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Solutions for Homework 2
1.
3.1.1 The graph of
f
(
x
) =
1
+
rx
+
x
2
is a parabola, opening upwards. The minimum occurs where
f
′
(
x
) =
0, that is, at
x
0
=
−
r
2
.
At this point, the value of
f
is
f
(
x
0
) =
1
−
r
2
2
+
r
2
4
=
1
−
r
2
4
.
If

r

<
2, then
f
(
x
0
)
>
0, therefore
f
(
x
)
>
0 for all
x
, and there are no fixed points. If

r

>
2,
then
f
(
x
0
)
<
0, and therefore there are two fixed points. They are easy to compute:
f
(
x
) =
0
⇔
x
=
−
r
±
√
r
2
−
4
2
.
We will write
x
±
=
−
r
±
√
r
2
−
4
2
.
(1)
Since the graph of
f
is a parabola opening upwards, it must be true that
f
′
(
x
−
)
<
0 and
f
′
(
x
+
)
>
0, therefore
x
−
is stable and
x
+
is unstable. Let us plot the bifurcation diagram. This requires
plotting
x
−
and
x
+
as functions of
r
. We could do that, but life becomes easier if we solve the
equation
1
+
rx
±
+
x
2
±
=
0
for
r
:
r
=
−
x
±
−
1
x
±
.
The function
r
(
x
) =
−
x
−
1
x
is really easy to plot:
1
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3
2
1
0
1
2
3
4
4
3
2
1
0
1
2
3
4
x
r
Now the graph of
x
as a function of
r
is obtained simply by reversing
x
and
r
in the above
plot. Here is what you get when you do that, remembering that the smaller of the two fixed
points is stable, and the larger is unstable:
4
3
2
1
0
1
2
3
4
4
3
2
1
0
1
2
3
4
r
x
This is the bifurcation diagram. There are two saddlenode bifurcations, one at
r
=
−
2 and the
other at
r
=
2. I’ll leave it to you to plot the qualitatively different vector fields.
3.1.3 Let’s first think about the fixed point:
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 Spring '09
 Dr.Borgers

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