# Runge Kutta method for coupled oscillator system.

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Heya :) on 10 Nov 2020
Commented: Heya :) on 12 Nov 2020
I am trying to solve these equations with the help of Runge Kutta Method. I am not sure whether I am writing the code correctly or we can use this method for coupled also getting this error (mentioned at the end of the code). Please help me to refine my code.
close all; clear all; clc;
%initializing x,y,t
t(1)=0;
x1(1)=1;
y1(1)=1;
x2(1)=2;
y2(1)=2;
%value of constants
a=0.1;
b=0.3;
omega=4;
Cnp=0.2;
G=1;
h=0.1; %step size
t=0:h:50;
%ode
p=@(t,x1,y1,x2,y2) (a-x1^2-y1^2)*x1-omega*y1+G*Cnp(x2-x1);
q=@(t,x1,y1,x2,y2) (a-x1^2-y1^2)*y1+omega*x1+G*Cnp(y2-y1);
%loop
for i=1:(length(t)-1)
k1=p(t(i),x1(i),y1(i),x2(i),y2(i));
l1=q(t(i),x1(i),y1(i),x2(i),y2(i));
k2=p(t(i)+h/2,(x1(i)+(h/2)*k1),(y1(i)+(h/2)*l1),(x2(i)+(h/2)*k1),(y2(i)+(h/2)*l1));
l2=q(t(i)+h/2,(x1(i)+(h/2)*k1),(y1(i)+(h/2)*l1),(x2(i)+(h/2)*k1),(y2(i)+(h/2)*l1));
k3=p(t(i)+h/2,(x1(i)+(h/2)*k2),(y1(i)+(h/2)*l2),(x2(i)+(h/2)*k2),(y2(i)+(h/2)*l2));
l3=q(t(i)+h/2,(x1(i)+(h/2)*k2),(y1(i)+(h/2)*l2),(x2(i)+(h/2)*k2),(y2(i)+(h/2)*l2));
k4=p(t(i)+h,(x1(i)+k3*h),(y1(i)+l3*h),(x2(i)+k3*h),(y2(i)+l3*h));
l4=q(t(i)+h,(x1(i)+k3*h),(y1(i)+l3*h),(x2(i)+k3*h),(y2(i)+l3*h));
x(i+1) = x(i) + h*(k1+2*k2+2*k3+k4)/6;
y(i+1) = y(i) + h*(l1+2*l2+2*l3+l4)/6;
end
%draw
plot(t,x,'r')
xlabel('X','fontsize',14,'fontweight','bold')
ylabel('y','fontsize',14,'fontweight','bold')
set(gca,'Color','k')
****Error***
Array indices must be positive integers or logical values.
Error in coupled>@(t,x1,y1,x2,y2)(a-x1^2-y1^2)*x1-omega*y1+G*Cnp(x2-x1) (line 17)
p=@(t,x1,y1,x2,y2) (a-x1^2-y1^2)*x1-omega*y1+G*Cnp(x2-x1);
Error in coupled (line 25)
k2=p(t(i)+h/2,(x1(i)+(h/2)*k1),(y1(i)+(h/2)*l1),(x2(i)+(h/2)*k1),(y2(i)+(h/2)*l1));

Alan Stevens on 10 Nov 2020
Do you mean like this
%initializing x,y,t
h=0.1; %step size
t=0:h:50;
x1 = zeros(1,numel(t)); y1 = x1; x2 = x1; y2 = x1;
x1(1)=1;
y1(1)=1;
x2(1)=2;
y2(1)=2;
%value of constants
a=0.1;
b=0.3;
omega=4;
Cnp=0.2;
G=1;
%ode
p=@(x1,y1,x2,y2) (a-x1^2-y1^2)*x1-omega*y1+G*Cnp*(x2-x1);
q=@(x1,y1,x2,y2) (a-x1^2-y1^2)*y1+omega*x1+G*Cnp*(y2-y1);
%loop
for i=1:(length(t)-1)
k1=p(x1(i),y1(i),x2(i),y2(i));
l1=q(x1(i),y1(i),x2(i),y2(i));
k2=p((x1(i)+(h/2)*k1),(y1(i)+(h/2)*l1),(x2(i)+(h/2)*k1),(y2(i)+(h/2)*l1));
l2=q((x1(i)+(h/2)*k1),(y1(i)+(h/2)*l1),(x2(i)+(h/2)*k1),(y2(i)+(h/2)*l1));
k3=p((x1(i)+(h/2)*k2),(y1(i)+(h/2)*l2),(x2(i)+(h/2)*k2),(y2(i)+(h/2)*l2));
l3=q((x1(i)+(h/2)*k2),(y1(i)+(h/2)*l2),(x2(i)+(h/2)*k2),(y2(i)+(h/2)*l2));
k4=p((x1(i)+k3*h),(y1(i)+l3*h),(x2(i)+k3*h),(y2(i)+l3*h));
l4=q((x1(i)+k3*h),(y1(i)+l3*h),(x2(i)+k3*h),(y2(i)+l3*h));
x1(i+1) = x1(i) + h*(k1+2*k2+2*k3+k4)/6;
y1(i+1) = y1(i) + h*(l1+2*l2+2*l3+l4)/6;
x2(i+1) = x2(i) + h*(k1+2*k2+2*k3+k4)/6;
y2(i+1) = y2(i) + h*(l1+2*l2+2*l3+l4)/6;
end
%draw
plot(t,x1,'r',t,x2,'y')
xlabel('t','fontsize',14,'fontweight','bold')
ylabel('x1 & x2','fontsize',14,'fontweight','bold')
set(gca,'Color','k')
legend('x1','x2','TextColor','w')
figure
plot(t,y1,'r',t,y2,'y')
xlabel('t','fontsize',14,'fontweight','bold')
ylabel('y1 & y2','fontsize',14,'fontweight','bold')
set(gca,'Color','k')
legend('y1','y2','TextColor','w')
##### 3 CommentsShow 1 older commentHide 1 older comment
Alan Stevens on 12 Nov 2020
That is just setting aside storage space for those variables. It makes the loop more efficient as space doesn't then have to be found for their new elements every iteration of the loop.
Heya :) on 12 Nov 2020
Thank you :)