并行Robin-Robin迭代法解决混合有限元二阶椭圆问题

0 下载量 135 浏览量 更新于2024-07-16 1 收藏 254KB PDF 举报
"王锋和曾玉平在‘Huge An Optimal Parallel Robin-Robin Iterative Method for the Mixed Finite Element Discretization of the Second Order Elliptic Problems’中提出了一个针对混合有限元离散二阶椭圆问题的并行优化Robin-Robin迭代方法。这种方法是基于非重叠区域分解,利用界面的Robin边界条件来实现子区域间的通信。" 在计算数学领域,二阶椭圆问题是一个基础且重要的问题,常常出现在各种物理和工程模型中。混合有限元方法是解决这类问题的有效工具,它通过引入多种类型的未知函数来离散原问题,从而获得更高效的数值解。然而,当问题的规模增大时,直接求解离散后的大型线性系统变得极其困难,这就需要采用迭代方法。 Robin-Robin迭代方法是区域分解方法的一种,它将连续域划分为多个子区域,并在子区域间使用特定的Robin边界条件进行交互。这种边界条件结合了Dirichlet(狄利克雷)和Neumann(诺伊曼)边界条件的特点,使得子区域间的耦合得以有效地处理。在非重叠区域分解中,Robin-Robin方法可以避免直接通信,降低了计算复杂度。 本文中,作者提出了一种并行优化的Robin-Robin迭代策略,特别针对混合有限元离散的二阶椭圆问题。他们提供了一种选择Robin参数和松弛参数的方法,并基于混合有限元的诺伊曼到狄利克雷算子的特征值估计和分块算子矩阵的谱理论,证明了该算法的收敛性。值得注意的是,该算法的收敛速度独立于网格尺寸和系数的跳跃,这意味着即使在不规则或具有大系数变化的域上,算法也能保持稳定且快速的收敛性能。 为了验证理论结果,文章还包含了一些数值实验。这些实验进一步证实了所提出的并行Robin-Robin迭代方法在解决大规模混合有限元问题时的效率和准确性,展示了其在实际应用中的潜力,特别是在需要并行计算的高维或大规模问题中。 关键词:计算数学,Robin-Robin迭代方法,区域分解,混合有限元 这篇论文属于O241.82(偏微分方程的数值解法)和O246(有限元法)的分类范畴,对于理解和改进混合有限元方法的并行计算策略具有重要意义。

The starting configuration of this puzzle is a row of cells, with disks located on cells through . The goal is to move the disks to the end of the row using a constrained set of actions. At each step, a disk can only be moved to an adjacent empty cell, or to an empty cell two spaces away if another disk is located on the intervening square. Given these restrictions, it can be seen that in many cases, no movements will be possible for the majority of the disks. For example, from the starting position, the only two options are to move the last disk from cell to cell , or to move the second-to-last disk from cell to cell . 1. [15 points] Write a function solve_identical_disks(length, n) that returns an optimal solution to the above problem as a list of moves, where length is the number of cells in the row and n is the number of disks. Each move in the solution should be a twoelement tuple of the form (from, to) indicating a disk movement from the cell from to the cell to. As suggested by its name, this function should treat all disks as being identical. Your solver for this problem should be implemented using a breadth-first graph search. The exact solution produced is not important, as long as it is of minimal length. Unlike in the previous two sections, no requirement is made with regards to the manner in which puzzle configurations are represented. Before you begin, think carefully about which data structures might be best suited for the problem, as this choice may affect the efficiency of your search

2023-06-06 上传

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L). To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river. Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N). FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks. Input Line 1: Three space-separated integers: L, N, and M Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position. Output Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks Sample Inputcopy Outputcopy 25 5 2 2 14 11 21 17 4 Hint Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).

2023-07-24 上传