本文研究了以下含有中性标量场及规范位势的非线性薛定谔方程组

(1.1) $ \begin{equation} \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. \end{equation} $

其中$ N $ : $ {{\Bbb R}} ^{2}\rightarrow {{\Bbb R}} $是中性标量场, $ p>2 $且

$ h_{u}(|x|) = \int_{0}^{|x|} s u^{2}(s)\, {\rm d}s, $

$ 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 $和$ 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

$ \begin{eqnarray*} {\cal L}^R\left(\phi, N, A_{0}, A_{1}, A_{2}\right)& = & \overline{D_{\mu} \phi} D^{\mu} \phi+\frac{1}{2 q} \partial_{\mu} N \partial^{\mu} N-\frac{1}{c^{2}}|\phi|^{2}\left(N+\frac{1}{\kappa c} v^{2}\right)^{2}\\ & &-\frac{q}{2 c^{2}}\left(|\phi|^{2}+\kappa c N\right)^{2} -\frac{1}{4 q} F_{\mu \nu} F^{\mu \nu}+\frac{\kappa}{4} \varepsilon^{\mu \nu \alpha} F_{\mu \nu} A_{\alpha} \end{eqnarray*} $

可以简化为拉格朗日密度

(1.2) $ \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:

(1.3) $ \begin{equation} -\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, \end{equation} $

(1.4) $ \begin{equation} {\bf A} \cdot \nabla\left(u^{2}\right) = 0, \end{equation} $

(1.5) $ \begin{equation} -\Delta N+\kappa^{2} q^{2} N+q\left(1+\frac{\kappa q}{2 m}\right) u^{2} = 0, \end{equation} $

(1.6) $ \begin{equation} \kappa\left(\partial_{1} A_{2}-\partial_{2} A_{1}\right) = e u^{2}, \end{equation} $

(1.7) $ \begin{equation} \kappa m \partial_{2} A_{0} = e^{2} A_{1} u^{2}, \end{equation} $

(1.8) $ \begin{equation} \kappa m \partial_{1} A_{0} = -e^{2} A_{2} u^{2}. \end{equation} $

当$ 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|) $,

(1.9) $ \begin{equation} 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|). \end{equation} $

如文献[1] 有

(1.10) $ \begin{equation} h(x) = h_u(x) = \int^{|x|}_0su^2(s)\, {\rm d}s. \end{equation} $

因此可以得到

(1.11) $ \begin{equation} A_0(x) = \frac{e^3}{mk^2} \int^\infty_{|x|}\frac{u^2(s)}sh_u(s)\, {\rm d}s, \end{equation} $

$ A_1^2(x)+A_2^2(x) = \frac{e^2}{\kappa^2}\frac 1{|x|^2}h^2_u(x) = \frac{e^2}{\kappa^2}\frac 1{|x|^2}\bigg(\int^{|x|}_0su^2(s)\, {\rm d}s\bigg)^2. $

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.

在$ 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 $和$ 2<p\leq 4 $的情况. 对于$ p>4 $, 我们得到以下结果.

定理1.1  假设$ p>4 $, 且存在$ e^{*}, q^{*}>0 $, 使得$ 0<e<e^{*} $, $ 0<q<q^{*} $, 则问题(1.1) 有两个非平凡解.

我们将用变分法证明定理1.1. 首先, 对于任意给定的$ u\in H^1({{\Bbb R}} ^{2}) $, 由(1.1) 式中的第二个方程可以得到

(1.12) $ \begin{equation} -\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}. \end{equation} $

问题(1.12) 有唯一解如下

(1.13) $ \begin{equation} 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, \end{equation} $

其中积分核$ G^\mu(x) $是Yukawa势, 即

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

因此, 可以把$ N $看作是$ u $的函数. 那么可以通过找相应泛函的全局极小值点和山路定理刻画的临界点来证明定理1.1.

对于$ 2< p \leq 4 $的情况, 我们得到以下结果.

定理1.2   假设$ 2< p \leq 4 $, 当耦合常数$ e $充分小时, 则问题(1.1) 有一个非平凡的解.

虽然问题(1.1) 对应泛函$ J_{e}(u) $具有山路几何结构, 但很难验证$ (P S)_{c} $序列的有界性. 为了克服这一困难, 我们将泛函$ J_{e}(u) $截断, 并应用山路定理.

薛定谔泊松方程和Chern-Simon薛定谔方程有类似的结果, 详见文献[11, 12]和[13].

在本文第2节中, 证明了相关泛函满足$ (PS)_c $条件. 在第3节和第4节中, 将分别证明在$ p>4 $及$ 2<p\leq4 $的情况下, 问题(1.1) 解的存在性.

本节中, 我们将在$ H^1_r({{\Bbb R}} ^2)\times H^1_r({{\Bbb R}} ^2) $空间, 证明问题(1.1) 对应的泛函

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

满足$ (P S)_{c} $条件, 即对于任意使得$ J_e(u_n) $有界, 且$ J^{\prime}_e(u_n) $趋于0的序列$ \{u_n\}\subset H^1_r({{\Bbb R}} ^2) $包含收敛的子列, 其中$ H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $是$ H^{1}\left({{\Bbb R}} ^{2}\right) $包含径向对称函数的子空间, 且由文献[19], 可知$ H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $紧嵌入$ L^p({{\Bbb R}} ^{2}) $. 由(1.12) 式可得

(2.2) $ \begin{equation} \|\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. \end{equation} $

因此, 将(2.2) 式代入(2.1) 式, 我们可以得到简化的泛函

(2.3) $ \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} $

根据Strauss不等式^[19]有

(2.4) $ \begin{equation} |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). \end{equation} $

对于任意$ u \in H_{r}^{1}\left({{\Bbb R}} ^{2}\right), $有

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

成立, 由文献[1] 和[7], 我们得到以下结果.

引理2.1  如果$ u \in H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $是泛函$ J_{e} $的临界点, 那么$ \left(u, N_{u}\right) $是$ {\cal E}_{e} $的临界点, 因此$ u $是方程(1.1) 的解.

可以验证$ J_{e} \in C^{1} $, 而且

$ J_{e}^{\prime}(u) = \partial_u {\cal E}_{e}\left(u, N_{u}\right)+\partial_{N} {\cal E}_{e}\left(u, N_{u}\right) N_{u}^{\prime} = \partial_{u} {\cal E}_{e}(u, N). $

令

$ H(u) = \int_{{{\Bbb R}} ^{2}} \frac{u_{n}^{2}}{|x|^{2}}\left(\int_0^{{|x|}}\tau|u_{n}|^{2} {\rm d}\tau\right)^{2}\, {\rm d}x. $

在文献[3] 中, 证明了以下结论.

命题2.1   假设在空间$ H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $中, 当$ n \rightarrow \infty $时, 序列$ \left\{u_{n}\right\} $弱收敛到函数$ u $, 则对于任意$ \varphi\in H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $, 当$ n\to\infty $时, 有$ H(u_n), \, \langle H'(u_n), \varphi\rangle $及$ \langle H'(u_n), u_n \rangle $收敛到$ H(u), \, \langle H'(u), \varphi \rangle $和$ \langle H'(u), u \rangle $. 此外

$ \lim\limits _{n \rightarrow \infty} \int_{{{\Bbb R}} ^{2}}\left(\int_{|x|}^{\infty} \frac{u_{n}^{2}(\tau) \int_0^{{\tau}}\left|u_{n}\right|^{2}}{\tau} {\rm d}\tau\right) u_{n}^{2}\, {\rm d}x = \int_{{{\Bbb R}} ^{2}}\left(\int_{|x|}^{\infty} \frac{u^{2}(\tau) \int_0^{{\tau}}|u|^{2}}{\tau} {\rm d}\tau\right) u^{2}\, {\rm d}x . $

在文献[1] 中证明了如下结果.

引理2.2   对于任意$ u \in H^{1}\left({{\Bbb R}} ^{2}\right), $存在唯一的$ N_{u} \in H^{1}\left({{\Bbb R}} ^{2}\right) $是方程(1.1) 第二个方程的解. 此外

$ \rm (i) $$ N_{u} \leq 0 $ a.e. in $ {{\Bbb R}} ^{2} $;

$ \rm (ii) $映射$ u \mapsto N_{u} $是$ 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) $若$ u \in H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $, 则$ N_{u} \in H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $.

定义

$ I(u) = \int_{{{\Bbb R}} ^{2}}N_u(x)u^2(x)\, {\rm d}x. $

命题2.2   假设在空间$ H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $中, 当$ n\to\infty $时, $ \left\{u_{n}\right\} $弱收敛到函数$ u $, 则对于任意$ \varphi\in H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $, 当$ n\to\infty $时, 有$ I(u_n), \, \langle I'(u_n), \varphi \rangle $, $ \langle I'(u_n), u_n \rangle $收敛到$ I(u), \, \langle I'(u), \varphi\rangle $, $ \langle I'(u), u \rangle $.

证  我们仅证明当$ n\to\infty $时$ I(u_n)\to I(u) $, 其它项可以同样地证明. 因为

$ I(u_n)-I(u) = \int_{{{\Bbb R}} ^{2}}(N_{u_{n}}u_{n}^{2}-N_{u_n}u^2)\, {\rm d}x + \int_{{{\Bbb R}} ^{2}}(N_{u_{n}}u^{2}-N_uu^2)\, {\rm d}x. $

由(1.13) 式, 富比尼定理和引理2.2, 有

$ \begin{eqnarray*} &&\bigg|\int_{{{\Bbb R}} ^{2}}(N_{u_{n}}(x)-N_u(x))u^{2}(x)\, {\rm d}x \bigg| \\ &\leq& q(1+\frac{\kappa q}{2m} )\int_{{{\Bbb R}} ^{2}} \int_{{{\Bbb R}} ^{2}} G^{\kappa q}(x-y)| u_n^{2}(y)-u^{2}(y)| u^{2}(x)\, {\rm d}y {\rm d}x \\ &\leqslant& C \int_{{{\Bbb R}} ^{2}}\left|u_n^{2}(y)-u^{2}(y)\right| \int_{{{\Bbb R}} ^{2}}\left|G^{\kappa q}(x-y) u^{2}(x)\right|\, {\rm d}{x}{\rm d}y\\ &\leqslant &C \int_{{{\Bbb R}} ^{2}}\left|u_n^{2}(y)-u^{2}(y)\right|\left(\int_{{{\Bbb R}} ^{2}}\left|G^{\kappa q}(x-y)\right|^{s}\, {\rm d}x\right)^{\frac{1}{s}}\left(\int_{{{\Bbb R}} ^{2}}|u|^{r} \, {\rm d}x\right)^{\frac{2}{r}}\, {\rm d}y, \end{eqnarray*} $

其中$ r>2, s = \frac r{r-2} $. 由文献[7] 中的引理4.1, 对于$ p\geq 1 $有

$ \|G^\mu\|_{L^p}\leq C\mu^{-\frac{2}{p}}, $

又因为$ H_r^1( {{\Bbb R}} ^2 ) $紧嵌入$ L^p( {{\Bbb R}} ^2 ) $, $ p \geq 2 $, 故有$ u_{n} $在$ L^{p}( {{\Bbb R}} ^2 ) $强收敛到$ u $, 因此, 当$ n\to\infty $时

$ \bigg|\int_{{{\Bbb R}} ^{2}}(N_{u_{n}}(x)-N_u(x))u^{2}(x)\, {\rm d}x \bigg| \leq C(\kappa q)^{-\frac{2}{s}}\int_{{{\Bbb R}} ^{2}}|u_n(y)-u(y)|^2\, {\rm d}y\rightarrow 0, $

类似地, 当$ n\to\infty $时, 有

$ \begin{eqnarray*} &&\bigg|\int_{{{\Bbb R}} ^{2}}(N_{u_{n}}(x)u_n^{2}(x)-N_{u_{n}}(x)u^{2}(x))\, {\rm d}x \bigg| \\ & = &\int_{{{\Bbb R}} ^{2}}|N_{u_{n}}(x)||u_n^2(x)-u^2(x)| \, {\rm d}x\\ &\leq& q(1+\frac{\kappa q}{2m} )\int_{{{\Bbb R}} ^{2}} \int_{{{\Bbb R}} ^{2}} |G^{\kappa q}(x-y)u_n^2(y)|dy|u_n^2(x)-u^2(x)|{\rm d}x\\ &\leq &C \int_{{{\Bbb R}} ^{2}}\bigg(\int_{{{\Bbb R}} ^{2}}|G^{\kappa q}(x-y)|^s{\rm d}y\bigg)^\frac{1}{s} \bigg(\int_{{{\Bbb R}} ^{2}}|u_n(y)|^r{\rm d}y\bigg)^\frac{2}{r}|u_n^2(x)-u^2(x)|\, {\rm d}x\\ &\leq &C (\kappa q)^{-\frac{2}{s}}\left(\int_{{{\Bbb R}} ^{2}}|u_n|^{r} \, {\rm d}x\right)^{\frac{2}{r}}\int_{{{\Bbb R}} ^{2}}|u_n(x)-u(x)|^2\, {\rm d}y\rightarrow 0. \end{eqnarray*} $

其中$ r>2, s = \frac r{r-2} $. 命题2.2证毕.

命题2.3   假设$ p>2 $, $ \left\{u_{n}\right\} \subset H_r^1({{\Bbb R}} ^2) $是泛函$ J_{e} $有界的$ (P S)_{c} $序列, 即当$ n \rightarrow \infty $时, 泛函$ J_{e} $满足

$ J_{e}\left(u_{n}\right) \rightarrow c, \quad J_{e}^{\prime}\left(u_{n}\right) \rightarrow 0, $

则$ \left\{u_{n}\right\} $包含一个收敛子列.

证  因为$ \{u_{n}\} $在$ 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

(2.6) $ \begin{equation} 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}. \end{equation} $

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

(2.7) $ \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.

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\} $在$ H_r^1({{\Bbb R}} ^{2}) $是一致有界的.

证  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

$ \begin{eqnarray*} c&\geq &J_{e}(u_n)\geq \min\Big\{\frac{1}{4m}, \frac{\omega}{2}\Big\}\|u_n\|^2-\frac{1}{4\kappa^2q}\left (1+\frac{\kappa q}{2 m}\right)^2\|u_n\|^4_{L^4}+\frac{q}{p^2 m^{2}}\|u_n\|^p_{L^p}\\ & &+\frac{e^{4}}{4 m \kappa^{2}} \int_{{{\Bbb R}} ^{2}} \frac{u^{2}(|x|) h_{u}^{2}(|x|)}{|x|^{2}}\, {\rm d}x\\ &\geq & C\|u_n\|^2-C_\varepsilon\int_{{{\Bbb R}} ^{2}} |u_n|^2{\rm d}x-\varepsilon\int_{{{\Bbb R}} ^{2}} |u_n|^p{\rm d}x +\frac{q}{p^2 m^{2}}\int_{{{\Bbb R}} ^{2}} |u_n|^p\, {\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. \end{eqnarray*} $

选取$ \varepsilon = \frac{q}{2p^2m^2} $, 可得

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

我们证明$ \|u_n\|^2_{L^2({{\Bbb R}} ^{2})} $一致有界, 即证$ \|u_n\|^2_{L^2({{\Bbb R}} ^{2})}\leq C $, C与$ n $无关. 用反证法, 假设当$ n \rightarrow \infty $时有

$ \|u_n\|^2_{L^2({{\Bbb R}} ^2)}\rightarrow \infty. $

令$ 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 \left\|u_{n}\right\|^{2}_{L^2}\|v_n\|^2+\frac{q}{2 p^2 m^{2}}\|u_{n}\|^p_{L^2} \int_{{{\Bbb R}} ^{2}}\left|v_{n}\right|^{p}\, {\rm d}x + \|u_n\|^6_{L^2}\int_{{{\Bbb R}} ^{2}} \frac{v_{n}^{2}(|x|) h_{v_n}^2(|x|)}{|x|^{2}}\, {\rm d}x\\ & \leq& c+C \|u_n\|^2_{L^2}\int_{{{\Bbb R}} ^{2}}\left|v_{n}\right|^{2}\, {\rm d}x. \end{eqnarray*} $

即

(3.1) $ \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} $

从而$ \|v_n\|\leq C $, 又因为$ H_r^1( {{\Bbb R}} ^2 ) $紧嵌入$ L^p( {{\Bbb R}} ^2 ) $, 因此我们可以假设

$ v_{n} \rightharpoonup v \quad \mbox{ in } \quad H_r^1({{\Bbb R}} ^2), \quad v_{n} \rightarrow v \quad \mbox{ in } \quad L^{p}\left({{\Bbb R}} ^{2}\right), \, \, \ p \geq 2, \quad v_{n} \rightarrow v \quad \mbox{ a.e. } \quad {{\Bbb R}} ^{2} $

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

(3.2) $ \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} $

则当$ n\to\infty $时, 由(3.1) 式有

$ \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 \rightarrow 0. $

于是可以得到

(3.3) $ \begin{equation} \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. \end{equation} $

故有

(3.4) $ \begin{equation} \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. \end{equation} $

从而得到

$ A_{j}(v) v = 0, \quad \mbox{ a.e. } \quad {{\Bbb R}} ^{2}, \quad \partial_{2}\left(A_{1}(v) v^{2}\right) = 0, \quad \partial_{1}\left(A_{2}(v) v^{2}\right) = 0 \quad \mbox{ a.e. } \quad {{\Bbb R}} ^{2}. $

这说明

(3.5) $ \begin{equation} 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, \end{equation} $

且

(3.6) $ \begin{equation} 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. \end{equation} $

结合(3.5), (3.6) 式可以得到

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^{2}}\left(\partial_{2} A_{1}(v)-\partial_{1} A_{2}(v)\right) v^{2}\, {\rm d}x+2 \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*} $

由(1.6) 式有

(3.7) $ \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} $

因此$ v = 0 $ a.e. $ {{\Bbb R}} ^{2}, $这与

$ 1 = \lim\limits_{n\to\infty}\int_{{{\Bbb R}} ^{2}}\left|v_{n}\right|^{2}\, {\rm d}x = \int_{{{\Bbb R}} ^{2}}|v|^{2}\, {\rm d}x. $

矛盾. 引理3.1证毕.

定理1.1的证明   考虑泛函

$ I_{0}(u) = \left.J_{e}\right|_{e, q = 0}(u) = \int_{{{\Bbb R}} ^{2}} \Big(\frac{1}{4{m}}|\nabla u|^{2}+\frac{w}{2}|u|^{2}\Big)\, {\rm d}x+\frac{1}{4} \int_{{{\Bbb R}} ^{2}} N_u u^{2}\, {\rm d}x. $

因为$ N_u \leq 0 $, 当$ t_0 $充分大且$ u_0\in H_r^1({{\Bbb R}} ^{2}) $固定时, 可以证明$ I_0(t_0u_0)< 0 $. 如果$ \|u\|< r $充分小, 则$ I_0(u)\geq \alpha > 0 $. 由$ J_{e} $对于$ e, q $的连续性, 存在$ e^{*}, q^{*}>0 $, 当$ 0<e<e^{*} $, $ 0<q<q^{*} $时, 若$ \|u\| = r $有$ J_{e}(u) \geq \alpha $, $ J_{e}(t_0u_0)<0 . $因为$ J_{e} $满足$ (P S)_c $条件, 由山路定理, 泛函$ J_{e}(u) $有临界点$ u_{c} $, 且$ J_{e}\left(u_{c}\right)>0 $.

下面, 我们考虑当$ 0<e<e^{*} $, $ 0< q <q^* $时的极小化问题

$ m_e = \inf\limits_{u\in H_r^1({{\Bbb R}} ^{2})}J_e(u). $

因为$ J_{e}(t_0u_0)<0 $, 由引理3.1可得$ -\infty<m_{e}<0 $. 另一方面, 由Ekland变分原理可知存在一个数列$ \left\{u_{n}\right\} \subset H_r^1({{\Bbb R}} ^{2}) $, 当$ n \rightarrow \infty $时

$ J_{e}\left(u_{n}\right) \rightarrow m_{e}, \quad J_{e}^{\prime}\left(u_{n}\right) \rightarrow 0. $

再由引理3.1及命题, 可得$ \left\{u_{n}\right\} $在$ H_r^1({{\Bbb R}} ^{2}) $中有界, 且有收敛的子列. 因此存在$ u_e $使得$ m_{e} = J_{e}\left(u_{e}\right) $, $ J_{e}^{\prime}\left(u_{e}\right) = 0 . $定理1.1证毕.

在泛函$ J_{e} $中, 两个非局部项之间互相影响, 使得很难验证泛函$ J_{e} $的几何结构及其Palais-Smale序列的有界性. 为了克服这些困难, 如文献[1, 10], 我们引入了截断函数$ \chi \in C^{\infty}({{\Bbb R}} +, {{\Bbb R}} ) $,

$ \left\{\begin{array}{ll} \chi(s) = 1, & s \in[0, 1], \\ 0 \leq \chi \leq 1 , & s \in[1, 2], \\ \chi(s) = 0, & s \in[2, +\infty[, \\ \left\|\chi^{\prime}\right\|_{\infty} \leq 2. \end{array}\right. $

对于任意$ T>0 $, 定义以下截断泛函$ J_{e, T} $:

$ \begin{eqnarray*} J_{e, T}(u):& = & \frac{1}{4 m}\|\nabla u\|_{L^2}^{2}+\frac{\omega}{2}\|u\|_{L^2}^{2}+\frac{e^{4}}{4 m \kappa^{2}} K_{T}(u) \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} u^{2}{\rm d}x+\frac{q}{p^2 m^{2}}\|u\|^{p}_{L^p}, \, \end{eqnarray*} $

其中

$ K_{T}(u): = \chi\left(\frac{\|u\|^{2}}{T^{2}}\right). $

可以验证$ J_{e, T} $是$ C^{1} $的, 对于确定的$ T, e>0 $, 若$ \bar{u} \in H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $是泛函$ J_{e, T} $的临界点, 且满足$ \|\bar{u}\|<T $, 则$ \bar{u} $是泛函$ J_{e} $的临界点, 再由引理$ 2.2, $可知$ \left(\bar{u}, N_{\bar{u}}\right) $是(1.1) 的解.

对于任意$ e, T>0 $, 截断的泛函$ J_{e, T} $都满足山路定理的几何假设, 更准确地说, 我们有

引理4.1   泛函$ J_{e, T} $满足以下性质:

$ {\rm (i)} $存在$ \rho>0 $, $ \alpha>0 $, 当$ \|u\| = \rho $时, 使得$ J_{e, T}(u) \geq \alpha $;

$ {\rm (ii)} $存在$ \bar{u} \in H_{r}^{1}\left({{\Bbb R}} ^{2}\right), $当$ \|\bar{u}\|>\rho $时, 使得$ J_{e, T}(\bar{u})<0, $其中$ \alpha $, $ \rho $与$ e, T $无关.

证   (i) 若$ \|u\| $充分小, 我们有

$ \begin{eqnarray*} J_{e, T}(u) &\geqslant & \frac{1}{4 m}\|\nabla u\|_{L^2}^{2}+\frac{\omega}{2}\|u\|_{L^2}^{2}-\frac{1}{4 \kappa^2q}\left(1+\frac{\kappa q}{2 m}\right)^2\|u\|_{L^4}^{4} +\frac{q}{p^2 m^{2}}\|u\|_{L^p}^{p}\geq\alpha>0, \end{eqnarray*} $

(ii) 对于任意$ u \in H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $, 当$ t>0 $充分大时, $ K_T(tu) = 0 $. 则由引理2.2可以推导出

$ \begin{eqnarray*} J_{e, T}(t u) & = &t^{2} \Big(\frac{1}{4 m}\|\nabla u\|_{L^2}^{2}+\frac{\omega}{2}\|u\|_{L^2}^{2}\Big)+\frac{e^{4} t^{6}}{4 m{\kappa}^{2}}K_{T}(t u) \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) t^{4} \int_{{{\Bbb R}} ^{2}}|N u| u^{2}\, {\rm d}x+\frac{q t^{p}}{p^2 m^{2}} \int_{{{\Bbb R}} ^{2}}|u|^{p}\, {\rm d}x, \end{eqnarray*} $

这说明当$ t\to\infty $时, 则$ J_{e, T}\rightarrow -\infty $.

对于任意$ e, T>0, $定义泛函$ J_{e, T} $对应的山路值$ c_{e, T} $, 即

$ {c_{e,T}}: = \mathop {\inf }\limits_{t \in [0,1]} \mathop {\sup }\limits_{\gamma (t) \in {\Gamma _{e,T}}} {J_{e,T}}(\gamma (t)),$

其中

$ \Gamma_{e, T}: = \left\{\gamma \in C\left([0, 1], H_{r}^{1}\left({{\Bbb R}} ^{2}\right)\right): \gamma(0) = 0, \|\gamma(1)\|>\rho, J_{e, T}(\gamma(1))<0\right\}, $

则$ c_{e, T}\geq\alpha . $由山路定理, 泛函$ J_{e, T} $在$ H_{r}^{1}\left({{\Bbb R}} ^{2}\right) $中存在Palais-Smale序列$ \left\{\left(u_{e, T}\right)_{n}\right\} $, 即当$ n \rightarrow +\infty $时有

$ J_{e, T}\left(\left(u_{e, T}\right)_{n}\right) \rightarrow c_{e, T}, \quad J_{e, T}^{\prime}\left(\left(u_{e, T}\right)_{n}\right) \rightarrow 0. $

下面证明, 若$ e $充分小则Palais-Smale序列$ \left\{u_{n}\right\}: = \left\{\left(u_{e, T}\right)_{n}\right\} $一致有界.

引理4.2   存在与$ e $无关的$ \bar{T}>0 $, $ e^{*}: = e(\bar{T})>0 $, 使得当$ 0<e<e^{*} $时有

$ \mathop {\lim \sup }\limits_n \left\|u_{n}\right\|<\bar T.$

证  我们用反证法: 假设对于任意$ T>0 $, 存在$ e>0 $使得

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

因为

$ \begin{eqnarray*} \langle J_{e, T}^{\prime}(u_{n}), u_n \rangle& = &\int_{{{\Bbb R}} ^{2}} \left(\frac{1}{2 m}|\nabla u_{n}|^{2}+\omega|u|^{2}\right)\, {\rm d}x+\frac{3 e^{4}}{2 m \kappa^{2}} K_{T}\left(u_{n}\right) \int_{{{\Bbb R}} ^{2}} \frac{u_{n}^{2}(|x|) h_{u_n}^{2}(|x|)}{|x|^{2}}\, {\rm d}x\\ &&+\frac{e^4}{2m\kappa^2T^2}K_T^{\prime}(u_n)\|u_n\|^2\int_{{{\Bbb R}} ^{2}}\frac{u_n^2(|x|)h^2_{u_n}(|x|)}{|x|^2}\, {\rm d}x\\ &&+\Big(1+\frac{\kappa q}{2m}\Big)\int_{{{\Bbb R}} ^{2}}N_{u_n}u_n^2\, {\rm d}x+\frac{q}{p m^2}\int_{{{\Bbb R}} ^{2}}|u_n|^p\, {\rm d}x, \end{eqnarray*} $

我们有

$ \begin{eqnarray*} &&c_{e, T}+o(1)\|u_n\|\\ & = &J_{e, T}(u_n)-\frac{1}{p}\langle J_{e, T}^{\prime}(u_n), u_n\rangle\\ & = &\int_{{{\Bbb R}} ^{2}}\Big(\frac{1}{2}-\frac{1}{p}\Big) \Big(\frac{1}{2m}|\nabla u_n|^2+\omega|u|^2\Big)\, {\rm d}x +\frac{e^4}{2m\kappa^2}\Big(\frac{1}{2}-\frac{3}{p}\Big)K_T\int_{{{\Bbb R}} ^{2}}\frac{u^2(|x|)h_u(|x|)^2}{|x|^2}\, {\rm d}x\\ &&-\frac{e^4}{2pm\kappa^2}K^{\prime}_T\|u_n\|^2\int_{{{\Bbb R}} ^{2}}\frac{u^2(|x|)h^2_u(|x|)}{|x|}\, {\rm d}x + \Big(\frac{1}{4}-\frac{1}{p}\Big)\Big(1+\frac{\kappa q}{2m}\Big)\int_{{{\Bbb R}} ^{2}}N_uu^2\, {\rm d}x\\ &\geq& C \|u_n\|^2-\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_u(|x|)^2}{|x|^2}\, {\rm d}x. \end{eqnarray*} $

因此

(4.2) $ \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} $

现在, 估计$ c_{e, T} $. 由引理2.2可得

$ \begin{eqnarray*} c_{e, T} &\leq & \max _{t \geq 0} J_{e, T}(t \bar{u}) \\ &\leq & \max _{t \geq 0}\left[t^{2} \int_{{{\Bbb R}} ^{2}} \Big(\frac{1}{4 m}|\nabla \bar{u}|^{2}+\frac{\omega}{2}|u|^{2}\Big)\, {\rm d}x-\frac{1}{4}\left(1+\frac{\kappa q}{2 m}\right) t^{4} \int_{{{\Bbb R}} ^{2}}\left|N_{\bar{u}}\right| \bar{u}^{2}\, {\rm d}x + \frac{q t^{p}}{p m^{2}} \int_{{{\Bbb R}} ^{2}}|\bar{u}|^{p}\, {\rm d}x \right]\\ &&+\frac{e^4}{4m\kappa^2}\max _{t \geq 0} \bigg[t^6\chi\Big(\frac{t^2\|\bar u\|^2}{T^2}\Big)\int_{{{\Bbb R}} ^{2}}\frac{{\bar u}^2(|x|)h^2_{\bar u}(|x|)}{|x|^2}\, {\rm d}x \bigg]\\ &\leq& A_{1}+A_{2}(T). \end{eqnarray*} $

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

$ A_2(T)\leq\frac{e^4}{4m\kappa^2}\frac{8T^6}{\|\bar u\|^6}\int_{{{\Bbb R}} ^{2}}\frac{{\bar u}^2(|x|)h^2_{\bar u}(|x|)}{|x|^2}\, {\rm d}x \leq Ce^4T^6. $

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

(4.3) $ \begin{equation} c_{e, T} \leq A_{1}+C e^{4} T^{6}. \end{equation} $

此外, 由(2.5) 式, 我们有

(4.4) $ \begin{equation} \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}. \end{equation} $

因为$ \|\chi^{\prime} \|_\infty\leq 2 $, $ K_T(u) = \chi (\frac{\|u\|}{T^2}) $, 可以推出

(4.5) $ \begin{equation} \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}. \end{equation} $

结合(4.2) 式, (4.3) 式及(4.5) 式可以得到

$ C\|u_n\|^2 + o(1)\|u_n\|\leq A_1+ Ce^4T^6. $

另一方面, (4.1) 式说明

$ C\|u_n\|^2+o(1)\|u_n\|\geq CT^2-T. $

于是

$ CT^2-T\leq A_1+Ce^4T^6, $

如果$ e = e(T) $足够小, 而$ T>0 $足够大, 就会产生矛盾.

定理1.2的证明   由引理4.1, 4.2, 可以得到泛函$ J_{e, \bar T} $有有界的Palais-Smale序列$ \{u_n\} $, 则$ J_{e, \bar{T}}\left(u_{n}\right) = J_{e}\left(u_{n}\right) $. 再由命题, 可得$ u_{n} $在$ H_r^{1}\left({{\Bbb R}} ^{2}\right) $中强收敛到$ u_{0} $, 故$ u_{0} $是泛函$ J_{e} $对应值$ c_{e, \bar{T}} $的临界点, 且$ u_{0} $非平凡. 因此$ \left(u_{0}, N_{u_{0}}\right) $是方程组(1.1) 的解. 证毕.