本文在有序Banach空间$ (E, \|\cdot\|) $中,利用上下解方法与混合单调迭代技巧研究下述具有阻尼弹性系统初值问题

(1.1) $ \begin{equation} \left\{ \begin{array}{ll} u''(t)+\rho Bu'(t)+Au(t) = f(t, u(t), u(t)), 0<t<a, \\ u(0) = u_0\in D(A), u'(0) = u_1\in E, \end{array} \right. \end{equation} $

$ L $ -拟mild解的存在性,其中$ A:D(A)\subset E\rightarrow E $为$ E $中稠定闭线性算子.

1982年, Chen和Russell^[1]在具有内积Hilbert空间$ H $中研究了具有阻尼二阶线性弹性系统

(1.2) $ \begin{equation} u''(t)+Bu'(t)+Au(t) = 0, \end{equation} $

其中$ A $与$ B $在$ H $中都是正定自伴算子.

在$ H\times H $中方程(1.2)可推得如下一阶方程

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}\left( \begin{array}{c} A^{\frac{1}{2}}u \\ u' \\ \end{array} \right) = \left( \begin{array}{cc} 0 & A^{\frac{1}{2}} \\ -A^{\frac{1}{2}} &-B \\ \end{array} \right) \left( \begin{array}{c} A^{\frac{1}{2}}u \\ u' \\ \end{array} \right). \end{eqnarray*} $

设$ V = D(A^{\frac{1}{2}}), {\cal H} = V\times H $,那么在$ {\cal H} $中方程(1.2)等价于下述方程

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}\left( \begin{array}{c} A^{\frac{1}{2}}u \\ u' \\ \end{array} \right) = {\cal A}_B \left( \begin{array}{c} A^{\frac{1}{2}}u \\ u' \\ \end{array} \right), \end{eqnarray*} $

其中

$ {\cal A}_B = \left( \begin{array}{cc} 0 & I\\ -A &-B \\ \end{array} \right), $

$ D({\cal A}_B) = D(A)\times [D(A^{\frac{1}{2}})\cap D(B)]. $

Chen和Rusell^[1]推测,若满足如下条件

$ D(A^{\frac{1}{2}})\subset D(B), $

当$ \beta_1, \beta_2>0 $时,满足下列不等式之一:

$ \beta_1(A^{\frac{1}{2}}v, v)\leq (Bv, v)\leq\beta_2(A^{\frac{1}{2}}v, v), v\in D(A^{\frac{1}{2}}), $

$ \beta_1(Av, v)\leq(B^2v, v)\leq\beta_2(Av, v), v\in D(A), $

则$ {\cal A}_B $是$ {\cal H} $中解析半群的无穷小生成元.关于这两个推测的完全证明最终由Huang^[2-3]给出.

另外,在文献[4-10]中,根据算子$ {\cal A}_B $或$ \overline{{\cal A}_B} $的预解算子显式矩阵表示,对$ 0 <\alpha \leq1 $,取算子$ A^\alpha $的比较算子$ B $,讨论了算子$ {\cal A}_B $或$ \overline{{\cal A}_B} $生成解析半群或可微半群的充分条件.

在文献[11-12]中,范等人在Banach空间$ E $中研究了如下具有结构阻尼线性弹性系统和半线性弹性系统解的渐近稳定性

(1.3) $ \begin{equation} \left\{ \begin{array}{ll} u''(t)+\rho Au'(t)+A^2u(t) = f(t, u(t)), 0<t<a, \\ u(0) = u_0\in D(A), u'(0) = u_1\in E, \end{array} \right. \end{equation} $

其中$ \rho $为阻尼系数, $ A:D( A)\subset E\rightarrow E $为扇形算子, $ -A $生成指数稳定的解析半群$ S(t)(t\geq0) $, $ f \in C (J\times E ), u_0 \in D(A ), u_1 \in E $为已知元.

在文献[13]中,范等人分别运用不动点定理和上下解单调迭代技巧获得了具有结构阻尼弹性系统(1.3) mild解的存在性.但该理论仍有待发展到非线性情况.

另一方面,上下解方法与单调迭代技巧是一种有效而灵活的方法,在文献[19]中,李将单调迭代方法推广到有序Banach空间中的发展方程.然而,在上述工作基础上又基于文献[19]的思想和方法,本文通过耦合上、下拟解及混合单调迭代技巧,获得了初值问题(1.1)的耦合最小与最大$ L $ -拟mild解的存在性,本文所得的结果,改进和推广了文献[11-13, 20]中的结果.

设$ (E, \|\cdot\|) $为有序Banach空间，其正元锥$ P = \{x\in E: x\geq 0\} $为正规锥,正规常数为$ N $.设$ C(J, E) $为定义在$ J $上取值于$ E $的全体连续函数按范数$ { }\|u\|_{C} = \max_{t\in J}\|u(t)\| $构成Banach空间.对任意常数$ a>0 $,设$ J = [0, a] $.显然, $ C(J, E) $为有序Banach空间,其正元锥$ P' = \{u\in E|u(t)\geq 0, t\in J\} $为正规锥.符号$ D(L) $表示算子$ L $的定义域.

对$ E $中任意有界集合$ D $, $ \alpha(D) $表示$ D $的Kuratowski非紧性测度.Kuratowski非紧性测度的定义及其相关性质可见文献[14-15].

对任意$ B\subset C(J, E) $和$ t\in J $,令$ B(t) = \{u(t) : u\in B\}\subset E $.若$ B $为有界集，则$ B(t) $也为有界集且$ \alpha(B(t))\leq \alpha(B) $.

引理2.1^[17]  设$ E $为Banach空间且$ D $为$ E $中有界集,则存在可数集$ D_0\subset D $,使得$ \alpha(D)\leq 2\alpha(D_0) $.

引理2.2  ^[16]设$ E $为Banach空间且$ D = \{u_n\}\subset C([b_1, b_2], E) $为有界可数集,对常数$ -\infty<b_1< b_2<+\infty $,使得

$ \alpha\Big(\Big\{\int_{b_1}^{b_2} u_n(t){\rm d}t: n\in {\mathbb N}\Big\}\Big) \leq 2\int_{b_1}^{b_2}\alpha(D(t)){\rm d}t. $

引理2.3  ^[14]设$ D $为$ C([b_1, b_2], E) $中等度连续有界集,则$ \alpha(D) $等度连续且

$ \alpha(D) = \max\limits_{t\in [b_1, b_2]}\alpha(D(t)). $

引理2.4  ^[15]设$ E $为实Banach空间, $ \Omega_0\subset E $为有界凸闭集, $ Q:\Omega_0\rightarrow \Omega_0 $为凝聚算子,则$ Q $在$ \Omega_0 $中至少存在一个不动点.

引理2.5  ^[18]设$ f\in C(J, E) $且$ A $生成$ C_0 $ -半群$ \{T(t)\}_{t\geq0} $,则非齐次柯西问题

(2.1) $ \begin{equation} \left\{ \begin{array}{ll} u'(t) = Au(t)+f(t), t\in J, \\ u(0) = u_0\in D(A) \end{array} \right. \end{equation} $

有一个mild解$ u $满足积分方程

$ u(t) = T(t)u_0+\int_0^t T(t-s)f(s){\rm d}s, t\in J. $

基于文献[13]的思想和方法.首先考虑下列线性阻尼弹性系统

(2.2) $ \begin{equation} \left\{ \begin{array}{ll} u''(t)+\rho Bu'(t)+Au(t) = h(t), t\in J, \\ u(0) = u_0\in D(A), u'(0) = u_1\in E, \end{array} \right. \end{equation} $

其中$ A:D(A)\subset E\to E $与$ B:D(B)\subset E\rightarrow E $为复Banach空间$ E $中稠定闭线性算子且$ h:J \rightarrow E $.

二阶发展方程

(2.3) $ \begin{equation} u''(t)+\rho Bu'(t)+Au(t) = h(t) \end{equation} $

可写为

(2.4) $ \begin{equation} \Big(\frac{\rm d}{{\rm d}t}+E_1(\rho)\Big)\Big(\frac{\rm d}{{\rm d}t}+E_2(\rho)\Big)u = h(t), t>0. \end{equation} $

即

(2.5) $ \begin{equation} \frac{{\rm d}^2u}{{\rm d}t^2}+\Big(E_1(\rho)+E_2(\rho)\Big)\frac{{\rm d}u}{{\rm d}t}+E_1(\rho)E_2(\rho)u = h(t). \end{equation} $

由(2.3)式及(2.5)式推得

(2.6) $ \begin{equation} E_1(\rho)+E_2(\rho) = \rho B, E_1(\rho)E_2(\rho) = A. \end{equation} $

再由(2.6)式,得到

(i)若$ C(\rho) = \rho^2B^2-4B = L^2(\rho)>0, $则有

(2.7) $ \begin{eqnarray} &&E_1(\rho) = \frac{\rho B-\sqrt{\rho^2B^2-4A}}{2} = \frac{\rho B-L(\rho)}{2}, \\ &&E_2(\rho) = \frac{\rho B+\sqrt{\rho^2B^2-4A}}{2} = \frac{\rho B+L(\rho)}{2}, \end{eqnarray} $

(ii)若$ C(\rho) = \rho^2B^2-4B = L^2(\rho) = 0, $则有

$ E_1(\rho) = E_2(\rho) = \frac{\rho B}{2}. $

(iii)若$ C(\rho) = \rho^2B^2-4B = L^2(\rho)<0, $则有

(2.8) $ \begin{eqnarray} &&E_1(\rho) = \frac{\rho B-\sqrt{\rho^2B^2-4A}}{2} = \frac{\rho B-{\rm i}L(\rho)}{2}, \\ && E_2(\rho) = \frac{\rho B+\sqrt{\rho^2B^2-4A}}{2} = \frac{\rho B+{\rm i}L(\rho)}{2}. \end{eqnarray} $

注2.1  为了研究方程$ (1.1) $解的存在性,我们利用上面的线性算子,把$ A $和$ B $结合起来: $ C(\rho) = \rho^2B^2-4A = L^2(\rho) $满足$ D(C(\rho)) = D(B^2)\cap D(A) $.在下面讨论中,我们讨论以下情况: $ C(\rho)<0 $,其中$ L(\rho):D(L(\rho))\subset E\rightarrow E $为稠定闭线性算子.显然,对$ C(\rho)>0 $与$ C(\rho) = 0 $的情形,在文献[11-12]中,已有相关的结论.

引理2.6  设$ L(\rho):D(L(\rho))\subset E\rightarrow E $为$ E $中稠定闭线性算子,使得$ u_0\in D(L(\rho))\cap D(B) $, $ C(\rho) = \rho^2B^2-4A = L^2(\rho)<0 $且$ -E_1(\rho) $和$ -E_2(\rho) $分别生成$ E $中$ C_0 $ -半群$ T_1(t)(t\geq0) $与$ T_2(t)(t\geq0), h\in C(J, E) $,则方程$ (2.2) $有唯一解

$ \begin{eqnarray*} u(t)& = &T_2(t)u_0+\int_0^tT_2(t-s)T_1(s)(u_1+E_2(\rho)u_0){\rm d}s\\ &&+\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)h(\tau){\rm d}\tau{\rm d}s, \end{eqnarray*} $

其中$ E_1(\rho) $, $ E_2(\rho) $由$ (2.7) $式所定义.

证  令

$ \frac{{\rm d}u}{{\rm d}t}+E_2(\rho)u = v(t), t\in J, $

这意味着

$ v_0: = v(0) = u_1+E_2(\rho)u_0. $

因此,我们将线性弹性系统$ (2.2) $写为Banach空间$ E $中两个抽象柯西问题

(2.9) $ \begin{eqnarray} \left\{ \begin{array}{ll} { } \frac{{\rm d}v}{{\rm d}t}+E_1(\rho)v = h(t), t\in J, \\ v(0) = v_0 \end{array} \right. \end{eqnarray} $

与

(2.10) $ \begin{eqnarray} \left\{ \begin{array}{ll} { } \frac{{\rm d}u}{{\rm d}t}+E_2(\rho)u = v(t), t\in J, \\ u(0) = u_0. \end{array} \right. \end{eqnarray} $

显然,对$ -E_1(\rho) $及$ -E_2(\rho) $来说,问题(2.9)与(2.10)分别是线性非齐次初值问题.

因此,由算子半群理论(见文献[11]),易知$ -E_1(\rho) $与$ -E_2(\rho) $分别为$ C_0 $ -半群的无穷小生成元,这意味着初值问题(2.9)与(2.10)适定.

再由引理2.6推得,若$ h\in C(J, E) $,则初值问题(2.9)有一个mild解$ v $满足下列积分方程

(2.11) $ \begin{eqnarray} v(t) = T_1(t)v_0+\int_0^tT_1(t-s)h(s){\rm d}s. \end{eqnarray} $

同理,若$ v\in C(J, E) $,则初值问题(2.10)有一个mild解$ u $满足下列积分方程

(2.12) $ \begin{eqnarray} u(t) = T_2(t)u_0+\int_0^tT_2(t-s)v(s){\rm d}s. \end{eqnarray} $

将(2.11)式代入(2.12)式,推得

$ \begin{eqnarray*} u(t)& = &T_2(t)u_0+\int_0^tT_2(t-s)T_1(s)(u_1+E_2(\rho)u_0){\rm d}s\\ &&+\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)h(\tau){\rm d}\tau{\rm d}s. \end{eqnarray*} $

引理2.6证毕.

定义2.1  设$ -E_1(\rho) $与$ -E_2(\rho) $分别生成$ E $中$ C_0 $ -半群$ T_1(t)\ (t\geq0) $与$ T_2(t)\ (t\geq0) $,对任意$ f\in C(J\times E\times E, E) $,则积分方程

$ \begin{eqnarray*} u(t)& = &T_2(t)u_0+\int_0^tT_2(t-s)T_1(s)(u_1+E_2(\rho)u_0){\rm d}s\\ &&+\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)f(\tau, u(\tau), u(\tau)){\rm d}\tau{\rm d}s \end{eqnarray*} $

的连续解称为初值问题$ (1.1) $的mild解,其中$ E_1(\rho) $, $ E_2(\rho) $由(2.7)式定义.

定义2.2  设$ L\geq0 $为常数.若函数$ v_0, w_0\in C(J, E)\cap C^2(J, E)\cap C(J, E_1) $满足下述不等式

(2.13) $ \begin{equation} \begin{array}{c} v_0''(t)+\rho Bv_0'(t)+Av_0(t)\leq f(t, v_0(t), w_{0}(t))+L(v_0(t)-w_0(t)), t\in J', \\ v_0(0)\leq v_{0}, v_0'\leq v_1, \end{array} \end{equation} $

(2.14) $ \begin{equation} \begin{array}{c} w_0''(t)+\rho Bw_0'(t)+Aw_0(t)\geq f(t, w_0(t), v_{0}(t))+L(w_0(t)-v_0(t)), t\in J', \\ w_0(0)\geq w_{0}, w_0'\geq w_1, \end{array} \end{equation} $

则称$ v_0, w_0 $是初值问题$ (1.1) $的耦合$ L $ -拟下解与耦合$ L $ -拟上解.当$ (2.13) $式与$ (2.14) $式中仅取等号成立时,称$ (v_0, w_0) $是初值问题$ (1.1) $的耦合$ L $ -拟解对.进一步,当$ u_0: = v_0 = w_0 $,则称$ u_0 $是初值问题$ (1.1) $的解.

定义2.3  设$ T(t)(t \geq0) $为$ E $中$ C_0 $ -半群,如$ T(t) $在$ (0, +\infty) $内按算子范数连续,则称$ T(t)(t\geq0) $为$ E $中等度连续半群.

定义2.4  设$ T(t)(t\geq 0) $为有序Banach空间$ E $中$ C_0 $ -半群.若对任意$ x\geq \theta, t\geq 0 $,都有$ T(t)x\geq \theta $,则称$ T(t)(t\geq 0) $为$ E $中正$ C_0 $ -半群.

对于$ v, w \in C(J, E) $且$ v \leq w $,序区间$ \{u \in C(J, E) | v \leq u \leq w\} $记为$ [v, w] $,序区间$ \{u \in E | v(t) \leq u(t) \leq w(t), t \in J\} $记为$ [v(t), w(t)] $.另记$ {\cal L}(E) $为$ E $中全体有界线性算子组成的Banach空间.为了方便,我们记

$ \begin{eqnarray*} M_1 = \sup\limits_{t\in J}\|T_1(t)\|_{{\cal L}(E)}, M_2 = \sup\limits_{t\in J}\|T_2(t)\|_{{\cal L}(E)}. \end{eqnarray*} $

定理3.1  设$ E $为有序Banach空间,其正元锥$ P $为正规锥,存在稠定闭线性算子$ L(\rho):D(L(\rho))\subset E\rightarrow E $,使得$ u_0\in D(L(\rho))\cap D(B) $, $ C(\rho) = \rho^2B^2-4A = L^2(\rho)<0, $$ BL(\rho) = L(\rho)B $且$ -E_1(\rho) $及$ -E_2(\rho) $分别生成$ E $中正$ C_0 $半群, $ f\in C(J\times E\times E, E) $.若初值问题$ (1.1) $存在耦合$ L $ -拟下解$ v_0\in (J, E) $及耦合$ L $ -拟上解$ w_0\in (J, E) $满足$ v_0\leq w_0 $且非线性项$ f $满足下列条件

(H1)存在常数$ M>0 $, $ L\geq0 $,使得对$ \forall t\in J $, $ v_0(t)\leq u_1\leq u_2\leq w_0(t) $, $ v_0(t)\leq v_2\leq v_1\leq w_0(t) $,有

$ f(t, u_2, v_2)-f(t, u_1, v_1)\geq -M(u_2-u_1)-L(v_1-v_2). $

(H2)存在常数$ L_1>0 $,使得对$ \forall t\in J $,有

$ \alpha(\{f(t, u_n, v_n)\})\leq L_1(\alpha(\{u_n\})+\alpha(\{v_n\})), $

其中$ \{u_n\}\subset[v_0(t), w_0(t)] $, $ \{v_n\}\subset[v_0(t), w_0(t)] $为递增序列或递减序列.

则对任意$ u_1\in E $,初值问题$ (1.1) $在$ v_0 $与$ w_0 $之间存在最小与最大耦合$ L $ -拟解.

证  定义如下算子$ Q:[v_0, w_0]\times[v_0, w_0]\to C(J, E) $：

(3.1) $ \begin{eqnarray} Q(u, v)(t)& = &T_2(t)u_0+\int_0^tT_2(t-s)T_1(s)(u_1+E_2(\rho)u_0){\rm d}s \\ &&+\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)[f(\tau, u(\tau), v(\tau))+(M+L)u(\tau)-Lv(\tau)]{\rm d}\tau{\rm d}s. \end{eqnarray} $

首先,证明$ Q $为连续算子.事实上,令$ u_n, v_n\in C(J, E) $,使得$ u_n\rightarrow u, v_n\rightarrow v $.由非线性项$ f $关于第二变量的连续性可知,对任意$ s\in J $,推得

(3.2) $ \begin{eqnarray} f(s, u_n(s), v_n(s))\rightarrow f(s, u(s), v(s)), n\rightarrow\infty. \end{eqnarray} $

即对任意$ \epsilon>0, $存在$ N, $使得$ n>N, $有

(3.3) $ \begin{eqnarray} \|f(s, u_n(s), v_n(s))-f(s, u(s), v(s))\|\leq \epsilon. \end{eqnarray} $

进而,推得

$ \begin{eqnarray*} \|Q(u_n, v_n)(t)-Q(u, v)(t)\| &\leq& M_1M_2 \int_{0}^t\int_0^s (\|f(\tau, u_n(\tau), v_v(\tau))-f(\tau, u(\tau), v(\tau))\|\\ &&+(M+L)\|u_n(\tau)-u(\tau)\|+L\|v_n(\tau)-v(\tau)\|){\rm d}\tau{\rm d}s\\ &\leq &M_1M_2a^2\|f(\tau, u_n(\tau), v_v(\tau))-f(\tau, u(\tau), v(\tau))\|\\ &&+(M+L)\|u_n(\tau)-u(\tau)\|+L\|v_n(\tau)-v(\tau)\|. \end{eqnarray*} $

因此,当$ n>N $,有

$ \|Q(u_n, v_n)-Q(u, v)\|_{C}\leq (M_1M_2a^2+M+2L)\epsilon, $

这意味着算子$ Q:[v_0, w_0]\times [v_0, w_0]\to C(J, E) $连续.再由定义2.1知,初值问题(1.1)的耦合$ L $ -拟mild解等价于算子$ Q $的耦合不动点.

下证$ Q:[v_0, w_0]\times [v_0, w_0]\rightarrow C(J, E) $为混合单调算子且满足$ v_0\leq Q(v_0, w_0), Q(w_0, v_0)\leq w_0 $.事实上,对$ \forall t\in J, v_0(t)\leq u_1(t)\leq u_2(t)\leq w_0, v_0(t)\leq v_2(t)\leq v_1(t)\leq w_0(t), $由假设条件(H1),有

$ f(t, u_1(t), v_1(t))+(M+L)u_1(t)-Lv_1(t)\leq f(t, u_2(t), v_2(t))+(M+L)u_2(t)-Lv_2(t). $

再由算子$ T_1(t) $与$ T_2(t) $的正性,推得

$ \begin{eqnarray*} &&\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)[f(\tau, u_1(\tau), v_1(\tau))+(M+L)u_1(\tau)-Lv_1(\tau)]{\rm d}\tau{\rm d}s\\ &\leq&\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)[f(\tau, u_2(\tau), v_2(\tau))+(M+L)u_2(\tau)-Lv_2(\tau)]{\rm d}\tau{\rm d}s. \end{eqnarray*} $

因此, $ Q(u_1, v_1)\leq Q(u_2, v_2), $推得$ Q $为连续混合单调算子.令

$ h(t) = v_0''(t)+\rho Bv_0'(t)+Av_0(t), $

由(2.13)式及定义2.1,对$ h\in C(J, E) $, $ h(t)\leq f(t, v_0, w_0)+(L+M)v_0-Lw_0, t\in J, $推得

$ \begin{eqnarray*} v_0(t)& = &T_2(t)v_0+\int_0^tT_2(t-s)T_1(s)(u_1+E_2(\rho)v_0){\rm d}s+\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)h(\tau){\rm d}\tau{\rm d}s\\ &\leq& T_2(t)v_0+\int_0^tT_2(t-s)T_1(s)(u_1+E_2(\rho)v_0){\rm d}s\\ &&+\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)[f(\tau, v_0(\tau), w_0(\tau)) +(M+L)v_0(\tau)-Lw_0(\tau)]{\rm d}\tau{\rm d}s\\ & = &Q(v_0, w_0)(t), t\in J. \end{eqnarray*} $

这说明, $ v_0\leq Q(v_0, w_0) $.类似地,可以证明$ Q(w_0, v_0)\leq w_0 $.因此, $ Q: [v_0, w_0]\times [v_0, w_0]\rightarrow [v_0, w_0] $为连续混合单调算子.

在$ [v_0, w_0] $中作两个迭代序列$ \{v_n\} $与$ \{w_n\} $如下:

(3.4) $ \begin{equation} v_n = Q(v_{n-1}, w_{n-1}), w_n = Q(w_{n-1}, v_{n-1}), n = 1, 2, \cdots. \end{equation} $

由算子$ Q $的混合单调性推得

(3.5) $ \begin{equation} v_0\leq v_1\leq v_2\leq\dots\leq v_n\leq \dots \leq w_n\leq \dots\leq w_2\leq w_1\leq w_0. \end{equation} $

下证$ \{v_n\} $与$ \{w_n\} $在$ J $上一致收敛.

为了叙述方便,记$ B = \{v_n: n\in {\mathbb N}\}+\{w_n: n\in {\mathbb N}\} $, $ B_1 = \{v_{n}: n\in {\mathbb N}\} $, $ B_2 = \{w_{n}: n\in {\mathbb N}\} $, $ B_{10} = \{v_{n-1}: n\in {\mathbb N}\} $, $ B_{20} = \{w_{n-1}: n\in {\mathbb N}\} $.则有$ B_1 = Q(B_{10}, B_{20}) $, $ B_2 = Q(B_{20}, B_{10}) $.

再由非紧性测度的性质及$ B_{10} = B_1\bigcup\{v_0\} $, $ B_{20} = B_2\bigcup\{w_0\} $,对任意$ t\in J $,有$ \alpha(B_{10}(t)) = \alpha(B_1(t)) $, $ \alpha(B_{20}(t)) = \alpha(B_2(t)) $.设$ \varphi(t): = \alpha(B(t)), t\in J $,因此,由引理2.2, (H3)及(3.2)式,推得

$ \begin{eqnarray*} \varphi(t)& = &\alpha(B(t))\\ & = &\alpha(B_1(t)+B_2(t))\\ & = &\alpha(Q(B_{10}, B_{20})(t)+Q(B_{20}, B_{10})(t))\\ & = &\alpha\Big(\Big\{T_2(t)u_0+\int_0^tT_2(t-s)T_1(s)(u_1+E_2(\rho)u_0){\rm d}s\\ &&+\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)[f(\tau, v_{n-1}(\tau), w_{n-1}(\tau))+(M+L)v_{n-1}(\tau)\\ &&-Lw_{n-1}(\tau)]{\rm d}\tau{\rm d}s +T_2(t)u_0+\int_0^tT_2(t-s)T_1(s)(u_1+E_2(\rho)u_0){\rm d}s\\ &&+\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)[f(\tau, w_{n-1}(\tau), v_{n-1}(\tau))\\ && +(M+L)w_{n-1}(\tau)-Lv_{n-1}(\tau)]{\rm d}\tau{\rm d}s\Big\}\Big) \\ &\leq& 2M_1M_2a\int_0^t\alpha\Big(\Big\{f(\tau, v_{n-1}(\tau), w_{n-1}(\tau)) +f(\tau, w_{n-1}(\tau), v_{n-1}(\tau))\\ &&+M(v_{n-1}(\tau)-w_{n-1}(\tau))\Big\}\Big){\rm d}\tau \\ &\leq&2M_1M_2a\int_0^t(2L_1+M)(\alpha(B_{10}(s))+\alpha(B_{20}(s))){\rm d}s\\ &\leq&4M_1M_2a(2L_1+M)\int_0^t\varphi(s){\rm d}s. \end{eqnarray*} $

再应用Bellman不等式,对任意$ t\in J $,推出$ \varphi(t) = 0 $.因此,

$ \int_0^t\varphi(s){\rm d}s\equiv0, $

再由上面的不等式,有$ \varphi(t)\leq0 $,结合非紧性测度的性质,亦有$ \varphi(t)\equiv0, t\in J $.特别地,

$ \alpha(B_{10}(t_1)) = 0, \alpha(B_{20}(t_1)) = 0, $

这意味着$ B_{10}(t_1) $与$ B_{20}(t_1) $在$ E $中相对紧.

因此,对任意$ t\in J, $推得$ \{v_n(t)\}+\{w_n(t)\} $相对紧,进而$ \{v_n(t)\}, \{w_n(t)\} $相对紧.再结合单调条件(3.5)易知, $ \{v_n(t)\} $及$ \{w_n(t)\} $相对紧.即

$ { }\lim\limits_{n\to \infty}v_n(t) = \underline{u}(t), t\in J. $

类似地推得,

$ { } \lim\limits_{n\to \infty}w_n(t) = \overline{u}(t), t\in J. $

再由(3.4)式及Lebesgue控制收敛定理可知

$ \underline{u} = Q\underline{u}, \overline{u} = Q\overline{u}. $

再应用单调条件(3.5),推得$ v_0\leq \underline{u}\leq \overline{u}\leq w_0 $.再由$ Q $的单调性易知, $ \underline{u} $与$ \overline{u} $分别为算子$ Q $在$ [v_0, w_0] $中的最大、最小耦合不动点,因此, $ \underline{u} $与$ \overline{u} $分别为初值问题(1.1)在$ [v_0, w_0] $上的最小、最大耦合$ L $ -拟mild解.定理3.1证毕.

在定理3.2中,如果$ E $为弱序列完备的Banach空间,那么条件(H2)自然成立.事实上,由文献[19]中的定理2.2可知,任何单调序有界序列是列紧的.设$ \{u_n\} $为$ E $中的单调递增序列, $ \{v_n\} $为$ E $中的单调递减序列,再由条件(H2)与(H1),易知$ \{f(t, u_n, v_n)\} $亦为单调序有界序列.再由非紧性测度的性质,推得

$ \alpha(\{f(t, u_n, v_n)\})\leq\alpha(\{f(t, u_n, v_n)+Mu_n-Lv_n\})+M\alpha(\{u_n\})+L\alpha(\{v_n\}) = 0. $

因此,条件(H2)成立.故由定理3.1得到下述推论.

推论3.1  设$ E $为弱序列完备的Banach空间,其正元锥$ P $为正规锥,存在稠定闭线性算子$ L(\rho):D(L(\rho))\subset E\rightarrow E $,使得$ u_0\in D(L(\rho))\cap D(B) $, $ C(\rho) = \rho^2B^2-4A = L^2(\rho)<0, $$ BL(\rho) = L(\rho)B $且$ -E_1(\rho) $与$ -E_2(\rho) $分别生成$ E $中正$ C_0 $ -半群, $ f\in C(J\times E\times E, E) $.若初值问题(1.1)存在耦合$ L $ -拟下解$ v_0\in (J, E) $及耦合$ L $ -拟上解$ w_0\in (J, E) $满足$ v_0\leq w_0 $且非线性项$ f $满足假设条件$ \rm(H1) $与$ \rm(H2) $,则对任意$ u_1\in E $,初值问题$ (1.1) $在$ v_0 $和$ w_0 $之间存在最小与最大耦合$ L $ -拟mild解.

下面我们讨论初值问题(1.1)在最小与最大耦合$ L $ -拟解之间mild解的存在性.若用下述条件代替(H2):

(H2) $ ^* $存在常数$ L_1>0 $,使得对任意$ t\in J $,有

$ \alpha(f, D_1\times D_2)\leq L_1(\alpha(D_1)+\alpha(D_2)), $

其中$ D_1 = \{v_n\} $与$ D_2 = \{w_n\} $为$ [v_0(t), w_0(t)] $中的可数集.

定理3.2  设$ E $为有序Banach空间,其正元锥$ P $为正规锥,存在稠定闭线性算子$ L(\rho):D(L(\rho))\subset E\rightarrow E $，使得$ u_0\in D(L(\rho))\cap D(B) $, $ C(\rho) = \rho^2B^2-4A = L^2(\rho)<0, $$ BL(\rho) = L(\rho)B $且$ -E_1(\rho) $及$ -E_2(\rho) $分别生成$ E $中正$ C_0 $ -半群, $ f\in C(J\times E\times E, E) $.若初值问题$ (1.1) $存在耦合$ L $ -拟下解$ v_0\in (J, E) $及耦合$ L $ -拟上解$ w_0\in (J, E) $满足$ v_0\leq w_0 $且非线性项$ f $满足假设条件(H1), (H2)$ ^* $,则对任意$ u_1\in E $,初值问题$ (1.1) $在$ v_0 $和$ w_0 $之间存在最小与最大耦合$ L $ -拟解$ \underline{u} $及$ \overline{u} $且在$ \underline{u} $与$ \overline{u} $之间至少有一个mild解.

证  易证(H2)$ ^* $$ \Rightarrow $ (H2).因此,由定理3.1,可推得初值问题$ (1.1) $在$ v_0 $和$ w_0 $之间存在最小与最大耦合$ L $ -拟解$ \underline{u} $及$ \overline{u} $.下证,在$ \underline{u} $与$ \overline{u} $之间存在mild解.令算子$ {\cal F}u = Q(u, u) $,显然算子$ {\cal F} : [v_0, w_0]\rightarrow [v_0, w_0] $连续且初值问题$ (1.1) $的mild解等价于算子$ {\cal F} $的不动点.下证算子$ {\cal F} : [v_0, w_0]\rightarrow [v_0, w_0] $等度连续.对任意$ u\in [v_0, w_0] $及$ 0< t' < t''\leq a $,有

$ \begin{eqnarray*} &&\|({\cal F}u)(t'')-({\cal F}u)(t')\|\\ &\leq& \|T_2(t'')u_0-T_2(t')u_0\| +\Big\|\int_{0}^{t'}\Big[T_2(t''-s)-T_2(t'-s)\Big]T_1(s)(u_1+E_2(\rho)u_0){\rm d}s\Big\|\\ & &+\Big\|\int_{t'}^{t''} T_2(t''-s)T_1(s)(u_1+E_2(\rho)u_0){\rm d}s\Big\|\\ &&+\Big\|\int_{0}^{t'}\int_0^s[T_2(t''-s)-T_2(t'-s)]\times T_1(s-\tau)[f(\tau, u(\tau), u(\tau))+Mu(\tau)]{\rm d}\tau{\rm d}s\Big\|\\ &&+\Big\|\int_{t'}^{t''}\int_0^sT_2(t''-s)\times T_1(s-\tau)[f(\tau, u(\tau), u(\tau))+Mu(\tau)]{\rm d}\tau{\rm d}s\Big\|\\ &: = &I_1+I_2+I_3+I_4+I_5, \end{eqnarray*} $

其中

$ I_1 = \|T_2(t'')u_0-T_2(t')u_0\|, $

$ I_2 = \Big\|\int_{0}^{t'}\Big[T_2(t''-s)-T_2(t'-s)\Big]T_1(s)(u_1+E_2(\rho)u_0){\rm d}s\Big\|, $

$ I_3 = \Big\|\int_{t'}^{t''} T_2(t''-s)T_1(s)(u_1+E_2(\rho)u_0){\rm d}s\Big\|, $

$ I_4 = \Big\|\int_{0}^{t'}\int_0^s[T_2(t''-s)-T_2(t'-s)]\times T_1(s-\tau)[f(\tau, u(\tau), u(\tau))+Mu(\tau)]{\rm d}\tau{\rm d}s\Big\|, $

$ I_5 = \Big\|\int_{t'}^{t''}\int_0^sT_2(t''-s)\times T_1(s-\tau)[f(\tau, u(\tau), u(\tau))+Mu(\tau)]{\rm d}\tau{\rm d}s\Big\|. $

事实上,我们仅需要证明$ I_1, I_2, I_3, I_4 $, $ I_5\rightarrow 0 $, $ t''-t'\to 0 $.

注意到,对任意$ t\geq0 $,函数$ T_2(t)u_0 $是连续函数.因此, $ T_2(t)u_0 $在$ J $一致连续,进而推得$ { }\lim_{t''\rightarrow t'}I_1 = 0. $

对$ I_2 $,有

$ \begin{eqnarray*} I_2&\leq&\int_{0}^{t'} \Big\|T_2(t''-s)-T_2(t'-s)\Big\|_{{\cal L}(E)}\times\|T_1(s)\|_{{\cal L}(E)}\|u_1+E_2(\rho)u_0\|{\rm d}s\\ &\leq &M_1\|u_1+E_2(\rho)u_0\|\int_{0}^{t'} \Big\|T_2(t''-s)-T_2(t'-s)\Big\|_{{\cal L}(E)}{\rm d}s. \end{eqnarray*} $

因此,由函数$ t\mapsto \|T_1(t)\| $及$ t\mapsto \|T_2(t)\| $的连续性,对任意$ t\in J $,推得$ { }\lim_{t''\rightarrow t'}I_2 = 0. $

对$ I_4 $,有

$ \begin{eqnarray*} I_4&\leq&\int_{0}^{t'}\int_0^s\Big\|T_2(t''-s)-T_2(t'-s)\Big\|_{{\cal L}(E)}\\ &&\times \|T_1(s-\tau)\|_{{\cal L}(E)}\|f(\tau, u(\tau), u(\tau))+Mu(\tau)\|{\rm d}\tau{\rm d}s\\ &\leq &M_1a\|f(\tau, u(\tau), u(\tau))\|\times\int_{0}^{t'}\Big\|T_2(t''-s)-T_2(t'-s)\Big\|_{{\cal L}(E)}{\rm d}s. \end{eqnarray*} $

推得, $ { }\lim_{t''\rightarrow t'}I_4 = 0. $

对$ I_3, I_5 $,得

$ I_3\leq M_1M_2\|u_1+E_2(\rho)u_0\|\cdot|t''-t'|, $

$ I_5\leq M_1M_2\|f(\tau, u(\tau), u(\tau))+Mu(\tau)\|\cdot|t''-t'|. $

因此可推得, $ { } \lim_{t''\rightarrow t'}I_3 = \lim_{t''\rightarrow t'}I_5 = 0. $

故推得$ \|({\cal F}u)(t'')-({\cal F}u)(t')\|\rightarrow 0 $, $ t''-t'\to 0 $,这意味着$ {\cal F}:[v_0, w_0]\to [v_0, w_0] $等度连续.

对任意$ D \subset [v_0, w_0] $,算子$ {\cal F}(D) \subset [v_0, w_0] $有界且等度连续,再由引理2.1,易知存在可数集$ D_0 = \{u_n\}\subset D $,使得

(3.6) $ \begin{eqnarray} \alpha({\cal F}(D))\leq 2\alpha({\cal F}(D_0)). \end{eqnarray} $

因为$ {\cal F}(D_0)\subset {\cal F}(D) $有界且等度连续,可由引理2.3,推得

(3.7) $ \begin{eqnarray} \alpha({\cal F}(D_0)) = \max\limits_{t\in J} \alpha({\cal F}(D_0)(t)). \end{eqnarray} $

对任意$ t\in J $,由引理2.2,条件(H2)及(3.1)式,得

(3.8) $ \begin{eqnarray} \alpha({\cal F}(D_0)(t)) & = & \alpha\Big(\Big\{T_2(t)u_0+\int_0^tT_2(t-s)T_1(s)(u_1+E_2(\rho)u_0){\rm d}s \\ &&+\int_0^t\int_0^sT_2(t-s)T_1(s-\tau)[f(\tau, u_n(\tau), u_n(\tau))+Mu_n(\tau)]{\rm d}\tau {\rm d}s\Big\}\Big) \\ &\leq & 2M_1M_2a\int_{0}^t \alpha(\{f(\tau, u_n(\tau), u_n(\tau))+Mu_n(\tau)\}){\rm d}\tau \\ &\leq & 2M_1M_2a\int_{0}^t (2L_1+M)\alpha(D_0(s)){\rm d}s \\ &\leq & 2M_1M_2(2L_1+M)a^2\alpha(D). \end{eqnarray} $

因此,由(3.6)式及(3.8)式,有

$ \begin{eqnarray*} \alpha({\cal F}(D))\leq 4M_1M_2(2L_1+M)a^2\alpha(D). \label{e3.21} \end{eqnarray*} $

这就证明了算子$ {\cal F}:[v_0, w_0]\to [v_0, w_0] $为凝聚映射.再由引理2.4,易知算子$ {\cal F} $在$ [v_0, w_0] $中至少有一个不动点.即$ u $是初值问题(1.1)在$ [v_0, w_0] $中的mild解.

最后,因为$ u = {\mathcal F}u = Q(u, u), v_0 \leq u \leq w_0 $,由算子$ Q v_1 = Q(v_0, w_0) \leq Q(u, u) \leq Q(w_0, v_0) = w_1 $的混合单调性.类似地推得, $ v_2 \leq u \leq w_2 $.总之, $ v_n \leq u \leq w_n $, $ n\rightarrow\infty $,推得$ \underline{u} \leq u \leq \overline{u}. $因此,初值问题(1.1)在$ \underline{u} $与$ \overline{u} $之间至少有一个mild解.定理3.2证毕.

注3.1  由文献[18]可知,解析半群与可微半群都是等度连续半群.在偏微分方程的应用中,如抛物线方程与强阻尼波方程,其对应的解半群是解析半群,因此,定理$ 3.1 $的结论具有广泛的适用性.

接下来,我们讨论初值问题$ (1.1) $在区间$ [v_0, w_0] $上mild解的存在唯一性.若用下述条件代替条件(H2):

(H4)存在常数$ C, L>0 $,使得对任意$ t\in J $及$ v_0(t)\leq u_1\leq u_2\leq w_0(t), v_0(t)\leq v_2\leq v_1\leq w_0(t) $,有

$ f(t, u_2, v_2)-f(t, u_1, v_1)\leq C(u_2-u_1)+L(v_1-v_2). $

定理3.3  设$ E $为有序Banach空间,其正元锥$ P $为正规锥,存在稠定闭线性算子$ L(\rho):D(L(\rho))\subset E\rightarrow E $,使得$ u_0\in D(L(\rho))\cap D(B) $, $ C(\rho) = \rho^2B^2-4A = L^2(\rho)<0, $$ BL(\rho) = L(\rho)B $且$ -E_1(\rho) $及$ -E_2(\rho) $分别生成$ E $中正$ C_0 $ -半群, $ f\in C(J\times E\times E, E) $.若初值问题$ (1.1) $存在耦合$ L $ -拟下解$ v_0\in (J, E) $及耦合$ L $ -拟上解$ w_0\in (J, E) $满足$ v_0\leq w_0 $且非线性项$ f $满足假设条件(H1)及(H4),则对任意$ u_1\in E $,初值问题$ (1.1) $在$ v_0 $和$ w_0 $之间存在唯一mild解.

证  首先证明由条件(H1),推得条件(H4).对任意$ t\in J $,设$ \{u_n\}\subset [v_0(t), w_0(t)] $为递增序列及$ \{v_n\}\subset [v_0(t), w_0(t)] $为递减序列.对$ m, n\in {\mathbb N} $满足$ m>n $,可由条件(H1)及(H4),有

$ \theta\leq f(t, u_m, v_m)-f(t, u_n, v_n)\leq C(u_m-u_n)+L(v_n-v_m). $

再由上式及锥$ P $的正规性,有

$ \begin{eqnarray*} \|f(t, u_m, v_m)-f(t, u_n, v_n)\|&\leq & N\|C(u_m-u_n)+L(v_n-v_m)\|\\ &\leq &NC\|u_m-u_n\|+N L\|v_n-v_m\|. \end{eqnarray*} $

应用上述不等式及非紧性测度的性质推得

$ \alpha(\{f(t, u_n, v_n)\})\leq NC\alpha(\{u_n\})+NL\alpha(\{v_n\})\leq L_1(\alpha(\{u_n\})+\alpha(\{v_n\})), $

其中$ L_1 = N(C+L) $.若$ \{u_n\} $为递减序列及$ \{v_n\} $为递增序列,故上述不等式亦成立.因此,条件(H2)成立.再由定理3.1,推得初值问题(1.1)在$ v_0 $与$ w_0 $之间存在最小、最大耦合$ L $ -拟解$ \underline{u} $及$ \overline{u} $.

对任意$ t\in J $,由(3.1)式及假设条件(H4),有

$ \begin{eqnarray*} \theta&\leq&\overline{u}(t)-\underline{u}(t) = Q(\overline{u}, \underline{u})(t)-Q(\underline{u}, \overline{u})(t)\\ & = &\int_0^t\int_0^s T_2(t-s)T_1(s-\tau)[f(\tau, \overline{u}(\tau), \underline{u}(\tau))-f(\tau, \underline{u}(\tau), \overline{u}(\tau))]{\rm d}\tau{\rm d}s\\ &\leq& 2M_1M_2(C+L)a \int_0^t(\overline{u}(s)-\underline{u}(s)){\rm d}s. \end{eqnarray*} $

再由上式及锥$ P $的正规性,有

$ \|\overline{u}(t)-\underline{u}(t)\|\leq2M_1M_2a(C+L)\int_0^t\|\overline{u}(s)-\underline{u}(s)\|{\rm d}s. $

应用上式及Bellman不等式,推得

$ \underline{u}(t)\equiv\overline{u}(t). $

因此, $ \widehat{u}: = \underline{u} = \overline{u} $是初值问题$ (1.1) $在$ v_0 $与$ w_0 $之间的唯一mild解.定理3.3证毕.

设$ \Omega\subset {\mathbb R}^N $为有界开区域,其边界$ \partial\Omega $充分光滑.又设$ E = L^2(\Omega) $,那么$ E $按$ L^2 $ -范数构成Banach空间.

为了说明主要结果的可行性,考虑下列阻尼弹性系统

(4.1) $ \begin{eqnarray} \left\{ \begin{array}{ll} { } \frac{\partial^2u(t, x)}{\partial t^2}+2\gamma\frac{\partial u(t, x)}{\partial t}+(\gamma^2+\Delta^2) u(t, x) = f(t, x, u(t, x), u(t, x)), (t, x)\in J\times \Omega, \\ [2mm] \Delta u(t, x) = u(t, x) = 0, (t, x)\in J\times\partial\Omega, \\ { } u(0, x) = u_0(x), \frac{\partial }{\partial t}u(0, x) = u_1(x), x\in\Omega, \end{array} \right. \end{eqnarray} $

其中$ \gamma = \rho>0 $是常数,函数$ f:J\times \Omega\times E\times E\rightarrow E $是连续函数且$ \Delta $为Laplace算子, $ J = [0, 1] $,在$ E $中定义线性算子$ A $与$ B $满足

$ Au = (\gamma^2I+\Delta^2) u, u\in D(A) = D(\Delta^2) = \{u\in H^4(\Omega):\Delta u = u = 0\ \mbox{on}\ \partial\Omega\}, $

$ Bu = 2 u, u\in D(B) = L^2(\Omega). $

显然, $ C(\rho) = \rho^2B^2-4A = -4\Delta^2<0 $,其中$ Lu = -2\Delta u $.显见$ BL = LB $.进而,对$ \rho>0 $,有

$ E_1(\gamma) = \gamma I+{\rm i}\Delta, E_2(\gamma) = \gamma I-{\rm i}\Delta, $

由山路定理知, $ T(t) = {\rm e}^{-{\rm i}t\Delta} $是$ C_0 $ -半群且$ -\Delta $在$ L^p(\Omega) $上是自伴算子.

注意到

$ T(t) = {\rm e}^{-{\rm i}t\Delta}, T(-t) = {\rm e}^{{\rm i}t\Delta}, t\geq0, $

分别是$ C_0 $ -半群在$ L^p(\Omega) $上.因此,根据算子半群理论(见文献[18]),易知$ -E_1(\gamma) $与$ -E_2(\gamma) $在$ L^2(\Omega) $上分别生成指数稳定的等度连续$ C_0 $ -半群$ T_1(t) = {\rm e}^{-t\gamma}T(t) $与$ T_2(t) = {\rm e}^{-t\gamma}T(-t), t\geq0 $且对任意$ t\geq0 $,有

$ \|T_1(t)\|\leq {\rm e}^{-\gamma t}, \|T_2(t)\| = {\rm e}^{-\gamma t}. $

设$ u(t) = u(t, \cdot) $, $ f(t, u(t)) = f(t, \cdot, u(t, \cdot), u(\cdot, t)) $,则初值问题(4.1)可写成初值问题(1.1).为了研究问题(4.1),需要如下假设：

(1) $ u_0\in D(L)\cap D(B), u_1\in L^p(\Omega) $.

定理4.1  如果条件(1), (H1)及(H2)$ ^* $成立,则初值问题$ (4.1) $有一个mild解$ u\in C(J, L^p(\Omega)). $

证  令$ v_0 = u, w_0 = u, $易知$ v_0, w_0 $是初值问题(1.1)的耦合$ L $ -拟解.由假设条件(1)保证条件(H1), (H2)及(H4)成立.因此,可由定理3.3,推得初值问题(4.1)有唯一mild解.定理4.1证毕.

本文通过利用混合单调迭代技术和算子半群理论,主要研究了Banach空间中具有阻尼弹性系统$ L $ -拟mild解的存在性.本文所得结果改进并推广了文献[11-13]中已有的结果.在今后的工作中,将进一步研究该类问题解的渐近稳定性,以及所对应半群的解析性和指数稳定性.