complete solution manual for nonlinear systems

4 There is a unique equilibrium point at the origin, which js unstable focus.. Al trajectories atartil1g inside the stable limit cycle, except the origin, approach it as i tends to infin

Zn+m-1 = Zn+m Zn+m = u 'JJ = :i:1

periodicity, 0.4067 + 2mr and 1.6398 + 2mr are also roots for n = :H,:1:2, Ea.ch root z1 = z gives an

equilibrium point (z, 0, 0.4075/ Binz)

(c) With E, = constant, the model reduces to

Z1 = X2

z2 = ~ - ix2 - ZE,sinx1

which is a pendulum equation with an input torque

• 1.9 (a) Let Z1 = t/>L, Z2 = Ve,

O= Cz => -cA-1 Bsine = G(O)Bine = 0

G(O)~O => sine=O => e=±mr, n=0,1,2,· and z=O

(c) For G(s) = 1/(,,-, + 1), take A= -1/'r, B = 1/T and C = 1 Then

1 1 z=- -z+-sme e=-z

• 1.14 The equation of motion is

where k1 , k2 , and l:3 are positive constants Let :c = v, u = F, and w = sin 8

V = mdii(Lcos8) + mg= mdt(-L8sin8) +mg= -mL8sin6-m.tA"" cosl+mg

Substituting V and H in the 8-equation yields

=: -mL26(1in.8)2 -mL2e2sin9coslJ+mgLsin6

-ml,j cod - mL21(cod)2 + mL2e2 siu6C088

= -mL2i[(sin8)3 + (cos8)2) + mgLsin9- mLficos9

6 = Ll~B) [(m + M)mgLsin9-mLcos8(F +mL.82 sin8-k1i))

.6.{~i) [(m + M)mgLsinz1 - mLcou:1(u + mL~ sinz1 - kz.))

x

ii= Ll~B) [-m2L2gsin8cos8+(l+mL2)(F+m.l.82sin9-kti)]

.6.(~i) [-m2 L2gsinz1 cosz1 +(I+ mL2)(u + mLzi sinz1 - b,.)J

-mLfcma9-mL2i(e019)2 + mL2f' sinlcosl

= v-mL2i[(ain9)~ + (co8')2J-mLftccosl

fie z: ~~B) [-mLucos9 +(I+ mL2)(mL82 sin8-kxc)]

~(:i) [-mLucosx1 +(I+ mL2)(mL~sinz1 - b 3)]

( d) Take u = constant Setting the derivatives z, = 0, we obtain z2 = z, = 0 and

0 = (m + M)u + mkL:ca cosz1

0 ;:: -mLucosz1 - k(I + mL2)xs Eliminating x3 between the two equations yields

u[(m + M)(/ + mL2) - m2 L 2 cos2 x1 ] = u 4(:ri) = 0 Since 4(x1) > 0, equilibrium can be maintained only at u = 0 Then, zs = O Thus, the system has an equilibrium set {x, =:rs= x• = O}

• 1.17 (a) Let z1 = i1, X2 = i8, and %3 = w

(b) Take "• = v = constant and "I = u

(c) 'lake v1 = v, = constant and "• = u A constant field voltage implies that (at steady state) i1 =

V1/R1 ~ 11 = constant Hence, the model reduces to ihe aec:ond-oder linear model

= L(i,) - L~ 2 ti> (11) di/ dL

= L(s,) 1 ( v - Ri) + £2(s,) a(l ef> + Lo s,/a)211

= _! [-Rz3 + Looz2za +u]

(b) The equilibrium equations a.re

Set M 1 = r, ~, = 1,., and tl = V., Then

I - (2mg(a+r}2)1/2 - Loa • v =RI

The solutions ofthis equation are giftm by the intersection of the curve (w,/k)2 with the curve t/>{wi), which

is shown in Figure 1.29 of the text From the figure, it is clear that there is only one intersection point

z2 = Pl = : (to1 -w,) = : [kiv'z1 -:z:2 - k:av'ii]

(b) The equilibrium equa.tions are

· From the second equation, we have

which is shown in Figure 1.29 of the text From the figure, it is clear that there is only one intersection point

(c) Whenµ= µ,,.z2/(km + z2 + k1zl}, the equilibrium equations are

from the fint equation, !1 = 0 or !!2 is the root of d =-µ(!!2), Since d < ma.x.2~o{µ(z2)}, the equation

d = µ(.:f2) has two roots Denote these roots by !2a and!» Subetituting z1 = 0 in the second equilibrium equation yields ~2 = :i:21 Substituting !2 = !2 in the second equilibrium equation yields

d(z21 - !2a) - µ(!!2a)!1/Y = 0 ~ z1 = Y{z21 - !2a)

since µ(:!2a} = d Similarly, substituting f2 = fn in the second equilibrium equation yields !1 = Y(:i:21

-!2111) Thus, there are three equilibriUUl points at

(Y(z21-z2.),f2c), (Y(z21-!2•),!n), and (0,z21)

-12-

(2)

0 = X1 (1 + Z2), 0 = -.:i::2 + xi + Z1X2 - xf

0 = z1(l +%2) => z1 =O or z2 = -1

%1 = 0 => 0 = -%2 + ~ => :Z:2 = 0 or X2 = 1 X2 = -1 :::> 0 = 2 - X1 - z? => Z1 = 1

There are three equilibrium points at (O, 0), (O, 1), uid (1, -1} Determine the type of eacli point using

From the second equation, z2 = 0 or x2 = 2(1 + x1),

z2 = 0 => X1 = 0 or x1 = 1 x2 = 2(1 + :r:i} => 0 = (x1 + 3):r:1 => X1 = 0 or :1:1 = -3 There are four equilibrium points at (0,0), (1,0), (0, 2), and (-3, -4) Notice that we have assumed 1 +:r:1 :/:,

O; otherwise the equation would not be well defined

(-1,-1)

Z1 = 0 => Z2 = 0, Z2 = 2 => zf ,;: 1 Z1 = 1 or - 1 There are three equilibrium points at (0,0), {1,2), and (-1,2) Determine the type of each point using linearlzaacm

There is a UDique equilibrium point at (0, 0) Determine its type by linearization

& (O.O} = -3(x1 - z2) 2 - 0.5 -1 + 3(z1 - z2)2 + 0.5 (O.O) = -0.5

The eigenvalues are -(1/4) :I: j./7/4 Hence, (0, 0) is stable focus

• 2.3 (1) The system has three equilibrium points: (O,O) is a Btable focus, {1,-1) and (-1,1) are saddle points The phase portrait is shown in Figure 2.1 The stable trajectories of the saddle form a lobe around the stable focus All trajectories inside the lobe converge to the stable focus All trajectories outside it diverge to infinity

(2) The system has three equilibrium points: (0, O) is a saddle, (0, 1) is unstable node, and (1, -1) is a stable focus The phase portrait is showu in Figure 2.2 The z2-axis is a trajectory itself since z1 (t) = 0 => z1 (t) = 0 The x2-axis is a separatrix All trajectories in the right half converge to the stable focus All trajectories in the left have diverge to infinity On the z2-a:x:is itself, trajectories starting at z 2 < 1 converge to the origin,

while trajectories starting at z2 > 1 diverge to infinity

:i:1(0 i!E O => :i1(t) = O Hence, trajectories starting in {z1 :2:: O} ltay there for all time The phaae trait ill shown in Figure 2.3 Notice that the z1-axis ia a trajectory since :i:2(t) !E O * z2(t) = O It ia a

por-separam,c that divides the half plane {z1 ~ O} into two quarters All trajectories starting in the quarter {z1 > 0, z2 > O} converge to the stable Dode (0, 2) All uajectoriea nartiDg in the quarter {z1 > 0, z2 < O}

diverge to infinity 'lrajectories starting on the z2-axis approach the stable node (0, 2) if z2(0) > 0 and

diverge to infinity if z2(0) < 0 'lrajectoriea starting on the z1•axil with z1(0) > 0 approach the addle (1,0)

(4) There is a unique equilibrium point at the origin, which js unstable focus The phase portrait is shown in Figure 2.4 There are two limit cycles The inner cycle is stable while the outer one ii unstable

Al trajectories atartil1g inside the stable limit cycle, except the origin, approach it as i tends to infinity

Trajectories starting in the region between the two limit cycles approach the stable Umit cycle Trajectories

starting outside the unstable limit cycle diverge to infinity

(5) The system has an equilibrium set at the unit circle and unstable focus at the origin The phue portrait

is shown in Figure 2.5 All trajectoriel, except the origin, approach the unit circle at t tends to infinity (8) The system has three equilibrium points: a saddle at (0,0) and stable nodes at (1,1) and (-1,-1) The phase portrait is shown in Figure 2.6 The stable trajectories of the saddle lie on the line z1 + :r2 = 0All irajectorie& to the right ol this line converge to the stable node (1, 1) and all trajectories to its left converge to the stable node (-1, -1) Trajectories on the line :r1 + z2 = 0 itself converge to the origin

x, Figure 2.3: Exercise 2.3(3)

Figure 2.4: Exercise 2.3(4)

-1

~ -1 0

x, 1 2

Figure 2.5: Exercise 2.3(5) Figure 2.6: Extrcise 2.3(6)

• 2.4 (.1) The system bu three equilibrium points at (0,0),-(a,O), and (-a,0), where a is the root of

a = tan(a/2) =t- a Aj 2.3311 The Jacobian matrix is

lk = 1 - 2/[1 + (~1 + z2)2] -2/[1 + (z1 + z2)2J

8/1 = [ 0 1 ] * A1,2 = -1, -1 /jz (O,O) -1 -2

Although we have multiple eigenvalues, we can conclude that the origin is a stable node because /(z) is an

analytic function of z in the neighborhood of the origin

:t2 3311,0) = f o.~2 -o.!10s ] => >.1,2 :r: 0.6892, -1 ~ (2.3311,0) is a saddle

Similarly, (-2.3311,0} is a saddle The phase portrait is Bhown in Figure 2.7 with the arrowheads The

stable trajectories of the two saddle points forms two aeparatrices, which divide the plane into three regions

AD trajectories in the middle region converge to the origin u t tends to infinity All trajectories in the outer

regions diverge to infioity

(2)

0 = x1(2 - x2}, 0 = 2x? -:&2 From the first equation, z1 = 0 or x2 = 2

on the :i:raxis, They form a separatrix that divides the plane in two halves 'bajectories iD the right half converge to the stable focus (1, 2) and thoee in the half converge to the stable focus (-1, 2)

(3) There is a unique equilibrium point at the origin

(0,0)

The phase portrait is shown in Figure 2 7 with the arrowheads There is a stable limit cycle a.round the

origin All trajectoriea, except the origin, approach the limit cycle u t teuds to mfinlty

(4) The equilibrium points are gmm by the real roota of the equation

0 = 1/4 - 2112 + II

where :t1 = ti and :t2 = l-11 It can be seen that the equatiou hu four roots at,:; 0, 1, (-1:v'S)/2 Bence,

there are four equilibrium points at (0, 1), (1,0), ((3 - /5)/2, (3- -v'S)/2), and ((3 + /5)/2, (3 + /5)/2)

The following table shows the Jacobian matrix and the type of each point

The phase portrait is shown in Figure 2 7 with the arrowheads The stable trajectories of the Addle points divide the plane into two regions The region that contains the stable focus has the feature that all trajectories inside it converge to the stable focus All trajectories in the other region diverge to infinity

a(z) = (zf + zl)(ln /4 +zip' /J(z) = In v'zf + 4

Noting that lims-+0X;%Jc:t(z) = 0 for i,; = 1, 2 and lim.~ /J(z) = o it can be seen that

a,, [ -1 o ] th _1 1 od

& .-o = 0 _ 1 => e ongm 18 a Hau e n e

(b) Transform the state equation into the polar coordinates

r = {zt +~, 8 = tan-1 (::)

to obtain

r = -r => r(t) = r0e-'

(c) /(z) is continuously differentiable, but not analytic, in the neighborhood of z = O See the discussion

on page 54 of the text

• 2.6 (a) The equilibrium points are the real roots of

0 = -z1 + a:r:2 -bz1z2 + zi, 0 =-(a+ b)z1 + bzf - z1z2

From the second equation we have

The eigenvalues of A are

.\ = -1 ::I: y'l - 4b(a + b)

2

The equilibrium point { ~~t:>, -J:tf>) is a stable focus if 4b(a+b) > 1, a stable node if O < 4b(a+b) <

1, and a saddle if b(a + b} < 0

The variOUI cues are summarized in the following table

(O,O) {0,-a) ( 1:"t:', -lw) )

b > 0, 44· a+b) > 1, 4b(a+b) > 1 stable focus saddle stable focus

b > 0, 4a a+b) > 1, 4b a+b )<1 stable locus saddle liable node

b > o, 4<I a+b) < 1, ~ a+b I> l stable node saddle stable focua

b > 0, 4a a+b) < 1, 4b o+ b 1<1 stable node saddle stable node

b < 0, 4 + b > 0, 4a a+b) > 1 stable fOCU8 stable node saddle

b < 0, a + b > 0, 4a a+b) < 1 stable node stable node saddle

b < 0, a+ b < 0, 4h a+b) > 1 saddle stable node stable focus

b < 0, a+ b < O, 4a a +b) < l saddle stable node stable node

If any one of the above conditions holds with equality rather than inequality, we end up with multiple eigenvalues or eigenvalues with uro real parts, in which cue linearization fails to determine the type of the equilibrium point of the DODlinear system

(c) The phase portraits of the three cases are shOWD in Figures 2.8 through 2.10

i a = 1, = 1 The equilibrium points are

in the right half aprPO&Cbing (1, -1) and all trajectories in the left half approacl,ing (0,0)

ii a= 1, b == -½, The equilibrium points are

(O, 0) stable focus

(0,-1) stable node

The linearizat.ion at tbe saddle is A, where (1 + b2)A • [ -~:~:~ ~!~:~ ] The stable eigenvector

is [ -~::: ] and the UDStable eigenvector is [ ~ : : : ] They are uaed to geuerate the stable and UJL1tab1e trajectories of the saddle The stable trajectories form a separatrix in the form of a lobe All trajectories outside the Jobe approac:h (0, -l)i all trajectories inside the lobe approach (0, 0)

ill a= 1, b: -2 The equilibrium points are

unstable eigenvector is [ ~ : : ; ] They are med to generate the stable and unstable trajectories of the saddle The stable trajectories form a separatrix m the form of a lobe All trajectories outside the lobe approach (O -1); all trajectories inside the lobe approach (l, ¼)

has a unique equilibrium point at the origin Linearization at the origin yields A = [ _ ~ ! ] whose

etgeovalues are 0.5 ¼ 0.866j Hence the origin is unstable focus The phase portrait is shown in Figure 2.11

There are three limit cyclea The inner limit cycle is stab~ the middle one is UD8ta.ble1 and the outer one is

stable AU trajectories starting inside the middle limit cycle, other than the origin, approach the inner limit

cycle as t tends to infinity AU trajectories starting outside the middle limit cycle approach the outer limit

cycle as t tends to ioftnity Trajectories starting at the unstable focus or on the unatable limit cycle remain

there

• 2.8 (a) The equilibrium points are the real roots of

O=z2,

(2,0) and (-2,0) are saddle points

(b) The phase portrait can be sketched by constructing a vector field diagram and using the

informa-tion about the equilibrium points, espedal]y the directions of the stable and unstable trajectories &t the

aaddJe point.s The stable and umtable eigenvectors of the linearization at the saddle points are

(b) At the equilibrium points, we have

From the first equation, z2 = vc,, and from the second one, z1 =(Kc+ K1v1 + K,,v~)/KJ' This is the only equilibrium point Linearization at· the equilibrium point yields

[ 0 -1 ]

A= K,/m -(K1 + 2Katid + K,,)/m whose eigenvalues a.re

(d) The eigenvalues of the linearization are -o.1s1:;:1:o.2646;; hence the equilibrium point is a stable focus

The phue portrait is lhown in Figure 2.13 All trajectories approach the stable focus Notice the inc:reued ovenboot compared with the previous cue For example, starting at the iDitial state (z1 = 15, z2 = 10),

the speed reaches about 36 m/ ,u before approaching the steady•nate of 30 m/ ,u

(e) The phue portrait is shown in Figure 2.14 The local behavior near the equilibrium point is not affected lince saturation will not be effective However, far from the equilibrium point we can aee that the state of the integrator, z1, '8.bs large values during saturation, resulting in au increaaed overshoot

• 2.10 (a) Using the same aca1ing u in Example 2.1, the state equation is giW!D by

i-1 = O.S{-h(z1) + z2), i2 = 0.2(-:1 - 0.2z:, + 0.2}

where h(z1 ) is given in Example 2.1 The equilibrium points are the intenection points of the curves

z2 = h(z1) and :2 = 1 : 52:1 , Figure 2.15 shows that there is a unique equilibrium point Uling the "roots" commUJd of MATLAB, the equilibrium point was determined to be~= (0.057,0.7151)

8/ = [ -0.5h'(z1 ) 0.5 ]

lJz -0.2 -0.04

lJ f I = [ - 1.046l _ 00·_504 ] => Eigenvalues= -0.9343, -0.1518 => stable node

In •=t -0.2

(b) The phase portrait is shown in Figure 2.16 All trajectories approach the stable node This circuit is

known as "monostable" becaUle it hu one steady-state operating point

0 20 40 90 ID 100

~ Figure 2.12: Exercise 2.9(c) Figure 2.13: Exercise 2.9(d)

O L - - - '

0 100 200 300

Figure 2.14: Exercise 2.D(e)

• 2,11 {a) Using the same sc.aling as in Example 2.1, the state equation is given by

t1 a 0.5[-h(z1) + :t,], ZJ = 0.2(-zi - 0.2z2 + 0.4) ·

where h(!:1 ) i8 given in Example 2.1 The equilibrium points are the iDienectioD points of the curve1

z2 = h(~1) and z2 s 2-5%1 Figure 2.17 sbowa that there i8 a unique equilibrium point Using the "roots" oornrnanA of MATLAB, the equilibrium point , determined to be%= (0.2582,0.7091)

:! '·=- = [ 1!t.~3

-~-~ ] => Eigell~ues = 1 7618, 0.0155 ~ uutable node (b) The phase portrait is shown iD Figure 2.18 The circuit has a stable limn cycle All trajectories, acept the amstant solution at the equilibrium point, approach the limit cycle Thia circuit is known u "anable."

L

Figure 2.15: Exercise 2.10: equilibrium point Figure 2.16: Exercise 2.10: phase portrait

,t

Figure 2.17: Exercise 2.11: equilibrium point Figure ~-18: Exercise 2.11: phase portrait

• 2.12 (a) Note that T 12 = T21 = I .J-.n1~ = _L llil = 1 and Tu = T22 = 0 _L "Jlu = L Jtii = 0 Hence the state equation is given by

z1 = h(z1)[:r2 - 2f}(z1)] ~ fi(z), z2 = h(z2)[z1 - 27J(z2)J ~ h(z)

where h(z) = '> car(wz/2) and 71(:r) = g-1(z) = (2/,r.\) tan(.-z/2) Equilibrium ,om.ta are the intersection

pointaofthecurvesz2 = 217(2:1) andz1 = 2T](z2), Notethatf1'(0) = l/ \and71'(z) = (1/.\)sec2(1rz/2) ~ 1/'> Therefore, for ,\ 5 2, the two curves intersect only at the origin (0, 0) For \ > 2, there are three intersection

points &t (0, 0), (a, a) and (-a, -a) where O < a < 1 depends on \ This fact can be seen be sketching the curves and uaiDg aymmetry; see Figure 2.19 '!he partial derivatives of Ii and /2 are given by

For ,\ < 2, the unique equilibrium point at (O, 0) is a stable node For \ > 2, the equilibrium point (0, 0) is

a saddle For A > 2 there are two other equilibrium pob:"8 at (a, a) and (-a, -a)

~JI = h(a) [ - 2,r~a) _ 2 1,(a) ] => Eigenvalues= h(a)[-2q'(a) ± 1)

Figure 2.19: Exercise 2.12: equilibrium point Figure 2.20: Exercise 2.12: phase portrait

• 2.lS (a) Using Kirchhoff's Voltage law, we obtain

The system has a Wlique equilibrium point at the origin The Jacobian at the origin is given by

:1 = f ~l ~~22:] => Eigenvalues =0.117±0.993j

21 ==O;sa :mO

Hence the origin is an UD8table focus The phase portrait is shown in Figures 2.21 and 2.22 using two different scales The system has two limit cycles The inner limit cycle is stable, while the outer one is unstable AU trajectories starting inside the outer limit cycle, except the origin, approach the inner one All trajectories starting outside the outer limit cycle diverge to infinity

• 2,14 The system is given by

where

:i:1 = z2

±2 = -kz1 - c:i:2 - 11(:r1, z2)

for lz2I > 0 for :r2 = 0&jz1 I :$ µ.mg/k

for %2 = 0&lz1I > µ.mg/k

Figure 2.21: Exercise 2.13 Figure 2.22: Exercise 2.13

For x2 > 0, the state equation is given by

:i:1 = :r2 :i:2 = -kx1 - c:r2 - µ1img

while, for x2 < 0, it iB given by

:i:1 = x2

:i:2 = -kx1 - c:r2 + µ1emg

In each half, we C&D determine the trajectories by studying the respective linear equation Let us start with

.z2 > O The linear state equation has an equilibrium point at (-µ1,mg/k,0) Shifting the equilibrium to the origin, we obtain a linear state equation with the matrix [ ~k ~c ] , whose characteristic equation is ,\2 + c,\ + k = 0, where k ~ c are positive constants The equilibrium point is a stable focus when 4k > cl

and a stable node when 4k $ d' We shall continue our discussion assuming 4k > cl 'Irajectories would tend

to spiral toward the equilibrium point (-µ1,mg/k,O) It will not actually spiral toward the point becaUBe

the equa.tion is valid only for z2 > 0 Thus, for any point in the upper half, we can solve the linear equation

to find the trajectory that should spiral toward the equilibrium point, but follow the trajectory only UJ1til it hits the :i: 1-axis For z2 < 0, we have a similar situation·e:x:cept tnat trajectories tend t.o spiral toward the

point (µ1,mg/k, 0) On the x1 -axis itself, we should distinguish between two regions If a trajectory bits the

x1-axis within the interval [-µ.mg/k,µ.mg/k), it will rest at equilibrium If it bits outside this interval,

it will have z2 '# O and will continue motion Notice that trajectories reaching the 2:1-axis in the interval

x1 > µ,mg/k will be coming from the upper half of the plane and will continue their motion into the lower

half By symmetry, trajectories reaching the :r1-axis in the interval z1 < -µ mg I k will be coming from the lower half of the plane and will continue their motion into the upper half Thus, a trajectory starting far from the origin, will spiral toward the origin, until it hits the xi-axis within the interval (-µ.mg/k, µ.mg/k]

The phase portrait is sketched in Figure 2.23

• 2.15 The solution of the state equation

to u = 1 aod the curve in the upper half corresponds to u = -1 We will refer to these curves a., the switching curves To move any point in the plane to the origin, we can switch between ±1 For example,

to move the point A to the origm, we apply u = -1 until the trajectory hits the switclling curve, then we

switch to u = 1 Similarly, to move the point B to the origin, we apply u = 1 until the trajectory hits the switching curve, then we switch to u = -1 When the trajectory reaches the origin we can keep it there by

switching to u = 0 which makes the origin an equilibrium point

From the first equation, we have x 1 = 0 or x1 = 1-az2 Substitution of :1:1 = 0 in the second equation results

in x2 = 0 Substitution of X1 = 1-a:i:2 in the second equation results in :r:2(1- az2 - x2 ) = 0 which yields

Hence, (cl;, r:f:a) is a stable node if 1 - 2b + b2 - 4ab > 0 and a st.able focus if l - 2b + b2 - 4ab < O

The phase portrait is shown in Figure 2.27 The equilibrium point (½, ½) is a stable focus that attracts all trajectories in the first quadrant except trajectories on the xi-axis or the :z:2 axis Trajectories on the x1-axis move on it approaching the saddle point at (1,0) Motion on the x1-ax:is corresponds to the case when there are no predators, iD which case the prey population settles at x1 = 1 Motion on the z2: a.xis corresponds

to the case when there are no preys, in which the case the predator population settles at :r:2: = O; i.e., the predators vanish In the presence of both preys and predators, their populations reach a balance at the equilibrium point ( ½, ½)

• 2.l'T (1) Assume & > 0 and let X1 = y, z2 - 'ii, and V(:c) = xr + z~

f(z) · VV(z) = 2ez~(l - :cf -:z:~) = 2e~(l - V)

1.5 2

x,

Figure 2.27: Exercise 2.16

Hence, J(z) · VV(z) ~ 0 for V(x) ~ 1 In particular, all trajectories starting in M = {V(z) ~ 1} stay in M

for all future time M contains only one equilibrium point at the origin Linearization at the origin yields the matrix [ ~l ! ] Bence, the origin is unstable node or unstable focus By the Poincare-Bendixson criterion, there is a periodic orbit in M

(2) Let V(z) = ~ + z~

/(z) · VV{z) = 2z~(2 - 3zf - 2x~) = 4z~(l - zf - ~) - 2z~z~ ~ 4z~(i - zf - z~)

Hence, J(z) · VV(z) ;$ 0 for~+ zi 2:: l In particular, all trajectories starting in M = {V(z) ;$ 1} stay

in M for all future time M contains only one equilibrium point at the origin Linearization at the origin yields the matrix [ ~l ! ] , whose eigenvalues are 1 and 1 Since /(z) is an analytic function of x, we

conclude that the origin iB unstable node By the Poincare-Bendixson criterion, there is a periodic orbit in M

(4) The equilibrium points are the roots of

0 = x1 + x2 - z1 max{lz1I, lx2I},

From the first equation, we have z2 = z1 (max{lxd, lz21} -1) Substitution in the second equation results in

0 = -2x1 - xi (max{lz1 I, lx2I} - 1)2 = -z1[2 + (max{lz1 I, lz2I} - 1)2) ~ :t1 = 0 ~ z2 = 0

Hence there is a unique equilibrium point at the origin Lineariz&tion at the origin yields the matrix

A = [ ~2 ! ] whose eigenvalues are 1 ± jl.4142 Hence, the origin is an unstable focus Now consider

V(x) = x? +~

VV · f = 2x1 (x1 + X2 - 1:1 max{lx1I, lx21}) + 2z2(-2x1 + x2 - x2 ma.x:{lx1I, lz21})

= 2x~ - 2:t1.z2 + 2z~ - 2(z~ + x~) max{lx1I, lz21}

= 2 [:i? Px - Jlxll~ max{lx1 I, lz2IH, where P = [ -~.S -~-5 ]

The matrix P is positive definite with maximum eigenvalue 1.5 Therefore,

VV · f ~ 2(1.Sllxll~ -Uxll~max{lz1l,lz21}] < 0, for ma.x:{lz1l,lx21} > 1.5

Thus, by choosing c large enough, VV · f will be negative on the surface {V(z) = c} Hence, all trajectories

starting in the set M = {V(x) ~ c} stay in M for all future time and M contains a single equilibrium point

which is unstable focus It follows from the Poincare-Bendixson's criterion that there is a periodic orbit in

M

• 2.18

(a)

V = z2±2 + g(z1)±1 = -x2g(z1) + X2g(x1) = 0 (b) For small c > 0, the equation V(z) = c defines a closed curve that encloses the origin Since V = 0, a

trajectory starting on the curve must remain on the curve for all t Moreover, from :t1 = z2, we see that the

trajectory can only move in the clockwise direction Hence, a trajectory starting at any point on the closed

curve V(z) = c must move around the curve until it comes back to the starting point Thus, the trajectory

is a periodic orbit

(c) Extension of (b) because V(z) = c is a closed curve for all c > 0

(d) V(z) = ½zi +G(z1) = coustant At z = (A,0), V = G(A) Thus

2 - /_A vf2[G(A)-G(y)J - Jo v'G(A) - G(y)

where we h&ve used the fact that G(z1) is an even function

(f) We can generate the trajectories using the equation in part (d) For each value of A, we solve the equation to find x 2 as a function of z1 The function G(x1) has a minimum, a maximum, or a point of

inflection at each equilibrium point of the system In particular, It has a miDimum at z1 = 0 corresponding

to the equilibrium point at the origin Starting from small values of A, the equation will have a solution

defining a closed orbit As we increase the value of A, the equation will continue to define a closed orbit until

A reaches the level of a maximum point of G(z1) Fbr values of A higher than the maximum, the curves

will not be closed Depending on the shape of G(x1 }; the equation may have multiple solutions defining

trajectories in different parts of the plane The conditions of part(c) ensure that G{xi) will have a global minimum at z1 = 0

• 2.19 The phase portraits can be generated by solving the equation of the previous exercise either icaJly or using a computer We will only give the function G(xi) and calculate the period of the trajectory through (1,0)

G(y) = lo" zs dz = is,4

G(y) -+ oo as 1111 -+ oo and zg(z) ~ 0 for all z Hencet every solution is periodic

By Bendixson 's criterion, there are no periodic orbits

(2) The equilibrium points are the roots of

0 = z1(-l + xf + x:), 0 = :z:2(-l +x~ + z;)

The system has an isolated equilibrium point at the origin and a continuum of equilibrium points on the

unit circle ~ + ~ = 1 It can be checked that the origin is a stable node 'Iransform the system into the

polar coordinates z1 = r oos 8, x2 = r sin 9 It can be verified that

(6) The equilibrium points are the roots of

The equilibrium point are (±mr,O) for n = 0, 1,2, · · · Linearization at the equilibrium points yields the matrix ( ~ ~ ] where a = ±1 Hence, all equilibrium points are saddles Since, by Corollary 2.1, a periodic orbit must enclose equilibrium points such that N - S = 1, we conclude that there are no periodic orbits

Hence, trajectories on the boundary of D must move into D, which shows that trajectories starting in D

cannot leave it

closed orbits through any point in D

• 2.22 (a) The value of :i:2 on the zi-axis ia .i'2 = l,zf ~ 0 Thus, trajectories starting in D cannot leave it

Hence, the origin is a stable focus for µ < 0 and unstable focus for µ > O The phase portrait for different

values of µ is shown in Figure 2.29 For µ < O, there is a stable focus at the origin and unstable limit cycle around the origin The size of the limit cycle shrinks as µ tends to zero For µ > 0, the origin is an unstable focus and the limit cycle disappears Hence, there is a subcritical Hopf bifurcation atµ= O

(3) The equilibrium points are the real roots of

There is a saddle-node bifurcation at µ = O

(4) The equilibrium points a.re the real roots of

0 = X2, 0 = -(1 + µ.2).z1 + 2µ:t2 + JJ,X~ - X~X2

• 2.24 Suppose M does not contain an equilibrium point Then, by the Poincar~Bendixson criterion, there

is a periodic orbit in M But, by Corollary 2.1, the periodic orbit must contain an equilibrium point: A contradiction Thus, M contains an equilibrium point

• 2.25 Verifying Lemma 2.3 by examining the vector fields is simple, but requires drawing several sketches Hence, it is skipped

• 2.26

(1) Linearization at the origin yields [ ~ ~l ] Hence, the origin is not hyperbolic The index of the origin is zero This can be ea.sily seen by noting that Ji = xf is always nonnegative Clearly, the vector field cannot make a full rotation as we encircle the origin because this will require Ji to be negative

(2) Linearization at the origin yields [ ~ ~ ] Hence, the origin is not hyperbolic The index of the origin

is two This can be seen by sketching the vector field along a closed curve around the origin

• 2.27 (1) The equilibrium points are the real roots of

Thus, there is superaitical pitchfork bifurcation at µ = 0

(2) The equilibrium points are the real roots of

0 = -zf + z2, 0 = -(1 + µ2)x1 + 2µz2 - µx: + 2(z2 - µzi) 3

x2 = x: ~ 0 = X1 {-1 + (xJ -µ)[µ + 2zUx~ - µ)21}

For all values ofµ, there is an equilibrium point at (O, 0) Atµ= 0, there are two other equilibrium points

at (a, a3 ) and (-a, -a3 ), where a8 = 0.5 It can be checked that these two equilibrium points are saddles

By continuous dependence of the roots of a polynomial equation on its parameters, we see that there i! a

range of values of µ around zero for which these two saddle points will persist We will limit our attention

to such values of µ and study local bifurcation at µ = O

{J:z (0,0) = -(1 + 1,i) 2µ => \1,2 = µ ± J

Hence, the origin is a stable focus for µ < 0 and unstable focus for µ > O The phase portrait for different

values of µ is shown in Figure 2.29 For µ < O, there is a stable focus at the origin and unstable limit cycle around the origin The size of the limit cycle shrinks as µ tends to zero For µ > 0, the origin is an unstable focus and the limit cycle disappears Hence, there is a subcritical Hopf bifurcation atµ= O

(3) The equilibrium points are the real roots of

There is a saddle-node bifurcation at µ = O

(4) The equilibrium points a.re the real roots of

0 = X2, 0 = -(1 + µ.2).z1 + 2µ:t2 + JJ,X~ - X~X2

X2 = 0 => 0 = -(1 + µ2)2:1 + µx~

For all values ofµ, there is an equilibrium point at (0, 0) For µ > 0 there are two other equilibrium points

at (a,O) and (-a,O), where a= v'(l + µ2}/µ

Fig-µ increases, until they reach ±2 atµ= 1 Then they move again toward infinity For p = 0.2, the phase portrait is shown in a larger area that includes the saddle points

0.25 < µ < 0.933 Unstab]e focus

0.933 < µ Unstable node Thus, there is a supercritical pitchfork bifurcation atµ = 0 We also examine /J = 0.25, where the equi-librium points (y'ii,0) and (-vP",0) change from stable focus to unstable focus The phase portraits for

µ = 0.24 and JJ = 0.26 are shown in Figure 2.31 As µ crosses 0.25 new stable limit cycles are created around

the points (.Jji,O) and (-v'Ji,O) Thus, there is a supercritical Hopf bifurcation atµ= 0.25

-x1

0.5 1

Hence, the origin is a stable node forµ< 0 and a saddle forµ> D

:!t,,,O) = [ ~µ -(,/+ l) ] => \1,2 = -1, -µ

Hence, (µ, O) is a saddle forµ < 0 and a stable node forµ > 0 There is a tra.nscritical bifurcation atµ= O

• 2.28 (a) The equilibrium points are the real roots of

Clearly there is an equilibrium point at the origin (O, 0) By plotting the two curves for different values of AT

(see Figure 2.32), it can be seen that the origin is the only intersection point In fact, the two curves touch each other asymptotically as AT ~ oo Thus, we conclude that the origin is the only equilibrium point Next

we use linearization to determine the type of the equilibrium point

Choosing r = c > 4-r, we conclude that on the circler== c, r < o Hence, vector fields on r = c point to the

inside of the circle Thus, the set M = {r $ c} has the property that every trajectory starting in M stays

in M lot all future time Moreover, Mis closed, bounded, and cont.aim only one equilibrium point which is

unstable focus By the Poincare-Beodixson criterion, we conclude that there is a periodic orbit in M

(b) The phase portrait is shown in Figure 2.33 The origin is an unstable focus and there is a stable limit cycle around it All trajectories, except the trivial solution z = 0, approach the limit cycle asymptoti-

From the second equation we have x1 :::: 0 or x2 = 1 + xf The first equation cannot be satisfied with z1 = 0

Substitution of x2 = 1 + xf in the first equation results in x1 = a/5 Thus, there is a unique equilibrium point at ((a/5),_1 + (a/5)2) The Jacobian matrix is given by

8/

- 1+:!' 1

&x ((a/S),l+(a/S)3) - 1 + (a/5)2 2b(a/5)2 -b(a/5) - 1 + (a/5)2