9.1 The Phase Plane - · PDF filedt ’(x(t);y(t)) = @ ... 9.2.3. (a) stable, biaxial node; (b) real, negative 9.2.4. (a) stable, coaxial node; (b) real, repeated, negative 9.2.5

154 Chapter 9 Hints, Answers, and Solutions

9.1 The Phase Plane

9.1.1–4. The particular trajectories are highlighted in the phase portraits below.

1. 2.

3. 4.

9.1.5. Shown below is one possibility with x(t) and y(t) periodic. There are many othersbecause the same parametric curve could be traced out from different initial pointsand with different speeds.

t

2

4

6

8

10

12 xy

9.1.6. Shown below is one possibility with x(t) and y(t) approaching sinusoidal functions ast → ∞. There are many others because the same parametric curve could be tracedout from different initial points and with different speeds.

9.1 Hints, Answers, and Solutions 155

9.1.7. Shown below is one possibility with x(0) = y(0) = 0 and x(t) and y(t) each ap-proaching a sinusoidal function with amplitude 2 as t→∞.

t

-2

-1

1

2y

x

9.1.8. Shown below is one possibility with x(t) and y(t) each approaching a periodic func-tion with amplitude 1 as t→ −∞.

9.1.9. Since x(t) and y(t) appear to be sinusoidal, the parametric curve is an ellipse.

2 4x

-1

1

y

10. 11. 12.


9.1.13.

2 4 6 8 10 12 14t

-2

-1

1

2x

2 4 6 8 10 12 14 t

-2

-1

1

2

y

9.1.14.

0 2 4 6 8 10 12t

0.25

1

2

3

4x

0 2 4 6 8 10 12t

1

2

3

4y

9.1.15. (0, 0), (1, 0), and (3, 0) are unstable; (2, 0) is neutrally stable.

9.1.16. (0, 0) and (3, 0) are asymptotically stable; (1, 0) and (2, 0) are unstable.

9.1.17. The first two plots in Figure 9 correspond to the two inner, periodic orbits in Figure8, respectively. The third plot corresponds to the clockwise orbit comprising theupper portion of outer circle, and the fourth corresponds to the counterclockwiseorbit comprising the lower portion of outer circle.

9.1.18. dydx = −x

y =⇒ y dy = −x dx =⇒ 12y

2 = − 12x

2 = C =⇒ x2 + y2 = C. Orbits traceout circles centered at the origin. The orientation is clockwise, since (e.g.) x′ > 0when y > 0.

9.1.19. dydx = x

y =⇒ y dy = x dx =⇒ 12y

2 = 12x

2 = C =⇒ y2 − x2 = C. These arehyperbolas. The phase portrait is shown below.

-3 -2 -1 1 2 3

-2

-1

1

2


9.1.20. dydx = −xy

xy =⇒ dydx = −1 =⇒ y = −x+ C. The phase portrait is shown below.

-3 -2 -1 1 2 3

-2

-1

1

2

9.1.21. dydx = −xy

y =⇒ dydx = −x =⇒ y = − 1

2x2 + C. The phase portrait is shown below.

-3 -2 -1 1 2 3

-2

-1

1

2

9.1.22. dydx = −y2+2x2y

xy =⇒ dydx = − y

x + 2x =⇒ y′ + 1xy = 2x2. Use the integrating factor

x to obtain the solution y = 23x

2 + Cx . The phase portrait is shown below.

-3 -2 -1 1 2 3

-2

-1

1

2

9.1.23. dydx = − sin x

y =⇒ y dy = − sinx dx =⇒ 12y

2 = cosx+ C =⇒ y2 = 2 cosx+ C

9.1.24. Implicitly differentiate x2 + y2 = 1 with respect to x to obtain dydx = −x

y . Now using

x2 + y2 = 1 in the right-hand side of the slope equation, dydx = 1−(x+y)2

2y2(2−x2−y2) , we have

dy

dx=

1− (x+ y)2

2y2(2− x2 − y2)=

1− (x2 + 2xy + y2)2y2(2− (x2 + y2))

= −2xy2y2

= −xy.

Thus the slope field is tangent to the unit circle at each point on it. No single orbittraces out the entire unit circle, because (±1, 0) are equilibrium points.


9.1.25. Rearrange the slope equation into (2x+ x2)dx+ 2ydy − (xdy + ydx) = 0. Integrateto get x3 + 3x2 + 3y2 − 6xy = C.

9.1.26.d

dtϕ(x(t), y(t)) =

∂ϕ

∂x

dx

dt+∂ϕ

∂y

dy

dt= −∂ϕ

∂x

∂ϕ

∂y+∂ϕ

∂y

∂ϕ

∂x= 0 =⇒ ϕ(x(t), y(t)) = C.

9.2 Phase Portraits of Homogeneous Linear Systems

9.2.1. (a) unstable, saddle point; (b) real, opposite sign

9.2.2. (a) stable, spiral point; (b) nonreal, negative real part

9.2.3. (a) stable, biaxial node; (b) real, negative

9.2.4. (a) stable, coaxial node; (b) real, repeated, negative

9.2.5. (a) unstable, biaxial node; (b) real, positive

9.2.6. (a) unstable, spiral point; (b) nonreal, positive real part

9.2.7. From the given eigenvectors, the straight line orbits trace out y = ±2x. Since theeigenvalues are both negative, the origin is a stable node.

9.2.8. From the given eigenvectors, the straight line orbits trace out y = ±2x. Since theeigenvalues are real with opposite signs, the origin is a stable node.


9.2.9. From the given eigenvectors, the straight line orbits trace out y = x and y = − 12x.

Since the eigenvalues are both positive, the origin is an unstable node.

9.2.10. From the given eigenvectors, the straight line orbits trace out only y = x. Since therepeated eigenvalue is negative, the origin is a stable coaxial node.

9.2.11. Let A =(a b

c d

). From the given sign diagram, we know a, c > 0 while b, d < 0.

Along the positive x axis, the system x′ = ax+by, y′ = cx+dy reduces to x′ = ax > 0and y′ = cx > 0, i.e. x and y are both increasing. Along the negative x axis, we haveprecisely the opposite behavior. On the other hand, along the positive y axis, thesystem reduces to x′ = by < 0 and y′ = dy < 0, i.e. x and y are both decreasing.Along the negative y axis, we have precisely the opposite behavior. Since the realparts of the eigenvalues of A were negative by assumption, the orbits spiral in towardsthe origin as shown in the phase portrait below.

Note that if the real parts of the eigenvalues of A were positive, the phase portraitwould look the same except that the orbits would spiral outward from the origin.

9.2.12. Let A =(a b

c d

). From the given sign diagram, we know a, c > 0 while b, d < 0.

Along the positive x axis, the system x′ = ax+by, y′ = cx+dy reduces to x′ = ax > 0


and y′ = cx < 0, i.e. x is increasing while y is decreasing. Along the negative x axis,we have precisely the opposite behavior. On the other hand, along the positive yaxis, the system reduces to x′ = by > 0 and y′ = dy < 0, i.e. x is increasing whiley is decreasing. Along the negative y axis, we have precisely the opposite behavior.Since the real parts of the eigenvalues of A were negative by assumption, the orbitsspiral in towards the origin as shown in the phase portrait below.


9.2.13. Let A =(a b

c d

). From the given sign diagram, we know a, c, d < 0 while b > 0.

Along the positive x axis, the system x′ = ax+by, y′ = cx+dy reduces to x′ = ax < 0and y′ = cx < 0, i.e. x and y are both decreasing. Along the negative x axis, wehave precisely the opposite behavior. On the other hand, along the positive y axis,the system reduces to x′ = by > 0 and y′ = dy < 0, i.e. x is increasing while y isdecreasing. Along the negative y axis, we have precisely the opposite behavior. Sincethe real parts of the eigenvalues of A were negative by assumption, the orbits spiralin towards the origin as shown in the phase portrait below.


9.2.14. Let A =(a b

c d

). From the given sign diagram, we know a, b, d < 0 while c > 0.

Along the positive x axis, the system x′ = ax+by, y′ = cx+dy reduces to x′ = ax < 0and y′ = cx > 0, i.e. x is decreasing while y is increasing. Along the negative x axis,we have precisely the opposite behavior. On the other hand, along the positive y


axis, the system reduces to x′ = by < 0 and y′ = dy < 0, i.e. x and y are bothdecreasing. Along the negative y axis, we have precisely the opposite behavior. Sincethe real parts of the eigenvalues of A were negative by assumption, the orbits spiralin towards the origin as shown in the phase portrait below.


9.2.15. (a, b)

(c, d)

9.2.16. The eigenvalues of A are −1 + k2 ±√k2 − 8k + 4. The real parts of the eigenvalues

are plotted below as a function of the parameter k.

-4 -2 2 4 6 8 10

-6

-4

-2

2

4

6

k

Re λ

Notice the eigenvalues are nonreal when 4 − 2√

3 < k < 4 + 2√

3, corresponding tothe straight part of the graph above. The following can be inferred about the type


of the equilibrium point as k changes:

k < 0 : saddle point

0 < k < 4− 2√

3 : stable biaxial node

k = 4− 2√

3 : stable coaxial node

4− 2√

3 < k < 2 : stable spiral point

k = 2 : center

2 < k < 4 + 2√

3 : unstable spiral point

k = 4 + 2√

3 : unstable coaxial node

k > 4 + 2√

3 : unstable biaxial node

9.2.17. (a) The eigenvalues of A are 12

(k ±√k2 − 4

). The real parts of the eigenvalues are

plotted below as a function of the parameter k.

-4 -2 2 4

-3

-2

-1

1

2

3

(b) Real and distinct when |k| > 2; real and repeated when k = ±2; imaginary whenk = 0; nonreal complex when |k| < 2.

(c) The following can be inferred about the type of the equilibrium point as k changes:

k < −2 : stable biaxial node

k = −2 : stable coaxial node

−2 < k < 0 : stable spiral point

k = 0 : center

0 < k < 2 : unstable spiral point

k = 2 : unstable coaxial node

k > 2 : unstable biaxial node

9.2.18. (a) The eigenvalues of A are 12

(k − 1±

√k2 + 2k − 7

). The real parts of the eigen-

values are plotted below as a function of the parameter k.

-6 -4 -2 2 4 6

-4

-2

2

4


(b) Real and distinct when k < −1− 2√

2 or k > −1 + 2√

2; real and repeated whenk = −1 ± 2

√2; imaginary when k = 1; nonreal complex when −1 − 2

√2 < k <

−1 + 2√

2.(c) The following can be inferred about the type of the equilibrium point as k changes:

k < −1− 2√

2 : stable biaxial node

k = −1− 2√

2 : stable coaxial node

−1− 2√

2 < k < 1 : stable spiral point

k = 1 : center

1 < k < −1 + 2√


k = −1 + 2√

2 : unstable coaxial node

−1 + 2√

2 < k < 2 : unstable biaxial node

k > 2 : saddle point

9.2.19. |A− λI | = λ2 − (a11 + a22)λ+ |A|; hence the eigenvalues are

12

(a11 + a22 ±

√(a11 + a22)2 − 4|A|

).

SinceA has a repeated eigenvalue, it follows that (a11+a22)2 = 4|A|, and the repeatedeigenvalue is 1

2 (a11 + a22). Now, |A−λI | = λ2− (a11 + a22 + ε)λ+ |A|+ a22ε; henceits eigenvalues are

λ1,2 =12

(a11 + a22 + ε±

√(a11 + a22 + ε)2 − 4(|A|+ a22ε)

).

Using (a11 + a22)2 = 4|A|, we simplify these as follows:

λ1,2 =12

(a11 + a22 + ε±

√(a11 + a22)2 + 2(a11 + a22)ε+ ε2 − 4(|A|+ a22ε)

)=

12

(a11 + a22 + ε±

√2(a11 + a22)ε+ ε2 − 4a22ε

)=

12

(a11 + a22 + ε±

√ε(2(a11 − a22) + ε

)).

From this last form, the stated results follow by consideration of cases.

9.2.20. A reasonable sketch of the phase portrait can be made based on the eigenvalues,nullclines, and straight-line orbits. This information is summarized below.

Eigenvalues: 12 (1±

√5).

Nullclines: y = 0, y = −x.Straight-line orbits: m = 1

2 (1±√

5) =⇒ y = 12 (1±

√5)x.

The phase portrait is shown below. Nullclines are dashed. Straight-line orbits arethe thicker lines.



Eigenvalues: −2,−1. Nullclines: y = 0, y = − 23x.

Straight-line orbits: m = −2,−1 =⇒ y = −2x, y = −x.The phase portrait is shown below. Nullclines are dashed. Straight-line orbits arethe thicker lines.


Eigenvalues: ±2. Nullclines: y = 13x, y = −x.

Straight-line orbits: m = − 13 , 1 =⇒ y = − 1

3x, y = x.The phase portrait is shown below. Nullclines are dashed. Straight-line orbits arethe thicker lines.

9.2.23. Eigenvalues: −1± 2i. Nullclines: y = x, y = −4x.There are no straight-line orbits, since m = −4−m

−1+m has no real solutions.


9.2.24. Eigenvalues: −1,−1. Nullclines: y = 34x, y = x.

Straight-line orbits: m = 12 =⇒ y = 1

2x.

9.2.25. Eigenvalues: 0, 2. Nullclines: y = x, on which every point is an equilibrium point.Straight-line orbits: dy

dx = −1 everywhere except on y = x. So all nontrivial orbitstrace out straight lines with slope −1.

9.3 Phase Portraits of Nonlinear Systems

9.3.1. (a) We want to find all simultaneous solutions of x(y−x−2) = 0 and x2−y = 0. Beginby solving x2−y = 0 for y and then substituting the result into x(y−x−2) = 0.That gives x(x2− x− 2) = x(x+ 1)(x− 2) = 0, whose solutions are x = 0,−1, 2.Now using y = x2 again, we get equilibrium points (0, 0), (−1, 1), and (2, 4).

(b) J (x, y) =(−2− 2x+ y x

2x −1

). Then

J (0, 0) =(−2 00 −1

)has eigenvalues − 2, 1,

J (−1, 1) =(

1 −1−2 −1

)has eigenvalues ±

√3,

J (2, 4) =(−2 24 −1

)has eigenvalues

12

(−3±√

33).

(c) (0, 0) is a stable node; (−1, 1) and (2, 4) are saddle points.


(e)

-6 -4 -2 2 4 6

-4

-2

2

4

9.3.2. (a) We want to find all simultaneous solutions of x(1 − y) = 0 and −y(1 − x) = 0.First, x(1−y) = 0 if x = 0 or y = 1. If x = 0, then−y(1−x) = −y = 0 =⇒ y = 0.This gives the equilibrium point (0, 0). If y = 1, then y′ = −y(1 − x) = x − 1 =0 =⇒ x = 1. This gives the equilibrium point (1, 1).

(b) J (x, y) =(

1− y −xy −1 + x

). Then

J (0, 0) =(

1 00 −1

)has eigenvalues ± 1,

J (1, 1) =(

0 −11 0

)has eigenvalues ± i.

(c) (0, 0) is a saddle point; (1, 1) is a possible center or spiral point.(e)

-6 -4 -2 2 4 6

-4

-2

2

4

9.3.3. (a) Solving x(1− y) = 0 and x− y = 0 gives (0, 0) and (1, 1).

(b) J (x, y) =(

1− y −x1 −1

). Then

J (0, 0) =(

1 01 −1


J (1, 1) =(

0 −11 −1

)has eigenvalues

12

(−1± i√

3).

(c) (0, 0) is a saddle point; (1, 1) is stable spiral point


(e)

-4 -2 2 4

-3

-2

-1

1

2

3

9.3.4. (a) Solving y = 0 and x(4− x) = 0 gives (0, 0) and (4, 0).

(b) J (x, y) =(

0 14− 2x 0

). Then

J (0, 0) =(

0 14 0


J (4, 0) =(

0 1−4 0

)has eigenvalues ± 2i.

(c) (0, 0) is a saddle point; (4, 1) is possible center or spiral point.(e)

-6 -4 -2 2 4 6

-4

-2

2

4

9.3.5. (a) Solving 2− x2 − y2 = 0 and x− y = 0 gives (−1,−1) and (1, 1).

(b) J (x, y) =(−2x −2y

1 −1

). Then

J (−1,−1) =(

2 21 −1

)has eigenvalues

12

(1±√

17),

J (1, 1) =(−2 −21 −1

)has eigenvalues

12

(3± i√

7).

(c) (−1,−1) is a saddle point; (1, 1) is stable spiral point.


(e)

-4 -2 2 4

-3

-2

-1

1

2

3

9.3.6. (a) Solving 2− x2 − y2 = 0 and y(x− y) = 0 gives (−1,−1), (1, 1), and (±√

2, 0).

(b) J (x, y) =(−2x −2yy x− 2y

). Then

J (−1,−1) =(

2 2−1 1

)has eigenvalues

12

(3± i√

7),

J (1, 1) =(−2 −21 −1

)has eigenvalues

12

(−3± i√

7),

J (−√

2, 0) =

(√8 0

0 −√

2

)has eigenvalues −

√2,√

8,

J (√

2, 0) =

(−√

8 00

√2

)has eigenvalues −

√8,√

2.

(c) (±√

2, 0) are saddle points; (−1,−1) is an unstable spiral point; (1, 1) is a stablespiral point.

(e)

-4 -2 2 4

-3

-2

-1

1

2

3

9.3.7. (a) x2 + y2 − 2 = 0 and x2 − y2 = 0 at (1,±1) and (−1,±1).

(b) J (x, y) =(

2x 2y2x −2y

). Then

J (−1,−1) =(−2 −2−2 2

)and J (1, 1) = −J (−1,−1) have eigenvalues ±

√8,

J (−1, 1) =(−2 2−2 −2

)and J (1,−1) = −J (−1, 1) have eigenvalues −2± 2i.


(c) (−1,−1) and (1, 1) are saddle points; (−1, 1) is a stable spiral point; (1,−1) isan unstable spiral point.

(e)

-4 -2 2 4

-3

-2

-1

1

2

3

9.3.8. (a) −xy + 3 = 0 and −x2y + 2x+ 1 = 0 at (1, 3).

(b) J (x, y) =(−y −x

2− 2xy −x2

). J (1, 3) =

(−3 −1−4 −1

)has eigenvalues − 2±

√5.

(c) (1, 3) is a saddle point.(e)

-4 -2 2 4

-3

-2

-1

1

2

3

9.3.9. (a) x2y + y + 1 = 0 and −x2y + x− y = 0 at (1, 1/2).

(b) J (x, y) =(

2xy 1 + x2

1− 2xy −1− x2

). J (1, 1/2) =

(1 20 −2

)has eigenvalues − 2, 1.

(c) (1, 1/2) is a saddle point.

(e)

-4 -2 2 4

-3

-2

-1

1

2

3


9.3.10. The Jacobian is J (x, y) =(

0 1−k cosx 0

). For odd n, we get J (±nπ, 0) =

(0 1k 0

),

which has eigenvalues ±√k. Hence these equilibrium points are saddle points. Each

odd n corresponds to the unstable position at the top of the pendulum’s arc.

9.3.11. The x equation implies that y = 0 at every equilibrium point. Since ρ(0) = 0, itfollows from the y equation that the equilibrium points are (±nπ, 0), n = 0, 1, 2, . . .

The Jacobian is J (x, y) =(

0 1−k cosx −ρ′(y)

). For odd n, we get

J (±nπ, 0) =(

0 1k −ρ′(0)

)which has eigenvalues

−ρ′(0)±√ρ′(0)2 + 4k

2.

Since the eigenvalues are real and of opposite sign, these equilibrium points are saddlepoints. For even n, we get

J (±nπ, 0) =(

0 1−k −ρ′(0)

)which has eigenvalues

−ρ′(0)±√ρ′(0)2 − 4k

2.

If ρ′(0) is small relative to k, we get a pair of complex conjugate eigenvalues withnegative real parts. Therefore the equilibrium points are stable spiral points. If ρ′(0)is large, we get two negative real eigenvalues. Therefore the equilibrium points arestable nodes.As in the undamped case, each odd n corresponds to the unstable position at thetop of the pendulum’s arc, and each even n corresponds to the stable position at thebottom of the pendulum’s arc. When ρ′(0) is small relative to k, damped oscillationsoccur, corresponding to a stable spiral point. For larger values of ρ′(0), dampingprevents oscillation and causes the equilibrium point to be a stable node.

9.3.12. The x equation implies that y = 0 at every equilibrium point. The y equation thenimplies that x = 0 at every equilibrium point. Thus the sole equilibrium point is(0, 0). The Jacobian is

J (x, y) =(

0 1−1− 2kxy −k(x2 − 1)

),

and so

J (0, 0) =(

0 1−1 k

), which has eigenvalues

12

(k ±

√k2 − 4

).

A plot of the real parts of the eigenvalues (for k > 0) is shown below.

1 2 3 4

1

2

3

k

Re λ


The eigenvalues are nonreal when 0 < k < 2, corresponding to the straight part ofthe graph above. The following can be inferred about the type of the equilibriumpoint as k changes:

0 < k < 2 : unstable spiral point,

k = 2 : unstable coaxial node,

k > 2 : unstable biaxial node.

9.3.13. The equilibrium point is (k2, 1/k). The Jacobian is

J (x, y) =(−y2 −2xyy2 2xy − k

),

and so

J (k2, 1/k) =(−1/k2 −2k1/k2 k

), which has eigenvalues − 1− k3 ±

√k6 − 6k3 + 1

2k2.

A plot of the real parts of the eigenvalues (for k > 0) is shown below.

0.5 1 1.5 2 2.5 3

-4

-3

-2

-1

1

2

k

Re λ

The eigenvalues are nonreal when 3 − 2√

2 < k3 < 3 + 2√

2. The following can beinferred about the type of the equilibrium point as k changes:

0 < k3 < 3− 2√

2 : stable node

3− 2√

2 < k3 < 1 : stable spiral point

k = 1 : possible center

1 < k3 < 3 + 2√


k3 > 3 + 2√

2 : unstable node

Phase portraits for each case are shown below.Stable node if 0 < k3 < 3− 2

√2. Stable spiral point if 3− 2

√2 < k3 < 1.


Possible center if k = 1. Unstable spiral point if 1 < k3 < 3 + 2√

2.

Unstable node if k3 > 3 + 2√

2.

9.3.14. The nonnegative equilibrium points are (0, 0) and (1, 1).

J (x, y) =(

1− 2xy −x2

y x− 2y

).

J (0, 0) =(

1 00 0

)has eigenvalues 0, 1.

J (1, 1) =(−1 −11 −1

)has eigenvalues − 1± i.

So (0, 0) is unstable, and (1, 1) is a stable spiral. The phase portrait is shown below.

1 2 3 4

1

2

3

4


9.3.15. The nonnegative equilibrium points are (0, 0), (1, 1), and (0, 2).

J (x, y) =

(1− 3x2y −x3

−4xy(1+x2)2

21+x2 − 2y

).

J (0, 0) =(

1 00 2


J (1, 1) =(−2 −1−1 −1

)has eigenvalues

12

(−3±√

5).

J (0, 2) =(

1 00 −2


So (0, 0) is an unstable node; (1, 1) and (0, 2) are saddle points.

1 2 3 4

1

2

3

4

9.3.16. The equilibrium points are (0, 0), (1, 0), (2, 0), and (3, 0).

J (x, y) =(

0 15x4 − 28x3 + 51x2 − 34x+ 6 0

).

J (0, 0) =(

0 16 0

)has eigenvalues ±

√6.

J (1, 0) =(

0 10 0


J (2, 0) =(

0 1−2 0

)has eigenvalues ± i

√2.

J (3, 0) =(

0 112 0

)has eigenvalues ±

√12.

So (0, 0) and (3, 0) are saddle points; (2, 0) is a center; the nature of (1, 0) is notdetermined.

9.3.17. (0, 0) and (3, 0) are stable spiral points, and (2, 0) is a saddle point. (1, 0) has char-acteristics of a stable node and a saddle point, consistent with the fact that theeigenvalues of J (1, 0) are −1 and 0.


9.3.18. The equilibrium points are (0, 0), (0, 2), (5/7, 4/7), (6/7, 2/7), (2, 0), and (3, 0). TheJacobian here is quite complicated, so a computer will be useful. It turns out that

J (0, 0) =(−6 00 1

)has eigenvalues − 6, 1,

J (0, 2) =(−6/13 0−2 −1

)has eigenvalues − 1,−6/13,

J (5/7, 4/7) =(−1/14 −2/7−4/7 −2/7

)has eigenvalues

128

(−5±√

137),

J (6/7, 2/7) =(

6/35 24/35−2/7 −1/7

)has eigenvalues

170

(1± i√

839),

J (2, 0) =(

2/5 8/50 −1

)has eigenvalues − 1, 2/5,

J (3, 0) =(−3/10 −6/5

0 −2

)has eigenvalues − 2,−3/10.

So (0, 0), (5/7, 4/7), and (2, 0) are saddle points; (0, 2) and (0, 3) are stable nodes;(6/7, 2/7) is an unstable spiral point.

9.3.19. (a) The equilibrium points are (0, 0) and (5/2, 0).

J (x, y) =(

0 6− 4y−10 + 8x+ 2y −1 + 2x+ 8y

).

J (0, 0) =(

0 6−10 −1

)has eigenvalues

12

(−1± i√

239).

J (5/2, 0) =(

0 610 4


So (0, 0) is a stable spiral point while (5/2, 0) is a saddle point.

(b)

-1 1 2 3 4

-2

-1

1

2

(c) First, (x − 1/2)2 + y2 = 1 =⇒ 2(x − 1/2) + 2y dydx = 0 =⇒ dy

dx = 1−2x2y . Also,

the equation of the circle is equivalent to 4x2 + 4y2 = 3 + 4x. Use this in theright-hand side of the slope equation as follows:

4x2 + 2xy + 4y2 − 10x− y6y − 2y2

= · · · = 3− 6x+ 2xy − y6y − 2y2

=(3− y)(1− 2x)

2y(3− y)=

1− 2x2y

.

Therefore the slope field is tangent to the circle at each point on it.


9.3.20. (a) The x equation implies that either x = 0 or y = 0 at every equilibrium point,and y equation implies that y = ±x at every equilibrium point. Thus the onlyequilibrium point is (0, 0).

J (x, y) =(−2y −2x2x −2y

); thus J (0, 0) =

(0 00 0

).

(c) First, x2 + y2 − 2ax = 0 =⇒ 2x + 2y dydx − 2a = 0 =⇒ dy

dx = a−xy . Also, on

these circles we have y2 = 2ax− x2. So the right hand side of the slope equationbecomes

−x2 − y2

2xy=−x2 + 2ax− x2

2xy=−2x(x− a)

2xy=a− xy

.

-4 -2 2 4

-4

-2

2

4

9.3.21. (a) The orbits follow the level curves of the surface.(b) Nearby orbits trace out closed curves around (x0, y0) so this point is a center

point.

9.3.22. (a) The equilibrium points are (0, 0) and (±1/√

2, 0).

J (x, y) =(

0 14− 24x2 0

).

J (0, 0) =(

0 14 0

)has eigenvalues ± 2.

J (±1/√

2, 0) =(

0 1−8 0

)has eigenvalues ± i

√8.

(b) Solve the separable equation to get the implicit family 4x4 − 4x2 + y2 = C.

(c) Solving ∂∂x (4x4−4x2 +y2) = ∂

∂y (4x4−4x2 +y2) = 0 yields (0, 0) and (±1/√

2, 0).Now apply the second derivative test for functions of two variables: Let D(x, y) =ϕxxϕyy − (ϕxy)2. Since D(0, 0) < 0, we know ϕ(0, 0) is neither a max nor a min.On the other hand D(±1/

√2, 0) = 32 > 0 and ϕxx(±1/

√2, 0) = 16 > 0 so a

local min occurs at these points.A surface plot and a plot of the level curves are shown below.


-1-0.5

0

0.5

1 -1

-0.5

0

0.5

1

y-1

0

1

φ

--0.5

0

0.5x-1 -0.5 0 0.5 1

x-1

-0.5

0

0.5

1

y

(d) With the help of the identity sin 2θ = 2 sin θ cos θ, it is straightforward to checkthat (sin θ, sin 2θ) satisfies 4x4 − 4x2 + y2 = 0. The phase portrait (including theLissajous curve) is shown below.

-1.5 -1 -0.5 0.5 1 1.5

-2

-1

1

2

9.3.23. (a) Solve the separable equation to get the implicit family y2 − 2 cosx = C.(b) Apply the second derivative test for functions of two variables: Let D(x, y) =

ϕxxϕyy − (ϕxy)2. Here, D(x, y) = 4 cosx. When n is even, D(±nπ, 0) = 4 > 0and ϕxx(±nπ, 0) = 2 > 0; so a local min occurs at each of these points.

9.4 Limit Cycles

1. (a) p′ = 2xx′ + 2yy′ = 2x(y + x(1− x2 − y2)) + 2y(−x+ y(1− x2 − y2))= 2(x2 + y2)(1− x2 − y2) = 2p(1− p)

(b) Suppose r2 ≤ p(0) ≤ 1. Then by (a), p(t) is nondecreasing for t ≥ 0, and p(t)→ 1as t→∞. On the other hand, if 1 ≤ p(0) ≤ R2, then p(t) is nonincreasing for allt ≥ 0, and p(t)→ 1 as t→∞.

(c) To see that (0, 0) is the only equilibrium point of the system, look at yf(x, y)−xg(x, y). As in (b) , if 0 < p(0) < 1 then p(t) is increasing for all t ≥ 0 andp(t) → 1 as t → ∞. Therefore any orbit initially near the origin will move awayfrom the origin.

(d) We can conclude that there exists a periodic orbit of the system in Ar,R.


9.4.2. Take M = (x, y) | x2 + y2 = r2 ≤ 1 and let p(t) = x(t)2 + y(t)2 as in Problem1. Since p′(t) ≤ 0 when x2 + y2 ≤ 1, the hypotheses of Theorem 2 are satisfied.Therefore M is a forward invariant region. None ofthese disks contain a periodicorbit since (0, 0) is asymptotically stable. See Figure 3b in the section.

9.4.3. (a) p′ = 2xx′ + 2yy′ = 2x(ay + xϕ) + 2y(−ax+ yϕ) = 2(x2 + y2)ϕ = 2pϕ(b) Notice that if R2 ≤ p(t) ≤ r2, then p′(t) < 0. Therefore every closed disk centered

at the origin with radius r ≥ R is a forward invariant region.(c) Solve f(x, y) = 0 and g(x, y) = 0 simultaneously to see that (0, 0) is the only

equilibrium point of the system. Moreover, p′(t) > 0 when (x(t), y(t)) is near theorigin. Thus any orbit initially near the origin will move away from the origin.

(d) We can conclude that there exists a periodic orbit of the system in the diskx2 + y2 ≤ R2.

9.4.4. Assume that a, b, c are positive. Let p = bx2 + ay2. Then p′ = 2(bx2 + acy2)ϕ. LetR ≥ R be such that x2 + y2 ≥ R2 whenever bx2 + acy2 ≥ R2. Then p′ < 0 wheneverx2 + y2 ≥ R2. Thus every closed disk centered at (0, 0) with radius r ≥ R is aforward-invariant region. All of the same conclusions as in Problem 3 can now bereached.

9.4.5. (a) If x(t) ≥ 0 and y(t) = 0, then y′(t) = bx(t) + β ≥ 0. If x(t) = 0 and y(t) ≥ 0,then x′(t) = py(t) + α ≥ 0. Therefore, no orbit can leave the first quadrant.

(b) First observe that x′ + y′ = (a + b)x + (p + q)y + α + β. Comparing the linesx+ y = k and (a+ b)x+ (p+ q)y + α+ β = 0 and using the fact that a+ b < 0,p+ q < 0, α+ β ≥ 0, and the assumption on k, we see that x+ y = k lies above(a + b)x + (p + q)y + α + β = 0 in the first quadrant. Thus x′ + y′ < 0 whenx+ y = k.

(c) By (a) and (b), the triangular region is a forward invariant region when k ≥ 4/3.Solving f(x, y) = 0 and g(x, y) = 0 simultaneously, we see that (1/2, 2/3) isthe only equilibrium point, and it is a stable spiral point. Therefore every firstquadrant orbit approaches (1/2, 2/3) as t → ∞. The phase portrait is shownbelow.

0.2 0.4 0.6 0.8 1 1.2 1.4

0.2

0.4

0.6

0.8

1

1.2

1.4


9.4.6. (a) Solve f(x, y) = g(x, y) = 0 to see that (0, 0) and (3/2, 6/5) are the only equilib-rium points of the system with nonnegative entries. Now

J (x, y) =(−2 + 2xy 1/4 + x2

1− 2xy 1− x2

),

J (0, 0) =(−2 1/41 1

)has eigenvalues

12

(−1±√

10),

J (3/2, 6/5) =(

8/5 5/2−13/5 −5/4

)has eigenvalues

140

(7± i√

7151).

(b) (x+ 2y)′ = −x2y + 9y/4 = (9/4− x2)y(c) For each fixed y > 0, the minimum value of x2y−2x+y/4 occurs where 2xy−2 =

0, i.e., at x = 1/y. Therefore, x2y − 2x + y/4 ≥ 1/y − 2/y + y/4 = y2−44y for all

y > 0 and all x. Therefore, if y > 2, then

dy

dx=−x2y + x+ y

x2y − 2x+ y/4≤ (−x2y + x+ y) 4y

y2 − 4≤ 4y2 + 1

y2 − 4,

where the last inequality comes from maximizing −x2y+ x+ y (for fixed y > 0).(d) Observe that 4y2+1

y2−4 is decreasing for y > 2 with limit 4 as y → ∞. Thus, for ally > 5 its value is less than 101/21, which is less than 5.

(e, f)

9.4.7. The only equilibrium point is (1, 3), which is an unstable spiral point. A forward-invariant region containing (1, 3) is

(x, y) | x ≥ 0, y ≥ 0, y ≤ 94

(x+ 5), x+ y ≤ 334

.

This follows from:(i) x′ = 1 when x = 0, and y′ = 3x ≥ 0 when y = 0 and x ≥ 0.

(ii) (x+ y)′ = 1− x ≤ 0 if x ≥ 1.

(iii) For fixed y > 0, −x2y + 3x is maximized by x = 32y ; so y′ ≤ 9

4y for all y > 0.

(iv) For fixed y > 0, x2y−4x+1 is minimized by x = 2y ; so x′ ≥ 1−4/y for all y > 0.

(v) By (iii) and (iv), dydx ≤

9/(4y)1−4/y = 9

4y−16 ; thus dydx ≤

94 if y ≥ 5.

(vi) The line x+ y = c intersects the line y = 94 x+ 5 at x = 1, if c = 33/4.

9.4.8. (a) (i) x′ = a > 0 when x = 0, and y′ = b > 0 when y = 0.(ii) (x+ y)′ = a+ b− x ≤ 0 if x ≥ a+ b.


(iii) y′ ≤ b and x′ ≥ a− 14y = 4ay−1

4y for all (x, y) in the first quadrant; thus dydx ≤

4by4ay−1

for all (x, y) in the first quadrant. Therefore, dydx ≤

2b/a1 = 2b/a for all x > 0 and

y > 12a .

(iv) The line x + y = c intersects the line y = 2ba x + 1

2a at x = a + b, if c =2(a+2b)(a+b)+1

2a .From (i)–(iv) it follows that the region in the first quadrant bounded by the linesy = 2b

a x+ 12a and x+ y = 2(a+2b)(a+b)+1

2a is forward invariant.

9.4.10. There is no equilibrium point, and hence no periodic orbit.

9.4.11. Every periodic orbit must enclose the origin. Therefore, since the first quadrant is aforward-invariant region (by Theorem 2), no periodic orbit can exist.

9.4.12. The only equilibrium point is (1, 1); so if there is a periodic orbit, it must enclose(1, 1).

9.4.13. Every periodic orbit must enclose a point on the line y = x, which consists en-tirely of equilibrium points. This implies that a periodic orbit must pass through anequilibrium point, which is impossible. Therefore, no periodic orbit can exist.

9.4.14. The system has no periodic orbit. One example is x′ = −xy, y′ = (1− x)2(1− y).

9.4.15. (a) Let p be the period of the orbit. By Green’s theorem,∫∫

Ω(fx + gy) dx dy =∫ p

0(−g x′ + f y′)dt =

∫ p

0(−g f + f g)dt = 0.

(b) fx + gy = −1 + cos y − 1 + sinx ≤ 0 for all x, y and not identically zero on anyregion; therefore, for any region Ω,

∫∫Ω

(fx + gy) dx dy 6= 0.(c) With y = x′, the equation becomes the system x′ = y, y′ = −ρ(x)y−µ(x). Thus

fx + gy = −ρ(x). The result now follows.

9.4.16. (a) Clearly (0, 0), (±1, 0) are equilibrium points of this system.

J (x, y) =

(−x(2− 6x2 + 3x4 + 2y2) 1− 2(x2 − 1)y

1− 3x2 + 2xy − 2x3y x2 − 12x

4 − 3y2

).

J (0, 0) =(

0 11 0

)has eigenvalues ± 1.

J (1, 0) =(

1 1−2 1

2

)has eigenvalues

14

(3± i√

31 ).

J (−1, 0) =(−1 1−2 1

2

)has eigenvalues

14

(−1± i√

23 ).

Therefore, (0, 0) is a saddle point, (1, 0) is an unstable spiral point, and (−1, 0)is a stable spiral point.

(b) Let us consider the behavior of orbits crossing the x axis near (±1, 0). First notethat on the x axis, we have y′ = x(1 − x2). Now, x > 1 implies y′ < 0 while0 < x < 1 implies y′ > 0. Therefore, the flow around the spiral point (1, 0) isclockwise. Similarly, −1 < x < 0 implies y′ < 0 while x < −1 implies y′ > 0.Therefore the flow around the spiral point (−1, 0) is clockwise.


(d)

-1.5 -1 -0.5 0.5 1 1.5

-1

-0.5

0.5

1

9.4.17. (a) The equilibrium points are (0, 0), (0, 1), (− 12 ,

12 ), ( 1

2 ,12 ). The x-nullcline is the

circle x2 + (y − 12 )2 = ( 1

2 )2, and the y-nullclines are the lines x = 0 and y = 12 .

(b) Suppose that y(0) = 12 . Then y(t) = 1

2 for all t, and so x′ = x2 − 14 for all t.

Therefore, x′(t) ≤ 0 for all t if x(0) ≤ 12 and x′(t) > 0 for all t if x(0) > 1

2 .(c) Differentiate the ellipse equation implicitly with respect to x.(d) d

dtϕ = yy′ − (2xyx′ + x2y′) − y2y′ + xx′ = (x − 2xy)x′ + (y − x2 − y2)y′ =x(1 − 2y)(x2 + y2 − y) + (y − x2 − y2)x(1 − 2y) = 0. Observe that ϕ(x, 1

2 ) isconstant and that implicit differentiation of ϕ(x, y) = C (with respect to x)yields the slope equation in (c).

(e) ϕx = ϕy = 0 and ϕxxϕyy − ϕ2xy = 2y(12y − 6) − 24x2 > 0 at each of (0, 0) and

(0, 1). Therefore, those equilibrium points are surrounded by periodic orbits.(f)

-1 1

1

9.4.18. Consider a closed level curve Γ of ϕ that surrounds (x∗, y∗). The gradient ∇ϕ =(ϕx, ϕy) is orthogonal to Γ and points out of the region enclosed by Γ (since ϕ

increases in the direction of ∇ϕ). The vector (f, g) is tangent to any orbit, andthe cosine of the angle θ between (f, g) and ∇ϕ at any point has the same sign as∇ϕ · (f, g) = ϕxf +ϕyg, which we assume is nonpositive on Γ. Thus θ ≥ π/2, whichimplies that (f, g) is either tangent to Γ or points into the region enclosed by Γ.

9.4.19. ϕxf+ϕyg = 2xy−2yx = 0. Therefore, disks centered at (0, 0) are forward-invariant.(In fact, orbits are circular.)

9.4.20. ϕxf + ϕyg = 2x(y − x) + 2y(−x− y) = −2(x2 + y2) ≤ 0. Therefore, disks centeredat (0, 0) are forward-invariant.

9.4.21. ϕxf + ϕyg = 2x(y2 − x3)− 2xy2 = −2x4 ≤ 0. Therefore, disks centered at (0, 0) areforward-invariant. Also, (0, 0) is asymptotically stable.


9.5 Beyond the Plane

9.5.1. Solving f(x, y, z) = g(x, y, z) = h(x, y, z) = 0 yields only (0, 0, 0). Now

J (x, y, z) =

−a a 0b− z −1 −xy x −c

=⇒ J (0, 0, 0) =

−a a 0b −1 00 0 −c

,

which has eigenvalues

−c, 12

(−(a+ 1)±√

(a− 1)2 + 4ab).

Since a > 0, 0 ≤ b < 1, and c > 0, all the eigenvalues are real and negative. Hencethe equilibrium point is asymptotically stable.

9.5.2. Solving f(x, y, z) = g(x, y, z) = h(x, y, z) = 0 yields (0, 0, 0), (−9.30,−9.30, 27), and(9.30, 9.30, 27). The Jacobian is

J (x, y, z) =

−10 10 028− z −1 −xy x −3.2

.

Then

J (0, 0, 0) has eigenvalues λ = −22.83, 11.83,−3.2,

J (−9.30,−9.30, 27) has λ = −14.20, .002± 11.03i,

J (9.30, 9.30, 27) has λ = −14.20, .002± 11.03i.

At each equilibrium point, J has an eigenvalue with positive real part. Thereforeeach equilibrium point is unstable.

9.5.3. Solving f(x, y, z) = g(x, y, z) = h(x, y, z) = 0 yields (0, 0, 0), (−9.35,−9.35, 27), and(9.35, 9.35, 27). The Jacobian is

J (x, y, z) =

−10 10 028− z −1 −xy x −3.24

.

Then

J (0, 0, 0) has eigenvalues − 22.83, 11.83,−3.24,

J (−9.35,−9.35, 27) has eigenvalues − 14.23,−.006± 11.09i,

J (9.35, 9.35, 27) has eigenvalues − 14.23,−.006± 11.09i.

At each of the two nonzero equilibrium points, J has eigenvalues with negative realpart. Therefore each such equilibrium point is asymptotically stable.


9.5.4. (a) d`dt = 0; so `(x, y, z) is a conserved quantity.

(b) d`dt ≤ 0 always and the origin is a minimum of `. Therefore ` is a Lyapunovfunction.

5. (a) d`dt = −2(x2 + x2z2 + y2z2); so `(x, y, z) is not a conserved quantity.

(b) d`dt ≤ 0 always and the origin is a minimum of `. Therefore ` is a Lyapunovfunction.

9.5.6. (a) d`dt = −2x2z − 2y2z + 2xyz2; so `(x, y, z) is not a conserved quantity.

(b) d`dt = −2x2z − 2y2z + 2xyz2 = −2z(x + y)2 > 0 whenever z < 0. Therefore ` isnot a Lyapunov function.

9.5.7. (a) d`dt = 2x2 − 2xy + 2x2z − 2xyz; so `(x, y, z) is not a conserved quantity.

(b) d`dt = 2x2 − 2xy + 2x2z − 2xyz = 2x(x − y)(1 + z), which is positive whenever0 < y < x and z > −1. Therefore ` is not a Lyapunov function.

9.5.8. d`dt = `xx

′ + `yy′ = −g(x)h(x, y)f(y) + f(y)h(x, y)g(x) = 0.

9.5.9. Take h(x, y) = x2 + y3, f(y) = y, and g(x) = −2x. Then use (a) to get `(x, y) =x2 + 1

2y2. Then (0, 0) is a minimum of ` and d`

dt = 2xx′ + yy′ = 0. Therefore, ` isa Lyapunov function. Now apply the Lyapunov stability theorem to conclude that(0, 0) is a stable equilibrium point.

9.5.10. Write the system as x′ = k1xy(

a−yy

), y′ = −k2xy

(b−x

x

). With

h(x, y) = xy, f(y) = k1

(a− yy

), and g(x) = −k2

(b− xx

),

apply (a) to get

`(x, y) = −k2x− k1y + bk2 lnx+ ak1 ln y.

Now we show that ` is a Lyapunov function. First, `x(x, y) = −k2 + bk21x and

`y(x, y) = −k1 + ak11y , which are each 0 at (b, a). Also,

`xx`yy − (`xy)2 = (−bk2

x2)(−ak1

y2)− 0 =

abk1k2

x2y2,

which is positive at (b, a), while `xx(b, a) < 0. Therefore, ` has a local minimum at(b, a). Finally,

d`

dt=(−k2 + bk2

1x

)x′ +

(−k1 + ak1

1y

)y′

=(−k2 + bk2

1x

)k1x(a− y) +

(−k1 + ak1

1y

)(−k2y(b− x))

= k1k2

(− x(a− y) + b(a− y) + y(b− x)− a(b− x)

)= 0.

Therefore, ` is a Lyapunov function at (b, a), which implies that (b, a) is stable.

9.5.11. dϕdt = ϕxx

′ + ϕyy′ + ϕzz

′. Replace x′, y′, z′ with the values from the given system.Simplification gives dϕ

dt = 0.

Documents

9.1 The Phase Plane - · PDF filedt ’(x(t);y(t)) = @ ... 9.2.3. (a) stable, biaxial node; (b) real, negative 9.2.4. (a) stable, coaxial node; (b) real, repeated, negative 9.2.5