Place a wire mesh surface across the upstream end of

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 wire-mesh \surface" across the upstream end of the tube|in, say, the form of a hemisphere. Through each subregion of the mesh ows some number of gallons-per-minute, 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 ux-preserving. The above description is meant to motivate the de nition below, where cross-section 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 inward-pointing 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 line-segment. 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 rst-return 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 cross-section L is \transverse" to th...
View Full Document

This note was uploaded on 01/26/2012 for the course MTG 6401 taught by Professor Staff during the Fall '09 term at University of Florida.

Ask a homework question - tutors are online