MATH 304 -- Ordinary Differential Equations

Equilibrium points for non-linear systems

November 8, 2004

The first-order system

    dx/dt = x^2-4

    dy/dt = -x^2*y+x^2-y+1

has two equlibrium points:  (2,1)  and (-2,1)

Let's use Maple's phaseportrait command

to plot some solutions to visualize what is

going on:

>    eqns:={diff(x(t),t)=x(t)^2-4,diff(y(t),t) =-x(t)^2*y(t)+x(t)^2-y(t)+1};

eqns := {diff(x(t),t) = x(t)^2-4, diff(y(t),t) = -x(t)^2*y(t)+x(t)^2-y(t)+1}

>    with(DEtools):

>    phaseportrait(eqns,[x(t),y(t)],t=0..0.15,[seq(seq([x(0)=i,y(0)=j],j=-3..3),i=-3..3)],linecolor=black,arrows=slim);

[Maple Plot]

From this evidence it looks like the equlibrium point at (-2,1) is a sink  (since all

solutions that start nearby seem to be tending toward that point).  On the other

hand, (2,1) is a saddle point (since some solutions approach, but others

tend away from that point).  

In fact the linearization of the system at (-2,1) is the 1st order linear system

du/dt = -4*u  ,   dv/dt = -5*v ,  where   u = x + 2,   v = y - 1

which has a sink at ( u,v ) = (0,0). Similarly, at (2,1) we get

du/dt = 4*u  ,   dv/dt = 5*v ,  where u = x - 2,   v = y - 1,

which has a saddle point at ( u,v ) = (0,0).

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