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 Commenti

Jan
Jan il 14 Nov 2021
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
Jan
Jan il 15 Nov 2021
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?
Jan
Jan il 16 Nov 2021
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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Jan
Jan il 14 Nov 2021
Modificato: Jan il 15 Nov 2021

0 voti

The test contains cos(dt), but your code only cos(t).

3 Commenti

Ghofran Khaled
Ghofran Khaled il 14 Nov 2021
Modificato: Ghofran Khaled il 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
Jan
Jan il 18 Nov 2021
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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il 14 Nov 2021

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Jan
il 18 Nov 2021

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