考虑下列含临界指数椭圆方程

\begin{equation} \left\{ \begin{array}{*{35}{l}} -\text{div}({{\left| x \right|}^{-2a}}\nabla u)-\mu {{\left| x \right|}^{-2(a+1)}}u=f(x){{\left| x \right|}^{-bp}}{{\left| u \right|}^{p-2}}u+\lambda g(x),\ x\in {{\mathbf{R}}^{N}}/\{0\},\\ u\in H_{a}^{1}({{\mathbf{R}}^{N}}),\\ \end{array} \right. %\eqno(1) \end{equation}

(1.1)

其中$N \geq 3,\lambda > 0,p = 2N / (N - 2 + 2(b - a))$是临界Caffarelli-Kohn-Nirenberg指数, $0 \le a < (N - 2) / 2,a \le b < a + 1,0 \le \mu < \mu _1 = ((N - 2(a + 1)) / 2)^2.$ $H_a^1 ({\rm {\bf R}}^N)$表示$C_0^\infty ({\rm {\bf R}}^N)$ 的完备化空间,它的范数是

$$ \left\| u \right\|_1 = \left( {\int_{{\rm {\bf R}}^N} {\left| x \right|^{ - 2a}\left| {\nabla u} \right|^2{\rm d}x} } \right)^{1 / 2}, $$

在方程(1.1)中,设$(H_a^1 ({\rm {\bf R}}^N))'$表示$H_a^1 ({\rm {\bf R}}^N)$的对偶空间. $g(x) \in (H_a^1 ({\rm {\bf R}}^N))'\backslash \{0\}$ 是${\rm {\bf R}}^N$ 上的一个连续的函数. $f(x)$是${\rm {\bf R}}^N$上的有界函数,满足条件

$(A)~f\in {{L}^{\infty }}({{\mathbf{R}}^{N}}),\underset{\left| x \right|\to 0}{\mathop{\lim }}\,f(x)=\underset{\left| x \right|\to \infty }{\mathop{\lim }}\,f(x)={{f}_{0}}>0,f(x)\ge {{f}_{0}},a.e.x\in {{\mathbf{R}}^{N}}.$

近来,关于含临界指数椭圆方程解的存在性得到了广泛的研究,当$a=b=0,\mu = 0$时, Tarantello在文献[1]中证明了,当

$$\int_\Omega {\lambda gu {\rm d}x < \frac{4}{N - 2}} \Big (\frac{N - 2}{N + 2}\Big)^{(N + 2) / 4}(\left\| {\nabla u} \right\|_2 )^{(N + 2) / 2},$$

方程(1.1)存在两个正解. 曹在文献[2]中应用山路引理研究了方程

$$\left\{ \begin{array}{*{35}{l}} -\text{div}({{\left| \nabla u \right|}^{p-2}}\nabla u)={{\left| u \right|}^{{{p}^{*}}-2}}u+\lambda h(x),\ x\in {{\mathbf{R}}^{N}},\\ u\in D_{a}^{1}({{\mathbf{R}}^{N}}) \\ \end{array} \right.$$

解的存在性. 王在文献[3]中研究了方程(1.1)中$a=0$的情况. 相似的方法可见文献[4, 5, 6, 8, 9]. 文献[10, 11, 12, 13, 15]利用Nehari流形的方法得到了椭圆方程的多解性. 本文应用Nehari 流形和变分方法,在一定条件下,方程(1.1) 存在两个非平凡解.

设空间$E$表示$C_0^\infty ({\rm {\bf R}}^N)$的完备化空间,它的范数是

$$ \left\| u \right\| = \left( {\int_{{\rm {\bf R}}^N} {(\left| x \right|^{ - 2a}\left| {\nabla u} \right|^2\mbox{ - }\mu \left| x \right|^{ - 2(a + 1)}u^2){\rm d}x} } \right)^{1 / 2}. $$

由Caffarelli-Kohn-Nirenberg不等式

$$ \left( {\int_{{\rm {\bf R}}^N} {\left| x \right|^{ - bp}\left| {u} \right|^p{\rm d}x} } \right)^{1 / p} \le C(a,b)\left( {\int_{{\rm {\bf R}}^N} {\left| x \right|^{ - 2a}\left| {\nabla u} \right|^2{\rm d}x} } \right)^{1 / 2} $$

及Hardy 不等式

$$ \int_{{\rm {\bf R}}^N} {\left| x \right|^{ - 2(a + 1)}u^2{\rm d}x} \le \mu _1 ^{ - 1}\int_{{\rm {\bf R}}^N} {\left| x \right|^{ - 2a}\left| {\nabla u} \right|^2{\rm d}x} $$

知道范数$\left\| \cdot \right\|_1$与$\left\| \cdot \right\|$等价.

对于$u \in E,u \ne 0,$ 设

$$ S =\inf \frac{\int_{{\rm {\bf R}}^N} {(\left| x \right|^{ - 2a}\left| {\nabla u} \right|^2{\rm d}x} - \mu \left| x \right|^{ - 2(a + 1)}u^2\mbox{)d}x}{\left( {\int_{{\rm {\bf R}}^N} {\left| x \right|^{ - bp}\left| u \right|^p{\rm d}x} } \right)^{2 / p}}, $$

记

$$\left\| u \right\|_{b,p} = \left( {\left. {\int_{{\rm {\bf R}}^N} {\left| x \right|^{ - bp}\left| u \right|^p{\rm d}x} } \right)} \right.^{1 / p},$$

则有

$\begin{equation} \left\| u \right\|_{b,p}^p \le \left\| u \right\|^pS^{ - p / 2}. \end{equation}$

(1.2)

$S$的达到函数

$$v_\varepsilon (x) = \frac{(2p(\mu _1 - \mu )\varepsilon ^2)^{\beta_{1}} }{\left| x \right|^\gamma (\varepsilon ^2 + \left| x \right|^{\sqrt {\mu _1 - \mu } / {\beta_{1}} })^{2{\beta_{1}} }},$$

其中$\gamma = \sqrt {\mu _1 } - \sqrt {\mu _1 - \mu }, \beta_{1} = (N - 2(1 + a - b)) / 4(1 + a - b),$ 且$v_\varepsilon (x)$是下列方程的唯一正解

$$ - \mbox{div}(\left| x \right|^{ - 2a}\nabla u) - \mu \left| x \right|^{ - 2(a + 1)}u = \left| x \right|^{ - bp}\left| u \right|^{p - 2}u,u \in E. $$

因此有

$$ \left\| {v_\varepsilon } \right\|^{2}= \left\| {v_\varepsilon } \right\|_{b,p}^p = S^{p / (p - 2)}, $$

详见文献[7].

下面给出弱解的定义.

定义1.1   $u\in E$是方程(1.1)的一个弱解是指任意$\phi \in E$满足

$$ \int_{{\rm {\bf R}}^N} {(\left| x \right|^{ - 2a}\nabla u\nabla \phi - \mu \left| x \right|^{ - 2(a + 1)}u\phi ){\rm d}x = } \int_{{\rm {\bf R}}^N} {(f(x)\left| x \right|^{ - bp}\left| u \right|^{p - 2}u\phi + \lambda g(x)\phi } ){\rm d}x. $$

下面给出本文的主要结果.

定理1.2   假设条件(A)满足,$0 < \lambda < C_1 ,$ 则方程(1.1)至少存在一个非平凡解. 其中 $C_1$的定义将在下文中引理2.3 中给出.

定理1.3   假设条件(A)满足,则存在$\Lambda > 0,$ 当$0 < \lambda < \Lambda < C_1 ,$ 方程(1.1)至少存在两个非平凡解.

定义能量泛函

$$\begin{align} & J(u)=\frac{1}{2}{{\left\| u \right\|}^{2}}-\frac{1}{p}\int_{{{\mathbf{R}}^{N}}}{f(x){{\left| x \right|}^{-bp}}{{\left| u \right|}^{p}}}\text{d}x-\lambda \int_{{{\mathbf{R}}^{N}}}{g(x)u}\text{d}x \\ & \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}{{\left\| u \right\|}^{2}}-\frac{1}{p}A(u)-K(u),\\ \end{align}$$

其中

$$A(u) = \int_{{\rm {\bf R}}^N} {f(x)\left| x \right|^{ - bp}\left| u \right|^p} {\rm d}x,~K(u) = \lambda \int_{{\rm {\bf R}}^N} {g(x)u} {\rm d}x.$$

考虑Nehari流形

$$N_{\lambda} = \left\{ {u} \right. \in E\backslash \{(0,0)\}\left| {\left. {\left\langle {J'(u),u} \right\rangle = 0} \right\} } \right.,$$

因此$u \in N_{\lambda} $当且仅当

$$\left\langle {J'(u),u} \right\rangle = \left\| {u} \right\|^2 - A(u)- K(u) = 0.$$

我们有下列结论.

引理2.1   在$N_{\lambda}$里,$J(u)$是强制的、有下界的.

证   对于$u \in N_{\lambda}$,

$$\begin{align} & J(u)=\frac{1}{2}{{\left\| u \right\|}^{2}}-\frac{1}{p}A(u)-K(u) \\ & =\frac{1}{2}{{\left\| u \right\|}^{2}}-\frac{1}{p}\left( \left. {{\left\| u \right\|}^{2}}-K(u) \right) \right.-K(u) \\ & =(\frac{1}{2}-\frac{1}{p}){{\left\| u \right\|}^{2}}-(1-\frac{1}{p})K(u). \\ \end{align}$$

设$\left\| g \right\|_{E'}$表示$g(x)$在$E$的对偶空间的范数,于是

$\begin{equation} K(u) = \lambda \int_{{\rm {\bf R}}^N} {g(x)u} {\rm d}x \le \lambda \left\| g \right\|_{E'} \left\| u \right\|, \end{equation}$

(2.1)

因此

$\begin{equation} J(u) \ge \Big(\frac{1}{2} - \frac{1}{p}\Big)\left\| u \right\|^2 -\Big (1 - \frac{1}{p}\Big)\lambda \left\| g \right\|_{E'} \left\| u \right\|. \end{equation}$

(2.2)

于是$J(u)$是强制的、有下界的.

定义

$$ M(u) = \left\langle {J'(u),u} \right\rangle. $$

对于$u \in N_{\lambda}$,

$\begin{align} & \left\langle {M}'(u),u \right\rangle =2{{\left\| u \right\|}^{2}}-pA(u)-K(u) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(2-p)A(u)+K(u) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\left\| u \right\|}^{2}}-(p-1)A(u) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(2-p){{\left\| u \right\|}^{2}}+(p-1)K(u). \\ \end{align}$

(2.3)

把$N_{\lambda}$分成三个部分

$$\begin{align} & N_{\lambda }^{+}=\left\{ u \right.\in {{N}_{\lambda }}\left| \left. \left\langle {M}'(u),u \right\rangle >0 \right\} \right.,\\ & N_{\lambda }^{0}=\left\{ u \right.\in {{N}_{\lambda }}\left| \left. \left\langle {M}'(u),u \right\rangle =0 \right\} \right.,\\ & N_{\lambda }^{-}=\left\{ u \right.\in {{N}_{\lambda }}\left| \left. \left\langle {M}'(u),u \right\rangle <0 \right\} \right.. \\ \end{align}$$

于是,我们有下列结论.

引理2.2   假设$u_0 $是$J$在$N_{\lambda}$里的一个极小值点,且$u_0 \notin N_{\lambda}^{0}$,则$J'(u_0) = 0$,即 $u_0 $是$J(u)$的一个临界点.

证    此证明类似于文献[13],这里略去证明.

引理2.3   假设

$$0 < \lambda < C_1 = \left( {\left. {\frac{S^{p / 2}}{(p - 1)\left\| f \right\|_\infty }} \right)} \right.^{\frac{1}{p - 2}}\frac{p - 2}{\left\| g \right\|_{E'} (p - 1)},$$

则$N_{\lambda}^{0} = \emptyset .$

证   假设当$0 < \lambda < C_1 ,N_{\lambda}^{0} \notin \emptyset ,$ 于是对于$u \in N_{\lambda}^{0},$ 由(1.2)式得到

$\begin{equation} A(u) = \int_{{\rm {\bf R}}^N} {f(x)\left| x \right|^{ - bp}\left| u \right|^p} {\rm d}x \le \left\| f \right\|_\infty \left\| u \right\|^pS^{ - p / 2}, \end{equation}$

(2.4)

由(2.1),(2.3)和(2.4)式得到

$$\left( {\left. {\frac{S^{p / 2}}{(p - 1)\left\| f \right\|_\infty }} \right)} \right.^{\frac{1}{p - 2}} \le \left\| u \right\| \le \frac{p - 1}{p - 2}\lambda \left\| g \right\|_{E'}.$$

因此

$$ \lambda \geq C_1 = \left( {\left. {\frac{S^{p / 2}}{(p - 1)\left\| f \right\|_\infty }} \right)} \right.^{\frac{1}{p - 2}}\frac{p - 2}{\left\| g \right\|_{E'} (p - 1)},$$

矛盾,因此,结论成立.

由引理2.3,$N_{\lambda} = N_{\lambda}^{+} \cup N_{\lambda}^{-} ,$ 由引理2.1,我们可以定义

$$\xi = \mathop {\inf }\limits_{u \in N_{\lambda} } J(u),~ \xi ^ + = \mathop {\inf }\limits_{u \in N_{\lambda}^{+} } J(u),~ \xi ^ - = \mathop {\inf }\limits_{u \in N_{\lambda}^{-} } J(u).$$

引理2.4   假设$0 < \lambda < C_1 ,$ 则$\xi \le \xi ^ + < 0;$ 假设$0 < \lambda < \frac{1}{2}C_1 ,$ 则 存在$C_0 = C_0 (p,S,\lambda ,\left\| f \right\|_\infty ,\left\| g \right\|_{E'} ) > 0,$ 使得$\xi ^ - > C_0 .$

证 设$u \in N_{\lambda}^{+}$,由(2.3)式

$$A(u) < \frac{1}{p - 1}\left\| u \right\|^2,$$

因此

$$\begin{align} & J(u)=(\frac{1}{2}-1){{\left\| u \right\|}^{2}}+(1-\frac{1}{p})A(u) \\ & \ \ \ \ \ \ \ \ <[(\frac{1}{2}-1)+(1-\frac{1}{p})\frac{1}{p-1}]{{\left\| u \right\|}^{2}}<0,\\ \end{align}$$

于是$\xi \le \xi ^ + < 0.$

(2)~ 设$u \in N_{\lambda}^{-}$,由(2.1),(2.3)和(2.4)式得到$A(u) > 0$,且

$$\frac{1}{p - 1}\left\| u \right\|^2 < A(u) \le \left\| f \right\|_\infty \left\| u \right\|^pS^{ - p / 2},$$

于是

$$\left\| u \right\| > \left( {\left. {\frac{S^{p / 2}}{(p - 1)\left\| f \right\|_\infty }} \right)} \right.^{\frac{1}{p - 2}}.$$

由(2.2)式

$$\begin{align} & J(u)\ge (\frac{1}{2}-\frac{1}{p}){{\left\| u \right\|}^{2}}-(1-\frac{1}{p})\lambda {{\left\| g \right\|}_{{{E}'}}}\left\| u \right\| \\ & \ \ \ \ \ \ \ =\left\| u \right\|[(\frac{1}{2}-\frac{1}{p})\left\| u \right\|-\lambda (1-\frac{1}{p}){{\left\| g \right\|}_{{{E}'}}}] \\ & \ \ \ \ >{{\left( \left. \frac{{{S}^{p/2}}}{(p-1){{\left\| f \right\|}_{\infty }}} \right) \right.}^{\frac{1}{p-2}}}[(\frac{1}{2}-\frac{1}{p}){{\left( \frac{{{S}^{p/2}}}{(p-1){{\left\| f \right\|}_{\infty }}} \right)}^{\frac{1}{p-2}}}-\lambda (1-\frac{1}{p}){{\left\| g \right\|}_{{{E}'}}}]. \\ \end{align}$$

因此,当$0 < \lambda < \frac{1}{2}C_1$时,

$$J(u) > C_0 = C_0 = C_0 (p,S,\lambda ,\left\| f \right\|_\infty ,\left\| g \right\|_{E'} ) > 0, \xi ^ - > C_0 . $$

证毕.

对于每个$u \in E$,且$A(u)>0,$ 记

$$ t_0 = \bigg( \frac{\left\| u \right\|^2}{(p - 1)A(u)} \bigg) ^{\frac{1}{p - 2}} > 0, $$

引理2.5   对于$u \in E,0 < \lambda < C_1 ,$ 且$A(u) > 0,$ 有下列结论

(1) 假如$K(u) \le 0$,则存在唯一的$t_1 > t_0$使得$(t_1 u) \in N_{\lambda}^{-} ,$ 且

$$ J(t_1 u) = \mathop {\sup }\limits_{t \ge 0} J(tu). $$

(2) 假如$K(u) > 0$,则存在唯一的$0 < t_2 < t_0 < t_1$使得$t_1 u \in N_{\lambda}^{-} , t_2 u \in N_{\lambda}^{+} ,$ 且

$$ J(t_2 u) = \mathop {\inf }\limits_{0 \le t \le t_0 } J(tu),~ J(t_1 u) = \mathop {\sup }\limits_{t \ge 0} J(tu). $$

证   设$u \in E$,且$A(u) > 0$,

$$\begin{align} & \left\langle {J}'(tu),tu \right\rangle ={{t}^{2}}{{\left\| u \right\|}^{2}}-{{t}^{p}}A(u)-tK(u) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =t[t{{\left\| u \right\|}^{2}}-{{t}^{p-1}}A(u)-K(u)]. \\ \end{align}$$

设

$$h(t) = t\left\| u \right\|^2 - t^{p - 1}A(u),t \ge 0.$$

显然$h(0) = 0,t \to + \infty ,h(t) \to - \infty ,$

$$h'(t) = \left\| u \right\|^2 - (p - 1)t^{p - 2}A(u).$$

令$h'(t) = 0,$ 得到$t = t_0,$ 当$t \in [0,t_0 ),h'(t) > 0,$ 当$t \in [t_0 ,\infty ), h'(t) < 0,$ 因此$h(t)$在$t = t_0$达到最大值,由(2.4)式

$\begin{align} & h({{t}_{0}})={{t}_{0}}{{\left\| u \right\|}^{2}}-t_{0}^{p-1}A(u) \\ & \ \ \ \ \ \ \ \ \ \ =\left\| u \right\|[{{\left( \left. \frac{1}{p-1} \right) \right.}^{\frac{1}{p-2}}}-{{\left( \left. \frac{1}{p-1} \right) \right.}^{\frac{p-1}{p-2}}}]{{\left( \left. \frac{{{\left\| u \right\|}^{p}}}{A(u)} \right) \right.}^{\frac{1}{p-2}}} \\ & \ \ \ \ \ \ \ \ \ \ge \left\| u \right\|\left( \left. \frac{p-2}{p-1} \right) \right.{{\left( \left. \frac{{{S}^{p/2}}}{(p-1){{\left\| f \right\|}_{\infty }}} \right) \right.}^{\frac{1}{p-2}}}. \\ \end{align}$

(2.5)

(1) 假如$K(u) \le 0,$ 则存在唯一的$t_1 > t_0$使得$h(t_1 ) = K(u),$ 且$h'(t_1 ) < 0$,由(2.3)式得到

$$\begin{align} & \left\langle {J}'({{t}_{1}}u),({{t}_{1}}u) \right\rangle =t_{1}^{q}[h({{t}_{1}})-K(u)]=0,\\ & \left\langle {M}'(u),u) \right\rangle =2{{\left\| u \right\|}^{2}}-pA(u)-K(u),\\ \end{align}$$

且

$$\begin{align} & \left\langle {M}'({{t}_{1}}u),({{t}_{1}}u) \right\rangle =2t_{1}^{2}{{\left\| u \right\|}^{2}}-pt_{1}^{p}A(u)-{{t}_{1}}K(u) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =t_{1}^{2}{{\left\| u \right\|}^{2}}-(p-1)t_{1}^{p}A(u) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =t_{1}^{2}[{{\left\| u \right\|}^{2}}-(p-1)t_{1}^{p-2}A(u)] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =t_{1}^{2}{h}'({{t}_{1}})<0. \\ \end{align}$$

于是$t_1 u \in N_{\lambda}^{-} .$ 又$h(t_0 ) = t_0 \left\| u \right\|^2 - t_0^{p - 1} A(u)$,

$$ \frac{\rm d}{{\rm d}t}J(t_1 u) = t_1 \left\| u \right\|^2 - t_1^{p - 1} A(u) - K(u) = h(t_1 ) - K(u) = 0.$$

当$t \in [0,t_1 ),\frac{\rm d}{{\rm d}t}J(tu) > 0;$ 当$t \in (t_1 ,+ \infty ), \frac{\rm d}{{\rm d}t}J(tu) < 0$,因此

$$ J(t_1 u) = \sup \limits_{t \ge 0} J(tu). $$

(2) 假如$K(u) > 0,0 < \lambda < C_1 ,$ 由(2.1)和(2.5)式得到

$$\begin{align} & h(0)=0 < K(u)\le \lambda {{\left\| g \right\|}_{{{E}'}}}\left\| u \right\| \\ & \ \ \ \ \ \ \ \le \left( \left. \frac{p-2}{p-1} \right) \right.{{\left( \left. \frac{{{S}^{p/2}}}{(p-1){{\left\| f \right\|}_{\infty }}} \right) \right.}^{\frac{1}{p-2}}}\left\| u \right\|\le h({{t}_{0}}). \\ \end{align}$$

因此存在唯一的$t_2$和$t_1$,当$0 < t_2 < t_0 < t_1 $时,$h(t_2 ) = h(t_1 ) = K(u),$ 且$h'(t_2 ) > 0 > h'(t_1 ).$ 于是$t_1 u \in N_{\lambda}^{-},t_2 u \in N_{\lambda}^{+} $,当$t_2 < t < t_1 ,J(t_2 u) \le J(tu) \le J(t_1 u),$ 当$0 < t < t_2 ,J(t_2 u) \le J(tu),$ 于是

$$J(t_2 u) = \mathop {\inf }\limits_{0 \le t \le t_0 } J(tu),~ J(t_1 u) = \mathop {\sup }\limits_{t \ge 0} J(tu).$$

证毕.

定义3.1   序列$\{u_n \} \subset E$叫做一个$(PS)_c$列, 假如存在$c \in {\rm {\bf R}}$有

$$I(u_n ) \to c,~ I'(u_n ) \to 0,n \to \infty .$$

引理3.2   假设条件(A)满足,$0 < \lambda < C_1 ,$ 则泛函$J(u)$存在一个$(PS)_\xi$列$\{u_n \} \subset N_{\lambda} $; 假设条件(A)满足,$0 < \lambda < {(q/2)}C_1 ,$ 则泛函$J(u)$存在一个$(PS)_{\xi^{-}}$列$\{u_n \} \subset N_{\lambda}^{-}$.

证   此证明类似于文献[15],这里略去证明.

引理3.3   假设条件(A)满足,$0 < \lambda < C_1 ,$ 则泛函$J(u)$有一个极小值点$u_0^ + \in N^{+}_\lambda,$ 且满足

(1) $J(u_0^ + ) = \xi = \xi ^ + < 0$.

(2) $u_0^ + $是方程(1.1)的一个非平凡解.

证   由引理3.2知道,泛函$J(u)$存在一个$(PS)_\xi$列$\{u_n \} \subset N_{\lambda} $ 使得

$$J(u_n ) \to c,J'(u_n ) \to 0,n \to \infty .$$

由引理2.1,在$N_{\lambda}$里面,$J(u)$是强制的,在$E$里,因此$\{u_n \}$是有界的, 由紧嵌入定理,存在一个子列(不妨仍记作$\{u_n \}$)以及$u_0^ + \in E,$ 且在$E$里及在$L^p({\rm {\bf R}}^N,\left| x \right|^{ - bp})$里, $u_n \rightharpoonup u_0^ +$,在${\rm {\bf R}}^N$里,$u_n \rightarrow u_0^ + $ a.e., 于是由(2.1)式得到当$n \to \infty,K(u_n ) \to K(u_0^ + ),$ 于是, $u_0^ + $是方程(1.1)的一个解. 此外由于$\{u_n \} \subset N_{\lambda}$,于是

$$J(u_n ) = \Big(\frac{1}{2} - \frac{1}{p}\Big)\left\| {u_n } \right\|^2 - \Big(1 - \frac{1}{p}\Big)K(u_n ) \ge - \Big(1 - \frac{1}{p}\Big)K(u_n ).$$

让$ n \to \infty$,由引理2.4,$\xi < 0$,

$$0 > \xi \ge -\Big (1 - \frac{1}{p}\Big)K(u_0^ + ),$$

于是$K(u_0^ +) > 0$,$u_0^ +$是方程(1.1)的一个非平凡解.

现在我们证明: 在$E$里$u_n \rightarrow u_0^ + $且$J(u_0^ + ) = \xi$. 注意到$\{u_n \} \subset N_{\lambda}$,由Fatou引理有

$$\begin{align} & \xi \le J(u_{0}^{+})=(\frac{1}{2}-\frac{1}{p}){{\left\| u_{0}^{+} \right\|}^{2}}-(1-\frac{1}{p})K(u_{0}^{+}) \\ & \ \ \ \le \underset{n\to \infty }{\mathop{\lim }}\,\inf [(\frac{1}{2}-\frac{1}{p}){{\left\| {{u}_{n}} \right\|}^{2}}-(1-\frac{1}{p})K({{u}_{n}})] \\ & \ \ =\underset{n\to \infty }{\mathop{\lim }}\,\inf J({{u}_{n}})=\xi . \\ \end{align}$$

因此$J(u_0^ + ) = \xi,\lim \limits_{n \to \infty } \left\| {u_n } \right\|^2 = \left\| {u_0^ + } \right\|^2.$ 设$v_n = u_n - u_0^ +$,于是$v_n \rightharpoonup 0$. 由Brezis-lemma 引理(参见文献[5])推出

$$\left\| {u_n } \right\|^2 = \left\| {v_n } \right\|^2 + \left\| {u_0^ + } \right\|^2 + o_n (1),$$

因此$u_n \rightarrow u_0^ + $,此外$u_0^ + \in N^{+}_\lambda.$ 如果$u_0^ + \in N^{-}_\lambda,$ 由(2.3)(2.4)式得到$A(u_0^ +)>0$,由引理2.5,存在唯一的$t_{2}$和$t_{1}$ 使得$t_1 u_0^ + \in N_{\lambda}^{-} ,t_2 u_0^ + \in N_{\lambda}^{+} ,$ 且$t_2 < t_1 = 1.$ 由于当$t \in (t_2 ,t_1 ),\frac{\rm d}{{\rm d}t}J(tu_0^ + ) > 0,$ 于是存在$\tau \in (t_2 ,t_1 )$使得

$$\xi \le \xi ^ + \le J(t_2 u_0^ + ) < J(\tau u_0^ + ) < J(t_1 u_0^ + ) = \xi,$$

矛盾,因此$u_n \to u_0^ +$且$J(u_0^ + ) = \xi ,$ 由引理2.2, 知道$u_0^ + $是方程(1.1)的一个非平凡解. 证毕.

由引理3.3,完成定理1.2的证明.

引理4.1   对于$c \in {\rm {\bf R}}$,若序列$\{u_n \} \subset E$是 泛函$J$的一个$(PS)_c$列,且$\{u_n\}\rightharpoonup u \in E$,则$J'(u)= 0$, 并且存在常数$C_2 = C_2 (p) > 0$ 使得$J(u) \ge - C_2 \lambda ^2\left\| g \right\|_{E'}.$

证   对于$c \in {\rm {\bf R}}$,若序列$\{u_n \} \subset E$是泛函$J$ 的一个$(PS)_c$ 列,且$\{u_n\}\rightharpoonup u \in E$,容易看出$J'(u) = 0, \left\langle {J'(u),u} \right\rangle = 0,u \in N$,由(2.2)式及Young表达式得到

$$\begin{align} & J(u)\ge (\frac{1}{2}-\frac{1}{p}){{\left\| u \right\|}^{2}}-(1-\frac{1}{p})\lambda {{\left\| g \right\|}_{{{E}'}}}\left\| u \right\| \\ & \ \ \ \ \ \ \ \ge (\frac{1}{2}-\frac{1}{p}){{\left\| u \right\|}^{2}}-(\frac{1}{2}-\frac{1}{p}){{\left\| u \right\|}^{2}}-{{C}_{2}}{{\lambda }^{2}}{{\left\| g \right\|}_{{{E}'}}} \\ & \ \ \ \ \ \ =-{{C}_{2}}{{\lambda }^{2}}{{\left\| g \right\|}_{{{E}'}}}. \\ \end{align}$$

其中正常数$C_2 = C_2 (p).$

引理4.2   对于$c \in {\rm {\bf R}}$,若序列$\{u_n \} \subset E$是泛函$J$的一个$(PS)_c$列,则在$E$里,$\{u_n \}$有界.

证   由$(PS)_c$列的定义,存在$c \in {\rm {\bf R}}$,有

$$ J(u_n ) \to c,J'(u_n ) \to 0,n \to \infty . $$ $$ J(u_n ) = \frac{1}{2}\left\| {u_n } \right\|^2 - \frac{1}{p}A(u_n ) - K(u_n ) = c + o_n (1), $$ $$ \left\langle {J'(u_n ),u_n } \right\rangle = \left\| {u_n } \right\|^2 - A(u_n ) - K(u_n ) = o_n (1). $$

于是

$$ c + o_n (1) = J(u_n ) - \frac{1}{p}\left\langle {J'(u_n ),u_n } \right\rangle = \Big(\frac{1}{2} - \frac{1}{p}\Big)\left\| {u_n } \right\|^2 - \Big(1 - \frac{1}{p}\Big)K(u_n ). $$

由(2.1)式

$$ c + o_n (1) + \Big(1 - \frac{1}{p}\Big)\lambda \left\| g \right\|_{E'} \left\| {u_n } \right\| \ge\Big (\frac{1}{2} - \frac{1}{p}\Big)\left\| {u_n } \right\|^2, $$

当$n \to \infty ,\left\| {u_n } \right\| \to \infty$,则由上式得出矛盾, 故$\{u_n \}$有界.

引理4.3   对于$c \in {\rm {\bf R}}$,若序列$\{u_n \} \subset E$是泛函$J$的一个$(PS)_c$列,则当

$$c < c^\ast = \Big(\frac{1}{2} - \frac{1}{p}\Big)(S^{p / 2}f_0^{ - 1} )^{2 / (p - 2)} - C_2 \lambda ^2\left\| g \right\|_{E'}$$

时,$u_n (x) \to u(x).$

证 由$(PS)_c$列的定义,存在$c \in {\rm {\bf R}}$,有

$$J(u_n ) \to c,J'(u_n ) \to 0,n \to \infty .$$

由引理4.2,在$E$里,$\{u_n \}$有界,则在$E$里存在弱收敛的子列, 不妨仍记为$\{u_n \}$,存在$u(x) \in E$,在$E$里及在$L^p({\rm {\bf R}}^N, \left| x \right|^{ - bp})$里,$u_n \rightharpoonup u$. 在${\rm {\bf R}}^N$里, $u_n \rightarrow u$~a.e.,$\left\langle {J'(u),\phi } \right\rangle = 0,u$ 是方程(1.1)的一个解,$J'(u) = 0$,且$K(u_n ) \to K(u).$

设$v_n = u_n - u$,于是$v_n \rightharpoonup 0$. 由Brezis-lemma引理(文献[5])推出

$$\begin{align} & {{\left\| {{u}_{n}} \right\|}^{2}}={{\left\| {{v}_{n}} \right\|}^{2}}+{{\left\| u \right\|}^{2}}+{{o}_{n}}(1),\\ & A({{u}_{n}})=A({{v}_{n}})+A(u)+{{o}_{n}}(1),\\ \end{align}$$

因此

$$\begin{align} & J({{u}_{n}})=J(u)+\frac{1}{2}{{\left\| {{v}_{n}} \right\|}^{2}}-\frac{1}{p}A({{v}_{n}})+{{o}_{n}}(1),\\ & \left\langle {J}'({{u}_{n}}),{{u}_{n}} \right\rangle ={{\left\| {{v}_{n}} \right\|}^{2}}-A({{v}_{n}})+{{o}_{n}}(1). \\ \end{align}$$

由条件(A)假设

$$\mathop {\lim }\limits_{n \to \infty } \left\| {v_n } \right\|^2 = \mathop {\lim }\limits_{n \to \infty } A(v_n ) = f_0 \lim\limits_{n \to \infty } \left\| {v_n } \right\|_{b,p}^p = l,$$

由(1.2)式$l \le f_0 l^{p / 2}S^{ - p / 2},l = 0$,或者$l \ge (S^{p / 2}f_0^{ - 1} )^{2 / (p - 2)}.$ 若$l \ge (S^{p / 2}f_0^{ - 1} )^{2 / (p - 2)},$

$$J(u_n ) - \frac{1}{2}\left\langle {J'(u_n ),u_n } \right\rangle = J(u) + \Big(\frac{1}{2} - \frac{1}{p}\Big)A(v_n ) + o_n (1).$$

让$n \to \infty$,由引理4.1知道

$$c = J(u) + \Big(\frac{1}{2} - \frac{1}{p}\Big)l \ge\Big (\frac{1}{2} - \frac{1}{p}\Big)(S^{p / 2}f_0^{ - 1} )^{2 / (p - 2)} - C_2 \lambda ^2\left\| g \right\|_{E'} ,$$

矛盾,故结论成立.

引理4.4   若条件(A)满足,则存在$u \in E\backslash \{0\}$以及$\Lambda > 0,$ 当$0 < \lambda ^2\left\| g \right\|_{E'} < \Lambda$时

$$\mathop {\sup }\limits_{t \ge 0} J(tu) < c^\ast = \Big(\frac{1}{2} - \frac{1}{p}\Big)(S^{p / 2}f_0^{ - 1} )^{2 / (p - 2)} - C_2 \lambda ^2\left\| g \right\|_{E'},$$

特别有$\xi ^ - < c^\ast .$

证 考虑下列函数

$$\rho _\varepsilon (x) = \left\{ {\begin{array}{ll} v_\varepsilon (x),&g(x) \ge 0,x \in {\rm {\bf R}}^N, \\ v_\varepsilon (x - x_0 ),~~&\exists x_0 \in {\rm {\bf R}}^N,g(x_0 ) \ge 0,x \in {\rm {\bf R}}^N,\\ - v_\varepsilon (x),&g(x) \le 0,x \in {\rm {\bf R}}^N. \end{array}} \right.$$

则存在$\varepsilon _0 > 0$使得

$\begin{equation} \int_{{\rm {\bf R}}^N} {g(x)\rho _\varepsilon (x)} {\rm d}x > 0. \end{equation}$

(4.1)

这是因为,若$g(x) \geq 0$或$g(x) \leq 0$,(4.1)式显然成立. 若$\exists x_0 \in {\rm {\bf R}}^N,g(x_0 ) \ge 0,$则由$g(x)$的连续性, $\exists r > 0$,使得$g(x) > 0,x \in B_r (x_0 )$,由$v_\varepsilon (x - x_0 )$的定义,(4.1)式显然成立.

考虑下列函

数 $$\begin{align} & G(t)=J(t{{\rho }_{\varepsilon }})=\frac{{{t}^{2}}}{2}{{\left\| {{\rho }_{\varepsilon }} \right\|}^{2}}-\frac{{{t}^{p}}}{p}A({{\rho }_{\varepsilon }})-tK({{\rho }_{\varepsilon }}),\\ & \ \ \ \ \ \ \ H(t)=\frac{{{t}^{2}}}{2}{{\left\| {{\rho }_{\varepsilon }} \right\|}^{2}}-\frac{{{t}^{p}}}{p}{{f}_{0}}\left\| {{\rho }_{\varepsilon }} \right\|_{b,p}^{p}. \\ \end{align}$$

由$\lim \limits_{t \to + \infty } H(t) = - \infty , \lim\limits_{t \to 0^ + } H(t) > 0,$ 当$t \ge 0,$ $ \sup H(t)$ 在某个$t_\varepsilon > 0$达到,由$H'(t) = 0$得到

$$t_\varepsilon = \bigg(\frac{\left\| {\rho _\varepsilon } \right\|^2}{f_0 \left\| {\rho _\varepsilon } \right\|_{b,p}^p }\bigg)^{\frac{1}{p - 2}}.$$

由$S$的定义

,

$\begin{equation} \max \limits_{t \ge 0} H(t) = H(t_\varepsilon ) = \Big(\frac{1}{2} - \frac{1}{p}\Big)(S^{p / 2}f_0^{ - 1} )^{2 / (p - 2)}, \end{equation}$

(4.2)

取$\lambda$足够小,存在$\Lambda $,当$0 < \lambda < \Lambda ,$

$$G(0) = 0 < \Big(\frac{1}{2} - \frac{1}{p}\Big)(S^{p / 2}f_0^{ - 1} )^{2 / (p - 2)} - C_2 \lambda ^2\left\| g \right\|_{E'}.$$

由$G(t)$的连续性,存在$t_1 > 0,$ 当$0 < t < t_1 ,$

$$G(t) < \Big(\frac{1}{2} - \frac{1}{p}\Big)(S^{p / 2}f_0^{ - 1} )^{2 / (p - 2)} - C_2 \lambda ^2\left\| g \right\|_{E'}.$$

由(4.2)式

$$\sup\limits_{t \ge 0} G(t) < \Big(\frac{1}{2} - \frac{1}{p}\Big)(S^{p / 2}f_0^{ - 1} )^{2 / (p - 2)} - t_1 K(\rho _\varepsilon ),$$

取$\lambda$足够小,使得

$$ - t_1 K(\rho _\varepsilon ) < - C_2 \lambda ^2\left\| g \right\|_{E'},$$

于是

$$\begin{align} & \underset{t\ge 0}{\mathop{\sup }}\,G(t)<(\frac{1}{2}-\frac{1}{p}){{({{S}^{p/2}}f_{0}^{-1})}^{2/(p-2)}}-{{C}_{2}}{{\lambda }^{2}}{{\left\| g \right\|}_{{{E}'}}},\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underset{t\ge 0}{\mathop{\sup }}\,J(t{{\rho }_{\varepsilon }})<{{c}^{*}}. \\ \end{align}$$

容易看出$A(\rho _\varepsilon) > 0,$ 由引理2.4及上式,当$0 < \lambda < \Lambda < C_1 ,$ 存在$\tau _0 > 0$使得$\tau _0 \rho _\varepsilon \in N_{\lambda}^{-} ,$ 且$\xi ^ - \le J(\tau _0 \rho_\varepsilon ) \le \sup\limits_{\tau \ge 0} J(\tau \rho_\varepsilon ) < c^\ast $.

引理4.5   若条件(A)满足,则存在$u \in E\backslash \{0\}$以及$\Lambda > 0,$ 当$0 < \lambda ^2\left\| g \right\|_{E'} < \Lambda$时, 则泛函$J(u)$有一个极小值点$u_0^ - \in N^{-},$ 且满足

(1)~ $J(u_0^ - ) = \xi ^ - $.

(2)~ $u_0^ - $是方程(1.1)的一个正解.

证   由引理3.2知道,当$0 < \lambda < \Lambda < C_1 ,$ 泛函$J(u)$存在一个$(PS)_{\xi ^ - }$列$\{u_n \} \subset N^{-}$,由引理4.3, 引理4.4和引理2.4,存在$u_0^ - \in N_{\lambda}^{-} ,J(u_0^ - ) = \xi ^ - > 0,$ 与引理3.3同样讨论知道,$u_0^ -$是方程(1.1)的一个非平凡解.

定理1.3 的证明   由引理3.3及引理4.5,存在$\Lambda > 0$, 当$0 < \lambda ^2\left\| g \right\|_{E'} < \Lambda$时,$u_0^ + \in N_{\lambda}^{+} , u_0^ - \in N_{\lambda}^{-} $是方程(1.1)的两个非平凡解, 且$N_{\lambda}^{+}\cap N_{\lambda}^{-} = \emptyset ,$ 说明$u_0^ + $和$u_0^ - $不同. 证毕.