How to find fixed points in nonlinear differential equations?

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Michael 2014-1-24

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评论： Myriam Ghali 2018-11-19

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.

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Answer 1

Mischa Kim 2014-1-24

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编辑：Mischa Kim 2014-1-24

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:

From the two equations compute dx/dy. Choose initial conditions [x0; y0] and with dx/dy compute the trajectory.
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.

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Mischa Kim 2014-1-24

在 MATLAB Online 中打开

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,:);