Logistic differential equation--how to solve this one?

Question

The population of the world was about 5.3 billion in 1990.  Birth rates in the 1990s ranged from 35 to 40 billion per year and death rates ranged from 15 to 20 million per year.  Let's assume that the carrying capacity for world population is 100 billion.

a. Write the logistic differential equation for these data. (Because the initial population is small compared to the carrying capacity, you can take k to be an estimate of the initial relative growth rate.

b. Use the logistic model to estimate the world population in the year 2000.

schmiso · Accepted Answer

a.

dP/dt = r∙P∙(1 - P/K)

K is the capacity

Using P units of billions

K = 100

The initial growth rate ist the difference of birth rate and death rates.

Taking the arithmetic average of these rate you find for initial rate:

(dP/dt)₀ = 0.0375 - 0.0175 = 0.02 (unit billions per year)

Since K is much larger than initial population you can ignore second term in differential equation for initial conditions:

(dP/dt)₀ ≈ r∙P₀

=>

r = (dP/dt)₀ /P₀ = 0.02 / 5.3 ≈ 3.7736×10⁻³ (unit 1/year)

hence:

dP/dt = 3.7736×10⁻³ ∙ P∙(1 - P/100)

with P in billions

and t in years

b.

let x = P/K

=>

dxdt = r∙x∙(1 - x)

<=>

1/[x∙(1 - x)] dx = r dt

<=>

[ 1/x + 1/(1-x) ] dx = r dt

<=>

∫ [ 1/x + 1/(1-x) ] dx = r dt

<=>

ln(x) - ln(1 - x) = r∙t + c

<=>

ln( x/(1 - x) ) = r∙t + c

<=>

x/(1 - x) = C∙e^(r∙t) (with c=e^c)

<=>

P/(K - P) = C∙e^(r∙t)

apply initial condition:

P(t=0) = P₀

=>

C = P₀/(K - P₀)

=>

P/(K - P) = [P₀/(K - P₀)]∙e^(r∙t)

<=>

P∙(K - P₀) = (K - P)∙P₀∙e^(r∙t)

<=>

P∙(K - P₀ + P₀∙e^(r∙t)) = K∙P₀∙e^(r∙t)

<=>

P = K∙P₀∙e^(r∙t) / ( K + P₀∙(e^(r∙t) - 1) )

For this particular model:

P = 530∙e^(3.7736×10⁻³∙t) / ( 100 + 5.3∙(e^(3.7736×10⁻³∙t) - 1) )

in year 2000, i.e for t=10

P = 530∙e^(3.7736×10⁻³∙10) / ( 100 + 5.3∙(e^(3.7736×10⁻³∙10) - 1) )

= 530∙1.03846 / ( 100 + 5.3∙0.03846) - 1) )

= 5.493

Logistic differential equation--how to solve this one?

1 Antwort