Lemma 2.1
Under assumption (F2) and (F3), we can obtain:
-
(i)
\(\frac{f(s)}{s}\)
is nonincreasing in
\((-\infty,0)\), and nondecreasing in
\((0,+\infty)\).
-
(ii)
\(f(s)s\ge2F(s)\ge0\), \(\forall s\in\mathbb{R}\).
-
(iii)
\(\frac{F(s)}{s^{2}}\)
is nonincreasing in
\((-\infty,0)\), and nondecreasing in
\((0,+\infty)\).
The proof is elementary, so we omit here.
Lemma 2.2
Under assumptions (V0), (Q0), and (F1)–(F3),
$$ I(u)\ge I(tu)+\frac{1-t^{4}}{4}\bigl\langle I'(u),u \bigr\rangle +\frac {(1-t^{2})^{2}}{4}\lVert u\rVert^{2},\quad \forall u\in H^{1}\bigl(\mathbb{R}^{3}\bigr), t\ge0. $$
(2.1)
Proof
Note that
$$ I(u)=\frac{1}{2}\lVert u\rVert^{2}+ \frac{1}{4} \int_{\mathbb{R}^{3}} \phi _{u} (x)u^{2}\,dx- \frac{1}{2} \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac {Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)F\bigl(u(x)\bigr)\,dx\,dy $$
(2.2)
and
$$ \bigl\langle I'(u),u\bigr\rangle = \lVert u \rVert^{2} + \int_{\mathbb{R}^{3}} \phi_{u} (x)u^{2}\,dx- \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac {Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)f\bigl(u(x)\bigr)u(x)\,dx\,dy. $$
(2.3)
Thus, from (2.2) and (2.3), one has
$$\begin{aligned} I(u)-I(tu)&=\frac{1-t^{2}}{2}\lVert u\rVert^{2}+\frac{1-t^{4}}{4} \int_{\mathbb{R}^{3}} \phi_{u} (x)u^{2}\,dx \\ &\quad {}-\frac{1}{2} \int_{\mathbb{R}^{3}} \int _{\mathbb{R}^{3}} \frac{Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)F\bigl(u(x)\bigr)\,dx\,dy \\ &\quad {} +\frac {1}{2} \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac {Q(y)F(tu(y))}{|x-y|^{\mu}} Q(x)F\bigl(tu(x)\bigr)\,dx\,dy \\ &=\frac{1-t^{2}}{4}\bigl\langle I'(u),u\bigr\rangle + \frac{(1-t^{2})^{2}}{4}\lVert u\rVert^{2} \\ &\quad {}+ \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \biggl[\frac{1-t^{4}}{4}\frac{Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)f \bigl(u(x)\bigr) \bigl(u(x)\bigr) \\ &\quad {} +\frac{1}{2}\frac{Q(y)F(tu(y))}{|x-y|^{\mu}} Q(x)F\bigl(tu(x)\bigr)- \frac{1}{2}\frac{Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)F\bigl(u(x)\bigr) \biggr]\,dx\,dy. \end{aligned}$$
Define a function
$$\begin{aligned} h(t,u)&= \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \biggl[\frac{1-t^{4}}{4} \frac{Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)f \bigl(u(x)\bigr) \bigl(u(x)\bigr)+\frac{1}{2}\frac {Q(y)F(tu(y))}{|x-y|^{\mu}} Q(x)F \bigl(tu(x)\bigr) \\ &\quad {} -\frac{1}{2}\frac {Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)F\bigl(u(x)\bigr)\biggr]\,dx \,dy, \quad \forall t\ge0, u\in H^{1}\bigl(\mathbb{R}^{3} \bigr). \end{aligned}$$
By (Q0), Lemma 2.1, and elementary computations, we have
$$\begin{aligned} h'(t)&= \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \biggl[\frac {Q(y)F(tu(y))}{|x-y|^{\mu}}Q(x)f\bigl(tu(x)\bigr)u(x) \,dx\,dy \\ &\quad {}-t^{3}\frac {Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)f\bigl(u(x)\bigr)u(x)\biggr]\,dx \,dy \\ &=t^{3} \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac {Q(x)Q(y)u^{2}(x)u^{2}(y)}{|x-y|^{\mu}} \biggl[\frac{F(tu(y))}{(t^{2}(u(y))} \biggl( \frac{f(tu(x))}{tu (x)}-\frac{f(u(x))}{u(x)} \biggr) \\ &\quad {} +\frac {f(u(x))}{u(x)} \biggl(\frac{F(tu(y))}{tu^{2}(y)}-\frac {F(u(y))}{u^{2}(y)} \biggr) \biggr]\,dx\,dy \textstyle\begin{cases} \ge0,t\ge1,\\ < 0,0< t< 1, \end{cases}\displaystyle \end{aligned}$$
which yields \(h(t)\ge h(1)=0\). Therefore, we have (2.1) holds. □
Corollary 2.3
Under assumptions (V0), (Q0), and (F1)–(F3), for any
\(u\in \mathcal{N}\),
$$ I(u)=\max_{t\ge0} I(tu). $$
(2.4)
Lemma 2.4
(Hardy–Littlewood–Sobolev inequality [8])
Let
\(s,r>1\), and
\(0<\mu<N\)
with
\(\frac{1}{s}+\frac{\mu}{N}+\frac {1}{r}=2\), \(f\in L^{s}(\mathbb{R}^{N})\)
and
\(h\in L^{r}(\mathbb{R}^{N})\). Then there exists a sharp constant
\(C(s,N,\mu,r)\)
independent of
f, h
such that
$$ \int_{\mathbb{R}^{N}} \int_{\mathbb{R}^{N}} \frac{f(x)h(y)}{|x-y|^{\mu}}\le C(s,N,\mu,r)\lVert f \Vert_{s}\lVert h\rVert_{r}. $$
In the sequel, we set \(N=3\) and \(r=\frac{6}{6-\mu}\) (we take the same value for r in the following part of this paper). It follows from \((F1)\) that \(2< rq<2^{*}\). So, for any \(u\in H^{1}(\mathbb{R}^{3})\), by an elementary computation, we have
$$ \bigl\lVert F(u) \bigr\rVert _{r}\le C\bigl(\lVert u \rVert_{2}^{2-\frac{\mu}{3}}+\lVert u\rVert_{rq}^{q} \bigr). $$
(2.5)
By (Q0), (2.5), and Hardy–Littlewood–Sobolev inequality, we can obtain
$$ \biggl\vert \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac {Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)F\bigl(u(x)\bigr)\,dx\,dy \biggr\vert \le C \bigl\lVert F(u) \bigr\rVert _{r}^{2}\le C \bigl(\lVert u\rVert_{2}^{4-\frac{2\mu}{3}}+\lVert u\rVert_{rq}^{2q} \bigr). $$
(2.6)
Similarly,
$$ \biggl\vert \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac {Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)f\bigl(u(x)\bigr)u(x)\,dx\,dy \biggr\vert \le C \bigl\lVert F(u) \bigr\rVert _{r}^{2}\le C \bigl(\lVert u\rVert_{2}^{4-\frac{2\mu}{3}}+\lVert u\rVert_{rq}^{2q} \bigr). $$
(2.7)
To show \(\mathcal{N}\neq\emptyset\) in our situation, we define a set as follows:
$$ \mathcal{J}:= \textstyle\begin{cases} H^{1}(\mathbb{R}^{3})\setminus \{0\},&\mbox{if }\mu< 1; \\ \{u\in H^{1}(\mathbb{R}^{3}):\int_{\mathbb{R}^{3}}\phi_{u}(x)u^{2}\,dx \\ \quad {}-\int _{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}\frac{Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)f(u(x))u(x)\,dx\,dy< 0 \},&\mbox{if }1\le\mu< 3. \end{cases} $$
Lemma 2.5
Under assumptions (V0), (Q0), and (F1)–(F4), (i) \(\mathcal {N}\subset\mathcal{J}\ne\emptyset\); (ii) for any
\(u\in\mathcal {J}\), there exists unique
\(t(u)>0\)
such that
\(t(u)u\in \mathcal {N}\).
Proof
(i) It is easy to see that \(\mathcal{J}\neq\emptyset\) if \(\mu<1\). Next, we consider that \(1\le\mu<3\). From Lemma 2.4 and Sobolev imbedding theorem, there exists \(C>0\) such that \(\int_{\mathbb{R}^{3}} \phi_{u} (x)u^{2}\,dx\le C\lVert u\rVert^{4}\) for any \(u\in H^{1}(\mathbb{R}^{3})\). Fix \(v\in H^{1}(\mathbb{R}^{3})\) and \(v(x)>0 \) for any \(x\in\mathbb{R}^{3}\) and set \(v_{t} (x)=v(tx)\) for \(t>0\). By (Q0), one has
$$\begin{aligned} & \int_{\mathbb{R}^{3}} \phi_{(t^{2}v_{t})} (x) \bigl(t^{2}v_{t} \bigr)^{2}\,dx- \int_{\mathbb {R}^{3}} \int_{\mathbb{R}^{3}}\frac{Q(y)F(t^{2}v_{t}(y))}{|x-y|^{\mu}} Q(x)f\bigl(t^{2}v_{t}(x) \bigr)t^{2}v_{t}(x)\,dx\,dy \\ &\quad =t^{3} \int_{\mathbb{R}^{3}} \phi_{v} (x)v^{2} \,dx-t^{\mu-6} \int_{\mathbb {R}^{3}} \int_{\mathbb{R}^{3}} \frac{Q(t^{-1}y)F(t^{2}v(y))}{|x-y|^{\mu}} Q\bigl(t^{-1}x\bigr)f \bigl(t^{2}v(x)\bigr)t^{2}v(x)\,dx\,dy \\ &\quad \le t^{3} \biggl(C\lVert v\rVert^{4}-Q_{\infty}^{2} \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}}\frac{F(t^{2}v(y))}{(t^{2}v(y))^{\frac{9-\mu}{4}}}\frac {f(t^{2}v(x))t^{2}v(x)}{(t^{2}v(x))^{\frac{9-\mu}{4}}} \frac{[v(x)v(y)]^{\frac{9-\mu}{4}}}{|x-y|^{\mu}} \biggr)\,dx\,dy, \end{aligned}$$
(2.8)
where \(Q_{\infty}=\inf_{x\in\mathbb{R}^{3}}Q(x)\). Since \(1\le\mu<3\), then \(\sigma=\frac{9-\mu}{4}\) in (F4), and so
$$ \frac{F(t^{2}v(y)))}{(t^{2}v(y))^{\frac{9-\mu}{4}}}\to+\infty \quad \mbox{and}\quad \frac {f(t^{2}v(x)))t^{2}v(x))}{(t^{2}v(x))^{\frac{9-\mu}{4}}}\to+\infty \quad \mbox{as }t\to +\infty, $$
(2.9)
which, together with (Q0), (2.8), and (2.9), implies
$$\begin{aligned}& \int_{\mathbb{R}^{3}} \phi_{(t^{2}v_{t})} (x) \bigl(t^{2}v_{t} \bigr)^{2}\,dx- \int_{\mathbb {R}^{3}} \int_{\mathbb{R}^{3}}\frac{Q(y)F(t^{2}v_{t}(y))}{|x-y|^{\mu}} Q(x)f\bigl(t^{2}v_{t}(x) \bigr)t^{2}v_{t}(x)\,dx\,dy \\& \quad \to -\infty \quad \mbox{as } t\to+\infty. \end{aligned}$$
Thus taking \(u=t^{2}v_{t}\) for t large enough, we have \(u\in\mathcal {J}\), and so \(\mathcal {J}\ne\varnothing\) in the case \(1\le\mu<3\). From (2.3), it is easy to see that \(\mathcal {N}\subset\mathcal {J}\).
(ii) First, we prove the existence of \(t(u)\). Since \(\sigma=2\) in (F4) if \(\mu<1\), then through a standard argument, the existence of \(t(u)\) can be proved easily. After that, we consider the case \(1\le\mu <3\). Let \(u\in\mathcal {J}\) be fixed and define a function \(g(t)=\langle I'(tu),tu\rangle\) on \([0,+\infty)\). By (2.3), Lemma 2.4, and Sobolev imbedding theorem, one has
$$\begin{aligned} g(t)&=t^{2}\lVert u\rVert^{2} +t^{4} \int_{\mathbb{R}^{3}} \phi_{u} (x)u^{2}\,dx- \int _{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac{Q(y)F(tu(y))}{|x-y|^{\mu}} Q(x)f\bigl(tu(x)\bigr)tu(x)\,dx \,dy \\ &\ge t^{2}\lVert u\rVert^{2}-C\bigl(t^{4-\frac{2\mu}{3}}\lVert u \rVert _{2}^{4-\frac{2\mu}{3}}+t^{2q}\lVert u\rVert_{rq}^{2q} \bigr) \\ &\ge t^{2}\lVert u\rVert^{2}-C\bigl(t^{4-\frac{2\mu}{3}}\lVert u \rVert ^{4-\frac{2\mu}{3}}+t^{2q}\lVert u\rVert^{2q}\bigr). \end{aligned}$$
(2.10)
By (2.3) and Lemma 2.1, we have
$$\begin{aligned} g(t)&=t^{2}\lVert u\rVert^{2}+t^{4} \int_{\mathbb{R}^{3}} \phi_{u} (x)u^{2}\,dx \\ &\quad {}- \int _{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac{Q(y)F(tu(y))}{|x-y|^{\mu}} Q(x)f\bigl(tu(x)\bigr)tu(x)\,dx \,dy \\ &\le t^{2}\lVert u\rVert^{2}+t^{4} \int_{\mathbb{R}^{3}} \phi_{u} (x)u^{2}\,dx \\ &\quad {}-t^{4} \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac {Q(x)Q(y)u^{2}(x)u^{2}(y)}{|x-y|^{\mu}}\frac{F(tu(y))}{(tu(y))^{2}} \frac {f(tu(x))}{tu (x)}\,dx\,dy \\ &\le t^{2}\lVert u\rVert^{2}+t^{4} \int_{\mathbb{R}^{3}} \phi_{u} (x)u^{2}\,dx \\ &\quad {}-t^{4} \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac {Q(x)Q(y)u^{2}(x)u^{2}(y)}{|x-y|^{\mu}}\frac{F(u(y))}{(u(y))^{2}} \frac {f(u(x))}{u (x)}\,dx\,dy \\ &\le t^{2}\lVert u\rVert^{2}+t^{4}\biggl( \int_{\mathbb{R}^{3}} \phi_{u} (x)u^{2}\,dx \\ &\quad {}- \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}}\frac {Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)f\bigl(u(x)\bigr)u(x)\,dx\,dy \biggr),\quad \forall t\ge1. \end{aligned}$$
(2.11)
It follows from (2.10) and (2.11) that \(g(t)>0\) for \(t>0\) small, and \(g(t)<0\) for t large due to \(u\in\mathcal {J}\). Therefore, there exists \(t_{0}=t(u)>0\) such that \(g(t_{0})=0\) and \(t(u)u\in\mathcal {N}\).
Next, we prove that \(t(u)\) is unique for any u
\(\in\mathcal {J}\). For any given \(u\in\mathcal {N}\), let \(t_{1}, t_{2}>0\) such that \(g(t_{1})=g(t_{2})=0\). Jointly with (2.1), we have
$$\begin{aligned} I(t_{1}u)&\ge I(t_{2}u)+\frac{t_{1}^{4}-t_{2}^{4}}{4t_{1}^{4}}\bigl\langle I'(t_{1}u),t_{1}u\bigr\rangle + \frac{(t_{1}^{2}-t_{2}^{2})^{2}}{4t_{1}^{2}}\lVert u\rVert^{2} \\ &=I(t_{2}u)+\frac{(t_{1}^{2}-t_{2}^{2})^{2}}{4t_{1}^{2}}\lVert u\rVert^{2}, \end{aligned}$$
(2.12)
and
$$\begin{aligned} I(t_{2}u)&\ge I(t_{1}u)+\frac{t_{2}^{4}-t_{1}^{4}}{4t_{2}^{4}}\bigl\langle I'(t_{2}u),t_{2}u\bigr\rangle + \frac{(t_{2}^{2}-t_{1}^{2})^{2}}{4t_{2}^{2}}\lVert u\rVert^{2} \\ &=I(t_{1}u)+\frac{(t_{2}^{2}-t_{1}^{2})^{2}}{4t_{2}^{2}}\lVert u\rVert^{2}. \end{aligned}$$
(2.13)
Equations (2.12) and (2.13) imply \(t_{1}=t_{2}\). Hence, \(t(u)>0\) is unique for any \(u\in\mathcal {J}\). □
Lemma 2.6
Under assumptions (V0), (Q0), and (F1)–(F4), then
$$ \inf_{u\in\mathcal {N}} I(u):=c=\inf_{u\in\mathcal{J},u\ne0} \max_{t\ge 0}I(tu)>0. $$
(2.14)
The proof is standard, so we omit here.
Lemma 2.7
Under assumptions (V0), (Q0), and (F1)–(F4), there exist a constant
\(c_{*}\in(0,c]\)
and a sequence
\(\{u_{n}\}\subset H^{1}({\mathbb {R}^{3}})\)
satisfying
$$ I(u_{n})\to c_{*}, \qquad \bigl\lVert I'(u_{n}) \bigr\rVert \bigl(1+ \lVert u_{n} \rVert\bigr) \to0. $$
(2.15)
Proof
It follows from Lemma 2.4 and Sobolev imbedding theorem that
$$\begin{aligned} I(u)&=\frac{1}{2}\lVert u\rVert^{2}+\frac{1}{4} \int_{\mathbb{R}^{3}} \phi _{u} (x)u^{2}\,dx- \frac{1}{2} \int_{\mathbb{R}^{3}} \int_{\mathbb{R}^{3}} \frac {Q(y)F(u(y))}{|x-y|^{\mu}} Q(x)F\bigl(u(x)\bigr)\,dx\,dy. \\ &\ge\frac{1}{2}\lVert u\rVert^{2}-C\bigl(\lVert u \rVert_{2}^{4-\frac{2\mu }{3}}+\lVert u\rVert_{rq}^{2q} \bigr) \\ &\ge\frac{1}{2}\lVert u\rVert^{2}-C\bigl(\lVert u \rVert^{4-\frac{2\mu }{3}}+\lVert u\rVert^{2q}\bigr). \end{aligned}$$
(2.16)
From (F1), we know that there exist \(\delta_{0}>0\) and \(\rho_{0} >0\) such that
$$ I(u)\ge\rho_{0},\qquad \lVert u\rVert= \delta_{0}. $$
(2.17)
In view of Lemmas 2.5 and 2.6, we may choose \(v_{k}\in \mathcal{N}\subset\mathcal{J}\) such that
$$ c-\frac{1}{k}< I(v_{k})< c+\frac{1}{k}, \quad k\in\mathbb{N}. $$
(2.18)
Using Lemma 2.2 and (2.17), we can obtain that \(I(tv_{k})>0\) for small \(t>0\) and \(I(tv_{k})<0\) for large \(t>0\) due to \(v_{k}\in\mathcal {N}\). Since \(I(0)=0\), then the mountain pass lemma implies that there exists a sequence \(\{u_{k,n}\}\subset H^{1}(\mathbb{R}^{3})\) satisfying
$$ I(u_{k,n})\to c_{k},\qquad \bigl\lVert I'(u_{k,n}) \bigr\rVert \bigl(1+ \lVert u_{k,n} \rVert\bigr)\to 0, \quad k\in\mathbb{N}, $$
(2.19)
where \(c_{k}\in[\rho_{0},\sup_{t\ge0}I(tv_{k})]\). As a result of Corollary 2.3, we have \(I(v_{k})=\sup_{t\ge0}I(tv_{k})\). Then, by (2.18) and (2.19), one has
$$ I(u_{k,n})\to c_{k}\in \biggl[\rho_{0},c+ \frac{1}{k} \biggr),\qquad \bigl\lVert I'(u_{k,n}) \bigr\rVert \bigl(1+ \lVert u_{k,n} \rVert\bigr)\to0,\quad k\in\mathbb{N}. $$
(2.20)
Choose a sequence \(\{n_{k}\}\subset\mathbb{N}\) such that
$$ I(u_{k,n_{k}})\to c_{k}\in \biggl[\rho_{0},c+ \frac{1}{k} \biggr),\qquad \bigl\lVert I'(u_{k,n_{k}}) \bigr\rVert \bigl(1+ \lVert u_{k,n_{k}} \rVert\bigr)< \frac{1}{k} , \quad k \in\mathbb{N}. $$
(2.21)
Let \(u_{k}=u_{k,n_{k}}\), \(k\in\mathbb{N}\). Then, going if necessary to a subsequence, we have
$$ I(u_{n})\to c_{*}\in[\rho_{0},c], \qquad \bigl\lVert I'(u_{n}) \bigr\rVert \bigl(1+ \lVert u_{n} \rVert\bigr) \to0. $$
□
Lemma 2.8
Under assumptions (V0), (Q0), and (F1)–(F4), any sequence
\(\{u_{n}\} \subset H^{1}(\mathbb{R}^{3})\)
satisfying (2.15) is bounded in
\(H^{1}(\mathbb{R}^{3})\).
Proof
By Lemma 2.2, one has
$$ c_{*}+o(1)=I(u_{n})-\frac{1}{4}\bigl\langle I'(u_{n}),u_{n} \bigr\rangle \ge\frac {1}{4}\lVert u_{n}\rVert^{2}. $$
This shows that the sequence \(\{u_{n}\}\) is bounded in \(H^{1}(\mathbb{R}^{3})\). □
Similar to the proof of [31, Lemma 2.7], we can obtain the following lemma.
Lemma 2.9
Under assumptions (V0), (Q0), and (F1)–(F4), if
\(u_{0}\in\mathcal{N}\)
and
\(I(u_{0})=c\), then
\(u_{0}\)
is a critical point of
I.