practical06_sol - Practical Session 6 Solutions Exercises...

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Practical Session 6 — Solutions Exercises 1. (a) The estimate of y (1) is 2 . 714081 , which is accurate to within 0.2%. By comparison, Euler’s method requires about 400 steps to achieve comparable accuracy. Second order Runge-Kutta is quite accurate when applied to this problem, in particular given that the ODE is unstable. (b) See Figure 1. 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 0 0.2 0.4 0.6 0.8 1 y x "ode_exp_rk2.txt" exp(x) Figure 1: Result for 1. (b). 2. (a) For the ODE we have (see Q2 of prac 5 for a more detailed explanation) d N d t = d N dN dt d t dN dt = 1 N * 1 a aN (1 - N/N * ) = N (1 - N ) . (1) For the analytic solution, dividing both sides by N * gives the required form. (b) Equilibrium values are found by setting the RHS of the ODE equal to zero, i.e. are the solutions to N (1 - N ) = 0 . Hence there are two equilibrium values, N = 0 or N = 1 , corresponding to extinction or a population equal to the carrying capacity. The carrying capacity is the maximum sustainable population. (c) Instability is defined by ∂f/∂ N > 0 , where f = N (1 - N ) is the RHS of the ODE. This leads to the condition N < 1 2 . 1
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(f) The plot is Figure 2 (the analytic solution is shown by the solid curve, the Euler’s method solution is indicated by vertical crosses, and the Runge-Kutta solution is indicated by angled crosses). You might observe that the population estimates approach the carrying capacity, and that the Runge-Kutta estimates are much more accurate (closer to the ana-
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This note was uploaded on 09/29/2009 for the course COSC 1002 taught by Professor Wheatland during the Three '09 term at University of Sydney.

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practical06_sol - Practical Session 6 Solutions Exercises...

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