We have two possible solutions (intersections of null-clines. dt = bv + muv = g(u, v). du = au nuv = f (u, v),
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1 Let us apply the approach presented above to the analysis of population dynamics models. 9. Lotka-Volterra predator-prey model: phase plane analysis. Earlier we introduced the system of equations for prey population u and predator population v (a, b, n, m > 0 are const.): du = au nuv = f (u, v), dv = bv + muv = g(u, v). The steady states are solutions of the system: f (ū, v)=aū nū v = ū(a n v)=0, g(ū, v)= b v + mū v = v( b + mū)=0. We have two possible solutions (intersections of null-clines f (u, v)=0 and g(u, v)=0: (a) ū 1 = 0, v 1 = 0; (b) ū 2 = b/m, v 2 = a/n. 54 Let us characterize the equilibrium points. The Jacobian matrix that must be evaluated at the steady states has the form: f u (ū, v) f v (ū, v) A(ū, v)= g u (ū, v) g v (ū, v). b + mū = a n v nū. m v 55
2 (a) For steady state (ū 1, v 1 )=(0, 0), we have: A(ū 1, v 1 )= a 0. 0 b Thus, det A = ab < 0, and no matter what tr A is, the steady state (0, 0) is a saddle. (b) For steady state (ū 2, v 2 )=(b/m, a/n), we have: A(ū 2, v 2 )= 0 nb m am n 0. Thus, tr A = 0, det A =+ab > 0, and the steady state (b/m, a/n) is a center. Global behavior of trajectories on the phase plane =? Equation for trajectories: du f (u, v) = dv g(u, v), u(v 0)=u 0, 56 or dv g(u, v) = du f (u, v), v(u 0)=v 0. How to convert phase trajectory information into solution curves? 57
3 What are the implications? Let us find out more about behavior of trajectories near the steady states. We write for the perturbations: u(t)=ū + x(t), v(t)= v + y(t). In the case of the saddle point, (ū 1, v 1 )=(0, 0), the systems of equation for x and y, dx = ax, dy = by, leads to the following equation for the trajectories: dy dx = b y a x. This can be easily integrated to obtain: y = C x b/a, 58 where C is an arbitrary constant of integration. Different choices of C following from the initial conditions produce different (non-intersecting) trajectories on the phase plane! In the case of the center, (ū 2, v 2 )=(b/m, a/n), the systems of equation for x and y, dx = nb m y, dy = am n x, leads to the following equation for the trajectories: dy dx = am2 bn 2 x y = N x y, 59
4 where N > 0 is a known constant. We integrate the equation to obtain: y 2 + Nx 2 = C, where C is the constant of integration. Evidently, for different choices of C > 0 the above equation describes a family of (non-intersecting) elliptic trajectories! Important note on structural stability. We characterized steady states of nonlinear systems and their stability properties using linearization procedure. Will the stability properties be the same for all types of the steady states for both nonlinear and corresponding linear equations? That is, whether some small perturbations can change the 60 character of a steady state? It turns out that all the steady states that we discussed, but one, are structurally stable! The center in the linearized system, however, in the original nonlinear system may either be a center, or a stable or unstable focus! What is the situation with the center steady state in our original predator-prey system? We need to study the trajectories of the original system: dv v(mu b) = du u(a nv). Integration using separation of variables: v a e nv = Cu b e mu, where the constant of integration C is defined by initial conditions. If trajectory is a spiral, it intersects a given line (e.g., null-cline) at an infinite number of points. If trajectory is a 61
5 closed curve, it intersects a given line (e.g., null-cline) at most at two points. Let us, e.g., fix u. Then the equation becomes v a e nv = const. Solutions of the above equation produce points of intersection of the original solution with a vertical straight line. How many intersections with a straight line? Let us show that geometrically for v a e nv we have the following picture: 62 Indeed, the unique point of maximum v = v is found by setting the first derivative of the function v a e nv with respect to v to zero: v a e nv =(av a 1 nv a )e nv =(a nv)v a 1 e nv = 0, and thus, v = a n. This allows us to complete the phase portrait of the Lotka-Volterra system (not only locally, but globally). 10. Competition model: phase plane analysis. Let us re-write previously derived competition model 63
6 du = k 1u α 1 u 2 β 1 uv, dv = k 2v α 2 v 2 β 2 uv, in a slightly different form: du = k 1u 1 umu vnv, dv = k 2v 1 vmv unu, where M u = k 1 /α 1, M v = k 2 /α 2, N u = k 2 /β 2, N v = k 1 /β 1. PLAN: null-clines steady states steady states classification for various cases phase portraits for various cases + analysis + conclusions! Null-clines: du = 0: u = 0; 1 u M u v N v = 0; 64 dv = 0: v = 0; 1 v M v u N u = 0. Steady states (intersections of null-clines): (a) ū 1 = 0, v 1 = 0, always exists; (b) ū 2 = 0, v 2 = M v, always exists (M v is a carrying capacity for v); (c) ū 3 = M u, v 3 = 0, always exists (M u is a carrying capacity for u); (d) Non-trivial steady state, ū 4 = N um u (M v N v ) MuM v N u N v, v 4 = N v M v (M u N u ) MuM v N u N v, is the intersection of the straight lines, 1 u/m u v/n v = 0 and 1 v/m v u/n u = 0, that must occur in the first quadrant due to a natural biological constraint: ū > 0, v > 0. Evidently, this steady 65
7 state does not always exist since two straight lines may not necessarily intersect in the first quadrant! Conditions for existence of steady state (d): we must have simultaneously Mv > N v, M u > N u or Mv < N v, M u < N u. Thus, we arrive at the following 4 different possible cases: (1) M v < N v, M u > N u (2) M v > N v, M u < N u 66 (3) M v > N v, M u > N u (4) M v < N v, M u < N u Characterization of steady states. The Jacobian matrix: A(ū, v)= k 1 1 2ū Mu v N v k 1 ū N v v k 2 k v ū N u Mv N u. Let us evaluate the Jacobian matrix at various steady states: (a) For (ū 1, v 1 )=(0, 0), 67
8 A(0, 0)= k 2 k Here tra = k 1 + k 2 > 0, det A = k 1 k 2 > 0, and (tra/2) 2 det A =(k 1 k 2 ) 2 /4 > 0, and thus, this trivial steady state is always an unstable node. Alternatively, we may notice that λ 1 = k 1 > 0, λ 2 = k 2 > 0 (i.e., λ s are the elements standing on the main diagonal of A in the case where A is upper triangular, lower triangular, or diagonal matrix). We will use similar argument for the two steady states that we consider next. (b) For (ū 2, v 2 )=(0, M v ), A(0, M v )= k 1 1 M v N v 0 k 2 M v N u k In the lower triangular matrix A, the diagonal elements correspond to λ 1 = k 1 1 M v N v, λ 2 = k 2 < 0. For M v > N v, we have that λ 1 < 0, and thus, in this case the steady state is a stable node. For M v < N v, we have that λ 1 > 0, and thus, in this case the steady state is a saddle (unstable). (c) For (ū 3, v 3 )=(M u,0), A(M u,0)= k 1 k 1 M u N v 0 k 2 1 M u N u. In the upper triangular matrix A, the diagonal elements correspond to λ 1 = k 1 < 0, λ 2 = k 2 1 M u. N u 69
9 For M u > N u, we have that λ 2 < 0, and thus, in this case the steady state is a stable node. For M u < N u, we have that λ 2 > 0, and thus, in this case the steady state is a saddle (unstable). (d) For (ū 4, v 4 ), we have (see the system of equations from which we determined ū 4 > 0, v 4 > 0): ū 4 k 1 Mu A(ū 4, v 4 )= k 1 ū 4 N v k 2 v 4 N u k 2 v 4. Mv For the above matrix, tra = k 1 ū 4 Mu k 2 v 4 Mv < 0, ū 4 v 4 det A = k 1 k 2 1 M um v. MuM v N u N v We recall that the steady state (ū 4, v 4 ) exists only if 70 Mv > N v M u > N u or if Mv < N v M u < N u. In the former case det A < 0, and so, the steady state is a saddle, while in the latter case det A > 0, and the steady state is a stable node (it can be shown that (tra/2) 2 det A > 0). Let us now construct the phase portraits for various cases: (1) M v < N v, M u > N u (2) M v > N v, M u < N u 71
10 (3) M v > N v, M u > N u (4) M v < N v, M u < N u Principle of Competitive Exclusion! 11. SIR model: phase plane analysis. Let us repeat the schematic representation: S I R 72 The system of equations for S, I, and R (susceptible, infected, and recovered) has the form: ds = αis, =+αis β I, dr =+β I, S(0)=S 0, I(0)=I 0, R(0)=0 without immunization. Here α>0 is the infection rate constant, β>0isthe removal rate constant of infectives. We recall that at any instant of time S(t)+I(t)+R(t)=S 0 + I 0 = N = const, where N is the total population (assume no deaths, births, etc.). We also note that in the SIR system the equation for R is decoupled from a closed system of 2 equations for S and I (and when I is known, R can be obtained by integration). Thus, for SIR model we analyze the system of 2 equations: 73
11 ds = αis, =(αs β)i. The relation S(t)+I(t)+R(t)=N means that the solutions of the above system will always lie inside the triangle S(t)+I(t) N, S(t) 0, I(t) 0 (since R 0), and, without immunization, the initial conditions will lie of the line S 0 + I 0 = N: 74 Null-clines: ds = 0: I = 0; S = 0; = 0: I = 0; S = β α = σ. New feature!: Here the steady states are not isolated points, but the whole I = 0, S 0 half-line. This means that as t we have that I(t) 0, and S(t) S( ) > 0 =const (that depends on initial conditions). Let us characterize the signs of velocity vectors: ds < 0 : for all I > 0, S > 0. < 0 : for I > 0, S < β α = σ, > 0 : for I > 0, S > β α = σ. 75
12 We may have the following situations: Some terminology: σ = β/α is called relative removal rate; 1/σ is the infection s contact rate. The equation for I may be re-written as = β αs β 1 I. What happens at the beginning of the infection (at t = 0) depends on the sign of the term in the brackets in the above equation, i.e., on whether αs 0 β 1 > 0 or αs 0 β 1 < The quantity (common notation that has nothing to do with recovered!) R 0 = αs 0 β or, if I 0 N, so that S 0 N, R 0 = αn β is called the basic reproduction rate of the infection. This number shows how many secondary infections are produced from one primary infection in the initial population. Interpretation of the term 1/β: it is the characteristic time showing for how long the individual infection lasts. Evidently, if R 0 > 1, then / > 0, and we have an epidemic. How do we define an epidemic (number of new infections is increasing, or via total fraction of the original population infected, etc.)? If we assume that / > 0 corresponds to epidemic, and / < 0 corresponds to no epidemic, then we get the following implication for immunization policies: the 77
13 fraction of population, p, that must be immunized (with fraction (1 p) not immunized) is defined by the relation α(1 p)n β =(1 p)r 0 < 1, and thus, p > 1 1 R 0, where R 0 is the original basic reproduction rate of the infection. Why does the disease die out: from the lack of infectives or from the lack of susceptibles? SIRS model: phase plane analysis. We recall the schematic representation for SIRS: S I R Corresponding model system has the form (here 1/γ is the characteristic time it takes for the recovered to become susceptible again): ds = αis + γr, =+αis β I, dr =+β I γr, S(0)=S 0, I(0)=I 0, R(0)=0 without immunization. Once again, we recall that the quantity S(t)+I(t)+R(t)=N = S 0 + I 0 =const is conserved. Thus, we may express 79
14 R(t)=N I(t) S(t). Substituting this into equations for S and I, we obtain: ds = αis + γ(n I S)= f (S, I), =+αis β I = g(s, I), S(0)=S 0, I(0)=I 0. Null-clines: ds = 0: αis + γ(n I S)=0; = 0: I = 0; S = β α = σ. The functional representation of S null-cline can be written as 80 γ(n S) I = αs + γ. It follows from the above expression that this null-cline intersects the S- and I-axes at the points (N,0) and (0, N), respectively. Also, it is immediately seen that in the first quadrant this curve lies below the straight line I = N S. We have either one or two steady states (intersections of null-clines) in the first quadrant: (a) ( S 1, Ī 1 )=(N,0), (b) β γ(n β/α) ( S2, Ī 2 )=,. α β + γ 81
15 We note that the second (non-trivial) steady state exists only if the following condition is satisfied: αn β = R 0 > 1. Let us characterize the steady states when R 0 < 1 and when R 0 > 1. The Jacobian matrix has the form: (αī + γ) (α S + γ) A( S, Ī)= αī α S β. For R 0 = αn/β < 1 only one steady state, ( S 1, Ī 1 )=(N,0), exists and γ (αn + γ) A( S 1, Ī 1 )= 0 αn β. From the above (taking into account that the above matrix is upper triangular), we obtain: λ 1 = γ<0, λ 2 = αn β<0. 82 Thus, this unique steady state is a stable node. The phase portrait in this case: For R 0 = αn/β > 1 we have two steady states. (a) For ( S 1, Ī 1 )=(N,0) we have the same Jacobian matrix as in the previous case, but now (due to condition on R 0 ) λ 1 = γ<0, λ 2 = αn β>0. So, this steady state is a saddle. (b) For ( S 2, Ī 2 ), with S 2 > 0, Ī 2 > 0, we obtain 83
16 (αī 2 + γ) (β + γ) A( S 2, Ī 2 )=. αī 2 0 Since tra( S 2, Ī 2 )= (αī 2 + γ) < 0, and det A( S 2, Ī 2 )=γβ(r 0 1) > 0, the non-trivial steady state is either stable node or stable focus (depending on the numerical values of parameters). The phase portrait in the case of oscillatory approach to ( S2, Ī 2 ): 84
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