catalogued.nb 1 ME 163 A Catalogue of Equilibria for Sstems of Two Linear Autonomous First Order Equations Introduction In this notebook, we look at solutions of equations of the following form: d = a + b, d = c + d. Here a, b, c and d are constants, and, unless otherwise specified, we assume that These equations have solutions of the form a d - b c 0. J N = Ju v N ert, where r, u, and v satisf (a - r) u + b v = 0, c u + (d - r) v = 0. For these two homogeneous linear equations to have a non-trivial solution, the determinant of the coefficient arra must vanish. Thus r - Ha + dl r + Had - bcl = 0. Each of the two roots of this quadratic determines a solution. The solution is completed b solving the equations for u and v for each value of r. Because the equations are homogeneous, the vector (u,v) is determined onl to within a constant multiple, which ma be chosen at convenience. The two solutions can then be assembled into the general solution of the problem:
catalogued.nb J N = C 1 J u 1 v 1 N e r 1 t + C J u v N e r t. The stabilit of the equilibrium is determined b the eigenvalues r 1 and r. If the are negative or if the are comple with negative real parts, then the solutions deca as t ö, and the equilibrium is called strictl stable. If either or both of the eigenvalues is positive (or if the real part is positive for a pair of comple conjugate eigenvalues), then there are growing solutions in the neighborhood of the equilibrium and it is unstable. If the eigenvalues are pure imaginar so that the solution is oscillator, the equilibrium is called stable (but not strictl stable). It is usual and useful to classif the possible equilibria on the basis of the eigenvalues, and this is done in Table One below. We have omitted some sub-categories that are sometimes included. Character of Eigenvalues Name of Equilibrium Stabilit real, both negative node strictl stable real, both positive node unstable real, one negative one positive saddle unstable comple conjugate, negative real part spiral strictl stable comple conjugate, positive real part spiral unstable comple conjugate, zero real part center stable Table One -- Classification of Equilibria In the remainder of this notebook, we analze a sequence of eamples, with emphasis on constructing in each case a good phase portrait. Ttaken together, the eamples illustrate the range of possibilities for linear second order autonomous sstems. Before starting the eamples, we discuss briefl the software to be used here. Software Used in This Notebook The pictures in this notebook have been constructed with DnPac. DnPac is a dnamical sstems package which runs in Mathematica. It is available on the Mac Lab server, and also on the Internet at the following URL: http://www.me.rochester.edu:8080/~clark/dnpac.html. In this notebook, we simpl use the package, without much eplanation, to generate the pictures we want. If ou are interested in learning how to use the package, download it from the server or the net. It comes with full documentation and 17 tutorials. The Eamples In[99]:= ssid Mathematica., DnPac 10.58, 3ê31ê000 ü Defining the Equation for DnPac In ever case, we are integrating equations of the form d/ = a + b, d/ = c + d. The state vector for the sstem is {,}, which we declare now for DnPac: In[300]:= setstate@8, <D;
catalogued.nb 3 The parameters in the problem are the four coefficients {a,b,c,d}, which are declared as follows: In[301]:= setparm@8a, b, c, d<d; Before carring out an solutions, we will have to assign values to all four of these parameters. The slope vector is the vector of right-hand sides, and we now define it for DnPac: In[30]:= slopevec = 8a + b, c + d <; Finall, we name our sstem LinEq: In[303]:= ssname = "LinEq"; We can check what we have defined at an time b tping ssreport: In[30]:= ssreport SYSTEM DEFINITION H10.58L Sstem name ssname = LinEq State vector statevec = 8, < State units stateunits = 8, < Slope vector slopevec = 8a * + b *, c * + d * < Parameter vector parmvec = 8a, b, c, d< Parameter values parmval = 8a, b, c, d< Parameter units vector parmunits = 8,,,< Time unit timeunit = Sstem Tpe = differential equation As we work each eample, we will first set values for the coefficients, then choose a set of initial conditions, and finall ask DnPac to carr out integrations and plot the solutions. We begin with nodes, which are sstems with two real eigenvalues of the same sign. ü Stable Nodes For our first case, we take the equations d = -, d = -,
catalogued.nb corresponding to a = -1, b = 0, c = 0, and d = -. We set these values for DnPac: In[305]:= parmval = 8-1, 0, 0, -<; For our first task, we ask DnPac to classif this equilibrium, using the function classifd, which takes the location of the equilibrium as its argument: In[306]:= classifd@80, 0<D Abbreviations used in classifd. L = linear, NL = nonlinear, R = repeated root. Z1 = one zero root, Z = two zero roots. This message printed once. strictl stable - node DnPac tells us that this is a strictl stable node. We can get the eigenvalues as follows: In[307]:= eigval@80, 0<D Out[307]= 8-, -1< We see that the eigenvalues are - and -1 -- both real and both negative, as the should be for a stable node. We can also get the eigenvectors and eigenvalues together: In[308]:= eigss@80, 0<D Out[308]= 88-, -1<, 880, 1<, 81, 0<<< The eigenvector {0,1} goes with the eigenvalue -, and the eigenvector {1,0} goes with the eigenvalue -1. Now its time to construct a picture. We will use a plotting window of {-5,5} in both and. We specif the window for Dnpac. In[309]:= plrange = 88-5, 5<, 8-5, 5<<; We set the image size to 380. In[310]:= imsize = 380; We also ask for one arrow at the mid-point of each solution curves to indicate the positive time direction. In[311]:= arrowflag = True; In[31]:= arrowvec = 81 ê <; We check our graphics settings with plotreport:
catalogued.nb 5 In[313]:= plotreport PLOTTING PARAMETERS Aspect ratio asprat = Automatic Color list for RGBColor colorvec = 8Black< Dashing pattern for D dashvec = 88<< Bo ratio for 3D borat = Automatic Viewpoint for 3D viewpt = 813, -, 0< Plot range for D plrange = 88-5, 5<, 8-5, 5<< Plot range for 3D plrange3d = All Point size ptsize = 0.005 Poincaré map dot poincptsize = 0.008 Line thickness lnthick = 0.00 Grid points in direction field dirpt = 0 Point connection flag pointcon = True Aes flag aon = True Aes origin aorg = Automatic Aes label flag alabon = True Frame flag frameon = False Frame label flag framelabon = True Plot label flag labon = True Plot label shift labshift = 0 Total digits in plot label totdig = Digits to right of decimal in label decdig = Flag for scientific notation in labels sciflag = False Gra level background gralev = 0.9 Displa flag displa = True Graphics image size imsize = 380 Graph background color backcolor = White Arrows on flag arrowflag = True Fractional arrow arclength arrowvec = 9 1 ÅÅÅÅ =
catalogued.nb 6 Because the solutions go inward to the origin, we start our integration on the frame of the plotting window. We will use a set of 16 initial conditions, spaced more or less uniforml around the frame. We define the initial conditions now. In[31]:= initlist = 885, 0<, 85,.5<, 85, 5<, 8.5, 5<, 80, 5<, 8-.5, 5<, 8-5, 5<, 8-5,.5<, 8-5, 0<, 8-5, -.5<, 8-5, -5<, 8-.5, -5<, 80, -5<, 8.5, -5<, 85, -5<, 85, -.5<<; For the numerical integration, we must specif the initial time t0, the integration step size h, and the number of integration steps, nsteps. In[315]:= t0 = 0.0; h = 0.01; nsteps = 500; Now we use the DnPac function portrait to get a phase portrait. In[316]:= node1 = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8-1.00, 0.00, 0.00, -.00< - - - - The arrows help b showing the flow direction. The graph is labeled b the values of the parameters used in the integration. Notice that the solutions are all tangent to the -ais at the origin, ecept for the solutions which start on the -ais -- the remain on the -ais as the approach the origin. This can be understood b looking at the analtical form of the general solution, namel
catalogued.nb 7 J N = C 1 J 0 1 N e- t + C J 1 0 N e-t. We see that when t gets ver large, the first term gets much smaller than the second term, so that for an solution in which C 0, we have J N º C J 1 0 N e-t as t ö, which shows that all such solutions become asmptoticall tangent to the -ais. Let's look at a second eample of a stable node. We consider the equations d = - -, We set the parameter values for DnPac. In[317]:= parmval = 8-, -, 1, -1<; We classif the equilbrium: In[318]:= classifd@80, 0<D strictl stable - node d = -. DnPac tells us that this is a strictl stable node. We get the eigenvalues: In[319]:= eigval@80, 0<D Out[319]= 8-3, -< We see that the eigenvalues are -3 and - -- both real and both negative, as the should be for a stable node. We can also get the eigenvectors and eigenvalues together: In[30]:= eigss@80, 0<D Out[30]= 88-3, -<, 88-, 1<, 8-1, 1<<< The eigenvector {-,1} goes with the eigenvalue -3, and the eigenvector {-1,1} goes with the eigenvalue -. We can predict from this that as the orbits approach the origin, the will do so tangent to the vector {-1,1}, ecept for a single orbit along the line {-,1}. We construct the phase portrait using the parameters defined for the preceding eample, after a slight modification of the list of initial conditions. In[31]:= initlist = 885, 0<, 85,.5<, 85, 5<, 80, 5<, 8-.5, 5<, 8-5, 5<, 8-5,.5<, 8-5, 0<, 8-5, -.5<, 8-5, -5<, 80, -5<, 8.5, -5<, 85, -5<, 85, -.5<<;
catalogued.nb 8 In[3]:= node = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 -.00, -.00, 1.00, -1.00< - - - - We see clearl that all of the solutions approach the equilibrium point tangent to the direction {-1,1}, ecept the single solution along the direction {-,1}. Now we look at a few unstable nodes. Our first eample is d =- + 5, In[33]:= parmval = 8-, 5, -3, 6<; In[3]:= classifd@80, 0<D unstable - node d =-3 + 6.
catalogued.nb 9 In[35]:= eigval@80, 0<D Out[35]= 81, 3< Thus we have two positive eigenvalues, so this is an unstable node, as classifd has alread told us. We must change our integration strateg slightl, because now orbits starting on the frame of our plotting window will in general leave the window and head off to infinit. To get orbits connecting the origin with the plotting frame, we integrate backwards in time, so that we start on the plotting frame and go toward the origin. We do this b assigning a negative value to the time step. In[36]:= t0 = 0.0; h =-0.01; nsteps = 500; We use the following initial conditions: In[37]:= initlist = 885, 5<, 8, 5<, 81, 5<, 83, 5<, 80, 5<, 8-5, -5<, 8-3, -5<, 8-1, -5<, 8-, -5<, 80, -5<, 85, 3<, 8-5, -3<<; In[38]:= node3 = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 -.00, 5.00, -3.00, 6.00< - - - - The arrows point awa from the origin, consistent with our finding that this is an unstable equilibrium. Our net node illustrates the case of a repeated root.
catalogued.nb 10 d = +, We set the parameter values. In[39]:= parmval = 81, 1, -1, 3<; d We classif the sstem and find the eigenvalues and eigenvectors. In[330]:= classifd@80, 0<D unstable - spiral-node transition HRL In[331]:= eigval@80, 0<D Out[331]= 8, < In[33]:= eigss@80, 0<D Out[33]= 88, <, 881, 1<, 80, 0<<< =- + 3. Note that DnPac has called this a spiral-node transition. The eplanation for this is the following. For a node there are two real roots. For a spiral, there are two comple conjugate roots of the form a ± ib. Now we can imagine that we change a parameter of the sstem so that the eigenvalues change. Suppose the parameter change causes b to decrease to zero and a to remain non-zero. Just as b reaches zero, we have a repeated real root, which we normall call a node. But we see that it could equall well be regarded as the limit of a spiral as the oscillation frequenc goes to zero. Thus this case of a repeated real root is eactl in between a spiral and a node, and so it is called a spiral-node transition. We see that our repeated root comes with onl one independent eigenvector, namel {1,1}. (See the net eample for a repeated root with two independent eigenvectors.) Thus the solution will contain te t terms as well as e t terms. We can verif this b asking DSolve for the analtical solution. In[333]:= Out[333]= DSolve@8'@tD == @td + @td, '@td == -@td + 3@tD<, 8@tD, @td<, td 88@tD Ø- t H-C@1D + tc@1d - tc@dl, @td Ø- t HtC@1D - C@D - tc@dl<< The form verifies our prediction. As we did in the last eample, we integrate this one backwards in time, so that orbits starting on the plotting window frame will move toward the origin. In[33]:= initlist = 885, 0<, 85,.5<, 85, 5<, 8.5, 5<, 80, 5<, 8-.5, 5<, 8-5, 5<, 8-5, 0<, 8-5, -.5<, 8-5, -5<, 8-.5, -5<, 80, -5<, 8.5, -5<, 85, -5<<;
catalogued.nb 11 In[335]:= node = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 1.00, 1.00, -1.00, 3.00< - - - - As we go back in time, all of the solutions will approach the origin tangent to {1,1}, which is the onl eigenvector. We do one last node eample. It is slightl artificial but interesting nonetheless. The equations are d = -, We set the parameter values. In[336]:= parmval = 8-, 0, 0, -<; d =-. The artificial feature is that the equations are uncoupled. We can solve for independentl of, and independentl of. Of course we can also solve them as a sstem, which we do now. First we classif the equilibrium and then find the eigenvalues and eigenvectors.
catalogued.nb 1 In[337]:= classifd@80, 0<D strictl stable - spiral-node transition HRL In[338]:= eigss@80, 0<D Out[338]= 88-, -<, 880, 1<, 81, 0<<< We get a repeated eigenvalue but two independent eigenvectors, so there are no te t terms in the solution, which we verif with DSolve: In[339]:= Out[339]= DSolve@8'@tD == -@td, '@td == -@td<, 8@tD, @td<, td 88@tD Ø -t C@1D, @td Ø -t C@D<< No t terms, as predicted. Now we construct the phase portrait. Because the sstem is stable, we integrate forward in time, starting on the frame of our plotting window. In[30]:= initlist = 885, 0<, 85,.5<, 85, 5<, 8.5, 5<, 80, 5<, 8-.5, 5<, 8-5, 5<, 8-5,.5<, 8-5, 0<, 8-5, -.5<, 8-5, -5<, 8-.5, -5<, 80, -5<, 8.5, -5<, 85, -5<, 85, -.5<<; We restore the time direction to forward b taking the time step h positive. In[31]:= t0 = 0.0; h = 0.01; nsteps = 500;
catalogued.nb 13 In[3]:= node5 = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 -.00, 0.00, 0.00, -.00< - - - - The graph, which is startling at first glance, makes perfect sense if ou remember that [t]/[t] = C[1]/C[] = constant for an solution, so all of the orbits are straight lines through the origin. Sometimes the terminolog for nodes is refined to distinguish some of the various special cases discussed above, such as the repeated root with one eigenvector and the repeated root with two eigenvectors. However, there is a serious lack of uniformit in this terminolog in the literature, so it is of limited usefulness and we do not consider it here. ü Spirals The equations for our net eample are d = - + 3, d Å = -3 -.
catalogued.nb 1 We define the equation for DnPac and then classif it. In[33]:= parmval = 8-1, 3, -3, -1<; In[3]:= classifd@80, 0<D strictl stable - spiral We have a stable spiral. The eigenvalues are In[35]:= eigval@80, 0<D Out[35]= 8-1 - 3 Â, -1 + 3 Â< The eigenvectors will be comple: In[36]:= eigss@80, 0<D Out[36]= 88-1 - 3 Â, -1 + 3 Â<, 88Â, 1<, 8-Â,1<<< Although our goal here is to produce phase portraits, we take a small detour to show how to get real-valued solutions for this equation (something that DSolve can't manage). We start b giving names to the eigenvalues and eigenvectors. In[37]:= 8r1, r< = 8First@First@eigss@80, 0<DDD, Last@First@eigss@80, 0<DDD< Out[37]= 8-1 - 3 Â, -1 + 3 Â< In[38]:= 8vec1, vec< = 8First@Last@eigss@80, 0<DDD, Last@Last@eigss@80, 0<DDD< Out[38]= 88Â, 1<, 8-Â, 1<< Now we get a real valued solution b taking the real part of the comple-valued solution: In[39]:= Out[39]= firstsol = CompleEpand@Re@vec1 Ep@r1 tddd 8 -t Sin@3 td, -t Cos@3tD< We get the second solution as the imaginar part of the comple-valued solution: In[350]:= Out[350]= secondsol = CompleEpand@Im@vec1 Ep@r1 tddd 8 -t Cos@3 td, - -t Sin@3tD< Now we construct the phase portrait. We choose initial conditions around the frame of the plotting window, which is still {{-5,5},{-5,5}}. In[351]:= initlist = 885, 0<, 8.5, 5<, 8-.5, 5<, 8-5,.5<, 8-5, 0<, 8-.5, -5<, 80, -5<, 85, -.5<<;
catalogued.nb 15 In[35]:= spiral1 = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8-1.00, 3.00, -3.00, -1.00< - - - - We look at one last eample of a spiral, this time unstable. The equations are d = - 3, d = 3 +. Notice that these are the same as the equations for the previous eample, ecept the right-hand sides have the opposite sign. An interesting wa to produce these equations is to change the direction of time in the previous equations. That transformation induces a minus sign on each left hand side which can then be transferred to the right hand side. Given that observation, we epect to get the same picture as above, but with the arrows reversed in direction. Let's tr it. In[353]:= parmval = 81, -3, 3, 1<; In[35]:= eigval@80, 0<D Out[35]= 81-3 Â, 1+ 3 Â<
catalogued.nb 16 The eigenvalues indicate an unstable spiral. In[355]:= eigss@80, 0<D Out[355]= 881-3 Â, 1 + 3 Â<, 88-Â, 1<, 8Â, 1<<< We use the same initial conditions as in the last eample. In[356]:= initlist = 885, 0<, 8.5, 5<, 8-.5, 5<, 8-5,.5<, 8-5, 0<, 8-.5, -5<, 80, -5<, 85, -.5<<; We must now integrate backwards in time to get to the origin from the plot window frame, so we make the time step negative. In[357]:= t0 = 0.0; h =-0.01; nsteps = 500; In[358]:= spiral = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 1.00, -3.00, 3.00, 1.00< - - - - As predicted, we get the same picture with the arrows reversed.
catalogued.nb 17 ü Centers There is an interesting connection between spirals and centers, which can be seen b looking again at how these objects are defined b the properties of their eigenvalues. A stable spiral has eigenvalues of the form a ± ib, where a < 0. An unstable spiral has eigenvalues of the form a ± ib, where a > 0. A center has eigenvalues of the form ± ib. Thus we see that a center is a kind of transition sstem between stable and unstable spirals. Let's look at this a little more, b introducing a sstem with a parameter e which can be varied. The sstem is d =, d = - - e. (You might recognize this as the sstem form of a damped oscillator with m = 1, k = 1, and b = e.) We define the sstem for DnPac: In[359]:= In[360]:= Clear@eD parmval = 80, 1, -1, -e<; We now look at three cases: e = 0.5, 0, and -0.5. In each case we will classif the equilibrium. In[361]:= e=0.5; classifd@80, 0<D strictl stable - spiral In[36]:= eigval@80, 0<D Out[36]= 8-0.5 + 0.9686 Â, -0.5-0.9686 Â< In[363]:= e=0.0; classifd@80, 0<D stable HLL, indeterminate HNLL - center In[36]:= eigval@80, 0<D Out[36]= 80. + 1. Â, 0.- 1. Â< In[365]:= e=-0.5; classifd@80, 0<D unstable - spiral In[366]:= eigval@80, 0<D Out[366]= 80.5 + 0.9686 Â, 0.5-0.9686 Â< We see eactl the transition we talked about above. As e goes from positive to negative, the sstem goes from a stable spiral to an unstable spiral, with the transition occurring at e = 0, for which the sstem is a center. Although we are focusing mostl on centers here, we plot all three cases just to show the transition. We start with the stable spiral. We specif the list of initial conditions, on the frame of the plotting window, and we specif forward integration in time. In[367]:= initlist = 88.5, 5<, 8-.5, 5<, 8-.5, -5<, 80, -5<<;
catalogued.nb 18 In[368]:= t0 = 0.0; h = 0.01; nsteps = 000; In[369]:= e=0.5; In[370]:= precenter = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 0.00, 1.00, -1.00, -0.50< - - - - Let's look now at the transition case, the center when e = 0. In[371]:= e=0.0; In[37]:= initlist = 885, 0<, 8, 0<, 83, 0<, 8, 0<, 81, 0<<; In[373]:= t0 = 0.0; h = 0.01; nsteps = 630; This particular diagram looks better with two arrows, so we ask for arrows at the 1/8 and 5/8 points of each orbit. In[37]:= arrowvec = 81 ê 8, 5 ê 8<;
catalogued.nb 19 In[375]:= center = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 0.00, 1.00, -1.00, 0.00< - - - - time. Finall we look at the unstable spiral for e = -0.5. We start on the plotting window frame and integrate backwards in In[376]:= e=-0.5; In[377]:= initlist = 88.5, 5<, 8-.5, 5<, 8-.5, -5<, 8.5, -5<<; In[378]:= t0 = 0.0; h =-0.01; nsteps = 000; In[379]:= arrowvec = 81 ê <;
catalogued.nb 0 In[380]:= postcenter = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 0.00, 1.00, -1.00, 0.50< - - - - This is the unstable spiral associated with e < 0. We do one last center, unrelated to the sequence we just did. The equations are d = -, We define the sstem for DnPac: In[381]:= parmval = 8, -,, -<; We classif the sstem and look at the eigenvalues: d = -.
catalogued.nb 1 In[38]:= classifd@80, 0<D stable HLL, indeterminate HNLL - center In[383]:= eigval@80, 0<D Out[383]= 8- Â, Â< This is a center. The motion is periodic with angular frequenc. We construct the portrait, and we ask for two arrows. In[38]:= arrowvec = 81 ê 8, 5 ê 8<; In[385]:= initlist = 88.5, 0<, 81.5, 0<, 80.5, 0<, 8, 0<, 81, 0<<; In[386]:= t0 = 0.0; h = 0.01; nsteps = 315; In[387]:= center = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8.00, -.00,.00, -.00< - - - - We see that rotation is the opposite wa for this center. Is there an eas wa to tell which wa the rotation is going to be? There are several was. Perhaps the best is to calculate the curl of the direction field, namel G - ÅÅÅÅÅÅ F. If it is positive, then the rotation is counterclockwise. If it is negative, the rotation is clockwise.
catalogued.nb ü Saddles Our final categor is saddles --equilibria for which there is one negative and one positive eigenvalue. Our first eample is ver simple and somewhat artificial, but also tpical in most was. d = -, d =. These are ver similar to our first node eample, and artificial in the same wa: the equations are uncoupled and can be solved independentl of one another. Nevertheless, the make a useful two-dimensional eample. We define the equation for DnPac and then look at the eigenvalues. In[388]:= parmval = 8-1, 0, 0, <; In[389]:= eigval@80, 0<D Out[389]= 8-1, < In[390]:= classifd@80, 0<D unstable - saddle Both eigval and classifd tell us that this is a saddle point. We get the eigenvectors too. In[391]:= eigss@80, 0<D Out[391]= 88-1, <, 881, 0<, 80, 1<<< Thus the -ais goes with the eigenvalue -1, and the -ais goes with the eigenvalue. We will see from the phase portrait that these are special straight line solutions of the equations. We construct the phase portrait now. We revert to one arrow. We use a somewhat different technique here. We integrate both directions in time, starting with initial points along the diagonals = and = -. We also use range checking and stop the integration as soon as it goes out of the bo. In[39]:= arrowvec = 81 ê <; In[393]:= In[39]:= In[395]:= rangeflag = True; bothdirflag = True; ranger = plrange; In[396]:= initlist = 881., 1.<, 8, <, 8.8,.8<, 83.5, 3.5<, 8-1., 1.<, 8-, <, 8-.8,.8<, 8-3.5, 3.5<, 8-1., -1.<, 8-, -<, 8-.8, -.8<, 8-3.5, -3.5<, 81., -1.<, 8, -<, 8.8, -.8<, 83.5, -3.5<, 85, 0<, 8-5, 0<, 80, 0.05<, 80, -0.05<<; In[397]:= t0 = 0.0; h = 0.01; nsteps = 00;
catalogued.nb 3 In[398]:= saddle = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8-1.00, 0.00, 0.00,.00< - - - - For ever starting point ecept along the -ais, the solutions eventuall head off to inifinit. In ever neighborhood of the origin, there are starting points from which solutions go to infinit. Thus the equilibrium at the origin is unstable. Solutions starting on the -ais head toward the equilibrium at the origin. However, such solutions are unstable, because the slightest perturbation will knock them off the -ais onto an orbit which is heading for infinit. The -ais is called the stable manifold of the saddle point, and the -ais is called the unstable manifold of the saddle point. These manifolds divide the phase plane into four disjoint quadrants. Because solutions cannot cross, a solution remains in the quadrant in which it starts. We do one more saddle point, this time for the more tpical case when the equations are coupled. Our sstem is d = + 3, d = 3 +. We define the sstem for DnPac, and then get the classification, the eigenvalues and the eigenvectors.
catalogued.nb In[399]:= parmval = 81, 3, 3, 1<; In[00]:= classifd@80, 0<D unstable - saddle In[01]:= eigval@80, 0<D Out[01]= 8-, < In[0]:= eigss@80, 0<D Out[0]= 88-, <, 88-1, 1<, 81, 1<<< Now we construct the phase portrait moving our arrow location to the 3/5 point of the orbit. In[03]:= arrowvec = 83 ê 5<; We see that the stable manifold (belonging to eigenvalue -) is the line along {-1,1} -- i.e., = -. The unstable manifold (belonging to eigenvalue ) is the line along {1,1} -- i.e., =. Now the picture. First the initial condition list. In[0]:= initlist = 881.0, 0<, 8.0, 0<, 83.0, 0<, 8, 0<, 8-1.0, 0<, 8-.0, 0<, 8-3.0, 0<, 8-, 0<, 81, 1<, 8-1, -1<, 81, -1<, 8-1, 1<, 80, 1.0<, 80,.0<, 80, 3<, 80, <, 80, -1.0<, 80, -.0<, 80, -3<, 80, -<<;
catalogued.nb 5 In[05]:= saddle = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 1.00, 3.00, 3.00, 1.00< - - - - The previous two saddles are atpical in that the stable and unstable manifolds meet at right angles. Let's look at one last saddle that is more tpical in that respect. The equations are d = 3 +, d = - -. We enter the parameters for DnPac, classif the sstem and then get the eigenvalues and eigenvectors. In[06]:= parmval = 83,, -, -<; In[07]:= classifd@80, 0<D unstable - saddle
catalogued.nb 6 In[08]:= eigss@80, 0<D Out[08]= 88-1, <, 88-1, <, 8-, 1<<< Thus the stable manifold, belongin to the eigenvalue -1, is along the vector {-1,}. The unstable manifold, belongin to the eigenvalue, is along the vector {-,1}. We first construct a portrait with just these two manifolds. We continue to integrate both directions in time, and we start the integrations on these manifolds. In[09]:= initlist = 88-1, <, 81, -<, 8, -1<, 8-, 1<<; In[10]:= saddle3man = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 3.00,.00, -.00, -.00< - - - - With this picture, it is somewhat easier to pick other initial conditions to fill in the rest of the picture. In[11]:= initlist = 881, 1<, 8, <, 83, 3<, 8, <, 8-1, -1<, 8-, -<, 8-3, -3<, 8-, -<, 8, -<, 83, -3<, 8, -<, 8-, <, 8-3, 3<, 8-, <<;
catalogued.nb 7 In[1]:= saddle3fill = portrait@initlist, t0, h, nsteps, 1, D; LinEq 8a, b, c, d<=8 3.00,.00, -.00, -.00< - - - - Now we get our final picture with a Show command:
catalogued.nb 8 In[13]:= saddle3 = Show@saddle3man, saddle3filld; LinEq 8a, b, c, d<=8 3.00,.00, -.00, -.00< - - - - This concludes our catalogue. These pictures are important in understanding the behavior of linear sstems, and are conceptual building blocks for thinking about more complicated nonlinear sstems.