Contracting Axially Symmetric Hypersurfaces by Powers of the $\sigma _k$-Curvature

Abstract

In this paper, we investigate the contracting curvature flow of closed, strictly convex axially symmetric hypersurfaces in $\mathbb {R}^{n+1}$ and $\mathbb {S}^{n+1}$ by $\sigma _k^\alpha$, where $\sigma _k$ is the k-th elementary symmetric function of the principal curvatures and $\alpha \ge 1/k$. We prove that for any $n\ge 3$ and any fixed k with $1\le k\le n$, there exists a constant $c(n,k)>1/k$ such that if $\alpha$ lies in the interval [1/k, c(n, k)], then we have a nice curvature pinching estimate involving the ratio of the biggest principal curvature to the smallest principal curvature at every point of the flow hypersurface, and we prove that the properly rescaled hypersurfaces converge exponentially to the unit sphere. In the case $1<k\le n \le k^2$, we can choose $c(n,k)=1/(k - 1)$. Our results provide an evidence for the general convergence result without initial curvature pinching conditions.

Appendix A: Sturm’s Theorem and the Computer Algorithm

In this section, we will prove that for each n and any fixed k with $1\le k\le n$, there exists a constant $c_0(n,k) >\frac{1}{k}$ such that the polynomial Q defined by (3.14) is non-positive for any $x>0$ and $\alpha = c_0(n,k)$. This is needed in the proof of Theorem 3.2. Note that in order to make sure that Q defined by (3.14) is non-positive for any $x>0$ and $\alpha = c_0(n,k)$, the highest coefficient of Q needs to be non-positive, which implies that

$$\begin{aligned} \frac{1}{k} \le c_0(n,k) \le \frac{1}{k-1}\quad \forall {k}\ge 2. \end{aligned}$$

(A.1)

In the case $1<k\le n \le k^2$, we prove that $c_0(n,k) =\frac{1}{k-1}$ satisfies that Q is non-positive for any $x>0$ and $\alpha = c_0(n,k)$, see Proposition  A.1. For general case, we prove that for each n and any fixed k with $1\le k\le n$, we can find a constant $c_0(n,k) >\frac{1}{k}$ such that Q is non-positive for any $x>0$ and $\alpha = c_0(n,k)$. Although we cannot write down $c_0(n,k)$ in term of an explicit function of n and k, $c_0(n,k)$ can be precisely determined by applying Sturm’s theorem, see Proposition A.2. We list some of the values of $c_0(n,k)$ (see (A.3)). We also give some estimates of the constant $c_0(n,k)$. We prove that $c_0(n,1)\ge 1+\frac{7}{n}$ and $c_0(n,k) \ge \frac{1}{k}+\frac{k}{(k-1)n}$ for $k\ge 2$ and $n\ge k^2$, see Proposition  A.3 and Proposition A.4. The estimate for the case with $k\ge 2$ and $n\ge k^2$ is optimal in the sense that when $n=k^2$, the lower bound $\frac{1}{k}+\frac{k}{(k-1)n}$ equals $\frac{1}{k-1}$, which is an upper bound for $c_0(n,k)$ (see (A.1)).

First, in the case $1<k\le n \le k^2$, we have the following result.

Proposition A.1

If $1<k\le n \le k^2$, $\alpha =\frac{1}{k-1}$, then all the coefficients of $Q=Q(x)$ defined by (3.14) are non-positive, which implies that Q is non-positive for any $x>0$.

Proof

We denote the coefficients of $Q=Q(x)$ by $c_i$, i.e., we write

$$\begin{aligned} Q(x)=c_6 x^6+c_5 x^5+ c_4 x^4+c_3 x^3+ c_2 x^2 +c_1 x^1 +c_0. \end{aligned}$$

We will prove that if $1<k\le n \le k^2$, $\alpha =\frac{1}{k-1}$, then $c_i\le 0$, $i=0,1,\ldots ,6$.

If $1<k\le n \le k^2$, $\alpha =\frac{1}{k-1}$, it is obvious that $c_6=0$ and $c_0$ is non-positive. We will estimate $c_1$ to $c_5$ one by one. For $c_1$, if $1<k\le n \le k^2$, $\alpha =\frac{1}{k-1}$, then we have that

$$\begin{aligned} \begin{aligned} c_1&= \frac{(n-k)^2(n-6k^2+8k)}{k-1}\le \frac{(n-k)^2(k^2-6k^2+8k)}{k-1}\\&=\frac{(n-k)^2(-5k^2+8k)}{k-1}\le 0. \end{aligned} \end{aligned}$$

For $c_2$, note that from the expression of $c_2$ we know that $c_2=0$ if $n=k$, hence we only need to consider that case that $k+1\le n\le k^2$. If $2<k+1\le n \le k^2$, $\alpha =\frac{1}{k-1}$, then we have

$$\begin{aligned} \begin{aligned} c_2&= (k-n)(\alpha ^2 k^2(k-2n+3)+\alpha k(6k^2-k(3n+22)+12n+3)+3k(n+1)-4n)\\&=\frac{k-n}{(k-1)^2}(2k^2(3k^2-12k+11)+n(k+1)(3k-4))\\&\le \frac{k-n}{(k-1)^2}(2k^2(3k^2-12k+11)+(k+1)(k+1)(3k-4))\\&= \frac{k-n}{k-1}(6k^3-15k^2+9k+4)<0. \end{aligned} \end{aligned}$$

For $c_3$, we can regard $c_3$ as a quadratic polynomial of n. If $1<k\le n \le k^2$, $\alpha =\frac{1}{k-1}$, then the coefficient of $n^2$ is $-2(\alpha k-1)(2\alpha k-3)=\frac{2(k-3)}{(k-1)^2}$, which is non-negative when $k\ge 3$, which means $c_3$ is a convex function of n when $k\ge 3$. We first consider the case $k\ge 3$. In this case, it suffices to prove that both $c_3|_{n=k}$ and $c_3|_{n=k^2}$ are non-positive. We have that

$$\begin{aligned} \begin{aligned} c_3|_{n=k}=-\frac{k^2(k-2)(2k-3)}{k-1}<0,~~c_3|_{n=k^2}=-\frac{k^3(9k-16)}{k-1} < 0. \end{aligned} \end{aligned}$$

In the case $k=2$, we have $\alpha =1$ and $c_3 =-2n(n-2)\le 0$.

Similarly, for $c_4$, we can regard it as a quadratic polynomial of n. Note that the coefficient of $n^2$ is $2(\alpha k-1)^2$, which means that $c_4$ is a convex function of n. For any k and $\alpha$, the maximum of $c_4$ is attained at either $n=k$ or $n=k^2$. We only need to prove that both $c_4|_{n=k}$ and$c_4|_{n=k^2}$ are non-positive. If $1<k\le n \le k^2$, $\alpha =\frac{1}{k-1}$, then we have

$$\begin{aligned} c_4|_{n=k}=-\frac{5k^2(k-2)}{k-1}\le 0,~~c_4|_{n=k^2}=-\frac{5k^3}{k-1}\le 0. \end{aligned}$$

For $c_5$, if $1<k\le n \le k^2$, $\alpha =\frac{1}{k-1}$, then we have $c_5=\frac{4k(n-k^2)}{(k-1)^2}\le 0$.

Therefore, we have proved that if $1<k\le n \le k^2$, $\alpha =\frac{1}{k-1}$, then $c_i \le 0$, for $i=0,1,\ldots ,6$. Consequently, we obtain that the polynomial Q defined by (3.14) is non-positive for any $x>0$ if $1<k\le n \le k^2$ and $\alpha =\frac{1}{k-1}$. $\square$

For general cases, we apply Sturm’s theorem to prove the following proposition.

Proposition A.2

For each n and any fixed k with $1\le k\le n$, we can find a constant $c_0(n,k) >\frac{1}{k}$ such that Q defined by (3.14) is non-positive for any $x>0$ and $\alpha = c_0(n,k)$.

Proof

First, we define a standard sequence of a polynomial $p(x)\in \mathbb {R}[x]$ of positive degree (cf. Chapter 5.2 of [32]) by applying Euclid’s algorithm to p(x) and $p'(x)$:

$$\begin{aligned} \begin{aligned} p_0(x)&:= p(x), \\ p_1(x)&:= p'(x), \\ p_2(x)&:= -\text {rem}(p_0,p_1), \\ p_3(x)&:= -\text {rem}(p_1,p_2), \\&\,\,\,\,\vdots \\ 0&\,= -\text {rem}(p_{m-1},p_m), \end{aligned} \end{aligned}$$

(A.2)

where $\text {rem}(p_i,p_j)$ is the polynomial remainder of the polynomial long division of $p_i$ by $p_j$. We call the above sequence of polynomials the standard Sturm sequence of p(x). We will apply the following theorem.

Sturm’s theorem (cf. [32]) Let $p_0(x), \ldots , p_m(x)$ be the standard Sturm sequence of a polynomial $p(x)\in \mathbb {R}[x]$ with positive degree. Assume that [a, b] is an interval such that $p(a)p(b)\ne 0$, and let $\sigma (\xi )$ denote the number of sign changes (ignoring zeroes) in the sequence

$$\begin{aligned} \{p_0(\xi ),p_1(\xi ),\ldots ,p_m(\xi )\}, \end{aligned}$$

then the number of distinct roots of p(x) in (a, b) is $\sigma (a)-\sigma (b)$.

For each n and any fixed k with $1\le k\le n$, it is obvious that $Q<0$ if $x=0$. Therefore, in order to prove Proposition A.2, by applying Sturm’s theorem, we only need to prove that for each n and any fixed k with $1\le k\le n$, we can find a constant $c_0(n,k) >\frac{1}{k}$ such that the number $\sigma (0)- \sigma (\infty )$ of the polynomial $Q(x, k, n, c_0(n,k)$ equals 0. We will describe how to use the computer program Mathematica to help us to find the constant $c_0(n,k)$, by using a method of bisection and applying Sturm’s theorem. We can use Mathematical algorithm to run the following procedure: For each n and fixed k with $1\le k\le n$, we fix an arbitrary precision $\delta$ and set the initial data as follows.

$$\begin{aligned} \alpha ^{\text {initial}}_{\min } =\frac{1}{k},\quad \alpha ^{\text {initial}}_{\max }=6 ~~\text {for}~~ k=1,\quad \alpha ^{\text {initial}}_{\max }=\frac{1}{k-1}+\delta ~~\text {for}~~ k\ge 2. \end{aligned}$$

Whenever $\alpha _{\max }-\alpha _{\min } \ge \delta$, we do the following loop:

• (1) Set $\alpha _{test}=\frac{1}{2}(\alpha _{\max }+\alpha _{\min })$.

• (2) Use Euclid’s algorithm to compute the Sturm sequence for the polynomial $Q(x, k, n, \alpha _{test})$.

• (3) Compute $\sigma (0)- \sigma (\infty )$ of the polynomial $Q(x, k, n, \alpha _{test})$, if $\sigma (0)- \sigma (\infty )=0$, then we set $\alpha _{\min }=\alpha _{test}$, otherwise, we set $\alpha _{\max }=\alpha _{test}$.

Once the loop ends, we obtain two constants $\alpha _{\min }(n,k)$ and $\alpha _{\max }(n,k)$ which satisfy that $\alpha _{\max }(n,k)-\alpha _{\min }(n,k)$ is less than the given precision $\delta$, and the number $\sigma (0)- \sigma (\infty )$ of the polynomial $Q(x,k,n,\alpha _{\min }(n,k))$ equals 0.

Therefore, $c_0(n,k) =\alpha _{\min }(n,k)$ is the constant we seek for. $\square$

We list some values of $c_0(n,k)$ for some specific k, n, with precision $\delta =0.01$:

$$\begin{aligned} \begin{aligned}&c_0(3,1)=3.64\ldots ,~ c_0(4,1)=2.93\ldots ,~ c_0(5,1)=2.56\ldots ,~ c_0(6,1)=2.33\ldots ,\\&c_0(7,1)=2.17\ldots ,~ c_0(8,1)=2.05\ldots ,~ c_0(9,1)=1.96\ldots ,~ c_0(10,1)=1.89\ldots ,\\&c_0(11,1)=1.83\ldots ,~ c_0(12,1)=1.78\ldots ,~ c_0(3,2)= c_0(4,2)=1. \end{aligned} \end{aligned}$$

(A.3)

In the following, we give more details of how to apply Sturm’s theorem by proving the following estimate for $c_0(n,1)$.

Proposition A.3

Let $p_0(x), \ldots , p_m(x)$ be the Sturm sequence given by (A.2) for the polynomial $p(x)=Q(x,1,n, \alpha )$ defined by (3.14) with regard to x, and let $\sigma (\xi )$ denote the number of sign changes (ignoring zeroes) in the sequence

$$\begin{aligned} \{p_0(\xi ),p_1(\xi ),\ldots ,p_m(\xi )\}. \end{aligned}$$

If $n > 12$ and $\alpha =1+\frac{7}{n}$, then $\sigma (0)- \sigma (\infty ) = 0$, which means that $Q(x,1,n, \alpha )$ has no root on $(0,\infty )$ if $\alpha =1+\frac{7}{n}$. Consequently, we obtain that $Q(x,1,n, 1+\frac{7}{n}) < 0$ for all $x\in (0,\infty )$ and $n>12$, since $Q<0$ when $x=0$. If $3\le n\le 12$, one can easily check that $c_0(n,1)\ge 1+\frac{7}{n}$. Therefore, we have $c_0(n,1)\ge 1+\frac{7}{n}$.

Proof

When $n > 12$ and $\alpha =1+\frac{7}{n}$, we can divide the standard Sturm sequence of $Q(x,1,n, 1+\frac{7}{n})$ by some suitable positive functions to obtain a simpler Sturm sequence, as Sturm’s theorem only concerns the sign of each term in the Sturm sequence. We still denote the simpler Sturm sequence by

$$\begin{aligned} \{p_0(x),p_1(x),\ldots ,p_m(x)\}. \end{aligned}$$

Although the expression of $p_i(x)$ might be very complicated, we only need to know the signs of $p_i(0)$ and $p_i(\infty )$, that is, the signs of the zero-order terms and the coefficients of the highest order terms of $p_i(x)$. We divide the zero-order terms and the highest order terms of $p_i(x)$ by some suitable positive functions, and denote the remaining terms by $Z_i$ and $I_i$, respectively. Therefore, $Z_i$ has the same sign as the zero-order term of $p_i(x)$ and $I_i$ has the same sign as the coefficient of the highest order term of $p_i(x)$. We have

$$\begin{aligned} \begin{aligned} Z_0=&7 - 19 n + 15 n^2 - n^3 - 2 n^4,~Z_1=1,\\ Z_2=&2744 n - 12348 n^2 + 18172 n^3 - 8453 n^4 - 1794 n^5 + 1535 n^6 + 144 n^7,\\ Z_3=&16672544 n - 60658864 n^2 + 78969576 n^3 - 38201184 n^4 - 2317896 n^5 \\&+ 6372732 n^6 - 576295 n^7 - 270278 n^8 + 6081 n^9 + 3584 n^{10},\\ Z_4=&2529924096 - 11497601472 n + 20565314112 n^2 - 18321051392 n^3 + 8896937056 n^4\\&- 2856559664 n^5 + 619264184 n^6 + 142496512 n^7 - 32213404 n^8- 29801085 n^9 \\&- 21819236 n^{10} + 4720867 n^{11} + 725886 n^{12} - 296460 n^{13} - 40000 n^{14},\\ Z_5=&-165288374272 + 822014504192 n - 1439948464640 n^2 + 635221775104 n^3 \\&+ 1099756498624 n^4 - 1442680610560 n^5 + 317014594400 n^6 + 331846621568 n^7\\&- 142328426016 n^8 - 34283941676 n^9 + 15836869820 n^{10} + 3675343420 n^{11}\\&- 667512847 n^{12} - 214699395 n^{13} + 31704806 n^{14} + 13123168 n^{15} + 994304 n^{16},\\ Z_6=&330576748544 - 422075670016 n - 593003666752 n^2 + 717369012864 n^3 \\&+ 326600077888 n^4 - 375952652096 n^5 - 61529189456 n^6 \\&+ 54892083792 n^7 + 9999889760 n^8 - 2733091200 n^9 - 549429077 n^{10}\\&+ 149191472 n^{11} + 43911424 n^{12} + 2985984 n^{13}= I_6,\\ I_0=&-1, ~I_1=-1,~ I_2=-192080 + 263424 n - 60368 n^2 - 13272 n^3 - 53 n^4 + 144 n^5, \\ I_3=&90354432 - 180708864 n + 96693072 n^2 - 9686320 n^3 + 6803552 n^4 - 3159968 n^5 \\&+ 71104 n^6 + 240196 n^7 + 8186 n^8 - 2400 n^9,\\ I_4=&2529924096 - 16557449664 n + 37348649856 n^2 - 36713556544 n^3 + 13721909824 n^4\\&- 526931104 n^5 + 1003345392 n^6 - 687439728 n^7 - 136384936 n^8 + 21404941 n^9 \\&+ 7869536 n^{10} - 1740356 n^{11} - 780868 n^{12} - 68800 n^{13},\\ I_5=&165288374272 - 822014504192 n + 1463561089536 n^2 - 741689414144 n^3 \\&- 920500835072 n^4 + 1357950741952 n^5 - 451152740416 n^6 - 132175466304 n^7 \\&+ 73039703968 n^8 + 20211061820 n^9 - 9966788912 n^{10} - 1997678860 n^{11} \\&+392467304 n^{12} + 110406478 n^{13} - 19057514 n^{14} - 7572032 n^{15} - 594432 n^{16}. \end{aligned} \end{aligned}$$

Note that for sufficiently large n, the signs of $\{Z_i\}$ are $-,+,+,+,-,+,+,$ and the signs of $\{I_i\}$ are $-,-,+,-,-,-,+.$ This implies that $\sigma (0)= \sigma (\infty )=3$. So it remains to show that for any $i\in \{0,1,\ldots ,6\}$, all real roots of $Z_i$ and $I_i$ are not greater than 12. Since $Z_i$ and $I_i$ are unary polynomials, this can be proved directly by applying Sturm’s theorem. For example, we show how to apply Sturm’s theorem to $I_2$:

$$\begin{aligned} I_2=-192080 + 263424 n - 60368 n^2 - 13272 n^3 - 53 n^4 + 144 n^5. \end{aligned}$$

First, we can obtain a Sturm sequence by using Euclid’s algorithm (A.2) and removing some positive coefficients to get a simpler Sturm sequence $\{q_i\}$. Then we have

$$\begin{aligned} \begin{aligned} q_0(n)=&-192080 + 263424 n - 60368 n^2 - 13272 n^3 - 53 n^4 + 144 n^5, \\ q_1(n)=&263424 - 120736 n - 39816 n^2 - 212 n^3 + 720 n^4, \\ q_2(n)=&169381632 - 188065528 n + 33126282 n^2 + 4780729 n^3, \\ q_3(n)=&-14501462505796+11364288885852 n-795070863791 n^2, \\ q_4(n)=&11296812839226538 - 7895204048274613 n,\\ q_5(n)=&-1. \end{aligned} \end{aligned}$$

(A.4)

Therefore the signs of $\{q_i(12)\}$ are $+,+,+,+,-,-,$ and the signs of $\{q_i(\infty )\}$ are $+,+,+,-,$

$-,-.$ This implies that $\sigma (12)= \sigma (\infty )=1$. By applying Sturm’s theorem, $I_2$ has no real root greater than 12. In a similar way, we obtain that all real roots of $Z_i$ and $I_i$ are not greater than 12. Hence, we obtain that the difference of the numbers of sign changes $\sigma (0)- \sigma (\infty )$ equals 0 for $Q(x,1,n, 1+\frac{7}{n})$ regarded as a polynomial of x, hence we obtain that $Q(x,1,n, 1+\frac{7}{n}) < 0$ for all $x\in (0,\infty )$, since $Q<0$ when $x=0$. $\square$

When $k\ge 2$, we have the following estimate for $c_0(n,k)$.

Proposition A.4

If $k\ge 2$, $n\ge k^2$, then we have $c_0(n,k) \ge \frac{1}{k}+\frac{k}{(k-1)n}$.

Proof

First, from Proposition A.1, when $n=k^2$, we have $c_0(n,k) =\frac{1}{k-1}= \frac{1}{k}+\frac{k}{(k-1)n}$. Second, we can use the method of bisection as described in the proof of Proposition A.2 to estimate $c_0(n,2)$ and $c_0(n,3)$, and obtain that $c_0(n,2)\ge \frac{1}{2}+\frac{2}{n}$, $\forall ~n\in [4,44]$; $c_0(n,3)\ge \frac{1}{3}+\frac{3}{2n}$, $\forall ~n\in [9,15]$. In the remaining cases, i.e., $k=2, n>44$, or $k=3,~n>15$, or $k\ge 4, ~n\ge k^2+1$, in order to prove Proposition A.4, we only need to show that

$$\begin{aligned} Q\left( x,k,n,\frac{1}{k}+\frac{k}{(k-1)n}\right) \le 0 \quad \forall {x}>0. \end{aligned}$$

For convenience, we reduce $Q(x,k,n,\frac{1}{k}+\frac{k}{(k-1)n})$ to a simpler polynomial:

$$\begin{aligned} Q\left( x,k,n,\frac{1}{k}+\frac{k}{(k-1)n}\right) =\frac{1}{n^2(k-1)^2} \sum _{i=0}^{6}a_i x^i, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} a_0=&-n(n-k)^3(k-1)(2n(k-1)+k^2),\\ a_1=&-n(n-k)^2(k-1)(n(5k^2-12k+6)+k^2(4k-6)),\\ a_2=&-6n^4(k-1)^2+n^3(k-1)(-3k^3+22k^2-30k+6)\\&+n^2 k (-3k^4+9k^3+11k^2-21k+6)+n k^3(6k^3-29k^2+26k-9)+k^5(k+3), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} a_3=&-2n^3(k-1)(5k^2-12k+6)+n^2(k^5-5k^4-17k^3+37k^2-22k+2)\\&+n k^2(-4k^4+37k^3-45k^2+32k-4)+2k^4(-2k^2-5k+1),\\ a_4=&-6n^3(k-1)^2+n^2(15k^3-37k^2+30k-6)\\&+n k^2(k^4-22k^3+37k^2-42k+12)+6k^4(k^2+2k-1),\\ a_5=&(n-k^2)(n(-5k^3+17k^2-18k+6)+4k^4+6k^3-6k^2),\\ a_6=&-(k-1) (n-k^2) ((k+2) k^2+2 (k-1) n). \end{aligned} \end{aligned}$$

When $k\ge 2$, we have $5k^2-12k+6>0$, and it is obvious that $a_0<0$, $a_1< 0$ and $a_6< 0$ for $n\ge k^2+1$. Fix any $j\in \{2,3,4\}$, we regard $a_j$ as a polynomial of n, and we have that $a_j \rightarrow -\infty$ as $n \rightarrow \infty$. If we set $n=k^2+1$, then $a_j$ is a unary polynomial of k, and we can prove that $a_j|_{n=k^2+1}< 0$ for $k\ge 2$ directly by applying Sturm’s theorem. Finally, we can apply Sturm’s Theorem to show that $a_j$ (with regard to n) has no roots in $[k^2+1,\infty )$, hence $a_j< 0$, if $j\in \{2,3,4\}$, $k\ge 2$, and $n\ge k^2+1$. For $a_5$, we need to discuss three cases.

• (i) If $k=2$, then $a_5=-2(n-4)(n-44)$. Hence, if $n>44$, then $a_5< 0$.

• (ii) If $k=3$, then $a_5=-6(n-9)(5n-72)$. Hence, if $n>15$, then $a_5< 0$.

• (iii) If $k\ge 4$, since $n\ge k^2+1$, we have

  $$\begin{aligned} \begin{aligned} a_5&\le (n-k^2)((k^2+1)(-5k^3+17k^2-18k+6)+4k^4+6k^3-6k^2)\\&=-(n-k^2)(k(k-4)+2)(5k^3-k^2+3k-3)<0. \end{aligned} \end{aligned}$$

Therefore, we have proved that all the coefficients $a_0,\ldots ,a_6$ are non-positive when $k=2, n>44$, or $k=3,~n>15$, or $k\ge 4, ~n\ge k^2+1$. Consequently, we obtain that

$$\begin{aligned} Q\left( x,k,n,\frac{1}{k}+\frac{k}{(k-1)n}\right) \le 0\quad \forall {x}>0, \end{aligned}$$

when $k=2, n>44$, or $k=3,~n>15$, or $k\ge 4, ~n\ge k^2+1$. This completes the proof of Proposition A.4. $\square$