Existence and multiplicity of solutions for a class of sublinear Schrödinger-Maxwell equations

Abstract

In this paper I consider a class of sublinear Schrödinger-Maxwell equations, and new results about the existence and multiplicity of solutions are obtained by using the minimizing theorem and the dual fountain theorem respectively.

1 Introduction and main result

Consider the following semilinear Schrödinger-Maxwell equations:

$$\{\begin{array}{c}-\mathrm{\Delta }u+V(x)u+\varphi u=f(x,u),\text{in }{R}^3,\hfill \\ -\mathrm{\Delta }\varphi =u^2,{lim}_{|x|\to \mathrm{\infty }}\varphi (x)=0,\text{in }{R}^3.\hfill \end{array}$$

(1)

Such a system, also known as the nonlinear Schrödinger-Poisson system, arises in an interesting physical context. Indeed, according to a classical model, the interaction of a charge particle with an electromagnetic field can be described by coupling the nonlinear Schrödinger and the Maxwell equations (we refer to [1, 2] for more details on the physical aspects and on the qualitative properties of the solutions). In particular, if we are looking for electrostatic-type solutions, we just have to solve (1).

In recent years, system (1), with $V(x)\equiv 1$ or being radially symmetric, has been widely studied under various conditions on f; see, for example, [3–11]. Since (1) is set on $R^3$, it is well known that the Sobolev embedding $H^1(R^3)\hookrightarrow L^s(R^3)$ ($2\le s\le 2^{\ast }=6$) is not compact, and then it is usually difficult to prove that a minimizing sequence or a sequence that satisfies the $(\mathit{PS})$ condition, briefly a Palais-Smale sequence, is strongly convergent if we seek solutions of (1) by variational methods. If $V(x)$ is radial (for example, $V(x)\equiv 1$), we can avoid the lack of compactness of Sobolev embedding by looking for solutions of (1) in the subspace of radial functions of $H^1(R^3)$, which is usually denoted by $H_r^1(R^3)$, since the embedding $H_r^1(R^3)\hookrightarrow L^s(R^3)$ ($2<s<6$) is compact. Specially, Ruiz [11] dealt with (1) under the assumption that $V(x)\equiv 1$ and $f(u)=u^p$ ($1<p<5$) and got some general existence, nonexistence and multiplicity results.

Moreover, in [12] the authors considered system (1) with periodic potential $V(x)$, and the existence of infinitely many geometrically distinct solutions was proved by the nonlinear superposition principle established in [13].

There are also some papers treating the case with nonradial potential $V(x)$. More precisely, Wang and Zhou [14] got the existence and nonexistence results of (1) when $f(u)$ is asymptotically linear at infinity. Chen and Tang [15] proved that (1) has infinitely many high energy solutions under the condition that $f(x,u)$ is superlinear at infinity in u by the fountain theorem. Soon after, Li, Su and Wei [16] improved their results.

Up to now, there have been few works concerning the case that $V(x)$ is nonradial potential and $f(x,u)$ is sublinear at infinity in u. Very recently, Sun [17] treated the above case based on the variant fountain theorem established in Zou [18].

Theorem 1.1 [17]

Assume that the following conditions hold:

($V_1^{\prime }$) $V\in C(R^3,R)$ satisfies ${inf}_{x\in R^3}V(x)\ge a>0$, where $a>0$ is a constant. For every $M>0$, $meas\{x\in R^3:v(x)\le M\}<\mathrm{\infty }$.

($H_1$) $F(x,u)=a(x){|u|}^r$, where $F(x,u)={\int }_0^{u}f(x,y)dy$, $a:R^3\to R^{+}$ is a positive function such that $a\in L^{\frac{2}{2-r}}(R^3)$ and $1<r<2$.

Then problem (1) has infinitely many nontrivial solutions $\{(u_k,{\varphi }_k)\}$ satisfying

$$\frac{1}{2}{\int }_{R^3}({|\mathrm{\nabla }{u}_k|}^2+V(x)u_k^2)dx-\frac{1}{4}{\int }_{R^3}{|\mathrm{\nabla }{\varphi }_k|}^{2}dx+\frac{1}{2}{\int }_{R^3}{\varphi }_k{u}_k^{2}dx-{\int }_{R^3}F(x,u_k)dx\to 0^{-}$$

as $k\to \mathrm{\infty }$.

In the present paper, based on the dual fountain theorem, we can prove the same result under a more generic condition, which generalizes the result in [17]. Our first result can be stated as follows.

Theorem 1.2 Assume that V satisfies

($V_1$) $V\in C(R^3,R)$ and ${inf}_{x\in R^3}V(x)>0$;

and f satisfies the following conditions.

($W_1$) There exist constants $\delta >0$, $r_1\in (1,2)$ and a function $a_1\in L^{\frac{2}{2-r_1}}(R^3,[0,+\mathrm{\infty }))$ such that

$$|f(x,u)|\le a_1(x){|u|}^{r_1-1}$$

for all $x\in R^3$ and $|u|\le \delta$;

($W_2$) There exist constants $M>0$, $r_2\in (1,2)$ and a function $a_2\in L^{\frac{2}{2-r_2}}(R^3,[0,+\mathrm{\infty }))$ such that

$$|f(x,u)|\le a_2(x){|u|}^{r_2-1}$$

for all $x\in R^3$ and $|u|\ge M$;

($W_3$) For every $m>\delta$, there exist a constant $r_3\in (1,2)$ and a function $b_m\in L^{\frac{2}{2-r_3}}(R^3,[0,+\mathrm{\infty }))$ such that

$$|f(x,u)|\le b_m(x)$$

for all $x\in R^3$ and $|u|\le m$;

($W_4$) There exist constants $r_4\in (1,2)$, $\eta >0$ and $\zeta >0$ such that

$$F(x,u)\ge \eta {|u|}^{r_4}$$

for all $x\in \mathrm{\Omega }$ and $|u|\le \zeta$, where $meas\{\mathrm{\Omega }\}>0$, $F(x,u):={\int }_0^{u}f(x,y)dy$;

($W_5$) $F(x,-u)=F(x,u)$ for all $x\in R^3$ and $u\in R$.

Then problem (1) has infinitely many nontrivial solutions $\{(u_k,{\varphi }_k)\}$ satisfying

$$\frac{1}{2}{\int }_{R^3}({|\mathrm{\nabla }{u}_k|}^2+V(x)u_k^2)dx-\frac{1}{4}{\int }_{R^3}{|\mathrm{\nabla }{\varphi }_k|}^{2}dx+\frac{1}{2}{\int }_{R^3}{\varphi }_k{u}_k^{2}dx-{\int }_{R^3}F(x,u_k)dx\to 0^{-}$$

as $k\to \mathrm{\infty }$.

By Theorem 1.2, we obtain the following corollary.

Corollary 1.3 Assume that L satisfies ($V_1$) and W satisfies

($W_6$) $F(x,u)=a(x){|u|}^r$, where $F(x,u)={\int }_0^{u}f(x,y)dy$, $1<r<2$ is a constant and $a:R^3\to R$ is a function such that $a\in L^{\frac{2}{2-r}}(R^3)$ and $a(x)>0$ for $x\in \mathrm{\Omega }$, where $meas\{\mathrm{\Omega }\}>0$.

Then problem (1) has infinitely many nontrivial solutions $\{(u_k,{\varphi }_k)\}$ satisfying

$$\frac{1}{2}{\int }_{R^3}({|\mathrm{\nabla }{u}_k|}^2+V(x)u_k^2)dx-\frac{1}{4}{\int }_{R^3}{|\mathrm{\nabla }{\varphi }_k|}^{2}dx+\frac{1}{2}{\int }_{R^3}{\varphi }_k{u}_k^{2}dx-{\int }_{R^3}F(x,u_k)dx\to 0^{-}$$

as $k\to \mathrm{\infty }$.

Remark 1.4 In Theorem 1.2, infinitely many solutions for problem (1) are obtained under the symmetry condition ($W_5$) by using the dual fountain theorem. As a special case of Theorem 1.2, Corollary 1.3 generalizes and improves Theorem 1.1. To show this, it suffices to compare ($V_1^{\prime }$) and ($V_1$), ($H_1$) and ($W_6$). Firstly, it is clear that ($V_1$) is really weaker than ($V_1^{\prime }$). Secondly, in ($H_1$) a is assumed to be positive, while in ($W_6$) we assume that a is indefinite.

Moreover, under all the conditions of Theorem 1.2 except ($W_5$) we obtain an existence result.

Theorem 1.5 Assume that L satisfies ($V_1$) and W satisfies ($W_1$), ($W_2$), ($W_3$), ($W_4$). Then problem (1) possesses a nontrivial solution.

Remark 1.6 In Theorem 1.5 we obtain the existence of solutions for problem (1) under the assumption that $f(x,u)$ is indefinite and without any coercive assumptions respect to V such as ($V_1^{\prime }$). There are functions V and f which satisfy Theorem 1.5, but do not satisfy the corresponding results in [2–16]. For example,

$$V(x)\equiv 1,f(x,u)=\tilde{a}(x){|u|}^{\frac{3}{2}}$$

(2)

and

$$\tilde{a}(x)=\{\begin{array}{cc}{(-1)}^n{n}^3(|x|-n)\hfill & \text{for }n\le |x|\le n+\frac{1}{n^2},\hfill \\ 0\hfill & \text{else},\hfill \end{array}$$

(3)

in which $n\ge 3$. It is clear that $\tilde{a}\in C(R^3,R)$ is indefinite. Denoting by π the area of the unit ball in $R^3$, we obtain

$$\begin{array}{rcl}{\int }_{R^3}{\tilde{a}}^4(x)dx& =& \sum _{n=3}^{\mathrm{\infty }}({\int }_n^{n+\frac{1}{n^2}}{n}^{12}{r}^2{(r-n)}^{4}dr+{\int }_{n+\frac{1}{n^2}}^{n+\frac{2}{n^2}}{n}^{12}{r}^2{(n+\frac{2}{n^2}-r)}^{4}dr)\pi \\ =& \pi \sum _{n=3}^{\mathrm{\infty }}2n^{12}{\int }_0^{\frac{1}{n^2}}{r}^{6}dx\\ =& \frac{2\pi }{7}\sum _{n=3}^{\mathrm{\infty }}{n}^{-2}\\ <& \mathrm{\infty },\end{array}$$

(4)

which means that $\tilde{a}\in L^{\frac{2}{2-\frac{3}{2}}}(R^3)$. So, (2) satisfies our results, but does not satisfy the results in [3–17].

2 Preliminary results

In order to establish our results via critical point theory, we firstly describe some properties of the space $H^1(R^3)$, on which the variational functional associated with problem (1) is defined. Define the function space

$$H^1\left(R^3\right):=\{u\in L^2\left(R^3\right):\mathrm{\nabla }u\in {\left(L^2\left(R^3\right)\right)}^3\}$$

equipped with the norm

$${\parallel u\parallel }_{H^1}:={\left({\int }_{R^3}({|\mathrm{\nabla }u|}^2+u^2)dx\right)}^{1/2}$$

and the function space

$$D^{1,2}\left(R^3\right):=\{u\in L^{2^{\ast }}:\mathrm{\nabla }u\in {\left(L^2\left(R^3\right)\right)}^3\}$$

with the norm

$${\parallel u\parallel }_{D^{1,2}}={\left({\int }_{R^3}{|\mathrm{\nabla }u|}^{2}dx\right)}^{1/2}.$$

Let

$$E:=\{u\in H^1\left(R^3\right):{\int }_{R^3}V(x)u^{2}dx<+\mathrm{\infty }\}$$

equipped with the inner product

$$(u,v)={\int }_{R^3}(\mathrm{\nabla }u\cdot \mathrm{\nabla }v+V(x)uv)dx$$

and the corresponding norm

$${\parallel u\parallel }^2=(u,u).$$

Note that the following embeddings

$$E\hookrightarrow L^s\left(R^3\right),2\le s\le 2^{\ast },D^{1,2}\left(R^3\right)\hookrightarrow L^{2^{\ast }}\left(R^3\right)$$

are continuous, where $2^{\ast }=6$ is the critical exponent for the Sobolev embeddings in dimension 3. Therefore, there exist constants $C_p$ and $C_{\ast }$ such that

$${\parallel u\parallel }_{L^p}\le C_p\parallel u\parallel ,{\parallel u\parallel }_{L^{2^{\ast }}}\le C_{\ast }{\parallel u\parallel }_{D^{1,2}}$$

(5)

for all $u\in E$. Here $L^p(R^3)$ ($2\le p\le 2^{\ast }$) denotes the Banach spaces of a function on $R^3$ with values in R under the norm

$${\parallel u\parallel }_{L^p}={\left({\int }_{R^3}{|u(x)|}^{p}dx\right)}^{1/p}.$$

Let

$$L_a^r\left(R^3\right):=\{u:R^3\to R:{\int }_{R^3}a(x){|u|}^{r}dx<+\mathrm{\infty }\},$$

where $a(x)>0$ for a.e. $x\in R^3$. Then $L_a^r(R^3)$ is a Banach space with the norm

$${\parallel u\parallel }_{L_a^r}={({\int }_{R^3}a(x){|u|}^{r}dx)}^{1/r}.$$

Lemma 2.1 Suppose that assumption ($V_1$) holds. Then the embedding of E in $L_a^r(R^3)$ is compact, where $r\in (1,2)$, $a\in L^{\frac{2}{2-r}}(R^3)$ is positive for a.e. $x\in R^3$.

Proof For any bounded set $K\subset E$, there exists a positive constant $M_0$ such that $\parallel u\parallel \le M_0$ for all $u\in K$. We claim that K is precompact in $L_a^r(R^3)$. In fact, since $a\in L^{\frac{2}{2-r}}(R^3)$, for any $\epsilon >0$, there exists $T_{\epsilon }>0$ such that

$${\left({\int }_{|x|\ge T_{\epsilon }}a{(x)}^{\frac{2}{2-r}}dx\right)}^{(2-r)/2}<\epsilon .$$

For any $u,v\in K$, applying the Hölder inequality for r such that $\frac{r}{2}+\frac{2-r}{2}=1$ and the first inequality in (5), we have

$$\begin{array}{rcl}{\int }_{|x|\ge T_{\epsilon }}a(x){|u-v|}^{r}dx& \le & {\left({\int }_{|x|\ge T_{\epsilon }}a{(x)}^{\frac{2}{2-r}}dx\right)}^{(2-r)/2}{\left({\int }_{|x|\ge T_{\epsilon }}{|u-v|}^{2}dx\right)}^{r/2}\\ \le & {\parallel u-v\parallel }_{L^2}^r{\left({\int }_{|x|\ge T_{\epsilon }}a{(x)}^{\frac{2}{2-r}}dx\right)}^{(2-r)/2}\\ \le & C_2^r{\parallel u-v\parallel }^r\epsilon \\ \le & 2C_2^r{M}_0^r\epsilon .\end{array}$$

(6)

Besides, since $E(B_{T_{\epsilon }}(0))\subset H^1(B_{T_{\epsilon }}(0))$ is compactly embedded in $L_a^r(B_{T_{\epsilon }}(0))$, where $B_{T_{\epsilon }}(0)=\{x\in R^3:|x|\le T_{\epsilon }\}$, there are $u_1,u_2,\dots ,u_m\in K$ such that for any $u\in K$,

$${\int }_{|x|\le T_{\epsilon }}a(x){|u-u_i|}^{r}dx<\epsilon .$$

(7)

Now it follows from (6) and (7) that K is precompact in $L_a^r(R^3)$. Obviously, we have E is compact embedded in $L_a^r(R^3)$, where $r\in (1,2)$, $a\in L^{\frac{2}{2-r}}(R^3)$ is positive for a.e. $x\in R^3$. □

Lemma 2.2 Assume that assumptions ($V_1$), ($W_1$), ($W_2$) and ($W_3$) hold and $u_n\rightharpoonup u$ in E. Then

$$f(x,u_n)\to f(x,u)$$

in $L^2(R^3)$.

Proof Assume that $u_n\rightharpoonup u$ in E. Then, by Lemma 2.1,

$$u_n\to u$$

in $L_a^r(R^3)$, where $r\in (1,2)$, $a\in L^{\frac{2}{2-r}}(R^3)$ is positive for a.e. $x\in R^3$. Passing to a subsequence if necessary, it can be assumed that

$$\sum _{n=1}^{\mathrm{\infty }}{\parallel u_n-u\parallel }_{L_a^r}<+\mathrm{\infty }.$$

It is clear that

$$h_k(x):=\sum _{n=1}^k|u_n(x)-u(x)|\in L_a^r\left(R^3\right)$$

(8)

and

$${\parallel h_g-h_l\parallel }_{L_a^r}\le \sum _{n=l}^g{\parallel u_n-u\parallel }_{L_a^r}$$

(9)

for all $g>l\in N^{+}$. Since $\{u_n\}$ is a Cauchy sequence in $L_a^r(R^3)$, so by (9) we know that $\{h_k\}$ is also a Cauchy sequence in $L_a^r(R^3)$. Therefore, by the completeness of $L_a^r(R^3)$, there exists $h\in L_a^r(R^3)$ such that $h_k\to h$ in $L_a^r(R^3)$. Now we show that

$$h_k(x)\le h(x)$$

(10)

for all $k\in N^{+}$ and almost every $x\in R^3$. If not, there exist $k_0\in N^{+}$ and $S\subset R^3$, with $meas\{S\}>0$, such that

$$h_{k_0}(x)>h(x)$$

for all $x\in S$. Then there exist a constant $c>0$ and $S_0\subset S$, with $meas\{S_0\}>0$, such that

$$h_{k_0}(x)\ge h(x)+c$$

for all $x\in S_0$. By the definition of $h_k$, we have

$$h_k(x)\ge h_{k_0}(x)\ge h(x)+c$$

for all $k\ge k_0$ and $x\in S_0$. Therefore, one has

$$\begin{array}{rcl}{\int }_{R^3}a(x){|h_k-h|}^{r}dx& \ge & {\int }_{S_0}a(x){|h_k-h|}^{r}dx\\ \ge & c^r{\int }_{S_0}a(x)dx.\end{array}$$

Letting $k\to \mathrm{\infty }$, we get

$$0\ge c^r{\int }_{S_0}a(x)dx,$$

which contradicts the fact that $a(x)>0$ for a.e. $x\in R^3$. Now we have proved (10). It follows from ($W_2$) that there exists $M>0$ such that

$$|f(x,u)|\le a_2(x){|u|}^{r_2-1}$$

(11)

for all $x\in R^3$ and $|u|\ge M$. By ($W_1$), there exists $\delta >0$ such that

$$|f(x,u)|\le a_1(x){|u|}^{r_1-1}$$

(12)

for all $x\in R^3$ and $|u|\le \delta$, which together with ($W_3$) shows there exists $b_M\in L^{\frac{2}{2-r_3}}(R^3)$ such that

$$|f(x,u)|\le a_1(x){|u|}^{r_1-1}+\frac{b_M(x)}{{\delta }^{r_3-1}}{|u|}^{r_3-1}$$

(13)

for all $x\in R^3$ and $|u|\le M$. Combining (11) and (13), we have

$$|f(x,u)|\le a_1(x){|u|}^{r_1-1}+a_2(x){|u|}^{r_2-1}+\frac{b_M}{{\delta }^{r_3-1}}{|u|}^{r_3-1}$$

(14)

for all $x\in R^3$ and $u\in R$. Hence, by (10) one has

$$\begin{array}{rcl}|f(x,u_n)-f(x,u)|& \le & a_1(x)({|u_n|}^{r_1-1}+{|u|}^{r_1-1})+a_2(x)({|u_n|}^{r_2-1}+{|u|}^{r_2-1})\\ +\frac{b_M(x)}{{\delta }^{r_3-1}}({|u_n|}^{r_3-1}+{|u|}^{r_3-1})\\ \le & a_1(x)({|u_n-u|}^{r_1-1}+2{|u(x)|}^{r_1-1})+a_2(x)({|u_n-u|}^{r_2-1}+2{|u|}^{r_2-1})\\ +\frac{b_M(x)}{{\delta }^{r_3-1}}({|u_n-u|}^{r_3-1}+2{|u|}^{r_3-1})\\ \le & a_1(x)({|h|}^{r_1-1}+2{|u|}^{r_1-1})+a_2(x)({|h|}^{r_2-1}+2{|u|}^{r_2-1})\\ +\frac{b_M(x)}{{\delta }^{r_3-1}}({|h|}^{r_3-1}+2{|u|}^{r_3-1})\end{array}$$

for all $n\in N$ and $x\in R^3$. It follows that

$$\begin{array}{rcl}{|f(x,u_n)-f(x,u)|}^{2}dx& \le & 6a_1^2(x)({|h|}^{2(r_1-1)}+4{|u|}^{2(r_1-1)})dx\\ +6a_2^2(x)({|h|}^{2(r_2-1)}+4{|u|}^{2(r_2-1)})dx\\ +\frac{6b_M^2(x)}{{\delta }^{2(r_3-1)}}({|h|}^{2(r_3-1)}+4{|u|}^{2(r_3-1)})dx\\ =:& \varrho (x)\end{array}$$

(15)

for all $n\in N$. By the Hölder inequality, we have

$$\begin{array}{rcl}{\int }_{R^3}{a}_1^2(x){|h|}^{2(r_1-1)}dx& \le & {\left({\int }_{R^3}{a}_1{(x)}^{\frac{2}{2-r_1}}dx\right)}^{\frac{2-r_1}{r_1}}{({\int }_{R^3}{a}_1(x){|h|}^{r_1}dx)}^{\frac{2(r_1-1)}{r_1}}\\ =& {\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}^{\frac{2}{r_1}}{\parallel h\parallel }_{L_{a_1}^{r_1}}^{2(r_1-1)}\\ <& \mathrm{\infty }.\end{array}$$

(16)

Similarly, we can prove

$$\begin{array}{r}{\int }_{R^3}{a}_1^2(x){|u|}^{2(r_1-1)}dx<\mathrm{\infty },{\int }_{R^3}{a}_2^2(x){|h|}^{2(r_2-1)}dx<\mathrm{\infty },\\ {\int }_{R^3}{a}_2^2(x){|u|}^{2(r_2-1)}dx<\mathrm{\infty },\end{array}$$

(17)

also

$${\int }_{R^3}{b}_M^2(x){|h|}^{2(r_3-1)}dx<\mathrm{\infty },{\int }_{R^3}{b}_M^2(x){|u|}^{2(r_3-1)}dx<\mathrm{\infty }.$$

(18)

It follows from (15), (16), (17) and (18) that

$$\varrho \in L^1\left(R^3\right),$$

which together with Lebesgue’s convergence theorem shows

$${\int }_{R^3}{|f(x,u_n)-f(x,u)|}^{2}dx\to 0$$

(19)

as $n\to \mathrm{\infty }$. Now we have proved the lemma. □

In the proof of Theorem 1.2, the following lemma is needed.

Lemma 2.3 Assume that $G\subset R^3$ is an open set. Then, for any closed set $H\subset G$, there exists a function $\phi \in C_0^{\mathrm{\infty }}(R^3)$ such that $\phi (x)=0$ for all $x\in R^3\setminus G$, $\phi (x)=1$ for all $x\in H$ and $0\le \varphi (x)\le 1$ for all $x\in G\setminus H$.

Proof Letting

$$\tilde{\alpha }(x)=\{\begin{array}{cc}{e}^{\frac{1}{{|x|}^2-1}},\hfill & |x|<1,\hfill \\ 0,\hfill & |x|\ge 1,\hfill \end{array}$$

then $\tilde{\alpha }\in C_0^{\mathrm{\infty }}(R^3)$ and $supp\tilde{\alpha }=B_1(0)$. For any given $\epsilon >0$, defining α and ${\alpha }_{\epsilon }$ as follows,

$$\alpha (x)=\frac{\tilde{\alpha }(x)}{{\int }_{R^3}\tilde{\alpha }(x)dx},{\alpha }_{\epsilon }(x)=\frac{1}{{\epsilon }^3}\alpha \left(\frac{x}{\epsilon }\right),$$

one has ${\alpha }_{\epsilon }\in C_0^{\mathrm{\infty }}(R^3)$, $supp{\alpha }_{\epsilon }=\{x:|x|\le \epsilon \}$ and ${\int }_{R^3}{\alpha }_{\epsilon }(x)dx=1$. Denoting

$$d_0=\underset{x\in H,y\in \partial G}{inf}d(x,y)$$

and

$$G_{\theta }:=\{x\in G,d(x,\partial G)\ge \theta \},$$

it is clear that $d_0>0$ and $H\subset G_{d_0}$. Lastly, we define

$$\psi (x)=\{\begin{array}{cc}1,\hfill & x\in G_{\frac{d_0}{2}},\hfill \\ 0,\hfill & x\in R\setminus G_{\frac{d_0}{2}}\hfill \end{array}$$

and

$$\phi (x)={\int }_{R^3}\psi (x-y){\alpha }_{\frac{d_0}{4}}(y)dy,$$

then $\phi (x)=1$ for all $x\in H$ and $\phi (x)=0$ for all $x\in G_{\frac{d_0}{4}}$. Moreover, by the definition of ${\alpha }_{\epsilon }$, we have $\phi \in C_0^{\mathrm{\infty }}(R^3)$ and $0\le \phi (x)\le 1$. □

Since E is a Hilbert space, then there exists a basis $\{v_n\}\subset X$ such that $X=\overline{{⨁}_{j\ge 1}{X}_j}$, where $X_j=span\{v_j\}$. Letting $Y_k={⨁}_{j=1}^k{X}_j$, $Z_k=\overline{{⨁}_{j\ge k}{X}_j}$, now we show the following lemma, which will be used in the proof of Theorem 1.2.

Lemma 2.4 Suppose $r\in (1,2)$ and $a\in L^{\frac{2}{2-r}}(R^3)$, then we have

$${\beta }_k(a,r):=\underset{u\in Z_k,\parallel u\parallel =1}{sup}{\parallel u\parallel }_{L_a^r}\to 0$$

as $k\to \mathrm{\infty }$.

Proof It is clear that $0<{\beta }_{k+1}(a,r)\le {\beta }_k(a,r)$, so there exists $\beta (a,r)\ge 0$ such that

$${\beta }_k(a,r)\to \beta (a,r)$$

(20)

as $k\to \mathrm{\infty }$. By the definition of ${\beta }_k(a,r)$, there exists $u_k\in Z_k$ with $\parallel u_k\parallel =1$ such that

$${\parallel u_k\parallel }_{L_a^r}>\frac{{\beta }_k(a,r)}{2}.$$

(21)

Since ${\{u_k\}}_{k\in N}$ is bounded, then there exists $u\in E$ such that

$$u_k\rightharpoonup u$$

as $k\to \mathrm{\infty }$. Now, since $\{v_j\}$ is a basis of E, it follows that for all $j\in N$,

$$\begin{array}{rcl}0& =& (u_k,v_j)\mathrm{\forall }k>j\\ \to & (u,v_j)\end{array}$$

as $k\to \mathrm{\infty }$, which shows that $u=0$. By Lemma 2.1 we have

$$u_k\to 0$$

in $L_a^r(R^3)$ for all $r\in (1,2)$ and $a\in L^{\frac{2}{2-r}}(R^3)$, which together with (20) and (21) implies that $\beta (a,r)=0$ for all $r\in (1,2)$ and $a\in L^{\frac{2}{2-r}}(R^3)$. □

We obtain the existence of a solution for problem (1) by using the following standard minimizing argument.

Lemma 2.5 [19]

Let E be a real Banach space and $\mathrm{\Phi }\in C^1(E,R)$ satisfying the $(\mathit{PS})$ condition. If Φ is bounded from below,

$$c:=\underset{E}{inf}\mathrm{\Phi }$$

is a critical value of Φ.

In order to prove the multiplicity of solutions, we will use the dual fountain theorem. Firstly, we introduce the definition of the ${(\mathit{PS})}_c^{\ast }$ condition.

Definition 2.6 Let $\mathrm{\Phi }\in C^1(E,R)$ and $c\in R$. The function Φ satisfies the ${(\mathit{PS})}_c^{\ast }$ condition if any sequence $\{u_{n_j}\}\in E$, such that

$$\mathrm{\Phi }(u_{n_j})\to c,\mathrm{\Phi }{|}_{Y_{n_j}}^{\prime }(u_{n_j})\to 0\text{as }{n}_j\to \mathrm{\infty },$$

contains a subsequence converging to a critical point of Φ.

Now we show the following dual fountain theorem.

Lemma 2.7 [20]

If $\mathrm{\Phi }(-u)=\mathrm{\Phi }(u)$ and for every $k\ge k_0$, there exists ${\rho }_k>{\gamma }_k>0$ such that

• (i) $a_k:={inf}_{u\in Z_k,\parallel u\parallel ={\rho }_k}\mathrm{\Phi }(u)\ge 0$,

• (ii) $b_k:={max}_{u\in Y_k,\parallel u\parallel ={\gamma }_k}\mathrm{\Phi }(u)<0$,

• (iii) $d_k:={inf}_{u\in Z_k,\parallel u\parallel ={\rho }_k}\mathrm{\Phi }(u)\to 0$ as $k\to \mathrm{\infty }$.

Moreover, if $\mathrm{\Phi }\in C^1(X,R)$ satisfies the ${(\mathit{PS})}_c^{\ast }$ condition for all $c\in [d_{k_0},0)$, then Φ has a sequence of critical points $\{u_k\}$ such that $\mathrm{\Phi }(u_k)\to 0^{-}$ as $k\to \mathrm{\infty }$.

3 Proof of theorems

Define the functional $I:E\times D^{1,2}(R^3)\to R$ by

$$I(u,\varphi )=\frac{1}{2}{\parallel u\parallel }^2-\frac{1}{4}{\int }_{R^3}{|\mathrm{\nabla }\varphi |}^{2}dx+\frac{1}{2}{\int }_{R^3}\varphi u^{2}dx-{\int }_{R^3}F(x,u)dx.$$

(22)

It is easy to know that I exhibits a strong indefiniteness, namely it is unbounded both from below and from above on an infinitely dimensional subspace. This indefiniteness can be removed using the reduction method described in [1], by which we are led to study a variable functional that does not present such a strong indefinite nature.

Now we recall this method. For any $u\in E$, consider the linear functional $T_u:D^{1,2}(R^3)\to R$ defined as

$$T_u(v)={\int }_{R^3}{u}^{2}vdx.$$

By the Hölder inequality and using the second inequality in (5), we have

$$\begin{array}{rcl}{\int }_{R^3}{u}^{2}vdx& \le & {\parallel u^2\parallel }_{L^{6/5}}{\parallel v\parallel }_{L^6}\\ \le & {\parallel u\parallel }_{L^{12/5}}{\parallel v\parallel }_{L^6}\\ \le & C_{12/5}{C}_{\ast }{\parallel u\parallel }^2{\parallel v\parallel }_{D^{1,2}}.\end{array}$$

So, $T_u$ is continuous on $D^{1,2}(R^3)$. Set

$$\mu (u,v)={\int }_{R^3}\mathrm{\nabla }u\cdot \mathrm{\nabla }vdx$$

for all $u,v\in D^{1,2}(R^3)$. Obviously, $\mu (u,v)$ is bilinear, bounded and coercive. Hence, the Lax-Milgram theorem implies that for every $u\in E$, there exists a unique ${\varphi }_u\in D^{1,2}(R^3)$ such that

$$T_u(v)=\mu ({\varphi }_u,v)$$

for any $v\in D^{1,2}(R^3)$, that is,

$${\int }_{R^3}{u}^{2}vdx={\int }_{R^3}\mathrm{\nabla }{\varphi }_u\cdot \mathrm{\nabla }vdx$$

for any $v\in D^{1,2}(R^3)$. Using integration by parts, we get

$${\int }_{R^3}\mathrm{\nabla }{\varphi }_u\cdot \mathrm{\nabla }vdx=-{\int }_{R^3}v\mathrm{\Delta }{\varphi }_{u}dx$$

for any $v\in D^{1,2}(R^3)$, therefore

$$-\mathrm{\Delta }{\varphi }_u=u^2$$

(23)

in a weak sense. We can write an integral expression for ${\varphi }_u$ in the form

$${\varphi }_u=\frac{1}{4\pi }{\int }_{R^3}\frac{u^2(y)}{|x-y|}dy$$

for any $u\in C_0^{\mathrm{\infty }}(R^3)$ (see [21], Theorem 1); by density it can be extended for any $u\in E$ (see Lemma 2.1 of [22]). Clearly, ${\varphi }_u\ge 0$ and ${\varphi }_{-u}={\varphi }_u$ for all $u\in E$.

It follows from (23) that

$${\int }_{R^3}{\varphi }_u{u}^{2}dx={\int }_{R^3}{\varphi }_u(-\mathrm{\Delta }{\varphi }_u)dx={\int }_{R^3}{|\mathrm{\nabla }{\varphi }_u|}^{2}dx,$$

(24)

and by the Hölder inequality, we have

$$\begin{array}{rcl}{\parallel {\varphi }_u\parallel }_{D^{1,2}}^2& =& {\int }_{R^3}{\varphi }_u{u}^{2}dx\\ \le & {\left({\int }_{R^3}{\varphi }_u^{6}dx\right)}^{1/6}{\left({\int }_{R^3}{|u|}^{\frac{12}{5}}\right)}^{5/6}\\ =& C_{\ast }{\parallel {\varphi }_u\parallel }_{D^{1,2}}{\parallel u\parallel }_{L^{12/5}}^2,\end{array}$$

and it follows that

$${\parallel {\varphi }_u\parallel }_{D^{1,2}}\le C_{\ast }{\parallel u\parallel }_{L^{12/5}}^2.$$

(25)

Hence,

$${\int }_{R^3}{\varphi }_u{u}^{2}dx\le {C_{\ast }}^2{\parallel u\parallel }_{L^{12/5}}^4\le {C_{\ast }}^2{C}_{12/5}^4{\parallel u\parallel }^4:=C{\parallel u\parallel }^4.$$

(26)

So, we can consider the functional $\mathrm{\Phi }:E\to R$ defined by $\mathrm{\Phi }(u)=I(u,{\varphi }_u)$. By (24), the reduced functional takes the form

$$\mathrm{\Phi }(u)=\frac{1}{2}{\parallel u\parallel }^2+\frac{1}{4}{\int }_{R^3}{\varphi }_u{u}^{2}dx-{\int }_{R^3}F(x,u)dx.$$

(27)

By (12), we have

$$|F(x,u)|\le \frac{a_1(x)}{r_1}{|u|}^{r_1}$$

(28)

for all $x\in R^3$ and $|u|\le \delta$, where $r_1\in (1,2)$ and $a_1\in L^{\frac{2}{2-r_1}}(R^3)$. Let $u\in E$, then $u\in C^0(R^3)$, the space of continuous function u on $R^3$, such that $u(x)\to 0$ as $|x|\to \mathrm{\infty }$. Therefore there exists $T_1>0$ such that

$$|u(x)|\le \delta$$

(29)

for all $|x|>T_1$. Hence, one has

$$\begin{array}{rl}{\int }_{|x|>T_1}|F(x,u)|dx& \le {\int }_{|x|>T_1}\frac{a_1(x)}{r_1}{|u(x)|}^{r_1}dx\\ \le \frac{1}{r_1}{\left({\int }_{|x|\ge T_1}{a}_1{(x)}^{\frac{2}{2-r_1}}dx\right)}^{(2-r_1)/2}{\left({\int }_{|x|\ge T_1}{|u(x)|}^{2}dx\right)}^{r_1/2}\\ \le \frac{1}{r_1}{\left({\int }_{|x|\ge T_1}{a}_1{(x)}^{\frac{2}{2-r_1}}dx\right)}^{(2-r_1)/2}{\parallel u\parallel }_{L^2}^{r_1}\\ \le \frac{1}{r_1}{C}_2^{r_1}{\parallel u\parallel }^{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}\\ <\mathrm{\infty },\end{array}$$

which together with (26) shows that Φ is well defined. Furthermore, it is well known that Φ is a $C^1$ functional with derivative given by

$$〈{\mathrm{\Phi }}^{\prime }(u),v〉={\int }_{R^3}[(\mathrm{\nabla }u\cdot \mathrm{\nabla }v)+V(x)uv+{\varphi }_{u}uv-f(x,u)v]dx.$$

It can be proved that $(u,\varphi )\in E\times D^{1,2}(R^3)$ is a solution of problem (1) if and only if $u\in E$ is a critical point of the functional Φ and $\varphi ={\varphi }_u$; see, for instance, [1].

Lemma 3.1 Under conditions ($V_1$), ($W_1$), ($W_2$), ($W_3$), Φ satisfies the ${(\mathit{PS})}_c^{\ast }$ condition.

Proof Assume that $\{u_{n_j}\}\subset E$ is a sequence such that

$$\mathrm{\Phi }(u_{n_j})\to c,\mathrm{\Phi }{|}_{Y_{n_j}}^{\prime }(u_{n_j})\to 0\text{as }{n}_j\to \mathrm{\infty }.$$

Then there exists $\sigma >0$ such that

$$|\mathrm{\Phi }(u_{n_j})|\le \sigma ,{\parallel \mathrm{\Phi }{|}_{Y_{n_j}}^{\prime }(u_{n_j})\parallel }_E^{\ast }\le \sigma$$

for all $n_j\in N$.

Firstly, we show that $\{u_{n_j}\}$ is bounded. By (14), we have

$$|F(x,u)|\le \frac{a_1(x)}{r_1}{|u|}^{r_1}+\frac{a_2(x)}{r_2}{|u|}^{r_2}+\frac{b_M(x)}{r_3{\delta }^{r_3-1}}{|u|}^{r_3}$$

(30)

for all $u\in R$ and $x\in R^3$, which together with ${\int }_{R^3}{\varphi }_{u_{n_j}}{u}_{n_j}^{2}dx\ge 0$ implies

$$\begin{array}{rcl}{\parallel u_{n_j}\parallel }^2& =& 2\mathrm{\Phi }(u_{n_j})-\frac{1}{2}{\int }_{R^3}{\varphi }_{u_{n_j}}{u}_{n_j}^{2}dx+2{\int }_{R^3}F(x,u_{n_j})dx\\ \le & 2\sigma +\frac{2}{r_1}{\int }_{R^3}{a}_1(x){|u_{n_j}|}^{r_1}dx+\frac{2}{r_2}{\int }_{R^3}{a}_2(x){|u_{n_j}|}^{r_2}dx\\ +\frac{2}{r_3{\delta }^{r_3-1}}{\int }_{R^3}{b}_M(x){|u_{n_j}|}^{r_3}dx\\ \le & 2\sigma +\frac{2}{r_1}{\left({\int }_{R^3}{a}_1{(x)}^{\frac{2}{2-r_1}}dx\right)}^{(2-r_1)/2}{\left({\int }_{R^3}{|u_{n_j}|}^{2}dx\right)}^{r_1/2}\\ +\frac{2}{r_2}{\left({\int }_{R^3}{a}_2{(x)}^{\frac{2}{2-r_2}}dx\right)}^{(2-r_2)/2}{\left({\int }_{R^3}{|u_{n_j}|}^{2}dx\right)}^{r_2/2}\\ +\frac{2}{r_3{\delta }^{r_3-1}}{\left({\int }_{R^3}{b}_M{(x)}^{\frac{2}{2-r_3}}dx\right)}^{(2-r_3)/2}{\left({\int }_{R^3}{|u_{n_j}|}^{2}dx\right)}^{r_3/2}\\ \le & 2\sigma +\frac{2}{r_1}{C}_2^{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}{\parallel u_{n_j}\parallel }^{r_1}+\frac{2}{r_2}{C}_2^{r_2}{\parallel a_2\parallel }_{L^{\frac{2}{2-r_2}}}{\parallel u_{n_j}\parallel }^{r_2}\\ +\frac{2}{r_3{\delta }^{r_3-1}}{C}_2^{r_3}{\parallel b_M\parallel }_{L^{\frac{2}{2-r_3}}}{\parallel u_{n_j}\parallel }^{r_3}.\end{array}$$

(31)

Noting that $r_i<2$ for all $i=1,2,3$, so $\parallel u_{n_j}\parallel$ is bounded.

By the fact that $\{u_{n_j}\}$ is bounded in E, there exists $u\in E$ and a constant $d>0$ such that

$$\underset{n_j\in N}{sup}\parallel u_{n_j}\parallel \le d,\parallel u\parallel \le d$$

(32)

and

$$u_{n_j}\rightharpoonup u$$

in E as $n_j\to \mathrm{\infty }$. It is obvious that

$$〈{\mathrm{\Phi }}^{\prime }(u_{n_j})-{\mathrm{\Phi }}^{\prime }(u),u〉\to 0$$

(33)

and

$${\varphi }_{u}u(u_{n_j}-u)\to 0$$

(34)

as $n_j\to \mathrm{\infty }$. On the other hand, by ($V_1$), (32) and Lemma 2.2, one has

$$\begin{array}{rcl}\left|{\int }_{R^3}(f(x,u_{n_j})-f(x,u))u_{n_j}dx\right|& \le & {\parallel f(x,u_{n_j})-f(x,u)\parallel }_{L^2}{\parallel u_{n_j}\parallel }_{L^2}\\ \le & C_2{\parallel f(x,u_{n_j})-f(x,u)\parallel }_{L^2}\parallel u_{n_j}\parallel \\ \le & C_{2}d{\parallel f(x,u_{n_j})-f(x,u)\parallel }_{L^2}\\ \to & 0\end{array}$$

(35)

as $n_j\to \mathrm{\infty }$, which implies

$$〈{\mathrm{\Phi }}^{\prime }(u_{n_j})-{\mathrm{\Phi }}^{\prime }(u),u_{n_j}〉\to 0$$

(36)

as $n_j\to \mathrm{\infty }$. Summing up (33) and (36), we have

$$〈{\mathrm{\Phi }}^{\prime }(u_{n_j})-{\mathrm{\Phi }}^{\prime }(u),u_{n_j}-u〉\to 0$$

(37)

as $n_j\to \mathrm{\infty }$. By the Hölder inequality and (25), one gets

$$\begin{array}{rcl}{\int }_{R^3}{\varphi }_{u_{n_j}}{u}_{n_j}(u_{n_j}-u)dx& \le & {\parallel {\varphi }_{u_{n_j}}{u}_{n_j}\parallel }_{L^2}{\parallel u_{n_j}-u\parallel }_{L^2}\\ \le & {\parallel {\varphi }_{u_{n_j}}\parallel }_{L^6}{\parallel u_{n_j}\parallel }_{L^3}{\parallel u_{n_j}-u\parallel }_{L^2}\\ \le & C_{\ast }{\parallel {\varphi }_{u_{n_j}}\parallel }_{D^{1,2}}{\parallel u_{n_j}\parallel }_{L^3}{\parallel u_{n_j}-u\parallel }_{L^2}\\ \le & {C_{\ast }}^2{\parallel u_{n_j}\parallel }_{L^{12/5}}^2{\parallel u_{n_j}\parallel }_{L^3}{\parallel u_{n_j}-u\parallel }_{L^2}\\ \le & {C_{\ast }}^2{C}_{12/5}^2{C}_3{C}_2{\parallel u_{n_j}\parallel }^3\parallel u_{n_j}-u\parallel \\ \le & 2{C_{\ast }}^2{C}_{12/5}^2{C}_3{C}_2{d}^4\\ <& \mathrm{\infty }.\end{array}$$

Then by Lebesgue’s convergence theorem, we have

$${\int }_{R^3}{\varphi }_{u_{n_j}}{u}_{n_j}(u_{n_j}-u)dx\to 0$$

as $n_j\to \mathrm{\infty }$, which together with (34) implies

$${\int }_{R^3}({\varphi }_{u_{n_j}}{u}_{n_j}-{\varphi }_{u}u)(u_{n_j}-u)dx\to 0$$

(38)

as $n_j\to \mathrm{\infty }$. By Lemma 2.2 and (32), we get

$$\begin{array}{rcl}|{\int }_{R^3}(f(x,u_{n_j})-f(x,u))(u_{n_j}-u)dx|& \le & {\parallel f(x,u_{n_j})-f(x,u(x))\parallel }_{L^2}{\parallel u_{n_j}-u\parallel }_{L^2}\\ \le & C_2{\parallel f(x,u_{n_j})-f(x,u)\parallel }_{L^2}\parallel u_{n_j}-u\parallel \\ \le & 2C_{2}d{\parallel f(x,u_{n_j})-f(x,u)\parallel }_{L^2}\\ \to & 0\end{array}$$

as $n_j\to \mathrm{\infty }$. Moreover, an easy computation shows that

$$\begin{array}{rcl}〈{\mathrm{\Phi }}^{\prime }(u_{n_j})-{\mathrm{\Phi }}^{\prime }(u),u_{n_j}-u〉& =& {\parallel u_{n_j}-u\parallel }^2+{\int }_{R^3}({\varphi }_{u_{n_j}}{u}_{n_j}-{\varphi }_{u}u)(u_{n_j}-u)dx\\ -{\int }_{R^3}(f(x,u_{n_j})-f(x,u))(u_{n_j}-u)dx.\end{array}$$

Consequently, $\parallel u_{n_j}-u\parallel \to 0$ as $n_j\to \mathrm{\infty }$. Φ satisfies the ${(\mathit{PS})}_c^{\ast }$ condition. □

Remark 3.2 Under conditions ($V_1$), ($W_1$), ($W_2$), ($W_3$), Φ satisfies the $(\mathit{PS})$ condition. Assume that $\{u_n\}\subset E$ is a sequence such that $I(u_n)$ is bounded and

$$I^{\prime }(u_n)\to 0$$

as $n\to \mathrm{\infty }$. Then there exists $\sigma >0$ such that

$$|I(u_n)|\le \sigma ,{\parallel I^{\prime }(u_n)\parallel }_E^{\ast }\le \sigma$$

for all $n\in N$. The rest of the proof is the same as that of Lemma 3.1.

Proof of Theorem 1.2 For any $k\in N$, we take k disjoint open sets $\{{\mathrm{\Omega }}_i|i=1,\dots ,k\}$ such that

$$\bigcup _{i=1}^k{\mathrm{\Omega }}_i\subset \mathrm{\Omega }.$$

For any $\epsilon >0$ and ${\mathrm{\Omega }}_i$, there exist a closed set $H_i$ and an open set $G_i$ such that $H_i\subset {\mathrm{\Omega }}_i\subset G_i$ and

$$meas\{G_i\setminus {\mathrm{\Omega }}_i\}<\epsilon ,meas\{{\mathrm{\Omega }}_i\setminus H_i\}<\epsilon .$$

For every $G_i$ ($i=1,\dots ,k$), by Lemma 2.3 there exists ${\phi }_i\in C_0^{\mathrm{\infty }}(G_i,R)$ such that ${\phi }_i{|}_{H_i}=1$ and $0\le {\phi }_i\le 1$. Letting $v_i=\frac{{\phi }_i}{\parallel {\phi }_i\parallel }$, can be extended to be a basis $\{v_n\}\subset X$. Therefore $X=\overline{{⨁}_{j\ge 1}{X}_j}$, where $X_j=span\{v_j\}$. Now we define $Y_k:={⨁}_{j=1}^k{X}_j$, $Z_k:=\overline{{⨁}_{j\ge k}{X}_j}$.

By Lemma 3.1, $\mathrm{\Phi }\in C^1(E,R)$ satisfies the ${(\mathit{PS})}_c^{\ast }$ condition and $\mathrm{\Phi }(u)=\mathrm{\Phi }(-u)$. Hence, to prove Theorem 1.2, we should just show that Φ has the geometric property (i), (ii) and (iii) in Lemma 2.7.

• (i) By Lemma 2.4

  $${\beta }_k(a,r)=\underset{u\in Z_k,\parallel u\parallel =1}{sup}{\parallel u\parallel }_{L_a^r}\to 0$$

as $k\to \mathrm{\infty }$ for $r\in (1,2)$ and $a\in L^{\frac{2}{2-r}}(R^3)$. In view of (30) and the fact that ${\int }_{R^3}{\varphi }_u{u}^{2}dx\ge 0$, we have

$$\begin{array}{rcl}\mathrm{\Phi }(u)& =& \frac{1}{2}{\parallel u\parallel }^2+\frac{1}{4}{\int }_{R^3}{\varphi }_u{u}^{2}dx-{\int }_{R^3}F(x,u)dx\\ \ge & \frac{1}{2}{|u|}^2-{\int }_{R^3}F(x,u)dx\\ \ge & \frac{1}{2}{\parallel u\parallel }^2-\frac{2}{r_1}{\int }_{R^3}{a}_1(x){|u|}^{r_1}dx-\frac{2}{r_2}{\int }_{R^3}{a}_2(x){|u|}^{r_2}dx\\ -\frac{2}{r_3{\delta }^{r_3-1}}{\int }_{R^3}{b}_M(x){|u|}^{r_3}dx\\ \ge & \frac{1}{2}{\parallel u\parallel }^2-\frac{2{\parallel u\parallel }_{L_{a_1}^{r_1}}^{r_1}}{r_1}-\frac{2{\parallel u\parallel }_{L_{a_2}^{r_2}}^{r_2}}{r_2}-\frac{2{\parallel u\parallel }_{L_{a_3}^{r_3}}^{r_3}}{r_3{\delta }^{r_3-1}}\\ \ge & \frac{1}{2}{\parallel u\parallel }^2-\frac{2{\beta }_k{(a_1,r_1)}^{r_1}}{r_1}{\parallel u\parallel }^{r_1}-\frac{2{\beta }_k{(a_2,r_2)}^{r_2}}{r_2}{\parallel u\parallel }^{r_2}-\frac{2{\beta }_k{(b_M,r_3)}^{r_3}}{r_3{\delta }^{r_3-1}}{\parallel u\parallel }^{r_3}.\end{array}$$

(39)

Let $r:=min\{r_1,r_2,r_3\}$, ${\beta }_k:=max\{{\beta }_k(a_1,r_1),{\beta }_k(a_2,r_2),{\beta }_k(b_M,r_3)\}$, $C^{\prime }:=max\{\frac{2}{r_1},\frac{2}{r_2},\frac{2}{r_3{\delta }^{r_3-1}}\}$, then ${\beta }_k\to 0$ as $k\to \mathrm{\infty }$. Hence, we have

$$\mathrm{\Phi }(u)\ge \frac{1}{2}{\parallel u\parallel }^2-3C^{\prime }{\beta }_k^r{\parallel u\parallel }^r$$

(40)

when $\parallel u\parallel \le 1$ and ${\beta }_k\le 1$. Now we can choose ${\rho }_k={(12{\beta }_k^r{C}^{\prime })}^{1/(2-r)}$, then ${\rho }_k\to 0$ as $k\to \mathrm{\infty }$. When k is large enough, we have ${\rho }_k\le 1$, ${\beta }_k\le 1$, which together with (40) shows

$$a_k:=\underset{u\in Z_k,\parallel u\parallel ={\rho }_k}{inf}\mathrm{\Phi }(u)\ge \frac{1}{4}{\rho }_k^2>0.$$

• (ii) For any $u\in Y_k$, there exists ${\lambda }_i=1,2,\dots ,k$ such that

  $$u=\sum _{i=1}^k{\lambda }_i{v}_i.$$

Then we have

$$\begin{array}{rcl}{\parallel u\parallel }_{L^{r_4}}^{r_4}& =& {\int }_{R^3}{|u(x)|}^{r_4}dx\\ =& \sum _{i=1}^k{|{\lambda }_i|}^{r_4}{\int }_{{\mathrm{\Omega }}_i}{|v_i(x)|}^{r_4}dx+\sum _{i=1}^k{|{\lambda }_i|}^{r_4}{\int }_{G_i\setminus {\mathrm{\Omega }}_i}{|v_i(x)|}^{r_4}dx\\ =& \sum _{i=1}^k{|{\lambda }_i|}^{r_4}{\int }_{{\mathrm{\Omega }}_i}{|v_i(x)|}^{r_4}dx+\sum _{i=1}^k{|{\lambda }_i|}^{r_4}{\int }_{G_i\setminus {\mathrm{\Omega }}_i}\frac{{|{\phi }_i(x)|}^{r_4}}{{\parallel {\phi }_i\parallel }^{r_4}}dx\\ \le & \sum _{i=1}^k{|{\lambda }_i|}^{r_4}{\int }_{{\mathrm{\Omega }}_i}{|v_i(x)|}^{r_4}dx+\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_4}}{{\parallel {\phi }_i\parallel }^{r_4}}meas\{G_i\setminus {\mathrm{\Omega }}_i\}\\ \le & \sum _{i=1}^k{|{\lambda }_i|}^{r_4}{\int }_{{\mathrm{\Omega }}_i}{|v_i(x)|}^{r_4}+\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_4}}{{\parallel {\phi }_i\parallel }^{r_4}}\epsilon \end{array}$$

(41)

and also

$$\begin{array}{rcl}{\parallel u\parallel }^2& =& {\int }_{R^3}[{|\mathrm{\nabla }u|}^2+V(x)u^2]dx\\ =& \sum _{i=1}^k{\lambda }_i^2{\int }_{G_i}[{|\mathrm{\nabla }{v}_i|}^2+V(x)v_i^2]dx\\ =& \sum _{i=1}^k{\lambda }_i^2{\parallel v_i\parallel }^2\\ =& \sum _{i=1}^k{\lambda }_i^2.\end{array}$$

(42)

Since all the norms of a finite dimensional space are equivalent, there is a constant $\tilde{C}$ such that

$$\tilde{C}\parallel u\parallel \le {\parallel u\parallel }_{L^{r_4}}$$

for all $u\in Y_k$. By (30), one has

$$F(x,{\lambda }_i{v}_i)\ge -\frac{a_1(x)}{r_1}{|{\lambda }_i{v}_i|}^{r_1}-\frac{a_2(x)}{r_2}{|{\lambda }_i{v}_i|}^{r_2}-\frac{b_M(x)}{r_3{\delta }^{r_3-1}}{|{\lambda }_i{v}_i|}^{r_3}.$$

Therefore, we have

$$\begin{array}{l}\sum _{i=1}^k{\int }_{G_i\setminus {\mathrm{\Omega }}_i}F(x,{\lambda }_i{v}_i)dx\\ \ge -\sum _{i=1}^k{\int }_{G_i\setminus {\mathrm{\Omega }}_i}\frac{{|{\lambda }_i|}^{r_1}}{r_1}{a}_1(x){|v_i|}^{r_1}dx-\sum _{i=1}^k{\int }_{G_i\setminus {\mathrm{\Omega }}_i}\frac{{|{\lambda }_i|}^{r_2}}{r_2}{a}_2(x){|v_i|}^{r_2}dx\\ -\sum _{i=1}^k{\int }_{G_i\setminus {\mathrm{\Omega }}_i}\frac{{|{\lambda }_i|}^{r_3}}{r_3{\delta }^{r_3-1}}{b}_M(x){|v_i|}^{r_3}dx\\ \ge -\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_1}}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}{\left({\int }_{G_i\setminus {\mathrm{\Omega }}_i}{|v_i|}^{2}dx\right)}^{r_1/2}\\ -\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_2}}{r_2}{\parallel a_2\parallel }_{L^{\frac{2}{2-r_2}}}{\left({\int }_{G_i\setminus {\mathrm{\Omega }}_i}{|v_i|}^{2}dx\right)}^{r_2/2}\\ -\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_3}}{r_3{\delta }^{r_3-1}}{\parallel b_M\parallel }_{L^{\frac{2}{2-r_3}}}{\left({\int }_{G_i\setminus {\mathrm{\Omega }}_i}{|v_i|}^{2}dx\right)}^{r_3/2}\\ \ge -\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_1}}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}{\left({\int }_{G_i\setminus {\mathrm{\Omega }}_i}\frac{{|{\phi }_i|}^2}{{\parallel {\phi }_i\parallel }^2}dx\right)}^{r_1/2}\\ -\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_2}}{r_2}{\parallel a_2\parallel }_{L^{\frac{2}{2-r_2}}}{\left({\int }_{G_i\setminus {\mathrm{\Omega }}_i}\frac{{|{\phi }_i|}^2}{{\parallel {\phi }_i\parallel }^2}dx\right)}^{r_2/2}\\ -\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_3}}{r_3{\delta }^{r_3-1}}{\parallel b_M\parallel }_{L^{\frac{2}{2-r_3}}}{\left({\int }_{G_i\setminus {\mathrm{\Omega }}_i}\frac{{|{\phi }_i|}^2}{{\parallel {\phi }_i\parallel }^2}dx\right)}^{r_3/2}\\ =-\frac{1}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_1}}{{\parallel {\phi }_i\parallel }^{r_1}}{(meas\{G_i\setminus {\mathrm{\Omega }}_i\})}^{r_1/2}\\ -\frac{1}{r_2}{\parallel a_2\parallel }_{L^{\frac{2}{2-r_2}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_2}}{{\parallel {\phi }_i\parallel }^{r_2}}{(meas\{G_i\setminus {\mathrm{\Omega }}_i\})}^{r_2/2}\\ -\frac{1}{r_3{\delta }^{r_3-1}}{\parallel b_M\parallel }_{L^{\frac{2}{2-r_3}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_3}}{{\parallel {\phi }_i\parallel }^{r_3}}{(meas\{G_i\setminus {\mathrm{\Omega }}_i\})}^{r_3/2}\\ \ge -\frac{1}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_1}}{{\parallel {\phi }_i\parallel }^{r_1}}{\epsilon }^{r_1/2}-\frac{1}{r_2}{\parallel a_2\parallel }_{L^{\frac{2}{2-r_2}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_2}}{{\parallel {\phi }_i\parallel }^{r_2}}{\epsilon }^{r_2/2}\\ -\frac{1}{r_3{\delta }^{r_3-1}}{\parallel b_M\parallel }_{L^{\frac{2}{2-r_3}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_3}}{{\parallel {\phi }_i\parallel }^{r_3}}{\epsilon }^{r_3/2}.\end{array}$$

(43)

For any $u\in Y_k$ with $\parallel u\parallel ={\sum }_{i=1}^k{\lambda }_i^2={\gamma }_k$, we can choose ${\gamma }_k$ small enough such that $|{\lambda }_i{v}_i(x)|<\zeta$ for all $x\in R^3$ and $i=1,\dots ,k$, which together with ($W_4$) implies

$$F(x,{\lambda }_i{v}_i)\ge \eta {|{\lambda }_i{v}_i|}^{r_4}$$

(44)

for all $x\in {\mathrm{\Omega }}_i$ and $i=1,\dots ,k$. Combining (24), (41), (42), (43) and (44), we have

$$\begin{array}{rcl}\mathrm{\Phi }(u)& =& \frac{1}{2}{\parallel u\parallel }^2+\frac{1}{4}{\int }_{R^3}{\varphi }_u{u}^{2}dx-{\int }_{R^3}F(x,u)dx\\ =& \frac{1}{2}{\parallel u\parallel }^2+\frac{C}{4}{\parallel u\parallel }^4-\sum _{i=1}^k{\int }_{G_i}F(x,{\lambda }_i{v}_i)dx\\ \le & \frac{1}{2}{\parallel u\parallel }^2-\sum _{i=1}^k[{\int }_{G_i\setminus {\mathrm{\Omega }}_i}F(x,{\lambda }_i{v}_i)dx+{\int }_{{\mathrm{\Omega }}_i}F(x,{\lambda }_i{v}_i)dx]\\ \le & \frac{1}{2}{\parallel u\parallel }^2+\frac{C}{4}{\parallel u\parallel }^4+\frac{1}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_1}}{{\parallel {\phi }_i\parallel }^{r_1}}{\epsilon }^{r_1/2}\\ +\frac{1}{r_2}{\parallel a_2\parallel }_{L^{\frac{2}{2-r_2}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_2}}{{\parallel {\phi }_i\parallel }^{r_2}}{\epsilon }^{r_2/2}+\frac{1}{r_3{\delta }^{r_3-1}}{\parallel b_M\parallel }_{L^{\frac{2}{2-r_3}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_3}}{{\parallel {\phi }_i\parallel }^{r_3}}{\epsilon }^{r_3/2}\\ -\eta \sum _{i=1}^k{|{\lambda }_i|}^{r_4}{\int }_{{\mathrm{\Omega }}_i}{|v_i|}^{r_4}dx\\ =& \frac{1}{2}{\parallel u\parallel }^2+\frac{C}{4}{\parallel u\parallel }^4+\frac{1}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_1}}{{\parallel {\phi }_i\parallel }^{r_1}}{\epsilon }^{r_1/2}\\ +\frac{1}{r_2}{\parallel a_2\parallel }_{L^{\frac{2}{2-r_2}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_2}}{{\parallel {\phi }_i\parallel }^{r_2}}{\epsilon }^{r_2/2}+\frac{1}{r_3{\delta }^{r_3-1}}{\parallel b_M\parallel }_{L^{\frac{2}{2-r_3}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_3}}{{\parallel {\phi }_i\parallel }^{r_3}}{\epsilon }^{r_3/2}\\ -\eta ({\parallel u\parallel }_{L^{r_4}}^{r_4}-\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_4}}{{\parallel {\phi }_i\parallel }^{r_4}}\epsilon )\\ \le & \frac{1}{2}{\parallel u\parallel }^2+\frac{C}{4}{\parallel u\parallel }^4-\eta {\tilde{C}}^{r_4}{\parallel u\parallel }^{r_4}+\frac{1}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_1}}{{\parallel {\phi }_i\parallel }^{r_1}}{\epsilon }^{r_1/2}\\ +\frac{1}{r_2}{\parallel a_2\parallel }_{L^{\frac{2}{2-r_2}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_2}}{{\parallel {\phi }_i\parallel }^{r_2}}{\epsilon }^{r_2/2}+\frac{1}{r_3{\delta }^{r_3-1}}{\parallel b_M\parallel }_{L^{\frac{2}{2-r_3}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_3}}{{\parallel {\phi }_i\parallel }^{r_3}}{\epsilon }^{r_3/2}\\ +\eta \sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_4}}{{\parallel {\phi }_i\parallel }^{r_4}}\epsilon \\ =& \frac{1}{2}\sum _{i=1}^k{\lambda }_i^2+\frac{C}{4}{\left(\sum _{i=1}^k{\lambda }_i^2\right)}^2-\eta {\tilde{C}}^{r_4}{\left(\sum _{i=1}^k{\lambda }_i^2\right)}^{r_4/2}+\frac{1}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_1}}{{\parallel {\phi }_i\parallel }^{r_1}}{\epsilon }^{r_1/2}\\ +\frac{1}{r_2}{\parallel a_2\parallel }_{L^{\frac{2}{2-r_2}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_2}}{{\parallel {\phi }_i\parallel }^{r_2}}{\epsilon }^{r_2/2}+\frac{1}{r_3{\delta }^{r_3-1}}{\parallel b_M\parallel }_{L^{\frac{2}{2-r_3}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_3}}{{\parallel {\phi }_i\parallel }^{r_3}}{\epsilon }^{r_3/2}\\ +\eta \sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_4}}{{\parallel {\phi }_i\parallel }^{r_4}}\epsilon \\ =& \frac{1}{2}{\gamma }_k^2+\frac{C}{4}{\gamma }_k^4-\eta {(\tilde{C}{\gamma }_k)}^{r_4}+\frac{1}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_1}}{{\parallel {\phi }_i\parallel }^{r_1}}{\epsilon }^{r_1/2}\\ +\frac{1}{r_2}{\parallel a_2\parallel }_{L^{\frac{2}{2-r_2}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_2}}{{\parallel {\phi }_i\parallel }^{r_2}}{\epsilon }^{r_2/2}+\frac{1}{r_3{\delta }^{r_3-1}}{\parallel b_M\parallel }_{L^{\frac{2}{2-r_3}}}\sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_3}}{{\parallel {\phi }_i\parallel }^{r_3}}{\epsilon }^{r_3/2}\\ +\eta \sum _{i=1}^k\frac{{|{\lambda }_i|}^{r_4}}{{\parallel {\phi }_i\parallel }^{r_4}}\epsilon \\ \le & {\gamma }_k^2+\frac{C}{4}{\gamma }_k^4-\eta {(\tilde{C}{\gamma }_k)}^{r_4}\end{array}$$

for all $u\in Y_k$ with $\parallel u\parallel ={\gamma }_k$, when ε and ${\gamma }_k$ are both small enough. Since $r_4<2$, we can choose ${\gamma }_k<{\rho }_k$ small enough such that

$$b_k:=\underset{u\in Y_k,\parallel u\parallel ={\gamma }_k}{max}\mathrm{\Phi }(u)<0.$$

• (iii) By (40), for any $u\in Z_k$ with $\parallel u\parallel ={\rho }_k$, we have

  $$\mathrm{\Phi }(u)\ge -3C^{\prime }{\beta }_k^r{\parallel u\parallel }^r.$$

Therefore

$$0\ge \underset{u\in Z_k,\parallel u\parallel \le {\rho }_k}{inf}\mathrm{\Phi }(u)\ge -3C^{\prime }{\beta }_k^r{\rho }_k^r.$$

Since ${\beta }_k,{\rho }_k\to 0$ as $k\to \mathrm{\infty }$, we have

$$d_k:=\underset{u\in Z_k,\parallel u\parallel \le {\rho }_k}{inf}\mathrm{\Phi }(u)\to 0$$

as $k\to \mathrm{\infty }$.

Hence, by Lemma 2.7, we obtain that problem (1) has infinitely many solutions $\{(u_k,{\varphi }_k)\}$ satisfying

$$\frac{1}{2}{\int }_{R^3}({|\mathrm{\nabla }{u}_k|}^2+V(x)u_k^2)dx-\frac{1}{4}{\int }_{R^3}{|\mathrm{\nabla }{\varphi }_k|}^{2}dx+\frac{1}{2}{\int }_{R^3}{\varphi }_k{u}_k^{2}dx-{\int }_{R^3}F(x,u_k)dx\to 0^{-}$$

as $k\to \mathrm{\infty }$. □

Proof of Theorem 1.5 Similar to (31), there exist constants $k_i>0$, $i=1,2,3$, such that

$$\mathrm{\Phi }(u)\ge \frac{1}{2}{\parallel u\parallel }^2-\sum _{i=1}^3{k}_i{\parallel u\parallel }^{r_i}$$

(45)

for all $u\in E$. Since $1<r_i<2$, it follows from (45) that the functional Φ is bounded from below. By Lemma 2.5 and Remark 3.2, Φ possesses a critical point u satisfying

$$\mathrm{\Phi }(u)=\underset{E}{inf}\mathrm{\Phi },{\mathrm{\Phi }}^{\prime }(u)=0.$$

It remains to show that u is nontrivial. For every $\epsilon >0$, there exist an open set G and a closed set H such that $H\subset \mathrm{\Omega }\subset G$ and

$$meas\{G\setminus \mathrm{\Omega }\}<\epsilon ,meas\{\mathrm{\Omega }\setminus H\}<\epsilon .$$

By Lemma 2.3, there exists a function $\phi \in C_0^{\mathrm{\infty }}(R^3)$ such that $0\le \phi (x)\le 1$ and $\phi {|}_H(x)=1$, $\phi {|}_{R\setminus G}(x)=0$, then $\phi \in E$. Choosing $0<\lambda <min\{\delta ,\zeta \}$, then $|\lambda \phi (x)|<\delta$ for all $x\in R^3$, which together with (28) shows

$$F(x,\lambda \phi (x))\ge -\frac{a_1(x)}{r_1}{|\lambda \phi (x)|}^{r_1}$$

for all $x\in R^3$. Therefore, one has

$$\begin{array}{rcl}{\int }_{G\setminus H}F(x,\lambda \phi )dx& \ge & -{\int }_{G\setminus H}\frac{{\lambda }^{r_1}}{r_1}{a}_1(x){\phi }^{r_1}dx\\ \ge & -\frac{{\lambda }^{r_1}}{r_1}{\left({\int }_{G\setminus H}{a}_1{(x)}^{\frac{2}{2-r_1}}dx\right)}^{(2-r_1)/2}{\left({\int }_{G\setminus H}{\phi }^{2}dx\right)}^{r_1/2}\\ \ge & -\frac{{\lambda }^{r_1}}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}{\left({\int }_{G\setminus H}1dx\right)}^{r_1/2}\\ \ge & -\frac{{\lambda }^{r_1}}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}{(meas\{G\setminus H\})}^{r_1/2}\\ \ge & -\frac{{\lambda }^{r_1}}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}{(2\epsilon )}^{r_1/2}.\end{array}$$

(46)

In view of $\lambda <\zeta$, we have $|\lambda \phi (x)|<\zeta$ for all $x\in R^3$, which together with ($W_4$) implies

$$F(x,\lambda \phi )\ge \eta {|\lambda \phi |}^{r_4}$$

(47)

for all $x\in \mathrm{\Omega }$. It follows from (24), (46), (47) that

$$\begin{array}{rcl}\mathrm{\Phi }(\lambda \phi )& =& \frac{{\lambda }^2}{2}{\parallel \phi \parallel }^2+\frac{1}{4}{\int }_{R^3}{\varphi }_{\lambda \phi }{(\lambda \phi )}^{2}dx-{\int }_{R^3}F(x,\lambda \phi )dx\\ \le & \frac{{\lambda }^2}{2}{\parallel \phi \parallel }^2+C{\lambda }^4{\parallel \phi \parallel }^4-{\int }_{R^3}F(x,\lambda \phi )dx\\ \le & \frac{{\lambda }^2}{2}{\parallel \phi \parallel }^2+C{\lambda }^4{\parallel \phi \parallel }^4-{\int }_{G}F(x,\lambda \phi )dx\\ =& \frac{{\lambda }^2}{2}{\parallel \phi \parallel }^2+C{\lambda }^4{\parallel \phi \parallel }^4-[{\int }_{H}F(x,\lambda \phi )dx+{\int }_{G\setminus H}F(x,\lambda \phi )dx]\\ \le & \frac{{\lambda }^2}{2}{\parallel \phi \parallel }^2+C{\lambda }^4{\parallel \phi \parallel }^4-{\lambda }^{r_4}{\int }_H\eta {|\phi |}^{r_4}dx+\frac{{\lambda }^{r_1}}{r_1}{\parallel a_1\parallel }_{L^{\frac{2}{2-r_1}}}{(2\epsilon )}^{r_1/2}\\ \le & {\lambda }^2{\parallel \phi \parallel }^2+C{\lambda }^4{\parallel \phi \parallel }^4-{\lambda }^{r_4}\eta meas\{H\}\\ <& 0\end{array}$$

when ε and λ are both small enough. Since $\mathrm{\Phi }(0)=0$, then $u\ne 0$. Hence, $(u,{\varphi }_u)$ is a nontrivial solution of problem (1). □