Math Insight

A graphical approach to finding equilibria of discrete dynamical systems

Math 201, Spring 2017
Name:
ID #:
Due date:
Table/group #:
Group members:
Total points: 3
  1. Consider the dynamical system written in function iteration form \begin{align*} q_{t+1} &= f(q_t)\\ q_0 &= q_0, \end{align*} where $f(x)=x^2-x$. Let's begin by reviewing the the analytic approach to finding its equilibria. (Analytic approach means manipulating equations.) Then, we'll introduce a graphical approach to finding the equilibria.

    1. Analytic equilibria. Equilibria are values of $q_t$ where $f(q_t)=q_t$, i.e., find values of $q_t$ that when plugged into the function $f(x)=x^2-x$, we get $q_t$ back out again. The values $q_t$ must satisfy the equation $$q_t^2 - q_t = q_t.$$ To stress that we are talking about equilibria, let's substitute $E$ for $q_t$, so the equation becomes


      One thing we cannot do is divide both sides of the equation by $E$. Why can't we do that? We have to remember that we cannot divide both sides of the equation by
      and it is possible that $E$ might be equal to
      .

      Instead, we need to set the equation equal to zero by subtracting
      from both sides of the equation. The resulting equation is


      $=0$.
      Factor out the $E$ from the left hand size, obtaining:

      $=0$
      Since we have the product of two factors equal to zero, one of them must be zero, so we can conclude that

      $=0$ or
      $=0$.
      Therefore, the equilibria are
      $E=$
      (separate answers by commas).
    2. Graphical equilibria. A second way to find equilibria is via a plot of $q_{t+1}$ versus $q_t$ (i.e., have $q_t$ on the $x$-axis and $q_{t+1}$ on the $y$-axis). On the below axes, graph the function $q_{t+1}=f(q_t)$. (Recall that the function is $f(x)=x^2-x$.) Using a different color (or different line thickness), also plot the diagonal line $q_{t+1}=q_t$. (Online, drag the blue points so the thick blue curve is the graph of $q_{t+1}=f(q_t)$ and drag the green points so the thin green line is the graph of $q_{t+1}=q_t$.)

      Feedback from applet
      function:
      intersections:
      line:

      Find the points where the diagonal and the graph of $f$ intersect and draw them on the graph. (Online, change the $n_i$ slider to indicate the number of intersections and drag the points that appear to the proper locations.) What are the coordinates of the points? If we label the points $A$ and $B$, the points are:

      $A =$
      and $B=$

      At those points $(q_{t},q_{t+1})$, both $q_{t+1}=f(q_t)$ and $q_{t+1}=q_t$. This means that $q_{t}=f(q_t)$ (i.e., $f$ didn't change the value $q_t$). The value of $q_t$ at those points are the equilibria $E$ for our dynamical system. The equilibria are

      $E=$
      .

  2. For the dynamical system \begin{align*} x_{n+1} &= f(x_n)\\ x_0 &= a, \end{align*} where $f(x) = x^3-8x$, the function $f$ along with the diagonal are plotted on a graph of $x_{n+1}$ versus $x_n$.
    Feedback from applet
    Equilibria:
    Number of equilibria:
    Point:

    1. Find the equilibria using the above graph. Label the points you used to determine the equilibria and give the resulting value of the equilibria. (Online, move the $n_e$ slider to indicate how many equilibria you found and drag the resulting points to the locations of the equilibria.)

      Remember the values of the equilibria are the first components of the points you found.
      The equilibria are: $E=$

      (Separate multiple values by commas.)

      Need help?

    2. To find the equilibria $x_n=x_{n+1}=E$ analytically, you must solve the equation (in terms of $E$)


      $=0$.
      You should be able to factor the left hand side of the equation into three factors. (Remember you can factor $a^2-b^2 = (a-b)(a+b)$.) Therefore, you get three equations for $E$. The equilibria are:
      $E=$

      (Separate multiple values by commas.)

    3. Starting with the initial condition $x_0 = -3$, calculate $x_1=$
      , $x_2=$
      , and $x_3 =$
      .
    4. If we start with the initial condition $x_0 = 1$, use the graph to approximate $x_1=f(x_0)$. Label the point on the graph that you used to estimate this value of $x_1$. (Online, check the “show point” box and move the resulting blue point to the location to estimate $x_1$.) $x_1=$

      Check the accuracy of your estimate by calculating $f(x_0)$ from the analytic expression for $f$.

  3. For the dynamical system \begin{align*} y_{t+1} &=g(y_t)\\ y_0 &= y_0, \end{align*} the function $g$ along with the diagonal are plotted on a graph of $y_{t+1}$ versus $y_t$.
    Feedback from applet
    Equilibria:
    Number of equilibria:
    Point:
    1. Find the equilibria using the above graph. Label the points you used to determine the equilibria and give the resulting values of the equilibria. (Online, move the $n_e$ slider to indicate how many equilibria you found and drag the resulting points to the locations of the equilibria.)

      The equilibria are: $E=$

    2. Starting with $y_0 = 1$, calculate $y_1 =$
      , $y_2 =$
      , and $y_3 =$
    3. If we start with $y_0 = -3$, use the graph to approximate $y_1=g(y_0)$. Label the point on the graph that you used to estimate this value of $y_1$. (Online, check the “show point” box and move the resulting blue point to the location to estimate $y_1$.)
      $y_1=$

Thread navigation

Math 201, Spring 2017