Proof. Let Q be the unique positive radial solution of (1.3). For $\gamma>0$ , define

$$ \begin{align*} \varphi_\gamma(x):=\frac{\gamma Q (\gamma x)}{\|Q\|_{2}}, \end{align*} $$

so that $\|\varphi _\gamma (x)\|^2_{2}=1$ . Since $\|\nabla Q\|^2_{2}=\tfrac 12\|Q\|^4_{4}=a_*$ (by (2.2)), then,

(2.3) $$ \begin{align} E_a(\varphi_\gamma)&=\frac{1}{2}\int_{\mathbb{R}^2}|\nabla \varphi_\gamma|^2\,dx-\frac{a}{4}\int_{\mathbb{R}^2}|\varphi_\gamma|^4\,dx+\frac{a^2}{6}\int_{\mathbb{R}^2}|\varphi_\gamma|^6\,dx\nonumber\\ &=\frac{\gamma^2}{2}\bigg(1-\frac{a}{a_*} \bigg)+\frac{a^2\gamma^{4}}{6a_*^3}\int_{\mathbb{R}^2}|Q|^6\,dx. \end{align} $$

By letting $\gamma \to 0^+$ , we deduce that $e(a)\leq 0$ .

To prove that $e(a)=0$ if and only if $0<a\leq a_*$ , we just need to show that for ${0<a\leq a_*}$ , we have $E_a(\varphi )\geq 0$ for any $\varphi \in H^1(\mathbb {R}^2)$ . We deduce from the Gagliardo– Nirenberg inequality (2.1) that

$$ \begin{gather*} \begin{aligned} E_a(\varphi)=\frac{1}{2}\int_{\mathbb{R}^2}|\nabla\varphi|^2\,dx-\frac{a}{4}\int_{\mathbb{R}^2}|\varphi|^4\,dx+\frac{a^2}{6}\int_{\mathbb{R}^2}|\varphi|^6\,dx \geq\frac{a^2}{6}\int_{\mathbb{R}^2}|\varphi|^6\,dx\geq0. \end{aligned} \end{gather*} $$

Thus, $e(a)=0$ if and only if $0<a\leq a_*$ .

Next, we claim that for any $a>a_*$ , we have $e(a)<0$ . By (2.3),

$$ \begin{gather*} \begin{aligned} e(a)\leq\frac{\gamma^2}{2}\bigg(1-\frac{a}{a_*} \bigg)+\frac{a^2\gamma^{4}}{6a_*^3}\int_{\mathbb{R}^2}|Q|^6\,dx =:-A(a-a_*)\gamma^2+B\gamma^{4}, \end{aligned} \end{gather*} $$

where

$$ \begin{align*}A=\frac{1}{2a_*}>0\quad\text{and}\quad B=\frac{a^2}{6a_*^3}\int_{\mathbb{R}^2}|Q|^6\,dx>0.\end{align*} $$

Now, let $\gamma =C_0(a-a_*)^{1/2}$ , taking $C_0$ small enough so that $AC_0^2-BC_0^4> 0$ . Then,

(2.4) $$ \begin{align} e(a)\leq-( AC_0^2-BC_0^4)(a-a_*)^{2}\lesssim-(a-a_*)^{2}<0 \end{align} $$

for any $a>a_*$ . This completes the proof of the Lemma 2.1.

Proof of Theorem 1.1.

Part (2) of Theorem 1.1 comes from [Reference Carles and Sparber1], or it can be proved by the standard concentration-compactness principle [Reference Lions12].

Next, we prove that there is no minimiser for (1.2) with $0<a\leq a_*=\|Q\|_{2}^2$ . Suppose that there exists a minimiser $\varphi _a$ with $0<a\leq a_*$ . As pointed out in Section 1, we can assume $\varphi _a$ to be nonnegative. We deduce from the Gagliardo–Nirenberg inequality (2.1) and $e(a)=0$ that

$$ \begin{align*} \frac{1}{2}\int_{\mathbb{R}^2}|\nabla \varphi_a|^2\,dx=\frac{a}{4} \int_{\mathbb{R}^2}|\varphi_a|^4\,dx \end{align*} $$

and

$$ \begin{align*} \int_{\mathbb{R}^2}|\varphi_a|^6\,dx=0 \end{align*} $$

for $0<a\leq a_*$ . This implies $\varphi _a=0$ a.e., which is a contradiction with $\| \varphi _a \|^2_2 =1$ . This completes the proof of the first part of Theorem 1.1.

To prove the stated properties of the energy $e(a)$ , note that Lemma 2.1 implies that $e(a)=0$ for $0<a\leq a_*=\|Q\|_{2}^2$ . We have already shown that $e(a)\lesssim -(a-a_*)^{2}$ for ${a>a_*}$ in (2.4), hence it remains to show that $\lim _{a\searrow a_*}e(a)=e(a_*)=0$ for $a>a_*$ . This will complete the proof of Theorem 1.1.

Proof. In view of (2.4), we just need to prove the lower bound. First, for any ${\varphi \in H^1(\mathbb {R}^2)}$ , by using the Hölder’s inequality and Young’s inequality with $\epsilon $ ,

(3.4) $$ \begin{align} \int_{\mathbb{R}^2}|\varphi|^4\,dx&\leq\bigg(\int_{\mathbb{R}^2}|\varphi|^2\,dx \bigg)^{{1}/{2}} \bigg( \int_{\mathbb{R}^2}|\varphi|^6\,dx\bigg)^{{1}/{2}}\nonumber\\ &\leq \frac{3(a-a_*)}{8a^2}\int_{\mathbb{R}^2}|\varphi|^2\,dx+ \frac{2a^2(a-a_*)^{-1}}{3}\int_{\mathbb{R}^2}|\varphi|^6\,dx. \end{align} $$

Then, for any $\varphi (x)\in H^1(\mathbb {R}^2)$ with $\|\varphi \|^2_{2}=1$ , by using the Gagliardo–Nirenberg inequality (2.1) and (3.4),

$$ \begin{gather*} \begin{aligned} E_a(\varphi) &=\frac{1}{2}\int_{\mathbb{R}^2}|\nabla \varphi|^2\,dx-\frac{a}{4}\int_{\mathbb{R}^2}|\varphi|^4\,dx+\frac{a^2}{6}\int_{\mathbb{R}^2}|\varphi|^6\,dx\\ &\geq-\frac{a-a_*}{4}\int_{\mathbb{R}^2}|\varphi|^4\,dx+\frac{a^2}{6}\int_{\mathbb{R}^2}|\varphi|^6\,dx\\ &\gtrsim-(a-a_*)^{2}. \end{aligned} \end{gather*} $$

This completes the proof of Lemma 3.1.

Proof. From the Gagliardo–Nirenberg inequality (2.1),

(3.6) $$ \begin{align} \frac{1}{2}\int_{\mathbb{R}^2}|\nabla \varphi_a(x)|^2\,dx-\frac{a}{4}\int_{\mathbb{R}^2}|\varphi_a(x)|^4\,dx\geq-\frac{a-a_*}{4}\int_{\mathbb{R}^2}|\varphi_a(x)|^4\,dx. \end{align} $$

However, by the definition of $e(a)$ and Lemma 3.1,

(3.7) $$ \begin{align} \frac{1}{2}\int_{\mathbb{R}^2}|\nabla \varphi_a(x)|^2\,dx-\frac{a}{4}\int_{\mathbb{R}^2}|\varphi_a(x)|^4\,dx\leq E_a(\varphi_a)=e(a)\lesssim-(a-a_*)^2. \end{align} $$

From the inequalities (3.6) and (3.7),

$$ \begin{gather*} \begin{aligned} \int_{\mathbb{R}^2}|\varphi_a(x)|^4\,dx\gtrsim(a-a_*). \end{aligned} \end{gather*} $$

Moreover, by the Gagliardo–Nirenberg inequality (2.1) and (3.7),

$$ \begin{align*} \int_{\mathbb{R}^2}|\nabla \varphi_a(x)|^2\,dx\gtrsim\int_{\mathbb{R}^2}|\varphi_a(x)|^4\,dx\gtrsim(a-a_*). \end{align*} $$

From (3.1), (3.2) and Lemma 3.1,

(3.8) $$ \begin{align} \frac{a^2}{6}\int_{\mathbb{R}^2}|\varphi_a|^6\,dx=-e(a)\sim(a-a_*)^2. \end{align} $$

By Hölder’s inequality together with (3.7) and (3.8),

$$ \begin{gather*} \begin{aligned} \int_{\mathbb{R}^2}|\nabla \varphi_a(x)|^2\,dx\lesssim\int_{\mathbb{R}^2}|\varphi_a(x)|^4\,dx\lesssim\bigg( \int_{\mathbb{R}^2}|\varphi_a|^6\,dx\bigg)^{{1}/{2}} \lesssim(a-a_*). \end{aligned} \end{gather*} $$

This completes the proof of the lemma.

Proof of Theorem 1.3.

First, (1.4) in Theorem 1.3 comes from (3.5). Next, we prove (1.5). Note that $\{w_\tau \}$ is radially symmetric, since $\varphi _a$ is radially symmetric (see Remark 1.2). By (3.9), $\{w_\tau \}$ is uniformly bounded in $H_{\mathrm {rad}}^1(\mathbb {R}^2)$ and there exists a subsequence $\{w_{\tau _k}\}$ such that $w_{\tau _k}\rightharpoonup w_0$ weakly in $H_{{\mathrm { rad}}}^1(\mathbb {R}^2)$ , where $H_{{\mathrm {rad}}}^1( \mathbb {R}^2)$ denotes the Sobolev space of radial $H^1( \mathbb {R}^2)$ functions. For $2<p<+\infty $ , the embedding $H_{{\mathrm {rad}}}^1(\mathbb {R}^2)\hookrightarrow L^p(\mathbb {R}^2)$ is compact, so $w_{\tau _k}\to w_0$ strongly in $L^p(\mathbb {R}^2)$ . This implies that

(3.12) $$ \begin{align} \int_{\mathbb{R}^2}|w_{\tau_k}|^4\,dx\to\int_{\mathbb{R}^2} |w_0|^4\,dx\quad\text{and}\quad\int_{\mathbb{R}^2}|w_{\tau_k}|^6\,dx\to\int_{\mathbb{R}^2}|w_0|^6\,dx. \end{align} $$

By the Pohozaev identity (3.2), $w_{\tau _k}(x)$ satisfies

$$ \begin{gather*} \begin{aligned} \int_{\mathbb{R}^2}|\nabla w_{\tau_k}|^2\,dx-\frac{a_k}{2}\int_{\mathbb{R}^2}|w_{\tau_k}|^4\,dx+\frac{2a_k^2}{3}\int_{\mathbb{R}^2}|w_{\tau_k}|^6\,dx=0 \end{aligned} \end{gather*} $$

and it follows from (3.12) that

(3.13) $$ \begin{align} \lim_{k\to\infty}\int_{\mathbb{R}^2}|\nabla w_{\tau_k}|^2\,dx= \frac{a_*}{2}\int_{\mathbb{R}^2}|w_{0}|^4\,dx+\frac{2a_*^2}{3}\int_{\mathbb{R}^2}|w_{0}|^6\,dx. \end{align} $$

By passing to the weak limit $\tau _k\to 0^+$ in (3.10), we see that $w_0(x)$ satisfies

(3.14) $$ \begin{align} -\Delta w_0(x)=-\beta^2 w_0(x)+a_*w_0^3(x)-a_*^2w_0^5(x). \end{align} $$

We also have the Pohozaev identity (see [Reference Carles and Sparber1]),

(3.15) $$ \begin{align} \beta^2\int_{\mathbb{R}^2}|w_0 |^2\,dx-\frac{a_*}{2}\int_{\mathbb{R}^2}|w_0 |^4\,dx+\frac{a_*^2}{3}\int_{\mathbb{R}^2}|w_0|^6\,dx=0, \end{align} $$

where $\beta ^2\in (0,3/16)$ since $w_0\not = 0$ (see [Reference Carles and Sparber1]). From (3.14) and (3.15),

(3.16) $$ \begin{align} \int_{\mathbb{R}^2}|\nabla w_{0}|^2\,dx=\frac{a_*}{2}\int_{\mathbb{R}^2}|w_0|^4\,dx-\frac{2a_*^2}{3}\int_{\mathbb{R}^2}|w_0|^6\,dx. \end{align} $$

It follows from (3.13) that

(3.17) $$ \begin{align} \lim_{k\to\infty}\int_{\mathbb{R}^2}|\nabla w_{\tau_k}|^2\,dx=\int_{\mathbb{R}^2}|\nabla w_0|^2\,dx. \end{align} $$

However, by (3.11) together with (3.12) and (3.17),

$$ \begin{gather*} \begin{aligned} \lim_{k\to\infty} \int_{\mathbb{R}^2}|w_{\tau_k}|^2\,dx&=\lim_{k\to\infty}\frac{1}{\tau_k^2\lambda_{a_k}}\bigg( \int_{\mathbb{R}^2}|\nabla w_{\tau_k}|^2\,dx-a_k \int_{\mathbb{R}^2}|w_{\tau_k}|^4\,dx+ a_k^2\int_{\mathbb{R}^2}|w_{\tau_k}|^6\,dx\bigg) \\ &=-\frac{1}{\beta^2}\bigg( \int_{\mathbb{R}^2}|\nabla w_{0}|^2\,dx-a_* \int_{\mathbb{R}^2}|w_{0}|^4\,dx+a_*^2 \int_{\mathbb{R}^2}|w_{0}|^6\,dx\bigg)\\ &=\int_{\mathbb{R}^2}|w_0|^2\,dx. \end{aligned} \end{gather*} $$

Combining this with (3.17) shows $w_{\tau _k}\to w_0$ strongly in $H^1(\mathbb {R}^2)$ and this yields (1.5).

By (3.8) and (3.12),

(3.18) $$ \begin{align} \lim_{k\to\infty}\tau_k^4e(a_k)&=-\lim_{k\to\infty} \frac{\tau_k^4a_k^2}{6}\int_{\mathbb{R}^2}|\varphi_{a_k}|^6\,dx \nonumber\\ &=-\lim_{k\to\infty}\frac{a_k^2}{6}\int_{\mathbb{R}^2}|w_{\tau_k}|^6\,dx =-\frac{a_*^2}{6}\int_{\mathbb{R}^2}|w_{0}|^6\,dx. \end{align} $$

Finally, by applying the same argument as we used before to (3.18), we can take a subsequence $\{\tau _k\}$ with $\tau _k\to +\infty $ as $k\to +\infty $ , such that

(3.19) $$ \begin{align} \liminf_{a\searrow a_*}\tau^4e(a)=\lim_{k\to\infty}\tau_k^4e(a_k)=-\frac{a_*^2}{6}\int_{\mathbb{R}^2}|w_{0}|^6\,dx, \end{align} $$

where $w_0(x)$ satisfies (3.15) with some $\beta ^2\in (0,3/16)$ and $\int _{\mathbb {R}^2}|w_0(x)|^2\,dx=1$ . However, taking the test function $\phi _\tau =\tau ^{-1}w_0(\tau ^{-1}x)$ in $E_a(\cdot )$ , we deduce from (3.16) that

$$ \begin{gather*} \begin{aligned} \tau^4e(a)\leq \tau^4 E_a(\phi_\tau)&=\frac{1}{2}\int_{\mathbb{R}^2}|\nabla w_{0}|^2\,dx-\frac{a}{4}\int_{\mathbb{R}^2}|w_0|^4\,dx+\frac{a^2}{6}\int_{\mathbb{R}^2}|w_0|^6\,dx\\ &=\frac{a_*-a}{4}\int_{\mathbb{R}^2}|w_0|^4\,dx+\frac{a^2-2a_*^2}{6}\int_{\mathbb{R}^2}|w_0|^6\,dx. \end{aligned} \end{gather*} $$

This means that

$$ \begin{gather*} \begin{aligned} \limsup_{a\searrow a_*}\tau^4e(a)\leq-\frac{a_*^2}{6}\int_{\mathbb{R}^2}|w_{0}|^6\,dx. \end{aligned} \end{gather*} $$

Combining this with (3.19) gives (1.6). This completes the proof of Theorem 1.3.