Help with lorenz equation

I used a function to call it to get the lorenz solution and then plot it

this is my code

if true
  % syms s t 
[t1 x1]=ode45(@lorenz,[0,10],[1;0;0])
[t2 x2]=ode45(@lorenz,[0,10],[1;0;0])
[t3 x3]=ode45(@lorenz,[0,10],[1;0;0])
figure(1)
plot(t1,x1)
figure(2)
plot(t2,x2)
figure(3)
plot(t3,x3)
figure(4)
plot(x1,x2)
figure(5)
plot(x1,x3)
figure(6)
plot(x2,x3)
end 
and the function was the following

if true
  % function [xdot]=lorenz(t,x)
t=0:0.001:10
xdot=[10*(x(2)-x(1));-1*x(1)*x(3)-x(2);x(1)*x(2)-(8/3)*x(3)]
end
end

any advice on how to resolve ; it is not running and just looking to continue my approach just to get it solved , thank you

Answer 1

[Star Strider]

You’re close. Be sure you have the correct initial conditions.

This should get you started:

lorenz = @(t,x) [10*(x(2)-x(1));-1*x(1)*x(3)-x(2);x(1)*x(2)-(8/3)*x(3)];    % Anonymous Function

[T,X] = ode45(lorenz, [0 10], [1; 0; 0]);

figure(1)
plot(T, X(:,1))
grid
xlabel('Time')

figure(2)
plot(X(:,1), X(:,2))
grid
axis equal

[diana betancur]

thank you , can I still use my function or is better to use anonymous ? thank you once again

[Star Strider]

My pleasure.

You can do either. I used an anonymous function because I find them easier to work with.

[diana betancur]

still is not running with the code

[Star Strider]

The code I posted in my Answer runs without error in R2017a.

In the Wikipedia article on the Lorenz system, the MATLAB simulation has the initial conditions vector as [1 1 1], and the correct version of the Lorenz system, that being:

lorenz = @(t,x) [10*(x(2)-x(1)); x(1).*(28-x(3))-x(2); x(1)*x(2)-(8/3)*x(3)];    % Anonymous Function

Try that. It reproduces the plot in the Wikipedia article.