5th Order Runge Kutta

Hey,

I've been looking at Runge Kutta methods and was wondering how we would write a 5th order runge kutta method. I've stumbled upon lot of work online about how to write to the 4th order but not the 5th?

I've been looking at the code below...

% It calculates ODE using Runge-Kutta 4th order method
% Author Ido Schwartz
% Originally available form: http://www.mathworks.com/matlabcentral/fileexchange/29851-runge-kutta-4th-order-ode/content/Runge_Kutta_4.m
% Edited by Amin A. Mohammed, for 2 ODEs(April 2016)

clc;                                               % Clears the screen
clear all;

h=0.1;                                             % step size
x = 0:h:1;                                         % Calculates upto y(1)
y = zeros(1,length(x)); 
z = zeros(1,length(x)); 
y(1) = 3;                                          % initial condition
z(1) = 1;                                          % initial condition
% F_xy = @(t,r) 3.*exp(-t)-0.4*r;                  % change the function as you desire
F_xyz = @(x,y,z) z;                                  % change the function as you desire
G_xyz = @(x,y,z) 6*y-z;

for i=1:(length(x)-1)                              % calculation loop
    k_1 = F_xyz(x(i),y(i),z(i));
    L_1 = G_xyz(x(i),y(i),z(i));
    k_2 = F_xyz(x(i)+0.5*h,y(i)+0.5*h*k_1,z(i)+0.5*h*L_1);
    L_2 = G_xyz(x(i)+0.5*h,y(i)+0.5*h*k_1,z(i)+0.5*h*L_1);
    k_3 = F_xyz((x(i)+0.5*h),(y(i)+0.5*h*k_2),(z(i)+0.5*h*L_2));
    L_3 = G_xyz((x(i)+0.5*h),(y(i)+0.5*h*k_2),(z(i)+0.5*h*L_2));
    k_4 = F_xyz((x(i)+h),(y(i)+k_3*h),(z(i)+L_3*h)); % Corrected        
    L_4 = G_xyz((x(i)+h),(y(i)+k_3*h),(z(i)+L_3*h));
    %k_5
    %L_5
    y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;  % main equation
    z(i+1) = z(i) + (1/6)*(L_1+2*L_2+2*L_3+L_4)*h;  % main equation

end

Anyone have any ideas in what the 5th order equations should be and how to implement it to this code?

Many Thanks Chris

[krishna kadiyam]

bro..have u gotten the solution? if u have gotten pls share with me also

[Chris Potts]

http://maths.cnam.fr/IMG/pdf/RungeKuttaFehlbergProof.pdf

just follow this method, it's not too difficult

[krishna kadiyam]

thank u man!!!

Answer 1

[James Tursa]

E.g., 5th Order Runge-Kutta-Fehlberg coefficients can be found here:

https://en.wikipedia.org/wiki/Runge%25E2%2580%2593Kutta%25E2%2580%2593Fehlberg_method

Answer 2

[Suganya V.P]

ATLAB AnswersToggle Sub Navigation 5th Order Runge Kutta 134 views (last 30 days) Show older comments Chris Potts Chris Potts on 30 Jan 2018 ⋮ Commented: krishna kadiyam on 5 Feb 2018 Accepted Answer: James Tursa Hey, I've been looking at Runge Kutta methods and was wondering how we would write a 5th order runge kutta method. I've stumbled upon lot of work online about how to write to the 4th order but not the 5th? I've been looking at the code below... % It calculates ODE using Runge-Kutta 4th order method % Author Ido Schwartz % Originally available form: http://www.mathworks.com/matlabcentral/fileexchange/29851-runge-kutta-4th-order-ode/content/Runge_Kutta_4.m % Edited by Amin A. Mohammed, for 2 ODEs(April 2016) clc; % Clears the screen clear all; h=0.1; % step size x = 0:h:1; % Calculates upto y(1) y = zeros(1,length(x)); z = zeros(1,length(x)); y(1) = 3; % initial condition z(1) = 1; % initial condition % F_xy = @(t,r) 3.*exp(-t)-0.4*r; % change the function as you desire F_xyz = @(x,y,z) z; % change the function as you desire G_xyz = @(x,y,z) 6*y-z; for i=1:(length(x)-1) % calculation loop k_1 = F_xyz(x(i),y(i),z(i)); L_1 = G_xyz(x(i),y(i),z(i)); k_2 = F_xyz(x(i)+0.5*h,y(i)+0.5*h*k_1,z(i)+0.5*h*L_1); L_2 = G_xyz(x(i)+0.5*h,y(i)+0.5*h*k_1,z(i)+0.5*h*L_1); k_3 = F_xyz((x(i)+0.5*h),(y(i)+0.5*h*k_2),(z(i)+0.5*h*L_2)); L_3 = G_xyz((x(i)+0.5*h),(y(i)+0.5*h*k_2),(z(i)+0.5*h*L_2)); k_4 = F_xyz((x(i)+h),(y(i)+k_3*h),(z(i)+L_3*h)); % Corrected L_4 = G_xyz((x(i)+h),(y(i)+k_3*h),(z(i)+L_3*h)); %k_5 %L_5 y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h; % main equation z(i+1) = z(i) + (1/6)*(L_1+2*L_2+2*L_3+L_4)*h; % main equation end