MAT1193 – 6a Derivatives and the stability of discrete time dynamical systems We already talked about equilibrium states for dynamical systems. An equilibrium state is a state where all the factors leading to change in the system are in balance so the state does not change in time. Suppose we have DTDS with h(x) as an updating function and st is the state at time t. This means that st+1=h(st). If the system is in an equilibrium state, then updating the system does not lead to any change. In other words if s* is an equilibrium state and st = s*, then st+1 is also equal to s* leading to the condition s* = h(s*). There are two very different types of equilibrium states. These are best illustrated by considering a pendulum that is free to swing a full 360 degrees. There are two equilibrium states for this system. If the pendulum is pointing straight down, it obviously won’t move and the system is in equilibrium. But the system is also is at an equilibrium state when the pendulum is pointing straight up,, since any tendency to move counterclockwise is exactly balanced by the tendency to move in a clockwise direction. While both of these states are equilibrium states, something very different happen when the pendulum is moved by a small angle. In the case where the pendulum is pointing down, the updating function will make the system move back toward the equlibrium state (straight down). We say that this equilibrium is stable. However, if the pendulum is pointing straight up and then moved by a small angle, the updating function will make the angle move away from the equilbrium state. We say that the state of pointing straight up is an unstable equilibrium. So even if the pendulum points straight up, any of the little jiggle that is comes from living in the real world will cause the pendulum to move away from the equilibrium, swing down, and eventually ending up at the stable equilibrium state of pointing downward. Given the importance of the concept of stability, we want to be able to tell which equilibria are stable and which are unstable. We’ll do this both from the perspective of the graph of the updating function, as well as by it’s formula. So let’s start by examining the formula for the updating function h(x) = x2 which means that st+1 = h(st) = (st)2. What are the equilibrium values? Let s* be an equilibrium. Then s* = (s*)2. From this it’s clear that s*=0 is one equilibrium. If s*≠0, we can divide both sides by s* and get 1=s*. Suppose we’re near one of the two states and we ‘bump’ the state by changing it by a small amount. So we’ll assume that our initial condition is some small distance away from our equilibrium, s0 -­‐ s* = Δs0. Now we’ll apply the updating function to find s1 = h(s0), and find out how far that value is away from the equilibrium: Δs1 = s1 -­‐ s* . The main question we’re interested in is whether the updating function moves the state ‘back’ closer the equilibrium, or does it moves it further away from the equilibrium. In other words, we want to know whether Δs1 is Δs
smaller than or bigger than Δs0. The way we’ll do that is to look at the ratio 1 . If Δs0
that ratio is bigger than 1, then Δs1 is bigger than Δs0, the state at time 1 (=s1) is €
Δs1
Δs0
ratio is smaller than 1, then Δs1 is smaller than Δs0, the state at time 1 (=s1) is closer to s* than where it started, and the equilibrium is stable. further away from s* than where it started, and so the equilibrium is unstable. If Now we plug in the definitions of Δs1: €
Δs1 s1 − s * h(s0 ) − s * h(s * +Δs0 ) − s *
=
=
=
Δs0
Δs0
Δs0
Δs0
Pulling one more trick out of our sleeve, we use the fact that since s* is an equilibrium, s*=h(s*). Substituting this in we find €
Δs1 h(s * +Δs0 ) − h(s*)
=
Δs0
Δs0
The stuff on the right hand side of the equation should look familiar, since €
h(s * +Δs0 ) − h(s*)
ΔS 0 → 0
Δs0
hʹ′(s* ) = lim
so if Δs0 is small
€
€
Δs1 h(s * +Δs0 ) − h(s*)
=
≈ hʹ′(s*) Δs0
Δs0
That means that as long as the initial change in state (=Δs0) is small, we can use the derivative of the updating function, evaluated at the equilibrium state, to determine if that equilibrium is stable or unstable: If h’(s*)>1 then Δs1>Δs0 and the equilibrium is unstable If h’(s*)<1 then Δs1<Δs0 and the equilibrium is stable Going back to our example with h(x) = x2, we already found the equilibrium states s*=0 and s*=1. Using the power rule to take the derivative, we find that h’(x) = 2x. Now we evaluate the derivative at the first equilibrium s*=0 and find that h’(s*) = h’(0) = 2*0=0. Since 0<1, the equilibrium s*=0 is a stable equilibrium. At the other equilibrium, s*=1 and h’(s*) = h’(1) = 2*1=1. Since 2>1 this is an unstable equilibrium. Now let’s examine the same system by looking at the graph of the updating function. 1.6
1.4
st+1=h(st)
1.2
1
0.8
0.6
0.4
0.2
0
−0.2
0
0.5
st
1
The equilibrium values s*=0 and s*=1 can be determined by finding where the graph of the updating function (thick line) intersects the dashed diagonal. At the upper equilibrium, s*=1, the tangent line is steeper than the diagonal (slope>1), so the updating function passes from underneath the diagonal to above. If you cobweb the graph near this equilibrium you will see that trajectories move away from s*=1. This equilibrium is unstable. In contrast, at s*=0 the slope of the tangent line is <1, and cobwebbing reveals that trajectories get closer to the equilibrium. This equilibrium is stable. To wrap things up, we note that if Δs0 is a small number, then Δs1=h’(s*) Δs0 should also be pretty small. But then we can repeat the argument to get Δs2 = h'(s*)Δs1 . Repeating this over and over, we have an updating function for Δst: Δst +1 = h'(s*)Δst . Since h’ evaluated at the equilibrium is a constant value, this is the exponential growth/decay problem we solved earlier: Δst +1 = rΔst . ISince whether we have €
growth or decay depends on whether r is greater than or less than 1, we have shown €
that near an equilibrium, the distance from that equilibrium shows exponential growth or decay depending on whether h’( s*) is bigger or smaller than 1. €