Is Shur's arm lemma true for nonconvex arcs?

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Exercise 8
1.16 Shur's Arm Lemma
The following result describes how the distance between the end points of a
planar curve is effected by its curvature:
Theorem 7 (Shur's Arm Lemma). Let a1, az: [0, L] → R³ be unit speed
C' curves such that the union of each a, with the line segment from a,(0) to
a,(L) is a conver curve. Suppose that for almost all t e [0, L], K, (t) is well
defined, e.g., a, is piecewise C², and
Ki(t) 2 K2(t)
for almost all t e (0, L]. Then
(n1(0), an(1) < dist (02(0), a{(1).
dist
az(L)
a(0)
a,(L)
az0)
az(6)
a(to)
Proof. After a rigid motion we may assume that the segment a1 (0)a (L) is
parallel to the r-axis and a is rotating counterclockwise, see the picture
below. Then there exists to € [0, L) such that a (to) is horizontal. After a
rigid motion, we may assume that a, (to) is horizontal as well. Now let 6, be
the angle that a makes with the positive direction of the r-axis measured
counterclockwise. Then 6, € [-x, 7] (for 0 this follows from convexity of
a1, and for 02, this follows from the assumption that k2 < K1). Further note
that
10,(t)| = |0,(t) – 0,(to)| =
Thus |61(t)| > |62(t)|, and, since |0,(t)| € [0, #], it follows that
cos (61(t)| < cos |02(t)|-
Finally note that, if we set e := (1,0), then
||a1(L) – a1(0)|| = (a1(L) – a1(0), e1)
cos |0,(t)| dt
cos |02(t)| dt
(az(L) – az(0), e,)< ||a2(L) – az(0)|l-
Exercise 8. Is Shur's arm lemma true for nonconvex arcs?
Transcribed Image Text:1.16 Shur's Arm Lemma The following result describes how the distance between the end points of a planar curve is effected by its curvature: Theorem 7 (Shur's Arm Lemma). Let a1, az: [0, L] → R³ be unit speed C' curves such that the union of each a, with the line segment from a,(0) to a,(L) is a conver curve. Suppose that for almost all t e [0, L], K, (t) is well defined, e.g., a, is piecewise C², and Ki(t) 2 K2(t) for almost all t e (0, L]. Then (n1(0), an(1) < dist (02(0), a{(1). dist az(L) a(0) a,(L) az0) az(6) a(to) Proof. After a rigid motion we may assume that the segment a1 (0)a (L) is parallel to the r-axis and a is rotating counterclockwise, see the picture below. Then there exists to € [0, L) such that a (to) is horizontal. After a rigid motion, we may assume that a, (to) is horizontal as well. Now let 6, be the angle that a makes with the positive direction of the r-axis measured counterclockwise. Then 6, € [-x, 7] (for 0 this follows from convexity of a1, and for 02, this follows from the assumption that k2 < K1). Further note that 10,(t)| = |0,(t) – 0,(to)| = Thus |61(t)| > |62(t)|, and, since |0,(t)| € [0, #], it follows that cos (61(t)| < cos |02(t)|- Finally note that, if we set e := (1,0), then ||a1(L) – a1(0)|| = (a1(L) – a1(0), e1) cos |0,(t)| dt cos |02(t)| dt (az(L) – az(0), e,)< ||a2(L) – az(0)|l- Exercise 8. Is Shur's arm lemma true for nonconvex arcs?
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