Neumann boundary value problem for general curvature flow with forcing term

Abstract

We consider the evolution of a strictly convex hypersurface by a class of general curvature. We prove that given some Neumann boundary condition, the flow exists for all time and converges to a solution with prescribed general curvature that satisfies the Neumann boundary condition. Our method also works for the corresponding elliptic setting.

1 Introduction

In this paper, we study the deformation of a strictly convex graph over a bounded, convex domain $\Omega \subset \mathbb {R}^n, n\ge 2,$ to a convex graph with prescribed general curvature and Neumann boundary condition.

More precisely, let $\Sigma (t)=\{X:=(x, u(x, t))| (x, t)\in \Omega \times [0, T)\},$ we study the long time existence and convergence of the following flow problem

$$\begin{aligned} \left\{ \begin{aligned}&\dot{u}=w\left( f(\kappa [\Sigma (t)])-\Phi (x, u)\right)&\text{ in } \Omega \times [0, T)\\&u_\nu =\varphi (x, u)&\qquad \text{ on } \partial \Omega \times [0, T)\\&u|_{t=0}=u_0&\text{ in } \Omega , \end{aligned}\right. \end{aligned}$$

(1.1)

where $\Phi , \varphi : \bar{\Omega }\times \mathbb {R}\rightarrow \mathbb {R}$ are smooth functions, $\nu$ denotes the outer unit normal to $\partial \Omega ,$ $w=\sqrt{1+|Du|^2},$ $\kappa [\Sigma (t)]=(\kappa _1, \ldots , \kappa _n)$ denotes the principal curvatures of $\Sigma (t),$ and $u_0: \bar{\Omega }\rightarrow \mathbb {R},$ the initial hypersurface, is a smooth, strictly convex function over $\Omega .$

To guarantee that as long as the flow exits, $\Sigma (t)$ stays convex, the curvature function f has to satisfy some structure conditions. Accordingly, the function f is assumed to be defined in the convex cone $\Gamma _n^+\equiv \{\lambda \in \mathbb {R}^n: \text{ each } \text{ component } \lambda _i>0 \}$ in $\mathbb {R}^n$ and satisfying the following conditions:

$$\begin{aligned} f_i(\lambda )\equiv \frac{\partial f(\lambda )}{\partial \lambda _i}>0\,\,\text{ in } \Gamma _n^+, \quad 1\le i\le n, \end{aligned}$$

(1.2)

and

$$\begin{aligned} f \text{ is } \text{ a } \text{ concave } \text{ function. } \end{aligned}$$

(1.3)

In addition, f will be assumed to satisfy some more technical assumptions. These include

$$\begin{aligned}&f>0\,\,\text{ in } \Gamma _n^+, f=0 \quad \text{ on } \partial \Gamma _n^+, \end{aligned}$$

(1.4)

$$\begin{aligned}&f(1, \ldots , 1)=1, \end{aligned}$$

(1.5)

and

$$\begin{aligned} f \text{ is } \text{ homogeneous } \text{ of } \text{ degree } \text{ one. } \end{aligned}$$

(1.6)

Moreover, for any $C>0$ and every compact set $E\subset \Gamma ^+_n,$ there is $R=R(E, C)>0$ such that

$$\begin{aligned} f(\lambda _1, \ldots , \lambda _{n-1}, \lambda _n+R)\ge C, \quad \forall \lambda \in E. \end{aligned}$$

(1.7)

An example of functions satisfying all assumptions above is given by $f=\frac{1}{2}\left[ H_n^\frac{1}{n}+(H_n/H_l)^{\frac{1}{n-l}}\right] ,$ where $H_l$ is the normalized l-th elementary symmetric polynomial. However, we point out that the pure curvature quotient $(H_n/H_l)^{\frac{1}{n-l}}$ does not satisfy (1.7).

For a graph of u,  the induced metric and its inverse matrix are given by

$$\begin{aligned} g_{ij}=\delta _{ij}+u_iu_j\quad \text{ and } \quad g^{ij}=\delta _{ij}-\frac{u_iu_j}{w^2}, \end{aligned}$$

(1.8)

where $w=\sqrt{1+|Du|^2}.$ Following [2], the principle curvature of graph u are eigenvalues of the symmetric matrix $A[u]=[a_{ij}]:$

$$\begin{aligned} a_{ij}=\frac{\gamma ^{ik}u_{kl}\gamma ^{lj}}{w},\quad \text{ where } \gamma ^{ik}=\delta _{ij}-\frac{u_iu_k}{w(1+w)}. \end{aligned}$$

(1.9)

The inverse of $\gamma ^{ij}$ is denoted by $\gamma _{ij},$ and

$$\begin{aligned} \gamma _{ij}=\delta _{ij}+\frac{u_iu_k}{1+w}. \end{aligned}$$

(1.10)

Geometrically $[\gamma _{ij}]$ is the square root of the metric, i.e. $\gamma _{ik}\gamma _{kj}=g_{ij}.$

Now, for any positive definite symmetric matrix A,  we define the function F by

$$\begin{aligned} F(A)=f(\lambda (A)), \end{aligned}$$

where $\lambda (A)$ denotes the eigenvalues of A. We will use the notation

$$\begin{aligned} F^{ij}(A)=\frac{\partial F}{\partial a_{ij}},\,\,F^{ij, kl}=\frac{\partial ^2 F}{\partial a_{ij}\partial a_{kl}}(A). \end{aligned}$$

The matrix $[F^{ij}(A)]$ is symmetric and has eigenvalues $f_1, \ldots , f_n.$ By (1.2), $[F^{ij}(A)]$ is positive definite. Moreover, by (1.3), F is a concave function of A, that is

$$\begin{aligned} F^{ij, kl}(A)\xi _{ij}\xi _{kl}\le 0, \end{aligned}$$

for any $n\times n$ symmetric matrix $[\xi _{ij}].$

We rewrite Eq. (1.1) as following,

$$\begin{aligned} \left\{ \begin{aligned}&\dot{u}=w\left[ F\left( \frac{\gamma ^{ik}u_{kl}\gamma ^{lj}}{w}\right) -\Phi (x, u)\right]&\text{ in } \Omega \times [0, T)\\&u_\nu =\varphi (x, u)&\text{ on } \partial \Omega \times [0, T)\\&u|_{t=0}=u_0&\text{ in } \Omega . \end{aligned} \right. \end{aligned}$$

(1.11)

We will prove

Theorem 1.1

Let $\Omega$ be a smooth bounded, strictly convex domain in $\mathbb {R}^n.$ Let $\Phi , \varphi : \bar{\Omega }\times \mathbb {R}\rightarrow \mathbb {R},$ be smooth functions satisfying

$$\begin{aligned}&\Phi >0 \quad \mathrm{and }\quad \Phi _z\ge 0, \end{aligned}$$

(1.12)

$$\begin{aligned}&\varphi _z\le c_\varphi <0. \end{aligned}$$

(1.13)

Let $u_0$ be a smooth, strictly convex function that satisfies the compatibility condition on $\partial \Omega$:

$$\begin{aligned} \left. \nu ^iu_i-\varphi (x, u)\right| _ {t=0}=0. \end{aligned}$$

(1.14)

We also assume

$$\begin{aligned} f(\kappa [\Sigma _0])-\Phi (x, u_0)\ge 0, \end{aligned}$$

(1.15)

where $\Sigma _0=\{(x, u_0(x))|x\in \Omega \}.$ Then there exists a solution $u\in C^\infty (\bar{\Omega }\times (0, \infty ))$ of Eq. (1.11). Moreover, as $t\rightarrow \infty ,$ the function u(x, t) smoothly converges to a smooth limit function $u^\infty ,$ such that $u^\infty$ solves the Neumann boundary value problem

$$\begin{aligned} \left\{ \begin{aligned}&F\left( \frac{\gamma ^{ik}u^\infty _{kl}\gamma ^{lj}}{w}\right) =\Phi (x, u^\infty )&\mathrm{in }\,\Omega \\&u^\infty _\nu =\varphi (x, u^\infty )&\mathrm{on }\,\partial \Omega ,\\ \end{aligned} \right. \end{aligned}$$

(1.16)

where $\nu$ is the outer unit normal of $\partial \Omega .$

Remark 1.2

As it is explained in [11], in view of the compatibility assumption (1.14), the short time existence for Eq. (1.11) follows from Theorem 5.3 in [6] and the implicit function theorem. Moreover, the solution $u(\cdot , t)$ approaches $u_0$ in $C^2(\bar{\Omega })$ as $t\rightarrow 0,$ this implies $\dot{u}$ is continuous up to $t=0.$

By applying short time existence theorem, we know that the flow exists for $t\in [0, T^*),$ for some $T^*>0$ very small. In the following sections, we fix $T<T^*,$ and establish the uniform $C^2$ bounds for the solution u of (1.11) in (0, T]. Since our estimates are independent of T,  repeating this process we obtain the longtime existence of Eq. (1.11).

Neumann boundary problem has attracted lots of attetions through the years. In particular, real Monge–Ampère equations in bounded uniformly convex domains are solved with Neumann boundary conditions by Lions, Trudinger, and Urbas in [8]. There, they built the foundation for $C^2$ a priori estimates of Neumman boundary problem, which departs completely from that of the Dirichlet problem. By adapting and developing the techniques in [8], Jiang et al. [5] proved the classical solvability of a generalized Monge–Ampère type equation with Neumann boundary condition. Recently, Ma and Qiu proved the existence of the solution to Hession equations with Neumann boundary condition in their beautiful paper [9], which confirms a longstanding conjecture by Trudinger.

The Neumann boundary problems for parabolic equations have been widely studied as well. For example, mean curvature flow with Neumann boundary condition have been studied in [1, 3, 10, 14]; Guass curvature flow with Neumann boundary condition have been studied in [12, 13].

Our paper is oganized as follows: In Sect. 2 we prove the uniform estimate for $\dot{u},$ which also implies the convexity of $u(\cdot , t).$ This estimate is used in Sect. 3 to derive the bounds for |u| and |Du|. Section 4 is the most important section, in which we derive the $C^2$ estimates for u. Finally, in Sect. 5, we combine all results above to prove the convergence of the solution of (1.11) as $t\rightarrow \infty$.

2 Speed estimates

Lemma 2.1

As long as a smooth convex solution of (1.11) exists, we have

$$\begin{aligned} \min \{\min \limits _{t=0}\dot{u}, 0\}\le \dot{u}\le \max \{\max \limits _{t=0}\dot{u}, 0\}. \end{aligned}$$

(2.1)

Proof

If $(\dot{u})^2$ achieves a positive local maximum at $(x, t)\in \partial \Omega \times (0, T],$ then by (1.13) at this point we would have

$$\begin{aligned} (\dot{u})^2_\nu =2\dot{u}\dot{u}_\nu =2(\dot{u})^2\varphi _z<0, \end{aligned}$$

(2.2)

which leads to a contradiction. Thus, we assume $(\dot{u})^2$ achieves its maximum at an interior point. Now let’s denote

$$\begin{aligned} \tilde{G}(D^2u, Du, u)=wF\left( \frac{\gamma ^{ik}u_{kl}\gamma ^{lj}}{w}\right) -w\Phi (x, u) \end{aligned}$$

and $r=(\dot{u})^2.$ A straightforward calculation gives us

$$\begin{aligned} \dot{r}=\tilde{G}^{ij}r_{ij}-2\tilde{G}^{ij}\dot{u}_i\dot{u}_j+\tilde{G}^sr_s+2\tilde{G}_ur, \end{aligned}$$

(2.3)

where $\tilde{G}^{ij}=\frac{\partial \tilde{G}}{\partial u_{ij}},$ $\tilde{G}^s=\frac{\partial \tilde{G}}{\partial u_{s}},$ and $\tilde{G}_u=\frac{\partial \tilde{G}}{\partial u}.$ Since

$$\begin{aligned} \tilde{G}_u:=\frac{\partial \tilde{G}}{\partial u}=-w\Phi _u\le 0, \end{aligned}$$

(2.4)

we have

$$\begin{aligned} \dot{r}-\tilde{G}^{ij}r_{ij}-\tilde{G}^sr_s\le 0. \end{aligned}$$

(2.5)

By the maximum principle we know that a positive local maximum of $(\dot{u})^2$ can not occur at an interior point of $\Omega \times (0, T].$ Therefore, we proved this Lemma. $\square$

Lemma 2.2

A solution of (1.11) satisfies $\dot{u}>0$ for $t>0$ if $0\not \equiv \dot{u}\ge 0$ for $t=0.$

Proof

Differentiating

$$\begin{aligned} \dot{u}=\tilde{G}(D^2u, Du, u), \end{aligned}$$

(2.6)

with respect to t we get,

$$\begin{aligned} \frac{d}{dt}u_t=\tilde{G}^{ij}(u_t)_{ij}+\tilde{G}^s(u_t)_s+\tilde{G}_uu_t. \end{aligned}$$

(2.7)

Then, for any constant $\lambda$ we have

$$\begin{aligned} \frac{d}{dt}(u_te^{\lambda t})=\tilde{G}^{ij}(u_te^{\lambda t})_{ij}+\tilde{G}^s(u_te^{\lambda t})_s+\tilde{G}_u(u_te^{\lambda t})+\lambda u_te^{\lambda t}. \end{aligned}$$

(2.8)

Now we fix $t_0>0,$ and choosing a constant $\lambda$ such that $\lambda +\tilde{G}_u>0$ for $(x, t)\in \bar{\Omega }\times [0, t_0].$ If $u_te^{\lambda t}=0$ at some interior point $(x_1, t_1)\in \Omega \times (0, t_0],$ then by the strong maximum principle we would have $u_te^{\lambda t}$ vanishes identically in $\bar{\Omega }\times [0, t_0]$ , which leads to a contradiction.

Assuming $u_te^{\lambda t}=0$ at a boundary point $(x_1, t_1)\in \partial \Omega \times (0, t_0],$ then we would have

$$\begin{aligned} (u_te^{\lambda t})_\nu =\varphi _z(u_te^{\lambda t})=0. \end{aligned}$$

(2.9)

This contradicts the Hopf Lemma. $\square$

Remark 2.3

Lemma 2.2 implies that, if we start from a strictly convex hypersurface $\Sigma _0$ that satisfies the inequality (1.15), then as long as the flow exists, the flow surfaces $\Sigma (t)$ are strictly convex and satisfying $f(\kappa [\Sigma (t)])-\Phi (x, u)>0.$

3 $C^0$ and $C^1$ estimates

Recall that $u_\nu =\varphi (x, u)$ on $\partial \Omega ,$ the strict convexity of u and the fact that $\varphi (\cdot , z)\rightarrow -\infty$ uniformly as $z\rightarrow \infty$ implies that u is uniformly bounded from above. By Lemma 2.2 we also have,

$$\begin{aligned} u(x, t)=u(x, 0)+\int _0^t\dot{u}(x, \tau )d\tau \ge u(x, 0). \end{aligned}$$

(3.1)

This yields u is bounded from below. To conclude, we have

Theorem 3.1

($C^0$ estimates) Under our assumption (1.15) on $u_0,$ a solution of equation (1.11) satisfies

$$\begin{aligned} |u|\le C_0, \end{aligned}$$

(3.2)

where $C_0=C_0(u_0, \varphi ).$

Theorem 3.2

($C^1$ estimates) For a convex solution u of Eq. (1.11), the gradient of u remains bounded during the evolution,

$$\begin{aligned} |Du|\le C_1, \end{aligned}$$

(3.3)

where $C_1=C_1(|u|_{C^0}, \Omega , \varphi ).$

Proof

The proof is the same as Theorem 2.2 in [8], for readers convenience we include it here. By the convexity of u we have for any $t\in [0, T]$

$$\begin{aligned} \max \limits _{\Omega }|Du(\cdot , t)|=\max \limits _{\partial \Omega }|Du(\cdot , t)|. \end{aligned}$$

(3.4)

Let $x_0\in \partial \Omega$ and let $\tau$ be a direction such that $\nu \cdot \tau =0$ at $x_0.$ Let $B=B_R(z)$ be an interior ball at $x_0,$ L be the line through $x_0$ in the direction of $-\nu ,$ and L intersects $\partial B$ at $y_0.$ Then $z=\frac{1}{2}(x_0+y_0),$ we also let $y\in \partial B$ be the unique point such that $\frac{y-z}{|y-z|}=\tau .$

Now let $\omega$ be an affine function such that $\omega (x_0)=u(x_0, t)$ and $D\omega =Du(x_0, t).$ Then $\omega \le u(x, t),\,\,x\in \Omega$ and

$$\begin{aligned} \begin{aligned} \omega (z)&=\omega (x_0)+D\omega (x_0)\cdot (z-x_0)\\&=u(x_0, t)+Du(x_0, t)\cdot \frac{z-x_0}{|z-x_0|}\cdot |z-x_0|\\&\ge u(x_0, t)-M_1R, \end{aligned} \end{aligned}$$

(3.5)

where we assume $|\varphi (x, u)|\le M_1$ in $\bar{\Omega }\times [-C_0, C_0].$ Therefore,

$$\begin{aligned} D_\tau u(x_0, t)=D_\tau \omega (x_0)=\frac{\omega (y)-\omega (z)}{|y-z|}\le \frac{u(y, t)-u(x_0, t)+M_1R}{R}\le \frac{2C_0}{R}+M_1. \end{aligned}$$

(3.6)

Since $\tau ,$ $x_0,$ and t are arbitrary, we are done. $\square$

4 $C^2$ estimates

First of all, we will list some evolution equations that will be used later. Since the calculations are straightforward, we will only state our results here.

Lemma 4.1

Let u be a solution to the general curvature flow (1.11). Then we have the following evolution equations:

• (i) $\frac{d}{dt}g_{ij}=-2(F-\Phi )h_{ij},$

• (ii) $\frac{d}{dt}\mathbf {n}=-g^{ij}(F-\Phi )_i\tau _j,$

• (iii) $\frac{d}{dt}\mathbf {n}^{n+1}=-g^{ij}(F-\Phi )_iu_j,$

• (vi) $\frac{d}{dt}h^j_i=(F-\Phi )^j_i+(F-\Phi )h^k_ih^j_k$,

where $g_{ij}, h_{ij}$ are the first and second fundamental forms, $\mathbf {n}$ is the upward unit normal to $\Sigma (t),$ $\mathbf {n}^{n+1}=\left<\mathbf {n}, e^{n+1}\right>,$ and $h_i^j=g^{jk}h_{kj}.$

4.1 $C^2$ interior estimates

In this subsection, we will prove the following theorem.

Theorem 4.2

Let $\Sigma (t)=\{(x, u(x, t))| x\in \Omega , t\in [0, T]\}$ be the flow surfaces, where u(x, t) satisfies Eq. (1.11) and

$$\begin{aligned} \mathbf {n}^{n+1}\ge 2a>0\quad \text{ on } \Sigma (t),\quad \forall t\in [0, T]. \end{aligned}$$

For $X\in \Sigma (t),$ let $\kappa _{\max }(X)$ be the largest principle curvature of $\Sigma (t)$ at X. Then

$$\begin{aligned} \max \limits _{\bar{\Omega }_T}\frac{\kappa _{\max }}{\mathbf {n}^{n+1}-a}\le C_2(\Phi , |u|_{C^1})\left( 1+\max \limits _{\partial \Omega _T}\kappa _{\max }\right) , \end{aligned}$$

(4.1)

where $\Omega _T=\Omega \times (0, T].$

Proof

Let’s consider

$$\begin{aligned} M_0=\max \limits _{\bar{\Omega }_T}\frac{\kappa _\text {max}}{\mathbf {n}^{n+1}-a}, \end{aligned}$$

we assume $M_0>0$ is attained at an interior point $(x_0, t_0)\in \Omega \times (0, T].$ We can choose a local coordinate at $(x_0, t_0)$ such that $\kappa _1=\kappa _\text {max},$ $h^j_i=\kappa _i\delta _{ij},$ and $g_{ij}=\delta _{ij}.$

At $(x_0, t_0),$ $\psi =\frac{h_1^1}{\mathbf {n}^{n+1}-a}$ achieves its local maximum. Therefore, at this point we have

$$\begin{aligned} \frac{h^1_{1i}}{h_1^1}-\frac{\nabla _i\mathbf {n}^{n+1}}{\mathbf {n}^{n+1}-a}=0. \end{aligned}$$

(4.2)

Moreover, by Lemma 4.1

$$\begin{aligned} \begin{aligned} \frac{\partial }{\partial t}\psi&=\frac{\dot{h_1^1}}{\mathbf {n}^{n+1}-a}-\frac{h_1^1\dot{\mathbf {n}}^{n+1}}{(\mathbf {n}^{n+1}-a)^2}\\&=\frac{1}{\mathbf {n}^{n+1}-a}\left\{ \nabla _{11}F-\nabla _{11}\Phi +(F-\Phi )\kappa _1^2\right\} +\frac{h_1^1}{(\mathbf {n}^{n+1}-a)^2}(F-\Phi )_iu_i. \end{aligned} \end{aligned}$$

(4.3)

Since

$$\begin{aligned} \nabla _{11}\Phi= & {} \Phi _{x_1x_1}(x, u)+2\Phi _{x_1z}u_1+\Phi _zu_{11}, \end{aligned}$$

(4.4)

$$\begin{aligned} \nabla _{11}u= & {} \left<X, e_{n+1}\right>_{11}=\left<h_{11}\mathbf {n}, e_{n+1}\right>=h_{11}\mathbf {n}^{n+1}, \end{aligned}$$

(4.5)

$$\begin{aligned} \nabla _{ii}\mathbf {n}^{n+1}= & {} \nabla _i\left<-h_{ik}\tau _k, e_{n+1}\right> =\left<-h_{iik}\tau _k, e_{n+1}\right>-h^2_{ii}\left<\mathbf {n}, e_{n+1}\right>, \end{aligned}$$

(4.6)

and

$$\begin{aligned} \begin{aligned} \nabla _{11}F&=F^{ij}h_{ij11}+F^{ij,rs}h_{ij1}h_{rs1}\\&=F^{ij}(h_{11ij}-h^2_{11}h_{ij}+h_{ik}h_{kj}h_{11})+F^{ij, rs}h_{ij1}h_{rs1}. \end{aligned} \end{aligned}$$

(4.7)

In view of Eqs. (4.3), (4.6), and (4.7), we get at $(x_0, t_0)$

$$\begin{aligned} \begin{aligned} 0&\le \frac{\partial }{\partial t}\psi -F^{ii}\nabla _{ii}\psi \\&=\frac{1}{\mathbf {n}^{n+1}-a}\left\{ F^{ii}h_{ii11}+F^{ij, rs}h_{ij1}h_{rs1}-\nabla _{11}\Phi +(F-\Phi )\kappa _1^2\right\} \\&+\frac{h_{11}}{(\mathbf {n}^{n+1}-a)^2}(F-\Phi )_iu_i-\frac{F^{ii}h_{11ii}}{\mathbf {n}^{n+1}-a}+\frac{h_{11}}{(\mathbf {n}^{n+1}-a)^2}F^{ii}\mathbf {n}^{n+1}_{ii}\\&=\frac{1}{\mathbf {n}^{n+1}-a}F^{ii}(h^2_{ii}h_{11}-h^2_{11}h_{ii})+\frac{F^{ij, rs}h_{ij1}h_{rs1}}{\mathbf {n}^{n+1}-a}\\&-\frac{\nabla _{11}\Phi }{\mathbf {n}^{n+1}-a}+\frac{(F-\Phi )\kappa _1^2}{\mathbf {n}^{n+1}-a}+\frac{h_{11}}{(\mathbf {n}^{n+1}-a)^2}(F-\Phi )_iu_i\\&+\frac{h_{11}}{(\mathbf {n}^{n+1}-a)^2}F^{ii}\left( -\nabla _kh_{ii}u_k-h^2_{ii}\mathbf {n}^{n+1}\right) .\\ \end{aligned} \end{aligned}$$

(4.8)

By our assumptions (1.6) and (1.3), we know that at $(x_0, t_0),$

$$\begin{aligned} F^{ii}h_{ii}=f_i\kappa _i=F \end{aligned}$$

(4.9)

and

$$\begin{aligned} F^{ij, rs}h_{ij1}h_{rs1}\le 0. \end{aligned}$$

(4.10)

Substituting (4.4) and (4.5) into (4.8), then combining with (4.9) and (4.10) we get

$$\begin{aligned} \begin{aligned} 0&\le \frac{\partial }{\partial t}\psi -F^{ii}\nabla _{ii}\psi \\&\le \frac{-ah_{11}}{(\mathbf {n}^{n+1}-a)^2}f_i\kappa _i^2-\frac{\Phi \kappa _1^2}{\mathbf {n}^{n+1}-a} +\frac{C}{\mathbf {n}^{n+1}-a}\\&\quad -\,\frac{\Phi _z\kappa _1\mathbf {n}^{n+1}}{\mathbf {n}^{n+1}-a}-\frac{\kappa _1}{(\mathbf {n}^{n+1}-a)^2} (\Phi _i+\Phi _zu_i)u_i,\\ \end{aligned} \end{aligned}$$

(4.11)

which implies,

$$\begin{aligned} 0\le \frac{-a\kappa _1}{(\mathbf {n}^{n+1}-a)^2}f_i\kappa _i^2-\frac{\left( \inf \limits _{\bar{\Omega }\times [-C_0, C_0]}\Phi \right) \kappa _1^2}{\mathbf {n}^{n+1}-a}+C\kappa _1, \end{aligned}$$

(4.12)

thus

$$\begin{aligned} {\kappa _1}\le C=C(\Phi , |u|_{C^1}). \end{aligned}$$

(4.13)

Note that the constants C in (4.12) and (4.13) also depend on a;  since our choice of a depends on $|u|_{C^1},$ we omit the dependency on a. Therefore, we conclude that

$$\begin{aligned} \max \limits _{\bar{\Omega }_T}\frac{\kappa _{\text {max}}}{\mathbf {n}^{n+1}-a}\le C_2\left( 1+\max \limits _{\partial \Omega _T}\kappa _{\text {max}}\right) . \end{aligned}$$

(4.14)

$\square$

4.2 $C^2$ boundary estimates

We use $\nu$ for the outer unit normal of $\partial \Omega$ and $\tau$ for a direction that tangential to $\partial \Omega .$ By the exactly same argument as Lemma 4.1 of [13] we have

Lemma 4.3

(Mixed $C^2$ estimates at the boundary) Let u be the solution of our flow Eq. (1.11). Then the absolute value of $u_{\tau \nu }$ remains a priori bounded on $\partial \Omega$ during the evolution.

Now we consider the function

$$\begin{aligned} V(x, \xi , t):=u_{\xi \xi }-2(\xi \cdot \nu )\xi _i'(D_i\varphi -D_kuD_i\nu ^k), \end{aligned}$$

(4.15)

where $\xi \in \mathbb {S}^{n-1}$ is a unit vector and $\xi '=\xi -(\xi \cdot \nu )\nu .$ By Theorem 4.2, we may assume $V(x, \xi , t)$ achieves its maximum at $(x_0, t_0)\in \partial \Omega \times (0, T],$ otherwise, we would be done.

We will devide it into 3 cases.

• (i) $\xi$ is tangential We will compute the second tangential derivatives of the boundary condition. The proof is the same as in [9], for readers convenience, we will include it here. Following the notation in [9], we denote $c^{ij}=\delta _{ij}-\nu ^i\nu ^j.$ Differentiating the boundary condition with respect to the tangential direction twice we obtain

  $$\begin{aligned} u_{li}\nu ^l=c^{ij}D_j\varphi -c^{ij}u_lD_j\nu ^l+\nu ^i\nu ^j\nu ^lu_{lj}, \end{aligned}$$

  and

  $$\begin{aligned} u_{lip}\nu ^l=c^{pq}D_q\left( c^{ij}D_j\varphi -c^{ij}u_lD_j\nu ^l+\nu ^i\nu ^j\nu ^lu_{lj}\right) +\nu ^p\nu ^q\nu ^lu_{liq}-c^{pq}u_{li}D_q\nu ^l. \end{aligned}$$

  Summing with $\xi ^i\xi ^p$ yields

  $$\begin{aligned} u_{\xi \xi \nu }&=-2\xi ^p\xi ^iu_{li}D_p\nu ^l-u_l\xi ^pD_{ip}\nu ^l\xi ^i +u_{\nu \nu }\sum \limits _{i}\xi ^pD_p\nu ^i\xi ^i\\&\quad -\,\sum \limits _{i}\xi ^p\xi ^i\nu ^jD_p\nu ^iD_j\varphi +\varphi _zu_{\xi \xi } +\xi ^p\xi ^i\varphi _{ip}\\&\quad +\,\varphi _{zz}u^2_{\xi }+2u_{\xi }\xi ^i\varphi _{zi}, \end{aligned}$$

  where we have used $\sum _j\nu ^jD_l\nu ^j=0.$ Therefore, at $(x_0, t_0)$ we have

  $$\begin{aligned} D_{\xi \xi \nu }u\le -2(D_i\nu ^k)D_{jk}u\xi _i\xi _j+(D_i\nu ^j)\xi _i\xi _jD_{\nu \nu }u+\varphi _zD_{ij}u\xi _i\xi _j+C, \end{aligned}$$

  (4.16)

  where $C=C(\Vert u\Vert _{C^1}, \Vert \partial \Omega \Vert _{C^3}, \Vert \varphi \Vert _{C^2}).$ Next, since V attains its maximum at $(x_0, t_0)$, we get

  $$\begin{aligned} 0\le D_\nu V=u_{\xi \xi \nu }-a_kD_{k\nu }u-(D_\nu a_k)D_ku-D_\nu b, \end{aligned}$$

  (4.17)

  where $a_k=2(\xi \cdot \nu )(\varphi _z\xi '_k-\xi '_iD_i\nu ^k)$ and $b=2(\xi \cdot \nu )\xi '_k\varphi _{k}.$ Thus, applying Lemma 4.3

  $$\begin{aligned} u_{\xi \xi \nu }\ge a_\nu D_{\nu \nu }u-C(\Vert \varphi \Vert _{C^2}, \Vert u\Vert _{C^1}, \Vert \partial \Omega \Vert _{C^3})=-C, \end{aligned}$$

  (4.18)

  where we have used $a_{\nu }=0.$ Combine with (4.16) and condition (1.13) yields

  $$\begin{aligned} -2(D_i\nu ^k)D_{jk}u\xi _i\xi _j+(D_i\nu ^j)\xi _i\xi _ju_{\nu \nu } +c_{\varphi }D_{ij}u\xi _i\xi _j+C\ge -C. \end{aligned}$$

  (4.19)

  By virtue of the uniformly convexity of the domain $\Omega$, we have $[D_i\nu ^k]\ge c_0I,$ for some $c_0>0.$ This gives

  $$\begin{aligned} D_{\xi \xi }u(x_0, t_0)\le C(1+D_{\nu \nu }u(x_0, t_0)). \end{aligned}$$

  (4.20)

• (ii) $\xi$ is non-tangential We write $\xi =\alpha \tau +\beta \nu ,$ where $\alpha =\xi \cdot \tau ,$ $\beta =\xi \cdot \nu \ne 0.$ Then

  $$\begin{aligned} \begin{aligned} D_{\xi \xi }u&=\alpha ^2D^2_{\tau \tau }u+\beta ^2D_{\nu \nu }u+2\alpha \beta D_{\tau \nu }u\\&=\alpha ^2D_{\tau \tau }u+\beta ^2D_{\nu \nu }u+V'(x, \xi ),\\ \end{aligned} \end{aligned}$$

  (4.21)

  where $V'=2(\xi \cdot \nu )\xi '_i(D_i\varphi -D_kuD_i\nu ^k).$ Thus we get,

  $$\begin{aligned} \begin{aligned} V(x_0, \xi , t_0)&=\alpha ^2 V(x_0, \tau , t_0)+\beta ^2 V(x_0, \nu , t_0)\\&\le \alpha ^2 V(x_0, \xi , t_0)+\beta ^2 V(x_0, \nu , t_0),\\ \end{aligned} \end{aligned}$$

  (4.22)

  which yeilds

  $$\begin{aligned} u_{\xi \xi }(x_0, t_0)\le C(1+u_{\nu \nu }(x_0, t_0)). \end{aligned}$$

  (4.23)

• (iii) Double normal $C^2$-estimates at the boundary Let’s recall our evolution equation

  $$\begin{aligned} \left\{ \begin{aligned} \dot{u}&=w\left[ F\left( \frac{\gamma ^{ik}u_{kl}\gamma ^{lj}}{w}\right) -\Phi (x, u)\right] \\ u_\nu&=\varphi (x, u) \end{aligned}\right. \end{aligned}$$

  (4.24)

  In the following we denote

  $$\begin{aligned} G(D^2u, Du)=F\left( \frac{\gamma ^{ik}u_{kl}\gamma ^{lj}}{w}\right) , \end{aligned}$$

  then G satisfies similar structure conditions to those of F. We have

  $$\begin{aligned} G^{ij}:=\frac{\partial G}{\partial u_{ij}}=\frac{1}{w}F^{kl}\gamma ^{ik}\gamma ^{lj}, \end{aligned}$$

  (4.25)

  and it’s easy to see that

  $$\begin{aligned} \frac{1}{w^3}\sum F^{ii}\le \sum G^{ii}\le \frac{1}{w}\sum F^{ii}. \end{aligned}$$

  (4.26)

  By a straightforward calculation we get (for details see the proof of Lemma 2.3 in [4]),

  $$\begin{aligned} G^s:=\frac{\partial G}{\partial u_s}=-\frac{u_s}{w^2}F-\frac{2}{w(1+w)}F^{ij}a_{ik}(wu_k\gamma ^{sj}+u_j\gamma ^{ks}), \end{aligned}$$

  (4.27)

  where we have used $\sum f_i\kappa _i=f(\kappa ).$ Since $[a_{ij}]$ is positive definite, we obtain

  $$\begin{aligned} \sum |G^i|\le CF\le \tilde{C}_0. \end{aligned}$$

  (4.28)

  Now, let

  $$\begin{aligned} \Omega _\mu :=\{x\in \bar{\Omega }: 0<d(x)=\text {dist}(x, \partial \Omega )<\mu \}. \end{aligned}$$

  Consider $q(x)=-d(x)+Nd^2(x),$ then $q\in C^\infty$ in $\Omega _\mu$ for some constant $\mu$ satisfies $\mu \le \tilde{\mu }$ and $N\mu \le \frac{1}{8},$ where $\tilde{\mu }$ is a small constant depending on $\Omega .$ Since

  $$\begin{aligned} -Dd(y_0)=\nu (x_0) \end{aligned}$$

  where $x_0\in \partial \Omega$ and $\text {dist}(y_0, \partial \Omega )=\text {dist}(x_0, y_0),$ q satisfies the following properties in $\Omega _\mu :$

  $$\begin{aligned} -\mu +N\mu ^2\le q\le 0\quad \text{ and } \quad \frac{1}{2}\le |Dq|\le 2. \end{aligned}$$

  (4.29)

  Moreover, $\frac{Dq}{|Dq|}=\nu$ in $\Omega _\mu ,$ here $\nu$ is the unit outer normal to the boundary $\partial \Omega$. Next, let

  $$\begin{aligned} M=\max \limits _{\partial \Omega \times [0, T]}u_{\nu \nu } \end{aligned}$$

  (4.30)

  and $Q(x, t)=Q(x)=(A+\frac{1}{2}M)q(x)$ in $\Omega _\mu ,$ where $\mu , A, N$ are positive constants to be chosen later. We consider the following function

  $$\begin{aligned} P(x, t):=Du\cdot Dq-\varphi -Q \end{aligned}$$

  (4.31)

Lemma 4.4

For any $(x, t)\in \bar{\Omega }_\mu \times [0, T],$ if we choose $A, N>0$ large, $\mu >0$ small, then we have $P(x, t)\ge 0.$

Proof

First, let’s assume P(x, t) attains its minimum at $(x_0, t_0)\in \Omega _\mu \times (0, T].$ Let’s choose a local coordinate such that $a_{ij}(x_0, t_0)=\kappa _i(x_0, t_0)\delta _{ij}, 1\le i, j\le n.$ Then at this point we have $F^{ij}=\frac{\partial f}{\partial \kappa _i}\delta _{ij}.$ Differentiating P twice, we get

$$\begin{aligned} P_i=\sum _lu_{li}q_l+\sum _lu_lq_{li}-\varphi _i-Q_i, \end{aligned}$$

(4.32)

and

$$\begin{aligned} P_{ij}=\sum _lu_{lij}q_l+2\sum _lu_{li}q_{lj}+\sum _lu_lq_{lij}-\varphi _{ij}-Q_{ij}. \end{aligned}$$

(4.33)

Moreover,

$$\begin{aligned} \begin{aligned} P_t&=Du_t\cdot Dq-\varphi _t\\&=\sum \limits _{l}[w(F-\Phi )]_lq_l-\varphi _zu_t=\sum \limits _{l}[w(F-\Phi )]_lq_l-\varphi _zw(F-\Phi ).\\ \end{aligned} \end{aligned}$$

(4.34)

Therefore, at $(x_0, t_0)$ we have

$$\begin{aligned} \begin{aligned}&\frac{1}{w}P_t-G^{ij}P_{ij}\\&\quad =\frac{1}{w}[w(F-\Phi )]_lq_l-\varphi _z(F-\Phi )-G^{ij}\left( \sum _lu_{lij}q_l+2\sum _lu_{li}q_{lj}\right. \\&\qquad \left. +\,\sum _lu_lq_{lij}-\varphi _{ij}\right) +\left( A+\frac{1}{2}M\right) G^{ij}q_{ij}\\&\quad =\sum \limits _{l}\frac{1}{w}[w(F-\Phi )]_lq_l-\varphi _z(F-\Phi )-G^{ij}\sum _lu_{lij}q_l\\&\qquad -\,2\sum \limits _{l}G^{ij}u_{li}q_{lj}-\sum \limits _{l}G^{ij}u_lq_{lij}+G^{ij}\varphi _{ij} +\left( A+\frac{1}{2}M\right) G^{ij}q_{ij}. \end{aligned} \end{aligned}$$

(4.35)

This implies at $(x_0, t_0)$

$$\begin{aligned} \begin{aligned} 0&\ge \frac{1}{w}P_t-G^{ij}P_{ij}\\&=\sum \limits _{l}\frac{(F-\Phi )}{w}\cdot \frac{u_su_{sl}q_l}{w} +\sum \limits _{l}F_lq_l-\sum \limits _{l}\Phi _lq_l-\varphi _z(F-\Phi )\\&\quad -\,G^{ij}\sum _lu_{lij}q_l-2\sum \limits _{l}G^{ij}u_{li}q_{lj}-\sum _lG^{ij}u_lq_{lij}\\&\quad +\,G^{ij}(\varphi _{x_ix_j}+2\varphi _{x_iz}u_j+\varphi _{zz}u_iu_j+\varphi _zu_{ij}) +\left( A+\frac{1}{2}M\right) G^{ij}q_{ij}.\\ \end{aligned} \end{aligned}$$

(4.36)

Since $G(D^2u, Du)=F$ we have

$$\begin{aligned} G^{ij}u_{ijl}+G^su_{sl}=F_l, \end{aligned}$$

(4.37)

which gives us

$$\begin{aligned} F_lq_l-G^{ij}u_{ijl}q_l=G^su_{sl}q_l. \end{aligned}$$

(4.38)

By (4.28) and (4.29) we have

$$\begin{aligned} |G^su_{sl}q_l|\le \tilde{C}_1(M+1). \end{aligned}$$

(4.39)

Furthermore, by the speed estimate (2.1), height estimate (3.2), and the gradient estimate (3.3), we obtain

$$\begin{aligned} |\Phi _lq_l|+\left| \frac{F-\Phi }{w}\cdot \frac{u_su_{sl}q_l}{w}+\varphi _zG^{ij}u_{ij}\right| \le \tilde{C}_2 M. \end{aligned}$$

(4.40)

Now, by the convexity of $\partial \Omega ,$ we may assume

$$\begin{aligned} \kappa [2k_0\delta _{\alpha \beta }]\le \kappa [-d_{\alpha \beta }]\le \kappa [k_1\delta _{\alpha \beta }],\,\,1\le \alpha , \beta \le n-1, \end{aligned}$$

(4.41)

for some $k_0, k_1 >0$ depending on $\partial \Omega .$ Thus, in $\Omega _\mu$ we have

$$\begin{aligned} \kappa [(k_1+3N)\delta _{ij}]\ge \kappa [q_{ij}]=\kappa [-d_{ij}+2Ndd_{ij}+2Nd_id_j]\ge \kappa [k_0\delta _{ij}], \end{aligned}$$

(4.42)

where $1\le i, j\le n$ and $\kappa [A]$ denotes the eigenvalue of the matrix A. This gives

$$\begin{aligned} |2G^{ij}u_{li}q_{lj}|\le \tilde{C}_3(k_1+3N), \end{aligned}$$

(4.43)

where $\tilde{C}_3$ depends on F. Next, an easy calculations yields

$$\begin{aligned} q_{ijl}=-d_{ijl}+2Nd_ld_{ij}+2Ndd_{ijl}+4Nd_{il}d_j, \end{aligned}$$

(4.44)

which implies

$$\begin{aligned} |q_{ijl}|\le C(|\partial \Omega |_{C^3})+6Nk_1. \end{aligned}$$

(4.45)

Therefore,

$$\begin{aligned} |G^{ij}u_lq_{lij}|\le \left( C(|\partial \Omega |_{C^3})+6Nk_1\right) C_1\sum G^{ii}, \end{aligned}$$

(4.46)

where we have used $G^{ij}=\frac{1}{w}F^{kl}\gamma ^{ik}\gamma ^{lj}\le C\sum F^{kk}\le C\sum G^{ii}.$ Consequently, we have

$$\begin{aligned} |G^{ij}u_lq_{lij}+G^{ij}(\varphi _{x_ix_j}+2\varphi _{x_iz}u_j+\varphi _{zz}u_iu_j)|\le (\tilde{C}_4+\tilde{C}_5Nk_1)\sum G^{ii}. \end{aligned}$$

(4.47)

To conclude, we obtained

$$\begin{aligned} \begin{aligned} 0&\ge \frac{1}{w}P_t-G^{ij}P_{ij}\\&\ge -\tilde{C}_2M-\tilde{C}_1(M+1)-\tilde{C}_3(k_1+3N)-(\tilde{C}_4+\tilde{C}_5 Nk_1)\sum G^{ii}\\&+\left( \frac{A}{2}+\frac{M}{4}\right) k_0\sum G^{ii}+\left( \frac{A}{2}+\frac{M}{4}\right) G(D^2 q, Du), \end{aligned} \end{aligned}$$

(4.48)

Note that here we used the inequality $G^{ij}(D^2u,Du)q_{ij}\ge G(D^2 q, Du),$ which follows from the concavity of f. By Lemma 2.2 of Guan and Spruck [4], we may choose N sufficiently large such that

$$\begin{aligned} \frac{1}{4}G(D^2 q, Du)\ge \tilde{C}_1+\tilde{C}_2+1, \end{aligned}$$

(4.49)

then we choose A such that

$$\begin{aligned} \frac{k_0}{2}A\sum G^{ii}>\tilde{C}_3(k_1+3N)+(\tilde{C}_4+N\tilde{C}_5k_1)\sum G^{ii}, \end{aligned}$$

(4.50)

here we have used $\sum G^{ii}\ge c_0(|u|_{C^1})>0,$ which follows from (4.26) and the assumptions (1.3), (1.6) of f. Substituting (4.49) and (4.50) into (4.48) we get

$$\begin{aligned} \frac{1}{w}P_t-G^{ij}P_{ij}>0 \end{aligned}$$

(4.51)

at $(x_0, t_0),$ which leads to a contradiction.

Finally, note that for any $(x, t)\in \partial \Omega \times [0, T]$ we have

$$\begin{aligned} P(x, t)=0. \end{aligned}$$

For $(x, t)\in \partial \Omega _\mu \setminus \partial \Omega \times [0, T]$ we have

$$\begin{aligned} P(x, t)\ge -\tilde{C}_6+(A+\frac{1}{2}M)\cdot \frac{1}{2}\mu >0, \end{aligned}$$

when $A\ge \frac{2\tilde{C}_6}{\mu }.$ Moreover, when $A\ge \tilde{C}_7=\tilde{C}_7(|u_0|_{C^2}, |\varphi |_{C^1}, |\partial \Omega |_{C^2}),$ we have for $x\in \Omega _\mu$

$$\begin{aligned} P(x, 0)\ge 0. \end{aligned}$$

Thus, choosing

$$\begin{aligned} A=\frac{2[\tilde{C}_3(k_1+2N)+\tilde{C}_4+N\tilde{C}_5k_1]}{k_0c_0}+\frac{2\tilde{C}_6}{\mu }+\tilde{C}_7 \end{aligned}$$

we have $P(x, t)\ge 0$ in $\Omega _\mu \times [0, T].$ Here $c_0=\min \{1, \sum G^{ii}\}>0.$ $\square$

Theorem 4.5

Let $\Omega$ be a smooth bounded, strictly convex domain in $\mathbb {R}^n,$ u is a smooth solution of (1.11), $\nu$ is the outer unit normal vector of $\partial \Omega .$ Then we have

$$\begin{aligned} \max \limits _{\partial \Omega \times [0, T]}u_{\nu \nu }\le C. \end{aligned}$$

(4.52)

Proof

Assume $(z_0, t_0)\in \partial \Omega \times [0, T]$ is the maximum point of $u_{\nu \nu }$ on $\partial \Omega \times [0, T].$ By Lemma 4.4 we have

$$\begin{aligned} \begin{aligned} 0&\ge P_\nu (z_0, t_0)=\left( \sum _lu_{l\nu }q_l+u_lq_{l\nu }-\varphi _\nu \right) -\left( A+\frac{1}{2}M\right) q_\nu \\&\ge u_{\nu \nu }-C(|u|_{C^1}, N, |\partial \Omega |_{C^2}, |\varphi |_{C^1})-\left( A+\frac{1}{2}M\right) ,\\ \end{aligned} \end{aligned}$$

(4.53)

Therefore we have,

$$\begin{aligned} \max \limits _{\partial \Omega \times [0, T]}u_{\nu \nu }\le C+\frac{1}{2}M. \end{aligned}$$

(4.54)

Inequality (4.52) follows from (4.54) and the assumption (4.30). $\square$

5 Convergence to a stationary solution

Let us go back to our original problem (1.1), which is a scalar parabolic differential equation defined on the cylinder $\Omega _T=\Omega \times [0, T]$ with initial value $u_0.$ In view of a priori estimates, which we have estimated in the preceding sections, we know that

$$\begin{aligned}&|D^2 u|\le C, \end{aligned}$$

(5.1)

$$\begin{aligned}&|Du|\le C, \end{aligned}$$

(5.2)

and

$$\begin{aligned} |u|\le C. \end{aligned}$$

(5.3)

Therefore,

$$\begin{aligned} F \text{ is } \text{ uniformly } \text{ elliptic. } \end{aligned}$$

Moreover, since F is concave, we can apply the results of Chapter 14 in [7] to obtain uniform $C^{2,\alpha }$ estimates for u. Then standard Schauder estimates imply uniform bounds for u in $C^k, k\ge 0.$ Therefore, a smooth solution of (1.1) exists for all $t\ge 0.$

Lemma 5.1

If a solution of the flow Eq. (1.1) exists for all $t\ge 0.$ Moreover, the initial surface satisfies (1.15). Then the solution converges uniformly to a solution of the Neummann boundary problem

$$\begin{aligned} \left\{ \begin{aligned}&F\left( \frac{\gamma ^{ik}u^\infty _{kl}\gamma ^{lj}}{w}\right) =\Phi (x, u^\infty )&\mathrm{in }\,\Omega ,\\&u^\infty _\nu =\varphi (x, u^\infty )&\mathrm{on }\,\partial \Omega .\\ \end{aligned} \right. \end{aligned}$$

(5.4)

Proof

By integrating the flow equation with respect to t we get

$$\begin{aligned} u(x, t^*)-u(x, 0)=\int _0^{t^*}w(F-\Phi )dt. \end{aligned}$$

(5.5)

In particular, by (5.3) we have

$$\begin{aligned} \int _0^{\infty }w(F-\Phi )dt<\infty \,\,\forall x\in \Omega . \end{aligned}$$

(5.6)

Hence for any $x\in \Omega$ there existes a sequence $t_k\rightarrow \infty$ such that $F-\Phi \rightarrow 0.$ On the other hand, by Lemmas 2.1 and 2.2 we know $u(x, \cdot )$ is monotone increasing and bounded. Therefore, u(x, t) converges uniformly to $u^\infty .$ By virtue of our a priori estimates, we also know that $u^\infty$ is of class $C^\infty (\bar{\Omega })$. Moreover, it’s easy to see that $u^\infty$ is a stationary solution of our problem, i.e., $f(\kappa [\Sigma ^\infty ])=\Phi (x, u^\infty )$ and $u^\infty _\nu =\phi (x, \infty ).$ $\square$