how I create improved lorenz code

hi everyone
I want to create improved lorenz code , but its doesnt work well
could someone help me , please
this is my code
function dy= lorenz(t,y)
a = 8/3;
b = 25;
c = 10;
d = 5.3;
e = 17.5;
f = 10.5;
g = 13.3;
h = 14;
if cos(5.3*t)>=0
p=1;
else
p=-1;
end
dy=zeros(3,1);
dy(1)=a*y(1)+p*y(2)*y(3);
dy(2)=y(3)*b-y(3)*c*cos(d*t) - b*y(2)+y(2)+c*cos(d*t);
dy(3)= -1*p*y(1)*y(2)-(e*cos(d*t)+f)*y(2)+(g-h*y(3)*(cos(d*t)));
end

6 Comments

If you do see a problem, it is a good idea to explain it clearly. This helps to solve the problem. After reading "its doesnt work well", we have to guess, what the problem is.
The graphic appears incorrectly
when initial parameters y=[0.9613 0.5201 0.1314] and t=[0 300]
the graph must be like below
but its apear like this . I dont know why
I've implemented it by my own and get an equivalent trajectory.
@Jan I didn't understand well, did it work for you correctly? What is the code you used?
If I follow the instructions from scratch, I get the same output as you. So why to you assume that the diagram of the top is the correct solution? It looks like the standard Lorenz attractor, but the formula is something different.
@Jan Thank you very much for helping me

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 Accepted Answer

Jan
Jan on 14 Nov 2021
Edited: Jan on 15 Nov 2021
The test contains cos(dt), but your code only cos(t).

3 Comments

Ghofran Khaled
Ghofran Khaled on 14 Nov 2021
Edited: Ghofran Khaled on 14 Nov 2021
I fixed it and it still doesn't look right
@Jan Please can you help me this took me a week and I don't know where I went wrong I'm out of time
I still do not see a convincing reason to assume, that there is any problem. I've written some code to solve the answer and get the same output as you (as far as I can see). Why do you think that your code is wrong?

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Asked:

on 14 Nov 2021

Commented:

Jan
on 18 Nov 2021

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