ExercisesSolutions

Denotes tumor cells and e denotes effector cytotoxic

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denotes tumor cells and E denotes effector (cytotoxic im- mune) cells. The T nullclines are found to be at T = 0 E = a (1 - bT ) cT 1 3 , and the E nullcline is at E = s ( g + T ) ( d + mT )( g + T ) - pT . These nullclines are plotted in Figure 10. There are 2 possible equilibria for this system. One is located at ( T, E ) = (0 , s d ), while the other is at E = a (1 bT ) cT 1 3 given that T is a solution of the nonlinear equation a (1 bT ) cT 1 3 = s ( g + T ) ( d + mT )( g + T ) pT , which can be solved numerically using, for example, the Newton-Raphson Method. The Jacobian for the system at ( T, E ) = (0 , s d ) is singular, hence the stability of the equilibrium point cannot be determined using linear stability analysis. From Maple plots, it appears that this equilibrium point is a saddle with an extremely unstable manifold. The stability of the other equi- librium point is also difficult to determine, however using Maple, it is found to be a stable node. The parameter values used to determine the stability of these points are the same as those used in the Kuznetsov paper [KMTP94]. The qualitative behaviour of the model differs slightly to the one presented in class due to the T 2 3 term in the Von Bertalanffy equation. The equilibria at A and B in the model presented in class still exist in this model, however due to the behaviour of the term mention above, the two equilibria at C and D no longer exist. 9
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0 100000 200000 300000 400000 500000 E –400 –200 200 400 600 800 1000 T Figure 10: The T nullclines (thick line) and E nullcline (thin line) for Exercise 4. 10
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Exercises for Qualitative Analysis Module 1. Purpose: To study non-linear centers. Notes: Define a reversible system to be any second-order system that is invariant under t → - t and y → - y . For example, any system ofthe form ˙ x = f ( x, y ) ˙ y = g ( x, y ) , where f is odd in y and g is even in y (i.e., f ( x, - y ) = - f ( x, y ) and g ( x, - y ) = g ( x, y )) is reversible. Theorem: (Nonlinear centers for reversible systems) Suppose the origin x * = 0 is a linear center for the continously differentiable system ˙ x = f ( x, y ) ˙ y = g ( x, y ) , and suppose that the system is reversible. Then sufficiently close to the origin, all trajectories are closed curves. from[Str94], Example 6.6.1 Exercise: Show that the system ˙ x = y - y 3 ˙ y = - x - y 2 has a nonlinear center at the origin, and plot the phase portrait. Solution: For the system ˙ x = y - y 3 = f ( x, y ) ˙ y = - x - y 2 = g ( x, y ) the fixed point x is (0 , 0). Compute the Jacobian for the system and evaluate it at the fixed point, A = bracketleftbigg 0 1 - 3 y 2 - 1 - 2 y bracketrightbigg vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle (0 , 0) = bracketleftbigg 0 1 - 1 0 bracketrightbigg . The eigenvalues of A are i and - i ; purely imaginary eigenvalues are characteristic of a center; thus x = 0 is a center. To show that the system is reversible, consider f ( x, - y ) = - y - ( - y ) 3 = - y + y 3 = - ( y - y 3 ) = - f ( x, y ) g ( x, - y ) = - x - ( - y ) 2 = - x - y 2 = g ( x, y ) which illustrates that f is odd in y and g is even in y . Thus all trajectories sufficiently close to the origin are closed curves. The phase portrait is shown in Figure 11.
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