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Unformatted text preview: oincare section of a flow Imagine a large tube submerged horizontally in a river, through which water ows in some complicated way. Place a wiremesh \surface" across the upstream end of the tubein, say, the form
of a hemisphere. Through each subregion of the mesh ows some number of gallonsperminute,
the \ ux through the surface", which therefore induces a measure on this surface. If we place a
mesh also across the downstream end, we get a map from the upstream surface to the downstream
surface simply by watching molecules of water ow from the one to the other. Since the ow
preserves volume (water being incompressible), this \induced map" is uxpreserving.
The above description is meant to motivate the de nition below, where crosssection
is a
\surface", vol( ) = 0, which is \transverse" to the ow in an appropriate sense. Let (0;t]( ) be
the set of cueballs swept out as ows for t seconds. (More generally, for any subset W R of time, let
W ( ) denote the union of t , over all t
W .) The ux of is the limiting rate that the volume of
(0;t] ( ) grows. By this we mean
2 ux( ) := tlim 1 vol (0;t]( ) :
&0 t (10) In order to show that this limit exists in 0; 1], we employ a type of argument which is often
useful in dynamics: subadditivity. The function V t] := vol (0;t] is subadditive because
V t] + V s] = vol( (0;t] ) + vol( (t;t+s]
vol( (0;t+s] ) ) V t + s] : Now x a positive t. Given any smaller positive time s, let N be the integer such that Ns
1
(N 1)s. By subadditivity, V s] N V N s]. Thus
1 V s]
s 1 Ns V N s] 1 V t]
Ns
N 11
V t] :
N
t Math. Intelligencer, vol.17 no.1, (1995), 8{16. t> 10 J.L. King Sending s & 0 along any sequence sends the associated N to in nity, and so liminf s&0 1 V s]
s
dominates 1 V t]. Taking a supremum over positive t shows that the limit in (10) always exists
t
in 0; 1], and shows that
1
ux( ) = sup vol (0;t] ( ) :
(11a)
t
t>0 Figure 17 illustrates how this induced measure, ux( ), will be used to prove the conservativity
result of this article, theorem 14.
How to visualize the induced measure. An important special case is when we have a curve
which is a subarc of the table's boundary @ , and is the set of inwardpointing cueballs with
footpoint on . We can get an explicit integral for vol( ) by describing a cueball v on the boundary
in terms of \relative angle". Write v = (v; ), where 2 ( 2 ; 2 ) denotes the angle that v makes
relative to the inward normal of @ at v. Figure 12 shows an example in which is a linesegment.
ρ t Σ Boundary of Γ Figure 12 In a xed relative direction , assume that the points of can be moved a distance t
without encountering the cushion. Then the shaded parallelogram shows the location of the
footpoints of those v in of relative angle , after they have owed for at most t seconds. The area of the above parallelogram is t cos( ) times the length of . Multiplying by 1=t and
then integrating over gives
ZZ
ux( ) =
cos( ) d dv ;
(11b)
(v; )2 where \dv" denotes arclength measure along @ and \d " is arclength on 2 ; 2 .
It is routine to check that this formula remains valid for a general arc by rst approximating
the arc by line segments and then sending t & 0. This last step uses that has a rstreturn
function R : ! 0; 1] which is everywhere positive, where R (v) := sup t>0 (0;t)(v) is disjoint from : It turns out that for an arbitrary set L of cueballs, the condition RL > 0 is a reasonable de nition
that crosssection L is \transverse" to th...
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