text:
\section{Introduction}
When addressing standard issues of control theory such as motion
planning and stabilization, one may adopt an approach based on optimal
control, e.g., Hamilton-Jacobi type methods and shooting
algorithms. One is then immediately facing intrinsic difficulties due
to the possible presence of singular trajectories. It is therefore
important to characterize these trajectories, by studying in
particular their existence, optimality status, and the related
computational aspects.
In this paper, we provide answers to the aforementioned questions for
control-affine systems, under generic assumptions, and then
investigate consequences in optimal control and its applications.

Let $M$ be a smooth (i.e.\ $C^{\infty}$) manifold of dimension $n$.
Consider the control-affine system
$$
(\Sigma)\qquad\dot{x}=f_0(x)+\sum_{i=1}^m u_if_i(x),
$$
where $x\in M$, $m$ is a positive integer, $(f_0,\ldots,f_m)$ is a
$(m+1)$-tuple of smooth vector fields on $M$, and the control
$u=(u_1,\ldots,u_m)$ takes values in an open subset $\Omega$ of
$\mathbb{R}^m$.
For $x_0\in M$ and $T>0$, a control $u\in L^\infty([0,T],\Omega)$ is
said to be \textit{admissible} if the trajectory $x(\cdot,x_0,u)$ of
$(\Sigma)$ associated to $u$ and starting at $x_0$ is well defined on
$[0,T]$. On the set ${\cal U}_{x_0,T}$ of
admissible controls, define the {\em end-point
  mapping} by
$$E_{x_0,T}(u) := x(T,x_0,u).$$
It is classical that ${\cal U}_{x_0,T}$ is an open subset of
$L^\infty([0,T],\Omega)$ and that $E_{x_0,T}:{\cal U}_{x_0,T} \rightarrow
M$ is a smooth map.

\begin{definition}
A control $u\in {\cal U}_{x_0,T}$ is said to be {\em singular} if $u$
is a critical point of the end-point mapping $E_{x_0,T}$,
i.e.\ its differential at $u$, $DE_{x_0,T}(u)$, is not surjective.
A trajectory $x(\cdot,x_0,u)$ is said to be {\em singular} if
$u$ is singular and {\em of corank one} if the codimension in the tangent
space of the range of $E_{x_0,T}(u)$ is equal to one.
\end{definition}

In other words, a control $u\in {\cal U}_{x_0,T}$ is singular if the
linearized system along the trajectory $x(\cdot,x_0,u)$ is not
controllable on $[0,T]$.
Singular trajectories appear as singularities in the set of
trajectories of $(\Sigma)$ joining two given points, and hence,
they play a crucial role in variational problems associated to
$(\Sigma)$ and in optimal control, as described next.

Let $x_0$ and $x_1$ be two points of $M$, and $T>0$.
Consider the following optimal control problem:
among all the trajectories of $(\Sigma)$ steering $x_0$ to $x_1$,
determine a trajectory minimizing the {\em cost}
\begin{equation}\label{coutintro}
C_{U,\alpha,g}(T,u)=\int_0^T\Big(\frac{1}{2}
u(t)^TU(x(t))u(t)+\alpha(x(t))^Tu(t)
+ g(t,x(t)) \Big) dt,
\end{equation}
where $\alpha=(\alpha_1,\ldots,\alpha_m)\in C^{\infty}(M,\mathbb{R}^m)$,
$g\in C^\infty(\mathbb{R}\times M)$, and
$U$ takes values in the set of symmetric positive definite $m\times
m$ matrices.

According to the Pontryagin Maximum Principle (see \cite{P}), for every
optimal trajectory $x(\cdot):=x(\cdot,x_0,u)$, there exists a nonzero
pair $(\lambda(\cdot),\lambda^0)$, where $\lambda^0$ is
a nonpositive real number and $\lambda(\cdot)$ is an
 absolutely continuous function on $[0,T]$ (called {\em
adjoint vector}) with $\lambda(t) \in T^*_{x(t)}M$,
such that, almost everywhere on $[0,T]$,
\begin{equation}
\label{hamil}
\begin{split}
&\dot{x}(t)=\frac{\partial H}{\partial
  \lambda}(t,x(t),\lambda(t),\lambda^0,u(t)),\\
&\dot{\lambda}(t)=-\frac{\partial H}{\partial x}(t,x(t),\lambda(t),\lambda^0,u(t)),\\
&\frac{\partial H}{\partial u}(t,x(t),\lambda(t),\lambda^0,u(t))=0,
\end{split}
\end{equation}
where
$$H(t,x, \lambda,\lambda^0,u):=\sum_{i=1}^mu_i\langle
\lambda,f_i(x)\rangle+\lambda^0 \left( \frac{1}{2}
u^TU(x)u+\alpha(x)^Tu+ g(t,x) \right)$$
is the {\em Hamiltonian} of the system.
An {\em extremal} is a $4$-tuple
$(x(\cdot),\lambda(\cdot),\lambda^0,u(\cdot))$ solution
of the system of equations (\ref{hamil}). The extremal is said to be
{\em normal} if $\lambda^0\neq 0$ and {\em abnormal} if $\lambda^0=0$.

The relevance of singular trajectories in optimal control lies in the
fact that they are exactly the projections of abnormal extremals.
Note that a singular trajectory may be the projection of several
abnormal extremals, and also of a normal extremal.
A singular trajectory is said to be {\em strictly abnormal} if it is
not the projection of a normal extremal.
Notice that a singular trajectory is of corank one if and only if it
admits a unique (up to scalar normalization) abnormal extremal lift;
it is strictly abnormal and of corank one if and only if it admits a
unique extremal lift which is abnormal.

For a normal extremal, it is standard to adopt the normalization
$\lambda^0=-1$, and to derive the control $u$ as the feedback function
of $(x,\lambda)$
\begin{equation}\label{controlenormal}
u(t) =  \begin{pmatrix}u_1(t)\\ \vdots \\ u_m(t)\end{pmatrix}
=  U(x(t))^{-1} \begin{pmatrix} h_1(t)-\alpha_1(x(t))
  \\ \vdots \\ h_m(t)-\alpha_m(x(t))\end{pmatrix},
\end{equation}
for every $t\in[0,T]$, where
$ h_i(t) := \langle \lambda(t), f_i(x(t)) \rangle$, for $i=1,\ldots,m$.
In particular, normal extremals are smooth on $[0,T]$.

For abnormal extremals, the situation is much more involved, since
equations (\ref{hamil}) do not provide directly an expression for
abnormal controls. Abnormal extremals may be nonsmooth, and it is not
always possible to determine an explicit expression for
singular controls. Indeed, it follows from (\ref{hamil}) that
\begin{equation}\label{hiegalzero}
h_i(\cdot)\equiv 0 \ \textrm{on}\ [0,T],\ i=1,\ldots,m,
\end{equation}
along every abnormal extremal. At that point, in order to compute the
singular control, one usually differentiates iteratively
(\ref{hiegalzero}) with respect to $t$, until the control appears
explicitly (in an affine way). To recover the control, an
invertibility property is then required, which may not hold in
general.

In this paper, we prove that, in a generic context, such an
invertibility property is obtained with a minimal number of
differentiations (cf.\ Theorem \ref{thmA}). This is the concept of {\em
  minimal order}, defined in Definition \ref{defordre}.
Here, genericity means that the $(m+1)$-tuple $(f_0,\ldots,f_m)$
belongs to an open and dense subset of the set of vector fields
equipped with the Whitney topology. The corank one property is also
proved to hold generically. We obtain similar results in the driftless
case (cf.\ Theorem \ref{srthmA}).

In a preliminary step for deriving the above results, we establish a
theorem of independent interest, asserting that any trajectory of a
generic control-affine system satisfies $\dot{x}=0$ almost everywhere
on the set where the vector fields are linearly dependent
(cf.\ Theorem \ref{thmrang}).

\medskip

When considering optimal control problems, singular minimizing
trajectories may exist, and play a major role, since
they are not dependent on the specific minimization problem.
The issue of such minimizing trajectories was already well known in
the classical theory of calculus of variations (see for instance
\cite{bliss,Young}) and proved to be a major focus, during the forties, when
the whole domain eventually developed into optimal control theory (cf
\cite{Bonnard}).
The optimality status of singular trajectories was chiefly investigated in
\cite{BK1,trelatCOCV} in relation to control-affine systems with $m=1$, in
\cite{AS,LS,Montgomery,trelatCOCV} regarding driftless systems with
$m=2$ and in \cite{AS2,Sar} for general nonlinear control systems.

In this paper, we prove that, for generic systems with $m\geq 2$ (and
$m\geq 3$ in the driftless case), and for a fixed cost
$C_{U,\alpha,g}$, there does not exist minimizing singular
trajectories (cf.\ Corollaries \ref{cor1112} and \ref{srcor1112}).
We also prove that, given a fixed system $(\Sigma)$, singular
trajectories are strictly abnormal, generically with respect to the
cost (\ref{coutintro}) (cf.\ Propositions \ref{propstrictB11} and
\ref{srpropstrictB}).
We then show how the abovementioned results can be used to derive
regularity results for the value function and in the theory of
Hamilton-Jacobi equations, which in turn have applications for
stabilization and motion planning.

\medskip

This paper is organized as follows.
Section \ref{sec2} is devoted to the statement of the main results,
firstly in the control-affine case, and secondly in the driftless
case. The consequences are detailed in Section \ref{sec3}, and
proofs are provided in Section \ref{sec4}.

\section{Statement of the main results}\label{sec2}
Let $M$ be a smooth, $n$-dimensional manifold.
Throughout the paper, $VF(M)$ denotes the set of smooth vector fields
on $M$, endowed with the $C^\infty$ Whitney topology.

\subsection{Trajectories of control-affine systems}
Let $T$ be a positive real number. Consider the control-affine system
\begin{equation}
\label{affine}
\dot{x}(t)=f_0(x(t))+\sum_{i=1}^mu_i(t)f_i(x(t)),
\end{equation}
where $(f_0,\ldots,f_m)$ is an $(m+1)$-tuple of smooth
vector fields on $M$, and the set of admissible controls
$u=(u_1,\ldots,u_m)$ is an open subset of $L^\infty([0,T],\Omega)$.

For every trajectory $x(\cdot):=x(\cdot,x_0,u)$ of (\ref{affine}),
define $I_{\mathrm{dep}}(x(\cdot))$ as the closed subset of $[0,T]$
\begin{equation}\label{Idep}
I_{\mathrm{dep}}(x(\cdot)) := \{ t\in[0,T]\ \vert\
\mathrm{rank}\{f_0(x(t)),\ldots,f_m(x(t))\}  < m+1 \} .
\end{equation}

Note that, on the open subset of $\mathbb{R}^n$ where
$\mathrm{rank}\{f_0,\ldots,f_m\}  = m+1$, there is a one-to-one
correspondence between trajectories and controls.
In contrast, on $I_{dep}(x(\cdot))$, there is no uniqueness of the
control associated to $x(\cdot)$; in particular, $x(\cdot)$ may be
associated to both singular and nonsingular controls.
This fact emphasizes the following result, which describes, in a generic
context, trajectories on the subset of $\mathbb{R}^n$ where
$\mathrm{rank}\{f_0,\ldots,f_m\}  < m+1$.

\begin{theorem}\label{thmrang}
Let $m<n$ be a nonnegative integer.
There exists an open and dense subset $O_{m+1}$ of
$VF(M)^{m+1}$ so that, if the $(m+1)$-tuple $(f_0,\ldots,f_m)$
belongs to $O_{m+1}$, then every trajectory $x(\cdot)$ of the associated
control-affine system
$
\dot{x}=f_0(x)+\sum_{i=1}^mu_if_i(x)
$
verifies
\begin{equation}\label{xpointIdep}
\dot{x}(t)=0,\ \textrm{for a.e.}\ t\in I_{\mathrm{dep}}(x(\cdot)).
\end{equation}
In addition, for every integer $N$, the set  $O_{m+1}$ can be chosen
so that its complement has codimension greater than $N$.
\end{theorem}

\begin{remark}\label{remidep}
At the light of the previous result, one can choose the admissible
control $u$ on $I_{dep}(x(\cdot))$ such that, for every $t\in
I_{dep}(x(\cdot))$, $u(t)$ consists of any $m$-tuple
$(\alpha_1,\ldots,\alpha_m)$ so that
$$
f_0(x(t))+\sum_{i=1}^m \alpha_if_i(x(t))=0.
$$
In particular, on any subinterval of $I_{dep}(x(\cdot))$, the
trajectory $x(\cdot)$ is constant, and the control can be
chosen constant as well.
\end{remark}

\begin{remark}\label{remtrivial}
A trajectory $x(\cdot)$ is said to be {\em trivial} if it reduces to a
point; otherwise it is said to be {\em nontrivial}.
It is clear that, if $I_{dep}(x(\cdot))\neq [0,T]$, then
$\dot{x}(t)\neq 0$ for $t\notin I_{dep}(x(\cdot))$ and $x(\cdot)$ is
nontrivial.

Let $x(\cdot)$ be a trajectory  of a control-affine system
associated to an $(m+1)$-tuple of $O_{m+1}$. As a consequence of
Theorem \ref{thmrang}, $x(\cdot)$ is trivial if and only if
$I_{dep}(x(\cdot))=[0,T]$.
\end{remark}

\subsection{Singular trajectories}
Recall that a singular trajectory $x(\cdot)$ is the projection of an abnormal
extremal $(x(\cdot),\lambda(\cdot))$.
For $t\in[0,T]$ and $i,j\in\{0,\ldots,m\}$, we define
$$
h_i(t):=\langle \lambda(t),f_i(x(t)\rangle, \qquad
h_{ij}(t):=\langle \lambda(t),[f_i,f_j](x(t))\rangle .
$$
Along an abnormal extremal, we have for every $t\in[0,T]$,
\begin{equation}\label{s-affine}
h_0(t) = \textrm{constant},\ \ h_i(t)=0, \ \ i=1,\ldots,m.
\end{equation}
Differentiating (\ref{s-affine}), one gets, almost everywhere on
$[0,T]$,
\begin{equation}\label{sss-affine}
h_{i0}(t)+\sum_{j=1}^m h_{ij}(t)u_j(t)=0,\quad i\in\{0,\ldots,m\}.
\end{equation}

\begin{definition}
Along an abnormal extremal $(x(\cdot),\lambda(\cdot),u(\cdot))$ of the system
(\ref{affine}), the {\em Goh matrix} $G(t)$
at time $t\in [0,T]$ is
the $m\times m$ skew-symmetric matrix given by
\begin{equation}\label{goh-aff}
G(t):=\big(h_{ij}(t)\big)_{1\leq i,j\leq m}.
\end{equation}
\end{definition}

Since $G(t)$ is skew-symmetric, $\mathrm{rank}\ G(t)$ is
even, and Equation (\ref{sss-affine}) rewrites, almost everywhere on
$[0,T]$,
\begin{equation}\label{Au=b}
G(t) u(t) = b(t),
\end{equation}
with $b(t):=-(h_{i0}(t))_{1\leq i\leq m}$.

Note that, if $G(t)$ is invertible, then $u(t)$ is uniquely determined
by Equation (\ref{Au=b}). This only occurs for $m$ even.

If $m$ is odd, $G(t)$ is never invertible. However, a similar
construction is derived as follows. Define
\begin{equation}\label{Gbar}
\overline{G}(t):=\big(h_{ij}(t)\big)_{0\leq i,j\leq m}.
\end{equation}
Since $\overline{G}(t)$ is skew-symmetric, the determinant of
$\overline{G}(t)$ is the square of a polynomial $\overline{P}(t)$ in
the $h_{ij}(t)$ with degree $(m+1)/2$, called the {\em Pfaffian} (see
of $\overline{G}(t)$ \cite{artin}).
 From Equation (\ref{sss-affine}), $\overline{G}(t)$ is not
invertible, and thus, along the
extremal, $\overline{P}(t)=0$. After differentiation, one gets, almost
everywhere on $[0,T]$,
\begin{equation}\label{sss-affine2}
\{\overline{P},h_0\}(t)+\sum_{i=1}^m
u_j(t)\{\overline{P},h_j\}(t)=0.
\end{equation}
Define the $(m+1)\times m$ matrix $\widetilde{G}(t)$ as
$G(t)$ augmented with the row
$(\{\overline{P},h_j\}(t))_{1\leq j\leq m}$, and the $(m+1)$-dimensional
vector $\widetilde{b}(t)$ as $b(t)$ augmented with the coefficient
$-\{\overline{P},h_0\}(t)$.
Then, from Equations (\ref{Au=b}) and (\ref{sss-affine2}), there
holds, almost everywhere on $[0,T]$,
\begin{equation}\label{Atildeu=b}
\widetilde{G}(t)u(t)=\widetilde{b}(t).
\end{equation}
If $\widetilde{G}(t)$ is of rank $m$, then $u(t)$ is uniquely determined
by Equation (\ref{Atildeu=b}).

These facts, combined with Remark \ref{remidep}, motivate the following
definition.

\begin{definition}\label{defordre}
If $m$ is even (resp.\ odd),
a singular trajectory $x(\cdot)$ is said to be {\em of minimal order}
if:
\begin{itemize}
\item[(i)] $\dot{x}(t)=0$, for almost every $t\in I_{dep}(x(\cdot))$;
\item[(ii)] it admits an abnormal extremal lift such that, for almost every
$t\in [0,T]\setminus I_{dep}$, $\mathrm{rank}\
G(t)=m$ if $m$ is even, resp., $\mathrm{rank}\ \widetilde{G}(t)=m$ if
$m$ is odd.
\end{itemize}
\end{definition}

On the opposite, for arbitrary $m$,
a singular trajectory is said to be a {\em Goh trajectory} if it
admits an abnormal extremal
lift along which the Goh matrix is identically equal to $0$.

\begin{theorem}\label{thmA}
Let $m<n$ be a positive integer.
There exists an open and dense subset $O_{m+1}$ of
$VF(M)^{m+1}$ so that, if the $(m+1)$-tuple
$(f_0,\ldots,f_m)$ belongs to $O_{m+1}$, then every nontrivial
singular trajectory of the associated control-affine system
$
\dot{x}(t)=f_0(x(t))+\sum_{i=1}^mu_i(t)f_i(x(t)),
$
is of minimal order and of corank one.
In addition, for every integer $N$, the set  $O_{m+1}$ can be chosen
so that its complement has codimension greater than $N$.
\end{theorem}

\begin{corollary}\label{corGoh2}
With the notations of Theorem~\ref{thmA} and if $m\geq 2$,
there exists an open and dense subset $O_{m+1}$ of $VF(M)^{m+1}$
so that every control-affine system defined with an
$(m+1)$-tuple of $O_{m+1}$ does not admit nontrivial Goh singular
trajectories.
\end{corollary}

\subsection{Minimizing singular trajectories}\label{subsec2.3}
We keep here the notations of the previous sections.
Consider the control-affine system
\begin{equation}\label{CAsyst}
\dot{x}(t)=f_0(x(t))+\sum_{i=1}^mu_i(t)f_i(x(t)),
\end{equation}
and the quadratic cost given by
\begin{equation}\label{coutUg}
C_{U,g}(T,u)=\frac{1}{2}\int_0^T \Big( u(t)^TU(x(t))u(t) + g(t,x(t)) \Big) dt,
\end{equation}
where $U\in\mathcal{S}^+_m(M)$ and $g\in C^\infty(\mathbb{R}\times M)$. Here,
$\mathcal{S}^+_m(M)$ denotes the set of smooth mappings $x\mapsto U(x)$ on
$M$, taking values in the set $\mathcal{S}^+_m$ of $m\times m$ real
positive definite matrices.

For $x_0\in M$ and $T>0$, define the optimal control problem
\begin{equation}\label{opt}
\inf \{ C_{U,g}(T,u)\ \vert\ E_{x_0,T}(u)=x \}.
\end{equation}
We next state two sets of genericity results, depending whether the
cost or the control system is fixed.

\subsubsection{Genericity w.r.t.\ the control system, with a fixed cost}
$\ $

\begin{proposition}\label{propstrictA}
Fix $U\in\mathcal{S}^+_m(M)$ and $g\in C^\infty(\mathbb{R}\times M)$.
There exists an open and dense subset $O_{m+1}$ of
$VF(M)^{m+1}$ such that
every nontrivial singular trajectory of
a control-affine system defined by an $(m+1)$-tuple
of $O_{m+1}$ is strictly abnormal for the optimal
control problem (\ref{opt}).
\end{proposition}

Corollary~\ref{corGoh2} together with Proposition~\ref{propstrictA}
yields the next corollary.

\begin{corollary}\label{cor1112}
Fix $U\in\mathcal{S}^+_m(M)$ and $g\in C^\infty(\mathbb{R}\times M)$.
Let $m\geq 2$ be an integer.
There exists an open and dense subset $O_{m+1}$ of $VF(M)^{m+1}$ so that
the optimal control problem
(\ref{opt}) defined with an
$(m+1)$-tuple of $O_{m+1}$ does not admit nontrivial minimizing singular
trajectories.
\end{corollary}

\begin{remark}
In both previous results, the set  $O_{m+1}$ can be chosen
so that its complement has arbitrary codimension.
\end{remark}

\subsubsection{Genericity w.r.t.\ the cost, with a fixed control system}
We endow $\mathcal{S}^+_m(M)$ with the Whitney topology.
An $(m+1)-$ tuple $(f_0,\ldots,f_m)$ of $VF(M)^{m+1}$ is said to verify the
{\em Lie Algebra Rank Condition} if the Lie algebra generated by
$f_0,\ldots,f_m$ is of dimension $n$ at every point of $M$.

\begin{proposition}\label{propstrictB}
Fix $(f_0,\ldots,f_m)\in VF(M)^{m+1}$ so that the
Lie Algebra Rank Condition is satisfied and the zero control $u\equiv
0$ is not singular. Let $g\in C^\infty(\mathbb{R}\times M)$. Then, there
exists an open and dense subset $\mathcal{A}_m$ of
$\mathcal{S}^+_m(M)$ such that every nontrivial singular trajectory of
the control-affine system associated to the $(m+1)$-tuple
$(f_0,\ldots,f_m)$ is strictly abnormal for the optimal
control problem (\ref{opt}) defined with $U\in \mathcal{A}_m$
and $g$.
\end{proposition}

Assuming that the zero control $u\equiv 0$ is not singular is
a necessary hypothesis. Indeed, the fact that a control $u$ is
singular is a property of the sole
$(m+1)$-tuple $(f_0,\ldots,f_m)$ and is independent of the cost.
On the other hand, every trajectory $x:=x(\cdot,x_0,0)$ associated
to the zero control
is always the projection of the normal extremal $(x(\cdot),
0,-1,0)$ of any optimal
control problem (\ref{opt}). As a consequence, if the zero
control is singular, such a trajectory $x(\cdot,x_0,0)$ cannot be
strictly abnormal.

In order to handle the case of a singular zero control, it is
therefore necessary to consider more general quadratic costs such as
\begin{equation}\label{coutUg11}
C_{U,\alpha,g}(T,u)=\int_0^T\Big(\frac{1}{2}
u(t)^TU(x(t))u(t)+\alpha(x(t))^Tu(t)
+ g(t,x(t)) \Big) dt,
\end{equation}
where $U\in\mathcal{S}^+_m(M)$, $\alpha\in C^{\infty}(M,\mathbb{R}^m)$
and $g\in C^\infty(\mathbb{R}\times M)$.

\begin{proposition}\label{propstrictB11}
Fix $(f_0,\ldots,f_m)\in VF(M)^{m+1}$ satisfying the
Lie Algebra Rank Condition and $g\in C^\infty(\mathbb{R}\times M)$.
Then, there exists an open and dense subset $\mathcal{B}_m$ of
$\mathcal{S}^+_m(M)\times C^{\infty}(M,\mathbb{R}^m)$ such that every
nontrivial singular trajectory of
the control-affine system associated to the $(m+1)$-tuple
$(f_0,\ldots,f_m)$ is strictly abnormal for the optimal
control problem (\ref{CAsyst})-(\ref{coutUg11}) defined with
$(U,\alpha)\in \mathcal{B}_m$ and $g$.
\end{proposition}

\subsection{Driftless control-affine systems}

Let $T$ be a positive real number. Consider the driftless
control-affine system
\begin{equation}
\label{sraffine}
\dot{x}(t)=\sum_{i=1}^mu_i(t)f_i(x(t)),
\end{equation}
where $(f_1,\ldots,f_m)$ is an $m$-tuple of smooth
vector fields on $M$, and the set of admissible controls
$u=(u_1,\ldots,u_m)$ is an open subset of $L^\infty([0,T],\Omega)$.

For every trajectory $x(\cdot):=x(\cdot,x_0,u)$ of (\ref{affine}),
define $I_{\mathrm{dep}}(x(\cdot))$ as the closed subset of $[0,T]$
$
I_{\mathrm{dep}}(x(\cdot)) := \{ t\in[0,T]\ \vert\
\mathrm{rank}\{f_1(x(t)),\ldots,f_m(x(t))\}  < m \} .
$

\begin{theorem}\label{srthmrang}
Let $m\leq n$ be a positive integer.
There exists an open and dense subset $O_{m}$ of
$VF(M)^{m}$ so that, if the $m$-tuple $(f_1,\ldots,f_m)$
belongs to $O_{m}$, then every trajectory $x(\cdot)$ of the associated
driftless control-affine system
$
\dot{x}=\sum_{i=1}^m u_i f_i(x)
$
verifies
$
\dot{x}(t)=0,\ \textrm{for a.e.}\ t\in I_{\mathrm{dep}}(x(\cdot)).
$
In addition, for every integer $N$, the set  $O_{m}$ can be chosen
so that its complement has codimension greater than $N$.
\end{theorem}

\subsubsection{Singular trajectories}
Let $x(\cdot)$ be a singular trajectory, projection of an abnormal
extremal $(x(\cdot),\lambda(\cdot))$. Similarly to the previous section,
we define, for $t\in[0,T]$ and $i,j\in\{1,\ldots,m\}$,
$$
h_i(t):=\langle \lambda(t),f_i(x(t)\rangle, \qquad
h_{ij}(t):=\langle \lambda(t),[f_i,f_j](x(t))\rangle .
$$
For every $t\in[0,T]$,
\begin{equation}\label{srs-affine}
h_i(t)=0, \ \ i=1,\ldots,m.
\end{equation}
Differentiating (\ref{srs-affine}), one gets, almost everywhere on
$[0,T]$,
\begin{equation}\label{srsss-affine}
\sum_{j=1}^m h_{ij}(t)u_j(t)=0,\quad i\in\{1,\ldots,m\}.
\end{equation}

\begin{definition}
Along an abnormal extremal $(x(\cdot),\lambda(\cdot),u(\cdot))$ of the system
(\ref{affine}), the {\em Goh matrix} $G(t)$
at time $t\in [0,T]$ is
the $m\times m$ skew-symmetric matrix given by
\begin{equation}\label{srgoh-aff}
G(t):=\big(h_{ij}(t)\big)_{1\leq i,j\leq m}.
\end{equation}
\end{definition}

Since $G(t)$ is skew-symmetric, $\mathrm{rank}\ G(t)$ is
even, and Equation (\ref{srsss-affine}) rewrites, almost everywhere on
$[0,T]$,
\begin{equation}\label{srAu=b}
G(t) u(t) = 0.
\end{equation}

Note that, if $\mathrm{rank}\ G(t)=m-1$,
one can deduce from~(\ref{srAu=b})  an
expression for $u(t)$, up to time reparameterization.
This only occurs for $m$ odd.

If $m$ is even, $\mathrm{rank}\ G(t)$ is always smaller than
$m-1$. However, a similar
construction is derived as follows. The determinant of
$G(t)$ is the square of the  Pfaffian $P(t)$, and, along the
extremal, $P(t) \equiv 0$. After differentiation, one gets, almost
everywhere on $[0,T]$,
\begin{equation}\label{srsss-affine2}
\sum_{i=1}^m
u_j(t)\{ P,h_j\}(t)=0.
\end{equation}
Define the $(m+1)\times m$ matrix $\widetilde{G}(t)$ as
$G(t)$ augmented with the row
$(\{ P,h_j\}(t))_{1\leq j\leq m}$.
Then, from Equations (\ref{srAu=b}) and (\ref{srsss-affine2}), there
holds, almost everywhere on $[0,T]$,
\begin{equation}\label{srAtildeu=b}
\widetilde{G}(t)u(t)=0.
\end{equation}
If $\widetilde{G}(t)$ is of rank $m-1$, one can deduce
from~(\ref{srAtildeu=b})  an
expression for $u(t)$, up to time reparameterization.

\begin{definition}\label{srdefordre}
If $m$ is odd (resp.\ even), a singular trajectory $x(\cdot)$ is said
to be {\em of minimal order}
if:
\begin{itemize}
\item[(i)] $\dot{x}(t)=0$, for almost every $t\in I_{dep}(x(\cdot))$;
\item[(ii)] it admits an abnormal extremal lift such that, for almost every
$t\in [0,T]\setminus I_{dep}$, $\mathrm{rank}\
G(t)=m-1$ if $m$ is odd, resp., $\mathrm{rank}\ \widetilde{G}(t)=m-1$ if
$m$ is even.
\end{itemize}
\end{definition}

On the opposite, for arbitrary $m$,
a singular trajectory is said to be a {\em Goh trajectory} if it
admits an abnormal extremal
lift along which the Goh matrix is identically equal to $0$.

\begin{theorem}\label{srthmA}
Let $m$ be an integer such that $2 \leq m \leq n$.
There exists an open and dense subset $O_{m}$ of $VF(M)^{m}$ so that,
if the $m$-tuple $(f_1,\ldots,f_m)$ belongs to $O_{m}$, then every nontrivial
singular trajectory of the associated driftless control-affine system
$\dot{x}(t)=\sum_{i=1}^mu_i(t)f_i(x(t))$
is of minimal order and of corank one.
In addition, for every integer $N$, the set  $O_{m}$ can be chosen
so that its complement has codimension greater than $N$.
\end{theorem}

\begin{corollary}\label{srcorGoh2}
With the notations of Theorem~\ref{srthmA} and if $m\geq 3$,
there exists an open and dense subset $O_{m}$ of $VF(M)^{m}$ so that
every driftless control-affine system defined with an
$m$-tuple of $O_{m}$ does not admit nontrivial Goh singular
trajectories.
\end{corollary}

\subsubsection{Minimizing singular trajectories}\label{subsubsec2.4.2}
Consider the optimal control problem
associated to the driftless control-affine system
\begin{equation}\label{srCAsyst}
\dot{x}(t)=\sum_{i=1}^mu_i(t)f_i(x(t)),
\end{equation}
with the quadratic cost given by
\begin{equation}\label{srcoutUg}
C_{U,g}(T,u)=\frac{1}{2}\int_0^T \Big( u(t)^TU(x(t))u(t) + g(t,x(t))
\Big) dt,
\end{equation}
where $U\in\mathcal{S}^+_m(M)$ and $g\in C^\infty(\mathbb{R}\times M)$.

For $x_0\in M$ and $T>0$, define the optimal control problem
\begin{equation}\label{sropt}
\inf \{ C_{U,g}(T,u)\ \vert\ E_{x_0,T}(u)=x \}.
\end{equation}
We next state genericity results with respect to the control system, with a
fixed cost.

\begin{proposition}\label{srpropstrictA}
Fix $U\in\mathcal{S}^+_m(M)$ and $g\in C^\infty(\mathbb{R}\times M)$.
There exists an open and dense subset $O_{m}$ of
$VF(M)^{m}$ such that every nontrivial singular trajectory of
a driftless control-affine system defined by an $m$-tuple
of $O_{m}$ is strictly abnormal for the optimal
control problem (\ref{sropt}).
\end{proposition}

Corollary~\ref{srcorGoh2} together with Proposition~\ref{srpropstrictA}
yields the next corollary.

\begin{corollary}\label{srcor1112}
Fix $U\in\mathcal{S}^+_m(M)$ and $g\in C^\infty(\mathbb{R}\times M)$.
Let $m\geq 3$ be an integer.
There exists an open and densee subset $O_{m}$ of $VF(M)^{m}$ so that the optimal control problem
(\ref{sropt}) defined with an
$m$-tuple of $O_{m}$ does not admit nontrivial minimizing singular
trajectories.
\end{corollary}

\begin{remark}
In both previous results, the set  $O_{m}$ can be chosen
so that its complement has arbitrary codimension.
\end{remark}

We also have have a genericity result with respect to the cost, with a
fixed control system.

\begin{proposition}\label{srpropstrictB}
Fix $(f_1,\ldots,f_m)\in VF(M)^{m}$ so that the
Lie Algebra Rank Condition is satisfied. Let $g\in C^\infty(\mathbb{R}\times
M)$. Then, there
exists an open and dense subset $\mathcal{A}_m$ of
$\mathcal{S}^+_m(M)$ such that every nontrivial singular trajectory of
the driftless control-affine system associated to the $m$-tuple
$(f_1,\ldots,f_m)$ is strictly abnormal for the optimal
control problem (\ref{sropt}) defined with $U\in \mathcal{A}_m$
and $g$.
\end{proposition}

\begin{remark}
In the driftless case, the control $u\equiv 0$ is always singular, but
corresponds to a trivial trajectory. Therefore, in opposition to the
control-affine case, it is not necessary to add the linear term
$\alpha(x)^Tu$ in the cost.
\end{remark}

\section{Consequences}
\label{sec3}
\subsection{Regularity of the value function}
Consider the optimal control problem (\ref{opt}), associated to the
control-affine system (\ref{CAsyst}) and the cost (\ref{coutUg}). The value
function is defined by
\begin{equation}\label{opt1}
S_{x_0,T}(x) := \inf \{ C_{U,g}(T,u)\ \vert\ E_{x_0,T}(u)=x \},
\end{equation}
for every $x\in\mathbb{R}^n$ (with, as usual,
$\inf\emptyset:=-\infty$). We assume in the sequel that all data are
analytic.

The regularity of $S_{x_0,T}$ is closely related to the existence of
nontrivial minimizing singular trajectories starting from $x_0$.
It is proved in \cite{trelatJDCS} that, in the absence of minimizing
singular trajectories, the value function is continuous and
subanalytic (see e.g.\ \cite{Hironaka} for a definition of a subanalytic
function).
For driftless control-affine systems and $g\equiv 0$, the value
function coincides with the square of a sub-Riemannian distance
(see \cite{Be} for an introduction to sub-Riemannian geometry). In
particular, in this case, the value function is always continuous, but
the trivial trajectory $x(\cdot)\equiv x_0$ is always minimizing and
singular. Moreover, if there is no nontrivial minimizing singular
trajectories, then the value function is subanalytic outside $x_0$
(see \cite{Ag,AgS}). This situation holds for generic distributions of
rank greater than or equal to three (see \cite{AG,CJT}).

The results of Section \ref{subsec2.3} have the following
consequence on the regularity of $S_{x_0,T}$.

\begin{corollary}
With the notations of Corollary \ref{cor1112}, and
if in addition the functions $g$, $U$, and
the vector fields of the $(m+1)$-tuple in $O_{m+1}$
are analytic, then the associated value function $S_{x_0,T}$ is continuous
and subanalytic on its domain of definition.
\end{corollary}

\begin{remark}
If there exists a nontrivial minimizing trajectory, the value function
may fail to be subanalytic or even continuous.
For example, consider the control-affine system
in $\mathbb{R}^2$ given by
\begin{equation*}
\dot{x}(t)=1+y(t)^2,\quad
\dot{y}(t)=u(t),
\end{equation*}
and the cost
$C(T,u)=\int_0^T u(t)^2dt$.
The trajectory $(x(t)=t,y(t)=0)$, associated
to the control $u=0$, is a nontrivial minimizing singular trajectory,
and the value function $S_{(0,0),T}$ has the asymptotic expansion,
near the point $(T,0)$,
\begin{equation*}
S_{(0,0),T}(x,y)= \frac{1}{4}\frac{y^4}{x-T} +
\frac{y^4}{x-T}\,\mathrm{exp}\left(-\frac{y^2}{x-T}\right)
+\mathrm{o}\left(\frac{y^4}{x-T}\,\mathrm{exp}\left(-\frac{y^2}{x-T}\right)\right)
\end{equation*}
(see \cite{trelatJDCS} for details).
Hence, it is not continuous, nor subanalytic, at the point $(T,0)$.
\end{remark}

In the driftless control-affine case, by using results of Section
\ref{subsubsec2.4.2}, we derive the following similar consequence.

\begin{corollary}
With the notations of Corollary \ref{srcor1112}, and
if in addition the functions $g$, $U$, and
the vector fields of the $m$-tuple in $O_{m}$
are analytic, then the associated value function $S_{x_0,T}$ is
subanalytic outside $x_0$.
\end{corollary}

\subsection{Regularity of viscosity solutions of Hamilton-Jacobi equations}
Assume that the assumptions of the previous subsection hold.
It is standard (see \cite{Evans,Lions}) that the value function
$v(t,x)=S_{x_0,t}(x)$ is a viscosity solution of the Hamilton-Jacobi
equation
\begin{equation}\label{hjbeq}
\frac{\partial v}{\partial t} + \mathcal{H}\left(x,\frac{\partial v}{\partial
    x}\right)= g(t,x),
\end{equation}
where
$
\mathcal{H}(x,p) =
\langle p,f_0(x)\rangle + \frac{1}{2}\sum_{i,j=1}^m
(U^{-1}(x))_{ij} \langle p,f_i(x)\rangle \langle p,f_j(x)\rangle  $.

Conversely, the viscosity solution of (\ref{hjbeq}) with analytic
Dirichlet-type conditions is subanalytic, as soon as the corresponding
optimal control does not admit minimizing singular trajectories
(see \cite{trelatHJB}). Using the results of the previous sections,
this situation holds generically if $m\geq 2$ (and, similarly for
driftless control-affine systems, if $m\geq 3$).

As a consequence, the analytic singular set $\mathrm{Sing}(v)$ of the
viscosity solution $v$, i.e., the subset of $\mathbb{R}^n$ where $v$ is not
analytic, is a (subanalytic) stratified manifold of codimension
greater than or equal to one (see \cite{Tamm} for more details on the
subject). Since $\mathrm{Sing}(v)$ is also the locus
where characteristic curves intersect, the abovementioned property
turns out to be instrumental for the global convergence of numerical
schemes for Equation (\ref{hjbeq}) (see \cite{Evans}). Indeed, the
analytic singular set must be as ``nice'' as possible in order to
integrate energy functions on the set of characteristic curves.

\subsection{Applications to stabilization and motion planning}
For a driftless control-affine system verifying the Lie
Algebra Rank Condition, there exist general stabilizing strategies
stemming from dynamic programming. As usual, the stabilizing feedback
is computed using the gradient of the value function $S$ for a suitable
optimal control problem. Of course this is only possible outside the
singular set $\mathrm{Sing}(S)$, and one must device another
construction for the feedback on $\mathrm{Sing}(S)$.
Let us mention two such strategies, the first one providing an hybrid
feedback (see \cite{PrieurTrelatSIAM}), and the second one a
smooth repulsive stabilizing (SRS) feedback (see \cite{Rifford1,Rifford2}).
Both strategies crucially rely on the fact that $\mathrm{Sing}(S)$ is
a stratified manifold of codimension greater than or equal to one.

As seen before, the latter fact holds generically in the analytic
category for $m\geq 3$.

\medskip

On the other hand, the absence of singular minimizing trajectories is
the basic requirement for the convergence of usual algorithms in
optimal control (such as direct or indirect methods, see
e.g. \cite{Betts,Pesch}). We have proved that
this situation holds generically for control-affine systems if $m\geq
2$, and for driftless control-affine systems if $m\geq 3$.

As a final application,
consider a driftless control-affine system verifying the Lie
Algebra Rank Condition. According to Proposition \ref{srpropstrictB}, it is
possible to choose a (generic) cost function $C_{U,g}$ such that all
singular trajectories are strictly abnormal. Combining that fact with
\cite[Theorem 1.1]{RiffordTrelatMA}, we deduce that there exists a
dense subset $N$ of $\mathbb{R}^n$ such that every point of $N$ is reached by
a unique minimizing trajectory, which is moreover nonsingular.
As a consequence, a shooting method with target in $N$ will converge.
That fact may be used for solving (at least approximately) motion
planning problems.

\section{Proofs of the results}
\label{sec4}
\subsection{Proof of Theorem~\ref{thmrang} and of Theorem~\ref{srthmrang}}
Every trajectory of the control-affine system
$
\dot{x}=f_0(x)+\sum_{i=1}^mu_if_i(x)
$
is also a trajectory of the driftless control system
$
\dot{x}=\sum_{i=0}^mu_if_i(x),
$
with $u_0\equiv 1$. Therefore, Theorem~\ref{thmrang} follows from
Theorem~\ref{srthmrang}, whose proof is provided next.

Let $x(\cdot)=x(\cdot,x_0,u)$ be a trajectory of the
driftless control system
$\dot{x}=\sum_{i=1}^m u_if_i(x)$, with $2\leq m\leq n$.
Consider the set $I_{\mathrm{dep}}(x(\cdot))$ defined by
(\ref{Idep}). We argue by contraposition, and assume that
$I_{\mathrm{dep}}(x(\cdot))$ contains a subset $I$ of positive measure
such that $\dot{x}(t)\neq 0$ for $t\in I$.
Since Lebesgue points of $u$ are of full Lebesgue measure, we assume
that $u$ is continuous on $I$.

Up to considering a subset of $I$, and relabeling the $f_i$'s, we
assume that, for every $t\in I$:
\begin{itemize}
\item[(i)] there exists $1\leq k<m$ such that
$$\mathrm{rank}\{f_1(x(t)),\ldots,f_m(x(t))\}=k;$$
\item[(ii)] $f_1(x(t)),\ldots,f_k(x(t))$ are linearly independent, and
  thus, there exist real numbers $\alpha_i^j(t)$, $i=1,\ldots,k$,
  $j=k+1,\ldots,m$, such that
\begin{equation*}
f_j(x(t)) = \sum_{i=1}^k \alpha_i^j(t) f_i(x(t)),\quad j=k+1,\ldots,m.
\end{equation*}
Therefore,
$\displaystyle
\dot{x}(t) = \sum_{i=1}^k \delta_i(t) f_i(x(t)),
$
where $\displaystyle\delta_i(t):=u_i(t)+\sum_{j=k+1}^m \alpha_i^j(t)u_j(t)$;
\item[(iii)] $\delta_1(t)\neq 0$.
\end{itemize}

\begin{remark}\label{remrangouvert}
Up to reducing $I$, we furthermore assume that $I$ is contained in an
open interval $\mathcal{I}$ on which
$\mathrm{rank}\{f_1(x(t)),\ldots,f_k(x(t))\}=k$.
\end{remark}

Set $\mathrm{ad}^0g(h)=h$, where $g,h\in\mathrm{VF}(M)$,
and $\mathrm{ad}^kg(h)=[g,\mathrm{ad}^{k-1}g(h)]$ for $k\geq 1$.
The length of the iterated Lie bracket $[f_{i_1},[f_{i_2},[\cdots,
f_{i_k}]\cdots]$ of $f_1,\ldots,f_m$ is the integer $k$.

\begin{proposition}\label{lemroutine}
Let $N$ be a positive integer. There exists a subset $J_N\subset I$ of
positive measure such that, for every $t\in J_N$, and every
$\ell\in\{1,\ldots,N\}$,
\begin{equation}\label{rout0}
\delta_1(t)^{\ell-1} \mathrm{ad}^{\ell-1}f_1(f_m) (x(t)) =
h_t^\ell(x(t)) + R_t^\ell(x(t)),
\end{equation}
where
\begin{itemize}
\item $h_t^\ell(x(t))\in\mathrm{Span}\{f_1(x(t)),\ldots,f_k(x(t))\}$,
\item $R_t^\ell$ is a linear combination of iterated Lie
  brackets of $f_1,\ldots,f_m$, of length smaller than $\ell-1$, and
  of iterated Lie brackets of $f_1,\ldots,f_k$, of length smaller than
  or equal to $\ell$.
\end{itemize}
\end{proposition}

\begin{proof
For $t\in I$, let $F_t\in VF(M)$ be the vector field defined by
$$
F_t(x):=\sum_{i=1}^k \delta_i(t) f_i(x).
$$
Notice that $\dot{x}(t)=F_t(x(t))$, for $t\in I$.
For the argument of Proposition~\ref{lemroutine}, we need the
following lemma.

\begin{lemma}\label{lem-recursif}
Consider a set $J\subset I$ of positive measure and $h\in VF(M)$ so
that
$h(x(t))\in \mathrm{Span}\{f_1(x(t)),\ldots,f_k(x(t))\}$ on
$J$,
i.e., for every $t\in J$, there exist real numbers $\beta_i(t)$,
$i=1,\ldots,k$, such that
\begin{equation}\label{h00}
h(x(t)) = \sum_{i=1}^k \beta_i(t) f_i(x(t)).
\end{equation}
For $t\in J$, define $g_t\in VF(M)$ by
$$
g_t(x):= h(x)- \sum_{i=1}^k \beta_i(t) f_i(x).
$$
Then, there exists a set $J'\subset J$ of positive measure such
that
\begin{equation}\label{eq00}
[F_t,g_t](x(t))\in \mathrm{Span}\{f_1(x(t)),\ldots,f_k(x(t))\} \quad \hbox{ on
}J'.
\end{equation}
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lem-recursif}]
 Using Remark \ref{remrangouvert},
there exist $e_j\in VF(M)$, $k+1\leq j\leq n$, so
that, for every $t\in \mathcal{I}$, the vectors
$f_1(x(t))$, $\dots$, $f_k(x(t))$, $e_{k+1}(x(t))$, $\dots$, $e_n(x(t))$ span
$T_{x(t)}M$. Thus,
there exist $n$ smooth functions $b_i$, $1\leq
i\leq n$, defined on $M$, such that
$$
h(x)=\sum_{i=1}^kb_i(x)f_i(x)+\sum_{i=k+1}^nb_i(x)e_i(x),
$$
for $x$ in an open neighborhood of $x(\mathcal{I})$.
For $i=1,\dots, n$, define $\beta_i(t):=b_i(x(t))$, for $t\in \mathcal{I}$
(this notation is consistent with (\ref{h00})).
The $\beta_i's$ are absolutely continuous on $\mathcal{I}$ and
differentiable everywhere on $J$. For $i=k+1,\dots, n$, there holds
$\beta_i\equiv 0$ on $J$ and therefore, it follows that
$\dot\beta_i\equiv 0$ on a subset $J'\subset J$ of full
measure (cf. \cite[Lemma p.~177]{Ru}).

For $t\in J$, using that $g_t(x(t))=0$, and $F_t(x(t))=\dot{x}(t)$, it
holds
\begin{eqnarray*}
[F_t,g_t](x(t)) &=& dg_t\circ F_t(x(t)) \\
&=& \sum_{i=1}^k (db_i(x(t)).\dot{x}(t)) f_i(x(t)) +
 \sum_{i=k+1}^n (db_i(x(t)).\dot{x}(t)) e_i(x(t)) \\
&=& \sum_{i=1}^k \dot{\beta}_i(t) f_i(x(t)) +
 \sum_{i=k+1}^n \dot{\beta}_i(t) e_i(x(t)).
\end{eqnarray*}
On $J'$, the second sum of the right-hand side of the last equation
vanishes, and the lemma follows.
\end{proof}

Applying Lemma \ref{lem-recursif} to $h=f_m$ and $J=I$, we get
\begin{equation*}
[F_t,g_t^1](x(t))\in \mathrm{Span}\{f_1(x(t)),\ldots,f_k(x(t))\} \quad \hbox{ on
}J_1,
\end{equation*}
where $J_1\subset I$ and
$g_t^1 := f_m-\sum_{i=1}^k \alpha_i^m(t)f_i$.

Set $h_t^1=[F_t,g_t^1]$. We next iterate the above procedure, for
$1\leq\ell\leq N$. Assume that the vector fields $h_t^\ell$,
$g_t^\ell$, and the set $J_\ell$ of positive measure are defined, such
that
$h_t^\ell(x(t))\in \mathrm{Span}\{f_1(x(t)),\ldots,f_k(x(t))\}$ on
$J_\ell$.
For every $t\in J_\ell$, let $\beta_i^\ell(t)$,
$i=1,\ldots,k$, be the real numbers such that
\begin{equation*}
h_t^\ell(x(t)) = \sum_{i=1}^k \beta_i^\ell(t) f_i(x(t)),
\end{equation*}
and define $g_t^{\ell+1}\in VF(M)$ by
$
g_t^{\ell+1}:= h_t^\ell- \sum_{i=1}^k \beta_i^\ell(t) f_i.
$
Set $ h_t^{\ell+1} := [F_t,g_t^{\ell+1}]$.
Applying Lemma \ref{lem-recursif}, there exists a subset
$J_{\ell+1}\subset J_\ell$ of positive measure such that
$h_t^{\ell+1}(x(t))\in \mathrm{Span}\{f_1(x(t)),\ldots,f_k(x(t))\}$
on $J_{\ell+1}$.

For $t\in J_N$, and for $\ell=1,\ldots,N$, we express $h_t^\ell(x(t))$
using iterated Lie brackets of $f_1,\ldots,f_m$, and an easy induction
yields (\ref{rout0}).
\end{proof}

Combining Proposition \ref{lemroutine} with routine transversality
arguments (see for instance \cite{BK} and \cite{CJT}), it follows that
the $(m+1)$-tuple $(f_0,\ldots,f_m)$ belongs to a closed subset of
$VF(M)^{m+1}$ of codimension greater than or equal to
$N$. Theorem~\ref{srthmrang} follows.

\begin{remark}\label{remtransverse}
The fact that $f_1(x(t))\neq 0$ is essential in order
to derive, from (\ref{rout0}), an infinite number of independent
relations, and then to apply the abovementioned transversality
arguments.
\end{remark}

\subsection{Proof of Theorem \ref{thmA}}
The minimal order and corank one properties are proved separately in
the following lemmas.

\begin{lemma}\label{keylemma1}
There exists an open and dense subset $O^1_{m+1}$ of $VF(M)^{m+1}$ so that,
if the $(m+1)$-tuple $(f_0,\ldots,f_m)$ belongs to $O^1_{m+1}$, then every
singular trajectory of the associated control-affine system
$
\dot{x}(t)=f_0(x(t))+\sum_{i=1}^mu_i(t)f_i(x(t))
$
is of minimal order.
In addition, for every integer $N$, the set $O^1_{m+1}$ can be chosen
so that its complement has codimension greater than $N$.
\end{lemma}

\begin{lemma}\label{keylemma2}
There exists an open and dense subset $O_{m+1}$ of $O^1_{m+1}$ so that,
if the $(m+1)$-tuple $(f_0,\ldots,f_m)$ belongs to $O_{m+1}$, then every nontrivial
singular trajectory of the associated control-affine system
$
\dot{x}(t)=f_0(x(t))+\sum_{i=1}^mu_i(t)f_i(x(t))
$
is of corank one.
In addition, for every integer $N$, the set  $O^2_{m+1}$ can be chosen
so that its complement has codimension greater than $N$.
\end{lemma}

The conclusion of Theorem \ref{thmA} with the set $O_{m+1}$ whose
existence is stated above.

\subsubsection{Proof of Lemma \ref{keylemma1}}
 From Theorem \ref{thmrang}, there exists an open and dense subset $O_{m+1}^{11}$
of $VF(M)^{m+1}$ such that, if $(f_0,\ldots,f_m)\in O_{m+1}^{11}$, then
every trajectory $x(\cdot)$ of $\dot{x}=f_0(x)+\sum_{i=1}^m u_if_i(x)$
verifies Item (i) of Definition \ref{defordre}.

It is therefore enough to show the existence of an open and dense subset
$O_{m+1}^{12}$ of $VF(M)^{m+1}$ such that, if $(f_0,\ldots,f_m)\in
O_{m+1}^{12}$, then every singular trajectory $x(\cdot)$ of
$\dot{x}=f_0(x)+\sum_{i=1}^m u_if_i(x)$ verifies Item (ii) of
Definition \ref{defordre}.
Then, by choosing $O^1_{m+1}:=O_{m+1}^{11}\cap O_{m+1}^{12}$, the
conclusion of
Lemma~\ref{keylemma1} follows.

Consider a singular trajectory $x(\cdot):=x(\cdot,x_0,u)$ of
$\dot{x}=f_0(x)+\sum_{i=1}^m u_if_i(x)$, admitting
an abnormal extremal $(x(\cdot),\lambda(\cdot))$. Assume that there
exists $J\subset [0,T]\setminus I_{dep}(x(\cdot))$ of
positive measure such that
$G(t)$ is not of rank $m$ if $m$ is even, respectively,
$\widetilde{G}(t)$ is not of rank $m$ if $m$ is odd.
We will show that the $(m+1)$-tuple  $(f_0,\ldots,f_m)$ belongs to a
subset of arbitrary codimension in $VF(M)^{m+1}$ whose complement
contains an open and dense subset.

Note that, on $[0,T]\setminus I_{dep}(x(\cdot))$, the vector fields
$f_0(x(t)),\ldots, f_m(x(t))$ are linearly independent.
The remaining part of the argument consists of reformulating the
problem in order to follow the chain of arguments in the proof of
\cite[Theorem 2.4]{CJT}
concerning the case of everywhere linearly independent vector fields.
For that purpose, we distinguish the cases $m$ even and $m$ odd.

Assume first that $m$ is even. As in (\ref{Gbar}), define, for $t\in
J$,
$
\overline{G}(t):=\big(h_{ij}(t)\big)_{0\leq i,j\leq m}.
$
 From (\ref{Au=b}), we have, for $t\in J$,
$$
\overline{G}(t) =
\begin{pmatrix}
  0 & \big(G(t)u(t)\big)^T \\
  -G(t)u(t) & G(t)
\end{pmatrix} .
$$
Since the ranks of both $\overline{G}(t)$ and $G(t)$ are even,
they must be equal, for $t\in J$, and hence,
the rank of $\overline{G}(t)$ is smaller than $m$ on $J$.
This is exactly the starting point of the proof of
\cite[Lemma 3.8]{CJT}.
The machinery of \cite{CJT} then applies and we deduce that the
$(m+1)$-tuple  $(f_0,\ldots,f_m)$ belongs to a subset of arbitrary
codimension in $VF(M)^{m+1}$ whose complement contains an open and
dense subset $O_{m+1}^2$ of $VF(M)^{m+1}$.

Assume next that $m$ is odd.
Define the $(m+2)\times (m+1)$ matrix $\widehat{G}(t)$ as
$\overline{G}(t)$ augmented in the last row with
$(\{\overline{P},h_j\}(t))_{0\leq j\leq m}$.

\begin{lemma}\label{lemrank}
With the notations above,
$
\mathrm{rank}\ \widehat{G}(t) \leq \mathrm{rank}\ \widetilde{G}(t)+1.
$
\end{lemma}

\begin{proof
It amounts to show that, $\xi\in\ker\widetilde{G}(t)$ implies
$(0,\xi)\in\ker\widehat{G}(t)$. This follows from the fact that
if $\widetilde{G}(t)\xi=0$, then $G(t)\xi=0$, and thus $\xi$ is
orthogonal to the range of $G(t)$ since $G(t)$ is skew-symmetric.
\end{proof}

Using Lemma \ref{lemrank}, the rank of $\widehat{G}(t)$ is less than
$m+1$ on $J$.
This is exactly the starting point of the proof of \cite[Lemma
3.9]{CJT}. The machinery of \cite{CJT} then applies and we deduce that
the $(m+1)$-tuple  $(f_0,\ldots,f_m)$ belongs to a subset of arbitrary
codimension in $VF(M)^{m+1}$ whose complement contains an open and
dense subset $O_{m+1}^{12}$ of $VF(M)^{m+1}$.

\subsubsection{Proof of Lemma \ref{keylemma2}}
We argue by contraposition.
Consider a nontrivial singular trajectory $x(\cdot):=x(\cdot,x_0,u)$ of
$\dot{x}=f_0(x)+\sum_{i=1}^m u_if_i(x)$, with $(f_0,\ldots,f_m)\in
O_{m+1}^1$. Assume that $x(\cdot)$ admits two abnormal extremal
lifts $(x(\cdot),\lambda^{[1]}(\cdot))$ and
$(x(\cdot),\lambda^{[2]}(\cdot))$ such that, for some $t_0 \in [0,T]$,
$\lambda^{[1]}(t_0)$ and $\lambda^{[2]}(t_0)$ are linearly independent.
By linearity, $\lambda^{[1]}(\cdot)$ and $\lambda^{[2]}(\cdot)$ are linearly independent
everywhere on $[0,T]$.
Since $x(\cdot)$ is nontrivial, it follows from
Remark~\ref{remtrivial} that there exists a nonempty subinterval $J$
of $[0,T]\setminus I_{dep}(x(\cdot))$.
We are now in a position to exactly follow the arguments of \cite{CJT}
corresponding to the corank one property, i.e., \cite[Lemma 4.4]{CJT}.

\subsection{Proof of Theorem~\ref{srthmA}}
We start with the proof of the statement dealing with the minimal
order property.

 From Theorem~\ref{srthmrang}, there exists an open and dense subset
$O_{m}^{1}$ of $VF(M)^{m}$ such that, if $(f_1,\ldots,f_m)\in
O_{m}^{1}$, then
every trajectory $x(\cdot)$ of $\dot{x}=\sum_{i=1}^m u_if_i(x)$
verifies Item (i) of Definition~\ref{srdefordre}.

It is therefore enough to show the existence of an open and dense subset
$O_{m}^{2}$ of $VF(M)^{m}$ such that, if $(f_1,\ldots,f_m)\in
O_{m}^{2}$, then every singular trajectory $x(\cdot)$ of
$\dot{x}=\sum_{i=1}^m u_if_i(x)$ verifies Item (ii) of
Definition~\ref{srdefordre}.
Then, by choosing $O_{m}:=O_{m}^{1}\cap O_{m}^{2}$, the
statement dealing with the minimal
order property in Theorem~\ref{srthmA} follows.

Consider a singular trajectory $x(\cdot):=x(\cdot,x_0,u)$ of
$\dot{x}=\sum_{i=1}^m u_i f_i(x)$ admitting
an abnormal extremal $(x(\cdot),\lambda(\cdot))$. Assume that there
exists $J\subset [0,T]\setminus I_{dep}(x(\cdot))$ of
positive measure such that
$G(t)$ is not of rank $m-1$ if $m$ is odd, respectively,
$\widetilde{G}(t)$ is not of rank $m-1$ if $m$ is even.
Following exactly the proofs of Lemmas~3.8 and 3.9 in~\cite{CJT}, the
$m$-tuple  $(f_1,\ldots,f_m)$ belongs to a
subset of arbitrary codimension in $VF(M)^{m}$ whose complement
contains an open and dense subset.

We proceed similarly for an argument of the  statement dealing with
the corank one property.

\subsection{Proofs of Propositions~\ref{propstrictA}
  and \ref{srpropstrictA}}
We only treat the control-affine case, the argument for the driftless
control-affine case being identical. We argue by contraposition.
Consider a nontrivial singular trajectory $x(\cdot):=x(\cdot,x_0,u)$ of
$\dot{x}=f_0(x)+\sum_{i=1}^m u_if_i(x)$, with $(f_0,\ldots,f_m)\in
VF(M)^{m+1}$. Assume that $x(\cdot)$ admits on the one
part a normal extremal lift $(x(\cdot),\lambda^{[n]}(\cdot))$ and on
the other part an abnormal extremal lift
$(x(\cdot),\lambda^{[a]}(\cdot))$.

Let us introduce some notations.
For $k\in \mathbb{N}$, let $L=l_1 \cdots l_k$ be a multi-index of
$\{0,\dots,m\}$. The length of $L$ is $|L|=k$ and $f_{L}$ is the
vector field defined by
$$
f_L:=[[\dots[f_{l_1},f_{l_2}], \dots],f_{l_k}].
$$
A multi-index $L=j l \cdots l$ with $k$ consecutive occurrences
of the index $l$ is denoted as $L=j l^k$.

For every multi-index $L$ of $\{0,\ldots,m\}$ and $t\in [0,T]$, set
$$
h_L^{[n]}(t) = \langle\lambda^{[n]}(t),f_L(x(t))\rangle, \
\textrm{and}\ \
h_L^{[a]}(t) = \langle\lambda^{[a]}(t),f_L(x(t))\rangle.
$$
After time differentiation, we have on $[0,T]$,
\begin{eqnarray}
\frac{d}{dt} h_{L}^{[n]}(t)
  = \sum_{l=1}^m u_l(t) h_{Ll}^{[n]}(t)\label{eq:00},\\
\frac{d}{dt} h_{L}^{[a]}(t)
  = \sum_{l=1}^m u_l(t) h_{Ll}^{[a]}(t)\label{eq:11}.
\end{eqnarray}
Recall that, according to the Pontryagin Maximum Principle, there holds
\begin{equation}\label{inconnus}
u(t) = \begin{pmatrix}u_1(t)\\ \vdots \\ u_m(t)\end{pmatrix}
=  U(x(t))^{-1} \begin{pmatrix} h_1^{[n]}(t)
  \\ \vdots \\ h_m^{[n]}(t)\end{pmatrix},
\end{equation}
and, for every $t\in [0,T]$,
\begin{equation}\label{eqanor}
h_0^{[a]}(t)=\textrm{constant}, \ h_l^{[a]}(t) = 0 ,
\end{equation}
for every $l\in\{1,\ldots,m\}$, and $t\in[0,T]$. Since the trajectory
$x(\cdot)$ is nontrivial, there exists an open interval
$J\subset[0,T]$ and $i\in\{0,\ldots,m\}$ such that
$u_{i}(\cdot)f_{i}(x(\cdot))$ is never vanishing (with the
convention $u_0\equiv 1$). Fix $j\in\{0,\ldots,m\}\setminus\{i\}$.
Differentiating $s$ times (with $s\geq 1$) the relation $h_j^{[a]}(t)
= \mathrm{constant}$ with respect to $t\in J$, one gets, by using
(\ref{eq:00}), (\ref{eq:11}), and (\ref{inconnus}), that
\begin{equation}\label{eq:induc}
0 = \frac{d^s}{dt^s} h_{j}^{[a]}(t)
  = (u_i(t))^s h^{[a]}_{ji^s}(t) + R_s(t),
\end{equation}
where $R_{s}(t)$ is polynomial in $h_L^{[n]}(t)$ and $h_K^{[a]}(t)$, $\vert
L\vert\leq s$, $\vert K\vert\leq s+1$, with $K$ different from
$ji^s$ and $iji^{s-1}$.
Fix $t\in J$. Since $u_i(t)\neq 0$ and $f_i(x(t))\neq 0$,
we are in a position to apply routine transversality arguments.
It follows that
the $(m+1)$-tuple $(f_0,\ldots,f_m)$ belongs to a closed subset of
$VF(M)^{m+1}$ of arbitrary codimension. Proposition \ref{propstrictA}
follows.

\subsection{Proofs of Propositions~\ref{propstrictB},
  \ref{propstrictB11}, and~\ref{srpropstrictB}}
We first prove Proposition \ref{propstrictB11} and argue by contraposition.
Consider a nontrivial singular trajectory $x(\cdot):=x(\cdot,x_0,u)$ of
$\dot{x}=f_0(x)+\sum_{i=1}^m u_if_i(x)$. Assume that $x(\cdot)$ admits
on the one part a normal extremal lift
$(x(\cdot),\lambda^{[n]}(\cdot))$ and on the other part an abnormal
extremal lift $(x(\cdot),\lambda^{[a]}(\cdot))$.

 From the Pontryagin Maximum Principle, there holds, for
$l=1,\ldots,m$,
$$ u_l(t) = \sum_{p=1}^m Q^{lp}(x(t))  \beta_p(x(t)),\qquad
\beta_p(x(t)):=h^{[n]}_p(t)-\alpha_p(x(t)),$$
where the $Q^{lp}(x)$ and the $\alpha_p(x)$ are respectively the
coefficients of $U^{-1}(x)$ and of $\alpha(x)$. Note that
the $u_l$'s are smooth functions of the time.

Since the trajectory $x(\cdot)$ is nontrivial, there exists an open
interval $J\subset[0,T]$ such that $\dot x$ is never vanishing on $J$
and one of the two following cases holds.
\medskip

\underline{Case $1$:} $u\equiv 0$ on $J$.

In that case, $\dot x(t)=f_0(x(t))$ for $t\in J$,
and $f_0(x(\cdot))$ is never vanishing on $J$. Moreover, for
$p=1,\ldots,m$, $\beta_p\equiv 0$ on $J$,
i.e. $\alpha_p(x(t))=h^{[n]}_p(t)$ for $t\in J$. By differentiating
the latter relation with respect to the time, we deduce that, for all
$N\geq 0$, $t\in J$ and $p=1,\dots,m$,
$$
L_{f_0}^{N}\alpha_p(x(t))=L_{f_0}^{N}h^{[n]}_p(x(t)),
$$
where $L_{f_0}$ denotes the Lie derivative with respect to the vector
field $f_0$.
Applying routine transversality arguments, it follows that
$\alpha$ belongs to a closed subset of $C^{\infty}(M,\mathbb{R}^m)$
of arbitrary codimension.

\medskip

\underline{Case $2$:} $u$ is never vanishing on $J$.

Using (\ref{s-affine}) and the Lie Algebra Rank Condition, there exist
a multi-index $L$, an index $j_0\in\{0,\ldots,m\}$, and a subinterval of $J$
(still denoted $J$), such that
$$h_L^{[a]}(t)=\mathrm{constant},\ \textrm{and}\
h_{Lj_0}^{[a]}(t)\neq 0,$$
for every $t\in J$.
Differentiating $h_L^{[a]}$ on $J$, one gets
\begin{equation}\label{formder}
\begin{split}
0 = \frac{d}{dt}h_L^{[a]}(t) &=
 h_{L0}^{[a]}(t) + \sum_{l=1}^m u_l(t)  h^{[a]}_{Ll}(t) \\
& =  h_{L0}^{[a]}(t) + \sum_{1\leq l\leq p\leq m}
 c_{lp}(t)  Q^{lp}(x(t)) ,
\end{split}
\end{equation}
where $c_{ll}(t) :=  \beta_l(t) h^{[a]}_{Ll}(t) $, and
$ c_{lp}(t) := \beta_p(t) h^{[a]}_{Ll}(t) +  \beta_l(t)
h^{[a]}_{Lp}(t)$ if $l< p$.

\begin{lemma}
Up to reducing the interval $J$, there exist indices $j$ and $l$ in
$\{1,\ldots,m\}$ such that $c_{lj}(t)$ or $c_{jl}(t)$ is never
vanishing on $J$.
\end{lemma}

\begin{proof}
If $j_0=0$, then $h^{[a]}_{L0}(t)\neq 0$, and it follows from
(\ref{formder}) that there exist $l,j\in\{1,\ldots,m\}$ such that
$c_{lj}(t)\neq 0$. Otherwise, take $j:=j_0$. In that case, one of the
$\beta_p$'s
does not vanish on $J$ since $u$ is not zero. First, assume that
$\beta_{j}(t)$ is not identically equal to zero on $J$; then, up to
reducing $J$,
$c_{jj}(t)$ is never vanishing on $J$. Otherwise, there exists $l\neq
j$ such that, up to reducing $J$, $\beta_l$ is never vanishing on $J$
and thus similarly for $c_{lj}$ (or $c_{jl}$).
\end{proof}

For $t\in J$, let $F_t\in VF(M)$ be the vector field defined by
$$F_t(x) := f_0(x)+ \sum_{i=1}^m u_i(t) f_i(x).$$
Notice that $F_t(x(t))=\dot{x}(t)\neq 0$.
For all $N\geq 0$ and $t\in J$, we get, by taking the $(N+1)$-th time
derivative of $h_L^{[a]}$ on $J$,
$$
0 = \frac{d^{N+1}}{dt^{N+1}}h_L^{[a]}(t) =
 c_{lj}(t) L^N_{F_t} Q^{jl}(x(t))
+ R_N(t),
$$
where $R_N(t)$ is a linear combination of $L^s_{F_t}
Q^{pi}(x(t))$, with $s\leq N$, $p\leq i$ in $\{1,\dots,m\}$ and
$s<N$ if  $(p,i)=(j,l)$, and of
$L^s_{f_r}Q^{pi}(x(t))$, with $s<N$, $p\leq i$ in $\{1,\ldots,m\}$,
and $r\in\{0,\ldots,m\}$.
Applying routine transversality arguments,
it follows that $(U,\alpha)$ belongs to a closed subset of
$\mathcal{S}^+_m(M)\times C^{\infty}(M,\mathbb{R}^m)$
 of arbitrary codimension. Proposition
\ref{propstrictB11} is proved.

To show Propositions~\ref{propstrictB} and~\ref{srpropstrictB}, simply
notice that the
argument of Case $2$ with $\alpha=0$ applies with suitable
modifications.

redpajama_set_name: RedPajamaArXiv


As usual, the sun took hold of the день in Danville, giving warm rays of bright happiness throughout the whole Tri-state area. Though the morning was very misleading for today, the Flynn-Fletcher boys were hard at work, hammering down on the day's project.
"A Flying Machine!" Phineas stretched his arms wide, loosing grip and falling backward. Lucky he was attached to the rope, или else his head would have slammed to the ground. Now he was upside down, dangling and bouncing slightly. "Well, I promise that wont happen again," He chuckled, turning upright and shaking his head.
Isabella giggled, "What do Ты need help with?"
"Come on, I'll Показать you," He took her hand and held onto her waist as he was pulled up again. Isabella let out a slight yelp as they flew upward so quickly. "Here." Phineas let go and guided her to some other screws that needed to be hooked in. "I'll be right back to you, I just need to assist Ferb for a moment. For now, we just need help with those screws."
"Got it!" Picking up the electric отвертка that sat Далее to her feet, Isabella quickly hooked them in.
"What are Ты doing?" She demanded, stomping up to her little brothers. "What is this?"
"Well," Phineas began, smiling and feeling proud, adding a slight chuckle. "It makes your Кости lighter for thirty минуты so you're able to fly. It's actually quite simple, all Ты do is insert the manipulator into-"
"I don't care!" Candace interrupted, crossing her arms. Her blue eyes blazed with cold, angry огонь as she growled, "I'm. Telling. Mom."
"Well, she's happier than she was yesterday," Phineas прокомментировал(-а) to Ferb. Ferb nodded and they got briskly back to work.
Isabella glanced at their project. "Are Ты sure? It looks perfect to me."
Just when he was about to answer, Phineas's mother came out with Candace and Lawrence fallowing behind. "Phineas, Ferb," She called. "I'm going to the antique store with your father. Candace is coming along, I think she needs some time away from here..." Linda rolled her eyes and smiled. "We wont be long, Ты kids be good!"
The sound of the Flying Machine starting up edged Phineas, making him even еще eager to try it out. Running over, he accidentally tripped over something, almost panicking once he realized what it was. It was the manipulator! "Oh, no..." Phineas gasped at the Flying machine began to smoke, sparks flying everywhere as it shook with steaming rage. "Everyone, get back!" He demanded with a yelp, quickly stumbling up and running to the safety of their oak tree. Ferb wasn't far behind as Phineas tried to climb up as quick as he could. Fear rattled him like a maraca every step higher he went into the tree. His auspiciousness made him feel regretful and stupid for not installing the main important part.
But еще important that that was the fact that Isabella was no where to be seen. "Wait here!" Phineas told Ferb, almost falling down as he tried to locate her. "Isabella, where are you?" He called, looking everywhere over the lawn.
"No time for explanations, we must hurry!" Almost dragging Isabella by the hand, they were barely six feet away from the дерево when it happened.
Once realizing his brother was awake and okay, Ferb jumped down from the дерево and rushed over. "You're alright?" He asked.
"Do Ты know where she is?"
"Isabella?" Phineas's voice rose to locate her. "Are Ты hurt?"
The sight of blood-tipped трава made Phineas's eyes widen. "Oh, no...No, no, no!" Speeding up quicker than a cheetah, he zipped around the дерево and skidded to a halt, his words caught in his throat, as if he was choking. The image he saw was unbelievable; it just couldn't be true!
"Told me what?" Phineas prompted, eager to know what she was going to say. "If it's about how I left Ты and this happened, I know, I'm sorry-"
"Shh," With her eyes еще than halfway closed, she put her finger to his lips. "Let me finish."
"Phineas," She croaked, a single tear running slowly down her perfect face. "I've"-she stopped to cough again-"I've noticed th-that Ты probably care a l-lot for me, an-nd that I really like y-you..." While hearing her heartbreaking stutters, Phineas held her hands once еще as she lay down on the cold grass, her breath slowing.
"And I'm sorry that it took Ты to get hurt for me to realize," Phineas apologized, suddenly becoming even sadder. "Please..." He begged, closing his eyes to try to stop the waterfall of tears. "Don't leave me...I'll miss Ты terribly..."
Gray rain started to patter the ground, hissing secrets as it splashed to the earth. Isabella's blood began to wash slowly away, trickling down the grass. She hugged him back, whispering groggily into his ear, "Phineas, I Любовь you, and never, ever forget it."
"I know, and trust me, I never will forget it," Closing his eyes and just letting the tears roll down his face, he replied smoothly, "And I Любовь you, too."
"Phineas," He whispered as he set a hand on the twelve-year-old's shoulder. "I know Ты loved her, and now that she's gone-"
Clearly Phineas didn't get it. "Phineas," Ferb said, "It's too late. We've Остаться в живых her."
Phineas weakly smiled. "You know," He began. "People die, but real Любовь is forever."


Our crisp refreshing, very easy drinking Mexican-style lager, inspired by the crossing of the King River and Ovens River meeting place, now the township of Wangaratta. Using pilsner malt for a clean, refined base with flaked maize to lighten the body, its a perfect quencher on a hot day.
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304 Pearce Drive is a $400,000, 3,814 square foot, 4 bedroom, 4.0 bath home on a 0.40 acre lot located in Jamestown, NC.
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What was your first Spyderco?
A Delica first, then I found out about the Paramilitary 2 and have been hooked ever since. I'm a big fan of Spyderco. I know......to much information! I'll be quiet now.
"What was your first Spyderco?"
That's a good question. I purchased my first Spyderco in 1999 at a local dealer, but I don't remember the model name. I believe it was a Jess Horn Lightweight.
A Chapparal FRN arrived today. It's the first pocketknife I ever bought. I did not know how I'd be using it, so I wanted one unobtrusive but competent. I barely know it's in my pocket, so the first test passed!
My boy is getting a less expensive knife as a gift for Xmas but that one's very different: Ruike P801. Obviously I'll be playing with that one as well to get a feel for a larger, heavier (but more mainstream size) knife compared to mine. Regardless, I know I want to also get a PM2 someday.
Another gift will be for my daughter, a SAK Spartan, in Massdrop's "City Street" colorway. As a kid I was given a small SAK of some kind but it's been gone a long time.
Spyderco UKPK in S30v and G10. Bought it the day after I attended a Spyderco meet-up in Leeds UK. Been EDCing a UKPK ever since.
Zdp189 delica sprint. Hated the steel. Factory edge was prone to microchipping. So, I sharpened them out and sold it. Glad that experience didn't turn me off to spyderco knives though.
Blue weave Domino, probably 2015. I wasn't sure what i'd think of the Spydie Hole at the time, so the flipper seemed a safer way to dip my toe in.


text:
U.S. Takes Control of Group B in 2009 CONCACAF Gold Cup With 2-0 Victory Against
Jul 10, 2009 by Rodolfo Macias	in Sports
WASHINGTON, D.C. — Second half goals by Santino Quaranta and Brian Ching four minutes apart fired the U.S. to a 2-0 victory against Honduras and the top spot in Group B before an enthusiastic crowd of 26,079 at RFK Stadium.
The U.S. currently sits in first place in Group B as the only team with six points after two games. Honduras and Haiti each have three points, while Grenada sits in last place with zero. A win or a tie for the U.S. against Haiti in their final group match on Saturday, July 11, in Foxborough, Mass., would clinch the top spot in the group. With a five-goal advantage over second place Haiti in goal differential, the U.S. could also still finish first or second in the group even with a first-ever loss in group play in the Gold Cup. Their current all-time record stands at 22-0-3 through 10 regional championship appearances.
"We knew that this would be a good challenge tonight," said U.S. Men's National Team head coach Bob Bradley. "Honduras is a good team and this was the game that was going to go a long ways towards determining who would finish first in the group. I thought the response of our team was quite good. Games like this there's got to be a good collective effort from start to finish, you have to have the ability at certain points to bring on subs who help and in all ways I thought it was a good solid team effort."
For each of the goal scorers, the match marked a return to international action after a layoff – although Quaranta's was by far the most notable and longest. The 24-year-old, who was a member of the U.S. Gold Cup champion team in 2005 and made his most recent appearance more than three years ago, provided a storybook return by netting the game-winner – and first career international goal – in his home stadium in front of family and friends.
Quaranta broke the deadlock after a nice build-up that started on the right sideline when Chad Marshall reacted quickly to step around his man and win the ball just across the midfield line. The ball found the feet of Heath Pearce and he squared the ball for Benny Feilhaber, who entered the game only 10 minutes earlier. Feilhaber played a quick combination with Ching and then found his fellow substitute Charlie Davies with his back to goal about 16 yards out. Holding off his man, Davies smartly guided the ball to his right toward a charging Quaranta near the top of the box and the D.C. United midfielder directed a hard, first-time shot low and to the left of goalkeeper Donis Escobar.
"It was a very special night for me, personally," said Quaranta. "But to be back wearing the jersey was probably more special. Just to be back with the team and to listen to the anthem was very emotional for me because it's been such a fun and long road back. It was a special night."
The goal took the wind out of the Hondurans' sails, and off a restart in the 79th-minute the U.S. doubled the lead. Ching, who hadn't played for the U.S. since April 1 due to a hamstring strain, drew a foul trying to turn with the ball 40 yards from goal on the left side of midfield, and after Honduras brought on Georgie Welcome for Carlos Costly, Feilhaber quickly restarted play and switched the ball to Steve Cherundolo who had acres of space on the right side. Cherundolo served a cross into the six-yard box, where Ching out jumped his defender and headed the ball off Nery Medina's shoulder and into the top of the net.
The goal was Ching's 10th of his career, becoming the 18th player in U.S. history to tally double digits in his career. It was also Ching's second career Gold Cup goal, with the last coincidentally coming in the second group match of the 2007 Gold Cup.
The two goals were well deserved as the U.S. controlled most of the pace throughout the match even though both teams had chances during the scoreless first half in front of a festive atmosphere charged by partisan fans supporting both sides.
Bradley brought in three new faces to the starting lineup from the Group B against Grenada on July 4 by Quaranta starting on the right, Ching starting at forward, and Michael Parkhurst stepping into the back line.
The U.S. got the first chance in the 15th minute when a series of passes and a nice flick from Freddy Adu put Robbie Rogers through to the baseline. He chipped a ball to the penalty spot, where a charging Quaranta put a diving header just wide. Seven minutes later, Kyle Beckerman played Ching through the back line but was shouldered off as he attempted to shoot and slipped the ball past the goalkeeper but off target before it was cleared.
After the U.S. had grabbed control of possession, finding space on the wings for the combination of Robbie Rogers and Pearce on the left and Cherundolo on the right making advances, Honduras began claiming the momentum in the last 15 minutes of the first half with pressure on the U.S.
In the 30th minute, the U.S adverted disaster after a bad giveaway in midfield providing a quick counter for Honduras. In behind the U.S. defense and all alone, Walter Martinez had only Troy Perkins to beat. Choosing to go near post, his effort struck the outside of the side netting.
The U.S. had two more scoring opportunities in the 34th and 38th minute. Adu's golden chance came as the U.S. countered off a Honduran free kick. Ching sent Pearce off to the races down the left wing, where he picked out Adu alone 10 yards out, but his header went straight into the hands of Escobar. Four minutes later, Rogers hit a bouncing volley over the crossbar on a back-post cross by Quaranta.
The second half opened with neither team staking claim, although as the game carried on the U.S. began playing downhill. Bradley went to his bench in the 64th minute and brought on a pair of veterans from the team that finished second at the Confederations Cup, bringing on Davies and Feilhaber in favor of Adu and Logan Pause, respectively.
The hustle of Davies saw the U.S. nearly steal a goal in the 71st minute. As Honduran goalkeeper Donis Escobar received a back pass and set up to clear, Davies closed quickly and blocked the kick, sending the ball towards the Honduran goal. Mercifully for Honduras, it lacked the pace to cross the line and Escobar was able to collect.
Four minutes later, Quaranta put the U.S. ahead and Ching followed suit to put the game away. After the second goal, Kenny Cooper replaced Ching and the U.S., through the work of Feilhaber's fresh legs, kept possession to finish off the game. Cooper ended up with three shots, but none could beat Escobar.
In stoppage time, Perkins was called upon one final time as Honduras had a free kick from 30 yards out. As Honduras pushed their attackers forward for a cross at the far post, Martinez attempted a surprise shot, but Perkins pushed the blast wide of his left post to preserve his second shutout in three appearances.
Following tonight's match, both Adu and Feilhaber will join their clubs in Europe to begin preseason training for the upcoming season.
The U.S. improved to 12-3-3 at RFK Stadium, and closes out Group B play at against Haiti on July 11 in another familiar location, Foxborough, Mass., where the U.S. is 16-1-4 all time, including a 6-0-1 record in the Gold Cup at Gillette Stadium. Kickoff on Saturday is 7 p.m. ET live on Fox Soccer Channel, TeleFutura and the Futbol de Primera Radio Network.
Haiti currently sits in second place in the group with three points and a plus-one goal differential, following a 2-0 victory against Grenada in the first match of the evening. Honduras is third with three points and a minus-one goal differential, and faces Grenada on Saturday. Tickets for the Foxborough doubleheader are available at goldcup.org.
Match: United States vs. Honduras
Competition: CONCACAF Gold Cup – Group Stage
Venue: RFK Stadium – Washington, D.C.
Kickoff: 9 p.m. ET
Attendance: 26, 079
Weather: 71 degrees, mostly cloudy
Scoring Summary: 1 2 F
USA 0 2 2
HON 0 0 0
USA – Santino Quaranta (Charlie Davies) 75 minute
USA – Brian Ching (Steve Cherundolo) 79
USA: 1-Troy Perkins; 6-Steve Cherundolo (capt.), 4-Chad Marshall, 14-Michael Parkhurst, 2-Heath Pearce; 20-Santino Quaranta, 8-Logan Pause (25- Benny Feilhaber, 64), 5-Kyle Beckerman, 7-Robbie Rogers; 11-Brian Ching (17-Kenny Cooper, 82), 19-Freddy Adu (9-Charlie Davies, 64)
Subs not used: 10-Stuart Holden, 16-Jay Heaps, 18-Luis Robles, 21-Brad Evans
Head Coach: Bob Bradley
HON: 22-Donis Escober; 5-Erick Norales, 11-Mariano Acevedo, 2-Osman Chavez, 16-Nery Medina; 14-Carlos Palacios, 23-Roger Espinoza (7-Rigoberto Padilla, 70), 18-Melvin Valladares, 13-Carlos Costly (24-Georgie Welcome, 79); 10-Marvin Chavez (8-Allan Lalin, 65), 15-Walter Martínez (capt.)
Subs not used: 3-David Molina, 4-Johnny Palacios, 12-Ricardo Canales, 21-Luis Ramos
Head Coach: Reynaldo Rueda
Stats Summary: USA / HON
Shots: 16 / 6
Shots on Goal: 6 / 2
Saves: 2 / 3
Corner Kicks: 8 / 2
Fouls: 15 / 8
Offside: 1 / 8
Misconduct Summary:
HON – Nery Medina (caution) 38th minute
USA – Heath Pearce (caution) 40
USA – Michael Parkhurst (caution) 90
HON – Walter Martínez (caution) 93+
Referee: Courtney Campbell (JAM)
Assistant Referee 1: Ricardo Morgan (JAM)
Assistant Referee 2: José Luis Camargo (MEX)
Fourth Official: Walter Quesada (CRC)
ussoccer.com Man of the Match:
Santino Quaranta
Search: ussoccer.com is the official website of U.S. Soccer, the governing body of soccer in the United States
About the author Rodolfo Macias
San Antonio Newspaper, Inc. Copyright 2015 San Antonio Newspaper. All rights reserved.

redpajama_set_name: RedPajamaCommonCrawl


MyApartmentMap is a housing search web site with pet friendly listings all over the USA. Katie Reardon created this web site after searching for an apartment that would accept her Jack Russell Daisy, and her Beagle Jerry. As a pet owner she found it difficult to find pet friendly apartments, and wanted to help other pet owners find a home quick and easy.
Apartments listings all over the US and you can search on pet friendly listings.
Zillow.com and Hotpads.com both allow you to select Pet Friendly rental housing in their filters.
FidoFriendly.com is a wonderful resource for pet friendly travelers. Looking for a pet friendly hotel or off leash beach where you are visiting. They can help.


text:
UK weather set to be be hotter than Greece and Portugal this weekend
The long-range forecasts looks very promising
Emma Grimshaw
A lovely weekend of weather is forecast (Image: Oli Scarff/Getty Images)
Summer is finally on the horizon as forecasters's long-range predictions show the mercury surging at the end of the month.
Experts say during the last weeks of April the country will bask in temperatures higher than average for this time of year.
It's good news for millions who are staying at home and can enjoy the sunshine in their gardens and balconies, reports the Mirror.
And for those without either option, they can soak up the good weather during their daily exercise outdoors.
Temperatures across the south east of England are set to be warmer than Greece and Portugal this weekend.
Check the weather forecast near you, powered by In Your Area:
Met Office chiefs say the heatwave will be brief – but people should expect the sun to return later this month. They predict temperatures to be "trending above average by the end of the month" in the southern half of the UK.
The Met Office forecast adds: "More dominant settled conditions may begin to develop as we head towards the end of April.
"This dry and fine weather has the potential to gradually push northwestwards across much of the UK."
Frosts will become "infrequent" by the end of April, the Met Office's three-month prediction system forecasts, with "below-average" rainfall between April and June.
It should remain dry and cloudly across Bristol today (Friday) with temperatures peaking at 11C. The mercury is set to rise to 17C on Sunday but our city will miss out on the highs expected on the east coast.
UK weather forecast
Variable cloud and sunny spells, best in the north and east, whilst showers are expected in western, central and northern parts, these still wintry in the far north.
Outlook for Saturday to Monday:
Mostly dry over the weekend with sunny spells. Becoming warm, but also windy. Outbreaks of rain moving east later Sunday and into Monday with sunshine and showers following.


Give input to the development projects from a technical perspective in bringing know how regarding construction, material, production, quality and customer.
Responsible of the documentation and the drawings.
Secure they are updated and at correct quality level according to Organization Documentation Guideline and standards.
Responsible for choosing test methods and assist in developing test methods in Co-operation with Test Lab and PR&C.
Responsible for planning and executing quality checks.
Support production implementation in cooperation with Purchasing.
Contribute to product assessments and analyze production risk assessments.
Take initiatives to product and supplier improvements together with Product developers and product engineers and Purchasing Technicians.
Be updated within applicable Organization Standards, specifications, working methods and IT tools.
Keep an up-dated picture of customers needs and demands around the Globe.
Supporting external users in handling PLUS.
Experience in product development and implementation of product specifications in production environment.
Understanding of customers needs and expectations towards the key and secondary functions of different products (e.g. safety, durability, ergonomics etc.).


You do not know fix out of service laminate? About this problem you can read in this article.
Possible my advice may seem unusual, however nonetheless for a start there meaning wonder: whether it is necessary general repair out of service laminate? may logical will purchase new? I personally inclined according to, has meaning for a start learn, how is a new laminate. For it possible just make appropriate inquiry finder, eg, rambler or yandex.
If you still decided own hands repair, then the first thing has meaning learn how repair laminate. For it one may use rambler or yandex, or try find response desired question on popular forum or community.
I think you do not nothing spent time and this article could help you make fix laminate. In the next article I will write how fix leather jacket or leather jacket.


CEPLAS is the only German Cluster of Excellence focusing exclusively on plant science. Arable land, water and nutrients become scarce, while the climate is changing and the world's population is further increasing. To ensure a sustainable supply with food, feed, and plant-based raw materials, innovative strategies for crop improvement are required.
CEPLAS addresses these grand challenges through world-class fundamental research on complex plant traits of agronomic relevance that impact on yield and adaption to limited resources. This knowledge is essential for designing and breeding plants that react in a predictable way to future challenges.
Additionally, CEPLAS has established a technology platform on Plant Metabolism and Metabolomics Facility.


text:
Best answer: How is hydropower regulated?
How is hydroelectric power regulated?
What is hydropower policy?
Is hydropower sustainable or not?
Is hydroelectric power illegal?
What is a FERC license?
When was the Federal Power Act passed?
How many hydroelectric dams are in Scotland?
What percentage of electricity is generated by hydropower in India?
Most dams in California are regulated by the Federal Energy Regulatory Commission.
On 26.8. 1998, the Government has accorded approval to the policy on Hydro Power Development. The object of the Policy is to prevent a decline in hydro share and to undertake measures for the exploitation of vast hydro-electric potential in the country especially in the North and North Eastern Regions.
Hydropower is a climate-friendly energy source, generating power without producing air pollution or toxic by-products. Using hydropower avoids approximately 200 million metric tons of carbon pollution in the U.S. each year – equal to the output of over 38 million passenger cars.
As outlined on FERC's website, any hydroelectric project must apply for a license or an exemption from licensing if the project is or will be: located on a navigable waterway of the U.S.; occupying U.S. lands; utilizing surplus water or water power from a U.S. government dam; or.
The Federal Energy Regulatory Commission (Commission or FERC) licenses and relicenses hydroelectric projects, including new hydrokinetic technologies, and wants you to know how to participate in the licensing process.
Enacted in 1935, Part II of the Federal Power Act (FPA) has been the basis for federal regulation of interstate electricity transmission and wholesale electricity sales for over eighty years.
THIS IS INTERESTING: What is the most essential use of geothermal energy?
Scotland has two pumped-storage hydro-electric power stations, which pump water back up to a storage reservoir during periods of off-peak demand.
As of 31 March 2020, India's installed utility-scale hydroelectric capacity was 46,000 MW, or 12.3% of its total utility power generation capacity.


Every week there seems to be a new revision of how many Internet of Things things there will. So to put an end to this I am now publishing my own version which is future proof: "By 2050 the number of connected devices will be huge ™". In any case, we all realise that the Internet of Things will create a data tsunami of unimaginable size and that the opportunity lies in making sense of all that data, in finding the signals in all of that noise.
Today IBM announces the creation of a new global headquarters for Watson IoT in Munich where one thousand data scientists, researchers and programmers will work in this domain. Yes indeed, the name is Watson IoT. Watson is course of Jeopardy fame but in the mean time it (he?) has evolved into a cognitive computing solution with deep knowledge in many industries. The only way to take advantage of all the sensor data will be with a system that can learn and reason on a massive scale while presenting an easy, natural way for humans to interact with it.
interact with the system in human language. Using natural language processing to help understand the intent of human language by linking it to other sources of data to put it into context in specific situations.
look at data coming from devices, sensors and other sources to understand what is 'normal', what is a deviation and what is the trend in order to propose actions when necessary.
find correlations and anomalies in video feeds and image snapshots to identify scenes and patterns.
discover potential issues in a certain domains by reading maintenance logs, blog comments, etc. using text analytics.
Here is the video of the announcement.
Next Next post: I'm an AoT newbie and it feels great!


Turkey's parliament on Monday ratified changes to its bylaws to be implemented under the new presidential system.
The changes were backed by all party groups represented in parliament.
The revised bylaws cover a wide range of issues, from when ministers take their oath to how the budget bill is to be debated.
The changes stipulate that parliament is to debate presidential decrees within three months after they are signed. If it is unable to do so, the decree will be automatically abolished.
Parliament can also extend the state of emergency for four months in line with presidential decrees demanding its extension.
Members of the cabinet will also be able to attend the general assembly during the budget debate.
However, they can only attend other bill debates with the permission of the parliament speaker.
In an April 2017 referendum, Turkish voters approved the switch from a parliamentary to a presidential system.


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Nagle Energy Solutions (NES) announced today its patent-pending garage ventilation control system is delivering a 97.2% reduction in the energy consumed by the mechanical ventilation system at Main Street Cupertino Lofts garage in Cupertino, CA. The NES system shaves more than 324,000 kilowatt-hours (kWh) a year from the property's baseline of energy consumption, providing a recurring, operational-cost savings exceeding $66,000 a year, not accounting for future utility rate increases.
Continue reading Garage Ventilation Control System Delivers 97.2% Energy Savings at Silicon Valley Commercial Property at STL.News.


text:
Q: xamarin.forms: retrieving existing data from .db file In Xamarin Forms 4.3, I'm following this "Local Databases" doc: https://learn.microsoft.com/en-us/xamarin/xamarin-forms/data-cloud/data/databases
This doc explains how to create and use a new sqlite db, but how do I go about in loading and displaying data in an existing .db file with the end purpose of transferring it into a new local sqlite database?
using SQLite;

namespace TruckExample.Models
{
    public class Truck
    {
        [PrimaryKey, AutoIncrement]
        public int TruckNo { get; set; }
        public string TruckName { get; set; }
    }
}
//////////////////////////////////////////////
using System.Collections.Generic;
using System.Threading.Tasks;
using SQLite;
using TruckExample.Models;

namespace TruckExample.Data
{
    public class TruckDatabase
    {
        readonly SQLiteAsyncConnection _database;

        public TruckDatabase(string dbPath)
        {
            _database = new SQLiteAsyncConnection(dbPath);
            _database.CreateTableAsync<Truck>().Wait();
        }

        public Task<List<Truck>> GetTrucksAsync()
        {
            return _database.Table<Truck>().ToListAsync();
        }
    }
}
////////////////////////////////////////////////
using System;
using Xamarin.Forms;
using Xamarin.Forms.Xaml;
using TruckExample.Data;
using System.IO;

[assembly: XamlCompilation(XamlCompilationOptions.Compile)]
namespace TruckExample
{
    public partial class App : Application
    {
        static TruckDatabase database;

        public static TruckDatabase Database
        {
            get
            {
                if(database == null)
                {
                    database = new TruckDatabase(Path.Combine(Environment.GetFolderPath(Environment.SpecialFolder.LocalApplicationData), "Truck.db"));
                }

                return database;
            }
        }

        public App()
        {
            InitializeComponent();
            MainPage = new NavigationPage(new MainPage());
        }
   }
}
/////////////////////////////////////////
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

using Xamarin.Forms;
using Xamarin.Forms.Xaml;

namespace TruckExample
{
    [XamlCompilation(XamlCompilationOptions.Compile)]
    public partial class MainPage : ContentPage
    {
        public MainPage ()
        {
            InitializeComponent ();
        }

        protected override async void OnAppearing()
        {
            base.OnAppearing();
            listView.ItemsSource = await App.Database.GetTrucksAsync();
            Debug.Write(App.Database.GetTrucksAsync());

        }
    }
}

A: For others, helpful resources:
Implementation by Hameed Kunkanoor: "Creating a sqlite databse and storing your data in your android and ios application in Xamarin Forms" https://medium.com/@hameedkunkanoor/creating-a-sqlite-databse-and-storing-your-data-in-your-android-and-ios-application-in-xamarin-2ebaa79cdff0
(Refer to his git if needed)
Explanation/response by Matthewrdev: Use a local database in Xamarin
To improve the implementation -- Brandon Minnick: "Xamarin: Efficiently Using a SQLite Database" https://codetraveler.io/2019/11/26/efficiently-initializing-sqlite-database/

redpajama_set_name: RedPajamaStackExchange


text:
Daydream Island Resort & Spa, Whitsundays and Resort View
Daydream Island Resort & Spa, (clockwise from left): Suite Living Area, Superior King, Superior Twin, Serenity Suite and Deluxe King
Daydream Island Resort & Spa, Restaurant
Daydream Island Resort & Spa, Aeriel Island View, Pool View, Western Shore, Whitehaven Beach and Lovers Cove
Daydream Island Resort
from €125 per person per night
Located in the Whitsundays on Australia's Great Barrier Reef, travel to the award winning Daydream Island Resort and step into a tropical paradise. Accessible via a 30 minute ferry ride from Hamilton Island or The Port of Airlie, this vibrant resort is home to an abundance of exciting activities and facilities to make your holiday an unforgettable one. The resort offers refurbished modern rooms and suites overlooking the gardens, pools and ocean.
The Living Reef, the resort's 200m man-made outdoor aquarium wraps around the central building, is a big highlight. Holding 1.5 million litres of water, over 100 species of marine life, colourful corals and fantastic activities such as joining the team of marine biologists who live on the island and feed baby stingrays, this attraction will be sure to thrill the whole family.
The children will love this island resort, with fun-filled children's programs, three large, lagoon style swimming pools, and tasty food and drinks. Families are well accommodated here, with the option of interconnected rooms and suites, featuring bright contemporary décor and breath-taking views.
With beautiful beaches, gym facilities, swimming pools, brilliant non-motorised water sports, rainforest walks, and a range of delicious restaurants including three new restaurants and bars, you'll never get bored at this wonderful island retreat.
Located in the Whitsundays on Australia's Great Barrier Reef. This tropical resort is nestled on Daydream Island, one of seven Molle Group isles that form a sub-group of the beautiful Whitsunday Islands, along the North Eastern coast of Queensland.
30 minutes by ferry from Hamilton Island Airport and from The Port of Airlie
There are 277 rooms and suites, all featuring:
Ensuite and shower
Bar size fridge and mini bar
Complimentary tea and coffee facilities
Choose between:
Standard – choose between either one king bed or two queen beds, these contemporary rooms are conveniently located near the centre of the resort with views of the tropical gardens. Sleep a maximum of four people, including two adults and two children.
Superior – With the same amenities as the Standard Garden, plus a spacious balcony. Available in Pool View, Poolside or Ocean View. Choose between either two queen beds or one king bed. Sleeps a maximum of four people, including two adults and two children.
Deluxe Ocean Terrace – Airy rooms located on the eastern side of the Driftwood Wing, with uninterrupted views of the ocean towards the Whitsunday Islands. Choose between either two queen beds or one king bed. Sleeps a maximum of four people, including two adults and two children.
Deluxe Serenity Twin – Colourful suites with two queen beds, a sitting area, and a balcony with gorgeous views. Families can also interconnect this suite with a Daydreamer Suite. Sleeps a maximum of four people, including two adults and two children.
Serenity Suite – One expansive suite which is all about the spectacular Lovers Cove View. Featuring a terrace with outdoor furniture, and a walk in shower. Sleeps a maximum of two adults and two children.
Family Rooms – Two spacious resort rooms with an interconnecting door, choose from a Superior Pool View, Pool Sode or Ocean View. Sleeps a maximum of four guests, including two adults and two children.
Inkstone Kitchen & Bar – Inspired Australian dishes, serving the best in steak and seafood. Open for breakfast 06:30am-10:00am, lunch from 11:30am-3:00pm and dinner from 5:30pm-9:00pm.
Infinity – Enjoy an Asian Fusion dining experience, along with expansive views of the Whitsunday Islands. Open daily from 5:30pm-9:00pm
Graze – A modern buffet offering a range of international cuisines with live cooking stations. Open daily for breakfast from 06:30am-10:00am and dinner from 5:30pm-9:00pm.
Barefoot Bar – Centrally located amongst the gardens the Barefoot Bar is an outdoor venue where you can relax with a drink, snack or a gourmet burger. Open daily.
Tonic Bar – spanning from the atrium out towards the living reef you can find the Tonic Bar. As the island's speciality Gin bar, it's a great place to meet for a G&T before or after dinner.
Three large, lagoon style swimming pool, one features Barefoot, the swim-up bar.
Living Reef – the resort's outdoor aquarium
Multiple water sports are available, including diving and snorkelling.
Children aged 0-12 years can eat and stay free of charge if sharing a room with adults and utilising existing bedding.
Kids Club – Open daily; choose between a morning (8:30am-12 midday) or afternoon (1:00pm-4:30pm) session or 3 hour babysitting services. The 4-12 year's program includes wildlife experiences such as fish feeding, coral collecting, rainforest wildlife walks and talks, beach sports, scavenger hunts, arts and crafts, movies, kite making and much more. Must be pre-booked.
Living Reef – Located by the entrance of the resort, this free from living reef wraps 200m around the building holding 1.5 million litres of water. Home to over 100 species of marine fish, coral, starfish, sea cucumbers and crabs. There are free educational feeding sessions, Stingray Splash Experience, wading knee deep into the reef to say hello and guided fish feeding experiences are all available to book.
Enjoy a relaxing massage or facial in one of the two spa suites.
Stinger season takes place between October and March every year, during this time a stinger suit must be worn when swimming in the Whitsundays
Hamilton Island Palm Bungalows
Hamilton Island Reef View Hotel
Coral Sea Marina Resort, Airlie Beach


text:
<?php

namespace Doctrine\Tests\Models\References;

use Doctrine\ODM\PHPCR\Mapping\Annotations as PHPCRODM;

/**
 * @PHPCRODM\Document(referenceable=true)
 */
class ParentWithChildrenTestObj
{
    /** @PHPCRODM\Id(strategy="parent") */
    public $id;
    /** @PHPCRODM\ParentDocument */
    public $parent;
    /** @PHPCRODM\Nodename */
    public $nodename;
    /** @PHPCRODM\Field(type="string") */
    public $name;
    /** @PHPCRODM\Children() */
    public $children;

    public function getParentDocument()
    {
        return $this->parent;
    }

    public function setParentDocument($parent)
    {
        $this->parent = $parent;
    }
}

redpajama_set_name: RedPajamaGithub


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Located just behind the House, our gymnasium is the perfect place for children's birthday parties, dances, basketball or large meetings. The open basketball court makes this space quite versatile. Use of toys and gym equipment appropriate for all age groups is included with your rental. Kids enjoy basketball, parachute, scooters, ride-on toys, bowling, obstacle courses and more. Tables and chairs are also available for your use. Why not bring in your own magician, musician, or even rent our moon bounce?
$200 for 3 hours; $50 per each additional hour.
Moon Bounce rental: $75 for the duration of your gym party.
Up to 10 6-foot rectangular tables and 50 chairs are readily available. Additional tables and chairs may be available for an additional cost if available and with at least 2 weeks notice.
Call for date and time availability.
We have a no-smoking policy throughout our facilities.
No alcohol on the premises unless served by an insured bartending service.
The gym comfortably holds 50-75 children.