Solving large-scale continuous-time algebraic Riccati equations by doubling

Solving large-scale continuous-time algebraic Riccati equations by doubling
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通过加倍求解大规模连续时间代数 Riccati 方程

DOI:
10.1016/j.cam.2012.06.006
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发表时间:
2013
影响因子:
2.4
通讯作者:
Weng, Peter Chang-Yi
Weng, Peter Chang-Yi
中科院分区:
数学2区
文献类型:
--
作者:
Li, Tiexiang;Chu, Eric King-wah;Lin, Wen-Wei;Weng, Peter Chang-Yi

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考虑具有数值低阶解的大规模代数Riccati方程的解。对于离散时间情况,采用了保持结构的加倍算法,A的迭代不显式计算,而是采用递归形式[公式:见文本],Dk(1)和Dk(2)是低阶的,Sk−1是小维的。对于连续时间情况,在应用加倍之前,将首先用Cayley变换处理代数Riccati方程。由于n是代数方程的维数,所得到的算法每次迭代的计算复杂度为O(n),不需要任何内部迭代,并且本质上是二次收敛的。本文将给出一些数值结果。例如,在第5.2节中,例3,维数n=20209,解X中有2.04亿个变量,在MacBook Pro上使用MATLAB在45秒内求解,机器精度为0(10−16)。
We consider the solution of large-scale algebraic Riccati equations with numerically low-ranked solutions. For the discrete-time case, the structure-preserving doubling algorithm has been adapted, with the iterates for A not explicitly computed but in the recursive form [Formula: see text] , with Dk(1)and Dk(2)being low-ranked and Sk−1being small in dimension. For the continuous-time case, the algebraic Riccati equation will be first treated with the Cayley transform before doubling is applied. With n being the dimension of the algebraic equations, the resulting algorithms are of an efficient O(n) computational complexity per iteration, without the need for any inner iterations, and essentially converge quadratically. Some numerical results will be presented. For instance in Section 5.2, Example 3, of dimension n=20209 with 204 million variables in the solution X, was solved using MATLAB on a MacBook Pro within 45 s to a machine accuracy of O(10−16).
DOI: 10.1109/9.863597
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