In this section, we will find the range of c where the \((\mathrm{PS})_{c}\) condition holds for the functional \(I_{\lambda}\). For convenience, we give some notations.
Notations
-
The dual space of a Banach space E will be denoted by \(E^{\ast}\).
-
\(B_{r}:=\{x\in\mathbb{R}^{N}:|x|\leq r\}\) is the ball in \(\mathbb{R}^{N}\).
-
c, \(c_{i}\) represent various positive constants, the exact values of which are not important.
-
Let \(C_{0}^{\infty}(\mathbb{R}^{N})\) denote the collection of smooth functions with compact support.
-
\(o(1)\) denotes \(o(1)\rightarrow0\) as \(n\rightarrow\infty\).
-
\(S_{\alpha,\beta}\) is the best Sobolev embedding constant defined by
$$ S_{\alpha,\beta}= \inf_{u,v \in H^{1}(\mathbb{R}^{N})} \frac{\int_{\mathbb{R}^{N}}(|\nabla u|^{2}+|\nabla v|^{2})}{(\int_{\mathbb {R}^{N}}|u|^{\alpha}|v|^{\beta})^{\frac{2}{\alpha+\beta}}}. $$
(3.1)
We can obtain
$$ S_{\alpha,\beta}= \biggl(\biggl(\frac{\alpha}{\beta}\biggr)^{\frac{\beta}{\alpha +\beta}}+\biggl( \frac{\beta}{\alpha}\biggr)^{\frac{\alpha}{\alpha+\beta}}\biggr)S, $$
where S is the best Sobolev embedding constant defined by
$$ S= \inf_{u \in H^{1}(\mathbb{R}^{N})} \frac{\int_{\mathbb {R}^{N}}|\nabla u|^{2}}{(\int_{\mathbb{R}^{N}}|u|^{2^{*}})^{\frac{2}{2^{*}}}}. $$
Based on the assumptions of Theorem 2 in [27], we can show that \(I_{\lambda}\in C^{1}(E,\mathbb{R})\) and the critical points of \(I_{\lambda}\) are weak solutions of the problem (2.1).
Recall that we say that a sequence \(\{(u_{n},v_{n})\} \subset E\) is a \((\mathrm{PS})\) sequence at level c (\((\mathrm{PS})_{c}\) sequence, for short) if \(I_{\lambda}(u_{n},v_{n})\rightarrow c\) and \(I_{\lambda}'(u_{n},v_{n})\rightarrow0\). \(I_{\lambda}\) is said to satisfy the \((\mathrm{PS})_{c}\) condition if any \((\mathrm{PS})_{c}\) sequence contains a convergent subsequence.
Lemma 3.1
If the sequence
\(\{(u_{n},v_{n})\}\subset E\)
is a
\((\mathrm{PS})_{c}\)
sequence for
\(I_{\lambda}\), then we find that
\(c\geq0\)
and
\(\{(u_{n},v_{n})\}\)
is bounded in the space
E.
Proof
We have
$$\begin{aligned}& I_{\lambda}(u_{n},v_{n})-\frac{1}{\mu}I_{\lambda }'(u_{n},v_{n}) (u_{n},v_{n}) \\& \quad = \frac{1}{2}\bigl\Vert (u_{n},v_{n})\bigr\Vert _{E}^{2}-\lambda\int_{\mathbb {R}^{N}} \bigl(F(x,u)+G(x,v)\bigr)-\frac{\lambda}{\alpha+\beta} \int_{\mathbb {R}^{N}}K(x)|u_{n}|^{\alpha}|v_{n}|^{\beta} \\& \qquad {}-\frac{1}{\mu}\biggl[\bigl\Vert (u_{n},v_{n}) \bigr\Vert _{E}^{2}-\lambda\int_{\mathbb {R}^{N}} \bigl(f(x,u_{n})u_{n}+g(x,v_{n})v_{n} \bigr)-\lambda\int_{\mathbb {R}^{N}}K(x)|u_{n}|^{\alpha}|v_{n}|^{\beta} \biggr] \\& \quad = \biggl(\frac{1}{2}-\frac{1}{\mu}\biggr)\bigl\Vert (u_{n},v_{n})\bigr\Vert _{E}^{2}+ \lambda\int_{\mathbb{R}^{N}}\biggl(\frac{1}{\mu }\bigl(f(x,u_{n})u_{n}+g(x,v_{n})v_{n} \bigr)-F(x,u_{n})-G(x,v_{n})\biggr) \\& \qquad {}+\biggl(\frac{1}{\mu}-\frac{1}{\alpha+\beta}\biggr)\lambda\int _{\mathbb {R}^{N}}K(x)|u_{n}|^{\alpha}|v_{n}|^{\beta}. \end{aligned}$$
Together with (K0), (H3), and \(2<\mu<2^{\ast}\), we get
$$I_{\lambda}(u_{n},v_{n})-\frac{1}{\mu}I_{\lambda }'(u_{n},v_{n}) (u_{n},v_{n})\geq\biggl(\frac{1}{2}- \frac{1}{\mu}\biggr)\bigl\Vert (u_{n},v_{n})\bigr\Vert _{E}^{2}. $$
By the fact that \(I_{\lambda}(u_{n},v_{n})\rightarrow c\) and \(I_{\lambda }'(u_{n},v_{n})\rightarrow0\), we easily obtain the desired conclusion. □
Lemma 3.2
There exists a subsequence
\(\{ (u_{n_{j}},v_{n_{j}})\}\)
such that for any
\(\varepsilon>0\), there is
\(r_{\varepsilon}>0\)
with
\(r\geq r_{\varepsilon}\),
$$\lim_{j\rightarrow\infty}\sup\int_{B_{j}\setminus B_{r}} \bigl(|u_{n_{j}}|^{d}+|v_{n_{j}}|^{d}\bigr)\leq \varepsilon, $$
where
\(2\leq d<2^{\ast}\).
Proof
Together with Lemma 3.1, the \((\mathrm{PS})_{c}\) sequence \(\{(u_{n},v_{n})\}\) for \(I_{\lambda}\) is bounded in E. So, we assume \((u_{n},v_{n})\rightharpoonup (u,v)\) in E, \(u_{n}\rightarrow u, v_{n}\rightarrow v\) a.e. in \(\mathbb {R}^{N}\) and \((u_{n},v_{n})\rightarrow(u,v)\) in \(L_{\mathrm{loc}}^{d}(\mathbb {R}^{N})\times L_{\mathrm{loc}}^{d}(\mathbb{R}^{N})\) for any \(2\leq d<2^{\ast}\). Note that, for each \(j\in\mathbb{N}\), we have
$$ \int_{B_{j}}\bigl(|u_{n}|^{d}+|v_{n}|^{d} \bigr)\rightarrow\int_{B_{j}}\bigl(|u|^{d}+|v|^{d} \bigr). $$
Thus, there exists \(n_{0}\in\mathbb{N}\) such that
$$ \int_{B_{j}}\bigl(|u_{n}|^{d}+|v_{n}|^{d}-|u|^{d}-|v|^{d} \bigr)< \frac{1}{j} $$
for all \(n\geq n_{0}+1\). Without loss of generality, we may choose \(n_{j}=n_{0}+j\) such that
$$ \int_{B_{j}}\bigl(|u_{n_{j}}|^{d}+|v_{n_{j}}|^{d}-|u|^{d}-|v|^{d} \bigr)< \frac{1}{j}. $$
It is easy to see there is \(r_{\varepsilon}\) satisfying
$$ \int_{\mathbb{R}^{N}\setminus B_{r}}\bigl(|u|^{d}+|v|^{d}\bigr)< \varepsilon \quad \mbox {for all } r \geq r_{\varepsilon}. $$
Since
$$ \int_{B_{j}\setminus B_{r}}\bigl(|u_{n_{j}}|^{d}+|v_{n_{j}}|^{d} \bigr)< \frac{1}{j}+\int_{\mathbb{R}^{N}\setminus B_{r}}\bigl(|u|^{d}+|v|^{d} \bigr) +\int_{B_{r}}\bigl(|u|^{d}-|u_{n_{j}}|^{d}+|v|^{d}-|v_{n_{j}}|^{d} \bigr) $$
and
$$ (u_{n},v_{n})\rightarrow(u,v) \quad \mbox{in } L_{\mathrm{loc}}^{d}\bigl(\mathbb {R}^{N}\bigr)\times L_{\mathrm{loc}}^{d}\bigl(\mathbb{R}^{N}\bigr), $$
and the lemma follows. □
Let \(\eta\in C^{\infty}(\mathbb{R}^{+},[0,1])\) be a smooth function satisfying \(0\leq\eta(t)\leq1\), \(t\geq0\). \(\eta(t)=1\) if \(t\leq1\) and \(\eta(t)=0\) if \(t\geq2\). Define \(\tilde{u}_{j}(x)=\eta (2|x|/j)u(x)\) and \(\tilde{v}_{j}(x)=\eta(2|x|/j)v(x)\), then
$$ \tilde{u}_{j}\rightarrow u,\qquad \tilde{v}_{j} \rightarrow v \quad \mbox{in } E_{\lambda}\mbox{ as } j\rightarrow\infty. $$
(3.2)
Lemma 3.3
One has
$$\lim_{j\rightarrow\infty}\biggl\vert \int_{\mathbb {R}^{N}} \bigl(f(x,u_{n_{j}})-f(x,u_{n_{j}}-\tilde{u}_{j})-f(x, \tilde {u}_{j})\bigr)\varphi\biggr\vert =0 $$
and
$$\lim_{j\rightarrow\infty}\biggl\vert \int_{\mathbb {R}^{N}} \bigl(g(x,v_{n_{j}})-g(x,v_{n_{j}}-\tilde{v}_{j})-g(x, \tilde{v}_{j})\bigr)\psi\biggr\vert =0 $$
uniformly in
\((\varphi,\psi)\in E\)
with
\(\|(\varphi,\psi)\|_{E}\leq1\).
Proof
Note that (3.2) and local compactness of the Sobolev embedding imply that for any \(r>0\),
$$\lim_{j\rightarrow\infty}\biggl\vert \int_{\mathbb {B}^{r}} \bigl(f(x,u_{n_{j}})-f(x,u_{n_{j}}-\tilde{u}_{j})-f(x, \tilde {u}_{j})\bigr)\varphi\biggr\vert =0 $$
uniformly in \(\|\varphi\|\leq1\). For any \(\varepsilon>0\), it follows from
$$\int_{\mathbb{R}^{N}\setminus B_{r}}\bigl(|u|^{d}+|v|^{d}\bigr)< \varepsilon $$
that
$$\lim_{j\rightarrow\infty}\sup\int_{B_{j}\setminus B_{r}}|\tilde {u_{j}}|^{d}\leq\int_{\mathbb{R}^{N}\setminus B_{r}}|u|^{d} \leq\varepsilon \quad \mbox{for all } r \geq r_{\varepsilon}. $$
By using Lemma 3.2 and the assumption (H2), we get
$$\begin{aligned}& \lim_{j\rightarrow\infty}\sup\biggl\vert \int_{\mathbb {R}^{N}} \bigl(f(x,u_{n_{j}})-f(x,u_{n_{j}}-\tilde{u}_{j})-f(x, \tilde {u}_{j})\bigr)\varphi\biggr\vert \\& \quad = \lim_{j\rightarrow\infty}\sup\biggl\vert \int_{B_{j}\setminus B_{r}} \bigl(f(x,u_{n_{j}})-f(x,u_{n_{j}}-\tilde{u}_{j})-f(x, \tilde{u}_{j})\bigr)\varphi \biggr\vert \\& \quad \leq c_{2}\lim_{j\rightarrow\infty}\sup\biggl\vert \int _{B_{j}\setminus B_{r}}\bigl(\vert u_{n_{j}}\vert +| \tilde{u}_{j}|\bigr)|\varphi|\biggr\vert \\& \qquad {} +c_{3}\lim_{j\rightarrow\infty}\sup\biggl\vert \int _{B_{j}\setminus B_{r}}\bigl(|u_{n_{j}}|^{q-1}+| \tilde{u}_{j}|^{q-1}\bigr)|\varphi|\biggr\vert \\& \quad \leq c_{2}\lim_{j\rightarrow\infty}\sup\bigl(\Vert u_{n_{j}}\Vert _{L^{2}(B_{j}\setminus B_{r})}+\|\tilde{u}_{j} \|_{L^{2}(B_{j}\setminus B_{r})}\bigr)\| \varphi\|_{2} \\& \qquad {}+c_{3}\lim_{j\rightarrow\infty}\sup\bigl( \|u_{n_{j}}\|_{L^{q}(B_{j}\setminus B_{r})}^{q-1}+\|\tilde{u}_{j} \|_{L^{q}(B_{j}\setminus B_{r})}^{q-1}\bigr)\|\varphi\| _{q} \\& \quad \leq c_{4}\varepsilon^{\frac{1}{2}}+c_{5} \varepsilon^{\frac{q-1}{q}}, \end{aligned}$$
which implies that
$$\lim_{j\rightarrow\infty}\biggl\vert \int_{\mathbb {R}^{N}} \bigl(f(x,u_{n_{j}})-f(x,u_{n_{j}}-\tilde{u}_{j})-f(x, \tilde {u}_{j})\bigr)\varphi\biggr\vert =0. $$
Similar to this proof, we can prove that the other result is correct. □
Lemma 3.4
Passing to a subsequence, we have
$$I_{\lambda}(u_{n}-\tilde{u}_{n},v_{n}- \tilde{v}_{n})\rightarrow c-I_{\lambda}(u,v) $$
and
$$I_{\lambda}'(u_{n}-\tilde{u}_{n},v_{n}- \tilde{v}_{n})\rightarrow0\quad \textit{in } E^{\ast}. $$
Proof
Together with the fact that \((u_{n},v_{n})\rightharpoonup(u,v)\), \((\tilde {u}_{n},\tilde{v}_{n})\rightarrow(u,v)\) in E, we get
$$\begin{aligned}& I_{\lambda}(u_{n}-\tilde{u}_{n},v_{n}- \tilde{v}_{n}) \\& \quad = I_{\lambda}(u_{n},v_{n})-I_{\lambda}( \tilde{u}_{n},\tilde{v}_{n}) \\& \qquad {} +\frac{\lambda}{\alpha+\beta}\int_{\mathbb {R}^{N}}K(x) \bigl(|u_{n}|^{\alpha}|v_{n}|^{\beta}-|u_{n}- \tilde{u}_{n}|^{\alpha }|v_{n}-\tilde{v}_{n}|^{\beta}-| \tilde{u}_{n}|^{\alpha}|\tilde {v}_{n}|^{\beta} \bigr) \\& \qquad {}+\lambda\int_{\mathbb{R}^{N}}\bigl(F(x,u_{n})-F(x,u_{n}- \tilde {u}_{n})-F(x,\tilde{u}_{n})\bigr) \\& \qquad {}+\lambda\int_{\mathbb{R}^{N}}\bigl(G(x,v_{n})-G(x,v_{n}- \tilde {v}_{n})-G(x,\tilde{v}_{n})\bigr)+o(1). \end{aligned}$$
Similar to the proof of the Brézis-Lieb lemma [28], we easily get
$$\begin{aligned}& \lim_{n\rightarrow\infty}\int_{\mathbb{R}^{N}}K(x) \bigl(|u_{n}|^{\alpha }|v_{n}|^{\beta}-|u_{n}- \tilde{u}_{n}|^{\alpha}|v_{n}-\tilde{v}_{n}|^{\beta }-| \tilde{u}_{n}|^{\alpha}|\tilde{v}_{n}|^{\beta} \bigr)=0, \\& \lim_{n\rightarrow\infty}\int_{\mathbb {R}^{N}}\bigl(F(x,u_{n})-F(x,u_{n}- \tilde{u}_{n})-F(x,\tilde{u}_{n})\bigr)=0 \end{aligned}$$
and
$$\lim_{n\rightarrow\infty}\int_{\mathbb {R}^{N}}\bigl(G(x,v_{n})-G(x,v_{n}- \tilde{v}_{n})-G(x,\tilde{v}_{n})\bigr)=0. $$
Observing the fact that \(I_{\lambda}(u_{n},v_{n})\rightarrow c\) and \(I_{\lambda}(\tilde{u}_{n},\tilde{v}_{n})\rightarrow I_{\lambda}(u,v)\), we obtain
$$I_{\lambda}(u_{n}-\tilde{u}_{n},v_{n}- \tilde{v}_{n})\rightarrow c-I_{\lambda}(u,v). $$
In addition, for any \((\varphi,\psi)\in E\), we get
$$\begin{aligned}& I_{\lambda}'(u_{n}-\tilde{u}_{n},v_{n}- \tilde{v}_{n}) (\varphi,\psi) \\& \quad = I_{\lambda}'(u_{n},v_{n}) ( \varphi,\psi)-I_{\lambda}'(\tilde {u}_{n}, \tilde{v}_{n}) (\varphi,\psi) \\& \qquad {} +\frac{\lambda\alpha}{\alpha+\beta}\int_{\mathbb {R}^{N}}K(x) \bigl(|u_{n}|^{\alpha-2}u_{n}|v_{n}|^{\beta}-|u_{n}- \tilde {u}_{n}|^{\alpha-2}(u_{n}-\tilde{u}_{n})|v_{n}- \tilde{v}_{n}|^{\beta} \\& \qquad {}-|\tilde {u}_{n}|^{\alpha-2} \tilde{u}_{n}|\tilde{v}_{n}|^{\beta}\bigr)\varphi \\& \qquad {} +\frac{\lambda\beta}{\alpha+\beta}\int_{\mathbb {R}^{N}}K(x) \bigl(|u_{n}|^{\alpha}|v_{n}|^{\beta-2}v_{n}-|u_{n}- \tilde {u}_{n}|^{\alpha}|v_{n}-\tilde{v}_{n}|^{\beta-2}(v_{n}- \tilde{v}_{n}) \\& \qquad {}-|\tilde {u}_{n}|^{\alpha}| \tilde{v}_{n}|^{\beta-2}\tilde{v}_{n}\bigr)\psi \\& \qquad {} +\lambda\int_{\mathbb{R}^{N}}\bigl(f(x,u_{n})-f(x,u_{n}- \tilde {u}_{n})-f(x,\tilde{u}_{n})\bigr)\varphi \\& \qquad {} +\lambda\int_{\mathbb{R}^{N}}\bigl(g(x,v_{n})-g(x,v_{n}- \tilde {v}_{n})-g(x,\tilde{v}_{n})\bigr)\psi. \end{aligned}$$
It is standard to check
$$\lim_{n\rightarrow\infty}\int_{\mathbb{R}^{N}}K(x) \bigl(|u_{n}|^{\alpha -2}u_{n}|v_{n}|^{\beta}-|u_{n}- \tilde{u}_{n}|^{\alpha-2}(u_{n}-\tilde {u}_{n})|v_{n}-\tilde{v}_{n}|^{\beta}-| \tilde{u}_{n}|^{\alpha-2}\tilde {u}_{n}| \tilde{v}_{n}|^{\beta}\bigr)\varphi=0 $$
and
$$\lim_{n\rightarrow\infty}\int_{\mathbb{R}^{N}}K(x) \bigl(|u_{n}|^{\alpha }|v_{n}|^{\beta-2}v_{n}-|u_{n}- \tilde{u}_{n}|^{\alpha}|v_{n}-\tilde {v}_{n}|^{\beta-2}(v_{n}-\tilde{v}_{n})-| \tilde{u}_{n}|^{\alpha}|\tilde {v}_{n}|^{\beta-2} \tilde{v}_{n}\bigr)\psi=0 $$
uniformly in \(\|(\varphi,\psi)\|_{E}\leq1\). By the fact of Lemma 3.3 and \(I_{\lambda}'(u_{n},v_{n})\rightarrow0\), we complete the proof of Lemma 3.4. □
Set \(u_{n}^{1}=u_{n}-\tilde{u}_{n}\) and \(v_{n}^{1}=v_{n}-\tilde{v}_{n}\), then \(u_{n}-u=u_{n}^{1}+(\tilde{u}_{n}-u)\) and \(v_{n}-v=v_{n}^{1}+(\tilde{v}_{n}-v)\). We easily get \((u_{n},v_{n})\rightarrow(u,v)\) in E if and only if \((u_{n}^{1},v_{n}^{1})\rightarrow(0,0)\) in E.
Observe that
$$\begin{aligned}& I_{\lambda}\bigl(u_{n}^{1},v_{n}^{1} \bigr)-\frac{1}{2}I_{\lambda }'\bigl(u_{n}^{1},v_{n}^{1} \bigr) \bigl(u_{n}^{1},v_{n}^{1}\bigr) \\& \quad = \biggl(\frac{1}{2}-\frac{1}{\alpha+\beta}\biggr)\lambda\int _{\mathbb {R}^{N}}K(x)\bigl\vert u_{n}^{1}\bigr\vert ^{\alpha}\bigl\vert v_{n}^{1}\bigr\vert ^{\beta} \\& \qquad {} +\lambda\int_{\mathbb{R}^{N}}\biggl(\frac {1}{2}\bigl(f \bigl(x,u_{n}^{1}\bigr)u_{n}^{1}+g \bigl(x,v_{n}^{1}\bigr)v_{n}^{1}\bigr)-F \bigl(x,u_{n}^{1}\bigr)-G\bigl(x,v_{n}^{1} \bigr)\biggr) \\& \quad \geq \frac{\lambda}{N}K_{0}\int_{\mathbb{R}^{N}}\bigl\vert u_{n}^{1}\bigr\vert ^{\alpha }\bigl\vert v_{n}^{1}\bigr\vert ^{\beta}, \end{aligned}$$
where \(K_{0}=\inf_{x\in\mathbb{R}^{N}}K(x)>0\). In connection with \(I_{\lambda}(u_{n}^{1},v_{n}^{1})\rightarrow c-I_{\lambda}(u,v)\) and \(I_{\lambda}'(u_{n}^{1},v_{n}^{1})\rightarrow0\) in \(E^{\ast}\), we get
$$ \int_{\mathbb{R}^{N}}\bigl\vert u_{n}^{1} \bigr\vert ^{\alpha}\bigl\vert v_{n}^{1}\bigr\vert ^{\beta}\leq\frac {N(c-I_{\lambda}(u,v))}{\lambda K_{0}}+o(1). $$
(3.3)
In addition, by (K0) and (H2), for any \(b>0\), there is a constant \(C_{b}>0\) such that
$$\begin{aligned}& \int_{\mathbb{R}^{N}}\bigl(K(x)\bigl\vert u_{n}^{1} \bigr\vert ^{\alpha}\bigl\vert v_{n}^{1}\bigr\vert ^{\beta }+f\bigl(x,u_{n}^{1} \bigr)u_{n}^{1}+g\bigl(x,v_{n}^{1} \bigr)v_{n}^{1}\bigr) \\& \quad \leq b\bigl(\bigl\Vert u_{n}^{1}\bigr\Vert _{2}^{2}+\bigl\Vert v_{n}^{1}\bigr\Vert _{2}^{2}\bigr)+C_{b}\int _{\mathbb {R}^{N}}\bigl\vert u_{n}^{1}\bigr\vert ^{\alpha}\bigl\vert v_{n}^{1}\bigr\vert ^{\beta}. \end{aligned}$$
Let \(V_{b}(x):=\max\{V(x),b\}\), where b is the positive constant in the assumption (V0). Since the set \(\nu^{b}:=\{x\in\mathbb{R}^{N}:V(x)< b\}\) has a finite Lebesgue measure and \((u_{n}^{1},v_{n}^{1})\rightarrow(0,0)\) in \(L_{\mathrm{loc}}^{2}(\mathbb{R}^{N})\times L_{\mathrm{loc}}^{2}(\mathbb{R}^{N})\), we have
$$ \int_{\mathbb{R}^{N}}V(x) \bigl(\bigl\vert u_{n}^{1}\bigr\vert ^{2}+\bigl\vert v_{n}^{1}\bigr\vert ^{2}\bigr)=\int _{\mathbb {R}^{N}}V_{b}(x) \bigl(\bigl\vert u_{n}^{1}\bigr\vert ^{2}+\bigl\vert v_{n}^{1}\bigr\vert ^{2}\bigr)+o(1). $$
(3.4)
Thus
$$\begin{aligned}& S_{\alpha,\beta}\biggl(\int_{\mathbb{R}^{N}}\bigl\vert u_{n}^{1}\bigr\vert ^{\alpha }\bigl\vert v_{n}^{1}\bigr\vert ^{\beta}\biggr)^{\frac{2}{\alpha+\beta}} \\& \quad \leq \int_{\mathbb{R}^{N}}\bigl(\bigl\vert \nabla u_{n}^{1}\bigr\vert ^{2}+\bigl\vert \nabla v_{n}^{1}\bigr\vert ^{2}\bigr) \\& \quad = \int_{\mathbb{R}^{N}}\bigl(\bigl\vert \nabla u_{n}^{1}\bigr\vert ^{2}+\bigl\vert \nabla v_{n}^{1}\bigr\vert ^{2}+\lambda V(x)\bigl\vert u_{n}^{1}\bigr\vert ^{2}+\lambda V(x) \bigl\vert v_{n}^{1}\bigr\vert ^{2}\bigr)-\int _{\mathbb {R}^{N}}\lambda V(x) \bigl(\bigl\vert u_{n}^{1} \bigr\vert ^{2}+\bigl\vert v_{n}^{1}\bigr\vert ^{2}\bigr) \\& \quad = \lambda\int_{\mathbb{R}^{N}}\bigl(K(x)\bigl\vert u_{n}^{1}\bigr\vert ^{\alpha}\bigl\vert v_{n}^{1}\bigr\vert ^{\beta }+f\bigl(x,u_{n}^{1} \bigr)u_{n}^{1}+g\bigl(x,v_{n}^{1} \bigr)v_{n}^{1}\bigr) \\& \qquad {}-\lambda\int_{\mathbb {R}^{N}}V_{b}(x) \bigl(\bigl\vert u_{n}^{1}\bigr\vert ^{2}+ \bigl\vert v_{n}^{1}\bigr\vert ^{2}\bigr)+o(1) \\& \quad \leq \lambda C_{b}\int_{\mathbb{R}^{N}}\bigl\vert u_{n}^{1}\bigr\vert ^{\alpha }\bigl\vert v_{n}^{1}\bigr\vert ^{\beta}+\lambda b\bigl(\bigl\Vert u_{n}^{1}\bigr\Vert ^{2}+\bigl\Vert v_{n}^{1}\bigr\Vert ^{2}\bigr) -\lambda V_{b}(x) \bigl(\bigl\Vert u_{n}^{1}\bigr\Vert ^{2}+\bigl\Vert v_{n}^{1}\bigr\Vert ^{2}\bigr)+o(1) \\& \quad \leq \lambda C_{b}\int_{\mathbb{R}^{N}}\bigl\vert u_{n}^{1}\bigr\vert ^{\alpha }\bigl\vert v_{n}^{1}\bigr\vert ^{\beta}+o(1). \end{aligned}$$
Together with (3.3), we have
$$\begin{aligned} S_{\alpha,\beta}&\leq\lambda C_{b}\biggl(\int_{\mathbb {R}^{N}} \bigl\vert u_{n}^{1}\bigr\vert ^{\alpha}\bigl\vert v_{n}^{1}\bigr\vert ^{\beta} \biggr)^{1-\frac{2}{\alpha+\beta }}+o(1) \\ &\leq \lambda C_{b}\biggl(\frac{N(c-I_{\lambda}(u,v))}{\lambda K_{0}} \biggr)^{\frac {2}{N}}+o(1) \\ & = \lambda^{1-\frac{2}{N}}C_{b}\biggl(\frac{N}{K_{0}} \biggr)^{\frac {2}{N}}\bigl(c-I_{\lambda}(u,v)\bigr)^{\frac{2}{N}}+o(1). \end{aligned}$$
Set \(\alpha_{0}=S_{\alpha,\beta}^{\frac{N}{2}}C_{b}^{-\frac {N}{2}}N^{-1}K_{0}\). This implies \(\alpha_{0}\lambda^{1-\frac {N}{2}}\leq c-I_{\lambda}(u,v)+o(1)\).
Lemma 3.5
Assume that (V0), (K0), and (H1)-(H3) are satisfied. Then, for any
\((\mathrm{PS})_{c}\), the sequence
\(\{(u_{n},v_{n})\}\)
for
\(I_{\lambda}\), there exists a constant
\(\alpha _{0}>0\) (independent of
λ) such that the functional
\(I_{\lambda}(u,v)\)
satisfies the
\((\mathrm{PS})_{c}\)
condition for all
\(c< \alpha _{0}\lambda^{1-\frac{N}{2}}\).
Proof
We can check that, for any \((\mathrm{PS})_{c}\) sequence \(\{(u_{n},v_{n})\}\subset E\) with \((u_{n},v_{n})\rightharpoonup(u,v)\), either \((u_{n},v_{n})\rightarrow (u,v)\) or \(c-I_{\lambda}(u,v)\geq\alpha_{0}\lambda^{1-\frac{N}{2}}\).
On the contrary, if \((u_{n},v_{n})\nrightarrow(u,v)\), this shows
$$\lim\inf_{n\rightarrow\infty}\bigl\Vert (u_{n},v_{n}) \bigr\Vert _{E}>0 $$
and
$$c-I_{\lambda}(u,v)>0. $$
Based on the above mentioned conclusion, we easily find that the functional \(I_{\lambda}(u,v)\) satisfies the \((\mathrm{PS})_{c}\) condition for all \(c<\alpha_{0}\lambda^{1-\frac{N}{2}}\). □