Two-step processes by one-step methods of order 3 and of order 4

Hisayoshi Shintani
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引用次数: 10

Abstract

and the initial condition y(χo) = jo, where /(#, y) is assumed to be a sufficiently smooth function. We are concerned with the case where the equation (1.1) is integrated numerically by one-step methods of order 3 and of order 4. It is well known that the one-step methods of order 3 such as Kutta method and those of order 4 such as Runge-Kutta method require three and four evaluations of the derivative respectively. It is also known that, if the same step-size is used twice in succession, an approximate value of the truncation error can be obtained by integrating again with the double step-size. This method of approximating the truncation error requires, per two steps of integration, eight and eleven evaluations of f(x, y) for any one-step method of order 3 and for that of order 4 respectively. In our previous paper C19], it has been shown that there exists a onestep formula of order 4 such that, after two steps of integration with the same step-size, only one additional evaluation of the derivative makes/it possible to approximate the truncation error. In that formula, however, four values of f(χ, y) evaluated in the first step of integration are not used explicitly in the second step. Thus there remains a possibility of reducing the number of evaluations of the derivative by utilizing all the values of the derivative computed already. In this paper, it is shown that there exist one-step integration formulas of order 3 and those of of order 4 such that approximate values zx and z2 of y(χo + h) and y(xo-\-2h) and an approximation to their truncation errors can be obtained with five and seven evaluations of f(x, y) respectively. Finally two numerical examples are presented.
用3阶和4阶的一步法进行两步过程
初始条件y(χ 0) = jo,其中/(#,y)是一个足够光滑的函数。我们所关心的是用3阶和4阶的一步法对方程(1.1)进行数值积分的情况。众所周知,库塔法等3阶的一步法和龙格-库塔法等4阶的一步法分别需要求3次和4次导数。我们还知道,如果连续两次使用相同的步长,将截断误差与两次步长再次积分即可得到一个近似值。这种近似截断误差的方法需要在每两步积分中分别对3阶和4阶的单步方法求8次和11次f(x, y)。在我们之前的论文[19]中,已经证明了存在一个4阶的一步公式,使得在相同步长的两步积分之后,只需要对导数进行一次额外的求值就可以近似截断误差。然而,在该公式中,在第一步积分中计算的f(χ, y)的四个值在第二步中没有明确使用。因此,仍然有可能通过利用已经计算的导数的所有值来减少导数的评估次数。本文证明了存在3阶和4阶的一步积分公式,使得分别求5次和7次f(x, y),就可以求得y(χo + h)和y(xo-\-2h)的近似值zx和z2及其截断误差的近似值。最后给出了两个数值算例。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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