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ODE  Bifurcation diagrams

It Started, The Bifurcation of Time

Saddlenode bifurcation
Transcription
I'll introduce now the idea of a 'bifurcation diagram'. Bifurcation diagrams are way study differential equations which may depend on some extra parameter. Now you've seen such a thing already. When I first introduced the Logistic equation I also discussed the addition of a harvesting term: this 'h'. So this differential equation would describe a population, 'x' governed by this Logistic model. But then I have a 'h' and that represents some constant rate of harvesting applied to the population, 'x'. Now the way to study such a thing, and how it depends on the parameter, 'h' is essentially to draw a lot of phaselines. And we draw them for various values of the parameter, 'h'. So let's look first at the phaseline for 'h = 0'. So it looks like this: and the equilibria are at x = 0 and x = 10. And how I find those? Well, the equilibria are where dx/dt = 0. So I take this righthand side and I set it equal to '0'. And I remember that I've set h = 0 in order to draw this phaseline, So I have (1/10)*x*(10  x)  0 = 0 and that means x = 10 and x = 0. So those are the two equilibria. And then I can test of the sign of 'dx/dt' above x = 10 by plugging it in this righthand side again with 'h = 0'. And I find it's negative, so the solution goes down. Inbetween x = 0 and 10, it's positive, so the solution goes up, And below x = 0 dx/dt is negative, so the solution goes down. How about the phaseline for 'h = 1'? I did 'h = 0' right here, so I'll move over one unit and do 'h = 1' right here. Now it looks roughly the same, except the two equilibria have moved in a little bit more And how I know that? Well I set h = 1 in this equation and I get (1/10)*x*(10  x)  1 = 0, and solve for the two x values  which are the equilibria. And those guys are at: x = 5 + 10^(1/2) and x = 5  10^(1/2). So you see that they've moved in a little closer just inched in toward.. '5' Okay, how about 'h = 2'? At h = 2 something kind of special happens: I only have one equilibrium value. And above and below that equilibrium dx/dt will be negative. I no longer have an area where dx/dt is positive, like on the two previous phaselines. And, where do I get all this information? Well, I set 'h = 2' in the righthand side of this equation, and I get (1/10)*x*(10  x)  2 = 0. There's only one route to that equation: that's 'x = 5' and I can test that this quantity is negative for x > 5, and negative for x < 5. I can keep drawing more phaselines, for instance 'h = 3' I'll find there are no equilibria and, everywhere, dx/dt is negative. And, again, I find that by simply plugging in 'h = 3' into the righthand side of the equation, setting it equal to '0', and finding there are no realroots 'x' and hence no equilibria. In fact that quantity on the righthand side is always negative hence the arrow is pointing down everywhere. Okay I've taken away all the calculations What we're left with is the beginning of a bifurcation diagram. I've drawn phaselines for various values of 'h' And we can see in those phaselines the bifurcation  or the 'splitting' of equilibria. For instance, at 'h= 3', we have no equilibrium values. At 'h = 2', one equilibrium appears. Then, at 'h = 1' we have two equilibria. At 'h = 0' I still have two [equilbria] Now, you can imagine drawing in other phaselines, in between these: say at 'h = 0.5' or 'h = 1.5' (etc.) And we'd (sort of) get a 'smooth' picture of the way that these equilibrium values split and begin to move apart. So I can draw that smooth line in (it's a parabola in fact.) And how do I know it's a parabola? Well, it's exactly described by this equation: (1/10)*x*(10  x)  h = 0. And that is a parabola in the x,h plane So if I draw in these axes, here, make this the 'haxis' and these are the various values of the 'h' that I've chosen, h = 0... h = 1... h = 2 (etc) and this is the xaxis represented by the vertical position on the phaseline, then that's the parabola: it's a sidewaysparabola given by this equation. So that pretty much completes our picture. We now have a bifurcation diagram, which is the bifurcation diagram for the example equation that I gave, and that is the Logistic equation with some harvesting parameter, 'h'. And we can use that bifurcation diagram to study how this Logistic model depends on the harvesting, 'h'. And if you remember, that was our original goal: to study how an ODE depends on a parameter. And in fact we can answer some questions about 'harvestingrates'. Now let's suppose (just to make things a little more real) that 'x' represents a population of fish let's say in some lake. So that means, without any harvesting, that the population of fish is governed by a Logistic model, given here. Now, 'h' would represent some fishingrate: some rate at which I catch fish and remove them from the lake. Now my question for you would be: what is the maximum sustainable fishingrate? At what rate do I know that I can fish without killing off the fish population entirely?
Science and technology
 Bifurcation theory, the study of sudden changes in dynamical systems
 Bifurcation, of an incompressible flow, modeled by squeeze mapping the fluid flow
 River bifurcation, the forking of a river into its tributaries
 Bifurcation lake, a lake that flows into two different drainage basins
 Bifurcated bonding, a single hydrogen atom participates in two hydrogen bonds
Other uses
 Bifurcation (law), the division of issues in a trial
See also
 Aortic bifurcation, the point at which the abdominal aorta bifurcates into the left and right common iliac arteries
 Tracheal bifurcation, or the carina of trachea (Latin: bifurcatio tracheae)
 Bifurcation diagram
 Bifurcate merging, a kinship system
 False dilemma or bifurcation
 Tongue bifurcation (disambiguation)
 Fork (disambiguation)
 All pages with titles beginning with Bifurcation
 All pages with titles containing Bifurcation