Reverse Isoperimetric Inequalities in R3

REVERSE ISOPERIMETRIC INEQUALITIES IN R3

DISSERTATION

Presented in Partial Fulﬁllment of the Requirements for the Degree Doctor of

Philosophy in the Graduate School of the Ohio State University

Andrew Gard, BS

Graduate Program in Mathematics

The Ohio State University

2012

Dissertation Committee:

Dr. Fangyang Zheng, Advisor

Dr. Ulrich Gerlach

Dr. Bo Guan

Dr. Chris Hans c Copyright by

Andrew Gard

2012 ABSTRACT

We formulate conditions under which the classical isoperimetric inequality in R3 can be reversed. Restricting our attention to surfaces with rotational symmetry, we enforce bounds on curvature and overall size (both in the appropriate technical senses) and show that these suﬃce to guarantee the existence of shapes of minimal volume for given ﬁxed surface area. In the fundamental case where both principal curvatures are bounded, we construct the surface of minimal volume.

ii For my family.

iii ACKNOWLEDGMENTS

The ﬁrst thanks of course go to my advisor, Dr. Fangyang Zheng, who brought the seed of this project to me and suggested how I could make it grow. His support and advice have been invaluable throughout this process. I also wish to thank professors

Chris Hans, Bo Guan and Ulrich Gerlach for generously taking the time to review this document and serve at my defense. I recognize that this is not a small commitment; all I can do in return is to attempt to be worthy of their eﬀorts.

Without the support of friends and family, this project would have driven me mad. Thanks particularly go to Isabel and Jenny for the coﬀee, Moy and Justin for the debates, and Bob and Billy for the nights oﬀ. Thanks also to the denizens of

MW200 past and present for the never-ending sharing of mathematical ideas. We’ll tease out the truth yet.

Finally and most importantly, my wife Kristina, too-often a math widow of late, has my inﬁnite gratitude and love, as does my son Clayton, who still thinks that math just means adding things up. You may be right after all, son.

iv VITA

1974 ...... Born in Hammond Indiana, USA

1999 ...... B.Sc. in Mathematics, The Ohio State University

2004-Present ...... Graduate Teaching Associate, The Ohio State University

FIELDS OF STUDY

Major Field: Mathematics

Specialization: Diﬀerential Geometry

v TABLE OF CONTENTS

Abstract ...... ii

Dedication ...... iii

Acknowledgments ...... iv

Vita...... v

List of Figures ...... vii

CHAPTER PAGE

1 Introduction ...... 1

1.1 A Little History ...... 1 1.2 Preliminaries ...... 3

2 Existence: |κθ|, |κσ| ≤ 1...... 6 2.1 Formulation of the Problem ...... 6 2.2 Existence of the Minimal Shape ...... 15

3 Construction of the Minimal Shape: |κθ|, |κσ| ≤ 1 ...... 17 4 Existence: |H| ≤ 1 ...... 40

4.1 Formulation of the problem ...... 40 4.2 Existence of the Minimal Shape ...... 55

Bibliography ...... 58

vi LIST OF FIGURES

FIGURE PAGE

1.1 The Howard/Treibergs result ...... 2

2.1 A typical element of FA0 ...... 7

2.2 The threshold σT ...... 8 2.3 From axis to cap ...... 12

3.1 The neighborhood to be deleted ...... 28

3.2 The optimal σmin ...... 31 3.3 Swapping pinwheels while preserving ∆x ...... 32

3.4 Connecting a local minimum ...... 33

3.5 Minimal area when β = π/2 ...... 34

3.6 Yet another simpliﬁction ...... 35

3.7 The last simpliﬁcation ...... 37

3.8 Nearing a hollow sphere ...... 39

3.9 Adding area without adding volume ...... 39

4.1 Cap comparison ...... 46

4.2 A typical |H| ≡ 1 sleeve ...... 50

4.3 A typical |H| ≡ 1 half-balloon ...... 53

vii CHAPTER 1

INTRODUCTION

1.1 A Little History

The isoperimetric inequality says, roughly, that among domains of of ﬁxed boundary area the one having greatest volume is the n-sphere. Analytically,

1−1/n A(∂Ω) V (Ω) ≤ 1/n (1.1.1) nωn n Where Ω is any domain in R , V is n-dimensional volume, A is (n − 1)-dimensional n boundary area, and ωn is the volume of the unit ball B . Equality holds exactly when Ω = Bn.

In the plane, this result was known to the ancient Greeks and proved by Steiner in 1838. Countless new proofs have emerged over the years since, drawing on nearly every branch of mathematics and extending the result in every conceivable direction, most notably to manifolds and abstract measure spaces. See [3] for a somewhat recent survey.

The opposite inequality remains almost completely unaddressed.

In a sense this is logical. There is in general no minimum volume for fixed surface area (or maximum surface area for fixed volume). Restrictions must be imposed before the question becomes meaningful. Yet the conditions that we must set turn out to be so simple and obvious - bounds on curvature and overall size - that there need be no unbearable loss of elegance in the statement of the problem. 1 Moreover, the question is not without physical interpretation or application. One needs to look no further than the dimple on the bottom of a bottle of wine to see the commercial desirability of volume reduction. In the human body, the red blood cell carries oxygen on its exterior, and so naturally seems to desire greater surface area for relatively fixed volume. In a sense, nature has already addressed this problem.

In 1995, Howard and Treibergs [8] proved a reverse isoperimetric inequality for simple plane curves, ﬁnding for perimeter 2π ≤ L < 14π/3 and curvature |κ| ≤ 1 the relation L − 2π A ≥ π + 4 sin (1.1.2) 4 where L represents perimeter length. The extremal shape resembles a peanut:

Figure 1.1: The Howard/Treibergs result

As L increases, the concave portions of the peanut pinch inward until finally meeting when L = 14π/3. The reverse-isoperimetric problem ceases to be meaningful beyond this threshhold. √ Figure 1.1 is also extremal for the dual problem, in which A ∈ [π, π + 2 3] is √ fixed and L is maximized. Here the threshhold value A = π + 2 3 corresponds to the situation where L can be made arbitrarily large by pinching the concave parts of the peanut and stretching the convex parts away from one another like taffy.

2 In higher dimensions the reverse isoperimetric problem rapidly becomes more diﬃcult, with multiple formulations possible. In this dissertation, we address the rotationally-symmetric three-dimensional case, showing that the problem is well- deﬁned when either mean curvature or both principal curvatures are bounded and surface area A0 is less than a determined threshhold. In the principal curvature case, we construct the optimal shape.

1.2 Preliminaries

Most of this dissertation will be derived from ﬁrst principles, but we will occasionally require some fundamental ideas from local theory of smooth manifolds. Here we establish those results.

1 2 3 Let X be a diﬀerentiable map from a domain D in the (x , x )-plane into R .

By considering the partial derivatives of X we obtain the tangent space TpX =

Span{X1,X2} of X at point p and the ﬁrst fundamental form I : TpX × TpX → R of the surface X(D):   hX ,X i hX ,X i  1 1 1 2  (gij) =   hX2,X1i hX2,X2i

3 where the inner product is inherited from R . This matrix is non-singular whenever

X1 and X2 are linearly independent (so-called regular points). For convenience we set

ij −1 (g ) = (gij) and g = det(gij). This Riemannian metric is the basis for all intrinsic calculations of length, angle, and area. Since all the surfaces in this dissertation will be C1 or better, the reliability of this metric will never be at issue. X ∧ X The unit normal n = 1 2 is now well-deﬁned and independent of our choice |X1 ∧ X2| coordinates (x1, x2) up to sign. The second fundamental form II is then given in the

3 basis {X1,X2} by:     hX , ni hX , ni −hX , n i −hX , n i  11 12   1 1 1 2  (hij) =   =   (1.2.1) hX21, ni hX22, ni −hX2, n1i −hX2, n2i

Closely related to II is the shape operator S (or Weingarten map) on TpX, deﬁned by S(u) = −Dun. Like the second fundamental form, the it tracks changes in the unit

3 normal and thus gives information about the ways in which X(D) curves locally in R . The shape operator satisﬁes the important properties II(u, v) = hS(u), vi = hu, S(v)i.

The ﬁrst equality is a matter of deﬁnition; let us verify the second.

j Since hXi, ni = 0, we have hXi, nji = −hn,Xiji = hXj, nii. Writing u = u Xj and

j v = v Xj, where summation is understood to take place whenever an index occurs in both a lower and upper position, we obtain:

i j i j hS(u), vi = −hu ni, v Xji = −u v hni,Xji (1.2.2)

i j i j = −u v hXi, nji = −hu Xi, v nji = hu, S(v)i (1.2.3)

In light of the fact that 0 = Dwhn, ni = 2hDwn, ni, S is a map from TpX to itself

j and we may write S(Xi) = Si Xj. Direct computation shows that in fact:

j X kj Si = hkig (1.2.4)

X jk ni = − hijg Xk (1.2.5)

The eigenvectors of S represent the directions of minimal and maximal changing of n.

The corresponding eigenvalues are called the principal curvatures of X(D). The determinant and trace of S are called the Gauss and mean curvatures, respectively. The former can be shown to be depend only on (gij) and its derivative (Gauss’s Theorema egregium), and hence measures the way in whichthe surface curves intrinsically.

Now suppose that X is rotationally symmetric. Orienting so that the axis of rotation is the y-axis, we have the standard parameterization

X(θ, s) = (x(s) cos θ, y(s), x(s) sin θ) (1.2.6) 4 where σ(s) = (x(s), y(s)) is a unit-speed curve in the xy-plane and x(s) ≥ 0 ∀s. With

0 0 0 respect to the basis {Xθ,Xs}, the unit normal is n = (−y cos θ, x , −y sin θ) and the ﬁrst and second fundamental forms are given by:     x2 0 xy0 0     (gij) =   and (hij) =   (1.2.7) 0 1 0 x0y00 − x00y0

Line 1.2.4 yields  y0  0  x  S =   (1.2.8) 0 x0y00 − x00y0 The principal curvatures of X are just the diagonal entries. Note that the second of these is just the signed curvature of the plane curve σ(s). We choose our labels accordingly: y0 κ = x0y00 − x00y0 and κ = (1.2.9) σ θ x

It is worth noting at the outset that κθ is formally defined even when σ fails to be twice-differentiable, a fact which will occasionally simplify our computations. On the other hand, κσ requires the usual smoothness before it is well-defined, even formally. Our first task must be to address this difficulty.

5 CHAPTER 2

EXISTENCE: |κθ|, |κσ| ≤ 1

2.1 Formulation of the Problem

It is obvious that some regularity condition must be imposed if the idea of constrained curvature is to make any sense. However, we also desire that our class of surfaces be closed under uniform convergence, lest we arrive at a sequences of surfaces approach- ing an optimal limit not contained in that class. In particular, it is too much to require that the surfaces in question be C2 or even piecewise C2, as such conditions may be lost under uniform convergence. Thus it is necessary to tread carefully when talking about curvature. We make the following deﬁnitions:

ϑ(s) − ϑ(t) y0(s) κ (s, t) = and κ (s) = (2.1.1) σ s − t θ x(s) n 2 1 0 K = σ(s) = (x(s), y(s)) : [0,L(σ)] ,→ R+, σ ∈ C , |σ | ≡ 1, o |κσ|, |κθ| ≤ 1 ∀s, t ∈ [0,L] (2.1.2)

n 0 0 FA0 = σ ∈ K : x(0) = x(L) = y(0) = y (0) = y (L) = 0, Z L o 2π x(s)ds = A0 (2.1.3) 0 where ϑ is the angle between σ0 and the positive x-axis (measured continuously from

2 2 s = 0), R+ = {(x, y) ∈ R : x ≥ 0}, and L(σ) is the length of σ. Thus in K each σ0 is absolutely continuous, hence a.e differentiable and equal to the integral of that derivative. Rotation about the y-axis then yields a 2-manifold with surface area 6 Z L Z L 2 0 A(σ) = 2π x(s) ds, volume V (σ) = π [x(s)] y (s) ds and principal curvatures 0 0 defined at least a.e and bounded in absolute value by 1 wherever defined. FA0 consists of curves of regularity class K that are well-behaved at the axis of rotation and have prescribed surface area A0. See Figure 2.1.

Figure 2.1: A typical element of FA0

We seek to minimize V (σ) among elements of FA0 .

It is clear that this problem is only well-deﬁned for small values of A0. Figures

2.1 and 2.2 illustrate the reason: large values of A0 allow sequences {σi} ⊂ FA0 of curves with descending V (σi) converging to non-embedded limits. Such degeneracy, however, is impossible for smaller values of A0 (in the extreme case of A0 = 4π, for

example, FA0 consists of only one element: the unit semicircle). Our strategy, then, is to require that A0 < AT , where AT is the least surface area possible among such degenerate curves. The following theorem, proved over the remainder of the section, establishes threshhold.

2π √ Theorem 2.1.1. A = (5π 3 + 6). This threshold is realized only for the non- T 3 embedded curve σT in Figure 2.2. Explicitly,

7   sin(s), −1 + cos(s) s ∈ [0, π/3]   √ 5π 5π σT (s) = 3 + cos s + , sin s + s ∈ [π/3, 2π] (2.1.4)  6 6  π π  cos s + , 1 − sin s + s ∈ [2π, 7π/3]  6 6

Figure 2.2: The threshold σT

Proof. Since the threshold curve is non-embedded but can be made so through a deformation that adds any > 0 to its surface area, σT must possess a pinched neck

0 0 where σT (s1) = σT (s2) and σT (s1) = −σT (s2), 0 ≤ s1 < s2 ≤ L. Of the two arcs

σT |[0,s1] and σT |[s2,0], we may replace the one contributing more surface area with the one contributing less to ensure that the two agree. Thus σT has the shape of a balloon.

We proceed with a sequence of lemmas. We begin by extending some basic facts about smooth curves with bounded curvature to members of K.

Lemma 2.1.2. (Dubins) Let σ be a C1 curve with |σ0(s)−σ0(t)| ≤ |s−t| ∀s, t ∈ [a, b], where |b − a| ≤ π, and let γ be a unit circle. Assume both σ and γ are parameterized by arc length. Then,

|σ0(b) − σ0(a)| ≤ |γ0(b) − γ0(a)| (2.1.5)

8 hσ0(b), σ0(a)i ≥ hγ0(b), γ0(a)i (2.1.6)

3 Equality holds exactly when σ is an arc of a unit circle. In particular, in R , if ∂ σ(0) = γ(0) = 0 and σ0(0) = γ0(0) = , then σ(s) lies outside the interiors of the ∂x unit balls with centers (0, 1) and (0, −1) for s ≤ π.

2 Proof. Note that this lemma, as stated, is not restricted to R . As a result, it makes no mention of ϑ. Although the following proof would require only trivial modiﬁcations to make use of that formulation, we opt instead for the more expansive equivalence of Lemma 2.1.3, below.

Following Dubin’ original work (see [5]), we use the fact that the curve σ0 lies on the surface of Bn. The condition |σ0(s)−σ0(t)| ≤ |s−t| ∀s, t ∈ [a, b] implies that σ00 exists Z b on [a, b] as an L1 function bounded in magnitude by 1. σ0 has length |σ00(τ)| dτ ≤ a |b − a|, a bound obviously shared by the minimal geodesic connecting the points σ(a) and σ(b). Since |b − a| is exactly the length of γ0, the shortest great-circular arc connecting σ0(a) with σ0(b) is shorter than the shortest great-circular arc connecting

γ0(a) with γ0(b), and both have lengths at most π. Under such circumstances, the chord connecting σ0(a) and σ0(b) must also be shorter than the chord connecting γ0(a) and γ0(b). In other words, |σ0(b) − σ0(a)| ≤ |γ0(b) − γ0(a)|, which is exactly equation

(2.1.5). Equation (2.1.6) is now clear as well.

Next assume that |σ0(b) − σ0(a)| = |γ0(b) − γ0(a)|, so the smallest great circular arc connecting σ0(a) and σ0(b) has length exactly |b − a|. Then the arc length of σ0 is no less than |b − a|. In the last paragraph, we observed that this arc length was no greater than |b − a|, so in fact the two are equal. Thus σ0 is a minimal geodesic on the surface of Bn connecting σ0(a) and σ0(b), from which we conclude that it must be a great-circular arc. Thus σ is an arc of a unit circle, as claimed.

9 0 0 σ (b) − σ (a) Remark Dubins gives the name average curvature to the quantity , b − a terminology which is both natural and inaccurate, as can be quickly confirmed by carrying out the computations for a unit circle. For a twice-differentiable unit-speed plane curve, the actual average value of κ on an interval I is given by |κσ(a, b)| as defined in line 2.1.1. We have, however, the following equivalency:

2 1 Lemma 2.1.3. Let σ :[a, b] → R be C . The following bounds on average plane curvature are equivalent:

|ϑ(t) − ϑ(s)| (1) ≤ K ∀s, t ∈ [a, b] (2.1.7) t − s

|σ0(t) − σ0(s)| (2) ≤ K ∀s, t ∈ [a, b] (2.1.8) t − s

Proof. (1) =⇒ (2): This implication holds even without the quantiﬁers. Without loss of generality, take s = 0, σ(0) = 0, ϑ(0) = 0, and ϑ(t) ≥ 0, so condition (1) becomes ϑ(t)/t ≤ K. We compute:

|σ0(t) − σ0(s)| |(cos ϑ(t), sin ϑ(t)) − (1, 0)| p2 − 2 cos ϑ(t) = = t − s t t s ϑ(t) 2 − 2 cos ϑ(t) ϑ(t) = ≤ ≤ K t ϑ2(t) t

(2) =⇒ (1): Suppose condition (1) fails for some values of s and t, which we take to be the endpoints a and b respectively . Without loss of generality, assume

ϑ(a) < ϑ(b). We now claim that (1) fails for an entire subinterval, speciﬁcally:

ϑ(t) − ϑ(s) > K ∀s, t ∈ [c, d] ⊂ [a, b] (2.1.9) t − s

Writing f(t) := ϑ(t) − Kt, the problem becomes to show (f(t) − f(s)) > 0 ∀s, t ∈

[c, d] ⊂ [a, b] under the assumption that f(b) − f(a) > 0. This is essentially the

Riesz Rising Sun Lemma, however: G = {s ∈ [a, b]: f(s) > inf f(t)} is an open t∈[a,s] 10 [ set which can be written G = (ai, bi). On any such interval, the desired condition (f(t) − f(s)) > 0 is satisﬁed.

By reducing the interval slightly, we get a [c, d] on which (ϑ(t) − ϑ(s)) > (K +

)(t − s) for all suﬃciently small > 0 . Fix s, t ∈ [c, d] and reorient once more so that s = 0, σ(0) = 0, ϑ(s) = 0, and ϑ(t) > 0. Then s r |σ0(t) − σ0(s)| ϑ(t) 2 − 2 cos ϑ(t) ϑ2(t) = ≥ (K + ) 1 − > K t − s t ϑ2(t) 12

The last inequality is valid for suﬃciently small t, which is all that we require.

We now make a fundamental distinction:

Deﬁnition 2.1.4. Let σ be the generating curve for a rotationally symmetric C1 manifold. By the cap of σ we mean any chosen subarc connecting points where x0 = 1 and x0 = −1 which lacks intervening horizontal tangents. Thus on the cap, y(s) is monotone and x(s) includes a local maximum. The sleeves are the remaining arcs, if such exist.

To prove the Theorem 2.1.1 it suﬃces to separately show that σT minimizes surface area A among all possible caps and all sleeves.

The ﬁrst of these tasks is short. For any ﬁxed x0 ≥ 0, letσ ˜ be the unit semicircular cap beginning at x(s0) = x0, and let σ be any other cap with the same

0 0 starting horizontal. By the Dubins Lemma, x (s) ≥ x˜ (s) on [s0, s0 + π]. Thus σ contributes more surface area thanσ ˜. Moreover, since σ must be longer thanσ ˜ and |y0(s)| ≤ |y˜0(s)| within π/2 of either horizontal, σ must traverse a greater total y-distance thanσ ˜, a fact which will be relevant shortly.

We now turn our attention to the sleeves of σT .

∂ Proposition 2.1.5. Let σ ∈ K satisfy x(0) = y(0) = 0, y(l) = h, σ0(0) = σ0(l) = . ∂x Then: 11 h 1) If h ≤ 2, then A(σ) ≥ 2φ sin φ, where φ = cos−1(1− ), as illustrated in Figure 2

2.3. Equality holds if and only if σ =σ ˜h consists of two complimentary unit circular arcs.

  sin s, 1 − cos s s ∈ [0, φ] σ˜h(s) = (2.1.10)  2 sin φ − sin(2φ − s), h − 1 + cos(2φ − s s ∈ [φ, 2φ]

2) If h > 2, then A(σ) > π.

Φ y = h

sinΦ, 1 - cosΦ

H L

Σ s

H L

Figure 2.3: From axis to cap

Proof. Consider the case h ≤ 2. By the Dubins Lemma, the shortest arc from (0, 0) ∂ to the line y = h/2 with σ0(0) = isσ ˜ | . Similarly, any shortest arc from height ∂x h [0,φ] ∂ h to h/2 with initial tangent of − is a translated reparametrization ofσ ˜ | . ∂x h [φ,2φ]

Thus any σ satisfying the hypotheses of the lemma is longer thanσ ˜h. We must consider two cases:

12 0 Case 1 : ∃s0 ∈ (0, 2φ): x (s0) = 0. Let us assume s0 to be the largest such value.

By the Dubins Lemma, s0 ≥ π/2 ≥ φ. Moreover, for all s ≤ φ, x(s) ≥ x˜(s) and x0(s) ≥ x˜0(s). Thus:

Z s0 Z φ Z φ

A(σ|[0,s0]) = 2π x(s) ds ≥ 2π x(s) ds ≥ 2π x˜(s) ds = A(˜σh|[0,φ]) 0 0 0 (2.1.11) ˜ Next, observe that σ deﬁnes a function y = f(x) on [x(s0), x(l)]. Let y = f(x) correspond to the upper half of the unit circle with center (x(s0) + 1, y(s0)). Then 0 ˜0 x(l) ≥ x(s0) + 1 and f (x) ≥ f (x) on [x(s0), x(s0) + 1], so:

Z x(l) Z x(s0)+1 q p 0 2 ˜0 2 A(σ[s0,l]) = 2π x 1 + (f (x)) dx ≥ 2π x 1 + (f (x)) dx (2.1.12) x(s0) x(s0) The rightmost expression represents the surface area generated by a unit-circular arc connecting a vertical at x(s0) to the horizontal. Since x(s0) ≥ 1, this is greater than

A(˜σh|[φ,2φ]), which consists of a part of such an arc translated toward the axis. Thus A(σ) ≥ A(˜σ), completing the proof of case 1.

Case 2 : x0(s) 6= 0 on [0, 2φ], so σ deﬁnes a function y = f(x) on [0, x(l)]. Let s1 = inf{s : x(s) < x˜(s)} and set x1 = x(s1) =x ˜(s1). By the Dubins Lemma, 0 ˜0 ˜ s1 ≥ φ, x1 ≥ sin φ, and f (x1) ≥ f (x1), where y = f(x) is the function deﬁned byσ ˜ 0 ˜0 on [0, 2φ]. Now f (x) ≥ f (x) ∀x ∈ [x1, x(l)] (Dubins again) and we have:

A(σ) ≥ A(σ|s∈[0,s1]) + A(σ|x∈[x1,2 sin φ]) Z 2 sin φ p 0 2 ≥ A(˜σ|s∈[0,s1]) + 2π x 1 + (f (x)) dx x1 Z 2 sin φ q ˜0 2 ≥ A(˜σ|s∈[0,s1]) + 2π x 1 + (f (x)) dx = A(˜σ) x1

The second part of the lemma is proved by comparing σ toσ ˜h with h = 2 (for which A = π), progressing exactly as above.

Now to put the pieces together. We take it as obvious that if x(s) lacks a local maximum between its self-intersection (so that σ doubles back on itself), then its 13 surface area will be substantially larger than AT . Thus we consider only cases where the sleeves intersect one another.

Deﬁne A(φ1, φ2) = φ1 sin φ1 + φ2 sin φ2, which tracks the contribution to surface area of two optimal arms reaching heights h1 and −h2, where hi = 2 − 2 cos φi as in the previous lemma. From our discussion of the cap, we know that if σ is not embedded, then h1 + h2 ≥ 2. Once we show that A is minimized for φ1 = φ2 = π/3

(so h1 = h2 = 1), the proposition will be proved, since it is clear that any other choice of φi would necessitate a greater starting x-value for one end of the cap, which we already know can be approximated by a semicircular arc.

Set h1 + h2 = 2d, so 2 − cos φ1 − cos φ2 = d. The method of Lagrange multipliers yields the following system:

  φi cos φi = (λ − 1) sin φi (2.1.13)  2 − cos φ1 − cos φ2 = d

−1 Two solutions are immediately apparent: φi = 0, φj = cos (1 − d), i 6= j. We easily verify that φ1 = φ2 is a third for an appropriate choice of λ and d. To see that this list is exhaustive, set f(t) = (λ − 1) sin t − t cos t for t ∈ [0, π/2]. If λ < 1, f(t) < 0.

Since f 0(t) = (λ − 2) cos t + t sin t, if λ > 2 then f 0(t) > 0. Thus f can have nonzero roots only if λ ∈ (1, 2). For λ in this range, f 00(t) = t cos t + (3 − λ) sin t > 0 and so f has at most one root in (0, π/2). Hence if both φ1 and φ2 are nonzero, they must be equal. When φ1 = φ2, A is clearly least when d = 1, which corresponds to sleeves

−1 of length φi = π/3. We verify directly that A(π/3, π/3) < A(0, cos (1 − d)) = A(cos−1(1 − d), 0), completing the proof of Theorem 2.1.1.

14 2.2 Existence of the Minimal Shape

We have already ﬁnished the heavy lifting. By taking A0 < AT in the deﬁnition of

FA0 , we guarantee that every σ ∈ FA0 must be embedded. Thus we need only show that this family is closed. To begin, we require an upper bound on the length of the

generating curves in FA0 .

A(σ) Lemma 2.2.1. ∀σ ∈ F , L(σ) ≤ + π − 1. A0 2π

0 0 Proof. |κθ| ≤ 1, i.e. |y (s)| ≤ x(s) ∀s. In particular, if x = 0 (for example at a local minimum of x) then x ≥ 1. Thus x(s) < 1 is possible only in neighborhoods of s = 0 and s = L. Let s1 = inf{s ∈ [0,L]: x(s) ≥ 1} and s2 = sup{s ∈ [0,L]: x(s) ≥ 1}, so σ deﬁnes y as a function of x on both [0, s1] and [s2,L]. Let y = f(x) be either of these functions, and let y = g(x) be a unit circle tangent to σ at the origin. Then

|f 0(x)| ≤ |g0(x)| for x ∈ [0, 1], and:

Z s1 Z L Z 1 p 0 2 L(σ|x<1) = ds + ds ≤ 2 1 + (g (x)) dx = π (2.2.1) 0 s2 0

Z Z A(σ) = 2π x(s) ds + x(s) ds (2.2.2) x<1 x≥1 Z ≥ 2π + 2π x ds x≥1

≥ 2π + 2πL(σ|x≥1)

≥ 2π[1 + L(σ) − π] (2.2.3) which proves the lemma.

Now for our ﬁrst major result.

Theorem 2.2.2. For any A0 < AT , there exists an element σmin ∈ FA0 with

V (σmin) ≤ V (σ) ∀σ ∈ FA0 .

15 Proof. Reparameterize each element σ(s) ∈ FA0 with s(t) = L(σ)t to achieve a common domain [0, 1] upon which |σ0(t)| ≡ L(σ). The Lipshitz condition on σ0 then becomes |σ0(b) − σ0(a)| ≤ L(σ)|b − a|, which is certainly preserved under convergence

1 0 in the C norm kσk = kσku +kσ ku, as are the endpoint conditions and the properties

0 |y | ≤ x, and A(σ) = A0. Finally, since A0 < AT , convergence in FA0 must preserve

1 embeddedness. C ([0, 1]) is complete, so FA0 is as well for any A0 < AT .

Let {σk} be a sequence of reparameterized arcs in FA0 with V (σk) & inf{V (σ): A σ ∈ F }. Since |σ0 (b)−σ0 (a)| ≤ L|b−a| and L ≤ 0 +π−1, both {σ } and {σ0 } are A0 k k 2π k k bounded and equicontinuous. By Arzela’s Theorem, {σk} has a subsequence {σki }

1 that is Cauchy with respect to the C norm. Since FA0 is complete, ∃σmin ∈ FA0 with σk ⇒ σmin and V (σk) & V (σmin).

16 CHAPTER 3

CONSTRUCTION OF THE MINIMAL SHAPE: |κθ|, |κσ| ≤ 1

Having established the existence of an optimal ﬁgure, we may now legitimately ask which characteristics such an extremum must possess. Ultimately, we will be able to rule out every shape but one.

The curvature constraint K is said to be binding in a neighborhood U if at least one of the inequalities |κθ| ≤ 1 and |κσ| ≤ 1 turns out to be an equality throughout U. The following theorem establishes that binding is universal for any optimal solution

σ. It is worth noting that the proof below uses the speciﬁc constraint K only as a matter of convenience; there is no reason to think the result will be any diﬀerent when K is altered or weakened.

Theorem 3.0.3. Suppose that σ generates a surfaces of minimal volume among

elements of FA0 . Then the curvature constraint K is binding everywhere.

We again proceed with a series of propositions, beginning with a pair of standard results from the calculus of variations.

3 Proposition 3.0.4. For any relatively compact surface X ∈ R , the ﬁrst variations of enclosed volume V (X) and surface area A(X) are given by: Z Z δ1V (ξ) = ξ dA and δ1A(ξ) = −2 Hξ dA (3.0.1) X X where ξ = hW, ni is the normal component of the variational vector ﬁeld W and n is taken to be the exterior unit normal. 17 Proof. We follow [3] and [10].

3 To establish the volume formula, consider a family of smooth maps Φt :Ω ⊂ R →

3 dΦt R where ∂Ω = X and Φ0 is the identity map. With this convention, W := dt t=0 is the variational vector ﬁelfd on X extended to all of Ω. Viewing Φt as a change of parameter, we obtain:

Z Z Z d d d δ1V = dV = JΦt dV = JΦt dV (3.0.2) dt t=0 Φt(Ω) dt t=0 Ω Ω dt t=0

3 where dV is the standard volume element in R and JΦt is the Jacobian determinant. We now compute: d ∂ ∂Φi ∂ ∂Φi (J ) = t = t (3.0.3) dt Φt ij ∂t ∂xj ∂xj ∂t ∂Φi d ∂W i At t = 0, t = W i, so (J ) = . We now apply the formula ∂t dt Φt ij ∂xj d dM det M = det M· tr M−1 (3.0.4) dt dt which is valid for non-singular M to obtain:

d d ∂W i J = det J · tr J −1 J = tr = div W (3.0.5) Φt Φ0 Φ0 Φt j dt t=0 dt t=0 ∂x

Z Z Z d δ1V = JΦt dV = div W dV = hW, nˆi dA (3.0.6) Ω dt t=0 Ω ∂Ω Z = ξ dA (3.0.7) X as claimed.

We change perspective somewhat to prove the area formula. To begin, extend the variational vector field W smoothly to a neighborhood of X, so Xt = X(u, v) + tW (u, v) is defined for small t, where u and v are local coordinates in U. The metric coefficients of Xt are then given by:

18 ∂X ∂X ∂X ∂W E = t , t = E + 2t , + [t2] t ∂u ∂u ∂u ∂u ∂X ∂X ∂X ∂W ∂X ∂W F = t , t = F + t , + , + [t2] t ∂v ∂u ∂u ∂v ∂v ∂u ∂X ∂X ∂X ∂W G = t , t = G + 2t , + [t2] t ∂v ∂v ∂v ∂v where [t2] represents the nonlinear part of the above coeﬃcients. Now:

2 gt = EtGt − Ft ∂X ∂W ∂X ∂W = EG − F 2 + 2t E , + G , − ∂v ∂v ∂u ∂u ∂X ∂W ∂X ∂W −F , + , + [t2] (3.0.8) ∂u ∂v ∂v ∂u

Since this is a local computation, we may take (u, v) to be isothermal parameters 1 ∂X for the surface X = X. That is, let E = G = λ2 and F = 0 by taking e = 0 1 λ ∂u 1 ∂X and e = . The upshot is: 2 λ ∂v

4 2 2 gt = λ + 2λ t hXu,Wui + hWv,Wvi + [t ] (3.0.9)

T ⊥ T T Write W = W + W = W + ξn, so Wu = Wu + ξun + ξnu, hXu,Wui =

T hXu,Wu i + hXu, ξnui, and

T T hXu,Wui + hXv,Wvi = hXu,Wu i + hXu, ξnui + hXv,Wv i + hXv, ξnvi

T = div W + ξ hXu, nui + hXv, nvi (3.0.10)

1 Note that under isothermal coordinates, mean curvature is given by H = − (hX , n i+ 2λ2 u u

19 Z T hXv, nvi). Moreover, since W is compactly supported, div W dA = 0. Thus: U Z Z d d √ dAt = gt dudv (3.0.11) dt t=0 U U dt t=0 Z 1 d = √ gt dudv (3.0.12) U 2 g dt t=0 Z = ξ (hXu,Wui + hWv,Wvi) dudv (3.0.13) U Z = −2 λ2ξH dudv (3.0.14) U Z = −2 ξH dA (3.0.15) U For compact manifolds, a partition of unity extends the result to variations supported on all of X

The following technical result shows that functions satisfying δ1A(ξ) = 0 necessarily give rise to variations preserving area.

Proposition 3.0.5. Let ξ be a smooth real-valued function with supp(ξ) ⊂ U ⊂ X(D) Z Z and ξH dA = 0 on U. If H dA 6= 0 for some V ⊂ U (so H is not identically U V ∂X zero as a distribution), then there exists a variation X of X with t = ξn which t ∂t preserves surface area for small t. Z Proof. Let µ : U → R be a second smooth function, this one satisfying µH dA 6= 0. U Deﬁne a 2-parameter variation Xs,t = X + (tξ + sµ)n, and notice that ∂A(X ) Z s,t = −2 µH dA 6= 0 (3.0.16) ∂s U

Applying the implicit function theorem to the relation A(Xs,t)−A(X) = 0, we obtain a neighborhood of zero where s is a function of t and surface area is constant. Write Xt = X + tξ + s(t)µ n, a 1-parameter variation of X ﬁxing A for small t. ∂X ∂A(X ) To complete the proof, we need only check that t = ξn. Since s,t = 0, ∂t ∂t

∂A ∂s ∂A + = 0 (3.0.17) ∂s (0,0) ∂t t=0 ∂t (0,0) 20 ∂A R ∂s ∂t (0,0) 2 U ξH dA = − = − = 0 (3.0.18) ∂t ∂A 2 R µH dA t=0 ∂s (0,0) U ∂Xt ∂s = ξ + µ n = ξn (3.0.19) ∂t t=0 ∂t t=0 as desired.

Z Remark Identical logic shows that for any ξ with ξ dA = 0, there exists a U ∂X variation X preserving volume for small t with t = ξn. In this case, the stipulation t ∂t H 6= 0 is unnecessary.

Taken together, the last two propositions imply that if a manifold X has minimal volume for ﬁxed surface area, then for any smooth function ξ with supp(ξ) = U, where U is an open set upon which binding does not occur and H is not zero a.e.,

if δ1A(ξ) = 0 then δ1V (ξ) = 0. Thus any optimal element σmin ∈ FA0 must have constant mean curvature (understood as a distribution) on any open set where binding does not occur and H 6= 0. Moreover, only one such constant is possible. Let us restrict our attention to one neighborhood, write H = H0, and see what conclusions we can draw. We begin by adapting some results from the theory of stable surfaces of constant mean curvature.

Proposition 3.0.6. Let kBk be the norm of the second fundamental form of X, and let all other notation be as before. For any normal variation W = ξn, δ2A(ξ) is given by: Z 2 2 δ2A(ξ) = −ξ∆ξ − kBk ξ dA (3.0.20) U

∂H 2 Proof. In view of Proposition 3.0.4, our goal is to establish 2 = ∆ξ+kBk . As ∂t t=0 before, we view this as a local problem with a neighborhood U ⊂ X parameterized by

2 isothermal coordinates so that hXu,Xvi = 0 and hXu,Xui = hXv,Xvi = λ . Drawing

21 inspiration from Barboso & do Carmo (see [2]), we make use of the following three identities: −1 hD E D Ei H = nu ∧ Xv, n + Xu ∧ nv, n (3.0.21) 2|Xu ∧ Xv|

2 2 2 kBk = 4H − 2(auuavv − auv) (3.0.22)

1 ∂nu ∂nv ∆ξ = − 2 ∧ Xv + Xu ∧ , n (3.0.23) λ ∂t t=0 ∂t t=0 where the aij’s are deﬁned, as usual, by nu = auuXu + auvXv = −(SuuXu + SuvXv) and nv = avuXu + avvXv = −(SvuXu + SvvXv). Using this notation, the right side of the ﬁrst equation is interpreted as follows:

D E D E nu ∧ Xv, n = auuXu ∧ Xv, n = auu|Xu ∧ Xv|

D E D E Xu ∧ nv, n = Xu ∧ avvXv, n = avv|Xu ∧ Xv| 1 Since H = (S + S ), the ﬁrst identity is clear. Note the standard identiﬁcation 2 uu vv 3 of Xu ∧ Xv with the positively-oriented orthogonal vector in R having magnitude

|Xu ∧ Xv|.

2 2 2 2 The second identity is trivial, as kBk = auu + avv + 2auv and 2H = auu + avv, both by deﬁnition. The third requires a bit of work. ∂n D∂n E To begin, note that is a tangent vector, as hn, ni = 1 implies , n = 0. ∂t ∂t Then:

∂n 1 ∂Xu ∂Xv = ∧ Xv + Xu ∧ + (normal components) ∂t |Xu ∧ Xv| ∂t ∂t ∂X ∂X ∂X At t = 0, = ξn, so u = ξ n + ξn and v = ξ n + ξn . Hence: ∂t ∂t u u ∂t v v

∂n 1 h = 2 (ξun + ξauuXu + ξauvXv) ∧ Xv + ∂t t=0 λ i +Xu ∧ (ξvn + ξavuXu + ξavvXv) + normal components 1 1 = (ξ n ∧ X + ξ X ∧ n) = − (ξ X + ξ X ) (3.0.24) λ2 u v v u λ2 u u v v

Since the above is a tangent vector, all normal components have cancelled by this last line. Here we have used the identiﬁcations n ∧ Xv = −Xu and Xu ∧ n = −Xv. Now, at t = 0:

∂n 1 u = − (ξ X + ξ X + ξ X + ξ X ) ∂t λ2 uu u u uu vu v v vu 1 = − (ξ X + ξ X ) + (normal components) (3.0.25) λ2 uu u vu v ∂n ξ u ∧ X = − uu X ∧ X + (tangential components) ∂t v λ2 u v

= −ξuun + (tangential components) (3.0.26)

∂n Similarly, X ∧ v = −ξ n + (tangential components). Putting everything to- u ∂t vv gether, we get our third identity:

1 ∂nu ∂nv 1 − 2 ∧ Xv + Xu ∧ , n = − 2 (−ξuu − ξvv) = ∆ξ λ ∂t t=0 ∂t t=0 λ

We are ﬁnally ready to prove the proposition.

∂H 1 D ∂Xv ∂Xu E 2 = − nu ∧ + ∧ nv, n + ∂t |Xu ∧ Xv| ∂t ∂t D∂n ∂n E D ∂nE + u ∧ X + X ∧ v , n + n ∧ X + X ∧ n , + ∂t v u ∂t u v u v ∂t 1 1 D E ∂ D E + 3 nu ∧ Xv + Xu ∧ nv, n Xu ∧ Xv,Xu ∧ Xv 2 |Xu ∧ Xv| ∂t

23 where

∂ D E D∂X ∂X E X ∧ X ,X ∧ X = 2 u ∧ X + X ∧ v ,X ∧ X ∂t u v u v ∂t v u ∂t u v D E = 2 (ξun + ξnu) ∧ Xv + Xu ∧ (ξvn + ξnv) ,Xu ∧ Xv D E = 2 ξauuXu ∧ Xv + ξavvXu ∧ Xv,Xu ∧ Xv

2 = 2ξ(auu + avv)|Xu ∧ Xv|

2 = −4ξH|Xu ∧ Xv| . (3.0.27)

This implies that

∂H 1 hD Ei 2 = − nu ∧ (ξvn + ξnv) + (ξun + ξnu) ∧ nv, n + ∂t |Xu ∧ Xv| 2ξH D E +∆ξ − nu ∧ Xv + Xu ∧ nv, n (3.0.28) |Xu ∧ Xv| 1 D E = − n ∧ ξn + ξn ∧ n , n + ∆ξ + 4ξH2 λ2 u v u v 2ξ D E = − (a X + a X ) ∧ (a X + a X ), n + ∆ξ + 4ξH2 λ2 uu u uv v uv u vv v 2ξ D E = − (a a − a2 )X ∧ X , n + ∆ξ + 4ξH2 λ2 uu vv uv u v 2 2 = −2ξ(auuavv − auv) + ∆ξ + 4ξH

= ∆ξ + kBk2ξ (3.0.29)

Proposition 3.0.6 also holds for non-orthogonal variations of X, though we shall have no need of such a result. See [2] for the messy details.

Proposition 3.0.7. For any orthogonal variation W = ξn, if H = H0 is constant on U ⊂ X and 1 Z kBk2 dA < 2π (3.0.30) 2 U then δ2A(ξ) > 0 on U.

24 This proposition, essentially due to Ruchert [11], can be interpreted to mean that

3 surfaces of constant mean curvature in R , like minimal surfaces, are stable in the small. His proof apparently exists only in German, so we reconstruct it here.

2 Proof. We continue to assume isothermal coordinates on U, so gij = δijλ . Define Z Z 1 2 ˆ 1 2 a new metricg îj := kBk gij, and note that dA = kBk dA. We must now 2 U 2 U show that on U, the Gaussian curvature Kˆ satisfies Kˆ ≤ 1:

1 1 1 kBk2Kˆ = − ∆ log kBkλ 2 λ2 2 1 1 1 = − ∆ log λ − ∆ log kBk λ2 λ2 2 1 = K − ∆ log kBk2 (3.0.31) 2λ2 1 Writing w = kBk2 and making use of complex coordinates, this yields: 2 2 w3Kˆ = w2K + (w w − ww ) (3.0.32) λ2 z z zz

Let L = hXuu, ni,M = hXuv, ni, and N = hXvv, ni be the coeﬃcients of the second fundamental form, as usual. It is well-known (and easily veriﬁable) that the function 1 Ω = (L − N) − iM is holomorphic whenever X is a surface of constant mean 2 curvature parameterized with isothermal coodinates, as is the case here. Since |Ω| = 1 1 1 (L2 − 2LN + N 2) + M 2, w = (L2 + 2M 2 + N 2), and H = (L + N) we have: 4 2λ4 0 2λ2 |Ω|2 w = + H2 = λ−4ΩΩ + H2 (3.0.33) λ4 0 0 −5 2 −4 wz = −4λ λz|Ω| + λ ΩzΩ

−5 2 −4 wz = −4λ λz|Ω| + λ ΩΩz

−6 2 −5 2 −5 −5 −4 wzz = 20λ λzλz|Ω| − 4λ λzz|Ω| − 4λ λzΩΩz − 4λ λzΩzΩ + λ ΩzΩz

25 and hence:

2 w3Kˆ = w2K + (w w − ww ) (3.0.34) λ2 z z zz 2 16λ λ |Ω|4 4λ Ω Ω|Ω|2 4λ ΩΩ |Ω|2 |Ω|2|Ω |2 = w2K + z z − z z − z z + z − λ2 λ10 λ9 λ9 λ8 20λ λ |Ω|2w 4λ |Ω|2w 4λ ΩΩ w 4λ Ω Ωw Ω Ω w − z z + zz + z z + z z − z z λ6 λ5 λ5 λ5 λ4 |Ω|2 16λ λ |Ω|2 4λ ΩΩ 4λ Ω Ω |Ω |2 = w2K − 2 w − z z − z z − z z + z − λ4 λ8 λ7 λ7 λ6 8w|Ω|2 − (λ λ − λλ ) λ8 z z zz 2H 16λ λ |Ω|2 4λ ΩΩ 4λ ΩΩ = w2K − 0 z z − z z − z z + |Ω |2 − λ6 λ2 λ λ z 2w|Ω|2 4 − · (λ λ − λλ ) λ4 λ4 z z zz 2 2 2|Ω| 2H0 4λzΩ = wK w − − − Ωz (3.0.35) λ4 λ6 λ 2 2 2 2H0 4λzΩ = w(2H − w) − − Ωz 0 λ6 λ 2 2 3 2 |Ω| 2H0 4λzΩ = w − 4H w − − Ωz 0 λ4 λ6 λ ≤ w3 which proves that Kˆ ≤ 1.

We now require a comparison result of Barbosa & do Carmo:

Lemma 3.0.8. Let D ⊂ M be simply connected and let the Gaussian curvature K of

∗ ∗ M satisfy K ≤ K0. Let D be a geodesic disk having A(D ) = A(D) and contained

∗ in a surface of constant curvature K0. Then λ1(D) ≥ λ1(D ), where λ1 is the ﬁrst eigenvalue of the Laplacian on the appropriate domain.

Proof. See [1].

In the present context, this implies that since Kˆ (U) ≤ K(S2), we must have

2 λ1(U) ≥ λ1(W ), where W ⊂ S satisﬁes A(W ) = A(U). Here the requirement that 26 Z 1 Z dAˆ = kBk2 dA < 2π is essential, allowing us to compare to the unit half- U 2 U sphere. It is standard that λ1(W ) ≥ 4π/A > 2 whenever A(W ) < 2π (for a proof, see [1]), and that: R ˆ 2 ˆ U |∇ξ| dA λ1(U) ≤ (3.0.36) R 2 ˆ U ξ dA ˆ with equality exactly when ξ is an eigenfunction for λ1 . Here ∇ is just the gradient with respect to metricg ˆij. Putting this all together:

R ˆ 2 ˆ U |∇ξ| dA 2 < λ1(W ) ≤ λ1(U) ≤ (3.0.37) R 2 ˆ U ξ dA Z |∆ˆ ξ|2 − 2ξ2 dAˆ > 0 U Z −ξ∆ξ − kBk2ξ2 dA > 0 (3.0.38) U from which we conclude that δ2A(X) > 0.

The significance of Proposition 3.0.7 in the present context is that on a sufficiently small neighborhood of X where binding does not occur, any arbitrary variation fixing Z V (X) must increase A(X). By taking any nontrivial ξ with ξ dA = 0 and applying U the remark following Propostion 3.0.5, we may always obtain a variation fixing V .

It suffices, then, to find another variation somewhere on X that reduces both A(X) and V (X). A glance at the first variational formulas in Proposition 3.0.4 makes it clear that if X includes any neighborhood where binding does not occur and H0 < 0, then any ξ < 0 at all will do the trick. Henceforth let us assume that no such neighborhoods exists.

See ﬁgure 3.1. The idea is to locate and then delete a pair of neighborhoods of the sort shown, translating the loose ends inward and reconnecting them in a C1 fashion.

Such surgery reduces both A(X) and V (X). We will need the following two lemmas:

27 Figure 3.1: The neighborhood to be deleted

0 Lemma 3.0.9. Let σ ∈ FA0 have x(0) = x (0) = 0 and x(b) = 1, where b ≤ π/2.

0 If there exists an s3 ∈ (0, b) with |y (s3)| > x(s3) then there also exist s1 and s2 with

0 0 0 ≤ s1 < s2 ≤ s3 such that |σ (s2) − σ (s1)| > s2 − s1. That is, if σ violates the constraint on rotational curvature κθ at s3, then it also violates the constraint on average plane curvature somewhere before s3.

Proof. If σ does not violate the constraint on average plane curvature on [0, s3], then

0 0 0 0 the Dubins Lemma applies, and hσ (0), σ (s)i ≥ hσ˜ (0), σ˜ (s)i for all s ∈ [0, s3], where σ˜ is a unit circle withx ˜(0) = 0 andx ˜0(0) = 1. In particular, x0(s) ≥ x˜0(s), x(s) ≥ x˜(s),

0 0 and |y (s)| ≤ |y˜ (s)| for all s ≤ s3, implying that at s3, |κθ(σ)| ≤ |κθ(˜σ)| = 1. Thus σ does not violate the constraint on rotational curvature at s3, proving the lemma.

Lemma 3.0.10. Let σ ∈ K. If there exist values s = a and s = b for which

|σ0(b) − σ0(a)| = |b − a|, that is, for which the constraint on average plane curvature is binding, then in fact it is binding on every subinterval [c, d] ⊂ [a, b].

Proof. Since σ ∈ K, |σ0(d) − σ0(c)| ≤ d − c for any [c, d] ⊂ [a, b]. Suppose this inequality is strict on [c, d]. Then:

b − a = |σ0(b) − σ0(a)| ≤ |σ0(b) − σ0(d)| + |σ0(d) − σ0(c)| + |σ0(c) − σ0(a)|

≤ |σ0(b) − σ0(d)| + (d − c) + |σ0(c) − σ0(a)| (3.0.39)

(b − d) + (c − a) < |σ0(b) − σ0(d)| + |σ0(c) − σ0(a)| (3.0.40)

28 which implies that either (b − d) < |σ0(b) − σ0(d)| or (c − a) < |σ0(c) − σ0(a)|. This contradiction proves the lemma.

Lemma 3.0.9 implies that when binding occurs anywhere other than along the line x = 1, it must due to the constraint on average plane curvature. In such a case, there exist a and b with |σ0(b) − σ0(a)| = b − a. Lemma 3.0.10 then implies that this binding obtains throughout the entire interval [a, b]. Thus σ consists of countably many intervals with either |κσ| = 1, H = H0, H = 0 or x = 1. In particular, we can deal with the curve piecewise and ignore the distinction between average plane curvature and κ.

Let σ1 and σ2 be the portions of σ extending from the axis of rotation to the nearest vertical tangent. On each of these arcs, we seek to locate a neighborhoods like the one in Figure 3.1, i.e., one in which κσ changes sign (or is identically zero) and in which the beginning and ending tangents agree. Let us call such neighborhoods pinwheels. Deleting such pinwheels and reconnecting the arcs in the obvious way provides the desired reductions in both volume and surface area.

Suppose the upper arc σ1 does not have a pinwheel neighborhood. In between its π horizontal and vertical tangents, κ (a, b) = − , so κ < 0 on [a, b]. Since we σ 2(b − a) also have κθ ≤ 0 and H0 > 0, there is no possibility that H = H0 or H = 0. Thus

κ ≡ −1 and σ1 is just a quarter-circle.

Identical logic applies to σ2. Each of the two arcs, therefore, is either part of a unit circle or includes a pinwheel. We immediately rule out the possibility that σ includes only one pinwheel total, for in such a case the arc on which it lay would not be able to connect to its counterpart’s vertical end at x = 1. We also dismiss as degenerate the case where σ is a semicircle, for then A0 = 4π and the standard

isoperimetric inequality implies that σ is the sole member of FA0 . Deleting pinwheel neighborhoods (which can be taken as small as we like) on both

29 σ1 and σ2 and reconnecting the curve by shifting the remaining arcs inward yields a new arc which generates less surface area and volume than the original. This is most easily seen via the formulations Z Z V (X) = π x2(s)y0(s) ds and A(X) = 2π x(s) ds (3.0.41) σ σ where we notice that for corresponding s-values, the above surgery reduces x and

ﬁxes y0.

A small diﬃculty arises in the case where the condition |κθ| ≤ 1 is binding somewhere along the arc joining σ1 and σ2, for then an inward shift would increase |κθ| beyond the permissible threshhold. Let σ3 be that connecting arc. Since σ3 does not

0 connect to the axis of rotation, κθ can only be binding when x = 1 and x = 0. Be- tween such a point and the local x-maxima where σ3 ends, we are guaranteed of points where |x0| reaches local maxima. These provide additional pinwheel neighborhoods.

Deleting them relieves the need to shift all of σ3 inward. Thus unless σ generates a unit sphere, there is always a surgery to reduce A and

V by arbitrarily small amounts. When combined with the previously constructed variation reducing V and increasing A, this eliminates the possibility that the optimal surface X includes any neighborhood where binding does not occur. Thus the generating arc consists entirely of arcs with |κσ| = 1 or x = 1. This completes the proof of Theorem 3.0.3.

The problem now is simply to piece these arcs together in the most eﬃcient manner possible.

Theorem 3.0.11. The arc σmin ∈ FA0 that generates the surface X of minimal volume consists of two unit quarter-circles connected by a vertical line segment. Thus

X has the shape of a hot dog. 30 Figure 3.2: The optimal σmin

We will prove this by performing a sequence of surgeries, each reducing both V and

A in such a way that 2∆V > ∆A. Such surgeries will be called beneficial. Eventually, we will arrive at a figure with a vertical tangent at x = 1 that we may stretch vertically to restore the original surface area. This final operation has 2∆V = ∆A, implying an overall reduction in volume.

We begin by eliminating a key sort of joint:

Proposition 3.0.12. Consider an arc σ1 consisting of a pinwheel whose convex part is closer to the axis of rotation, with center at x = a > cos θ, central declination

θ, and half-length β (see Figure 3.3, where the interior of X lies to the left of each arc). Replacing σ1 with a concave-convex pinwheel σ2 with half-length α chosen to preserve ∆x, and then reconnecting the remaining arcs in the obvious way results in a beneﬁcial reduction of surface area and volume.

The restriction a > cos θ amounts only to a requirement that σ1 not connect directly to the axis of rotation. That situation is dealt with separately in Proposition

3.0.14.

Proof. This is a direct calculation. Note that since the alterations in question do

0 not alter x(s) or y (s) away from σ1, the changes to V and A depend entirely on

31 Θ

Α x=a

Figure 3.3: Swapping pinwheels while preserving ∆x

the eﬀect within the neighborhood. Also note that we may consider arbitrarily small neighborhoods, so it is the case ∆x → 0 that concerns us.

1 ∆x = cos θ − cos(θ + β) = cos(θ + β) − cos(θ + β + α) 2

2 cos(θ + β) = cos θ + cos(θ + β + α) (3.0.42) dα 2 sin(θ + β) − sin(θ + β + α) = dβ sin(θ + β + α)

dα = 1 (3.0.43) dβ ∆x=0 Here we assume that θ 6= 0 and that α, β are small enough so that the above derivative is well-defined. In particular, near β = 0 we may write α = α(β). Fixing θ and a and defining ∆A(α(β), β) = A(σ1) − A(σ2), we find ∆A = 4πa(β − α), and: d sin(θ + β) − sin(θ + β + α) d ∆A = 8πa hence = 0 dβ sin(θ + β + α) dβ ∆x=0

d2 2 sin2(θ + β) cos(θ + β + α) − sin2(θ + β + α) cos(θ + β) ∆A = 8πa dβ2 sin3(θ + β + α)

2 d Hence 2 ∆A = 8πa cot θ. dβ ∆x=0

32 Z Z Similarly, deﬁne ∆V = x2y0 ds− x2y0 ds. A long but elementary calculation σ1 σ2 then yields

2 ∂ ∂ 2 ∆V = 0 and 2 ∆V = 4πa csc θ (3.0.44) ∂β β=0 ∂ β β=0 from whence we conclude lim(2∆V − ∆A) = 8π cot θ(a sec θ − 1). We have assumed β→0 that sec θ > a−1, so 2∆V − ∆A > 0 for suﬃciently small β.

Proposition 3.0.12 imlpies, in particular, that any portion of σ connecting a local minimum of x(s) to a local maximum can consist only of a vertical pinwheel; otherwise it includes neighborhoods of the above type which can all be beneﬁcially replaced.

We next establish that such vertical pinwheels can simply be removed. Note that because of the threshhold AT , these can occur only where rotational curvature κθ is negative.

Figure 3.4: Connecting a local minimum

Proposition 3.0.13. Removing a vertical pinwheel of the type in Figure 3.4 is ben- eﬁcial, i.e. 2∆V > ∆A.

The surgery in question destroys continuity, of course. This will be recovered shortly.

33 Proof. As in the previous proposition, we take the center of the pinwheel to lie at x = a and denote its half-length by β. Another direct calculation yields:

A = 4πaβ (3.0.45) 2 V = π 2a2 sin β + 2 sin β − 2β cos β − sin3 β (3.0.46) 3 1 1 (2V − A) = a2 sin β − aβ + sin β − β cos β − sin3 β (3.0.47) 4π 3 Claim that each of the two grouped quantities in line 3.0.47 is positive. In the ﬁrst case, the constraint on rotational curvature κθ implies a ≥ 2 − cos β and hence:

a sin β − β ≥ 2 sin β − sin β cos β − β (3.0.48)

Call the right hand side of this inequality g(β). Then g(0) = 0 and g0(β) = 2 cos β(1− cos β) > 0 on (0, π/2), so g(β) ≥ 0 there.

Figure 3.5: Minimal area when β = π/2

Similar logic applies to the second quantity above: it is zero at β = 0 and increasing on (0, π/2). This proves the proposition for β < π/2.

34 Among arcs having a β = π/2 vertical pinwheel, the one with minimal area includes only an additional pair of unit quarter circles and a straight horizontal to the axis of rotation (see Figure 3.5), a fact which can easily be proved using the methods of Theorem 2.1.1. For this curve, A = 2π(4 + 3π) > AT , implying that β ≥ π/2 is impossible.

We are thus able to remove all sections of the cap between the ﬁrst and last vertical tangents. At present, these may not connect, however, and so we continue to deal with them individually.

Proposition 3.0.12 implies that there can be at most one concavity between each of these maxima and the axis of rotation (since any convex arc lying to the left of a concave arc must connect directly to the axis of rotation). The next two propositions show that a beneﬁcial surgery is possible in each possible case.

Proposition 3.0.14. Replacing the convex-concave-convex curve σ1 of Figure 3.6 with the quarter-circle σ2 is beneﬁcial.

Σ2

Σ1

Figure 3.6: Yet another simpliﬁction

35 Proof. Noting that 0 < θ + β < π/2, we compute:

A1 = 2π 2β cos θ + sin θ + 2(θ + β)(cos θ − cos(θ + β)) (3.0.49)

2 2 V1 = π 8 cos θ sin(θ + β) − 2 cos θ sin θ − sin θ + 2 sin(θ + β) − 2β cos θ +

+2(θ + β)(cos θ − cos(θ + β)) − 8 cos θ cos(θ + β) sin(θ + β) + 1 2 + sin3 θ − sin3(θ + β) + 2 cos2(θ + β) sin(θ + β) (3.0.50) 3 3 π A = 2π sin θ and V = π sin θ − sin3 θ (3.0.51) 2 2 3

Let ∆V = V1 − V2 and ∆A = A1 − A2. Then:

2∆V − ∆A = 2π 8 cos2 θ sin(θ + β) − 2 cos2 θ sin θ − 2 sin θ + 2 sin(θ + β) − 2 −4β cos θ − 8 cos θ cos(θ + β) sin(θ + β) + sin3 θ − 3 2 − sin3(θ + β) + 2 cos2(θ + β) sin(θ + β) 3 2 2 (2∆V − ∆A)β = 4π 4 cos θ cos(θ + β) + cos(θ + β) − 2 cos θ + 4 cos θ sin (θ + β) −

−4 cos θ cos2(θ + β) + cos(θ + β) − cos(θ + β) sin2(θ + β) − −3 cos(θ + β) sin2(θ + β) = 8π cos θ − cos(θ + β) 2 cos θ cos(θ + β) + 2 sin2(θ + β) − 1

≥ 0

For β = 0, 2∆V − ∆A = 0, so for β ≥ 0 we conclude 2∆V − ∆A ≥ 0.

Proposition 3.0.15. Replacing the concave-convex shape pictured with a single quarter- circle yields 2∆V − ∆A > 0.

Proof. σ1 consists of a horizontal pinwheel of half-length β and a unit quart-circle.

36 Figure 3.7: The last simpliﬁcation

Once more we compute:

A1 = 2π 1 + (π + 2β) sin β (3.0.52) 2 2 V = π − 2 cos β + cos3 β − 2 cos β sin2 β − (π + 2β) sin β) (3.0.53) 1 3 3

A2 = 2π (3.0.54) 2 V = π (3.0.55) 2 3 4 2 2∆V − ∆A = 2π − + 2 cos β − cos3 β + 2 cos β sin3 β := h(β) (3.0.56) 3 3

Noting that β ∈ [0, π/3) by the threshhold AT , we observe that h(0) = 0, h(π/3) = 2π/3, and

h0(β) = 4 sin β(2 cos2 β − 1) (3.0.57) which implies that h lacks a local minimum on the interval in question and hence must remain positive there. This proves Proposition 3.0.15.

It is clear that h(β) ≥ 0 for values of β well beyond π/3, so the conclusion holds for some trans-threshhold values of A0. In such cases, the hot dog of Figure 3.2 is still optimal even when compared to non-embedded shapes, so long as volume is viewed as a functional rather than a physical property.

Taken together, the above propositions reduce σ to a semicircle while reducing volume and surface area in a ratio of better than one to two. This shape has a vertical 37 tangent at x = 1 which can be extended by length l to restore σ to the original surface area A0. The resulting changes to V and A are

∆V = πl and ∆A = 2πl

Hence 2∆V = ∆A. Overall, we have reduced V while maintaining A. This completes the proof of Theorem 3.0.11.

Remark Although Theorem 3.0.11 deals only with the question of minimizing volume for ﬁxed surface area, it essentially solves the dual problem of maximizing area for ﬁxed volume V0 as well. In such a case, the threshhold curve σT remains unchanged, a fact which can be easily proved with the methods of Theorem 2.1.1. The variational arguments in Propositions 3.0.4 and 3.0.6 then imply that for an extremal curve σmax with V (σmax) = V0 < V (σT ), binding must occur everywhere. Let A0 be the surface area of the hot dog with this V0. If σ is not itself a hot dog, then since it must include a pair of pinwheels, the arguments of section 4 provide a way of reducing both A and V until either A = A0 and V < V0 (contradicting Theorem 3.0.11), or there are no longer any pinwheels remaining to shorten. In the latter case, we again arrive at a hot dog and the same contradiction.

The minimal volume problem considered here is preferable to its dual only in its potential for generalization. In particular, when we consider a bound on mean curvature rather than principal curvatures, there is no surface of maximal area for any ﬁxed volume V0 > 0. The reason is illustrated in Figure 3.8. This shape can be pinched in and extended at vertically near x = 1 as much as desired.

In fact, a similar issue occurs in the principal curvature case of the maximal area problem, but only for large trans-threshhold values of V0. See Figure 3.9. A pinched

38 Figure 3.8: Nearing a hollow sphere

neck tangent to the line x = 1 (point P in the ﬁgure) can be extended indeﬁnitely without adding any volume.

Figure 3.9: Adding area without adding volume

39 CHAPTER 4

EXISTENCE: |H| ≤ 1

4.1 Formulation of the problem

3 We now wish to consider rotationally-symmetric surfaces in R with mean curvature H bounded by one. As in the previous case, we must allow for non-C2 surfaces, so

H should be interpreted in the weak sense. Speciﬁcally, we continue to require that

00 the generating curve σ have absolutely continuous tangent vector, so that σ and κσ Z t Z t 00 0 0 are deﬁned a.e with σ (τ) dτ = σ (t) − σ (s) and κσ(τ) dτ = ϑ(t) − ϑ(s), where s s ϑ is again the angle that σ0 makes with the positive x-axis, measured continuously from s = 0. We then deﬁne the average mean curvature of σ on [s, t] as:

1 Z t 1 Z t H(s, t) = H(τ) dτ = (κσ + κθ) dτ (4.1.1) t − s s 2(t − s) s

0 00 00 0 0 When σ is unit speed, we may orient σ so that κσ = x y − x y and κθ = y /x to get the following convenient formulations:

1 Z t y0 H(s, t) = x0y00 − x00y0 + dτ (4.1.2) 2(t − s) s x 1 Z t y0 = ϑ(t) − ϑ(s) + dτ (4.1.3) 2(t − s) s x We are now ready to set the scope of our problem. Let

n 2 1 0 o K = σ(s) : [0,L(σ)] ,→ R+, σ ∈ L2, |σ | ≡ 1, |H(s, t)| ≤ 1 ∀s, t ∈ [0,L] Z L n 0 0 o GA0 = σ ∈ K : x(0) = x(L) = y(0) = y (0) = y (L) = 0, 2π x(s)ds = A0 0 40 2 2 1 1 where R+ = {(x, y) ∈ R : x ≥ 0} and L2 is the space of C maps with absolutely continuous tangent vectors. Thus GA0 consists of generating arcs for rotationally- symmetric surfaces with prescribed surface area A0 and average mean curvature

bounded everywhere by 1. By further stipulating that y(0) > y(L) ∀σ ∈ GA0 , we indirectly specify that the sphere has negative principal curvatures.

Our aim is to minimize volume V (σ) among embedded elements of GA0 . As

was the case for FA0 , this problem is only well-deﬁned if we limit the size of A0.

Otherwise there may well be a sequence {σi} with decreasing volume converging to a non-embedded limit. In particular, for A0 ≥ 8π, we can have shapes like the one in Figure 3.8, where the point at which x(s) takes its minimum can be drawn arbitrarily close to the axis (increasing κθ while making κσ very negative) so that σ approaches a pair of overlapping semicircles.

We will rule out such cases by ﬁnding a non-embedded element σT ∈ K with x(0) = x(L) = 0 and y0(0) = y0(L) = 0 of least surface area. In particular, we show that A ≥ AT = 8π for any such non-embedded ﬁgure, thereby establishing that the hollow sphere described above is such a σT . The following proposition is interesting in its own right.

Proposition 4.1.1. Suppose h(s) : [0, b] → R is Lipshitz continuous, and let (x0, y0) ∈ 2 0 0 1 R and (x0, y0) ∈ S be ﬁxed. Then there is a unique unit-speed curve σ(s) = 2 1 (x(s), y(s)) : [0, b] → R+ in L2 that generates a surface with average mean curva- 0 0 0 ture H(0, s) = h(s) ∀s ∈ [0, b] and which has σ(0) = (x0, y0) and σ (0) = (x0, y0).

Proof. To begin, we restrict our attention to σ ∈ C2 and consider the diﬀerential equation

2H(s)x(s) = y00(s)x0(s)x(s) − y0(s)x00(s)x(s) + y0(s) (4.1.4) where H(s) is continuous, (x0)2 + (y0)2 = 1, and σ(0) and σ0(0) are as speciﬁed.

41 Following Kenmotsu [10], we deﬁne Z(s) = x(s)(x0(s) + iy0(s)) and reduce (4.1.4) to the elementary ﬁrst order linear equation

Z0(s) − 2iH(s)Z(s) − 1 = 0 (4.1.5)

The general solution is most conveniently written as

Z(s) = (F (s) + iG(s) + c)(F 0(s) − iG0(s)) (4.1.6)

where c = c1 + ic2 is a complex constant and Z s Z u F (s) := sin 2H(t) dt du (4.1.7) 0 0 Z s Z u G(s) := cos 2H(t) dt du (4.1.8) 0 0 Since |Z(s)| = x(s), the general solution to (4.1.4) is

p 2 2 x(s) = (F (s) + c1) + (G(s) + c2) (4.1.9) Z s (G(t) + c )F 0(t) − (F (t) + c )G0(t) y(s) = 2 1 dt (4.1.10) 0 x(t) 0 0 Application of the given initial conditions yields c1 = −x0y0 and c2 = x0x0. This

0 formulation applies even when x0 = 0, provided of course that x0 = 1 as required by the bound on average mean curvature.

Note that this solution makes sense even when H(s) is merely L1. In fact, H ∈ L1

1 whenever σ ∈ L2, which is the regularity required by the lemma. It therefore suffices to confirm that σ(s) = (x(s), y(s)) as defined in lines (4.1.9) and (4.1.10) remains valid and unique for h(s) as in the statement of the lemma. Specifically, we check that σ satisfies the integral equation: 1 Z s y0(t) h(s) = ϑ(s) − ϑ(0) + dt (4.1.11) 2s 0 x(t) Since h(s) is Lipshitz, there exists an essentially bounded function H(s) ∈ L1 defined Z s 1 1 a.e on [0,L(σ)] such that h(s) = H(t) dt. Similarly, since σ ∈ L2, ϑ(s) is 2s 0 differentiable a.e allowing us write the weak differential equation 42 Z s y0 0 = 2H − y00x0 − y0x00 + dt ∀s ∈ [0,L(σ)] 0 x Or, equivalently.

Z s 0 = (2Hx − y00x0x − y0x00x + y0) dt ∀s ∈ [0,L(σ)] (4.1.12) 0 which is obviously satisﬁed by σ(s) as deﬁned in lines (4.1.9) and (4.1.10). To prove uniqueness, take Z(s) as before to get a Volterra integral equation:

Z s 0 = (Z0(t) − 2iH(t)Z(t) − 1) dt ∀s ∈ [0,L(σ)] (4.1.13) 0 Z s Z(s) = Z(0) + 2i (H(t)Z(t) + 1) dt ∀s ∈ [0,L(σ)] (4.1.14) 0

The rest is standard (see for example chapter 2 of [7]). We consider the Banach

1 space X of bounded C functions from [0,L(σ)] to C with the norm

kϕk := max e−3Ms|ϕ(s)| (4.1.15) [0,l(σ)] where M = kHk∞. This is equivalent to the standard sup norm on BC([0,L(σ)]). Deﬁne a linear transformation T on X by

Z s T (ϕ)(s) := Z(0) + 2i (H(t)ϕ(t) + 1) dt 0

For any ϕ, ψ ∈ X and s ∈ [0,L(σ)], we have:

Z s

|T (ϕ)(s) − T (ψ)(s)| = 2H(t)(ϕ(t) − ψ(t)) dt 0 Z s Z s ≤ 2M |ϕ(t) − ψ(t)| dt ≤ 2M e3Mtkϕ − ψk dt 0 0 2 2 = (e3Ms − 1)kϕ − ψk ≤ (e3ML(σ) − 1)kϕ − ψk 3 3

Hence,

2 2 e−3ML(σ)|T (ϕ)(s) − T (ψ)(s)| ≤ (1 − e−3ML(σ))kϕ − ψk ≤ kϕ − ψk 3 3 43 This holds for all s ∈ [0,L(σ)], including the one maximizing the left hand side. In short, 2 kT (ϕ)(s) − T (ψ)(s)k ≤ kϕ − ψk 3 The Banach Fixed Point Theorem implies that T has exactly one ﬁxed point in X.

1 Thus in the smaller class L2, the known solution

p 0 2 0 2 x(s) = (F (s) − x0y0) + (G(s) + x0x0) (4.1.16) Z s (G(t) + x x0 )F 0(t) − (F (t) − x y0 )G0(t) y(s) = 0 0 0 0 dt (4.1.17) 0 x(t) is unique.

This analog of the Fundamental Theorem of Plane Curves allows us to reconstruct any plane curve from its starting element and average mean curvature h(s), provided only that h(s) is diﬀerentiable a.e with h0(s) bounded. In the present context, it is the explicit representation of σ given in (4.1.16) and (4.1.17) that is most signiﬁcant.

Lemma 4.1.2 (Cap Lemma). Let σ˜ be the unit-speed H ≡ −1 arc beginning at

0 0 0 0 x˜(0) = x0 ≥ 0 and having initial tangent vector σ˜ (0) = (x0, y0), where x0 > 0 and

0 y0 ≤ 0 (so that κθ ≤ 0), and let σ be any other element of K with the same initial conditions. Then x0(s) ≥ x˜0(s) (hence also y0(s) ≥ y˜0(s)) ∀s ∈ [0, π/4].

Proof. It suﬃces to show (x2)0 ≥ (˜x2)0 on [0, π/4]. Using (4.1.16), we obtain

2 0 0 0 0 0 (x ) = 2(F − x0y0)F + 2(G + x0x0)G (4.1.18)

0 0 0 0 0 0 = 2(FF + GG ) + 2x0(x0G − y0F ) (4.1.19)

0 0 where x0 > 0, x0 ≥ 0 and y0 ≤ 0. Since σ ∈ K, we have H ≥ −1, which implies

0 ≥ F (s) ≥ F˜(s) 0 ≥ F 0(s) ≥ F˜0(s) ∀s ∈ [0, π/4] (4.1.20)

G(s) ≥ G˜(s) ≥ 0 G0(s) ≥ G˜0(s) ≥ 0 ∀s ∈ [0, π/2] (4.1.21) 44 0 0 0 0 0 ˜0 0 ˜0 Clearly x0G − y0F ≥ x0G − y0F , so the second summand in line (4.1.19) is greater for σ than forσ ˜ on [0, π/4]. We need only prove a similar result for the ﬁrst summand, which can be succinctly written as 2h(F,G), (F 0,G0)i.

Formulas (4.1.7) and (4.1.8) are suggestive. Let us deﬁne γ(s) = (F (s),G(s)) andγ ˜ = (F˜(s), G˜(s)). In the (F,G)-plane,γ ˜ is a circle of radius 1/2, while γ is a

C1 curve with absolutely continuous tangent vector and average curvature bounded everywhere by 2. For any given s ∈ [0, π/4], orient γ andγ ˜ so that γ(s) =γ ˜(s) =

(0, 0) and γ0(s) =γ ˜0(s) = ∂/∂y. This rigid motion shifts γ(0) to (−F (s)G0(s) +

G(s)F 0(s), −F (s)F 0(s) − G(s)G0(s)). The Dubins Lemma applies and says that the y-value of γ is less than the y-value ofγ ˜ for s ∈ [0, π/2]. That is,

−F (s)F 0(s) − G(s)G0(s) ≤ −F˜(s)F˜0(s) − G˜(s)G˜0(s)) D E D E (F (s),G(s)), (F 0(s),G0(s)) ≥ (F˜(s), G˜(s)), (F˜0(s), G˜0(s))

which proves the lemma. Note the essential hypothesis is that onσ ˜, both κr and

κθ have the same sign. As the proof suggests, this condition can be removed at the expense of the uniform interval s ∈ [0, π/4].

In order to make future calculations less unwieldy, we will frequently consider Z b Z b surface area energy E = x2 ds rather than surface area A = 2π x ds. Of a a course, Z b 2 2 E(σ1) − E(σ2) > 0 =⇒ (x1 − x2) ds > 0 a Z b =⇒ inf{x1 + x2} (x1 − x2) ds > 0 a Z b =⇒ (x1 − x2) ds > 0 a

=⇒ A(σ1) − A(σ2) > 0 so E(σ1) > E(σ2) implies that A(σ1) > A(σ2). This fact simpliﬁes the following corollary substantially. 45 Corollary 4.1.3. Let σ be the unit-speed H ≡ −1 cap beginning at x˜(0) = x0 ≥ 0 (so x˜0(0) = 1) and let σ be any other cap in K with the same starting x-value. Then

A(σ) ≥ A(˜σ). That is, the cap of least area for ﬁxed starting x0 consists of an arc of an H ≡ −1 nodary.

Figure 4.1: Cap comparison

Proof. The Cap Lemma implies that A(σ) ≥ A(˜σ) for the half-cap beginning at x0, and that σ reaches its vertical tangent at a greater x-value thanσ ˜. See Figure 4.1.

We now extend this result to the entire cap.

Write xend for the x-value of the endpoint of σ. Obviously, if xend = x0, the Cap Lemma applies equally well to the second half of σ as to the ﬁrst and we are done. If xend > x0, it suﬃces to show that the surface area contributed by a nodary half-cap is increasing in x0. It is here convenient to introduce a standard form for unit-speed nodaries:

√ Z s 1 2 1 + B sin 2t σB(s) = 1 + B + 2B sin 2s , − √ dt (4.1.22) 2 2 0 1 + B + 2B sin 2t

46 where B > 1. This is obtained from lines (4.1.16) and (4.1.17) by assuming a √ 2 1 + B 0 B vertical tangent at s = π/4 and then setting x0 = , x = √ and 2 0 1 + B2 0 1 y = −√ . Obviously, x0 is increasing in B. The ﬁrst half of the cap then 0 1 + B2 1 1 π corresponds to the interval I = − sin−1 , , upon which 2 B 4 1 Z E(σ ) = (1 + B2 + 2B sin 2s) ds (4.1.23) B 4 I √ ∂ B 1 B2 − 1 E(σ ) = π + 2 sin−1 + > 0 (4.1.24) ∂B B 8 B B2 so the surface area energy and hence the surface area itself is increasing in B for the relevant values of B. This veriﬁes the corollary in the case where xend > x0.

Now suppose xend < x0. Reversing parameterization so that x(0) = xend, we consider σ on s ∈ [0, smax] andσ ˜ on [0, s˜max], where smax ands ˜max correspond to the

0 0 points where the caps reach their (unique) maxima. Then at x = x0, x < x˜ = 1.

On the interval [x0, x˜(˜smax)), neitherσ ˜ nor σ can become vertical, so each deﬁnes a ˜ ˜0 0 function there. Calling these f(x) and f(x) respectively, we have 0 = f (x0) < f (x0) ˜0 0 and f > f near x =x ˜(˜smax). Let x1 be the smallest value in [x0, x˜(˜smax)]) where equality holds, and let s1 ands ˜1 be the corresponding s-values. Finally, let s0 be the s-value on σ where x = xend. We have

A(σ|s∈[0,smax]) = A(σ|s∈[0,s0]) + A(σ|x∈[x0,x1]) + A(σ|s∈[s1,smax]) Z x1 q Z smax > x 1 + (f 0(x))2 dx + x(s) ds x0 s1 r Z x1 2 Z s˜max ≥ x 1 + f˜0(x) dx + x˜(s) ds x0 s˜1

= A(˜σ|x∈[x0,xmax]

This proves the ﬁnal case and veriﬁes the corollary.

Another immediate consequence of the Cap Lemma is that if σ has a self-intersection at the axis of rotation, say at s = s0, then x(s) ≥ sin 2s for s ∈ [s0 − π/4, s0 + π/4]. 47 Similar considerations apply to the intervals [0, π/4] and [L(σ)−π/4,L(σ)], so A(σ) ≥

8π. We thus need only consider arcs with self-intersection away from x = 0. Let us call such a shape a balloon. Z π 1 π The hollow sphere with A = 8π has E = 2 sin2 s ds = s − sin 2s = π. 0 2 0 Thus our goal is to show that any balloon must necessarily be have E ≥ π. We do this

ﬁrst in the case where σ includes a local x-minimum to the left of a self-intersection and then in the case where it includes a local x-maximum to the right of it. These two possibilities are obviously exhaustive.

Lemma 4.1.4. Suppose that a curve σ(s) ∈ GA0 includes a local x-minimum at x0 = x(s0) and a horizontal tangent at a point where x ≥ x0. Then E(σ) > π.

Proof. On one side of s = s0, σ may connect directly to the axis of rotation, while on the other it must include the cap. We estimate E(σ) using only these two elements.

The surface area energy contributed by the cap of σ is at least that of an H ≡ −1 nodary starting from a horizontal tangent at x = xh. where B > 1 and s ∈ I = 1 1 π 1 1 [− sin−1( ), + sin−1( )]. The surface area energy is given by 2 B 2 2 B Z 1 2 E(σB) = (1 + B + 2B sin 2s) ds (4.1.25) 4 I 1 π 1 √ = (1 + B2) + sin−1 + 2 B2 − 1 (4.1.26) 4 2 B

It is easy to verify that E(σB) is increasing in B. Moreover, σB has its horizontals 1√ at x = B2 − 1, so x is increasing in B as well. Since x ≤ x , we can approximate h 2 h 0 h q 2 the surface area of the cap of an arbitrary σ from below using σB for Bx0 = 4x0 + 1. Line (4.1.26) then becomes: !! π 2 2 −1 1 E(σB ) = 2πx + 4x0 + π + 2(1 + 2x ) sin (4.1.27) x0 4 0 0 p 2 1 + 4x0 r π This quantity is increasing in x and equal to π when x = . Thus we 0 0 π + 1 need only be concerned with smaller values of x0. 48 We now consider the sleeve of σ which includes the local x-minimum. On this

0 sleeve, σ moves from x = 0 to x = 1, where y = 1, then back to x = x0. Call this

interval [0, s0]. The Cap Lemma applies, and σ|[0,s0] has more surface area energy

than the unit-circular arc γx0 from x = 0 to x = x0 with length between π/2 and π. We have:

−1 Z π−sin x0 q 2 1 2 −1 E(γx0 ) = sin s ds = π + x0 1 − x0 − sin x0 (4.1.28) 0 2

Now E(σ) ≥ E(σ ) + E(γ ) := ε(x). A quick computation shows ε(0) = π and Bx0 x0

2 2 ! ! 0 2x0 + 1 x0 −1 1 ε (x) = 1 − + πx0 − + 2x sin 4x2 + 1 p 2 p 2 0 1 − x0 1 + 4x0 2 q 2x0 x0 2 ≥ + π 1 − x − x0 (4.1.29) 4x2 + 1 p 2 0 0 1 − x0 r π ≥ 0 for x ∈ 0, (4.1.30) π + 1

Thus ε0(x) ≥ 0 and hence ε(x) ≥ π. This proves the lemma.

In order to address the case where σ includes a local x-maximum to the right of its self-intersection, we require some information about its sleeves.

Lemma 4.1.5 (Sleeve Lemma). Let σ˜ be the |H| ≡ 1 pinwheel connecting horizontal

(x0 = 1) tangents at y = 0 and y = h ≤ 1, and let σ be any other element of K with these same end conditions. Then A(˜σ) ≤ A(σ). Moreover, σ ends at a greater x-value than σ˜.

Proof. In light of Lemma (4.1.4), the hypothesis that x0 = 1 at the ends of the sleeve implies that σ has no vertical tangent and hence can be described as a function of x.

Suppose there exist values of s with x(s) < x˜(s), which is the only case relevant to the second part of the lemma (xend ≥ x˜end), since x(s) can never reach a local maximum. Let s1 = inf{s : x(s) < x˜(s)} and note that s1 ≥ s0 by the Cap Lemma,

49 y = h

Figure 4.2: A typical |H| ≡ 1 sleeve

where s0 is the point onσ ˜ where H changes sign, as in Figure 4.2. Then A(˜σ|[0,s1]) ≥ 0 ˜0 A(σ|[0,s1]). Write x1 = x(s1) =x ˜(s1), and note that f (x1) ≥ f (x1). 0 ˜0 Suppose there exists some x3 > x1 with f (x3) < f (x3). Let x2 = sup{x < x3 : 0 ˜0 0 ˜0 0 ˜0 f (x3) = f (x3), so f (x2) = f (x2) and f (x) < f (x) on (x2, x3). Reparameterizing, let s2 be the point on both σ andσ ˜ where x2 = x(s2) =x ˜(s2), and let s3 be the

0 0 point on σ where x(s3) = x3. On (s2, s3), we have y < y˜ and x > x˜, while also ˜ ˜ ϑ(s3) < ϑ(s3) and ϑ(s2) = ϑ(s2). Taken together,

1 Z s3 y0 H(s2, s3) = ϑ(s3) − ϑ(s2) + ds 2(l − s1) s2 x Z s3 0 1 ˜ ˜ y˜ < ϑ(s3) − ϑ(s2) + ds 2(l − s1) s2 x˜ ˜ = H(s2, s3) = −1

This violates the constraint on average mean curvature and implies that f 0(x) ≥ f˜0(x) ˜ for all x ≥ x1 in the domain of f. In particular, f(x) reaches a horizontal only after ˜ f(x), implying xend ≥ x˜end and so proving the second part of the lemma. In light of the formulation, Z b p 0 2 A(σ|[a,b]) = x 1 + (f (x)) dx (4.1.31) a we also have that A(σ|x>x1 ) > A(˜σ|x>x1 ). This proves the ﬁrst part of the lemma whenever there exists an s with x(s) < x˜(s). 50 Now suppose there is no such s. The lemma is trivial unless the length l of σ is less than the length ˜l ofσ ˜, so we make that supposition as well. In particular, we

0 0 0 have x(l) > x˜(l) and 0 = y (l) < y˜ (l), hence κθ(l) < κ˜θ(l). Let s1 = sup{s : y (s) ≥ Z l Z l 0 0 0 y˜ (s)} < l. Noting that h = y ds ≥ y˜ ds, we ﬁnd s1 > 0, which immediately 0 0 generalizes to s1 > s0 by the Cap Lemma.

0 0 On [s1, l], y (s) < y˜ (s) and x(s) > x˜(s), so κθ(s) < κ˜θ(s). On the other hand, ˜ ˜ ϑ(s1) = ϑ(s1) and ϑ(l) < ϑ(l). Thus

1 Z l H[s1, l] = ϑ(l) − ϑ(s1) + κθ ds 2(l − s1) s1 Z l 1 ˜ ˜ < ϑ(l) − ϑ(s1) + κ˜θ ds 2(l − s1) s1 ˜ = H[s1, l] = −1

Another contradiction. This completes the proof of the Sleeves Lemma.

Next we give an explicit representation of the sleeveσ ˜ described above.

Lemma 4.1.6. The |H| ≡ 1 curve of the Sleeve Lemma (see Figure 4.2) consists r B2 − 1 of a unit circular arc σ (s) = (sin s, 1 − cos s) on [0, s ], where s = sin−1 0 0 0 8 for some B ∈ (1, 3], and part of a nodary σB(s) as described in equation (4.1.22), 1 3 + B2 vertically shifted and deﬁned on the interval [s , s ], where s = − sin−1 1 2 1 2 4B 1 1 and s = − sin−1 . 2 2 B

Proof. An |H| ≡ 1 arc beginning at the axis of rotation can only be a unit-circular arc; this can be veriﬁed directly from the deﬁnition of H or simply by taking B = 1 in equation (4.1.22). The parameterization σ0(s) = (sin s, 1 − cos s) is simplest and most natural, however, so we use that.

0 The condition x(s) = y (s) must be satisﬁed by σB(s) at the point where it

51 connects to the unit-circlular arc. Applying this requirement to line (4.1.22), we obtain:

1 (1 + B2 + 2B sin 2s) = −(1 + B sin 2s) 2 3 + B2 sin 2s = − 4B r B2 − 1 Let s be the solution in (−π/4, π/4), at which point x = . On a unit circle 1 8 r B2 − 1 with x(s) = sin s, this x-value corresponds to s = sin−1 . Clearly, inter- 0 8 sections with the unit circle occur only when B ∈ (1, 3], with B = 1 corresponding to the degenerate case of where the nodary begins at the axis of rotation (and as a result is itself a circular arc). The fact that a nodary reaches its horizontals when 1 1 1 sin 2s = − , so s = − sin−1( ) as claimed, can be veriﬁed directly. B 2 2 B

We are now ready to rule out the possibility that x(s) reaches a local maximum in the interval beginning and ending at the self-intersection of σ. Let smax be the location

of such a maximum, set xmax = x(smax), and deﬁne the half balloons σ1 = σ|[0,smax]

and σ2 = σ|[smax,L]. We may assume that y(s) is monotone increasing on the sleeve

0 of σ1 (and decreasing on the sleeve of σ2), for if ever y < 0 on some interval (a, b) ⊂

[0, smax], we can simply replace y|(a,b) with −y|(a,b) while keeping x(s) unchanged, then shift the pieces vertically to restore continuity. This operation preserves surface area A while possibly reducing y(s1), thus only increasing the overlap of σ1 and σ2.

On the cap of σ, we have −1 < κθ < 0. Thus since |H| ≤ 1, average plane curvature is bounded below by −2, and by the Dubins Lemma σcap lies outside of the circle of radius 1/2 with the same starting point and tangent. Thus if σ fails to be embedded, the sleeves together must traverse a y-distance of at least one unit. In particular, at least one of the sleeves must reach height h ≥ 1/2. We now show that this is only possible for trans-threshhold values of surface area energy E.

52 Figure 4.3: A typical |H| ≡ 1 half-balloon

For any given h, the |H| ≡ 1 sleeve minimizes both E and xend. Thus we can estimate E(σ) from below with an |H| ≡ 1 half-balloon like that in Figure (4.3), where B is chosen so as to preserve xend. Using equation (4.1.22), we may arrive at expressions for the energy E(B) of the half-balloon and height h(B) attained by its sleeve. The vertical of the nodary occurs at occurs at s2 = π/4 regardless of B, so:

Z s0 1 Z π/4 E(B) = sin2 s ds + (1 + B2 + 2B sin 2s) ds 0 4 s1 r 1 B2 − 1 π 1 3 + B2 = sin−1 + (1 + B2) + (1 + B2) sin−1 2 8 16 8 4B Z π/4 1 + B sin 2t h(B) = 1 − cos s0 − √ dt 2 s1 1 + B + 2B sin 2t

In particular,

E(1) = π/4 and E(2.3) > 3π/4

As we might expect, both E and h are monotone in B:

53 Z π/4 dE ds0 2 1 2 ds1 1 = sin s0 − (1 + B + 2B sin 2s1) + (2B + 2 sin 2s) ds dB dB 4 dB 4 s1 B B2 − 1 (B2 − 1) 3 − B2 = · − + p(B2 − 1)(9 − B2) 8 8 2Bp(B2 − 1)(9 − B2) 1 Z π/4 + (2B + 2 sin 2s) ds 4 s1 r 2 Z π/4 3 2 B − 1 1 = (B − 1) 2 + (2B + 2 sin 2s) ds > 0 8 9 − B 4 s1 Z π/4 ∂h ds0 ds1 1 + B sin 2s1 ∂ 1 + B sin 2t = sin s0 + √ − √ dt 2 2 ∂B dB dB 1 + B + 2B sin 2s1 s1 ∂B 1 + B + 2B sin 2t r B B2 − 1 (3 − B2) 1 + B sin 2s = · + · √ 1 + p 2 2 p 2 2 2 (B − 1)(9 − B ) 8 2B (9 − B )(B − 1) 1 + B + 2B sin 2s1 Z π/4 B cos2 2t + 2 3/2 dt s1 (1 + B + 2B sin 2t) 3(B2 − 1) r 2 Z π/4 B cos2 2t = · 2 + 2 3/2 dt > 0 8B 9 − B s1 (1 + B + 2B sin 2t)

In particular, if either σi reaches a height of h(2.3) or greater, then the total surface area energy of σ is larger than π. Such cases are ruled out.

It is now suﬃcient to verify that h(2.3) < 1/2, since each half of the cap must traverse a y-distance of at least 1/2 by the Cap Lemma. This can be done by approx- imating the nodary portion of the |H| ≡ 1 sleeve using a circular arc of appropriate radius. Observe that κθ is decreasing on the relevant part:

y0 −2(1 + B sin 2s) κ = = θ x 1 + B2 + 2B sin 2s

dκ 4B(1 − B2) cos 2s θ = ≤ 0 ∀s ∈ [s , π/4] ds (1 + B2 + 2B sin 2s)2 1

Since H ≡ −1, this implies that κσ is increasing (and |κr| is decreasing) on the right portion of the |H| ≡ 1 sleeve. In short, the nodary curves more near the axis of rotation, just as Figure (4.3) suggests. To approximate h(2.3) from above,

54 then, we replace the nodary with the osculating circle at the end of the sleeve, where 1√ x = B2 − 1, κ = 0 and κ = −2. The radius of this circle is 1/2. 2 θ σ r !! 1 B2 − 1 h(B) ≤ 1 + 1 − cos sin−1 2 8 r 3 3 9 − B2 = − 2 4 2 r 3 3 9 − (2.3)2 h(2.3) ≤ − ≈ .48 < .5 2 4 2

Thus in order for σ to reach suﬃcient height to allow its sleeves to intersect, it must exceed the threshhold for surface area energy. We have proved the following:

Theorem 4.1.7. For A0 < AT = 8π, every element of GA0 is embedded.

4.2 Existence of the Minimal Shape

One last preliminary result is needed in order to establish an existence result about

elements of minimal volume in GA0 .

Lemma 4.2.1. Fix A0 < AT . There exists xm > 0 such that no element σ ∈ GA0 can have a local x-minimum for x ≤ xm. Thus for any such σ, x < xm only in neighborhoods of s = 0 and s = L(σ).

1 Proof. By corollary 4.1.3, we may estimate A(σ) from below using the unit-circular 2 arc γ that starts at x = 0, reaches a vertical at x = 1 and then arches back to x = xm for any chosen xm < 1. We compute:

−1 Z π−sin xm p 2 A(γ) = 2π x(s) ds = 2π 1 + 1 − xm 0

55 s A 2 Noting that A ∈ [4π, 8π), set x = 1 − 0 − 1 . Then x > 0 and A(γ) > 0 m 4π m 1 A . This proves the lemma. 2 0 |y0| 1 The signiﬁcance of Lemma 4.2.1 is twofold. First, |κθ| = ≤ away from x xm the beginning and end of σ. The Cap Lemma applies to the remaining pieces, where |y˜0| the comparison arcσ ˜ is just a unit circle with center on the x-axis, so |κ | ≤ = 1 θ x˜ there. Hence within GA0 , κθ is uniformly bounded, which in turn implies a uniform bound on average plane curvature κσ as well:

Z t 0 1 y 1 ≥ |H(a, b)| = ϑ(t) − ϑ(s) + ds 2(t − s) s x Z t 0 |ϑ(t) − ϑ(s)| 1 y ≤ 2 + ds| t − s (t − s) s x 1 ≤ 2 + xm

Second, Lemma 4.2.1 provides an upper bound on the length L(σ) of the gener-

ating arcs in GA0 . With notation as before, we have Z Z A(σ) = 2π x(s) ds + x(s) ds x≤xm x>xm h i ≥ 2π xm + xmL(σ|x≥xm ) (4.2.1)

L(σ) = L(σ|x

A0 ≤ L(˜σ|x

Now for the payoﬀ:

Theorem 4.2.2. For any A0 < AT , there exists an element σmin ∈ GA0 with

V (σmin) ≤ V (σ) ∀σ ∈ GA0 .

56 Proof. Reparameterize each σ(s) ∈ GA0 via s(t) = L(σ)t to acheive a common domain

0 [0, 1] upon which |σ (t)| ≡ L(σ) ≤ Lmax. The bound H(s, t) then becomes:

Z t2 0 1 ϑ(t2) − ϑ(t1) y 1 ≥ |H(t1, t2)| = + dτ (4.2.3) 2L t2 − t1 t1 x which is preserved under convergence in the C1 norm, as are the ﬁxed conditions on

σ(0) and σ(L) and the stipulation that A(σ) = A0. Because κσ is uniformly bounded

0 0 in GA0 , there exists a constant M such that |σ (b) − σ (a) ≤ M|b − a| ∀a, b ∈ [0, 1].

1 1 This bound respects convergence in C , hence in GA0 , so does the condition σ ∈ L2.

1 Finally, since A0 < AT , embeddedness must also be preserved. Since C [0, 1] is

complete, we conclude that GA0 is complete as well.

Now let {σk(t)} be a sequence in GA0 , parameterized as above, with V (σk) &

0 0 inf{V (σ): σ ∈ GA0 }. Since L(σk) ≤ Lmax and |σk(b) − σk(a) ≤ M|b − a| ∀a, b ∈ [0, 1],

0 both {σk} and {σk} are bounded and equicontinuous. By Arzela’s Theorem, {σk}

1 has a subsequence {σki } that is Cauchy with respect to the C norm. Since GA0 is complete, ∃σmin ∈ GA0 with σki ⇒ σmin and V (σki ) & V (σmin).

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