Plane Curves, Convex Curves, and Their Deformation

Home , Convex curve, Torsion of a curve

37^ A/81 ~7S~//

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

Johanna M. Debrecht, B.A.

Denton, Texas

August, 1998 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. 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. 37^ A/81 ~7S~//

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

Johanna M. Debrecht, B.A.

Denton, Texas

August, 1998 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 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 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 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 CHAPTER 1

INTRODUCTION

The main focus of this paper is the study of convex, plane curves undergoing deformation 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, 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 advantage 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) 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 situations 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 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. 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 76 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 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 €

an [a, b], then 3 a convergent subsequence, {£«*}> d 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

Vk > N, => 17 (tnk) — 7 (i*)| = \xQ — 7(t*)| < e. That is, by definition, y(t*) = x0. 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 = tUm ^ ^-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, 8

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 \d&\ \4£\ 1^1 \ dv \ \du 1 dv 1 du 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. • Definition 2.1.7 (Regular Curve Length) The length of a regular curve, 7: [a, b] —

R3 is given by mi*. 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) 7^ 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 10

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)

= 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, 0 (since 7 is a regular curve). dt 11

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

= yj (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 IJo 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) = eln(1+^) ^cos (ln (* + ) >sin (lQ (2 + ) , i) • 12

Example 2.1.12 Let 0(t) = (21, t2,0), which is a parabola. Then dfi/dt = (2,21,0) and dp = V4 + 4*2 = 2Vl +12. 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, 13 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. 14

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. 15

2.2 The Frenet-Serret Apparatus

These five notions |/c, r,T,iV, 51 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 16

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+(§, AfB. (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 differentiating 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). 17

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 Then the FVenet-Serret equations for the curve /? are the following.

1. F = KVN

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 apparatus.

da 5 12 = ((-jgSins,-coss, • — sin • s 1\ ds 18

da I 25 . 144 sin2 s + cos2 s + —— sin2 s = 1. ds - Vim81 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

, da / 5 . 12 . \ T(s) = ^=(-l3SmS'-COSS'l3SmV

So then we can compute K by first finding T'.

rt,, x ( 5 .12 \ 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 .12 \ ^(s)=TW = r iscos 8>sm iscos 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 60 . 60 . = (_lsin3s-^co»»»)?-(- r-r sin s cos s + —r sin s cos sT 13z 13^ )

2 2 2 = (-±sin2s_lcos S)r+(-Asm S-^cos s)

Hence, => B (s) = (—^,0, —Finally, by the definition for T,

T (s) = — (s), iV (s)^ = 0 fsince B' (s) = 0). 19

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 20

» J k Thus, B (s) = x' (s) y' (s) 0

£isL 0 /x"2+y"2 ^x"2+y"2 x' (s) y" (s) _ x" (s) y'(s) = (o-o)r-(o-o)j*+ k y/x"2 + y'a y/x"2 + y'n

2 2 ) it . ( W-'W ) k ((see Lemma 3.1.1)y yfx'/x"u + yy" / \\x'y"-x"y'\J = (0,0,±1).

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. 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 curvature. 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)

21 22

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 = yj (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 23

{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'). 24

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* Ml - I*(»)| =±sw-

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 = tn

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. 25

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-

P metrized by arc length, then a is also closed (and regular) with period L = J0 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.) 26

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 27 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 ( v' (s) _ 1 x'y" - x"y' »(>) = \z'(s)/ 1 + 4 X12 i+($02 *

XV X = x x k xa + ya 'y" ~ "v' = 00» provided x' (s) ^ 0. (3.3) 28

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 (fj)dfj,. J So

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. 29

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 6(L) - 0(O)=0(L)= f k (fi) d/i = rV(s)ds+ /"V(a)ds + f & (s) ds JO J 0 J Si J S2 lim arctan ~~ *im arctan arctan ( 5-»S7 \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- jj^ -> +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. 30

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 _HL r*i+i »(L)-9( 0) = = ff(s)ds i=0 J** i=0 31

= £ lim arctan (Krrlt i — limarcta n S—>8 \x (s) J \x'(s)J i=0 t+1 3-^57 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. PSi psr«t+ i 6' = lim arctan (

n 1. = — — lim arcta\x'(s)Jn s-ysf \x'(s)J 2 , ,im ^Mf>') _ 1

= — lim arctan \x'(s)J 3~>Si+1 \x'(s)J 2 + lim arctan + 0

\X'(s)J s^s-+1 \X'(s)J 32

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

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. 33

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^. tm ds+ m Lh arcuW))}n C{"* arctan {Wj)} * arctan + arctan - '™- ^ (?$) - 0® J&, G® 34

0 — lim arctan 5—YS;

= ^lirn arctan (*g) -

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

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) >

1 0 (y'/x') > 0 and as s s{ from the right, y' (s) > 0, x (s) < 0 =*• (y'/x') < 0. (See Figure 3.8.) Then at Si, we have {arctM (?8)} * arctan + arctan lim arctan S—>S- (£§) - G?§) it. 8S S — lim arctan(ti l\ 35

= — — lim arctan "**•(«)- 7T m-A—m*2 = lim arctan

That is, the point Si adds tt to the value of Q{L).

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'. 36

Thus, finally, putting all of these results together, we have

= E lim arctan (~ lim arctan (^44) 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 lim arctan f ^4) - lim arctan (^441 i=0 (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 37 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) 38 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) = , we let p (u, u) = t (u) and p (0, L) = —t (0) = —t (L) |<5(v) — a(w)|

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.) 39

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' then dP dQ &o dv dvdu' du dudv' Since a (u, v) is C2 on A, we have

2 d o _ dV dvdu dudv

Hence, Green's Theorem becomes

f do da f f ( d2a dPo \ Ltodu+fodv = JjAd^~a^)duiv = 0- (34) "V*" it 0 Now, in general, for some curve c(s) = (u (s), v (s)), where a < s

J^Pdu + Qdv =f J ^P(u(s) ,v(s))^ + Q(u(s) ,v(s))^ ds

= [ lp(u (s) > V (s)) u' (s) + Q (u (s), v (s)) v' (s)] ds. J a

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. 40

On 71

Let u = s =£• du = ds > for 0 < s < L. v = s dv = ds Thus do do , fL — du + -z-dv = da (i A f a<7 (l i\ ? 1 s,s 1 ds. (3.5) du dv J0 an*'T I '' J/ 71 On 72

Let u = s du — ds • for 0 < s < L. v = L dv = 0 Thus,

do do , -^-du+ — dv J du dv *—jf !<••*>*• J 79 (3.6) On 73

Let u = 0 du = 0 for 0 < s < L. v = s dv = ds 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:

du+ dv du+ dv+ iu+ iv+ iu+ dv 0 ~ L^ a^ = /, fa to Jv % % f^ % % - However, this equation can be true if and only if the following is true.

I [a^(s's) + a?(s-s> ds = I ^3'^ds + l §;(«,>)d> (3.8) 41

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 becomes

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)] !2 :2 ^ x (a)] +[^°(L)-2/(s)] _ 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) 42

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 L x (s) y' (s) - x' (s) y (s) = IIo x2 (s) + y2 (s)

Proof. Suppose a = (x (s),y (s)) and 3 {si}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 arctan (y (s) /x (s)) at s = 0 (or at s = L) is of the form arctan (0/0), we can apply L'Hopital's rule:

lims_0+ arctan (fg) = lims_0+ arctan = arctan (f) = arctan (0) = 0 43

and lims_^- arctan (§ff) = 0. (3.9)

L x (s) y' (s) - x' (s) y (s) Then ds= ds I x2 (s) + y2 (s) tC - 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.)

(y{s)\ i So, lim arctan = and lim arctan (V{*)\ _ JH (3.10) 5—>S~ U(')/ 2 S-¥$t \x{s)) 2"

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^ = — ^ and lim arctan (V(*)\ = £ (3.11) \x(s)J 2 2'

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 44

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 (^41 • 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 larctan (f$)}'+f' {arctan GS)}' = lim arctan _ arctan areta s-t-sT G&!) J;,™ (^}) + _ " (f||) lim arctan [y(s)\ \x(s)J 45

+ lim arCtan = %• — lim arctan (\x(s)J) s->,r+1 {\x(s)J~ f2 8-*8,*-1 = lim arctan arctan f^+0. \x(s)J s->sUlim \x(s)J 5-4«i+1 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 {si}1

Furthermore, by the placement of a, we know that both Si and sn must be of type case I. Thus, we have

-E lim arctan (\ — lim arctan ( 2=0 *-K"+i \x(s)J s->sf \x{s)J n lim arctan (— lim arctan (^7-^ (since s = s i) -E «-wr \x{s)J s-*sf \x(s)J 0 n+ z=0 n lim arctan (- lim arctan (^7^ (by equations 3.9). \x{s)J s-nt \x{s)J -Ei= 1 From equation 3.10, we see that a case I point adds a net of tx to the sum. Likewise, 46 from equation 3.11, a case II point adds a net of —it to the sum. Thus,

^ J, . £ [ l,m arctan - lim arcta„ ( 2 2 i=l X 8 X S J0 x (s) + y (s) nt f -»T \ ( )J -+4 \ ( )J = ^ (—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. 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.

47 48

Definition 4.1.2 A function / (t) is called monotone increasing if t\ / (<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. 49

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 50

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, sj and [s2, L]. Since a is

closed, t(L) = t (0). Therefore, the tangent lines at a (si) and a (s2) not only have 51 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. 52

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, 53 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.) 54

Figure 4.6: The circle and the curve a.

—T The circle (3 can be parametrized by:

-y/r2 — x2 0 < s < S2 13 (s) = (z (s), w (s)) where z(s) = x (s) and w (s) = < 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' 55 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

I xdy = I ldxdy = Area (D) JdD JD Similarly, take

and <3 = 0 ==^ ^ = 0. ox

Then

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 56

By the Cauchy-Schwarz Inequality,

< |(^)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. 57

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 actually 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 (xy'). (4.6) 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 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,

iti 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.

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. 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 61

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 {u, t) = kN = k (tt, t) N (it, t). (5.1) ~dt

Figure 5.1: The curve is deformed in the direction of the arrows. 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

ru| drf(u) = s(u) = f du. Jo du

As 7 (ti, t) is deformed over time, however, we get

du. (5.2) (M) = / |s;(M)

Then

'dx . ' ^7 + du (u,t) a^(M)

Hence, by substitution back into equation 5.2

By + (v,t) dv. dv

So, ds_ dj (u,() (5.3) du du

Thus, the tangent vector field to 7 at a fixed time, t, is given by di f(u) ^ ^ ==» T(u) = dds ds 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

dT(u) d'y dj (it) f' (u) = kN k (u) N (u) du du 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, 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

S T(s,t) = -gj ^ 7 unit_8peed Qg ( ) dy 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_ 'a?, J > ds ds 'J ds'1 '' Hence,

2? = <3 ^ & = md 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 9 = v w 1{ du-g;di a; ' ^" = „(«,t) = _£ = In addition, d_ 1 d ds (5.5) \Afi!(M))2 + (g(u,t))2 90 So the arclength parameter is ds = vdu. That is, since

<(«,*)=£ >(.,() dw and ~=v(u,t) ds = v du. 64

So by equation 2.3 in Chapter 2, we have

dT dN g^^vkN 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.

2 v2 = #7 / dj dy du Xdu' du. d So = 2/—?! ?l\ -0/ &Y dl & \9tdu,du/~ \dudi^/ (u^tare independent)

d d ±ft+k JL du du dN 2{k~fa,vf) (sinceTlN)

2kv v (^du'^ ~ (~ kT,T^ (by equation 5.6)

—2k2v2

That is,

chai rule 2?>— ^ by above „l9 , dt ~ = ~2kv- Therefore,

— - -k2„ dt~ kv- 65

Lemma 5.1.2

dL = — I k2 ds, 10/iere L is the length of the curve 7. dt Jo

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) dL 2ir Qy dt / — 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

d_d_ _ d_d_ - o d dtds dsdt

Proof.

d_d_ (substitution from equation 5.5) dtds s(:s) d_ 1 d2 1 dv 8_ 1 d_d_ [!(;)] du^ v dtdu v2 dt du^ vdtdu 1 d d [-i (-*••)] 9u + v8tdu (by Lemma 5JJ)

tf_d_ 1 d_d_ l2 (I d \ 1 d d , v du vdtdu = k [vd^)+ vdudt (U> * mdePendent)

kd~s + Wsdi (b« Wt'on 5.5) 66

Lemma 5.1.4

dT_dkfi 8N dk- ~oT ~Z~N, ana —— = T dt ds dt ds

Proof. Recall that T = dj/ds. So,

dT _ d_ (&f\ _ ^7 2^7 , dt at + aJ iemma 5i

- I (£J)+ 1H+**-£*+*!?+«• / -A ^ = ds \ ) + ^ ^ (by Theorem 2.2.1) = ^-N -k2T + k2T=—N. us ds

Since N is a unit vector, => ^N,N^ = 1. 5o

at ((Ar"iV})= §i(1) = 0 "*=*•2 (w'") = 0 ^ ^ = o-

As a vector, dff/dt can be written as a linear combination off} 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 \df,N/= °*==* \aN + bT,N^ =0<=> (aN,N^ = 0 <=> a = 0.

So, dN , dt ~ T• (5.7) Then, since N _L T,

<*f>- 0 -> !<**)_!(0)-0 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- an , 69 "SatT asa " -7dsT — k

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 ^'Sin 6 W = (" Sin 6 W cos 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.) 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 , + fk3 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 m m ~2'•

Proof. It is known that

1 [2* ( dy dx\ 2 Jo [x»l~yai)du- 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, _dy_ x v).(?y du ydu {X,y) \du' du)- Recall that did d d V ds vdu V ds du So, combining the last several results, we get

dy dx (dy dx\ ( dy dx\ x

= (®»y) • (-f) iv = ^7,-vN^ = — {^7, vfi\ .

Hence,

77iis implies that

OA 1 /" 5 /- dt = — ^ I dty,vN)du (since u, t are independent) r2ir - --} v + J + ,v )l 2 Jo &' *) (*& *) 0 w du

= -\L [*» - +(-?.»(~|^) «*)] ««» = -irikv-(^)+(^-aif)}d"- 70

Now consider the following integral:

jT £(***)*.

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 du dt -\ J*" [to - (* vk?N)kv + (y, ~duT +k f +k ;i -ir[ -(^)du (^ ) { ' —^ J ^kv — ^7, vk2N^ + k (T' vkN^ du

-iI' [to-(7-+ (g.if) + Ivk'N}] du /»z7r . . ^ '27rZ7Tr = — - 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. 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=kN- (^) 72

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

2 f(ui,u2,t) = \j(ui,t) -7(u2,t)| (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 off.

Proof. By the chain rule,

d f d = ~dt -7(w2,*),7(«i,t)-7(w2,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^(ui,t)--f(ti2,t),f («!,t)^ U2 as? 2 ( aI ^ aZ ^ ' ^^

+2 (7 (mi, t) - 7 («2, <), 1^- (tii, <) 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,

df2 = 2^(Ul,t)-fKt),^(«llf)-gKt)

= 2^7(^1,*) ~7(ti2,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

+ = 2 + 2 ; u t fs[ fsf ( y( i' ) -i(u2,t),kN(uut)^ + 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- 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 parametrized 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, kf. 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 dd. ds ~kg~ ,fc/l " ds by integration we have, for s > s0,

= |»/W-«/(».)!.

For s < s0, this implies -«-£§—r f»-m 75

B T=(x',y) D

Figure 5.2: Position of the curves / and g.

[°Mj_ < = -\0f(s)-ef(So)\ J So dS 1*mi = -\ef(s0)-6f(s)\ < -I- r* Js ds Multiplying through by — 1, we get

Og («) > \9f (so) - Of (5)| = 16f (s) - 9f (s0)|

Thus, Vs between 0 and L,

IMs)l>!Ms)-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.) 76

L Thus, dist (AB) = f^x'(s) ds = /0 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) >

2 [| sin (| |si — s2|)] , 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)

2 be a unit speed curve, (parametrized by arclength). Let g : |si — s2| =4» R 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

c jjjc) ~ - Thus, |A: (u, t)| = |A:^|

2 2 2 dist {A, B) < dist (C, D) *=> dist (A, B) < dist (C, D) = I7 fo) - 7 (s2) |

= / (si> s2) (by definition).

Recall that arclength = radius x central angle (in radians). That is, s = rd. Thus

1 c 0 |si - s | = -9 c |si - s | = 9 <=$• - |s - s | = r. 2 c 2 1 x 2 2 77

Y (st)=B A =y(sl)

D = y(Sj

C = y(s,)

Figure 5.3: g is formed with an arc of the circle and chord AB.

Also,

-dist (A, B)

dist (A, B) = ^ sin |sx - s2|)

Therefore,

2 2 . (c f (si, s2) > dist {A, B) = -sin 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) | |Sl - 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,

Sin =* (| 1*1 ~ S2|) = 0 <«=• 81 = 82. 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 -s2| = \ 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)> -sin(||s1-s2|n .

In this set, however,

ISi_S2i = :=>sin(£.£)=sin(|) = 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 off, f (sx, s2,0) >

0, since for t = 0, we have 7, the initial curve.) In other words, /(si,s2,0) =

It(si>0) - 7(^2,0)1 # 0, since 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

d ?L-At-A ± 9 _ d[f(si,s2,t) +et] dt 1 dt dt df det - i+9T=A/-4+£- 79

Observe that

&f &9 A &5 &9 \ r a ~ as? 94~ 94^ f ~ 9' So | = A,-4 + ..

Now let 0 < 5 < to. (So (5 < (2/c)2 and 5 < p.) Suppose g achieves the value

1 1 TO - S (> 0) on D = (5 x 5 x [0,T)) \ 1£. Let

2 Since t £ [0, T), which is bounded below, =$• t0 € [0,T). D is a closed subset of R

(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

On a small interval about the point t0,

- S-3-(iIfc)'" «>

Also,

2 ^ ^ = (7 («i, 0-7 («2, t), -f (u2, t)}

d2g = 2 (f K, t), -T (u , t)) + 0 = -2 (f ( , t), -f («2, <)) = ±2. dsids$ 2 ttl 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,

929 d 9 9 929 aq = 1 * >2 r ds\ ds2 ~ V ds\ ds2 By equation 5.12, we have

2 l&g&g u SPg \ &9 dPg ds\ ds% ~ y \ds1ds2) ds\ds2 dsids2 Furthermore, since

&9 d2g = ±2 = 21±2| = 4. dsids: dsids? Putting all the pieces together, we get the following equation

— >4. ds\ ds2 V ds\ 3s| ds\ds2 But by assumption, da da ^ = Aj-4 + e and - < 0.

So for Ag > 4 =!• (dg/dt) = A5-4 + 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, ^

s / ( i> 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. 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 curvature function, up to translation. Suppose further that 7 is given by 7 = (x (9) ,y (9)), where /n\ fe 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. 82

Also, dx cos 9 , dy smd x{e) = M = W) and 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 ofj = |t' (s)| = \9' (s)\ = k (s). 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

2 dk , 2 d k ., a;=kw+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 (smce km = Ts m = Ts) dk , (dk\2 k = fr + {de) ' (515)

M ^

dedk d (d$\ (de\2 d (dk\ , dsdd' d6\ds) + \ds) dO\de) ' y the cham rule^ 2 = 96 dk M 2^_M(dk\ 2&k ds'de'de dd2 ds\de) de2 ,{dk\2 &k t2 516 = k\a»)+kW < > 84

Recall from Lemma 5.1.6 that

dk dPk ,, ¥ = a? + *-

Then by equation 5.15

2 2 dk (dk\ cPk 3 /dk\ jod^k 3 8f v»/ = d? \ae)+kW+ (>>y equation 5.16) dk ,o

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 (Sl 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* (?#) = ft ($) = 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. 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 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 deformation 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^. 87

Also, because the curve 7 is convex, then the breadth must satisfy 2rm < < 2rc;r; that is, the breadth of the curve must be between the diameters of the inscribed and circumscribed circles. This implies rm < (j, —N^ < rCjr. 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%~vi£)du=-\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)is-

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. 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,-Ny 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 Hds 89

on the left hand side, and

LiA L A1 2 2 _ 2 (Li + L2) A 2LA 7T 7T 7T 7T ow the right hand side. Combining them, we obtain the desired result s iM~ ) 7T

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 rL 2 L f1 k ds 10 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 -jj / <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< (jf (7,-.iv}2 (fe) ' (£k2ds

~ <*-*>'*) CM- 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 £ DS £ (F ) QM v CM =$• < J k2ds = J k2 ds.

Thus, we have showed that a closed, convex, C2, plane curve, satisfies the isoperimetric 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. 91

By Lemma 5.1.2 , we know that 7 has the following relationship

rL dL=_f k2ds. dt Jo

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

2 2 d (L \ 2L^A-^L L L dL A dA dt (?) -

Therefore,

eds-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., lim^r fcmaxAmin = 1- In other words, over the evolution process, the curvature is approaching a constant 92 value. The development of this result involves establishing "a priori" estimates for the solution of the nonlinear heat equation

dk l2&k f3

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. 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.