2019
DOI: 10.1002/mma.5936
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Two‐derivative Runge‐Kutta methods with optimal phase properties

Abstract: In this work, we consider two-derivative Runge-Kutta methods for the numerical integration of first-order differential equations with oscillatory solution.We construct methods with constant coefficients and special properties as minimum phase-lag and amplification errors with three and four stages. All methods constructed have fifth algebraic order. We also present methods with variable coefficients with zero phase-lag and amplification errors. In order to examine the efficiency of the new methods, we use four… Show more

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Cited by 11 publications
(6 citation statements)
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“…Example The Orbital Problem in Kalogiratou et al 21 : righty1left=y1+11000cos(x),y1(0)=1,y1(0)=0,righty2left=y2+11000sin(x),y2(0)=0,y2(0)=999510000,x[0,10].$$ {\displaystyle \begin{array}{cc}\hfill {y}_1^{{\prime\prime} }& =-{y}_1+\frac{1}{1000}\cos (x),{y}_1(0)=1,{y}_1^{\prime }(0)=0,\hfill \\ {}\hfill {y}_2^{{\prime\prime} }& =-{y}_2+\frac{1}{1000}\sin (x),{y}_2(0)=0,{y}_2^{\prime }(0)=\frac{9995}{10000},x\in \left[0,10\right].\hfill \end{array}} $$ The exact solution is y1false(xfalse)=cosfalse(xfalse)+120000.1emxsinfalse(xfalse),$$ {y}_1(x)=\cos (x)+\frac{1}{2000}\kern0.1em x\sin (x), $$ y2false(xfalse)=sinfalse(xfalse)120000.1emxcosfalse(xfalse).$$ {y}_2(x)=\sin (x)-\frac{1}{2000}\kern0.1em x\cos (x). $$ Now we consider w=1.0…”
Section: Numerical Examplesmentioning
confidence: 99%
“…Example The Orbital Problem in Kalogiratou et al 21 : righty1left=y1+11000cos(x),y1(0)=1,y1(0)=0,righty2left=y2+11000sin(x),y2(0)=0,y2(0)=999510000,x[0,10].$$ {\displaystyle \begin{array}{cc}\hfill {y}_1^{{\prime\prime} }& =-{y}_1+\frac{1}{1000}\cos (x),{y}_1(0)=1,{y}_1^{\prime }(0)=0,\hfill \\ {}\hfill {y}_2^{{\prime\prime} }& =-{y}_2+\frac{1}{1000}\sin (x),{y}_2(0)=0,{y}_2^{\prime }(0)=\frac{9995}{10000},x\in \left[0,10\right].\hfill \end{array}} $$ The exact solution is y1false(xfalse)=cosfalse(xfalse)+120000.1emxsinfalse(xfalse),$$ {y}_1(x)=\cos (x)+\frac{1}{2000}\kern0.1em x\sin (x), $$ y2false(xfalse)=sinfalse(xfalse)120000.1emxcosfalse(xfalse).$$ {y}_2(x)=\sin (x)-\frac{1}{2000}\kern0.1em x\cos (x). $$ Now we consider w=1.0…”
Section: Numerical Examplesmentioning
confidence: 99%
“…Definition 1. Given the method in ( 2)-( 4), an interval I = (− hs , 0) is said to be an interval of absolute stability if for all h ∈ I, it is |λ 1,2 | < 1, where λ 1,2 are the solutions of the equation in (19).…”
Section: Stability Analysismentioning
confidence: 99%
“…Definition 2. An interval (− hp , 0) corresponding to the RKN method in Equations ( 2)-( 4) is said to be an interval of periodicity if for every h ∈ (− hp , 0), |λ 1,2 | = 1, with λ 1 = λ 2 , where λ 1,2 are the roots of the equation in (19).…”
Section: Stability Analysismentioning
confidence: 99%
“…An advantage of these methods is that the number of algebraic order conditions is significantly reduced in comparison with the classical Runge-Kutta methods of the same order, thereby allowing the construction of high-order schemes with only a few stages. Following this, implicit TDRK methods were derived in [13,4], general linear TDRK methods were presented in [1,2,3,11], and recently TDRK methods with optimal phase properties were constructed in [32]. In the case if a good estimate of frequency is known in advance, one can further improve the numerical solution of these methods by incorporating the exponential fitting idea [22,8,29,15] in order to integrate exactly the system who solutions are linear combinations of e ˘iωx (ω is a prescribed frequency).…”
Section: Introductionmentioning
confidence: 99%
“…In particular, with s " 2 stages we obtain superconvergent schemes of order p " 2s and p " 2s `1, and with s " 3 stages, a method of order p " 2s is derived. This is a significant improvement since previous EFTDRK methods (either explicit or fully implicit) [12,32,4,13] only achieve orders p " 2s for s " 2 stages and p " 2s ´1 for s " 3 stages.…”
Section: Introductionmentioning
confidence: 99%