考虑下面非线性的schrödinger-Maxwell方程

$ \begin{equation} \left\{\begin{array}{cc} -\triangle u+V(x)u+K(x)\phi(x)u = f(x, u)- g(x, u), \ x\in R^{3}, \\ -\triangle\phi = K(x)u^{2}.\ x\in R^{3}, \ \ \end{array} \right. \end{equation} $

(1.1)

其中$ F(x, u) = \int^{u}_{0}f(x, s)ds $, $ G(x, u) = \int^{u}_{0}g(x, s)ds $, 此系统又称为schrödinger-Poisson系统, 它是物理学家们在求解非线性schrödinger方程和一未知电场相互作用的解时得到的.关于$ (1.1) $式的具体推导过程及更详尽的物理意义, 文献[3, 4]有较详细的论述.

自从系统$ (1.1) $在文献[3]中提出以后, 很多学者对系统解的问题产生了兴趣并进行了研究.在文献[1]中, Sun用变化的喷泉定理得出了一类次线性项schrödinger-Maxwell方程的无限多个解.在文献[2]中, Li, Su, Wei用变化的喷泉定理得出了非线性schrödinger-Maxwell方程的无穷多个高能解的存在性.受文献[2]的启发, 本文在文献[2]的非线性项$ f(x, u) $ (见本文定理1.1的(f1)–(f4)条件)的基础上, 添加了一个次线性项$ g(x, u) $ (见定理1.1的(g)条件), 将文献[1]中的次线性项$ 0<p<1 $, 推广到本文的$ q\in(2, 3) $, 运用山路定理, 证出了系统$ (1.1) $的基态解.在文献[5]中, Liu, Guo, Zhang用变分法得出了系统的基态解的存在性.在文献[6]中, Sun, Ma用新方法证得了系统的基态解.在文献[7]中, 方立婉, 黄文念, 谢苏静运用山路定理得出了一类非线性schrödinger-Maxwell方程在$ R^3 $的基态解, 关于系统$ (1.1) $的更多的结论见文献[8-12].

定理1.1   假设$ V, K $和非线性项$ f, g $满足如下条件:

(V)  $ V\in C(R^{3}, R) $且$ \inf\limits_{x\in R^3}V(x)> 0 $.进一步地, $ \forall \ M >0 $, 有$ {\rm meas} \{x\in R^{3}:V(x)\leq M\}<\infty $.

(K)  $ K\in L^{\infty}(R^{3}, R) $, 且$ K(x)\geq0 $, $ \forall {x\in R^3} $.

(f1)  $ f\in C(R^{3}\times R, R) $且存在$ a>0 $, $ p\in(2, 6) $, 使得$ |f(x, u)| \leq a(1+|u|^{p-1}) $, $ \forall \ (x, u)\in R^3\times R $.

(f2)  $ \lim\limits_{|u|\rightarrow0}\frac{F(x, u)}{|u|^2} = 0 $对于所有$ x\in R^3 $一致成立.

(f3)  $ \lim\limits_{|u|\rightarrow\infty}\frac{F(x, u)}{|u|^4} = +\infty $对于所有$ x\in R^3 $一致成立.

(f4)  $ \forall \ (x, u)\in R^{3}\times R $, $ 0< F(x, u)\leq\frac{1}{4}f(x, u)u $.

(g)  $ G(x, u) = \frac{1}{4-q}b(x)|u|^{4-q} $, $ q\in(2, 3) $, $ b(x)\geq0 $, $ \forall \ {x\in R^3} $, $ b(x)\in L^{\frac{4}{q}}(R^{3}, R) $.

则系统$ (1.1) $存在一个解$ u\in E $使得$ \Phi(u) = \inf\limits_{\mathcal{N}}\Phi(u)>0 $. $ {\mathcal{N}} $定义见式$ (2.9) $.

对于任意的$ 1\leq r<\infty $, $ L^r{(R^3)} $表示通常意义下的Lebesgue空间, 其范数为

$ \|u\|_{r} = \left( \int_{R^3}|u|^{r}\right)^{\frac{1}{r}}. $

$ {H}^{1}(R^{3}) $则表示通常意义下的Sobolev空间, 其范数取为

$ \|u\|_{{H}^{1}} = \left\{ \int_{R^3} \left[|\nabla{u}|^2+|u|^2\right]\mathrm{d}x\right\}^{1/2}. $

定义如下空间

$ \mathcal{D}^{1, 2} = \{u\in L^{2^{*}}(R^{3})\mid \nabla u\in L^{2}(R^{3})\}, $

其范数取为

$ \|u\|_{\mathcal{D}^{1, 2}} = \left( \int_{R^3}|\nabla u|^{2}\right)^{\frac{1}{2}}. $

$ 2^{*} = 6 $为三维空间的临界Sobolev嵌入指数, 则$ \mathcal{D}^{1, 2}(R^{3})\hookrightarrow L^{2^{*}}(R^{3}) $, $ C_{1} $是最佳嵌入正的常数, 即

$ \begin{equation} \|u\|_{6}\leq C_{1}\|u\|_{\mathcal{D}^{1, 2}}. \end{equation} $

(2.1)

由Lax-Milgram定理(见文献[13]), $ \forall \ u\in\ H^{1}(R^3) $, $ \exists $唯一的$ \phi_u \in \mathcal{D}^{1, 2}(R^{3}) $使得

$ \begin{equation} -\triangle\phi_u = K(x)u^{2}. \end{equation} $

(2.2)

进一步地有

$ \phi_{u} = \frac{1}{4\pi} \int_{R^{3}}\frac{K(y)u^{2}(y)}{|x-y|}dy. $

从而$ \phi_{u}\geq0 $. $ \forall u\in H^{1}(R^{3}) $, 由式$ (2.1), (2.2) $和Hölder不等式知

$ \|\phi_{u}\|_{\mathcal{D}^{1, 2}}^{2} = \int_{R^{3}} K(x)\phi_{u}u^{2}dx\leq \|K\|_{\infty} \|\phi_{u}\|_{6}\|u\|_{\frac{12}{5}}^{2}\leq C_{1} \|K\|_{\infty} \|\phi_{u}\|_{\mathcal{D}^{1, 2}}\|u\|_{\frac{12}{5}}^{2}. $

从而

$ \begin{equation} \|\phi_{u}\|_{\mathcal{D}^{1, 2}}\leq C_{1} \|K\|_{\infty}\|u\|_{\frac{12}{5}}^{2}, \end{equation} $

(2.3)

且令$ C_2 = C_{1}^{2} \|K\|_{\infty}^2 $

$ \begin{equation} \int_{R^{3}} K(x)\phi_{u}u^{2}dx\leq C_{2}\|u\|_{\frac{12}{5}}^{4}. \end{equation} $

(2.4)

令

$ \begin{equation} E = \left\{u\in H^{1}(R^3) : \int_{R^3} |\nabla u|^{2}+V(x)|u|^2\mathrm{d}x< +\infty \right\} \end{equation} $

(2.5)

和

$ \|u\| = \left\{ \int_{R^3} \left[|\nabla{u}|^2+V(x)|u|^2\right]\mathrm{d}x\right\}^{1/2}, \ \ \ \ \forall \ u\in E. $

(V) 条件显示$ \|\cdot\| $和$ \|\cdot\|_{H^1} $在$ E $上是两个等价范数, 由Sobolev嵌入定理, $ \exists C_3>0 $, 使得

$ \begin{equation} \|u\|_{r}\le C_{3}\|u\|, \ \forall \ u\in E, \ \ 2\le r\le 2^*. \end{equation} $

(2.6)

在空间$ E\times\mathcal{D}^{1, 2} $中定义一个泛函$ I $,

$ \begin{eqnarray} I(u, \phi)& = &\frac{1}{2} \int_{R^{3}}(|\nabla u|^{2}+V(x)u^{2})dx-\frac{1}{4} \int_{R^{3}}|\nabla\phi|^{2}dx+\frac{1}{2} \int_{R^{3}} K(x)\phi u^{2}dx\\ &&- \int_{R^{3}}F(x, u)dx + \int_{R^{3}}G(x, u)dx, \end{eqnarray} $

(2.7)

则$ I $是有意义的, 且$ I\in C^{1}(E\times\mathcal{D}^{1, 2}) $, 同时, $ I $的每个临界点就是系统$ (1.1) $的一个解(这里是指弱解).由式$ (2.2) $得出

$ \begin{equation} \Phi(u) = \frac{1}{2}\|u\|^2+\frac{1}{4} \int_{R^{3}} K(x)\phi_{u} u^{2}dx- \int_{R^{3}}F(x, u)dx+ \int_{R^{3}}G(x, u)dx. \end{equation} $

(2.8)

易知$ \Phi\in C^{1}(E, R) $且

$ \langle\Phi'(u), v\rangle = \int_{R^{3}}(\nabla u \nabla v+V(x)uv)dx+ \int_{R^{3}} K(x)\phi_{u} uvdx- \int_{R^{3}}f(x, u)vdx+ \int_{R^{3}}g(x, u)vdx. $

由临界点理论及变分法知:当$ u\in E $为泛函$ \Phi $的一个临界点, 则$ (u, \phi_u) $为系统$ (1.1) $的一组解.对应的Nehari流形为

$ \begin{equation} \mathcal{N} = \{u\in E\backslash\{0\} : \langle\Phi'(u), u\rangle = 0\}. \end{equation} $

(2.9)

定义2.1   (见文[14])设$ E $是Banach空间, $ \Phi\in C^1(E, R) $, $ c\in R $.泛函$ \Phi $满足(PS)$ _c $条件是指: $ \forall \ \{u_n\}\subset E $使得

$ \Phi(u_n)\rightarrow c, \ \ \Phi'(u_n)\rightarrow 0 $

有一个收敛的子列.

定理2.2   (见文[15])设$ E $是Banach空间, $ E^* $为其对偶空间, 若$ \Phi\in C^1(E, R) $且满足

$ \Phi(v)<\Phi(0) = 0<\rho\leq\inf\limits_{\|u\| = r}\Phi(u), $

其中$ \rho>0, r>0 $, 以及$ v\in E $且$ \|v\|>r $.令$ c\geq\rho $且$ c = \inf\limits_{\gamma\in\Gamma}\max\limits_{s\in[0, 1]}\Phi(\gamma(s)), $是从$ 0 $到$ v $的连续曲线.其中$ \Gamma = \{\gamma\in C([0, 1], E):\gamma(0) = 0, \gamma(1) = v\} $, 则存在$ \{u_n\}\subset E $, 使得$ \Phi(u_n)\rightarrow c>0 $且$ \Phi'(u_n)\rightarrow0 $, 若$ \Phi $满足(PS)$ _c $条件, 则$ c\geq\rho $是$ \Phi $的临界值.

引理3.1  在定理$ 1.1 $的假设条件下, 则

(ⅰ)  $ \exists\; r >0, \rho >0 $, 使得$ \rho\leq\inf\limits_{\|u\| = r}\Phi(u) $;

(ⅱ)  $ \exists\; v\in E $, 满足$ \|v\|>r $, 使得$ \Phi(v)<0 $.

证   (ⅰ)由(f2)条件知$ \forall \ \varepsilon>0, \ \exists \ \alpha>0 $, 使得$ \forall \ x\in R^3, \ 0\leq|u|\leq\alpha $, 有$ \frac{|F(x, u)|}{|u|^2}\leq\frac{\varepsilon}{2} $, 所以

$ \begin{equation} |F(x, u)| \leq\frac{\varepsilon}{2}|u|^2. \end{equation} $

(3.1)

进一步得出

$ \begin{equation} |f(x, u)|\leq\varepsilon|u|. \end{equation} $

(3.2)

由(f1)条件知$ \forall \ x\in R^3, \ |u|\geq\alpha $, 这里$ C = a((\frac{1}{\alpha})^{p-1}+1) $,

$ \begin{eqnarray} &&|f(x, u)|\leq{} a+a|u|^{p-1} \leq{}a((\frac{1}{\alpha})^{p-1}+1)|u|^{p-1} \leq{}C|u|^{p-1}, \\ &&|f(x, u)|\leq C|u|^{p-1}, \end{eqnarray} $

(3.3)

因此

$ \begin{equation} |F(x, u)|\leq \frac{C}{p}|u|^{p}. \end{equation} $

(3.4)

结合式$ (3.2), (3.3) $, $ \forall \ x\in R^3, \ u \in R $, 有

$ \begin{equation} |f(x, u)|\leq C|u|^{p-1}+\varepsilon|u|. \end{equation} $

(3.5)

结合式(3.1), (3.4), $ \forall \ x\in R^3, \ u \in R $, 有

$ \begin{equation} |F(x, u)|\leq \frac{C}{p}|u|^{p}+\frac{\varepsilon}{2}|u|^2. \end{equation} $

(3.6)

结合(g)条件和式$ (2.6) $, 有

$ \Phi (u) =\frac{1}{2}\| u\|^{2}+\frac{1}{4} \int_{R^{3}}K(x) \phi_{u}{u^2}dx- \int_{R^{3}}F(x, u)dx+ \int_{R^{3}}\frac{1}{4-q}b(x)|u|^{4-q}dx\\ \geq{}\frac{1}{2}\|u\|^2-\frac{\varepsilon}{2}\|u\|_{2}^2-\frac{C}{p}\|u\|_{p}^{p} \geq{}\frac{1}{2}\|u\|^2-\frac{C_{3}^2\varepsilon}{2}\|u\|^{2}-\frac{CC_{3}^p}{p}\|u\|^{p}. $

取$ \varepsilon $充分小, 因此$ \exists \ r >0, \ \rho >0 $, 使得$ \rho\leq\inf\limits_{\|u\| = r}\Phi(u) $.

(ⅱ)  由(f3)条件知$ \forall \ M >0, \ \ \exists \ \delta>0 $, 使得$ |u|\geq\delta $时,

$ \begin{equation} F(x, u)\geq M|u|^{4}. \end{equation} $

(3.7)

由(f2)条件知$ \forall \ x\in R^3, \ 0\leq|u|\leq \alpha $, 有

$ \begin{equation} \frac{|F(x, u)|}{|u|^2}\leq \frac{1}{2}, \end{equation} $

(3.8)

从而

$ \begin{equation} F(x, u)\geq -\frac{1}{2}|u|^2. \end{equation} $

(3.9)

由(f1)条件知$ \exists \ M_{1} > 0 $, 使得$ \forall \ x\in R^3, \ \ \alpha \leq|u|\leq \delta $, 有

$ \begin{equation} |\frac{f(x, u)u}{u^2}|\leq \frac{|a(1+|u|^{p-1})u|}{|u|^2} \leq M_{1}, \end{equation} $

(3.10)

从而

$ \begin{equation} F(x, u)\geq -\frac{M_{1}}{2}|u|^2. \end{equation} $

(3.11)

结合式$ (3.7) $, $ (3.9) $, $ (3.11) $, $ \forall \ M>0, \ \exists \ M_{1}>0 $使得$ \forall \ x\in R^3, \ u\in R $, 取$ \tilde{M} = \frac{M_{1}+1}{2}+ M|u|^4 $, 有

$ \begin{equation} F(x, u)\geq -\tilde{M}|u|^2+M|u|^{4}. \end{equation} $

(3.12)

由式$ (2.4) $, $ (2.6) $, $ 2<q<3 $, 知

$ \begin{eqnarray*} \Phi(u)& = {}&\frac{1}{2}\|u\|^2+\frac{1}{4} \int_{R^{3}} K(x)\phi_{u} u^{2}dx- \int_{R^{3}}F(x, u)dx+ \int_{R^{3}}\frac{1}{4-q}b(x)|u|^{4-q}dx\\ &\leq{}&\frac{1}{2}\|u\|^{2}+\frac{1}{4}C_2 C_3^4 \|u\|^{4}+\tilde{M} C_3 ^2\|u\|^{2}-M \|u\|_4^4+\frac{1}{4-q}\|b\|_{\frac{4}{q}}C_3^{4-q}\|u\|^{4-q}, \\ \Phi(te)&\leq&\frac{1}{2}t^2\|e\|^2+\frac{1}{4}C_2 C_3^4 t^4 \|e\|^{4}+\tilde{M} C_3^2 t^2 \|e\|^{2}-Mt^4 \|e\|_4^4+\frac{1}{4-q}\|b\|_{\frac{4}{q}}t^{4-q}C_3^{4-q}\|e\|^{4-q}. \end{eqnarray*} $

取M充分大, 则$ \Phi(te)\leq0 $, 取$ v = te $, 当$ t $充分大时, $ \|v\|>r $, 使得$ \Phi(v)\leq0 $.

引理3.2  在定理$ 1.1 $的条件下, $ \Phi $的(PS)$ _c $序列有界.

证  因为$ \Phi(u_n)\rightarrow c $, 所以$ \exists \ M_2 >0 $, 使得$ \Phi(u_n)\leq M_2 $.又因$ \Phi'(u_n)\rightarrow0 $, 所以$ \forall \ \delta>0 $, $ \|\Phi'(u_n)\|\leq \delta $,

$ \begin{eqnarray*} \Phi(u_n)& = &\frac{1}{2}\|u_n\|^2+\frac{1}{4 } \int_{R^3}K(x)\phi_{u_{n}} u_{n}^2dx- \int_{R^3}F(x, u_n)dx+ \int_{R^{3}}\frac{1}{4-q}b(x)|u_n|^{4-q}dx, \\ \|u_n\|^{2}& = {}&2\Phi(u_n)-\frac{1}{2} \int_{R^3}K(x)\phi_{u_{n}}u_{n}^2dx+2 \int_{R^{3}}F(x, u_n)dx-2 \int_{R^{3}}\frac{1}{4-q}b(x)|u_n|^{4-q}dx\\ &\leq{}& 2M_2-\frac{1}{2} \int_{R^3}K(x)\phi_{u_{n}} u_{n}^2dx+2 \int_{R^{3}}F(x, u_n)dx-2 \int_{R^{3}}\frac{1}{4-q}b(x)|u_n|^{4-q}dx. \end{eqnarray*} $

由Cauchy-Schwartz不等式

$ |\langle\Phi'(u_n), u_n\rangle| = |\|u_n\|^{2}+ \int_{R^3}K(x)\phi_{u_{n}}u_{n}^2dx- \int_{R^{3}}f(x, u_n)u_n dx+ \int_{R^{3}}b(x)|u_n|^{4-q}dx| \leq \delta\|u_n\|, $

所以

$ \|u_n\|^{2}+ \int_{R^3}K(x)\phi_{u_{n}}u_{n}^2dx- \int_{R^{3}}f(x, u_n)u_n dx + \int_{R^{3}}b(x)|u_n|^{4-q}dx\geq -\delta\|u_n\|, $

即

$ \int_{R^{3}}f(x, u_n)u_n dx- \int_{R^3}K(x)\phi_{u_{n}} u_{n}^2dx - \int_{R^{3}}b(x)|u_n|^{4-q}dx\leq \|u_n\|^{2}+\delta\|u_n\|. $

由(f4)条件知$ 2F(x, u_n)\leq \frac{1}{2}f(x, u_n)u_n, $因为$ 2< q <3 $, 所以$ 1<\frac{2}{4-q}<2 $.因此

$ \begin{eqnarray*} \|u_n\|^{2}&\leq& 2M_2+\frac{1}{2}( \int_{R^{3}}f(x, u_n)u_n dx- \int_{R^3}K(x)\phi_{u_{n}} u_{n}^2dx- \int_{R^{3}}b(x)|u_n|^{4-q}dx)\\ &\leq& 2M_2+\frac{1}{2}\|u_n\|^{2}+\frac{\delta}{2} \|u_n\|, \end{eqnarray*} $

即$ \{u_n\} $有界.

引理3.3  在定理$ 1.1 $的条件下, $ \Phi $的有界(PS)$ _{c} $序列收敛, 即$ \Phi(u_n)\rightarrow c, \Phi'(u_n)\rightarrow 0 $, 则$ \{u_n\} $存在一个收敛的子列.

证  因为$ \{u_n\}\subset E $有界, 所以$ \exists \ u\in E $, 使得$ u_{n}\rightharpoonup u, $由(V)条件知, 在$ L^{t}(R^{3}), t\in[2, 6) $中, 有

$ \begin{equation} u_{n}\rightarrow u. \end{equation} $

(3.13)

因为

$ \begin{array}{l} \parallel {u_n} - u{\parallel ^2} = \langle \Phi '({u_n}) - \Phi '(u),{u_n} - u\rangle + \int_{{R^3}} {(f(} x,{u_n})\;\;\;\;\;\;\;\;\;\;\;\;\;\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; - (f(x,u))({u_n} - u)dx - \int_{{R^3}} {(g(} x,{u_n}) - (g(x,u))({u_n} - u)dx\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; - \int_{{R^3}} K (x)({\phi _{{u_n}}}{u_n} - {\phi _u}u)({u_n} - u)dx, \end{array} $

显然, 当$ n\rightarrow\infty $时,

$ \begin{equation} \langle\Phi'(u_{n})-\Phi'(u), u_{n}- u\rangle\rightarrow 0. \end{equation} $

(3.14)

由式$ (3.5) $可知

$ \begin{array}{l} \int_{{R^3}} {(f(} x,{u_n}) - f(x,u))({u_n} - u)dx\\ \le \int_{{R^3}} \varepsilon (|{u_n}| + |u|) + C(|{u_n}{|^{p - 1}} + |u{|^{p - 1}})({u_n} - u))dx \le \\ \varepsilon (\parallel {u_n}{\parallel _2} + \parallel u{\parallel _2})\parallel {u_n} - u{\parallel _2} + C(|{u_n}|_p^{p - 1} + |u|_p^{p - 1})\parallel {u_n} - u{\parallel _p}. \end{array}$

因为在$ L^s(R^3), \ \ 2 \leq s <6 $中, 有$ u_{n}\rightarrow u, \ \ $所以$ \int_{R^3}(f(x, u_{n})-f(x, u))(u_{n}-u)dx \rightarrow 0. $由Hölder不等式和(g)条件知

$ \begin{split} & \int_{R^3}(g(x, u_{n})-g(x, u))(u_{n}-u)dx\\ \leq{}& \int_{R^3} ( b(x)(|u_{n}|^{3-q}+|u|^{3-q})|u_{n}-u|)dx\\ \leq{}&( \int_{R^3}|b(x)|^{\frac{4}{3}}|u_{n}|^{\frac{4(3-q)}{3}}dx)^{\frac{3}{4}}+( \int_{R^3}|b(x)|^{\frac{4}{3}}|u|^{\frac{4(3-q)}{3}}dx)^{\frac{3}{4}}) \|u_{n}-u\|_{4}\\ \leq{}& \|b\|_{\frac{4}{q}}(\|u_n\|_{4}^{3-q}+\|u\|_{4}^{3-q})\|u_{n}-u\|_{4}. \end{split} $

因为在$ L^s(R^3), \ \ 2 \leq s <6 $中, 有$ u_{n}\rightarrow u, \ \ $所以$ \int_{R^3}(g(x, u_{n})-g(x, u))(u_{n}-u)dx \rightarrow 0. $由Hölder不等式,

$ \begin{split} \int_{R^3}K(x)\phi_{u_{n}} u_{n}(u_{n}-u)dx \leq {}& \|K\|_{\infty}\|\phi_{u_{n}}u_{n}\|_{2}\|u_{n}-u\|_{2}\\ \leq {}& \|K\|_{\infty}\|\phi_{u_n}\|_{6}\|u_{n}\|_{3}\|u_{n}-u\|_{2}\\ \leq{}& C_{1}\|K\|_{\infty}\|\phi_{u_{n}}\|_{\mathcal{D}^{1, 2}}\|u_{n}\|_{3}\|u_{n}-u\|_{2}\\ \leq{}& C_{1}^2 \|K\|_{\infty}^2\|u_{n}\|_{\frac{12}{5}}^{2} \|u_{n}\|_{3}\|u_{n}-u\|_{2}. \end{split} $

所以

$ \begin{equation} \int_{R^3}K(x)\phi_{u_{n}} u_{n}(u_{n}-u)dx \rightarrow 0. \end{equation} $

(3.15)

类似的

$ \begin{equation} \int_{R^3}K(x)\phi_{u} u(u_{n}-u)dx \rightarrow 0. \end{equation} $

(3.16)

因此

$ \begin{equation} \|u_{n}-u\| \rightarrow 0. \end{equation} $

(3.17)

则已证$ \forall \ c $, $ \Phi $的$ (PS)_{c} $条件成立.因此至少存在一个非零解$ u\in E, \ \ \Phi(u) = c, \ \ \Phi'(u) = 0 $.由式(2.9)知, 显然$ \mathcal{N} $非空, $ \forall \ u\in \mathcal{N} $, 有

$ \begin{equation} \begin{split} 0 = \langle \Phi'(u), u\rangle{}& = \|u\|^2 + K(x) \int_{R^3}\phi_u u^2 dx- \int_{R^3} f(x, u)udx+ \int_{R^3} b(x)|u|^{4-q}dx\\ {}&\geq\|u\|^2-\varepsilon C_{3}^2 \|u\|^{2}-CC_{3}^p\|u\|^p. \end{split} \end{equation} $

(3.18)

由$ \mathcal{N} $的定义, 显然$ u\not\equiv0 $, 取$ \varepsilon $充分小, 则

$ \|u\| \geq \sqrt[p-2]{\frac{1-\varepsilon C_{3}^2}{CC_{3}^p}}, $

又因为$ 2< q <3 $, 所以$ \frac{1}{2}<\frac{1}{4-q}<1 $.

$ \begin{equation} \begin{split} \Phi(u){}& = \Phi(u)-\frac{1}{4}\langle\Phi'(u), u \rangle\\ {}& = \frac{1}{4}\|u\|^{2}+ \int_{R^3}(\frac{1}{4}f(x, u)u-F(x, u))dx+ \int_{R^3}(G(x, u)-\frac{1}{4}g(x, u)u)dx\\ {}&\geq\frac{1}{4}\|u\|^2+(\frac{1}{4-q}-\frac{1}{4}) \int_{R^3} b(x)|u|^{4-q}dx\\ {}&\geq \frac{1}{4}\|u\|^2.\\ \end{split} \end{equation} $

(3.19)

所以$ \Phi $在$ \mathcal{N} $上是下有界的, 且$ \Phi $具有强制性, 即$ (\|u\|\rightarrow\infty, \ \ \Phi(u)\rightarrow \infty) $因此定义

$ m = \inf\limits_{\mathcal{N}}\Phi>0. $

引理3.4  (见文[14])设$ r>0 $, 如果$ \{u_{n}\}\subset H^{1}(R^{3}) $有界, 且

$ \lim \limits_{n \rightarrow\infty}\sup \limits_{y\in R^{3}} \int_{B_{r}(y)}|u_{n}|^{2}dx = 0, $

则$ \forall \ s\in(2, 6) $, 在$ L^{s}(R^{3}) $中有$ u_{n}\rightarrow 0 $.

定理1.1的证明  设$ \{u_{n}\}\subset \mathcal{N} $是$ \Phi $的一个极小化序列, 则$ \{u_{n}\} $有界, 而且通过一个恰当的$ Z^3 $变换, $ \exists \ u \in \mathcal{N} $, 使得$ u_n \rightharpoonup u $和$ \Phi(u) = s = \inf\limits_{\mathcal{N}}\Phi. $

证  由$ \Phi $具有强制性, 显然, $ \{u_{n}\} $有界.在子列意义条件下, 有$ u_n \rightharpoonup u $.假设引理$ 3.4 $成立, 结合式$ (3.5) $, 当$ n\rightarrow\infty $时, 有$ \int_{R^3} f(x, u_n)u_{n} = o(\|u_n\|) $.从而有

$ \begin{align*} o(\|u_{n}\|)& = \langle\Phi'(u_{n}), u_{n}\rangle = \|u_n\|^{2}+ \int_{R^{3}}K(x)\phi_{u_{n}}u_{n}^{2}dx- \int_{R^{3}}f(x, u_{n})u_{n}dx+ \int_{R^{3}} b(x)|u_n|^{4-q}dx\\ &\geq\|u_{n}\|^{2}-o(\|u_{n}\|). \end{align*} $

所以$ \|u_{n}\|\rightarrow 0 $, 这与式$ (3.18) $相矛盾, 因此$ \exists \ r >0, \{y_{n}\}\subset Z^{3} $, 满足

$ \lim \limits_{n \rightarrow\infty}\sup \limits_{y\in R^{3}} \int_{B_{r}(y)}|u_{n}|^{2}dx>0. $

因此$ u_{n}\rightharpoonup u\neq0 $, 则$ \Phi'(u) = 0 $.因此$ u\in\mathcal{N} $.显然$ \Phi(u)\geq s $.由(f4)条件, Fatou's引理, $ \|\cdot\| $的弱连续和$ \{u_{n}\} $有界得

$ \begin{align*} s+o(1)& = \Phi(u_{n})-\frac{1}{4}\langle\Phi'(u_{n}), u_{n}\rangle\\ & = \frac{1}{4}\|u_n\|^{2}+ \int_{R^3}(\frac{1}{4}f(x, u_{n})u_{n}-F(x, u_{n})dx+ \int_{R^3}(G(x, u_{n})-\frac{1}{4}g(x, u_{n})u_{n})dx\\ &\geq\frac{1}{4}\|u\|^{2}+ \int_{R^3}(\frac{1}{4}f(x, u)u-F(x, u)dx+ \int_{R^3}(G(x, u)-\frac{1}{4}g(x, u) u) dx+o(1)\\ & = \Phi(u)-\frac{1}{4}\langle\Phi'(u), u\rangle+o(1) = \Phi(u)+o(1). \end{align*} $

所以$ \Phi(u)\leq s $, 从而$ \Phi(u) = s = \inf \limits_{\mathcal{N}}\Phi>0. $