Chapter III: Nonconvex ancient solutions to Curve Shortening Flow
3.4 Convergence
In this section, we obtain uniform curvature bounds on a sequence of βreally old solutionsβ {π(π‘)} and extract a subsequence of solutions that converges locally smoothly to an ancient solution of curve shortening flow (CSF).
There exists aπ such that for any πβ > π, the curvatures |π π| of the βreally old solutionsβ{π(π‘)}πare bounded independently of πon the interval[βπββ1/4,βπβ]. The strategy to obtain these bounds is as follows: a) decompose an element π(π‘)of this sequence into the union of several graph representations, b) use theπΏ1bound on the error to obtainπΏβestimates for these graphs, and c) apply the standard estimates for divergence-form quasilinear parabolic equations to establish a uniform curvature bound.
Let πβ > 0 be a large positive number, which may be increased as necessary throughout this section. The obvious candidates for the sequence of βreally old so- lutionsβ{π(π‘)}are the CSF solutions defined on[βπ ,βπβ]with the initial condition
π(βπ) =β (βπ). At any timeπ‘ β [βπ ,βπβ], the unsigned area enclosed by the curves
β(π‘) and π(π‘)is bounded by the quantity
β« π‘
βπ
β«
β(π)
|πβπ |ππ π π .
By the results of the previous sections, this quantity is in πΏ1 and givenπ > 0, we can findπβ > 0 such that
(πβ) =
β« βπβ
ββ
β«
β(π)
|πβπ |ππ π π < π .
This estimate gives a uniform bound on the unsigned area between π(π‘)andβ(π‘)for any sufficiently large π andπ‘ <βπβ, whenever defined.
For simplicity of calculation, we consider an alternative sequence of βreally old solutionsβ {π(π‘)}, called βsquare-profile approximationsβ, which do not satisfy the initial conditionπ(βπ) =β (βπ). They are constructed as follows: Let[πββ(π‘), π+β(π‘)]
be the interval of polar angles that contain the approximate solution at a given time π‘. Let π = (π) be the central branch of the Yin-Yang foliation, so the Yin-Yang solution is given by the two branchesπ = (π+π/2βπ‘)andπ = (πβπ/2βπ‘). We define π(π‘) to be the solution of curve shortening which at timeπ‘ =βπ is given by
β’ Two arms of the Yin-Yang soliton π = (πβ π/2βπ‘) = (π βπ/2+ π), and π =(π+π/2βπ‘) = (π+π/2+ π) truncated atπ =π+β(βπ);
β’ a straight line segment connecting the two arms of the Yin-Yang soliton. This segment is part of the rayπ =+β (βπ).
Notice that the square-profile approximationsπ(π‘)enclose the CSF solutions starting fromβ(βπ), and that the area bounded by these solutions stays constant along the flow, for allπ‘ β [βπ ,βπβ]. The area π΄π between π(βπ)andβ(βπ) is small and goes to zero as π β β. Thus, the area between the old solution π(π‘) with the βsquare initial dataβ and the approximate solutionβ(π‘)is bounded by
Area(π(π‘),β(π‘)) β©½ π΄π+
β« π‘
βπ
β«
β(π)
|πβπ |ππ π π .
In order to improve these area bounds toπΏβbounds, we will use the geometry of the
π(π‘) and several properties of CSF. In particular, we often appeal to the maximum principle and the following Sturmian property for intersections of curve shortening flows.
Consider two CSF solutionsπΎ
0, πΎ
1: [π
1, π
2) Γ [0,1] β2, for which
π πΎ0(π‘) β©πΎ
1(π‘) =π πΎ
1(π‘) β©πΎ
0(π‘) =β holds for anyπ‘ β [π
1, π
2). Then the number of intersections ofπΎ
0(π‘)andπΎ
1(π‘)is a finite and non-increasing function ofπ‘ β (π
1, π
2). It decreases whenever πΎ
0andπΎ
1
have a tangency.
There is a useful related theorem for inflections points.
LetπΎ : [π
1, π
2) Γπ1 β2be a solution of CSF. Then, for anyπ‘ β (π
1, π
2), πΎ(π‘) has at most a finite number of inflection points, and this number does not increase with time. In fact, it drops whenever the curvatureπ has a multiple zero.
While the curves π(π‘) are not convex, we do have a one sided curvature bound. If π is the curvature of a counterclockwise oriented parametrization π of the curves
π(π‘), then
π β β¨π , ππ β©> 0.
Assuming that the parametrization π is normal (ππ‘ β₯ ππ ), the curvature evolves by
π π‘ =π π π +π 3.
A short computation using ππ‘ =ππ π andβ₯ππ β₯ =1 shows that
ππ‘βπ2
π
β₯πβ₯2=2β¨ππ‘, πβ© β β¨ππ π , πβ© β2β₯ππ β₯2 =β2.
Differentiating with respect to arclength, using the commutator[ππ‘, ππ ] =π 2ππ , and alsoππ β₯πβ₯2=2β¨π , ππ β©we get
ππ‘β¨π , ππ β©=π2
π β¨π , ππ β© +π 2β¨π , ππ β©.
Hence π and β¨π , ππ β©satisfy the same linear equation. Therefore π = π β β¨π , ππ β© also satisfies
ππ‘ =ππ π +π 2π .
The quantityπvanishes on the rotating soliton (see the appendix).
The square-profile initial curves π(βπ) consist of two arcs. One is the Yin-Yang soliton, so on this arc we haveπ = 0. The other arc is the radial line segment on the ray π = πβ+(βπ). On this segment we clearly have π = 0. Since we orient π counterclockwise, π and ππ are parallel with opposite directions; i.e. β¨π , ππ β© > 0.
Hence π > 0 on the line segment. Finally, the initial curve π(βπ) is not smooth, having two corners where the line segment and Yin-Yang arms meet. If one rounds these corners off by replacing them with small circle arcs with radius π βͺ 1, then the curvature of these arcs will beπ = πβ1 β« 1, so thatπ >0 on the circular arcs, provided π is sufficiently small. The resulting curve has π = 0 on the Yin-Yang arms, andπ >0 on the line segment, as well as the small circular arcs. The solution to CS starting from the modified initial curve therefore hasπ > 0. Letting π β 0 we conclude thatπ >0 also holds on π(π‘).
With Theorem 3.4, we can decompose the solutions π(π‘) into exactly two graphs over the polar angle parameter.
For anyπ‘ β (βπ ,βπβ], there is an interval[β
π(π‘),+
π (π‘)]such that the curveπ(π‘)can be written as the union of two graphs of polar functions,π β
π(π , π‘) andπ +
π(π , π‘)defined forπ β [βπ(π‘),+
π (π‘)]. The functionsπ‘ β¦β πβ
π(π‘)andπ‘ β¦β π+
π(π‘) are strictly increasing and decreasing, respectively.
By the maximum principle, the βreally old solutionsβ π(π‘)will be contained inside of the Yin-Yang curve. The Sturmian property, Theorem 3.4, tells us that the number of intersections of π(π‘)and the rays π =π
0 βis non-increasing, and only decreases when there is a tangency. This implies that the desired graph decomposition exists.
These two graphs are bounded above and below by the branches of the Yin-Yang soliton on their polar interval of definition, [β
π(π‘),+
π (π‘)].
Similarly, we can always write each π(π‘)as a union of two graphs taking values inπ, the polar angle. Recall that the images of(πβπ‘)forπ‘ β [βπ, π)foliate the punctured plane2\ {0}. See Figure 3.2.
For all π‘, π(π‘) can be decomposed into two graphs of two functions which take leaves of the foliation as inputs and have their range in the set of polar angles. More specifically, forπ βͺ 0 there existπ¦π ,
1, π¦π ,
2: (ββ, π] β (β1
2π,1
2π) and functions ΞΒ±π :{(π‘ , π¦) :π‘ < π , π¦π ,
1(π‘) < π¦ < π¦π ,
2(π‘)} β such that the very old solution π(π‘)is the union of the two curves
π(ΞΒ±π(π‘ , π¦), π‘ , π¦) = ΞΒ±π(π‘ , π¦) βπ‘+π¦ 1(π)
((3.7)).
The initial square-profile curve π(βπ)is tangent to the graphs of (πΒ±π/2+ π)and intersects the graphs of (π + π¦ + π), π¦ β (βπ/2, π/2) twice: once at the origin and once on the line segment connecting the two branches of π(βπ). Then, by the Sturmian theorem, for all subsequent π‘ > βπ, π(π‘) can be split into two graphs corresponding to the βupperβ and βlowerβ intersection points with the leaves of the Yin-Yang foliation. At each timeπ‘, these graphs split at two unique leaves of the foliation, marked by values π¦π ,
1(π‘), π¦π ,
2(π‘) β (βπ/2, π/2), so that π(π‘)is tangent to the curves{π = (π+π¦π ,
1(π‘) βπ‘)}and{π =(π+π¦π ,
2(π‘) βπ‘)}. We know that these two points are unique since a greater number of tangencies would introduce more than two intersection points for other curves {π = (π+π¦ βπ‘)}. We call the coordinate system(π¦, π) β (βπ, π) Γ (0,β)the βYin-Yang polar coordinate systemβ and denote the two functions giving the upper and lower graphs comprising π(π‘) by Ξβπ(π¦, π‘) andΞ+π(π¦, π‘)respectively, defined on the interval (π¦π ,
1(π‘), π¦π ,
2(π‘)) β (βπ/2, π/2). There exist π < 0 and πΆ > 0 such that π+
π(π‘) β©½ πβ+(π‘) + πΆ for all π β and all π‘ β [βπ ,βπ].
Assume thatπ < π/16.
For anyπ‘ β [βπ , π] at which π+
π(π‘) >+β (π‘) we consider the area π(π‘) of the βreally old solutionβ π(π‘) inside the polar interval [β+(π‘), π+
π(π‘)], whereπ+
π(π‘) and +β(π‘) are the endpoints of the intervals of definition of the approximate solutionβ(π‘)and π(π‘) respectively. This area measures the βtailβ of the π(π‘)that may form between the tip of π(π‘) and the tip ofβ(π‘). Note that the areaπ(π‘)is bounded above by the error
π(π‘) β©½ π΄π + (πβ) = π΄π+
β« βπβ
ββ
β«
β(π)
|π βπ |ππ π π < π . To calculate this area, first consider the functionΞπ‘ , π(π¦) :=max{Ξ+
π(π¦, π‘) βπβ+(π‘),0} over the interval (π¦π ,
1(π‘), π¦π ,
2(π‘)). Then in the (π , π¦) βYin-Yang coordinates,β we can integrate to find the area:
π(π‘) =
β« π¦π ,2(π‘) π¦π ,
1(π‘)
β« Ξπ‘ , π(π¦)+πβ+ π+β
(π+π¦βπ‘)det(π·π)ππ ππ¦,
whereπ : (0,β) Γ (βπ/2, π/2) β (0,β) Γ (0,β) is the coordinate transformation given byπ(π , π¦) = (π ,(π+π¦βπ‘)). Clearly, detπ =β² (π+π¦βπ‘), so
π(π‘) =
β« π¦π ,2(π‘) π¦π ,
1(π‘)
β« Ξπ‘ , π(π¦)+π+β
πβ+
(π+π¦βπ‘)β²(π+π¦βπ‘)ππ ππ¦β
β« π¦π ,2(π‘) π¦π ,
1(π‘)
Ξπ‘ , π(π¦)ππ¦,
by the asymptotic expansions in (3.1).
We argue that given a smallπΏ >0, it is possible to pick an angleπ
0independent of π such that the measure|{π¦:Ξπ‘ , π(π¦) > π
0}| < πΏ. Indeed, it follows from π0
{π¦ :Ξπ‘ , π(π¦) > π
0} β©½
β« π¦π ,2(π‘) π¦π ,
1(π‘)
Ξπ‘ , π(π¦)ππ¦ < π that ifπ
0< π
πΏ, then|{π¦:Ξπ‘ , π(π¦) > π
0}| < πΏ holds for allπ‘ , π. The two points intersection of π(π‘)with the rayπ =πβ+(π‘) +π
0are πΒ±(π‘) = π Β±
π(πβ(π‘) +π
0, π‘). LetπΎ(π‘)be the arc on π(π‘) on whichπ β₯ π+β(π‘) +π
0, and whose endpoints therefore are πΒ±(π‘). Consider the area π΄(π‘) of the region enclosed by πΎ(π‘) and the line segment connectingπΒ±(π‘). This area changes because the arcπΎ(π‘)moves, and also because the line segmentπβπ+ moves. The rate of change is therefore the sum of
ββ«
πΎ(π‘)π ππ and the rate at which the segmentπβπ+ sweeps out area.
Figure 3.3: Left: The arcπΎ(π‘). Right: the angles π½Β±
Ifπ: πΎ(π‘) βis the tangent angle along the arc (i.e.ππ =1(π)), then the curvature integral is
β«
πΎ(π‘)
π ππ =ππβ(π‘)βππ+(π‘). The line segmentπ+πβmoves with angular velocityπ π‘π
πβ+(π‘)and therefore adds area to the region enclosed byπΎ(π‘)at the rate
1 2 n
π +
π
2
β π β
π
2oπ π+β(π‘) π π‘
in whichπ Β±
π are evaluated atπ =πβ+(π‘) +π
0. Our construction of the cap implies that πβ+(π‘) =βπ‘+π(1), and that this relation may be differentiated: π π‘ππβ+(π‘) =β1+π(π‘). The radii π Β±
π(πβ+(π‘) +π
0, π‘) are given in terms of their Yin-Yang coordinates π¦Β±(π‘) via
π Β±
π(π+β(π‘) +π
0, π‘) =(πβ+(π‘) +π
0βπ‘+π¦Β±). It follows that atπβ+(π‘) +π
0
1 2 n
π +
π
2
β π β
π
2o
=(πβ+(π‘) +π
0βπ‘+π¦+)2β (πβ+(π‘) +π
0βπ‘+π¦β)2
= β²(π¦+βπ¦β) in which ,β²are evaluated at πβ(π‘) +π
0+π¦Λ for some Λπ¦ β [π¦β, π¦+] that is provided by the mean value theorem. The asymptotics of imply thatβ²= 1+π(1) < 2. Our choice ofπ
0was such that 0 < π¦+βπ¦β β©½ πΏ. Hence
1 2 n
π +
π
2
β π β
π
2oπ πβ+(π‘) π π‘
β©½ 2πΏ.
In total, the rate at which the area π΄(π‘) enclosed by the arc πΎ(π‘) grows is bounded by
ππ΄
π π‘ β©½ β ππβ(π‘) βππ+(π‘) +2πΏ.
We estimate the change in tangent angle across the arc πΎ(π‘). Let π½+ be the coun- terclockwise angle from the ray π = πβ+(π‘) +π
0 to the tangent ππ to πΎ at π+, and similarly, let π½β be the counterclockwise angle from the same ray to the tangent to πΎ atπβ (see Figure 3.3). We have 0< π½+ < π < π½βandππβ βππ+ =π½ββπ½+. Recall thatπ β β¨π , ππ β© > 0 along π(π‘). Since π = ππ andβ¨π , ππ β© = 1
2 π
ππ π2, where π = β₯πβ₯, it follows thatπβ 1
2π2increases as one traverses πΎfrom π+ toπβ. Thus, at any point with polar coordinates(π , π)onπΎone has
π > ππ++1 2
(π2βπ2
π+).
The lowest value π has on πΎ occurs at the point πβ, and we have just shown that π2
π+βπ2
πβ β©½ 2πΏ. Henceπ > ππ+β2πΏon the entire arcπΎ. It follows that ifπ½+ > 3
4π, then the angle between the tangent toπΎand the rayπ π+ (π is the origin) will always be at least 34π β2πΏ, i.e. more than 58π, provided we chooseπΏ < π
16.
Consider the line β through π+ whose angle with π π+ is 58π. The euclidean distance between πβ and π+ is ππ+ βππβ β©½ 2πΏ/(ππβ +ππ+) β©½ πΆ πΏ|π‘|β1/2, since ππ+ > ππβ β³ |π‘|1/2.
At this scale the Yin-Yang leaves will be almost straight lines near πΒ±, so that the lineβ then intersects the Yin-Yang leaf with π¦ = π¦β(π‘) at a pointπβ(π‘), also at a distanceπ(πβ, πβ) β² πΏ|π‘|β1/2.
π+
π(π‘) βπβ+(π‘) βπ
0 βΌ π(πβ, πβ)
ππβ β² πΏ|π‘|β1βͺ π
0. Hence the largest polar angle onπΎ will be at most
π+
π(π‘) β©½ π+β(π‘) +π
0+πΏ|π‘|β1 β©½ πβ+(π‘) +2π
0.
Thus we find that if|π‘|is sufficiently large, then eitherππ(π‘) < πβ(π‘), or elseπ½+ < 5
8π. In the latter case the area enclose byπΎ(π‘) decreases faster than
ππ΄
π π‘ β©½ βπ½β+π½++2πΏ β©½ βπ+ 5 8
π+2πΏ=β3 8
π+2πΏ <βπ 4 , again, assumingπΏ < π/16.
We now finally prove that π+
π(π‘) βπ+β(π‘) is uniformly bounded for all π‘ β [βπ , π] and π.
At π‘ = βπ we have π+
π(π‘) < π+β(π‘) + π
0, by definition of the initial curve π(βπ). Hence, if at any timeπ‘
1 < π one hasπ+
π(π‘) > πβ+(π‘) +2π
0, then there is a largest interval (π‘
2, π‘
3) β π‘
1on which π+
π(π‘) > πβ+(π‘) +2π
0. In particular, atπ‘ =π‘
2one has π+
π(π‘) =π+β(π‘) +2π
0.
Define the arcπΎ(π‘) as above. Its enclosed area is at mostπ, where we may assume thatπ < π/4. During the time interval(π‘
2, π‘
3)the area decreases at a rate of at least π/4, and therefore the lengthπ‘
3βπ‘
2of the time interval cannot exceedπ/(π/4) =1.
At timeπ‘ =π‘
2we hadπ+
π(π‘) =π+β(π‘) +2π
0. Sinceπ+
π(π‘)is a nonincreasing function, we have throughout(π‘
2, π‘
3) π+
π(π‘) βπ+β(π‘) +2π
0 β©½ π+
π(π‘
2) βπβ+(π‘) +2π
0
β©½ π+
π(π‘
2) βπβ+(π‘
2) +2π
0+π+β(π‘) βπ+β(π‘
2)
β©½ πβ+(π‘
3) βπβ+(π‘
2). Since π π‘ππβ+(π‘) =1+π(1) we find that
π+
π(π‘) βπ+β(π‘) +2π
0 β©½ 1+π(1) < 2
for allπ‘ β (π‘
2, π‘
3).
To summarize, we can now decompose every very-old solution π(π‘), forπ‘ β [βπββ 1/2,βπβ] into four graphs in two different coordinate systems, π Β±
π(π , π‘) in polar coordinates, andΞΒ±
π(π¦, π‘)in Yin-Yang polar coordinates.
[Curvature bounds] There existπ < 0 such that for anyπβ² < π the lengths πΏπ(π‘) and curvatures of π(π‘)are uniformly bounded for all πand allπ‘ β [πβ², π β2]. The length bounds follow from the fact that in (π¦, π) coordinates each π(π‘) is contained in a uniformly bounded rectangle |π¦| β©½ π/2, βπ‘ β©½ π β©½ πβ+(π‘) +2π
0+2, and the fact that π(π‘) decomposes into four segments on each of which both π¦ and π are monotone.
Consider a given value πβ² < π. Assume that our Lemma fails, and that along some subsequence π the maximal curvature of π(π‘) with π‘ β [πβ², π β2] becomes unbounded.
Forπ‘ β [πβ²β2, π] the lengthsπΏπ(π‘)of π(π‘) are uniformly bounded by someπΏ >0.
It follows that
β« πβ²β1 πβ²β2
β«
π(π‘)
π 2ππ π π‘ =
βπΏπ(π‘)πβ²β1
πβ²β2
< πΏ . Therefore, there is a sequenceπ‘π β [πβ²β2, πβ²β1]such that
β«
π(π‘π)
π 2ππ < πΏ .
By a Sobolev embedding theorem this implies that the curves π(π‘π) are uniformly πΆ1,1/2, i.e. they are continuously differentiable, and their tangent angles ππ are uniformly HΓΆlder continuousβin fact, for any two points at arclength coordinates π 1, π
2in π(π‘π)one has
|π(π
2) βπ(π
1) | =
β« π 2
π 1
π ππ β©½ β
π 2βπ
1
βοΈβ« π 2
π 1
π 2ππ β©½
β πΏ
β π 2βπ
1.
It follows that all π(π‘π) are uniformly locally Lipschitz curves. Now consider the solutions to curve shortening with π(π‘π)as initial data, i.e. considerΛπ(π‘) =π (π‘π +π‘). These solutions all exist for 0 β©½ π‘ β©½ π βπ‘π β©Ύ π βπβ². Supposing that along some subsequence ofπ‘π the curvatures of the Λπ are not bounded for 1 β©½ π‘ β©½ π βπβ², we pass to a further subsequence for which the initial curves Λπ(0) converge in πΆ1 to some limit curveΛβ. The enclosed areas of theΛπ(0)then also converge, and hence, by Graysonβs theorem (Grayson, 1986) the evolution by Curve Shorteningβ(π‘)starting
from β exists for 0 β©½ π‘ β©½ π βπβ². By continuous dependence on initial data it follows that the solutionsΛπ(π‘)converge inπΆβtoβ(π‘)on any time interval[πΏ, πβπβ²] withπΏ > 0. This implies that the curvatures of the Λπ(π‘) are uniformly bounded for π‘ β [1, πβπβ²], which then implies that the curvatures of π(π‘)are uniformly bounded after all forπ‘ β [π‘π +1, π‘π +π βπβ²] β [πβ², πβ2].
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