Sample Solutions of Assignment 10 for MATH3270A for ODE
November 18,2013
a
Contact Mr. XIAO Yao(yxiaomath.cuhk.edu.hk)
directly if you have any questions on the
solutions
1. Consider the following Lienard equation:
d2 x
dx
+ c(x) + g(x) = 0
2
dt
dt
where c(0) = g(0) = 0. (Note that damped or undamped pendulum equations are special cases
of Lienard equation.)
(a) Write the Lienard equation as a system of two first order equations by introducing the
variable y =
dx
.
dt
(b) Show that (0, 0) is a critical point and that the system is almost linear near (0, 0).
0
(c) Show that if c(0) > 0, g (0) > 0, then (0, 0) is asymptotically stable, and that if c(0) < 0
0
or g (0) < 0, then the critical point is unstable.
Answer: a). Let y = x0 , then from the original equation, we get
x0 = y
y 0 = −c(x)y − g(x).
b). Since g(0) = 0, then (0,0) is the critical point of the system. Since −c(x)y − g(x) is
continuously differentiable, the system is almost linear in the neighborhood of (0,0).
c). The corresponding linear system near (0, 0) is
0 0
1
u
u
=
v
v0
−g 0 (0) −c(0)
Hence
p
c2 (0) − 4g 0 (0)
λ1 =
p2
−c(0) − c2 (0) − 4g 0 (0)
λ1 =
2
−c(0) +
0
Case A: c(0) > 0, g (0) > 0
0
If 4 = c2 (0) − 4g > 0, then λ2 < λ1 < 0 and (0,0) is an asymptotical stable critical point.
0
If 4 = c2 (0) − 4g < 0, then λ2 = α + βi, λ1 = α − βi where α = − c(0)
< 0 and (0,0) is an
2
asymptotical stable critical point.
0
If 4 = c2 (0) − 4g = 0, then λ2 = λ1 = − c(0)
< 0 and (0,0) is an asymptotical stable critical
2
point.
1
2
Case B: c(0) < 0
0
If 4 = c2 (0) − 4g > 0, then λ1 > λ2 > 0 and (0,0) is an unstable critical point.
0
If 4 = c2 (0) − 4g < 0, then λ2 = α + βi, λ1 = α − βi where α = − c(0)
> 0 and (0,0) is an
2
unstable critical point.
0
If 4 = c2 (0) − 4g = 0, then λ2 = λ1 = − c(0)
> 0 and (0,0) is unstable critical point.
2
0
Case C: g (0) < 0
In this case, λ1 > 0 > λ2 > 0 and (0,0) is an unstable critical point.
2. In each of the following ODEs, construct a suitable Liapunov functions of the form ax2 + cy 2 ,
where a, c are to be determined. Then show that the critical point (0, 0) is of the indicated type.
dy
dx
= −x3 + xy 2 ,
= −2x2 y − y 3 , asymp. stable.
(a)
dt
dt
dx
1
dy
(b)
= − x3 + 2xy 2 ,
= −y 3 , asymp. stable
dt
2
dt
dx
dy
(c)
= −x3 + 2y 3 ,
= −2xy 2 , stable.
dt
dt
dx
3
3 dy
(d)
=x −y ,
= 2xy 2 + 4x2 y + 2y 3 , unstable
dt
dt
Answer: a). (0,0) is an isolated critical point of the system. Let
V (x, y) = ax2 + cy 2
then
dV
= −2ax4 − 2cy 4 + (2a − 4c)x2 y 2
dt
Let a = 2, c = 1, then V (x, y) = 2x2 + y 2 is positive definite and
dV
= −4x4 − 2y 4
dt
is negative definite. Hence, (0,0) is an asymptotical stable critical point.
b). (0,0) is an isolated critical point of the system. Let
V (x, y) = ax2 + cy 2
then
dV
= −ax4 − 2cy 4 − 4ax2 y 2
dt
Let a = 1, c = 2, then V (x, y) = x2 + 2y 2 is positive definite and
dV
= −x4 − 4x2 y 2 − 4y 4 = −(x2 + 2y 2 )2
dt
is negative definite. Hence, (0,0) is an asymptotical stable critical point.
3
c). (0,0) is an isolated critical point of the system. Let
V (x, y) = ax2 + cy 2
then
dV
= −2ax4 + 4axy 3 − 4cxy 3
dt
Let a = 1, c = 1, then V (x, y) = x2 + y 2 is positive definite and
dV
= −2x4
dt
is negative semidefinite. Hence, (0,0) is a stable critical point.
d). (0,0) is an isolated critical point of the system. Let
V (x, y) = ax2 + cy 2
then
dV
= 2ax4 + 4cy 4 + (−2a + 4c)x2 y 2
dt
Let a = 2, c = 1, then V (x, y) = 2x2 + y 2 is positive definite and
dV
= 4x4 + 8x2 y 2 + 4y 4 = (x2 + y 2 )2
dt
is positive definite. Hence, (0,0) is an unstable critical point.
3. Consider the following system of equations
dx
dy
= y − xf (x, y),
= −x − yf (x, y)
dt
dt
By constructing Liaponov function of the type c(x2 + y 2 ), show that of f (x, y) > 0 in some neighborhood of (0, 0), then (0, 0) is asymptotically stable, and if f (x, y) < 0 in some neighborhood
of (0, 0), then (0, 0) is an unstable critical point.
Answer: (0,0) is an isolated critical point of the system. Let
V (x, y) = cx2 + cy 2
then
dV
= −2c(x2 + y 2 )f (x, y)
dt
Let c = 1, then V (x, y) = x2 + y 2 is positive definite and
dV
= −2(x2 + y 2 )f (x, y)
dt
If f (x, y) > 0 in some neighborhood of (0,0), then dV
is negative definite. Hence, (0,0) is an
dt
asymptotical stable critical point.
If f (x, y) < 0 in some neighborhood of (0,0), then
unstable critical point.
dV
dt
is positive definite. Hence, (0,0) is an
4
4. Consider the following 2nd order ODE;
d2 x
+ g(x) = 0
dt2
where g(x) satisfies: g(0) = 0, xg(x) > 0 for x 6= 0, |x| < k. (A typical example is g(x) =
sin x, k = π2 .)
(a) Write the equation as a system of two first order equations by introducing the variable
.
y = dx
dt
(b) Show that (0, 0) is a critical point.
(c) Show that
Z x
1 2
g(s)ds, −k < x < k
V (x, y) = y +
2
0
is positive definite, and use this result to show that (0, 0) is stable.
Answer: a). Let y = x0 then
x0 = y
y 0 = −g(x)
b). Since g(0) = 0, (0,0) is a critical point of the system.
c). Let
Z x
1 2
V (x, y) = y +
g(s)ds, −k < x < k
2
0
Then V (0, 0) = 0.
If −k < x < 0, from xg(x) > 0, we get g(x) < 0. So
Z x
Z 0
g(s)ds = −
g(s) > 0ds
0
x
If 0 < x < k, from xg(x) > 0, we get g(x) > 0. So
Z x
g(s)ds > 0
0
Hence, V (x, y) is positive definite in (−k, k) × (−1, 1)
dV
≡0
dt
i.e. dV
is negative semidefinite.
dt
From above argument, (0,0) is a stable critical point of the system.
P541:Each of Problems 1 through 6 can be interpreted as describing the interaction of two species
with populations x and y.In each of these problems,carry out the following steps.
(a)Draw a direction field and describe how solutions seem to behave.
(b)Find the critical points.
5
(c)For each critical point find the corresponding linear system.Find the eigenvalues and eigenvectors of the linear system;classify each critical point as to type,and determine whether it is
asymptotically stable,stable,or unstable.
(d)Sketch the trajectories in the neighbirhood of each critical point.
(e)Compute and plot enough trajectories of the given system to show clearly the behavior of the
solutions.
(f)Determine the limiting behavior of x and y as t → ∞ and interpret the results in terms of the
populations of the two species.
3.dx/dt = x(1.5 − 0.5x − y)
dy/dt = y(1.75 − y − 1.125x)
5.dx/dt = x(1 − x − y)
dy/dt = y(2 − y − x)
Answer: 3.F (x, y) = x(1.5 − 0.5x − y), G(x, y) = y(1.75 − y − 1.125x),and thus the critical
points are
F (x, y) = 0
⇒
Gx, y = 0
x=0
or
y=0
3
2
x=0
or
y = 1.75
x=3
or
y=0
x = 0.4
y = 1.3
At (0, 0),the coefficient matrix is
,and the eigenvalues are λ1 = 23 , λ2 = 74 ,this point is
0 74
1
0
1
2
a node, unstable; hence the corresponding eigenvectors are ξ =
,ξ =
;
0
1
1
−4
0
At (0, 74 ),the coefficient matrix is
,and the eigenvalues are λ1 = − 14 , λ2 = 47 ,this
63
− 32
− 74
1
1
point is a node,asymptotically stable;hence the corresponding eigenvectors are ξ =
, ξ2 =
− 21
16
0
1
3
− 2 −3
,and the eigenvalues are λ1 = − 32 , λ2 = − 13
,this
At (3, 0),the coefficient matrix is
8
0 − 13
8
1
point is a node, asymptotically stable; hence the corresponding eigenvectors are ξ 1 =
, ξ2 =
0
24
;
1
√
−1± 27
− 15 − 25
2 13
10
At ( 5 , 10 ),the coefficient matrix is
,and the eigenvalues are λ =
,this point
4
2
− 117
−
80
5
is a saddle point,unstable;
0
5.F (x, y) = x(1 − x − y), G(x, y) = y(2 − y − x),and thus the critical points are
F (x, y) = 0
⇒
Gx, y = 0
x=0
or
y=0
x=0
or
y=2
x=1
y=0
6
At (0, 0),the coefficient matrix is
,and the eigenvalues are λ1 = 1, λ2 = 2,this point is a
1
0
1
2
node, unstable; hence the corresponding eigenvectors are ξ =
,ξ =
;
0
1
−1 0
At (0, 2),the coefficient matrix is
,and the eigenvalues are λ1 = −1, λ2 = −2,this
−2 −2
1
1
point is a node,asymptotically stable;hence the corresponding eigenvectors are ξ =
, ξ2 =
−2
0
1
−1 −1
At (1, 0),the coefficient matrix is
,and the eigenvalues are λ1 = −1, λ2 = 1,this point
0
1
1
−2
1
2
is a saddle point, unstable; hence the corresponding eigenvectors are ξ =
,ξ =
.
0
1
1 0
0 2
P542 8. Two species of fish that compete with each other for food, but do not prey on each
other, are bluegill and redear. Suppose that a pond is stocked with bluegill and redear, and let
x and y be the populations of bluegill and redear, respectively, at time t. Suppose further that
the competition is modeled by the equations
dx
= x(1 − σ1 x − α1 y),
dt
dy
= y(2 − σ2 x − α2 y).
dt
(a) If 2 /α2 > 1 /σ1 and 2 /σ2 > 1 /α1 , show that the only equilibrium populations in the
pond are no fish, no redear, or no bluegill. What will happen?
(b) If 1 /σ1 > 2 /α2 and 1 /α1 > 2 /σ2 , show that the only equilibrium populations in the pond
are no fish, no redear, or no bluegill. What will happen?
Answer: . (a) From
x(1 − σ1 x − α1 y) = 0, y(2 − σ2 x − α2 y) = 0,
and the condition 2 /α2 > 1 /σ1 and 2 /σ2 > 1 /α1 , we have
x = 0, y = 0; x = 0, y = 2 /σ2 ; x = 1 /σ1 , y = 0.
For x = 0, y = 0, the linear system is
0 x
1 0
x
=
.
y
0 2
y
the critical point is Node, US.
For x = 0, y = 2 /σ2 , the linear system is
0 1 − σ22 α1 0
u
u
=
,
v
− σ22 α2 −2
v
7
where u = x, v = y − 2 /σ2 , the critical point is Node, AS.
For x = 1 /σ1 , y = 0, the linear system is
0 −1 − σ11 α1
u
u
=
,
1
v
0 2 − σ1 α2
v
where u = x − 1 /σ1 , v = y, the critical point is Saddle, US.
Hence x −→ 0, y −→ σ22 when t −→ ∞.
(b) Using the same method as in (a), we have x −→
1
,y
σ1
−→ 0 when t −→ ∞.
9. Consider the competition between bluegill and redear mentioned in Problem 8. Suppose that
2 /α2 > 1 /σ1 and 1 /α1 > 2 /σ2 , so as shown in the text, there is a stable equilibrium point at
which both species can coexist. It is convenient to rewrite the equations of Problem 8 in terms
of the carrying capacities of the pond for bluegill (B = 1 /σ1 ) in the absence of redear and for
redear (R = 2 /σ2 ) in the absence of bluegill.
(a). Show that the equation of Problem 8 take the form
1
γ1
dx
= 1 x(1 − x − y),
dt
B
R
1
γ2
dy
= 2 y(1 − y − x),
dt
R
R
where γ1 = α1 /σ1 and γ2 = α2 /σ2 . Determine the coexistence equilibrium point (X, Y ) in terms
of B, R, γ1 and γ2 .
(b). Now suppose that a fisherman fishes only for bluegill with the effect that B is reduced.
What effect does this have on the equilibrium populations? Is it possible, by fishing, to reduce
the population of bluegill to such a level that they will die out?
Answer: (a) By computation, we can easily show that
dx
1
γ1
= 1 x(1 − x − y),
dt
B
R
dy
1
γ2
= 2 y(1 − y − x),
dt
R
R
where γ1 = α1 /σ1 and γ2 = α2 /σ2 .
Let 1 − B1 x −
γ1
y
R
= 0 and 1 − R1 y −
γ2
x
R
= 0, we have x =
(B−γ1 R)
,y
(1−γ1 γ2 )
=
(R−γ2 B)
.
(1−γ1 γ2 )
(b) From the condition, we have σ1 σ2 > α1 α2 , that is, 1 − γ1 γ2 > 0. If B is reduced then x
reduced and y is increased. If B become less that γ1 R, then x −→ 0 and y −→ R as t −→ ∞.
10. Consider the system (2) in the text, and assume that σ1 σ2 − α1 α2 = 0.
(a) Find all the critical points of the system. Observe that the result depends on whether
σ1 2 − α2 1 is zero.
8
(b) If σ1 2 − α2 1 > 0, classify each critical point and determine whether it is asymptotically
stable, stable, or unstable. Then do the same if σ1 2 − α2 1 < 0.
(c) Analysis the nature of the trajectory when σ1 2 − α2 1 = 0.
Answer: (a) If σ1 2 − α2 1 6= 0, then the critical points are x = 0, y = 0; x = 0, y = 2 /σ2 ; x =
1 /σ1 , y = 0.
If σ1 2 −α2 1 = 0, then the critical points are x = 0, y = 0 and all points on the line σ1 x+α1 y = 1 .
(b) If σ1 2 − α2 1 > 0, for x = 0, y = 0, the linear system is
0 x
1 0
x
=
,
y
0 2
y
Node, US.
For x = 0, y = 2 /σ2 , the linear system is
0 1 − σ22 α1 0
u
u
=
,
−2
v
α
−2
v
σ2 2
where u = x, v = y − 2 /σ2 . From the condition σ1 2 − α2 1 > 0, the critical point is Node, AS.
For x = 1 /σ1 , y = 0, the linear system is
0 −1 − σ11 α1
u
u
,
=
1
0 2 − σ1 α2
v
v
where u = x − 1 /σ1 , v = y, the critical point is Saddle, US.
Use the same ideal, when σ1 2 −α2 1 < 0, we have x = 0, y = 0 ⇒ Node, US. X = 0, Y = 2 /σ2 ⇒
Saddle, US. x = 1 /σ1 , y = 0 ⇒ Node, AS.
(c) when σ1 2 − α2 1 = 0, we have (0,0) is unstable node. For σ1 x + α1 y = 0, the points are
nonisolated stable.
p543 11.Consider the system (3) in Example 1 of the text.Recall that this system has an asymptotically stable critical point at (0.5, 0.5),corresponding to the stable coexistence of the two
population species.Now suppose that immigration or emigration occurs at the constant rates of
δa and δb for the species x and y,respectively.In this case Eqs.(3) are replaced by
dx/dt = x(1 − x − y) + δa, (i)
dy/dt = y(0.75 − y − 0.5x) + δb.
The question is what effect this has on the location of the stable equilibrium point.
(a)To find the new critical point,we must solve the equations
x(1 − x − y) + δa = 0, (ii)
y(0.75 − y − 0.5x) + δb = 0.
One way to proceed is to assume that x and y are given by power series in the parameter δ;thus
x = x0 + x 1 δ + · · · ,
y = y0 + y1 δ + · · · .(iii)
9
Substitute Eqs.(iii) into Eqs.(ii) and collect terms according to powers of δ.
(b)From the constant terms(the terms not involving δ),show that x0 = 0.5 and y0 = 0.5,thus
confirming that in the absence of immigration or emigration,the critical point is (0.5, 0.5).
(c)From the terms that are linear in δ,show that
x1 = 4a − 4b, y1 = −2a + 4b.(iv)
(d)Suppose that a > 0 and b > 0 so that immigration occurs for both species.Show that the
resulting equilibrium solution may represent an increase in both populations,or an increase in
one but a decrease in the other.Explain intuitively why this is a reasonable result.
Answer: (a)Substitute (iii) into (ii) we get:


x0 (x0 + y0 − 1) = 0



 x0 (x1 + y1 ) + (x0 + y0 − 1)x1 = a
y0 (0.5x0 + y0 − 0.75) = 0


y0 (y1 + 0.5x0 ) + y1 (0.5x0 + y0 − 0.75) = b


 x = y = 0(k > 1)
k
k
Thus we get:




x0 = 0
x0 = 0
x0 = 1
x = 0.5










 0

y0 = 0
y0 = 0.75
y0 = 0
y0 = 0.5
or
or
or
x
=
−a
x
=
4a
x
=
a
+
4b
x1 = 4a − 4b




1
1
1




 y = −4b
 y = 4b
 y = −4b
 y = −2a + 4b
1
1
1
1
3
3
(b) So there are four critical points (−aδ, − 43 δ), (−4aδ, 0.75 + 43 bδ), (1 + (a + 4b)δ, −4bδ), (0.5 +
(2a − 2b + 0.25)δ, 0.5 + (2b − 0.25)δ),rule out the vanishing term,and when δ = 0,the critical point
is (0.5, 0.5)
(c) As is shown in (a)
(d)As the equilibrium is x = 0.5 + (4a − 4b)δ, y = 0.5 + 2(2b − a)δ. When b < a < 2b,then both
species increase;when a > 2b or a < b,then one species increase and the other decrease.
p551-553 6. In this problem we examine the phase difference between the cyclic variations of
the predator and prey populations as given by Eqs.(24) of this section. Suppose we assume that
K > 0 and that t is measured from the time that the prey population (x) is a maximum; then
φ = 0.
√
(a)Show that the predator population (y) is a maximum at t = π/2 ac =
period of the oscillation.
T
,
4
where T is the
(b)When is the prey population increasing most rapidly? decreasing most rapidly? a minimum?
(c)Answer the same questions for predator population.
(d)Draw a typical elliptic trajectory enclosing the point (c/γ, a/α),and mark on it the points
found in parts (a),(b),(c).
10
Answer: From
√
c
c
+ K cos( act + φ),
γ γ
r
√
a c
a
K sin( act + φ),
y= +
α α a
√
and φ = 0 we can get the following results: when t = π/2 ac = T4 , y is a maximum.
x=
when t = 0, T, 2T, · · · , the prey population increasing most rapidly. When t =
T 5T
, ,···,
4 4
the
predator population is a maximum.
when t = T2 , 3T
, · · · , the prey population is a minimum. When t =
2
population is a minimun.
3T 7T
, 4 ,···,
4
the predator
P552,11.Consider the system
x0 = x(1 − σx − 0.5y), y 0 = y(−0.75 + 0.25x),
where σ > 0.Observe that this system is a modification of the system (2) in Example 1.
(a)Find all of the critical points.How does their loacation change as σ increases from zero?Observe
that there is a critical point in the interior of the first quadrant only if σ < 1/3.
(b)Determine the type and stability property in the interior of the first quadrant change (c)Draw
a direction field and phase portrait for a value of σ between zero and σ1 ;for a value of σ between
σ1 and 1/3.
Answer: (a)F (x, y) = x(1 − σx − 0.5y), G(x, y) = y(−0.75 + 0.25x))
F (x, y) = 0
x=0
x=3
x = σ1
⇒
or
or
G(x, y) = 0
y=0
y = 2 − 6σ
y=0
So the critical points are (0, 0), (3, 2 − 6σ), ( σ1 , 0).and only if 2 − 6σ > 0 so that there is a critical
point in the first quadrant,i.e.,σ < 13 1
0
(b)At (0, 0),the coefficient matrix is
,thus λ1 = −2, λ2 = −0.75,it’s a saddle point
0 −0.75
and unstable.
−3σ
−1.5
At (3, 2−6σ), the coefficient matrix is
,λ2 +3σλ+1.5(0.5−1.5σ) = 0,hence
0.5 − 1.5σ
0
√
∆ = 9σ 2 + 9σ − 3,when 0 < σ < − 21 + 621 ,∆ < 0,thus λ = a ± ib where a < 0,so it’s spriral
√
point which is asymptotically stable;when − 21 + 621 < σ < 31 , ∆ > 0 thus λ1,2 < 0,so it’s an
√
asymtotically stable node.And σ1 = − 12 + 621 1
−1 − 2σ
,λ1 = −1, λ2 = 14 ( σ1 − 3) > 0,thus it’s a saddle
At ( σ1 , 0),the coefficient matrix is
1
0 4σ
− 34
point which is unstable.
p553,12.Consider the systerm
dx/dt = x(a − σx − αy),
dy/dt = y(−c + γx),
11
where a, σ, α, c and γ are positive constants.
(a)Find all critical points of the given system.How does their location change as σ increases from
zero?Assume that a/σ > c/γ,that is σ < aγ/c.Why is this assumption necessary?
(b)Determine the nature and stability characterristics of each critical point.
(c)Show that there is a value of σ between zero and aγ/c where the critical point in the interior
of the first quadrant changes from a spiral point to a node.
(d)Describe the effect on the two populations as σ increases from zero to aγ/c.
Answer: (a)F (x, y) = x(a − σx − αy), G(x, y) = y(−c + γx)
x = γc
F (x, y) = 0
x=0
x = σa
⇒
or
or
cσ
G(x, y) = 0
y=0
y=0
y = αa − γα
Since x and y denote the population so y =
a
α
−
cσ
γα
>0⇒
a
σ
>
c
γ
to make sense.
(b)Fx = a − 2σx
− αy, Fy = αx, Gx = γy, Gy = −c + γx.
a 0
At (0, 0),A =
,thus λ1 = a > 0, λ2 = −c < 0,so it’s a unstable saddle point.
0 −c
aα
−a
a
σ
At ( σ , 0),A =
,λ1 = −a < 0, λ2 = −c + aγ
> 0,so it’s a unstable saddle point.
σ
0 −c + aγ
σ
αc
− cσ
2
2
2
cσ
γ
γ
),A =
.λ2 + cσ
λ − ac + c γσ = 0 ∆ = γc 2 σ 2 − 4 cγ σ + 4ac.With respect
At ( γc , αa − γα
aγ
cσ
γ
− α 0
α
3
0
to this σ equation,∆ = 16 γc 2 (c − a).
Case 1: If c < a,∆0 < 0,then ∆ > 0,and thus λ have two real solutions such that λ − 1 < 0 <
λ2 ,thus it’s a unstable saddle point.
Case 2: If c > a,then ∆ < 0 for 0 < σ < 2γ +
√
c2 − ac,then λ1,2 = a ± ib where a < 0,hence
√
it’s a asymptotically stable spiral point;∆ > 0 for 2γ + 2γc c2 − ac < σ < aγ/c,then λ1 < 0 < λ2 ,
thus it’s a unstable saddle point.
(c)By the work of (b),the value is 2γ +
(d)Indicated by (b).
2γ
c
√
2γ
c
c2 − ac
p562-563 In each of Problems1 through 4, construct a suitable Liapunov functions of the form
ax2 + cy 2 , where a, c are to be determined.Then show that the critical point (0, 0) is of the
indicated type.
dx
1
dy
2.
= − x3 + 2xy 2 ,
= −2y 3 , asymp. stable
dt
2
dt
dx
dy
3.
= −2x3 + 2y 3 ,
= −2xy 2 , stable.
dt
dt
dx
dy
4.
= 2x3 − y 3 ,
= 2xy 2 + 4x2 y + 2y 3 , unstable
dt
dt
2. (0,0) is an isolated critical point of the system. Let
V (x, y) = ax2 + cy 2
12
then
dV
= −ax4 − 4cy 4 + 4ax2 y 2
dt
Let a = 1, c = 1, then V (x, y) = x2 + y 2 is positive definite and
dV
= −x4 + 4x2 y 2 − 4y 4 = −(x2 − 2y 2 )2
dt
is negative definite. Hence, (0,0) is an asymptotical stable critical point.
3. (0,0) is an isolated critical point of the system. Let
V (x, y) = ax2 + cy 2
then
dV
= −4ax4 + 4axy 3 − 4cxy 3
dt
Let a = 1, c = 1, then V (x, y) = x2 + y 2 is positive definite and
dV
= −4x4
dt
is negative semidefinite. Hence, (0,0) is a stable critical point.
4. (0,0) is an isolated critical point of the system. Let
V (x, y) = ax2 + cy 2
then
dV
= 4ax4 + 4cy 4 + 8cx2 y 2 + (4c − 2a)xy 3
dt
Let a = 2, c = 1, then V (x, y) = 2x2 + y 2 is positive definite and
dV
= 8x4 + 8x2 y 2 + 4y 4
dt
is positive definite. Hence, (0,0) is an unstable critical point.
5. Consider the following system of equations
dx
dy
= y − xf (x, y),
= −x − yf (x, y)
dt
dt
By constructing Liaponov function of the type c(x2 + y 2 ), show that of f (x, y) > 0 in some neighborhood of (0, 0), then (0, 0) is asymptotically stable, and if f (x, y) < 0 in some neighborhood
of (0, 0), then (0, 0) is an unstable critical point.
Answer: (0,0) is an isolated critical point of the system. Let
V (x, y) = cx2 + cy 2
then
dV
= −2c(x2 + y 2 )f (x, y)
dt
Let c = 1, then V (x, y) = x2 + y 2 is positive definite and
dV
= −2(x2 + y 2 )f (x, y)
dt
13
If f (x, y) > 0 in some neighborhood of (0,0), then
dV
dt
is negative definite. Hence, (0,0) is an
dV
dt
is positive definite. Hence, (0,0) is an
asymptotical stable critical point.
If f (x, y) < 0 in some neighborhood of (0,0), then
unstable critical point.
p563, 7. By introducing suitable dimensionless variables, we can write the system of nonlinear
equations for the damped pendulum[Eqs.(8) of Section 9.3] as
dx/dt = y, dy/dt = −y − sinx
(a) Show that the origin is a critical point.
(b) Show that while V (x, y) = x2 + y 2 is positive definite, V˙ (x, y) takes on both positive and
negative values in any domain containing the origin, so that V is not a Liapunov function.
(c) Using the energy function V (x, y) = 12 y 2 + (1 − cos x) mentioned in Problem 6(b), show that
the origin is a stable critical point. Since there is damping in the system, we can expect that
the origin is asymptotically stable. However, is not possible to draw this conclusion using this
Liapunov function.
(d) To show asymptotic stability it is necessary to construct a better Liapunov function than the
one used in part(c). Show that V (x, y) = 21 (x + y)2 + x2 + 21 y 2 is such a Liapunov function, and
conclude the origin is an asymptotically stable critical point.
Answer: Set F (x, y) = y, G(x, y) = −y − sin x.
(a) Since F (0, 0) = G(0, 0) = 0, the origin is a critical point.
(b) It is immediate that V (x, y) = x2 + y 2 is positive definite.
V˙ (x, y) = 2xy + 2y(−y − sin x) = 2y(x − sinx − y).
For any r such that 0 < r < 1, V˙ (0, r/2) < 0, setting a = (r/2) − sin(2/r), then V˙ (r/2, a/2) < 0.
since r is arbitrary, V˙ (x, y) takes on both positive and negative values in any domain containing
the origin.
(c)V˙ (x, y) = (sin x)y + y(−y − sin y) = −y 2 is negative semidefinite, hence the origin is a stable
critical point.
(d)By taylor’s formula, sin x = x − αx3 /3!,where α depends on x but o < α < 1 for −π/2 < x <
π/2. V˙ (x, y) = (3x + y)y − (x + 2y)(y + sin x), substituting sin x = x − αx3 /3! into V˙ , we get
V˙ (x, y) = −x2 − y 2 + (α/3!)(x4 + 2x3 y). Letting x = r cos θ, y = r sin θ, V˙ (r, θ) = −r2 [1 − h(r, θ)],
where h = (α/3!)r2 (cos4 θ + cos3 α sin α). If r < 1, |h| ≤ (1/3!)(1 + 2) ≤ 1/2. Hence V˙ is negative
definite and the origin is an asymptotically stable critical point.
14
In Problems 10 and 11 we will prove part of Theorem 9.3.2: If the critical point (0,0) of the
almost linear system
dx/dt = a11 x + a12 y + F1 (x, y), dy/dt = a21 x + a22 y + G1 (x, y) (i)
is an asymptotically stable critical point of the corresponding linear system
dx/dt = a11 x + a12 y, dy/dt = a21 x + a22 y (iI)
then it is an asymptotically stable critical point of the almost linear system(i). Problem 12 deals
with the corresponding result for instability. 10. Consider the linear system(ii)
(a) Since (0,0) is an asymptotically stable critical point, show that a11 + a12 < 0 and a11 a22 −
a12 a21 > 0.
(b) Constructa Liapuno function V (x, y) = Ax2 + Bxy + Cy 2 such that V is positive definite
and V˙ is negative definite. One way to ensure that V˙ is negative definite is to choose A, B, and
C so that V˙ = −x2 − y 2 Show that this leads to the result
a221 + a222 + (a11 a22 − a12 a21 )
a12 a22 + a11 a21
a2 + a212 + (a11 a22 − a12 a21 )
,B =
, C = − 11
,
2∆
∆
2∆
where ∆ = (a11 + a22 )(a11 a22 − a12 a21 ).
A=−
(c) Using the result of part(a), show that A > 0 and then show that
(a211 + a212 + a221 + a222 )(a11 a22 − a12 a21 ) + 2(a11 a22 − a12 a21 )2
> 0.
∆2
Thus, by Theorem 9.6.4, V is positive definite.
4AC − B 2 =
Answer: (a)Let p = a11 + a12 and q = a11 a22 − a12 a21 , by Table 9.1.1 and Problem 21 of Section
9.1, we see that we must have p < 0 and q > 0 for the origin to be an asymptotically stable
critical point.
(b) Setting A, B, C and ∆ as suggested, computation shows that
V˙ = (2Ax + By)(a11 x + a12 y) + (2Cy + Bx)(a21 x + a22 y) = −x2 − y 2
(c) By (a), we see that A > 0, computation shows that
4AC − B 2 =
(a211 + a212 + a221 + a222 )(a11 a22 − a12 a21 ) + 2(a11 a22 − a12 a21 )2
>0
∆2
as require.
11. In this problem we show that the Liapunov function constructed in the preceding problem
is also a Liapunov function for the almost linear system(i). We must show that there is some
region containing the origin for which V˙ is negative definite.
(a) Show that
V˙ (x, y) = −(x2 + y 2 + (2Ax + By)F1 (x, y) + (Bx + 2Cy)G1 (x, y).
(b) Recall that F1 (x, y)/r → 0 and G1 (x, y)/r → 0 as r → 0. This means that, given any
> 0, there exists a circle r = R about the origin such that for 0 < r < R, |F1 (x, y) < r and
15
|G1 (x, y) < r. Letting M be the maximum of |2A|, |B|, and |2C|, show by introducing polar
coordinates that R can be chosen so that V˙ (x, y) < 0 for r < R.
Answer: (a) A direct computation using the definitions immediate gives the result.
(b) Converting to polar coordinate, for r 6= 0,
V˙ = −r2 + r(2A cos θ + B sin θ)F1 + r(2C sin θ + B cos θ)G1
= r2 (−1 + (A cos θ + B sin θ)(F1 /r) + r(2C sin θ + B cos θ)(G1 /r).
Since the system is almost linear, there is an R such that
|(A cos θ + B sin θ)(F1 /r) + r(2C sin θ + B cos θ)(G1 /r)| < (1/2),
and hence for r < R
1
V˙ < − r2
2
Hence V˙ is negative definite on the domain r < R.
12. In this problem we prove a part of Theorem 9.3.2 related to instability.
(a) Show that if a11 + a22 > 0 and a11 a22 − a12 a21 > 0, then the critical point (0,0) of the linear
system is unstable.
(b) The same result holds for the almost linear system(i). As in Problems 10 and 11, construct a
positive definite function V such that V˙ (x, y) = x2 + y 2 and hence is positive definite, and then
invoke Theorem 9.6.2.
Answer: (a). We consider the linear system
0 x
a11 a12
x
=
y
a21 a22
y
Let V (x, y) = Ax2 + Bxy + Cy 2 , in which A, B, C and ∆ are the same as in Problem 10. Based
on the hypothesis, the coefficients A and B are negative. Therefore, except for the origin, V (x, y)
is negative on each of the coordinate axes. Along each trajectory,
V˙ = (2Ax + By)(a11 x + a12 y) + (2Cy + Bx)(a21 x + a22 y) = −x2 − y 2
Hence V˙ (x, y) is negative definite. Theorem 9.6.2 asserts that the origin is an unstable critical
point.
(b). We now consider the system(i).Let V (x, y) = Ax2 + Bxy + Cy 2 where
a221 + a222 + (a11 a22 − a12 a21 )
a12 a22 + a11 a21
a2 + a212 + (a11 a22 − a12 a21 )
,B = −
, C = 11
,
2∆
∆
2∆
and ∆ = (a11 + a22 )(a11 a22 − a12 a21 ). Based on the hypothesis, A, B ¿ 0. Except for the origin,
A=
V (x, y) is positive on each of the coordinate axes. Along each trajectory,
V˙ = x2 + y 2 + (2Ax + By)F1 (x, y) + (2Cy + Bx)G1 (x, y).
16
Converting to polar coordinate, for r 6= 0,
V˙ = r2 + r(2A cos θ + B sin θ)F1 + r(2C sin θ + B cos θ)G1
= r2 (1 + (A cos θ + B sin θ)(F1 /r) + r(2C sin θ + B cos θ)(G1 /r).
Since the system is almost linear, there is an R such that
|(A cos θ + B sin θ)(F1 /r) + r(2C sin θ + B cos θ)(G1 /r)| < (1/2),
and hence for r < R
1
V˙ > r2
2
˙
Hence V is positive definite on the domain r < R.By Theorem 9.6.2, the origin is an unstable
critical point.