Systems of Differential Equations

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1 WWW Problems and Solutions 5.1 Chapter 5 Sstems of Differential Equations Section 5.1 First-Order Sstems www Problem 1. (From Scalar ODEs to Sstems). Solve each linear ODE; construct and solve an equivalent first-order sstem of ODEs. For each first-order sstem of linear ODEs below find an equivalent single ODE, solve it, and then use these solutions to construct all solutions of the sstem. (a) 4 = 0 (b) + 9 = 0 (c) = 0 (d) 1 = 2, 2 = (e) 1 = 2, 2 = 16 1 (f) = 0 (a) The characteristic polnomial r 2 4for 4 = 0 has roots ±2, so the general solution is = C 1 e 2t + C 2 e 2t.If 1 = and 2 =, the equivalent sstem is 1 = 2, 2 = 4 1 with general solution 1 = = C 1 e 2t + C 2 e 2t, 2 = = 2C 1 e 2t + 2C 2 e 2t. (c) The characteristic polnomial r 2 + 5r + 4for = 0 has roots 4, 1, so the general solution is = C 1 e 4t + C 2 e t. The equivalent sstem is 1 = 2, 2 = with general solution 1 = = C 1 e 4t + C 2 e t, 2 = = 4C 1 e 4t C 2 e t. (e) Let 1 =, so 2 = 1 = and = 2 = 16 1 = 16. The scalar ODE that is equivalent to the sstem, 1 = 2, 2 = 16 1,is + 16 = 0. The characteristic polnomial is r , which has roots ±4i. The general solution is = C 1 cos4t + C 2 sin 4t. The general solution of the sstem is 1 = = C 1 cos4t + C 2 sin 4t, 2 = = 4C 1 sin 4t + 4C 2 cos4t. Section 5.2 Properties of Sstems www Problem 1. Verif that each IVP satisfies the hpotheses of Theorem for all values of t and the state variables. Find formulas for the solutions. For (a) (e) plot the nullclines and the nine orbits corresponding to all possible combinations of (0), (0) = a, b = 1, 0, 2 in the rectangle 3, 3. Identif an orbits that are equilibrium points or ccles. (a) =, = 2 [Hint: Note that = 0.] (b) = 2, = 4 (c) =, = 9 (d) = 3, = 3 [Hint: Write as d/d = 3 / 3.]

2 5.2 Chapter 5/Sstems of Differential Equations (e) = 3, = (f) =, = 26 2, z = z/2; (0) = (0) = z(0) = 1. Sketch the projection of the orbit in each coordinate plane. Use the ranges 1, 4 3, 1 z 15, and the time interval 0 t 10. The hpotheses of the Fundamental Theorem are satisfied in each part for all points in state space and for all t since the rate functions in each case are polnomials in the state variables, and all polnomials are continuous and continuousl differentiable. So, each of the initial value problems has a unique maimall etended solution defined on some time interval containing t 0 = 0. The solutions are defined for all real t unless specificall noted otherwise. The origin in state space is the onl equilibrium point since the right-hand sides vanish onl at that point. The - and -nullclines are the dashed curves. The are not alwas visible because sometimes orbits lie on top of them. Some of the nine orbits coincide with one another, and so fewer than nine orbits ma be visible in the pictures. The 9 initial points to use are ( 1, 1), ( 1, 0), ( 1, 2), (0, 1), (0, 0), (0, 2), (2, 1), (2, 0),and(2, 2). (a) The sstem is =, = 2. The corresponding scalar IVP, = 0, (0) = a, (0) = (0) = b, has characteristic polnomial r 2 + 2r + 1 with 1as a double root. The general solution has the form = C 1 e t + C 2 te t. Appling the initial conditions, we obtain = e t (a + (a + b)t). Then = = e t (b (a + b)t). See Fig. 1(a), where the -nullcline is = 0, and the -nullcline is = /2; (0, 0) is the equilibrium point. (c) The sstem is =, = 9. The equivalent scalar IVP + 9 = 0, (0) = a, (0) = (0) = b has characteristic polnomial r with roots ±3i. The general solution has the form = C 1 cos 3t + C 2 sin 3t. Appling the initial conditions, we obtain = a cos3t + (b sin 3t)/3, and so = = 3a sin3t + b cos3t. All orbits aside from the equilibrium point (0, 0) are ccles. See Fig. 1(c), where the - and -nullclines are, respectivel, = 0and = 0. (e) The ODEs = 3, = are uncoupled and ma be solved directl. Separating variables, we have 3 d = dt, (0) = a, which leads to a 2 2 = 2t, or = a(1 + 2a 2 t) 1/2,wheret > 1/2a 2 (or if a = 0, t < ). In addition, the solution of the IVP, =, (0) = b, is = be t for the same t-interval. See Fig. 1(e), where the - and-nullclines are, respectivel, = 0and = 0; (0, 0) is an equilibrium point.

3 5.3/ Models of Interacting Species 5.3 Problem 5.2.1(a). Problem 5.2.1(c). Problem 5.2.1(e). Section 5.3 Models of Interacting Species www Problem 5. (Knocking Out the Competition). The sstem = (2 2) + H, = (2 2 ) models competition where the -species can be restocked (H > 0) at a rate proportional to the population. Using a numerical solver to plot solution curves, find the restocking coefficient H of minimal magnitude H 0 so that the -species becomes etinct regardless of its initial population. Plot orbits for values of H below, at, and above H 0 and eplain what ou see. [Hint: Find the value H 0 of the restocking coefficient that has the propert that as H increases through H 0, the equilibrium point inside the population quadrant eits across an ais.] The sstem is = (2 + H 2), = (2 2 ). The equilibrium point determined b the intersection of the -nullcline 2 + H 2 = 0andthe-nullcline 2 2 = 0 has coordinates = (2 H)/3, = 2(1 + H)/3. This point eits the first quadrantas H increases through the value 2. The other equilibrium points are (0, 0), (0, 2),and(2 + H, 0). For0< H < 2, it seems that we have a model of competitive eclusion (see Fig. 5, Graph 1 where H = 1). At H = 2, the -species seems to win the competition (see Fig. 5, Graph 2 where H = 2). For H > 2, the -species wins (see Fig. 5, Graph 3 where H = 3). Problem 5.3.5, Graph 1. Problem 5.3.5, Graph 2. Problem 5.3.5, Graph 3.

4 5.4 Chapter 5/Sstems of Differential Equations Section 5.4 Predator-Pre Models www Problem 3. (Linearizing to Estimate the Periods of Ccles). To linearize a planar sstem = f (, ), = g(, ) about an equilibrium point ( 0, 0 ), epand each rate function in a Talor series about ( 0, 0 ) and discard all terms higher than first order. The result is called the linearization of the original sstem at ( 0, 0 ). Let s appl this process to the predator-pre rate equations = a + b, = c d in order to estimate the periods of the population ccles near the equilibrium point (c/d, a/b) inside the population quadrant. (a) Show that the sstem of linearized rate equations for the predator-pre sstem is = bc/d ac/d, = ad/b + ac/b. (b) Show that for certain values of ω and A/B, = c/d + A cos ωt, = a/b + B sin ωt solves the linearized sstem in part (a). What are these values? (c) Set a = b = c = d = 1 and plot orbits and component graphs of the nonlinear and the linearized sstems, using the common initial data 0 = 1, 0 = 1, 1.1, 1.3, 1.5, 1.9. Plot over 0 t 20. Eplain the graphs and compare the periods of the ccles of the two sstems using a common initial point. The -sstem is = f (, ) = a + b, = g(, ) = c d.let 1 = c/d and 1 = a/b so ( 1, 1 ) is the equilibrium point. (a) To linearize we first find the partial derivatives of f (, ) and g(, ) with respect to both and. These are f/ = a + b, f/ = b, g/ = d,and g/ = c d. So with 1 = c/d and 1 = a/b,wehave = f/ ( 1, 1 )( 1 ) + f/ ( 1, 1 )( 1 ) = ( a + b(a/b))( c/d) + b(c/d)( a/b) = bc/d ac/d and = g/ ( 1, 1 )( 1 ) + g/ ( 1, 1 )( 1 ) = d(a/b)( c/d) + (c d(c/d))( a/b) = ad/b + ac/b. The linearized sstem is = bc/d ac/d, = ad/b + ac/b. (c) See Fig. 3(c), Graph 1 for the graphs of the predator-pre orbits and component graphs of the nonlinear Lotka-Volterra sstem, = ( 1 + ), = (1 ). Graph 2 shows the orbits and component graphs for the corresponding linearized sstem. Note from the -componentgraphshowthe period of the predator-precclesof the nonlinear sstem increases with amplitude, but the period of the linearized ccles is fied at 2π/ ac (= 2π if a = c = 1). The initial points used for the graphs in both figures are 0 = 1, 0 = 1, 1.1, 1.3, 1.5, and 1.9.

5 5.5/ The Possum Plague: A Model in the Making 5.5 Problem 5.4.3(c), Graph 1. t Problem 5.4.3(c), Graph 2. t Section 5.5 The Possum Plague: A Model in the Making www Problem 1. (Outbreak of a Measles Epidemic: Kermack McKendrick SIR Model). We ma model the evolution of an epidemic in a fied population b dividing the population into three distinct classes: S = Susceptibles, those who have never had the illness and can catch it I = Infectives, those who are infected and are contagious R = Recovered, those who had the illness and have recovered Assume that the disease is mild (everone eventuall recovers) the disease confers immunit on the recovered, and that the diseased are infective until the recover. (a) Argue that the evolution of the epidemic ma be modeled b the nonlinear sstem of first-order ODEs S = asi, I = asi bi, R = bi, where the parameter b represents the reciprocal of the period of infection and a represents the reciprocal of the level of eposure of a tpical person. [Hint: Compare with models for predatorpre interaction.] (b) Show that the onset of an epidemic can onl occur if the susceptible population is large enough. Specificall, find the threshold value for S above which more people are infected each da than recover. (c) Suppose that German measles lasts for four das. Suppose also that the tpical susceptible person meets 0.3% of the infected population each da and that the disease is transmitted in 1 out of ever 6 contacts with an infected person. Find the values of the parameters a and b in the SIR model. How small must the susceptible population be for this illness to fade awa without becoming an epidemic? Verif b plotting the component graph for I(t) for our choice of I(0) and for values of S(0) that are 50% above and 50% below the threshold value found in part (b). Plot over 0 t 30 and discuss what ou see. (d) Suppose that another illness has parameter values a = and b = 0.08, and suppose that 100 infected individuals are introduced into a population. Investigate how the spread of the infection depends on the size of the population b plotting S-, I-, and R-component graphs for 0 t 50, I(0) = 100, R(0) = 0, and values of

6 5.6 Chapter 5/Sstems of Differential Equations S(0) ranging from 0 to 2000 in increments of 500. How does the value of S(0) affect the speed with which the epidemic runs its course? (e) Using a = and b = 0.08, find I as a function of S, I 0,andS 0, and plot I(t) against S(t) for various values of S 0 and I 0,0< S 0 < 1600, 0 < I 0 < Zooming near the origin, look at the long-term behavior. Interpret our graphs in terms of the model. [Hint: di/ds = 1 + b(as) 1.] (a) Since the parameter b represents the reciprocal of the period of infection, the rate equation for the population of recovered is R = bi. The rate equation for the population of susceptibles is S = asi because the spread of the disease depends upon contact between susceptibles and infectives, and so a mass action term such as asi is appropriate, where 1/a measures the level of eposure of the average susceptible to the average infective. Since we assume that the total population is constant, then S + I + R =constant, which implies that I = (S + R ) = asi bi. (c) We have a = (1/6) = Since the duration of the illness in a person is four das, b = 1/4 = It is shown in part (b) that the epidemic occurs if S > 2b/a = / = See Fig. 1(c) for graphs of I(t), wherei(0) = 50, S 0 = 500, 1000, 1500 (solid, long dashes, and short dashes, respectivel). Note that an epidemic occurs onl when the susceptible population is large enough (in the plotted cases, S(0) = 1000, 1500). (e) Because di = [b/(as) 1]dS, the general solution is I = (b/a) ln S S + C. According to the initial condition, we know that I = (b/a) ln(s/s 0 ) + (S 0 S) + I 0. See Fig. 1(e), Graph 1 where a = 0.001, b = The four curves (from top down) correspond to SI-initial data (1500, 50), (1250, 35), (1000, 20), (800, 15); the time span is 0 t 50. The S-ais is a line of equilibrium points, and it seems that each SI-orbit starts (at t = ) at a point (S 1, 0) where S 1 is large, and ends (as t + ) at(s 2, 0) where S 2 is small. This is plausible since as a disease runs its course the susceptible pool diminishes and the number of infectives first increases and then diminishes. Fig. 1(e), Graph 1 is deceptive because it suggests that as time goes on the number of susceptibles tends to 0. Fig. 1(e), Graph 2 zooms on the left-hand ends of the SI-orbits, and we see that as t + the separate orbits actuall approach different points on the S-ais. Infectives I Infectives I Infectives I Problem 5.5.1(c). t Susceptibles S Problem 5.5.1(e), Graph 1. Susceptibles S Problem 5.5.1(e), Graph 2.