Nonlocal estimates for the volume preserving mean curvature flow and applications

Abstract

We obtain estimates on nonlocal quantities appearing in the volume preserving mean curvature flow (VPMCF) in the closed, Euclidean setting. As a result we demonstrate that blowups of finite time type I singularities of VPMCF are ancient solutions to mean curvature flow (MCF), prove that monotonicity methods may always be applied at these finite times and obtain information on the asymptotics of the flow. In the case of type II singularities, asymptotic flows corresponding to ’Hamilton’s rescaling procedures’ are eternal solutions of the MCF.

1 Introduction

A key question for the study of Volume Preserving Mean Curvature Flow (VPMCF) is the extent to which it has similar properties to the Mean Curvature Flow (MCF). It is now known that the extra nonlocal term in VPMCF causes significant differences. For example, Cabezas-Rivas and Miquel [13] prove that key properties used in many MCF methods do not hold for VPMCF, such as preservation of mean convexity (along with most other curvature conditions bar convexity). This demonstrates the need to a VPMCF specific methods, and in this paper we begin to develop some of these. In this paper, we give evidence that there are still significant links between singularities of these two flows: We show that blowups of type I finite time singularities of VPMCF are indeed ancient solutions to MCF (compare with [13, point (c), p 288]). Finite time type II singularities of the VPMCF look like eternal solutions of MCF after a suitable parabolic blowup procedure (first used by Hamilton [29]). This is particularly interesting in the light of the significant progress has been made in recent years in the classification of such singularities – see for example [3, 8, 14, 15].

In order to study these blowup solutions we needed to get control over the (nonlocal) Lagrange parameter in the equation. We found links between $L^2$-bounds on the (nonlocal) averaged mean curvature and the extrinsic diameter along the flow. These estimates play an important role in the monotonicity formula, and essentially allow initial results on the analysis of singularities to be proven without assuming any further bounds on $\overline{H}$.

The VPMCF of n dimensional hypersurfaces in $\mathbb {R}^{n+1}$ is the $L^2$-gradient flow of the area functional under the constraint that the $(n+1)$-dimensional enclosed volume is fixed. This flow was introduced by Gage [25] for curves and by Huisken [30] for hypersurfaces. We consider the closed setting, meaning that $M^n$ is a smooth, orientable, compact n-dimensional manifold without boundary. Then a smooth family of immersions $X:M^n\times [0,T)\rightarrow \mathbb {R}^{n+1}$ is a solution of the VPMCF if

$$\begin{aligned} \left\{ \begin{aligned} \frac{ d X}{dt}&= -(H-\overline{H})\nu \ ,{} & {} \text {on}\,M^n \times [0,T),\\ X(\cdot , t)&= X_0 \ {}{} & {} \text { on }\,M^n, \end{aligned}\right. \end{aligned}$$

where $X_0: M^n \rightarrow \mathbb {R}^{n+1}$ is a given immersion. Here, $\nu$ is a unit normal along $X(\cdot , t) =:M_t$, $h_{ij} = - \langle \partial _i\partial _jX,\nu \rangle$ are the local coefficients of the second fundamental form A, $g_{ij} = \left\langle X_i , X_j \right\rangle$ is the induced metric and $H = \sum g^{ij}h_{ij}$ is the mean curvature. Furthermore, $\overline{H} =\frac{\int H d\mu _t}{|M_t|}$ is the Lagrange parameter of the above mentioned geometric variational problem. Thus, $\frac{d}{dt}|M_t|\le 0$ and $\frac{d}{dt}V(t)=0$ is satisfied along the flow (for the precise definition of the enclosed volume V(t) see Sect. 2). Note that the formulation of the VPMCF does not depend on the choice of the unit normal. If it is possible to choose between an “inner” and “outer” unit normal with respect to $M_t$, then we choose the outer unit normal, so that $H>0$ for round spheres.

We collect some known results about the VPMCF. Gage (for $n=1$) and Huisken (for $n\ge 2$) proved that convex solutions of the VPMCF stay convex and do not develop singularities. They converge smoothly to a round sphere enclosing the same amount of volume as the initial hypersurface. In [23], Escher and Simonett proved that hypersurfaces being $h^{1+\beta }$ close to a round sphere converge to (possibly another) round sphere with the same enclosed volume as the initial surface. Li obtained related results in [37]. For example, he proved convergence to a round sphere if the integrated trace-free second fundamental form $\overset{\circ }{A}_{ij} = h_{ij} - \frac{H}{n}g_{ij}$ of the initial surface is small enough in $L^2$. The needed smallness depends on $\max |A|(0),$ $\overline{H}(0)>0$ and $V(0)>0$, see [37, Theorem 1]. Further results about the VPMCF in a non-Euclidean ambient manifold can be found for example in [2, 11, 12, 21, 33, 44].

Athanassenas, later with Kandanaarachchi, studied in [4,5,6] the VPMCF of a rotational symmetric surface with and without boundary conditions. For example, in [6] it is shown that a height bound on the generating curve for the rotational symmetric surfaces prevents the flow from developing a singularity. The hypersurfaces then converge to a sphere (or a half-sphere in the considered boundary setting). Note that rotational symmetry (additionally to convexity) is one the of few properties that is known to be preserved under the VPMCF.

For curves, we expect stronger results compared to the hypersurface case because the main part of the Lagrange parameter has a geometric interpretation – the term $\int \kappa ds$ is scaling invariant and measures the turn of the angle of a curve. In the closed setting, this is just $2\pi m$ with $m\in {\mathbb {N}}$ (the index of a closed curve). So an $L^\infty$-in-time bound on $\overline{\kappa } = \frac{\int \kappa ds_t}{L_t}$ is immediate if one controls the length from below (e.g. by an isoperimetric inequality). Inspired by [16], Escher and Ito proved in [22] the existence of singularities of the VPMCF for curves under some conditions on the initial curve. One scenario of the initial curve leading to a singularity is that the enclosed volume of the initial curve with $\frac{1}{2\pi }\int \kappa ds \ge 1$ is negative. Inspired by [22], the second author proved in [40] an analogous result for curves in the (volume preserving) free boundary setting. In this setting, $\int \kappa ds$ is not bounded a priori, but still, an $L^\infty$-in-time bound proven earlier in [39] is used in the proof of existence of singularities. In [39], the second author had introduced the (Neumann) free boundary setting for the VPMCF in detail (for curves and surfaces) and proved convergence to a stationary solution for (non-closed) curves under some conditions on the initial curve. The most advanced result about closed curves can be found in [19]. There, Dittberner was able to derive a comparison principle between the intrinsic and extrinsic distance – based on previous work of Huisken [32] for the MCF – to show convergence to a sphere of a closed curve under the condition that the initial curve is embedded and satisfies $\int _{x}^y \kappa ds \ge -\pi$ for all x, y. This can be seen as the analogue of Grayson’s theorem – which says that the MCF first makes every embedded, closed curve convex and then shrinks it to a point while becoming more and more round, see [24, 27]. Without assuming $\int _{x}^y \kappa ds \ge -\pi$ for all x, y we cannot expect that such a strong result holds because there is an example, already suggested by Gage [25] and also studied by Dittberner [19] where this condition is violated and a self-intersections develops after starting the VPMCF. Numerical computations in fact indicate that then a singularity appears for this initial curve, see Mayer [42]. For curves, it was also shown recently, that a star-shaped, centrosymmetric set keeps this properties and converges to a round sphere, see [26].

Whether star-shapedness is perserved in general seems to be an open questions. In the work of Kim and Kwon certain approximating solutions of the VPMCF are considered. Using them, it is proven that a strong version of star-shapedness is preserved [34]. In [43], Mugnai, Seis and Spadaro presented a distributional formulation of the VPMCF using the setting of Caccioppoli sets. They show global-in-time existence of their weak flow. A phase field method of the VPMCF was studied for example in the work of Takasao [46, 47], see also the references therein. In which generality these weak flows or limits of approximating sets can be applied to any initially smoothly immersed surfaces in unclear to the authors. An immersed surface does not need to bound a domain, see also the definition of Alexandrov immersions below. Recently, Laux showed in in [36] that a (strong) solution where the hypersurfaces bound a domain agrees with a distributional solution he defines (as long as the smooth one exists).

We now point out the structure of the paper and formulate the main results. Due to the lack of preserved quantities under the VPMCF we need to work in a quite general setting. In Sect. 3, we study the connection between (extrinsic) diameter bounds and $L^2$-estimates of $\overline{H}(t)=\frac{\int H d\mu _t}{|M_t|}$. One of our results is the following:

$$\begin{aligned} \begin{array}{cc} &{}\textit{If the initial immersion has non-vanishing enclosed volume } V_0\not =0, \textit{ then there} \\ &{}\textit{are constants}~c \textit{ and } C \textit{ only depending on } M_0 \textit{ such that} \\ &{}\int _0^t \overline{H}^2 (\tau ) d\tau \le C(1+ t^2) e^{c\sqrt{t}}. \end{array} \end{aligned}$$

We provide an example of a convex, embedded curve where the diameter is growing for a short time (note the contrast to the classical MCF). In Sect. 4, we re-prove a monotonicity formula for the VPMCF already shown in [38] and study consequences for finite time singularities. In the monotonicity formula it is important that the $L^2$-norm of $\overline{H}$ is appearing in form of $\exp (-\frac{1}{2}\int _0^t\overline{H}^2(\tau ) d\tau )$ as a multiplicative factor in front of the usual integrated Gaußian in Huisken’s monotonicity formula [31]. Consequently, the $L^2$-control from Sect. 3 allows us to get properties of asymptotic flows appearing after suitable parabolic blowups at finite time singularities. One result is:

Parabolic blowups of VPMCF about finite time singularities of type I ($\max |A|^2 \le \frac{C}{T-t}$) produce properly immersed, ancient, non-compact homothetically

shrinking solutions of MCF.

Finite type II blowups are eternal solutions of the MCF after Hamilton’s parabolic rescaling.

In Sect. 4.3, we illustrate that an $L^\infty$-bound on $\overline{H}$ in fact implies that the diameter of $|M_t|$ along the VPMCF always stays uniformly bounded for all times.

Section 5 contains statements about the VPMCF with infinite life-span and uniform diameter bound. We introduce a new “extended isoperimetric ratio” by

$$\begin{aligned} {\mathcal {I}}(M_t): = \frac{n+1}{n}\overline{H} \frac{V(t)}{|M_t|}. \end{aligned}$$

Note that ${\mathcal {I}}$ of the round sphere is equal to one. We motivate this definition with the Alexandrov-Fenchel inequalities proven for k-convex, star-shaped domains in [28]. The main result of Sect. 5 is the following:

$$\begin{aligned} \begin{array}{cc} &{}\textit{If } V_0\not =0 \textit{ and } M_t \textit{ satisfies a uniform diameter bound along the VPMCF with } T=\infty ,\\ &{}\textit{then, for each } t_i\rightarrow \infty , \textit{ there exists a subsequence (not relabeled) such that} \\ {} &{}\textit{either } {\mathcal {I}}(M_{t_i})\rightarrow 0 \textit{ or } {\mathcal {I}}(M_{t_i})\rightarrow 1. \end{array} \end{aligned}$$

As a corollary of the Alexandrov-Fenchel inequalities [28] we get that

$$\begin{aligned} \begin{array}{cc} &{}\textit{Any mean convex, star-shaped solution of the VPMCF with uniformly bounded curvature}\\ &{}\textit{exists for all times and converges to a round sphere.} \end{array} \end{aligned}$$

In Sect. 6, we remind the reader of the notion of an Alexandrov immersion, which is—roughly speaking—an n-dimensional immersion that bounds an $(n+1)$-dimensional manifold that is immersed in ${\mathbb {R}}^{n+1}$ (see Definition 1 for the precise definition). Alexandrov immersions bound an $(n+1)$-dimensional “domain” that is allowed to have a certain kind of self-overlaps. The definition goes back to Alexandrov [1] in his work about closed surfaces of constant mean curvature in Euclidean space.¹ It has also been used successfully for example by Brendle in his work about minimal tori in ${\mathbb {S}}^3$, see [7]. Using maximum principle arguments in a “one-sided situation” we prove the following result:

$$\begin{aligned} \begin{array}{cc} &{}\textit{The property of being Alexandrov immersed is preserved under the VPMCF as long as } \\ &{}\overline{H}\ge 0~\textit{(using the outer unit normal for the definition of } \overline{H}). \end{array} \end{aligned}$$

Unfortunately, we found an (even embedded) example where $\overline{H}\ge 0$ is lost along the VPMCF. We explain this example in Appendix A.

2 Notation

We complement the definitions from the introduction with the formula of the signed volume on the immersion $M_t$

$$\begin{aligned}V(t):= {{\text {Vol}}}(M_t) = \frac{1}{n+1}\int _{M_t} \left\langle X , \nu \right\rangle d\mu .\end{aligned}$$

The defining property of VPMCF is that it preserves this volume, that is,

$$\begin{aligned}{{\text {Vol}}}^{n+1}(M_t)={{\text {Vol}}}^{n+1}(M_0)=:V_0\ .\end{aligned}$$

In particular if $\Omega _t\subset \mathbb {R}^{n+1}$ is a domain with smooth boundary $\partial \Omega _t=M_t$, then the divergence theorem implies that ${{\text {Vol}}}^{n+1}(M_t)={\mathcal {L}}^{n+1}(\Omega _t)$ where ${\mathcal {L}}^{n+1}$ is the $(n+1)$–dimensional Lebesgue measure.

We will regularly use the extrinsic diameter of the immersion $M_t$ defined by

$$\begin{aligned} {\text {diam}}(M_t):={\text {diam}}_{{{\text {ext}}}}(M_t) = \max _{(x,y)\in M^n\times M^n}|X(x,t) - X(y,t)|\ . \end{aligned}$$

(1)

This is distinct to the intrinsic metric diameter of $M_t$, ${{\text {diam}}}_{{{\text {int}}}}(M_t)$ where we have the inequality ${\text {diam}}(M_t)\le {{\text {diam}}}_{{{\text {int}}}}(M_t)$.

3 Diameter bounds and $L^2$-bounds of the Lagrange parameter

Lemma 1

Let $X:M^n\times [0,T) \rightarrow {\mathbb {R}}^{n+1}$ be a closed VPMCF for $t\in [0,T)$. Then the extrinsic diameter (as in Eq. 1) is Lipschitz continuous and therefore differentiable almost everywhere in [0, T). For a time of differentiability t it satisfies

$$\begin{aligned} \frac{d}{dt} {\text {diam}}(M_t) = \langle -(H(x,t) -\overline{H}(t)) \nu (x,t) + (H(y,t) - \overline{ H }(t)) \nu (y,t), \tfrac{X(x,t) - X(y,t)}{|X(x,t) - X(y,t)|}\rangle , \end{aligned}$$

where $(x,y)\in M^n\times M^n$ are points where the maximum of the diameter is attained.

Proof

First we note that the map $(x,y,t) \mapsto |X(x,t) - X(y,t)|^2$ is smooth and $M^n\times M^n$ is compact. Thus, Mantegazza’s version of ’Hamilton’s trick’² [41, Lemma 2.1.3] tells us that $t\mapsto \max _{(x,y)\in M^n\times M^n}|X(x,t) - X(y,t)|^2$ is Lipschitz and therefore differentiable almost everywhere in [0, T). For such a time t, the derivative can be computed as

$$\begin{aligned} \frac{d}{dt} \max _{(x,y)\in M^n\times M^n}|X(x,t) - X(y,t)|^2 = \frac{\partial }{\partial t} |X(x,t) - X(y,t)|^2\big |_{(x,y) = (x^*,y^*)}, \end{aligned}$$

where $(x^*, y^*)$ is a pair such that the maximum in $\max _{(x,y)\in M^n\times M^n}|X(x,t) - X(y,t)|^2$ is attained. Note that $|X(x^*,t) - X(y^*,t)|^2>0$, otherwise $X(\cdot ,t)$ is just a point which contradicts the fact that $X(\cdot , t)$ is an immersion. So, when t is such a time of differentiability of $u(t):=\max _{(x,y)\in M^n\times M^n}|X(x,t) - X(y,t)|^2$, then also ${\text {diam}}(M_t)$ is differentiable and $\frac{d}{dt} {\text {diam}}(M_t) = \frac{\partial _t u(t)}{2 {\text {diam}}(M_t)}$. This shows that $t\mapsto {\text {diam}}(M_t)$ is differentiable almost everywhere in [0, T) and

$$\begin{aligned} \frac{d}{dt} {\text {diam}}(M_t)&= \frac{1}{|X(x,t) - X(y,t)|} \langle \partial _t X(x,t) - \partial _t X(y,t), X(x^*,t) - X(y,t) \rangle \\&= \langle -(H(x,t) -\overline{H}(t)) \nu (x,t) + (H(y,t) - {{\bar{H}}}(t)) \nu (y,t), \tfrac{X(x,t) - X(y,t)}{|X(x,t) - X(y,t)|}\rangle , \end{aligned}$$

where $(x,y)\in M^n\times M^n$ are points where the maximum in u(t) und thus also in the diameter is attained. $\square$

Lemma 2

Let $X:M^n\times [0,T) \rightarrow {\mathbb {R}}^{n+1}$ be a closed VPMCF for $t\in [0,T)$. Then the extrinsic diameter satisfies for almost every time $t\in [0,T)$

$$\begin{aligned} \frac{d}{dt} {\text {diam}}(M_t)&\le \overline{H} \left[ \langle \nu ^x, \tfrac{X(x,t) - X(y,t)}{|X(x,t) - X(y,t)|}\rangle - \langle \nu ^y, \tfrac{X(x,t) - X(y,t)}{|X(x,t) - X(y,t)|}\rangle \right] - \frac{4n}{{\text {diam}}(M_t)}\\&\le 2 |\overline{H}| - \frac{4n}{{\text {diam}}(M_t)}. \end{aligned}$$

Proof

We work again with the smooth squared distance function $\varphi : M^n \times M^n \times [0,T) \rightarrow \mathbb {R}$, $\varphi (x,y,t): = |X(x,t) - X(y,t)|^2$ as in the proof of Lemma 1. We write $(z_1, \ldots , z_{2n}) := (x_1, \ldots , x_n ,y_1, \ldots , y_n)$ if needed. We leave out the time dependence in the notation from now on. We calculate the second derivatives of $\varphi$. But we only want to use them for x and y such that the maximum in ${\text {diam}}(M_t)^2= \max _{(x,y)\in M^n\times M^n}|X(x,t) - X(y,t)|^2$ is attained. Thus, we compute

$$\begin{aligned}\partial _{x_i}\varphi = 2\left\langle \partial _{x_i} X(x) , X(x)-X(y) \right\rangle , \qquad \partial _{y_i}\varphi = -2\left\langle \partial _{y_i}X(y) , X(x)-X(y) \right\rangle \end{aligned}$$

$$\begin{aligned} \nabla ^2_{x^i x^j}\varphi&= -2h_{ij}(x) \left\langle \nu (x) , X(x)-X(y) \right\rangle +2g_{ij}(x)\\ \nabla ^2_{x^i y^j}\varphi&=-2\left\langle \partial _{x_i} X (x) , \partial _{y_j}X(y) \right\rangle \\ \nabla ^2_{y^i y^j}\varphi&= 2h_{ij}(y) \left\langle \nu (y) , X(x)-X(y) \right\rangle +2g_{ij}(y) \end{aligned}$$

At almost every time t, ${\text {diam}}(M_t)^2$ is differentiable, and by Hamilton’s trick [41, Lemma 2.1.3], we know that

$$\begin{aligned}\frac{ d }{dt} {\text {diam}}(M_t)^2 = 2 \langle -(H(x) -\overline{H}) \nu (x) + (H(y) - {{\bar{H}}}) \nu (y), {X(x) - X(y)}\rangle \end{aligned}$$

for x and y attaining the maximum. At such points we have that $0=\partial _{x_i}\varphi = \partial _{y_i}\varphi$ and $0\ge \nabla ^2_{z^iz^j} \varphi$. This particularly implies that $X(x) -X(y)$ is normal to the surface in x and y, so the tangent spaces at x and y are parallel and we may take local orthonormal coordinates so that $\partial _{x_i} X(x)=\partial _{y_i}X(y)=e_i$. Furthermore,

$$\begin{aligned}0\ge \sum _i \nabla ^2_{e_i^x-e_i^y,e_i^x-e_i^y}\varphi = -2 \left\langle \nu (x) , X(x)-X(y) \right\rangle H(x)+2\left\langle \nu (y) , X(x)-X(y) \right\rangle H(y) +8n\end{aligned}$$

which implies

$$\begin{aligned}\frac{ d }{dt}{\text {diam}}^2(M_t)\le 2\overline{H} [\left\langle \nu (x) , X(x)-X(y) \right\rangle -\left\langle \nu (y) , X(x)-X(y) \right\rangle ]-8n.\end{aligned}$$

Since ${\text {diam}}(M_t) = |X(x,t) - X(y,t)|>0$, we get

$$\begin{aligned} \begin{aligned} \frac{ d }{dt}{\text {diam}}(M_t)&\le \overline{H} \left[ \left\langle \nu (x) , \frac{X(x)-X(y)}{|X(x)-X(y)|} \right\rangle -\left\langle \nu (y) , \frac{X(x)-X(y)}{|X(x)-X(y)|} \right\rangle \right] -\frac{4n}{{\text {diam}}(M_t)}\\&=:2\left( \overline{H} \sigma -\frac{2n}{{\text {diam}}(M_t)}\right) \\&\le 2\left( |\overline{H}|-\frac{2n}{{\text {diam}}(M_t)}\right) \end{aligned} \end{aligned}$$

(2)

as stated in the lemma. Here, $\sigma \in \{1,0,-1\}$ depends on the configuration of normals and positions. $\square$

Remark 1

Lemma 2 implies that the diameter along the VPMCF is bounded on every bounded time interval [0, T), $T<\infty$, if one is able to show a bound $\sup _{t\in [0,T)}|\overline{H}(t)| \le c$. Unfortunately, such a bound is hard to get in a very general setting. See Proposition 15 for a weakening of this condition.

Remark 2

The above calculation is to some extent optimal in bounding diameter, even in the case of convex curves. Indeed, even in the convex setting, the nonlocal term can force the diameter to increase initially, before converging back to a circle. See Example 1 for details.

Definition 1

An immersion $X:M^n\rightarrow \mathbb {R}^{n+1}$ is Alexandrov immersed if there exists an $(n+1)$-dimensional manifold $\overline{\Omega }$ with $\partial \overline{\Omega } = M^n$ and an immersion $G:\overline{\Omega } \rightarrow \mathbb {R}^{n+1}$ such that $G|_{\partial \overline{\Omega }}$ also parametrises ${{\text {Im}}}(X)$.

Remark 3

For an Alexandrov immersion, there is obviously a natural notion of an an inner unit normal ${{\tilde{\nu }}}$ (the one where $G^*({{\tilde{\nu }}})$ shows into $\Omega$) and an outer unit normal $-{{\tilde{\nu }}}$.

Corollary 3

Let $X:M^n\times [0,T) \rightarrow {\mathbb {R}}^{n+1}$ be a closed VPMCF for $t\in [0,T)$ that is an Alexandrov immersion for all $t\in [0,T)$. Then we choose $\nu$ to be the outer unit normal. In this case

$$\begin{aligned} \left\langle \nu (x) , \frac{X(x)-X(y)}{|X(x)-X(y)|} \right\rangle -\left\langle \nu (y) , \frac{X(x)-X(y)}{|X(x)-X(y)|} \right\rangle = 2 \end{aligned}$$

for x(t) and y(t) such that the maximum in the definition of the diameter is attained. As a consequence, we have that

$$\begin{aligned} \frac{ d }{dt}{\text {diam}}(M_t) \le 2\left( \overline{H} - \frac{2n}{{\text {diam}}(M_t)}\right) \end{aligned}$$

for almost every $t\in [0,T)$.

Proof

Since there are no points $X({{\tilde{x}}})$ that are further apart from X(y) in ${\mathbb {R}}^{n+1}$ than X(x), the outer unit normal $\nu (x)$ must agree with $\frac{X(x)-X(y)}{|X(x)-X(y)|}$. For X(y), an analogous argument works. $\square$

Corollary 4

Let $X:M^n\times [0,T) \rightarrow {\mathbb {R}}^{n+1}$ be a closed VPMCF that is Alexandrov-immersed for all $t\in [0,T)$ and $\overline{H}\le 0$ on $t\in [0,T)$, where $\overline{H}$ is computed with respect to the outer unit normal. Then $T\le \frac{{\text {diam}}(M_0)^2}{8n}$.

Proof

The assumption $\overline{H}\le 0$ and Corollary 3 imply

$$\begin{aligned} \frac{ d }{dt}{\text {diam}}(M_t) \le -\frac{4n}{{\text {diam}}(M_t)}, \ \ \ \ t\in [0,T) \end{aligned}$$

which is equivalent to

$$\begin{aligned} {\text {diam}}(M_t)^2 \le {\text {diam}}(M_0)^2 - 8nt \ \ \text {for } t\in [0,T) \end{aligned}$$

via integration. So we have that

$$\begin{aligned} \limsup _{t \rightarrow T} {\text {diam}}(M_t)^2 \le {\text {diam}}(M_0)^2 - 8nT \end{aligned}$$

which leads to a contradiction for $T> \frac{{\text {diam}}(M_0)^2}{8n}$. $\square$

Remark 4

This above statement should be interpreted in the following way: If we know the flow exists for quite some time and stays Alexandrov immersed, then $\overline{H} \le 0$ is not preserved. For more on this see also Theorem 16.

Lemma 5

Let $X:M^n\times [0,T) \rightarrow {\mathbb {R}}^{n+1}$ be a VPMCF with enclosed volume $V_0 \not =0$. Then we can express $\overline{H}$ as

$$\begin{aligned} \overline{H} = \frac{\int _M\left\langle \partial _t X , X-x \right\rangle d\mu +n|M_t|}{(n+1)V_0} \end{aligned}$$

(3)

for any $x=x(t) \in {\mathbb {R}}^{n+1}$.

Proof

We have that $\Delta |X-x|^2= -2H\left\langle \nu , X-x \right\rangle +2n$ and so by the divergence theorem

$$\begin{aligned} |M_t| = \frac{1}{n}\int _M H\left\langle \nu , X-x \right\rangle d\mu \ . \end{aligned}$$

(4)

The formula for the signed volume enclosed by $M_t$ reads $V(t) =\frac{1}{n+1}\int _{M_t}\left\langle X-x , \nu \right\rangle d\mu$ for any $x=x(t)\in {\mathbb {R}}^{n+1}$. We compute

$$\begin{aligned}\int _M\left\langle X_t , X-x \right\rangle d\mu = \overline{H}\int _M \left\langle \nu , X-x \right\rangle d\mu -\int _M H\left\langle \nu , X-x \right\rangle d\mu \end{aligned}$$

and so

$$\begin{aligned}\overline{H} = \frac{\int _M\left\langle \partial _t X , X-x \right\rangle d\mu +\int _M H\left\langle \nu , X-x \right\rangle d\mu }{\int _M \left\langle \nu , X-x \right\rangle d\mu }=\frac{\int _M\left\langle \partial _t X , X-x \right\rangle d\mu +n|M_t|}{(n+1)V_0}\end{aligned}$$

using the volume preserving property. $\square$

Proposition 6

Suppose that $(M_t)_{t\in [0,T)}$ satisfies $V_0\ne 0$ and $\text {diam}(M_t)<R (t)$ with $R(s) \le R(t)$ for $s<t\le T$. Then we have for $t\in [0,T)$,

$$\begin{aligned}\int _0^t \overline{H}^2(\tau ) d\tau \le \frac{ 2|M_0|^2}{V_0^2(n+1)^2}(R(t)^2+n^2t) \end{aligned}$$

Proof

First we choose $x(t)\in {\mathbb {R}}^{n+1}$ to be any point such that $M_t\subset B_R(x(t))$. We have that

$$\begin{aligned}\frac{ d }{dt}|M_t| = -\int _M H(H-\overline{H})d\mu =-\int _M (H-\overline{H})^2d\mu = -\int _M|\partial _t X|^2d\mu .\end{aligned}$$

Formula (3) implies

$$\begin{aligned} |\overline{H}| \le \frac{1}{(n+1)|V_0|}\left( \sqrt{\int _M|\partial _t X|^2d\mu }\sqrt{\int _M|X-x|^2d\mu }+n|M_0|\right) . \end{aligned}$$

and thus

$$\begin{aligned}\overline{H}^2\le \frac{2}{(n+1)^2V_0^2}\left( -\frac{ d }{dt}|M_t|\int _M |X-x|^2d\mu +n^2|M_0|^2\right) \ .\end{aligned}$$

Estimating $|X-x|^2\big |_{s}\le \text {diam}^2\left( M_s\right) \le R^2(s) \le R^2(t)$ for $s<t$ and using $|M_s|\le |M_0|$ again, we know now

$$\begin{aligned} \int _0^t\overline{H}^2ds \le \frac{2}{V_0^2(n+1)^2}|M_0|^2(R^2(t)+ n^2t)\ . \end{aligned}$$

$\square$

Proposition 7

Let $X:M^n\times [0,T) \rightarrow {\mathbb {R}}^{n+1}$ be a closed VPMCF for $t\in [0,T)$ with enclosed volume $V_0\not =0$. Then there are constants c, C only depending on $M_0$ such that

$$\begin{aligned} {\text {diam}}(M_t) \le C(1+t) e^{c\sqrt{t}} \end{aligned}$$

for $t\in [0,T)$

Proof

We have

$$\begin{aligned}\frac{ d }{dt}{\text {diam}}(M_t)\le 2|\overline{H}|-\frac{4 n}{{\text {diam}}(M_t)}\le 2|\overline{H}|\end{aligned}$$

and from (3)

$$\begin{aligned} |\overline{H}|&\le \frac{1}{(n+1)|V_0|}\left( \sqrt{-\frac{ d }{dt}|M_t|}\sqrt{|M_0|} {\text {diam}}(M_t)+n|M_0|\right) \\&: = c_1 + c_2 {\text {diam}}(M_t)\sqrt{-\frac{ d }{dt}|M_t|}. \end{aligned}$$

We pick

$$\begin{aligned} \log \psi = -2c_2\int _0^t \sqrt{-\frac{ d }{dt}|M_t|}\ge -2c_2 \sqrt{t} (|M_0|-|M_t|)^{\frac{1}{2}} \ge -2c_2 \sqrt{t} |M_0|^{\frac{1}{2}} \end{aligned}$$

(5)

so that

$$\begin{aligned}\frac{1}{\psi }\frac{ d }{dt}({\text {diam}}(M_t) \psi ) = \frac{ d }{dt}{\text {diam}}(M_t)-2c_2 {\text {diam}}(M_t)\sqrt{-\frac{ d }{dt}|M_t|}\le 2c_1\end{aligned}$$

and thus (using that $\psi \le 1$ and (5))

$$\begin{aligned} {\text {diam}}(M_t)&\le \frac{1}{\psi (t)}{\text {diam}}(M_0)+2c_1\frac{1}{\psi (t)}\int _0^t\psi (\tau ) d\tau \\&\le \left( {\text {diam}}(M_0) + 2c_1 t\right) \frac{1}{\psi (t)} \\&\le C(1+t) e^{2c_2|M_0|^{\frac{1}{2}}\sqrt{t}}. \end{aligned}$$

$\square$

Corollary 8

Let $X:M^n\times [0,T) \rightarrow {\mathbb {R}}^{n+1}$ be a closed VPMCF for $t\in [0,T)$ with enclosed volume $V_0\not =0$. Then there are constants $c,C>0$ only depending on n, $V_0$ and $M_0$ such that

$$\begin{aligned}\int _0^t \overline{H}^2(s) ds \le C (1+t^2) e^{c\sqrt{t}}. \end{aligned}$$

Proof

Put together Propositions 6 and 7. $\square$

Remark 5

• (1) In general, we can not expect to have a bound of the form $\int _0^t \overline{H}^2(s) ds \le C$, where C is independent of t because round spheres are (stationary) solutions of the VPMCF, and they satisfy $\int _0^t \overline{H}^2(s) ds = c^2 t$. We do not know whether the $e^{\sqrt{t}}$-growth can be improved (e.g. to a linear growth in t) in this very general setting.

• (2) Integrability conditions of a Lagrange multiplier was successfully used earlier, see for example the work of Rupp [45] on the volume preserving Willmore flow. But there, the fixed-enclosed-volume constraint is of lower order compared to the order of the flow, therefore Rupp was able to get strong results with the integrability.

3.1 Optimality of diameter bounds

In this subsection we consider a curve $\gamma :{\mathbb {S}}^1 \rightarrow {\mathbb {R}}^2$, where $\partial _s$ is differentiation with respect to arclength and $\nu = -J \partial _s \gamma$ is the rotation of the tangent $\partial _s\gamma$ by $\frac{\pi }{2}$ in negative direction. In order to be consistent with the definitions for surfaces, we use the definition $\kappa = -\langle \partial _s^2 \gamma , \nu \rangle$. The area preserving curve shortening flow reads

$$\begin{aligned} \partial _t \gamma = -(\kappa -\overline{\kappa })\nu \end{aligned}$$

for $\overline{\kappa } : = \frac{\int \kappa ds}{L(\gamma _t)}$. Here, $ds = |\partial _x \gamma |dx$ denotes integration with respect to arclength and $L(\gamma _t)$ is the length of $\gamma _t: =\gamma (\cdot ,t)$ for the family $\gamma :{\mathbb {S}}^1\times [0,T) \rightarrow {\mathbb {R}}^2$. Of course, as $\gamma$ is closed and $\int \kappa ds$ is the turn of the tangents, we have that $\overline{\kappa } = \frac{2\pi m}{L(\gamma _t)}$. If $\gamma _0$ is embedded, then $m=1$.

On a general curve we have that

$$\begin{aligned} \kappa _0(x) = \kappa _0(y), \ \ \frac{\gamma _0(x) -\gamma _0(y)}{|\gamma _0(x) -\gamma _0(y)|} = \nu _0(x), \ \ \frac{\gamma _0(x) -\gamma _0(y)}{|\gamma _0(x) -\gamma _0(y)|} = - \nu _0(y), \end{aligned}$$

(6)

where $x, y \in {\mathbb {S}}^1$ are such that the maximum in ${\text {diam}}(\gamma _0)$ is attained. Note that for a positively oriented curve $\gamma _0$, $\nu _0$ is the outer unit normal. The proof of Lemma 2 shows that

$$\begin{aligned} 0&\ge -2 \langle \nu _0(x), \gamma _0(x) - \gamma _0(y) \rangle \kappa _0(x) + 2 \langle \nu _0(y), \gamma _0(x) - \gamma _0(y) \rangle \kappa _0(y) + 8. \end{aligned}$$

Then (6) implies that

$$\begin{aligned} \langle \nu _0(x), \gamma _0(x) - \gamma _0(y) \rangle&= |\gamma _0(x) -\gamma _0(y)| = {\text {diam}}(\gamma _0)\\ \langle \nu _0(y), \gamma _0(x) - \gamma _0(y) \rangle&= -|\gamma _0(x) -\gamma _0(y)| = -{\text {diam}}(\gamma _0) \end{aligned}$$

and thus

$$\begin{aligned} \kappa _0(x) \ge \frac{2}{{\text {diam}}(\gamma _0)}. \end{aligned}$$

(7)

This inequality can be interpreted geometrically: The curvature at the points where the maximum is attained cannot be smaller than $\frac{2}{{\text {diam}}(\gamma _0)}$ – otherwise the straight line realising the maximal distance would be attained at other points (tilting that line would increase the distance). Bearing this is mind, we construct a somehow sharp example.

We now provide an example demonstrating, somewhat counter-intuitively, that even on a convex curve, (which we know will converge to a round circle in infinite time [30]), the diameter will initially increase while the curvature at extremal points initially decreases.

Fig. 1

An embedded, convex initial curve for which the diameter increases

Example 1

In this example we show that $\frac{d}{dt} {\text {diam}}(\gamma _t)\big |_{t=0}>0$ can occur, even on closed convex curves. We construct an explicit example as in Figure 1 as follows: First we set ${\text {diam}}(\gamma _0) = 1$. We put a circle with radius $r: =\frac{1}{2}$ at the midpoint of the line realising the maximal distance (w.l.o.g. $\{0\}\times [0,1]\subset {\mathbb {R}}^2$). So we use very short arcs of the circle at the points $(0,1) = \gamma _0(x)$ and $(0,0) = \gamma _0(y)$. Thus, in x and y, the curvature is precisely equal to 2. We close this curve (smoothly) by straight lines (going from south to north and the other way round at the end of the arcs). The diameter is clearly realised by the points x and y.

In Example 1 we have equality in Eq. (7). With Lemma 1, we know that

$$\begin{aligned} \frac{d}{dt} {\text {diam}}(\gamma _t)\big |_{t=0}&= \langle -2(\kappa _0(x) - \overline{\kappa }_0)\nu _0(x), \frac{\gamma _0(x) - \gamma _0(y)}{|\gamma _0(x) - \gamma _0(y)|} \rangle = -4 (1- \frac{\pi }{L(\gamma _0)}) \end{aligned}$$

taking into account our situation. The length of $\gamma _0$ can be chosen to be $L(\gamma _0) = 2+\epsilon$. The curve only needs to satisfy $L(\gamma _0)<\pi$ to get an example where $\frac{d}{dt} {\text {diam}}(\gamma _t)\big |_{t=0} >0$, in particular $\kappa _0(x) - \overline{\kappa }(0) <0$.

There is something else worth noticing about the above example: The evolution equation of the curvature along the area preserving curve shortening flow is

$$\begin{aligned} \partial _t \kappa = \partial _s^2 \kappa + (\kappa - \overline{\kappa }) \kappa ^2. \end{aligned}$$

We use this for our initial curve $\gamma _0$ at the points x and y (where the maximal distance is attained). Since $\kappa \equiv 2$ on a small neighborhood around x and y, we know $\partial _s^2 \kappa (x) = 0 = \partial _s^2 \kappa (y)$ and hence

$$\begin{aligned} \partial _t \kappa (x) \big |_{t=0} = \partial _t \kappa (y) \big |_{t=0} = (\kappa _0(x) - \overline{\kappa }(0)) \kappa ^2(x)<0. \end{aligned}$$

This means that, at the points $\gamma _0(x)$ and $\gamma _0(y)$, the curve becomes even a bit flatter, and the diameter grows a bit. This is surprising, because eventually, this curve converges to a round circle. This example shows that the short-time behavior of the flow can be somehow counterintuitive.

4 Finite time singularities and monotonicity results

4.1 Monotonicity formula

We derive a monotonicity formula for VPMCF and demonstrate various corollaries following a similar course to [20]. This has been calculated by the second author in [38] already, inspired by [5]. Note that also [46] contains a version of a monotonicity formula.

The evolution equation of the area measure is

$$\begin{aligned}\frac{ d }{dt}d \mu =-(H-\overline{H})H d \mu \ .\end{aligned}$$

We now set $\Phi = \Phi _{(x_0,t_0)}$ the standard (appropriately rescaled version of the) “ambient backward heat kernel”,

$$\begin{aligned}\Phi _{(x_0,t_0)} (x,t):= \frac{1}{(4\pi (t_0-t))^\frac{n}{2}}e^{-\frac{|x-x_0|^2}{4(t_0-t)}}, \ \ x\in {\mathbb {R}}^{n+1}, t<t_0, \end{aligned}$$

where in the following calculation we will suppress the subscript for simplicity of notation.

Proposition 9

(Monotonicity formula) Suppose that $M_t$ satisfies smooth VPMCF. We define

$$\begin{aligned}\psi (t):=e^{-\frac{1}{2}\int _0^t \overline{H}^2(s) ds}\end{aligned}$$

and let $\phi$ be a $C^2$-function that is either defined on $M^n \times [0,T)$ or on a neighborhood of $M_t$ in ${\mathbb {R}}^{n+1}$. Then we have that

$$\begin{aligned} \frac{ d }{dt} \left( \psi \int _{M_t} \phi \Phi d \mu \right)&= \psi \int _{M_t} \Phi \left( \frac{d}{dt} -\Delta \right) \phi d \mu +\psi \int _{\partial M_t} \Phi \nabla _\mu \phi -\phi \nabla _\mu \Phi d \mu ^\partial \\&\qquad -\frac{\psi }{2}\int _{M_t}\left[ \left| (H-\overline{H})-\frac{\left\langle \nu , x-x_0 \right\rangle }{2(t_0-t)}\right| ^2+\left| H-\frac{\left\langle \nu , x-x_0 \right\rangle }{2(t_0-t)}\right| ^2 \right] \\&\qquad \times \phi \Phi d \mu \ . \end{aligned}$$

Proof

The proof can be found in [38], see also [39, Remark before Proposition 4.10]. For the convenience of the reader we repeat it here. For general functions $\phi$ and $\Phi :\mathbb {R}^{n+1}\times [0,T) \rightarrow \mathbb {R}$ we have that

$$\begin{aligned} \frac{ d }{dt}\int _{M_t} \phi \Phi d \mu&= \int _{M_t} \Phi \left( \frac{d}{dt} -\Delta \right) \phi +\text {div}(\Phi \nabla \phi -\phi \nabla \Phi ) \\&\quad +\phi \left( \frac{ d \Phi }{dt}+\Delta \Phi \right) -H(H-\overline{H})\phi \Phi d \mu \\&= \int _{M_t} \Phi \left( \frac{d}{dt} -\Delta \right) \phi +\text {div}(\Phi \nabla \phi -\phi \nabla \Phi )d \mu \\&\qquad +\int _{M_t}\Bigg [\frac{1}{\Phi }\Bigg (\left( \frac{\partial }{\partial t}+\overline{\Delta }\right) \Phi -\overline{\nabla }^2_{\nu \nu } \Phi \\&\qquad -(H-\overline{H})\overline{\nabla }_\nu \Phi -H\overline{\nabla }_\nu \Phi \Bigg )-H(H-\overline{H})\Bigg ]\phi \Phi d \mu . \end{aligned}$$

For our choice of $\Phi$, we get that

$$\begin{aligned}\frac{\partial \Phi }{\partial t} = \left( \frac{n}{2(t_0-t)}-\frac{|x-x_0|^2}{4(t_0-t)^2}\right) \Phi ,\qquad \overline{\nabla }_V \Phi = -\frac{\left\langle V , x-x_0 \right\rangle }{2(t_0-t)}\Phi , \end{aligned}$$

$$\begin{aligned}\overline{\nabla }^2_{VW} \Phi = \left( \frac{\left\langle V , x-x_0 \right\rangle \left\langle W , x-x_0 \right\rangle }{4(t_0-t)^2}-\frac{\left\langle V , W \right\rangle }{2(t_0-t)}\right) \Phi \ .\end{aligned}$$

In particular we have

$$\begin{aligned}\left( \frac{\partial }{\partial t}+\overline{\Delta }\right) \Phi =\frac{-1}{2(t_0-t)}\Phi , \qquad \overline{\nabla }^2_{\nu \nu } \Phi = \left( \left( \frac{\left\langle \nu , x-x_0 \right\rangle }{2(t_0-t)}\right) ^2-\frac{1}{2(t_0-t)}\right) \Phi \end{aligned}$$

and so the contents of the square brackets in the last integrand above are

$$\begin{aligned}&-\left( \frac{\left\langle \nu , x-x_0 \right\rangle }{2(t_0-t)}\right) ^2+(H-\overline{H})\frac{\left\langle \nu , x-x_0 \right\rangle }{2(t_0-t)}+H\frac{\left\langle \nu , x-x_0 \right\rangle }{2(t_0-t)}-H(H-\overline{H})\\&\quad =-\frac{1}{2}\left| (H-\overline{H})-\frac{\left\langle \nu , x-x_0 \right\rangle }{2(t_0-t)}\right| ^2-\frac{1}{2}\left| H-\frac{\left\langle \nu , x-x_0 \right\rangle }{2(t_0-t)}\right| ^2+\frac{1}{2}\overline{H}^2\ . \end{aligned}$$

We have chosen $\psi (t)=e^{-\frac{1}{2}\int _0^t \overline{H}^2(s) ds}$ so that $\frac{ d \psi }{dt}= -\frac{1}{2} \overline{H}^2$ which yields

$$\begin{aligned} \frac{ d }{dt}\int _{M_t} \psi \phi \Phi d \mu&= \int _{M_t}\psi \Phi \left( \frac{d}{dt} -\Delta \right) \phi d \mu +\psi \int _{\partial M_t} \Phi \nabla _\mu \phi -\phi \nabla _\mu \Phi d \mu ^\partial \\&\qquad -\frac{\psi }{2}\int _{M_t}\left[ \left| (H-\overline{H})-\frac{\left\langle \nu , x-x_0 \right\rangle }{2(t_0-t)}\right| ^2+\left| H-\frac{\left\langle \nu , x-x_0 \right\rangle }{2(t_0-t)}\right| ^2 \right] \phi \Phi d \mu \ . \end{aligned}$$

$\square$

Remark 6

We have that

and so we may rewrite

Remark 7

Consider a parabolic rescaling of $M_t$, ${\widetilde{M}}_s$ where $t=\lambda ^2 s+t_0$, and ${M}_t = \lambda {\widetilde{M}}_s +x_0$. Then $\overline{H}({\widetilde{M}}_s)=\lambda \overline{H}(M_t)$ and we have that

$$\begin{aligned}\log {\widetilde{\psi }}(s) = \int _0^s\overline{H}({\widetilde{M}}_{{\hat{s}}})^2d{\hat{s}}=\int _0^s\lambda ^{2}\overline{H}({M}_{t({{\hat{s}}})})^{2}d{\hat{s}}=\int _{t_0}^{t}\overline{H}({M}_{{\hat{t}}})^{2}d{\hat{t}}=\log \psi (t)\ ,\end{aligned}$$

which implies that our object to differentiate is invariant under parabolic rescaling.

4.2 VPMCF blowups are ancient MCFs

We will use the following MCF terminology.

Definition 2

Given any X(x, t) satisfying VPMCF the parabolic rescaling about a point $p\in \mathbb {R}^{n+1}$ at time T given by

$$\begin{aligned}X^\lambda _p(x,\tau ) := \lambda (X(x, \lambda ^{-2} \tau +T)-p)\end{aligned}$$

for $\tau \le 0$. Equivalently we may write this as

$$\begin{aligned}M^{\lambda , p}_\tau =\lambda (M_{\lambda ^{-2}\tau +T}-p)\ .\end{aligned}$$

A parabolic rescaling of VPMCF gives another solution of VPMCF (although with a different fixed volume).

Definition 3

Let $M_t$ be a mean curvature flow on the time interval [0, T), and suppose that the second fundamental form blows up at time T that is, $\limsup _{t\rightarrow T} \sup _{M_t}|A| = \infty$. Then we say that this is a singularity of VPMCF where

• (i) the singularity is of type I if there exists a $C>0$ such that

  $$\begin{aligned}\sup _{M_t}|A|^2 \le \frac{C}{T-t}\end{aligned}$$

  for all $t<T$ and

• (ii) the singularity is of type II otherwise.

Definition 4

Let $M_t$ be a mean curvature flow on the time interval [0, T). Then

• (i) Suppose we are given a point $p \in \mathbb {R}^{n+1}$ and a sequence $\lambda _i\rightarrow \infty$. Then a type I (or centred) blow up sequence is the sequence of solutions to VPMCF given by

  $$\begin{aligned}M^i_\tau = \lambda _i(M_{\lambda _i^{-2}\tau + T}-p)\end{aligned}$$

  for $\tau \in [-\lambda _i^2 T,0)$.

• (ii) Assuming that the second fundamental form blows up as $t\rightarrow T$, then we follow Hamilton [29] and define a type II blow up sequence by a sequence times $t_i$ and points $x_i\in M_{t_i}$ such that

  $$\begin{aligned}\lambda _i:=\lambda _i^2 (T - i^{-1} - t_i)=\max _{(x,t)\in M^n \times [0,T-i^{-1}]}|A(x,t)|^2(T-i^{-1}-t)\end{aligned}$$

  Then the type II blow up along $(x_i,t_i)$ is given by

  $$\begin{aligned}M^i_\tau = \lambda _i(M_{t_i+\lambda _i^{-2}\tau }-x_i)\end{aligned}$$

  for $\tau \in [-\lambda _i^2 t_i, \lambda _i^2(T-i^{-1}-t_i))$

Remark 8

Once we have uniformly bounded curvature, $\overline{H}$ is bounded and Shi-type estimates hold to all orders, as in [20, Proposition 3.22]. As a result, if a type I singularity occurs, then any type I blowup sequence will converge locally smoothly to a smooth flow for $\tau \in (-\infty , 0)$ (which is possibly empty depending on the p). This will be referred to as an ancient flow. Later in Lemma 11 we will see that this is a self similar solution of MCF. If we have a type II singularity then the type II blowup sequence converges smoothly to a flow for $\tau \in (-\infty , \infty )$ [29].

We are now in a position to prove the following:

Theorem 10

Suppose that a singularity of VPMCF occurs at a finite time. Then

• (a) If the singularity is type I then any locally smoothly converging subsequence of the type I blow up sequence is an ancient solution to MCF.

• (b) If the singularity if type II then any locally smoothly converging subsequence of a type II blow up sequence yields an eternal solution to MCF.

Proof

We first consider the type I blow up case: By dilating we may assume wlog that $|A|^2(x,t)<\frac{1}{T-t}$. By Corollary 8, we know that for any finite T

$$\begin{aligned}\int _0^T \overline{H}^2(t) dt<C\ .\end{aligned}$$

By continuous dependence of an integral on it’s domain of definition (Lebesgue’s differentiation theorem) we have that there exist $T_i\rightarrow T$ such that

$$\begin{aligned} \int ^T_{T_j} \overline{H}^2 dt = j^{-1} \end{aligned}$$

(8)

and therefore, writing $t = \lambda _i^{-2}(\tau +T)$ then

$$\begin{aligned}j^{-1}=\int ^T_{T_j} \overline{H}^2(t) dt =\int ^0_{-\lambda ^2(T-T_j)}\lambda ^{-2}\overline{H}^2(\tau )d\tau =\int ^0_{-\lambda _i^2(T-T_j)}(\overline{H}^i)^2(\tau )d\tau \ .\end{aligned}$$

where $\overline{H}^i$ is the average mean curvature of $M_\tau ^i$ of the type I blowup. Due to the type I singularity hypothesis, the second fundamental form of $M^i_\tau$ satisfies

$$\begin{aligned}|A^i(x,\tau )|^2\le \frac{1}{|\tau |}\ .\end{aligned}$$

Furthermore, we have that

(9)

Given any $a<b<0$, if $\lambda _i$ is so large that $[a,b]\subset [-\max \{\lambda _i^2(T-T_j), \lambda _i^2T\}, 0)$ then we have that there is a constant $C_1=C_1(a,b)$ such that

$$\begin{aligned}\overline{H}^i<C_1(a,b)j^{-1}\end{aligned}$$

where we used the estimate (9). Therefore, for any smoothly converging subsequence, $\overline{H}^i\rightarrow 0$ uniformly on [a, b] and so the flow converges smoothly on [a, b] to a solution to MCF,

$$\begin{aligned}\frac{ d X^\infty }{dt} = -H^\infty \nu ^\infty \ .\end{aligned}$$

Note that the above argument implies that for any subsequence such that the blow up sequence converges in $C^2$, we still get a mean curvature flow.

The argument for the type II blow up is similar: Suppose that we have a converging subsequence of a type II blow up. Then, defining $T_j$ as in (8) then substituting $t=t_i+\lambda _i^{-2}\tau$

$$\begin{aligned}j^{-1}=\int ^T_{T_j} \overline{H}^2(t) dt \ge \int ^{\lambda _i^2(T-t_i-i^{-1})}_{\lambda ^2_i(T_j-t_i)}\lambda ^{-2}_i\overline{H}^2(\tau )d\tau =\int ^{\lambda _i^2(T-t_i-i^{-1})}_{\lambda ^2_i(T_j-t_i)}\overline{H}^2_i(\tau )d\tau \ .\end{aligned}$$

By the type II hypothesis we know that $\lambda _i^2(T-t_i-i^{-1})\rightarrow \infty$, see for example [41, p. 89], while we may assume that for i large enough $T_j<t_i$ and so $\lambda _i^2(T_j-t_i)\rightarrow -\infty$. This time, by our choice of blow up sequence, $|A^i(x,t)|\le 1$ on the time interval $[-\lambda _i^2 t_i, \lambda _i^2(T-i^{-1}-t_i))$ and so, estimating similarly to (9), on this time interval $|\frac{ d }{dt}\overline{H}|\le 4$. As a result, given any finite time interval $[a,b]\subset \mathbb {R}$, if i is large enough so that $[a,b]\subset [\lambda _i^2(T_j-t_i), \lambda _i^2(T-t_i-i^{-1})]$ then there exists a $C_2=C_2(b-a)$ such that $\overline{H}^i<C_2 j^{-1}$. As we know that the type II sequence converges locally smoothly, we see that on [a, b] we $\overline{H}^i\rightarrow 0$ uniformly and again the limit is a solution to MCF. $\square$

Lemma 11

Suppose that a singularity of VPMCF occurs at a finite time T which is of type I. Then the asymptotic limit flow from the above theorem is a (possibly trivial) properly immersed, non-compact homothetically shrinking solution of the Mean Curvature Flow.

Proof

The proof works as in the case of type I singularities of the Mean Curvature Flow. We sketch the proof for the convenience of the reader. Note that the Monotonicity formula implies

$$\begin{aligned} \frac{d}{dt}\left( \psi (t)\int _{M_t}\Phi _{(x_0,T)}(x,t)\,d\mu ^n_t\right) \le 0, \end{aligned}$$

thus the limit $\lim _{t\rightarrow T}\Lambda _{(x_0,T)}(t)$ exists, where

$$\begin{aligned} \Lambda _{(x_0,T)}(t):=\psi (t)\int _{M_t}\Phi _{(x_0,T)}(x,t)\,d\mu ^n_t\,. \end{aligned}$$

By parabolic rescaling we have that

$$\begin{aligned} \Lambda _{(x_0,T)}(t)&=\psi (t)\int _{M_{t}}\Phi _{(x_0,T)}(x,t)\,d\mu _{t} \nonumber =\psi _i(\tau )\int _{M^i_\tau }\Phi _{(0,0)}(y,\tau )\,d\mu ^i_{\tau } =:\Lambda _{(0,0)}^i(\tau )\,, \end{aligned}$$

where $t = T + \frac{\tau }{\lambda _i^2}$. We apply the rescaled monotonicity formula and estimate for $\tau _1<\tau _2$

$$\begin{aligned} 0&\le \frac{1}{2}\int _{\tau _1}^{\tau _2}\psi _i(\tau ) \int _{M^i_\tau }\left| H^i+\frac{\langle y,\nu \rangle }{2\tau }\right| ^2\Phi _{0,0}\,d\mu ^n_\tau d\tau \\&\le \frac{1}{2}\int _{\tau _1}^{\tau _2} \psi _i(\tau )\int _{M^i_\tau } \left\{ \left| H^i - \overline{H}^i+\frac{\langle y,\nu \rangle }{2\tau }\right| ^2 +\left| H^i+\frac{\langle y,\nu \rangle }{2\tau }\right| ^2\right\} \times \Phi _{0,0}\,d\mu ^n_\tau d\tau \\&\le \Lambda _{(0,0)}^i(\tau _1)-\Lambda _{(0,0)}^i(\tau _2)&\nonumber \\&=\Lambda _{(x_0,T)}\!\left( T+\frac{\tau _1}{\lambda _i^2}\right) -\Lambda _{(x_0,T)}\!\left( T+\frac{\tau _2}{\lambda _i^2}\right)&\end{aligned}$$

for all i. Since

$$\begin{aligned} T+\frac{\tau _l}{\lambda _i^2}\rightarrow T \end{aligned}$$

for $i\rightarrow \infty$ and $l=1,2$, the right-hand side of the above converges to 0 for $i\rightarrow \infty$. The flows $\big (\big (M^i_\tau \big )_{\tau \in [\tau _1,\tau _2]}\big )_{i\in {\mathbb {N}}}$ converge smoothly along a subsequence and on compact subsets of ${\mathbb {R}}^{n+1}$ to a smooth flow $\left( M^\infty _\tau \right) _{\tau \in [\tau _1,\tau _2]}$. Let $R>0$. There exists a $i_0\in {\mathbb {N}}$ so that for all $i\ge i_0$, $M_\tau ^i\cap B_R(0)$ can be parametrised over $M_\tau ^\infty \cap B_R(0)$. That is, there exist immersions $Y_i:M_\tau ^\infty \cap B_R(0)\rightarrow {\mathbb {R}}^{n+1}$ with

$$\begin{aligned} M_\tau ^i\cap B_R(0)=Y_i(M_\tau ^\infty \cap B_R(0)) \end{aligned}$$

and $Y_i\rightarrow {\text {id}}$ for $i\rightarrow \infty$. From the above we have that

$$\begin{aligned}&\Lambda _{(x_0,T)}\!\left( T+\frac{\tau _1}{\lambda _i^2}\right) -\Lambda _{(x_0,T)}\!\left( T+\frac{\tau _2}{\lambda _i^2}\right) \\&\ge \frac{1}{2}\int _{\tau _1}^{\tau _2}\psi _i(\tau ) \int _{M^i_\tau \cap B_R(0)}\left| H^i+\frac{\langle y,\nu \rangle }{2\tau }\right| ^2\Phi _{0,0}\,d\mu ^n_\tau d\tau \ . \end{aligned}$$

Taking a $\liminf$ and applying Fatou’s lemma,

$$\begin{aligned} 0&=\liminf _{i\rightarrow \infty } \int _{\tau _1}^{\tau _2} \int _{M^i_\tau \cap B_R(0)} \frac{\psi _i(\tau )}{2}\left| H^i+\frac{\langle y,\nu ^i\rangle }{2\tau }\right| ^2 \Phi _{0,0}\,d\mu ^i_{\tau }d\tau \nonumber \\&=\liminf _{i\rightarrow \infty } \int _{\tau _1}^{\tau _2} \int _{M^i_\tau \cap B_R(0)} \frac{\psi _i(\tau )}{2}\left| H^i+\frac{\langle y,\nu ^i\rangle }{2\tau }\right| ^2 \Phi _{0,0}\,\sqrt{\det (DY_i)}\,dxd\tau \nonumber \\&\ge \int _{\tau _1}^{\tau _2}\int _{M^\infty _\tau \cap B_R(0)} \liminf _{i\rightarrow \infty }\left( \frac{\psi _i(\tau )}{2} \left| H^i+\frac{\langle Y_i,\nu ^i\rangle }{2\tau }\right| ^2 \Phi _{0,0}\sqrt{\det (DY_i)}\right) \,dx d\tau \nonumber \\&= \frac{1}{2} \int _{\tau _1}^{\tau _2}\int _{M^\infty _\tau \cap B_R(0)} \left| H^\infty +\frac{\langle y,\nu ^\infty \rangle }{2\tau }\right| ^2 \Phi _{0,0}\,d\mu ^\infty _{\tau }d\tau \,, \end{aligned}$$

where we have used that $\overline{H}^i \rightarrow 0$ uniformly as shown in the proof of Theorem 10, and thus $\psi _i(\tau ) \rightarrow 1$. Since $R>0$ was arbitrary, we deduce

$$\begin{aligned} \int _{\tau _1}^{\tau _2}\int _{M^\infty _\tau }\left| H^\infty +\frac{\langle y,\nu ^\infty \rangle }{2\tau }\right| ^2\Phi _{0,0}\,d\mu _\tau d\tau =0\,. \end{aligned}$$

Since the convergence is smooth, and sending $\tau _1\rightarrow -\infty$ and $\tau _2\rightarrow 0$ yields

$$\begin{aligned} \left| H^\infty +\frac{\langle y,\nu ^\infty \rangle }{2\tau }\right| ^2=0 \end{aligned}$$

for every $\tau \in (-\infty ,0)$ and every $y\in M^\infty _\tau$, so $M^\infty _\tau$ is a homothetically shrinking (possibly trivial) solution of the Mean Curvature Flow.

It remains to show that the limit flow is proper and non-compact. The rescaled monotonicity formula and the $L^2$-bound on $\overline{H}$ implies

$$\begin{aligned} \int _{M_\tau ^i} \Phi _{(0,0)}(y,\tau ) d\mu _\tau ^n \le C \end{aligned}$$

for all $\tau \in (-\infty ,0)$ and for all i. Fatou’s lemma on ambient balls yields

$$\begin{aligned} C&\ge \int _{M_\tau ^\infty \cap B_R(0)} \Phi _{(0,0)} d\mu ^\infty _\tau \\&\ge \frac{1}{(-4\pi \tau )^{\frac{n}{2}}} \exp \left( -\frac{R^2}{-4\tau }\right) |M^\infty _\tau \cap B_R(0)|, \end{aligned}$$

which implies that $M^\infty _\tau$ is proper. The enclosed volume of the rescaled flows is $V^i = \lambda ^{n+1}_i V_0$. Since $V_0\not =0$, this clearly blows up as $\lambda _i\rightarrow \infty$. If the limit flow were compact, it would be contained in a ball with finite radius (for each $\tau$). But because of the properness of the limiting hypersurface $M^\infty _\tau$ there can only be finite volume contained in that ball which is a contradition. Thus, the limiting ancient flow is non-compact. $\square$

Remark 9

See [10] for examples of non-compact self-shrinkers of arbitrary genus.

4.3 Density estimates and the clearing out lemma

In this section we demonstrate that the monotonicity arguments in [20, Chapter 4] may be modified to the case of VPMCF with essentially minor additional assumptions depending only on the $L^2$ norm of $\overline{H}$. Throughout this section we will assume that our flow is smooth and properly immersed up to the final, possibly singular, time.

From the Monotonicity formula we have that

$$\begin{aligned} \frac{ d }{dt} \left( \psi \int _{M_t} \phi \Phi _{(x_0,t_0)} d \mu \right)&\le \psi \int _{M_t} \Phi _{(x_0,t_0)}\left( \frac{d}{dt} -\Delta \right) \phi d \mu \ . \end{aligned}$$

and from Corollary 8 we have that $1=\psi (0)\ge \psi (t)\ge \epsilon (t)>0$ for some strictly positive function $\epsilon (t)$, and by the dominated convergence theorem, $\psi (t)$ is continuous.

Definition 5

The space-time track of a flow is defined by

$$\begin{aligned}{\mathcal {M}}:=\cup _{t\in [0,T)} M_t\times \{t\} \subset \mathbb {R}^{n+1}\times [0,T)\ .\end{aligned}$$

The Gaussian density on ${\mathcal {M}}$ is determined by

$$\begin{aligned}\Theta ({\mathcal {M}}, x_0, t_0):=\lim _{t \nearrow t_0}\int _{M_t} \Phi _{(x_0,t_0) }d\mu _t\end{aligned}$$

where, as usual, integration also counts multiplicities in ${\mathcal {M}}$.

Indeed, by the monotonicity formula (with $\phi =1$) we see that

$$\begin{aligned}\int _{M_t} \Phi _{(x_0,t_0) }d\mu _t\le \frac{1}{\psi (t)}\int _{M_0} \Phi _{(x_0,t_0) }d\mu _0=: \frac{C_0(M_0)}{\psi (t)}\end{aligned}$$

and so by dominated convergence theorem, the density converges everywhere. It is also useful to define a localised density, and so we consider the localisation function

$$\begin{aligned}\phi _{(x_0,t_0),\rho }(x,t):=\left( 1-\frac{|x-x_0|^2+2n(t-t_0)}{\rho ^{2}}\right) ^3_+\end{aligned}$$

then we have that on the support of $\phi _{(x_0,t_0),\rho }(x,t)$,

$$\begin{aligned}\left( \frac{d}{dt} -\Delta \right) \phi _{(x_0,t_0),\rho }(x,t)=-\frac{6}{\rho ^2}\overline{H}\left\langle x-x_0 , \nu \right\rangle \phi _{(x_0,t_0),\rho }^\frac{2}{3}\le \frac{6}{\rho }|\overline{H}|\end{aligned}$$

and so by the monotonicity formula

$$\begin{aligned}\frac{ d }{dt} \left( \psi (t)\int _{M_t} \phi _{(x_0,t_0),\rho }\Phi _{(x_0,t_0)} d \mu \right) \le \frac{6}{\rho }|\overline{H}|\int _{M_t}\psi (t)\Phi _{(x_0,t_0)}d\mu \le \frac{6}{\rho }|\overline{H}|C_0(M_0) \end{aligned}$$

where we apply the monotonicity formula to estimate the integral in terms of $M_0$. In particular, we may estimate that for any $t_2<t_3$,

$$\begin{aligned}&\psi (t_3)\int _{M_{t_3}} \phi _{(x_0,t_0),\rho }\Phi _{(x_0,t_0)} d \mu \nonumber \\&\quad \le \psi (t_2)\int _{M_{t_2}} \phi _{(x_0,t_0),\rho }\Phi _{(x_0,t_0)} d \mu +6C_0(M_0)\sqrt{\frac{t_3-t_2}{\rho ^2}}\sqrt{\int _{t_2}^{t_3} \overline{H}^2dt}\ . \end{aligned}$$

(10)

With some abuse of notation, we define localised monotonicity to be

$$\begin{aligned}\Theta ({\mathcal {M}}, x_0, t_0):=\lim _{t \nearrow t_0}\int _{M_t} \phi _{(x_0,t_0),\rho }\Phi _{(x_0,t_0) }d\mu _t\ ,\end{aligned}$$

and note that, by (10), this limit always exists and we may estimate that for $t\in (t_0-\rho ^2, t_0)$

$$\begin{aligned} \Theta ({\mathcal {M}}, x_0, t_0)\le \frac{\psi (t)}{\psi (t_0)}\int _{M_{t}} \phi _{(x_0,t_0),\rho }\Phi _{(x_0,t_0)} d \mu +6\frac{C_0(M_0)}{\psi (t_0)}\sqrt{\frac{t_0-t}{\rho ^2}}\sqrt{\int _{t}^{t_0} \overline{H}^2dt}. \end{aligned}$$

(11)

Lemma 12

(Upper semicontinuity of $\Theta$) $\Theta ({\mathcal {M}}, x_0, t_0)$ is upper semi continuous in time and space. Explicitly, if $x_j \rightarrow x_0$, $t_j \nearrow t_0$ then

$$\begin{aligned}\limsup _{j\rightarrow \infty } \Theta ({\mathcal {M}}, x_j, t_j)\le \Theta ({\mathcal {M}}, x_0, t_0)\ .\end{aligned}$$

Proof

We have that for any $t\in (t_0-\rho ^2,t_0)$, if j is large enough then

$$\begin{aligned}\Theta ({\mathcal {M}}, x_j, t_j)\le \frac{\psi (t)}{\psi (t_j)}\int _{M_{t}} \phi _{(x_j,t_j),\rho }\Phi _{(x_j,t_j)} d \mu +6\frac{C_0(M_0)}{\psi (t_j)}\sqrt{\frac{t_j-t}{\rho ^2}}\sqrt{\int _{t}^{t_j} \overline{H}^2dt}\ . \end{aligned}$$

so, taking a limsup,

$$\begin{aligned}\limsup _{j\rightarrow \infty }\Theta ({\mathcal {M}}, x_j, t_j)\le \frac{\psi (t)}{\psi (t_0)}\int _{M_{t}} \phi _{(x_0,t_0),\rho }\Phi _{(x_0,t_0)} d \mu ++6\frac{C_0(M_0)}{\psi (t_0)}\sqrt{\frac{t_0-t}{\rho ^2}}\sqrt{\int _{t}^{t_0} \overline{H}^2dt} \end{aligned}$$

and limiting $t\rightarrow t_0$ yields the required inequality (as $\psi$ is continuous). $\square$

Corollary 13

For any point $(x_0,t_0)$ reached by a smooth properly immersed flow,

$$\begin{aligned}\Theta ({\mathcal {M}},x_0,t_0)\ge 1\end{aligned}$$

Proof

On any smooth manifold with $x_j\in M_{t_j}$, $\Theta ({\mathcal {M}},x_j,t_j)=1$, so the statement follows from upper semicontinuity. $\square$

Proposition 14

(The clearing out lemma) Suppose that $M_t$ is smooth and properly immersed for $t\in [t_0-\rho _0^2,t_0)$. We suppose that for positive constants $C_0$ and L

$$\begin{aligned}\int _{t_0-\rho _0^2}^{t_0} \overline{H}^2 dt\le L \qquad \psi (t_0-\rho _0^2)\int _{M_{t_0-\rho ^2_0}} \Phi _{(x_0,t_0)} d\mu \le C_0\ .\end{aligned}$$

Then there exists a constant $\beta _0=\beta _0(n,C_0,L)$ such that for any $\beta \in (0,\beta _0)$ there exists a constant $\theta (\beta , n)$ such that for any $\rho \in (0,\rho _0)$

$$\begin{aligned}\frac{{\mathcal {H}}^n(M_{t_0-\beta \rho ^2}\cap B_{\rho })}{\rho ^n}\ge \theta e^{-^\frac{L}{2}}\ .\end{aligned}$$

Proof

By Corollary 13 and (11) we have that for $t\in (t_0-\sigma ^2,t_0)$ where $\sigma \in (0,\rho _0)$,

$$\begin{aligned}1\le \frac{\psi (t)}{\psi (t_0)}\int _{M_{t}} \phi _{(x_0,t_0),\sigma }\Phi _{(x_0,t_0)} d \mu +6\frac{C_0(M_0)}{\psi (t_0)}\sqrt{\frac{t_0-t}{\sigma ^2}}\sqrt{\int _{t}^{t_0} \overline{H}^2dt}\ .\end{aligned}$$

Rewriting, and setting $t_0-t=\alpha \sigma ^2$ for some $\alpha \in (0,1]$,

$$\begin{aligned}&1-6\frac{C_0(M_0)}{\psi (t_0)}\sqrt{\frac{t_0-t}{\sigma ^2}}\sqrt{\int _{t}^{t_0} \overline{H}^2dt}\\&\quad \le \frac{\psi (t)}{\psi (t_0)}\int _{M_{t}} \left( 1-\frac{|x-x_0|^2+2n(t-t_0)}{\sigma ^{2}}\right) ^3_+\frac{1}{(4\pi (t_0-t))^\frac{n}{2}}e^{-\frac{|x-x_0|^2}{4(t_0-t)}} d \mu \\&\quad \le \frac{\psi (t)}{\psi (t_0)}\frac{(1+2n\alpha )^3}{(4\pi \alpha \sigma ^2)^\frac{n}{2}}{\mathcal {H}}^n(M_{t_0-\alpha \sigma ^2}\cap B_{\sqrt{1+2n\alpha } \sigma }(x_0))\ . \end{aligned}$$

We wish to change constants from $(\alpha , \sigma )$ to $(\beta , \rho )$ by setting $\rho = \sqrt{1+2n \alpha } \sigma$ and $\beta \rho ^2 = \alpha \sigma ^2$. For this to be possible, we need $\beta <\frac{1}{2n}$ at which point we have that

$$\begin{aligned} \beta =\frac{\alpha }{1+2n\alpha }{} & {} \alpha =\frac{\beta }{1-2n\beta }{} & {} \sigma =\sqrt{1-2n\beta }\rho \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{(4\pi \beta )^\frac{n}{2}}{(1+2n\alpha )^3}\left( 1-6\sqrt{\alpha }C_0\sqrt{L}e^\frac{L}{2}\right)&\le e^\frac{L}{2}\frac{{\mathcal {H}}^n(M_{t_0-\beta \rho ^2}\cap B_{\rho })}{\rho ^n}\ . \end{aligned}$$

To get a positive density we want $\alpha = \min \{\frac{e^{-L}}{144C_0^2 L},1\}$ which corresponds to

$$\begin{aligned}\beta \le \min \left\{ \frac{1}{144C_0^2Le^{L}+2n}, \frac{1}{1+2n}\right\} =:\beta _0\ .\end{aligned}$$

Finally we obtain,

$$\begin{aligned} {{\theta }}(n,\beta )e^{-^\frac{L}{2}}:=\frac{1}{2}(4\pi \beta )^\frac{n}{2}(1-2n\beta )^3e^{-^\frac{L}{2}}&\le \frac{{\mathcal {H}}^n(M_{t_0-\beta \rho ^2}\cap B_{\rho }(x_0))}{\rho ^n}\ . \end{aligned}$$

$\square$

4.4 A condition for a uniform diameter bound

In this section we apply the clearing out lemma to obtain conditions under which we may ensure a uniform curvature bound.

Proposition 15

Suppose that there exists constants $h, C>0$ such that for all $t\in [0, T-h)$

$$\begin{aligned} \int _t^{t+h}\overline{H}^2 dt <C\ . \end{aligned}$$

(12)

Then there exists a constant $R=R(C, n, M_0)$ such that for all $t\in [0,T)$

$$\begin{aligned}{\text {diam}}(M_t)<R\ .\end{aligned}$$

Proof

We apply the clearing out lemma (Proposition 14) with $\rho _0=\sqrt{h}$ on any time interval of the form $[t,t+h)$. On such an interval by assumption we have the $L^2$-bound on $\overline{H}$, and the second assumption is automatically fulfilled for some $C_0=C_0(M_0)$ by the monotonicity formula. Therefore we have: for $\beta =\frac{\beta _0(n, C, M_0)}{2}$ there is a ${\tilde{\theta }}=\theta (\beta , n)e^{-\frac{C}{2}}$ such that ${\mathcal {H}}^n(M_{t_0-\beta \rho ^2}\cap B_{\rho })\ge {\tilde{\theta }}\rho ^n$.

Suppose that $t>h\beta$. We have that $|M_t|\le |M_0|$ and there exists $\frac{\rho _0}{4}\le \rho <\rho _0$ such that ${\text {diam}}(M_t)=2N\rho$ for $N\in {\mathbb {N}}$. Then there must be at least N disjoint ambient balls of radius $\rho$ whose centers lie on $M_t$. We use the area of $M_t$ in these balls as lower bounds. We see that (as $t>\beta \rho ^2$)

$$\begin{aligned}|M_0|>|M_{t-\beta \rho ^2}|>N{{\tilde{\theta }}} \rho ^n=\tfrac{1}{2}{{\tilde{\theta }}} \rho ^{n-1}{\text {diam}}(M_t) .\end{aligned}$$

As $\rho \ge \frac{\rho _0}{4}$ and ${{\tilde{\theta }}}$ is fixed, we see that ${\text {diam}}(M_t)\le 2 {{\tilde{\theta }}}^{-1}\rho _0^{1-n}4^{n-1}|M_0|$.

If $t<h \beta$, we may instead apply Proposition 7 to see that ${\text {diam}}(M_t)\le C(M_0, h, \beta )$. $\square$

Remark 10

The condition (12) in the above Proposition may also be replaced by

$\int _t^{t+h}\int _{M_t} H^2d\mu dt <C$ or equivalently $|\frac{ d }{dt}|M_t||<C$. Clearly a uniform bound on $\overline{H}$ also implies this condition.

5 Long time behaviour of VPMCF

In this section we consider the flow under the assumptions that

• a) the flow exists for all time, and

• b) the diameter is uniformly bounded, that is, there exists an $R_0>0$ such that

  $$\begin{aligned} {\text {diam}}(M_t)\le R_0 \end{aligned}$$

  (13)

  for all $t\in [0,\infty )$.

By Proposition 15 we can replace (13) with (12). Both of the above assumptions are therefore implied by a curvature bound on the flow which is uniform in time.

Definition 6

We define a “Newton inequality-like” extended isoperimetric ratio by the scale invariant quantity

$$\begin{aligned}{\mathcal {I}}(M_t) = \frac{n+1}{n}\overline{H}\frac{{\text {Vol}}^{n+1}(M_t)}{|M_t|}\ .\end{aligned}$$

Remark 11

This isoperimetric ratio appears to be a natural one to consider: If we think of the Alexandrov–Fenchel inequalities as an “averaged” version of the Mclaurin inequalities for symmetric polynomials, then the above would be a ratio similar to an “averaged” Newton inequalities. Importantly, the Newton inequalities hold without any assumptions on the curvature cone.

Remark 12

Denoting $\omega _{n}$ for the area of the n-sphere, if we write $v_{n+1}=\frac{n+1}{\omega _{n}}{{\text {Vol}}}^{n+1}(M_t)$, $v_n=|M_t|$ and $v_{n-1}=\frac{1}{n}\int _{M_t} H d\mu$ then, while $M_t$ is mean convex and starshaped, then it bounds some domain $\Omega _t$ and we know that, by the Alexandrov–Fenchel inequalities, $v_{n+1}^\frac{1}{n+1}\le v_{n}^\frac{1}{n}\le v_{n-1}^\frac{1}{n-1}$ with equalities if and only if $M_t$ is a sphere (see [28] – here we are writing, $v_{n+1-k}=\frac{V_{n+1-k}(\Omega _t)}{V_{n+1-k}(B)}$ in that paper). Applying these we have that

$$\begin{aligned}1\le \frac{v_{n+1}}{v_n^\frac{n+1}{n}}\le {\mathcal {I}}(M_t)=\frac{v_{n-1}v_{n+1}}{v_n^2}\le \frac{v_{n-1}}{v_n^\frac{n-1}{n}}=\frac{1}{{\mathcal {I}}_1(\Omega _t)^{n-1}}\ ,\end{aligned}$$

where ${\mathcal {I}}_1(\Omega _t)$ is the first Quermass integral [28, equation (6)]. Equalities hold in the above if and only if $M_t$ is a sphere. Therefore if $M_t$ is starshaped and mean convex with ${\mathcal {I}}(M_t)=1$, $M_t$ is a round sphere.

Under the above assumptions we may improve the assumptions of Sect. 3 to get the following:

Theorem 16

Suppose that $M_t$ satisfies VPMCF for all $t>0$ with $V_0\ne 0$ and has uniformly bounded diameter as in (13). Then for any sequence $t_i\rightarrow \infty$ there exists a subsequence (not relabeled) such that either

$$\begin{aligned}{\mathcal {I}}(M_{t_i})\rightarrow 1 \qquad \text { or }\qquad {\mathcal {I}}(M_{t_i})\rightarrow 0\end{aligned}$$

as $i\rightarrow \infty$. Equivalently, there exists a subsequence such that the extended isoperimetric ratio goes to that of a constant mean curvature surface or of a minimal surface.

Corollary 17

Let X be a mean convex embedded starshaped solution of the VPMCF with uniformly bounded curvature for all times. Then $X(\cdot ,t)$ converges smoothly to a round sphere with enclosed volume $V(X(\cdot ,0))$ for $t\rightarrow \infty$.

Proof of Corollary 17

We first note that the uniform curvature bound implies that the diameter is bounded by Proposition 15. The area remains also bounded due to the gradient flow property. This allows us to use Shi-type estimates as sketched in Remark 8 to get uniform estimates on all derivatives of curvature.

We now pick a sequence of points and times $(p_i, t_i)$ such that $M_{t_i}=:M_i$ is starshaped over $p_i$. We denote $X_i:S^n\rightarrow \mathbb {R}^{n+1}$ to be the immersion for $M_{i}$ translated so that $p_i$ is moved to the origin. We may now apply the compactness theorem of Breuning [9] and get a limiting immersion ${{\tilde{X}}}:{{\tilde{M}}}\rightarrow {\mathbb {R}}^{n+1}$ and smooth subconvergence of $X_i$ to ${{\tilde{X}}}$ on compact subsets of ${{\tilde{M}}}^n$ with the following properties: a) ${{\tilde{M}}}^n$ is connected and a topological sphere because the domain enclosed by $X_i$ is starshaped around 0 for all i – so ${{\tilde{M}}}^n$ can be assumed to be $M^n={\mathbb {S}}^n$ (we compose with appropriate diffeomorphisms if necessary, see [9]); b) in fact, the convergence is uniform because $M^n$ is compact.

For the convenience of the reader, we sketch the above compactness result under our stronger assumptions (which include having a sequence of immersions on a fixed manifold): Due to the curvature bound there exists an $r>0$ such that for any i, $M_i$ may be locally written as graphs over a ball $B_{3r}\subset T_p M_i$ for any $p\in M_i$. As a result we claim that there exists an N such that for all i, $M_i$ may be covered by N graphs over $B_r$ tangent balls. Take all such balls of radius $B_{\frac{r}{5}}$ then by the Vitali covering lemma there exists a countable subcollection of balls of radius r with centres $p^i_1,p^i_2, \ldots$ such that $B_r\subset T_{p_i}M_i$ cover $M_i$ and the graphs over $B_{\frac{r}{5}}$ are disjoint. Each of the $B_{\frac{r}{5}}$ graphs contribute a fixed amount of area and so there can only be a finite number N given by our (uniform-in-i) area bound. Writing $p_1^i, \ldots , p_N^i$ for the tangent space centres on $M_i$, using that $M_i$ stays in a bounded region given by the diameter bound, and by compactness of $S^n=Gr(\mathbb {R}^n;\mathbb {R}^{n+1})$, we may choose a subsequence (also labelled with i) so that these centres and the tangent spaces converge to a sequence of points $p_1^\infty , \ldots , p_N^\infty$, so that graphs over $B_{2r}$ cover a smooth manifold $M_\infty$. After taking further subsequences, Arzelà–Ascoli implies the convergence is smooth and that for i large enough, images of the $X_i$ may be written as smooth local graphs over the $p_i^\infty$ converging to $M_\infty$. However we immediately see that for i large enough $M_\infty$ has the same topology as $M_i$, which due to starshapedness must be a topological sphere.

As $\left\langle \frac{x}{|x|} , \nu \right\rangle \ge 0$ on the $M_i$, we get ${\mathcal {I}}(M_i)\ge 1$, which implies ${\mathcal {I}}(M_\infty )\ge 1$. By applying Theorem 16, we conclude ${\mathcal {I}}(M_\infty )=1$. Therefore, as in Remark 12, we know that $M_\infty$ is a round sphere. Finally, picking a $t_{i_0}$ so that $M_{i_0}$ is (smoothly) close enough to $M_\infty$, and restarting the flow from $t_{i_0}$ we may now apply [23] to see that the flow converges exponentially and smoothly to a round sphere. But now by smooth convergence it must be $M_\infty$, which is a round sphere at finite distance from $p_{i_0}$. So the translating procedure can be reversed and the original immersions $X(\cdot ,t)$ converge to a round sphere for $t\rightarrow \infty$.

Remark 13

• If we have a VPMCF that stays Alexandrov immersed for all times and $|A|\le C$ uniformly, then for each $t_j\rightarrow \infty$, the flow subconverges smoothly to a closed limiting surface with constant mean curvature ($H(\cdot ,{{\tilde{t}}}_j)$ must converge to $\overline{H}_\infty$ see e.g. [30]). In this Alexandrov immersed case the limiting object must be a sphere – and thus ${\mathcal {I}}(M_\infty )=1$. This follows from Alexandrov’s theorem: A closed Alexandrov immersion having constant mean curvature must be a sphere [1]. We can also improve subconvergence to convergence by [23] again.

• In general, if no singularity appears, i.e. $T=\infty$ and $|A|\le C$ (uniformly in t), then ${\mathcal {I}}(M_{t_j})\rightarrow 0$ cannot happen because any limiting surface satisfies $|M_\infty |>0$, so ${\mathcal {I}}(M_{{{\tilde{t}}}_j})\rightarrow 0$ would imply we found a closed minimal surface as limiting object which is clearly impossible. At this point we do not know whether ${\mathcal {I}}(M_{\infty })=1$ implies that $M_{\infty }$ is a sphere. We only know this for the starshaped case and for Alexandrov immersions.

In order to show Theorem 16 we first improve our estimates on $\overline{H}^2$ in the two cases of ${\mathcal {I}}(M_t)\le 1$ (where ${\mathcal {I}}(\mathbb {S}^n)=1$) and ${\mathcal {I}}(M_t)\ge 1$.

Proposition 18

Suppose that $M_t$ as in (13) and $V_0\ne 0$. Let $\alpha \in (0,1)$, and suppose that on the time interval $[t_1, t_2]$ we may estimate

$$\begin{aligned}\overline{H}(n+1)V_0\le \alpha n |M_t|\ .\end{aligned}$$

Then we have that

$$\begin{aligned} \int _{t_1}^{t_2}|\overline{H}|^2dt&\le \max \left( \frac{\alpha ^2}{(1-\alpha )^2}, 1\right) \frac{R^2_0 }{2(n+1)^2V_0^2}|M_{t_1}|^2\ . \end{aligned}$$

Note that $\frac{\alpha ^2}{(1-\alpha )^2 }\ge 1$ if and only if $\alpha \ge \frac{1}{2}$.

Proof

In Proposition 6, we estimated $\overline{H}$ using Young’s inequality. Here we estimate more carefully using the above assumption.

Using (4), we note that,

$$\begin{aligned} \begin{aligned} \int _{M_t} \left\langle X_t , X-x \right\rangle d \mu&=\overline{H}\int _M\left\langle \nu , X-x \right\rangle d \mu -\int _M H\left\langle \nu , X-x \right\rangle d \mu \\&=\overline{H}(n+1)V_0-n|M_t| \\&\le - n(1-\alpha ) |M_t| \end{aligned} \end{aligned}$$

(14)

We set $b=n|M_t|$ and $a=-\int _{M_t} \left\langle X_t , X-x \right\rangle d \mu \ge 0$. Then we consider two cases. First, we assume that $a<b$. We use $a= \alpha a + (1-\alpha ) a \ge (1-\alpha )b$ which implies $|b-a|=b-a \le \frac{\alpha }{1-\alpha } a = \frac{\alpha }{1-\alpha } |a|$. On the other hand, if $a>b$, then we know that $|b-a| = a-b < a=|a|$ because $b>0$. Putting this together, we get that $|b-a|\le \max (1,\frac{\alpha }{1-\alpha })|a|$ and thus

$$\begin{aligned}\left| \int _{M_t} \left\langle X_t , X-x \right\rangle d \mu +n|M_t|\right| \le \max \left( \frac{\alpha }{1-\alpha },1\right) \left| \int _M \left\langle X_t , X-x \right\rangle d \mu \right| \end{aligned}$$

and

$$\begin{aligned}|\overline{H}|\le \max \left( \frac{\alpha }{1-\alpha },1\right) \frac{\left| \int _M \left\langle X_t , X-x \right\rangle d \mu \right| }{(n+1) |V_0|}\ .\end{aligned}$$

Estimating as before we have

$$\begin{aligned} |\overline{H}|^2&\le \max \left( \frac{\alpha ^2}{(1-\alpha )^2},1\right) \frac{\int _M |H-\overline{H}|^2d \mu \int _M |X|^2d \mu }{(n+1)^2V_0^2}\\&\le -\max \left( \frac{\alpha ^2}{(1-\alpha )^2},1\right) \frac{R_0^2}{2(n+1)^2V_0^2}\frac{ d }{dt}|M_t|^2\ . \end{aligned}$$

And so

$$\begin{aligned}\int _{t_1}^{t_2}|\overline{H}|^2dt \le \max \left( \frac{\alpha ^2}{(1-\alpha )^2},1\right) \frac{R^2_0}{2(n+1)^2V_0^2}[|M_{t_1}|^2-|M_{t_2}|^2]\ .\end{aligned}$$

$\square$

We may now repeat this but for the opposite inequality of the (extended) isoperimetric (or Alexandrov-Fenchel) inequality.

Proposition 19

Suppose that $M_t$ has ${\text {diam}}(M_t)\le R$ and $V_0\ne 0$. If on the time interval $[t_1, t_2]$ we may estimate

$$\begin{aligned}\overline{H}(n+1)V_0\ge (1+\beta ) n |M_t|\ ,\end{aligned}$$

for some $\beta >0$, then we have that

$$\begin{aligned}\int _{t_1}^{t_2}|\overline{H}|^2 dt \le \frac{R^2(1+\beta ^{-1})^2}{2(n+1)^2V_0^2}|M_{t_1}|^2 \ .\end{aligned}$$

Proof

We note that from the assumption,

$$\begin{aligned}\int _{M_t} \left\langle X_t , X-x \right\rangle d \mu&=\overline{H}(n+1)V_0-n|M_t| \\&\ge \beta n |M_t|\ . \end{aligned}$$

This time, may directly estimate using the triangle inequality that

$$\begin{aligned}|\overline{H}|\le \frac{(1+\beta ^{-1})\left| \int _M \left\langle X_t , X-x \right\rangle d \mu \right| }{(n+1)|V_0|}\ .\end{aligned}$$

so, as above

$$\begin{aligned}|\overline{H}|^2\le -\frac{R^2_0(1+\beta ^{-1})^2}{2(n+1)^2V_0^2}\frac{ d }{dt}|M_t|^2\ ,\end{aligned}$$

and so

$$\begin{aligned}\int _{t_1}^{t_2}|\overline{H}|^2 dt \le \frac{R^2_0(1+\beta ^{-1})^2}{2(n+1)^2V_0^2}[|M_{t_1}|^2-|M_{t_2}|^2]\ .\end{aligned}$$

$\square$

Proof of Theorem 16

Suppose not. Then there exists a $T, \epsilon >0$ such that for all $t>T$, ${\mathcal {I}}(M_t)\in (-\infty , -\epsilon )\cup (\epsilon ,1-\epsilon )\cup (1+\epsilon ,\infty )$. By continuity we have that ${\mathcal {I}}(M_t)$ is in one of the intervals $(-\infty , -\epsilon )$, $(\epsilon ,1-\epsilon )$ or $(1+\epsilon ,\infty )$ for all $t>T$. We particularly have that $\overline{H}(n+1) V_0 = n|M_t| {\mathcal {I}}(M_t)$ is either not larger than $(1-\epsilon )n|M_t|$ or not less than $(1+\epsilon )n|M_t|$. As a result, by applying Propositions 18 and 19 we have one of

$$\begin{aligned}\int _T^\infty \overline{H}^2dt \le \max \left( \frac{\epsilon ^2}{(1-\epsilon )^2},1\right) \frac{R^2_0}{2(n+1)^2V_0^2}|M_{t_1}|^2\ .\end{aligned}$$

or

$$\begin{aligned}\int _T^\infty \overline{H}^2dt \le \frac{R^2_0(1+\epsilon ^{-1})^2}{2(n+1)^2V_0^2}|M_{t_1}|^2 \ .\end{aligned}$$

Using the isoperimetric inequality, we have that $|\overline{H}|= \frac{n|M_t|}{(n+1)|V_0|}|{\mathcal {I}}(M_t)|\ge c>0$ in both cases. This is clearly impossible.

Corollary 20

If a solution exists for all $t\ge 0$ and for all this time $\overline{H}(n+1)V_0\le \alpha n|M_t|$ for some $\alpha \in (0,1)$ then there exist $t_i$ such that $\overline{H}(t_i)\rightarrow 0$ as $i\rightarrow \infty$.

6 Barriers and preservation of Alexandrov immersions

In Corollary 4 we already saw that the property of being Alexandrov immersed and having $\overline{H}\le 0$ is not preserved for long under the VPMCF. Now we will study under which conditions the property of being Alexandrov immersed is preserved for all times. We will also describe how barriers for the VPMCF can be constructed.

We start by introducing some notation. We will use descriptive superscripts to distinguish between manifold, e.g. $M^\text {outer}_t$, $M^\text {Inner}$ and so on. For example, we let $f^*:[0,T) \rightarrow \mathbb {R}$ be some smooth function.

Definition 7

We will say a flow $M_t^*$ satisfies Forced Mean Curvature Flow with forcing term $f^*$ (and write this as FMCF($f^*$)) if it has a time dependent parametrisation $X^*:N^n \times [0,T)\rightarrow \mathbb {R}^{n+1}$ (and chosen normal $\nu ^*$) so that

$$\begin{aligned}\left\langle \frac{ d X^*}{dt}+H^*\nu ^* , \nu ^* \right\rangle = f^*\ .\end{aligned}$$

We will use this to produce barriers for the flow.

Definition 8

Suppose that $M_t^1$ satisfies FMCF($f^1$) and $M_t^2$ satisfies FMCF($f^2$). Suppose that there is an open ball $B_{r}(x)$ so that $M_t^1$ has non-trivial intersection with $B_{r}(x)$ so that $B_{r}(x)\setminus M_t^1$ is made up of two connected components.

We define the component which the normal points away from to be the local interior of $M_t^1$, and the component the normal points into to be the local exterior of $M_t^1$.

We will say that $M^2_t$ is locally inside $M_t^1$ if it is contained in the local interior of $M_t^1$. We will say that $M^2_t$ is locally outside $M_t^1$ if it is contained in the local exterior of $M_t^1$

Lemma 21

Suppose that $M_t^\text {out}$ satisfies FMCF($f^\text {out}$) and $M_t^\text {in}$ satisfies FMCF($f^\text {in}$). Suppose that in open some ball $B_{2r}(x)$, $M^\text {in}_0$ is locally inside $M^\text {out}_0$. Given $\sigma \in \{1,-1\}$, suppose that

$$\begin{aligned}f^\text {out}-\sigma f^\text {in}\ge 0\ .\end{aligned}$$

Then $M^\text {in}_t\cap B_{r}(x)$ cannot intersect $M^\text {out}_t\cap B_{r}(x)$ for the first time with $\left\langle \nu ^\text {in} , \nu ^\text {out} \right\rangle =\sigma$ at an interior point.

Proof

Step 1: The graph equation. We describe the evolution of a graphical piece of a surface moving by FMCF($f^\text {gr}$):

We consider a solution $M^\text {gr}_t$ to FMCF($f^\text {gr}$) locally as a graph of the function graph function $u^\text {gr}$ in direction $e_{n+1}$ (possibly after rotation). Then we calculate with respect to unit normal $\mu ^\text {gr}$ showing upwards in the graphical setting (the graphical normal $\mu ^\text {gr}$ may not be the same normal as the a priori chosen $\nu ^\text {gr}$)

$$\begin{aligned}&v^\text {gr}=\sqrt{1+|Du^\text {gr}|^2}, \qquad \mu ^\text {gr} = v^{-1}_\text {gr}(-Du^\text {gr} + e_{n+1}),\\&g_{ij} = \delta _{ij} + D_i u^\text {gr} D_j u^\text {gr},\qquad g^{ij} = \delta ^{ij} -v^{-2}_\text {gr} D^i u^\text {gr} D^j u^\text {gr}, \end{aligned}$$

and

$$\begin{aligned} h_{ij}^\text {gr} =-v^{-1}_\text {gr} D^2_{ij} u^\text {gr}, \qquad H=-v^{-1}_\text {gr}(\delta ^{ij} -v^{-2}_\text {gr} D^i u^\text {gr} D^j u^\text {gr})D^2_{ij} u^\text {gr}\end{aligned}$$

and so

$$\begin{aligned}\left\langle u^\text {gr}_t e_{n+1} - v^{-1}_\text {gr}(\delta ^{ij} -v^{-2} D^i u^\text {gr} D^j u^\text {gr})D^2_{ij} u^\text {gr} \nu ^\text {gr} , \nu ^\text {gr} \right\rangle =f^\text {gr}\left\langle \nu ^\text {gr} , \mu ^\text {gr} \right\rangle \end{aligned}$$

or

$$\begin{aligned}u_t^\text {gr} = (\delta ^{ij} -v^{-2}_\text {gr} D^i u^\text {gr} D^j u^\text {gr})D^2_{ij} u^\text {gr} + v^\text {gr} f^\text {gr}\left\langle \nu ^\text {gr} , \mu ^\text {gr} \right\rangle \ .\end{aligned}$$

Here (while hypersurface is graphical), $\left\langle \nu ^\text {gr} , \mu ^\text {gr} \right\rangle \in \{1,-1\}$.

Step 2: Proof of the statement.

Suppose not. Suppose that the first point of intersection is $(p_0, t_0)$ for some $t_0>0$. Then after rotation we may assume that $\nu ^\text {out}(p_0,t_0)=e_{n+1}$ and so $\nu ^\text {in}(p_0,t_0)=\pm e_{n+1}$. For some $\epsilon >0$ we can write both $M^\text {out}_t$ and $M^\text {in}_t$ graphically on $B_\epsilon (p_0)\times (t_0-\epsilon ,t_0]$ where

$$\begin{aligned}u_t^\text {out} = (\delta ^{ij} -v^{-2}_\text {out} D^i u^\text {out} D^j u^\text {out})D^2_{ij} u^\text {out} + v^\text {out} f^\text {out}\ ,\end{aligned}$$

and

$$\begin{aligned}u_t^\text {in} = (\delta ^{ij} -v^{-2}_\text {in} D^i u^\text {in} D^j u^\text {in})D^2_{ij} u^\text {in} + v^\text {in} f^\text {in}\left\langle \nu ^\text {in} , \mu ^\text {in} \right\rangle \ .\end{aligned}$$

Due to our definitions we have that $w:=u^\text {out}-u^\text {in}>0$ for $t<t_0$ and

$$\begin{aligned} w_t&= a^{ij}D^2_{ij} w +{\hat{b}}^i D_i w +\sqrt{1+|Du^\text {out}|^2}f^\text {out} - \sqrt{1+|Du^\text {in}|^2}\left\langle \nu ^\text {in} , \mu ^\text {in} \right\rangle f^\text {in}\\&= a^{ij}D^2_{ij} w +{\hat{b}}^i D_i w +\sqrt{1+|Du^\text {out}|^2}(f^\text {out}-\left\langle \nu ^N , \mu ^N \right\rangle f) \\&\qquad +(\sqrt{1+|Du^\text {out}|^2}- \sqrt{1+|Du^\text {in}|^2})\left\langle \nu ^\text {in} , \mu ^\text {in} \right\rangle f^\text {in}\\&= a^{ij}D^2_{ij} w +{\hat{b}}^i D_i w +\sqrt{1+|Du^\text {out}|^2}(f^\text {out}-\left\langle \nu ^\text {in} , \mu ^\text {in} \right\rangle f)\\&\qquad +\frac{(D^iu^\text {out}+D^iu^\text {in})(D^iu^\text {out}-D^iu^\text {in})}{\sqrt{1+|Du^\text {in}|^2}+ \sqrt{1+|Du^\text {in}|^2}}\left\langle \nu ^\text {in} , \mu ^\text {in} \right\rangle f^\text {in}\\&= a^{ij}D^2_{ij} w +b^i D_i w +\sqrt{1+|Du^\text {out}|^2}(f^\text {out}-\left\langle \nu ^\text {in} , \mu ^\text {in} \right\rangle f^\text {in}) \end{aligned}$$

where $a^{ij}$ and $b^i$ are the usual terms and we have computed the additional terms to needless accuracy. We know that $\left\langle \nu ^\text {in} , \mu ^\text {in} \right\rangle =\left\langle \nu ^\text {in}(p_0,t_0) , \nu ^\text {out}(p_0,t_0) \right\rangle =\sigma$. Therefore as $f^\text {out}-\left\langle \nu ^\text {in} , \mu ^\text {in} \right\rangle f^\text {in}\ge 0$, we have a contradiction to the strong maximum principle. $\square$

Lemma 22

Suppose that $M_t^1$ satisfies FMCF($f^1$) and $M_t^\text {out}$ satisfies FMCF($f^\text {out}$). Suppose that in a ball $B_r(x)$ we have that $M_0^1$ is locally outside $M_0^\text {in}$. Suppose that for some $\sigma \in \{1,-1\}$

$$\begin{aligned}f^1-\sigma f^\text {out}\le 0\end{aligned}$$

Then $M^1_t\cap B_{r}(x)$ cannot intersect $M^\text {out}_t\cap B_{r}(x)$ for the first time with $\left\langle \nu ^1 , \nu ^ \text {out} \right\rangle =\sigma$ at an interior point.

Proof

Suppose not. This time, we may take $e_{n+1} = \nu ^1(p_0,t_0)=\sigma \nu ^\text {out}(p_0,t_0)$ (as otherwise one would have been locally inside the other). This time $w=u^1-u^\text {out}<0$ and

$$\begin{aligned} w_t&= a^{ij}D^2_{ij} w +b^i D_i w +\sqrt{1+|Du^1|^2}(f^1-\sigma f^\text {out})\ . \end{aligned}$$

Again we get a contradiction to the strong maximum principle. $\square$

Proposition 23

(Outer Barriers) Suppose that $M^\text {out}_t$ is a solution for FMCF($f^\text {out}$) which is embedded for all time and bounds the compact region $\Lambda (t)$, and $\nu ^N$ points out of $\Lambda (t)$. Suppose that $M_t$ satisfies VPMCF and $M_0\subset \overset{\circ }{\Lambda (0)}$ and for all $t\in [0,T)$

$$\begin{aligned}f^\text {out}\ge |\overline{H}| .\end{aligned}$$

Then for all $t\in [0,T)$, $M_t \subset \Lambda (t)$.

Proof

Suppose not. We have that for $\sigma \in \{1,-1\}$,

$$\begin{aligned}f^\text {out} - \sigma \overline{H}\ge f^\text {out}-|\overline{H}|\ge 0\end{aligned}$$

and so by Lemma 21, at a first time of intersection the two normals cannot be multiples of each other, a contradiction. $\square$

Proposition 24

(Inner Barriers) Suppose that $N_t$ is a solution for FMCF(f) which is embedded for all time and bounds the compact region $\Lambda (t)$, and $\nu ^N$ points out of $\Lambda (t)$. Suppose that $M_0\cap \overline{\Lambda (0)}=\emptyset$, and for all $t\in [0,T)$

$$\begin{aligned}f\le -|\overline{H}| .\end{aligned}$$

Then for all $t\in [0,T)$, $M_t\cap \overline{\Lambda (0)}=\emptyset$.

Proof

Suppose not. For any $\sigma \in \{1,-1\}$

$$\begin{aligned}f^1-\sigma \overline{H}\le f^1+ |\overline{H}|\le 0\end{aligned}$$

and so by Lemma 22, at a first time of intersection the two normals cannot be multiples of each other, a contradiction. $\square$

We need the following corollary from our previous paper:

Corollary 25

([35, Corollary 5]) A compact flowing manifold with bounded curvature may only loose the property of being Alexandrov immersed at time T if there exist points $p and q(t)\in M^n$ so that $|X(p,t)-X(q(t),t)|$ goes to zero with $\left\langle \nu (p,t) , \nu (q(t),t) \right\rangle =-1$ and where $\left\langle \nu (p) , X(q(t),t)-X(p,t) \right\rangle <0$ and $\left\langle \nu (q,t) , X(p,t)-X(q(t),t) \right\rangle <0$ for $T-\delta<t<T$.

Theorem 26

Let $M_0$ be Alexandrov immersed $M_t$ the VPMCF starting from $M_0$. As long as $\overline{H} \ge 0$, the flow stays Alexandrov immersed.

Proof

From Corollary 25 we know that at a time where Alexandrov immersion property is lost two disjoint pieces of the flow must intersect, with the normals are in opposite directions. In Lemma 21 this corresponds to $\sigma =-1$ while $f^\text {out}=f^{\text {in}}=\overline{H}$, and so we immediately see that this cannot happen. $\square$

A natural question to ask is when $\overline{H}\ge 0$ is preserved if we know that $M_t$ is Alexandrov immersed. This is not always true without further assumptions – see the Example 2 below.

Appendix A: The Trilobite - an example in which \({\overline{H}}\ge 0\) is lost

A natural hope, given the previous section, is to attempt to show that under Alexandrov immersed VPMCF, $\overline{H}(t)\ge 0$ is preserved. The following example demonstrates that this is not the case without further assumptions on the flow.

The evolution equation for $\overline{H}$ is given by

So, for surfaces ($n=2$), if $\overline{H}(0)=0$, then

using the Gauss equations. We now describe initial data such that $\int _M H d\mu =0$ and $\int _M KH d\mu >0$, meaning that $\overline{H}\ge 0$ is not preserved.

Our example is constructed from rotationally symmetric pieces. If $\gamma$ is the profile curve (parametrised by arclength) of a rotationally symmetric surface (rotated about the y-axis) then we have that the curvatures are given by

$$\begin{aligned}\kappa _1 = \kappa , \qquad \kappa _2 =\frac{\left\langle {\dot{\gamma }} , e_y \right\rangle }{x}=\frac{{\dot{y}}}{x}, \qquad H=\frac{{\dot{y}}}{x}+\kappa , \qquad K=\frac{{\dot{y}}\kappa }{x}\end{aligned}$$

and we may see that $d\mu = 2\pi x ds$.

Therefore we have that (for $I=[a,b]$)

$$\begin{aligned} \int _M H d\mu = 2\pi \int _\gamma x\kappa ds +2\pi ( y(b)-y(a)), \qquad \int _M HK d\mu = 2\pi \int _\gamma {\dot{y}} \kappa ^2 + \kappa \frac{{\dot{y}}^2}{x} ds \ . \end{aligned}$$

(15)

We calculate the above for various profiles.

On a hemisphere $E_\rho$ of radius $\rho$: We have that $\kappa _1=\kappa _2=\rho ^{-1}$ and so in this case $H=2\rho ^{-1}$, $K = \rho ^{-2}$,

$$\begin{aligned}\int _{E_\rho } H = 4\pi \rho , \qquad \int _{E_\rho } HK d\mu = 4\pi \rho ^{-1}\ .\end{aligned}$$

On a cylinder $C_{\rho , l}$ of radius $\rho$ and length l: We have that $\kappa _1=0$, $\kappa _2=\rho ^{-1}$ and so in this case $H=\rho ^{-1}$, $K = 0$,

$$\begin{aligned}\int _{C_{\rho , l}} H = 2\pi l, \qquad \int _{C_{\rho , l}} HK d\mu = 0 \ .\end{aligned}$$

On a rotated arc $I^{\theta _1, \theta _2}_{(\tilde{x},{{\tilde{y}}}), \rho }\subset \mathbb {R}_+$ centred at $({{\tilde{x}}},{{\tilde{y}}})$ through angles $[\theta _1, \theta _2]$ from the horizontal: Parametrising as $\gamma (s) =(\rho \cos (\rho ^{-1} s)+\tilde{x},\rho \sin (\rho ^{-1} s)+{{\tilde{y}}})$ for $s\in [\rho \theta _1,\rho \theta _2]$, we have that $\kappa =\rho ^{-1}$ so

$$\begin{aligned}\int _{I^{\theta _1, \theta _2}_{({{\tilde{x}}},{{\tilde{y}}}), \rho }} H&= 2\pi \rho ^{-1} \int _{\rho \theta _1}^{\rho \theta _2} \rho \cos (\rho ^{-1} s)+{{\tilde{x}}} ds+2\pi \rho (\sin (\theta _2)-\sin (\theta _1))\\&=4\pi \rho [\sin ({{\theta }}_2)-\sin ({{\theta }}_1)]+2\pi {{\tilde{x}}}({{\theta }}_2-{{\theta }}_1) \end{aligned}$$

while

$$\begin{aligned} \int _{I^{\theta _1, \theta _2}_{({{\tilde{x}}},{{\tilde{y}}}), \rho }} HK d\mu&= 2\pi \rho ^{-1}\int _{\rho \theta _1}^{\rho \theta _2} \cos (\rho ^{-1} s)\rho ^{-1}+\frac{\cos ^2(\rho ^{-1} s)}{\rho \cos (\rho ^{-1} s)+{{\tilde{x}}}} ds\\&= 2\pi \int _{\theta _1}^{\theta _2}\frac{\cos ^2(t)}{\rho \cos (t)+{{\tilde{x}}}} dt+2\pi \rho ^{-1}[\sin ({{\theta }}_2)-\sin ({{\theta }}_1)]\ . \end{aligned}$$

As $\int _{\theta _1}^{\theta _2} \cos ^2(t)dt=\frac{1}{4}[\sin (2{{\theta }}_2)-\sin (2{{\theta }}_1)] +\frac{1}{2}[{{\theta }}_2-{{\theta }}_1]$ we may estimate the integral as

$$\begin{aligned}{} & {} \pi \frac{\frac{1}{2}[\sin (2{{\theta }}_2)-\sin (2{{\theta }}_1)] +[{{\theta }}_2-{{\theta }}_1]}{{{\tilde{x}}}+\rho }\le 2\pi \int _{\theta _1}^{\theta _2}\frac{\cos ^2(t)}{\rho \cos (t)+{{\tilde{x}}}} dt \\{} & {} \quad \le \pi \frac{\frac{1}{2}[\sin (2{{\theta }}_2)-\sin (2{{\theta }}_1)] +[{{\theta }}_2-{{\theta }}_1]}{{{\tilde{x}}}-\rho }\end{aligned}$$

Fig. 2

The construction of the Trilobite example in pictorial form

Example 2

(The Trilobite) We construct a family of embedded surfaces in $\mathbb {R}^3$ as follows. We start with a construction that is $C^1$ everywhere and smooth away from 1 dimensional glueing points (Fig. 2).

Step 1 - Capped cylinders, $Q_{\rho ,l}$ We start with a hemisphere of radius $\rho$, $E_\rho$, attach this to a long cylinder $C_{\rho , l}$ and then attach this to the plane using a quarter circle $I_{(2\rho ,-\rho ),\rho }^{(\frac{\pi }{2}, \pi )}$ of radius $\rho$, as shown. We will call this entire union $Q_{\rho ,l}$. In our constuction, we will choose out normals to point into the cylinders meaning that they contribute negatively to both integrals. We calculate

	$\int Hd\mu$	$\int HK d\mu$
$E_\rho ^1$	$-4\pi \rho$	$-4\pi \rho ^{-1}$
$C_{\rho , l}$	$-2\pi l$	0
$I_{(2\rho ,-\rho ),\rho }^{(\frac{\pi }{2}, \pi )}$	$2\pi \rho (\pi -2)$	$\ge \frac{\pi ^2}{6\rho } -2\pi \rho ^{-1}$

We get that

$$\begin{aligned}\int _{Q_{\rho ,l}} H d\mu = 2\pi [-(4-\pi )\rho -l], \qquad \int _{Q_{\rho ,l}} KH d\mu \ge -6\pi \rho ^{-1} \end{aligned}$$

Step 2 - Attach n capped cylinders to a disk We now attach n of these to a flat disk of radius $2n \rho$ which we may always do (e.g. in a line). We will refer to this modified disk as $P_{\rho ,l,n}$ where we note that the integrals over H and HK on this disk are just n times those on $Q_{\rho ,l}$.

Step 3 - Glue into a final rotationally symmetric surface We now attach this to the rotationally symmetric surface given by taking another third of a circle $I^{-\frac{\pi }{6}, \frac{\pi }{2}}_{(2n\rho ,-l),l}$ before closing the surface with a cone of slope $\sqrt{3}$ which we will denote $O_{\sqrt{3}}$. We round off the point of the cone using a spherical cap of radius r denoted $A_r$.

Step 4 - A careful choice of constants and smoothing In the above, altering l in the cylinders $Q_{\rho , l}$ changes only $\int H$, and by increasing n sufficiently, we will see below see that this may be used to ensure that in the above we have $\int H=0$. Unfortunately, the caps and joins add a significant negative quantity to $\int HK$. However $A_r$ has only negligable $\int H$ contribution while adding an arbitrarily large amount to $\int HK$. Balancing these (see calculations below) mean we can ensure that $\int H=0$ while $\int HK$ can be made arbitrarily large. However, we would like a smooth manifold - to do this we smooth only locally to the joins, perturbing $\int H$, $\int HK$ by an arbitrarily small amount. Away from the joins, the interior of the cylinders are still cylinders and so shortening or lengthening one of these slightly will again restore $\int H=0$.

We now do the accountancy and calculate the following contributions to the integrals:

	$\int Hd\mu$	$\int HK d\mu$
$P_{\rho ,l,n}$	$2n\pi [-(4-\pi )\rho -l]$	$\ge -6n\pi \rho ^{-1}$
$I^{-\frac{\pi }{6}, \frac{\pi }{2}}_{(2n\rho ,-l),l}$	$6\pi l+\frac{8}{3}n\pi ^2\rho$	positive
$O_{\sqrt{3}}$	$2n\rho +l-r$	0
$A_r$	$2\pi (2-\sqrt{3}) r$	$2\pi (2-\sqrt{3}) r^{-1}$

so

$$\begin{aligned}&\int H d\mu = 2\pi [(3+(2\pi )^{-1}-n)l+(\frac{11}{3}\pi +\pi ^{-1} -4) n\rho +(2-\sqrt{3}-(2\pi )^{-1})r]\\&\quad \int HK d\mu \ge 2\pi (2-\sqrt{3}) r^{-1}-6n\pi \rho ^{-1}\ . \end{aligned}$$

Therefore we may pick (for example) $\rho =1$, $n\ge 4$ (e.g. $n=7$), and choose

$$\begin{aligned}l(r)=\frac{(\frac{11}{3}\pi +\pi ^{-1} -4) n\rho +(2-\sqrt{3}-(2\pi )^{-1})r}{n-3-(2\pi )^{-1}}\ .\end{aligned}$$

Then, by setting r to sufficiently small (compared to $n^{-1}\rho$) we may make $\int HK d\mu$ arbitrarily large.