Calculus Online: Lab 4

Welcome to Lab 4 of Math 101, Sections 203, 204, 207 and 209



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We had some technical problems with our server on Monday. As a result, the due date for this lab has been extended until 11:59 P.M. on 17 March 1999.


Instructions

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Qualitative Analysis of Differential Equations

In this section we are going to study the solutions of differential equations of the form

\[ \frac{dy}{dt} = f(y) \]

By now you know, at least in principle, how to solve a separable differential equation like this. We can compute the integrals

\[ \int_0^t\frac{1}{f(y(T))} \frac{dy}{dT}dT = \int_0^t 1 dT \]

to obtain an algebraic expression relating y(t) and t, namely

\[ \int_{y(0)}^{y(t)}\frac{1}{f(Y)} dY = t \] .

However, there are a couple of drawbacks to this approach.

First, finding antiderivatives to help us evaluate the definite integral can be difficult, if not impossible. Then if we are able to evaluate the integral, we need to solve for y(t) as a function of t and this can be difficult as well. (Remember how much work it was to solve the logistic equation?)

Furthermore, we are usually not particularly interested in the exact solutions of these equations. For example, if we are modelling the growth of a biological population with a differential equation, we don't expect that the resulting numerical answers to be all that accurate since the differential equation is based on a number of assumptions and simplifications.

Generally, we are more interested in questions like, "will the species become extinct?" or "will the population grow if its food source increases?" In this lab, we are going to look at some ways to determine the general behaviour of the solutions of a differential equation without having to calculate its solutions directly.

We have seen in class that there are some important solutions to a differential equation that we can find fairly easily---the constant solutions which we called critical points. Remember that for a differential equation $ \frac{dy}{dt} = f(y) $ , a constant solution is one which satisfies

\[ \frac{dy}{dt} = f(y) = 0. \]

Consider, for example, the differential equation $ \frac{dy}{dt} =-y^3-y $ so here we have $ f(y)=-y^3-y $ . The plot on the right shows a graph of this function. From the graph we can see that f(y) = 0 only when y = 0.

From the graph we see more than this. Whenever $ y > 0 $ , we have $ \frac{dy}{dt} = f(y) < 0 $ , so y is decreasing. Also, when $ y < 0, $ we see that y is increasing. We have drawn arrows on the axis to indicate this behaviour.

From the arrows on the diagram we can see one more thing. Solutions starting near y = 0 get closer to y = 0 as time goes on. We say that the critical point y = 0 is stable.

If, on the other hand, the situation is like the one to the right, we see that solutions that start near the steady solution y = 0 get farther away as time progresses. In this case we say that y = 0 is an unstable critical point.

Question 1a (2 points)

Which of the following plots could be the graph of a solution of the differential equation

\[ \frac{dy}{dt} = -y^3 - y. \]

a)

b)

c)

d)

e)


Now we know how to determine something about the general behaviour of the solutions of some differential equations without having to solve the system. Another reasonable thing to do is to see how the general behaviour changes as we change the environment of the system.

Many differential equations arising from models contain parameters that describe variable but fixed conditions of the system being modelled. These can be things like the mass of a weight, the temperature of the ocean, or the amount of food available to a species. Sometimes, the general behaviour of solutions of a differential equation changes radically when these parameters change even a small amount. Critical points can vanish, new ones can appear, and the stability of equilibrium solutions can change from stable to unstable (or vice versa). We are going to analyze solutions of

\[ \frac{dy}{dt}=-\frac{y^3}3 +ay \]

as the parameter a changes.

Question 1b (6 points)

We are going to make a plot showing how the number and stability of the steady solutions of $ \frac{dy}{dt}=-\frac{y^3}3 +ay $ depend on a using the plot below. On the left hand side is a plot of $ -\frac{y^3}3 +ay $ . You can change the value of a by dragging the red dot on that diagram. On the right hand side is an empty plot. You can put dots representing stable solutions and squares representing unstable solutions into the plot by dragging on the dot and square above the plot. For example, if y = 0 is a stable equilibrium when a = -0.2 (and it is), then you would put a dot at the point (-0.2, 0). (We have done this one for you.)

For the values of a equal to -1.0, -0.6, -0.2, 0.2, 0.6, and 1.0, place appropriate boxes or squares in the plot on the right corresponding to each of the steady solutions of the differential equation for that parameter value.

When you were creating your plot you should have noticed something special happens around a = 0. When $ a < 0 $ , there is only one steady critical point. When $ a > 0, $ there are three critical points. The change that occurs when a = 0 is called a bifurcation.

The main idea here is that the solutions to the differential equation behave quite differently for different values of the parameter a. But you might be wondering to yourself why we might be interested in critical points or bifurcations. The next section shows how these ideas have significant meaning in a model problem.

Spruce Budworms

In this section we are going to look at the growth of the spruce budworm, a species of insect which is considered a pest by the logging industry in Ontario where it consumes valuable softwood resources. Our objective is to consider a simple model for the population of the budworm and to see how a bifurcation that occurs in this model has real meaning about the behaviour of the population of the species. The variable we will work with is B, the population density of the budworms in the forest at a given time. So B is the average number of budworms per hectare in the forest.

We have seen before that the logistic equation,

\[ \frac{dB}{dt} = k B ( 1 - \frac BM) \]

where k and M are constants, gives a model for the growth of a population B. Now we are going to introduce another effect into the model, that of predation. We will introduce the predation rate P(B) which describes the rate at which budworms are eaten by birds.

A few things seem pretty clear. First, if there are no budworms, the predation rate should be zero; that is, P(0) = 0. Also, the more budworms there are, the more will be eaten. This means that P(B) should be an increasing function. Finally, there are only so many budworms that a bird can eat in a day. Consequently, we expect that the predation rate should level off at some point.

To take this features into account, we will adopt the model

\[ P(B)=p \frac{B^2}{a^2+B^2}. \]

The diagram to the right shows a plot of this function. The parameter p is the maximum level of predation and the parameter a describes how quickly this maximum is achieved as the budworm population grows.

The rate of change in the budworm population is the usual rate of change given by the logistic equation minus the rate at which they are consumed by predators. So the model we have arrived at for the spruce budworm population is

\[ \frac{dB}{dt}=kB(1-\frac{B}{M}) - \frac{pB^2}{a^2+B^2} \] .

That's a pretty complicated looking equation which could be very difficult to solve. So we will use the qualitative methods that we discussed in Question 1.

Our first step is to simplify this equation. If we define a new variable b = sB with an appropriately chosen value for the constant s, the equation becomes

\[ \frac{db}{dt}= Rb(1-\frac{b}{Q}) - \frac{b^2}{1+b^2} \] .

We won't go into the details of this transformation here (you can do it for yourself if you want). The variable b is a measure of the budworm population. The parameter Q is a fixed number similar to the carrying capacity of the budworm population.

Most importantly, the parameter R is a measure of how quickly the budworm population grows. In young forests, R is small but as the forest matures, R becomes a larger number. We are going to look at what happens to solutions of the differential equation as we change R.

To begin, we notice that b = 0 is always a solution. We can factor the right hand side of the differential equation to get

\[ \frac{db}{dt}= b\left( R(1-\frac{b}{Q}) - \frac{b}{1+b^2} \right) \] .

This make senses: if there are no budworms, there will never be any budworms.

To analyze the system further, we see that any other constant solutions happen at values of b where

\[ R(1-\frac bQ) - \frac{b}{1+b^2} \]
is zero. We can think of this expression as a balance between the growth of the species g(b) = R(1-b/Q) and predation on the species p(b) = b/(1+b2). So equilibria happen when g(b) - p(b)=0.

In the diagram below you see a plot of g(b) - p(b). You can use the dot on the diagram to change the value of R between 0.1 and 2. As you do this, you will see that the number of constant solutions change. This means that bifurcations occur as we change the value of R.

Question 2a (4 points)

There are two bifurcations that occur between R = 0.1 and R = 2.

Enter the value of R at which these bifurcations appear into the boxes below.

Using the diagram below, we can see what happens to the budworm population over time. You can change the initial budworm population by moving the dot on the b axis. Then press the Start button to see a plot develop that shows the budworm population. You can pause the simulation using the Stop button and continue it with the Resume button. (The Stop button appears when the simulation is running and the Resume button appears when it is stopped). You can reset the simulation and start over with the Reset button. At any time, you can change the value of R in the diagram.

Go ahead an try out the simulator before going on to the experiment.

Now we are going to do an experiment to show some interesting behaviour in this system.
  1. Reset the simulation by pressing the Reset button.
  2. Move the initial condition dot to a value near 0.5.
  3. Set the value of R to 0.56.
  4. Click on the Start button and wait for the system to reach its new equilibrium. You will see that the system has reached equilibrium when the plot becomes constant. Notice the equilibrium value.
  5. Click on the Stop button and increase the value of R just a bit to 0.57.
  6. Start the simulation again with the Resume button. The system starts up again with a new value of R and an initial condition of the old equilibrium. Wait for the system to come to its new equilibrium. Notice that the equilibrium value is not much different than when R is 0.56.
  7. Click on the Stop button and increase the value of R just a bit more to 0.58.
  8. Start the simulation again with the Resume button. This time, the system will take longer to come to equilibrium. Be patient; the change starts slowly but becomes obvious as time progresses. Why did the value of b change so much?
  9. Click on the Stop button and decrease the value of R back down to 0.57.
  10. Start the simulation again with the Resume button. Wait for the system to reach equilibrium. It doesn't go back down to the low levels. You have an outbreak of spruce budworms on your hands!

Now go back to the plot of g(b)-p(b) and see if anything interesting happens in that plot around the value of R = 0.57 that might explain the phenomenon you have just seen. How does this relate to your answers in Question 2a?

Question 2b (2 points)

We have seen in the experiment that if R is increased to 0.58 the budworm population explodes, and that it doesn't come down when R is lowered to 0.57. So the question is, what is the largest value of R that will bring the population back down again to, say, a value of b less than 2?

There is an easy way and a hard way to answer this question. You can go back to the simulator and experiment like you did to see the population explode. (This is the hard way). You can also think about what change made the population explode and see from the plot of g(b) - p(b) the value of R that will reverse this change.

Enter the value of R at which the the population is brought under control into the box below.



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