Classification of convex ancient free boundary curve shortening flows in the disc.

Define

$$f(\lambda,\theta)\doteqdot\cos\theta\tanh(\lambda\cos\theta)-\sin\theta\cot(\lambda\sin\theta))\,.$$

Observe that

$$\lim_{\lambda\searrow 0}f(\lambda,\theta)=-\infty\,,\;\;\lim_{\lambda\nearrow\frac{\pi}{2\sin\theta}}f(\lambda,\theta)=\cos\theta\tanh(\tfrac{\pi}{2}\cot\theta)>0\,,$$

and

(2)

$$\frac{\partial f}{\partial\lambda}=\cos^{2}\theta(1-\tanh^{2}(\lambda\cos\theta))+\sin^{2}\theta(1+\cot^{2}(\lambda\sin\theta))>0\,.$$

The first claim follows.

Next observe that

	$\displaystyle\frac{\partial f}{\partial\theta}={}$	$\displaystyle-\sin\theta\tanh(\lambda\cos\theta)-\lambda\cos\theta\sin\theta\operatorname{sech}^{2}(\lambda\cos\theta)$
		$\displaystyle-\cos\theta\cot(\lambda\sin\theta)+\lambda\sin\theta\cos\theta\csc^{2}(\lambda\sin\theta)\,.$

Given $\theta\in(0,\frac{\pi}{2})$, we obtain, at the unique zero $\lambda\in(0,\frac{\pi}{2\sin\theta})$ of $f(\cdot,\theta)$,

	$\displaystyle\frac{\partial f}{\partial\theta}={}$	$\displaystyle-\sin\theta\tan\theta\cot(\lambda\sin\theta)-\lambda\cos\theta\sin\theta(1-\tan^{2}\theta\cot^{2}(\lambda\sin\theta))$
		$\displaystyle-\cos\theta\cot(\lambda\sin\theta)+\lambda\sin\theta\cos\theta\csc^{2}(\lambda\sin\theta)$
	$\displaystyle={}$	$\displaystyle-\sec\theta\cot(\lambda\sin\theta)\left(1-\lambda\sin\theta\cot(\lambda\sin\theta)\right).$

Since $Y\cot Y<1$ for $Y\in(0,\frac{\pi}{2})$, this is less than zero. The second claim follows. ∎

By the strong maximum principle, the two families of curves can never develop contact at an interior point. Since the families are monotonic, they cannot develop boundary contact at a boundary point $(\cos\theta,\sin\theta)$ with $|\theta|\leq\theta_{\alpha}$. On the other hand, since $\lambda\leq\lambda(\theta_{\alpha})$, (2) implies that

$$f(\lambda,\theta_{\alpha})\leq f(\lambda_{\alpha},\theta_{\alpha})=0\,,$$

and hence, by the argument of Lemma 2.2,

$$f(\lambda,\theta)\leq 0\;\;\text{for}\;\;\theta\geq\theta_{\alpha}\,.$$

So the Hopf boundary point lemma implies that no boundary contact can develop for $\theta\geq\theta_{\alpha}$ either. ∎

Existence of a maximal solution to curve shortening flow out of $\Gamma^{\rho}$ which meets $\partial B^{2}$ orthogonally was proved by Stahl [22, Theorem 2.1]. Stahl also proved that this solution remains convex and locally uniformly convex and shrinks to a point on the boundary of $B^{2}$ at the final time (which is finite) [21, Proposition 1.4]. We obtain $\{\Gamma_{t}^{\rho}\}_{t\in[\alpha_{\rho},0)}$ by time-translating Stahl’s solution.

By uniqueness of solutions $\Gamma^{\rho}_{t}$ remains reflection symmetric about the $y$-axis for $t\in(\alpha_{\rho},0)$, so the final point is $(0,1)$.

The reflection symmetry also implies that $\kappa^{\rho}_{s}=0$ at the point $p_{t}\doteqdot\Gamma^{\rho}_{t}\cap\{x=0\}$ for all $t\in[\alpha_{\rho},0)$. By [21, Proposition 2.1], $\kappa^{\rho}_{s}=\kappa^{\rho}>0$ at the boundary point $q_{t}\doteqdot\partial\Gamma^{\rho}_{t}\cap\{x>0\}$ for all $t\in(\alpha_{\rho},0)$. Applying Sturm’s theorem [4] to $\kappa^{\rho}_{s}$, we thus find that $\kappa^{\rho}_{s}>0$ on $\Gamma^{\rho}_{t}\cap B^{2}\cap\{x>0\}$ for all $t\in(\alpha_{\rho},0)$.

Since $\mathrm{C}_{\theta_{\rho}}\subset\Omega^{\rho}$, the final property follows from Proposition 2.1. ∎

To prove the lower bound for $\overline{\kappa}$, it suffices to show that the circle $\mathrm{C}_{\overline{\theta}(t)}$ (see (1)) lies locally below $\Gamma_{t}$ near $q_{t}$. If this is not the case, then, locally around $q_{t}$, $\Gamma_{t}$ lies below $\mathrm{C}_{\overline{\theta}(t)}$ and hence $\kappa(q_{t})\leq\tan\overline{\theta}(t)$. But then we can translate $\mathrm{C}_{\overline{\theta}(t)}$ downwards until it touches $\Gamma_{t}$ from below in an interior point at which the curvature must satisfy $\kappa\geq\tan\overline{\theta}(t)$. This contradicts the unique maximization of the curvature at $q_{t}$.

The estimate (4) now follows by integrating the inequality

$$\frac{d}{dt}\sin\overline{\theta}=\cos\overline{\theta}\,\overline{\kappa}\geq\sin\overline{\theta}$$

between any initial time $t$ and the final time $0$ (at which $\overline{\theta}=\frac{\pi}{2}$ since the solution contracts to the point $(0,1)$).

The upper bound for $\underline{y}$ follows from convexity and the boundary condition $\overline{y}=\sin\overline{\theta}$. To prove the lower bound, we will show that the circle $\mathrm{C}_{\overline{\theta}(t)}$ lies nowhere above $\Gamma_{t}$. Suppose that this is not the case. Then, since $\mathrm{C}_{\overline{\theta}(t)}$ lies locally below $\Gamma_{t}$ near $q_{t}$, we can move $\mathrm{C}_{\overline{\theta}(t)}$ downwards until it is tangent from below to a point $p^{\prime}_{t}$ on $\Gamma_{t}\cap\{x\geq 0\}$, at which we must have $\kappa\geq\tan\overline{\theta}(t)$. But then, since $\kappa_{s}\geq 0$ in $\{x>0\}$, we find that $\kappa\geq\tan\overline{\theta}(t)$ for all points between $p^{\prime}_{t}$ and $q_{t}$. But this implies that this whole arc (including $p^{\prime}_{t}$) lies above $\mathrm{C}_{\overline{\theta}(t)}$, a contradiction.

To prove the upper bound for $\underline{\kappa}$, fix $t$ and consider the circle $C$ centred on the $y$-axis through the points $p_{t}$ and $q_{t}$. Its radius is $r(t)$, where

$$r\doteqdot\frac{\cos^{2}\overline{\theta}+(\sin\overline{\theta}-\underline{y})^{2}}{2(\sin\overline{\theta}-\underline{y})}\,.$$

We claim that $\Gamma_{t}$ lies locally below $C$ near $p_{t}$. Suppose that this is not the case. Then, by the symmetry of $\Gamma_{t}$ and $C$ across the $y$-axis, $\Gamma_{t}$ lies locally above $C$ near $p_{t}$. This implies two things: first, that

$$\kappa(p_{t})\geq r^{-1},$$

and second, that, by moving $C$ vertically upwards, we can find a point $p^{\prime}_{t}$ (the final point of contact) which satisfies

$$\kappa(p^{\prime}_{t})\leq r^{-1}\,.$$

These two inequalities contradict the (unique) minimization of $\kappa$ at $p_{t}$. We conclude that

$$\underline{\kappa}\leq\frac{2(\sin\overline{\theta}-\underline{y})}{\cos^{2}\overline{\theta}+(\sin\overline{\theta}-\underline{y})^{2}}\leq\tan\overline{\theta}$$

due to the lower bund for $\underline{y}$. ∎

For each $\rho>0$, consider the old-but-not-ancient solution $\{\Gamma^{\rho}_{t}\}_{t\in[\alpha_{\rho},0)}$, $\Gamma^{\rho}_{t}=\partial\Omega^{\rho}_{t}$, constructed in Lemma 2.5. By (4), $\Omega^{\rho}_{t}$ contains $\mathrm{C}_{\omega(t)}\cap B^{2}$, where $\omega(t)\in(0,\frac{\pi}{2})$ is uniquely defined by

$$\frac{1-\cos\omega(t)}{\sin\omega(t)}=\mathrm{e}^{t}\,.$$

If we represent $\Gamma^{\rho}_{t}$ as a graph $x\mapsto y^{\rho}(x,t)$ over the $x$-axis, then convexity and the boundary condition imply that $|y^{\rho}_{x}|\leq\tan\omega$. Since $\omega(t)$ is independent of $\rho$, Stahl’s (global in space, interior in time) Ecker–Huisken type estimates [22] imply uniform-in-$\rho$ bounds for the curvature and its derivatives. So the limit

$$\{\Gamma^{\rho}_{t}\}_{t\in[\alpha_{\rho},0)}\to\{\Gamma_{t}\}_{t\in(-\infty,0)}$$

exists in $C^{\infty}$ (globally in space on compact subsets of time) and the limit $\{\Gamma_{t}\}_{t\in(-\infty,0)}$ satisfies curve shortening flow with free boundary in $B^{2}$. On the other hand, since $\{\Gamma^{\rho}_{t}\}_{t\in(\alpha_{\rho},0)}$ contracts to $(0,1)$ as $t\to 0$, (the contrapositive of) Proposition 2.1 implies that $\Gamma^{\rho}_{t}$ must intersect the closed region enclosed by $\mathrm{C}_{\theta^{+}(t)}$ for all $t<0$. It follows that $\Gamma_{t}$ must intersect the closed region enclosed by $\mathrm{C}_{\theta^{+}(t)}$ for all $t<0$. Since each $\Gamma_{t}$ is the limit of convex boundaries, each is convex. It follows that $\Gamma_{t}$ converges to $(0,1)$ as $t\to 0$ and, by [22, Corollary 4.5], that $\Gamma_{t}$ is locally uniformly convex for each $t$. ∎

It suffices to prove that $\frac{\kappa}{\cos\theta}\geq\lambda\tan(\lambda y)$ on each of the old-but-not-ancient solutions $\{\Gamma^{\lambda}_{t}\}_{t\in[\alpha_{\lambda},0)}$. Note that equality holds on the initial timeslice $\Gamma^{\lambda}_{\alpha_{\lambda}}=\mathrm{A}^{\lambda}_{t_{\lambda}}$.

Given any $\mu<\lambda$, set $u\doteqdot\mu\tan(\mu y)$ and $v\doteqdot x_{s}=\cos\theta=\left\langle\nu,e_{2}\right\rangle$. Observe that

$$u_{s}=\mu^{2}\sec^{2}(\mu y)\sin\theta\,,\;\;(\partial_{t}-\Delta)u=-2\mu^{2}\sec^{2}(\mu y)\sin^{2}\theta u\,,$$

$$v_{s}=-\kappa\sin\theta\;\;\text{and}\;\;(\partial_{t}-\Delta)v=\kappa^{2}v\,.$$

At an interior maximum of $\frac{uv}{\kappa}$ we observe that

$$\frac{\nabla\kappa}{\kappa}=\frac{\nabla u}{u}+\frac{\nabla v}{v}$$

and hence

	$\displaystyle 0\leq(\partial_{t}-\Delta)\frac{uv}{\kappa}={}$	$\displaystyle\frac{uv}{\kappa}\left(\frac{(\partial_{t}-\Delta)u}{u}-2\left\langle\frac{\nabla u}{u},\frac{\nabla v}{v}\right\rangle\right)$
(8)		$\displaystyle={}$	$\displaystyle 2\mu^{2}\sec^{2}(\mu y)\sin^{2}\theta\left(1-\frac{uv}{\kappa}\right).$

At a (without loss of generality right) boundary maximum of $\frac{uv}{\kappa}$, we have $y_{s}=y$ and $\kappa_{s}=\kappa$, and hence

	$\displaystyle\left(\frac{uv}{\kappa}\right)_{s}={}$	$\displaystyle\frac{uv}{\kappa}\left(\frac{u_{s}}{u}+\frac{v_{s}}{v}-\frac{\kappa_{s}}{\kappa}\right)$
	$\displaystyle={}$	$\displaystyle\frac{uv}{\kappa}\left(\frac{\sec^{2}(\mu y)\mu y}{\tan{\mu y}}-\kappa\frac{y}{v}-1\right)$
(9)		$\displaystyle\leq{}$	$\displaystyle\left(\frac{uv}{\kappa}-1\right)\tan(\mu y)\mu y\,.$

We may now conclude that $\max_{\overline{\Gamma}^{\lambda}_{t}}\frac{uv}{\kappa}$ remains less than one. Indeed, if $\frac{uv}{\kappa}$ ever reaches $1$, then there must be a first time $t_{0}>0$ and a point $x_{0}\in\overline{\Gamma}_{t}$ at which this occurs (note that $uv/\kappa$ is continuous on $\overline{\Gamma}_{t}$ up to the initial time). The point $x_{0}$ cannot be an interior point, due to (8), and it cannot be a boundary point, due to (9) and the Hopf boundary point lemma. We conclude that

$$\frac{\kappa}{\cos\theta}\geq\mu\tan(\mu y)$$

on $\{\Gamma^{\lambda}_{t}\}_{t\in[\alpha_{\lambda},0)}$ for all $\mu<\lambda$. Now take $\mu\to\lambda$. ∎

We will prove the estimate for each old-but-not-ancient solution $\{\Gamma^{\lambda}_{t}\}_{t\in(\alpha_{\lambda},0)}$. We first prove a crude gradient estimate of the form

(12)

$$|\kappa_{s}|\leq 2\kappa$$

for $t$ sufficiently negative. It will suffice to prove that

(13)

$$|\kappa_{s}|-\kappa+\left\langle\gamma,\nu\right\rangle\leq 0\,.$$

Indeed, since $\left\langle\gamma,\nu\right\rangle_{s}=\kappa\left\langle\gamma,\tau\right\rangle$ has the same sign as the $x$-coordinate, we may estimate, as in (7),

(14)

$$|\!\left\langle\gamma,\nu\right\rangle\!|\leq|\!\left\langle\gamma,\nu\right\rangle\!|_{x=0}\leq\lambda^{-2}\kappa|_{x=0}=\lambda^{-2}\min_{\Gamma_{t}}\kappa\leq\kappa\,.$$

For $\lambda$ sufficiently close to $\lambda_{0}$, we have $\kappa|_{t=\alpha_{\lambda}}<1/2$. Denote by $T^{\lambda}$ the first time at which $\kappa$ reaches $1/2$. Since $\kappa$ is continuous up to the initial time $\alpha_{\lambda}$, we have $T^{\lambda}>\alpha_{\lambda}$. We claim that (13) holds for $t<T^{\lambda}$. Indeed, it is satisfied on the initial timeslice $\Gamma^{\lambda}_{\alpha_{\lambda}}=\mathrm{A}^{\lambda}_{t_{\lambda}}$ since

$$\kappa_{s}^{2}-\kappa^{2}=\lambda^{2}\left(\cos^{2}\theta\sin^{2}\theta-\sin^{2}\theta-a^{2}_{\lambda}\right)=-\lambda^{2}(\sin^{4}\theta+a_{\lambda}^{2})\leq 0\,,$$

whereas $\left\langle\gamma,\nu\right\rangle\leq 0$. We will show that

$$f_{\varepsilon}\doteqdot|\kappa_{s}|-\kappa+\left\langle\gamma,\nu\right\rangle-\varepsilon\mathrm{e}^{t-\alpha_{\lambda}}$$

remains negative up to time $T^{\lambda}$. Suppose, to the contrary, that $f_{\varepsilon}$ reaches zero at some time $t<T^{\lambda}$ at some point $p\in\overline{\Gamma}_{t}$. Since $|\kappa_{s}|-\kappa+\left\langle\gamma,\nu\right\rangle$ vanishes at the boundary, $p$ must be an interior point. Since $\kappa_{s}$ vanishes at the $y$-axis, and the curve is symmetric, we may assume that $x(p)>0$. At such a point,

	$\displaystyle 0\leq(\partial_{t}-\Delta)f_{\varepsilon}={}$	$\displaystyle\kappa^{2}(4\kappa_{s}-\kappa+\left\langle\gamma,\nu\right\rangle)-2\kappa-\varepsilon\mathrm{e}^{t-\alpha_{\lambda}}$
	$\displaystyle={}$	$\displaystyle\kappa^{2}(3[\kappa-\left\langle\gamma,\nu\right\rangle]+4\varepsilon\mathrm{e}^{t-\alpha_{\lambda}})-2\kappa-\varepsilon\mathrm{e}^{t-\alpha_{\lambda}}\,.$

Recalling (14) and estimating $\kappa\leq\frac{1}{2}$ yields

$$0\leq 6\kappa^{3}-2\kappa+(4\kappa^{2}-1)\varepsilon\mathrm{e}^{t-\alpha_{\lambda}}<0\,,$$

which is absurd. So $f_{\varepsilon}$ does indeed remain negative, and taking $\varepsilon\to 0$ yields (12) for $t<T^{\lambda}$.

Since $\mathrm{Length}(\Gamma_{t}\cap\{x\geq 0\})\leq 1$, integrating (12) yields

$$\overline{\kappa}\leq\mathrm{e}^{2}\underline{\kappa}\;\;\text{for}\;\;t<T^{\lambda}\,.$$

Recalling (3) and (4), this implies that

$$\overline{\kappa}\leq\mathrm{e}^{2}\frac{\mathrm{e}^{t}}{\sqrt{1-\mathrm{e}^{2t}}}\;\;\text{for}\;\;t<T^{\lambda}\,.$$

Taking $t=T^{\lambda}$ we find that $T^{\lambda}\geq T$, where $T$ is independent of $\lambda$, so we conclude that

$$\overline{\kappa}\leq C\mathrm{e}^{t}\;\;\text{for}\;\;t<T\,,$$

where $C$ and $T$ do not depend on $\lambda$. ∎

Consider the old-but-not-ancient solution $\{\Gamma^{\lambda}_{t}\}_{t\in(-\infty,0)}$. By (11), we can find $C<\infty$ and $T>-\infty$ such that

	$\displaystyle(\partial_{t}-\Delta)\frac{\kappa}{y}={}$	$\displaystyle\kappa^{2}\frac{\kappa}{y}+2\left\langle\nabla\frac{\kappa}{y},\frac{\nabla y}{y}\right\rangle$
	$\displaystyle\leq{}$	$\displaystyle C\mathrm{e}^{2t}\frac{\kappa}{y}+2\left\langle\nabla\frac{\kappa}{y},\frac{\nabla y}{y}\right\rangle\;\;\text{for}\;\;t<T\,.$

Since, at a boundary point,

$$\left(\frac{\kappa}{y}\right)_{s}=\frac{\kappa_{s}}{y}-\frac{\kappa}{y}\frac{y_{s}}{y}=0\,,$$

the Hopf boundary point lemma and the ode comparison principle yield

$$\max_{\Gamma^{\lambda}_{t}}\frac{\kappa}{y}\leq C\max_{\Gamma^{\lambda}_{\alpha_{\lambda}}}\frac{\kappa}{y}\;\;\text{for}\;\;t\in(\alpha_{\lambda},T)\,.$$

But now

$\displaystyle(\partial_{t}-\Delta)\frac{\kappa}{y}\leq{}$

$\displaystyle C\mathrm{e}^{2t}\max_{\Gamma^{\lambda}_{\alpha_{\lambda}}}\frac{\kappa}{y}+2\left\langle\nabla\frac{\kappa}{y},\frac{\nabla y}{y}\right\rangle\;\;\text{for}\;\;t<T\,,$

and hence, by ode comparison,

$\displaystyle\max_{\Gamma^{\lambda}_{t}}\frac{\kappa}{y}\leq\max_{\Gamma^{\lambda}_{\alpha_{\lambda}}}\frac{\kappa}{y}\left(1+C\mathrm{e}^{2t}\right)\;\;\text{for}\;\;t\in(\alpha_{\lambda},T)\,.$

Since, on the initial timeslice $\Gamma^{\lambda}_{\alpha_{\lambda}}=\mathrm{A}_{t_{\lambda}}^{\lambda}$,

$$\frac{\kappa}{y}=\frac{\lambda\tan(\lambda y)}{y}\cos\theta\,,$$

the claim follows upon taking $\lambda\to\lambda_{0}$. ∎

Set $A(t)\doteqdot\operatorname{area}(\Omega_{t})$. Integrating the variational formula for area yields

$$A(t)=\int_{t}^{0}\!\!\!\int_{\Gamma_{t}}d\theta\,,$$

where $\theta$ is the turning angle. Since convexity ensures that the total turning angle $\int_{\Gamma_{t}}d\theta$ is increasing and $A(t)\leq\pi$ for all $t$, we find that

$$\int_{\Gamma_{t}}d\theta\to 0\,\,\text{as}\;\;t\to-\infty\,.$$

Monotonicity of the flow, the free boundary condition and convexity now imply that the enclosed regions $\Omega_{t}$ satisfy

$$\overline{\Omega}_{t}\to B^{2}\cap\{y\geq 0\}\,\,\text{as}\;\;t\to-\infty$$

in the Hausdorff topology.

If we now represent $\Gamma_{t}$ graphically over the $x$-axis, then convexity and the boundary condition ensure that the height and gradient are bounded by the height at the boundary. Stahl’s estimates [22] now give bounds for $\kappa$ and its derivatives up to the boundary depending only on the height at the boundary. We then get smooth subsequential convergence along any sequence of times $t_{j}\to-\infty$. The claim follows since any sublimit is the horizontal segment. ∎

Given any $\omega\in(0,\pi)$, we define the halfspace

$$H_{\omega}=\{(x,y):(x,y)\cdot(\sin\omega,-\cos\omega)>0\}$$

and denote by $R_{\omega}$ the reflection about $\partial H_{\omega}$. We first claim that, for every $\omega$, there exists $t=t_{\omega}$ such that

$$(R_{\omega}\cdot\Gamma_{t})\cap(\Gamma_{t}\cap H_{\omega})=\emptyset\;\;\text{for all}\;\;t<t_{\omega}\,.$$

Assume that the claim is not true. Then there exists $\omega\in(0,\pi)$, a sequence of times $t_{i}\to-\infty$, and a sequence of pairs of points $p_{i},q_{i}\in\Gamma_{t_{i}}$ such that $R_{\omega}(p_{i})=q_{i}$. This implies that the line passing through $p_{i}$ and $q_{i}$ is parallel to the vector $(\sin\omega,-\cos\omega)$, so the mean value theorem yields for each $i$ a point $r_{i}$ on $\Gamma_{t_{i}}$ where the normal is parallel to $(\cos\omega,\sin\omega)$. This contradicts Lemma 3.1. ∎