Project on Runge Kutta method 1. INTRODUCTION 3 1.4 Runge Kutta Order 1 Method We consider the Runge Kutta order 1 method here because it is very short and easy. There is …

Project on ? Runge Kutta method

Nguyen Quan Ba Hong1

Doan Tran Nguyen Tung2

Nguyen An Thinh3

Students at Faculty of Math and Computer Science

Ho Chi Minh University of Science, Vietnam

email. [email protected]

email. [email protected]

blog. http://hongnguyenquanba.wordpress.com 4

October 2, 2016

1Student ID. 14111032Student ID. 14113523Student ID. 14112894Copyright c© 2016 by Nguyen Quan Ba Hong, Student at Ho Chi Minh University

of Science, Vietnam. This document may be copied freely for the purposes of educationand non-commercial research. Visit my site http://hongnguyenquanba.wordpress.

com to get more.

http://hongnguyenquanba.wordpress.com



Contents

List of Figures iii

1 Introduction 11.1 Initial Value Problems . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Mesh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 General Runge Kutta method . . . . . . . . . . . . . . . . . . . . 21.4 Runge Kutta Order 1 Method . . . . . . . . . . . . . . . . . . . . 3

2 Runge Kutta order 2 method 42.1 Derivation of Runge Kutta Order 2 Method . . . . . . . . . . . . 4

2.1.1 Runge Kutta Order 2 Formula . . . . . . . . . . . . . . . 42.1.2 Taylor Series Expansion Formula . . . . . . . . . . . . . . 52.1.3 Derivation of System of Equations . . . . . . . . . . . . . 52.1.4 Solutions of System of Equations . . . . . . . . . . . . . . 62.1.5 Some Cases . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Matlab Implementation . . . . . . . . . . . . . . . . . . . . . . 7

3 Runge Kutta order 3 method 83.1 Derivation of Runge Kutta Order 3 Method . . . . . . . . . . . . 8


3.2 Matlab Implementation . . . . . . . . . . . . . . . . . . . . . . 19

4 Runge Kutta Order 4 Method 204.1 Derivation of Runge Kutta Order 4 Method . . . . . . . . . . . . 20


4.2 Matlab Implementation . . . . . . . . . . . . . . . . . . . . . . 334.2.1 Curtiss-Hirchfelder Equation . . . . . . . . . . . . . . . . 33

4.2.1.1 Subroutine s.m . . . . . . . . . . . . . . . . . . . 334.2.1.2 Subroutine f.m . . . . . . . . . . . . . . . . . . . 334.2.1.3 Subroutine x.m . . . . . . . . . . . . . . . . . . . 34

i

CONTENTS ii

4.2.1.4 Main Routine RK4 . . . . . . . . . . . . . . . . 344.2.1.5 Results . . . . . . . . . . . . . . . . . . . . . . . 36

4.2.2 Brusselator Equation . . . . . . . . . . . . . . . . . . . . . 374.2.2.1 Subroutine f.m . . . . . . . . . . . . . . . . . . . 374.2.2.2 Subroutine x.m . . . . . . . . . . . . . . . . . . . 384.2.2.3 Main Routine RK4.m . . . . . . . . . . . . . . . 384.2.2.4 Results . . . . . . . . . . . . . . . . . . . . . . . 43

4.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.2.3.1 Subroutine s2.m . . . . . . . . . . . . . . . . . . 444.2.3.2 Subroutine f2.m . . . . . . . . . . . . . . . . . . 444.2.3.3 Subroutine x2.m . . . . . . . . . . . . . . . . . . 444.2.3.4 Main Routine RK4ex2.m . . . . . . . . . . . . . 444.2.3.5 Results . . . . . . . . . . . . . . . . . . . . . . . 50

List of Figures

4.1 Numerical Solutions. . . . . . . . . . . . . . . . . . . . . . . 364.2 Absolute Errors. . . . . . . . . . . . . . . . . . . . . . . . . . 374.3 Relative Errors. . . . . . . . . . . . . . . . . . . . . . . . . . 384.4 Numerical Solutions. . . . . . . . . . . . . . . . . . . . . . . 434.5 Dependency of y2 with respect to y1. . . . . . . . . . . . . 444.6 Absolute Errors. . . . . . . . . . . . . . . . . . . . . . . . . . 454.7 Relative Errors. . . . . . . . . . . . . . . . . . . . . . . . . . 464.8 Numerical Solutions. . . . . . . . . . . . . . . . . . . . . . . 504.9 Numerical Solutions. . . . . . . . . . . . . . . . . . . . . . . 514.10 Dependency of y2 with respect to y1. . . . . . . . . . . . . 514.11 Absolute Errors. . . . . . . . . . . . . . . . . . . . . . . . . . 524.12 Relative Errors. . . . . . . . . . . . . . . . . . . . . . . . . . 52

iii

Chapter 1

Introduction

1.1 Initial Value Problems

The Runge Kutta methods are an important family of iterative methods for theapproximation of solutions of ODE’s, that were developed around 1900 by theGerman mathematicians Carl Runge (1856-1927) and Martin. W. Kutta (1867-1944). Modern developments are mostly due to John Butcher in the 1960s.

We consider initial value problems expressed in autonomous form. Startingwith the non-autonomous, we assume that f (x, y) is a continuous function withdomain [a, b]× Rn where t ∈ [a, b] and y ∈ Rn.

Consider the initial value problem

dy (t)

dt= f (t, y (t)) (1.1)

y (x0) = y0 (1.2)

where

y (t) = (y1 (t) , y2 (t) , . . . , yn (t))T

(1.3)

f : [a, b]× Rn → Rn (1.4)

We assume that

‖f (t, y1)− f (t, y2)‖L2(Rn) ≤ L‖y1 − y2‖L2(Rn) (1.5)

for all t ∈ [a, b] , y1 ∈ Rn, y2 ∈ Rn.Thus the IVP (1.1) has a unique solution.For convenience, we write (1.1) briefly as

yt = f (1.6)

Most efforts to increases the order of the Runge Kutta methods have beenaccomplished by increasing the number of Taylor’s series terms used and thusthe number of functional evaluations, e.g. Butcher 1987, Gear 1971. The use ofhigher order derivative terms has been proposed for stiff problems, e.g. Rosen-brock 1963, Enright 1974. Our method add higher order derivative terms to theRunge Kutta ki terms (i > 1) to achieve a higher order of accuracy.

We are interested in a numerical approximation of the continuously differ-entiable solution y (t) of the IVP (1.1) over the time interval t ∈ [a, b].

1

CHAPTER 1. INTRODUCTION 2

1.2 Mesh

We subdivide the interval [a, b] into M equal subintervals and select the meshpoints tj

tj = a+ jh, j = 0, 1, . . . ,M (1.7)

where

h =b− aM

(1.8)

is called a step size.

1.3 General Runge Kutta method

The family of explicit Runge Kutta (RK) methods of the mth stage is given by

y(n+1) = y(n) + h

s∑i=1

biki (1.9)

where

ki = f (τi, ηi) , i = 1, 2, . . . , s (1.10)

and

τi = tn + cih (1.11)

ηi = yn + h

i−1∑j=1

aijkj (1.12)

= yn + h

i−1∑j=1

aijf (τj , ηj) (1.13)

We use the notation

fn := f (tn, yn) (1.14)

To specify a particular method, we need to provide the integer s (the numberof stages), and the coefficients ci, i = 2, . . . , s, aij , 1 ≤ j < i ≤ m and bi, i =1, 2, . . . , s.

These data are usually arranged in a so-called Butcher tableau (after JohnC. Butcher)

0c2 a21c3 a31 a32...

......

. . .

cs as1 as2 · · · ass−1b1 b2 · · · bs−1 bs

(1.15)

CHAPTER 1. INTRODUCTION 3

1.4 Runge Kutta Order 1 Method

We consider the Runge Kutta order 1 method here because it is very short andeasy. There is no need to represent this into a separated chapter.

For s = 1, (1.10) becomes

k1 = f (tn, xn) (1.16)

and (1.10) becomes

y(n+1) = y(n) + hb1k1 (1.17)

= y(n) + hb1fn (1.18)

On the other hand, the Taylor expansion yields

y(n+1) = y(n) + hyt|tn +O(h2)

(1.19)

= y(n) + hfn +O(h2)

(1.20)

Comparing (1.18) and (1.20), we obtain easily

b1 = 1 (1.21)

Hence, The Butcher table in this case has the following form

01

(1.22)

Remark 1.1. The Runge Kutta order 1 method is equivalent to the explicitEuler’s method. Note that the Euler’s method is of the first order of accuracy.Hence, we get the name Runge Kutta method of the ‘first order’ as above.

Chapter 2

Runge Kutta order 2method

2.1 Derivation of Runge Kutta Order 2 Method

To set up the Runge Kutta order 2 method, we need to do 4 steps.

1. Write down Runge Kutta order 2 formula described by (1.9) and (1.10).

2. Write down Taylor series expansion.

3. Compare the coefficients of two formulas above to obtain a system ofequations.

4. Solve the system of equation or find some its solutions.

Remark 2.1. This process is also applied for Runge Kutta method of higherorders. Therefore, we can regard it as the standard process for derivation of thegeneral Runge Kutta method in this context.

After these steps, with some solutions of the derived system of equations,we can simulate some IVPs easily.

2.1.1 Runge Kutta Order 2 Formula


k1 = fn (2.1)

k2 = f (tn + c2h, yn + ha21fn) (2.2)

and (1.9) becomes

y(n+1) = y(n) + hb1k1 + hb2k2 (2.3)

= y(n) + hb1fn + hb2f (tn + c2h, yn + ha21fn) (2.4)

Now we write down the Taylor series expansion O(h2)

for k2

k2 = f (tn + c2h, yn + ha21fn) (2.5)

= fn + c2hft + ha21fnfy +O(h2)

(2.6)

4

CHAPTER 2. RUNGE KUTTA ORDER 2 METHOD 5

Inserting (2.6) into (2.4), we obtain

y(n+1) = y(n) + hb1fn + hb2[fn + c2hft + ha21fnfy +O

(h2)]

(2.7)

= y(n) + h (b1 + b2) fn + h2b2c2ft + h2b2a21fnfy +O(h3)

(2.8)

2.1.2 Taylor Series Expansion Formula

We need to compute ytt for Taylor series expansion below.

ytt =df

dt(2.9)

=∂f

∂t

∂t

∂t+∂f

∂y

∂y

∂t(2.10)

= ft + ffy (2.11)

Now we write down the Taylor series expansion of y in the neighborhood oftn with O

(h3).

y(n+1) = y(n) + hyt|tn +h2

2ytt|tn +O

(h3)

(2.12)

= y(n) + hfn +h2

2ytt|tn +O

(h3)

(2.13)

= y(n) + hfn +h2

2(ft + ffy)|tn +O

(h3)

(2.14)

= y(n) + hfn +h2

2ft

∣∣∣∣tn

+h2

2fn fy|tn +O

(h3)

(2.15)

Remark 2.2. (Important) From here to later, if nothing is misundertood,we can abbreviate notation ftαyβ for

ftαyβ∣∣tn

(2.16)

in the Taylor series expansion.For example, under this abbreviation, (2.15) can be rewritten briefly as

y(n+1) = y(n) + hfn +h2

2ft +

h2

2fnfy +O

(h3)

(2.17)

This abbreviation reduces the complexity of the formulas in this context. Asyou will see later, this abbreviation is really essential.

2.1.3 Derivation of System of Equations

Return to our problem, as usual, comparing (2.8) and (2.17), we obtain

hfn : 1 = b1 + b2 (2.18)

h2ft :1

2= b2c2 (2.19)

h2fnfy :1

2= b2a21 (2.20)


Hence, we obtain the system of equations

b1 + b2 = 1 (2.21)

b2c2 =1

2(2.22)

b2a21 =1

2(2.23)

2.1.4 Solutions of System of Equations

We can solve the above system of equations easily. Here, the authors representtwo solutions. The idea behind these two solutions will be use later for higherorder Runge Kutta method.

Solution 1. The system involves four unknowns in three equations. Tak-ing b2 = α is free variable. Due to (2.23) we must have α 6= 0. Then we caneasily obtain the general solution for (2.21)-(2.23) is

b2 = α (2.24)

b1 = 1− α (2.25)

c2 = a21 =1

2α(2.26)

where α is an arbitrary real number.Butcher tableau in this case becomes

012α

12α

1− α α(2.27)

�

Solution 2. Due to (2.21) and (2.22), we must have c2 = a21. So, we cantake

c2 = a21 = β (2.28)

as a free variable. And the remaining is very easy, we obtain

b1 = 1− 1

2β(2.29)

b2 =1

2β(2.30)

c2 = a21 = β (2.31)

Butcher tableau in this case becomes

0β β

1− 12β

12β

(2.32)

This Butcher tableau appears in [5]. �

Remark 2.3. Since Runge Kutta order 2 is still simple, you can not see thedifferences between two solutions. With Runge Kutta of higher order, you willsee that the first choice of free variables is very important in entire solution.


2.1.5 Some Cases

We discuss two useful choices

Case α =1

2. In this case, (2.26) becomes

b2 =1

2(2.33)

b1 =1

2(2.34)

c2 = a21 = 1 (2.35)

The corresponding Butchet tableau reads

01 1

1

2

1

2

(2.36)

Thus, in this case the Runge Kutta method of second order takes the form

y(n+1) = y(n) +h

2[fn + f (tn + h, yn + hfn)] (2.37)

and is equivalent to the Heun’s method.

Case α = 1. In this case, (2.26) becomes

b2 = 1 (2.38)

b1 = 0 (2.39)

c2 = a21 =1

2(2.40)

The corresponding Butcher tableau reads

01

2

1

20 1

(2.41)

In this case Runge Kutta method of second order can be written as

y(n+1) = y(n) + hf

(tn +

h

2, yn +

h

2fn

)(2.42)

and is called the RK2 method.

Remember α ∈ R∗, so there is infinite many choices of solution for (2.21)-(2.23).

2.2 Matlab Implementation

See section 4.2 for general code.

Chapter 3

Runge Kutta order 3method




k1 = fn (3.1)

k2 = f (tn + c2h, yn + ha21fn) (3.2)

k3 = f (tn + c3h, yn + ha31k1 + ha32k2) (3.3)

= f (tn + c3h, yn + ha31fn + ha32f (tn + c2h, yn + ha31k1 + ha32k2)) (3.4)

and (1.9) becomes

y(n+1) = y(n) + hb1k1 + hb2k2 + hb3k3 (3.5)


for k2.

k2 = f (tn + c2h, yn + ha21fn) (3.6)

= fn + c2hft + ha21fnfy + h2c222ftt + h2c2a21fnfty + h2

a2212f2nfyy (3.7)

+O(h3)

(3.8)

And we also write down the Taylor series expansion O(h3)

for k3.

k3 = f (tn + c3h, yn + ha31fn + ha32k2) (3.9)

= fn + c3hft + h (a31fn + a32k2) fy (3.10)

+ h2c232ftt + c3h

2 (a31fn + a32k2) fty + h2(a31fn + a32k2)

2

2fyy (3.11)

+O(h3)

(3.12)

8


Inserting (3.8) into (3.12)


= fn + c3hft + ha31fnfy (3.14)

+ ha32fy (fn + hc2ft + ha21fnfy) (3.15)

+ h2c232ftt + c3h

2 (a31fn + a32fn) fty (3.16)

+ h2(a231f

2n + a232f

2n + 2a31a32f

2n

)2

fyy (3.17)

+O(h3)

(3.18)

Collecting terms respect to exponents of h

k3 = fn + h (c3ft + a31fnfy + a32fyfn) (3.19)

+ h2

a32c2ftfy + a21a32fnf

2y +

c232ftt

+c3a31fnfty + c3a32fnfty +a2312f2nfyy

+a2322f2nfyy + a31a32f

2nfyy

(3.20)

+O(h3)

(3.21)

Inserting (3.12) and (3.21) into (3.5)

y(n+1) = y(n) + hb1fn (3.22)

+ hb2

fn + hc2ft + ha21fnfy + h2c222ftt

+h2c2a21fnfty + h2a2212f2nfyy

(3.23)

+ hb3

fn + h (c3ft + a31fnfy + a32fyfn)

+h2


2y +

c232ftt

+c3a31fnfty + c3a32fnfty +a2312f2nfyy

+a2322f2nfyy + a31a32f

2nfyy

(3.24)

+O(h4)

(3.25)


y(n+1) = y(n) + h (b1fn + b2fn + b3fn) (3.26)

+ h2 (b2c2ft + a21b2fnfy + b3c3ft + a31b3fnfy + a32b3fyfn) (3.27)

+ h3

b2c22

2ftt + a21b2c2fnfty +

a221b22

f2nfyy

+a32b3c2ftfy + a21a32b3fnf2y +

b3c23

2ftt

+a31b3c3fnfty + a32b3c3fnfty +a231b3

2f2nfyy

+a232b3

2f2nfyy + a31a32b3f

2nfyy

(3.28)

+O(h4)

(3.29)



We need to compute yttt for Taylor series expansion below.

yttt =d

dt(ft + ffy) (3.30)

= (ft + ffy)t + f(ft + ffy)y (3.31)

= ftt + ftfy + ffty + ffty + ff2y + f2fyy (3.32)

= ftt + ftfy + 2ffty + ff2y + f2fyy (3.33)


(h4).

y(n+1) = y(n) + hyt +h2

2ytt +

h3

6yttt +O

(h4)

(3.34)

= y(n) + hfn +h2

2(ft + fnfy) (3.35)

+h3

6

(ftt + ftfy + 2fnfty + fnf

2y + f2nfyy

)+O

(h4)

(3.36)


Comparing (3.29) and (3.36)

hfn : 1 = b1 + b2 + b3 (3.37)

h2ft :1

2= b2c2 + b3c3 (3.38)

h2fnfy :1

2= a21b2 + a31b3 + a32b3 (3.39)

h3ftt :1

6=b2c

22

2+b3c

23

2(3.40)

h3ftfy :1

6= a32b3c2 (3.41)

h3fnfty :1

3= a21b2c2 + a31b3c3 + a32b3c3 (3.42)

h3fnf2y :

1

6= a21a32b3 (3.43)

h3f2nfyy :1

6=a221b2

2+a231b3

2+a232b3

2+ a31a32b3 (3.44)


Hence, we obtain the system of 8 equations with 8 unknowns.

b1 + b2 + b3 = 1 (3.45)

b2c2 + b3c3 =1

2(3.46)

b2a21 + b3 (a31 + a32) =1

2(3.47)

b2c22 + b3c

23 =

1

3(3.48)

b3a32c2 =1

6(3.49)

b2c2a21 + b3c3 (a31 + a32) =1

3(3.50)

b3a32a21 =1

6(3.51)

b2a221 + b3(a31 + a32)

2=

1

3(3.52)


We now solve the above system of equations in two different ways.

Solution 1. We take b2 = α, b3 = β as two free variables, then

b1 = 1− α− β (3.53)

our task remains to solve

αc2 + βc3 =1

2(3.54)

αa21 + β (a31 + a32) =1

2(3.55)

αc22 + βc23 =1

3(3.56)

βa32c2 =1

6(3.57)

αc2a21 + βc3 (a31 + a32) =1

3(3.58)

βa32a21 =1

6(3.59)

αa221 + β(a31 + a32)2

=1

3(3.60)

We can solve c2 and c3 by using two equations

αc2 + βc3 =1

2(3.61)

αc22 + βc23 =1

3(3.62)

Since (3.59), we must have β 6= 0.


Then we can take c3 =1

2β− αβc2 from (3.61) and insert into (3.62) to obtain

the following equation respect to c2.

12α (α+ β) c22 − 12αc2 + 3− 4β = 0 (3.63)

Consider two following cases.

1. Case α (α+ β) = 0. Consider two subcases.

(a) Case α = 0. Then we have immediately

β =3

4(3.64)

c3 =2

3(3.65)

b1 =1

4(3.66)

and the system of equations remains

a31 + a32 =2

3(3.67)

a32c2 =2

9(3.68)

3

4a32a21 =

1

6(3.69)

To solve this remaining system of equations, we can take a32 = γ 6= 0as a free variable. Then

a31 =2

3− γ (3.70)

c2 =2

9γ(3.71)

a21 =2

9γ(3.72)

Hence, we obtain the solutions

b1 =1

4(3.73)

b2 = 0 (3.74)

b3 =3

4(3.75)

c2 =2

9γ(3.76)

c3 =2

3(3.77)

a21 =2

9γ(3.78)

a31 =2

3− γ (3.79)

a32 = γ (3.80)


where γ is an arbitrary nonzero real number, in this subcase.

Butcher tableau becomes

02

9γ

2

9γ2

3

2

3− γ γ

1

40

3

4

(3.81)

(b) Case α+ β = 0, α 6= 0. Then we have immediately

b1 = 1 (3.82)

b2 = α (3.83)

b3 = −α (3.84)

c2 =1

3+

1

4α(3.85)

and the system of equations remains

c2 − c3 =1

2α(3.86)

a21 − (a31 + a32) =1

2α(3.87)

c22 − c23 =1

3α(3.88)

a32c2 = − 1

6α(3.89)

c2a21 − c3 (a31 + a32) =1

3α(3.90)

a32a21 = − 1

6α(3.91)

a221 − (a31 + a32)2

=1

3α(3.92)

We again have immediately

c3 =1

3− 1

4α(3.93)

a32 = − 2

4α+ 3(3.94)

a21 =1

3+

1

4α(3.95)

a31 =16α2 + 24α− 9

12α (4α+ 3)(3.96)


Hence, we obtain the solution

b1 = 1 (3.97)

b2 = α (3.98)

b3 = −α (3.99)

c2 =1

3+

1

4α(3.100)

c3 =1

3− 1

4α(3.101)

a21 =1

3+

1

4α(3.102)

a31 =16α2 + 24α− 9

12α (4α+ 3)(3.103)

a32 = − 2

4α+ 3(3.104)

where α is an arbitrary nonzero real number, in this subcase.

Butcher tableau becomes

01

3+

1

4α

1

3+

1

4α1

3− 1

4α

16α2 + 24α− 9

12α (4α+ 3)− 2

4α+ 31 α −α

(3.105)

2. Case α (α+ β) 6= 0. (3.63) is a quadratic equation respect to c2.

Computing the determinant of (3.63)

∆′ = 36α2 + 12α (α+ β) (4β − 3) (3.106)

= 12αβ (4α+ 4β − 3) (3.107)

Hence, we have to make the assumption

αβ (4α+ 4β − 3) ≥ 0 (3.108)

so that (3.63) has roots in this case.

Under this assumption, (3.63) have two roots

c2 =3α±

√3αβ (4α+ 4β − 3)

6α (α+ β)(3.109)

Consider two subcases respect to c2.

(a) Case c2 =3α+

√3αβ (4α+ 4β − 3)

6α (α+ β). We easily solve the remaining


system of equations to get

b1 = 1− α− β (3.110)

b2 = α (3.111)

b3 = β (3.112)

c2 =3α+

√3αβ (4α+ 4β − 3)

6α (α+ β)(3.113)

c3 =3β −

√3αβ (4α+ 4β − 3)

6β (α+ β)(3.114)

a21 =3α+

√3αβ (4α+ 4β − 3)

6α (α+ β)(3.115)

a31 =1

2β−

3α+√

3αβ (4α+ 4β − 3)

6β (α+ β)(3.116)

− α (α+ β)

β(

3α+√

3αβ (4α+ 4β − 3)) (3.117)

a32 =α (α+ β)

β(

3α+√

3αβ (4α+ 4β − 3)) (3.118)

Butcher tableau reads all obtained coefficients.

(b) Case c2 =3α−

√3αβ (4α+ 4β − 3)

6α (α+ β). We also easily solve the re-

maining system of equations to get

b1 = 1− α− β (3.119)

b2 = α (3.120)

b3 = β (3.121)

c2 =3α−

√3αβ (4α+ 4β − 3)

6α (α+ β)(3.122)

c3 =3β +

√3αβ (4α+ 4β − 3)

6β (α+ β)(3.123)

a21 =3α−

√3αβ (4α+ 4β − 3)

6α (α+ β)(3.124)

a31 =1

2β−

3α−√

3αβ (4α+ 4β − 3)

6β (α+ β)(3.125)

− α (α+ β)

β(

3α−√

3αβ (4α+ 4β − 3)) (3.126)

a32 =α (α+ β)

β(

3α−√

3αβ (4α+ 4β − 3)) (3.127)

Butcher tableau reads all obtained coefficients.

We have solved the system of equations (3.45)-(3.52) completely. �


Remark 3.1. In the first solution, we have used b2 and b3 as two free vari-ables. This choice makes square roots appear in the solutions. This is quiteeasy to understand. Because of choice of b2, b3 as free variables, we have tosolve a quadratic equation. This quadratic equation make square roots appearobviously.

Now, we solve our system of equation by alternative ways. The idea is verysimple. It is just a matter of the first choice. More explicitly, instead of choosingb2, b3 as two free variables, we will choose c2, c3 as two free variables. Let us seethe differences between two solutions through the following second one.

Solution 2. We take c2 = α, c3 = β as two free variables and focus onthe following two equations of our system of equations.

b2c2 + b3c3 =1

2(3.128)

b2c22 + b3c

33 =

1

3(3.129)

We consider two cases respect to c2 and c3.

1. Case c2 = c3 = α. Due to (3.57), we must have α 6= 0. Then the abovesub-system of equations becomes

b2 + b3 =1

2α(3.130)

b2 + b3 =1

3α2(3.131)

Hence

c2 = c3 = α =2

3(3.132)

The remaining system of equations is

b1 + b2 + b3 = 1 (3.133)

b2 + b3 =3

4(3.134)

b2a21 + b3 (a31 + a32) =1

2(3.135)

b3a32 =1

4(3.136)

b3a32a21 =1

6(3.137)

b2a221 + b3(a31 + a32)

2=

1

3(3.138)

We obtain immediately

b1 =1

4(3.139)

a21 =2

3(3.140)


and the remaining system of equations is

b2 + b3 =3

4(3.141)

2

3b2 + b3 (a31 + a32) =

1

2(3.142)

b3a32 =1

4(3.143)

4

9b2 + b3(a31 + a32)

2=

1

3(3.144)

We now choose b3 = γ then

b2 =3

4− γ (3.145)

a31 =2

3− 1

4γ(3.146)

a32 =1

4γ(3.147)

Therefore,

c2 =2

3(3.148)

c3 =2

3(3.149)

b1 =1

4(3.150)

b2 =3

4− γ (3.151)

b3 = γ (3.152)

a21 =2

3(3.153)

a31 =2

3− 1

4γ(3.154)

a32 =1

4γ(3.155)

where γ is an arbitrary nonzero real number, is the solution of (3.45)-(3.52)in this case.

2. Case c2 6= c3. Due to (3.49) and (3.51), we have immediately

a21 = c2 = α (3.156)

Due to (3.128) and (3.129), we obtain

b2 =3β − 2

6α (β − α)(3.157)

b3 =3α− 2

6β (α− β)(3.158)


Hence,

b1 =6αβ − 3α− 3β + 2

6αβ(3.159)

The remaining system of equations is

3β − 2

6α (β − α)α+

3α− 2

6β (α− β)(a31 + a32) =

1

2(3.160)

3α− 2

6β (α− β)a32α =

1

6(3.161)

3β − 2

6 (β − α)α+

3α− 2

6 (α− β)(a31 + a32) =

1

3(3.162)

3β − 2

6α (β − α)α2 +

3α− 2

6β (α− β)(a31 + a32)

2=

1

3(3.163)

We easily solve this and obtain

a21 = α (3.164)

a31 = β − β (α− β)

α (3α− 2)(3.165)

a32 =β (α− β)

α (3α− 2)(3.166)

Therefore,

c2 = α (3.167)

c3 = β (3.168)

b1 =6αβ − 3α− 3β + 2

6αβ(3.169)

b2 =3β − 2

6α (β − α)(3.170)

b3 =3α− 2

6β (α− β)(3.171)

a21 = α (3.172)

a31 = β − β (α− β)

α (3α− 2)(3.173)

a32 =β (α− β)

α (3α− 2)(3.174)

where α 6= β, α 6= 2

3are two arbitrary nonzero real numbers, is the solution

of (3.45)-(3.52) in this case.

We have solved (3.45)-(3.52) completely. �

Remark 3.2. In the second solution,nth roots do not appear in the solu-tion because we does not need to solve any polynomial equations. Since thisway is more easy and effective, it will be used for Runge Kutta method of higherorders.


3.1.5 Some Cases

We consider some cases of Runge Kutta order 3 method respect to some solu-tions of its associated system of equations.

Case α =2

3, β =

1

6.

We use case 2.(b) above to obtain

01

2

1

21 −1 2

1

6

2

3

1

6

(3.175)

This Butcher tableau appears in [6] - Kutta’s third order method and in [4], p.82.

Case α =3

8, β =

3

8.

02

3

2

32

30

2

31

4

3

8

3

8

(3.176)

This Butcher tableau appears in [4], p.82.

Case α =3

4, β =

1

4.

02

3

2

30 −1 1

03

4

1

4

(3.177)



See section 4.2 for general code.

Chapter 4

Runge Kutta Order 4Method




k1 = fn (4.1)

k2 = f (tn + c2h, yn + ha21fn) (4.2)


k4 = f (tn + c4h, yn + ha41fn + ha42k2 + ha43k3) (4.4)

and (1.9) becomes

y(n+1) = y(n) + hb1k1 + hb2k2 + hb3k3 + hb4k4 (4.5)


for k2.

k2 = f (tn + c2h, yn + ha21fn) (4.6)

= fn + hc2ft + ha21fnfy + h2c222ftt + h2a21c2fnfty + h2

a2212f2n (4.7)

+ h3c326fttt + h3

a21c22

2fnftty + h3

a221c22

f2nftyy + h3a3216f3nfyyy +O

(h4)(4.8)

= fn + h (c2ft + a21fnfy) + h2(c222ftt + a21c2fnfty +

a2212f2n

)(4.9)

+ h3(c326fttt +

a21c22

2fnftty +

a221c22

f2nftyy +a3216f3nfyyy

)+O

(h4)

(4.10)

20


We write down the Taylor series expansion O(h4)

for k3.


= fn + hc3ft + h (a31fn + a32k2) fy (4.12)

+ h2c232ftt + h2c3 (a31fn + a32k2) fty +

1

2h2(a31fn + a32k2)

2fyy (4.13)

+c33h

3

6fttt +

h3c232

(a31fn + a32k2) ftty +h3c3

2(a31fn + a32k2)

2ftyy (4.14)

+h3

6(a31fn + a32k2)

3fyyy +O

(h4)

(4.15)

Inserting (4.10) into (4.15)

k3 = fn + hc3ft + ha31fnfy (4.16)

+ ha32fy

[fn + h (c2ft + a21fnfy) + h2

(c222ftt + a21c2fnfty +

a2212f2n

)](4.17)

+h2c23

2ftt + h2c3a31fnfty + h2c3a32fty [fn + h (c2ft + a21fnfy)] (4.18)

+h2

2a231f

2nfyy +

h2

22a31a32fnfyy [fn + h (c2ft + a21fnfy)] (4.19)

+h2

2a232fyy[fn + h (c2ft + a21fnfy)]

2(4.20)

+c33h

3

6fttt +

h3c232

(a31fn + a32fn) ftty (4.21)

+h3c3

2

(a231f

2n + 2a31a32f

2n + a232f

2n

)ftyy (4.22)

+h3

6

(a331f

3n + 3a231a32f

3n + 3a31a

232f

3n + a332f

3n

)fyyy +O

(h4)

(4.23)


k3 = fn + h (c3ft + a31fnfy + a32fnfy) (4.24)

+ h2

a32c2ftfy + a21a32fnf2y +

c232ftt + a31c3fnfty

+c3a32fnfty +a2312f2nfyy + a31a32f

2nfyy +

a2322f2nfyy

(4.25)

+ h3

a32c22

2fyftt + a32a21c2fnfyfty +

a221a322

f2nfy

+a32c2c3ftfty + a21a32c3fnfyfty + a31a32c2fnftfyy+a21a31a32f

2nfyfyy + a232c2fnffyyt + a21a

232f

2nfyfyy

+c336fttt +

a31c23

2fnftty +

a32c23

2fnftty +

a231c32

f2nftyy

+a31a32c3f2nftyy +

a232c32

f2nftyy +a3316f3nfyyy

+a231a32

2f3nfyyy +

a31a232

2f3nfyyy +

a3326f3nfyyy

(4.26)

+O(h4)

(4.27)


We continue to write down the Taylor series expansion O(h4)

for k4.

k4 = f (tn + c4h, yn + ha41fn + ha42k2 + ha43k3) (4.28)

= fn + c4hft + h (a41fn + a42k2 + a43k3) fy (4.29)

+c24h

2

2ftt + c4h

2 (a41fn + a42k2 + a43k3) fty (4.30)

+1

2h2(a41fn + a42k2 + a43k3)

2fyy +

c34h3

6fttt (4.31)

+c242h3 (a41fn + a42k2 + a43k3) ftty (4.32)

+c4h

3

2(a41fn + a42k2 + a43k3)

2ftyy (4.33)

+1

6h2(a41fn + a42k2 + a43k3)

3fyyy +O

(h4)

(4.34)


Inserting (4.15) and (4.27) to (4.34)

k4 = fn + hc4ft + ha41fnfy (4.35)

+ ha42fy

fn + h (c2ft + a21fnfy)

+h2(c222ftt + a21c2fnfty +

a2212f2n

) (4.36)

+ ha43fy

fn + h (c3ft + a31fnfy + a32fnfy)

+h2


2y +

c232ftt

+a31c3fnfty + c3a32fnfty +a2312f2nfyy

+a31a32f2nfyy +

a2322f2nfyy

(4.37)

+ h2c242ftt + h2a41c4fnfty + h2a42c4fty [fn + h (c2ft + a21fnfy)] (4.38)

+ h2a43c4fty [fn + h (c3ft + a31fnfy + a32fnfy)] (4.39)

+ h2a2412f2nfyy + h2

a2422fyy[fn + h (c2ft + a21fnfy)]

2(4.40)

+ h2a2432fyy[fn + h (c3ft + a31fnfy + a32fnfy)]

2(4.41)

+ h2a41a42fnfyy [fn + h (c2ft + a21fnfy)] (4.42)

+ h2a42a43fyy [fn + h (c2ft + a21fnfy)]× (4.43)

× [fn + h (c3ft + a31fnfy + a32fnfy)] (4.44)

+ h2a41a43fnfyy [fn + h (c3ft + a31fnfy + a32fnfy)] (4.45)

+ h3c346fttt + h3

c242

(a41fn + a42fn + a43fn) ftty (4.46)

+ h3c42

(a241f

2n + a242f

2n + a243f

2n + 2a41a42f

2n

+2a42a43f2n + 2a41a43f

2n

)ftyy (4.47)

+ h3

a3416f3n +

a3426f3n +

a3436f3n +

a241a422

f3n

+a241a43

2f3n +

a242a412

f3n +a242a43

2f3n +

a243a412

f3n

+a243a42

2f3n + a41a42a43f

3n

fyyy (4.48)

+O(h4)

(4.49)



k4 = fn + h (c4ft + a41fnfy + a42fnfy + a43fnfy) (4.50)

+ h2


2y + a43c3ftfy + a31a43fnf

2y

+a32a43fnf2y +

c242ftt + c4a41fnfty + c4a42fnfty

+c4a43fnfty +a2412f2nfyy +

a2422f2nfyy +

a2432f2nfyy

+a41a42f2nfyy + a42a43f

2nfyy + a41a43f

2nfyy

(4.51)

+ h3

a42c22

2fyftt + a21a42c2fnfyfty +

a42a221

2f2nfyfyy

+a43a32c2ftf2y + a21a32a43fnf

3y +

a43c23

2fyftt

+a31a43c3fnfyfty + a32a43c3fnfyfty +a231a43

2f2nfyfyy

+a31a32a43f2nfyfyy +

a232a432

f2nfyfyy + a42c2c4ftfty

+a21a42c4fnfyfty + a43c3c4ftfty + a31a43c4fnfyfty+a32a43c4fnfyfty + a242c2fnftfyy + a21a

242f

2nfyfyy

+a243c3fnftfyy + a31a243f

2nfyfyy + a32a

243f

2nfyfyy

+a41a42c2fnftfyy + a21a41a42f2nfyfyy + a42a43c3fnftfyy

+a31a42a43f2nfyfyy + a32a42a43f

2nfyfyy + a42a43c2fnftfyy

+a21a42a43f2nfyfyy + a41a43c3fnftfyy + a31a41a43f

2nfyfyy

+a32a41a43f2nfyfyy +

c346fttt +

a41c24

2fnftty +

a42c24

2fnftty

+a43c

24

2fnftty +

a241c42

f2nftyy +a242c4

2f2nftyy

+a243c4

2f2nftyy + a41a42c4f

2nftyy + a42a43c4f

2nftyy

+a41a43c4f2nftyy +

a3416f3nfyyy +

a3426f3nfyyy

+a3436f3nfyyy +

a241a422

f3nfyyy +a241a43

2f3nfyyy

+a41a

242

2f3nfyyy +

a242a432

f3nfyyy +a41a

243

2f3nfyyy

+a42a

243

2f3nfyyy + a41a42a43f

3nfyyy

(4.52)

+O(h4)

(4.53)


Inserting (4.10), (4.27) and (4.53) into (4.5)

y(n+1) (4.54)

= y(n) + h (b1fn + b2fn + b3fn + b4fn) (4.55)

+ h2(b2c2ft + a21b2fnfy + b3c3ft + a31b3fnfy + a32b3fnfy+b4c4ft + a41b4fnfy + a42b4fnfy + a43b4fnfy

)(4.56)

+ h3

b2c22

2ftt + b2c2a21fnfty +

a221b22

f2nfyy + a32b3c2ftfy

+a32b3a21fnfyfy +b3c

23

2ftt + c3b3a31fnfty + c3b3a32fnfty

+a231b3

2f2nfyy + a31a32b3fnfnfyy +

a232b32

f2nfyy + a42b4c2ftfy

+a42b4fya21fnfy + a43b4c3ftfy + a43a31b4fnf2y + +a43a32b4fnf

2y

+b4c

24

2ftt + c4b4a41fnfty + c4b4a42ftyfn + c4b4a43ftyfn

+a241b4

2f2nfyy +

a242b42

f2nfyy +a243b4

2f2nfyy + a41a42b4fyyf

2n

+a42a43b4fyyf2n + a41a43b4fyyf

2n

(4.57)

+ h4

b2c32

6fttt +

b2c22a212

fnftty +b2c2a

221

2f2nftyy +

a321b26

f3nfyyy

+a32b3c

22

2fttfy + a32b3c2a21fnftyfy +

a32a221b3

2f2nfyyfy

+c3a32b3c2ftfty + c3a32b3a21fnfyfty + a31a32b3c2fnftfyy+a31a32a21b3fnfnfyfyy + a232b3fnc2ftfyy + a232b3fna21fnfyfyy

+b3c

33

6fttt +

b3c23 (a31 + a32)

2fnftty +

b3c3a231

2f2nftyy

+b3c3a31a32fnfnftyy +b3c3a

232

2f2nftyy +

b3a331

6f3nfyyy

+b3a

231a322

f3nfyyy +b3a31a

232

2f3nfyyy +

b3a332

6f3nfyyy

+a42b4c

22

2fyftt + a42b4c2a21fnfyfty +

a42a221b4

2f2nfyfyy

+a43a32b4c2ftf2y + a43a32a21b4fnf

3y +

a43b4c23

2fyftt

+a43b4c3a31fnfyfty + a43b4c3a32fnfyfty +a43a

231b4

2f2nfyfyy

+a43a31a32b4f2nfyfyy +

a43a232b4

2f2nfyfyy + c4a42b4c2ftfty

+c4a42a21b4fnfyfty + c4a43c3b4ftfty + c4a43a31b4fnfyfty+c4a43a32b4fnfyfty + a242b4c2fnftfyy + a242a21b4f

2nfyfyy

+a243b4c3fnftfyy + a243a31b4f2nfyfyy + a243a32b4f

2nfyfyy

+a41a42b4c2fnftfyy + a41a21a42b4f2nfyfyy + a42a43b4c3fnftfyy

+a42a43a31b4f2nfyfyy + a42a43a32b4f

2nfyfyy + a42a43b4c2fnftfyy

+a42a43a21b4f2nfyfyy + a41a43b4c3fnftfyy + a41a43a31b4f

2nfyfyy

+a41a43a32b4f2nfyfyy +

b4c34

6fttt +

b4c24a412

fnftty +b4c

24a422

fnftty

+b4c

24a432

fnftty +c4b4a

241

2f2nftyy +

c4b4a242

2f2nftyy +

c4b4a243

2f2nftyy

+c4b4a41a42f2nftyy + c4a42a43b4f

2nftyy + c4a41a43b4f

2nftyy

+a341b4

6f3nfyyy +

a342b46

f3nfyyy +a343b4

6f3nfyyy +

a241a42b42

f3nfyyy

+a241a43b4

2f3nfyyy +

a242a41b42

f3nfyyy +a242a43b4

2f3nfyyy

+a243a41b4

2f3nfyyy +

a243a42b42

f3nfyyy + a41a42a43b4f3nfyyy

(4.58)

+O(h5)

(4.59)



We need to compute ytttt for Taylor series expansion below.

ytttt =d

dt

(ftt + ftfy + 2ffty + ff2y + f2fyy

)(4.60)

=(ftt + ftfy + 2ffty + ff2y + f2fyy

)t

(4.61)

+ f(ftt + ftfy + 2ffty + ff2y + f2fyy

)y

(4.62)

= fttt + fyftt + ftfty + 2ftfty + 2fftty + ftf2y + 2ffyfty (4.63)

+ 2fftfyy + f2ftyy + fftty + ffyfty + fftfyy + 2ffyfty (4.64)

+ 2f2ftyy + ff3y + 2f2fyfyy + 2f2fyfyy + f3fyyy (4.65)

= fttt + fyftt + 3ftfty + 3fftty + ftf2y + 5ffyfty + 3fftfyy (4.66)

+ 3f2ftyy + ff3y + 4f2fyfyy + f3fyyy (4.67)


(h5).

yn+1 = yn + hfn +h2

2(ft + fnfy) (4.68)

+h3

6

(ftt + ftfy + 2fnfty + fnf

2y + f2nfyy

)(4.69)

+h4

24

(fttt + fyftt + 3ftfty + 3fnftty + ftf

2y + 5fnfyfty

+3fnftfyy + 3f2nftyy + fnf3y + 4f2nfyfyy + f3nfyyy

)(4.70)

+O(h5)

(4.71)


Compare the coefficients of (4.68) and (4.59)


hfn : 1 = b1 + b2 + b3 + b4 (4.72)

h2ft :1

2= b2c2 + b3c3 + b4c4 (4.73)

h2fnfy :1

2= b2a21 + b3a31 + b3a32 + a41b4 + a42b4 + a43b4 (4.74)

h3ftt :1

6=b2c

22

2+b3c

23

2+b4c

24

2(4.75)

h3ftfy :1

6= a32b3c2 + a42b4c2 + a43b4c3 (4.76)

h3fnfty :1

3= b2c2a21 + c3b3a31 + c3b3a32 (4.77)

+ c4b4a41 + c4b4a42 + c4b4a43 (4.78)

h3fnf2y :

1

6= a32b3a21 + a42b4a21 + a43a31b4 + a43a32b4 (4.79)

h3f2nfyy :1

6=a221b2

2+a231b3

2+ a31a32b3 +

a232b32

+a241b4

2(4.80)

+a242b4

2+a243b4

2+ a41a42b4 + a42a43b4 + a41a43b4 (4.81)

h4fttt :1

24=b2c

32

6+b3c

33

6+b4c

34

6(4.82)

h4fyftt :1

24=a32b3c

22

2+a42b4c

22

2+a43b4c

23

2(4.83)

h4ftfty :1

8= c3a32b3c2 + c4a42b4c2 + c4a43c3b4 (4.84)

h4fnftty :1

8=b2c

22a212

+b3c

23 (a31 + a32)

2+b4c

24a412

(4.85)

+b4c

24a422

+b4c

24a432

(4.86)

h4ftf2y :

1

24= a43a32b4c2 (4.87)

h4fnfyfty :5

24= a32b3c2a21 + c3a32b3a21 + a42b4c2a21 (4.88)

+ c4a42a21b4 + c4a43a31b4 + a43b4c3a31 (4.89)

+ a43b4c3a32 + c4a43a32b4 (4.90)

h4fnftfyy :1

8= a31a32b3c2 + a232b3c2 + a242b4c2 + a243b4c3 (4.91)

+ a41a42b4c2 + a42a43b4c3 + a42a43b4c2 + a41a43b4c3 (4.92)

h4f2nftyy :1

8=b2c2a

221

2+b3c3a

231

2+ b3c3a31a32 +

b3c3a232

2(4.93)

+c4b4a

241

2+c4b4a

242

2+c4b4a

243

2+ c4b4a41a42 (4.94)

+ c4a42a43b4 + c4a41a43b4 (4.95)

h4fnf3y :

1

24= a43a32a21b4 (4.96)


h4f2nfyfyy :1

6=a32a

221b3

2+ a31a32a21b3 + a232b3a21 +

a42a221b4

2(4.97)

+a43a

231b4

2+ a43a31a32b4 +

a43a232b4

2+ a242a21b4 (4.98)

+ a243a31b4 + a243a32b4 + a41a21a42b4 + a42a43a31b4 (4.99)

+ a42a43a32b4 + a42a43a21b4 + a41a43a31b4 (4.100)

+ a41a43a32b4 (4.101)

h4f3nfyyy :1

24=a321b2

6+b3a

331

6+b3a

231a322

+b3a31a

232

2+b3a

332

6(4.102)

+a341b4

6+a342b4

6+a343b4

6+a241a42b4

2+a241a43b4

2(4.103)

+a242a41b4

2+a242a43b4

2+a243a41b4

2(4.104)

+a243a42b4

2+ a41a42a43b4 (4.105)


Hence, we obtain the system of equations

b1 + b2 + b3 + b4 = 1 (4.106)

b2c2 + b3c3 + b4c4 =1

2(4.107)

b2a21 + b3 (a31 + a32) + b4 (a41 + a42 + a43) =1

2(4.108)

b2c22 + b3c

23 + b4c

24 =

1

3(4.109)

b3a32c2 + b4 (a42c2 + a43c3) =1

6(4.110)

b2c2a21 + b3c3 (a31 + a32) + b4c4 (a41 + a42 + a43) =1

3(4.111)

b3a32a21 + b4 [a42a21 + a43 (a31 + a32)] =1

6(4.112)

b2a221 + b3(a31 + a32)

2+ b4(a41 + a42 + a43)

2=

1

3(4.113)

b2c32 + b3c

33 + b4c

34 =

1

4(4.114)

b3a32c22 + b4

(a42c

22 + a43c

23

)=

1

12(4.115)

b3c3a32c2 + b4c4 (a42c2 + a43c3) =1

8(4.116)

b2c22a21 + b3c

23 (a31 + a32) + b4c

24 (a41 + a42 + a43) =

1

4(4.117)

b4a43a32c2 =1

24(4.118){

b3a32a21 (c2 + c3)+b4 [a42a21 (c2 + c4) + a43 (a31 + a32) (c3 + c4)]

}=

5

24(4.119)

b3a32c2 (a31 + a32) + b4 (a42c2 + a43c3) (a41 + a42 + a43) =1

8(4.120)

b2c2a221 + b3c3(a31 + a32)

2+ b4c4(a41 + a42 + a43)

2=

1

4(4.121)

b4a43a32a21 =1

24(4.122) b3a32a21 [a21 + 2 (a31 + a32)]

+b4a42a21 [a21 + 2 (a41 + a42 + a43)]+b4a43 (a31 + a32) [a31 + a32 + 2 (a43 + a42 + a41)]

=1

3(4.123)

b2a321 + b3(a31 + a32)

3+ b4(a41 + a42 + a43)

3=

1

4(4.124)


We now solve the above system of equations.We take

c2 = α (4.125)

c3 = β (4.126)

c4 = γ (4.127)


as three free variables.Then, our system of equations becomes

b1 + b2 + b3 + b4 = 1 (4.128)

αb2 + βb3 + γb4 =1

2(4.129)

b2a21 + b3 (a31 + a32) + b4 (a41 + a42 + a43) =1

2(4.130)

α2b2 + β2b3 + γ2b4 =1

3(4.131)

αb3a32 + b4 (αa42 + βa43) =1

6(4.132)

αb2a21 + βb3 (a31 + a32) + γb4 (a41 + a42 + a43) =1

3(4.133)

b3a32a21 + b4 [a42a21 + a43 (a31 + a32)] =1

6(4.134)

b2a221 + b3(a31 + a32)

2+ b4(a41 + a42 + a43)

2=

1

3(4.135)

α3b2 + β3b3 + γ3b4 =1

4(4.136)

α2b3a32 + b4(α2a42 + β2a43

)=

1

12(4.137)

αβb3a32 + γb4 (αa42 + βa43) =1

8(4.138)

α2b2a21 + β2b3 (a31 + a32) + γ2b4 (a41 + a42 + a43) =1

4(4.139)

αb4a43a32 =1

24(4.140){

b3a32a21 (α+ β)+b4 [a42a21 (α+ γ) + a43 (a31 + a32) (β + γ)]

}=

5

24(4.141)

αb3a32 (a31 + a32) + b4 (αa42 + βa43) (a41 + a42 + a43) =1

8(4.142)

αb2a221 + βb3(a31 + a32)

2+ γb4(a41 + a42 + a43)

2=

1

4(4.143)

b4a43a32a21 =1

24(4.144) b3a32a21 [a21 + 2 (a31 + a32)]

+b4a42a21 [a21 + 2 (a41 + a42 + a43)]+b4a43 (a31 + a32) [a31 + a32 + 2 (a43 + a42 + a41)]

=1

3(4.145)

a321b2 + b3(a31 + a32)3

+ b4(a41 + a42 + a43)3

=1

4(4.146)

We now need to focus on three special equations of the above system.

αb2 + βb3 + γb4 =1

2(4.147)

α2b2 + β2b3 + γ2b4 =1

3(4.148)

α3b2 + β3b3 + γ3b4 =1

4(4.149)


Case (α = β) ∨ (β = γ) ∨ (γ = α).Reader do this case as exercises.

Case α, β, γ are pairwise distinct.Solving this system of three equations respect to b2, b3, b4, we obtain

b2 = − 4β + 4γ − 6βγ − 3

12α (α− β) (α− γ)(4.150)

b3 = − 4γ + 4α− 6γα− 3

12β (β − γ) (β − α)(4.151)

b4 = − 4α+ 4β − 6αβ − 3

12γ (γ − α) (γ − β)(4.152)

Then we use (4.128) to obtain

b1 =4 (α+ β + γ)− 6 (αβ + βγ + γα) + 12αβγ − 3

12αβγ(4.153)

Now using b1, b2, b3, b4, we need to focus on the following three equations ofthe remaining system.

αb3a32 + b4 (αa42 + βa43) =1

6(4.154)

α2b3a32 + b4(α2a42 + β2a43

)=

1

12(4.155)

αβb3a32 + γb4 (αa42 + βa43) =1

8(4.156)

Solving this system of three equations respect to a32, a42, a43, we obtain

a32 =β (4γ − 3) (α− β)

2α (4α+ 4γ − 6αγ − 3)(4.157)

a42 =γ (α− γ)

(4β2 − 5β + 3α+ 2γ − 4αγ

)2α (6α2β − 4α2 − 6αβ2 + 3α+ 4β2 − 3β)

(4.158)

a43 =γ (1− 2α) (γ − α) (γ − β)

β (6α2β − 4α2 − 6αβ2 + 3α+ 4β2 − 3β)(4.159)

Now, we only need to solve a21, a31, a41. To do this, we need to focus on thefollowing three equations of the remaining system.

b2a21 + b3a31 + b4a41 =1

2− b3a32 − b4a42 − b4a43 (4.160)

αb2a21 + βb3a31 + γb4a41 =1

3− βb3a32 − γb4a42 − γb4a43 (4.161)

(b3a32 + b4a42) a21 + b4a43a31 =1

6− b4a43a32 (4.162)

We solve this system of three equations respect to a21, a31, a41 by the followingMatlab routine because the formulas are quite long.

format long


syms a b c; % a=alpha, b=beta, c=gamma

A = [a b c; a^2 b^2 c^2; a^3 b^3 c^3];

v=[1/2;1/3;1/4];

B = A^-1*v;

b2 = B(1);

b3 = B(2);

b4 = B(3);

A1 = [a*b3 a*b4 b*b4;

a^2*b3 a^2*b4 b^2*b4;

a*b*b3 c*b4*a c*b4*b];

v1 = [1/6;1/12;1/8];

B1 = A1^-1*v1;

a32 = B1(1)

a42 = B1(2)

a43 = B1(3)

A2 = [b2 b3 b4;a*b2 b*b3 c*b4;(b3*a32+b4*a42) b4*a43 0];

v2 = [1/2-b3*32-b4*a42-b4*a43;

1/3-b*b3*a32-c*b4*a42-c*b4*a43;

1/6-b4*a43*a32];

B2 = A2^-1*v2;

a21 = B2(1)

a31 = B2(2)

a41 = B2(3)

This Matlab routine returns the formulas of a21, a31, a41 in terms of α, β, γ.Hence, we obtain the solutions of our system of equations in terms of α, β, γ.Checking whether this solutions satisfies the remaining system of equations isjust computation matter. Although those computations is quite complicated byhand, checking by Matlab is quite easy and there is no benefit to representthose computations here. You should check yourself.

In practice, we don’t need to know the final three formulas of a21, a31, a41 interms of α, β, γ. For each 3 tuples α, β, γ, we can easily solve the other unknownsby the process described above.

We have solved (4.106)-(4.124) completely. �

4.1.5 Some Cases

Case α =1

2, β =

1

2, γ = 1. Using the above process, we obtain

01

2

1

21

20

1

21 0 0 1

1

6

1

3

1

3

1

6

(4.163)



Case α =1

3, β =

2

3, γ = 1.

01

3

1

32

3−1

31

1 1 −1 11

8

3

8

3

8

1

8

(4.164)

This Butcher tableau appears in [5].


4.2.1 Curtiss-Hirchfelder Equation

In this subsection, we will use Runge Kutta order 4 method to solve numericallythe Curtiss-Hirchfelder equation.

Definition 4.1. The Curtiss-Hirchfelder equation is an ordinary differentialequation (ODE) which has the following form

dy

dt= −50 (y − cos t) (4.165)

y (0) = 1 (4.166)

In general,

dy

dt= f (t, y) (4.167)

y (x0) = y0 (4.168)

The following Matlab routines aims at

1. Approximating solution of (4.165) by using Runge Kutta order 4 method.

2. Compute absolute errors and relative errors and plot the obtained numer-ical solutions and errors.

4.2.1.1 Subroutine s.m

This subroutine provides the exact solution of (4.165).

function s = s(t)

s = 50/2501*(50*cos(t)+sin(t)) + exp(-50*t)/2501;

4.2.1.2 Subroutine f.m

This subroutine provides the function f in RHS of (4.165). Users can modifythese functions later.

function f = f(t,y)

f = -50*y + 50*cos(t);


4.2.1.3 Subroutine x.m

This subroutine provides the Runge Kutta order 4 method.

function x = x(a,b,c,h,t,y)

k1 = f(t,y);

k2 = f(t+c(2)*h,y+h*a(2,1)*k1);

k3 = f(t+c(3)*h,y+h*(a(3,1)*k1+a(3,2)*k2));

k4 = f(t+c(4)*h,y+h*(a(4,1)*k1+a(4,2)*k2+a(4,3)*k3));

x = h*(b(1)*k1+b(2)*k2+b(3)*k3+b(4)*k4);

4.2.1.4 Main Routine RK4

close all

clear all

clc

format long

tic

%% Initial.

A1(1) = 1;

A2(1) = 1;

% A3(1) = 1;

N=1000;

h=25/N;

t = 0:h:25;

%% Coefficients

a1 = [0 0 0 0 ;

1/2 0 0 0 ;

0 1/2 0 0 ;

0 0 1 0];

b1 = [1/6 1/3 1/3 1/6];

c1 = [0 1/2 1/2 1];

a2 = [0 0 0 0 ;

1/3 0 0 0 ;

-1/3 1 0 0 ;

1 -1 1 0];

b2 = [1/8 3/8 3/8 1/8];

c2 = [0 1/3 2/3 1];

%% Numerical Solution.

for n=1:N

A1(n+1) = A1(n) + x(a1,b1,c1,h,t(n),A1(n));

A2(n+1) = A2(n) + x(a2,b2,c2,h,t(n),A2(n));

end

%% Plot Numerical Solution


figure(1)

hold on

plot(t,s(t),’b’);

plot(t,A1,’r’);

legend(’Exact Solution’,’Numerical Solution’);

title(’Numerical Solution (Butcher Table 1)’);

figure(2)

hold on

plot(t,s(t),’b’);

plot(t,A2,’r’);

legend(’Exact Solution’,’Numerical Solution’);

title(’Numerical Solution (Butcher Table 2)’);

%% Absolute Error.

display(’Absolute Error 1’)

ae1 = h*sum(abs(A1-s(t)))

display(’Absolute Error 2’)

ae2 = h*sum(abs(A2-s(t)))

figure(3)

hold on

plot(t,abs(A1-s(t)),’g’);

plot(t,abs(A2-s(t)),’b’);

legend(’a1 b1 c1’,’a2 b2 c2’);

title(’Absolute Error’);

%% Relative Error.

display(’Relative Error 1’)

re1 = h*sum(abs((A1-s(t))./s(t)))

display(’Relative Error 2’)

re2 = h*sum(abs((A2-s(t))./s(t)))

figure(4)

hold on

plot(t,abs((A1-s(t))./A1),’g’);

plot(t,abs((A2-s(t))./A2),’b’);


title(’Relative Error’);

toc


4.2.1.5 Results

Figure 4.1: Numerical Solutions.


Figure 4.2: Absolute Errors.

4.2.2 Brusselator Equation

Definition 4.2. The Brusselator equation is the following system of equations

dy1dt

= 1− 4y1 + y21y2 (4.169)

dy2dt

= 3y1 − y21y2 (4.170)

In general,

dy1dt

= f1 (t, y1, y2) (4.171)

dy2dt

= f2 (t, y1, y2) (4.172)

In vector form y = (y1, y2),

dy

dt= f (t, y) (4.173)

4.2.2.1 Subroutine f.m

This subroutine provides f1 and f2 in the RHS of (4.171) and (4.172). Userscan modify these functions later.

function f = f(t,y)

f(1) = 1-4*y(1)+y(1)^2*y(2);

f(2) = 3*y(1)-y(1)^2*y(2);


Figure 4.3: Relative Errors.

4.2.2.2 Subroutine x.m

This subroutine provides the Runge Kutta order 4 method.

function x = x(a,b,c,h,t,y)

k1 = f(t,y);

k2 = f(t+c(2)*h,y+h*a(2,1)*k1);

k3 = f(t+c(3)*h,y+h*(a(3,1)*k1+a(3,2)*k2));

k4 = f(t+c(4)*h,y+h*(a(4,1)*k1+a(4,2)*k2+a(4,3)*k3));

x = h*(b(1)*k1+b(2)*k2+b(3)*k3+b(4)*k4);

4.2.2.3 Main Routine RK4.m

clear all

close all

clc

format long

tic

%% Initial for Reference Solutions.

N0 = 10^6;

h0= 20/N0;

t0 = 0:h0:20;

B1 = zeros(N0,2);


B1(1,1) = 1.5;

B1(1,2) = 3;

B2 = zeros(N0,2);

B2(1,1) = 1.5;

B2(1,2) = 3;

%% Initial for Numerical Solutions.

N = 10^4;

h = 20/N;

t = 0:h:20;

A1 = zeros(N,2);

A1(1,1) = 1.5;

A1(1,2) = 3;

A2 = zeros(N,2);

A2(1,1) = 1.5;

A2(1,2) = 3;

A3 = zeros(N,2);

A3(1,1) = 1.5;

A3(1,2) = 3;

% Step

step = N0/N;

%% Runge Kutta initials.

% Coefficients

a1 = [0 0 0 0 ;

1/2 0 0 0 ;

0 1/2 0 0 ;

0 0 1 0];

b1 = [1/6 1/3 1/3 1/6];

c1 = [0 1/2 1/2 1];

a2 = [0 0 0 0 ;

1/3 0 0 0 ;

-1/3 1 0 0 ;

1 -1 1 0];

b2 = [1/8 3/8 3/8 1/8];

c2 = [0 1/3 2/3 1];

%% Reference Solutions.

for n=1:N0

B1(n+1,:) = B1(n,:) + x(a1,b1,c1,h0,t0(n),B1(n,:));

B2(n+1,:) = B2(n,:) + x(a2,b2,c2,h0,t0(n),B2(n,:));

end

%% Numerical Solutions, Absolute Errors and Relative Errors.


ae1 = zeros(N+1,2);

ae2 = zeros(N+1,2);

re1 = zeros(N+1,2);

re2 = zeros(N+1,2);

for n=1:N

% Numerical Solutions

A1(n+1,:) = A1(n,:) + x(a1,b1,c1,h,t(n),A1(n,:));

A2(n+1,:) = A2(n,:) + x(a2,b2,c2,h,t(n),A2(n,:));

% Absolute Errors

ae1(n,:) = h.*abs(A1(n,:)-B1(step*(n-1)+1,:));

ae2(n,:) = h.*abs(A2(n,:)-B2(step*(n-1)+1,:));

% ae1(n,1) = abs((A1(n,1)-s(1,t));

% ae1(n,2) = abs((A1(n,2)-s(2,t));

% ae2(n,1) = abs((A2(n,i)-s(1,t));

% ae2(n,2) = abs((A2(n,i)-s(2,t));

% Relative Errors

re1(n,:) = h.*abs((A1(n,:) ...

-B1(step*(n-1)+1,:))./B1(step*(n-1)+1,:));

re2(n,:) = h.*abs((A2(n,:) ...

-B2(step*(n-1)+1,:))./B2(step*(n-1)+1,:));

% re1(n,1) = abs((A1(n,1)-s(1,t))./s(i,t));

% re1(n,2) = abs((A1(n,2)-s(2,t))./s(i,t));

% re2(n,1) = abs((A2(n,i)-s(1,t))./s(i,t));

% re2(n,2) = abs((A2(n,i)-s(2,t))./s(i,t));

end

% Absolute Errors

display(’Absolute Error of Table 1’)

ae1_y1 = h*sum(ae1(:,1))

ae1_y2 = h*sum(ae1(:,2))


ae2_y1 = h*sum(ae2(:,1))

ae2_y2 = h*sum(ae2(:,2))

% Relative Errors

display(’Relative Error of Table 1’)

re1_y1 = h*sum(re1(:,1))

re1_y2 = h*sum(re1(:,2))


re2_y1 = h*sum(re2(:,1))

re2_y2 = h*sum(re2(:,2))



figure(1)

subplot(2,1,1)

hold on

% plot(t,s(1,t),’b’)

plot(t0,B1(:,1),’b’)

plot(t,A1(:,1),’r’)

legend(’Exact/Reference’,’Numerical Solution’);

title(’Numerical Solution for Y1 (Butcher Table 1)’);

subplot(2,1,2)

hold on


plot(t0,B1(:,2),’b’)

plot(t,A1(:,2),’r’)



figure(2)

subplot(2,1,1)

hold on


plot(t0,B2(:,1),’b’)

plot(t,A2(:,1),’r’)



subplot(2,1,2)

hold on


plot(t0,B2(:,2),’b’)

plot(t,A2(:,2),’r’)



%% Plot: Dependency of y2 with respect to y1.

figure(3)

hold on

plot(A1(:,1),A1(:,2),’b’);

plot(A2(:,1),A2(:,2),’g’);

% plot(A3(:,1),A3(:,2),’r’);


title(’Dependency of y2 with respect to y1’);

%% Plot Absolute Errors.

figure(4)

subplot(2,1,1)

hold on

plot(t,ae1(:,1),’b’);


plot(t,ae2(:,1),’g’);

% plot(t,ae3(:,1),’r’);


title(’Absolute Errors Y1’);

subplot(2,1,2)

hold on

plot(t,ae1(:,2),’b’);

plot(t,ae2(:,2),’g’);

% plot(t,ae3(:,2),’r’);



%% Plot Relative Errors.

figure(5)

subplot(2,1,1)

hold on

plot(t,re1(:,1),’b’);

plot(t,re2(:,1),’g’);

% plot(t,re3(:,1),’r’);


title(’Relative Errors Y1’);

subplot(2,1,2)

hold on

plot(t,re1(:,2),’b’);

plot(t,re2(:,2),’g’);

% plot(t,re3(:,2),’r’);



toc


4.2.2.4 Results



Figure 4.5: Dependency of y2 with respect to y1.

4.2.3 Examples

4.2.3.1 Subroutine s2.m

function s2 = s2(t)

s2 = [3*exp(2*t)-2*exp(3*t);-3*exp(2*t)+4*exp(3*t)];

4.2.3.2 Subroutine f2.m

function f2 = f2(t,y)

% f2=[y(1)-y(2),2*y(1)+4*y(2)];

f2(1) = y(1)-y(2);

f2(2) = 2*y(1)+4*y(2);

4.2.3.3 Subroutine x2.m

function x2 = x2(a,b,c,h,t,y)

k1 = f2(t,y);

k2 = f2(t+c(2)*h,y+h*a(2,1)*k1);

k3 = f2(t+c(3)*h,y+h*(a(3,1)*k1+a(3,2)*k2));

k4 = f2(t+c(4)*h,y+h*(a(4,1)*k1+a(4,2)*k2+a(4,3)*k3));

x2 = h*(b(1)*k1+b(2)*k2+b(3)*k3+b(4)*k4);

4.2.3.4 Main Routine RK4ex2.m

clear all



close all

clc

format long

tic

% %% Initial for Reference Solutions.

% N0 = 10^6;

% h0= 20/N0;

% t0 = 0:h0:20;

%

% B1 = zeros(N0,2);

% B1(1,1) = 1;

% B1(1,2) = 1;

%

% B2 = zeros(N0,2);

% B2(1,1) = 1;

% B2(1,2) = 1;

%% Initial for Numerical Solutions.

N = 10^4;

h = 1/N;

t = 0:h:1;

A1 = zeros(N+1,2);



A1(1,1) = 1;

A1(1,2) = 1;

A2 = zeros(N+1,2);

A2(1,1) = 1;

A2(1,2) = 1;

A3 = zeros(N+1,2);

A3(1,1) = 1;

A3(1,2) = 1;

% Step

% step = N0/N;

%% Runge Kutta initials.

% Coefficients

a1 = [0 0 0 0 ;

1/2 0 0 0 ;

0 1/2 0 0 ;

0 0 1 0];

b1 = [1/6 1/3 1/3 1/6];

c1 = [0 1/2 1/2 1];

a2 = [0 0 0 0 ;


1/3 0 0 0 ;

-1/3 1 0 0 ;

1 -1 1 0];

b2 = [1/8 3/8 3/8 1/8];

c2 = [0 1/3 2/3 1];

% %% Reference Solutions.

% for n=1:N0

% B1(n+1,:) = B1(n,:) + x(a1,b1,c1,h0,t0(n),B1(n,:));

% B2(n+1,:) = B2(n,:) + x(a2,b2,c2,h0,t0(n),B2(n,:));

% end

s=s2(t)’;

%% Numerical Solutions, Absolute Errors and Relative Errors.

ae1 = zeros(N+1,2);

ae2 = zeros(N+1,2);

re1 = zeros(N+1,2);

re2 = zeros(N+1,2);

for n=1:N

% Numerical Solutions

A1(n+1,:) = A1(n,:) + x2(a1,b1,c1,h,t(n),A1(n,:));

A2(n+1,:) = A2(n,:) + x2(a2,b2,c2,h,t(n),A2(n,:));

% Absolute Errors

% ae1(n,:) = h.*abs(A1(n,:)-B1(step*(n-1)+1,:));

ae1 = (abs(A1-s));

% ae2(n,:) = h.*abs(A2(n,:)-B2(step*(n-1)+1,:));

ae2 = (abs(A2-s));

% ae1(n,1) = abs((A1(n,1)-s(1,t));

% ae1(n,2) = abs((A1(n,2)-s(2,t));

% ae2(n,1) = abs((A2(n,i)-s(1,t));

% ae2(n,2) = abs((A2(n,i)-s(2,t));

% Relative Errors

% re1(n,:) = h.*abs((A1(n,:)-B1(step*(n-1)+1,:))./B1(step*(n-1)+1,:));

re1 = (abs((A1-s)./s));

% re2(n,:) = h.*abs((A2(n,:)-B2(step*(n-1)+1,:))./B2(step*(n-1)+1,:));

re2 = (abs((A2-s)./s));

% re1(n,1) = abs((A1(n,1)-s(1,t))./s(i,t));

% re1(n,2) = abs((A1(n,2)-s(2,t))./s(i,t));

% re2(n,1) = abs((A2(n,i)-s(1,t))./s(i,t));

% re2(n,2) = abs((A2(n,i)-s(2,t))./s(i,t));

end

% Absolute Errors


ae1_y1 = h*sum(ae1(:,1))


ae1_y2 = h*sum(ae1(:,2))


ae2_y1 = h*sum(ae2(:,1))

ae2_y2 = h*sum(ae2(:,2))

% Relative Errors


re1_y1 = h*sum(re1(:,1))

re1_y2 = h*sum(re1(:,2))


re2_y1 = h*sum(re2(:,1))

re2_y2 = h*sum(re2(:,2))


figure(1)

subplot(2,1,1)

hold on


plot(t,s(:,1),’b’)

plot(t,A1(:,1),’r’)



subplot(2,1,2)

hold on


plot(t,s(:,2),’b’)

plot(t,A1(:,2),’r’)



figure(2)

subplot(2,1,1)

hold on


plot(t,s(:,1),’b’)

plot(t,A2(:,1),’r’)



subplot(2,1,2)

hold on


plot(t,s(:,2),’b’)

plot(t,A2(:,2),’r’)




%% Plot: Dependency of y2 with respect to y1.

figure(3)

hold on

plot(A1(:,1),A1(:,2),’b’);

plot(A2(:,1),A2(:,2),’g’);

% plot(A3(:,1),A3(:,2),’r’);

legend(’Butcher Table 1’,’Butcher Table 2’);

title(’Dependency of y2 with respect to y1’);

%% Plot Absolute Errors.

figure(4)

subplot(2,1,1)

hold on

plot(t,ae1(:,1),’b’);

plot(t,ae2(:,1),’g’);

% plot(t,ae3(:,1),’r’);



subplot(2,1,2)

hold on

plot(t,ae1(:,2),’b’);

plot(t,ae2(:,2),’g’);

% plot(t,ae3(:,2),’r’);



%% Plot Relative Errors.

figure(5)

subplot(2,1,1)

hold on

plot(t,re1(:,1),’b’);

plot(t,re2(:,1),’g’);

% plot(t,re3(:,1),’r’);



subplot(2,1,2)

hold on

plot(t,re1(:,2),’b’);

plot(t,re2(:,2),’g’);

% plot(t,re3(:,2),’r’);



toc


4.2.3.5 Results




Figure 4.10: Dependency of y2 with respect to y1.




Alphabetical Index

Aabsolute errors 33autonomous form 1

BBrusselator equation 37Butcher table 3Butcher tableau 2

CCurtiss-Hirchfelder equation 33

EEuler’s method 3explicit Euler’s method 3

Ffamily of explicit Runge Kutta

methods 2

Iinitial value problems 1iterative method 1

KKutta’s third order method 19

Mmesh 2

Nnon-autonomous form 1

Oorder of accuracy 1

Rrelative errors 33Runge Kutta method 1Runge Kutta method of second order

7Runge Kutta order 1 method 3Runge Kutta order 2 method 4Runge Kutta order 3 method 8Runge Kutta order 4 method 20, 33

Sstep size 2stiff problems 1

TTaylor’s series 1

53

Bibliography

[1] Richard L. Burden, J. Douglas Faires, Numerical Analysis, Ninth Edition.

[2] David Goeken, Olin Johnson, Fifth-order Runge-Kutta with higher orderderivative approximations, Electronic Journal of Differential Equations,Conference 02, 1999, pp. 1-9.

[3] J. C. Butcher, On fifth and sixth order explicit Runge-Kutta methods: Or-der conditions and Order barriers, Canadian applied mathematics quar-terly, Volume 17, Number 3, Fall 2009.

[4] John Butcher, Runge-Kutta methods for ordinary differential equations,The University of Auckland New Zealand, COE Workshop on NumericalAnalysis, Kyushu University, May 2005.

[5] https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods

[6] https://en.wikipedia.org/wiki/List_of_Runge%E2%80%93Kutta_

methods

54

https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods

https://en.wikipedia.org/wiki/List_of_Runge%E2%80%93Kutta_methods

https://en.wikipedia.org/wiki/List_of_Runge%E2%80%93Kutta_methods

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