Exercises 5.4
(c)
To find critical points for the system in (a), we solve
v
= 0
,
−
x
+
1
λ
−
x
= 0
⇒
v
= 0
,
x
2
−
λx
+ 1 = 0
⇒
v
= 0
,
x
=
λ
±
√
λ
2
−
4
2
.
For 0
< λ <
2,
λ
2
−
4
<
0 and so both roots are complex numbers. However, for
λ >
2
there are two distinct real solutions,
x
1
=
λ
−
√
λ
2
−
4
2
and
x
2
=
λ
+
√
λ
2
−
4
2
,
and the critical points are
λ
−
√
λ
2
−
4
2
,
0
and
λ
+
√
λ
2
−
4
2
,
0
.
(d)
The phase plane diagrams for
λ
= 1 and
λ
= 3 are shown in Figures B.40 and B.41 in
the answers section of the text.
(e)
From Figures B.40 we conclude that, for
λ
= 1, all solution curves approach the vertical
line
x
= 1(=
λ
). This means that the bar is attracted to the magnet. The case
λ
= 3
is more complicated. The behavior of the bar depends on the initial displacement
x
(0)
and the initial velocity
v
(0) =
x
(0). From Figure B.41 we see that (with
v
(0) = 0) if
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Critical Point, Complex Numbers, Velocity, Bar association, real solutions, distinct real solutions, phase plane diagrams

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