1. No category

Sample Solutions of Assignment 10 for MAT3270B
1. For the following ODEs, (a) determine all critical points; (b) find the corresponding
linear system near each critical point; (c) find the eigenvalues of each linear system; (d)
state if the system is almost linear at each critical point; (e) conclude the types and
stability of each critical point.
= 1 − y, dy
= x2 − y 2
a). dx
dt
dt
dx
b). dt = 2x + y + xy 3 , dy
= x − 2y − xy
dt
dy
dx
c). dt = (1 + x) sin y, dt = 1 − x − cos y
d). dx
= y + x(1 − x2 − y 2 ), dy
= −x + y(1 − x2 − y 2 )
dt
dt
dy
dx
e). dt = x(1.5 − x − 0.5y), dt = y(2 − y − 0.75x)
f). dx
= x(1 − x − y), dy
= y(1.5 − y − x)
dt
dt
Answer: a). From
F =
1−y
=0
G = x2 − y 2 = 0
we know that all the critical points are (−1, 1), (1, 1). Since the F, G are continuous
differentiable, the system is almost linear near the critical points.
µ
¶ µ
¶
Fx Fy
0 −1
=
Gx Gy
2x −2y
The linear system near critical point (−1, 1) is
du
= −v
dt
dv
= −2u − 2v
dt
√
√
where u = −1 + x, v = 1 + y. The eigenvalues of the linear system are −1 − 3, −1 + 3,
hence the critical point (−1, 1) is saddle point and unstable. The linear system near
critical point (1, 1) is
du
= −v
dt
dv
= 2u − 2v
dt
where u = 1 + x, v = 1 + y. The eigenvalues of the linear system are −1 − i, −1 + i and
the real parts of the eigenvalues are negative, hence the critical point (1, 1) is spiral point
and is asymptotically stable.
1
2
b). From
F = 2x + y + xy 3 = 0
G=
we know that all the critical
continuous differentiable, the
µ
Fx
Gx
x − 2y − xy
=0
points are (0, 0), (−1.19345, −1.4797). Since the F, G are
system is almost linear near the critical points.
¶ µ
¶
Fy
2 + y 3 1 + 3xy 2
=
Gy
1 − y −2 − x
The linear system near critical point (0, 0) is
du
= 2u + v
dt
dv
= u − 2v
dt
√ √
where u = x, v = y. The eigenvalues of the linear system are − 5, 5, hence the critical
point (0, 0) is saddle point and unstable.
The linear system near critical point (−1.19345, −1.4797) is
du
= −1.2399u − 8.8393v
dt
dv
= 2.4797u − 0.80655v
dt
where u = −1.19345 + x, v = −1.4797 + y. The eigenvalues of the linear system are
−1.0232±4.1125i, hence the critical point (−1.19345, −1.4797) is spiral point and asymptotically stable.
c). From
F =
(1 + x) sin y
=0
G = 1 − x − cos y = 0
we know that all the critical points are (0, 2kπ), (2, (2k + 1)π). Since the F, G are
continuous differentiable, the system is almost linear near the critical points.
¶ µ
¶
µ
sin y (1 + x) cos y
Fx Fy
=
Gx Gy
−1
sin y
The linear system near critical point (2, (2k + 1)π) is
du
= −3v
dt
dv
= −u
dt
3
√ √
where u = 2 + x, v = (2k + 1)π + y. The eigenvalues of the linear system are − 3, 3,
hence the critical point (2, (2k + 1)π) is saddle point and unstable. The linear system
near critical point (0, 2kπ) is
du
= v
dt
dv
= −u
dt
where u = x, v = 2kπ + y. The eigenvalues of the linear system are −i, i, hence the
critical point (0, 2kπ) is center point and stable of the corresponding linear system. Then
(0, 2kπ) maybe the spiral point or center point of the nonlinear system. The stability of
(0, 2kπ) is indeterminate.
d). From
F =
y + x(1 − x2 − y 2 )
=0
G = −x + y(1 − x2 − y 2 ) = 0
we know that all the critical point is (0, 0). Since the F, G are continuous differentiable,
the system is almost linear near the critical points.
µ
¶ µ
¶
Fx Fy
1 − 3x2 − y 2
1 + 2xy
=
Gx Gy
−1 − 2xy 1 − x2 − 3y 2
The linear system near critical point (2, (2k + 1)π) is
du
= u+v
dt
dv
= −u + v
dt
where u = 2 + x, v = (2k + 1)π + y. The eigenvalues of the linear system are 1 ± i and
and the real parts of the eigenvalues are positive, hence the critical point (0, 0) is spiral
point and unstable.
e). From
F = x(1.5 − x − 0.5y) = 0
G = y(2 − y − 0.75x) = 0
we know that all the critical point is (0, 0), (1.5, 0), (0, 2), (0.8, 1.4). Since the F, G are
continuous differentiable, the system is almost linear near the critical points.
¶ µ
¶
µ
1.5 − 2x − 0.5x
−0.5x
Fx Fy
=
Gx Gy
−0.75y
2 − 2y − 0.75x
4
The linear system near critical point (0, 0) is
du
= 1.5u
dt
dv
= 2v
dt
where u = x, v = y. The eigenvalues of the linear system are 1.5, 2, hence the critical
point (0, 0) is node and unstable.
The linear system near critical point (1.5, 0) is
du
= −1.5u − 0.75v
dt
dv
= 0.8725v
dt
where u = 1.5 + x, v = y. The eigenvalues of the linear system are −1.5, 0.8725, hence
the critical point (1.5, 0) is saddle point and unstable.
f). From
F =
x(1 − x − y)
=0
G = y(1.5 − y − x) = 0
we know that all the critical point are (0, 0), (1, 0) and (0, 1.5).
Since the F, G are continuous differentiable, the system is almost linear near the critical
points.
µ
¶ µ
¶
Fx Fy
1 − 2x − y
−x
=
Gx Gy
−y
1.5 − x − 2y
The linear system near critical point (0, 0) is
du
= u
dt
dv
= 1.5v
dt
where u = x, v = y. The eigenvalues of the linear system are 1, 1.5, hence the critical
point (0, 0) is node and unstable.
The linear system near critical point (1, 0) is
du
= −u − v
dt
dv
= 0.5v
dt
where u = 1 + x, v = y. The eigenvalues of the linear system are 1, −0.5, hence the
critical point (1, 0) is saddle point and unstable.
5
The linear system near critical point (0, 1.5) is
du
= −0.5u
dt
dv
= −1.5u − 1.5v
dt
where u = x, v = 1.5 + y. The eigenvalues of the linear system are −0.5, −1.5, hence the
critical point (0, 0) is node and asymptotically stable.
2. Consider the following two systems of ODEs:
(1). dx
= y + x(x2 + y 2 ), dy
= −x + y(x2 + y 2 )
dt
dt
(2). dx
= y − x(x2 + y 2 ), dy
= −x − y(x2 + y 2 )
dt
dt
(a). Show that (0,0) is a critical point for (1) and (2) and furthermore, is a center of the
corresponding linear system.
(b). Show that each system is almost linear at (0,0).
(c). Using polar coordinates, x = r cos θ, y = r sin θ, find out the corresponding ODEs
for (r, θ). Then show that (0,0) is asymptotical stable for (2) while (0,0) is unstable for (1).
Answer:
(1)Let
F (x, y) = y + x(x2 + y 2 )
G(x, y) = −x + y(x2 + y 2 )
Obviously, (0, 0) is a critical point of the system.
The linear system near (0, 0) is
µ ¶ µ
¶µ ¶
d
x
0 1
x
=
y
−1
0
y
dt
The eigenvalues of this linear system are λ1 = i, λ2 = −i. So (0, 0) is a center of the
linear system.
Let x = r cos θ, y = r sin θ then
x(x2 + y 2 )
= r2 cos θ → 0, r → 0
r
y(x2 + y 2 )
= r2 sin θ → 0, r → 0
r
Hence, the system is almost linear.
6
Under above polar coordinate, the original system can be written as
dr
= r3 > 0
dt
dθ
= −1
dt
Hence, (0,0) is an unstable critical point of the original system.
(2)Let
F (x, y) = y − x(x2 + y 2 )
G(x, y) = −x − y(x2 + y 2 )
Obviously, (0, 0) is a critical point of the system.
The linear system near (0, 0) is
µ ¶ µ
¶µ ¶
d
x
0 1
x
=
−1 0
y
dt y
The eigenvalues of this linear system are λ1 = i, λ2 = −i. So (0, 0) is a center of the
linear system.
Let x = r cos θ, y = r sin θ then
−x(x2 + y 2 )
= −r2 cos θ → 0, r → 0
r
−y(x2 + y 2 )
= −r2 sin θ → 0, r → 0
r
Hence, the system is almost linear.
Under above polar coordinate, the original system can be written as
dr
= −r3 < 0
dt
dθ
= −1
dt
Hence, (0,0) is an asymptotically stable critical point of the original system.
3. Consider the following undamped pendulum equation
Let x = θ, y =
dθ
dt
d2 θ
+ ω 2 sin θ = 0
2
dt
to obtain the system of equations
dx
dy
= y,
= −ω 2 sin x.
dt
dt
7
(a). Find out all critical points, and their corresponding linear systems, and types and
stability of the corresponding linear systems.
(b). Show that the system is almost linear. What can you say about the type and
stabilities of the critical points?
dy
(c). Solving the corresponding equation for dx
, show that the trajectories are given by
ω 2 (1 − cos x) +
y2
= c.
2
Find the trajectory passing through (0, v).
Answer: Let
x = θ, y =
then
dθ
dt
dx
dθ
=
=y
dt
dt
dy
d2 θ
= 2 = −ω 2 sin θ = −ω 2 sin x
dt
dt
Let
F (x, y) = y
G(x, y) = −ω 2 sin x
The critical points of the system are (2kπ, 0), ((2k + 1)π, 0), k is integer.
The linear system near (2kπ, 0) is
µ ¶ µ
¶µ ¶ µ
¶µ
¶
d
x
Fx (0, 0) Fy (0, 0)
x
0 1
x−1
=
=
Gx (0, 0) Gy (0, 0)
y
−ω 2 0
y−1
dt y
The eigenvalues of this linear system are λ1 = iω, λ2 = −iω. So (2kπ, 0) is a stable center
of the linear system.
The linear system near ((2k + 1)π, 0) is
µ ¶ µ
¶µ ¶ µ
¶µ
¶
d
x
Fx (0, 0) Fy (0, 0)
x
0 1
x−1
=
=
Gx (0, 0) Gy (0, 0)
y
ω2 0
y−1
dt y
The eigenvalues of this linear system are λ1 = −ω, λ2 = ω. So ((2k+1)π, 0) is an unstable
saddle of the linear system.
(b). The original system is almost linear at each critical point.
From the stability of the linear system, ((2k + 1)π, 0) are unstable saddles of the original
system. The stability of (2kπ, 0) is indeterminate.
(c). We get
sin x
dy
= −ω 2
dx
y
8
Hence,
1
ω 2 (1 − cos x) + y 2 = c.
2
From the initial conditions (0, v), c = 21 v 2 and the trajectory is
1
1
ω 2 (1 − cos x) + y 2 = v 2 .
2
2
4. Consider the following Lienard equation
d2 x
dx
+ c(x) + g(x) = 0
2
dt
dt
where g(0) = 0. (Note that damped or undamped pendulum equations are special case
of Lienard equation.)
.
a). Write it as a system of two first order equations by introducing y = dx
dt
b). Show that (0,0) is a critical point and that the system is almost linear in the neighborhood of (0,0).
0
c). Show that if c(0) > 0, g (0) > 0 then the critical point is asymptotically stable, and
0
that if c(0) < 0 or g (0) < 0, then the critical point is unstable.
Answer: a).Let
y=
then
dx
dt
dy
d2 x
= 2
dt
dt
From the original equation, we get
dy
= −c(x)y − g(x).
dt
Let
µ ¶
x
X =
y
We get
µ
X
0
=
0
1
0 −c(x)
¶
µ
X+
0
−g(x)
¶
9
b). Since g(0) = 0, then (0,0) is the critical point of the system.
Let
c(x) = c(0) + η1 (x)
where η1 (x) → 0, x → 0
0
g(x) = g (0)x + η2 (x)
where η1 (x)/x → 0, x → 0
The system can written as
µ
0
X =
0
1
0
−g (0) −c(0)
¶
µ
X+
0
−η1 (x)y − η2 (x)
¶
Take
GT = (0, −η1 (x)y − η2 (x))
Then ||G|| → 0, (x, y) → (0, 0). Hence the system is almost linear in the neighborhood of
(0,0).
c).
¯
¯ −λ
1
0
det(A − λI) = ¯¯
−g (0) −c(0) − λ
Hence
¯
¯
¯ = λ2 + c(0)λ + g 0 (0) = 0
¯
p
c2 (0) − 4g 0 (0)
2
p
−c(0) − c2 (0) − 4g 0 (0)
λ1 =
2
λ1 =
−c(0) +
0
Case A: c(0) > 0, g (0) > 0
0
If 4 = c2 (0) − 4g (0) > 0, then λ2 < λ1 < 0 and (0,0) is an asymptotically stable critical
point.
If 4 = c2 (0) − 4g (0) < 0, then λ2 = α + βi, λ1 = α − βi where α = − c(0)
< 0 and
2
(0,0) is an asymptotically stable critical point.
0
If 4 = c2 (0) − 4g (0) = 0, then λ2 = λ1 = − c(0)
< 0 and (0,0) is an asymptotically
2
stable critical point.
0
Case B: c(0) < 0
0
If 4 = c2 (0) − 4g (0) > 0, then λ1 > λ2 > 0 and (0,0) is an unstable critical point.
10
If 4 = c2 (0) − 4g (0) < 0, then λ2 = α + βi, λ1 = α − βi where α = − c(0)
> 0 and
2
(0,0) is an unstable critical point.
0
If 4 = c2 (0) − 4g (0) = 0, then λ2 = λ1 = − c(0)
> 0 and (0,0) is unstable critical point.
2
0
0
Case C: g (0) < 0
In this case, λ1 > 0 > λ2 and (0,0) is an unstable critical point.
5. 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.
a.
b.
c.
d.
dx
dt
dx
dt
dx
dt
dx
dt
= −x3 + xy 2 , dy
= −2x2 y − y 3 , asymptotically stable.
dt
= 12 x3 + 2xy 2 , dy
= −y 3 , asymptotical stable.
dt
= −x3 + 2y 3 , dy
= −2xy 2 , stable.
dt
= x3 − y 3 , dy
= 2xy 2 + 4x2 y + 2y 3 , unstable.
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 asymptotically 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
11
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 asymptotically stable critical point.
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.
6. Consider the following system of the equations
dy
dx
= y − xf (x, y),
= −x − yf (x, y)
dt
dt
By construction Liapunov function of the type c(x2 + y 2 ), show that of f (x, y) > 0 in
some neighborhood of (0, 0), the (0, 0) is asymptotically stable, and if f (x, y) < 0 in some
neighborhood of (0, 0), the (0, 0) is an unstable critical point.
12
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)
dt
is an asymptotically stable critical point.
If f (x, y) < 0 in some neighborhood of (0,0), then dV
is positive definite. Hence, (0,0)
dt
is an unstable critical point.
7. Consider the following 2nd order of 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 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
V (x, y) = y +
g(s)ds, −k < x < k
2
0
is positive definite, and use this result to show that (0,0) is stable.
Answer:
a). Let
y=
dx
dt
then
dy
d2 x
= 2 = −g(x)
dt
dt
13
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)ds > 0
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.
8. Do the same as in Problem 7 by consider the following Lienard equation:
d2 x
dx
+ c(x) + g(x) = 0
2
dt
dt
0
where c(0) = 0, c (0) ≥ 0 and g(x) satisfies: g(0) = 0, xg(x) > 0 for x 6= 0, |x| < k.
Answer: a). Let
y=
then
dx
dt
d2 x
dy
= 2 = −c(x)y − g(x)
dt
dt
b). Since c(0) = 0, g(0) = 0, (0,0) is a critical point of the system.
c). Let
Z x
1 2
g(s)ds, −k < x < k
V (x, y) = y +
2
0
Then V (0, 0) = 0.
If −k < x < 0, from xg(x) > 0, we get g(x) < 0.
14
So
Z
Z
x
0
g(s)ds = −
0
g(s)ds > 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
= −c(x)y 2
dt
i.e. dV
is negative semidefinite.
dt
From above argument, (0,0) is a stable critical point of the system.
9. Consider the following undamped pendulum equation:
d2 x dx
+
+ sin x = 0
dt2
dt
Show that V (x, y) = 12 (x + y)2 + x2 + 12 y 2 is a Liapunov function and conclude that (0,0)
is asymptotically stable.
Answer:
Let
y=
dx
dt
then
dy
d2 x
= 2 = −y − g(x)
dt
dt
Obviously, (0,0) is a critical point of the system.
Let
1
1
V (x, y) = (x + y)2 + x2 + y 2
2
2
It is easy to check V (x, y) is positive definite and
dV
= −y 2 + 2xy − 2y sin x − x sin x
dt
Since sin x = x − α6 x3 , where α depends on x and 0 < α < 1,
dV
α
α
= −y 2 − x2 + x3 y + x4
dt
3
6
15
Set x = r cos θ, y = r sin θ
dV
α
α
= −r2 [1 − r2 ( cos3 θ sin θ + cos4 θ)] < −r2 (1 − r2 )
dt
3
6
1
Hence, if r = (x2 + y 2 ) 2 < 1, then dV
is negative definite.
dt
From above argument, (0,0) is an asymptotically stable critical point of the system.
10. Consider the following Lienard equation:
d2 x dx
+
+ g(x) = 0
dt2
dt
where g(x) satisfies: g(0) = 0, xg(x) > 0 for x 6= 0, |x| < k.
By choosing A appropriately, show that
Z x
1 2
V (x, y) = y + Ayg(x) +
g(s)ds
2
0
is a Liapunov function and conclude that (0,0) is asymptotically stable.
Answer: We first note that, since g(0) = 0, xg(x) > 0 for x 6= 0, |x| < k,
Z x
g(s)ds > 0, if x > 0
0
We also suppose that g(x) is Lipschitz continuous, then,
|g 0 (x)| < M,
if |x| < r <
k
2
Without lose generalization, we assume M > 1.
Let y = x0 , then we get
x0 = y
y 0 = −g(x) − y
Let V (x, y) = 12 y 2 + A yg(x) +
Rx
g(s)ds, here A is a undetermined number. Then
Z x
1 2
y + A yg(x) +
g(s)ds
V (x, y) =
2
0
Z x
1 2
2 2
≥
y −A g +
g(s)ds
4
0
Z x
1 2
y +
g(s)(1 − 2A2 g 0 (s))ds
=
4
0
0
16
Hence in a small ball B(0, r), if we choose |A| small enough say |A| < δ1 =
1 − 2A2 g 0 (x) >
√1
2 M
then,
1
2
Hence, in ball B(0, r),
Z
1 2 1 x
V (x, y) > y +
g(s)ds > 0, if (x, y) 6= (0, 0).
4
2 0
On the other hand,
W (x, y) = V˙ (x, y) = Ag 0 (x)y 2 − y 2 − Ag 2 (x) − Ag(x)y
1
1
≤ Ag 0 (x)y 2 − y 2 − Ag 2 (x) + ( Ag 2 (x) + Ay 2 )
2
2
1
1
≤ −(1 − A − Ag 0 (x))y 2 − Ag 2 (x)
2
2
In the ball B(0, r), if we choose |A| small enough, say |A| < δ2 = 2M1+1 , then
1
1
W (x, y) = V˙ (x, y) ≤ − y 2 − δ2 g 2 (x) < 0, if (x, y) 6= (0, 0)
2
2
Then, if we choose A such that |A| < min(δ1 , δ2 ) then, in the ball B(0, r), we have
V (x, y) > 0,
V˙ (x, y) < 0
If (x, y) 6= (0, 0). Hence the critical point (0, 0) is asymptotically stable.