Answers 1

Chapter 10

Section 10.1

1.

−5 0 5

0

5

10

(0,0)

(0,5)

(2,0)

(−4 9)

(0, 0) is a nodal source. (0, 5) is a nodal sink. (2, 0) is a saddle. (−4, 9) is asaddle.

3.

−1 0 1 2−1

0

1

2

(0,0)

(0,1)

(1,0)

(2/3,2/3)

(0, 0) is a source. (0, 1) is a saddle. (1, 0) is a saddle. (2/3, 2/3) is a nodal sink.

5.

2 Answers

0 1 2 3

0

1

2

(0,0)

(3,5/4)

(0, 0) is a saddle. (3, 5/4) cannot be classified.

7.

−8 −4 0 4 8−2

0

2

(kπ, 0) is a spiral sink if k is even and a saddle if k is odd.

9.

Answers 3

−5 0 5

−5

0

5

10

x

y

−1 0 1

−1

−0.5

0

0.5

1

u

v

11.

−0.5 0 0.5 1 1.5

−0.5

0

0.5

1

1.5

x

y

4 Answers

−1 0 1

−1

−0.5

0

0.5

1

u

v13.

−1 0 1 2 3 4

−1

−0.5

0

0.5

1

1.5

2

x

y

−1 0 1

−1

−0.5

0

0.5

1

u

v

15.

Answers 5

−5 0 5

−3

−2

−1

0

1

2

3

x

y

−1 0 1

−1

−0.5

0

0.5

1

u

v

17. (b)

−1 −0.5 0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

1.5

2

x

y

(0, 0)

(0, 4/3)

(1, 0)

(1/4, 3/4)

6 Answers

−1 −0.5 0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

1.5

2

u

v

(−1/4, −3/4)

(0, 0)

(3/4 −3/4)

u′ = (x − 1/4)′

= x ′

= (1 − x − y)x= [1 − (1/4 + u)− (3/4 + v)](1/4 + u)= −(u+ v)(1/4 + u)= −u/4 − v/4 − U62 − uv.

v′ = (y − 3/4)′

= y ′

= (4 − 7x − 3y)y

= [4 − 7(1/4 + u)− 3(3/4 + v)](3/4 + v)= −(7u+ 3v)(3/4 + v)= −21u/4 − 9v/4 − 7uv − 3v2

17. (c) If we throw away the rerms having degree 2 or more we get the linearization.19. (c) Integrating we get θ(t) = −t + C. Hence θ is decreasing at a uniform rate.Hence the solution curve is spiraling clockwise. Since r ′ = αr3, we see that if α < 0,r is decreasing, and the curve is spiraling into the origin. If α > 0, r is increasingand the curve is spiraling outward.21.

x ′1 = (a1 − b1x1 + c1x2)x1

x ′2 = (a2 − b2x2 + c2x1)x2

where a1 < 0, b1 = 0, c1 > 0, a2 > 0, b2 > 0, and c2 > 0.23.

x ′1 = (a1 − b1x1 + c1x2)x1

x ′2 = (a2 − b2x2 + c2x1)x2

where a1 < 0, b1 = 0, c1 > 0, a2 < 0, b2 = 0, and c2 > 0.25.

x ′1 = (a1 − b1x1 + c1x2)x1

x ′2 = (a2 − b2x2 + c2x1)x2

Answers 7

where a1 > 0, b1 > 0, c1 > 0, a2 > 0, b2 > 0, and c2 > 0.

27. The model is

x ′1 = (a1 + b11x1 + b12x2 + b13x3)x1

x ′2 = (a2 + b21x1 + b22x2 + b23x3)x2

x ′3 = (a3 + b31x1 + b32x3 + b33x3)x3.

The constant aj is the reproductive rate of the population xj in the abasence of theothers. The constant bjj is negative if there is a logistic limit, and zero if not. Fori �= j the constant bij measures the effect of xj on the reproduction of xi . If bij > 0the effect is to increase the reproductive rate, and if bij < 0 the reproductive rate isdiminished.

Section 10.2

1. (0, 0) is a saddle. (±1, 1) are spiral sinks.

3. (0, 0) is a nongeneric sink. (−5, 1) is a saddle.

5. (0, 0, 0) is a sink. (2/3, 4/9, 2/9) is unstable.

7. (0, 0, 0)is unstable.

9. The origin is unstable.

11. The origin is asymptotically stable.

13. 0 is also an eigenvalue, so it is not possible to classify the equilibrium point usingthe Jacobian.

15. The origin is asymptotically stable.

17. 0 is unstable. c+ and c− are asymptotically stable. Solutions starting at (2, 1, 1)T

and −2,−1, 1)T .

0 1 2 3 4 5

−10

−4

0

46

10

t

x, y, & z

8 Answers

0 1 2 3 4 5

−10

−4

0

46

10

t

x, y, & z

19. All three equilibrium points are unstable. Solutions starting at (8, 8, 27)T and(−8,−8, 27)T .

0 10 20 30 40−10−x

0

0

x0

z0

50

t

x, y, & z

0 10 20 30 40−10−x

0

0

x0

z0

50

t

x, y, & z

21. (b) The equilbrium point is x = (5/4, 1/2, 3/4)T . It is asymptotically stable.

Section 10.31. If x(t) solves the logistic equation x ′ = (2 − x)x and y(t) = 0 then x and yare a solution to the system in the exercise. Every point in the x-axis is contained

Answers 9

in such a solution curve, which stays in the x-axis. Hence the x-axis is invariant.Similarly the functions x(t) = 0 and y(t) defined by y ′ = (3 − y)y are solutionsto the system in the exercise, and their solution curves exhaust the y-axis, makingit invariant. A solution curve starting in one of the four quadrants must stay in thatquadarant, because to get out it has to cross one of the axes. It cannot do so becausethe unique solution curve through any point in the axis must be entirely contained inthe axis.3. If x(t) solves the logistic equation x ′ = (1 − x)x and y(t) = 0 then x and yare a solution to the system in the exercise. Every point in the x-axis is containedin such a solution curve, which stays in the x-axis. Hence the x-axis is invariant.Similarly the functions x(t) = 0 and y(t) defined by y ′ = (6 − 3y)y are solutionsto the system in the exercise, and their solution curves exhaust the y-axis, makingit invariant. A solution curve starting in one of the four quadrants must stay in thatquadarant, because to get out it has to cross one of the axes. It cannot do so becausethe unique solution curve through any point in the axis must be entirely contained inthe axis.5. The axes are invariant, so no solution curves can cross them. Along the line y = 3we have x ′ = (2 − x − y)x = −(1 + x)x < 0, so x is decreasing. Thus the solutioncurve is moving into S. Along the line x = 3 we have y ′ = (3 − 3x − y)y =−(6 + y)y < 0, so y is decreasing. Again the solution curves are moving into S.Consequently no solution curves cross the boundary of S going out of S. Hence S isinvariant.7. The axes are invariant, so no solution curves can cross them. Along the line y = 3we have x ′ = (1 − x − y)x = −(2 + x)x < 0, so x is decreasing. Thus the solutioncurve is moving into S. Along the line x = 4 we have y ′ = (6 − 2x − 3y)y =−(2 + 3y)y < 0, so y is decreasing. Again the solution curves are moving into S.Consequently no solution curves cross the boundary of S going out of S. Hence S isinvariant.9.

0 1 20

1

2

3

I

II

III

IV

The origin is a nodal source. (2, 0) is a nodal sink. (0, 3) is also a nodal sink.(1/2, 3/2) is a saddle. All solution curves flow to one of the sinks at (2, 0) and (0, 3),with the exception of the two stable solution curves for the saddle at (1/2, 3/2).

10 Answers

11.

0 1 2 3

0

1

2

I

II

III

The origin is a nodal source. (2, 0) is a saddle. (0, 3) is a nodal sink. All solutioncurves in the positive quadrant flow to the sink at (0, 3).

13.

−2 −1 0 1 2

0

1

2

(1, 1) is a saddle. (−1, 1) is a spiral source.

15.

Answers 11

−1 0 1

−1

0

1

The origin is a saddle. (1, 1) and (−1,−1) are nodal sinks. All solutions convergeto one of the two sinks, except for the two stable solutions for the saddle point.17. (b) All solutions tend to ∞.17. (c) All solutions tend to the equilibrium point

u = 1 + a1 − ab and v = 1 + b

1 − ab .

Section 10.41. r ′ = r(1 − r), so r = 1 is an attracting limit cycle.3. r ′ = r(4 − r2), so r = 2 is an attracting limit cycle.5. r ′ = (r− 2)(r− 1), so r = 1 is an attracting limit cycle, while r = 2 is a repellinglimit cycle.

−2 0 2

−2

−1

0

1

2

x

y

7. r ′ = r(r2 − 1)2, so r = 1 is an unstable limit cycle.

12 Answers

−1 0 1

−1

0

1

x

y

9. r ′ = (x2 + 2y2)(1 − r2)/r. r2θ ′ = −(x2 − xy − y2) + xyr2. Notice that theunit circle is an attractive, invariant set. The equilibrium points are (0, 0), a spiralsource, (1/

√2, 1/

√2) and (−1/

√2,−1/

√2) which are degenerate in the sense that

the Jacobian has 0 determinant. All solutions are attracted to one of these latter twoequilibrium points, but they are not sinks!

−1 0 1

−1

−0.5

0

0.5

1

x

y

11. (b) r ′ > 0 on r = 1/2 and r ′ < 0 on r = 1.

11. (c) The only equilibrium points of the system is (0, 0).

11. (d)

Answers 13

−2 −1 0 1 2−1.5

−1

−0.5

0

0.5

1

1.5

x

y

13. Compute that rr ′ = r2 −[3x4 +3x2y2 +y4]. Prove that r4 ≤ 3x4 +3x2y2 +y4 ≤3r4. Then the annulus R defined by 1/3 ≤ r ≤ 1 is invariant.

−2 −1 0 1 2−1.5

−1

−0.5

0

0.5

1

1.5

x

y

15.

−2 −1 0 1 2

−2

−1

0

1

2

x

y

17.

14 Answers

−0.5 0 0.5 1 1.5

0

0.5

1

x

y

19.

−1 −0.5 0 0.5

−0.5

0

0.5

x

y

21.

−1 0 1

0

0.2

0.4

0.6

0.8

1

x

y

23. (b)

Answers 15

−4 −2 0 2 4

−4

−2

0

2

4

x

y

23. (d)

−4 −2 0 2 4

−4

−2

0

2

4

x

y

25.

−5 0 5

−5

0

5

x

y

Section 10.5

16 Answers

1. E(y, v) = y2 + v2/2

3. E(y, v) = y3/3 − y2/2 − v2/2

5. E(y, v) = y2/2 − y4/4 − v2/2

7. E(x, y) = ex − ey9. E(x, y) = cos x − y2/2

11. The conserved quantity is E(y, v) = y2 − v2/2.

−5 0 5−5

0

5

y

v

1

1

5

5

5511

1111

11

11

11

20

20

20 20

20

20

36

36

36

36

36

36

13. The conserved quantity is E(y, v) = y3/3 − y2/2 − v2/2.

−5 0 5−5

0

5

y

v

−1/6

−1/6

−1/6

−30

−30

−30 −10

−10

−10−10

−5

−5

−5

−5

0

0

0

5

5

510

10

1030

30

30

15. The conserved quantity is E(y, v) = y2/2 − y4/4 − v2/2.

Answers 17

−5 0 5−5

0

5

y

v 00−1

−1

−1

−5

−5

−5

−5

−10

−10

−10

−10

−10

−40

−40

−40

−40

17. The conserved quantity is E(x, y) = ex − ey .

−2 −1 0 1 2

−2

−1

0

1

2

x

y

−2

−2−2−2

−1

−1

−1−1

0

0

0

0 1

1

1

2

2

2

19. The conserved quantity is E(x, y) = cos x − y2/2.

−10 −5 0 5 10

−3

−2

−1

0

1

2

3

x

y

−4 −4 −4 −4

−4 −4 −4 −4

−2 −2 −2 −2 −2

−2 −2 −2 −2 −2

−1

−1 −1

−1

−1 −1

−1

−1

−1 −1

−1

−1

0

00

0

0

00

00.0.7 0.7

0.7

21. E′ = −v2

18 Answers

23. E′ = −v2

25. (b)

−2 0 2

−3

−2

−1

0

1

2

3

xy

0.5

0.5

0.5

0.5

0.5

0.5

0.5 0.50.5

Section 10.61.

y

v

3.

y

v

Answers 19

5.

y

v

7. Min at y = 1/2, U(y) = −y + y2,

y

v

−3 3

y

U (y)

−3 3

9. Min at y = 4, Max at y = 0, U(y) = −2y2 + (1/3)y3,

20 Answers

y

v

−2 7

y

U (y)

−2 7

11. Minima at y = kπ , k odd, maxima at y = kπ , k even. U(y) = cos y,

y

v

−2π 2π

y

U (y)

2π 2π

13. Separatrix 12v

2 − 14y

4 + 12y

2 = 14 , Saddles at (−1, 0) and (1, 0), center at (0, 0).

Answers 21

−2 2

−2

2

y

v

15. Separatrix 12v

2 − cos y = 1, Saddles at (−π, 0) and (π, 0), centers at (−2π, 0),(0, 0), (2π, 0) on [−2π, 2π ].

−3.1416 0 3.1416−3

3

y

v

17. β > 0 implies hard spring; β < 0 yields a soft spring.19. (a) Equilibrium points: (kπ, 0), k an integer.19. (b) Spiral sink: (kπ, 0), k even. Saddles: (kπ, 0), k odd.

−10 −5 0 5 10

−3

−2

−1

0

1

2

3

θ

ω

22 Answers

21. Hint: If dH/dt = 0, then equating coefficients leads to a system of threeequations in A, B, and C.23. H(x, y) = −2xy − y3 + x3

25. H(x, y) = y3 + x3

27. H(x, y) = −xy + y2 + x2

29. Not Hamiltonian31. E = 0, so E is conserved.

Section 10.71. Positive definite3. None5. Negative semidefinite7. None9. Positive definite13. dV/dt = −2y4

15. dV/dt = (2 − x)y3 + y2 + 4xy − x2

19. Hint: Argue that V (x, y) = −2(x2 + y2) is negative definite.21. Hint: Show that V (x, y) = −2(x2(1 − y)+ y2(1 − x)).Section 10.85. In either of these two cases all solution curves approach the equilibrium point at(1, 0).

Section 10.9

Section 10.10