In this paper, a study is made on the possibility of developing a nonlinear four-step method based on contraharmonic mean. The study is done since the four-step methods always give higher order than popular methods like Numerov and classical Runge-Kutta methods. A detailed study of consistency, stability, convergence and interval of periodicity has been done to convince ourselves of using this new method. The numerical results shows that the method is more accurate than the existing one

Phang, Chang

Yaacob, Nazeeruddin

English

Universiti Teknologi Malaysia Institutional Repository

Matematika, 2003, Jilid 19, bil. 1, hlm. 47–56c©Jabatan Matematik, UTM.New Nonlinear Four-Step Method fory′′ = f(t, y)Nazeeruddin Yaacob & Phang ChangDepartment of MathematicsUniversity Teknologi Malaysia81310 Skudai,Johor, MalaysiaAbstractIn this paper, a study is made on the possibility of developing a nonlinearfour-step method based on contraharmonic mean. The study is done since thefour-step methods always give higher order than popular methods like Numerovand classical Runge-Kutta methods. A detailed study of consistency, stability,convergence and interval of periodicity has been done to convince ourselves ofusing this new method. The numerical results shows that the method is moreaccurate than the existing one.Keywords Contraharmonic mean, interval of periodicity, second order initialvalue problem.Abstrak Dalam kertas kerja ini, satu kajian telah dijalankan tentang kemungk-inan untuk membangunkan satu kaedah empat-langkah taklinear berdasarkanmin kontraharmonik. Kajian ini telah dijalankan kerana kaedah empat-langkahselalunya memberikan peringkat yang lebih tinggi daripada kaedah terkenalseperti kaedah Numerov dan kaedah Runge-Kutta klasik. Kajian yang ter-perinci tentang kekonsistenan, kestabilan, penumpuan dan selang berkala telahmeyakinkan kita penggunaan kaedah baru ini. Keputusan berangka menun-jukkan keputusan yang lebih jitu daripada kaedah yang sedia ada.Katakunci Min kontraharmonik, selang berkala, masalah nilai awal peringkatkedua.1 IntroductionFor years, much work has been devoted to developing a better formulae to solve secondorder initial value problems of the type y′′ = f(t, y) with solution which is periodic oroscillating with known frequency. In recent years, multistep methods for this problem have48 Nazeeruddin Yaacob & Phang Changbeen particularly been of concern. It is so because single step methods are inefficient sincethey do not make full use of the available information. It seems plausible that more accuracycan be obtained if the value of yn+1 is made to depend not only on yn , but also on yn−1and yn−2 .The same is true for the more recent method proposed by Simos [4] and Ixaruand Rizea [3].The new nonlinear method is developed here based on some important hypotheses byLambert and Watson [2] and Sommeijer, Van der Houwen and Neta [1].2 Development of Nonlinear Four-Step Method basedon Contraharmonic MeanTo develop a nonlinear four-step method, we use the formula of a linear multistep methodas (1) and make a slightly modification to suit nonlinear four-step method as (2).k∑j=0ajyn+j = h2k∑j=0cjψj(f), k ≤ 2, (1)k∑j=0ajyn+j = h215∑j=0cjψj(f), k = 4, (2)where aj and cj are constants and are the function formed by taking nonlinear combi-nation of f at some fixed discrete points of special second order IVP y′′ = f(t, y) which ischaracterized by the polynomials ρ and σ , whereρ(ζ) =k∑j=0ajξj , (3)σ(ζ) =k∑j=0cjξj . (4)Throughout this paper, we refer to the work of Lambert and Watson [2] for linearmultistep methods as a guide. We shall assume that of nonlinear multistep methods satisfythe following hypotheses:(i) ak = 1, |a0 + a5| 6= 0,∑15j=1 |cj | 6= 0.(ii) ρ and σ have no common factors.(iii) ρ(1) = ρ′(1) = 0, ρ′′(1) = 2σ(1) ; this is necessary and sufficient for the method to beconsistent, that is, to have order at least one.(iv) The method (ρ, σ) is zero-stable; that is, all the roots of ρ lie in or on the unit circle,and those on the unit circle having multiplicity not greater than two.New Nonlinear Four-Step Method for y′′ = f(t, y) 49If we intend to have the four-step formula to be based on contraharmonic mean, thenwe have to form the nonlinear combination as in (5).4∑j=0ajyn+j = h215∑j=1cjψj (5)whereψ1 = fn+4, ψ2 = fn+3, ψ3 = fn+2, ψ4 = fn+1, ψ5 = fn,ψ6 =f2n+4 + f2n+3fn+4 + fn+3, ψ7 =f2n+3 + f2n+2fn+3 + fn+2, ψ8 =f2n+2 + f2n+1fn+2 + fn+1,ψ9 =f2n+1 + f2nfn+1 + fn, ψ10 =f2n+4 + f2nfn+4 + fn, ψ11 =f2n+3 + f2n+1fn+3 + fn+1,ψ12 =f2n+4 + f2n+1fn+4 + fn+1, ψ13 =f2n+4 + f2n+2fn+4 + fn+2, ψ14 =f2n+3 + f2nfn+3 + fn,ψ15 =f2n+2 + f2nfn+2 + fn.To fulfill the hypothesis (i) and properties of symmetric methods by Sommeijer, Vander Houwen and Neta [1] , we shall let the left hand side of equation (5) asyn+4 + q1yn+3 + q2yn+1 + yn.We may then write (5) asyn+4 + q1yn+3 + q2yn+1 + yn = h2(c1fn+4 + c2fn+3 + c3fn+2 + c4fn+1 + c5fn+c6f2n+4 + f2n+3fn+4 + fn+ 3+ c7f2n+3 + f2n+2fn+3 + fn+ 2+ c8f2n+2 + f2n+1fn+2 + fn+ 1+c9f2n+1 + f2nfn+1 + fn+ c10f2n+4 + f2nfn+4 + fn+ c11f2n+3 + f2n+1fn+3 + fn+ 1+c12f2n+4 + f2n+1fn+4 + fn+ 1+ c13f2n+4 + f2n+2fn+4 + fn+ 2+ c14f2n+3 + f2nfn+3 + fn+c15f2n+2 + f2nfn+2 + fn(6)Using Taylor’s series we expand both sides of equation (5), and compare each coefficient.We obtain the system of equations as follows:50 Nazeeruddin Yaacob & Phang ChangUsing MATHEMATICA, we get the solution for system of equations in free parameterq2 as follows:c1 =170001124000000+q2480, c2 =58000096000000− q220, c3 =9249791000000− 97q2240c4 =19332832000000− q220,c5 =169993124000000+q2480, c6 = −1250000, c7 =11000000, c8 =411000000,c9 =162500, c10 =194000000, c11 =1200000, c12 = −7600000, c13 =1100000,c14 = −1200000, c15 = −1100000, and q1 = −1−q22.Thus, we obtain a new method for solving y′′ = f(t, y) . We call it a nonlinear 4-stepmethod based on contraharmonic mean. ( We abreviate it as 4-step CoM method ).yn+4 − (1 +q22)yn+3 + q2yn+2 − (1 +q22)yn+1 + yn= h2(( 170001124000000+q2480)fn+4 + (58000096000000− q220)fn+3 + (9249791000000− 97q2240)fn+2+ (19332832000000−q220)fn+1 + (169993124000000+q2480)fn − (1250000)(f2n+4 + f2n+3fn+4 + fn+3)+ (11000000)(f2n+3 + f2n+2fn+3 + fn+2)+ (411000000)(f2n+2 + f2n+1fn+2 + fn+1)+ (162500)(f2n+1 + f2nfn+1 + fn)+ (194000000)(f2n+4 + f2nfn+4 + fn)+ (1200000)(f2n+3 + f2n+1fn+3 + fn+1)− ( 7600000)(f2n+4 + f2n+1fn+4 + fn+1)+ (1100000)(f2n+4 + f2n+2fn+4 + fn+2)+ (1200000)(f2n+3 + f2nfn+3 + fn)− ( 1100000)(f2n+2 + f2nfn+2 + fn)) (7)New Nonlinear Four-Step Method for y′′ = f(t, y) 513 Determination of Interval of Periodicity for 4-stepCoM MethodThe first and the second characteristic polynomial for 4-step CoM method may be writtenas followsρ(ζ) = ζ4 −(1 +q22)ζ3 + q2ζ2 −(1 +q22)ζ + 1 (8)σ(ζ) =(170001124000000+q2480)ζ4 +(58000096000000− q220)ζ3 +(9249791000000− 97q2240)ζ2+(19332832000000−q220)ζ +(169993124000000+q2480)−(1250000)(ζ8 + ζ6ζ4 + ζ3)+(11000000)(ζ6 + ζ4ζ3 + ζ2)+(411000000)(ζ4 + ζ2ζ2 + ζ)+(162500)(ζ2 + 1ζ + 1)+(194000000)(ζ8 + 1ζ4 + 1)+(1200000)(ζ6 + ζ2ζ3 + ζ)−(7600000)(ζ8 + ζ2ζ4 + ζ)+(1100000)(ζ8 + ζ4ζ4 + ζ2)+(1200000)(ζ6 + 1ζ3 + 1)−(1100000)(ζ4 + 1ζ2 + 1)(9)Next, we consider the equationφ(ζ;H2) = ρ(ζ) +H2σ(ζ). (10)The objective is to find H20 such that |ζj | = 1 where ζj are zeroes of the equation φ(ζ;H2) =0 , and H2 ∈ (0, H20 ) . This is done by fixing some values of H2 which will give values ofφ = 0 in the equation (10). We do this until the required H20 is found. We present some ofthe values obtained in Table 1 using q2 = 2 (where the method is consistent).From the Table 1, it is observed that the value of H20 is approximately 0.55. This is truefor H2 ∈ (0, 0.55) , we have |ζj | = 1 . Hence, the interval of periodicity for 4-step CoMmethod is (0,0.55).4 Numerical ResultsWe shall test the new method using two different values of free parameter for two differentproblems. We compare our results to those of Method V developed by Lambert and Watson[2] given as follows:yn+4 − (2 + q)yn+3 + (2 + 2q)yn+2 − (2 + 2q)yn+1 + yn= h2[(18 + q24)fn+4 +(26 + 3q30)fn+3 +(14− 97q120)fn+2+(26− 3q30)fn+1 +(18 + q240)fn](11)52 Nazeeruddin Yaacob & Phang ChangTable 1: Determination of H20H2 ζj |ζj |0.50 −0.999998 0.999998−0.707107± 0.707106i 1.−0.504766± 0.863261i 1.−6.93165± 0.999997i 0.9999970.500002±0.866025i 1.0.707106±0.707107i 1.0.877584±0.479422i 1.0.55 −0.999997 0.999997−0.707107± 0.707106i 1.−0.505909± 0.862593i 1.−8.58288± 0.999997i 0.9999970.500002± 0.866025i 1.0.707106± 0.707107i 1.0.852528± 0.522681i 1.0.56 −0.999997 0.999997−0.707107± 0.707106i 1.−0.506155± 0.862449i 1.000001−8.94191± 0.999996i 0.9999960.500002± 0.866025i 1.0.707106± 0.707108i 1.0.84726± 0.531179i 1.New Nonlinear Four-Step Method for y′′ = f(t, y) 53Problem 1 Consider the problem of solvingy′′ = −4y, y(0) = 0, y′(0) = 2where the exact solution of this problem is y(t) = sin(2t).Using the 4-step CoM formula as in (7), we solve Problem 1 using q2 = 1/1000 withstep size h = 0.1. The absolute errors for the numerical solution near the zeroes of y(t) aregiven in Table 2.Table 2: Absolute errors in Method V and 4-step CoM methodusing step size h = 0.1 for Problem 1t Exact solution Absolute errors Absolute errorsy(t) near zero (Method V) (4-step CoM method)pi −8.3089402817e-02 4.6211810859e-07 6.5705635603e-064pi 6.7208072525e-02 1.8615250752e-06 2.1496595322e-0716pi 7.2493920780e-02 7.6815485629e-06 3.5145162423e-0625pi −7.9548542875e-02 1.1634989843e-05 6.0222911229e-0632pi −6.1890250722e-02 1.4912883813e-05 4.4873629581e-0736pi 5.3289163091e-03 1.6806657479e-05 4.1541678249e-0640pi 7.2523985782e-02 1.8628661728e-05 1.8771784017e-0650pi 4.0723376706e-02 2.3326315267e-05 8.2667880686e-06Problem 2: Undamped Duffing problemWe use 4-step CoM method with q2 = 2 and Method V to find the numerical solutionof the following initial value problem, which is the nonlinear undamped Duffing equationy′′ + y + y3 = B cosΩtforced by a harmonic function where B = 0.002 and Ω = 1.01 . The exact solution computedby the Galerkin method with a precision 10−12 of the coefficients is given byy(t) = A1 cosΩt+A3 cos 3Ωt+A5 cos 5Ωt+A7 cos 7Ωt+A9 cos 9ΩtwhereA1 = 0.200179477536,A3 = 0.000246946143,A5 = 0.000000304014,A7 = 0.000000000374,A9 = 0.000000000000.(This problem are also solved by U.Anantha Krishnaiah [6]). We present the results inTable 3.54 Nazeeruddin Yaacob & Phang ChangTable 3: Absolute error in Method V and 4-step CoM methodusing step size h = pi/10 for Problem 2t Exact solution Absolute errors Absolute errorsnear zero (Method V) (4-step CoM method)4pi 1.9883085347e-01 3.0174707690e-03 9.0133968507e-044.2pi 1.4489947717e-01 3.1834901583e-03 8.4771739226e-034.8pi −1.7782049510e-01 4.0003985992e-03 7.7119748608e-047.9pi 1.9998640851e-01 4.6872008091e-03 1.4612857199e-0311.8pi 1.9375073508e-01 5.9496054289e-03 4.8996022141e-0312.7pi −1.7131986058e-01 5.4529363933e-03 6.6239472623e-0315.8pi 1.9866753458e-01 1.0080427819e-02 1.1368402356e-0326.1pi 8.4424476916e-02 2.0912004348e-02 4.6788378483e-0327.4pi 1.0381052847e-01 2.2980506875e-02 2.1040142289e-055 ConsistencyThe 4-step CoM method is said to be consistent if and only if ρ′(1) = ρ(1) = 0 andρ′′(1) = 2σ(1).Proof:ρ(ζ) = ζ4 − (1 + q22)ζ3 + q2ζ2 − (1 +q22)ζ + 1=⇒ ρ(1) = (1)4 − (1 + q22)(1)3 + q2(1)2 − (1 +q22)(1) + 1 = 0and ρ′(ζ) = 4ζ3 − 3(1 +q22)ζ2 + 2q2ζ − (1 +q22)ρ′(1) = 4(1)3 − 3(1 +q22)(1)2 + 2q2(1)− (1 +q22) = 0.So ρ′(1) = ρ(1) = 0.Next, we haveρ′′(ζ) = 12ζ2 − 6(1 +q22)ζ + 2q2=⇒ ρ′′(1) = 12(1)2 − 6(1 + q22)(1) + 2q2 = 4− 2q2New Nonlinear Four-Step Method for y′′ = f(t, y) 55andσ(1) =(170001124000000+q2480)(1)4 +(58000096000000− q220)(1)3 +(9249791000000− 97q2240)(1)2+(19332832000000− q220)(1) +(169993124000000+q2480)−(1250000)((1)8 + (1)6(1)4 + (1)3)+(11000000)((1)6 + (1)4(1)3 + (1)2)+(411000000)((1)4 + (1)2(1)2 + (1))+(162500)((1)2 + 1(1) + 1)+(194000000)((1)8 + 1(1)4 + 1)+(1200000)((1)6 + (1)2(1)3 + 1)−(7600000)((1)8 + (1)2(1)4 + 1)+(1100000)((1)8 + (1)4(1)4 + (1)2)−(1200000)((1)6 + 1(1)3 + 1)−(1100000)((1)4 + 1(1)2 + 1))= 2− q2 (12)Since ρ′′(1) = 2σ(1) , the 4-step CoM method is consistent.6 Zero-stabilityThe multistep method is said to be zero-stable if no root of the first characteristic polynomialρ(ζ) has modulus greater than one, and every root of modulus one has multiplicity notgreater than two.Proof:Sinceρ(ζ) = ζ4 − 2ζ3 + 2ζ2 − 2ζ + 1 = 0,we get ζ = i,−i, 1 and 1. Thus |ζ| = 1. So(i) No modulus of root > 1, and(ii) Every root of modulus one has multiplicity exactly 2.Hence the 4-step CoM method is zero-stable.7 ConvergenceThe necessary and sufficient conditions for a multistep method to be convergent are that itbe consistent and zero-stable.Proof:The nonlinear multistep method based on contraharmonic mean is shown to be(i) consistent, and(ii) zero-stable.Thus, the 4-step CoM method is convergent.56 Nazeeruddin Yaacob & Phang Chang8 ConclusionIn this paper, we have pioneered in using nonlinear mean to develop the nonlinear 4-stepmethod for y′′ = f(t, y). Finally, we shall determine the agreement of the hypothesesmentioned in Section 1.(i) ak = 1, |a0|+ |c5| 6= 0,∑15j=1 |cj | 6= 0(ii) ρ and σ have no common factors.(iii) ρ(1) = ρ′(1) = 0, ρ′′(1) = 2σ(1);This is a necessary and sufficient conditions for the method (ρ, σ) to be consistent,that is, to have order at least one.(iv) The method (ρ, σ) is zero-stable;That is all the roots of ρ lie in or on the unit circle, and those on the unit circle havingmultiplicity not greater than two.Proof:(i) a4 = 1, |a0| = 1, c5 =169993124000000+q2480.So |a0|+ |c5| 6= 0 andσ(1) = ( 170001124000000+2480) + (58000096000000− 220) + (9249791000000− 97(2)240)+ (19332832000000− 220) + (169993124000000+2480)− 1250000+11000000+411000000+162500+194000000+1200000−7600000+1100000− 1200000− 1100000= 2 6= 0(ii), (iii) and (iv) are verified in Sections 5 to 7.Furthermore, from the Table 2 and 3, we notice that this new nonlinear four-step methodis generally more accurate than the Method V.References[1] B. P. Sommeijer, P. J. Van der Houwen and B. Neta, Symmetric Linear MultistepMethods for Second Order Differential Equations with Periodic Solutions, Journal onApplied Numerical Mathematics, 2(1986), 69-77.[2] J. D. Lambert and I.A. Watson, Symmetric Multistep Methods for Periodic InitialValue Problems, J. Inst. Math Applics., 18 (1976), 189-202.New Nonlinear Four-Step Method for y′′ = f(t, y) 57[3] L.G. Ixaru and M. Rizea, Four Step Methods for y′′ = f(x, y), Journal of Computa-tional and Applied Mathematics, 79 (1997), 87-99.[4] T.E. Simos, An Explicit Four-Step Phase-Fitted Method for the Numerical Integrationof Second-Order Periodic Initial Value Problems, J. Comput. Appl. Math., 58 (1995),125-136.[5] N.Yaacob, New Nonlinear Runge-Kutta Methods for Solving Initial Value Problems,PhD. Thesis, UTM, (1996).[6] U.Anantha Krishnaiah, Adaptive Methods for Periodic Initial Value Problem of SecondOrder Differential Equations, Journal of Computational and Applied Mathematics, 8(1982), 101-104.

New nonlinear four-step method for y"=f(t,y)

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New nonlinear four-step method for y"=f(t,y)

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