Predator - Prey Model Trajectories and the nonlinear conservation law

Transcription

1 Predator - Prey Model Trajectories and the nonlinear conservation law James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University October 28, 2013

2 Outline 1 Drawing Trajectories Again 2 Only Quadrant One Is Biologically Relevant Trajectories on the y + Axis Trajectories on the x + Axis 3 The Nonlinear Conservation Law

3 Abstract This lecture talks about the trajectories of the predator prey model and the nonlinear conservation law this model satisfies.

4 Drawing Trajectories Again We drew trajectories for the linear system models already without a lot of background discussion. Now we ll go over it again in more detail. How do we draw trajectories? We use the algebraic signs of x and y to determine this. For example, in Region I, the sign of x is negative and the sign of y is positive. Thus, x decreases and y increases in this region. If we graphed (x(t), y(t)) in the x y plane for all t > 0, we would plot a y versus x curve. We would have y = f (x) for some function of x. Note by the chain rule dy dt = f (x) dx dt. Hence, as long as x is not zero (and this is true in Region I!), we have the slope of the curve y = f (x) is given by df dx (t) = y (t) x (t).

5 Drawing Trajectories Again Since our pair (x, y) is the solution to a differential equation, we expect that x and y both are continuously differentiable with respect to t. So if we draw the curve for y vs x in the x y plane, we do not expect to see a corner in it (as a corner means the derivative fails to exist). So we can see three possibilities: a straight line as x equals y at each t meaning the slope is always the same, a curve that is concave up or a curve that is concave down. We illustrate this three possibilities in the next figure. When we combine trajectories from one region with another, we must attach them so that we do not get corners in the curves. This is how we can determine whether or not we should use concave up or down or straight in a given region.

6 Drawing Trajectories Again y axis x = 0 y = a b x = 0 II (, ) I (,+) III (+, ) IV (+,+) In this figure, we show the three trajectory types for the signs of region I x axis y = 0 y = 0 x = c d

7 Only Quadrant One Is Biologically Relevant To analyze this nonlinear model, we need a fact from more advanced courses. For these kinds of nonlinear models, trajectories that start at different initial conditions can not cross. Assumption We can show, in a more advanced course, that two distinct trajectories to the Predator - Prey model x = a x b x y y = c y + d x y x(0) = x 0 y(0) = y 0 can not cross.

8 Only Quadrant One Is Biologically Relevant Trajectories on the y + Axis Let s look at a trajectory that starts on the positive y axis. We therefore need to solve the system x = a x b x y y = c y + d x y x(0) = 0 y(0) = y 0 > 0 It is easy to guess the solution is the pair (x(t), y(t)) with x(t) = 0 always and y(t) satisfying y = c y(t). Hence, y(t) = y 0 e ct and y decays nicely down to 0 as t increases.

9 Only Quadrant One Is Biologically Relevant Trajectories on the x + Axis If we start on the positive x axis, we want to solve x = a x b x y y = c y + d x y x(0) = x 0 > 0 y(0) = 0 Again, it is easy to guess the solution is the pair (x(t), y(t)) with y(t) = 0 always and x(t) satisfying x = a x(t). Hence, x(t) = x 0 e at and the trajectory moves along the positive x axis always increasing as t increases.

10 Only Quadrant One Is Biologically Relevant Trajectories on the x + Axis Since trajectories can t cross other trajectories, this tells us a trajectory that begins in Quadrant I with a positive (x 0, y 0 ) can t hit the x axis or the y axis in a finite amount of time because it it did, we would have two trajectories crossing. This is good news for our biological model. Since we are trying to model food and predator interactions in a real biological system, we always start with initial conditions (x 0, y 0 ) that are in Quadrant One. It is very comforting to know that these solutions will always remain positive and, therefore, biologically realistic.

11 Only Quadrant One Is Biologically Relevant Trajectories on the x + Axis In fact, it doesn t seem biologically possible for the food or predators to become negative, so if our model permitted that, it would tell us our model is seriously flawed! Hence, for our modeling purposes, we need not consider initial conditions that start in Regions V - IX. Indeed, if you look at our pictures, you can see that a solution trajectory could only hit the y axis from Region II. But that can t happen as if it did, two trajectories would cross! Also, a trajectory could only hit the x axis from a start in Region III. Again, since trajectories can t cross, this is not possible either. So, a trajectory that starts in Quadrant I, stays in Quadrant I kind of has a Las Vegas feel doesn t it?.

12 The Nonlinear Conservation Law Let s analyze a trajectory that starts in QI. Assume we start in Region II and the resulting trajectory hits the y = a b line at some time t. At that time, we will have x (t ) = 0 and y(t ) < 0. We show this situation in next figure. y axis x = 0 II (, ) y = a b x = 0 x axis y = 0 III (+, ) (x 0, y 0 ) (x(t ), a b ) y = 0 x = c d I (,+) IV (+,+) We show a trajectory that starts in Region II and terminates on the y = a b line at the point shown.

13 The Nonlinear Conservation Law Look at the Predator - Prey model dynamics for 0 t < t. Since all variables are positive and their derivatives are not zero for these times, we can look at the fraction y (t)/x (t). y (t) x (t) = y(t) ( c + d x(t)) x(t) (a b y(t)). To make this easier to understand, let s do a specific example. x (t) = 2 x(t) 5 x(t) y(t) y (t) = 6 y(t) + 3 x(t) y(t) Rewrite as y /x : y (t) x (t) = = 6y(t) + 3x(t)y(t) 2x(t) 5x(t)y(t) y(t)( 6 + 3x(t)) x(t)(2 5y(t)).

14 The Nonlinear Conservation Law Put all the y stuff on the left and all the x stuff on the right: y (t) 2 5y(t) y(t) = x (t) 6 + 3x(t). x(t) Rewrite as separate pieces: 2 y (t) y(t) 5y (t) = 6 x (t) x(t) + 3x (t). Integrate both sides from 0 to t: 2 t 0 y (s) t t y(s) ds 5 y x (s) t (s) ds = x(s) ds + 3 x (s) ds. 0

15 The Nonlinear Conservation Law Do the integrations: everything is positive so we don t need absolute values in the ln s t t t t 2 ln(y(s)) 5y(s) = 6 ln(x(s)) + 3x(s). y 0 0 Evaluate: ( ) y(t) 2 ln 5(y(t) y 0 ) = 6 ln 0 ( x(t) x ) + 3(x(t) x 0 ). Put ln s on the left and other terms on the right: ( ) ( ) x(t) y(t) 6 ln + 2 ln = 3(x(t) x 0 ) + 5(y(t) y 0 ). x 0 y 0

16 The Nonlinear Conservation Law Combine ln terms: ( (x(t) ) ) ( 6 (y(t) ) ) 2 ln + ln = 3(x(t) x 0 ) + 5(y(t) y 0 ). x 0 y 0 Combine ln terms again: ( (x(t) ) 6 ( ) ) y(t) 2 ln = 3(x(t) x 0 ) + 5(y(t) y 0 ). x 0 y 0 Exponentiate both sides: ( ) x(t) 6 ( ) y(t) 2 = e 3(x(t) x 0)+5(y(t) y 0 ). x 0 y 0 Simplify the exponential term: ( ) x(t) 6 ( ) y(t) 2 = e3x(t) x 0 y 0 e 3x 0 e 5y(t) e 5y 0

17 The Nonlinear Conservation Law Put all function terms on the left and all constant terms on the right: (x(t)) 6 (y(t)) 2 e 3x(t) e 5y(t) = (x 0) 6 e 3x 0 Define the functions f and g by f (x) = x 6 /e 3x. g(y) = y 2 /e 5y. Then we can rewrite our result as (y 0 ) 2 e 5y 0 f (x(t)) g(y(t)) = f (x 0 ) g(y 0 ). We did this analysis for Region II, but it works in all the regions. So for the entire trajectory, we know f (x(t)) g(y(t)) = f (x 0 ) g(y 0 ).

18 The Nonlinear Conservation Law The equation f (x(t)) g(y(t)) = f (x 0 ) g(y 0 ). for f (x) = x 6 /e 3x and g(y) = y 2 /e 5y, is called the Nonlinear Conservation Law or NLCL for the Predator - Prey model x (t) = 2 x(t) 5 x(t) y(t) y (t) = 6 y(t) + 3 x(t) y(t) In the text, we also do this derivation for the general Predator - Prey model so it is a bit harder to follow as it uses a, b, c and d instead of numbers. Also, we discuss what happens as t approaches the t value where the trajectory crosses the nullcline. So you should read that to get the full picture.

19 The Nonlinear Conservation Law The equation f (x(t)) g(y(t)) = f (x 0 ) g(y 0 ). for f (x) = x c /e dx and g(y) = y a /e by, is called the Nonlinear Conservation Law or NLCL for the general Predator - Prey model x (t) = a x(t) b x(t) y(t) y (t) = c y(t) + d x(t) y(t)

20 The Nonlinear Conservation Law Homework 67 For the following Predator - Prey models, derive the nonlinear conservation law as discussed in the book and this lecture x (t) = 100 x(t) 25 x(t) y(t) y (t) = 200 y(t) + 40 x(t) y(t) x (t) = 1000 x(t) 250 x(t) y(t) y (t) = 2000 y(t) + 40 x(t) y(t) x (t) = 900 x(t) 45 x(t) y(t) y (t) = 100 y(t) + 50 x(t) y(t)