Predator - Prey Model Trajectories and the nonlinear conservation law

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1 Predaor - Prey Model Trajecories and he nonlinear conservaion law James K. Peerson Deparmen of Biological Sciences and Deparmen of Mahemaical Sciences Clemson Universiy Ocober 28, 213 Ouline Drawing Trajecories Again Only Quadran One Is Biologically Relevan Trajecories on he y + Axis Trajecories on he x + Axis The Nonlinear Conservaion Law

2 Absrac This lecure alks abou he rajecories of he predaor prey model and he nonlinear conservaion law his model saisfies. We drew rajecories for he linear sysem models already wihou a lo of background discussion. Now we ll go over i again in more deail. How do we draw rajecories? We use he algebraic signs of x and y o deermine his. For example, in Region I, he sign of x is negaive and he sign of y is posiive. Thus, x decreases and y increases in his region. If we graphed (x(), y()) in he x y plane for all >, we would plo a y versus x curve. We would have y = f (x) for some funcion of x. Noe by he chain rule dy d = f (x) dx d. Hence, as long as x is no zero (and his is rue in Region I!), we have he slope of he curve y = f (x) is given by df dx () = y () x ().

3 Since our pair (x, y) is he soluion o a differenial equaion, we expec ha x and y boh are coninuously differeniable wih respec o. So if we draw he curve for y vs x in he x y plane, we do no expec o see a corner in i (as a corner means he derivaive fails o exis). So we can see hree possibiliies: a sraigh line as x equals y a each meaning he slope is always he same, a curve ha is concave up or a curve ha is concave down. We illusrae his hree possibiliies in he nex figure. When we combine rajecories from one region wih anoher, we mus aach hem so ha we do no ge corners in he curves. This is how we can deermine wheher or no we should use concave up or down or sraigh in a given region. y axis x = y = a b x = II (, ) I (,+) III (+, ) IV (+,+) In his figure, we show he hree rajecory ypes for he signs of region I x axis y = y = x = c d

4 To analyze his nonlinear model, we need a fac from more advanced courses. For hese kinds of nonlinear models, rajecories ha sar a differen iniial condiions can no cross. Assumpion We can show, in a more advanced course, ha wo disinc rajecories o he Predaor - Prey model x = a x b x y y = c y + d x y x() = x y() = y can no cross. Le s look a a rajecory ha sars on he posiive y axis. We herefore need o solve he sysem x = a x b x y y = c y + d x y x() = y() = y > I is easy o guess he soluion is he pair (x(), y()) wih x() = always and y() saisfying y = c y(). Hence, y() = y e c and y decays nicely down o as increases.

5 If we sar on he posiive x axis, we wan o solve x = a x b x y y = c y + d x y x() = x > y() = Again, i is easy o guess he soluion is he pair (x(), y()) wih y() = always and x() saisfying x = a x(). Hence, x() = x e a and he rajecory moves along he posiive x axis always increasing as increases. Since rajecories can cross oher rajecories, his ells us a rajecory ha begins in Quadran I wih a posiive (x, y ) can hi he x axis or he y axis in a finie amoun of ime because i i did, we would have wo rajecories crossing. This is good news for our biological model. Since we are rying o model food and predaor ineracions in a real biological sysem, we always sar wih iniial condiions (x, y ) ha are in Quadran One. I is very comforing o know ha hese soluions will always remain posiive and, herefore, biologically realisic.

6 In fac, i doesn seem biologically possible for he food or predaors o become negaive, so if our model permied ha, i would ell us our model is seriously flawed! Hence, for our modeling purposes, we need no consider iniial condiions ha sar in Regions V - IX. Indeed, if you look a our picures, you can see ha a soluion rajecory could only hi he y axis from Region II. Bu ha can happen as if i did, wo rajecories would cross! Also, a rajecory could only hi he x axis from a sar in Region III. Again, since rajecories can cross, his is no possible eiher. So, a rajecory ha sars in Quadran I, says in Quadran I kind of has a Las Vegas feel doesn i?. Le s analyze a rajecory ha sars in QI. Assume we sar in Region II and he resuling rajecory his he y = a b line a some ime. A ha ime, we will have x ( ) = and y( ) <. We show his siuaion in nex figure. y axis x = II (, ) y = a b x = x axis y = III (+, ) (x, y ) (x( ), a b ) y = x = c d I (,+) IV (+,+) We show a rajecory ha sars in Region II and erminaes on he y = a b line a he poin shown.

7 Look a he Predaor - Prey model dynamics for <. Since all variables are posiive and heir derivaives are no zero for hese imes, we can look a he fracion y ()/x (). y () x () = y() ( c + d x()) x() (a b y()). To make his easier o undersand, le s do a specific example. x () = 2 x() 5 x() y() y () = 6 y() + 3 x() y() Rewrie as y /x : y () x () = = 6y() + 3x()y() 2x() 5x()y() y()( 6 + 3x()) x()(2 5y()). Pu all he y suff on he lef and all he x suff on he righ: y () 2 5y() y() Rewrie as separae pieces: = x () 6 + 3x(). x() 2 y () y() 5y () = 6 x () x() + 3x (). Inegrae boh sides from o : y (s) 2 y(s) ds 5 y x (s) (s) ds = 6 x(s) ds + 3 x (s) ds.

8 Do he inegraions: everyhing is posiive so we don need absolue values in he ln s 2 ln(y(s)) 5y(s) = 6 ln(x(s)) + 3x(s). Evaluae: ( ) y() 2 ln 5(y() y ) = 6 ln y ( x() x ) + 3(x() x ). Pu ln s on he lef and oher erms on he righ: ( ) ( ) x() y() 6 ln + 2 ln = 3(x() x ) + 5(y() y ). x y Combine ln erms: ( (x() ) ) 6 ( (y() ) ) 2 ln x + ln y = 3(x() x ) + 5(y() y ). Combine ln erms again: ( (x() ) 6 ( ) ) y() 2 ln x y = 3(x() x ) + 5(y() y ). Exponeniae boh sides: ( ) x() 6 ( ) y() 2 = e 3(x() x)+5(y() y). x y Simplify he exponenial erm: ( ) x() 6 ( ) y() 2 = e3x() x y e 3x e 5y() e 5y

9 Pu all funcion erms on he lef and all consan erms on he righ: (x()) 6 (y()) 2 e 3x() e 5y() = (x)6 (y ) 2 e 3x e 5y Define he funcions f and g by f (x) = x 6 /e 3x. g(y) = y 2 /e 5y. Then we can rewrie our resul as f (x()) g(y()) = f (x ) g(y ). We did his analysis for Region II, bu i works in all he regions. So for he enire rajecory, we know f (x()) g(y()) = f (x ) g(y ). The equaion f (x()) g(y()) = f (x ) g(y ). for f (x) = x 6 /e 3x and g(y) = y 2 /e 5y, is called he Nonlinear Conservaion Law or NLCL for he Predaor - Prey model x () = 2 x() 5 x() y() y () = 6 y() + 3 x() y() In he ex, we also do his derivaion for he general Predaor - Prey model so i is a bi harder o follow as i uses a, b, c and d insead of numbers. Also, we discuss wha happens as approaches he value where he rajecory crosses he nullcline. So you should read ha o ge he full picure.

10 The equaion f (x()) g(y()) = f (x ) g(y ). for f (x) = x c /e dx and g(y) = y a /e by, is called he Nonlinear Conservaion Law or NLCL for he general Predaor - Prey model x () = a x() b x() y() y () = c y() + d x() y() Homework 67 For he following Predaor - Prey models, derive he nonlinear conservaion law as discussed in he book and his lecure x () = 1 x() 25 x() y() y () = 2 y() + 4 x() y() x () = 1 x() 25 x() y() y () = 2 y() + 4 x() y() x () = 9 x() 45 x() y() y () = 1 y() + 5 x() y()

Predator - Prey Model Trajectories and the nonlinear conservation law

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