MATH 1014 N 3.0 W2015 APPLIED CALCULUS II - SECTION P. Perhaps the most important of all the applications of calculus is to differential equations.

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1 MATH 1014 N 3.0 W2015 APPLIED CALCULUS II - SECTION P Stewart Chapter 9 Differential Equations Perhaps the most important of all the applications of calculus is to differential equations. 9.1 Modeling with Differential Equations In describing the process of modeling in we can talk about formulating a mathematical model of a real-world problem through either: Intuitive reasoning about the phenomenon A physical law based on evidence from experiments The model often takes the form of a differential equation.this is an equation that contains an unknown function and some of its derivatives. 1st Order, 2nd order etc (degree, linear or non-linear). MODELS OF POPULATION GROWTH One model for the growth of a population is based on the assumption that the population grows at a rate proportional to the size of the population. Exponential growth. dp/ = kp (1) That is a reasonable assumption for a population of bacteria or animals under ideal conditions, such as: Unlimited environment. Adequate nutrition. Absence of predators. Immunity from disease. If we rule out a population of 0, then with dp/ = kp, P(t) > 0 for all t. If k > 0, then Equation 1 shows that: P (t) > 0 for all t. This means that the population is always increasing. Any exponential function of the form P(t) = Cekt is a solution of Equation 1. We will see later that there is no other solution. This is the General Solution. 1

2 Allowing C to vary through all the real numbers, we get the family of solutions P(t) = Cekt, whose graphs are shown (for k > 0). However, populations have only positive values. So, we are interested only in the solutions with C > 0.Also, we are probably concerned only with values of t greater than the initial time t = 0. 2

3 Putting t = 0, we get: P(0) = Cek(0) = C. The constant C is the initial population, P(0). This in an INITIAL CONDITION which is applied to the GENERAL SOLUTION in order to find the UNIQUE solution satisfying this requirement. Equation 1 is appropriate for modeling population growth under ideal conditions. However, we have to recognize that a more realistic model must reflect the fact that a given environment has limited resources. See for example Limits to Growth (1972 Club of Rome Report see for example or Wikipedia. Many populations start by increasing in an exponential manner. However, the population levels off when it approaches its carrying capacity K (or decreases toward K if it ever exceeds K.) A simple expression that incorporates these assumptions is given by the equation dp Pö æ = kp ç1 - è Kø (2) Equation 2 is called the logistic differential equation. It was proposed by the Dutch mathematical biologist Pierre-François Verhulst in the 1840s as a model for world population growth. Also works for spread of disease or rumours. We could rearrange this as dp/ = mp(ptot-p), m = k/ptot If P is the infected population and Ptot is the total population. In Section 9.4, we will develop techniques that enable us to find explicit solutions of the logistic equation. (Using partial fractions). For now, we can deduce qualitative characteristics of the solutions directly from Equation 2. 3

4 We first observe that the constant functions P(t) = 0 and P(t) = K are solutions. dp Pö æ = kp ç 1 - è Kø If the initial population P(0) lies between 0 and K, then the right side of Equation 2 is positive. So, dp/ > 0 and the population increases. However, if the population exceeds the carrying capacity (P > K), then 1 P/K is negative. So, dp/ < 0 and the population decreases. One equilibrium is stable, one unstable. 4

5 MODEL FOR MOTION OF A SPRING We consider the motion of an object with mass m at the end of a vertical spring. Equilibrium position balances force of gravity, or lying on a frictionless plane, etc. Hooke s Law. If a spring is stretched (or compressed) x units from its natural length, it exerts a force proportional to x: restoring force = -kx where k is a positive constant (the spring constant). If we ignore any external resisting forces (due to air resistance or friction) then, by Newton s Second Law, we have: 2 d x m 2 = - kx 5

6 This is an example of a second-order differential equation. It involves second derivatives. We can rewrite as d 2 x -k = x 2 m or d 2x 2 x 0 2 since k and m are > 0. Solutions? x = A sin ωt + B cos ωt In fact this is the general solution, A and B are arbitrary constants. How could they be determined? Initial conditions on x and dx/. GENERAL DIFFERENTIAL EQUATIONS In general, a differential equation is an equation that contains an unknown function and one or more of its derivatives. Ordinary and partial differential equations, ODEs (single independent variable) and PDEs. For ODEs indendent variable could be t, or x, or anything. In general, solving a differential equation is not an easy matter. There is no systematic technique that enables us to solve all differential equations Direction Fields and Euler s Method Suppose we are asked to sketch the graph of the solution of the initialvalue problem y = x + y; y(0) = 1. Suppose we don t know a formula for the solution. So, how can we possibly sketch its graph? The equation y = x + y tells us that the slope at any point (x, y) on the graph (called the solution curve) is equal to the sum of the x- and ycoordinates of the point. In particular, as the curve passes through the point (0, 1), its slope there must be = 1. 6

7 As a guide to sketching the rest of the curve, let s draw short line segments at a number of points (x, y) with slope x + y. 7

8 Now, we can sketch the solution curve through the point (0, 1) by following the direction field as in this figure. How about the solution with y = -1 at x = 0? dy/dx = x + y. A solution is y = -x -1 dy/dx = -1 etc. A particular solution. 8

9 9.3 Separable Equations A separable equation is a first-order differential equation in which the expression for dy/dx can be factored as a function of x times a function of y. In other words, it can be written in the form dy/dx = g(x)f(y) OR g(x)/h(y) The name separable comes from the fact that the expression on the right side can be separated into a function of x and a function of y. But the separation has to be as a product (not a sum). To solve this equation, we rewrite it in the differential form h(y) dy = g(x) dx (1) so that: All y s are on one side of the equation. All x s are on the other side. Then, we integrate both sides of the equation to give, ò h( y ) dy = ò g ( x) dx This defines y implicitly as a function of x. In some cases, we may be able to solve for y in terms of x. We can use the Chain Rule to justify this procedure. If h and g satisfy Equation 2, then d dx d dy ( ) ò h( y) dy = = g ( x) ( ò h( y) dy ) dy dx and d dx h( y ) ( ò g ( x) dx ) dy = g ( x) dx Thus, Equation 1 is satisfied. 9

10 Example 1: Solve the differential equation dy x 2 = 2 dx y Find the solution of this equation that satisfies the initial condition y(0) = 2. y = x + 3C 3 3 y = x Family of solutions above. Specific solution with y(0) = 2 shown in red. Mixing Problems A typical mixing problem involves a tank of fixed capacity filled with a thoroughly mixed solution of some substance, such as salt. A solution of a given concentration enters the tank at a fixed rate. The mixture, thoroughly stirred, leaves at a fixed rate, which may differ from the entering rate. But rate is usually the same - as in a chemostat, or a lake etc. PS 9.3: a lot of word problems, Do some! 10

11 9.4 Models for Population Growth One of the models for population growth we considered in Section 9.1 was based on the assumption that the population grows at a rate proportional to the size of the population: dp/ = kp. (1) In general, it seems reasonable that the growth rate should be proportional to the size but there may be other factors. Equation 1 is a separable differential equation. dp ò P = ò k ln P = kt + C P =e kt + C =e e C kt P = Ae kt A is the initial value of the function. P0. P (t ) = P0 e kt Variations: 1) What if k is negative? 2)We can account for emigration (or harvesting ) at a fixed rate, m, from a population by modifying Equation 1 as dp = kp - m 11

12 LOGISTIC MODEL A population often increases exponentially in its early stages, but levels off eventually and approaches its carrying capacity because of limited resources. A model for this is dp Pö æ = kp ç1 - è Kø (4) dp/ becomes negative if P ever exceeds its carrying capacity K (the maximum population that the environment is capable of sustaining in the long run). Suppose dp P ö æ = 0.08P ç 1 è 1000 ø Growth rate of 8% per year, max sustainable population Direction field is: 12

13 Note that the logistic equation is autonomous (dp/ depends only on P, not on t). So, the slopes are the same along any horizontal line. Here, we use the direction field to sketch solution curves with initial populations P(0) = 100, P(0) = 400, P(0) = 1300 Can we find an analytic solution? It is separable: Partial Fractions dp Pö æ = kp ç1 - è Kø dp ò P(1 - P / K ) = ò k K 1 1 = + P( K - P) P K - P 13

14 So 1 ö æ1 ò çè P + K - P ø dp = ò k ln P - ln K - P = kt + C K -P ln = - kt - C P K -P = e - kt -C = e - C e - kt P K -P = Ae - kt P Which gives K P= - kt 1 + Ae as our solution to the Logistic Equation. A depends on initial conditions. K - P0 = Ae0 = A P0 As an example for dp P ö æ = 0.08P ç 1 è 1000 ø with P0 = 100 we can ask when is P = 900. (t = 54.9 years) 14

15 Belgium population: A dubious example: Many countries that formerly experienced exponential growth are now finding that their rates of population growth are declining and the logistic model provides a better model. But a base of 9840 has been added not really valid as far as I can see? 15

16 A better example is protozoa growth see textbook for details. Other models: Logistical growth with harvesting. Minimum viable population (m) dp Pö æ = kp ç1 - - c è Kø dp æ P öæ m ö = kp ç1 - ç 1 - è K øè P ø Still separable equations. 16

17 9.5 Linear Equations A first-order linear differential equation is one that can be put into the form dy (1) dx + P( x) y = Q( x) where P and Q are continuous functions on a given interval. This type of equation occurs frequently in various sciences. An example of a linear equation is can be written in the form xy + y = 2x because, for x 0, it 1 y '+ y = 2 x (2) Notice that this differential equation is not separable. It s impossible to factor the expression for y as a function of x times a function of y. However, we can still solve the equation by noticing, by the Product Rule, that xy + y = (xy). So, we can rewrite the equation as: (xy) = 2x. If we now integrate both sides, we get: xy = x2 + C or y = x + C/x If the differential equation had been in the form of Equation 2, we would have had to initially multiply each side of the equation by x. We call this an Integrating Factor, generally denoted by I(x).. It turns out that every first-order linear differential equation can be solved in a similar fashion by multiplying both sides of Equation 1 by a suitable function I(x). 17

18 As a general method we try to find I so that the left side of Equation 1, dy + P ( x) y = Q( x) dx when multiplied by I(x), becomes the derivative of the product I(x)y, so I(x)(y + P(x)y) = (I(x)y) (3) If we can find such a function I, then Equation 1 becomes: (I(x)y) = I(x)Q(x) Integrating both sides, we would have: I(x)y = I(x)Q(x) dx + C So, the solution would be: y ( x) = 1 é ù I ( x ) Q ( x ) dx + C ò û I ( x) ë So how to find I(x)? To find such an I, we expand Equation 3 and cancel terms: So I(x)y + I(x)P(x)y = (I(x)y) = I (x)y + I(x)y I(x)P(x) = I (x) This is a separable ODE: di ò I = ò P( x) dx ln I = ò P ( x) dx the Integrating Factor is P ( x ) dx ò I = Ae and we take A = 1. 18

19 LINEAR EQUATIONS To summarize: To solve the linear differential equation y + P(x)y = Q(x) multiply both sides by the integrating factor P ( x ) dx ò Ix = e and integrate both sides. Example 1: Solve the differential equation 2 Linerar Equn, P(x) = 3x, 3x ò I ( x) = e 2 dx dy + 3x 2 y = 6 x 2 dx =e x3 3 d x3 (e y ) = 6 x 2 e x dx So Equn becomes RHS happens to be easily integrated, e y = ò 6 x e dx = 2e + C x3 2 x3 y = 2 + Ce x3 - x3 CHECK that it satisfies the equation. Solutions look like: 19

20 Example 2: Find the solution of the initial-value problem x2y + xy = y '+ y = 2 x x x>0 y(1) = 2 x>0 The integrating factor is: I(x) = e (1/x) dx = eln x = x. So 1 xy '+ y = x Integrating, or 1 ( xy ) ' = x 1 xy = ò dx = ln x + C x Use IC to determine C = 2> So ln x + 2 y= x PS 9.5: Maybe try 31,

21 9.6 Predator-Prey Systems In this section, we will learn about models that take into account the interaction of two species in the same habitat. Examples of prey and predators include: Rabbits and wolves in an isolated forest Food fish and sharks Aphids and ladybugs Bacteria and amoebas We let R(t) be the number of prey (R for rabbits) and W(t) be the number of predators (W for wolves) at time t. In the absence of predators, and with an ample food supply we would support exponential growth of the prey, that is, dr = kr with k > 0. In the absence of prey, we assume that the predator population would decline at a rate proportional to itself, that is, dw = - rw with r > 0. 21

22 With both species present, we assume that: The principal cause of death among the prey is being eaten by a predator. Alternately k could be the difference between natural birth and death rates. The birth and survival rates of the predators depend on their available food supply namely, the prey. We also assume that the two species encounter each other at a rate that is proportional to both populations and is, therefore, proportional to the product RW. The more there are of either population, the more encounters there are likely to be. A system of two differential equations that incorporates these assumptions is dr = kr - arw dw = -rw + brw where k, r, a, and b are positive constants. The term arw decreases the natural growth rate of the prey. The term brw increases the natural growth rate of the predators. 22

23 The equations in (1) are known as the predator-prey equations, or the Lotka-Volterra equations.they were proposed as a model to explain the variations in the shark and food-fish populations in the Adriatic Sea by the Italian mathematician Vito Volterra ( ). A solution of this system of equations is a pair of functions R(t) and W(t) that describe the populations of prey and predator as functions of time. As the system is coupled (R and W occur in both equations), we can t solve one equation and then the other. We have to solve them simultaneously. Unfortunately, it is usually impossible to find explicit formulas for R and W as functions of t. However, we can use graphical methods to analyze the equations. Suppose that populations of rabbits and wolves are described by the Lotka-Volterra equations with: k = 0.08, a = 0.001, r = 0.02, b = The time t is measured in months. Units of a, b, k, r? dr = kr - arw dw = -rw + brw 23

24 Are there equilibrium solutions? And are they stable or unstable, or neutral? Both R and W will be constant if both derivatives are 0. That is, R = R( W) = 0 W = W( R) = 0 R = W = 0 is one (unstable) equilibrium. The equations are also satisfied if 0.08 W= = R= = This means that 1000 rabbits are just enough to support a constant wolf population of 80. The wolves aren t too many which would result in fewer rabbits. They aren t too few which would result in more rabbits. Since the system is autonomous, we can eliminate t : dw dw -0.02W RW = = dr dr 0.08R RW 24

25 This is a separable ODE. We can solve (implicitly) and obtain a relationship between W and R. We can also draw a direction field, and some solution curves Note the equilibrium point (R,W) = (1000, 40). It is a neutral equilibrium. When we represent solutions of a system of differential equations as here, we refer to the RW-plane as the phase plane. A phase trajectory is a path traced out by solutions (R, W) as time goes by. 25

26 A phase portrait, as shown, consists of Equilibrium points and Typical phase trajectories. But we need direction arrows. Which way do the trajectories go? A little thought shows that we move counterclockwise around the phase trajectory. 26

27 These Lotka-Volterra equations have periodic solutions An important part of the modeling process is to interpret our mathematical conclusions as real-world predictions and test them against real data. 27

28 Not a count of actual animals, but of furs traded, implying a proportionality with populations. One way to possibly modify the Lotka-Volterra equations is to assume that, in the absence of predators, the prey grow according to a logistic model with carrying capacity K. dr æ Rö = kr ç1 - - arw è Kø dw = - rw + brw Models have also been proposed to describe and predict population levels of two species that compete for the same resources (e.g. or cooperate for mutual benefit. 28

29 Taylor, P.A. and P.J. LeB. Williams, 1975: Theoretical studies on the co-existence of competing species under continuous flow conditions. Can. J. of Microbiology, 21, In continuous-flow environments, such as a chemostat, we find that, in theory, stable mixed populations of competing species obeying Michaelis-Menten growth kinetics (extended to more than one limiting substrate) cannot coexist unless the number of growth-limiting substrates is equal to or greater than the number of species. Additional restrictions on the relative values of the growth-yield constants and input-substrate concentrations must also be satisfied. Examples of the theoretical growth of mixed cultures after initial inoculation in stable and unstable cases are presented. The ecological consequences of these findings are briefly discussed. A little mathematics can go a long way if you know how to use it. 29