4 Insect outbreak model

Transcription

1 4 Insect outbreak model In this lecture I will put to a good use all the mathematical machinery we discussed so far. Consider an insect population, which is subject to predation by birds. It is a very well known and very well documented fact that often it is possible to observe so-called outbreaks of the insect population, when the number of insects increase from almost undetectable to extremely high. Here is the question: Is it possible to suggest a minimalist mathematical model in the form of one parameterdependent ODE, which would exhibit this phenomenon? Assume that the dynamics of the insect population is governed by the ODE of the form Ṅ = F N) P N), where Nt) is the population size at time t, F is the growth rate of the population in the absence of predators, and P describes the effect of the birds on the rate of change of the population. I am free to choose any reasonable functional form for F and P. Since I am looking for a model as simple as possible, let me take the logistic equation for F : ) F N) = r B N 1 NKB, where r B > 0 is the per capita growth rate when N is close to zero, and K B is the population carrying capacity. I know that if there are no predators, then Nt) K B as t for any initial condition N 0 > 0. To choose P, I note that it is reasonable to expect that if N then P N) B = const since the birds can eat only some amount of insect. It is usually said that function P in this case should exhibit a saturation effect. Another qualitative feature of P is that it should decrease to zero faster than the linear function. The rational for this is that if the insect population is low, then birds prefer looking for food somewhere else. Putting these two assumptions together, I obtain that the graph of P should look similar to the one shown in the figure. PN) Figure 1: Qualitative form of P N) in the insect outbreak model Math 484/684: Mathematical modeling of biological processes by Artem Novozhilov artem.novozhilov@ndsu.edu. Fall 2015 N 1

2 It is possible to pick various analytical expressions for P, but one of the simplest with the desired properties is possibly P N) = BN 2 A 2 + N 2, where B > 0 is the saturation constant and A > 0 determines the inflection point of the graph of P N). Putting everything together, ) Ṅ = r B N 1 NKB BN 2 A 2 + N 2. 1) Model 1) has four parameters. It is usually quite a difficult task to fully describe the behavior of solutions depending on more than two parameters. Luckily for our case, it is frequently possible to reduce the number of model parameters by properly rescaling the variables. Let me make the change of variables and introduce new parameters x = N A, τ = Bt A, r = Ar B B, q = K B A. Note that new variables and parameters are dimensionless I will explain how to actually figure out a possible change of the variables at the end of the lecture). In the new variables I have ẋ = rx 1 x ) x2 =: fx, r, q). 2) q 1 + x2 To find equilibria of 2) I need to solve fx, r, q) = 0. Obviously, ˆx = 0 is always an equilibrium, which is unstable for any parameter values r, q actually, f 0, r, q) = r > 0). Other possible equilibria, if they exist, should satisfy r 1 x ) = x q 1 + x 2. Finding the zeros of the last expression boils down to solving a cubic equation, which, albeit formally possible, usually leads to messy expressions, whose use is dubious at best. It is better to look for the equilibria graphically. The equilibria are the points of intersections of g 1 x) = r1 x/q) and g 2 x) = u/1 + x 2 ). Since g 2 x) does not have any parameters, I can plot it once and for all. g 1 x) defines a straight line, which intersect x-axis at the point q and y-axis at the point r. By changing r for a fixed q I can see that it is possible to have additionally from one to three equilibria see the figure), which I will denote ˆx 1, ˆx 2, ˆx 3. How to determine stability of these equilibria? I can use the standard technique by analyzing the sign of the derivative f evaluated at the equilibria, but an easier method is to note that if all the equilibria are hyperbolic, then any two equilibria without other equilibria between them must be such that one is unstable and another one is asymptotically stable. I know that ˆx = 0 is unstable. Therefore, if there is only one additional equilibrium, say ˆx 1 and it is hyperbolic i.e., f ˆx 1 ) 0), then it must be asymptotically stable. If there are all three equilibria then ˆx 1 and ˆx 3 are asymptotically stable and ˆx 2, which is between them, is unstable see the figure). In the same figure I can also see that the bifurcations in the system happen when either two equilibria collide and disappear, or when two equilibria appear as the parameters change. This is 2

3 g1x), g2x) r 1 r 2 g 1 x) = r 1 x ) q r 3 g 2 x) = x 1+x 2 x q Figure 2: Nontrivial equilibria of 2) given as the points of intersection of g 1 x) and g 2 x). It is possible to have one r 1 and r 3 cases) or three r 2 case) equilibria. For some parameter values the graph of g 1 x) touches the graph of g 2 x) and there are two equilibria, one of which is non-hyperbolic exactly the fold bifurcation. It is interesting to find the bifurcation values of r, q. They can be found as the solutions to the following system fx, r, q) = rx 1 x ) x2 q 1 + x 2 = 0, df x, r, q) = r 1 2x ) 2x dx q 1 + x 2 ) 2 = 0. The first equation here is the condition for x to be an equilibrium, and the second equation is the condition for this equilibrium to be non-hyperbolic. By solving this system with respect to r and q I get 2x 3 r = 1 + x 2 ) 2, q = 2x3 x 2 1. fx) 0 ˆx 1 ˆx 2 ˆx 3 x Figure 3: The case of bistability in model 2). asymptotically stable Equilibria 0, ˆx 2 are unstable, and ˆx 1 and ˆx 3 are 3

4 The last expressions is a parametrically defined curve on the plane q, r). The shape of this curve can be analyzed by the usual means I used Mathematica to plot it, see the figure). Note that there is a singularity on this curve, which is called the cusp singularity. In the area inside the curve the model 2) has three nontrivial equilibria, and outside the curve there is only one equilibrium different from zero. Everywhere on the curve, except for the cusp point, crossing of this curve corresponds to the fold bifurcation, when either two equilibria appear or disappear. D 1 2 C r B q A Figure 4: The bifurcation curve in the space of parameters r, q) of model 2) Model 2) and, correspondingly, 1)) exhibits the phenomenon of hysteresis. Hysteresis can be defined as the dependence of the system state not only on the current environment but also on the previous history how the system found itself in this particular state. To illustrate this phenomenon with the help of the model 2), consider the surface in the space r, q, x), which is defined by the equilibrium condition fx, r, q) = 0. This surface is shown in the figure below, here for each fixed r, q) the x-coordinate on the surface gives the equilibria of 2). The previous figure can be obtained by the projection of this surface onto r, q) plane. Now I assume that parameter q is fixed, but somehow I can change the value of parameter r in the model. I start at the point A see the figures), where there is only one nontrivial equilibrium ˆx 1 this corresponds to the case in Figure 3 for r 1 ). This point is asymptotically stable, and the insect population settles at this point. Now I assume that r starts growing such that I reach point B in Figure 4. For an observer not much changed, since the population is still at the equilibrium ˆx 1, which is slightly greater than before, but quite close. On the other hand, for the whole system actually I have that the fold bifurcation happened, and aside of ˆx 1 two new equilibria ˆx 2 and ˆx 3 appeared, the former is unstable, and the latter one is asymptotically stable. If I increase r to the point C in Figure 4, I have observed continuous growth of ˆx 1 -equilibrium, in which the population resided. At this point actually another bifurcation happens, points ˆx 1 and ˆx 2 collide and disappear. What is going to happen with the population? It must find another stable state, which is actually present in the system and given by ˆx 3. Note however, that the coordinate of ˆx 3 is much larger than that of ˆx 1, and at this point the system experiences the outbreak! Now the insect population at ˆx 3 -equilibrium for any larger values 4

5 q C D B C ˆx A B r Figure 5: This surface is defined by the condition fx, r, q) = 0 in model 2) of r point D). The path for the population state in Figure 5 is given by A, lower B, lower C, upper C, D. Now I assume that somehow I am able to reduce the value of r, the insect population will start declining, but staying still at the point ˆx 3, now I will need to reduce the parameter value to the point B to actually make sure that the insect population is again at ˆx 1. Hence, the path on Figure 5 when we reduce the values of r is given by D, upper C, upped B, lower B, A. 4.1 Non-dimensional variables In the mathematical model of the insect outbreak a change of variables was presented, which allowed me to reduce the number of parameters. Here is the explanation how to actually find such a change of variables. In physics this process is frequently referred to as non-dimensionalization. As an example consider the logistic equation in the form Ṅ = rn 1 N ), K where r > 0 and K > 0 are parameters of the problem. The former is the rate of growth and the latter is the carrying capacity. Nt) is the size of population at time moment t. Let me assume that here I am talking about elephants, hence Nt) is the number of elephants in the population at the time moment t. What are the dimensions of the parameters in this problem? First note that on the left hand side I have the derivative Ṅ, i.e., the rate of change, which means that the units on the left 5

6 [ ] have to be [Ṅ] = elephants time the square parenthesis mean dimensions ), which implies that the right hand side has to have the same units, and I have no choice other than consider that [K] = [elephants] and [r] = [ 1 time]. Now I make the substitutions Nt) = Anτ), t = T τ, where τ is a new independent variable, nτ) is a new dependent variable, and A and T are constants to be determined. Using the chain rule hence A T dnt) dt = A dnτ) dt dn dτ = ran 1 An ) K Since I am free to choose A and T, I can set = A dnτ) dτ dτ dt = A T dnτ) dτ = dn dτ = rt N 1 An ). K T = 1, [T ] = [time], A = K, [A] = [elephants], r and the change of variables nτ) = Nt) K, τ = rt, reduces the logistic equation to the form without any parameters: ṅ = n1 n), where now ṅ means the derivative with respect to τ. Note that the new variables are dimensionless: nτ) is the proportion of the population at time τ, and time is measured in some abstract nondimensional units. It is usually possible to suggest several alternative changes of the variables that lead to dimensionless form. Which one to choose is up to the researcher. For example, in the insect outbreak model the choice was made so that all the remaining parameters are in the linear function, whose behavior is easy to figure out. Here is another example for the system of two autonomous equations: Ṅ = an bnp, P = dp + cnp. Here N, P are dependent variables, that depend on the independent variable t time), and a, b, c, d are parameters, all of which are supposed to be nonnegative. Let Nt) = Axτ), P t) = Byτ), t = T τ be my change of variables such that A, B, T are some constants I would like to determine. Using again the chain rule, I find or, after canceling, Aẋ = aat x babt xy, Bẏ = dt By + cabt xy, ẋ = at x bbt xy, ẏ = dt y + cat xy. 6,

7 Putting at = 1, bbt = 1, cat = 1, I find that the change of variables and parameters leads to xτ) = cnt) a, yτ) = bp t) a, τ = at, α = d a, ẋ = x xy, ẏ = αy + xy, which has only one dimensionless parameter. You should convince yourself that the change yields xτ) = cnt) d, yτ) = bp t) a, τ = at, γ = d a, ẋ = x xy, ẏ = γyx 1), which is in some cases more convenient for subsequent analysis. 7