How to find fixed points in nonlinear differential equations?

I'm struggling with such problem that I need to find fixed points, and then sketch the nullclines,the vector field and a phase portrait. I cannot handle finding fixed points of those two differential equations in one point: dx/dt=2xy dy/dt=y^2-x^2 How to get those fixed points and then sketch this phase portrait? I know how should it look but have no idea of how to put my hands on it.

Answer 1

[Mischa Kim]

Hello Michael, by definition the state derivate at a fixed point is equal to the zero vector. In your case that means x'= 0 and y'= 0. One obvious fixed point is at x = y = 0. There are various ways of getting the phase diagram:

1. From the two equations compute dx/dy. Choose initial conditions [x0; y0] and with dx/dy compute the trajectory.
2. Alternatively you could use the differential equations to calculate the trajectory. Pick initial conditions [x0; y0] and use, e.g. ode45 to do the integration.

Once you have a trajectory you can get the phase portrait by picking different initial conditions and repeat 1. or 2.

[Michael]

Oh I see now.From the other equation y=x so it means that when I pick for example x0=1 then y0=1? And in ode45 which would be my fixed points f.ex. ode45('function', range of Xes, fixed points)?

[Mischa Kim]

Almost. What I mean is that

• at a fixed point the derivatives are zero. That is xy = 0 and y^2 - x^2 = 0. The only solution is x = 0 and y = 0. That is your fixed point. No need to use MATLAB for that part of the problem.
• Use

tspan = 0:0.1:10;  %you might want to adapt the range for t
[T,Z] = ode45(@myfunc, tspan, [x0 y0]); %myfun is your DE function
x = Z(1,:);
y = Z(2,:);

[Myriam Ghali]

Could you please explain how you code for myfunc ?