2.7. Explicit Solutions to Dierential Equations 109 power (kw) 20 15 10 5 0 08:00 10:00 12:00 14:00 16:00 time 18:00 2.7 Explicit Solutions to Dierential Equations In the very rare case in which an antiderivative function can be found, we can use the antiderivative function to create explicit solutions to some simple dierential equations. In Chapter 1, we said that for every well-behaved vector field, the integral curve exists. This is the red curve, and its existence is guaranteed by the fundamental theorem on the existence and uniqueness of solutions to ordinary dierential equations. But we also said that while the red curve is known to exist, the equation for the red curve is generally unknown and unknowable, in the sense that most dierential equations do not have solutions in terms of elementary functions. We will now deal with one of the few cases in which the equation for the red curve is known. This is called an explicit solution to the dierential equation X = f (X), and we can actually write out the function X(t), and then show that ! " d X(t) = f X(t) dt We will study two of the simplest dierential equations, X = kX and X = kX. These equations have explicit solutions. Suppose an individual in a population of size X gives birth, on average, to bospring per unit time (i.e., the population has per capita birth rate b). The population has per capita death rate d, per capita immigration rate i , and per capita emigration rate e, all assumed to be constants. In this case, the per capita population growth rate r = b+ i d e is constant, and we can write the dierential equation (2.4) X = r X What behavior follows from this dierential equation? Of course, we can integrate it numerically, using SageMath. But in this case, there is an explicit solution to the dierential equation. We saw earlier that d kt e = ke kt dt so if X(t) = e kt , then X (t) = d d kt X(t) = e = ke kt = kX dt dt Therefore, the function X(t) = e kt solves the dierential equation X = kX. 110 Derivatives and Integrals If we plot both the numerical integration of X = kX and the explicit solution X(t) = e kt , we see that they agree (Figure 2.26). Indeed, e kt is the equation for the true red curve that solves the dierential equation. The discrete points are the numerical approximation (blue line) to the red curve. 4 3 X X(t) 2 1 0 10 20 30 t Figure 2.26: Left: state space and vector field for X = 0.05X. Right: time series plots showing e 0.05t (red) and Euler integration of X = 0.05X (blue). Initial condition X(0) = 1. Exercise 2.7.1 Use SageMath to plot e kt for three dierent values of k, say k = 0.1, k = 1, and k = 5. The type of growth that corresponds to these equations is called exponential growth. How fast is it? The Rate of Exponential Growth In the year 1256, the Arab scholar ibn Khallikan wrote down the story of the inventor of chess and the Indian king who wished to reward him for his invention. The inventor asked that the king place one grain of wheat (or in other versions of the legend, rice) on the first square of the chessboard, two on the second, four on the third, and so on, doubling the number of grains with each succeeding square, until the 64 squares were filled. The king thought this a very meager reward, but the inventor insisted. To the king's shock, it turned out that there was no way he could give the inventor that much grain, even if he bankrupted the kingdom. Let's plot X(t) = # of grains of rice at time t. If we plot the number of grains, starting at t = 0 with 1 grain on the first square, then the resulting first few iterations look like Figure 2.27. The red points lie exactly on an exponential curve (black), namely # of grains X(4) = 16 15 10 X(3) = 8 X(2) = 4 X(1) = 2 X(0) = 1 5 0 1 2 3 4 time Figure 2.27: The red dots are calculations of X(t) = 2t for t = 0, 1, 2, 3 and 4. The black curve is the continuous function X(t) = 2t , which exhibits exponential growth. 2.7. Explicit Solutions to Dierential Equations 111 X(t) = e (ln 2)t because e (ln 2)t = 2t . You can think of the smooth curve e (ln 2)t as representing a process in which growth occurs smoothly all the time, whereas the points represent a process in which growth happens and/or is measured only at the time points t = 0, 1, 2, 3, . . . This latter process is called a discrete-time process and it will be studied in Chapters 5 and 6. Exercise 2.7.2 How many grains of wheat would there be on the last square of the chessboard? Exercise 2.7.3 Your employer oers you a choice: be paid $1 million for thirty days of work or receive $0.01 on the first day and double your earnings each day. Which do you pick and why? Exercise 2.7.4 Use SageMath's built-in dierential equation solver or your own implementation of Euler's method to simulate equation (2.4) on page 109 for at least three dierent values of r . Mind-Blowing Fact of the Day A sheet of paper is about 0.1 mm thick. If you folded such a sheet of paper in half 50 times, the resulting stack would reach 3/4 of the way to the sun! It would take light 6.3 minutes to travel this distance. The dierential equation X = r X has the solution X = X0 e r t . Exponential Decay The dierential equation X = kX models a process in which a constant fraction k of X is removed or dies at any given time, for example, a liquid that evaporates at a constant rate k, or a chemical species that decays into another at a constant rate k X A In population dynamics, in a population whose the per capita death rate d exceeded the per capita birth rate b, growth would be governed by X = kX where k = d b. What is the behavior predicted by this model? The vector field is always negative, and the arrows get smaller and smaller as we get closer to zero (Figure 2.28 left). The dierential equation X = kX has an explicit solution: X(t) = X0 e kt Exercise 2.7.5 Verify that X(t) = X0 e kt solves the dierential equation X = kX. 112 Derivatives and Integrals X(0) = 1 1 0.8 X X 0.6 X(1) = 0.5 0.4 X(2) = 0.25 0.2 1 2 time 3 4 5 X(3) = 0.125 X(4) = 0.0625 X(5) = 0.03125 Figure 2.28: Vector field and time series for X = (ln 0.5) X The graph of X(t) depits what is called exponential decay (Figure 2.28 right). In this case, the dierential equation is X = (ln 0.5) X, and the explicit solution is X(t) = e (ln 0.5)t . Just as in exponential growth, we can imagine exponential decay it as a continuous process governed by a dierential equation, or as a discrete process in which, say, half of what is left is removed every night at midnight. When Models Break Down Over short periods of time, populations can be modeled as growing or decaying exponentially. However, if a population truly underwent exponential growth, it would never stop growing. An exponentially growing population of aardvarks, elephants, or naked mole rats would reach the mass of the Earth, then the solar system and, eventually, the universe. Clearly, something must prevent real populations from growing exponentially, and if we want a population growth model to be useful over more than just the short term, the exponential model must be modified to incorporate processes that stop the population from growing. Similarly, exponential decay can't be pushed too far. If every night, your roommate removes half of the chocolate left in the fridge, then after a month of this, the fraction of chocolate left would be 2130 10110 , which is smaller than a molecule of chocolate. The point here is not to bash exponential growth or decay as a model. It fails in a particularly spectacular way, but all models fail at some point. A perfect model would be as complex as the system being modeled and therefore useless. Rather, the lesson is to always do a sanity check when working with models. When using mathematical models in biology, keep an open mind some strange phenomena first discovered in models have been found in the real worldbut use your biological knowledge to judge when a model's predictions are suciently wrong to require the model to be modified in order to be useful for the task at hand. There is no universal recipe for doing this, which is why modeling is often called an art. Exercise 2.7.6 Name two biological factors or processes that might stop a population from growing. Exercise 2.7.7 A population of 20 million bacteria is growing continuously at a rate of 5% an hour. How many bacteria will there be in 24 hours? 2.7. Explicit Solutions to Dierential Equations 113 Exercise 2.7.8 A pollutant breaks down at the rate of 2% a year. What fraction of the current amount will be present in 20 years? Let X(0) = 1. Further Exercises 2.7 1. The population of an endangered species is declining at a rate of 2.5% a year. If there are currently 4000 individuals of this species, how many will there be in 20 years? 2. If money in a bank account earns an interest rate of 1.5%, compounded continuously, and the initial balance is $1000, how much money will be in the account in ten years? 3. Radioactive iodine, used to treat thyroid cancer, has a half-life of eight days. Find its decay constant, r . (Hint: You'll need to use natural logarithms.) 4. Mary is going to have an outdoor party in 10 days. She wants to have her backyard pond covered in water lilies before the party, so she goes to the nursery to buy some water lilies. Mary gives the clerk the dimensions of her pond, and the clerk, knowing the growth rate of the water lilies that he stocks, calculates that if she purchases a single water lily, it will produce a population of ten thousand lilies that will completely cover the surface of the pond in 20 days. Mary reasons that if she buys two water lilies instead of one, she can meet her goal of having the pond surface covered in 10 days. Is there anything wrong with Mary's reasoning? How many water lilies will Mary need to buy to meet her goal? 5. You have $10,000 and can put it either in an account bearing 3.9% interest compounded monthly or one bearing 4% interest compounded annually. If the money will be in the account for five years, which one should you choose? 6. The economic activity of a country is often quantified as the gross domestic product (GDP), which is the sum of private and government consumption, investments, and net exports (the value of exports minus the value of imports). For a developed country such as the United States, economists might see a GDP growth rate of 3% a year as reasonable. However, production and consumption create some pollution. By how much would pollution per dollar of GDP have to decline for pollution levels 50 years from now to be the same as current levels, assuming a 3% annual growth rate of GDP? In 75 years? In 100 years? (Hint: Let the current pollution level be 1 and find out what the future pollution level would be). 7. If you want to approximate the time it takes an exponentially growing quantity to double, you can divide 70 by the percentage growth rate. For example, a population growing at 2% a year has a doubling time of about 35 years. Find an exact equation for doubling time and explain why the rule of 70 works as an approximation. (Hint: Start by trying a few concrete examples.) Chapter Equilibrium Behavior 3.1 When X Is Zero A major clue to the behavior of dynamical systems is given by the existence and location of equilibrium points. These are points in state space at which the system does not change. More formally, an equilibrium point of a dierential equation X = f (X) is a point X0 at which f (X0 ) = 0. Since a dierential equation specifies a vector field, we can also say that such a point is an equilibrium point of the vector field X = f (X). So far, we have studied dierential equation models almost entirely by simulating them. Simulation is a powerful tool. Indeed, it is sometimes the only available one. But it can also lead us astray. To see one example of how, consider the following modification of the logistic equation: X X )( 1) (3.1) k a We will delay discussion of the biological meaning of this equation until page 123. One way to study this equation is to pick some values for r , a, and k and an initial condition X(0) and numerically integrate the resulting equation. Figure 3.1 does this for three values of X(0). X = r X(1 200 150 X 100 50 0 4 8 12 time Figure 3.1: Simulations of equation (3.1) with r = 0.1, k = 100, a = 5 and initial conditions X(0) = 10 (red), X(0) = 20 (black), X(0) = 200 (blue). Looking at Figure 3.1, we might think we understand how the model behaves. If the population starts below k, it grows slowly, speeds up, and then gradually reaches k. If it starts above k, it gradually declines to that level. The new equation appears to behave just like the logistic. c Springer International Publishing AG 2017 115 A. Garfinkel et al., Modeling Life, DOI 10.1007/978-3-319-59731-7_3 3 116 Equilibrium Behavior But what happens if we try one more value for X(0)? Suppose we start with an initial condition X(0) = 4 and all parameters as before. Now the population declines instead of growing (Figure 3.2). 100 X 50 0 10 20 30 time Figure 3.2: An additional simulation of equation (3.1) with initial condition X(0) = 4 (green, arrow). Note that the population declines. The trick here, of course, is that the value we chose for a was 5. All initial values in Figure 3.1 were above a, while the one in Figure 3.2 is below a. Equation (3.1) models a situation in which populations fail to grow below a certain threshold size, namely a. We will return to this model later in this section. It would be very useful to be able to figure out that that other behavior was possible, since we can't run simulations from every initial condition. There is such a method. In one dimension, it tells us the whole behavior of the system, and even in higher dimensions, it gives us very important landmarks that determine system behavior. The first thing we need to do is to find the points where the system is not changing, that is, the equilibrium points. 3.2 Equilibrium Points in One Dimension As we learned in calculus, the derivative of a constant is zero. This is true because a derivative is a rate of change, and the value of a constant function doesn't change. Looking at the same issue geometrically, we note that the graph of a constant function is a horizontal line, and the slope of a horizontal line is zero. The converse is also true. If the derivative of a function at some point is zero, the value of the function is not changing at that point. In the context of dierential equations, such points are called equilibrium points (or fixed points or constant steady states). Equilibria are very important to the dynamics of all models, and they are especially important in single-variable models. The dynamics of such models are very limitedthe state value can either grow without limit or go to an equilibrium point. There are also special cases, in which the state variable is something like the position of a runner on a closed track, or the position of the hand on a clock, where moving in the same direction can bring you back to your starting point. In those cases, oscillations are possible. (See, for example, the angular variable for the pendulum in Chapter 6.) Finding Equilibria How do we find the equilibria of a dierential equation? We know that at an equilibrium point, the derivative is zero. Since what a dierential equation gives us is the derivative, all we have to do is set the equation equal to zero and solve for the state variable. 3.2. Equilibrium Points in One Dimension 117 For example, the logistic equation, dX X = r X(1 ) dt k is a common model for population growth that we've already encountered. To find its equilibria, we need to find the values of X for which X 0 = r X(1 ) k These can be found either by multiplying the expression out and then solving the resulting algebraic equation, or by looking thoughtfully at the right-hand side and seeing that it is the product of two terms. The only way the product of several quantities can be zero is for at least one of those quantities itself to be zero. Looking at the logistic equation shows that it is equal to zero if X=0 so X = 0 is one equilibrium point. If X isn't zero, the population could still be at equilibrium if 1 X =0 k This occurs when Xk = 1, implying that X = k is another equilibrium point. Have we found all the equilibria of the logistic equation? Multiplying it out would give a term with X 2 , and since a quadratic equation has at most two distinct solutions, we are done. To find the equilibrium points of a dierential equation X = f (X), set X = 0 and solve the resulting equation to find the values of X that make X = 0. Exercise 3.2.1 Consider a population of organisms that reproduce by cloning and have genotypes A and a with per capita growth rates rA and ra . If we denote the fraction of the population having genotype a by Y , the equation describing how the prevalence of genotype a changes is dY = (ra rA )Y (1 Y ) dt a) What does the quantity 1 Y represent? b) Using reasoning similar to what we used for the logistic equation, find the equation's equilibria. Explain how you know you have found all of them. Stability of Equilibrium Points Having found a model's equilibrium points, we next want to know whether the system will stay at these equilibria if perturbed. We might also be interested in knowing whether the system can spontaneously reach a particular equilibrium if it did not start there. These are questions about the stability of equilibria. A simple way of thinking about stability is illustrated in Figure 3.3. The picture on the left illustrates a stable equilibriumif the ball in the cup is given a slight push, it will return to the bottom of the cup. In the picture on the right, the ball is on a hilltop and will roll away, never to return, with even a tiny push. 118 Equilibrium Behavior Figure 3.3: Stable (left) and unstable (right) equilibria. Returning to the language of dynamical systems, if an equilibrium is stable, the system returns to it after a small perturbation. If it is unstable, even a tiny perturbation will send the system to a dierent equilibrium or a trajectory of infinite growth. Stability Analysis 1: Sketching the Vector Field In one dimension, a model's vector field can be used to completely figure out whether equilibria are stable. We start by drawing a line to represent the system's state space. Then we need to find the equilibrium points and mark them on this line. Let's use the logistic equation as an example: X ) k As we saw above, the equilibrium points of this equation are X = 0 and X = k (Figure 3.4). X = r X(1 X=0 X=k X Figure 3.4: Equilibrium points of the logistic equation. Note that the points divide the line into intervals. The next task is to determine whether the equilibrium points are stable or unstable. In other words, we need to find out how the state point would move if nudged o an equilibrium point: toward it or away from it. In one dimension this is easy; all we have to do is figure out whether the point moves to the left or to the right, which means that we need to find out whether the sign of the vector field is positive or negative. Since the point moves to the left when the state variable is decreasing, leftward movement corresponds to a negative value of X ; similarly, rightward movement corresponds to a positive value of X . Thus, we need to figure out whether X is positive or negative on each interval and draw change vectors accordingly, as in Figure 3.5. In this case, only the direction of the vectors matters; we don't have to worry about their length. A drawing of a model's state space showing equilibrium points, a few representative change vectors, and key trajectories is called a phase portrait of the model. Figure 3.5 is a phase portrait of the logistic equation. (Often, in one dimension, we omit drawing the trajectories and show only the change vectors. The trajectories are obvious.) X=0 X=k X Figure 3.5: A phase portrait of the logistic equation showing unstable (left) and stable (right) equilibria. 3.2. Equilibrium Points in One Dimension 119 To create Figure 3.5, it was necessary to find the sign of the logistic equation on various intervals of the state space. Luckily, this can be done without any algebraic calculations. Instead, we take advantage of the following observations: 1) The parameter r , the population's per capita growth rate in the absence of intraspecific competition, is positive. 2) X is a population size, so it must be nonnegative (positive or zero). 3) The carrying capacity k is also a population size and must be positive. X = rX !"#$ non negative (1 X ) k To find the sign of X , we first note that r X is always nonnegative, so the sign of 1 Xk determines the sign of X . When X < k, 1 Xk is positive, and when X > k, 1 Xk is negative. Therefore, X is positive on the interval 0 < X < k and negative on the interval X > k. This gives the phase portrait in Figure 3.5. Once the phase portrait is drawn, stability becomes obvious. An equilibrium point is stable if the vector field would move the system back to the equilibrium if it was nudged o; if the vector field would carry the system away from the equilibrium point, that equilibrium point is unstable. If vectors on one side of the equilibrium point toward it and those on the other side point away from it, we say the equilibrium is semistable. Semistability is an unusual situation that we will not devote much attention to. Exercise 3.2.2 Draw phase portraits to confirm each of the above statements. Stability Analysis 1 (Continued): The Method of Test Points In the logistic equation example, it was easy to sketch a phase portrait of the system simply by looking at the sign of each term in the equation and multiplying the signs. However, there are many models for which this won't work, or at least won't be as simple. For example, consider a population that undergoes logistic growth but also has 10% of individuals removed every year, say by fishing. If r = 0.2 and k = 1000, the change equation for this system is X ) 0.1X 1000 This system's equilibria are X = 0 and X = 500. X = 0.2X(1 Exercise 3.2.3 Confirm that the equilibria given above are correct. As before, we can draw the state space and mark the equilibria at X = 0 and X = 500, dividing the line into intervals. However, it's no longer obvious how to find the sign of X in each interval. For reasons that we will soon explain, it is enough to pick one point in each interval and find the sign of X at that point. In the region between 0 and 500, we might choose X = 100. Then X = 0.2100(1100/1000)0.1100 = 8, so the change vectors point to the right. Above 500, we can use X = 1000. Then, X = 0.2 1000 (1 1000/1000) 0.1 1000 = 100, so the change vectors point to the left. This means that the equilibrium at X = 0 is unstable and the one at X = 500 is stable (Figure 3.6). 120 Equilibrium Behavior 0 X 500 Figure 3.6: Phase portrait for the logistic growth with harvesting example. Exercise 3.2.4 Find the equilibria of X = 0.1X(1 determine their stability. X 800 ) 0.05X and use test points to You may wonder what allows us to use only one point in each interval. How do we know that the sign of X won't change between adjacent equilibrium points? The dierential equations that we deal with are nearly always continuous functions. Informally, saying that a function is continuous just means that you can draw its graph without lifting your pen from the paper. If a function is continuous, it can't jump from one value to anotherit has to pass through all the values in between. (This is called the intermediate value theorem.) This matters for our purposes, because when a continuous function goes from positive to negative or vice versa, it has to pass through zero. Since the function in question is X , every value at which it is zero is an equilibrium point. But we've already found and plotted all the equilibria! Thus, X can't change sign between equilibria, and we can use test points to perform graphical stability analysis. Exercise 3.2.5 Draw several functions (by hand or using SageMath) to convince yourself that a continuous function can't change sign without passing through zero. Stability Analysis 2: Linear Stability Analysis Drawing vector fields to determine stability works wonderfully in 1D, somewhat in 2D, badly in 3D, and not at all in higher dimensions. The more general way to find the stability of an equilibrium point is to use linear approximation. Here we will illustrate this method for a onedimensional vector field, but in Chapter 6, we will see it in its full glory in n dimensions. We begin by making a new kind of plot. Since X is a function of X, we can plot this function in a graph. We will put on the X axis the state space X, and on the Y axis we put the vector field X , which is f (X). Note the places where the graph of X intersects the X axis, that is, the line X = 0. The intersection points are equilibrium points. Exercise 3.2.6 Why is this true? If we make this plot for the logistic vector field, X = X(1 X ) k we get Figure 3.7. As we already know, there are two equilibria, at X = 0 and X = k. The one at X = 0 is unstable, while the one at X = k is stable. 3.2. Equilibrium Points in One Dimension 121 Now we can connect the vector field X = f (X) to the graph of f (X). When the graph is above the X axis, X is positive, which means that X is increasing, and when the graph is below the X axis, X is negative, which means that X is decreasing. Look at Figure 3.7. The equilibrium point at X = 0 occurs when f (X) goes from negative to positive. If f (X) goes from negative to positive, it is increasing. This means that the tangent to f (X) at X = 0 has a positive slope, as shown in Figure 3.8. X' = f(X) 0 k X Figure 3.7: Vector field plot for logistic vector field. The black curve shows X at each point X in state space. The points in X at which the curve intersects the horizontal axis (X = 0) are equilibrium points of the vector field. Here they are at X = 0 and X = k. X' = f(X) 0 k X Figure 3.8: Graphical linear stability analysis. The slope of the tangent lines at the equilibrium points determines the stability of the equilibrium point: positive slopes imply unstable equilibrium points, and negative slopes imply stable equilibrium points. Let's put this another way. This is the most important way, and it generalizes to n dimensions. Since f (X) is a function of X, we can dierentiate it like any other function of X. This gives us the derivative of f (X) with respect to X, d( dX dX df (X) dt ) = = dX dX dX We know that the derivative of a function at a point gives us the slope of the tangent to the graph of the function at that point. Clearly, if the slope of the tangent is positive, then the function is going from negative to positive. Consider the equilibrium point at X = 0. The slope of the tangent to f (X) at X = 0 (the red line passing through X = 0) is positive, so X is going from negative to positive. But that means that the change vectors to the left of the equilibrium point to the left, and the change vectors to the right of the equilibrium point to the right. In other words, X = 0 must be an unstable equilibrium point! 122 Equilibrium Behavior Now let's look at the equilibrium point X = k. Here the slope of the tangent (the red line passing through X = k) is negative, which means that X is going from positive to negative. Therefore, the change vectors to the left of the equilibrium point to the right, while the change vectors to the right of the equilibrium point to the left. In other words, X = k must be a stable equilibrium point.1 We have discovered a deep truth: the stability of an equilibrium point of a vector field is determined by the linear approximation to the vector field at the equilibrium point. This principle, called the Hartman-Grobman theorem, enables us to use linearization to determine the stability of equilibria. The 1D version of the Hartman-Grobman theorem, also called the principle of linearization, says that if the slope of the linear approximation to a vector field at an equilibrium point is positive, then the equilibrium point is unstable, and if the slope is negative, the equilibrium point is stable. Exercise 3.2.7 Sketch graphs of two functions, as in Figure 3.7. (No equations are needed.) For each function, which we'll refer to as f (X), sketch the vector field of X = f (X). Mark the equilibrium points and indicate their stability. Calculating the Linear Approximation Since the derivative of a function is the slope of the linear approximation to the function, this method of using derivatives to learn about stability is called linear stability analysis. It works (X) (X) is not equal to zero. If dfdX = 0, graphical methods are required. whenever dfdX We can actually calculate these linear approximations by calculating the derivative. In the example above (with r = 1 for simplicity), X ) k we can calculate the derivative of f (X) at the point X as f (X) = X(1 df (X) 2X =1 dX k At X = 0, that yields and at X = k, % df (X) %% = +1 dX %X=0 % df (X) %% = 1 dX %X=k So the equilibrium point at X = 0 is unstable, and the equilibrium point at X = k is stable. 1A rare. semistable equilibrium can also occur when the function touches zero without changing sign, but this is 3.2. Equilibrium Points in One Dimension 123 At an equilibrium point X (pronounced \"X-star,\" a common notation for equilibria): % (1) If df (X) % is an unstable equilibrium. is positive, then X (2) If dX X=X df (X) dX |X=X is negative, then X is a stable equilibrium. Exercise 3.2.8 Find the equilibria of the dierential equation N N )( 1) 1000 50 and use linear stability analysis to find their stability. Then, use the graphical method to check your results. N = 0.1N(1 Exercise 3.2.9 Do the same thing for the model Y = (1 32 Y )Y (1 Y ). Exercise 3.2.10 Suppose we try to evaluate the stability of the X = 0 equilibrium point of the vector field X = 2X 2 X a) Perform a linear stability analysis at the point X = 0. What is the character of this equilibrium point according to this analysis? b) Suppose we did a test point analysis for confirmation and chose two test points, X = 1 and X = +1. When we calculate the change vectors X at these two points, we see that the change vector at X = 1 is positive, % X %X=1 = 2(1)2 (1) = 3 and the change vector at X = +1 is also positive, % X %X=+1 = 2(+1)2 (+1) = 1 Explain why this test point method conflicts with the linear stability analysis. What have we done wrong? (Hint: Plot the X function.) Example: The Allee Eect In some species, a minimal number of animals is necessary to ensure the survival of the group. For example, some animals, such as African hunting dogs, require the help of others to bring up their young. As a result, their reproductive success declines at low population levels, and a population that's too small may go extinct. This decline in per capita population growth rates at low population sizes is called the Allee eect. As an example of the Allee eect, consider the strategy employed by Elon Musk, the developer of the Tesla electric car. Musk announced that he would give away, free of charge, all the patents that his company held on electric cars. These patents are valuable. Why would he give them away? Because he realized that for electric cars to succeed, they require substantial infrastructure: tax benefits, dedicated highway lanes, and public recharging networks. None of these would happen if there was only one electric car company. In other words, if there were only one electric car company, there would soon be no electric car companies. A critical mass is necessary. 124 Equilibrium Behavior We can model the Allee eect by adding another term to the logistic equation. The modified equation becomes X X X = r X(1 )( 1) k a We already saw this model in equation (3.1). Let's carry out the full analysis of this equation. Equilibrium Points We start by finding the equilibrium points. Setting X = 0, we solve X X )( 1) k a by realizing that the product of three terms can be 0 only when at least one of them is 0. So we have three choices: X = 0, X = k, and X = a (Figure 3.9). 0 = r X(1 0 a k X Figure 3.9: State space for the Allee eect model, with its three equilibrium points (black dots). Stability 1: Method of Test Points We can determine the stability of each of the three by choosing appropriate test points. If we choose values of X in the three intervals 0 < X < a, a < X < k, and k < X, and calculate the change vectors X , we see the direction of flow (Figure 3.10). 0 a k X Figure 3.10: By drawing representative change vectors on state space, we can easily see the stability of the system's equilibrium points. Clearly, X = 0 and X = k are stable equilibrium points and X = a is unstable. (To see the time series of these flow simulations, see Figure 3.2 on page 116). Exercise 3.2.11 Judging by the phase portrait, what is the biological meaning of a? Exercise 3.2.12 Choose values for r , a, and k. Find the model's equilibria and use test points to determine their stability. Stability 2: Linear Stability Analysis Finally, let's confirm this with linear stability analysis. First, we graph X as a function of X (the black curve in Figure 3.11). We see that the linear approximation to X at the equilibrium point X = 0 has a negative slope, the linear approximation at X = a has a positive slope, and the linear approximation at X = k has a negative slope. Therefore, by the principle of linearization, 3.2. Equilibrium Points in One Dimension 125 the equilibrium point at X = 0 is stable, the equilibrium point at X = a is unstable, and the equilibrium point at point X = k is stable. X' 0 a k X X Figure 3.11: Graphical linear stability analysis for the Allee eect model. We can confirm this by a calculation. In order to calculate dX dX , the easiest method is to multiply out the terms in X X r r r X = r X(1 )( 1) = X 3 + X 2 + X 2 r X k a ak a k and then use the power rule to dierentiate X , giving dX 3r 2r 2r = X2 + X + X r dX ak a k If we then plug the three values X = 0, X = a, and X = k into this expression, we get dX %% = r % dX X=0 a dX %% = r (1 ) % dX X=a k k dX %% = r (1 ) % dX X=k a Since we are assuming that a is less than k, these values are negative, positive, and negative. Therefore, we have confirmed the results obtained by looking at the vector field. Exercise 3.2.13 Carry out the same analysis for the example you chose in Exercise 3.2.12. Example: Game Theory Models in Evolution and Social Theory Much significant modeling has been done using models of the dynamics of games, with applications to evolution and also to social theory. Game theory was introduced into evolution as a way of talking about how dierent genes might succeed or fail in various environments. For example, suppose you are a bird that can have a gene for oily feathers or a gene for dry feathers. Which is better? It depends! If it is going to rain, then you would definitely prefer oily feathers that can shed the rain, but if it is going to be dry, then you would prefer dry feathers. So we can 126 Equilibrium Behavior view the choice as a game: you, as the bird, are the gambler. You can bet on \"oily feathers\" or \"dry feathers.\" Every individual in the population makes such a bet. The croupier spins the wheel, and it comes up \"rain,\" with probability X, or \"dry,\" with probability 1 X. When it comes up \"rain,\" she pays the bet on \"oily feathers\" and rakes in the chips from the bet on \"dry feathers,\" and when the wheel comes up \"dry,\" she does the opposite. In social theory, game theory models are used to explain how various patterns of behaviors can evolve in society, such as how cooperation develops among self-interested individuals (for example, in the game called \"prisoners' dilemma\"). These games are all described by dierential equations. We will study a simple model here. The basic idea is really already familiar to you. We will imagine two strategies, call them A and B; we will use the letters A and B as state variables to represent the numbers of people (or animals) currently playing each strategy. Which strategy you play is determined by whether you have the A or B genotype.2 Thus, these are basically population dynamics models like the shark-tuna model of Chapter 1. The basic idea is to write A = rA A B = rB B where rA and rB are the reproductive rates of individuals carrying the two genotypes. Only now rA and rB are not going to be constant, but will vary: the reproductive rate will be a direct outcome of success in previous encounters. More specifically, rA is proportional to A's success in recent encounters, and rB is proportional to B's success in recent encounters: rA A's success in recent encounters rB B's success in recent encounters (The sign \"\" is read \"proportional to.\") The Replicator Equation Instead of looking at the raw numbers of individuals playing A or B, we will look at the fraction of the population that each group represents. These are A B X= and Y = A+B A+B Let's form a dierential equation for X by dierentiating this expression. & ' A X = A+B Now we need the quotient rule, which gives us & ' A (A + B)A A(A + B) = A+B (A + B)2 + BA AB AA AA = (A + B)2 But A = rA A and B = rB B, so X = 2 For rA AB rB BA (A + B)2 simplicity, we assume that all individuals are haploid. 3.2. Equilibrium Points in One Dimension Recall that X = 127 A B and Y = = 1 X, so this gives us A+B A+B X = (rA rB )X(1 X) This is called the replicator equation. Payos Now we need to find a model for the reproductive rates rA and rB . We said that reproductive rate previous success But what is previous success? It consists in the success of encounters with individuals of the same genotype and encounters with individuals of the other genotype. So the payo to rA is & ' & ' the payo for the payo for X + Y X-X encounters X-Y encounters What is the payo for these encounters? It varies from game to game! A number of dierent games have been proposed as evolutionary models. Here we will study one of them. Hawks and Doves We will now apply stability analysis to a classic problem in the evolution of behavior. This example will illuminate why dierent genotypes can persist in a population. Suppose that an animal population consists of individuals of two genotypes, \"hawks\" (A) and \"doves\" (B). These individuals compete for access to a resource, such as mates or food. Hawks always fight when they encounter a competitor, while doves share the resource equally on encountering another dove and bow out on encountering a hawk. Fighting carries a substantial cost for the loser. In this example, the cost of losing a fight is 3, so its payo value is 3, while the value of the resource gained is +2. All hawks have the same fighting ability, so the probability of a hawk winning a fight with another hawk is 50%. Therefore, the expected payo to a hawk when it encounters another hawk is a 50% chance of +2 and a 50% chance of 3, giving a total expected value of 0.5 (+2) + 0.5 (3) = 0.5. Doves never fight. When a dove encounters another dove, they split the resource, whose value is still +2, so the outcome for a dove encountering another dove is +1. When a hawk encounters a dove, the hawk takes the resource, but the dove doesn't risk fighting. Therefore, the benefit to the hawk is +2, while the dove incurs neither a cost nor a benefit. The costs and benefits of various encounters are summarized in the payo table (or payo matrix) in Table 3.1. Hawk(A) Dove(B) Hawk(A) Dove(B) (0.5, 0.5) (0, +2) (+2, 0) (+1, +1) Table 3.1: Payo table describing the costs and benefits to participants in hawk-dove interactions. 128 Equilibrium Behavior We want to know what will happen to the prevalence of hawk and dove genotypes over time. We will denote the fraction of the population consisting of hawks as X. Since all individuals are either hawks or doves, the fraction of the population that consists of doves is 1 X. Next, we define the per capita growth rate of each genotype as the sum of the outcomes of its interactions with members of the same and the other genotype. For example, if rA is the per capita growth rate of hawks, then rA = 0.5 ! "# $ payo when a hawk encounters another hawk X !"#$ + frequency of encountering another hawk 2 !"#$ 1 !"#$ payo when a hawk encounters a dove (1 X) ! "# $ frequency of encountering a dove = 2 2.5 X Similarly, the per capita growth rate of doves is rB = 0 !"#$ payo when a dove encounters a hawk X !"#$ frequency of encountering a hawk + payo when a dove encounters another dove (1 X) ! "# $ frequency of encountering another dove =1X Substituting for rA and rB in the replicator equation, which we derived earlier, gives X = dX = (rA rB )X(1 X) dt ' ( = 2 2.5X (1 X) X(1 X) = (1 1.5X)X(1 X) So the hawk-dove dierential equation is X = (1 1.5X)X(1 X) (3.2) where X is the fraction of the population who are hawks. Exercise 3.2.14 Explain the values in the payo table of Table 3.1. Exercise 3.2.15 Derive a similar equation for the payo table. Hawk(A) Dove(B) Hawk(A) Dove(B) (1, 1) (0, +3) (+3, 0) (+0.5, +0.5) Equilibrium Points How will this system behave? Let's begin by finding equilibrium points. If we set X = 0, two of the equilibria of this equation, X = 0 and X = 1, can be immediately found by inspection of the equation. We find the third one by solving the equation 1 1.5X = 0 3.2. Equilibrium Points in One Dimension 129 which gives the third equilibrium point (Figure 3.12), X= X=0 2 3 2 X= 3 X X=1 Figure 3.12: Equilibrium points of the hawk-dove dierential equation. Stability 1: Method of Test Points To find the stability of these equilibria, we need to know the sign of X on the intervals 0 < X < 23 and 23 < X < 1. One easy way to do this it to pick a value in each interval and plug it into the hawk-dove dierential equation (equation (3.2)). For the interval 0 < X < 23 , we can use X = 0.5. Then X > 0, 1X > 0, and 11.50.5 > 0, so X > 0 in the left-hand interval. For the interval 23 < X < 1, we can use the point X = 0.8. For this value of X, X and 1 X remain positive, but 1 1.5 0.8 < 0, so X < 0. Thus, the method of test points tells us that X = 0 and X = 1 are unstable equilibrium points, while X = 23 is stable (Figure 3.13). X=0 2 X= 3 X X=1 Figure 3.13: Stability of equilibria for the hawk-dove model, by the method of test points. Stability 2: Linear Stability Analysis To use linear stability analysis, we first plot X = f (X), giving us the black curve in Figure 3.14. Note the three places the curve intersects the X = 0 axis, representing the three equilibrium points. The tangents to the curve at the three points are shown in red. Their slopes are obviously positive, negative, and positive, indicating that the equilibrium points are unstable, stable, and unstable. X' 0 2 3 1 X X Figure 3.14: Linear stability analysis of the hawk-dove dierential equation. 130 Equilibrium Behavior Finally, we confirm the linear stability analysis by calculating the sign of the derivatives at the three equilibrium points. First, we multiply out the X equation, X = (1 1.5X)X(1 X) to give X = 1.5X 3 2.5X 2 + X Then we use the polynomial rule to dierentiate this expression, dX d = (1.5X 3 2.5X 2 + X) dX dX = 4.5X 2 5X + 1 evaluated at X0 = 0, evaluated at X0 = 23 , and evaluated at X0 = 1, dX %% = +1 % dX X0 =0 dX %% 1 = % dX X0 = 23 3 dX %% = +0.5 % dX X0 =1 Therefore, we have confirmed that the three equilibrium points are unstable, stable, and unstable. Exercise 3.2.16 Redo this stability analysis for the model you obtained in Exercise 3.2.15. Use both methods. So the overall conclusion of our analysis of the hawk-dove game is that the two populations, hawks and doves, will evolve to a stable equilibrium at X = 23 . In other words, the population will evolve to a stable state in which there are two hawks for every dove. Notice that this conclusion is far from obvious. This is why we model. It would be very easy to wave our hands, consult our personal intuition, and say \"Oh, the hawks will prevail; it will be all hawks,\" or \"Oh, the hawks will kill each other and the doves will prevail.\" It turns out that neither scenario is true. The model predicts the coexistence of the two genotypes, in the ratio 2: 1. Other evolutionary games include \"stag hunt,\" which is a model of group collective behavior, \"prisoners' dilemma,\" which is a model of cooperation and competition, and \"rock/paper/scissors,\" which is a model of cyclic population dynamics. 3.2. Equilibrium Points in One Dimension Further Exercises 3.2 1. A kayaker is paddling directly into the wind but the kayak keeps veering either left or right. a) Use your physical intuition to explain why the kayaker is having diculty going straight. b) Describe this situation in terms of equilibria and stability. (Hint: Sketch a vector field. No equations are necessary.) 2. The spread of a genetic mutation in a population of mice can be modeled by the dierential equation P = 2P (1 P ) (1 3P ) where P is the fraction of the mice that have the new gene. (This means that 0 P 1.) a) Find the equilibrium points of this model and determine the stability of each one. b) If 10% of the mice have the new gene (so P = 0.1) initially, what fraction of the population will have the new gene in the long run? c) What if the initial fraction is 90% of the mice? 3. The von Bertalany growth model, which can be used to model the growth of individual organisms, is given by L = r (k L) where L is the length of the organism, and r and k are positive constants. Find the equilibrium point(s) for this model and determine their stability. How large will the organism eventually grow? 4. The Gompertz growth model, which is sometimes used to model the growth of tumors, is given by & & '' X X = X k ln X(0) where X is the mass of the tumor, and k and are positive constants. Find the equilibrium points for this model and determine the stability of each one. How large will the tumor eventually grow? 5. (Modified from Strogatz) Consider a system of two chemical compounds, A and X. One molecule of A and one of X react to produce two molecules of X, with rate constant 0.1. Also, two molecules of X can react to form one molecule of X and one molecule of A, with rate constant 0.05. 2X A+X 131 132 Equilibrium Behavior The amount of A is much larger than that of X, so its concentration can be thought of as a constant, 2. a) Write a dierential equation for the concentration of X. (Hint: Look back at the predator-prey and disease models studied in earlier sections.) b) Find the equilibria of this system and describe their stability. 6. This problem will look at equilibria in chemistry more generally. You may find it helpful to review Section 1.4 on page 34. kf a) In the chemical equation A B, what do kf and kb mean in dynamical terms? kb b) Write models for the following chemical reactions: k 1. A B k 2. A + B C kf 3. A B kb c) Look back at all the models you just wrote. Do you notice anything unusual about the equations? d) Use the observation you just made to help you find an expression for the equilibkf rium of the reaction A B. (Solve for kkbf .) The expression you get is called an equilibrium constant. kb kf e) Do the same thing for A + B C + D. kb k f) Write a model for the reaction 2A B. (Hint: Keep the coecients in mind.) k f g) Write a model for 2A + B C + 3D and find the equilibrium constant. (Hint: kb How many molecules of each substance are coming together in each reaction?) k f h) Write a model for aA + bB cC + dD and find the equilibrium constant. If the kb result doesn't look familiar, it should after you take more chemistry. 7. Is it possible for a one-dimensional system to have two stable equilibria without an unstable one between them? Explain. (Hint: Try drawing the situation.) 8. How could you use simulation (numerical integration) to determine whether an equilibrium point of a dierential equation is stable or unstable? 9. In the text, we said that linear stability analysis fails if df dX |X0 = 0. Here, we will see why. a) All of the following dierential equations have an equilibrium point at X = 0. By looking at the vector field, determine the stability of this equilibrium point for each equation. a) X = X 3 b) X = X 3 c) X = X 2 d) X = X 2 3.3. Equilibrium Points in Higher Dimensions b) Now find % df % dX X=0 133 for each function. What do you notice? 10 The text said that semistable equilibrium points are rare. Here, we will see why. a) X = X 2 has an equilibrium point at X = 0. Determine the stability of this equilibrium point. b) Use graphical methods to find the equilibria of X = X 2 + a for at least one positive and two negative values of a. For each value of a, determine the stability of the equilibria. c) Use your findings to explain why semistable equilibria rarely occur in real life. 3.3 Equilibrium Points in Higher Dimensions In one dimension, the only possible types of long-term behavior are perpetual growth and movement toward an equilibrium point. In multivariable systems, much more complex behaviors are possible, but equilibria are still important, both as forms of behavior and as landmarks that help determine system behavior. Finding Equilibrium Points The definition of an equilibrium point in a multivariable system is a point at which all changes vanish. In order to find the equilibrium points of a system of dierential equations in several variables, we solve for values of the state variables at which all the equations are equal to zero. An equilibrium point of the dierential equation, X = f1 (X, Y, . . . , Z) Y = f2 (X, Y, . . . , Z) .. . Z = fn (X, Y, . . . , Z) is a point (X , Y , . . . , Z ) for which f1 (X , Y , . . . , Z ) = 0 f2 (X , Y , . . . , Z ) = 0 .. . fn (X , Y , . . . , Z ) = 0 Exercise 3.3.1 Find the equilibrium point of the system of equations X = 0.5X, Y = Y . 134 Equilibrium Behavior Types of Equilibrium Points in Two Dimensions Equilibrium Points Without Rotation One way to make equilibrium points in 2D is to take two 1D equilibria and put them together. Recall from Chapter 1 that if we have two 1D spaces X and Y , then we can make the 2D space X Y , called the Cartesian product of X and Y , which is the set of all pairs (x, y ) with x in X and y in Y . Geometrically, this corresponds to to using X and Y as the two perpendicular axes in our new 2D space. We will now use a similar technique, mixing and matching pairs of state points and change vectors to generate a series of 2D phase portraits. Look at Figure 3.15. For every point in the state space X, there is a change vector in the tangent space X , and for every point in the state space Y , there is a change vector in the tangent space Y . Since both spaces are one-dimensional, both the state and the change vectors can be thought of simply as real numbers. Now let's say that X = X and Y = Y . Suppose X = 1 and Y = 2. Then, in this particularly simple example, at the point (1,2), we have (X , Y ) = (1, 2). We obtain the whole vector field in the same way, mixing and matching. Exercise 3.3.2 What is the change vector at the point (3, 4)? We can also look at the whole vector field at once. For example, let's take the 1D phase portrait for X = X. This has an unstable equilibrium point at X = 0. Then we take a second 1D phase portrait, for Y = Y . This has a second unstable equilibrium point at Y = 0. If we take these two 1D phase portraits and join them together, we get a 2D unstable equilibrium point at (X, Y ) = (0, 0) (Figure 3.15). Y Y X X X'=X Y'=Y X'=X, Y'=Y Figure 3.15: Unstable node. This type of unstable equilibrium is called an unstable node (Figure 3.16). Y Y X X'=X, Y'=Y X X'=X, Y'=2Y Figure 3.16: Unstable nodes. 3.3. Equilibrium Points in Higher Dimensions 135 Similarly, we can take a stable equilibrium point in X and combine it with a stable equilibrium point in Y to get a stable equilibrium in 2D, called a stable node (Figure 3.17). Y Y X X X'= X, Y'= Y X'= X, Y'= 2Y Figure 3.17: Stable nodes. Stable and unstable nodes are essentially similar to stable and unstable equilibrium points in one dimension, not exhibiting any really new features. Another type of equilibrium point can be created by taking a stable equilibrium point in X and an unstable equilibrium point in Y (or vice versa) and joining them to make a new kind of equilibrium point. This new type of equilibrium point, called a saddle point, is more interesting (Figure 3.18). Y Y X X'= X, Y'= Y X X'= X, Y'= Y Figure 3.18: Saddle points. Left: X-axis is unstable, Y -axis is stable. Right: X-axis is stable, Y -axis is unstable. A saddle point has a stable direction and an unstable direction. The typical state point will move under the influence of both, that is, it will move in the stable direction (toward the unstable axis), as well as in the unstable direction (away from the stable axis). The only way to approach the equilibrium point in the long run is to start exactly on the stable axis. Since the typical trajectory does not lie exactly on the stable axis, a saddle point is considered unstable. Exercise 3.3.3 Sketch time series (for both X and Y ) corresponding to two trajectories in Figure 3.18. Nodes and saddle points are important examples of 2D equilibrium points. We should mention that sometimes it is possible to get nonisolated equilibria. For example, there may be a line completely made up of equilibrium points. Such situations are mathematically pathological and require special handling. 136 Equilibrium Behavior Equilibrium Points with Rotation So far, we have been taking two 1D equilibrium points and joining them together to make a 2D equilibrium point. Now we will consider a new kind of equilibrium point that is irreducibly two-dimensional, not made up of two one-dimensional systems. These equilibrium points all involve rotation, which is impossible in one dimension because there is no room for it. Recall the spring with friction: X = V V = X V It has an equilibrium point at (X, V ) = (0, 0). What kind of equilibrium point is this? If we plot a trajectory, it looks like Figure 3.19, left. Notice that the point (0, 0) meets the definition of a stable equilibrium point: if we perturb the system a little bit from the equilibrium point, it returns to it. So (0, 0) is a stable equilibrium point of this system. It is called a stable spiral. Similarly, if we consider the spring with \"negative friction,\" X = V V = X + V we get the equilibrium point in Figure 3.19, middle, which is called an unstable spiral. Y Y X X'= Y, Y'= X Y Y X X X'= Y, Y'= X Y X'= Y, Y'= X Figure 3.19: Equilibrium points in 2D with rotation. Left: stable spiral. Middle: unstable spirals. Right: center. Finally, there is one more kind of 2D equilibrium point. We saw it in the predator-prey model and the frictionless spring: X = V V = X Here, the equilibrium point (Figure 3.19, right) is clearly not stable, but neither is it clearly unstable. A small perturbation from the equilibrium point does not go far away, and neither does it return to the equilibrium point. Instead, it hangs around the neighborhood of the equilibrium point and oscillates in a new trajectory nearby. This type of equilibrium point is called a neutral equilibrium point or a center . We have now classified all the equilibrium points that can occur robustly in a 2D system. 3.3. Equilibrium Points in Higher Dimensions 137 Equilibrium Points in n Dimensions The generalization to n dimensions is straightforward: to make an n -dimensional equilibrium point, we simply take as many 1D equilibrium points as we like (stable or unstable nodes), and as many 2D equilibrium points as we like (stable or unstable spirals or centers), and mix and match them to make an n -dimensional equilibrium point (of course, the total number of dimensions has to add up to n ). These equilibrium points will be studied systemically in Chapter 6. They are all the equilibrium points of linear vector fields in n dimensions. Here let's look at an example in three dimensions. Let's take an unstable spiral in X and Y , and a stable node in Z, giving us a 3D unstable equilibrium point. A trajectory near this equilibrium point will spiral out in the X-Y plane, while it heads toward Z = 0 (Figure 3.20). Z Z Y X Y X Figure 3.20: Left: unstable equilibrium point in 3D, composed of one stable dimension (Z) and a 2D unstable spiral in X and Y . Right: a trajectory near this equilibrium point. Further Exercises 3.3 1. Consider the following Romeo and Juliet model, in which (as usual) R represents Romeo's love for Juliet, and J represents Juliet's love for Romeo (recall that these variables can be positive or negative): R = J 0.1R J = R a) Verify that this system has one equilibrium point and it is at the origin. b) Sketch the vector field for this system, using eight to ten change vectors. c) What can you say about the equilibrium point at the origin? d) Plot the vector field in SageMath. Can you determine the type of equilibrium point at the origin now? 138 Equilibrium Behavior e) Choose some initial conditions and use SageMath to simulate this system and plot (at least) one trajectory. Can you determine the type of equilibrium point at the origin now? 2. Repeat the same analysis as in the previous problem, but with the following dierential equations: R = J J = R + 0.1J 3. Create a SageMath interactive that allows you to explore the eects of parameters on the vector field of the Romeo-Juliet system R = aR +bJ, J = cR +dJ. Use parameter values ranging between 2 and 2 in steps of 0.5, using the syntax a = (2, 2, 0.5) in your function definition. (This will allow you to control parameter values more precisely.) Then, do the following exercises, supplementing the vector field with simulations when necessary. a) Set b and c to zero and d to 1. Classify the equilibrium point at (0, 0) for a < 1, a = 1, 1 < a < 0, a = 0, and a > 0. Do you get the same results if you switch the roles of a and d? b) Set a and d to zero and manipulate b and c. What happens to the equilibrium when both a and d are negative? When both are positive? When they are of opposite signs? c) How is each type of equilibrium point you found in the previous part aected by manipulating b and c? 3.4 Multiple Equilibria in Two Dimensions We have now seen all the types of simple equilibrium points that can occur in two dimensions. (Later, we will see that these are exactly the linear equilibrium points.) A typical nonlinear vector field will have multiple equilibrium points. Example: Competition Between Deer and Moose Consider two populations of deer and moose, which compete with each other for food. The deer population is denoted by D, and the moose population is denoted by M. If there were no environmental limitations, the deer population would grow at a per capita rate 3, and the moose population would grow at a per capita rate 2. Each animal competes for resources within its own species, giving rise to the D2 and M 2 intraspecies crowding terms. In addition, deer compete with moose and vice versa, although the impact of the deer on the moose is only 0.5, giving rise to the cross species term 0.5MD in the M equation, while the impact of the moose on the deer is harsher, and has value 1, giving rise to the MD term in the D equation. These assumptions make up the Lotka-Volterra competition model. D = 3D MD D2 M = 2M 0.5MD M 2 (3.3) LS 30A: MATHEMATICS FOR LIFE SCIENTISTS FALL 2017 - LECTURE 1 HOMEWORK 8 (Due MONDAY 11/27 at the beginning of the lecture in La Kretz 110) IMPORTANT Please be so kind as to: 1) at the top of the first sheet, write \"LS 30A-1 HOMEWORK 8\" and YOUR TA SECTION, as well as your NAME and your STUDENT ID NUMBER; 2) staple your sheets together; seriously, please do it; 3) come to the lecture on Monday as early as you can so that we can collect your homework and still begin the lecture on time. ASSIGNMENT WARNING Section 2.7 FE 5, FE 6, FE 7 Section 3.2 Ex 3.2.14, Ex 3.2.15, Ex 3.2.16; FE 1, FE 2, FE 3, FE 4 Section 3.3 Ex 3.3.1, Ex 3.3.2, Ex 3.3.3; FE 1, FE 2, FE 3 READ Section 3.3 (as you are doing the exercises) Lecture Notes 18, 19, 20 (as a review) The Further Exercises from Section 3.3 are CoCalc problems, and some of them may be time-consuming. You should allow plenty of time for completing them