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Further methods and applications

13.2 Logistic growth (and other models)

9 problems · hints, answers and solutions shown beside each one

Spread of infection. In the model for the spread of a disease (starting page  in the text), we used the fact that the total population is constant (S(t)+I(t)=N=S(t)+I(t)=N=constant) to eliminate S(t)S(t) and analyze a differential equation for I(t)I(t) on its own.

Carry out a similar analysis, but eliminate I(t)I(t). Then analyze the differential equation you get for S(t)S(t) to find its steady states and behaviour, practicing the qualitative analysis discussed in this chapter.

Social media. Sally Sweetstone has invented a new social media App called HeadSpace, which instantly matches compatible mates according to their changing tastes and styles. Users hear about the App from one another by word of mouth and sign up for an account. The account expires randomly, with a half-life of 11 month. Suppose y1(t)y_1(t) are the number of individuals who are not subscribers and y2(t)y_2(t) are the number of are subscribers at time tt. The following model has been suggested for the evolving subscriber population

dy1dt=by2ay1y2,dy2dt=ay1y2by2.\begin{align*} \frac{dy_1}{dt}&=b y_2-a y_1 y_2, \\ \frac{dy_2}{dt}&=a y_1 y_2 -b y_2.\\ \end{align*}
  1. Explain the terms in the equation. What is the value of the constant bb?

  2. Show that the total population P=y1(t)+y2(t)P=y_1(t)+y_2(t) is constant.

    Note: this is a conservation statement.

  3. Use the conservation statement to eliminate y1y_1. Then analyze the differential equation you obtain for y2y_2.

  4. Use your model to determine whether this newly launched social media will be successful or whether it will go extinct.

Answer

(c) Steady states at at y2=0y_2=0 and y2=Pb/ay_2=P-b/a. (d) Social media persists if Pa/b>1Pa/b>1.

A bimolecular reaction. Two molecules of AA can react to form a new chemical, BB. The reaction is reversible

so that BB also continually decays back into 22 molecules of AA. The differential equation model proposed for this system is

dadt=μa2+2βbdbdt=μ2a2βb,\begin{align*} \frac{da}{dt}&= -\mu a^2 +2 \beta b\\ \frac{db}{dt}&= \frac{\mu}{2} a^2 -\beta b, \end{align*}

where a(t),b(t)>0a(t), b(t)>0 are the concentrations of the two chemicals.

  1. Explain the factor 22 that appears in the differential equations and the conservation statement. Show that the total mass M=a(t)+2b(t)M=a(t)+2b(t) is constant.

  2. Use the techniques in this chapter to investigate what happens in this chemical reaction, to find any steady states, and to explain the behaviour of the system

Answer

(b) Stable steady state at a=β2μ(1+1+4μM/β)\displaystyle a=\frac{\beta}{2 \mu}\left(-1+\sqrt{1+4\mu M/\beta}\right)

Scaling time in the logistic equation. Consider the scaled logistic equation 12.1.3

dydt=ry(1y).\diff{y}{t}=ry(1-y).

Recall that rr has units of 1/time, so 1/r1/r is a quantity with units of time. Now consider scaling the time variable in the displayed equation by defining t=s/rt=s/r. Then ss carries no units (ss is “dimensionless”).

Substitute this expression for tt in the displayed equation and find the differential equation so obtained (for dy/dsdy/ds).

Answer

dyds=y(1y)\diff{y}{s}=y(1-y)

The Law of Mass Action. The Law of Mass Action in Section 12.1 led to the assumption that the rate of a reaction involving two types of molecules (A and B) is proportional to the product of their concentrations, kabk \cdot a \cdot b.

Explain why the sum of the concentrations, k(a+b)k \cdot (a+b), would not make for a sensible assumption about the rate of the reaction.

Glucose solution in a tank. A tank that holds 11 liter is initially full of plain water. A concentrated solution of glucose, containing 0.250.25 gm/cm3^{3} is pumped into the tank continuously, at the rate 1010 cm3^{3}/min and the mixture (which is continuously stirred to keep it uniform) is pumped out at the same rate.

Let G(t)G(t) be the amount of glucose in the tank after ttminutes. Write a differential equation for GG, and give its initial condition.

How much glucose is in the tank after a long time?

Answer

The initial value problem is dGdt=52G100\diff{G}{t}=\frac52 - \frac{G}{100}, G(0)=0G(0)=0. After a long time, there is 250 gm of glucose in the tank.

Pollutant in a lake. A lake of constant volume VV gallons contains Q(t)Q(t) pounds of pollutant at time tt evenly distributed throughout the lake. Water containing a concentration of kk pounds per gallon of pollutant enters the lake at a rate of rr gallons per minute, and the well-mixed solution leaves at the same rate.

  1. Set up a differential equation that describes the way that the amount of pollutant in the lake changes.

  2. Determine what happens to the pollutant level after a long time if this process continues.

  3. If k=0k=0 find the time TT for the amount of pollutant to be reduced to one half of its initial value.

Answer
  1. Q(t)=krQVr=rV[QkV]\displaystyle Q'(t) = kr - \frac{Q}{V}r = - \frac{r}{V} [ Q - kV];

  2. Q=kVQ=kV;

  3. T=Vln2/rT= V \ln 2/r.

A sugar solution. Sugar dissolves in water at a rate proportional to the amount of sugar not yet in solution. Let Q(t)Q(t) be the amount of sugar undissolved at time tt. The initial amount is 100100 kg and after 44 hours the amount undissolved is 7070 kg.

  1. Find a differential equation for Q(t)Q(t) and solve it.

  2. How long does it take for 5050 kg to dissolve?

Answer
  1. dQdt=kQ\displaystyle \frac{dQ}{dt}=kQ; Q(t)=1000.7t/4Q(t) = 100\cdot 0.7^{t/4}

  2. 4log2log(0.7)7.77  hr\frac{-4\log 2}{\log(0.7)} \approx 7.77~\text{ hr}. Note you should be able to get the exact value without the use of a calculator.

Leaking water tank. A cylindrical tank with cross-sectional area AA has a small hole through which water drains. The height of the water in the tank y(t)y(t) at time tt is given by:

y(t)=(y0kt2A)2y(t) = \left(\sqrt{y_{0}} - \frac{kt}{2A}\right)^{2}

where k,y0k, y_{0} are constants.

  1. Show that the height of the water, y(t)y(t), satisfies the differential equation

    dydt=kAy.\frac{dy}{dt} = -\frac{k}{A}\sqrt{y}.
  2. What is the initial height of the water in the tank at time t=0t=0 ?

  3. At what time is the tank be empty ?

  4. At what rate is the volume of the water in the tank changing when t=0t=0?

Answer
  1. not provided

  2. y0y_{0}

  3. t=2Ay0k\displaystyle t=\frac{2A\sqrt{y_{0}}}{k}

  4. ky0-k\sqrt{y_{0}}

From the UBC Math 100 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.