本文研究如下一阶脉冲微分包含非线性边界问题

(1.1) $ \begin{equation} \left\{ \begin{array}{ll} x'(t)\in F(t, x(t)), \quad & t\in J', \\ \Delta x(t_{k}) = I_{k}(x(t_{k})), & k = 1, \cdots , m, \\ g(x(0), x(T)) = 0, \end{array} \right. \end{equation} $

其中$ J' = J\backslash\{t_{1}, \cdots , t_{m}\} $, $ J = [0, T], \ T>0 $, $ 0<t_{1}<t_{2}<\cdots<t_{m}< T $, $ \Delta x(t_{k}) = x(t_{k}^{+})-x(t_{k}^{-}) $, $ x(t_{k}^{+}) = \lim\limits_{\varepsilon\rightarrow0^{+}}x(t_{k}+\varepsilon) $, $ x(t_{k}^{-}) = \lim\limits_{\varepsilon\rightarrow0^{+}}x(t_{k}-\varepsilon) $, $ F:J\times {\mathbb R}\rightarrow P({\mathbb R}) $是一个多值映射, $ P({\mathbb R}) $是$ {\mathbb R} $的所有非空子集组成的集合, $ I_{k}\in C({\mathbb R}, {\Bbb R} )\ (k = 1, \cdots , m) $, $ g:{\Bbb R} ^{2}\rightarrow {\Bbb R} $是一个单值映射.

脉冲微分包含理论在生物和社会宏观系统、信息科学、控制系统、通讯、生命科学等领域中都有重要的应用.对于一个实际应用较强的系统来说，研究脉冲微分包含解的存在性是研究系统其它性质的首要基础.讨论脉冲微分包含解的存在性常见的方法有不动点理论,参见文献[7-8, 12],还有其他方法,参见文献[1, 3, 5].在文献[4]中,作者仅讨论了周期边界条件$ x(0) = x(T) $的一阶脉冲微分包含,即$ g(x, y) = x-y = 0 $.对于反周期边界条件$ x(0) = -x(T) $的一阶脉冲微分包含,即$ g(x, y) = x+y = 0 $,文献[4]中的结论不能应用.

受上述工作的启发,我们讨论问题(1.1)解的存在性,其主要方法是Martelli不动点定理结合上下解反向方法.本文结构如下:第二节给出本文应用的定义和引理.第三节我们证明系统(1.1)解的存在性结果,最后第四节我们给出一些推论.

下面我们引进一些记号,定义以及一些引理.

设$ X $是一个Banach空间, $ Z $是$ X $的子集.记$ P(X) = \{Z\subset X\mid Z\neq\emptyset\} $, $ P_{cv}(X) = \{Z\subset P(X)\mid Z\ \mbox{是凸的} \} $, $ P_{cp}(X) = \{Z\subset P(X)\mid Z\ \mbox{是紧的}\} $, $ P_{cv, cp}(X) = P_{cv}(X)\cap P_{cp}(X) $,以此类推.例如,设$ X = {\mathbb R} $,我们有记号$ P({\Bbb R} ) $, $ P_{cv}({\Bbb R} ) $, $ P_{cp}({\Bbb R} ) $和$ P_{cv, cp}({\Bbb R} ) $.

设$ PC(J, {\Bbb R} ) = \{x:J\rightarrow {\mathbb R}| x(t)\ \mbox{除}\ t_{k}\mbox{外处处连续}, \mbox{其中}\ x(t_{k}^{-}), \ x(t_{k}^{+})\ \mbox{存在且} \ x(t_{k}^{-}) = x(t_{k}), \ k = 1, \cdots , m \} $,其范数为$ \|x\|_{PC} = \sup\{|x(t)|: t\in J\ \} $,则$ PC(J, {\Bbb R} ) $是一个Banach空间. $ L^{1}(J, {\Bbb R} ) = \{x:J\rightarrow {\mathbb R}|\ |x|:J\rightarrow[0, +\infty)\rm\ \mbox{勒贝格可积} \} $,其范数为$ \|x\|_{L^{1}} = \int_{0}^{T} |x(t)|{\rm d}t $,则$ L^{1}(J, {\Bbb R} ) $是一个Banach空间. $ AC(J, {\Bbb R} ) $是所有绝对连续函数$ x:J\rightarrow {\mathbb R} $组成的空间.

定义2.1  多值映射$ F:J\times {\mathbb R}\rightarrow P({\Bbb R} ) $称为$ L^{1} $-Carathéodory,如果: (ⅰ)对每个$ x\in{\mathbb R} $, $ t\rightarrow F(t, x) $是可测的; (ⅱ)几乎对所有的$ t\in J $, $ x\rightarrow F(t, x) $是上半连续的; (ⅲ)对每个$ \rho>0 $,存在$ \varphi_{\rho}\in L^{1}(J, [0, +\infty)) $满足

$ \|F(t, x)\|_{P({\Bbb R} )} = \sup\{|v|: v\in F(t, x)\}\leq\varphi_{\rho}(t), \ \forall\ |x|\leq\rho, \ {\rm a.e.}\ t\in J. $

定义2.2  函数$ \alpha, \ \beta\in PC(J, {\Bbb R} )\cap AC(J', {\Bbb R} ) $称为问题(1.1)的反向上下解,如果$ \beta(t)\leq\alpha(t) $, $ t\in J $,且存在$ v_{1}, \ v_{2}\in L^{1}(J, {\Bbb R} ) $满足

$ \left\{ \begin{array}{ll} v_{1}(t)\in F(t, \alpha(t)), &{\rm a.e.}\ t\in J, \\ \alpha'(t)\geq v_{1}(t), \ &{\rm a.e.}\ t\in J', \\ \Delta \alpha(t_{k})\geq I_{k}(\alpha(t_{k})), \ &k = 1, \cdots , m, \\ g(\alpha(0), \alpha(T))\geq0 \end{array} \right. $

和

$ \left\{ \begin{array}{ll} v_{2}(t)\in F(t, \beta(t)), &{\rm a.e.}\ t\in J, \\ \beta'(t)\leq v_{2}(t), \ &{\rm a.e.}\ t\in J', \\ \Delta \beta(t_{k})\leq I_{k}(\beta(t_{k})), \ &k = 1, \cdots , m, \\ g(\beta(0), \beta(T))\leq0. \end{array} \right. $

定义2.3  $ x\in PC(J, {\Bbb R} )\cap AC(J', {\Bbb R} ) $称为系统(1.1)的解,如果存在$ v\in L^{1}(J, {\Bbb R} ) $满足$ v(t)\in F(t, x(t)) $, $ \rm{a.e.} $$ t\in J $, $ x'(t) = v(t) $, $ \rm{a.e.} $$ t\in J' $, $ \Delta x(t_{k}) = I_{k}(x(t_{k})) $, $ k = 1, \cdots , m $,且$ g(x(0), x(T)) = 0 $.

引理2.1^[10]  设$ X $是Banach空间, $ F:J\times X\rightarrow P_{cv, cp}(X) $是$ L^{1} $-Carathéodory多值映射, $ S_{F, x} = \{f\in L^{1}(J, X)| \ f(t)\in F(t, x(t)), {\rm a.e.}\ t\in J\}\neq\emptyset $, $ \Gamma:L^{1}(J, X)\rightarrow C(J, X) $是线性连续映射,那么映射$ \Gamma\circ S_{F}:C(J, X)\rightarrow P_{cv, cp}(C(J, X)), \ u\mapsto (\Gamma\circ S_{F})(x): = \Gamma( S_{F, x}) $在$ C(J, X)\times C(J, X) $中是闭图.

引理2.2 (Martelli不动点定理^[11])  设$ X $是Banach空间, $ G:X\rightarrow P_{cv, cp}(X) $是上半连续的凝聚映射.如果集合$ \Re = \{x\in X:\mbox {对某些}\ \lambda>1, \ \lambda x\in G(x)\} $有界,则$ G $有不动点.

注2.1  (ⅰ)如果多值映射$ F $是全连续的,并且是非空紧值的,则$ F $是上半连续的当且仅当$ F $有闭图(i.e., $ x_{n}\rightarrow x^{\ast}, \ y_{n}\rightarrow y^{\ast}, \ y_{n}\in F(x_{n}) $可推出$ y^{\ast}\in F(x^{\ast})) $; (ⅱ)如果多值映射$ F $是全连续的,则$ F $是凝聚的.可参见文献[9].

设$ J_{0} = [0, t_{1}] $, $ J_{k} = (t_{k}, t_{k+1}] $, $ k = 1, \cdots , m $, $ t_{m+1} = T $.

定义2.4^[2]  函数族$ S $称为拟等度连续,如果对每个$ \varepsilon>0 $存在$ \delta>0 $满足:如果$ x\in S $, $ k = 0, 1, \cdots , m $,则$ \|x(t_{1})-x(t_{2})\|<\varepsilon, \ \forall\ t_{1}, \ t_{2}\in J_{k}, \ |t_{1}-t_{2}|<\delta. $

引理2.3 (紧性标准^[2])  集合$ S\in PC(J, R^{n}) $是相对紧的当且仅当(ⅰ) $ S $有界, i.e.,对每个$ x\in S $和某些$ c>0 $, $ \|x\|<c $; (ⅱ) $ S $拟等度连续的.

定义2.5  设$ X $是Banach空间,多值映射$ F $称为全连续的,如果对每个有界子集$ U\subseteq X $, $ F(U) $是相对紧的.

下面我们给出本文的主要结果以及证明.

定理3.1  假设下列条件成立.

(H1) $ F:J\times {\Bbb R} \rightarrow P_{cv, cp}({\Bbb R} ) $是$ L^{1} $-Carathéodory多值映射.

(H2) $ \alpha, \beta\in PC(J, {\Bbb R} )\cap AC(J', {\Bbb R} ) $满足定义2.2.

(H3) $ I_{k}\in C({\Bbb R} , {\Bbb R} ) $, $ k = 1, \cdots , m $.

(H4)当$ (x, y)\in[\beta(0), \alpha(0)]\times[\beta(T), \alpha(T)] $, $ g(x, y) $是连续的单值映射,且关于$ y $非增.

则系统(1.1)至少有一个解$ x $,且满足$ \beta(t)\leq x(t)\leq \alpha(t) $, $ t\in J $.

证  我们把系统(1.1)转换成不动点问题.考虑如下问题

(3.1) $ \begin{equation} \left\{ \begin{array}{ll} x'(t)+x(t)\in F_{1}(t, x(t)), \ \ t\in J', \\ \Delta x(t_{k}) = I_{k}(\tau(t_{k}, x(t_{k}))), \ k = 1, \cdots , m, \\ x(0) = \tau(0, x(0)-g(\tau(0, x), \tau(T, x))), \end{array} \right. \end{equation} $

其中$ F_{1}(t, x) = F(t, \tau(t, x))+\tau(t, x) $, $ \tau:\ C(J, {\Bbb R} )\rightarrow C(J, {\Bbb R} ) $的定义如下

$ \tau(t, x) = \left\{ \begin{array}{ll} \alpha(t), \ \ x(t)>\alpha(t), \\ x(t), \ \ \beta(t)\leq x(t)\leq \alpha(t), \\ \beta(t), \ \ x(t)<\beta(t). \end{array} \right. $

显然,如果$ x $是问题(3.1)的解,且$ \beta(t)\leq x(t)\leq \alpha(t) $, $ \beta(0)\leq x(0)-g(\tau(0, x), \tau(T, x))\leq \alpha(0) $,则$ x $是系统(1.1)的解.通过直接计算,我们知道问题(3.1)的解是下列算子$ N:PC(J, {\Bbb R} )\rightarrow P(PC(J, {\Bbb R} )) $的不动点,

$ \begin{array}{lll} N(x) = \biggl\{h\in PC(J, {\Bbb R} ): h(t) = x(0)+ \int^{t}_{0}[v(s)+\tau(s, x)-x(s)]{\rm d}s\\ \ \ \ \ \ \ \ \ \ \ \ \ \ + \sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x(t_{k}))), \ v\in \widetilde{S}_{F, x}\biggl\}, \end{array} $

其中

$ \widetilde{S}_{F, x} = \{v\in S_{F, \tau(t, x)}: v(t)\leq v_{1}(t), \ {\rm a.e.}\ t\in A_{1}, \ v(t)\geq v_{2}(t), \ {\rm a.e.}\ t\in A_{2}\}, $

$ S_{F, \tau(t, x)} = \{v\in L^{1}(J, {\Bbb R} ):v(t)\in F(t, \tau(t, x)), \ {\rm a.e.}\ t\in J\}, $

$ A_{1} = \{t\in J: \beta(t)\leq\alpha(t)<x(t)\}, \ A_{2} = \{t\in J: x(t)<\beta(t)\leq\alpha(t)\}. $

注意到对每个$ x\in C(J, {\Bbb R} ) $, $ S_{F, x} $非空(见文献[10]),所以$ \widetilde{S}_{F, x} $非空.

下面我们将应用引理2.2证明$ N $有不动点,其证明过程分成5步.

第1步  对每个$ x\in PC(J, {\Bbb R} ) $, $ N(x) $是凸的.

的确,如果$ h_{1}, \ h_{2}\in N(x) $,则存在$ \overline{v}_{1}, \ \overline{v}_{2}\in \widetilde{S}_{F, x} $使得

$ h_{i}(t) = x(0)+ \int^{t}_{0}[\overline{v}_{i}(s)+\tau(s, x)-x(s)]{\rm d}s+ \sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x(t_{k}))), \ i = 1, 2. $

设$ 0\leq l\leq1 $,则对每个$ t\in J $,我们有

$ \begin{array}{lll} [l h_{1}+(1-l)h_{2}](t) = x(0)+ \int^{t}_{0}[l\overline{v}_{1}(s)+(1-l)\overline{v}_{2}(s)+\tau(s, x)-x(s)]{\rm d}s\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x(t_{k}))). \end{array} $

由于$ \widetilde{S}_{F, x} $是凸的(因为在(H1)中$ F $有凸值),则$ l h_{1}+(1-l)h_{2}\in N(x) $,所以$ N(x) $是凸的.

第2步  $ N $是全连续的.

首先,我们将证明$ N $映有界集到$ PC(J, {\Bbb R} ) $中的有界集.设$ q $是正数, $ B_{q} = \{x\in PC(J, {\Bbb R} ): \|x\|_{PC}<q\} $是有界集,且$ x\in B_{q} $.则对每个$ h\in N(x) $,存在$ v\in \widetilde{S}_{F, x} $满足

(3.2) $ \begin{equation} h(t) = x(0)+ \int^{t}_{0}[v(s)+\tau(s, x)-x(s)]{\rm d}s+ \sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x(t_{k}))). \end{equation} $

注意到问题(3.1)的边界条件和$ \tau $的定义,我们有

(3.3) $ \begin{equation} \beta(0)\leq x(0)\leq\alpha(0), \end{equation} $

(3.4) $ \begin{equation} \beta(t)\leq \tau(t, x)\leq\alpha(t). \end{equation} $

设$ \rho_{1} = \max(q, \ \sup\limits_{t\in J}|\alpha(t)|, \ \sup\limits_{t\in J}|\beta(t)|) $,则有$ |\tau(t, x)|\leq\rho_{1} $.由条件(H1)可知,存在$ \varphi_{\rho_{1}}\in L^{1}(J, [0, +\infty)) $满足

(3.5) $ \begin{equation} \sup\{|v|: v\in F(t, \tau(t, x))\}\leq\varphi_{\rho_{1}}(t). \end{equation} $

如果$ x\in B_{q} $,由$ I_{k} $的连续性和(3.4)式知,存在$ c_{k}>0 $, $ k = 1, \cdots , m $满足$ |I_{k}(\tau(t_{k}, x(t_{k})))|\leq c_{k} $.所以,由(3.3)和(3.5)式,我们有

$ \begin{array}{llll} |h(t)|\leq|x(0)|+ \int_{0}^{T}[|v(s)|+|\tau(s, x)|+|x(s)|]{\rm d}s +\sum\limits_{0<t_{k}<t}|I_{k}(\tau(t_{k}, x(t_{k})))|\\ \ \ \ \ \ \ \ \ \leq \max(|\alpha(0)|, |\beta(0)|)+\|\varphi_{\rho_{1}}\|_{L^{1}}+T(\rho_{1}+q) +\sum\limits^{m}_{k = 1}c_{k}: = K, \end{array} $

于是$ \|N(x)\|_{PC}\leq K $.

其次,我们将证明$ N $映有界集到$ PC(J, {\Bbb R} ) $中的拟等度连续集.设$ u_{1}, u_{2}\in J_{k} $, $ k = 0, 1, $$ \cdots , m $, $ u_{1}<u_{2} $, $ x\in B_{q} $, $ h\in N(x) $.则有

$ \begin{array}{lll} |h(u_{2})-h(u_{1})|\leq \int_{u_{1}}^{u_{2}}\varphi_{\rho_{1}}(s){\rm d}s+(u_{2}-u_{1})(\rho_{1}+q)+ \sum\limits_{u_{1}< t_{k}<u_{2}}c_{k}. \end{array} $

当$ u_{2}\rightarrow u_{1} $,上面不等式的右端趋于0,由定义2.4知$ N(B_{q}) $是拟等度连续的.再由引理2.3和定义2.5, $ N $是全连续的,因此是凝聚映射(见注2.1(ⅱ)).

第3步  $ N $有闭图.

设$ x_{n}\rightarrow x^{\ast} $, $ h_{n}\in N(x_{n}) $,且$ h_{n}\rightarrow h^{\ast} $.我们将证明$ h^{\ast}\in N(x^{\ast}) $. $ h_{n}\in N(x_{n}) $意味着存在$ v_{n}\in \widetilde{S}_{F, x_{n}} $满足

$ h_{n}(t) = x_{n}(0)+ \int^{t}_{0}[v_{n}(s)+\tau(s, x_{n})-x_{n}(s)]{\rm d}s+ \sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x_{n}(t_{k}))). $

下面我们需要证明存在$ v^{\ast}\in \widetilde{S}_{F, x^{\ast}} $,对每个$ t\in J $满足

$ h^{\ast}(t) = x^{\ast}(0)+ \int^{t}_{0}[v^{\ast}(s)+\tau(s, x^{\ast})-x^{\ast}(s)]{\rm d}s+ \sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x^{\ast}(t_{k}))). $

因为$ x_{n}\rightarrow x^{\ast} $, $ h_{n}\rightarrow h^{\ast} $, $ \tau $和$ I_{k} $ ($ k = 1, 2, \cdots , m $)是连续的,当$ n\rightarrow\infty $我们有

(3.6) $ \begin{eqnarray} &&\biggl\|h_{n}(t)-x_{n}(0)- \int^{t}_{0}[\tau(s, x_{n})-x_{n}(s)]{\rm d}s-\sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x_{n}(t_{k})))\\ &&- \left[h^{\ast}(t)-x^{\ast}(0)- \int^{t}_{0}[\tau(s, x^{\ast})-x^{\ast}(s)]{\rm d}s-\sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x^{\ast}(t_{k})))\right]\biggl\|_{PC}\rightarrow0, \end{eqnarray} $

考虑线性连续算子$ \Gamma: L^{1}(J, {\Bbb R} )\rightarrow C(J, {\Bbb R} ) $,有

$ v\mapsto \Gamma(v)(t) = \int^{t}_{0}v(s){\rm d}s. $

注意到$ \widetilde{S}_{F, x} $非空,由引理2.1知, $ \Gamma\circ \widetilde{S_{F}} $是闭图.而且,

(3.7) $ \begin{equation} h_{n}(t)-x_{n}(0)- \int^{t}_{0}[\tau(s, x_{n})-x_{n}(s)]{\rm d}s-\sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x_{n}(t_{k})))\in \Gamma(\widetilde{S}_{F, x_{n}}). \end{equation} $

因为$ x_{n}\rightarrow x^{\ast} $,由(3.6)和(3.7)式知,存在$ v^{\ast}\in \widetilde{S}_{F, x^{\ast}} $满足

$ h^{\ast}(t)-x^{\ast}(0)- \int^{t}_{0}[\tau(s, x^{\ast})-x^{\ast}(s)]{\rm d}s-\sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x^{\ast}(t_{k}))) = \int^{t}_{0}v^{\ast}(s){\rm d}s. $

由第1步到第3步可知, $ N $是具有凸闭值的,全连续的,且上半连续的多值映射.

第4步  集合$ \Re = \{x\in PC(J, {\Bbb R} ):\mbox {对某些}\ \lambda>1, \ \lambda x\in G(x)\} $有界.

设$ x\in\Re $,则对某些$ \lambda>1 $, $ \lambda x\in N(x) $.所以,对每个$ t\in J $,有

$ x(t) = \lambda^{-1}\biggl[x(0)+ \int^{t}_{0}[v(s)+\tau(s, x)-x(s)]{\rm d}s+ \sum\limits_{0<t_{k}<t}I_{k}(\tau(t_{k}, x(t_{k})))\biggl], \ v\in \widetilde{S}_{F, x}. $

设$ \rho_{2} = \max(\sup\limits_{t\in J}|\alpha(t)|, \sup\limits_{t\in J}|\beta(t)|) $,由(3.4)式得$ |\tau(t, x)|\leq\rho_{2} $.由条件(H1)知存在$ \varphi_{\rho_{2}}\in L^{1}(J, [0, +\infty)) $满足

(3.8) $ \begin{equation} \sup\{|v|: v\in F(t, \tau(t, x))\}\leq\varphi_{\rho_{2}}(t). \end{equation} $

由条件(H3) $ I_{k} $中的连续性和(3.4)式,存在$ c'_{k}>0 $, $ k = 1, \cdots , m $使得$ |I_{k}(\tau(t_{k}, x(t_{k})))|\leq c'_{k} $.所以由(3.3)和(3.8)式,对每个$ t\in J $我们有

$ \begin{array}{llll} |x(t)|\leq|x(0)|+ \int_{0}^{t}[|v(s)|+|\tau(s, x)|+|x(s)|]{\rm d}s +\sum\limits_{0<t_{k}<t}|I_{k}(\tau(t_{k}, x(t_{k})))|\\ \ \ \ \ \ \ \ \ \leq \max(|\alpha(0)|, |\beta(0)|)+\|\varphi_{\rho_{2}}\|_{L^{1}}+T\rho_{2} + \int_{0}^{t}|x(s)|{\rm d}s+ \sum\limits^{m}_{k = 1}c'_{k}. \end{array} $

设$ K_{0} = \max(|\alpha(0)|, |\beta(0)|)+\|\varphi_{\rho_{2}}\|_{L^{1}}+T\rho_{2} +\sum\limits^{m}_{k = 1}c'_{k} $.应用Gronwall引理(参见文献[6, p36]),对每个$ t\in J $,我们有

$ |x(t)|\leq K_{0} e^{t}. $

所以

$ \|x\|_{PC}\leq K_{0} e^{T}. $

这表明集合$ \Re $有界.应用引理2.2, $ N $有一个不动点,此不动点也是系统(3.1)的一个解.

第5步  系统(3.1)的解$ x $满足

(3.9) $ \begin{equation} \beta(t)\leq x(t)\leq \alpha(t), \ t\in J, \end{equation} $

(3.10) $ \begin{equation} \beta(0)\leq x(0)-g(\tau(0, x), \tau(T, x))\leq \alpha(0). \end{equation} $

首先我们证明(3.9)式.设$ x $是(3.1)式的一个解,我们证明对所有的$ t\in J $, $ x(t)\leq \alpha(t) $.应用反证法证明,假设在点$ s_{0}\in J $处, $ x-\alpha $取得最大正值.由(3.3)式知, $ s_{0}\neq0 $,我们只需考虑$ s_{0}\in(0, T] $.于是存在$ s_{1}\in(0, s_{0}) $, $ s_{1}\neq t_{k}\ (k = 1, 2, \cdots , m) $使得

$ 0<x(t)-\alpha(t)\leq x(s_{0})-\alpha(s_{0}), \ t\in[s_{1}, s_{0}]. $

于是当$ t\in[s_{1}, s_{0}] $, $ \alpha(t)< x(t) $,所以$ \tau(t, x) = \alpha(t) $, $ t\in A_{1} $.此时存在$ v\in \widetilde{S}_{F, x} $满足$ v(t)\in F(t, \alpha(t)) $, $ v(t)\leq v_{1}(t) $使得

$ \begin{eqnarray*} \alpha(s_{0})-\alpha(s_{1})&\leq & x(s_{0})-x(s_{1}) = \int^{s_{0}}_{s_{1}}[v(s)+\alpha(s)-x(s)]{\rm d}s\\ & <& \int^{s_{0}}_{s_{1}}v(s){\rm d}s\leq\int^{s_{0}}_{s_{1}}v_{1}(s){\rm d}s\leq\int^{s_{0}}_{s_{1}}\alpha'(s){\rm d}s\\ & = &\alpha(s_{0})-\alpha(s_{1}). \end{eqnarray*} $

这是矛盾的.因此,对所有的$ t\in J $, $ x(t)\leq\alpha(t) $.同样我们可以证明$ \beta(t)\leq x(t) $, $ t\in J $,在此省略证明过程.因此(3.9)式成立.

其次我们证明系统(3.1)的解$ x $满足(3.10)式.假设

(3.11) $ \begin{equation} \beta(0)>x(0)-g(\tau(0, x), \tau(T, x)). \end{equation} $

由系统(3.1)的边界条件和$ \tau $的定义,我们有

(3.12) $ \begin{equation} x(0) = \beta(0). \end{equation} $

由(3.9)式和$ \tau $的定义,我们有

(3.13) $ \begin{equation} \tau(0, x) = x(0), \ \tau(T, x) = x(T). \end{equation} $

从(3.11)到(3.13)式,我们有

$ g(\beta(0), x(T)) = g(\tau(0, x), \tau(T, x))>0. $

因为在条件(H4)中$ g $关于第二个变量非增,且$ \beta(T)\leq x(T) $,我们有

$ g(\beta(0), \beta(T))\geq g(\beta(0), x(T))>0, $

这与定义2.2的$ g(\beta(0), \beta(T))\leq0 $矛盾.因此我们有

(3.14) $ \begin{equation} \beta(0)\leq x(0)-g(\tau(0, x), \tau(T, x)). \end{equation} $

同样我们可以证明

(3.15) $ \begin{equation} x(0)-g(\tau(0, x), \tau(T, x))\leq \alpha(0). \end{equation} $

(3.14)和(3.15)式表明(3.10)式成立.

由第1步到第5步可知,系统(3.1)的解$ x $也是系统(1.1)的解.证毕.

注3.1  如果在系统(1.1)中$ g(x(0), x(T)) = x(0)-x(T) $, i.e., $ x(0) = x(T) $,则$ g $满足条件(H4),此时系统(1.1)变成一个脉冲微分包含周期边值问题.

下面我们给出系统(1.1)不同的反向上下解定义,都可以得到系统(1.1)解的存在性结果.

定义4.1  $ \alpha, \ \beta\in PC(J, {\Bbb R} )\cap AC(J', {\Bbb R}) $称为问题(1.1)的反向相关上下解,如果$ \beta(t)\leq\alpha(t) $, $ t\in J $,且存在$ v_{1}, \ v_{2}\in L^{1}(J, {\Bbb R} ) $满足

$ \left\{ \begin{array}{ll} v_{1}(t)\in F(t, \alpha(t)), & {\rm a.e.}\ t\in J, \\ \alpha'(t)\geq v_{1}(t), \ & {\rm a.e.}\ t\in J', \\ \Delta \alpha(t_{k})\geq I_{k}(\alpha(t_{k})), \ &k = 1, \cdots , m, \\ g(\alpha(0), \beta(T))\geq0 \end{array} \right. $

和

$ \left\{ \begin{array}{ll} v_{2}(t)\in F(t, \beta(t)), &{\rm a.e.}\ t\in J, \\ \beta'(t)\leq v_{2}(t), \ &{\rm a.e.}\ t\in J', \\ \Delta \beta(t_{k})\leq I_{k}(\beta(t_{k})), \ &k = 1, \cdots , m, \\ g(\beta(0), \alpha(T))\leq0. \end{array} \right. $

推论4.1  假设(H1), (H3)和如下条件成立.

(H5) $ \alpha, \beta\in PC(J, {\Bbb R} )\cap AC(J', {\Bbb R}) $满足定义4.1.

(H6)当$ (x, y)\in[\beta(0), \alpha(0)]\times[\beta(T), \alpha(T)] $, $ g(x, y) $是连续的单值映射,且关于$ y $非减.

则系统(1.1)至少有一个解$ x $,且满足$ \beta(t)\leq x(t)\leq \alpha(t) $, $ t\in J $.

证明过程类似定理3.1的证明,在此我们省略.

注4.1  如果在系统(1.1)中$ g(x(0), x(T)) = x(0)+x(T) $, i.e., $ x(0) = -x(T) $,则$ g $满足条件(H6),此时系统(1.1)变成一个脉冲微分包含反周期边值问题.

定义4.2  $ \alpha, \ \beta\in PC(J, {\Bbb R} )\cap AC(J', {\Bbb R}) $称为问题(1.1)的反向相关上下解,如果$ \beta(t)\leq\alpha(t) $, $ t\in J $,且存在$ v_{1}, \ v_{2}\in L^{1}(J, {\Bbb R} ) $满足

$ \left\{ \begin{array}{ll} v_{1}(t)\in F(t, \alpha(t)), & {\rm a.e.}\ t\in J, \\ \alpha'(t)\geq v_{1}(t), \ &{\rm a.e.}\ t\in J', \\ \Delta \alpha(t_{k})\geq I_{k}(\alpha(t_{k})), \ &k = 1, \cdots , m, \\ g(\alpha(0), \beta(T))\leq0 \end{array} \right. $

和

$ \left\{ \begin{array}{ll} v_{2}(t)\in F(t, \beta(t)), &{\rm a.e.}\ t\in J, \\ \beta'(t)\leq v_{2}(t), \ &{\rm a.e.}\ t\in J', \\ \Delta \beta(t_{k})\leq I_{k}(\beta(t_{k})), \ &k = 1, \cdots , m, \\ g(\beta(0), \alpha(T))\geq0. \end{array} \right. $

推论4.2  假设条件(H1), (H3), (H4)和如下条件成立.

(H7) $ \alpha, \beta\in PC(J, {\Bbb R} )\cap AC(J', {\Bbb R}) $满足定义4.2.

则系统(1.1)至少有一个解$ x $,且满足$ \beta(t)\leq x(t)\leq \alpha(t) $, $ t\in J $.

证  我们把系统(1.1)转换成不动点问题.考虑如下问题

(4.1) $\left\{ \begin{array}{ll} x'(t)+x(t)\in F_{1}(t, x(t)), \ \ t\in J', \\ \Delta x(t_{k}) = I_{k}(\tau(t_{k}, x(t_{k}))), \ k = 1, \cdots , m, \\ x(0) = \tau(0, x(0)+g(\tau(0, x), \tau(T, x))), \end{array} \right.$

其中$ F_{1} $, $ \tau $的定义同(3.1)式.推论4.2剩下的证明类似定理3.1的证明,在此我们省略.

定义4.3  $ \alpha, \ \beta\in PC(J, {\Bbb R} )\cap AC(J', {\Bbb R} ) $称为问题(1.1)的反向上下解,如果$ \beta(t)\leq\alpha(t) $, $ t\in J $,且存在$ v_{1}, \ v_{2}\in L^{1}(J, {\Bbb R} ) $满足

$ \left\{ \begin{array}{ll} v_{1}(t)\in F(t, \alpha(t)), &{\rm a.e.}\ t\in J, \\ \alpha'(t)\geq v_{1}(t), \ &{\rm a.e.}\ t\in J', \\ \Delta \alpha(t_{k})\geq I_{k}(\alpha(t_{k})), \ &k = 1, \cdots , m, \\ g(\alpha(0), \alpha(T))\leq0 \end{array} \right. $

和

$ \left\{ \begin{array}{ll} v_{2}(t)\in F(t, \beta(t)), &{\rm a.e.}\ t\in J, \\ \beta'(t)\leq v_{2}(t), \ &{\rm a.e.}\ t\in J', \\ \Delta \beta(t_{k})\leq I_{k}(\beta(t_{k})), \ &k = 1, \cdots , m, \\ g(\beta(0), \beta(T))\geq0. \end{array} \right. $

推论4.3  假设条件(H1), (H3), (H6)和如下条件成立.

(H8) $ \alpha, \beta\in PC(J, {\Bbb R} )\cap AC(J', {\Bbb R} ) $满足定义4.3.

则系统(1.1)至少有一个解$ x $,且满足$ \beta(t)\leq x(t)\leq \alpha(t) $, $ t\in J $.

证明过程类似推论4.2的证明,在此我们省略.