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REU 2004

REU 2004. Population Models Day 1 Competing Species. REU’04—Day 1. Often know how populations change over time (e.g. birth rates, predation, etc.), as opposed to knowing a ‘population function’ Differential Equations! Knowing how population evolves over time

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REU 2004

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  1. REU 2004 Population Models Day 1 Competing Species

  2. REU’04—Day 1 • Often know how populations change over time (e.g. birth rates, predation, etc.), as opposed to knowing a ‘population function’ Differential Equations! • Knowing how population evolves over time w/ initial population  population function

  3. Example – Hypothetical rabbit colony lives in a field, no predators. Let x(t) be population at time t; Want to write equation for dx/dt Q: What is the biggest factor that affects dx/dt? A: x(t) itself! more bunnies  more baby bunnies

  4. 1st Model—exponential, MalthusianSolution: x(t)=x(0)exp(at)

  5. Critique • Unbounded growth • Non integer number of rabbits • Unbounded growth even w/ 1 rabbit! Let’s fix the unbounded growth issue dx/dt = ????

  6. Logistic Model • dx/dt = ax(1-x/K) K-carrying capacity we can change variables (time) to get dx/dT = x(1-x/K) • Can actually solve this DE

  7. Solutions: • Critique: • Still non-integer rabbits • Still get rabbits with x(0)=.02

  8. Suppose 2 species x(t) rabbits and y(t) deer compete for the same food source. dx/dt = dy/dt = Ax(1-x/K) -Cxy By(1-y/W) -Dxy Or…. (after changes of coordinates…) dx/dt = x(1-x-ay) dy/dt = y(b-by-cx)

  9. Analysis of one case dx/dt = x(1-x-2y) dy/dt = y(2-2y-5x) Equlibria: (0,0) , (0,1), (1,0), (1/4,3/8) Jacobian:

  10. Jacobians • J(0,0) = • J(1,0) = • J(0,1) = • J(1/4,3/8)= Evals 1, 2 so unstable node!—evect : [1,0], [0,1] Evals both negative– stable node evect: [1,0], [1,1] Evals both negative– stable node evects: [0,1],[1,-5] Evals , -3/2,1/2 –saddleevects: uns [1,-1.5], stab [1,2.5]

  11. Nullcline Analysis

  12. XPP Phase portrait

  13. 3 competing species • A first step->May-Leonard modeldx/dt= x(1-x-ay-bz)dy/dt= y(1-bx-y-az)dz/dt= z(1-ax-by-z)for all parameter values, characterize the behavior of all solutions.

  14. What is the goal? • Goal is to complete a phase portrait for all parameter values • In higher dimension, phase portraits won’t work, so we want to describe be able to describe the fate of a solution with a given initial condition, (e.g. goes to a fixed point, goes to a periodic, etc.)

  15. Some tools • http://vortex.bd.psu.edu/~jpp/talk10-2-2003/hirsch.pdf • Stable/Center manifold Theorem

  16. Project 1 • Describe the phase portrait for all values of the general 3 competing species model and possibly 4 competing species • More tricks to come! Stokes’ Theorem!, Invariant sets!

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