29.3. SEPARABLE DIFFERENTIAL EQUATIONS, STABILITY 553

Is there a systematic way to figure this out without doing lots of computer generatedpictures? The answer is yes! Furthermore, it is very easy to do. Consider the right side ofthe equation. If you graph the function z = f (y) , you get something which looks like theright side of the above.

Look at the graph. When y ∈ (0,T ) , you have the slope of the graph is negative andso, from the equation, dy

dt is negative and so t → y(t) is decreasing. (Remember calculus.)If y ∈ (T,K) , then the graph is positive and so dy

dt is positive which requires that t → y(t)is increasing. When y ∈ (K,∞) , the graph is negative and so t → y(t) is decreasing. ThusT is unstable, K is stable while 0 is also stable. I have not considered the case wherey < 0 because this is not too interesting in the example which typically describes y as apopulation of something. However, you can see from the graph that if y < 0, then t→ y(t)is increasing.

In general, you can consider y′= f (y) and the equilibrium points. The following pictureis descriptive of the situation. Such an equation is called autonomous because the functionon the right depends only on y and not on t.

•unstable

•stable

Pieces of the graph of f

Proposition 29.3.10 Suppose f is continuous with continuous derivative and that f (y0) =0, f ′ (y0)< 0. Then y0 is asymptotically stable.

Proof: By continuity of f ′, there is δ > 0 such that for y∈ (y0−δ ,y0 +δ ) = I, f ′ (y)≤−2η ,η > 0. Thus

f (y) = f (y0)+ f ′ (y0)(y− y0)+o(y− y0)

= f ′ (y0)(y− y0)+o(y− y0)

Then if y1 ∈ I, and if y(t) is the solution to the equation y′ = f (y) having this initialcondition, then

y(t)− y0 = y1− y0 +∫ t

0f (y(s))ds

= y1− y0 +∫ t

0f ′ (y0)(y(s)− y0)ds+

∫ t

0o(y(s)− y0)ds

We can also assume δ is small enough that |o(y− y0)| < η |y− y0| for y ∈ I. Say y1 > y0.Then by assumption, t → y(t) is decreasing since y′ = f (y) < 0 and so if y(t) fails toconverge to y0, there would exist ε > 0 which is the limit of y(t)− y0. Then

ε +∫ t

0− f ′ (y0)(y(s)− y0)ds≤ y1− y0 +η

∫ t

0(y(s)− y0)

Thusε +

∫ t

0ηε ≤ y1− y0