Runge-Kutta 4th order method

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Mariam Gasra
Mariam Gasra 2019-5-5
编辑： Torsten 2024-5-5
% It calculates ODE using Runge-Kutta 4th order method
% Author Ido Schwartz
clc; % Clears the screen
clear;
h=5; % step size
x = 0:h:100; % Calculates upto y(3)
Y = zeros(1,length(x));
y(1) = [-0.5;0.3;0.2];
% initial condition
F_xy = @(t,r) 3.*exp(-t)-0.4*r; % change the function as you desire
for i=1:(length(x)-1) % calculation loop
k_1 = F_xy(x(i),y(i));
k_2 = F_xy(x(i)+0.5*h,y(i)+0.5*h*k_1);
k_3 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k_2));
k_4 = F_xy((x(i)+h),(y(i)+k_3*h));
y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h; % main equation
end
display(Y(i+1));
if i run the programme i get answer =0;
how can i solve this problem if i have three initial condition -0.5 ,0.3 and 0.2
while x=0:5:100
and how i can plot the answer with respect to x?
  2 个评论
Naveen
Naveen 2024-5-5
what can be done if functions are very large?
Torsten
Torsten 2024-5-5
编辑：Torsten 2024-5-5
Define them as a function instead of a function handle:
https://uk.mathworks.com/help/matlab/ref/function.html

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采纳的回答

David Wilson
David Wilson 2019-5-6
编辑：MathWorks Support Team 2023-4-18
First up, you will need a much smaller step size to get an accurate solution using this explicit RK4 (with no error control). I suggest h = 0.05. Validate using say ode45 (which does have error control).
Then you will need to run your ode above three separate times, once starting from y(1) = -0.5, again with y(1) = 0.3, etc.
Then finally plot the result with plot(x,y,'o-').
h=0.05; % step size
x = 0:h:100; % Calculates upto y(3)
y = zeros(1,length(x));
%y(1) = [-0.5;0.3;0.2];
y(1) = -0.5; % redo with other choices here.
% initial condition
F_xy = @(t,r) 3.*exp(-t)-0.4*r; % change the function as you desire
for i=1:(length(x)-1) % calculation loop
k_1 = F_xy(x(i),y(i));
k_2 = F_xy(x(i)+0.5*h,y(i)+0.5*h*k_1);
k_3 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k_2));
k_4 = F_xy((x(i)+h),(y(i)+k_3*h));
y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h; % main equation
end
% validate using a decent ODE integrator
tspan = [0,100]; y0 = y(1);
[tx, yx] = ode45(F_xy, tspan, y0);
plot(x,y,'o-', tx, yx, '--')
  4 个评论
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Walter Roberson
Walter Roberson 2022-11-14
the code would need to be adjusted slightly if the ode function has more than one state (and so returns a vector.)

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更多回答（6 个）

Sandip Das
Sandip Das 2021-7-28
%Published in 25 July 2021
%Sandip Das
clc;
clear all;
dydt=input('Enter the function : \n');
t0=input('Enter the value of t0 : \n');
y0=input('Enter the value of y0 : \n');
tn=input('Enter the value of t for which you want to find the value of y : \n');
h=input('Enter the step length : \n');
i=0;
while i<tn
k_1 = dydt(t0,y0);
k_2 = dydt(t0+0.5*h,y0+0.5*h*k_1);
k_3 = dydt((t0+0.5*h),(y0+0.5*h*k_2));
k_4 = dydt(((t0)+h),(y0+k_3*h));
nexty = y0 + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;
y0=nexty
t0=t0+h
i=i+h;
end
fprintf('The value of y at t=%f is %f',t0,y0);
mahmoud mohamed abd el kader
function [x,y] = rk4th(dydx,xo,xf,yo,h)
x = xo:h:xf ;
y = zeros(1,length(x));
y(1)= yo ;
for i = 1:(length(x)-1)
k_1 = dydx(x(i),y(i));
k_2 = dydx(x(i)+0.5*h,y(i)+0.5*h*k_1);
k_3 = dydx((x(i)+0.5*h),(y(i)+0.5*h*k_2));
k_4 = dydx((x(i)+h),(y(i)+k_3*h));
y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;
end
dydx = @(x,y) 3.*exp(-x)-0.4*y;
%[x,y] = rk4th(dydx,0,100,-0.5,0.5);
%plot(x,y,'o-');
end
  3 个评论
显示 1更早的评论
soham roy
soham roy 2022-12-8
What modifications do we need to make in this code to solve 3 ODEs with different initial conditions?
Walter Roberson
Walter Roberson 2022-12-8
@soham roy
y = zeros(1,length(x));
would change to
y = zeros(length(x), length(y0));
and below that, each y(INDEX) would be replaced with y(INDEX,:)

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Mj
Mj 2020-11-7
Hello everyone!
I have to solve this second order differential equation by using the Runge-Kutta method in matlab:
can anyone help me please? and how can i plot the figure?(a against e)
d2a/de2=(((((2+c2)*(Fu^2))/(1+c2))+1)*(a^c2)-((2+c2/1+c2)*(Fu^2/a))-a^(2+(2*c2)))/(((2+c2)*Fu^2)/(1+c2)*(3+c2));
Fu=1
c2=0 , 0.5 , 1 (there are 3 values for c2)
initial conditions are: a=0.8 , d_a=
David Wilson
David Wilson 2021-5-6
Ran in:
Wow, you haven't given us too much to go on, so that makes a real challenge.
First up, your 2nd order ODE is needlessly complex given that Fu=1, and c2 =0 say. (I'm not sure what the other valuesare for, Are you solving this 3 seprate times? (Be good to know if that is the case.)
If you have the symbolic toolbox, it makes it easy to simplify your problem to something doable. First up, I'm going to try and solve it analytically.
syms Fu c2 real
syms a(t)
f2 = (((((2+c2)*(Fu^2))/(1+c2))+1)*(a^c2)-((2+c2/1+c2)*(Fu^2/a))-a^(2+(2*c2)))/(((2+c2)*Fu^2)/(1+c2)*(3+c2));
f2_a = subs(f2,Fu,1)
f2_a(t) = 
f2_b = subs(f2_a,c2,0) % subs c2 for 0
f2_b(t) = 
Da = diff(a);
D2a = diff(a,2);
% Now attempt to solve analytically
dsolve(D2a == f2_b, a(0) == 0.8, Da(0) == 1)
Warning: Unable to find symbolic solution.
ans = [ empty sym ]
Well that didn't work, but no real suprise there.
Let's try a numerical method:
syms Fu c2 real
syms a real
f2 = (((((2+c2)*(Fu^2))/(1+c2))+1)*(a^c2)-((2+c2/1+c2)*(Fu^2/a))-a^(2+(2*c2)))/(((2+c2)*Fu^2)/(1+c2)*(3+c2));
f2_a = subs(f2,Fu,1); f2_b = subs(f2_a,c2,0); pretty(f2_b)
2 1 a 1 - - -- - --- 2 6 3 a
We need to encode this as a system of 2 ODES. (Convert to Cauchy form)
aprime = @(t,a) [a(2); ...
0.5 - a(1).^2/6 - 1./(a(1)*3)]
Now we are ready to solve the ODE. I'll use ode45, and guess a t-span, and guess one of the initial conditions since you forgot to help us out there.
aprime = @(t,a) [a(2); ...
0.5 - a(1).^2/6 - 1./(a(1)*3)]
a0 = [0.8; 0]
[t,a] = ode45(aprime, [0,4], a0)
plot(t,a)
Amr Mohamed
Amr Mohamed 2021-5-9
how can we write the code for this problem :
  2 个评论
Moneeb Ur Rehman
Moneeb Ur Rehman 2021-5-27
get the y on other side, integrate then to find 1st derivative. Now apply R.k method to solve. Hope you understood;

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monsef
monsef 2023-7-17
y=x^2-2yx
h=0.2
y0=0
x0=1
wriet program im mathlab
  1 个评论
Ahmed J. Abougarair
Ahmed J. Abougarair 2024-3-24
clc;
clear all;
F = @(t,y) 4*exp(0.8*t)-0.5*y
t0=input('Enter the value of t0 : \n');
y0=input('Enter the value of y0 : \n');
tn=input('Enter the value of t for which you want to find the value of y : \n');
h=input('Enter the step length : \n');
i=0;
while i<tn
k_1 = F(t0,y0);
k_2 = F(t0+0.5*h,y0+0.5*h*k_1);
k_3 = F((t0+0.5*h),(y0+0.5*h*k_2));
k_4 = F(((t0)+h),(y0+k_3*h));
nexty = y0 + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;
y0=nexty;
t0=t0+h;
i=i+h;
end
fprintf('The value of y at t=%f is %f \n',t0,y0)
% validate using a decent ODE integrator
tspan = [0,1]; Y0 = 2;
[tx,yx] = ode45(F, tspan, Y0);
fprintf('The true value of y at t=%f is %f \n',tspan(end),yx(end))
Et= (abs(yx(end)-y0)/yx(end))*100;
fprintf('The value of error Et at t=%f is %f%% \n',tspan(end),Et)

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