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Research article

Existence of axially symmetric solutions for a kind of planar Schrödinger-Poisson system

  • Received: 13 March 2021 Accepted: 10 May 2021 Published: 18 May 2021
  • MSC : 35J20, 35J62, 35Q55

  • In this paper, we study the following kind of Schrödinger-Poisson system in R2

    {Δu+V(x)u+ϕu=K(x)f(u),   xR2,Δϕ=u2,                                       xR2,

    where fC(R,R), V(x) and K(x) are both axially symmetric functions. By constructing a new variational framework and using some new analytic techniques, we obtain an axially symmetric solution for the above planar system. Our result improves and extends the existing works.

    Citation: Qiongfen Zhang, Kai Chen, Shuqin Liu, Jinmei Fan. Existence of axially symmetric solutions for a kind of planar Schrödinger-Poisson system[J]. AIMS Mathematics, 2021, 6(7): 7833-7844. doi: 10.3934/math.2021455

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  • In this paper, we study the following kind of Schrödinger-Poisson system in R2

    {Δu+V(x)u+ϕu=K(x)f(u),   xR2,Δϕ=u2,                                       xR2,

    where fC(R,R), V(x) and K(x) are both axially symmetric functions. By constructing a new variational framework and using some new analytic techniques, we obtain an axially symmetric solution for the above planar system. Our result improves and extends the existing works.



    We consider the following planar Schrödinger-Poisson system:

    {Δu+V(x)u+ϕu=K(x)f(u),   xR2,Δϕ=u2,                                xR2, (1.1)

    where K, V and f satisfy the following basic assumptions:

    (V1) VC(R2,(0,)), V(x)=V(x1,x2)=V(|x1|,|x2|),  xR2 and lim inf|x|V(x)>0;

    (K1) KC(R2,(0,)), K(x)=K(x1,x2)=K(|x1|,|x2|),  xR2 and lim inf|x|K(x)>0;

    (F1) f(u)=o(|u|) as u0;

    (F2) fC(R,R), there exists c0>0 and p>2 such that |f(u)|c0(1+|u|p1),   uR.

    It is pointed out that (V1) and (K1) imply that V(x) and K(x) are both axially symmetric functions. As shown in [1], axially symmetric functions are widely existing in real world, but axially symmetric functions are less used in the existing works because of the lack of compact embedding from the subspace of H1(RN) to Ls(RN) for N2, where the elements of the subspace are axially symmetric functions. In recent years, the following nonlinear Schrödinger-Poisson equations have gained more attentions:

    {Δu+V(x)u+μϕu=f(x,u),   xRN,Δϕ=u2,                              xRN, (1.2)

    where μR{0}, VC(RN,(0,)) and fC(RN×R,R). It is easy to see that system (1.1) is a special form of system (1.2).

    From [2], we know that system (1.2) comes from semiconductor theory and quantum mechanics theory. Recently, the N=2 problem has attracted a lot of attention in relation to optical propagation in certain media. Paredes, Olivieri and Michinel [3] gave a detail review on nonlinear Schrödinger-Poisson systems. They mainly reviewed two kinds of Schrödinger-Poisson systems: nonlinear optics in thermo-optical media and the 1+2D Schrödinger-Poisson model; ultralight axion dark matter and 3D Schrödinger-Poisson systems. Specially, they presented families of stationary solutions, discussed the implications of the simulation of propagation dynamics and discussed some numerical methods to solve the system of time-dependent partial differential equations. In the physical aspects, the solution ϕ of Δϕ=u2 in system (1.2) can be solved by ϕ=ΓNu2, where

    ΓN(x)={ln|x|2π,      N=2,|x|2NN(2N)ωN,  N3,

    is the fundamental solutions of the Laplacian, is the convolution in RN and ωN is the volume of the unit N-ball. With this formal inversion, an integro-differential equation is obtained as follows

    Δu+V(x)u+μ(ΓNu2)u=f(x,u),  xRN. (1.3)

    When N=2 and μ0, there are only a few works dealing with system (1.2) or (1.3). Chen, Chen and Tang [4] investigated system (1.2) in the periodic and asymptotically periodic cases using the non-Nehari manifold method derived from [5]. Bernini and Mugnai [6] have rewritten a nonlinear planar Schrödinger-Poisson system as a nonlinear Hartree equation and obtained an existence result of radially symmetric solutions when V(x) is a positive constant and μ=1. If f(x,u)=f(u) and μ>0, the authors in [7,8] dealt with periodic case and constructed a variational setting for (1.3). Recently, the author in [9] improved and extended the main results obtained in [8] with V(x)=1 and more general nonlinearity f(u). Very recently, Chen and Tang [10] dealt with axially symmetric potential instead of the periodic case and developed a natural constraint function space for system (1.2). More recently, Chen and Tang [1] considered the case that the nonlinearity is sub-cubic growth at infinity. As pointed out in [1] that this case is more difficult and the methods used in [10] is no longer available since it is not sure whether {un} are bounded in H1(RN). Motivated by [1], Wen, Chen and Rădulescu [11] studied system (1.1) with V(x)=0 and K(x) is a axially symmetric function and obtained a main result.

    When μ=0, system (1.2) reduces to Schrödinger equations. Many researchers investigated Schrödinger equations and obtained many existence results of nontrivial solutions, see [12,13,14,15,16,17,18,19,20] and references therein. It is pointed out that critical point theory is very important for studying Schrödinger equations and other kinds of elliptic systems [21,22,23,24]. However, most of the existing works of Schrödinger equations or Schrödinger-Poisson equations are dealt with one of the following two cases: i) infxRNV(x)>0; ii) V(x) and K(x) vanish at infinity. There is a question: what will happen if infxR2V(x)>0 and lim inf|x|K(x)>0 in system (1.1). Moreover, the methods handling the case N=3 are no longer available for N=2 since the integral Γ2=ln|x|2π is sign-changing and unbounded, which causes the functional associated with system (1.1) is not well-defined on H1(R2) even if VL(R2) and infxR2V(x)>0. As far as we known, there seems no related works in the case of infxR2V(x)>0 and lim inf|x|K(x)>0. In this paper, motivated by the aforementioned works, we will give a positive answer and obtain an axially symmetric solution for system (1.1) by establishing a new variational setting and using some new analytic tricks.

    To present our result, the following assumptions are needed.

    (V2) VC1(R2,R), tt2[2V(tx)V(tx)(tx)] is nondecreasing on (0,) for all xR2;

    (K2) KC1(R2,R), K(x)x0, t4K(tx)K(tx)(tx) is nonincreasing on (0,) for all xR2;

    (F3) the function f(u)uF(u)u3 is nondecreasing on both (,0) and (0,), where and in the sequel, F(u)=u0f(s)ds.

    The main result is as follows.

    Theorem 1.1. Suppose that V, K and f satisfy (V1), (V2), (K1), (K2) and (F1)–(F3). Then (1.1) possesses an axially symmetric solution ˉu satisfying

    Φ(ˉu)=infuMΦ(u)=infmaxuE{0}  t>0Φ(t2ut)  with  M:={uE{0}:I(u):=2Φ(u),uP(u)=0},

    where ut=ut(x)=u(tx), the definitions of Φ, E and P will be given in the next section.

    In the next section, we will construct a variational setting and give some preliminaries. In Section 3, we give the proof of Theorem 1.1. Throughout this paper, and \|\cdot\|_{s} denote the norms of H^{1}({ \mathbb{R}}^{2}) and L^{s}({ \mathbb{R}}^{2}) for 1\leq s\leq\infty , respectively. C_{i} are different positive constants in different places.

    The following bilinear forms are given as

    \begin{equation*} \label{2.1} (u, v)\mapsto A_{1}(u, v) = \frac{1}{2\pi}\int_{{ \mathbb{R}}^{2}}\int_{{ \mathbb{R}}^{2}}\ln (2+|x-y|)u(x)v(y)dxdy, \end{equation*}
    \begin{equation*} \label{2.2} (u, v)\mapsto A_{2}(u, v) = \frac{1}{2\pi}\int_{{ \mathbb{R}}^{2}}\int_{{ \mathbb{R}}^{2}}\ln \left(1+\frac{2}{|x-y|}\right)u(x)v(y)dxdy, \end{equation*}

    and

    \begin{equation*} \label{2.3} (u, v)\mapsto A_{0}(u, v) = A_{1}(u, v)-A_{2}(u, v) = \frac{1}{2\pi}\int_{{ \mathbb{R}}^{2}}\int_{{ \mathbb{R}}^{2}}\ln (|x-y|)u(x)v(y)dxdy, \end{equation*}

    where u, v:{ \mathbb{R}}^{2}\rightarrow { \mathbb{R}} are measurable functions. Since u, v are measurable functions, A_{1}(u, v) , A_{2}(u, v) and A_{0}(u, v) are well defined in Lebesgue sense. From the Hardy-Littlewood-Sobolev inequality [25] and 0\leq\ln(1+t)\leq t for t\geq0 , we have

    \begin{equation} |A_{2}(u, v)|\leq \frac{1}{\pi}\int_{{ \mathbb{R}}^{2}}\int_{{ \mathbb{R}}^{2}}\frac{1}{|x-y|}|u(x)v(y)|dxdy\leq C_{1}\|u\|_{4/3}\|v\|_{4/3}, \end{equation} (2.1)

    where C_{1} is a positive constant. In order to obtain the existence of ground state solution for system (1.1), we develop a new variational framework for system (1.1). The working function space is

    E: = X\cap H^{1}_{as} = \left\{u\in H^{1}_{as}({ \mathbb{R}}^{2}):\int_{{ \mathbb{R}}^{2}}[V(x)+\ln (2+|x|)]u^{2}(x)dx < \infty\right\},

    where

    X = \left\{u\in H^{1}({ \mathbb{R}}^{2}):\int_{{ \mathbb{R}}^{2}}[V(x)+\ln (2+|x|)]u^{2}(x)dx < \infty\right\},

    and

    H^{1}_{as} = \{u\in H^{1}({ \mathbb{R}}^{2}):u(x): = u(x_{1}, x_{2}) = u(|x_{1}|, |x_{2}|), \ \forall\ x\in{ \mathbb{R}}^{2}\}.

    Under (V1) and (K1), it is easy to see that the space E is a suitable constraint to study system (1.1). The norm of E is given by

    \begin{equation} \|u\|_{E}: = (\|u\|^{2}+\|u\|^{2}_{*})^{\frac{1}{2}}, \end{equation} (2.2)

    where

    \|u\|^{2} = \int_{{ \mathbb{R}}^{2}}[|\nabla u|^{2}+V(x)u^{2}(x)]dx, \ \ \forall\ u\in X,
    \|u\|^{2}_{*} = \int_{{ \mathbb{R}}^{2}}\ln (2+|x|)u^{2}(x)dx, \ \ \forall\ u\in X.

    The energy functional of system (1.1) on E is given by

    \begin{equation} \Phi(u) = \frac{1}{2}\int_{{ \mathbb{R}}^{2}}[|\nabla u|^{2}+V(x)u^{2}(x)]dx+\frac{1}{4}A_{0}(u^{2}, u^{2})-\int_{{ \mathbb{R}}^{2}}K(x)F(u)dx. \end{equation} (2.3)

    From (F1), (F2) and [1,(2.9)], we have \Phi\in C^{1}(X, { \mathbb{R}}) and the embedding X\hookrightarrow L^{s}({ \mathbb{R}}^{2}) is compact for s\in[2, \infty) , moreover,

    \begin{equation*} \label{2.7} \langle\Phi'(u), v\rangle = \int_{{ \mathbb{R}}^{2}}(\nabla u \nabla v+V(x)uv)dx+A_{0}(u^{2}, uv)-\int_{{ \mathbb{R}}^{2}}K(x)f(u)vdx. \end{equation*}

    Now, the Pohožaev functional associated to (1.1) is defined as follows:

    \begin{eqnarray*} \label{2.8} \mathcal {P}(u)&: = & \frac{1}{2}\int_{{ \mathbb{R}}^{2}}[\nabla V(x)x+2V(x)]u^{2}(x)dx-\int_{{ \mathbb{R}}^{2}}F(u)\nabla K(x)\cdot xdx\\ & &-2\int_{{ \mathbb{R}}^{2}}K(x)F(u)dx+A_{0}(u^{2}, u^{2})+\frac{1}{8\pi}\|u\|^{4}_{2}. \end{eqnarray*}

    Similar to [1], any solution u of (1.1) satisfies \mathcal {P}(u) = 0 . The following constraint is defined as:

    \begin{equation*} \label{2.9} \mathscr{M} = \{u\in E\backslash\{0\}:I(u) = 2\langle\Phi'(u), u\rangle-\mathcal {P}(u) = 0\}, \end{equation*}

    where

    \begin{eqnarray} I(u)& = &2\langle\Phi'(u), u\rangle-\mathcal {P}(u)\\ & = & 2\|\nabla u\|^{2}_{2}-\frac{1}{8\pi}\|u\|^{4}_{2}-2\int_{{ \mathbb{R}}^{2}}[f(u)u-F(u)]K(x)dx+\int_{{ \mathbb{R}}^{2}}F(u)\nabla K(x)\cdot xdx\\ & &+\frac{1}{2}\int_{{ \mathbb{R}}^{2}}[2V(x)-\nabla V(x) \cdot x]u^{2}(x)dx+A_{0}(u^{2}, u^{2}), \ \ \forall\ u\in E. \end{eqnarray} (2.4)

    Similar to [1,11], the following lemmas are obtained.

    Lemma 2.1. Assume that (V1), (K1), (F1) and (F2) hold. If u is a critical point of \Phi restricted to E , then u is a critical point of \Phi on X .

    Lemma 2.2. Assume that (V1) and (V2) hold. Then

    \begin{equation*} \label{2.11} A_{1}(u^{2}, v^{2})\geq \frac{1}{8\pi}\|u\|_{2}^{2}\|v\|_{*}^{2}, \ \ \forall\ u, v\in E, \end{equation*}

    and there exists a constant \gamma > 0 such that

    \begin{equation*} \label{2.12} \gamma\|u\|_{H^{1}}^{2}\leq 2\|\nabla u\|^{2}_{2}+\frac{1}{2}\int_{{ \mathbb{R}}^{2}}[2V(x)-\nabla V(x)\cdot x]u^{2}dx+A_{1}(u^{2}, u^{2}), \ \ \forall\ u, v\in E. \end{equation*}

    Lemma 2.3. Assume that (V1), (V2), (K1), (K2), (F1)–(F3) hold. Then for all t > 0 , u\in{ \mathbb{R}} and x\in{ \mathbb{R}}^{2} ,

    \begin{eqnarray} g(t, x, u)&: = &\frac{1}{t^{2}}F(t^{2}u)K(t^{-1}x)+\frac{1-t^{4}}{2}[f(u)u-F(u)]K(x)\\ & & -\frac{1-t^{4}}{4}F(u)\nabla K(x)\cdot x-F(u)K(x)\geq0. \end{eqnarray} (2.5)

    Lemma 2.4. Assume that (V2) holds. Then

    \begin{equation} \alpha(t, x): = (1+t^{4})V(x)+\frac{1-t^{4}}{2}\nabla V(x)\cdot x-2t^{2}V(t^{-1}x)\geq0, \ \ \forall\ x\in{ \mathbb{R}}^{2}, \ t > 0. \end{equation} (2.6)

    In this section, we first establish an energy estimate inequality related to \Phi(u) , \Phi(t^{2}u_{t}) and I(u) , where

    \begin{eqnarray} \Phi(t^{2}u_{t})& = &\frac{t^{4}}{2}\|\nabla u\|_{2}^{2}+\frac{1}{2}\int_{{ \mathbb{R}}^{2}}t^{2}V(t^{-1}x)u^{2}dx+\frac{t^{4}}{4}A_{0}(u^{2}, u^{2})\\ & & -\frac{t^{4}\ln t}{8\pi}\|u\|_{2}^{4}-\int_{{ \mathbb{R}}^{2}}\frac{1}{t^{2}}F(t^{2}u)K(t^{-1}x)dx, \ \ \forall\ u\in E, \ t > 0. \end{eqnarray} (3.1)

    Lemma 3.1. Assume that (V1), (V2), (K1), (K2) and (F1)–(F3) hold. Then

    \begin{equation} \Phi(u)\geq \Phi(t^{2}u_{t})+\frac{1-t^{4}}{4}I(u)+\frac{1-t^{4}+4t^4\ln t}{32\pi}\|u\|_{2}^{4}, \ \ \forall\ t > 0, \ u\in E, \end{equation} (3.2)

    and

    \begin{equation} \Phi(u)\geq \frac{1}{4}I(u)+\frac{1}{32\pi}\|u\|_{2}^{4}, \ \ \forall\ u\in E. \end{equation} (3.3)

    Proof. From (2.3), (2.5), (2.6) and (3.1), we have

    \begin{eqnarray} \Phi(u)-\Phi(t^{2}u_{t})& = & \frac{1-t^{4}}{2}\|\nabla u\|_{2}^{2}+\frac{1-t^{4}}{4}A_{0}(u^{2}, u^{2})+\frac{1}{2}\int_{{ \mathbb{R}}^{2}}[V(x)-t^{2}V(t^{-1}x)]u^{2}dx\\ & &+\frac{t^{4}\ln t}{8\pi}\|u\|_{2}^{4}+\int_{{ \mathbb{R}}^{2}}\frac{1}{t^{2}}F(t^{2}u)K(t^{-1}x)dx-\int_{{ \mathbb{R}}^{2}}F(u)K(x)dx\\ & = &\frac{1-t^{4}}{4}I(u)+\frac{1}{4}\int_{{ \mathbb{R}}^{2}}\left[(1+t^{4})V(x)+\frac{1-t^{4}}{2}\nabla V(x)\cdot x-2t^{2}V(t^{-1}x)\right]u^{2}dx\\ & & +\frac{1-t^{4}+4t^4\ln t}{32\pi}\|u\|_{2}^{4}+\int_{{ \mathbb{R}}^{2}}\left\{\frac{1}{t^{2}}F(t^{2}u)K(t^{-1}x)+\frac{1-t^{4}}{2}[f(u)u-F(u)]K(x)\right.\\ & &\left.-\frac{1-t^{4}}{4}F(u)\nabla K(x)\cdot x-F(u)K(x)\right\}dx\\ &\geq& \frac{1-t^{4}}{4}I(u)+\frac{1-t^{4}+4t^4\ln t}{32\pi}\|u\|_{2}^{4}, \ \ \forall\ t > 0, \ u\in E. \end{eqnarray} (3.4)

    Now, we prove that K is bounded in { \mathbb{R}}^2 . From (K1), we know that K\in C({ \mathbb{R}}^{2}, (0, \infty)) , then K is bounded in bounded domain. Now we need to prove that K is also bounded at infinity. From (K2), we have

    \begin{eqnarray} t\mapsto K(t^{-1}x)\ \ \text{is nondecreasing for}\ \ t > 0. \end{eqnarray} (3.5)

    For every fixed x\in{ \mathbb{R}}^2 , if letting t\rightarrow0 , then |t^{-1}x|\rightarrow \infty and if letting t\rightarrow \infty , then |t^{-1}x|\rightarrow0 . Hence, from (K1) and (3.5), we have K(\infty) = \lim_{t\rightarrow0}K(t^{-1}x)\leq\lim_{t\rightarrow \infty}K(t^{-1}x) = K(0) < \infty , which implies that K is also bounded at infinity. Hence, K is bounded in { \mathbb{R}}^2 , that is

    \begin{equation} K\in L^{\infty}({ \mathbb{R}}^2). \end{equation} (3.6)

    From (V2) and (2.6), we have

    \begin{equation} (1+t^{4})V(x)+\frac{1-t^{4}}{2}\nabla V(x)\cdot x\geq 2t^{2}V(t^{-1}x)\geq0, \ \ \forall\ t > 0, \ x\in { \mathbb{R}}^{2}. \end{equation} (3.7)

    Let t\rightarrow0 in (3.7), we obtain

    \begin{equation} 2V(x)+\nabla V(x)\cdot x\geq0, \ \ \forall\ x\in { \mathbb{R}}^{2}. \end{equation} (3.8)

    From (K1), (K2), (2.3), (2.4), (3.8) and (2.5) when t\rightarrow0 with the boundedness of K(x) , we get

    \begin{eqnarray} \Phi(u)-\frac{1}{4}I(u)& = &\frac{1}{32\pi}\|u\|_{2}^{4}+\frac{1}{8}\int_{{ \mathbb{R}}^{2}}[2V(x)+\nabla V(x)\cdot x]u^{2}dx\\ & &+\frac{1}{2}\int_{{ \mathbb{R}}^{2}}[f(u)u-3F(u)]K(x)dx-\frac{1}{4}\int_{{ \mathbb{R}}^{2}}F(u)\nabla K(x)\cdot xdx\\ &\geq&\frac{1}{32\pi}\|u\|_{2}^{4}, \ \ \forall\ u\in E. \end{eqnarray} (3.9)

    It follows from (3.4) and (3.9) that Lemma 3.1 holds.

    From (3.2) and the fact that 1+4t^{4}\ln t-t^{4}\geq0 for t > 0 , the following corollary is obtained.

    Corollary 3.2. Assume that (V1), (V2), (K1), (K2) and (F1)–(F3) hold. Then \Phi(u) = \max_{t > 0}\Phi(t^{2}u_{t}) for all u\in\mathscr{M} .

    Lemma 3.3. Assume that (V1), (V2), (K1), (K2) and (F1)–(F3) hold. Then for any u\in E\backslash\{0\} , there exists a constant t_{u} > 0 such that t_{u}^{2}u_{t_{u}}\in\mathscr{M} .

    Proof. Fix u\in E\backslash\{0\} and define \xi(t): = \Phi(t^{2}u_{t}) on (0, \infty) . From (2.4) and (3.1), one has

    \begin{eqnarray*} \label{3.10} \xi'(t) = 0 &\Leftrightarrow& 2t^{3}\|\nabla u\|^{2}_{2}+t^{3}A_{0}(u^{2}, u^{2})+\frac{t}{2}\int_{{ \mathbb{R}}^{2}}[2V(t^{-1}x)-\nabla V(t^{-1}x)\cdot t^{-1}x]u^{2}dx\nonumber\\ & & -\frac{4t^{3}\ln t+t^{3}}{8\pi}\|u\|^{4}_{2}+\frac{1}{t^{3}}\int_{{ \mathbb{R}}^{2}}F(t^{2}u)\nabla K(t^{-1}x)\cdot (t^{-1}x)dx\nonumber\\ & &-\frac{2}{t^{3}}\int_{{ \mathbb{R}}^{2}}[f(t^{2}u)t^{2}u-F(t^{2}u)]K(t^{-1}x)dx = 0\nonumber\\ &\Leftrightarrow& I(t^{2}u_{t}) = 0\Leftrightarrow t^{2}u_{t}\in \mathscr{M}, \ \ \forall\ t > 0. \end{eqnarray*}

    From (F1)–(F3), we have

    \begin{equation} \beta(t, u): = \frac{1-t^{4}}{2}f(u)u+\frac{t^{4}-3}{2}F(u)+\frac{1}{t^2}F(t^2u). \end{equation} (3.10)

    By (3.10), we obtain

    \begin{equation} \lim\limits_{t\rightarrow0}\beta(t, u) = \frac{1}{2}[f(u)u-3F(u)]\geq0. \end{equation} (3.11)

    From (3.11), we get

    \begin{equation} \frac{F(u)}{u^{3}}\ \ \text{is nondecreasing on}\ \ (-\infty, 0)\cup(0, +\infty). \end{equation} (3.12)

    From (3.12), we have

    \begin{eqnarray} \frac{F(t^2u)}{t^6} = \frac{F(t^2u)}{(t^2u)^3}\cdot u^3\leq \frac{F(u)}{u^3}\cdot u^3 = F(u), \ \ \text{for}\ 0 < t < 1, \end{eqnarray} (3.13)

    and

    \begin{eqnarray} \frac{F(t^2u)}{t^6} = \frac{F(t^2u)}{(t^2u)^3}\cdot u^3\geq \frac{F(u)}{u^3}\cdot u^3 = F(u), \ \ \text{for}\ t > 1. \end{eqnarray} (3.14)

    From (F3), we have

    \begin{eqnarray} \frac{f(t^2u)t^2u-F(t^2u)}{(t^2u)^3}\cdot u^3\leq \frac{f(u)u-F(u)}{u^3}\cdot u^3 = f(u)u-F(u), \ \ \text{for}\ 0 < t < 1, \end{eqnarray} (3.15)

    and

    \begin{eqnarray} \frac{f(t^2u)t^2u-F(t^2u)}{(t^2u)^3}\cdot u^3\geq \frac{f(u)u-F(u)}{u^3}\cdot u^3 = f(u)u-F(u), \ \ \text{for}\ t > 1. \end{eqnarray} (3.16)

    By (V2), we have

    \begin{equation} 2V(x)-\nabla V(x)x\geq0, \ \ \forall\ x\in{ \mathbb{R}}^{2}, \end{equation} (3.17)
    \begin{equation} t^{-2}[2V(t^{-1}x)-\nabla V(t^{-1}x)t^{-1}x]\geq 2V(x)-\nabla V(x)x, \ \ \forall\ 0 < t < 1, \ x\in{ \mathbb{R}}^{2}, \end{equation} (3.18)

    and

    \begin{equation} t^{-2}[2V(t^{-1}x)-\nabla V(t^{-1}x)t^{-1}x]\leq 2V(x)-\nabla V(x)x, \ \ \forall\ t > 1, \ x\in{ \mathbb{R}}^{2}. \end{equation} (3.19)

    By (K2), we obtain

    \begin{equation} -2K(x)\leq\nabla K(x)x\leq 2K(x), \ \ \forall\ x\in{ \mathbb{R}}^{2}. \end{equation} (3.20)

    It follows from (K1), (K2), (3.5), (3.6), (3.13), (3.15), (3.17), (3.18) and (3.20) that

    \begin{eqnarray} \frac{\xi'(t)}{t^{3}}& = & 2\|\nabla u\|^{2}_{2}+A_{0}(u^{2}, u^{2})+\frac{1}{2}\int_{{ \mathbb{R}}^{2}}t^{-2}[2V(t^{-1}x)-\nabla V(t^{-1}x)t^{-1}x]-\frac{4\ln t+1}{8\pi}\|u\|_{2}^{4}\\ &&+\frac{1}{t^6}\int_{{ \mathbb{R}}^{2}}F(t^2u)\nabla K(t^{-1}x)\cdot (t^{-1}x)dx-\frac{2}{t^6}\int_{{ \mathbb{R}}^{2}}[f(t^2u)t^2u-F(t^2u)]K(t^{-1}x)dx\\ &\geq&2\|\nabla u\|^{2}_{2}+A_{0}(u^{2}, u^{2})-\frac{4\ln t+1}{8\pi}\|u\|_{2}^{4}-2\int_{{ \mathbb{R}}^{2}}F(u)K(x)dx-2\int_{{ \mathbb{R}}^{2}}[f(u)u-F(u)]K(x)dx\\ & = &2\|\nabla u\|^{2}_{2}+A_{0}(u^{2}, u^{2})-\frac{4\ln t+1}{8\pi}\|u\|_{2}^{4}-2\int_{{ \mathbb{R}}^{2}}K(x)f(u)udx\\ &\geq& 2\|\nabla u\|^{2}_{2}+A_{0}(u^{2}, u^{2})-\frac{4\ln t+1}{8\pi}\|u\|_{2}^{4} -2\|K\|_{\infty}\int_{{ \mathbb{R}}^{2}}f(u)udx, \ \ \forall\ 0 < t < 1. \end{eqnarray} (3.21)

    From (K1), (K2), (3.5), (3.6), (3.11), (3.14), (3.16), (3.17), (3.19) and (3.20), we have

    \begin{eqnarray} \frac{\xi'(t)}{t^{3}}& = & 2\|\nabla u\|^{2}_{2}+A_{0}(u^{2}, u^{2})+\frac{1}{2}\int_{{ \mathbb{R}}^{2}}t^{-2}[2V(t^{-1}x)-\nabla V(t^{-1}x)t^{-1}x]-\frac{4\ln t+1}{8\pi}\|u\|_{2}^{4}\\ &&+\frac{1}{t^6}\int_{{ \mathbb{R}}^{2}}F(t^2u)\nabla K(t^{-1}x)\cdot (t^{-1}x)dx-\frac{2}{t^6}\int_{{ \mathbb{R}}^{2}}[f(t^2u)t^2u-F(t^2u)]K(t^{-1}x)dx\\ &\leq& 2\|\nabla u\|^{2}_{2}+A_{0}(u^{2}, u^{2})-\frac{4\ln t+1}{8\pi}\|u\|_{2}^{4}+\frac{1}{2}\int_{{ \mathbb{R}}^{2}}[2V(x)-\nabla V(x)x]u^{2}dx\\ &&+2\int_{{ \mathbb{R}}^{2}}F(u)K(t^{-1}x)dx-2\int_{{ \mathbb{R}}^{2}}[f(u)u-F(u)]K(x)dx\\ &\leq&2\|\nabla u\|^{2}_{2}+A_{0}(u^{2}, u^{2})+\frac{1}{2}\int_{{ \mathbb{R}}^{2}}[2V(x)-\nabla V(x)x]u^{2}dx\\ & & -\frac{4\ln t+1}{8\pi}\|u\|_{2}^{4} +2\|K\|_{\infty}\int_{{ \mathbb{R}}^{2}}F(u)dx, \ \ \forall\ t > 1. \end{eqnarray} (3.22)

    Then, from (3.21) and (3.22), for t\in(0, 1) small enough, one has \xi'(t) > 0 and for t > 1 large enough, \xi'(t) < 0 . Hence, there exists t_{u} > 0 such that \xi'(t_{u}) = 0 and t_{u}^{2}u_{t_{u}}\in\mathscr{M} .

    From Corollary 3.2 and Lemma 3.3, we get the following lemma.

    Lemma 3.4. Assume that (V1), (V2), (K1), (K2) and (F1)–(F3) hold. Then

    \begin{equation} \inf\limits_{u\in\mathscr{M}}\Phi(u): = c = \inf\limits_{u\in E\backslash\{0\}}\max\limits_{t > 0}\Phi(t^{2}u_{t}). \end{equation} (3.23)

    Lemma 3.5. Assume that (V1), (V2), (K1), (K2) and (F1)–(F3) hold. Then c = \inf_{u\in\mathscr{M}}\Phi(u) > 0 .

    Proof. By a standard argument, by (F1), (F2) and I(u) = 0 for u\in \mathscr{M} , one can easily show that there exists \sigma > 0 such that \|u\|_{H^{1}}\geq\sigma , \forall \ u\in \mathscr{M} . Let \{u_{n}\}\subset\mathscr{M} be such that \Phi(u_{n})\rightarrow c . We consider two cases:

    Case 1. \inf_{n\in{ \mathbb{N}}}\|u_{n}\|_{2}: = \sigma_{1} > 0 . From (3.3), we get

    \begin{equation*} \label{3.18} c+o(1) = \Phi(u_{n})\geq\Phi(t_{n}^2(u_{n})_{t_{n}})\geq\frac{1}{32\pi}\|u_{n}\|_{2}^{4}\geq\frac{1}{32\pi}\sigma_{1}^{4}. \end{equation*}

    Case 2. \inf_{n\in{ \mathbb{N}}}\|u_{n}\|_{2}: = 0 . Since \|u\|_{H^{1}}\geq\sigma for all u\in \mathscr{M} , passing to a subsequence, one obtains

    \begin{equation} \|u_{n}\|_{2}\rightarrow0, \ \ \|\nabla u_{n}\|_{2}\geq\frac{\sigma}{2}. \end{equation} (3.24)

    From (2.1) and the Gagliardo-Nirenberg inequality, we get

    \begin{equation} 0\leq A_{2}(u_{n}^{2}, u_{n}^{2})\leq C_{1}\|u_{n}\|_{8/3}^{4}\leq C_{2}\|u_{n}\|_{2}^{3}\|\nabla u_{n}\|_{2}, \ \ \|u_{n}\|_{p}^{p}\leq C_{3}\|u_{n}\|_{2}^{2}\|\nabla u_{n}\|_{2}^{p-2}. \end{equation} (3.25)

    From (3.24), we have

    \begin{equation} \frac{|\ln(\|\nabla u_{n}\|_{2})|}{\|\nabla u_{n}\|_{2}^{2}}\leq C_{4}. \end{equation} (3.26)

    From (F1) and (F2), there exists a positive constant C_{5} such that

    \begin{equation} |F(t_{n}^{2}u_{n})|\leq |t_{n}^{2}u_{n}|^2+C_{5}|t_{n}^{2}u_{n}|^p, \ \ \forall\ t_{n} > 0, \ u_{n}\in{ \mathbb{R}}. \end{equation} (3.27)

    Let t_{n} = \|\nabla u_{n}\|_{2}^{-1/2} . Since I(u_{n}) = 0 and the boundedness of K , from (3.1), (3.24), (3.25), (3.26), (3.27) and Corollary 3.2, we have

    \begin{eqnarray*} \label{3.22} c+o(1)& = &\Phi(u_{n})\geq\Phi(t^{2}_{n}(u_{n})_{t_{n}})\nonumber\\ & = &\frac{t_{n}^{4}}{2}\|\nabla u_{n}\|_{2}^{2}+\frac{t_{n}^{4}}{4}[A_{1}(u_{n}^{2}, u_{n}^{2})-A_{2}(u_{n}^{2}, u_{n}^{2})]-\frac{t_{n}^{4}\ln t_{n}}{8\pi}\|u_{n}\|_{2}^{4}\nonumber\\ & &+\frac{t_{n}^{2}}{2}\int_{{ \mathbb{R}}^{2}}V(t_{n}^{-1}x)u_{n}^{2}dx-\frac{1}{t_{n}^{2}}\int_{{ \mathbb{R}}^{2}}K(t_{n}^{-1}x)F(t_{n}^{2}u_{n})dx\nonumber\\ &\geq&\frac{t_{n}^{4}}{2}\|\nabla u_{n}\|_{2}^{2}-\frac{t_{n}^{4}}{4}A_{2}(u_{n}^{2}, u_{n}^{2})-\frac{t_{n}^{4}\ln t_{n}}{8\pi}\|u_{n}\|_{2}^{4}-\frac{\|K\|_{\infty}}{t_{n}^{2}}\int_{{ \mathbb{R}}^{2}}[|t_{n}^{2}u_{n}|^{2}+C_{5}|t_{n}^{2}u_{n}|^{p}]dx\nonumber\\ &\geq&\frac{t_{n}^{4}}{2}\|\nabla u_{n}\|_{2}^{2}-\frac{C_{2}t_{n}^{4}}{4}\|u_{n}\|_{2}^{3}\|\nabla u_{n}\|_{2}-\frac{t_{n}^{4}\ln t_{n}}{8\pi}\|u_{n}\|_{2}^{4}\nonumber\\ & &-t_{n}^{2}\|K\|_{\infty}\|u_{n}\|^{2}_{2}-C_{6}\|K\|_{\infty}t_{n}^{2p-2}\|u_{n}\|^{2}_{2}\|\nabla u_{n}\|^{p-2}_{2}\nonumber\\ & = &\frac{1}{2}-\frac{C_{2}\|u_{n}\|^{3}_{2}}{4\|\nabla u_{n}\|_{2}}+\frac{\ln(\|\nabla u_{n}\|_{2})}{16\pi\|\nabla u_{n}\|_{2}^{2}}\|u_{n}\|^{4}_{2}-\frac{\|K\|_{\infty}\|u_{n}\|^{2}_{2}}{\|\nabla u_{n}\|_{2}}-\frac{C_{6}\|K\|_{\infty}\|u_{n}\|^{2}_{2}}{\|\nabla u_{n}\|_{2}}\nonumber\\ & = &\frac{1}{2}+o(1). \end{eqnarray*}

    It follows from the above two cases that c = \inf_{u\in\mathscr{M}}\Phi(u) > 0 .

    Lemma 3.6. Assume that (V1), (V2), (K1), (K2) and (F1)–(F3) hold. Then c is achieved. Moreover, if \bar{u}\in \mathscr{M} and \Phi(\bar{u}) = c , then \bar{u} is a critical point of \Phi in E .

    Proof. Let \{u_{n}\}\subset\mathscr{M} be such that \Phi(u_{n})\rightarrow c . From (3.9) and I(u_{n}) = 0 , it yields

    \begin{equation} c+o(1) = \Phi(u_{n})-\frac{1}{4}I(u_{n})\geq\frac{1}{32\pi}\|u_{n}\|_{2}^{4}. \end{equation} (3.28)

    From (3.28), we know that \{\|u_{n}\|_{2}\} is bounded. It is needed to prove that \{\|\nabla u_{n}\|_{2}\} is bounded too. Arguing by indirectly, assume that \|\nabla u_{n}\|_{2}\rightarrow \infty as n\rightarrow \infty . Let t_{n} = (2\sqrt{c}/\|\nabla u_{n}\|_{2})^{1/2} , then t_{n}\rightarrow0 as n\rightarrow \infty . Hence, t_{n}^{4}\ln t_{n}\rightarrow0 as n\rightarrow \infty . From (K1), (F1), (F2), (2.1), (3.1), (3.25), (3.27), the boundedness of K and Corollary 3.2, we have

    \begin{eqnarray} c+o(1)& = &\Phi(u_{n})\geq\Phi(t^{2}_{n}(u_{n})_{t_{n}})\\ & = &\frac{t_{n}^{4}}{2}\|\nabla u_{n}\|_{2}^{2}+\frac{t_{n}^{4}}{4}[A_{1}(u_{n}^{2}, u_{n}^{2})-A_{2}(u_{n}^{2}, u_{n}^{2})]-\frac{t_{n}^{4}\ln t_{n}}{8\pi}\|u_{n}\|_{2}^{4}\\ & &+\frac{t_{n}^{2}}{2}\int_{{ \mathbb{R}}^{2}}V(t_{n}^{-1}x)u_{n}^{2}dx-\frac{1}{t_{n}^{2}}\int_{{ \mathbb{R}}^{2}}K(t_{n}^{-1}x)F(t_{n}^{2}u_{n})dx\\ &\geq&\frac{t_{n}^{4}}{2}\|\nabla u_{n}\|_{2}^{2}-\frac{t_{n}^{4}}{4}A_{2}(u_{n}^{2}, u_{n}^{2})-\frac{\|K\|_{\infty}}{t_{n}^{2}}\int_{{ \mathbb{R}}^{2}}F(t_{n}^{2}u_{n})dx+o(1)\\ &\geq&\frac{t_{n}^{4}}{2}\|\nabla u_{n}\|_{2}^{2}-\frac{t_{n}^{4}}{4}A_{2}(u_{n}^{2}, u_{n}^{2})-\frac{\|K\|_{\infty}}{t_{n}^{2}}\int_{{ \mathbb{R}}^{2}}[|t_{n}^{2}u_{n}|^{2}+C_{5}|t_{n}^{2}u_{n}|^{p}]dx+o(1)\\ &\geq&\frac{t^{4}}{2}\|\nabla u_{n}\|_{2}^{2}-\frac{C_{2}t_{n}^{4}}{4}\|u_{n}\|_{2}^{3}\|\nabla u_{n}\|_{2} -t_{n}^{2}\|K\|_{\infty}\|u_{n}\|^{2}_{2}\\ & &-C_{6}\|K\|_{\infty}t_{n}^{2p-2}\|u_{n}\|^{2}_{2}\|\nabla u_{n}\|^{p-2}_{2}+o(1)\\ & = &2c-\frac{C_{2}c\|u_{n}\|^{3}_{2}}{\|\nabla u_{n}\|_{2}}-\frac{C_{3}(2\sqrt{c})^{p-1}\|K\|_{\infty}\|u_{n}\|^{2}_{2}}{\|\nabla u_{n}\|_{2}}+o(1)\\ & = &2c+o(1), \end{eqnarray} (3.29)

    a contradiction, hence, we have that \{\|\nabla u_{n}\|_{2}\} is bounded too, so \{u_{n}\} is bounded in H^{1}({ \mathbb{R}}^{2}) . Thanks to (K1), (F1), (F2), (2.1) and (2.2), we know that \{\|u_{n}\|\} and A_{1}(u_{n}^{2}, u_{n}^{2}) are both bounded. From [1,Lemma 3.5], one has

    \begin{equation} \limsup\limits_{n\rightarrow \infty}\|u_{n}\|_{2} > 0, \end{equation} (3.30)

    which together with Lemma 2.2 shows that \{\|u_{n}\|_{*}\} is bounded. Then \{u_{n}\} is bounded in E . Passing to a subsequence, one may assume that u_{n}\rightharpoonup \bar{u} in E , u_{n}\rightarrow \bar{u} in L^{s}({ \mathbb{R}}^{2}) for s\in[2, \infty) , u_{n}\rightarrow \bar{u} a.e. on { \mathbb{R}}^{2} . By a standard argument, we have

    \begin{equation} I(\bar{u})\leq\liminf\limits_{n\rightarrow \infty}I(u_{n}) = 0. \end{equation} (3.31)

    It follows from (3.30) and (3.31) that \bar{u}\neq0 . From Lemma 3.3, there exists \bar{t} > 0 such that \bar{t}^{2}\bar{u}_{\bar{t}}\in \mathscr{M} and \Phi(\bar{t}^{2}\bar{u}_{\bar{t}})\geq c . Hence, by (2.2), (2.4), (3.30), Fatou's Lemma, Lebesgue's dominated convergence theorem and the fact 1+4t^{4}\ln t-t^{4}\geq0 for t > 0 , we obtain

    \begin{eqnarray} c& = &\lim\limits_{n\rightarrow \infty}\left[\Phi(u_{n})-\frac{1}{4}I(u_{n})\right]\\ & = &\lim\limits_{n\rightarrow \infty}\left\{\frac{1}{32\pi}\|u_{n}\|_{4}^{2}+\frac{1}{8}\int_{{ \mathbb{R}}^{2}}[2V(x)+\nabla V(x)\cdot x]u_{n}^{2}dx\right.\\ & &+\left.\frac{1}{2}\int_{{ \mathbb{R}}^{2}}\{[f(u_{n})u_{n}-3F(u_{n})]K(x)-F(u_{n})\nabla K(x)\cdot x\}dx\right\}\\ &\geq&\frac{1}{32\pi}\|\bar{u}\|_{4}^{2}+\frac{1}{8}\int_{{ \mathbb{R}}^{2}}[2V(x)+\nabla V(x)\cdot x]\bar{u}^{2}dx\\ & &+\frac{1}{2}\int_{{ \mathbb{R}}^{2}}[f(\bar{u})\bar{u}-3F(\bar{u})]K(x)dx-\frac{1}{4}\int_{{ \mathbb{R}}^{2}}F(\bar{u})\nabla K(x)\cdot xdx\\ & = &\Phi(\bar{u})-\frac{1}{4}I(\bar{u}) \geq\Phi(\bar{t}^{2}\bar{u}_{\bar{t}})-\frac{\bar{t}^{4}}{4}I(\bar{u}) \geq c-\frac{\bar{t}^{4}}{4}I(\bar{u})\geq c. \end{eqnarray} (3.32)

    From (3.32), we have I(\bar{u}) = 0 and \Phi(\bar{u}) = c . Similar to [1,Lemma 4.11], we can obtain that \bar{u} is a critical point of \Phi in E .

    From Lemmas 2.1, 3.4 and 3.6, it is easy to get Theorem 1.1. The proof is complete.

    This paper studies a kind of Schrödinger-Poisson system in { \mathbb{R}}^{2} . By constructing a new variational framework and using some new analytic techniques, we obtain an axially symmetric solution. Specially speaking, we first establish an energy estimate inequality related to \Phi(u) , \Phi(t^{2}u_{t}) and I(u) , which is useful in the proof of the main result. Secondly, we prove that for any u\in E\backslash\{0\} , there exists a constant t_{u} > 0 such that t_{u}^{2}u_{t_{u}}\in\mathscr{M} , and then prove \inf_{u\in\mathscr{M}}\Phi(u): = c = \inf_{u\in E\backslash\{0\}}\max_{t > 0}\Phi(t^{2}u_{t}) and c = \inf_{u\in\mathscr{M}}\Phi(u) > 0 . Finally, we prove that c is achieved. Moreover, if \bar{u}\in \mathscr{M} and \Phi(\bar{u}) = c , then \bar{u} is a critical point of \Phi in E . Our result improves and extends some existing works.

    This work is supported by the National Natural Science Foundation of China (No. 11961014, No. 61563013) and Guangxi Natural Science Foundation (2018GXNSFAA281021, 2018GXNSFBA281019, 2019GXNSFAA245031). The authors would like to thank the editors and referees for their useful suggestions and comments.

    All authors declare no conflict of interest in this paper.



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