Topic 30 Notes Jeremy Orloff

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1 Toic 30 Notes Jeremy Orloff 30 Alications to oulation biology 30.1 Modeling examles Volterra redator-rey model The Volterra redator-rey system models the oulations of two secies with a redatorrey relationshi. The equations are where a, b,, q are all ositive constants. x = ax xy = x(a y) rey y = by + qxy =y( b + qx) redator, Notice, if y = 0, then there is no redator and the rey oulation grows exonentially. If x = 0, then there is no rey and the redator oulation decays exonentially. It is easy to find that there are two critical oints Critical oints: (0, 0), q, a ). Volterra s Princile: Looking at the critical oint (b/q, a/) we see: If you increase a (the growth rate of rey) this leaves x (the rey oulation) unchanged but increases y (the redator oulation). More recisely, this describes the behavior of the critical oint, i.e. increasing a increases the y comonent of the critical oint. Likewise, increasing b (the decay rate of the redator) leaves, y unchanged, but increases x (the rey oulation). More recisely, increasing b increases the x comonent of the critical oint. Let s analyze this system by linearizing near the critical oints. [ ] a 0 J(0, 0) =. This has eigenvalues = a, b, with eigenvectors = 0 b [ ] 1, 0 [ ] 0. 1 The linearized system is a saddle. It is structurally stable, so the nonlinear system also has a saddle at (0, 0), (See lot below). J q, a ) [ ] 0 b = q qa. This has eigenvalues ±i ab. 0 The linearized system is a center. This not structurally stable, so nonlinear system is either a center or a siral sink or source. Since J q, a ) [ ] [ ] 1 0 = qa we know that the center or siral turns counterclockwise. 0 It turns out the nonlinear system has a center. The roof of this is given below. Here is a Matlab diagram 1

2 30 APPLICATIONS TO POPULATION BIOLOGY 2 Volterra redator-rey Fancier redator-rey Examle Consider the following redator-rey oulation model x =3x x 2 xy y = y y 2 + xy. (a) Which one of the variables reresents the redator oulation and which the rey? (b) Describe the oulation growth of each secies in the absence of the other. (c) Analyze the critical oints and sketch use that to sketch a hase ortrait. (d) Describe what haens to the oulations over time. answer: (a) We see that the resence of y, i.e. y > 0 decreases the growth rate of x and the resence of x increases the growth rate of y. Therefore, x reresents the rey oulation and y the redator. (b) If y = 0 then x = 3x x 2. This is a logistic oulation model with carrying caacity 3. Likewise, if x = 0 then y = y y 2 is a logistic oulation model. So, in the absence of the other, each oulation stabilizes at the carrying caacity of its logistic model. (c) Finding the critical oints is relatively easy. The two eqautions are x = 3x x 2 xy = x(3 x y) = 0 y = y y 2 + xy = y(1 y + x) = 0 In each equation one of the factors must be 0. This gives four critical oints (0, 0), (0, 1), (3, 0), (1, 2). [ ] 3 2x y x We comute the Jacobian =. Next we linearize at each critical y 1 2y + x oint. You should do this yourself. The results are comiled in the following table.

3 30 APPLICATIONS TO POPULATION BIOLOGY 3 Critical oints (0, 0) (0, 1) (3, 0) (1, 2) [ ] [ ] [ ] [ ] J λ 3, 1 2, 1 3, 4 ( 3 ± 7 i)/2 linear tye source saddle saddle siral sink [ ] [ ] [ ] [ ] [ ] [ ] v,,, Not needed Structural stability stable stable stable stable By considering the tangent vector at (u, v) = (1, 0) we see the siral sink at (1, 2) turns in the counterclockwise direction. Hand sketch of hase lane Matlab lot of hase lane (d) As long as both oulations are initially ositive, the model redicts they will go asymtotically to the stable equilibrium at (1,2) Proof the Volterra redator-model has closed trajectories You are not resonsible for the following roof. Claim: In the Volterra redator-rey model the critical oint at q, a ) is a center. More recisely, every trajectory with initial condition (x 0, y 0 ) in the first quadrant is a closed loo in the first quadrant that circles the critical oint. Proof: Because the ositive x and y axis are trajectories existence and uniqueness imlies a trajectory that starts in the first quadrant must stay there i.e., it can t cross out of the quadrant. To understand the trajectory in more detail we use the following trick. dy dx = dy/dt y( b + qx) = dx/dt x(a y).

4 30 APPLICATIONS TO POPULATION BIOLOGY 4 This is a searable equation: dy (a y) y = dx ( b + qx) x a ln y y = b ln x + qx + c. (*) This is an imlicit equation for a trajectory. Each value of c corresonds to a different trajectory. w x, y A B C y Q w=a ln y y w= b ln x+qx+c D E F w vs. x and w vs. y P (b/q, a/) S R Phase lane trajectory Now we have to show that the grah of the imlicit function defined in (*) is a closed loo. Using techniques we can show that the grahs of w = b ln x + qx + c and w = a ln y y are as shown above. Equation (*) tells us that a oint (x, y) is on the trajectory if the w = a ln(y) y curve and w = b ln(x) + qx + c curve are at the same hight. We can translate this to the hase lane trajectory as follows: Draw any horizontal line in the first grah. Its oints of intersection with the two curves give x and y coordinates of oints on the trajectory. Let A 1, B 1, etc. be the first coordinate of A, B, etc. Then the oints A, B and C in the first lot corresond to the oints P = (A 1, B 1 ) and R = (C 1, B 1 ) in the second lot. The oints D, E and F in the first lot corresond to the oints S = (E 1, D 1 ) and Q = (E 1, F 1 ) in the second lot. Now ay attention, the closed loo corresonds to the following ath along the two curves in the first grah. As the horizontal line goes down from its eak, its intersection oints travel from A to E along the x curve and from B to F along the y curve. This means that both x and y are increasing since they are the first coordinates of their resective curves. So this corresonds to the trajectory from P to Q on the second grah. Continuing, here s a table describing the closed trajectory (roving it s closed). horizontal line x, y curves x, y trajectory to to bottom A, B to E, F increase, increase P to Q bottom to to E, F to C, B increase, decrease Q to R to to bottom C, B to E, D decrease, decrease R to S bottom to to E, D to A, B decrease, increase S to P Easier, indirect argument There is an easier indirect argument that the trajectory must be closed. x

5 30 APPLICATIONS TO POPULATION BIOLOGY 5 Since, in the left-hand grah above, each horizontal line intersects each curve in at most 2 oints there are at most 2 oints on a trajectory with the same y-value. This means the trajectory can not be a siral. Hence it must be a center.