Intro to Mathematical Biology
Intro to Mathematical Biology MATH 345
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Bellow s model Amy xn1 1 or more general xn1 f13n Observation 1 If A lt 1 then the species is doomed to extinc tion why Observation 2 If b lt 1 then f13 is an increasing function then the sequence mm is increasing or decreasing Equilibrium fixed point constant solution If 1 satisfies 1 f13 then 1 is an equilibrium of the difference equation xn1 f13n An equilibrium is where the graphs of y 1 and y f13 intersect Monotone Dynamics If the function f13 is increasing then for the sequence generated by xn1 f13n every solution either decreases to an equilib rium possibly oo or increases toward an equilibrium possibly Example a xn1 9xn with O lt b g 1 Mm 1 513 Use calculus to show g x gt O for all 13 b xn1 2am sin xn see the cobweb diagram Stability of an equilibrium suppose that 513 is an equilibrium of xn1 f13n If we start the iteration sequence from any initial point x0 Close to 513 we always have Iimngt00 mm 513 then 513 is stable otherwise it is unstable Monotone one if the graph y f13 cuts the graph of y 1 at an equilibrium 1 513 from above then it is stable if the cut is from below it is unstable How do we do it more mathematically Linearization x w fx f xx 22 linear approximation first order Taylor expansion IF 1 is close to m So near an equilibrium m xn1 fan aws f xxn 22 22 f xxn 22 or xn1 513 m f xxn 513 A linear equation Let gm 2 am 513 Then yn1 f xyn If f 13 gt 1 then exponential growth unstable If 0 lt f 13 lt 1 then exponential decay stable If f 13 lt 1 then exponential growing oscillation unstable If 1 lt f 13 lt 0 then exponential decaying oscillation stable Linearization Theorem Suppose that 513 is an equilibrium of xn1 f13n If f 13 lt 1 then 513 is stable if f 13 gt 1 then 513 is unstable What happens when f 13 1 Example xn1 2 mm 513 and xn1 2 am 513 the equilibrium 513 O can be either stable or unstable Amy Question 1 Find all possible equilibria 2 Determine the stability of all equilibria One more example Ricker s model xn1 lune x same question Bifurcation a qualitative change in the mathematical system and it can be reflected from the asymptotic behavior of solutions 1 Number of equilibrium changes 2 Stability of equilibria changes A3371 Bifurcation dia ram 2 g inn H 1 513 O lt A lt 1 only equilibrium x0 O which is stable for all initial values A gt 1 equilibria x0 O and 9 A 11b130 O is unstable and how about 513 fxb b 1AI bifurcation will occur when f 13 1 or f 13 1 A 1 is where f 13 1 a new equilibrium emerges A b then m is always stable 2 is where f 13 1 what happens here if b lt 2 Periodic solution If for some integer p xnp 2 am for all n then am is a periodic solution for xn1 f13n with period p pr 1 then it is equilibrium period two solution 0 39 gt 1 fo7E 0 39 gt 2 f1 ffo 0 so x0 is an equilibrium of the map ff13 Question A try to calculate the period 2 solutions for xn1 xquot 1x Chain rule ffx f fxf x If 513 f13 and f 13 1 then the derivative of ff13 is 1 So a new equilibrium for 1 ff13 which is a period 2 solution for 1 f13 emerges Flip bifurcation period doubling bifurcation When f 13 1 the equilibrium loses the stability and a stable period 2 so lution emerges when bgt 2 x is a flip bifurcation pomt for xn1 xquot A b 9 2 14 33 What happens next when A gt b 2 Brief description at next bifurcation point another period doubling bifurcation occurs and a period 4 solution emerges then period 8 period 16 and this a cascade of period doubling bifurcations But all this happens for A less than a finite number M eventraly chaos occurs Leslie matrix model 81ml 817712 Slmk l Slmk 52 o o o o 53 o o L O O O O o o 5k 0 zn z1nz2n zknT is the population vector with zin being the population of the i th age group at time n z1n is the population of the youngest group and zkn is that of the oldest group Let 51gt be the survival rate from age i 1 to i Oltsilt1 for 1 i k let mi be the average reproduction rate per capita for age 139 mi gt 0 only for a g 139 g 3 fertile period Consider solving the equation take n 4 for simplicity k 901 tota birth rate 2 miuim 1 i1 u10 811901 l11901 11201 52511901 1 l2bn 1 11301 5352511901 2 l3bn 2 for n 2 5 11401 545352511901 3 41901 3 Define l1 2 51554 535251 to be the survival function which is the fraction of population surviving form birth to age 139 Renewal equation k k 901 1 Z limibn 1 or 901 Z l m bn i1 i1 fi limi is the net maternity function the number of offsprings produced by each female of age 139 k 901 Z f bn z39 k th order linear difference equation 1 IL characteristic equation for eigenvalue V f k l mm fk eigenvalues A1gt2 Ak 901 C1gt71L CQAEL ck 11101 can be solved by 901 since 11101 libn i Life Table http en wikipedia orgwikiLifetable In actuarial science a life table also called a mortality table or actuarial table is a table which shows for a person at each age what the probability is that they die before their next birthday From this starting point a number of statistics can be derived and thus also included in the table 1 the probability of surviving any particular year of age 2 remaining life expectancy for people at different ages 3 the proportion of the original birth cohort still alive Life tables are usually constructed separately for men and for women because of their substantially different mortality rates Other characteristics can also be used to distinguish different risks such as smoking status occupation socio economic class and others Life tables are also used in biology Example 1 Semelparous reproduction Semelparous organisms reproduce only once in their lifetime such as annual plants Often they die shortly after reproduction i ll quot81 5f f 0 0 6 Lifetable Lesliematrix 12 0 O 2 12 0 12 0 O 13 0 3 16 6 13 1 Renewal equation A3 1 l Mia Eigenvalues A1 1 A273 li l Harro Bernardelli Population waves 1941 Journal of the Burma Research Society oscillation of Burmese population from 1901 to 1931 Example 2 US population in 1966 N Keyfitz W Flieger 1971 Population Facts and Methods of Demography z 1 2 3 4 5 slmi 0 000102 008515 030574 040002 Si 099670 099837 099870 099672 2 6 7 8 9 10 slmi 028061 015260 006420 001483 000089 Si 099607 099472 099240 098867 098274 5 year age classes Population growth from life table A1 10498 growth rate per capita 498 eigenvector stable population distribution 012520118901131010750102000968 0091700867o0817o0765 Reaction diffusion equation Example 1 Genetics drifting model Lecture 20 dp E 8191 p where pt is the fraction of one Allele at generation 75 Now consider a species randomly dispersing in an unbounded habitat and we look for one particular gene Let p13t be the density of the species which possesses an advantageous allele Then Fisher equation or diffusive logistic equation 1937 2 D 8191 p Phenomenon the advantageous gene initializes at one location and it will spread to other places Example 3 Epidemic model Lecture 18 19 St SI Rt 2 Oz Phenomenon The locally infected group reaches a high then returns to zero but nearby region will have the peak of infected people in a slightly later time so the peak of infected propagates from center to surrounding areas That is an epidemic wave Here consider the case of rabies population of foxes Rabies is lethal so that removed class is dead thus no diffusion Healthy foxes tend to stay in their own territory but rabid will travel at random and attack other foxes Thus only infected class will diffuse The Black Death also known as the Black Plague was a devastating pandemic that first struck Europe in the mid late 14th century 13471350 killing between a third and two thirds of Europe s population Almost si multaneous epidemics occurred across large portions of West Asia and the Middle East during the same period indicating that the European outbreak was actually part of a multi regional pandemic Including Middle Eastern lands India and China the Black Death killed at least 75 million people http en wikipedia orgwikiBlackDeath Traveling wave solution a solution with fixed spatial shape but moving when time evolves Suppose the shape is wx when t O and the speed of moving is c then the solution is uat 11133 ct ie when t 1 ua1 2 1113 c when t 2 ua 2 11133 26 etc 2 WE a 15 fuxt format of traveling wave solution u13t wx ct then wz here we use z 1 ct satisfies dw d2w CE w fw which is an ordinary differential equation w cw fw O a 2nd order differential equation 2nd order differential equation iS a lst Order system 111 cw fw 0 Let 2 111 then 111 v and 12 w cw fw cv fw wzv vz cv fOJJ We can do phase plane analysis Traveling wave solution of Fisher s equation Max 7 Z 8215 t uxt1 m 2 let W t W Ct 9t w cw I w1 w 0 let vzw U121 12 cv w1 w Equilibrium points 00 and 10 we look for a solution sat isfying im2gtoowz 0 and IimzooowCz 1 road to the advantageous gene Harvesting Example 5 Constant yield harvesting dP kp lt1 5 h dt N Mathematical Analysis P 1 Nondimensionalization u N s kt d h uu1 u H H dt kN 2 Bifurcation a subcritical saddle node bifurcation occurs at H 025 or h 025kN 3 Qualitative analysis when 0 lt H lt 025 H 025 and H gt 025 4 Analytic method solve the equation you could Example 5 Cont Constant yield harvesting Biological interpretation 1 When 0 lt H lt 025 there are two equilibrium points P1 gt P2 gt O for PO gt P2 limtgtOOPt P1 and for O lt PO lt P2 P05 lt O for t gt to P1 is smaller than N that means the carrying capacity decreases because of harvesting the behavior of solu tions with PO gt P2 is similar to that of logistic equation if the initial population is less than P2 then the population becomes extinct in finite time 2 When H gt 025 there is no equilibrium points and for any initial population the population becomes extinct in finite time 3 H 025 or h 025kN is called Maximum Sustainable Yield MSY dP P Example 6 Constant effort harvesting 7 kP 1 7 qEP q catchability coefficient E effort Mathematical Analysis 1 Nondimensionalization u 2 ve 2 kt d E lu1 u Hu HL t 2 Bifurcation transcritical bifurcation at H 1 3 A different question for which H we can get the maximum yield Hu ma Em mm eflm Yield effort curve Realistic fishery models How to measure fishing effort time spent on fishing Catchability catch per unit of fishing effort tons per 100 hours Data available for analysis total catch fishing effort catch po sition size of fish but not the population of fish fishery stock Yield effort curve can be approximated from data of total catch yield and fishing effort Fishery policy increase effort until the yield falls then decrease effort until the yield falls again the maximum sustainable yield then can be determined Spruce Budworm in Eastern Canada and Northern Minnesota Ludwig Jones and Holling Qualitative analysis of insect out break systems the spruce budworm and forest Journal of Ani mal Ecology 1978 47 315 332 October 6 1977 issue of Nature Magazine Vol 269 pg 471 The extensive bibliography at the end of the article provides an excellent reading list for those interested in pursuing other realistic examples of mathematical modeling Spruce budworm Spruce budworm distribution Background Spruce budworm Choristoneura fumiferana is a serious pest in eastern Canada and northern Minnesota The spruce budworm crawls upon and consumes the leaves of coniferous trees Exces sive consumption can damage and kill the host The budworms themselves are eaten primarily by birds who eat many other in sects as well The budworms prefer larger trees A key factor in determining the spruce budworm population is the leaf sur face area per tree Larger trees have larger leaf surface areas resulting in larger spruce budworm populations The Canadians had observed that the spruce budworm popu lation underwent irruptions approximately every 40 years For 8 unknown reasons the budworm population would explode dev astating the pineries and then return to their previous manage able levels The loss of timber represented a significant cost to the Canadian wood products industry and various management techniques pesticide application for example were tried without success In an effort to understand the cycles of spruce budworm popu lations and with an eye toward developing inexpensive and ef fective management of the problem several scientists at the University of British Columbia R Morris D Ludwig D Jones and CS Holling studied the problem and produced a series of mathematical models They simplified the problem by exploiting a separation of time scales the budworm population evolves on a fast time scale they can increase their density fivefold in a year so their have a characteristic time scale of months whereas the trees grow and die on a slow time scale they can completely replace their foliage in about 7 10 years and their life span in the absence of budworms is 100 150 years Thus as far as the budworm dynamics are concerned the forest variables may be treated as constants Assumptions 1 In the absence of predation the population satisfies logistic growth d szPlt1 gt hP dt N 2 The effort of predators saturates at high prey densities ie there is an upper limit to the rate of budworm mortality due to the predation 3 If the spruce budworm density is low birds will opt for some other prey which most likely lives in other parts of the trees like beetles P So lim hP a gt 0 lim L O or hP m bP2 near P gtoo P P gtoo P O 10 Holling s type III model dP P DCP2 kP lt1 gt dt N A2 P2 the natural growth rate as in the logistic model E the carrying capacity as in the logistic model Q is a measure of predation efficiency If birds are good at catching spruce budworms this number will be larger than if birds often miss the budworm they are attacking Q is the bird population considered a constant in this model A is called the switching value that is the population at which predators begin showing increased interest in harvesting bud worms 11 Through observation it had been observed that two of the model parameters A and N are directly dependent on the average leaf surface area per tree Letting S be the average leaf surface area then A 058 N 48 As the finishing touch of the modeling we denote B 2 CD Then dP P BP2 dt kP 1 lt N A2P2 12 Mathematical Analysis 1 Nondimensionalization39 Q P Bt 2 8 2 A A d 2 ZQQ19gt c2 d5 b 1 I Q2 where az bzg B A 2 Equilibrium points we Q2 Q Q 1Q2gtQ Ooralt1 bgt1Q2 A Real test for your algebra a bifurcation problem c2 HQ2 in How many times will ag 2 alt1 gt and 9Q tersect Three two or one The border line is given by 2133 b 2133 a 7 1 5122 1 5132 Since a b and A 053 N 2 48 where S is the 28 average leaf surface area then b m 8 and a m g 14 81 When b 8 the curve is 0 es mum There are two saddle node bifurcation points we call a1 lt 619 Biological interpretation 1 When the forest is young so 8 is small and a lt a1 then there is only one small positive equilibrium point which is a sink So the budworm population is controlled by the birds the equilibrium is kept in a low level which we call refuge 2 When the forest grows a passes a1 then there are three positive equilibrium points two of them are sinks the refuge and a much larger one which we call outbreak level Outbreak level of budworm is dangerous for the forest But since when the forest grows the budworm is kept at refuge level then it cannot jump to the outbreak level Thus the forest is still in good shape since budworm is still in low level 15 Half life P kP PO Po the time T so that PT P02 Example Polonium 210 has a half life of 140 days 1 If a sample has a mass of 50 mg find the mass after 75 days 2 Find the mass after 100 days Example Determining an infusion rate An asthmatic patient is given a continuous infusion of theophylline to relax and open the air passages in his lungs The desired steady state level of theophylline in the patients blood stream is 15 mgL The average half life of theophylline is about 4 hours and the patient has 56 liters of blood Find the infusion required to maintain the desired steady state level P Linear moder t aP b PO P0 solution in a form of Pt P PO Pe t where P is the equilibrium solution Direction field slope field dP E ft P If the function P05 is a solution of the equation and if its graph passes through the point tOPO where P0 yt0 then the differential equation says that the derivative dPdt at t to is given by the number ftOPO Direction field at each point selected we draw a minitangent line we call it slope mark whose slope is ftP Matlab program dfield httpmathriceedu dfie1d Java Version run onlinel httpmathriceeduquotdfielddfpphtm1 Autonomous Equation dP E fP The direction field of an autonomous equation does not depend on 25 so the slope marks at t1P and t2P are same All solutions are parallel in P direction in the sense that if P05 is a solution then Ptc is also a solution So we only need to draw slope marks for a fixed 25 Phase line Phase line is a simplified direction field for au tonomous eduation only On a vertical line we indicate the positive derivative by an arrow pointing up and negative deriva tive by an arrow pointing down dP Equilibrium solutions E fP If for some number a fa 0 then Pt a is an equilibrium solution constant solution How to draw the phase line 1 Draw the P Iine usually vertical 2 Find and mark the equilibrium points 3 Find the intervals of P values for which fP gt O and draw arrows pointing up in these intervals 4 Find the intervals of P values for which fP lt O and draw arrows pointing down in these intervals Example 1 1 Draw the phase line of P P1 P2 P 2 Determine the long time behavior of the solutions with PO 03 and PO 14 Stability of an equilibrium point Suppose that y 2 yo is an equilibrium point of y fy yo is a sink if any solution with initial condition close to yo tends toward yo as 75 increase yo is a source if any solution with initial condition close to yo tends toward yo as 75 decrease yo is a node if it is neither a sink nor a source Linearization Theorem Suppose that y 2 yo is an equilibrium point of y fy o if f y0 lt 0 then yo is a sink exponential stable 0 if f y0 gt 0 then yo is a source exponential unstable o if f y0 0 then yo can be any type but in addition if f y0 gt O or f y0 lt 0 then yo is a node Model 3 Allee effect Assumptions 1 If the population is too large then the growth is negative 2a If the population is too small then the growth is negative Example The fox squirrel a small mammal native to the Rocky Mountains If the population is too small fertile adults run the risk of not being able to find suitable mates so the growth is negative ewe aw dt K M r 2 growth rate per capita coefficient K carrying capacity M sparsity constant and O lt M lt K Basic growth models P Logistic model P 7P 1 E P P growth rate per capita 7 lt1 E decreasmg maximum growth rate at zero population compensatory Allee effect d P 7P 1 1 dt K M P P P growth rate per capita 7 lt1 gt lt 1 increasing then P K M decreasnng maximum growth rate at a positive population depensatory critical depensation growth is negative for small population Paradiso Purgatory and Inferno model xloii l 51 d 15d O1d x1n 13201 1 075d 1 d O1d 13201 13301 1 075d 05d 021 d 13301 x1O 100 x2O 2 200 x3O 300 Shorter form 5501 1 A 5501 f0 100 200 300 The solution A3301 1 A2330quot 2 A3330quot 3 Anx0 here Aquot is the power of matrix A which is not easy to calculate but easy for Matlab Observation from Matlab simulation d 02 1 All population grow 2 The total population grows exponentially logP has a linear growth 3 The fraction of each population in the total population tends to a constant d 08 1 All population decay 2 The total population decays exponentially 3 The fraction of each population in the total population tends to a constant A stage Structured model 3201 1 lt 025 01515 gt 5101 13O200O Observation exponential growth the numbers ofjuveniles and adults are almost same after a few time units 100 n100 nlOO For 5130 lt 100 then A 100 gt 11 100 100 100 AltlOOgt 11ltIOOgtI so the matrix power could bejust the power of a scalar number Linear algebra can explain all these Math 211 A crash course of linear algebra The numbers A satisfying Ax 2 Am are called eigenvalues The corresponding M72 0 is called eigenvector associated with the eigenvalue An n x n matrix has exactly n eigenvalues they could be same but with different eigenvectors If v is an eigenvector so is co c is a constant Example 2 x 2 matrix Find the eigenvalues and eigenvectors of A if g gt Eigenvalue and eigenvector of 2 x 2 matrix a b a A a a A b a O c d y y 39 c d A y O 39 solve a Ad A bc 2 O characteristic equation for eigen values A199 Example Fibonacci model Eigenvalues eigenvectors for higher dimensional matrices use Matlab to solve Solution of 5101 1 Axn A is a k x k matrix k c dfvl CQAEL UQ ckAka Z cikgbvi i1 where M are eigenvalues vi are eigenvectors and a are con stants The eigenvalues can be ordered so that gt1 2 gt2 2 2 Ak and A1 is called dominant eigenvalue 3301 m clk lvl for large n Univariate difference equation Nn1 fNn Recap of mathematical tools see Appendix A1 A2 Linear equation Nn1 ANn solution Nn 2 NOW Equilibrium solution of N fN Periodic solution Nn 2 Nn for some p p is the period Linearization at equilibrium N linearized equation Nn1 N f NNn N stable if f N lt 1 unstable if f N gt 1 Behavior of linearized sequence If f N gt 1 exponential growth monotonically unstable If 0 lt f N lt 1 exponential decay monotonically stable If f N lt 1 growing oscillation oscillatorin unstable If 1 lt f N lt O decaying oscillation oscillatorily stable Graphing tools can be implemented via Matlab plotting the recursive sequence cobwebbing bifurcation diagram Bifurcation a qualitative change in the mathematical system and it can be reflected from the asymptotic behavior of solutions 1 Number of equilibrium changes 2 Stability of equilibria changes Bifurcation diagram of xn1 fxnu the curve of equilibria in mm space Use solid curve for stable equilibria and dotted curve for unstable equilibria Example 1 Saddle node blue sky bifurcation xn1 2 mm u mg 2 Transcritical bifurcation xn1 2 mm uxn mg 3 Pitchfork bifurcation xn1 2 mm iixn 323 all these three satisfying f 13 1 at the bifurcation point 4 Period doubling bifurcation xn1 mle mm at r 3 f 13 1 but ffa 1 so that the equilibrium becomes an unstable one and a stable period two solution emerges form the bifurcation before bifurcation stable pattern is a period one equilibrium after bifurcation a stable pattern is period two solution 23371 1 3371 Example Beverton Holt with harvesting xn1 2 consider the bifurcation with parameter h 3 Biological modeling Malthus model Nn1 ANn solution Nn NOW based on assumption the ratio of parents to offsprihgs is a constant n n1 gt Nonlinear model density dependent Nn1 ANnSNn SUV is the survival rate of the population which depends on the size of population and ASUV is the per capita reproduction rate How to determine SN 1 From certain assumptions to derive the mathematical forms 2 Available data suggests the shape of function then find func tions with such shape curve fitting see Lecture 2 New assumption the ratio NtNHl is an increasing function of the population size the parents have fewer children if the current population is larger Principle of modeling Everything should be made as simple as possible but not simpler Einstein Our choice the ratio NnNn1 is a increasing linear function Beverton Holt model Nnil i 1 1RNn or Nn1 R137 Nn R K 1 TNT Beverton Holt recruitment curve R maximum growth rate per capita K carrying capacity when the population size is over K it decreasgs 33m 1n Nondimensionalized equation xn1 Math 345 Notes Nondimensionalization J unping Shi The form of a solution of a differential equation can depend critically on the units one chooses for the various quantities involved Frequently these choices can lead to substantial problems when numerical approximation techniques such as Euler s method are applied These dif culties can be controlled or avoided by proper scaling We describe a technique that changes variables so that the new variables are dimensionless This technique will lead to a simple form of the equation with fewer parameters It makes clear that the parameters can interact in the equation and a simpler combined parameter can suffice for more than one parameter We illustrate this technique which is called Nondimensionalization with an example Example Consider the following model of an outbreak of the spruce budworm dP P BPZ 7kP177 77 P0P 1 dt lt N A2P2 H 0 H We give a step by step approach to nondimensionalize this initial value problem Step 1 List all of the variables parameters and their dimensions For the dimensions we use 739 for time and p for population in number of worms Variable Dimension Parameter Dimension 25 739 k 1 739 P p N p A p B p 739 P0 p Step 2 Take each variable and create a new variable by dividing by the combination of parameters that has the same dimension in order to create a dimensionless variable Not that there is not always a unique way to do that so some experimentation may be necessary Here we create P Bt 7 3 7 A7 A We can use our table of dimensions above to check that these new variables are now dimensionless LL Step 3 Now use the chain rule to derive a new differential equation dPidP du 13 du B du 7777iA Bi dt du 13 dt 13 A 13 2 2 The term kP 1 7 becomes kAu lt1 7 and the term simpli es to 15722 Thus the equation simpli es to du Au Buz B7 kA 1 7 7 7 ds 7quot lt N gt 1 u Dividing by B gives Lug 1L L ds B NA 1u2 Lotka Volterra system of predator prey dR dF R FR bF dFR dt a C dt R05 the population of prey Ft the population of predator Original Lotka Volterra problem the proportion of predatory fish increases and the proportion of prey fish decreases during lst world war while the fishing is reduced equation with fishing E is fishing effort pq catchability coef ficients for prey and predator dR dF Equilibrium solution REFE TqEgt qEFE Pro ortion P p pERE decreasing with respect to E Summary of Lotka Volterra predator prey model 1 Solutions are periodic solutions and the orbit circles around the equilibrium 2 The equilibrium represents the average population over dif ferent seasons 3 Volterra s principle if we kills or removes both prey and preda tors in proportion to their population sizes then the average prey population increases while the average predator population de creases 4 The model is flawed as no stable structure equilibrium or pe riodic solution in the model thus it cannot be used to explain the stable oscillations in predator prey interaction It cottoy cushion scale insect prey left vedalia beetle predator right An application of Volterra Principle on pest control In 1868 the cottony cushion scale insect was introduced accidentally from Australia to America and it became a serious pest of citrus in the 1880 s In one of the classical success stories for biological control cottony cushion scales were completely controlled by the introduction of a ladybeetle called the vedalia beetle When the insecticide DDT was discovered it was applied in the hope of further control But instead the result was an increase in the scale insect population 3 Designing better models of predator prey More general predation equation prey P predator Q 62 1kPlt1 gt hPQ hP is the functional response of predators to prey population Assumptions 1 hO 0 when the population is zero there is nothing to harvest 2 h P 2 0 when there is more can be harvested then the harvest rate is higher In Lotka Volterra model MP 2 kP linear predation AP Hollin s t e 11 functional res onse h P 9 YD D 1 BP dPzPlt gt APQ dt N 1 BP Assumptions The number of predator is assumed to be con stant and they cannot consume more preys when P is large It takes the predator a certain amount of time to kill and eat each prey So suppose that in one hour the predator a wolf can catch AP number of prey rabbits it is proportional to P since when P is larger the wolf has better chance to meet rabbits but it needs T hour to handle and eat each rabbit caught So for all AP rabbits it takes ATP hours and in fact the wolf spends 1 ATP hours on these AP rabbits So in 1 hour the wolf AP actually only eats rabbits We use B 2 AT as a new 1 ATP parameter in the equation Holling s type I model Cit fzkplt1 gt h13 where MP 2 LP when 0 g P g PO and MP 2 aPO when P gt PO Assumptions The number of predator is assumed to be con stant and they cannot consume more preys when P is large Type I is unusual except for the case of filter feeding crustacea feeding on algar cells Holling s type 111 model dP P AP kP 1 dt N 1 BPquot spruce budworm model lecture 8 Predator prey model with Holling type II predation saturating interaction non dimensionalized d d uu1 mU m EZ CU F pm ds u 1 d5 u 1 u nullccline u O 2 1 muu 1 v nullcline v O u p c Equilibrium points 00 771 170 c pP c mc p C 19 62 Case 1 p lt m 1c Two equilibrium points at 00 and Tn 10 and Tn 10 is a sink which attracts any initial values Case 2 p gt m 1c Three equilibrium points at 00 in10 me f W C m C p C2 Stability of uovo A brief theory of periodic orbits Theorem 1 Inside a periodic orbit there is at least one equilib rium point Corollary 2 Let 0340 be an equilibrium point If there is a stable or unstable orbit of this equilibrium going to infinity as t gt oo then there is no periodic orbit around this equilibrium point Corollary 3 Let 0340 be an equilibrium point If the direction of vector field on the nullclines near 0340 is not clockwise or counterclockwise then there is no periodic orbit around this equilibrium point Theorem 4 Poincar Bendixon If there is a donut region a region without equilibrium point that the vector field is pointing inside at any point on the boundary then there is a periodic orbit in the donut region Corollary 5 If in a region there is only one equilibrium point which is a source or spiral source and the vector field is pointing inside at any point on the boundary then there is a periodic orbit around this equilibrium point A typical case the solutions from outside are spiraling in and the equilibrium point is a spiral source Theorem 6 Hopf bifurcation If there is a bifurcation occurring that an equilibrium point changes from a spiral sink to spiral source and the solutions from outside are always spiraling in then there is a periodic orbit around the spiral source Hopf bifurcation is similar to flip bifurcation for difference equa tion 39 HON B furcat on d uu1 mu m z cv I pm d5 u I l7 d5 u I ll Stability Of u0v0 Cp c mp mc c J 1919 0 p p m I 1c O Cp c mp mc Eigenvalue equation A2 pp 0 A I Ep c mc O p p C Spiral sink or sink p c mp mc lt O or m gt P p c pc 11 Spiral source or source p c mp mcgt O or O lt m lt p c pc c c p but near p 110220 IS a spiral snnk and all solutions pc pc tend to u0vo in an oscillating fashing when m passes Conclusion m is where Hopf bifurcation occurs When mgt p C C but p C C from u0vo still tends toward 110220 spiral inward thus a periodic solution with small oscillation emerges around u0vo and it is a limit cycle near 110220 becomes a spiral source but all solutions away Examples p 1 c 05 the Hopf bifurcation point is m 13 m 033333 12 Half life P kP PO Po the time T so that PT P02 Example Polonium 210 has a half life of 140 days 1 If a sample has a mass of 50 mg find the mass after 75 days 2 Find the mass after 100 days Example Determining an infusion rate An asthmatic patient is given a continuous infusion of theophylline to relax and open the air passages in his lungs The desired steady state level of theophylline in the patients blood stream is 15 mgL The average half life of theophylline is about 4 hours and the patient has 56 liters of blood Find the infusion required to maintain the desired steady state level P Linear moder t aP b PO P0 solution in a form of Pt P PO Pe t where P is the equilibrium solution Direction field slope field dP E ft P If the function P05 is a solution of the equation and if its graph passes through the point tOPO where P0 yt0 then the differential equation says that the derivative dPdt at t to is given by the number ftOPO Direction field at each point selected we draw a minitangent line we call it slope mark whose slope is ftP Matlab program dfield httpmathriceedu dfie1d Java Version run onlinel httpmathriceeduquotdfielddfpphtm1 Autonomous Equation dP E fP The direction field of an autonomous equation does not depend on 25 so the slope marks at t1P and t2P are same All solutions are parallel in P direction in the sense that if P05 is a solution then Ptc is also a solution So we only need to draw slope marks for a fixed 25 Phase line Phase line is a simplified direction field for au tonomous eduation only On a vertical line we indicate the positive derivative by an arrow pointing up and negative deriva tive by an arrow pointing down dP Equilibrium solutions E fP If for some number a fa 0 then Pt a is an equilibrium solution constant solution How to draw the phase line 1 Draw the P Iine usually vertical 2 Find and mark the equilibrium points 3 Find the intervals of P values for which fP gt O and draw arrows pointing up in these intervals 4 Find the intervals of P values for which fP lt O and draw arrows pointing down in these intervals Example 1 1 Draw the phase line of P P1 P2 P 2 Determine the long time behavior of the solutions with PO 03 and PO 14 Stability of an equilibrium point Suppose that y 2 yo is an equilibrium point of y fy yo is a sink if any solution with initial condition close to yo tends toward yo as 75 increase yo is a source if any solution with initial condition close to yo tends toward yo as 75 decrease yo is a node if it is neither a sink nor a source Linearization Theorem Suppose that y 2 yo is an equilibrium point of y fy o if f y0 lt 0 then yo is a sink exponential stable 0 if f y0 gt 0 then yo is a source exponential unstable o if f y0 0 then yo can be any type but in addition if f y0 gt O or f y0 lt 0 then yo is a node Model 3 Allee effect Assumptions 1 If the population is too large then the growth is negative 2a If the population is too small then the growth is negative Example The fox squirrel a small mammal native to the Rocky Mountains If the population is too small fertile adults run the risk of not being able to find suitable mates so the growth is negative ewe aw dt K M r 2 growth rate per capita coefficient K carrying capacity M sparsity constant and O lt M lt K Basic growth models P Logistic model P 7P 1 E P P growth rate per capita 7 lt1 E decreasmg maximum growth rate at zero population compensatory Allee effect d P 7P 1 1 dt K M P P P growth rate per capita 7 lt1 gt lt 1 increasing then P K M decreasnng maximum growth rate at a positive population depensatory critical depensation growth is negative for small population