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Unformatted text preview: Fall Semester 04-05 Akila Weerapana LECTURE 19: Phase Diagram Analysis I. INTRODUCTION Todays lecture discusses how to analyze the dynamics of a system of non-linear difference or differential equations using algebraic solutions and using phase diagrams. The phase diagrams in the two variable case are more complicated than in the single variable case because we have to keep track of the motion of two variables. However, this phase diagram technique will become extremely useful when we are dealing with a system of non-linear equations. It will also be useful when we do not want to explicitly solve a system of linear differential equations, but instead prefer to do comparative static analysis with the dynamics of the model. We will first work with a phase diagram for a system of linear differential equations and then draw the phase diagram for a system of linear difference equations. II. PHASE DIAGRAMS FOR STABLE SYSTEMS Example 1 Consider the following system of first order differential equations y t =- 4 y t + 2 x t + 32 x t =- 2 x t The solution to this system is y t = Be- 2 t + De- 4 t + 8 x t = Be- 2 t We can see that this is a stable system. Now look at the phase diagram. The locus of points corresponding to x = 0 is easy, that is given by the equation x t = 0, i.e. the y-axis. The locus of points corresponding to y = 0 is 4 y t = 2 x t + 32. The intersection of the two lines represents the set of points for which x = 0 and y = 0. The two lines intersect at x = 0 and y = 8. To the right of the x = 0 line, x > 0 so x < 0, i.e. x decreases over time (a left arrow in the diagram). Points to the left of the x = 0 locus correspond to x increasing over time (a right arrow in the diagram). Similarly, to the right of the y = 0 line, x is increasing and y is decreasing so y t =- 4 y t + 2 x t + 32 > 0 In other words y is increasing over time (an up arrow in the diagram), while points to the left of the line are associated with decreasing y over time (a down arrow in the diagram). The phase diagram would look as follows.- 6 ? y t x t y t = 0 x t = 0 (0 , 8) ?- 6- ? 6 Notice that wherever we start, the dynamics of the system push us directly towards the steady state....
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This note was uploaded on 01/11/2012 for the course ECON 200 taught by Professor Riley during the Fall '10 term at UCLA.
- Fall '10