MATH 485 T EST 2 Due May 9th, 2016 Your Name: I certify that the work on this test is mine and mine alone (please sign): Directions: You get one point for signing the certi...cation above. Then do two of the three problems below. Note that you must do the problem completely and may not do parts of all three. Score: 1. 2. Signed Certi...cation Total 1 /25 1. (12pts) Recall the logistic map given by the equation xn+1 = axn (1 xn ) : (1) This map has two ...xed points, and for certain values of a; there is an attractive ...xed point, while for other values of a neither ...xed point is attractive and more complicated behavior must ensue. In this problem we will alter this map by replacing one of the xn on the right side of the equation with the new value xn+1 : The following is an implicit map: xn+1 = axn (1 xn+1 ) (2) since it does not explicitly give xn+1 as a function of xn ; i.e., the iteration is determined by solving an equation of the form xn+1 = F (xn ; xn+1 ) : a. Show that both maps (1) and (2) have the same ...xed points. b. Show that map (2) has one attractive ...xed point no matter which a is chosen. c. The extra credit below shows that both maps are approximations to the logistics dierential equation dx = ax (1 x) : dt Knowing that both maps are approximations to the logistics dierential equation as well as the results from parts a and b, which map do you think is a better approximation to the logistics dierential equation? Explain. Extra credit. Show that both maps are approximations of the logistic dierential equation dx = ax (1 x) : dt Hint: A dierential equation dx = F (x) dt can be approximated by approximating the derivative and the function, e.g., dx = dt xn+1 xn (for some value h) and F (x) = F (xn ) or F (xn+1 ) or something like that. By h approximating the logistic equation as : yn+1 yn = cyn (1 h yn ) you should be able to rearrange and rescale the variables to get map (1). The argument for map (2) is similar. 2. (12pts) Consider the system of dierential equations dx = x (a + bx + cy) dt dy = y (e + f x + gy) : dt 2 where a; b; c; e; f; g are parameters. This is called the generalized Lotka-Volterra Equations a. (2pts) Give the signs of the parameters (positive/negative/zero) that result in the predatory/prey model. b. (2pts) Give the signs of the parameters (positive/negative/zero) that result in the two competing species model. c. (4pts) If we suppose that a and e are positive and b is negative, ...nd positive values of A; B; and C such that taking X = Ax, Y = By; and T = Ct transforms the system into dX = X (1 X + hY ) dT dY = Y (1 + kX + mY ) dT for parameters h; k; m: Also relate the parameters h; k; m to the original parameters in the system. d. (4pts) Note that (X; Y ) = (1; 0) is a ...xed point of the system. Linearize near the ...xed point (1; 0) and give conditions on the parameter k that would indicate that it is possible for the population represented by X to drive the population represented by Y to extinction (this may depend on the initial values of X and Y; but ...nd a value of k such that some positive initial values of X and Y lead to extinction of the population represented by Y ). Justify your answer. 3