Solutions for homework 12

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1 y Soluions for homework Secion Nonlinear sysems: The linearizaion of a nonlinear sysem Consider he sysem y y y y y (i) Skech he nullclines Use a disincive marking for each nullcline so hey can be disinguished (ii) Use analysis o find he equilibrium poins for he sysem Label each equilibrium poin on your skech wih is coordinaes (iii) Use he Jacobian o classify each equilibrium poin (spiral source, nodal sink, ec) Soluion (i) The and y nullclines are or y + ( -nullcline ) y or y + ( y-nullcline ) y y y y y (ii) The equilibrium poin are he inersecion of he nullclines: (, ), (, ), (, ), ()

2 y y y y y y (iii) The Jacobian is (A) A (, ) his is y J(, y) y y J(, ), so he characerisic equaion λ T r(j)λ + de(j) λ + λ yields he eigenvalues: one posiive, one negaive λ, λ, and herefore (, ) is a saddle poin (B) A (, ) his is ( J ),, so he characerisic equaion λ T r(j)λ + de(j) λ + λ 9

3 yields he eigenvalues: boh negaive λ and herefore (, ) is a nodal sink (C) A (, ) his is so he characerisic equaion 577, λ + 6, J(, ), λ T r(j)λ + de(j) λ + λ yields he eigenvalues: one negaive, one posiive λ, λ, and herefore (, ) is a saddle poin (D) A (, ) his is J(, ), so he characerisic equaion yields he eigenvalues: boh posiive and herefore (, ) is a source 7 Consider he sysem λ T r(j)λ + de(j) λ λ + y λ, λ, y sin y (i) Skech he nullclines Use a disincive marking for each nullcline so hey can be disinguished (ii) Use analysis o find he equilibrium poins for he sysem Label each equilibrium poin on your skech wih is coordinaes (iii) Use he Jacobian o classify each equilibrium poin (spiral source, nodal sink, ec) Soluion (i) The and y nullclines are y ( -nullcline ) y sin ( y-nullcline )

4 y y y y sin() (ii) The equilibrium poin are he inersecion of he nullclines: sin() k kπ( for any ineger k ), y, ie, (kπ, ), k Z y y sin()

5 y (iii) The Jacobian is (A) A ((l + )π, ) his is ( ) J(, y) cos() J((l + )π, ) so he characerisic equaion λ T r(j)λ + de(j) λ + λ yields he eigenvalues: one negaive, one posiive λ 5 68, λ + and herefore ((l + )π, ) is a saddle (B) A (lπ, ) his is J(lπ, ) so he characerisic equaion λ T r(j)λ + de(j) λ + λ + yields he eigenvalues: comple, wih negaive real par λ i, λ + i, and herefore (lπ, ) is a spiral sink y y cos() y 5 68,

6 y y 5 Use your numerical solver o compare he phase porrai of he nonlinear sysem y y sin y wih ha of is linearizaion near he equilibrium poin (π, ) y y sin() y Soluion The linearizaion a (π, ) is he sysem u Ju, u u y y y

7 Secion 9 Linear sysems wih consan coefficiens: Phase plane porrais For he mari 5 A 5 use p(λ) λ T λ+d, where T r(a) and D de(a), o compue he characerisic polynomial Then use p(λ) de(a λi) o calculae he characerisic polynomial a second ime and compare he resuls Soluion λ T r(a)λ + de(a) λ + 5 and de λ 5 ( λ)( λ) + 5 λ λ The general soluion of y Ay is y() C e ( ) + C e Wihou he help of a compuer or calculaor, skech he half-line soluions generaed by each eponenial erm of he soluion Then, skech a rough approimaion of a soluion in each region deermined by he half-line soluions Use arrows o indicae he direcion of moion on all soluions Classify he equilibrium poin as a saddle, a nodal sink, or a nodal source Soluion Boh eigenvalues are real and posiive, herefore he origin is a nodal source The general soluion of y Ay is y() C e + C e Wihou he help of a compuer or calculaor, skech he half-line soluions generaed by each eponenial erm of he soluion Then, skech a rough approimaion of a soluion in each region deermined by he 7

8 half-line soluions Use arrows o indicae he direcion of moion on all soluions Classify he equilibrium poin as a saddle, a nodal sink, or a nodal source Soluion Boh eigenvalues are real and negaive, herefore he origin is a nodal sink 5 The general soluion of y Ay is y() C e + C e 5 Wihou he help of a compuer or calculaor, skech he half-line soluions generaed by each eponenial erm of he soluion Then, skech a rough approimaion of a soluion in each region deermined by he half-line soluions Use arrows o indicae he direcion of moion on all soluions Classify he equilibrium poin as a saddle, a nodal sink, or a nodal source Soluion The eigenvalues are boh real posiive, hence he origin is a nodal source 7 Verify ha he equilibrium poin a he origin is a cener by showing ha he real pars of he sysem s comple eigenvalues are zero Calculae and skech he vecor generaed by he RHS of he sysem a he poin (,) Use his o help skech he ellipic soluion rajecory for he sysem passing hrough he poin (, ) Draw arrows on he soluion, indicaing he direcion of moion Use you numerical solver o check your resul y y Soluion The characerisic equaion is λ +9, hence he eigenvalues are λ, ±i, hence he origin is a cener When y (, ) T, he RHS is (, ) T, easily seen from he pplane plo Calculae he eigenvalues o deermine wheher he equilibrium poin is a spiral sink or a source Calculae and skech he vecor generaed by he righ-hand side of he sysem a he poin (, ) Use his o help skech he soluion rajecory for he sysem passing hrough he poin (,) Draw arrows on he soluion, indicaing he direcion of moion Use your numerical solver o check your resul y y 5 Soluion The characerisic polynomial is λ T r(a)λ + de(a) λ λ + and he eigenvalues are λ, ± i 8

9 y Since he real par is posiive, he equilibrium poin is a spiral source The vecor generaed by RHS a is + y y 5 + y Calculae he eigenvalues o deermine wheher he equilibrium poin is a spiral sink or a source Calculae and skech he vecor generaed by he righ-hand side of he sysem a he poin (, ) Use his o help skech he soluion rajecory for he sysem passing hrough he poin (,) Draw arrows on he soluion, indicaing he direcion of moion Use your numerical solver o check your resul y y Soluion The characerisic polynomial is and he eigenvalues are λ T r(a)λ + de(a) λ + λ + λ, ± i Since he real par is negaive, he equilibrium poin, ie he origin is a spiral sink The vecor generaed by RHS a is ( ) 9

10 Secion 99 Linear sysems wih consan coefficiens: Inhomogeneous Linear Sysems Use he variaion of parameers echnique o find he general soluion o y 5 6 e y + e Soluion The general soluion is The characerisic polynomial is y y h + y p e A y + e A e As f(s)ds λ λ +, hence he eigenvalues are λ, λ, and he corresponding eigenvecors ( ( v, v ) ) The general soluion corresponding o he homogeneous par is The fundamenal mari is Y () and herefore he eponenial mari y h c e λ v + c e λ v c e + c e () e e e e, Y, Y e A Y ()Y e e e e, e + e 6e + 6e e e e e Then e As e f(s) s + e s 6e s + 6e s e s e s e s e s e s e s + e s 6 + 6e s 9 + e s e s + e s 6 5e s, and e As f(s)ds Therefore he paricular soluion is ( 9 + e s y p e A e As f(s)ds 6 5e s ) ds 9 e e 5

11 e + e 6e + 6e 9 e + e e e e 6 + 5e 5 9e e + e 6e + 5e 5e () Finally, by () and () y c e +c e 9e + e + e 6e + 5e 5e c e +(c +5)e 9e + e 6e + 5e Use he variaion of parameers echnique o find he general soluion o ( y 6 y + ) Soluion The characerisic equaions is λ λ hence he eigenvalues are λ, λ, wih corresponding eigenvecors ( ( v, v ), ) and he fundamenal mari ( e Y () e ), Y, Y Then e A Y ()Y e e 6 + 6e e e + e Now and e As f(s)ds y p () e A e As f(s)ds e s 6 + 6e s e s + e s ds Finally, y() y h ()+y p () c +c e + 5e s 6 + e s ds 5e e + e 6 + 6e 5e e e + e 6 e e 5 + 5e e c +(c +5)e + Use he echnique of undeermined coefficiens o find a paricular soluion for he sysem ( y 6 8 y + 6 ) 5 6

12 Soluion The soluion is y y h + y p, where (λ, ±) y h c e + c e Since he forcing is consan, y p mus saisfy ( ( ( 6 8 u +, ) 6) v ) hence 5 Consider he sysem y y p 5 y + sin Because he firs derivaive of he sine funcion is a cosine funcion, we migh hazard a guess ha involves boh sines and cosines Use he mehod of undeermined coefficiens and he guess y p a cos + b sin, where a (a, a ) T and b (b, b ) T o find a paricular soluion of he sysem Soluion Since y p is a paricular soluion, i has o saisfy he equaion, and herefore a sin + b cos a cos + b sin + a sin + b cos a cos + b sin sin Taking and π we obain he sysem b a, b a a, a b, a b b +, which gives a + b, a, b, and finally ( ) y p cos + sin