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On the invariant sphere theorem with emphasis in the round sphere $\mathbb S^n$

Part of: General theory in ordinary differential equations Qualitative theory

Published online by Cambridge University Press:  15 October 2026

Jaume Llibre Open the ORCID record for Jaume Llibre [Opens in a new window]  and
Luis Fernando Mello Open the ORCID record for Luis Fernando Mello [Opens in a new window]
Jaume Llibre
Affiliation:
Universitat Autònoma de Barcelona , Spain e-mail: jaume.llibre@uab.cat
Luis Fernando Mello*
Affiliation:
Universidade Federal de Itajubá , Brazil
*
e-mail: lfmelo@unifei.edu.br
Rights & Permissions [Opens in a new window]

Abstract

This article discusses vector fields in $\mathbb R^n$, $n \geq 3$, that can be written as the sum of a radial repelling vector field and an attracting nonlinear homogeneous vector field of odd degree. The nontrivial dynamics for such vector fields is restricted to an invariant topological sphere. A necessary and sufficient condition for this invariant topological sphere to be the unit sphere centered at the origin of coordinates is given. Examples exhibiting distinct dynamics on these invariant spheres are presented.

Keywords

Polynomial vector fieldlimit cycleinvariant sethomogeneous nonlinearity

MSC classification

Primary: 34C25: Periodic solutions 34A26: Geometric methods in differential equations 34A34: Nonlinear equations and systems, general

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Type
Article
Information
Canadian Mathematical Bulletin , First View , pp. 1 - 12
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Copyright
© The Author(s), 2026. Published by Cambridge University Press on behalf of Canadian Mathematical Society

1 Introduction and statement of the main results

The theorem below is known as invariant sphere theorem and can be found in [Reference Field3, p. 388].

Theorem 1.1 Let U be an n-dimensional real vector space, $n \geq 2$ n greater than or equals 2 , and $Q: U \to U$ upper Q colon upper U right arrow upper U be a homogeneous polynomial map of degree $2 d + 1 (d> 0)$ 2 d plus 1 left parenthesis d greater than 0 right parenthesis . Assume that, for some choice $(\cdot , \cdot )$ left parenthesis dot comma dot right parenthesis of positive definite inner product on U, we have

$$ \begin{align*} (Q(x), x) < 0, \quad \text{all} \quad x \neq 0. \end{align*} $$ left parenthesis upper Q left parenthesis x right parenthesis comma x right parenthesis less than 0 comma all x not equals 0 period

For $\mu \in \mathbb R$ mu element of double struck upper R , consider the differential equation

(1.1) $$ \begin{align} x' = F(x) = \mu \, x + Q(x). \end{align} $$ x prime equals upper F left parenthesis x right parenthesis equals mu x plus upper Q left parenthesis x right parenthesis period

Then, for every $\mu> 0$ mu greater than 0 , there exists a unique $(n-1)$ left parenthesis n minus 1 right parenthesis -dimensional sphere $S(\mu ) \subset U \setminus \{0\}$ upper S left parenthesis mu right parenthesis subset of upper U minus StartSet 0 EndSet which is invariant by the flow of (1.1). Furthermore, $S(\mu )$ upper S left parenthesis mu right parenthesis is embedded as a topological submanifold of U and the bounded component of the complement of $S(\mu )$ upper S left parenthesis mu right parenthesis contains the origin of U. Finally, $S(\mu )$ upper S left parenthesis mu right parenthesis is globally attracting in the sense that every trajectory of (1.1) through $x \neq 0$ x not equals 0 is forward asymptotic to $S(\mu )$ upper S left parenthesis mu right parenthesis .

The following remarks are immediate consequences of Theorem 1.1.

  1. (i) As the topological sphere $S(\mu )$ upper S left parenthesis mu right parenthesis is invariant and globally attracting, the nontrivial dynamics of (1.1) with $\mu> 0$ mu greater than 0 are contained in $S(\mu )$ upper S left parenthesis mu right parenthesis . In particular, all the invariant sets, equilibrium points different from the equilibrium point at the origin, and periodic orbits are contained in $S(\mu )$ upper S left parenthesis mu right parenthesis .

  2. (ii) Consider the linear involution $\phi : U \to U$ phi colon upper U right arrow upper U , $\phi (x) = - x$ phi left parenthesis x right parenthesis equals negative x . The set of fixed points of $\phi $ phi is $M = \{0\}$ upper M equals StartSet 0 EndSet , the origin of U. By a direct calculation it follows that $D \phi (F(x)) = F(\phi (x))$ upper D phi left parenthesis upper F left parenthesis x right parenthesis right parenthesis equals upper F left parenthesis phi left parenthesis x right parenthesis right parenthesis , for all $x \in U$ x element of upper U . So, the phase portrait of the vector field F in (1.1) is symmetric with respect to the origin.

  3. (iii) From the previous observation, in particular, the topological sphere $S(\mu )$ upper S left parenthesis mu right parenthesis is symmetric with respect to the origin.

  4. (iv) Through a scaling, one can always assume $\mu = 1$ mu equals 1 in equation (1.1). We will consider this condition throughout this article.

From this point on, we will specialize the inner product space $(U, (\cdot , \cdot ))$ left parenthesis upper U comma left parenthesis dot comma dot right parenthesis right parenthesis as the Euclidean space $\mathbb R^n$ double struck upper R Superscript n , that is, the real vector space $\mathbb R^n$ double struck upper R Superscript n equipped with the Euclidean structure induced by the standard Euclidean inner product $\left < \cdot , \cdot \right>$ left angle bracket dot comma dot right angle bracket . The unit sphere is defined by

$$\begin{align*}\mathbb S^{n-1} = \left\{ x = (x_1, \ldots, x_n) \in \mathbb R^n: \| x \|^2 = x_1^2 + \cdots + x_n^2 = 1 \right\}. \end{align*}$$ double struck upper S Superscript n minus 1 Baseline equals StartSet x equals left parenthesis x 1 comma ellipsis comma x Subscript n Baseline right parenthesis element of double struck upper R Superscript n Baseline colon StartMetric x EndMetric squared equals x 1 squared plus midline horizontal ellipsis plus x Subscript n Superscript 2 Baseline equals 1 EndSet period

The theorem below complements Theorem 1.1 by providing a necessary and sufficient condition for the topological sphere $S(\mu = 1)$ upper S left parenthesis mu equals 1 right parenthesis to be the unit sphere $\mathbb S^{n-1}$ double struck upper S Superscript n minus 1 .

Theorem 1.2 Consider the differential equation

(1.2) $$ \begin{align} x' = F(x) = x + Q(x), \end{align} $$ x prime equals upper F left parenthesis x right parenthesis equals x plus upper Q left parenthesis x right parenthesis comma

where $x \in \mathbb R^n$ x element of double struck upper R Superscript n , $n \geq 2$ n greater than or equals 2 , and $Q: \mathbb R^n \to \mathbb R^n$ upper Q colon double struck upper R Superscript n Baseline right arrow double struck upper R Superscript n is a homogeneous polynomial map of degree $m = 2 d + 1$ m equals 2 d plus 1 , $d> 0$ d greater than 0 . The differential equation (1.2) has the unit sphere $\mathbb S^{n-1}$ double struck upper S Superscript n minus 1 invariant if and only if

$$ \begin{align*} \left< x, Q(x) \right> = - \left(\| x \|^2 \right)^{d + 1}. \end{align*} $$ left angle bracket x comma upper Q left parenthesis x right parenthesis right angle bracket equals minus left parenthesis StartMetric x EndMetric squared right parenthesis Superscript d plus 1 Baseline period

From Theorem 1.1 it follows that, under the hypotheses of Theorem 1.2, the invariant sphere $\mathbb S^{n-1}$ double struck upper S Superscript n minus 1 is a global attractor except for the equilibrium point $x = 0$ x equals 0 .

We now restrict our attention to the case $n=3$ n equals 3 , that is, we study the differential equation (1.2) in the Euclidean space $\mathbb R^3$ double struck upper R cubed . The next theorem describes how the orbits of the differential equation (1.2) tend to the invariant sphere $\mathbb S^2$ double struck upper S squared .

Theorem 1.3 Consider the differential equation (1.2) in $\mathbb R^3$ double struck upper R cubed . Assume that the unit sphere $\mathbb S^2$ double struck upper S squared is invariant by the flow of the differential equation (1.2). Consider $\gamma (t) \in \mathbb S^2$ gamma left parenthesis t right parenthesis element of double struck upper S squared a solution of (1.2) with $t \in \mathbb R$ t element of double struck upper R . Define the ruled surface

$$\begin{align*}C_{\gamma} = \big\{ \lambda \, \gamma(t): \forall \, \lambda, t \in \mathbb R \big\}. \end{align*}$$ StartLayout 1st Row upper C Subscript gamma Baseline equals left brace lamda gamma left parenthesis t right parenthesis colon for all lamda comma t element of double struck upper R right brace period EndLayout

Then, $C_{\gamma }$ upper C Subscript gamma is invariant by the flow of (1.2).

The following corollary follows directly from the previous theorem.

Corollary 1.4 The following statements hold:

  1. (a) If $p \in \mathbb S^2$ p element of double struck upper S squared is an equilibrium point of the differential equation (1.2), then the straight line joining p to the origin is invariant by the flow of (1.2).

  2. (b) Consider the polynomial vector field F defined in (1.2). Then, the restriction of the Poincaré compactification of F to the infinite sphere is topologically equivalent to the restriction of F to the invariant sphere $\mathbb S^2$ double struck upper S squared .

Some results concerning the planar analog of the sphere theorem, namely, the existence of an invariant circle, can be found in [Reference Alarcón, Castro and Labouriau1] (see also [Reference Llibre and Zhang8]).

The class of polynomial differential systems whose nonlinearities are homogeneous polynomials of the same degree has been studied by several authors (see, for instance, [Reference Carbonell and Llibre2, Reference Gasull, Yu and Zhang4–Reference Llibre and Zhang7]).

The article is organized as follows. Theorems 1.2 and 1.3 are proved in Section 2. Section 3 presents some examples that illustrate the studies conducted in this article.

2 Proofs of Theorems 1.2 and 1.3

We begin with the proof of Theorems 1.2.

Proof Define $f: \mathbb R^n \to \mathbb R$ f colon double struck upper R Superscript n Baseline right arrow double struck upper R , $f(x) = \| x \|^2 - 1$ f left parenthesis x right parenthesis equals StartMetric x EndMetric squared minus 1 . Firstly, assume

$$\begin{align*}\left< x, Q(x) \right> = - \left(\| x \|^2 \right)^{d + 1}. \end{align*}$$ left angle bracket x comma upper Q left parenthesis x right parenthesis right angle bracket equals minus left parenthesis StartMetric x EndMetric squared right parenthesis Superscript d plus 1 Baseline period

Then,

$$ \begin{align*} \left< \nabla f(x), x + Q(x) \right> & = 2 \left( \left< x, x \right> + \left< x, Q(x) \right> \right) = 2 \left( \| x \|^2 - \left(\| x \|^2 \right)^{d + 1} \right) \\ & = - 2 \, \| x \|^2 \left( \left(\| x \|^2 \right)^d - 1 \right) \\ & = - 2 \, \| x \|^2 \left( \left( 1 + \| x\|^2 + \cdots + \left( \| x \|^2 \right)^{d - 1} \right) \left( \| x \|^2 - 1 \right) \right) \\ & = - 2 \, \| x \|^2 \left( 1 + \| x\|^2 + \cdots + \left( \| x \|^2 \right)^{d - 1} \right) \, f(x). \end{align*} $$ StartLayout 1st Row 1st Column left angle bracket nabla f left parenthesis x right parenthesis comma x plus upper Q left parenthesis x right parenthesis right angle bracket 2nd Column equals 2 left parenthesis left angle bracket x comma x right angle bracket plus left angle bracket x comma upper Q left parenthesis x right parenthesis right angle bracket right parenthesis equals 2 left parenthesis StartMetric x EndMetric squared minus left parenthesis StartMetric x EndMetric squared right parenthesis Superscript d plus 1 Baseline right parenthesis 2nd Row 1st Column Blank 2nd Column equals minus 2 StartMetric x EndMetric squared left parenthesis left parenthesis StartMetric x EndMetric squared right parenthesis Superscript d Baseline minus 1 right parenthesis 3rd Row 1st Column Blank 2nd Column equals minus 2 StartMetric x EndMetric squared left parenthesis left parenthesis 1 plus StartMetric x EndMetric squared plus midline horizontal ellipsis plus left parenthesis StartMetric x EndMetric squared right parenthesis Superscript d minus 1 Baseline right parenthesis left parenthesis StartMetric x EndMetric squared minus 1 right parenthesis right parenthesis 4th Row 1st Column Blank 2nd Column equals minus 2 StartMetric x EndMetric squared left parenthesis 1 plus StartMetric x EndMetric squared plus midline horizontal ellipsis plus left parenthesis StartMetric x EndMetric squared right parenthesis Superscript d minus 1 Baseline right parenthesis f left parenthesis x right parenthesis period EndLayout

Therefore, $\mathbb S^{n-1} = f^{-1}(\{0\})$ double struck upper S Superscript n minus 1 Baseline equals f Superscript negative 1 Baseline left parenthesis StartSet 0 EndSet right parenthesis is invariant by the flow of (1.2).

Now assume that the sphere $\mathbb S^{n-1} = f^{-1}(\{0\})$ double struck upper S Superscript n minus 1 Baseline equals f Superscript negative 1 Baseline left parenthesis StartSet 0 EndSet right parenthesis is invariant by the flow of (1.2). Consider the change of variables

$$\begin{align*}x = r \, u, \quad r = \| x \|, u \in \mathbb S^{n - 1}. \end{align*}$$ x equals r u comma r equals StartMetric x EndMetric comma u element of double struck upper S Superscript n minus 1 Baseline period

In these new coordinates, the differential equation (1.2) writes as

(2.1) $$ \begin{align} u' = r^{m - 1} \big( Q(u) - \left< u, Q(u) \right> u \big), \quad r' = r \, \big(1 + r^{m - 1} \left< u, Q(u) \right>\big). \end{align} $$ StartLayout 1st Row u prime equals r Superscript m minus 1 Baseline left parenthesis upper Q left parenthesis u right parenthesis minus left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket u right parenthesis comma r prime equals r left parenthesis 1 plus r Superscript m minus 1 Baseline left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket right parenthesis period EndLayout

Indeed, from (1.2) and the above change of variables, it follows that

(2.2) $$ \begin{align} r' \, u + r \, u' = r \, u + r^m Q(u). \end{align} $$ r prime u plus r u Superscript prime Baseline equals r u plus r Superscript m Baseline upper Q left parenthesis u right parenthesis period

Taking the inner product of both sides of the above equation with u, we obtain

$$ \begin{align*} \left<r' \, u + r \, u', u \right>& = r' \left< u, u \right> + r \left< u', u \right> = r', \\ \left<r \, u + r^m Q(u), u \right>& = r \left< u, u \right> + r^m \left< u, Q(u) \right> = r + r^m \left< u, Q(u) \right>. \end{align*} $$ StartLayout 1st Row 1st Column left angle bracket r prime u plus r u Superscript prime Baseline comma u right angle bracket 2nd Column equals r prime left angle bracket u comma u right angle bracket plus r left angle bracket u prime comma u right angle bracket equals r Superscript prime Baseline comma 2nd Row 1st Column left angle bracket r u plus r Superscript m Baseline upper Q left parenthesis u right parenthesis comma u right angle bracket 2nd Column equals r left angle bracket u comma u right angle bracket plus r Superscript m Baseline left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket equals r plus r Superscript m Baseline left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket period EndLayout

Therefore,

$$\begin{align*}r' = r + r^m \left< u, Q(u) \right> = r \, \big( 1 + r^{m - 1} \left< u, Q(u) \right> \big), \end{align*}$$ StartLayout 1st Row r prime equals r plus r Superscript m Baseline left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket equals r left parenthesis 1 plus r Superscript m minus 1 Baseline left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket right parenthesis comma EndLayout

which is the second equation of (2.1). Using this equation in (2.2), we get

$$ \begin{align*} r \, u' & = r \, u + r^m Q(u) - r' \, u = r \, u + r^m Q(u) - \left( r + r^m \left< u, Q(u) \right> \right) \, u \\ & = r^m Q(u) - r^m \left< u, Q(u) \right> u = r^m \left( Q(u) - \left< u, Q(u) \right> u \right), \end{align*} $$ StartLayout 1st Row 1st Column r u prime 2nd Column equals r u plus r Superscript m Baseline upper Q left parenthesis u right parenthesis minus r prime u equals r u plus r Superscript m Baseline upper Q left parenthesis u right parenthesis minus left parenthesis r plus r Superscript m Baseline left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket right parenthesis u 2nd Row 1st Column Blank 2nd Column equals r Superscript m Baseline upper Q left parenthesis u right parenthesis minus r Superscript m Baseline left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket u equals r Superscript m Baseline left parenthesis upper Q left parenthesis u right parenthesis minus left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket u right parenthesis comma EndLayout

from which we obtain the first equation of (2.1).

It follows from the hypothesis and the second equation in (2.1) that

(2.3) $$ \begin{align} \left< u, Q(u) \right> = -1, \quad \forall \, u \in \mathbb S^{n - 1}. \end{align} $$ left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket equals negative 1 comma for all u element of double struck upper S Superscript n minus 1 Baseline period

The expression on the left-hand side of (2.3) is a homogeneous polynomial function of degree $2 (d + 1)$ 2 left parenthesis d plus 1 right parenthesis , that is,

(2.4) $$ \begin{align} \left< u, Q(u) \right> = \sum_{i_1 + \cdots + i_n = 2 (d+1)} a_{i_1, \ldots, i_n} \, u_1^{i_1} \ldots u_n^{i_n}, \end{align} $$ left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket equals sigma summation Underscript i 1 plus midline horizontal ellipsis plus i Subscript n Baseline equals 2 left parenthesis d plus 1 right parenthesis Endscripts a Subscript i 1 comma ellipsis comma i Sub Subscript n Subscript Baseline u 1 Superscript i 1 Baseline ellipsis u Subscript n Superscript i Super Subscript n Superscript Baseline comma

with $u = (u_1, \ldots , u_n) \in \mathbb S^{n-1}$ u equals left parenthesis u 1 comma ellipsis comma u Subscript n Baseline right parenthesis element of double struck upper S Superscript n minus 1 and $a_{i_1, \ldots , i_n} \in \mathbb R$ a Subscript i 1 comma ellipsis comma i Sub Subscript n Subscript Baseline element of double struck upper R . Evaluating (2.4) at appropriate points, exploring the symmetries of $\mathbb S^{n-1}$ double struck upper S Superscript n minus 1 and using (2.3), it follows that

$$\begin{align*}\left< u, Q(u) \right> = - \big( u_1^2 + \cdots + u_n^2 \big)^{d + 1} = - (\| u \|^2)^{d + 1}, \end{align*}$$ left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket equals minus left parenthesis u 1 squared plus midline horizontal ellipsis plus u Subscript n Superscript 2 Baseline right parenthesis Superscript d plus 1 Baseline equals minus left parenthesis StartMetric u EndMetric squared right parenthesis Superscript d plus 1 Baseline comma

as we wanted to demonstrate.

We prove the previous equality when the homogeneous polynomial map Q has the degree $3$ 3 and the differential equation is in $\mathbb R^3$ double struck upper R cubed . The proof when the homogeneous polynomial map Q has an arbitrary order degree in $\mathbb R^n$ double struck upper R Superscript n is similar.

From equations (2.3) and (2.4), one has

$$\begin{align*}h(u_1, u_2, u_3) = \left< u, Q(u) \right> = \sum_{i + j + k = 4} a_{i, j, k} \, u_1^{i} \, u_2^{j} \, u_3^{k} = -1, \end{align*}$$ h left parenthesis u 1 comma u 2 comma u 3 right parenthesis equals left angle bracket u comma upper Q left parenthesis u right parenthesis right angle bracket equals sigma summation Underscript i plus j plus k equals 4 Endscripts a Subscript i comma j comma k Baseline u 1 Superscript i Baseline u 2 Superscript j Baseline u 3 Superscript k Baseline equals negative 1 comma

for all $u = (u_1, u_2, u_3) \in \mathbb S^2$ u equals left parenthesis u 1 comma u 2 comma u 3 right parenthesis element of double struck upper S squared . Evaluating h at $(1,0,0), (0,1,0), (0,0,1) \in \mathbb S^2$ left parenthesis 1 comma 0 comma 0 right parenthesis comma left parenthesis 0 comma 1 comma 0 right parenthesis comma left parenthesis 0 comma 0 comma 1 right parenthesis element of double struck upper S squared , we obtain

(2.5) $$ \begin{align} a_{4,0,0} = a_{0,4,0} = a_{0,0,4} = -1. \end{align} $$ a Subscript 4 comma 0 comma 0 Baseline equals a Subscript 0 comma 4 comma 0 Baseline equals a Subscript 0 comma 0 comma 4 Baseline equals negative 1 period

Evaluating h at the points $(\sqrt {3}/2, \pm 1/2, 0), (1/2, \pm \sqrt {3}/2, 0) \in \mathbb S^2$ left parenthesis StartRoot 3 EndRoot divided by 2 comma plus or minus 1 divided by 2 comma 0 right parenthesis comma left parenthesis 1 divided by 2 comma plus or minus StartRoot 3 EndRoot divided by 2 comma 0 right parenthesis element of double struck upper S squared , we obtain

(2.6) $$ \begin{align} a_{1,3,0} = a_{3,1,0} = 0. \end{align} $$ a Subscript 1 comma 3 comma 0 Baseline equals a Subscript 3 comma 1 comma 0 Baseline equals 0 period

Evaluating h at the points $(\sqrt {3}/2, 0, \pm 1/2), (1/2, 0, \pm \sqrt {3}/2) \in \mathbb S^2$ left parenthesis StartRoot 3 EndRoot divided by 2 comma 0 comma plus or minus 1 divided by 2 right parenthesis comma left parenthesis 1 divided by 2 comma 0 comma plus or minus StartRoot 3 EndRoot divided by 2 right parenthesis element of double struck upper S squared , we obtain

(2.7) $$ \begin{align} a_{1,0,3} = a_{3, 0,1} = 0. \end{align} $$ a Subscript 1 comma 0 comma 3 Baseline equals a Subscript 3 comma 0 comma 1 Baseline equals 0 period

Evaluating h at the points $(0, \sqrt {3}/2, \pm 1/2), (0, 1/2, \pm \sqrt {3}/2) \in \mathbb S^2$ left parenthesis 0 comma StartRoot 3 EndRoot divided by 2 comma plus or minus 1 divided by 2 right parenthesis comma left parenthesis 0 comma 1 divided by 2 comma plus or minus StartRoot 3 EndRoot divided by 2 right parenthesis element of double struck upper S squared , we obtain

(2.8) $$ \begin{align} a_{0,1 ,3} = a_{0, 3, 1} = 0. \end{align} $$ a Subscript 0 comma 1 comma 3 Baseline equals a Subscript 0 comma 3 comma 1 Baseline equals 0 period

Evaluating h at $(1/2, 1/2, \pm 1/\sqrt {2}), ( 1/4, 1/2, \pm \sqrt {11}/4), (\pm 1/\sqrt {2}, 1/2, 1/2) \in \mathbb S^2$ left parenthesis 1 divided by 2 comma 1 divided by 2 comma plus or minus 1 divided by StartRoot 2 EndRoot right parenthesis comma left parenthesis 1 divided by 4 comma 1 divided by 2 comma plus or minus StartRoot 11 EndRoot divided by 4 right parenthesis comma left parenthesis plus or minus 1 divided by StartRoot 2 EndRoot comma 1 divided by 2 comma 1 divided by 2 right parenthesis element of double struck upper S squared , we obtain

(2.9) $$ \begin{align} a_{1, 2, 1} = a_{2, 1, 1} = a_{1, 1, 2} = 0. \end{align} $$ a Subscript 1 comma 2 comma 1 Baseline equals a Subscript 2 comma 1 comma 1 Baseline equals a Subscript 1 comma 1 comma 2 Baseline equals 0 period

Finally, evaluating h at $(\sqrt {3}/2, 1/2, 0), ( 1/2, 0, \sqrt {3}/2), (0, 1/2, \sqrt {3}/2) \in \mathbb S^2$ left parenthesis StartRoot 3 EndRoot divided by 2 comma 1 divided by 2 comma 0 right parenthesis comma left parenthesis 1 divided by 2 comma 0 comma StartRoot 3 EndRoot divided by 2 right parenthesis comma left parenthesis 0 comma 1 divided by 2 comma StartRoot 3 EndRoot divided by 2 right parenthesis element of double struck upper S squared , we obtain

(2.10) $$ \begin{align} a_{2, 2, 0} = a_{2, 0, 2} = a_{0, 2, 2} = -2. \end{align} $$ a Subscript 2 comma 2 comma 0 Baseline equals a Subscript 2 comma 0 comma 2 Baseline equals a Subscript 0 comma 2 comma 2 Baseline equals negative 2 period

Collecting the coefficients obtained in (2.5)–(2.10), it follows that

$$ \begin{align*} h(u_1, u_2, u_3) & = -u_1^4 - 2 u_1^2 u_2^2 - u_2^4 - 2 u_1^2 u_3^2 - 2 u_2^2 u_3^2 - u_3^4 \\ & = -(u_1^2 + u_2^2 + u_3^2)^2 = -(\| u \|^2)^2.\\[-32pt] \end{align*} $$ StartLayout 1st Row 1st Column h left parenthesis u 1 comma u 2 comma u 3 right parenthesis 2nd Column equals minus u 1 Superscript 4 Baseline minus 2 u 1 squared u 2 squared minus u 2 Superscript 4 Baseline minus 2 u 1 squared u 3 squared minus 2 u 2 squared u 3 squared minus u 3 Superscript 4 Baseline 2nd Row 1st Column Blank 2nd Column equals minus left parenthesis u 1 squared plus u 2 squared plus u 3 squared right parenthesis squared equals minus left parenthesis StartMetric u EndMetric squared right parenthesis squared period EndLayout

We now present the proof of Theorem 1.3.

Proof Consider $p_0 = \gamma (t_0) \in \mathbb S^2$ p 0 equals gamma left parenthesis t 0 right parenthesis element of double struck upper S squared with $t_0 \in \mathbb R$ t 0 element of double struck upper R an arbitrary point. The tangent plane of the ruled surface $C_{\gamma }$ upper C Subscript gamma at $p_0$ p 0 , $T_{p_0} C_{\gamma }$ upper T Subscript p 0 Baseline upper C Subscript gamma , is generated by $\{ p_0, F(p_0) \}$ StartSet p 0 comma upper F left parenthesis p 0 right parenthesis EndSet . So, the normal vector to $T_{p_0} C_{\gamma }$ upper T Subscript p 0 Baseline upper C Subscript gamma is

$$\begin{align*}n(p_0) = p_0 \times F(p_0) = p_0 \times (p_0 + Q(p_0)) = p_0 \times Q(p_0), \end{align*}$$ StartLayout 1st Row n left parenthesis p 0 right parenthesis equals p 0 times upper F left parenthesis p 0 right parenthesis equals p 0 times left parenthesis p 0 plus upper Q left parenthesis p 0 right parenthesis right parenthesis equals p 0 times upper Q left parenthesis p 0 right parenthesis comma EndLayout

where $\times $ times denotes the cross product of two vectors in $\mathbb R^3$ double struck upper R cubed . It follows that $\left < n(p_0), F(p_0) \right> = 0$ left angle bracket n left parenthesis p 0 right parenthesis comma upper F left parenthesis p 0 right parenthesis right angle bracket equals 0 .

Consider $p \in C_{\gamma }$ p element of upper C Subscript gamma an arbitrary point. So, there exist $\lambda \neq 0$ lamda not equals 0 and $p_0 \in \mathbb S^2$ p 0 element of double struck upper S squared such that $p = \lambda \, p_0$ p equals lamda p 0 . As $C_{\gamma }$ upper C Subscript gamma is a ruled surface, it follows that $n(p)$ n left parenthesis p right parenthesis is parallel to $n(p_0)$ n left parenthesis p 0 right parenthesis . Thus,

$$ \begin{align*} \left< n(p_0), F(p) \right> & = \left< p_0 \times Q(p_0), \lambda \, p_0 + Q(\lambda \, p_0) \right> \\ & = \left< p_0 \times Q(p_0), \lambda \, p_0 \right> + \left< p_0 \times Q(p_0), \lambda^m \, Q(p_0) \right> = 0, \end{align*} $$ StartLayout 1st Row 1st Column left angle bracket n left parenthesis p 0 right parenthesis comma upper F left parenthesis p right parenthesis right angle bracket 2nd Column equals left angle bracket p 0 times upper Q left parenthesis p 0 right parenthesis comma lamda p 0 plus upper Q left parenthesis lamda p 0 right parenthesis right angle bracket 2nd Row 1st Column Blank 2nd Column equals left angle bracket p 0 times upper Q left parenthesis p 0 right parenthesis comma lamda p 0 right angle bracket plus left angle bracket p 0 times upper Q left parenthesis p 0 right parenthesis comma lamda Superscript m Baseline upper Q left parenthesis p 0 right parenthesis right angle bracket equals 0 comma EndLayout

that is, $n(p_0)$ n left parenthesis p 0 right parenthesis is orthogonal to $F(p)$ upper F left parenthesis p right parenthesis . Therefore, $n(p)$ n left parenthesis p right parenthesis is orthogonal to $F(p)$ upper F left parenthesis p right parenthesis , implying that $F(p)$ upper F left parenthesis p right parenthesis is tangent to the ruled surface $C_{\gamma }$ upper C Subscript gamma . As $p \in C_{\gamma }$ p element of upper C Subscript gamma is an arbitrary point, we conclude that $C_{\gamma }$ upper C Subscript gamma is invariant by the flow of F.

3 Some examples

In this section, we will present some examples exploring the results contained in Theorems 1.1 and 1.2.

Example 3.1 Consider the differential equation (1.2) with the homogeneous polynomial map of degree $3$ 3

(3.1) $$ \begin{align} Q: \mathbb R^3 \to \mathbb R^3, \quad Q(x, y, z) = \big( -x^3 -2 x z^2, -2 x^2 y -\!2 y z^2 -\!y^3, -z^3 \big). \end{align} $$ StartLayout 1st Row upper Q colon double struck upper R cubed right arrow double struck upper R cubed comma upper Q left parenthesis x comma y comma z right parenthesis equals left parenthesis minus x cubed minus 2 x z squared comma minus 2 x squared y minus 2 y z squared minus y cubed comma minus z cubed right parenthesis period EndLayout

As

$$\begin{align*}\left< (x, y, z), Q(x, y, z) \right> = - \left( x^2 + y^2 + z^2 \right)^2 < 0, \quad \forall (x, y, z) \neq (0, 0, 0), \end{align*}$$ left angle bracket left parenthesis x comma y comma z right parenthesis comma upper Q left parenthesis x comma y comma z right parenthesis right angle bracket equals minus left parenthesis x squared plus y squared plus z squared right parenthesis squared less than 0 comma for all left parenthesis x comma y comma z right parenthesis not equals left parenthesis 0 comma 0 comma 0 right parenthesis comma

it follows from Theorem 1.2 that the unit sphere $\mathbb S^2$ double struck upper S squared is invariant by the flow of (1.2).

Note that the coordinate planes $x = 0$ x equals 0 , $y = 0$ y equals 0 , and $z = 0$ z equals 0 are invariant. Consequently, the coordinate axes are invariant as well. Hence, there exist six equilibrium points on $\mathbb S^2$ double struck upper S squared

$$\begin{align*}\begin{array}{lll} p_1 = (1, 0, 0), & p_2 = (-1, 0, 0), & p_3 = (0, 1, 0), \\ p_4 = (0, -1, 0), & p_5 = (0, 0, 1), & p_6 = (0, 0, -1), \end{array} \end{align*}$$ StartLayout 1st Row 1st Column p 1 equals left parenthesis 1 comma 0 comma 0 right parenthesis comma 2nd Column p 2 equals left parenthesis negative 1 comma 0 comma 0 right parenthesis comma 3rd Column p 3 equals left parenthesis 0 comma 1 comma 0 right parenthesis comma 2nd Row 1st Column p 4 equals left parenthesis 0 comma negative 1 comma 0 right parenthesis comma 2nd Column p 5 equals left parenthesis 0 comma 0 comma 1 right parenthesis comma 3rd Column p 6 equals left parenthesis 0 comma 0 comma negative 1 right parenthesis comma EndLayout

and the equilibrium point $(0,0,0)$ left parenthesis 0 comma 0 comma 0 right parenthesis . By a direct calculation, the equilibrium points $p_1$ p 1 and $p_2$ p 2 are saddles, $p_3$ p 3 and $p_4$ p 4 are unstable nodes, and $p_5$ p 5 and $p_6$ p 6 are stable nodes for the restriction of the differential equation (1.2) on the invariant sphere $\mathbb S^2$ double struck upper S squared (see Figure 1).

Figure 1

Some orbits of the differential equation (1.2) with Q given in (3.1) on the unit invariant sphere $\mathbb S^2$ double struck upper S squared . The six equilibrium points are draw in red. For interpretation of the references to color in this figure legend, the reader is referred to the web version of the article.

A green 3D sphere with black orbital lines intersecting at right angles. Six red equilibrium points are located at the intersections on the top, bottom, front, back, left, and right poles. Arrows indicate flow direction along the paths.

Example 3.2 Consider the differential equation (1.2) with the homogeneous polynomial map of degree $3$ 3

(3.2) $$ \begin{align} Q: \mathbb R^3 \to \mathbb R^3, \quad Q(x, y, z) = \left( Q_1(x,y,z), Q_2(x,y,z), Q_3(x,y,z) \right), \end{align} $$ StartLayout 1st Row upper Q colon double struck upper R cubed right arrow double struck upper R cubed comma upper Q left parenthesis x comma y comma z right parenthesis equals left parenthesis upper Q 1 left parenthesis x comma y comma z right parenthesis comma upper Q 2 left parenthesis x comma y comma z right parenthesis comma upper Q 3 left parenthesis x comma y comma z right parenthesis right parenthesis comma EndLayout

where

$$ \begin{align*} Q_1(x,y,z) & = - x^3-\dfrac{3 x^2 y}{4}-2 x y^2-y^3+y z^2, \\ Q_2(x,y,z) & = \dfrac{3 x^3}{4}+x y^2-\dfrac{x z^2}{4}-y^3, \\ Q_3(x,y,z) & = -2 x^2 z-\dfrac{3 x y z}{4}-2 y^2 z-z^3. \end{align*} $$ StartLayout 1st Row 1st Column upper Q 1 left parenthesis x comma y comma z right parenthesis 2nd Column equals minus x cubed minus StartFraction 3 x squared y Over 4 EndFraction minus 2 x y squared minus y cubed plus y z squared comma 2nd Row 1st Column upper Q 2 left parenthesis x comma y comma z right parenthesis 2nd Column equals StartFraction 3 x cubed Over 4 EndFraction plus x y squared minus StartFraction x z squared Over 4 EndFraction minus y cubed comma 3rd Row 1st Column upper Q 3 left parenthesis x comma y comma z right parenthesis 2nd Column equals minus 2 x squared z minus StartFraction 3 x y z Over 4 EndFraction minus 2 y squared z minus z cubed period EndLayout

By a direct calculation, we get

$$\begin{align*}\left< (x, y, z), Q(x, y, z) \right> = - \left( x^2 + y^2 + z^2 \right)^2 < 0, \quad \forall (x, y, z) \neq (0, 0, 0). \end{align*}$$ left angle bracket left parenthesis x comma y comma z right parenthesis comma upper Q left parenthesis x comma y comma z right parenthesis right angle bracket equals minus left parenthesis x squared plus y squared plus z squared right parenthesis squared less than 0 comma for all left parenthesis x comma y comma z right parenthesis not equals left parenthesis 0 comma 0 comma 0 right parenthesis period

It follows from Theorem 1.2 that the unit sphere $\mathbb S^2$ double struck upper S squared is invariant by the flow of (1.2).

The differential equation (1.2) has three equilibrium points: one is $(0,0,0)$ left parenthesis 0 comma 0 comma 0 right parenthesis and the other two are

$$\begin{align*}p_1 = (0, 0, 1), \quad p_2 = (0, 0, -1), \end{align*}$$ p 1 equals left parenthesis 0 comma 0 comma 1 right parenthesis comma p 2 equals left parenthesis 0 comma 0 comma negative 1 right parenthesis comma

on the invariant sphere $\mathbb S^2$ double struck upper S squared . By a direct calculation, the equilibrium points $p_1$ p 1 and $p_2$ p 2 are both unstable foci for the restriction of the differential equation (1.2) on the invariant sphere $\mathbb S^2$ double struck upper S squared .

Note that the coordinate plane $z = 0$ z equals 0 is invariant. Hence, the equator of $\mathbb S^2$ double struck upper S squared is a limit cycle. Since the third component of the vector field F in (1.2) is negative for $0 < z < 1$ 0 less than z less than 1 and positive for $-1 < z < 0$ negative 1 less than z less than 0 , the equator of $\mathbb S^2$ double struck upper S squared is a stable limit cycle (see Figure 2 for an illustration).

Figure 2

Some orbits of the differential equation (1.2) with Q given in (3.2) on the unit invariant sphere $\mathbb S^2$ double struck upper S squared . The unstable foci $(0,0,1)$ left parenthesis 0 comma 0 comma 1 right parenthesis and $(0,0,-1)$ left parenthesis 0 comma 0 comma negative 1 right parenthesis are draw in red. The equator of the invariant sphere $\mathbb S^2$ double struck upper S squared is a stable limit cycle and is draw in blue. Two positive orbits corresponding to the initial conditions $(1/2,0,\pm \sqrt {3}/2)$ left parenthesis 1 divided by 2 comma 0 comma plus or minus StartRoot 3 EndRoot divided by 2 right parenthesis are drawn in black. For interpretation of the references to color in this figure legend, the reader is referred to the web version of the article.

Phase portrait on a green sphere showing red dots at the North and South poles. A blue equatorial limit cycle and two black dashed orbits spiral toward the equator from the upper and lower hemispheres.

Example 3.3 Consider $Q: \mathbb R^3 \to \mathbb R^3$ upper Q colon double struck upper R cubed right arrow double struck upper R cubed ,

(3.3) $$ \begin{align} Q(x, y, z) = \left( - x^3, - y^3, - z^3 \right), \end{align} $$ upper Q left parenthesis x comma y comma z right parenthesis equals left parenthesis minus x cubed comma minus y cubed comma minus z cubed right parenthesis comma

a homogeneous polynomial map of degree $3$ 3 in the differential equation (1.2). Note that

$$\begin{align*}\left< (x, y, z), Q(x, y, z) \right> = - \left( x^4 + y^4 + z^4 \right) < 0, \quad \forall (x, y, z) \neq (0, 0, 0). \end{align*}$$ left angle bracket left parenthesis x comma y comma z right parenthesis comma upper Q left parenthesis x comma y comma z right parenthesis right angle bracket equals minus left parenthesis x Superscript 4 Baseline plus y Superscript 4 Baseline plus z Superscript 4 Baseline right parenthesis less than 0 comma for all left parenthesis x comma y comma z right parenthesis not equals left parenthesis 0 comma 0 comma 0 right parenthesis period

It follows from Theorem 1.1 that there exists a unique two-dimensional topological sphere $S(\mu = 1) \subset \mathbb R^3 \setminus \{(0, 0, 0)\}$ upper S left parenthesis mu equals 1 right parenthesis subset of double struck upper R cubed minus left brace left parenthesis 0 comma 0 comma 0 right parenthesis right brace invariant by the flow of (1.2). This topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis is embedded as a topological submanifold of $\mathbb R^3$ double struck upper R cubed , and the bounded component of the complement of $S(1)$ upper S left parenthesis 1 right parenthesis contains the origin of $\mathbb R^3$ double struck upper R cubed . Furthermore, $S(1)$ upper S left parenthesis 1 right parenthesis is globally attracting, that is, every trajectory of (1.2) with initial condition $(x,y,z) \neq (0,0,0)$ left parenthesis x comma y comma z right parenthesis not equals left parenthesis 0 comma 0 comma 0 right parenthesis converges to $S(1)$ upper S left parenthesis 1 right parenthesis . By Theorem 1.2, the topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis is not the unit sphere $\mathbb S^2$ double struck upper S squared .

Note that the planes $x = \pm 1$ x equals plus or minus 1 , $y = \pm 1$ y equals plus or minus 1 , and $z = \pm 1$ z equals plus or minus 1 are invariant. So, the topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis is precisely the cube with vertices at the points

$$\begin{align*}\begin{array}{llll} p_1 = (1, 1, 1), & p_2 = (-1, 1, 1), & p_3 = (-1, -1, 1), & p_4 = (1, -1, 1), \\ p_5 = (1, 1, - 1), & p_6 = (-1, 1, - 1), & p_7 = (-1, -1, -1), & p_8 = (1, -1, -1). \end{array} \end{align*}$$ StartLayout 1st Row 1st Column p 1 equals left parenthesis 1 comma 1 comma 1 right parenthesis comma 2nd Column p 2 equals left parenthesis negative 1 comma 1 comma 1 right parenthesis comma 3rd Column p 3 equals left parenthesis negative 1 comma negative 1 comma 1 right parenthesis comma 4th Column p 4 equals left parenthesis 1 comma negative 1 comma 1 right parenthesis comma 2nd Row 1st Column p 5 equals left parenthesis 1 comma 1 comma negative 1 right parenthesis comma 2nd Column p 6 equals left parenthesis negative 1 comma 1 comma negative 1 right parenthesis comma 3rd Column p 7 equals left parenthesis negative 1 comma negative 1 comma negative 1 right parenthesis comma 4th Column p 8 equals left parenthesis 1 comma negative 1 comma negative 1 right parenthesis period EndLayout

In addition to the six invariant planes described above, there are three more invariant planes: $x = 0$ x equals 0 , $y = 0$ y equals 0 , and $z = 0$ z equals 0 . The 9 invariant planes determine 27 invariant straight lines; so, there are 27 equilibrium points, one of which is $(0,0,0)$ left parenthesis 0 comma 0 comma 0 right parenthesis and the other 26 are on the invariant cube: the vertices denoted by $p_i$ p Subscript i , $i = 1, \ldots , 8$ i equals 1 comma ellipsis comma 8 ; the midpoints of each of the 12 edges of the cube denoted by $q_i$ q Subscript i , $i= 1, \ldots , 12$ i equals 1 comma ellipsis comma 12 ; and the centers of each of the 6 faces of the cube denoted by $r_i$ r Subscript i , $i = 1, \ldots , 6$ i equals 1 comma ellipsis comma 6 . By a direct calculation, the equilibrium points $p_i$ p Subscript i , $i = 1, \ldots , 8$ i equals 1 comma ellipsis comma 8 are stable nodes, the equilibrium points $q_i$ q Subscript i , $i = 1, \ldots , 12$ i equals 1 comma ellipsis comma 12 are saddles, and the equilibrium points $r_i$ r Subscript i , $i = 1, \ldots , 8$ i equals 1 comma ellipsis comma 8 are unstable nodes for the restriction of the differential equation (1.2) on the invariant cube (see Figure 3).

Figure 3

Restriction of the vector field F given in (1.2) with Q given in (3.3) to each face of the topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis (cube). The nine equilibrium points are draw in red. For interpretation of the references to color in this figure legend, the reader is referred to the web version of the article.

A vector field plot on a light green square face with nine red equilibrium points and black directional arrows. See long description.
Figure 3 Long description

The vector field is displayed on a square light green surface. Nine red circular dots represent equilibrium points. One dot is located at the exact center of the square. Four dots are located at the corners of the square, and four dots are located at the midpoints of each of the four edges. Black arrows represent the vector field F. From the central red dot, arrows flow outward in all directions toward the edges. Along the horizontal and vertical centerlines, the arrows point directly away from the center toward the edge midpoints. In the four quadrants created by these centerlines, the arrows curve symmetrically, originating from the central region and terminating or converging toward the perimeter equilibrium points. The density of the arrows is uniform across the surface, illustrating a source-like behavior from the center point.

Example 3.4 Consider the differential equation (1.2) with the homogeneous polynomial map of degree $3$ 3

(3.4) $$ \begin{align} Q: \mathbb R^3 \to \mathbb R^3, \quad Q(x, y, z) = \big( Q_1(x, y, z), Q_2(x, y, z), Q_3(x, y, z) \big), \end{align} $$ StartLayout 1st Row upper Q colon double struck upper R cubed right arrow double struck upper R cubed comma upper Q left parenthesis x comma y comma z right parenthesis equals left parenthesis upper Q 1 left parenthesis x comma y comma z right parenthesis comma upper Q 2 left parenthesis x comma y comma z right parenthesis comma upper Q 3 left parenthesis x comma y comma z right parenthesis right parenthesis comma EndLayout

where

$$ \begin{align*} Q_1 (x, y, z) & = \dfrac{1}{2} \left( -x^3 - x^2 y - 2 x y^2 - y^3 - x^2 z + x y z - 3 y^2 z - 2 x z^2 - 3 y z^2 - 2 z^3 \right), \\ Q_2 (x, y, z) & = \dfrac{1}{2} \left( x^2 y + x y^2 - y^3 + 2 x^2 z + x y z - 2 y^2 z + x z^2 - 2 y z^2 - 2 z^3 \right), \\ Q_3 (x, y, z) & = \dfrac{1}{2} \left( x^3 - 3 x^2 y + 2 x y^2 + 2 y^3 + x^2 z + 2 x y z + y^2 z + 2 x z^2 + 2 y z^2 - z^3 \right). \end{align*} $$ StartLayout 1st Row 1st Column upper Q 1 left parenthesis x comma y comma z right parenthesis 2nd Column equals one half left parenthesis minus x cubed minus x squared y minus 2 x y squared minus y cubed minus x squared z plus x y z minus 3 y squared z minus 2 x z squared minus 3 y z squared minus 2 z cubed right parenthesis comma 2nd Row 1st Column upper Q 2 left parenthesis x comma y comma z right parenthesis 2nd Column equals one half left parenthesis x squared y plus x y squared minus y cubed plus 2 x squared z plus x y z minus 2 y squared z plus x z squared minus 2 y z squared minus 2 z cubed right parenthesis comma 3rd Row 1st Column upper Q 3 left parenthesis x comma y comma z right parenthesis 2nd Column equals one half left parenthesis x cubed minus 3 x squared y plus 2 x y squared plus 2 y cubed plus x squared z plus 2 x y z plus y squared z plus 2 x z squared plus 2 y z squared minus z cubed right parenthesis period EndLayout

As

$$\begin{align*}\left< (x, y, z), Q(x, y, z) \right> = - \dfrac{1}{2} \left(x^4 + x^2 y^2 + y^4 + x^2 z^2 + y^2 z^2 + z^4 \right) < 0, \end{align*}$$ left angle bracket left parenthesis x comma y comma z right parenthesis comma upper Q left parenthesis x comma y comma z right parenthesis right angle bracket equals minus one half left parenthesis x Superscript 4 Baseline plus x squared y squared plus y Superscript 4 Baseline plus x squared z squared plus y squared z squared plus z Superscript 4 Baseline right parenthesis less than 0 comma

for all $(x, y, z) \neq (0, 0, 0)$ left parenthesis x comma y comma z right parenthesis not equals left parenthesis 0 comma 0 comma 0 right parenthesis , it follows from Theorem 1.1 that there exists a unique two-dimensional topological sphere $S(\mu = 1) \subset \mathbb R^3 \setminus \{(0, 0, 0)\}$ upper S left parenthesis mu equals 1 right parenthesis subset of double struck upper R cubed minus left brace left parenthesis 0 comma 0 comma 0 right parenthesis right brace which is invariant by the flow of (1.2). This topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis is embedded as a topological submanifold of $\mathbb R^3$ double struck upper R cubed , and the bounded component of the complement of $S(1)$ upper S left parenthesis 1 right parenthesis contains the origin of $\mathbb R^3$ double struck upper R cubed . Furthermore, $S(1)$ upper S left parenthesis 1 right parenthesis is globally attracting, that is, every trajectory of (1.2) through $(x, y, z) \neq (0, 0, 0)$ left parenthesis x comma y comma z right parenthesis not equals left parenthesis 0 comma 0 comma 0 right parenthesis tends to $S(1)$ upper S left parenthesis 1 right parenthesis . By Theorem 1.2, the invariant topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis is not the unit sphere $\mathbb S^2$ double struck upper S squared .

By a direct calculation, in addition to the equilibrium point at the origin, there are two equilibrium points on the topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis

$$\begin{align*}p_1 = (0.66899, -1.37908, 0.26803), \quad p_2 = (-0.66899, 1.37908, -0.26803), \end{align*}$$ p 1 equals left parenthesis 0.66899 comma negative 1.37908 comma 0.26803 right parenthesis comma p 2 equals left parenthesis negative 0.66899 comma 1.37908 comma negative 0.26803 right parenthesis comma

with five decimals round-off precision. The equilibrium points $p_1$ p 1 and $p_2$ p 2 are both stable foci for the restriction of the differential equation (1.2) on the invariant topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis . Therefore, even without explicitly knowing the invariant topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis , there exists at least one limit cycle on it (see Figure 4).

Figure 4

Some orbits of the differential equation (1.2) with Q given in (3.4). The stable foci $p_1$ p 1 and $p_2$ p 2 are draw in red. These two equilibrium points belong to the invariant topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis . For interpretation of the references to color in this figure legend, the reader is referred to the web version of the article.

A phase portrait showing two sets of spiraling orbits in blue and black converging toward two red stable foci. See long description.
Figure 4 Long description

A phase portrait of a differential equation on a white background.

* On the left, a blue trajectory begins at a black dot at the bottom center and curves upward and to the left, forming a series of concentric, tightening loops. These loops spiral inward toward a central red dot, identified as the stable focus P sub 1.

* On the right, a black trajectory begins at a black dot at the top center and curves downward and to the right. It forms a similar series of tightening loops that spiral inward toward a second red dot, identified as the stable focus P sub 2.

* Both sets of orbits exhibit asymptotic behavior as they approach their respective equilibrium points on the invariant topological sphere S super 2 open parenthesis 1 close parenthesis.

Example 3.5 Consider the differential equation (1.2) initially in $\mathbb R^2$ double struck upper R squared with the homogeneous polynomial map of degree $3$ 3

$$ \begin{align*} \overline{Q}: \mathbb R^2 \to \mathbb R^2, \quad \overline{Q}(x, y) = \big( (-x + y) (x^2 + y^2), - (x + y) (x^2 + y^2) \big). \end{align*} $$ StartLayout 1st Row ModifyingAbove upper Q With quotation dash colon double struck upper R squared right arrow double struck upper R squared comma ModifyingAbove upper Q With quotation dash left parenthesis x comma y right parenthesis equals left parenthesis left parenthesis negative x plus y right parenthesis left parenthesis x squared plus y squared right parenthesis comma minus left parenthesis x plus y right parenthesis left parenthesis x squared plus y squared right parenthesis right parenthesis period EndLayout

It follows that

$$\begin{align*}\left< (x, y), \overline{Q}(x, y) \right> = - \left( x^2 + y^2 \right)^2 < 0, \quad \forall (x, y) \neq (0, 0). \end{align*}$$ left angle bracket left parenthesis x comma y right parenthesis comma ModifyingAbove upper Q With quotation dash left parenthesis x comma y right parenthesis right angle bracket equals minus left parenthesis x squared plus y squared right parenthesis squared less than 0 comma for all left parenthesis x comma y right parenthesis not equals left parenthesis 0 comma 0 right parenthesis period

By Theorem 1.2, the unit sphere $\mathbb S^1$ double struck upper S Superscript 1 is invariant by the flow of (1.2). In this case, the sphere is the unit circle. The origin $(0,0)$ left parenthesis 0 comma 0 right parenthesis is the unique equilibrium point of the differential equation (1.2). Thus, each orbit of the differential equation (1.2) by an initial condition different from the origin has the unit circle $\mathbb S^1$ double struck upper S Superscript 1 as the $\omega $ omega -limit set.

Now consider the differential equation (1.2) in $\mathbb R^4$ double struck upper R Superscript 4 defined by

(3.5) $$ \begin{align} \begin{array}{lll} x' = x + (-x + y) (x^2 + y^2), & y' = y - (x + y) (x^2 + y^2), \\ z' = z + (-z + w) (z^2 + w^2), & w' = w - (z + w) (z^2 + w^2), \end{array} \end{align} $$ StartLayout 1st Row 1st Column x prime equals x plus left parenthesis negative x plus y right parenthesis left parenthesis x squared plus y squared right parenthesis comma 2nd Column y prime equals y minus left parenthesis x plus y right parenthesis left parenthesis x squared plus y squared right parenthesis comma 2nd Row 1st Column z prime equals z plus left parenthesis negative z plus w right parenthesis left parenthesis z squared plus w squared right parenthesis comma 2nd Column w prime equals w minus left parenthesis z plus w right parenthesis left parenthesis z squared plus w squared right parenthesis comma EndLayout

that is, the differential equation (1.2) is defined in $\mathbb R^4$ double struck upper R Superscript 4 with the homogeneous polynomial map of degree $3$ 3

$$ \begin{align*} Q: \mathbb R^4 \to \mathbb R^4, \quad Q(x, y, z, w) = \left( \overline{Q}(x, y), \overline{Q}(z, w) \right). \end{align*} $$ StartLayout 1st Row upper Q colon double struck upper R Superscript 4 Baseline right arrow double struck upper R Superscript 4 Baseline comma upper Q left parenthesis x comma y comma z comma w right parenthesis equals left parenthesis ModifyingAbove upper Q With quotation dash left parenthesis x comma y right parenthesis comma ModifyingAbove upper Q With quotation dash left parenthesis z comma w right parenthesis right parenthesis period EndLayout

By a direct calculation, we get

$$\begin{align*}\left< (x, y, z, w), Q(x, y, z, w) \right> = - \left(\left( x^2 + y^2 \right)^2 + \left( z^2 + w^2 \right)^2 \right) < 0, \end{align*}$$ left angle bracket left parenthesis x comma y comma z comma w right parenthesis comma upper Q left parenthesis x comma y comma z comma w right parenthesis right angle bracket equals minus left parenthesis left parenthesis x squared plus y squared right parenthesis squared plus left parenthesis z squared plus w squared right parenthesis squared right parenthesis less than 0 comma

for all $(x, y, z, w) \neq (0, 0, 0, 0)$ left parenthesis x comma y comma z comma w right parenthesis not equals left parenthesis 0 comma 0 comma 0 comma 0 right parenthesis . It follows from Theorem 1.1 that there exists a unique three-dimensional topological sphere $S(\mu = 1) \subset \mathbb R^4 \setminus \{(0, 0, 0, 0)\}$ upper S left parenthesis mu equals 1 right parenthesis subset of double struck upper R Superscript 4 Baseline minus left brace left parenthesis 0 comma 0 comma 0 comma 0 right parenthesis right brace which is invariant by the flow of (1.2). This topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis is embedded as a topological submanifold of $\mathbb R^4$ double struck upper R Superscript 4 , and the bounded component of the complement of $S(1)$ upper S left parenthesis 1 right parenthesis contains the unique equilibrium point at the origin of $\mathbb R^4$ double struck upper R Superscript 4 . Furthermore, $S(1)$ upper S left parenthesis 1 right parenthesis is globally attracting, that is, every trajectory of (1.2) through $(x, y, z, w) \neq (0, 0, 0, 0)$ left parenthesis x comma y comma z comma w right parenthesis not equals left parenthesis 0 comma 0 comma 0 comma 0 right parenthesis tends to $S(1)$ upper S left parenthesis 1 right parenthesis . It follows from Theorem 1.2 that the invariant topological sphere $S(1)$ upper S left parenthesis 1 right parenthesis is not the unit sphere $\mathbb S^3$ double struck upper S cubed .

Consider the initial condition $x_0 = (0, 1, 0, 1)$ x 0 equals left parenthesis 0 comma 1 comma 0 comma 1 right parenthesis . By a direct calculation, the function

$$\begin{align*}\gamma : \mathbb R \to \mathbb R^4, \quad \gamma (t) = \left( \sin(t), \cos(t), \sin(t), \cos(t) \right), \end{align*}$$ StartLayout 1st Row gamma colon double struck upper R right arrow double struck upper R Superscript 4 Baseline comma gamma left parenthesis t right parenthesis equals left parenthesis sine left parenthesis t right parenthesis comma cosine left parenthesis t right parenthesis comma sine left parenthesis t right parenthesis comma cosine left parenthesis t right parenthesis right parenthesis comma EndLayout

is the solution of the initial value problem defined by differential equation (1.2) and $x_0$ x 0 . The image of $\gamma $ gamma is a periodic orbit.

Consider the torus

$$\begin{align*}\mathcal T = \left\{ (x, y, z, w) \in \mathbb R^4 : x^2 + y^2 = 1, \, z^2 + w^2 = 1 \right\}, \end{align*}$$ script upper T equals left brace left parenthesis x comma y comma z comma w right parenthesis element of double struck upper R Superscript 4 Baseline colon x squared plus y squared equals 1 comma z squared plus w squared equals 1 right brace comma

the Cartesian product of two circles $\mathbb S^1 \times \mathbb S^1$ double struck upper S Superscript 1 Baseline times double struck upper S Superscript 1 . By the above calculations, it follows that

$$\begin{align*}\gamma (t) \in \mathcal T \subset S(1), \quad \forall t \in \mathbb R. \end{align*}$$ gamma left parenthesis t right parenthesis element of script upper T subset of upper S left parenthesis 1 right parenthesis comma for all t element of double struck upper R period

The construction presented in Example 3.5, where the differential equation (1.2) in $\mathbb R^4$ double struck upper R Superscript 4 is obtained from the product of two differential equations in $\mathbb R^2$ double struck upper R squared with separate variables, provides a simple example of an invariant topological sphere that does not coincide with the sphere $\mathbb S^3$ double struck upper S cubed .

In the next example, we slightly modify the differential equation (3.5) in order to obtain $\mathbb S^3$ double struck upper S cubed as the invariant sphere.

Example 3.6 Consider the differential equation (1.2) in $\mathbb R^4$ double struck upper R Superscript 4 defined by

$$ \begin{align*} \begin{aligned} x' & = x + (-x + y) (x^2 + y^2) -2 x z^2, \\ y' & = y - (x + y) (x^2 + y^2) -2 y w^2, \\ z' & = z + (-z + w) (z^2 + w^2) -2 y^2 z, \\ w' & = w - (z + w) (z^2 + w^2) -2 x^2 w, \end{aligned} \end{align*} $$ StartLayout 1st Row 1st Column x prime 2nd Column equals x plus left parenthesis negative x plus y right parenthesis left parenthesis x squared plus y squared right parenthesis minus 2 x z squared comma 2nd Row 1st Column y prime 2nd Column equals y minus left parenthesis x plus y right parenthesis left parenthesis x squared plus y squared right parenthesis minus 2 y w squared comma 3rd Row 1st Column z prime 2nd Column equals z plus left parenthesis negative z plus w right parenthesis left parenthesis z squared plus w squared right parenthesis minus 2 y squared z comma 4th Row 1st Column w prime 2nd Column equals w minus left parenthesis z plus w right parenthesis left parenthesis z squared plus w squared right parenthesis minus 2 x squared w comma EndLayout

that is, the differential equation (1.2) is defined in $\mathbb R^4$ double struck upper R Superscript 4 with the homogeneous polynomial map of degree $3$ 3 , $Q: \mathbb R^4 \to \mathbb R^4$ upper Q colon double struck upper R Superscript 4 Baseline right arrow double struck upper R Superscript 4 ,

$$ \begin{align*} Q(x, y, z, w) = \left( Q_1(x, y, z, w), Q_2(x, y, z, w), Q_3(x,y,z,w), Q_4(x, y, z, w) \right), \end{align*} $$ StartLayout 1st Row upper Q left parenthesis x comma y comma z comma w right parenthesis equals left parenthesis upper Q 1 left parenthesis x comma y comma z comma w right parenthesis comma upper Q 2 left parenthesis x comma y comma z comma w right parenthesis comma upper Q 3 left parenthesis x comma y comma z comma w right parenthesis comma upper Q 4 left parenthesis x comma y comma z comma w right parenthesis right parenthesis comma EndLayout

where

$$ \begin{align*} Q_1(x, y, z, w) & = (-x + y) (x^2 + y^2) -2 x z^2,\\ Q_2(x, y, z, w) & = - (x + y) (x^2 + y^2) -2 y w^2, \\ Q_3(x, y, z, w) & = (-z + w) (z^2 + w^2) -2 y^2 z, \\ Q_4(x, y, z, w) & = - (z + w) (z^2 + w^2) -2 x^2 w. \end{align*} $$ StartLayout 1st Row 1st Column upper Q 1 left parenthesis x comma y comma z comma w right parenthesis 2nd Column equals left parenthesis negative x plus y right parenthesis left parenthesis x squared plus y squared right parenthesis minus 2 x z squared comma 2nd Row 1st Column upper Q 2 left parenthesis x comma y comma z comma w right parenthesis 2nd Column equals minus left parenthesis x plus y right parenthesis left parenthesis x squared plus y squared right parenthesis minus 2 y w squared comma 3rd Row 1st Column upper Q 3 left parenthesis x comma y comma z comma w right parenthesis 2nd Column equals left parenthesis negative z plus w right parenthesis left parenthesis z squared plus w squared right parenthesis minus 2 y squared z comma 4th Row 1st Column upper Q 4 left parenthesis x comma y comma z comma w right parenthesis 2nd Column equals minus left parenthesis z plus w right parenthesis left parenthesis z squared plus w squared right parenthesis minus 2 x squared w period EndLayout

By a direct calculation, we obtain

$$\begin{align*}\left< (x, y, z, w), Q(x, y, z, w) \right> = - \left( x^2 + y^2 + z^2 + w^2 \right)^2 < 0, \end{align*}$$ left angle bracket left parenthesis x comma y comma z comma w right parenthesis comma upper Q left parenthesis x comma y comma z comma w right parenthesis right angle bracket equals minus left parenthesis x squared plus y squared plus z squared plus w squared right parenthesis squared less than 0 comma

for all $(x, y, z, w) \neq (0, 0, 0, 0)$ left parenthesis x comma y comma z comma w right parenthesis not equals left parenthesis 0 comma 0 comma 0 comma 0 right parenthesis . It follows from Theorem 1.2 that the unit sphere $\mathbb S^3$ double struck upper S cubed is invariant by the flow of (1.2) and it is globally attracting. The bounded component of the complement of $\mathbb S^3$ double struck upper S cubed contains the unique equilibrium point at the origin of $\mathbb R^4$ double struck upper R Superscript 4 .

Data availability statement

No data were used for the research described in the article.

Acknowledgments

The authors thank their colleague Antonio Carlos Fernandes for the conversations regarding the subject studied here. The second author is grateful for the warm hospitality of the Universitat Autònoma de Barcelona during the development of this article.

Author contributions

All authors have collaborated equally in the preparation of this manuscript. The author ordering is alphabetical.

Competing interests

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this article.

Footnotes

The first author is partially supported by the Agencia Estatal de Investigación of Spain grant PID2022-136613NB-100. The second author is partially supported by Fundação de Amparo à Pesquisa do Estado de Minas Gerais (Grant Nos. APQ-02153-23, APQ-05207-23, and RED-00133-21).

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Figure 0

Figure 1 Some orbits of the differential equation (1.2) with Q given in (3.1) on the unit invariant sphere S2$\mathbb S^2$. The six equilibrium points are draw in red. For interpretation of the references to color in this figure legend, the reader is referred to the web version of the article.

Figure 1

Figure 2 Some orbits of the differential equation (1.2) with Q given in (3.2) on the unit invariant sphere S2$\mathbb S^2$. The unstable foci (0,0,1)$(0,0,1)$ and (0,0,−1)$(0,0,-1)$ are draw in red. The equator of the invariant sphere S2$\mathbb S^2$ is a stable limit cycle and is draw in blue. Two positive orbits corresponding to the initial conditions (1/2,0,±3/2)$(1/2,0,\pm \sqrt {3}/2)$ are drawn in black. For interpretation of the references to color in this figure legend, the reader is referred to the web version of the article.

Figure 2

Figure 3 Restriction of the vector field F given in (1.2) with Q given in (3.3) to each face of the topological sphere S(1)$S(1)$ (cube). The nine equilibrium points are draw in red. For interpretation of the references to color in this figure legend, the reader is referred to the web version of the article.Figure 3 long description.

Figure 3

Figure 4 Some orbits of the differential equation (1.2) with Q given in (3.4). The stable foci p1$p_1$ and p2$p_2$ are draw in red. These two equilibrium points belong to the invariant topological sphere S(1)$S(1)$. For interpretation of the references to color in this figure legend, the reader is referred to the web version of the article.Figure 4 long description.