d}x\right)^2. \end{align*}

These results imply

(2.15)\begin{equation} S(u_n)=S(w_n)\rightarrow d_N, \ \mathrm{as}\ n\rightarrow\infty. \end{equation}

We obtain that $\{u_n\}$ is also a minimizing sequence of $S$. It follows from (2.15) that $\{u_n\}$ is bounded in $H^1(\mathbb {R}^2)$. Therefore, there exist a subsequence of $u_n$ (denoted by itself) and $u\in H^1(\mathbb {R}^2)$ such that $u_n\rightharpoonup u$ weakly in $H^1(\mathbb {R}^2)$ and $u_n\rightarrow u$ strongly in $L^{p}(\mathbb {R}^2)$. By the Fatou's lemma,

\begin{align*} 0& = \lim_{n\rightarrow\infty}V(u_n)=\displaystyle\lim_{n\rightarrow\infty}\left(\frac{1}{p}\int_{\mathbb{R}^2}|u_n|^{p}\,{\rm d}x-\frac{\omega}{2}\int_{\mathbb{R}^2}|u_n|^2\,{\rm d}x\right)\\ & \leq \frac{1}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x-\frac{\omega}{2}\int_{\mathbb{R}^2}|u|^2\,{\rm d}x\\ & = V(u). \end{align*}

By the weak lower semicontinuity of the $L^2$ norm,

\[ T(u)\leq\liminf_{n\rightarrow\infty}T(u_n). \]

Therefore,

\[ V(u)\geq0, \quad \mathrm{and} \quad S(u)\leq d_N. \]

We claim that $V(u)=0$. Indeed, if $V(u)>0$, then $u\neq 0$. So there exists $\lambda \in (0,\,1)$ such that $v=\lambda u$ satisfies $V(v)=0$. Thus $v\in N$. Furthermore,

\begin{align*} \frac{1}{2}\int_{\mathbb{R}^2}|\nabla v|^2\,{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R}^2}|\nabla v|^2\,{\rm d}x\right)^2& = \displaystyle\frac{\lambda^2}{2}\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+\frac{b}{4}\lambda^4\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & < \displaystyle\frac{1}{2}\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2. \end{align*}

Noting that $d_N=\lim _{n\to \infty }S(u_n)$ and $V(u_n)=0$, then we have

\begin{align*} d_N& = \displaystyle\lim_{n\to\infty}S(u_n)\\ & = \displaystyle\lim_{n\to\infty}\left\{\frac{1}{2}\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x\right)^2-V(u_n)\right\}\\ & = \displaystyle\lim_{n\to\infty}\left\{\frac{1}{2}\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x\right)^2\right\}\\ & \geq \displaystyle\frac{1}{2}\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2. \end{align*}

From the above information, we get

\begin{align*} S(v)& =\frac{1}{2}\int_{\mathbb{R}^2}|\nabla v|^2\,{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R}^2}|\nabla v|^2\,{\rm d}x\right)^2\\& <\frac{1}{2}\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\leq d_N. \end{align*}

This implies that $S(v)< d_N$. This contradicts (2.13) since $S(v)\geq d_N$. Therefore, $V(u)=0$. This implies that

\[ \lim_{n\rightarrow\infty}V(u_n)=V(u). \]

Consequently,

\[ \int_{\mathbb{R}^2}|u|^2\,{\rm d}x=\lim_{n\rightarrow\infty}\int_{\mathbb{R}^2}|u_n|^2\,{\rm d}x=\gamma, \]

and so $u$ satisfies (2.13).

Step 2. We claim that every solutions of (2.13) belongs to $A$.

Indeed, consider a solution $u$ of (2.13). There exists a Lagrange multiplier $\lambda$ such that

(2.16)\begin{equation} -\left(1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)\Delta u=\lambda(|u|^{p-2}u-\omega u). \end{equation}

For any $\varphi \in H^1(\mathbb {R}^2)$, it follows from (2.16) that

\[ \left(1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)\int_{\mathbb{R}^2}\nabla u\nabla\varphi \,{\rm d}x=\lambda\int_{\mathbb{R}^2}(|u|^{p-2}u-\omega u)\varphi \,{\rm d}x. \]

Taking the test function $\varphi =u$, we obtain

(2.17)\begin{equation} T(u)=\left(1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x=\lambda\int_{\mathbb{R}^2}(|u|^{p}-\omega |u|^2)\,{\rm d}x. \end{equation}

Noting that

\begin{align*} S(u)& =\displaystyle\frac{1}{2}\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & = d_N, \end{align*}

we deduce that

\[ \int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x=\frac{-1+\sqrt{1+4bd_N}}{b}, \]

so that

\[ T(u)=\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+b\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2=\frac{1+4bd_N-\sqrt{1+4bd_N}}{b}. \]

It follows from $V(u)=0$ and $\int _{\mathbb {R}^2}|u|^2\,{\rm d}x=\gamma$ that

\[ \int_{\mathbb{R}^2}|u|^{p}=\frac{p\gamma\omega}{2}. \]

Together with (2.17), there holds

\[ \frac{1+4bd_N-\sqrt{1+4bd_N}}{b}=\lambda\frac{(p-2)\gamma\omega}{2}=\lambda\frac{(p-2)\omega}{2}\cdot\frac{2(4bd_N+1)-2\sqrt{4bd_N+1}}{\omega b(p-2)}, \]

and so $\lambda =1$. Therefore, $u$ satisfies problem (1.1). So $A\neq \emptyset$.

Step 3. We claim that $u$ satisfies problem (2.13) if and only if $u\in G$.

Consider any solution $u$ of (2.13) and any $v\in A$. That is,

\[ -\left(1+b\int_{\mathbb{R}^2}|\nabla v|^2\,{\rm d}x\right)\Delta v=|v|^{p-2}v-\omega v. \]

By Pohozaev's identity, we get $V(v)=0$. This means $v\in N$, and

\begin{align*} \int_{\mathbb{R}^2}|v|^{p}\,{\rm d}x& =\frac{p\omega}{2}\int_{\mathbb{R}^2}|v|^2\,{\rm d}x,\\ T(v)& =\int_{\mathbb{R}^2}|v|^{p}\,{\rm d}x-\omega\int_{\mathbb{R}^2}|v|^2\,{\rm d}x=\frac{(p-2)\omega}{2}\int_{\mathbb{R}^2}|v|^2\,{\rm d}x. \end{align*}

From (2.2), there holds

(2.18)\begin{equation} S(v)=\frac{(p-2)\omega\int_{\mathbb{R}^2}|v|^2\,{\rm d}x}{8}+\frac{\sqrt{1+2b(p-2)\omega\int_{\mathbb{R}^2}|v|^2\,{\rm d}x}-1}{8b}. \end{equation}

Since $v\in N$, by the definition of $d_N$, we have

\[ S(v)\geq S(u)=d_N, \]

which implies that $u\in G\neq \emptyset$.

Assume further that $v\in G$. Since $u\in G$ also, we have $S(u)=S(v)$. Noting that $S(u)=d_N$ and $\int _{\mathbb {R}^2}|u|^2\,{\rm d}x=\gamma =\frac {2(4bd_N+1)-2\sqrt {4bd_N+1}}{\omega b(p-2)}$, we obtain

\[ d_N=\frac{(p-2)\gamma\omega}{8}+\frac{\sqrt{1+2b(p-2)\gamma\omega}-1}{8b}. \]

Applying (2.18), we have

\begin{align*} & \frac{(p-2)\gamma\omega}{8}+\frac{\sqrt{1+2b(p-2)\gamma\omega}-1}{8b}\\ & \quad =\frac{(p-2)\omega\int_{\mathbb{R}^2}|v|^2\,{\rm d}x}{8}+\frac{\sqrt{1+2b(p-2)\omega\int_{\mathbb{R}^2}|v|^2\,{\rm d}x}-1}{8b}, \end{align*}

which implies that $\gamma =\int _{\mathbb {R}^2}|v|^2\,{\rm d}x,$ which means that $v$ satisfies (2.13). Hence, the proof is complete.

Proof. Proof of theorem 1.1

Consider $u\in G$, it follows that

\[ -\left(1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)\Delta u+\omega u=|u|^{p-2}u, \]

and so

(2.19)\begin{equation} -\Delta u+\frac{\omega}{\overline{b}} u=\frac{a(x)}{\overline{b}}u, \end{equation}

where

\[ a(x)=|u(x)|^{p-2}, \quad \overline{b}=\left(1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right). \]

According to lemma 2.1, $u\in C^2(\mathbb {R}^2)$ and $a(x)\rightarrow 0$ as $|x|\rightarrow \infty$. Applying Theorem 2 in [Reference Gidas, Ni and Nirenberg10], we obtain that there exists a positive, spherically symmetric solution $\varphi$ of Eq. (2.19) and $y\in \mathbb {R}^2$ such that $u(\cdot )=\varphi (\cdot -y)$. Note that $\varphi$, being radially symmetric, satisfies the ordinary differential equation

\[ \left(1+b\int_0^\infty4\pi r^2u_r^2dr\right)\left(u_{rr}+\frac{1}{r}u_r\right)+\omega u(r)-u^{p-1}(r)=0. \]

Since $\int _0^\infty 4\pi r^2u_r^2dr$ is independent of choice of $u$, we deduce that $1+b\int _0^\infty 4\pi r^2u_r^2dr$ is a positive constant. As a result, we obtain that $\varphi$ is unique by applying the uniqueness results in [Reference Kwong18] and [Reference Li, Luo, Peng, Wang and Xiang19]. The proof is now complete.

Proof. Let $u\in H^1(\mathbb {R}^2)$, $u\neq 0$, and let $u_\lambda =\mathcal {P}(\lambda,\,u)$. We have

\begin{align*} f(\lambda)& :=S(u_\lambda) =\frac{\lambda^2}{2}\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+\frac{b\lambda^4}{4}\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & \quad +\frac{\omega}{2}\int_{\mathbb{R}^2}|u|^2\,{\rm d}x-\frac{\lambda^{p-2}}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x. \end{align*}

Then

\[ f'(\lambda)=\lambda\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+b\lambda^3\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-\frac{(p-2)\lambda^{p-3}}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x, \]

so that

\[ \frac{d}{d\lambda}S(\mathcal{P}(\lambda,u))=\frac{1}{\lambda}Q(\mathcal{P}(\lambda,u)), \]

Consequently, property (vi) holds. It follows from $f'(\lambda )=0$ that

(3.1)\begin{equation} \int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+b\lambda^2\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-\frac{(p-2)\lambda^{p-4}}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x=0. \end{equation}

Thus

\begin{align*} f''(\lambda)& =\displaystyle\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+3b\lambda^2\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & \quad\displaystyle-\frac{(p-2)(p-3)\lambda^{p-4}}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & ={-}(p-4)\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x-b(p-6)\lambda^2\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & <\displaystyle0, \end{align*}

which implies that (3.1) has a unique positive solution $\lambda ^*(u)$ (by Rello's theorem). In particular, $f$ achieves its maximum at $\lambda ^*(u)$. Therefore, properties (i), (ii) and (v) follow.

(iii) Let $\lambda ^*(u)$ be the solution of (3.1), namely

(3.2)\begin{equation} \int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+b(\lambda^*(u))^2\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-\frac{(p-2)(\lambda^*(u))^{p-4}}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x=0. \end{equation}

If $Q(u)<0$. Then

\[ \int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+b\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x<0, \]

which implies that

\[ b\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2<\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x. \]

If $\lambda ^*(u)\geq 1$. By (3.2), we deduce that

\begin{align*} 0& > \int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+b\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & \quad -\displaystyle\left[\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+b(\lambda^*(u))^2\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-\frac{(p-2)(\lambda^*(u))^{p-4}}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\right]\\ & =\displaystyle\frac{p-2}{p}[(\lambda^*(u))^{p-4}-1]\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x+b(1-(\lambda^*(u))^2)\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & \geq \displaystyle\frac{p-2}{p}[(\lambda^*(u))^{p-4}-1]\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x+(1-(\lambda^*(u))^2)\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & = \displaystyle\frac{p-2}{p}[(\lambda^*(u))^{p-4}-(\lambda^*(u))^2]\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & = \displaystyle\frac{p-2}{p}(\lambda^*(u))^2[(\lambda^*(u))^{p-6}-1]\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x, \end{align*}

which is impossible. This implies that if $Q(u)<0$ then $\lambda ^*(u)<1$. Now we claim that $Q(u)<0$ when $\lambda ^*(u)<1$. Indeed, $\lambda ^*(u)<1$ leads to

\[ b\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\leq\frac{(p-2)(\lambda^*(u))^{p-6}}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x<\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x. \]

Therefore,

\begin{align*} Q(u)& =\displaystyle\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+b\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & =\displaystyle\frac{(p-2)(\lambda^*(u))^{p-4}}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x-b(\lambda^*(u))^2\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & \quad +b\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & =\displaystyle\frac{p-2}{p}[(\lambda^*(u))^{p-4}-1]\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x+b(1-(\lambda^*(u))^2)\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & <\displaystyle\frac{p-2}{p}[(\lambda^*(u))^{p-4}-1]\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x+(1-(\lambda^*(u))^2)\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & =\displaystyle\frac{p-2}{p}[(\lambda^*(u))^{p-4}-(\lambda^*(u))^2]\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & =\displaystyle\frac{p-2}{p}(\lambda^*(u))^2[(\lambda^*(u))^{p-6}-1]\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & < 0. \end{align*}

Thus, property (iii) follows.

(iv) If $\lambda ^*(u)=1$, it follows from (3.2) that $u\in M$. On the other hand, assume $u\in M$, then

\[ \begin{cases} \displaystyle\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+b\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x=0,\\ \displaystyle\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+b(\lambda^*(u))^2\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2-\frac{(p-2)(\lambda^*(u))^{p-4}}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x=0, \end{cases} \]

so that

(3.3)\begin{equation} \frac{p-2}{p}[1-(\lambda^*(u))^{p-4}]\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x+b((\lambda^*(u))^2-1)\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2=0. \end{equation}

If $\lambda ^*(u)>1$, using (3.3), we have

\begin{align*} 0& =\displaystyle\frac{p-2}{p}[1-(\lambda^*(u))^{p-4}]\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x+b((\lambda^*(u))^2-1)\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & <\displaystyle b[1-(\lambda^*(u))^{p-4}]\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2+b((\lambda^*(u))^2-1)\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & =\displaystyle b(\lambda^*(u))^2(1-(\lambda^*(u))^{p-6})\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & < 0. \end{align*}

Consequently, $\lambda ^*(u)\leq 1$. Similarly, we can reach a contradiction for the case of $\lambda ^*(u)<1$. As a result, $\lambda ^*(u)=1$.

Finally, property (vii) follows easily from the definition of Schwarz's symmetrization. For (viii), given $\lambda >0$, the operator $u\mapsto \mathcal {P}(\lambda,\,u)$ is linear and strongly continuous $H^1(\mathbb {R}^2)\rightarrow H^1(\mathbb {R}^2)$. Therefore, it is also weakly continuous. The $L^{p}(\mathbb {R}^2)$ continuity is immediate. Thus (viii) follows. This ends the proof.

Proof. The set $M$ is nonempty from lemma 3.1 (i). Let $u\in H^1(\mathbb {R}^2)$ such that $Q(u)<0$. By lemma 3.1 (ii) (iii), $f''(x)\leq 0$ on $(\lambda ^*(u),\, 1)$. Therefore,

\[ f(x)=f(1)+f'(1)(x-1)+\frac{f''(\xi)}{2}(x-1)^2\leq f(1)+f'(1)(x-1), \]

which means

\[ f(1)\geq f(\lambda^*(u))+f'(1)(1-\lambda^*(u)). \]

Applying lemma 3.1 (vi), we obtain

\begin{align*} S(u)& \geq f(\lambda^*(u))+(1-\lambda^*(u))Q(u)\\ & \geq f(\lambda^*(u))+Q(u). \end{align*}

Noting that $\lambda ^*(u)\in M$, we have $f(\lambda ^*(u))=S(\lambda ^*(u))\geq m$, and so

\[ S(u)\geq m+Q(u), \]

which completes the proof.

Proof. Proof of theorem 1.2

We proceed in three steps.

Step 1. We claim that the minimization problem (3.4) has a solution.

Since $M\neq \emptyset$, $S$ has a minimizing sequence $\{v_n\}$. In particular, $Q(v_n)=0$ and $S(v_n)\rightarrow m$. Let $w_n=|v_n|^*$, and $u_n=\mathcal {P}(\lambda ^*(w_n),\,w_n)$. It follows from lemma 3.1 (i) that $u_n\in M$. Furthermore, it follows from lemma 3.1 (vii) that $u_n=|\mathcal {P}(\lambda ^*(w_n),\,v_n)|^*$. Therefore,

\begin{align*} S(u_n)& \leq S(\mathcal{P}(\lambda^*(w_n),v_n))\leq S(\mathcal{P}(\lambda^*(v_n),v_n)) \ (\mathrm{by~Lemma~} 3.1\ (\mathrm{v}))\\ & \leq S(v_n) \ (\mathrm{since} \ \lambda^*(v_n)=1). \end{align*}

In particular, $\{u_n\}$ is a nonnegative, spherically symmetric, nonincreasing minimizing sequence of $S$. Furthermore, note that as $n\rightarrow \infty$

\begin{align*} m\displaystyle\leftarrow S(u_n)-\frac{1}{p-2}Q(u_n)& =\displaystyle\left(\frac{1}{2}-\frac{1}{p-2}\right)\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x \\ & \quad +b\left(\frac{1}{4}-\frac{1}{p-2}\right)\left(\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x\right)^2+\frac{\omega}{2}\int_{\mathbb{R}^2}|u_n|^2\,{\rm d}x\\ & \geq \displaystyle\left(\frac{1}{2}-\frac{1}{p-2}\right)\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x+\frac{\omega}{2}\int_{\mathbb{R}^2}|u_n|^2\,{\rm d}x, \end{align*}

which implies that $\{u_n\}$ is bounded in $H^1(\mathbb {R}^2)$. Since $Q(u_n)=0$, then

(3.5)\begin{equation} \frac{p}{p-2}\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x\leq\int_{\mathbb{R}^2}|u_n|^{p}\,{\rm d}x. \end{equation}

Therefore, by the Gagliardo–Nirenberg inequality and the boundedness of $\{u_n\}$ in $L^2(\mathbb {R}^2)$, there exists a constant $C>0$ such that

\[ \int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x\leq C\left(\int_{\mathbb{R}^2}|\nabla u_n|^2\,{\rm d}x\right)^{\frac{p-2}{2}}. \]

We obtain that $|\nabla u_n|_{L^2}$ is bounded from below, and so it follows from (3.5) that

(3.6)\begin{equation} \int_{\mathbb{R}^2}|u_n|^{p}\,{\rm d}x\geq c, \ \mathrm{for~some}\ c>0. \end{equation}

Therefore, there exist $v\in H^1(\mathbb {R}^2)$ and a subsequence, which we still denoted by $\{u_n\}$, such that $u_n\rightharpoonup v$ in $H^1(\mathbb {R}^2)$ weakly and in $L^{p}(\mathbb {R}^2)$ strongly ($v\neq 0$, by (3.6)). Therefore, we may define $\zeta =\mathcal {P}(\lambda ^*(v),\,v)=\lambda ^*(v)v(\lambda ^*(v)x)$. By lemma 3.1 (i), it follows directly that $\zeta \in M$, and hence

\[ S(\zeta)\geq m. \]

On the other hand, according to lemma 3.1 (viii), we know that $\mathcal {P}(\lambda ^*(v),\,u_n)\rightharpoonup \zeta$ in $H^1(\mathbb {R}^2)$ weakly and in $L^{p}(\mathbb {R}^2)$ strongly. Therefore,

\begin{align*} S(\zeta)& \leq \displaystyle\liminf_{n\rightarrow\infty} S(\mathcal{P}(\lambda^*(v),u_n)) \\ & \leq \displaystyle\liminf_{n\rightarrow\infty} S(\mathcal{P}(\lambda^*(u_n),u_n))\\ & = \displaystyle\liminf_{n\rightarrow\infty} S(u_n)=m \ (\mathrm{since}\ \lambda^*(u_n)=1). \end{align*}

Consequently, we obtain $S(\zeta )=m$. Thus, $\zeta$ is the solution of (3.4).

Step 2. We claim that every solution of (3.4) satisfies equation (1.1).

Consider any solution $u$ of (3.4). We have

\[ S(u)=m. \]

Then $\langle S'(u),\,u\rangle _{H^{-1},\,H^1}=0$, where $S'$ is the gradient of the $C^1$ functional $S$, i.e.,

\[ S'(u)={-}\left(1+b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)\Delta u+\omega u-|u|^{p-2}u. \]

Notice that

\[ Q'(u)={-}\left(2+4b\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)\Delta u-(p-2)|u|^{p-2}u. \]

It follows from $u\in M$ that

\begin{align*} \langle Q'(u),u\rangle_{H^{{-}1},H^1}& = \displaystyle2\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x+4b\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & \quad -(p-2)\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x -pQ(u)\\ & =\displaystyle-(p-2)\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x-(p-4)b\left(\int_{\mathbb{R}^2}|\nabla u|^2\,{\rm d}x\right)^2\\ & < 0. \end{align*}

Finally, since $u$ solves (3.4), there exists a Lagrange multiplier $\lambda$ such that

\[ S'(u)=\lambda Q'(u). \]

Thus,

\[ 0=\langle S'(u),u\rangle_{H^{{-}1},H^1}=\lambda \langle Q'(u),u\rangle_{H^{{-}1},H^1}. \]

Noting that $\langle Q'(u),\,u\rangle _{H^{-1},\,H^1}<0$, we obtain $\lambda =0$. Consequently, $S'(u)=0$, which means that $u$ solves problem (1.1).

Step 3. Conclusion.

Consider

(3.7)\begin{equation} l=\min\{S(u): u\in A\}. \end{equation}

Let $u\in G$ be such that $S(u)=l$. Now we claim that $u\in M$. Indeed, since $u$ is a solution of equation (1.1), we have $V(u)=0$ and

\[ T(u)=\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x-\omega\int_{\mathbb{R}^2}|u|^2\,{\rm d}x. \]

Thus

\begin{align*} Q(u)& = \displaystyle T(u)-\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & = \displaystyle\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x-\omega\int_{\mathbb{R}^2}|u|^2\,{\rm d}x-\frac{p-2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x\\ & =\displaystyle\frac{2}{p}\int_{\mathbb{R}^2}|u|^{p}\,{\rm d}x-\omega\int_{\mathbb{R}^2}|u|^2\,{\rm d}x\\ & =\displaystyle\frac{1}{2}V(u)\\ & = 0. \end{align*}

Therefore, $u\in M$, which implies that $S(u)\geq m$. In particular

(3.8)\begin{equation} l=S(u)\geq m. \end{equation}

Consider now a solution $v$ of (3.4). Since $S(v)=m$ and $v\in A$ (by Step 2), it follows from (3.7) that $m\geq l$. Combining with (3.8), we obtain $m=l$. The equivalence of the two problems follows easily. The proof is complete.