$\displaystyle=o(1)\|(-t_{n})\star(\phi,\psi)\|_{H^{1}(\mathbb{R}^{3})\times
H^{1}(\mathbb{R}^{3})}$
$\displaystyle=o(1)\|(\phi,\psi)\|_{H^{1}(\mathbb{R}^{3})\times
H^{1}(\mathbb{R}^{3})},$

therefore $(\widetilde{u}_{n},\widetilde{v}_{n})$ is a radial Palais-Smale
sequence of
$\widetilde{J}_{\beta}|_{\mathrm{T}_{b_{1},r}\times\mathrm{T}_{b_{2},r}}$ and
hence a radial symmetric Palais-Smale sequence of
$\widetilde{J}_{\beta}|_{\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}}$ at level
$\sigma_{\beta}(b_{1},b_{2}).$ By the same arguments as Lemma 4.8, we can
prove prove the $H^{1}$ convergence of the Palais-Smale sequence. We should
point out that there are several differences form the proof of Lemma 4.8.
First, to eliminate the disappearance of the solutions, we use
$m^{-}(b_{1},b_{2})>0$ to get a contradiction. To eliminate the semi-trivial
solution, we use Brézis-Lieb lemma and Lemma 4.11 to get a contradiction.

Hence,

$\displaystyle m^{-}_{\beta}(b_{1},b_{2})$
$\displaystyle=\lim_{n\rightarrow+\infty}J_{\beta}(\widetilde{u}_{n},\widetilde{v}_{n})=\lim_{n\rightarrow+\infty}J_{\beta}(w_{n},\sigma_{n})+J_{\beta}(u,v)$
$\displaystyle=\lim_{n\rightarrow+\infty}\left[\frac{1}{6}\int_{\mathbb{R}^{3}}(|\nabla
w_{n}|^{2}+|\nabla\sigma_{n}|^{2})dx-\frac{\beta}{4}\int_{\mathbb{R}^{3}}w_{n}^{2}\sigma_{n}dx\right]+J_{\beta}(u,v)$
$\displaystyle\geq\lim_{n\rightarrow+\infty}\bigg{[}\frac{1}{6}\int_{\mathbb{R}^{3}}(|\nabla
w_{n}|^{2}+|\nabla\sigma_{n}|^{2})dx-\frac{|\beta|}{4}\left(\int_{\mathbb{R}^{3}}|w_{n}|^{3}dx\right)^{\frac{2}{3}}\left(\int_{\mathbb{R}^{3}}|\sigma_{n}|^{3}dx\right)^{\frac{1}{3}}\bigg{]}+J_{\beta}(u,v)$
$\displaystyle\geq\lim_{n\rightarrow+\infty}\frac{1}{6}\int_{\mathbb{R}^{3}}(|\nabla
w_{n}|^{2}+|\nabla\sigma_{n}|^{2})dx+J_{\beta}(u,v)\geq
m^{-}_{\beta}(b_{1},b_{2}).$

Thus, $J_{\beta}(u,v)=m^{-}_{\beta}(b_{1},b_{2})\ \text{and}\
(\widetilde{u}_{n},\widetilde{v}_{n})\rightarrow(u,v)$ in
$H^{1}(\mathbb{R}^{3})\times H^{1}(\mathbb{R}^{3}).$ ∎

######  Lemma 4.13.

We have

$m^{-}_{\beta}(b_{1},b_{2})=\inf_{(u,v)\in\mathrm{T}_{b_{1},r}\times\mathrm{T}_{b_{2},r}}\max_{t\in\mathbb{R}}J_{\beta}(t\star(u,v)).$

###### Proof.

On the one hand

$m^{-}_{\beta}(b_{1},b_{2})=\inf_{(u,v)\in\mathcal{P}^{-}_{b_{1},b_{2}}}J_{\beta}(u,v)=J_{\beta}(\widetilde{u},\widetilde{v}).$

Then, by Lemma 4.4, we have

$m^{-}_{\beta}(b_{1},b_{2})=J_{\beta}(\widetilde{u},\widetilde{v})=\max_{t\in\mathbb{R}}J_{\beta}(t\star(\widetilde{u},\widetilde{v}))\geq\inf_{(u,v)\in\mathrm{T}_{b_{1},r}\times\mathrm{T}_{b_{2},r}}\max_{t\in\mathbb{R}}J_{\beta}(t\star(u,v)).$

On the other hand, for any
$(u,v)\in\mathrm{T}_{b_{1},r}\times\mathrm{T}_{b_{2},r},$ we have
$t_{u,v}\star(u,v)\in\mathcal{P}^{-}_{b_{1},b_{2}}$ and hence

$\max_{t\in\mathbb{R}}J_{\beta}(t\star(u,v))=J_{\beta}(t_{u,v}\star(u,v))\geq
m^{-}_{\beta}(b_{1},b_{2}).$

∎

Before giving the estimate of $m^{-}_{\beta}(b_{1},b_{2})$, we would like to
study the dependence of $m^{-}_{\beta}(b_{1},b_{2})$ on $\beta$.

######  Lemma 4.14.

For any $0<\beta_{1}<\beta_{2}$, then $m^{-}_{\beta_{2}}(b_{1},b_{2})\leq
m^{-}_{\beta_{1}}(b_{1},b_{2})\leq m^{-}_{0}(b_{1},b_{2}).$

###### Proof.

By Lemma 4.13

$\displaystyle
m^{-}_{\beta_{2}}(b_{1},b_{2})\leq\max_{t\in\mathbb{R}}J_{\beta_{2}}(t\star(\widetilde{u}_{\beta_{1}},\widetilde{v}_{\beta_{1}}))$
$\displaystyle\leq\max_{t\in\mathbb{R}}J_{\beta_{1}}(t\star(\widetilde{u}_{\beta_{1}},\widetilde{v}_{\beta_{1}}))$
$\displaystyle=J_{\beta_{1}}(\widetilde{u}_{\beta_{1}},\widetilde{v}_{\beta_{1}})=m^{-}_{\beta_{1}}(b_{1},b_{2}),$
$m^{-}_{\beta_{1}}(b_{1},b_{2})\leq\max_{t\in\mathbb{R}}J_{\beta_{1}}(t\star(\widetilde{u}_{0},\widetilde{v}_{0}))\leq\max_{t\in\mathbb{R}}J_{0}(t\star(\widetilde{u}_{0},\widetilde{v}_{0}))=J_{0}(\widetilde{u}_{0},\widetilde{v}_{0})=m^{-}_{0}(b_{1},b_{2}).$

∎

###### Proof of Theorem 1.3.

Let us consider a minimizing sequence $(u_{n},v_{n})$ for
$J_{\beta}(u,v)\mid_{A_{R_{0}}}$ , we assume that
$(u_{n},v_{n})\in\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}$ is radial
decreasing for every $n$. Furthermore, by Lemma 4.4, for every $n$ we can take
$s_{u_{n},v_{n}}\ast(u_{n},v_{n})\in\mathcal{P}^{+}_{b_{1},b_{2}}$ such that
$(\int_{\mathbb{R}^{3}}(|\nabla(s_{u_{n},v_{n}}\star
u_{n})|^{2}+|\nabla(s_{u_{n},v_{n}}\star v_{n})|^{2})dx)^{\frac{1}{2}}\leq
R_{0}$ and

$\displaystyle J_{\beta}(s_{u,v}\star(u_{n},v_{n}))$
$\displaystyle=\min\bigg{\\{}J_{\beta}(t\star(u_{n},v_{n}):t\in\mathbb{R}\
\text{and}\left(\int_{\mathbb{R}^{3}}(|\nabla(t\star
u_{n})|^{2}+|\nabla(t\star
v_{n})|^{2})dx\right)^{\frac{1}{2}}<R_{0}\bigg{\\}}$
$\displaystyle<J_{\beta}(u_{n},v_{n}).$

Thus, we obtain a new minimizing sequence
$\\{w_{n},\sigma_{n}\\}=\\{s_{u_{n},u_{n}}\star v_{n},s_{u_{n},v_{n}}\star
v_{n}\\}$ with
$(w_{n},\sigma_{n})\in\mathrm{T}_{b_{1,r}}\times\mathrm{T}_{b_{2,r}}\cap\mathcal{P}^{+}_{a,\mu}$
radially decreasing for every $n$. By Lemma 4.5, we have
$\left(\int_{\mathbb{R}^{3}}(|\nabla
w_{n}|^{2}+|\nabla\sigma_{n}|^{2})dx\right)^{\frac{1}{2}}\leq R_{0}$ for every
$n$ and hence by Ekeland’s variational principle in a standard way, we know
the existence of a new minimizing sequence for $\\{u_{n},v_{n}\\}\subset
A_{R_{0}}$ for $m_{\beta}^{+}(b_{1},b_{2})$ with
$\|(u_{n},v_{n})-(w_{n},\sigma_{n})\|\rightarrow 0$ as $n\rightarrow+\infty$,
which is also a Palais-Smale sequence for $J_{\beta}(u,v)$ on
$\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}$. By the boundedness of
$\\{(w_{n},\sigma_{n})\\}$, $\|(u_{n},v_{n})-(w_{n},\sigma_{n})\|\rightarrow
0$, Brézis-Lieb lemma and Sobolev embedding theorem, we have

$P_{b_{1},b_{2}}(u_{n},v_{n})=P_{b_{1},b_{2}}(w_{n},\sigma_{n})+o(1)\rightarrow
0\ \ \text{and}\ u^{-}_{n},v^{-}_{n}\rightarrow 0\ a.e.\text{in}\
\mathbb{R}^{3}.$

Hence $(u_{n},v_{n})$ satisfies all the assumptions of Lemma 4.8, as a
consequence, up to a subsequence $(u_{n},v_{n})\rightarrow(u,v)$ strongly in
$H^{1}(\mathbb{R}^{3})\times H^{1}(\mathbb{R}^{3})$ and $(u,v)$ is an interior
local minimizer for $J_{\beta}(u,v)|_{A_{R_{0}}}$. Finally, we prove that
$(u,v)$ is in $\mathcal{P}^{+}_{b_{1},b_{2}}(u,v)$. i.e.
$\beta\int_{\mathbb{R}^{3}}u^{2}vdx>0$. Indeed, from
$(u_{n},v_{n})\in\mathcal{P}^{+}_{b_{1},b_{2}}(u,v)$, we have
$\beta\int_{\mathbb{R}^{3}}u^{2}vdx\geq 0$. Assume
$\beta\int_{\mathbb{R}^{3}}u^{2}vdx=0$, then from Pohozaev identity, we have

$\displaystyle\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+|\nabla
v|^{2})dx=\frac{3}{4}\int_{\mathbb{R}^{3}}(\mu_{1}u^{4}+\mu_{2}v^{4}+2\rho
u^{2}v^{2})dx.$

Thus, from (2.1)-(2.3), we have

$\displaystyle\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+|\nabla v|^{2})dx$
$\displaystyle=\frac{3}{4}\int_{\mathbb{R}^{3}}\left(\mu_{1}u^{4}+\mu_{2}v^{4}+2\rho
u^{2}v^{2}\right)dx$
$\displaystyle\leq\frac{3\mu_{1}}{4}C^{4}_{3,4}b_{1}\|\nabla
u\|^{3}_{L^{2}(\mathbb{R}^{3})}+\frac{3\mu_{2}}{4}C^{4}_{3,4}b_{2}\|\nabla
v\|^{3}_{L^{2}(\mathbb{R}^{3})}$ $\displaystyle\quad+\frac{3}{4}\rho
C^{4}_{3,4}b^{\frac{1}{2}}_{1}b^{\frac{1}{2}}_{2}\left[\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{3})}+\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{3})}\right]^{\frac{3}{2}}$
$\displaystyle\leq\frac{3}{4}C^{4}_{3,4}(\mu_{1}b_{1}+\mu_{2}b_{2}+\rho
b^{\frac{1}{2}}_{1}b^{\frac{1}{2}}_{2})\left[\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{3})}+\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{3})}\right]^{\frac{3}{2}},$

so,

$\displaystyle\left[\|\nabla u\|^{2}_{L^{2}(\mathbb{R}^{3})}+\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{3})}\right]^{\frac{1}{2}}\geq\frac{4}{3C^{4}_{3,4}(\mu_{1}b_{1}+\mu_{2}b_{2}+\rho
b^{\frac{1}{2}}_{1}b^{\frac{1}{2}}_{2})}.$ (4.37)

From Corollary 4.1 and the definition of $R_{0}$ in Lemma 4.2, we have

$\displaystyle\left[\|\nabla u\|^{2}_{L^{2}(\mathbb{R}^{3})}+\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{3})}\right]^{\frac{1}{2}}<R_{0}\leq\widetilde{t}=\frac{2}{3C^{4}_{3,4}(\mu_{1}b_{1}+\mu_{2}b_{1}+\rho
b^{\frac{1}{2}}_{1}b^{\frac{1}{2}}_{2})}.$ (4.38)

a contradiction. Thus $\beta\int_{\mathbb{R}^{3}}u^{2}vdx>0$ and
$(u,v)\in\mathcal{P}^{+}_{b_{1},b_{2}}(u,v)$. From Lemma 4.5, we know that
$(u,v)$ is a ground normalized solution. From Lemma 4.12, we get a second
critical point of mountain pass type for
$J_{\beta}(u,v)\mid_{\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}}$.

Next, we prove the second part of Theorem 1.3, that is the limit behavior of
the ground state solution as $\beta\rightarrow 0.$

For $b_{1},b_{2}>0$ fixed, from the proof Lemma 4.2, we can deduce that when
$\beta\rightarrow 0$, then $R_{0}(b_{1},b_{2},\rho,\beta)\rightarrow 0$. By
corollary 4.1, when $(\widehat{u},\widehat{v})$ is the ground normalized
solution obtained in Theorem 1.3, then

$\big{(}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+|\nabla
v|^{2})dx\big{)}^{\frac{1}{2}}<R_{0}(b_{1},b_{2},\rho,\beta)\rightarrow 0,$

and

$\displaystyle 0>$ $\displaystyle m^{+}_{\beta}(b_{1},b_{2})=J_{\beta}(u,v)$
$\displaystyle\geq\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+|\nabla
v|^{2})dx-\mathcal{D}_{1}\big{(}\int_{\mathbb{R}^{3}}|\nabla
u|^{2}dx\big{)}^{\frac{3}{2}}-\mathcal{D}_{2}\big{(}\int_{\mathbb{R}^{3}}|\nabla
v|^{2}dx\big{)}^{\frac{3}{2}}$
$\displaystyle\quad-\rho\mathcal{D}_{3}\big{(}\int_{\mathbb{R}^{3}}(|\nabla
u|^{2}+|\nabla
v|^{2})dx\big{)}^{\frac{3}{2}}-|\beta|\mathcal{D}_{4}\big{(}\int_{\mathbb{R}^{3}}(|\nabla
u|^{2}+|\nabla v|^{2})dx\big{)}^{\frac{3}{4}}$ $\displaystyle\geq
h\Big{(}\big{(}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+|\nabla
v|^{2})dx\big{)}^{\frac{1}{2}}\Big{)}\rightarrow 0,$

thus

$m^{+}_{\beta}(b_{1},b_{2})\rightarrow 0.$

Denote
$\\{(\widetilde{u}_{\beta},\widetilde{v}_{\beta}):0<\beta<\overline{\beta}\\}$
with $\overline{\beta}$ small enough. By the same arguments as Lemma 4.8, we
can obtain the $H^{1}$ convergence. First, to eliminate the disappearance of
the solutions, we use Lemma 4.14 and
$0<m^{-}_{\beta}(b_{1},b_{2})<m^{-}_{0}(b_{1},b_{2})$ to get a contradiction.
To eliminate the semi-trivial solution, we use Brézis-Lieb lemma and energy
comparison to get a contradiction.

Hence,

$\displaystyle m^{-}_{0}(b_{1},b_{2})$ $\displaystyle=\lim_{\beta\rightarrow
0}J_{\beta}(\widetilde{u}_{\beta},\widetilde{v}_{\beta})=\lim_{\beta\rightarrow
0}J_{\beta}(w_{\beta},\sigma_{\beta})+\lim_{\beta\rightarrow
0}J_{\beta}(\widetilde{u},\widetilde{v})$
$\displaystyle=\lim_{\beta\rightarrow
0}\left[\frac{1}{6}\int_{\mathbb{R}^{3}}(|\nabla
w_{\beta}|^{2}+|\nabla\sigma_{\beta}|^{2})dx-\frac{\beta}{4}\int_{\mathbb{R}^{3}}w_{\beta}^{2}\sigma_{\beta}dx\right]+\lim_{\beta\rightarrow
0}J_{\beta}(\widetilde{u},\widetilde{v})$
$\displaystyle\geq\lim_{\beta\rightarrow
0}\frac{1}{6}\int_{\mathbb{R}^{3}}(|\nabla
w_{\beta}|^{2}+|\nabla\sigma_{\beta}|^{2})dx+\lim_{\beta\rightarrow
0}J_{\beta}(\widetilde{u},\widetilde{v})\geq m^{-}_{0}(b_{1},b_{2}).$

Thus, $\lim_{\beta\rightarrow
0}J(\widetilde{u}_{\beta},\widetilde{v}_{\beta})=m^{-}_{0}(b_{1},b_{2})\
\text{and}\
(\widetilde{u}_{\beta},\widetilde{v}_{\beta})\rightarrow(\widetilde{u},\widetilde{v})$
in $H^{1}(\mathbb{R}^{3})\times H^{1}(\mathbb{R}^{3}).$ ∎

##  5 Proof of Theorem 1.4

In this section, we give a refined upper of $m^{+}_{\beta}(b_{1},b_{2})$ and
search for
$(\lambda_{1,b_{1},b_{2}},\lambda_{2,b_{1},b_{2}},u_{b_{1},b_{2}},v_{b_{1},b_{2}})$
solving

$\begin{cases}-\Delta u+\lambda_{1}u=\beta uv&\text{in}\ \ \mathbb{R}^{3},\\\
-\Delta v+\lambda_{2}v=\frac{\beta}{2}u^{2}&\text{in}\ \
\mathbb{R}^{3},\end{cases}$ (5.1)

satisfying the additional conditions

$\displaystyle\int_{\mathbb{R}^{3}}u^{2}dx=b^{2}_{1}\ \text{and}\
\int_{\mathbb{R}^{3}}v^{2}dx=b^{2}_{2}.$ (5.2)

Denote

$J_{0}(u,v)=\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+|\nabla
v|^{2})dx-\frac{\beta}{2}\int_{\mathbb{R}^{3}}u^{2}vdx,$

on the constraint $\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}$.

$\displaystyle\mathcal{P}^{0}_{b_{1},b_{2}}(u,v):=\Big{\\{}$
$\displaystyle(u,v)\in\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}:P^{0}_{b_{1},b_{2}}(u,v)=0\Big{\\}},$

where

$P^{0}_{b_{1},b_{2}}(u,v)=\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+|\nabla
v|^{2})dx-\frac{3}{4}\beta\int_{\mathbb{R}^{3}}u^{2}vdx.$

Then, the solution of (5.1)-(5.2) can be found as minimizers of

$m^{+}_{0}(b_{1},b_{2})=\inf_{\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}}J_{0}(u,v)>-\infty.$

If $(u_{0},v_{0})$ is the unique positive solution of (1.12), then
$(u_{0},v_{0})=(\sqrt{2}\beta^{-1}w,\beta^{-1}w)$, and $w$ is the unique
positive solution of (1.13).

Let

$\theta_{1}=\frac{2\beta^{2}b_{1}^{\frac{6}{5}}b_{2}^{\frac{4}{5}}}{16^{\frac{2}{5}}\|w\|^{2}_{L^{2}(\mathbb{R}^{3})}},\
\theta_{2}=\frac{\beta^{2}b_{1}^{\frac{8}{5}}b_{2}^{\frac{2}{5}}}{16^{\frac{1}{5}}\|w\|^{2}_{L^{2}(\mathbb{R}^{3})}},\
L_{1}=\frac{2\beta^{4}b_{1}^{\frac{14}{5}}b_{2}^{\frac{6}{5}}}{16^{\frac{3}{5}}\|w\|^{4}_{L^{2}(\mathbb{R}^{3})}},\
\
L_{2}=\frac{4\beta^{4}b_{1}^{\frac{12}{5}}b_{2}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{4}_{L^{2}(\mathbb{R}^{3})}},$
$\lambda_{1,b_{1},b_{2}}=\frac{4\beta^{4}b_{1}^{\frac{12}{5}}b_{2}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{4}_{L^{2}(\mathbb{R}^{3})}},\
\ \
\lambda_{2,b_{1},b_{2}}=\frac{\beta^{4}b_{1}^{\frac{16}{5}}b_{2}^{\frac{4}{5}}}{16^{\frac{2}{5}}\|w\|^{4}_{L^{2}(\mathbb{R}^{3})}},$

we have following lemma.

######  Lemma 5.1.

(5.1)-(5.2) has a unique positive solution

$\big{(}\lambda_{1,b_{1},b_{2}},\lambda_{2,b_{1},b_{2}},L_{1}u_{0}(\theta_{1}x),L_{2}v_{0}(\theta_{2}x)\big{)}.$

Moreover,

$\displaystyle
m^{+}_{0}(b_{1},b_{2})=J_{0}(L_{1}u_{0}(\theta_{1}x),L_{2}v_{0}(\theta_{2}x))=-\frac{1}{6}\Big{[}\frac{4\beta^{4}b_{1}^{\frac{22}{5}}b_{2}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1}^{\frac{16}{5}}b_{2}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}.$

###### Proof.

By elementary calculation, we have

$(\lambda_{1,b_{1},b_{2}},\lambda_{2,b_{1},b_{2}},L_{1}u_{0}(\theta_{1}x),L_{2}v_{0}(\theta_{2}x))$

is the unique positive solution of (5.1)-(5.2). Furthermore, we have

$\displaystyle m^{+}_{0}(b_{1},b_{2})$
$\displaystyle=J_{0}(L_{1}u_{0}(\theta_{1}x),L_{2}v_{0}(\theta_{2}x))$
$\displaystyle=-\frac{1}{6}\Big{[}\frac{L^{2}_{1}}{\theta_{1}}\int_{\mathbb{R}^{3}}|\nabla
u_{0}|^{2}dx+\frac{L^{2}_{2}}{\theta_{2}}\int_{\mathbb{R}^{3}}|\nabla
v_{0}|^{2}dx\Big{]}$
$\displaystyle=-\frac{1}{6}\Big{[}2\frac{L^{2}_{1}}{\theta_{1}}\beta^{-2}+\frac{L^{2}_{2}}{\theta_{2}}\beta^{-2}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}$
$\displaystyle=-\frac{1}{6}\Big{[}\frac{4\beta^{4}b_{1}^{\frac{22}{5}}b_{2}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1}^{\frac{16}{5}}b_{2}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}.$

∎

######  Lemma 5.2.

We have

$m^{-}_{\beta}(b_{1},b_{2})<m^{+}_{0}(b_{1},b_{2})=-\frac{1}{6}\Big{[}\frac{4\beta^{4}b_{1}^{\frac{22}{5}}b_{2}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1}^{\frac{16}{5}}b_{2}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}.$

###### Proof.

Since

$\displaystyle J_{0}(u,v)$
$\displaystyle=\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+|\nabla
v|^{2})dx-\frac{\beta}{2}\int_{\mathbb{R}^{3}}u^{2}vdx$
$\displaystyle\geq\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+|\nabla
v|^{2})dx-\frac{|\beta|}{2}\Big{[}\big{(}\frac{2}{3}b^{\frac{3}{2}}_{1}+\frac{1}{3}b^{\frac{3}{2}}_{2}\big{)}C^{3}_{3,3}\big{[}\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{3})}+\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{3})}\big{]}^{\frac{3}{4}}\Big{]}$
$\displaystyle=g\Big{(}\big{(}\int_{\mathbb{R}^{3}}(|\nabla u|^{2}+|\nabla
v|^{2})dx\big{)}^{\frac{1}{2}}\Big{)},$

where

$g(t)=\frac{1}{2}t^{2}-\frac{|\beta|}{2}\Big{(}\frac{2}{3}b^{\frac{3}{2}}_{1}+\frac{1}{3}b^{\frac{3}{2}}_{2}\Big{)}C^{3}_{3,3}t^{\frac{3}{2}}.$

It is easy to see that $g(t)<0$ if $t\in(0,\widetilde{t})$ and $g(t)>0$ if
$t\in(\widetilde{t},+\infty)$, where
$\widetilde{t}=\Big{[}|\beta|\big{(}\frac{2}{3}b^{\frac{3}{2}}_{1}+\frac{1}{3}b^{\frac{3}{2}}_{2}\big{)}C^{3}_{3,3}\Big{]}^{2}$.
Since
$J_{0}(L_{1}u_{0}(\theta_{1}x),L_{2}v_{0}(\theta_{2}x))=m^{+}_{0}(b_{1},b_{2})<0$,
we get

$\Big{[}\int_{\mathbb{R}^{3}}(|\nabla L_{1}u_{0}(\theta_{1}x)|^{2}+|\nabla
L_{2}v_{0}(\theta_{2}x)|^{2})dx\Big{]}^{\frac{1}{2}}<\widetilde{t}.$

From Lemma 5.1, the definition of $R_{0}$ in Lemma 4.2 and the definition of
$\widetilde{t}$, we have $\widetilde{t}<R_{0}$,

$\|L_{1}u_{0}(\theta_{1}x)\|^{2}_{L^{2}(\mathbb{R}^{3})}=b^{2}_{1},\
\|L_{2}v_{0}(\theta_{2}x)\|^{2}_{L^{2}(\mathbb{R}^{3})}=b^{2}_{2},$

and

$\displaystyle\Big{[}\int_{\mathbb{R}^{3}}(|\nabla
L_{1}u_{0}(\theta_{1}x)|^{2}+|\nabla
L_{2}v_{0}(\theta_{2}x)|^{2})dx\Big{]}^{\frac{1}{2}}<\widetilde{t}<R_{0}.$

Therefore, we have

$\displaystyle m^{-}_{\beta}(b_{1},b_{2})$
$\displaystyle=\inf_{A_{R_{0}}}J_{\beta}(u,v)\leq
J_{\beta}(L_{1}u_{0}(\theta_{1}x),L_{2}v_{0}(\theta_{2}x))$
$\displaystyle\quad<J_{0}(L_{1}u_{0}(\theta_{1}x),L_{2}v_{0}(\theta_{2}x))=m_{0}(b_{1},b_{2})$
$\displaystyle=-\frac{1}{6}\Big{[}\frac{4\beta^{4}b_{1}^{\frac{22}{5}}b_{2}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1}^{\frac{16}{5}}b_{2}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}.$

∎

###### Proof of Theorem 1.4.

Let $b_{1,k},\ b_{2,k}\rightarrow 0^{+}$ with $b_{1,k}\backsim b_{2,k}$ as
$k\rightarrow+\infty$ and $(u_{b_{1,k},b_{2,k}},v_{b_{1,k},b_{2,k}})\in
A_{R_{0}}$ be a positive minimizer of $m(b_{1,k},b_{2,k},R_{0})$
for each $k\in\mathbb{N}$. From

$\displaystyle P_{b_{1},b_{2}}(u_{b_{1,k},b_{2,k}},v_{b_{1,k},b_{2,k}})$
$\displaystyle=\int_{\mathbb{R}^{3}}(|\nabla u_{b_{1,k},b_{2,k}}|^{2}+|\nabla
v_{b_{1,k},b_{2,k}}|^{2})dx-\frac{3}{4}\int_{\mathbb{R}^{3}}\left(\mu_{1}u_{b_{1,k},b_{2,k}}^{4}+\mu_{2}v_{b_{1,k},b_{2,k}}^{4}+2\rho
u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}^{2}\right)dx$
$\displaystyle\quad-\frac{3}{4}\beta\int_{\mathbb{R}^{N}}u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}dx=0,$

we have

$\displaystyle J_{\beta}(u_{b_{1,k},b_{2,k}},v_{b_{1,k},b_{2,k}})$
$\displaystyle=\frac{1}{6}\int_{\mathbb{R}^{3}}(|\nabla
u_{b_{1,k},b_{2,k}}|^{2}+|\nabla
v_{b_{1,k},b_{2,k}}|^{2})dx-\frac{\beta}{4}\int_{\mathbb{R}^{N}}u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}dx$
$\displaystyle=-\frac{1}{6}\int_{\mathbb{R}^{3}}(|\nabla
u_{b_{1,k},b_{2,k}}|^{2}+|\nabla v_{b_{1,k},b_{2,k}}|^{2})dx$
$\displaystyle\quad+\frac{1}{4}\int_{\mathbb{R}^{3}}\big{(}\mu_{1}u_{b_{1,k},b_{2,k}}^{4}+\mu_{2}v_{b_{1,k},b_{2,k}}^{4}+2\rho
u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}^{2}\big{)}dx$
$\displaystyle<-\frac{1}{6}\Big{[}\frac{4\beta^{4}b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}.$

Thus

$\displaystyle\Big{[}\frac{4\beta^{4}b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}\leq\int_{\mathbb{R}^{3}}(|\nabla
u_{b_{1,k},b_{2,k}}|^{2}+|\nabla v_{b_{1,k},b_{2,k}}|^{2})dx$

and

$\displaystyle\int_{\mathbb{R}^{3}}(|\nabla u_{b_{1,k},b_{2,k}}|^{2}+|\nabla
v_{b_{1,k},b_{2,k}}|^{2})dx<\left[\frac{3|\beta|}{2}\Big{[}\big{(}\frac{2}{3}b^{\frac{3}{2}}_{1,k}+\frac{1}{3}b^{\frac{3}{2}}_{2,k}\big{)}C^{3}_{3,3}\Big{]}\right]^{4}.$

Therefore

$\displaystyle\Big{[}\frac{4\beta^{4}b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}$ (5.3)
$\displaystyle\leq\int_{\mathbb{R}^{3}}(|\nabla
u_{b_{1,k},b_{2,k}}|^{2}+|\nabla
v_{b_{1,k},b_{2,k}}|^{2})dx<\left[\frac{3|\beta|}{2}\Big{[}\big{(}\frac{2}{3}b^{\frac{3}{2}}_{1,k}+\frac{1}{3}b^{\frac{3}{2}}_{2,k}\big{)}C^{3}_{3,3}\Big{]}\right]^{4}.$

When $b_{1,k}\backsim b_{2,k}$, we can get

$\displaystyle\int_{\mathbb{R}^{3}}(|\nabla u_{b_{1,k},b_{2,k}}|^{2}+|\nabla
v_{b_{1,k},b_{2,k}}|^{2})dx\backsim
b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}+b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}.$
(5.4)

From (2.1)-(2.2) and (5.3), we have

$\displaystyle\frac{1}{4}\int_{\mathbb{R}^{3}}\big{(}\mu_{1}u_{b_{1,k},b_{2,k}}^{4}+\mu_{2}v_{b_{1,k},b_{2,k}}^{4}+2\rho
u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}^{2}\big{)}dx$ (5.5)
$\displaystyle\leq\frac{1}{4}\int_{\mathbb{R}^{3}}\Big{(}(\mu_{1}+\rho)u_{b_{1,k},b_{2,k}}^{4}+(\mu_{2}+\rho)v_{b_{1,k},b_{2,k}}^{4}\Big{)}dx$
$\displaystyle\leq\Big{[}\frac{\mu_{1}+\rho}{4}C^{4}_{3,4}b_{1,k}+\frac{\mu_{2}+\rho}{4}C^{4}_{3,4}b_{2,k}\Big{]}\left[\|\nabla
u_{b_{1,k},b_{2,k}}\|^{2}_{L^{2}(\mathbb{R}^{3})}+\|\nabla
v_{b_{1,k},b_{2,k}}\|^{2}_{L^{2}(\mathbb{R}^{3})}\right]^{\frac{3}{2}}$
$\displaystyle\leq\Big{[}\frac{\mu_{1}+\rho}{4}C^{4}_{3,4}b_{1,k}+\frac{\mu_{2}+\rho}{4}C^{4}_{3,4}b_{2,k}\Big{]}\left[\frac{3|\beta|}{2}\Big{[}\big{(}\frac{2}{3}b^{\frac{3}{2}}_{1,k}+\frac{1}{3}b^{\frac{3}{2}}_{2,k}\big{)}C^{3}_{3,3}\Big{]}\right]^{6}\rightarrow
0$

as $k\rightarrow+\infty.$ Since

$m^{-}_{\beta}(b_{1,k},b_{2,k})<-\frac{1}{6}\Big{[}\frac{4\beta^{4}b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})},$

we obtain

$\displaystyle-\frac{1}{6}\Big{[}\frac{4\beta^{4}b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}$ (5.6)
$\displaystyle>m^{-}_{\beta}(b_{1,k},b_{2,k})=J_{\beta}(u_{b_{1,k},b_{2,k}},v_{b_{1,k},b_{2,k}})$
$\displaystyle\geq\inf_{\mathrm{T}_{b_{1,k}}\times\mathrm{T}_{b_{2,k}}}\Big{\\{}\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla
u|^{2}+|\nabla
v|^{2})dx-\frac{\beta}{2}\int_{\mathbb{R}^{3}}u^{2}vdx\Big{\\}}$
$\displaystyle\quad-\frac{1}{4}\int_{\mathbb{R}^{3}}\big{(}\mu_{1}u_{b_{1,k},b_{2,k}}^{4}+\mu_{2}v_{b_{1,k},b_{2,k}}^{4}+2\rho
u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}^{2}\big{)}dx$
$\displaystyle=-\frac{1}{6}\Big{[}\frac{4\beta^{4}b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}$
$\displaystyle\quad-\frac{1}{4}\int_{\mathbb{R}^{3}}\big{(}\mu_{1}u_{b_{1,k},b_{2,k}}^{4}+\mu_{2}v_{b_{1,k},b_{2,k}}^{4}+2\rho
u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}^{2}\big{)}dx$
$\displaystyle\geq-\frac{1}{6}\Big{[}\frac{4\beta^{4}b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}+\frac{\beta^{4}b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}}{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}\Big{]}\|\nabla
w\|^{2}_{L^{2}(\mathbb{R}^{3})}$ (5.7)
$\displaystyle\quad-\Big{[}\frac{\mu_{1}+\rho}{4}C^{4}_{3,4}b_{1,k}+\frac{\mu_{2}+\rho}{4}C^{4}_{3,4}b_{2,k}\Big{]}\left[\frac{3|\beta|}{2}\Big{[}\big{(}\frac{2}{3}b^{\frac{3}{2}}_{1,k}+\frac{1}{3}b^{\frac{3}{2}}_{2,k}\big{)}C^{3}_{3,3}\Big{]}\right]^{6}.$

When $b_{1,k}\backsim b_{2,k}$, we can get

$\displaystyle m^{-}_{\beta}(b_{1,k},b_{2,k})\backsim
b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}+b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}$
(5.8)

and

$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{3}}(|\nabla
u_{b_{1,k},b_{2,k}}|^{2}+|\nabla
v_{b_{1,k},b_{2,k}}|^{2})dx-\frac{\beta}{2}\int_{\mathbb{R}^{N}}u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}dx\backsim
b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}+b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}.$
(5.9)

The Lagrange multipliers rule implies the existence of some
$\lambda_{1,k},\lambda_{2,k}\in\mathbb{R}$ such that

$\displaystyle\int_{\mathbb{R}^{3}}(\nabla
u_{b_{1,k},b_{2,k}}\nabla\varphi)dx+\int_{\mathbb{R}^{3}}(\lambda_{1,k}u_{b_{1,k},b_{2,k}}\varphi)dx$
$\displaystyle=\int_{\mathbb{R}^{3}}(\mu_{1}u_{b_{1,k},b_{2,k}}^{3}\varphi+\rho
v_{b_{1,k},b_{2,k}}^{2}u_{b_{1,k},b_{2,k}}\varphi)dx+\beta\int_{\mathbb{R}^{3}}u_{b_{1,k},b_{2,k}}v_{b_{1,k},b_{2,k}}\varphi,$
$\displaystyle\int_{\mathbb{R}^{3}}(\nabla
v_{b_{1,k},b_{2,k}}\nabla\psi)dx+\int_{\mathbb{R}^{3}}(\lambda_{2,k}v_{b_{1,k},b_{2,k}}\psi)dx$
$\displaystyle=\int_{\mathbb{R}^{3}}(\mu_{2}v_{b_{1,k},b_{2,k}}^{3}\psi+\rho
u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}\psi)dx+\frac{\beta}{2}\int_{\mathbb{R}^{3}}u_{b_{1,k},b_{2,k}}^{2}\psi,$

for each $\varphi,\psi\in H^{1}(\mathbb{R}^{3})$. Taking
$\varphi=u_{b_{1,k},b_{2,k}}$ and $\psi=v_{b_{1,k},b_{2,k}}$, we have

$\displaystyle\lambda_{1,k}b_{1,k}^{2}$
$\displaystyle=-\int_{\mathbb{R}^{3}}|\nabla
u_{b_{1,k},b_{2,k}}|^{2}dx+\int_{\mathbb{R}^{3}}(\mu_{1}u_{b_{1,k},b_{2,k}}^{4}+\rho
v_{b_{1,k},b_{2,k}}^{2}u^{2}_{b_{1,k},b_{2,k}})dx$ (5.10)
$\displaystyle+\beta\int_{\mathbb{R}^{3}}u^{2}_{b_{1,k},b_{2,k}}v_{b_{1,k},b_{2,k}},$
$\displaystyle\lambda_{2,k}b_{2,k}^{2}$
$\displaystyle=-\int_{\mathbb{R}^{3}}|\nabla
v_{b_{1,k},b_{2,k}}|^{2}dx+\int_{\mathbb{R}^{3}}(\mu_{1}v_{b_{1,k},b_{2,k}}^{4}+\rho
v_{b_{1,k},b_{2,k}}^{2}u^{2}_{b_{1,k},b_{2,k}})dx$ (5.11)
$\displaystyle+\frac{\beta}{2}\int_{\mathbb{R}^{3}}u^{2}_{b_{1,k},b_{2,k}}v_{b_{1,k},b_{2,k}}.$

Since $P_{b_{1},b_{2}}(u_{b_{1,k},b_{2,k}},v_{b_{1,k},b_{2,k}})=0,$ we get

$\displaystyle\int_{\mathbb{R}^{3}}(|\nabla u_{b_{1,k},b_{2,k}}|^{2}+|\nabla
v_{b_{1,k},b_{2,k}}|^{2})dx-\frac{3}{4}\beta\int_{\mathbb{R}^{N}}u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}dx$
(5.12)
$\displaystyle-\frac{3}{4}\int_{\mathbb{R}^{3}}\left(\mu_{1}u_{b_{1,k},b_{2,k}}^{4}+\mu_{2}v_{b_{1,k},b_{2,k}}^{4}+2\rho
u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}^{2}\right)dx=0.$

Hence, from (5.10), (5.11), (5.12) and $b_{1,k}\backsim b_{2,k}$, we have

$\displaystyle-\int_{\mathbb{R}^{3}}(|\nabla u_{b_{1,k},b_{2,k}}|^{2}+|\nabla
v_{b_{1,k},b_{2,k}}|^{2})dx+\frac{3\beta}{2}\int_{\mathbb{R}^{3}}u^{2}_{b_{1,k},b_{2,k}}v_{b_{1,k},b_{2,k}}$
$\displaystyle\leq\lambda_{1,k}b_{1,k}^{2}+\lambda_{2,k}b_{2,k}^{2}=\int_{\mathbb{R}^{3}}(|\nabla
u_{b_{1,k},b_{2,k}}|^{2}+|\nabla v_{b_{1,k},b_{2,k}}|^{2})dx$
$\displaystyle\quad-\frac{1}{2}\int_{\mathbb{R}^{3}}\left(\mu_{1}u_{b_{1,k},b_{2,k}}^{4}+\mu_{2}v_{b_{1,k},b_{2,k}}^{4}+2\rho
u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}^{2}\right)dx,$

combing with (5.4) and (5.9), we get

$\lambda_{1,k}b_{1,k}^{2}+\lambda_{2,k}b_{2,k}^{2}\backsim
b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}+b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}},$

thus $\lambda_{1,k}\backsim b_{1,k}^{4}$ and $\lambda_{2,k}\backsim
b_{1,k}^{4}$ when $b_{1,k}\backsim b_{2,k}$.

Denote

$\theta_{1,k}=\frac{2\beta^{2}b_{1,k}^{\frac{6}{5}}b_{2,k}^{\frac{4}{5}}}{16^{\frac{2}{5}}\|w\|^{2}_{L^{2}(\mathbb{R}^{3})}},\
\theta_{2,k}=\frac{\beta^{2}b_{1,k}^{\frac{8}{5}}b_{2,k}^{\frac{2}{5}}}{16^{\frac{1}{5}}\|w\|^{2}_{L^{2}(\mathbb{R}^{3})}},\
L_{1,k}=\frac{2\beta^{4}b_{1,k}^{\frac{14}{5}}b_{2,k}^{\frac{6}{5}}}{16^{\frac{3}{5}}\|w\|^{4}_{L^{2}(\mathbb{R}^{3})}},\
\
L_{2,k}=\frac{4\beta^{4}b_{1,k}^{\frac{12}{5}}b_{2,k}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{4}_{L^{2}(\mathbb{R}^{3})}}.$

Define

$\widetilde{u}_{b_{1,k},b_{2,k}}=L^{-1}_{1,k}u_{b_{1,k},b_{2,k}}(\theta^{-1}_{1,k}x),\
\
\widetilde{v}_{b_{1,k},b_{2,k}}=L^{-1}_{2,k}v_{b_{1,k},b_{2,k}}(\theta^{-1}_{2,k}x),$

then

$\displaystyle\int_{\mathbb{R}^{3}}(|\nabla\widetilde{u}_{b_{1,k},b_{2,k}}|^{2}+|\nabla\widetilde{v}_{b_{1,k},b_{2,k}}|^{2})dx$
(5.13)
$\displaystyle=\frac{\theta_{1,k}}{L^{2}_{1,k}}\int_{\mathbb{R}^{3}}|\nabla
u_{b_{1,k},b_{2,k}}|^{2}dx+\frac{\theta_{2,k}}{L^{2}_{2,k}}\int_{\mathbb{R}^{3}}|\nabla
v_{b_{1,k},b_{2,k}}|^{2}dx$
$\displaystyle\leq\Big{[}\frac{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}{2\beta^{6}b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}}+\frac{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}{2\beta^{6}b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}}\Big{]}\int_{\mathbb{R}^{3}}(|\nabla
u_{b_{1,k},b_{2,k}}|^{2}+|\nabla v_{b_{1,k},b_{2,k}}|^{2})dx$
$\displaystyle\leq\Big{[}\frac{16^{\frac{4}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}{2\beta^{6}b_{1,k}^{\frac{22}{5}}b_{2,k}^{\frac{8}{5}}}+\frac{16^{\frac{2}{5}}\|w\|^{6}_{L^{2}(\mathbb{R}^{3})}}{2\beta^{6}b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{14}{5}}}\Big{]}\left[\frac{3|\beta|}{2}\Big{[}\big{(}\frac{2}{3}b^{\frac{3}{2}}_{1,k}+\frac{1}{3}b^{\frac{3}{2}}_{2,k}\big{)}C^{3}_{3,3}\Big{]}\right]^{4},$
$\displaystyle\int_{\mathbb{R}^{3}}(|\widetilde{u}_{b_{1,k},b_{2,k}}|^{2}+|\widetilde{v}_{b_{1,k},b_{2,k}}|^{2})dx$
$\displaystyle=\frac{\theta_{1,k}^{3}}{L^{2}_{1,k}}\int_{\mathbb{R}^{3}}|u_{b_{1,k},b_{2,k}}|^{2}dx+\frac{\theta_{2,k}^{3}}{L^{2}_{2,k}}\int_{\mathbb{R}^{3}}|v_{b_{1,k},b_{2,k}}|^{2}dx$
(5.14)
$\displaystyle=\frac{2\|w\|^{2}_{L^{2}(\mathbb{R}^{3})}+\|w\|^{2}_{L^{2}(\mathbb{R}^{3})}}{\beta^{2}},$
$\displaystyle\int_{\mathbb{R}^{3}}\Big{(}\frac{\theta_{1,k}^{3}}{L^{4}_{1,k}}\mu_{1}u_{b_{1,k},b_{2,k}}^{4}+\mu_{2}\frac{\theta_{2,k}^{3}}{L^{4}_{2,k}}v_{b_{1,k},b_{2,k}}^{4}\Big{)}dx$
(5.15)
$\displaystyle\quad<\int_{\mathbb{R}^{3}}\left(\mu_{1}\widetilde{u}_{b_{1,k},b_{2,k}}^{4}+\mu_{2}\widetilde{v}_{b_{1,k},b_{2,k}}^{4}+2\rho\widetilde{u}_{b_{1,k},b_{2,k}}^{2}\widetilde{v}_{b_{1,k},b_{2,k}}^{2}\right)dx$
$\displaystyle=\int_{\mathbb{R}^{3}}\Big{(}\frac{\theta_{1,k}^{3}}{L^{4}_{1,k}}\mu_{1}u_{b_{1,k},b_{2,k}}^{4}+\mu_{2}\frac{\theta_{2,k}^{3}}{L^{4}_{2,k}}v_{b_{1,k},b_{2,k}}^{4}+2\rho
u_{b_{1,k},b_{2,k}}^{2}v_{b_{1,k},b_{2,k}}^{2}\Big{)}dx$
$\displaystyle\quad\leq\int_{\mathbb{R}^{3}}\Big{(}\frac{\theta_{1,k}^{3}}{L^{4}_{1,k}}(\mu_{1}+\rho)u_{b_{1,k},b_{2,k}}^{4}+(\mu_{2}+\rho)\frac{\theta_{2,k}^{3}}{L^{4}_{2,k}}v_{b_{1,k},b_{2,k}}^{4}\Big{)}dx.$

By the definition of $\theta_{1,k},\ \theta_{2,k},\ L_{1,k},\ L_{2,k}$, it is
easy to see that

$\displaystyle\int_{\mathbb{R}^{3}}\big{(}\mu_{1}\widetilde{u}_{b_{1,k},b_{2,k}}^{4}+\mu_{2}\widetilde{v}_{b_{1,k},b_{2,k}}^{4}+2\rho\widetilde{u}_{b_{1,k},b_{2,k}}^{2}\widetilde{v}_{b_{1,k},b_{2,k}}^{2}\big{)}dx\rightarrow+\infty\
\text{as}\ k\rightarrow+\infty.$ (5.16)

From (5.13)- (5.14), we know that
$(\widetilde{u}_{b_{1,k},b_{2,k}},\widetilde{v}_{b_{1,k},b_{2,k}})$ is bounded
in $H^{1}(\mathbb{R}^{3})\times H^{1}(\mathbb{R}^{3}).$ Then, we have

$\big{(}\widetilde{u}_{b_{1,k},b_{2,k}}(x),\widetilde{v}_{b_{1,k},b_{2,k}}(x)\big{)}\rightharpoonup(\bar{u},\bar{v})\neq(0,0),$

for some $(\bar{u},\bar{v})\in H^{1}(\mathbb{R}^{3})\times
H^{1}(\mathbb{R}^{3})$. Thus, we see that
$(\widetilde{u}_{b_{1,k},b_{2,k}}(x),\widetilde{v}_{b_{1,k},b_{2,k}}(x))$
satisfies

$\begin{cases}-\Delta\widetilde{u}_{b_{1,k},b_{2,k}}+\frac{\lambda_{1,k}}{\theta^{2}_{1,k}}\widetilde{u}_{b_{1,k},b_{2,k}}=\mu_{1}\frac{L^{2}_{1,k}}{\theta^{2}_{1,k}}\widetilde{u}_{b_{1,k},b_{2,k}}^{3}+\rho\frac{L^{2}_{2,k}}{\theta^{2}_{1,k}}\widetilde{v}_{b_{1,k},b_{2,k}}^{2}\widetilde{u}_{b_{1,k},b_{2,k}}+\beta\frac{L_{2,k}}{\theta^{2}_{1,k}}\widetilde{u}_{b_{1,k},b_{2,k}}\widetilde{v}_{b_{1,k},b_{2,k}},\\\
-\Delta\widetilde{v}_{b_{1,k},b_{2,k}}+\frac{\lambda_{2,k}}{\theta^{2}_{2,k}}\widetilde{v}_{b_{1,k},b_{2,k}}=\mu_{2}\frac{L^{2}_{2,k}}{\theta^{2}_{2,k}}\widetilde{v}_{b_{1,k},b_{2,k}}^{3}+\rho\frac{L^{2}_{1,k}}{\theta^{2}_{2,k}}\widetilde{u}_{b_{1,k},b_{2,k}}^{2}\widetilde{v}_{b_{1,k},b_{2,k}}+\frac{\beta}{2}\frac{L^{2}_{1,k}}{L_{2,k}\theta^{2}_{2,k}}\widetilde{u}_{b_{1,k},b_{2,k}}^{2}.\end{cases}$
(5.17)

By the definition of $\theta_{1,k},\ \theta_{2,k},\ L_{1,k},\ L_{2,k}$, we
have

$\frac{L^{2}_{1,k}}{\theta^{2}_{1,k}}=\frac{\beta^{4}b_{1,k}^{\frac{16}{5}}b_{2,k}^{\frac{4}{5}}}{16^{\frac{2}{5}}\|w\|^{4}_{L^{2}(\mathbb{R}^{3})}}\rightarrow
0,\
\frac{L^{2}_{2,k}}{\theta^{2}_{2,k}}=\frac{\beta^{4}b_{1,k}^{\frac{8}{5}}b_{2,k}^{\frac{12}{5}}}{16^{\frac{1}{5}}\|w\|^{4}_{L^{2}(\mathbb{R}^{3})}}\rightarrow
0,$
$\frac{L^{2}_{2,k}}{\theta^{2}_{1,k}}=\frac{4\beta^{4}b_{1,k}^{\frac{12}{5}}b_{2,k}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{4}_{L^{2}(\mathbb{R}^{3})}}\rightarrow
0,\ \
\frac{L^{2}_{1,k}}{\theta^{2}_{2,k}}=\frac{4\beta^{4}b_{1,k}^{\frac{12}{5}}b_{2,k}^{\frac{8}{5}}}{16^{\frac{4}{5}}\|w\|^{4}_{L^{2}(\mathbb{R}^{3})}}\rightarrow
0,\ \frac{L_{2,k}}{\theta^{2}_{1,k}}=1,\ \
\frac{L^{2}_{1,k}}{L_{2,k}\theta^{2}_{2,k}}=1,$

and there exists $\lambda^{\ast}_{1}>0$ and $\lambda^{\ast}_{2}>0$ such that

$\frac{\lambda_{1,k}}{\theta^{2}_{1,k}}\rightarrow\lambda^{\ast}_{1},\
\frac{\lambda_{2,k}}{\theta^{2}_{2,k}}\rightarrow\lambda^{\ast}_{2}\
\text{as}\ k\rightarrow+\infty.$

Therefore, $(\bar{u},\bar{v})$ solves that

$\begin{cases}-\Delta u+\lambda^{\ast}_{1}u=\beta uv&\text{in}\ \
\mathbb{R}^{3},\\\ -\Delta
v+\lambda^{\ast}_{2}v=\frac{\beta}{2}u^{2}&\text{in}\ \
\mathbb{R}^{3}.\end{cases}$ (5.18)

From Theorem 1.2 of [49], we know that when
$\lambda^{\ast}_{1}=\lambda^{\ast}_{2}$, then

$\big{(}\sqrt{2}\lambda^{\ast}_{1}\beta^{-1}w((\lambda^{\ast}_{1})^{\frac{1}{2}}x),\beta^{-1}\lambda^{\ast}_{1}w((\lambda^{\ast}_{1})^{\frac{1}{2}}x)\big{)}$

is the unique positive solution of above system, where $w$ is the unique
positive solution of

$-\Delta u+u=u^{2},\ u\in H^{1}(\mathbb{R}^{3}).$

Texting (5.17) and (5.18) with $\widetilde{u}_{b_{1,k},b_{2,k}}-\bar{u}$,
$\widetilde{v}_{b_{1,k},b_{2,k}}-\bar{v}$ respectively, we obtain

$\displaystyle\int_{\mathbb{R}^{3}}|\nabla$
$\displaystyle(\widetilde{u}_{b_{1,k},b_{2,k}}-\bar{u})|^{2}dx+\int_{\mathbb{R}^{3}}(\frac{\lambda_{1,k}}{\theta^{2}_{1,k}}\widetilde{u}_{b_{1,k},b_{2,k}}-\lambda^{\ast}_{1}\bar{u})(\widetilde{u}_{b_{1,k},b_{2,k}}-\bar{u})$
(5.19)
$\displaystyle-\beta\int_{\mathbb{R}^{3}}(\widetilde{u}_{b_{1,k},b_{2,k}}\widetilde{v}_{b_{1,k},b_{2,k}}-\bar{u}\bar{v})(\widetilde{u}_{b_{1,k},b_{2,k}}-\bar{u})$
$\displaystyle=\int_{\mathbb{R}^{3}}|\nabla(\widetilde{u}_{b_{1,k},b_{2,k}}-\bar{u})|^{2}dx+\lambda^{\ast}_{1}\int_{\mathbb{R}^{3}}|\widetilde{u}_{b_{1,k},b_{2,k}}-\bar{u}|^{2}=o_{k}(1),$
$\displaystyle\int_{\mathbb{R}^{3}}|\nabla$
$\displaystyle(\widetilde{v}_{b_{1,k},b_{2,k}}-\bar{v})|^{2}dx+\int_{\mathbb{R}^{3}}(\frac{\lambda_{2,k}}{\theta^{2}_{2,k}}\widetilde{v}_{b_{1,k},b_{2,k}}-\lambda^{\ast}_{2}\bar{v})(\widetilde{v}_{b_{1,k},b_{2,k}}-\bar{v})$
(5.20)
$\displaystyle-\frac{\beta}{2}\int_{\mathbb{R}^{3}}(\widetilde{u}^{2}_{b_{1,k},b_{2,k}}-\bar{u}^{2})(\widetilde{v}_{b_{1,k},b_{2,k}}-\bar{v})$
$\displaystyle=\int_{\mathbb{R}^{3}}|\nabla(\widetilde{v}_{b_{1,k},b_{2,k}}-\bar{v})|^{2}dx+\lambda^{\ast}_{2}\int_{\mathbb{R}^{3}}|\widetilde{v}_{b_{1,k},b_{2,k}}-\bar{v}|^{2}=o_{k}(1).$

Therefore

$\displaystyle\frac{2(\lambda^{\ast}_{1})^{\frac{1}{2}}\|w\|^{2}_{L^{2}(\mathbb{R}^{3})}}{\beta^{2}}=\|\sqrt{2}\lambda^{\ast}_{1}\beta^{-1}w((\lambda^{\ast}_{1})^{\frac{1}{2}}x)\|^{2}_{L^{2}(\mathbb{R}^{3})}=\|\bar{u}\|^{2}_{L^{2}(\mathbb{R}^{3})}$
$\displaystyle=\lim_{k\rightarrow+\infty}\int_{\mathbb{R}^{3}}|\widetilde{u}_{b_{1,k},b_{2,k}}|^{2}dx=\lim_{k\rightarrow+\infty}\frac{\theta_{1,k}^{3}}{L^{2}_{1,k}}\int_{\mathbb{R}^{3}}|u_{b_{1,k},b_{2,k}}|^{2}dx=\frac{2\|w\|^{2}_{L^{2}(\mathbb{R}^{3})}}{\beta^{2}},$
$\displaystyle\frac{(\lambda^{\ast}_{1})^{\frac{1}{2}}\|w\|^{2}_{L^{2}(\mathbb{R}^{3})}}{\beta^{2}}=\|\beta^{-1}\lambda^{\ast}_{1}w((\lambda^{\ast}_{1})^{\frac{1}{2}}x))\|^{2}_{L^{2}(\mathbb{R}^{3})}=\|\bar{v}\|^{2}_{L^{2}(\mathbb{R}^{3})}$
$\displaystyle=\lim_{k\rightarrow+\infty}\int_{\mathbb{R}^{3}}|\widetilde{v}_{b_{1,k},b_{2,k}}|^{2}dx=\lim_{k\rightarrow+\infty}\frac{\theta_{1,k}^{3}}{L^{2}_{1,k}}\int_{\mathbb{R}^{3}}|v_{b_{1,k},b_{2,k}}|^{2}dx=\frac{\|w\|^{2}_{L^{2}(\mathbb{R}^{3})}}{\beta^{2}},$

therefore

$\lambda^{\ast}_{1}=\lambda^{\ast}_{2}=1.$

From (5.17), (5.18) ,(5.19) and (5.20), we have that

$(\widetilde{u}_{b_{1,k},b_{2,k}},\widetilde{v}_{b_{1,k},b_{2,k}})\rightarrow(\sqrt{2}\beta^{-1}w,\beta^{-1}w)\
\ \text{in}\ H^{1}(\mathbb{R}^{3})\times H^{1}(\mathbb{R}^{3}).$

Moreover, as the limit function $(\sqrt{2}\beta^{-1}w,\beta^{-1}w)$ is
independent of the sequence that we choose, which implies that the convergence
is true for the whole sequence. Therefore

$\big{(}\widetilde{u}_{b_{1,k},b_{2,k}},\widetilde{v}_{b_{1,k},b_{2,k}}\big{)}=\big{(}L^{-1}_{1,k}u_{b_{1,k},b_{2,k}}(\theta^{-1}_{1,k}(x)),L^{-1}_{2,k}v_{b_{1,k},b_{2,k}}(\theta^{-1}_{2,k}(x))\big{)}\rightarrow\big{(}\sqrt{2}\beta^{-1}w,\beta^{-1}w\big{)},$

in $\ H^{1}(\mathbb{R}^{3})\times H^{1}(\mathbb{R}^{3})$ as
$b_{1,k},b_{2,k}\to 0$ and $b_{1,k}\backsim b_{2,k}$.

∎

Next, we prove Theorem 1.5. In order to prove Theorem 1.5, we first give
following lemma.

######  Lemma 5.3.

Let $\overline{b}_{1},\ \overline{b}_{2},\ \tau_{1},\ \tau_{2}>0.$ There
exists
$\widetilde{\beta}=\widetilde{\beta}(\overline{b}_{1}+\tau_{1},\overline{b}_{2}+\tau_{2})>0$
such that if $0<b_{1}\leq\overline{b}_{1}$, $0<b_{2}\leq\overline{b}_{2}$ and
$0<\beta<\widetilde{\beta}$, then

(i)
$2R^{2}_{0}(\overline{b}_{1}+\tau_{1},\overline{b}_{2}+\tau_{2},\rho,\beta)<R^{2}_{1}(\overline{b}_{1},\overline{b}_{2},\rho,\beta)$,

(ii) for any $d_{i}>0,\ c_{i}>0,i=1,2$ such that
$d^{2}_{i}+c^{2}_{i}=b_{1}^{i},\ i=1,2$, then
$R^{2}_{0}(c_{1},c_{2},\rho,\beta)+R^{2}_{0}(d_{1},d_{2},\rho,\beta)\leq
R^{2}_{1}(b_{1},b_{2},\rho,\beta).$

###### Proof.

By Lemma 4.2,
$0<R_{0}=R_{0}(b_{1},b_{2},\rho,\beta)<R_{1}=R_{1}(b_{1},b_{2},\rho,\beta)$
are the roots of

$g(t,b_{1},b_{2},\rho,\beta)=\frac{1}{2}t^{\frac{1}{2}}-\frac{1}{4}(\mathcal{D}_{1}+\mathcal{D}_{2}+\rho\mathcal{D}_{3})t^{\frac{3}{2}}-\frac{1}{2}|\beta|\mathcal{D}_{4}=\varphi(t,b_{1},b_{2},\rho)-\frac{1}{2}|\beta|\mathcal{D}_{4}$

and the existence of $R_{0}$ and $R_{1}$ is guaranteed by the condition

$\displaystyle\beta\left(2b^{\frac{3}{2}}_{1}+b^{\frac{3}{2}}_{2}\right)C^{3}_{3,3}C^{2}_{3,4}\sqrt{\mu_{1}b_{1}+\mu_{2}b_{2}+\rho
b^{\frac{1}{2}}_{1}b^{\frac{1}{2}}_{2}}<\frac{2\sqrt{6}}{3}.$ (5.21)

Let $\overline{b}_{1},\ \overline{b}_{2},\ \tau_{1},\ \tau_{2}>0$ and consider
the range of $\beta>0$ such that (5.21) is satisfied with
$b_{1}=\overline{b}_{1}+\tau_{1},\ b_{2}=\overline{b}_{2}+\tau_{2}$. Taking
the limit as $\beta\rightarrow 0^{+}$, by the continuity we have that
$R_{0}(\overline{b}_{1}+\tau_{1},\overline{b}_{2}+\tau_{2},\rho,\beta)\rightarrow
0$ and $R_{1}(\overline{b}_{1}+\tau_{1},\overline{b}_{2}+\tau_{2},\rho,\beta)$
converge to the only positive root of
$\varphi(t,\overline{b}_{1}+\tau_{1},\overline{b}_{2}+\tau_{2},\rho)$.
Particularity, for every $\overline{b}_{1},\ \overline{b}_{2},\ \tau_{1},\
\tau_{2}>0$ fixed , there exists
$\widetilde{\beta}=\widetilde{\beta}(\overline{b}_{1}+\tau_{1},\overline{b}_{2}+\tau_{2})>0$
such that

$\displaystyle
2R^{2}_{0}(\overline{b}_{1}+\tau_{1},\overline{b}_{2}+\tau_{2},\rho,\beta)<R^{2}_{1}(\overline{b}_{1}+\tau_{1},\overline{b}_{2}+\tau_{2},\rho,\beta)\
\ \text{whenever}\ 0<\beta<\widetilde{\beta}.$ (5.22)

Let $0<b_{1}\leq\overline{b}_{1}+\tau_{1},$
$0<b_{2}\leq\overline{b}_{2}+\tau_{2}$ and $0<\beta<\widetilde{\beta}$. Under
the condition of (5.21), we have that

$\partial_{t}g(t,b_{1},b_{2},\rho,\beta)=\partial_{t}\varphi(t,b_{1},b_{2},\rho).$

It is easy to check that $\varphi(t,b_{1},b_{2},\rho)$ has a unique critical
point on $(0,+\infty)$, which is a strict maximum point in
$\overline{t}=\overline{t}(b_{1},b_{2},\rho)$ with
$0<R_{0}(b_{1},b_{2},\rho)<\overline{t}<R_{1}(b_{1},b_{2},\rho)$, therefore

$\partial_{t}g(R_{0}(b_{1},b_{2},\rho),b_{1},b_{2},\rho,\beta)=\partial_{t}\varphi(R_{0}(b_{1},b_{2},\rho),b_{1},b_{2},\rho)>0.$

By the implicit function theorem, we know that $R_{0}(b_{1},b_{2},\rho,\beta)$
is a locally unique $C^{1}$ function of $((b_{1},b_{2},\rho,\beta))$ with

$\frac{\partial_{t}R_{0}(b_{1},b_{2},\rho,\beta)}{\partial
b_{1}}=-\frac{\partial_{b_{1}}g(R_{0}(b_{1},b_{2},\rho,\beta),b_{1},b_{2},\rho,\beta)}{\partial_{t}g(R_{0}(b_{1},b_{2},\rho),b_{1},b_{2},\rho,\beta)}>0,$
$\frac{\partial_{t}R_{0}(b_{1},b_{2},\rho,\beta)}{\partial
b_{2}}=-\frac{\partial_{b_{2}}g(R_{0}(b_{1},b_{2},\rho),b_{1},b_{2},\rho,\beta)}{\partial_{t}g(R_{0}(b_{1},b_{2},\rho,\beta),b_{1},b_{2},\rho,\beta)}>0.$

Similarly, we can proof that $R_{1}(b_{1},b_{2},\rho,\beta)$ is a locally
unique $C^{1}$ function of
$((b_{1},b_{2},\rho,\beta))$ with

$\frac{\partial_{t}R_{1}(b_{1},b_{2},\rho,\beta)}{\partial
b_{1}}=-\frac{\partial_{b_{1}}g(R_{1}(b_{1},b_{2},\rho,\beta),b_{1},b_{2},\rho,\beta)}{\partial_{t}g(R_{1}(b_{1},b_{2},\rho),b_{1},b_{2},\rho,\beta)}<0,$
$\frac{\partial_{t}R_{1}(b_{1},b_{2},\rho,\beta)}{\partial
b_{2}}=-\frac{\partial_{b_{2}}g(R_{1}(b_{1},b_{2},\rho),b_{1},b_{2},\rho,\beta)}{\partial_{t}g(R_{1}(b_{1},b_{2},\rho,\beta),b_{1},b_{2},\rho,\beta)}<0.$

So, $R_{0}$ is monotone increasing in $b_{1},b_{2}$ and $R_{1}$ is monotone
decreasing in $b_{1},b_{2}$. By (5.22) and the monotonicity of $R_{1}$, we
have
$2R^{2}_{0}(\overline{b}_{1}+\tau_{1},\overline{b}_{2}+\tau_{2},\rho,\beta)<R^{2}_{1}(\overline{b}_{1},\overline{b}_{2},\rho,\beta)$.
For any $d_{i}>0,\ c_{i}>0,i=1,2$ such that $d^{2}_{i}+c^{2}_{i}=b_{1}^{i},\
i=1,2$, then

$\displaystyle
R^{2}_{0}(c_{1},c_{2},\rho,\beta)+R^{2}_{0}(d_{1},d_{2},\rho,\beta)<2R^{2}_{0}(b_{1},b_{2},\rho,\beta)<2R^{2}_{0}(\overline{b}_{1}+\tau_{1},\overline{b}_{2}+\tau_{2},\rho,\beta)$
$\displaystyle<2R^{2}_{1}(\overline{b}_{1},\overline{b}_{2},\rho,\beta)<R^{2}_{1}(b_{1},b_{2},\rho,\beta).$

∎

###### Proof of Theorem 1.5.

Suppose that there exists a $\epsilon_{0}>0$ , a sequence of initial date
$\\{u^{0}_{n},v^{0}_{n}\\}\subset H^{1}(\mathbb{R}^{3})\times
H^{1}(\mathbb{R}^{3})$ and a sequence $\\{t_{n}\\}\subset\mathbb{R}^{+}$ such
that the solution $(u_{n},v_{n})$ of (1.14) with the initial date
$(u_{n}(0,\cdot),v_{n}(0,\cdot))=(u_{n}(\cdot),v_{n}(\cdot))$ satisfies

$dist_{H^{1}(\mathbb{R}^{3})\times
H^{1}(\mathbb{R}^{3})}((u^{0}_{n},v^{0}_{n}),\mathcal{M}_{b_{1},b_{2}})<\frac{1}{n}$

and

$dist_{H^{1}(\mathbb{R}^{3})\times
H^{1}(\mathbb{R}^{3})}((u_{n}(t_{n},\cdot),v_{n}(t_{n},\cdot)),\mathcal{M}_{b_{1},b_{2}})\geq\epsilon_{0}.$

Without loss of generality, we may assume that
$\\{(u^{0}_{n},v^{0}_{n})\\}\subset\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}},$
since

$dist_{H^{1}(\mathbb{R}^{3})\times
H^{1}(\mathbb{R}^{3})}((u^{0}_{n},v^{0}_{n}),\mathcal{M}_{b_{1},b_{2}})\rightarrow
0\ \text{as}\ \ n\rightarrow+\infty.$

So, $\|u^{0}_{n}\|_{L^{2}(\mathbb{R}^{3})}:=b_{1n}\rightarrow b_{1}$
$\|v^{0}_{n}\|_{L^{2}(\mathbb{R}^{3})}:=b_{2n}\rightarrow b_{2}$ and
$J_{\beta}(u^{0}_{n},v^{0}_{n})\rightarrow m_{\beta}(b_{1},b_{2})$. By (i) of
Lemma 5.3 and the continuity, we can deduce that

$\int_{\mathbb{R}^{3}}(|\nabla u^{0}_{n}|^{2}+|\nabla
v^{0}_{n}|^{2})dx<R_{0}(b_{1}+\tau_{1},b_{2}+\tau_{2},\rho,\beta)<R_{1}(b_{1n}+\tau_{1},b_{2n}+\tau_{2},\rho,\beta)$

for every $n$ large enough. Since

$\int_{\mathbb{R}^{3}}(|\nabla u^{0}_{n}|^{2}+|\nabla
v^{0}_{n}|^{2})dx\in[R_{0}(b_{1n},b_{2n},\rho,\beta),R_{1}(b_{1n},b_{2n},\rho,\beta)],$

we have $J_{\beta}(u^{0}_{n},v^{0}_{n})\geq 0$. Thus, we can deduce that

$\int_{\mathbb{R}^{3}}(|\nabla u^{0}_{n}|^{2}+|\nabla
v^{0}_{n}|^{2})dx<R_{0}(b_{1n},b_{2n},\rho,\beta)<R_{0}(b_{1}+\tau_{1},b_{2}+\tau_{2},\rho,\beta).$

Since $(u^{0}_{n},v^{0}_{n})\in A_{R_{0}(b_{1n},b_{2n},\rho,\beta)}$, if
$(u_{n}(t_{n},\cdot),v_{n}(t_{n},\cdot)$ exist from
$A_{R_{0}(b_{1n},b_{2n},\rho,\beta)}$ there exists $t_{n}\in(0,T_{\max})$ such
that

$\int_{\mathbb{R}^{3}}(|\nabla u_{n}(t_{n},\cdot)|^{2}+|\nabla
v_{n}(t_{n},\cdot)|^{2})dx=R_{0}(b_{1n},b_{2n},\rho,\beta);$

however, $J_{\beta}(u_{n}(t_{n},\cdot),v_{n}(t_{n},\cdot)\geq h(R_{0})=0$
contract with the conservation of energy. Therefore, the solutions starting in
$A_{R_{0}(b_{1n},b_{2n},\rho,\beta)}$ are globally defined in time and satisfy

$\int_{\mathbb{R}^{3}}(|\nabla u_{n}(t_{n},\cdot)|^{2}+|\nabla
v_{n}(t_{n},\cdot)|^{2})dx<R_{0}(b_{1n},b_{2n},\rho,\beta)<R_{0}(b_{1}+\tau_{1},b_{2}+\tau_{2},\rho,\beta)$

for every $t_{n}\in(0,+\infty)$. Then by the conservation laws of energy and
mass

$\\{(u_{n}(t_{n},\cdot),v_{n}(t_{n},\cdot))\\}$

is a minimizing sequence. Thus,$\\{(u_{n}(t_{n},\cdot),v_{n}(t_{n},\cdot))\\}$
is a minimizing sequence of $m^{r}_{\beta}(b_{1},b_{2})$. Then according the
proof of Theorem 1.3 there exists $(u_{0},v_{0})\in\mathcal{M}_{b_{1},b_{2}}$
such that $(u_{n}(t_{n},\cdot),v_{n}(t_{n},\cdot))\rightarrow(u_{0},v_{0})$ in
$H^{1}(\mathbb{R}^{3})\times H^{1}(\mathbb{R}^{3})$, which contradicts to

$dist_{H^{1}(\mathbb{R}^{3})\times
H^{1}(\mathbb{R}^{3})}((u_{n}(t_{n},\cdot),v_{n}(t_{n},\cdot)),\mathcal{M}_{b_{1},b_{2}})\geq\epsilon_{0}.$

∎

##  6 Proof of Theorems 1.6 and 1.7

In this section, we first give some preliminaries. For $N=4$, we have the
Sobolev inequality

$\mathcal{S}\|u\|_{L^{4}(\mathbb{R}^{4})}^{2}\leq\|\nabla
u\|_{L^{2}(\mathbb{R}^{4})}^{2},\quad\forall u\ \in D^{1,2}(\mathbb{R}^{4}),$

where $D^{1,2}(\mathbb{R}^{4})$ is the completion of
$C_{c}^{\infty}(\mathbb{R}^{4})$ with respect to the norm
$||u||_{D^{1,2}}:=\|\nabla u\|_{2}$. For $2<p<4$, the Gagliardo-Nirenberg
inequality (see [39]) is

$\|u\|_{L^{p}(\mathbb{R}^{4})}\leq C_{4,p}\|\nabla
u\|_{L^{2}(\mathbb{R}^{4})}^{\gamma_{p}}\|u\|_{L^{2}(\mathbb{R}^{4})}^{1-\gamma_{p}},\quad\forall\
u\in H^{1}(\mathbb{R}^{4}),$ (6.1)

where $C_{4,p}>0$ is a constant and $\gamma_{p}=\frac{2(p-2)}{p}$. If $p=4$,
$\gamma_{4}=1$, then $\mathcal{S}=|C_{4,p}|^{-2}$.

We have the following already known results.

######  Lemma 6.1.

([28, Corollary B.1]) Suppose $(u,v)\in H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})$ ($N\geq 3$) is a nonnegative solution of (1.1)-(1.2),
then $(u,v)$ is a smooth solution.

######  Lemma 6.2.

([28, Lemma 2.3]) Suppose $\mu_{1},\mu_{2},\beta>0$ and $(u,v)\in
H^{1}(\mathbb{R}^{4})\times H^{1}(\mathbb{R}^{4})$ is a nonnegative solution
of (1.1)-(1.2), then $u>0$ implies $\lambda_{1}>0$; $v>0$ implies
$\lambda_{2}>0$.

From [12, 38], we get the least energy solutions to (1.16). Note that (1.16)
has semi-trivial solutions $(\mu^{-\frac{1}{2}}_{1}U_{\varepsilon},0)$ and
$(0,\mu^{-\frac{1}{2}}_{2}U_{\varepsilon})$. Here, we are only interested in
nontrivial solutions of (1.16), this is
$(\sqrt{k_{1}}U_{\varepsilon},\sqrt{k_{2}}U_{\varepsilon})$, where
$U_{\varepsilon}$ is defined in (1.15),
$k_{1}=\frac{\rho-\mu_{2}}{\rho^{2}-\mu_{1}\mu_{2}}$ and
$k_{2}=\frac{\rho-\mu_{1}}{\rho^{2}-\mu_{1}\mu_{2}}$. The main results in this
aspect are summarized below. Next, we turn to the related limiting elliptic
system (1.16).

######  Lemma 6.3.

([12, Theorems 1.5 and 4.1]) For $\mu_{1},\mu_{2}>0$, if
$0<\rho<\min\\{\mu_{1},\mu_{2}\\}$ or $\rho>\max\\{\mu_{1},\mu_{2}\\}$, then
any positive least energy solution $(u,v)$ of (1.16) must be of the form

$(u,v)=\big{(}\sqrt{k_{1}}U_{\varepsilon},\sqrt{k_{2}}U_{\varepsilon}\big{)}.$

Solutions to (1.16) correspond to critical points of the functional

$I(u,v)=\int_{\mathbb{R}^{4}}\frac{1}{2}\big{(}|\nabla u|^{2}+|\nabla
v^{2}|\big{)}dx-\frac{1}{4}\big{(}\mu_{1}|u|^{4}+\mu_{2}|v|^{4}+2\rho|u|^{2}|v|^{2}\big{)}dx,\
\ $

$u,v\in D^{1,2}(\mathbb{R}^{4}).$ From [38, Lemma 2.1], if
$0<\rho<\min\\{\mu_{1},\mu_{2}\\}$ or $\rho>\max\\{\mu_{1},\mu_{2}\\}$,
$(u_{0},v_{0})$ is a least energy solution of (1.16), we can deduce that

$I(u_{0},v_{0})=\frac{k_{1}+k_{2}}{4}\mathcal{S}^{2}\ \ \text{and}\ \
\mathcal{S}_{\mu_{1},\mu_{2},\rho}=\sqrt{k_{1}+k_{2}}\mathcal{S}.$

We first study the case $\beta<0$, $\mu_{i}>0(i=1,2)$ and $\rho>0$.

Proof of Theorem 1.11. If $\beta<0$, let $(u,v)$ be a constraint critical
point of $J_{\beta}$ on $\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}$ and
$u,v>0$. Therefore, $(u,v)$ solves (1.1)-(1.2) for some
$\lambda_{1},\lambda_{2}\in\mathbb{R}$. By the Pohozaev identity
$P_{b_{1},b_{2}}(u,v)=0$, we have

$\displaystyle\lambda_{1}\|u\|^{2}_{L^{2}(\mathbb{R}^{4})}+\lambda_{2}\|v\|^{2}_{L^{2}(\mathbb{R}^{4})}$
$\displaystyle=\frac{4-N}{4}\big{(}\mu_{1}\|u\|^{4}_{L^{4}(\mathbb{R}^{4})}+\mu_{2}\|v\|^{4}_{L^{4}(\mathbb{R}^{4})}+2\rho\|uv\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{)}$
$\displaystyle\quad+\frac{6-N}{4}\beta\int_{\mathbb{R}^{N}}|u|^{2}vdx$

For $N=4,5$, then one of the $\lambda_{1},\lambda_{2}$ is negative. With out
of generality, we assume $\lambda_{1}<0$. It follows from Lemma 6.1 that
$(u,v)$ is smooth, and is in $L^{\infty}(\mathbb{R}^{N})\times
L^{\infty}(\mathbb{R}^{N})$; thus, $|\Delta u|,|\Delta v|\in
L^{\infty}(\mathbb{R}^{N})$ as well, and stand gradient estimates for the
Poisson equation (see formula (3,15) in [14]) imply that $|\nabla u|,|\nabla
v|\in L^{\infty}(\mathbb{R}^{N})$. Combining the fact that $u,v\in
L^{2}(\mathbb{R}^{N})$, we get $u(x),v(x)\to 0$ as $|x|\to\infty$. Thus, we
have

$-\Delta u=\big{(}-\lambda_{1}+\beta v+\mu_{1}u^{2}+\rho
v^{2}\big{)}u\geq\frac{-\lambda_{1}}{2}u>0\ \ $

for $|x|>R_{0}$, with $R_{0}>0$ large enough, and then $u$ is superharmonic at
infinity. From the Hadamard three spheres theorem [37, Chapter 2], this
implies that the function $m(r):=\min\limits_{|x|=r}u(x)$ satisfies

$m(r)\geq\frac{m(r_{1})\big{(}r^{2-N}-r_{2}^{2-N}\big{)}+m(r_{2})\big{(}r_{1}^{2-N}-r^{2-N}\big{)}}{r_{1}^{2-N}-r_{2}^{2-N}}\qquad\forall
R_{0}<r_{1}<r<r_{2}.$

Since $u$ decays at infinity, we have that $m(r_{2})\to 0$ as
$r_{2}\to+\infty$, it is not difficult to see that $r\mapsto r^{N-2}m(r)$ is
monotone non-decreasing for $r>R_{0}$. Moreover, $m(r)>0$ for every $r>0$
because $u>0$ in $\mathbb{R}^{N}$. Thus,

$m(r)\geq m(R_{0})R_{0}^{N-2}\cdot r^{2-N}\qquad\forall r>R_{0}.$

If $N=4$, it follows from $u\in H^{1}(\mathbb{R}^{N})$ that $u\in
L^{\frac{N}{N-2}}(\mathbb{R}^{N})$. We deduce that

$\begin{split}\|u\|_{L^{\frac{N}{N-2}}(\mathbb{R}^{N})}^{\frac{N}{N-2}}\geq
C\int_{R_{0}}^{+\infty}|m(r)|^{\frac{N}{N-2}}r^{N-1}dr\geq
C\int_{R_{0}}^{+\infty}\frac{1}{r}dr=+\infty,\end{split}$

with $C>0$. This is a contradiction. If $N=5$, the fact that $u\in
H^{1}(\mathbb{R}^{N})$ does not imply that $u\in
L^{\frac{N}{N-2}}(\mathbb{R}^{N})$ or that $u\in L^{p}(\mathbb{R}^{N})$ for
some $p\in(0,\frac{N}{N-2}]$. But, imposing such condition as an assumption,
we still reach a contradiction.

If $N\geq 4$, and $u\in H^{1}(\mathbb{R}^{N})$ is a radial function by [4,
Radial Lemma A.II], there exist $C>0$ and $R_{1}>0$ such that

$|u(x)|\leq C|x|^{-\frac{N-1}{2}}\ \ \text{for}\ \ |x|\geq R_{1}.$

If $\beta<0$, let $(u,v)$ be a non-trivial radial solution of (1.1)-(1.2).
Similarly, we assume $\lambda_{1}<0$. Setting $p(x)=-\beta v-\mu_{1}u^{2}-\rho
v^{2}$, then

$-\Delta u+p(x)u=-\lambda_{1}u.$ (6.2)

Therefore, for $N\geq 4$,

$\lim_{n\to+\infty}|x||p(x)|\leq\lim_{n\to+\infty}\big{[}C|x|^{\frac{3-N}{2}}+C|x|^{2-N}\big{]}=0.$

By Kato’s result [24], i.e. Schrödinger operator $H=-\Delta+p(x)$ has no
positive eigenvalue with an $L^{2}$-eigenfunction if $p(x)=o(|x|^{-1})$, then
(6.2) has no solution. We get (1.1)-(1.2) has no non-trivial radial solution
for $\beta<0$. ∎

######  Lemma 6.4.

If $0<\beta b_{1}<\\!\frac{3}{2|C_{4,3}|^{3}}$ and $0<\beta
b_{2}<\\!\frac{3}{|C_{4,3}|^{3}}$, for all
$(u,v)\in\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}$, there exists $t_{(u,v)}$
such that $t_{(u,v)}\star(u,v)\in\mathcal{P}_{b_{1},b_{2}}$. $t_{(u,v)}$ is
the unique critical point of the function $\Psi_{u,v}$ and is a strict maximum
point at positive level. Moreover: $(1)$ $\Psi_{u,v}^{\prime\prime}(0)<0$ and
$P_{b_{1},b_{2}}(u,v)<0$ iff $t_{(u,v)}<0$.
$(2)$ $\Psi_{u,v}$ is strictly increasing in $(-\infty,t_{(u,v)})$.
$(3)$ The map $(u,v)\mapsto t_{(u,v)}\in\mathbb{R}$ is of class $C^{1}$.

###### Proof.

We are therefore considering the defocusing nonlinearity mass-critical. We
rewrite (4.2) as

$\displaystyle\Psi_{u,v}(s)$
$\displaystyle=\frac{e^{2s}}{2}\int_{\mathbb{R}^{4}}\big{(}|\nabla
u|^{2}+|\nabla
v|^{2}-\beta|u|^{2}v\big{)}dx-\frac{e^{4s}}{4}\int_{\mathbb{R}^{4}}\big{(}\mu_{1}|u|^{4}+\mu_{2}|v|^{4}+2\rho|u|^{2}|v|^{2}\big{)}dx$
$\displaystyle\geq\frac{e^{2s}}{2}\Big{[}\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{(}1-\frac{2\beta|C_{4,3}|^{3}}{3}b_{1}\big{)}+\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{(}1-\frac{\beta|C_{4,3}|^{3}}{3}b_{2}\big{)}\Big{]}$
$\displaystyle\quad-\frac{e^{4s}}{4}\Big{(}\mu_{1}\|u\|^{4}_{L^{4}(\mathbb{R}^{4})}+\mu_{2}\|v\|^{4}_{L^{4}(\mathbb{R}^{4})}+2\rho\|uv\|^{2}_{L^{2}(\mathbb{R}^{4})}\Big{)}.$

$s\star(u,v)\in\mathcal{P}_{b_{1},b_{2}}$ if and only if
$\Psi^{\prime}_{u,v}(s)=0$, it’s easy to see that if $\big{[}\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{(}1-\frac{2\beta|C_{4,3}|^{3}}{3}b_{1}\big{)}+\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{(}1-\frac{\beta|C_{4,3}|^{3}}{3}b_{2}\big{)}\big{]}$
is positive, then $\Psi_{u,v}(s)$ has a unique critical point $t_{(u,v)}$,
which is a strict maximum point at positive level. Therefore, under the
condition of $0<b_{1}<\\!\frac{3}{2\beta|C_{4,3}|^{3}}$ and
$0<b_{2}<\\!\frac{3}{\beta|C_{4,3}|^{3}}$, we know that
$\int_{\mathbb{R}^{4}}\big{(}|\nabla u|^{2}+|\nabla
v|^{2}-\beta|u|^{2}v\big{)}dx>0$. If $(u,v)\in\mathcal{P}_{b_{1},b_{2}}$, then
$t_{(u,v)}$ is a maximum point, we have that
$\Psi^{\prime\prime}_{u,v}(t_{(u,v)})\leq 0$. We claim that
$\Psi^{\prime\prime}_{u,v}(t_{(u,v)})<0$. By contradiction, this is
$\Psi^{\prime}_{u,v}(0)=\Psi^{\prime\prime}_{u,v}(0)=0$, then necessarily
$\int_{\mathbb{R}^{4}}\big{(}\mu_{1}|u|^{4}+\mu_{2}|v|^{4}+2\rho|u|^{2}|v|^{2}\big{)}dx=0$,
which is not possible because
$(u,v)\in\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}$. Thus,
$\Psi_{u,v}^{\prime\prime}(0)<0$.

As in the proof of Lemma 4.4 shows that the map
$(u,v)\in\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}\mapsto
t_{(u,v)}\in\mathbb{R}$ is of class $C^{1}$. Finally,
$\Psi^{\prime}_{u,v}(s)<0$ if and only if $s>t_{(u,v)}$, then
$P_{b_{1},b_{2}}(u,v)=\Psi^{\prime}_{u,v}(0)<0$ if and only if $t_{(u,v)}<0$.
∎

######  Lemma 6.5.

If $0<\beta b_{1}<\\!\frac{3}{2|C_{4,3}|^{3}}$ and $0<\beta
b_{2}<\\!\frac{3}{|C_{4,3}|^{3}}$, then the set $\mathcal{P}_{b_{1},b_{2}}$ is
a $C^{1}$-submanifold of codimension 1 in
$\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}$, and it is a $C^{1}$-submanifold
of codimension 3 in $H^{1}(\mathbb{R}^{4})\times H^{1}(\mathbb{R}^{4})$.

###### Proof.

The proof is similar to that of Lemma 4.3. ∎

######  Lemma 6.6.

Assume $\mu_{i},b_{i}>0(i=1,2)$ and $\rho,\beta>0$. Let $0<\beta
b_{1}<\\!\frac{3}{2|C_{4,3}|^{3}}$ and $0<\beta
b_{2}<\\!\frac{3}{|C_{4,3}|^{3}}$, then

$m_{\beta}(b_{1},b_{2}):=\inf_{(u,v)\in\mathcal{P}_{b_{1},b_{2}}}J_{\beta}(u,v)>0.\\\
$

###### Proof.

If $(u,v)\in\mathcal{P}_{b_{1},b_{2}}$, then by Gagliardo-Nirenberg and the
Sobolev inequalities, we have

$\displaystyle\|\nabla u\|^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{4})}$
$\displaystyle=\mu_{1}\|u\|^{4}_{L^{4}(\mathbb{R}^{4})}+\mu_{2}\|v\|^{4}_{L^{4}(\mathbb{R}^{4})}+2\rho\|uv\|^{2}_{L^{2}(\mathbb{R}^{4})}+\beta\int_{\mathbb{R}^{4}}|u|^{2}vdx$
$\displaystyle\leq\mathcal{S}^{2}_{\mu_{1},\mu_{2},\rho}\big{[}\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{]}^{2}+\frac{2\beta|C_{4,3}|^{3}}{3}b_{1}\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{4})}+\frac{\beta|C_{4,3}|^{3}}{3}b_{2}\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{4})}.$

Moreover, $0<\beta b_{1}<\\!\frac{3}{2|C_{4,3}|^{3}}$ and $0<\beta
b_{2}<\\!\frac{3}{|C_{4,3}|^{3}}$, and $\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla v\|^{2}_{L^{2}(\mathbb{R}^{4})}\neq
0$ (since $(u,v)\in\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}$), we get

$\inf_{(u,v)\in\mathcal{P}_{b_{1},b_{2}}}\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla v\|^{2}_{L^{2}(\mathbb{R}^{4})}\geq
C>0.$

So

$\displaystyle m_{\beta}(b_{1},b_{2})$
$\displaystyle=\inf_{(u,v)\in\mathcal{P}_{b_{1},b_{2}}}J_{\beta}(u,v)$
$\displaystyle=\inf_{(u,v)\in\mathcal{P}_{b_{1},b_{2}}}\frac{1}{4}\big{[}\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{4})}-\beta\int_{\mathbb{R}^{4}}|u|^{2}vdx\big{]}$
$\displaystyle\geq\inf_{(u,v)\in\mathcal{P}_{b_{1},b_{2}}}\frac{1}{4}\big{(}1-\frac{2\beta|C_{4,3}|^{3}b_{1}}{3}\big{)}\|\nabla
u\|^{2}_{L^{2}(\mathbb{R}^{4})}+\frac{1}{4}\big{(}1-\frac{\beta|C_{4,3}|^{3}b_{2}}{3}\big{)}\|\nabla
v\|^{2}_{L^{2}(\mathbb{R}^{4})}$ $\displaystyle\geq C>0.$

∎

From the above lemmas, then
$J_{\beta}\big{|}_{\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}}$ has a mountain
pass geometry. We need an estimate from above on
$m_{\beta}(b_{1},b_{2})=\inf\limits_{(u,v)\in\mathcal{P}_{b_{1},b_{2}}}J_{\beta}(u,v)$.

######  Lemma 6.7.

Let $\beta,\mu_{1},\mu_{2}>0$,
$\rho\in\big{(}0,\min\\{\mu_{1},\mu_{2}\\}\big{)}\cup\big{(}\max\\{\mu_{1},\mu_{2}\\},\infty\big{)}$.
If $0<\beta b_{1}<\\!\frac{3}{2|C_{4,3}|^{3}}$ and $0<\beta
b_{2}<\\!\frac{3}{|C_{4,3}|^{3}}$, then

$0<m_{\beta}(b_{1},b_{2})<\frac{k_{1}+k_{2}}{4}\mathcal{S}^{2},$

where $k_{1}=\frac{\rho-\mu_{2}}{\rho^{2}-\mu_{1}\mu_{2}}$ and
$k_{2}=\frac{\rho-\mu_{1}}{\rho^{2}-\mu_{1}\mu_{2}}$.

###### Proof.

From (1.15),
$U_{\varepsilon}=\frac{2\sqrt{2}\varepsilon}{\varepsilon^{2}+|x|^{2}}$, taking
a radially decreasing cut-off function $\xi\in C_{0}^{\infty}(\mathbb{R}^{4})$
such that $\xi\equiv 1$ in $B_{1}$, $\xi\equiv 0$ in $\mathbb{R}^{4}\backslash
B_{2}$, and let $W_{\varepsilon}(x)=\xi(x)U_{\varepsilon}(x)$. We have (see
[23, Lemma 7.1]),

$\|\nabla
W_{\varepsilon}\|^{2}_{L^{2}(\mathbb{R}^{4})}=\mathcal{S}^{2}+O(\varepsilon^{2}),\quad\quad\|W_{\varepsilon}\|^{4}_{L^{4}(\mathbb{R}^{4})}=\mathcal{S}^{2}+O(\varepsilon^{4}),$
(6.3)

and

$\|W_{\varepsilon}\|^{3}_{L^{3}(\mathbb{R}^{4})}=O(\varepsilon),\quad\quad\|W_{\varepsilon}\|^{2}_{L^{2}(\mathbb{R}^{4})}=O(\varepsilon^{2}|\ln\varepsilon|).$
(6.4)

Setting

$\big{(}\overline{W}_{\varepsilon},\overline{V}_{\varepsilon}\big{)}=\Big{(}a_{1}\frac{W_{\varepsilon}}{\|W_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}},a_{2}\frac{W_{\varepsilon}}{\|W_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}}\Big{)},$

then
$(\overline{W}_{\varepsilon},\overline{V}_{\varepsilon})\in\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}$.
From Lemma 6.4, there exists $\tau_{\varepsilon}\in\mathbb{R}$ such that
$\tau_{\varepsilon}\star(\overline{W}_{\varepsilon},\overline{V}_{\varepsilon})\in\mathcal{P}_{b_{1},b_{2}}$,
this implies that,

$m_{\beta}(b_{1},b_{2})\leq
J_{\beta}\big{(}\tau_{\varepsilon}\star(\overline{W}_{\varepsilon},\overline{V}_{\varepsilon})\big{)}=\max_{t\in\mathbb{R}}J_{\beta}\big{(}t\star(\overline{W}_{\varepsilon},\overline{V}_{\varepsilon})\big{)},$

and

$\displaystyle\big{[}\|\nabla\overline{W}_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}^{2}+\|\nabla\overline{V}_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}^{2}\big{]}$
$\displaystyle=e^{2\tau_{\varepsilon}}\big{[}\mu_{1}\|\overline{W}_{\varepsilon}\|_{L^{4}(\mathbb{R}^{4})}^{4}+\mu_{2}\|\overline{V}_{\varepsilon}\|_{L^{4}(\mathbb{R}^{4})}^{4}$
(6.5)
$\displaystyle\quad+2\rho\|\overline{W}_{\varepsilon}\overline{V}_{\varepsilon}\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{]}+\int_{\mathbb{R}^{4}}\beta|\overline{W}_{\varepsilon}|^{2}\overline{V}_{\varepsilon}dx.$

Then,

$\displaystyle e^{2\tau_{\varepsilon}}$
$\displaystyle\big{[}\mu_{1}\|\overline{W}_{\varepsilon}\|_{L^{4}(\mathbb{R}^{4})}^{4}+\mu_{2}\|\overline{V}_{\varepsilon}\|_{L^{4}(\mathbb{R}^{4})}^{4}+2\rho\|\overline{W}_{\varepsilon}\overline{V}_{\varepsilon}\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{]}$
$\displaystyle=\big{[}\|\nabla\overline{W}_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}^{2}+\|\nabla\overline{V}_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}^{2}\big{]}-\int_{\mathbb{R}^{4}}\beta|\overline{W}_{\varepsilon}|^{2}\overline{V}_{\varepsilon}dx$
$\displaystyle\geq\big{(}1-\frac{2\beta|C_{4,3}|^{3}b_{1}}{3}\big{)}\|\nabla\overline{W}_{\varepsilon}\|^{2}_{L^{2}(\mathbb{R}^{4})}+\big{(}1-\frac{\beta|C_{4,3}|^{3}b_{2}}{3}\big{)}\|\nabla\overline{V}_{\varepsilon}\|^{2}_{L^{2}(\mathbb{R}^{4})}.$

Therefore, for $\varepsilon$ small enough, we have

$\displaystyle
J_{\beta}\big{(}\tau_{\varepsilon}\star(\overline{W}_{\varepsilon},\overline{V}_{\varepsilon})\big{)}$
$\displaystyle=\frac{e^{2\tau_{\varepsilon}}}{2}\big{(}\|\nabla\overline{W}_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}^{2}+\|\nabla\overline{V}_{\varepsilon}\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{)}-\frac{e^{2\tau_{\varepsilon}}}{2}\int_{\mathbb{R}^{4}}|\overline{W}_{\varepsilon}|^{2}\overline{V}_{\varepsilon}dx$
$\displaystyle\quad-\frac{e^{4\tau_{\varepsilon}}}{4}\big{(}\mu_{1}\|\overline{W}_{\varepsilon}\|_{L^{4}(\mathbb{R}^{4})}^{4}+\mu_{2}\|\overline{V}_{\varepsilon}\|_{L^{4}(\mathbb{R}^{4})}^{4}+2\rho\|\overline{W}_{\varepsilon}\overline{V}_{\varepsilon}\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{)}$
$\displaystyle\leq\max_{s_{1},s_{2}>0}\frac{s^{2}_{1}+s^{2}_{2}}{2}\|\nabla
W_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}^{2}-\frac{\big{(}\mu_{1}s^{4}_{1}+\mu_{2}s^{4}_{2}+2\rho
s^{2}_{1}s^{2}_{2}\big{)}}{4}\|W_{\varepsilon}\|_{L^{4}(\mathbb{R}^{4})}^{4}$
$\displaystyle\quad-\frac{e^{2\tau_{\varepsilon}}}{2}\int_{\mathbb{R}^{4}}|\overline{W}_{\varepsilon}|^{2}\overline{V}_{\varepsilon}dx,$

where
$s_{1}=\frac{e^{\tau_{\varepsilon}}b_{1}}{\|W_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}}$
and
$s_{2}=\frac{e^{\tau_{\varepsilon}}b_{2}}{\|W_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}}$.
Define

$f(s_{1},s_{2})=\frac{s^{2}_{1}+s^{2}_{2}}{2}\|\nabla
W_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}^{2}-\frac{(\mu_{1}s^{4}_{1}+\mu_{2}s^{4}_{2}+2\rho
s^{2}_{1}s^{2}_{2})}{4}\|W_{\varepsilon}\|_{L^{4}(\mathbb{R}^{4})}^{4}.$

Using that, for all $0<s_{1},s_{2}$,

$\max_{s_{1},s_{2}>0}f(s_{1},s_{2})\leq\frac{k_{1}+k_{2}}{4}\mathcal{S}^{2}+O(\varepsilon^{2}).$

Finally,

$\displaystyle\frac{e^{2\tau_{\varepsilon}}}{2}\int_{\mathbb{R}^{4}}|\overline{W}_{\varepsilon}|^{2}\overline{V}_{\varepsilon}dx$
$\displaystyle=\frac{e^{2\tau_{\varepsilon}}}{2}\frac{b^{2}_{1}b_{2}}{\|W_{\varepsilon}\|^{3}_{L^{2}(\mathbb{R}^{4})}}\int_{\mathbb{R}^{4}}|W_{\varepsilon}|^{3}dx$
$\displaystyle\geq\frac{C}{\|W_{\varepsilon}\|_{L^{2}(\mathbb{R}^{4})}}\int_{\mathbb{R}^{4}}|W_{\varepsilon}|^{3}dx$
$\displaystyle\geq C|\ln\varepsilon|^{-\frac{1}{2}},$

hence, we deduce that

$\max_{t\in\mathbb{R}}J_{\beta}\big{(}t\star(\overline{W}_{\varepsilon},\overline{V}_{\varepsilon})\big{)}<\frac{k_{1}+k_{2}}{4}\mathcal{S}^{2}.$

∎

Now, we are ready to show that the infimum is attained by nontrivial positive
radial functions.

######  Lemma 6.8.

Let $\mu_{i},b_{i}>0(i=1,2)$, and
$\rho\in\big{(}0,\min\\{\mu_{1},\mu_{2}\\}\big{)}\cup\big{(}\max\\{\mu_{1},\mu_{2}\\},\infty\big{)}$.
If $0<\beta b_{1}<\\!\frac{3}{2|C_{4,3}|^{3}}$, $0<\beta
b_{2}<\\!\frac{3}{|C_{4,3}|^{3}}$ and

$0<m_{\beta}(b_{1},b_{2})<\frac{k_{1}+k_{2}}{4}\mathcal{S}^{2},$

where $k_{1}=\frac{\rho-\mu_{2}}{\rho^{2}-\mu_{1}\mu_{2}}$ and
$k_{2}=\frac{\rho-\mu_{1}}{\rho^{2}-\mu_{1}\mu_{2}}$, then
$m_{\beta}(b_{1},b_{2})$ can be achieved by some function
$(u_{b_{1},b_{2}},v_{b_{1},b_{2}})\in\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}$
which is real valued, positive, radially symmetric and radially decreasing.

###### Proof.

By Lemma 4.12, we can also find a radial Palais-Smale sequence for
$J_{\beta}\big{|}_{\mathrm{T}_{b_{1}}\times\mathrm{T}_{b_{2}}}$ at level
$m_{\beta}(b_{1},b_{2})$ such that $P(u_{n},v_{n})\to 0$ and
$u^{-}_{n},v^{-}_{n}\to 0$ a.e. in $\mathbb{R}^{4}$. The rest of proof is
similar to that of Lemma 5.5 in [31], and just needs a slight modification.. ∎

######  Lemma 6.9.

Let $\mu_{i},\beta>0(i=1,2)$, and
$\rho\in\big{(}0,\min\\{\mu_{1},\mu_{2}\\}\big{)}\cup\big{(}\max\\{\mu_{1},\mu_{2}\\},\infty\big{)}$.
If $0<b_{1}<\frac{3}{2\beta|C_{4,3}|^{3}}$ and
$0<b_{2}<\frac{3}{\beta|C_{4,3}|^{3}}$, for any ground state
$(u_{b_{1},b_{2}},v_{b_{1},b_{2}})$ of (1.1)-(1.2), then

$J_{\beta}(u_{b_{1},b_{2}},v_{b_{1},b_{2}})=m_{\beta}(b_{1},b_{2})\to\frac{k_{1}+k_{2}}{4}\mathcal{S}^{2}\
\ \text{as}\ \ (b_{1},b_{2})\to(0,0).$

Moreover, there exists $\varepsilon_{b_{1},b_{2}}>0$ such that

$\big{(}\sigma_{1}u_{b_{1},b_{2}}(\sigma_{1}x),\sigma_{1}v_{b_{1},b_{2}}(\sigma_{1}x)\big{)}\to(\sqrt{k_{1}}U_{\varepsilon_{0}},\sqrt{k_{2}}U_{\varepsilon_{0}})\
\ \text{in}\ D^{1,2}(\mathbb{R}^{4})\times D^{1,2}(\mathbb{R}^{4}),$

for some $\varepsilon_{0}>0$ as $(b_{1},b_{2})\to(0,0)$ up to a subsequence,
where $k_{1}=\frac{\rho-\mu_{2}}{\rho^{2}-\mu_{1}\mu_{2}}$ and
$k_{2}=\frac{\rho-\mu_{1}}{\rho^{2}-\mu_{1}\mu_{2}}$.

###### Proof.

By $P_{b_{1},b_{2}}(u_{b_{1}},v_{b_{2}})=0$, we have

$\displaystyle\frac{k_{1}+k_{2}}{4}\mathcal{S}^{2}$
$\displaystyle>J_{\beta}(u_{b_{1},b_{2}},v_{b_{1},b_{2}})=\frac{1}{4}\big{(}\|\nabla
u_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}-\beta\int_{\mathbb{R}^{4}}|u_{b_{1},b_{2}}|^{2}v_{b_{1},b_{2}}dx\big{)}$
$\displaystyle\geq\frac{1}{4}\big{(}1-\frac{2\beta
b_{1}}{3}|C_{4,3}|^{3}\big{)}\|\nabla
u_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}+\frac{1}{4}\big{(}1-\frac{\beta
b_{2}}{3}|C_{4,3}|^{3}\big{)}\|\nabla
v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})},$

then $\\{(u_{b_{1},b_{2}},v_{b_{1},b_{2}})\\}$ is bounded in
$H^{1}(\mathbb{R}^{4})\times H^{1}(\mathbb{R}^{4})$. Using the Hölder and
Gagliardo-Nirenberg inequality again,

$\beta\int_{\mathbb{R}^{4}}|u_{b_{1},b_{2}}|^{2}v_{b_{1},b_{2}}dx\leq\beta|C_{4,3}|^{3}\|\nabla
u_{b_{1},b_{2}}\|^{\frac{4}{3}}_{L^{2}(\mathbb{R}^{4})}\|\nabla
v_{b_{1},b_{2}}\|^{\frac{2}{3}}_{L^{2}(\mathbb{R}^{4})}b^{2}_{1}b_{2}\to 0,$

as $(b_{1},b_{2})\to(0,0)$. Therefore, it follows from
$P(u_{b_{1},b_{2}},v_{b_{1},b_{2}})=0$ that

$\displaystyle\|\nabla u_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}=\mu_{1}\|u_{b_{1},b_{2}}\|^{4}_{L^{4}(\mathbb{R}^{4})}+\mu_{2}\|v_{b_{1},b_{2}}\|^{4}_{L^{4}(\mathbb{R}^{4})}+2\rho\|u_{b_{1},b_{2}}v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}+o(1).$

From (1.17), we have

$\displaystyle\sqrt{k_{1}+k_{2}}\mathcal{S}\big{(}\mu_{1}\|u_{b_{1},b_{2}}\|^{4}_{L^{4}(\mathbb{R}^{4})}+\mu_{2}\|v_{b_{1},b_{2}}\|^{4}_{L^{4}(\mathbb{R}^{4})}+2\beta\|u_{b_{1},b_{2}}v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{)}^{\frac{1}{2}}\leq||\nabla
u_{b_{1},b_{2}}||^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}.$

Thus, we distinguish the two cases

$\displaystyle\text{either}\ \ (i)\ \ ||\nabla
u_{b_{1},b_{2}}||^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}\to 0\ \ \text{or}(ii)\ \
||\nabla u_{b_{1},b_{2}}||^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}\to l>0.$

We claim that $(i)$ is impossible. Indeed, if $l=0$, by
$\Psi^{\prime\prime}_{u_{b_{1},b_{2}},v_{b_{1},b_{2}}}(0)<0$,

$\displaystyle\|\nabla u_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}$
$\displaystyle<2\big{[}\mu_{1}\|u_{b_{1},b_{2}}\|^{4}_{L^{4}(\mathbb{R}^{4})}+\mu_{2}\|v_{b_{1},b_{2}}\|^{4}_{L^{4}(\mathbb{R}^{4})}$
$\displaystyle\quad+2\rho\|u_{b_{1},b_{2}}v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{]}+\int_{\mathbb{R}^{4}}\beta|u_{b_{1},b_{2}}|^{2}v_{b_{1},b_{2}}dx,$

we obtain a contradiction

$\displaystyle\big{(}1-\frac{2\beta b_{1}}{3}|C_{4,3}|^{3}\big{)}\|\nabla
u_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}+\big{(}1-\frac{\beta
b_{2}}{3}|C_{4,3}|^{3}\big{)}\|\nabla
v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}$
$\displaystyle<\frac{2}{\mathcal{S}^{2}_{\mu_{1},\mu_{2},\rho}}\big{[}\|\nabla
u_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}\big{]}^{2}.$

The claim is proved. Therefore, $l>0$, we obtain that
$J_{\beta}(u_{b_{1},b_{2}},v_{b_{1},b_{2}})\to\frac{k_{1}+k_{2}}{4}\mathcal{S}^{2}$
as $(b_{1},b_{2})\to(0,0)$.

Thus, we know that

$\displaystyle||\nabla u_{b_{1},b_{2}}||^{2}_{L^{2}(\mathbb{R}^{4})}+\|\nabla
v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}\to(k_{1}+k_{2})\mathcal{S}^{2},$
$\displaystyle\mu_{1}\|u_{b_{1},b_{2}}\|^{4}_{L^{4}(\mathbb{R}^{4})}+\mu_{2}\|v_{b_{1},b_{2}}\|^{4}_{L^{4}(\mathbb{R}^{4})}+2\beta\|u_{b_{1},b_{2}}v_{b_{1},b_{2}}\|^{2}_{L^{2}(\mathbb{R}^{4})}\to(k_{1}+k_{2})\mathcal{S}^{2},$

as $(b_{1},b_{2})\to(0,0)$. It follows that, up to a subsequence,
$\\{(u_{b_{1},b_{2}},v_{b_{1},b_{2}})\\}$ is a minimizing sequence of the
minimizing problem (1.17). From Lemma 6.8, $(u_{b_{1},b_{2}},v_{b_{1},b_{2}})$
is radially symmetric. By [47, Theorem 1.41] or [26, Lemma 3.5], up to a
subsequence, there exists $\sigma_{1}:=\sigma_{1}(a_{1},a_{2})$ such that for
some $\varepsilon_{0}>0$,

$\big{(}\sigma_{1}u_{b_{1},b_{2}}(\sigma_{1}x),\sigma_{1}v_{b_{1},b_{2}}(\sigma_{1}x)\big{)}\to\big{(}\sqrt{k_{1}}U_{\varepsilon_{0}},\sqrt{k_{2}}U_{\varepsilon_{0}}\big{)}$

in $D^{1,2}(\mathbb{R}^{4})\times D^{1,2}(\mathbb{R}^{4})$, as
$(b_{1},b_{2})\to(0,0)$. ∎

Proof of Theorem 1.5. The proof is finished when we combine Lemma 6.8 and
Lemma 6.9. ∎

Acknowledgements:  The research of J. Wei is partially supported by NSERC of
Canada. X. Luo was supported by the National Natural Science Foundation of
China (Grant No. 11901147) and the Fundamental Research Funds for the Central
University of China(Grant No. JZ2020HGTB0030). Maoding Zhen was supported by
the Fundamental Research Funds for the Central Universities (Grant No.
JZ2021HGTA0177).

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