SDSU

 

Math 121 - Calculus for Biology I
Spring Semester, 2004
More Applications of Nonlinear Dynamical Systems - Worked Examples

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San Diego State University -- This page last updated 20-Mar-04


More Applications of Nonlinear Dynamical Systems - Worked Examples

  1. Ricker's Growth Model
  2. Hassell's Model
  3. Model for Cellular Division with Inhibition

This section applies the techniques of qualitative analysis to population models using Ricker's and Hassell's growth models. We also introduce a model for cellular division control by chalones, which is similar mathematically to Hassell's growth model.

Ricker's Growth Model

Example 1: Let Pn be the population of fish in any year n. Consider Ricker's model for population growth given by the equation

Pn+1 = R(Pn) = aPnexp(-bPn),

where a = 7 and b = 0.02. Sketch a graph of the updating function 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 Pn+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 Pc = 50. Thus, there is a critical point for the updating function at

(Pc, R(Pc)) = (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 Pe for Pn+1 and Pn in the Ricker's model. The resulting equation is given by

Pe = 7Pe exp(-0.02Pe).

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

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

This gives the other equilibrium Pe = 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 Pe = 0. We see that,

R '(0) = 7e0(1 - 0) = 7 > 1,

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

Next we consider Pe = 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, Pe = 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 2: 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 Pn+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 Pc = 50, which gives a maximum at

(Pc, R(Pc)) = (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 Pe for Pn+1 and Pn in the Ricker's model. The resulting equation is given by

Pe = 9Pe exp(-0.02Pe).

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

The stability analysis uses the same techniques as before, but the behavior changes at the upper equilibrium. Near Pe = 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 Pe = 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.

 

 

Hassell's model

Example 3: Suppose that a population of insects is measured weekly and satisfies Hassell's model. Assume it follows the discrete dynamical model given by the equation

 

where n is measured in weeks.

a. Assume that p0 = 200 and find the population for the next four weeks, p1, p2, and p3. Graph a simulation of the model for 10 weeks.

b. Find the p-intercepts and the horizontal asymptote for H(p) and sketch a graph of the updating function 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.) Include the identity map on the graph.

c. Determine all equilibria, pe, for this model and discuss their stability.

Solution:

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

Below is the simulation of the first 10 weeks of the insect population

b. For the updating function, the only intercept for H(p) is (0, 0), while the horizontal asymptote is H = 1000. 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). Thus, the population (after the initial population) must always remain below Pn = 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 pe = pn+1 = pn, so we solve

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

1 + 0.02 pe = 20.

Solving this equation gives the other equilibrium,

pe = 950.

To determine the stability of the equilibria, we must differentiate H(p). From the quotient rule,

Notice that H '(p) > 0, so H(p) is always increasing. We expect the behavior of the model near the zero equilibrium to grow exponentially away from 0. Since

H '(0) = 20 > 1,

the equilibrium pe = 0 is unstable with solutions monotonically growing away from it.

At pe = 950, we can easily compute the derivative obtaining

Thus, our properties on the stability of discrete dynamical models give pe = 950 as being stable with all solutions monotonically approaching this equilibrium.

 

Example 4: (Hassell's Model) Suppose that a study shows that a population, Pn, of butterflies satisfies the dynamic model given by the following equation:

 

where n is measured in weeks. Let P0 = 200, then find P1 and P2. Find the intercepts, all extrema of H(P), and any asymptotes for P > 0.

Determine the equilibria and analyze the behavior of the solution near the equilibria.

Solution: The iterations are found in the standard way with

P1= H(200) = 16200/(1.4)4 = 4271.

P2= H(4271) = 43.

Thus, we see dramatic population swings with this model, suggesting an instability.

Analyzing H(P), we find that the only intercept is (0, 0) and that there is a horizontal asymptote with H = 0 (since the power of the denominator exceeds the power of the numerator). To find where the maximum occurs, we differentiate the function.

Solving H '(P) = 0, we have 1 - 0.006P = 0 or P = 500/3 = 166.7. With H(500/3) = 4271.5, the maximum occurs at (166.7, 4271.5). A graph of the updating function with the identity function is shown below.

 

As always, the equilibria are found by solving Pe = H(Pe), which is equivalent to solving

Pe(1 + 0.002Pe)4 = 81Pe.

So either Pe = 0 or (1 + 0.002Pe)4 = 81, which gives 1 + 0.002Pe = 3 or Pe = 1000. The stability of these equilibria can be determined by examining the derivative at the equilibria. At Pe = 0, H '(0) = 81, which implies from our rules that the solutions monotonically grow away from 0. At Pe = 1000, H '(1000) = 81(-5)/243 = -5/3. This implies that the solution near this equilibrium oscillates and goes away from the equilibrium. In fact, this model produces a period 4 solution with the solution asymptotically oscillating from 163 to 4271 to 42 to 2453. A simulation of this model is shown below.

 

 

Model for Cellular Division with Inhibition

Example 5: In the quotient rule of differentiation, a model for mitosis control by chalones was introduced. This model suggests that a biochemical agent known as a chalone is released by cell to inhibit the mitosis of nearby cells, preventing the over crowding of cells. Early models of cancer speculated that a break down in this control would lead to canceer. Consider the mitotic model given by the equation

Find the equilibria for this model and determine the behavior of the population near the equilibria. Also, start with an initial population of P0 = 10, and simulate this model for 20 mitotic divisions.

Solution: To find the equilibria, we let Pe = Pn = Pn+1 in the discrete dynamical model above. We find that

Thus, either Pe = 0 or Pe = 100, which is what is predicted from the lecture notes.

To analyze the behavior near the equilibria, we differentiate f(P), which gives

Since f '(0) = 2 > 1, solutions of the model near the zero equilibrium are unstable and grow monotonically away from 0. At Pe = 100, we see that f '(100) = -4/(1 + 1)2 = -1. Thus, this equilibrium right on the border of the stability region. The solutions will oscillate and slowly approach the equilibrium.

The simulation below, starting with P0 = 10, shows this behavior.