POPULATION DYNAMICS: TWO SPECIES MODELS; Susceptible Infected Recovered (SIR) MODEL Next logical step: consider dynamics of more than one species. We start with models of 2 interacting species. We consider, so-called, box models where species are assumed to be well mixed. Models that include spatial movement of species are no discussed in this course. (Ordinary vs. partial diff. equations.) If they co-exist in the same environment: rate of change of u=+growth effect of predator-prey encounters rate of change of v= decay+ effect of predator-prey encounters How to describe the effect of encounters? Law of Mass Action! 1. Lotka-Volterra predator-prey model: heuristic derivation. Consider 2 species, prey u, and predator v. Population of prey without predator grows (a>0 is a const.): = au; u(0)=u 0; population of predator without prey decays (b>0 is a const.): = bv; v(0)= v 0. According to Law of Mass Action the probability of encounters of 2 species is proportional to the proct of population densities of these species. Proportionality coefficients depend of various factors. Thus, we arrive at 24 25
the system of equations (a, b, n, m>0 are const.): = au nuv, = bv+muv; u(0)=u 0, v(0)= v 0. Possible model modifications: logistic growth for prey (instead of exponential), etc. 2. Competition model: derivation. Heuristic derivation: consider two species that consume the same resource. Assume that each species population in the absence of the other is described by the logistic equation (here u and v are population densities of the two species): = k 1u α 1 u 2, = k 2v α 2 v 2. When the other species is present: rate of change of u=+growth competition between u competition between u and v rate of change of v = + growth competition between v competition between u and v Thus, using once again the Law of Mass Action, we write: = k 1u α 1 u 2 β 1 uv, = k 2v α 2 v 2 β 2 uv; u(0)=u 0, v(0)= v 0. Another way to derive: Consider a well stirred batch reactor. Let u and v be the population densities of two types of bacteria, and c be the food concentration. The same food is consumed by both types of bacteria. Assume that growth rate coefficient for each bacteria type is a linear function of c: K i = K i (c)=κ i c (i= 1, 2). Then we have a system of equations 26 27
=κ 1cu, =κ 2cv, dc = a 1κ 1 cu a 2 κ 2 cv, where 1/a i (i= 1,2) are, so-called, yield factors. Initial conditions are u(0)=u 0, v(0)= v 0, c(0)= c 0. The above system can be reced to two equations as follows. Let us multiply the first equation by a 1, the second by a 2, and add the three equations. We obtain, a 1 + a 2 + dc = d(a 1u+ a 2 v+c) = 0. Integrating this equation, we get for any t: a 1 u(t)+ a 2 v(t)+ c(t)=const, and from the initial conditions: Conservation of mass! From the above we express c(t)= a 1 u 0 + a 2 v 0 + c 0 a 1 u(t) a 2 v(t)=a a 1 u a 2 v, and substitute this expression in the original equations for u and v to obtain the system: =κ 1cu=κ 1 u(a a 1 u a 2 v) = k 1 u α 1 u 2 β 1 uv; =κ 2cv=κ 2 v(a a 1 u a 2 v) = k 2 v α 2 v 2 β 2 uv; where k i =κ i A,α 1 =κ 1 a 1,α 2 =κ 2 a 2,β 1 =κ 1 a 2, and β 2 =κ 2 a 1. In the future, for the analysis, we will write this system in yet another form. 3. SIR model. It is common to start with a schematic representation: a 1 u(t)+ a 2 v(t)+ c(t)= a 1 u 0 + a 2 v 0 + c 0. 28 29
S I R Which processes affect the rates of change of respective populations? What are the assumptions? Law of Mass Action! For box models it does not matter whether we use the population densities or actual populations: numerical values of coefficients will be different but qualitative behavior is going to be the same! Let us use notations S, I, and R for susceptible, infected, and recovered. Then, di ds = αis, =+αis β I, dr =+β I, S(0)=S 0, I(0)= I 0, R(0)=0 without immunization. Remark 1. It can be easily checked that in this system the total population is conserved: at any instant of time S(t)+ I(t)+R(t)=S 0 + I 0 = N total = const. Remark 2. If deaths are included in the model, we get the system: ds = αis δ 1S, di =+αis β I δ 2I, dr =+β I δ 1R, S(0)=S 0, I(0)= I 0, R(0)=0. We note that the SIR system is, in fact, a combination of a closed system of 2 equations for S and I (this system can be solved independently of R) and the differential relation for R (i.e., when I is known, R is obtained by simple integration). So, we actually have to analyze the system of 2 equations: 30 31
di ds = αis, =+αis β I. If recovered can become susceptible again we arrive at the, so-called, SIRS model. 4. SIRS model. Schematic representation: dr =+β I γr, S(0)=S 0, I(0)= I 0, R(0)=R 0 in general case. Using the fact that in this model (without deaths) the total population, N, is once again conserved, we can rece the above 3-dimensional system to a system of 2 equations. We express R in terms of S and I as follows. For N= S 0 + I 0 + R 0 = const known, we have: R(t)= N I(t) S(t). Substituting this into equations for S and I we, finally, obtain: S I R Corresponding model system will now be in the form: ds = αis+γ(n I S), di =+αis β I, ds = αis+γr, di =+αis β I, S(0)=S 0, I(0)= I 0. 4. General systems of two nonlinear differential equations. It can be easily seen that all the models 32 33
introced above can be written in the following general form: = f(u, v), = g(u, v), u(0)=u 0, v(0)= v 0 known. somehow known: u = u(t), v = v(t). Consider a (u, v) plane, which we call a phase plane. Then the point with coordinates(u(t), v(t)) (where time t is changing) will trace a curve on this plane. Functions f(u, v) and g(u, v) describe the rules of species interactions and behavior. Our goal is to describe possible types of solutions: we want to know when the populations will grow to certain values, go extinct, oscillate, etc., and how will these types of solutions depend on numerical values of model parameters? Let us extend the approach that worked previously for scalar (single) nonlinear differential equations to systems of two (and later, to systems of three and more) differential equations. 5. Model solutions and phase plane. Geometry of the model system. Assume that the solution of a system is 34 35
If the solution goes to a steady state (i.e., as t, (u(t), v(t)) (ū, v)) then the point on the phase plane corresponding to a solution will eventually stop. If the solution oscillates (i.e., the values of(u(t), v(t)) periodically repeat themselves with certain period T) then the point on the phase plane will trace a closed loop. The velocity vector w related to motion of a point (u(t), v(t)) on the plane (this vector is tangential to the, so-called, trajectory of the moving point, it shows the direction of point s motion and its length shows how fast the point is moving) is defined as follows: w=, =(f(u, v), g(u, v)). The above expression means that the system of differential equations specifies the, so-called, vector field on the phase plane: with every point(u, v) we associate a vector ( f(u, v), g(u, v)). These vectors show the direction and speed of motion of a solution point, that is currently located at position(u, v), as time increases. Sometimes it is convenient to show only the direction of motion at every point of the phase plane and not how fast the solution changes along the phase trajectory. Then instead of velocity vector field one may show the, so-called, direction field: all the vectors associated with different locations in the phase plane will have the same length and will possibly differ only by direction. To normalize the vectors with coordinates(f(u, v), g(u, v)) (i.e., to make them all be of the same length L) the following formula may be used: the new normalized vectors will have coordinates (F(u, v), G(u, v)) 36 37
L f(u, v) =, f 2 (u, v)+ g 2 (u, v) Lg(u, v). f 2 (u, v)+ g 2 (u, v) The prescribed initial condition hits one of the trajectories on the phase plane and follows it as time increases. For us it is important to know where it will go. If the initial condition corresponds to a steady state, then the solution will stay at this steady state forever. It turns out that if the initial condition is not at a steady state, only a few possibilities may occur on the phase plane: (a) solution tends to a stable steady state, (b) moves away (to infinity), (c) belongs to a limit cycle, (d) tends to a limit cycle, (e) moves away from a limit cycle, (f) what else? Same as in the case of the scalar equation, steady states (ū, v) are the constant solutions that satisfy the following system of equations (since dū/= 0 and d v/= 0): 0= f(ū, v), 0= g(ū, v). In terms of behavior on the phase plane we have that if the the point corresponds to a steady state, it will not move since the velocity vector at this point has zero entries: ( f(ū, v), g(ū, v)) =(0, 0) (and thus, no direction of motion is defined, and the speed of motion is zero). Important! No chaos for continuous systems of 2 autonomous differential equations, i.e., no chaos on the phase plane! We need 3 equations to proce chaos. 38 39
Our goal will be to characterize possible types of solution behavior in the vicinity of the steady states, which will lead to identification of several distinct types of steady states (also called equilibrium points). Then, using some additional information on general behavior of phase trajectories away from steady states (e.g., the fact that trajectories can only intersect at steady states), we will be able to qualitatively characterize the global behavior of solutions (not only near the steady states). In what follows we will extensively use the idea of a null-cline. The curves in the phase plane whose(u, v) coordinates satisfy the equation f(u, v)=0 are called u null-clines. Special feature: solution trajectories intersect these null-clines vertically (since on these curves the u-component of velocity vectors is zero). Similarly, curves in the phase plane whose(u, v) coordinates satisfy the equation g(u, v) = 0 are called v null-clines. Their special feature: solution trajectories intersect these null-clines horizontally (since on these curves the v-component of velocity vectors is zero). Evidently, the steady states correspond to points of intersection of u null-clines and v null-clines. 6. Linearization procere for systems of two nonlinear differential equations. Given a system of equations = f(u, v), = g(u, v), and a steady state(ū, v) satisfying 0= f(ū, v), 0= g(ū, v). Let us perturb this steady state, i.e., we consider initial 40 41
conditions u(0)=ū+α(0), v(0)= v+β(0) with α(0), β(0) 1. We want to find out what happens to α(t), β(t) at t increases. We substitute u(t)=ū+α(t), into the system to obtain: d(ū+α) = dα v(t)= v+β(t), = f(ū+α, v+β) = f(ū, v)+ f u (ū, v)α+ f v (ū, v)β+... ; d( v+β) = dβ = g(ū+α, v+β) = g(ū, v)+ g u (ū, v)α+ g v (ū, v)β+.... Here f u, g u are the derivatives of f and g with respect to u, f v, g v are the derivatives of f and g with respect to v. Thus, approximately, we obtain dα = f u(ū, v)α+ f v (ū, v)β; 42 dβ = g u(ū, v)α+ g v (ū, v)β, andα=0,β= 0 correspond to the steady state solution of the above linear system. = Importance of linear equations! To find out how perturbationsα(t) andβ(t) behave, we first need to discuss solutions of systems of 2 linear constant coefficient differential equations. 7. Linear systems of two equations with constant coefficients and equivalent second order equation. Let us write down such system in a special form (here we assume that x and y are the dependent variables): d x = a 11x+ a 12 y, d y = a 21x+ a 22 y. Evidently, the point in the(x, y)-phase plane with coordinates x= 0, y= 0 is an equilibrium solution of this system. 43
Constant coefficients a i j may be written as a matrix (called Jacobian matrix): A= a 11 a 12 a 21 a 22. For the future classification of steady states the following quantities are important: trace of matrix A: tra= a 11 + a 22 ; determinant of matrix A: det A= a 11 a 22 a 12 a 21. There are several approaches to finding solutions x(t), y(t). Here, for simplicity, we use the fact that every system of 2 first order equations (i.e., system of 2 equations containing only first order derivatives) is equivalent to some scalar second order equation (i.e., one differential equation that contains both first and second order derivatives). Let us derive an equivalent second order equation, e.g., for x (equation for y will be the same). We begin with differentiating the first equation of the system with respect to t: d 2 x 2 = a d x 11 + a d y 12. Then we substitute here d y/ from the second equation: d 2 x 2 = a d x 11 + a 12(a 21 x+ a 22 y), and finally, we eliminate y using the first equation of the original system: y=(d x/ a 11 x)/a 12. Collecting the terms, we obtain the following equation for x: d 2 x 2 (a 11+ a 22 ) d x +(a 11a 22 a 12 a 21 )x= 0. We note that this equation may be re-written as: d 2 x d x 2 (tra) +(det A) x= 0. The initial conditions for the original system are x(0)= x 0, y(0)= y 0. The second order equation requires 2 conditions for x. Substituting x 0 and y 0 into the first 44 45
equation of the original system we see that the condition for y may be converted into the second condition for x: d x (0)= a 11x 0 + a 12 y 0 known. If x(t) is known (as a solution of the second order equation with corresponding 2 initial conditions), then y(t) is obtained from y= d x/ a 11x a 12. The original system and the above second order equation are equivalent in the following sense: if x is the solution of one, then it is the solution of the other (the same is true for y). 8. Possible solutions: characterization of steady states. We start with the observation that x= 0 (and thus, y= 0) is one of the solutions of the second order equation (which was expected since(0,0) is the steady state of the original system). Assume that the initial conditions are chosen close to(0,0): x(0)= x 0, y(0)= y 0, with x 0 and y 0 not equal to zero simultaneously. What will happen to x(t) and y(t)? Based on our previous experience with scalar first order equations, we seek solution of the second order equation in the form: x= e λt, whereλis yet unknown. Substituting this into the equation, we obtain: λ 2 e λt (tra)λe λt +(deta)e λt = 0. Canceling e λt 0, we arrive at the CHARACTERISTIC EQUATION forλ: λ 2 (tra)λ+det A= 0. This is a quadratic equation, and thus, it has 2 solutions. So, as a result, we will have not one x(t), but two: x 1 (t) and x 2 (t). Since the second order differential equation is linear, the linear combination of these two solutions, C 1 x 1 (t)+ C 2 x 2 (t), where C 1 and C 2 are arbitrary 46 47
constants of integration, is also a solution. Thus, the so-called general solution of the linear second order equation can be written as follows: (a) For det A> 0, and tr A> 0:λ 1 > 0,λ 2 > 0, and the steady state(0, 0) is unstable. x(t)= C 1 x 1 (t)+ C 2 x 2 (t). Two constants of integration are defined by 2 known initial conditions x(0) and d x/d t(0). Let us findλ 1,2 and corresponding x 1,2 for different possible cases. From equation forλ, we obtain: λ 1,2 = tra 2 ± tra 2 det A. 2 Steady state of this type is called an unstable node. (b) For det A> 0, and tr A< 0:λ 1 < 0,λ 2 < 0, and the steady state(0, 0) is (asymptotically) stable. Case 1: If tra 2 deta> 0, 2 we have 2 distinct realλ s; x 1 (t)=exp(λ 1 t), x 2 (t)=exp(λ 2 t), and Steady state of this type is called a stable node. x(t)= C 1 e λ 1t + C 2 e λ 2t. (c) For det A< 0, and tr A either positive or negative: 48 49
λ 1 < 0,λ 2 > 0, and and the steady state(0,0) is unstable. Figures are similar to those presented earlier for stable and unstable node. Case 3: If tra 2 deta< 0, Steady state of this type is called a saddle. Case 2: If tra 2 deta= 0, 2 we have 2 identical realλ s:λ=tra/2; x 1 (t)=exp(λt), x 2 (t)= t exp(λt), and 2 we have 2 distinct complexλ s:λ 1,2 =σ±iδ, i= 1, where σ= tra 2, δ= deta tra 2 ; and x 1 (t)=exp(σt) sin(δt), x 2 (t)=exp(σt) cos(δt). So, x(t)= C 1 exp(σt) sin(δt)+ C 2 exp(σt) cos(δt). 2 x(t)= C 1 e λt + C 2 t e λt. (a) For tr A> 0= λ>0: unstable node. (a) For tr A= 0 we have pure oscillations:σ=0, and x(t)= C 1 sin(δt)+ C 2 cos(δt), whereδ= det A. The steady state(0, 0) is stable (but not asymptotically stable). (b) For tr A< 0= λ<0: stable node. 50 51
Steady state of this type is called a center. Steady state of this type is called a stable focus. (b) For tr A> 0 we have oscillations with growing amplitude. The steady state(0,0) is unstable. Let us summarize our findings using the following scheme: Steady state of this type is called an unstable focus. (c) For tr A< 0 we have oscillations with decreasing amplitude. The steady state(0,0) is (asymptotically) stable. What happens when det A= 0 in the linear case/ nonlinear case? 52 53