Solutions for Homework 3
1.
3.4.16 a) The potential is
V
(
x
) =
−
rx
+
x
3
3
.
You could add a constant
C
, but that would not make any essential difference; see below. Here
is the graph of
V
for
r
=
−
2,
r
=
0, and
r
=
2:
4
2
0
2
4
5
0
5
r=2
4
2
0
2
4
5
0
5
r=0
4
2
0
2
4
5
0
5
r=2
The fixed points of the differential equation are the stationary points of
V
, that is, the values
x
for which
V
′
(
x
) =
0. Local minima are stable fixed points, since in a local minimum,
V
′
(
x
)
changes from negative to positive, and therefore
−
V
′
(
x
)
changes from positive to negative.
Local maxima are unstable fixed points. Stationary points that are neither local maxima, nor
local minima are semistable points.
The potential
V
is defined to be the
negative
of an antiderivative of the righthand side of
the differential equation. Why the minus sign? The answer is this: If we were to leave out the
minus sign, simply writing the equation in the form ˙
x
=
dV
/
dx
, then stable fixed points would
be local maxima of
V
, and unstable ones would be local minima. This is unintuitive, if you
think of a little ball rolling on the graph of
V
!
Back to our example. There are no stationary points of
V
for
r
<
0, exactly one for
r
=
0
(namely, 0), and two for
r
>
0 (namely,
±
√
r
). If we had added a constant
C
to
V
, the graph
of
V
would have shifted upwards or downwards, but the stationary points would have remained
the same.
b) The potential is
V
(
x
) =
−
r
x
2
2
+
x
3
3
.
Here is the graph of
V
for
r
=
−
2,
r
=
0, and
r
=
2:
1
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4
2
0
2
4
5
0
5
r=2
4
2
0
2
4
5
0
5
r=0
4
2
0
2
4
5
0
5
r=2
For any
r
negationslash
=
0, there are two stationary points of
V
, and therefore two fixed points of the dif
ferential equation, namely
x
=
0 and
x
=
r
. For
r
=
0, there is only one, namely
x
=
0. For
r
<
0,
x
=
r
is a local maximum of
V
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 Spring '09
 Dr.Borgers
 Critical Point, Optimization, Fermat's theorem

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