Extended Runge Kutta-like formulae

Applied Numerical Mathematics 56 006 584 605 www.elsevier.com/locate/apnum Extended Runge Kutta-like formulae Xinyuan Wu a Jianlin Xia b a State Key Laboratory for Novel Software Technology Department of Mathematics Nanjing University Nanjing 009 PR China b Department of Mathematics University of California Berkeley CA 9470 USA Available online 6 January 006 Abstract In this paper we present a new family of extended Runge Kutta formulae in which just like in Enright s methods it is assumed that the user will evaluate both f and f readily when solving the autonomous system y = fy numerically. This means that we introduce some new parameters in the extended Runge Kutta-like formulae in order to enhance the order of accuracy of the solutions using evaluations of both f and f instead of the evaluations of f only. Moreover if f is approximated by a difference quotient of past and current evaluations of f the order of convergence can be retained. The resulting two-step Runge Kutta method can be regarded as replacing the function evaluations of f with approximations of f. Specifically the proposed formulae with f are more efficient for cases where f is not more expensive to evaluate than f and the proposed derivative-free formulae are more attractive for use when past values of f are available. Furthermore error estimates and step-choose strategies are considered for the derivative-free extended Runge Kutta methods. 005 IMACS. Published by Elsevier B.V. All rights reserved. MSC: 65L05 Keywords: Extended Runge Kutta-like method; Two-step Runge Kutta method; Absolute stability. Introduction This paper is mainly concerned with the autonomous system of ordinary differential equations y t = f yt. and attempts to solve initial value problems based on. using step-by-step numerical methods. Throughout the paper we assume that fyhas derivatives to the desired order in a domain D in R q and we assume that fy fy L y y holds for all y y D where L is the Lipschitz constant. Many efforts have been made to improve the order of Runge Kutta methods by means of increasing the number of terms in the Taylor series expansion. This increases the number of function evaluations accordingly. The direct use of * Corresponding author. E-mail address: xywu@nju.edu.cn X. Wu. The research of this author was done partially while visiting Mathematics Institute of Tübingen University Germany. The author is indebted to Prof. Christian Lubich and Dr. Achim Schädle for their hospitality and helpful discussion. 068-974/$0.00 005 IMACS. Published by Elsevier B.V. All rights reserved. doi:0.06/j.apnum.005..008

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 585 the Jacobian matrix in an integrator for stiff problems has been proposed by many authors see [ 57] etc.. In a recent paper [6] Goeken et al. proposed a class of Runge Kutta methods using higher derivatives and presented new third and fourth order numerical methods. Specifically f is embedded in f. This motivates a new family of extended Runge Kutta-like formulae of the form where y n+ = y n + h k j m j= j = f y n + h b j k j + h s= a js k s m j= c j k j. k j j = f y n + h s= b js k s j =...m. which also can be thought of as extended explicit Enright-like methods. Obviously with c j = 0 j =...min. the methods reduce to classical Runge Kutta methods y n+ = y n + h m b j k j j= where j k j = f y n + h a js k s j =...m. s= We note also that if a js = b js j =...m s =...j in. then we have where y n+ = y n + h k j m j= j = f y n + h b j k j + h s= a js k s m j= c j k j.4 k j j = f y n + h s= a js k s j =...m..5 Moreover if k j in. and.4 is approximated by a difference quotient of the past and the current evaluations of f namely k j fy n + h j s= b jsk s where k n j = f j y n + h a js k n s s= fy n + h j s= b jsk h n s and k n 0 = fy n then we will obtain two-step Runge Kutta methods which can be regarded as derivative-free extended Runge Kutta methods. The rest of the paper is organized as follows. In Section third order formulae with two function evaluations are derived and derivative-free formulae are provided. In Section fourth order formulae with three function evaluations are developed and derivative-free extended Runge Kutta methods are presented. In Section 4 extended fifth order Runge Kutta-like formulae and the order conditions of the derivative-free extended Runge Kutta formula of fifth order are considered. The stability analysis is given in Section 5. Numerical experiments are provided in Section 6 to illustrate that these new formulae are capable of achieving the order claimed and can be very effective. In Section 7 error estimates and step-choose strategies are discussed for the derivative-free extended Runge Kutta methods.

586 X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605. Third-order formulae with two function evaluations Formula. with m = is trivial. Now let us consider. with m = namely y n+ = y n + h b k + b k + h c k + c k. where k = fy n k = f y n + ha k k = f y n k = f y n + hb k.. Our task is to determine the coefficients of the extended Runge Kutta methods. to get third order methods. Employing the Taylor series expansion and comparing the terms of order and with those of the true solution we arrive at the following conditions b + b = b a + c + c = b a + c b = 6 c b = 6. It follows from the last two equations of. that a b = 0. So we have. b + b = b a + c + c = c b = 6 b a = 0..4a.4b.4c.4d.4c means that c 0 and b 0 so.4c can be rewritten as b =. 6c In order to choose the best possibilities from the varieties of possible third order formulae let a = b = 0 in.4b. It therefore follows from.4a.4d that b = c = c b = c 0 6c where c is a free parameter and c 0. From.. and.5 we obtain the following formulae [ y n+ = y n + hf n + h c f n + c f y n + 6c f n where f n = fy n and f n = f yy n f y n. Letting c = it follows that b = including only two function evaluations per step y n+ = y n + hf n + h f y n + hf n with the local truncation error given by.5 ].6 6c = and we have the best formula.7

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 587 Tth= yt + h {yt + hf yt + h f yt + hf yt } = yt + h {yt + hf yt + h [ fyt+ h f yf + h ] f f yy + O h 5} = yt + y h + h y + h 6 y + h4 4 y4 [ {yt + hf + h f y f + h fyy f + fy f + h fyyy f + f yy f y f ] + O h 5} 8 = h4 [ fyyy f + 6f y f yy f + fy 7 f ] + O h 5..8 Next we consider the derivative-free patterns corresponding to the formulae.. For this purpose it is enough to approximate f to some given accuracy by employing the current and previous evaluations of f such as k = f y n fy n fy n h k = f y n + hb fy n fy n + hb fy n fy n + hb fy n. h This motivates a class of two-step Runge Kutta formulae as follows y n+ = αy n + αy n + h [ b + c f y n + b f y n + ha fy n + c f y n + hb fy n c fy n c f y n + hb fy n ].9 where <α<. Employing Taylor series expansion and comparing the terms of order and with those of the true solution we obtain the following conditions α + b + b = α + b a + c + c = α c c b = 6.0 α 6 + b a c +c + c b = 6. From this we get b + b = + α b a + c + c = α c b c c = + α b a = 0. Letting b = 0 we have b = + α c = α c b = 5 α c 0. c This results in a class of two-step Runge Kutta methods of third order as follows.a.b.c.d.a.b.c y n+ = αy n + αy n + h [ + α + c f n c f n + c fyn + hb f n fy n + hb f n ].

588 X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 where c and b are determined by.b and.c. Letting α = 0 b = 0 a = 0 and c = wehavec = 0 and b = 5 6. This leads to the following formula [ y n+ = y n + h f n + f y n + 5 6 hf n f y n + 5 ] 6 hf n.4 which is a two-step Runge Kutta method of third order and only needs two function evaluations of f per step. Letting α = 0b = 0a = 0 and c = in. then c = and b = 5. Thus we have [ y n+ = y n + h f n + f n + f y n + 5 hf n f y n + 5 ] hf n..5 Let α = 0b = 0a = 0 and c = 5 48 in. then c = 48 and b = 4 5. Hence we get y n+ = y n + h [ 47f n + f n + 5 f y n + 4 48 5 hf n f y n + 4 ] 5 hf n..6 If α = 0b = 0a = 0 and c = 5 4 are chosen in. then c = 4 and b = 5 consequently we have y n+ = y n + h [ f n + f n + 5 f y n + 4 5 hf n f y n + ] 5 hf n.7 y n+ = y n + h [ 4f n + 5f n + 9 f y n + 0 48 9 hf n f y n + 0 ] 9 hf n..8 It is easy to see that all of the formulae.5.6.7 and.8 are of third order with only two function evaluations of f per step.. Fourth-order formulae with three function evaluations Extended Runge Kutta-like methods.. with m = are of the following form y n+ = y n + h b k + b k + b k + h c k + c k + c k. where k = fy n k = f y n + ha k k = f y n k = f y n + hb k k = f y n + ha k + ha k k = f y n + hb k + hb k.. In order to determine the coefficients of formulae. we use Taylor s series expansion and compare the terms of order and 4 with those of the true solution. We then obtain b + b + b = b a + b a + a + c + c + c = b a + b a + a + c b + c b + b = b a a + c b + c b + b = 6. b a + b a + a + c b + b + b = 4 b a a a + a + a + c b + c b + b = 4 c b a = 4. Here we have five free parameters resulting from twelve unknowns with seven equations. Those parameters are b b c a and c. The specific formula of interest is y n+ = y n + hf n + 6 h f n + h f y n + y hf n + 4 hf n.4 with the local truncation error Tth= yt n+ y n+ = h5 4fyyyy fn 4 880 + 8f yf yyy fn + f yy f n + 69f y f yyfn + 4f y 4 f n..5

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 589 This is because yt n+ = yt n + hf n + h f yf n + h fyy fn 6 + f y f h n 4 + 4fy f yy fn 4 + f yyyfn + f y f n + h5 fyyyy fn 4 0 + 7f yyyf y fn + 4f yy f n + f yyfy f n + f y 4 f n + O h 6 and y n+ = y n + hf n + h 6 f n y + h f n + hf y n + h 4 f n { = yt n + hf n + h 6 f yf n + h f y f n + h [ f n + h 4 f yf n + ] hfn f f yy y 4 + f yyf n + [ h f n + h 4 f yf n + ] hfn f yy f yyyf n + f y f yy 4 + [ ] hfn fyyyyf n + 4f y f yyy + f } yy + O h 6 6 = yt n + hf n + h f yf n + h f 6 y f n + f yy fn h 4 [ + f 4 y f n + f yy f y fn ] h 4 [ + fyyy fn 4 + f yf yy fn ] [ + h 5 fyy fy 9 f n + f yy f n + fyyy f y fn 48 + f yyfy f n + fyyyy fn 4 44 + 4f yyyf y fn + f yy f n ] + O h 6 = yt n + hf n + h f yf n + h f 6 y f n + f yy fn h 4 [ + f 4 y f n + f yy f y fn ] h 4 [ + fyyy fn 4 + f yf yy fn ] + h5 [ 4fyyyy fn 4 576 + 8f yyyf y fn + 5f yy f n + 9f yyfy f n ] + O h 6. Now let us consider the derivative-free formula corresponding to the formula.. It suffices to approximate f to some given accuracy by employing the current and previous evaluations of f. This motivates the following formula y n+ = α y n + αy n + h [ b f n + b fy n + ha f n + b f y n + ha f n + ha f ] y n + ha f n + h [ c f n f n + c fyn + hb f n fy n + hb f n + c f yn + hb f n + hb fy n + ha f n f y n + hb f n + hb fy n + ha f n ]..6 In order to determine the coefficients of.6 we use Taylor series expansion. Comparing the terms of order and 4 with those of the true solution we obtain α + b + b + b = α + b a + b a + a + c + c + c = α + b a + b a + a + c b + c b + b c + c + c = α + b a a + c b + c b + b c + c + c = α 4 + b a + b a + a + c b b + c b + b b + b + c + c + c = 4 α + b a a a + a + a + 6c b b c b + b b a + 6c b + b + 4c + c + c = α 4 c b c b + b b a + c + c + c = 4..7 In.7 we have six free parameters resulting from twelve unknowns with seven equations for each α <α<. In order to be able to derive a method of order four with only three function evaluations of f weseta = b

590 X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 a = b and a = b in.7. In this way we may obtain many desirable formulae and all of them are of fourth order with only three function evaluations of f. For example the special formulae are [ y n+ = y n + h 5 48 f 8 7f n + f n + 5 y n + 4 5 hf n y n+ = [ 5 9y n 4y n + h + 9 00 f y n 90 84 00 f y n+ = y n + h 48 48 f y n + 4 5 hf n 548 f 5 f n + 84 00 f 9 hf n + 90 9 hf 9 00 f f y n + 0 9 hf n [ 47f n + f n + 5 + 5 48 f y n hf n + hf y n + 4 5 hf n y n + 4 ] 5 hf n.8 y n hf n + hf y n + 0 y n + 0 9 hf n 9 hf n y n 90 9 hf n + 90 y n + 4 5 hf y n + 5 hf n 9 hf y n + 0 9 hf n ].9 f y n + 45 hf y n + ] 5 hf n.0 y n+ = y n + h 50 + 5 f [ 79f n + 7f n + 94 f y n + 5 y n + 5 88 hf y n + 5 hf n f Now letting a = 6+47 4 00 a = 757+799 4 600 a = 49 4 00 b = 0 b = 0 c = 0 α= 9 0 then we can obtain b = 0 c = 079 4800 b = 40 570+80 4 5854 b = 4006 47 4 5854 c = 48 4800 from.7 and then we have formula hf n y n + 5 88 hf y n+ = 9 0 y n 9 { 0 y n + h 0 f n + 079 4800 f n f n + 48 4800 + 4006 47 4 hf 5854 y n + 6 + 47 4 hf n 00 f y n + 5 hf n [ f y n + 5 ] hf n.. y n + 40 570 + 80 4 hf n 5854

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 59 f y n + 40 570 + 80 4 5854 + 4006 47 4 hf 5854 hf n y n + 6 + 47 4 hf n ]}.. 00 All of these formulae.8. are of fourth-order formulae with only three function evaluations per step. 4. Extended fifth order Runge Kutta-like formulae and order conditions of two-step formulae Extended Runge Kutta-like methods.. with m = 4 are as follows y n+ = y n + h b k + b k + b k + b 4 k 4 + h c k + c k + c k + c 4 k 4 4. where k = fy n k = f y n + ha k k = f y n + ha k + ha k k 4 = f y n + ha 4 k + ha 4 k + ha 4 k k = f y n k = f y n + hb k k = f y n + hb k + hb k k 4 = f y n + hb 4 k + hb 4 k + hb 4 k. 4. We use Taylor series expansion. Comparing the terms of order 4 and 5 with those of the true solution we can obtain twelve equations with twenty unknowns. In this system of nonlinear equations we let b = 0 b = 0 b 4 = 0 c = 0b 4 = 0 b 4 = 0 and b = a then this system is reduced as b = 0 c + c + c 4 = 0 b c + b 4 c 4 6 = 0 b c + b 4 c 4 6 = 0 a b 4 c 4 + a b 4 c 4 4 = 0 b c + a b 4 c 4 + a b 4 c 4 + b4 c 4 6 = 0 b c + b 4 c 4 4 = 0 a a b 4 c 4 0 = 0 a b 4c 4 + a a b 4 c 4 + a a b 4 c 4 + a b 4c 4 + a b4 c 4 + a b4 c 4 0 = 0 a b 4c 4 + a a b 4 c 4 + a b 4c 4 + a b4 c 4 + a b4 c 4 0 = 0 b c + a b 4c 4 + a a b 4 c 4 + a b 4c 4 + a b4 c 4 + a b4 c 4 + b 4 c 4 0 7 = 0 6 b c + 6 b 4 c 4 0 = 0. Then we can obtain two particular solutions. The first solution is b = c = 6 b = 5 b = 4 c = 50 + c 4 = 5 + 68 68 5 4 5 a = 4 5 b 4 = + 0

59 X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 and the second solution is b = c = + c = 50 c 4 = 5 + 6 68 68 b = 5 + b = + 4 5 4 5 a = 4 5 b 4 =. 0 Consequently we have produced two extended Runge Kutta-like formulae [ y n+ = y n + hf y n + h 6 + 5 + f y n + h + 68 0 + h 4 5 f y n + h 5 ] fy n 4 f y n + 50 + f y n + h 5 68 4 f y n + h 5 4 5 fy n fy n 4. and [ + y n+ = y n + hf y n + h 6 5 + f 68 + h 4 5 f f y n + 50 f y n + h 5 + fy n 68 4 y n + h f y n + h + 0 5 4 fy n 5 ]. 4.4 y n + h 5 + fy n 4 Next we consider the derivative-free formulae corresponding to the formulae 4. that is the following two-step Runge Kutta formulae y n+ = αy n + αy n + hb k + b k + b k + b 4 k 4 + c q + c q + c q + c 4 q 4 <α< where k = fy n k = fy n + ha k k = fy n + ha k + ha k k 4 = fy n + ha 4 k + ha 4 k + ha 4 k q = f n f n q = fy n + hb f n fy n + hb f n q = fy n + hb f n + hb fy n + ha f n fy n + hb f n + hb fy n + ha f n q 4 = fy n + hb 4 f n + hb 4 fy n + ha f n + hb 4 fy n + ha f n + ha fy n + ha f n fy n + hb 4 f n + hb 4 fy n + ha f n + hb 4 fy n + ha f n + ha fy n + ha f n. 4.5 4.6

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 59 In order to determine the coefficients of formulae 4.5 we use Taylor series expansion and compare the terms of order 4 and 5 with those of the true solution. This leads to the following equations α + b + b + b + b 4 = α + c + c + c + c 4 + b a + b a + b a + b 4 a 4 + b 4 a 4 + b 4 a 4 = α 6 c + c + c + c 4 + b a a + b 4 a a 4 + b 4 a a 4 + b 4 a a 4 + c b + c b + c b + c 4 b 4 + c 4 b 4 + c 4 b 4 = 6 b a + b a + b a + b 4a4 + b 4a4 + b 4a4 α 6 c + c + c + c 4 + b a a + b 4 a 4 a 4 + b 4 a 4 a 4 + b 4 a 4 a 4 + c b + c b + c b + c 4 b 4 + c 4 b 4 + c 4 b 4 = 6 α 4 + c 6 + c 6 + c 6 + c 4 6 + b 4 a a a 4 c b c b c b + c a b c 4b 4 c 4b 4 + c 4 a b 4 c 4b 4 + c 4 b b 4 + c 4 a b 4 = 4 b a a + b 4a 4 a + b a a + b 4 a4 a + b b a a + b 4 a 4 a 4 a + b 4 a 4 a 4 a + c b a + c 4 b 4 a + b 4 b a4 + b 4a a4 + c b + c b + c b + c 4b 4 + c 4b 4 + c 4b 4 + α 6 + c + c + c + c 4 + b 4b a 4 + b 4a a 4 + b 4 b a a 4 + b 4 b a 4 a 4 + b 4 a a 4 a 4 + b 4 b a 4 a 4 + b 4 a a 4 a 4 c b c b c b + c b b c 4 b 4 c 4 b 4 + c 4 b 4 b 4 c 4 b 4 + c 4 a b 4 + c 4 a b 4 + c 4 b 4 b 4 + c 4 b 4 b 4 = 6 6 b a + 6 b a + 6 b a + 6 b 4a4 + 6 b 4a4 + 6 b 4a4 + b a a + b 4a 4 a4 + b 4a 4 a4 + b 4a 4 a4 + c b + c b + c b + c 4b 4 + c 4b 4 + c 4b 4 + α 4 + c 6 + c 6 + c 6 + c 4 6 + b a a + b 4a 4 a 4 + b 4a 4 a 4 + b 4a 4 a 4 + b 4 a 4 a 4 a 4 c b c b c b + c b b c 4b 4 c 4b 4 + c 4 b 4 b 4 c 4b 4 + c 4 b 4 b 4 + c 4 b 4 b 4 = 4 0 α c 4 c 4 c 4 c 4 4 + 6 c b + 6 c b + 6 c b c a b + 6 c 4b 4 + 6 c 4b 4 c 4a b 4 + 6 c 4b 4 c 4b b 4 c 4a b 4 + c 4 a a b 4 = 0 b a a + b 4a4 a + b 4a a 4 a + c b a + c 4b 4 a + b 4 a a4 a + c b a + c 4 b4 a + b 4 a a 4a + b 4 b a a 4 a + b 4 a a 4 a 4 a + b 4 a a 4 a 4 a + b 4 a a 4 a 4 a 7 c b a + c b b a 7 c 4b 4 a + c 4 b 4 b 4 a + c 4 a b 4 a + c 4 b 4 b 4 a + b 4a a 4 + b 4a a 4 + b 4a a a 4 c b c b c b c 4b 4 c 4b 4 c 4 b 4 + c 4a b 4 + c 4a b 4 α 0 c 4 c 4 c 4 c 4 4 + 6 c b + 6 c b + 6 c b c b b + 6 c 4b 4 + 6 c 4b 4 c 4 b 4 b 4 + c 4 a b 4 + c 4 a b 4 + 6 c 4b 4 7 c 4a b 4 7 c 4a b 4 + c 4 a a b 4 c 4 b 4 b 4 + c 4 a b 4 b 4 + c 4 a b 4 b 4 c 4 b 4 b 4 + c 4 a b 4 b 4 + c 4 a b 4 b 4 = 0

594 X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 b a a + b 4a 4 a + b a a a + b 4a 4 a 4 a + b 4a 4 a 4 a + c b a + c 4b 4 a + c b a + c 4 b 4 a c b a + c b b a c 4 b 4 a + c 4 b 4 b 4 a + c 4 b 4 b 4 a + b 4a a 4 + b 4a a 4 + b 4a a a 4 c b c b c b c 4b 4 c 4b 4 c 4b 4 + c 4 a b 4 + c 4a b 4 α 0 c 6 c 6 c 6 c 4 6 + b 4a a 4a 4 + b 4 a a 4a 4 + b 4 a a a 4 a 4 + b 4a a 4a 4 + b 4a a 4a 4 + b 4 a a a 4 a 4 + c b + c b + c b c b b + c 4b 4 + c 4b 4 c 4 b 4 b 4 + c 4a b 4 + c 4a b 4 + c 4b 4 c 4 a b 4 c 4 a b 4 + c 4 a a b 4 c 4 b 4 b 4 + c 4 a b 4 b 4 + c 4 a b 4 b 4 c 4 b 4 b 4 + c 4 a b 4 b 4 + c 4 a b 4 b 4 = 0 6 b a a + 6 b 4a 4 a + c b a + c 4b 4 a + b a a + b 4a4 a + b a a a + b 4 a 4 a4 a + b 4a 4 a4 a + c b a + c 4 b4 a + b a a a + b 4a4 a 4a + b 4 a4 a 4a + b 4 a 4 a 4 a 4 a c b a + c b b a c 4 b 4 a + c 4 b 4 b 4 a + c 4 b 4 a 4 a + b 4a a4 + b 4a a4 + c b + c b + c b + c 4b4 + c 4b4 + c 4b4 + b 4 a a 4 a 4 + b 4a a 4 a 4 + b 4a a 4 a 4 + b 4a a 4 a 4 5 4 c b 5 4 c b 5 4 c b + c b b 5 4 c 4a 4 5 4 c 4b 4 + c 4b 4 b4 5 4 c 4b4 + c 4 a b4 + c 4a b4 + c 4b 4 b4 + c 4b 4 b4 0 7α 7c 4 7c 4 7c 4 7c 4 4 + 6 b 4a a 4 + 6 b 4a a 4 + b 4a a a 4 + b 4a a4 a 4 + b 4a a4 a 4 + b 4a a 4 a 4 + b 4a a 4 a 4 + b 4a a a 4 + b 4 a a 4 a 4 a 4 + b 4 a a 4 a 4 a 4 + 7 6 c b + 7 6 c b + c b b + 7 6 c b 5 c b b + 7 6 c 4b 4 + c 4b 4 b 4 + 7 6 c 4b 4 5 c 4b 4 b 4 + c 4a b 4 + c 4a b 4 + c 4b 4 b 4 + c 4b 4 b 4 + 7 6 c 4b 4 c 4 a b 4 c 4 a b 4 + c 4 a a b 4 5 c 4b 4 b 4 + c 4 a b 4 b 4 + c 4 a b 4 b 4 5 c 4b 4 b 4 + c 4 a b 4 b 4 + c 4 a b 4 b 4 + c 4 b 4 b 4 b 4 = 0 7 4 b a 4 + 4 b a 4 + 4 b a 4 + 4 b 4a4 4 + 4 b 4a4 4 + 4 b 4a4 4 + 6 b a a + 6 b 4a 4 a4 + 6 b 4a 4 a4 + 6 b 4a 4 a4 + 6 c b + 6 c b + 6 c b + 6 c 4b 4 + 6 c 4b 4 + 6 c 4b 4 + 4 b a a + 4 b 4a 4 a 4 + 4 b 4a 4 a 4 + 4 b 4a 4 a 4 + b 4a 4 a 4 a 4 4 c b 4 c b 4 c b + c b b 4 c 4b 4 4 c 4b 4 + c 4b 4 b 4 4 c 4b 4 + c 4b 4 b 4 + c 4b 4 b 4 α 0 c 4 c 4 c 4 c 4 4 + 6 b a a + 6 b 4a 4 a 4 + 6 b 4a 4 a 4 + 6 b 4a 4 a 4 + b 4a 4 a 4 a 4 + b 4a 4 a 4a 4 + 6 c b + 6 c b + c b b + 6 c b c b b + 6 c 4b 4 + c 4b 4 b 4 + 6 c 4b 4 c 4b 4 b 4 + c 4b 4 b 4 + c 4b 4 b 4 + 6 c 4b 4 c 4b 4 b 4 c 4b 4 b 4 + c 4 b 4 b 4 b 4 = 0. There is a great deal of tedious manipulations involved in deriving the above twelve identities with twenty one unknowns. The existence of solution for the above system of equations will be shown in the following text. Letting b = 0 b 4 = 0 c = 0 c = 0 b 4 = b 4 = then from the above system we can obtain b = 6 + 4 c = 6 9 + 4 α = + 4 b = 0 a = 600 757 + 799 4 a = 00 49 4 a = 00 6 + 47 4 b 4 = 0 9 + 4 c 4 = 6 6 9 4.

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 595 Consequently We have the following formula of fifth order with only four function evaluations of f y n+ = + 4 y n + + 4 y n +h{ 6 + 4f n + 6 9 + 4 f n f n Letting + 6 6 9 4 [fy n + 0 9 + 4 hf n + h fy n + 00 6 + 47 4 hf n h fy n + 600 757 + 4 hf n + 00 49 4 hf y n + 00 6 + 47 4hf n fy n + 0 9 + 4 hf n + h fy n + 00 6 + 47 4 hf n h fy n + 600 757 + 4 hf n + 00 49 4 hf y n + 00 6 + 47 4 hf n ]}. a 4 = b 4 a 4 = b 4 = 8 a 4 = b 4 = 8 4.7 b = 0 b 4 = 0 c = 0 α= 0 we can figure out b = b = 0 c = 7+5 6 c = 5 8 008 c 4 = 5 48 a = 97+45 50 a = 989 90755 80850 a = 496 97+45 6795 b = 4+ 0 b 4 = 4+ 0. Therefore we have the corresponding derivative-free formula of fifth order with only five function evaluations per step as follows: y n+ = y n + h{f n + 7+5 6 f n f n + 5 8 008 [fy n + h 4+ 0 f n fy n + h 4+ 0 f n ] + 5 48 [fy n + h 4+ 0 f n h 8 fy n h 97+45 50 f n + h 8 fy 4.8 n + h 989 90755 80850 f n +h 496 97+45 6795 fy n h 97+45 50 f n fy n + h 4+ 0 f n h 8 fy n h 97+45 50 f n + h 8 fy n + h 989 90755 80850 f n +h 496 97+45 6795 fy n h 97+45 50 f n ]}.

596 X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 Likewise we can obtain another derivative-free formula of fifth order with only five function evaluations per step for which it is enough to list the corresponding coefficients α = 0 b = b = 0 b = 0 b 4 = 0 c = 7 5 6 c = 5+8 008 c = 0 c 4 = 5 48 a = 97+45 50 a = 989+90755 80850 a = 49697+45 6795 a 4 = b 4 a 4 = b 4 = 8 a 4 = b 4 = 8 b = 4 0 b 4 = 4 0. 5. Stability analysis The stability regions of the formulae.7 and. using one function evaluation of f are the same as the classical third order and fourth order Runge Kutta methods respectively because they have the same stability polynomials. Formulae.4.5.6.7 and.8 have the same stability polynomial that is Φξμ = ξ + μ + 5 μ ξ + μ + 56 μ where μ = hλ. Let b = bμ = + μ + 5 μ and c = cμ = μ + 56 μ. We have Φξμ = ξ + bξ + c. To find the region of absolute stability we employ Schur criterion see [90]. To do this we define Φξμ = cξ + bξ + Φ ξ μ = Φ0 μφξ μ Φ0μ Φξμ ξ where c and b are the complex conjugates of c and b. Clearly Φ ξ μ has degree at least. Then by a theorem of Schur [90] Φξμ is a Schur polynomial if and only if Φ0μ > Φ0μ and Φ ξ μ is a Schur polynomial. This results in the following conditions c < and b c b < c. Namely we have μ + 5 μ 6 < and

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 597 + μ + 5 μ + μ + 56 μ + μ + 5 μ < 4 μ + 5. 6 μ For formula.8 the stability polynomial is Φξμ = ξ + bμξ + cμ where bμ = + μ + 5 μ + 6 μ cμ= μ + 5 μ + 6 μ. Suppose that ξ and ξ are the zeros of the stability polynomial. From Schur criterion ξ < and ξ < if and only if μ + 5 μ + 6 μ < μ + 5 μ + μ 6 μ + + 5 μ + 6 μ μ + 5 μ + μ 6 < μ + 5 μ + 6 μ. For formula.9 the stability polynomial is Φξμ = ξ + bμξ + cμ where 9 bμ = 5 + 0 μ + 60 μ + 6 μ and cμ = 4 5 + 9 0 μ + 9 60 μ + 6 μ. From Schur criterion ξ < and ξ < if and only if cμ < and b cμ bμ < cμ. Similarly we can discuss the stability polynomial for formulae. 4.7 and 4.8. The stability regions of these formulae.4.5.6.7.8.8.9.0.. 4.7 and 4.8 are sketched in Figs. 5 respectively. The corresponding intervals of absolute stability are listed in Table. The intervals of absolute stability of formula 4.5 is the same as that of the classical Runge Kutta formula of fifth order. 6. Numerical experiments In this section in order to demonstrate the new formulae are of the claimed order several equations have been solved with the new third fourth and fifth order methods on scalar autonomous equations see Table from [6] and system of autonomous equations. For scalar autonomous equations we list in Table the relative errors of the numerical solutions by the formulae.7.4.7..8. 4. 4.7 and 4.8 with different step Table The intervals of absolute stability Formulae.7.4..8.9. 4.7 4.8 Intervals.0 0.6 0.78 0.5 0. 0.4 0. 0.88 0

598 X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 Fig.. The region of absolute stability for formulae.4.5.6.7 and.8. Fig.. The region of absolute stability for formulae.8.0 and.. sizes h = 0.00 0.004 0.008 0.06 and 0.0. For two-step Runge Kutta formulae two starting values y 0 and y are needed. For convenience y is provided by the true solution yh. For an autonomous system of differential equations examples we consider first the Van der Pol equation with parameter λ see [] x λ x x + x = 0 x0 = α x 0 = 0. 6.

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 599 Fig.. The region of absolute stability for formula.9. Fig. 4. The region of absolute stability for formula 4.7. Since it is known that 6. has a unique periodic solution for each λ we consider the problem as that of determining the constant α for which the solution of the initial value problem defined by 6. is periodic. Let T represent the period of the solution then we note that x T = α x T = 0.

600 X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 Fig. 5. The region of absolute stability for formula 4.8. Fig. 6. The region of absolute stability for formula.. Table Test problems Function Solution Initial value y0 y = y y = e t y = y y = t+ y = y 4 y 80 y = 0 +9e t 4

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 60 Table The relative errors of numerical solutions at t = 0 Eq. h.7.4.5.6.7.. 0.00 6.68E 9.67E 8.67E 8 9.0E 8 9.0E 8.68E.5E 0 0.004 5.5E 8.4E 7.06E 7 7.E 7 7.E 7 4.8E.E 9 0.008 4.9E 7.7E 6.4E 6 5.8E 6 5.8E 6 6.87E 0.70E 8 0.06.46E 6.8E 5.8E 6 4.69E 5 4.69E 5.E 8 6.4E 7 0.0.80E 5.E 4.90E 4.8E 4.8E 4.79E 7.E 5 0.00.E.E 0 7.84E.57E 0.99E 0.69E 4 4.95E 0.004.58E 0 9.69E 0 6.7E.05E 9.9E 9.06E.4E 0.008.08E 9 7.74E 9 5.0E 0.6E 8.90E 8 4.94E.54E 0 0.06.68E 8 6.8E 8 4.0E 9.9E 7.5E 7 7.98E 4.9E 9 0.0.6E 7 4.9E 7.E 8.0E 6.9E 6.0E 9 8.0E 8 0.00.9E 7.88E 4.6E 7.48E 7.E.89E 5.E 0.004.54E 6.E.E 5.97E 5.78E.E 5.55E 0.008.E 5.06E.66E 4.77E 0 4.6E 0 4.66E 5 5.80E 0.06 9.85E 4.04E 0.E 0.8E 9.68E 9 6.77E 4 4.0E 0.0 7.88E 0.E 9.75E 0.0E 8.9E 8.07E 7.E Eq. h.8.9.0. 4. 4.7 4.8 0.00.8E 6.77E.8E.8E 5.6E 5.44E 4 8.4E 5 0.004.E 0.09E 9.E 0.E 0.0E 4.E.7E 0.008.55E 9.78E 8.55E 9.55E 9 9.0E 6.8E 5.49E 0.06 5.7E 8.96E 7 5.7E 8 5.7E 8.95E.0E 0.77E 0 0.0 9.5E 7 5.E 6 9.5E 7 9.5E 7 9.58E 0 7.6E 9 5.7E 9 0.00 9.0E 4.E 8.57E 4 8.5E 4 6.6E 6 4.07E 5.05E 5 0.004.4E.8E.4E.4E.76E 4.45E 4 5.09E 6 0.008.8E.98E 0.8E.5E.5E.67E 4.44E 4 0.06.64E 0 4.97E 9.64E 0.4E 0 4.6E 5.7E.4E 0.0 5.78E 9 8.57E 8 5.78E 9 5.4E 9.8E.68E 0 4.68E 0.00.44E 5 5.44E 5 8.0E 6 8.0E 6 8.0E 6 0.00 8.0E 6 0.004.55E 5.68E 4.60E 5.60E 5.8E 5.60E 6.4E 5 0.008.9E 4.0E.7E 4.58E 4.E 5 4.E 5 6.0E 6 0.06 4.40E 4.4E 4.4E 4.E 5.6E 4.40E 5 6.0E 6 0.0 7.07E 6.9E 7.07E 6.78E 8.78E 8.6E 5 8.6E 5 Table 4 The numerical results by formula. for Van der Pol equation λ T x 0 = α h N Numerical x T Theoretical xt 50 4.5.00945 0.00 45.00956.00945 0 9.59.0485 0.00 959.0485.0485 Let y = x and y = x. The second order Van der Pol equation can be transformed into the following autonomous system { y = y y = λy y y 6. with initial values y 0 = α y 0 = 0. For the numerical solutions generated by formula. see Table 4. Secondly we consider the well known nonlinear system of ordinary differential equation { y = 00y + 000y y = y 6. y + y with the initial values y 0 = y 0 =. For the numerical solutions at t = T = 5 and t = T = 50 generated by formula..7 and the classical forth order Runge Kutta method RK see Table 5.

60 X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 Table 5 Numerical results by formulae..7 4.8 and RK for initial value problem 6. Formulae T h N Y Numerical Theoretical Relative error.4 5 0.00 5000 y.9874988e.9874985e.685e 8 y.8879486497e.8879486486e 7.78E 50 0.00 5000 y.700779e 44.7007598E 44 5.08E 7 y.9874984798e.9874985e.8e.7 5 0.00 5000 y.9874968e.9874985e 8.588E 8 y.8879485e.8879486e.04e 9 50 0.00 5000 y.7007e 44.7007598E 44 7.07E 7 y.987498e.9874985e.670e 8 4.8 5 0.00 5000 y.9874985e.9874985e.685e y.88794864964e.88794864964e 5.9E 5 50 0.00 5000 y.70075985e 44.70075985E 44.4E 9 y.987498479687e.98749847969e.5e 4 RK 5 0.00 5000 y.987499e.9874985e.6e 8 y.887949e.887949e 7.78E 50 0.00 5000 y.7007975e 44.7007598E 44.0E 6 y.9874985e.9874985e.8e 7. Error estimates and step-choosing strategies From a content point of view practical error estimates are necessary so that the step size h is chosen sufficiently small to achieve a prescribed tolerance of the local error and the step size h is chosen large enough to avoid unnecessary computational cost. So we should derive suitable error estimates and step-choosing strategies to ensure that the proposed formulae in this paper could be used in an adaptive style. This is straightforward for the extended Runge Kutta-like formulae. such as the special formulae.7 with the local truncation error.8 and.4 with the local truncation error.5 since these formulae are all one-step methods. For automatic step size control of one-step methods it is well known that whenever a starting step size h has been chosen the algorithm computes two steps of size h and one step of size h then an error estimate can be obtained. For a detailed description of this issue see E. Hairer S.P. Norsett and G. Wanner [8]. However for the class of derivative-free formulae the situation is different because these formulae are special two-step methods and require equal step sizes for the starting values any change in step size necessitates recalculating a new starting value at that point. In the special situation this will be done by calling a Runge Kutta sub-algorithm. Obtaining a strategy for error estimation for these two-step methods requires some effort. It must handle a multitude of details not only step size selection and error control but also temporary storage management communication with other programs etc. Let us consider formula pair.7 and.0 developed in Sections and respectively. We will present a procedure which automatically adjusts the step-size in order to achieve a prescribed tolerance of the local error. Since we cannot generally determine the global error of the methods we work instead with the local truncation error of the method. We use.7 a formula of third order to demonstrate a variable-step version of the two step formula. Here we employ formula.0 to estimate the local error of formula.7 We observe that for formula.0 three evaluations of f are required per step but two of them are the same as that of formula.7. Thus a clear advantage to this formula pair is that only three evaluations of f are required per step whereas arbitrary classical Runge Kutta methods of order three and four used together would require seven evaluations of f per step three evaluations for the three-order method and an additional four evaluations for the fourth order method. The local error of formula.7 and.0 can be expressed as d n h = yt n+ y n+ = Ch 4 and ˆd n h = yt n+ ŷ n+ = Ĉh 5

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 60 respectively. In order to estimate the local error of formula.7 we calculate yt n+ y n+ = yt n+ ŷ n+ +ŷ n+ y n+. Since the order of ˆd nh is higher than that of dn h namely the most significant portion of yt n+ y n+ must be attributed to ŷ n+ y n+ this can be reduced to yt n+ y n+ ŷ n+ y n+. 7. Now suppose that a tolerance eps > 0 is prescribed and we want the local error at every step no more than eps namely yt n+ y n+ eps h. If we can check the local error at every step then we can control the global error. With the aid of 7. we can choose h such that ŷ n+ y n+ eps h where eps is a prescribed tolerance. It is followed that yt n+ y n+ = 5 [ kn k n + q n q n p n p n ] h 48 from.7 and.0 under the assumption of y n = yt n and ŷ n = yt n where k i = f i p i = f y i + hf i q i = f y i + 4h5 5 f y i + h 5 f i i = n n are available at every step. For simplicity we only consider the cases where the step-size is double or halved. Suppose that we have y n and y n with step-size h we want to advance the next step. If ŷ n+ y n+ eps h then we accept y n+ and we still use this step-size h. Otherwise h is halved and y n+ is re-calculated. Besides if the estimate of the error is lower than eps to some extent then double the step-size. In fact if y n+ is already accepted and the estimate of the error is lower than eps 6 that is ŷ n+ y n+ < eps h 6 then h is doubled. The reason we choose eps 6 is that formula.0 is a fourth-order formula the error should be 4 = 6 times that of the former. If a step change from h to h is required prior to the above step the approximation of ỹ n+ = yt n + h has to be supplied. Obviously the most straightforward is that we can employ a classical Runge Kutta formula of order three or order four to obtain the approximate of ỹ n+ = yt n + h. Likewise we can consider the adaptive version for the other two formula pairs of.6.8.5. and.8.9 developed in Sections and respectively. To illustrate the technique of the adaptive version we present an example in which the error estimate and stepchanging strategies are used. In practice we should control the halved step-size to ensure that it will not circulate infinitely. That is to say we should prescribed a lower limit for the step-size such as h min.ifh<h min then the program is stopped and minimum h exceeded is printed out. We also prescribe a maximum step-size h max. We solve problem { y = y t [0 0] 7. y0 = by the adaptive version with eps =.e 5 h 0 = 0. h min = 0.00 and h max = 0.. The initial value of h is chosen as h 0 = 0.. We employ the classical Runge Kutta method of order four to supply the starting value y and ỹ n+ whenever a step change from h to h is required. The numerical results are listed in Table 6. It should be noticed that we also try to test the adaptive version for formula pair.7.0 via employing the Runge Kutta Fehlberg method [4] to supply the starting value y and ỹ n+ whenever a step change from h to h is required. The numerical results are almost the same as those in Table 6.

604 X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 Table 6 Numerical results by the adaptive version for initial value problem 7. h i t i y i y i yt i ŷ i y i h i 0.0500 0.500 0.860708056785 8.0045777e 08 9.000000000e 06 0.050 0.750 0.89457099597 7.890508e 08 9.000000000e 06 0.050 0.000 0.887077454680.468805e 08.5566e 06 0.0500 0.500 0.77880080560765.56440e 08 9.000000000e 06 0.050 0.750 0.7595745679.04984e 08 9.000000000e 06 0.050 0.000 0.740889468.85489e 08.00469668e 06 0.0500 0.500 0.70468806475508.496660e 08 9.000000000e 06 0.050 0.750 0.687895450080.4906965e 08 9.000000000e 06 0.050 0.4000 0.670997769 6.87094877e 08.894455e 06 0.0500 0.4500 0.676808799460 6.67797e 08 9.000000000e 06 0.050 0.4750 0.688499446048 6.004576e 08 9.000000000e 06 0.050 0.5000 0.60650558504.008476e 07.640665e 06 0.0500 0.5500 0.5769497567449 9.47059969e 08 9.000000000e 06 0.050 0.5750 0.567047764860 9.0940e 08 9.000000000e 06 0.050 0.6000 0.5488509944.68998856e 07.4855544e 06 0.0500 0.6500 0.504565746748.9959e 07 9.000000000e 06 0.050 0.6750 0.5095604069.605858e 07 9.000000000e 06 0.050 0.7000 0.49658557006.467850754e 07.478809e 06 0.0500 0.7500 0.47664497.8479e 07 9.000000000e 06 0.050 0.7750 0.46070646098.4889790e 07 9.000000000e 06 0.050 0.8000 0.4498808080.676465e 07.59096e 06 0.0500 0.8500 0.4744779607.568799e 07 9.000000000e 06 0.0500 0.9000 0.406569067688 5.95977e 07 9.468470e 06 0.0500 0.9500 0.867400980 9.847494e 07.898860e 06 0.0500.0000 0.67878076095.569489e 06.456e 06 0.0500.0500 0.499606659.6464457e 06.949690e 06 0.0500.000 0.8695806755.95605e 06.74885e 06 0.0500.500 0.664595784.7667e 06.054e 06 0.0500.000 0.09889869.99955e 06.50555e 06 0.0500.500 0.865060.5856468e 06.096779e 06 0.0500.000 0.75909889.75850e 06.048755e 06 0.0500.500 0.59768.899507686e 06 9.94047459e 07 0.0500.4000 0.4659995447.04407e 06 9.44005760e 07 0.0500.4500 0.4567578670.0679e 06 8.97960089e 07 0.0500.5000 0.6946777.850658e 06 8.546795e 07 0.0500.5500 0.4468059.90775555e 06 8.5087754e 07 0.0500.6000 0.0896967679.487869e 06 7.78840e 07 0.0500.6500 0.9046560087.949879e 06 7.5867546e 07 0.0500.7000 0.86800997970.445570e 06 6.990580e 07 0.0500.7500 0.777049855.4449699e 06 6.650479e 07 0.0500.8000 0.659548078.45547808e 06 6.779056e 07 0.000.9000 0.495655067959.04046e 06 9.000000000e 06 0.0500.9500 0.47075464.9608874e 06 9.000000000e 06 0.0500.0000 0.567796.96646649e 06.044654505e 06 0.00000000 8.00000000 0.000000050699.76665579e 0.655088e 0.00000000 8.00000000 0.0000000055.749758e 0.77497e 0.00000000 8.500000000 0.00000000904849.8896500e 0.45069449e 0.00000000 8.700000000 0.0000000074060.5696659e 0.876699e 0.00000000 8.900000000 0.0000000060669.0607984e 0 9.786978e 0.00000000 9.00000000 0.000000004968.084765e 0 7.9540508e 0.00000000 9.00000000 0.00000000406079 8.98595785e 6.504489e 0.00000000 9.500000000 0.0000000068 7.45895e 5.867998e 0.00000000 9.700000000 0.00000000707 6.89975e 4.64464e 0.00000000 9.900000000 0.00000000657 5.5747e.569744e 0.00000000 0.000000000 0.000000000469 4.64657047e.976887e

X. Wu J. Xia / Applied Numerical Mathematics 56 006 584 605 605 8. Conclusion A family of new extended Runge Kutta-like formulae has been presented in this paper. The new formulae exploit the use of first order derivatives f. In particular if f is approximated by the difference quotients of past and current evaluations of f then the two-step Runge Kutta methods have been developed which can be regarded as derivativefree extended Runge Kutta methods. Specifically the proposed new formulae with f such as.7. 4. and 4.4 are more efficient for cases where f is not more expensive to evaluate than f. For example with a linear system of equations y = Ay f = Ay requires the same time to compute as does f = Ay. And the proposed derivativefree formulae such as.4.8.8. 4.7 and 4.8 are more attractive because the use of historical values of f is cheaper than evaluating f. As an example for formula pair.7.0 developed in Sections and respectively we derive suitable error estimate of the local error and give step-choosing strategies therefore the formula.7 can be used in an adaptive style. This procedure consists of using an extended Runge Kutta method with local truncation error of order four with only three evaluations to estimate the local error in a extended Runge Kutta method of order three per step. A clear advantage to this technique is that only three evaluations of f are required per step whereas arbitrary classical Runge Kutta methods of order three and four used together would require six evaluations of f per step. Acknowledgements The authors are grateful to the anonymous referees for helpful and useful comments on the presentation and substance of this paper. References [] P.C. Chakrivat M.S. Kamew Stiffly stable second multi-step methods with higher order and improved stability regions BIT 98 75 8. [] Y.F. Chang G. Corliss ATOMFT: Solving ODEs and DAEs using Taylor series Comput. Math. Appl. 8 994 09. [] W.H. Enright Second derivative multi-step methods for stiff ordinary differential equations SIAM J. Numer. Anal. 974. [4] E. Fehlberg Klassische Runge Kutta Formeln Vierter und niedrigerer Ordnung mit Schrittweiten Kotrolle und ihre Anwendung auf Wärmeleitungs Problems Computing 6 970 6 7. [5] C.W. Gear Numerical Initial Value Problems in Ordinary Differential Equations Prentice-Hall Englewood Cliffs NJ 97. [6] D. Goeken O. Johnson Runge Kutta with higher derivative approximations Appl. Numer. Math. 9 000 49 57. [7] E. Hairer G. Wanner Solving Ordinary Differential Equations II Springer Berlin 99. [8] E. Hairer S.P. Norsett G. Wanner Solving Ordinary Differential Equations I Springer Berlin 987. [9] J.D. Lambert Computational Methods in Ordinary Differential Equations J. Wiley London 974. [0] J.J.H. Miller On the location of zeros of certain classes of polynomials with applications to numerical analysis J. Inst. Math. Appl. 8 97 97 406. [] H.H. Rosenbrock Some general implicit processes for the numerical solution of ordinary differential equations Comput. J. 96 9 0. [] X.Y. Wu A six-order a-stable explicit one-step method for stiff systems Comput. Math. Appl. 5 9 998 59 64. [] H.Y. Xu X.Y. Wu A class of second derivate multi-step methods for solving stiff ODEs Nanjing Daxue Xuebao J. Nanjing Univ. 6 990 75 8.