MATH 215/255
Fall 2014
Assignment 9
§3.9, §8.1, §8.2
Solutions to selected exercises can be found at the end of the draft chapter of Lebl’s book
on nonlinear systems.
• 3.9.8: Consider the equation
→
−
x0 =
1/t −1
1 1/t
→
−
x +
t2
−t
.
Check that a complementary solution is
t sin t
t cos t
→
−
x c = c1
+ c2
.
−t cos t
t sin t
Use it to find a particular solution, by variation of parameters.
t sin t
Answer.
To check the complementary solution, it is enough to check that
−t cos t
t cos t
are solutions (since they are linearly independent). This is the case
and
t sin t
since
0
t sin t
sin t + t cos t
t sin t
1/t −1
,
=
=
−t cos t
t sin t − cos t
−t cos t
1 1/t
0
1/t −1
t cos t
cos t − t sin t
t cos t
=
=
.
1 1/t
t sin t
t cos t + sin t
t sin t
Therefore the fundamental matrix and its inverse are
1 sin t − cos t
sin t cos t
, X(t)−1 =
X(t) = t
.
− cos t sin t
sin t
t cos t
T
→
−
Writing f (t) = t2 −t , the particular solution given by the method of variation
of parameters is then
Z
→
−
−
→
xp = X(t) X(t)−1 f (t)dt
2 Z 1 sin t − cos t
t
= X(t)
dt
sin t
−t
t cos t
Z t sin t + cos t
= X(t)
dt
t cos t − sin t
−t cos t + 2 sin t
= X(t)
.
t sin t + 2 cos t
A final computation yields the particular solution:
2t
−
→
xp =
.
t2
• 8.1.3 (modified): Find the critical points and the Jacobian matrix at any point for
the following systems.
0 0 0 2
x
x2 − y 2
x
−y
x
x +y
a)
=
, b)
=
, c)
=
.
y0
x2 + y 2 − 1
y0
3x + yx2
y0
y2 + x
Answer.
a) We solve
(
x2 = y 2
x2 + y 2 = 1
(
√
x = ±1/ 2
√
⇔
y = ±1/ 2
(
x = ±y
⇔
2y 2 = 1
and therefore the critical points are
±
√1 , ± √1
2
2
2x −2y
2x
2y
. The Jacobian at (x, y) is
.
b) We solve −y = 0, 3x + yx2 = 0 and find easily that (0, 0) is the only critical point.
The Jacobian at (x, y) is
0
−1
.
J(x, y) =
3 + 2xy x2
c) We solve
(
y = −x2
y2 + x = 0
(
y = −x2
⇔
x4 + x = 0
(
y = −x2
⇔
x(x + 1)(x2 + x + 1) = 0,
P
j j n−1−j valid for n
where we used the special identity an + bn = (a + b) n−1
j=0 (−1) a b
odd (with n = 3, a = x, b = 1). The polynomial x2 + x + 1 has negative discriminant
−3 and as such it has no real roots. The only critical points are therefore (0, 0) and
(−1, −1). The Jacobian at (x, y) is
2x 1
J(x, y) =
.
1 2y
2
• 8.1.4: For the following systems, find the linearization at (0, 0).
0 0 x
x + 2y + x2 − y 2
x
−y
a)
=
,
b)
=
,
y0
2y − x2
y0
x − y3
0 x
ax + by + f (x, y)
∂f ∂g ∂g
c)
=
, where f, g, ∂f
∂x , ∂y , ∂x , ∂y are zero at (0, 0).
y0
cx + dy + g(x, y)
Answer.
In each case it is enough to compute the Jacobian at (0, 0), since the
linearization is the system
0
u
u
= J(0, 0)
.
v
v
1 + 2x 2 − 2y
1
J(0, 0) =
a) J(x, y) =
−2x
2
0
0
0
−1
J(0, 0) =
b) J(x, y) =
2
1
1 −3y
"
#
∂f
a + ∂f
a
∂x b + ∂y
c) J(x, y) =
J(0, 0) =
∂g
∂g
c
c + ∂x d + ∂y
2
2
.
−1
.
0
b
.
d
• 8.1.101: Find the critical points and linearizations of the following systems.
a) x0 = sin πy + (x − 1)2 , y 0 = y 2 − y
b) x0 = x + y + y 2 , y 0 = x
c) x0 = (x − 1)2 + y, y 0 = y + x2 .
Answer.
Again we only need to find the critical points and to compute the
Jacobians at each of these points.
a) We have to solve
(
sin πy + (x − 1)2 = 0
y2 = y
(
(x − 1)2 = 0
⇔
y = 0 or 1.
Therefore (1, 0) and (1, 1) are the only critical points. We have
2(x − 1) π cos πy
0
π
0 −π
J(x, y) =
, J(1, 0) =
, J(1, 1) =
.
0
2y − 1
0 −1
0
1
b) We solve x = 0, y(y + 1) = 0 to find that the critical points are (0, 0) and (0, −1).
The Jacobians are
1 1 + 2y
1 1
1 −1
J(x, y) =
, J(0, 0) =
, J(0, −1) =
.
1
0
1 0
1
0
3
c) We solve y = −x2 , x2 − 2x + 1 − x2 = 0 to find that ( 21 , − 14 ) is the only critical
point. The Jacobians are
2(x − 1) 1
−1 1
1
1
J(x, y) =
, J( 2 , − 4 ) =
.
2x 1
1 1
• 8.1.103: The concepts of critical points and linearization also generalize to higher
dimensions, by adding more functions and more variables to the Jacobian matrix. For
the following system of 3 equations, find the critical points and their linearizations:
 0  

x
x + z2
 y0  =  z2 − y  .
z0
z + x2
Answer.
We solve

2

x + z = 0
z2 − y = 0


z + x2 = 0

4

x + x = 0
⇔ y = x4


z = −x2 .
The real solutions to x+x4 = 0 were found in 8.1.3 c) to be x = 0 and x = −1. Therefore the critical points are (0, 0, 0) and (−1, 1, −1). The three-dimensional Jacobian
is


1
0 2z
J(x, y, z) =  0 −1 2z  .
2x
0 1
Inserting the values of critical points, we obtain




1
0
0
1
0 −2
0  , J(−1, 1, −1) =  0 −1 −2  .
J(0, 0, 0) =  0 −1
0
0
1
−2
0
1
• 8.1.104: Write down the non-autonomous system of 2 equations x0 = f (x, y, t),
y 0 = g(x, y, t) (where x = x(t), y = y(t)) as an autonomous system of 3 equations,
using variables u, v, w.
Answer.
system
Introduce variables u = x, v = y, ω = t to obtain the 3-dimensional
0 

u
f (u, v, ω)
 v  =  g(u, v, ω)  .
ω
1

4
• 8.2.1: For the systems below, find and classify the critical points (this includes a
discussion of stability).
0 0 2
0 x
−x + 3x2
x
x + y2 − 1
x
yex
a)
=
, b)
=
, c)
=
.
y0
−y
y0
x
y0
y − x + y2
Answer.
a) We solve y = 0, x(3x − 1) = 0 to find that (0, 0) and ( 13 , 0) are the
critical points. We have
−1 + 6x
0
−1
0
1
0
1
J(x, y) =
, J(0, 0) =
, J( 3 , 0) =
.
0 −1
0 −1
0 −1
The matrix J(0, 0) is diagonal, so its eigenvalues are the diagonal entries −1, −1,
which are both real and negative. Therefore we have a sink at (0, 0), which is stable.
The matrix J( 13 , 0) is also diagonal with two real eigenvalues of opposite signs: 1 and
−1. Therefore we have a saddle-point at ( 31 , 0), which is unstable.
b) We solve x = 0, y 2 = 1 to find that the critical points are (0, ±1). We have
0 −2
2x 2y
0 2
.
, J(0, −1) =
J(x, y) =
, J(0, 1) =
1
0
1 0
1 0
√
At (0, 1) we have a discriminant equation λ2 − 2 = 0, hence two real eigenvalues ± 2
of opposite signs. Therefore we have a saddle-point at (0, 1), which is unstable.
At (0, −1) we √
have a discriminant equation λ2 + 2 = 0, hence two purely imaginary
eigenvalues ±i 2. This corresponds to a center and therefore we cannot determine
stability.
c) Since ex never vanishes, (0, 0) is the only critical point. We have
x
0 1
ye
ex
.
, J(0, 0) =
J(x, y) =
−1 1
−1 1 + 2y
The determinant equation is
−λ
1
−1 1 − λ
= λ2 − λ + 1 = 0.
√
The roots are 1±i2 3 , and they are complex with positive real part. Therefore we have
a spiral source at (0, 0), which is unstable.
• 8.2.3 (modified): Find the critical point of the following system, and determine
whether the system is almost-linear at it:
0 x
−x2
=
.
y0
−y 2
5
Answer.
There is clearly a unique critical point (0, 0), and the Jacobians are
−2x
0
0 0
J(x, y) =
, J(0, 0) =
.
0 −2y
0 0
Since J(0, 0) is (clearly) not invertible, the system is not almost-linear at (0, 0) (note:
the terminology ”almost-linear” in not standard in the field of ODEs, and is used only
as a commodity in this course).
• 8.2.7: Consider the system x0 = f (x, y), y 0 = g(x, y). Suppose that g(x, y) > 1 for
all (x, y). Are there any critical points? What can we say about the trajectories as t
goes to infinity?
Answer.
There is no critical point since g(x, y) never vanishes. The derivative of
the y-coordinate of solutions (x, y) is always larger than 1, therefore the trajectories
go to infinity towards the right in the xy plane.
• 8.2.101: For the systems belows, find and classify the critical points.
0 0 0 xy
x
y − y2 + x
x
−x + x2
x
.
=
, c)
=
, b)
=
a)
x+y−1
y0
−x
y0
y
y0
Answer.
a) The critical points are (0, 0) and (1, 0), and the Jacobians are
1 0
−1 0
−1 + 2x 0
.
, J(1, 0) =
, J(0, 0) =
J(x, y) =
0 1
0 1
0
1
At (0, 0) we have two real eigenvalues ±1 of opposite signs, and therefore a saddlepoint, which is unstable. At (1, 0) we have two real positive eigenvalues 1, 1, and
therefore a source, which is unstable.
b) The critical points are (0, 0) and (0, 1), and the Jacobians are
1 −1
1 1
1 1 − 2y
.
, J0, 1) =
J(x, y) =
, J(0, 0) =
−1 0
−1
0
−1
0
At (0, 0) the eigenvalues are
√
1±i 3
, so we have a spiral source, which is unstable.
√2
√
1± 5
(note that 1 − 5 < 0), and therefore we have a
2
At (0, 1) the eigenvalues are
saddle-point, which is unstable.
c) The critical points are (0, 1) and (1, 0), and the Jacobians are
y x
1 0
0 1
J(x, y) =
, J(0, 1) =
, J(1, 0) =
.
1 1
1 1
1 1
At (0, 1) the eigenvalues
are 1, 1 and we have a source, which is unstable. At (1, 0)
√
1± 5
the eigenvalues are 2 and we have a saddle-point, which is unstable.
6