A parameterized curve γ is closed with a period of T, if ∀t∈R,∀k∈N.γ(t)=γ(t+kT)
A parameterized curve γ 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 is ±2π.
Vertex
A vertex of a plane curve γ is a point where dtdκs=0. Note that this definition is independent of parameterizations.
Example Every point on the circle γ(t)=(Rcost,Rsint) is a vertex.
proof. We known that κs=R−1 for the circle γ, so that dtdκs=0 everywhere.
Circumscribed circle
For a simple closed curve γ:R→R2, we can define a circumscribed circle C(c,R)={x∈R2:∥x−c∥=R} that contains γ(R) and attains the smallest possible radius.
Claim existence
proof. By Jordon's Theorem, since γ is simple, closed. Its interior is bounded. There exists some circles C(c,R) that bounds γ(R).
Then, we need to consider whether inf{R} exists.
Take a sequence cn,Rn s.t. Rn↘inf{R}=R∞. Note that ∥c−γ(s)∥≤Rn, take a convergent subsequence cnk s.t. limk→∞cnk=c∞.
We claim that circle C(c∞,R∞) contains γ(R). The idea of the claim is that if we assume C∞=C(c∞,R∞) does not contain γ(R), then there are some part of γ(R) "sticks out" of C∞, i.e.
Let D:=∥γ(s0)−c∞∥+∥c∞−cnk∥,ϵk:=∥c∞−cnk∥. so that
∥γ(s0)−cnk∥≥D−ϵk
Take k large enough s.t. ϵk<2D−R∞ and Rnk−R∞<2D−R∞. Therefore,
∥γ(s0)−cnk∥≥D−2D−R∞=2D−R∞+R∞>Rnk
Then, by the construction of our sequence, this is a contradiction.
Claim. uniqueness
Assume not unique, then exists C1=(c1,R),C2=(c2,R) (R must be the same since minimum) s.t. ∥γ(s)−c1∥<R and ∥γ(s)−c2∥<R. Therefore, γ(R)∈int(C1)∩int(C2). Then, we can construct new circle according to the picture, where c=2c1+c2 and by triangle inequality, R′<R
Lemmas
Fact 1C(c,R) and γ(R) has at least one intersection. Assume not, then we can keep c unmoved, note that ∀s.∥γ(s)−c∥<R so that we can shrink R′<R.
Fact 2C(c,R) and γ(R) has at least two intersections. Assume not, by fact 1, we take s0 to be the only intersection. Then move c along the direction of γ(s0)−c by arbitrarily small ϵ, then it has no intersection, which violates Fact 1.
Fact 3 If C(c,R) and γ(R) has exactly two intersections, then the line segment between intersection points is a diameter. If not, then move c along the direction normal to the line.
Fact 4γ(R) and C has the same tangent at their point of intersections. Assume there are at least two intersections p1,...,pn. First, we want to show that at all intersections, the orientation is the same. Assume γ is oriented the same as C as intersection point pk, to show that all intersection points have the same intersection, it's sufficient to show that of pk+1. Note that the curve γ([pk,pk+1]) and the arc of circle between pk and pk+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+1 to pk, and it forms a simple closed curve with γ([pk,pk+1]). Therefore, they must oriented the same.
Fact 5.1 If γ's image is within int(C) in a neighborhood of the intersection point, then the curvature of γ,κs(si)≥R−1. Wlog assume c=(0,0). At point of intersection, we have that pk=γ(sk), then note that ∥γ(sk)∥2=γ(sk)⋅γ(sk)=R2 must attain its maximum (since other points are ∥γ(s)∥≤R by circumscribed circle). Then, we have
dsd(γ(sk)⋅γ(sk))=2γ′(sk)⋅γ(sk)=0
Therefore, we have that t(sk) is perpendicular to the radius at pk and is the same tangent vector.
Fact 5.2 If γ's image is within ext(C) in a neighborhood of the intersection point, then the curvature of γ,κs(si)≤R−1.
The idea is similar to Fact 5.1 while here γ(sk)⋅γ(sk) attains its local minimum, so that
ds2d2(γ(sk)⋅γ(sk))κs(sk)≥0≤R−1
Theorem (Four Vertex Theorem)
For a simple closed curve γ:R→R2, it has at least 4 vertex.
Let γ be a simple closed curve, and C be its circumscribed curve, pk=γ(sk) and pk+1=γ(sk+1) be two conservative intersection points. Connecting the two points with the line segment l=pkpk+1 and move the the center of the circle along the direction normal to l until the last moment it touches γ. Then, the γ([sk,sk+1]) is all in ext(C), by Fact 5.2, there exists at least one point qk s.t. κs(qk)≤R−1. Note that there are two opposite directions to move c.
Therefore, we have that the curve passing through pk,qk,pk+1 must have a local minimum since it goes down from ≥R−1 to ≤R−1 and then back to ≥R−1. Therefore, we have at least n local minimum. Then, consider the curve passing through qk,pk,qk+1, similarly we can obtain n local maximum.
Note that the only possible case is that κs at pk,q,pk+1 are all R−1. In this case, there is infinitely many points of γ in the neighborhood of pk, that is coincided with the arc of the circle. Since C is the circumscribed circle, either γ([pk,pk+1] coincides with the arc of the circle all the way, or κ(qk)<R−1
Example : Limacon
Let γ:R→R2 be defined as
γ(t)=((1+2cost)cost,(1+2cost)sint)
Note that γ is closed (but not simple) with a period of 2π.