DOC PREVIEW
HARVARD MATH 19 - Lecture 28

This preview shows page 1 out of 4 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Math 19. Lecture 28Periodic SolutionsT. JudsonFall 20051 An Improved Predator-Prey ModelIf p(t) be the number of prey and q(t) is the number of predators at time t,then we can mo del a predator-prey system asdpdt=23p1 −p4−pq1 + p(1)dqdt= sq1 −qp, (2)where s > 0.• If there are no predators, our system (1) is just logistic growth:dpdt=23p1 −p4.We have a stable equilibrium at p = 4.• The existence of predators decreases dp/dt by pq/(1 + q).– When p is small,pq1 + p≈ pq.This tells us that the dp/dt is dependent on predator-prey inter-action.1– When p is large,pq1 + p≈ q.In other words, food is abundant and the death rate is only de-pendent on the number of predators.• In (2), we have the standard logistic equation if p is constant:dqdt= sq1 −qp.A lion can only eat s o much! Thus, this equation models the fact t hatthe carrying capacity for the predator is proportional to the number ofprey.2 The Phase Plane• The p null clines are p = 0 and q = (2/3)(1 − p/4)(1 + p).• The q null clines are q = 0 and q = p.• The equilibrium points for p > 0 are(p, q) = (1, 1)(p, q) = (4, 0).3 Stability• At p = 1, q = 1,A =1/12 −1/2s −s.In this case,tr(A) =112− sdet(A) =512s.This point is a stable equilibrium point if s > 1/12 and unstable ifs < 1/12.2• At p = 4, q = 0,A =−2/3 −4/50 s.In this case, det A = −2 s/3 < 0. Therefore, this point is not stable.4 A Repelling Equilibrium Poi ntAn equilibrium point is a repelling equilibrium point if whenever a non-equilibrium solution is close to the equilibrium solution at t, it moves furtheraway as t increases. The equilibrium point p = 1 and q = 1 is repelling ifs < 1/12. If p(t) and q(t) are both near 1, thenpqis almost a solution toddtp − 1q − 1=1/12 −1/2s −sp − 1q − 1=(p − 1 )/12 − (q − 1)/12s(p − 1) − s(q − 1).These solutions grow exponentially with time. An equilibrium point is re-pelling if tr(A) > 0 and det(A) > 0.5 Basin of Attractio nA basin o f attraction or a trapping region is a region V in the (p, q)-planewhere no solutionp(t)q(t)of our predator-system that enters V ever leaves V . We claim that the squareregionV = {(p, q) : 0 < p < 4, 0 < q < 4}is a basin of attraction.36 Poincar´e-Bendixs on The oremConsider the systemdpdt= f(p, q)dqdt= g(p, q),and supp ose that a region V is a basin of at t raction in the (p, q)-plane. If Vcontains a single equilibrium point that is repelling, then t he system has aperiodic solution that is inside V for all t.7 Periodic SolutionsBy the Poincar´e-Bendixson Theorem, our predator-prey system has a peri-odic solution if s < 1/12.8 StabilityOur periodic solution is stable in the following sense. Starting inside the pe-riodic solution, a trajectory will spiral out towards the stable orbit. Startingoutside the periodic solution, a trajectory will spiral in towards the stableorbit.Readings and References• C. Taubes. Modeling Differe ntial Equations in Biology. Prentice Hall,Upper Saddle River, NJ, 2001. Chapter 23.• “Snowshoe Hare Populations: Squeezed from Below and Above,” pp.382–385• “Impact of Food and Predation on the Snowshoe Hare Cycle,”


View Full Document

HARVARD MATH 19 - Lecture 28

Download Lecture 28
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture 28 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture 28 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?