Hi
Can sb help me please?
I should draw 3D-Plot for a first order differential system, but I could find just one example which was very complicated for me to learn!
a=1; b=2; c=a*(b^2-1/(b^2));
f = @(t,x)[x(2)*x(3)-a*x(1)
x(1)*(x(3)-c)-a*x(2)
1-x(1)*x(2)];
t=linspace(0,50);
[t,x]=ode45(f,t,[1 0 1]);
now I should draw the 3D trajectory (x1-x2-x3 diagram) and also indicate them in the course of time !
can someone help with it?
Thank you very much in advance

4 Comments

madhan ravi
madhan ravi on 6 Apr 2019
In 3D plane what does x1 ,x2, x3 & t what do they represent? Only three variables can be represented in x,y & z plane.
Yes, you are right.
I put the parameter time aside, and want to draw just x1 and x2 and x3 in a 3D space.
But what is the code/command to draw x1-x2-x3 in a differential system?
madhan ravi
madhan ravi on 7 Apr 2019
simply use plot3()
Thanks a lot

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 Accepted Answer

Star Strider
Star Strider on 6 Apr 2019
In our universe, there are three large dimensions (although string theory posits 11 in all). That means that you can only plot three spatial coordinates, and if you want to plot time as well, the only way is to add colour to the various regions.
This slight variation on your code does that:
a=1; b=2; c=a*(b^2-1/(b^2));
f = @(t,x)[x(2)*x(3)-a*x(1)
x(1)*(x(3)-c)-a*x(2)
1-x(1)*x(2)];
t=linspace(0,50,1000);
[t,x]=ode45(f,t,[1 0 1]);
cm = jet(numel(t));
figure
hold all
for k = 1:size(x,1)-1
hl = plot3([x(k,1),x(k+1,1)], [x(k,2),x(k+1,2)], [x(k,3),x(k+1,3)]);
set(hl, 'LineStyle','-.', 'Color',cm(k,:));
end
hold off
grid on
view(50, 30)
The jet (link) colormap (link) (that some object to using) colours the times here from earliest (blue) to latest (red). You can use any map (link) you want. Substitute that name into the ‘cm’ assignment to change it. I am not aware of any other way to depict time in a 3D plot that does not include time as an independent variable on one of the axes.

3 Comments

oh thank you very much!
Now I can analyse and change the commands step by step to see what happens and learn it.
Thank you a lot again
darova
darova on 7 Apr 2019
accept the answer then
Star Strider
Star Strider on 7 Apr 2019
@Saeed Ahmadzadeh — As always, my pleasure!
@darova — Thank you!

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