Equilibria occur where the curves intersect The nullclines for the competition

# Equilibria occur where the curves intersect the

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Equilibria occur where the curves intersect. The nullclines for the competition model are only straight lines: The dX dt = 0 has X = 0 or the Y -axis preventing solutions in X from becoming negative. The dY dt = 0 has Y = 0 or the X -axis preventing solutions in Y from becoming negative. The other two nullclines are straight lines with negative slopes passing through the positive quadrant, X > 0 and Y > 0. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (44/68)

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Introduction Linear Applications of Systems of 1 st Order DEs Nonlinear Applications of Systems of DEs Model of Glucose and Insulin Control Glucose Tolerance Test Competition Model Nullclines 3 Example 1 : Consider the competition model : dX dt = 0 . 1 X - 0 . 01 X 2 - 0 . 02 XY, dY dt = 0 . 2 Y - 0 . 03 Y 2 - 0 . 04 XY. Nullclines where dX dt = 0 are 1 X = 0. 2 0 . 1 - 0 . 01 X - 0 . 02 Y = 0 or Y = 5 - 0 . 5 X . Nullclines where dY dt = 0 are 1 Y = 0. 2 0 . 2 - 0 . 03 Y - 0 . 04 X = 0 or Y = 20 3 - 4 3 X . Equilibria occur at intersections of a nullcline with dX dt = 0 and one with dY dt = 0. The 4 equilibria are (0 , 0), ( 0 , 20 3 ) , (10 , 0), and (2 , 4). Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (45/68)
Introduction Linear Applications of Systems of 1 st Order DEs Nonlinear Applications of Systems of DEs Model of Glucose and Insulin Control Glucose Tolerance Test Competition Model Linearization Linearization : The competition model is below: dX dt = 0 . 1 X - 0 . 01 X 2 - 0 . 02 XY = f 1 ( X, Y ) , dY dt = 0 . 2 Y - 0 . 03 Y 2 - 0 . 04 XY = f 2 ( X, Y ) , and the linearization about the equilibria is found by evaluating the Jacobian matrix at the equilibria: J ( X, Y ) = ∂f 1 ( X,Y ) ∂X ∂f 1 ( X,Y ) ∂Y ∂f 2 ( X,Y ) ∂X ∂f 2 ( X,Y ) ∂Y ! = 0 . 1 - 0 . 02 X - 0 . 02 Y - 0 . 02 X - 0 . 04 Y 0 . 2 - 0 . 06 Y - 0 . 04 X ! . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (46/68)

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Introduction Linear Applications of Systems of 1 st Order DEs Nonlinear Applications of Systems of DEs Model of Glucose and Insulin Control Glucose Tolerance Test Competition Model Linearization and Equilibria Linearization : Consider the extinction equilibrium , ( X e , Y e ) = (0 , 0), the Jacobian satisfies: J (0 , 0) = 0 . 1 0 0 0 . 2 ! . This has eigenvalues λ 1 = 0 . 1 ( ξ 1 = [1 , 0] T ) and λ 2 = 0 . 2 ( ξ 1 = [0 , 1] T ). This is an unstable node , as we’d expect for low populations. At the X e carrying capacity equilibrium , ( X e , Y e ) = (10 , 0), the Jacobian satisfies: J (10 , 0) = - 0 . 1 - 0 . 2 0 - 0 . 2 ! . This has eigenvalues λ 1 = - 0 . 1 ( ξ 1 = [1 , 0] T ) and λ 2 = - 0 . 2 ( ξ 1 = [2 , 1] T ). This is a stable node . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (47/68)
Introduction Linear Applications of Systems of 1 st Order DEs Nonlinear Applications of Systems of DEs Model of Glucose and Insulin Control Glucose Tolerance Test Competition Model Linearization and Equilibria Linearization : At the Y e carrying capacity equilibrium , ( X e , Y e ) = (0 , 20 / 3), the Jacobian satisfies: J (0 , 20 / 3) = - 0 . 03333 0 - 0 . 2667 - 0 . 2 !

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