Entry
Math: Chaos theory: Equation: Math: Which equations are candidates for chaos? [Physics: Dynamics]
Nov 29th, 2003 11:37
Knud van Eeden,
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--- Knud van Eeden --- 29 November 2003 - 08:29 pm -------------------
Math: Chaos theory: Equation: Math: Which equations are candidates for
chaos? [Physics: Dynamics]
---
Conditions for chaos.
We now return to the idea of chaotic oscillations in nonlinear systems.
There is one key condition which must be met for such oscillations to
be at all possible: the associated phase space must be at least
three-dimensional.
To put the matter another way, chaos cannot occur in a 2-dimensional
autonomous system.
That is, chaos can NOT occur in a system of 2 first order differential
equations, in 2 variables:
+--------------+
| |
| . |
| x = f(x,y) |
+--------------+
| . |
| y = g(x,y) |
| |
+--------------+
The reason for this is an important result called the
Poincar<e'>-Bendixon theorem, which we now describe.
Imagine first that we have determined the equilibrium points, if any,
of the above system (11.4), i.e. per definition the points for which
both
f(x,y) = 0
and
g(x,y) = 0
Now suppose that a phase path starts at some point and cannot leave a
certain bounded region of the x, y plane. Then the Poincar<e'>-Bendixon
theorem says that the phase path must eventually either
+----------------------------------------------------------------+
|a) terminate at an equilibrium point |
| |
| . . > . |
| . figure: 'o' is an equilibrium point|
| .. . |
| . O . |
| . . |
| . . |
+----------------------------------------------------------------+
or
+----------------------------------------------------------------+
|b) return to the original point, giving a closed path |
| |
| O |
| . . |
| . > |
| . |
| . . figure: 'o' is an equilibrium point|
| . . |
| . . |
| . |
+----------------------------------------------------------------+
or
+----------------------------------------------------------------+
|c) approach a limit cycle |
| |
| o . . > . . . |
| . . . . |
| . . . . |
| . . . figure: 'o' is an equilibrium point|
| . . . . |
| . . . . |
| . . . |
| . |
+----------------------------------------------------------------+
In particular then, chaotic solutions are ruled out.
While the above theorem is certainly not obvious, one severe constraint
which helps to bring the theorem about can be seen quite easily.
For it follows immediately from (11.4) that a slope in the (x,y) plane,
calculated from the given differential equations is:
. dy
y ---- +---------------+
___ = dt | dy g(x,y)|
. ------ =| ---- = ------+
x dx | dx f(x,y)|
+---------------+
----
dt
for example:
2
dy 3 x + y
---- = ---------
dx 2 x - y
which assigns a unique (that is 0 or 1 value, but not more) of the
phase path slope dy/dx to each point (x,y) of the plane (because when
you choose (x,y), e.g. (1,1), you choose only one point in the plane.
And if you calculate the right side of (11.5), the value is completely
determined by this choice of that x and y point, and gives only one
value.
+---------------------------------------------------------------------+
|This because: |
|1) f(x,y) is unique |
| |
|2) g(x,y) is unique |
| |
|3) from (1) and (2) follows that the quotient f(x,y)/g(x,y) is unique|
| (-with the exception when f(x,y) = 0 |
| |
| (because then you can give g(x,y) any value, e.g. 0/3=0 but |
| also 0/5=0 |
| or 0/10 = 0|
| ...) |
| |
| -with the exception when f(x,y) = 0 and also g(x,y) = 0 (because |
| then the value is also not determined uniquely, 0/0=3 can be |
| chosen because 0 = 0.3 and 0/0=4 can be chosen, because 0=0.4|
| etc.) |
+---------------------------------------------------------------------+
So only 1 slope.
The only exception is when the point in question is an equilibrium
point, at which both f(x,y) and g(x,y) are zero.
So because every point has only one slope, a phase path
+---------------------------------------------------------------------+
| so this is possible: |
| |
| . |
| . |
| . |
| o (x,y) |
| . figure: only one slope per given point (x,y)|
| . possible if only 2 variables x,y |
| . |
| |
+---------------------------------------------------------------------+
+---------------------------------------------------------------------+
| and so this is NOT possible: |
| |
| . . |
| . . |
| . . figure: more than one slope in a 2 variable plane|
| . o (x,y) per given point (x,y) |
| . . . is not possible if only 2 variables x,y |
| . |
| . |
| |
+---------------------------------------------------------------------+
in a two-dimensional autonomous system can not cross itself (with the
exception at a equilibrium point).
This prohibits, at a stroke, a system with a two-dimensional phase path
behaving like in figure 11.2 (where you see a picture of points in the
(x,y) plane wih a lot of crossings, in other words a lot of
non-equilibrium points having more than 1 slope at a given point (x,y).
+-----------------------------------------------------------+
| . . . |
| . |
| . . . |
| . . . . |
| .. . . ... . . . |
| . . . .... . . |
| . . . .. . |
|. . . . . . |
| . . . . . |
+-----------------------------------------------------------+
---
[book: source: Acheson, David - from Calculus to Chaos, an
Introduction to Dynamics - 1997 - p. 156]
---
[kn, zoe, wo, 12-01-2000 21:55:57]
So, if you read the above, how to go about to prove that systems of 3
differential equations with 3 variables and higher are candidates for
chaos?
You do this by showing that this equations do indeed allow more than 1
slope per given point (x,y) in the plane.
Now the general form of the 3 first order differential equations, in 3
variables is:
+--------------+
| . |
| x = f(x,y,z) | (12)
+--------------+
| . |
| y = g(x,y,z) |
+--------------+
| . |
| z = h(x,y,z) |
+--------------+
for example:
.
x = 2 * x + y - 3 z (13)
.
y = x - y + 2 y.z
. 2
z = x + y - 6 z
If you choose a point (x,y), say (1,1) in the plane and fill this in in
(13), you get the following equations:
.
x = 2 * 1 + 1 - 3 z
.
y = 1 - 1 + 2 1.z
. 2
z = 1 + 1 - 6 z
or
.
x = 3 - 3 z
.
y = 2 z
.
z = 2 - 6 z
so if you look in the xy plane for the
slope:
.
y g(x,y,z) 2 z
--- = -------- = ---------
. f(x,y,z) 3 - 3 z
x
then you see, because of this free to
choose 3th variable z, after fixing
that first 2 variables, you can give 1
point x,y a lot of different slopes
(e.g. when you choose z=2, you have the
slope 2.2/3-3.2 = -4/3,
when you choose z=3, you have the
slope 2.3/3-3.3 = -6/6
etc...)
A simular reasoning can of course also be applied for first order
systems of differential equations in 4 variables, 5 variables, 6
variables, etc...
So a system of 3 first order differential equations in 3 variables is
the
minimum system in which chaos can occur.
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Internet: see also:
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Math: Chaos theory: Overview: Can you give me an overview of chaos
theory?
http://www.faqts.com/knowledge_base/view.phtml/aid/26929/fid/867
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