642:527
SOLUTIONS: ASSIGNMENT 8
FALL 2007
Section 7.4: 7 (a) By introducing
y
=
x
′
we can reformulate the given equation as
x
′
=
y,
y
′
=
2
P
ml
x
bracketleftbigg
1
−
parenleftBig
x
l
parenrightBig
2
bracketrightbigg
−
1
/
2
−
k
m
x.
The right hand sides vanish when
x
=
y
= 0 so the origin is indeed a critical point. To
linearize there we simply drop the factor [1
−
(
x/l
)
2
]
−
1
/
2
in the second equation to obtain
z
′
=
A
z
,
z
=
bracketleftbigg
x
y
bracketrightbigg
,
A
=
bracketleftbigg
0
1
2
P/
(
ml
)
−
k/m
0
bracketrightbigg
.
This system has eigenvalues
λ
±
=
±
radicalbig
2
P/
(
ml
)
−
k/m
, so it is a center if
P < k/
2
l
and
a saddle point if
P > k/
2
l
. For
P < k/
2
l
the mass will oscillate around its equilibrium
poition at the origin, but the saddle point is unstable so for
P > k/
2
l
the mass will leave
the neighborhood of the origin  this is “buckling”.
FURTHER DISCUSSION (not asked for in problem):
In the buckling regime, that is,
when
P > k/
2
l
, there are two additional singular points, located at
x
=
±
x
c
, where
x
c
=
l
radicalbig
1
−
(
kl/
2
P
)
2
. One easily finds that these are centers, so that after the buckling
the system can oscillate around these new singular points.
There is also a “figure 8”
trajectory through the saddle point at the origin and trajectories which go around all
three critical points.
(b) The value
k/lm
is a bifurcation point for
P
: when
P
passes through this value the
nature of the phase plane changes.
(c) Omitted.
Section 7.5: 4 (a) We write
x
=
r
cos
θ
,
y
=
r
sin
θ
, and remember that
r
and
θ
are functions
of
t
. Then
x
′
=
r
′
cos
θ
−
rθ
′
sin
θ
and
y
′
=
r
′
sin
θ
+
rθ
′
cos
θ
, and the equations become
r
′
cos
θ
−
rθ
′
sin
θ
=
ǫr
cos
θ
+
r
sin
θ
−
r
3
cos
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 Fall '07
 SPEER
 Critical Point, Polar coordinate system, dx, saddle point, Stationary point, hessian matrix

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