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This is the 4d Lorenz equation. how to solve it on Matlab?

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Nix Jr
Nix Jr 2021-7-25
编辑： Yongzhen Mi 2021-7-26
dx/dt=y+y^2-a*y*z;
dy/dt=-z^2+b*y*z-u;
dz/dt=x*y;
du/dt=-c*x;
  2 个评论
Nix Jr
Nix Jr 2021-7-25
Here has the function with code. But I didn't understand what's the problem. If you don't mind, would you please help me to fix the problem?
...............................................................function..........................................
function df = test(~, x)
% HELP: Lorenz Functions
%dx/dt=y+y^2-a*y*z;
%dy/dt=-z^2+b*y*z-u;
%dz/dt=x*y;
%du/dt=-c*x;
u=10; a=8/3; b=7; c=.5;
% ICs: x(0)=5; y(0)=5; z(0)=5; % Your ICs
df=[x(2)+([x(2)]^2)-a*x(2).*x(3); ...
-([x(3)]^2)+b*x(2).*x(3)-x(4);...
x(1).*x(2)...
-c*x(1)];
end
...................................................................................main code...........................
ICs=[5, 5, 5]; % Your data
t=[0, 20]; % Your simulation space
OPTs = odeset('reltol', 1e-6, 'abstol', 1e-8); % Optional ODE options set up
[time, fOUT]=ode45(@test, t, ICs, OPTs);
close all
figure
plot3(fOUT(:,1), fOUT(:,2), fOUT(:,3), grid
xlabel('x(t)'), ylabel('y(t)'), zlabel('z(t)')
title('LORENZ functions x(t) vs. y(t) vs. z(t)')
axis tight
figure plot3(fOUT(:,1), fOUT(:,2), fOUT(:,3),
figure
subplot(311)
plot(time, fOUT(:,1), 'b','linewidth', 3), grid minor
title 'LORENZ functions x(t), y(t), z(t)', xlabel 'time', ylabel 'x(t)'
subplot(312)
plot( time', fOUT(:,2), 'r', 'linewidth', 2 ), grid minor
xlabel 'time', ylabel 'y(t)'
subplot(313)
plot(time, fOUT(:,3),'k', 'linewidth', 2), grid minor, xlabel 'time', ylabel 'z(t)'
figure
plot(fOUT(:,1), fOUT(:,2), 'b', 'linewidth', 1.5)
grid minor, title('LORENZ functions'), xlabel('x(t)'), ylabel 'y(t)'
axis square
figure
plot(fOUT(:,1), fOUT(:,3), 'k', 'linewidth', 1.5)
grid minor, title('LORENZ functions'), xlabel('x(t)'), ylabel 'z(t)'
axis square
figure
plot(fOUT(:,2), fOUT(:,3), 'm', 'linewidth', 1.5)
grid minor, title('LORENZ functions'), xlabel('y(t)'), ylabel 'z(t)'
axis square

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回答（1 个）

Yongzhen Mi
Yongzhen Mi 2021-7-26
Hi, Nix Jr. I think the error is that you defined four variables in your differential equations, but you provided only three initial values to the ode45 function. Therefore, ICs should be something like this [5, 5, 5, 5]' (column vector better).
In addition, you can transfer parameters into the target function by using [time, fOUT]=ode45(@(x)test(~,x,a,b,c), t, ICs, OPTs);. In this way, the parameters a, b, and c can be claimed in the main function once, instead of being defined again and again when the target function is called.
  2 个评论
Nix Jr
Nix Jr 2021-7-26
Thanks for you suggestion but this process doesn't work. or might be I didn't get your process.
Yongzhen Mi
Yongzhen Mi 2021-7-26
编辑：Yongzhen Mi 2021-7-26
Take this as an example:
a = 5;
b = 20;
c = 1;
d = 0.1;
k = 0.1;
e = 20.6;
h = 1;
F = @(t,Y) [a*(Y(2)-Y(1))-e*Y(4);Y(1)*Y(3)-h*Y(2);b-Y(1)*Y(2)-c*Y(3);k*Y(2)-d*Y(4)];
CI = [3.2 8.5 3.5 2.0]';
T = linspace(0,100,3000);
[t,Y]=ode45(F,T,CI);
plot3(Y(:,1),Y(:,2),Y(:,3))
I drew some pictures and the attractors could be observed. Please try again and hopefully this will work.

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