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Four Vertex Theorem

Simple Closed Curves

A parameterized curve γ\gamma is closed with a period of TT, if tR,kN.γ(t)=γ(t+kT)\forall t\in \mathbb R, \forall k\in\mathbb N. \gamma(t) = \gamma(t+kT)

A parameterized curve γ\gamma is simple if it has no self-intersections.

Jordan's Theorem

Claim Any simple closed curve in the plane has a connected bounded "interior" and an connected unbounded "exterior".

Hopf's Umlaufsatz Theorem

The total signed curvature of a simple closed curve in R2\mathbb R^2 is ±2π\pm 2\pi.

Vertex

A vertex of a plane curve γ\gamma is a point where dκsdt=0\frac{d\kappa_s}{dt} = 0. Note that this definition is independent of parameterizations.

Example Every point on the circle γ(t)=(Rcost,Rsint)\gamma(t) = (R \cos t, R\sin t) is a vertex.

proof. We known that κs=R1\kappa_s = R^{-1} for the circle γ\gamma, so that dκsdt=0\frac{d\kappa_s}{dt} = 0 everywhere.

Circumscribed circle

For a simple closed curve γ:RR2\gamma:\mathbb R\rightarrow\mathbb R^2, we can define a circumscribed circle C(c,R)={xR2:xc=R}C(c, R) = \{x\in\mathbb R^2: \|x - c\| = R\} that contains γ(R)\gamma(\mathbb R) and attains the smallest possible radius.

Claim existence

proof. By Jordon's Theorem, since γ\gamma is simple, closed. Its interior is bounded. There exists some circles C(c,R)C(c,R) that bounds γ(R)\gamma(\mathbb R).

Then, we need to consider whether inf{R}\text{inf}\{R\} exists.

Take a sequence cn,Rnc_n, R_n s.t. Rninf{R}=RR_n\searrow \text{inf}\{R\} = R_\infty.
Note that cγ(s)Rn\|c-\gamma(s)\| \leq R_n, take a convergent subsequence cnkc_{n_k} s.t. limkcnk=c\lim_{k\rightarrow\infty} c_{n_k} = c_\infty.

We claim that circle C(c,R)C(c_\infty, R_\infty) contains γ(R)\gamma(\mathbb R).
The idea of the claim is that if we assume C=C(c,R)C_\infty = C(c_\infty, R_\infty) does not contain γ(R)\gamma(\mathbb R), then there are some part of γ(R)\gamma(\mathbb R) "sticks out" of CC_\infty, i.e.

s0R.γ(s0)c>R\exists s_0\in\mathbb R. \|\gamma(s_0) - c_\infty\| > R_\infty

By triangle inequality we have

γ(s0)cγ(s0)cnkγ(s0)c+ccnk\|\gamma(s_0) - c_\infty\| \leq \|\gamma(s_0) - c_{n_k}\| \leq \|\gamma(s_0) - c_\infty\| + \|c_\infty - c_{n_k}\|

Let D:=γ(s0)c+ccnk,ϵk:=ccnkD:= \|\gamma(s_0) - c_\infty\| + \|c_\infty - c_{n_k}\|, \epsilon_k := \|c_\infty - c_{n_k}\|. so that

γ(s0)cnkDϵk\|\gamma(s_0) - c_{n_k}\| \geq D - \epsilon_{k}

Take kk large enough s.t. ϵk<DR2\epsilon_k < \frac{D-R_\infty}{2} and RnkR<DR2R_{n_k} - R_\infty < \frac{D-R_\infty}{2}. Therefore,

γ(s0)cnkDDR2=DR2+R>Rnk\|\gamma(s_0) - c_{n_k}\| \geq D - \frac{D-R_\infty}{2} = \frac{D-R_\infty}{2} + R_\infty > R_{n_k}

Then, by the construction of our sequence, this is a contradiction.

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Claim. uniqueness

Assume not unique, then exists C1=(c1,R),C2=(c2,R)C_1 = (c_1, R), C_2 = (c_2, R) (RR must be the same since minimum) s.t. γ(s)c1<R\|\gamma(s) - c_1\| < R and γ(s)c2<R\|\gamma(s) - c_2\| < R. Therefore, γ(R)int(C1)int(C2)\gamma(\mathbb R) \in \text{int}(C_1)\cap \text{int}(C_2). Then, we can construct new circle according to the picture, where c=c1+c22c = \frac{c_1 + c_2}{2} and by triangle inequality, R<RR' < R

Lemmas

Fact 1 C(c,R)C(c, R) and γ(R)\gamma(\mathbb R) has at least one intersection.
Assume not, then we can keep cc unmoved, note that s.γ(s)c<R\forall s. \|\gamma(s) - c\| < R so that we can shrink R<RR' < R.

Fact 2 C(c,R)C(c, R) and γ(R)\gamma(\mathbb R) has at least two intersections.
Assume not, by fact 1, we take s0s_0 to be the only intersection. Then move cc along the direction of γ(s0)c\gamma(s_0)-c by arbitrarily small ϵ\epsilon, then it has no intersection, which violates Fact 1.

Fact 3 If C(c,R)C(c, R) and γ(R)\gamma(\mathbb R) has exactly two intersections, then the line segment between intersection points is a diameter.
If not, then move cc along the direction normal to the line.

Fact 4 γ(R)\gamma(\mathbb R) and CC has the same tangent at their point of intersections.
Assume there are at least two intersections p1,...,pnp_1,...,p_n.
First, we want to show that at all intersections, the orientation is the same. Assume γ\gamma is oriented the same as CC as intersection point pkp_k, to show that all intersection points have the same intersection, it's sufficient to show that of pk+1p_{k+1}. Note that the curve γ([pk,pk+1])\gamma([p_k, p_{k+1}]) and the arc of circle between pkp_k and pk+1p_{k+1} will form a simple closed curve, by Jordon's theorem, it's bounded and connected.
Then, consider any simple curve within the enclosed region from pk+1p_{k+1} to pkp_k, and it forms a simple closed curve with γ([pk,pk+1])\gamma([p_k, p_{k+1}]). Therefore, they must oriented the same.

Fact 5.1 If γ\gamma's image is within int(C)\text{int}(C) in a neighborhood of the intersection point, then the curvature of γ,κs(si)R1\gamma, \kappa_s(s_i) \geq R^{-1}. Wlog assume c=(0,0)c = (0,0). At point of intersection, we have that pk=γ(sk)p_k = \gamma(s_k), then note that γ(sk)2=γ(sk)γ(sk)=R2\|\gamma(s_k)\|^2 = \gamma(s_k)\cdot\gamma(s_k)= R^2 must attain its maximum (since other points are γ(s)R\|\gamma(s)\| \leq R by circumscribed circle). Then, we have

dds(γ(sk)γ(sk))=2γ(sk)γ(sk)=0\frac{d}{ds}(\gamma(s_k)\cdot \gamma(s_k)) = 2\gamma'(s_k)\cdot\gamma(s_k) = 0

Therefore, we have that t(sk)\mathbf t(s_k) is perpendicular to the radius at pkp_k and is the same tangent vector.

In addition, we have that

d2ds2(γ(sk)γ(sk))02(γ(sk)γ(sk)+γ(sk)γ(sk))02(κs(sk)pkns(sk)+1)0\begin{align*} \frac{d^2}{ds^2}(\gamma(s_k)\cdot \gamma(s_k)) &\leq 0\\ 2(\gamma''(s_k)\cdot\gamma(s_k) + \gamma'(s_k)\cdot\gamma'(s_k))&\leq 0\\ 2(\kappa_s(s_k)p_k \cdot \mathbf n_s(s_k) + 1) &\leq 0 \end{align*}

Then note ns(sk)\mathbf n_s(s_k) is parallel and directly opposite to pkp_k, since pk=R,pkns(sk)=R\|p_k\| = R, p_k \cdot \mathbf n_s(s_k) = -R

d2ds2(γ(sk)γ(sk))0(κs(sk)R+1)0κs(sk)R1\begin{align*} \frac{d^2}{ds^2}(\gamma(s_k)\cdot \gamma(s_k)) &\leq 0\\ (-\kappa_s(s_k)R + 1) &\leq 0\\ \kappa_s(s_k) &\geq R^{-1} \end{align*}

Fact 5.2 If γ\gamma's image is within ext(C)\text{ext}(C) in a neighborhood of the intersection point, then the curvature of γ,κs(si)R1\gamma, \kappa_s(s_i) \leq R^{-1}.

The idea is similar to Fact 5.1 while here γ(sk)γ(sk)\gamma(s_k)\cdot \gamma(s_k) attains its local minimum, so that

d2ds2(γ(sk)γ(sk))0κs(sk)R1\begin{align*} \frac{d^2}{ds^2}(\gamma(s_k)\cdot \gamma(s_k)) &\geq 0\\ \kappa_s(s_k) &\leq R^{-1} \end{align*}

Theorem (Four Vertex Theorem)

For a simple closed curve γ:RR2\gamma:\mathbb R\rightarrow\mathbb R^2, it has at least 4 vertex.

Let γ\gamma be a simple closed curve, and CC be its circumscribed curve, pk=γ(sk)p_k = \gamma(s_k) and pk+1=γ(sk+1)p_{k+1} = \gamma(s_{k+1}) be two conservative intersection points. Connecting the two points with the line segment l=pkpk+1l = \overline{p_k p_{k+1} } and move the the center of the circle along the direction normal to ll until the last moment it touches γ\gamma. Then, the γ([sk,sk+1])\gamma([s_k, s_{k+1}]) is all in ext(C)\text{ext}(C), by Fact 5.2, there exists at least one point qkq_k s.t. κs(qk)R1\kappa_s(q_k) \leq R^{-1}. Note that there are two opposite directions to move cc.

Therefore, we have that the curve passing through pk,qk,pk+1p_{k}, q_k, p_{k+1} must have a local minimum since it goes down from R1\geq R^{-1} to R1\leq R^{-1} and then back to R1\geq R^{-1}. Therefore, we have at least nn local minimum. Then, consider the curve passing through qk,pk,qk+1q_k, p_k, q_{k+1}, similarly we can obtain nn local maximum.

Note that the only possible case is that κs\kappa_s at pk,q,pk+1p_k, q, p_{k+1} are all R1R^{-1}. In this case, there is infinitely many points of γ\gamma in the neighborhood of pkp_k, that is coincided with the arc of the circle. Since CC is the circumscribed circle, either γ([pk,pk+1]\gamma([p_k, p_{k+1]} coincides with the arc of the circle all the way, or κ(qk)<R1\kappa(q_k) < R^{-1}

Example : Limacon

Let γ:RR2\gamma:\mathbb R\rightarrow\mathbb R^2 be defined as

γ(t)=((1+2cost)cost,(1+2cost)sint)\gamma(t) = ((1+2\cos t)\cos t, (1 + 2\cos t)\sin t)

Note that γ\gamma is closed (but not simple) with a period of 2π2\pi.

Claim Limacon only has 2 vertices.

proof.

γ(t)=(sint2sin2t,cost+2cos2t)γ(t)=sin2t+4sintsin2t+4sin22t+cos2t+4costcos2t+4cos22t=5+8costsin2t+4cost8costsin2t)=5+4cost\begin{align*} \gamma'(t) &= (-\sin t - 2\sin 2t, \cos t + 2\cos 2t)\\ \|\gamma'(t) \| &= \sqrt{\sin^2 t + 4\sin t\sin 2t + 4\sin^2 2t + \cos^2 t + 4\cos t\cos 2t + 4\cos^2 2t}\\ &= \sqrt{5 + 8\cos t \sin^2 t + 4\cos t - 8\cos t\sin^2 t)}\\ &= \sqrt{5 + 4\cos t} \end{align*}

Then, let γ(t)=15+4cost(sint2sin2t,cost+2cos2t)\gamma'(t) = \frac{1}{\sqrt{5 + 4\cos t} }(-\sin t - 2\sin 2t, \cos t + 2\cos 2t) and define the turning angle by

cosφ=sint2sin2t5+4cost.sinφ=cost+2cos2t5+4cost\cos\varphi = \frac{-\sin t - 2\sin 2t}{\sqrt{5 + 4\cos t} }. \sin\varphi = \frac{\cos t + 2\cos 2t}{ {\sqrt{5 + 4\cos t} } }

Because we only interested in κs=φ\kappa_s = \varphi', observe that

ddtsinφ=ddφsinφdφdtddtsinφ=(sint4sin2t)5+4cost(cost+2cos2t)(4sint)125+4cost5+4cost=sint(24cos2t+42cost+9)(5+4cost)3/2φ(t)=ddtsinφcosφ=sint(24cos2t+42cost+9)(5+4cost)3/2sint2sin2t5+4cost=9+6cost5+4costκs=dφds=9+6cost5+4cost15+4cost=9+6cost(5+4cost)3/2dκsdt=12sint(2+cost)(5+4cost)5/2\begin{align*} \frac{d}{dt}\sin \varphi &= \frac{d}{d\varphi}\sin\varphi\frac{d\varphi}{dt}\\ \frac{d}{dt}\sin \varphi &= \frac{(-\sin t - 4\sin 2t)\sqrt{5+4\cos t} - (\cos t + 2\cos 2t)(-4\sin t)\frac{1}{2\sqrt{5+4\cos t} } }{5+4\cos t}\\ &= \frac{-\sin t(24\cos^2 t + 42\cos t +9)}{(5+4\cos t)^{3/2} }\\ \varphi'(t) &= \frac{\frac{d}{dt}\sin \varphi}{\cos \varphi} \\ &= \frac{\frac{-\sin t(24\cos^2 t + 42\cos t +9)}{(5+4\cos t)^{3/2} } }{\frac{-\sin t - 2\sin 2t}{\sqrt{5 + 4\cos t} } }\\ &= \frac{9+6\cos t}{5+4\cos t}\\ \kappa_s &= \frac{d\varphi}{ds} = \frac{9+6\cos t}{5+4\cos t} \frac{1}{\sqrt{5+4\cos t} }\\ &= \frac{9+6\cos t}{(5+4\cos t)^{3/2} }\\ \frac{d\kappa_s}{dt} &= \frac{12\sin t (2+\cos t)}{(5+4\cos t)^{5/2} } \end{align*}

Therefore, dκsdt=0\frac{d\kappa_s}{dt} = 0 only when t=0,πt=0, \pi for t[0,2π)t\in [0, 2\pi)