Scientific Computing I Part I Module 3: Population Modelling – Continuous Models ODE Models Michael Bader Lehrstuhl Informatik V Winter 2007/2008 Discrete vs. Contiuous Models Model of Malthus (1798) Only one species: birth rate γ (number of births per time interval) 1 dp dt = F ( p , t ,... ) proportional to size of population death rate δ proportional to size of population 2 p ( t ) =? thus: constant growth (or decay) rate: r = γ − δ 3 Modelling: constant growth rate discrete model: continuous model: p ( t ) ∈ N individuals p : R → R , p ( t ) =? dp dt = r · p Move to Continuous Models: growth within a time interval easier(?) type of mathematical problem: differential equations, calculus p ( t + δ t ) = p ( t )+ r · p ( t ) · δ t analytical solutions available(?) Model of Malthus – Differential Equation Model of Malthus – Solutions The model of Malthus describes exponential growth or Written as an ordinary differential equation: decay of a population: ˙ p ( t ) = r · p ( t ) 3 Requires initial condition (population at start) 2,5 p ( 0 ) = p 0 2 p 1,5 Analytical solution: 1 p ( t ) = p 0 e rt 0,5 0 0 2 4 6 8
Model of Verhulst (19th century) Model of Verhulst – Saturation solve initial value problem: Objective: p ( t ) = α − β p ( t ) , ˙ p ( 0 ) = p 0 model populations that approach saturation value solution: Assumptions: p ∞ = α p ( t ) = p ∞ + e − β t ( p 0 − p ∞ ) , growth/death rate depend on population size; β assume linear dependency: 3 g ( t ) = g 0 − g 1 · p ( t ) d ( t ) = d 0 + d 1 · p ( t ) 2,5 2 p 1,5 leads to differential equation: 1 p ( t ) = g ( t ) − d ( t ) = ( g 0 − d 0 ) ˙ − ( g 1 + d 1 ) p ( t ) 0,5 � �� � � �� � =: α 0 =: β 0 2 4 6 8 Model of Verhulst – Logistic Growth Logistic Growth other formulation � � 1 − p ( t ) p ( t ) = α ˙ p ( t ) β saturation model does no longer model exponential growth solution: β idea: let growth/death rate decrease linearly with p ( t ) = ( 1 − e − α t )+ β p 0 e − α t size of population but keep growth/death rate proportional to 3 population size 2,5 leads to differential equation: 2 ˙ p ( t ) = ( α − β p ( t )) p ( t ) p 1,5 1 0,5 0 0 2 4 6 8 Logistic Growth with Threshold Example – The Passenger Pigeon extended version of Verhulst’s model: beginning of the 19th century, estimated � �� � 1 − p ( t ) 1 − p ( t ) population in North America: four billion p ( t ) = α ˙ p ( t ) β δ hunting diminished its number below a critical threshold (late 1880s) solutions ( β = 2 , δ = 4): The last passenger pigeon died on September, 1st 1914. 5 4 3 p 2 1 0 0 2 4 6 8
Scientific Computing I Part II Module 3: Population Modelling – Continuous Models (Parts II and III) More Than One Species – Systems of ODE Michael Bader Lehrstuhl Informatik V Winter 2007/2008 A Linear Model First Example: Arms Race similar to Verhulst’s saturation model additional growth term proportional to other armament of two (hostile) countries species our suspicion: a 12 > 0, a 21 > 0 leads to system of differential equations: Observation: ˙ p ( t ) = b 1 + a 11 p ( t )+ a 12 q ( t ) long-time behaviour depends on size of parameters q ( t ) ˙ = b 2 + a 21 p ( t )+ a 22 q ( t ) steady-state solutions exist typically: solutions exist that show unlimited growth b 1 > 0 , b 2 > 0 (growth term) a 11 < 0 , a 22 < 0 (saturation) a 12 , a 21 ? Second Example: Competition A Non-Linear Model similar to Verhulst’s logistic growth model additional growth term proportional to other two species sharing a common natural habitat species competition: a 12 < 0, a 21 < 0 leads to system of differential equations: Observation: ˙ p ( t ) = ( b 1 + a 11 p ( t )+ a 12 q ( t )) p ( t ) long-time behaviour depends on size of parameters ˙ q ( t ) = ( b 2 + a 21 p ( t )+ a 22 q ( t )) q ( t ) steady-state solutions exist some scenarios are physically incorrect! typically: (negative population size) b 1 > 0 , b 2 > 0 (growth term) a 11 < 0 , a 22 < 0 (saturation) a 12 , a 21 ?
The Non-Linear Competition Model Competition – Steady State system of differential equations: � √ √ � 5 24 − 5 3 3 ˙ p ( t ) = 2 + 8 p ( t ) − 24 q ( t ) p ( t ) � √ √ � two species sharing a common natural habitat 8 + 3 7 2 − 3 3 8 p ( t ) − 7 3 ˙ q ( t ) = 8 q ( t ) q ( t ) competition: a 12 < 0, a 21 < 0 solution for p 0 = 1 4 , q 0 = 3: Possible Scenarios: 4 steady-state one species dies out (extinction) 3 no obvious nonsense p 2 1 0 2 4 6 8 10 t Competition – Extinction Predator-Prey system of differential equations: � � 71 8 − 23 12 p ( t ) − 25 ˙ p ( t ) = 12 q ( t ) p ( t ) two species: predator p and prey q � � 73 8 − 25 12 p ( t ) − 23 ˙ q ( t ) = 12 q ( t ) q ( t ) predator eats prey: a 12 > 0 solution for p 0 = 1 4 , q 0 = 1 4 : prey is eaten by predator: a 21 < 0 Possible Scenarios: stable oscillations 4 p one species dies out (what happens with the other, 3 then?) 2 1 0 0 2 4 6 8 10 t Predator-Prey by Lotka & Volterra Open Questions system of differential equations: � � − 1 1 ˙ Methods to Analyse a Given Model? p ( t ) = 2 + 200 q ( t ) p ( t ) � 1 � 5 − 1 q ( t ) ˙ = 50 p ( t ) q ( t ) predict approximate solution or shape of solution? predict possible steady states? solution for p 0 = 6, q 0 = 50: predict critical points? (species on edge of extiction?) 160 Methods to Improve Modeling? 120 predict failure of the model? p 80 tune parameters to model a specific situation? 40 0 20 40 60 80 100 t
Analysing the Slope of a Solution Example: Model of Malthus Part III ˙ p ( t ) = α p ( t ) Discussion and Analysis of ODE Models for a sensible solution: p ( t ) > 0 α decides slope of solution: α > 0: growing population (accelerated growth) α < 0: receding population (decelerated reduction) Points of Equilibrium Critical Points Example: Model of Verhulst (saturation) Observation on Logistic Growth: p ( t ) = α − β p ( t ) ˙ constant solution p ( t ) = β , if p ( 0 ) = β constant solution p ( t ) = 0, if p ( 0 ) = 0 equilibrium: ˙ p ( t ) = 0 equilibrium at p = β is reached for nearly all initial only, if p ( t ) = α β conditions ⇒ attractive (stable) equilibrium Example: Logistic Growth equilibrium at p = 0 is not reached for any other � � 1 − p ( t ) initial conditions (“repulsive”) p ( t ) = α ˙ p ( t ) β ⇒ unstable equilibrium constant solution, if p ( t ) = β or p ( t ) = 0 Critical Points – Derivatives Direction Field plot derivatives vs. time and size of population: Examine derivatives: Example: Logistic Growth critical point p = ¯ � � p 1 − p ( t ) p ( t ) = α ˙ p ( t ) attractive equilibrium (asymptotically stable): β p < 0 ˙ for p = ¯ p + ε 3 p > 0 ˙ for p = ¯ p − ε 2,5 p(t) unstable equilibrium: 2 1,5 p > 0 ˙ for p = ¯ p + ε 1 p < 0 ˙ for p = ¯ p − ε 0,5 otherwise: saddle point 0 0 2 4 6 8 10 t
Direction Field (2) Identifying Critical Points attractive equilibrium: Example: Logistic Growth with Threshold � �� � 1 − p ( t ) 1 − p ( t ) ˙ p ( t ) = α p ( t ) β δ 5 unstable equilibrium 4 p(t) 3 2 saddle point 1 0 0 2 4 6 8 10 t Critical Points in 2D Direction Field for a System of ODE Example: Arms Race system of differential equations example: 2D system of differential equations: equilibrium: ˙ p = 0, ˙ q = 0 ˙ p ( t ) = b 1 + a 11 p ( t )+ a 12 q ( t ) p ( t ) ˙ = b 1 + a 11 p ( t )+ a 12 q ( t ) = 0 ˙ q ( t ) = b 2 + a 21 p ( t )+ a 22 q ( t ) ˙ q ( t ) = b 2 + a 21 p ( t )+ a 22 q ( t ) = 0 natural exension: 3D plot: t vs. p vs. q solution of a linear system of equations: 1D direction field for p vs. t or q vs. t not sufficient: what values to chose for q (or p resp.)? a 11 p ( t )+ a 12 q ( t ) = − b 1 but: stationary problem ⇒ independent of t a 21 p ( t )+ a 22 q ( t ) = − b 2 thus: plot directions depending on p and q in most cases one critical point critical line, if system matrix is singular 2D Direction Field – Arms Race Arms Race – unlimited growth system of differential equations: system of differential equations: 2 − p ( t )+ 1 3 1 2 − 3 ˙ ˙ p ( t ) = 2 q ( t ) p ( t ) = 4 p ( t )+ q ( t ) 0 + 1 − 5 4 + p ( t ) − 3 q ( t ) ˙ = 2 p ( t ) − q ( t ) q ( t ) ˙ = 4 q ( t ) direction field – with critical point at ( 2 , 1 ) : direction field – with critical point at ( 2 , 1 ) : 2 5 4 q q 1,5 3 1 2 0,5 1 0 0 0 1 2 3 4 0 1 2 3 4 5 p p
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