9.3-4: Phase Plane Portraits - Colorado State University

1 : Phase Plane PortraitsClassification of 2d Systems:x =Ax,A=[a bc d]:T=a+d,D=ad bc,p( ) = 2 T +DCase A:T2 4D >0 real distinct eigenvalues 1,2= (T T2 4D)/2 General Solution:(v1,v2: eigenvectors)x(t) =c1e 1tv1+c2e 2tv2L1,2: Full lines generated byv1,2 Half line trajectories :ifc2= 0 x(t) =c1e 1tv1 trajectory is half lineH1+={x= v1| >0}ifc1>0H1 ={x= v1| <0}ifc1<0 Same forH2 ifc1= 0,c2>0 or<0 The 4 half line trajectories separate4 regions ofR2y x v1 v2 H1+ H1 H2 H2+ 2v2 v2 v1 2v1 v2 2v2 v1 2v1 x y Phase portrait:Sketch trajectories . Indicatedirection of motionby arrows point-ing in the direction of increasingtDirection of Motion on Half LineTrajectories: If 1>0 thenx(t) =c1e 1tv1 moves out to fort (outwards arrow onH1+) approaches0fort If 1<0 thenx(t) =c1e 1tv1 approaches0fort (inwards arrow onH1+) moves out to fort 1 Subcases of Case ASaddle 1>0> 2 Half line trajectoriesy x L1 L2 Generic TrajectoriesL1 L2 x y Generic trajectoryineach region approaches L1fort L2fort Nodal source 1> 2>0 Half line trajectoriesy x L2 L1 fast slow Generic TrajectoriesL1L2 y x : fast escape to Generic trajectoryis parallel toL1fort tangent toL2fort Nodal sink 1< 2<0 Half line trajectoriesy x L2 L1 fast slow Generic TrajectoriesL1L2 y x : fast approach to0 Generic trajectoryis parallel toL1fort tangent toL2fort 2 Phase Portraits and Time Plots for Cases A (pplane6)SaddleEx.

2 :A=[1 42 1] 1= 3 v1= [2,1]T 2= 3 v2= [ 1,1]Tx =x+4y, y =2x y 505 505xyTime Plots for thick trajectory 30 20 100102030tx and yxyNodal SourceEx.:A=[3 11 3] 1= 4 v1= [1,1]T 2= 2 v2= [ 1,1]Tx =3x+y, y =x+3y 505 505xyTime Plots for thick trajectory 2 1 and yxyNodal SinkEx.:A=[ 3 1 1 3] 1= 4 v1= [1,1]T 2= 2 v2= [ 1,1]Tx = 3x y, y = x 3y 505 505xyTime Plots for thick trajectory and yxy3 Case B:T2 4D <0 = +i ; =T /2, = 4D T2/2 complex eigenvectorv=u+iwcomplex no half line solutionsGeneral Solution:x(t) =e t[c1(ucos t wsin t) +c2(usin t+wcos t)]Subcases of Case BCenter: =0 x(t) periodic trajectories areclosed curvesy x Spiral Source: >0 growing oscillations trajectories areoutgoing spiralsyx Spiral Sink: <0 decaying oscillations trajectories areingoing spiralsyx Direction of Rotation:Atx= [1,0]T:y =c. If{c >0 counterclockwisec <0 clockwiseBorderline Case:Center ( = 0) is border between spiral source ( >0) and spiral sink ( <0).}

3 4 Phase Portraits and Time Plots for Cases B (pplane6)CenterEx.:A=[4 102 4] = 2i v=[2 +i1]x =4x 10y, y =2x 4y 505 2 1012xyTime Plots for thick trajectory 505 5 4 3 2 1012345tx and yxySpiral SourceEx.:A=[ 1 1 ] = +i v=[1i]x = +y, y = x+ 1 1 Plots for thick trajectory 20 15 10 505 2 10123tx and yxySpiral SinkEx.:A=[ 1 ] = +i v=[1i]x = +y, y = x 1 1 Plots for thick trajectory 505101520 4 3 2 10123tx and yxy5 Degenerate Node: Borderline Case Spiral/Node AssumeT2 4D= 0 single eigenvalue =T /2 Assume generic case: (A I)6= 0 single eigenvectorv Let (A I)w=v General solution:x(t) =c1e tv+c2e t(w+tv) only two half line solutions on straight line generated byvDegenerate Nodal Source:T >0borderline case{nodal sourcespiral source x y Degenerate Nodal Sink:T <0borderline case{nodal sinkspiral sink x y 6 Saddle Node: Borderline Case Node/Saddle AssumeD= 0,T6= 0 eigenvalues 1= 0, 2=T Letv1,v2be the eigenvectors General solution:x(t) =c1v1+c2e 2tv2 line of equilibrium points generated byv1 infinitely many half line solutions on straight linesparallel to line generated byv2 Unstable Saddle Node:T >0borderline case{nodal sourcesaddle x y 1=0 2>0 Stable Saddle Node.}}}

4 T <0borderline case{nodal sinksaddle x y 1=0 2<0 : The(T,D) Plane : =T/2 T2 4D/2 five Generic Cases: ifD <0 saddle ifD >0 and T >0 source T <0 sink T2>4D node T2<4D spiralBorderline Cases: ifT= 0 andD >0 center ifD=0,T6=0 saddle-node ifT >0 unstable ifT <0 stable ifT2=4D,A6=(T /2)I, and T >0 d. nodal source T <0 d. nodal sinkOther Special Case:A= I, 6= 0 only half line solutions from origin Name:{unstablestable}star if{ >0 <0}spiral source degeneratenodal source spiral sink degenerate nodal sink nodalsourcenodal sink saddle unstablesaddle node stablesaddle nodeT D D=T2/4 center Ex.:A=[8 5 10 7]{D= 6} saddleEx.:A=[ 2 01 1] {D= 2, T= 3T2 4D= 1} nodal sinkEx.:A=[ 10 255 10] {D= 25T= 0} centerc= 5>0 counterclockwisedirection of rotation8 Typical Homework and Exam a matrixA=[a bc d],classifythe type of Phase the case of centers and spirals you may also be asked to determinethedirection of a matrixA=[a bc d],sketchthe Phase sketch should show all special trajectories and a few each trajectory the direction of motion should be indicated by an arrow.}

5 In the case of centers, sketch a few closed trajectories with the rightdirection of rotation. For spirals, one generic trajectory issufficient. In the case of saddles or nodes, the sketch should include all half linetrajectories and a generic trajectory in each of the four regions separatedby the half line trajectories . The half line trajectories should be sketchedcorrectly, that is, you have to compute eigenvalues as well aseigenvectors. In the case of nodes you should also distinguish between fast (doublearrow) and slow (single arrow) motions (see ). , find the general solution (or a solution to an IVP), classify thephase portrait, and sketch the Phase