18.03 Class 8, Feb 19, 2010 Autonomous equations [1] Logistic equation [2] Phase line [3] Extrema, points of inflection Announcements: Final Tuesday, May 18, 9:00 - 12:00, Johnson Track Hour exam next Wednesday: Rooms to be announced Monday. Please go to the hour you are registered for. 50 minutes. [1] Back to qualitative study of differential equations. I'll use (t,y) today. The general first order equation is y'(t) = F(t,y) Autonomous ODE: y'(t) = g(y) . Autonomous means conditions are constant in time, though they may depend on the current value of y . Eg [Natural growth/decay] Constant growth rate: so y' = k0 y . Eg Bank account with interest rate NOT depending on time but possibly depending upon current balance, and constant savings rate: y' = I(y) y + q . Extended example: Population model with variable growth rate k(y) depending on the current population but NOT ON TIME; so y' = k(y) y. Suppose that when y is small the growth rate is approximately k0 , but that there is a maximal sustainable population p , and as y gets near to p the growth rate decreases to zero. When y > p , the growth rate becomes negative; the population declines back to the maximal sustainable population. In the simplest version of this, when you graph of k(y) against y you get a straight line with vertical intercept k0 and horizontal intercept p : k(y) = k0 (1 - (y/p)). so k(0) = k0 , and k(p) = 0 . When y > p , k(y) < 0 . The Logistic Equation is y' = k0 (1 - (y/p)) y = g(y) . This is more realistic than Natural Growth when you want to account for limits to growth. It is nonlinear but it IS autonomoous.Autonomous equations are always separable, but we aim for a qualitative grasp of solutions. I sketched the nullcline: y = 0 and y = p . Clearly all the isoclines will be collections of horizontal straight lines - the differential equation is constant in time. To get a clear idea of the other isoclines, I will draw a graph of g(y) as a function of y . It's a parabola opening downward, meeting the horizontal axis at y = 0 and y = p . This says that for y < 0 the slopes are negative for 0 < y < p the slopes are positive for y > p the slopes are positive. I drew more isoclines and some solution curves. The bounded solutions are "logistic curves" or "S-curves." The represent the population drifting from just above one equilibrium (no oryx) towards the stable population (p oryx). If the population exceeds the maximal stable population, it falls back towards it. The max stable population is a "stable equilibrium;" the zero population is "unstable." Values of y for which g(y) = 0 are called "critical points" of the differential equation y' = g(y) . The are also called "equilibria." [2] Phase line Since the direction field is constant horizontally, its essential content can be conveyed more efficiently. Draw a vertical line. Mark on it the equilibria, where g(y) = 0. In between them, draw an upward pointing arrow if g(y) > 0 and a downward pointing arrow if g(y) < 0 . This simple diagram tells you roughly how the system behaves. It's called the "phase line." Question 8.1. In the autonomous equation y' = g(y) , where g(y) has a graph which I sketch, looking like g(y) = y^3 - y , is the rightmost critical point 1. Stable 2. Unstable 3. Can't tell, could be either Blank: don't know This can be made clear by sketching the phase line. Ans: Unstable. In terms of the graph of g(y), the stable equilibria occur when g'(p) < 0 , unstable when g'(p) > 0. Now, the Kenyan government wants to establish a game preserve on which it will allow the hunting of oryx. It wants to sell licenses to kill 250 oryx each year and wants to know how big a preserve to establish to guarantee that this is sustainable. It's known that the population (measured in kilo-oryx, without hunting) follows the logistic equation y' = (a-y)yHere a is the stable population of oryx, in the absence of hunting. It is proportional to the area of the preserve. With our 1/4 kilo-oryx per year hunt, then, we have y' = g(y) =(a-y)y - 1/4 = - ( y^2 - ay + 1/4 ) I invoked the Mathlet <Phase Lines> to visualize what happens. For small a , the population crashes. When a is increased to a critical size, a single equilibrium appears, which is "semistable." You can watch the graph of g(y) at the lower left of the applet; it's a parabola, opening downward, and it rises as a increases till at a = 1 it has a single root. We can solve for where the critical points are, using the quadratic formula: y = a/2 +- (1/2)(sqrt(a^2 - 1)) There is just one exactly when the square root is zero, ie a = 1 (or a = -1 ). So the Kenyan Chamber of Commerce recommends a preserve of this size, a = 1. Is this wise? if the population falls every so slightly below the critical population (which seems to be y = 1/2), it crashes. When you increase a further, things get better. There are two critical points. The farther they are from each other, the more stable the population system is. I returned a to zero and opened the Bifurcation Diagram. and watched what happened as a increased. The bifurcation diagram puts all the phase lines, for various a , together. [3] In an autonomous equation y' = g(y) , the conditions represented by the ODE are constant in time. Direction fields are constant in the horizontal direction. Consequently, any horizontal (time) translate of a solution is another solution. A "time translate" of a function y(t) is a function y(t-t_0): the graph gets shifted horizontally (to the right) by t_0 units. Ex. Solutions of y' = k_0 y exhibit three different behaviors, illustrated by y = e^{k_0 t} , y = 0 , y = - e^{k_0 t} Any solution is a horizontal translate of one of them: any solution is either y = e^{k_0 (t - t_0)} , y = 0 (whose only time-translate is itself) , or y = - e^{{k_0 (t - t_0)} Question 8.2. Solutions of autonomous equations can have astrict local maximum. (A strict local maximum for f(t) is a time t = a such that f(a) > f(t) for all t near but not equal to a . ) 1. True 2. False Extreme points occur where y' = 0 , i.e. where g(y) = 0. These are the constant solutions, and they don't have strict maxima or minima. So it's true, solutions can't have strict
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