Some Basic Modeling with Differential Equations

Transcription

1 Some Basic Modeling with Differential Equations S. F. Ellermeyer Kennesaw State University October 6, What is a Mathematical Model? A mathematical model is an equation or set of equations that attempt to give a mathematical description of some real phenomenon. Example 1 Consider the statement: The number of McDonalds franchises in any given city is proportional to the population of that city. A mathematical model corresponding to this statement is M = kp (1) where M stands for the number of McDonalds franchises in a city, P stands for the population of that city and k isaconstant(calledaconstantofproportionality). The mathematical model (1) is an algebraic model. It is the simplest kind of mathematical model. Example 2 Consider the statement: At any given time, the rate of growth of the population of the United States is proportional to the population of the United States. A mathematical model corresponding to this statement is dt = kp (2) 1

2 where P stands for the population of the United States at time t, andk is a constant of proportionality. Note that P is a function of time (t) and that /dt is the derivative of the function P at time t. Since the derivative of a function at time t tells us the rate at which that function is changing at time t, the above model captures what is being said in the statement. The model (2) is an ordinary differential equation. It is more complicated than thealgebraicmodelinthepreviousexamplebecauseitinvolvesanunknown function, P, and also involves the derivative of P. Example 3 Newton s Law of Cooling asserts that if an object (say a ball of metal) is placed in surroundings that remain at a constant temperature, then therateofchangeintemperatureoftheobjectatanytime,t, isproportional to the difference between the temperature of the object at that time and the temperature of the surroundings. Let us write down a mathematical model of Newton s Law of Cooling: Assuming that the temperature of the surroundings always remains constant at the fixed temperature T s,wecanlett stand for the temperature of the object at time t. The difference between the temperature of the object at any time t and the temperature of the surroundings is T s T. (Note that this difference is positive is the object is cooler than its surroundings and is negative if the object is warmer than its surroundings.) The rate at which the temperature of the object is changing at any time t is dt/dt. Thus, a mathematical model of Newton s Law of Cooling is dt dt = k (T s T ) (3) where k is a constant of proportionality. The model (3) is an ordinary differential equation. Note that the constant of proportionality, k, mustbeassumed to be a positive constant in order for this model to make sense. In particular, if the object is cooler than its surroundings, then T s T > 0 and we know that the object will warm up over time (meaning that T will increase over time). We thus want our model to reflect the fact that dt/dt > 0 when T s T > 0. This will be the case only if k>0. Likewise, if k>0, then our model will reflect the fact that if T s T<0 (meaning that the object is warmer than its surroundings, then dt/dt < 0 (meaning that the object cools over time). 2

3 2 Basic Guidelines for Modeling The basic procedure for formulating a mathematical model is as follows: 1. Understand the phenomenon you are trying to model. 2. Give names to all independent variables, dependent variables, and parameters that are necessary to formulate a model of the phenomenon. 3. Write down your model. Some basic things to look for in trying to formulate a model are phrases like rate of change and is proportional to. The phrase rate of change means that a derivative will be involved and your model will be a differential equation. The phrase is proportional to means that there will be a parameter (a constant of proportionality) in your model. The sentence A is proportional to B has the mathematical translation A = kb. In modeling, it is very important to state what all of the variables and parameters being used in the model stand for. For example, if you just write down the differential equation dy dt =5y B without explaining what y and B stand for, then nobody will know what it is you are trying to model. Somebody who is familiar with differential equations would know that y is a function (the unknown of the differential equation) and would probably guess that B is a parameter, but they would not know what the model is attempting to describe. 3

4 3 Two Basic Models of Population Growth The two most basic mathematical models of population growth are the Malthusian model and the Logistic model. By basic, we mean that they are based on very simple assumptions and do not take into account the many complex factors that determine population dynamics in reality. Nonetheless, one must understand the Malthusian and logistic models before more realistic (and hence more complicated) models of population dynamics can be understood. 3.1 The Malthusian Model The Malthusian model of population growth is the one given in Example 2. The only hypothesis of this model is that a population grows at a rate proportional to itself. The model is dt = kp where P (which is a function of time, t) stands for the population at time t, and k is a constant of proportionality. Let us use the Malthusian model to model the growth of the population of the United States since the year 1790: In order to do this, we will use the fact (taken from actual census data) that the U. S. population in 1790 was 3.9 million. Letting t =0be the year 1790 and letting P stand for the U. S. population (in millions) at any time t after 1790, we obtain the initial value problem = kp dt (4) P (0) = 3.9. (5) The initial condition (5) states that the population in 1790 was 3.9 million. The differential equation (4) states that the population always grows at a rate proportional to itself. From our previous work, we know that the solution of the initial value problem (4,5) is the function P =3.9e kt. (6) 4

5 We could use the solution (6) to make predictions about the U.S. population after 1790 if we knew the value of k. In order to assign a value to k, we will use the fact (taken from census data) that the U.S. population in 1800 was 5.3 million. In other words, P (10) = 5.3. Substituting this information into the solution (6) then gives 5.3 =3.9e 10k and solving for k gives k = 1 µ ln Sincewenowhaveadefinite value for k, we can use the solution (6) to predict the U. S. population at any time after For example, the predicted U.S. population in 1900 is P (110) = 3.9e 110k 3.9e 110( ) 114 million and the predicted U.S. population in 1980 is P (190) = 3.9e 190k 3.9e 190( ) 1, 324 million. Figure 1 and the accompanying table show the Malthusian predicted U. S. population and the actual U. S. population according to census data taken at ten year intervals from 1790 to Observe that the Malthusian model predicts the U. S. population fairly accurately up to about the year 1860 (which is t = 70), but predicts a much larger population (than the real population) for years after It is not surprising that the model does not predict an accurate population very far into the future because the model was based on only a very simple hypothesis: that the population is always growing at a rate proportional to itself. 5

6 Actual and Malthusian Predicted U. S. Population Year actual predicted Year actual predicted Figure 1: Actual and Malthusian-Predicted U. S. Population 3.2 The Logistic Model The Logistic model of population growth is based on the following hypotheses: 6

7 1. Due to space limitations, limited food supply, etc., any population has a carrying capacity. The carrying capacity, which is a positive constant denoted by N, is a certain population level such that if the population is less than N, then the population will increase; if the population is greater than N, then the population will decrease; and if the population is exactly equal to N, then the population will neither increase nor decrease. 2. When the population is very small (well below carrying capacity), it will grow at a rate roughly proportional to itself (like in the Malthusian model). Amodelwhichembodiestheabovehypothesesis µ dt = kp 1 P (7) N where the constant k is assumed to be positive. The important features of the model (7), showing that the model embodies the modeling hypotheses, are: 1. If P>0and P<N,then1 P/N > 0, and hence µ dt = kp 1 P > 0 N (indicating that the population is increasing when the population is less than its carrying capacity). 2. If P>N,then1 P/N < 0, and hence µ dt = kp 1 P < 0 N (indicating that the population is decreasing when the population is greater than its carrying capacity). 3. If P = N, then1 P/N =0, and hence µ dt = kp 1 P =0 N (indicating that the population does not change when the population is exactly at its carrying capacity). 7

8 4. If P>0but very small (meaning that P 0), then 1 P/N 1 and hence µ dt = kp 1 P kp N (indicating approximate Malthusian type growth of a small population). Let us examine what happens when we attempt to model the U. S. population using the Logistic model. We will again start with the given information that the U. S. population in 1790 was 3.9 million. We must also hypothesize what we think the carrying capacity of the United States is. This is hard to do, but let us assume that the carrying capacity is double the 1990 U.S. population of 249 million. Thus, we will take N =. Based on our modeling assumptions, and again taking the year 1790 to be t =0, the Logistic model for the U.S. population is the initial value problem dt = kp µ 1 P (8) P (0) = 3.9. (9) The first step in finding the solution of the logistic model is to find the general solution of the differential equation (8). To do this, we first write the differential equation as dt = kp ( P ) and then (separating variables) as P ( P ) = kdt. This gives us Z Z P ( P ) = Using the partial fraction decomposition kdt. P ( P ) = 1 P 1 P, 8

9 we obtain Z 1 Z P Z 1 P = and performing the integrations then gives us kdt ln P ln P = kt + C. We now use algebra to solve for P : First, we use properties of the logarithm to write µ P ln = kt + C. P Since P is assumed to be less than (because the population will always stay below carrying capacity), we see that P < 0 and hence P = P. Also, since the population is always positive, we have P = P. This gives us µ P ln = kt + C. P Exponentiating both sides of the above equation gives us P P = ekt+c or P P = Bekt (where we have written B in place of e C ). We now have P = Be kt ( P ) or P + Be kt P =Be kt or P 1+Be kt =Be kt or P = Bekt 1+Be kt. We can divide both numerator and denominator of the above right hand side by Be kt to obtain P = Ae kt +1 9

10 (where we have written A in place of 1/B). To find the value of the constant, A, that will make our solution satisfy the initial condition (9), we observe that we want to have 3.9 = Ae k(0) +1 which is the same as 3.9 = A +1 Solving this equation for A, weobtain A +1= 3.9 which gives us A = = Therefore, the (approximate) logistic model solution for the U. S. population is P = e kt +1. (10) An immediate observation that we can make about this solution is that lim P = lim t t e kt +1 = 0+1 =, indicating that the logistic model predicts that as time goes on (meaning that t ), the population will approach the carrying capacity value of million. In order to use the solution (10) to estimate the U. S. population at any particular time after 1790, we need to know the value of k. Aswedidwith the Malthusian model, we will assign a value to k by applying the given information that P (10) = 5.3 to the solution (10). This gives us 5.3 = e 10k +1. Solving the above equation for k gives k Therefore we have P for all t 0. (11) e t +1 10

11 Thegraphofthesolution(11)alongwithaplotoftheactualU.S. population data between 1790 and 1990 is shown in Figure 2. Note that the logistic model seems to be a better predictor of U. S. population than the Malthusian model, although, of course, it is far from perfect. Actual and Logistic Predicted U. S. Population Year actual predicted Year actual predicted Figure 2: Actual and logistic predicted U. S. Populations 11

12 4 Newton s Law of Cooling Newton s Law of Cooling, introduced in Example 3, asserts that dt dt = k (T s T ) where T is the temperature of some object at time t, andt s is the temperature (assumed to be constant) of the surroundings in which the object whose changing temperature is being monitored is situated. The parameter, k, is a constant of proportionality. The following example illustrates a typical application of Newton s Law of Cooling. Example 4 Suppose that a cup of hot chocolate has initial temperature of 150 F and suppose that this hot chocolate is sitting in a room whose (constant) temperature is 60 F. Also, suppose that it is observed that the hot chocolate has cooled to 130 F after one minute. Assuming that Newton s Law of Cooling is obeyed, find the function, T, that gives the temperature of the hot chocolate at any time t 0. How long does it take for the hot chocolate to cool to 100 F? Solution 5 We wish to solve the initial value problem dt dt = k (60 T ) T (0) = 150. To find the general solution of the above differential equation, we separate variables to obtain dt 60 T = kdt and then integrate to obtain Z Z dt 60 T dt = Evaluation of the above integrals gives kdt. ln 60 T = kt + C. Since the T is always greater than 60 in this problem, we observe that 60 T = T 60. Thus ln (T 60) = kt + C 12

13 or ln (T 60) = kt + C (where we have written C in place of C, which doesn t matter since this is an arbitrary constant to be determined later). Exponentiation of both sides of the above equation gives or (writing D in place of e C ) T 60 = e kt+c T =60+De kt. In order to satisfy the initial condition, T (0) = 150, wemusthave 150 = 60 + De k(0), which tells us that D =90and hence that T =60+90e kt for all t 0. (12) We are also given that T (1) = 130 and this is the information that we will use to determine the value of k. In particular, we have 130 = e k which, when solved for k, gives µ 9 k =ln Using the function (12) with the value k =0.2513, we would like find the time, t, atwhicht (t) =100. To do this, we solve the equation e kt =100. The solution of this equation is t = 1 µ ln We conclude that it takes a little more than three minutes for the hot chocolate to cool to 100 F. The graph of the function (12) over the time interval 0 t 10 is shown in Figure 3. 13

14 Figure 3: Graph of T =60+90e t 14