5
$\begingroup$

Updated:

A polynomial $P(x,y)\in \mathbb{R}[x,y]$ is called an elliptic polynomial if its last homogeneous part does not vanish on $\mathbb{R}^2\setminus\{0\}$.The two answers to this post provide a proof for the following theorem:

Theorem: If $p$ is an elliptic polynomial whose last homogeneous part is positive definitive, then for $c$ sufficiently large , $p^{-1}(c)$ is a simple closed curve. Moreover if the centroid of interior of $p^{-1}(c)$ is denoted by $e_c$ then $e_c$ is convergent as $c$ goes to $+\infty$. The limit $\lim_{c\to \infty} e_c$ can be written in terms of coefficients of $p$. If we drop the ellipticity condition then this convergence result is not necessarily true.

The previous version of the post:

Is there a polynomial function $P:\mathbb{R}^2 \to \mathbb{R}$ with the following property?

For sufficiently large $c>0$, $P^{-1}(c)$ is a simple closed curve $\gamma_c$, homeomorphic to $S^1$, but as $c$ goes to $+\infty$. the centroid $e_c$ of the interior of $\gamma_c$ does not converge to any point of $\mathbb{R}^2$.

Motivation: The answer is negative if we consider this question for polynomials $p:\mathbb{R} \to \mathbb{R}$ whose eventual level sets are $2$-pointed set, i.e. $S^0$.(Namely a polynomial of even degree). The motivation comes from line -3, item III, page 4 of Taghavi - On periodic solutions of Liénard equations, which can be generalized to every even degree polynomial with one variable.

$\endgroup$

2 Answers 2

7
+800
$\begingroup$

Concerning homogeneous polynomials: Let $P(x,y)=\sum_{j=0}^n a_j x^j y^{n-j}$ be such a polynomial, of degree $n$ such that $C:=P^{-1}(\{c\})$ is a simple closed curve for all large enough $c>0$.

If $n$ is odd, then every line through the origin will have at most one point of intersection with $C$. So, then $C$ cannot be a simple closed curve for any real $c$ -- because every line through any point interior to a simple closed curve must intersect the curve in at least two points.

It remains to consider the case when $n$ is even. Then $C$ is symmetric about the origin, and hence so is the interior of $C$. Then the centroid of the interior is the origin, and it does not depend on the level $c$.


Consider now the case of an elliptic polynomial \begin{equation*} P(z)=P(x,y)=\sum_{j=0}^n a_j x^j y^{n-j}+\sum_{j=0}^{n-1}b_j x^j y^{n-1-j}+K|z|^{n-2} \end{equation*} of (necessarily even) degree $n$, where $z:=(x,y)$ and $K=O(1)$ (as $|z|\to\infty$). The ellipticity here is understood as the following condition: \begin{equation*} \min_{|z|=1}\sum_{j=0}^n a_j x^j y^{n-j}>0. \end{equation*}

For any $d_*\in(0,1)$ and any real $D>0$, let $\mathcal P_{n,d_*,D}$ denote the set of all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ such that $d_n\ge d_*$ and $\sum_{j=0}^n|d_j|\le D$. Then it is not hard to see that there is a real $c_*(n,d_*,D)>0$, depending only on $n,d_*,D$, such that for any polynomial $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ and for all real $c\ge c_*(n,d_*,D)$ the equation $p(x)=c$ has exactly two roots $x_\pm:=x_\pm(c)$ such that $x_-<0<x_+$ and, moreover, \begin{equation*} x_\pm=\pm\Big(\frac c{d_n}\Big)^{1/n}-(1+o(1))\frac{d_{n-1}}{nd_n} \tag{1} \end{equation*} uniformly over all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$; here and in the sequel the asymptotic relations are for $$c\to\infty,$$ unless otherwise specified. This uniformity can be obtained by refining this reasoning.
Moreover, without loss of generality (wlog), \begin{equation*} \text{for all $p\in\mathcal P_{n,d_*,D}$ and all real $c\ge c_*(n,d_*,D)$ we have $p'(x_\pm)\ne0$.} \tag{1.5} \end{equation*} Indeed, because (1) holds uniformly over all $p\in\mathcal P_{n,d_*,D}$, wlog \begin{equation*} |x_\pm|\ge\Big(\frac cD\Big)^{1/n}-2\frac D{nd_*}\to\infty, \tag{1.6} \end{equation*} so that $|x_\pm|\to\infty$ uniformly over all $p\in\mathcal P_{n,d_*,D}$. Also, taking any polynomial $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ and writing $p'(x)=\sum_{j=1}^n d_j jx^{j-1}$, we see that for $|x|\ge1$ \begin{equation*} \frac{|p'(x)|}{|x|^{n-1}}\ge nd_n-\sum_{j=1}^{n-1} |d_j| j|x|^{j-n} \ge nd_*-n D |x|^{-1}\underset{x\to\infty}\longrightarrow nd_*>0. \end{equation*} So, by (1.6), wlog (1.5) holds indeed.

Let us now turn back to the elliptic polynomial $P(x,y)$. For each real $t$ consider the polynomial \begin{equation*} p_t(r):=P(r\cos t,r\sin t). \end{equation*} By the ellipticity of the polynomial $P(x,y)$, there exist $d_*\in(0,1)$ and a real $D>0$ such that $p_t\in\mathcal P_{n,d_*,D}$ for all real $t$. Take now any real $c\ge c_*(n,d_*,D)$. Then, by the paragraph right above, for each real $t$ the equation $p_t(r)=c$ has exactly two roots $r_\pm(t):=r_\pm(c;t)$ such that $r_-(t)<0<r_+(t)$ and, moreover,
\begin{equation*} r_\pm(t)=\pm\Big(\frac c{a(t)}\Big)^{1/n}-(1+o(1))\frac{b(t)}{na(t)} \end{equation*} uniformly in real $t$, where \begin{equation*} a(t):=\sum_{j=0}^n a_j \cos^jt\, \sin^{n-j}t,\quad b(t):=\sum_{j=0}^{n-1} b_j \cos^jt\, \sin^{n-1-j}t. \end{equation*} Moreover, $\frac{dp_t(r)}{dr}|_{r=r_\pm(t)}\ne0$. So, by the implicit function theorem, the functions $r_\pm$ are continuous (in fact, infinitely smooth). Also, the functions $r_\pm$ are periodic with period $2\pi$, since for each real $t$ we have $p_{t+2\pi}=p_t$ and the values $r_\pm(t)$ of the the functions $r_\pm$ at $t$ are uniquely determined by the polynomial $p_t$. Furthermore, for all real $r$ and $t$ we have $p_{t+\pi}(r)=p_t(-r)$, which implies $r_+(t+\pi)=-r_-(t)$. So, letting $$z_\pm(t):=r_\pm(t)(\cos t,\sin t),$$ we see that $z_\pm(t+2\pi)=z_\pm(t)$ and $z_+(t)=z_-(t-\pi)$ for all real $t$. So, the $c$-level curve of $P(x,y)$ is \begin{align*} C=P^{-1}(\{c\})&=\{z_+(t)\colon t\in\mathbb R\}\cup\{z_-(t)\colon t\in\mathbb R\} \\ &=\{z_+(t)\colon t\in[0,2\pi)\}\cup\{z_-(t)\colon t\in[0,2\pi)\} \\ &=\{z_+(t)\colon t\in[0,2\pi)\} \\ &=\{z_+(t)\colon t\in[0,\pi)\}\cup\{z_-(t-\pi)\colon t\in[\pi,2\pi)\} \\ &=\{z(t)\colon t\in[0,2\pi)\}, \end{align*} where \begin{equation*} z(t):=R(t)(\cos t,\sin t), \quad R(t):= \begin{cases} r_+(t)>0&\text{ for }t\in[0,\pi],\\ |r_-(t-\pi)|>0&\text{ for }t\in[\pi,2\pi]. \end{cases} \end{equation*} So, the level curve $C$ is closed and simple, and its interior is \begin{equation*} I(c):=\{r\,(\cos t,\sin t)\colon0\le r<R(t)\}. \end{equation*}

The main idea for the elliptic polynomial case is to consider, for all real $c\ge c_*(n,d_*,D)$, the two opposite infinitesimal sectors of the interior $I(c)$ of the simple closed curve $C=P^{-1}(\{c\})$ between the rays $t$ and $t+dt$ and between the rays $t+\pi$ and $t+\pi+dt$, where $t$ is the polar angle in the interval $[0,\pi)$. The centroid of the union of these two sectors of $I(c)$ is at (signed) distance \begin{equation*} d(t)\sim \frac23\,\Big(r_+(t)\frac{|r_+(t)|^2}{|r_+(t)|^2+|r_-(t)|^2} +r_-(t)\frac{|r_-(t)|^2}{|r_+(t)|^2+|r_-(t)|^2}\Big) \tag{2} \end{equation*} from the origin. Formula (2) follows because (i) the centroid of an infinitesimal sector of radius $r>0$ between the rays $t$ and $t+dt$ is at distance $\frac23\,r$ from the origin, (ii) the area of this sector is $\frac12\,r^2\,dt$, and (iii) the centroid of the union of the two sectors is the weighted average of the centroids of the two sectors, with weights adding to $1$ and proportional to the areas of the sectors, and thus proportional to the squared radii of the sectors.

Simplifying (2), we get
\begin{equation*} d(t)\sim-\frac{2b(t)}{na(t)}. \end{equation*} Averaging now over all the pairs of opposite infinitesimal sectors, we see that the centroid converges to \begin{align*} &-\int_0^\pi dt\,\frac{2b(t)}{na(t)}(\cos t,\sin t)\frac12\,\Big(\frac c{a(t)}\Big)^{2/n} \Big/\int_0^\pi dt\,\frac12\,\Big(\frac c{a(t)}\Big)^{2/n} \\ &=-\int_0^\pi dt\,\frac{2b(t)}{na(t)}(\cos t,\sin t)\Big(\frac1{a(t)}\Big)^{2/n} \Big/\int_0^\pi dt\,\Big(\frac1{a(t)}\Big)^{2/n}. \tag{3} \end{align*}


I have checked this result numerically for $P(x,y)=x^4 + y^4 + 3 (x - y)^4 + y^3 + x y^2 + 10 x^2$, getting the centroid $\approx(-0.182846, -0.245149)$ for $c=10^4$ and $\approx(-0.189242,-0.25)$ for the limit (as $c\to\infty$) given by (3). From the above reasoning, one can see that the distance of the centroid from its limit is $O(1/c^{1/n})$; so, the agreement in this numerical example should be considered good, better than expected.


One may also note that in general the level sets $P^{−1}([0,c])$ will not be convex, even if $P$ is a positive elliptic homogeneous polynomial. E.g., take $P(x,y)=(x−y)^2(x+y)^2+h(x^4+y^4)$ for a small enough $h>0$. Here is the picture of this level set for $c=1$ and $h=1/10$:

enter image description here

Clearly, the shape of this level set does not depend on $c>0$.

This non-convexity idea can be generalized, with $$P(x,y)=P_{k,h}(x,y) :=\prod_{j=0}^{2k-1}\Big(x\cos\frac{\pi j}k-y\sin\frac{\pi j}k\Big)^2+h(x^{4k}+y^{4k})$$ for natural $k$ and real $h>0$. Here is the picture of the curve $P_{k,h}^{-1}(\{1\})$ for $k=5$ and $h=(3/10)^{4k}$:

enter image description here

$\endgroup$
26
  • 1
    $\begingroup$ I have added the case of elliptic polynomials. $\endgroup$ Feb 28, 2020 at 20:38
  • 1
    $\begingroup$ @AliTaghavi : The convexity is not needed because we now have the uniformity: (1) holds uniformly over all polynomials in $\mathcal P_{n,d_*,D}$ if $c\ge c_*(n,d_*,D)$. So, if $c>0$ is large enough, then for all $t\in[0,2\pi)$ at once the polynomials $P(r\cos t,r\sin t)-c$ in $r$ have exactly two roots, $r_\pm(t)$, satisfying condition (2a). I have now inserted the previously missing qualification "for all large enough $c>0$" into the sentence "The main idea for the elliptic polynomial case is ...". $\endgroup$ Mar 5, 2020 at 3:31
  • 1
    $\begingroup$ @AliTaghavi : I have added details on why the $c$-level curve for an elliptic polynomial is necessarily simple and closed for all large enough $c>0$. I have not stated or used any convexity properties. $\endgroup$ Mar 8, 2020 at 3:12
  • 1
    $\begingroup$ @AliTaghavi : I am glad you liked the answer. Please let me know if more clarifications are needed in some places. $\endgroup$ Mar 24, 2020 at 0:05
  • 1
    $\begingroup$ @AliTaghavi : I have now added details on why without loss of generality $p'(x_\pm)\ne0$ for all $p\in\mathcal P_{n,d_*,D}$ and all real $c\ge c_*(n,d_*,D)$ -- this statement is now formula (1.5). $\endgroup$ Mar 24, 2020 at 21:31
5
+150
$\begingroup$

$(y-x^2)^2+x^2\phantom{aaaaaaaaaaaaaaaaaaaaa}$

$\endgroup$
6
  • $\begingroup$ Thanks for your answer. Is there a homogenous example or a polynomial whose last homogenous part os elliptic(non degenerate)? $\endgroup$ Feb 27, 2020 at 13:29
  • $\begingroup$ In case of convergence of centroid to a point q, how can one write q in terms of coefficoent of the polynomial p?(as in one variable). BTW it seems that the elliptic assumption is a reasonable generalization of 1 variable case. $\endgroup$ Feb 27, 2020 at 15:27
  • $\begingroup$ You actually send the centroid to infinity along the y axis. I think your example is based on choosing a volum preserving diffeomorphism as change of cordinate. But is there really a homogenuos or elliptic example? $\endgroup$ Feb 27, 2020 at 17:26
  • $\begingroup$ Apart from above questions, is there an example for which the centroid is bounded but is not convergent?(very irregular behaviour)? $\endgroup$ Feb 27, 2020 at 17:50
  • $\begingroup$ To be honest, before your answer I was interested in a kind of non degeneracy conditioñs on the last homogeneous part please see the comment conversations on this post $\endgroup$ Feb 27, 2020 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.