C2 Differential Equations : Computational Modeling and Simulation Instructor: Linwei Wang

Transcription

1 C2 Differential Equations : Computational Modeling and Simulation Instructor: Linwei Wang

2 Part IV Dynamic Systems

3 Equilibrium: Stable or Unstable? Equilibrium is a state of a system which does not change A differential equation / a system of differental equations: setting a derivative (all derivatives) to zero. Example 1: Logistic model Equilibrium N * : r 0 N * (1" N * Example 2: System of differential equation dn dt = r 0 N(1! N K ) K ) = 0 # N * = 0,N * = K " dx $ = f (x, y) dt # $ dy = g(x, y) % dt " # $ g(x *, y * ) = 0 Equilibrium x *,y * : f (x *, y * ) = 0

4 Equilibrium: Stable or Unstable? Equilibrium is a state of a system which does not change A differential equation / a system of differential equations: setting a derivative (all derivatives) to zero. A difference equation (discrete time) Example: Ricker s model (discrete-time analog of logistic model) Equilibrium N * : N t +1 = f (N t ) = N t exp[r(1" N t K )] N * = N * exp[r(1" N * K )] " N * = 0,N * = K

5 Equilibrium: Stable or Unstable? An equilibrium is considered stable (for simplicity we will consider asymptotic stability only) if the system always returns to it after small disturbances. If the system moves away from the equilibrium after small disturbances, then the equilibrium is unstable. One differential equation: stable equilibria are characterized by a negative first derivative, df/dn (negative feedback), whereas unstable equilibria are characterized by a positive one (positive feedback). dn dt = r 0 N(1" N K )

6 Equilibrium: Stable or Unstable? An equilibrium is considered stable (for simplicity we will consider asymptotic stability only) if the system always returns to it after small disturbances. System of differential equations: stable equilibria are characterized by negative real parts of ALL eigenvalues of the Jacobian matrix. If there is one eigenvalue with a positive real part, the equilibrium point is unstable #"f % "x A = % % "g $ %"x "f & "y ( ( "g ( "y'(

7 Equilibrium: Stable or Unstable? An equilibrium is considered stable (for simplicity we will consider asymptotic stability only) if the system always returns to it after small disturbances One difference equation: The equilibrium point is stable if and only if Example: Ricker s model "1 < df dn N= N * <1 N t +1 = f (N t ) N t +1 = f (N t ) = N t exp[r(1" N t K )] N t +1 = f (N t ) = N t exp[r(1" N t K )] df = (1" rn t dn t K )exp[r(1" N t K )] * N t = N ### $ 1" r Ricker s model has a stable equilibrium point N * if 0<r<2

8 Equilibrium: Stable or Unstable? An equilibrium is considered stable (for simplicity we will consider asymptotic stability only) if the system always returns to it after small disturbances. One difference equation

9 Equilibrium: Stable or Unstable? An equilibrium is considered stable (for simplicity we will consider asymptotic stability only) if the system always returns to it after small disturbances. System of difference equations: ALL eigenvalues of the Jacobian matrix have to lie within the circle with radius = 1 in the complex plain Find equilibrium points Approximate Jacobian matrix Calculate eigenvalues of the Jacobian matrix

10 Equilibrium: Stable or Asymptotically Stable An equilibrium is considered stable if the solution starting close enough to the equilibrium remain close forever If for every ε >0, there exists a! =!(") > 0 such that, if x(0)! x e <!, Then x(t)! x e <! for every t>=0 An equilibrium is considered to be asymptotically stable if the solution starting close enough to the equilibrium not only remain close but also converge to the equilibrium. If it is stable and if there exists! > 0 such that if x(0)! x e <!, then lim x(t)# x e = 0 t!"

11 Growth of - Population - Pollution - Cancer cell Growth and Decay Processes - Poverty - Investment - - Nonlinear: limiting nature of growth - Coupled / intertwined

12 Discrete Growth Processes Growth of money in a saving account Annual rate of interest r% The interest is paid & compound n times a year Deposite can be made at any time, but start to earn interest from the beginnng of the next time period y(k): total funds at the end of the kth period u(k): total deposits made during the kth period y(k) = y(k "1) + (r /n)y(k "1) + u(k) = (1+ r /n)y(k "1) + u(k)

13 Discrete Growth Processes Mortage Total debt: d dollars Monthly mortage interest: r Monthly payment: p What should be the value of p if the debt is to be paid off by N payments? y(k): balance at the end of the kth period u(k): amout of the kth period = p y(k) = y(k "1) + ry(k "1) " u(k) = (1+ r)y(k "1) " u(k) = (1+ r)y(k "1) " p, y(0) = d

14 Discrete Growth Process Growth of rabbit population A pair of rabbits is born to each pair of adult rabbits at the end of every month A newborn pair produces first offspring at two months of age A pair of rabbits keep producing forever according to the above rules y(k): number of rabbit pairs at the end of kth month y(k) = y(k "1) + y(k " 2)

15 Continuous Growth Population growth Assumption: when a population is introduced into an environment, where it is not already found, growth occurs at a rate that is directly proportional to the size of the population dn dt = kn, N(0) = N 0 " N(t) = N 0 e kt A large number of growth / decay phenomena follows this equation (physical, chemical, biological, medical, social sicence)

16 Example 1: Drug Dose Dose level of a drug Concentration level in the blood y(t) The rate of change of concentration in the blood is directly proportationally to the amount present Ingestion of medicine is added to the body with a constant amount y 0 at intervals of time T dy dt = "ky # y = y 0e "kt y(t) = y 0 e "kt + y 0 y(2t) = (y 0 e "kt + y 0 )e "kt + y 0! y(nt) = (y 0 + y 0 e "kt +"+ y 0 e "(n"1)kt )e "kt + y 0 = y 0(1" e "(n +1)kT ) 1" e "kt

17 Example 1: Drug Dose To achieve constant drug concentration y(nt) = y 0 (1" e"(n +1)kT ) 1" e "kt t " # What is the initial dose plan? (y init for first dose, y 0 afterwards) y(nt) = (y 0 + y 0 e "kt +!+ y init e "(n"1)kt )e "kt + y 0 = y 0 (1" e"nkt ) 1" e "kt + y init e "nkt = y 0 1" e "kt y init = y 0 1" e "kt

18 Example 2: Body Burden Body burden: study the effect of environmental pollutants on human health; defined as the level of pollutants in the organs, blood or other body fluids of an organism x(t): amount excreted through urine and faeces in (0,t) y(t): amount consumed via food and beverages in (0,t) z(t): amount per unit volume of blood at t u(t): amount inspired via lungs in (0,t) v(t): amount stored in body tissues in (0,t)

19 Example 2: Body Burden x(t): amount excreted through urine and faeces in (0,t) y(t): amount consumed via food and beverages in (0,t) z(t): amount per unit volume of blood at t u(t): amount inspired via lungs in (0,t) v(t): amount stored in body tissues in (0,t) dz(t) dt = "( dy(t) dt + du(t) dt # dx(t) dt # dv(t) ) dt Generally, the rate of change in the tissue is proportional to the amount in the blood dz dt + bz = "( d(y + u) dt # dx dt )

20 Example 3: Mathematical Economics Building steel mills in a developing country S a (t): Quantity of steel produced / available at t S c (t): Capacity of mills at t Assume all other materials & labors are abundant, the rate of steel production is propotational to the mill capacity Because building mills needs steel, the rate of mill building (mill capacity) is proportional to available steels " ds c $ dt # $ ds a % $ dt = k c S a, S c (0) = c 1 = k a S c! bs a, S a (0) = c 2

21 Limits To Growth Population growth: competition for food and resources lower birth rate, higher death rate dn dt = kn = (b " m)n m = m 1 N dn dt = b(1" N M )N, M = b /m 1 N(t) = (MN 0 /(M " N 0 ))e bt 1+ (N 0 /(M " N 0 ))e bt Small t: N(t) = MN 0 (M " N 0 ) ebt Large t: N(t) " M

22 Limits To Growth Population growth: In practice, environmental factors are much more complicated much harder to collect data for validation dn dt = k(t)n + Q(t) Enviromental factors Abnormal growth pa7ern, such as sudden natural calamity, hun<ng habits of predator, seasonal factor, etc

23 Competition Among Species Biological species Political parties Business enterprises Coupled reacting chemical components

24 Competition Among Species Ecological niche Two species S 1 and S 2 with ecological niches 1 and 2, respectively dn 1 dt = (b 1 " m 1 N 1 )N 1, N 1 (0) = N 10 dn 2 dt = (b 2 " m 2 N 2 )N 2, N 2 (0) = N 20 In the initial growth stage: Coexist only happens if b 1 =b 2! N 1 /N 2 = (N 10 /N 20 )exp(b 1 " b 2 )t

25 Competition Among Species Possible explanations Geographical isolation Ecological differentiation: different specialization Interbreeding & merging Apply to business and other problems as well Revision to the model Add interaction between species dn 1 dt dn 2 dt = m 1 (M 1 " N 1 "# 12 N 2 )N 1, N 1 (0) = N 10 = m 2 (M 2 " N 2 "# 21 N 1 )N 2, N 2 (0) = N 20

26 Predator-Prey Systems Consumer-resource Prey-plant Parasite-host Tumor cells (virus) immune systems Susceptible infectious

27 Predator-Prey Systems Lotka-Vloterra model Prey: H Predator: P Assumption: In the absence of predator, prey population grows exponentially In the absence of prey, predator population declines exponentially Interaction: prey decreases and predator increases at a rate proportional to the frequency of predator-prey encounters H = bh " shp H(0) = H 0 P = "dp + eshp P(0) = P 0

28 Predator-Prey Systems Equibilirium points # H(b " sp) = 0 $ % P("d + esh) = 0 Stability # J = b " sp "sh & % ( $ esp "d + esh' (H *,P * ) = (0,0),(H *,P * ) = ( d es,b s ) # J(0,0) = b 0 & % ( " 1 = b," 2 = #d $ 0 "d' # J( d es, b 0 " d & s ) = % e ( % e ( % 0 ( $ b ' " = ±i ed be

29 Predator-Prey Systems (H *,P * ) = ( d es,b s ) Stable, but not asymptotically stable Since a model is not a precise descrip<on of a system, qualita<ve predic<on should not be altered by slight modifica<ons!

30 Predator-Prey Systems Change of assumption: In the absence of predator, prey population grows by the logistic model # % H = bh(1" H $ K ) " shp H(0) = H 0 & % P = "dp + eshp P(0) = P 0

31 Predator-Prey Systems Equibilirium points # % H(b(1" H $ K ) " sp) = 0 & % P("d + esh) = 0 Stability # J = b " 2bH & % " sp "sh K ( $ % esp "d + esh' ( (H *,P * ) = (0,0) (H *,P * ) = (K,0) (H *,P * ) = ( d es, b s (1" d esk )) # J(0,0) = b 0 & % ( $ 0 "d' # J(K,0) = "b "sk & % ( " 1 = #b," 2 = esk # d $ 0 "d + esk' J( d es,b s (1" d esk )) =?

32 Predator-Prey Systems (H *,P * ) = ( d es,b s (1" d esk )) Structurally stable! b d < 4 Kes d (Kes d "1)

33 Predator-Prey Systems More realistic assumptions: Intra-species competition Relation between the predator s consumption rate and prey density Efficiency of converting prey to predator # H = r H H(1" H K )" a HP H % b + H $ P = a HP P &% b + H " cp # H % = r H H(1" H K ) " shp $ % P = r p P(1" P & ch ) Consump<on rate is increasing func<on of prey density but approaching a limit at high density Leslie s model

34 Predator-Prey Systems More realistic assumptions: Fluctuating environment Time-delay for reproduction N = rn(t)(1" N(t " T) ) K(t) Any more?

35 Epidemic Model Kermack-McKendrick epidemic model Population is divdied into three disjoint groups Suseptible (S) individuals are capable of contracting the diseases and become infective Infective (I) individuals are capable of trasmitting the diseases to others Removed (R) individuals have had the diseases and are dead, recovered & permanently immune, or are isolated until recovery & permanent immunity occur Rules The rate of change in S is proportional to the contacts between S & I I are removed at a rate proportional to their number Total number of population is constant

36 Epidemic Model # S = "isi % $ I = isi " ri % & R = ri Implications Initial number of S vs. occurance of epidemic: critical value of susceptible population In the case of a deadly diseases, an epidemic has less chance to occur!

37 Exercise Harvesting renewable resources, e.g. fish Use logistic population with an addiitonal mortality term (harvesting yield, proportional (catch rate) to the population). Find equilibrium population, determine the catch rate for stable equiblibrium, and determine the corresponding maximum yield N = rn(1" N K ) " cn Equilibrium point & stable condition Maximum yield N * = K(1" c r ),c < r c = r /2,(cN) max = rk /4

38 Exercise Now assume constant yield H 0, analyze the equilibrium points and what happens when H 0 approaches rk/4 N = rn(1" N K ) " H 0 N * 1 = 1 2 K(1" 1" 4 H 0 rk ) N * 2 = 1 2 K(1+ 1" 4 H 0 rk )