On the rigidity theorems for Lagrangian translating solitons in pseudo-Euclidean space III

Abstract

Let f be a smooth strictly convex solution of

$$\begin{aligned} \det \left( \frac{\partial ^{2}f}{\partial x_{i}\partial x_{j}}\right) =\exp \left\{ \sum _{i=1}^n- a_i\frac{\partial f}{\partial x_{i}} +\sum _{i=1}^n b_ix_i+c\right\} \end{aligned}$$

defined on $\mathbb {R}^{n}$, where $a_i$, $b_i$ and c are constants, then the graph $M_{\nabla f}$ of $\nabla f$ is a space-like translating soliton for mean curvature flow in pseudo-Euclidean space $\mathbb {R}^{2n}_{n}$ with the indefinite metric $\sum dx_idy_i$. In this paper, we classify the entire solutions of the PDE above for dimension $n=1$ and show every entire classical strictly convex solution $(n\ge 2)$ must be a quadratic polynomial under a decay condition on the hessian $(D^2f)$.

Appendix

Proposition 5.1

Let f be an entire smooth strictly convex solution of (1.3) $(n\ge 2)$. If the function $\Phi$ has an upper bound, then the graph hypersurface $M=\{(x, f(x))\}$ is complete with respect to the Calabi metric.

Proof

Let $p\in M$ be any fixed point. Up to an affine transformation of $\mathbb {R}^{n+1}$, we may assume that p has coordinates $(0,\ldots ,0,0)$ and

$$\begin{aligned} f(0)=0,\;\;\;\;\frac{\partial f}{\partial x^i}(0)=0,\;\;\;\;1\le i\le n. \end{aligned}$$

The key point of the proof of Proposition 5.1 is to estimate $\frac{\Vert \nabla f\Vert ^2}{\left( 1+f\right) ^{2}}$. We shall show that it is bounded if $\Phi$ is bounded. To estimate $\frac{\Vert \nabla f\Vert ^2}{\left( 1+f\right) ^{2}}$, we consider the following function

$$\begin{aligned} F(x):=\exp \left\{ \frac{-m}{C-f}+\Psi \right\} \frac{\Vert \nabla f\Vert ^2}{\left( 1+f\right) ^{2}} \end{aligned}$$

(5.1)

defined on the section

$$\begin{aligned} S_{f}(0,C):=\left\{ x\in R^n\;\big |\; f(x)<C\right\} , \end{aligned}$$

where

$$\begin{aligned} \Psi :=\exp \{\Phi \}, \end{aligned}$$

and m is a positive constant to be determined later. Clearly, F attains its supremum at some interior point $p^*$ of $S_{f}(0,C)$. We can assume that $\Vert \nabla f\Vert >0$ at $p^*$. Choose a local orthonormal frame field of the Calabi metric $e_1,\ldots ,e_n$ on M such that, at $p^*,\;f_{,1}=\Vert \nabla f\Vert >0,\;f_{,i}=0\;(i\ge 2)$. Then, at $p^*$,

$$\begin{aligned} F_{,i}=0, \end{aligned}$$

(5.2)

$$\begin{aligned} \sum F_{,ii}\le 0. \end{aligned}$$

(5.3)

Now we calculate both expressions (5.2) and (5.3) explicitly. By (5.2) and (5.3), we have

$$\begin{aligned} 2\sum _j f_{,j}f_{,ji}+\left( -g f_{,i}-2\frac{f_{,i}}{1+f}+\Psi _{,i}\right) \sum _j (f_{,j})^2=0, \end{aligned}$$

(5.4)

$$\begin{aligned} 2 \sum _{i,j} (f_{,ij})^2+2\sum f_{,j}f_{,jii}+2\sum _{i,j} \left( -g f_{,i}-2\frac{f_{,i}}{1+f}+ \Psi _{,i}\right) f_{,j}f_{,ji} \end{aligned}$$

$$\begin{aligned} +\left[ -g^{\prime }\sum (f_{,i})^{2}-g \Delta f+ 2\frac{\sum (f_{,i})^2}{(1+f)^{2}}-2\frac{\Delta f}{1+f} +\Delta \Psi \right] \sum _j (f_{,j})^{2}\le 0, \end{aligned}$$

(5.5)

where

$$\begin{aligned} g:=\frac{m}{(C-f)^2},\quad \quad g^\prime :=\frac{2m}{(C-f)^3}. \end{aligned}$$

Let us simplify (5.5). From (5.4) we have

$$\begin{aligned} 2f_{,1i}=\left( g f_{,i}+2\frac{f_{,i}}{1+f} -\Psi _{,i}\right) f_{,1}. \end{aligned}$$

(5.6)

Applying the inequality of Schwarz, we get

$$\begin{aligned} 2\sum (f_{,ij})^2\ge 2\left( \frac{n}{n-1}-\delta \right) (f_{,11})^2+4\sum _{i>1}(f_{,1i})^2-\frac{2}{\delta (n-1)^2}(\Delta f)^2, \end{aligned}$$

(5.7)

for any $0<\delta <1$. We insert (5.6) and (5.7) into (5.5) and obtain

$$\begin{aligned}&\left[ 2\left( \frac{n}{n-1}-\delta \right) -4\right] (f_{,11})^2+2\sum f_{,j}f_{,jii}-\frac{2}{\delta (n-1)^2}(\Delta f)^2\nonumber \\&\quad +\left[ -g^{\prime }(f_{,1})^2-g\Delta f +2\frac{(f_{,1})^{2}}{(1+f)^{2}}-2\frac{\Delta f}{1+f}+\Delta \Psi \right] (f_{,1})^2\le 0. \end{aligned}$$

(5.8)

Now we calculate $\Delta \Psi$. From Proposition 3.1 we obtain

$$\begin{aligned} \Delta \Phi \ge \frac{\delta }{\rho ^2}\sum (\rho _{,ij})^2 - C. \end{aligned}$$

(5.9)

Then we get

$$\begin{aligned} \Psi _{,i}=\Psi \Phi _{,i}, \end{aligned}$$

(5.10)

and

$$\begin{aligned} \Delta \Psi \ge \Psi \sum (\Phi _{,i})^2 + \Psi \frac{\delta }{\rho ^2}\sum (\rho _{,ij})^2 - C. \end{aligned}$$

(5.11)

Let us now compute the term $\sum f_{,j}f_{,jii}$ in (5.8). An application of the Ricci identity shows that

$$\begin{aligned} 2\sum _{i,j} f_{,j}f_{,jii}=2\sum _{j} f_{,j}(\Delta f)_{,j}+2\sum _{i,j} R_{ij}f_{,i}f_{,j}. \end{aligned}$$

Note

$$\begin{aligned} \Delta f=n+\frac{n+2}{2} \sum (\log \rho )_{,i}f_{,i}, \end{aligned}$$

we get

$$\begin{aligned} 2\sum&f_{,j}f_{,jii}=(n+2)\sum _{i,j}\left( f_{,ij}\frac{\rho _{,i}}{\rho }+ f_{,i}\frac{\rho _{,ij}}{\rho }- f_{,i}\frac{\rho _{,i}\rho _{,j}}{\rho ^2}\right) f_{,j}+2 R_{11}(f_{,1})^2 \nonumber \\&=(n+2)\left( \frac{\rho _{,11}}{\rho }(f_{,1})^2- \frac{(\rho _{,1})^2}{\rho ^2}(f_{,1})^2 +\sum _i f_{,1i}f_{,1}\frac{\rho _{,i}}{\rho }-\sum _k A_{11k}\frac{\rho _{k}}{\rho }(f_{,1})^2\right) \nonumber \\&\quad +2\sum _{m,l} A_{ml1}^2(f_{,1})^2 \nonumber \\&\ge (n+2)\frac{\rho _{,11}}{\rho }(f_{,1})^2- C(n,\delta ) \Phi (f_{,1})^2 \nonumber \\&\quad -\delta \sum (f_{,1i})^2 +(2-\delta )\sum A_{ml1}^2(f_{,1})^2, \end{aligned}$$

(5.12)

where $\delta$ is a positive constant as before, and $C(n,\delta )$ is a positive constant depending only on n and $\delta$. From the structure equation we get

$$\begin{aligned} f_{,ij}=f_{ij}+A_{ij1}f_{,1}, \end{aligned}$$

and

$$\begin{aligned} \sum _{i,j} (f_{,ij})^2=\sum _{i,j} A_{ij1}^2f_{,1}^2+n+(n+2)\frac{\rho _{,1}f_{,1}}{\rho }. \end{aligned}$$

(5.13)

A combination of (5.7) and (5.13) gives

$$\begin{aligned} 2\sum f_{,j}f_{,jii}\ge&(2-6\delta )\sum (f_{,1i})^2-\frac{2}{\delta (n-1)}(\Delta f)^2 \nonumber \\&+(n+2)\frac{\rho _{,11}}{\rho }(f_{,1})^2-C(n,\delta ) \Phi (f_{,1})^2-2n\nonumber \\ \ge&(2-6\delta )\sum (f_{,1i})^2-\frac{2}{\delta (n-1)}(\Delta f)^2-\delta \Psi \sum \frac{(\rho _{,ij})^2}{\rho ^2}(f_{,1})^2\nonumber \\&-\frac{(n+2)^2}{4\delta }(f_{,1})^2- C(n,\delta ) \Phi (f_{,1})^2-2n. \end{aligned}$$

(5.14)

We insert (5.11), (5.14) into (5.8) and use (5.6):

$$\begin{aligned}&\left( \frac{1}{2(n-1)}{-}2\delta \right) \left( g f_{,1}+2\frac{f_{,1}}{1+f}{-}\Psi _{,1}\right) ^2(f_{,1})^2 {-}\frac{4}{(n{-}1)\delta }(\Delta f)^2{-} C(n,\delta )\Phi (f_{,1})^2\nonumber \\&\quad - \frac{(n+2)^2}{4\delta }(f_{,1})^2+\Psi \sum (\Phi _{,i})^2(f_{,1})^2-C(n,\delta )(f_{,1})^2-2n\nonumber \\&\quad +\left[ -g^{\prime }(f_{,1})^{2}-g \Delta f +2\frac{(f_{,1})^{2}}{(1+f)^{2}}-2\frac{\Delta f}{1+f}\right] (f_{,1})^{2} \le 0. \end{aligned}$$

(5.15)

We choose the following values for $\delta$ and m:

$$\begin{aligned} \delta :=\frac{1}{16(n-1)},\quad \quad m:=40\left( 3n+\exp \{N\}\right) C, \end{aligned}$$

where N is an upper bound of $\Phi$ on $\mathbb {R}^n$. To simplify the expression we denote

$$\begin{aligned} a_1:=\frac{1}{2(n-1)}-2\delta ,\;\;\;\;a_2:=\frac{4}{(n-1)\delta }, \end{aligned}$$

$$\begin{aligned} a_3:= C(n,\delta )N +\frac{(n+2)^2}{4\delta }+ C(n,\delta ),\quad a_4:=\frac{a_1}{1+a_1\exp \{N\}}. \end{aligned}$$

Recall that $\Psi _{,i}=\Psi \Phi _{,i}$. Then

$$\begin{aligned} a_1\left( gf_{,1}+2\frac{f_{,1}}{1+f}-\Psi _{,1}\right) ^2+ \Psi \sum (\Phi _{,i})^2 \ge a_4\left( g f_{,1}+ 2\frac{f_{,1}}{1+f}\right) ^2. \end{aligned}$$

(5.16)

Inserting (5.16) into (5.15) we get

$$\begin{aligned}&a_4\left( g f_{,1}+2\frac{f_{,1}}{1+f}\right) ^{2} (f_{,1})^{2}-a_2(\Delta f)^2-a_3 (f_{,1})^2\nonumber \\&\quad +\left( -g^{\prime } (f_{,1})^2-g\Delta f+2\frac{(f_{,1})^2}{(1+f)^{2}}-2\frac{\Delta f}{1+f}\right) (f_{,1})^{2}-2n\le 0. \end{aligned}$$

(5.17)

Multiply both sides of (5.17) by $\frac{1}{(1+f)^{2}}$. Then we obtain

$$\begin{aligned}&a_4\left( g f_{,1}+ 2 \frac{f_{,1}}{1+f}\right) ^2 \frac{(f_{,1})^2}{(1+f)^{2}}-a_2 \left( \frac{\Delta f}{1+f}\right) ^2-a_3 \frac{(f_{,1})^2}{(1+f)^{2}}-g^{\prime } \frac{(f_{,1})^4}{(1+f)^{2}}\nonumber \\&\quad -g \Delta f \frac{(f_{,1})^2}{(1+f)^{2}} +2\frac{(f_{,1})^4}{(1+f)^{4}}-2 \left( \frac{\Delta f}{1+f}\right) \frac{(f_{,1})^2}{(1+f)^{2}}-2n\le 0. \end{aligned}$$

(5.18)

Using $g'\le \frac{a_4}{20}g^2$, to further estimate (5.18), we have the following three inequalities:

$$\begin{aligned} g^{\prime }\frac{(f_{,1})^4}{(1+f)^{2}}\le & {} \frac{a_4}{10} \left( gf_{,1}+2\frac{f_{,1}}{1+f}\right) ^{2} \frac{(f_{,1})^2}{(1+f)^{2}},\\ \left( \frac{\Delta f}{1+f}\right) ^2\le & {} 2n^2+2n^2\Phi \frac{(f_{,1})^2}{(1+f)^2},\\ 2 \left( \frac{\Delta f}{1+f}\right) \frac{(f_{,1})^2}{(1+f)^2}\le & {} 2n \frac{(f_{,1})^2}{(1+f)^2}+\frac{(f_{,1})^4}{(1+f)^4}+ n^2\Phi \frac{(f_{,1})^2}{(1+f)^2}. \end{aligned}$$

We use the inequality of Schwarz and obtain

$$\begin{aligned} g\Delta f \frac{(f_{,1})^2}{(1+f)^{2}}&=ng\frac{(f_{,1})^2}{(1+f)^{2}} +\frac{n+2}{2}g\frac{\rho _{,1}}{\rho } \frac{(f_{,1})^3}{(1+f)^{2}}\\&\le \frac{5n^2}{2a_4}+\frac{a_4}{10}g^2\frac{(f_{,1})^4}{(1+f)^{4}}+ \frac{10(n+2)^2}{16a_4}\Phi \frac{(f_{,1})^2}{(1+f)^{2}}+\frac{a_4}{10}g^2 \frac{(f_{,1})^4}{(1+f)^{2}}\\&\le \frac{a_4}{5} \left( g f_{,1}+2\frac{f_{,1}}{1+f}\right) ^2 \frac{(f_{,1})^2}{(1+f)^{2}}+\frac{5n^2}{2a_4} +\frac{5n^2}{2a_4}\Phi \frac{(f_{,1})^2}{(1+f)^{2}}. \end{aligned}$$

We insert these inequalities into (5.18) and get

$$\begin{aligned} \frac{(f_{,1})^4}{(1+f)^{4}} -a\frac{(f_{,1})^2}{(1+f)^{2}}-b\le 0, \end{aligned}$$

(5.19)

where we use the abbreviations:

$$\begin{aligned} a:=\left( 2n^2a_2+ \frac{5n^2}{2a_4}+n^2\right) N+2n+a_3, \quad b:=2n^2a_2+\frac{5n^2}{2a_4}+2n. \end{aligned}$$

The left hand term in (5.19) is a quadratic expression in $\frac{(f_{,1})^2}{(1+f)^{2}}$. If one considers its zeroes it follows that

$$\begin{aligned} \frac{(f_{,1})^2}{(1+f)^{2}}\le a+\sqrt{b}. \end{aligned}$$

Thus from (5.1) we get, with our special choice of $\delta$ and m:

$$\begin{aligned} F\le \exp \{\exp \{N\}\}(a+\sqrt{b}), \end{aligned}$$

which holds at $p^*$, where F attains its supremum. Hence, at any interior point of $S_f(0,C)$, we have

$$\begin{aligned} \frac{\Vert \nabla f\Vert ^2}{(1+f)^{2}}\le \exp \{\exp \{N\}\}(a+\sqrt{b})\exp \left\{ \frac{40(3n+\exp \{N\})C}{C-f}\right\} . \end{aligned}$$

(5.20)

Let $C\rightarrow \infty$ then

$$\begin{aligned} \frac{\Vert \nabla f\Vert ^2}{(1+f)^{2}}\le \exp \left\{ \exp \{N\}+40(3n+\exp \{N\})\right\} (a+\sqrt{b}):=Q, \end{aligned}$$

(5.21)

where Q is a constant.

Using the gradient estimate (5.21) we can prove that M is complete with respect to the Calabi metric, namely: for any unit speed geodesic, starting from p,

$$\begin{aligned} \sigma :[0,S]\rightarrow M \end{aligned}$$

we have

$$\begin{aligned} \frac{df}{ds}\le \Vert \nabla f\Vert \le \sqrt{Q}(1+f). \end{aligned}$$

It follows that

$$\begin{aligned} s\ge \frac{1}{\sqrt{Q}}\int ^{x_{n+1}(\sigma (S))}_{0}\frac{df}{1+f}. \end{aligned}$$

(5.22)

Since

$$\begin{aligned} \int ^{\infty }_{0}\frac{df}{1+f}=\infty \end{aligned}$$

and $f:\Omega \rightarrow \mathbb R$ is proper (i.e., the inverse image of any compact set is compact), (5.22) implies that M is complete with respect to the Calabi metric. This completes the proof of Proposition 5.1. $\square$