1.2. Introduction to Modeling. P (t) = r P (t) (b) When r > 0 this is the exponential growth equation.

Transcription

1 G. NAGY ODE January 9, Introduction to Modeling Section Objective(s): Review of Exponential Growth. The Logistic Population Model. Competing Species Model. Overview of Mathematical Models Review of the Exponential Growth Equation. Remarks: (a) Last class we obtained the differential equation P (t) = r P (t) as the continuum limit of a discrete eqauation. (b) When r > 0 this is the exponential growth equation. (c) When r < 0 this is the exponential decay equation. (d) The solutions are P (t) = P 0 e rt. P P = r P (t) P (t) dt = r dt ln( P ) = rt + c 0, where c 0 R is an arbitrary integration constant, and we ln( P ) = P /P. Then, P (t) = ±e rt+c 0 = ±e c 0 e rt, denote P 0 = ±e c 0 P (t) = P 0 e rt, P 0 R. Here P (0) = P 0 e 0 = P 0. (e) In recitation we also solved y (t) = r y(t) + b, r, b const. Rewrite the differential equation as follows, y (t) ry(t) + b = 1 1 r y (t) dt y(t) + (b/r) = u = y + (b/r) dt 1 du = y r dt. du u = dt 1 r ln( u ) = t + c 0 ln y + (b/r) = rt + c 1 y + (b/r) = e rt+c1 = e rt e c1 y(t) + (b/r) = (±e c1 ) e rt y(t) = c 2 e rt b r.

2 2 G. NAGY ODE january 9, The Logistic Population Model. Remark: The exponential growth equation describe a population with unlimited food resources. We now describe a population system that has finite food resources. If the population is small enough: If the population is large enough If P (t) is small enough, then P (t) r P (t), r > 0. If P (t) is large enough, then P (t) < r P (t), r > 0. Definition The logistic equation for the function P, which depends on the independent variable t, is P (t) = r P (t) 1 P (t) ", (1.2.1) r > 0 is the growth constant and > 0 is the carrying capacity. Example 1.2.1: Suppose the function P is solution to the logistic equation P (t) = r P (t) 1 P (t) ". (a) For what values of P is the population in equilibrium that is, time independent? (b) For what values of P is the population increasing in time? (c) For what values of P is the population decreasing in time? Solution: (a) If P is an equilibrium solution, then P (t) = onstant. Therefore, P (t) = P = 0. But P is solution of the logistic equation, that means 0 = P = r P 1 P " P = 0 or P = Pc. So, there are only two equilibrium solutions, the trivial case of no population P = 0, or the case when the population is equal the carrying capacity P =. (b) Any solution of the logistic equation is increasing when P (t) > 0, but the logistic equation implies that r P (t) 1 P (t) " > 0 P (t) 1 P (t) " > 0.

3 G. NAGY ODE January 9, Since P (t) 0, the population is going to be increasing for 1 P (t) " > 0 0 < P (t) <. (c) Any solution of the logistic equation is decreasing when P (t) < 0. But a calculation similar to the one in the previous part implies that the population is going to be decreasing for P (t) >. People say that in this case The population exceeds the carrying capacity of the environment. We can sketch a qualitative graph of the solutions to the logistic equation. P Equil. Sol. 0 Equil. Sol. t Figure 1. Qualitative graphs of solutions P of the logistic equation.

4 4 G. NAGY ODE january 9, Competing Species Model. Example 1.2.2: Write a simple model to describe how rabbits and sheep populations evolve in time when they compete on the grass on a particular piece of land. Solution: One way to find a simple model for these competing species is to start with the case that they do not compete at all. Each species has its own, finite, food resources. Let s call R(t) and S(t) the populations of rabbits and sheep at the time t. Then the independent species population model with finite food resources is nothing more than two logistic equations, one for each species, R = r r R 1 R " R c S = r s S 1 S ", S c where r r, r s are the growth rates and R c, S c are the carrying capacities for each species. The next step is to introduce competition terms into this model. Consider the first equation above for the rabbit population. The fact that sheep are eating the same grass as the rabbits will decrease R, so we need to add a term of the form α R(t) S(t) on the right hand side, that is R = r r R 1 R " α R S, R c with α > 0. The new term must depend on the product of the populations RS, because the product is a good measure of how often the species meet when they compete for the same grass. If the rabbit population is very small, then the interaction must be also small. Something similar must occur for the sheep population, so we get S = r s S 1 S " β R S, S c with β > 0.

5 G. NAGY ODE January 9, Definition The competing species equation for the functions x and y, which depend on the independent variable t, are x = r x x 1 x " α x y x c y = r y y 1 y " β x y, y c where the constants r x, r y and x c, x c are positive and α, β are nonnegative. Remarks: (a) In general, the solutions to competing species equation cannot be written in terms simple functions such as exponentials, trigonometric functions, and polynomials. (b) However, we will be able to describe the qualitative behavior solutions. of the Example 1.2.3: The following systems are models of the populations of pair of species that either compete for resources (an increase in one species decreases the growth rate in the other) or cooperate (an increase in one species increases the growth rate in the other). For each of the following systems identify the independent and dependent variables, the parameters, such as growth rates, carrying capacities, measure of interactions between species. Do the species compete of cooperate? (a) (b) dx dt = c x 2 dx 1x c 1 b 1 xy K 1 dt = x x xy Solution: (a) dy dt = c y 2 2y c 2 b 2 xy. K 2 (b) dy dt = 2 y y xy. The species compete. t is the independent variable. x, y are the dependent variables. c 1, c 2 are the growth rates. K 1, K 2 are the carrying capacities. B 1, b 2 are the competition coefficients. The species coperate. t is the independent variable. x, y are the dependent variables. 1, 2 are the growth rates. 5, 12 (not 6) are the carrying capacities. 5, 2 are the competition coefficients.

6 6 G. NAGY ODE january 9, Overview of Mathematical Models. Remarks: (a) A mathematical model is description of a situation in physics, engineering, ecology, biology, etc., using mathematical concepts and language. (b) We do not produce a perfect copy of the real-life situation, but rather to capture certain features one could say the essential features that govern the behavior of the system. Main Steps to Construct a Mathematical Model: Step 1. Clearly state all the assumptions. Step 2. Describe all the independent variables, dependent variables, parameters to be used in the model. Step 3. Use your assumptions to derive equations relating the variables and parameters. Step 4. Analyze the predictions of the model. Consider the population models we studied in this section. The step 1 could include: Does the system have unlimited food supplies? If yes, we would end up with the exponential growth equation. If not, we would end up with the logistic equation. The step 2 could include: Do we study only one species, or are we studying two or more species? is time the independent variable? The step 3 is how we obtained the differential equation for each of the population models. Step 4 is the analysis of the solutions, which we carried over for the exponential growth and the logistic population models.

7 G. NAGY ODE January 9, Remark: Some famous differential equations: Example 1.2.4: (a) Newton s Law: Mass times acceleration equals force, ma = f, where m is the particle mass, a = d 2 x/dt 2 is the particle acceleration, and f is the force acting on the particle. Hence Newton s law is the differential equation m d2 x dt 2 (t) = f (t, x(t), x (t)). where the unknown is the position of the particle in space, x(t), at the time t. Remark: This is a second order Ordinary Differential Equation (ODE). (b) Radioactive Decay: The amount u of a radioactive material changes in time as follows, du (t) = k u(t), k > 0, dt where k is a positive constant representing radioactive properties of the material. Remark: This is a first order ODE. (c) The Heat Equation: The temperature T in a solid material changes in time and in one space dimension according to the equation T t (t, x) = k 2 T (t, x), k > 0, x2 where k is a positive constant representing thermal properties of the material. Remark: This is a first order in time and second order in space Partial Differential Equation (PDE). (d) The Wave Equation: A wave perturbation u propagating in time t and in one space dimension x through the media with wave speed v > 0 is 2 u t 2 (t, x) = v2 2 u (t, x). x2 Remark: This is a second order in time and in space PDE.