Basic Assumption: population dynamics of a group controlled by two functions of time

Transcription

1 opulaion Models Basic Assumpion: populaion dynamics of a group conrolled by wo funcions of ime Birh Rae β(, ) = average number of birhs per group member, per uni ime Deah Rae δ(, ) = average number of deahs per group member, per uni ime If () = populaion a ime, hen, beween and + : Toal birhs β Toal deahs δ Change in populaion is he difference: (β δ) = = (β δ) = (β δ) (ake limi ) Differen models depend on choices/observaions/predicions of birh and deah raes Naural Growh Suppose β and δ consan Naural growh (β δ > ) or decay (β δ < ) equaion: = (β δ) = () = e (β δ) where is he populaion a ime = β > δ β = δ β < δ

2 Growh rae proporional o populaion Suppose δ = and β = k wih k > consan: no deah, while birh rae increases wih populaion. Then = (β δ) = k2 = 2 d = k d = + = k = () = k () as (in finie ime!) k Ridiculous model for long erm: ypically large populaion reduces birh rae in real-world siuaions. k The Logisic equaion Assume growh rae β δ a linear funcion of : β δ = k(m ()) where M, k are consan. Definiion. The Logisic Differenial Equaion is = k(m ). The logisic equaion is separable, hence k d = (M ) d = M + M d = Mk = ln (M ) (M ) (parial fracions). = () = M + (M )e Mk 2

3 Examples.. A populaion of = 25 fish has a consan deah rae δ =.5 per fish, per year and a birh rae β() =.5. per fish, per year small = many offspring large = reduced birh rae Then = (β δ) = (.) =.( ), so ha k =. and M = in he logisic model. The soluion is M 75 () = + (M )e Mk 25 5 = e = e (fish) 5 5 (years) 2. Suppose ha = 5 wih he same birh and deah raes as before (so M = and k =.) Then () = M + (M )e = 5 3 = Mk 5 5e 3 e (fish) 5 5 ex 5 5 (years) In boh cases sabilizes a M = for very large 3

4 Oher applicaions of he Logisic Equaion Limied environmen β δ proporional o M = room for expansion Compeiion for resources β consan and δ proporional o Disease/Rumor spread rae of spread proporional o produc of hose ha have/know and hose ha don (i.e. and M ) Example: Doomsday versus Exincion Suppose number of birhs per uni ime proporional o # females # males (= rae of random encouners) bu deah rae consan. Then β = k 2 / = k (rae relaive o populaion), from which = k2 δ = k(m ) a reverse logisic equaion (coefficien of 2 posiive) The soluion is () = M + (M )e Mk which gives hree possibiliies... < M opulaion decreases owards zero as M < > opulaion increases o as km ln M = M opulaion consan M = M > Example (From Chemisry). KNO 3 dissolves in mehanol; he mass x grams dissolved a ime seconds saisfies he differenial equaion dx =.8x.4x2 (a) Wha is he maximum amoun of sal ha will ever dissolve? (b) If x() = 5 g, how long will i ake for an addiional 5 g o dissolve? Since.8x.4x 2 = 25 x(2 x), his is he logisic equaion wih k = M = 2 g. Thus he maximum dissolved solue is 2 g Assuming we keep adding more han 2 g, he exra will immediaely sar o precipiae ou of soluion 25 and sable soluion 4

5 The soluion o he logisic equaion wih x = 5, k = 25, and M = 2 g is Hence x() = x M x + (M x )e = km 5 + 5e = 2 4/5 + 3e 4/5 x() = g + 3e 4/5 = 2 e 4/5 = 3 = 5 ln s 4 Sabiliy of Equilibrium Soluions Limiing opulaion Original Equaion k > 2 Doomsday/Exincion Reverse Equaion k < M M () M sable () unsable () M unsable () sable If iniial condiion = () is close o M, hen in our examples:. () moves back owards M 2. () moves away from M Logisic Equaion wih Harvesing Suppose he populaion () of fish in a lake obeys he logisiic equaion wih k > and sable populaion M Suppose also ha h addiional fish are removed per uni ime. Obain new differenial equaion: = k(m ) h Criical poins: use quadraic formula o find he roos of he RHS as a funcion of : = M M 2 4h k There are hree possibiliies 2 = M 2 2 M 2 4h k 5

6 Two real roos In his case he harvesing rae is low: h < 4 km2 = > 2 > The differenial equaion is = k( )( 2 ) wih phase diagram + Soluion Direcion 2 Sabiliy Unsable Sable 2 The sable populaion of fish is now < M, so in he long run he populaion of fish should sabilize bu a a lower level han wih no fishing. 2 Repeaed real roos The harvesing rae is criical: h = 4 km2 = = 2 = M 2 > The differenial equaion is = k( M/2)2 wih phase diagram Soluion Direcion Sabiliy = 2 = M 2 Semi-sable 6

7 = 2 Soluions eiher decrease o M/2, are consan = M/2, or go exinc. If iniial populaion of fish was he original sable populaion = M, hen we expec he populaion o decrease owards = M/2. This is a dangerous siuaion (for he fish!) if anyhing unexpeced happens o ip he populaion under M/2, hen he model predics exincion. 3 Complex roos The harvesing rae is high: h > 4 km2 =, 2 / R There are no real equilibrium soluions and < for all For any iniial populaion, exincion in finie ime is ineviable 7