systems of nonlinear differential equations chapter 11
TRANSCRIPT
Systems of Nonlinear Differential Equations
CHAPTER 11
Ch11_2
Contents
11.1 Autonomous Systems11.2 Stability of Linear Systems11.3 Linearization and Local Stability11.4 Autonomous Systems as Mathematical Models11.5 Periodic Solutions, Limit Cycles, and Global Sta
bility
Ch11_3
11.1 Autonomous Systems
IntroductionA system of first-order differential equations is called autonomous, when it can be written as
(1)
),,,(
),,,(
),,,(
21
2122
2111
nnn
n
n
xxxgdt
dx
xxxgdt
dx
xxxgdtdx
Ch11_4
Example 1
The above is not autonomous, since the presence of t on the right-hand side.
)sin(
3
212
221
1
txxdt
dx
txxdtdx
Ch11_5
Example 2
Consider
If we let x = , y = , then
is a system of first-order.
0sin2
2
lg
dt
d
xlg
y
yx
sin
Ch11_6
Vector Field Interpretation
A plane autonomous system can be written as
The vector V(x, y) = (P(x, y), Q(x, y)) defines a vector field of the plane.
),(
),(
yxQdtdy
yxPdtdx
Ch11_7
Example 3
A vector field for the steady-state flow of a fluid around a cylinder of radius 1 is given by
where V0 is the speed of the fluid far from the cylinder.
222222
22
0 )(
2,
)(1),(
yx
xy
yx
yxVyxV
Ch11_8
Example 3 (2)
If a small cork is released at (−3, 1), the path X(t) = (x(t), y(t)) satisfies
subject to X(0) = (−3, 1). See Fig 11.1.
2220
222
22
0
)(
2
)(1
yx
xyV
dtdy
yx
yxV
dtdx
Ch11_9
Fig 11.1
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Types of Solutions
(i) A constant solution x(t) = x0, y(t) = y0 (or X(t) = X0 for all t). The solution is called a critical or stationary point, and the constant solution is called an equilibrium solution. Notice that X(t) = 0 means
0),(
0),(
yxQ
yxP
Ch11_11
(ii) A solution defines an arc – a plane curve that does not cross itself. See Fig 11.2(a). Referring to Fig11.2(b), it can not be a solution, since there would be two solutions starting from point P.
Fig 11.2
Ch11_12
(iii) A periodic solution – is called a cycle. If p is the period, then X(t + p) = X(t). See Fig 11.3.
Ch11_13
Example 4
Find all critical points of the following:(a) (b) (c)
Solution (a)
then y = x. There are infinitely many critical points.
yxy
yxx
0
0
yx
yx
yxy
yxx
2
22 6
)2.060(05.0
)100(01.0
xyyy
yxxx
Ch11_14
Example 4 (2)
(b)
Since x2 = y, then y2 + y – 6 = (y + 3)(y – 2) = 0. If y = – 3, then x2 = – 3, there are no real solutions. If y = 2, then . The critical points are and .
0
062
22
yx
yx
2x )2,2()2,2(
Ch11_15
Example 4 (3)
(c)From 0.01x(100 – x – y) = 0, we have x = 0 or x + y = 100. If x = 0, then 0.05y(60 – y – 0.2x) = 0 becomes y(60 – y) = 0. Thus y = 0 or y = 60, and (0, 0) and (0, 60) are critical points.If x + y = 100, then 0 = y(60 – y – 0.2(100 – y)) = y(40 – 0.8y). We have y = 0 or y = 50. Thus (100, 0) and (50, 50) are critical points.
Ch11_16
Example 5
Determine whether the following system possesses a periodic solution. In each case, sketch the graph pf the solution satisfying X(0) = (2, 0).(a) (b)
Solution(a) In Example 6 of Section 10.2, we have shown
yxy
yxx
2
82
yxy
yxx
2/1
2
tctcy
ttcttcx
2sin)2cos(
)2sin22cos2()2sin22cos2(
21
21
Ch11_17
Example 5 (2)
Thus every solution is periodic with period . The solution satisfying X(0) = (2, 0) is
x = 2 cos 2t + 2 sin 2t, y = – sin 2tSee Fig 11.4(a).
Ch11_18
Example 5 (3)
(b) Using the similar method, we have
Since the presence of et, there are no periodic solutions.The solution satisfying X(0) = (2, 0) is
See Fig 11.4(b).
)cos()sin(
)sin2()cos2(2
1
21
tectecy
tectecxtt
tt
teytex tt sin,cos2
Ch11_19
Fig 11.4(b)
Ch11_20
Changing to Polar Coordinates
Please remember that the transformations are
r2 = x2 + y2 and = tan–1(y/x),
dtdy
xdtdx
yrdt
ddtdy
ydtdx
xrdt
dr2
1,
1
Ch11_21
Example 6
Find the solution of the following system
satisfying X(0) = (3, 3).
Solution
22
22
yxyxy
yxxyx
2)]()([1
ryrxyxryxrdt
dr
1)]())([(12 yrxxxryy
rdtd
Ch11_22
Example 6 (2)
Since (3, 3) is in polar coordinates, then X(0) = (3, 3) becomes and (0) =π/4.Using separation of variables, we have the solution is
for r 0. Applying the initial conditions, we have
)4/,23( 23)0( r
21
,1
ctct
r
4,
6/21
t
tr
Ch11_23
Example 6 (3)
The graph of
is shown in Fig 11.5.
4/6/21
r
Ch11_24
Fig 11.5
Ch11_25
Example 7
Consider the system in polar coordinates:
Find and sketch the solutions satisfying X(0) = (0, 1) and X(0) = (3, 0) in rectangular coordinates.
Solution By separation of variables, we have
1 ),3(5.0 dtd
rdtdr
25.0
1 ,3 ctecr t
Ch11_26
Example 7 (2)
If X(0) = (0, 1), then r(0) = 1 and (0) = /2. Thus c1 = –2, c2 =/2. The solution curve is the spiral
. Notice that as t → ， increases without bound and r approaches 3.
If X(0) = (3, 0), then r(0) = 3 and (0) = 0. Thus c1 = c2 = 0 and r = 3, = t. We have the solution is x = r cos = 3 cos t and y = r sin = 3 sin t. It is a periodic solution. See Fig 11.6.
)2/(5.023 er
Ch11_27
Fig 11.6
Ch11_28
11.2 Stability of Linear Systems
Some Fundamental QuestionsSuppose X1 is a critical point of a plane autonomous system and X = X(t) is a solution satisfying X(0) = X0. Our questions are when X0 is near X1:
(i) Is limt X(t) = X1?
(ii) If the answer of (i) is “no”, does it remain close to X1 or move away from X1?
See Fig 11.7
Ch11_29
Fig 11.7
Ch11_30
Referring to Fig11.7(a) and (b), we call the critical point locally stable.
However, if an initial value results in behavior similar to (c) can be found in nay given neighborhood, we call the critical point unstable.
Ch11_31
Stability Analysis
Considerx = ax + by y = cx + dy
We have the system matrix as
To ensure that X0 = (0, 0) is the only critical point, we assume the determinant = ad – bc 0.
dc
baA
Ch11_32
Then det (A – I) = 0 becomes2 − + = 0
where = a + d. Thus
2/)4( 2
Ch11_33
Example 1
Find the eigenvalues of the system
in terms of c, and use a numerical solver to discover the shape of solutions corresponding to the case c = ¼ , 4, 0 and −9.
ycxy
yxx
Ch11_34
Example 1 (2)
Solution Since the coefficient matrix is
then we have = −2, and = 1 – c. Thus
1
11
c
cc 1
2)1(442
Ch11_35
Example 1 (3)
If c = ¼ , = −1/2 and −3/2. Fig 11.8(a) shows the phase portrait of the system.
When c = 4, = 1 and 3. See Fig 11.8(b).
Ch11_36
Example 1 (4)
When c = 0, = −1. See Fig 11.8(c).When c = −9, = −1 3i. See Fig 11.8(d).
Ch11_37
Case I: Real Distinct Eigenvalues
According to Sec 10.2, the general solution is
(a) Both eigenvalues negative: Stable NodeIt is easier to check that under this condition,
X(t) 0 as t See Fig 11.9.
)(
)()(
2211
2211
121
21
tt
tt
ecce
ecect
KK
KKX
Ch11_38
Fig 11.9
Ch11_39
(b) Both eigenvalues positive: Unstable NodeIt is easier to check that under this condition,
|X(t)| becomes unbounded as t See Fig 11.10
Ch11_40
Fig 11.10
Ch11_41
(c) Eigenvalues have opposite signs (2 < 0 < 1): Saddle Point
On if c1 = 0, will approach 0 along the line determined by
K2 as t . This unstable solution is called a saddle point. See Fig 11.11.
Ch11_42
Fig 11.11
Ch11_43
Example 2
Classify the critical point (0, 0) of each system X = AX as either a stable node, an unstable node, or a saddle point.(a) (b)
Solution (a) Since the eigenvalues are 4, −1, (0, 0) is a saddle point. The corresponding eigenvectors are respectively
12
32A
1915
610A
2
31K
1
12K
Ch11_44
Example 2 (2)
If X(0) lies on the line y = −x, then X(t) approaches 0. For any other initial conditions, X(t) will become unbounded in the direction determined by K1. That is, y = (2/3)x serves an asymptote. See Fig 11.12.
Ch11_45
Fig 11.12＞
Ch11_46
(b) Since the eigenvalues are − 4, −25, (0, 0) is a stable node. The corresponding eigenvectors are respectively
See Fig 11.13.
1
11K
5
22K
Ch11_47
Fig 11.13
Ch11_48
Case II: A Repeated Real Eigenvalue
According to Sec 10.2, we have the following conditions.
(a) Two linearly independent eigenvectorsThe general solution is
If 1 < 0, then X(t) approaches 0 along the line determined by c1K1 + c2K2 and the critical point is called a degenerate stable node. Fig 11.14(a) shows the graph for 1 < 0 and the arrows are reversed when 1 > 0, and is called a degenerate unstable node.
ttt eccecect 111 )()( 22112211 KKKKX
Ch11_49
Fig 11.14
Ch11_50
(b) A single linearly independent eigenvectorWhen we only have a single eigenvector, the general solution is
If 1 < 0, then X(t) approaches 0 in one of directions determined by K1(See Fig 11.14(b)). This critical point is again called a degenerate stable node.If 1 > 0, this critical point is again called a degenerate unstable node.
)(
))(
21
112
1211
1
111
PKK
P(KKX
tc
tc
cte
etecect
t
ttt
Ch11_51
Case III: Complex eigenvalues (2 – 4 < 0)
(a) Pure imaginary roots (2 – 4 < 0, = 0)We call this critical point a center. See Fig 11.15
Ch11_52
(b) Nonzero real part (2 – 4 < 0, 0)real part > 0: unstable spiral point (Fig 11.16(a))real part < 0: stable spiral point (Fig 11.16(b))
Ch11_53
Example 3
Classify the critical point (0, 0) of each system
Solution (a) The characteristic equation
2 + 6 + 9 = ( + 3)2= 0 so (0, 0) is a degenerate stable node.
(b) The characteristic equation 2 + 1 = 0
so (0, 0) is a center.
11
21 (b)
92
183 (a) AA
Ch11_54
Example 4
Classify the critical point (0, 0) of each system
for positive constants.
Solution (a) = −0.01, = 2.3789, 2 − 4 < 0: (0, 0) is a stable spiral point.
ydycd
xabxa
ˆˆ
ˆˆ (b)
02.110.1
10.301.1 (a) AA
Ch11_55
Example 4 (2)
(b)
spiral stableor
stable, degenerate satble,either :0,1 if
point saddle a:0,1 if
)1(ˆˆ
,0)ˆˆ(
bc
bc
bcyxad
ydxa
Ch11_56
For a linear plane autonomous system X’ = AX with det A 0,
let X = X(t) denote the solution that satisfies the initial condition
X(0) = X0, where X0 0.
(a) limt→X(t) = 0 if and only if the eigenvalues of A have negative real parts. This will occur when ∆ > 0 and < 0.
(b) X(t) is periodic if and only if the eigenvalues of A are pure imaginary. This will occur when ∆ > 0 and = 0.
(c) In all other cases, given any neighborhood of the region, there is at least one X0 in the neighborhood for which X(t)becomes unbounded as t increases.
THEOREM 11.1Stability Criteria for Lonear Systems
Ch11_57
11.3 Linearization and Local Stability
Let X1 be a critical point of an autonomous system, and let X = X(t) denote the solution that satisfies the initial condition X(0) = X0, where X0 X1. We saythat X1 is a stable critical point when, given any radiusρ > 0, there is a corresponding radius r > 0 such that if the initial position X0 satisfies │X0 – X1│< r, thenthe corresponding solution X(t) satisfies │X(t) – X1│< ρ for all t > 0. If, in addition limt→X(t) = X1 whenever │X0 – X1│< r, we call X1 an asymptotically stable critical point.
DEFINITION 11.1Stable Critical Points
Ch11_58
This definition is shown in Fig 11.20(a). To emphasize that X0 must be selected close to X1, we also use the terminology locally stable critical point.
Ch11_59
Let X1 be a critical point of an autonomous system, and let X = X(t) denote the solution that satisfies the initial condition X(0) = X0, where X0 X1. We saythat X1 is an unstable critical point if there is a diskof radius ρ > 0 with the property that, for any r > 0, there is at least one initial position X0 satisfies│X0 – X1│< r, yet the corresponding solution X(t) satisfies │X(t) – X1│ ρ for all t > 0.
DEFINITION 11.2Unstable Critical Points
Ch11_60
If a critical point X1 is unstable, no matter how small the neighborhood about X1, an initial position X0 can be found that the solution will leave some disk at some time t. See Fig 11.20(b).
Ch11_61
Example 1
Show that (0, 0) is a stable critical point of the system
Solution In Example 6 of Sec 11.1, we have shown
r = 1/(t + c1), = t + c2 is the solution. If X(0) = (r0, 0), then
r = r0/(r0 t + 1), = t +0
Note that r < r0 for t > 0, and r approaches (0, 0) as t increase. Hence the critical point (0, 0) is stable and is in fact asymptotically stable. See Fig 11.21.
22
22
'
'
yxyxy
yxxyx
Ch11_62
Fig 11.21
Ch11_63
Example 2
Consider the plane system
Show that (x, y) = (0, 0) is an unstable critical point.
1
)3(05.0
dtd
rrdtdr
Ch11_64
Example 2 (2)
SolutionSince x = r cos and y = r sin , we have
Since dr/dt = 0.05r(3-r), then r = 0 implies dr/dt = 0. Thus when r = 0, we have dx/dt = 0, dy/dt = 0. We conclude that (x, y) = (0, 0) is a critical point.
sincos
cossin
dtdr
dtd
rdtdy
dtdr
dtd
rdtdx
Ch11_65
Example 2 (3)
Solving the given differential equation with r(0) = r0 and r0 0, we can have
No matter how close to (0, 0) a solution starts, the solution will leave (0, 0). Thus (0, 0) is an unstable critical point. See Fig 11.22.
ec
rrc
ecr
tt
t
31
3lim
Since ./)3( where
1
3
15.00
000
15.00
Ch11_66
Fig 11.22
Ch11_67
Linearization
If we write the systems in Example 1 and 2 as X = g(X). The process to find a liner term A(X – X1) that most closely approximates g(X) is called linearization.
Ch11_68
Let x1 be a critical point of the autonomous differentialequation x = g(x), where g is differentiable at x1.(a) If g(x1) < 0, then x1 is an asymptotically stable critical point.(b) If g(x1) > 0, then x1 is an unstable critical point.
THEOREM 11.2Stability Criteria for Linear Systems
Ch11_69
Example 3
predict the behavior of solutions near these two critical points. Since
Therefore x = /4 is an asymptotically stable critical point but x = 5/4 is unstable. See Fig 11.23.
.sincos'
of points critical are 4
5 and
4Both
xxx
xx
02)45(',02)4('
cossin)('
gg
xxxg
Ch11_70
Fig 11.23
Ch11_71
Example 4
Without solving explicitly, analyze the critical points of the system x = (r/K)x(K – x), where r and K are positive constants.
SolutionWe have two critical points x = 0 and x = K. Since
Therefore x = K is an asymptotically stable critical point but x = 0 is unstable.
0)(',0)0('
)2()('
rKgrg
xKKr
xg
Ch11_72
Jacobian Matrix
An equation of the tangent plane to the surface z = g(x, y) at X1 = (x1, y1) is
Similarly, when X1 = (x1, y1) is a critical point, then P (x1, y1) = 0, Q (x1, y1) = 0.
We have
)()(),( 1),(1),(11 1111yy
yg
xxxg
yxgz yxyx
)()(),('
)()(),('
1),(1),(
1),(1),(
1111
1111
yyyQ
xxxQ
yxQy
yyyP
xxxP
yxPx
yxyx
yxyx
Ch11_73
The original system X = g(X) may be approximated by X = A(X – X1), where
This matrix is called the Jacobian Matrix at X1 and is denoted by g(X1).
),(),(
),(),(
1111
1111
yxyx
yxyx
yQ
xQ
yP
xP
A
Ch11_74
Let X1 be a critical point of the autonomous differentialequation X’ = g(X), where P(x, y) and Q(x, y) havecontinuous first partials in a neighborhood of X1.(a) If the eigenvalues of A = g’(X1) have negative real part, then X1 is an asymptotically stable critical point.(b) If A = g’(X1) has an eigenvalue with positive real part, then X1 is an unstable critical point.
THEOREM 11.3Stability Criteria for PlaneAutonomous Systems
Ch11_75
Example 5
Classify the critical points of each system.(a) x’ = x2 + y2 – 6 (b) x’ = 0.01x(100 – x – y) y’ = x2 – y y’ = 0.05y(60 – y – 0.2x)
Solution(a)
12
22)('
),2 ,2( and )2 ,2( are points critical The
x
yxXg
Ch11_76
Example 5 (2)
Since the determinant of A1 is negative, A1 has a positive real eigenvalue. Therefore unstable. A1 has a positive determinant and a negative trace. Both the eigenvalues have negative real parts. Therefore
is stable.
122
422))2 ,2(('
122
422))2 ,2(('
2
1
gA
gA
)2,2(
)2,2(
Ch11_77
Example 5 (3)
(b) The critical points are (0, 0), (0, 60), (100, 0), (50, 50). The Jacobiam matrix is
36.0
04.0))60 ,0(('
30
01))0 ,0(('
)2.0260(05.001.0
01.0)2100(01.0)('
2
1
gA
gA
Xgxyy
xyx
Ch11_78
Example 5 (4)
Checking the signs of the determinant and trace of each matrix, we conclude that (0, 0) is unstable; (0, 60) is unstable; (100, 0) is unstable; (50, 50) is stable.
5.25.0
5.05.0))50,50(('
20
11))0,100(('
4
3
gA
gA
Ch11_79
Example 6
Classify each critical point of the system in Example 5(b).
Solution For the matrix A1 corresponding to (0, 0), = 3, = 4, 2 – 4 = 4. Therefore (0, 0) is an unstable node. The critical points (0, 60) and (100, 0) are saddles since < 0 in both cases. For A4, > 0, < 0, (50, 50) is a stable node.
Ch11_80
Example 7
Consider the system x + x – x3 = 0. We havex = y,
y = x3 – x. Find and classify the critical points.
Solution
(-1,0). 0), (1,0), (0,
are points critical the,0)1( 23 xxxx
Ch11_81
Example 7 (2)
The corresponding matrices are
doubt.in is 0) (0, of status
theand are of seigenvalue The points.
saddleboth are 0) (-1, and 0) (1, ,0det Since
02
10))0,1(('))0,1(('
01
10))0,0(('
1
2
2
1
i
A
A
ggA
gA
Ch11_82
Example 8
Use the phase-plane method to classify the sole critical point (0, 0) of the system
x = y2
y = x2
Solution The determinant of the Jacobian matrix
is 0 at (0, 0), and so the nature of (0, 0) is in doubt.
02
20)('
x
yXg
Ch11_83
Example 8 (2)
Using the phase-plane method, we get
Fig 11.26 shows a collection of solution curves. The critical point (0, 0) is unstable.
.or ), ,0((0) If
or ,
//
3 30
330
330
3322
2
2
yxyyxyy
cxydxxdyy
y
xdtdxdtdy
dxdy
X
Ch11_84
Fig 11.26
Ch11_85
Example 9
Use the phase-plane method to determine the nature of the solutions to x + x − x3 = 0 in a neighborhood of (0, 0).
Solution
yxx
dtdxdtdy
dxdy
xxdtdyydtdx
3//
.3/ then /let weIf3
3
.2
)1( thus
,242
or ,)3(
0
222
2423
cx
y
cxxy
dxxxydy
Ch11_86
Example 9 (2)
Note that y = 0 when x = −x0 and the right-hand side is positive when −x0 < x < x0. So each x has two corresponding values of y. The solution X = X(t) that satisfies X(0) = (x0, 0) is periodic, and (0, 0) is a center. See Fig 11.27.
2))(2(
2)1(
2)1(
,2
)1(then ,10 ),0((0) If
220
20
2220
222
220
000
xxxxxxy
xcx,x
X
Ch11_87
Fig 11.27
Ch11_88
11.4 Autonomous Systems as Mathematical Models
Nonlinear Pendulum Consider the nonlinear second-order differential equation
When we let x = , y = , then we can write
0sin2
2
lg
dt
d
xlg
y
yx
sin
Ch11_89
The critical points are (k, 0) and the Jacobian matrix is
If k = 2n + 1, < 0, and so all critical points ((2n +1), 0) are saddle points. Particularly, the critical point (, 0) is unstable as expected. See Fig 11.28.
0)1(
10))0,((' 1
lgk kg
Ch11_90
Fig 11.28
Ch11_91
When k = 2n, the eigenvalues are pure imaginary, and so the nature of these critical points remains in doubt. Since we assumed that there are no damping forces, we expect that all the critical points ((2n, 0) are centers. From
)cos(cos2
then ),0 ,((0) If
cos2
then ,sin
//
02
0
2
xxlg
yx
cxlg
yy
ylg
dtdxdtdy
dxdy
X
Ch11_92
Note that y = 0 when x = −x0, and that (2g/l)(cos x – cos x0) > 0 for |x| < |x0| < . Thus each such x has two corresponding values of y, and so the solution X = X(t) that satisfies X(0) = (x0, 0) is periodic. We may conclude that (0, 0) is a center. See Fig 11.29.
Ch11_93
Fig 11.29
Ch11_94
Example 1
A pendulum is an equilibrium position with = 0 is given an initial velocity of 0 rad/s. Determine under what conditions the resulting motion is periodic.
SolutionThe initial condition is X(0) = (0, 0).
)2
1(cos2
thatfollowsit ),(cos2
From
20
2
2
gl
xlg
y
cxlg
y
Ch11_95
Example 1 (2)
To establish that the solution X(t) is periodic it is sufficient to show that there are two x-intercepts x = x0 between − and and that the right-hand side is positive for |x| < |x0|. Each such x then has two corresponding values of y.
If y = 0, cos x = 1 – (l/2g)02, and this equation has two
solutions x = x0 between − and , provided 1 – (l/2g)0
2 > −1. Note that (l/2g)(cos x – cos x0) is positive for |x| < |x0|. The restriction on the initial angular velocity may be written as
lg
20
Ch11_96
Nonlinear Oscillations: The Sliding Bead
See Fig 11.30. The tangential force F has the magnitude mg sin , thus Fx = − mg sin cos . Since tan = f (x), then
')]('[1
)('" :law sNewton' From
, and , force damping a is thereAssume
)]('[1
)('cossin
2
2
xxf
xfmgmx
dtdx
DD
xf
xfmgmgF
x
x
Ch11_97
and the corresponding plane autonomous system is
If X1 = (x1, y1) is a critical point, then y1 = 0 and f (x1) = 0. The bead must be at rest at a point on the wire where the tangent line is horizontal.
ymxf
xfgy
yx
2)]('[1
)(''
'
Ch11_98
The Jacobian matrix at X1 is
).("44 ),(" ,
so and ,)("
10)('
12
22
1
11
xgfm
xgfm
mxgf
Xg
Ch11_99
We can make the following conclusions.
(i) f ”(x1) < 0 :A relative maximum occurs at x = x1 and since < 0, an unstable saddle point occurs at X1 = (x1, 0).
(ii) f ”(x1) > 0 and > 0:A relative minimum occurs at x = x1 and since < 0 and > 0, X1 = (x1, 0) is a stable critical point. If 2 > 4gm2f (x1), the system is overdamped and the critical point is a stable node.
Ch11_100
If 2 < 4gm2f (x1), the system is underdamped and the critical point is a stable spiral point. The exact nature of the stable critical point is still in doubt if 2 = 4gm2f (x1).
(iii) f ”(x1) > 0 and the system is undamped ( = 0):In this case the eigenvalues are pure imaginary, but the phase plane method can be used to show that the critical point is a center. Thus solutions with X(0) = (x(0), x(0)) near X1 = (x1, 0) are periodic.
Ch11_101
Example 2
A 10-gram bead slides along the graph z = sin x. The relative minima at x1 = −/2 and x2 = 3/2 are stable critical points. See Fig 11.31.
Ch11_102
Fig 11.32
Fig 11.32 shows the motions when the critical points are stable spiral points.
Ch11_103
Fig 11.33
Fig 11.33 shows a collection of solution curves for the undamped case.
Ch11_104
Lotka-Volterra Predator-Prey Model
Recall the predator-prey model:
.0/
/0))/,/(('
and 0
0))0,0(('
then),( and )0,0( are points critical The
)('
)('
2
1
bac
cbdbacd
d
a
d/c,a/b
dcxydycxyy
byaxbxyaxx
gA
gA
Ch11_105
Fig 11.34
The critical point (0, 0) is a saddle point. See Fig 11.34.
Ch11_106
Since A2 has the pure imaginary eigenvalues, the critical point may be a center. Since
0
1
))((
or ,lnln
,
then,)()(
ceyex
cxdcxbyya
dyx
dcxdy
ybya
byaxdcxy
dxdy
byacxd
Ch11_107
Fig 11.35
Typical graphs are shown in Fig 11.35.
Ch11_108
1. If y = a/b, the equation F(x)G(y) = c0 has exactly two solutions xm and xM that satisfy xm < d/c < xM.
2. If xm < x1 < xM and x = x1, then F(x)G(y) = c0 has exactly two solutions y1 and y2 that satisfy y1 < a/b < y2.
3. If x is outside the interval [xm, xM], then F(x)G(y) = c
0 has no solutions.
The graph of a typical periodic solution is shown in Fig 11.36.
Ch11_109
Fig 11.36
Ch11_110
Example 3
If we let a = 0.1, b = 0.002, c = 0.0025, d = 0.2, the critical point in the first quadrant is (d/c, a/b) = (80, 50), and we know it is a center. See Fig 11.37.
Ch11_111
Fig 11.37
Ch11_112
Lotka-Volterra Competition Model
Consider the model:
This system has critical points at (0, 0), (K1, 0) and (0, K2).
(1) )('
)('
2122
2
1211
1
xyKyKr
y
yxKxKr
x
Ch11_113
Example 4
Consider the model
Find and classify all critical points.
Solution Critical points are (0, 0), (50, 0), (0, 100), (20, 40). Since 1221 = 2.25 > 1, and so the critical point (20, 40) is a saddle point. The Jacobian matrix is
)0.3100(001.0'
)75.050(004.0'
xyyy
yxxx
Ch11_114
Example 4 (2)
05.00
15.02.0))0,50(('
1.00
02.0))0,0(('
003.0002.01.0003.0
003.0003.0008.02.0)('
2
1
gA
gA
Xgxyy
xyx
1.03.0
01.0))100,0(('
04.006.0
12.008.0))40,20(('
4
3
gA
gA
Ch11_115
Example 4 (3)
Therefore (0, 0) is unstable, whereas both (50, 0) and (0, 100) are stable nodes and (20, 40) is a saddle point.
Ch11_116
11.5 Periodic Solutions, Limit Cycles and Global Stability
If a plane autonomous system has a periodic solutionX = X(t) in a simply connected region R, then the systemhas at least one critical point inside the correspondingsimple closed curve C. If there is a single critical point inside C, then that critical point cannot be a saddle point.
THEOREM 11.4Cycles and Critical Points
If a simply connected region R either contains no criticalpoints of a plane autonomous system or contains a single saddle point, then there are no periodic solutions in R.
COROLLARY
Ch11_117
Example 1
Show that the plane autonomous system x’ = xyy’ = −1 – x2 – y2
has no periodic solutions.
Solution If (x, y) is a critical point, then either x = 0 or y = 0. If x = 0, then −1 – y2 = 0, y2 = –1. Likewise, y = 0 implies x2
= –1. Thus this system has no critical points and has no periodic solutions.
Ch11_118
Example 2
Show that
has no periodic solutions in the first quadrant.
Solution From Example 4 of Sec 11.4, we knew only (20, 40) lies in the first quadrant and (20, 40) is a saddle point. By the corollary, there are no periodic solutions in the first quadrant.
)0.3100(001.0'
)75.050(004.0'
xyyy
yxxx
Ch11_119
If div V = P/y + Q/ y does not change sign in a connected region R, then the plane autonomous systemhas no periodic solution in R.
THEOREM 11.5Bendixson Negative Criterion
Ch11_120
Example 3
Investigating possible periodic solutions of each system.
Solution
)2(' ),2(' (b)
2' ,42' (a)2222
3223
yxyxyyxxyx
yyxyxyyxyxx
solutions. periodic no are there
so and,332121
// div (a)222
yxx
yQxPV
Ch11_121
Example 3 (2)
If R is the interior of the given circle, div V > 0 and so there are no periodic solutions inside the disk. Note that div V < 0 on the exterior of the circle. If R is any simply connected subset of the exterior, then there are no periodic solutions in R. If there is a periodic solution in the exterior, it must enclose the circle x2 + y2 = 1.
)(44
)32()32( div (b)22
2222
yx
yxyx
V
Ch11_122
Example 4
The sliding bead in Sec 11.4 satisfies
Show that there are no periodic solutions.
Solution
')]('[1
)('" 2 x
xf
xfmgmx
0div
)]('[1
)('',' 2
myQ
xP
ymxf
xfgyyx
V
Ch11_123
If (x, y) has continuous first derivatives in a simply
connected region R and does not change
sign in R, then the plane autonomous system has no
periodic solution in R.
THEOREM 11.6Dulac Negative Criterion
yQ
xP
)()(
Ch11_124
Example 5
Show that
has no periodic solutions.
Solution
'')(" 22 xxxxx
)()12(
)()(
,),( Letting
.','
22
22
yxyxbeyaye
yQ
xP
eyxδ
yxyxyyx
byaxbyax
byax
Ch11_125
Example 5 (2)
If we set a = −2, b = 0, then
which is always negative. The system has no periodic solutions.
byaxeyQ
xP
)()(
Ch11_126
Example 6
Use (x, y) = 1/(xy) to show
have no periodic solutions in the first quadrant.
)('
)('
2122
2
1211
1
xyKyKr
y
yxKxKr
x
Ch11_127
Example 6 (2)
Solution
For (x, y) in the first quadrant, the last expression is always negative.
)1
()1
()()(
)(),(
2
2
1
1
212
2
212
1
1
1
xKr
yKr
yQ
xP
xy
xK
Kr
Qyx
yK
Kr
P
Ch11_128
Fig 11.40 shows two standard types of invariant regions.
A region R is called an invariant region for a plane autonomous system if, whenever X0 is in R, theX = X(t) satisfying X(0) = X0 remains in R.
DEFINITION 11.3Invariant Region
Ch11_129
Fig 11.40
Ch11_130
If n(x, y) denote a normal vector on the boundary thatpoint inside the region, then R will be an invariant region for the plane autonomous system provided V(x, y)‧n(x, y) 0 for all points (x, y) on the boundary.
THEOREM 11.7Normal Vector and Invariant Regions
Ch11_131
Example 7
Find a circular region with center at (0, 0) that serves an invariant region for the system
Solution For the circle x2 + y2 = r2, n = (−2x, −2y) is a normal vector that points toward the interior of the circle. Since
we may conclude that Vn 0 on the circle x2 + y2 = r2. By Theorem 11.7, the circular region x2 + y2 r2 serves as an invariant region for the system for any r > 0.
3
3
'
'
yxy
xyx
)(2)2,2(),( 4433 yxyxyxxy nV
Ch11_132
Example 8
Find an annular region bounded by circles that serves as an invariant region for the system
Solution As in Example 7, the normal vector n1 = (−2x, −2y) points inside the circle x2 + y2 = r2, while the normal vector n2 = − n1 points outside the circle.
522
522
)(5'
)(5'
yyxyyxy
xyxxyxx
)5(2 66421 yxrr nV
Ch11_133
Example 8 (2)
If r = 1, V‧n1 = 8 – 2(x6 + y6) 0. If r = 1/4, V n‧ 1 – 2(r2 – 5r4) < 0 and so V n‧ 2 > 0. The annular 1/16 x2 + y2 1 is an invariant region.
Ch11_134
Example 9
The Van der Pol equation is a nonlinear second-order differential equation that arise in electronics,
Fig 11.41 shows the vector field for = 1, together with the curves y = 0 and (x2 – 1)y = −x along which the vectors are vertical and horizontal, respectively.
xyxy
yx
)1('
'2
Ch11_135
Fig 11.41
It is not possible to find a simple invariant region whose boundary consists of lines or circles.
Ch11_136
Let R be an invariant region for a plane autonomous system and suppose that R has no critical points on its boundary.
(a) If R is a Type I region that has a single unstable node or an unstable spiral point in its interior, then there is at least one periodic solution in R.
(b) If R is a Type II region that contains no critical points of the system, then there is at least one periodic solution in R.
In either of the two cases, if X = X(t) is a nonperiodic solutionin R, then X(t) spirals toward a cycle that is a solution to the solution to the system. This periodic solution is called a limit cycle.
THEOREM 11.8Poincare-Bendixson I
Ch11_137
Example 10
Use Theorem 11.8 to show that
has at least one periodic solution.
SolutionWe first construct an invariant region bounded by circles. If n1 = (−2x, −2y) then
)()1('
)()1('2222
2222
yxxyxyxy
yxyyxxyx
)1(2 221 rr nV
Ch11_138
Example 10 (2)
If we let r = 2 and then r = ½, we conclude that the annular region R: ¼ x2 + y2 4 is invariant. If (x1, y1) is a critical point, then V‧n1 = (0, 0)‧n1 = 0. Therefore r = 0 or r = 1. If r = 0, then (x1, y1) = (0, 0) is a critical point.
If r = 1, the system reduces to −2y = 0, 2x = 0 and we have reached a contradiction. Therefore (0, 0) is the only critical point and is not in R. Thus the system has at least one periodic solution in R.
Ch11_139
Example 11
Show that the Van der Pol equations has a periodic solution when > 0.
Solution We found that the only critical point is (0, 0) and the Jacobian matrix is
44 ,1 ,
then,1
10))0 ,0(('
22
g
Ch11_140
Example 11 (2)
Since > 0, the critical point is either an unstable spiral point or an unstable node. Bu part (i) of Theorem 11.8 the system has at least one periodic solution in R. See Fig 11.42.
Ch11_141
Fig 11.42
Ch11_142
Let R be a Type I invariant region for a plane autonomous system that has no periodic solution in R.
(a) If R has a finite number of nodes or spiral points, then given any initial position X0 in R, limt→X(t) = X1 for some critical point X1.
(b) If R has a single stable node or stable spiral point X1 in its interior and no critical points on its boundary, the limt→X(t) = X1 for all initial position X0 in R.
THEOREM 11.8Poincare-Bendixson II
Ch11_143
Example 12
Investigate global stability for the system in Example 7.
Solution It is not hard to show that the only critical point is (0, 0) and the Jacobian matrix is
3
3
'
'
yxy
xyx
1 ,0 and ,01
10))0 ,0(('
g
Ch11_144
Example 12 (2)
(0, 0) may be either a stable or an unstable spiral. Theorem 11.9 guarantees that
The critical point is therefore a globally stable spiral point. See Fig 11.43.
plane. in the 0)(position initialany for )0 ,0()(lim
havemust wepoint, criticalonly theis 0) (0, Since
. points critical somefor )(lim 11
XX
XXX
t
t
t
t
Ch11_145
Fig 11.43
