CONTINUOUS DYNAMICAL SYSTEMS I
Math 21b, O. Knill
Homework: Section 9.1: 4,8,10,26,32,24*,46* until Tuesday
CONTINUOUS DYNAMICAL SYSTMES. A differential
equation
d
dt
~x
=
f
(
~x
) defines a dynamical system. The solu
tions is a curve
~x
(
t
) which has the
velocity vector
f
(
~x
(
t
))
for all
t
. One often writes ˙
x
instead of
d
dt
x
. So, we have the
problem that we know a formula for the tangent at each
point.
The aim is to find a curve
~x
(
t
) which starts at a
given point
~v
=
~x
(0).
IN ONE DIMENSION. A system ˙
x
=
g
(
x, t
) is the general differential equation in one dimensions. Examples:
•
If
˙
x
=
g
(
t
)
, then
x
(
t
) =
R
t
0
g
(
t
)
dt
. Example: ˙
x
= sin(
t
)
, x
(0) = 0 has the solution
x
(
t
) = cos(
t
)

1.
•
If
˙
x
=
h
(
x
)
, then
dx/h
(
x
) =
dt
and so
t
=
R
x
0
dx/h
(
x
) =
H
(
x
) so that
x
(
t
) =
H

1
(
t
). Example:
˙
x
=
1
cos(
x
)
with
x
(0) = 0 gives
dx
cos(
x
) =
dt
and after integration sin(
x
) =
t
+
C
so that
x
(
t
) = arcsin(
t
+
C
).
From
x
(0) = 0 we get
C
=
π/
2.
•
If
˙
x
=
g
(
t
)
/h
(
x
)
, then
H
(
x
) =
R
x
0
h
(
x
)
dx
=
R
t
0
g
(
t
)
dt
=
G
(
t
) so that
x
(
t
) =
H

1
(
G
(
t
)).
Example:
˙
x
= sin(
t
)
/x
2
, x
(0) = 0 gives
dxx
2
= sin(
t
)
dt
and after integration
x
3
/
3 =

cos(
t
) +
C
so that
x
(
t
) =
(3
C

3 cos(
t
))
1
/
3
. From
x
(0) = 0 we obtain
C
= 1.
Remarks:
1) In general, we have no closed form solutions in terms of known functions. The solution
x
(
t
) =
R
t
0
e

t
2
dt
of ˙
x
=
e

t
2
for
example can not be expressed in terms of functions exp
,
sin
,
log
,
√
·
etc but it can be solved using Taylor series: because
e

t
2
= 1

t
2
+
t
4
/
2!

t
6
/
3!+
. . .
taking coefficient wise the antiderivatives gives:
x
(
t
) =
t

t
3
/
3+
t
4
/
(32!)

t
7
/
(73!)+
. . .
.
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 Spring '03
 JUDSON
 Math, Differential Equations, Linear Algebra, Algebra, Derivative, Stability theory

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