This has eigenvalues � 1 03333 ξ 1 1 1 6 T and � 2 2 ξ 1 0 1 T This is a stable

This has eigenvalues ? 1 03333 ξ 1 1 1 6 t and ? 2 2

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. This has eigenvalues λ 1 = - 0 . 03333 ( ξ 1 = [1 , - 1 . 6] T ) and λ 2 = - 0 . 2 ( ξ 1 = [0 , 1] T ). This is a stable node . At the cooperative equilibrium , ( X e , Y e ) = (2 , 4), the Jacobian satisfies: J (2 , 4) = - 0 . 02 - 0 . 04 - 0 . 16 - 0 . 12 ! . This has eigenvalues λ 1 = - 0 . 1643 ( ξ 1 = [1 , 3 . 609] T ) and λ 2 = 0 . 02434 ( ξ 1 = [1 , - 1 . 1085] T ). This is a saddle node . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (48/68)
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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 Phase Portrait The figure below was generated with pplane8 and shows that Example 1 exhibits competitive exclusion with all solutions going to either the carrying capacity equilibria , ( X e , Y e ) = ( 0 , 20 3 ) or ( X e , Y e ) = (10 , 0). Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (49/68)
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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 Example/Equilibria Example 2 : Consider the competition model : dX dt = 0 . 1 X - 0 . 02 X 2 - 0 . 01 XY, dY dt = 0 . 2 Y - 0 . 04 Y 2 - 0 . 03 XY. Nullclines where dX dt = 0 are 1 X = 0. 2 0 . 1 - 0 . 02 X - 0 . 01 Y = 0 or Y = 10 - 2 X . Nullclines where dY dt = 0 are 1 Y = 0. 2 0 . 2 - 0 . 04 Y - 0 . 03 X = 0 or Y = 5 - 0 . 75 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 , 5), (5 , 0), and (4 , 2). Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (50/68)
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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 Linearization : The competition model is below: dX dt = 0 . 1 X - 0 . 02 X 2 - 0 . 01 XY = f 1 ( X, Y ) , , dY dt = 0 . 2 Y - 0 . 04 Y 2 - 0 . 03 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 . 04 X - 0 . 01 Y - 0 . 01 X - 0 . 03 Y 0 . 2 - 0 . 08 Y - 0 . 03 X ! . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (51/68)
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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 ) = (5 , 0), the Jacobian satisfies: J (5 , 0) = - 0 . 1 - 0 . 05 0 0 . 05 ! . This has eigenvalues λ 1 = - 0 . 1 ( ξ 1 = [1 , 0] T ) and λ 2 = 0 . 05 ( ξ 1 = [1 , - 3] T ). This is a saddle node . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (52/68)
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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 : At the Y e
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