# Geometry of Level sets of elliptic polynomials in two real variables

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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.

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 $$x_-<0 and, moreover, $$\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 $$r_-(t)<0 and, moreover,
$$\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

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 $$h=1/10$$:

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 $$h=(3/10)^{4k}$$:

• I have added the case of elliptic polynomials. – Iosif Pinelis Feb 28 at 20:38
• @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 ...". – Iosif Pinelis Mar 5 at 3:31
• @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. – Iosif Pinelis Mar 8 at 3:12
• @AliTaghavi : I am glad you liked the answer. Please let me know if more clarifications are needed in some places. – Iosif Pinelis Mar 24 at 0:05
• @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). – Iosif Pinelis Mar 24 at 21:31

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

• Thanks for your answer. Is there a homogenous example or a polynomial whose last homogenous part os elliptic(non degenerate)? – Ali Taghavi Feb 27 at 13:29
• 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. – Ali Taghavi Feb 27 at 15:27
• 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? – Ali Taghavi Feb 27 at 17:26
• Apart from above questions, is there an example for which the centroid is bounded but is not convergent?(very irregular behaviour)? – Ali Taghavi Feb 27 at 17:50
• 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 – Ali Taghavi Feb 27 at 17:59