Math Insight

The stability of equilibria for discrete dynamical systems

Math 201, Spring 2017
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Total points: 3
  1. As a warm up for stability of nonlinear discrete dynamical systems, recall how to determine stability for the linear dynamical systems, such as a system with slope $R$: \begin{align*} x_{n+1} &= Rx_n\\ x_0 &= a. \end{align*}
    1. If $R \ne 1$, what is the one equilibrium? $E=$
    2. What is the solution of the dynamical system? $x_n=$
      Need help?
    3. For what values of the slope $R$ does this solution get smaller (i.e., closer to zero) with time? I.e., for what values of $R$ does multiplying by $R$ lead to smaller numbers? Be sure to consider both positive and negative $R$.

      The solution will get smaller with time if
      $\lt R \lt$

      For what values of the slope $R$ does this solution get larger (i.e., further from zero) with time? I.e., for what values of $R$ does multiplying by $R$ lead to larger numbers? Be sure to consider both positive and negative $R$.

      The solution will get larger with time if $R \gt$
      or if $R \lt$

    4. For what values of the slope $R$ is the equilibrium stable?
      $\lt R \lt$

      For what values of the slope $R$ is the equilibrium unstable? $R \gt$
      or $R \lt$

      Need help?
    5. We can rewrite the condition even more simply using absolute values. The equilibrium is stable for $|R| \lt$
      . The equilibrium is unstable for $|R| \gt $
      .

  2. Consider the dynamical system $x_{n+1} = f(x_n)$, where $f(x)=x^2$.
    1. What are the two equilibria of the dynamical system? $E=$
      (Enter in increasing order, separated by commas.)
    2. What are the slopes of the tangent lines to $f(x)$ at the two equilibria?
      For the first equilibrium, $f'($
      $)=$

      For the second equilibrium, $f'($
      $)=$
    3. The slopes of the tangent lines, $f'(E)$, play the role of the slope $R$, above. Translate the above conclusions about stability in terms of the slope of the tangent line.
      An equilibrium $E$ is stable if
      $\lt f'(E) \lt$
      , which we can rewrite as $|f'(E)| \lt$

      An equilibrium $E$ is unstable if $f'(E) \gt$
      or if $ f'(E) \lt$
      , which we can rewrite as $|f'(E)| \gt$
      .
      (We've just stated the stability theorem for discrete dynamical systems.)

      Based on the tangent line slopes, determine the stability of the equilibria.
      The first equilibrium, $E=$
      is
      .
      The second equilibrium, $E=$
      is

    4. On the following graph of $f(x)=x^2$ and the diagonal, sketch the tangent lines at the equilibria. Cobweb to verify your conclusions regarding equilibrium stability. (To verify stability, cobweb with initial conditions near the equilibria.)
      Feedback from applet
      Points on diagonal:
      Points on function:
      Need help?

  3. The dynamical system \begin{align*} x_{t+1}&=f(x_t)\\ x_0 &= x_0 \end{align*} with $f(x)=-\frac{1}{8}x^3+\frac{3}{4}x^2+\frac{3}{8}x$ has three equilibria: $E=0$, and $E=1$ and $E=5$.
    1. Determine the stability of the equilibria analytically. For stable equilibria, indicate whether the solution spirals toward the equilibrium or moves steadily toward the equilibrium. For unstable equilibrium, indicate whether the solution spirals away from the equilibrium or moves steadily away from the equilibrium.

      For the equilibrium $E=0$, since $f'(0)=$
      and $|f'(0)|$
      $1$, the equilibrium $E=0$ is
      . Moreover, since $f'(0)$ is
      , the solution
      the equilibrium.

      For the equilibrium $E=1$, since $f'(1)=$
      and $|f'(1)|$
      $1$, the equilibrium $E=1$ is
      . Moreover, since $f'(1)$ is
      , the solution
      the equilibrium.

      For the equilibrium $E=5$, since $f'(5)=$
      and $|f'(5)|$
      $1$, the equilibrium $E=5$ is
      . Moreover, since $f'(5)$ is
      , the solution
      the equilibrium.

    2. Demonstrate the stability of the equilibria by cobwebbing. The right panel is a zoomed in view to use for the lower equilibria.

  4. For a dynamical system in difference form, such as \begin{align*} x_{n+1} - x_n &= \frac{1}{2} \left(x_{n} - 2\right) \left(x_{n} - 1\right)\\ \end{align*} one needs to convert to function iteration form before determining stability.
    1. First, find the equilibria. $E=$

      (Enter in increasing order, separated by commas.)
    2. Next, add $x_n$ to both sides to convert to function iteration form.
      $x_{n+1}=$
      Need help?
    3. Then, calculate the derivative of the right hand side to determine stability of the equilibria.

      The derivatives at the equilibria are:

      Enter in the derivatives in the same order as the equilibria, above, separated by commas.

      The stability of the equilibria are:

      Enter stable or unstable for each equilibrium, separated by commas, in the same order as above. For example, if the first equilibrium is stable and the second equilibrium is unstable, enter stable, unstable.

      Need help?

  5. A population of termites would be increasing at the rate of 10% per year, except that due to a scarcity of rotten wood, there is only enough food to support 10000 termites. Therefore, if $m_t$ is the number of termites in year $t$, the population evolves according to the dynamical system \begin{align*} m_{t+1} - m_{t} = 0.1 m_t\left(1 - \frac{ m_t}{ 10000 }\right). \end{align*}

    Calculate the equilibria of the dynamical system and their stability. If the initial population is $m_0=100$ termites, what will happen to the population after a long time? If the initial population is $m_0=15000$, what will happen to the population after a long time?

    Equilibria: $E=$

    Stability of equilibria:

    In both cases, separate answers by commas. Enter the equilibria in increasing order and enter the stability results in the same order. Enter stable for a stable equilibrium and unstable for an unstable equilibrium. For example, if the first equilibrium is stable and the second equilibrium is unstable, enter stable, unstable.

    For $m_0=100$, the population
    toward
    .
    For $m_0=15000$, the population
    toward
    .

    Need help?

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Math 201, Spring 2017