Skip to content

MathBio: Gearing up for Test 2

If you a looking for problems to work on to prepare for Midterm 2, this post will be the place to look.  This will be updated several times over the next few days.

Challenge problems for discrete-time systems.

Solutions to the important parts of the Horseshoe Crab project are written up here:

Horseshow crabs 1a   Horseshoe crabs 1b

For more practice, I recommend 3.3 #5 from the book.

Challenge problems for continuous-time systems.

The first and most important place to find problems for continuous time systems is in my supplemental lecture notes:  Systems of ODEs (pdf).  See in particular problems 4, 5, 6 and 7.  Solutions are included already for 4-6.  I will post a solution to 7 on Monday.

Challenge problem on epidemics.

\dot S = - \alpha SI + \gamma R

\dot I = \alpha SI - \beta I

\dot R = \beta I - \gamma R

Set \beta = 2, \gamma = 1 and impose the constraint that S + I + R = 1.  Rewrite the system as a two dimensional system (in S and I only) and determine a condition for \alpha such that there exists an endemic equilibrium (meaning an equilibrium (S_*,I_*) where S_* > 0 and I_* > 0.  Classify the type of the endemic equilibrium.

SOLUTION

There is an endemic equilibrium if \alpha > 2.  CHECK me on this next part!!!  I’m getting that the endemic equilibrium is a stable spiral if \alpha \in (2.0294, 35.9706)

Challenge Problem for Nullcline Analysis

Consider the following system of ODEs which are from the class of Niche Competition models introduced in class.  (Both species exhibit logistic growth models when alone and have negative mutual interactions.) In this case, we have one unknown parameter a.

\dot x = x ( 2 -x - y)

\dot y = y (a - 2x - y)

  • Sketch the x and y nullclines in three cases:  a = 1, 3, 5.  In each case, note whether there is an equilibrium in the first quadrant (a co-existence equilibrium).
  • For what values of a does this system have a coexistence equilibrium?
  • When the coexistence equilibrium exists, what is its type?

SOLUTION

The x- and y- nullclines for this problem are pictured below.

Nullclines of the Challenge problem

Nullclines of the Challenge problem

There is only a coexistence equilibrium if a \in (2,4). This equilibrium has coordinates (x_*,y_*) = (a-2, 4-a).  After computing the Jacobian and evaluating at the fixed point, we see that the trace of the Jacobian is -2 for all values of a, while the determinant is -(a-2)(4-a) which is always negative.  We conclude that, when it exists, the coexistence equilibrium is always a saddle node.

 

7 Comments Post a comment
  1. William Vebert #

    I wasn’t sure where to post this, but wanted it to be noted. Problem 3.3 #3 has a typo (either in the solutions or the statement). The second equation should have a y withy the -s coefficient rather than an x.

    March 30, 2014
  2. William Vebert #

    I think there may be a typo at the top of page 10 in your supplemental notes, Dr. McKinley. In the case of complex roots, should the general solution have the t next to the second vector? Maybe I wrote it down wrong when we discussed it in class.

    April 1, 2014
    • Scott Alister McKinley #

      Will, you are right. There is a typo there. There should be no t where you are seeing it.

      April 1, 2014
  3. Franco #

    Will there be a post with the solutions to these questions so we may compare against what we did?

    April 1, 2014
    • Scott Alister McKinley #

      Solutions to the last two challenge problems are now posted along with the solutions to the parts of the Horseshoe Crab problem that I graded but did not cover in class.

      April 1, 2014
  4. William Vebert #

    I got slightly different bounds for alpha in the epidemic problem (2.303, 31.697). Here is a link to the work that I did: http://tinypic.com/r/s4oy77/8 (I don’t feel like fiddling with the LaTeX right now/sorry it’s grainy). I did not actually draw the nullclines on this page, but did on another to check where the intersections were.

    It’s quite possible I messed up somewhere in the Jacobian. Did anybody else get something different?

    April 1, 2014
  5. Marcus Reis #

    No, I got the same answer William, I think the posted solution is just a tad off, checked your work too and couldn’t find any mistakes.

    April 2, 2014

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s