29.3. SEPARABLE DIFFERENTIAL EQUATIONS, STABILITY 551
The bottom axis is the t axis. Note how all the integralcurves in the picture approach K as t increases. This is whyK is called a stable equilibrium point.
Definition 29.3.8 Consider the equation dydt = f (y) . Then y0
is called an equilibrium point if f (y0) = 0. Note that the so-lution to the initial value problem y′ = f (y) ,y(t0) = y0 is justy = y0. An equilibrium point is stable if whenever y1 is closeenough to y0, it follows that the solution to the initial value
problem y′ = f (y) , y(0) = y1 stays close to y0 for all t > 0. It is asymptotically stable ifwhenever y1 is close enough to y0, it follows that for y the solution to the initial value prob-lem, y′ = f (y) , y(0) = y1 satisfies limt→∞ y(t) = y0. The equilibrium point y0 is unstableif there are initial conditions close to y0 but the solution does not stay close to y0. That is,there exists ε > 0 such that for any δ > 0 there is y1 with |y1− y0| < δ but the solution toy′ = f (y) ,y(0) = y1 has the property that for some t > 0, |y(t)− y0| ≥ ε . An equilibriumpoint y0 is semi-stable if it is stable from one side and unstable from the other.
Now observe that y = 0 is the solution which results if you begin with the initial con-dition y(0) = 0. If there is nothing to start with, it can’t grow. However, if you have anyother positive number for y(0) , then you see that the solution curve approaches the stablepoint K. You can see this, not just by looking at the picture but also by taking the limit ast→ ∞ in the above formulae.
One of the interesting things about this equation is that it is possible to determine K themaximum capacity, by taking measurements at three equally spaced times. Suppose youdo so at times t,2t,3t and obtain y1,y2,y3 respectively. Assume you are in the region wherey < K. In an actual experiment, this is where you would be. Let λ ≡ ert . Then from theabove formula for y, you have the equations
KCλ = y1 (Cλ +1) ,KCλ2 = y2
(Cλ
2 +1),KCλ
3 = y3
(Cλ
3 +1)
Then divide the second equation by λ and compare with the first. This shows that λ =y2/y1. Next divide the top equation by Cλ and the last by Cλ
3. This yields
K = y1
(1+
1Cλ
)= y3
(1+
1
Cλ3
)Now it becomes possible to solve for C. This yields
C =
(y3
1y3− y21y2
2)
y1y32− y3
2y3
Then substitute this in to the first equation. This obtains
K
((y3
1y3− y21y2
2)
y1y32− y3
2y3
)y2
y1= y1
((y3
1y3− y21y2
2)
y1y32− y3
2y3
(y2
y1
)+1
)Then you can solve this for K. After some simplification, it yields
y22y3− y2
1y3
y22− y1y3
= K
29.3. SEPARABLE DIFFERENTIAL EQUATIONS, STABILITY 55115) VoyeyeoyeoyoviyVi yyy yr The bottom axis is the ¢ axis. Note how all the integralAWW NNN ANNA : . .it SSS SS curves in the picture approach K as ¢ increases. This is whyZYE34077077 Kiscalleda stable equilibrium point.LV 4447S dy54/1 7474777 7 7 Definition 29.3.8 Consider the equation = f(y). Then yoZS/E TE EEE, vey oo tCLEC Is called an equilibrium point if f (yo) = 0. Note that the so-giticscre--- lution to the initial value problem y' = f (y) ,y (to) = yo is just0 5 10 y=~yo. An equilibrium point is stable if whenever y, is closeenough to yo, it follows that the solution to the initial valueproblem y' = f (y), y(0) = y1 stays close to yo for allt > 0. It is asymptotically stable ifwhenever y is close enough to yo, it follows that for y the solution to the initial value prob-lem, y' = f (y), y(0) = y1 satisfies lim,_,.. y(t) = yo. The equilibrium point yo is unstableif there are initial conditions close to yo but the solution does not stay close to yo. That is,there exists € > 0 such that for any 6 > 0 there is y, with |y, —yo| < 6 but the solution toy’ = f (y),y(0) = y has the property that for some t > 0,|y(t) —yo| > €. An equilibriumpoint yo is semi-stable if it is stable from one side and unstable from the other.Now observe that y = 0 is the solution which results if you begin with the initial con-dition y (0) = 0. If there is nothing to start with, it can’t grow. However, if you have anyother positive number for y (0), then you see that the solution curve approaches the stablepoint K. You can see this, not just by looking at the picture but also by taking the limit ast — co in the above formulae.One of the interesting things about this equation is that it is possible to determine K themaximum capacity, by taking measurements at three equally spaced times. Suppose youdo so at times ¢, 2t, 3t and obtain yj, y2, y3 respectively. Assume you are in the region wherey < K. In an actual experiment, this is where you would be. Let A = e”. Then from theabove formula for y, you have the equationsKCA =y| (CA+1),KCA2 = yp (cr? + 1) .KCM3 =y3 (ca? + 1)Then divide the second equation by A and compare with the first. This shows that 2 =y2/y1. Next divide the top equation by CA and the last by CA°. This yields] ]k=n (14g) =»(1+q3)Now it becomes possible to solve for C. This yields(viys — yiy3)Yiy3 — Y3y3Then substitute this in to the first equation. This obtains(vivs 193) \x2 [ (viv3 —yty3) (9K\ 3S J Se | a7 Sd Iyi —y3y3 J yiyz—y3y3 \y1Then you can solve this for K. After some simplification, it yieldsY3¥3 —YTY3 _5 KYa — Y1Y3
