# Flows of piecewise analytic vector fields in convex polytope decompositions

Tianqi Wu

###### Abstract.

We prove that for a convex polytope decomposition of a domain in
$\mathbb{R}^{n}$, an integral curve of a piecewise analytic vector field is
chopped by the decomposition into finitely many pieces. As a consequence, we
prove the finiteness of the number of the edge flips in a discrete Yamabe
flow.

###### Contents

  1. 1 Introduction
    1. 1.1 Setup and the main theorem
    2. 1.2 Finiteness of the edge flips in the discrete Yamabe flow
    3. 1.3 Notations and a convex geometric estimate
  2. 2 Local finiteness
  3. 3 $\lambda$-series and convergence
    1. 3.1 $n$-dim analytic functions
    2. 3.2 $\lambda$-series
    3. 3.3 Convergence of a $\lambda$-series
  4. 4 Limiting behavior for analytic vector fields
    1. 4.1 Formal $\lambda$-series solutions to ODEs
    2. 4.2 Construction of $\lambda$-series solutions
    3. 4.3 Dominating functions
    4. 4.4 Convergence of $\lambda$-series solutions
  5. 5 Limiting behavior for piecewise analytic vector fields
    1. 5.1 Preliminary estimates
    2. 5.2 Proof of Theorem 1.3

##  1\. Introduction

Gu et al. [4] introduced a discrete Yamabe flow with edge flips, which are
combinatorial changes of triangulations. It would be theoretically interesting
and practically useful to know if there are only finitely many edge flips
during the flow.

In this paper we prove that for a convex polytope decomposition of a domain in
$\mathbb{R}^{n}$, an integral curve of a piecewise analytic vector field is
chopped by the decomposition into finitely many pieces. As a consequence, we
prove the finiteness of the edge flips in a discrete Yamabe flow mentioned
above.

###  1.1. Setup and the main theorem

(a) A subset $H$ of $\mathbb{R}^{n}$ is called a _half space_ if

$H=\\{x\in\mathbb{R}^{n}:a_{1}x_{1}+...+a_{n}x_{n}\geq b\\}$

for some nonzero $(a_{1},...,a_{n})\in\mathbb{R}^{n}$ and $b\in\mathbb{R}$. A
subset $D$ of $\mathbb{R}^{n}$ is called a _convex polytope_ if $D$ has
nonempty interior and is the intersection of finitely many half spaces.

(b) Given $U\subseteq\mathbb{R}^{n}$, a _vector field_ on $U$ is a function
from $U$ to $\mathbb{R}^{n}$.

(c) Given $U\subseteq\mathbb{R}^{n}$, a function $f$ on $U$ is called
_analytic_ if there exists an open superset $\tilde{U}$ of $U$ and an analytic
function $\tilde{f}$ on $\tilde{U}$ such that $\tilde{f}\equiv f$ on $U$.

(d) Given $M\subseteq\mathbb{R}^{n}$, a flow on $M$ is a continuous function
from $[T,\infty)$ to $M$ for some $T\in\mathbb{R}$.

Here is our main theorem.

######  Theorem 1.1.

Let $U$ be an open domain in $\mathbb{R}^{n}$ and $U\subseteq\cup_{i}D_{i}$ be
a finite closed cover such that each $D_{i}$ is a convex polytope. Suppose
$M\ni 0$ is an analytic submanifold in $U$ and $V(x)$ is a $C^{1}$ vector
field on $U$ such that

  1. (i)

$V(x)$ is analytic on each $D_{i}\cap U$,

  2. (ii)

$DV(x)(T_{x}M)\subseteq T_{x}M$ for all $x\in M$,

  3. (iii)

$V(0)=0$, and

  4. (iv)

$DV(0)|_{T_{0}M}$ is diagonalizable and has only negative eigenvalues.

Suppose $x(t)$ is a flow on $M$ such that

  1. (i)

$x^{\prime}(t)=V(x(t))$ on $[0,\infty)$, and

  2. (ii)

$\lim_{t\rightarrow\infty}x(t)=0$.

Then there exists a sequence of finitely many increasing numbers
$0=t_{0}<t_{1}<...<t_{n-1}<t_{n}=\infty$ such that for all
$i\in\\{1,...,n\\}$, $x([t_{i-1},t_{i}))\subseteq D_{j}$ for some $j$.

To prove Theorem 1.1, by compactness we only need to prove Theorems 1.2 and
1.3, describing the local behavior and the limiting behavior respectively.

######  Theorem 1.2.

Let $U$ be an open domain in $\mathbb{R}^{n}$ and $U\subseteq\cup_{i}D_{i}$ be
a finite closed cover such that each $D_{i}$ is a convex polytope. Suppose
$V(x)$ is a $C^{1}$ vector field on $U$ and is analytic on each $D_{i}\cap U$.
If $x^{\prime}(t)=V(x(t))$ near $0$, then there exists $\epsilon>0$ such that
$x([0,\epsilon))\subseteq D_{i}$ for some $i$.

######  Theorem 1.3.

Let $U$ be an open domain in $\mathbb{R}^{n}$ and $U\subseteq\cup_{i}D_{i}$ be
a finite closed cover such that each $D_{i}$ is a convex polytope. Let $M\ni
0$ be an $m$-dim analytic submanifold in $U$. Suppose $V(x)$ is a $C^{1}$
vector field on $U$ such that

  1. (i)

$V(x)$ is analytic on each $D_{i}\cap U$,

  2. (ii)

$DV(x)(T_{x}M)\subseteq T_{x}M$ for all $x\in M$,

  3. (iii)

$V(0)=0$, and

  4. (iv)

$DV(0)|_{T_{0}M}$ is diagonalizable and has $m$ negative eigenvalues
$\lambda_{1}\geq...\geq\lambda_{m}$.

Suppose $x(t)$ is a flow on $M$ such that

  1. (i)

$x^{\prime}(t)=V(x(t))$ on $[0,\infty)$, and

  2. (ii)

$\lim_{t\rightarrow\infty}x(t)=0$.

Then there exists $T>0$ such that $x([T,\infty))\subseteq D_{j}$ for some $j$.

###  1.2. Finiteness of the edge flips in the discrete Yamabe flow

As a consequence of Theorem 1.1, we prove the finiteness of the number of edge
flips in the discrete Yamabe flow introduced in [4]. One may refer [4] for
backgrounds. Other related work on discrete conformality and geodesic
triangulations could be found in [17, 10, 14, 11, 1, 15, 5, 9, 7, 16, 8, 6,
13, 12, 4, 3, 2]. Here we give a very brief explanation.

A discrete conformal class in [4] is parametrized by $\mathbb{R}^{n}$ and has
a natural finite cell decomposition $\mathbb{R}^{n}=\cup_{i}U_{i}$. A discrete
Yamabe flow $u(t)$ is an integral curve of a $C^{1}$ vector field $F(u)$ on
$\mathbb{R}^{n}$, satisfying that

(a) $F(u)$ is analytic on each $U_{i}$,

(b)
$F(u)\in\mathbf{1}^{\perp}_{n}=\\{(x_{1},...,x_{n})\in\mathbb{R}^{n}:x_{1}+...+x_{n}=0\\}$
for all $u$, and

(c) $DF(u)$ is always symmetric and negative definite as a transformation on
$\mathbf{1}_{n}^{\perp}$.

(d) $F(\bar{u})=0$ for some $\bar{u}\in u(0)+\mathbf{1}_{n}^{\perp}$.
A discrete Yamabe flow $u(t)$ always exists on $[0,\infty)$ and satisfies that

(a) $u(t)-u(0)\in\mathbf{1}_{n}^{\perp}$.

(b) $u^{\prime}(t)=F(u(t))$, and

(c) $\lim_{t\rightarrow\infty}u(t)=\bar{u}$.
Furthermore, the analytic change of coordinates
$(u_{1},...,u_{n})\mapsto(e^{-2u_{1}},...,e^{-2u_{n}})$ will map each cell
$U_{i}$ to $D_{i}\cap\mathbb{R}^{n}_{>0}$ for some convex polytope $D_{i}$.

If we let $M$ be the image of $u(0)+\mathbf{1}_{n}^{\perp}$ under this
coordinate change, then Theorem 1.1 implies the finiteness of the number of
switches in the new coordinates.

###  1.3. Notations and a convex geometric estimate

(a) Given $x\in\mathbb{R}^{n}$ (or $\mathbb{Z}^{n}$), denote
$x=(x_{1},...,x_{n})$.

(b) Given $x\in\mathbb{R}^{n}$ (or $\mathbb{Z}^{n}$), denote
$|x|=|x|_{1}=|x_{1}|+...+|x_{n}|$.

(c) Given $x\in\mathbb{R}^{n}$ (or $\mathbb{Z}^{n}$), denote
$|x|_{2}=\sqrt{x_{1}^{2}+...+x_{n}^{2}}$.

(d) Given $x\in\mathbb{R}^{n}$ (or $\mathbb{Z}^{n}$), denote
$|x|_{\infty}=\max_{i}|x_{i}|$.

(e) Given $x\in\mathbb{R}^{n},I\in\mathbb{Z}^{n}$, denote
$x^{I}=x_{1}^{I_{1}}...x_{n}^{I_{n}}$.

(f) $\mathbf{1}_{n}$ denotes $(1,...,1)\in\mathbb{R}^{n}$.

(g) Given $U\subseteq\mathbb{R}^{n}$ and $p\in\mathbb{R}^{n}$, denote
$d(p,U)=\inf_{q\in U}|p-q|_{2}$.

(h) Given a half space $H\subseteq\mathbb{R}^{n}$, $d_{s}(x,H)$ denotes the
singed distance from $x$ to $H$. To be specific, $d_{s}(\cdot,H)$ is a linear
function on $\mathbb{R}^{n}$ such that $d_{s}(x,H)=d(x,H)$ whenever
$d(x,H)>0$.

######  Lemma 1.4.

Suppose $D=\cap_{i=1}^{m}H_{i}$ is a convex polytope in $\mathbb{R}^{n}$, and
$H_{i}$’s are half spaces. Then for all $p\in int(D)$ and $x\notin D$ we have
that

$\frac{\min_{i}d(p,\partial H_{i})}{|x-p|_{2}}\cdot
d(x,D)\leq\max_{i}d(x,H_{i})\leq d(x,D).$

###### Proof.

The second part of the inequality is obvious. Denote $q$ as the intersection
of $\partial D$ and the straight arc from $p$ to $x$. Assume $q\in\partial
H_{j}$ and then

$\frac{|x-q|_{2}}{|x-p|_{2}}=\frac{d(x,H_{j})}{d(p,\partial
H_{j})+d(x,H_{j})}\leq\frac{d(x,H_{j})}{d(p,\partial
H_{j})}\leq\frac{\max_{i}d(x,H_{i})}{\min_{i}d(p,\partial H_{i})}.$

Notice that $|x-q|_{2}\geq d(x,D)$ and we are done. ∎

##  2\. Local finiteness

###### Proof of Theorem 1.2.

Without loss of generality, assume $x(0)=0$. It suffices to show that for any
$D_{i}$ there exists $\epsilon>0$ such that $x([0,\epsilon])\subseteq D_{i}$
or $x([0,\epsilon])\cap D_{i}=\\{0\\}$.

Fix $D=D_{i}$. Let $V_{*}$ be an analytic vector field on an open superset of
$D\cap U$ such that $V_{*}(x)=V(x)$ on $D\cap U$. Let $y(t)$ be the local
analytic solution to

$\left\\{\begin{array}[]{ll}y(0)=0\\\
y^{\prime}(t)=V_{*}(y(t))\end{array}\right..$

If there exists $\epsilon>0$ such that $y([0,\epsilon])\subseteq D$, then by
Picard’s uniqueness theorem $x(t)=y(t)\in D$ on $[0,\epsilon]$.

So we may assume that for all $\epsilon>0$, $y([0,\epsilon])\not\subseteq D$.
Suppose $D=\cap_{j=1}^{m}H_{j}$ where $H_{j}$’s are half spaces. For a fixed
$j$, $d(y(t),H_{j})=0$ for all small $t$ or $d(y(t),H_{j})=at^{k}+o(t^{k})$
for some $a>0$ and $k\in\mathbb{Z}_{\geq 1}$. Since we cannot have that
$d(y(t),H_{j})=0$ for all small $t$ and for all $j$, we have that

$\max_{j}d(y(t),H_{j})=at^{k}+o(t^{k})$

for some $a>0$ and $k\in\mathbb{Z}_{\geq 1}$. By Lemma 1.4, there exists
constants $a_{1},a_{2}>0$ such that $a_{1}t^{k}\leq d(y(t),D)\leq a_{2}t^{k}$
for small $t$.

We will prove $x(t)=y(t)+o(t^{k})$, and then $x(t)\notin D$ for small $t>0$.
We will prove $x(t)=y(t)+o(t^{j})$ inductively for all $j=0,1,...,k$.
Obviously $x(t)=y(t)+o(1)$. Now we assume $x(t)=y(t)+o(t^{j})$ and $j\leq
k-1$. Denote $y_{*}(t)$ as the closest point to $y(t)$ in $D$, and then
$y_{*}(t)=y(t)+o(t^{j})=x(t)+o(t^{j})$ and

$V(x(t))=V(y_{*}(t))+o(t^{j})=V_{*}(y_{*}(t))+o(t^{j})$
$=V_{*}(y(t))+o(t^{j})=y^{\prime}(t)+o(t^{j})$

and

$x(t)=\int_{0}^{t}V(x(s))ds=y(t)+o(t^{j+1}).$

∎

##  3\. $\lambda$-series and convergence

###  3.1. $n$-dim analytic functions

An $n$-dim analytic function $f(x)$ near $0$ in $\mathbb{R}^{m}$ can be
written as

$f(x)=\sum_{I\in\mathbb{Z}^{m}_{\geq 0}}b_{I}x^{I}.$

where $b_{I}\in\mathbb{R}^{n}$ and

$|b_{I}|\leq M^{|I|+1}$

for some constant $M>0$.

Suppose $f(x)=\sum_{I}b_{I}x^{I}$ is an $n$-dim analytic function near $0$ in
$\mathbb{R}^{m}$, and $g(x)=\sum_{J}c_{J}x^{J}$ is an $m$-dim analytic
function near $0$ in $\mathbb{R}^{k}$ with $c_{0}=0$. Then $(f\circ g)(x)$ is
an $n$-dim function near $0$ in $\mathbb{R}^{k}$, defined as

$(f\circ g)(x)=f(g(x)).$

Furthermore, $f\circ g$ is analytic near $0$ and

$(f\circ
g)(x)=f(g(x))=\sum_{I}b_{I}(\sum_{J}c_{J}x^{J})^{I}=\sum_{J}d_{J}x^{J}$

where

(3.1)
$d_{J}=\sum_{I}b_{I}\sum_{J_{i,j}:\sum_{i=1}^{m}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}(c_{J_{i,j}})_{i}.$

Here we let $d_{0}=b_{0}$ as a convention and denote $(c_{J_{i,j}})_{i}$ as
the $i$-th component of $c_{J_{i,j}}$. Notice that the summation in equation
(3.1) is well-defined since it contains only finitely many nonzero terms. This
is because $c_{0}=0$ and if $|I|>|J|$ then
$\sum_{i=1}^{m}\sum_{j=1}^{I_{i}}J_{i,j}=J$ forces some $J_{i,j}$ to be $0$.

###  3.2. $\lambda$-series

$P(t)=\sum_{i=0}^{k}a_{i}t^{i}$ is called an _$n$ -dim polynomial_ if
$a_{i}\in\mathbb{R}^{n}$ for all $i$. Given $\lambda\in\mathbb{R}^{m}_{<0}$,
we define an _$n$ -dim $\lambda$-series_ as a formal expression

$x(t)=x(t;\lambda)=\sum_{J\in\mathbb{Z}^{m}_{\geq 0}}P_{J}(t)e^{\lambda\cdot
Jt}$

where each $P_{J}(t)$ is an $n$-dim polynomial. To be rigorous, such an
$n$-dim $\lambda$-series could be represented by a map $J\mapsto P_{J}$ from
$\mathbb{Z}_{\geq 0}^{m}$ to the space of $n$-dim polynomials. However, we
will always represent a $\lambda$-series as the above infinite summation for
better intuition. The _formal derivative_ of a $\lambda$-series is also a
$\lambda$-series naturally defined by

$(\sum_{J\in\mathbb{Z}^{m}_{\geq 0}}P_{J}(t)e^{\lambda\cdot
Jt})^{\prime}=\sum_{J\in\mathbb{Z}^{m}_{\geq
0}}(P_{J}^{\prime}(t)+\lambda\cdot JP_{J}(t))e^{\lambda\cdot Jt}.$

If $f(x)=\sum_{I\in\mathbb{Z}^{n}_{\geq 0}}b_{I}x^{I}$ is an analytic function
in $n$ variables and

$x(t)=\sum_{J\in\mathbb{Z}^{m}_{\geq 0}}P_{J}(t)e^{\lambda\cdot Jt}$

is an $n$-dim $\lambda$-series with $P_{0}(t)=0$, we can heuristically
formally expand $f(x(t))$ as

$f(x(t))=\sum_{I\in\mathbb{Z}^{n}_{\geq
0}}b_{I}\bigg{(}\sum_{J\in\mathbb{Z}^{m}_{\geq 0}}P_{J}(t)e^{\lambda\cdot
Jt}\bigg{)}^{I}=\sum_{J\in\mathbb{Z}^{m}_{\geq 0}}Q_{J}(t)e^{\lambda\cdot Jt}$

where

(3.2) $Q_{J}(t)=\sum_{I\in\mathbb{Z}_{\geq
0}^{n}}b_{I}\sum_{J_{i,j}:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}(P_{J_{i,j}})_{i}(t).$

Here we let $Q_{0}=b_{0}$ as a convention and denote $(P_{J_{i,j}})_{i}(t)$ as
the $i$-th component of the $n$-dim polynomial $P_{J_{i,j}}(t)$. So given such
analytic function $f(x)$ and $\lambda$-series $x(t)$, we define $(f\circ
x)(t)$ as a formal $\lambda$-series

$(f\circ x)(t)=\sum_{J}Q_{J}(t)e^{\lambda\cdot Jt}$

where $Q(t)$ is defined as in equation (3.2). Notice that the summation in
equation (3.2) is well-defined since it contains only finitely many nonzero
terms. This is because $P_{0}(t)=0$ and if $|I|>|J|$ then
$\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J$ forces some $J_{i,j}$ to be $0$.
Also notice that equation (3.2) is similar to equation (3.1). This is because
a $\lambda$-series could be viewed as a power series with $m$ variables
$e^{\lambda_{1}t},...,e^{\lambda_{m}t}$ and polynomial-valued coefficients.

######  Proposition 3.1.

Suppose $f(x)=\sum_{I}b_{I}x^{I}$ is a $k$-dim analytic function in $n$
variables, $g(x)=\sum_{I}c_{I}x^{I}$ is an $n$-dim analytic function in $m$
variables with $c_{0}=0$, and $x(t)=\sum_{I}P_{I}(t)e^{\lambda\cdot Jt}$ is an
$m$-dim $\lambda$-series with $P_{0}=0$. Then

$(f\circ(g\circ x))(t)=((f\circ g)\circ x)(t).$

###### Proof.

This can be shown by a routine but tedious computation, using equations (3.1)
and (3.2). ∎

###  3.3. Convergence of a $\lambda$-series

We have a simple sufficient condition for a $\lambda$-series to converge.

######  Proposition 3.2.

Given $\lambda\in\mathbb{R}^{m}_{<0}$ and a $\lambda$-series
$x(t)=\sum_{J}P_{J}(t)e^{\lambda\cdot Jt}$, suppose there exists $T>0$ and
$q\in\mathbb{R}_{>0}$ such that $|P_{J}(t)|\leq t^{q|J|}$ for all
$J\in\mathbb{Z}^{m}_{\geq 0}$ and $t\geq T$. Then

(a) the series $x(t)$ converges absolutely and uniformly for sufficiently
large $t$, and

(b) for any $a\in\mathbb{R}$, as a real-valued function

$x(t)=\sum_{J\in\mathbb{Z}^{m}_{\geq 0}:\lambda\cdot J\geq
a}P_{J}(t)e^{\lambda\cdot Jt}+o(e^{at})$

as $t\rightarrow\infty$.

###### Proof.

(a) Without loss of generality, we may assume that $T$ is so large that
$t^{q}e^{\lambda_{i}t}$ is decreasing and less than $1$ on $[T,\infty)$ for
all $i=1,...,m$.

For $t\in[T,\infty)$

$\sum_{|J|_{\infty}>n}|P_{J}(t)e^{\lambda\cdot
Jt}|\leq\sum_{|J|_{\infty}>n}t^{q|J|}e^{\lambda\cdot
Jt}\leq\sum_{|J|_{\infty}>n}T^{q|J|}e^{\lambda\cdot JT}$
$=\sum_{J}T^{q|J|}e^{\lambda\cdot JT}-\sum_{|J|_{\infty}\leq
n}T^{q|J|}e^{\lambda\cdot JT}$
$=\prod_{i=1}^{m}\left(\sum_{j=0}^{\infty}(T^{q}e^{\lambda_{i}T})^{j}\right)-\prod_{i=1}^{m}\left(\sum_{j=0}^{n}(T^{q}e^{\lambda_{i}T})^{j}\right)$

which is independent on $t$ and goes to $0$ as $n$ goes to infinity, since for
all $i$

$\sum_{j=0}^{n}(T^{k}e^{\lambda_{i}T})^{j}\rightarrow\sum_{j=0}^{\infty}(T^{q}e^{\lambda_{i}T})^{j}=\frac{1}{1-T^{q}e^{\lambda_{i}T}}<\infty$

as $n\rightarrow\infty$.

(b) Pick a large $n$ such that $\lambda\cdot J<a$ for all $J$ with
$|J|_{\infty}>n$. Then

$\sum_{\lambda\cdot J<a}P_{J}(t)e^{\lambda\cdot
Jt}=\sum_{|J|_{\infty}>n}P_{J}(t)e^{\lambda\cdot Jt}+o(e^{at}).$

Then we finish the proof by showing that for sufficiently large $t$

$\sum_{|J|_{\infty}>n}P_{J}(t)e^{\lambda\cdot
Jt}\leq\sum_{|J|_{\infty}>n}t^{q|J|}e^{\lambda\cdot Jt}$
$=\sum_{J}t^{q|J|}e^{\lambda\cdot Jt}-\sum_{|J|_{\infty}\leq
n}t^{q|J|}e^{\lambda\cdot Jt}$
$=\prod_{i=1}^{m}\left(\sum_{j=0}^{\infty}(t^{q}e^{\lambda_{i}t})^{j}\right)-\prod_{i=1}^{m}\left(\sum_{j=0}^{n}(t^{q}e^{\lambda_{i}t})^{j}\right)$
$=\sum_{i=1}^{m}O((t^{q}e^{\lambda_{i}t})^{n+1})=o(e^{at}).$

∎

######  Proposition 3.3.

Given $\lambda\in\mathbb{R}^{m}_{<0}$ and an $n$-dim $\lambda$-series
$x(t)=\sum_{J}P_{J}(t)e^{\lambda\cdot Jt}$ with $P_{0}(t)=0$, suppose there
exists $T>1$ and $q\in\mathbb{R}_{>0}$ such that $|P_{J}(t)|\leq t^{q|J|}$ for
all $J\in\mathbb{Z}^{m}_{\geq 0}$ and $t\geq T$. If $f(x)=\sum_{I}b_{I}x^{I}$
is a $1$-dim analytic function on a neighborhood of $0$ in $\mathbb{R}^{n}$,
then there exists $r>0$ such that the $\lambda$-series

$(f\circ x)(t)=\sum_{J}Q_{J}(t)e^{\lambda\cdot Jt}.$

satisfies that $|Q_{J}(t)|\leq t^{r|J|}$ for all $J$ and $t\geq T$.
Furthermore, for sufficiently large $t$,

(3.3) $({f\circ x})(t)=f(x(t)).$

Here $(f\circ x)(t)$ and $x(t)$ are viewed as two functions induced by taking
the limit of the corresponding $\lambda$-series. Equation (3.3) means that the
limit of the $\lambda$-series $(f\circ x)(t)$ is equal to the analytic
function $f$ evaluated at the limit of the $\lambda$-series $x(t)$.

###### Proof.

Without loss of generality, assume $f(0)=0$. Suppose $t\geq T$ and by equation
(3.2),

$|Q_{J}(t)|\leq\tilde{Q}_{J}(t):=\sum_{I\in\mathbb{Z}_{\geq
0}^{n}}|b_{I}|\sum_{J_{i,j}:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}|(P_{J_{i,j}})_{i}(t)|$
$=\sum_{I\in\mathbb{Z}_{\geq 0}^{n}}|b_{I}|\sum_{J_{i,j}\neq
0:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}|(P_{J_{i,j}})_{i}(t)|$
$\leq\sum_{I\in\mathbb{Z}_{\geq 0}^{n}}|b_{I}|\sum_{J_{i,j}\neq
0:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}t^{q|J_{i,j}|}$
$=t^{q|J|}\sum_{I\in\mathbb{Z}_{\geq 0}^{n}}|b_{I}|\sum_{J_{i,j}\neq
0:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}1$

where

$c_{J}:=\sum_{I\in\mathbb{Z}_{\geq 0}^{n}}|b_{I}|\sum_{J_{i,j}\neq
0:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}1$

is the coefficient of $x^{J}$ in the local analytic function

$\sum_{I}|b_{I}|(\mathbf{1}_{n}\sum_{i=1}^{\infty}x^{i})^{I}=f_{*}(\frac{x}{1-x}\mathbf{1}_{n})$

where $f_{*}(x)=\sum_{I}|b_{I}|x^{I}$ is analytic near $0$.

So $c_{J}\leq R^{|J|}$ for some $R>0$, and for all $J\in\mathbb{Z}^{m}_{\geq
0}$ and $t\geq T$

$|Q_{J}(t)|\leq\tilde{Q}_{J}(t)\leq R^{|J|}t^{q|J|}\leq t^{(q+\log_{T}R)|J|}.$

Suppose $t$ is sufficiently large so that $x(t)$ and $(f\circ x)(t)$ and
$\sum_{J}\tilde{Q}_{J}e^{\lambda\cdot Jt}$ all converge absolutely and $f$ is
analytic near $x(t)$. To prove $({f\circ x})(t)=f(x(t))$ we need to show that

$\sum_{J}Q_{J}(t)e^{\lambda\cdot
Jt}=\sum_{I}b_{I}(\sum_{J}P_{J}(t)e^{\lambda\cdot Jt})^{I}.$

Let

$A_{N,M}=\sum_{|I|\leq N}b_{I}(\sum_{J:|J|\leq M}P_{J}(t)e^{\lambda\cdot
Jt})^{I}$

and

$B_{N}=\sum_{J:|J|\leq N}Q_{J}(t)e^{\lambda\cdot Jt}.$

If $N\leq M$, then

$|A_{N,M}-B_{N}|=\left|\sum_{J:|J|>N}e^{\lambda\cdot Jt}\sum_{I:|I|\leq
N}b_{I}\sum_{J_{i,j}\neq
0:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J,|J_{i,j}|\leq
M}\prod_{i,j}(P_{J_{i,j}})_{i}(t)\right|$
$\leq\sum_{J:|J|>N}\tilde{Q}_{J}(t)e^{\lambda\cdot Jt}$

which is independent on $M$ and goes to $0$ as $N$ goes to infinity. Let
$M\rightarrow\infty$ and then $N\rightarrow\infty$, and then we get

$|\sum_{I}b_{I}(\sum_{J}P_{J}(t)e^{\lambda\cdot
Jt})^{I}-\sum_{J}Q_{J}(t)e^{\lambda\cdot Jt}|=0.$

∎

##  4\. Limiting behavior for analytic vector fields

###  4.1. Formal $\lambda$-series solutions to ODEs

Let $V(x)=\sum_{I}{b_{I}}x^{I}$ be an analytic vector field on a neighborhood
of $0$ in $\mathbb{R}^{m}$ such that $V(0)=0$ and

$\Lambda:=DV(0)=\text{diag}(\lambda_{1},...,\lambda_{m})$

for some $\lambda=(\lambda_{1},...,\lambda_{m})\in\mathbb{R}^{m}_{<0}$. A
$\lambda$-series

$x(t)=\sum_{J}P_{J}(t)e^{\lambda\cdot Jt}$

with $P_{0}(t)=0$ formally satisfies $x^{\prime}=V(x)$ if and only if for all
$J\neq 0$,

$P_{J}^{\prime}+\lambda\cdot
JP_{J}=\sum_{I}b_{I}\sum_{J_{i,j}:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}(P_{J_{i,j}})_{i},$

which is equivalent to that

(4.1) $\left(\frac{d}{dt}-(\Lambda-\lambda\cdot J)\right)P_{J}=Q_{J}$

where

(4.2) $Q_{J}=\sum_{I:|I|\geq
2}b_{I}\sum_{J_{i,j}:\sum_{i=1}^{m}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}(P_{J_{i,j}})_{i}.$

###  4.2. Construction of $\lambda$-series solutions

Given $u\in\mathbb{R}$ and a 1-dim polynomial $Q(t)$, define

$(\frac{d}{dt}-u)^{-1}Q(t)=-u^{-1}(1+u^{-1}\frac{d}{dt}+u^{-2}\frac{d^{2}}{dt^{2}}+...)Q(t),$

if $u\neq 0$ and

$(\frac{d}{dt}-u)^{-1}Q(t)=\int_{0}^{t}Q(s)ds$

if $u=0$. It is straightforward to verify that

$(\frac{d}{dt}-u)(\frac{d}{dt}-u)^{-1}Q(t)=Q(t).$

Given an $m$-dim diagonal matrix $U=\text{diag}(u_{1},...,u_{m})$ and an
$m$-dim polynomial $Q(t)$,

$P(t)=(\frac{d}{dt}-U)^{-1}Q(t)$

is defined to be such that

$P_{i}(t)=(\frac{d}{dt}-u_{i})^{-1}Q_{i}(t)$

for all $i=1,...,m$. Clearly we have that

$(\frac{d}{dt}-U)(\frac{d}{dt}-U)^{-1}Q(t)=Q(t).$

Denote

$\vec{e}_{1}=(1,0,...,0)\in\mathbb{R}^{m},$ $\vdots$
$\vec{e}_{m}=(0,...,0,1)\in\mathbb{R}^{m}.$

Given any $c\in\mathbb{R}^{m}$, we construct a formal $\lambda$-series
solution

$x(t;c)=\sum_{J}P_{J}(t;c)e^{\lambda\cdot Jt}$

to $x^{\prime}=V(x)$ as follows.

(4.3) $P_{0}(t;c)=0,$ (4.4) $P_{\vec{e}_{i}}(t;c)=c_{i}\vec{e}_{i},$ (4.5)
$P_{J}(t;c)=\left(\frac{d}{dt}-(\Lambda-\lambda\cdot
J)\right)^{-1}Q_{J}(t;c),$

if $|J|\geq 2$ where

(4.6) $Q_{J}(t;c)=\sum_{I\in\mathbb{Z}_{\geq 0}^{m}:|I|\geq
2}b_{I}\sum_{J_{k,l}:\sum_{i=1}^{m}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}(P_{J_{i,j}})_{i}(t;c).$

It is easy to verify such $x(t;c)$’s are formal $\lambda$-series solutions to
$x^{\prime}=V(x)$ parameterized by $c\in\mathbb{R}^{m}$.

###  4.3. Dominating functions

Given a $1$-dim polynomial in one variable

$P(t)=\sum_{i=0}^{q}a_{i}t^{i},$

we denote

$P^{*}(t)=\left(\sum_{i=0}^{q}a_{i}t^{i}\right)^{*}=\sum_{i=0}^{q}|a_{i}t^{i}|$

as a $1$-dim function in one variable. Given an $n$-dim polynomial in one
variable $P(t)=(P_{1}(t),...,P_{n}(t))$, then denote

$\deg(P)=\max_{i}\deg(P_{i})$

and

$P^{*}(t)=\max_{i}P_{i}^{*}(t).$

It is routine to verify the following property.

######  Proposition 4.1.

Suppose $t\in\mathbb{R}$ and $a\in\mathbb{R}$ and $P,\tilde{P}$ are two
$n$-dim polynomials in one variable. Then

(a) $(P+\tilde{P})^{*}(t)\leq P^{*}(t)+\tilde{P}^{*}(t)$,

(b) $(aP(t))^{*}=|a|P^{*}(t)$, and

(c) $(P(t)\tilde{P}(t))^{*}\leq P^{*}(t)\tilde{P}^{*}(t)$ if $n=1$.

######  Lemma 4.2.

Suppose $u\geq 2$ and $Q(t)$ are $1$-dim polynomial and

$P(t)=(\frac{d}{dt}-u)^{-1}Q(t).$

Then

$P^{*}(t)\leq Q^{*}(t)\quad\text{ for all }\quad t\geq\frac{2\deg(Q)}{u}.$

###### Proof.

Assume $Q(t)=a_{q}t^{q}+...+a_{1}t+a_{0}$ and denote

$\mathcal{A}=(\frac{d}{dt}-u)^{-1}=-u^{-1}(1+u^{-1}\frac{d}{dt}+u^{-2}\frac{d^{2}}{dt^{2}}+...).$

If $0\leq i\leq q$ and $t\geq{2q}/{u}\geq 2i/u$,

$(\mathcal{A}(t^{i}))^{*}=u^{-1}t^{i}(1+\frac{i}{ut}+\frac{i(i-1)}{(ut)^{2}}+...)\leq\frac{1}{2}t^{i}(1+\frac{1}{2}+\frac{1}{2^{2}}+...)\leq
t^{i}$

and then

$P^{*}(t)=\left(\sum_{i=0}^{q}a_{i}\mathcal{A}(t^{i})\right)^{*}\leq\sum_{i=0}^{q}|a_{i}|(\mathcal{A}(t^{i}))^{*}\leq\sum_{i=1}^{q}|a_{i}|t^{i}=Q^{*}(t).$

∎

We have a similar result for $n$-dim polynomials as a direct consequence.

######  Corollary 4.3.

Suppose $U=\text{diag}(u_{1},...,u_{n})$ is a diagonal matrix such that
$u_{i}\geq 2$ for all $i$. If $P(t),Q(t)$ are $n$-dim polynomials such that

$(\frac{d}{dt}-U)P(t)=Q(t),$

then

$P^{*}(t)\leq Q^{*}(t)\quad\text{ for all }\quad
t\geq\frac{2\deg(Q)}{\min_{i}u_{i}}.$

###  4.4. Convergence of $\lambda$-series solutions

######  Lemma 4.4.

Let $n\in\mathbb{Z}_{>0}$ and $M>0$ and $a_{J}\in\mathbb{R}$ for all
$J\in\mathbb{Z}_{\geq 0}^{m}$. $a_{J}$ is defined inductively as follows.

(4.7) $\displaystyle a_{0}=0$ (4.8) $\displaystyle a_{J}=1$ if $|J|=1$ (4.9)
$\displaystyle a_{J}=\sum_{I\in\mathbb{Z}^{n}_{\geq 0}:|I|\geq
2}M^{|I|}\sum_{J_{i,j}:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}a_{J_{i,j}}$
if $|J|\geq 2$.

Then there exists $R>0$ such that $|a_{J}|\leq R^{|J|}$.

###### Proof.

Consider the following analytic function $F(x_{1},...,x_{m},y)$ of $(m+1)$
variables.

$F(x_{1},...,x_{m},y)=\frac{1}{(1-My)^{n}}-(Mn+1)y+(x_{1}+...+x_{m}).$

We have that $F(0)=1$ and $F^{\prime}_{y}(0)=-1\neq 0$. Then by the analytic
implicit function theorem, there exists an analytic function
$f(x)=\sum_{J}b_{J}x^{J}$ near $0\in\mathbb{R}^{m}$ such that $f(0)=0$ and

$F(x_{1},...,x_{m},f(x_{1},...,x_{m}))=1,$

i.e.,

(4.10)
$1-(x_{1}+...+x_{m})+(Mn+1)f=\frac{1}{(1-Mf)^{n}}=(1+Mf+(Mf)^{2}+...)^{n}.$

It suffices to show that $b_{J}=a_{J}$ for all $J$. Clearly we have that
$b_{0}=0=a_{0}$ and $b_{J}=1=a_{J}$ for all $J$ with $|J|=1$. Then it suffices
to show that

$b_{J}=\sum_{I\in\mathbb{Z}^{n}_{\geq 0}:|I|\geq
2}M^{|I|}\sum_{J_{i,j}:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}b_{J_{i,j}}$

for all $J$ with $|J|\geq 2$. Now we fix $J\in\mathbb{Z}^{m}_{\geq 0}$ with
$|J|\geq 2$. The coefficient of $x^{J}$ on the left-hand side of equation
(4.10) is

$(Mn+1)b_{J}.$

Denote $C_{J}(P)$ as the coefficient of $x^{J}$ in the power series $P$. Then
the coefficient of $x^{J}$ on the right-hand side of equation (4.10) is

$\sum_{I\in\mathbb{Z}^{n}_{\geq
1}}C_{J}\left(\prod_{i=1}^{n}(Mf)^{I_{i}}\right)=\sum_{I\in\mathbb{Z}^{n}_{\geq
1}}M^{|I|}C_{J}\left(\prod_{i=1}^{n}f^{I_{i}}\right)$

where

$C_{J}\left(\prod_{i=1}^{n}f^{I_{i}}\right)=C_{J}\left(\prod_{i=1}^{n}(\sum_{J}b_{J}x^{J})^{I_{i}}\right)=\sum_{J_{i,j}:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}b_{J_{i,j}}.$

It remains to show that

$Mnb_{J}=\sum_{I\in\mathbb{Z}^{n}_{\geq
0}:|I|=1}M^{|I|}\sum_{J_{i,j}:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}b_{J_{i,j}}.$

This is a consequence of that for all $I$ with $|I|=1$

$\sum_{J_{i,j}:\sum_{i=1}^{n}\sum_{j=1}^{I_{i}}J_{i,j}=J}\prod_{i,j}b_{J_{i,j}}=b_{J}.$

∎

######  Theorem 4.5.

Let $U\ni 0$ be an open set in $\mathbb{R}^{m}$ and $V$ be an analytic vector
field on $U$ such that $V(0)=0$ and

$DV(0)=\text{diag}(\lambda_{1},...,\lambda_{m})$

for some $\lambda=(\lambda_{1},...,\lambda_{m})\in\mathbb{R}^{m}_{<0}$.
Suppose the $n$-dim $\lambda$-series

$x(t)=\sum_{J\in\mathbb{Z}^{m}_{\geq 0}}P_{J}(t)e^{\lambda\cdot Jt}$

formally satisfies that $P_{0}=0$ and $x^{\prime}(t)=V(x(t))$ as
$\lambda$-series. Then we have the following.

(a) There exists $T>0$ and $r>0$ such that $|P_{J}(t)|\leq t^{r|J|}$ for all
$J\in\mathbb{Z}^{m}_{\geq 0}$ and $t\geq T$.

(b) $x(t),x^{\prime}(t)$ converge and satisfy $x^{\prime}(t)=V(x(t))$ as
functions for sufficiently large $t$.

(c) If $\lambda_{0}>\max_{i}\lambda_{i}$ and $C\in\mathbb{R}^{m}$ is nonzero
and $x(t;c)$ denotes the formal solution defined as in Section 4.2, as
$t\rightarrow\infty$ we have

$x(t;c+C)-x(t;c)=\sum_{i=1}^{m}C_{i}e^{\lambda_{i}t}\vec{e}_{i}+o(e^{(\lambda_{C}+\lambda_{0})t})$

where $\lambda_{C}=\max\\{\lambda_{i}:i\in\\{1,...,m\\},C_{i}\neq 0\\}$.

###### Proof.

(a)

Without loss of generality, we may assume
$0>\lambda_{1}\geq...\geq\lambda_{m}.$ Denote
$\Lambda=\text{diag}(\lambda_{1},...,\lambda_{m})=DV(0)$ and suppose

$V(x)=\Lambda x+\sum_{I\in\mathbb{Z}^{m}_{\geq 0}:|I|\geq 2}b_{I}x^{I}.$

Since formally $x^{\prime}(t)=V(x(t))$ we have that

$P_{J}^{\prime}(t)+\lambda\cdot JP_{J}(t)=\Lambda P_{J}(t)+Q_{J}(t)$

where

$Q_{J}(t)=\sum_{I\in\mathbb{Z}_{\geq 0}^{m}:|I|\geq
2}b_{I}\sum_{J_{k,l}:\sum_{k=1}^{m}\sum_{l=1}^{I_{k}}J_{k,l}=J}\prod_{k,l}(P_{J_{k,l}})_{k}(t).$

So

$Q_{J}^{*}(t)\leq\sum_{I\in\mathbb{Z}_{\geq 0}^{m}:|I|\geq
2}|b_{I}|\sum_{J_{k,l}:\sum_{k=1}^{m}\sum_{l=1}^{I_{k}}J_{k,l}=J}\prod_{k,l}(P^{*}_{J_{k,l}})(t).$

and

(4.11) $\left(\frac{d}{dt}-(\Lambda-\lambda\cdot J)\right)P_{J}=Q_{J}$

Here $\Lambda-\lambda\cdot J$ is a diagonal matrix whose smallest diagonal
entry is

$(\lambda_{m}-\lambda\cdot J).$

Pick $s\in\mathbb{Z}_{\geq 0}$ sufficiently large such that for all $J$ with
$|J|\geq s$,

$\lambda_{m}-\lambda\cdot J\geq\frac{-\lambda_{1}|J|}{2}\geq 2.$

Then for any $J$ with $|J|\geq s$, $\deg(P_{J})=\deg(Q_{J})$.

Let $p\in\mathbb{R}_{\geq 0}$ be a constant such that

$\deg(P_{J})\leq p|J|\quad\text{ and }\quad\deg(Q_{J})\leq p|J|$

for all $J\in\mathbb{Z}^{m}_{\geq 0}$ with $|J|<s$. Then by induction it is
straightforward to see that

$\deg(P_{J})\leq p|J|\quad\text{ and }\quad\deg(Q_{J})\leq p|J|$

for all $J$. By Corollary 4.3, for $J\in\mathbb{Z}^{m}_{\geq 0}$ with $|J|\geq
s$ and

$t\geq\frac{4p}{-\lambda_{1}}=\frac{2p|J|}{-\lambda_{1}|J|/2}\geq\frac{2\deg(Q_{J})}{\lambda_{m}-\lambda\cdot
J}$

we have $P_{J}^{*}(t)\leq Q_{J}^{*}(t)$.

Pick a sufficiently large $q$ such that

$P_{J}^{*}(t)\leq t^{q|J|}$

for all $t\geq 2$ and $J\in\mathbb{Z}^{m}_{\geq 0}$ with $|J|\leq s$. Suppose
$|b_{I}|\leq M^{|I|}$ for some $M>0$ and all $I\in\mathbb{Z}^{m}_{\geq 0}$.
Let $a_{J}$ be defined as in Lemma 4.4. By induction it is straightforward to
see that $P_{J}^{*}(t)\leq a_{J}t^{q|J|}$ for all

$t\geq\max\\{\frac{4p}{-\lambda_{1}},2\\}$

and $J\in\mathbb{Z}^{m}_{\geq 0}$. By Lemma 4.4 there exists $R>0$ such that
$a_{J}\leq R^{|J|}$. So

$|P_{J}(t)|_{\infty}\leq P^{*}_{J}(t)\leq R^{|J|}t^{q|J|}\leq
2^{|J|\log_{2}R}\cdot t^{q|J|}\leq t^{(q+\log_{2}R)|J|}$

for all $J$ and

$t\geq\max\\{\frac{4p}{-\lambda_{1}},2\\}.$

(b) By Propositions 3.2 (a) and 3.3, $x(t),V(x(t))$ converge uniformly. Since
any partial sum of $x^{\prime}(t)=V(x(t))=(V\circ x)(t)$ is the derivative of
the corresponding partial sum of $x(t)$ as functions, $x^{\prime}(t)=V(x(t))$
is the derivative of $x(t)$ as functions.

(c) By Proposition 3.2 (b), we only need to compare the coefficients of
$e^{\lambda\cdot Jt}$ for nonzero $J$ satisfying $\lambda\cdot
J\geq\lambda_{C}+\lambda_{0}$. Such $J$ could either be $\vec{e}_{i}$ or
satisfy that $J_{i}=0$ if $C_{i}\neq 0$. The first case is immediate from the
definition. In the second case of $J$, we can straightforwardly show that
$P_{J}(t;c)=P_{J}(t,c+C)$ by induction. ∎

##  5\. Limiting behavior for piecewise analytic vector fields

###  5.1. Preliminary estimates

######  Lemma 5.1.

Let $U\ni 0$ be an open domain in $\mathbb{R}^{n}$ and $U\subset\cup_{i}D_{i}$
be a closed finite cover. Suppose $V(x)$ is a $C^{1}$ vector field on $U$ such
that $V(x)$ is analytic on each $D_{i}\cap U$. Then

$V(x)-V(y)=DV(0)(x-y)+O((|x|+|y|)|x-y|).$

###### Proof.

On a small neighborhood $U_{0}$ of $0$, $DV(x)$ is Lipschitz continuous on
each $D_{i}\cap U_{0}$, since $V(x)$ is analytic on $D_{i}\cap U_{0}$. Since
$DV(x)$ is continuous on $U_{0}$, it is easy to see that $DV(x)$ is Lipschitz
continuous on $U_{0}$. So on $U_{0}$ we may assume that

$|DV(x)-DV(0)|\leq C|x|.$

for some constant $C>0$. Then

$V(x)-V(y)-DV(0)(x-y)$
$=\left(\int_{0}^{1}DV(tx+(1-t)y)(x-y)dt\right)-DV(0)(x-y)$
$=\int_{0}^{1}\bigg{(}DV(tx+(1-t)y)-DV(0)\bigg{)}(x-y)dt$
$\leq\left(\int_{0}^{1}C|(tx+(1-t)y)|_{2}dt\right)|x-y|_{2}$
$=O((|x|+|y|)|x-y|).$

∎

######  Lemma 5.2.

Suppose $V(x)$ is a $C^{1}$ vector field on a neighborhood $U$ of $0$ in
$\mathbb{R}^{m}$, such that

$V(x)=\Lambda x+O(|x|^{2})$

where

$\Lambda=\text{diag}(\lambda_{1},...,\lambda_{m})$

and

$0>\lambda_{1}\geq...\geq\lambda_{m}.$

If $x^{\prime}(t)=V(x(t))$ on $[0,\infty)$ and $x(t)=o(1)$ as
$t\rightarrow\infty$, then $x(t)=O(e^{\lambda_{1}t})$ as $t\rightarrow\infty$.

###### Proof.

As $t\rightarrow\infty$,

$(|x|_{2}^{2})^{\prime}=2x\cdot x^{\prime}=2x\cdot V(x)$ $=2x\cdot(\Lambda
x+O(|x|^{2}))\leq 2\lambda_{1}|x|_{2}^{2}+O(|x|^{3}).$

If $\lambda_{0}>\lambda_{1}$,

$(e^{-2\lambda_{0}t}|x|_{2}^{2})^{\prime}=-2\lambda_{0}e^{-2\lambda_{0}t}|x|_{2}^{2}+(2\lambda_{1}|x|_{2}^{2}+O(|x|^{3}))e^{-2\lambda_{0}t}$
$\leq(2\lambda_{1}-2\lambda_{0})e^{-2\lambda_{0}t}|x|_{2}^{2}+e^{-2\lambda_{0}t}O(|x|)^{3}.$

So $(e^{-2\lambda_{0}t}|x|_{2}^{2})^{\prime}<0$ if $t$ is sufficiently large,
and then

$e^{-2\lambda_{0}t}|x|_{2}^{2}=O(1)$

and

$|x|=O(e^{\lambda_{0}t}).$

So

$(e^{-2\lambda_{1}t}|x|_{2}^{2})^{\prime}=-2\lambda_{1}e^{-2\lambda_{1}t}|x|_{2}^{2}+(2\lambda_{1}|x|_{2}^{2}+O(|x|^{3}))e^{-2\lambda_{1}t}$

$=e^{-2\lambda_{1}t}O(|x|)^{3}=O(e^{(3\lambda_{0}-2\lambda_{1})t}).$

So if $\lambda_{0}=3\lambda_{1}/4>\lambda_{1}$

$e^{-2\lambda_{1}t}|x|_{2}^{2}=|x(0)|_{2}^{2}+\int_{0}^{t}O(e^{\lambda_{1}s/4})ds=|x(0)|_{2}^{2}+O(1)=O(1)$

and $|x|=O(e^{\lambda_{1}t})$. ∎

######  Lemma 5.3.

Suppose $V(x)$ is a $C^{1}$ vector field on a neighborhood $U$ of $0$ in
$\mathbb{R}^{m}$, such that

$DV(0)=\text{diag}(\lambda_{1},...,\lambda_{m})$

where

$0>\lambda_{1}\geq...\geq\lambda_{m}.$

Suppose

$x^{\prime}(t)=V(x(t)),\quad y^{\prime}(t)=V(x(t))$

on $[0,\infty)$ and

$x(t)=o(1),\quad y(t)=o(1)$

as $t\rightarrow\infty$. If $y(t)-x(t)=O(e^{\lambda_{0}t})$ for some
$\lambda_{0}<\lambda_{m}$, then $x(t)=y(t)$.

###### Proof.

Let $f(s)=sy+(1-s)x$. Then

$V(y)-V(x)=V(f(1))-V(f(0))=\int_{0}^{1}(V\circ f)^{\prime}(s)ds$

and

$|V(y)-V(x)|_{2}\leq\int_{0}^{1}|(V\circ f)^{\prime}(s)|_{2}ds$
$=\int_{0}^{1}|DV(f(s))|_{2}\cdot|f^{\prime}(s)|_{2}ds=\max_{0\leq s\leq
1}|DV(f(s))|_{2}\cdot|y-x|_{2}\leq\frac{\lambda_{0}+\lambda_{m}}{2}|y-x|_{2}$

if $|x|,|y|$ are sufficiently small. So for $t$ sufficiently large,

$(|y-x|_{2}^{2})^{\prime}=2(y-x)\cdot(V(y)-V(x))\geq-2|y-x|_{2}\cdot|V(y)-V(x)|_{2}\geq(\lambda_{0}+\lambda_{m})|y-x|_{2}^{2}.$

So $|y-x|_{2}^{2}\geq e^{(\lambda_{0}+\lambda_{m})t}|y(0)-x(0)|_{2}^{2}$. So
we must have $y(0)=x(0)$ if $y(t)-x(t)=O(e^{\lambda_{0}t})$.

∎

######  Lemma 5.4.

Suppose $a<0,b<0,a\neq b$, and $x(t),h(t)$ are two functions on $[T,\infty)$
for some $T\in\mathbb{R}$, such that

$x^{\prime}(t)=ax(t)+h(t)$

and

$h(t)=o(e^{bt})$

as $t\rightarrow\infty$. Then

$x(t)=Ce^{at}+o(e^{bt})$

as $t\rightarrow\infty$ for some constant $C$.

###### Proof.

If $a<b$,

$x(t)=e^{at}\int_{T}^{t}e^{-as}h(s)ds+Ce^{at}$
$=e^{at}\int_{T}^{t}e^{-as}\cdot o(e^{bs})ds+Ce^{at}$
$=e^{at}\int_{T}^{t}o(e^{(b-a)s})ds+Ce^{at}$ $=e^{at}\cdot
o(e^{(b-a)t})+Ce^{at}$ $=o(e^{bt})+Ce^{at}$

for some constant $C$.

If $a>b$,

$x(t)=e^{at}\int_{\infty}^{t}e^{-as}h(s)ds+Ce^{at}$
$=e^{at}\int_{\infty}^{t}e^{-as}\cdot o(e^{bs})ds+Ce^{at}$
$=e^{at}\int_{\infty}^{t}o(e^{(b-a)s})ds+Ce^{at}$ $=e^{at}\cdot
o(e^{(b-a)t})+Ce^{at}$ $=o(e^{bt})+Ce^{at}$

for some constant $C$.

∎

###  5.2. Proof of Theorem 1.3

###### Proof of Theorem 1.3.

It suffices to prove that for any cell $D$ in the finite cover, $x(t)\in D$
for sufficiently large $t$ or $x(t)\notin D$ for sufficiently large $t$. Let
us fix a cell $D$ and assume $0\in D$ without loss of generality.

By a linear transformation, we may assume that $T_{0}M$ is tangent to the
$(x_{1}...x_{m})$-plane and

$\Lambda:=DV(0)|_{T_{0}M}=\text{diag}(\lambda_{1},...,\lambda_{m}).$

Without loss of generality, we may assume that

$M=\\{(x_{1},...,x_{n}):(x_{1},...,x_{m})\in
U_{0},(x_{m+1},...,x_{n})=f(x_{1},...,x_{m})\\}$

for some open set $U_{0}\subset\mathbb{R}^{m}$ and analytic function $f$ on
$U_{0}$. Then $f(0)=0$ and $Df(0)=0$. Denote

$\bar{x}(t)=(x_{1}(t),...,x_{m}(t)),\text{ and}$
$\bar{V}(x)=(V_{1}(x),...,V_{m}(x)).$

Then

$\bar{x}^{\prime}(t)=\bar{V}(\bar{x},f(\bar{x}))=:\tilde{V}(\bar{x}).$

where $\tilde{V}(\bar{x})$ is $C^{1}$ in a neighborhood of $0$ in
$\mathbb{R}^{m}$. Furthermore,

$D\tilde{V}(0)=\left(\Lambda,\frac{\partial\bar{V}}{\partial(x_{m+1},...,x_{n})}(0)\right)(Id_{m},Df(0))^{T}=\Lambda.$

So by Lemma 5.2 $\bar{x}(t)=O(e^{\lambda_{1}t})$ and then
$x(t)=O(e^{\lambda_{1}t})$.

Suppose

$p_{m}(x_{1},...,x_{n})=(x_{1},...,x_{m})$

is the projection from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$. Then
$\tilde{V}(\bar{x})$ is analytic on $p_{m}(M\cap D_{i})$ for each cell $D_{i}$
in the finite cover. Then by Lemma 5.1,

(5.1)
$\tilde{V}(\bar{x})-\tilde{V}(\bar{y})=\Lambda(\bar{x}-\bar{y})+O((|x|+|y|)|x-y|).$

Let $V_{*}$ be the analytic vector field on an open superset of $D\cap U$ such
that $V_{*}(x)=V(x)$ on $D\cap U$. Denote

$\bar{V}_{*}(x)=(V_{*,1}(x),...,V_{*,m}(x))$

and

$\tilde{V}_{*}(\bar{x})=\bar{V}_{*}(\bar{x},f(\bar{x})).$

Given $c\in\mathbb{R}^{m}$, let $\bar{y}(t;c)$ be the formal $\lambda$-series
solution to $\bar{y}^{\prime}=\tilde{V}_{*}(\bar{y})$ as constructed in
Section 4.2.

Pick $\lambda_{0}\in(\lambda_{1},0)$ such that $k\lambda_{0}\neq\lambda_{i}$
for all $k\in\mathbb{Z}_{\geq 0}$ and $i=1,...,m$. Let

$A=\\{k\in\mathbb{Z}_{\geq 0}:\text{there exists $c\in\mathbb{R}^{m}$ such
that }\bar{x}(t)-\bar{y}(t;c)=o(e^{k\lambda_{0}t})\text{ as
$t\rightarrow\infty$}\\}$

$A$ is nonempty since $0\in A$. Suppose $D=\cap_{i}H_{i}$ where $H_{i}$’s are
half spaces, and denote $y(t;c)=(\bar{y}(t;c),f(\bar{y}(t;c)))$ as a $n$-dim
$\lambda$-series.

(a) If $A$ has no maximum, then we can find $k\in A$ and $c\in\mathbb{R}^{m}$
such that $k\lambda_{0}<\lambda_{m}$ and
$\bar{x}(t)-\bar{y}(t;c)=o(e^{k\lambda_{0}t})$. Then by Lemma 5.3
$\bar{x}(t)=\bar{y}(t;c)$ as $\mathbb{R}^{m}$-valued functions. Then by
Proposition 3.3 $x(t)=y(t;c)$ as $\mathbb{R}^{n}$-valued functions. Then for
each $H_{i}$, the signed distance $d_{s}(y(t;c),H_{i})$ is a $\lambda$-series
and converge for sufficiently large $t$. Furthermore, by Proposition 3.3 the
limit of the $\lambda$-series $d_{s}(y(t;c),H_{i})$ is indeed the signed
distance from $y(t;c)$ to $H_{i}$. For each $H_{i}$, $d(y(t;c),H_{i})\equiv 0$
for sufficiently large $t$ or $d(y(t;c),H_{i})=at^{q}e^{rt}+o(at^{q}e^{rt})$
for some $a>0$, $q\in\mathbb{Z}_{\geq 0}$ and $r<0$. So $x(t)=y(t;c)\notin D$
for sufficiently large $t$ or $x(t)=y(t;c)\in D$ for sufficiently large $t$.

(b) If $k=\max A$, let $c$ be such that

$\bar{x}(t)=\bar{y}(t;c)+o(e^{k\lambda_{0}t}).$

If

$d(y(t;c),H_{i})=at^{q}e^{rt}+o(at^{q}e^{rt})$

for some $H_{i}$ and $a>0$ and $q\in\mathbb{Z}_{\geq 0}$ and $r\geq
k\lambda_{0}$, then

$d(x(t),H_{i})=at^{q}e^{rt}+o(at^{q}e^{rt}).$

If not, $d(y(t;c),H_{i})=o(e^{k\lambda_{0}t})$ for all $H_{i}$, and then

$d(y(t;c),D)\leq d(y(t;c),0)=O(e^{\lambda_{1}t})$

and by Lemma 1.4

$d(y(t;c),D)=o(e^{k\lambda_{0}t}).$

Denote $y_{*}(t)$ as the closest point to $y(t;c)$ in $D$. Then

$y_{*}(t)=O(y(t))=O(e^{\lambda_{1}t})$

and

$x(t)=y(t;c)+o(e^{k\lambda_{0}t})=y_{*}(t)+o(e^{k\lambda_{0}t}),$

and by Lemmas 5.1 and 5.2

$(\bar{x}-\bar{y})^{\prime}=\tilde{V}(\bar{x})-\tilde{V}_{*}(\bar{y})$
$=(\tilde{V}(\bar{x})-\tilde{V}(\bar{y}_{*}))+(\tilde{V}_{*}(\bar{y}_{*})-\tilde{V}_{*}(\bar{y}))$
$=\Lambda(\bar{x}-\bar{y}_{*})+O((|\bar{x}|+|\bar{y}_{*}|)|\bar{x}-\bar{y}_{*}|)+\Lambda(\bar{y}_{*}-\bar{y})+O((|\bar{y}_{*}|+|\bar{y}|)|\bar{y}_{*}-\bar{y}|)$
$=\Lambda(\bar{x}-\bar{y})+O(e^{\lambda_{1}t}\cdot
e^{k\lambda_{0}t})=\Lambda(\bar{x}-\bar{y})+o(e^{(k+1)\lambda_{0}t}).$

Then by Lemma 5.4 for all $i=1,...,m$,

$\bar{x}_{i}(t)-\bar{y}_{i}(t;c)=C_{i}e^{\lambda_{i}t}+o(e^{(k+1)\lambda_{0}t})$

for some constant $C_{i}$. Let $C=(C_{1},...,C_{m})$, and then $C$ is nonzero
by the maximality of $k$. Since
$\bar{x}(t)-\bar{y}(t;c)=o(e^{k\lambda_{0}t})$, $C_{i}=0$ if
$\lambda_{i}>k\lambda_{0}$. So

$\lambda_{C}:=\max\\{\lambda_{i}:i\in\\{1,...,m\\},C_{i}\neq
0\\}<k\lambda_{0}.$

By Theorem 4.5 (c),

$\bar{y}(t;c+C)=\bar{y}(t;c)+\sum_{i=1}^{m}C_{i}e^{\lambda_{i}t}\vec{e}_{i}+o(e^{(\lambda_{C}+\lambda_{0})t})$
$=\bar{x}(t;c)+o(e^{(\lambda_{C}+\lambda_{0})t})=\bar{x}(t;c)+o(e^{(k+1)\lambda_{0}t}).$

This contradicts the maximality of $k$. ∎

## References

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