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//**************** 2D Bifurcations - February 5th, 2019 *********************//
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- Well, pray and hope and pray, I suppose
- (in other news, the projector clock is extremely confused and thinks it's Monday. No, projector clock! You have been lied to - by yourself!)
- What we're going to do today:
- Go through an example of this 2D analysis
- Talk about bifurcations in 2D
- IMPORTANT ANNOUNCEMENT: We don't have class on Thursday (since Professor Vuduc will be out of town), so instead, work on your projects during that time - a TA will be having office hours instead of class. Keep working!
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- So, last time, we left the comfortable world of 1D linear systems and began looking at 2D linear systems - and then briefly say how we can use the Jacobian to at least kind of guesstimate the behavior of systems around their fixed point
- Essentially, the Jacobian does this by linearizing the equation components to get us an approximation for points near the critical point we're looking at
- To restate our 2nd-order stability cases:
- Real part (lam0), real part(lam1) > 0 => Repeller
- Re(lam0), Re(lam1) < 0 => Attractor
- "If you didn't do the exercise from last time on your own, you should do it, just to make sure you do know how to do it"
- So: example time. Let's say we have a field (a big, beautiful, grass-green field), and within the field are two mutual enemies locked in a struggle for survival: sheep and rabbits!
- By the way, this is the Lotka-Volterra population model for competing species, taken from Strogatz chapter 6.4
- So, rabbits produce faster than sheep, but the sheep are bullies: they can push the rabbits away to get the grass all by themselves
- To model this population, we'll basically use a "logistic - competition factor" equation
- For the sake of time, let's say that we know the coefficients in advance:
drabbits/dt = rabbits*(3 - rabbits - 2*sheep)
dsheep/dt = sheep*(2 - sheep - rabbits)
- So, the higher either population is, the lower the growth rate of either one (in equation form, the multiplication of sheep*rabbits shows this competition-style relationship)
- So, to analyze this system, we need to 1) Find the fixed points 2)
- There are 4 fixed points for this particular system; to save time, let's just give them here:
[0,0], [0,2], [3,0], [1,1]
- Next, we'll solve for the Jacobian:
[3-2*rabbits-2*sheep, -2*rabbits ]
[-sheep, 2-2*sheep-rabbits]
- Plugging in our first fixed point, we'd get:
A(0, 0) = [3 0]
[0 2]
- This gets us a trace of 5 and a determinant of 6, which suggest the point is UNSTABLE
- Continuing to do this for the rest of the points, we see that [0,2] and [3,0] are stable, and [1,1] is a saddle
- Because this is a nonlinear system, we can NOT draw the whole graph, since we only know the local behavior immediately next to the critical points - so instead, we can only label the likely field immediately next to the critical points
- However, things probably still flow smoothly, so we can *tentatively* connect those lines to suggest the most probable behavior of the system
- What we can infer from this drawing, though, is that there's not a single fixed point where the population stays for any given region (?)
- Instead, there are 2 "basins of attraction" around the stable points where solutions are likely to drift
- Now, onto the "fun" stuff - 2D bifurcations! (this sentence was said by the Professor entirely without irony)
- As you remember, in 1D, a saddle node bifurcation was where a single fixed point split into 2 (or vice-versa) as the model's parameters changed
- In 2D, the "normal form" for a saddle node bifuraction is something like this:
x0' = u - x0^2
x1' = -x1
- This system is already decoupled (the two equations aren't related), so it should be pretty clear that x0' is just a 1D saddle-node bifurcation waiting to happen
- In this 2D system, the critical points will be of the form:
x* = [+-sqrt(u), 0]
- So, u > 0 has 2 critical points (one is stable, one is unstable), u = 0 will only have one semi-stable critical point, and u < 0 will have none (solutions drift to the left everywhere)
- As an example, let's steal an exercise from Strogatz Chapter 8 (Griffith '71 model) and look at a model of gene control, where there's a liquid containing some quantity of protein and a gene that "codes" to that protein, creating more of it in exchange for itself (thereby changing the concentration) (I think?)
- The model itself looks like:
x0(t) = protein concentration
x1(t) = gene concentration
x0' = -ax0 + x1
x1' = x0^2 / (1 + x0^2) - b*x1
- So, this is a non-linear, 2D, highly coupled system - neat!
- First things first, let's try to plot the "null clines" of this graph:
f0' = 0 => x1 = ax
f1' = 0 => x1 = (1/b) * x^2/(1 + x^2)
- So, f0's null cline is just a linear slope through the origin, while f1 is a sort of logistic-y curve
- The two functions can have up to 3 intersection points (at the origin, and then twice more), which means we can have up to 3 critical points for this system
- In general, this model will be a stable point at the origin, a saddle in the middle, and then another stable point farther off, meaning that the best chance we have of keeping both genes and proteins around is to have a high concentration of each
- There were two other bifurcations we talked about in the 1D case (transcritical and pitchfork); here're their 2D normal forms:
- For transcritical:
x0' = u*x0 - x0^2
x1' = -x1
- For pitchforks, there are 2 forms (in this form, the bisection point for both is at u=0):
- The 'supercritical' form (where 1 stable point turns into an unstable point w/ 2 new stable points to the side when u > 0):
x0' = ux0 - x0^3
- The 'subcritical' form:
x0' = ux0 + x0^3
- An example of a supercritical pitchfork:
x0' = ux0 + x1 + sin(x0)
x1' = x0 - x1
- To show that this bifurcation exists, and finding the 'u' value where the bifurcation happens, is pretty hard without some hints, (hint: Taylor expansion for the sine of x is 'x - x^3/3! + ...', and approximate it with just these first 2 terms)
- ...so, um, yeah. Class just sort of ends here. Bye.