Next lemma can be viewed as a generalization of Struwe’s monotonicity trick [12] and is the main tool for obtaining a bounded Palais-Smale sequence.
Lemma 3.1
(see [13] or [14])
Let
X
be a Banach space equipped with a norm
\(\Vert \cdot \Vert _{X}\), and let
\(J\subset\mathbb{R}^{+}\)
be an interval. We consider a family
\(\{\Phi_{\mu}\}_{\mu\in J}\)
of
\(C^{1}\)-functionals on
X
of the form
$$\Phi_{\mu}(u)=A(u)-\mu B(u),\quad\forall\mu\in J, $$
where
\(B(u)\geq0\)
for all
\(u\in X\)
and such that either
\(A(u)\rightarrow +\infty\)
or
\(B(u)\rightarrow+\infty\)
as
\(\Vert u \Vert _{X}\rightarrow+\infty\). We assume that there are two points
\(v_{1}\), \(v_{2}\)
in
X
such that
$$c_{\mu}=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]} \Phi_{\mu}\bigl(\gamma(t)\bigr)>\max\bigl\{ \Phi_{\mu}(v_{1}), \Phi_{\mu}(v_{2})\bigr\} , $$
where
$$\Gamma=\bigl\{ \gamma\in C\bigl([0,1],X\bigr):\gamma(0)=v_{1}, \gamma(1)=v_{2}\bigr\} . $$
Then, for almost every
\(\mu\in J\), there is a bounded
\((PS)_{c_{\mu}}\)
sequence for
\(\Phi_{\mu}\), that is, there exists a sequence
\(\{u_{n}\}\subset X\)
such that
-
(1)
\(\{u_{n}\}\)
is bounded in
X,
-
(2)
\(\Phi_{\mu}(u_{n})\rightarrow c_{\mu}\),
-
(3)
\(\Phi'_{\mu}(u_{n})\rightarrow0\)
in
\(X^{*}\), where
\(X^{*}\)
is the dual of
X.
Remark 3.2
In [13], it is also proved that, under the assumptions of Lemma 3.1, the map \(\mu\mapsto c_{\mu}\) is left-continuous.
In the paper, we set \(X:=E\), \(\Vert \cdot \Vert _{X}:= \Vert \cdot \Vert \) and \(J:=[\frac{1}{2},1]\). Let us define \(I_{\mu}:E\rightarrow\mathbb{R}\) by \(I_{\mu}(u)=A(u)-\mu B(u)\), where
$$\begin{aligned}& A(u)=\frac{a}{2} \int_{\mathbb{R}^{N}} \vert \nabla u \vert ^{2}\,dx+ \frac{b}{4} \biggl( \int_{\mathbb{R}^{N}} \vert \nabla u \vert ^{2}\,dx \biggr)^{2}+\frac{W}{2} \int_{\mathbb{R}^{N}}u^{2}\,dx- \int_{\mathbb{R}^{N}}hu\,dx, \\& B(u)= \int_{\mathbb{R}^{N}}K(u)\,dx. \end{aligned}$$
Then \(I_{1}(u)=I(u)\). By \((k_{1})\)-\((k_{3})\) and \((h_{1})\), it is obvious that \(I_{\mu}\in C^{1}(E,\mathbb{R})\), \(B(u)\geq0\) for all \(u\in E\) and \(A(u)\geq\frac{\min\{a,W\}}{2} \Vert u \Vert ^{2}-C \vert h \vert _{2} \Vert u \Vert \rightarrow+\infty\) as \(\Vert u \Vert \rightarrow+\infty\).
Lemma 3.3
Assume that
\((k_{1})\)-\((k_{3})\)
and
\((h_{1})\)
hold. Then there exist
\(\rho >0\), \(\alpha>0\)
and
\(m_{0}>0\)
such that
\(I_{\mu}(u)\vert_{\Vert u\Vert=\rho}\geq\alpha\)
for all
h
satisfying
\(\vert h\vert_{2}< m_{0}\)
and for all
\(\mu\in J\).
Proof
First, we consider \(N=2\). It follows from \((k_{1})\) and \((k_{2})\) that, for all \(t\in\mathbb{R}\), we have
$$\begin{aligned} \bigl\vert K(t) \bigr\vert \leq\frac{W+d}{4} \vert t \vert ^{2}+C \vert t \vert ^{p}. \end{aligned}$$
(3.1)
By (3.1), the Hölder inequality and the Sobolev inequality, for all \(\mu\in J\) and \(u\in E\), one has
$$\begin{aligned} I_{\mu}(u) \geq&\frac{a}{2} \int_{\mathbb{R}^{2}} \vert \nabla u \vert ^{2}\,dx+ \frac{W}{2} \int_{\mathbb{R}^{2}}u^{2}\,dx- \int_{\mathbb {R}^{2}}K(u)\,dx- \int_{\mathbb{R}^{2}}hu\,dx \\ \geq&\frac{a}{2} \int_{\mathbb{R}^{2}} \vert \nabla u \vert ^{2}\,dx+ \frac{W}{2} \int_{\mathbb{R}^{2}}u^{2}\,dx-\frac{W+d}{4} \int_{\mathbb {R}^{2}}u^{2}\,dx-C \int_{\mathbb{R}^{2}} \vert u \vert ^{p}\,dx- \vert h \vert _{2} \vert u \vert _{2} \\ \geq&\frac{\min\{2a,W-d\}}{4} \Vert u \Vert ^{2}-C_{1} \Vert u \Vert ^{p}-C_{2} \vert h \vert _{2} \Vert u \Vert \\ =& \Vert u \Vert \biggl(\frac{\min\{2a,W-d\}}{4} \Vert u \Vert -C_{1} \Vert u \Vert ^{p-1}-C_{2} \vert h \vert _{2} \biggr). \end{aligned}$$
Let \(g_{1}(t)=\frac{\min\{2a,W-d\}}{4}t-C_{1}t^{p-1}\) for \(t\geq0\). Since \(p>2\), we know that there exists a constant \(\rho>0\) such that \(\max_{t\geq 0}g_{1}(t)=g_{1}(\rho)>0\). Choose \(m_{0}=\frac{1}{2C_{2}}g_{1}(\rho)\), then there exists \(\alpha>0\) such that \(I_{\mu}(u)\vert_{\Vert u\Vert=\rho}\geq\alpha\) for all h satisfying \(\vert h\vert_{2}< m_{0}\).
Next when \(N=3\), it follows from \((k_{1})\) and \((k_{2})\) that, for all \(t\in\mathbb{R}\), we have
$$\begin{aligned} \bigl\vert K(t) \bigr\vert \leq\frac{W+d}{4} \vert t \vert ^{2}+C \vert t \vert ^{6}. \end{aligned}$$
(3.2)
By (3.2), the Hölder inequality and the Sobolev inequality, for all \(\mu\in J\) and \(u\in E\), one has
$$\begin{aligned} I_{\mu}(u) \geq&\frac{a}{2} \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx+ \frac{W}{2} \int_{\mathbb{R}^{3}}u^{2}\,dx- \int_{\mathbb {R}^{3}}K(u)\,dx- \int_{\mathbb{R}^{3}}hu\,dx \\ \geq&\frac{a}{2} \int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2}\,dx+ \frac{W}{2} \int_{\mathbb{R}^{3}}u^{2}\,dx-\frac{W+d}{4} \int_{\mathbb {R}^{3}}u^{2}\,dx-C \int_{\mathbb{R}^{3}} \vert u \vert ^{6}\,dx- \vert h \vert _{2} \vert u \vert _{2} \\ \geq&\frac{\min\{2a,W-d\}}{4} \Vert u \Vert ^{2}-C_{3} \Vert u \Vert ^{6}-C_{4} \vert h \vert _{2} \Vert u \Vert \\ =& \Vert u \Vert \biggl(\frac{\min\{2a,W-d\}}{4} \Vert u \Vert -C_{3} \Vert u \Vert ^{5}-C_{4} \vert h \vert _{2} \biggr). \end{aligned}$$
Let \(g_{2}(t)=\frac{\min\{2a,W-d\}}{4}t-C_{3}t^{5}\) for \(t\geq0\), we know that there exists a constant \(\rho>0\) such that \(\max_{t\geq0}g_{2}(t)=g_{2}(\rho)>0\). Choose \(m_{0}=\frac {1}{2C_{4}}g_{2}(\rho)\), then there exists \(\alpha>0\) such that \(I_{\mu}(u)\vert_{\Vert u\Vert =\rho}\geq\alpha\) for all h satisfying \(\vert h\vert_{2}< m_{0}\). □
Lemma 3.4
Assume that
\((k_{1})\)-\((k_{3})\)
and
\((h_{1})\)
hold. Then
\(-\infty< c:=\inf\{ I(u): \Vert u \Vert \leq\rho\}<0\), where
ρ
is given by Lemma 3.3.
Proof
Since \(h\in L^{2}(\mathbb{R}^{N})\) and \(h\not\equiv0\), then for \(\varepsilon=\frac{ \vert h \vert _{2}}{2}\), there exists \(\phi \in C^{\infty}_{0}(\mathbb{R}^{N})\) such that \(\vert h-\phi \vert _{2}<\varepsilon\). Thus
$$\begin{aligned} \int_{\mathbb{R}^{N}}\bigl(h^{2}-h\phi\bigr)\,dx\leq \int_{\mathbb {R}^{N}} \bigl\vert h^{2}-h\phi \bigr\vert \,dx \leq \vert h-\phi \vert _{2} \vert h \vert _{2}< \varepsilon \vert h \vert _{2}, \end{aligned}$$
and then
$$\begin{aligned} \int_{\mathbb{R}^{N}}h\phi \,dx\geq \vert h \vert _{2}^{2}- \varepsilon \vert h \vert _{2}=\frac{ \vert h \vert _{2}^{2}}{2}>0. \end{aligned}$$
Hence
$$\begin{aligned} I(t\phi)\leq\frac{at^{2}}{2} \int_{\mathbb{R}^{N}} \vert \nabla\phi \vert ^{2}\,dx+ \frac{bt^{4}}{4} \biggl( \int_{\mathbb{R}^{N}} \vert \nabla\phi \vert ^{2}\,dx \biggr)^{2}+\frac{Wt^{2}}{2} \int_{\mathbb {R}^{N}}\phi^{2}\,dx-t \int_{\mathbb{R}^{N}}h\phi \,dx< 0 \end{aligned}$$
for \(t>0\) small enough. Then we get \(c=\inf\{I(u): \Vert u \Vert \leq\rho\}<0\). \(c>-\infty\) is obvious. □
In order to prove the compactness, we define \(g(t)=k(t)-dt\), \(\forall t\in\mathbb{R}\). Then, by \((k_{1})\) and \((k_{2})\), we get that
$$\begin{aligned} \lim_{t\rightarrow0^{+}}\frac{g(t)}{t}=0, \end{aligned}$$
(3.3)
and when \(N=2\),
$$\begin{aligned} \lim_{t\rightarrow+\infty}\frac{g(t)}{t^{p-1}}=0, \end{aligned}$$
(3.4)
when \(N=3\),
$$\begin{aligned} \lim_{t\rightarrow+\infty}\frac{g(t)}{t^{5}}=0. \end{aligned}$$
(3.5)
Lemma 3.5
Suppose that
\((k_{1})\)-\((k_{3})\), \((h_{1})\)
and
\((h_{3})\)
hold. Assume that
\(\{ u_{n}\}\subset E\)
is a bounded Palais-Smale sequence of
\(I_{\mu}\)
for each
\(\mu\in J\). Then
\(\{u_{n}\}\)
has a convergent subsequence in
E.
Proof
Since \(\{u_{n}\}\) is bounded in E and \(E\hookrightarrow L^{s}(\mathbb{R}^{3})\), \(\forall s\in(2,6)\), \(E\hookrightarrow L^{s}(\mathbb {R}^{2})\), \(\forall s\in(2,+\infty)\) are compact (see [15]), up to a subsequence, we can assume that there exists \(u\in E\) such that \(u_{n}\rightharpoonup u\) in E, \(u_{n}\rightarrow u\) in \(L^{s}(\mathbb{R}^{3})\), \(\forall s\in(2,6)\), \(u_{n}\rightarrow u\) in \(L^{s}(\mathbb{R}^{2})\), \(\forall s\in(2,+\infty)\), \(u_{n}(x)\rightarrow u(x)\) a.e. in \(\mathbb{R}^{N}\).
By (3.3) and (3.4), for any \(\varepsilon>0\), we have
$$\begin{aligned} \bigl\vert g(t) \bigr\vert \leq\varepsilon \vert t \vert +C_{\varepsilon} \vert t \vert ^{p-1},\quad \forall t\geq0. \end{aligned}$$
(3.6)
Then, by (3.6) and the Hölder inequality, one has
$$\begin{aligned}& \biggl\vert \int_{\mathbb{R}^{2}}g(u_{n}) (u_{n}-u)\,dx \biggr\vert \\& \quad\leq\varepsilon \int_{\mathbb{R}^{2}} \vert u_{n} \vert \vert u_{n}-u \vert \,dx+C_{\varepsilon}\int_{\mathbb{R}^{2}} \vert u_{n} \vert ^{p-1} \vert u_{n}-u \vert \,dx \\& \quad\leq\varepsilon \vert u_{n} \vert _{2} \vert u_{n}-u \vert _{2}+C_{\varepsilon}\biggl( \int_{\mathbb{R}^{2}} \vert u_{n} \vert ^{p}\,dx \biggr)^{\frac{p-1}{p}} \vert u_{n}-u \vert _{p} \\& \quad\leq C\varepsilon+o_{n}(1). \end{aligned}$$
Similarly, we can obtain that
$$\begin{aligned} \biggl\vert \int_{\mathbb{R}^{2}}g(u) (u_{n}-u)\,dx \biggr\vert =o_{n}(1). \end{aligned}$$
By (3.3) and (3.5), for any \(\varepsilon>0\), we have
$$\begin{aligned} \bigl\vert g(t) \bigr\vert \leq\varepsilon\bigl( \vert t \vert + \vert t \vert ^{5}\bigr)+C_{\varepsilon} \vert t \vert ^{3},\quad \forall t\geq0. \end{aligned}$$
(3.7)
Hence, by (3.7) and the Hölder inequality, one has
$$\begin{aligned}& \biggl\vert \int_{\mathbb{R}^{3}}g(u_{n}) (u_{n}-u)\,dx \biggr\vert \\& \quad\leq\varepsilon \int_{\mathbb{R}^{3}} \vert u_{n} \vert \vert u_{n}-u \vert \,dx+\varepsilon \int_{\mathbb{R}^{3}} \vert u_{n} \vert ^{5} \vert u_{n}-u \vert \,dx+C_{\varepsilon}\int _{\mathbb{R}^{3}} \vert u_{n} \vert ^{3} \vert u_{n}-u \vert \,dx \\& \quad\leq\varepsilon \vert u_{n} \vert _{2} \vert u_{n}-u \vert _{2}+\varepsilon \biggl( \int_{\mathbb{R}^{3}} \vert u_{n} \vert ^{6}\,dx \biggr)^{\frac{5}{6}} \vert u_{n}-u \vert _{6} +C_{\varepsilon}\biggl( \int_{\mathbb{R}^{3}} \vert u_{n} \vert ^{\frac {9}{2}}\,dx \biggr)^{\frac{2}{3}} \vert u_{n}-u \vert _{3} \\& \quad\leq C\varepsilon+o_{n}(1). \end{aligned}$$
Similarly, we can obtain that
$$\begin{aligned} \biggl\vert \int_{\mathbb{R}^{3}}g(u) (u_{n}-u)\,dx \biggr\vert =o_{n}(1). \end{aligned}$$
Hence when \(N=2\) or 3, one has
$$\begin{aligned} \biggl\vert \int_{\mathbb{R}^{N}} \bigl(g(u_{n})-g(u) \bigr) (u_{n}-u)\,dx \biggr\vert =o_{n}(1). \end{aligned}$$
It is clear that
$$\bigl\langle I'_{\mu}(u_{n})-I'_{\mu}(u),u_{n}-u \bigr\rangle =o_{n}(1) $$
and
$$b \biggl( \int_{\mathbb{R}^{N}}\bigl( \vert \nabla u \vert ^{2}- \vert \nabla u_{n} \vert ^{2}\bigr)\,dx \biggr) \int_{\mathbb{R}^{N}}\bigl(\nabla u,\nabla(u_{n}-u) \bigr)\,dx=o_{n}(1). $$
Note that
$$\begin{aligned} \bigl\langle I_{\mu}'(u_{n})-I'_{\mu}(u),u_{n}-u \bigr\rangle =& \biggl(a+b \int_{\mathbb {R}^{N}} \vert \nabla u_{n} \vert ^{2}\,dx \biggr) \int_{\mathbb {R}^{N}} \bigl\vert \nabla(u_{n}-u) \bigr\vert ^{2}\,dx \\ &{}+(W-\mu d) \int_{\mathbb{R}^{N}} \vert u_{n}-u \vert ^{2}\,dx \\ &{}-b \biggl( \int_{\mathbb{R}^{N}}\bigl( \vert \nabla u \vert ^{2}- \vert \nabla u_{n} \vert ^{2}\bigr)\,dx \biggr) \int_{\mathbb{R}^{N}}\bigl(\nabla u,\nabla(u_{n}-u)\bigr)\,dx \\ &{}-\mu \int_{\mathbb{R}^{N}} \bigl(g(u_{n})-g(u) \bigr) (u_{n}-u)\,dx \\ \geq& \min\{a,W-\mu d\} \Vert u_{n}-u \Vert ^{2} \\ &{}-b \biggl( \int_{\mathbb{R}^{N}}\bigl( \vert \nabla u \vert ^{2}- \vert \nabla u_{n} \vert ^{2}\bigr)\,dx \biggr) \int_{\mathbb{R}^{N}}\bigl(\nabla u,\nabla(u_{n}-u)\bigr)\,dx \\ &{}-\mu \int_{\mathbb{R}^{N}} \bigl(g(u_{n})-g(u) \bigr) (u_{n}-u)\,dx. \end{aligned}$$
Therefore we get that \(\Vert u_{n}-u \Vert \rightarrow0\) as \(n\rightarrow\infty\). □
Proof of the first solution of Theorem 1.2
By Lemma 3.4 and Ekeland’s variational principle [11], there exists a sequence \(\{u_{n}\}\subset E\) such that \(\Vert u_{n} \Vert \leq\rho\), \(I(u_{n})\rightarrow c\) and \(I'(u_{n})\rightarrow 0\) as \(n\rightarrow\infty\). From Lemma 3.5 with \(\mu=1\), there exists \(u_{0}\in E\) such that \(u_{n}\rightarrow u_{0}\) in E and then \(I'(u_{0})=0\) and \(I(u_{0})=c<0\). Put \(u_{0}^{-}:=\max\{-u_{0},0\}\), one has
$$ \begin{aligned}[b] 0&=\bigl\langle I'(u_{0}),u_{0}^{-} \bigr\rangle \\ &=-a \int_{\mathbb{R}^{N}} \bigl\vert \nabla u_{0}^{-} \bigr\vert ^{2}\,dx-b \int _{\mathbb{R}^{N}} \vert \nabla u_{0} \vert ^{2}\,dx \int_{\mathbb {R}^{N}} \bigl\vert \nabla u_{0}^{-} \bigr\vert ^{2}\,dx-W \int_{\mathbb{R}^{N}} \bigl\vert u_{0}^{-} \bigr\vert ^{2}\,dx \\ &\quad{}- \int_{\mathbb{R}^{N}}hu_{0}^{-}dx, \end{aligned} $$
(3.8)
which implies that \(u_{0}^{-}=0\) and then \(u_{0}\geq0\). By the strong maximum principle, we get \(u_{0}>0\). □
For ρ and α in Lemma 3.3, we have following result.
Lemma 3.6
Assume that
\((k_{1})\)-\((k_{3})\)
and
\((h_{1})\)
hold. Then
-
\((*)\)
:
-
\(\exists v_{2}\in E\)
with
\(\Vert v_{2} \Vert >\rho\)
such that
\(I_{\mu}(v_{2})<0\), \(\forall\mu\in J\).
-
\((**)\)
:
-
\(c_{\mu}=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}I_{\mu}(\gamma (t))>\max\{I_{\mu}(0),I_{\mu}(v_{2})\}\), \(\forall\mu\in J\), where
$$\begin{aligned} \Gamma=\bigl\{ \gamma\in C\bigl([0,1],E\bigr):\gamma(0)=0,\gamma(1)=v_{2} \bigr\} . \end{aligned}$$
Proof
It follows from \((k_{3})\) that, for any \(L>0\), there exists \(C_{L}>0\) such that, for all \(t\geq0\), one has
$$ K(t)\geq Lt^{2}-C_{L}. $$
(3.9)
Fix \(0\leq w\in C_{0}^{\infty}(\mathbb{R}^{N})\) with \(\operatorname {supp}w\subset B_{1}:=\{x\in\mathbb{R}^{N}: \vert x \vert <1\}\) and \(w\not\equiv0\). Define \(w_{t}(x)=tw(\frac{x}{t^{2}})\) for \(t>0\), then
$$\begin{aligned} \operatorname{supp}w_{t}=\bigl\{ t^{2}y: y\in\operatorname{supp}w\bigr\} . \end{aligned}$$
By direct computation, we have
$$\begin{aligned}& \int_{\mathbb{R}^{N}} \vert \nabla w_{t} \vert ^{2}\,dx=t^{2N-2} \int _{\mathbb{R}^{N}} \vert \nabla w \vert ^{2}\,dx, \\& \int_{\mathbb{R}^{N}}w_{t}^{2}\,dx=t^{2N+2} \int_{\mathbb{R}^{N}}w^{2}\,dx \end{aligned}$$
and, by (3.9),
$$\begin{aligned} \int_{\mathbb{R}^{N}}K(w_{t})\,dx =& \int_{\operatorname{supp}w_{t}}K(w_{t})\,dx \\ \geq& L \int_{\operatorname{supp}w_{t}}w_{t}^{2}\,dx-C_{L} \int_{\operatorname{supp}w_{t}}\,dx \\ \geq& Lt^{2N+2} \int_{\operatorname{supp}w}w^{2}\,dx-C_{L} \int_{\{t^{2}y:y\in B_{1}\} }\,dx \\ =&Lt^{2N+2} \int_{\mathbb{R}^{N}}w^{2}\,dx-C_{L}Ct^{2N}. \end{aligned}$$
Therefore
$$\begin{aligned}& I_{\mu}(w_{t}) \\& \quad =\frac{a}{2} \int_{\mathbb{R}^{N}} \vert \nabla w_{t} \vert ^{2}\,dx+\frac{b}{4} \biggl( \int_{\mathbb{R}^{N}} \vert \nabla w_{t} \vert ^{2}\,dx \biggr)^{2}+\frac{W}{2} \int_{\mathbb{R}^{N}}w_{t}^{2}\,dx\\& \qquad {}-\mu \int _{R^{N}}K(w_{t})\,dx- \int_{\mathbb{R}^{N}}hw_{t}\,dx \\& \quad \leq\frac{at^{2N-2}}{2} \int_{\mathbb{R}^{N}} \vert \nabla w \vert ^{2}\,dx+ \frac{bt^{4N-4}}{4} \biggl( \int_{\mathbb{R}^{N}} \vert \nabla w \vert ^{2}\,dx \biggr)^{2}+\frac{Wt^{2N+2}}{2} \int_{\mathbb {R}^{N}}w^{2}\,dx \\& \qquad {}-\frac{Lt^{2N+2}}{2} \int_{\mathbb{R}^{N}}w^{2}\,dx+ C_{L}Ct^{2N} \end{aligned}$$
for all \(\mu\in J\). When \(N=2\), we choose \(L=2W\). When \(N=3\), we choose \(L=2W+b\frac{ (\int_{\mathbb{R}^{N}} \vert \nabla w \vert ^{2}\,dx )^{2}}{\int_{\mathbb{R}^{N}}w^{2}\,dx}\). Then \(I_{\mu}(w_{t})\rightarrow -\infty\) as \(t\rightarrow+\infty\). Hence there exists \(t'>0\) such that \(v_{2}:=w_{t'}\) with \(\Vert v_{2} \Vert >\rho\) and \(I_{\mu}(v_{2})<0\), \(\forall\mu\in J\). This completes the proof of \((*)\).
By Lemma 3.3 and the definition of \(c_{\mu}\), for all \(\mu\in J\), we have
$$0< \alpha\leq c_{1}\leq c_{\mu}\leq c_{\frac{1}{2}}\leq\max _{t\in [0,1]}I_{\frac{1}{2}}(tv_{2})< +\infty. $$
Therefore, by \(I_{\mu}(0)=0\) and \(I_{\mu}(v_{2})<0\), we obtain the proof of \((**)\). □
So far we have verified all the conditions of Lemma 3.1. Then there exists \(\{\mu_{j}\}\subset J\) such that
-
(i)
\(\mu_{j}\rightarrow1^{-}\) as \(j\rightarrow\infty\), \(\{u_{n}^{j}\}\) is bounded in E;
-
(ii)
\(I_{\mu_{j}}(u_{n}^{j})\rightarrow c_{\mu_{j}}\) as \(n\rightarrow \infty\);
-
(iii)
\(I'_{\mu_{j}}(u_{n}^{j})\rightarrow0\) as \(n\rightarrow\infty\).
Using (i)-(iii) and Lemma 3.5, there exists \(u_{j}\in E\) such that \(u_{n}^{j}\rightarrow u_{j}\) in E as \(n\rightarrow\infty\) and then \(I_{\mu_{j}}(u_{j})=c_{\mu_{j}}\) and \(I'_{\mu_{j}}(u_{j})=0\). Hence, from \(I_{\mu_{j}}(u_{j})=c_{\mu_{j}}\) and \(\langle I'_{\mu _{j}}(u_{j}),u_{j}\rangle=0\), we get respectively
$$\begin{aligned}& \begin{aligned}[b] &\frac{a}{2} \int_{\mathbb{R}^{N}} \vert \nabla u_{j} \vert ^{2}\,dx+\frac{b}{4} \biggl( \int_{\mathbb{R}^{N}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{2}+\frac{W}{2} \int_{\mathbb{R}^{N}}u_{j}^{2}\,dx \\ &\quad-\mu_{j} \int _{\mathbb{R}^{N}}K(u_{j})\,dx- \int_{\mathbb{R}^{N}}hu_{j}\,dx=c_{\mu_{j}}, \end{aligned} \end{aligned}$$
(3.10)
$$\begin{aligned}& \begin{aligned}[b] & a \int_{\mathbb{R}^{N}} \vert \nabla u_{j} \vert ^{2}\,dx+b \biggl( \int _{\mathbb{R}^{N}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{2}+W \int _{\mathbb{R}^{N}}u_{j}^{2}\,dx \\ &\quad-\mu_{j} \int_{\mathbb{R}^{N}}k(u_{j})u_{j}\,dx- \int_{\mathbb {R}^{N}}hu_{j}\,dx=0. \end{aligned} \end{aligned}$$
(3.11)
Next, for obtaining \(\{u_{j}\}\) is bounded in E, we need the following lemma (Pohozaev type identity). The proof is similar to Lemma 2.6 in [16], and we omit its proof in here.
Lemma 3.7
Suppose that
\((h_{1})\)
and
\((h_{2})\)
hold. If
\(I'_{\mu}(u)=0\), we have
$$ \begin{aligned} &\frac{a(N-2)}{2} \int_{\mathbb{R}^{N}} \vert \nabla u \vert ^{2}\,dx+ \frac{b(N-2)}{2} \biggl( \int_{\mathbb{R}^{N}} \vert \nabla u \vert ^{2}\,dx \biggr)^{2}+\frac{NW}{2} \int_{\mathbb{R}^{N}}u^{2}\,dx \\ &\quad{}-N\mu \int_{\mathbb{R}^{N}}K(u)\,dx-N \int_{\mathbb{R}^{N}}hu\,dx- \int _{\mathbb{R}^{N}}\bigl(\nabla h(x),x\bigr)u\,dx=0. \end{aligned} $$
Since \(I'_{\mu_{j}}(u_{j})=0\), by Lemma 3.7, we get that
$$ \begin{aligned}[b] &\frac{a(N-2)}{2} \int_{\mathbb{R}^{N}} \vert \nabla u_{j} \vert ^{2}\,dx+\frac{b(N-2)}{2} \biggl( \int_{\mathbb{R}^{N}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{2}+\frac{NW}{2} \int_{\mathbb{R}^{N}}u_{j}^{2}\,dx \\ &\quad{}-N\mu_{j} \int_{\mathbb{R}^{N}}K(u_{j})\,dx-N \int_{\mathbb {R}^{N}}hu_{j}\,dx- \int_{\mathbb{R}^{N}}\bigl(\nabla h(x),x\bigr)u_{j}\,dx=0. \end{aligned} $$
(3.12)
Lemma 3.8
Assume that
\((k_{1})\)-\((k_{3})\)
and
\((h_{1})\)-\((h_{3})\)
hold. Then
\(\{u_{j}\}\)
is bounded in
E.
Proof
It follows from (3.10) and (3.12) that
$$ a \int_{\mathbb{R}^{N}} \vert \nabla u_{j} \vert ^{2}\,dx+\frac {b(4-N)}{4} \biggl( \int_{\mathbb{R}^{N}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{2}+ \int_{\mathbb{R}^{N}}\bigl(\nabla h(x),x\bigr)u_{j}\,dx=Nc_{\mu_{j}}. $$
(3.13)
Be similar to (3.8), by \(I'_{\mu_{j}}(u_{j})=0\), we obtain \(u_{j}\geq0\).
Firstly, we consider \(N=2\). From (3.13) and \(c_{\mu_{j}}\leq c_{\frac{1}{2}}\), we get
$$ \begin{aligned}[b] a \int_{\mathbb{R}^{2}} \vert \nabla u_{j} \vert ^{2}\,dx &\leq a \int_{\mathbb{R}^{2}} \vert \nabla u_{j} \vert ^{2}\,dx+\frac {b}{2} \biggl( \int_{\mathbb{R}^{2}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{2} \\ &\quad{}-2c_{\mu_{j}}+2c_{\mu_{j}} \\ &=- \int_{\mathbb{R}^{2}}\bigl(\nabla h(x),x\bigr)u_{j}\,dx+2c_{\mu_{j}}. \end{aligned} $$
(3.14)
Since \((\nabla h(x),x)\geq0\), by (3.14) and \(u_{j}\geq0\), one has \(\{\int_{\mathbb{R}^{2}} \vert \nabla u_{j} \vert ^{2}\,dx\}\) is bounded. Next we prove \(\{\int_{\mathbb{R}^{2}}u_{j}^{2}\,dx\}\) is bounded. Inspired by [14], we suppose by contradiction that \(\lambda _{j}:= \vert u_{j} \vert _{2}\rightarrow+\infty\). Define \(w_{j}:=u_{j}(\lambda_{j}x)\), then
$$ \int_{\mathbb{R}^{2}} \vert \nabla w_{j} \vert ^{2}\,dx= \int_{\mathbb {R}^{2}} \vert \nabla u_{j} \vert ^{2}\,dx\leq C $$
and
$$ \int_{\mathbb{R}^{2}} \vert w_{j} \vert ^{2}\,dx= \frac{1}{\lambda _{j}^{2}} \int_{\mathbb{R}^{2}} \vert u_{j} \vert ^{2}\,dx=1. $$
(3.15)
Hence \(\{w_{j}\}\) is bounded in E. Up to a subsequence, we may assume that \(w_{j}\rightharpoonup w\) in E, \(w_{j}\rightarrow w\) in \(L^{s}(\mathbb {R}^{2})\), \(\forall s\in(2,+\infty)\), \(w_{j}\rightarrow w\) in \(L_{\mathrm{loc}}^{s}(\mathbb{R}^{2})\), \(\forall s\in[1,+\infty)\), \(w_{j}(x)\rightarrow w(x)\) a.e. in \(\mathbb{R}^{2}\). By \(I'_{\mu_{j}}(u_{j})=0\), one has
$$ - \biggl(a+b \int_{\mathbb{R}^{2}} \vert \nabla w_{j} \vert ^{2}\,dx \biggr)\frac{1}{\lambda_{j}^{2}}\triangle w_{j}+(W-d \mu_{j})w_{j}=\mu_{j}g(w_{j})+h( \lambda_{j}x). $$
(3.16)
For any \(v\in C_{0}^{\infty}(\mathbb{R}^{2})\), one has
$$ \biggl\vert \int_{\mathbb{R}^{2}}h(\lambda_{j}x)v\,dx \biggr\vert \leq \vert v \vert _{2} \biggl( \int_{\mathbb{R}^{2}} \bigl\vert h(\lambda _{j}x) \bigr\vert ^{2}\,dx \biggr)^{\frac{1}{2}} =\frac{1}{\lambda_{j}} \vert v \vert _{2} \vert h \vert _{2}\rightarrow0 $$
(3.17)
and by the Lebesgue dominated convergence theorem, we have
$$ \biggl\vert \int_{\mathbb{R}^{2}}g(w_{j})v\,dx- \int_{\mathbb{R}^{2}}g(w)v\,dx \biggr\vert \leq C \int_{\operatorname{supp}v} \bigl\vert g(w_{j})-g(w) \bigr\vert \,dx \rightarrow0. $$
(3.18)
Hence by (3.16)-(3.18), we have \((W-d)w=g(w)\) in \(\mathbb{R}^{2}\), from which we get that \(w=0\). Indeed, since 0 is an isolated solution of \((W-d)z=g(z)\), \(w=0\). Therefore by (3.6), (3.15) and (3.16), one has
$$ \begin{aligned} W-d&=(W-d) \int_{\mathbb{R}^{2}} \vert w_{j} \vert ^{2}\,dx \\ &\leq \biggl(a+b \int_{\mathbb{R}^{2}} \vert \nabla w_{j} \vert ^{2}\,dx \biggr)\frac{1}{\lambda_{j}^{2}} \int_{\mathbb{R}^{2}} \vert \nabla w_{j} \vert ^{2}\,dx+(W-d\mu_{j}) \int_{\mathbb{R}^{2}} \vert w_{j} \vert ^{2}\,dx \\ &=\mu_{j} \int_{\mathbb{R}^{2}}g(w_{j})w_{j}\,dx+ \int_{\mathbb{R}^{2}}h(\lambda _{j}x)w_{j}\,dx \\ &\leq\varepsilon \int_{\mathbb{R}^{2}} \vert w_{j} \vert ^{2}\,dx+C_{\varepsilon}\int_{\mathbb{R}^{2}} \vert w_{j} \vert ^{p}\,dx+ \frac{1}{\lambda_{j}} \vert h \vert _{2} \vert w_{j} \vert _{2} \\ &\leq C\varepsilon+o_{n}(1), \end{aligned} $$
which implies a contradiction. Hence \(\{\int_{\mathbb{R}^{2}} \vert u_{j} \vert ^{2}\,dx\}\) is bounded and then \(\{u_{j}\}\) is bounded in E.
Secondly, for \(N=3\), we have a simple proof. From (3.13), \((h_{2})\) and \(c_{\mu_{j}}\leq c_{\frac{1}{2}}\), we get
$$ \begin{aligned}[b] a \int_{\mathbb{R}^{3}} \vert \nabla u_{j} \vert ^{2}\,dx &\leq a \int_{\mathbb{R}^{3}} \vert \nabla u_{j} \vert ^{2}\,dx+\frac {b}{4} \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{2}-3c_{\mu_{j}}+3c_{\mu_{j}} \\ &=- \int_{\mathbb{R}^{3}}\bigl(\nabla h(x),x\bigr)u_{j}\,dx+3c_{\mu_{j}} \\ &\leq \bigl\vert \bigl(\nabla h(x),x\bigr) \bigr\vert _{2} \vert u_{j} \vert _{2}+3c_{\frac{1}{2}} \\ &\leq C \vert u_{j} \vert _{2}+3c_{\frac{1}{2}}. \end{aligned} $$
(3.19)
We prove directly \(\{\int_{\mathbb{R}^{3}}u_{j}^{2}\,dx\}\) is bounded. Similar to (3.19), we obtain
$$ \begin{aligned}[b] \frac{b}{4} \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{2} &\leq a \int_{\mathbb{R}^{3}} \vert \nabla u_{j} \vert ^{2}\,dx+\frac {b}{4} \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{2}-3c_{\mu_{j}}+3c_{\mu_{j}} \\ &\leq C \vert u_{j} \vert _{2}+3c_{\frac{1}{2}}. \end{aligned} $$
(3.20)
By the Hölder inequality, we have
$$\begin{aligned} \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{3}\leq C \biggl( \int_{\mathbb{R}^{3}}u_{j}^{2}\,dx \biggr)^{\frac{3}{4}}+C. \end{aligned}$$
(3.21)
By (3.3) and (3.5), for all \(t\in\mathbb{R}\), one has
$$ \bigl\vert g(t)t \bigr\vert \leq\frac{W-d}{2} \vert t \vert ^{2}+C \vert t \vert ^{6}. $$
(3.22)
From (3.11), (3.21), (3.22), \(\mu_{j}\leq 1\) and \(D^{1,2}(\mathbb{R}^{3})\hookrightarrow L^{6}(\mathbb{R}^{3})\), it follows that
$$\begin{aligned} (W-d) \int_{\mathbb{R}^{3}}u_{j}^{2}\,dx \leq& a \int_{\mathbb{R}^{3}} \vert \nabla u_{j} \vert ^{2}\,dx+b \biggl( \int_{\mathbb{R}^{3}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{2}+(W-\mu_{j}d) \int_{\mathbb{R}^{3}}u_{j}^{2}\,dx \\ =&\mu_{j} \int_{\mathbb{R}^{3}}g(u_{j})u_{j}\,dx+ \int_{\mathbb{R}^{3}}hu_{j}\,dx \\ \leq&\frac{W-d}{2} \int_{\mathbb{R}^{3}}u_{j}^{2}\,dx+C \int_{\mathbb {R}^{3}}u_{j}^{6}\,dx+ \vert h \vert _{2} \biggl( \int_{\mathbb {R}^{3}}u_{j}^{2}\,dx \biggr)^{\frac{1}{2}} \\ \leq&\frac{W-d}{2} \int_{\mathbb{R}^{3}}u_{j}^{2}\,dx+C \biggl( \int_{\mathbb {R}^{3}} \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{3}+ \vert h \vert _{2} \biggl( \int_{\mathbb{R}^{3}}u_{j}^{2}\,dx \biggr)^{\frac{1}{2}} \\ \leq&\frac{W-d}{2} \int_{\mathbb{R}^{3}}u_{j}^{2}\,dx+C \biggl( \int_{\mathbb {R}^{3}}u_{j}^{2}\,dx \biggr)^{\frac{3}{4}} +C+ \vert h \vert _{2} \biggl( \int_{\mathbb{R}^{3}}u_{j}^{2}\,dx \biggr)^{\frac{1}{2}}, \end{aligned}$$
which implies that \(\{\int_{\mathbb{R}^{3}}u_{j}^{2}\,dx\}\) is bounded. Combining with (3.19), we get that \(\{u_{j}\}\) is bounded in E. □
Proof of the second solution of Theorem 1.2
By \(I_{\mu_{j}}(u_{j})=c_{\mu_{j}}\), \(I'_{\mu_{j}}(u_{j})=0\), \(\mu_{j}\rightarrow 1^{-}\) and Remark 3.2, we get \(I(u_{j})\rightarrow c_{1}\) and \(I'(u_{j})\rightarrow0\) as \(n\rightarrow +\infty\). By Lemmas 3.5 and 3.8, there exists \(v_{0}\in E\) such that \(u_{j}\rightarrow v_{0}\) in E as \(n\rightarrow+\infty\) and then \(I(v_{0})=c_{1}>0\), \(I'(v_{0})=0\). Be similar to (3.8), we get \(v_{0}\geq0\). By the strong maximum principle, one has \(v_{0}>0\). □