A quick explanation of stability?

[dillonmhudson]

I have the following DE:
y'=y*(1-y^2)
And understand that y=0, and y=+/-1 are critical points

I am confused as to which are stable/unstable and why.

TIA

 

[HallsofIvy]

Yes, y'= y(1- y)(1+ y) so y= 0, 1, and -1 give "equilibrium solutions". If, at any time, y is equal to one of those, its derivative is 0 so it doesn't change- those are constant solutions.

Now, look at what happens for other initial values of y. Suppose, for some t, y< -1.
Then all three of y and 1+ y are negative bur 1- y is positive so their product is positive. That is, y'> 0 so y is increasing, moving toward -1.

Suppose y lies between -1 and 0. Then y is still negative but both 1- y and 1+ y are positive. Their product is negative so now y' is negative. Now y is decreasing but that means it is still movin g toward -1.

If y is some value close to -1, whether above or below, y(t) moves toward -1. That is what "stable equilibrium" means- if y is close to the equilibrium point but not exactly equal to it, y(t) will stay close or even move toward it.

Of course, since y(t) is decreasing for -1< y< 0, it is moving away from 0. If y is between 0 and 1, now all three of y, 1- y and 1+ y are positive so their product, and therefore y' is positive. y is increasing and so is moving away from 0. That is, if y is close to the equilibrium value, 0, it will move away from that value. That means that y= 0 is an unstable equilibrium- values near there do NOT stay near it.

Of course, for 0< x< 1, since y is increasing, it is moving toward y= 1. And, if y is larger than 1, y and 1+ y are positive but now 1- y is negative. Their product is negative which means y' is negative. y is decreasing and again moving toward 1. That is, if y is close to 1 but not exactly at 1, whether above or below 1, it will move toward 1. That is a "stable" equilibrium.

Another way of thinking about this- if you look at the anti-derivative of $y(1- y^2)$, $(1/2)y^2- (1/4)y^4$ you will see a quartic graph that has a minimum at x= 1, a maximum at x= 0, and a second minimum at x=-1. If you think of that as representing a hill between two valleys, if you were to set a ball on the hill at x= 0, it will not immediately roll away because it is balanced there (so it is in "equilibrium"). But the slightest disturbance would knock it off that hill and start it rolling downward. That is an "unstable" equilibrium. If, instead, you set the ball at either of the two valleys, at x=-1 or x= 1, A slight disturbance might start the ball rocking but it would roll back to the bottom of the valley. Those are "stable" equilibria.

 

[dillonmhudson]

Wow thanks so much for the helpful reply!