设$ H $为实Hilbert空间, $ C\subset H $是非空闭凸子集, $ \langle \cdot , \cdot \rangle $和$ \left\|\cdot \right\| $分别表示$ H $中的内积和范数. 设$ x\in H $, $ P_{C}(x) $表示向量$ x $在$ C $上的投影. 设$ F:H\rightarrow H $是一个映射. 本文考虑如下变分不等式问题: 寻找点$ x^*\in C $, 使得

(1.1) $ \begin{eqnarray} \langle F(x^*), x-x^*\rangle \geq 0, \quad \forall x\in C. \end{eqnarray} $

记$ S $为问题$ (1.1) $的解集. 近年来, 变分不等式模型在经济金融、管理、博弈论及非线性规划等领域中有着广泛的应用, 许多学者对变分不等式的理论和算法进行了研究, 也推动着变分不等式的发展, 见文献[1-8]. Korpelevich^[9]最早提出了求解变分不等式的外梯度算法:

(1.2) $ \begin{eqnarray} y_{n} = P_{C}(x_{n}-\lambda F(x_{n})), \; x_{n+1} = P_{C}(x_{n}-\lambda F(y_{n})), \end{eqnarray} $

其中映射$ F $是单调的且Lipschitz连续的, $ L $为Lipschitz常数, $ \lambda\in(0, \frac{1}{L}) $. 然而该算法每次迭代需要计算两次到$ C $上的投影. 若集合$ C $结构不够简单，则在$ C $上的投影难以计算, 从而会影响算法的效率和适用性. 为了克服这一缺点, He^[3]在1997年提出了收缩投影算法, 该算法每次迭代只需要计算一次到$ C $上的投影, 具体算法如下:

(1.3) $ \begin{eqnarray} \left\{ \begin{array}{ll} \ y_{n} = P_{C}(x_{n}-\lambda F(x_{n})), \\ \ d({x_{n}, y_{n}}) = x_{n}-y_{n}-\lambda (F(x_{n})-F(y_{n})), \\ \ x_{n+1} = x_{n}-\gamma \beta_{n}d({x_{n}, y_{n}}), \forall n\geq 0, \end{array} \right. \end{eqnarray} $

其中映射$ F $是单调的且Lipschitz连续的, $ \lambda\in(0, \frac{1}{L}), \gamma \in (0, 2), \beta_{n}: = \frac{\langle x_{n}-y_{n}, d({x_{n}, y_{n}})\rangle }{\left\|d({x_{n}, y_{n}})\right\|^2} $. 后来, 惯性方法由于其良好的加速性质得到了学者们的广泛关注^[10-11]. 据此, Dong等^[10]提出了带惯性项的收缩投影算法, 同时在映射$ F $为单调且Lipschitz连续的条件下证明了该算法的弱收敛性定理. 进一步, Thong等^[11]结合黏度方法得到了如下改进的惯性收缩投影算法:

(1.4) $ \begin{eqnarray} \left\{ \begin{array}{ll} \ w_{n} = x_{n}+\alpha_{n}(x_{n}-x_{n-1}), \\ \ y_{n} = P_{C}(w_{n}-\lambda F(w_{n})), \\ \ d_n = w_{n}-y_{n}-\lambda (F(w_{n})-F(y_{n})), \\ \ x_{n+1} = \beta_nf(x_{n})+(1-\beta_n)(w_n-\theta_{n}d_n), \end{array} \right. \end{eqnarray} $

其中映射$ F $是单调的且Lipschitz连续的, $ \lambda \in(0, \frac{1}{L}), \theta_{n}: = (1-\lambda L)\frac{\left\|w_{n}-y_{n}\right\|^2}{\left\|d_n\right\|^2} $. Thong等证明了由算法$ (1.4) $产生的无穷序列强收敛到变分不等式问题的解.

对于算法(1.2)–(1.4), 其步长$ \lambda $都与Lipschitz常数$ L $有关, 而Lipschitz常数很难计算或估计. 因此, 为了避免估算Lipschitz常数, 学者们通常采用一类自适应方法得到步长. 2019年, Yang和Liu^[12]提出了一种无需搜索得到步长的算法:

(1.5) $ \begin{eqnarray} \left\{ \begin{array}{ll} \ y_n = P_{C}(x_{n}-\lambda_n F(x_{n})), \\ \ z_n = y_n-\lambda_n (F(x_n)-F(y_n)), \\ \ x_{n+1} = \alpha_nf(x_n)+(1-\alpha_n)z_n, \end{array} \right. \end{eqnarray} $

其中

(1.6) $ \begin{eqnarray} \lambda_{n+1}: = \left \{ \begin{array}{ll} { } \min \left\{\frac{\mu \left\|x_{n}-y_{n}\right\|}{\left\|F(x_{n})-F(y_{n})\right\|}, \lambda_{n}\right\}, \, \, &\mbox{若}\, \, F(x_{n})-F(y_{n})\neq 0, \\ \lambda_{n} , \, \, &\mbox{其它, } \end{array} \right. \end{eqnarray} $

这里映射$ F $是单调的且Lipschitz连续的($ L $未知), $ \lambda_0 >0, \mu \in (0, 1). $

最近, Gibali等^[13]给出了一种黏度投影算法, 在其算法中步长$ \tau_n $是最大的$ \tau \in \{\lambda, \lambda l, \lambda l^2, \cdots \} $满足$ \tau \left\|F(x_n)-F(y_n)\right\|\leq \mu \left\|x_n-y_n\right\|, $其算法为

(1.7) $ \begin{eqnarray} \left\{ \begin{array}{ll} \ y_{n} = P_{C}(x_{n}-\tau_n F(x_{n})), \\ \ z_n = x_n-\gamma \eta_n d_n, \\ { } \ \eta_n = (1-\mu )\frac{\left\|x_n-y_n\right\|^2}{\left\|d_n\right\|^2}, \\ \ d_n = x_n-y_n-\tau_n(F(x_n)-F(y_n)), \\ \ x_{n+1} = \alpha_nf(x_n)+(1-\alpha_n)z_n, \end{array} \right. \end{eqnarray} $

Gibali等在映射$ F $是单调且Lipschitz连续的($ L $未知)条件下证明了算法的强收敛性. 然而映射$ F $单调这一假设比较强. 事实上, 许多函数不满足单调的假设. 因此, 如何在较弱的条件下证明算法的收敛性是本文讨论的重点.

本文受文献[11-13]的启发, 在映射是伪单调和Lipschitz连续($ L $未知)的假设下, 证明了由算法产生的无穷序列强收敛到变分不等式问题的解, 并用数值实验验证了算法的适用性. 本文所得结果改进和推广了文献[11-13]中相应结果.

记$ x_n\rightharpoonup x(n\to \infty ) $为$ \{x_n\} $弱收敛于$ x $, $ x_n\to x(n\to \infty ) $为$ \{x_n\} $强收敛于$ x $. 对于任意的$ x\in H $, 有

$ \left\|x-P_{C}(x)\right\|\leq \left\|x-y\right\|, \; \forall y\in C. $

对于任意的$ x, y\in H $, 有

$ \left\|x+y\right\|^2 = \left\|x\right\|^2+2\langle x, y\rangle +\left\|y\right\|^2, $

$ \left\|x+y\right\|^2\leq \left\|x\right\|^2+2\langle y, x+y\rangle, $

$ \left\|\alpha x+(1-\alpha)y\right\|^2 = \alpha \left\|x\right\|^2+(1-\alpha )\left\|y\right\|^2-\alpha (1-\alpha )\left\|x-y\right\|^2. $

定义2.1^[8]  设$ F:H\to H $是映射, 如果对于任意的$ x, y \in H $, 有

$ \langle F(x), y-x\rangle \geq 0 \Rightarrow \langle F(y), y-x\rangle \geq 0, $

则称$ F $是伪单调的.

定义2.2  设$ F:H\to H $是映射, 如果对于任意的$ x, y\in H, $有

$ \left \|F(x)-F(y)\right\|\leq L\left \|x-y\right\|, $

则称$ F $是Lipschitz连续的, 其中Lipschitz常数为$ L>0. $若$ L = 1 $, 则称$ F $是非扩张的; 若$ L\in (0, 1), $则称$ F $是压缩的.

引理2.1^[2]  设$ C\subset H $是非空闭凸子集. 对于任意的$ x\in H $, $ z\in C $, 有

$ z = P_C(x) \; \Longleftrightarrow \; \langle x-z, z-y\rangle \geq 0, \; \forall y\in C. $

引理2.2^[2]  设$ C\subset H $是非空闭凸子集. 对于任意的$ x\in H $, 有

$ {\rm (i)} $$ \left \|P_C(x)-P_C(y)\right\|^2\leq \langle P_C(x)-P_C(y), x-y\rangle , \forall y\in H $;

$ {\rm (ii)} $$ \left \|P_C(x)-y\right\|^2\leq \left\|x-y\right\|^2-\left\|x-P_C(x)\right\|^2, \forall y\in C $.

引理2.3^[14]  设$ \{a_{n_{j}}\} $是非负实序列$ \{a_{n}\} $的子列, 且子列满足对于任意的$ j\in {\Bbb N} $, 有$ a_{n_{j}} < a_{n_{j+1}} $成立, 则存在单调递增的整数子列$ \{m_{k}\} $满足: $ \lim\limits_{k \to \infty}m_{k} = \infty $; 而且当$ k\in {\Bbb N} $充分大时有: $ a_{m_{k}}\leq a_{m_{k+1}}, a_{k}\leq a_{m_{k+1}}. $事实上$ m_{k} $是集合$ \{1, 2, \cdots, k\} $中满足$ a_{n} < a_{n+1} $的最大的整数.

引理2.4^[15]  设$ \{a_{n}\} $是非负实序列, 满足

$ a_{n+1}\leq (1-\gamma_{n})a_{n}+\gamma_{n}b_{n}, \; \forall n\geq 0, $

其中$ \{\gamma_{n}\}\subset (0, 1) $且$ \{b_{n}\} $为$ {{\Bbb R}} $中的一个数列满足

$ {\rm (i)} $$ \sum\limits_{n = 0}^\infty \gamma_{n} = \infty $; $ {\rm (ii)} $$ {\limsup\limits_{n \to \infty}}b_{n}\leq 0 $, 则$ {\lim\limits_{n \to \infty}}a_{n} = 0. $

本文假设

$ (C_1) $$ S\neq \emptyset $;

$ (C_2) $映射$ F:H\to H $是伪单调且Lipschitz连续的, 但Lipschitz常数$ L $未知;

$ (C_3) $映射$ F:H\to H $满足: 若$ \{u_{n}\}\subset C $, $ u_{n}\rightharpoonup z(n\to \infty ) $, 则$ \left\|F(z)\right\|\leq {\liminf\limits_{n \to \infty}}\left\|F(u_{n})\right\| $;

$ (C_4) $序列$ \{\alpha_{n}\}, \{\beta_{n}\}\subset (0, 1) $满足

$ {\lim\limits_{n \to \infty}}\frac{\alpha_{n}} {\beta_{n}}\left\|u_{n}-u_{n-1}\right\| = 0, \; {\lim\limits_{n \to \infty}}\beta_{n} = 0, \; \sum\limits_{n = 1}^\infty \beta_{n} = \infty . $

设$ f:H\rightarrow {H} $是压缩映射, 压缩参数$ \rho \in(0, 1). $本文提出如下的算法:

算法3.1  选取初始点$ u_{0} $, $ u_{1}\in H $, $ \lambda_{0}> 0 $, $ \mu \in (0, 1) $, $ \gamma\in (0, 2) $.

步骤1  令$ w_{n} = u_{n}+\alpha_{n}(u_{n}-u_{n-1}) $, 计算

$ q_{n} = P_{C}(w_{n}-\lambda_{n}F(w_{n})), $

如果$ q_{n} = w_{n} $或$ F(q_{n}) = 0, $则停止; $ q_{n} $是变分不等式问题(1.1) 的解. 否则, 转到步骤2.

步骤2  计算

$ h_{n} = w_{n}-\gamma \theta_{n}d_{n}, $

其中

$ d_{n}: = w_{n}-q_{n}+\lambda_{n}(F(q_{n})-F(w_{n})), \; \; \; \; \theta_{n}: = \frac{\langle w_{n}-q_{n}, d_{n}\rangle }{\left\|d_{n}\right\|^2}, $

并计算

$ \lambda_{n+1}: = \left \{ \begin{array}{ll} { } \min \left\{\frac{\mu \left\|w_{n}-q_{n}\right\|}{\left\|F(w_{n})-F(q_{n})\right\|}, \lambda_{n}\right\}, \, \, &\mbox{若}\, \, F(w_{n})-F(q_{n})\neq 0, \\ \lambda_{n}, \, \, &\mbox{否则}. \end{array} \right. $

步骤3  计算

$ u_{n+1} = \beta_{n}f(u_{n})+(1-\beta_{n})h_{n}. $

令$ n: = n+1 $，转到步骤1.

引理3.1  假设条件$ (C_2) $成立, 则算法$ 3.1 $产生的步长$ \{\lambda _{n}\} $是一个非增序列, 且满足

$ {\lim\limits_{n \to \infty}\lambda _{n}} = \lambda \geq \left\{ \lambda_{0}, \frac{\mu }{L} \right\}. $

证  该引理的证明与文献[12]中引理$ 3.1 $的证明类似, 故不再赘述.

引理3.2  设$ \{d_{n}\} $是由算法$ 3.1 $产生的无穷序列, 则$ d_{n} = 0 $当且仅当$ w_{n} = q_{n}. $

证

$ \begin{eqnarray*} \left\|d_{n}\right\| & = &\left\|w_{n}-q_{n}-\lambda_{n}(F(w_{n})-F(q_{n}))\right\|\\ &\geq& \left\|w_{n}-q_{n}\right\|-\lambda_{n}\left\|F(w_{n})-F(q_{n})\right\|\\ &\geq &\left\|w_{n}-q_{n}\right\|-\mu \frac{\lambda_{n}}{\lambda_{n+1}}\left\|w_{n}-q_{n}\right\|\\ & = &(1-\mu \frac{\lambda_{n}}{\lambda_{n+1}})\left\|w_{n}-q_{n}\right\|. \end{eqnarray*} $

类似可得

$ \left\|d_{n}\right\|\leq (1+\mu \frac{\lambda_{n}}{\lambda_{n+1}})\left\|w_{n}-q_{n}\right\|. $

因此

$ (1-\mu \frac{\lambda_{n}}{\lambda_{n+1}})\left\|w_{n}-q_{n}\right\|\leq \left\|d_{n}\right\|\leq (1+\mu \frac{\lambda_{n}}{\lambda_{n+1}})\left\|w_{n}-q_{n}\right\|. $

故$ d_{n} = 0 $当且仅当$ w_{n} = q_{n}. $

注3.1  由引理$ 3.2 $可知, 若$ d_{n} = 0 $, 则算法停止, 故$ q_{n} $是变分不等式问题$ (1.1) $的解.

引理3.3  假设条件$ (C_1) $–$ (C_3) $成立, $ \{w_{n}\} $是由算法$ 3.1 $产生的无穷序列. 如果存在$ \{w_{n}\} $的子序列$ \{w_{n_{k}}\} $满足$ \{w_{n_{k}}\} $弱收敛于$ z\in H $且$ {\lim\limits_{n \to \infty }\left\|w_{n_{k}}-q_{n_{k}}\right\|} = 0, $则$ z\in S $.

证  由$ q_{n_{k}} = P_{C}(w_{n_{k}}-\lambda_{n}F(w_{n_{k}})) $可得

$ \langle w_{n_{k}}-\lambda _{n_{k}}F(w_{n_{k}})-q_{n_{k}}, x-q_{n_{k}}\rangle \leq 0, \; \forall x\in C. $

上式等价于

$ \frac{1}{\lambda_{n_{k}}}\langle w_{n_{k}}-q_{n_{k}}, x-q_{n_{k}}\rangle \leq \langle F(w_{n_{k}}), x-q_{n_{k}}\rangle , \; \forall x\in C, $

即

(3.1) $ \begin{eqnarray} \frac{1}{\lambda_{n_{k}}}\langle w_{n_{k}}-q_{n_{k}}, x-q_{n_{k}}\rangle+\langle F(w_{n_{k}}), q_{n_{k}}-w_{n_{k}}\rangle \leq \langle F(w_{n_{k}}), x-w_{n_{k}}\rangle, \; \forall x\in C. \end{eqnarray} $

因为$ \{w_{n}\} $的子序列$ \{w_{n_{k}}\} $弱收敛于$ z\in H $, 所以$ \{w_{n_{k}}\} $是有界的. 由$ F $的Lipschitz连续性可知$ \{F(w_{n_{k}})\} $也是有界的. 又因为$ \left \|w_{n_{k}}-q_{n_{k}}\right\|\to 0 $, 所以$ \{q_{n_{k}}\} $也是有界的. 在$ (3.1) $式中, 令$ k\to \infty $, 有

(3.2) $ \begin{equation} {\liminf\limits_{k \to \infty}} \langle F(w_{n_{k}}), x-w_{n_{k}}\rangle \geq 0, \; \forall x\in C . \end{equation} $

注意到

(3.3) $ \begin{equation} \langle F(q_{n_k}), x-q_{n_k}\rangle = \langle F(q_{n_k})-F(w_{n_k}), x-w_{n_k}\rangle +\langle F(w_{n_{k}}), x-w_{n_{k}} \rangle +\langle F(q_{n_{k}}), w_{n_{k}}-q_{n_{k}} \rangle . \end{equation} $

由于$ \lim\limits_{k\to \infty}\left\|w_{n_{k}}-q_{n_{k}}\right\| = 0 $且$ F $在$ H $上是Lipschitz连续的, 故

$ \lim\limits_{k\to \infty}\left\|F(w_{n_{k}})-F(q_{n_{k}}) \right\| = 0. $

结合$ (3.2) $和$ (3.3) $式得

$ {\liminf\limits_{k \to \infty}}\langle F(q_{n_{k}}), x-q_{n_{k}}\rangle \geq 0, \; \forall x\in C . $

下证$ z\in S $. 选择非负递减序列$ \{\epsilon_{k}\}, $使得$ {\lim\limits_{k \to \infty}}\epsilon_{k} = 0. $对于每一个$ \epsilon_{k}, $记$ N_{k} $为使得下式成立的最小正整数

(3.4) $ \begin{equation} \langle F(q_{n_{j}}), x-q_{n_{j}}\rangle +\epsilon _{k}\geq 0, \; \forall j\geq N_{k}. \end{equation} $

由于$ \{\epsilon_{k}\} $是递减序列, 易知$ N_{k} $是递增序列. 则对于每一个$ k, $由$ \{q_{N_{k}}\}\subset C $可知$ F(q_{N_{k}})\neq 0 $, 令$ v_{N_{k}} = \frac{F(q_{N_{k}})}{\left\|F(q_{N_{k}})\right\|^2} $. 易知$ \langle F(q_{N_{k}}), v_{N_{k}}\rangle = 1. $对于每一个$ k, $由$ (3.4) $式知$ \langle F(q_{N_{K}}), x+\epsilon _{k}v_{N_{k}}-q_{N_{k}}\rangle \geq 0 $. 再由$ F $的伪单调性知

$ \langle F(x+\epsilon _{k}v_{N_{k}}), x+\epsilon _{k}v_{N_{k}}-q_{N_{k}}\rangle \geq 0. $

上式等价于

(3.5) $ \begin{eqnarray} \langle F(x), x-q_{N_{k}}\rangle \geq \langle F(x)-F(x+\epsilon _{k}v_{N_{k}}), x+\epsilon _{k}v_{N_{k}}-q_{N_{k}}\rangle -\epsilon _{k}\langle F(x), v_{N_{k}}\rangle . \end{eqnarray} $

下证$ {\lim\limits_{k \to \infty}}\epsilon _{k}v_{N_{k}} = 0 $. 由$ w_{n_{k}}\rightharpoonup z $, $ k\to \infty $和$ {\lim\limits_{k \to \infty }}\left\|w_{n_{k}}-q_{n_{k}}\right\| = 0 $知$ q_{N_{k}}\rightharpoonup z $, $ k\to \infty. $又由$ \{q_{n}\}\subset C $得$ z \in C $. 根据$ (C_3) $知

$ 0<\left\|F(z)\right\|\leq {\liminf\limits_{k \to \infty}}\left\|F(q_{n_{k}})\right\|. $

又因为$ \{q_{N_{k}}\} \subset \{q_{n_{k}}\} $以及$ \epsilon _{k} \to \infty (k\to \infty ), $故

$ 0\leq {\limsup\limits_{n \to \infty }}\left\|\epsilon _{k}v_{N_{k}}\right\| = {\limsup\limits_{n \to \infty}}\left (\frac{\epsilon _{k}}{\left\|F(q_{n_{k}})\right\|}\right )\leq \frac{{\limsup\limits_{n \to \infty }}\epsilon _{k}}{{\lim\limits_{k \to \infty}}\left\|F(q_{n_{k}})\right\|} = 0. $

所以$ {\lim\limits_{k \to \infty }}\epsilon _{k}v_{N_{k}} = 0. $又当$ k\to \infty $时, 由$ (3.5) $式知$ {\liminf\limits_{k \to \infty }}\langle F(x), x-q_{N_{k}}\rangle \geq 0. $对任意的$ x\in C $, 有

$ \langle F(x), x-z\rangle = {\lim\limits_{k \to \infty }}\langle F(x), x-q_{N_{k}}\rangle = {\liminf\limits_{k \to \infty }}\langle F(x), x-q_{N_{k}}\rangle \geq 0. $

又由$ F $的伪单调性得

$ \langle F(z), x-z\rangle \geq 0, \; \forall x\in C. $

即$ z\in S $.

定理3.1  假设条件$ (C_1) $–$ (C_4) $成立, $ \{u_{n}\} $是由算法$ 3.1 $所产生的无穷序列, 则$ \{u_{n}\} $强收敛于某一个$ p\in S, $其中$ p = P_{S}\circ f(p). $

证  该定理的证明分为以下六步进行.

第一步  首先证明

(3.6) $ \begin{equation} \left\|h_{n}-p\right\|^2\leq \left\|w_{n}-p\right\|^2-\frac{2-\gamma}{\gamma}\left\|w_{n}-h_{n}\right\|^2. \end{equation} $

事实上

(3.7) $ \begin{eqnarray} \langle w_{n}-p, d_{n}\rangle & = &\langle w_{n}-q_{n}, d_{n}\rangle+\langle q_{n}-p, d_{n}\rangle\\ & = &\langle w_{n}-q_{n}, d_{n}\rangle+\langle q_{n}-p, w_{n}-q_{n}-\lambda_{n}(F(w_{n})-F(q_{n}))\rangle. \end{eqnarray} $

由$ q_{n} = P_C(w_{n}-\lambda_nF(w_{n})) $得$ \langle w_{n}-q_{n}-\lambda_{n}F(w_{n}), q_{n}-p\rangle\geq 0. $由于$ p\in S, $则$ \langle F(p), q_{n}-p\rangle \geq0. $由$ F $的伪单调性得$ \langle F(q_{n}), q_{n}-p\rangle\geq0. $因此, $ \langle q_{n}-p, w_{n}-q_{n}-\lambda_{n}(F(w_{n})-F(q_{n}))\rangle\geq0. $再根据$ \{\theta_{n}\} $的定义以及$ (3.7) $式有

(3.8) $ \begin{equation} \langle w_{n}-p, d_{n}\rangle \geq \langle w_{n}-q_{n}, d_{n}\rangle = \theta_{n}\left\|d_{n}\right\|^2. \end{equation} $

由此可得

$ \begin{eqnarray*} \left\|h_{n}-p\right\|^2& = &\left\|w_{n}-\gamma \theta_{n}d_{n}-p\right\|^2\\ & = &\left\|w_{n}-p\right\|^2-2\gamma\theta_{n}\langle w_{n}-p, d_{n}\rangle+\gamma^2\theta_{n}^2\left\|d_{n}\right\|^2\\ &\leq& \left\|w_{n}-p\right\|^2-\frac{2-\gamma}{\gamma}\left\|\gamma\theta_{n}d_{n}\right\|^2\\ & = &\left\|w_{n}-p\right\|^2-\frac{2-\gamma}{\gamma}\left\|w_{n}-h_{n}\right\|^2. \end{eqnarray*} $

第二步  证明序列$ \{w_{n}\} $是有界的. 由$ (3.6) $式知

(3.9) $ \begin{eqnarray} \left\|h_{n}-p\right\|\leq \left\|w_{n}-p\right\|. \end{eqnarray} $

注意到

(3.10) $ \begin{eqnarray} \left\|w_{n}-p\right\| & = &\left\|w_{n}-u_{n}+u_{n}-p\right\|{}\\ &\leq& \left\|w_{n}-u_{n}\right\|+\left\|u_{n}-p\right\|{}\\ & = &\alpha_{n}\left\|u_{n}-u_{n-1}\right\|+\left\|u_{n}-p\right\|{}\\ & = &\beta_{n}\frac{\alpha_{n}}{\beta_{n}}\left\|u_{n}-u_{n-1}\right\|+\left\|u_{n}-p\right\|. \end{eqnarray} $

因为$ \frac{\alpha_{n}}{\beta_{n}}\left\|u_{n}-u_{n-1}\right\|\to 0 $, 所以存在$ M_{1}>0 $使得

(3.11) $ \begin{equation} \frac{\alpha_{n}}{\beta_{n}}\left\|u_{n}-u_{n-1}\right\|\leq M_{1}, \;\forall\;n\geq 1. \end{equation} $

由$ (3.9) $–$ (3.11) $式知

(3.12) $ \begin{equation} \left\|h_{n}-p\right\|\leq \left\|w_{n}-p\right\|\leq \left\|u_{n}-p\right\|+\beta_{n}M_{1}. \end{equation} $

另一方面

$ \begin{eqnarray*} \left\|u_{n+1}-p\right\|\nonumber & = &\left\|\beta_{n}f(u_{n})+(1-\beta_{n})h_{n}-p\right\|\nonumber\\ & = &\left\|\beta_{n}(f(u_{n})-p)+(1-\beta_{n})(h_{n}-p)\right\|\nonumber\\ &\leq& \beta_{n}\left\|f(u_{n})-p\right\|+(1-\beta_{n})\left\|h_{n}-p\right\|\nonumber\\ &\leq &\beta_{n}\left\|f(u_{n})-f(p)\right\|+\beta_{n}\left\|f(p)-p\right\|+(1-\beta_{n})\left\|h_{n}-p\right\|\nonumber\\ &\leq &\beta_{n}\rho \left\|u_{n}-p\right\|+\beta_{n}\left\|f(p)-p\right\|+(1-\beta_{n})\left\|h_{n}-p\right\|.\nonumber \end{eqnarray*} $

将$ (3.12) $式代入上式得

$ \begin{eqnarray*} \left\|u_{n+1}-p\right\|\nonumber &\leq &(1-(1-\rho)\beta_{n})\left\|u_{n}-p\right\|+(1-\beta_{n})\beta_{n} M_{1}+\beta_{n}\left\|f(p)-p\right\|\nonumber\\ &\leq &(1-(1-\rho)\beta_{n})\left\|u_{n}-p\right\|+\beta_{n} M_{1}+\beta_{n}\left\|f(p)-p\right\|\nonumber\\ &\leq &(1-(1-\rho)\beta_{n})\left\|u_{n}-p\right\|+(1-\rho )\beta_{n}\frac{M_{1}+\left\|f(p)-p\right\|}{1-\rho } \nonumber\\ &\leq &\max\{ \left\|u_{n}-p\right\|, \frac{M_{1}+\left\|f(p)-p\right\|}{1-\rho }\}\nonumber\\ &\leq &\cdots \leq \max\{ \left\|u_{0}-p\right\|, \frac{M_{1}+\left\|f(p)-p\right\|}{1-\rho }\}.\nonumber \end{eqnarray*} $

因此, $ \{u_{n}\} $是有界的. 故$ \{h_{n}\} $, $ \{w_{n}\} $和$ \{f(u_{n})\} $也是有界的.

第三步  证明

(3.13) $ \begin{equation} \frac{2-\gamma}{\gamma}(1-\beta_{n})\left\|h_{n}-w_{n}\right\|^2\leq \left\|u_{n}-p\right\|^2-\left\|u_{n+1}-p\right\|^2+\beta_{n}M_{4}, \end{equation} $

其中$ M_{4}>0 $. 事实上

(3.14) $ \begin{eqnarray} \left\|u_{n+1}-p\right\|^2 & = &\left\|\beta_{n}f(u_{n})+(1-\beta_{n})h_{n}-p\right\|^2\\ & = &\left\|\beta_{n}(f(u_{n})-p)+(1-\beta_{n})(h_{n}-p)\right\|^2\\ & = &\beta_{n}^2\left\|f(u_{n})-p\right\|^2+(1-\beta_{n})^2\left \|h_{n}-p\right\|^2+2\beta_{n}(1-\beta_{n})\langle f(u_{n})-p, h_{n}-p\rangle\\ &\leq& \beta_{n}^2(\left\|f(u_{n})-f(p)\right\|+\left\|f(p)-p\right\|)^2+(1-\beta_{n})^2\left \|h_{n}-p\right\|^2\\ &&+2\beta_{n}(1-\beta_{n})\left\|f(u_{n})-p\right\|\cdot \left\|h_{n}-p\right\|\\ &\leq& \beta_{n}^2(\rho\left\|u_{n}-p\right\|+\left\|f(p)-p\right\|)^2+(1-\beta_{n})^2\left \|h_{n}-p\right\|^2\\ &&+2\beta_{n}(1-\beta_{n})\left\|f(u_{n})-p\right\|\cdot \left\|h_{n}-p\right\|\\ &\leq& \beta_{n}\left\|u_{n}-p\right\|^2+(1-\beta_{n})\left \|h_{n}-p\right\|^2+\beta_{n}(\left\|f(p)-p\right\|^2\\ &&+2\left\|f(p)-p\right\|\cdot\left\|u_{n}-p\right\|+2\left\|f(u_{n})-p\right\|\cdot\left\|h_{n}-p\right\|). \end{eqnarray} $

记

$ M_{2} = \left\|f(p)-p\right\|^2+2\left\|f(p)-p\right\|\cdot\left\|u_{n}- p\right\|+2\left\|f(u_{n})-p\right\|\cdot\left\|h_{n}-p\right\|, $

则$ M_{2}>0. $

由$ (3.6) $和$ (3.14) $式得

(3.15) $ \begin{equation} \left\|u_{n+1}-p\right\|^2\leq \beta_{n}\left\|u_{n}-p\right\|^2+(1-\beta_{n})\left\|w_{n}-p\right\|^2-\frac{2-\gamma}{\gamma}(1-\beta_{n})\left\|w_{n}-h_{n}\right\|^2+\beta_{n}M_{2}. \end{equation} $

又由$ (3.12) $式得

(3.16) $ \begin{eqnarray} \left\|w_{n}-p\right\|^2 &\leq &(\left\|u_{n}-p\right\|+\beta_{n}M_{1})^2{}\\ & = &\left\|u_{n}-p\right\|^2+\beta_{n}(2M_{1}\left\|u_{n}-p\right\|+\beta_{n}M_{1}^2){}\\ &\leq &\left\|u_{n}-p\right\|^2+\beta_{n}M_{3}, \end{eqnarray} $

其中$ M_{3} = 2M_{1}\left\|u_{n}-p\right\|+\beta_{n}M_{1}^2 > 0 $. 将$ (3.16) $式代入$ (3.15) $式得

$ \begin{eqnarray*} \left\|u_{n+1}-p\right\|^2&\leq&\beta_{n}\left\|u_{n}-p\right\|^2+(1-\beta_{n})\left\|u_{n}-p\right\|^2\\ &&+(1-\beta_{n})\beta_{n}M_{3}-\frac{2-\gamma}{\gamma}(1-\beta_{n})\left\|w_{n}-h_{n}\right\|^2+\beta_{n}M_{2}\\ &\leq&\left\|u_{n}-p\right\|^2+\beta_{n}M_{3}-\frac{2-\gamma}{\gamma}(1-\beta_{n})\left\|w_{n}-h_{n}\right\|^2+\beta_{n}M_{2}. \end{eqnarray*} $

上式等价于

$ \frac{2-\gamma}{\gamma}(1-\beta_{n})\left\|w_{n}-h_{n}\right\|^2\leq \left\|u_{n}-p\right\|^2-\left\|u_{n+1}-p\right\|^2+\beta_{n}M_{4}, $

其中$ M_{4}: = M_{2}+M_{3}. $

第四步  证明

(3.17) $ \begin{equation} \left\|w_{n}-q_{n}\right\|^2\leq \frac{(1+\mu \frac{\lambda_{n}}{\lambda_{n+1}})^2}{\gamma^2(1-\mu \frac{\lambda_{n}}{\lambda_{n+1}})^2}\left\|h_{n}-w_{n}\right\|^2. \end{equation} $

事实上

(3.18) $ \begin{eqnarray} \langle w_{n}-q_{n}, d_{n}\rangle & = &\langle w_{n}-q_{n}, w_{n}-q_{n}-\lambda_{n}(F(w_{n})-F(q_{n}))\rangle{}\\ & = &\left\|w_{n}-q_{n}\right\|^2- \lambda_{n}\langle w_{n}-q_{n}, F(w_{n})-F(q_{n})\rangle{}\\ &\geq&\left\|w_{n}-q_{n}\right\|^2-\lambda_{n}\left\|w_{n}-q_{n}\right\|\cdot \left\|F(w_{n})-F(q_{n})\right\|{}\\ & = &\left\|w_{n}-q_{n}\right\|^2-\mu \frac{\lambda_{n}}{\lambda_{n+1}}\left\|w_{n}-q_{n}\right\|^2{}\\ & = &(1-\mu \frac{\lambda_{n}}{\lambda_{n+1}})\left\|w_{n}-q_{n}\right\|^2. \end{eqnarray} $

结合$ (3.8) $和$ (3.18) $式得

$ \langle w_{n}-q_{n}, d_{n}\rangle = \theta_{n}\left\|d_{n}\right\|^2\geq (1-\mu \frac{\lambda_{n}}{\lambda_{n+1}})\left\|w_{n}-q_{n}\right\|^2. $

上式等价于

(3.19) $ \begin{equation} \left\|w_{n}-q_{n}\right\|^2\leq\frac{\theta_{n}\left\|d_{n}\right\|^2}{1-\mu \frac{\lambda_{n}}{\lambda_{n+1}}} = \frac{\left\|\theta_{n}d_{n}\right\|^2}{1-\mu \frac{\lambda_{n}}{\lambda_{n+1}}}\cdot \frac{1}{\theta_{n}} = \frac{\left\|w_{n}-h_{n}\right\|^2}{\gamma^2(1-\mu \frac{\lambda_{n}}{\lambda_{n+1}})}\cdot \frac{1}{\theta_{n}}. \end{equation} $

又由引理$ 3.2 $知

$ \left\|d_{n}\right\|\leq (1+\mu \frac{\lambda_{n}}{\lambda_{n+1}})\left\|w_{n}-q_{n}\right\|. $

故

(3.20) $ \begin{equation} \left\|d_{n}\right\|^2\leq (1+\mu \frac{\lambda_{n}}{\lambda_{n+1}})^2\left\|w_{n}-q_{n}\right\|^2. \end{equation} $

由$ (3.18) $和$ (3.20) $式知

(3.21) $ \begin{equation} \theta_{n} = \frac{\langle w_{n}-q_{n}, d_{n}\rangle}{\left\|d_{n}\right\|^2}\geq \frac{(1-\mu \frac{\lambda_{n}}{\lambda_{n+1}})\left\|w_{n}-q_{n}\right\|^2} {(1+\mu \frac{\lambda_{n}}{\lambda_{n+1}})^2\left\|w_{n}-q_{n}\right\|^2} = \frac{1-\mu \frac{\lambda_{n}}{\lambda_{n+1}}}{(1+\mu \frac{\lambda_{n}}{\lambda_{n+1}})^2}. \end{equation} $

再由$ (3.19) $和$ (3.21) $式得

$ \left\|w_{n}-q_{n}\right\|^2\leq \frac{(1+\mu \frac{\lambda_{n}}{\lambda_{n+1}})^2}{(1-\mu \frac{\lambda_{n}}{\lambda_{n+1}})^2}\left\|\theta_{n}d_{n}\right\|^2 = \frac{(1+\mu \frac{\lambda_{n}}{\lambda_{n+1}})^2} {\gamma^2(1-\mu \frac{\lambda_{n}}{\lambda_{n+1}})^2}\left\|h_{n}-w_{n}\right\|^2. $

第五步  证明

(3.22) $ \begin{eqnarray} \left\|u_{n+1}-p\right\|^2 &\leq& (1-(1-\rho )\beta_{n})\left\|u_{n}-p\right\|^2{}\\ &&+(1-\rho )\beta_{n}\left[\frac{M}{1-\rho}\cdot \frac{\alpha_{n}} {\beta_{n}}\left\|u_{n}-u_{n-1}\right\|+\frac{2}{1-\rho}\langle f(p)-p, u_{n+1}-p\rangle \right], {\qquad} \end{eqnarray} $

其中$ M = 2\left\|u_{n}-p\right\|+\alpha_{n}\left\|u_{n}-u_{n-1}\right\|>0 $. 注意到

(3.23) $ \begin{eqnarray} \left\|w_{n}-p\right\|^2 & = &\left\|u_{n}+\alpha_{n}(u_{n}-u_{n-1})-p)\right\|^2{}\\ & = &\left\|u_{n}-p\right\|^2+2\alpha_{n}\langle u_{n}-p, u_{n}-u_{n-1}\rangle+\alpha_{n} ^2\left\| u_{n}-u_{n-1}\right\|^2{}\\ &\leq& \left\|u_{n}-p\right\|^2+2\alpha_{n}\left\|u_{n}-p\right\|\cdot \left\|u_{n}-u_{n-1}\right\|+\alpha_{n} ^2\left\| u_{n}-u_{n-1}\right\|^2{}\\ & = &\left\|u_{n}-p\right\|^2+\alpha_{n}\left\|u_{n}-u_{n-1}\right\|(2\left\|u_{n}-p\right\|+\alpha_{n}\left\|u_{n}-u_{n-1}\right\|){}\\ & = &\left\|u_{n}-p\right\|^2+M\alpha_{n}\left\|u_{n}-u_{n-1}\right\|. \end{eqnarray} $

另一方面

(3.24) $ \begin{eqnarray} \left\|u_{n+1}-p\right\|^2 & = &\left\|\beta_{n}f(u_{n})+(1-\beta_{n})h_{n}-p\right\|^2{}\\ & = &\left\|\beta_{n}(f(u_{n})-f(p))+(1-\beta_{n})(h_{n}-p)+\beta_{n}(f(p)-p)\right\|^2{}\\ &\leq& \left\|\beta_{n}(f(u_{n})-f(p))+(1-\beta_{n})(h_{n}-p)\right\|^2+2\beta_{n}\langle f(p)-p, u_{n+1}-p\rangle{}\\ & = & \beta_{n}\left\|f(u_{n})-f(p)\right\|^2+(1-\beta_{n})\left\|h_{n}-p\right\|^2{}\\ &&-\beta_{n}(1-\beta_{n})\left\|f(u_{n})-f(p)-h_{n}+p\right\|^2+2\beta_{n}\langle f(p)-p, u_{n+1}-p\rangle{}\\ &\leq& \beta_{n}\left\|f(u_{n})-f(p)\right\|^2+(1-\beta_{n})\left\|h_{n}-p\right\|^2+2\beta_{n} \langle f(p)-p, u_{n+1}-p\rangle{}\\ &\leq& \beta_{n}\rho^2\left\|u_{n}-p\right\|^2+(1-\beta_{n})\left\|h_{n}-p\right\|^2+2\beta_{n} \langle f(p)-p, u_{n+1}-p\rangle{}\\ &\leq& \beta_{n}\rho\left\|u_{n}-p\right\|^2+(1-\beta_{n})\left\|w_{n}-p\right\|^2+2\beta_{n} \langle f(p)-p, u_{n+1}-p\rangle . \end{eqnarray} $

由$ (3.23) $和$ (3.24) $式得

$ \begin{eqnarray*} \left\|u_{n+1}-p\right\|^2 &\leq& (1-(1-\rho )\beta_{n})\left\|u_{n}-p\right\|^2+M\alpha_{n}(1-\beta_{n})\left\|u_{n}-u_{n-1}\right\|\\ &&+2\beta_{n}\langle f(p)-p, u_{n+1}-p\rangle \\ &\leq& (1-(1-\rho )\beta_{n})\left\|u_{n}-p\right\|^2\\ &&+(1-\rho)\beta_{n}\left[\frac{M}{1-\rho}\frac{\alpha_{n}}{\beta_{n}}\left\|u_{n}-u_{n-1}\right\|+\frac{2}{1-\rho}\langle f(p)-p, u_{n+1}-p\rangle \right]. \end{eqnarray*} $

第六步  证明序列$ \left\{\left\|u_{n}-p\right\|^2\right\} $收敛到零, 考虑两种情形.

情形1  若存在$ N\in {\Bbb N} $使得

$ \left\|u_{n+1}-p\right\|^2\leq \left\|u_{n}-p\right\|^2, \; \forall n\geq N. $

这就意味着$ {\lim\limits_{n \to \infty}}\left\|u_{n}-p\right\| $存在. 由$ (3.13) $式和$ \lim\limits_{n \to \infty}\beta_{n} = 0 $得

(3.25) $ \begin{equation} {\lim\limits_{n \to \infty}}\left\|h_{n}-w_{n}\right\| = 0. \end{equation} $

又因为

(3.26) $ \begin{eqnarray} \left\|u_{n+1}-u_{n}\right\| & = &\left\|\beta_{n}f(u_{n})+(1-\beta_{n} )h_{n}-u_{n}\right\|{}\\ & = &\left\|\beta_{n}f(u_{n})-\beta_{n}u_{n}+\beta_{n}u_{n}+(1-\beta_{n} )h_{n}-u_{n}\right\|{}\\ &\leq& \beta_{n}\left\|f(u_{n})-u_{n}\right\|+(1-\beta_{n} )\left\|h_{n}-u_{n}\right\|. \end{eqnarray} $

另一方面,

$ \left\|w_{n}-u_{n}\right\| = \alpha_{n}\left\|u_{n}-u_{n-1}\right\| = \beta_{n}\cdot \frac{\alpha_{n}}{\beta_{n}}\left\|u_{n}-u_{n-1}\right\|\to 0. $

由$ (3.25) $式得$ \left\|h_{n}-u_{n}\right\|\to 0. $再由$ (3.26) $式知$ \left\|u_{n+1}-u_{n}\right\|\to 0 $. 联合$ (3.17) $和$ (3.25) $式得

(3.27) $ \begin{equation} \left\|w_{n}-q_{n}\right\|\to 0. \end{equation} $

又由于$ \left\{{u_{n}} \right\} $是有界的, 也即存在$ \left\{{u_{n}} \right\} $的子序列$ \left\{{u_{n_{k}}} \right\} $弱收敛于某一点$ z\in H $满足

$ {\limsup\limits_{n \to \infty}}\langle f(p)-p, u_{n}-p\rangle = {\lim\limits_{k \to \infty}}\langle f(p)-p, u_{n_{k}}-p\rangle = \langle f(p)-p, z-p\rangle. $

同理, 因为$ \left\{{w_{n}} \right\} $也是有界的, 所以存在$ \left\{{w_{n}}\right\} $的子序列$ \left\{{w_{n_{k}}}\right\} $弱收敛于某一点$ z\in H $且根据$ (3.27) $式有

$ {\lim\limits_{n \to \infty}}\left\|w_{n_{k}}-q_{n_{k}}\right\| = 0. $

故由引理$ 3.3 $知$ z\in S $. 再根据$ p $的定义和引理$ 2.1 $有

$ {\limsup\limits_{n \to \infty}}\langle f(p)-p, u_{n} -p\rangle = \langle f(p)-p, z-p\rangle\leq0. $

因此

(3.28) $ \begin{eqnarray} {\limsup\limits_{n \to \infty}}\langle f(p)-p, u_{n+1}-p\rangle &\leq& {\limsup\limits_{n \to \infty}}\langle f(p)-p, u_{n+1}-u_{n}\rangle +{\limsup\limits_{n \to \infty}}\langle f(p)-p, u_{n}-p\rangle{}\\ & = &\langle f(p)-p, z-p\rangle \leq0. \end{eqnarray} $

最后由引理$ 2.4 $、$ (3.22) $和$ (3.28) $式得$ {\lim\limits_{n \to \infty}\left\|u_{n}-p\right\|} = 0 $, 即$ u_{n}\to p, n\to \infty $.

情形2  若存在序列$ \left\{\left\|u_{n}-p\right\|^2\right\} $的子序列$ \left\{\left\|u_{n_{j}}-p\right\|^2\right\} $满足

$ \left\|u_{n_{j}}-p\right\|^2<\left\|u_{n_{j}+1}-p\right\|^2, \; \forall j\in {\Bbb N}. $

在这种情况下, 由引理$ 2.3 $可知存在一个非减序列$ \left\{m_{k}\right\} $使得$ {\lim\limits_{k \to \infty}m_{k}} = \infty $且有如下不等式成立:

(3.29) $ \begin{align} \left\|u_{m_{k}}-p\right\|^2\leq \left\|u_{m_{k}+1}-p\right\|^2, \; \forall k\in {\Bbb N}, \end{align} $

(3.30) $ \begin{align} \left\|u_{k}-p\right\|^2\leq \left\|u_{m_{k}}-p\right\|^2. \end{align} $

则由$ (3.13) $和$ (3.29) $式得

$ \frac{2-\gamma}{\gamma}\cdot (1-\beta_{m_{k}})\left\|h_{m_{k}}-w_{m_{k}}\right\|^2\leq \left\|u_{m_{k}}-p\right\|^2-\left\|u_{m_{k}+1}-p\right\|^2+\beta_{m_{k}}M_{4}\leq \beta_{m_{k}}M_{4}. $

因此

$ \lim\limits_{n \to \infty}\left\|h_{m_{k}}-w_{m_{k}}\right\| = 0. $

类似于情形$ 1 $证明可得

(3.31) $ \begin{align} \left\|u_{m_{k}+1}-u_{m_{k}}\right\|\to 0 \end{align} $

以及

(3.32) $ \begin{align} {\limsup\limits_{n \to \infty}}\langle f(p)-p, u_{m_{k}+1}-p\rangle\leq0. \end{align} $

联立$ (3.22) $和$ (3.29) $式有

$ \begin{eqnarray*} \left\|u_{m_{k}+1}-p\right\|^2 &\leq& (1-(1-\rho )\beta_{m_{k}})\left\|u_{m_{k}+1}-p\right\|^2\\ &&+(1-\rho )\beta_{m_{k}}\left[\frac{2}{1-\rho} \langle f(p)-p, u_{m_{k}+1}-p\rangle +\frac{M}{1-\rho }\cdot\frac{\alpha_{m_{k}}}{\beta_{m_{k}}}\left\|u_{m_{k}}-u_{m_{k}-1}\right\|\right]. \end{eqnarray*} $

由此可得

(3.33) $ \begin{align} \left\|u_{m_{k}+1}-p\right\|^2\leq \frac{2}{1-\rho} \langle f(p)-p, u_{m_{k}+1}-p\rangle +\frac{M}{1-\rho }\cdot\frac{\alpha_{m_{k}}}{\beta_{m_{k}}}\left\|u_{m_{k}}-u_{m_{k}-1}\right\|. \end{align} $

故由$ (3.31) $、$ (3.32) $和$ (3.33) $式知

$ {\limsup\limits_{k \to \infty}}\left\|u_{m_{k}+1}-p\right\|\leq0. $

再由$ (3.29) $和$ (3.30) $式得$ {\limsup\limits_{k \to \infty}}\left\|u_{k}-p\right\|\leq0 $, 即$ u_{k}\to p, k\to \infty $.

注3.2  定理$ 3.1 $从以下三个方面改进了文献[11]中定理$ 1 $:

$ (1) $$ F $的单调性推广为伪单调性.

$ (2) $定理$ 3.1 $的证明不要求Lipschitz常数$ L $的任何信息, 而文献[11]中定理$ 1 $要求步长$ \lambda \in (0, \frac{1}{L}) $.

$ (3) $定理$ 3.1 $中$ r\in (0, 2) $, 而文献[11]中定理$ 1 $的$ r = 1 $.

注3.3  定理$ 3.1 $推广了文献[12]中定理$ 3.1 $到伪单调性情形, 同时增加了惯性项使算法的收敛速度更快.

注3.4  定理$ 3.1 $推广了文献[13]中定理$ 3.2 $到伪单调性情形, 同时$ \theta_{n} $的定义比文献[13]中的更加简洁并且增加了惯性项, 使收敛速度更快.

本节给出数值实验, 所有代码均在MATLAB R2019a和Windows10系统下运行. 计算机基本参数为AMD Ryzen 5 3550H with Radeon Vega Mobile Gfx 2.10 GHz和16GB内存. 为了体现本文算法的数值效果, 将算法3.1 (记为NIPCM算法)与文献[11]的IPCM算法、文献[12]的Y.L算法和文献[13]的VPTA算法进行比较, 其中"Iter"代表程序循环次数, "Time" 代表程序的CPU运行时间, 单位为秒.

例1  设$ H = L^2([0, 1]) $, 内积

$ \langle x, y\rangle : = \int_0^1 x(t)y(t)\, {\rm d}t, {\quad} \forall x, y\in H. $

范数

$ \left\|x\right\|: = \left(\int_0^1|x(t)|^2\, {\rm d}t\right)^\frac{1}{2}, {\quad} \forall x\in H. $

令$ C: = \{x\in H: \left\| x\right\|\leq 1\} $为单位球. 设$ F:C\rightarrow H $满足$ (Fx)(t) = \max \{0, x(t)\}, $故$ F $在$ C $上为单调Lipschitz连续的. 进一步, 由变分不等式的解集$ S = \{0\} $可得

$ P_{C}(x) = \left \{\begin{array}{ll} { } \frac{x}{\left\|x\right\|_{L^2}}, \, \, &\mbox{若}\, \, \left\|x\right\|_{L^2}>1, \\ x , \, \, &\mbox{若}\, \, \left\|x\right\|_{L^2}\leq 1. \end{array} \right. $

选取$ \mu = 0.85 $, $ \gamma = 1.99 $, $ u_1(t) = f(t) = 0.9t $和不同的初始点$ u_{0}(t) $, 其中$ t\in [0, 1] $为等差分布的数列, 取终止条件为$ \left\|u_{n+1}-u_{n}\right\|\leq 10^{-30} $对算法进行测试, 结果见表 1和图 1–5 (取$ u_0(t) = \frac{1}{600}(t^2-e^{-t}) $).

例2  设算子$ F:{{\Bbb R}} ^m\to {{\Bbb R}} ^m $满足

$ F(x) = Mx+q, $

其中$ q\in {{\Bbb R}} ^{m} $, 同时

$ M = BB^T+H+D, $

其中$ B\in {{\Bbb R}} ^{m\times m}, H\in {{\Bbb R}} ^{m\times m} $是斜对称矩阵, $ D\in {{\Bbb R}} ^{m\times m} $为对角元素非负的对角矩, 可行集为$ C = \{x\in {{\Bbb R}} ^m|Qx\leq b\}, $$ Q\in {{\Bbb R}} ^{m\times m}, $$ b\in {{\Bbb R}} ^m, $$ q $, $ b $, $ Q $, $ B $, $ H $在$ (2, -2) $中随机产生, $ D $中的对角元素在$ (0, 2) $中产生. 我们选取$ \gamma = 1 $, $ \mu = 0.05, $$ \alpha_{n} = \frac{2-\gamma}{1.01\gamma } $和初始点$ u_0 = (1, 1, \cdots , 1)^T\in {{\Bbb R}} ^m $, 终止条件为$ \left\|u_n\right\|<\epsilon $, 测试结果见表 2和图 6.

注4.1  从数值实验的结果来看, 我们有如下结论:

$ {\rm (i)} $ NIPCM算法、VPTA算法、Y.L算法和IPCM算法均收敛于变分不等式问题的解, 但NIPCM算法受初始点取值的影响最小, 见表 1.

$ {\rm (ii)} $从迭代次数、程序CPU运行时间和算法精度来看, NIPCM算法、Y.L算法和VPTA算法均优于IPCM算法, 尤其是NIPCM算法充分体现了迭代步数较少的优势, 而VPTA算法和Y.L算法稳定性较弱, 这可能会对该算法的收敛性造成影响, 见图 1–5及表 1.

$ {\rm (iii)} $在适当参数的选取下, 初始值$ u_{0}(t) $和$ f(x) $的取值对NIPCM算法的影响较小(见表 1和图 6), 但终止条件精度的选取对NIPCM算法的影响较大(见表 2).

$ {\rm (iv)} $ NIPCM算法优于VPTA算法、Y.L算法和IPCM算法.