Notes. October 19, Continuous Time Consider the logistic equation which was introduced by Pierre-François Verhulst in 1845 [4]:

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1 Notes October 19, Single species models 1.1 Logistic growth Continuous Time Consider the logistic equation which was introduced by Pierre-François Verhulst in 1845 [4]: ( dx (t) = rx(t) 1 x(t) ), x(0) = x 0, (1) dt Figure 1: Pierre-François Verhulst Here, r and are positive constants known as the maximal per capita growth rate and carrying capacity (why you think do they have these names?), and x(t) is a measure of the amount of some biological population at time t. The units of x(t) are often biomass (e.g. measured in grams), or densities (e.g. measured in grams per liter). Finally, we assume that the initial time is t = 0, and x 0 represents the initial value of x(t), a non-negative number. Our goal is to understand how x(t) will evolve over time, and in particular what happens asymptotically, i.e. when t. Ideally, we seek to solve the logistic equation, which is just a specific example of a differential equation to which an initial condition is supplied. We will see later that this is indeed possible, but we can already learn quite a bit from more qualitative considerations. Let us first think what (1) really says? On the left side we have a derivative representing the rate of change of the amount of the population level x(t), or in other words the net growth rate of the population. The equation is a prescription for that rate of change, and expresses that it depends quadratically on the population level x(t). Indeed, if we would plot the function Introduction to Mathematical Modeling in Biology, Instructor: Patrick De Leenheer, deleenhp@math.oregonstate.edu. 1

2 f(x) = rx(1 x/), we would get a parabola with zeros at x = 0 and x =, achieving its maximum r/4 at x = /2. The zeros at x = 0 and x = are interesting, because the equation says that there, the rate of change of the equation is zero. In other words, if we were to initialize the population at those levels, nothing would happen! We call these steady states, or equilibria. When x(t) belongs to (0, ), the rate of change is positive, hence the population level increases, and when x(t) >, the rate of change is negative, and thus the population shrinks. A natural question to ask is: How much will the population level increase or decrease, in the respective cases where x(t) belongs to (0, ) or to (, )? A rather natural guess would be that they approach as t. This can indeed be proved using elementary properties about solutions of differential equations, combined with facts from calculus (can you provide the details?), but let s postpone a proof until later. We point out an important property that any reasonable continuous-time population model should have: Every solution x(t) that starts from a non-negative level x 0, remains non-negative for future times t > 0. After all, the equation is supposed to represent the population level of some biological species, and these levels cannot be negative. To prove this, first note that if x 0 = 0, then x(t) = 0 for all times, establishing the claim in this special case. Indeed, recall that x = 0 is a steady state, and thus the corresponding solution remains there for all future times. Before proceeding, we recall a very important fact from the theory of differential equations. It says that the solutions of the differential equations we shall consider here always exist, are unique, and are differentiable functions of time. Now continuing our proof, suppose that for some x 0 > 0, the corresponding solution x(t) would be negative for some future time t = T > 0. Then continuity of the function x(t), would imply that there is some t between t = 0 and t = T such that x(t ) = 0. But this would imply that there are multiple solutions through x = 0, contradicting uniqueness of solutions. We return to a proof of the fact that if x 0 is in (0, ), then the corresponding solution x(t) converges to (you should try to prove the same result in case x 0 belongs to (, )). We already know that x(t) is an increasing function of time, and thus if we can show that it is bounded above, then we know it must converge. We claim that x(t) is bounded above by, i.e. x(t) < for all t > 0. Indeed, if this were not the case, then by continuity of x(t), there would be some time T > 0 such that x(t ) =. But then this would contradict the uniqueness of solutions result from the theory of differential equations. Thus, we know that there is some L such that lim t x(t) = L. We now claim that L =. If not, then L <, and since x(t) is increasing with x(0) = x 0 > 0, there must hold that L > 0. Denoting the right hand side of the equation (1) by f(x(t)), it follows from continuity of f(x), and of x(t) that lim t f(x(t)) = f(l) > 0, where the last inequality comes from the fact that L belongs to (0, ). By continuity of L, we can find an interval [L ɛ, L + ɛ] for some ɛ > 0, such that f(x) > 0 for all x in this interval. Let m > 0 be the minimum of f on this interval, i.e. f(x) m for all x in [L ɛ, L + ɛ]. Now, since x(t) converges to L, there is some T such that for all t T, x(t) belongs to the interval [L ɛ, L + ɛ], and thus f(x(t)) m for all t T. Consequently, for all t T : x(t) = x(t ) + t T f(x(s))ds x(t ) + m(t T ), which contradicts that x(t) for all t > 0, since the right hans side converges to + as t. Let s move on and investigate the second derivative of a solution x(t), i.e. the rate of change of the net growth rate of the population. A short calculation, and using the shorthand notation ẋ for dx/dt(t) and ẍ for d 2 x/dt 2 (t), we get: ẍ = rẋ (1 2 ) ( x = r 2 x 1 x ) ( 1 2 ) x, where we use (1) to show the second equality. Thus, ẍ = 0 if Moreover, x = 0 or (the steady states), or if x = If x(t) belongs to (0, /2), then x(t) is increasing (because we know that ẋ > 0), and since ẍ > 0 as well, the rate of change of the population s growth rate is positive, hence the population grows at an increasing pace. 2

3 2. If x(t) belongs to (/2, ), then x(t) is increasing (because we know that ẋ > 0), but since ẍ < 0, the rate of change of the population s growth rate is negative, hence the population grows at a decreasing pace. 3. If x(t) belongs to (, ), then x(t) is decreasing (because we know that ẋ < 0), and since ẍ > 0, the rate of change of the population s growth rate is positive, hence the population declines more slowly. Putting all our knowledge about the solutions x(t) of the logistic equation together, allows us to understand the plots below. Figure 2: Some solutions of the logistic equation where N represents x, a = 1 represents r, and = 100. For comparison, the solution to the exponential growth model dn dt (t) = N(t), N(0) = N 0 = 10 is given in the dashed line. This solution is N(t) = N 0 e t. Next, we consider a more quantitative approach to understand (1). We are fortunate that equation (1) is a separable equation (consider your ODE textbook in case you forgot what this means), and thus it can be solved explicitly. By separating variables, and integrating: Solving (do this!), gives: and rearranging gives: x(t) x 0 dy y ( 1 y x(t) ln 1 x(t) x(t) = ) = 1 x0 x 0 t 0 rdτ. = rt, x 0 x 0 e rt +1 [The formula even makes sense for x 0 = 0 by slightly re-writing it. Do this.] From (2), we note many if the properties we already know: 1. The steady states are x = 0 and x =. 2. All solutions are non-negative for all times. 3. lim t x(t) = if x 0 > We know that when x 0 belongs to (0, /2), then the solution x(t) has a single inflection point when it reaches /2. With a straightforward calculation we can show when this happens, namely for: ( ) x ln 0 x 0 t = r (2) 3

4 An alternative solution method that transforms the ODE into a linear one: When x > 0, we define a new variable: u = 1 x, and transform equation (1): u = ẋ x 2 = r ( 1 x ) ( = ru 1 1 ), x u or: u = ru + r, u(0) = u 0 = 1 x 0, which is a linear equation whose solution follows from the variation of constants formula (see your ODE textbook): t u(t) = e rt u 0 + e r(t s) r ds, which after some algebra simplifies to: Returning to x via x = 1/u, we get: u(t) = e rt u (1 e rt ) 0 x(t) = x 0 x 0 e rt +1, which is the same as (2). Fitting to data Consider the data by Georgii Frantsevich Gause, see our textbook on p. 11,12 and 13. The graphical representation on p.12 represents the density of a population of Paramecium Figure 3: Georgii Frantsevich Gause aurelia, measured in number of individuals per volume of 0.5 cm 3. It suggests that the growth of this population might be modeled using a logistic equation (1). The question is to determine the values of r and that would best fit the data. A rough visual estimate is that should be in the range of , but guessing r is more complicated. How can we find appropriate values for r and? Many fitting methods rely in some way or another on a linear least squares method, which you probably studied in your linear algebra course. The idea is this. One is given a system of linear equations of the form Az = b, 4

5 where A is a given n m matrix, with n > m (often n is much larger than m), b is a given column vector of size n, and z is an unknown vector of size m. One wishes to solve this for the vector z in an optimal way ( optimal refers to the fact that the solution z has the property that the Euclidean distance between the vectors Az and b, i.e. Az b 2, is minimized; this explains the chosen terminology of least squares method ), and it turns out that this optimal solution is given by the following expression: z = (A T A) 1 A T b, (3) where A T denotes the transposed of the matrix A, obtained by writing the columns of A as rows instead. Our challenge is to rewrite our fitting problem as a linear problem of the form Az = b, where z is the vector of unkowns, which ideally would be the vector with components r and. The data of our problem can be summarized as the points (t i, x(t i )), with i = 1,..., n (the table on p. 11 of [5] shows there are 25 data points, i.e. n = 25). If our data points would match perfectly with logistic growth, we would have that for all i = 1,..., 25: x(t i ) = x 0 x 0 e rti +1, by formula (2). It is obviously impossible to reformulate this as a system of linear equations of the form Az = b, when choosing z to be the 2 vector with components r and, because clearly the relation is not linear in r and. However, not everything is lost. Indeed, as long as x(t) 0 (i.e. as long as we don t consider the zero steady state solution) the logistic equation (1) can be rewritten as: ẋ(t) x(t) = r r x(t). Let us define the vector z = ( r r and in the spirit of Euler s method, approximate the derivative ẋ(t i ) by a difference quotient: ) ẋ(t i ) = x(t i+1) x(t i ) t i+1 t i. Thus, we can write Az = b, where 1 x(t 1 ) ẋ(t 1 )/x(t 1 ) 1 x(t 2 ) A =.. and b = ẋ(t 2 )/x(t 2 )., 1 x(t n ) ẋ(t n )/x(t n ) where in the vector b we use the formula above for the derivatives ẋ(t i ), in order to express these in terms of the available data points (t i, x(t i )). The least-squares solution z, given in (3), then takes the form: ( ) 1 z n x 1 = x 1 x 2 A T b, 2 where x is the n-vector having components x(t i ), x 1 = n i=1 x i is the l 1 -norm of the vector x, and x 2 = ( n ) 1/2 i=1 x2 i is the l 2 -norm of the vector x. HW problem for all students: Use Gause s data, see p. 11 in [5], and determine z, and from this the least-square values for r and. Make sure to clearly indicate what the vectors x and b are that you use in your calculation. In particular, be careful in saying what the first and last components of these vectors are. HW problem for all students: Develop a fitting method for the geometric or exponential growth model: ẋ = rx, x(0) = x 0 (4) Start by solving this equation. Then assume that n data points (t i, x(t i )) for i = 1,..., n are given to you by a biologist, who wishes to fit the data to this model, and wants you to determine the 5

6 optimal value for r. Suppose that the biologist s data looks like Gause s data? Argue qualitatively why a fit to a linear model as (4), is probably not a very good idea. HW for math grad students only: When is the 2 2 matrix in the last formula invertible? To answer this more generally, reconsider the general least-squares problem Az = b where A is an n m matrix with n > m. Show that the matrix A T A is invertible if and only if the matrix A has (full column) rank equal to m. In this case, there is a unique least-squares solution z given by the formula (3). Applied to our particular problem, the invertibility condition is that the vector x and the n-vector consisting of ones, are linearly independent. What requirement does this impose on the data set (t i, x(t i ))? In other words, what would a data set look like, for which the matrix A T A is not invertible? Discrete Time Next we consider scenarios where time is discrete, say enumerated by whole numbers t = 0, 1, 2,... A famous discrete-time model for logistic growth is the Beverton-Holt model, first proposed in 1957 in the context of fisheries [1]: µx t x t+1 = 1 + µ 1 x, x 0 > 0 given (5) t Here, x t represents the number of individuals of a population at time t (e.g. fish stock), and the model above says that the next population level x t+1 depends only on the current level x t. There are two positive parameters µ and, and we assume that µ > 1. Note that the growth factor between consecutive time points (i.e. the ratio of two consecutive population levels) is given by µ 1 + µ 1 x, t and therefore decreases with x t. The maximal growth factor is achieved when x t = 0, and equals µ. Thus, the interpretation of µ is similar (but, as we will see later on, not the same!) as in the continuous-time logistic growth model (1). Similarly, > 0 will be called the carrying capacity because as we will soon see, all positive solutions of (5) converge to as t. The Beverton-Holt model (5) has the form µx t x t+1 = f(x t ), with f(x t ) = 1 + µ 1 x, t and for general discrete-time models of the form x t+1 = f(x t ), we call x a fixed point if f(x ) = x. Fixed points of discrete-time models are similar to steady states of continuous-time models, because if the system is initialized at a fixed point at time zero, it will stay there for all future times. Let s find the fixed points for the Beverton-Holt model. A simple calculation (do it) shows that there are two fixed points, namely: x 1 = 0 and x 2 =. Now consider the graph of the function f that appears in the right-hand side of the Beverton-Holt model (5). It is depicted in Figure 2.15 (b) on p.27 of [5]. Note that f is an increasing function, and its graph intersects the 45 o line precisely in the two fixed points. More generally, when studying models of the form x t+1 = f(x t ), you can reverse this, and find all the fixed points by simply plotting the function f, and determining where its intersects the 45 o line. Question: Does the Beverton-Holt model have the required property that all solutions starting from a non-negative x 0, remain non-negative for all future times? (As we know, any reasonable model of population growth should have this property). Can you prove why? To understand qualitatively what will happen to solution sequences x 0, x 2, x 2,... of the Beverton-Holt model (5), we can cobbweb. Doing this using Figure 2.15 (b) on p.27 of [5], it won t take much to convince yourself that all positive solution sequences converge to as t. Can we prove this? We will see that the key property here is that f(x) is monotonically increasing in x, that is: If x y, then f(x) f(y). 6

7 If we take any x 0 0, then either x 0 f(x 0 ), or x 0 f(x 0 ). Case 1: x 0 f(x 0 ). Iterated use of the monotonicity of f shows that x 0 x 1 = f(x 0 ) x 2 = f(f(x 0 )) x 3 = f(f(f(x 0 )))..., and thus x n = f n (x 0 ), where the latter denotes the n-fold composition of f with itself, must be a non-decreasing sequence. Case 2: x 0 f(x 0 ). Similarly, iterated use of the monotonicity of f shows that x n must be a non-increasing sequence. If we can prove that the solution sequences are bounded, then we know they must converge. For the Beverton-Holt model, we see that Case 1 occurs if x 0 belongs to [0, ], and Case 2 if x 0 belongs to [x 0, ). We shall show that in Case 2, all (non-increasing) solution sequences are bounded below by (you should prove a similar result in Case 1). This is pretty obvious: f(x) for all x, as can be seen from Figure 2.15 (b) on p.27 of [5]. Thus, any solution sequence starting at some x 0, must converge to, say L. We show next that L = : Continuity of L, and the fact that x n+1 = f(x n ) yields: L = lim t x t+1 = lim t f(x t ) = f(l), and hence L must be a fixed point. But there is only one fixed point in [, ), namely L =, which concludes the proof. HW problem (all students): Suppose that f : R R is continuous and increasing, and has finitely many fixed points x 1 < x 2 < < x n. Prove that every solution sequence of the model x t+1 = f(x t ), either converge to one of the fixed points, or to + or to. (Hint: Make plots of the function f first, and identify various scenarios by cobbwebbing) Question (not part of the HW): Does this result remains valid if f is assumed to be decreasing, rather than increasing? Let s return to the Beverton-Holt model (5), but take a more quantitative approach. Can we solve the equation explicitly? Recall the idea to solve the continuous-time logistic equation (1) which clearly shares qualitative properties with the solutions of the Beverton-Holt model, namely that all positive solutions converge monotonically to the carrying capacity. The trick there was the reciprocal transformation: so let s try this here as well. Then u = 1 x, u t+1 = 1 x t+1 = 1 µx t + µ 1 µ, or u t+1 = 1 µ u t + µ 1 µ, u(0) = u 0, which is a linear difference equation that is easily solved: Write down u 1, u 2, u 3 in terms of u 0 and the model parameters, and you will discover the following general formula for u t (do this): u t = 1 µ t u t 1 1 µ µ i µ, or i=0 u t = 1 µ t u 0 1 Returning to x t, we find after some algebra (do it): ( ) 1 µ t 1. x t = x 0, t = 0, 1, 2,... (6) 1 x 0 µ + 1 t From this we see immediately that lim t x t = if x 0 > 0 (recall that µ > 1), confirming what we had already learned from the qualitative analysis earlier. 7

8 Let us now compare the solution of the (continuous-time) logistic equation (2), to the above solution (6) of the (discrete-time) Beverton-Holt model. Note that if we set µ = e r, then the solutions expressions coincide. In other words, if we would sample a positive solution of the logistic equation (1) at integer times, we would find the solution sequence of the Beverton-Holt model (5) starting from the same initial condition x 0, provided µ and r are related as in the formula above. 1.2 Linearization An important tool in the analysis of discrete or continuous time models is the linearization principle. The idea is to approximate the function that describes the system dynamics (the net growth rate function for continuous time models, and the next generation population number for discrete time models) by a linear function at steady states (continuous time) or at fixed points (discrete time). Of course, we cannot expect that the dynamics of this approximated model will yield the same solutions of the original model. Nevertheless, the approximation is often enough to conclude whether the steady state or fixed point under consideration is asymptotically stable, or unstable. Discrete time Given is a smooth function f : X X, where X is some subset of R (in most of our models X = R + ), and suppose that x t+1 = f(x t ), t = 0, 1, 2,..., and x 0 given, (7) is a discrete time model that has a fixed point at x (i.e. f(x ) = x ). Definition We say that x is a stable fixed point of (7) if for any ɛ > 0, there is a δ > 0 such that if x 0 x < δ, then x t x < ɛ. 2. We say that x is an unstable fixed point of (7) if it is not a stable fixed point of (7). 3. We say that x is a locally asymptotically stable fixed point of (7) if it is a stable fixed point, and if in addition there exists an r > 0, such that if x 0 x < r, then lim t x t = x. 4. We say that x is a globally asymptotically stable fixed point of (7) if it is a stable fixed point, and if in addition lim t x t = x, for any x 0 in R +. Note the subtle difference between a stable fixed point, and an asymptotically stable fixed point. Every asymptotically fixed point is necessarily stable, but the converse is not true. To see this, consider the following example: x t+1 = x t, which has a stable fixed point at x = 0, that is not asymptotically stable. (prove this) Define u t = x t x. Thus, u t is the difference between the current state x t and the fixed point x. Then, u t satisfies the following equation: u t+1 = x t+1 x = f(x t ) x = f(u t + x ) f(x ) f (x )u t. In the last step we approximated the difference by the first order term in a Taylor series expansion of f near the fixed point x. Thus, the dynamics for the deviations from the fixed point has been approximated by a linear equation: u t+1 = f (x )u t, which is easily solved, when an initial condition u 0 is supplied: u t = (f (x )) t u 0, t = 0, 1, 2,... Thus, as t, this difference converges to 0 if f (x ) < 1, but grows unbounded, possible in an oscillatory manner, if f (x ) > 1. The linearization principle allows us to conclude that under these two respective conditions, the fixed point x of the original model x t+1 = f(x t ) is locally asymptotically stable, respectively unstable. 8

9 Theorem 1. (Principle of Linearization) Let x be a fixed point of (7). Then 1. x is a locally asymptotically stable fixed point of (7), if f (x ) < x is an unstable fixed point of (7), if f (x ) > 1. It is worth emphasizing two points concerning the Principle of Linearization: 1. It only provides information about the nature of solutions that start near the fixed point x, i.e. it provides local information only. Other methods are needed to understand the dynamics globally. In particular, other methods are needed to conclude the global asymptotic stability of fixed points x. 2. It does not yield any information in the intermediate case where f (x ) = 1, i.e. where either f (x ) = 1 or f (x ) = 1. In fact, one can construct examples of models having a fixed point x with f (x ) = 1 and such that x is either stable, unstable or asymptotically stable. HW problem (both for undergrads and grad students): Reconsider the Beverton-Holt model, and use the Principle of Linearization to determine whether the fixed points at 0 and are asymptotically stable, or unstable. Continuous time Given is a smooth function f : X X, where X is some subset of R (in most of our models X = R + ), and suppose that dx dt (t) = f(x(t)), t > 0 and x(0) = x 0 given, (8) is a continuous time model that has a steady state at x (i.e. f(x ) = 0). The definitions of a stable, unstable, locally and globally asymptotically stable fixed point of a discrete time system given above, carry over verbatim for a steady state of a continuous time system, by simply replacing x t by x(t), where the latter denotes the solution of the model (8). To motivate the Principle of Linearization for continuous time models, we take our cue from the discrete time model case, and define the difference u(t) = x(t) x, between the solution of (8), and the steady state x. It is easy to show (do it) that: du dt (t) f (x )u(t), Replacing by =, we obtain a linear differential equation which is easily solved: u(t) = e f (x )t u 0, where u 0 = x 0 x. Thus, as t, the difference u(t) converges to 0 if f (x ) < 1, but grows unbounded, if f (x ) > 0. Once again, the linearization principle allows us to conclude that under these two respective conditions, the steady state x of the original model ẋ = f(x) is locally asymptotically stable, respectively unstable. Theorem 2. (Principle of Linearization) Let x be a steady state of (8). Then 1. x is a locally asymptotically stable steady state of (8), if f (x ) < x is an unstable fixed point of (8), if f (x ) > 0. HW problem (both for undergrads and grad students): Reconsider the logistic equation, and use the Principle of Linearization to determine whether the steady states at 0 and are asymptotically stable, or unstable. 9

10 Figure 4: Warder Clyde Allee 1.3 Non-logistic growth models Allee effect: The Allee effect is named after the ecologist Warder Clyde Allee, who, through experimental work in the 1930-ies with goldfish observed that the fish grew more rapidly when there were more individuals in the tank. According to Wikipedia, The Allee effect is a phenomenon in biology characterized by a correlation between population size or density and the mean individual fitness (often measured as per capita population growth rate) of a population or species. The last statement is a bit vague. First, it does not say if the correlation is positive or negative. However, referring back to Allee s experimental work, it seems fair to say that the correlation is meant to be positive: A higher population size gives rise to a higher population growth rate. Secondly, the statement contains the word fitness, one of the favorite notions of biologists, which may have different meanings depending on the context. Let this serve as a warning which may save you a lot of headaches: You should always try to find out what fitness really means when talking to biologists: Ask them to properly define the concept, before proceeding with the discussion. Fortunately, in the above definition, fitness is defined as the per capita growth rate of the population, a precise notion for continuous time population growth models. Let s reconsider the logistic equation (1), and determine if it exhibits an Allee effect. The per capita growth rate equals r x, which decreases with x. This is contrary to what the Allee effect is supposed to be! Therefore, the logistic equation would not be an appropriate model to explain Allee s experiments of growing goldfish populations. This raises the problem of how to modify the logistic equation, so that the modified model does exhibit an Allee effect. Here is one such model: dx (1 dt (t) = rx(x ɛ) x ), x(0) = x 0. (9) Here, the parameters r and are positive, and ɛ is assumed to satisfy: 0 < ɛ <<, i.e., ɛ is positive, but small, and much smaller than. Formally, the model (9) is obtained from the logistic model by multiplying r(1 x/) by the factor (x ɛ). The per capita growth rate is now a quadratic function of x: ( r(x ɛ) 1 x ) The per capita growth rate is increasing in x, if x belongs to the interval ( ) 0, ɛ+ 2 (note that the right endpoint is precisely the midpoint of the interval [ɛ, ]). However, for x in (0, ɛ), the 10

11 population actually declines, and thus we can only claim that an Allee effect exists for this model when the population level is in the interval: ( ɛ, ɛ + ) 2 Indeed, in this case, the population grows, and it does so more quickly for larger population levels. HW problem (all students): Explain why it still makes sense to call the carrying capacity, but that it does no longer make sense to call r the maximal per capita growth rate. To answer this, note that the parameter ɛ presents a threshold level for the population level: Below it, the population is not viable and will go extinct. What happens above it? Find all the steady states of the model, and apply the Linearization Principle. Describe what happens to all solutions x(t) of (9) as t. HW problem (all students): How would you define the Allee effect in the context of a discrete-time population model x t+1 = f(x t )x t, where f is the (population-dependent) relative growth rate, i.e. the ratio between consecutive population levels. [The function f should necessarily be a non-negative function.] Modify the Beverton-Holt model so that it exhibits an Allee effect in the sense of your suggested definition. Quadratic map and Ricker equation (Illustrate linearization, and point out qualitative difference between discrete and continuous time: oscillations, chaos...) The quadratic map: ( x t+1 = rx t 1 x ) t, t = 0, 1, 2,..., has received, and continuous to receive a lot of attention in the literature, both in mathematical biology, and more generally in dynamical systems. We will not spend much time on this model, but highlight some of its astounding features. First, we shall simplify the analysis a bit by assuming that = 1. This assumption means, in fact, no loss of generality because x t can always be scaled by, and the scaled variable can easily be shown to satisfy the difference equation above but with = 1. The graph of the function is a parabola with zeros at x = 0 and x = 1, see Figure 2.4 on p.15 in the text for the case r = 2.8. One problem is that the quadratic function f(x) = rx(1 x) takes negative values when x > 1. Thus, if x is supposed to represent a (scaled) number of individuals of some population (e.g. x = 1 may represent 1000 individual cells in some cellular population), it makes no sense to assume that x > 1, because in the next time step, the model would predict a negative population number. To remedy this, one often assumes that x belongs to [0, 1], but to ensure that the next iterate remains in [0, 1], we need to impose that the maximum of the parabola, which is achieved for x = 1/2 and equals r/4, does not exceed 1. Thus, we assume that: 0 < r 4, and restrict the analysis of the model to x in [0, 1]. Let us determine the non-negative fixed points. It is easy to see that x 1 = 0 is always a fixed point. A second fixed point x 2 = 1 1 r exists in the interval [0, 1], provided that r > 1 (verify this). Thus, if 0 < r 1, then there is only one fixed point x 1 = 0. Verify that this fixed point is globally attracting, i.e. that lim t x t = 0, for all x 0 in [0, 1]. Thus, any interesting dynamics can only occur if we restrict the value of r to the smaller interval: 1 < r 4, an assumption we make henceforth. Now let us apply the Principle of Linearization to both fixed points. We have that: f (x 1) = r > 1 and f (x 2) = 2 r, 11

12 and x 1 = 0 is always unstable. The fixed point x 2 is asymptotically stable if r belongs to (1, 3), but unstable of r belongs to (3, 4). The latter range for r leads to the most interesting dynamical behavior, see Figure 2.10 on p.22 in the text. For r = 3.2 panel (b) shows a periodic solution with period 2, or a 2-cycle for short. Panel (d) shows a 4-cycle when r = 3.55, and (e) is so-called chaotic behavior when r = If you wish to understand the mechanism underlying these phenomena a little bit better, you may (but don t have to) read the material concerning the so-called bifurcation diagrams in Figures 2.12 and 2.13 which display the route the chaos via a cascade of so-called period doubling bifurcations of the various cycles the model exhibits. Figure 5: William E. Ricker The Ricker model was proposed by Bill Ricker in 1954 in the context of fisheries, just like the 1957 Beverton-Holt model which dates back to the same period. It has the following form: x t+1 = e r(1 x t ) xt, t = 0, 1, 2,..., where r > 0 and > 0 are model parameters. The model expresses that the ratio of two consecutive population levels equals e r(1 x t ), and therefore decreases exponentially with the current population level. The maximal ratio is achieved when x t equals zero, and equals e r. This suggests an idea that is similar to the Beverton- Holt model and the quadratic model, but deviates from those in the details of how exactly the ratio decreases, namely exponentially. These details have far-reaching consequences. First note that the Ricker model does not suffer from the problem exhibited by the quadratic model where future populations could be negative. This is why in the quadratic model we had to restrict the values of x to [0, 1], and impose conditions on the value of r. The Ricker model never has this problem, and therefore future populations levels are always non-negative. Let us consider the fixed points of the Ricker model. There are always exactly two fixed points: The Principle of Linearization shows that x 1 = 0 and x 2 =. f (x 1) = e r > 1 and f () = 1 r, and thus x 1 = 0 is always unstable, whereas x 2 = is asymptotically stable if r belongs to (0, 2), and unstable if r > 2. When r increases in the interval (2, ), The Ricker model can be shown to exhibit a s similar route towards chaotic behavior via period doubling bifurcations, just like the quadratic model did. We conclude the discussion about single species population models with two remarks: 12

13 1. Both the quadratic model and the Ricker model are difference equations in which the right hand side is a function that is called unimodal or single-hump shaped. When you encounter models of this type, you should always proceed with caution, because these are the kinds of models that may exhibit chaotic dynamics, via a cascade of period doubling bifurcations. Models in which the right hand side is monotonically increasing cannot show this kind of complex behavior. Indeed, in a previous homework problem (see p.7) you have proved that all solution sequences of such models converge monotonically to a fixed point, or to +, or to. 2. It is worth pointing out that another striking distinction exists between single species models in continuous time, and single species models in discrete time: All solutions of (timeindependent) continuous-time models of the form ẋ = f(x) where f : X X is smooth on some subset X of R, behave in a very simple way, namely they converge monotonically to a steady state, or to +, or to (can you sketch a proof of this fact?). Note that this is the case without any additional assumptions regarding the monotonicity properties of f: The function f does not have to be increasing, or does not even have to be a monotone function on its domain! This is in stark contrast to what we have learned from the last two discrete time models we considered (quadratic and Ricker), which not only may exhibit periodic solutions, but also chaotic behavior. Periodic solutions never occur in single-variable continuous time models. We shall see later that they may occur in 2-variable continuous-time models. In other words, we need at least two populations of interacting species to observe oscillatory behavior. Another way of looking at this is as follows: Suppose a biologist presents data of a population that exhibits a periodic pattern in time. Then a single species population model in continuous time can never capture this behavior: Either you must choose a discrete time model, or you must poll the biologist about another possible species that is interacting with the one for which data is available. 1 2 Models with two interacting species Many models (but not all!) of two biologically interacting species are characterized by the type of interaction occurring between the two species. These are of a competitive, cooperative (or mutualistic) or predator-prey nature. In a competitive, respectively cooperative interaction, the presence of each species is harmful, respectively beneficial to the other. In a predator-prey interaction, one species benefits from the other s presence (the predator), whereas the other does not (the prey). First, we discuss some general mathematical methods that are useful when studying these models. 2.1 Linearization principle for 2 by 2 systems Discrete Time Let x t be a vector in R 2 whose entries denote the population levels of 2 interacting species at time t. Many 2 species population models take the form: x t+1 = f(x t ), (10) where f : R 2 + R 2 + is a given smooth function. Fixed points of these models are solutions, which are 2-vectors x that solve the algebraic system: x = f(x ). The notions of stability, instability and local or global asymptotic stability of a fixed point of system (10) carry over verbatim from Definition 1 of single species models, to the 2 (and even n) species context here. The only difference is that we need to replace the model (7) by (10). Interestingly, the Principle of Linearization, see Theorem 1 also carries over: 1 A third possibility would be to consider time-dependent continuous time models ẋ = f(t, x), where the time dependence of f may reflect diurnal or seasonal fluctuations in the environment for instance. 13

14 Theorem 3. (Principle of Linearization for 2 species models) Let x be a fixed point of (10). 1. x is a locally asymptotically stable fixed point of (10), if λ < 1 for every eigenvalue of the Jacobian matrix J(x ). 2. x is an unstable fixed point of (10), if λ > 1 for some eigenvalue of the Jacobian matrix J(x ). A new concept, namely the Jacobian matrix J(x ) appears in the above statement. It is obtained as follows. Consider the matrix that contains the partial derivatives of the 2-variable vector function ( ) f1 (x f(x 1, x 2 ) = 1, x 2 ). f 2 (x 1, x 2 ) This matrix is called the Jacobian matrix of f at x, and it is defined as follows: ) J(x) = ( f1 x 1 f 2 x 2 f 2 f 1 x 1 x 2 When this matrix is evaluated at the fixed point x, we obtain the Jacobian matrix mentioned in the Principle of Linearization above. Recall from Linear Algebra that a 2 2 matrix has 2 (possibly complex) eigenvalues, which by definition are the roots of the characteristic equation of the matrix, i.e. of the equation det (λi J(x )) = 0. To see how the Jacobian matrix arises in the Principle of Linearization above, we mimic the approach we took for fixed points of single variable models. Let x be a fixed point of (10). Define u t = x t x, which is the difference between the current state x t and the fixed point x. Then, u t satisfies the following equation: u t+1 = x t+1 x = f(x t ) x = f(u t + x ) f(x ) J(x )u t, by using the first order term approximation of f in a Taylor series expansion near the fixed point x. Thus, the dynamics for the deviations from the fixed point has been approximated by a linear equation: u t+1 = J(x )u t, which is easily solved, when an initial condition u 0 is supplied: u t = (J(x )) t u 0, t = 0, 1, 2,... From the theory of linear difference equations follows that as t, this difference converges to 0 if λ < 1 for both eigenvalues of J(x ), but that some solutions of this equation grow unbounded, possible in an oscillatory manner, if λ > 1 for at least one eigenvalue of J(x ). (Can you reconstruct a proof of these statements?) Jury conditions Question: When are all the eigenvalues of a real n n matrix A strictly inside the unit disk of the complex plane? Answer in case n = 2: Let tr(a) and det(a) denote the trace 2 and determinant of A respectively. Then the eigenvalues of A are strictly inside the unit disk if and only if tr(a) < 1 + det(a) < 2 (11) 2 the trace of a square matrix is the sum of the diagonal entries of the matrix 14

15 Proof. If λ 1 and λ 2 are the (possibly complex) eigenvalues of the real 2 2 matrix A, then its characteristic polynomial can be written as: det(λi A) = λ 2 tr(a)λ + det(a) = (λ λ 1 )(λ λ 2 ) (12) [If you have never seen the first equality,perform the calculation, and convince yourself that the coefficients of the characteristic polynomial of a 2 2 matrix are indeed tr(a) and det(a) respectively.] Consequently, tr(a) = λ 1 + λ 2 and det(a) = λ 1 λ 2, that is, the sum of the eigenvalues of A equals the trace of A, and the product of the eigenvalues equals the determinant of A. Necessity: Suppose that λ 1 and λ 2 are strictly inside the unit disk, i.e. λ 1 < 1 and λ 2 < 1. Then det(a) = λ 1 λ 2 < 1, from which the right inequality in (11) follows. If λ 1 and λ 2 are real, then they are both between 1 and +1. Consequently: which by (12) implies that ( 1 λ 1 )( 1 λ 2 ) > 0 and (1 λ 1 )(1 λ 2 ) > 0, ( 1) 2 tr(a)( 1) + det(a) > 0 and 1 2 tr(a)(1) + det(a) > 0, (13) but these two inequalities are equivalent to the left inequality in (11). If λ 1 and λ 2 are complex, they are complex conjugate, and hence and (1 λ 1 )(1 λ 2 ) = (1 λ 1 )(1 λ 1 ) = 1 λ 1 2 > 0, ( 1 λ 1 )( 1 λ 2 ) = ( 1 λ 1 )( 1 λ 1 ) = 1 λ 1 2 > 0, from which (13) follows once again. Sufficiency: Assume that (11) holds. We need to prove that the eigenvalues λ 1 and λ 2 of A are strictly inside the unit disk. If λ 1 and λ 2 are real, then the left inequality in (11) is equivalent to (13), and this in turn is equivalent to: ( 1 λ 1 )( 1 λ 2 ) > 0 and (1 λ 1 )(1 λ 2 ) > 0, This implies that λ 1 and λ 2 are on the same side of 1 and of +1 respectively. There are 3 possibilities: Either they are both to the left of 1, or to the right of +1, or between 1 and +1. The first two cases would imply that λ 1 λ 2 = det(a) > 1, which would contradict the right inequality in (11). Thus λ 1 and λ 2 are between 1 and +1, and hence they belong to the interior of the unit disk. If λ 1 and λ 2 are complex, then their product λ 1 λ 2 = λ 1 2 = det(a) < 1 by the right inequality in (11), hence λ 1 = λ 2 < 1. This concludes the proof. Remark: To check if all eigenvalues of a real n n matrix are strictly inside the unit disk, there exist the so-called Jury test, see [6] Continuous Time Let x(t) be a vector in R 2 whose entries denote the population levels of 2 interacting species at time t. Many 2 species population models take the form: dx (t) = f(x(t)), (14) dt where f : R 2 + R 2 + is a given smooth function. Steady states of these models are by definition vector solutions x that solve the algebraic system: f(x ) = 0. The notions of stability, instability and local or global asymptotic stability of a fixed point of system (14) carry over verbatim from Definition 1 of single species models, to the 2 (and even n) species context here. The only difference is that we need to replace the model (7) by (14). The Principle of Linearization, see Theorem 3 also carries over, to the 2 species case: 15

16 Theorem 4. (Principle of Linearization for 2 species models) Let x be a steady state of (14). 1. x is a locally asymptotically stable steady state of (14), if Re(λ) < 1 for every eigenvalue of the Jacobian matrix J(x ). 2. x is an unstable steady state of (14), if Re(λ) > 1 for some eigenvalue of the Jacobian matrix J(x ). You should try to sketch a proof of this result! Routh-Hurwitz test Question: When are all the eigenvalues of a real n n matrix A in the open left half plane of the complex plane (i.e. when is the real part of all eigenvalues negative?)? Answer in case n = 2: The eigenvalues of A are in the open left half plane if and only if tr(a) < 0 and det(a) > 0 (15) Homework problem (all students): Prove this. Remark: To check if all eigenvalues of a real n n matrix are strictly in the left half plane of the complex plane, there is the so-called Routh-Hurwitz test, see [7]. 2.2 Host-parasitoid system (an example of a predator-prey system) Discrete time See text p.44-48: 1. The Nicholson-Bailey model (2.91) (2.92): Summary: There are two fixed points: with respective Jacobian matrices: H n+1 = k e apn H n P n+1 = c ( 1 e apn) H n (H, P ) = (0, 0) and ( H, P ) = ( ) k and ( k ln k ac(k 1), ln k ), a ( ) 1 k ln k c(k 1) k c(k 1) ln k k 1 Since k > 1 by assumption, (0, 0) is an unstable fixed point, and so is ( H, P ) because the 2nd Jury condition in (11) fails (check). Some follow-up analysis: We next investigate whether the eigenvalues of J( H, P ) are real or complex because oscillatory behavior near the fixed point would arise if they are complex. (but also if they are real, provided one of them is negative. Do you understand why?) Since eigenvalues are roots of the characteristic equation, which is a quadratic equation in this case, we can come up with a simple condition to check this. Indeed, the sign of the discriminant associated to this equation determines the answer, and the discriminant, as a function of k > 1, is given by: ( ) 2 D(k) = (tr(j( H, P ))) 2 4det(J( H, P k 1 ln k )) = 4 ln k k 1 Verify this. Note that lim D(k) = 0, k 1+ confirming that there is a double eigenvalue when k = 1. This eigenvalue equals 1, hence is located on the unit circle of the complex plane. Also note that: lim D(k) =. k 16

17 Verify this. Therefore, for all large k, the discriminant D(k) is negative, and oscillatory behavior occurs near the positive fixed point. Similarly, dd(k) lim < 0. k 1+ dk Verify this as well. Consequently, for k > 1 but near 1, D(k) will be negative, and thus oscillatory behavior occurs near the positive fixed point. A natural question would be to ask if D(k) < 0 for all k > 1, not just for large k s and for k s near 1. Extra credit for any student who can either prove that this is indeed the case, or who can prove that it is not necessarily true. 2. The Beddington model (2.97) (2.98): H n+1 = e Hn r(1 ) e ap n H n P n+1 = c ( 1 e apn) H n where r > 1 and > 0, and the other parameters are like in the Nicholson-Bailey model. How does this model deviate from the Nicholson-Bailey model? Ask yourself: How does the host grow in the absence of the parasite? Do you recognize this growth mechanism? Do you remember the features of it? How does it compare to the host s growth of the Nicholson- Bailey model when there are no parasites. 3. An intermediate model (with Beverton-Holt dynamics for the host in the absence of the parasite.) Has this model been named in the literature? (Extra credit for any student who can answer this question.) H n+1 = µ 1 + µ 1 H e apn H n n P n+1 = c ( 1 e apn) H n where µ > 1 and > 0, and the other parameters are like in the Nicholson-Bailey model. Spoiler alert: This model will feature prominently on the first Midterm! Continuous time (The Lotka-Volterra predator-prey model, or HW?) The Rosenzweig-Macarthur model: Smith s notes-time permitting? References [1] Beverton, R. J. H., and Holt, S. J., On the Dynamics of Exploited Fish Populations, Fishery Investigations Series II Volume XIX, Ministry of Agriculture, Fisheries and Food, [2] Gause, Georgii Frantsevich, The struggle for existence, 1934, (a more recent version was published by Dover, New York, in 2003). [3] Ricker, W. E., Stock and Recruitment, Journal of the Fisheries Research Board of Canada 11, , [4] Verhulst, Pierre-François, Recherches mathmatiques sur la loi d accroissement de la population [Mathematical Researches into the Law of Population Growth Increase]. Nouveaux Mémoires de l Académie Royale des Sciences et Belles-Lettres de Bruxelles 18: 1-42, [5] de Vries, Gerda, and Hillen, Thomas, and Lewis, Mark, and Müller, Johannes, and Schönfisch, Birgitt, A course in mathematical biology, Quantitative modeling with mathematical and computational methods, SIAM [6] stability criterion [7] stability criterion 17