differential equations

HAND-IN HOMEWORK #2 (due Thursday, September 12)

PART 2:

Read the accompanying pages about controlling a staph infection. Then answer the following questions.

(a) Find a solution formula for the equation (4.4).

(b) Use your answer in (a) to find an exact value for the critical dosage dcr (which was estimated by the computer simulations to lie betweeen 1.5 gm and 3.0 gm). Find a formula for the cure time tc when the dosage d > dcr.

(c) A natural expectation would be that higher dosages d > dc would result in shorter cure times. Does your formula for the cure time in (b) justify this expectation?

1

4. APPLICATIONS 55

4. Applications

4.1. Bacterial Cell Growth. When placed in an environment of abundant resources (nutrients, space, etc.) cell cultures typically grow in such a way that their per capita rate of change is constant. Mathematically, this means the number of cells x = x(t) at time t satisfies the differential equation

x0 = rx

where the constant r > 0 is the “per capita growth rate”. Often a particular microorganism’s growth rate is described by the time it takes the number of cells in the culture to double. This time δ is called the “doubling time” (or “generation time”) and it is related to the growth rate according to the formula

r = ln2

δ .

For more detailed discussion of these topics and of population growth models see Sec.6, Chapter 3.

As an example, the doubling time of the bacterium Staphylococcus aureus is approximately δ = 30 minutes, which corresponds to a per capita growth rate of

r = ln2

30 = 0.02310 (per minute).

The growth of a culture of S. aureus initially consisting of 106 cells is described by the initial value problem

x0 = 0.02310x(4.1)

x(0) = 1.

Here x is measured in units of 106 cells. According to Theorem 1.1, this initial value problem has a unique solution

x = x(t). A slope field and a solution graph (drawn using Heun’s Algorithm with step size s = 0.05) appear in Fig. 2.9. Notice the number of cells grows rapidly, following a seemingly exponential-like curve. Indeed, the solution formula for the initial value problem

x = e0.02310t

shows the growth is indeed exponential.

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5

10

15

20

t

x

Figure 2.9. The slope field of the differential equation x0 = 0.02310x and the solution of the initial value problem (4.1) drawn using Heun’s Algorithm with step size s = 0.05.

S. aureus is a common cause of bacterial skin infection (particularly in patients with HIV). The rapid exponential growth of a staph infection can be a serious problem if left untreated. Our modeling application involves determining the effect

56 1. FIRST ORDER EQUATIONS

of a medical treatment that removes staph cells from the patient at a certain rate h > 0 (cells/minute). We set up our mathematical model (i.e., perform the Model Derivation Step in Fig. 2.1 of the Introduction ) by applying the inflow-outflow rule (2.1) to the staph cell population numbers. This leads to the differential equation

(4.2) x0 = 0.02310x−h More specifically, suppose a milligram (mg) of antibiotic in a particular patients

kills staph cells at a rate of 104 per minute. Then a dosage of d mg kills a total of 104d staph cells per minute. In units of 106 cells, we have

(4.3) h = 104

106 d = 0.01d.(per minute)

Suppose, for the moment, that this removal rate h remains constant in time, as might be the case for example if the antibiotic were continuously administered intravenously. We want to know what dosages d, if any, will eliminate the staph infection from the patient, and if so in what amount of time.

The antibiotic kill rate h in (4.3) leads to the initial value problem

x0 = 0.02310x−0.01d(4.4) x(0) = 1.

for the number of staph cell x = x(t). Our next goal is to perform the Model Solution Step in the Modeling Cycle. What we want to learn from the solution x = x(t) is whether or not it continues to increase or whether it decreases and eventually equals 0. The answer will presumably depend on the dosage d.

One way to obtain answers to our questions would be from a formula for the solution x(t). We will learn how to find such a formula in Chapter 2. Here, however, we will investigate the solution by means of the methods developed in Sec. 2 and 2.2.

Fig. 2.10 shows slope fields and solution graphs, for a selection of dosages d, obtained by a computer. These graphs indicate the existence of a critical dosage level dcr above which the staph infection is eliminated and below which it is not. From Fig. 2.10 this critical dose lies between 1.5 gm and 3.0 gm. Further computer explorations, using other values of d, suggest this critical value is approximately dcr = 2.31 gm.

Another way to determine the critical value is to reason as follows. For d < dcr, the staph infection increases (x0 > 0) and for d > dcr it decreases (x0 < 0). Therefore, at the critical dose d = dcr the infection should do neither, but instead remain constant. From the initial value problem (4.4), we see that x remains at x(0) = 1, and hence x0 = 0, means

0.02310−0.01dcr = 0 or

dcr = 2.31.

At the critical dose dcr the staph infection remains constant, but at a higher dose d > dcr our computer studies indicate that x(t) = 0 at some finite time tc. This (“cured”) time tc = tc(d) when the infection is eliminated depends on d, as Fig. 2.10 shows. The higher the dose, the quicker the staph is eliminated; that is, tc(d) is a decreasing function.

4. APPLICATIONS 57

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0.5

1.0

1.5

t

x d = 1.5

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0.5

1.0

1.5

t

x d = 2.0

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0.5

1.0

1.5

t

x d = 2.5

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0.5

1.0

1.5

t

x d = 3.0

Figure 2.10. The slope field of the differential equation x0 = 0.02310x−0.01d and the solution of the initial value problem (4.1) for selected values of the antibiotic dose d.

We emphasize that computer explorationsdo not “prove” our conclusions about the existence of a critical dosage and the dependence of tc on d. This is because, when doing computer studies, we can calculate only a finite number of solutions for only a finite selection of dosages d. An advantage of a solution formula, if available (or, if not, other methods of analysis) is that these conclusions can be rigorously established. (See Exercise 6.36 in Chapter 2).

Often antibiotics are not continuously administered to a patient, but a dose is applied by pill or injection. In this case, the effect of the antibiotic is not constant, but decreases over time. To account for this change we return to the model equa- tion (4.2) to see what adjustments must be made (this is the Model Modification Step of the Modeling Cycle). To proceed we need information concerning how the effectiveness of the antibiotic changes over time, so that we can derive a formula for the staph removal rate h.

Suppose, for example, the effectiveness of the antibiotic decreases exponentially so that

h = 0.01de−at

Under this model assumption, the initial effectiveness of the antibiotic is 0.01d (cells/minute), but the effectiveness decreases over time with an exponential decay rate of a > 0. Suppose it is observed that the effectiveness decreases by 50% every hour. This allows us to calculate a. In 60 minutes, h is decreased by a fraction of 1/2 and therefore

e−a60 = 0.5.

or

58 1. FIRST ORDER EQUATIONS

a = 0.01155.

These assumptions lead us to a new initial problem for a staph infection starting with 106 cells:

x0 = 0.02310x−0.01de−0.01155t(4.5) x(0) = 1.

(Recall x is measured in units of 106.) Again we ask: what dosages d, if any, will eliminate the staph infection? Fig. 2.11 shows the slope field and the solution of the initial value problem

(4.5) for some selected values of the dose d. These samples suggest that this initial value problem also has a critical dosage dcr below which the treatment does not eliminate the staph infection. The particular examples in Fig. 2.11 indicate that dcr lies between 2.5 gm and 4.0 gm. (See Exercise 6.37 in Chapter 2.)

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0.5

1.0

1.5

t

x d = 2.5

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0.5

1.0

1.5

t

x d = 3.3

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0.5

1.0

1.5

t

x d = 3.5

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0.5

1.0

1.5

t

x d = 4.0

Figure 2.11. The slope field of the differential equation x0 = 0.02310x−0.01de−0.01155t and the solution of the initial value prob- lem (4.5) for selected values of the antibiotic dose d.

An interesting difference between the intravenous treatment modeled by (4.4) and the pill or injection treatment modeled by (4.5) occurs for doses below the critical level dcr. Unlike the intravenous treatment, the pill or injection treatment can show an initial improvement (x initially decreases in Fig. 2.11 for d = 2.5 and 3.3) even though the infection ultimately “bounces back” and grows unabated. Thus, one must guardagainst amistakenconclusion, basedon its early effectiveness, that the treatment will result in a cure.