Chapter 3 Modeling With First-Order Differential Equations

1. Chapter 3 Modeling With First-Order Differential Equations 1

2. INTRODUCTION • Mathematical models for population growth, radioactive decay, velocity of a falling body, etc.. • The chapter concludes the systems of first-order differential equations as mathematical models. • With the methods developed above, we can now solve some of the linear differential equations in section 3.1. And solve the nonlinear differential equations in section 3.2. 2

3. 3.1. Linear Equations • The standard form of linear equation is by the integration factor enables us write the DE as The equation can solved by integrating both sides of this last form. • Here it includes exponential growth and decay, half-life, carbon dating, Newton’s law of cooling, mixtures, series circuits, transient term, steady-state term. 3

4. Growth and Decay • Growth and Decay The initial value problem where k is a constant of proportionality, serves as a model for diverse phenomena involving either growth and decay. • In physics and chemistry, the equation means is directly proportional to the amount of x, like U-238 by radioactivity into Th-234. • Knowing the population at some arbitrary initial time , we can then use the solution of the equation to predict the population in the future-that is , at times t > . The constant of proportionality k can be determined from the solution of the initial-value problem, using a subsequent measurement of x at a time 4

5. Bacterial Growth • Example 1. Bacterial Growth • A culture initially has number of bacteria. At t=1h, the number of bacteria is measures to be If the rate of growth is proportional to the number of bacteria p(t) present at time t, determined the time necessary for the number of bacteria to triple. • <sol> • Then multiply the both side of integration factor and • integrating gives • therefore At t=0 so 5

6. And at t=1, so , To find the time bacteria has tripled , so 6

7. Growth constant and Decay-constant • The figure 3.2 shown the exponential function increase as t • increases for k>0, and decreases as t increases for k<0. • k is either a growth constant (k>0) or a decay constant(k<0) • Half-life • In physics the half-life is a measure of the stability of a radioactive substance. • The longer half-life of a substance, the more stable it is. 7

8. Half-life of Plutonium • Example 2. Half-Life of Plutonium • A breeder reactor converts relatively stable uranium 238 into the isotope plutonium 239. After 15 years it is determined that 0.043% of the initial amount of plutonium has disintegrated. Find the half-life of this isotope if the rate of disintegration is proportional to the amount remaining. <sol> By the equation The solution of the initial-value problem is To find the decay constant k, we use means Half-life is the corresponding value of time which 8

9. Solving for t gives yields 9

10. Carbon Dating • Example 3. Age of a fossil • A fossilized bone is found to contain one-thousandth the original amount of C-14. Determine the age of the fossil. (The half-life of C-14is approximately 5600 years) • <sol> To determine the value of the decay constant k, We use the fact that From We get k= Therefore And we have so Thus 10

11. Newton’s Law of Cooling • The mathematical formulation of Newton’s empirical law of cooling of an object is given by the linear first-order differential equation • K is a constant of proportionality • T(t) is the temperature of the object for t >0 • is the ambient temperature-the temperature of the medium around the object. Here we assume that is constant. 11

12. Cooling of a Cake • Example 4. Cooling of a Cake • When a cake is removed from an oven, its temperature is measured at 300°F. Three minutes later its temperature is 200ºF. How ling will it take for the cake to cool off to a room temperature of 70°F? • <sol> Then solve the initial-value problem Separating variables, • Integration yields so • When t=0, T=300, so 300=70+ gives =230 • Therefore 12

13. The measurement T(3)=200 leads to • So • Thus • Since we know that there is no finite solution to • T(t)=70. • But we can see the figure cake will approximately 70 about one-half • hour. 13

14. Mixtures • The mixing of two fluids sometimes gives rise to a linear first-order equation. • We assumed that the rate which the amount of salt in the mixing tank changes was a net rate. 14

15. Mixture of two salt solutions • Example 5. Mixture of two salt solutions • Recall that the large tank considered in Section 1.3 initially held 300 gallons of brine solution. Salt was entering and leaving the tank; a brine solution was being pumped into the tank at the rate of 3gal/min, mixed with the solution there, and then pumped out of the tank at the rate of 3gal/min. The concentration of the entering solution was 2 lb/gal, and so salt was entering the tank at the rate and leaving the tank at the rate From these data and We get equation . Let us pose this equation: If 50 pounds of salt was dissolved in the initial 300 gallons, how much salt would be in the tank after a long time. 15

16. <sol>We solve the initial-value problem • Note that A(0)=50 is the initial amount of liquid. • And the integrating factor of the linear differential equation is • We can write the equation as • Then derive the general solution • When t=0, A=50, so we find that c=-550. • Thus the amount of salt in the tank at any time t is given by 16

17. Series circuits • Series circuits • In figure 3.5, the sum of the voltage drop across the inductor and the voltage drop across the resistor is the same as the impressed voltage on the circuit. • Thus the linear differential equation for the current i(t), is • In figure 3.6, the voltage drop across a capacitor with capacitance C is given by q(t)/c. And derive • By the relation I=dq/dt, the above equation becomes the linear equation 17

18. Series Circuit • Example 6. Series Circuit • A 12-volt battery is connected to a series circuit in which the inductance is ½ henry and the and the resistance is 10 ohms. Determine the current i if the initial current is zero. • <sol>We must solve subject to i(0)=0 We multiply the equation by 2 and read the integration factor • Then obtain • Integrating each side derive • Now i(0)=0 implies 0=6/5 +c or c=-6/5 18

19. Remarks • Remarks • The solution of initial-value problem in Example 1described the population is a continuous function that takes on all real numbers in the interval at any time t >0. • Moreover, we would not expect the population to grow continuously- that is, there may be intervals of time • over which there is no growth at all. • The left figure is a more realistic description of population than just use a exponential continuous function. 19

20. 3.2. Nonlinear equations • In this session, we turn to the solution of nonlinear first-order mathematical models. And in this session, we solve the most differential equations by separation of variables. • Nonlinear equation includes population models, relative growth rate, density-dependent hypothesis, logistic differential equation, logistic function, technique of integration. 20

21. Population Dynamics • The model for exponential growth begins with the assumption that for some K >0. In this model the relative, or specific, growth rate defined by is assumed to be a constant K. • The assumption that the rate at which a population grows(or declines) is dependent only on the present and not on any time-dependent mechanisms such as seasonal phenomena can be stated as It is called the density-dependent hypothesis. 21

22. Logistic Equation • Suppose when an environment is capable of sustaining no more than a fixed number K of its population. The quantity K is called the carrying capacity, hence f(k)=0 and we simply let f(0)=r. • The left figure shows three function f that satisfy the above f(k)=0 and f(0)=r two conditions. The simplest assumption that we can make is that f(p) is linear-that is • And Or we can write the equation as • The equation is known as the logistic equation when a>0 and b>0. And its solution is called the logistic function. The graph of a logistic function is called a logistic curve. 22

23. When the population is very large, the linear equation does not provide a very accurate model. Overcrowded conditions, or with the pollution environment, can have an inhibiting effect on population growth. • The equation is bounded as If we rewrite as the nonlinear term can be interpreted as an “inhibition” or “competition” term. And in most applications the positive constant a is much larger than the constant b. 23

24. Solution of the Logistic Equation • One method of solving is separation of variables. Decomposing the left side of into partial fractions and integrating gives So And it follows If and so, after substituting and simplifying, the solution becomes 24

25. Graph of P(t) • From equation we see that • The dashed line P=a/2b shown in left two figure corresponds • to the ordinate of a point of inflection of the logistic curve. • We differentiate by the product rule: 25

26. From calculus the points where are possible points of inflection, hence is the only possible ordinate value at which the concavity of the graph can change. • For and When the initial value satisfies the graph of assumes the shape of an S, as seen in Figure(a). When the inflection point occurs at a negative value of t, as shown in figure(b). 26

27. Logistic Growth • Example 1. Logistic Growth • Suppose a student carrying a flu virus returns to an isolated college campus of 1000 students. If it is assumed that the rate at which the virus spreads is proportional not only to the number x of infected students but also to the number of students not infected, determine the number of infected students after 6 days if it is further observed that after 4 days x(4)=50. • <sol> We must solve the initial-value problem • By making the identifications a=1000k, b=k, from above • page, we derive • By x(4)=50, we determine k from • And • Finally students. 27

28. Modifications of the Logistic Equation • There are many variations of the logistic equation. For example, the differential equation could serve,when h>0 is a constant, the DE can be solved by separation of variables. • The rate h could be a function of time t or could be population dependent. • In the latter instance, the model would look like for example, a reasonable model for the population of the community would be • Another equation of the form is is a modification of the logistic equation known as the Gompertz differential equation. 28

29. Chemical Reaction • Suppose that a gram s of chemical A is combined with b grams of chemical B. If there are M parts of A and N parts of B formed in the compound and X(t) is the number of grams of chemical C formed, and the number of grams of chemical A and B remaining at time t are,respectively, • The mass action law states the rate at which the two substances react is proportional to the product of the amounts of A and B that are untransformed at time t: • If we factor out M/(M+N) from the first factor and N/(M+N) from the second and introduce a constant proportionality K>0, we have the form 29

30. Second-order Chemical Reaction • Example 2. Second-order Chemical Reaction • A compound C is formed when chemicals A and B are combined. The resulting reaction between the two chemicals is such that for each gram of A, 4 grams of B is used. It is observes that 30 grams of the compound C is formed in 10 minutes. Determine the amount of C at any time if the rate of the reaction if proportional to the amounts of A and B remaining and if initially there are 50 grams of A and 32 grams of B. How much of the compound C is present at 15 minutes? Interpret the solution as • <sol>Let X(t) denote the number of the compound C present at any time t. Clearly X(0)=0g, X(10)=30g, In general, for X grams of C we must use • The amounts of A and B remaining at any time are then • respectively. 30

31. The rate at which compound C is formed satisfies We factor 1/5 from the first term and 4/5 from the second and introduce the constant of proportionality: By separation of variables and partial fractions we can write Integrating gives When t=0, x=0, Using x=30g at t=10, we find With this information, we solve the equation for X: This means that 40 grams of compound C is formed, leaving 31

32. Remarks • Remarks • The indefinite integral can be evaluated in terms of logarithm, the inverse hyperbolic tangent, or the inverse hyperbolic cotangent. • For example, of the two results: 32

33. Systems of linear and nonlinear differential equation • In this section we are going to discuss mathematical models based on the some of the topics explored in the preceding two sections. • The mathematical models that we are discuss are the systems of first-order differential equation, but we are not going to develop any methods for solving these systems. We shall examine solution methods for systems in Chapter 8, and higher-order differential equations in Chapter 4 and 7. 33

34. Linear/Nonlinear Systems • Up to now, all the mathematical models that we gave considered have been single differential equations. But if there are two interacting and perhaps competing species living in the same environment, then a model for their populations X(t) and Y(t) might be a system of two first-order differential equations such as When are linear in the variables x and y-that is, they have the forms When the coefficients could only depend on t-then the above DE is said to linear system,and is said to nonlinear when DE is not linear. 34

35. Radioactive Series • When a substance decays by radioactivity it usually doesn’t just transmute in one step into a stable substance, the half-live of the various elements in a radioactive series can range from billions of years to a fraction of a second. • Suppose a radioactive series is described by are the decay constant for substances X and Y, respectively, and Z is a stable element. X(t), y(t), z(t) denote amounts of substance X, Y, Z, respectively, remaining at time t. The decay of element X is described by whereas the rate at which the second element Y decays is the net rate Z is a stable element, it is simply the decay of element Y: The elements is the linear system of three first-order equations 35

36. Mixtures • Consider the two tanks shown in figure. Let us suppose for the sake of discussion that tank A contains 50 gallons of water in which 25 pounds of salt is dissolved. Suppose tank B contains 50 gallons of pure water. Liquid is pumped in and out of the tanks as indicated in the figure; the mixture exchanged between the two tanks and the liquid pumped out of tank B are assumed to be well stirred. We wish to construct a mathematical model that describes the number of pounds of salt in tanks A and B, respectively, at time t. • <pf>We see that the net rate of change of for tank A is 36

37. <pf>Similarly, for tank B the net rate of change of is Thus we obtain the linear system Observe that the foregoing system is accompanied by the initial conditions 37

38. A Predator-Prey Model • Let x(t) and y(t) denote, respectively, the fox and the rabbit populations at any time t. If there were no rabbits, then one might expect that the foxes, lacking an adequate food supply, would decline in number according to When rabbits are present in the environment, these two species per unit time is jointly proportional to their populations x and y- that is, proportional to the product xy. • Thus when rabbits are present there is a supply food, so foxed are added to the system at a rate bxy, b>0. It gives a model for the fox population: 38

39. If there were no foxes, then the rabbits would, with an added assumption of unlimited food supply, grow at a rate that is proportional to the number of rabbits present at time t: • But when foxes are present, a model for rabbit population is decreased by cxy, c>0-that is, decrease by the rate at which the rabbits are eaten during their encounters with the foxes: • It constitute a system of nonlinear differential equations whereas a, b, c, and d are positive constants.This famous system of equations is known as the Lotka-Volterra predator-prey model. 39

40. Predator-Prey Model • Example 1. Predator-Prey Model • Suppose represents a predator-prey model. Since we are dealing with populations, we have The figure shows the population curves of the predators(fox) and prey(rabbit) for this model superimposed on the same coordinate axes. Observe that the model seems to predict that both populations x(t) and y(t) are periodic in time. As the number of prey decreases, the predator population decreases because of a diminished food supply, but attendant to a decrease in the number of predators is an increase in the number of prey; this in turn gives rise to an increased number of predators, which brings decrease in the number of prey. 40

41. Competition Models • Suppose two different species of animals occupy the same ecosystem, let us assume that the rate at which each population grows is given by respectively. Since the two species compete, each of these rates is diminished simply by the influence, or existence, of the other population. Thus a model for the two populations is given by the linear system: where a, b, c ,and d are positive constants. • And each growth should be reduced by a rate proportional to the number of interactions between the two species: 41

42. Inspection shows that this nonlinear system is similar to the Lotka-Volterra predator-prey model. It indicate that the population of each species in isolation grows exponentially, with rates indicating that each population grows logistically: when the new rates are decreased by rates proportional to the number of interactions, we obtain another nonlinear model where all coefficients are positive. The above equation are called competition models. 42

43. Networks • An electrical network having more than one loop also gives rise to simultaneous differential equations. In the figure, the current 43

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45. Example • Example. • The left figure is the network containing a resistor, an inductor, and a capacitor, and we can describe the system of differential equations by 45