09/29/11 - Autonomous ODEs

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09/29/11 - Autonomous ODEs

			General first-order ODEs have the form y' = f(t,y).

			First-order ODEs of the form y' = f(t) can be solved simply with integration.

			First-order ODEs of the form y' = f(y) are called autonomous because they depend explicitly only on y. There is no explicit dependence on the independent variable t.

			An interesting property of autonomous ODEs is that their direction fields do not vary in the horizontal (t) direction.

			Hence, a solution with initial condition y(3) = 1 is just a time-shifted version of a solution with initial condition y(0) = 1.

			The roots of the function y' = f(y) = 0 correspond to equilibrium values (where the solution y approaches a constant), since no change occurs once y' = 0 (zero rate of change).

			Since there is no change in the direction field or the solution as the initial condition is shifted in time, graphs of autonomous solutions can be reduced to one dimensional representations, called phase lines, which show only the vertical change.

			Application of Autonomous First-order ODEs

			The duck population x around a hunting lodge is modeled by the ODE where H is the constant harvesting rate. Answer the following questions without solving the ODE.

			a) If each hunter is allowed to shoot up to 20 ducks per year, how many licenses can be issued per year so that the duck population has a chance of survival (x' > 0)?

			Note that the ODE above is autonomous: x' = f(x). So plot the function f(x) versus x in Graphing Calculator.

				Click on the "small mountain" icon to zoom out: Notice that the value of this function (vertical height) is the value of the rate of change of x (the fish population) as a function of x (the fish population). This graph tells us there are two points where the rate of change of the fish population is zero: near x = 0 and near x = 1000. To the left of x = 0, the function is negative, telling us the rate of change of the population (x') is negative, i.e., the population is decreasing (as if x could be negative). To the right of x = 1000, the function is again negative, again telling us the rate of change of the population (x') is negative, i.e., the population is decreasing. Between x = 0 and x = 1000, the population is positive, meaning the population is increasing. The rate of change of the population is positive but small for x a little above 0 and for x a little less than 1000. The population is growing fastest when the population x is near 500. We would describe the points where the curve crosses the x-axis as the two equilibrium points for this population. At those two points the function (which happens to be the rate of change of the population) is zero, so the population will neither increase nor decrease. The equilibrium point on the left (near x = 0) is unstable, because to the left of that point the function (rate of change) is negative, meaning the population will decrease (x will move to the left, away from the equilibrium point) and for points to the right of the equilibrium point near x = 0, the function (rate of change) is positive, meaning the population will increase (x will move to the right, away from the equilibrium point). The equilibrium point on the right (near x = 1000) is stable, because to the left of that point the function (rate of change) is positive, meaning the population will increase (x will move to the right, back toward the equilibrium point) and to the right of that point, the function (rate of change) is negative, meaning the population will decrease (x will move to the left, back toward the equilibrium point). Now create a slider for the parameter H so that we can assess the effect of harvesting ducks. Type H=slider(0,400) into Graphing Calculator on a new line: When you press RETURN (or ENTER) the graph should appear: Start sliding the blue dot (slider) to the right to increase the value of H: Note that the function moves down (as you would expect when you subtract a constant from a function. Also note that this moves the two equilibrium values towards the center. In the graph above, the left equilibrium point has moved to about x = 60 and the right equilibrium point has moved to about x = 940. That means with that level of harvesting (H = 60), if the population drops below about x = 60 (or above x = 940), the rate of growth of the duck population turns negative and the duck population goes extinct. As we continue to move the slider to the right (harvesting more ducks), the equilibrium points approach each other until finally we get:

				b) Suppose N licenses are issued, where N is the maximal number found in part (a). What values of the initial duck population lead to total extinction of the species? Explain.

			Homework: