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# hw5 - u = 0.2 and ρ = 3.01 3.05 and 3.1 respectively In...

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Numerical Analysis in Engineering ME 140A, Fall 2007 Homework #5 Due: Thursday Nov 15, 8 am. November 8, 2007 Consider the logistic difference equation: u n +1 = ρu n (1 - u n ) . (1) 1. For ρ = 3 . 2 plot the solution for the initial conditions u 0 = 0.2, 0.4, 0.6 and 0.8. Observe that in each case the solution approaches a steady oscillation between the same two values. This illustrates that the long-term behavior of the solution is independent of the initial conditions. 2. The solutions to the logistic equation change from convergent sequences to periodic oscillations of period two as the parameter ρ passes through the value 3. To see more clearly how this happens, carry out the following calculations: a) Plot the solution for u 0 = 0.2 and ρ = 2.9, 2.95 and 2.99, respectively. In each case, estimate how many iterations it takes for the solution to get to within 0.01% of the limiting value. b) Plot the solution for
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Unformatted text preview: u = 0.2 and ρ = 3.01, 3.05 and 3.1, respectively. In each case, estimate how many iterations it takes for the solution to get to within 0.01% of the limiting values of the steady state oscillation. 3. By plotting the solution for diﬀerent values of ρ , estimate the value of ρ for which the solution changes from an oscillation of period two to one of period four. In the same way, estimate the value of ρ for which the solution changes from period four to period eight. Ten bonus points will be awarded to the person who gets closest to the accurate value for each of the two estimates. 4. Explore the range 1 ≤ ρ ≤ 4, with steps Δ ρ = 0 . 01. For each value of ρ , and for u = 0.2, calculate u 1 to u 500 . Then plot a dot for each of the values u 401 to u 500 , for each value of ρ , in a ( ρ, u )-plot....
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