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Mathematical Problems in Engineering
Volume 2011 (2011), Article ID 810324, 18 pages
http://dx.doi.org/10.1155/2011/810324
Research Article

Solution of Higher-Order ODEs Using Backward Difference Method

Department of Mathematics, Faculty of Science, UPM, Selangor Darul Ehsan, 43400 Serdang, Malaysia

Received 9 November 2010; Revised 25 March 2011; Accepted 13 May 2011

Academic Editor: Francesco Pellicano

Copyright © 2011 Mohamed Bin Suleiman et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The current numerical technique for solving a system of higher-order ordinary differential equations (ODEs) is to reduce it to a system of first-order equations then solving it using first-order ODE methods. Here, we propose a method to solve higher-order ODEs directly. The formulae will be derived in terms of backward difference in a constant stepsize formulation. The method developed will be validated by solving some higher-order ODEs directly with constant stepsize. To simplify the evaluations of the integration coefficients, we find the relationship between various orders. The result presented confirmed our hypothesis.

1. Introduction

Differential equations constantly arise in various branches of science and engineering. Many of these problems are in the form of higher-order ordinary differential equations (ODEs). A few examples where these problems can be found are, in the motion of projectiles, the bending of a thin clamped beam and population growth.

The popular practice for solving a system of higher-order ODEs is by reducing it to a system of first-order equations and then solving with first-order methods. These methods worked, so that methods for solving higher-order ODEs have been disregarded as robust codes. However, the work by Krogh [1], Suleiman [2], Majid and Suleiman [3], and Omar and Suleiman [4] has revived the interest in solving higher-order ODEs directly and the theoretical development of the methods.

Related works for solving higher-order ODEs can be found in Collatz [5], Gear [6], Krogh [1, 7], and Suleiman [2]. Krogh [7] proposed the direct integration (DI) method for nonstiff problems using modified divided difference while Suleiman [2] proposed the DI method using the standard divided difference. In this paper, we will derive the constant stepsize backward difference formulae of solving higher-order ODEs up to third order. The main reason for developing the constant stepsize formulae is that, in developing the theory on convergence and stability, the approach is through constant stepsize formulation. Another reason is that it is possible to use this formula in conjunction with other similar formulae as in Majid and Suleiman [3] to develop a code for variable stepsize and order.

The advantage of such a code is that the integration or differentiation constants are calculated only once at the start of the first step of integration, whereas other formulations calculate the constants at every step.

In this paper, we will focus only on nonstiff ODEs of the form 𝑦(𝑑)𝑌=𝑓𝑥,,(1.1)𝑌(𝑥)=𝑓𝑥,𝑦,𝑦,𝑦,,𝑦(𝑑1),̃𝜂=𝜂,𝜂,𝜂,,𝜂(𝑑1),(1.2) where 𝑌(𝑎)=̃𝜂 in the interval 𝑎𝑥𝑏 and 𝑑 is the order of the ODE.

Without loss of generality, we will be considering the scalar equation in (1.1).

This paper will be organized as follows. First, the integration coefficients of the explicit constant stepsize backward difference formulation of the DI method will be derived. Then, the coefficients of the implicit method are formulated and their relationship with the explicit coefficients is shown. We start the derivation with the coefficients of the first-order system, which is given in Henrici [8]. Next, the second-order coefficients are derived and their relationship with the corresponding first-order coefficients is given, likewise the relationship of the coefficients for the second- and third-order systems. Finally, the method developed using backward difference will be validated numerically.

2. The Formulation of the Predict-Evaluate-Correct-Evaluate (PECE) Multistep Method in Its Backward Difference Form (MSBD) for Nonstiff Higher-Order ODEs

The code developed will be using the PECE mode. The predictor and corrector will have the following form:

predictor:𝑝𝑟𝑦(𝑑𝑡)𝑛+1=𝑡1𝑖=0𝑖𝑦𝑖!𝑛(𝑑𝑡+1)+𝑑𝑘1𝑖=0𝛾(𝑑𝑡),𝑖𝑖𝑓𝑛,𝑡=1,2,,𝑑,(2.1) corrector:𝑦(𝑑𝑡)𝑛+1=𝑡1𝑖=0𝑖𝑦𝑖!𝑛(𝑑𝑡+1)+𝑑𝑘1𝑖=0𝛾(𝑑𝑡),𝑖𝑖𝑓𝑛+1,𝑡=1,2,,𝑑.(2.2) The corrector will be reformulated, so that it will be in terms of the predictor. The reformulated corrector can be written as𝑦(𝑑𝑡)𝑛+1=𝑝𝑟𝑦(𝑑𝑡)𝑛+1+𝛾(𝑑𝑡),𝑘𝑘𝑝𝑟𝑓𝑛+1,𝑡=1,2,,𝑑,(2.3) where 𝑝𝑟𝑓𝑛+1 indicates 𝑓𝑛+1 evaluated using predicted values. The integration coefficient 𝛾(𝑑𝑡),𝑖 and 𝛾(𝑑𝑡),𝑖 will be derived using the method of generating function. Finally, the constant stepsize algorithm will be constructed and validated with some test problems and examples from physical situations.

3. Derivation up to Third-Order Explicit Integration Coefficients

Integrating (1.1) once yields 𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑥𝑛+1𝑥𝑛𝑓𝑥,𝑦,𝑦,𝑦𝑑𝑥.(3.1) Let 𝑃𝑛(𝑥) be the interpolating polynomial which interpolates the 𝑘 values (𝑥𝑛,𝑓𝑛),(𝑥𝑛1,𝑓𝑛1),,(𝑥𝑛𝑘+1,𝑓𝑛𝑘+1), then𝑃𝑛(𝑥)=𝑘1𝑖=0(1)𝑖𝑖𝑠𝑖𝑓𝑛.(3.2) Next, approximating 𝑓 in (3.1) with 𝑃𝑛(𝑥) and letting𝑥=𝑥𝑛+𝑠or𝑠=𝑥𝑥𝑛(3.3) gives us𝑦𝑥𝑛+1𝑥=𝑦𝑛+10𝑘1𝑖=0(1)𝑖𝑖𝑠𝑖𝑓𝑛𝑑𝑠,(3.4) or𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑘1𝑖=0𝛾1,𝑖𝑖𝑓𝑛,(3.5) where𝛾1,𝑖=(1)𝑖10𝑖𝑠𝑑𝑠.(3.6) The generating function 𝐺1(𝑡) for the coefficients 𝛾1,𝑖 is defined as follows:𝐺1(𝑡)=𝑖=0𝛾1,𝑖𝑡𝑖.(3.7) Substituting 𝛾1,𝑖 in (3.6) in 𝐺1(𝑡) gives𝐺1(𝑡)=𝑖=0(𝑡)𝑖10𝑖𝐺𝑠𝑑𝑠,1(𝑡)=10(1𝑡)(𝑠)𝐺𝑑𝑠,1(𝑡)=10𝑒𝑠log(1𝑡)𝑑𝑠(3.8) which leads to𝐺1(𝑡)=(1𝑡)11log(1𝑡)log(1𝑡).(3.9) Equation (3.9) can be written as𝑖=0𝛾1,𝑖𝑡𝑖𝑡log(1𝑡)=(1𝑡)(3.10)or𝛾1,0+𝛾1,1𝑡+𝛾1,2𝑡2+𝛾1,3𝑡3𝑡+𝑡+22+𝑡33+𝑡44+=𝑡1+𝑡+𝑡2+𝑡3+.(3.11) Hence, the coefficients of 𝛾1,𝑘 are given by𝑘𝑖=0𝛾1,𝑖𝛾𝑘𝑖+1=1,1,𝑘=1𝑘1𝑖=0𝛾1,𝑖𝑘𝑖+1𝑘=1,2,,𝛾1,0=1.(3.12)

4. Second-Order ODE Formulae

Integrate (1.1) twice for second-order ODEs where 𝑑=2. Integrating once leads to the same coefficients as given in (3.6). Integrating twice yields𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑦𝑥𝑛+2𝑘𝑖=0𝛾2,𝑖𝑖𝑓𝑛.(4.1) Substituting 𝑥 with 𝑠 gives𝛾2,𝑖=(1)𝑖10(1𝑠)𝑖1!𝑠𝑑𝑠.(4.2) The generating function 𝐺2(𝑡) of the coefficients 𝛾2,𝑖 is defined as follows𝐺2(𝑡)=𝑖=0𝛾2,𝑖𝑡𝑖.(4.3) Substituting (4.2) into 𝐺2(𝑡) above gives𝐺2(𝑡)=10(1𝑠)𝑒1!𝑠log(1𝑡)𝑑𝑠.(4.4) Substituting 𝐺1(𝑡) into (4.4) yields𝐺21(𝑡)=11!log(1𝑡)1!𝐺1(𝑡)log(1𝑡).(4.5) Equation (4.5) can be written as𝑖=0𝛾2,𝑖𝑡𝑖1log(1𝑡)=1!11!𝐺1(𝑡)(4.6)or𝛾2,0+𝛾2,1𝑡+𝛾2,2𝑡2𝑡+𝑡+22+𝑡33=1+𝛾1!1+1!1,0+𝛾1,1𝑡+𝛾1,2𝑡2+.(4.7) Hence the coefficients of 𝛾2,𝑘 in relation to coefficients of the previous order 𝛾1,𝑘 are given by𝑘𝑖=0𝛾2,𝑖𝑘𝑖+1=𝛾1,𝑘+1,𝛾2,0=𝛾1,1,𝛾2,𝑘=𝛾1,𝑘+1𝑘𝑖=0𝛾2,𝑖𝑘𝑖+1,𝑘=1,2,.(4.8)

5. Third-Order Formulae

Next, the case of the third-order ODE where 𝑑=3 will be considered. In the case of 𝑦,𝑦, the corresponding coefficients are 𝛾1,𝑖,𝛾2,𝑖 as in (3.6) and (4.2). For the solution 𝑦(𝑥), integrating three times yields𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑦𝑥𝑛+2𝑦2!𝑥𝑛+𝑥𝑛+1𝑥𝑛𝑥𝑛+1𝑥(2)𝑓(2)!𝑥,𝑦,𝑦,𝑦𝑑𝑥(5.1) or in the backward difference formulation, given by𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑦𝑥𝑛+2𝑦2!𝑥𝑛+3𝑘𝑖=0𝛾3,𝑖𝑖𝑓𝑛,(5.2) where𝛾3,𝑖=(1)𝑖10(1𝑠)2!2𝑖𝑠𝑑𝑠.(5.3) The generating function 𝐺3(𝑡) of the coefficients 𝛾3,𝑖 is defined as follows:𝐺3(𝑡)=𝑖=0𝛾3,𝑖𝑡𝑖.(5.4)Substituting (5.3) into 𝐺3(𝑡) above yields𝐺3(𝑡)=10(1𝑠)2!2𝑒𝑠log(1𝑡)𝑑𝑠.(5.5) As in (4.4), we now substitute 𝐺2(𝑡) in (5.5) which gives𝐺31(𝑡)=12!log(1𝑡)2!𝐺2(𝑡)log(1𝑡).(5.6) Equation (5.6) can be written as𝑖=0𝛾3,𝑖𝑡𝑖1log(1𝑡)=2!12!𝐺2(𝑡)(5.7)

or𝛾3,0+𝛾3,1𝑡+𝛾3,2𝑡2𝑡+𝑡+22+𝑡33=1+𝛾2!1+2!2,0+𝛾2,1𝑡+𝛾2,2𝑡2+.(5.8) Hence, the coefficients of 𝛾3,𝑘 in relation to coefficients of the previous order 𝛾2,𝑘 are given by𝑘𝑖=0𝛾3,𝑖𝑘𝑖+1=𝛾2,𝑘+1,𝛾3,0=𝛾2,1,𝛾3,𝑘=𝛾2,𝑘+1𝑘𝑖=0𝛾3,𝑖𝑘𝑖+1,𝑘=1,2,.(5.9)

6. Derivation up to the Third-Order Implicit Integration Coefficients

Integrating (1.1) once yields 𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑥𝑛+1𝑥𝑛𝑓𝑥,𝑦,𝑦,𝑦𝑑𝑥.(6.1) Let 𝑃𝑛(𝑥) be the interpolating polynomial which interpolates the 𝑘 values (𝑥𝑛,𝑓𝑛),(𝑥𝑛1,𝑓𝑛1),,(𝑥𝑛𝑘+1,𝑓𝑛𝑘+1):𝑃𝑛(𝑥)=𝑘𝑖=0(1)𝑖𝑖𝑠𝑖𝑓𝑛+1.(6.2) As in the previous derivation, we choose𝑥=𝑥𝑛+1+𝑠,or𝑠=𝑥𝑥𝑛+1.(6.3) Replacing 𝑥by 𝑠 yields 𝑦𝑥𝑛+1𝑥=𝑦𝑛+0𝑘1𝑖=0(1)𝑖𝑖𝑠𝑖𝑓𝑛+1𝑑𝑠.(6.4) Simplify 𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑘𝑖=0𝛾1,𝑖𝑖𝑓𝑛,(6.5) where𝛾1,𝑖=(1)𝑖01𝑖𝑠𝑑𝑠.(6.6) The generating function 𝐺1(𝑡) of the coefficients𝛾1,𝑖 is defined as follows:𝐺1(𝑡)=𝑖=0𝛾1,𝑖𝑡𝑖(6.7)

or𝐺1(𝑡)=𝑖=0(𝑡)𝑖01𝑖𝐺𝑠𝑑𝑠,1(𝑡)=01(1𝑡)(𝑠)𝐺𝑑𝑠,1(𝑡)=01𝑒𝑠log(1𝑡)𝑑𝑠(6.8) which leads to𝐺11(𝑡)=log(1𝑡)(1𝑡)log(1𝑡).(6.9) For the case 𝑑=2, the approximate solution of 𝑦 has the form𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑦𝑥𝑛+𝑥𝑛+1𝑥𝑛𝑥𝑛+1𝑥(1)!(1)𝑓𝑥,𝑦,𝑦,𝑦𝑑𝑥.(6.10) The coefficients are given by𝛾2,𝑖=(1)𝑖01(𝑠)𝑖1!𝑠𝑑𝑠,(6.11) where 𝛾2,𝑖 are the coefficients of the backward difference formulation of (6.11) which can be represented by𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑦𝑥𝑛+2𝑘𝑖=0𝛾2,𝑖𝑖𝑓𝑛.(6.12) The generating function 𝐺2(𝑡) for the coefficients 𝛾2,𝑖 is defined as follows:𝐺2(𝑡)=𝑖=0𝛾2,𝑖𝑡𝑖.(6.13) Substituting (6.11) into 𝐺2(𝑡) above,𝐺2(𝑡)=01(𝑠)𝑒1!𝑠log(1𝑡)𝑑𝑠.(6.14) Solving (6.14) with the substitution of (3.8) produces the relationship𝐺21(𝑡)=1!(1𝑡)log(1𝑡)1!𝐺1(𝑡)log(1𝑡).(6.15) Integrating (1.1) thrice yields𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑦𝑥𝑛+2𝑦2!𝑥𝑛+𝑥𝑛+1𝑥𝑛𝑥𝑛+1𝑥(2)𝑓(2)!𝑥,𝑦,𝑦,𝑦𝑑𝑥.(6.16) The coefficients are given by𝛾3,𝑖=(1)𝑖01(𝑠)2!2𝑖𝑠𝑑𝑠,(6.17) where 𝛾3,𝑖 are the coefficients of the backward difference formulation of (6.17) which can be represented by𝑦𝑥𝑛+1𝑥=𝑦𝑛+𝑦𝑥𝑛+2𝑦2!𝑥𝑛+3𝑘𝑖=0𝛾3,𝑖𝑖𝑓𝑛.(6.18)

The generating function 𝐺3(𝑡) of the coefficients 𝛾3,𝑖 is defined as follows:𝐺3(𝑡)=𝑖=0𝛾3,𝑖𝑡𝑖.(6.19)

Substituting (6.17) into 𝐺3(𝑡) above yields𝐺3(𝑡)=01(𝑠)2!2𝑒𝑠log(1𝑡)𝑑𝑠.(6.20) Solving (6.11) with the substitution of (6.20) produces the relationship𝐺31(𝑡)=2!(1𝑡)log(1𝑡)2!𝐺2(𝑡)log(1𝑡).(6.21)

7. The Relationship between the Explicit and Implicit Integration Coefficients

Calculating the integration coefficients directly is time consuming when large numbers of integration are involved. A more efficient way of computing the coefficients is by obtaining a recursive relationship between the coefficients. With this recursive relationship, we are able to obtain the implicit integration coefficient with minimal time consumption. The relationship between the explicit and implicit coefficients is expressed below.

For first-order coefficients,𝐺11(𝑡)=log(1𝑡)1𝑡log(1𝑡).(7.1)

It can be written as𝐺11(𝑡)=(1𝑡)1(1𝑡)log(1𝑡)log(1𝑡).(7.2) By substituting 𝐺11(𝑡)=1(1𝑡)log(1𝑡)log(1𝑡)(7.3) into (7.2), we have 𝐺1(𝑡)=(1𝑡)𝐺1(𝑡),𝑖=0𝛾1,𝑖𝑡𝑖=(1𝑡)𝑖=0𝛾1,𝑖𝑡𝑖.(7.4) Expanding the equation yields𝛾1,0+𝛾1,1𝑡+𝛾1,2𝑡2=1+1+𝑡+𝑡2𝛾+1,0+𝛾1,1𝑡+𝛾1,2𝑡2,𝛾+1,0+𝛾1,1𝑡+𝛾1,2𝑡2+1+𝑡+𝑡2=𝛾+1,0+𝛾1,1𝑡+𝛾1,2𝑡2.+(7.5) This gives the recursive relationship𝑘𝑖=0𝛾1,𝑖=𝛾1,𝑘.(7.6) For second-order coefficient, 𝐺21(𝑡)=11!log(1𝑡)1!𝐺1(𝑡)log(1𝑡).(7.7) It can be written as𝐺2(𝑡)=(1𝑡)11!log(1𝑡)1!𝐺1(𝑡)(1𝑡)log(1𝑡).(7.8) Substituting (7.4) into the equation above gives𝐺2(𝑡)=(1𝑡)11!log(1𝑡)1!(1𝑡)𝐺1(𝑡)(1𝑡)log(1𝑡)(7.9)or𝐺2(𝑡)=(1𝑡)11!log(1𝑡)1!𝐺1(𝑡)log(1𝑡).(7.10) Substituting (4.5) into (7.10) gives𝐺2(𝑡)=(1𝑡)𝐺2(𝑡),𝑖=0𝛾2,𝑖𝑡𝑖=(1𝑡)𝑖=0𝛾2,𝑖𝑡𝑖.(7.11) Expanding the equation, we have𝛾2,0+𝛾2,1𝑡+𝛾2,2𝑡2=1+1+𝑡+𝑡2𝛾+2,0+𝛾2,1𝑡+𝛾2,2𝑡2,𝛾+2,0+𝛾2,1𝑡+𝛾2,2𝑡2+1+𝑡+𝑡2=𝛾+2,0+𝛾2,1𝑡+𝛾2,2𝑡2.+(7.12) The above gives the relationship𝑘𝑖=0𝛾2,𝑖=𝛾2,𝑘.(7.13) For third-order coefficient, we have𝐺31(𝑡)=12!log(1𝑡)2!𝐺2(𝑡)log(1𝑡).(7.14) It can be written as𝐺3(𝑡)=(1𝑡)12!log(1𝑡)2!𝐺2(𝑡)(1𝑡)log(1𝑡).(7.15) Substituting (7.10) into (7.15) gives𝐺3(𝑡)=(1𝑡)12!log(1𝑡)2!(1𝑡)𝐺2(𝑡)(1𝑡)log(1𝑡)(7.16)

or𝐺3(𝑡)=(1𝑡)12!log(1𝑡)2!𝐺2(𝑡)log(1𝑡).(7.17) Substituting 𝐺31(𝑡)=12!log(1𝑡)2!𝐺2(𝑡)log(1𝑡)(7.18)into (6.15) leads to𝐺3(𝑡)=(1𝑡)𝐺3(𝑡),𝑘𝑖=0𝛾3,𝑖𝑡𝑖=(1𝑡)𝑖=0𝛾3,𝑖𝑡𝑖.(7.19) Expanding the equation into,𝛾3,0+𝛾3,1𝑡+𝛾3,2𝑡2=1+1+𝑡+𝑡2𝛾+3,0+𝛾3,1𝑡+𝛾3,2𝑡2,𝛾+3,0+𝛾3,1𝑡+𝛾3,2𝑡2+1+𝑡+𝑡2=𝛾+3,0+𝛾3,1𝑡+𝛾3,2𝑡2+(7.20)

which leads to a recursive relationship 𝑘𝑖=0𝛾3,𝑖=𝛾3,𝑘.(7.21)

Tables 1 and 2 are a few examples of the explicit and implicit integration coefficients.

tab1
Table 1: The explicit integration coefficients for 𝑘 from 0 to 6.
tab2
Table 2: The implicit integration coefficients for 𝑘 from 0 to 6.

8. Numerical Results

For error calculations, we will be using the three error tests, namely, absolute error, relative error, and mixed error tests. The error formula is given by,Error=Max𝑥𝑛𝑦𝑥𝑛𝑦𝑛𝑥𝐴+𝐵𝑦𝑛,(8.1) where 𝐴=1, 𝐵=0 gives the absolute error test, 𝐴=0, 𝐵=1 gives the relative error test, and 𝐴=1, 𝐵=1 gives the mixed error test.

In (8.1), 𝑦(𝑥𝑛) is the exact solution for the problem considered and 𝑦𝑛 the computed solution. In a general code when the exact solution is not available for the relative error, 𝑦(𝑥𝑛) is replaced by 𝑦𝑛 the computed value.

When 𝑦(𝑥𝑛)𝑦𝑛 is small, the error in (8.1) will approximate the absolute error. However, when it is large, the mixed error test will approximate the relative error. The numerical results give the three errors.

The following notations hold MAX ABS: maximum error when using absolute error test, MAX MIX: maximum error when using mixed error test, MAX REL: maximum error when using relative error test, : step size selected.

For the choice of problems to be tested, we choose four linear problems consisting of a second- and a third-order problem. The third problem is a mix system of second- and first-order equations and the fourth problem is a system of three second-order equations. Our reason for choosing the linear problems is that if the formulae are correct, then they should solve linear problems. The choice of system of equations is to raise the degree of difficulty of solving the problems. The rest of the problems are nonlinear, which occur in physical situations. The choices of the physical problems are those with exact solutions known. We give our comments on the numerical results right after the numerical Tables 3, 4, 5, 6, 7, 8, and 9.

tab3
Table 3
tab4
Table 4
tab5
Table 5
tab6
Table 6
tab7
Table 7
tab8
Table 8
tab9
Table 9

Problem 1. 𝑦=2𝑦𝑦,𝑦(0)=0,𝑦(0)=1,0𝑥64.(8.2)

Solution 1. 𝑦=𝑥𝑒𝑥.(8.3)

Source: Krogh [7].

This is a linear equation used by Krogh [7] to test his code. The solution increases exponentially to a maximum value of 𝑦(64)=64𝑒64 which is considered large and therefore not suitable for absolute error test and hence the large values of the error.

Problem 2. 𝑦=2𝑦4,0𝑥30,𝑦(0)=1,𝑦(0)=2,𝑦(0)=6.(8.4)

Solution 2. 𝑦(𝑥)=𝑥2+𝑒2𝑥.(8.5)

Source: Omar and Suleiman [4].

This is a third-order problem with an exponential solution. The difference between Problems 1 and 2 is that one is third order and the other is second order. Again, absolute error test does not work for the same reason given above.

Problem 3. 𝑦1=2𝑦15𝑦2+3,𝑦2=𝑦1+2𝑦2,𝑦0𝑥16𝜋,1(0)=0,𝑦1(0)=0,𝑦2(0)=1.(8.6)

Solution 3. 𝑦1𝑦(𝑥)=2cos𝑥+6sin𝑥26𝑥,2(𝑥)=2cos𝑥+2sin𝑥+3.(8.7)

Source: Bronson [9].

For this problem, all error tests worked well.

Problem 4. 𝑦1=1𝑦2𝑦3,𝑦2=𝑦3𝑦1,𝑦3=𝑦1+𝑦2,𝑦0𝑥4𝜋,1(0)=0,𝑦1(0)=1,𝑦2(0)=0,𝑦2(0)=0,𝑦3(0)=1,𝑦3(0)=1.(8.8)

Solution 4. 𝑦1𝑦(𝑥)=sin𝑥,2𝑦(𝑥)=cos𝑥1,2(𝑥)=sin𝑥cos𝑥.(8.9)

Source: Bronson [9].

This problem does not work for relative error test because of the small value of the solution for certain values of 𝑥.

Problem 5. 𝑦1=𝑥𝑦+1𝑥2𝑦+1𝑥,1𝑥50,𝑦(1)=2621ln2(2)+99104,𝑦(1)=40521ln(2)13,𝑦3(1)=+4267ln(2).(8.10)

Solution 5. 𝑥𝑦(𝑥)=28𝑥2ln2332133+1ln(2)326𝑥21ln2ln(2)+3326.(8.11)

Source: Russel and Shampine [10].

This problem is the symmetrical bending of a laterally loaded circular plate.

The numerical results of this problem show the failure to control the error using relative error test. This is because the solution is zero when 𝑥=2.

Problem 6. 𝑦1𝑦=1𝑟3,𝑦1(0)=1,𝑦1(𝑦0)=0,2𝑦=2𝑟3,𝑦2(0)=0,𝑦2𝑦(0)=1,𝑟=21+𝑦221/2,0𝑥16𝜋.(8.12)

Solution 6. 𝑦1(𝑥)=cos𝑥,𝑦2(𝑥)=sin𝑥.(8.13)

Source: Shampine and Gordon [11].

This problem is Newton’s equations of motion for the two-body problem.

Again, relative error test does not work too well for this problem because 𝑦𝑛 is very small at certain points 𝑥𝑛.

Problem 7. 𝑦=2𝑦3,0𝑥5,𝑦(0)=1,𝑦(0)=1.(8.14)

Solution 7. 1𝑦(𝑥)=𝑥+1.(8.15)

Source: Robert Jr. [12].

For this problem, all error tests worked well.

All the numerical results show that the errors in the mixed error mode give a reliable error estimate for all the problems given. The absolute error mode failed to give meaningful error results for Problems 1 and 2. This is because the value of 𝑦𝑛 increases as 𝑥 increases and this becomes large. Similarly, for Problems 4, 5, and 6, the relative error failed to give an acceptable result because 𝑦𝑛 is small.

The research work done shows that the method developed for solving higher-order ODEs directly using the backward difference is successful. We recommend that, for multistep method, the error control procedure should use the mixed error test. This research suggests the potential of this work developing a robust code for solving higher-order ODEs directly.

Acknowledgments

This paper has been supported by the UPM Graduate Research Fellowship (GRF) and Science Fund.

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