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The following system of odes models nonlinear

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The following system of ODEs models nonlinear chemical reactions y 1 = - αy 1 + βy 2 y 3 y 2 = αy 1 - βy 2 y 3 - γy 2 2 y 3 = γy 2 2 where α = 4 × 10 2 , β = 10 4 , and γ = 3 × 10 7 . Starting with initial conditions y 1 (0) = 1 and y 2 (0) = y 3 (0) = 0 , integrate this ODE from t = 0 to t = 3 . You may use either a library routine or an ODE solver of your own design. Try both stiff and non-stiff methods, and experiment with various error tolerances. Compare the efficiencies of the stiff and non-stiff methods as a function of error tolerance. Solution: The following results were found using MATLAB ODE23 (nonstiff method) The following results were found using MATLAB ODE23s (stiff method) From these results, it can be seen that the stiff method seems to be more efficient than the non-stiff method, requiring less time steps and generally less CPU time. 24
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Error tolerance t steps Failures CPU time (secs) 1e-12 2921 10 2.703 1e-10 2683 11 2.624 1e-8 2633 11 2.644 1e-6 2623 9 2.994 1e-4 2623 32 0.47 Error tolerance t steps Failures CPU time (secs) 1e-12 1690 4 5.387 1e-10 266 3 0.941 1e-8 50 3 0.43 1e-6 19 3 0.141 1e-4 15 3 0.741 25
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Exercises for Bifurcation Module 1. Purpose: To study saddle node bifurcations in a model. From [Str94], Example 8.1.1 . Exercise: The following system has been discussed by [Gri71] as a model for a genetic control system. The activity of a certain gene is assumed to be directly induced by two copies of the protein for which it codes. In other words, the gene is stimulated by its own product, potentially leading to an autocatalytic feedback process. In dimensionless form, the equations are ˙ x = - ax + y ˙ y = x 2 1 + x 2 - by where x and y are proportional tothe concentrations of the protein and the messenger RNA from which it is translated, respectively, and a, b > 0 are parameters that govern the rate of degradation of x and y . Show that the system has three fixed points when a < a c , where a c is to be determined. Show that two of these fixed points coalesce in a saddle-node bifurcation when a = a c . Then sketch the phase portrait for a < a c , and give a biological interpretation. Solution: The fixed points of a system are found by solving ˙ x = 0 and ˙ y = 0. For this particular system, solve: y = ax and y = 1 b x 2 1 + x 2 , ax = x 2 b (1 + x 2 ) for x = 0 , 1 ± 1 - 4 a 2 b 2 2 ab which illustrates the required three fixed points. Notice that if 1 - 4 a 2 b 2 = 0, or rather 2 ab = 1, the two latter fixed points coalesce into one. Thus we can determine a c = 1 / 2 b , below which value three fixed points exist. At a = a c , the non-zero fixed point is x = 1 / 2 ab . The presence of the saddle-node bifurcation can be confirmed by evaluating the Jacobian matrix, A = parenleftbigg - a 1 2 x (1+ x 2 ) 2 - b parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle x = x * = ab - 2 x (1 + ( x ) 2 ) 2 at the fixed point. Since two fixed points coalesce at a = a c , and the eigenvalues of A are both real positive and negative, there exists a saddle-node bifurcation. The phase portrait for a < a c is shown in Figure 22.
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