Department of Mathematics

IDEA: Internet Differential Equations Activities


In calculus you were introduced to the most basic population growth model, the exponential growth model, described by the differential equation

x' = ax.

In this model x is the population of a given species (often a bacteria or similar microorganism), and thus x' represents the change in population per unit time. If time is measured in days then the parameter a, known as the growth rate, is positive and has units of 1/days.

Problem 1. Compute the general solution to the exponential growth model described above. For each x0 >= 0 describe the asymptotic behavior of the initial value problem x' = ax, x(0) = x0. Why do you think that this model is often used to describe the growth of a bacteria colony but is rarely used to describe the population growth of elephants?

In more sophisticated models the growth rate is assumed to be a function of the population x. If the growth rate is denoted by r(x) then the differential equation describing population growth take the form

x' = x r(x).

Here r(x) = a has units of 1/days. In the exponential growth model we assumed that r(x) = a. The next logical step in developing our model is to assume that r(x) is a linear function of the form r(x) = a-bx where a and b are nonnegative constants. Note that r(x) is a decreasing function. If the population is small then the growth rate is large and if the population is large then the growth rate is small, or even negative. The x-intercept of r(x) is at x=a/b = K. K is known as the carrying capacity because at x = K the growth rate is zero and hence the population remains unchanged at a population of K.

When r(x) = a-bx the differential equation becomes

x' = x (a-bx)

and is known as the logistic growth model.

Problem 2. It is often desirable to nondimensionalize a differential equation in order to reduce the number of parameters. Because the carrying capacity limits the population size we want to introduce a new variable y so that the carrying capacity at y is at 1. If we let y = x/K then when x = K (the carrying capacity in x-coordinates), y = 1. In other words y represents the fraction of the population relative to the carrying capacity. Use this change of coordinates to show that the new differential equation in y-coordinates is y' = a y (1 -y).

Problem 3. Verify analytically that y(t) = 0 and y(t) = 1 are solutions to this differential equation. Because these solutions are independent of time they are known as equilibria. Describe the stability of each of these points. Describe the asymptotic behavior of all solutions with initial conditions between 0 and 1. Interpret the behavior of the solutions in ecological terms. Is this model better than the exponential growth model described above? Why or why not?

Problem 4. Suppose that r(x) is a differentiable function such that r(x) > 0 for 0 <= x < 1, r(1) = 0, and r'(1) < 0. Show that the qualitative behavior exhibited by the differential equation x' = x r(x) in the region 0 <= x <= 1 is identical to that described in problem 3. What does this say about our assumption of linearity?

Of course, most populations do not live a life unmolested by outside influences. Next we will examine the dynamics of a population that is affected by harvesting. Let's consider a plant population that grows according to the logistic growth model and is harvested by cattle. We add a harvesting term to the logistic differential equation to give the new differential equation

x' = x (a-bx) - h(x)

where h(x) represents the harvest rate of plants.

The harvest rate is traditionally modeled by a function of the form

h(x) = qsx/(x + .05 d)

where q is the maximum amount of vegetation eaten per head of cattle per day, d is the plant population at which the animal is 95% full, and s is the number of head of cattle stocked. Thus h(x) has units of vegetation/day. Because the parameters qand d are innate properties of the cattle, the only parameter that cattle managers can control is s, the number of cattle allowed to graze.

Problem 5. Graph h(x) and provide an interpretation of its properties in terms of the model. In particular, what happens as the value of s is changed? Note that in the graph below the horizontal axis is x, the number of plants, and the vertical axis is h(x), the amount of vegetation harvested per day.

Putting everything together gives us the differential equation

x' = x (a - bx) - qsx/(x + .05d).

Problem 6. Using the change of coordinates given in problem 2 (y = x/K) and by introducing a nondimensional time T = at show that the nondimensional version of this differential equation is

dy/dT = y (1 - y) - s1 y/(b1 y + .05)

where s1 = qs/ad and b1 = k/d. (Hint: dy/dT = dy/dx dx/dt dt/dT. Use the equations y = x/K and T = at to compute dy/dx and dt/dT.)

The parameter s1 is known as the dimensionless stocking variable (why?) and it is our goal to understand what happens to the plant population y as the stocking variable changes.

Problem 7. Recall that the first term of this differential equation g(y) = y (1 - y) is the nondimensional growth rate and the second term c(y)=s1y/(b1+.05) is the nondimensional consumption rate. Graph g(y) and c(y) on the same set of axes for various values of s1. Explain why an intersection of these two curves corresponds to an equilibrium of the differential equation. Show how you can use the graphs of g(y) and c(y) to determine the stability of the critical points. Interpret this in terms of the model.

Problem 8. The applet below plots solutions to the differential equation derived in problem 6 with b1 = 4/3. You can change the parameter s1 using the second set of up/down buttons. For s1 = .1, .2, .3, .4, .5 describe the asymptotic behavior of all solutions (be sure to take multiple initial conditions to get a complete picture!). Describe how increasing the stocking rate s1 affects the plant population. Does this make sense?

Problem 9. Compute the bifurcation diagram for this differential equation. In other words, plot the value of the critical points as a function of s. Indicate curves of stable critical points using solid lines and curves of unstable critical points using dashed lines. Interpret this diagram in terms of the model.

The Bureau of Land Management

With the advent of HTML5, Javascript is now ready for prime time for mathematical applications. There are new Javascript demos illustrating how we might use interactive web objects to help students learn Calculus.

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