Mathematics > Solutions Guide > University of California, Los Angeles LIFESCIENC 30A. Homework 7 Solutions. All Exercises. (All)

University of California, Los Angeles LIFESCIENC 30A. Homework 7 Solutions. All Exercises.

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LS 30A Homework 7 Solutions Exercise 3.2.4 Find the equilibria of X0 = 0:1X(1 - 800 X ) - 0:05X and use test points to determine their stability. You need to solve X0 = 0: Thus, the equilibrium... point X∗ = 0 is unstable (the arrows around it are pointing away from it), while the equilibrium point X∗ = 400 is stable (the arrows around it are pointing at it). Note that since the state space of this model is R>0, there is no interval to the left of 0. Exercise 3.2.6 Why is this true? Equilibrium points are by definition the points X∗ where X0(X∗) = 0. But, the points where the graph of X0 intersects the X-axis are precisely the points where X0 = 0. Hence, they are precisely the equilibrium points. Exercise 3.2.8 Find the equilibria of the differential equation 1 LS 30A Homework 7 Solutions to find that the equilibrium points are N∗ = 0; N∗ = 50, and N∗ = 1000. Further,. Using \graphical methods" would mean plugging the given expression for N0 into CoCalc (say) and seeing what the graph looks like. For example, you would see that, to the left of N = 50, N0 < 0, and to the right of N = 50, N0 > 0. So, you would conclude that N = 50 is an unstable equilibrium Exercise 3.2.10 of the vector field Suppose we try to evaluate the stability of the X = 0 equilibrium point X0 = 2X2 - X 1. Perform a linear stability analysis at the point X = 0. What is the character of this equilibrium point according to this analysis? 2. Suppose we did a test point analysis for confirmation and chose two test points, X = -1 and X = +1. When we calculate the change vectors X0 at these two points, we see that 2 LS 30A Homework 7 Solutions the change vector at X = -1 is positive, X0jX=-1 = 2(-1)2 - (-1) = 3 and the change vector at X = +1 is also positive, X0jX=+1 = 2(+1)2 - (+1) = 1 Explain why this test point method conflicts with the linear stability analysis. What have we done wrong? (Hint: Plot the X0 function.) 1. 4 , and X = 1 would lead to a successful analysis. Exercise 3.2 FE 1 A kayaker is paddling directly into the wind but the kayak keeps veering either left or right. 1. Use your physical intuition to explain why the kayaker is having difficulty going straight. 2. Describe this situation in terms of equilibria and stability. (Hint: Sketch a vector field. No equations are necessary.) 1. The kayak keeps veering either left or right because anytime the nose of the kayak is not pointing directly at the wind, the wind blows on the side of the kayak and pushes it further to the left or right. 2. The nose of the kayak pointing directly into the wind represents an unstable equilibrium point. If the nose is pointing directly into the wind, the wind is pushing on both sides of the kayak equally and the direction is not changed. However, whenever the kayak gets just a little out of alignment, the direction of the kayak begins to diverge away from the equilibrium point. Exercise 3.2 FE 2 The spread of a genetic mutation in a population of mice can be modeled by the differential equation P 0 = 2P · (1 - P ) · (1 - 3P ) where P is the fraction of the mice that have the new gene. (This means that 0 ≤ P ≤ 1.) 1. Find the equilibruim points of this model and determine the stability of each one. 3 LS 30A Homework 7 Solutions 2. If 10% of the mice have the new gene (so P = 0:1) initially, what fraction of the population will have the new gene in the long run? 3. What if the initial fraction is 90% of the mice? 2. The initial condition P = 0:1 is in the basin of attraction of the stable equilibrium point P∗ = 1 3 . Thus, in the long run, approximately 1 3 of the population will have the new gene. 3. The initial condition P = 90% = 0:9 is also in the basin of attraction of the stable equilibrium point P∗ = 1 3 . Thus, in the long run, approximately 1 3 of the population will have the new gene. Exercise 3.2 FE 5 Consider a system of two chemical compounds, A and X. One molecule of A and one of X react to produce two molecules of X, with rate constant 0.1. Also two molecules of X can react to form one molecule of X and one molecule of A, with rate constant 0.05. A + X 2X The amount of A is much larger than that of X, so its concentration can be thought of as a constant, 2. a) Write a differential equation for the concentration of X. (Hint: Look back at the predator-prey and disease models studied in earlier sections.) b) Find the equilibria of this system and describe their stability. Exercise 3.3.2 What is the change vector at the point (3; -4)? X0 = X and Y 0 = Y in this vector field. Therefore the change vector at point P = (3; -4) is (X0; Y 0) = (3; -4:) Exercise 3.3.3 Sketch time series (for both X and Y ) corresponding to two trajectories in Figure 3.18. The plots below show time series for the system X0 = X; Y 0 = -Y with four different initial conditions: (X0; Y0) = (-0:1; 0:1); (0:1; 0:1); (-0:1; -0:1); (0:1; -0:1) in reading order. The plots below show time series for the system X0 = -X; Y 0 = Y with four different initial conditions: (X0; Y0) = (-0:1; 0:1); (0:1; 0:1); (-0:1 order. 5 LS 30A Homework 7 Solutions Exercise 3.3 FE 1 Consider the following Romeo and Juliet model, in which (as usual) R represents Romeo’s love for Juliet, and J represents Juliet’s love for Romeo (recall that these variables can be positive or negative): R0 = J - 0:1R J0 = -R a) Verify that this system has one equilibrium point and it is at the origin. d) Plot the vector field in SageMath. Can you determine the type of equilibrium point at the origin now? e) Choose some initial conditions and use SageMath to simulate this system and plot (at least) one trajectory. Can you determine the type of equilibrium point at the origin now? a) In order for J0 = 0, we must have R = 0. But, then, in order for R0 = 0, we must have J - (0:1)(0) = 0, so R = 0 also. Thus, (0; 0) is the only equilibrium point of this model. d) 6 LS 30A Homework 7 Solutions e) starting at point (R; J) = (1; 1) the trajectory in the vector field is depicted below. Since the trajectory of the point (1; 1) spirals in towards (0; 0), we can infer that the equilibrium point at (0; 0) is a stable spiral. Exercise 3.3 FE 2 Repeat the same analysis as in the previous problem, but with the following differential equations: a) Verify that this system has one equilibrium point and it is at the origin. d) Plot the vector field in SageMath. Can you determine the type of equilibrium point at the origin now? e) Choose some initial conditions and use SageMath to simulate this system and plot (at least) one trajectory. Can you determine the type of equilibrium point at the origin now? a) Suppose R0 = 0; this means that J = 0. Now suppose, in addition, that J0 = 0; this means that -R + (0:1)(0) = 0, so that R = 0. Thus, (0; 0) is the only equilibrium point of this model. d/e) The equilibrium point at (0; 0) is an unstable spiral. 8 [Show More]

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