Project on Runge Kutta method

Transcription

1 Project on Runge Kutta method Nguyen Quan Ba Hong Doan Tran Nguyen Tung Nguyen An Thinh 3 Students at Faculty of Math and Computer Science Ho Chi Minh University of Science, Vietnam nguyenquanbahong@gmail.com dtrngtung@live.com blog. 4 October, 06 Student ID. 403 Student ID Student ID Copyright c 06 by Nguyen Quan Ba Hong, Student at Ho Chi Minh University of Science, Vietnam. This document may be copied freely for the purposes of education and non-commercial research. Visit my site com to get more.

2 Contents List of Figures iii Introduction. Initial Value Problems Mesh General Runge Kutta method Runge Kutta Order Method Runge Kutta order method 4. Derivation of Runge Kutta Order Method Runge Kutta Order Formula Taylor Series Expansion Formula Derivation of System of Equations Solutions of System of Equations Some Cases Matlab Implementation Runge Kutta order 3 method 8 3. Derivation of Runge Kutta Order 3 Method Runge Kutta Order 3 Formula Taylor Series Expansion Formula Derivation of System of Equations Solutions of System of Equations Some Cases Matlab Implementation Runge Kutta Order 4 Method 0 4. Derivation of Runge Kutta Order 4 Method Runge Kutta Order 4 Formula Taylor Series Expansion Formula Derivation of System of Equations Solutions of System of Equations Some Cases Matlab Implementation Curtiss-Hirchfelder Equation Subroutine s.m Subroutine f.m Subroutine x.m i

3 CONTENTS ii Main Routine RK Results Brusselator Equation Subroutine f.m Subroutine x.m Main Routine RK4.m Results Examples Subroutine s.m Subroutine f.m Subroutine x.m Main Routine RK4ex.m Results

4 List of Figures 4. Numerical Solutions Absolute Errors Relative Errors Numerical Solutions Dependency of y with respect to y Absolute Errors Relative Errors Numerical Solutions Numerical Solutions Dependency of y with respect to y Absolute Errors Relative Errors iii

5 Chapter Introduction. Initial Value Problems The Runge Kutta methods are an important family of iterative methods for the approximation of solutions of ODE s, that were developed around 900 by the German mathematicians Carl Runge (856-97) and Martin. W. Kutta ( ). Modern developments are mostly due to John Butcher in the 960s. We consider initial value problems expressed in autonomous form. Starting with the non-autonomous, we assume that f (x, y) is a continuous function with domain [a, b] R n where t [a, b] and y R n. Consider the initial value problem where We assume that dy (t) = f (t, y (t)) dt (.) y (x 0 ) = y 0 (.) y (t) = (y (t), y (t),..., y n (t)) T (.3) f : [a, b] R n R n (.4) f (t, y ) f (t, y ) L (R n ) L y y L (R n ) (.5) for all t [a, b], y R n, y R n. Thus the IVP (.) has a unique solution. For convenience, we write (.) briefly as y t = f (.6) Most efforts to increases the order of the Runge Kutta methods have been accomplished by increasing the number of Taylor s series terms used and thus the number of functional evaluations, e.g. Butcher 987, Gear 97. The use of higher order derivative terms has been proposed for stiff problems, e.g. Rosenbrock 963, Enright 974. Our method add higher order derivative terms to the Runge Kutta k i terms (i > ) to achieve a higher order of accuracy. We are interested in a numerical approximation of the continuously differentiable solution y (t) of the IVP (.) over the time interval t [a, b].

6 CHAPTER. INTRODUCTION. Mesh We subdivide the interval [a, b] into M equal subintervals and select the mesh points t j t j = a + jh, j = 0,,..., M (.7) where is called a step size. h = b a M (.8).3 General Runge Kutta method The family of explicit Runge Kutta (RK) methods of the mth stage is given by where s y (n+) = y (n) + h b i k i (.9) i= k i = f (τ i, η i ), i =,,..., s (.0) and τ i = t n + c i h (.) i η i = y n + h a ij k j (.) j= i = y n + h a ij f (τ j, η j ) (.3) j= We use the notation f n := f (t n, y n ) (.4) To specify a particular method, we need to provide the integer s (the number of stages), and the coefficients c i, i =,..., s, a ij, j < i m and b i, i =,,..., s. These data are usually arranged in a so-called Butcher tableau (after John C. Butcher) 0 c a c 3 a 3 a c s a s a s a ss b b b s b s (.5)

7 CHAPTER. INTRODUCTION 3.4 Runge Kutta Order Method We consider the Runge Kutta order method here because it is very short and easy. There is no need to represent this into a separated chapter. For s =, (.0) becomes and (.0) becomes On the other hand, the Taylor expansion yields k = f (t n, x n ) (.6) y (n+) = y (n) + hb k (.7) = y (n) + hb f n (.8) y (n+) = y (n) + hy t tn + O ( h ) (.9) Comparing (.8) and (.0), we obtain easily = y (n) + hf n + O ( h ) (.0) b = (.) Hence, The Butcher table in this case has the following form 0 (.) Remark.. The Runge Kutta order method is equivalent to the explicit Euler 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.

8 Chapter Runge Kutta order method. Derivation of Runge Kutta Order Method To set up the Runge Kutta order method, we need to do 4 steps.. Write down Runge Kutta order formula described by (.9) and (.0).. Write down Taylor series expansion. 3. Compare the coefficients of two formulas above to obtain a system of equations. 4. Solve the system of equation or find some its solutions. Remark.. This process is also applied for Runge Kutta method of higher orders. Therefore, we can regard it as the standard process for derivation of the general Runge Kutta method in this context. After these steps, with some solutions of the derived system of equations, we can simulate some IVPs easily... Runge Kutta Order Formula For s =, (.0) becomes and (.9) becomes k = f n (.) k = f (t n + c h, y n + ha f n ) (.) y (n+) = y (n) + hb k + hb k (.3) = y (n) + hb f n + hb f (t n + c h, y n + ha f n ) (.4) Now we write down the Taylor series expansion O ( h ) for k k = f (t n + c h, y n + ha f n ) (.5) = f n + c hf t + ha f n f y + O ( h ) (.6) 4

9 CHAPTER. RUNGE KUTTA ORDER METHOD 5 Inserting (.6) into (.4), we obtain y (n+) = y (n) + hb f n + hb [ fn + c hf t + ha f n f y + O ( h )] (.7) = y (n) + h (b + b ) f n + h b c f t + h b a f n f y + O ( h 3) (.8).. Taylor Series Expansion Formula We need to compute y tt for Taylor series expansion below. y tt = df dt = f t t t + f y y t (.9) (.0) = f t + ff y (.) Now we write down the Taylor series expansion of y in the neighborhood of t n with O ( h 3). y (n+) = y (n) + hy t tn + h y tt tn + O ( h 3) (.) = y (n) + hf n + h y tt tn + O ( h 3) (.3) = y (n) + hf n + h (f t + ff y ) tn + O ( h 3) (.4) = y (n) + hf n + h f t + h tn f n f y tn + O ( h 3) (.5) Remark.. (Important) From here to later, if nothing is misundertood, we can abbreviate notation f t α y for β f t α y β tn (.6) in the Taylor series expansion. For example, under this abbreviation, (.5) can be rewritten briefly as y (n+) = y (n) + hf n + h f t + h f nf y + O ( h 3) (.7) This abbreviation reduces the complexity of the formulas in this context. As you will see later, this abbreviation is really essential...3 Derivation of System of Equations Return to our problem, as usual, comparing (.8) and (.7), we obtain hf n : = b + b (.8) h f t : = b c (.9) h f n f y : = b a (.0)

10 CHAPTER. RUNGE KUTTA ORDER METHOD 6 Hence, we obtain the system of equations b + b = (.) b c = b a = (.) (.3)..4 Solutions of System of Equations We can solve the above system of equations easily. Here, the authors represent two solutions. The idea behind these two solutions will be use later for higher order Runge Kutta method. Solution. The system involves four unknowns in three equations. Taking b = α is free variable. Due to (.3) we must have α 0. Then we can easily obtain the general solution for (.)-(.3) is b = α (.4) b = α (.5) c = a = (.6) α where α is an arbitrary real number. Butcher tableau in this case becomes 0 α α α α (.7) Solution. Due to (.) and (.), we must have c = a. So, we can take c = a = β (.8) as a free variable. And the remaining is very easy, we obtain Butcher tableau in this case becomes 0 β This Butcher tableau appears in [5]. b = β (.9) b = β (.30) c = a = β (.3) β β β (.3) Remark.3. Since Runge Kutta order is still simple, you can not see the differences between two solutions. With Runge Kutta of higher order, you will see that the first choice of free variables is very important in entire solution.

11 CHAPTER. RUNGE KUTTA ORDER METHOD 7..5 Some Cases We discuss two useful choices Case α =. In this case, (.6) becomes b = (.33) The corresponding Butchet tableau reads b = (.34) c = a = (.35) 0 (.36) Thus, in this case the Runge Kutta method of second order takes the form y (n+) = y (n) + h [f n + f (t n + h, y n + hf n )] (.37) and is equivalent to the Heun s method. Case α =. In this case, (.6) becomes b = (.38) b = 0 (.39) c = a = (.40) The corresponding Butcher tableau reads 0 0 (.4) In this case Runge Kutta method of second order can be written as ( y (n+) = y (n) + hf t n + h, y n + h ) f n (.4) and is called the RK method. Remember α R, so there is infinite many choices of solution for (.)-(.3).. Matlab Implementation See section 4. for general code.

12 Chapter 3 Runge Kutta order 3 method 3. Derivation of Runge Kutta Order 3 Method 3.. Runge Kutta Order 3 Formula For s = 3, (.0) becomes k = f n (3.) k = f (t n + c h, y n + ha f n ) (3.) k 3 = f (t n + c 3 h, y n + ha 3 k + ha 3 k ) (3.3) = f (t n + c 3 h, y n + ha 3 f n + ha 3 f (t n + c h, y n + ha 3 k + ha 3 k )) (3.4) and (.9) becomes y (n+) = y (n) + hb k + hb k + hb 3 k 3 (3.5) Now we write down the Taylor series expansion O ( h 3) for k. k = f (t n + c h, y n + ha f n ) (3.6) = f n + c hf t + ha f n f y + h c f tt + h c a f n f ty + h a f nf yy (3.7) + O ( h 3) (3.8) And we also write down the Taylor series expansion O ( h 3) for k 3. k 3 = f (t n + c 3 h, y n + ha 3 f n + ha 3 k ) (3.9) = f n + c 3 hf t + h (a 3 f n + a 3 k ) f y (3.0) + h c 3 f tt + c 3 h (a 3 f n + a 3 k ) f ty + h (a 3f n + a 3 k ) f yy (3.) + O ( h 3) (3.) 8

13 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD 9 Inserting (3.8) into (3.) k 3 = f (t n + c 3 h, y n + ha 3 f n + ha 3 k ) (3.3) = f n + c 3 hf t + ha 3 f n f y (3.4) + ha 3 f y (f n + hc f t + ha f n f y ) (3.5) + h c 3 f tt + c 3 h (a 3 f n + a 3 f n ) f ty (3.6) ( ) a + h 3 fn + a 3fn + a 3 a 3 fn f yy (3.7) + O ( h 3) (3.8) Collecting terms respect to exponents of h k 3 = f n + h (c 3 f t + a 3 f n f y + a 3 f y f n ) (3.9) a 3 c f t f y + a a 3 f n fy + c 3 f tt + h +c 3 a 3 f n f ty + c 3 a 3 f n f ty + a 3 f nf yy (3.0) + a 3 f nf yy + a 3 a 3 fnf yy + O ( h 3) (3.) Inserting (3.) and (3.) into (3.5) y (n+) = y (n) + hb f n (3.) f n + hc f t + ha f n f y + h c f tt + hb +h c a f n f ty + h a f nf yy f n + h (c 3 f t + a 3 f n f y + a 3 f y f n ) a 3 c f t f y + a a 3 f n fy + c 3 f tt + hb 3 +h +c 3 a 3 f n f ty + c 3 a 3 f n f ty + a 3 f nf yy + a 3 (3.3) (3.4) f nf yy + a 3 a 3 fnf yy + O ( h 4) (3.5) Collecting terms respect to exponents of h y (n+) = y (n) + h (b f n + b f n + b 3 f n ) (3.6) + h (b c f t + a b f n f y + b 3 c 3 f t + a 3 b 3 f n f y + a 3 b 3 f y f n ) (3.7) b c f tt + a b c f n f ty + a b f nf yy +a 3 b 3 c f t f y + a a 3 b 3 f n f + h 3 y + b 3c 3 f tt +a 3 b 3 c 3 f n f ty + a 3 b 3 c 3 f n f ty + a 3b 3 f (3.8) nf yy + a 3b 3 f nf yy + a 3 a 3 b 3 fnf yy + O ( h 4) (3.9)

14 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD Taylor Series Expansion Formula We need to compute y ttt for Taylor series expansion below. y ttt = d dt (f t + ff y ) (3.30) = (f t + ff y ) t + f(f t + ff y ) y (3.3) = f tt + f t f y + ff ty + ff ty + ff y + f f yy (3.3) = f tt + f t f y + ff ty + ff y + f f yy (3.33) Now we write down the Taylor series expansion of y in the neighborhood of t n with O ( h 4). y (n+) = y (n) + hy t + h y tt + h3 6 y ttt + O ( h 4) (3.34) = y (n) + hf n + h (f t + f n f y ) (3.35) + h3 ( ftt + f t f y + f n f ty + f n fy + f ) ( 6 nf yy + O h 4 ) (3.36) 3..3 Derivation of System of Equations Comparing (3.9) and (3.36) hf n : = b + b + b 3 (3.37) h f t : = b c + b 3 c 3 (3.38) h f n f y : = a b + a 3 b 3 + a 3 b 3 (3.39) h 3 f tt : 6 = b c + b 3c 3 (3.40) h 3 f t f y : 6 = a 3b 3 c (3.4) h 3 f n f ty : 3 = a b c + a 3 b 3 c 3 + a 3 b 3 c 3 (3.4) h 3 f n f y : 6 = a a 3 b 3 (3.43) h 3 f nf yy : 6 = a b + a 3b 3 + a 3b 3 + a 3 a 3 b 3 (3.44)

15 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD Hence, we obtain the system of 8 equations with 8 unknowns. b + b + b 3 = (3.45) b c + b 3 c 3 = b a + b 3 (a 3 + a 3 ) = b c + b 3 c 3 = 3 b 3 a 3 c = 6 b c a + b 3 c 3 (a 3 + a 3 ) = 3 b 3 a 3 a = 6 b a + b 3 (a 3 + a 3 ) = 3 (3.46) (3.47) (3.48) (3.49) (3.50) (3.5) (3.5) 3..4 Solutions of System of Equations We now solve the above system of equations in two different ways. Solution. We take b = α, b 3 = β as two free variables, then b = α β (3.53) our task remains to solve αc + βc 3 = αa + β (a 3 + a 3 ) = αc + βc 3 = 3 βa 3 c = 6 αc a + βc 3 (a 3 + a 3 ) = 3 βa 3 a = 6 αa + β(a 3 + a 3 ) = 3 (3.54) (3.55) (3.56) (3.57) (3.58) (3.59) (3.60) We can solve c and c 3 by using two equations αc + βc 3 = αc + βc 3 = 3 (3.6) (3.6) Since (3.59), we must have β 0.

16 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD Then we can take c 3 = β α β c from (3.6) and insert into (3.6) to obtain the following equation respect to c. Consider two following cases. α (α + β) c αc + 3 4β = 0 (3.63). Case α (α + β) = 0. Consider two subcases. (a) Case α = 0. Then we have immediately b = 4 β = 3 4 c 3 = 3 (3.64) (3.65) (3.66) and the system of equations remains a 3 + a 3 = 3 (3.67) a 3 c = a 3a = 6 (3.68) (3.69) To solve this remaining system of equations, we can take a 3 = γ 0 as a free variable. Then Hence, we obtain the solutions a 3 = 3 γ (3.70) c = 9γ a = 9γ (3.7) (3.7) b = 4 (3.73) b = 0 (3.74) b 3 = 3 4 c = 9γ c 3 = 3 a = 9γ (3.75) (3.76) (3.77) (3.78) a 3 = 3 γ (3.79) a 3 = γ (3.80)

17 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD 3 where γ is an arbitrary nonzero real number, in this subcase. Butcher tableau becomes 0 9γ 3 9γ 3 γ γ (3.8) (b) Case α + β = 0, α 0. Then we have immediately and the system of equations remains b = (3.8) b = α (3.83) b 3 = α (3.84) c = 3 + 4α (3.85) We again have immediately c c 3 = α a (a 3 + a 3 ) = α c c 3 = 3α a 3 c = 6α c a c 3 (a 3 + a 3 ) = 3α a 3 a = 6α a (a 3 + a 3 ) = 3α c 3 = 3 4α a 3 = 4α + 3 a = 3 + 4α a 3 = 6α + 4α 9 α (4α + 3) (3.86) (3.87) (3.88) (3.89) (3.90) (3.9) (3.9) (3.93) (3.94) (3.95) (3.96)

18 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD 4 Hence, we obtain the solution b = (3.97) b = α (3.98) b 3 = α (3.99) c = 3 + 4α (3.00) c 3 = 3 4α (3.0) a = 3 + 4α (3.0) a 3 = 6α + 4α 9 α (4α + 3) (3.03) a 3 = 4α + 3 (3.04) where α is an arbitrary nonzero real number, in this subcase. Butcher tableau becomes α 3 4α 3 + 4α 6α + 4α 9 α (4α + 3) 4α + 3 α α (3.05). Case α (α + β) 0. (3.63) is a quadratic equation respect to c. Computing the determinant of (3.63) = 36α + α (α + β) (4β 3) (3.06) = αβ (4α + 4β 3) (3.07) Hence, we have to make the assumption αβ (4α + 4β 3) 0 (3.08) so that (3.63) has roots in this case. Under this assumption, (3.63) have two roots c = 3α ± 3αβ (4α + 4β 3) 6α (α + β) (3.09) Consider two subcases respect to c. (a) Case c = 3α + 3αβ (4α + 4β 3). We easily solve the remaining 6α (α + β)

19 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD 5 system of equations to get b = α β (3.0) b = α (3.) b 3 = β (3.) c = 3α + 3αβ (4α + 4β 3) 6α (α + β) c 3 = 3β 3αβ (4α + 4β 3) 6β (α + β) (3.3) (3.4) a = 3α + 3αβ (4α + 4β 3) (3.5) 6α (α + β) a 3 = β 3α + 3αβ (4α + 4β 3) (3.6) 6β (α + β) α (α + β) ( β 3α + ) (3.7) 3αβ (4α + 4β 3) α (α + β) a 3 = ( β 3α + ) (3.8) 3αβ (4α + 4β 3) Butcher tableau reads all obtained coefficients. (b) Case c = 3α 3αβ (4α + 4β 3). We also easily solve the remaining system of equations to 6α (α + β) get b = α β (3.9) b = α (3.0) b 3 = β (3.) c = 3α 3αβ (4α + 4β 3) 6α (α + β) c 3 = 3β + 3αβ (4α + 4β 3) 6β (α + β) (3.) (3.3) a = 3α 3αβ (4α + 4β 3) (3.4) 6α (α + β) a 3 = β 3α 3αβ (4α + 4β 3) (3.5) 6β (α + β) α (α + β) ( β 3α ) (3.6) 3αβ (4α + 4β 3) α (α + β) a 3 = ( β 3α ) (3.7) 3αβ (4α + 4β 3) Butcher tableau reads all obtained coefficients. We have solved the system of equations (3.45)-(3.5) completely.

20 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD 6 Remark 3.. In the first solution, we have used b and b 3 as two free variables. This choice makes square roots appear in the solutions. This is quite easy to understand. Because of choice of b, b 3 as free variables, we have to solve a quadratic equation. This quadratic equation make square roots appear obviously. Now, we solve our system of equation by alternative ways. The idea is very simple. It is just a matter of the first choice. More explicitly, instead of choosing b, b 3 as two free variables, we will choose c, c 3 as two free variables. Let us see the differences between two solutions through the following second one. Solution. We take c = α, c 3 = β as two free variables and focus on the following two equations of our system of equations. b c + b 3 c 3 = b c + b 3 c 3 3 = 3 (3.8) (3.9) We consider two cases respect to c and c 3.. Case c = c 3 = α. Due to (3.57), we must have α 0. Then the above sub-system of equations becomes Hence b + b 3 = α (3.30) b + b 3 = 3α (3.3) c = c 3 = α = 3 (3.3) The remaining system of equations is b + b + b 3 = (3.33) b + b 3 = 3 4 b a + b 3 (a 3 + a 3 ) = b 3 a 3 = 4 b 3 a 3 a = 6 b a + b 3 (a 3 + a 3 ) = 3 (3.34) (3.35) (3.36) (3.37) (3.38) We obtain immediately b = 4 a = 3 (3.39) (3.40)

21 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD 7 and the remaining system of equations is We now choose b 3 = γ then b + b 3 = b + b 3 (a 3 + a 3 ) = b 3 a 3 = b + b 3 (a 3 + a 3 ) = 3 (3.4) (3.4) (3.43) (3.44) b = 3 4 γ (3.45) a 3 = 3 4γ a 3 = 4γ (3.46) (3.47) Therefore, c = 3 c 3 = 3 b = 4 (3.48) (3.49) (3.50) b = 3 4 γ (3.5) b 3 = γ (3.5) a = 3 a 3 = 3 4γ a 3 = 4γ (3.53) (3.54) (3.55) where γ is an arbitrary nonzero real number, is the solution of (3.45)-(3.5) in this case.. Case c c 3. Due to (3.49) and (3.5), we have immediately Due to (3.8) and (3.9), we obtain a = c = α (3.56) b = 3β 6α (β α) b 3 = 3α 6β (α β) (3.57) (3.58)

22 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD 8 Hence, b = The remaining system of equations is 6αβ 3α 3β + 6αβ 3β 6α (β α) α + 3α 6β (α β) (a 3 + a 3 ) = 3α 6β (α β) a 3α = 6 3β 6 (β α) α + 3α 6 (α β) (a 3 + a 3 ) = 3 3β 6α (β α) α + 3α 6β (α β) (a 3 + a 3 ) = 3 We easily solve this and obtain (3.59) (3.60) (3.6) (3.6) (3.63) Therefore, a = α (3.64) β (α β) a 3 = β α (3α ) β (α β) a 3 = α (3α ) (3.65) (3.66) c = α (3.67) c 3 = β (3.68) 6αβ 3α 3β + b = 6αβ b = 3β 6α (β α) b 3 = 3α 6β (α β) (3.69) (3.70) (3.7) a = α (3.7) β (α β) a 3 = β α (3α ) β (α β) a 3 = α (3α ) (3.73) (3.74) where α β, α are two arbitrary nonzero real numbers, is the solution 3 of (3.45)-(3.5) in this case. We have solved (3.45)-(3.5) completely. Remark 3.. In the second solution,nth roots do not appear in the solution because we does not need to solve any polynomial equations. Since this way is more easy and effective, it will be used for Runge Kutta method of higher orders.

23 CHAPTER 3. RUNGE KUTTA ORDER 3 METHOD Some Cases We consider some cases of Runge Kutta order 3 method respect to some solutions of its associated system of equations. Case α = 3, β = 6. We use case.(b) above to obtain (3.75) This Butcher tableau appears in [6] - Kutta s third order method and in [4], p.8. Case α = 3 8, β = This Butcher tableau appears in [4], p.8. Case α = 3 4, β = (3.76) This Butcher tableau appears in [4], p (3.77) 3. Matlab Implementation See section 4. for general code.

24 Chapter 4 Runge Kutta Order 4 Method 4. Derivation of Runge Kutta Order 4 Method 4.. Runge Kutta Order 4 Formula For s = 4, (.0) becomes and (.9) becomes k = f n (4.) k = f (t n + c h, y n + ha f n ) (4.) k 3 = f (t n + c 3 h, y n + ha 3 f n + ha 3 k ) (4.3) k 4 = f (t n + c 4 h, y n + ha 4 f n + ha 4 k + ha 43 k 3 ) (4.4) y (n+) = y (n) + hb k + hb k + hb 3 k 3 + hb 4 k 4 (4.5) Now we write down the Taylor series expansion O ( h 4) for k. k = f (t n + c h, y n + ha f n ) (4.6) = f n + hc f t + ha f n f y + h c f tt + h a c f n f ty + h a f n (4.7) + h 3 c3 6 f ttt + h 3 a c f nf tty + h 3 a c f nf tyy + h 3 a3 6 f nf 3 yyy + O ( h 4) (4.8) ( ) c = f n + h (c f t + a f n f y ) + h f tt + a c f n f ty + a f n (4.9) ( c + h f ttt + a c ) f nf tty + a c f nf tyy + a3 6 f nf 3 yyy + O ( h 4) (4.0) 0

25 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD We write down the Taylor series expansion O ( h 4) for k 3. k 3 = f (t n + c 3 h, y n + ha 3 f n + ha 3 k ) (4.) = f n + hc 3 f t + h (a 3 f n + a 3 k ) f y (4.) + h c 3 f tt + h c 3 (a 3 f n + a 3 k ) f ty + h (a 3 f n + a 3 k ) f yy (4.3) + c3 3h 3 6 f ttt + h3 c 3 (a 3f n + a 3 k ) f tty + h3 c 3 (a 3f n + a 3 k ) f tyy (4.4) + h3 6 (a 3f n + a 3 k ) 3 f yyy + O ( h 4) (4.5) Inserting (4.0) into (4.5) k 3 = f n + hc 3 f t + ha 3 f n f y (4.6) [ ( )] c + ha 3 f y f n + h (c f t + a f n f y ) + h f tt + a c f n f ty + a f n (4.7) + h c 3 f tt + h c 3 a 3 f n f ty + h c 3 a 3 f ty [f n + h (c f t + a f n f y )] (4.8) + h a 3f nf yy + h a 3a 3 f n f yy [f n + h (c f t + a f n f y )] (4.9) + h a 3f yy [f n + h (c f t + a f n f y )] (4.0) + c3 3h 3 6 f ttt + h3 c 3 (a 3f n + a 3 f n ) f tty (4.) + h3 c 3 ( a 3 fn + a 3 a 3 fn + a 3fn ) ftyy (4.) + h3 6 ( a 3 3 fn 3 + 3a 3a 3 fn 3 + 3a 3 a 3fn 3 + a 3 3fn 3 ) fyyy + O ( h 4) (4.3) Collecting terms respect to exponents of h k 3 = f n + h (c 3 f t + a 3 f n f y + a 3 f n f y ) (4.4) + h a 3 c f t f y + a a 3 f n fy + c 3 f tt + a 3 c 3 f n f ty (4.5) +c 3 a 3 f n f ty + a 3 f nf yy + a 3 a 3 fnf yy + a 3 f nf yy a 3 c f yf tt + a 3 a c f n f y f ty + a a 3 f nf y +a 3 c c 3 f t f ty + a a 3 c 3 f n f y f ty + a 3 a 3 c f n f t f yy +a a 3 a 3 f nf y f yy + a 3c f n ff yyt + a a 3fnf y f yy + h 3 + c3 3 6 f ttt + a 3c 3 f nf tty + a 3c 3 f nf tty + a 3c 3 f nf tyy (4.6) +a 3 a 3 c 3 fnf tyy + a 3c 3 f nf tyy + a3 3 6 f nf 3 yyy + a 3a 3 f 3 nf yyy + a 3a 3 f 3 nf yyy + a3 3 6 f nf 3 yyy + O ( h 4) (4.7)

26 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD We continue to write down the Taylor series expansion O ( h 4) for k 4. k 4 = f (t n + c 4 h, y n + ha 4 f n + ha 4 k + ha 43 k 3 ) (4.8) = f n + c 4 hf t + h (a 4 f n + a 4 k + a 43 k 3 ) f y (4.9) + c 4h f tt + c 4 h (a 4 f n + a 4 k + a 43 k 3 ) f ty (4.30) + h (a 4 f n + a 4 k + a 43 k 3 ) f yy + c3 4h 3 6 f ttt (4.3) + c 4 h3 (a 4 f n + a 4 k + a 43 k 3 ) f tty (4.3) + c 4h 3 (a 4f n + a 4 k + a 43 k 3 ) f tyy (4.33) + 6 h (a 4 f n + a 4 k + a 43 k 3 ) 3 f yyy + O ( h 4) (4.34)

27 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 3 Inserting (4.5) and (4.7) to (4.34) k 4 = f n + hc 4 f t + ha 4 f n f y (4.35) f n + ( h (c f t + a f n f y ) ) + ha 4 f y c +h f tt + a c f n f ty + a f n (4.36) f n + h (c 3 f t + a 3 f n f y + a 3 f n f y ) a 3 c f t f y + a a 3 f n fy + c 3 f tt + ha 43 f y +h +a 3 c 3 f n f ty + c 3 a 3 f n f ty + a 3 f nf (4.37) yy +a 3 a 3 fnf yy + a 3 f nf yy + h c 4 f tt + h a 4 c 4 f n f ty + h a 4 c 4 f ty [f n + h (c f t + a f n f y )] (4.38) + h a 43 c 4 f ty [f n + h (c 3 f t + a 3 f n f y + a 3 f n f y )] (4.39) + h a 4 f nf yy + h a 4 f yy[f n + h (c f t + a f n f y )] (4.40) + h a 43 f yy[f n + h (c 3 f t + a 3 f n f y + a 3 f n f y )] (4.4) + h a 4 a 4 f n f yy [f n + h (c f t + a f n f y )] (4.4) + h a 4 a 43 f yy [f n + h (c f t + a f n f y )] (4.43) [f n + h (c 3 f t + a 3 f n f y + a 3 f n f y )] (4.44) + h a 4 a 43 f n f yy [f n + h (c 3 f t + a 3 f n f y + a 3 f n f y )] (4.45) + h 3 c3 4 6 f ttt + h 3 c 4 (a 4f n + a 4 f n + a 43 f n ) f tty (4.46) + h 3 c ( ) 4 a 4 fn + a 4fn + a 43fn + a 4 a 4 fn +a 4 a 43 fn + a 4 a 43 fn f tyy (4.47) a f n 3 + a3 4 6 f n 3 + a f n 3 + a 4a 4 fn 3 + h 3 + a 4a 43 fn 3 + a 4a 4 fn 3 + a 4a 43 fn 3 + a 43a 4 f 3 n f yyy (4.48) + a 43a 4 fn 3 + a 4 a 4 a 43 fn 3 + O ( h 4) (4.49)

28 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 4 Collecting terms respect to exponents of h k 4 = f n + h (c 4 f t + a 4 f n f y + a 4 f n f y + a 43 f n f y ) (4.50) a 4 c f t f y + a a 4 f n fy + a 43 c 3 f t f y + a 3 a 43 f n f y + h +a 3 a 43 f n fy + c 4 f tt + c 4 a 4 f n f ty + c 4 a 4 f n f ty (4.5) +c 4 a 43 f n f ty + a 4 f nf yy + a 4 f nf yy + a 43 f nf yy +a 4 a 4 fnf yy + a 4 a 43 fnf yy + a 4 a 43 fnf yy a 4 c f yf tt + a a 4 c f n f y f ty + a 4a f nf y f yy +a 43 a 3 c f t fy + a a 3 a 43 f n fy 3 + a 43c 3 f yf tt +a 3 a 43 c 3 f n f y f ty + a 3 a 43 c 3 f n f y f ty + a 3a 43 f nf y f yy +a 3 a 3 a 43 fnf y f yy + a 3a 43 f nf y f yy + a 4 c c 4 f t f ty +a a 4 c 4 f n f y f ty + a 43 c 3 c 4 f t f ty + a 3 a 43 c 4 f n f y f ty +a 3 a 43 c 4 f n f y f ty + a 4c f n f t f yy + a a 4f nf y f yy +a 43c 3 f n f t f yy + a 3 a 43fnf y f yy + a 3 a 43f nf y f yy +a 4 a 4 c f n f t f yy + a a 4 a 4 f nf y f yy + a 4 a 43 c 3 f n f t f yy +a 3 a 4 a 43 fnf y f yy + a 3 a 4 a 43 f nf y f yy + a 4 a 43 c f n f t f yy + h 3 +a a 4 a 43 fnf y f yy + a 4 a 43 c 3 f n f t f yy + a 3 a 4 a 43 fnf y f yy +a 3 a 4 a 43 fnf y f yy + c3 4 6 f ttt + a 4c 4 f nf tty + a 4c 4 f nf tty + a 43c 4 f nf tty + a 4c 4 f nf tyy + a 4c 4 f nf tyy + a 43c 4 f nf tyy + a 4 a 4 c 4 fnf tyy + a 4 a 43 c 4 fnf tyy +a 4 a 43 c 4 fnf tyy + a3 4 6 f nf 3 yyy + a3 4 6 f nf 3 yyy + a f nf 3 yyy + a 4a 4 f 3 nf yyy + a 4a 43 f 3 nf yyy + a 4a 4 f 3 nf yyy + a 4a 43 f 3 nf yyy + a 4a 43 f 3 nf yyy + a 4a 43 f 3 nf yyy + a 4 a 4 a 43 fnf 3 yyy (4.5) + O ( h 4) (4.53)

29 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 5 Inserting (4.0), (4.7) and (4.53) into (4.5) y (n+) (4.54) = y (n) + h (b f n + b f n + b 3 f n + b 4 f n ) (4.55) ( ) + h b c f t + a b f n f y + b 3 c 3 f t + a 3 b 3 f n f y + a 3 b 3 f n f y (4.56) +b 4 c 4 f t + a 4 b 4 f n f y + a 4 b 4 f n f y + a 43 b 4 f n f y b c f tt + b c a f n f ty + a b f nf yy + a 3 b 3 c f t f y +a 3 b 3 a f n f y f y + b 3c 3 f tt + c 3 b 3 a 3 f n f ty + c 3 b 3 a 3 f n f ty + a 3b 3 + h 3 f nf yy + a 3 a 3 b 3 f n f n f yy + a 3b 3 f nf yy + a 4 b 4 c f t f y +a 4 b 4 f y a f n f y + a 43 b 4 c 3 f t f y + a 43 a 3 b 4 f n fy + +a 43 a 3 b 4 f n fy + b 4c 4 f tt + c 4 b 4 a 4 f n f ty + c 4 b 4 a 4 f ty f n + c 4 b 4 a 43 f ty f n + a 4b 4 f nf yy + a 4b 4 f nf yy + a 43b 4 f nf yy + a 4 a 4 b 4 f yy fn +a 4 a 43 b 4 f yy fn + a 4 a 43 b 4 f yy fn (4.57) b c 3 6 f ttt + b c a f n f tty + b c a f nf tyy + a3 b 6 f nf 3 yyy + a 3b 3 c f tt f y + a 3 b 3 c a f n f ty f y + a 3a b 3 f nf yy f y +c 3 a 3 b 3 c f t f ty + c 3 a 3 b 3 a f n f y f ty + a 3 a 3 b 3 c f n f t f yy +a 3 a 3 a b 3 f n f n f y f yy + a 3b 3 f n c f t f yy + a 3b 3 f n a f n f y f yy + b 3c f ttt + b 3c 3 (a 3 + a 3 ) f n f tty + b 3c 3 a 3 f nf tyy +b 3 c 3 a 3 a 3 f n f n f tyy + b 3c 3 a 3 f nf tyy + b 3a f 3 nf yyy + b 3a 3a 3 f nf 3 yyy + b 3a 3 a 3 f nf 3 yyy + b 3a f 3 nf yyy + a 4b 4 c f y f tt + a 4 b 4 c a f n f y f ty + a 4a b 4 f nf y f yy +a 43 a 3 b 4 c f t fy + a 43 a 3 a b 4 f n fy 3 + a 43b 4 c 3 f y f tt +a 43 b 4 c 3 a 3 f n f y f ty + a 43 b 4 c 3 a 3 f n f y f ty + a 43a 3b 4 f nf y f yy + h 4 +a 43 a 3 a 3 b 4 fnf y f yy + a 43a 3b 4 f nf y f yy + c 4 a 4 b 4 c f t f ty +c 4 a 4 a b 4 f n f y f ty + c 4 a 43 c 3 b 4 f t f ty + c 4 a 43 a 3 b 4 f n f y f ty +c 4 a 43 a 3 b 4 f n f y f ty + a 4b 4 c f n f t f yy + a 4a b 4 fnf y f yy +a 43b 4 c 3 f n f t f yy + a 43a 3 b 4 fnf y f yy + a 43a 3 b 4 fnf y f yy +a 4 a 4 b 4 c f n f t f yy + a 4 a a 4 b 4 fnf y f yy + a 4 a 43 b 4 c 3 f n f t f yy +a 4 a 43 a 3 b 4 fnf y f yy + a 4 a 43 a 3 b 4 fnf y f yy + a 4 a 43 b 4 c f n f t f yy +a 4 a 43 a b 4 fnf y f yy + a 4 a 43 b 4 c 3 f n f t f yy + a 4 a 43 a 3 b 4 fnf y f yy +a 4 a 43 a 3 b 4 fnf y f yy + b 4c f ttt + b 4c 4a 4 f n f tty + b 4c 4a 4 f n f tty + b 4c 4a 43 f n f tty + c 4b 4 a 4 f nf tyy + c 4b 4 a 4 f nf tyy + c 4b 4 a 43 f nf tyy +c 4 b 4 a 4 a 4 f nf tyy + c 4 a 4 a 43 b 4 fnf tyy + c 4 a 4 a 43 b 4 fnf tyy + a3 4b 4 6 f nf 3 yyy + a3 4b 4 6 f nf 3 yyy + a3 43b 4 6 f nf 3 yyy + a 4a 4 b 4 f 3 nf yyy + a 4a 43 b 4 f 3 nf yyy + a 4a 4 b 4 f 3 nf yyy + a 4a 43 b 4 f 3 nf yyy + a 43a 4 b 4 f 3 nf yyy + a 43a 4 b 4 f 3 nf yyy + a 4 a 4 a 43 b 4 fnf 3 yyy (4.58) + O ( h 5) (4.59)

30 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD Taylor Series Expansion Formula We need to compute y tttt for Taylor series expansion below. y tttt = d ( ftt + f t f y + ff ty + ffy + f ) f yy dt = ( f tt + f t f y + ff ty + ffy + f ) f yy + f ( f tt + f t f y + ff ty + ff y + f f yy ) t y (4.60) (4.6) (4.6) = f ttt + f y f tt + f t f ty + f t f ty + ff tty + f t f y + ff y f ty (4.63) + ff t f yy + f f tyy + ff tty + ff y f ty + ff t f yy + ff y f ty (4.64) + f f tyy + ff 3 y + f f y f yy + f f y f yy + f 3 f yyy (4.65) = f ttt + f y f tt + 3f t f ty + 3ff tty + f t f y + 5ff y f ty + 3ff t f yy (4.66) + 3f f tyy + ff 3 y + 4f f y f yy + f 3 f yyy (4.67) Now we write down the Taylor series expansion of y in the neighborhood of t n with O ( h 5). y n+ = y n + hf n + h (f t + f n f y ) (4.68) + h3 ( ftt + f t f y + f n f ty + f n fy + f ) 6 nf yy (4.69) ( ) + h4 fttt + f y f tt + 3f t f ty + 3f n f tty + f t fy + 5f n f y f ty 4 +3f n f t f yy + 3fnf tyy + f n fy 3 + 4fnf y f yy + fnf 3 (4.70) yyy + O ( h 5) (4.7) 4..3 Derivation of System of Equations Compare the coefficients of (4.68) and (4.59)

31 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 7 hf n : = b + b + b 3 + b 4 (4.7) h f t : = b c + b 3 c 3 + b 4 c 4 (4.73) h f n f y : = b a + b 3 a 3 + b 3 a 3 + a 4 b 4 + a 4 b 4 + a 43 b 4 (4.74) h 3 f tt : 6 = b c + b 3c 3 + b 4c 4 (4.75) h 3 f t f y : 6 = a 3b 3 c + a 4 b 4 c + a 43 b 4 c 3 (4.76) h 3 f n f ty : 3 = b c a + c 3 b 3 a 3 + c 3 b 3 a 3 (4.77) + c 4 b 4 a 4 + c 4 b 4 a 4 + c 4 b 4 a 43 (4.78) h 3 f n f y : 6 = a 3b 3 a + a 4 b 4 a + a 43 a 3 b 4 + a 43 a 3 b 4 (4.79) h 3 f nf yy : 6 = a b + a 3b 3 + a 3 a 3 b 3 + a 3b 3 + a 4b 4 (4.80) + a 4b 4 + a 43b 4 + a 4 a 4 b 4 + a 4 a 43 b 4 + a 4 a 43 b 4 (4.8) h 4 f ttt : 4 = b c b 3c b 4c (4.8) h 4 f y f tt : 4 = a 3b 3 c + a 4b 4 c + a 43b 4 c 3 (4.83) h 4 f t f ty : 8 = c 3a 3 b 3 c + c 4 a 4 b 4 c + c 4 a 43 c 3 b 4 (4.84) h 4 f n f tty : 8 = b c a h 4 f n f y f ty : + b 4c 4a 4 + b 3c 3 (a 3 + a 3 ) + b 4c 4a 43 + b 4c 4a 4 (4.85) (4.86) h 4 f t fy : 4 = a 43a 3 b 4 c (4.87) 5 4 = a 3b 3 c a + c 3 a 3 b 3 a + a 4 b 4 c a (4.88) + c 4 a 4 a b 4 + c 4 a 43 a 3 b 4 + a 43 b 4 c 3 a 3 (4.89) + a 43 b 4 c 3 a 3 + c 4 a 43 a 3 b 4 (4.90) h 4 f n f t f yy : 8 = a 3a 3 b 3 c + a 3b 3 c + a 4b 4 c + a 43b 4 c 3 (4.9) h 4 f nf tyy : 8 = b c a + a 4 a 4 b 4 c + a 4 a 43 b 4 c 3 + a 4 a 43 b 4 c + a 4 a 43 b 4 c 3 (4.9) + b 3c 3 a 3 + b 3 c 3 a 3 a 3 + b 3c 3 a 3 (4.93) + c 4b 4 a 4 + c 4b 4 a 4 + c 4b 4 a 43 + c 4 b 4 a 4 a 4 (4.94) + c 4 a 4 a 43 b 4 + c 4 a 4 a 43 b 4 (4.95) h 4 f n fy 3 : 4 = a 43a 3 a b 4 (4.96)

32 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 8 h 4 f nf y f yy : 6 = a 3a b 3 h 4 f 3 nf yyy : + a 3 a 3 a b 3 + a 3b 3 a + a 4a b 4 (4.97) + a 43a 3b 4 + a 43 a 3 a 3 b 4 + a 43a 3b 4 + a 4a b 4 (4.98) + a 43a 3 b 4 + a 43a 3 b 4 + a 4 a a 4 b 4 + a 4 a 43 a 3 b 4 (4.99) + a 4 a 43 a 3 b 4 + a 4 a 43 a b 4 + a 4 a 43 a 3 b 4 (4.00) + a 4 a 43 a 3 b 4 (4.0) 4 = a3 b + b 3a b 3a 3a 3 + b 3a 3 a 3 + b 3a a3 4b a 4a 4 b 4 + a 43a 4 b 4 + a3 4b a3 43b a 4a 43 b 4 + a 4a 4 b 4 + a 43a 4 b 4 + a 4a 43 b 4 (4.0) (4.03) (4.04) + a 4 a 4 a 43 b 4 (4.05)

33 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 9 Hence, we obtain the system of equations b + b + b 3 + b 4 = (4.06) b c + b 3 c 3 + b 4 c 4 = b a + b 3 (a 3 + a 3 ) + b 4 (a 4 + a 4 + a 43 ) = b c + b 3 c 3 + b 4 c 4 = 3 b 3 a 3 c + b 4 (a 4 c + a 43 c 3 ) = 6 b c a + b 3 c 3 (a 3 + a 3 ) + b 4 c 4 (a 4 + a 4 + a 43 ) = 3 b 3 a 3 a + b 4 [a 4 a + a 43 (a 3 + a 3 )] = 6 b a + b 3 (a 3 + a 3 ) + b 4 (a 4 + a 4 + a 43 ) = 3 b c 3 + b 3 c b 4 c 3 4 = 4 b 3 a 3 c ( + b 4 a4 c + a 43 c ) 3 = b 3 c 3 a 3 c + b 4 c 4 (a 4 c + a 43 c 3 ) = 8 b c a + b 3 c 3 (a 3 + a 3 ) + b 4 c 4 (a 4 + a 4 + a 43 ) = 4 b 4 a 43 a 3 c = { } 4 b3 a 3 a (c + c 3 ) = 5 +b 4 [a 4 a (c + c 4 ) + a 43 (a 3 + a 3 ) (c 3 + c 4 )] 4 b 3 a 3 c (a 3 + a 3 ) + b 4 (a 4 c + a 43 c 3 ) (a 4 + a 4 + a 43 ) = 8 b c a + b 3 c 3 (a 3 + a 3 ) + b 4 c 4 (a 4 + a 4 + a 43 ) = 4 b 4 a 43 a 3 a = 4 b 3 a 3 a [a + (a 3 + a 3 )] +b 4 a 4 a [a + (a 4 + a 4 + a 43 )] +b 4 a 43 (a 3 + a 3 ) [a 3 + a 3 + (a 43 + a 4 + a 4 )] = 3 b a 3 + b 3 (a 3 + a 3 ) 3 + b 4 (a 4 + a 4 + a 43 ) 3 = 4 (4.07) (4.08) (4.09) (4.0) (4.) (4.) (4.3) (4.4) (4.5) (4.6) (4.7) (4.8) (4.9) (4.0) (4.) (4.) (4.3) (4.4) 4..4 Solutions of System of Equations We now solve the above system of equations. We take c = α (4.5) c 3 = β (4.6) c 4 = γ (4.7)

34 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 30 as three free variables. Then, our system of equations becomes b + b + b 3 + b 4 = (4.8) αb + βb 3 + γb 4 = b a + b 3 (a 3 + a 3 ) + b 4 (a 4 + a 4 + a 43 ) = α b + β b 3 + γ b 4 = 3 αb 3 a 3 + b 4 (αa 4 + βa 43 ) = 6 αb a + βb 3 (a 3 + a 3 ) + γb 4 (a 4 + a 4 + a 43 ) = 3 b 3 a 3 a + b 4 [a 4 a + a 43 (a 3 + a 3 )] = 6 b a + b 3 (a 3 + a 3 ) + b 4 (a 4 + a 4 + a 43 ) = 3 α 3 b + β 3 b 3 + γ 3 b 4 = 4 α ( b 3 a 3 + b 4 α a 4 + β ) a 43 = αβb 3 a 3 + γb 4 (αa 4 + βa 43 ) = 8 α b a + β b 3 (a 3 + a 3 ) + γ b 4 (a 4 + a 4 + a 43 ) = 4 αb 4 a 43 a 3 = { } 4 b3 a 3 a (α + β) = 5 +b 4 [a 4 a (α + γ) + a 43 (a 3 + a 3 ) (β + γ)] 4 αb 3 a 3 (a 3 + a 3 ) + b 4 (αa 4 + βa 43 ) (a 4 + a 4 + a 43 ) = 8 αb a + βb 3 (a 3 + a 3 ) + γb 4 (a 4 + a 4 + a 43 ) = 4 b 4 a 43 a 3 a = 4 b 3 a 3 a [a + (a 3 + a 3 )] +b 4 a 4 a [a + (a 4 + a 4 + a 43 )] +b 4 a 43 (a 3 + a 3 ) [a 3 + a 3 + (a 43 + a 4 + a 4 )] = 3 a 3 b + b 3 (a 3 + a 3 ) 3 + b 4 (a 4 + a 4 + a 43 ) 3 = 4 (4.9) (4.30) (4.3) (4.3) (4.33) (4.34) (4.35) (4.36) (4.37) (4.38) (4.39) (4.40) (4.4) (4.4) (4.43) (4.44) (4.45) (4.46) We now need to focus on three special equations of the above system. αb + βb 3 + γb 4 = α b + β b 3 + γ b 4 = 3 α 3 b + β 3 b 3 + γ 3 b 4 = 4 (4.47) (4.48) (4.49)

35 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 3 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 b, b 3, b 4, we obtain 4β + 4γ 6βγ 3 b = α (α β) (α γ) 4γ + 4α 6γα 3 b 3 = β (β γ) (β α) 4α + 4β 6αβ 3 b 4 = γ (γ α) (γ β) (4.50) (4.5) (4.5) Then we use (4.8) to obtain b = 4 (α + β + γ) 6 (αβ + βγ + γα) + αβγ 3 αβγ (4.53) Now using b, b, b 3, b 4, we need to focus on the following three equations of the remaining system. αb 3 a 3 + b 4 (αa 4 + βa 43 ) = 6 α ( b 3 a 3 + b 4 α a 4 + β ) a 43 = αβb 3 a 3 + γb 4 (αa 4 + βa 43 ) = 8 (4.54) (4.55) (4.56) Solving this system of three equations respect to a 3, a 4, a 43, we obtain β (4γ 3) (α β) a 3 = α (4α + 4γ 6αγ 3) a 4 = γ (α γ) ( 4β 5β + 3α + γ 4αγ ) α (6α β 4α 6αβ + 3α + 4β 3β) γ ( α) (γ α) (γ β) a 43 = β (6α β 4α 6αβ + 3α + 4β 3β) (4.57) (4.58) (4.59) Now, we only need to solve a, a 3, a 4. To do this, we need to focus on the following three equations of the remaining system. b a + b 3 a 3 + b 4 a 4 = b 3a 3 b 4 a 4 b 4 a 43 (4.60) αb a + βb 3 a 3 + γb 4 a 4 = 3 βb 3a 3 γb 4 a 4 γb 4 a 43 (4.6) (b 3 a 3 + b 4 a 4 ) a + b 4 a 43 a 3 = 6 b 4a 43 a 3 (4.6) We solve this system of three equations respect to a, a 3, a 4 by the following Matlab routine because the formulas are quite long. format long

36 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 3 syms a b c; % a=alpha, b=beta, c=gamma A = [a b c; a^ b^ c^; a^3 b^3 c^3]; v=[/;/3;/4]; B = A^-*v; b = B(); b3 = B(); b4 = B(3); A = [a*b3 a*b4 b*b4; a^*b3 a^*b4 b^*b4; a*b*b3 c*b4*a c*b4*b]; v = [/6;/;/8]; B = A^-*v; a3 = B() a4 = B() a43 = B(3) A = [b b3 b4;a*b b*b3 c*b4;(b3*a3+b4*a4) b4*a43 0]; v = [/-b3*3-b4*a4-b4*a43; /3-b*b3*a3-c*b4*a4-c*b4*a43; /6-b4*a43*a3]; B = A^-*v; a = B() a3 = B() a4 = B(3) This Matlab routine returns the formulas of a, a 3, a 4 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 is just computation matter. Although those computations is quite complicated by hand, checking by Matlab is quite easy and there is no benefit to represent those computations here. You should check yourself. In practice, we don t need to know the final three formulas of a, a 3, a 4 in terms of α, β, γ. For each 3 tuples α, β, γ, we can easily solve the other unknowns by the process described above. We have solved (4.06)-(4.4) completely Some Cases Case α =, β =, γ =. Using the above process, we obtain This Butcher tableau appears in [4], p (4.63)

37 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 33 Case α = 3, β = 3, γ = This Butcher tableau appears in [5] (4.64) 4. Matlab Implementation 4.. Curtiss-Hirchfelder Equation In this subsection, we will use Runge Kutta order 4 method to solve numerically the Curtiss-Hirchfelder equation. Definition 4.. The Curtiss-Hirchfelder equation is an ordinary differential equation (ODE) which has the following form In general, The following Matlab routines aims at dy = 50 (y cos t) dt (4.65) y (0) = (4.66) dy = f (t, y) dt (4.67) y (x 0 ) = y 0 (4.68). Approximating solution of (4.65) by using Runge Kutta order 4 method.. Compute absolute errors and relative errors and plot the obtained numerical solutions and errors Subroutine s.m This subroutine provides the exact solution of (4.65). function s = s(t) s = 50/50*(50*cos(t)+sin(t)) + exp(-50*t)/50; 4... Subroutine f.m This subroutine provides the function f in RHS of (4.65). Users can modify these functions later. function f = f(t,y) f = -50*y + 50*cos(t);

38 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD Subroutine x.m This subroutine provides the Runge Kutta order 4 method. function x = x(a,b,c,h,t,y) k = f(t,y); k = f(t+c()*h,y+h*a(,)*k); k3 = f(t+c(3)*h,y+h*(a(3,)*k+a(3,)*k)); k4 = f(t+c(4)*h,y+h*(a(4,)*k+a(4,)*k+a(4,3)*k3)); x = h*(b()*k+b()*k+b(3)*k3+b(4)*k4); Main Routine RK4 close all clear all clc format long tic %% Initial. A() = ; A() = ; % A3() = ; N=000; h=5/n; t = 0:h:5; %% Coefficients a = [ ; / ; 0 / 0 0 ; 0 0 0]; b = [/6 /3 /3 /6]; c = [0 / / ]; a = [ ; / ; -/3 0 0 ; - 0]; b = [/8 3/8 3/8 /8]; c = [0 /3 /3 ]; %% Numerical Solution. for n=:n A(n+) = A(n) + x(a,b,c,h,t(n),a(n)); A(n+) = A(n) + x(a,b,c,h,t(n),a(n)); end %% Plot Numerical Solution

39 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 35 figure() hold on plot(t,s(t), b ); plot(t,a, r ); legend( Exact Solution, Numerical Solution ); title( Numerical Solution (Butcher Table ) ); figure() hold on plot(t,s(t), b ); plot(t,a, r ); legend( Exact Solution, Numerical Solution ); title( Numerical Solution (Butcher Table ) ); %% Absolute Error. display( Absolute Error ) ae = h*sum(abs(a-s(t))) display( Absolute Error ) ae = h*sum(abs(a-s(t))) figure(3) hold on plot(t,abs(a-s(t)), g ); plot(t,abs(a-s(t)), b ); legend( a b c, a b c ); title( Absolute Error ); %% Relative Error. display( Relative Error ) re = h*sum(abs((a-s(t))./s(t))) display( Relative Error ) re = h*sum(abs((a-s(t))./s(t))) figure(4) hold on plot(t,abs((a-s(t))./a), g ); plot(t,abs((a-s(t))./a), b ); legend( a b c, a b c ); title( Relative Error ); toc

40 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD Results Figure 4.: Numerical Solutions.

41 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD Brusselator Equation Figure 4.: Absolute Errors. Definition 4.. The Brusselator equation is the following system of equations In general, In vector form y = (y, y ), dy dt = 4y + yy (4.69) dy dt = 3y yy (4.70) dy dt = f (t, y, y ) (4.7) dy dt = f (t, y, y ) (4.7) dy dt = f (t, y) (4.73) 4... Subroutine f.m This subroutine provides f and f in the RHS of (4.7) and (4.7). Users can modify these functions later. function f = f(t,y) f() = -4*y()+y()^*y(); f() = 3*y()-y()^*y();

42 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 38 Figure 4.3: Relative Errors Subroutine x.m This subroutine provides the Runge Kutta order 4 method. function x = x(a,b,c,h,t,y) k = f(t,y); k = f(t+c()*h,y+h*a(,)*k); k3 = f(t+c(3)*h,y+h*(a(3,)*k+a(3,)*k)); k4 = f(t+c(4)*h,y+h*(a(4,)*k+a(4,)*k+a(4,3)*k3)); x = h*(b()*k+b()*k+b(3)*k3+b(4)*k4); Main Routine RK4.m clear all close all clc format long tic %% Initial for Reference Solutions. N0 = 0^6; h0= 0/N0; t0 = 0:h0:0; B = zeros(n0,);

43 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 39 B(,) =.5; B(,) = 3; B = zeros(n0,); B(,) =.5; B(,) = 3; %% Initial for Numerical Solutions. N = 0^4; h = 0/N; t = 0:h:0; A = zeros(n,); A(,) =.5; A(,) = 3; A = zeros(n,); A(,) =.5; A(,) = 3; A3 = zeros(n,); A3(,) =.5; A3(,) = 3; % Step step = N0/N; %% Runge Kutta initials. % Coefficients a = [ ; / ; 0 / 0 0 ; 0 0 0]; b = [/6 /3 /3 /6]; c = [0 / / ]; a = [ ; / ; -/3 0 0 ; - 0]; b = [/8 3/8 3/8 /8]; c = [0 /3 /3 ]; %% Reference Solutions. for n=:n0 B(n+,:) = B(n,:) + x(a,b,c,h0,t0(n),b(n,:)); B(n+,:) = B(n,:) + x(a,b,c,h0,t0(n),b(n,:)); end %% Numerical Solutions, Absolute Errors and Relative Errors.

44 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 40 ae = zeros(n+,); ae = zeros(n+,); re = zeros(n+,); re = zeros(n+,); for n=:n % Numerical Solutions A(n+,:) = A(n,:) + x(a,b,c,h,t(n),a(n,:)); A(n+,:) = A(n,:) + x(a,b,c,h,t(n),a(n,:)); % Absolute Errors ae(n,:) = h.*abs(a(n,:)-b(step*(n-)+,:)); ae(n,:) = h.*abs(a(n,:)-b(step*(n-)+,:)); % ae(n,) = abs((a(n,)-s(,t)); % ae(n,) = abs((a(n,)-s(,t)); % ae(n,) = abs((a(n,i)-s(,t)); % ae(n,) = abs((a(n,i)-s(,t)); % Relative Errors re(n,:) = h.*abs((a(n,:)... -B(step*(n-)+,:))./B(step*(n-)+,:)); re(n,:) = h.*abs((a(n,:)... -B(step*(n-)+,:))./B(step*(n-)+,:)); % re(n,) = abs((a(n,)-s(,t))./s(i,t)); % re(n,) = abs((a(n,)-s(,t))./s(i,t)); % re(n,) = abs((a(n,i)-s(,t))./s(i,t)); % re(n,) = abs((a(n,i)-s(,t))./s(i,t)); end % Absolute Errors display( Absolute Error of Table ) ae_y = h*sum(ae(:,)) ae_y = h*sum(ae(:,)) display( Absolute Error of Table ) ae_y = h*sum(ae(:,)) ae_y = h*sum(ae(:,)) % Relative Errors display( Relative Error of Table ) re_y = h*sum(re(:,)) re_y = h*sum(re(:,)) display( Relative Error of Table ) re_y = h*sum(re(:,)) re_y = h*sum(re(:,))

45 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 4 %% Plot Numerical Solution figure() subplot(,,) hold on % plot(t,s(,t), b ) plot(t0,b(:,), b ) plot(t,a(:,), r ) legend( Exact/Reference, Numerical Solution ); title( Numerical Solution for Y (Butcher Table ) ); subplot(,,) hold on % plot(t,s(,t), b ) plot(t0,b(:,), b ) plot(t,a(:,), r ) legend( Exact/Reference, Numerical Solution ); title( Numerical Solution for Y (Butcher Table ) ); figure() subplot(,,) hold on % plot(t,s(,t), b ) plot(t0,b(:,), b ) plot(t,a(:,), r ) legend( Exact/Reference, Numerical Solution ); title( Numerical Solution for Y (Butcher Table ) ); subplot(,,) hold on % plot(t,s(,t), b ) plot(t0,b(:,), b ) plot(t,a(:,), r ) legend( Exact/Reference, Numerical Solution ); title( Numerical Solution for Y (Butcher Table ) ); %% Plot: Dependency of y with respect to y. figure(3) hold on plot(a(:,),a(:,), b ); plot(a(:,),a(:,), g ); % plot(a3(:,),a3(:,), r ); legend( a b c, a b c ); title( Dependency of y with respect to y ); %% Plot Absolute Errors. figure(4) subplot(,,) hold on plot(t,ae(:,), b );

46 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 4 plot(t,ae(:,), g ); % plot(t,ae3(:,), r ); legend( a b c, a b c ); title( Absolute Errors Y ); subplot(,,) hold on plot(t,ae(:,), b ); plot(t,ae(:,), g ); % plot(t,ae3(:,), r ); legend( a b c, a b c ); title( Absolute Errors Y ); %% Plot Relative Errors. figure(5) subplot(,,) hold on plot(t,re(:,), b ); plot(t,re(:,), g ); % plot(t,re3(:,), r ); legend( a b c, a b c ); title( Relative Errors Y ); subplot(,,) hold on plot(t,re(:,), b ); plot(t,re(:,), g ); % plot(t,re3(:,), r ); legend( a b c, a b c ); title( Relative Errors Y ); toc

47 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD Results Figure 4.4: Numerical Solutions.

48 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 44 Figure 4.5: Dependency of y with respect to y Examples Subroutine s.m function s = s(t) s = [3*exp(*t)-*exp(3*t);-3*exp(*t)+4*exp(3*t)]; Subroutine f.m function f = f(t,y) % f=[y()-y(),*y()+4*y()]; f() = y()-y(); f() = *y()+4*y(); Subroutine x.m function x = x(a,b,c,h,t,y) k = f(t,y); k = f(t+c()*h,y+h*a(,)*k); k3 = f(t+c(3)*h,y+h*(a(3,)*k+a(3,)*k)); k4 = f(t+c(4)*h,y+h*(a(4,)*k+a(4,)*k+a(4,3)*k3)); x = h*(b()*k+b()*k+b(3)*k3+b(4)*k4); Main Routine RK4ex.m clear all

49 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 45 Figure 4.6: Absolute Errors. close all clc format long tic % %% Initial for Reference Solutions. % N0 = 0^6; % h0= 0/N0; % t0 = 0:h0:0; % % B = zeros(n0,); % B(,) = ; % B(,) = ; % % B = zeros(n0,); % B(,) = ; % B(,) = ; %% Initial for Numerical Solutions. N = 0^4; h = /N; t = 0:h:; A = zeros(n+,);

50 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 46 Figure 4.7: Relative Errors. A(,) = ; A(,) = ; A = zeros(n+,); A(,) = ; A(,) = ; A3 = zeros(n+,); A3(,) = ; A3(,) = ; % Step % step = N0/N; %% Runge Kutta initials. % Coefficients a = [ ; / ; 0 / 0 0 ; 0 0 0]; b = [/6 /3 /3 /6]; c = [0 / / ]; a = [ ;

51 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 47 / ; -/3 0 0 ; - 0]; b = [/8 3/8 3/8 /8]; c = [0 /3 /3 ]; % %% Reference Solutions. % for n=:n0 % B(n+,:) = B(n,:) + x(a,b,c,h0,t0(n),b(n,:)); % B(n+,:) = B(n,:) + x(a,b,c,h0,t0(n),b(n,:)); % end s=s(t) ; %% Numerical Solutions, Absolute Errors and Relative Errors. ae = zeros(n+,); ae = zeros(n+,); re = zeros(n+,); re = zeros(n+,); for n=:n % Numerical Solutions A(n+,:) = A(n,:) + x(a,b,c,h,t(n),a(n,:)); A(n+,:) = A(n,:) + x(a,b,c,h,t(n),a(n,:)); % Absolute Errors % ae(n,:) = h.*abs(a(n,:)-b(step*(n-)+,:)); ae = (abs(a-s)); % ae(n,:) = h.*abs(a(n,:)-b(step*(n-)+,:)); ae = (abs(a-s)); % ae(n,) = abs((a(n,)-s(,t)); % ae(n,) = abs((a(n,)-s(,t)); % ae(n,) = abs((a(n,i)-s(,t)); % ae(n,) = abs((a(n,i)-s(,t)); % Relative Errors % re(n,:) = h.*abs((a(n,:)-b(step*(n-)+,:))./b(step*(n-)+,:)); re = (abs((a-s)./s)); % re(n,:) = h.*abs((a(n,:)-b(step*(n-)+,:))./b(step*(n-)+,:)); re = (abs((a-s)./s)); % re(n,) = abs((a(n,)-s(,t))./s(i,t)); % re(n,) = abs((a(n,)-s(,t))./s(i,t)); % re(n,) = abs((a(n,i)-s(,t))./s(i,t)); % re(n,) = abs((a(n,i)-s(,t))./s(i,t)); end % Absolute Errors display( Absolute Error of Table ) ae_y = h*sum(ae(:,))

52 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 48 ae_y = h*sum(ae(:,)) display( Absolute Error of Table ) ae_y = h*sum(ae(:,)) ae_y = h*sum(ae(:,)) % Relative Errors display( Relative Error of Table ) re_y = h*sum(re(:,)) re_y = h*sum(re(:,)) display( Relative Error of Table ) re_y = h*sum(re(:,)) re_y = h*sum(re(:,)) %% Plot Numerical Solution figure() subplot(,,) hold on % plot(t,s(,t), b ) plot(t,s(:,), b ) plot(t,a(:,), r ) legend( Exact/Reference, Numerical Solution ); title( Numerical Solution for Y (Butcher Table ) ); subplot(,,) hold on % plot(t,s(,t), b ) plot(t,s(:,), b ) plot(t,a(:,), r ) legend( Exact/Reference, Numerical Solution ); title( Numerical Solution for Y (Butcher Table ) ); figure() subplot(,,) hold on % plot(t,s(,t), b ) plot(t,s(:,), b ) plot(t,a(:,), r ) legend( Exact/Reference, Numerical Solution ); title( Numerical Solution for Y (Butcher Table ) ); subplot(,,) hold on % plot(t,s(,t), b ) plot(t,s(:,), b ) plot(t,a(:,), r ) legend( Exact/Reference, Numerical Solution ); title( Numerical Solution for Y (Butcher Table ) );

53 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 49 %% Plot: Dependency of y with respect to y. figure(3) hold on plot(a(:,),a(:,), b ); plot(a(:,),a(:,), g ); % plot(a3(:,),a3(:,), r ); legend( Butcher Table, Butcher Table ); title( Dependency of y with respect to y ); %% Plot Absolute Errors. figure(4) subplot(,,) hold on plot(t,ae(:,), b ); plot(t,ae(:,), g ); % plot(t,ae3(:,), r ); legend( Butcher Table, Butcher Table ); title( Absolute Errors Y ); subplot(,,) hold on plot(t,ae(:,), b ); plot(t,ae(:,), g ); % plot(t,ae3(:,), r ); legend( Butcher Table, Butcher Table ); title( Absolute Errors Y ); %% Plot Relative Errors. figure(5) subplot(,,) hold on plot(t,re(:,), b ); plot(t,re(:,), g ); % plot(t,re3(:,), r ); legend( Butcher Table, Butcher Table ); title( Relative Errors Y ); subplot(,,) hold on plot(t,re(:,), b ); plot(t,re(:,), g ); % plot(t,re3(:,), r ); legend( Butcher Table, Butcher Table ); title( Relative Errors Y ); toc

54 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD Results Figure 4.8: Numerical Solutions.

55 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 5 Figure 4.9: Numerical Solutions. Figure 4.0: Dependency of y with respect to y.

56 CHAPTER 4. RUNGE KUTTA ORDER 4 METHOD 5 Figure 4.: Absolute Errors. Figure 4.: Relative Errors.

57 Alphabetical Index A absolute errors 33 autonomous form B Brusselator equation 37 Butcher table 3 Butcher tableau C Curtiss-Hirchfelder equation 33 E Euler s method 3 explicit Euler s method 3 F family of explicit Runge Kutta methods I initial value problems iterative method K Kutta s third order method 9 M mesh N non-autonomous form O order of accuracy R relative errors 33 Runge Kutta method Runge Kutta method of second order 7 Runge Kutta order method 3 Runge Kutta order method 4 Runge Kutta order 3 method 8 Runge Kutta order 4 method 0, 33 S step size stiff problems T Taylor s series 53