MATH10001 Mathematical Workshop Difference Equations part 2 Non-linear difference equations

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1 MATH10001 Mathematical Workshop Difference Equations part 2 Non-linear difference equations In a linear difference equation, the equation contains a sum of multiples of the elements in the sequence {y n }. In a non-linear difference equation, all these restrictions are lifted. Linear Non-linear y n+1 = y n, y n+1 = y 2 n, y n+2 = y n+1 + y n y n+2 = y n+1 y n, y n+2 = 4y n+1 + 3y n y n+2 = 4e y n, There is no standard method for finding solutions to non-linear difference equations. Indeed, it is an area, where numerical experiments are as important as theory in understanding the system. Fixed points of difference equations A simple technique that can be used to obtain a great deal of information about systems of non-linear difference equations is to use a fixed-point analysis. The idea is to find particular points for which the solution is fixed (does not change), in other words, y n+1 = y n, for all n N, for some N. Linear stability analysis of fixed points A fixed point is said to be linearly stable if the system moves back to the fixed point after it has been slightly perturbed. A fixed point is said to be linearly unstable if the system moves away from the fixed point after it has been slightly perturbed. In mathematical terms, we must consider what happens when y n = Y + ỹ n, where Y is a fixed point of the system and ỹ n 1 is a small perturbation to the system. The idea of the analysis is to find an explicit expression for ỹ n. If ỹ n 0 when n, then the system moves back to the fixed point at Y and we conclude that the fixed point is linearly stable. The word linear in the expression linear stability analysis refers to the following important assumption:

2 In a linear stability analysis we assume that the perturbation ỹ n is so small that we can neglect all terms of the form (ỹ n ) m where m 2. That is, we keep only terms linear in ỹ n. Example y n+1 = y 2 n. If Y is a fixed point we have Y = Y 2 and so Y = 0, 1. Let y n = Y + ỹ n. Then Y + ỹ n+1 = (Y + ỹ n ) 2 = Y 2 + 2Y ỹ n + ỹ 2 n. We neglect the term ỹ 2 n, assuming ỹ n is small and use the fact that Y = Y 2 to get ỹ n+1 = 2Y ỹ n. This is a linear difference equation with solution ỹ n = (2Y ) n ỹ 0. When Y = 0 we get ỹ n = 0 for all n 1 and so Y = 0 is a stable fixed point. When Y = 1 we get ỹ n = 2 n ỹ 0 as n and so Y = 1 is an unstable fixed point. Other methods for analysing non-linear difference equations Consider the logistic map y n+1 = ry n (1 y n ) where r > 0 is a constant and 0 y 0 1. We can rewrite this as y n+1 = f(y n ) where f is the function given by f(x) = rx(1 x). We can sketch this function for 0 x 1 (the vertical axis is multiples of r)

3 So for 0 r 4 the entire sequence y n will remain in the interval [0, 1] when y 0 [0, 1]. We can also draw an iteration diagram to examine the long term behaviour of the sequence. By using fixed point analysis and iteration diagrams we observe the following behaviour for the logistic map: When 0 r 1 there is one stable fixed point at Y = 0, so whatever the initial condition the sequence tends to 0. When 1 < r 3, the fixed point at Y = 0 becomes unstable and we get a new stable fixed point at (r 1)/r. When r > 3 this second fixed point also becomes unstable and the sequence cannot reach a constant value.

4 MATH10001 Mathematical Workshop Difference Equations Project Problems part 2 Problem 3 Consider the difference equation where a > 0 is a constant. y n+1 = { ay n if y n < 0.5 a(1 y n ) if y n 0.5 (i) Write the difference equation as y n+1 = f(y n ) for some function f and sketch the graph of f(x) for 0 x 1. For what values of a does the sequence remain in [0, 1] if y 0 [0, 1]? (ii) Consider the case when a = 0.3. Show that there is a single fixed point in this case. Determine whether this fixed point is stable or unstable. Let y 0 = 0.6 and calculate y 1, y 2, y 3 and y 4. Draw an iteration diagram starting at y 0 = 0.6. (iii) Now let a = 1.5. unstable. Find the fixed points and determine if they are stable or Let y 0 = 0.2. Calculate y 1, y 2, y 3 and y 4. Draw an iteration diagram starting at y 0 = 0.2. (iv) Let a = 4. Find all values of y 0 that give a sequence of period 2 ie. y n+2 = y n but y n+1 y n for all n 0. Project Report The assessment for this project is by an individual project report. The report should contain your solutions to the three problems and the (clearly labelled) graph that you produced as the homework for part 1. Your report should be well presented and the problem solutions should be clearly explained. Even though you have worked as a group on this project, the report should be all your own work. There are marks for the clarity as well as the correctness of your mathematical arguments. Please hand in your report to Alan Turing reception by 1pm on Friday 30th November. You should attach a cover sheet to your report and put your group number on the front page.

5 There are 25 marks for this project. (a) 16 marks for the solutions to the problems. (b) 3 marks for the homework. (c) 2 marks for the presentation of your report (clear explanations and layout). (d) 1/4 of the average mark for your group for (a) out of 4. Any student who does not attend one group session, without good reason, will get half the marks for (d). If a student misses both group sessions, without good reason, they will get 0 marks for (d). Please notify the School as soon as possible if you miss a session and fill in a Self Certification Form available from the Alan Turing Building reception.