Proof. By (1.9) and (1.10), $(\widetilde {v}_\mu, \widetilde {t}_\mu )$ is a solution of (1.11). In particular, $\widetilde {v}_\mu$ is a solution of (1.4) for $\lambda =1$ and $t=\widetilde {t}_\mu =\mu \widetilde {\lambda }_\mu ^{\frac {q\gamma _q-q}{2}}$. By the well-known Gidas–Ni–Nirenberg theorem [Reference Gidas, Ni and Nirenberg17], $\widetilde {v}_{\mu }$ is radial and decreasing for $r=|x|$ up to translations. Thus, without loss of generality, we may assume that $\widetilde {v}_{\mu }(0)=\max _{x\in \mathbb {R}^{N}}\widetilde {v}_{\mu }$. Recall that $\varepsilon _{\mu }^{\frac {1}{2}}\widetilde {u}_\mu (\varepsilon _{\mu } x)\to U_{\varepsilon _0}$ strongly in $D^{1,2}(\mathbb {R}^{3})$ for some $\varepsilon _0>0$ as $\mu \to 0$ up to a subsequence, by the classical elliptic regularity and the Sobolev embedding theorem, $\varepsilon _{\mu }^{\frac {1}{2}}\widetilde {u}_\mu (\varepsilon _{\mu } x)\to U_{\varepsilon _0}$ strongly in $C_{loc}^{1,\alpha }(\mathbb {R}^{3})$ for some $\alpha \in (0, 1)$ as $\mu \to 0$ up to a subsequence. In particular, $\varepsilon _{\mu }^{\frac {1}{2}}\widetilde {u}_\mu (0)\to U_{\varepsilon _0}(0)$ as $\mu \to 0$ up to a subsequence. Thus, by (2.1),

(2.5)\begin{equation} \widetilde{v}_{\mu}(0)=\widetilde{\lambda}_\mu^{-\frac{1}{4}}\widetilde{u}_\mu(0)\sim\widetilde{\lambda}_\mu^{-\frac{1}{4}}\varepsilon_\mu^{-\frac{1}{2}}\quad\text{as }\mu\to 0\text{ up to a subsequence}. \end{equation}

In the following, let us estimate $\widetilde {v}_{\mu }(0)$ and $\widetilde {t}_\mu$ as $\mu \to 0$. We begin with the estimate of $\widetilde {t}_\mu$. We first consider the case $2< q<3$. In this case, by (2.2), (2.3) and (2.4), $\widetilde {\lambda }_\mu \sim \varepsilon _\mu ^{2},$ which, together with (1.10), implies

\[\widetilde{t}_\mu\sim\varepsilon_\mu^{\frac{10-3q}{2}}(\varepsilon_\mu^{2})^{\frac{q-6}{4}}=\varepsilon_\mu^{2-q}\to+\infty\quad\text{as }\mu\to 0.\]

For $q=3$, by (2.2), (2.3) and (2.4),

\[\widetilde{\lambda}_\mu\sim\varepsilon_\mu^{2}\frac{\ln(\frac{1}{\sqrt{\widetilde{\lambda}_\mu}\varepsilon_\mu})}{\ln(\frac{1}{\varepsilon_\mu})}\gtrsim\varepsilon_\mu^{2}.\]

It follows that

\[\ln\left(\frac{1}{\varepsilon_\mu}\right)\lesssim\ln\left(\frac{1}{\sqrt{\widetilde{\lambda}_\mu}\varepsilon_\mu}\right)\lesssim\ln(\frac{1}{\varepsilon_\mu}).\]

Thus, we also have $\widetilde {\lambda }_\mu \sim \varepsilon _\mu ^{2}$ for $q=3$. By (1.10) and (2.2),

\[\widetilde{t}_\mu\sim\varepsilon_\mu^{\frac{1}{2}}\frac{1}{\ln(\frac{1}{\varepsilon_\mu})}(\varepsilon_\mu^{2})^{-\frac{3}{4}}=\varepsilon_\mu^{{-}1}\frac{1}{\ln(\frac{1}{\varepsilon_\mu})}\to+\infty\quad\text{as }\mu\to 0.\]

For $3< q<4$, by (2.2), (2.3) and (2.4), $\widetilde {\lambda }_\mu \sim \varepsilon _\mu ^{2}.$ Now, by (1.10),

\[\widetilde{t}_\mu\sim\varepsilon_\mu^{\frac{q-2}{2}}(\varepsilon_\mu^{2})^{\frac{q-6}{4}}=\varepsilon_\mu^{q-4}\to+\infty\quad\text{as }\mu\to 0.\]

Thus, for all $2< q<4$, we always have

(2.6)\begin{equation} \widetilde{\lambda}_\mu\sim\varepsilon_\mu^{2}\quad\text{and}\quad\widetilde{t}_\mu\sim \left\{\begin{aligned} & \varepsilon_\mu^{q-4},\quad 3< q<4,\\ & \varepsilon_\mu^{{-}1}\frac{1}{\ln(\frac{1}{\varepsilon_\mu})},\quad q=3,\\ & \varepsilon_\mu^{2-q},\quad 2< q<3, \end{aligned}\right. \end{equation}

as $\mu \to 0$. Now, by (2.2)–(2.4) and (2.5), we have

\[\widetilde{v}_{\mu}(0)\sim\left\{\begin{aligned} & \mu^{-\frac{2}{q-2}},\quad 3< q<4,\\ & \left(\frac{1}{\mu|\ln\mu|}\right)^{2},\quad q=3,\\ & \mu^{-\frac{2}{10-3q}},\quad 2< q<3. \end{aligned}\right.\]

It follows from (2.2) and (2.6) that

\[\widetilde{v}_{\mu}(0)\sim\left\{\begin{aligned} & \widetilde{t}_\mu^{\frac{1}{4-q}},\quad 3< q<4,\\ & \widetilde{t}_\mu\ln \widetilde{t}_\mu,\quad q=3,\\ & \widetilde{t}_\mu^{\frac{1}{q-2}},\quad 2< q<3.\end{aligned}\right.\]

Thus, by (2.6), $\widetilde {v}_{\mu }$ is a blow-up solution of (1.4) for $N=3$, $\lambda =1$, $2< q<4$ and $t=\widetilde {t}_\mu$. Note that by [Reference Ma and Moroz25, theorem 2.2], the ground-states of (1.4) for $\lambda =1$, say $\overline {v}_{t}$, satisfies $\|\overline {v}_{t}\|_\infty \sim t^{-\frac {1}{q-2}}$ as $t\to +\infty$. For $\mu >0$ sufficiently small, $\widetilde {v}_{\mu }$ is a second positive radial solution of (1.4) with $N=3$, $\lambda =1$, $2< q<4$ and $t>0$ sufficiently large.

Proof. We argue in the contrary by supposing that there exists $t_n\to 0$ as $n\to \infty$ such that $m(t_n)<\frac {1}{3}S^{\frac {3}{2}}$. Then, it is standard to show (cf.[Reference Alves, Souto and Montenegro6]) that $m(t_n)$ is attained by a positive and radial function, which is also a solution of (1.4) with $\lambda =1$, $N=3$ and $t=t_n$. We denote this solution by $v_{t_n}$. Since $t_n\to 0$ as $n\to \infty$, it is also standard to show that

(3.2)\begin{equation} \|\nabla v_{t_n}\|_2^{2}=\|v_{t_n}\|_6^{6}+o_n(1)=S^{\frac{3}{2}}+o_n(1)\quad\text{as }n\to\infty. \end{equation}

Thus, $\{v_{t_n}\}$ is a minimizing sequence of the Sobolev inequality. By Lions’ result (cf. [Reference Willem34, theorem 1.41]), up to a subsequence, there exists $\sigma _n>0$ such that for some $\varepsilon _*>0$,

\[w_{t_n}(x)=\sigma_n^{\frac{1}{2}}v_{t_n}(\sigma_n x)\to U_{\varepsilon_*}\text{ strongly in }D^{1,2}(\mathbb{R}^{3})\text{ as }n\to\infty.\]

Clearly, by direct computations, we know that $w_{t_n}$ satisfies the following equation:

(3.3)\begin{equation} -\Delta w_{t_n}+\sigma_n^{2}w_{t_n}=t_n\sigma_n^{3-\frac{q}{2}}w_{t_n}^{q-1}+w_{t_n}^{5}\quad\text{in }\mathbb{R}^{3}. \end{equation}

Since $v_{t_n}$ is positive and radial, $w_{t_n}$ is also positive and radial. Thus, by the boundedness of $\{w_{t_n}\}$ in $D^{1,2}(\mathbb {R}^{3})$, the Sobolev embedding theorem and Strusss radial lemma (cf. [Reference Berestycki and Lions8, lemma A.2]),

\[w_{t_n}\lesssim r^{-\frac{1}{2}}\quad\text{for all }r\geq 1\text{ uniformly as }n\to\infty.\]

On the other hand, since $w_{t_n}\to U_{\varepsilon _*}$ strongly in $D^{1,2}(\mathbb {R}^{3})$ as $n\to \infty$, by applying the Moser iteration in a standard way and using the Sobolev embedding theorem, we know that $w_{t_n}\to U_{\varepsilon _*}$ strongly in $C_{loc}^{1,\alpha }(\mathbb {R}^{3})$ as $n\to \infty$ for some $\alpha \in (0, 1)$. Thus,

\[w_{t_n}\lesssim (1+r)^{-\frac{1}{2}}\quad\text{for all }r\geq 0\text{ uniformly as }n\to\infty.\]

Now, we can adapt the ODE's argument in [Reference Atkinson and Peletier5, Reference Gazzola and Serrin18, Reference Knaap and Peletier21] as that in the proof of [Reference Wei and Wu33, lemma 4.1] to obtain

(3.4)\begin{equation} w_{t_n}\lesssim\frac{1}{(1+r^{2})^{\frac 12}}\quad\text{for all }r\geq 0\text{ uniformly as }n\to\infty. \end{equation}

On the other hand, since $N=3$, it is easy to check that $r^{-1}e^{-\sigma _nr}$ is a subsolution of $-\Delta u+\sigma _n^{2}u=0$ for $r\geq 1$. Thus, by the fact that $w_{t_n}\to U_{\varepsilon _*}$ strongly in $C_{loc}^{1,\alpha }(\mathbb {R}^{3})$ as $n\to \infty$ for some $\alpha \in (0, 1)$, we can use the maximum principle in a standard way to show that

\[w_{t_n}\gtrsim r^{{-}1}e^{-\sigma_nr}\quad \text{for }r\geq 1\text{ uniformly as }n\to\infty.\]

It follows that

(3.5)\begin{equation} \|w_{t_n}\|_q^{q}\gtrsim\int_1^{\frac{1}{\sigma_n}}r^{2-q}e^{{-}q\sigma_nr}{\rm d}r\sim\left\{ \begin{aligned} & \sigma_n^{q-3},\quad 2\leq q<3,\\ & |\ln\sigma_n|,\quad q=3,\\ & 1,\quad 3< q<6. \end{aligned}\right. \end{equation}

Since $t_n\to 0$ as $n\to \infty$, by (3.4), for $r\gtrsim (\frac {1}{\sigma _n})^{\frac 12}$, (3.3) reads as

\[-\Delta w_{t_n}+\frac{1}{4}\sigma_n^{2}w_{t_n}\leq 0\quad\text{in }\mathbb{R}^{3}.\]

Thus, by (3.4), we can use the maximum principle in a standard way again to obtain

\[w_{t_n}\lesssim r^{{-}1}e^{-\frac{\sigma_n}{4}r}\quad \text{for }r\gtrsim\left(\frac{1}{\sigma_n}\right)^{\frac 12}\text{ uniformly as }n\to\infty.\]

On the other hand, since $\|w_{t_n}\|_6^{6}=\|v_{t_n}\|_6^{6}=S^{\frac {3}{2}}+o_n(1)$, by (3.8) and the Hölder inequality,

\[\sigma_n^{2}\|w_{t_n}\|_2^{2}\lesssim t_n\sigma_n^{3-\frac{q}{2}}\|w_{t_n}\|_q^{q}\lesssim t_n\sigma_n^{3-\frac{q}{2}}\|w_{t_n}\|_2^{\frac{6-q}{2}},\]

which implies

\[\sigma_n\|w_{t_n}\|_2\lesssim t_n^{\frac{2}{q-2}}.\]

Since $w_{t_n}\to U_{\varepsilon _*}$ strongly in $D^{1,2}(\mathbb {R}^{3})$ as $n\to \infty$ and $U_{\varepsilon _*}\not \in L^{2}(\mathbb {R}^{3})$, by the Fatou lemma,

\[\liminf_{n\to\infty}\|w_{t_n}\|_2={+}\infty.\]

Thus, by $t_n\to 0$ as $n\to \infty$, we have $\sigma _n\to 0$ as $n\to \infty$. It follows from (3.4) once more that

(3.6)\begin{equation} \|w_t\|_q^{q}\lesssim 1+\int_1^{\frac{1}{\sigma_n}}r^{2-q}{\rm d}r+\int_{(\frac{1}{\sigma_n})^{\frac{1}{2}}}^{+\infty}r^{2-q}e^{-\frac{q}{4}\sigma_nr}{\rm d}r\sim \left\{\begin{aligned} & \sigma_n^{q-3},\quad 2\leq q<3,\\ & |\ln\sigma_n|,\quad q=3,\\ & 1,\quad 3< q<6. \end{aligned}\right. \end{equation}

Thus, by (3.5) and (3.6), we have

(3.7)\begin{equation} \|w_{t_n}\|_q^{q}\sim\left\{\begin{aligned} & \sigma_n^{q-3},\quad 2\leq q<3,\\ & |\ln\sigma_n|,\quad q=3,\\ & 1,\quad 3< q<6. \end{aligned}\right. \end{equation}

Note that as that of (1.8), by the Pohozaev identity, we have

(3.8)\begin{equation} \sigma_n^{2}\|w_{t_n}\|_2^{2}=(1-\gamma_q)t_n\sigma_n^{3-\frac{q}{2}}\|w_{t_n}\|_q^{q}. \end{equation}

Thus, by (3.7),

\[\sigma_n\sim \left\{\begin{aligned} & t_n\sigma_n^{\frac{q}{2}},\quad 2< q<3,\\ & t_n\sigma_n^{\frac{3}{2}}|\ln\sigma_n|,\quad q=3,\\ & t_n\sigma_n^{3-\frac{q}{2}},\quad 3< q<6, \end{aligned}\right.\]

which implies

(3.9)\begin{equation} t_n\sim \left\{\begin{aligned} & \sigma_n^{\frac{2-q}{2}},\quad 2< q<3,\\ & \sigma_n^{-\frac{1}{2}}\frac{1}{|\ln\sigma_n|},\quad q=3,\\ & \sigma_n^{\frac{q-4}{2}},\quad 3< q<6. \end{aligned}\right. \end{equation}

(3.9) contradicts the facts that $t_n,\sigma _n\to 0$ as $n\to \infty$ for $2< q\leq 4$. It follows that $m(t)\geq \frac {1}{3}S^{\frac {3}{2}}$ for $t>0$ sufficiently small in the case of $2< q\leq 4$. On the other hand, since $m(t)$ is the minimum of $\mathcal {E}_t(v)$ on the Nehari manifold $\mathcal {N}_{t}$, it is standard (cf. [Reference Wei and Wu33, lemma 3.3]) to use the fibering maps (3.1) to show that $m(t)$ is nonincreasing for $t>0$. Note that it is well known that $m(0)=\frac {1}{3}S^{\frac {3}{2}}$, thus, $m(t)\leq \frac {1}{3}S^{\frac {3}{2}}$ for all $t>0$. It follows that $m(t)=\frac {1}{3}S^{\frac {3}{2}}$ for $t>0$ sufficiently small in the case of $2< q\leq 4$.

Proof. The conclusion $\|v_t\|_q^{q}\lesssim 1$ as $t\to t_q^{*}$ is standard so we omit it. For the conclusion $\|v_t\|_q^{q}\gtrsim 1$ as $t\to t_q^{*}$, we argue in the contrary. Then there exists $t_n\to t_q^{*}$ as $n\to \infty$ such that $\|v_{t_n}\|_q^{q}\to 0$ as $n\to \infty$. Similar to that of (3.2), we also have

\[\|\nabla v_{t_n}\|_2^{2}=\|v_{t_n}\|_6^{6}+o_n(1)=S^{\frac{3}{2}}+o_n(1)\quad\text{as }n\to\infty.\]

Thus, $\{v_{t_n}\}$ is a minimizing sequence of the Sobolev inequality. By Lions’ result (cf. [Reference Willem34, theorem 1.41]), up to a subsequence, there exists $\sigma '_n>0$ such that for some $\varepsilon _*>0$,

\[w_n(x)=(\sigma'_n)^{\frac{1}{2}}v_{t_n}(\sigma'_n x)\to U_{\varepsilon_*}\text{ strongly in }D^{1,2}(\mathbb{R}^{3})\text{ as }n\to \infty.\]

Now, repeating the arguments for (3.9), we will arrive at

\[t_q^{*}\sim \left\{\begin{aligned} & (\sigma_n')^{\frac{2-q}{2}},\quad 2< q<3,\\ & (\sigma_n')^{-\frac{1}{2}}\frac{1}{|\ln\sigma'_n|},\quad q=3,\\ & (\sigma'_n)^{\frac{q-4}{2}},\quad 3< q<4. \end{aligned}\right.\]

This is impossible since $\sigma '_n\to 0$ as $n\to \infty$ by similar arguments as that used for $\sigma _n$ in the proof of lemma 3.1. Thus, we must have $\|v_t\|_q^{q}\gtrsim 1$ as $t\to t_q^{*}$.

Proof. We first prove that there is no ground-states of (1.4) for $0< t< t_q^{*}$ in the case of $2< q\leq 4$. Suppose the contrary that (1.4) has a ground-state for some $0< t< t_q^{*}$ in the case of $2< q\leq 4$. Then $m(t)$ is attained. Now, by use the fibering maps (3.1) in a standard way (cf. [Reference Wei and Wu33, lemma 3.3]), we have $m(t')< m(t)$ for all $t'>t$. It follows that $m(t')<\frac {1}{3}S^{\frac {3}{2}}$ for all $t'>t$, which contradicts the definition of $t_q^{*}$ given by (3.10). Thus, there is no ground-states of (1.4) for $0< t< t_q^{*}$ in the case of $2< q\leq 4$. It remains to prove that (1.4) has a ground-state for $t=t_q^{*}$ in the case of $2< q<4$, which is equivalent to prove that $m(t_q^{*})$ is attained for $2< q<4$. Let $v_t$ be a ground-state of (1.4), which is radial and positive for $t>t_q^{*}$ in the case of $2< q<4$ such that $t\to t_q^{*}$. By proposition 3.1, $\|v_t\|_q^{q}\gtrsim 1$ as $t\to t_q^{*}$. Since $v_t$ is radial, it is standard to show that $v_t\to v_{t_q^{*}}\not =0$ strongly in $H^{1}(\mathbb {R}^{3})$ as $t\to t_q^{*}$ up to a subsequence. Thus, $m(t_q^{*})$ is attained by $v_{t_q^{*}}$, which is also a ground-state of (1.4) for $t=t_q^{*}$ in the case of $2< q<4$.

Proof of theorem 1.2. It follows from propositions 2.1 and 3.2.

Proof of theorem 1.3. By [Reference Wei and Wu33, theorem 1.1] and [Reference Jeanjean and Le19, theorem 1.6], (1.7) has a normalized ground-state for $\mu >0$ sufficiently small. Thus, we only need to prove (1.7) has no normalized ground-states for $\mu >0$ sufficiently large, which is equivalent to show that $\overline {\mu }_{q,a}<+\infty$. Recall that $\gamma _q=\frac {N(q-2)}{2q}$, we always have $q>q\gamma _q$. It follows from (4.2) that $\mu _t\to 0$ as $t\to t_q^{**}$ for $N\geq 4$ since $t_q^{**}=0$ for $N\geq 4$. For $N=3$, we have $t_q^{**}=t_q^{*}>0$ and $\|v_t\|_q^{q}\sim 1$ as $t\to t_q^{*}$ by proposition 3.1. Thus, $\mu _t\lesssim 1$ as $t\to t_q^{**}$ for all $N\geq 3$. Since $v_t$ is a ground-state of (1.4) with the least energy $m(t)$ on the Nehari manifold $\mathcal {N}_t$, by standard arguments (cf. [Reference Chen and Zou11, lemma 2.2]),

(4.3)\begin{equation} m(t)=\frac{1}{N}S^{\frac{N}{2}}-\int_{t_q^{**}}^{t}\frac{1}{q}\|v_\tau\|_q^{q}d\tau\quad\text{for all }t>t_q^{**} \end{equation}

and

(4.4)\begin{equation} m'(t)={-}\frac{1}{q}\|v_t\|_q^{q}\quad\text{for a.e. }t>t_q^{**}. \end{equation}

As that of (1.8), by the Pohozaev identity, we have

(4.5)\begin{equation} \|\nabla v_t\|_2^{2}=\gamma_qt\|v_t\|_q^{q}+\|v_t\|_{2^{*}}^{2^{*}}\quad\text{and}\quad \|\nabla v_t\|_2^{2}=Nm(t). \end{equation}

Thus, by (4.3) and (4.4),

\[Nm(t)+q\gamma_qm'(t)t\geq 0\quad\text{ for a.e. }t>t_q^{**},\]

which implies $m(t)t^{\frac {N}{q\gamma _q}}$ is increasing for $t>t_q^{**}$. Now, let $t_0>t_q^{**}$ with $t_0-t_q^{**}>0$ sufficiently small such that $\mu _t\lesssim 1$ for $t< t_0$, then

(4.6)\begin{equation} m(t)\gtrsim t^{-\frac{N}{q\gamma_q}}\quad\text{for }t\geq t_0. \end{equation}

On the other hand, by the definition of $t_q^{**}$ given by (4.1), [Reference Alves, Souto and Montenegro6, theorem 1.2] and theorem 1.2, $m(t)<\frac {1}{N}S^{\frac {N}{2}}$ for $t>t_q^{**}$. Thus, it is standard to apply the classical elliptic estimates to show that $\|v_t\|_\infty \lesssim 1$ for all $t\geq t_0$. By (4.4) and (4.5),

\begin{align*} Nm(t)& =\|\nabla v_t\|_2^{2}\leqslant\left(1+O\left(\frac{1}{t}\right)\right)\gamma_q\|v_t\|_q^{q}t\\ & ={-}\left(1+O\left(\frac{1}{t}\right)\right)q\gamma_qm'(t)t\quad\text{ for a.e. }t\geq t_0, \end{align*}

which implies that for every $\varepsilon >0$ there exists $t_\varepsilon >0$ such that $m(t)\lesssim t^{-\frac {N}{q\gamma _q+\varepsilon }}$ for $t\geq t_\varepsilon$. It follows from (4.5) once more that

\[\|v_t\|_q^{q}\lesssim t^{-\frac{N}{q\gamma_q+\varepsilon}-1}\quad\text{for }t\geq t_\varepsilon.\]

Thus, by (4.5) and $\|v_t\|_\infty \lesssim 1$ for all $t\geq t_0$, we have

\[Nm(t)=\|\nabla v_t\|_2^{2}\leqslant \gamma_q\|v_t\|_q^{q}t+C_0t^{-\frac{N}{q\gamma_q+\varepsilon}-1} \quad\text{ for }t\geq t_\varepsilon,\]

which implies $m(t)t^{\frac {N}{q\gamma _q}}-C_1t^{-\frac {N}{q\gamma _q+\varepsilon }}$ is decreasing for $t\geq t_\varepsilon$. Therefore, $m(t)\lesssim t^{-\frac {N}{q\gamma _q}}$ for $t>0$ sufficiently large, which, together with (4.6), implies that

\[m(t)\sim t^{-\frac{N}{q\gamma_q}}\quad\text{as }t\to+\infty.\]

It follows from (4.5) and $\|v_t\|_\infty \lesssim 1$ for all $t\geq t_0$ that

\[\|v_t\|_q^{q}\sim t^{-\frac{N}{q\gamma_q}-1}\quad\text{as }t\to+\infty.\]

Since

\[\frac{2}{q-q\gamma_q}-\frac{N}{q\gamma_q}=\frac{2N(q-2-\frac{4}{N})}{(q-2)(2N-q(N-2))}<0 \quad\text{for } 2< q<2+\frac{4}{N},\]

by (4.2), $\overline {\mu }_{q,a}<+\infty$ for $2< q<2+\frac {4}{N}$.

Proof. The proof is standard so we only sketch it here. We first prove the existence of ground-states of (5.4). By the discussion in [Reference Bellazzini, Boussaid, Jeanjean and Visciglia7, 4.2 Symmetry of minimizers], we know that for the energy level $\mathfrak {m}(t)$, there exists a minimizing sequence $\{w_n\}$ on the Nehari manifold $\mathcal {M}_t$ such that $w_n$ is real and positive. Moreover, $w_n$ is radial and decreasing w.r.t. $(x_1, x_2)$ for all $x_3$ and $w_n$ is even and decreasing w.r.t. $x_3$ for all $(x_1, x_2)$. Since $\frac {10}{3}< p<6$, it is standard to use the fibering maps (5.6) to show that $\mathfrak {m}(t)>0$ on $\mathcal {M}_t$. Thus, by [Reference Bellazzini, Boussaid, Jeanjean and Visciglia7, lemma 3.4], there exists $\{z_n\}\in \mathbb {R}$ such that

\[w_n(x_1,x_2,x_3-z_n)\rightharpoonup w_0\not=0\quad\text{weakly in }X\text{ as }n\to\infty.\]

Since $\frac {10}{3}< p<6$, the fibering map of every $w\not =0$ in $X$, see (5.6), has a unique maximum point $s_0'$ and it interacts the Nehari manifold $\mathcal {M}_t$ only at the unique maximum point $s_0'$. Thus, we can use standard arguments (cf. [Reference Wei and Wu33, proposition 3.1]) to show that $w_0$ is a positive ground-state of (5.4). We next prove the convergent conclusion for $t\to +\infty$. Let $w_t$ be a positive ground-state of (5.4) for $t>0$. Since $V(x)\geq 0$, $t>0$ and $w_t$ is positive, we know that $w_t$ satisfies

(5.10)\begin{equation} -\Delta w_t+w_t\leq w_t^{p-1}\quad\text{in }\mathbb{R}^{3}. \end{equation}

By using the fibering maps (5.6) in a standard way (cf. [Reference Wei and Wu33, lemma 3.2]), we know that $\mathfrak {m}(t)$ is decreasing w.r.t. $t>0$. Thus, $\{w_t\}$ is bounded in $H^{1}(\mathbb {R}^{3})$. It follows from (5.10) and the classical elliptic estimates that

(5.11)\begin{equation} w_t\lesssim (1+|x|)^{{-}1}e^{-\frac{1}{2}|x|}\quad\text{in }\mathbb{R}^{3}\text{ for }t\geq 1. \end{equation}

Thus, by $V(x)=x_1^{2}+x_2^{2}$,

\[\int_{\mathbb{R}^{3}}V(x)w_t^{2}{\rm d}x\lesssim 1\quad\text{for all }t\geq 1,\]

which implies that

(5.12)\begin{equation} t^{{-}2}\int_{\mathbb{R}^{3}}V(x)w_t^{2}{\rm d}x=o_t(1)\quad\text{as }t\to+\infty. \end{equation}

Now, using the fibering maps (5.6) in a standard way, we know that $\mathfrak {m}(t)\geq \mathfrak {m}+o_t(1)$ as $t\to +\infty$, where

\[\mathfrak{m}=\inf_{v\in\mathcal{M}}\mathcal{J}(v)\]

with

\[\mathcal{J}(w)=\frac{1}{2}(\|\nabla w\|_2^{2}+\|w\|_2^{2})-\frac{1}{p}\|w\|_p^{p}\]

and

\[\mathcal{M}=\{w\in H^{1}(\mathbb{R}^{3})\backslash\{0\}\mid \mathcal{J}'(w)w=0\}.\]

On the other hand, it is well known that (5.9) has a unique (up to translations) positive radial solution $w_\infty$, which exponentially decays to zero at infinity. Thus, using $w_\infty$ as a test function and adapting the property of the fibering maps (5.6) in a standard way, we also have $\mathfrak {m}(t)\leq \mathfrak {m}+o_t(1)$ as $t\to +\infty$. It follows that $\mathfrak {m}(t)=\mathfrak {m}+o_t(1)$ as $t\to +\infty$, which implies that $\|w_t\|_p^{p}=\|w_\infty \|_p^{p}+o_t(1)$. Now, by standard arguments and the uniqueness of $w_\infty$, we can show that $w_t\to w_\infty$ strongly in $H^{1}(\mathbb {R}^{3})$ as $t\to +\infty$. We now turn to the proof of the convergent conclusion for $t\to 0$. For every $t>0$, let $w_t$ be a positive ground-state of (5.4), then by (5.2), $u_t$ is a positive solution of (5.7). Moreover, by direct calculations,

\[\mathcal{J}_t(w_t)=t^{\frac{p-6}{2(p-2)}}\mathcal{I}_t(u_t)\quad\text{and}\quad\mathcal{J}_t'(w_t)w_t=t^{\frac{p-6}{2(p-2)}}\mathcal{I}_t'(u_t)u_t.\]

Thus, $u_t$ is a positive ground-state of (5.7) for all $t>0$. On the other hand, by [Reference Bellazzini, Boussaid, Jeanjean and Visciglia7, lemma 2.1], Hölder and Sobolev inequalities,

\[\|u\|_p^{p}\lesssim\|u\|_2^{\frac{6-p}{2}}\|\nabla u\|_2^{\frac{3p-6}{2}}\lesssim\|u\|_X^{p}\quad\text{for all }u\in X.\]

Thus, by using the fibering maps (5.8) in a standard way, we know that $\mathbb {M}(0)>0$. By similar arguments as that used above to compare the energy levels $\mathbb {M}(0)$ and $\mathbb {M}(t)$, we can obtain that $\mathbb {M}(t)=\mathbb {M}(0)+o_t(1)$ as $t\to 0$. It follows that $\{u_t\}$ is bounded in $X$ and $\|u_t\|_p^{p}\sim 1$ as $t\to 0$. By [Reference Bellazzini, Boussaid, Jeanjean and Visciglia7, lemma 2.1], $\{u_t\}$ is also bounded in $H^{1}(\mathbb {R}^{3})$ as $t\to 0$. Now, by the Lions’ lemma (cf. [Reference Lions23, lemma I.1] or [Reference Willem34, lemma 1.21]), we can conclude that $\|u_t\|_2^{2}\sim 1$ as $t\to 0$. It follows from (5.2) that $\|w_t\|_2^{2}\sim t^{\frac {3p-10}{2(p-2)}}$ as $t\to 0$. We close this proof by showing the uniqueness of $w_t$ for $t>0$ sufficiently large. Let $w_t$ and $w_t'$ be two different positive ground-states of (5.4) and we define $\phi _t=\frac {w_t-w_t'}{\|w_t-w_t'\|_{L^{\infty }}(\mathbb {R}^{3})}$. Then by the Taylor expansion,

\[-\Delta\phi_t+\phi_t+t^{{-}2}V(x)\phi_t=(p-1)(w_t+\theta(w_t-w_t')))^{p-2}\phi_t,\quad\text{in }\mathbb{R}^{3},\]

where $\theta \in (0, 1)$. Since $V(x)\geq 0$, by (5.11),

\[-\Delta(\phi_t)^{2}+\frac{3}{2}(\phi_t)^{2}\leq 0,\quad\text{in }\mathbb{R}^{3}.\]

Thus, by the maximum principle, $|\phi _t|\lesssim e^{-\frac {1}{2}|x|}$ for $|x|\geq 1$. It is standard to show that $\phi _t\to \phi$ strongly in any compact sets as $t\to +\infty$ and

\[-\Delta\phi+\phi=(p-1)w_\infty^{p-2}\phi,\quad\text{in }\mathbb{R}^{3}.\]

Note that $w_t$ and $w_t'$ are radial w.r.t. $(x_1, x_2)$ for all $x_3$ and even w.r.t. $x_3$ for all $(x_1, x_2)$. Thus, $\phi _t$ is also radial w.r.t. $(x_1, x_2)$ for all $x_3$ and even w.r.t. $x_3$ for all $(x_1, x_2)$. Now, by the well-known nondegeneracy of $w_\infty$, we have $\phi _\infty \equiv 0$. It, together with $|\phi _t|\lesssim e^{-\frac {1}{2}|x|}$ for $|x|\geq 1$, contradicts $\|\phi _t\|_{L^{\infty }(\mathbb {R}^{3})}=1$. Therefore, $w_t$ is unique for $t>0$ sufficiently large.

Proof of theorem 1.4. By the uniqueness of $w_t$ given by proposition 5.1 for $t>0$ sufficiently large, say $t>T_*$. It is standard to show that $\int _{\mathbb {R}^{3}}V(x)w_t^{2}$ is continuous for $t>T_*$. Note that by proposition 5.1,

\[\left(\frac{6-p}{2p}\|w_t\|_p^{p}-2t^{{-}2}\int_{\mathbb{R}^{3}}V(x)w_t^{2}{\rm d}x\right)=\frac{6-p}{2p}\|w_\infty\|_p^{p}+o_t(1).\]

Thus, by $\frac {10}{3}< p<6$, for every $r<(T_*^{\frac {10-3p}{2(p-2)}}(\tfrac {6-p}{2p}\|w_{T_*}\|_p^{p}-2T_*^{-2}\int _{\mathbb {R}^{3}}V(x)w_{T_*}^{2}{\rm d}x))^{\frac {1}{2}}$, $f(r,t)=0$ has a solution $t_r>T_*$. This, together with [Reference Bellazzini, Boussaid, Jeanjean and Visciglia7, theorem 2], implies that (1.15) has a second positive normalized solution $u_{r,2}$ with a positive Lagrange multiplier $\lambda _{r,2}$. The asymptotic behaviour of $u_{r,2}$ and $\lambda _{r,2}$ is obtained by (5.2) and (5.13). It remains to show that $u_{r,2}$ is a mountain-pass solution of (1.15) for $r>0$ sufficiently small. As that in [Reference Bellazzini, Boussaid, Jeanjean and Visciglia7, remark 1.10], we introduce the mountain-pass level

\[\alpha(r)=\inf_{g\in\Gamma_r}\max_{t\in[0, 1]}\mathcal{Y}(g[t]),\]

where $\mathcal {Y}(u)=\frac {1}{2}\|u\|_{X}^{2}-\frac {1}{p}\|u\|_p^{p}$ and

\[\Gamma_r=\{g[s]\in C([0, 1], \mathcal{S}_r)\mid g[0]=u_{r,1}\quad\text{and}\quad \mathcal{Y}(g[1])<\mathcal{Y}(g[0])\}\]

with $u_{r,1}$ being a local minimizer of $\mathcal {Y}(u)$ in $\mathcal {S}_r$ found in [Reference Bellazzini, Boussaid, Jeanjean and Visciglia7] and $\mathcal {S}_r=\{u\in X\mid \|u\|_2^{2}=r^{2}\}$. Let

\[B_{\rho,X,t}=\{u\in X\mid \|u\|_{X,t}^{2}\leq \rho^{2}\},\]

where $\|u\|_{X,t}$ is a norm in $X$ given by

\[\|w\|_{X,t}=\left(\|\nabla w\|_2^{2}+\|w\|_2^{2}+t^{{-}2}\int_{\mathbb{R}^{3}}V(x)w^{2}{\rm d}x\right)^{\frac{1}{2}}.\]

Then by [Reference Bellazzini, Boussaid, Jeanjean and Visciglia7, lemma 2.1] and the Sobolev inequality, for a fixed $\rho >0$ sufficiently small, it can be proved by using $\frac {10}{3}< p<6$ in a standard way that

\[\mathfrak{m}(t)=\inf_{h\in\Theta}\max_{t\in[0, 1]}\mathcal{J}_t(h[s]),\]

where

\[\Theta=\{h[t]\in C([0, 1], X)\mid h[0]\in B_{\rho,X,t}\quad\text{and}\quad \mathcal{J}_t(h[1])<\frac{1}{4}\rho^{2}\}.\]

Now, for every $g[s]\in \Gamma _r$, we define $g^{*}[s]=\lambda _{r,2}^{-\frac {1}{p-2}}g[s](\lambda _{r,2}^{-\frac {1}{2}}x)$. Then

\[\mathcal{J}_{\lambda_{r,2}}(g^{*}[s])=\lambda_{r,2}^{\frac{p-6}{2(p-2)}}\left(\mathcal{Y}(g[s])+\frac{\lambda_{r,2}r^{2}}{2}\right).\]

By [Reference Bellazzini, Boussaid, Jeanjean and Visciglia7, theorem 1] and (1.16),

\[\|g^{*}[0]\|_{X,\lambda_{r,2}}^{2}\lesssim r^{2}\lambda_{r,2}^{\frac{p-6}{2(p-2)}}\sim\lambda_{r,2}^{{-}1}\to 0\quad\text{as }r\to 0.\]

Thus, $g^{*}[0]\in B_{\rho,X,\lambda _{r,2}}$ for $r>0$ sufficiently small and $\mathcal {J}_t(g^{*}[0])\to 0$ as $r\to 0$. By the definition of $g[t]$, we also have $\mathcal {J}_t(g^{*}[1])<\frac {1}{4}\rho ^{2}$. It follows that $g^{*}[t]\in \Theta$, which implies

\[\mathfrak{m}(\lambda_{r,2})\leq\lambda_{r,2}^{\frac{p-6}{2(p-2)}}\left(\alpha(r)+\frac{\lambda_{r,2}r^{2}}{2}\right).\]

On the other hand, the fibering map of $\mathcal {Y}(u)$ at $u_{r,2}$ is given by

\[\mathcal{T}_{u_{r,2}}(\tau)=\frac{\tau^{2}}{2}\|\nabla u_{r,2}\|_2^{2}+\frac{1}{2\tau^{2}}\int_{\mathbb{R}^{3}}V(x)u_{r,2}^{2}{\rm d}x-\frac{\tau^{p\gamma_p}}{p}\|u_{r,2}\|_p^{p}.\]

By direct calculations,

\[\mathcal{T}_{u_{r,2}}'(\tau)=\tau\|\nabla u_{r,2}\|_2^{2}-\frac{1}{\tau^{3}}\int_{\mathbb{R}^{3}}V(x)u_{r,2}^{2}{\rm d}x-\gamma_p\tau^{p\gamma_p-1}\|u_{r,2}\|_p^{p}\]

and

\[\mathcal{T}_{u_{r,2}}''(\tau)=\|\nabla u_{r,2}\|_2^{2}+\frac{3}{\tau^{4}}\int_{\mathbb{R}^{3}}V(x)u_{r,2}^{2}{\rm d}x-\gamma_p(p\gamma_p-1)\tau^{p\gamma_p-2}\|u_{r,2}\|_p^{p}.\]

Clearly, $\mathcal {T}_{u_{r,2}}'(1)=0$. Moreover, by (1.18), (5.12) and the Pohozaev identity of $w_\infty$,

\[\mathcal{T}_{u_{r,2}}''(1)=\lambda_{r,2}^{\frac{6-p}{2(p-2)}}(\gamma_p\|w_\infty\|_p^{p}(2-p\gamma_p)+o_r(1))<0\]

for $r>0$ sufficiently small. Now, let $h(\tau )=\tau ^{4}\|\nabla u_{r,2}\|_2^{2}-\gamma _p\tau ^{p\gamma _p+2}\|u_{r,2}\|_p^{p},$ then,

\[\max_{\tau\geq 0}h(\tau)=\left[\frac{4\|\nabla u_{r,2}\|_2^{2}}{\gamma_p(p\gamma_p+2)\|u_{r,2}\|_p^{p}}\right]^{\frac{4}{p\gamma_p-2}}\frac{p\gamma_p-2}{p\gamma_p+2}\|\nabla u_{r,2}\|_2^{2}>\int_{\mathbb{R}^{3}}V(x)u_{r,2}^{2}{\rm d}x.\]

It follows that there exists $\tau _r<1$ such that $\mathcal {T}_{u_{r,2}}'(\tau _r)=0$ and $\mathcal {T}_{u_{r,2}}''(\tau _r)>0$. We claim that $\tau _r\to 0$ as $r\to 0$. If not, then, there exists $r_n\to 0$ such that $\tau _{r_n}\gtrsim 1$ as $n\to \infty$. Without loss of generality, we may assume $\tau _r\gtrsim 1$ for all $r>0$ sufficiently small. By $\mathcal {T}_{u_{r,2}}'(\tau _{r})=0$, (1.18), (5.12) and the Pohozaev identity of $w_\infty$,

\begin{align*} \mathcal{T}_{u_{r,2}}''(\tau)& =\lambda_{r,2}^{\frac{6-p}{2(p-2)}}(4\|\nabla w_\infty\|_2^{2}-\gamma_p(p\gamma_p+2)\|w_\infty\|_p^{p}\tau_r^{p\gamma_p-2}+o_r(1))\\ & =\lambda_{r,2}^{\frac{6-p}{2(p-2)}}\gamma_p(4-(p\gamma_p+2)\tau_r^{p\gamma_p-2}+o_r(1))\|w_\infty\|_p^{p}, \end{align*}

which implies $\tau _r<(\tfrac {4}{p\gamma _p+2})^{\frac {1}{p\gamma _p-2}}<1$. Without loss of generality, we may assume that $\tau _r\to \tau _0$ as $r\to 0$. Then by $\mathcal {T}_{u_{r,2}}'(\tau _r)=0$, (5.12) and the fact that $w_\infty$ solves (1.17), we must have $\tau _0=0$. It is impossible. Thus, we must have $\tau _r\to 0$ as $r\to 0$. By (1.16) and (1.18),

\[\frac{1}{\tau_r^{4}\lambda_{r,2}^{2}}\left(\int_{\mathbb{R}^{3}}V(x)w_\infty^{2}{\rm d}x+o_r(1)\right)=\|\nabla w_\infty\|_2^{2}+o_r(1).\]

It follows from (1.16) and (1.18) that

\[\|(u_{r,2})_{\tau_r}\|_X^{2}=\tau_r^{2}\|\nabla u_{r,2}\|_2^{2}+\tau_r^{{-}2}\int_{\mathbb{R}^{3}}V(x)u_{r,2}^{2}{\rm d}x\sim\lambda_{r,2}^{\frac{10-3p}{2(p-2)}}\sim r^{2}\]

as $r\to 0$, where $(u_{r,2})_{\tau _r}=\tau _r^{\frac {3}{2}}u_{r,2}(\tau _r x)$. Thus, $(u_{r,2})_{\tau _r}\in B_{r\chi,X,1}$ for a fixed and large $\chi >0$. Since $B_{r\chi,X,1}$ is connected, we can find a continuous path $\Upsilon : [0, 1]$ with $\Upsilon (0)=u_{r,1}$ and $\Upsilon (1)=(u_{r,2})_{\tau _r}$. Now, we define

\[h^{**}[s]=\left\{\begin{aligned} & \Upsilon[(2s)],\quad 0\leq s\leq\frac 12,\\ & (u_{r,2})_{2(1-s)\tau_r+(2s-1)\tau_{r,*}}, \quad\frac 12\leq s\leq 1,\\ \end{aligned}\right.\]

where we choose $\tau _{r,*}>1$ such that $\mathcal {T}_{u_{r,2}}(\tau _{r,*})<\mathcal {Y}(u_{r,1})$. Note that $\mathcal {Y}(u)\lesssim r^{2}$ in $B_{r\chi,X,1}$ and $\mathcal {Y}(u_{r,2})\gtrsim 1$ by (1.16) and (1.18). Thus, for $r>0$ sufficiently small, $h^{**}[s]\in \Gamma _r$ and

\[\alpha(r)\leq\max_{0\leq s\leq 1}h^{**}[s]=\mathcal{T}_{u_{r,2}}(1)=\mathfrak{m}(\lambda_{r,2})\lambda_{r,2}^{\frac{6-p}{2(p-2)}}-\frac{\lambda_{r,2}r^{2}}{2}.\]

Therefore, $\mathfrak {m}(\lambda _{r,2})\lambda _{r,2}^{\frac {6-p}{2(p-2)}}-\frac {\lambda _{r,2}r^{2}}{2}=\alpha (r)$ and $u_{r,2}$ is a mountain-pass solution of (1.15) for $r>0$ sufficiently small.