# Math Insight

### Video: Introduction to finding equilibria of discrete dynamical systems graphically

This video is found in the pages

List of all videos

#### Transcript of video

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 \begin{align*} f(x) = x^2. \end{align*} The number $x$ could be the answer to the word problem if $x=f(x)$, i.e., if $x=x^2$. 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 $0 = x^2-x$. 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 $f(x)=x^2$, 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 $y=x^2$ 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 $y=x^2$ and $y=x$. Substituting $y=x$ into $y=x^2$ gives us $x=x^2$. 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 $x^2$.

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 \begin{align*} H_{t+1} &= H_t^2\\ \end{align*} where $H_t$ is our state variable. Here, we won't give the initial condition another name, but just leave it as $H_0$. We can rewrite the dynamical system in terms of our function $f(x)=x^2$ as \begin{align*} H_{t+1} &= f(H_t). \end{align*}

If we happened to start at a number $H_0$ that was the same as its square, then this evolution rule wouldn't change the value of $H_0$. We would find that $H_t$ would be equal to $H_0$ 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 $H_t=E$ and $H_{t+1}=E$ into the evolution rule to get the condition $E=f(E)$, or $E=E^2$. 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 \begin{align*} H_{t+1} &= f(H_t). \end{align*} 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 $H_t$, then we can label the vertical axis with the state variable at the next time step, i.e., with $H_{t+1}$. Now we have a plot of $H_{t+1}$ versus $H_t$.

The evolution rule of our dynamical system is that $H_{t+1}=f(H_t)$. We can plot this relationship by plotting the graph of $f$ on these axes. Since $f(H)=H^2$, 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., $H_{t+1}$. We just start with the value of $H_t$ 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 $H_{t+1}$.

For example, if $H_t=-2$, then we move up to the graph, and find that the graph of $f$ is at the height of $4$. Therefore $H_{t+1}=4$. 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 $H_{t+1}=H_t$, 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 $H_t$ 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 $H_{t+1}=H_t$ 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 $H_0=1$. To find the value of $H_1$, we could look up the value of the function above 1. We see that $f$ takes on the value 1 so that $H_1$, which is $f(H_0)$ or $f(1)$ is equal to $1$. If we repeated this process to find $H_2=f(H_1)$, we would again find that $H_2=1$. The dynamical system is not changing the value of $H$; it would stay at $H_t=1$ forever. The solution to the initial condition $H_0=1$ is $H_t=1$ for all $t$. We have found a constant solution, or an equilibrium.

The same reasoning would apply if we started with $H_0=0$. We would have the solution $H_t=0$, 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.