37^ a/81 ~7s~// plane curves, convex curves, and …/67531/metadc...the curve may not technically...
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PLANE CURVES, CONVEX CURVES,
AND THEIR DEFORMATION VIA
THE HEAT EQUATION
THESIS
Presented to the Graduate Council of the
University of North Texas in Partial
Fulfillment of the Requirements
For the Degree of
MASTER OF ARTS
By
Johanna M. Debrecht, B.A.
Denton, Texas
August, 1998
3 7 ^
A/81
~7S~//
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Debrecht, Johanna M., Plane Curves, Convex Curves, and Their Deformation
Via the Heat Equation. Master of Arts (Mathematics), August, 1998, 93 pp., 24
figures, bibliography, 5 titles.
We study the effects of a deformation via the heat equation on closed, plane
curves. We begin with an overview of the theory of curves in R3. In particular, we
develop the Frenet-Serret equations for any curve parametrized by arc length. This
chapter is followed by an examination of curves in R2, and the resultant adjustment
of the Frenet-Serret equations. We then prove the rotation index for closed, plane
curves is an integer and for simple, closed, plane curves is ±1. We show that a
curve is convex if and only if the curvature does not change sign, and we prove the
Isoperimetric Inequality, which gives a bound on the area of a closed curve with fixed
length. Finally, we study the deformation of plane curves developed by M. Gage and
R. S. Hamilton. We observe that convex curves under deformation remain convex,
and simple curves remain simple.
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Debrecht, Johanna M., Plane Curves, Convex Curves, and Their Deformation
Via the Heat Equation. Master of Arts (Mathematics), August, 1998, 93 pp., 24
figures, bibliography, 5 titles.
We study the effects of a deformation via the heat equation on closed, plane
curves. We begin with an overview of the theory of curves in R3. In particular, we
develop the Frenet-Serret equations for any curve parametrized by arc length. This
chapter is followed by an examination of curves in R2, and the resultant adjustment
of the Frenet-Serret equations. We then prove the rotation index for closed, plane
curves is an integer and for simple, closed, plane curves is ±1. We show that a
curve is convex if and only if the curvature does not change sign, and we prove the
Isoperimetric Inequality, which gives a bound on the area of a closed curve with fixed
length. Finally, we study the deformation of plane curves developed by M. Gage and
R. S. Hamilton. We observe that convex curves under deformation remain convex,
and simple curves remain simple.
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PLANE CURVES, CONVEX CURVES,
AND THEIR DEFORMATION VIA
THE HEAT EQUATION
THESIS
Presented to the Graduate Council of the
University of North Texas in Partial
Fulfillment of the Requirements
For the Degree of
MASTER OF ARTS
By
Johanna M. Debrecht, B.A.
Denton, Texas
August, 1998
3 7 ^
A/81
~7S~//
![Page 5: 37^ A/81 ~7S~// PLANE CURVES, CONVEX CURVES, AND …/67531/metadc...the curve may not technically lie in R2, by a suitable choice of coordinates, we can, without loss of generality,](https://reader035.vdocuments.net/reader035/viewer/2022071501/6120d737ac76cc71db69b3d7/html5/thumbnails/5.jpg)
ACKNOWLEDGMENTS
The writer gratefully acknowledges the patient and careful guidance of Dr. Joseph
Iaia, who directed the work on this paper. Gratitude is also expressed for the many
people who have given their encouragement, assistance, and support. These include
her advisor, committee members, family, and friends.
In particular, the writer wishes to express her deepest appreciation to her husband,
David, whose tireless support and encouragement made this achievement possible.
111
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TABLE OF CONTENTS
LIST OF FIGURES vi
1 INTRODUCTION 1
2 LOCAL CURVE THEORY 5
2.1 Basic Definitions and Concepts 5
2.2 The Frenet-Serret Apparatus 15
3 PLANE CURVE THEORY 21
3.1 Basic Definitions and Results for Plane Curves 21
3.2 Rotation Index for Closed Curves 26
3.3 Rotation Index for Simple Closed Curves 37
4 CONVEX CURVES AND THE ISOPERIMETRIC INEQUALITY 47
4.1 Convex Curves 47
4.2 The Isoperimetric Inequality 53
5 THE HEAT EQUATION 60
5.1 Curves in the Plane 61
5.2 Evolution of Simple Curves with Bounded Curvature 71
5.3 Convex Curves in the Plane 81
5.4 An Application of the Isoperimetric Inequality 85
BIBLIOGRAPHY 93
iv
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LIST OF FIGURES
2.1 A Moving Frame: | t , N, 2?j 14
3.1 A Non-simple Curve 26
3.2 Two Simple Curves 26
3.3 Definition of 9 in the Plane 27
3.4 The Rotation Index for the Unit Circle 28
3.5 No Change in the Sign of y'/x' 31
3.6 y' Changes Sign an Odd Number of Times . 33
3.7 Points where y' = 0, x' 0 34
3.8 The Sign of y'/x' Changes from Positive to Negative 35
3.9 The Sign of y'/x' Changes from Negative to Positive 35
3.10 The Vector Valued Function p(u,v) 38
3.11 The Region of A 38
3.12 Case I and II Type Points 43
3.13 Case III and IV Type Points 44
4.1 A Convex Curve 47
4.2 The Tangent Lines at A, B, and C 48
4.3 Proof That Line I is Parallel to Line li 49
4.4 Coinciding Tangent Lines 51
4.5 A Straight Line from Si to S2 52
4.6 The Circle (3 and the Curve a Between Two Tangent Lines 54
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4.7 The Circles O and O' and the Curve a 58
5.1 The Method of Deformation 61
5.2 Schur and Schmidt's Lemma 75
5.3 Convex Arc from a Circle 77
VI
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CHAPTER 1
INTRODUCTION
The main focus of this paper is the study of convex, plane curves undergoing defor-
mation by means of the heat equation. This topic was originally written about by M.
Gage and R. S. Hamilton in three journal articles, [Ga, Ga2, GaHa]. In order to fully
understand this topic, we begin with the basic theory of curves in R3. The majority
of the material in chapters 2-4 is based on the book by Millman and Parker, [MiPa].
After defining what we mean by a curve, we discuss the importance and advantages
of regular curves. The issue of how to represent the image of a curve is discussed
next. This is referred to as the parametrization of the curve. It turns out that
both the tangent vector field and the length of the curve are independent of the
parametrization. In other words, the tangent vector field and the length of a curve
are "geometric" quantities. Thus, we are able to represent it in any form we choose.
There are definite advantages to using the arc length to parametrize the curve. We
will largely assume that curves are parametrized by arc length, (we say such a curve
has unit speed), and any exceptions will be so noted. Thereafter, we define the
normal, and binormal vector fields, as well as the curvature and torsion of a curve.
After the basic definitions have been made, we develop the Frenet-Serret apparatus
for curves which have been parametrized by arc length. The Frenet-Serret apparatus
is the basic tool used in the study of curves. It involves three vector fields along the
given curve, namely the tangent, normal and binormal, and two real valued functions,
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the curvature and the torsion. In fact, the Frenet-Serret apparatus uniquely and
completely determines the geometry of the curve.
Having established the foundation for a study of curves, we restrict our focus
to plane curves; i.e., curves which lie in a plane. It is not always clear from the
parametrization of a curve whether or not the image lies in R3 or R2. (Though
the curve may not technically lie in R2, by a suitable choice of coordinates, we can,
without loss of generality, suppose that it does.) We will present an example in which
this is the case. Then, we will develop a theorem which states that certain conditions
in the Frenet-Serret apparatus are equivalent to ensuring that the curve lies in a
plane. That is, by merely examining the Frenet-Serret apparatus of the curve, we can
easily determine if it lies in a plane.
We then proceed to make "modifications" to the definitions for the tangent and
normal vector fields, and the curvature which are specific to plane curves. The ad-
vantage to making these changes, is that we can make well-defined definitions; that
is, the definitions are defined everywhere in the plane. (This is not the case for curves
in R3; in particular, the normal vector field is not defined when the curvature (in R3)
is zero.) From there, we continue by defining closed curves, simple curves, and the
period of a closed curve.
Our next focus is on the rotation index of a closed, plane curve. Roughly speaking,
the rotation index is an integer which represents the number of times the curve makes
a complete rotation (of 360°) before "closing up;" that is, before returning to an
arbitrarily chosen starting point. As we will discover, the rotation index is dependent
on the number of times the slope of the tangent line (to any point on the curve)
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changes sign, and upon the means by which the sign change occurs. Following this
theorem, we are lead to discover that the rotation index for a simple, closed, plane
curve is ±1. (A simple curve is one which does not intersect itself at any point
between the starting point and the length of the curve.)
The fourth chapter consists of two theorems, seemingly unrelated, and yet both
intertwined in the results developed in the final chapter dealing with the deformation
of convex curves via the heat equation. We begin by making a precise definition for
what we mean by a convex curve. This definition automatically implies that convex
curves are simple. Following this, we develop a sufficient and necessary condition for a
simple, closed, plane curve to be convex. In accordance with one's geometric intuition,
it turns out that this condition is the requirement that the curvature does not change
sign. Thereafter, we study a well known result, the Isoperimetric Inequality.
The Isoperimetric Inequality quantifies the limit placed on the area bounded by
a simple, closed, regular plane curve of fixed length. This particular topic has long
been a source of interest and study. Still today, we are interested in knowing how
to maximize the area of a region, given specific limitations on its perimeter, and
sometimes its shape. In addition, we will show that the geometric shape of fixed
length which bounds the maximum area is, in fact, a circle.
Finally, we begin the study of the major topic of this paper—the deformation
of plane curves via the heat equation. Like the Isoperimetric Inequality, the heat
equation has been a source of study for some time. Its behavior in particular situ-
ations is well known. Gage and Hamilton, in their journal article [GaHa], applied
a particular type of deformation to simple, closed, regular, plane curves, which lead
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to the development of the heat equation in describing the evolution of the curve.
The method chosen to deform the curve was both intuitively easy to understand and
mathematically simple to describe. Moreover, they were able to show that under this
type of deformation, subject to restrictions on the curvature, curves which did not
intersect themselves still did not intersect themselves while evolving. In addition,
when they restricted their attention to convex curves undergoing deformation, they
discovered that the evolving curves remained convex. Furthermore, they found that
these curves, in the sense of the Isoperimetric Inequality, begin to approach the shape
of a circle.
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CHAPTER 2
LOCAL CURVE THEORY
We begin with a study of curves in R3 . We will restrict our attention to those curves
which can be described by differentiable equations. In this way, we can avoid some
technicalities and pathologies that might otherwise arise. For most of this paper, we
will assume that curves are at least of class C3; that is, three times differentiable,
with the third derivative also continuous. Any exceptions will be specifically noted.
2.1 Basic Definitions and Concepts
While it is intuitively clear what we mean by a curve, it is considerably more difficult
to define it in a mathematically precise way. In addition, we would like to have some
way to distinguish a "smooth" curve from one that is "flat," or that has "corners."
A "flat" curve is simply a line, which is not geometrically interesting. On the other
hand, a curve with sharp "corners" could pose great difficulties, since functions are
generally not differentiable at such points. Thus, we make the following standard
definition for a curve in R3.
Definition 2.1.1 (Regular Curve) A regular curve in R3 is a function j: [a, 6] —»
R3 such that 7 6 Ck, for some A: > 1 and such that dj/dt ^ 0, Vt € [a, b].
Observe that if (dj/dt) = 0 over some interval, then the curve is constant there,
that is, it is actually a single point. Sharp corners could result from these "points,"
or from single values of t (not an interval) where (dj/dt) = 0. In addition, there is
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no well-defined tangent line at sharp corners (see Definition 2.1.4 below). Thus, by
requiring our curves to be regular curves, we avoid these difficulties.
Also note that by this definition, a curve is actually a function, and not the image
set. Two different curves (i.e. functions) could very well have the same image set.
We then say that the image set has two different parametrizations. Since we wish to
have only regular curves, we will want to be able to get a regular parametrization for
a given image set. This issue is discussed more fully later.
This definition does not require that a regular curve be one-to-one. Geometrically,
this means that the image may "cross," or intersect, itself. However, by the following
lemma, it cannot intersect itself infinitely often.
Lemma 2.1.2 Let 7 beaC1 curve on [a, b] in R3. Suppose 3 a sequence of points {£„}
in [a, 6] such that Vn € N, tn ^ tm unless n = m. Also, suppose Vn € N, 7 (tn) = x0,
(i.e., 7 intersects itself infinitely often at the point x0).
Then 7 is not a regular curve.
Proof. First note that since [a, b] is a compact interval in R, and Vn € N, tn €
[a, b], then 3 a convergent subsequence, {£«*}> and t* € [a, b] such that tnk -» t*.
We will show that (dj/dt) (t*) = 0. Since t* € [a, 6], then =>- 7 cannot be a regular
curve.
By assumption, 7 is differentiate on [a, 6], thus it is continuous on [a, b]. Let
£ > 0. By the continuity of 7, 35 > 0 such that if \t — f | <6, then | f( i) - 7(i*)| <
e. Since lim^oo tnk = t*, 3N G N such that VA; > N, |£nfc —t*| < S. But then
Vk > N, => 17 (tnk) — 7 (i*)| = \xQ — 7(t*)| < e. That is, by definition, y(t*) = x0.
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Now, let k ^ m > N. By assumption, tnk, tnm e [a, 6], and tnic ± tnm. Also,
7 (tnk) = X0 = 7 (tnm). By the definition of the derivative at t*, for k>N,
f ( 0 = t U m ^ ^ - f ( O = 0 . at tnk-t -*o tnk — t* t„k-t*-+Q tnk — t*
(Note that since Vfc, m € N, tnk ^ tUm, =$> VA: € N, except possibly at one value of
nk, then tnk ^ t*. Thus, for sufficiently large he N, tnk - t* ^ 0.)
Therefore, since (dy/dt) (t*) = 0 and t* € [a, 6], toen 7 is, definition, not a
regular curve. g
Now that we have a precise definition for a regular curve, we would like to interpret
this definition in some "real world" sense. It is customary to visualize a curve as the
path of a moving particle. We can then associate a more intuitive meaning to some
of the vector fields along the curve.
Definition 2.1.3 (Velocity Vector) The derivative dj/dt of a regular curve y(t)
at a particular time, t0, is its velocity vector at t0. The vector-valued function, dy/dt,
is the velocity vector field. The speed of a particle on the curve at time tQ is given by
the magnitude (length) of the velocity vector at ta, that is, speed = \(dj/dt) (£0)|.
Thus, the velocity vector at a particular point in time is in the direction the particle
was moving at that instant, with magnitude equal to the speed of the particle. Note
that requiring the curve to be regular implies that the speed is never zero. Thus, the
particle never stops moving, not even for a moment.
Definition 2.1.4 (Tangent Vector Field) The tangent vector field to a regular
curve, 7 (i), is the vector-valued function, T(t) = (dj/dt) / \dj/dt\. Geometrically,
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when evaluated at a particular time t0, the tangent vector field is the unit vector in the
direction that the particle was moving at that moment. Generally, the tangent vector
field is referred to as the tangent, or simply T. Note that the regularity condition
ensures that the tangent is defined at every point on the curve.
The tangent vector field turns out to be independent of the parametrization of
the image set. In other words, the tangent is a geometric property, depending only
on the geometric shape of the curve's image set.
Definition 2.1.5 (Reparametrization) Let 7: [a, 6] -> R3 be a C1 curve. Then a
reparametrization of 7 is a one-to-one, onto function g: [c, d\ —> [a, b] such that both
g and its inverse h: [a, 6] —» [c, d] are Ck for some k > 1.
Lemma 2.1.6 The tangent vector field is independent of the parametrization of the
image set for a particular regular curve.
Proof. Let 7 be a regular curve, and let g be a reparametrization of it. Define
P = 1°9- Suppose u0 = g (v0). Let T be the tangent vector field of 7 at u0, and S be
the tangent vector field of g at v0. Then by definition,
d/3 <0148. <*£ C dv du dv du dv
\4£\ \ dv \ \du
1^1 1 dv 1 du
\d&\ I dv I
= ( f ) (±1) = ± f .
Thus, if dg/dv > 0, then S = T, and if dg/dv < 0, then S = —T. The geometric
interpretation refers to whether 7 and /3 are traversed in the same or opposite direc-
tions. •
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Definition 2.1.7 (Regular Curve Length) The length of a regular curve, 7: [a, b] —
R3 is given by
m i * . A 1*1
The next lemma shows that the length of a curve turns out to be a geometric
invariant as well.
Lemma 2.1.8 Let a: [a, b] —>• R3 be a curve, and let g: [c, d] —• [a, 6] be a reparame-
trization, where j3: [c, d] —> R3 is defined by j3 — a og. Then the length of a is equal
to the length of (3.
Proof. We begin by showing the following claim.
Claim 2.1.9 /3 is a regular curve if a is regular.
Proof. Suppose that a is a regular curve. Thus, da/dt ^ 0 , Vi € [a, 6]. By the
chain rule, we have
0'(t) = a'(g(t)).g'(t).
By the definition of reparametrization, we know that g~l (g (t)) = t. Then by taking
the derivative with respect to t on both sides, we get that (<7-1)' (g(t)) • g' (t) = 1.
Thus, g' (t) 7̂ 0.
Then, by the equation for 0' (t) given above, since a' (g(t)) ± 0 ^ g' (t), then
(t) ^ 0. Therefore, by definition, 0 is a regular curve. I
Now, we will show that the lengths of ci and /? are equivalent. By the previous
definition, we have
l ( p ) = [ ' I3'(g(t))\\g'(t)\ dt. J c
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If g'(t)>0,=>£ \c?{g(t))\\g'(t)\dt = jf \& (g (t))\g' (t) dt
= f \at (u)| du = L (a) J a
(u — g (t) du = g' (t) dt)
and if sf (*) < 0, = > f K (g (t))| |g' (f)| dt = - J \St (g (t))| g' (f) dt pa
= — I |a' (u)\ du (as above)
= f |a' (m)| du = L (a) . J a
Since the length of a curve does not depend on the parametrization, it makes sense
to think about reparametrizing a curve by its length. The advantage to parametrizing
by arc length is that the velocity vector field is the same as the tangent vector field.
Then the original curve, say 7, will have a speed of one.
Definition 2.1.10 (Unit Speed Curve) A curve 7 (t): [a, 6] —»• R3 is called a unit
speed curve when Id'y/dt] = 1.
In order to reparametrize a curve by its arc length, note that for s equal to the
arc length, by definition 2.1.7, we can write
dt.
Thus,
J (t) = <?7 dt > 0 (since 7 is a regular curve).
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This implies that s (t) is invertible, and we will let t (s) be its inverse. (Note that this
is an abuse of notation, but it is the standard practice.) We then define /3 (t) to be
j(t (s)). Then since
9 (s) = f (t (s)) • a (s) =* 9 (s) = If (t (a))| I< (s)I = If (t (s))| = 1, *(*«)
—*
we have that j3 is a unit speed curve. Finally, we have that is 7 reparametrized by
arc length.
Hence, so long as we do not care about the particular parametrization used, we
can assume that any regular curve is parametrized by arc length. While in theory,
this is always possible, in practice it may be very difficult. In Example 2.1.11, we
find the arc length parametrization of a regular curve. Then, in Example 2.1.12, we
examine a curve for which the arc length parametrization would be difficult to obtain.
Example 2.1.11 Consider the curve a (t) = (et cos t, el sin t, el).
(da/dt) = (e4 cos t — el sin t, e* sin t + e* cos t, e') ^ 0 so a is regular. Then
= y j (e4 cos t — e* sin t)2 + (el sin t + el cos t)2 + (e1)2
— V e2t + e2t + e2t = V3e2t = et\/Z.
So s(t) = f 1 ^ dt = Vs f eldt = \ /3e ' ' = y/Se* - V3, Jo I J o 0
where s (t) is the arc length. Thus, solving for t, we get t = In (l + (s/\/3)) = g (s),
the arc length reparametrization of a. Then the unit speed reparametrization of a is
given by
0 (s) = e% (cos t, sin t, 1) = e l n(1 +^) ^cos (ln (* + ) >sin (lQ (2 + ) , i ) •
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Example 2.1.12 Let 0(t) = (21, t2,0), which is a parabola. Then dfi/dt = (2,21,0)
and
= V4 + 4*2 = 2Vl +12. dp dt
Thus
s(t) = J 2vT+~r^ dr = t\J 1 +t2 + In (t + Vl + t2^j .
Though the parabola is a rather straightforward geometric curve, it is very difficult to
find t = g(s) from this equation.
In addition, sometimes the integral s (t) = f* \dj/dt\ dt may not be elementary,
which can also make parametrization by arc length difficult. Nevertheless, it is a
common and relatively benign assumption to suppose that all regular curves have
been parametrized by arc length. Recall that this does not change the geometric
shape of the image set. Though the new parametrization is almost certainly not the
same as the original, it is a common abuse of notation to refer to it by the same name,
e.g. 7. Also, if a curve is parametrized by arc length, the usual variable of choice is
s. For a general curve, not necessarily parametrized by arc length, it is customary to
use the variable t. Furthermore, it is usual to talk about the shape of the curve when
really it is the shape of the image set. We will follow these conventions.
Recall that curves parametrized by arc length have unit speed. Thus, we have
s(t) = ^ dt = I = (2.1)
Hence, we can write T (s) = 7' (s) = 7'. Now we can discuss the notion of "curvature"
of the curve. That is, we wish to measure the amount of bending the curve undergoes
from one point to another. Intuitively, a straight line should have curvature of zero,
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while a circle should have a constant curvature. These considerations lead to the
following definition.
Definition 2.1.13 (Curvature) The curvature of a unit speed (regular) curve 7 (s)
is k (s) = T' (s) .
Then, since for a line the tangent is constant =*• K = 0. For a circle of radius r, we
can use the unit speed parametrization 7 = (r cos (s/r), r sin (s/r), 0). Then, after a
bit of work, we can see that K(S) = 1/r. Thus, the smaller the radius, the "faster"
the bending and the larger the curvature.
At this point, it is customary to define an orthonormal frame in which to describe
the curve. Unfortunately, the standard orthonormal frame {ei, e2,63} reflects the
geometry of R3 rather than the geometry of the curve. Again the standard approach
is to view the curve as it would appear to a particle traveling along it. The result is
a "moving frame" which travels along the curve with the particle. Consequently, the
frame is generally not fixed, but depends upon where on the curve one is; i.e., on the
arc length, s. To define this frame, we use the tangent, T, find another vector field
orthogonal to it, and use their cross product for the third. This leads to the following
definitions.
Definition 2.1.14 (Normal and Binormal Vector Fields) The principal normal
vector field to a unit speed (regular) curve 7, where the curvature K is nonzero, is
given by N (s) = f ' ( s ) /K (S) = f' (s) / T' (s) . Note that this implies that TV is a
unit vector field itself. The binormal vector field to 7 is given by B(s) =f (s) x N (s).
Since both T and N are unit vector fields and T _L N, then B is as well.
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Note that these definitions are only defined when «(s) ^ 0. When the curve is
traveled in a counter-clockwise direction, the N is assumed to point in the direction
which gives the | t , iV, £ j frame a right handed orientation. (See Figure 2.1.) Once
this frame is defined, it is possible to make the following definitions.
B A*
Figure 2.1: in R3
Definition 2.1.15 (Torsion) The osculating plane to a unit speed (regular curve)
7 (s), at a particular point along the arc length, 7 (sD), is the plane through the point
7(s0) which is perpendicular to B. Thus, the osculating plane is spanned by f and
N. The torsion, r, is then defined to be the real-valued function of arc length given
by
T(s) = -(S'(s),N(s)y
In the same way that the curvature k measures the "twisting" of the curve at a
particular point away from the tangent line at that point, the torsion r measures the
"twisting" of the curve out of the osculating plane.
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2.2 The Frenet-Serret Apparatus
These five notions |/c, r,T,iV, 5 1 provide the basis for the derivation of the Frenet-
Serret equations. These well known equations are used to study the geometry of
curves in R3, and actually uniquely determine the curve as well. (We will not con-
cern ourselves with uniqueness here, but a complete proof may be found in [MiPa,
pages 42-44].) We will derive the Frenet-Serret equations for a curve parametrized
by arc length, and state the Frenet-Serret equations for a curve which has not been
parametrized by arc length.
Theorem 2.2.1 (Frenet-Serret) Let a be a unit speed (regular) curve, with nonzero
curvature, (K^O), and Frenet-Serret apparatus j/c, r, T, N, B j . Then for all s
1. T ' ( s ) = K(S)N(S)
2. N ' ( s ) = - K ( S ) T { S ) +T(S)B(S)
3. B' (s) = —T (s) N (s).
Proof. By definition,
T'(s) N ( s ) = « = > T ( s ) = K(S)N(S)
K(S)
which completes the proof of 1. The proofs of 2 and 3 make use of the following
lemma. Lemma 2.2.2 Let v be a unit vector. Then the dot product v • if = {v, v1) = 0.
Proof. Suppose v = (a,b). So v1 = (a1,6'). Since v is a unit vector, then
|u| = y/a2 4- ft2 = 1 <$• a2 + b2 = 1. By differentiating both sides of the last equation,
we get 2aa! + 2bbf = 0 aa! + 66' = 0. Hence, v • if = (a, b) • (a', b') = aa' + bV = 0.
I
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Since { f , are all unit vectors and all mutually orthogonal, they form an
orthonormal frame. Thus any vector v can be written in terms of the three as follows:
v= (T,V}T+(N,V}N+(B,V}B.
In particular, since N' (s) is a vector, it can be written as
X' (s) = (T, JV') f+(/},&) N+(§, A f B . (2.2)
T _L N, so ^Tj — 0. Differentiating both sides of this equation, we get
(F\N) + (F,N') = 0<=*(F,N'} = ~(T',N}
= - (K (s) N (s), N (S)^ (by part I)
= -k(S).
Since N is a unit vector, by the lemma N • N> = 0. Thus, equation 2.2 has become
N' (s) = K (s) f + (b, N'^ B.
Since B ±N, then (B , N ^ = 0. Using the same method as above and differenti-
ating both sides, we derive
(B , + (B, iV'} = o «=* (B, N = -(B',N^=T (s) (by definition).
So N (5) = -K (s) T + r (s) B. This ends the proof of part 2.
Applying the same tricks to B' (s), we get
# ( « ) = S
— (T, B)T+ N (by the lemma).
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Since TLB, then ( t , B^j = 0. Differentiating, we get
(T,B,s) + (T',B} = 0 ^ (?,&) = -(?,&}
4=^ ( f , B'} = - (-k (s) N (s), Bj = 0 (since N _L B).
Then by definition, r = — (&', N^j = ^N, B'^j =$• B' (s) = —TN. This ends the
proof. I
Having derived the Frenet-Serret equations for curves parametrized by arc length,
we now state the Frenet-Serret equations for curves not parametrized by arc length.
These are found by reparametrizing by arc length, s, using the Frenet-Serret equations
in that situation, then resubstituting and writing everything back in terms of the
original variable, t.
Let /? be one such regular curve, and let v (t) = d(3/dt
equations for the curve /? are the following.
1. F = KVN
Then the FVenet-Serret
2- IT = -Ktrf +tvB (2-3)
3. f = -TVN
In the following example, we will compute the Frenet-Serret apparatus for a particular
curve.
Example 2.2.3 Consider the curve a (s) = cos s, ^ - sin s, -y§ cos s). We will
show that it is a unit speed, regular curve, and then compute its Frenet-Serret appa-
ratus.
da ds
( 5 • 1 2 • \ = ( - j g S i n s , - c o s s , — sins 1
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da ds
I 25 .
- V i m 8 1
144 sin2 s + cos2 s + —— sin2 s = 1.
169
So a is a unit speed curve. Since Vs, da/ds ^ (0,0,0), then a is also a regular curve.
Thus we may compute its Frenet-Serret apparatus, j/c (s), r (s), T (s) ,N (s),B(s) j .
By definition, we have
5 , da / 5 . 12 . \ T ( s ) = ^ = ( - l 3 S m S ' - C O S S ' l 3 S m V
So then we can compute K by first finding T'.
rt,, x ( 5 . 1 2 \ T'(s) = 1 cos s, sin s,— coss I
/c(s) = | f ' (s)| = cos2 s + sin2 s + cos2 s = 1.
—t
Next we find N.
# , , T'(s) { 5 . 1 2 \ ^ ( s ) = T W = r i s c o s 8>s m i s c o s 5 J
T/ien to find the binormal, we compute the cross product of T and N.
B (s) = T(s) x N (s) = —jjjsins —coss sins
—^cos s sins | | coss
= ( _ l s i n 3 s - ^ c o » » » ) ? - ( -60 . 60 . r - r sin s cos s + —r sin s cos s 13z 13̂ )T
= ( - ± s i n 2 s _ l c o s 2S ) r + ( - A s m 2
S - ^ c o s 2s )
Hence, => B (s) = (—^,0, — F i n a l l y , by the definition for T,
T (s) = — (s), iV (s)^ = 0 fsince B' (s) = 0).
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Recall from the definition of torsion r given earlier, that it measures the "twisting"
of the curve out of the osculating plane. (The osculating plane is spanned by T and
N.) Since r = 0 in the previous example, one might expect that a actually lies in a
plane. This is, in fact, the case, and we now prove a general theorem about curves of
this nature.
Theorem 2.2.4 Let a. (s) be a unit speed curve with K ^ 0. Then the following are
equivalent:
1. a is a plane curve. (That is, its image lies in a plane.)
2. B is a constant vector.
3. T = 0, Vs.
Proof. The equivalence of 2 and 3 may be readily seen by the Frenet-Serret —* —•
equation B = —TN. NOW, if we assume that a is a plane curve, then, without loss
of generality, we can suppose that a lies in the (x, y) plane (by appropriate choice of
coordinates). Thus, we can say
T(s) = tf(«) = (i '(»)>j/ '(«),0)#(0,0,0)
and t («) = ST (,) = (*" (»), y" (»), 0) # (0,0,0).
So, by the definition for B,
f'(s) B (s) = T(s)xN(s)=T(s)x
K{S)
= (*' («), v' W , 0) x 1 (x« (s), / (s), 0) V («))2 + ( y " (s))2 + o
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Thus, B (s) =
» J
x' (s) y' (s)
£isL /x"2+y"2 ^x"2+y"2
k
0
0
= (o-o)r-(o-o)j*+ x' (s) y" (s) _ x" (s) y'(s) y/x"2 + y'a y/x"2 + y'n k
yfx'u + y
= (0,0,±1).
) it . ( W - ' W ) k (see Lemma 3.1.1) /x"2 + y"2 / \\x'y"-x"y'\J ( y
77ms, JB IS a constant vector, so B' (s) = (0,0,0) = 0. Therefore, by definition,
T (») = - (5' (»), jv (»)) = - (5, J? (»)) = 5.
JVoiu, i/ we suppose that r = 0, Vs, then by the equivalence of 2 and 3, we know —•
tfiai B must be a constant vector. To show that a lies in a plane, it is sufficient to
find a vector, say v, such that for some point x0 on a, the dot product (a (s) — x0, v)
is identically zero. Examination of the Frenet-Serret equations leads us to let v be B
and x0 be a (0). Since B is a constant vector, then B' = 0. Thus,
(3 (s) -3(0), £ ) ' = (c? (s) , £ ) + ( a (s) - a (0), = ( f , g } + 0 = 0,
which implies that (^a (s) — a (0), B^ is a constant. Furthermore, when s = 0, then
this constant is clearly zero. Therefore, (a (s) — a (0), B^J = 0 and a must lie in a
plane. •
In the next chapter, we will restrict our focus to plane curves, rather than curves in
R3. After developing some elementary results, we will direct our attention to convex
plane curves.
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CHAPTER 3
PLANE CURVE THEORY
3.1 Basic Definitions and Results for Plane Curves
In the previous chapter, we examined curves by breaking them apart and studying
their behavior over small arc lengths. In this chapter, we take a more global approach.
In addition, we will focus our attention on curves in the plane. For simplicity, we will
assume that all curves have been written such that they lie in the (x, y) plane. Thus
we may view the curve as a function into R2, rather than K3. Recall that all curves
are assumed to be of at least class C3.
First, let us discover what happens to a regular plane curve with non-zero curva-
ture. We will need the following lemma.
Lemma 3.1.1 Let a (s) be a unit speed, plane curve given by (x (s), y (s)). Then
K=\xy x y \.
Proof. Since a is a unit speed curve, then \da/ds\ = 1. Thus,
=» yjx12 + yn = 1 <=> x'2 + y/2 = 1, (3.1)
and differentiating both sides gives 2x'x" + 2y'y" = 0
•<==> x'x" -I- y'y" = 0 <£=> x'x" = —y'y" (3-2)
Hence,
k = 15"| = y/x"2 + y'n = \/(xn + y'2) (x"2 + y"2) (by equation 3.1)
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yjxn (x,a + y'n) + yn (a;"2 + y"2) = yj(x'x")2 + (x'y")2 + (x"y')2 + (y'y")2
\J(x'x") {—y'y") + (x'y")2 + (x"y')2 + (y'y") (—x'x") (substitution from 3.2)
\J {x'y")2 — 2 (x'x" y'y") + (x"y')2 = y j (x'y" — x"y')2
I x'y" - x"y'\.
There axe other advantages to restricting our study to plane curves. In R3, the
normal vector field is only defined when « ^ 0. Furthermore, defining the tangent
vector field in R3 does not uniquely determine the other two vectors, N and B, to
form the orthonormal frame. In R2, however, we can give a definition of the normal
vector field which is everywhere defined, regardless of «, yet is largely consistent with
the definition in R3. In addition, this "new" definition for N will be uniquely fixed
once T is determined. Thus, B is uniquely determined as well.
Let us now define the tangent and normal vector fields in R2, and show their
relationship to the comparable concepts in R3. In addition, we will compare these
"new" definitions to the Frenet-Serret apparatus of the curve when viewed as a curve
in R3, which just happens to lie in the (x,y) plane. To distinguish the tangent and
normal vector fields in the plane from their counterparts in R3, we will use t(s) and
n (s). Note that since we are in the plane, we can associate B with the 2-axis.
Definition 3.1.2 (Tangent and Normal Vector Fields, and Curvature) Let
a(s) be a unit speed curve in R2. Then the tangent vector field to a is t(s) =
a' (s). The normal vector field to a is n (s), the unique unit vector field such that
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{F(s), n (s)} gives a right handed orthonormal basis of R2 for each s. The plane
curvature of a is k (s) = (s),n(s)).
Thus if a is a unit speed curve in the plane, then it is not hard to see that the
following relationships hold:
1. t* (s) = k (s) n (s); and
2. if a (s) = (x (s), y (s)) for real-valued functions x, y, then t (s) = (x' (s), y' (s))
and n (s) = (—y' (s), x' (s)).
In addition, we have the following relationships between the definitions in R2 and R3.
Lemma 3.1.3 Let a be a unit speed plane curve. Then
1. t(s) = T (s),
2. n (s) = ±N (s) at all points for which N (s) is defined,
3. K(S) = \k(s)\,
4• n(s) is differentiable, and
5. n' (s) = —k(s)t(s) Vs.
Proof. The first statement is trivially true, since by definition
t (s) = a' (s) = T (s) V (s) = f ' ( s ) .
To prove the third statement, recall the following relationships:
k « ( ? , a ) J i = (*', y') =S. ? = (x", y"); n = x').
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Thus
k = -x"y' + x'y" = x'y" - x"y'
=> K = |fc| by Lemma 3.1.1.
We can then use the third relationship to prove the second, provided K # 0.
N (s) =f ^ a — ^ (s) k (s) ™ (s) «(«) l*WI ~ I* M l - I*(»)| = ± s w -
In order to prove the remaining two statements, we differentiate both sides of the
equation, t-n = 0.
t • n + t-n' -0 <*=*• k + 1 • f? = 0 (by definition of k) <==> —k = t n
Now since t± n and we are in the plane, then t and n form a basis and any vector
can be written m terms of t and n. In particular, we can write n' = af+ bn, for
some real numbers a and b. However, n ± n', so by dotting both sides of the previous
equation by n, we get the following.
n' -n = (at) -n + (bn)-n <=> 0 = 6
Thus we have ff = at By using the same method, but dotting this time by t, we get
rf-f=(at)-f=a.
But we found above that I ff =-k, and sinee dot products are commutative, we have
that a = -k. Therefore, rT = -hi This proves statements 4 and 5, and completes
the proof.
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The sign of k indicates which way the curve a is bending. If a is curving in the
direction of n, then k is positive. If a is curving away from n, then k is negative.
Definition 3.1.4 (Closed Curve) A regular curve, j3 (t), is a closed curve if 3 a
positive constant p, such that Vi, 0(t) = /3 (t + p); i.e., /? is periodic. The period of
0 is the least such constant p. Thus, /? (0) = 0 (p).
The following lemma is easily established, and is stated here without proof. (A
complete proof may be found in [MiPa, page 54].)
Lemma 3.1.5 If P{t) is a closed (regular) curve with period p, and a (s) is 0 repara-
metrized by arc length, then a is also closed (and regular) with period L = J0P dt.
Observation 3.1.6 Thus for a unit speed curve, the period of the curve is also the
length.
Definition 3.1.7 (Simple Curve) A regular curve a (t) is simple if either of the
following is true:
1. a is a one-to-one function, or
2. a is a closed, periodic curve with period p, such that a (ti) = a (t2) if and only
if ti — t2 = np, for some integer n.
Geometrically, a non-periodic simple curve never intersects, or crosses, itself. For
a periodic simple curve, it does not intersect itself except at periodic intervals, where
it is essentially "retracing" itself. See the examples below.
Example 3.1.8 The figure-eight curve is not simple. (See Figure 3.1.)
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Figure 3.1: The figure-eight is not simple.
Figure 3.2: The circle and the ellipse are both simple curves.
Example 3.1.9 Circles and ellipses are closed, periodic simple curves. (See Fig-
ure 3.2.)
3.2 Rotation Index for Closed Curves
Let 5 be a closed, regular curve in the plane; without loss of generality we can assume
that a is unit speed, parametrized by arc length s. By studying how many times the
tangent vector, t, rotates as s goes from 0 to L, we can learn something about the
behavior of the curve itself. Since t(0) =t (L), we will show that this rotation must
be an integral multiple of 27r.
We can choose a right handed coordinate system in the plane such that the image
of a lies in the upper half plane, and place the starting point of a, a (0) at (0,0).
Thus, t(0) is horizontal. We will define 6 (s) to be the angle between the horizontal
axis and t, measured in the counter-clockwise direction. (See Figure 3.3.) Since a is
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continuous, then 6 (s) is also continuous for 0 < s < L, although 6 may be either less
than 0 or greater than 2ir. In fact 9 (L) may not be 0, though it must be an integral
multiple of 27r, (since a is a closed curve).
y*
a(0) = a (L)
Figure 3.3: Definition of 0
Definition 3.2.1 (Rotation Index) The rotation index of a closed, unit speed,
plane curve a (s) is given by the integer n$ = (0 (L) — 0 (0)) /27r = 9 (L) /27T.
Lemma 3.2.2 Let a (s) = (x (s), y (s)) be a closed, unit speed, plane curve. Then
0' (s) = k (s) for all s such that x' (s) ^ 0.
Proof. By the definition of 9, we have the relationship tan 9 = sin 9/ cos 9 =
y'(s)/x'(s), for all s such that x'(s) ^ 0. Thus, up to a multiple of 2ir, 9(s) =
arctan (y' (s) /x' (s)), Vx' (s) # 0.
Since a is unit speed, we have the relationship x12 + y12 = 1. Thus, if x' (s) = 0,
then y' (s) = ±1.
Then by Lemma 3.1.1,
1 d »(>) =
i + ( $ 0 2 *
( v' (s) _ 1 x'y" - x"y' \ z ' ( s ) / 1 + 4 X12
XVxa +
Xya
= x'y" ~ x"v' = k 00» provided x' (s) ^ 0. (3.3)
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Note that if x' (s) = 0 on a nonempty interval (c, d), then x (s) must be a constant.
Because a is unit speed, then y' (s) = ±1; without loss of generality, suppose y' (s) =
+1. Then a (s) = (x (s), y (s)) is a vertical straight line on (c, d), and thus k (s) = 0.
Moreover, 0 (s) is not changing over the interval, so 9' (s) = 0 over (c, d). Thus, we
have 9' (s) = k (5) over this interval.
If x' (s) = 0 at a single point (not over a nonempty interval), then we can take
the limits from the left and the right. Thus, since k and 9 are both continuous, we
can, without ambiguity define 6 to be the following:
e (s) = f k ( f j ) J So
dfj,.
Consider the following basic example.
Example 3.2.3 Let a (s) be the unit circle centered at (0,1), parametrized by arc
length in the counter-clockwise direction. (Then we can write a (s) = (cos s, 1 + sin s).)
So a (0) = (0,0) = a (L). Thus, 0 (0) = 0. Recall that although 9 (L) may not be 0,
it must be a multiple of 2n. Clearly, there are precisely two points where x' (s) = 0,
at a (sx) = (1,1) and a(s2) = (—1,1). (See Figure 3.4-)
-1 1
Figure 3.4: The rotation index for the unit circle.
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Note that there is a simple way to calculate the value of 6 (L) — 0(0) = 6 (L).
Since the radius is one, and the curvature of a circle, parametrized by arc length, is
l/r, then we know k = 1. Furthermore, we know that the length of a circle is its
circumference, 2ir. Thus we have 9 (L) = k = 2n. However, to demonstrate the
method which will be used to calculate 6 (L) in the proof of the next theorem, we can
compute this value as follows. By the Fundamental Theorem of Calculus and the rules
for improper integrals,
f*L psi ps2 pL
lim arctan 5-»S7
6(L) - 0 (O)=0(L)= f k (fi) d/i = r V ( s ) d s + / " V ( a ) d s + f & (s) ds JO J 0 J Si J S2
~~ *im a r c t a n arctan ( \x'(s)J s->o+ WOO/J W(«)/
- lim arctan ( y + [ lim arctan (^77-7^ - lim arctan ( y ^ ^ W \x'(s)J J I»L- W ( s ) / \x ' (s )J
as s -» y'{s) > 0 x' (s) > 0, =4- j j^ -> +00
as s -» 0+ y' (s) -» 0 x' (s) > 0, =*• ->• 0
as s —> s2, y' (s) < 0, x' (s) < 0, =>• —> +00
as s-^sf y' (s) > 0 x' (s) < 0, = * -» -00
as s L y' (s) ->• 0 x' (s) > 0, = > ^ ->• 0
as s -» 8% y' (s) < 0 x' (s) > 0, = • -> -00
7T ^ 7T 7T 7T = 2 — 2 2 2 ~
Thus, the rotation index for a circle, traversed counter-clockwise, is +1.
Theorem 3.2.4 Let a (s) be a closed, unit speed, plane curve of length L, parame-
trized by s, such that x', y' both have a finite number of zeroes. Let 6 be as defined
above. Then 6 (L) — 6 (0) = 2im, for some n € Z.
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Proof. Recall that at s = 0, 0(0) = 0 and a (0) = (0,0). Also, a is traversed
counter-clockwise, staying in the upper half plane. Thus, y > 0. Since y (0) = 0, y
has a minimum at s = 0. This implies that y' (0) = 0. Since a is unit speed,
xn + yn = 1; thus, we must have x' (0) = ±1. Since, by assumption, a is traversed
counter-clockwise, x' (0) = +1. In addition, a is periodic (since closed) with period
L, which implies that y' (L) = 0, x' (L) = 1. Hence, we have the following.
lim arctan ( | = 0 and lim arctan ( ^77-r | = 0. a—•0+ yx1 (S) J s-+L- \x' (s) J
As we unll see, the critical consideration in this proof is the number of times the
slope of the tangent line (i.e., y'/x') changes sign. For such a sign change to occur,
either y' or x' must become zero.
Observation 3.2.5 x' and y' cannot both be 0 at the same point, since this would
imply that xn (s) + y12 (s) = 0. However, as a is a unit speed curve, we must have
that x'2 + yn = 1 everywhere.
However, there could also be points where one of x' or y' is zero, but y'/x' does
not change sign. We need to consider the effect such a point might have. i4s we will
see, these points have no effect on the value of 9 (L) — 0 (0) at all. Before we proceed,
however, we would like to make the following definitions.
Since we supposed that x' (s) and y' (s) both have only a finite number of zeroes,
say m zeroes combined, then 3 0 = s0 < si < s2 < • • • < sm < sm+i = L, where
V 0 < i < m + 1, either x' (s») = 0 or y' (sj) = 0. Also, =*> Vs e (sh si+1), 0 <i <m,
x' (s) ^ 0 and y' (s) ^ 0; so
"> /*««+i _HL r*i+i »(L)-9( 0) = = ff(s)ds
i=0 J** i=0
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ds
= £
i=0
lim arctan ( K r r l i — lim arctan S—>8 t+1 3-̂ 57 \x'(s)J \xt (s) J
Having made these definitions, we will proceed with the proof that all uzeroes"
(either from x' ory1), across which the sign ofy'/x' does not change, can be eliminated
from consideration in the above calculation. We will give the proof for a point where
x' is zero; the proof for y' = 0 is similar. (See Figure 3.5.)
Figure 3.5: No change in the sign of y'fx' at the indicated points.
Claim 3.2.6 Suppose that Si, 1 < i < m, is a point such that x' (sj) = 0, and the
sign ofy'/x' is the same on the immediate left and on the immediate right of S{. That
is, the sign ofy'/x' is constant, non-zero, and the same on ( ) and (si,si+i).
Then any such point has no effect on the value of 9 (L) — 0 (0) = 6 (L).
Proof. Without loss of generality, suppose that the sign of y'/x' is positive on
(sj_i,sj) and (sj, Sj+i). Let us examine what happens at Sj.
r«t+i PSi ps
/ ff+J J Si-1 J Si
6' = lim arctan S-+S~
+ lim arctan
n 1. = — — lim arctan 2 ,
= — lim arctan
( < m _ l i f f i a r c t a „ ( s £ L )
\x'(s)J ,^+1 \x'(s)J
\x'(s)J s-ysf \x'(s)J , i m ^ M f > ' ) _ 1
\x'(s)J 3~>Si+1 \x'(s)J 2 + l i m a r c t a n + 0
\X'(s)J s^s-+1 \X'(s)J
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That is, the point Si adds nothing to the value of$(L). Hence, we can eliminate all
such points from consideration. (Note that this proof is not dependent on the number
of such points being finite. Thus, it is valid for an infinite number of these points.)
• We mil now assume that all such points have been removed from the set {s»}i<i<m,
and reordered into the set {sj}1<t<(1, where d <m, d G N. (We will still assume that
So = 0 and define s«i+i = L.) Thus, the only situations that merit consideration are
the ones where the sign of y'/x1 changes across some zero of x' or y'; i.e., at S{,
0 < i < d. Notice that the sign of y'/x' changes when precisely one (and only one) of
x' or y' changes sign across a zero.
Observation 3.2.7 Observe that the number of points where y' changes sign (and
x' does not) must be even. Recall that a remains in the upper half plane, and that
y (0) = 0. Furthermore, y' is of one sign on (0, si). Thus, we must have y' > 0 on
(0, Si). Otherwise, a would not remain in the upper half plane. Similarly, we must
have y' < 0 on (Sd,L). Hence, y' changes sign at s = 0 and there must be at least
one other point where y' changes sign. If not, then after leaving the origin, a would
never return; i.e., y' would stay greater than zero. This argument can be applied more
generally to an arbitrary (finite) number of zeroes ofy'. (See Figure 3.6.) Likewise, a
similar argument can be used to show that there are an even number of points where
x' changes sign (and y' does not). Therefore, the total number of points where a
sign change of y'/x' occurs must be even. Moreover, since SQ = 0 is identified with
Sd+i = L, (meaning both points represent (0,0)^, then there are precisely d+1 such
points => d + 1 is even.
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Figure 3.6: y' changes sign seven times, at the indicated points.
Let us examine what happens at these points S{, 0 < i < d. We will split these
into the following cases:
1. the sign ofy'/x' changes due to a change in sign of y' only,
2. the sign ofy'/x' changes from positive to negative due to a change in sign of x'
only, and
3. the sign ofy'/x' changes from negative to positive due to a change in sign of x'
only.
Let us consider the first case.
Claim 3.2.8 All of the points which fall into the first category have no effect on the
value of9(L).
Proof. Let Si be a point falling into the first category; i.e., y' (sj) = 0, x' (s{) ^ 0.
(See Figure 3.7.) Consider what happens at s^.
LhtmW))} ds+C{"*m{Wj)} * - '™arcun (?$) - arctan 0® + J & , a r c t a n G® - ̂arctan
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0 — lim arctan 5—YS;
= ^lirn arctan ( * g ) - <lim_arctan ( * g ) + 0
Thus, this point contributed a net of 0 to the value of 9 (L).
Figure 3.7: At the indicated points, y' = 0, x' ^ 0.
Thus, although points from the first case are included, in the set {si}0<j<d, their
combined net effect is 0. Now we mil examine points falling into the second group.
Claim 3.2.9 Each of the points which fall into the second category add ir to the
value of 6 (L).
Proof. Let Sj be a point falling into the second category. Again, this situation can
arise in a different way, but we prove only one case here. The other follows similarly.
Suppose that Si is a point such that as s -> s{ from the left, y' (s) > 0, x' (s) >
0 (y'/x') > 0 and as s s{ from the right, y' (s) > 0, x1 (s) < 0 =*• (y'/x ') < 0.
(See Figure 3.8.) Then at Si, we have
{ a r c t M ( ? 8 ) } * ( £ § ) - a r c t a n G ? § ) + i t . a r c t a n 8 S lim arctan
S—>S-
— lim arctan (tiSl\
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= — — lim arctan " * * • ( « ) -m-A—m* That is, the point Si adds tt to the value of Q{L).
7T 2
= lim arctan
Figure 3.8: y'/x' changes from positive to negative, due to a change in x'.
Finally, let us look at the last case. (See Figure 3.9.)
Claim 3.2.10 Each of the points in this situation add —7r to the value.
Proof. The proof of this case is almost identical to the previous claim. Though
these points arise from different situations, the equations are the same with the ex-
ception that the signs of n/2 are opposite in the fourth line. Thus, each such point
adds —7r to the value ofO(L). |
Figure 3.9: y'/x' changes from negative to positive, due to a change in x'.
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Thus, finally, putting all of these results together, we have
= E lim arctan ( ~ lim arctan ( ^ 4 4 ) V* (<0/ \x (s) J
Observe that since lim arctan ( ^-t41 = lim arctan ( *+l- \x'{s)J H«- V® /(«)/'
then we can change subscripts to get
- E i = 0
lim arctan f ^ 4 ) - lim arctan ( ^ 4 4 1 (s)J s_s+ \x'(s)J
Note that the number of bracketed terms which are 0 is equal to the number of places
where y' (only) changes sign. Since this number is even (by Observation 3.2.7), there
is an even number of bracketed terms which are 0 's. In addition, by Claims 3.2.9, 3.2.10
and Observation 3.2.7, the number of bracketed terms which are ±n is also even, say
21. Since there are an even number of summands (d+1), we can pair them up to
obtain
21 &(L) = Ui' where n{ = ir or n{ = -7r
t=i
= 2irn, for some integer n.
Observation 3.2.11 The previous theorem can be applied to a general curve j, where
7 may not necessarily have only a finite number of zeroes for either x' or y'. Though
we will not present a formal proof here, we will sketch it as follows.
Given a general curve 7, we can approximate ? by fk = (xt, yk), where both 4 and
l/t have only a finite number 0/zeroes. That is, we can find % such that lim^ fk = f
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and limfc_+oo jk — tS where the convergence of both limits is uniform. Care must be
taken to ensure that Xk and yk are appropriately differentiable.
Thus, by the previous theorem, 0k (L) = 2ivnk, for some nk € Z. Then as k —> oo,
0k (L) —• 0 (L). Thus, for large k, 0k (L) does not change. That is, 3k0 such that
for k > kQ, 0k (L) = 2nnk0. (Recall that nk is an integer.) Therefore, since 0(L) =
limjt^oo 0k (L) = 2irnk0, we obtain 0 (L) = 2-nn^.
The last theorem asserts that 0(L) — 0 (0) = 2itn, for some n € N. We are unable
to predict precisely what value n will take without knowing what the curve looks like,
or rather how many changes of sign of y'/x' it has, and of which type. However, as
we will see below, by restricting our study to simple closed curves, we can get a more
precise answer for 0 (L) —0(0).
3.3 Rotation Index for Simple Closed Curves
Recall that a simple plane curve is one which does not intersect itself. For a closed,
simple, plane curve, a, with period (or length) L, this means that Vsi, s2 € (0, L),
a (si) = a (s2) O Si = s2. By requiring a curve to be simple, we obtain a more exact
answer for the rotation index.
Theorem 3.3.1 The rotation index of a simple closed curve a (s), parametrized by
arc length, is ±1.
Proof. Let L be the length (or period) of a. We will assume that a is in the
upper half plane, a (0) = (0,0), and that a is traversed counter-clockmse. Let u, v be
points along a such that 0 < u < v < L. Now define a vector valued function p(u, v)
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to be the unit vector from a (u) in the direction of a (v) (see Figure 3.10); i.e., let
a (v) — a (u) p{u,v) =
|<5(v) — a(w)| , we let p (u, u) = t (u) and p (0, L) = —t (0) = —t (L)
Figure 3.10: p(u,v)
Then p is a C2 function in the region of A. (See Figure 3.11.) We can define a
C2 function on A, say a, where a (it, v) measures the angle between the x-axis and
p(u,v). Note that since p(u,u) = t(u), and by definition 6 (u) is the angle between
the x-axis and t (u), =£• a (u,u) = 6 (u).
(0,0) L
Figure 3.11: A
It can be shown that a (it, v) is Cl on the closure of A, and C2 on the interior.
(The Jordan Curve Theorem guarantees that a has an interior, A, and that dA = a.)
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While this is clear on the interior of A, care must be exercised along the diagonal
boundary.
Recall that Green's Theorem states that
Lpdu+Qdv=IL&~^)dudv' where dA represents the boundary itself, and A represents the region enclosed by the
boundary. Now if we let
P = Q = — du v dv'
dP dQ &o
then
dv dvdu' du dudv'
Since a (u, v) is C2 on A, we have
d2o _ d V
dvdu dudv
Hence, Green's Theorem becomes
f do da f f ( d2a dPo \ L t o d u + f o d v = J j A d ^ ~ a ^ ) d u i v = 0- ( 3 4 )
"V*" it
0
Now, in general, for some curve c(s) = (u (s), v (s)), where a < s <b,
J ^ P d u + Qdv = f J ^P(u(s) ,v(s))^ + Q(u(s) ,v(s))^
= [ lp(u (s) > V (s)) u' (s) + Q (u (s), v (s)) v' (s)] ds. J a
ds
Thus we can split up the boundary of A into pieces. We will let 71 be the diagonal
edge, 72 be the horizontal edge, and 73 be the vertical edge. We will examine each
edge separately.
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On 71
40
Let u = s =£• du = ds
dv = ds v = s > for 0 < s < L.
Thus
/ J 71
On 72
do do , fL
— du + -z-dv = du dv J0
Let u = s
v = L
da (i A f a<7 (l i\ ? a n * ' T 1 I s , s ' ' 1 ds. (3.5)
du — ds
dv = 0 • for 0 < s < L.
Thus,
J J 79
do do , -^-du+ — dv du dv * — j f ! < • • * > * •
(3.6)
On 73
Let u = 0
v = s
du = 0
dv = ds for 0 < s < L.
Thus,
/ I Qg, g^(0,3)ds. (3.7)
Combining equations 3.5, 3.6, and 3.7 with equation 3.4, we get the following:
0 ~ L^du+a^dv = / , fadu+todv+Jv %iu+%iv+f^ %iu+%dv-
However, this equation can be true if and only if the following is true.
I [ a ^ ( s ' s ) + a ? ( s - s > d s = I ^ 3 ' ^ d s + l § ; ( « , > ) d > (3.8)
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Working on the left side of equation 3.8, recall that a (it, u) = 6 (u) so =» a (s, s) =
0(s). Hence,
da . . da ,
So, by the Fundamental Theorem of Calculus, the left hand side of equation 3.8 be-
comes
f 9'(s)ds = d(L)-6(O) = 0(L). Jo
On the right hand side of equation 3.8, recoil that
a (u, u) = arctan ( + 2irn, \x(v)-x(u)J
for some neZ. Since we know a (s) = (x (s), y (s)) and 5(0) = 5 (L) = 0, we have
that x (0) = 0,x (L) = 0,y (0) = 0, and y (L) = 0. Thus we get
—(uv) =x> ^ ^ ^ ~ y ~ y' tx ~ x
^ ' [x(v)-x(u)¥+ [y(v)-y(u)]2 '
Therefore
~{sL) - X' ^ ~ V ~ y' ^ ~ x (s)]
^ x (a)]!2 + [ ^ ° ( L ) - 2 / ( s ) ] : 2
_ x (s) y' (s) - x' 00 y (s) x2 (s) + y2 (s) '
and ^-(u,v) = y'(v)lx(v)-xM1-x'(v)ly(v)-y(u)l V \x (v) ~ X (u)]2 + [y (v) — y (w)]2
So we find that
da <.\-x(s)y'{s)-x'{s)y(s) da. dvK'> x2{s) + y2{s)
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Thus, the right hand side of equation 3.8 becomes
fL x (s) y' (s) - x' (s)
Jo x2(s) + y2(s)
Note that
x (s) y1 (s) - x'(s) y (s) X2 (s) + y2 (s)
except where x (s) = 0, i.e. on the y-axis.
It is now necessary to establish what happens on the y-axis. Again, we will restrict
ourselves to the case where x(s) = 0 at a finite number of points, although it can be
shown for the infinite case as well.
Lemma 3.3.2 Suppose a simple, closed curve, a, touches the y-axis a finite number
of times. Then
I L x (s) y' (s) - x' (s) y (s) =
Io x2 (s) + y2 (s)
Proof. Suppose a = (x (s),y (s)) and 3 {si}0<i<n+1 such that V 0 < i < n + 1,
x (si) = 0, and for all other s € [0, L] \ {*<}„<<<„+!, x(s)^0.
Also suppose that Q = SQ < s\ < S2 < • • • < sn < sn+1 = L. By assumption of the
placement of a on the graph, we know that x (0) = 0 and y (0) = 0.
Since Vs E (0, L), y(s)> 0 and y (0) = 0, then y must have a (local) minimum
at s = 0. Thus y' (0) = 0. However, a is a unit speed curve, so Vs G (0, L),
xa (s) +yn (s) = 1. Hence => xa (0) = 1 x' (0) = ±1. (For a traversed in the
counter-clockwise direction, x' (0) = +1; otherwise, it is —1.)
Then, because a r c t a n (y (s) /x (s)) at s = 0 (or at s = L) is of the form a r c t a n (0/0),
we can apply L'Hopital's rule:
l i m s _ 0 + a r c t a n ( f g ) = l i m s _ 0 + a r c t a n = a r c t a n ( f ) = a r c t a n (0) = 0
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and
Then I lims_^- arctan (§ff) = 0.
L x (s) y' (s) - x' (s) y (s)
43
(3.9)
ds=tC ds-x2 (s) + y2 (s)
We will assume that a is being traversed in a counter-clockwise manner. Let us
examine what can occur at Si, for some 1 < i < n.
I. The curve a is passing from quadrant I into quadrant II. Thus, traveling from
the left, y (s) > 0, x (s) > 0, and traveling from the right, y (s) > 0, x (s) < 0.
(See Figure 3.12.)
So, lim arctan 5—>S~
(y{s)\ = i U ( ' ) / 2
and lim arctan S-¥$t
(V{*)\ _ JH \x{s)) 2"
(3.10)
II. The curve a is passing from quadrant II into quadrant I. Thus, traveling from
the left, y (s) > 0, x (s) < 0, and traveling from the right, y (s) > 0, x (s) > 0.
(See Figure 3.12.)
So, lim arctan (^7-r^ = — ̂ \x(s)J 2
and lim arctan (V(*)\ = £ 2'
(3.11)
Figure 3.12: The figure on the left is case I; on the right is case II.
III. The curve a remains in the closure of quadrant I, but becomes tangent to the
y-axis; i.e., a simply "bounces" off the y-axis. Thus, whether traveling from the
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left or the right, y (s) > 0, x (s) > 0. (See Figure 3. IS.)
So, lim arctan ( = £ = lim arctan ( . \x(s)J 2 \x(s)J
IV. Similarly to the previous case, the curve a remains in the closure of quadrant
II, becoming tangent to the y-axis; again, simply "bouncing" off the y-axis.
Thus, whether traveling from the left or the right, y (s) > 0 , x (s) < 0. (See
Figure 3.13.)
So, lim arctan ( = - | = lim arctan ( ^ 4 1 • s-yst \X(S)J 2 s_).s+ \x(s)J
Figure 3.13: The figure on the left is case III; on the right is case IV.
Claim 3.3.3 All points in cases III and IV can be ignored.
Proof. Without loss of generality, we will prove the result for points in case III;
the proof for case IV follows similarly. Let Si be a case III point, and observe what
happens at this point. Then
L la rc tan ( f $ ) } ' + f ' {arctan G S ) } ' G&! ) _ J;,™arctan ( ^ } ) + _areta" ( f | | )
[ y ( s ) \ \x(s)J
= lim arctan s-t-sT
lim arctan
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= %• — lim arctan 8-*8, *-1
= lim arctan 5-4«i+1
( ) + lim arCtan { ~ f \x(s)J s->,r+1 \x(s)J 2
l i m arctan f ^ + 0 . \x(s)J s->sU \x(s)J
That is, the point Si had no effect on the sum. Therefore, any such point can be
disregarded. I
We will now assume, without loss of generality, that all points falling into cases III
or IV have been removed from the points { s i } 1 < i < n . Since these points have been
removed from consideration, this clearly implies that the remaining points from cases I
and II alternate in type. That is, following a point of type case I, we have a point of
type case II. Note that by the assumptions on the placement of a and the fact that it
is a closed curve, a passes from quadrant I into quadrant II once more than from II
into I. (Otherwise, the curve will not be closed.) Thus n must be an odd number, with
case I occurring ((n — 1) /2) + 1 times, and case II occurring ((n — 1) /2) times.
Furthermore, by the placement of a, we know that both Si and sn must be of type
case I. Thus, we have
- E 2 = 0
n
-E z=0 n
-E i= 1
lim arctan ( \ — lim arctan ( *-K"+i \x(s)J s->sf \x{s)J
lim arctan ( — lim arctan ( ^ 7 - ^ «-wr \x{s)J s-*sf \x(s)J
lim arctan ( - lim arctan ( ^ 7 ^ \x{s)J s-nt \x{s)J
(since s0 = sn+i)
(by equations 3.9).
From equation 3.10, we see that a case I point adds a net of tx to the sum. Likewise,
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from equation 3.11, a case II point adds a net of —it to the sum. Thus,
^ J , . £ [ l,m arctan - lim arcta„ (
J0 x2(s) + y2(s) tf -»T \X(8)J -+4 \X(S)J i=l n
= ^ (—1),+17r = 7r ^since n is odd). i=1
(7/ <3 was traversed in the opposite direction, we would have one more case II type
point than case I type points, and the result would be —it.) I
Finally, to complete the proof of Theorem 3.3.1, for a simple curve a, we have
Thus, by definition, the rotation index of a simple, closed plane curve is ±1. I
One can do a similar approximation type argument to prove the previous theorem
in the case where a touches the y-axis an infinite number of times.
There is one other result which will be needed later.
Corollary 3.3.4 Suppose that a is a simple, closed, regular plane curve. Then the
tangent circular image t: [0, L] —> S1 is onto.
Proof. From the previous theorem, we know 6 (0) = 0 and 6 (L) = ±2ir. Suppose,
without loss of generality, that 6 (L) = 2n. Then given any 0 < 0O < 2ir, 3s0 such
that 6 (s0) = 60 by the Intermediate Value Theorem. I
This concludes our discussion of the basic properties of plane curves. In the next
chapter, we will develop and prove two specific results regarding plane curves; the
necessary and sufficient condition for a curve to be convex and the isoperimetric
inequality.
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CHAPTER 4
CONVEX CURVES AND THE ISOPERIMETRIC INEQUALITY
In this chapter, we will consider a necessary and sufficient condition for a curve to be
convex. Then we examine the Isoperimetric Inequality, which deals with the shape
of a geometric figure of fixed perimeter, and the area it encloses.
4.1 Convex Curves
Given any straight line /, we can divide R2 into two half planes, say Hi and H2, with
I as the dividing line. That is, Hi n H2 = I and Hi U #2 = R2- Then, if a curve lies
completely inside one of the half planes, we say that the curve lies on one side of I.
Definition 4.1.1 A regular curve a is convex if it lies on one side of each tangent
line. (See Figure 4.1.)
Figure 4.1: A convex curve is on one side of any tangent line.
The next theorem, which develops a necessary and sufficient condition for convex-
ity, requires the concept of monotonicity.
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Definition 4.1.2 A function / (t) is called monotone increasing if t\ <t2 f (t\) <
f (h)- f (t) is called monotone decreasing if t\ < t% =$• f (<i) > / (<2). We say that /
is monotone if it is either increasing or decreasing.
By the rules for the first derivative from Calculus, we know that a function / is
monotone if and only if f has constant sign. If / ' > 0 then / is monotone increasing,
while f < 0 implies that / is monotone decreasing.
Theorem 4.1.3 A simple, closed, regular plane curve a (s) is convex if and only if
the sign of k (s) does not change. That is, Vs, k(s) > 0 or k (s) < 0.
Proof. Since k (s) = d9/ds, this is equivalent to saying that 9 is monotone if and
only if a is convex.
Figure 4.2: The tangent lines at A, B, and C.
First, we prove the forward direction. Let 9 be monotone. Now, suppose to the
contrary that a is not convex. Then, by definition, 3 a point A such that a is on
both sides of the tangent line at A. Let the tangent line at A, t (^4), be I. Since a is
closed, 3 points B, C of a on opposite sides of I, which are farthest from I. Let the
tangent lines at B, C be known as li, I2 respectively. (See Figure 4-2.) Clearly the
tangent lines at A, B, and C must be distinct.
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Claim 4-1-4 h IIIII h-
Proof. Suppose, to the contrary, that this is not the case. Without loss of gen-
erality suppose that l\ ft I. (See Figure 4.3.) Now we construct a line through point
B, parallel to I, call it l[. By the construction ofl[, it is clear that it is not a tangent
line to a at any point. Thus, since it passes through B, there must be points of a on
both sides of l[.
Figure 4.3: Suppose I ^l \ .
Let C* be a point on a on the opposite side of l[ as A. Then the perpendicular
distance from line I to C* is equal to the distance from point A to point B, plus the
perpendicular distance from line l[ to C*, or
dJCl = AB + dJ&.
But then the point C» on a is farther away from line I than point B, which contradicts
the definition of B. Hence, => li || I. Similarly, it can be shown that I2 || I. I
Thus, by the claim, we have three parallel, distinct lines, lx, l2, and I. Since all
three lines are by definition tangent lines to a at particular points, they have a specific
direction associated with them. Then at least two of the three tangent lines must point
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in the same direction. So 3si < s2 such that t (si) = t*(s2) (i-e. their magnitudes and
directions are the same). Thus 0 (s2) = 0 (si) + 2irn, for some n € Z.
By Theorem 3.3.1, we know that n = ±1 or 0. Geometrically, if n = ±1, then a
has "rotated" between Si and s2. If n = 0, then either a has not rotated, or it has
"rotated back," so that the net change is 0. Upon further consideration, it should be
clear that the second possibility cannot occur due to the monotonicity of 0. Let us
take a look at the three possibilities separately.
If n = 0, then a has not rotated so => 6 (si) = 6{s2). By the monotonicity of
6, =$• Vs 6 (si,«2)j since Si < s < s2; then 9(si) < 9 (s) < 0(s2). Then by the
Squeeze Theorem and the equation above, 0(s) = 6 (si) = 0 (S2). In other words,
9 must be constant on the closed interval [si, s2].
Now consider the cases when n = ±1. Without loss of generality, suppose that
n = 4-1, so that 0 is monotone increasing. That is, 0 (52) = 0 (si) + 27r. Since a is
a simple, closed curve, then by Theorem 3.3.1, we know that 0 (L) — 0(0) = 2n. But
then
2ir = 0 (L) - 0 (0) = e(81)-0(O) + 0(82)-0(8i) + 0(L)-O{s2)
= 0(s1)-0(O) + 2ir + 0(L)-0(s2).
Now since 0 is monotone increasing, 0 < Si, and s2 < L, then
=> 0 (0) < 0 (si) and 0{s2)<0{L) =>• 0(si) — 0(O)>O and 0(L) — 0(s2)>O
==$• 0 (si) = 0 (0) and 0(L) = 0(s2).
Similarly, since 0 is monotone, then it is constant on [0, s j and [s2, L]. Since a is
closed, t(L) = t (0). Therefore, the tangent lines at a (si) and a (s2) not only have
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the same direction and magnitude, they actually coincide. (See Figure 4-4-) But
by Claim 4-l-4> these lines are distinct. This contradiction implies that a must be
convex.
Figure 4.4: The tangent lines at a (si) and a (S2) coincide.
Now we prove the reverse implication. That is, we suppose that a is convex, and
we show that this implies 9 is monotone.
Suppose for some 0 < si < s2 < L, 0(si) = 0(s2). Thus, t(s 1) = t(s2),
(magnitude and direction). By Corollary 3.3.4 from the previous chapter, we know
that t: [0, L\ —¥ S1 is onto. So 3s3 such that t (S3) = —t (si). Now, if the tangent lines
at si, S2, and S3 were all distinct, they would be parallel and one would be between the
other two. This implies there must exist a tangent line with points on a occurring on
both sides of the tangent line. However, the convexity of a rules out this possibility.
Hence, at least two of the tangent lines at {si, s2, S3} must be the same. We suppose
the tangent lines at S\ and s2 coincide. Now we need to establish the following lemma.
Lemma 4-1-5 <3 is a straight line from Si to s2.
Proof. Let a (si) = A, a(s2) = B. Suppose, to the contrary, that 3 some
point C, on line segment AB, such that C is not on the curve a. Let I' be the line
perpendicular to line segment AB, which passes through the point C. Now since a
is closed and convex, then V must intersect a at a minimum of two points. (See
Figure 4-5.) Thus I' cannot be a tangent line for any point s on a.
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Figure 4.5: Suppose C is not on the curve a.
This implies that I' has to cross a at a minimum of two points, say D, E, on the
same side of line segment AB. (Otherwise, the tangent line at A (equivalently, the
tangent line at B) crosses a, contradicting the convexity of a.) Let D be the point
closest to C. Then the tangent line at point D, t (D), has at least one of the points
A, B, E on opposite sides. (See Figure 4-5-) This contradicts the convexity of a.
Therefore, no such point C can exist, so a is a straight line from Si to s2- 1
By Lemma 4-1-5 then, 9 is constant on [si, §2]- To complete the proof, we suppose,
to the contrary, that 9 is not monotone. Then 3 some point s0, such that si < s0 <
S2 with 9 (s 1) = 6 (^2) (by the continuity of 9), but 9 (^2) 7^ 0 (so) • However, this
contradicts the previous lemma, and thus cannot occur.
Hence, we must have that 9 is monotone. Therefore, a simple, closed, regular
plane curve is convex if and only if the sign of k (s) does not change. •
At this point, we will briefly leave convex curves, and consider the area bounded by
simple, closed, regular plane curves, whether convex or not. As we will see, however,
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convexity does have an effect on the maximum area which can be contained by a
curve of fixed length.
4.2 The Isoperimetric Inequality
Given a fixed length (or perimeter) for a simple, plane curve, what shape will have
the greatest area? Even today, calculus and algebra students often consider a similar
question, namely how to find the rectangle, or triangle, with a fixed perimeter that
bounds the greatest possible area. We will now show that for simple, closed, regular
curves, the circle is the geometric shape that bounds the greatest area.
Theorem 4.2.1 (Isoperimetric Inequality) Let a be a simple, closed, regular plane
curve of length (perimeter) L. Let A be the area of the region bounded by a. Then
L2 > 47rA, with equality if and only if a is a circle. Thus, of all curves of fixed length,
the circle bounds the greatest area.
Proof. Let I2 be two parallel lines, tangent to a, with a bounded between them.
Let be a circle, radius r, between li, l2, and tangent to both lines, such that (5 does
not intersect a.
Now, choose the x, y coordinates for the plane with the origin at the center of 0,
and with the y-axis parallel to li, l2. Let the point where a is tangent to l\ be <3(0),
and the point where a is tangent to l2 be a(s2)- Since both the length and the area
enclosed by a are independent of parameterization, we may assume that a is a unit
speed curve (parametrized by arc length s). Thus, for a (s) = (x (s),y(s)), we have
x12 (s) + y12 (s) = 1 for all s. (See Figure 4-6.)
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2
Figure 4.6: The circle and the curve a.
—T
The circle (3 can be parametrized by:
13 (s) = (z (s), w (s)) where z(s) = x (s) and w (s) = < -y/r2 — x2 0 < s < S2
yjr2 — X2 S2 < s < L.
Observe that although j3 is C2, it may not be a regular curve, with this particular
parametrization.
The following lemma is proved by an application of Green's Theorem.
Lemma 4.2.2 Let a be a simple, closed (regular) plane curve, traversed counter-
clockwise. Then the area of the region it bounds, D, is given by
Area (D) = / xdy — — I y dx, J a J a
where x and y are the usual plane coordinates.
Proof. Recall the statement of Green's Theorem:
Lpdx+Qdy=L{^-^)dxdy'
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where D is the region bounded by a, and dD is the boundary of this region. Now we
make the following substitutions into Green's Theorem:
P . 0 - ^ = 0 ay
and Q = x =» = 1. ox
Thus, we get
Similarly, take
Then
I xdy = I ldxdy = Area (D) JdD JD
and <3 = 0 ==^ ^ = 0. ox
I ydx = I (—\)dxdy = —Area(D). JdD JD
Using Lemma 4-2.2 and the substitution y' = dy/ds, we get the following equation.
A = I xdy = I xy'ds J s Jo
Also, the area of (3, a circle, is clearly
7it2 = — f ydx — — f wz' ds = — f wx' ds. Jp Jo Jo
Thus, by adding the previous two equations together, we are led to the following.
A + irr2 = J (xy' — wx')ds< J \xy' — wx'\ds (4.1)
= / ds. Jo
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By the Cauchy-Schwarz Inequality,
< |(<yOII(-w>^)l = \j
= 1 • Vw2 + x2 = Vr2 = r (since a is unit speed).
Hence, putting it together, we get
A + 7rr2 < f rds = rL. (4.2) Jo
We now need to establish the following claim.
Claim 4-2.3 Suppose that a and b are both positive numbers. Then y/ab < | (a + b),
with equality if and only ifa = b.
Proof.
< - (a + b) <=$• 2y/ab < (a + b) <=$• ^2y/ab^j < (a + b)2
<=$> 4ab < a2 + 2ab + b2 <=> 0 < a2 — 2ab + b2 <*=>• 0 < (a — b)2
The last inequality is certainly true for any real numbers a and b, which implies the
original inequality is true. Furthermore,
Vab = - (a + ft) <=>• (a — b)2 = 0 4=> a — b.
Since both A and irr2 are areas, hence positive, we can apply the claim to get the
following:
VAnr2 < - (A + 7rr2) with equality <*=>• A = nr2.
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Combining the previous equation with equation 4-2 gives
VAnr2 < i (A + nr2) < \rL. (4.3)
Z A
Thus
r2L2
Anr2 < —— <=> An A < L2. (4.4) This establishes the Isoperimetric Inequality.
Now, we ivish to show that equality occurs in equation 4-4 if and only if a is a circle.
If a is a circle, the result is easily seen. Length for a circle is just the circumference,
2n r, and the area is n r2. So L2 = An A becomes (2nr)2 = An2r2 = An (nr2) = Air A.
To establish the forward direction, suppose L2 = An A. We want to show that a
must be a circle.
Note that equation 4-3 still holds. Then, by substituting nA = L2/A, it becomes
Y ^ \ ( A + nr2) < y <^=4- VAnr2 = i (A + nr2) = irL.
Then, by the Claim, A = nr2. Moreover, we have
A + nr2 = rL. (4.5)
Since this equation was derived by the Cauchy-Schwarz Inequality, then we must ac-
tually have equality
I (-<",*)> I = Ka/.iOl !(-«>,*)|.
The Cauchy-Schwarz Inequality then implies that they are linearly dependent; i.e.,
3c e R such that (-w, x) = c ( x y ' ) . (4.6)
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58
This implies that the inequality in equation 4-1 must actually be an equality. That is,
rL — A + irr2 = f (xy' — vox') ds by='6 f (ex72 + q/2) ds = cL c — r. Jo Jo
Then, in combination with equation 4-6, =$• x = ry'.
Because of the relationship, A = 7rr2, r must depend on A, and not on the choice
of l\. Now suppose we choose lines 1$, Z4 to be perpendicular to l\ and I2, with a
between I3 and I4, and tangent to each.
Then there exists a circle, say O', between I3 and I4, tangent to both, and having
radius r (since r is independent of the lines chosen). Let O be the similar circle
between li and I2, tangent to both. Now establish new coordinates, x and y, with the
origin at the center of O', where the y-axis of O is parallel to, and in the direction
of the x-axis, and the x-axis of O is parallel to, and in the direction of the -y-axis.
(See Figure 4-7.)
< O'
h
Figure 4.7: The circles O and O' and the curve a.
By performing operations similar to the ones done for O, we get the equation
x = ry' for O'. In addition, by virtue of the relationships between the two coordinate
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59
systems, 3 constants d and e such that x = y — d and y = e — x. Combining these
last three equations, we get y — d — x — ry' — —rx' (since y' — 0 — x'). Thus, by
substitution, we get
x2 + (y ~ d)2 = {ry')2 + {—rx')2 = r2yn + r2xn
= r2 {yn + xn) = r2,
ti
i
which is clearly the equation of a circle in the plane.
Therefore, a is a circle of radius r centered at (0, d) in the x, y coordinate system.
I
In the next chapter, we will apply both of the results developed in this chapter
to curves which are evolving over time. That is, we will study curves which are
undergoing a particular type of deformation (as time passes), and its effects on convex
curves and the area which they bound.
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CHAPTER 5
THE HEAT EQUATION
In the previous chapters, we have studied curves in R3, and curves in the plane,
with particular emphasis on convex curves. We also developed the Isoperimetric
Inequality, and found that the maximum area bounded by simple curves with fixed
length is achieved by circles, which are convex curves.
In this chapter, we will examine simple, closed, regular plane curves again. This
time, however, the actual shape, or rather the geometric image of the curve, will be
changing as a function of some variable. We will be associating this variable with
time, and saying that the curve is evolving over time. Instead of using the lowercase
symbols for the tangent and normal vectors in the plane, we will use the uppercase T
and N usually reserved for R3. This abuse of notation is adopted to avoid confusion
with other variables, most notably time, t.
Moreover, we will deform the curve in a particular fashion. Though the method
is somewhat arbitrarily chosen, it is an intuitively pleasing and simple idea (none
of which necessarily implies that it is a useful idea!) However, the equations used
to describe the deformation are analogous to the heat equation, which has been
extensively studied and about which much is known. For this reason, this method
of deformation is worthy of study. The material for this chapter was almost wholly
based on the journal articles by M. Gage and R. S. Hamilton, [Ga, Ga2, GaHa].
fin
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5.1 Curves in the Plane
Let 7 (u) = (xo (u), t/0 (u)) be a regular, closed, plane curve, (not necessarily convex),
where 0 < u < 2n. Recall that since 7 is closed, => 7(0) = 7 (27r). Furthermore,
along a closed curve, arclength s is unique only up to a constant. However, the
derivative of the curve with respect to arclength, dj/ds, is uniquely defined.
Although 7 is a function of it, over time we will "deform" the curve in the direction
of kN. Geometrically, for a curve traversed counter-clockwise, when the initial curve
is "dented in," the curvature is negative so 7 will be "pushed out" in the direction
of —N. When the initial curve is already "pushed out," the curvature is positive so
7 will be "pushed in" in the +N direction. (See Figure 5.1.) Thus, at time t, the
"new" curve 7 (u, t) is given by 7 (u, t) = (x (tt, t) ,y(u,t)). Then at time t = 0, the
"original" curve 7 can be written as 7 (it, 0) = (x (u, 0), y (u, 0)) = (rr0 (u), y0 (u)).
The precise formulation of the deformation is given by:
#7 ~dt
{u, t) = kN = k (tt, t) N (it, t). (5.1)
Figure 5.1: The curve is deformed in the direction of the arrows.
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62
Since 7 is a regular curve and a function of u , then the arclength s is also a
function of u . Thus by equation 2.1, for the initial curve we have
r u | d r f ( u ) = s ( u ) = f
Jo du d u .
As 7 (ti, t ) is deformed over time, however, we get
( M ) = / | s ; ( M ) d u . (5.2)
Then
^7 d u (u,t)
' d x . ' a ^ ( M ) +
Hence, by substitution back into equation 5.2
+ B y d v
( v , t ) d v .
So,
ds_ d u
d j d u
(u,() (5.3)
Thus, the tangent vector field to 7 at a fixed time, t , is given by
di f ( u ) ^ ^ ==» T ( u ) =
d d s d s u \
(Although technically, we should write T (u,t), since we are fixing the time t , it is a
common and acceptable abuse of notation to write T ( u ) . ) This implies
d T ( u ) du
f ' (u) = k N d'y du
k ( u ) N ( u ) dj (it)
du
We would prefer to have 7 as a function of arclength, in order to make it unit
speed, instead of as a function of the parameter u . To accomplish this goal, however,
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63
TO must first note that « depends on t, (as well as on «), whereas u does not depend
on t. Then, for 7 parametrized by arclength (y(s,t)), we have
^ 7 unit_8peed Qg (S) dy T(s,t) = -gj g ( . , « ) | i (5.4)
In the following, equation 5.1 gives the first equality, Theorem 2.2.1 gives the second,
and equation 5.4 gives the third.
d_ ds
' a ? , J > ds 'J ds'1 ''
Hence,
2 ? = <3 ^ & = m d dy^Py ds2 dt dt ds2 dt ds2
Thus, we see that x (s, t) and y (s, t) both satisfy the heat equation. There are some
advantages to using the heat equation. It has been studied for some time and there
are many results which are known to be true for it. In particular, we will later be
using a form of the Maximum Principle to prove results about the evolution of simple,
convex curves.
By the chain rule and from equation 5.3 above,
31_a=fds &y g du-g;di = a;v' w ^ " = „(«,t) = _£ =
91{
In addition,
d_ ds
1 d
\ A f i ! ( M ) ) 2 + (g(u , t ) ) 2 9 0
So the arclength parameter is ds = vdu. That is, since
(5.5)
<(«,*)=£ >(.,() dw and ~=v(u,t) ds = v du.
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64
So by equation 2.3 in Chapter 2, we have
dT dN g ^ ^ v k N and — = -vkf. (5.6)
These equations are the FVenet-Serret apparatus applied to this situation.
In order to consider the area bounded by one of these evolving curves, TO will first
need to develop quite a few smaller lemmas.
Lemma 5.1.1 dv/dt = -k2v
Proof.
v2 = #7 2 / dj dy Xdu' du. du
So = 2/—?! ?l\ -0/ d &Y dl & \9tdu,du/~ \dudi^/ (u^tare independent)
d±ft+kdJL
du du
dN 2{k~fa,vf) (sinceTlN)
2kv
—2k2v2
(^du'^ ~ (~vkT,T^ (by equation 5.6)
That is,
Therefore,
2?>— c h a i ^ r u l e by above „ l 9 , dt ~ = ~2kv-
— - -k2„ dt~ kv-
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65
Lemma 5.1.2
dL dt
= — I k2 ds, 10/iere L is the length of the curve 7. Jo
dL dt
Proof. By Lemma 3.1.5, we know that
r2v I dy r2* L (t) = J I — (u, t) du = J v(u,t)du, (definition of v)
/2ir Qy
— du (since u, t independent) p2 7T
= I {—k2v)du (by Lemma 5.1.1) Jo
= I {—k2) ds — — I (k2) ds (change of var., by def. of v) Jo Jo
Lemma 5.1.3
Proof.
d_d_ dtds
d_d_ _ d_d_ dtds dsdt
- o d
(substitution from equation 5.5)
d_ 1 d2
du^ v dtdu 1 dv v2 dt
s(:s) [!(;)]
[ - i ( -*• • ) ]
tf_d_ 1 d_d_ v du vdtdu
kd~s + Wsdi (b« Wt'on 5.5)
8_ 1 d_d_ du^ vdtdu
1 d d 9u + v8tdu (by Lemma 5 J J )
l2 (I d \ 1 d d , = k [vd^)+ vdudt (U> * mdePendent)
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66
Lemma 5.1.4
d T _ d k f i 8N dk-~oT ~Z~N, ana —— = T dt ds dt ds
Proof. Recall that T = dj/ds. So,
dT _ d_ (&f\ _ ^7 2^7 , at + aJ i e m m a 5 i
- I ( £ J ) + 1 H + * * - £ * + * ! ? + « • / -A ^
= ds \ ) + ^ ^ (by Theorem 2.2.1)
dt
= ^-N -k2T + k2T=—N. us d s
Since N is a unit vector, => ^N,N^ = 1. 5o
at ( ( A r " i V } ) = § i ( 1 ) = 0 "*=*•2 ( w ' " ) = 0 ^ ^ = o-
As a vector, d f f / d t can be written as a linear combination o f f } and T. (Recall that
in the plane, nothing is in the binormal direction, B). That is,
dN — = aN + bT.
We know that N A. T. Thus, by substitution,
/dN -A / ^ ^ v \ d f , N / = °*==* \ a N + bT,N^ = 0 < = > (aN,N^ = 0 <=> a = 0.
So,
dN , dt ~ T• (5.7)
Then, since N _L T,
< * f > - 0 - > ! < * * ) _ ! ( 0 ) - 0
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67
(bT,T^ + =0 (by equation 5.7) =$• 6+^ = 0
dA;
Resubstituting, we finally get
dN_ = _dk-. dt ds
Now, define 0 to be the angle between the tangent vector and the x-axis.
Lemma 5.1.5
86 dk , 69 "ST = a - a n " -7T — k at as ds
Proof. Since T is a unit vector, then by the definition of 9, we can write T =
(cos 9, sin 9). So,
^ = §t ^C°S 6 ^ ' S i n 6 W = ( " S i n 6 W c o s 6 W
= ^ ( -sm0(<) ,cos^(<)) . (5.8)
Since N ±T, (ti, f } = 0. Letting N = (a, b), we get
((a>&) > (cos9,sin0)) = 0 =*> acos0 + 6sin0 = 0
=£• a cos 9 = —fesin 9 =$• a = sin 9 and b = — cos 9
or a = - sin 9 and b = cos 9.
Since we have defined N to be such that it forms a right handed frame, then it must
be the latter. (This assumes, of course, that 7 is being traversed counter-clockwise.)
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68
Thus, N = (—sin0,cos0) and by Lemma 5.1.4
df dk~ dk . . a m
~dt=ds = ^ dk dO . .
==>• — = — (by equation 5.8). OS dt
Using the equation dT/ds = kN from the Frenet-Serret equations (Theorem 2.2.1),
we have
a / J cs fyp — (-sin0(s) ,cos0(s)) = —(cos0,smO) = — = kN = k(-sind,cosd)
dO Hence, ==> k = —.
as
Lemma 5.1.6
dk &k f , + k3
dt ds2
Proof.
dk ab^ve &o = dw 2de Lemma 5 l
dt dtds ds dt ds { " J
8 f)k fPk = + k2 (k) = -r-r- + k3 (by previous Lemma),
dsds ds2
Lemma 5.1.7 Let A be the area enclosed by a simple, closed, regular plane curve,
7 (s, t) = (x (s, t), y (s, t)). Then the evolution equation for the area is
dA
Proof. It is known that
m m ~2'•
1 [2* ( dy dx\ 2 Jo [x»l~yai)du-
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69
Also, by the definition of the normal vector in the plane and since T is a unit vector,
then
^(!)2+(!?)2=i and !<.,«)). Thus,
Recall that
x _dy_ du
v).(?y ydu {X,y) \du' du)-
d i d d d V
So, combining the last several results, we get
ds vdu V ds du
x dy dx (dy dx\ ( dy dx\
= (®»y) • ( - f ) iv = ^7,-vN^ = — {^7, vfi\ .
Hence,
A-\r ('s - *©du=s r (- •*» * • -* r <* *• 77iis implies that
OA dt
r2ir
&'v*) + (*&J*) + 0 , v w ) l d u
1 /" 5 / -
= — ̂ I dty,vN)du (since u, t are independent)
- - - } 2 Jo
= - \ L [ * » - + ( - ? . » ( ~ | ^ ) « * ) ] ««»
= -irikv-(^)+(^-aif)}d"-
du
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70
Now consider the following integral:
j T £(***)*.
Since 7, k, T are all 2ir periodic, by implicit integration by parts, we get
au°' r du -1 (%kf)du+i
— s: < du.
Since — 1 is a scalar, we have
r>2 7r dT
du du.
Thus by substitution we have
dA dt - \ J * " [to - ( * vk?N) + ( y , ~ T du
du
du -ir[kv-(^)+k(^f)+k{;i' —^ J ^kv — ̂ 7, vk2N^ + k (T' vkN^ du
- i I ' [ t o - ( 7 - + (g.if) + Ivk'N}] du
'27T /»z7r . . ^ rZ7r
= — - J ycv + (vT, du = — - J 2kv du = — J kvdu
pL pL
= — I kds (change of variable) = — I ff (s) ds
Jo Jo
= — J ^ds = — J dO = — 27r (since 7 is a simple, closed curve).
Therefore, dA/dt = —2ir.
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71
Observation 5.1.8 Before continuing, we should note that by a simple integration
of the evolution equation for the area, we get the following
A (t) = —27rt + A (0),
where A (0) is the area enclosed by the original curve 7, before any deformation.
Clearly, this deformation will not continue forever. At some later point in time, say
T, the curve will have shrunk to a point, and the enclosed area will be zero. At this
point the evolution must stop.
5.2 Evolution of Simple Curves with Bounded Curvature
Now that we have developed many of the lemmas needed, we want to consider how the
curvature of these "evolving" curves changes over time. Before we do this, however,
we need to recall the definition of a simple curve.
A simple curve does not intersect itself. That is, 7 is one-to-one for all t e (a, 6),
and if 7 is also closed, with length \b — a|, then 7 (a) = 7 (b). If the curve is not closed,
simple means that the curve never "crosses" itself. We will again be restricting our
study to closed curves. In this section, we will show that if the curvature of a simple
curve remains bounded during evolution, then the curve will remain simple. We will
start with the rigorous statement of this theorem, then develop a couple of necessary
lemmas before completing the proof of the theorem.
Theorem 5.2.1 Let 7: S1 x [0,T) —• R2 represent a one parameter family of closed
curves satisfying the evolution equation
dy -i = k N - ( ^ )
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If |A; (u,t)| < c, for some positive, real constant c, and if the initial curve 7 (•,()) is
simple, then 7 (•, t): S1 —> R2 is a simple curve for each t.
First, we will prove the following lemmas.
Consider the function / : Sl x S1 x [0, T) defined by
f(ui,u2,t) = \j(ui,t) -7(u 2 , t ) | 2 (5.10)
= (7 («i, t) - 7 (u2, t), 7 («i, *) - 7 («2, *)) •
Lemma 5.2.2 The function f satisfies the heat equation
at ' ds\ dsi '
52f f where A / = 25 the Laplacian o f f .
Proof. By the chain rule,
d f d ~dt = -7(w 2 ,* ) ,7(« i , t ) -7(w 2 , i ) )
= 2^7(«!,i) - 7 (u2,t),^ (uut) ~^(u2,t)^
= 2(^(u1,t)-;y(u2,t),kN(ui,t)-kN(u2,t(by 5.9). (5.11)
= ^(/y(ui,t)-j{u2,t),^-(ui,t)-^-(u2,t)
= 2 ^ ( u i , t ) - - f ( t i 2 , t ) , f («!,t)^
as? 2 ( aI ^ aZ ̂ U2' ̂ ^ +2 ( 7 (mi, t) - 7 («2, <), 1^- (tii, <)
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73
= 2 ( f (ui, t),f («!, + 2 ̂ 7 (tii, t) - 7 («2, *), («1} t)^
= 2 + 2^7(Mi,t)-7(M2 ,t),A:iV(«i,t)y
Similarly,
d f 2 = 2 ^ ( U l , t ) - f K t ) , ^ ( « l l f ) - g K t )
= 2^7(^1,*) ~ 7 ( t i 2 , t ) , - f (w2,t))
0
= * S = » ( £ ( * . « > - £ < « * * > . - * w > (
= 2 (ti2, <), - T (u2, t)^ + 2 ( j (tii,i) ~ 7 (^2,<), -fciV (ti2,
= 2 — 2 ̂ 7 (nx, t) — j (tz2, f ) , fciV (u2, t)^ .
Then, by adding the two results together, we get
f s [ + f s f = 2 + 2 ( ; y ( u i ' t ) - i ( u 2 , t ) , k N ( u u t ) ^ + 2
- 2 ^7 (tii, *) - 7 («2, *), fciV (ti2, t)^
= 4 + 2 ̂ 7 («!,<) — 7 (ti2,t), fciV (ui,t)^
- 2 ^7 (tii, *) - 7 («2, t), kN (u2,
= 4 + 2 (7 (tii, t) - 7 («2, t) , fcJV (tii, t) - kN (u2, t ) )
5 / , = 4 + — (by equation 5.11).
**+"• I = S + S - 4 -
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74
Now, suppose 3 c, a positive constant such that |A;| < c. We will show that for
small arc lengths, \si - s2| < 2/c, f{si,s2,t) = 0 & = s2. That is, "the curves
have no self intersections resulting from short kinks." [GaHa, Page 76]
Lemma 5.2.3 (A. Schur and E. Schmidt) Let g: [0, L] —> E2 be a curve para-
metrized by arclength, thus unit speed, traveling from point A to point B, such that
AB together with g forms a convex curve. Let f be a second curve of the same
length as g, say L, and with endpoints C and D. Assume f and g have continuous
tangents and piecewise continuous curvature. Assume g is traversed in the counter-
clockwise direction such that k > 0 (curvature is positive). Suppose the curvature at
each point of g, kg, is greater than the absolute value of the curvature at the cor-
responding point (at the same distance) on f , k f . That is, kg(s) > \kf (s)|. Then
dist (A, B) < dist (C, D).
Proof. Without loss of generality, position the curves such that AB and CD both
lie on the x-axis. (See Figure 5.2.)
Since g is convex over its length, then 3! point s0, where the tangent to g at sa,
g (s0), is parallel to the x-axis. ^4s 6 is the angle between the tangent and the x-axis,
=*- 0g (s0) = 0. Since
ds ~kg~ ,fc/l " dd. ds
by integration we have, for s > s0,
For s < s0, this implies
-«-£§—r f»-m
= | » / W - « / ( » . ) ! .
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B T=(x',y) D
Figure 5.2: Position of the curves / and g.
< [°Mj_
J So d S
= -\0f(s)-ef(So)\
1*mi < - I - r *
Js ds
= -\ef(s0)-6f(s)\
Multiplying through by — 1, we get
Og («) > \9f (so) - Of (5)| = 16f (s) - 9f (s0)|
Thus, Vs between 0 and L,
I M s ) l > ! M s ) - M * o ) | .
Since g is convex, for 0 < s < L, \9g (s)| < 7r, then -n < 0g (s) < n. (Otherwise, g
is not convex.) Recall that tan 6 = y'/x'. Thus, x' = (cos 6) / ^|T|) = COS 6, since T
is unit speed. (Similarly, y' = sin 6.) (See Figure 5.2.)
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76
Thus, dist (AB) = f^x'(s) ds = /0L cos 9G (s) ds = jjfcoslfl^s)! ds. Also note
that cosine is decreasing on the interval [0,7r]. SO since \9G (s)| > 19/ (s) — 9/ (sQ)| =>
cos 19g (s)| < cos 19/ (s) - 9/ (s0)|. Hence,
pL rL
dist (AB) = / cos \9G (s)| ds < J cos|0/ (s) — 9F (s0)| ds Jo Jo
= I cos (9/ (s) - 9F (s0)) ds. Jo
The last integral is the projection of the line segment CD onto the tangent to f (sQ)-
Thus, line segment CD (R.H.S.) is greater than or equal to the projection (dist(C'D'))}
and C'D' > AB. Therefore, CD > AB. I
Corollary 5.2.4 Suppose |A;(m, £)| < c, for some constant c > 0. Then f (sl5 s2) >
[ | sin ( | |si — s2|)]2, where f is as defined above. Note that although f is dependent
on t, the expression on the right is not.
Proof. Let si, s2 be arc lengths on the circle with radius r = 1/c. Let 7 (s, t)
be a unit speed curve, (parametrized by arclength). Let g : |si — s2| =4» R2 be the
arc along the circle between Si and s2- Hence, the length of g is |si — s2|- Then g,
traversed counter-clockwise, together with the chord sjs2 forms a convex curve. (See
Figure 5.3.) Since the curvature of a circle is 1 /r, then the curvature of g, kg is
jjjc) ~ c- Thus, |A: (u, t)| = |A:̂ | <c = kg. So, by Lemma 5.2.3, with 7 = / ,
dist {A, B) < dist (C, D) *=> dist2 (A, B) < dist2 (C, D) = I7 fo) - 7 (s2) |2
= / (si> s2) (by definition).
Recall that arclength = radius x central angle (in radians). That is, s = rd. Thus
1 c 0 |si - s 2 | = -9 c |si - s 2 | = 9 <=$• - |sx - s 2 | = r .
c 1 2
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77
g
Y (st)=B
Also,
A =y(sl)
C = y(s,)
D = y(Sj
Figure 5.3: g is formed with an arc of the circle and chord AB.
-dist (A, B)
dist (A, B) = ^ sin |sx - s2 |)
Therefore,
f (si, s2) > dist2 {A, B) = 2 . (c -s in c (51®. - *1)
Finally, we are ready to present the proof for Theorem 5.2.1.
Proof. Consider the following set E = {(Sl, s2, t) | |S l - s2 | < (*/<:)}. For all the
points in E, |sj — s2| < (ir/c) if and only if
- I . 7T C 7T 2 IS1 ~ ®2, < — * - = —.
Since c > 0, |si - s2 | > 0, and sinx = 0 if and only if x = rwr, n e Z,
= * S in ( | 1*1 ~ S2|) = 0 <«=• 81 = 82.
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78
So f (si,s2,t) = 0 <=>• si = S2, (follows from Corollary 5.2.4). Now restrict / to the
complementary domain D = (S1 x S1 x [0, T]) \ E.
Claim 5.2.5 On the boundary of D, f has a strictly positive minimum.
Proof. Clearly, the boundary of D is given by
{(Sl,S2,*)||Sl - s 2 | = \ 0 < t < t } U {(si,«2,0)||«i -S2\ >
On the first set, by Corollary 5.2.4,
"2 c 1 ^
f(si,s2,t)> - s i n ( | | s 1 - s 2 | n .
In this set, however,
I S i _ S 2 i = : = > s i n ( £ . £ ) = s i n ( | ) = 1
- » / ( W ) > g ) ( 1 ) 2 = ( J = | > 0 .
The second set has a strictly positive minimum, say p, because the initial curve is
simple. (That is, there are no self intersections; so by the definition o f f , f (sx, s2,0) >
0, since for t = 0, we have 7, the initial curve.) In other words, / ( s i , s 2 ,0) =
I t ( s i > 0 ) - 7(^2 ,0)1 # 0, s ince 7 is simple. (Note that Si = 0 and s = 2n are the
same point, identified with each other.)
Now, let m = min j (f)2 , p | - Then clearly m > 0. I
Now consider the function g (su s2, t) = / (si, s2, t) + et for some e > 0. Since by
Lemma 5.2.2, / satisfies
?L-At-A ± d9 _ d[f(si,s2,t) +et] dt 1 dt dt
df det - i + 9 T = A / - 4 + £ -
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79
Observe that
So
&f &9 A &5 &9 \ r a ~ as? 94~ 94^ f ~ 9'
| = A , - 4 + . .
Now let 0 < 5 < to. (So (5 < (2/c)2 and 5 < p.) Suppose g achieves the value
TO - S (> 0) on D = (51 x 51 x [0,T)) \ 1£. Let <Q = inf {t\g (si, s2,t) =m — S}.
Since t £ [0, T), which is bounded below, =$• t0 € [0,T). D is a closed subset of R2
(a Hausdorff space), and so D is compact. Since / and t are continuous, this implies
that g is continuous.
The continuity of g and the compactness of D, together with the boundary esti-
mate, ensure that the first occurrence of the value m - 6 is at some interior point, say
(si, S2, t). Thus, by definition of t0, the point (sT, s^, t0) is in the interior of D. (Recall
that on the boundary of D, f has a strictly positive minimum, which is > m. Hence,
g>f>m>0<&g>m on the boundary. So g^m — 8<m on the boundary.)
On a small interval about the point t0,
- S - 3 - ( i I f c ) ' " « >
Also,
So
d2g dsids$
^ ^ = 2 (7 («i, 0 - 7 («2, t), - f (u2, t)}
= 2 ( f K , t), -T (u2, t)) + 0 = - 2 ( f (ttl, t), -f («2, <)) = ±2.
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80
Geometrically, this means that at a minimum point, the tangent lines at si and s2
must be parallel. Then by an application of Claim 4.2.3,
a q = 929 1 d*9 >2 r 9 929
ds\ ds2 ~ V ds\ ds2
By equation 5.12, we have
l&g&g u SPg \ 2
ds\ ds% ~ y \ds1ds2) &9
ds\ds2
Furthermore, since
&9 dsids:
= ±2 d2g
dsids? = 21±2| = 4.
dPg
dsids2
Putting all the pieces together, we get the following equation
ds\ ds2 V ds\ 3s| — ds\ds2
>4.
But by assumption,
da da ^ = A j - 4 + e and - < 0.
So for Ag > 4 =!• (dg/dt) = A 5 - 4 + f > 4 - 4 + f > e > 0 , That is, (dg/dt) > 0,
which is a contradiction.
Thus, since 8 is arbitrary, we must have that g(sus2,t) > m on D. Hence
/ (si, s2, t) >m — et since / (sx, s2, t) = g (s1} s2, t) —et>m — et. Letting e —• 0, ^
/ (si> s2, t) > m > 0 on D. But by definition
f(si,s2,t) = py(si,£) - 7(s2,*)| > 0 on D.
= > 7 has no self-intersections Vt,
7 is simple Vt.
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81
5.3 Convex Curves in the Plane
For convex curves in the plane, we will use the angle 9, (between the tangent vector
field T and the rc-axis) as a parameter. The curvature k can then be written as a
function of 9. Now we consider which positive, 2it periodic functions can represent
the curvature function of a convex curve.
Lemma 5.3.1 Let 7 be a simple, closed, strictly convex plane curve. A positive 2ir
periodic function will represent the curvature function k (9) if and only if
r»27r _ _ _ a /»2*r
r*cos9jn r* sin 9 Jn n
L *(«> ~l *(») Proof. First we will prove the forward implication. Let k be the curvature function
of 7. Since 7 is closed, => f^T ds = 0. From Lemma 5.1.5, we have k = d9/ds.
So d9 = kds, => ds = d9/k(9). Then, with a change of variable from s to 9,
where s goes from 0 to L as 9 goes from 0 to 27r, we get T (9) = (cos 9, sin 9) from
T (s) = (x'(s), y' (s)). Thus
I fds=0^l To prove the reverse direction, suppose we are given an arbitrary function k,
(which is 2n periodic and positive), and the associated curved for which k is the curva-
ture function, up to translation. Suppose further that 7 is given by 7 = (x (9) ,y (9)),
where
/n\ f e cos/? P*1 sin/? y(9)-lw)d0-
Clearly, x (0) = 0 = y (0); by assumption, x (27r) = 0 — y (27r). So 7 is a closed curve.
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82
Also,
dx cos 9 , dy smd x { e ) = M = W ) a n d y { e ) = d§ = W)'
which implies
(I) +(I) = S ) +(!m) =(*M) (cos2#+siii2^=ipW-(5.13)
Then we can reparametrize to obtain
s (0, = / V W + ̂ ^ ^ (5.14)
Hence, we can now write 7 as a function of s, 7 (s) = (x(s) ,y(s)). By the chain
rule, we get
. dx dxdO dx = s = = mk(e) (* wofon 5.14)
dv and similarly, y' (s) = —k (9).
du
Then, this implies that
x>2 00 + yn (s) = W ~ 1 (bV equation 5.13).
However, this means that 7 is unit speed; so we can write the tangent at s as follows:
f(s) = f M = (*>W M) = (I;* (0), w)
= (cos 9 (s), sin 9 (s)) (since 7 is unit speed).
Then it follows that
-* dT dO T' (s) = — = — ( - sin 9, cos 6) = 9' (s) ( - sin 9, cos 9)
curvature o f j = | t ' (s)| = \9' (s)\ = k (s).
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83
Since cos, sin, and k are all 2n periodic functions, then for 6 between 0 and 2ir,
7 (s) is a one-to-one function onto S1. Hence, it must be a simple curve. Moreover,
since by assumption k is positive, then 7 is strictly convex.
That is, j is a simple, closed, strictly convex plane curve, with k as its curvature
function. •
Now we need to find an evolution equation for the curvature using 6 as a parameter
instead of s. We will use r as the time parameter. Note that (d/dt) ^ {d/dr) since
(d/dt) holds u fixed and (d/dr) holds 6 fixed. So we have k (6, r) instead of k (s, t).
Lemma 5.3.2
dk , 2 d2k .,
a ; = k w + k
Proof. By the chain rule
dk dk dk 89 dk dk dk di = d; + wdt = + <*» s-i-Q
dk , (dk\ dk ( . dk dd dk dk\ = d^ + { m ) kW ( s m c e k m = Ts m = T s )
dk , (dk\2
= fr + k{de) ' ( 5 1 5 )
<n d_ (dk\ _d$ d_ (dk\ _do d an ds2 ds\ds)~ ds do
M ^ d6_ dk ds dd \ds J ds d9
dedk d (d$\ (de\2 d (dk\ , dsdd' d6\ds) + \ds) dO\de) ' y the cham rule^
= 96 dk M 2 ^ _ M ( d k \ 2 2&k
ds'de'de dd2 ds\de) de2
,{dk\2 t2&k
= k \ a » ) + k W <516>
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84
Recall from Lemma 5.1.6 that
dk dPk ,, ¥ = a ? + * -
Then by equation 5.15
dk (dk\2 cPk 3 /dk\2 jod^k 3 8f v»/ = d? \ a e ) + k W + (>>y equation 5.16)
dk ,o<Pk
The following theorem is due to Gage and Hamilton [GaHa, pages 80-81], and is
stated here without proof.
Theorem 5.3.3 The curve shortening process for convex curves is equivalent to the
following initial value partial differential equation problem:
Find k\ S1 x [0,71) —y R satisfying
1. ke C2+a'1+a (S l x [0, T - e]) Ve > 0.
2. (.dk/dr) = A;2 (Pk/dB2) + k3.
3. k (9,0) = if) (0) where if) satisfies:
(a) iPeC^iS1),
(b) tp (0) > 0, and
(°) /o2* ( ? # ) = f t ( $ ) <u> = 0.
Now we wish to show that curves which are strictly convex remain strictly convex
during and after evolution. This follows from the evolution equation in Theorem 5.3.3
by means of the following lemma.
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85
Lemma 5.3.4 If k satisfies Theorem 5.3.3, then
kmin (r) = inf {k (0, r) |0 < 6 < 2tt}
is a nondecreasing function.
Proof. Let e > 0 be such that kmin (0) > e. (That is, r = 0.) Suppose, to the
contrary, at some later time t, that A;min(r) = /cmin (0) - e < fcmin (0). (Thus, k is
decreasing over the time interval (0, r)). Let r0 = inf {r|fcmin (r) = kmin (0) - e}.
Since k satisfies Theorem 5.3.3, k is continuous. Continuity, then, ensures that
this minimum is achieved at some point in the interval, say (90,t0). Since (60,t0) is
a local minimum,
dk ffik = * (00, To) < 0 = » — (0O, To) > 0.
Note that k (80, r0) > 0 since at r0, k^n (t0) = kmin (0) — s > 0, by assumption.
But this contradicts k satisfying the second requirement in Theorem 5.3.3, as it
implies
dk l2&k l3
Therefore, km\n (r) is a nondecreasing function. |
Then, since in Theorem 5.3.3, k (6,0) = ip(6) and > 0, =» k(6,r) > 0 Vr.
Thus, by definition, the curve is strictly convex both during and after evolution.
5.4 An Application of the Isoperimetric Inequality
Having established that strictly convex curves remain strictly convex, we turn our
attention to an application of the isoperimetric inequality for convex curves. This
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86
final section is based on the work by M. Gage [Ga], Specifically, we show that closed,
convex, C2 curves in the plane satisfy the isoperimetric inequality
L
k2ds, (5.17)
where L is again the length of the curve, k the curvature, and A the area enclosed
by the curve. Then we show that when the curve is deformed according to the defor-
mation equation 5.1, this inequality is equivalent to showing that the isoperimetric
ratio L2/A decreases. Since equality is achieved in the Isoperimetric Inequality (The-
orem 4.2.1) only when we have a circle, then in some sense the curve is becoming
more circular.
Let 7 be a closed, convex, C2, plane curve. We would like to show that 7 satisfies
inequality 5.17 above. First, we need to develop the following two lemmas. We will
use the Bonnesen inequality for a curve symmetric through the origin,
rL — A — nr2 > 0,
which holds for any number r satisfying rjn < r < rcjr. [Oss, Page 1] (rjn is the radius
of the largest circle contained inside the curve—the inscribed circle—and rCir is the
radius of the smallest circle which contains 7—the circumscribed circle.)
Lemma 5.4.1 Let 7 be a closed, convex, plane curve which is symmetric through the
origin. Then it satisfies the inequality
>2 .LA ,_$)) ds<—. (5.18)
Proof. Since the curve is symmetric through the origin, then the breadth (or width) of
the curve, b^^, in the direction determined by a normal, N (s), must be 2 ̂ 7, —N^.
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87
Also, because the curve 7 is convex, then the breadth must satisfy 2rm < < 2 r c ; r ;
that is, the breadth of the curve must be between the diameters of the inscribed and
circumscribed circles. This implies rm < ( j , —N^ < r C j r . Then, by the Bonnesen
inequality,
=• (7, -N^ L — A — TT ^ 7 , - iv}2^ > 0, for each s.
We now integrate this inequality with respect to s.
—N^ L — A — TT ^ 7 , —N^ ds> J Ods = 0
1 : ds — 7T r (7, -N^ ds> 0
L J ^7, -N^ ds - AL - 7r J ^7, -N^ ds > 0. (5.19)
Recall that in the proof of Lemma 5.1.7, we found by Green's Theorem that
A = \ l { x % ~ v i £ ) d u = - \ l for a simple, closed, plane curve a with parameter u. Since by assumption, the curve 7
has been parametrized by arc length, we may rewrite the previous equation as follows:
a = \ L { i ' n ) i s -
By combining this equation with inequality 5.19 above, we obtain the final result.
L (2A) — AL — 7r J ^7, —N^ ds> 0 <*=>• > J ^7, —N^ ds.
Lemma 5.4.2 For any closed, convex, C1, plane curve, it is possible to choose an
origin so that inequality 5.18 from the previous lemma is satisfied.
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88
Proof. By assumption, 7 is such a closed, convex, C1, plane curve. Now let 7 (s)
be any point on the curve. Then 3/ point, 7 (I) such that the chord 7 (s) 7 (s) bisects
the area bounded by 7.
Define a function, f , to be given by f (7 (s)) = (t(S) xf(s ) , n | , where T(s)
and T(s) are the tangents at 7 (s) and 7 (s), respectively, and n is a normal to
the plane (in the positive direction). Since 7 is C1, then f is continuous. Also,
f (7 ( s)) = ~f (7 (s)) by a property of cross products. From the Intermediate Value
Theorem, it follows that 3 some Si such that f (7 (si)) = 0 and that T (si) = — T (si).
Select the origin to be the midpoint of the chord from 7 (si) to 7 (s"i), and let the x-axis
lie along it.
To complete the proof, we will designate the portion of the curve lying above the
x-axis by 71 and the portion below the x-axis by 72. For the moment, consider only
the upper portion of the curve. Suppose that the length of the upper portion is L\.
Now reflect 71 across the origin to form a closed, convex curve, which is symmetric
about the origin. From Lemma 5.4.1, we note that
2 J ( i , - N y ds< 2 — • 7T
where 2Li is the length of 71 and its reflection. Because the x-axis bisects the original
enclosed area, then the area enclosed by 71 and its reflection is the same.
Similarly, applying the same process to 72 yields
2jT (<7,-iv)2 ds < 2 ^ .
Adding these two inequalities gives
2IL ̂ ds+L H d s
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89
on the left hand side, and
2 LiA L2A1 _ 2 (Li + L2) A 2LA 7T 7T 7T 7T
ow the right hand side. Combining them, we obtain the desired result
iM~s) 7T
rL k2ds
10
With the help of the previous lemmas, we axe now ready to show that closed,
convex, C2, plane curves satisfy the inequality 5.17.
Claim 5.4.3 Suppose 7 is a closed, convex, C1, pieeewise C2, plane curve. Then the
inequality
L f1
is satisfied.
Proof. Using the definition for the tangent, T, and one of the Frenet-Serret
equations, (Theorem 2.2.1), we can rewrite the following integral.
J (7, ~N)kds = £ -kty ds = £ ^ _?) ds
/
L - j j
<7, f ) ds = — ( f , f ) |o + / ( f , f ) ds L
= 0 + J Ids (since 7 is unit speed) = L
Then, using the Cauchy-Schwarz Inequality,
L = f (i,-^)kds< ( j f (7,-.iv}2 (fe) ' (£k2ds
~ < * - * > ' * ) C M -
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90
Since by assumption, 7 satisfies the requirements for Lemmas 5-4-1 and 5-4-2,
then we have
Finally, since 7 is piecewise C2, we combine the two results above to yield
L2 £ (F DS) QM£ v CM
=$• < J k2ds = J k2 ds.
Thus, we have showed that a closed, convex, C2, plane curve, satisfies the isoperi-
metric inequality, equation 5.17 above. Next, we prove that this is equivalent to
showing that the ratio L2/A decreases.
Claim 5.4.4 Let 7 (s, t) be a regular, closed, simple, plane curve which has been
parametrized by arc length s, and deformed according to the deformation equation 5.1.
Suppose 7 also satisfies the isoperimetric inequality
"L
°i<-L k ds,
where L is the length, and A the area bounded by the curve. This is equivalent to
showing the ratio L2/A decreases.
Proof. By the First Derivative rule from Calculus,
L2 . , . , . d (L2\ — is a decreasing function •<===>• — I — ) < 0. A at \ A J ~
Thus, it suffices to show that the first derivative is less than or equal to 0.
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By Lemma 5.1.2 , we know that 7 has the following relationship
rL dL=_f dt Jo
k2ds.
Furthermore, by Lemma 5.1.7, the evolution equation for the area is
dA —— = —27T. dt
Then, with substitution from the two preceding equations, we can take the derivative
of L2/A to get
d (L2\ 2L^A-^L2 L L dL A dA dt (?) -
Therefore,
e d s - T ^ ° ^ l eis^~A'
where the last inequality is true by assumption. I
Thus, we have shown that the ratio L2/A is decreasing. By our work with the
Isoperimetric Inequality in Chapter 4, we know that for any closed, simple, convex,
plane curve, L2 > Ait A with equality if and only if the curve is a circle. In a later
article [GaHa], Gage and Hamilton show that in fact this ratio is approaching the
value of 47r. In this sense, the curve is becoming more circular.
Furthermore, they support this conclusion by showing that the ratio of maximum
curvature to minimum curvature is approaching one; i.e., l im^r fcmaxAmin = 1- In
other words, over the evolution process, the curvature is approaching a constant
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value. The development of this result involves establishing "a priori" estimates for
the solution of the nonlinear heat equation
dk l2&k f 3
Recall that the curvature for a circle is a constant, given by 1/r, where r is the radius.
Again, we see that in some sense, the curve is becoming more circular.
In addition, we can rescale 7 (6, t) by
Then as t increases, the area of the region bounded by t] is always 7r. That is, the area
is kept constant. Gage [Ga2] was then able to show that, in fact, 77 ($, t) converges to
the unit circle ast —>T.
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BIBLIOGRAPHY
[Ga] M. Gage, An Isoperimetric Inequality With Applications to Curve Shortening,
Duke Math. J. 50 (1983) 1225-1229.
[Ga2] M. Gage, Curve Shortening Makes Convex Curves Circular, Invent. Math. 76
(1984) 357-364.
[GaHa] M. Gage, R. S. Hamilton, The Heat Equation Shrinking Convex Plane
Curves, J. Differential Geom. 23 (1986) 69-81.
[MiPa] R. S. Millman, G. D. Parker, Elements of Differential Geometry, Prentice-
Hall, Englewood Cliffs NJ, 1977.
[Oss] R. Osserman, Bonnesen-Style Isoperimetric Inequalities, Amer. Math.
Monthly 86 (1979) 1.
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