How to plot the trajectories of the equilibrium points

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Raja Emlik
Raja Emlik am 8 Apr. 2016
Kommentiert: Raja Emlik am 8 Apr. 2016
I want to plot the trajectories of the equilibrium points and to show the stability of them . This my code to find all the equilibrium points
r1=576; r2=23616/11; r3=39456/11;
a1=572; a2=4420; a3=5850;
b1=308; b2=1804; b3=2970;
c1=b1; c2=b1; c3=594;
syms x y z
vars = [x, y, z];
eqs=[(r1-a1*x-b1*y-c1*z)*x,(r2-a2*x-b2*y-c2*z)*y,(r3-a3*x-b3*y-c3*z)*z];
[xc, yc, zc] = solve(eqs(1), eqs(2), eqs(3));
[xc, yc, zc]
A = jacobian(eqs, vars);
disp('Matrix of linearized system:')
subs(A, vars, [xc(1), yc(1),zc(1)])
disp('eigenvalues:')
eig(ans)
disp('Matrix of linearized system:')
subs(A, vars, [xc(2), yc(2),zc(2)])
disp('eigenvalues:')
eig(ans)
disp('Matrix of linearized system:')
subs(A, vars, [xc(3), yc(3),zc(3)])
disp('eigenvalues:')
eig(double(ans))
disp('Matrix of linearized system:')
subs(A, vars, [xc(5), yc(5),zc(5)])
disp('eigenvalues:')
eig(ans)
disp('Matrix of linearized system:')
subs(A, vars, [xc(6), yc(6),zc(6)])
disp('eigenvalues:')
eig(ans)
disp('Matrix of linearized system:')
subs(A, vars, [xc(7), yc(7),zc(7)])
disp('eigenvalues:')
eig(ans)
disp('Matrix of linearized system:')
subs(A, vars, [xc(8), yc(8),zc(8)])
disp('eigenvalues:')
eig(ans)
there are 8 equilibrium points and one of them is negative so we do not care about it . Can you please help me how to plot these as the below graph. any help would be appreciated.

Antworten (1)

Mischa Kim
Mischa Kim am 8 Apr. 2016
Bearbeitet: Mischa Kim am 8 Apr. 2016
Raja, you are probably best off to solving differential equations of the system for different initial conditions (and the plot the solutions). You would pick the initial conditions close to the equilibrium points to get the figure you show above. See for example this answer.
  1 Kommentar
Raja Emlik
Raja Emlik am 8 Apr. 2016
the example seems different than mine as mine in 3D could you help me with that please ?

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