考虑如下具有$ \Phi $-Laplace算子的拟线性椭圆型方程正解的存在性、多重性和分歧性

(1.1) $ \begin{equation} \left\{ \begin{array}{ll} -\triangle_{\Phi}u = u^{\tau-1}+\lambda f(x, u), {\quad} &x\in\Omega, \\ u>0, &x\in\Omega, \\ u = 0, &x\in\partial\Omega, \end{array} \right. \end{equation} $

其中$ \Omega\subseteq{{\Bbb R}} ^{N} $是一个具有光滑边界的有界区域, $ \Delta_{\Phi}u = {\rm div}(\phi(|\nabla u|)\nabla u) $是$ \Phi $-Laplace算子, $ \Phi(t) = \int_{0}^{t}s\phi(s){\rm d}s $, $ f : \Omega \times {{\Bbb R}} \rightarrow {{\Bbb R}} $是Carathéodory函数, $ \lambda>0 $, $ \tau>1 $.

在过去几十年里,对具有$ \Phi $-Laplace算子的拟线性椭圆型方程解的存在性等相关问题得到广泛研究(参见文献[1-4]).同时, $ \Phi $-Laplace算子具有很强的物理背景,在偏微分方程、非牛顿流体、图像处理、等离子物理等领域有着广泛的应用.例如,对非牛顿流体问题,即当$ p>1 $, $ \Phi(t) = \frac{1}{p}|t|^{p} $时,问题(1.1)转化为$ p $-Laplace方程

(1.2) $ \begin{equation} \left\{ \begin{array}{ll} -\Delta_{p} u = f_{\lambda}(u), {\quad} &x\in\Omega, \\ u = 0, &x\in\partial\Omega. \end{array} \right. \end{equation} $

作者Azorero等^[5]利用局部极小化方法和山路定理研究了当$ f_{\lambda}(u) = |u|^{r-2} u+\lambda|u|^{p-2} u $, $ 1<p<r<p^{*} $, $ \lambda>0 $时问题(1.2)正解的存在性和多重性.

关于拟线性椭圆型方程解的分歧性的研究,一个具有代表性的结果是由Marano等^[6]给出.当$ 1<q<p<N $, $ p \in\left(q, q^{*}\right) $, $ \Phi(t) = \frac{1}{p}|t|^{p}+\frac{1}{q}|t|^{q} $时问题(1.1)转化为$ (p, q) $-Laplace方程

(1.3) $ \begin{equation} \left\{ \begin{array}{ll} -\triangle_{p}u-\triangle_{q}u = u^{\tau-1}+\lambda f(x, u), \; \; &x\in\Omega, \\ u>0, &x\in\Omega, \\ u = 0, &x\in\partial\Omega. \end{array} \right. \end{equation} $

作者利用临界点理论、截断技巧和比较原理研究了当$ 1<\tau<q<p<+\infty, \; \lambda>0 $时问题(1.3)正解的存在性和分歧性.

近年来,关于分歧性的研究成果还有很多(参见文献[7-8]),其中文献[8]是研究带有Robin边界条件的边值问题.

本文的目的是探究问题(1.1)的正解随参数$ \lambda $变化时解集的变化规律.

现在给出函数$ \phi $和函数$ f $的基本假设.函数$ \phi :(0, +\infty) \rightarrow(0, +\infty) $, $ \phi \in C^{1}(\Omega) $,且满足下列条件

($ \phi_{1} $)当$ t\rightarrow0 $时, $ t\phi(t)\rightarrow0 $;当$ t\rightarrow\infty $时, $ t\phi(t)\rightarrow\infty $;

($ \phi_{2} $) $ t\phi(t) $在$ (0, \infty) $上是严格增的;

($ \phi_{3} $)存在常数$ \ell, m\in(1, N) $,使得对于任意的$ t>0 $,成立

$ \ell-1 = \inf\limits _{t>0} \frac{(t \phi(t))^{\prime}}{\phi(t)} \leq \frac{(t \phi(t))^{\prime}}{\phi(t)} \leq \sup\limits_{t>0} \frac{(t \phi(t))^{\prime}}{\phi(t)} = m-1<\ell^{*}-1, \ell^{\ast} = \frac{N \ell}{N-\ell}; $

($ \phi_{4} $) $ t\mapsto\Phi(t^{\frac{1}{\ell}}) $在$ (0, +\infty) $上是凸函数.

进一步,假设$ f:\Omega\times{{\Bbb R}} \rightarrow{{\Bbb R}} $是Carathéodory函数,当$ t\leq0 $时, $ f(x, t) = 0 $, $ F(x, t): = \int^{t}_{0}f(x, s){\rm d}s $.此外, $ f $还满足以下条件

($ f_{1} $)存在常数$ \theta\in[\tau, \ell] $, $ r\in(m, \ell^{*}) $,使得

$ c_{1}t^{m-1}+c_{2}t^{\ell-1}\leq f(x, t)\leq c_{0}(t^{\theta-1}+t^{r-1}), \; \; \; \; \forall(x, t)\in\Omega\times{\Bbb R_{+}}, $

其中$ c_{2}>\frac{\overline{\lambda}_{1}\ell\|\phi_{1}\|^{\ell}_{\Phi}}{\lambda\|\phi_{1}\|^{\ell}_{\ell}} $, $ \overline{\lambda}_{1} $是算子$ -\Delta_{\Phi} $的第一特征值, $ \phi_{1} $是对应的特征函数;

$ (f_{2}) $$ \lim\limits_{t\rightarrow+\infty}\frac{F(x, t)}{t^{m}} = +\infty $对几乎所有的$ x\in\Omega $一致成立;

$ (f_{3}) $$ \lim\limits_{t\rightarrow+\infty}\frac{f(x, t)t-mF(x, t)}{t^{\beta}}\geq c_{3} $对几乎所有的$ x\in\Omega $一致成立,其中$ \beta>\tau $且

$ \frac{N(r-\ell)}{\ell}<\beta<\ell^{*}. $

由条件$ (f_{2}) $和$ (f_{3}) $可知, $ f(x, t) $在$ +\infty $处为$ (m-1) $次超线性增长.其次,我们知道在运用山路定理研究拟线性方程解的存在性时, (AR)条件起着关键作用.但$ \tau<\ell<m $, $ t\mapsto t^{\tau-1} $在$ +\infty $处为$ (\ell-1) $次线性增长,而$ t\mapsto f(x, t) $在$ +\infty $处为$ (m-1) $次超线性增长不满足一般的(AR)条件.另外,问题(1.1)具有凹凸非线性的特征,这一点增加了研究的难度.为了克服这些困难,需要采用适当的技巧.我们的方法主要基于临界点理论,结合截断技巧和比较原理.

在叙述本文的主要结果之前,先定义下列几类集合

$ {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) = \left\{u | u \in C_{0}^{1}(\overline{\Omega})_{+}: \; u(x)>0, \; \forall x \in \Omega;\; \frac{\partial u}{\partial n}(x)<0, \; \forall x \in \partial \Omega\right\}, $

其中$ n(x) $是边界$ \partial\Omega $上在$ x $点处的单位外法向量, $ C_{0}^{1}(\overline{\Omega})_{+} = \left\{u | u \in C_{0}^{1}(\overline{\Omega}): \; u\geq 0, \; x \in \overline{\Omega}\right\} $;记$ S(\Omega): = \{u:\Omega\rightarrow{{\Bbb R}} , \; u\mbox{是可测的}\} $,对$ u, v\in S(\Omega), \; \mbox{定义}[u, v]: = \{\omega\in S(\Omega):u\leq\omega\leq v\} $, $ [u): = \{\omega\in S(\Omega):u\leq\omega\} $;定义$ \Lambda = \{\lambda>0, \; E_{\lambda}\neq\emptyset\} $,其中$ E_{\lambda} = \{u|u\in W^{1, \Phi}_{0}(\Omega) $是问题(1.1)的正解$ \}; $$ K_{I}: = \{u\in W^{1, \Phi}_{0}(\Omega):I'(u) = 0\} $.

定理1.1  设$ (\phi_{1})-(\phi_{3}) $和$ (f_{1})-(f_{3}) $成立,则存在$ \lambda^{*}>0 $,使得下列结论成立:当$ \lambda \in\left(0, \lambda^{*}\right) $时,问题(1.1)至少存在两个解$ \hat{u}_{e}, \overline{u}_{e} \in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $,且$ \hat{u}_{e} \leq \overline{u}_{e} $;当$ \lambda = \lambda^{\ast} $时,问题(1.1)至少存在一个解$ u^{\ast}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $;当$ \lambda>\lambda^{\ast} $时,问题(1.1)无解.

定理1.2  设$ (\phi_{1})-(\phi_{4}) $和$ (f_{1})-(f_{3}) $成立,则当$ \lambda\in(0, \lambda^{*}] $时,问题(1.1)存在最小正解$ \overline{u}_{\lambda}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $.

定理1.3  设$ (\phi_{1})-(\phi_{4}) $和$ (f_{1})-(f_{3}) $成立,若$ \lambda\in(0, \lambda^{*}] $, $ \overline{u}_{\lambda} $为问题(1.1)的最小正解,则$ \lambda \mapsto \overline{u}_{\lambda} \in C_{0}^{1}(\overline{\Omega}) $是严格增和左连续的,即若$ \lambda_{1}<\lambda_{2} $,则$ \overline{u}_{\lambda_{2}}-\overline{u}_{\lambda_{1}} \in {\rm int}( C_{0}^{1}(\overline{\Omega})_{+}) $,且对任意的$ \lambda_{n} \rightarrow \lambda^{-} $$ (n\rightarrow\infty) $,有$ \overline{u}_{\lambda_{n}} \rightarrow \overline{u}_{\lambda} \in C_{0}^{1}(\overline{\Omega}) $.

记$ L_{\Phi}(\Omega) = \big\{u|u:\Omega\rightarrow {{\Bbb R}}\ \mbox{是可测的, }\ \int_{\Omega}\Phi(|u(x)|){\rm d}x<\infty\big\} $,在$ L_{\Phi}(\Omega) $上定义Luxemburg范数

$ \|u\|_{\Phi} = \inf\limits_{k}\left\{k>0\; \Big|\; \int_{\Omega}\Phi\left(\frac{|u(x)|}{k}\right){\rm d}x\leq1\right\}. $

记$ W^{1, \Phi}(\Omega) = \left\{u | u \in L_{\Phi}(\Omega), \; D_{i} u \in L_{\Phi}(\Omega), \; i = 1, \cdots , n\right\} $,在$ W^{1, \Phi}(\Omega) $上定义范数

$ \|u\|_{1, \Phi} = \|u\|_{\Phi}+\|\nabla u\|_{\Phi}. $

记$ W^{1, \Phi}_{0}(\Omega) $是$ C^{\infty}_{0}(\Omega) $在$ W^{1, \Phi}(\Omega) $中的闭包. $ L_{\Phi}(\Omega) $和$ W^{1, \Phi}(\Omega) $是可分、自反的Banach空间(参见文献[1]).

根据$ (\phi_{3}) $和$ {\rm Poincar\acute{e}} $不等式(参见文献[9]),有

$ \overline{\lambda}_{1}\int_{\Omega}\Phi(u){\rm d}x\leq\int_{\Omega} \Phi( | \nabla u|){\rm d}x, {\quad} u\in W^{1, \Phi}_{0}(\Omega). $

设$ d_{\Omega} = {\rm diam}(\Omega) $,则对于任意$ u\in W^{1, \Phi}_{0}(\Omega) $,有

$ \int_{\Omega}\Phi(u){\rm d}x\leq \int_{\Omega}\Phi\left(2d_{\Omega} |\nabla u|\right){\rm d}x, $

这样,得到$ \|u\|_{\Phi} \leq 2d_{\Omega}\|\nabla u\|_{\Phi} $,因此定义在$ W^{1, \Phi}_{0}(\Omega) $上的范数$ \|u\|: = \|\nabla u\|_{\Phi} $与$ \|\cdot\|_{1, \Phi} $等价.

设$ \Phi_{\ast}^{-1}(t) = \int_{0}^{t}{s^{-\frac{N+1}{N}}}\Phi^{-1}(s){\rm d}s $ ($ t\geq0 $),且当$ t<0 $时, $ \Phi_{\ast}(t) = \Phi_{\ast}(-t) $.如果对于所有的$ \nu>0 $,都有$ \lim\limits _{t \rightarrow \infty} \frac{\Upsilon(\nu t)}{\Phi_{\ast}(t)} = 0 $,则称N -函数$ \Upsilon $在无穷远处比$ \Phi_{\ast} $增长得更慢,记$ \Upsilon\ll\Phi_{\ast} $.如果$ \Upsilon\ll\Phi_{\ast} $,则$ W^{1, \Phi}_{0}(\Omega)\hookrightarrow\hookrightarrow L_{\Upsilon}(\Omega) $ (参见文献[9]).进一步,有$ W^{1, \Phi}_{0}(\Omega)\hookrightarrow L_{\Phi_{\ast}}(\Omega) $.在条件$ (\phi_{1})-(\phi_{3}) $下,还有如下连续嵌入: $ L^{m}(\Omega)\hookrightarrow L_{\Phi}(\Omega)\hookrightarrow L^{\ell}(\Omega) $, $ L_{\Phi_{\ast}}(\Omega)\hookrightarrow L^{\ell^{\ast}}(\Omega) $ (参见文献[10]).

注2.1  在条件$ (\phi_{3}) $下,可推得下式成立

$ \ell-2 \leq \frac{\phi^{\prime}(t) t}{\phi(t)} \leq m-2, \; \ell\leq\frac{\phi(t)t^{2}}{\Phi(t)}\leq m, \; t>0. $

下面给出本文需要的几个基本引理.

引理2.1^[2]  设$ (\phi_{1})-(\phi_{3}) $成立,对$ t\geq0 $,令

$ \eta_{0}(t) = \min\{t^{\ell-2}, t^{m-2}\}, \; \; \eta_{1}(t) = \max\{t^{\ell-2}, t^{m-2}\}, $

则对于任意$ \rho, t>0 $,成立$ \eta_{0}(t)\phi(\rho)\leq\phi(\rho t)\leq\eta_{1}(t)\phi(\rho) $.

引理2.2^[2]  设$ (\phi_{1})-(\phi_{3}) $成立,对$ t\geq0 $,令

$ \eta_{2}(t) = \min\{t^{\ell}, t^{m}\}, \; \; \eta_{3}(t) = \max\{t^{\ell}, t^{m}\}, $

则对于任意$ \rho, t>0 $, $ u\in L_{\Phi}(\Omega) $,成立

$ \eta_{2}(t)\Phi(\rho)\leq\Phi(\rho t)\leq\eta_{3}(t)\Phi(\rho), \; \; \eta_{2}(\|u\|_{\Phi})\leq\int_{\Omega}\Phi(u){\rm d}x\leq\eta_{3}(\|u\|_{\Phi}). $

众所周知,寻找问题(1.1)的非负弱解等价于求泛函

$ I(u) = \int_{\Omega}\Big(\Phi(|\nabla u|)-G(x, u)\Big){\rm d}x, \; \; u\in W^{1, \Phi}_{0}(\Omega) $

的临界点,其中$ G(x, t) = \int^{t}_{0}g(x, s){\rm d}s $, $ g(x, t) = (t^{+})^{\tau-1}+\lambda f(x, t^{+}) $.

引理2.3^[11]  设$ (\phi_{1})-(\phi_{3}) $和$ (f_{1})-(f_{3}) $成立,若$ u\in W^{1, \Phi}_{0}(\Omega) $是$ I $在$ C_{0}^{1}(\overline{\Omega}) $上的局部极小点,则存在常数$ \eta\in(0, 1) $,有$ u\in C^{1, \eta}(\overline{\Omega}) $,且$ u $也是$ I $在$ W^{1, \Phi}_{0}(\Omega) $上的局部极小点.

引理2.4^[11]  设$ (\phi_{1})-(\phi_{3}) $成立, $ b_{1}, b_{2}\in L^{\infty}(\Omega) $, $ b_{2}\geq b_{1} $,且集合$ S = \{x\in\Omega:b_{1}(x) = b_{2}(x)\} $的测度为零,若$ u_{1}, u_{2}\in W^{1, \Phi}_{0}(\Omega) $满足$ -\triangle_{\Phi}u_{i} = b_{i}, \; i = 1, 2, \; x\in\Omega $,则$ u_{2}-u_{1}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $.

引理2.5^[12]  设$ (\phi_{1})-(\phi_{3}) $成立,则$ -\triangle_{\Phi} $是$ (S_{+}) $型算子,即对任意给定的序列$ \{u_{n}\}\subseteq W^{1, \Phi}_{0}(\Omega) $,若$ u_{n}\rightharpoonup u $,且$ \limsup\limits_{n\rightarrow\infty}\langle-\triangle_{\Phi}u_{n}, u_{n}-u\rangle\leq0 $,则在$ W^{1, \Phi}_{0}(\Omega) $中有$ u_{n}\rightarrow u $.

本节的主要工作是证明定理1.1,先给出几个关键引理.

引理3.1  设$ (\phi_{1})-(\phi_{3}) $和$ (f_{1}) $成立,则对每一个$ \lambda>0 $,有$ E_{\lambda}\subseteq {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $.

证  设$ \lambda\in\Lambda $ (否则$ E_{\lambda} = \emptyset $),那么存在$ u\in W^{1, \Phi}_{0}(\Omega), \; u\geq0 $且$ u\neq0 $,使得

(3.1) $ \begin{equation} -\triangle_{\Phi}u = u^{\tau-1}+\lambda f(x, u). \end{equation} $

由文献[13]推得$ u\in L^{\infty}(\Omega) $.又由于

$\nabla(\phi(|\zeta|)\zeta)|\zeta|\leq(m-1)\Phi^{\prime}(|\zeta|), \; \; (\nabla(\phi(|\zeta|)\zeta)y, y)|\zeta|\geq(\ell-1)\Phi^{\prime}(|\zeta|)|y|^{2}$,

所以, $ \Phi $-Laplace算子满足文献[14]中的定理1.7条件,由该定理的结论有$ u\in C_{0}^{1}(\overline{\Omega})_{+}\backslash\{0\} $.

在(3.1)式两边同时加上$ L\phi(u)u $,得

$ -\triangle_{\Phi}u+L\phi(u)u = u^{\tau-1}+\lambda f(x, u)+L\phi(u)u: = \overline{h}(x, u). $

因为$ u\geq0 $且$ u\neq0, \; \lambda>0 $,若$ L>0 $且充分大,则有$ \overline{h}(x, u)\geq0 $,即$ -\triangle_{\Phi}u+L\phi(u)u\geq0 $.

定义$ H(t) = t^{2}\phi(t)-\Phi(t), t\in{{\Bbb R}} $,那么存在足够小$ \delta>0 $,使得^[12]

(3.2) $ \begin{equation} \int^{\delta}_{0}\frac{1}{H^{-1}(\Phi(s))}{\rm d}s = +\infty; \end{equation} $

利用Pucci-Serrin强极大值原理^[15],对任意的$ x\in\Omega $,有$ u>0 $;再由文献[15,定理5.5.1]的结论推得$ u\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $,故$ E_{\lambda}\subseteq {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $.证毕.

为了验证$ \Lambda\neq\emptyset $,考虑如下辅助问题

(3.3) $ \begin{equation} \left\{ \begin{array}{ll} -\triangle_{\Phi}u = g_{\lambda}(u), \; \; &x\in\Omega, \\ u>0, \; &x\in\Omega, \\ u = 0, &x\in\partial\Omega \end{array} \right. \end{equation} $

与问题(3.3)对应的能量泛函为$ I_{\lambda}:W^{1, \Phi}_{0}(\Omega)\rightarrow{{\Bbb R}} $,有

(3.4) $ \begin{equation} I_{\lambda}(u) = \int_{\Omega}\Big(\Phi(|\nabla u|)-G_{\lambda}(u)\Big){\rm d}x, \end{equation} $

其中

(3.5) $ \begin{equation} G_{\lambda}(t) = \int_{0}^{t}g_{\lambda}(s){\rm d}s, \; \; g_{\lambda}(t) = (t^{+})^{\tau-1}+\lambda c_{0}((t^{+})^{\theta-1}+(t^{+})^{r-1}). \end{equation} $

引理3.2  设$ (\phi_{1})-(\phi_{3}) $、$ (f_{1}) $和$ (f_{3}) $成立,则对每一个$ \lambda>0 $,能量泛函$ I_{\lambda} $满足$ \rm Cerami $条件.

证  给定$ \lambda>0 $,设序列$ \{u_{n}\}\subseteq W^{1, \Phi}_{0}(\Omega) $,满足

(3.6) $ \begin{equation} |I_{\lambda}(u_{n})|\leq A\; \; (A>0), \end{equation} $

(3.7) $ \begin{equation} (1+\|u_{n}\|)\|I_{\lambda}'(u_{n})\|\rightarrow0 \; \; \mbox{(在$W^{-1, \widetilde{\Phi}}_{0}(\Omega)$中), } \; n\rightarrow\infty. \end{equation} $

由(3.5)式易得,存在$ M^{\ast}>0 $,当$ t>M^{\ast} $, $ \sigma\in(m, r) $时,有

(3.8) $ \begin{equation} 0<\sigma G_{\lambda}(t)<g_{\lambda}(t)t. \end{equation} $

又由(3.4)式, (3.6)–(3.8)式和注2.1,当$ n $足够大时,有

$ A+1+o_{n}(1)\|u_{n}\|\geq(\sigma-m)\min\{\|u_{n}\|^{\ell}, \|u_{n}\|^{m}\}-c_{6}. $

由此推得$ \{u_{n}\} $在$ W^{1, \Phi}_{0}(\Omega) $中是有界的.因此,存在$ \{u_{n}\} $的子列(仍记为其本身),有

(3.9) $ \begin{equation} u_{n}\rightharpoonup u\ \mbox{(在$W^{1, \Phi}_{0}(\Omega)$中), } \end{equation} $

(3.10) $ \begin{equation} u_{n}\rightarrow u\ \mbox{(在$L_{\Phi}(\Omega)$中).} \end{equation} $

由(3.7)式知

$ |\langle I_{\lambda}'(u_{n}), h\rangle|\leq \frac{o_{n}(1)\|h\|}{1+\|u_{n}\|}, \; \; \forall h\in W_{0}^{1, \Phi}(\Omega), $

即

(3.11) $ \begin{equation} \Big|\int_{\Omega}\phi(|\nabla u_{n}|)\nabla u_{n}\nabla h {\rm d}x-\int_{\Omega}g_{\lambda}(u_{n})h {\rm d}x\Big|\leq\frac{o_{n}(1)\|h\|}{1+\|u_{n}\|}, \; \; \forall h\in W^{1, \Phi}_{0}(\Omega). \end{equation} $

取测试函数$ h = u_{n}-u $代入(3.11)式中,并利用Hölder不等式和插值不等式,那么存在$ t\in(0, 1) $, $ \ell<r<\ell^{\ast} $,使得

$ \begin{eqnarray*} &&\int_{\Omega}\phi(|\nabla u_{n}|)\nabla u_{n}\nabla(u_{n}-u){\rm d}x\\ &\leq&\|u_{n}\|^{\tau-1}\|u_{n}-u\|_{\tau}+\lambda c_{0}\Big(\|u_{n}\|^{\theta-1}\|u_{n} -u\|_{\theta}+\|u_{n}\|^{r-1}\|u_{n}-u\|^{t}_{\ell}\|u_{n}-u\|^{1-t}_{\ell^{*}}\Big)\\ && +\frac{o_{n}(1)c_{7}}{1+\|u_{n}\|}. \end{eqnarray*} $

在上式中,由于$ W^{1, \Phi}_{0}(\Omega)\hookrightarrow L_{\Phi_{\ast}}(\Omega) $, $ L_{\Phi_{\ast}}(\Omega)\hookrightarrow L^{\ell^{\ast}}(\Omega) $和$ W^{1, \Phi}_{0}(\Omega)\hookrightarrow\hookrightarrow L_{\Phi}(\Omega)\hookrightarrow L^{\ell}(\Omega) $,故当$ n\rightarrow\infty $时,有

(3.12) $ \begin{equation} \limsup\limits_{n\rightarrow\infty}\langle-\triangle_{\Phi}u_{n}, u_{n}-u\rangle\leq0. \end{equation} $

即算子$ -\triangle_{\Phi} $满足$ (S_{+}) $型条件,由引理2.5、(3.9)和(3.12)式推得在$ W^{1, \Phi}_{0}(\Omega) $中$ u_{n}\rightarrow u $.

引理3.3  设$ (\phi_{1})-(\phi_{3}) $和$ (f_{1}) $成立,则$ \Lambda\neq\emptyset $,且对每一个$ \lambda\in\Lambda $,有$ (0, \lambda]\subseteq\Lambda $.

证  首先,证明当$ \lambda>0 $且足够小,能量泛函$ I_{\lambda} $满足山路几何条件.

一方面,由引理2.2,对$ u\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $,当$ t\rightarrow\infty $时,有$ I_{\lambda}(tu)\rightarrow -\infty $.另一方面,令

$ \beta_{\lambda}(s) = c_{8}s^{\tau-\ell}+\lambda c_{9}(s^{\theta-\ell}+s^{r-\ell}), \ \beta^{\ast}_{\lambda}(s) = (c_{8}+\lambda c_{9})s^{\tau-\ell}+2\lambda c_{9}s^{r-\ell}, \; s>0. $

由$ \tau\leq\theta\leq\ell<m<r $可推得$ \lambda c_{9}s^{\theta-\ell}\leq\lambda c_{9}(s^{\tau-\ell}+s^{r-\ell}) $,从而$ 0<\beta_{\lambda}(s)\leq\beta^{\ast}_{\lambda}(s) $.

因为$ \lim\limits_{s\rightarrow 0^{+}}\beta^{\ast}_{\lambda}(s) = \lim\limits_{s\rightarrow +\infty}\beta^{\ast}_{\lambda}(s) = +\infty $,所以存在$ s_{0}>0 $,使得$ \beta^{\ast\prime}_{\lambda}(s_{0}) = 0 $.通过简单的计算,可得

$ s_{0} = \left(\frac{(c_{8}+\lambda c_{9})(\ell-\tau)}{2\lambda c_{9}(r-\ell)}\right)^{\frac{1}{r-\tau}} , $

且当$ \lambda\rightarrow0^{+} $时, $ s_{0}\rightarrow+\infty $和$ \beta^{\ast}_{\lambda}(s_{0})\rightarrow0 $,则可以找到$ \lambda_{0}>0 $,使得$ s_{0}\geq1 $且$ \beta^{\ast}_{\lambda}(s_{0})\leq\frac{1}{2} $.故当$ \lambda\in(0, \lambda_{0}) $, $ u\in\partial B_{s_{0}}(0) $时,有

$ I_{\lambda}(u)\geq \min\{\|u\|^{\ell}, \|u\|^{m}\}-c_{8}\|u\|^{\tau}_{\tau}-\lambda c_{9}\|u\|^{\theta}_{\theta}-\lambda c_{9}\|u\|^{r}_{r} \geq\frac{1}{2}\|u\|^{\ell}>0 = I_{\lambda}(0). $

由山路引理,便存在$ u_{\lambda}\in W^{1, \Phi}_{0}(\Omega) $,且$ u_{\lambda}\neq0 $,使得$ I_{\lambda}^{\prime}(u_{\lambda}) = 0 $,即对任意$ v\in W^{1, \Phi}_{0}(\Omega) $,有

(3.13) $ \begin{equation} \int_{\Omega}\phi(|\nabla u_{\lambda}|)\nabla u_{\lambda}\nabla v{\rm d}x = \int_{\Omega}\Big((u_{\lambda}^{+})^{\tau-1}+\lambda c_{0}((u_{\lambda}^{+})^{\theta-1}+(u_{\lambda}^{+})^{r-1})\Big)v{\rm d}x. \end{equation} $

取$ v = -u_{\lambda}^{-} $$ (u_{\lambda}^{-} = \max\{-u_{\lambda}, 0\}) $作为测试函数代入(3.13)式,有$ u_{\lambda}^{-} = 0 $,从而得到$ u_{\lambda}\geq0 $,再利用引理3.1,便有$ u_{\lambda}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $.

其次,考虑如下截断函数

$ \overline{g}_{\lambda}(x, u(x)) = \left\{ \begin{array}{ll} (u(x)^{+})^{\tau-1}+\lambda f(x, u(x)^{+}), \; \; &u(x)\leq u_{\lambda}(x), \\ u_{\lambda}(x)^{\tau-1}+\lambda f(x, u_{\lambda}(x)), &\mbox{其它}. \end{array} \right. $

此时能量泛函$ I(u) $成为$ \overline{I}_{\lambda}:W^{1, \Phi}_{0}(\Omega)\rightarrow{{\Bbb R}} $,有

(3.14) $ \begin{equation} \overline{I}_{\lambda}(u) = \int_{\Omega}\Big(\Phi(|\nabla u|)-\overline{G}_{\lambda}(x, u)\Big){\rm d}x, \end{equation} $

其中$ \overline{G}_{\lambda}(x, t) = \int^{t}_{0}\overline{g}_{\lambda}(x, s){\rm d}s $.

容易验证泛函$ \overline{I}_{\lambda}(u) $是强制且弱下半连续的,因此存在极小值点$ \overline{u}_{\lambda}\in W^{1, \Phi}_{0}(\Omega) $,使得对任意$ v\in W^{1, \Phi}_{0}(\Omega) $,成立

(3.15) $ \begin{equation} \int_{\Omega}\phi(|\nabla \overline{u}_{\lambda}|)\nabla \overline{u}_{\lambda}\nabla v{\rm d}x = \int_{\Omega}\overline{g}_{\lambda}(x, \overline{u}_{\lambda})v{\rm d}x. \end{equation} $

在(3.15)式中取$ v = -\overline{u}_{\lambda}^{-} $作为测试函数便得到$ \overline{u}_{\lambda}\geq0 $.利用$ (f_{1}) $和引理2.2,取$ u = \frac{t}{2}u_{\lambda} $,则当$ t $足够小时,我们有$ \overline{I}_{\lambda}(\frac{t}{2}u_{\lambda})<0 $.因此存在$ \widetilde{t}_{0}>0 $,使得$ \overline{I}_{\lambda}(\widetilde{t_{0}}u_{\lambda})<0 $.由$ \overline{I}_{\lambda}(\overline{u} _{\lambda}) = \min\limits_{v\in W^{1, \Phi}_{0}(\Omega)}\overline{I}_{\lambda}(v)\leq\overline{I} _{\lambda}(\widetilde{t}_{0}u_{\lambda})<0 = \overline{I}_{\lambda}(0) $知$ \overline{u}_{\lambda}\neq0 $.

取$ v = (\overline{u}_{\lambda}-u_{\lambda})^{+} $$ ((\overline{u}_{\lambda}-u_{\lambda})^{+} = \max\{(\overline{u}_{\lambda}-u_{\lambda}), 0\}) $作为测试函数代入(3.15)式中,由$ (f_{1}) $和(3.14)式,有

$ \int_{\Omega}\phi(|\nabla \overline{u}_{\lambda}|)\nabla \overline{u}_{\lambda}\nabla(\overline{u}_{\lambda}-u_{\lambda})^{+}{\rm d}x\leq\int _{\Omega}\phi(|\nabla u_{\lambda}|)\nabla u_{\lambda}\nabla(\overline{u}_{\lambda}-u_{\lambda})^{+}{\rm d}x. $

由$ (\phi_{2}) $可得$ \overline{u}_{\lambda}\leq u_{\lambda} $,故存在$ \lambda\in(0, \lambda_{0}) $,即$ \Lambda\neq\emptyset $.

最后,验证若$ \lambda\in\Lambda $,则$ (0, \lambda]\subseteq\Lambda $.

由$ \lambda\in\Lambda $,存在$ u_{\lambda}\in E_{\lambda} $,任取$ \gamma\in(0, \lambda] $,在$ u_{\lambda} $点处作与前面相似的截断$ \widetilde{g}_{\lambda} $以及$ \widetilde{G}_{\lambda} $,对应的能量泛函类似于(3.14)式,重复前面的证明过程,可得存在$ u_{\gamma}\in E_{\gamma} $,即$ \gamma\in\Lambda $.因此$ (0, \lambda]\subseteq\Lambda $.证毕.

利用上述引理结合引理2.4容易得到下面推论.

推论3.1  设$ (\phi_{1})-(\phi_{3}) $和$ (f_{1}) $成立,则对每一个$ \lambda\in\Lambda $,存在$ u_{\lambda}\in E_{\lambda} $,当$ \gamma\in(0, \lambda) $时,存在$ u_{\gamma}\in E_{\gamma} $满足$ u_{\lambda}-u_{\gamma}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $.

引理3.4  设$ 1<\ell<m<\infty $,有

(3.16) $ \begin{equation} h(m, \ell;t) = \frac{(m-1)t^{m-2}+(\ell-1)t^{\ell-2}}{(m-1)(t^{m-1}+t^{\ell-1})^{\frac{m-2}{m-1}}}, \end{equation} $

则$ \inf\limits_{t>0}h(m, \ell;t)>0 $和$ \inf\limits_{t>0}h(\ell, m;t)>0 $.

此引理的证明是初等的(参见文献[16]).

引理3.5  设$ (\phi_{1})-(\phi_{3}) $成立,则存在常数$ \omega>0 $,使得对一切$ u>0 $, $ \varphi\geq0 $,有

(3.17) $ \begin{equation} \phi(|\nabla u|)\nabla u\nabla\Big(\frac{\varphi^{m}}{u^{m-1}+u^{\ell-1}}\Big)\leq\frac{1}{\omega}(|\nabla\varphi|^{m}+|\nabla\varphi^{\frac{m}{\ell}}|^{\ell}), \; \; \forall x\in\Omega. \end{equation} $

证  首先,由简单的计算,有

$ \begin{eqnarray*} &&\phi(|\nabla u|)\nabla u\nabla\Big(\frac{\varphi^{m}}{u^{m-1}+u^{\ell-1}}\Big)\\ &\leq& m\phi(|\nabla u|)|\nabla u||\nabla\varphi|\frac{\varphi^{m-1}}{u^{m-1}+u^{\ell-1}}-\phi(|\nabla u|)|\nabla u|^{2}\varphi^{m}\frac{(m-1)u^{m-2}+(\ell-1)u^{\ell-2}}{(u^{m-1}+u^{\ell-1})^{2}}. \end{eqnarray*} $

利用Young不等式和引理3.4,当$ x\in\Omega\cap\{|\nabla u|>1\} $时,有

$ \begin{eqnarray*} &&\phi(|\nabla u|)\nabla u\nabla\Big(\frac{\varphi^{m}}{u^{m-1}+u^{\ell-1}}\Big)\\ & = &\frac{|\nabla\varphi|^{m}}{\omega^{m-1}}+\frac{\omega(m-1)\varphi^{m}\phi(|\nabla u|)|\nabla u|^{2}}{(u^{m-1}+u^{\ell-1})^{\frac{m}{m-1}}} ((\phi(1))^\frac{1}{m-1}-\frac{1}{\omega}(h(m, \ell;u))), \end{eqnarray*} $

其中$ h(m, \ell;u) $为(3.16)式所定义.由引理3.4,有$ \inf\limits_{u>0}h(m, \ell;u)>0 $,则存在足够小的常数$ \omega_{1}>0 $满足$ \omega_{1}(\phi(1))^{\frac{1}{m-1}}\leq \inf\limits_{u>0}h(m, \ell;u) $,于是

(3.18) $ \begin{equation} \phi(|\nabla u|)\nabla u\nabla\Big(\frac{\varphi^{m}}{u^{m-1}+u^{\ell-1}}\Big)\leq\frac{1}{\omega^{m-1}_{1}}|\nabla\varphi|^{m}, \; \; x\in\Omega\cap\{|\nabla u|>1\}. \end{equation} $

类似地,如果取$ \psi = \varphi^{\frac{m}{\ell}} $,则存在足够小的常数$ \omega_{2}>0 $满足$ \omega_{2}(\phi(1))^{\frac{1}{\ell-1}}\leq \inf\limits_{u>0}h(\ell, m;u) $,因此

(3.19) $ \begin{equation} \phi(|\nabla u|)\nabla u\nabla\Big(\frac{\psi^{\ell}}{u^{m-1}+u^{\ell-1}}\Big)\leq\frac{1}{\omega^{\ell-1}_{2}}|\nabla(\varphi^{\frac{m}{\ell}})|^{\ell}, \; \; x\in\Omega\cap\{|\nabla u|\leq1\}. \end{equation} $

取$ \omega = \min\{\omega_{1}^{m-1}, \omega_{2}^{\ell-1}\} $,结合(3.18)式和(3.19)式, (3.17)式得证.证毕.

引理3.6  设$ (\phi_{1})-(\phi_{3}) $和$ (f_{1}) $成立,若$ \lambda^{\ast} = \sup\Lambda $,则$ \lambda^{\ast}<\infty $.

证  给定$ \lambda\in\Lambda, \; u_{\lambda}\in E_{\lambda} $,不妨设$ \lambda>1 $ (否则$ \Lambda $有界).考虑如下截断函数

$ \begin{eqnarray*} z(x, u(x)) = \left\{ \begin{array}{ll} \lambda\big(c_{1}(u(x)^{+})^{m-1}+c_{2}(u(x)^{+})^{\ell-1}\big), \; \; &u(x)\leq u_{\lambda}(x), \\ \lambda\big(c_{1}(u_{\lambda}(x))^{m-1}+c_{2}(u_{\lambda}(x))^{\ell-1}\big), &\mbox{其它}. \end{array} \right. \end{eqnarray*} $

此时能量泛函$ I(u) $成为$ \Psi:W^{1, \Phi}_{0}(\Omega)\rightarrow{{\Bbb R}} $,有

(3.20) $ \begin{equation} \Psi(u) = \int_{\Omega}\Big(\Phi(|\nabla u|)-Z(x, u)\Big){\rm d}x, \end{equation} $

其中$ Z(x, t) = \int^{t}_{0}z(x, s){\rm d}s $.

重复引理3.3的证明方法,则存在极小值点$ u_{z} $,使得$ \Psi'(u_{z}) = 0 $,且$ u_{z}\in [0, u_{\lambda}] $.注意到$ u_{z} $, $ \phi_{1}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $,当$ t>0 $且足够小,则有$ t\phi_{1}\leq u_{z} $.又由$ \overline{\lambda}_{1}\int_{\Omega}\Phi(u) {\rm d}x\leq\int_{\Omega}\Phi(|\nabla u|){\rm d}x $,推得

$ \Psi(t\phi_{1})\leq\Big(\overline{\lambda}_{1}^{\ell}\|\phi_{1}\|^{\ell}_{\Phi} -\frac{\lambda c_{2}}{\ell}\|\phi_{1}\|^{\ell}_{\ell}\Big)t^{\ell}. $

因此

$ \Psi(u_{z}) = \min\limits_{v\in W^{1, \Phi}_{0}(\Omega)}\Psi(v)\leq \Psi(t\phi_{1})<0 = \Psi(0), $

则有$ u_{z}\neq0 $.于是方程

(3.21) $ \begin{equation} -\triangle_{\Phi}u = \lambda c_{1}|u|^{m-2}u+\lambda c_{2}|u|^{\ell-2}u \end{equation} $

存在正解$ u\in W^{1, \Phi}_{0}(\Omega) $.

现在,我们断言$ \lambda^{\ast}<\infty $.事实上,由引理3.1,有$ u\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $,从而可以选取$ \varphi\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $,使得$ \frac{\varphi}{u}\in L^{\infty}(\Omega) $ (参见文献[16]),故有$ \frac{\varphi^{m}}{u^{m-1}+u^{\ell-1}}\in W^{1, \Phi}_{0}(\Omega) $.由于$ u $是方程(3.21)的正解,取测试函数$ \varsigma = \frac{\varphi^{m}}{u^{m-1}+u^{\ell-1}} $,利用引理3.5,那么存在不依赖于$ u, \; \lambda $的常数$ \omega>0 $,使得

$ \begin{eqnarray*} \int_{\Omega}\phi(|\nabla u|)\nabla u\nabla\Big(\frac{\varphi^{m}}{u^{m-1}+u^{\ell-1}}\Big){\rm d}x&\leq& \frac{1}{\omega}\int_{\Omega\cap\{|\nabla u|>1\}}|\nabla\varphi|^{m}{\rm d}x+\frac{1}{\omega}\int_{\Omega\cap\{|\nabla u|\leq1\}}|\nabla\varphi^{\frac{m}{\ell}}|^{\ell}{\rm d}x\\ &\leq&\frac{1}{\omega}\int_{\Omega}|\nabla\varphi|^{m}{\rm d}x+\frac{1}{\omega}\int_{\Omega}|\nabla\varphi^{\frac{m}{\ell}}|^{\ell}{\rm d}x, \end{eqnarray*} $

则有

(3.22) $ \begin{equation} \lambda \min\{c_{1}, c_{2}\}\int_{\Omega}\varphi^{m}{\rm d}x\leq\frac{1}{\omega}\int_{\Omega}(|\nabla\varphi|^{m} +|\nabla\varphi^{\frac{m}{\ell}}|^{\ell}){\rm d}x. \end{equation} $

因此,由(3.22)式可推出$ \lambda^{\ast}<\infty $.证毕.

引理3.7  设$ (\phi_{1})-(\phi_{3}) $和$ (f_{1})-(f_{3}) $成立,则对每一个$ \lambda \in\left(0, \lambda^{*}\right) $,问题(1.1)至少存在两个解$ \hat{u}_{e}, \overline{u}_{e} \in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $,且$ \hat{u}_{e} \leq \overline{u}_{e} $.

证  令$ e\in (\lambda, \lambda^{\ast}) $,则由引理3.3,存在$ u_{e}\in E_{e}\subseteq {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $.作截断函数

$ \begin{eqnarray*} \widehat{g}_{\lambda}(x, u(x)) = \left\{ \begin{array}{ll} (u(x)^{+})^{\tau-1}+\lambda f(x, u(x)^{+}), \; \; &u(x)\leq u_{e}(x), \\ (u_{e}(x))^{\tau-1}+\lambda f(x, u_{e}(x)), &\mbox{其它}. \end{array} \right. \end{eqnarray*} $

此时能量泛函$ I(u) $成为

$ \widehat{I}_{\lambda}(u) = \int_{\Omega}\Big(\Phi(|\nabla u|)-\widehat{G}_{\lambda}(x, u)\Big){\rm d}x, \; u\in W^{1, \Phi}_{0}(\Omega), $

其中$ \widehat{G}_{\lambda}(x, t) = \int^{t}_{0}\widehat{g}_{\lambda}(x, s){\rm d}s $.类似引理3.3的证明方法,可以找到$ \widehat{u}_{e}\in W^{1, \Phi}_{0}(\Omega) $,使得

$ \widehat{u}_{e}(x)\in K_{\widehat{I}_{\lambda}}\subseteq[0, u_{e}(x)]\cap {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}). $

作另一截断函数

$ \begin{eqnarray*} g_{\lambda}^{\ast}(x, u(x)) = \left\{ \begin{array}{ll} (\widehat{u}_{e}(x))^{\tau-1}+\lambda f(x, \widehat{u}_{e}(x)), \; \; &u(x)\leq \widehat{u}_{e}(x), \\ (u(x))^{\tau-1}+\lambda f(x, u(x)), &\mbox{其它}. \end{array} \right. \end{eqnarray*} $

此时能量泛函$ I(u) $成为

$ I_{\lambda}^{\ast}(u) = \int_{\Omega}\Big( \Phi(|\nabla u|)-G_{\lambda}^{\ast}(x, u)\Big){\rm d}x, \; u\in W^{1, \Phi}_{0}(\Omega), $

其中$ G_{\lambda}^{\ast}(x, t) = \int^{t}_{0}g_{\lambda}^{\ast}(x, s){\rm d}s $.同样,可知$ K_{I_{\lambda}^{\ast}}\subseteq[\widehat{u}_{e}(x))\cap {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $,故假设$ K_{I_{\lambda}^{\ast}}\cap[0, u_{e}(x)] = \widehat{u}_{e}(x) $ (否则问题(1.1)存在另一个正解且大于$ \widehat{u}_{e}(x) $).

考虑截断函数

$ \begin{eqnarray*} \widetilde{g}_{\lambda}^{\ast}(x, u(x)) = \left\{ \begin{array}{ll} g_{\lambda}^{\ast}(x, u(x)), \; \; &u(x)\leq u_{e}(x), \\ g_{\lambda}^{\ast}(x, u_{e}(x)), &\mbox{其它}. \end{array} \right. \end{eqnarray*} $

此时能量泛函$ I(u) $成为

$ \widetilde{I}_{\lambda}^{\ast}(u) = \int_{\Omega}\Big(\Phi(|\nabla u|) -\widetilde{G}_{\lambda}^{\ast}(x, u)\Big){\rm d}x, \; u\in W^{1, \Phi}_{0}(\Omega), $

其中$ \widetilde{G}_{\lambda}^{\ast}(x, t) = \int^{t}_{0}\widetilde{g}_{\lambda}^{\ast}(x, s){\rm d}s $.类似地,存在$ \widetilde{u}^{\ast}_{e}\in W^{1, \Phi}_{0}(\Omega) $,使得

$ \widetilde{u}_{e}^{\ast}(x)\in K_{\widetilde{I}_{\lambda}^{\ast}}\subseteq[ \widehat{u}_{e}(x), u_{e}(x)]\cap {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}). $

注意到

$ I^{\ast}_{\lambda}(u)|_{[u_{1}, u_{2}]}: = \{I^{\ast}_{\lambda}(u)|u\in W^{1, \Phi}_{0}(\Omega), \; u_{1}\leq u\leq u_{2}\}, $

则有$ \widetilde{I}_{\lambda}^{\ast}(u) |_{[0, u_{e}]} = I^{\ast}_{\lambda}(u)|_{[0, u_{e}]} $,因此$ \widetilde{u}_{e}^{\ast}(x)\in K_{{I}_{\lambda}^{\ast}}\cap[0, u_{e}(x)] $,故有$ \widetilde{u}_{e}^{\ast} = \widehat{u}_{e} $.进一步,由引理2.3,推出$ \widehat{u}_{e} $是$ I^{\ast}_{\lambda} $在$ W^{1, \Phi}_{0}(\Omega) $上的局部极小点.

设$ K_{I_{\lambda}^{\ast}} $是有限的(否则问题(1.1)存在无限个正解且大于$ \widehat{u}_{e} $),因此存在$ \alpha\in (0, 1) $,使得$ I_{\lambda}^{\ast}(\widehat{u}_{e})<m_{\alpha}: = \inf\{I_{\lambda}^{\ast}(u):\|u-\widehat{u}_{e}\| = \alpha\} $.由$ (f_{2}) $知当$ t\rightarrow\infty $时, $ I_{\lambda}^{\ast}(tu)\rightarrow-\infty $.于是利用山路引理,则存在$ \overline{u}_{e}\in W^{1, \Phi}_{0}(\Omega) $满足$ \overline{u}_{e}\in K_{I_{\lambda}^{\ast}} $,且$ I_{\lambda}^{\ast}(\overline{u}_{e})\geq m_{\alpha} $.又因为$ K_{\widetilde{I}_{\lambda}^{\ast}} \subseteq[\widehat{u}_{e}(x), u_{e}(x)] $,便有$ \widehat{u}_{e}\leq\overline{u}_{e} $,且$ \widehat{u} _{e}\neq\overline{u}_{e} $.证毕.

引理3.8  设$ (\phi_{1})-(\phi_{3}) $和$ (f_{1})-(f_{3}) $成立,则$ \lambda^{\ast}\in\Lambda $.

证  设$ \{\lambda_{n}\}\subseteq(0, \lambda^{\ast}) $,且$ \lambda_{n}\uparrow\lambda^{\ast} $.由推论3.1,可以找到序列$ \{u_{n}\}\subseteq W^{1, \Phi}_{0}(\Omega) $,使得$ u_{n}\in E_{\lambda_{n}} $, $ u_{n}\leq u_{n+1} $,且满足

(3.23) $ \begin{equation} \int_{\Omega}\phi(|\nabla u_{n}|)\nabla u_{n}\nabla v{\rm d}x = \int_{\Omega}u_{n}^{\tau-1}v{\rm d}x+\lambda_{n}\int_{\Omega}f(x, u_{n})v{\rm d}x, \; \; \; \forall v\in W^{1, \Phi}_{0}(\Omega). \end{equation} $

由引理3.3,有$ I(u_{n})<0 $,即

(3.24) $ \begin{equation} \int_{\Omega}\Phi(|\nabla u_{n}|){\rm d}x-\frac{1}{\tau}\|u_{n}\|^{\tau}_{\tau}-\lambda_{n}\int_{\Omega}F(x, u_{n}){\rm d}x<0. \end{equation} $

对(3.23)式取测试函数$ v = u_{n} $,由(3.24)式和注2.1,有

$ \int_{\Omega}\phi(|\nabla u_{n}|)|\nabla u_{n}|^{2}{\rm d}x\leq \frac{m}{\tau}\|u_{n}\|^{\tau}_{\tau}+\lambda_{n}m\int_{\Omega}F(x, u_{n}){\rm d}x. $

通过计算可得

$ \lambda_{n}\int_{\Omega}(f(x, u_{n})u_{n}-mF(x, u_{n})){\rm d}x\leq(\frac{m}{\tau}-1)\|u_{n}\|^{\tau}_{\tau}. $

由于$ \lambda_{1}\leq\lambda_{n} $,因此

(3.25) $ \begin{equation} \int_{\Omega}(f(x, u_{n})u_{n}-mF(x, u_{n})){\rm d}x\leq\frac{1}{\lambda_{1}}(\frac{m}{\tau}-1)\|u_{n}\|^{\tau}_{\tau}, \; \; \; n\in{\Bbb N}. \end{equation} $

由$ (f_{1}) $和$ (f_{3}) $知$ f(x, u_{n})u_{n}-mF(x, u_{n})\geq c_{10}u_{n}^{\beta}-c_{11} $, $ (x, u_{n})\in\Omega\times{\Bbb R_{+}} $,从(3.25)式可得

$c_{10}\|u_{n}\|^{\beta}_{\beta}\leq \frac{1}{\lambda_{1}}(\frac{m}{\tau}-1)\|u_{n}\|^{\tau}_{\tau}+c_{12}\leq c_{13}\|u_{n}\|^{\tau}_{\beta}+c_{14}, \quad n\in{\Bbb N}$.

注意到$ \tau<\beta $,故$ \|u_{n}\|_{\beta}\leq c_{15}, \; n\in{\Bbb N} $.

下面将分两种情况证明$ \{u_{n}\} $在$ W^{1, \Phi}_{0}(\Omega) $中有界.

情形1  若$ r\leq\beta $,则$ \{u_{n}\}\subseteq L^{r}(\Omega) $有界,由条件$ (f_{1}) $、注2.1和引理2.2,有

$ \ell\min\{\|u_{n}\|^{\ell}, \|u_{n}\|^{m}\}\leq|\Omega|^{1-\frac{\tau}{r}}\|u_{n}\|^{\tau}_{r}+\lambda^{\ast}c_{0}\int_{\Omega}(1+2u_{n}^{r}){\rm d}x, $

因此可得$ \{u_{n}\} $在$ W^{1, \Phi}_{0}(\Omega) $中有界.

情形2  若$ \beta<r<\ell^{\ast} $,令$ t\in(0, 1) $满足

(3.26) $ \begin{equation} \frac{1}{r} = \frac{1-t}{\beta}+\frac{t}{\ell^{\ast}}. \end{equation} $

由插值不等式,得到$ \|u_{n}\|_{r}\leq\|u_{n}\|^{1-t}_{\beta}\|u_{n}\|^{t}_{\ell^{\ast}} $,从而$ \|u_{n}\|^{r}_{r}\leq c_{16}\|u_{n}\|^{tr}_{\ell^{\ast}} $,因此

$ \ell\min\{\|u_{n}\|^{\ell}, \|u_{n}\|^{m}\}\leq\ell\int_{\Omega}\Phi(|\nabla u_{n}|){\rm d}x\leq c_{17}(1+\|u_{n}\|^{tr}_{\ell^{\ast}})\leq c_{18}(1+\|u_{n}\|^{tr}). $

由条件$ (f_{3}) $和(3.26)式知$ tr<\ell $,故$ \{u_{n}\} $在$ W^{1, \Phi}_{0}(\Omega) $中有界.

于是,存在$ \{u_{n}\} $的子列(仍记为其本身),有

(3.27) $ \begin{equation} u_{n}\rightharpoonup u_{\lambda^{\ast}}\ \mbox{(在$W^{1, \Phi}_{0}(\Omega)$中), } \end{equation} $

(3.28) $ \begin{equation} u_{n}\rightarrow u_{\lambda^{\ast}}\ \mbox{(在$L_{\Phi}(\Omega)$中).} \end{equation} $

在(3.23)式中取$ v = u_{n}-u_{\lambda^{\ast}} $为测试函数并结合(3.28)式,有

(3.29) $ \begin{equation} \limsup\limits_{n\rightarrow\infty}\langle-\triangle_{\Phi}u_{n}, u_{n}-u_{\lambda^{\ast}}\rangle\leq0. \end{equation} $

注意到算子$ -\triangle_{\Phi} $满足$ (S_{+}) $型条件,由(3.27)式, (3.29)式和引理2.5,得到在$ W^{1, \Phi}_{0}(\Omega) $中$ u_{n}\rightarrow u_{\lambda^{\ast}} $,且$ 0\leq u_{n}\leq u_{\lambda^{\ast}} $.在(3.23)式中令$ n\rightarrow\infty $,推得

$ \langle-\triangle_{\Phi}u_{\lambda^{\ast}}, v\rangle = \int_{\Omega}(u_{\lambda^{\ast}})^{\tau-1}v {\rm d}x+\lambda^{\ast} \int_{\Omega}f(x, u_{\lambda^{\ast}})v {\rm d}x, \; \; \; \forall v\in W^{1, \Phi}_{0}(\Omega). $

因此$ u_{\lambda^{\ast}}\in E_{\lambda^{\ast}} $,即$ \lambda^{\ast}\in\Lambda $.证毕.

定理1.1的证明  根据引理3.7,对每一个$ \lambda\in(0, \lambda^{\ast}) $,问题(1.1)至少存在两个解$ \hat{u}_{e}, \overline{u}_{e} \in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $且$ \hat{u}_{e} \leq \overline{u}_{e} $;由引理3.8知$ \lambda^{\ast}\in\Lambda $,进一步,当$ \lambda = \lambda^{\ast} $时,问题(1.1)至少存在一个解$ u_{\lambda^{\ast}} $;最后,当$ \lambda>\lambda^{\ast} $时,易知问题(1.1)无解.证毕.

首先,考虑下列Dirichlet问题

(4.1) $ \begin{equation} \left\{ \begin{array}{ll} -\triangle_{\Phi}u = u^{\tau-1}, \; \; &x\in\Omega, \\ u>0, &x\in\Omega, \\ u = 0, &x\in\partial\Omega. \end{array} \right. \end{equation} $

易知(4.1)式对应的能量泛函为$ \widetilde{\Psi}(u) = \int_{\Omega} \big(\Phi(|\nabla u|)-\frac{1}{\tau}(u^{+})^{\tau}\big){\rm d}x $在$ W^{1, \Phi}_{0}(\Omega) $上是强制且弱下半连续的,则存在$ \widetilde{u}_{\lambda}\in W^{1, \Phi}_{0}(\Omega) $,使得$ \widetilde{\Psi}(\widetilde{u}_{\lambda}) = \inf\limits_{v\in W^{1, \Phi}_{0}(\Omega)} \widetilde{\Psi}(v) $,且$ \widetilde{\Psi}(\widetilde{u}_{\lambda})<0 = \widetilde{\Psi}(0) $,所以$ \widetilde{u}_{\lambda}\neq0 $.进一步,有$ \widetilde{\Psi}^{\prime}(\widetilde{u}_{\lambda}) = 0 $,即

(4.2) $ \begin{equation} \int_{\Omega}\phi(|\nabla\widetilde{u}_{\lambda}|)\nabla\widetilde{u}_{\lambda}\nabla v{\rm d}x = \int_{\Omega}(\widetilde{u}_{\lambda}^{+})^{\tau-1}v{\rm d}x, \; \; \; \forall v\in W^{1, \Phi}_{0}(\Omega). \end{equation} $

在(4.2)式中取$ v = -\widetilde{u}_{\lambda}^{-} $作为测试函数,得$ \widetilde{u}_{\lambda}\geq0 $,且$ \widetilde{u}_{\lambda}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $.

下面我们证明由(4.2)式给出的$ \widetilde{u}_{\lambda} $是问题(4.1)的唯一解,为此,我们先证明下列引理.

引理4.1  设$ (\phi_{1})-(\phi_{4}) $和$ (f_{1})-(f_{3}) $成立,则$ \widetilde{u}_{\lambda}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $是问题(4.1)的唯一解且对所有的$ \overline{u}\in E_{\lambda} $,有$ \widetilde{u}_{\lambda}\leq\overline{u} $.

证  定义$ Q:L^{1}(\Omega)\rightarrow{{\Bbb R}} \cup\{\infty\} $,有

$ Q(v) = \left\{ \begin{array}{ll} { } \int_{\Omega}\Phi(|\nabla v^{\frac{1}{\ell}}|){\rm d}x, \; \; &v\geq0, \; v^{\frac{1}{\ell}}\in W^{1, \Phi}_{0}(\Omega), \\ +\infty, &\mbox{其它}. \end{array} \right. $

我们断言映射$ v\mapsto\int_{\Omega}\Phi(|\nabla v^{\frac{1}{\ell}}|){\rm d}x $是凸的.事实上,若令$ v = ((1-t)v_{1}+tv_{2})^{\frac{1}{\ell}} $, $ t\in[0, 1] $, $ v_{1} $, $ v_{2}\in \overline{Q} = \{v|v\in L^{1}(\Omega), \; Q(v)<+\infty\} $,那么

$ |\nabla v|\leq\Big((1-t)|\nabla v_{1}^{\frac{1}{\ell}}|^{\ell}+t|\nabla v_{2}^{\frac{1}{\ell}}|^{\ell}\Big)^{\frac{1}{\ell}}. $

又因为$ \Phi(t) $是增函数且$ \Phi(t^{\frac{1}{\ell}}) $是凸的,从而$ \int_{\Omega}\Phi(|\nabla v^{\frac{1}{\ell}}|){\rm d}x $也是凸的.

由Fatou引理,易知$ Q $是弱下半连续的.若$ \widetilde{u}_{\lambda}^{\ast}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $是问题(4.1)的另一个解,对每一个$ h\in C_{0}^{1}(\Omega) $,当$ t $足够小时, $ \widetilde{u}^{\ell}_{\lambda}+th, \; \widetilde{u}^{\ast\ell}_{\lambda}+th\in\overline{Q} $,进一步,对所有的$ h\in W^{1, \Phi}_{0}(\Omega) $ (注意到$ C_{0}^{1}(\Omega) $在$ W^{1, \Phi}_{0}(\Omega) $中稠密).又由$ Q $是凸的,则$ Q' $是单调的,对$ \tau<\ell $,便有

$ 0\leq\frac{1}{\ell}\int_{\Omega}\Big(\frac{(\widetilde{u}_{\lambda}^{+})^{\tau-1}} {\widetilde{u}_{\lambda}^{\ell-1}}-\frac{({\widetilde{u}^{\ast+}_{\lambda}})^{\tau-1}}{{\widetilde{u}^{\ast(\ell-1)}_{\lambda}}}\Big) (\widetilde{u}^{\ell}_{\lambda}-\widetilde{u}^{\ast\ell}_{\lambda}){\rm d}x \leq0. $

因此$ \widetilde{u}_{\lambda} = \widetilde{u}^{\ast}_{\lambda}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $是方程(4.1)的唯一解.

对任意$ \overline{u}\in E_{\lambda} $,作截断函数

$ \widehat{z}(x, u(x)) = \left\{ \begin{array}{ll} ({u(x)^{+}})^{\tau-1}, \; \; &u(x)\leq\overline{u}(x), \\ u_{\lambda}(x)^{\tau-1}, &\mbox{其它}. \end{array} \right. $

此时能量泛函$ \widetilde{\Psi}(u) $成为

$ \widehat{\Psi}(u) = \int_{\Omega}\Big(\Phi(|\nabla u|)-\widehat{Z}(x, u(x))\Big){\rm d}x, \; u\in W^{1, \Phi}_{0}(\Omega), $

其中$ \widehat{Z}(x, t) = \int^{t}_{0}\widehat{z}(x, s){\rm d}s $.与引理3.3的证明类似,存在$ \widehat{u}_{\lambda}\in W^{1, \Phi}_{0}(\Omega) $,使得

$ \widehat{\Psi}(\widehat{u}_{\lambda}) = \inf\limits_{v\in W^{1, \Phi}_{0}(\Omega)}\widehat{\Psi}(v)<0 = \widehat{\Psi}(0), $

从而$ \widehat{u}_{\lambda}\neq0 $,且$ \widehat{u}_{\lambda}\in [0, \overline{u}] $.由唯一性,则有$ 0<\widetilde{u}_{\lambda} = \widehat{u}_{\lambda}\leq \overline{u} $.证毕.

定理1.2的证明  首先,我们有$ E_{\lambda} $是下有向集,即若$ u_{1}, \; u_{2}\in E_{\lambda} $,则存在$ \widetilde{u}_{0}(x)\in E_{\lambda} $,使得$ \widetilde{u}_{0}(x)\leq\widetilde{u}(x): = \min\{u_{1}, \; u_{2}\} $.由文献[17]知,可找到减序列$ \{u_{n}\}\subseteq E_{\lambda} $,使得$ \inf E_{\lambda} = \inf\{u_{n}:n\in {\Bbb N}\} $,且

$ \int_{\Omega}\phi(|\nabla u_{n}|)\nabla u_{n}\nabla v{\rm d}x = \int_{\Omega}u_{n}^{\tau-1}v{\rm d}x+\lambda\int_{\Omega}f(x, u_{n})v{\rm d}x, \; \; \; \forall v\in W^{1, \Phi}_{0}(\Omega). $

重复引理3.7的证明,存在$ \overline{u}_{\lambda}\in W^{1, \Phi}_{0}(\Omega) $,使得$ u_{n}\rightarrow\overline{u}_{\lambda} $ (在$ W^{1, \Phi}_{0}(\Omega) $中),由此可得

$ \int_{\Omega}\phi(|\nabla \overline{u}_{\lambda}|)\nabla \overline{u}_{\lambda}\nabla v{\rm d}x = \int_{\Omega}\overline{u}_{\lambda}^{\tau-1}v{\rm d}x+\lambda\int_{\Omega}f(x, \overline{u}_{\lambda})v{\rm d}x, \; \; \; \forall v\in W^{1, \Phi}_{0}(\Omega). $

由引理4.1,对任意的$ n\in {\Bbb N} $, $ \widetilde{u}_{\lambda}\leq u_{n} $,从而$ \widetilde{u}_{\lambda}\leq\overline{u}_{\lambda} $,即$ \overline{u}_{\lambda} = \inf E_{\lambda} $.证毕.

若$ \lambda_{1}, \lambda_{2}\in\Lambda $,且$ \lambda_{2}>\lambda_{1} $,由定理1.2知存在$ \overline{u}_{\lambda_{2}}\in E_{\lambda_{2}} $, $ \overline{u}_{\lambda_{1}}\in E_{\lambda_{1}} $,利用推论3.1,有$ \overline{u}_{\lambda_{2}}-\overline{u}_{\lambda_{1}}\in {\rm int}(C_{0}^{1}(\overline{\Omega})_{+}) $.

设$ \{\lambda_{n}\}\subseteq\Lambda $,且$ \lambda_{n}\leq\lambda_{n+1}, \; \lim\limits_{n\rightarrow\infty}\lambda_{n} = \lambda $,由正则性结果(参见文献[10]),存在$ \eta\in(0, 1) $和常数$ c_{19}>0 $ ($ c_{19} $与$ n $无关),使得$ \overline{u}_{\lambda_{n}}\in C_{0}^{1, \eta}(\overline{\Omega}) $,且$ \|\overline{u}_{\lambda_{n}}\|_{C_{0}^{1, \eta}(\overline{\Omega})}\leq c_{19} $.又由$ C_{0}^{1, \eta}(\overline{\Omega})\hookrightarrow C_{0}^{1}(\overline{\Omega}) $是紧的,因此存在子列(仍记为其本身)及$ \widetilde{u}_{\lambda}\in C_{0}^{1}(\overline{\Omega}) $,使得$ \overline{u}_{\lambda_{n}}\rightarrow \widetilde{u}_{\lambda} $$ (n\rightarrow\infty) $.

最后,我们断言$ \overline{u}_{\lambda} = \widetilde{u}_{\lambda} $.用反证法,若$ \overline{u}_{\lambda}\neq\widetilde{u}_{\lambda} $,则存在$ x_{0}\in\Omega $,使得$ \overline{u}_{\lambda}(x_{0})<\widetilde{u}_{\lambda}(x_{0}) $.对上述的$ \lambda_{n} $,当$ n $足够大时, $ \overline{u}_{\lambda}(x_{0})<\overline{u}_{\lambda_{n}}(x_{0}) $,而$ \lambda_{n}\leq\lambda $,这与定理1.3的第一个结论矛盾,因此$ \overline{u}_{\lambda} = \widetilde{u}_{\lambda} $.