Math Insight

Exponential growth and decay

Math 201, Spring 19
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Total points: 3
  1. Consider the dynamical system, with constant parameters $a$ and $c$: \begin{align*} z'(t) &= a z\\ z(0) &= c. \end{align*}
    1. Solve the dynamical system analytically (i.e., obtain a formula for $z(t)$).

      $z(t)=$

    2. For which values of $a$ does $z(t)$ exhibit exponential growth?

      For which values of $a$ does $z(t)$ exhibit exponential decay?
    3. An equilibrium is a constant solution, $z(t)=z_e$ for some number $z_e$. We find it by setting the change $z'(t)=0$ and substituting $z=z_e$. For $a \ne 0$, what is the equilibrium?
      $z_e=$

      For which values of $a$ is this equilibrium stable?

      For which values of $a$ is this equilibrium unstable?

    4. Solve the dynamical system graphically for both the case of exponential growth as follows.

      First, pick a value of $a$ that will lead to exponential growth. For that value of $a$, plot the right hand side of the dynamical system as a function of $z$. (I.e., plot the function $f(z)=az$, where you plot $z$ on the horizontal axis and $f$ on the vertical axis.)

      Feedback from applet
      sign of slope:
      vertical intercept:

      Using this graph to guide you, sketch the solution of $z$ versus $t$ on the below graph. Sketch solutions for initial conditions $c=1$, $c=0$, and $c=-1$.

      Feedback from applet
      Final values of curves:
      Initial conditions of curves:
      Number of curves:
      Speed profiles of curves:

      Even though for some initial conditions, $z(t)$ is decreasing, we still call it exponential growth, as it is shooting off to large, negative values.

    5. Next, pick a value of $a$ that will lead to exponential decay and plot the right hand side of the dynamical system as a function of $z$.
      Feedback from applet
      sign of slope:
      vertical intercept:

      Using this graph to guide you, sketch the solution of $z$ versus $t$ on the below graph. Sketch solutions for initial conditions $c=4$, $c=0$, and $c=-4$.

      Feedback from applet
      Final values of curves:
      Initial conditions of curves:
      Number of curves:
      Speed profiles of curves:

      Even though for some initial conditions, $z(t)$ is increasing, we still call it exponential decay, as the solution is decaying toward zero.

  2. Consider the dynamical system, with parameters $a$, $b$, and $c$: \begin{align*} \diff{ y }{t} &= a y +b\\ y(0) &= c. \end{align*}
    1. As long as $a \ne 0$, what is the equilibrium of the dynamical system?
      $y_e=$

      In this case, the only thing a nonzero value of $b$ does is shift the equilibrium away from zero.

    2. Solve the dynamical system graphically, for the following values of $a$ and $b$. (As above, first sketch the right hand side $f(y)=ay+b$, although we didn't include a graph for those sketches.)
      1. $a=1$, $b=1$. Use initial conditions $c=0$, $c=-1$, and $c=-2$.

        Feedback from applet
        Final values of curves:
        Initial conditions of curves:
        Number of curves:
        Speed profiles of curves:
      2. $a=1/2$, $b=-3/2$. Use initial conditions $c=4$, $c=3$, and $c=2$.

        Feedback from applet
        Final values of curves:
        Initial conditions of curves:
        Number of curves:
        Speed profiles of curves:
      3. $a=-2$, $b=-4$. Use initial conditions $c=3$, $c=-2$, and $c=-4$.

        Feedback from applet
        Final values of curves:
        Initial conditions of curves:
        Number of curves:
        Speed profiles of curves:
      4. $a=-1/4$, $b=1/2$. Use initial conditions $c=4$, $c=2$, and $c=-2$.

        Feedback from applet
        Final values of curves:
        Initial conditions of curves:
        Number of curves:
        Speed profiles of curves:
    3. Based on your experimentation with the above values of $a$ and $b$, determine how the stability of the equilibrium depends on $a$ and $b$. (If needed, try graphically solving the system for different values of $a$ and $b$ to further explore the dependence of stability on $a$ and $b$.)

      What is the condition on $a$ for the equilibrium to be stable?

      What is the condition on $a$ for the equilibrium to be unstable?

      Does the stability of the equilibrium depend on the value of $b$?

      Does the location of the equilibrium depend on the value of $b$?