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Models, Mathematical Analysis and Simulation, Part I, Fall 2005 Homework 5, Phase Portraits. (Score: 7.0) Three problems are to be solved in this homework assignment. The first consists in scaling of a homogeneous and a nonhonogeneous differential equation. The second is a study of a dynamical system with a simple bifurcation, and the third problem deals with predator-prey models. Hints for making phaseplanes with Matlab are annexed. Problem 1 (1.0). The differential equation for damped free oscillations is a second order homogeneous linear equation. It can be written in the following form: m d2u du + c + ku = 0. 2 dt dt Here the state variable u is the deviation from a stationary position and t is the time. The equation contains three parameters, i.e. the mass m, the damping constant c, and the spring constant k. We first introduce a dimensionless time τ by scaling of the original time variable t . After this is done, we derive a differential equation for u(τ) from the equation above for u(t ). One of the results is that the equation for nondimensionalized variables will often contain fewer parameters than the original equation. This fact will simplify the formal mathematical treatment of the problem, and it will also lead to improved insight about both the problem and its solution. We reparametrize by introducing two parameters as follows: The undamped (angular) fre√ quency ω0 is defined by ω0 = k/m and the critical damping c0 is defined by c0 = 2 km. Note that these new parameters are not dimensionless. We use them to define the dimensionless damping constant α = c/c0 and the dimensionless time τ = ω0 t . Your first task is to derive the differential equation for u(τ) and to show that it equals u + 2αu + u = 0. Observe that the scaling of the time has reduced the number of parameters from three to one. With forced oscillations we study the differential equation m d2u du + c + ku = F cos ωt . dt 2 dt Here there are two additional parameters, the amplitude F and the (angular) frequency ω of the external force. It is easy to nondimensionalize the driving frequency ω since we already have introduced the reference frequency ω0 . Thus we introduce a dimensionless driving frequency β. Furthermore 1 we introduce a dimensionless state variable y by proper choice of scale factor (this should contain F , of course). Your second task is to derive a differential equation for y(t ) In this case the scaling of both the dependent and the independent variables has reduced the number of parameters from five to two: α and β. Problem 2 (3.0). Consider the following linear system of differential equations: u =− 11 P1 u, where P is a real–valued parameter. (0.75) a) Determine the type of phase portrait as a function of P. (There are five different types.) Determine the slopes of the named manifolds (in those cases where they exist) as a function of P. (0.75) b) Determine the limit of u(t ) and of y(t )/x(t ) as t → ∞ in all five cases. Note that the result may depend on the initial point u(0) and on the value of P, Determine the bifurcation value P0 of P and describe the type of phase portrait on either side of it. (0.75) c) Use Matlab to plot one phase portrait of each of the five types, with your choice of P–values. Include the named manifolds when they exist. (0.75) d) There is a transition from one name to another of each named manifold when P passes from a value below P0 to a value above P0 . Describe this transition for each manifold. Describe in particular the boundary case of each of the manifolds when P = P0 . Problem 3 (3.0). (0.5) (a) Consider the two-species models of the form, x = a(x, y)x, ˙ y = b(x, y)y, ˙ (1) where x and y denote the populations of two species, and a and b denote the corresponding growth rates. Assume that a and b are smooth functions. We are now interested in a predator-prey model, where y denotes the predator and x denotes the prey population, which have the following (meaningful) assumptions: (i) If there is not enough prey, the predator population declines. (ii) An increase in the prey population increases the growth rate of the predator. (iii) If no predators are present, a small prey population will increase. 2 (iv) If the prey population goes beyond a certain size, it must decrease. (v) If the predator population increases, the growth rate of the prey population declines. Express the assumptions (i)–(v) in terms of a, b, x and y. Verify that the simple model with a(x, y) = α − βy − λx and b(x, y) = δx − γ − µy (α, β, γ, δ, λ, µ > 0) satisfy (i)–(v). (2.2) (b) Consider the predator-prey model of a), dx dt dy dt = (α − βy − λx)x , = (δx − γ − µy)y where α, β, γ, δ, λ and µ are positive constants. Show that x(t ), y(t ) > 0 for t > 0 if x(0), y(0) > 0. Determine the critical points of this system and determine their stability. Hint: to reduce the number of parameters, introduce the scaling, τ := ct , u := dx and v := ey. Choose the scalings to make the transformed equations have three parameters which are combinations of the original ones, and such that the Strang Predator-Prey model comes as a special case: u = (1 − v − Au)u , v = (u − C − Bv)v What is the condition on A, B, and C that there be a strictly positive critical point? Choose two sets of values, one to make the point a spiral, and one to make it a node. Hand in plots of the phase portrait of the linearized equations and original equations around the critical points which shows the difference between the solution of the linearized and “nonlinearized” equations. (0.3) (c) In his “proof” that the predator-prey populations are cyclic, Strang says that the curve a log u2 − bu2 = cu1 − d log u1 + constant, (2) is closed. Let u1 be fixed. Show that (2) has either 0, 1 or 2 solutions for u2 (depending on u1 ). 3 How to plot phase portraits with Matlab. There are some orbits in phase portraits that are more important, and carry more information, than others. Among the most important are the orbits on the so-called named manifolds. The term named manifolds is used to refer to the slow and fast manifolds of nodes and the stable and unstable manifolds of saddle points. We require of any plot of phase portraits that it includes orbits on all named manifolds. We show by treating a concrete example how you can use Matlab to plot the phase portrait of a linear system in the plane. The system of equations is written u’ = AA*u, where AA is a given 2 × 2-matrix and u is a column vector. The matrix is denoted in Matlab with two letters instead of one since it will be a so-called global variable, and it is customary in Matlab to denote global variables with long names with capital letters. The first step is to write an m-file that computes the left-hand side of the system of equations when the right–hand side is known. Let the name of the file be e.g. lin.m. The file may look like follows: % lin.m Linear diff eq system in two variables. function uprim = lin(t,u) global AA uprim = AA*u; The variable AA is global. It is declared global both in this m–file and in the main program. Its value is set in the main program. After you have saved this file, move to the main Matlab window and give the following commands: >> >> >> >> global AA AA = [-1 1;1 -3]; tspan = [0 5]; Va = [-1 1 -1 1]; axis(Va), axis equal, on, hold grid The second line sets the value AA. If you wish to work with another system you just feed in a new matrix AA. The third line above sets the time span — from initial value t0 = 0 to final value tf = 5 — of the independent variable t. If you later wish to run backwards in time, then all you have to do is to set tspan = [0,-5], that means t < 0. The command axis(Va) in the fourth line defines a window in the x-y-plane, whose corners are determined by Va = [xmin xmax ymin ymax]. axis equal sets equal units in both x- and y-directions. The command hold on allows you to plot several curves in the same figure without removing earlier plots. The command grid, finally, plots a grid in the plane. The main tool for plotting phase portraits is Matlab’s command ode23 or ode45. We start plotting one orbit of the phase portrait by giving the following commands: >> u0 = [1;1]; >> [t,u] = ode23(@lin,tspan,u0); plot(u(:,1),u(:,2)) 4 The initial point is denoted u0. The semicolon makes it a column vector. The second line contains the main command line. It will be used repeatedly in the sequel. Its first command asks Matlab to solve the system of equations in lin.m with the initial value u(t0) = u0. The solution is given in the form of a column vector of t-values in the interval from t0 to tf, and a matrix of corresponding values of u. The number of t-values, and their spacing, is determined by the program. It is used here with default values for tolerances etc. Read the help-file on ode23 (and ode45) to find out what they are and how to change them! You can check the number of steps the program has taken by giving the command size(t). The matrix of u-values has two columns. The first one contains a vector of values of x, and the second one contains a column of values of y. Check with size(u). The second command on the main command line plots the orbit. Here, u(:,1) and u(:,2) denote the first and second columns of the matrix u. When you wish to plot one more orbit, then you give a new initial point u0, after which you repeat the main command line. You do not have to retype any commands; the old ones can be recalled with the arrow-up key. Choose seven additional initial points uniformly distributed on the circumference of the window determined by Va. You might want to experiment with a loop like this1 n = 7; st = exp(i*2*pi*(1:n)/n); for s = st u0 = [real(s);imag(s)]; [t,u] = ode23(@lin,tspan,u0); plot(u(:,1),u(:,2)); end The phase portrait that you have produced so far shows eight orbits that all approach the origin with the same slope. This behavior is typical of the stable node. To complete the phase portrait you need to plot the orbits on the slow and fast manifolds. You find the orbits on the slow manifold by choosing initial points sufficiently far away from the origin. I suggest that you try u0 = [10;10] and then u0 = -u0. The orbits on the fast manifold are found in a different way. One way would be to experiment with a large number of initial points on the circumference of the window until you find one that gives you one of the orbits. An alternative is to start near the origin and run backwards in time. The initial point should then be chosen in a direction from the origin determined by the slope of the fast manifold. To determine the initial point, we need to determine eigenvalues and eigenvectors of the matrix AA: >> [V,D] = eig(AA) The matrix V contains two eigenvectors as columns, and the matrix D has the corresponding eigenvalues on its diagonal. The slopes of the two eigenvectors, sls and slf, are determined with the following commands. The last letter in the slope designation is s for the slow manifold and f for the fast one. 1A more advanced possibility consists of using Matlab script files. 5 >> v1 = V(:,1); >> v2 = V(:,2); sls = v1(2)/v1(1) slf = v2(2)/v2(1) Change the numbering if the slow manifold eigenvector is the second column of V! To plot an orbit on the fast manifold we choose an initial point whose coordinates are determined by the fast manifold. In addition we run time backwards: >> u0 = 0.02*v2; tspan = [0,-5]; After this we execute the main command line. You get the other orbit on the same manifold by changing the sign of u0 and then executing the main command line. When the phase portrait is finished and you are satisfied with its esthetic appearance, you may wish to equip it with a title. You can for example write as follows, replacing my name with yours: >> title(’Phase portrait by my name’) >> xlabel(’x’), ylabel(’y’) Finish off by giving the command print: >> print -deps figure.eps Even simpler, you may use the menu of the figure window (Export). Get your copy at the printer, and equip the orbits with arrows that show the directions they are traversed as time increases. Let us plot one more phase portrait. The window is erased with the command clf (clear figure). To prepare for a new phase portrait and reset the direction of the time axis to normal you give the following command: >> clf, axis(Va), axis equal, hold on, grid, tspan = [0,5]; Feed in a new matrix as follows: >> AA = [-1 4;4 -3]; Now plot the orbits for a number of initial points on the boundary of the window determined by Va. The orbits indicate that you are dealing with a saddle point. Use the command eig to determine eigenvalues and eigenvectors of the matrix AA. Determine the slopes sls and slu of the stable and unstable manifolds from the stable and unstable eigenvectors vs viz. vu. Use initial points u0 = 0.02*v and u0 = -u0, where v is replaced by vs and vu. Let time run forward to get the unstable manifold and backward to get the stable one. It should be noted that the rules that we have followed for plotting orbits on the named manifolds could be somewhat simplified if we are only concerned with linear systems. Our rules have the virtue that they can also be used when we are dealing with nonlinear systems. 6 We summarize the rules for plotting orbits on the named manifolds of stable nodes and of saddle points that have been used above: The orbits on the slow manifold of a stable node are plotted with initial points far away from the critical point, and with time running forward. All other orbits are plotted with initial points close to the critical point. The initial point is chosen on a straight line through the critical point whose slope is given by the corresponding eigenvector. The orbits on the unstable manifold of the saddle point are plotted with time running forward. The orbits on both the stable manifold of the saddle point and on the fast manifold of the stable node are plotted with time running backward. An unstable node is transformed into a stable one by letting time run backward. A note on using ode45 The initial value solver ode45 has a much higher order than ode23. On one hand, this is nice: the computation time is much lower than with ode23. On the other hand, the number of steps taken (e.g., size(t)) is much lower such that the phase plot may become a polygon with visible corners. The latter can be avoided by using an alternative “main command line”: >> >> >> >> sol = ode45(@lin,tspan,u0); t = linspace(tspan(1),tspan(end),101); u = deval(sol,t); plot(u(1,:),u(2,:)) % Note the transposition of u! For details please consult Matlab’s documentation. G OOD LUC K ! 7 ... View Full Document

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