Logistic Growth - Worked Examples

 

 

 

 

Example 1 (Logistic Growth model):

p_n+1 = 1.2p_n - 0.0004p_n²,

 

 

 

 

 

Solution:

p₁ = 1.2p₀ - 0.0004p₀² = 1.2(200) - 0.0004(200)² = 224

p₂ = 1.2(224) - .0004(224)² = 248.7

p₃ = 1.2(248.7) - 0.0004(248.7)² = 273.7

p_e = 1.2p_e - 0.0004p_e²

-0.2p_e + 0.0004p_e² = 0

p_e(0.0004p_e - 0.2) = 0

 

 

 

 

 

 

 

 

 

Example 2 (Growth Rate function):

p_n+1 = p_n + g(p_n) = p_n + 0.05 p_n (1 - 0.0001p_n),

where n is measured in hours

a. Assume that p₀ = 500 and find the population for the next three hours, p₁, p₂, and p₃.

b. Find the p-intercepts and the vertex for

g(p) = 0.05 p(1 - 0.0001p)

Sketch a graph of g(p).

c. By finding when the growth rate is zero, determine all equilibria for this model.

 

 

 

 

 

Solution:

a. We find the first three iterations by substituting each successive value into the given updating function. Thus,

p₁ = p₀ + g(p₀) = 500 + 0.05(500)(1 - 0.0001(500)) = 524,

p₂ = 524 + 0.05(524)(1 - 0.0001(524)) = 549,

p₃ = 549 + 0.05(549)(1 - 0.0001(549)) = 574.

 

 

b. Since g(p) = 0.05 p(1 - 0.0001p), the p-intercepts are found by solving

g(p) = 0.05 p(1 - 0.0001p) = 0,

which gives either p = 0 or 1 - 0.0001p = 0. The latter is equivalent to p = 10,000.

The vertex occurs halfway between the p-intercepts, so p = 5,000 and g(5000) = 0.05(5000)(1 - 0.0001(5000)) = 125.

The graph of g(p) is a standard parabola, which is shown below showing the p-intercepts and the vertex.

 

 

c. From the graph above, it is clear that the growth rate is zero at 0 and 10,000, so the equilibria occur at

p_e = 0 and 10,000.

 

 

 

 

 

Example 3 (Logistic Growth with Emigration): We extend the previous example to include the possibility that the population might be affected by immigration or emigration. Suppose that the growth rate for a population is given by

g(p) = 0.71 p - 0.001p² - 7.

This says that between each generation there is a 71% growth rate, while 0.001p_n² are lost due to crowding and 7 emigrate. The discrete dynamical model for this population model is given by

p_n+1 = p_n + g(p_n) = p_n + 0.71 p_n - 0.001p_n² - 7,

where n is measured in generations.

a. Assume that p₀ = 100 and find the population for the next three generations, p₁, p₂, and p₃.

b. Find the p -intercepts and the vertex for g(p) and sketch a graph of g(p).

c. By finding when the growth rate is zero, determine all equilibria for this model.

 

 

 

 

Solution:

a. As we did in the previous example, we iterate this discrete logistic model with emigration to obtain the next three generations. Thus,

p₁ = p₀ + g(p₀) = 100 + 0.71(100) - 0.001(100)² - 7 = 154,

p₂ = 154 + 0.71(154) - 0.001(154)² - 7 = 233,

p₃ = 233 + 0.71(233) - 0.001(233)² - 7 = 337.

 

 

 

b. Since g(p) = 0.71 p - 0.001p² - 7 = -0.001(p² - 710p + 7000), the p-intercepts are found by solving

g(p) = -0.001(p² - 710p + 7000) = -0.001(p - 10)(p - 700) = 0,

which gives either p = 0 or p = 700.

 

The vertex occurs halfway between the p-intercepts, so

p = 355 and g(355) = -0.001(345)(-345) = 119.

The graph of g(p) is a parabola, which is shown below showing the p-intercepts at p = 10 and 700 and the vertex.

 

 

c. From the graph above, it is clear that the growth rate is zero at 10 and 700, so the equilibria occur at

p_e = 10 and 700.

 

 

 

 

Simulation of Ricker's Model

• Let the parameter a vary and consider Ricker's model

P_n+1 = R(P_n) = aP_nexp(-0.01P_n),

• From the discussion above, the equilibria are

P_e = 0 and P_e = 100 ln(a)

 

 

 

 

 

• The product rule gives

R '(P) = aP (-0.01 e^-0.01P) + a e^-0.01P = a e^-0.01P(1 - 0.01P)

• The stability of P_e = 0 satisfies

R '(0) = a e⁰(1 - 0) = a

• Since a > 1, this equilibrium is unstable

 

 

 

 

• At P_e = 100 ln(a), the formula for the derivative gives

R '(100 ln(a)) = a e^-ln(a)(1 - ln(a)) = 1 - ln(a)

• For |R '(100 ln(a))| < 1, we must have a < 7.389

 

 

 

• Simulations for 50 generations of the Ricker's model with a = 5 and a = 8
• The initial population P₀ = 100
• For a = 5, the model tends towards the equilibrium P_e = 161
• For a = 8, the model oscillates between the two values of P = 139 and P = 277

 

 

 

 

 

 

 

 

 

 

Mitotic Model

 

 

 

 

 

 

 

 

 

Hassell's Model

• Other alternatives to the Logistic growth model and Ricker's model
• One applied to insect populations is Hassell's model
• One form of this model is

• A more general Hassell's model in the next section uses the chain rule
• The numerator of H(P_n) is the Malthusian growth model
• At low population densities, the population grows exponentially (a > 1)
• As the population increases, the denominator increases
• This slows the rate of growth

 

 

 

 

Graph the updating function, H(P_n), for P_n > 0

• The only intercept is (0, 0)
• The vertical asymptote for this function is outside the domain
• The horizontal asymptote is H = a/b
• Biologically, this implies that there is a maximum number in the next generation, no matter how large the population starts (which is reasonable considering limited resources)
• The population (after the initial population) must always remain below P_n = a/b
• If we differentiate H(P), the quotient rule gives

• Notice that H '(P) > 0, so H(P) is always increasing
• The graph starts at H(0) = 0, then increases and approaches the asymptote H = a/b
• Below let a = 20 and b = 0.01
• Graph the updating function and the identity map

 

 

 

 

 

 

 

Analysis of Hassell's Model

• Find the equilibria of this model by letting P_e = P_n = P_n+1

• Simplifying this last expression gives

 

 

 

 

 

Stability of the Equilibria

• Recall that the derivative of Hassell's model is

• Near the zero equilibrium

H '(0) = a > 1

• Solution is unstable and grows exponentially away from 0

 

• At P_e = (a - 1)/b, the derivative gives

• Thus, P_e = (a - 1)/b is always stable with solutions monotonically approaching this equilibrium
• Below is a simulation of 10 iterations for let a = 20 and b = 0.01 with P₀ = 200

 

 

 

 

 

Worked Examples

 

 

 

 

 

 

 

 

Analysis of the behavior of the chalone model

• When the population P_n is low, then the denominator of the model is insignificant

  • This approximates the Malthusian growth model
• Large populations increase the denominator

  • This causes a decline in production of new cells
• Determine equilibria of this model
• Find the stability of the equilibria

  • Depends on the value of the derivative of f(P) at the equilibria
  • Must differentiate the quotient given in f(P).

 

 

 

 

 

Example 4 (Hassell's model):

There is no reason that the discrete dynamical model should have a quadratic form and in fact, because a quadratic function goes negative for large values of p, this model becomes very unrealistic. Hassell modified this growth model (especially for insect populations) to use a rational function response (a rational updating function).

Consider the discrete dynamical model given by the equation

 

where n is measured in generations.

a. Assume that p₀ = 200 and find the population for the next four generations, p₁, p₂, p₃, and p₄.

b. Find the p -intercepts and the horizontal asymptote for H(p) and sketch a graph of H(p) for p > 0. (Note that the vertical asymptote occurs at p = -50, which is outside the biologically valid range as p should be greater than or equal to 0.)

c. By solving p_e = H(p_e), determine all equilibria, p_e, for this model.

 

 

 

 

Solution:

a. Once again, we iterate this nonlinear dynamical model by Hassell by substituting the value of p₀ = 200 into the model. The result is

 

 

 

b. The only intercept for H(p) is (0, 0), while the horizontal asymptote is H = 1000. Below is a graph of Hassell's updating function along with the identity map. The intersections give the equilibria.

 

 

 

c. The equilibria are found by setting p_e = p_n+1 = p_n, so we solve

One solution is clearly p_e = 0. Next we multiply both sides by (1 + 0.02 p_e)/p_e, which gives

1 + 0.02 p_e = 20.

Solving this equation gives the other equilibrium,

p_e = 950.

 

 

 

 

 

 

Ricker's Growth Model

Example 4: Let P_n be the population of fish in any year n. Consider Ricker's model for population growth given by the equation

P_n+1 = R(P_n) = aP_nexp(-bP_n),

where a = 7 and b = 0.02. Sketch a graph of R(P) with the identity function, showing the intercepts, all extrema, and any asymptotes. Find all equilibria of the model and describe the behavior of these equilibria. Let P₀ = 100, and simulate the model for 50 years, graphing the solution.

 

 

 

 

 

Solution: First, we see that the only intercept is the origin, (0, 0). Since the negative exponential dominates in the function R(P), there is a horizontal asymptote of P_n+1 = 0. To find the extrema, we differentiate R(P). Applying the product rule we obtain:

R '(P) = 7[P(-0.02e^-0.02P ) + e^-0.02P ] = 7e^-0.02P(1 - 0.02P).

This expression is zero only when 1 - 0.02P = 0 or P_c = 50. Thus, there is a critical point for the updating function at

(P_c, R(P_c)) = (50, 350 e^-1) ~ (50, 128.76).

The graph of the updating function along with the identity function is shown below. They intersect at the equilibria, which are calculated later.

To find the equilibria, we substitute P_e for P_n+1 and P_n in the Ricker's model. The resulting equation is given by

P_e = 7P_e exp(-0.02P_e).

One equilibrium is given by P_e = 0, so we can divide out P_e, leaving

1 = 7exp(-0.02P_e ) or exp(0.02P_e ) = 7.

This gives the other equilibrium P_e = 50 ln(7) ~ 97.3.

To find the behavior of the solution near each of these, we must substitute the value of the equilibrium into the expression for the derivative. First, we analyze the stability of P_e = 0. We see that,

R '(0) = 7e⁰(1 - 0) = 7 > 1,

which is unstable, and the population grows away from the equilibrium P_e = 0.

Next we consider P_e = 50 ln(7). Substituting into the formula for the derivative gives

R '(50 ln(7)) = 7e^{-0.02(50 ln(7))}(1 - 0.02(50 ln(7))) = 7e^-ln(7)(1 - ln(7)) = 1 - ln(7) ~ -0.95

Thus, -1 < R '(50 ln(7)) < 0, so we have a stable equilibrium point with solutions oscillating, but approaching the equilibrium, P_e = 50 ln(7).

The simulation with P₀ = 100 is shown in the graph below. The solution is slowly oscillating toward the equilibrium, as can be seen with 50 iterations.

 

 

 

 

 

 

 

Example 5: Repeat the previous problem with a = 9.

 

 

Solution: Many of the computations carry over from the example above. The only intercept is the origin, (0, 0), and there is a horizontal asymptote of P_n+1 = 0. The derivative is only slightly changed as the leading constant is the only variation, so

R '(P) = 9e^-0.02P(1 - 0.02P).

As in the previous example, the critical point satisfies P_c = 50, which gives a maximum at

(P_c, R(P_c)) = (50, 450 e^-1) ~ (50, 165.5).

The graph of the updating function along with the identity function is shown below with the important points labeled.

The equilibria are found just like the previous example by using P_e for P_n+1 and P_n in the Ricker's model. The resulting equation is given by

P_e = 9P_e exp(-0.02P_e).

The calculations are very similar giving the two equilibria, P_e = 0 and P_e = 50 ln(9) ~ 109.86.

The stability analysis uses the same techniques as before, but the behavior changes at the upper equilibrium. Near P_e = 0, we evaluate the derivative and obtain

R '(0) = 9e⁰(1 - 0) = 9 > 1,

which is unstable with the population growing away from this equilibrium.

Near P_e = 50 ln(9), we find that

R '(50 ln(9)) = 9e^{-0.02(50 ln(9))}(1 - 0.02(50 ln(9))) = 1 - ln(9) ~ -1.197

Thus, R '(50 ln(9)) < -1, so we have an unstable equilibrium point with solutions oscillating and moving away from the equilibrium. Below is a simulation with 30 iterations, starting with P₀ = 100. Clearly, the solution oscillates with increasing amplitude. The solution goes to a period 2 behavior, oscillating between 55 and 165.