MAT1193 – 3c DTDS – equilibria Lets start with an example. Consider the updating function st +1 = .3* (st − 0.5) +
1
1+ e
−6*(st −0.5)
Assume that the initial condition is s0=0.4. What will happen as the system evolves in time? We could start with just putting the values into the function and making a table: Start of time step Time=t State=st 0 0.4 1 .3243 2 .2058 3 .0579 4 -­‐.0668 5 -­‐.1378 6 -­‐.1700 7 -­‐.1834 8 -­‐.1887 9 -­‐.1908 €
End of time step State=st+1 Time=t+1 .3243 1 .2058 2 .0579 3 -­‐.0668 4 -­‐.1378 5 -­‐.1700 6 -­‐.1834 7 -­‐.1887 8 -­‐.1908 9 That does tell us that the values are decreasing over time, but what will happen as we keep going? Will the numbers keep decreasing forever? It’s hard to understand just from the function. So lets make a graph. In the graph on the left I have plotted the graph of the updating function – the function written at the top of the page that tells you how to get from the state at time t (=st) to the state at time t+1 (st+1). In the graph at the right, I have plotted the values from the table. This is just the first few points on the solution function -­‐ the function that takes time as input and spits out the state of the system at that time. The graphs here just plot the first 9 points in the solution – the values in the table we made above. Notice that the cobweb starts to ‘stack up’ at the place where the graph of the function crosses the diagonal line y=x (left). We can also see that the solution values start to ‘stack up’, with the output values not changing much as they approach a value somewhere near -­‐0.2. Let’s cobweb the same system but with a different initial condition s0=0.6. Here the graph of the function is above the diagonal at the initial condition and so the trajectory of output values increasesi as time goes on, but again the values begin to ‘stack up’ where the graph of the function crosses the diagonal. Cobwebbing for another initial condition s0=1.4, we see that the trajectory of points decreases and again stack up near the upper intersection of the graph and the diagonal line. So what do we learn from this example? One thing is by looking at the graph of the updating function and understanding cobwebbing, we can have a pretty good idea of the qualitative behavior of the system. When the updating function is above the diagonal, the output of the system increases over time and when the updating function is below the diagonal the values decrease. Under these conditions, we can predict the long term behavior of the system: generally it will accumulate at points where the updating function intersects the diagonal line y=x. What can we say about these special points? Suppose we take s0 = 0.5. Plugging this into the updating function we find s1 = .3* (0.5 − 0.5) +
1
1+ e −6*(0.5−0.5)
1
1
s1 = .3* 0 +
=
=
0.5
1+ e 0 2
€
In other words, if we start at 0.5 then we stay at 0.5 for the next time step. But of course if put in 0.5 into the updating function again, nothing changes. So if the system gets to where the state variable is equal to 0.5, the system does not change in time. We call s=0.5 an equilibrium state of the system. We use the word equilibrium to describe these points because they represent a state where any tendency for the state to increase in value is just balanced by any tendency for the state to decrease in value. Since these two tendencies are in equilibrium, the value of the state doesn’t change. In the book, these points are called fixed points, a term that is perhaps more descriptive. So how do we find the states where the system is in equilibrium? From the graph, it’s easy to look for the points where st = st+1 . Any point on the graph of the updating function has coordinates (st,st+1) since if st is the input to the updating function, st+1 is the output. But if st = st+1 then the point (st,st+1) also satisfies the equation y=x. That means that the equilibrium states of the system correspond to the intersection points of the updating function and the diagonal line y=x. In this example, there are three equilibria. How can we find the equilibrium states if we are only given the equation of the updating function? To do that we need to figure out how to translate the condition “an equilibrium state is a state that does not change in time” into a mathematical equation. The first step is to give the equilibrium state a name, then we can solve for it. So let’s let s* be the name of our equilibrium state and suppose that at time t the system is in that state. That translates as st = s*. Because s* is an equilibrium, if we apply the updating function the state of the system will still be s* (the state doesn’t change in time). So st+1 =s*. But now we know that both st = s* and st+1 =s* so we can substitute s* into both sides of the equation for the updating function and solve for s*. In the example we used above, the updating function is quite complicated and the equilibrium states are hard to solve for. Lets try an easier example: st +1 = 1.5 * st * (1 − st ) Now suppose that s* is an equilibrium state for this DTDS. Then we substitute in s* for both st+1 and st and get €
s* = 1.5s* (1 − s* ) Now we solve for s*. €
0 = −1.5(s* ) 2 +1.5s* − s*
0 = −1.5(s* ) 2 + 0.5s*
(
)
0 = s* −1.5s* − 0.5
€
In the last step we ‘factored out’ s* as a common factor of the terms -­‐1.5(s*)2 and 0.5s*. This is a common trick that will be used in this course. Now we have two numbers, s* and -­‐1.5s*-­‐0.5 that when multiplied together give us zero. The only way this can happen is if s* = 0 or -­‐1.5s*-­‐0.5 =0. If the latter statement is true, a little more algebra and we find s* = 0.5/1.5 = 0.33. So finally we find that there are two equilibrium states for this system, s* = 0 and s* = 0.33. We can check this in two ways. First we can plug these numbers into the updating function . If st=0, the st+1 = 1.5*0*(1-­‐0)=0, so the system doesn’t change. If st=0.33, then st+1 = 1.5*.33*(1-­‐0.33)=1.5*0.33*0.67=0.33, and the system doesn’t change. The other way to confirm these calculations is to look at the graph Indeed the graph of the updating function (blue) intersects the diagonal line at the points (0,0) and (.33,.33). That means that the equilibrium states of the system are 0 and 0.33. Note that some students make the mistake of answering the question “what are the equilibrium states of this system?” with (0,0) and (.33,.33). This answer is wrong because each equilibrium state is just a single state value, not 2 coordinates on a graph. Here we are using the graph to find the equilibrium states.