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We want to develop a method to find equilibria of discrete dynamical systems graphically.
But first, let's think about the more general problem of how to solve an equation graphically.
As a warm up, let's think of this word problem:
I'm thinking of a number.
It is the same as that number squared.
What are the numbers that I could possibly be thinking of?
It's not too difficult to just come up with the answers.
The number 1, if you square it, gives you 1 back.
So, I could have been thinking of the number 1.
Similarly, the number zero is the same as zero squared.
I must have been thinking of either 0 or 1.
But, let's imagine that you couldn't figure out the answer to this problem right away.
How do we set up a math problem to calculate the answer?
One method is to solve this problem analytically,
meaning we could write an equation with variables and then solve that equation.
Let f be the squaring function, i.e., define f of x equals x squared.
The number x could be the answer to the word problem if x=f(x), i.e., if x equals x squared.
Do you know how to solve this equation to find all possible values of x?
We could subtract x from both sides so that one side is zero.
We get the equation zero equals x squared minus x.
Next, we can factor by pulling out the common factor x.
Our condition becomes 0 = x(x-1).
We have a product of two factors equal to zero.
One of the factors must be zero. Either x=0 or x-1=0.
Our possible answers are x=0 or x=1.
Of course, we knew that already.
Another way to solve this problem is to use a graphical approach.
We could make a graph of y versus x.
Let's graph the function y=f(x) in green.
Since f of x equals x squared, we get a parabola.
Here's a neat trick.
We are looking for the condition where both the input and output of f are equal.
So, let's plot another line on the graph, the line where x and y are equal.
We can call this line "the diagonal," as it is a diagonal line through the origin.
The graph of y=f(x), or y equals x squared is the set of points (x,y) where y=f(x).
This means that if the point (x,y) is on the graph of f, then its y-coordinate is the square
of its x-coordinate.
The graph of the diagonal y=x is the set of points (x,y) where y=x.
This means that if the point (x,y) is on the graph of the diagonal,
then its y-coordinate is equal to its x-coordinate.
Notice there are two points where the two graphs intersect.
At these points, the input of f and the output of f are the same because the graph of f lies
along the diagonal.
Or, to think of it another way, for these points, we know that both y equals x suared
and y equals x.
Substituting y equals x into y equals x squared gives us x equals x squared.
That looks familar. That's the equation we solved earlier.
The points that line on both the graph of f and on the diagonal give us the solution
to our problem.
These points are (1,1) and (0,0).
Their x-coordinates are x=1 and x=0.
These numbers x are the same as their squares x squared.
The nice part about this graphical approach is that, once I have the graph of f, I don't
need its formula anymore.
If I just had the plot of f, along with a plot of the diagonal,
I can read right of the graph those values of x for which the input and output of f are
the same.
I just find the points of intersection and determine their x-coordinates.
How can we apply this idea to finding equilibria of dynamical systems?
Let's imagine we had a dynamical system that squared the value of the state variable at
each time step.
Let's write this dynamical system as H sub t+1 equals H sub t squared
where H sub t is our state variable.
Here, we won't give the initial condition another name, but just leave it as H naught.
We can rewrite the dynamical system in terms of our function f of x equal x squard
as H sub t+1 equals f of H sub t.
If we happened to start at a number H naught that was the same as its square,
then this evolution rule wouldn't change the value of H naught.
We would find that H sub t would be equal to H naught for all times t.
In other words, we would be at an equilibrium.
To find the equilibria analytically (i.e., with equations),
we plug in H sub t equals E and H sub t+1 equals E into the evolution rule
to get the condition E equals f of E, or E equals E squared.
We subtract E from both sides and factor to get the same solution we did before.
The equilibria are E=0 and E=1.
We can repeat our graphical solution method to develop a graphical method
to find the equilibria of H sub t+1 equals f of H sub t.
Let's make the same plot.
This time, though, rather than labeling the axes x and y,
let's label them with the state variable.
If we label the horizontal axis with the state variable at time t, i.e., with H sub t,
then we can label the vertical axis with the state variable at the next time step, i.e.,
with H sub t+1.
Now we have a plot of H sub t+1 versus H sub t.
The evolution rule of our dynamical system is that H sub t+1 equals f of H sub t.
We can plot this relationship by plotting the graph of f on these axes.
Since f of H equals H squared, the graph is the same parabola we plotted before.
The graph of f represents the evolution rule of our dynamical system.
If we know the value of H at any time step, say time step t,
we can use the graph of f to look up the value of H at the next time step, i.e., H sub t+1.
We just start with the value of H sub t on the horizontal axis and move up or down to
the graph of f.
The height of the resulting point gives us the value of H sub t+1.
For example, if H sub t=-2, then we move up to the graph,
and find that the graph of f is at the height of 4.
Therefore H sub t+1 equals 4. That makes sense.
The rule is to square the state variable at each time step, and the square of -2 is 4.
How do we use this graph to find equilibria?
We use the same trick as before. We plot the diagonal.
This time the formula for the diagonal is H sub t+1 equals H sub t,
but it looks just like the previous plot.
The diagonal shows us what the evolution rule would look like if nothing ever changed.
If we started at any value of H sub t on the horizontal axis
and went up or down to the diagonal,
we would always end up at a point whose vertical component
is exactly the same as the horizontal component we started with.
The rule implied by the formula H sub t+1 equals H sub t for the diagonal is a very
boring rule.
Our dynamical system, though, doesn't follow the rule from the diagonal.
It follows the rule represented by the graph of f.
However, at the points where the graph of f intersects the diagonal,
our dynamical system gives the same result as the diagonal.
These are the points where f does not change its inputs.
These points represent the equilibria.
If we label an equilibrium by E,
then the coordinates of the intersection points are (E,E).
To find the equilibria from the intersection points, we just read off one of the components.
In our case, we have two intersection points: (0,0) and (1,1).
From these two points, we see that the equilibria are E=0 and E=1.
Given the graph of f, we don't even need its formula to determine
that these are the equilibria.
Let's say we started with the initial condition H naught=1.
To find the value of H sub 1, we could look up the value of the function above 1.
We see that f takes on the value 1 so that H sub 1,
which is f of H naught or f of 1 is equal to 1.
If we repeated this process to find H sub 2 equals f of H sub 1,
we would again find that H sub 2 equals 1.
The dynamical system is not changing the value of H;
it would stay at H sub t equals 1 forever.
The solution to the initial condition H naught equals 1 is H sub t equals 1 for all t.
We have found a constant solution, or an equilibrium.
The same reasoning would apply if we started with H naught equals =0.
We would have the solution H sub t equals 0, another equilibrium.
Given the graph of f along with the graph of the diagonal,
it is a simple process to read off the equilibria of the discrete dynamical system.