1. Math 4B Systems of Differential Equations Population models Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2. An autonomous 2x2 system of ODEs has the form Nullclines are curves where f=0 (vertical) or g=0 (horizontal). The intersection of vertical and horizontal nullclines is an equilibrium point. To determine the stability near an equilibrium point, the system can be “linearized”. Essentially this means throwing out all the nonlinear terms, but the coordinates must first be shifted to match the equilibrium point. The linearized system will use the Jacobian matrix, defined below. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3. The following system is an example of a COMPETITION model. Two species, x and y, are competing for resources. Each species will grow logistically in the absence of the other, but when they interact, both populations are impacted negatively. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4. The following system is an example of a COMPETITION model. Two species, x and y, are competing for resources. Each species will grow logistically in the absence of the other, but when they interact, both populations are impacted negatively. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Equilibrium points occur where h- and v-nullclines intersect. Here we get 3 points - (0,0) (0,2) and (1,0)

5. The following system is an example of a COMPETITION model. Two species, x and y, are competing for resources. Each species will grow logistically in the absence of the other, but when they interact, both populations are impacted negatively. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

6. The following system is an example of a COMPETITION model. Two species, x and y, are competing for resources. Each species will grow logistically in the absence of the other, but when they interact, both populations are impacted negatively. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is:

7. The following system is an example of a COMPETITION model. Two species, x and y, are competing for resources. Each species will grow logistically in the absence of the other, but when they interact, both populations are impacted negatively. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is: Now plug in an equilibrium point to find the linearized system. Near (0,0): Eigenvalues are 1 and 2, so (0,0) is an unstable node.

8. The following system is an example of a COMPETITION model. Two species, x and y, are competing for resources. Each species will grow logistically in the absence of the other, but when they interact, both populations are impacted negatively. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is: Now plug in an equilibrium point to find the linearized system. Near (0,2): Eigenvalues are -1 and -2, so (0,2) is a stable node.

9. The following system is an example of a COMPETITION model. Two species, x and y, are competing for resources. Each species will grow logistically in the absence of the other, but when they interact, both populations are impacted negatively. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is: Now plug in an equilibrium point to find the linearized system. Near (1,0): Eigenvalues are -1 and 1, so (0,2) is an unstable saddle point.

10. The following system is an example of a COMPETITION model. Two species, x and y, are competing for resources. Each species will grow logistically in the absence of the other, but when they interact, both populations are impacted negatively. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Saddle Point Unstable Node Stable Node The long-term solution has species y surviving, and tending toward its equilibrium value of 2, and species x dying out (going to 0).

11. This system is another COMPETITION model. It looks very similar to the previous problem, but the results will be very different. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

12. This system is another COMPETITION model. It looks very similar to the previous problem, but the results will be very different. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Equilibrium points occur where h- and v-nullclines intersect. Here we get 4 points - (0,0) (0,2) (2,0) and (2/3,2/3)

13. This system is another COMPETITION model. It looks very similar to the previous problem, but the results will be very different. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is:

14. This system is another COMPETITION model. It looks very similar to the previous problem, but the results will be very different. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is: Now plug in an equilibrium point to find the linearized system. Near (0,0): Eigenvalue is 2 (repeated), so (0,0) is an unstable node.

15. This system is another COMPETITION model. It looks very similar to the previous problem, but the results will be very different. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is: Now plug in an equilibrium point to find the linearized system. Near (0,2): Eigenvalue is -2 (repeated), so (0,2) is a stable node.

16. This system is another COMPETITION model. It looks very similar to the previous problem, but the results will be very different. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is: Now plug in an equilibrium point to find the linearized system. Near (2,0): Eigenvalue is -2 (repeated), so (2,0) is a stable node.

17. This system is another COMPETITION model. It looks very similar to the previous problem, but the results will be very different. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is: Now plug in an equilibrium point to find the linearized system. Near (2/3,2/3): Eigenvalues are 2 and -2/3, so (2/3,2/3) is an unstable saddle point.

18. This system is another COMPETITION model. It looks very similar to the previous problem, but the results will be very different. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Unstable nodeStable Node Saddle Point The long-term solution tends toward one species or the other, depending on the starting point.

19. This is a Lotka-Volterra Predator-Prey model. In this model, one species preys on the other. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

20. This is a Lotka-Volterra Predator-Prey model. In this model, one species preys on the other. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Equilibrium points occur where h- and v-nullclines intersect. Here we get 2 points - (0,0) and (1,5/2) The next slides show the nullclines and some solution curves graphed using the PPlane application which can be found online at http://math.rice.edu/~dfield/dfpp.html http://math.rice.edu/~dfield/dfpp.html

21. h-nullcline: x=1 h-nullcline: y=0 v-nullcline: y=2.5 v-nullcline: x=0 Nullclines for this system

22. This is a Lotka-Volterra Predator-Prey model. In this model, one species preys on the other. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is:

23. This is a Lotka-Volterra Predator-Prey model. In this model, one species preys on the other. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is: Now plug in an equilibrium point to find the linearized system. Eigenvalues are 5 and -1. (0,0) is an unstable saddle point. Eigenvalues are ±√5 i (1,5/2) is a stable center *. * Complex roots with real part =0 are actually ambiguous for nonlinear systems, but the phase diagram shows the closed curves that indicate a stable solution.

24. Here is the phase- plane diagram for this predator-prey model. We are only concerned with solutions where x and y are positive or zero. (negative #s of rabbits don’t make sense)

25. A typical solution curve is shown. Notice that it forms a closed curve around the equilibrium point. The equilibrium point at (1,2.5) is a stable center, but it is NOT asymptotically stable. Equilibrium at (1,2.5)

26. Here is the same equation with “harvesting” (e.g., if we have rabbits and wolves, assume each species is hunted at the same rate). Notice that the solution is very similar to the previous case, except the equilibrium point has moved. New equilibrium at (1.1,2.45)

27. This is another Predator-Prey model. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

28. This is another Predator-Prey model. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Equilibrium points occur where h- and v-nullclines intersect. Here we get 2 points - (0,0) and (6,2)

29. This is another Predator-Prey model. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is:

30. This is another Predator-Prey model. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The Jacobian matrix for this system is: Now plug in an equilibrium point to find the linearized system. Eigenvalues are 2 and -3. (0,0) is an unstable saddle point. Eigenvalues are ±√6 i (6,2) is a stable center *. * Complex roots with real part =0 are actually ambiguous for nonlinear systems, but the phase diagram shows the closed curves that indicate a stable solution.

31. This is another Predator-Prey model. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Stable Center Saddle Point Solutions are stable orbits around the equilibrium at (6,2)