Ordinary Differential Equations

Transcription

1 Instructor: José Antonio Agapito Ruiz UCSC Summer Session II, 00 Math Ordinar Differential Equations Final Solution This report does not contain full answers to all questions of the final. Instead, ou will find the final is used as an excuse to get more familiar with Mathematica, version.. Not onl some relevant Mathematica code is included but also the use of notebooks and some of its nice features is shown. At the same time, I use Problem and to clarif some concepts to ou. I hope ou will find this useful. Problem. (a) Find the general solution of the given sstem of equations and describe the behavior of the solution as t. Also draw a direction field and plot a few trajectories of the sstem. Remember to justif our work. i x ' x ; Cleara a ; len Lengtha; SolveDeta IdentitMatrixlen 0,, Extra. Of course, ou can use built-in commands to perform desired computations. Here are some examples. Eigenvaluesa,

2 Eigensstema this command gives eigenvalues in the first column and its corresponding eigenvectors in the second column,,,,, CharacteristicPolnomiala, To find the general solution we tpe eq x 't x t x t, x 't x tx t; sol DSolveeq, x t, x t, t x t t t C t t C, x t t t C t t C It is ver instructive to compare this answer with the one we can obtain right awa from simple inspection b just looking at the eigenvalues and eigenvectors of the coefficient matrix a. The straightforward answer is x t C e t C e t, () where C and C are arbitrar constants. You ma want to appl the commands Simplif or FullSimplif to the answers given b Mathematica for x and x to get simpler expressions. I did that but I got the same answers as before FullSimplifsol x t t t C t C, x t t t C t C So, (since I don't know at the moment an other wa to do it with Mathematica), I performed the simplification b hand and obtained x tc C e t C C e t, which is equivalent to the expression we obtained in (). Remember, C and C are arbitrar constants. Finall, we draw a direction field and plot a few trajectories of the sstem for different initial conditions.

3 Cleara, b, eq, sol eq x 't x t x t, x 't x tx t; sol DSolveJoineq, x 0 a, x 0 b, x t, x t, t FullSimplif x t t ab a b t, x t t b t a t We let a var from - to in steps of, and then we give b first the value 0 and then the value -0: Clearsolset solset Tablesol, a,,, ; MapParametricPlotEvaluatex t, x t. solset. b #, t, 0, 6, PlotStle AbsoluteThickness0., DisplaFunction Identit &, 0, 0; Show%, DisplaFunction $DisplaFunction; These trajectories are not spiral. We can also plot a direction field:

4 Clearsol sol DSolveJoineq, x 0 0, x 0 0, x t, x t, t; Needs"Graphics`PlotField`" Block$DisplaFunction Identit, g PlotVectorField uv, uv, u, 0, 0, v, 0, 0, PlotPoints ; g ParametricPlotEvaluatex t, x t. sol, t, 0, 6; Showg, g, Axes True, AxesLabel "x ", "x "; 0 x x -0-0 Remark. Since both eigenvalues and of this sstem are negative, we know that the equilibrium 0, 0 is a node. Therefore, in the phase portrait, trajectories are converging to the equilibrium tangent to the phase line whose direction is the eigenvector v, that corresponds to the eigenvalue (we can see this from the general solution () when letting t ). The plots above confirm this. (ii) x ' x ; Cleara a ; len Lengtha; SolveDeta IdentitMatrixlen 0,,

5 Eigensstema,,,,, The general solution is Cleareq, sol eq x 't x t x t, x 't x tx t; sol DSolveeq, x t, x t, t x t 6 6 t 6t C t 6t C, x t 6 t 6t C 6 6 t 6 t C These formulae can be written as x tc e t C e t, () where C and C are arbitrar constants. We can also write the general solution of the sstem (ii) in terms of a different set of fundamental solutions, b simpl taking the real and imaginar parts of the corresponding fundamental solutions shown in (). Namel, FullSimplifsol x t C Cos t x t C Cos t C C Sin t, C C Sin t Recall, if the matrix of coefficients were then the general solution is of the form x tc e t Cos t C e Sin t t Sin t, Cos t () where and are the real and imaginar parts of the complex eigenvalue. The nice feature about formula () is that we can easil see that its corresponding phase portrait consists of circular trajectories if 0 or spirals otherwise. In our case, we don't have this standard matrix, but we can sa that the phase trajectories of sstem (ii) are closed curves alread since we do have 0. In fact, we can write the expressions for x and x obtained above with Mathematica as

6 x t Cost Sint Sint Sint Cost Sint C, C x tc Cost Sint Sint C Sint Cost Sint. () One can show from here that phase trajectories of sstem (ii) are ellipses. Now, we draw a direction field and plot a few trajectories of the sstem for different initial conditions. Cleara, b, eq, sol eq x 't x t x t, x 't x tx t; sol DSolveJoineq, x 0 a, x 0 b, x t, x t, t FullSimplif x t a Cos t x t b Cos t ab Sin t, ab Sin t We let a var from - to in steps of, and then we give b first the value 0 and then the value -0: Clearsolset solset Tablesol, a,,, ; MapParametricPlotEvaluatex t, x t. solset. b #, t, 0, 6, PlotStle AbsoluteThickness0., DisplaFunction Identit &, 0, 0; Show%, DisplaFunction $DisplaFunction; A plot of a phase trajector and the direction field can be obtained as follows 6

7 Clearsol sol DSolveJoineq, x 0 0, x 0 0, x t, x t, t; Needs"Graphics`PlotField`" Block$DisplaFunction Identit, g PlotVectorField uv, uv, u, 0, 0, v, 0, 0, PlotPoints ; g ParametricPlotEvaluatex t, x t. sol, t, 0, 6; Showg, g, Axes True, AxesLabel "x ", "x "; x x -0-0 Remark. Since the eigenvalues of this sstem are complex and conjugate with real part equal to zero, the equilibrium 0, 0 is a center (notice also that b definition 0, 0 is not a hperbolic point). As t, solutions sta bounded. The plots above confirm this. Finall, when plotting b hand our phase portrait, alwas tr some points to get the right direction. (iii) x ' x ; 7

8 Cleara a ; len Lengtha; SolveDeta IdentitMatrixlen 0,, Eigensstema,,,, 0, 0 We know that 0, 0 is not an eigenvector. This tells us that we need a generalized eigenvector to be able to write down the general solution. First, let's see what Mathematica gives us. Cleareq, sol eq x 't x t x t, x 't x tx t; sol DSolveeq, x t, x t, t x t t t C t t C, x t t t C t t C FullSimplifsol x t t t C t C, x t t t C t C We can write this as x t C e t 0 C C te t C e 0 t. If we let C C and then set K C and K C, we get x tk e t K te t e t 0. () 8

9 This is the form of the general solution we can get b hand, where 0, solution to the equation is the generalized eigenvector obtained as a A Iv v. In our case, v, is the eigenvector corresponding to the eigenvalue and v is a generalized eigenvector. Now, we draw phase trajectories and the direction field corresponding to this sstem. Cleara, b, eq, sol eq x 't x t x t, x 't x tx t; sol DSolveJoineq, x 0 a, x 0 b, x t, x t, t; TableFormFullSimplifsol x t t a tbt x t t a tbt We let a var from -6 to 6 in steps of, and then we give b first the value 0 and then the value -0: Clearsolset solset Tablesol, a, 6, 6, ; MapParametricPlotEvaluatex t, x t. solset. b #, t, 0, 6, PlotStle AbsoluteThickness0., DisplaFunction Identit &, 0, 0; Show%, DisplaFunction $DisplaFunction; And a plot for a phase trajector and the direction field is 9

10 Clearsol sol DSolveJoineq, x 0, x 0, x t, x t, t; Needs"Graphics`PlotField`" Block$DisplaFunction Identit, g PlotVectorFieldu v, uv, u,,, v,,, PlotPoints ; g ParametricPlotEvaluatex t, x t. sol, t, 0, 6; Showg, g, Axes True, AxesLabel "x ", "x "; x x (b) Solve the given value problem and describe the behavior of the solution as t. x ' 0 x, x 0 0 0

11 Cleara a 0 ; Eigensstema,,, 0,,,,, 0,,, Cleareq, sol eq x 't x tx tx t, x 't x tx t, x 't x tx tx t; sol DSolveJoineq, x 0, x 0 0, x 0, x t, x t, x t, t; TableFormFullSimplifsol x t t x t t t x t t As t, the solution x t goes to as well. (c) Find the general solution of the given sstem of equations. x ' x x t x ' x x e t Cleara a ; Eigensstema,,,,,

12 Cleareq, sol eq x 't x t x tt, x 't x t x texpt; sol DSolveeq, x t, x t, t FullSimplif x t t 0 t6 t C C 6 t C C, x t t t t tc C t C C This can be written as x t C C e t C C e t t et 6, which coincides with the form obtained b hand using the variation of parameters method when setting K C C and K C C. 7 Problem. (Predator-Pre model) We will stud a sstem x 'v x, where v is a nonlinear vector field. The particular sstem we are interested in is dx 9 xx dt d x dt (6) where 0 is a parameter. The variable x is the population of pre (rabbits) and is the population of predators (foxes). We want to understand what happens to these populations as t. Our job is to investigate the phase portrait of this sstem for various values of in the interval 0. We will follow the steps described in class (see also a nonlinear sstem analsis, written in Mathematica b Gavin LaRose from the Universit of Michigan). Step (Equilibrium points) We find the equilibrium points b setting both the first and the second components of the vector field v 9 xx, x equal to zero. Clear, x, Solve9 xx 0, x 0, x, x 0, 0, x 9, 0, 9, x Clearl, the second equilibrium point above makes sense when 0. For 0, we get

13 Clearx, Solve9x 0, x 0, x, x 0, 0, x, Step (Linearization. Analsis of eigenvalues for different values of ) We want to know what tpe of equilibrium points the are (hperbolic or non-hperbolic) If hperbolic, we know that we get a good approximation of the phase portrait of (6) b studing the linear sstem ' D v p at each hperbolic equilibrium point p. Thus, we can sa that p is like a node, source, center, saddle, and so on, according to the tpe of equilibrium 0, 0 is in the linear sstem ' D v p. We compute the Jacobian of the vector field v to then evaluate it at the equilibrium points. Here is a set of commands in Mathematica to perform this task Clear, x, v 9xttxt; v xtt; jacobif_list, x_list : OuterD, f, x; jac jacobiv, v, xt, t; MatrixFormSimplifjac 9 xt t xt t xt Again, the equilibrium points are equi Solvev 0, v 0, xt, t xt 0, t 0, xt 9, t 0, t 9, xt and the eigenvalues of their corresponding Jacobians are eig MapEigenvaluesjac. # &, equi, 9, 9, 9, 8, 8 Right awa we can see that independentl of, the equilibrium point 0, 0 is like a saddle. For the other two equilibrium points, we can tr several nonzero values for to see what we obtain.

14 ; equi xt 0, t 0, xt 9, t 0, t 7, xt eig, 9, 9, 7,, Looking at their eigenvalues, we see that all three equilibrium points are hperbolic: 9, 0 is a like a saddle and, is like a spiral with converging trajectories. 7 ; equi xt 0, t 0, xt 9, t 0, t, xt eig, 9, 9,, 6, 6 All three equilibrium points are hperbolic: 9, 0 is like a saddle and, is like a spiral with converging trajectories. ; equi xt 0, t 0, xt, t 0, t, xt eig, 9, 9,,, All three equilibrium points are hperbolic:, 0 is like a saddle and, is like a node.

15 ; equi xt 0, t 0, xt 9, t 0, t, xt eig, 9, 9,,, All three equilibrium points are hperbolic: 9, 0 is like a saddle and, is like a node..; equi xt 0, t 0, xt., t 0, t 0., xt eig, 9, 9, 0., 9., 0. Here, we have onl two equilibrium points: 0, 0 and, 0. And, 0 is not hperbolic ; equi xt 0, t 0, xt 9, t 0, t, xt eig, 9, 9,,, Again, all three equilibrium points are hperbolic: 9, 0 is like a node and, is like a saddle. When 0, we can independentl compute eigenvalues for the corresponding equilibrium points. Now, I denote v,0 and v,0 the first and second components of the vector field v.

16 Clearx,, equi v,0 9txt; v,0 xtt; jac jacobiv,0, v,0, xt, t; MatrixFormSimplifjac 9 t xt t xt equi Solvev,0 0, v,0 0, xt, t equilibrium points xt 0, t 0, xt, t eig MapEigenvaluesjac. # &, equi eigenvalues, 9,, Thus, for 0, we have two equilibrium points: 0, 0 is hperbolic, whereas, is not since its eigenvalues have real part equal to zero. Step (Direction fields, phase portraits) First, we plot direction fields of the sstem (6) for values of 0,,,,,, to get started. Clear Graphics`PlotField` TablePlotVectorField9 x x, x, x,,,,,, Axes True, AxesLabel "x", "", PlotLabel SequenceForm"",, PlotPoints 0,, 0, ; 6

17 0 - x - - x - 7

18 - x - - x - 8

19 - x - - x - 9

20 These plots give us alread a good picture of what is going on with stem (6) for different values of the parameter. B simple inspection we can guess where the equilibrium points are and the tpe of them (something we have done alread). It is alwas a good idea to plot the curves where x' and ' vanish (some authors call these curves nullclines). For instance, if ou are doing things b hand, ou can proceed b steps to construct a good sketch of the direction field of (6). For x'0, the resulting curves determine regions on the phase plane. In each region ou analze the horizontal component of the vector field v in (6). Along these curves, the vector field v is vertical since the horizontal component of v is zero (x'0). For '0, ou do a similar analsis but now focusing on the vertical components of v. Obviousl, here, the vector field v is horizontal along the corresponding curves where ' 0. Finall, ou overlap the two plots above to get a sketch of the direction field of the sstem. We do this for points where x' 0 and blue curves are points where ' 0. 0,,,,,. Red curves represent Clear h t. Solvev 0, t. xt x in terms of x 9 xx x v t. Solvev 0, t. xt x in terms of x 0 TableBlock$DisplaFunction Identit, hor Ploth, x, 0, 0, PlotStle RGBColor, 0, 0; hor GraphicsRGBColor, 0, 0, Line0,, 0, ; ver GraphicsRGBColor0, 0,, Line,,, ; ver Plotv, x, 0, 0, PlotStle RGBColor0, 0, ; equipts GraphicsPointSize.0, Point0, 0, Point 9 9, 0, Point, ; Showhor, hor, ver, ver, equipts, Axes True, PlotLabel SequenceForm"",, AxesLabel "x", "",,, ; 0

21 6 8 0 x x x - -6

22 x x The intersections of red and blue curves are the equilibrium points we have previousl obtained. Now, we plot some phase trajectories for different initial conditions.

23 Clear, eq, sol list of equations eq x't v, 't v ; 0 initial conditions iclist 0., 0.0,,,.,,, 0.0,.8,,, 0,.,,., 0.,, 0., 6, ; TableBlock$DisplaFunction Identit, getploteq_, x0_, 0_ : Modulesol, xs, s, sol NDSolveeq, x0 x0, 0 0, xt, t, t, 0, 0; xs, s xt, t. sol; ParametricPlotxs, s, t, 0, 0, DisplaFunction Identit; trajectories TablegetPloteq, iclisti,, iclisti,, i,, Lengthiclist; trajplot Showtrajectories, DisplaFunction $DisplaFunction, AspectRatio, PlotRange All, PlotLabel SequenceForm"",, AxesLabel "x", "",,, x

24 x x

25 x x Graphics, Graphics, Graphics, Graphics, Graphics

26 Finall, we can put all the plots together: direction fields, nullclines, equilibrium points and some phase trajectories. I leave this for ou as an exercise. You ma want to use again the command Table to plot all these graphs for different values of, all at once. Step (Other tpe of diagrams: x vs. t, vs. t) Here, just follow the pattern as in the Mathematica tutorial (Part b, pp ). Problem. (Extra credit) Solve Problem 8 of Section 7.8 of the textbook. The solution is similar to that of Problem 7 in the same section. Just follow the steps. 6