Research Article | Open Access

Volume 2011 |Article ID 810324 | https://doi.org/10.1155/2011/810324

Mohamed Bin Suleiman, Zarina Bibi Binti Ibrahim, Ahmad Fadly Nurullah Bin Rasedee, "Solution of Higher-Order ODEs Using Backward Difference Method", Mathematical Problems in Engineering, vol. 2011, Article ID 810324, 18 pages, 2011. https://doi.org/10.1155/2011/810324

# Solution of Higher-Order ODEs Using Backward Difference Method

Revised25 Mar 2011
Accepted13 May 2011
Published12 Jul 2011

#### 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 , Suleiman , Majid and Suleiman , and Omar and Suleiman  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 , Gear , Krogh [1, 7], and Suleiman . Krogh  proposed the direct integration (DI) method for nonstiff problems using modified divided difference while Suleiman  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  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 where in the interval and 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 . 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: corrector: The corrector will be reformulated, so that it will be in terms of the predictor. The reformulated corrector can be written as where indicates evaluated using predicted values. The integration coefficient and 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 Let be the interpolating polynomial which interpolates the values , then Next, approximating in (3.1) with and letting gives us or where The generating function for the coefficients is defined as follows: Substituting in (3.6) in gives which leads to Equation (3.9) can be written asor Hence, the coefficients of are given by

#### 4. Second-Order ODE Formulae

Integrate (1.1) twice for second-order ODEs where . Integrating once leads to the same coefficients as given in (3.6). Integrating twice yields Substituting with gives The generating function of the coefficients is defined as follows Substituting (4.2) into above gives Substituting into (4.4) yields Equation (4.5) can be written asor Hence the coefficients of in relation to coefficients of the previous order are given by

#### 5. Third-Order Formulae

Next, the case of the third-order ODE where will be considered. In the case of , the corresponding coefficients are as in (3.6) and (4.2). For the solution , integrating three times yields or in the backward difference formulation, given by where The generating function of the coefficients is defined as follows:Substituting (5.3) into above yields As in (4.4), we now substitute in (5.5) which gives Equation (5.6) can be written as

or Hence, the coefficients of in relation to coefficients of the previous order are given by

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

Integrating (1.1) once yields Let be the interpolating polynomial which interpolates the values : As in the previous derivation, we choose Replacing by yields Simplify where The generating function of the coefficients is defined as follows:

or which leads to For the case , the approximate solution of has the form The coefficients are given by where are the coefficients of the backward difference formulation of (6.11) which can be represented by The generating function for the coefficients is defined as follows: Substituting (6.11) into above, Solving (6.14) with the substitution of (3.8) produces the relationship Integrating (1.1) thrice yields The coefficients are given by where are the coefficients of the backward difference formulation of (6.17) which can be represented by

The generating function of the coefficients is defined as follows:

Substituting (6.17) into above yields Solving (6.11) with the substitution of (6.20) produces the relationship

#### 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,

It can be written as By substituting into (7.2), we have Expanding the equation yields This gives the recursive relationship For second-order coefficient, It can be written as Substituting (7.4) into the equation above givesor Substituting (4.5) into (7.10) gives Expanding the equation, we have The above gives the relationship For third-order coefficient, we have It can be written as Substituting (7.10) into (7.15) gives

or Substituting into (6.15) leads to Expanding the equation into,

which leads to a recursive relationship

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

 𝑘 0 1 2 3 4 5 6 𝛾 1 , 𝑘 1 1/2 5/12 3/8 251/720 95/288 19087/60480 𝛾 2 , 𝑘 1/2 1/6 1/8 19/180 3/32 863/10080 275/3456 𝛾 3 , 𝑘 1/6 1/24 7/240 17/720 41/2016 731/40320 8563/518400
 𝑘 0 1 2 3 4 5 6 𝛾 ∗ 1 , 𝑘 1 −1/2 −1/12 −1/24 −19/720 −3/160 −813/60480 𝛾 ∗ 2 , 𝑘 1/2 −1/3 −1/24 −7/360 −17/1440 −41/5040 −731/120960 𝛾 ∗ 3 , 𝑘 1/6 −1/8 −1/80 −1/180 −11/3360 −89/40320 −5849/3628800

#### 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, where , gives the absolute error test, , gives the relative error test, and , gives the mixed error test.

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

When 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.

 ℎ MAX ABS MAX MIX MAX REL 1 0 − 1 4.73734 + 026 1.18857 − 003 1.18857 – 003 1 0 − 2 4.70433 + 023 1.16697 − 006 1.16697 – 006 1 0 − 3 4.42360 + 020 1.18335 − 009 1.18335 − 009 1 0 − 4 4.05706 + 020 1.01668 − 009 1.01668 − 009 1 0 − 5 3.69529 + 021 9.26023 − 009 9.26024 − 009
 ℎ MAX ABS MAX MIX MAX REL 1 0 − 1 1.60072 + 023 1.40364 − 003 1.40364 − 003 1 0 − 2 1.52595 + 020 1.33620 − 006 1.33620 − 006 1 0 − 3 1.55425 + 017 1.36098 − 009 1.36098 − 009 1 0 − 4 1.27710 + 016 1.11807 − 010 1.11807 − 010 1 0 − 5 2.20767 + 017 1.93311 − 009 1.93311 − 009
 ℎ MAX ABS MAX MIX MAX REL 1 0 − 1 1.74071 − 002 1.68285 − 003 6.12179 − 003 1 0 − 2 2.32581 − 005 4.04838 − 006 5.34483 − 005 1 0 − 3 2.38274 − 008 4.39132 − 009 5.74568 − 007 1 0 − 4 5.33390 − 009 5.47762 − 010 9.42528 − 009 1 0 − 5 4.54132 − 008 4.54132 − 008 5.25218 − 007
 ℎ MAX ABS MAX MIX MAX REL 1 0 − 1 1.40347 + 001 1.55464 + 000 6.27279 + 002 1 0 − 2 1.78122 − 002 1.75004 − 002 1.60254 + 001 1 0 − 3 2.32649 − 005 2.32643 − 005 9.97054 − 001 1 0 − 4 2.44607 − 008 2.44600 − 008 9.82644 − 001 1 0 − 5 9.90539 − 010 9.90531 − 010 9.99724 − 001
 ℎ MAX ABS MAX MIX MAX REL 1 0 − 1 2.34537 − 001 4.96059 − 004 1.00000 + 000 1 0 − 2 2.85286 − 004 2.86491 − 007 1.00000 + 000 1 0 − 3 2.91084 − 007 6.19617 − 010 1.00083 + 000 1 0 − 4 3.00856 − 008 7.74497 − 013 1.80152 + 000 1 0 − 5 2.43305 − 007 1.09024 − 011 7.08729 + 000
 ℎ MAX ABS MAX MIX MAX REL 1 0 − 1 1.01336 − 004 8.29238 − 005 5.15346 − 002 1 0 − 2 8.33352 − 008 8.33254 − 008 9.08294 − 004 1 0 − 3 9.71863 − 011 9.71317 − 011 4.58972 − 006 1 0 − 4 5.45085 − 010 5.45075 − 010 3.45420 − 004 1 0 − 5 4.75544 − 009 4.75543 − 009 1.91598 − 002
 ℎ MAX ABS MAX MIX MAX REL 1 0 − 1 6.95439 − 002 5.63803 − 002 2.97860 − 001 1 0 − 2 8.31669 − 005 7.12977 − 005 4.99583 − 004 1 0 − 3 8.60027 − 008 7.37166 − 008 5.16016 − 007 1 0 − 4 8.66052 − 011 7.42330 − 011 5.19631 − 010 1 0 − 5 1.86645 − 012 1.55538 − 012 9.33222 − 012

Problem 1.

Solution 1.

Source: Krogh .

This is a linear equation used by Krogh  to test his code. The solution increases exponentially to a maximum value of which is considered large and therefore not suitable for absolute error test and hence the large values of the error.

Problem 2.

Solution 2.

Source: Omar and Suleiman .

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.

Solution 3.

Source: Bronson .

For this problem, all error tests worked well.

Problem 4.

Solution 4.

Source: Bronson .

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

Problem 5.

Solution 5.

Source: Russel and Shampine .

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 .

Problem 6.

Solution 6.

Source: Shampine and Gordon .

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 is very small at certain points .

Problem 7.

Solution 7.

Source: Robert Jr. .

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 increases as increases and this becomes large. Similarly, for Problems 4, 5, and 6, the relative error failed to give an acceptable result because 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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