Solution Notes on Problem set #3

1.

In 1926, the Lotka-Volterra predator-prey model was proposed as a means of modeling the populations of two co-existed species where one feeds upon the other.

$$\begin{eqnarray*}\frac{d p}{dt} & = & (a- b r) p \\ \frac{d r}{dt} & = & (-c + d p) r \end{eqnarray*}$$

where a, b, c and d are positive, and p and rare functions of time representing populations. Which function represents the predator and which the prey? Why?

This is a standard problem that can be found is almost all ODE textbooks. Examining the equations, we see that the p population grows if the r population is not present. The presence of rretards growth in the p population. Clearly, this is a description of a prey population. Similarly, r is a predator population.

2.

Under what conditions (in terms of the model coefficients) if any can both species co-exist forever?

While the trivial solution p=0 and r=0 is an equilibrium point, it does not describe any kind of co-existence. However, the other solution p=c/d and r=a/b describes a nontrivial co-existence.

3.

Study the local behavior near any equilibrium points? Sketch the phase portrait of the system for each type of stability. Bonus: Use Maple or Matlab to generate a phase portrait for specific examples of these cases.

If you linearize the system as demonstrated in class near the trivial solution, you obtain

$$\left[ \begin{array}{c} \frac{dp^{(1)}}{dt} \\ \frac{dr^{(1)... ...\left[ \begin{array}{c} p^{(1)} \\ r^{(1)} \end{array}\right]$$

The eigenvalues are a and -c, and the corresponding eigenvectors would be $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$ and $\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]$. Thus, we have a saddle point with growth along the p axis and decay along the raxis.

For the nontrivial equilibrium point, the local behavior near this point is described by

$$\left[ \begin{array}{c} \frac{dp^{(1)}}{dt} \\ \frac{dr^{(1)... ...\left[ \begin{array}{c} p^{(1)} \\ r^{(1)} \end{array}\right]$$

The eigenvalues are purely imaginary, $\pm \sqrt{ac} i$, and the local behavior is a center. Note: This does not mean that the equilibrium point is necessarily surround by closed orbits (why?), even though it is the case for this particular problem.

For the bonus, I set all the parameters to be one and let Maple do the work for me.

4.

A person saves $200 per week into a retirement account beginning at age 25. She plans to retire at age 60. If the retirement account earns 5.125% interest quarterly, develop a mathematical model for the balance in her account. How much money will she have saved when she retires? (Hint: Review geometric series and partial sums of geometric series.)

This problem is similar to a compound interest problem except for the sound term. If we denote the balance in quarter n as B_n, then we can describe the evolution of the balance as

B_n+1 = (1+r)B_n + q

where r is the quarterly interest rate of 0.05125/4 and qis the quarterly deposit rate. (By the way, there are 52 weeks in a year and 13 weeks in a quarter. Remember, a month has more than four weeks.) If we expand the recurrence relation above assuming that the employee makes one last deposit at the moment she retires, we can write the balance at the end of n quarters as a finite geometric sum

$$B_n = q\sum^n_{i=0} (1+r)^i = q\frac{1-r^{n+1}}{1-r}$$

If $q=\$2600$, r=0.05125/4 and $n=4 \cdot 35$, we see that she will have saved $\$ 1,003,239$ by the time she retires.

5.

Referring back to the previous problem, suppose her account earns 5% compounded continuously. Develop a mathematical model for the balance in her account. How is this model different from your previous model with quarterly interest? What is the connection between the two? How much money will she have saved when she retires?

Here, we replace the discrete problem with a continuous problem. The methodology is exactly the same. All discrete quantities are replaced with annual rates.

$$\frac{dB}{dt} = r B + q$$

where r is the annual interest rate and q is the annual deposit rate. Essentially, the continuous formulation assumes that the employee is constantly earning interest on the balance and constantly depositing small amounts of money into the retirement account. In this case, $q = \$10,400$ per year and r = 0.05. Solving the ODE, one obtains

P = 208000 (e^{0.05 t}-1).

After 35 years, she will have saved $\$988,957$ which is almost the same as what she would have earned with the higher quarterly rate. This is why financial institutions are required to provide consumers with the annualized percentage rate (APR). The APR is the aggregate rate over one year's time, and it removes ambiguity about when interest is compounded.