Phase line (mathematics)

Transcription

'Phase diagrams' are: "a way to visualize solutions to autonomous ordinary differential equations". And in this video we'll consider the case of just one dependent variable, x. And so an autonomous ordinary differential equation will have the form: dx/dt = f(x). Now what's 'special' about this form? What makes it 'autonomous'? The right-hand side does not include the independent variable (the independent variable is not used in the definition of the differential equation.) I'll show you how to construct phase diagrams through an example: Let's consider the differential equation: dx/dt = x*(1 - x) This is 'autonomous' because the independent variable 't' does not appear on the right-hand side. Now phase diagrams are kind of related to 'slope fields'. So I'll start by drawing the slope field for this differential equation (as I've done here.) I'll remind you how to draw the slope field: You pick a point (t,x) in the t-x plane, and you imagine that you have a solution passing through that point. The tangent-line to that solution will have slope 'dx/dt'. But the equation says that 'dx/dt' is the same as 'x*(1-x)'. So if you go to that point,and you draw in a little line which has slope x*(1-x) that will be a tangent-line the solution. If I do that for a bunch of points in this plane, I begin to get a picture of how the solutions will ..'flow'. Now notice a special property of this slope field: because the equation is autonomous, the right-hand side does not depend on 't' so the slope field will not change as I move from left to right. If I choose a particular 'x' and I evaluate the slope by evaluating 'x*(1 - x)' then I can move from left to right in other words I can change the value of 't' and I still have the same slope. Phase diagrams take special advantage of this left-to-right symmetry in the slope, in order to collapse this two-dimensional picture down into a single line. So here's our phase diagram, at least the beginning of it and since it's a line, we'll call it a 'phase-line'. Now, before we draw any more of this phase-line, let's sketch in a few solutions to this differential equation. So I go with the flow of the slope field and sketch in some solutions in orange. Now notice we have two particular solutions, which are special: and those are the solutions x(t) = 1 and x(t) = 0, 'constant' solutions. Those correspond to the points where dx/dt = 0 and the solution is not changing. For all of the other solutions, it will change with time. And so I mark those guys in here by filling in an orange 'circle' corresponding to the points x = 0 and x = 1. So these are the guys where dx/dt = 0: the solutions are not changing. What about elsewhere on the phase-line? Well, if I start out above x = 1 then my solution will decrease. If I start out between x = 0 and x = 1 my solution will increase. If I start out below x = 0 then my solution will decrease. So that corresponds to 'different signs' for the derivative: Above x = 1, dx/dt (by this formula) is negative. Between x = 0 and x = 1 (again, by this formula) dx/dt is positive. And below x = 0, dx/dt is negative. As a shorthand, we can mark those signs on our phase-line by drawing in little 'arrows': At these orange points the solution will 'stand still' (so I just draw a little 'circle') Now in between the orange points - say right here - the solution will increase. So I draw a little arrow going up. Above the top orange point - at x = 1 - the solution will decrease. So I draw an arrow going down Now (likewise) for below x = 0, I draw an arrow going down And that's it: that's our phase-line. So we have a line, with dots to indicate where the solution is standing still and arrows to indicate where it's moving, and in which direction it's moving. So let me give you a little vocabulary that will help you talk about the phase-line. And the first thing is that these points, - where the solution is standing still - these are called 'equilibrium-points' (in our case x = 1 and x = 0) 'Equilibrium' meaning that the solution is not changing. Now, these equilibrium points have different properties. This equilibrium point: at x = 1 we call a stable equilibrium and we call it 'stable' because both of the arrows are pointing-in. So you imagine that if you were to start off close to x = 1 (either about that or below it) you would 'tend toward' x = 1 and that's what makes it stable (it's like the bottom of a valley.) The other point: [at x = 0] the arrows are pointing away from that equilibrium and so it's called an 'unstable' equilibrium (imagine like the top of a mountain.) So if I start at anywhere near x = 0 then I'll move away from x = 0. Similarly, if I were to concoct a differential equation which had 'this' as part of its phase diagram. where an arrow was pointing-in and an arrow was pointing-out, I would also call that 'unstable' (because it is 'not-stable') Now I have a task for you: since this type of unstable equilibrium did not appear in our example, come up with an example of an equation that exhibits this type of unstable equilibrium. that is: one arrow going into the equilibrium and one arrow going out. Now, after you have figured out how to do that, I have a question: is there any phase-line any configuration of stable and unstable equilibria which 'could not' occur as the phase-line for a differential equation? In other words could I 'glue together' a whole bunch of these guys in some line which could not possibly be a phase-line for a differential equation? Or is it true that any picture I draw I can come up with an example of a differential equation which has that picture as its phase-line?