M469, Fall 2010, Practice Problems for the Final
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1 M469 Fall 00 Practice Problems for the Final The final exam for M469 will be Friday December 0 3:00-5:00 pm in the usual classroom Blocker 60 The final will cover the following topics from nonlinear systems of difference equations: parameter estimation non-dimensionalization fixed points and stability of fixed points m-cycles and stability of m-cycles The final will also cover the following topics from differential equations: compartment models population models parameter estimation nondimensionalization phase lines and phase planes nullcline analysis equilibrium points and stability maximum sustainable yield and recovery times for single equations Non-dimensionalize the Ricker-type predator-prey system of difference equations y t+ =ay t e by t y t+ =ry t e cy t Find all fixed points for the Malthusian-type competition model y t+ y t =r y t ay t y t y t+ y t =r y t by t y t and give the ranges of parameter values for which each is stable Assume all parameters are positive 3 For the system of difference equations y t+ = f y t with y + y f y = find an m-cycle that includes the point 0 ŷ = 0 y and determine whether or not the m-cycle is stable 4 For the system of difference equations with y t+ = f y t f y = y y find all fixed points and classify each as stable or unstable Also find all -cycles and classify each as stable or unstable
2 5 For the single-species population model dy dt = r y y K find all equilibrium points including negative ones and determine whether or not each is stable 6 Suppose that in the absence of fishermen the population of fish yt in a certain lake follows the single-species population model dy dt = r y y K and that fishing yield is added to the model as a percentage of population Determine the maximum sustainable yield for this population of fish and describe what will happen to the fish population if this yield is harvested 7 For the differential equation dy dt = y ; y0 = y 0 identify the stable equilibrium point there is exactly one and compute the recovery time to this point from y 0 = 0 8 Suppose saltwater flows into a tank at a constant rate of 6 L/min and out with a constant rate 8 L/min and that the tank initially contains 00 L of saltwater with 05 kg/l salt If the solution entering the tank is kg/l salt find an equation for the amount of salt in the tank at time t 9 Find a solution to the system of differential equations y = 4y + 7y ; 0 The Lotka-Volterra competition model is y 0 = y = y y ; y 0 = dy dt =r y y + s y K dy dt =r y s y + y K 0a Explain how you would use data {t j y j } N j= and linear regression to estimate values for r r s s K and K then explain how you would use nonlinear regression to refine your estimates 0b Obtain a dimensionless form for this equation The West Nile Virus is carried predominately by birds and mosquitoes: When an infected mosquito bites a susceptible bird the virus remains in the bird s blood for two or three days
3 long enough to infect a great number of mosquitoes Assume that when a bird recovers it becomes susceptible again That is it neither dies nor becomes immune Taking into account four variables number of uninfected mosquitoes mt number of infected mosquitoes nt number of uninfected birds bt and number of infected birds ct develop a model for the spread of West Nile Virus through a population of birds and mosquitoes While you should assume mosquitoes are born and die since the time horizon researchers are primarily concerned with is say one summer you should consider the total bird population bt+ct to be fixed Answer the following questions regarding the system of differential equations dy dt =y dy dt =y y a Find all equilibrium points for this system b Write down the phase plane equation for this system and find its general solution c Sketch three different integral curves for this system and place directional arrows on each d Determine the behavior as t for solutions starting from the following points: and 0 e Determine whether each equilibrium point for this problem is stable or unstable 3 If two populations are both produced at a constant rate for example certain cells such as the CD4 + helper T-cells of the immune system are created this way and each population is assumed to die off in proportion with the size of the other population we can describe the dynamics with the simple model 3a Non-dimensionalize this equation dy dt =a b y dy dt =a b y 3b Solve the phase plane equation for your non-dimensionalized version and sketch at least three different integral curves Use arrows to indicate trajectories and discuss the physical implications of your results 3c Carry out a nullcline phase plane analysis for this model and compare your results with b 3d Carry out a spectral stability analysis for this system and compare your results with b and c 3e No that s all I can think of 3
4 4 Carry out a nullcline phase plane analysis for the non-dimensionalized Lotka-Volterra competition model dy dt =y y ay dy dt =by cy y Assume the parameters a b and c are all positive and that the values a c and ac are all positive as well You need only work in the first quadrant where both populations are positive 5 Non-dimensionalize the Malthusian-type competition model dy dt =a y b y y dy dt =a y b y y and carry out a nullcline phase plane analysis for your non-dimensionalized equation Interpret your results in terms of the original system variables 6a Carry out a nullcline phase plane analysis for the non-dimensionalized mutualism model > and consider only positive popula- Consider separately the cases αβ tions dy dt = y y + αy y dy dt = y y + βy y < and αβ 6b Carry out an eigenvalue stability analysis for the model from a Again consider separately the cases αβ < and αβ > Compare your results with the nullcline analysis from a Solutions We begin by setting Y t = y t A ; Y t = y t C where A and C are constants with dimension biomass In terms of Y t and Y t our equations become AY t+ = aay t e bcy t CY t+ = ray t e ccy t 4
5 which we can re-write as Y t+ =ay t e bcy t Y t+ =r A C Y t e ccy t We see that a cannot be scaled away but we can eliminate both a and r by choosing C = b and A = Alternatively we could have eliminated c and r; we cannot eliminate both b rb and c We obtain the non-dimensionalized system Y t+ =ay t e Y t Y t+ =Y t e c b Y t Finally we would typically designate the combination c b as a single parameter say δ = c b The fixed points are solutions of 0 =r y ay y = y r ay 0 =r y by y = y r by From the first equation we have either y = 0 or y = r a If we substitute y = 0 into the second equation we obtain y = 0 and if we substitute y = r a into the second equation we obtain y = r b We conclude that there are two equilibrium points 0 0 and r b r a In order to check stability we set f y = + r y ay y + r y by y For 0 0 we have f + r ay y = ay by + r by f 0 0 = + r r with eigenvalues + r and + r Our stability condition is < + r j < < r j < 0 for j = and this is never satisfied for physical values of r r For r b r a we find f y = ar b br a with eigenvalues λ = ± r r Since + r r > for physical parameter values we conclude that this fixed point isn t stable either 3 First since we re given a point in the m-cycle we can compute the remaining points by iteration We find ŷ = fŷ = ; ŷ 0 3 = fŷ 0 0 = ; ŷ 4 = = ŷ 0 5
6 Ie we have a 3-cycle { 0 0 In order to check stability we compute 0 0 } f y = y + y y + y 0 We have f ŷ 3 f ŷ f ŷ = = The eigenvalues of this matrix are 4 and 0 so we conclude that this 3-cycle is unstable 4 The fixed points are solutions of the system y =y y = y Substituting the first equation into the second we have y = y y + y = 0 and by factoring we see that y = Substituting these values back into the first equation we conclude that we have two fixed points and In order to check if these are stable we compute For we have f y = 0 y 0 f = 0 0 with eigenvalues λ ± = ±i Clearly λ ± = > and we conclude that this fixed point is unstable For we have f = with eigenvalues λ ± = ± Clearly λ ± = > and we conclude that this fixed point is unstable 6
7 In order to find the -cycles we compute f f y = f f = y y Each point in a -cycle is a solution of the equation y = f f y which means we need to solve the system y = y y = y = y y = In particular we note the difference between this result and the result we obtained for fixed points For fixed points we had solutions and - but the values of y are determined by the values of y Here we have a solution so long as y takes either of the values or and independently y takes either of the same two values In this way we have four solutions The first two are fixed points so our -cycle is { } In order to check stability we compute f ŷ f ŷ = = The eigenvalues are λ = 4 from which we conclude instability 5 We find the equilibrium points by solving r ŷ ŷ K = 0 which has three solutions ŷ = K 0 K Either by plotting these points on a phase line or using our derivative condition we find that K and K are both asymptotically stable and 0 is unstable 6 While it would be reasonable to non-dimensionalize in order to streamline the calculations I ll proceed in this case with the original equation Incorporating fishing the model becomes dy dt = r y y K hy The equilibrium points for this model are the solutions to r ŷ ŷ K hŷ = ŷ[ r r h] ŷ = 0 ± K K K h ŷ r = 0 ±K h r 7
8 We are interested the positive equilibrium ŷ = K h for which our yield is r Y h = hŷ = Kh h r Clearly one requirement in this case is h r which keeps the argument of the square root positive We maximize this in the usual way by taking a derivative and setting it to 0 We have Y h = K[ Setting this to 0 we conclude h r + h r h r ] = K[ h h r r h r hmax = r 3 ] = K[ 3h r h r It s clear that this is a maximum because Y h > 0 for h < r and Y h < 0 for h > r 3 3 Consequently the optimal yield is Y hmax = K r 3 3 = rk 3 3 Finally the population approaches the equilibrium value ŷ = K h max = K r 3 = K 3 Using either a phase line or our derivative condition we can verify that this equilibrium point is asympotically stable 7 First the equilibrium points are ŷ = and it s clear either from a phase line or from our derivative condition that only ŷ = is stable Recalling that the recovery time T R is defined by we have yt R = + From class e yt R ŷ = e y 0 ŷ T R = ˆ + e 0 dy y For the integral in y we use partial fractions to show ] so that ˆ + e 0 ˆ dy + y = e 0 y = y y + y y + dy = ln y ln y = ln + e ln e = lne e 8
9 8 We begin by defining an appropriate variable in this case Our basic compartment analysis equation is and in this case we have dy dt yt = kg of salt at time t = incoming rate outgoing rate dy dt = 6 8y ; y0 = 0005 = 0 kg 00 t where we have observed that the volume of water in the tank is 00 t Simplified we have 9 This system has the form with dy dt = 4y ; y0 = 0 00 t A = d y dt = A y 4 7 The eigenvalues and eigenvectors for this matrix are λ = 3 v = λ = 5 v = 7 The general solution is yt = C e 3t + C e 5t 7 According to the initial conditions the constants C and C are solutions of the linear system =C + C = C + 7 C We find C = 3 and C 8 = so that 8 yt = 3 8 e 3t 8 e5t 7 0 For a we divide the first equation by y t giving dy y t dt = r r y r s y K K 9
10 We now use a central difference derivative approximation to approximate values of dy for dt {t j y j } j= N We fit this linearly as a first order linear polynomial in y and y We proceed similarly for the second equation Notice particularly that we obtain precisely enough information so that we can solve for all six parameters We refine our initial parameter values by using them as an initial guess for numerically minimizing the nonlinear error Er r s s K K = For b we define dimensionless variables N yt j ; r r s s K K y j j= τ = t A ; Y τ = y t B ; Y τ = y t C where A has dimension time and both B and C have dimension biomass Substituting these values into the system we find It s natural to make the choices to get It s natural to simplify notation by writing B A Y = r BY BY + s CY K C A Y = r CY s BY + CY K A = r ; B = K ; C = K Y = Y Y s K K Y Y = r r Y s K K Y Y a = s K K ; b = r r ; c = s K K This gives the dimensionless form we analyzed in class The simplest reasonable model is dm dt =k m k mc dn dt =k mc k 3 n db dt = k 4bn + k 5 c dc dt =k 4bn k 5 c 0
11 First equation: The term k m denotes the birth of mosquitoes with birth rate faster than death rate; the term k mc denotes uninfected mosquitoes infected by biting infected birds Second equation: The term k mc again denotes uninfected mosquitoes becoming infected; the term k 3 n denotes the death rate of infected mosquitoes no infected mosquitoes are born Third equation: The term k 4 bn denotes uninfected birds being infected by mosquitoes; the term k 5 c denotes the rate at which birds quit carrying the virus -3 days and consequently become uninfected Fourth equation: The term k 4 bn again denotes the number of birds becoming infected; k 5 c denotes the loss of infected birds when they cease carrying the virus Observe that it s clear from the last two equations that d bt + ct = 0 dt a In order to find the equilibrium points we need only solve the algebraic system ŷ =0 ŷ ŷ =0 from which we see that any point ŷ 0 is an equilibrium point ie the entire horizontal axis b The phase plane equation is dy dy = y y y = y so that parabolas opening upward c y = y + C y C>0 C=0 C<0 y Figure : Figure for Problem
12 d Since 0 0 is an equilibrium point if the system starts at this point it will remain at this point Starting at the point 0 the system will follow a parabola out to + + The interesting case is 0 for which the phase plane equation becomes y = y Since this corresponds with a parabola that opens upward from the point 0 its trajectory will hit the horizontal axis at which point it will stop since each point on the horizontal axis is an equilibrium point At this point y = 0 so y = ± According to the trajectory direction we see that the system approaches 0 e The critical observation to make here is that for each equilibrium point ŷ 0 for which ŷ < 0 nearby trajectories move toward it while for each such equilibrium point for which ŷ > 0 nearby trajectories move away from it In the case ŷ = 0 some trajectories move toward the equilibrium point and some away We conclude that points for which ŷ < 0 are stable not asymptotically stable while points for which ŷ 0 are unstable 3 For a we set τ = t A Y τ = y t B Y τ = y t C where A should be a constant with dimension time and B and C should both be constants with dimension biomass Our system becomes or B A Y = a b CY C A Y = a b BY Y = A B a b AC B Y Y = A C a b AB C Y One natural way to choose the constants A B and C is so that A B a = b AC B = and A C a = This gives C = a b A = a b a and B = a a b The non-dimensionalized system is where k = b b a a For b we return to lower case variables Y = Y Y = ky y = y y = ky
13 The phase plane equation is dy dy = ky y y dy = ky dy y y = y k y + C for some constant of integration C that can be determined by initial conditions We put this in standard form by completing the square on both y and y to obtain k y k y = C where C is a new constant that has subsumed values obtained in completing the square We recognize the integral curves now as hyperbolas centered at k with asymptotes y = ky and y = ky + See Figure y /k y Figure : Figure for Problem 3 According to this diagram we see that there is a line of initial values along which we approach the equilibrium point and that otherwise one species or the other will die out More k precisely y will die out for any populations initialized to the left of the line y = ky and y will die out for any populations initialized to the right of y = ky For c we note that the nullclines are: y -nullcline: y = y -nullcline: y = k The nullcline diagram is given in Figure 3 As with the phase plane diagram it s clear that one population will generally die out and for initial populations in quadrants II and IV we can identify this population but we have somewhat less information in quadrants I and III than we obtained from our full solution of the phase plane equation 3
14 y /k y Figure 3: Nullcline diagram for Problem 3 For d we set f y = y ky f y = 0 k 0 with eigenvalues λ = ±k Since k > 0 we conclude that the equilibrium point is k unstable This of course agrees with b and c More precisely the negative eigenvalue corresponds with the single stable direction we obtained in b while the positive eigenvalue corresponds with the purely unstable direction Although we didn t have time to discuss this in class those stable and unstable directions are precisely the eigenvectors associated with the stable and unstable eigenvalues respectively 4 First the nullclines satisfy so that we have: y y ay =0 by cy y =0 y -nullclines: y = 0 y = a y + a y -nullclines: y = 0 y = cy + See Figure 4 Notice that by assumption a > and c > Though it s natural to divide the first quadrant of the phase plane into four sections we see by following the trajectory arrows that the equilibrium point a ac c ac 4
15 y /a /c y Figure 4: Nullclines for Problem 4 will be approached from any pair of positive initial populations 5 To non-dimensionalize we set τ = t A ; for which the equations become Y τ = y t B ; Y τ = y t C B A Y = a BY b BCY Y C A Y = a CY b BCY Y and it s natural to choose A = a B = a b and C = a b and to define one combined parameter r = a a Returning to lower case coordinates we obtain the system The nullclines for this system are solutions of dy dt = y y y dy dt = ry y y y y =0 y -nullclines: y = 0 y = y r y =0 y -nullclines: y = r y = 0 The equilibrium points are 0 0 and r which in the original variables are 0 0 and a r b a b Referring to Figure 5 we see that if the initial populations are in Region II then 5
16 y II I III IV r y Figure 5: Figure for Problem 5 y will die out while if the initial conditions are in Region IV y will die out In Regions I and III there will be a stable curve passing through r so that y will die out for initial populations on one side and y will die out for initial populations on the other side 6 The nullclines satisfy and so we have: y y + αy =0 y y + β y =0 y -nullclines: y = 0 y = α y α y -nullclines: y = 0 y = β y + The two lines that do not correspond with axes intersect when α y α = β y + y = + α β α = α + αβ Clearly we have an intersection with positive populations if and only if > αβ which is Case The nullcline diagram for this case is given in Figure 6 In this case we see that if the system starts with any two positive populations then as t the populations will approach the equilibrium point α + + β αβ αβ 6
17 y y Figure 6: Figure for Problem 6 Case : αβ < The nullcline diagram for Case is given in Figure 7 In this case we see that if the system starts with any two positive populations both populations will grow without bound For b we set f y = y y + αy y y y + βy y f y + αy ŷ = αy βy y + βy We see from our nullcline diagram that the equilibrium points are 0 0 α + + β 0 0 αβ αβ where the fourth is only relevant in Case with αβ < In Case the diagonal y - nullcline intersects the diagonal y -nullcline at a value with negative populations; this is also clear algebraically For 00 we have f 0 0 = 0 0 with eigenvalues and We conclude that this equilibrium point is unstable Notice that the two unstable eigenvalues correspond with the two directions along which trajectories move directly away from 00 In particular the associated eigenvectors would give these directions For 0 we have f 0 = α 0 + β 7
18 y y Figure 7: Figure for Problem 6 Case : αβ > with eigenvalues and + β Since + β > 0 we conclude instability Notice that the presence of both a stable and an unstable eigenvalue corresponds with the nullcline observation that trajectories approach this equilibrium point along one direction and leave it along another In particular the associated eigenvectors would give these directions For 0 we have + α 0 f 0 = β with eigenvalues + α and Since + α > 0 we conclude instability Notice that the presence of both a stable and an unstable eigenvalue corresponds with the nullcline observation that trajectories approach this equilibrium point along one direction and leave it along another In particular the associated eigenvectors would give these directions For α+ +β = ŷ αβ αβ ŷ we recall that ŷ + αŷ = 0 ŷ + βŷ = 0 so f ŷ = ŷ + αŷ αŷ βŷ ŷ + βŷ The eigenvalues of this matrix are ŷ αŷ = βŷ ŷ λ = ŷ + ŷ ± ŷ + ŷ 4 αβŷ ŷ In Case we have > αβ so the radical is smaller than the negative value ŷ + ŷ This means both eigenvalues are negative and we can conclude asymptotic stability Notice 8
19 that the two stable eigenvalues correspond with the nullcline observation that there are two directions along which trajectories move directly toward this equilibrium point In particular the associated eigenvectors would give these directions 9
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