Question

Problem 5. (20 pts) Let f(y) be the real function f: R R depicted in Figurei, and consider the autonomous differential equation y(t) = f(y(t)). fly) у FIGURE 1. The function f(y) for Problem 4. (a) How many constant solutions does the above differential equation have ? (b) Study whether the behaviour of each of the constant solutions of the differential equation y(t) = f(y(t)) is stable, unstable or semistable. (c) Discuss the long-term behaviour of all solutions y(t) to this differential equation in terms of their initial condition y(0). (d) Consider also the autonomous differential equation: y(t) = f(y(t))+ 0.1. Compare the behavior of its solutions with the behavior of the solutions of y(t) = f(y(t). In particular, find how many constant solutions does the dif- ferential equation y(t) = f(y(t))+ 0.1 have.

Answer 1

y"(t) = f(y(t)) @ the constant the constant somnions are those values of y at which f(y)=0, ie the point a wehed fly) cuts yaxis. Total 6-times f(4) cuts y-axis so there are 6 constant sousons (say these are a, C2, C3, C4, C5 (6, starting from left, see in the fignue). 16. If yzc is the constant sourian, then 1 of f(y) <0 on the left of c and of (y)>0 .. on the night of then y= c is unstane. I . If fly) >0 on the left of c and of(y)<o on the right of c, then they=c is asymprorically stake. I stable. 6 of fly) >o ou both sides of c, a fly)<o on. both sides of a then the equilibrium solusion is semi stake. • Let's start with left most constant solution say so it is then ou left eft of y = c of y=c f(y) so f(ylso and on night fly) co, ac is stable so wsion. • Now, the second one from the left then on lift of y=6 flyco fly)>0 then & is unstable." is and say C2. ou right

from say it is С₂. Now the third cue then ou left of y = flyco & cz is stable the left f(y)>0 and • Now, the fourth one from the left, say it is ca then o u left of y = ca flygco and on sigut ' ce is unstable Now for the fifth one from the left, say it is then left and right nol y = 5, fly o & Co is same-stage. we • Now for sixth one from the left, say it is co then on left fly) and on right fly) co aco is stane. ©• 9f y(0) > <¢1, then fly) >0 yy'> z Lim y(t) 700 • 9f 700)=, then f (y) =0 Hy' =0 a c'm yct) with went • of yo) =k, where ack C C₂ then flyko , & yiko woode 2 limy(t) =cal toot

• $ ae) е , чьи річ. 1) = 4, • 3g to) = k , шаш <k <<3, 17)>° , 7' >0 - ua y) — ® • 9{ је ) = 4, #da ' (4) = (з – • 9 (0) = k, uш е, << < 04 , ал f() <o - С1+1 = • 3 yte) = 4, 4а, а 9 (+) = - • 9_yto) = k, lua 4<L <s , Кли + () >0 - Ом 1 ) =cs • 28 Я(0) - 4 и Діш Я (+) = cs • 9{ чо) = k , иш <<k <ec , Или fet>0,9 Hua ye) = to • 3 чо) - с., ал с 1 %) = 4 • 3 7са , +и + (1) го , и у %) = С. с then lim ult - Cs E9 до

| @9 y'et) = f(y(t)) +0,1 Since you can see a horizontal line f (y) = -1 80 -1 to,l = -0.9 lie to new horizontal live will be 0.9 graph will move upwend by oil. so the tuw constant so usions as and Co will be lost and no new constant solusion will be achieved. So a'ct)= flyct)) + 0.1 has & constant solutions the earlier one's only which are ci, c2,c₂, a.

Now the new solusions aco.ca C4 Take c - on left of c, do is stable ie if y=C), then f(y)>0 and f(y)<o take C2, on le if y=(2, then ou left of C₂ - f(y)<0 and on right f(y) >0 &c is unstasie Take Cz, leif = lg , then an left of C, f (4) >0 and ou right fly, eo is ez is stable. • Take Cy ie if y=(4, then on lef of cq f(y)<o and on right fly)>0 2 4 is unstable. the common souston all beaning the same way