In this section, we are going to prove the main statements of the paper.
Theorem 8.1
Let \((M^n,\sigma )\) be an asymptotically flat manifold with \(n\ge 3\) and \(u_0\in C^{0,1}(M)\) be uniformly spacelike such that
$$\begin{aligned} u_0(x)\rightarrow 0 \quad \text {for} \quad |x|\rightarrow \infty . \end{aligned}$$
Then there exists a unique uniformly spacelike solution of (2.7) with \(u(\cdot ,0)=u_0\) which exists for all times and satisfies \(u\in C^{2;1}\bigl (M\times (0,\infty ) \bigr )\cap C^0\bigl (M\times [0,\infty )\bigr )\). Moreover, u satisfies
$$\begin{aligned} u(x,t)\rightarrow 0 \quad \text {for} \quad |x|\rightarrow \infty . \end{aligned}$$
(8.1)
and this convergence is uniform in t. More precisely, there is a function \(b:M\rightarrow \mathbb {R}\) and a constant \(\kappa >0\), both only dependent on \(u_0\) and \(\sigma \), such that
-
(i)
\(|u(x,t)|\le b(x)\) for \((x,t)\in M\times (0,\infty )\),
-
(ii)
\(b(x)\rightarrow 0\) for \(|x|\rightarrow \infty \),
-
(iii)
\(|\nabla u(x,t)|_{\sigma } \le 1-\kappa \) for \((x,t)\in M\times (0,\infty )\).
Proof
By Lemma 7.5, the solutions \(u_R\) exist for all times and are uniformly spacelike in T and R. By standard theory for parabolic equations, we get uniform bounds on higher derivatives on \((\delta ,\infty )\) where the bounds only depend on \(u_0\), \(\sigma \) and \(\delta \) and are in particular independent of R. Therefore, by the Arzelà-Ascoli theorem, the family \(u_R\) subconverges in all derivatives to a solution u of (2.7) with initial data \(u_0\). The estimates (i)–(iii) are consequences of the estimates (i)–(iii) in Lemma 7.5 and Remark 7.6.
To show uniqueness, one uses (8.1) in the following sense: Suppose that v(x, t) is another uniformly spacelike solution of (2.7) which is defined up to a time T. Then by Theorem 5.4, v also satisfies (8.1) for each \(t\in [0,T)\). Therefore, \(u\pm \varepsilon \) can be used as a barrier for v on \(M\times [0,T)\). The maximum principle implies that for each \(\varepsilon >0\), we get \(u-\varepsilon \le v\le u+\varepsilon \). Uniqueness follows from \(\varepsilon \rightarrow 0\). \({\square }\)
Remark 8.2
Statement (8.1) also follows in dimensions \(n=1,2\) but we cannot conclude (i) and (ii) in these cases. These properties are however essential in the proof of the main theorem below. In the case \(n=1\), we give a separate proof that is done in a subsection below.
Theorem 8.3
Let u be the solution in Theorem 8.1. Then \(u\left( \cdot ,t\right) \xrightarrow [t\rightarrow \infty ]{}0\) uniformly in \(C^{l}\) for all \(l\in \mathbb {N}_{0}\). More precisely, for each \(l\in \mathbb {N}_0\), there exists a function \(v_l:[0,T)\rightarrow \mathbb {R}\) with \(v_l(t)\rightarrow 0\) as \(t\rightarrow \infty \) such that
$$\begin{aligned} \bigl \Vert u_t\bigr \Vert _{C^{l}(M)}\le v_l(t) \,. \end{aligned}$$
Proof
First we prove the statement for \(l=0\). Since u decays as \(x\rightarrow \infty \) by Theorem 8.1, we can apply the maximum principle to conclude that
$$\begin{aligned} v:[0,T)\rightarrow \mathbb {R}\,, \qquad v(t){:}{=} \sup _{x\in M} |u(x,t)| \,, \end{aligned}$$
is monotonically decreasing. If \(v(t)\rightarrow 0\) for \(t\rightarrow \infty \), the claim follows. Assume that \(v\left( t\right) \xrightarrow [t\rightarrow \infty ]{}\delta >0\). By Theorem 8.1 there exists \(R>0\) such that \(\left| u\left( x,t\right) \right| \le \frac{\delta }{2}\) for all \(\left| x\right| \ge R\) and \(t\ge 0\). Using the strict maximum principle once again in \(\overline{B_{R}\left( 0\right) }\times [0,\infty )\) yields that either v is strictly decreasing or \(u\left( \cdot ,t\right) \) is constant for all \(t\ge T_{0}\) for some \(T_0\ge 0\). In the second case we conclude (again by Theorem 8.1) that \(u\left( \cdot ,t\right) =0\) for all \(t\ge T_{0}\).
Let \(t_{k}\ge 0\) with \(t_{k}\xrightarrow [k\rightarrow \infty ]{}\infty \) be arbitrary. Due to the uniform estimates for u in \(C^{l}\) for \(l\in \mathbb {N}\) it follows by the Arzelà-Ascoli theorem (after choosing a subsequence also labelled \(\left( t_{k}\right) _{k\in \mathbb {N}}\)) that
$$\begin{aligned} u\left( \cdot ,t+t_{k}\right) \xrightarrow [k\rightarrow \infty ]{}\tilde{u}\left( \cdot ,t\right) \end{aligned}$$
uniformly and \(\tilde{u}\) is a solution of the same differential equation as u. Due to the uniform convergence we conclude that
$$\begin{aligned} \sup _{x\in M}\left| \tilde{u}\left( x,t\right) \right|&=\lim \limits _{k\rightarrow \infty }\sup \limits _{x\in M}\left| u\left( x,t+t_{k}\right) \right| =\lim \limits _{k\rightarrow \infty }v\left( t+t_{k}\right) =\delta >0 \end{aligned}$$
(8.2)
for all \(t\ge 0\). Because \(\left| u\left( x,t\right) \right| \xrightarrow [\left| x\right| \rightarrow \infty ]{}0\) uniformly in t the same holds for \(\tilde{u}\). Therefore, \(x\mapsto \tilde{u}(x,t)\) attains a maximum for each \(t\ge 0\). By the strict maximum principle, this is strictly decreasing unless \(\tilde{u}\) is constant, hence identically zero. However, this is a contradiction to (8.2).
For arbitrary \(l\in \mathbb {N}\), the statement follows immediately from the following interpolation inequality
$$\begin{aligned} \left\| \nabla f\right\| _{C^{0}\left( M \right) }^{2}\le c\left( n\right) \left\| f\right\| _{C^{0}\left( M\right) }\bigl \Vert f\bigr \Vert _{C^{2}\left( M\right) } \end{aligned}$$
since \(\left\| u\left( \cdot ,t\right) \right\| _{C^{0}\left( M\right) }\xrightarrow [t\rightarrow \infty ]{}0\) and the uniform bounds for the derivatives of u hold. \({\square }\)
The One-Dimensional Case
For \((M,\sigma )=(\mathbb {R},\delta )\), the mean curvature flow can be written in a particularly simple form which allows us to prove our main result with a different method. To be more precise, for maps \(u:\mathbb {R}\times [0,T)\rightarrow \mathbb {R}\), the graphical mean curvature flow (2.6) may equivalently be written as
$$\begin{aligned} \partial _t u = \frac{u''}{1-\bigl (u'\bigr )^2} = \frac{1}{2} \left[ \ln \frac{1+u'}{1-u'} \right] '\,. \end{aligned}$$
(8.3)
From Theorem 8.1 and Remark 8.2, we know already that (8.3) admits uniformly spacelike solutions that exist for all times, provided that the initial data is nice enough. In the following, we will show convergence to 0 for uniformly spacelike \(u_0\) and we will even get a convergence rate if \(u_0\in L^2(\mathbb {R})\).
Lemma 8.4
Let \(u:\mathbb {R}\times (0,T)\rightarrow \infty \) be a smooth solution of (8.3) (which is continuous up to \(t=0\)) with uniformly spacelike initial data \(u_0\in C^{0,1}(\mathbb {R})\cap L^2(\mathbb {R})\). Then \(u_t\in L^2(\mathbb {R})\) and \(\left\| u_t\right\| _{L^2}\le \left\| u_0\right\| _{L^{2}}\).
Proof
Choose a smooth cutoff function \(\psi \) with \(0\le \psi \le 1\),
$$\begin{aligned} \psi (x) = {\left\{ \begin{array}{ll} 1 \,, &{} x\in [-1,1] \,, \\ 0 \,, &{} x\notin (-2,2) \,, \end{array}\right. } \qquad \text {and} \qquad |\psi '|\le 2 \,. \end{aligned}$$
Let \(\varphi (x)=\psi (x/R)\) where \(R>0\). Then we get
$$\begin{aligned} \partial _t\int _{\mathbb {R}}\varphi ^2u^2\mathrm {d}x&= 2\int _{\mathbb {R}}\varphi ^2(\partial _t u) u\mathrm {d}x=\int _{\mathbb {R}}\varphi ^2u\left( \ln \frac{1+u'}{1-u'}\right) '\mathrm {d}x\\&\quad =-2\int _{\mathbb {R}}\varphi '\varphi u\left( \ln \frac{1+u'}{1-u'}\right) \mathrm {d}x -\int _{\mathbb {R}}\varphi ^2u'\left( \ln \frac{1+u'}{1-u'}\right) \mathrm {d}x\\&\quad \le \frac{1}{\delta }\int _{\mathbb {R}}(\varphi ')^2u^2\mathrm {d}x+\delta \int _{\mathbb {R}}\varphi ^2\left( \ln \frac{1+u'}{1-u'}\right) ^2\mathrm {d}x \\&\qquad -\int _{\mathbb {R}}\varphi ^2u'\left( \ln \frac{1+u'}{1-u'}\right) \mathrm {d}x , \end{aligned}$$
where \(\delta >0\) is fixed below. Also using \(x\ln \frac{1+x}{1-x} \ge x^2\) for \(|x|<1\), we further estimate
$$\begin{aligned} \partial _t\int _{\mathbb {R}}\varphi ^2u^2\mathrm {d}x&\le \frac{4}{\delta R^2}\int _{-2R}^{2R}u^2\mathrm {d}x+ C \delta \int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x -\int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x\\&\le \frac{4}{\delta R^2}\int _{-2R}^{2R}u^2\mathrm {d}x \,, \end{aligned}$$
where C depends on \(\sup |u'_t|\le 1-\varepsilon \) and \(\delta \in (0,C^{-1})\). Now, define
$$\begin{aligned} A(t,R)=\sup _{y\in \mathbb {R}}\int _{\mathbb {R}}u^2\varphi _{R,y}^2\mathrm {d}x \,, \end{aligned}$$
where \(\varphi _{R,y}(x)=\psi ((x-y)/R)\). Note that A(t, R) is bounded because u is. After integration in time, we get
$$\begin{aligned} \int _{\mathbb {R}}\varphi _{R,y}^2u_t^2\mathrm {d}x&\le \int _{\mathbb {R}}\varphi _{R,y}^2u_0^2\mathrm {d}x+\frac{4}{\delta R^2}\int _0^t \int _{-2R}^{2R}u_s^2\mathrm {d}x \mathrm {d}s \\&\le A(0,R) +\frac{8}{\delta R^2}\int _0^tA(s,R)\mathrm {d}s \end{aligned}$$
and taking the supremum over x on the left-hand side yields
$$\begin{aligned} A(t,R)\le A(0,R)+\frac{8}{\delta R^2}\int _0^tA(s,R)\mathrm {d}s \end{aligned}$$
and by the Gronwall inequality, we obtain
$$\begin{aligned} A(t,R)\le A(0,R)\exp \left( \frac{8t}{\delta R^2}\right) \end{aligned}$$
and the result follows by letting \(R\rightarrow \infty \). \({\square }\)
Now we extend the cutoff argument to get a uniform estimate of the \(H^1\)-norm.
Lemma 8.5
Let \(u:\mathbb {R}\times (0,T)\rightarrow \infty \) be a smooth solution of (8.3) with uniformly spacelike initial data \(u_0\in C^{0,1}(\mathbb {R})\cap L^2(\mathbb {R})\). Then \(u_t\in H^1(\mathbb {R})\) and
$$\begin{aligned} \left\| u_t\right\| _{L^2}^2+t \left\| u_t'\right\| _{L^2}^2\le \left\| u_0'\right\| _{L^2}^2 \end{aligned}$$
for all \(t>0\).
Proof
Let \(\varphi \) be as in the proof of Lemma 8.4. Then
$$\begin{aligned}&\partial _t \left( t \int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x\right) =\int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x+2t\int _{\mathbb {R}}\varphi ^2(u')(\partial _t u)'\mathrm {d}x\\&\quad =\int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x+2t\int _{\mathbb {R}}\varphi ^2u'\left( \frac{u''}{1-(u')^2}\right) '\mathrm {d}x\\&\quad =\int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x -2t\int _{\mathbb {R}}\varphi ^2\frac{(u'')^2}{1-(u')^2}\mathrm {d}x-4t\int _{\mathbb {R}}\varphi '\varphi u'\left( \frac{u''}{1-(u')^2}\right) \mathrm {d}x\\&\quad \le \int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x -2t\int _{\mathbb {R}}\varphi ^2\frac{(u'')^2}{1-(u')^2}\mathrm {d}x\\&\qquad +\frac{2t}{\delta }\int _{\mathbb {R}}(\varphi ')^2(u')^2\mathrm {d}x+2t\delta \int _{\mathbb {R}}\varphi ^2\frac{(u'')^2}{(1-(u')^2)^2}\mathrm {d}x\\&\quad \le \int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x+\frac{2t}{\delta }\int _{\mathbb {R}}(\varphi ')^2(u')^2\mathrm {d}x -2t(1-C\delta )\int _{\mathbb {R}}\varphi ^2\frac{(u'')^2}{1-(u')^2}\mathrm {d}x\\&\quad \le \int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x+\frac{8t}{R^2\delta }\int _{-2R}^{2R}(u')^2\mathrm {d}x -2t(1-C\delta )\int _{\mathbb {R}}\varphi ^2\frac{(u'')^2}{1-(u')^2}\mathrm {d}x \,, \end{aligned}$$
where C depends on \(\sup |u'_t|<1\). Now, pick \(\alpha \in (0,1)\). Then, combining the above estimate with the one from the Lemma 8.4, we obtain for any \(\delta \in (0,(1-\alpha )C^{-1})\)
$$\begin{aligned}&\partial _t\left( \int _{\mathbb {R}}\varphi ^2u^2\mathrm {d}x+\alpha t \int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x\right) \le \frac{4}{\delta R^2}\int _{-2R}^{2R}u^2\mathrm {d}x+\frac{8\alpha t}{R^2\delta }\int _{-2R}^{2R}(u')^2\mathrm {d}x\\&\qquad -(1-\alpha -C\delta )\int _{\mathbb {R}}\varphi ^2(u')^2\mathrm {d}x-2t(1-C\delta ) \alpha \int _{\mathbb {R}}\varphi ^2\frac{(u'')^2}{1-(u')^2}\mathrm {d}x\\&\quad \le \frac{4}{\delta R^2}\int _{-2R}^{2R}u^2\mathrm {d}x+\frac{8\alpha t}{R^2\delta }\int _{-2R}^{2R}(u')^2\mathrm {d}x \,. \end{aligned}$$
Analogously to the above, we define
$$\begin{aligned} A(t,R,\alpha )=\sup _{y\in \mathbb {R}}\left( \int _{\mathbb {R}}u^2\varphi _{R,y}^2\mathrm {d}x+\alpha t\int _{\mathbb {R}}(u')^2\varphi _{R,y}^2\mathrm {d}x\right) \,, \end{aligned}$$
where \(\varphi _{R,y}(x)\) as in the proof of the previous lemma. After integration in time, we get
$$\begin{aligned}&\int _{\mathbb {R}}\varphi _{R,y}^2u_t^2\mathrm {d}x+\alpha t\int _{\mathbb {R}}\varphi _{R,y}^2(u'_t)^2\mathrm {d}x \\&\quad \le \int _{\mathbb {R}}\varphi _{R,y}^2u_0^2\mathrm {d}x +\frac{4}{\delta R^2}\int _0^t \int _{-2R}^{2R}u_s^2\mathrm {d}x\mathrm {d}s + \frac{8\alpha }{R^2\delta }\int _0^ts\int _{-2R}^{2R}(u'_s)^2\mathrm {d}x\mathrm {d}s \\&\quad \le A(0,R,\alpha )+\frac{16}{\delta R^2}\int _0^tA(s,R,\alpha )\mathrm {d}s \end{aligned}$$
and taking the supremum over x on the left-hand side yields
$$\begin{aligned} A(t,R,\alpha )\le A(0,R,\alpha )+\frac{16}{\delta R^2}\int _0^tA(s,R,\alpha )\mathrm {d}s. \end{aligned}$$
By the Gronwall inequality, we obtain
$$\begin{aligned} A(t,R,\alpha )\le A(0,R,\alpha )\exp \left( \frac{16t}{\delta R^2}\right) \end{aligned}$$
and the result follows from letting \(R\rightarrow \infty \) and then letting \(\alpha \rightarrow 1\). \({\square }\)
Proposition 8.6
Let \(u:\mathbb {R}\times (0,T)\rightarrow \infty \) be a smooth solution of (8.3) with uniformly spacelike initial data \(u_0\in C^{0,1}(\mathbb {R})\cap L^2(\mathbb {R})\). Then \(u_t\rightarrow 0\) for \(t\rightarrow \infty \), uniformly in all derivatives. More precisely,
$$\begin{aligned} \left\| u^{(k)}\right\| _{L^{\infty }}\le C_k\cdot t^{-\frac{1}{4}} \end{aligned}$$
for some \(C\ge 0\) and for all \(t>0\).
Proof
By the one-dimensional Gagliardo–Nirenberg inequality and Lemma 8.5, we obtain
$$\begin{aligned} \left\| u\right\| _{L^{\infty }}\le C\left\| u'\right\| _{L^2}^{\frac{1}{2}}\left\| u\right\| _{L^2}^{\frac{1}{2}} = Ct^{-\frac{1}{4}}(t^{\frac{1}{2}}\left\| u'\right\| _{L^2})^{\frac{1}{2}}\left\| u\right\| _{L^2}^{\frac{1}{2}} \le Ct^{-\frac{1}{4}}\,. \end{aligned}$$
For higher derivatives, the assertion follows from short-time derivative estimates of the form
$$\begin{aligned} \left\| u^{(k)}_{t+1}\right\| _{L^{\infty }}\le C_k\cdot \left\| u_t\right\| _{L^{\infty }} \,, \end{aligned}$$
which finishes the proof. \({\square }\)
Remark 8.7
The convergence rate in Proposition 8.6 is the same as one gets for solutions of the 1-dimensional heat equation with initial data in \(L^2\). This can be seen as follows: If \(u_0\in L^2(\mathbb {R})\), the solution of the heat equation with initial data \(u_0\) is given by
$$\begin{aligned} u(x,t)=\int _{\mathbb {R}}K(y-x,t)u_0(y)\mathrm {d}y\,,\qquad K(x,t)=(4\pi t)^{-1/2}\cdot e^{-\frac{x^2}{4t}}\,, \end{aligned}$$
where K is the 1-dimensional heat kernel. A direct computation proves that the inequality \(\left\| K(\cdot ,t)\right\| _{L^2}\le C\cdot t^{-1/4}\) holds. Finally Young’s inequality for convolutions implies that
$$\begin{aligned} \left\| u(\cdot ,t)\right\| _{L^{\infty }}\le \left\| K(\cdot ,t)\right\| _{L^2}\left\| u_0\right\| _{L^2}\le C\cdot t^{-1/4}\cdot \left\| u_0\right\| _{L^2}\,, \end{aligned}$$
which is the estimate we claimed.
Theorem 8.8
Let \(u:\mathbb {R}\times (0,T)\rightarrow \infty \) be a smooth, uniformly spacelike solution of (8.3) with uniformly spacelike initial data \(u_0\in C^{0,1}(\mathbb {R})\) such that \(u_0(x)\rightarrow 0\) as \(|x|\rightarrow \infty \). Then, if \(t\rightarrow \infty \), \(u_t\rightarrow 0\) uniformly in all derivatives. More precisely, there exist functions \(v_k(t)\) with \(v_k(t)\rightarrow 0\) as \(t\rightarrow \infty \) such that
$$\begin{aligned} \left\| u^{(k)}\right\| _{L^{\infty }}\le v_k(t) \end{aligned}$$
for some \(C\ge 0\) and for all \(t>0\).
Proof
For each \(\varepsilon >0\), pick a compactly supported smooth function \(v_{0,\varepsilon }\) such that
$$\begin{aligned} -v_{0,\varepsilon }-\frac{\varepsilon }{2}\le u_0\le v_{0,\varepsilon }+\frac{\varepsilon }{2}\,. \end{aligned}$$
By the maximum principle, the corresponding solutions satisfy
$$\begin{aligned} -v_{\varepsilon }-\frac{\varepsilon }{2}\le u\le v_{\varepsilon }+\frac{\varepsilon }{2} \end{aligned}$$
for all times. As \(v_{\varepsilon }\rightarrow 0\), we conclude that \(|u|\le \varepsilon \) for all \(t\ge T(\varepsilon )\) and some large time \(T(\varepsilon )\), and the assertion for u follows for \(k=0\). For the higher derivatives, the claim follows from standard estimates. \({\square }\)