1 Introduction
Consider the following Kirchhoff-type equation:
$$\begin{align} \begin{aligned} -\left( 1 + \int_{\mathbb{R}^{3}} |\nabla u|^{2} \mathrm{d}x \right) \Delta u + \frac{A}{|x|^{\alpha}} u = f(u), \ \, x\in\mathbb{R}^{3}, \end{aligned} \end{align} $$
where
$\alpha \in (0,4)$
, and
$A>0$
is a parameter.
Equation
$(\mathcal {K})$
is related to the stationary analogue of the equation
$$ \begin{align} \begin{aligned} \begin{cases} u_{tt} - \left( 1+ \int_{\mathbb{R}^{3}} |\nabla u|^{2} \mathrm{d}x \right)\Delta u = f(u), \ \, &x\in\Omega, \\ u=0, &x\in\partial\Omega. \end{cases} \end{aligned} \end{align} $$
Equation (1.1) was first proposed by Kirchhoff [Reference Kirchhoff15] as an extension of the classical D’Alembert’s wave equation by considering the effects of the changes in the length of the string during the vibrations. Kirchhoff’s model takes into account the changes in length of the string produced by transverse vibrations. Nonlocal effect also finds its application in biological systems. After the celebrated work of Lions [Reference Lions17], which proposed an abstract functional analysis framework, much attention has been drawn to equation (1.1). For more mathematical and physical background on Kirchhoff-type equations, we refer to [Reference Arosio2, Reference Bernstein7, Reference Pohožaev21] and the references therein.
When
$\alpha =0$
, equation (1.1) turns into
$$ \begin{align} \begin{aligned} -\left( 1+ \int_{\mathbb{R}^{3}} |\nabla u|^{2} \mathrm{d}x \right) \Delta u+Au=f(u), \ \, x\in\mathbb{R}^{3}. \end{aligned} \end{align} $$
In the past decades, the existence and multiplicity of nontrivial solutions of equation (1.2) has been investigated intensively by many authors under different hypotheses on f. Several assumptions on f were considered, including subcritical and critical cases. To name a few, we refer to [Reference Alves, Corrêa and Ma1, Reference Cassani, Liu, Tarsi and Zhang11, Reference Perera and Zhang20] for subcritical case and [Reference Figueiredo and Santos14, Reference Naimen18, Reference Naimen19] for critical case.
More recently, the existence results of elliptic equations with singular potential began to attract much attention by many authors. Here, let us briefly recall some existing results.
Cao and Yan [Reference Cao and Yan9] considered an equation as follows:
$$ \begin{align} \begin{aligned} \begin{cases} -\Delta u - \frac{\mu}{|x|^{2}}u = |u|^{2^{*}-2}u + au, \ \, &x\in\Omega, \\ u=0, &x\in\partial\Omega, \end{cases} \end{aligned} \end{align} $$
where
$2^{*}=\frac {2N}{N-2}$
, and
$\Omega \subset \mathbb {R}^{N}$
is an open bounded domain which contains the origin. Under the assumptions of
$N\geqslant 7$
,
$\mu \in \left [0,\frac {(N-2)^{2}}{4}-4\right )$
, and
$a>0$
, Cao and Yan [Reference Cao and Yan9] proved equation (1.3) possesses infinitely many nontrivial solutions by using the perturbation arguments and variational methods. Subsequently, Chen and Zou [Reference Chen and Zou12] extended the result of [Reference Cao and Yan9] to the following two aspects. If
$N\geqslant 5$
and
$\mu \in \left (0,\left (\frac {N-2}{2}\right )^{2}-\left (\frac {N+2}{N}\right )^{2}\right )$
, equation (1.3) has a ground state solution for each fixed
$a>0$
. If
$N\geqslant 7$
and
$\mu \in \left (0,\frac {(N-2)^{2}}{4}-4\right )$
, equation (1.3) has infinitely many sign-changing solutions for each fixed
$a>0$
.
Su, Wang, and Willem [Reference Su, Wang and Willem24] dealt with a generalized version with the singular potential:
$$ \begin{align} \begin{aligned} \begin{cases} -\Delta u +V(|x|)u = Q(|x|)f(u), &u>0, \ \, x\in\mathbb{R}^{N}, \\ |u(x)|\rightarrow0, &\mathrm{as} \; |x|\rightarrow\infty, \end{cases} \end{aligned} \end{align} $$
where
$N\geqslant 2$
, V and f satisfy
-
(V 1)
$V\in C((0,\infty ),\mathbb {R})$
,
$V(t)\geqslant 0$
, and there exist
$\hat {a}$
and
$\hat {a}_{0}$
such that
$$ \begin{align*} \begin{aligned} \liminf_{t\rightarrow0} \frac{V(t)}{t^{\hat{a}_{0}}}>0, \quad \liminf_{t\rightarrow\infty} \frac{V(t)}{t^{\hat{a}}}>0. \end{aligned} \end{align*} $$
-
(Q 1)
$Q\in C((0,\infty ),\mathbb {R})$
,
$Q(t)\geqslant 0$
, and there exist
$\hat {b}$
and
$\hat {b}_{0}$
such that
$$ \begin{align*} \begin{aligned} \limsup_{t\rightarrow0} \frac{Q(t)}{t^{\hat{b}_{0}}}<\infty, \quad \limsup_{t\rightarrow\infty} \frac{Q(t)}{t^{\hat{b}}}<\infty. \end{aligned} \end{align*} $$
-
(F 1)
$f\in C(\mathbb {R},\mathbb {R})$
,
$f(0)=0$
, and there exists
$C>0$
such that where
$$ \begin{align*} \begin{aligned} f(u) \leqslant C(|t|^{p_{1}-1}+|t|^{p_{2}-1}), \ \, t\in\mathbb{R}, \end{aligned} \end{align*} $$
$2_{\alpha }^{*}:=2+\frac {4\alpha }{2N-2-\alpha }$
,
$2^{*}:=\frac {2N}{N-2}$
, and
$p_{1},p_{2}$
satisfy one of the following conditions:
$$ \begin{align*} \begin{aligned} \begin{cases} 2_{\alpha}^{*} <p_{1} \leqslant p_{2}<2^{*}, &\alpha\in(0,2), \\ 2^{*} <p_{1} \leqslant p_{2}<2_{\alpha}^{*}, &\alpha\in(2,2N-2), \\ 2^{*} <p_{1} \leqslant p_{2}, &\alpha\in[2N-2,\infty). \end{cases} \end{aligned} \end{align*} $$
-
(F 2) There exists
$\gamma>2$
such that
$\gamma F(t)\leqslant tf(t)$
, for any
$t\in \mathbb {R}$
, where
$F(t):=\int _{0}^{t}f(s)\mathrm {d}s$
. -
(F 3)
$F(t)>0$
, for any
$t\in \mathbb {R}$
. -
(F 4) f is odd.
Based on the parameters
$\hat {a},\hat {a}_{0},\hat {b},\hat {b}_{0}$
and an improved Strauss radial lemma [Reference Strauss23], Su, Wang, and Willem [Reference Su, Wang and Willem24] established some radial inequalities, and obtained the following continuous and compact embedding results, which is the basis of the following variational framework constructions.
Proposition 1.1 [Reference Su, Wang and Willem24]
Assume that
$N\geqslant 3$
and conditions
$(V_{1})$
–
$(Q_{1})$
hold. Then, the following continuous embeddings hold:
$$ \begin{align*} \begin{aligned} \begin{cases} W^{1,2}_{rad}(\mathbb{R}^{N},V) \hookrightarrow L^{r}(\mathbb{R}^{N}), \ \, r\in[2_{\alpha}^{*},2^{*}], \ \, \alpha\in(0,2); \\ W^{1,2}_{rad}(\mathbb{R}^{N},V) \hookrightarrow L^{r}(\mathbb{R}^{N}), \ \, r\in[2^{*},2_{\alpha}^{*}], \ \, \alpha\in(2,2N-2); \\ W^{1,2}_{rad}(\mathbb{R}^{N},V) \hookrightarrow L^{r}(\mathbb{R}^{N}), \ \, r\in[2^{*},\infty), \ \, \alpha\in[2N-2,\infty). \end{cases} \end{aligned} \end{align*} $$
Furthermore, the embeddings are compact if
$r\not =2_{\alpha }^{*}$
or
$r\not =2^{*}$
, where
$$ \begin{align*} \begin{aligned} L^{2}(\mathbb{R}^{N},V):= \left\{ u\in L^{2}(\mathbb{R}^{N}) \Big| \int_{\mathbb{R}^{N}} V(|x|)|u|^{2} \mathrm{d}x < +\infty \right\}, \end{aligned} \end{align*} $$
and
$W_{rad}^{1,2}(\mathbb {R}^{N},V):= D_{rad}^{1,2}(\mathbb {R}^{N}) \cup L^{2}(\mathbb {R}^{N},V)$
is the set of radial functions in
$W^{1,2}(\mathbb {R}^{N},V)$
.
As an application of Proposition 1.1, Su, Wang, and Willem [Reference Su, Wang and Willem24] got the existence and multiplicity of solutions to equation (1.4) via the variational arguments.
Li and Su [Reference Li and Su16] studied the following Kirchhoff-type equation:
$$ \begin{align} \begin{cases} -\Big[ 1+ \int_{\mathbb{R}^{3}} \big(|\nabla u|^{2} + V(x)u^{2}\big) \mathrm{d}x \Big] \Delta u+V(x)u = \lambda Q(x)f(u), &x\in\mathbb{R}^{3}, \\ u(x)\rightarrow0, &\mathrm{as} \; |x|\rightarrow\infty, \end{cases} \end{align} $$
where V, Q, and f satisfy
-
(V 2)
$V(x)\in C(\mathbb {R}^{3},\mathbb {R}^{+})$
is radial and there exists
$\bar {\bar {a}}\in \mathbb {R}$
such that
$\liminf \limits _{|x|\rightarrow \infty }V(x) |x|^{-\bar {\bar {a}}}>0$
. -
(Q 2)
$Q(x)\in C(\mathbb {R}^{3},\mathbb {R}^{+})$
is radial and there exists
$\bar {\bar {b}}\in \mathbb {R}$
such that
$\limsup \limits _{|x|\rightarrow \infty }Q(x) |x|^{-\bar {\bar {b}}}<\infty $
. -
(F 5)
$f\in C((-\delta ,\delta ),\mathbb {R})$
for some
$\delta>0$
and there exists
$q_{1}\in (4,6)$
such that where
$$ \begin{align*} \begin{aligned} \lim_{|t|\rightarrow0} \frac{F(t)}{|t|^{q_{1}}} = +\infty, \end{aligned} \end{align*} $$
$F(t)=\int _{0}^{t}f(\xi )\mathrm {d}\xi $
.
-
(F 6) There exists
$q_{2}\in (4,6)$
with
$q_{2}<q_{1}$
such that
$$ \begin{align*} \begin{aligned} \lim_{|t|\rightarrow0} \frac{f(t)t}{|t|^{q_{2}}} = 0. \end{aligned} \end{align*} $$
-
(F 7) There exist
$\beta \in (4,6)$
and
$\delta>0$
such that
$$ \begin{align*} \begin{aligned} 0< \beta F(t) \leqslant tf(t), \ \, 0<|t|\leqslant \delta. \end{aligned} \end{align*} $$
Li and Su [Reference Li and Su16] investigated the existence and multiplicity of nontrivial radial solutions for equation (1.5) via Proposition 1.1 and the variational methods. For more details and recent works of elliptic equations involving singular potential, we refer to [Reference Badiale, Benci and Rolando3–Reference Badiale and Rolando6, Reference Carrião, Demarque and Miyagaki10, Reference Conti, Crotti and Pardo13, Reference Rolando22] and the references therein.
For equation
$(\mathcal {K})$
, the case where the nonlinearity f has subcritical growth or supercritical growth can be discussed similar to Li and Su [Reference Li and Su16]. It is nature to ask whether there exists a solution to equation
$(\mathcal {K})$
with critical exponent? To the best of our knowledge, it seems that there is no affirmative answer in the literature. We are interested in some choice of f that violates the conditions
$(F_{1})$
and
$(F_{5})$
–
$(F_{6})$
as follows:
-
(F 8) if
$\alpha \in (0,2)$
, then
$f(u)= |u|^{2_{\alpha }^{*}-2}u + \lambda |u|^{q-2}u + |u|^{4}u $
, where
$\lambda>0$
is a parameter, and
$q\in (2_{\alpha }^{*},6)$
. -
(F 9) if
$\alpha \in (2,4)$
, then
$f(u)= |u|^{2_{\alpha }^{*}-2}u + \lambda |u|^{q-2}u + |u|^{4}u$
, where
$\lambda>0$
is a parameter, and
$q\in (6,2_{\alpha }^{*})$
.
Our first main result of the paper can be stated as follows.
Theorem 1.1 Assume that
$\alpha \in (0,2)$
and condition
$(F_{8})$
holds. Then there exists
$\Lambda>0$
such that for any
$\lambda>\Lambda $
, equation
$(\mathcal {K})$
admits a Nehari–Pohožaev-type ground state solution.
Remark 1.1 We summarize the main difficulties and novelties developed in this study.
(I). By Proposition 1.1, the presence of the embedding top and bottom indices makes the embeddings
$$ \begin{align*} \begin{aligned} W_{rad}^{1,2}(\mathbb{R}^{3},\alpha) \hookrightarrow L^{2_{\alpha}^{*}}(\mathbb{R}^{3}) \ \, \mathrm{and} \ \, W_{rad}^{1,2}(\mathbb{R}^{3},\alpha) \hookrightarrow L^{6}(\mathbb{R}^{3}) \end{aligned} \end{align*} $$
lack of compactness, which set an obstacle to prove the
$(PS)$
condition. In [Reference Li and Su16, Reference Su, Wang and Willem24], the authors overcame this difficulty for the case f satisfying subcritical growth. Clearly, their method is not suitable for our case. Motivated by [Reference Su25], we attempt to recover the compactness by the generalized version of Lions-type theorem (see Theorem 3.1) and taking advantage of the parameter
$\lambda $
.
(II). As one can see, for
$\alpha \in (0,\frac {4}{3})$
, we have
$2_{\alpha }^{*}<4$
. Then it seems much more complicated to get the boundedness of the
$(PS)$
sequences. In this paper, we adopt the method of Nehari–Pohožaev manifold to deal with this problem.
Remark 1.2 At the end of this section, we outline the sketch of our proof of Theorem 1.1. First, by using some new tricks and taking advantage of the parameter
$\lambda $
, we give an exact estimation of the least energy of the Nehari–Pohožaev manifold (see Lemma 3.5). According to the estimation, we show that the minimizing sequences are nonvanishing in
$L^{2_{\alpha }^{*}}(\mathbb {R}^{3})$
and
$L^{6}(\mathbb {R}^{3})$
. Second, we prove the minimizer of the energy functional on the corresponding Nehari–Pohožaev manifold can be achieved via the generalized version of Lions-type theorem (see Theorem 3.1). Finally, by applying the deformation arguments, we show that the minimizer is a ground state solution of equation
$(\mathcal {K})$
.
Our second main result of the paper can be stated as follows.
Theorem 1.2 Assume that
$\alpha \in (2,4)$
and condition
$(F_{9})$
holds. Then there exists
$\bar {\Lambda }>0$
such that for any
$\lambda>\bar {\Lambda }$
, equation
$(\mathcal {K})$
admits a Nehari–Pohožaev-type ground state solution.
Remark 1.3 It is notable that for
$\alpha \in [4,\infty )$
,
$6$
is the embedding bottom index. If the nonlinearity f has embedding bottom index, the study of this case becomes more complicated and challenging. At this stage, we are still working on this problem.
The remainder of this paper is organized as follows. In Section 2, we set up the variational framework for equation
$(\mathcal {K})$
, and some preliminaries are presented. In Sections 3 and 4, we give the proofs of Theorems 1.1 and 1.2, respectively.
2 Preliminaries and variational settings
Throughout this paper, we use symbols C,
$C_{i}\;(i=1,2,\dots )$
to denote different positive constants whose exact value is inessential.
Let
$C_{0}^{\infty }(\mathbb {R}^{3})$
be the collection of smooth functions with compact support. Let
$D^{1,2}(\mathbb {R}^{3})$
be the completion of
$C_{0}^{\infty }(\mathbb {R}^{3})$
under the seminorm
$$ \begin{align*} \begin{aligned} \|u\|_{D^{1,2}(\mathbb{R}^{3})} := \left( \int_{\mathbb{R}^{3}} |\nabla u|^{2} \mathrm{d}x \right)^{\frac{1}{2}}. \end{aligned} \end{align*} $$
Define
$$ \begin{align*} \begin{aligned} W^{1,2}(\mathbb{R}^{3},\alpha) &:= \left\{ u\in D^{1,2}(\mathbb{R}^{3}) \Big| \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)} ^{2} = \int_{\mathbb{R}^{3}} \frac{A}{|x|^{\alpha}}|u|^{2} \mathrm{d}x < +\infty \right\}\\ &= D^{1,2}(\mathbb{R}^{3}) \cap L^{2}(\mathbb{R}^{3},\alpha) \end{aligned} \end{align*} $$
with the norm
$$ \begin{align*} \begin{aligned} \|u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2} := \int_{\mathbb{R}^{3}} \big(|\nabla u|^{2} + \frac{A}{|x|^{\alpha}}|u|^{2}\big) \mathrm{d}x. \end{aligned} \end{align*} $$
Let
$W_{rad}^{1,2}(\mathbb {R}^{3},\alpha ) := D_{rad}^{1,2}(\mathbb {R}^{3}) \cap L^{2}(\mathbb {R}^{3},\alpha )$
denotes the radial subspace of
$W^{1,2}(\mathbb {R}^{3},\alpha )$
.
The following two inequalities play an important role in the estimation of the mountain pass energy.
$$ \begin{align} \mathcal{S}_{\alpha} \left( \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x \right) ^{\frac{2}{2_{\alpha}^{*}}} \leqslant \|u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2}, \ \, u\in W_{rad}^{1,2}(\mathbb{R}^{3},\alpha), \end{align} $$
and
$$ \begin{align} \mathcal{S} \left( \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \right) ^{\frac{1}{3}} \leqslant \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2}, \ \, u\in D^{1,2}(\mathbb{R}^{3}). \end{align} $$
We define the associated energy functional of equation
$(\mathcal {K})$
by
$$ \begin{align*} \begin{aligned} I(u) &= \frac{1}{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{1}{2} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{1}{4} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad {}- \frac{1}{2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{\lambda}{q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x - \frac{1}{6} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x, \end{aligned} \end{align*} $$
which is also well defined in
$W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
and
$I\in C^{1}(W_{rad}^{1,2}(\mathbb {R}^{3},\alpha ),\mathbb {R})$
with the derivative given by
$$ \begin{align*} \begin{aligned} \langle I'(u),\varphi \rangle &= \left( 1+\|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2}\right) \int_{\mathbb{R}^{3}} \nabla u\nabla\varphi \mathrm{d}x + A\int_{\mathbb{R}^{3}} \frac{u\varphi}{|x|^{\alpha}} \mathrm{d}x - \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}-2}u\varphi \mathrm{d}x \\ &\quad{}- \lambda \int_{\mathbb{R}^{3}} |u|^{q-2}u\varphi \mathrm{d}x - \int_{\mathbb{R}^{3}} |u|^{4}u\varphi \mathrm{d}x. \end{aligned} \end{align*} $$
3 The proof of Theorem 1.1
By standard arguments, we define the Pohožaev identity of equation
$(\mathcal {K})$
by
$$ \begin{align*} \begin{aligned} P(u) &= \frac{1}{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{3-\alpha}{2} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{1}{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}- \frac{3}{2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{3\lambda}{q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x - \frac{1}{2} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x. \end{aligned} \end{align*} $$
We define
$$ \begin{align*} \begin{aligned} J(u)= \frac{1}{2} \langle I'(u),u \rangle + P(u). \end{aligned} \end{align*} $$
Under direct calculation, one has
$$ \begin{align*} \begin{aligned} J(u) &= \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad {}- \frac{2_{\alpha}^{*}+6} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{\lambda(q+6)}{2q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x - \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x. \end{aligned} \end{align*} $$
To establish our main results, we introduce the following Nehari–Pohožaev manifold
$$ \begin{align*} \begin{aligned} \mathcal{M}= \big\{ u\in W_{rad}^{1,2}(\mathbb{R}^{3},\alpha)\backslash\{0\} \big| J(u)=0 \big\}. \end{aligned} \end{align*} $$
We establish a new inequality, which plays a key role in Lemma 3.2.
Lemma 3.1 Let
$0<a<b<\infty $
. Then, we have
$$ \begin{align*} \begin{aligned} at^{b}-bt^{a}+b-a\geqslant0, \ \, t\in(0,\infty). \end{aligned} \end{align*} $$
Proof Set
$$ \begin{align*} \begin{aligned} h(t)=at^{b}-bt^{a}+b-a. \end{aligned} \end{align*} $$
Then, we have
$$ \begin{align*} \begin{aligned} h'(t)=abt^{a-1}(t^{b-a}-1). \end{aligned} \end{align*} $$
It is easy to see that
$$ \begin{align*} \begin{aligned} \begin{cases} h'(t)<0,&t\in(0,1), \\ h'(t)=0,&t=1, \\ h'(t)>0,&t\in(1,\infty), \end{cases} \end{aligned} \end{align*} $$
and
$$ \begin{align*} \begin{aligned} \begin{cases} \lim\limits_{t\to0^{+}}h(t)=b-a, \\ h(1)=0. \end{cases} \end{aligned} \end{align*} $$
These imply that
$h(\cdot )$
achieves its minimum at
$t=1$
, and
$h(t)\geqslant 0$
for
$t\in (0,\infty )$
.▪
Lemma 3.2 Assume that all conditions described in Theorem 1.1 hold. Then, for any
$t>0$
,
$$ \begin{align*} \begin{aligned} I(u)\geqslant I(t^{\frac{1}{2}}u_{t}) + \frac{1-t^{4}}{4}J(u) + \frac{(t^{2}-1)^{2}}{4} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2}, \ \, u\in W_{rad}^{1,2}(\mathbb{R}^{3},\alpha), \end{aligned} \end{align*} $$
where
$u_{t}=u\left (\frac {x}{t}\right )$
.
Proof For
$u\in W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
and
$t>0$
, it follows that
$$ \begin{align} \begin{aligned} I(t^{\frac{1}{2}}u_{t}) &= \frac{t^{2}}{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{t^{4-\alpha}}{2} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{t^{4}}{4} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}- \frac{t^{\frac{2_{\alpha}^{*}}{2}+3}} {2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{\lambda t^{\frac{q}{2}+3}}{q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x - \frac{t^{6}}{6} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x. \end{aligned} \end{align} $$
From Lemma 3.1 and a direct calculation, one has
$$ \begin{align*} \begin{aligned} &I(u)-I(t^{\frac{1}{2}}u_{t})\\ &= \frac{1-t^{2}}{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{1-t^{4-\alpha}}{2} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{1-t^{4}}{4} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}- \frac{1-t^{\frac{2_{\alpha}^{*}}{2}+3}} {2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{\lambda(1-t^{\frac{q}{2}+3})}{q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x - \frac{1-t^{6}}{6} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \\ &= \frac{1-t^{4}}{4}J(u) + \frac{(t^{2}-1)^{2}}{4} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{(4-\alpha)t^{4}-4t^{4-\alpha}+\alpha}{8} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)} ^{2} \\ &\quad {}+ \frac{8t^{\frac{2_{\alpha}^{*}}{2}+3}- (2_{\alpha}^{*}+6)t^{4} + (2_{\alpha}^{*}-2)} {8\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x \\ &\quad {}+ \frac{\lambda[8t^{\frac{q}{2}+3} -(q+6)t^{4}+(q-2)]}{8q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x + \frac{2t^{6}-3t^{4}+1}{12} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \\ &\geqslant \frac{1-t^{4}}{4}J(u) + \frac{(t^{2}-1)^{2}}{4} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2}. \end{aligned} \end{align*} $$
The proof is completed.▪
Lemma 3.3 Assume that all conditions described in Theorem 1.1 hold. Then, for any
$u\in W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )\backslash \{0\}$
, there is a unique
$t_{u}>0$
such that
$t_{u}^{\frac {1}{2}}u_{t_{u}}\in \mathcal {M}$
. Moreover,
$I(t_{u}^{\frac {1}{2}}u_{t_{u}})= \max \limits _{t\geqslant 0} I(t^{\frac {1}{2}}u_{t})$
.
Proof For any
$u\in W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )\backslash \{0\}$
and
$t>0$
, we consider a function
$g(t)=I(t^{\frac {1}{2}}u_{t})$
. Clearly, from (3.1), we get
$$ \begin{align*} \begin{aligned} g'(t) &= t\|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{(4-\alpha)t^{3-\alpha}}{2} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)} ^{2} + t^{3}\|u\|_{D^{1,2}(\mathbb{R}^{3})} ^{4} \\ &\quad {}- \frac{(2_{\alpha}^{*}+6)t^{\frac{2_{\alpha}^{*}}{2}+2}} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{\lambda(q+6)t^{\frac{q}{2}+2}}{2q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x - t^{5}\int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x. \end{aligned} \end{align*} $$
Obviously, one can note that
$\lim \limits _{t\to 0^{+}}g(t)=0$
and
$\lim \limits _{t\to +\infty }g(t)=-\infty $
. Thus, there exists
$t_{u}>0$
corresponding to the maximum point of
$g(t)$
, namely
$g'(t_{u})=0$
and
$g(t_{u})=\max \limits _{t>0}g(t)$
. Moreover, under a direct calculation, it is easy to get
$$ \begin{align} \begin{aligned} J(t^{\frac{1}{2}}u_{t})=tg'(t). \end{aligned} \end{align} $$
So, from the above expression, we obtain
$t_{u}^{\frac {1}{2}}u_{t_{u}}\in \mathcal {M}$
and
$I(t_{u}^{\frac {1}{2}}u_{t_{u}}) = \max \limits _{t>0}I(t^{\frac {1}{2}}u_{t})$
.
In the following, we claim that the critical point of
$g(t)$
is unique. By the way of contradiction, we suppose that for any
$u\in W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )\backslash \{0\}$
, there exists
$t_{u}>\tilde {t}_{u}>0$
such that
$t_{u}^{\frac {1}{2}}u_{t_{u}},\tilde {t}_{u}^{\frac {1}{2}}u_{\tilde {t}_{u}}\in \mathcal {M}$
.
Then, we have
$$ \begin{align*} \begin{aligned} J(t_{u}^{\frac{1}{2}}u_{t_{u}}) &= t_{u}^{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{(4-\alpha) t_{u}^{4-\alpha}}{2} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + t_{u}^{4} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad {}- \frac{(2_{\alpha}^{*}+6)t_{u}^{\frac{2_{\alpha}^{*}}{2}+3}} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{\lambda(q+6)t_{u}^{\frac{q}{2}+3}}{2q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x - t_{u}^{6}\int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \\ &=0, \end{aligned} \end{align*} $$
and
$$ \begin{align*} \begin{aligned} J(\tilde{t}_{u}^{\frac{1}{2}} u_{\tilde{t}_{u}}) &= \tilde{t}_{u}^{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{(4-\alpha)\tilde{t}_{u}^{4-\alpha}}{2} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \tilde{t}_{u}^{4} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}- \frac{(2_{\alpha}^{*}+6)\tilde{t}_{u}^{\frac{2_{\alpha}^{*}}{2}+3}} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{\lambda(q+6)\tilde{t}_{u}^{\frac{q}{2}+3}}{2q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x - \tilde{t}_{u}^{6} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \\ &=0. \end{aligned} \end{align*} $$
Combining the above two equalities, one has
$$ \begin{align*} &\left( \frac{1}{t_{u}^{2}} - \frac{1}{\tilde{t}_{u}^{2}} \right) \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \left( \frac{1}{t_{u}^{\alpha}} - \frac{1}{\tilde{t}_{u}^{\alpha}} \right) \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} \\ &\quad= \frac{2_{\alpha}^{*}+6} {2\cdot 2_{\alpha}^{*}} \left( t_{u}^{\frac{2_{\alpha}^{*}}{2}-1} - \tilde{t}_{u}^{\frac{2_{\alpha}^{*}}{2}-1} \right) \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x + \frac{\lambda(q+6)}{2q} \left( t_{u}^{\frac{q}{2}-1} - \tilde{t}_{u}^{\frac{q}{2}-1} \right) \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x \\ &\qquad{}+ \left( t_{u}^{2} - \tilde{t}_{u}^{2} \right) \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x, \end{align*} $$
which yields a contradiction. This completes the proof.▪
Lemma 3.4 Let
$m:=\inf \limits _{u\in \mathcal {M}}I(u)$
, then
$m>0$
. Moreover,
$$ \begin{align*} \begin{aligned} \inf_{u\in\mathcal{M}}I(u) =m= \inf_{u\in W_{rad}^{1,2}(\mathbb{R}^{3},\alpha)\backslash\{0\}} \max_{t\geqslant 0}I(t^{\frac{1}{2}}u_{t})>0. \end{aligned} \end{align*} $$
Proof It follows from Lemma 3.3 that
$$ \begin{align*} \begin{aligned} m=\inf_{u\in W_{rad}^{1,2}(\mathbb{R}^{3},\alpha)\backslash\{0\}} \max_{t\geqslant 0}I(t^{\frac{1}{2}}u_{t}). \end{aligned} \end{align*} $$
In what follows, we show that
$m>0$
. For any
$u\in \mathcal {M}$
, one has
$$ \begin{align*} \begin{aligned} \|u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2} &\leqslant \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &= \frac{2_{\alpha}^{*}+6} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x + \frac{\lambda(q+6)}{2q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x + \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \\ &\leqslant C_{1}\|u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2_{\alpha}^{*}} + C_{2}\|u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{q} + C_{3}\|u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{6}, \end{aligned} \end{align*} $$
which gives that there exists
$C>0$
such that
$\|u\|_{W^{1,2}(\mathbb {R}^{3},\alpha )}\geqslant C$
. For any
$u\in \mathcal {M}$
and
$t>0$
, there holds
$$ \begin{align*} \begin{aligned} I(u) &= I(u)-\frac{2}{2_{\alpha}^{*}+6}J(u) \\ &= \frac{2_{\alpha}^{*}+2}{2(2_{\alpha}^{*}+6)} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{2_{\alpha}^{*}+2\alpha-2}{2(2_{\alpha}^{*}+6)} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{2_{\alpha}^{*}-2}{4(2_{\alpha}^{*}+6)} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}+ \frac{\lambda(q-2_{\alpha}^{*})}{q(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x + \frac{\lambda(6-2_{\alpha}^{*})}{6(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \\ &\geqslant \frac{2_{\alpha}^{*}+2\alpha-2}{2(2_{\alpha}^{*}+6)} \|u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2}, \end{aligned} \end{align*} $$
which implies
$m=\inf \limits _{u\in \mathcal {M}}I(u)>0$
. The proof is completed.▪
3.1 Estimation of m
The main feature of the functional I is that it satisfies the local compactness condition. Next, we give an estimation on the m.
Lemma 3.5 Assume that all conditions described in Theorem 1.1 hold. Then, there exists a
$\Lambda>0$
such that for all
$\lambda>\Lambda $
, we have
$$ \begin{align*} \begin{aligned} 0<m<m^{*} &:= \min\left\{ \frac{2_{\alpha}^{*}+2\alpha-2}{2(2_{\alpha}^{*}+6)} \left( \frac{2\cdot 2_{\alpha}^{*}}{2_{\alpha}^{*}+6} \right) ^{\frac{2}{2_{\alpha}^{*}-2}} \mathcal{S}_{\alpha} ^{\frac{2_{\alpha}^{*}}{2_{\alpha}^{*}-2}} , \right.\\ &\qquad\qquad\left. \frac{\mathcal{S}^{3} + \sqrt{\mathcal{S}^{6} + 4\mathcal{S}^{3}}} {6} + \frac{\mathcal{S}^{6} + 2\mathcal{S}^{3} + \mathcal{S}^{3} \sqrt{\mathcal{S}^{6} +4\mathcal{S}^{3}}} {24} \right\}. \end{aligned} \end{align*} $$
Proof We choose
$$ \begin{align*} \begin{aligned} \|v\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2} =1, \ \, \int_{\mathbb{R}^{3}} |v|^{q} \mathrm{d}x>0, \end{aligned} \end{align*} $$
and
$$ \begin{align*} \begin{aligned} \lim_{t\to+\infty} I(t^{\frac{1}{2}}v_{t}) = -\infty. \end{aligned} \end{align*} $$
Then,
$$ \begin{align*} \begin{aligned} \sup_{t\geqslant0} I(t^{\frac{1}{2}}v_{t}) = I(t_{v,\lambda}^{\frac{1}{2}} v_{t_{v,\lambda}}), \end{aligned} \end{align*} $$
for some
$t_{v,\lambda }>0$
. Note that Lemma 3.3,
$t_{v,\lambda }>0$
, satisfies
$$ \begin{align} \begin{aligned} &t_{v,\lambda}^{2} \|v\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{(4-\alpha) t_{v,\lambda}^{4-\alpha}}{2} \|v\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + t_{v,\lambda}^{4} \|v\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad= \frac{(2_{\alpha}^{*}+6)t_{v,\lambda}^{\frac{2_{\alpha}^{*}}{2}+3}} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |v|^{2_{\alpha}^{*}} \mathrm{d}x + \frac{\lambda(q+6)t_{v,\lambda}^{\frac{q}{2}+3}}{2q} \int_{\mathbb{R}^{3}} |v|^{q} \mathrm{d}x + t_{v,\lambda}^{6} \int_{\mathbb{R}^{3}} |v|^{6} \mathrm{d}x. \end{aligned} \end{align} $$
Thus, we get
$$ \begin{align*} \begin{aligned} t_{v,\lambda}^{2} \|v\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{(4-\alpha) t_{v,\lambda}^{4-\alpha}}{2} \|v\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + t_{v,\lambda}^{4} \|v\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \geqslant t_{v,\lambda}^{6} \int_{\mathbb{R}^{3}} |v|^{6} \mathrm{d}x. \end{aligned} \end{align*} $$
So we get
$\{t_{v,\lambda }\}$
is bounded.
We next claim that
$t_{v,\lambda }\to 0$
as
$\lambda \to +\infty $
. By the way of contradiction, we suppose that there exist
$\tilde {t}>0$
and a sequence
$\{\lambda _{n}\}$
with
$\lambda _{n}\to +\infty $
as
$n\to +\infty $
, such that
${t_{v,\lambda _{n}}\to \tilde {t}}$
as
$n\to +\infty $
. Passing the limit as
$n\to +\infty $
, it follows that
$$ \begin{align*} \begin{aligned} \frac{\lambda_{n}(q+6) \tilde{t}^{\frac{q}{2}+3}}{2q} \int_{\mathbb{R}^{3}} |v|^{q} \mathrm{d}x \to +\infty. \end{aligned} \end{align*} $$
Inserting this into (3.3), there holds
$$ \begin{align*} &\max\left\{\tilde{t}^{2}, \frac{(4-\alpha) \tilde{t}^{4-\alpha}}{2}\right\} \|v\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2} + \tilde{t}^{4} \|v\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{4} \\ &\quad\geqslant \tilde{t}^{2} \|v\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{(4-\alpha) \tilde{t}^{4-\alpha}}{2} \|v\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \tilde{t}^{4} \|v\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad= +\infty, \end{align*} $$
which leads to a contradiction. Hence, we get
$t_{v,\lambda }\to 0$
as
$\lambda \to +\infty $
. Combining this and (3.3), we have
$$ \begin{align*} \begin{aligned} &\lim_{\lambda\to+\infty} \sup_{t\geqslant0} I(t^{\frac{1}{2}}v_{t}) \\ &\quad= \lim_{\lambda\to+\infty} I(t_{v,\lambda}^{\frac{1}{2}}v_{t_{v,\lambda}}) \\ &\quad = \lim_{\lambda\to+\infty} \Bigg( \frac{t_{v,\lambda}^{2}}{2} \|v\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{t_{v,\lambda}^{4-\alpha}}{2} \|v\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{t_{v,\lambda}^{4}}{4} \|v\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\qquad\qquad\kern8pt {}- \frac{t_{v,\lambda}^{\frac{2_{\alpha}^{*}}{2}+3}} {2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |v|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{\lambda t_{v,\lambda}^{\frac{q}{2}+3}}{q} \int_{\mathbb{R}^{3}} |v|^{q} \mathrm{d}x - \frac{t_{v,\lambda}^{6}}{6} \int_{\mathbb{R}^{3}} |v|^{6} \mathrm{d}x\Bigg) \\ &\quad= \lim_{\lambda\to+\infty} \Bigg\{ t_{v,\lambda}^{2}\left( \frac{1}{2}-\frac{2}{q+6} \right) \|v\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + t_{v,\lambda}^{4-\alpha}\left( \frac{1}{2} - \frac{4-\alpha}{q+6} \right) \|v\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} \\ &\qquad{}+ t_{v,\lambda}^{4} \left( \frac{1}{4} - \frac{2}{q+6} \right) \|v\|_{D^{1,2}(\mathbb{R}^{3})}^{4} + t_{v,\lambda}^{\frac{2_{\alpha}^{*}}{2}+3} \left[ \frac{2_{\alpha}^{*}+6}{2_{\alpha}^{*}(q+6)} - \frac{1}{2_{\alpha}^{*}} \right] \int_{\mathbb{R}^{3}} |v|^{2_{\alpha}^{*}} \mathrm{d}x \\ &\qquad{}+ t_{v,\lambda}^{6} \left( \frac{2}{q+6} - \frac{1}{6} \right) \int_{\mathbb{R}^{3}} |v|^{6} \mathrm{d}x\Bigg\} \\ &\quad=0. \end{aligned} \end{align*} $$
Then, there exists
$0<\Lambda <+\infty $
, such that for any
$\lambda>\Lambda $
, we can arrive at
$$ \begin{align*} \begin{aligned} \sup_{t\geqslant0} I(t^{\frac{1}{2}}v_{t}) &< \min\left\{ \frac{2_{\alpha}^{*}+2\alpha-2}{2(2_{\alpha}^{*}+6)} \left( \frac{2\cdot 2_{\alpha}^{*}}{2_{\alpha}^{*}+6} \right) ^{\frac{2}{2_{\alpha}^{*}-2}} \mathcal{S}_{\alpha} ^{\frac{2_{\alpha}^{*}}{2_{\alpha}^{*}-2}}, \right.\\ &\qquad\qquad\left. \frac{\mathcal{S}^{3} + \sqrt{\mathcal{S}^{6} + 4\mathcal{S}^{3}}} {6} + \frac{\mathcal{S}^{6} + 2\mathcal{S}^{3} + \mathcal{S}^{3} \sqrt{\mathcal{S}^{6} +4\mathcal{S}^{3}}} {24} \right\}. \end{aligned} \end{align*} $$
This completes the proof.▪
3.2 Nonvanishing of the minimizing sequence
Lemma 3.6 Assume that all conditions described in Theorem 1.1 hold. Let
$\{u_{n}\}\subset \mathcal {M}$
be a bounded minimizing sequence of I and satisfies
$$ \begin{align*} \begin{aligned} I(u_{n})\to m\in(0,m^{*}) \ \, and \ \, J(u_{n})=0. \end{aligned} \end{align*} $$
Then, there exists a
$\Lambda>0$
such that for all
$\lambda>\Lambda $
, we have
$$ \begin{align*} \limsup_{n\rightarrow+\infty} \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x>0 \ \, and \ \, \limsup_{n\rightarrow+\infty} \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x >0. \end{align*} $$
Proof Let
$\{u_{n}\}$
be a bounded minimizing sequence of I. First, we give the proof of
$\limsup \limits _{n\rightarrow +\infty } \int _{\mathbb {R}^{3}} |u_{n}|^{2_{\alpha }^{*}} \mathrm {d}x>0$
. We argue by contradiction, and suppose that
$$ \begin{align} \limsup_{n\rightarrow+\infty} \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x=0. \end{align} $$
It follows from (3.4) and Hölder’s inequality that
$$ \begin{align} \limsup_{n\rightarrow+\infty} \int_{\mathbb{R}^{3}} |u_{n}|^{q} \mathrm{d}x &\leqslant C\limsup_{n\rightarrow+\infty} \left( \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x \right) ^{\frac{6-q}{6-2_{\alpha}^{*}}} \left( \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x \right) ^{\frac{q-2_{\alpha}^{*}}{6-2_{\alpha}^{*}}}\nonumber \\ &\leqslant C\limsup_{n\rightarrow+\infty} \left( \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x \right) ^{\frac{6-q}{6-2_{\alpha}^{*}}} \limsup_{n\rightarrow+\infty} \left( \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x \right) ^{\frac{q-2_{\alpha}^{*}}{6-2_{\alpha}^{*}}}\nonumber\\ &=0. \end{align} $$
In view of
$\{u_{n}\}$
is the minimizing sequence of I,
$$ \begin{align} m+o_{n}(1) &= \frac{1}{2} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{2} + \frac{1}{2} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2}\nonumber\\ &\quad{}+ \frac{1}{4} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{4} - \frac{1}{6} \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x, \end{align} $$
and
$$ \begin{align} \begin{aligned} o_{n}(1)= \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} - \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x. \end{aligned} \end{align} $$
Combined (3.6) with (3.7), we can deduce
$$ \begin{align} \begin{aligned} m+o_{n}(1) &= \frac{1}{3} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{2} + \frac{2+\alpha}{12} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{1}{12} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{4} \\ &\geqslant \frac{1}{3} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{2} + \frac{1}{12} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{4}. \end{aligned} \end{align} $$
Then, we can derive from (2.2) and (3.7) that
$$ \begin{align*} \begin{aligned} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{2} \geqslant \mathcal{S} \left( \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x \right) ^{\frac{1}{3}} \geqslant \mathcal{S} \left( \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{2} + \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{4} \right) ^{\frac{1}{3}}, \end{aligned} \end{align*} $$
which implies
$$ \begin{align} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{2} \geqslant \frac{\mathcal{S}^{3}+\sqrt{\mathcal{S}^{6}+4\mathcal{S}^{3}}}{2}. \end{align} $$
According to (3.8) and (3.9), we conclude
$$ \begin{align*} m\geqslant \frac{\mathcal{S}^{3} + \sqrt{\mathcal{S}^{6} + 4\mathcal{S}^{3}}} {6} + \frac{\mathcal{S}^{6} + 2\mathcal{S}^{3} + \mathcal{S}^{3} \sqrt{\mathcal{S}^{6} +4\mathcal{S}^{3}}} {24}, \end{align*} $$
which yields a contradiction with
$m<m^{*}$
. Hence, we get
$\limsup \limits _{n\rightarrow +\infty } \int _{\mathbb {R}^{3}} |u_{n}|^{2_{\alpha }^{*}} \mathrm {d}x>0$
.
Next, we claim
$\limsup \limits _{n\rightarrow +\infty } \int _{\mathbb {R}^{3}} |u_{n}|^{6} \mathrm {d}x>0$
. Otherwise, we can assume
$$ \begin{align} \limsup_{n\rightarrow+\infty} \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x =0. \end{align} $$
It follows from
$\{u_{n}\}$
is the minimizing sequence of I, (3.5), and (3.10) that
$$ \begin{align} m+o_{n}(1) &= \frac{1}{2} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{2} + \frac{1}{2} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2}\nonumber \\ &\quad{}+ \frac{1}{4} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})} ^{4} - \frac{1}{2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x, \end{align} $$
and
$$ \begin{align} \begin{aligned} 0= \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} - \frac{2_{\alpha}^{*}+6} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x. \end{aligned} \end{align} $$
In view of (3.11) and (3.12), we obtain
$$ \begin{align} \begin{aligned} m+o_{n}(1) &= \frac{2_{\alpha}^{*}+2}{2(2_{\alpha}^{*}+6)} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{2_{\alpha}^{*}+2\alpha-2}{2(2_{\alpha}^{*}+6)} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} \\ &\quad{}+ \frac{2_{\alpha}^{*}-2}{2(2_{\alpha}^{*}+6)} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\geqslant \frac{2_{\alpha}^{*}+2\alpha-2}{2(2_{\alpha}^{*}+6)} \|u_{n}\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2}. \end{aligned} \end{align} $$
In terms of (2.1) and (3.12), one can infer
$$ \begin{align*} \begin{aligned} \|u_{n}\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2} &\geqslant \mathcal{S}_{\alpha} \left( \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x \right) ^{\frac{2}{2_{\alpha}^{*}}} \geqslant \mathcal{S}_{\alpha} \left( \frac{2\cdot 2_{\alpha}^{*}}{2_{\alpha}^{*}+6} \right) ^{\frac{2}{2_{\alpha}^{*}}} \|u_{n}\|_{W^{1,2}(\mathbb{R}^{3},\alpha)} ^{\frac{4}{2_{\alpha}^{*}}}, \end{aligned} \end{align*} $$
which implies
$$ \begin{align} \|u_{n}\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2} \geqslant \left( \frac{2\cdot 2_{\alpha}^{*}}{2_{\alpha}^{*}+6} \right) ^{\frac{2}{2_{\alpha}^{*}-2}} \mathcal{S}_{\alpha} ^{\frac{2_{\alpha}^{*}}{2_{\alpha}^{*}-2}}. \end{align} $$
Taking into account (3.13) and (3.14), we can infer
$$ \begin{align*} m\geqslant \frac{2_{\alpha}^{*}+2\alpha-2}{2(2_{\alpha}^{*}+6)} \left( \frac{2\cdot 2_{\alpha}^{*}}{2_{\alpha}^{*}+6} \right) ^{\frac{2}{2_{\alpha}^{*}-2}} \mathcal{S}_{\alpha} ^{\frac{2_{\alpha}^{*}}{2_{\alpha}^{*}-2}}, \end{align*} $$
which contradicts with
$m<m^{*}$
. Hence, we obtain
$\limsup \limits _{n\rightarrow +\infty } \int _{\mathbb {R}^{3}} |u_{n}|^{6} \mathrm {d}x>0$
. This completes the proof.▪
3.3 The existence of ground state solution
Lemma 3.7 Assume that the assumptions of Theorem 1.1 hold. Let
$\{u_{n}\}\subset \mathcal {M}$
be a minimizing sequence at
$0<m<m^{*}$
. If
$u_{n}\rightharpoonup u$
in
$W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
, then
$J(u)=0$
.
Proof Because
$I(u_{n})\to m$
and
$J(u_{n})=0$
, one can easily get
$$ \begin{align*} \begin{aligned} m+o_{n}(1) &= \frac{1}{2} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{1}{2} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{1}{4} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}- \frac{1}{2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{\lambda}{q} \int_{\mathbb{R}^{3}} |u_{n}|^{q} \mathrm{d}x - \frac{1}{6} \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x, \end{aligned} \end{align*} $$
and
$$ \begin{align*} \begin{aligned} 0 &= \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}- \frac{2_{\alpha}^{*}+6} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x - \frac{\lambda(q+6)}{2q} \int_{\mathbb{R}^{3}} |u_{n}|^{q} \mathrm{d}x - \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x. \end{aligned} \end{align*} $$
It follows from the above expressions that
$$ \begin{align*} \begin{aligned} m &= \frac{2_{\alpha}^{*}+2}{2(2_{\alpha}^{*}+6)} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{2_{\alpha}^{*}+2\alpha-2}{2(2_{\alpha}^{*}+6)} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{2_{\alpha}^{*}-2}{4(2_{\alpha}^{*}+6)} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}+ \frac{\lambda(q-2_{\alpha}^{*})}{q(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u_{n}|^{q} \mathrm{d}x + \frac{6-2_{\alpha}^{*}}{6(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x \\ &\geqslant \frac{2_{\alpha}^{*}+2\alpha-2}{2(2_{\alpha}^{*}+6)} \|u_{n}\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}^{2}, \end{aligned} \end{align*} $$
which implies that
$\{u_{n}\}$
is uniformly bounded in
$W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
. Without loss of generality, up to a subsequence, there exists u such that
$$ \begin{align*} \begin{aligned} u_{n}\rightharpoonup u \; \mathrm{in} \; W^{1,2}_{rad}(\mathbb{R}^{3},\alpha), \;\; u_{n}\rightarrow u \; \mathrm{a.e.~in} \; \mathbb{R}^{3} \;\; \mathrm{and} \;\; u_{n}\rightarrow u \; \mathrm{in} \; L^{s}(\mathbb{R}^{3}) \; \mathrm{for} \; s\in(2_{\alpha}^{*},6). \end{aligned} \end{align*} $$
For
$u=0$
, we have
$J(u)=0$
. From now on, we assume
$u\not \equiv 0$
.
First, we show
$J(u)\geqslant 0$
. On the contrary, we suppose
$J(u)<0$
. In view of Lemma 3.3, there exists a unique
$t_{u}>0$
such that
$J(t_{u}^{\frac {1}{2}}u_{t_{u}})=0$
. We now claim
$t_{u}<1$
. In view of (3.2) and
$J(u)<0$
, we obtain
$$ \begin{align*} J(u)=g'(1)<0. \end{align*} $$
Moreover, it is easy to check that for
$t_{0}>0$
small enough, one has
$$ \begin{align*} g'(t_{0})>0. \end{align*} $$
By the Intermediate value theorem, it follows that there exists
$\bar {t}_{u}\in (t_{0},1)$
such that
$$ \begin{align*} g'(\bar{t}_{u})=0. \end{align*} $$
According to Lemma 3.3, we know
$\bar {t}_{u}$
is unique. Thus, we have
$$ \begin{align} t_{u} = \bar{t}_{u} \in(t_{0},1). \end{align} $$
Making use of (3.15), one has
$$ \begin{align*} m &\leqslant I(t_{u}^{\frac{1}{2}}u_{t_{u}}) = I(t_{u}^{\frac{1}{2}}u_{t_{u}}) - \frac{2}{2_{\alpha}^{*}+6} J(t_{u}^{\frac{1}{2}}u_{t_{u}}) \\ &= \frac{2_{\alpha}^{*}+2} {2(2_{\alpha}^{*}+6)} t_{u}^{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{2_{\alpha}^{*}+2\alpha-2} {2(2_{\alpha}^{*}+6)} t_{u}^{4-\alpha} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{2_{\alpha}^{*}-2} {4(2_{\alpha}^{*}+6)} t_{u}^{4} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}+ \frac{\lambda(q-2_{\alpha}^{*})} {q(2_{\alpha}^{*}+6)} t_{u}^{q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x + \frac{6-2_{\alpha}^{*}} {6(2_{\alpha}^{*}+6)} t_{u}^{6} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \\ &< \frac{2_{\alpha}^{*}+2} {2(2_{\alpha}^{*}+6)} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{2_{\alpha}^{*}+2\alpha-2} {2(2_{\alpha}^{*}+6)} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{2_{\alpha}^{*}-2} {4(2_{\alpha}^{*}+6)} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}+ \frac{\lambda(q-2_{\alpha}^{*})} {q(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x + \frac{6-2_{\alpha}^{*}} {6(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \\ &\leqslant \liminf_{n\rightarrow\infty} \bigg[ \frac{2_{\alpha}^{*}+2} {2(2_{\alpha}^{*}+6)} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \!+\! \frac{2_{\alpha}^{*}+2\alpha-2} {2(2_{\alpha}^{*}+6)} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} \!+\! \frac{2_{\alpha}^{*}-2} {4(2_{\alpha}^{*}+6)} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\qquad\qquad{}+ \frac{\lambda(q-2_{\alpha}^{*})} {q(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u_{n}|^{q} \mathrm{d}x + \frac{6-2_{\alpha}^{*}} {6(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x \bigg] \\ &= \liminf_{n\rightarrow\infty} \bigg[ I(u_{n}) - \frac{2}{2_{\alpha}^{*}+6} J(u_{n}) \bigg] \\ &\leqslant \lim_{n\rightarrow\infty} I(u_{n}) = m, \end{align*} $$
which yields a contradiction. Hence,
$J(u)\geqslant 0$
.
In what follows, we prove
$J(u)=0$
. Otherwise, if
$J(u)>0$
, then we have
$$ \begin{align} \begin{aligned} &I(u) - \frac{2}{2_{\alpha}^{*}+6} J(u) \\ &\quad= \frac{2_{\alpha}^{*}+2} {2(2_{\alpha}^{*}+6)} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{2_{\alpha}^{*}+2\alpha-2} {2(2_{\alpha}^{*}+6)} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{2_{\alpha}^{*}-2} {4(2_{\alpha}^{*}+6)} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\qquad{}+ \frac{\lambda(q-2_{\alpha}^{*})} {q(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x + \frac{6-2_{\alpha}^{*}} {6(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \\ &\quad>0. \end{aligned} \end{align} $$
Set
$v_{n}=u_{n}-u$
. By the Brésiz–Lieb lemma [Reference Brézis and Lieb8], we know
$$ \begin{align*} m+o_{n}(1)=I(u_{n})=I(v_{n})+I(u) +\frac{1}{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|v_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2}, \end{align*} $$
and
$$ \begin{align*} o_{n}(1)=J(u_{n})=J(v_{n})+J(u) + 2\|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|v_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2}, \end{align*} $$
which implies
$$ \begin{align} \begin{aligned} m &= I(u_{n}) - \frac{2}{2_{\alpha}^{*}+6} J(u_{n}) \\ &= \left[ I(v_{n}) - \frac{2}{2_{\alpha}^{*}+6} J(v_{n}) \right] + \left[ I(u) - \frac{2}{2_{\alpha}^{*}+6} J(u) \right] \\ &\quad{}+ \frac{2_{\alpha}^{*}-2}{2(2_{\alpha}^{*}+6)} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|v_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2}, \end{aligned} \end{align} $$
and
$$ \begin{align*} 0>J(v_{n})=-J(u) - 2\|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|v_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2}. \end{align*} $$
In terms of Lemma 3.3, there exists a unique
$t_{v_{n}}$
such that
$$ \begin{align} J(t_{v_{n}}^{\frac{1}{2}} (v_{n})_{t_{v_{n}}})=0 \ \, \mathrm{and} \ \, t_{v_{n}}\in(0,1). \end{align} $$
Then, we can derive from (3.17) and (3.18) that
$$ \begin{align*} m &\leqslant \liminf_{n\to+\infty} I(t_{v_{n}}^{\frac{1}{2}} (v_{n})_{t_{v_{n}}}) \\ &= \liminf_{n\to+\infty} \bigg[ I(t_{v_{n}}^{\frac{1}{2}} (v_{n})_{t_{v_{n}}}) - \frac{2}{2_{\alpha}^{*}+6} J(t_{v_{n}}^{\frac{1}{2}} (v_{n})_{t_{v_{n}}}) \bigg] \\ &= \liminf_{n\to+\infty} \bigg[ \frac{2_{\alpha}^{*}+2} {2(2_{\alpha}^{*}+6)} t_{v_{n}}^{2} \|v_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{2_{\alpha}^{*}+2\alpha-2} {2(2_{\alpha}^{*}+6)} t_{v_{n}}^{4-\alpha} \|v_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} \\ &\qquad\qquad{}+ \frac{2_{\alpha}^{*}-2} {4(2_{\alpha}^{*}+6)} t_{v_{n}}^{4} \|v_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} + \frac{\lambda(q-2_{\alpha}^{*})} {q(2_{\alpha}^{*}+6)} t_{v_{n}}^{q} \int_{\mathbb{R}^{3}} |v_{n}|^{q} \mathrm{d}x \\ &\qquad\qquad{}+ \frac{6-2_{\alpha}^{*}} {6(2_{\alpha}^{*}+6)} t_{v_{n}}^{6} \int_{\mathbb{R}^{3}} |v_{n}|^{6} \mathrm{d}x \bigg] \\ &< \liminf_{n\to+\infty} \bigg[ \frac{2_{\alpha}^{*}+2} {2(2_{\alpha}^{*}+6)} \|v_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \!+\! \frac{2_{\alpha}^{*}+2\alpha-2} {2(2_{\alpha}^{*}+6)} \|v_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} \!+\! \frac{2_{\alpha}^{*}-2} {4(2_{\alpha}^{*}+6)} \|v_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\qquad\qquad{}+ \frac{\lambda(q-2_{\alpha}^{*})} {q(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |v_{n}|^{q} \mathrm{d}x + \frac{6-2_{\alpha}^{*}} {6(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |v_{n}|^{6} \mathrm{d}x \bigg] \\ &= \liminf_{n\to+\infty} \bigg[I(v_{n}) - \frac{2}{2_{\alpha}^{*}+6} J(v_{n})\bigg] \\ &< \lim_{n\to+\infty} \bigg\{ \left[ I(v_{n}) - \frac{2}{2_{\alpha}^{*}+6} J(v_{n}) \right] + \left[ I(u)-\frac{2}{2_{\alpha}^{*}+6}J(u) \right] \\ &\qquad\qquad{}+ \frac{2_{\alpha}^{*}-2}{2(2_{\alpha}^{*}+6)} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|v_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \bigg\} \\ &= \lim_{n\to+\infty} \bigg[ I(u_{n}) - \frac{2}{2_{\alpha}^{*}+6} J(u_{n}) \bigg] \\ &= m, \end{align*} $$
which yields a contradiction. The proof is completed.▪
With some minor modifications to the proof of [Reference Su25, Theorem 1.3], one can deduce the following modified generalized version of Lions-type theorem.
Theorem 3.1 Let
$\alpha \in (0,2)$
and
$\{u_{n}\}\subset W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
be any bounded sequence satisfying
$$ \begin{align*} \begin{aligned} \limsup_{n\rightarrow+\infty} \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x>0 \ \, and \ \, \limsup_{n\rightarrow+\infty} \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x >0. \end{aligned} \end{align*} $$
Then, the sequence
$\{u_{n}\}$
converges weakly and a.e. to
$u\not \equiv 0$
in
$L_{loc}^{2}(\mathbb {R}^{3})$
.
Now, we are in a position to give the proof of Theorem 1.1.
Proof of Theorem 1.1
Let
$\{u_{n}\}\subset \mathcal {M}$
be a minimizing sequence of I. Obviously,
$\{u_{n}\}$
is bounded in
$W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
. By virtue of Lemmas 3.5–3.7 and Theorem 3.1, we obtain
$\{u_{n}\}$
converges weakly and a.e. to
$\tilde {u}\not \equiv 0$
in
$L^{2}_{loc}(\mathbb {R}^{3})$
.
Making use of the Brézis–Lieb lemma [Reference Brézis and Lieb8], it follows that
$$ \begin{align*} m &\leqslant I(\tilde{u}) \\ &= I(\tilde{u}) - \frac{2}{2_{\alpha}^{*}+6} J(\tilde{u}) \\ &= \frac{2_{\alpha}^{*}+2} {2(2_{\alpha}^{*}+6)} \|\tilde{u}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{2_{\alpha}^{*}+2\alpha-2} {2(2_{\alpha}^{*}+6)} \|\tilde{u}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{2_{\alpha}^{*}-2} {4(2_{\alpha}^{*}+6)} \|\tilde{u}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}+ \frac{\lambda(q-2_{\alpha}^{*})} {q(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |\tilde{u}|^{q} \mathrm{d}x + \frac{6-2_{\alpha}^{*}} {6(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |\tilde{u}|^{6} \mathrm{d}x \\ &\leqslant \lim_{n\rightarrow+\infty} \bigg[ \frac{2_{\alpha}^{*}+2} {2(2_{\alpha}^{*}+6)} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \!+\! \frac{2_{\alpha}^{*}+2\alpha-2} {2(2_{\alpha}^{*}+6)} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} \!+\! \frac{2_{\alpha}^{*}-2} {4(2_{\alpha}^{*}+6)} \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} \\ &\quad{}+ \frac{\lambda(q-2_{\alpha}^{*})} {q(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u_{n}|^{q} \mathrm{d}x + \frac{6-2_{\alpha}^{*}} {6(2_{\alpha}^{*}+6)} \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x \bigg] \\ &= \lim_{n\rightarrow+\infty} \bigg[ I(u_{n}) - \frac{2}{2_{\alpha}^{*}+6} J(u_{n}) \bigg] \\ &= \lim_{n\rightarrow+\infty} I(u_{n}) = m, \end{align*} $$
which implies
$I(\tilde {u})=m$
.
In the following, we shall show
$\|I'(\tilde {u})\|_{W^{-1,2}(\mathbb {R}^{3},\alpha )}=0$
. Otherwise, there exist
$\delta $
,
$\mu>0$
such that
$$ \begin{align*} \begin{aligned} \|I'(\tilde{u})\|_{W^{-1,2}(\mathbb{R}^{3},\alpha)}\geqslant \mu, \ \, u\in S_{3\delta}(\tilde{u}), \end{aligned} \end{align*} $$
where
$S_{3\delta }(\tilde {u})=: \left \{u\in W_{rad}^{1,2}(\mathbb {R}^{3},\alpha ) \Big | \|u-\tilde {u}\|_{W^{1,2}(\mathbb {R}^{3},\alpha )}\leqslant 3\delta \right \}$
. The fact that
$C^{\infty }_{0}(\mathbb {R}^{3})$
is dense in
$W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
yields
$$ \begin{align*} \begin{aligned} \lim_{t\to1} \|t^{\frac{1}{2}}\tilde{u}_{t}-\tilde{u}\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}=0. \end{aligned} \end{align*} $$
Then, there exists
$\delta _{1}\in (0,\frac {1}{4})$
such that
$$ \begin{align} \begin{aligned} |t-1|<\delta_{1}\Rightarrow \|t^{\frac{1}{2}}\tilde{u}_{t}-\tilde{u}\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}<\delta. \end{aligned} \end{align} $$
In terms of Lemma 3.2, there holds
$$ \begin{align} \begin{aligned} I(t^{\frac{1}{2}}\tilde{u}_{t}) &\leqslant I(\tilde{u}) - \frac{(t^{2}-1)^{2}}{4} \|\tilde{u}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \\ &= m- \frac{(t^{2}-1)^{2}}{4} \|\tilde{u}\|_{D^{1,2}(\mathbb{R}^{3})}^{2}, \ \, \forall\;t>0. \end{aligned} \end{align} $$
Let
$$ \begin{align*} \begin{aligned} 0<\varepsilon<\min \left\{ \frac{\|\tilde{u}\|_{D^{1,2}(\mathbb{R}^{3})}^{2}}{24} ,1,\frac{\mu\delta}{8} \right\}. \end{aligned} \end{align*} $$
By [Reference Willem26], there exists a deformation
$\eta \in C\big ([0,1]\times W_{rad}^{1,2}(\mathbb {R}^{3},\alpha ), W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )\big )$
such that
$$ \begin{align*} \begin{aligned} (a)&\quad \eta(1,u)=u, \ \, u\not\in I^{-1}\big([m-2\varepsilon,m+2\varepsilon]\big)\cap S_{2\delta}; \\ (b)&\quad \eta(1,I^{m+\varepsilon}\cap S_{\delta})\subset I^{m-\varepsilon}; \\ (c)&\quad I\big(\eta(1,u)\big)\leqslant I(u), \ \, u\in W_{rad}^{1,2}(\mathbb{R}^{3},\alpha); \\ (d)&\quad \|\eta(1,u)-u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)} \leqslant \delta, \ \, u\in W_{rad}^{1,2}(\mathbb{R}^{3},\alpha). \end{aligned} \end{align*} $$
By Lemma 3.3,
$I(t^{\frac {1}{2}}\tilde {u}_{t}) \leqslant I(\tilde {u})=m$
for
$t>0$
, then it follows from (3.19) and
$(b)$
that
$$ \begin{align} \begin{aligned} I\big(\eta(1,t^{\frac{1}{2}}\tilde{u}_{t})\big) \leqslant m-\varepsilon, \ \, t>0, \ \, |t-1|<\delta_{1}. \end{aligned} \end{align} $$
Moreover, putting together (3.20) with
$(c)$
, we can deduce
$$ \begin{align} \begin{aligned} I\big(\eta(1,t^{\frac{1}{2}}\tilde{u}_{t})\big) &\leqslant I(t^{\frac{1}{2}}\tilde{u}_{t}) \\ &\leqslant m- \frac{(t^{2}-1)^{2}}{4} \|\tilde{u}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \\ &\leqslant m- \frac{\delta_{1}^{2}}{4} \|\tilde{u}\|_{D^{1,2}(\mathbb{R}^{3})}^{2}, \ \, t>0, \ \, |t-1|\geqslant \delta_{1}. \end{aligned} \end{align} $$
Combining (3.21) with (3.22), one can infer
$$ \begin{align*} \begin{aligned} \max_{t\in [0.5,1.5]} I\big(\eta(1,t^{\frac{1}{2}}\tilde{u}_{t})\big)<m. \end{aligned} \end{align*} $$
Next, we prove
$\eta (1,t^{\frac {1}{2}}\tilde {u}_{t})\cap \mathcal {M}\not =\emptyset $
for some
$t\in [0.5,1.5]$
. Define
$$ \begin{align*} \begin{aligned} \Psi_{1}(t) = J(t^{\frac{1}{2}}\tilde{u}_{t}), \ \, \mathrm{and} \ \, \Psi_{2}(t) = J\big(\eta(1,t^{\frac{1}{2}}\tilde{u}_{t})\big), \ \, t>0. \end{aligned} \end{align*} $$
By Lemma 3.4 and the degree theory, we derive that
$$ \begin{align*} \begin{aligned} \mathrm{deg}(\Psi_{1}, (0.5, 1.5), 0)\not=0. \end{aligned} \end{align*} $$
It follows from (3.20) and
$(a)$
that
$\eta (1,t^{\frac {1}{2}}\tilde {u}_{t})=t^{\frac {1}{2}}\tilde {u}_{t}$
for
$t=0.5$
and
$t=1.5$
. Thus,
$$ \begin{align*} \begin{aligned} \mathrm{deg}(\Psi_{1}, (0.5, 1.5), 0) = \mathrm{deg}(\Psi_{2}, (0.5, 1.5), 0)\not=0. \end{aligned} \end{align*} $$
Hence,
$\Psi _{2}(t_{0}) = 0$
for some
$t_{0}\in (0.5, 1.5)$
, that is,
$\eta (1,t_{0}^{\frac {1}{2}}\tilde {u}_{t_{0}}) \in \mathcal {M}$
, which is a contradiction. The proof is completed.▪
4 Proof of Theorem 1.2
We study equation
$(\mathcal {K})$
again, and use the notations similarly in Section 3. We point out that the interval of parameter
$\alpha $
is different from Theorem 1.1.
Lemma 4.1 Assume that all conditions described in Theorem 1.2 hold. Then, the following statements are true.
$\mathrm {(i)}$
There exists
$0<\bar {\Lambda }<\infty $
such that for any
$\lambda>\bar {\Lambda }$
,
$$ \begin{align*} \begin{aligned} 0<m<\hat{m}^{*}, \end{aligned} \end{align*} $$
where
$$ \begin{align*} \begin{aligned} \hat{m}^{*}:= \min&\left\{ \frac{1}{12} \left[\frac{(\alpha-2)\mathcal{S}}{2}\right]^{\frac{3}{2}} , \frac{2_{\alpha}^{*}-2} {8\cdot 2_{\alpha}^{*}} \left[ \frac{2_{\alpha}^{*}(4-\alpha) \mathcal{S}_{\alpha}}{2_{\alpha}^{*}+6} \right]^{\frac{2_{\alpha}^{*}}{2_{\alpha}^{*}-2}} \right\}. \end{aligned} \end{align*} $$
$\mathrm {(ii)}$
Let
$\{u_{n}\}$
be a bounded minimizing sequence such that
$$ \begin{align*} \begin{aligned} I(u_{n})\to m\in(0,\hat{m}^{*}) \ \, and \ \, J(u_{n})=0. \end{aligned} \end{align*} $$
If
$u_{n}\rightharpoonup u$
in
$W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
, then
$J(u)=0$
.
$\mathrm {(iii)}$
Moreover, there exists
$0\not \equiv u\in W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
such that, up to a subsequence,
$$ \begin{align*} \begin{aligned} \|u_{n}-u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}\to0, \ \ \mathrm{as} \; n\to\infty. \end{aligned} \end{align*} $$
Proof Similar to Lemmas 3.5 and 3.7, we have (i) and (ii).
(iii) Let
$\{u_{n}\}$
be a bounded minimizing sequence such that
$$ \begin{align*} \begin{aligned} I(u_{n})\to m\in(0,\hat{m}^{*}) \ \, and \ \, J(u_{n})=0. \end{aligned} \end{align*} $$
Up to a subsequence, there exists u such that
$u_{n}\rightharpoonup u$
in
$W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
. From Lemma 4.1(ii), we have
$$ \begin{align} \begin{aligned} I(u) &= I(u)-\frac{1}{4}J(u) \\ &= \frac{1}{4} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} +\frac{\alpha}{8} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{2_{\alpha}^{*}-2} {8\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x \\ &\quad{}+ \frac{\lambda(q-2)}{8q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x + \frac{1}{12} \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x \\ &\geqslant 0. \end{aligned} \end{align} $$
Let
$w_{n}:=u_{n}-u$
. It follows from
$J(u_{n})=0$
and
$J(u)=0$
that
$$ \begin{align} \begin{aligned} &\lim_{n\rightarrow\infty} \left( \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|w_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4}\right. \\ &\qquad{}\left.+ 2\|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2}\right) \\ &\quad= \lim_{n\rightarrow\infty} \left( \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|u_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|u_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4}\right) \\ &\qquad{}- \left( \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|u\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{4}\right) \\ &\quad= \lim_{n\rightarrow\infty} \bigg[ \frac{2_{\alpha}^{*}+6} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u_{n}|^{2_{\alpha}^{*}} \mathrm{d}x + \frac{\lambda(q+6)}{2q} \int_{\mathbb{R}^{3}} |u_{n}|^{q} \mathrm{d}x + \int_{\mathbb{R}^{3}} |u_{n}|^{6} \mathrm{d}x\bigg] \\ &\qquad{}- \bigg[\frac{2_{\alpha}^{*}+6} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |u|^{2_{\alpha}^{*}} \mathrm{d}x + \frac{\lambda(q+6)}{2q} \int_{\mathbb{R}^{3}} |u|^{q} \mathrm{d}x + \int_{\mathbb{R}^{3}} |u|^{6} \mathrm{d}x\bigg] \\ &\quad= \lim_{n\rightarrow\infty} \bigg( \frac{2_{\alpha}^{*}+6} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |w_{n}|^{2_{\alpha}^{*}} \mathrm{d}x + \int_{\mathbb{R}^{3}} |w_{n}|^{6} \mathrm{d}x\bigg). \end{aligned} \end{align} $$
From
$0<\frac {4-\alpha }{2}<1$
and (4.2), we have
$$ \begin{align} \begin{aligned} &\lim_{n\rightarrow\infty} \left( 2\|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + 2\|w_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4}\right) \\ &\quad\geqslant \lim_{n\rightarrow\infty} \left( \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|w_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4}\right) \\ &\quad= \lim_{n\rightarrow\infty} \bigg( \frac{2_{\alpha}^{*}+6} {2\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |w_{n}|^{2_{\alpha}^{*}} \mathrm{d}x + \int_{\mathbb{R}^{3}} |w_{n}|^{6} \mathrm{d}x\bigg)- \lim_{n\rightarrow\infty} 2\|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2}. \end{aligned} \end{align} $$
It follows from (4.1) and (4.3) that
$$ \begin{align} \begin{aligned} m+o(1) &= I(u_{n}) \\ &= I(u)+I(w_{n}) + \frac{1}{2} \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \\ &\geqslant I(w_{n}) + \frac{1}{2} \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \\ &= \frac{1}{2} \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{1}{2} \|w_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \frac{1}{4}\|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4} - \frac{1}{2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |w_{n}|^{2_{\alpha}^{*}} \mathrm{d}x \\ &\quad{}- \frac{1}{6} \int_{\mathbb{R}^{3}} |w_{n}|^{6} \mathrm{d}x + \frac{1}{2} \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \\ &\geqslant \frac{2_{\alpha}^{*}-2} {8\cdot 2_{\alpha}^{*}} \int_{\mathbb{R}^{3}} |w_{n}|^{2_{\alpha}^{*}} \mathrm{d}x + \frac{1}{12} \int_{\mathbb{R}^{3}} |w_{n}|^{6} \mathrm{d}x. \end{aligned} \end{align} $$
In view of (2.1), (2.2), and
$0<\frac {4-\alpha }{2}<1$
, we have
$$ \begin{align} \begin{aligned} &\lim_{n\rightarrow\infty} \left( \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|w_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} + \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{4}\right. \\ &\qquad{}\left.+\ 2\|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} \|u\|_{D^{1,2}(\mathbb{R}^{3})}^{2}\right) \\ &\quad\geqslant \lim_{n\rightarrow\infty} \left( \frac{\alpha-2}{2} \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|w_{n}\|_{D^{1,2}(\mathbb{R}^{3})}^{2} + \frac{4-\alpha}{2} \|w_{n}\|_{L^{2}(\mathbb{R}^{3},\alpha)}^{2} \right) \\ &\quad\geqslant \lim_{n\rightarrow\infty} \frac{(4-\alpha)\mathcal{S}_{\alpha}}{2} \left( \int_{\mathbb{R}^{3}} |w_{n}|^{2_{\alpha}^{*}} \mathrm{d}x \right) ^{\frac{2}{2_{\alpha}^{*}}} + \lim_{n\rightarrow\infty} \frac{(\alpha-2)\mathcal{S}}{2} \left( \int_{\mathbb{R}^{3}} |w_{n}|^{6} \mathrm{d}x \right) ^{\frac{1}{3}}. \end{aligned} \end{align} $$
Putting (4.5) into (4.2), we get
$$ \begin{align*} &\frac{2_{\alpha}^{*}+6} {2\cdot 2_{\alpha}^{*}} \lim_{n\rightarrow\infty} \int_{\mathbb{R}^{3}} |w_{n}|^{2_{\alpha}^{*}} \mathrm{d}x + \lim_{n\rightarrow\infty} \int_{\mathbb{R}^{3}} |w_{n}|^{6} \mathrm{d}x \\ &\quad\geqslant \lim_{n\rightarrow\infty} \frac{(4-\alpha)\mathcal{S}_{\alpha}}{2} \left( \int_{\mathbb{R}^{3}} |w_{n}|^{2_{\alpha}^{*}} \mathrm{d}x \right) ^{\frac{2}{2_{\alpha}^{*}}} + \lim_{n\rightarrow\infty} \frac{(\alpha-2)\mathcal{S}}{2} \left( \int_{\mathbb{R}^{3}} |w_{n}|^{6} \mathrm{d}x \right) ^{\frac{1}{3}}, \end{align*} $$
which gives
$$ \begin{align} \begin{aligned} \frac{2_{\alpha}^{*}+6}{2_{\alpha}^{*}}X + 2Y \geqslant (4-\alpha)\mathcal{S}_{\alpha} X^{\frac{2}{2_{\alpha}^{*}}} + (\alpha-2)\mathcal{S} Y^{\frac{1}{3}}, \end{aligned} \end{align} $$
where
$X:=\lim \limits _{n\rightarrow \infty }\int _{\mathbb {R}^{3}} |w_{n}|^{2_{\alpha }^{*}} \mathrm {d}x$
and
$Y:=\lim \limits _{n\rightarrow \infty }\int _{\mathbb {R}^{3}} |w_{n}|^{6} \mathrm {d}x$
.
If
$0\not =X< \left [ \frac {2_{\alpha }^{*}(4-\alpha ) \mathcal {S}_{\alpha }}{2_{\alpha }^{*}+6} \right ]^{\frac {2_{\alpha }^{*}}{2_{\alpha }^{*}-2}}$
and
$0\not =Y< \left [\frac {(\alpha -2)\mathcal {S}}{2}\right ]^{\frac {3}{2}}$
, then we have
$$ \begin{align*} \begin{aligned} \frac{2_{\alpha}^{*}+6}{2_{\alpha}^{*}}X + 2Y < (4-\alpha)\mathcal{S}_{\alpha} X^{\frac{2}{2_{\alpha}^{*}}} + (\alpha-2)\mathcal{S} Y^{\frac{1}{3}}, \end{aligned} \end{align*} $$
which contradicts (4.6).
If
$X=0$
and
$Y\not =0$
, then we have
$2Y\geqslant (\alpha -2)\mathcal {S} Y^{\frac {1}{3}}$
, which gives
$$ \begin{align*} \begin{aligned} Y\geqslant \left[\frac{(\alpha-2)\mathcal{S}}{2}\right]^{\frac{3}{2}}. \end{aligned} \end{align*} $$
If
$X\not =0$
and
$Y=0$
, then we have
$\frac {2_{\alpha }^{*}+6}{2_{\alpha }^{*}}X \geqslant (4-\alpha )\mathcal {S}_{\alpha } X^{\frac {2}{2_{\alpha }^{*}}}$
, which gives
$$ \begin{align*} \begin{aligned} X\geqslant \left[ \frac{2_{\alpha}^{*}(4-\alpha) \mathcal{S}_{\alpha}}{2_{\alpha}^{*}+6} \right]^{\frac{2_{\alpha}^{*}}{2_{\alpha}^{*}-2}}. \end{aligned} \end{align*} $$
Hence, we have either
$Y\geqslant \left [\frac {(\alpha -2)\mathcal {S}}{2}\right ]^{\frac {3}{2}}$
or
$X\geqslant \left [ \frac {2_{\alpha }^{*}(4-\alpha ) \mathcal {S}_{\alpha }}{2_{\alpha }^{*}+6} \right ]^{\frac {2_{\alpha }^{*}}{2_{\alpha }^{*}-2}}$
.
Inserting
$Y\geqslant \left [\frac {(\alpha -2)\mathcal {S}}{2}\right ]^{\frac {3}{2}}$
into (4.4), we get
$$ \begin{align*} \begin{aligned} m\geqslant \frac{1}{12} \left[\frac{(\alpha-2)\mathcal{S}}{2}\right]^{\frac{3}{2}}. \end{aligned} \end{align*} $$
This contradicts
$m<\hat {m}^{*}$
.
Inserting
$X\geqslant \left [ \frac {2_{\alpha }^{*}(4-\alpha ) \mathcal {S}_{\alpha }}{2_{\alpha }^{*}+6} \right ]^{\frac {2_{\alpha }^{*}}{2_{\alpha }^{*}-2}}$
into (4.4), we get
$$ \begin{align*} \begin{aligned} m\geqslant \frac{2_{\alpha}^{*}-2} {8\cdot 2_{\alpha}^{*}} \left[ \frac{2_{\alpha}^{*}(4-\alpha) \mathcal{S}_{\alpha}}{2_{\alpha}^{*}+6} \right]^{\frac{2_{\alpha}^{*}}{2_{\alpha}^{*}-2}}. \end{aligned} \end{align*} $$
This contradicts
$m<\hat {m}^{*}$
.
From the above, we know that
$X=Y=0$
. Putting this result into (4.2), we have
$\lim \limits _{n\rightarrow \infty }\|u_{n}-u\|_{W^{1,2}(\mathbb {R}^{3},\alpha )} =0$
.
We now show
$u\not \equiv 0$
. If
$\lim \limits _{n\rightarrow \infty }\|u_{n}\|_{W^{1,2}(\mathbb {R}^{3},\alpha )}=0$
, then we have
$m=0$
. This contradicts
$m>0$
. Thus,
$\lim \limits _{n\rightarrow \infty }\|u_{n}\|_{W^{1,2}(\mathbb {R}^{3},\alpha )} \not =0$
. By virtue of Brézis–Lieb lemma [Reference Brézis and Lieb8], we get
$$ \begin{align*} \begin{aligned} \lim_{n\rightarrow\infty} \|u_{n}\|_{W^{1,2}(\mathbb{R}^{3},\alpha)} - \|u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)}= \lim_{n\rightarrow\infty} \|u_{n}-u\|_{W^{1,2}(\mathbb{R}^{3},\alpha)} =0, \end{aligned} \end{align*} $$
which gives
$\|u\|_{W^{1,2}(\mathbb {R}^{3},\alpha )}\not =0$
. Hence,
$u\not \equiv 0$
.▪
Proof of Theorem 1.2
Let
$\{u_{n}\}\subset \mathcal {M}$
be a minimizing sequence of I at
$m\in (0,m^{*})$
. It is easy to see that
$\{u_{n}\}$
is bounded in
$W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
. From Lemma 4.1(iii), there exists
$0\not \equiv u\in W_{rad}^{1,2}(\mathbb {R}^{3},\alpha )$
such that, up to a subsequence,
$\|u_{n}-u\|_{W^{1,2}(\mathbb {R}^{3},\alpha )} \to 0$
as
$n\to \infty $
. In view of Lemma 4.1(ii), we know
$J(u)=0$
. Similar to the proof of Theorem 1.1, we get
$\|I'(u)\|_{W^{-1,2}(\mathbb {R}^{3},\alpha )}=0$
.▪
