ExercisesSolutions

# Denotes tumor cells and e denotes effector cytotoxic

This preview shows pages 9–12. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern