MathBio: Homework 4, Due Wed Feb 26

One problem, due Wed Feb 26


(Logan and Wolesenksy, 3.3 #2)

In 1998, van der Meijden described the following model connecting the cinnabar moth and ragweed.  The life cycle of the moth is as follows.  It lives for one year, lays its eggs on the plant, and dies. The eggs hatch the next spring.  The number of eggs is proportional to the plant biomass the preceding year, or

E_{t+1} = aB_t.

The plant biomass the next year depends on the biomass the current year and the number of eggs according to

B_{t+1} = k e^{-cE_t / B_t}.

  • Explain why the last equation is a reasonable model (plant biomass vs eggs and biomass vs biomass).
  • What are the dimensions of k, c and a?
  • Rewrite the model equations in terms of the “rescaled” variables X_t = B_t / k and Y_t = E_t /ka.  Define b=ac.  What are the dimensions of X_t, \, Y_t and b?
  • Find the equilibrium and determine a condition on b for which the equilibrium is stable.
  • If the system is perturbed from equilibrium, describe how it returns to equilibrium in the stable case. (This means:  is the dominant eigenvalue real or complex?  If it is complex, this means there are decaying oscillations toward the equilibrium.)


One further note about the suggested problem, 3.3 #1.  It appears that there is a typo either in the statement of the problem or in the given solution.  If the statement of the problem is correct, then the projection matrix seen at the top of page 376 should be

\left( \begin{array}{cc} 2 & -2 \\ -1 & 1 \end{array}\right).

Otherwise, the model should be

\Delta P_t = - P_t - 2Q_t, \quad \Delta Q_t = - P_t.

As a hint, keep in mind that this is equivalent to our usual form

P_{t+1} = - 2 Q_t
Q_{t+1} = -P_t + Q_t.

For practice, it’s worth doing the problem both ways and comparing the outcomes.


UPDATE:  Here are the solutions to 3.3 #2 (b) – (e):  Solutions, HW4 (pdf)

3 thoughts on “MathBio: Homework 4, Due Wed Feb 26

  1. According to the back of the book, the nullclines are Q = P/2 where Δ P = 0, and P = 0 where Δ Q = 0. These nullclines only work using the original linear equations.

  2. Thank you for pointing that out, David. I have adjusted the wording of this post to accommodate either interpretation of the problem.

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