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MA354

MA354. Building Systems of Difference Equations T H 2:30pm– 3:45 pm Dr. Audi Byrne. Recall: Models for Population Growth. Very generally, P =  (increases in the population) –  (decreases in the population) Classically, P = “births” – “deaths” “Conservation Equation”.

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MA354

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  1. MA354 Building Systems of Difference Equations T H 2:30pm– 3:45 pm Dr. Audi Byrne

  2. Recall: Models for Population Growth • Very generally, P =  (increases in the population) –  (decreases in the population) • Classically, P = “births” – “deaths” “Conservation Equation”

  3. Practice: Writing Down Difference Equations

  4. Problem 1: Rabbit Population Model • Birth rate: Suppose that the birth rate of a rabbit population is 0.5*. • Death rate: Suppose that an individual rabbit has a 25% chance of dying each year. Write down a difference equation that describes the given dynamics of the bunny population. * E.g., for every two bunnies alive in year t approximately 1 bunny is born in year t+1.

  5. Problem 1: Rabbit Population Model • Rn+1 = • Rn+1 = • What is the long time behavior of the chicken population?

  6. Problem 3: Chicken Population Model • Birth rate: Suppose that 2000 chickens are born per year on a chicken farm. • Death rate: Suppose that the chicken death rate is 10% per year. Write down a difference equation that describes the given dynamics of the chicken population.

  7. Problem 2: Chicken Population Model • Cn+1= • Cn+1= • What is the long time behavior of the chicken population?

  8. Problem 2: Chicken Population Model • Cn+1= • Cn+1= • What is the long time behavior of the chicken population?

  9. Practice: Writing Down Systems of Difference Equations

  10. Systems of Difference Equations • Two or more populations interact with one another through birth or death terms. • Each population is given their own difference equation. • To find an equilibrium value for the system, all populations must simultaneously be in equilibrium.

  11. Example: Predator Prey Model W(t) – wolf population R(t) – rabbit population Without interaction: W(t+1) = (wb-wd) W(t) R(t+1) = (rb-rd) R(t) With predator/prey interaction: W(t+1) = (wb-wd) W(t) + k1W(t)R(t) R(t+1) = (rb-rd) R(t) – k2W(t)R(t)

  12. Example: Predator Prey Model W(t) – wolf population R(t) – rabbit population Without interaction: W(t+1) = (wb-wd) W(t) R(t+1) = (rb-rd) R(t) With predator/prey interaction: W(t+1) = (wb-wd) W(t) + k1W(t)R(t) R(t+1) = (rb-rd) R(t) – k2W(t)R(t)

  13. Example: Predator Prey Model W(t) – wolf population R(t) – rabbit population Without interaction: W(t+1) = (wb-wd) W(t) R(t+1) = (rb-rd) R(t) With predator/prey interaction: W(t+1) = (wb-wd) W(t) + k1W(t)R(t) R(t+1) = (rb-rd) R(t) – k2W(t)R(t) Population size just depends on independent births and deaths…

  14. Example: Predator Prey Model W(t) – wolf population R(t) – rabbit population Without interaction: W(t+1) = (wb-wd) W(t) R(t+1) = (rb-rd) R(t) With predator/prey interaction:

  15. Example: Predator Prey Model W(t) – wolf population R(t) – rabbit population Without interaction: W(t+1) = (wb-wd) W(t) R(t+1) = (rb-rd) R(t) With predator/prey interaction: W(t+1) = (wb-wd) W(t) + k1W(t)R(t)

  16. Example: Predator Prey Model W(t) – wolf population R(t) – rabbit population Without interaction: W(t+1) = (wb-wd) W(t) R(t+1) = (rb-rd) R(t) With predator/prey interaction: W(t+1) = (wb-wd) W(t) + k1W(t)R(t) R(t+1) = (rb-rd) R(t) – k2W(t)R(t)

  17. Example: Predator Prey Model W(t) – wolf population R(t) – rabbit population Without interaction: W(t+1) = (wb – wd) W(t) R(t+1) = (rb – rd) R(t) With predator/prey interaction: W(t+1) = (wb – wd) W(t) + k1W(t)R(t) R(t+1) = (rb – rd) R(t) – k2W(t)R(t) “mass-action” type interaction k1W(t)R(t)

  18. Example: Predator Predator Model W(t) – wolf population H(t) – hawk population

  19. Example: Predator Predator Model W(t) – wolf population H(t) – hawk population Without interaction: W(t+1) = (wb – wd) W(t) H(t+1) = (hb – hd) H(t)

  20. Example: Predator Predator Model W(t) – wolf population H(t) – hawk population Without interaction: W(t+1) = (wb – wd) W(t) H(t+1) = (hb – hd) H(t) With predator/predator interaction: W(t+1) = (wb – wd) W(t) – k1W(t)H(t) H(t+1) = (hb – hd) H(t) – k2W(t)H(t)

  21. Example: Predator/Predator/Prey Model W(t) – wolf population H(t) – hawk population R(r) – rabbit population System with interactions: W(t+1) = (wb-wd) W(t) – k1W(t)H(t) + k3W(t)R(t) H(t+1) = (hb-hd) H(t) – k2W(t)H(t) + k4W(t)R(t) R(t+1) = (hr-hr) R(t) – k5W(t)R(t) – k6H(t)R(t)

  22. Example: Predator/Predator/Prey Model W(t) – wolf population H(t) – hawk population R(r) – rabbit population System with interactions: W(t+1) = (wb – wd) W(t) – k1W(t)H(t) + k3W(t)R(t) H(t+1) = (hb – hd) H(t) – k2W(t)H(t) + k4W(t)R(t) R(t+1) = (hr – hr) R(t) – k5W(t)R(t) – k6H(t)R(t)

  23. Finding Equilibrium Valuesof Systems of Difference Equations

  24. Example: Predator Predator Model W(t) – wolf population H(t) – hawk population Dynamical System: W(t+1) = (wb-wd) W(t) - k1W(t)H(t) H(t+1) = (hb-hd) H(t) – k2W(t)H(t) W(t+1) = 1.2 W(t) – 0.001W(t)H(t) H(t+1) = 1.3 H(t) – 0.002W(t)H(t)

  25. Example: Predator Predator Model • What is the long time (equilibrium) behavior of the two populations? (Solution will be demonstrated on the board.)

  26. Example: A Car Rental Company • A rental company rents cars in Orlando and Tampa. It is found that 60% of cars rented in Orlando are returned to Orlando, but 40% end up in Tampa. Of the cars rented in Tampa, 70% are returned to Tampa and 30% are returned to Orlando. • Write down a system of difference equations to describe this scenario and decide how many cars should be kept in each city if there are 7000 cars in the fleet.

  27. Example: A Car Rental Company • What is the dynamical system describing this scenario? • What is the long time behavior of the two populations?

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