HW-10 Solution:

S10.1 - 1, 3, 5 ,19; S10.3 - 1, 5, 9, 13; S10.4 - 1, 2, 5

S10.1 - Q1

x′ = x(6− 3x)− 2xy

y′ = y(5− y)− xy

x(6− 3x)− 2xy = 0 ⇐⇒ x = 0 or 6− 3x− 2y = 0.

So, x-nullclines are x = 0 and 3x+ 2y = 6.

y(5− y)− xy = 0 ⇐⇒ y = 0 or 5− y − x = 0.

So, y-nullclines are y = 0 and x+ y = 5.

Equilibrium points are:

x = 0, y = 0 : (0, 0)

x = 0, x+ y = 5 : (0, 5)

3x+ 2y = 6, y = 0 : (2, 0)

3x+ 2y = 6, x+ y = 5 : (−4, 9).

The Jacobian is

J(x, y) =

6− 6x− 2y −2x

−y 5− 2y − x

.

J(0, 0) =

6 0

0 5

So (0, 0) is a nodal source.

J(0, 5) =

−4 0

−5 −5

The characteristic polynomial is p(λ) = λ2 + 9λ+ 20 = (λ+ 4)(λ+ 5).

p(λ) = 0 ⇐⇒ λ = −4 or − 5.

So (0, 5) is a nodal sink.

J(2, 0) =

−6 −4

0 3

The characteristic polynomial is p(λ) = λ2 + 3λ− 18 = (λ+ 6)(λ− 3).

p(λ) = 0 ⇐⇒ λ = −6 or 3.

So (2, 0) is a saddle point.

J(−4, 9) =

12 8

−9 −9

The characteristic polynomial is p(λ) = λ2 − 3λ− 36.

p(λ) = 0 ⇐⇒ λ =3±√

153

2= 7.685 or − 4.685.

So (−4, 9) is a saddle point.

10.1 - Q3

x′ = 2x− 2x2 − xy

y′ = 2y − xy − 2y2

2x− 2x2 − xy = 0 ⇐⇒ x = 0 or 2− 2x− y = 0.

So, x-nullclines are x = 0 and 2x+ y = 2.

2y − xy − 2y2 = 0 ⇐⇒ y = 0 or 2− x− 2y = 0.

So, y-nullclines are y = 0 and x+ 2y = 2.

Equilibrium points are:

x = 0, y = 0 : (0, 0)

x = 0, x+ 2y = 2 : (0, 1)

2x+ y = 2, y = 0 : (1, 0)

2x+ y = 2, x+ 2y = 2 : (2/3, 2/3).

The Jacobian is

J(x, y) =

2− 4x− y −x

−y 2− x− 4y

.

J(0, 0) =

2 0

0 2

So (0, 0) is a degenerate source.

The type of this equilibrium point for the non-linear system is unknown.

J(0, 1) =

1 0

−1 −2

The characteristic polynomial is p(λ) = λ2 + λ− 2 = (λ+ 2)(λ− 1).

p(λ) = 0 ⇐⇒ λ = −2 or 1.

So (0, 1) is a saddle point.

J(1, 0) =

−2 −1

0 1

The characteristic polynomial is p(λ) = λ2 + λ− 2 = (λ+ 2)(λ− 1).

p(λ) = 0 ⇐⇒ λ = −2 or 1.

So (1, 0) is a saddle point.

J(2/3, 2/3) =

−4/3 −2/3

−2/3 −4/3

The characteristic polynomial is p(λ) = λ2 +

8

3λ+

4

3.

p(λ) = 0 ⇐⇒ λ =(−8/3)±

√16/9

2= −2

3or − 2.

So (2/3, 2/3) is a nodal sink.

S10.1 - Q5

x′ = x(4y − 5)

y′ = y(3− x)

x(4y − 5) = 0 ⇐⇒ x = 0 or y = 5/4. So, x-nullclines are x = 0 and y = 5/4.

y(3− x) = 0 ⇐⇒ y = 0 or x = 3. So, y-nullclines are y = 0 and x = 3.

Equilibrium points are (0, 0) and (3, 5/4). The Jacobian is

J(x, y) =

4y − 5 4x

−y 3− x

.

J(0, 0) =

−5 0

0 3

. So (0, 0) is a saddle point.

J(3, 5/4) =

0 12

−5/4 0

The characteristic polynomial is p(λ) = λ2 + 15.

p(λ) = 0 ⇐⇒ λ = ±√

15i.

So (3, 5/4) is a linear center. We cannot classify this equilibrium point.

S10.1 - Q19

Part (a) r2 = x2 + y2.

Differentiate both sides by t.

2rdr

dt= 2x

dx

dt+ 2y

dy

dt

rdr

dt= x

dx

dt+ y

dy

dt

Next, consider the equation tan θ = y/x.

Differentiate both sides by t.

sec2(θ) · dθdt

=xdy

dt − ydxdt

x2

Note that sec2(θ) = r2/x2. So we have

r2

x2dθ

dt=

1

x2

(xdy

dt− y dx

dt

).

r2dθ

dt= x

dy

dt− y dx

dt

dθ

dt=

1

r2

(xdy

dt− y dx

dt

).

Part (b)

dx

dt= y + αx(x2 + y2)

dy

dt= −x+ αy(x2 + y2)

First we have

rdr

dt= x

dx

dt+ y

dy

dt

= xy + αx2(x2 + y2)− yx+ αy2(x2 + y2)

= α(x2 + y2)2 = αr4.

So,dr

dt= αr3.

Then we have

dθ

dt=

1

r2

(xdy

dt− y dx

dt

)

=1

r2

(x(−x+ αy(x2 + y2))− y(y + αx(x2 + y2))

)

=1

r2(−x2 − y2) = −1.

So,dθ

dt= −1.

Part (c)

1.dθ

dt= −1 =⇒ θ(t) = −t+ C for some constant C.

So θ is decreasing at a uniform rate. The solution is spiraling clockwise.

2. When α = 5 > 0, r′ = 5r3.

r is increasing and so the curve is spiraling out of the origin.

So the origin is a spiral source when α = 5.

3. When α = −5 < 0, r′ = −5r3.

r is decreasing and so the curve is spiraling into the origin.

So the origin is a spiral sink when α = −5.

S10.3 - Q1

(∗)

x′ = (2− x− y)x

y′ = (3− 3x− y)y

1. Suppose x = x(t) satisfies

x′ = (2− x)x and x(0) = x0 for some constant x0.

Then, (x(t), 0) is a solution to the system (∗).

Every point in the x-axis is contained in such a solution curve, which stays in the x-axis.

Hence, the x-axis is invariant.

2. Suppose y = y(t) satisfies

y′ = (3− y)y and y(0) = y0 for some constant y0.

Then, (0, y(t)) is a solution to the system (∗).

Every point in the y-axis is contained in such a solution curve, which stays in the y-axis.

Hence, the y-axis is invariant.

3. A solution curve starting in one of the four quadrants must stay in that quadrant, because to

get out it has to cross one of the axes. It cannot do so because the unique solution curve through

any point in the axis must be entirely contained in the axis.

S10.3 - Q5

(∗)

x′ = (2− x− y)x

y′ = (3− 3x− y)y

S = {(x, y) | 0 ≤ x ≤ 3 , 0 ≤ y ≤ 3}.

1. The x-axis and the y-axis are invariant. So, no solution curve can cross them.

2. On x = 3, x′ = (2− x− y)x = −3(1 + y) < 0.

So, x is decreasing, and the solution curve is moving into S.

3. On y = 3, y′ = (3− 3x− y)y = −9x < 0.

So, y is decreasing, and the solution curve is moving into S.

4. Since all the solution curve cross the boundary of S going into S, S is invariant.

S10.3 - Q9

(∗)

x′ = (2− x− y)x

y′ = (3− 3x− y)y

1. x-nullclines are x = 0 and x+ y = 2.

y-nullclines are y = 0 and 3x+ y = 3.

Equilibrium points are:

x = 0, y = 0 : (0, 0)

x = 0, 3x+ y = 3 : (0, 3)

x+ y = 2, y = 0 : (2, 0)

x+ y = 2, 3x+ y = 3 : (1/2, 3/2).

2. The Jacobian is

J(x, y) =

2− 2x− y −x

−3y 3− 3x− 2y

.

J(0, 0) =

2 0

0 3

So (0, 0) is a nodal source.

J(0, 3) =

−1 0

−9 −3

The characteristic polynomial is p(λ) = λ2 + 4λ+ 3 = (λ+ 1)(λ+ 3).

p(λ) = 0 ⇐⇒ λ = −1 or − 3.

So (0, 3) is a nodal sink.

J(2, 0) =

−2 −2

0 −3

The characteristic polynomial is p(λ) = λ2 + 5λ+ 6 = (λ+ 2)(λ+ 3).

p(λ) = 0 ⇐⇒ λ = −2 or − 3.

So (2, 0) is a nodal sink.

J(1/2, 3/2) =

−1/2 −1/2

−9/2 −3/2

The characteristic polynomial is p(λ) = λ2 + 2λ− 3/2.

p(λ) = 0 ⇐⇒ λ =−2±

√4 + 6

2= −1±

√10

2λ = 0.581 or − 2.5811.

So (1/2, 3/2) is a saddle point.

3. The flow of the solutions along the nullclines is shown by the arrows. This information shows

that regions I and III are invariant.

The solution curves flow from region II into regions I and III, except the one stable solution curve

for the saddle at (1/2, 3/2). The solution curves in region IV can flow directly to the sinks, or into

regions I and III, with the exception of the one stable solution curve for the saddle at (1/2, 3/2).

Finally, solution curves in the invariant region I flow to the sink at (0, 3) and the solution curves

in the invariant region III flow to the sink at (0, 2).

S10.3 - Q13 x′ = 1− y

y′ = y − x2

1. The x-nullcline is y = 1.

The y-nullcline is y = x2.

Equilibrium points are (1, 1) and (−1, 1).

2. The Jacobian is

J(x, y) =

0 −1

−2x 1

.

J(1, 1) =

0 −1

−2 1

The characteristic polynomial is p(λ) = λ2 − λ− 2 = (λ− 2)(λ+ 1).

p(λ) = 0 ⇐⇒ λ = 2 or − 1.

So (1, 1) is a saddle point.

J(−1, 1) =

0 −1

2 1

The characteristic polynomial is p(λ) = λ2 − λ+ 2.

p(λ) = 0 ⇐⇒ λ =1±√

1− 8

2=

1

2±√

7

2i

So (−1, 1) is a spiral source.

Phase portraits of S10.3 - Q9 and Q13 respectively.

S10.4 - Q1 x′ = y + x(1−√x2 + y2)

y′ = −x+ y(1−√x2 + y2)

r2 = x2 + y2. So we have,

rr′ = xx′ + yy′

= x(y + x(1−

√x2 + y2)

)+ y(− x+ y(1−

√x2 + y2)

)

= (x2 + y2) ·(

1−√x2 + y2

)

= r2(1− r).

∴ r′ = r(1− r).

For this differential equation, r = 1 is a stable equilibrium point.

Hence, r = 1 is an attracting limit cycle.

S10.4 - Q2 x′ = −y + x(√x2 + y2 − 3)

y′ = x+ y(√x2 + y2 − 3)

r2 = x2 + y2. So we have,

rr′ = xx′ + yy′

= x(− y + x(

√x2 + y2 − 3)

)+ y(x+ y(

√x2 + y2 − 3)

)

= (x2 + y2) ·(√

x2 + y2 − 3)

= r2(r − 3).

∴ r′ = r(r − 3).

For this differential equation, r = 3 is an unstable equilibrium point.

Hence, r = 3 is a repelling limit cycle.

S10.4 - Q5x′ = −y + x

(√x2 + y2 − 3 +

2√x2 + y2

)y′ = x+ y

(√x2 + y2 − 3 +

2√x2 + y2

)

r2 = x2 + y2. So we have,

rr′ = xx′ + yy′

= x(− y + x(r − 3 +

2

r))

+ y(x+ y(r − 3 +

2

r))

= r2(r − 3 +

2

r

)

= r3 − 3r2 + 2r.

∴ r′ = r2 − 3r + 2 = (r − 1)(r − 2).

Therefore, r = 1 is an attracting limit cycle, and r = 2 is an repelling limit cycle.