Systems of Differential Equations Autonomous Examples Math 4B Systems of Differential Equations Autonomous Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

An autonomous 2x2 system of DEs 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

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. There is an equilibrium point at (0,0). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. There is an equilibrium point at (0,0). Compute the Jacobian matrix at (0,0). No need to shift coodinates when the equilibrium is at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. There is an equilibrium point at (0,0). Compute the Jacobian matrix at (0,0). No need to shift coodinates when the equilibrium is at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. There is an equilibrium point at (0,0). Compute the Jacobian matrix at (0,0). No need to shift coodinates when the equilibrium is at the origin. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. There is an equilibrium point at (0,0). Compute the Jacobian matrix at (0,0). No need to shift coodinates when the equilibrium is at the origin. Note that we have basically thrown out the nonlinear terms to get the linearized system. 𝑥 ′ = −2 3 −1 1 𝑥 The linearized system near (0,0) is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. For this linearized system, find eigenvalues to determine stability. 𝑥 ′ = −2 3 −1 1 𝑥 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. For this linearized system, find eigenvalues to determine stability. 𝑥 ′ = −2 3 −1 1 𝑥 Eigenvalues are complex, with negative real part. A stable spiral. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. For this linearized system, find eigenvalues to determine stability. 𝑥 ′ = −2 3 −1 1 𝑥 Stable Spiral Eigenvalues are complex, with negative real part. A stable spiral. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Equilibrium points are at (0,0) and (-1,-1) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Equilibrium points are at (0,0) and (-1,-1) Compute the Jacobian matrix: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Equilibrium points are at (0,0) and (-1,-1) Compute the Jacobian matrix: At (0,0) the linearized system is: 𝑥 ′ = 1 1 −2 1 𝑥 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Equilibrium points are at (0,0) and (-1,-1) Compute the Jacobian matrix: At (0,0) the linearized system is: Eigenvalues: Unstable spiral 𝑥 ′ = 1 1 −2 1 𝑥 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Equilibrium points are at (0,0) and (-1,-1) Compute the Jacobian matrix: At (-1,-1) the linearized system is: 𝐸𝑖𝑔𝑒𝑛𝑣𝑎𝑙𝑢𝑒𝑠: 3± 33 2 𝑥 ′ = −1 −1 −2 4 𝑦 ; 𝑦 = 𝑥 − −1 −1 Saddle Point Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Unstable Spiral Equilibrium points are at (0,0) and (-1,-1) Saddle point Compute the Jacobian matrix: At (-1,-1) the linearized system is: 𝐸𝑖𝑔𝑒𝑛𝑣𝑎𝑙𝑢𝑒𝑠: 3± 33 2 𝑥 ′ = −1 −1 −2 4 𝑦 ; 𝑦 = 𝑥 − −1 −1 Saddle Point Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Equilibrium points are at (0,0) and (2/3,2/5) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Classify the stability of each. Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Equilibrium points are at (0,0) and (2/3,2/5) Compute the Jacobian matrix: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Equilibrium points are at (0,0) and (2/3,2/5) Compute the Jacobian matrix: At (0,0) the linearized system is: 𝑥 ′ = 1 −3 4 −6 𝑥 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Equilibrium points are at (0,0) and (2/3,2/5) Compute the Jacobian matrix: At (0,0) the linearized system is: Eigenvalues: -3 and -2 Real, negative: Stable Node 𝑥 ′ = 1 −3 4 −6 𝑥 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Equilibrium points are at (0,0) and (2/3,2/5) Compute the Jacobian matrix: At (2/3,2/5) the linearized system is: Eigenvalues: Real, opposite sign: Unstable Saddle Point 𝑥 ′ = 9 5 − 5 3 18 5 − 20 3 𝑦 ; 𝑦 = 𝑥 − 2 3 2 5 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the equilibrium point(s) for this system of differential equations. Classify the stability of each. Stable Node Saddle Point Equilibrium points are at (0,0) and (2/3,2/5) Compute the Jacobian matrix: At (2/3,2/5) the linearized system is: Eigenvalues: Real, opposite sign: Unstable Saddle Point 𝑥 ′ = 9 5 − 5 3 18 5 − 20 3 𝑦 ; 𝑦 = 𝑥 − 2 3 2 5 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Find the nulllclines and equilibrium points for this system of differential equations. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Nullclines are the curves that result when x’=0 or y’=0: Find the nulllclines and equilibrium points for this system of differential equations. Nullclines are the curves that result when x’=0 or y’=0: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Nullclines are the curves that result when x’=0 or y’=0: Find the nulllclines and equilibrium points for this system of differential equations. Nullclines are the curves that result when x’=0 or y’=0: Equilibrium points occur at the intersection of the nullclines. In this case that is at x=0,y=0 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 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Nullclines for this system v-nullcline: y=2x h-nullcline: y=x/4

Here is the phase plane diagram Here is the phase plane diagram. Notice the equilibrium point is where the nullclines intersect. Looks stable - all the arrows point toward it.

A typical solution trajectory is shown in blue A typical solution trajectory is shown in blue. Notice that is attracted to the equilibrium point, as expected.

Find the nulllclines and equilibrium points for this system of differential equations. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Nullclines are the curves that result when x’=0 or y’=0: Find the nulllclines and equilibrium points for this system of differential equations. Nullclines are the curves that result when x’=0 or y’=0: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Nullclines are the curves that result when x’=0 or y’=0: Find the nulllclines and equilibrium points for this system of differential equations. Nullclines are the curves that result when x’=0 or y’=0: Equilibrium points occur at the intersection of the nullclines. In this case that is at x=-1,y=1 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 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Nullclines for this system v-nullcline: y=1 h-nullcline: x=-1

Here is the phase plane diagram Here is the phase plane diagram. Notice the equilibrium point is where the nullclines intersect.

A typical solution trajectory. circle centered at (-1,1)

Note: This is a Lotka-Volterra Predator-Prey model. Find the nulllclines and equilibrium points for this system of differential equations. Note: This is a Lotka-Volterra Predator-Prey model. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Note: This is a Lotka-Volterra Predator-Prey model. Find the nulllclines and equilibrium points for this system of differential equations. Note: This is a Lotka-Volterra Predator-Prey model. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Note: This is a Lotka-Volterra Predator-Prey model. Find the nulllclines and equilibrium points for this system of differential equations. Note: This is a Lotka-Volterra Predator-Prey model. Equilibrium points occur when the nullclines intersect. Here we get 2 points - (0,0) and (1,2.5) 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 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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

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)

A typical solution curve is shown A typical solution curve is shown. Notice that it orbits around the equilibrium point. We would say that the equilibrium point at (1,2.5) is stable because the solution trajectories do not lead away from there. (even though the curves don’t go through the equilibirium point) Equilibrium at (1,2.5)

Here is the same equation with “harvesting” (e. g 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)