1. Systems of Nonlinear Differential Equations
CHAPTER 11
Systems of Nonlinear Differential Equations

2. Contents 11.1 Autonomous Systems 11.2 Stability of Linear Systems
11.3 Linearization and Local Stability
11.4 Autonomous Systems as Mathematical Models
11.5 Periodic Solutions, Limit Cycles, and Global Stability

3. 11.1 Autonomous Systems
Introduction A system of first-order differential equations is called autonomous, when it can be written as (1)

4. Example 1
The above is not autonomous, since the presence of t on the right-hand side.

5. Example 2
Consider If we let x = , y =  , then is a system of first-order.

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.

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.

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.

9. Fig 11.1

10. 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

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

12. (iii) A periodic solution – is called a cycle
(iii) A periodic solution – is called a cycle. If p is the period, then X(t + p) = X(t). See Fig 11.3.

13. Example 4 Find all critical points of the following: (a) (b) (c)
Solution (a) then y = x. There are infinitely many critical points. y x - = + 2 6

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

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.

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

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 2t See Fig 11.4(a).

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).

19. Fig 11.4(b)

20. Changing to Polar Coordinates
Please remember that the transformations are r2 = x2 + y2 and  = tan–1(y/x),

21. Example 6
Find the solution of the following system satisfying X(0) = (3, 3).
Solution

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

23. Example 6 (3)
The graph of is shown in Fig 11.5.

24. Fig 11.5

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

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.

27. Fig 11.6

28. 11.2 Stability of Linear Systems
Some Fundamental Questions Suppose 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

29. Fig 11.7

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.

31. Stability Analysis
Consider x = 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.

32. Then det (A – I) = 0 becomes 2 −  +  = 0 where  = a + d. Thus

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.

34. Example 1 (2)
Solution Since the coefficient matrix is then we have  = −2, and  = 1 – c. Thus

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).

36. Example 1 (4) When c = 0,  = −1. See Fig 11.8(c).
When c = −9,  = −1  3i. See Fig 11.8(d).

37. Case I: Real Distinct Eigenvalues
According to Sec 10.2, the general solution is
(a) Both eigenvalues negative: Stable Node It is easier to check that under this condition, X(t)  0 as t   See Fig 11.9.

38. Fig 11.9

39. (b) Both eigenvalues positive: Unstable Node
(b) Both eigenvalues positive: Unstable Node It is easier to check that under this condition, |X(t)| becomes unbounded as t   See Fig 11.10

40. Fig 11.10

41. (c) Eigenvalues have opposite signs. (2 < 0 < 1): Saddle Point
(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

42. Fig 11.11

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

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

45. Fig 11.12＞

46. (b) Since the eigenvalues are − 4, −25, (0, 0) is a stable node
(b) Since the eigenvalues are − 4, −25, (0, 0) is a stable node. The corresponding eigenvectors are respectively See Fig

47. Fig 11.13

48. Case II: A Repeated Real Eigenvalue
According to Sec 10.2, we have the following conditions.
(a) Two linearly independent eigenvectors The 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.

49. Fig 11.14

50. (b) A single linearly independent eigenvector When 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.

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

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))

53. Example 3
Classify the critical point (0, 0) of each system Solution (a) The characteristic equation 2 + 6 + 9 = ( + 3)2= so (0, 0) is a degenerate stable node.
(b) The characteristic equation 2 + 1 = so (0, 0) is a center.

54. Example 4
Classify the critical point (0, 0) of each system for positive constants.
Solution (a)  = −0.01,  = , 2 − 4 < 0: (0, 0) is a stable spiral point.

55. Example 4 (2)
(b)

56. For a linear plane autonomous system X’ = AX with det A  0,
THEOREM 11.1
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.
limt→X(t) = 0 if and only if the eigenvalues of A have negative real parts. This will occur when ∆ > 0 and  < 0.
X(t) is periodic if and only if the eigenvalues of A are pure imaginary. This will occur when ∆ > 0 and  = 0.
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.
Stability Criteria for Lonear Systems

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 say that X1 is a stable critical point when, given any radius ρ > 0, there is a corresponding radius r > 0 such that if the initial position X0 satisfies │X0 – X1│< r, then the 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 11.1
Stable Critical Points

58. This definition is shown in Fig 11. 20(a)
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.

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 say that X1 is an unstable critical point if there is a disk of 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 11.2
Unstable Critical Points

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).

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

62. Fig 11.21

63. Example 2
Consider the plane system Show that (x, y) = (0, 0) is an unstable critical point.

64. Example 2 (2)
Solution Since 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.

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

66. Fig 11.22

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.

68. THEOREM 11.2
Let x1 be a critical point of the autonomous differential equation 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.
Stability Criteria for Linear Systems

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

70. Fig 11.23

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.
Solution We have two critical points x = 0 and x = K. Since Therefore x = K is an asymptotically stable critical point but x = 0 is unstable.

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

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).

74. THEOREM 11.3
Let X1 be a critical point of the autonomous differential equation X’ = g(X), where P(x, y) and Q(x, y) have continuous 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.
Stability Criteria for Plane
Autonomous Systems

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)

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.

77. Example 5 (3)
(b) The critical points are (0, 0), (0, 60), (100, 0), (50, 50). The Jacobiam matrix is

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.

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.

80. Example 7
Consider the system x + x – x3 = 0. We have x = y, y = x3 – x. Find and classify the critical points.
Solution

81. Example 7 (2)
The corresponding matrices are

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.

83. Example 8 (2)
Using the phase-plane method, we get Fig shows a collection of solution curves. The critical point (0, 0) is unstable.

84. Fig 11.26

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

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

87. Fig 11.27

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

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

90. Fig 11.28

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

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

93. Fig 11.29

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.
Solution The initial condition is X(0) = (0, 0).

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)02 > −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

96. Nonlinear Oscillations: The Sliding Bead
See Fig The tangential force F has the magnitude mg sin , thus Fx = − mg sin  cos . Since tan  = f (x), then

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.

98. The Jacobian matrix at X1 is

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.

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.

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

102. Fig 11.32
Fig shows the motions when the critical points are stable spiral points.

103. Fig 11.33
Fig shows a collection of solution curves for the undamped case.

104. Lotka-Volterra Predator-Prey Model
Recall the predator-prey model:

105. Fig 11.34
The critical point (0, 0) is a saddle point. See Fig

106. Since A2 has the pure imaginary eigenvalues, the critical point may be a center. Since

107. Fig 11.35
Typical graphs are shown in Fig

108. If y = a/b, the equation F(x)G(y) = c0 has exactly two solutions xm and xM that satisfy xm < d/c < xM.
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.
If x is outside the interval [xm, xM], then F(x)G(y) = c0 has no solutions.
The graph of a typical periodic solution is shown in Fig

109. Fig 11.36

110. Example 3
If we let a = 0.1, b = 0.002, c = , 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

111. Fig 11.37

112. Lotka-Volterra Competition Model
Consider the model: This system has critical points at (0, 0), (K1, 0) and (0, K2).

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 1221 = 2.25 > 1, and so the critical point (20, 40) is a saddle point. The Jacobian matrix is

114. Example 4 (2)

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.

116. 11.5 Periodic Solutions, Limit Cycles and Global Stability
If a plane autonomous system has a periodic solution X = X(t) in a simply connected region R, then the system has at least one critical point inside the corresponding simple closed curve C. If there is a single critical point inside C, then that critical point cannot be a saddle point.
THEOREM 11.4
Cycles and Critical Points
If a simply connected region R either contains no critical points of a plane autonomous system or contains a single saddle point, then there are no periodic solutions in R.
COROLLARY

117. Example 1
Show that the plane autonomous system x’ = xy y’ = −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.

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.

119. THEOREM 11.5
If div V =  P/y + Q/ y does not change sign in a connected region R, then the plane autonomous system has no periodic solution in R.
Bendixson Negative Criterion

120. Example 3 Investigating possible periodic solutions of each system.

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.

122. Example 4
The sliding bead in Sec 11.4 satisfies Show that there are no periodic solutions.
Solution

123. If (x, y) has continuous first derivatives in a simply
THEOREM 11.6
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.
Dulac Negative Criterion

124. Example 5
Show that has no periodic solutions.
Solution

125. Example 5 (2)
If we set a = −2, b = 0, then which is always negative. The system has no periodic solutions.

126. Example 6
Use (x, y) = 1/(xy) to show have no periodic solutions in the first quadrant.

127. Example 6 (2)
Solution For (x, y) in the first quadrant, the last expression is always negative.

128. Fig 11.40 shows two standard types of invariant regions.
DEFINITION 11.3
A region R is called an invariant region for a plane autonomous system if, whenever X0 is in R, the X = X(t) satisfying X(0) = X0 remains in R.
Invariant Region
Fig shows two standard types of invariant regions.

129. Fig 11.40

130. THEOREM 11.7
If n(x, y) denote a normal vector on the boundary that point 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.
Normal Vector and Invariant Regions

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 Vn  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.

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.

133. Example 8 (2)
If r = 1, V‧n1 = 8 – 2(x6 + y6)  0. If r = 1/4, V‧n1  – 2(r2 – 5r4) < 0 and so V‧n2 > 0. The annular 1/16  x2 + y2  1 is an invariant region.

134. Example 9
The Van der Pol equation is a nonlinear second-order differential equation that arise in electronics, Fig 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.

135. Fig 11.41
It is not possible to find a simple invariant region whose boundary consists of lines or circles.

136. THEOREM 11.8
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 solution in 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.
Poincare-Bendixson I

137. Example 10
Use Theorem 11.8 to show that has at least one periodic solution.
Solution We first construct an invariant region bounded by circles. If n1 = (−2x, −2y) then

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.

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

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

141. Fig 11.42

142. THEOREM 11.8
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.
Poincare-Bendixson II

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

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

145. Fig 11.43

146. Thank You !