SMA 208: Ordinary differential equations I

Transcription

1 SMA 208: Ordinary differential equations I Modeling with First Order differential equations Lecturer: Dr. Philip Ngare (Contacts: pngare@uonbi.ac.ke, Tue 12-2 PM) School of Mathematics, University of Nairobi March 26, 2013 P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

2 Orthogonal trajectories An orthogonal trajectory of a family of a curve is a curve that intersects each curve of the family orthogonally, that is, at right angles. For instance, each member of the family y = mx of straight lines through the origin is an orthogonal trajectory of the family x 2 + y 2 = r 2 of concentric circles with center the origin. Example Find the orthogonal trajectories of the family of curves x = ky 2, where k is an arbitrary constant. P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

3 The curves x = ky 2 form a family of parabolas whose axis of symmetry is the x-axis. If we differentiate x = ky 2, we get 1 = 2ky dy dy dx or dx = 1 2ky. This differential equation depends on k, but we need an equation that is valid for all values of k simultaneously. We eliminate k, k = x y 2, so that the differential equation can be written as dy dx = 1 2ky = ( 1 ) 2 x y 2 y or dy dx = y 2x. This means that the slope of the tangent line at any point (x, y) on one of the parabolas is y = y/(2x). On an orthogonal trajectory the slope of the tangent line must be the negative reciprocal of this slope. Therefore the orthogonal trajectories must satisfy the differential equation dy dx = 2x y. The above equation is separable, hence ydy = 2xdx = y 2 2 = x 2 + c = x 2 + y 2 2 = c. Thus the orthogonal trajectories are the family of ellipses. P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

4 Example Find, in polar form, the orthogonal trajectories of the family of curves given by r = k sec θ. Solution Differentiation of r = k sec θ gives dr r sec θ dθ = k sec θ tan θ. k =, substituting this value to the above equation, dr dθ = r tan θ, r dθ dr = 1 tan θ. Therefore, the differential equation of the orthogonal family is dr rdθ = 1 dr tan θ or r = cot θdθ, whose solution is r sin θ = c. P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

5 Questions Find the orthogonal trajectories of the family of curves. Use a graphic device to draw several members of each family on a common screen. 1 y = kx 2 2 y = cx 5 3 x 2 y 2 = k 4 y = (x + k) 1 5 y = ke x 6 x 2 + 2xy y 2 = k 7 y 2 = kx 3 8 e x cos y = k P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

6 Linear Models Questions Read lecture 1 and attempt the following questions 1 (Bacterial Growth) A culture initially has P 0 number of bacteria. At t = 1h the number of bacteria is measured to be 3 2 P 0. If the rate of growth is proportional to the number of bacteria P(t) present at time t, determine the time necessary for the number of bacteria to triple. 2 (Cooling of a cake) When a cake is removed from an oven, its temperature is measured at F. Three minutes later its temperature is F. How long will it take for the cake to cool off to a room temperature of 70 0 F? 3 A tank contains 20 kg of salt dissolved in 5000L of water. Brine that contains 0.03kg of salt per liter of water enters the tank at a rate of 25L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt remains in the tank after half an hour? P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

7 Questions (4) Psychologist interested in learning theory study learning curves. A learning curve is the graph of a function P(t), the performance of someone learning a skill as a function of the training time t. The derivative dp/dt represents the rate at which performance improves. (a) When do you think P increases most rapidly? What happens to dp/dt as t increases? Explain. (b) If M is the maximum level of performance of which the learner is capable, explain why the differential equation dp dt = k(m P) k a positive constant. is a reasonable model for learning. (c) Solve this differential equation to find an expression for P(t). What is the limit of this expression? P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

8 Questions (5) A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is converted into other substances and removed from the bloodstream at a rate that is proportional to the concentration at that time. Thus a model for the concentration C = C(t) of the glucose in the bloodstream is dc dt = r kc where k is a positive constant. (a) Suppose that the concentration at time t = 0 at time t = 0 is C 0. Determine the concentration at any time t by solving the differential equation. (b) Assuming that C 0 < r/k, find lim t C(t) and interpret your answer. P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

9 Questions (6) A certain small country has $10 billion in paper currency in circulation, and each day $50 million comes into the country s banks. The government decides to introduce new currency by having the bank replace old bills. Let x = x(t) denote the amount of new currency in circulation at time t, with x(0) = 0. (a) Formulate a mathematical model in the form of an initial-value problem that represents the flow of the new currency into circulation. (b) Solve the initial-value problem in part (a). (c) How long will it take for the new bills to account for 90% of the currency in circulation? (7) The air in a room with volume 180m 3 contains 0.15% carbon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 2m 3 /min and the mixed air flows out at the same rate. Find the percentage of carbon dioxide in the room as a function of time. What happens in the long run? P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

10 Nonlinear Models An important class of first order equations are those in which the independent variable does not appear explicitly. Such equations are called autonomous and have the form dy dt = f (y). (Population dynamics) Let y = φ(t) be the population of the given species at time t. The simplest hypothesis is concerning the variation of population is that the rate of change of y is proportional to the current value of y, that is dy/dt = ry where the constant of proportionality r is called the rate of growth(if +ve) or decline (if -ve). Assume that r > 0 ( the population is growing) and subject to initial condition y = y 0 e rt. That is the population is growing exponentially for all time. This model has been observed to be reasonable for limited period of time. P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

11 Logistic Growth To take account of the fact that the growth rate actually depends on the population, we replace r by a function h(y), that is dy/dt = h(y)y. We now want to choose h(y) so that h(y) r > 0 when y is y is small, h(y) decreases as y grows larger and h(y) < 0 when y is sufficiently large. The simplest function having these properties is h(y) = r ay, where a is also positive constant. Thus, dy/dt = (r ay)y. This equation is known as the Verhulst equation or Logistic equation. It is often convenient to write the logistic equation in the equivalent form as dy dt = r ( 1 y ) k y where k = r/a. The constant r is called the intrinsic growth rate, that is, the growth rate in the absence of any limiting factors. We can easily show that y = dy dt = r ( 1 y k ) y. y 0 k y 0 +(k y 0 )e rt is a solution of P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

12 Remark The solution, y = y 0 k y 0 + (k y 0 )e rt (1) contains the equilibrium solution y = φ 1 (t) = 0 and y = φ 2 (t) = k corresponding to the initial condition y 0 = 0 and y 0 = k respectively. From equation (1), if y 0 = 0 then y(t) = 0 for all t. If y 0 > 0 and if we let t in equation (1), then we obtain k lim t y(t) = y 0 y 0 = k. Thus for each y 0 > 0 the solution approaches the y = φ 2 (t) = k asymptotically (infact, exponentially) as t. Therefore we say that the constant solution φ 2 (t) = k is an asymptotically stable equilibrium or critical point. This means that after a long time the population is close to the saturation level k regardless of the initial population size as long as it is positive. We notice that even solution that is very near to zero grows as t increases and we have seen approach k as t. We say that φ 1 (t) = 0 is an unstable equilibrium solution. P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

13 Questions 1 The Elk population in a( small mountain ) area is given by the logistic equation dp dt = 0.1P(t) 1 P(t) 300. If the initial Elk population is 120, find P as a function of t. What is the population in ten years? When does the population double? 2 Suppose that a given population can be divided into two parts: those who have a given disease and can infect others and those who do not have it but are susceptible. Let x be proportion of susceptible individual and y the proportion of infectious individual; then x + y = 1. Assume that the disease spread by contact between sick and well members of the population and that the rate of spread dy/dt is proportional to the number of such contacts. Further, assume that members of both groups move about freely among each other, so the number of contact is proportional to the product of x and y. Since x = 1 y, we obtain the initial value problem dy/dt = αy(1 y), y(0) = y 0 where α is a positive proportionality factor and y 0 is the initial proportion of infectious individuals. P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14

14 (a) Find the equilibrium points for the differential equation above and determine whether each is asymptotically stable or unstable. (b) Solve the initial value problem above and verify that the conclusions you reached in part (a) are correct. Show that y(t) 1 as t which means that ultimately the disease spreads through the entire population. (3) One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction y of the population who have heard the rumor and the fraction who have not heard the rumor. (a) Write a differential equation that is satisfied by y. (b) Solve the differential equation. (c) A small town has 100 inhabitants. At 8 AM, 80 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor? P. Ngare (University of Nairobi) SMA 208: Ordinary differential equations I March 26, / 14