## The Existence of Solutions for the Schrödinger-Chern-Simons-Higgs System

Deng Jin,

College of Mathematics and Statistics, Jiangxi Normal University, Nanchang 330022

 基金资助: 江西省教育厅研究生创新基金.  YJS2020053

 Fund supported: Postgraduate Innovation Fund of Educational Department in Jiangxi Province.  YJS2020053

Abstract

In this paper, we study the existence standing wave solutions for a nonlinear Schrödinger equation coupled with a neutral scalar field and a gauge field. We establish the existence result for the case that the exponent p of nonlinear term is greater than 2.

Keywords： Schrödinger equations ; Neutral scalar field ; Chern-Simons

Deng Jin. The Existence of Solutions for the Schrödinger-Chern-Simons-Higgs System. Acta Mathematica Scientia[J], 2021, 41(6): 1768-1778 doi:

## 1 引言

$$$\left\{\begin{array}{ll} { } -\frac{1}{2 m} \Delta u+\omega u+\frac{e^{4}}{2 m \kappa^{2}} \frac{h_{u}^{2} (|x|)}{|x|^{2}} u+\frac{e^{4}}{m \kappa^{2}} u \int_{|x|}^{\infty} \frac{u^{2}(s)}{s} h_{u}(s)\, {\rm d}s \\ { }{}{} +\left(1+\frac{\kappa q}{2 m}\right) N u+\frac{q}{ p m^{2}} |u|^{p-2}u = 0, & x\in {{\Bbb R}} ^{2}, \\ { } -\Delta N+\kappa^{2} q^{2} N+q\left(1+\frac{\kappa q}{2 m}\right) u^{2} = 0, & x\in{{\Bbb R}} ^{2}, \end{array}\right.$$$

$u: {{\Bbb R}} ^{2} \rightarrow {{\Bbb R}}$是径向对称的. 问题(1.1) 源于对$(2 + 1)$ -维Abelian Higgs模型的研究, 在该模型中, 带电的非相对论性物质场与仅包含Chern-Simons项的质量规范场在规范能量中相互作用, 详见文献[5]. 在物理上, 参数$m$, $\kappa $$q > 0 分别表示粒子的质量, Chern-Simons耦合常数和麦克斯韦耦合常数. 根据文献[5] 中的研究, (2 + 1) 维相对论拉格朗日密度 可以简化为拉格朗日密度 \begin{eqnarray} {\cal L}\left(\psi, N, A_{0}, A_{1}, A_{2}\right)& = &{\rm i} \bar{\psi} D_{t} \psi-\frac{1}{2 m}|{\bf D} \psi|^{2}+\frac{1}{2 q} \partial_{\alpha} N \partial^{\alpha} N-\frac{1}{2} \kappa^{2} q N^{2}-\frac{q}{p^2 m^{2} c^{2}}|\psi|^{p} {}\\ &&-\left(1+\frac{\kappa q}{2 m c}\right)|\psi|^{2} N+\frac{\kappa}{4} \varepsilon^{\mu \alpha \beta} A_{\mu} F_{\alpha \beta}, \end{eqnarray} 其中 \psi: {{\Bbb R}} \times {{\Bbb R}} ^{2} \rightarrow {\Bbb C} 是带电标量场, D_{t} = \partial_{t}+{\rm i} e A_{0}, \, {\bf D} = \nabla-{\rm i} e {\bf A},$$ \left(A_{0}, \, {\bf A}\right) = \left(A_{0}, -A_{1}, -A_{2}\right)$是规范场, $e>0$是耦合常数. 此外$F_{\mu \nu}: = \partial_{\mu} A_{v}-\partial_{\nu} A_{\mu}, $$\varepsilon^{\mu \nu \alpha} 是Levi-Civita张量. 对于静态驻波解 \psi(t, x) = e^{{\rm i}\omega t}u(x) , 即 A_\mu = A_\mu(x), \, N = N(x) , 假设 {\bf A} 是库仑规范的, 则 \partial_{1} A_{1}+\partial_{2} A_{2} = 0, 因此欧拉- 拉格朗日方程 {\cal L} 可以写成: $$-\frac{1}{2 m} \Delta u+\left[\frac{e^{2}}{2 m}|{\bf A}|^{2}+\left(\omega+e A_{0}\right)+\left(1+\frac{\kappa q}{2 m}\right) N\right] u+\frac{q}{p m^{2}} |u|^{p-2}u = 0,$$ $${\bf A} \cdot \nabla\left(u^{2}\right) = 0,$$ $$-\Delta N+\kappa^{2} q^{2} N+q\left(1+\frac{\kappa q}{2 m}\right) u^{2} = 0,$$ $$\kappa\left(\partial_{1} A_{2}-\partial_{2} A_{1}\right) = e u^{2},$$ $$\kappa m \partial_{2} A_{0} = e^{2} A_{1} u^{2},$$ $$\kappa m \partial_{1} A_{0} = -e^{2} A_{2} u^{2}.$$ p>2 时, 本文考虑问题(1.3)–(1.8) 径向解的存在性. 假设 A_0(x) = A_0(|x|) , $$A_{1}(x) = \frac{e}{\kappa} \frac{x_{2}}{|x|^{2}} h(|x|), \quad A_{2}(x) = -\frac{e}{\kappa} \frac{x_{1}}{|x|^{2}} h(|x|).$$ 如文献[1] 有 $$h(x) = h_u(x) = \int^{|x|}_0su^2(s)\, {\rm d}s.$$ 因此可以得到 $$A_0(x) = \frac{e^3}{mk^2} \int^\infty_{|x|}\frac{u^2(s)}sh_u(s)\, {\rm d}s,$$ 从而, 问题(1.3)–(1.8) 可以简化为问题(1.1). 如果 N = 0 , 问题(1.3)–(1.8) 是Chern-Simons薛定谔方程, 此类方程得到了广泛的研究. 已有各种存在性和不存在性结果. 特别地, p 的取值范围不同, 所得的结果不一样. 文献[3] 中, 对于 p> 4 的情况, 通过研究Nehari-Pohozaev流形上的极小化问题, 证明了解的存在性, 当 2<p<4 时, 极小化问题是限制在 L^2 球上考虑的. 而对于 p = 4 的情况, 通过研究刘维尔方程得到了自对偶的解, 详见文献[8] 和[9]. 进一步的结果可以参考文献[2-4, 6, 14-18] 及其中的参考文献. N\not\equiv0 的情况下, 文献[7] 中考虑了 e = 0 时问题(1.3)–(1.8) 的非平凡解的存在性, 以及当 q\rightarrow \infty 时的Chern-Simons极限. 当 p = 4 时, 文献[1] 中得到了进一步的存在性结果. 该文使用截断泛函的方法证明Palaiss-Smale序列的有界性, 且当 p = 4 时, 通过找相关泛函的临界点, 证明了问题(1.3)–(1.8) 解的存在性. 受这些工作的启发, 我们研究 p>2 时问题(1.1) 解的存在性. 我们以不同的方式处理 p>4$$ 2<p\leq 4$的情况. 对于$p>4$, 我们得到以下结果.

$$$-\Delta N+\kappa^{2} q^{2} N+q\left(1+\frac{\kappa q}{2 m}\right) u^{2} = 0, \quad x\in {{\Bbb R}} ^{2}.$$$

$$$N_u(x) = -q\left(1+\frac{\kappa q}{2 m}\right)\int_{{{\Bbb R}} ^{2}}G^{\kappa q}(x-y) u^{2}(y)\, {\rm d}y,$$$

$$$G^\mu(x) = \int_0^\infty\frac 1{4\pi t}e^{-(\frac{|x|^2}{4t}+\mu^2t)}\, {\rm d}y.$$$

$$$\|\nabla N\|^2_{L^2}+ \kappa^2q^2\|N\|_{L^2}^2 = -q\left(1+\frac{\kappa q}{2 m}\right)\int_{{{\Bbb R}} ^{2}} N u^{2}\, {\rm d}x.$$$

$\begin{eqnarray} J_e(u) = {{\cal E}_e}(u, N)& = &\int_{{{\Bbb R}} ^{2}} \Big(\frac{1}{4 m}|\nabla u|^{2}+\frac{w}{2}|u|^{2}\Big)\, {\rm d}x+\frac{e^{4}}{4 m \kappa^{2}} \int_{{{\Bbb R}} ^{2}} \frac{u^{2}(|x|) h_{u}^{2}(|x|)}{|x|^{2}}\, {\rm d}x{} \\ &&+\frac{1}{4}\left(1+\frac{\kappa q}{2 m}\right) \int_{{{\Bbb R}} ^{2}} N u^{2}\, {\rm d}x+\frac{q}{p^2 m^{2}}\int_{{{\Bbb R}} ^{2}}|u|^{p}\, {\rm d}x. \end{eqnarray}$

$$$|u(x)| \leq \frac{C}{\sqrt{|x|}}\|u\|_{H^{1}\left({{\Bbb R}} ^{2}\right)}, \quad u \in H_{r}^{1}\left({{\Bbb R}} ^{2}\right).$$$

$$$\int_{{{\Bbb R}} ^{2}} \frac{u^{2}(x) h_{u}^{2}(x)}{|x|^{2}}\, {\rm d}x \leq C\|u\|^{6}$$$

$\rm (ii)$映射$u \mapsto N_{u} $$C^{1} 的; \rm (iii)$$ \left|\int_{{{\Bbb R}} ^{2}} N_{u} u^{2}\, {\rm d}x \right| = -\int_{{{\Bbb R}} ^{2}} N_{u} u^{2}\, {\rm d}x \leq \frac{1}{\kappa^{2} q}\left(1+\frac{\kappa q}{2 m}\right)\|u\|_{L^4}^{4}$;

$\rm (iv) $$N_{t u} = t^{2} N_{u}, 对于任意 t \in {{\Bbb R}} ; \rm (v)$$ u \in H_{r}^{1}\left({{\Bbb R}} ^{2}\right)$, 则$N_{u} \in H_{r}^{1}\left({{\Bbb R}} ^{2}\right)$.

我们仅证明当$n\to\infty $$I(u_n)\to I(u) , 其它项可以同样地证明. 因为 由(1.13) 式, 富比尼定理和引理2.2, 有 其中 r>2, s = \frac r{r-2} . 由文献[7] 中的引理4.1, 对于 p\geq 1 又因为 H_r^1( {{\Bbb R}} ^2 ) 紧嵌入 L^p( {{\Bbb R}} ^2 ) , p \geq 2 , 故有 u_{n}$$ L^{p}( {{\Bbb R}} ^2 )$强收敛到$u$, 因此, 当$n\to\infty$

$\left\{u_{n}\right\}$包含一个收敛子列.

因为$\{u_{n}\} $$H_r^1({{\Bbb R}} ^2) 中有界, 且 H_r^1( {{\Bbb R}} ^2 ) 紧嵌入 L^p( {{\Bbb R}} ^2 ) , 我们可以假设 $$u_{n} \rightharpoonup u \quad \mbox{ in} \quad H_r^1({{\Bbb R}} ^2); \quad u_{n} \rightarrow u \quad \mbox{in} \quad L^{p}\left({{\Bbb R}} ^{2}\right), \, \, p \geq 2; \quad u_{n} \rightarrow u \quad \mbox{ a.e. } \quad {{\Bbb R}} ^{2}.$$ 由命题2.1和命题2.2, 可以得到 u 是泛函 J_e 的临界点, 即 J'_e(u) = 0 . 因此, 由(2.6) 式可以得到 \begin{eqnarray} &&\int_{{{\Bbb R}} ^{2}}\Big(\frac{1}{2m}|\nabla (u_n-u)|^2+\omega|u_n-u|^2\Big)\, {\rm d}x{}\\ & = &\langle J_e'(u_n)-J'_e(u), u_n-u\rangle +o(1){}\\ & = &\frac{3e^4}{2m\kappa^2}\langle H'(u_n)-H'(u), u_n-u\rangle +(1+\frac{\kappa q}{2m})\langle I'(u_n)-I'(u), u_n-u\rangle+o(1). \end{eqnarray} 然后再由命题2.1和2.2得到结果. 命题2.3证毕. ## 3 当 p>4 时解的存在性 在本节中, 我们考虑了 p>4 情况下问题(1.1) 解的存在性. 引理3.1 如果 \left\{u_{n}\right\} \subset H_r^1({{\Bbb R}} ^{2}) 使得 J_{e}\left(u_{n}\right) \leq c , 则 \left\{u_{n}\right\}$$ H_r^1({{\Bbb R}} ^{2})$是一致有界的.

因为$p>4$, 当$t\geq 0$时, 我们有$t^4\leq C_\varepsilon t^2 + \varepsilon t^p$, 其中$\varepsilon>0$待定. 由引理2.2, 有

$v_{n}(x) = \frac{u_{n}(x)}{\left\|u_{n}\right\|_{L^{2}\left({{\Bbb R}} ^{2}\right)}}$, 则$v_{n}(x)$满足

$\begin{eqnarray} &&C \|v_n\|^2+\frac{q}{2 p^2 m^{2}}\|u_{n}\|^{p-2}_{L^2} \int_{{{\Bbb R}} ^{2}}\left|v_{n}\right|^{p}\, {\rm d}x +\frac{e^{4}}{4 m \kappa^{2}} \|u_n\|^4_{L^2}\int_{{{\Bbb R}} ^{2}} \frac{v_{n}^{2}(x) h_{v_n}^2(|x|)}{|x|^{2}}\, {\rm d}x{}\\ &\leq &\frac{c}{\|u_n\|^2_{L^2}}+C \int_{{{\Bbb R}} ^{2}}\left|v_{n}\right|^{2}\, {\rm d}x, \end{eqnarray}$

$\int_{{{\Bbb R}} ^{2}} v^{2}\, {\rm d}x = 1$. 由命题2.1可得

$\begin{eqnarray} &&\lim\limits_{n\to\infty}\int_{{{\Bbb R}} ^{2}}\left(A_{1}^{2}\left(v_{n}\right) v_{n}^{2}+A_{2}^{2}\left(v_{n}\right) v_{n}^{2}\right) \, {\rm d}x {}\\ & = & \lim\limits_{n\to\infty}\int_{{{\Bbb R}} ^{2}} \frac{e^{2}}{4 m \kappa^{2}} \frac{h_{v_n}^{2}(|x|) v_{n}^{2}(x)}{|x|^{2}} {\rm d}x = \int_{{{\Bbb R}} ^{2}} \frac{e^{2}}{4 m \kappa^{2}} \frac{h_{v}^{2}(|x|) v^{2}(x)}{|x|^{2}}\, {\rm d}x. \end{eqnarray}$

$$$\lim\limits_{n\to\infty}\int_{{{\Bbb R}} ^{2}}\left(A_{1}^{2}\left(v_{n}\right) v_{n}^{2}+A_{2}^{2}\left(v_{n}\right) v_{n}^{2}\right) \, {\rm d}x = 0.$$$

$$$\int_{{{\Bbb R}} ^{2}}\left(A_{1}^{2}\left(v\right) v^{2}+A_{2}^{2}\left(v\right) v^{2}\right) \, {\rm d}x = 0.$$$

$$$0 = \int_{{{\Bbb R}} ^{2}} \partial_{2}\left(A_{1}(v) v^{2}\right)\, {\rm d}x = \int_{{{\Bbb R}} ^{2}} \partial_{2} A_{1}(v) v^{2}+A_{1} \partial_{2} v^{2}\, {\rm d}x,$$$

$$$0 = \int_{{{\Bbb R}} ^{2}} \partial_{1}\left(A_{2}(v) v^{2}\right)\, {\rm d}x = \int_{{{\Bbb R}} ^{2}} \partial_{1} A_{2}(v) v^{2}+A_{2} \partial_{1} v^{2}\, {\rm d}x.$$$

$\begin{eqnarray} \int_{{{\Bbb R}} ^{2}}|v|^{4}\, {\rm d}x& = &\frac{\kappa}{e} \int_{{{\Bbb R}} ^{2}}\left(\partial_{1} A_{2}(v)-\partial_{2} A_{1}(v)\right) v^{2}\, {\rm d}x {}\\ & = &\frac{2\kappa}{e} \int_{{{\Bbb R}} ^{2}}\left(v A_{2}(v) \partial_{1} v-v A_{1} \partial_{2} v\right)\, {\rm d}x = 0. \end{eqnarray}$

## 4 当$2<p<4$时解的存在性

我们用反证法: 假设对于任意$T>0$, 存在$e>0$使得

$$$\begin{array}{ll} \limsup _{n}\left\|u_{n}\right\| \geq T. \end{array}$$$

$\begin{eqnarray} C\|u_n\|^2+o(1)\|u_n\|&\leq &c_{e, T}+\frac{e^4}{2m\kappa^2}\Big(\frac{3}{p}-\frac{1}{2}\Big) K_T\int_{{{\Bbb R}} ^{2}}\frac{u^2(|x|)h^2_u(|x|)}{|x|}\, {\rm d}x{}\\ &&+\frac{e^4}{2pm\kappa^2}K_T^{\prime}\|u_n\|^2\int_{{{\Bbb R}} ^{2}}\frac{u^2(|x|)h^2_u(|x|)}{|x|^2}\, {\rm d}x. \end{eqnarray}$

$t\leq\frac{\sqrt{2}T}{\|\bar u\|}$, 则有

$t\geq\frac{\sqrt{2}T}{\|\bar u\|}$, $A_2(T) = 0$. 因此

$$$c_{e, T} \leq A_{1}+C e^{4} T^{6}.$$$

$$$\frac{e^{4}}{2 m \kappa^{2}}\left(\frac{3}{p}-\frac{1}{2}\right) K_{T}\left(u_{n}\right) \int_{{{\Bbb R}} ^{2}} \frac{u_{n}^{2}(|x|) h^2_{u_n}(|x|)}{|x|^{2}}\, {\rm d}x \leqslant Ce^{4} K_{T}\left(u_{n}\right)\left\|u_{n}\right\|^{6} \leqslant e^{4} T^{6}.$$$

$$$\frac{e^{4}}{2 p m \kappa^{2} T^{2}} K_{T}^{\prime}\left(u_{n}\right)\left\|u_{n}\right\|^{2} \int_{{{\Bbb R}} ^{2}} \frac{u_{n}^{2}\left(|x|\right) h^{2}_{u_{n}}(|x| )}{|x|^{2}}\, {\rm d}x \leqslant C e^{4} T^{6}.$$$

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