Math 1280
Practice Problems for Final Exam
Part 2 (Sections 6.6, 6.7, 6.8, and chapter 7)
Spring 2016
Solutions
1. Show that the given system has a nonlinear center at the origin.
ẋ = 9y − 5y 5 ,
Solution:
ẏ = −4x + x3 − xy 4
Denote f (y) = 9y − 5y 5 , g(y) = −4x + x3 − xy 4 .
f (−y) = −9y + 5y 5 = −f (y), f (y) is odd;
g(−y) = −4x + x3 − x(−y)4 = −4x + x3 − xy 4 = g(y), g(y) is even.
Therefore, the system is reversible.
It is easy to check that the origin is a FP because both right hand sides vanish when x = 0
and y = 0.
0 9 − 25y 4
Jacobian is A =
−4 + 3x2 − y 4
−4xy 3
0 9
A(0, 0) =
T = 0, ∆ = 36 > 0 Hence, there is a center at the origin.
−4 0
The system has a center at the origin and it is reversible.
Hence it has a nonlinear center at the origin.
2. Show that the given system is reversible but not conservative.
ẋ = 4y cos y + 3x sin y,
Hint: Show that the point 0, π2
Solution:
ẏ = 5 cos y + 4 sin x
is a FP and determine its type.
Denote f (y) = 4y cos y + 3x sin y, g(y) = 5 cos y + 4 sin x.
f (−y) = 4(−y) cos(−y) + 3x sin(−y) = −4y cos y − 3x sin y = −f (y), f (y) is odd;
g(−y) = 5 cos(−y) + 4 sin x = 5 cos y + 4 sin x = g(y), g(y) is even.
Therefore, the system is reversible.
It is easy to check that the point 0, π2
x = 0 and y = π2 .
is a FP because both right hand sides vanish when
"
Jacobian is A =
"
A(0, π2 ) =
3 sin y
4 cos y − 4y sin y + 3x cos y
#
4 cos x
−5 sin y
#
3 −2π
T = −1 < 0, ∆ = −15 + 8π > 0
4 −5
The FP 0, π2
is an attractor.
A conservative system can never have an attracting FP.
Therefore, the system cannot be conservative.
3. Consider the system
ẋ = x − y − x(x2 + y 2 ),
ẏ = x + y − y(x2 + y 2 )
(a) Write the system in polar coordinates
Solution:
ṙ =
1
1
1
(xẋ + y ẏ) = (x2 − xy − x2 r2 + xy + y 2 − y 2 r2 ) = (r2 − r4 )
r
r
r
ṙ = r(1 − r2 )
1
1
1
θ̇ = 2 (xẏ − y ẋ) = 2 (x2 + xy − xyr2 − xy + y 2 + xyr2 ) = 2 (r2 ) = 1
r
r
r
The system in polar coordinates is
ṙ = r(1 − r2 ),
θ̇ = 1
(b) Show that a limit cycle exists and find it.
Solution:
If r(1 − r2 ) = 0, then r = 0 or r = 1.
The circle r = 1 is the limit cycle.
4. Consider the system
ẋ = −x + y 2 ,
ẏ = y(2 + 2x − y 2 )
(a) Use index theory to show that the system has no closed orbits
Solution:
x-nullcline: x = y 2 ,
y-nullcline: y = 0, x = 21 y 2 − 1.
The curves x = y 2 and x = 12 y 2 − 1 don’t intersect.
Indeed, assume they intersect. Then we have y 2 = 12 y 2 − 1 or
impossible.
1 2
y
2
= −1 which is
Therefore, the only FP is the intersection of the curve x = y 2 and the line y = 0,
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which is the point (0, 0).
"
−1
2y
2 + 2x − 3y 2
−1 0
2y
#
Jacobian is A =
"
A(0, 0) =
0 2
∆ = −2 < 0
#
and the FP (0, 0) is saddle.
If there is a closed orbit C then it either encloses the saddle FP at the origin or does
not. An index of a saddle point is −1 and an index of a closed orbit according to
the index theory has to be +1. If C encloses the origin its index is −1. If C doesn’t
enclose it its index is 0. In either case the index of C is not +1. Therefore, the system
cannot have a closed orbit.
(b)
Use a method different from index theory and Dulac’s criterion to show that the
system has no closed orbits
Solution:
Denote f (x, y) = −x + y 2 , g(x, y) = y(2 + 2x − y 2 ).
Then fy = 2y = gx and the system is gradient.
Gradient systems have no closed orbits. Therefore, the system has no closed orbits.
(c)
Use Dulac’s criterion to show that the system has no closed orbits. You may use a
Dulac’s function g(x, y) = y1 .
Solution:
∂
=
∂x
∇ · (g(x, y)(ẋ, ẏ)) =
∂
∂
(g ẋ) +
(g ẏ)
∂x
∂y
x
∂
1
− +y +
(2 + 2x − y 2 ) = − − 2y.
y
∂y
y
Now let’s look at the x-axis. Suppose (x0 , 0) is a point on the axis. Let x(t) be a
solution to the equation ẋ = −x + y 2 with the initial condition x(0) = x0 , and let
y(t) = 0. Then (x(t), y(t)) = (x(t), 0) is a solution to the given system. This means
that the x-axis is a trajectory and cannot be crossed by another trajectory by existence
and uniqueness theorem. The x-axis divides the xy-plane into two parts: the upper
half-plane (y > 0) and the lower half-plane (y < 0).
In each halves ∇·(g(x, y)(ẋ, ẏ)) preserves its sign ( ∇·(g(x, y)(ẋ, ẏ)) < 0 when y > 0
and ∇ · (g(x, y)(ẋ, ẏ)) > 0 when y < 0 ). Each half is a simply connected region.
Hence in each half conditions for Dulac’s criterion hold and there are no closed orbits.
Closed orbits being trajectories cannot cross the x-axis.
Therefore, the system has no closed orbits in the xy-plane.
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5. For
ẋ = −y + 3xy 2 ,
ẏ = x − 4x2 y
use Lyapunov function to show that the system has no closed orbits.
Specifically, show that V (x, y) = x2 + y 2 is Lyapunov function for the system and hence
no closed orbits are allowed.
Solution:
Consider a Lyapunov function V (x, y) = x2 + y 2 .
1. V (0, 0) = 0.
2. V (x, y) > 0 for all (x, y) 6= 0.
3. V̇ = Vx ẋ + Vy ẏ = 2x(−y + 3xy 2 ) + 2y(x − 4x2 y) = −2xy + 6x2 y 2 + 2yx − 8x2 y 2 =
−2x2 y 2 < 0 for all (x, y), a < b.
V (x, y) = x2 + y 2 satisfies all requirements for a Lyapunov function
and the system has no closed orbits.
6. Determine if the given system is gradient. If it is, find its potential function V .
ẋ = 2xy 3 + 3y 2 cos x,
Solution:
ẏ = 3(x2 − 1)y 2 + 6y sin x
Denote f (x, y) = 2xy 3 + 3y 2 cos x, g(x, y) = 3(x2 − 1)y 2 + 6y sin x.
fy = 6xy 2 + 6y cos x, gx = 6xy 2 + 6y cos x.
fy = gx .
Therefore, the system is gradient.
Vx (x, y) = f (x, y) = 2xy 3 + 3y 2 cos x, V (x, y) = x2 y 3 + 3y 2 sin x + φ(y).
Vy (x, y) = g(x, y).
3x2 y 2 + 6y sin x + φ0 (y) = 3(x2 − 1)y 2 + 6y sin x.
R
φ0 (y) = −3y 2 , φ(y) = (−3y 2 ) dy = −y 3 + C.
V (x, y) = x2 y 3 + 3y 2 sin x − y 3 + C.
We can put C = 0.
V (x, y) = (x2 − 1)y 3 + 3y 2 sin x.
7. Consider the system
ẋ = 3x − 5y − x(x2 + 4y 2 ),
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ẏ = 5x + 3y − y(2x2 + y 2 )
(a) Classify the FP at the origin.
Solution:
Jacobian: A =
A(0, 0) =
3 −5
,
5
3
3 − 3x2 − 4y 2
5 − 4xy
−5 − 8xy
3 − 2x2 − 3y 2
T = 3, ∆ = 34, ∆ > T 2 .
(0, 0) is an unstable spiral.
(b) Switch to polar coordinates and write an equation for ṙ
Solution:
ṙ =
1
1
(xẋ + y ẏ) = (3x2 − 5xy − x4 − 4x2 y 2 + 5xy + 3y 2 − 2x2 y 2 − y 4 )
r
r
1
(3r2 − (x4 + 2x2 y 2 + y 4 ) − 4x2 y 2 )
r
1
1
=
3r2 − (x2 + y 2 )2 − 4r4 cos2 θ sin2 θ =
3r2 − r4 − r4 sin2 2θ
r
r
2
ṙ = r 3 − r2 − r2 sin 2θ
=
(c)
Determine the circle of maximum radius r1 , centered at the origin such that all
trajectories have a radially outward component on it.
Solution:
On this circle ṙ > 0.
To find its radius we require
3 − r2 − r2 sin2 2θ = 3 − 1 + sin2 2θ r2 > 0.
0 ≤ sin2 2θ ≤ 1, 1 ≤ 1 + sin2 2θ ≤ 2.
Then a sufficient condition is 3 − 2r2 > 0
q
q
3
Which gives r < 2 . We take r1 = 0.999 32 .
(d)
Determine the circle of minimum radius r2 , centered at the origin such that all
trajectories have a radially inward component on it.
Solution:
On this circle ṙ < 0.
To find its radius we require
3 − 1 + sin2 2θ r2 < 0.
As before, 1 ≤ 1 + sin2 2θ ≤ 2.
Then a sufficient condition is 3 − r2 < 0
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Which gives r >
√
√
3. We take r2 = 1.001 3.
(e) There is no FPs in the region r1 ≤ r ≤ r2 . You don’t have to show that.
Prove that the system has a limit cycle somewhere in the trapping region r1 ≤ r ≤ r2 .
Solution:
There is a source at the origin. All trajectories go inside the trapping
region through both circles of radii r1 and r2 centered at the origin. There are no FPs
inside the region. By Poincare-Bendixson theorem there must be a limit cycle inside
the region r1 ≤ r ≤ r2 .
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