642:527
SOLUTIONS: ASSIGNMENT 7
FALL 2009
NOTE:
A. There is a Maple worksheet on the “Assignments and solutions” web page
which shows the phase plane portraits for problems 2(b) and 2(k).
B. Vectors are denoted either by
boldface
or, for Greek letters,
underlining:
ζ
.
Review:
Consider
x
′
=
P
(
x, y
),
y
′
=
Q
(
x, y
). A point (
x
0
, y
0
) is a singular point for these
equations if
P
(
x
0
, y
0
) =
Q
(
x
0
, y
0
) = 0
.
(12.1)
To linearize near this singular point we make the the linear approximations
P
(
x, y
)
≈
P
x
(
x
0
, y
0
)(
x

x
0
) +
P
y
(
x
0
, y
0
)(
y

y
0
)
,
Q
(
x, y
)
≈
Q
x
(
x
0
, y
0
)(
x

x
0
) +
Q
y
(
x
0
, y
0
)(
y

y
0
)
(here we have used (12.1)). With
X
=
x

x
0
,
Y
=
y

y
0
, the linearized equations are
X
′
=
P
x
(
x
0
, y
0
)
X
+
P
y
(
x
0
, y
0
)
Y,
Y
′
=
Q
x
(
x
0
, y
0
)
X
+
Q
y
(
x
0
, y
0
)
Y.
(12.2)
Section 7.4 2. (a) The equations are
x
′
=
y
,
y
′
= 1

x
4
. To find the singular points we
set
y
= 0 and 1

x
4
= 0; since 1

x
4
= (1

x
)(1 +
x
)(1 +
x
2
) there are two solutions:
(1
,
0) and (

1
,
0). We analyze the phase plane near each.
(i) Near (1
,
0). Here
X
=
x

1,
Y
=
y
.
P
(
x, y
) =
y
is already linear and so it is its own
linear approximation. Then (12.2) becomes
X
′
=
Y,
Y
′
=

4
X.
From det(
A

λI
) =
vextendsingle
vextendsingle
vextendsingle
vextendsingle

λ
1

4

λ
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
λ
2
+ 4 = 0 we find the eigenvalues
λ
=
±
2
i
; this is a
center
in the linearized system. We can’t tell from this what it is in the true system: it
might be a center, a stable focus, or an unstable focus. But notice that there is a conserved
quantity,
F
(
x, y
) =
y
2
2

x
+
x
5
5
(this is just the total energy if we regard the equations as describing a particle, with
position
x
and velocity
y
, moving in a potential
V
(
x
) =

x
+
x
5
/
5). The level curves of
F
are then orbits and it is not hard to see that there are closed orbits circling (1
,
0)—so
this point stays a center.
(ii) Near (

1
,
0). Here
X
=
x
+ 1,
Y
=
y
. Here (12.2) becomes
X
′
=
Y,
Y
′
= 4
X.
From det(
A

λI
) =
vextendsingle
vextendsingle
vextendsingle
vextendsingle

λ
1
4

λ
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
λ
2

4 = 0 we find the eigenvalues
λ
=
±
2; this is
a
saddle point
in the linearized system. We find the special trajectories: for
λ
1
= 2, we
solve
bracketleftbigg

2
1
4

2
bracketrightbigg bracketleftbigg
ξ
η
bracketrightbigg
= 0 to find the eigenvector
ζ
(1)
=
bracketleftbigg
1
2
bracketrightbigg
; for
λ
2
=

2, we solve
1
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642:527
SOLUTIONS: ASSIGNMENT 7
FALL 2009
bracketleftbigg
2
1
4
2
bracketrightbigg bracketleftbigg
ξ
η
bracketrightbigg
= 0 to find the eigenvector
ζ
(2)
=
bracketleftbigg

1
2
bracketrightbigg
. The straightline trajectories for
the saddle point are
y
= 2
x
, which is oriented outward since
λ
1
>
0, and
y
=

2
x
, which is
oriented inward since
λ
2
<
0. The point (

1
,
0) is again a saddle point in the true system.
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 Fall '08
 Staff
 Math, Critical Point, Vectors, Web page, saddle point, Stationary point, hessian matrix, phase plane

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