Branciari^[1]首次引入了矩形度量空间的概念, 并给出了此空间中的Banach压缩映象定理.此后, 很多学者在此空间中研究了其他不同压缩条件下的不动点问题^[2-11], 同时也给出了Banach压缩定理和Kannan型不动点定理在矩形$b$ -度量空间中的一些应用.本文受到上述结论的启发, 将在矩形$b$ -度量空间中讨论耦合重合点和公共耦合不动点的存在性和唯一性问题, 得到了一类新的公共耦合不动点定理, 在很大程度上推广了相关文献^[12]的一些结果.

在介绍主要结果之前, 先介绍一些基本概念和已知结果.

定义 1.1 ^[1]设$X$是非空集, $d: X\times X\rightarrow [0, +\infty)$, 且$\forall x, y \in X$, 满足

(Rb1) $d(x, y)=0$当且仅当$x=y$;

(Rb2) $d(x, y)=d(y, x);$

(Rb3) $d(x, y)\leq s[d(x, u)+d(u, v)+d(v, y)]$, 其中$u, v\in X \setminus \{x, y\}$, 同时$u\neq v.$则称$(X, d)$为矩形$b$ -度量空间, 且$s\geq 1$为矩形$b$ -度量空间$(X, d)$的系数.

注 1.1 ^[13]每个度量空间都是矩形$b$ -度量空间, 每个矩形度量空间也为矩形$b$ -度量空间(这时$s=1$).反之不一定成立.

定义 1.2 ^[13]设$\{x_{n}\}$是矩形$b$ -度量空间$(X, d)$中的序列. $\{x_{n}\}$称为$X$中的Cauchy列, 如果$\lim\limits_{n, m\rightarrow\infty}d(x_{n}, x_{m})=0.$

定义 1.3 ^[13]设$\{x_{n}\}$是矩形$b$ -度量空间$(X, d)$中的序列. $\{x_{n}\}$称为$X$中的收敛列, 如果$\lim\limits_{n\rightarrow\infty}d(x_{n}, x)=0$

定义 1.4 ^[13]矩形$b$ -度量空间$(X, d)$称为完备的, 如果$X$中的每一Cauchy列都收敛于$X$中的某个点.

注 1.2 ^[13]矩形$b$ -度量空间$(X, d)$中的数列的极限点不一定唯一, 且收敛列不一定是Cauchy列.

定义 1.5 ^[14]称$(x, y)\in X \times X$是映象$F:X\times X\rightarrow X$的耦合不动点, 如果$F(x, y)=x, F(y, x)=y.$

定义 1.6 ^[15]称$(gx, gy)\in X \times X$是映象对$F:X\times X\rightarrow X$和$g:X\rightarrow X$的重合耦合点, 如果$F(x, y)=gx, F(y, x)=gy.$这时, 称$(x, y)\in X \times X$是映象对$F:X\times X\rightarrow X$和$g:X\rightarrow X$的耦合重合点.

定义 1.7 ^[15]称$(x, y)\in X \times X$是映象对$F:X\times X\rightarrow X$和$g:X\rightarrow X$的公共耦合不动点, 如果$F(x, y)=gx=x, F(y, x)=gy=y.$

定义 1.8 ^[16]设$X$为一非空集.映象对$F:X\times X\rightarrow X$和$g:X\rightarrow X$称为$\omega$ -相容的, 如果当$F(x, y)=gx$且$F(y, x)=gy$, 总有$gF(x, y)=F(gx, gy).$

定义 1.9 ^[17]设$X$是一非空集, $\preceq$是定义在$X$中的一偏序关系, 函数$T:X\times X\rightarrow X, g:X\rightarrow X$. $T$称为具有混合$g$ -单调性质, 如果$T(x, y)$关于$x$是$g$ -单调不减的, 关于$y$是$g$ -单调不增的, 即对任意的$x, y\in X$, 有

$ \begin{eqnarray*} &&x_{1}, x_{2}\in X, gx_{1}\preceq gx_{2}\Rightarrow T(x_{1}, y)\preceq T(x_{2}, y);\\ && y_{1}, y_{2}\in X, gy_{1}\preceq gy_{2}\Rightarrow T(x, y_{1})\succeq T(x, y_{2}). \end{eqnarray*} $

为方便起见, 下文中出现的函数$\psi$和$\phi$均指满足以下条件的函数^[19]:

(1) $\psi :[0, \infty)\rightarrow[0, \infty)$满足: 1) $\psi$是非减的且关于每个变元是连续的; 2) $\psi(t)=0$当且仅当$t=0.$

(2) $\phi :[0, \infty)\rightarrow [0, \infty)$满足: 1) $\phi$是下半连续的; 2) $\phi(t)=0\Leftrightarrow t=0.$

定理 2.1设$(X, d)$是一个矩形$b$-度量空间, 其系数$s> 1$, $ \preceq $是定义在$X$上的一偏序. $g:X\rightarrow X$为$X$上的自映象.映象$T:X\times X\rightarrow X$具有混合$g$ -单调性.且满足以下条件

(1) $T(X\times X)\subseteq g(X);$

(2) $\exists (x_{0}, y_{0})\in X\times X$使得$gx_{0}\preceq T(x_{0}, y_{0}), gy_{0}\succeq T(y_{0}, x_{0});$

(3) $\forall x, y, u, v \in X$, 如果$gx\preceq gu, gy\succeq gv$或者$gx\succeq gu, gy\preceq gv$, 则有

$ \begin{equation} \psi(sd(T(x, y), T(u, v)))\leq \psi(\max\left\{d(gx, gu), d(gy, gv)\right\})-\phi(\max\left\{d(gx, gu), d(gy, gv)\right\}), \end{equation} $

(2.1)

如果$g(X)$是$(X, d)$的完备子集, 则$g$和$T$在$X$中有重合耦合点.

证 根据条件(2)可得, 存在$ (x_{0}, y_{0})\in X\times X$, 使得$gx_{0}\preceq T(x_{0}, y_{0}), gy_{0}\succeq T(y_{0}, x_{0}), $由于$T(X\times X)\subseteq g(X), $所以存在$x_{1}, y_{1} \in X$, 使得$gx_{1}=T(x_{0}, y_{0})$, $gy_{1}=T(y_{0}, x_{0})$.类似的, 存在$x_{2}, y_{2} \in X$, 使得$gx_{2}=T(x_{1}, y_{1})$, $gy_{2}=T(y_{1}, x_{1})$.由于$gx_{0}\preceq T(x_{0}, y_{0}), gy_{0}\succeq T(y_{0}, x_{0})$, 可得到$gx_{0}\preceq gx_{1}, gy_{0}\succeq gy_{1}$.由于映象$T$具有混合$g$-单调性, 所以有

$ \begin{eqnarray*} &&gx_{1}=T(x_{0}, y_{0})\preceq T(x_{0}, y_{1})\preceq T(x_{1}, y_{1})=gx_{2};\\ && gy_{1}=T(y_{0}, x_{0})\succeq T(y_{0}, x_{1})\succeq T(y_{1}, x_{1})=gy_{2}. \end{eqnarray*} $

依此类推, 就可得到$X$中的两个序列$\{x_{n}\}$和$\{y_{n}\}$, 使得$gx_{n}=T(x_{n}, y_{n}), $ $gy_{n}=T(y_{n}, x_{n}), $并且$\{gx_{n}\}$和$\{gy_{n}\}$还满足

$ \left\{ \begin{array}{ll} gx_{0}\preceq gx_{1}\preceq gx_{2}\preceq \cdots \preceq gx_{n}\preceq gx_{n+1}\preceq \cdots ; \\ gy_{0}\succeq gy_{1}\succeq gy_{2}\succeq \cdots \succeq gy_{n}\succeq gy_{n+1}\succeq \cdots. \end{array} \right. $

(2.2)

在式(2.1)中, 取$(x, y)=(x_{n}, y_{n})$和$(u, v)=(x_{n+1}, y_{n+1})$, 并使用式(2.2)可得

$ \begin{eqnarray*} &&{\psi(sd(gx_{n+1}, gx_{n+2}))=\psi(d(T(x_{n}, y_{n}), T(x_{n+1}, y_{n+1})))}\nonumber\\ &\leq&\!\! \psi(\max\left\{d(gx_{n}, gx_{n\!+\!1}), d(gy_{n}, gy_{n\!+\!1})\right\})\!-\!\phi(\max\left\{d(gx_{n}, gx_{n\!+\!1}), d(gy_{n}, gy_{n\!+\!1})\right\}). \end{eqnarray*} $

即

$ \begin{eqnarray} &&{\psi(sd(gx_{n+1}, gx_{n+2}))}\nonumber\\ &\leq&\!\! \psi(\max\left\{d(gx_{n}, gx_{n\!+\!1}), d(gy_{n}, gy_{n\!+\!1})\right\})\!-\!\phi(\max\left\{d(gx_{n}, gx_{n\!+\!1}), d(gy_{n}, gy_{n\!+\!1})\right\}). \end{eqnarray} $

(2.3)

同理可得

$ \begin{eqnarray} &&{\psi(sd(gy_{n+1}, gy_{n+2}))}\nonumber\\ &\leq&\!\! \psi(\max\left\{d(gx_{n}, gx_{n\!+\!1}), d(gy_{n}, gy_{n\!+\!1})\right\})\!-\!\phi(\max\left\{d(gx_{n}, gx_{n\!+\!1}), d(gy_{n}, gy_{n\!+\!1})\right\}). \end{eqnarray} $

(2.4)

令

$ \begin{equation} d_{n}=\max\{d(gx_{n}, gx_{n+1}), d(gy_{n}, gy_{n+1})\}. \end{equation} $

(2.5)

由于$\max\{\psi(a), \psi(b)\}=\psi(\max\{a, b\}), \forall a, b\in[0, +\infty), $进而有

$ \begin{equation*} \psi(sd_{n+1})=\max\{\psi(sd(gx_{n+1}, gx_{n+2})), \psi(sd(gy_{n+1}, gy_{n+2}))\}. \end{equation*} $

再根据(2.3), (2.4), (2.5)式和上式可得

$ \begin{eqnarray*} &&{\psi(sd_{n+1}) }\nonumber\\ &\leq &\psi(\max\{d(gx_{n}, gx_{n+1}), d(gy_{n}, gy_{n+1})\})-\phi(\max\{d(gx_{n}, gx_{n+1}), d(gy_{n}, gy_{n+1})\})\nonumber\\ &= &\psi(d_{n})-\phi(d_{n}). \end{eqnarray*} $

又由函数$\phi :[0, \infty)\rightarrow [0, \infty)$, 因此由上式可得

$ \begin{equation} \psi(sd_{n+1})\leq \psi(d_{n})-\phi(d_{n})\leq \psi(d_{n}). \end{equation} $

(2.6)

由$\psi$是非减的, 可得

$ \begin{equation} sd_{n+1}\leq d_{n} . \end{equation} $

(2.7)

从而可知$\{d_{n}\}$单调递减的非负实数列, 因此存在$ r\in [0, \infty)$, 使得$\lim\limits_{n\rightarrow\infty}d_{n}=r.$式(2.6)两边令$n\rightarrow\infty$时, 取极限得$ \psi(sr)\leq\psi(r)-\phi(r)\leq \psi(r). $由$\psi$是非减的, 可得$sr\leq r$.又由$s> 1$, 所有当$r> 0$时, 出现矛盾, 进而可得$r=0.$于是有$\lim\limits_{n\rightarrow\infty}d_{n}=0.$由式(2.5)可知

$ \begin{equation} \lim\limits_{n\rightarrow\infty}d(gx_{n}, gx_{n+1})=\lim\limits_{n\rightarrow\infty}d(gy_{n}, gy_{n+1})=0. \end{equation} $

(2.8)

令$k=\frac{1}{s}$, 那么$0< k < 1$, 于是式(2.7)可化为$d_{n+1}\leq k d_{n}$, 进而可以得到

$ \begin{equation} d_{n+1}\leq k d_{n}\leq k^{2} d_{n-1} \leq \cdots\leq k^{n+1} d_{0}, n=0, 1, 2, \cdots . \end{equation} $

(2.9)

根据式(2.2), 在式(2.1)中取$(x, y)=(x_{n}, y_{n})$和$(u, v)=(x_{n+2}, y_{n+2})$, 可得

$ \begin{eqnarray} &&{\psi(sd(gx_{n+1}, gx_{n+3}))=\psi(d(T(x_{n}, y_{n}), T(x_{n+2}, y_{n+2})))}\nonumber\\ &\leq\!\! &\!\! \psi(\max\left\{d(gx_{n}, gx_{n\!+\!2}), d(gy_{n}, gy_{n\!+\!2})\right\})\!-\!\phi(\max\left\{d(gx_{n}, gx_{n\!+\!2}), d(gy_{n}, gy_{n\!+\!2})\right\}). \end{eqnarray} $

(2.10)

即

$ \begin{eqnarray} &&{\psi(sd(gx_{n+1}, gx_{n+3}))}\nonumber\\ &\leq\!\! &\!\! \psi(\max\left\{d(gx_{n}, gx_{n\!+\!2}), d(gy_{n}, gy_{n\!+\!2})\right\})\!-\!\phi(\max\left\{d(gx_{n}, gx_{n\!+\!2}), d(gy_{n}, gy_{n\!+\!2})\right\}). \end{eqnarray} $

(2.11)

同理可得

$ \begin{eqnarray} &&{\psi(sd(gy_{n+1}, gy_{n+3}))}\nonumber\\ &=\!\! &\!\!\psi(\max\left\{d(gx_{n}, gx_{n\!+\!2}), d(gy_{n}, gy_{n\!+\!2})\right\})\!-\!\phi(\max\left\{d(gx_{n}, gx_{n\!+\!2}), d(gy_{n}, gy_{n\!+\!2})\right\}). \end{eqnarray} $

(2.12)

令

$ \begin{equation} d^{*}_{n}=\max\{d(gx_{n}, gx_{n+2}), d(gy_{n}, gy_{n+2})\}. \end{equation} $

(2.13)

由于$\max\{\psi(a), \psi(b)\}=\psi(\max\{a, b\}), \forall a, b\in[0, +\infty), $进而有

$ \begin{equation*} \psi(sd^{*}_{n+1})=\max\{\psi(sd(gx_{n+1}, gx_{n+3})), \psi(sd(gy_{n+1}, gy_{n+3}))\}. \end{equation*} $

再根据(2.11), (2.12)和(2.13)式可得$\psi(sd^{*}_{n+1})\leq \psi(d^{*}_{n})-\phi(d^{*}_{n})\leq \psi(d^{*}_{n}).$又由$\psi$是非减的可知$sd^{*}_{n+1}\leq d^{*}_{n}, n=0, 1, 2, \cdots. $又由$k=\frac{1}{s}$, 可知$d^{*}_{n+1}\leq k d^{*}_{n}$, 依次类推可得到

$ \begin{equation} d^{*}_{n+1}\leq k d^{*}_{n}\leq k^{2} d^{*}_{n-1} \leq \cdots\leq k^{n+1} d^{*}_{0}, n=0, 1, 2, \cdots . \end{equation} $

(2.14)

接下来, 将证明$\{gx_{n}\}$和$\{gy_{n}\}$都是$gX$中的Cauchy列.

a) 当$p$是奇数时, 设$p=2m+1.$

事实上, 由性质(Rb3)可得

$ \begin{eqnarray*} &&{d(gx_{n}, gx_{n+2m+1})}\\ &\leq &s[d(gx_{n}, gx_{n+1})+d(gx_{n+1}, gx_{n+2})+d(gx_{n+2}, gx_{n+2m+1})]\\ &\leq &s[d(gx_{n}, gx_{n+1})+d(gx_{n+1}, gx_{n+2})]\\ &&+s^{2}[d(gx_{n+2}, gx_{n+3})+d(gx_{n+3}, gx_{n+4})+d(gx_{n+4}, gx_{n+2m+1})]\\ &\leq &s[d(gx_{n}, gx_{n+1})+d(gx_{n+1}, gx_{n+2})]+s^{2}[d(gx_{n+2}, gx_{n+3})+d(gx_{n+3}, gx_{n+4})]\\ &&+s^{3}[d(gx_{n+4}, gx_{n+5})+d(gx_{n+5}, gx_{n+6})+d(gx_{n+6}, gx_{n+2m+1})]\\ &\leq &\cdots \leq s[d(gx_{n}, gx_{n+1})+d(gx_{n+1}, gx_{n+2})]+s^{2}[d(gx_{n+2}, gx_{n+3})+d(gx_{n+3}, gx_{n+4})]\\ &&+s^{3}[d(gx_{n+4}, gx_{n+5})+d(gx_{n+5}, gx_{n+6})]+\cdots +s^{m}[d(gx_{n+2m-2}, gx_{n+2m-1})\\ &&+d(gx_{n+2m-1}, gx_{n+2m}) +d(gx_{n+2m}, gx_{n+2m+1})]. \end{eqnarray*} $

上式即

$ \begin{eqnarray} &&{d(gx_{n}, gx_{n+2m+1})}\nonumber\\ &\leq &s[d(gx_{n}, gx_{n+1})+d(gx_{n+1}, gx_{n+2})]+s^{2}[d_(gx_{n+2}, gx_{n+3})+d(gx_{n+3}, gx_{n+4})]\nonumber\\ &&+s^{3}[d(gx_{n+4}, gx_{n+5})+d(gx_{n+5}, gx_{n+6})]+\cdots+s^{m}[d(gx_{n+2m-2}, gx_{n+2m-1})\nonumber\\ &&+d(gx_{n+2m-1}, gx_{n+2m})]+s^{m}d(gx_{n+2m}, gx_{n+2m+1}). \end{eqnarray} $

(2.15)

同样道理可知

$ \begin{eqnarray} &&{d(gy_{n}, gy_{n+2m+1})}\nonumber\\ &\leq &s[d(gy_{n}, gy_{n+1})+d(gy_{n+1}, gy_{n+2})]+s^{2}[d(gy_{n+2}, gy_{n+3})+d(gy_{n+3}, gy_{n+4})]\nonumber\\ &&+s^{3}[d(gy_{n+4}, gy_{n+5})+d(gy_{n+5}, gy_{n+6})]+\cdots+s^{m}[d(gy_{n+2m-2}, gy_{n+2m-1})\nonumber\\ &&+d(gy_{n+2m-1}, gy_{n+2m})]+s^{m}d(gy_{n+2m}, gy_{n+2m+1}). \end{eqnarray} $

(2.16)

结合(2.9), (2.15)和(2.16)式, 又由$k=\frac{1}{s}$可得

$ \begin{eqnarray*} &&{\max\{d(gx_{n}, gx_{n+2m+1}), d(gy_{n}, gy_{n+2m+1})\}}\\ &\leq &sd_{n}+sd_{n+1}+s^{2}d_{n+2}+s^{2}d_{n+3}+s^{3}d_{n+4}+s^{3}d_{n+5}\\ &&+\cdots +s^{m}d_{n+2m-2}+s^{m}d_{n+2m-1}+s^{m}d_{n+2m}\\ &\leq &sk^{n}d_{0}+sk^{n+1}d_{0}+s^{2}k^{n+2}d_{0}+s^{2}k^{n+3}d_{0}+s^{3}k^{n+4}d_{0}+s^{3}k^{n+5}d_{0}\\ &&+\cdots +s^{m}k^{n+2m-2}d_{0}+s^{m}k^{n+2m-1}d_{0}+s^{m}k^{n+2m}d_{0}\\ &= &(sk^{n}+sk^{n+1}+s^{2}k^{n+2}+s^{2}k^{n+3}+s^{3}k^{n+4}+s^{3}k^{n+5}\\ &&+\cdots +s^{m}k^{n+2m-2}+s^{m}k^{n+2m-1}+s^{m}k^{n+2m})d_{0}\\ &= &(sk^{n}+s^{2}k^{n+2}+s^{3}k^{n+4}+\cdots +s^{m}k^{n+2m-2})\times d_{0}+s^{m}k^{n+2m}\times d_{0}\\ &&+(sk^{n+1}+s^{2}k^{n+3}+s^{3}k^{n+5}+\cdots +s^{m}k^{n+2m-1})\times d_{0}\\ &= &\left[\frac{(1-s^{m}k^{2m})(sk^{n}+sk^{n+1})}{1-sk^{2}}+s^{m}k^{n+2m}\right]d_{0}\\ &= &\left[\frac{(1-s^{m}\frac{1}{s^{2m}})(s\cdot \frac{1}{s^{n}}+s\frac{1}{s^{n+1}})}{1-s\cdot \frac{1}{s^{2}}}+s^{m}\cdot\frac{1}{s^{n+2m}}\right]d_{0}\\ &= &\left[\left(\frac{1}{s^{n-1}}+\frac{1}{s^{n+m-1}}\right)\left(\frac{s+1}{s-1}\right)+\frac{1}{s^{n+m}}\right]d_{0}. \end{eqnarray*} $

即

$ \begin{eqnarray}\label{2.17} \max\{d(gx_{n}, gx_{n\!+\!2m\!+\!1}), d(gy_{n}, gy_{n\!+\!2m\!+\!1})\}\leq\left[\left(\frac{1}{s^{n\!-\!1}}+\frac{1}{s^{n\!+\!m\!-\!1}}\right)\left(\frac{s\!+\!1}{s\!-\!1}\right)\!+\!\frac{1}{s^{n\!+\!m}}\right] d_{0}. \end{eqnarray} $

(2.17)

b) 当$p$是偶数时, 设$p=2m.$使用性质(Rb3)可得

$ \begin{eqnarray*} &&{d(gx_{n}, gx_{n+2m})}\\ &\leq &s[d(gx_{n}, gx_{n+1})+d(gx_{n+1}, gx_{n+2})+d(gx_{n+2}, gx_{n+2m})]\\ &\leq &s[d(gx_{n}, gx_{n+1})+d(gx_{n+1}, gx_{n+2})]\\ &&+s^{2}[d(gx_{n+2}, gx_{n+3})+d(gx_{n+3}, gx_{n+4})+d(gx_{n+4}, gx_{n+2m})]\\ &\leq &s[d(gx_{n}, gx_{n+1})+d(gx_{n+1}, gx_{n+2})]+s^{2}[d(gx_{n+2}, gx_{n+3})+d(gx_{n+3}, gx_{n+4})]\\ &&+\cdots +s^{m-1}[d(gx_{n+2m-4}, gx_{n+2m-3})+d(gx_{n+2m-3}, gx_{n+2m-2})]\\ &&+s^{m-1}d(gx_{n+2m-2}, gx_{n+2m})\\ &\leq &s[d(gx_{n}, gx_{n+1})+d(gx_{n+1}, gx_{n+2})]+s^{2}[d(gx_{n+2}, gx_{n+3})+d(gx_{n+3}, gx_{n+4})]\\ &&+\cdots +s^{m-1}[d(gx_{n+2m-4}, gx_{n+2m-3})+d(gx_{n+2m-3}, gx_{n+2m-2})]\\ &&+s^{m-1}d(gx_{n+2m-2}, gx_{n+2m}). \end{eqnarray*} $

即

$ \begin{eqnarray} &&{d(gx_{n}, gx_{n+2m})}\nonumber\\ &\leq &s[d(gx_{n}, gx_{n+1})+d(gx_{n+1}, gx_{n+2})]+s^{2}[d(gx_{n+2}, gx_{n+3})+d(gx_{n+3}, gx_{n+4})]\nonumber\\ &&+\cdots +s^{m-1}[d(gx_{n+2m-4}, gx_{n+2m-3})+d(gx_{n+2m-3}, gx_{n+2m-2})]\nonumber\\ &&+s^{m-1}d(gx_{n+2m-2}, gx_{n+2m}). \end{eqnarray} $

(2.18)

同样道理可知

$ \begin{eqnarray} &&{d(gy_{n}, gy_{n+2m})}\nonumber\\ &\leq &s[d(gy_{n}, gy_{n+1})+d(gy_{n+1}, gy_{n+2})] +s^{2}[d(gy_{n+2}, gy_{n+3})+d(gy_{n+3}, gy_{n+4})]\nonumber\\ &&+\cdots +s^{m-1}[d(gy_{n+2m-4}, gy_{n+2m-3})+d(gy_{n+2m-3}, gx_{n+2m-2})]\nonumber\\ &&+s^{m-1}d(gy_{n+2m-2}, gy_{n+2m}). \end{eqnarray} $

(2.19)

结合式(2.9), (2.14), (2.18)和(2.19)式, 又由$k=\frac{1}{s}$可得

$ \begin{eqnarray*} &&{\max\{d(gx_{n}, gx_{n+2m+2}), d(gy_{n}, gy_{n+2m+2})\}}\\ &\leq &sd_{n}+sd_{n+1}+s^{2}d_{n+2}+s^{2}d_{n+3}+\cdots +s^{m-1}d_{n+2m-4}+s^{m-1}d_{n+2m-3}+s^{m-1}d^{*}_{n+2m-2}\\ &\leq &sk^{n}d_{0}+sk^{n+1}d_{0}+s^{2}k^{n+2}d_{0}+s^{2}k^{n+3}d_{0}+s^{3}k^{n+4}d_{0}+s^{3}k^{n+5}d_{0}+\cdots +s^{m-1}k^{n+2m-4}d_{0}\\ &&+s^{m-1}k^{n+2m-3}d_{0}+s^{m-1}k^{n+2m-2} d^{*}_{0}\\ &= &[sk^{n}\!+\!sk^{n+1}\!+\!s^{2}k^{n+2}\!+\!s^{2}k^{n+3}\!+\!s^{3}k^{n+4}\!+\!s^{3}k^{n+5}\!+\!\cdots \!+\!s^{m-1}k^{n+2m-4}\!+\!s^{m-1}k^{n+2m-3}]d_{0}\\ &&+s^{m-1}k^{n+2m-2}d^{*}_{0}\\ &\leq &[sk^{n}+s^{2}k^{n+2}+s^{3}k^{n+4}+\cdots +s^{m-1}k^{n+2m-4}]d_{0}\\ &&+[sk^{n+1}+s^{2}k^{n+3}+s^{3}k^{n+5}+\cdots +s^{m-1}k^{n+2m-3}] d_{0}+s^{m-1}k^{n+2m-2}d^{*}_{0}\\ &\leq &\frac{(1-s^{m-1}k^{2m-2})(sk^{n}+sk^{n+1})}{1-sk^{2}}d_{0}+s^{m-1}k^{n+2m-2} d^{*}_{0}\\ &= &\frac{\left(1-s^{m-1}\cdot \frac{1}{s^{2m-2}}\right)\left(s\cdot\frac{1}{s^{n}}+s\cdot \frac{1}{s^{n+1}}\right)}{1-s\cdot \frac{1}{s^{2}}}d_{0}+s^{m-1}\cdot\frac{1}{s^{n+2m-2}} d^{*}_{0}\\ &= & \left[\left(\frac{1}{s^{n-1}}-\frac{1}{s^{m+n-2}}\right)\cdot\frac{s+1}{s-1}\right]d_{0}+\frac{1}{s^{n+m-1}} d^{*}_{0}. \end{eqnarray*} $

进而可知

$ \begin{eqnarray} \max\{d(gx_{n}, gx_{n\!+\!2m\!+\!2}), d(gy_{n}, gy_{n\!+\!2m\!+\!2})\} \leq\left[\left(\frac{1}{s^{n\!-\!1}}\!-\!\frac{1}{s^{m\!+\!n\!-\!2}}\right)\cdot\frac{s\!+\!1}{s\!-\!1}\right]d_{0}\!+\!\frac{1}{s^{n\!+\!m\!-\!1}} d^{*}_{0}. \end{eqnarray} $

(2.20)

由(2.17)和(2.20)式可得

$ \begin{eqnarray} \lim\limits_{n, m\rightarrow\infty}\max\{d(gx_{n}, gx_{n+p}), d(gy_{n}, gy_{n+p})\}=0. \end{eqnarray} $

(2.21)

从(2.21)式可知$\{gx_{n}\}$和$\{gy_{n}\}$是$g(X)$中的Cauchy列.

又因为$g(X)$在是矩形$b$ -度量空间$(X, d)$的完备子集, 所以存在$x, y \in X$, 使得

$ \begin{equation} \lim\limits_{n\rightarrow\infty}gx_{n}=gx, \lim\limits_{n\rightarrow\infty}gy_{n}=gy. \end{equation} $

(2.22)

又由(2.1)式可得

$ \begin{eqnarray}\label{2.23} &&{\psi(sd(T(x, y), gx_{n+1}))=\psi(sd(T(x, y), T(x_{n}, y_{n})))}\nonumber\\ &\leq &\psi(\max\left\{d(gx, gx_{n}), d(gy, gy_{n})\right\})-\phi(\max\left\{d(gx, gx_{n}), d(gy, gy_{n})\right\}). \end{eqnarray} $

(2.23)

使用三角不等式

$ \begin{equation}d(T(x, s), gx)\leq sd(T(x, s), gx_{n+1})+sd(gx_{n+1}, gx_n)+sd(gx_n, gx), \end{equation} $

(2.24)

在(2.24)式中, 令$n\rightarrow\infty$取极限,

$ \begin{eqnarray}d(T(x, s), gx)&\leq& \lim\limits_{n\rightarrow\infty}[sd(T(x, s), gx_{n+1})+sd(gx_{n+1}, gx_n)+sd(gx_n, gx)]\nonumber\\ &=&s\lim\limits_{n\rightarrow\infty}d(T(x, y), gx_{n+1}).\end{eqnarray} $

(2.25)

使用(2.22)和(2.25)式, 在(2.23)式两边令$n\rightarrow\infty$取极限, 可得

$ \begin{eqnarray*} \psi(d(T(x, y), gx))&\leq& \psi(s\lim\limits_{n\rightarrow\infty}d(T(x, y), gx_{n+1}))=\lim\limits_{n\rightarrow\infty}\psi(sd(T(x, y), gx_{n+1}))\\ &\leq&\psi(\max\left\{0, 0\right\})-\phi(\max\left\{0, 0\right\})=0. \end{eqnarray*} $

即$d(T(x, y), gx)=0$, 进而可知$T(x, y)=gx$.同理可得$T(y, x)=gy$.从而有$(gx, gy)$是$g$和$T$的重合耦合点, 证毕.

定理 2.2 设$(X, d)$是矩形$b$ -度量空间, 其系数$s> 1$, $ \preceq $是定义在$X$上的一个偏序. $g:X\rightarrow X$为$X$上的自映象.映象$T:X\times X\rightarrow X$具有混合$g$ -单调性.且满足以下条件

(1) $T(X\times X)\subseteq g(X);$

(2) $\exists (x_{0}, y_{0})\in X\times X$使得$gx_{0}\preceq T(x_{0}, y_{0}), gy_{0}\succeq T(y_{0}, x_{0});$

(3) $\forall x, y, u, v, a\in X$, 如果$gx\preceq gu, gy\succeq gv$或者$gx\succeq gu, gy\preceq gv$, 那么有

$ \begin{eqnarray*} &&\psi(d(T(x, y), T(u, v)))\\ &\leq& \psi(\frac{d(gx, gu)+d(gy, gv)}{2})-\phi(\frac{d(gx, gu)+d(gy, gv)}{2}). \end{eqnarray*} $

如果$g(X)$是$(X, d)$中的完备集, 则$g$和$T$在$X$中有重合耦合点.

证 与定理 2.1证明方法相同, 略去.

注 2.1 令定理 2.1中的自映象$g$为恒等映象$I$, 即有下面推论.

推论 2.1 设$(X, d)$是矩形$b$ -度量空间, 其系数$s> 1$, $ \preceq $是定义在$X$上的一个偏序.映象$T : X\times X \rightarrow X$.如果满足以下条件

(1) $\exists (x_{0}, y_{0})\in X\times X$使得$x_{0}\preceq T(x_{0}, y_{0}), y_{0}\succeq T(y_{0}, x_{0});$

(2) $\forall x, y, u, v \in X$, 如果$x\preceq u, y\succeq v$或者$x\succeq u, y\preceq v$, 那么有

$ \begin{eqnarray*} &&\psi(sd(T(x, y), T(u, v)))\\ &\leq& \psi(\max\left\{d(x, u), d(y, v)\right\})-\phi(\max\left\{d(x, u), d(y, v)\right\}), \end{eqnarray*} $

则$T$在$X$中有耦合不动点.

注 2.2 $\forall a, b, c\in [0, +\infty)$, 有

$ (a+b+c)^{k}\leq 3^{k-1}(a^{k}+b^{k}+c^{k}) (k=1, 2, \cdots). $

证 显然当$k=1, 2$时结论成立, 假设当$k=m$时结论成立, 即

$ (a+b+c)^{m}\leq 3^{m-1}(a^{m}+b^{m}+c^{m}), \forall x, y, z, w\in X. $

现证当$k=m+1$时结论也成立.事实上, 有

$ \begin{eqnarray*} &&{(a+b+c)^{m+1}\leq (a+b+c)(a+b+c)^{m}}\\ &\leq &(a+b+c)3^{m-1}(a^{m}+b^{m}+c^{m})\leq3^{m-1}(a+b+c)(a^{m}+b^{m}+c^{m})\\ &=&3^{m-1}[a^{m+1}+b^{m+1}+c^{m+1}+(a^{m}b+ab^{m})+(a^{m}c+ac^{m})+(b^{m}c+bc^{m})]\\ &\leq &3^{m-1}[a^{m+1}+b^{m+1}+c^{m+1}+(a^{m+1}+b^{m+1})+(a^{m+1}+c^{m+1})+(b^{m+1}+c^{m+1})]\\ &= &3^{m}(a^{m+1}+b^{m+1}+c^{m+1}). \end{eqnarray*} $

从而结论对一切自然数$k$成立.

例 2.1 设$X=\mathbb{R}$, $\forall x, y\in X$, 定义$d(x, y)=(x-y)^{2}$, 则$(X, d)$是一个系数$s=3$的矩形$b$ -度量空间.设$X$上的偏序关系$\preceq $定义如下: $x\preceq y\Leftrightarrow x\leq y.$定义函数$T:X \times X\rightarrow \mathbb{R}^{+}$和$g:X\rightarrow \mathbb{R}^{+}$分别如下

$ T(x, y)=\left\{ \begin{array}{ll} \frac{\sqrt{3}}{9}\ln(x-y+1),&x> y; \\ 0,&x\leq y, \end{array} \right. \, \, \, gx=e^{\frac{9x}{\sqrt{3}}}-1. $

易得$T(X\times X)\subseteq g(X)$, 且$T$具有$g$ -混合单调性.当

$ \left\{ \begin{array}{ll} g(x)=T(x, y), \\ g(y)=T(y, x) \end{array} \right. \Leftrightarrow \left\{ \begin{array}{ll} x=0, \\ y=0, \end{array} \right. \Rightarrow g(T(0, 0))=T(g0, g0). $

即$F$和$g$是$\omega$ -相容的.令$(x, y)=(0, 0)$时, 这时有$g0\preceq T(0, 0), g0\succeq T(0, 0)$.

取$x, y, u, v\in X$, 使得$x\preceq u$, $y\succeq v$, 即$x\leq u, y\geq v$.

定义函数$\psi, \phi: \mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$分别为$\psi(t)=\sqrt{t}, \, \, \, \phi(t)=\frac{1}{2}\sqrt{t}, \, \, t\in\mathbb{R}^{+}.$

下面分四种情况讨论.

1) 当$T(x, y)=\frac{\sqrt{3}}{9}\ln(x-y+1), T(u, v)=\frac{\sqrt{3}}{9}\ln(u-v+1)$时, 即$x> y, u> v$时, 有

$ \begin{eqnarray*} &&{\psi(3d(T(x, y), T(u, v)))=\sqrt{3d(T(x, y), T(u, v))}}\\ &\leq &\sqrt{3^{3}d(T(x, y), T(u, v))}= \sqrt{3^{3}(T(x, y)-T(u, v))^{2}}\\ &= & \sqrt{3^{3}(\frac{\sqrt{3}}{9}\ln(x-y+1)-\frac{\sqrt{3}}{9}\ln(u-v+1))^{2}}\\ &=& 3\sqrt{3}[\frac{\sqrt{3}}{9}\ln(x-y+1)-\frac{\sqrt{3}}{9}\ln(u-v+1)]=\ln(x-y+1)-\ln(u-v+1)\\ &=&\ln\frac{x-y+1}{u-v+1}=\ln\frac{x-y+1-(u-v+1)+u-v+1}{u-v+1}\\ &=&\ln[1+\frac{x-u+v-y}{u-v+1}]\leq \ln[1+(x-u)+(v-y)]\\ &\leq& \ln[1+(x-u)+(v-y)]\leq (e^{x}-e^{u})+(e^{v}-e^{y})\\ &\leq &\sqrt{2(e^{x}-e^{u})^{2}+2(e^{v}-e^{y})^{2}}\\ &\leq &\sqrt{2\max\{(e^{x}-e^{u})^{2}, (e^{v}-e^{y})^{2}\}}\\ &= &\sqrt{\frac{1}{4}\max\{(4e^{x}-4e^{u})^{2}, (4e^{v}-4e^{y})^{2}\}}\\ &= &\sqrt{\frac{1}{4}\max\{(gx-gu)^{2}, (gu-gy)^{2}\}}\\ &= &\frac{1}{2}\sqrt{\max\{d(gx, gu), d(gu, gy)\}}\\ &= &\sqrt{\max\{d(gx, gu), d(gu, gy)\}}-\frac{1}{2}\sqrt{\max\{d(gx, gu), d(gu, gy)\}}\\ &= &\psi(\max\{d(gx, gu), d(gu, gy)\})-\phi(\max\{d(gx, gu), d(gu, gy)\}). \end{eqnarray*} $

2) 当$T(x, y)=\frac{\sqrt{3}}{9}\ln(x-y+1), T(u, v)=0$时, 即$x> y, u\leq v$时, 有

$ \begin{eqnarray*} &&{\psi(3d(T(x, y), T(u, v)))=\sqrt{3d(T(x, y), T(u, v))}}\\ &\leq &\sqrt{3^{3}d(T(x, y), T(u, v))}= \sqrt{3^{3}(T(x, y)-T(u, v))^{2}}\\ &= & \sqrt{3^{3}(\frac{\sqrt{3}}{9}\ln(x-y+1)-0)^{2}}= 3\sqrt{3}\cdot \frac{\sqrt{3}}{9}\ln(x-y+1)=\ln(x-y+1)\\ &\leq &x-y\leq e^{x}-e^{y}= \sqrt{(e^{x}-e^{y})^{2}}\\ &=&\sqrt{(e^{x}-e^{u}+e^{u}-e^{v}+e^{v}-e^{y})^{2}}\\ &\leq &\sqrt{[(e^{x}-e^{u})+(e^{v}-e^{y})]^{2}}\\ &\leq &\frac{1}{4}\sqrt{[(4e^{x}-4e^{u})+(4e^{v}-4e^{y})]^{2}}\\ &\leq &\frac{1}{4}\sqrt{[(gx-gu)+(gv-gy)]^{2}}\\ &\leq &\frac{1}{4}\sqrt{2(gx-gu)^{2}+2(gv-gy)^{2}}\\ &\leq &\frac{1}{4}\sqrt{4\max\{(gx-gu)^{2}, (gv-gy)^{2}\}}\\ &= &\frac{1}{2}\sqrt{\max\{(gx-gu)^{2}, (gv-gy)^{2}\}}\\ &= &\sqrt{\max\{d(gx, gu), d(gu, gy)\}}-\frac{1}{2}\sqrt{\max\{d(gx, gu), d(gu, gy)\}}\\ &=&\psi(\max\{d(gx, gu), d(gu, gy)\})-\phi(\max\{d(gx, gu), d(gu, gy)\}). \end{eqnarray*} $

3) 当$T(u, v)=\frac{\sqrt{3}}{9}\ln(u-v+1), T(x, y)=0$时, 即$x\leq y, u> v$时, 同情况2)的证明类似.

4) 当$T(x, y)=0, T(u, v)=0$时, 即$x\leq y, u\leq v$时, 显然成立.

从而根据定理 2.1, 可证得$T$和$g$具有耦合重合点, 且$T(0, 0)=g0=0.$

设$X$是一个$(X, d)$矩形$b$ -度量空间, $\preceq$是$X$上的一个偏序关系, 在$X\times X$中规定一种可比较关系如下

$ (x, y)\preceq(z, t) \Leftrightarrow x\preceq z, y\succeq t, \forall (x, y), (z, t)\in X\times X. $

当$(x, y)\preceq(z, t)$或者$(x, y)\succeq(z, t)$时, 称$(x, y)$和$(z, t)$互为可比较.

定理 3.1 在定理 2.1的条件下, 假设$g$和$T$是$\omega$ -相容的, 且$g$和$T$的任意两个重合耦合点$(gx^{*}, gy^{*})$和$(gz^{*}, gt^{*})$都有相应的公共可比较点$(gu^{*}, gv^{*})\in X\times X$, 这时$g$和$T$有唯一的耦合公共不动点.

证 由定理 2.1可知$g$和$T$至少存在一个重合耦合点.不妨假设$(x, y)$, $(z, t)\in X\times X$是$g$和$T$的任意两个耦合重合点, 即$T(x, y)=gx, T(y, x)=gy$和$T(z, t)=gz$, $T(t, z)=gt.$现证$(gx, gy)=(gz, gt)$.

由已知条件可知, 存在$(u, v)\in X\times X$, 使得$(gu, gv)$分别和$(gx, gy)$, $(gz, gt)$都可比较.

a) 当可比较关系为

$ \begin{equation} (gx, gy)\preceq (gu, gv), \, \, (gt, gz)\preceq (gu, gv). \end{equation} $

(3.1)

令

$ \begin{eqnarray} u_{0}=u, v_{0}=v, x_{0}=x, y_{0}=y, z_{0}=z, t_{0}=t, \end{eqnarray} $

(3.2)

则$\exists (u_{1}, v_{1})\in X\times X, $使得

$ gu_{1}=T(u_{0}, v_{0}), gv_{1}=T(v_{0}, u_{0}). $

依次类推, 会得到两个数列$\{gu_{n}\}$和$\{gv_{n}\}$分别为

$ gu_{n+1}=T(u_{n}, v_{n}), \, \, gv_{n+1}=T(v_{n}, u_{n}), n= 1, 2, 3, \cdots. $

同样道理, 可以得到数列$\{gx_{n}\}$, $\{gy_{n}\}$和$\{gz_{n}\}$, $\{gt_{n}\}$分别为

$ \begin{eqnarray*} &&gx_{n+1}=T(x_{n}, y_{n}), \, \, gy_{n+1}=T(y_{n}, x_{n}), n=0, 1, 2, \cdots, \\ &&gz_{n+1}=T(z_{n}, t_{n}), \, \, gt_{n+1}=T(t_{n}, z_{n}), n= 0, 1, 2, \cdots. \end{eqnarray*} $

由于$(x, y)$是$g, T$的耦合重合点, 则有

$ \begin{eqnarray*} &&gx_{1}=T(x_{0}, y_{0})=T(x, y)=gx=gx_{0}, \, \, gy_{1}=T(y_{0}, x_{0})=T(y, x)=gy=gy_{0}, \\ &&gx_{2}=T(x_{1}, y_{1})=T(x_{0}, y_{0})=gx_{1}, gy_{2}=T(y_{1}, x_{1})=T(y_{0}, x_{0})=gy_{1}. \end{eqnarray*} $

依次类推可知$gx_{0}=gx_{1}=\cdots=gx_{n}=\cdots, gy_{0}=gy_{1}=\cdots=gy_{n}=\cdots.$即

$ \begin{eqnarray} gx_{n}= gx, gy_{n}= gy, \, \, \forall n=0, 1, 2, \cdots. \end{eqnarray} $

(3.3)

相同道理可知

$ \begin{equation} gz_{n}= gz, gt_{n}= gt, \, \, \forall n=0, 1, 2, \cdots. \end{equation} $

(3.4)

由于(3.1)和(3.2)式可得

$ (gx_{0}, gy_{0})\preceq (gu_{0}, gv_{0}), \, \, (gt_{0}, gz_{0})\preceq (gu_{0}, gv_{0}). $

由于$T$具有混合$g$ -单调性, 所以可得

$ \begin{eqnarray*} &&{gx_{1}=T(x_{0}, y_{0})\preceq T(u_{0}, y_{0})\preceq T(u_{0}, v_{0})=gu_{1}, };\\ &&gy_{1}=T(y_{0}, x_{0})\succeq T(v_{0}, x_{0})\succeq T(v_{0}, u_{0})=gv_{1}. \end{eqnarray*} $

即$(gx_{1}, gy_{1})\preceq (gu_{1}, gv_{1})$.同理可得$(gx_{2}, gy_{2})\preceq (gu_{2}, gv_{2}).$这样继续做下去, 可得

$(gx_{n}, gy_{n})\preceq (gu_{n}, gv_{n}), n=0, 1, 2, \cdots.$

类似方法, 可得

$(gt_{n}, gz_{n})\preceq (gu_{n}, gv_{n}), n=0, 1, 2, \cdots.$

结合(3.3)和(3.4)式可得

$ (gx, gy)\preceq (gu_{n}, gv_{n}), (gt, gz)\preceq (gu_{n}, gv_{n}), n=0, 1, 2, \cdots. $

于是, $gx\preceq gu_{n}, gy\succeq gv_{n}, n=0, 1, 2, \cdots$, 根据定理 2.1中的(2.1)式, 可得

$ \begin{eqnarray*} &&{\psi(sd(gx, gu_{n+1}))= \psi(sd(T(x, y), T(u_{n}, v_{n}))}\nonumber\\ &\leq &\psi(\max\left\{ d(gx, gu_{n}), d(gy, gv_{n})\right\})-\phi(\max\left\{ d(gx, gu_{n}), d(gy, gv_{n})\right\}). \end{eqnarray*} $

即

$ \begin{eqnarray} \psi(sd(gx, gu_{n+1}))\leq\psi(\max\left\{ d(gx, gu_{n}), d(gy, gv_{n})\right\})-\phi(\max\left\{ d(gx, gu_{n}), d(gy, gv_{n})\right\}). \end{eqnarray} $

(3.5)

同理可得

$ \begin{eqnarray} \psi(sd(gy, gv_{n+1}))\leq\psi(\max\left\{ d(gx, gu_{n}), d(gy, gv_{n})\right\})-\phi(\max\left\{ d(gx, gu_{n}), d(gy, gv_{n})\right\}). \end{eqnarray} $

(3.6)

现令

$ \begin{equation} \gamma_{n}=\max\left\{ d(gx, gu_{n}), d(gy, gv_{n})\right\}, \end{equation} $

(3.7)

联立(3.5), (3.6)和(3.7)式, 又由$\max\{\psi(a), \psi(b)\}=\psi(\max\{a, b\}), \forall a, b \in[0, +\infty), $可得到

$ \begin{eqnarray} \psi(s\gamma_{n+1})= \max\left\{\psi(sd(gx, gu_{n+1})), \psi(sd(gy, gv_{n+1})\right\}) \leq \psi( \gamma_{n})-\phi(\gamma_{n})\leq \psi( \gamma_{n}), \end{eqnarray} $

(3.8)

从而可得$ \gamma_{n+1}\leq \gamma_{n}$.所以可知$\{\gamma_{n}\}$单调递减的非负实数列, 存在$ r\in X$, 使得$r\geq 0$且$\lim\limits_{n\rightarrow\infty}\gamma_{n}=r$.因为$ \psi(s\gamma_{n+1})\leq \psi(\gamma_{n})-\phi(\gamma_{n}). $将上式两边令$n\rightarrow\infty$时, 取极限得$ \psi(sr)\leq \psi(r)-\phi(r). $当$r> 0$时, 出现矛盾.即$r=0.$于是有

$ \lim\limits_{n\rightarrow\infty}\gamma_{n}=\lim\limits_{n\rightarrow\infty}\max\left\{ d(gx, gu_{n}), d(gy, gv_{n})\right\}=0, $

进而可知

$ \begin{equation} \lim\limits_{n\rightarrow\infty}d(gx, gu_{n})=\lim\limits_{n\rightarrow\infty}d(gy, gv_{n})=0. \end{equation} $

(3.9)

同样道理可证得

$ \begin{equation} \lim\limits_{n\rightarrow\infty}d(gz, gu_{n})=\lim\limits_{n\rightarrow\infty}d(gt, gv_{n})=0. \end{equation} $

(3.10)

由(3.9)和(3.10)式可得到

$ \lim\limits_{n\rightarrow\infty}gu_{n}=gx=gz, \, \, \lim\limits_{n\rightarrow\infty}gv_{n}=gy=gt. $

即$(gx, gy)=(gz, gt)$.所以$g$和$T$具有唯一的重合耦合点.又因为$g$和$T$是$\omega$-相容的, 得到

$ \begin{equation} g(gx)=g(T(x, y))=T(gx, gy), \, \, g(gy)=g(T(y, x))=T(gx, gy). \end{equation} $

(3.11)

由于$\exists m, n \in X, $使得$gx=m, gy=n$, 那么(3.12)式可整理为

$ \begin{equation} gm=T(m, n), \, \, gn=T(n, m). \end{equation} $

(3.12)

因此$(m, n)$也是$g$和$T$的一个耦合重合点, 由重合耦合点的唯一性知$gm=gx=m, gn=gy=n.$又由(3.12)式可得$ m=gm=T(m, n), n=gn=T(n, m). $即证得$g$和$T$有耦合公共不动点.由于$g$和$T$的重合耦合点具有唯一性, 因此$g$和$T$的耦合公共不动点也具有唯一性.

b) 当可比较关系为$(gu, gv)\preceq (gx, gy), \, \, (gt, gz)\preceq (gu, gv).$

c) 当可比较关系为$(gx, gy)\preceq (gu, gv), \, \, (gu, gv)\preceq (gt, gz).$

d) 当可比较关系为$(gu, gv)\preceq (gx, gy), \, \, (gu, gv)\preceq (gt, gz).$同a)的证明方法类似, 上述三种情况同样也能证得$g$和$T$的耦合公共不动点也具有唯一性.

假设$X=C[a, b]$是定义在$[a, b]$上的连续函数全体. $\leq $是定义在$X$上的偏序关系, 定义

$(x, y)\preceq (u, v)\Leftrightarrow x\leq u, y\geq v.$

定义$d:X\times X\rightarrow \mathbb{R}^{+}$为

$ d(x, y)=\sup\limits_{r\in [0, 1]}|x(r)-y(r)|^{k}, \forall x, y\in X \, \, (k\geq 2). $

这时易知$(X, d)$是一个完备的矩形$b$ -度量空间, 由注2.2可知, 其系数$s=3^{k-1}$.

考虑以下积分方程组问题

$ \begin{eqnarray} \left\{ \begin{array}{ll} x(r)=\displaystyle\int_{a}^{b}S(r, t)(f(t, x(t))+g(t, y(t)))dt;\\ y(r)=\displaystyle\int_{a}^{b}S(r, t)(f(t, y(t))+g(t, x(t)))dt. \end{array} \right. \end{eqnarray} $

(4.1)

接下来, 假设下面几个条件成立.

(ⅰ) $S(r, t):[a, b]\times [a, b]\rightarrow\mathbb{R}^{+}$是连续函数.

(ⅱ) $f, g:[a, b]\times \mathbb{R}\rightarrow \mathbb{R}$是连续函数.

(ⅲ) $\exists (x_{0}, y_{0})\in X\times X$, 使得

$ x_{0}\leq \int_{a}^{b}S(r, t)(f(t, x_{0}(t))+g(t, y_{0}(t)))dt, \, \, \, y_{0}\geq \int_{a}^{b}S(r, t)(f(t, y_{0}(t))+g(t, x_{0}(t)))dt. $

(ⅳ) $\exists \mu, \nu$使得对于任意的$r\in [a, b], x, y\in \mathbb{R}$, 有

$ \left\{ \begin{array}{ll} |f(r, x)-f(r, y)|\leq \mu | x-y|, \\ |g(r, x)-g(r, y)|\leq \nu | x-y|. \end{array} \right. $

(ⅴ) $\parallel s \parallel_{\infty}=\sup\{S(r, t):r, t\in [a, b]\}$, 且$2^{k}\max\{\mu^{k}, \nu^{k}\}\parallel s\parallel_{\infty}^{k}\leq \frac{1}{3^{k}}$.

定理 4.1 由上述条件(i)-(v)成立, 则积分方程组(4.1)有唯一的解.

证 构造函数$T:X\times X\rightarrow X$和$g:X\rightarrow X$如下

$ T(x, y)(r)=\int_{a}^{b}S(r, t)(f(t, x(t))+g(t, y(t)))dt, \, \, \, \, gx(r)=x(r), \, \, \, \forall x, y\in X, r\in [a, b]. $

则由条件(ⅲ)可知$gx_{0}\preceq T(x_{0}, y_{0}), \, \, \, \, \, gy_{0} \succeq T(y_{0}, x_{0}), $且$g$和$T$具有$\omega$ -相容性.又由条件(iv)和(v)可得

$ \begin{eqnarray*} &&{\sup\limits_{r\in [a, b]}|T(x, y)-T(u, v)|}\\ &=&\sup\limits_{r\in [a, b]}|\int_{a}^{b}S(r, t)(f(t, x(t))+g(t, y(t)))dt-\int_{a}^{b}S(r, t)(f(t, u(t))+g(t, v(t)))dt|\\ &= &\sup\limits_{r\in [a, b]}|\int_{a}^{b}S(r, t)[(f(t, x(t))-f(t, u(t)))+(g(t, y(t))-g(t, v(t)))]dt|\\ &\leq &\sup\limits_{r\in [a, b]}\left(\int_{a}^{b}S(r, t)(| f(t, x(t))-f(t, u(t))| +| g(t, y(t))-g(t, v(t))|)dt\right)\\ &\leq &\sup\limits_{r\in [a, b]}\left(\int_{a}^{b}S(r, t)(\mu |x(t)-u(t)| +\nu | y(t)-v(t)|)dt\right)\\ &\leq &\max\{\mu, \nu\}\left(\sup\limits_{r\in [a, b]}\int_{a}^{b}S(r, t)(|x(t)-u(t)| + | y(t)-v(t)|)dt\right)\\ &\leq &\max\{\mu, \nu\}\sup\limits_{r\in [a, b]}\left(\left(\int_{a}^{b}S^{2}(r, t)dt\right)^{\frac{1}{2}}\left(\int_{a}^{b}(|x(t)-u(t)| + | y(t)-v(t)|)^{2}dt\right)^{\frac{1}{2}}\right)\\ &\leq &\max\{\mu, \nu\}\parallel s\parallel_{\infty}(\sup\limits_{r\in [a, b]}|x(r)-u(r)| + \sup\limits_{r\in [a, b]}| y(r)-v(r)|). \end{eqnarray*} $

于是由条件(ⅴ), 上式可整理为

$ \begin{eqnarray*} &&{\sup\limits_{r\in [a, b]}|T(x, y)-T(u, v)|^{k}}\\ &\leq &\max\{\mu^{k}, \nu^{k}\}\parallel s\parallel_{\infty}^{k}(\sup\limits_{r\in [a, b]}|x(r)-u(r)| + \sup\limits_{r\in [a, b]}| y(r)-v(r)|)^{k}\\ &\leq&2^{k}\max\{\mu^{k}, \nu^{k}\}\parallel s\parallel_{\infty}^{k}(\sup\limits_{r\in [a, b]}|x(r)-u(r)|^{k} + \sup\limits_{r\in [a, b]}| y(r)-v(r)|^{k})\\ &\leq &\frac{1}{3^{k}}(\sup\limits_{r\in [a, b]}|x(r)-u(r)|^{k} + \sup\limits_{r\in [a, b]}| y(r)-v(r)|^{k}). \end{eqnarray*} $

上式即为

$ \begin{eqnarray*} 3^{k-1}d(T(x, y), T(u, v))&\leq& \frac{1}{3}(\sup\limits_{r\in [a, b]}|x(r)-u(r)|^{k} + \sup\limits_{r\in [a, b]}| y(r)-v(r)|^{k})\\ &\leq &\frac{2}{3}\sup\limits_{r\in [a, b]}\max\{|x(r)-u(r)|^{k}, | y(r)-v(r)|^{k})\}. \end{eqnarray*} $

于是

$ \begin{eqnarray*} sd(T(x, y), T(u, v))&\leq& \frac{2}{3}\max\{d(gx, gu), d(gy, gv)\}\\ &=&\max\{d(gx, gu), d(gy, gv)\}-\frac{1}{3}\max\{d(gx, gu), d(gy, gv)\}. \end{eqnarray*} $

令函数$\psi, \phi: \mathbb{R}^{+}\rightarrow \mathbb{R}^{+} $分别为$\psi(t)=t, \phi(t)=\frac{t}{3}, \forall t\in \mathbb{R}^{+}$, 那么上式即为

$ \psi(sd(T(x, y), T(u, v)))\leq \psi(\max\{d(gx, gu), d(gy, gv)\})-\phi(\max\{d(gx, gu), d(gy, gv)\}). $

由此易知定理3.1的所有条件被满足, 于是, 存在$ (x^{*}, y^*)\in X\times x$, 使得

$T(x^{*}, y^{*})=gx^{*}=x^{*}, T(y^{*}, x^{*})=gy^{*}=y^{*}.$

即$(x^{*}, y^{*})$也为方程组(4.1)在$C[a, b]$上的唯一解.