Problem set 6: Solutions Math 207A, Fall x 0 2 x

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1 Problem se 6: Soluions Mah 7A, Fall 14 1 Skech phase planes of he following linear ssems: 4 a = ; 9 4 b = ; 9 1 c = ; 1 d = ; 4 e = ; f = 1 3 In each case, classif he equilibrium, =, as a saddle poin, node ec, deermine is sabili, and sa if i is hperbolic or non-hperbolic Soluion Phase planes are shown in Figure 1 The are drawn using pplane8m, which is available a hp://mahriceedu/~dfield/ a The eigenvalues and eigenvecors are λ = ±6i, r = ±3i The origin is a cener sable bu no asmpoicall sable, and nonhperbolic b The eigenvalues and eigenvecors are λ = ±6, r = ±3 The origin is a saddle poin unsable and hperbolic

2 c The eigenvalues and eigenvecors are 1 λ =, r = The eigenvalue λ = has algebraic muliplici and geomeric muliplici 1, and he mari is no diagonalizable The general soluion is [ ] 1 1 = c 1 e +c e + 1 The origin is an unsable node hperbolic d The eigenvalues and eigenvecors are 1 λ 1 =, λ = 4, r 1 = 1, r = The mari is singular, and here is a line of equilibria a, = c,c The origin is no an isolaed equilibrium unsable and non-hperbolic e The eigenvalues and eigenvecors are λ = 1±3i, r = 1±3i The origin is an unsable spiral poin hperbolic f The eigenvalues and eigenvecors are 1 λ 1 =, λ = 1, r 1 = 1, r = 1 The origin is a sable node asmpoicall sable and hperbolic

3 a b = A + B = C + D A = B = 4 C = D 9= = A + B = C + D A = B = 4 C = D 9 = c d = A + B = C + D A = B = 1 C = D = = A + B = C + D A = B = 1 C = D 4= e f = A + B = C + D A = B = C = D = = A + B = C + D A = B = C = D 1= 3 Figure 1: Phase planes for 1a 1f

4 Two n n linear ssems = A, = B are said o be differeniabl equivalen if here is a diffeomorphism ie, a differeniable map wih differeniable inverse h : R n R n such ha = h is a soluion of = B if and onl if is a soluion of = A Show ha if = A and = B are differeniabl equivalen, hen A and B have he same eigenvalues Is differeniable equivalence a useful wa o classif he qualiaive behavior of linear ssems? Eplain our answer Soluion Consider, more generall, wo n n nonlinear ssems = f, = g Suppose ha he change of variable = h maps he firs ssem ino he second We don indicae vecors eplicil Using he chain rule and he -equaion, we ge ha = Dh = Dh f, wherehelinearmapdh : R n R n ishederivaive ofh : R n R n a The mari of Dh is he Jacobian mari of h If ODEs have he same soluions, hen he have he same vecor fields, so i follows ha gh = Dh f 1 Now suppose ha is an equilbrium soluion for he -equaion, meaning ha f =, and le ȳ = h be he corresponding equilbrium of he -equaion The linearizaions of he,-equaions a =, = ȳ are: = A, A = Df ; = B, B = Dgȳ Differeniaing 1 wih respec o and using he chain rule, we ge ha Dgh Dh = D h f+dh Df; or, in componen noaion wih i = h i j, g i = i j f j, g i j j k = i j k f j + i j f j k, where we use he summaion convenion over repeaed j-indices

5 Seing = in and using he fac ha f =, we ge ha B Dh = Dh A Since h is a diffeomorphism, Dh is nonsingular, and B = Dh A Dh 1 This means ha he marices of A, B are similar; in paricular, he have he same eigenvalues Differeniable equivalence is oo fine a noion o provide a useful classificaion of he qualiaive dnamics of ODEs For eample, we would like o regard an wo planar linear ssems for which he origin is a saddle poin as equivalen However, if one equilibrium has eigenvalues λ = 1,1 and anoher has eigenvalues λ = 11,1, hen he corresponding ssems are no differeniabl equivalen This fac eplains wh we use he weaker noion of opological equivalence o classif dnamical ssems, even hough i is harder o work wih

6 3 Consider he following ssem of ODEs =, = + 3 a Find he equilibria b Linearize he ssem around he equilibria and classif hem c Skech he phase plane of he ssem d Discuss he asmpoic behavior of soluions as Indicae differen regions of he phase plane ha correspond o differen pes of asmpoic behavior Soluion a The equilibria saisf =, + =, which implies ha = and =, so, =, or, = 1,1 b The Jacobian mari of he ssem is The linearizaion a, is = The eigenvalues are λ = 1±i, so, is an unsable spiral poin The linearizaion a 1, 1 is 1 1 = 1 1 The eigenvalues are λ = ±, so 1,1 is a saddle poin c The phase plane is shown below The sable and unsable manifolds of he saddle poin are shown in green, and he nullclines in orange = and magena =

7 d The phase plane is divided ino wo pars b a curve ha consiss of he sable manifold of he saddle poin which includes he heeroclinic orbi from he spiral, o he saddle 1,1 and he rajecor from he spiral, such ha, 1/ as + Trajecories below his curve he region acuall winds around he spiral poin near, approach he righ par of he unsable manifold of he saddle poin, and have +, 1/ as + ; while poins above his curve approach he lef par of he unsable manifold of he saddle poin, and have, + as = =