Let $\xi = (\xi_1,\xi_2):\RR^2 \into \RR^2$ be a vector field of class $C^1$. Recall that the singular set of $\xi$ is the set $$Z(\xi):= \set{z \in \RR^2:\ \xi(z) = 0}.$$ The general theory of ordinary differential equations shows that, for every connected open $U \subseteq \RR^2 \setminus Z(\xi)$ and every $z \in U$, there exists a unique maximal connected $C^1$-manifold $L = L(z) \subseteq U$ such that $z \in L$ and $\xi(w) \in T_wL$ for every $w \in L$; such a manifold is called a leaf of $\xi$ in $U$. The leaves of $\xi$ in $U = \RR^2 \setminus Z(\xi)$ are simply called the leaves of $\xi$; the resulting family $$\F_\xi:= \set{L_z:\ z \in \RR^2 \setminus Z(\xi)}$$ is a partition of $\RR^2 \setminus Z(\xi)$ by the leaves of $\xi$, called the 1-dimensional foliation associated to $\xi$.
Definition
A cycle of $\xi$ is a compact leaf of $\xi$. A limit cycle of $\xi$ is a cycle $C$ of $\xi$ for which there exists a neighbourhood $U$ of $C$ such that $U$ contains no other cycle of $\xi$.
Exercise 1
- Show that every cycle of $\xi$ is a Jordan curve.
- Show that every leaf of $\xi$ is a submanifold of $\RR^2$, i.e., is embedded in $\RR^2$, not just immersed. (This uses the Jordan curve theorem as well as the flow-box theorem.)
A limit cycle $C$ is an “attractor” or “repellor” for $\xi$, in the sense that any other leaf $L$ of $\xi$ that comes sufficiently close to $C$ spirals toward or away from $C$. The general philosophy in the study of planar vector fields is that the arrangement of singular points and of limit cycles determines the phase portrait of the vector field, up to some reasonable equivalence relation (such being diffeomorphic, say).
Dulac’s Problem for $\xi$ is the following statement: the vector field $\xi$ has finitely many limit cycles. It is false in general even if $\xi$ is of class $C^\infty$; and it is a major theorem, proved independently by Ecalle and Ilyashenko, that it is true for real analytic $\xi$ that extend real-analytically to $S^2$.
By Exercise 1 and the Jordan curve theorem, the complement in $\RR^2$ of a cycle $C$ of $\xi$ has exactly two connected components $C_1$ and $C_2$, and the frontier of each of these is equal to $C$ (in this situation, $C$ is said to separate $\RR^2$).
Associated to $\xi$, we also define the vector field $\xi^\perp:\RR^2 \into \RR^2$ by $$\xi^\perp(z):= (-\xi_2(z),\xi_1(z)).$$
Exercise 2 (Orientation of cycles)
Show that, for any given cycle $C$ of $\xi$, there exists $i \in \{1,2\}$ such that $\xi^\perp(z)$ “points into $C_i$” for all $z \in C$. (This exercise includes the task of making sense of the expression in quotes.)
Definition
Let $M \subseteq \RR^2$ be a submanifold. For $z \in M$, we denote by $T_zM$ the tangent space of $M$ at $z$.
- $M$ is transverse to $\xi$ at $z \in M$ if $\xi(z) \notin T_zM$. We call $M$ transverse to $\xi$ if $M$ is transverse to $\xi$ at every $z \in M$.
- $M$ is tangent to $\xi$ at $z \in M$ if $\xi(z) \in T_zM$. We call $M$ tangent to $\xi$ if $M$ is tangent to $\xi$ at every $z \in M$.
Given an open set $U \subseteq \RR^2$ and a one-dimensional submanifold $V \subseteq U$, we say that $V$ separates $U$ if $U \setminus V$ has exactly two connected components and the frontier in $U$ of each of these is equal to $V$.
Lemma 1 (Rolle-Khovanskii)
Let $\,U \subseteq \RR^ \setminus Z(\xi)$ be open and simply connected, and let $L \subseteq U$ be a leaf of $\xi$ in $\,U$. Assume that $L$ separates $\,U$, and let $V \subseteq U$ be a one-dimensional $C^1$-submanifold that is transverse to $\xi$. Then $L \cap V$ contains at most one point.
Proof. Assume for a contradiction that $L \cap V$ contains two distinct points $z_0$ and $z_1$. Since $V$ is connected, there is a $C^1$-curve $\varphi:[0,1] \into V$ such that $\varphi(i) = z_i$ for $i=0,1$. Since $L$ separates $U$, the set $U \setminus L$ has two connected components $U_1$ and $U_2$.
The set $L \cap V$ is a discrete, closed subset of $V$; so we may assume, after reparametrizing if necessary, that $\varphi(t) \notin C$ for all $t \in (0,1)$. It follows that, after switching indices if necessary, we have $\varphi((0,1)) \subseteq U_1$.
Consider the continuous function $g:[0,1] \into \RR$ defined by $$g(t):= \varphi'(t) \cdot \xi^\perp(\varphi(t)).$$ Since $\varphi(t) \in U_1$ for all $t \in (0,1)$, the values $g(0)$ and $g(1)$ must have opposite signs; so by the Intermediate Value Theorem, there is $t \in (0,1)$ such that $g(t) = 0$. However, by transversality of $V$, we have $g(t) \ne 0$ for all $t \in [0,1]$, a contradiction. $\qed$
By definition, the lemma states that every leaf of $\xi$ in $U$ that separates $U$ is a Rolle leaf of $\xi$ in $U$. Denoting by $\RR_\xi$ the expansion of the real field by $\xi$, this means that every such leaf is definable in the pfaffian closure of $\RR_\xi$. Combining this with Exercise 1, we get:
Corollary 2 (see also Chris Miller’s module)
If $\,\RR_\xi$ is o-minimal, then so is the expansion of $\,\RR_\xi$ by all cycles of $\xi$. $\qed$
We assume from now on that $\RR_\xi$ is o-minimal.
Exercise 3
Show that the following sets are definable in $\RR_\xi$:
- $Z(\xi)$;
- for a definable submanifold $M \subseteq \RR^2$, the set Trans$_\xi(M)$ of all $z \in M$ at which $M$ is transverse to $\xi$;
- for a definable submanifold $M \subseteq \RR^2$, the set Tang$_\xi(M)$ of all $z \in M$ at which $M$ is tangent to $\xi$.
By o-minimality and Exercise 3, there is a $C^1$-cell decomposition $\C$ of $\RR^2$ that is definable in $\RR_\xi$ and compatible with the set $Z(\xi)$ and such that every $C \in \C$ is either transverse to $\xi$ or tangent to $\xi$. For any such decomposition $\C$, we set $$\C_{\textrm{trans}}:= \set{C \in \C:\ C \text{ is transverse to } \xi}.$$
We now fix such a $\C$ that is also a stratification; in particular, the union of all $C \in \C$ of dimension at most 1 is a closed subset of $\RR^2$. Hence the set $\bigcup\C_{\textrm{trans}} \cup Z(\xi)$ is a closed subset of $\RR^2$.
Definition
Let $U \subseteq \RR^2 \setminus Z(\xi)$ be open and $L$ be a leaf of $\xi$ in $U$. Then the time direction of $\xi$ induces a linear ordering $<_L$ of $L$.
Lemma 3
We can refine $\C$ so that for every open $C \in \C$ with $C \cap Z(\xi) = \emptyset$, the following holds:
- either every leaf of $\xi$ in $C$ is a vertical segment;
- or every leaf $L$ of $\xi$ in $C$ is the graph of a $C^1$-function $g_L:I_L \into \RR$, where $I_L \subseteq \RR$ is an open interval.
Proof. Refine $\C$ so that for each $C \in \C$, both $\xi_1$ and $\xi_2$ have constant sign on $C$. Now let $C \in \C$ be open such that $C \cap Z(\xi) = \emptyset$. If $\xi_1\rest{C} = 0$, then we are in Case 1; so we assume that $\xi_1\rest{C}$ has constant positive or negative sign.
Let $L$ be a leaf of $\xi$ in $C$; then $L$ is transverse to the constant vector field $(0,1)$. Since each vertical line separates the plane (i.e., its complement has exactly two connected components each of which has the line as its frontier), and since each vertical line is also a leaf of the vector field $(0,1)$ in $\RR^2$, it follows from Lemma 1 that $L$ is the graph of a $C^1$-function $g_L:I_L \into \RR$, as claimed. $\qed$
We assume from now on that $\C$ is as in Lemma 3.
Corollary 4
Let $C \in \C$ be open such that $C \cap Z(\xi) = \emptyset$, and let $L$ be a leaf of $\xi$ in $C$. Then $L$ separates $C$ (and hence is Rolle), and the frontier of $L$ contains at most two points.
Proof. Lemma 3 implies that $L$ spearates $C$, so by Corollary 2, $L$ is definable in the pfaffian closure of $\RR_\xi$. Combining this with Lemma 3 again gives the corollary. $\qed$
Exercise 4
Let $z \in \RR^2 \setminus Z(\xi)$. Show that there exists a neighbourhood $U$ of $z$ such that the leaf of $\xi$ in $U$ that contains $z$ is definable in the pfaffian closure of $\RR_\xi$.
Let now $z \in C \in \C_{\textrm{trans}}$, and let $L$ be the leaf of $\xi$ containing $z$. By Exercise 4, there is a unique $C’ \in \C$ and a unique leaf $L’$ of $\xi$ in $C’$ such that $C$ is contained in the frontier of $C’$ and $z$ is contained in the frontier of $L’$, and such that $$L’ \subseteq \set{z’ \in L:\ z’ >_L z}.$$ (In the case where $C = \{z\}$ is a singleton cell and $C’$ has dimension 1, we must have that $C’$ is tangent to $\xi$ and $L’ = C’$.) We define $$f_\xi(z):= \begin{cases} \text{the other point } z’ \text{ in } \fr(L’) &\text{if } |\fr(L’)| > 1 \text{ and } \xi(z’) \ne 0, \\ \infty &\text{otherwise.} \end{cases}$$ This is well defined by Corollary 4, and we extend $f_\xi$ to the set $$B:= \bigcup\C_{\textrm{trans}} \cup \{\infty\}$$ by setting $f_\xi(\infty):= \infty$.
Note that, by definition, we have $f_\xi:B \into B$.
Definition
We call the map $f_\xi$ the forward progression map of $\xi$ associated to $\C$.
By Lemma 3, every cycle of $\xi$ intersects the set $\bigcup\C_{\textrm{trans}}$. Therefore, a point $z \in B$ belongs to a cycle of $\xi$ if and only if there exists $n \in \NN$ such that $f_\xi^n(z) = z$, where $f_\xi^n$ denotes the $n$-th compositional iterate of $f$.
Proposition 5
There exists $N \in \NN$ such that for every $z \in B$, $z$ belongs to a cycle of $\xi$ if and only if $f_\xi^N(z) = z$.
Proof. By Lemma 1, each cycle of $\xi$ intersects each $C \in \C_{\textrm{trans}}$ in at most one point. Hence we can take $N :=$ lcm$\{2, \dots, |\C_{\textrm{trans}}|\}$. $\qed$
So for each $C \in \C_{\textrm{trans}}$, the set $$F_\xi(C):= \set{z \in C:\ f_\xi^N(z) = z}$$ represents the set of limit cycles of $\xi$ that intersect $C$.
Definition
Let $C \in \C_{\textrm{trans}}$ be of dimension 1. Then $\xi^\perp$ defines a linear ordering of $C$, which we denote by $<_C$. In this situation, the set $$IF_\xi(C):= \set{z \in F_\xi(C):\ z \text{ is } <_C-\text{isolated in } C}$$ represents the set of all limit cycles of $\xi$ that intersect $C$.
Corollary 6
The vector field $\xi$ has finitely many limit cycles if and only if $IF_\xi(C)$ is finite for every $C \in \C_{\textrm{trans}}$. $\qed$
We now define the structure $$\bo := (B; \{C:\ C \in \C_{\textrm{trans}}\}, \{<_C:\ C \in \C_{\textrm{trans}}\}, f_\xi).$$
Theorem 7 (Dolich-S)
Dulac’s Problem for $\xi$ holds if and only if for each $C \in \C_{\textrm{trans}}$, the structure induced on $C$ from $\bo$ is o-minimal.
In other words, proving Dulac’s Problem for $\xi$ is equivalent to proving that the forward progression map $\f_\xi$ is definable in some o-minimal expansion of the real line; it does not say anything about definability in any o-minimal expansion of the real field.